COMBINATORIAL AND GEOMETRIC RIGIDITY WITH ...COMBINATORIAL AND GEOMETRIC RIGIDITY WITH SYMMETRY CONSTRAINTS BY BERND SCHULZE A DISSERTATION SUBMITTED TO THE FACULTY OF GRADUATE STUDIES

COMBINATORIAL AND GEOMETRIC RIGIDITY WITH

SYMMETRY CONSTRAINTS

BY BERND SCHULZE

A DISSERTATION SUBMITTED TO THE FACULTY OF GRADUATE

STUDIES IN PARTIAL FULFILMENT OF THE REQUIREMENTS

FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

GRADUATE PROGRAM IN MATHEMATICS AND STATISTICS

YORK UNIVERSITY,

TORONTO, ONTARIO

MAY 2009

Abstract

In this thesis, we investigate the rigidity and flexibility properties of frame-

works consisting of rigid bars and flexible joints that possess non-trivial sym-

metries.

Using techniques from group representation theory, we first prove that

the rigidity matrix of a symmetric framework can be transformed into a

block-diagonalized form. Based on this result, we prove a generalization of

the symmetry-extended version of Maxwell’s rule given in [25] which can

be applied to both injective and non-injective realizations in all dimensions.

We then use this rule to prove that a symmetric isostatic (i.e., minimal in-

finitesimally rigid) framework must obey some very simply stated restrictions

on the number of joints and bars that are ‘fixed’ by various symmetry op-

erations of the framework. In particular, it turns out that a 2-dimensional

isostatic framework must belong to one of only six possible point groups. For

3-dimensional isostatic frameworks, all point groups are possible, although

restrictions on the placement of structural components still apply.

For three of the five non-trivial symmetry groups in dimension 2 that

allow isostatic frameworks, namely for the groups C2 and C3 generated by

a half-turn and a 3-fold rotation, respectively, and for the group Cs gener-

ated by a reflection, we establish symmetric versions of Laman’s Theorem

([33, 46]). More precisely, we show that the necessary conditions derived from

the symmetry-extended version of Maxwell’s rule, together with the Laman

conditions, are also sufficient for a framework whose joints are positioned as

generically as possible subject to the given symmetry conditions to be iso-

static. Symmetric versions of Henneberg’s Theorem ([40, 33]) and Crapo’s

iv

Theorem ([20, 33, 67]) for the groups C2, C3, and Cs are also established. For

the remaining two non-trivial symmetry groups in dimension 2 that allow

isostatic frameworks, we offer some analogous conjectures.

Finally, we derive sufficient conditions for the existence of a finite flex of a

symmetric framework. Finite flexes detected with these results have the nice

property that they preserve all of the symmetries of the given framework.

v

To my parents

vi

Acknowledgements

First, I would like to thank my supervisor Prof. Walter Whiteley for his

invaluable advice, guidance, and support throughout my time as his student.

His care and enthusiasm for my work as well as his vast knowledge and

exceptional insights into mathematics have always been a great source of

motivation and inspiration for me. The countless enlightening conversations

I have had with Prof. Walter Whiteley over the last few years have not only

played a crucial role in the writing of this thesis, but they have also nourished

my intellectual maturity which I will benefit from for a long time to come. In

addition, Prof. Walter Whiteley has always been available for me whenever

I faced any sort of trouble or had a question about my research or writing. I

simply could not have wished for a better or friendlier supervisor.

Secondly, I would like to express many thanks to Prof. Asia Weiss and

Prof. Mike Zabrocki for taking the time to read my thesis and serve on my

supervisory committee in the midst of all their other activities.

Thanks also to Prof. Ada Chan, Prof. Andy Mirzaian, and Prof. Meera

Sitharam for agreeing to be members of my examining committee.

Further, I would like to thank the organizers and participants of the AIM

workshop in Palo Alto in December 2007 and the BIRS workshop in Banff

in July 2008, particularly Prof. Robert Connelly, Prof. Simon Guest, and

Prof. Brigitte Servatius, for all the interesting and fruitful discussions.

A special thanks goes to my mother, father, and sister for their continued

support and encouragement throughout my studies at home and abroad.

vii

They have always been there for me.

Last, but not least I would like to thank my partner Krishna Wu for all

her care and love.

vii

Contents

1 Introduction 1

1.1 Background and motivation . . . . . . . . . . . . . . . . . . . 1

1.2 Outline of thesis . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Definitions and preliminaries 13

2.1 Graph theory terminology . . . . . . . . . . . . . . . . . . . . 13

2.2 Introduction to rigidity theory . . . . . . . . . . . . . . . . . . 17

2.2.1 Rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2.2 Infinitesimal rigidity . . . . . . . . . . . . . . . . . . . 19

2.2.3 Static rigidity . . . . . . . . . . . . . . . . . . . . . . . 23

2.2.4 Generic rigidity . . . . . . . . . . . . . . . . . . . . . . 28

2.2.5 Basic rigidity results . . . . . . . . . . . . . . . . . . . 31

2.3 Symmetry in frameworks . . . . . . . . . . . . . . . . . . . . . 38

3 A classification of symmetric frameworks 46

3.1 The classification . . . . . . . . . . . . . . . . . . . . . . . . . 48

viii

3.2 The notion of (S, Φ)-generic . . . . . . . . . . . . . . . . . . . 54

3.3 Of what types Φ can a framework be? . . . . . . . . . . . . . 68

3.4 When is a type Φ of a framework a homomorphism? . . . . . . 74

4 Using group representation theory to analyze symmetric

frameworks 79

4.1 Block-diagonalization of the rigidity matrix . . . . . . . . . . . 82

4.1.1 Basic definitions in group representation theory . . . . 82

4.1.2 The internal and external representation . . . . . . . . 85

4.1.3 The block-diagonalization . . . . . . . . . . . . . . . . 87

4.2 A symmetry-extended version of Maxwell’s rule . . . . . . . . 104

4.2.1 The necessary conditions . . . . . . . . . . . . . . . . . 106

4.2.2 The rule . . . . . . . . . . . . . . . . . . . . . . . . . . 118

4.2.3 Example and further remarks . . . . . . . . . . . . . . 123

4.3 Restrictions on the number of fixed joints and bars of sym-

metric isostatic frameworks . . . . . . . . . . . . . . . . . . . 127

4.3.1 Fixed versus geometrically unshifted . . . . . . . . . . 128

4.3.2 Isostatic frameworks in dimension 2 . . . . . . . . . . . 132

4.3.3 Isostatic frameworks in dimension 3 . . . . . . . . . . . 138

4.3.4 A remark on non-injective realizations . . . . . . . . . 149

4.4 Necessary conditions for independence and infinitesimal rigidity150

ix

5 Necessary and sufficient conditions for a graph to be (S, Φ)-

generically isostatic 154

5.1 Preliminary remarks and results . . . . . . . . . . . . . . . . . 155

5.2 Characterizations of (C3, Φ)-generically isostatic graphs . . . . 162

5.2.1 Symmetrized Henneberg moves and 3Tree2 partitions

for C3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

5.2.2 The main result for C3 . . . . . . . . . . . . . . . . . . 165

5.3 Characterizations of (C2, Φ)-generically isostatic graphs . . . . 188


for C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

5.3.2 The main result for C2 . . . . . . . . . . . . . . . . . . 191

5.4 Characterizations of (Cs, Φ)–generically isostatic graphs . . . . 210


for Cs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

5.4.2 The main result for Cs . . . . . . . . . . . . . . . . . . 215

5.5 Conjectures, algorithms, and further remarks . . . . . . . . . . 261

5.5.1 Dimension 2 . . . . . . . . . . . . . . . . . . . . . . . . 261

5.5.2 Dimension 3 . . . . . . . . . . . . . . . . . . . . . . . . 265

5.5.3 Independence and infinitesimal rigidity . . . . . . . . . 267

6 Symmetry as a sufficient condition for a flex 274

6.1 Alternate definitions of rigidity via the edge function . . . . . 276

x

6.2 Detection of symmetric flexes . . . . . . . . . . . . . . . . . . 278

6.3 Examples of flexible frameworks . . . . . . . . . . . . . . . . . 289

6.3.1 Examples in 2D . . . . . . . . . . . . . . . . . . . . . . 290

6.3.2 Examples in 3D . . . . . . . . . . . . . . . . . . . . . . 295

7 Further work 306

7.1 Rigidity of other types of symmetric structures . . . . . . . . . 307

7.1.1 Pinned frameworks . . . . . . . . . . . . . . . . . . . . 307

7.1.2 Body-bar structures . . . . . . . . . . . . . . . . . . . 309

7.1.3 Body-hinge and molecular structures . . . . . . . . . . 313

7.2 Coning, symmetry, and spherical frameworks . . . . . . . . . . 315

7.3 Symmetric global rigidity . . . . . . . . . . . . . . . . . . . . . 319

A Character tables of selected point groups 322

xi

List of Tables

4.1 Calculations of characters in the 2-dimensional symmetry-

extended Maxwell’s equation. . . . . . . . . . . . . . . . . . . . 133

4.2 Calculations of characters in the 3-dimensional symmetry-

extended Maxwell’s equation. . . . . . . . . . . . . . . . . . . . 139

xii

List of Figures

2.1 An invariant (b) and a non-invariant subgraph (c) of the

graph G under α = (v1 v2 v3)(v4 v5 v6) ∈ Aut(G). . . . . . . . . 16

2.2 A rigid (a) and a flexible (b) framework in the plane. The flex

shown in (c) takes the framework in (b) to the framework in

(d). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3 The arrows indicate the non-zero displacement vectors of an

infinitesimal rigid motion (a) and infinitesimal flexes (b, c) of

frameworks in R2. . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 (a), (b) The arrows indicate a tension (a) and a compression

(b) in a bar. (c) An equilibrium load on a non-degenerate

triangle. This load can be resolved by the triangle as shown

in (d). (e) An unresolvable equilibrium load on a degenerate

triangle: for any joint of this framework, tensions or com-

pressions in the bars cannot reach an equilibrium with the load

vector at this joint. . . . . . . . . . . . . . . . . . . . . . . . . 25

xiii

2.5 A Venn diagram showing the relationship between sets of var-

ious types of ‘generic’ configurations and the set of regular

points of a graph G (see Definition 2.2.22 in the end of Sec-

tion 2.2.5). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.6 The double banana satisfies the counts in Theorem 2.2.8 for

d = 3, but it is not generically 3-isostatic. . . . . . . . . . . . 34

2.7 Illustrations of the Vertex Addition Theorem (a) and the Edge

Split Theorem (b) in dimension 2. . . . . . . . . . . . . . . . . 35

2.8 Illustration of an X-replacement of a graph G. . . . . . . . . . 36

2.9 A proper (a) and a non-proper (b) 3Tree2 partition. . . . . . . 37

2.10 Symmetry elements corresponding to symmetry operations in

dimension 2: (a) a rotation Cm, m ≥ 2; (b) a reflection s; (c)

the identity Id. . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.11 Symmetry elements corresponding to symmetry operations in

dimension 3: (a) an improper rotation Sm, m ≥ 2; (b) a

rotation Cm, m ≥ 2; (c) a reflection s; (d) the identity Id. . . 42

3.1 2-dimensional realizations of (K3,3, Cs) of different types. . . . 50

3.2 2-dimensional realizations of (Gtp, C2) of different types. . . . . 51

3.3 3-dimensional realizations of (Gbp, Cs) of different types. . . . . 52

3.4 By the converse of Pascal’s Theorem, the joints of any real-

ization in R(K3,3,C2) or R(K3,3,Cs,Φb) lie on a conic section. . . . 55

3.5 A 3-dimensional realization of (K4, Cs) of type Υa (a) and of

type Υb (b). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

xiv

3.6 A realization of K3,3 that is (C2v, Φ)-generic, but not (Cs, Φa)-

generic, where Cs is the subgroup of C2v generated by sv and

Φa = Φ|Cs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.7 A realization of (Gt, C2) of type Θa and Θb (a) and a realization

of (Gbp, Cs) of type Ξa and Ξb (b). . . . . . . . . . . . . . . . . 69

3.8 Non-injective realizations with Aut(G, p) = id. . . . . . . . . 73

3.9 A graph G (a) and a realization (G, p) ∈ R(G,Cs) (b) for which

there does not exist a homomorphism Φ : Cs → Aut(G) so that

(G, p) is of type Φ. . . . . . . . . . . . . . . . . . . . . . . . . 76

3.10 A graph G (a) and a realization (G, p) ∈ R(G,C3) (b) for which

there does not exist a homomorphism Φ : C3 → Aut(G) so that

(G, p) is of type Φ. . . . . . . . . . . . . . . . . . . . . . . . . 77

4.1 A framework (K3, p) ∈ R(K3,Cs,Φ). . . . . . . . . . . . . . . . . 87

4.2 Illustration of the proof of Lemma 4.1.1 (i). . . . . . . . . . . 89

4.3 Illustration of the proof of Lemma 4.1.1 (i) in the case where

x is a reflection. . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.4 Illustration of the proof of Lemma 4.1.1 (ii). . . . . . . . . . . 90

4.5 (a, b) Vectors of the H ′e-invariant subspaces V

(A′)e (a) and

V(A′′)e (b) of R6; (c, d) vectors of the H ′

i-invariant subspaces

V(A′)i (c) and V

(A′′)i (d) of R3. . . . . . . . . . . . . . . . . . . 102

4.6 (a) A basis for the subspace V(A′)T ; (b) a basis for the subspace

V(A′′)T ; (c) a basis for the subspace R = V

(A′′)R . . . . . . . . . . 112

xv

4.7 (a) An infinitesimal flex of (K3,3, p) ∈ R(K3,3,C2v ,Φ) which is

symmetric with respect to B2 (the displacement vector at each

joint of (K3,3, p) remains unchanged under Id and sv and is

reversed under C2 and sh). (b) An unresolvable equilibrium

load on (K3,3, p) which is symmetric with respect to B2. (c) A

self-stress of (K3,3, p) which is symmetric with respect to A1

(the tensions and compressions in the bars of (K3,3, p) remain

unchanged under all symmetry operations in C2v). . . . . . . . 125

4.8 Geometrically unshifted bars in dimension 2: (a) a bar that is

geometrically unshifted by a half-turn C2; (b) possible place-

ment of a bar that is geometrically unshifted by a reflection

s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

4.9 Possible placement of a bar that is geometrically unshifted by:

(a) any rotation Cm, m ≥ 2 (in dimension 3); (b) a half-turn

C2 (in dimension 3) alone. . . . . . . . . . . . . . . . . . . . . 130

4.10 Possible placement of a bar that is geometrically unshifted by a

reflection s (in dimension 3): (a) lying in the reflection plane;

(b) lying perpendicular to the reflection plane. . . . . . . . . . 130

4.11 Possible placement of a bar that is geometrically unshifted by:

(a) any improper rotation Sm, m ≥ 2 (in dimension 3); (b)

an inversion i = S2 (in dimension 3) alone. . . . . . . . . . . 131

xvi

4.12 Examples, for each of the possible point groups, of small 2-

dimensional isostatic frameworks: (a) C1; (b) C2; (c) C3; (d)

Cs; (e) C2v; (f) C3v. For each of Cs and C3v, two examples

are given, where in (i) each mirror has a bar centered at and

perpendicular to the mirror line, whereas in (ii) each mirror

has a bar that lies in the mirror line. For C2v, the bar lying at

the center of C2 must lie in one mirror line and perpendicular

to the other. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.13 A regular octahedron (a), and a convex polyhedron (b) gen-

erated by capping every face of the original octahedron with a

twisted octahedron. The polyhedron in (b) has the rotational

but not the reflectional symmetries of the polyhedron in (a). If

a framework is constructed from either polyhedron by placing

bars along edges, and joints at vertices, the framework will be

isostatic. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

4.14 An icosahedron (a), and the second stellation of the icosahe-

dron (b). If a framework is constructed from either polyhedron

by placing bars along edges, and joints at vertices, the frame-

work will be isostatic. The framework (b) could be constructed

from the framework (a) by ‘capping’ each face of the original

icosahedron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

4.15 A series of ‘hats’ added symmetrically along a 3-fold axis of

an isostatic framework leaves the framework isostatic. . . . . . 146

xvii

4.16 3-dimensional frameworks with mirror symmetry satisfying

Maxwell’s original rule. (a) A framework which is not iso-

static, since bΦa(s) > jΦa(s); (b) a framework which is not

isostatic, since bΦb(s) < jΦb(s); (c) a framework which satis-

fies bΦc(s) = jΦc(s), but is not isostatic, because it contains the

frameworks depicted in (a) and (b) (Φa, Φb, Φc are uniquely

determined by the injective realizations). . . . . . . . . . . . . 147

4.17 3-dimensional frameworks with half-turn symmetry satisfying

Maxwell’s original rule. (a) An isostatic framework; (b) a

framework which is not isostatic, since bΦb(C2) = jΦb(C2) = 0;

(c) a framework which satisfies jΦc(C2) = 0 and bΦc(C2) = 2, but

is not isostatic, because it contains the non-isostatic framework

depicted in (b) (Φb and Φc are uniquely determined by the

injective realizations). . . . . . . . . . . . . . . . . . . . . . . . 148

4.18 Independent frameworks in R(G,C2,Φ) with jΦ(C2) = bΦ(C2) = 0

(a), jΦ(C2) = 0, bΦ(C2) = 2 (b), and jΦ(C2) = 1, bΦ(C2) = 0

(c). Infinitesimally rigid frameworks in R(G,C2,Φ) with jΦ(C2) =

bΦ(C2) = 0 (d), jΦ(C2) = 0, bΦ(C2) = 2 (e), and jΦ(C2) = 1,

bΦ(C2) = 0 (f). . . . . . . . . . . . . . . . . . . . . . . . . . . 152

5.1 Illustration of the proof of Lemma 5.1.2. . . . . . . . . . . . . 159

5.2 A (C3, Φ) vertex addition of a graph G, where Φ(C3) = γ and

Φ(C23) = γ2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

5.3 A (C3, Φ) edge split of a graph G, where Φ(C3) = γ and

Φ(C23) = γ2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

xviii

5.4 A (C3, Φ) ∆ extension of a graph G, where Φ(C3) = γ and

Φ(C23) = γ2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

5.5 (C3, Φ) 3Tree2 partitions of graphs, where Φ(C3) = γ and

Φ(C23) = γ2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

5.6 If a graph G satisfies the conditions in Theorem 5.2.1 (ii) and

has a vertex v of valence 3, then G is a graph of one of the

types depicted above. . . . . . . . . . . . . . . . . . . . . . . . 169

5.7 Construction of a (C3, Φ) 3Tree2 partition of G in the case

where G is a (C3, Φk−1) vertex addition of Gk−1. . . . . . . . . 175


where G is a (C3, Φk−1) edge split of Gk−1. . . . . . . . . . . . 176


where G is a (C3, Φk−1) ∆ extension of Gk−1. . . . . . . . . . . 177

5.10 The frame (G, p, q). . . . . . . . . . . . . . . . . . . . . . . . . 179

5.11 The frame (G, pt, qt). . . . . . . . . . . . . . . . . . . . . . . . 182




5.15 A realization of (G, C2) of type Φ (a) and a realization of

(G, Cs) of type Ψ (b). . . . . . . . . . . . . . . . . . . . . . . . 189

5.16 A (C2, Φ) vertex addition of a graph G, where Φ(C2) = γ. . . . 189

5.17 A (C2, Φ) edge split of a graph G, where Φ(C2) = γ. . . . . . . 190

xix

5.18 (C2, Φ) 3Tree2 partitions of graphs, where Φ(C2) = γ. The

edges in black color represent edges of the invariant trees. . . . 191


has a vertex v of valence 3, then G is a graph of one of the

types depicted above. . . . . . . . . . . . . . . . . . . . . . . . 195


where G is a (C2, Φk−1) vertex addition of Gk−1. The edges in

black color represent edges of the invariant tree T(k)0 . . . . . . . 201


where G is a (C2, Φk−1) edge split of Gk−1. The edges in black

color represent edges of the invariant trees. . . . . . . . . . . . 202

5.22 The frame (G, p, q). . . . . . . . . . . . . . . . . . . . . . . . . 204

5.23 The frame (G, pt, qt). . . . . . . . . . . . . . . . . . . . . . . . 207

5.24 A (Cs, Φ) single vertex addition of a graph G, where Φ(s) = σ. 210

5.25 A (Cs, Φ) single edge split of a graph G, where Φ(s) = σ. . . . 211

5.26 A (Cs, Φ) double vertex addition of a graph G, where Φ(s) = σ. 211

5.27 A (Cs, Φ) double edge split of a graph G, where Φ(s) = σ. . . . 212

5.28 A (Cs, Φ) X-replacement of a graph G, where Φ(s) = σ. . . . . 212

5.29 A (Cs, Φ) 3Tree2 ⊥ partition of a graph (a) and a (Cs, Φ)

3Tree2 ‖ partition of a graph (b), where Φ(s) = σ. . . . . . . . 214

xx


has a vertex v with NG(v) = v1, v2, v3 such that σ(vi, vj) 6=vi, vj for all i, j ⊆ 1, 2, 3, then G is a graph of one of

the types depicted above. . . . . . . . . . . . . . . . . . . . . . 221


has a vertex v with NG(v) = v1, v2, v3 such that σ(vi, vj) =

vi, vj for exactly one pair i, j ⊆ 1, 2, 3, then G is a graph

of one of the types depicted above. . . . . . . . . . . . . . . . . 230

5.32 If a graph G satisfies the conditions in Theorem 5.4.1 (ii),

has no vertex of valence two, no vertex of valence three that

is fixed by σ, and every 3-valent vertex v of G (except possibly

the vertices that are incident with the edge that is fixed by σ)

has the property that σ(u) = u for all u ∈ NG(v), then there

exists v ∈ V (G) with NG(v) = v1, v2, v3, σ(vi) = vi for all

i = 1, 2, 3, and valG(vi) = 4 for some i ∈ 1, 2, 3. . . . . . . . 235

5.33 Construction of a (Cs, Φ) ⊥ or (Cs, Φk−1) ‖ 3Tree2 partition of

G in the case where G is a (Cs, Φk−1) single vertex addition of

Gk−1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

5.34 Construction of a (Cs, Φ) ‖ 3Tree2 partition of G in the case

where G is a (Cs, Φk−1) single edge split of Gk−1. The edges in

black color represent edges of the invariant trees. . . . . . . . . 240

5.35 Construction of a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition of G

in the case where G is a (Cs, Φk−1) double vertex addition of

Gk−1 and v1, v2 /∈ V(T

(k−1)0

). . . . . . . . . . . . . . . . . . . . 241

xxi


in the case where G is a (Cs, Φk−1) double vertex addition of

Gk−1 and at least one of v1 or v2 is a vertex of T(k−1)0 . The

edges in black color represent edges of the invariant tree. . . . 242


in the case where G is a (Cs, Φk−1) double edge split of Gk−1,

v1, v2, σ(v1), σ(v2) ∈ E(T

(k−1)0

)and either v3 /∈ V

(T

(k−1)0

)

or v3 ∈ V(T

(k−1)0

)and σ(v3) 6= v3. The edges in black color

represent edges of the invariant trees. . . . . . . . . . . . . . . 243

5.38 Construction of a (Cs, Φ) ‖ 3Tree2 partition of G in the

case where G is a (Cs, Φk−1) double edge split of Gk−1,

v1, v2, σ(v1), σ(v2) ∈ E(T

(k−1)0

), v3 ∈ V

(T

(k−1)0

)and

σ(v3) = v3. The edges in black color represent edges of the

invariant trees. . . . . . . . . . . . . . . . . . . . . . . . . . . 245


where G is a (Cs, Φk−1) double edge split of Gk−1, v1, v2 ∈E

(T

(k−1)1

)and σ(v1), σ(v2) ∈ E

(T

(k−1)2

). The edges in black

color represent edges of the invariant tree. . . . . . . . . . . . 247


where G is a (Cs, Φk−1) X-replacement of Gk−1, v1, v2 ∈E

(T

(k−1)1

)and v3, v4 ∈ E

(T

(k−1)2

). . . . . . . . . . . . . . . . 248

xxii

5.41 Construction of a (Cs, Φ) ‖ 3Tree2 partition of G in the

case where G is a (Cs, Φk−1) X-replacement of Gk−1 and

v1, v2, v3, v4 ∈ E(T

(k−1)0

). The edges in black color rep-

resent edges of the invariant tree. . . . . . . . . . . . . . . . . 250

5.42 The frame (G, p, q) in Case 1 of the proof of Lemma 5.4.5. . . 253

5.43 The frame (G, pt, qt) in the case where 〈V1〉∩T0 is not connected.255

5.44 The frame (G, pt, qt) in the case where 〈V1〉∩T2 is not connected.256

5.45 The frame (G, pt, qt). . . . . . . . . . . . . . . . . . . . . . . . 258

5.46 The frame (G, p, q) in Case 2 of the proof of Lemma 5.4.5. . . 260

5.47 Two frameworks whose underlying graphs satisfy the condi-

tions of Case B.2.3 in the proof of Lemma 5.4.3 with respect

to the types Φ that are uniquely determined by the injective

realizations. Any symmetrized Henneberg’s sequence for any

of these two graphs needs to include a (Cs, Φi) X-replacement. . 261

5.48 A (C3v, Φ) 3Tree2 ⊥ partition of a graph (a) and a (C3v, Φ)

3Tree2 ‖ partition of a graph (b), where Φ(C3) = γ and Φ(s) =

σ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

5.49 A (Cs, Φ)-generic realization of the complete bipartite graph K4,6.266

5.50 A (C3, Φ) partial vertex addition of order 0 of a graph G (a),

a (C3, Φ) partial vertex addition of order 1 of a graph G (b),

and a (C3, Φ) ∆ addition of a graph G (c). . . . . . . . . . . . 269

xxiii

5.51 Independent frameworks in R(G,C3,Φ) with jΦ(C3) = 1. These

frameworks cannot be contained in an isostatic framework that

has the same joints and also C3 symmetry. . . . . . . . . . . . 271

5.52 Infinitesimally rigid frameworks in R(G,C3,Φ) with jΦ(C3) = 1.

These frameworks cannot contain an isostatic framework that

has the same joints and also C3 symmetry. . . . . . . . . . . . 273

6.1 Fully (S, Φ)-symmetric infinitesimal motions of frameworks:

(a) a fully (Cs, Φ)-symmetric infinitesimal rigid motion of

(K3, p) ∈ R(K3,Cs,Φ); (b) a fully (Cs, Φ)-symmetric infinitesimal

flex of (K3,3, p) ∈ R(K3,3,Cs,Φ); (c) a fully (C3, Φ)-symmetric in-

finitesimal flex of (Gtp, p) ∈ R(Gtp,C3,Φ). Since each of the above

frameworks is an injective realization, the type Φ is uniquely

determined in each case. . . . . . . . . . . . . . . . . . . . . . 283

6.2 A fully (Cs, Φ)-symmetric infinitesimal flex of the independent

framework (K2,2, p). . . . . . . . . . . . . . . . . . . . . . . . . 290

6.3 A fully (C2v, Φ)-symmetric infinitesimal flex of a (C2v, Φ)-

generic realization of K4,4. . . . . . . . . . . . . . . . . . . . . 292

6.4 Illustration of the proof that the joints of (K4,4, p) lie on a

conic section. . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

6.5 A fully (C2v, Ψ)-symmetric infinitesimal flex of a (C2v, Ψ)-

generic realization of K4,4. . . . . . . . . . . . . . . . . . . . . 295

6.6 Flexible octahedra: with point group C2 (a); with point group

Cs (b); with point group C2v (c). . . . . . . . . . . . . . . . . . 297

xxiv

6.7 An isostatic octahedron in R(G,Cs,Φd). . . . . . . . . . . . . . . 299

6.8 A fully (Cs, Φa)-symmetric infinitesimal flex of the framework

(G, p) ∈ R(G,Cs,Φa). . . . . . . . . . . . . . . . . . . . . . . . . 300

6.9 A fully (C2v, Φb)-symmetric infinitesimal flex of the framework

(G, p) ∈ R(G,C2v ,Φb). . . . . . . . . . . . . . . . . . . . . . . . . 301

6.10 A fully (Cs, Φ)-symmetric infinitesimal flex of a (Cs, Φ)-generic

realization of K4,6. . . . . . . . . . . . . . . . . . . . . . . . . 304

7.1 Isostatic pinned frameworks in the plane: (a) with point group

C4; (b) with point group C4v. . . . . . . . . . . . . . . . . . . . 308

7.2 3-dimensional body-bar frameworks modeling different types of

‘Steward platforms’: the non-symmetric body-bar framework in

(a) is isostatic; the body-bar framework in (b) is flexible due

to the presence of the 6-fold rotational symmetry, as predicted

by the necessary counts derived in [36]. . . . . . . . . . . . . . 311

7.3 3-dimensional body-hinge frameworks whose underlying multi-

graph is a hexagonal cycle: the non-symmetric body-hinge

framework in (a) is isostatic; the body-hinge framework in (b)

is flexible due to the half-turn symmetry, as predicted by the

counts derived in [36]. . . . . . . . . . . . . . . . . . . . . . . 314

7.4 Illustration of ‘symmetric coning’: both the framework (G, pH)

in Rd and the coned framework (G ∗ v, p∗) in Rd+1 have

mirror symmetry. . . . . . . . . . . . . . . . . . . . . . . . . . 316

xxv

7.5 Frameworks with mirror symmetry which are not globally rigid

in R2. The framework (G, p) in (a) is also not ‘symmetric

globally rigid’ in the sense of problem (2), since the framework

in (b) is another non-congruent realization of G in R2 with the

same edge lengths and the same mirror symmetry as (G, p);

the framework in (c), however, is ‘symmetric globally rigid’

within the set of all realizations of G in R2 with the same

mirror symmetry. . . . . . . . . . . . . . . . . . . . . . . . . . 320

xxvi

Chapter 1

Introduction

1.1 Background and motivation

The study of rigidity and flexibility has a rich history in what are cur-

rently a number of areas of engineering and mathematics, but historically

were connected in the work of many scientists who combined studies of en-

gineering, physics, and mathematics.

Early work which is now recognized as ‘rigidity theory’ included the con-

jecture of L. Euler about the rigidity of closed surfaces (e.g. polyhedra with

rigid faces) [23], the static and kinematic analysis of pin-jointed frameworks

for engineering structures (from the early 19th century) [49], and the large lit-

erature on linkages (frameworks with non-trivial - non-congruent - motions)

in the 19th century [52].

This wider range of work produced both counting rules (such as J.C.

Maxwell’s rules for built structures and M. Grubler’s counts for linkages) and

1

geometric analyses (such as reciprocal diagrams and various other tools for

analyzing the resolutions of forces in the structures, and analyzing possible

motions) [49, 52, 61, 62, 83].

By the end of the 19th century, there were detailed analyses of which

frameworks are ‘normally’ rigid (e.g. A.L. Cauchy’s famous theorem on the

uniqueness of convex triangulated polyhedra with fixed edge lengths [10, 22,

76]), as well as when certain frameworks that were ‘normally’ rigid became

flexible (e.g. R. Bricard’s work on flexible frameworks with the bars and

joints of an octahedron [9]).

The people working on engineering problems developed a number of prac-

tical methods for analyzing buildings (e.g. the Eiffel Tower) as well as numer-

ical and geometric rules of thumb for their design and construction. A whole

body of work, summarized by the engineer/mathematician L. Henneberg,

also developed explicit inductive techniques for generating rigid structures

[40]. Another stream, from A.F. Mobius through J. Plucker and F. Klein

to L. Cremona, investigated the projective geometry of static equilibria and

singular forms which lead to building failures, in fields with names such as

‘graphical statics’ [21, 83].

Pockets of work continued through the 20th century, including detailed

explorations of the design and analysis of mechanical linkages (mechanisms)

and some continuing geometric analyses of built structures such as built

trusses. However, much of the previous geometric and combinatorial the-

ory was submerged in the numerical analysis permitted by the developing

computers and the now ‘standard’ designs.

One notable exception to this decline in attention to the ’theory of rigid-

2

ity’ was the ongoing mathematical work on rigidity of polyhedral surfaces of

the Russian school of A.D. Alexandrov, N.V. Efimov, and A.V. Pogorelov,

and their analysis of the unique realizability of convex metrics [1].

In the past 40 years, there have been two major, parallel developments.

(A) A strong mathematical theory has flowered, refining old results

and techniques, with the splitting of key questions, techniques, and well-

developed results into combinatorial and geometric aspects.

These developments have brought in combinatorial methods from graph

theory, matroid theory, and associated combinatorial algorithms. The work

also refined geometric conditions that were sufficient for a shift in the rigidity

properties - projective conditions for the static/first-order kinematic theory,

and Euclidean and affine geometry for the theory of finite motions. The

static/first-order kinematic theory is expressed in the linear algebra of the

rigidity matrix, whose rank, row dependencies and column dependencies all

play key roles in the theory.

A highlight of these mathematical developments has been the basically

complete combinatorial theory of plane frameworks - most notably, the com-

binatorial characterization of rigid 2-dimensional ‘generic’ frameworks given

by G. Laman in 1970 [46]. Combined with strong results for certain classes

of structures in dimensions d ≥ 3, the work clarified the difficulties of char-

acterizing rigid ‘generic’ frameworks in 3- and higher-dimensional space.

During the 1970s, remarkable results have also been found in the theory

of flexible polyhedra: in 1975 H. Gluck was able to refine Euler’s famous

rigidity conjecture from 1766 [30], and in 1977 R. Connelly finally settled the

3

conjecture with his celebrated counterexample, the flexible ‘Connelly sphere’

[11].

(B) An expanding web of connections to problems in other fields where the

mathematical theory of rigidity makes substantial contributions to clarifying

and resolving central questions has arisen.

Starting with the connections with mechanical and civil engineering (hu-

man built structures), there are connections to the general theory of geomet-

ric constraints built into Computer Aided Design (CAD) and to computa-

tions for robotic motions. Less obvious, but very real, are the connections

to computer vision/recognition of geometric objects from geometric data.

In general, a sequence of problems in computational geometry have found

‘rigidity-type’ results and methods contribute to their understanding and

their solutions (as well as sometimes clarifying that the problems are ‘hard’).

In turn, computational geometry has contributed new problems and algo-

rithmic insights to the general theory of rigidity.

One step removed are the connections to scaled down natural structures,

such as granular materials which are sometimes modeled with packings of

spheres, abstracted as bar and joint frameworks. Over the last two decades,

there has also been a strong interest in applying rigidity theory to rapid

predictions of the rigid and flexible regions of large biomolecules, such as

proteins. Such predictions, even based on incomplete theories, are imple-

mented on the web because of the importance of flexibility and rigidity to

the function of biomolecules and the design of drugs to alter their functioning

[82].

Overall, both the elegant and expanding mathematical theory and the

4

growing network of applications have made rigidity theory a rich area for

research, and a source of new questions and new insights into a wide array

of pure and applied mathematical theories.

Symmetry is another central idea in geometry - and appears widely in

both natural structures such as crystals and biomolecules, and in structures

built by humans. One source of such symmetry is the efficiency of forming

the shape using multiple copies of a few key components.

Because the appearance of symmetry is wide-spread, sometimes ubiqui-

tous, there have recently been a series of papers by engineers and chemists,

which present criteria for rigidity and flexibility of symmetric frameworks.

The first of these breakthrough papers is due to the engineers R. Kangwai

and S. Guest: in 2000 they observed that the rigidity matrix of a symmet-

ric framework can be put into a block-diagonalized form using techniques

from group representation theory [44]. Using this result, some engineers and

chemists were able to make some further interesting and useful observations

concerning the rigidity of symmetric frameworks (see [25, 43], for example).

Many of these observations, however, are incomplete from a mathematical

point of view, since they are not presented with a mathematically precise

formulation nor with a thorough mathematical foundation or proofs. So

while this work has resulted in some important heuristics for engineers and

chemists to gain further insight into the rigidity properties of a symmetric

framework, there has not been a rigorous mathematical investigation of how

symmetry impacts the rigidity of frameworks. While this thesis was being

written, the mathematicians J.C. Owen and S.C. Power have also been work-

ing on grounding aspects of this theory [53]. In Chapter 4 we will describe

5

the one key area of overlap between their work and this thesis.

Again, the focus on classes of frameworks with given symmetries also has

both a combinatorial (graph automorphism) level and a geometric aspect

(point group, spatial isometries). In our study of these connections, we will

see some clearly combinatorial conditions about fixed vertices, edges, etc.

in the graph automorphisms which must show up in the geometry of the

realization.

When does this necessary geometry of symmetry from the graph auto-

morphisms and the geometry of reflections, rotations, and other isometries

force the rank of the rigidity matrix to drop? When does the geometry

of symmetry overlap with the geometry of the singular positions which are

traditionally expressed in projective form? We will see that under many cir-

cumstances, the addition of symmetry does not change the rigidity predicted

for more general, asymmetric realizations. In other circumstances there are

very simply stated added conditions.

The simplest example is the point group C3 in dimension 2 (Z3 as an

abstract group) which describes 3-fold rotational symmetry. We will see that

the combinatorial condition, once there is a group of graph automorphisms

associated with the point group C3, is simply that no vertices are fixed by

the automorphism corresponding to the 3-fold rotation (geometrically, no

vertices are placed on the center of rotation). This is a necessary condition

for any independent and rigid realization of the graph as a framework with

C3 symmetry (Chapter 4). It is also a sufficient condition for the most gen-

eral ‘C3-symmetric realizations’ to be isostatic (Chapter 5). The result is

striking in its simplicity: to test a ‘generic’ framework with C3 symmetry for

6

isostaticity, we just need to check the number of fixed vertices, as well as the

standard conditions for rigidity without symmetry.

More generally, the techniques in Chapter 4 work with counts of vertices

and edges, and counts of vertices and edges fixed by various elements of the

group. So the necessary conditions will always be of this type: vertices or

edges fixed by the automorphisms. The examination of when these necessary

conditions are also sufficient is the larger theme of Chapter 5, for an array

of plane symmetry groups. In fact, a collection of rigidity theory methods

which can now be called ’classical’ are symmetrized in Chapter 5 to establish

both necessary and sufficient conditions for realizations which are generic

with the given symmetry to be isostatic.

For general frameworks, an undercount of constraints becomes a predic-

tion of flexibility. For symmetric frameworks, we will show that there is an

extension that not only predicts finite motions, but predicts motions which

preserve the symmetries throughout their path (Chapter 6).

The interactions of combinatorics, geometry, and symmetry are rich. It is

no surprise that at almost every turn, we find not only fascinating and appeal-

ing results, but also possible extensions to explore as well as new questions

and new generalizations which can be conjectured and anticipated. We in-

vite the interested reader to join us in the exploration of these landscapes of

possibilities.

1.2 Outline of thesis

The thesis is organized as follows.

7

In Chapter 2, we give a brief introduction to rigidity, its linearized versions

infinitesimal and static rigidity, as well as generic (or combinatorial) rigidity.

We also introduce suitable mathematical definitions for the relevant terms

relating to symmetric structures that are frequently used in the chemistry

and engineering literature. In particular, we give a detailed mathematical

description of the Schoenflies notation for point groups in dimensions 2 and

3. We will be using this notation for all the examples throughout this thesis.

In Chapter 3, we introduce a natural classification of symmetric frame-

works. This classification is fundamental to all the results of this thesis. We

then define a symmetry-adapted notion of a ‘generic’ framework with respect

to this classification. This symmetrized notion of generic has the two funda-

mental properties that ‘almost all’ realizations in a given symmetry class are

generic and all generic realizations in this class share the same infinitesimal

rigidity properties. This classification therefore not only lays the foundation

for symmetrizing results in rigidity, infinitesimal rigidity, and static rigidity,

but it also allows us to develop a symmetry-adapted version of generic rigid-

ity theory in Chapter 5.

In the last two sections of the third chapter, we carefully examine the dif-

ficulties that arise in applying techniques from group representation theory

to the analysis of symmetric frameworks with non-injective configurations.

More precisely, in Section 3.3, we show that a framework with a non-injective

configuration can belong to more than one symmetry class, and we examine

how many distinct symmetry classes a given framework can possibly belong

to. In Section 3.4, we investigate under what conditions techniques from

group representation theory can be applied to the frameworks in a given

8

symmetry class.

All the results of the third chapter can be found in the manuscript [55]

which has been submitted for review.

Chapter 4 concerns the application of techniques from group representa-

tion theory to the rigidity analysis of symmetric frameworks. In Section 4.1,

we first give a complete self-contained mathematical proof that the rigid-

ity matrix of a symmetric framework can be block-diagonalized as described

by R. Kangwai and S. Guest in [44]. In Section 4.2, we use this result to

give a detailed proof for the symmetry-extended version of Maxwell’s rule

given by P. Fowler and S. Guest in [25]. This rule provides further necessary

conditions (in addition to Maxwell’s original condition from 1864 [49]) for a

symmetric framework (G, p) to be isostatic. While the rule in [25] is only

applicable to 2- or 3-dimensional frameworks with injective configurations,

we establish a more general result in this thesis, namely a rule that can be

applied to both injective and non-injective realizations in all dimensions.

The results of Sections 4.1 and 4.2 will constitute the main part of [56].

An alternate proof for the rule given in [25], as well as various general-

izations of this rule to other types of geometric constraint systems, is given

by J.C. Owen and S.C. Power in [53].

In Section 4.3, we show that the symmetry-extended version of Maxwell’s

rule can be used to prove that a symmetric isostatic framework must obey

some very simply stated restrictions on the number of joints and bars that are

‘fixed’ by various symmetry operations of the framework. In particular, these

restrictions imply that the symmetries of a 2-dimensional isostatic framework

must belong to one of only six possible point groups. For 3-dimensional iso-

9

static frameworks, all point groups are possible, although restrictions on the

placement of structural components still apply.

The main part of Section 4.3 is a mathematically explicit derivation of

the results presented without proof in [15]. This paper is joint work with R.

Connelly, P. Fowler, S. Guest, and W. Whiteley.

Finally, in Section 4.4, we use the results of the previous sections to also

establish necessary conditions for a symmetric framework to be independent

or infinitesimally rigid.

In Chapter 5, we present symmetric versions of some famous results

in generic rigidity theory. Given a graph G, Laman’s Theorem says that

Maxwell’s condition in 2D, i.e., |E(G)| = 2|V (G)| − 3, together with the

counts |E(H)| ≤ 2|V (H)| − 3 for all non-trivial subgraphs H of G, are nec-

essary and sufficient for all generic 2-dimensional realizations of G to be

isostatic. There are well known difficulties in extending this result to higher

dimensions (see [32, 33, 46], for example). Using the symmetry-adapted no-

tion of ‘generic’ introduced in Chapter 3, we establish symmetric versions of

Laman’s Theorem for three of the five non-trivial symmetry groups in dimen-

sion 2 that allow isostatic frameworks, namely for the groups C2 and C3 of

order 2 and 3 generated by a half-turn and a 3-fold rotation, respectively, and

for the group Cs of order 2 generated by a reflection. More precisely, we show

that for each of these groups, the conditions derived from the symmetry-

extended version of Maxwell’s rule, together with the Laman conditions, are

necessary and sufficient for realizations of G that are ‘generic’ within the

given symmetry class to be isostatic. These results were conjectured in [15].

Henneberg’s Theorem and Crapo’s Tree Covering Theorem are also fa-

10

mous combinatorial results that provide characterizations of generically 2-

isostatic graphs [20, 40, 33, 67, 68]. We show that for each of the symmetry

groups C2, C3 and Cs, there exist symmetric versions of these results as well.

The other two non-trivial symmetry groups in dimension 2 that allow iso-

static frameworks are the dihedral groups of order 4 and 6. For these groups,

we offer some analogous conjectures. To prove these conjectures with tech-

niques similar to the ones used for the results above, one has to consider an

unreasonably large number of cases.

In the final section of Chapter 5, we briefly discuss ‘symmetric-generically’

isostatic graphs in dimension 3.

The key results of the fifth chapter will be summarized in [57].

In Chapter 6, we study finite flexes of symmetric frameworks, i.e., flexes

that move the joints of a given framework on differentiable displacement

paths while holding the lengths of all bars fixed and changing the distance

between two unconnected joints. We prove that if a framework (G, p) is

‘generic’ within a given symmetry class and there exists a ‘fully-symmetric’

infinitesimal flex of (G, p) (i.e., the velocity vectors of the infinitesimal flex

remain unaltered under all symmetry operations of (G, p)), then (G, p) also

possesses a ‘symmetry-preserving’ finite flex, i.e., a flex which displaces the

joints of (G, p) in such a way that all the resulting frameworks have the same

symmetry as (G, p) (or possibly higher symmetry). This and other related

results are obtained by symmetrizing techniques described by L. Asimov and

B. Roth in [3] and by using the fact that the rigidity matrix of a symmetric

framework can be transformed into a block-diagonalized form as shown in

Chapter 4. As corollaries of these results, one obtains the results stated (but

11

not rigorously proven) in [35] and Proposition 1 in [43].

The finite flexes that can be detected with these symmetry-based methods

can in general not be found with the analogous non-symmetric methods.

The work of Chapter 6 will also be presented in [58].

Finally, in Chapter 7, we outline how the methods developed in this thesis

can be extended to analyze the rigidity and flexibility properties of various

other types of symmetric structures.

Several additional promising directions for future work are also presented

in this final chapter.

12

Chapter 2

Definitions and preliminaries

2.1 Graph theory terminology

We begin by establishing the graph theory vocabulary and notation we

will be using throughout this thesis.

Definition 2.1.1 A graph G is a finite nonempty set of objects called ver-

tices together with a (possibly empty) set of unordered pairs of distinct

vertices of G called edges. The vertex set of G is denoted by V (G) and the

edge set of G is denoted by E(G).

Definition 2.1.2 Two vertices u 6= v of a graph G are adjacent if u, v ∈E(G), and independent otherwise. A set S of vertices of G is independent if

every two vertices of S are independent.

Definition 2.1.3 Let G be a graph. The neighborhood NG(v) of a vertex

v ∈ V (G) is the set of all vertices that are adjacent to v and the elements of

13

NG(v) are called the neighbors of v.

Definition 2.1.4 Let G be a graph and e = u, v be an edge of G. Then

we say that u and e are incident, as are v and e. The valence valG(v) of

a vertex v ∈ V (G) is the number of edges of G that are incident with v.

Equivalently, valG(v) = |NG(v)|.

Definition 2.1.5 A graph is called complete if every two of its vertices are

adjacent. We write Kn for the complete graph on n vertices.

A graph G is called bipartite if the vertex set V (G) can be partitioned into

two sets X and Y (called partite sets) such that for every edge x, y ∈ E(G)

we have x ∈ X and y ∈ Y . A bipartite graph G with partite sets X and Y

is complete if x, y ∈ E(G) for all x ∈ X and y ∈ Y . We write Km,n for the

complete bipartite graph whose partite sets have cardinality m and n.

Definition 2.1.6 A graph H is a subgraph of a graph G if V (H) ⊆ V (G)

and E(H) ⊆ E(G), in which case we write H ⊆ G. A subgraph H of a graph

G is called spanning if |V (H)| = |V (G)|.

The simplest type of subgraph of a graph G is that obtained by deleting

a vertex or an edge from G. Let v be a vertex and e be an edge of G. Then

we write G − v for the subgraph of G that has V (G) \ v as its vertex

set and whose edges are those of G that are not incident with v. Similarly,

we write G− e for the subgraph of G that has V (G) as its vertex set and

E(G) \ e as its edge set. The deletion of a set of vertices or a set of edges

from G is defined and denoted analogously.

If u and v are independent vertices of G, then we write G +u, v for

14

the graph that has V (G) as its vertex set and E(G) ∪ u, v as its edge

set. The addition of a set of edges is again defined and denoted analogously.

Definition 2.1.7 Let G be a graph and U be a nonempty subset of V (G).

Then the subgraph 〈U〉 of G induced by U is the graph having vertex set U

and whose edges are those of G that are incident with two elements of U .

Definition 2.1.8 Let G1 and G2 be two graphs. The intersection G =

G1∩G2 is the graph with V (G) = V (G1)∩V (G2) and E(G) = E(G1)∩E(G2).

Similarly, the union G = G1 ∪G2 is the graph with V (G) = V (G1) ∪ V (G2)

and E(G) = E(G1) ∪ E(G2).

Definition 2.1.9 An automorphism of a graph G is a permutation α of

V (G) such that u, v ∈ E(G) if and only if α(u), α(v) ∈ E(G).

The automorphisms of a graph G form a group under composition which

is denoted by Aut(G).

Definition 2.1.10 Let H be a subgraph of a graph G and α ∈ Aut(G).

We define α(H) to be the subgraph of G that has α(V (H)

)as its vertex

set and α(E(H)

)as its edge set, where u, v ∈ α

(E(H)

)if and only if

α−1(u, v) = α−1(u), α−1(v) ∈ E(H).

We say that H is invariant under α if α(V (H)

)= V (H) and α

(E(H)

)=

E(H), in which case we write α(H) = H.

Example 2.1.1 The graph G in Figure 2.1 (a) has the automorphism

α = (v1 v2 v3)(v4 v5 v6). The subgraph H1 of G is invariant under α, but

the subgraph H2 of G is not, because α(E(H2)

) 6= E(H2).

15

...v3..v1

..v2

..v4

..v5

..v6

.G:

.(a)

...v3..v1

..v2

.H1:

.(b)

...v3..v1

..v2

.H2:

.(c)

Figure 2.1: An invariant (b) and a non-invariant subgraph (c) of the graph

G under α = (v1 v2 v3)(v4 v5 v6) ∈ Aut(G).

Definition 2.1.11 Let u and v be two (not necessarily distinct) vertices

of a graph G. A u-v path in G is a finite alternating sequence u =

u0, e1, u1, e2, . . . , uk−1, ek, uk = v of vertices and edges of G in which no vertex

is repeated and ei = ui−1, ui for i = 1, 2, . . . , k. A u-v path is called a cycle

if k ≥ 3 and u = v.

Let a u-v path P in G be given by u = u0, e1, u1, e2, . . . , uk−1, ek, uk = v

and let α ∈ Aut(G). Then we denote α(P ) to be the α(u)-α(v) path

α(u) = α(u0), α(e1), α(u1), α(e2), . . . , α(uk−1), α(ek), α(uk) = α(v) in G.

A vertex u is said to be connected to a vertex v in G if there exists a u−v

path in G. A graph G is connected if every two vertices of G are connected.

A graph with no cycles is called a forest and a connected forest is called

a tree.

A connected subgraph H of a graph G is a component of G if H = H ′

whenever H ′ is a connected subgraph of G containing H.

16

2.2 Introduction to rigidity theory

We now give a brief introduction to rigidity, its linearized versions in-

finitesimal and static rigidity, as well as generic rigidity, as we shall sym-

metrize results from each of these theories. The definitions and results listed

in this introduction are widely used in the rigidity theory literature so that

we will omit the proofs and leave more detailed explanations and illustrations

to be found in the references provided.

2.2.1 Rigidity

Definition 2.2.1 [32, 33, 81, 83] A framework (in Rd) is a pair (G, p), where

G is a graph and p : V (G) → Rd is a map with the property that p(u) 6= p(v)

for all u, v ∈ E(G). We also say that (G, p) is a d-dimensional realization

of the underlying graph G.

Given the vertex set V (G) = v1, . . . , vn of a graph G and a map p :

V (G) → Rd, it is often useful to identify p with a vector in Rdn by using the

order on V (G). In this case we also refer to p as a configuration of n points

in Rd.

Throughout this thesis we will simplify our notation by not differentiating

between an abstract vector and its coordinate vector relative to the canonical

basis.

Definition 2.2.2 Let (G, p) be a framework in Rd. A joint of (G, p) is

an ordered pair(v, p(v)

), where v ∈ V (G). A bar of (G, p) is an unordered

pair(

u, p(u)),(v, p(v)

)of joints of (G, p), where u, v ∈ E(G). We define

17

‖p(u)−p(v)‖ to be the length of the bar(

u, p(u)),(v, p(v)

), where ‖p(u)−

p(v)‖ is defined by the canonical inner product on Rd.

Note that we allow the map p of a framework (G, p) to be non-injective,

that is, two distinct joints(u, p(u)

)and

(v, p(v)

)of (G, p) may be located

at the same point p(u) = p(v) in Rd, provided that u and v are independent

vertices of G. However, if u, v ∈ E(G), then p(u) 6= p(v), and hence every

bar(

u, p(u)),(v, p(v)

)of (G, p) has a strictly positive length.

Definition 2.2.3 [32] Let (G, p) be a framework in Rd with V (G) =

v1, v2, . . . , vn. A motion of (G, p) is an indexed family of functions

Pi : [0, 1] → Rd, i = 1, 2, . . . , n, so that

(i) Pi(0) = p(vi) for all i;

(ii) Pi(t) is differentiable on [0, 1] for all i;

(iii) ‖Pi(t)− Pj(t)‖ = ‖p(vi)− p(vj)‖ for all t ∈ [0, 1] and vi, vj ∈ E(G).

A motion of a framework (G, p) displaces the joints of (G, p) on differen-

tiable displacement paths while preserving the lengths of all bars of (G, p).

Every framework has some trivial motions, namely those that correspond to

rigid motions of space (i.e., translations, rotations and their combinations).

Definition 2.2.4 A motion Pi of a framework (G, p) with V (G) =

v1, v2, . . . , vn is called a rigid motion if it preserves the distances between

every pair of joints of (G, p), that is, if ‖Pi(t)− Pj(t)‖ = ‖p(vi)− p(vj)‖ for

all t ∈ [0, 1] and all 1 ≤ i < j ≤ n.

18

Pi is called a flex if the distance between at least one pair of joints of

(G, p) is changed by Pi, that is, if ‖Pi(t)− Pj(t)‖ 6= ‖p(vi)− p(vj)‖ for all

t ∈ (0, 1] and some vi, vj /∈ E(G).

Definition 2.2.5 A framework (G, p) is called rigid if every motion of (G, p)

is a rigid motion. Otherwise (G, p) is called flexible.

.. .

.

.(a)

.. .

..

.(b)

.. .

.. . .

.(c)

.. .

. .

.(d)

Figure 2.2: A rigid (a) and a flexible (b) framework in the plane. The flex

shown in (c) takes the framework in (b) to the framework in (d).

Some alternate definitions of a rigid framework are common in the litera-

ture [3, 83] all of which are equivalent to Definition 2.2.5. We will introduce

some of these definitions in Chapter 6, where we examine the motions of

symmetric frameworks.

2.2.2 Infinitesimal rigidity

It is in general very difficult to determine whether a given framework is

rigid or not since it requires solving a system of quadratic equations. It is

therefore common to linearize this problem by differentiating the equations

in Definition 2.2.3 (iii). This gives rise to

Definition 2.2.6 [32, 33, 81, 83] Let (G, p) be a framework in Rd with

V (G) = v1, v2, . . . , vn. An infinitesimal motion of (G, p) is a function

19

u : V (G) → Rd such that

(p(vi)− p(vj)

) · (u(vi)− u(vj))

= 0 for all vi, vj ∈ E(G). (2.1)

An infinitesimal motion of a framework (G, p) is a set of displacement

vectors u(vi), one at each joint, that neither stretch nor compress the bars

of (G, p) at first order. More precisely, condition (2.1) says that for every

edge vi, vj ∈ E(G), the projections of u(vi) and u(vj) onto the line through

p(vi) and p(vj) have the same direction and the same length (see also Figure

2.3).

Definition 2.2.7 An infinitesimal motion u of a framework (G, p) with

V (G) = v1, v2, . . . , vn is called an infinitesimal rigid motion if there ex-

ists a rigid motion Pi of (G, p) such that for i = 1, 2, . . . , n, the vector

u(vi) is the derivative (at t = 0) of Pi. Otherwise, u is called an infinitesimal

flex of (G, p).

Remark 2.2.1 Let G be a graph with V (G) = v1, v2, . . . , vn and let u

be an infinitesimal motion of a d-dimensional realization (G, p) of G. If(p(vi) − p(vj)

) · (u(vi) − u(vj)) 6= 0 for some vi, vj /∈ E(G), then u is

an infinitesimal flex of (G, p). If the points p(v1), . . . , p(vn) span all of Rd

(in an affine sense), then the converse also holds, i.e., in this case, u is an

infinitesimal flex of (G, p) if and only if(p(vi)− p(vj)

) · (u(vi)− u(vj)) 6= 0

for some vi, vj /∈ E(G) or equivalently, u is an infinitesimal rigid motion of

(G, p) if and only if(p(vi)− p(vj)

) · (u(vi)− u(vj))

= 0 for all 1 ≤ i < j ≤ n

[32, 33, 81].

From now on, when we say that a set of points spans a space, then this

will always be in the affine sense.

20

...p1

..p2

.u1

.u2

.(a)

...p1

..p2

..p3

.u3.u1 = 0 .u2 = 0

.(b)

...p6

..p1

..p2

..p3

. .p4

. .p5

.u6

.u1

.u2

.u3.u4

.u5

.(c)

Figure 2.3: The arrows indicate the non-zero displacement vectors of an in-

finitesimal rigid motion (a) and infinitesimal flexes (b, c) of frameworks in

R2.

Definition 2.2.8 [32, 33, 81, 83] A framework (G, p) is infinitesimally rigid

if every infinitesimal motion of (G, p) is an infinitesimal rigid motion. Oth-

erwise (G, p) is said to be infinitesimally flexible.

The following theorem gives the main connection between rigidity and

infinitesimal rigidity. A proof of this result can be found in [3], [17] or [30],

for example.

Theorem 2.2.1 If a framework (G, p) is infinitesimally rigid, then (G, p) is

rigid.

Under certain conditions, rigidity and infinitesimal rigidity are equivalent.

We will give the relevant results in the end of Section 2.2.5 after we have

established the necessary definitions.

For a framework (G, p) whose underlying graph G has a vertex set that is

indexed from 1 to n, say V (G) = v1, v2, . . . , vn, we will frequently denote

21

p(vi) by pi for i = 1, 2, . . . , n. Similarly, for an infinitesimal motion u of

(G, p), we will frequently denote u(vi) by ui for all i. The kth component of

a vector x is denoted by (x)k.

The equations stated in Definition 2.2.6 form a system of linear equations

whose corresponding matrix is called the rigidity matrix. This matrix is

fundamental in the study of both infinitesimal and static rigidity.

Definition 2.2.9 [32, 33, 81, 83] Let G be a graph with V (G) =

v1, v2, . . . , vn and let p : V (G) → Rd. The rigidity matrix of (G, p) is

the |E(G)| × dn matrix

..R(G, p) =

...

0 . . . 0 pi − pj 0 . . . 0 pj − pi 0 . . . 0

...

.edge vi, vj,

.vi .vj.v1 .vn

that is, for each edge vi, vj ∈ E(G), R(G, p) has the row with (pi −pj)1, . . . , (pi−pj)d in the columns d(i−1)+1, . . . , di, (pj−pi)1, . . . , (pj−pi)d

in the columns d(j − 1) + 1, . . . , dj, and 0 elsewhere.

Remark 2.2.2 The rigidity matrix is defined for arbitrary pairs (G, p),

where G is a graph and p : V (G) → Rd is a map. If (G, p) is not a frame-

work, then there exists a pair of adjacent vertices of G that are mapped to

the same point in Rd under p and every such edge of G gives rise to a zero-row

in R(G, p).

If we identify an infinitesimal motion of a d-dimensional framework (G, p)

with a column vector in Rd|V (G)| (by using the order on V (G)), then the

22

kernel of R(G, p) is the space of infinitesimal motions of (G, p). It is well

known that the infinitesimal rigid motions arising from d translations and(

d2

)rotations of Rd form a basis for the space of infinitesimal rigid motions

of (G, p), provided that the points p1, . . . , pn span an affine subspace of Rd

of dimension at least d − 1 [33, 81]. Thus, for such a framework (G, p), we

have nullity(R(G, p)

) ≥ d +(

d2

)=

(d+12

)and (G, p) is infinitesimally rigid

if and only if nullity(R(G, p)

)=

(d+12

)or equivalently, rank

(R(G, p)

)=

d|V (G)| − (d+12

).

Theorem 2.2.2 [3, 30] A framework (G, p) in Rd is infinitesimally rigid if

and only if either rank(R(G, p)

)= d|V (G)|−(

d+12

)or G is a complete graph

Kn and the points p(v), v ∈ V (G), are affinely independent.

Remark 2.2.3 Let 1 ≤ m ≤ d and let (G, p) be a framework in Rd. If

(G, p) has at least m + 1 joints and the points p(v), v ∈ V (G), span an

affine subspace of Rd of dimension less than m, then (G, p) is infinitesimally

flexible (recall Figure 2.3 (b)). In particular, if (G, p) is infinitesimally rigid

and |V (G)| ≥ d, then the points p(v), v ∈ V (G), span an affine subspace of

Rd of dimension at least d− 1.

2.2.3 Static rigidity

We now also give a brief introduction to the static approach to rigidity.

The intuitive test for static rigidity of a framework (G, p) is to apply an

external load to (G, p) (i.e., a set of forces, one to each joint) and investigate

whether there exists a set of tensions and compressions in the bars of (G, p)

that reach an equilibrium with this load at the joints (see also Figure 2.4).

23

Of course only loads which do not correspond to a translation or rotation of

space can possibly be resolved in this way.

Definition 2.2.10 [21, 68, 76, 81] Let (G, p) be a framework in Rd with

V (G) = v1, v2, . . . , vn. A load on (G, p) is a function l : V (G) → Rd, where

for i = 1, 2, . . . , n, the vector l(vi) represents a force applied to the joint(vi, pi

)of (G, p).

A load l on (G, p) is called an equilibrium load if l satisfies

(i)∑n

i=1 li = 0;

(ii)∑n

i=1

((li)j(pi)k − (li)k(pi)j

)= 0 for all 1 ≤ j < k ≤ d,

where li denotes the vector l(vi) for each i.

The physical intuition for conditions (i) and (ii) in Definition 2.2.10 is

the following: condition (i) rules out loads that would produce a translation

of (G, p) and (ii) says that there is no net rotational twist of (G, p).

Definition 2.2.11 [21, 68, 76, 81] Let l be an equilibrium load on a frame-

work (G, p) in Rd with V (G) = v1, v2, . . . , vn. A resolution of l by (G, p)

is a function ω : E(G) → R such that at each joint(vi, pi

)of (G, p) we have

∑

j with vi,vj∈E(G)

ωij(pi − pj) + li = 0,

where ωij denotes ω(vi, vj) for all vi, vj ∈ E(G).

The scalars ωij represent tensions (ωij < 0) and compressions (ωij > 0)

in the bars of (G, p), so that the bar forces reach an equilibrium with li at

each joint(vi, pi

).

24

.. .

.(a)

. .

.(b)

.. .

.

.

.(c)

.. .

.

.

.(d)

.. ..

.

.(e)

Figure 2.4: (a), (b) The arrows indicate a tension (a) and a compression (b)

in a bar. (c) An equilibrium load on a non-degenerate triangle. This load can

be resolved by the triangle as shown in (d). (e) An unresolvable equilibrium

load on a degenerate triangle: for any joint of this framework, tensions or

compressions in the bars cannot reach an equilibrium with the load vector at

this joint.

Definition 2.2.12 [21, 68, 76, 81] A framework (G, p) is statically rigid if

every equilibrium load on (G, p) has a resolution by (G, p).

Note that if we identify l and ω with a column vector in Rdn and R|E(G)|,

respectively, then (after changing the sign of l) the equations in Definition

2.2.11 can be written in a compact form in terms of the rigidity matrix

R(G, p) as

R(G, p)T ω = l.

Let (vh, ph) and (vk, pk) be two joints of (G, p). Then it is easy to see that

the column vector Fhk, where

(Fhk)T = (0, . . . , 0, ph − pk, 0, . . . , 0, pk − ph, 0, . . . , 0),

is an equilibrium load on (G, p). Further, if vi, vj ∈ E(G), then (Fij)T

is the row vector of R(G, p) that corresponds to vi, vj and Fij is clearly

25

resolved by the bar (vi, pi), (vj, pj) of (G, p). Note that if (G, p) is statically

rigid, then Fhk has a resolution by (G, p) for every pair (vh, ph), (vk, pk) of

joints of (G, p) (even if vh, vk /∈ E(G)).

If the points p1, . . . , pn span all of Rd, then the converse also holds, since

in this case, the vectors Fhk, 1 ≤ h < k ≤ n, generate the entire space of

equilibrium loads on (G, p) (see [76]). This space is a subspace of Rdn of

dimension dn− (d+12

)(defined by the equations in Definition 2.2.10).

Thus, if we want to test such a framework (G, p) for static rigidity, we

need to investigate whether the rows of R(G, p) generate a space of dimension

dn− (d+12

), that is, the entire space of equilibrium loads on (G, p). In other

words, we need to investigate whether

rank(R(G, p)T

)= dn−

(d + 1

2

).

So, the essential information for both infinitesimal and static rigidity of

a framework (G, p) is comprised by the rigidity matrix R(G, p). While in

infinitesimal rigidity, we investigate the column space and column rank of

R(G, p), in static rigidity, we investigate the row space and row rank of

R(G, p). In the light of these remarks, the following fundamental facts do

not come as a surprise.

Theorem 2.2.3 [54] The load Fhk on a framework (G, p) has no resolution

by (G, p) if and only if there exists an infinitesimal motion u of (G, p) with

(ph − pk) · (uh − uk) 6= 0.

Theorem 2.2.4 [40, 54, 73] A framework (G, p) is infinitesimally rigid if

and only if (G, p) is statically rigid.

26

Theorem 2.2.4 allows us to use the terms infinitesimally rigid and stati-

cally rigid interchangeably.

Definition 2.2.13 [21, 76, 81] Given a framework (G, p), a function ω :

E(G) → R is called a stress of (G, p). Equivalently, a stress of (G, p) is a

resolution of a load on (G, p). A resolution of the zero-load is called a self-

stress of (G, p). So, if we identify a stress of (G, p) with a vector in R|E(G)|,

then a vector ω ∈ R|E(G)| is a self-stress of (G, p) if R(G, p)T ω = 0.

The framework (G, p) is said to be independent if the rigidity matrix

R(G, p) has linearly independent rows. Equivalently, (G, p) is independent if

(G, p) has no non-zero self-stress. Otherwise, (G, p) is said to be dependent.

Definition 2.2.14 A framework (G, p) is isostatic if it is infinitesimally (or

statically) rigid and independent.

The rows of the rigidity matrix of an isostatic framework (G, p) form a

basis for the space of equilibrium loads on (G, p), provided that the points

p(v), v ∈ V (G), span all of Rd.

Theorem 2.2.5 [33, 83] For a d-dimensional realization (G, p) of a graph

G with |V (G)| ≥ d, the following are equivalent:

(i) (G, p) is isostatic;

(ii) (G, p) is infinitesimally rigid and |E(G)| = d|V (G)| − (d+12

);

(iii) (G, p) is independent and |E(G)| = d|V (G)| − (d+12

);

27

(iv) (G, p) is minimal infinitesimally rigid, i.e., (G, p) is infinitesimally

rigid and the removal of any bar results in a framework that is not

infinitesimally rigid.

2.2.4 Generic rigidity

Generic rigidity is concerned with the infinitesimal (or equivalently, static)

rigidity of ‘almost all’ geometric realizations of a given graph. The following

standard definition of ‘generic’ specifies what we mean by ‘almost all’.

Definition 2.2.15 Let Kn be the complete graph on n vertices with

V (Kn) = v1, v2, . . . , vn. For each i = 1, 2, . . . , n, we introduce a d-tuple

p′i =((p′i)1, . . . , (p

′i)d

)of variables and let

R(n, d) =

...

0 . . . 0 p′i − p′j 0 . . . 0 p′j − p′i 0 . . . 0

...

be the matrix that is obtained from the rigidity matrix R(Kn, p) of a d-

dimensional realization (Kn, p) by replacing each (pi)j ∈ R with the variable

(p′i)j. We call R(n, d) the d-dimensional indeterminate rigidity matrix of

Kn.

Definition 2.2.16 [32, 33] Let V = v1, v2, . . . , vn and p : V → Rd be a

map. Further, let Kn be the complete graph with V (Kn) = V .

We say that p is generic if the determinant of any submatrix of R(Kn, p)

is zero only if the determinant of the corresponding submatrix of R(n, d) is

(identically) zero.

A framework (G, p) is said to be generic if p is a generic map.

28

There are two fundamental facts regarding this definition of generic. First,

the set of all non-generic maps p of a finite set V = v1, v2, . . . , vn to Rd

is a closed set of measure zero [33]. To see this, identify p with a vector in

Rdn and observe that the determinant of every submatrix of R(Kn, p) is a

polynomial in the variables (p′i)j. If such a polynomial is not identically zero,

then, by general algebraic geometry, it is non-zero for an open dense set of

p ∈ Rdn. Since R(Kn, p) has only finitely many minors, the set of generic

p ∈ Rdn is still an open dense subset of Rdn.

Secondly, the infinitesimal rigidity properties are the same for all generic

realizations of a graph G, as the following result shows:

Theorem 2.2.6 [32, 33, 83] For a graph G and a fixed dimension d, the

following are equivalent:

(i) (G, p) is infinitesimally rigid (independent, isostatic) for some map

p : V (G) → Rd;

(ii) every d-dimensional generic realization of G is infinitesimally rigid (in-

dependent, isostatic).

It follows that for generic frameworks, infinitesimal (and static) rigidity

is purely combinatorial, and hence a property of the underlying graph. This

gives rise to the following definition of infinitesimal rigidity for graphs:

Definition 2.2.17 A graph G is called generically rigid (independent, iso-

static) in dimension d or generically d-rigid (d-independent, d-isostatic) if

d-dimensional generic realizations of G are infinitesimally rigid (independent,

isostatic).

29

An easy but often useful observation concerning generic frameworks is

that if a framework (G, p) in Rd is generic, then the joints of (G, p) are in

general position, that is, for 1 ≤ m ≤ d, no m + 1 joints of (G, p) lie in an

m− 1-dimensional affine subspace of Rd [33].

In Chapter 3, we introduce a natural classification of symmetric frame-

works and introduce a symmetry-adapted notion of generic with respect to

this classification.

Remark 2.2.4 There are some other notions of a ‘generic’ realization of

a graph G that are commonly used in rigidity theory (see, for example,

[14, 48, 80]). One such notion of generic is obtained by replacing the matri-

ces R(Kn, p) and R(n, d) in Definition 2.2.16 with the rigidity matrix R(G, p)

and the matrix R(G)(n, d) which is obtained from R(n, d) by deleting those

rows that do not correspond to edges of G, respectively [80]. We shall refer

to this definition of generic as G-generic.

Clearly, if a map p : V (G) → Rd is generic, then p is also G-generic.

Moreover, like the set of all generic realizations of a graph G, the set of all

G-generic realizations of G is also a dense open subset of Rdn. However, the

fact that the property of being generic (in the sense of Definition 2.2.16) is

invariant under addition or deletion of edges in G makes this definition a

more convenient one for our purposes than the definition of G-generic.

A configuration p ∈ Rdn is frequently also defined to be ‘generic’ if the

coordinates of p are algebraically independent over Z, i.e., if there does

not exist a polynomial h(x1, . . . , xdn) with integer coefficients such that

h((p1)1, . . . , (pn)d

)= 0 [14, 48]. We refer to this definition of generic as

A-generic.

30

The set of all A-generic realizations of G is a dense, but not an open

subset of Rdn. The definition of A-generic is therefore not a very suitable

definition of generic for our purposes.

The relationship between all of these different types of generic realizations

of a given graph G is illustrated in Figure 2.5 by means of a Venn diagram.

.

.A-generic configurations

.generic configurations

.G-generic configurations

.regular points of G

Figure 2.5: A Venn diagram showing the relationship between sets of various

types of ‘generic’ configurations and the set of regular points of a graph G

(see Definition 2.2.22 in the end of Section 2.2.5).

2.2.5 Basic rigidity results

In this section, we give a number of important results (as well as some

definitions) in rigidity theory that we will later extend to frameworks that

are realized with certain symmetries.

In 1864, J. C. Maxwell gave a necessary, but not sufficient condition for

a 2- or 3-dimensional framework to be isostatic [49]. The following theorem

is the d-dimensional version of this condition.

31

Theorem 2.2.7 (Maxwell’s rule, 1864) Let (G, p) be a d-dimensional re-

alization of a graph G with |V (G)| ≥ d. If (G, p) is isostatic then

|E(G)| = d|V (G)| −(

d + 1

2

).

Let (G, p) be a framework in Rd with the property that the points p(v),

v ∈ V (G), span an affine subspace of Rd of dimension at least d− 1, so that

the space of infinitesimal rigid motions of (G, p) has dimension(

d+12

). Also,

let the vector space of infinitesimal motions of (G, p) be denoted by I(p)

and the vector space of self-stresses of (G, p) be denoted by Ω(p). Then the

equation in Maxwell’s rule can be written in its extended form as

|E(G)| − d|V (G)| = dim(Ω(p)

)− dim(I(p)

).

So, if |E(G)| − (d|V (G)| − (

d+12

))= k > 0, then we can conclude that (G, p)

has at least k linearly independent self-stresses and if |E(G)| − (d|V (G)| −

(d+12

))= −k < 0, then (G, p) has at least k linearly independent infinitesimal

flexes (see also [33]).

Note that Maxwell’s rule is an immediate consequence of Theorem 2.2.5

which gives both necessary and sufficient conditions for a framework to be

isostatic. However, the advantage of Maxwell’s rule is that it provides a

purely combinatorial necessary condition for (G, p) to be isostatic, and this

condition can easily be verified since it only requires a simple count of the

edges and vertices of G. In Chapter 4, we will use techniques from group

representation theory to extend Maxwell’s rule to frameworks that possess

non-trivial symmetries.

In addition to the condition in Theorem 2.2.7, there exist further nec-

essary conditions for a graph G to be generically d-isostatic. The following

32

result includes necessary conditions for all non-trivial subgraphs of G.

Theorem 2.2.8 [32, 33] Let G be a graph that is generically d-isostatic with

|V (G)| ≥ d. Then

(i) |E(G)| = d|V (G)| − (d+12

);

(ii) |E(H)| ≤ d|V (H)| − (d+12

)for all H ⊆ G with |V (H)| ≥ d.

Clearly, generic rigidity is a combinatorial concept for all dimensions.

However, a combinatorial characterization has only been found for dimen-

sions 1 and 2: for d ∈ 1, 2, the counts stated in the previous theorem are

also sufficient for a graph to be generically d-isostatic.

For dimension 1, this says that a graph G is generically 1-isostatic if and

only if G is a tree, and G is generically 1-rigid if and only if G is connected

[33].

For dimension 2, the sufficiency of the counts in Theorem 2.2.8 for a graph

to be generically 2-isostatic was proven by G. Laman in 1970.

Theorem 2.2.9 (Laman, 1970) [46] A graph G with |V (G)| ≥ 2 is gener-

ically 2-isostatic if and only if

(i) |E(G)| = 2|V (G)| − 3;

(ii) |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G with |V (H)| ≥ 2.

Various proofs of Laman’s Theorem can be found in [32], [33], [67], and

[79], for example. In Chapter 5, we establish symmetric versions of Laman’s

Theorem for non-trivial symmetry groups.

33

Throughout this thesis, we will refer to the conditions (i) and (ii) in

Theorem 2.2.9 as the Laman conditions.

For dimension d ≥ 3, the counts in Theorem 2.2.8 are not sufficient

for generic d-rigidity. The most famous example to demonstrate this for

dimension 3 is the so-called double banana (see Figure 2.6) [32, 33, 68].

.. ..

.

.

.

.

.

Figure 2.6: The double banana satisfies the counts in Theorem 2.2.8 for

d = 3, but it is not generically 3-isostatic.

There are some inductive construction techniques that preserve the

generic rigidity properties of a graph. These construction techniques can

be used to prove theorems such as Laman’s Theorem, to analyze graphs

for generic rigidity, and to characterize generically 1-isostatic and 2-isostatic

graphs. For all dimensions d, they provide a tool to generate classes of gener-

ically d-isostatic graphs.

Definition 2.2.18 [68, 81] Let G be a graph, U ⊆ V (G) with |U | = d

and v /∈ V (G). Then the graph G with V (G) = V (G) ∪ v and E(G) =

E(G) ∪ v, u|u ∈ U

is called a vertex d-addition (by v) of G.

Theorem 2.2.10 (Vertex Addition Theorem) [32, 33, 68, 81] A vertex

d-addition of a generically d-isostatic graph is generically d-isostatic. Con-

versely, deleting a vertex of valence d from a generically d-isostatic graph

results in a generically d-isostatic graph.

34

Definition 2.2.19 [68, 81] Let G be a graph, U ⊆ V (G) with |U | = d+1 and

u1, u2 ∈ E(G) for some u1, u2 ∈ U . Further, let v /∈ V (G). Then the graph

G with V (G) = V (G)∪v and E(G) =(E(G)\u1, u2

)∪v, u|u ∈ U

is called an edge d-split (on u1, u2; v) of G.

Theorem 2.2.11 (Edge Split Theorem) [32, 33, 68, 81] An edge d-split

of a generically d-isostatic graph is generically d-isostatic. Conversely, if one

deletes a vertex v of valence d + 1 from a generically d-isostatic graph, then

one may add an edge between one of the pairs of vertices adjacent to v so

that the resulting graph is generically d-isostatic.

.

.

. .

.

.

.

.

.

.. .

.

.

.

..

.(a)

.

.

.(b)

.for some pair

Figure 2.7: Illustrations of the Vertex Addition Theorem (a) and the Edge

Split Theorem (b) in dimension 2.

In 1911, L. Henneberg gave the following characterization of generically

2-isostatic graphs.

Theorem 2.2.12 (Henneberg, 1911) [40] A graph is generically 2-

isostatic if and only if it may be constructed from a single edge by a sequence

of vertex 2-additions and edge 2-splits.

For a proof of Henneberg’s Theorem, see [33] or [68], for example.

35

There exist a few additional inductive construction techniques that

are frequently used in rigidity theory. One of these techniques, the X-

replacement, will play a pivotal role in proving the symmetrized version of

Laman’s Theorem for symmetry groups consisting of the identity and a single

reflection.

Definition 2.2.20 [68, 81] Let G be a graph, u1, u2, u3, u4 be four distinct

vertices of G with u1, u2, u3, u4 ∈ E(G), and let v /∈ V (G). Then the

graph G with V (G) = V (G)∪v and E(G) =(E(G)\u1, u2, u3, u4

)∪v, ui|i ∈ 1, 2, 3, 4

is called an X-replacement (by v) of G .

...

. .

...

. .

.

Figure 2.8: Illustration of an X-replacement of a graph G.

Theorem 2.2.13 (X-Replacement Theorem) [68, 81] An X-replacement

of a generically 2-isostatic graph is generically 2-isostatic.

The reverse operation of an X-replacement performed on a generically

2-isostatic graph does in general not result in a generically 2-isostatic graph.

For more details and some additional inductive construction techniques, we

refer the reader to [68].

Another way of characterizing generically 2-isostatic graphs is due to H.

Crapo and uses partitions of a graph into edge disjoint trees.

36

Definition 2.2.21 [20, 47, 67] A 3Tree2 partition of a graph G is a partition

of E(G) into the edge sets of three edge disjoint trees T0, T1, T2 such that each

vertex of G belongs to exactly two of the trees.

A 3Tree2 partition is called proper if no non-trivial subtrees of distinct

trees Ti have the same span (i.e., the same vertex sets).

.. .

.

.

.

.

.(a)

. .. .

.

.

.

.

.(b)

Figure 2.9: A proper (a) and a non-proper (b) 3Tree2 partition.

Remark 2.2.5 If a graph G has a 3Tree2 partition, then it satisfies |E(G)| =2|V (G)| − 3. This follows from the presence of exactly two trees at each

vertex of G and the fact that for every tree T we have |E(T )| = |V (T )| − 1.

Moreover, note that a 3Tree2 partition of a graph G is proper if and only if

every non-trivial subgraph H of G satisfies the count |E(H)| ≤ 2|V (H)| − 3

[47].

Theorem 2.2.14 (Crapo, 1989) [20] A graph G is generically 2-isostatic

if and only if G has a proper 3Tree2 partition.

Symmetrized versions of Crapo’s Theorem are discussed in Chapter 5.

Finally, as promised, we give some results which assert that under the

right conditions, rigidity and infinitesimal rigidity are equivalent. In Chapter

6, we establish symmetric analogs to these theorems.

37

Definition 2.2.22 Let G be a graph with n vertices and let d ≥ 1 be an

integer. A point p ∈ Rdn is said to be a regular point of G if there exists a

neighborhood Np of p in Rdn so that rank(R(G, p)

) ≥ rank(R(G, q)

)for

all q ∈ Np.

A framework (G, p) is said to be regular if p is a regular point of G.

Theorem 2.2.15 [3] Let G be a graph with n vertices and let (G, p) be a

d-dimensional framework. If p ∈ Rdn is a regular point of G, then (G, p) is

infinitesimally rigid if and only if (G, p) is rigid.

If a framework (G, p) is generic or independent, then (G, p) is clearly also

regular (see also Figure 2.5), so that we immediately obtain the following

results.

Corollary 2.2.16 If a framework (G, p) is generic, then (G, p) is infinites-

imally rigid if and only if (G, p) is rigid.

Corollary 2.2.17 If a framework (G, p) is independent, then (G, p) is in-

finitesimally rigid if and only if (G, p) is rigid.

2.3 Symmetry in frameworks

In this section we establish the concept of a symmetric framework and

give mathematically precise definitions of terms relating to symmetry which

might have different meanings in different contexts. In the literature about

symmetric structures it is common to systematize the notion of symmetry

38

by introducing the concept of a symmetry operation and its corresponding

symmetry element [6, 19, 37]. We begin with our definitions of these terms.

First, recall that an isometry of Rd is a map x : Rd → Rd such that

‖x(a)− x(b)‖ = ‖a− b‖ for all a, b ∈ Rd.

Definition 2.3.1 Let (G, p) be a framework in Rd. A symmetry operation

of (G, p) is an isometry x of Rd such that for some α ∈ Aut(G), we have

x(p(v)

)= p

(α(v)

)for all v ∈ V (G).

A symmetry operation x of a framework (G, p) carries (G, p) into a frame-

work (G, xp) which is ‘geometrically indistinguishable’ from (G, p). In other

words, up to the labeling of the vertices of the underlying graph G, the frame-

works (G, p) and (G, x p) are the same.

Definition 2.3.2 Let x be a symmetry operation of a framework (G, p) in

Rd. The symmetry element corresponding to x is the affine subspace Fx of

Rd which consists of all points a ∈ Rd such that x(a) = a.

Since we only consider finite graphs, it follows directly from Definition

2.3.1 that a symmetry operation cannot be a translation. This implies in

particular that a symmetry element is always non-empty. In fact, it is

easy to see that if x is a symmetry operation of a framework (G, p) with

V (G) = v1, . . . , vn, then the point 1n

∑ni=1 pi must be fixed by x. Figures

2.10 and 2.11 depict the possible symmetry elements in dimensions 2 and 3.

Note that distinct symmetry operations of a framework may have the

same corresponding symmetry element. For example, distinct rotational

39

symmetry operations of a 3-dimensional framework may share the same ro-

tational axis.

The set of all symmetry operations of a given framework forms a group

under composition. We adopt the following vocabulary from chemistry and

crystallography:

Definition 2.3.3 Let (G, p) be a framework. Then the group which consists

of all symmetry operations of (G, p) is called the point group of (G, p).

For a systematic method to find the point group of a given framework,

see [6, 19, 37], for example.

If P is the point group of a d-dimensional framework, then, as noted

above, there exists a point in Rd which is fixed by every symmetry operation

in P . Note that if the origin of Rd is fixed by x ∈ P , then x is an orthogonal

linear transformation of Rd. So, if the origin of Rd is fixed by every sym-

metry operation in P , then P is a subgroup of the orthogonal group O(Rd)

consisting of all orthogonal linear transformations of Rd.

Definition 2.3.4 A subgroup of the orthogonal group O(Rd) is called a

symmetry group (in dimension d).

Given a d-dimensional framework (G, p), the framework (G, T p), where

T is a translation of Rd, clearly has the same rigidity properties as (G, p).

Therefore, for our purposes we may wlog restrict our attention to frameworks

whose point groups are symmetry groups. In this thesis, unless otherwise

specified, the point group of every framework is assumed to be a symmetry

group.

40

We use the Schoenflies notation to denote symmetry operations and sym-

metry groups in dimensions 2 and 3, as this is one of the standard notations

in the literature about symmetric structures [2, 4, 6, 19, 37]. Another mo-

tivation for using the Schoenflies notation in this thesis is to be consistent

with the notation in the papers [15, 25, 34, 35, 36], for example.

In the plane, the three kinds of possible symmetry operations are the

identity Id, rotations Cm about the origin by an angle of 2πm

, where m ≥ 2,

and reflections s in lines through the origin. The symmetry elements corre-

sponding to these symmetry operations are shown in Figure 2.10.

.

.(a) .(b) .(c)

Figure 2.10: Symmetry elements corresponding to symmetry operations in

dimension 2: (a) a rotation Cm, m ≥ 2; (b) a reflection s; (c) the identity

Id.

In the Schoenflies notation we differentiate between the following four

types of symmetry groups in dimension 2: C1, Cs, Cm and Cmv, where m ≥ 2.

C1 denotes the trivial group which only contains the identity Id. Cs de-

notes any symmetry group in dimension 2 that consists of the identity Id and

a single reflection s. For m ≥ 2, Cm denotes any cyclic symmetry group of

order m which is generated by a rotation Cm, and Cmv denotes any symmetry

group in dimension 2 that is generated by a pair Cm, s. So, as abstract

groups, any group Cs is the cyclic group Z2, any group Cm is the cyclic group

41

Zm, and any group Cmv is the dihedral group of order 2m.

In 3-space, there are the following symmetry operations: the identity Id,

rotations Cm about axes through the origin by an angle of 2πm

, where m ≥ 2,

reflections s in planes through the origin, and improper rotations Sm fixing

the origin, where m ≥ 3. An improper rotation Sm is a rotation Cm followed

by the reflection s whose symmetry element is the plane through the origin

that is perpendicular to the axis of Cm. The axis of Cm is called the improper

rotation axis of Sm. By convention, S1 and S2 are treated separately, since S1

is simply a reflection s and S2 is the inversion in the origin which is denoted

by i.

.

.(a) .(b) .(c) .(d)

Figure 2.11: Symmetry elements corresponding to symmetry operations in

dimension 3: (a) an improper rotation Sm, m ≥ 2; (b) a rotation Cm, m ≥ 2;

(c) a reflection s; (d) the identity Id.

This gives rise to the following families of possible symmetry groups in

dimension 3: C1, Cs, Ci, Cm, Cmv, Cmh, Dm, Dmh, Dmd, S2m, T , Td, Th, O, Oh,

I, and Ih, where m ≥ 2.

Analogous to the notation in dimension 2, C1 again denotes the trivial

group that only contains the identity Id, Cm denotes any symmetry group

in dimension 3 that is generated by a rotation Cm, where m ≥ 2, and Cs

denotes any symmetry group in dimension 3 that consists of the identity Id

42

and a single reflection s.

Ci is the symmetry group which consists of the identity and the inversion

i of R3.

Cmv denotes any symmetry group that is generated by a rotation Cm and

a reflection s whose symmetry element contains the rotational axis of Cm.

Similarly, a symmetry group Cmh is generated by a rotation Cm and the re-

flection s whose symmetry element is perpendicular to the axis of Cm. It

follows that every symmetry group Cmh contains an improper rotation Sm.

Note that if a symmetry group S contains an improper rotation Sm, where

m is odd, then S must also contain both the rotation Cm whose symmetry

element is the improper rotation axis of Sm and the reflection s whose mirror

plane is perpendicular to the axis of Cm. Therefore, for odd m, a symmetry

group S is of type Cmh if and only if S is generated by an improper rotation

Sm. The abstract groups corresponding to Cmv and Cmh are the dihedral

group Dm of order 2m and the group Zm × Z2 (which is isomorphic to Z2m

if m is odd), respectively.

The symbol Dm is used to denote a symmetry group in dimension 3 that

is generated by a rotation Cm and another 2-fold rotation C2 whose rota-

tional axis is perpendicular to the one of Cm. As an abstract group, Dm is

again the dihedral group of order 2m.

Symmetry groups of the types Dmh and Dmd are generated by the gener-

ators Cm and C2 of a group Dm and by a reflection s. In the case of Dmh, the

symmetry element of s is the plane that is perpendicular to the Cm axis and

contains the origin (and hence contains the rotational axis of C2), whereas in

the case of Dmd, the symmetry element of s is a plane that contains the Cm

43

axis and forms an angle of πm

with the C2 axis (i.e., the symmetry element of

s bisects the angle between adjacent half-turn axes created by rotating the

C2 axis about the Cm axis). As abstract groups, Dmh is the group Dm × Z2

(which is isomorphic to D2m if m is odd) and Dmd is the group D2m.

If a symmetry group S in dimension 3 is generated by an improper rota-

tion Sk, where k is even, say k = 2m, then S is denoted by S2m. The abstract

group that corresponds to S2m is of course the group Z2m.

The remaining seven types of symmetry groups in dimension 3 are related

to the Platonic solids and are placed into three divisions: the tetrahedral

groups T , Td and Th, the octahedral groups O and Oh, and the icosahedral

groups I and Ih.

A symmetry group whose elements are all the rotational symmetry oper-

ations of a regular tetrahedron is denoted by T , and a symmetry group that

consists of all the symmetry operations of a regular tetrahedron is denoted

by Th. Td denotes a symmetry group that is generated by the elements of a

group T and those three reflections whose symmetry elements each contain

two of the three axes that correspond to half-turns in T . The abstract groups

corresponding to T , Td and Th are A4, S4 and A4 × Z2, respectively, where

A4 is the alternating group and S4 the symmetric group on 4 elements.

O denotes a symmetry group that consists of all rotational symmetry

operations of a regular octahedron (or, equivalently, a regular cube), and Oh

denotes a symmetry group that consists of all the symmetry operations of

a regular octahedron. Similarly, a symmetry group whose elements are all

rotational symmetry operations of a regular icosahedron (or, equivalently,

a regular dodecahedron) is denoted by I, and the symbol Ih is used for a

44

symmetry group that consists of all the symmetry operations of a regular

icosahedron. The abstract groups corresponding to O, Oh, I and Ih are S4,

S4×Z2, A5 and A5×Z2, respectively, where Ai is the alternating group and

Si the symmetric group on i elements for i = 4, 5.

45

Chapter 3

A classification of symmetric

frameworks

Recall from Definition 2.3.1 that a symmetry operation of a framework

(G, p) imposes geometric constraints on (G, p) by taking into account the

combinatorial structure of the underlying graph G. So, these kinds of sym-

metry constraints have both a geometric and a combinatorial aspect. As

the results of this thesis will show, it turns out that both of these aspects

of the symmetry constraints play an important role in a symmetry-based

rigidity analysis of a symmetric framework. The classification of symmetric

frameworks we introduce in Section 3.1 is motivated by this fact.

The usual starting point in most applications of rigidity of symmetric

frameworks is that one is given a symmetric structure with a non-trivial point

group (such as a biomolecule, for example) whose rigidity and flexibility prop-

erties are to be examined. This approach is used in [25, 26, 27, 35, 43, 44, 45],

for example. The same approach will be used in this thesis.

46

Alternatively, one could start with a graph G and a subgroup A of Aut(G),

and then consider the possible geometric realizations of G that satisfy the

symmetry constraints imposed by A. This approach is used in [53], for ex-

ample.

Note that although these two approaches have a different starting point,

they both result in the same geometric and combinatorial conditions for the

final theorems.

As we will see in Sections 3.3 and 3.4, if (G, p) is an injective realization

of a graph G, and S is a subgroup of the point group of (G, p), then the

map Φ : S → Aut(G) which turns each isometry x ∈ S into a symmetry

operation of (G, p) by assigning an appropriate graph automorphism to x is

not only uniquely determined, but it is also a group homomorphism. In this

case, symmetry-based techniques can be applied to the rigidity analysis of

(G, p) in a unique way (for a fixed group S).

In their studies of symmetric frameworks, engineers and chemists usu-

ally restrict their attention to injective realizations. While this is a rea-

sonable assumption for most applications (atoms of biomolecules or joints

of 3-dimensional physical structures never coincide, for example), there are

occasions where we do want to analyze frameworks with non-injective config-

urations (if we want to model a linkage in the plane with overlapping joints,

for example). So, in order to obtain more general mathematical results and a

more complete theory, we develop the mathematical foundation for the rigid-

ity of symmetric frameworks in such a way that it also allows us to analyze

symmetric frameworks with non-injective configurations.

We will see in Sections 3.3 and 3.4 that if (G, p) is a non-injective real-

47

ization of G, then one may have several choices for a graph automorphism

to turn an isometry in S into a symmetry operation of (G, p). The subtle

difficulties that can occur in such a case, as well as their consequences for the

application of symmetry-based techniques to the rigidity analysis of (G, p),

are carefully examined in these sections.

All the key results in this chapter are original and are contained in the

manuscript [55] which has been submitted for review.

3.1 The classification

In order to symmetrize results in rigidity theory, particularly results in

generic rigidity theory, we first of all need an appropriate classification of

symmetric frameworks. Naturally, we require that frameworks in the same

class have the same underlying graph. This classification should also be such

that ‘almost all’ frameworks within a given class share the same infinitesimal

rigidity properties, so that we can develop a symmetrized version of generic

rigidity theory with respect to this classification.

Definition 3.1.1 Let G be a graph and S be a symmetry group in dimension

d. Then R(G,S) is the set of all d-dimensional realizations of G whose point

group is either equal to S or contains S as a subgroup. An element of R(G,S)

is said to be a realization of the pair (G,S).

Theorem 3.1.1 Let (G, p) be a d-dimensional realization of a graph G and

S be a symmetry group in dimension d. Then (G, p) ∈ R(G,S) if and only if

there exists a map Φ : S → Aut(G) such that x(p(v)

)= p

(Φ(x)(v)

)for all

48

v ∈ V (G) and all x ∈ S.

Proof. It follows immediately from the definitions that (G, p) ∈ R(G,S) if

and only if S is a subgroup of the point group of (G, p) if and only if every

element of S is a symmetry operation of (G, p) if and only if for every x ∈ S,

there exists an automorphism αx of G that satisfies x(p(v)

)= p

(αx(v)

)for

all v ∈ V (G). ¤

Remark 3.1.1 Note that the set R(G,S) can possibly be empty. For example,

there clearly exists no realization of (K2, C3), where C3 is a symmetry group

in dimension 2.

Theorem 3.1.1 gives rise to the following natural classification of the

frameworks within a set R(G,S).

Definition 3.1.2 Let S be a symmetry group, (G, p) be a framework in

R(G,S), and Φ be a map from S to Aut(G). Then (G, p) is said to be of type

Φ if the following equations hold:

x(p(v)

)= p

(Φ(x)(v)

)for all v ∈ V (G) and all x ∈ S.

The set of all realizations of (G,S) which are of type Φ is denoted by R(G,S,Φ).

Given a graph G and a symmetry group S in dimension d, different choices

of types Φ : S → Aut(G) frequently lead to very different geometric types

of realizations of (G,S). This is because a type Φ forces the joints and bars

of a framework in R(G,S,Φ) to assume certain geometric positions in Rd. We

give a few examples for small symmetry groups in dimensions 2 and 3 to

demonstrate this.

49

..

.p5 ..p3

..p6

..p1 . .p2

..p4

.(a)

...p6

..p1

..p2

. .p3

..p4

..p5

.(b)

Figure 3.1: 2-dimensional realizations of (K3,3, Cs) of different types.

Example 3.1.1 Figure 3.1 shows two realizations of (K3,3, Cs) of different

types, where K3,3 is the complete bipartite graph with partite sets v1, v2, v3and v4, v5, v6, and Cs = Id, s is a symmetry group in dimension 2 gen-

erated by a reflection. The framework in Figure 3.1 (a) is a realization of

(K3,3, Cs) of type Φa, where Φa : Cs → Aut(K3,3) is defined by

Φa(Id) = id

Φa(s) = (v1 v2)(v5 v6)(v3)(v4),

and the framework in Figure 3.1 (b) is a realization of (K3,3, Cs) of type Φb,

where Φb : Cs → Aut(K3,3) is defined by

Φb(Id) = id

Φb(s) = (v1 v4)(v2 v5)(v3 v6).

Note that for any framework (K3,3, p) in the set R(K3,3,Cs,Φa), the points p3

and p4 must lie in the symmetry element corresponding to s (i.e., in the

mirror line of s), because s(p(vi)

)= p

(Φa(s)(vi)

)= p(vi) for i = 3, 4. This

says in particular that for any framework (K3,3, p) in R(K3,3,Cs,Φa), the entire

50

undirected line segment p3p4 which corresponds to the bar(v3, p3), (v4, p4)

of (K3,3, p) must lie in the mirror line of s. We shall immediately become less

formal and say that the bar(v3, p3), (v4, p4)

lies in the mirror line of s.

Similarly, for any framework (K3,3, p) in R(K3,3,Cs,Φb), the bars(v1, p1), (v4, p4)

,(v2, p2), (v5, p5)

and

(v3, p3), (v6, p6)

must be perpen-

dicular to and centered at the mirror line of s.

.

..p3

..p6

..p5

..p2

..p1

. .p4

.(a)

.

..p3

..p5

..p6

..p2

..p1

. .p4

.(b)

Figure 3.2: 2-dimensional realizations of (Gtp, C2) of different types.

Example 3.1.2 Figure 3.2 depicts two realizations of (Gtp, C2) of different

types, where Gtp is the graph of a triangular prism and C2 = Id, C2 is the

half-turn symmetry group in dimension 2. The framework in Figure 3.2 (a)

is a realization of (Gtp, C2) of type Ψa, where Ψa : C2 → Aut(Gtp) is defined

by

Ψa(Id) = id

Ψa(C2) = (v1 v4)(v2 v6)(v3 v5).

and the framework in Figure 3.2 (b) is a realization of (Gtp, C2) of type Ψb,

where Ψb : C2 → Aut(Gtp) is defined by

Ψb(Id) = id

Ψb(C2) = (v1 v4)(v2 v5)(v3 v6).

51

It follows from the definitions of Ψa and Ψb that for any framework (Gtp, p)

in R(Gtp,C2,Ψa), the bar(v1, p1), (v4, p4)

must be centered at the ori-

gin (which is the center of the half-turn C2), whereas for any framework

(Gtp, p) in R(Gtp,C2,Ψb), all three bars(v1, p1), (v4, p4)

,

(v2, p2), (v5, p5)

,

and(v3, p3), (v6, p6)

must be centered at the origin.

.

..

..

.. ....

.p1 .p2

.p3

.p4

.p5

.(a)

... ..

..

.. ..

.p1 .p2

.p3

.p4 .p5

.(b)

Figure 3.3: 3-dimensional realizations of (Gbp, Cs) of different types.

Example 3.1.3 Finally, Figure 3.3 depicts two realizations of (Gbp, Cs) of

different types, where Gbp is the graph of a triangular bipyramid and Cs =

Id, s is a symmetry group in dimension 3. The framework in Figure 3.3

(a) is an element of R(Gbp,Cs,Ξa), where Ξa : Cs → Aut(Gbp) is defined by

Ξa(Id) = id

Ξa(s) = (v1 v2)(v3)(v4)(v5),

and the framework in Figure 3.3 (b) is an element of R(Gbp,Cs,Ξb), where

52

Ξb : Cs → Aut(Gbp) is defined by

Ξb(Id) = id

Ξb(s) = (v1 v2)(v4 v5)(v3).

For any framework (Gbp, p) in R(Gbp,Cs,Ξa) or R(Gbp,Cs,Ξb), the bar(v1, p1), (v2, p2)

must be perpendicular to and centered at the mirror plane

of s. Further, for any framework (Gbp, p) in R(Gbp,Cs,Ξa), the joints (vi, pi),

i = 3, 4, 5, must lie in the mirror plane of s, whereas for a framework (Gbp, p)

in R(Gbp,Cs,Ξb), only the joint (v3, p3) must have this property and the joints

(v4, p4) and (v5, p5) must be mirror images of each other with respect to s.

Remark 3.1.2 Given a non-empty set R(G,S), it is possible that R(G,S,Φ) = ∅for some map Φ : S → Aut(G).

Consider, for example, the non-empty set R(K2,C2), where C2 = Id, C2is the half-turn symmetry group in dimension 2, and let I : C2 → Aut(K2) be

the map which sends both Id and C2 to the identity automorphism of K2. If

(K2, p) ∈ R(K2,C2,I), then both joints of (K2, p) must be located at the origin

(which is the center of C2). This contradicts Definition 2.2.1 of a framework,

and hence we have R(K2,C2,I) = ∅.

We will see in the next section that ‘almost all’ frameworks within a set

of the form R(G,S,Φ) share the same infinitesimal rigidity properties. This

will allow us to develop a symmetrized version of generic rigidity theory with

respect to the classes R(G,S,Φ) of symmetric frameworks.

53

3.2 The notion of (S, Φ)-generic

Given a graph G, a non-trivial symmetry group S in dimension d clearly

imposes restrictions on the possible geometric positions of realizations of

(G,S) in Rd. For most groups S, these restrictions are in fact so strong that

they force the joints of any realization in the set R(G,S) to lie in non-generic

positions. In some situations, the realizations in R(G,S) are even forced to be

non-regular, as the following examples demonstrate.

Every realization of (K3, C2), where C2 is a half-turn symmetry group in

dimension 2 or 3, must be a degenerate triangle and is therefore non-regular

(recall Figure 2.3 (b)).

For a less trivial example, consider the complete bipartite graph K3,3 and

the symmetry group C2 in dimension 2. As shown in Figure 3.4 (a), the joints

of any realization (K3,3, p) in R(K3,3,C2) can be labeled in such a way that for

the resulting hexagon p1 p2 . . . p6, there exists a pair of opposite sides which

intersect in the origin. If all three pairs of opposite sides of this hexagon

are extended to their points of intersection, then the half-turn symmetry of

(K3,3, p) guarantees that these three points are collinear. Therefore, by the

converse of Pascal’s Theorem, the joints of (K3,3, p) must lie on a conic sec-

tion. It is well known that 2-dimensional realizations of K3,3 whose joints lie

on a conic section are in fact non-regular [7, 71, 75].

This shows that our notion of generic (without symmetry) is clearly not

suitable once we restrict our attention to symmetric frameworks that lie

within a set of the form R(G,S).

Note also that for a graph G, a symmetry group S, and two distinct maps

Φ and Ψ from S to Aut(G), it is possible that all realizations in R(G,S,Φ) are

54

infinitesimally flexible, whereas ‘almost all’ realizations in R(G,S,Ψ) are iso-

static.

For example, consider again the complete bipartite graph K3,3, a symme-

try group Cs in dimension 2, and the types Φa and Φb from Example 3.1.1.

K3,3 is known to be a generically 2-isostatic graph and the pure condition

(see [71]) for K3,3 says that a 2-dimensional realization (K3,3, p) is infinites-

imally flexible if and only if the joints of (K3,3, p) lie on a conic section. It

follows (again from the converse of Pascal’s Theorem) that every realization

in R(K3,3,Cs,Φb) is infinitesimally flexible (see also Figure 3.4 (b)), whereas

‘almost all’ realizations in R(K3,3,Cs,Φa) are isostatic.

.

..p5

..p2

..p1

. .p6

..p4

. .p3

.(a)

.. .p5..p2

..p1

..p6

..p4

..p3

.(b)

Figure 3.4: By the converse of Pascal’s Theorem, the joints of any realization

in R(K3,3,C2) or R(K3,3,Cs,Φb) lie on a conic section.

Therefore, in order to define a modified, symmetry-adapted notion of

generic for a set C ⊆ R(G,S) of symmetric frameworks in such a way that ‘al-

most all’ realizations within C are generic and all generic realizations within

C share the same infinitesimal rigidity properties, we need to restrict C to

a set of the form R(G,S,Φ).

55

Let G be a graph with V (G) = v1, v2, . . . , vn, S be a symmetry group

in dimension d, and Φ be a map from S to Aut(G). We will define a

symmetry-adapted notion of generic for the set R(G,S,Φ) in an analogous way

as we defined generic in Definition 2.2.16. This requires the definition of a

symmetry-adapted indeterminate rigidity matrix for R(G,S,Φ). The following

observations lay the foundation for the definition of such a matrix.

Recall that for every framework (G, p) in the set R(G,S,Φ), the equations

stated in Definition 3.1.2 are satisfied, that is, we have x(p(vi)

)= p

(Φ(x)(vi)

)

for all i = 1, 2, . . . , n and all x ∈ S. Since every element of S is an orthogonal

linear transformation, we may identify each x ∈ S with its corresponding

orthogonal matrix Mx that represents x with respect to the canonical basis of

Rd. Therefore, for each x ∈ S, the equations in Definition 3.1.2 corresponding

to x form a system of linear equations which can be written as

M(x)

p1

p2

...

pn

= PΦ(x)

p1

p2

...

pn

,

where

M(x) =

Mx 0 . . . 0

0 Mx. . .

...

.... . . . . . 0

0 . . . 0 Mx

,

and PΦ(x) is the dn × dn matrix which is obtained from the permutation

matrix corresponding to Φ(x) by replacing each 1 by a d× d identity matrix

56

and each 0 by a d× d zero matrix. Equivalently, we have

(M(x) −PΦ(x)

)

p1

p2

...

pn

= 0.

We denote Lx,Φ = ker(M(x) − PΦ(x)

)and U =

⋂x∈S Lx,Φ. Then U is a

subspace of Rdn which may be interpreted as the space of all those (possibly

non-injective) configurations of n points in Rd that possess the symmetry

imposed by S and Φ. In particular, if (G, p) is a framework in R(G,S,Φ),

then the configuration p is an element of U . Therefore, if we fix a basis

BU = u1, u2, . . . , uk of U , then every framework (G, p) ∈ R(G,S,Φ) can be

represented uniquely by the k × 1 coordinate vector of p relative to BU .

We are now ready to define the symmetry-adapted indeterminate rigidity

matrix for R(G,S,Φ).

Definition 3.2.1 Let G be a graph with V (G) = v1, v2, . . . , vn, Kn be the

complete graph with V (Kn) = V (G), S be a symmetry group in dimension

d, and Φ be a map from S to Aut(G). Further, let BU = u1, u2, . . . , ukbe a basis of U =

⋂x∈S Lx,Φ. The symmetry-adapted indeterminate rigidity

matrix for R(G,S,Φ) (corresponding to BU) is the matrix RBU(n, d) which

is obtained from the indeterminate rigidity matrix R(n, d) by introducing

a k-tuple (t′1, t′2, . . . , t

′k) of variables and replacing the dn variables (p′i)j of

R(n, d) as follows.

For each i = 1, 2, . . . , n and each j = 1, . . . , d, we replace the variable

(p′i)j in R(n, d) by the linear combination t′1(u1)ij + t′2(u2)ij + . . . + t′k(uk)ij .

57

Remark 3.2.1 Let (G, p) ∈ R(G,S,Φ) and BU = u1, u2, . . . , uk be a basis

of U =⋂

x∈S Lx,Φ. Then

p1

p2

...

pn

= t1u1 + . . . + tkuk, for some t1, . . . , tk ∈ R.

So, if for i = 1, . . . , k, the variable t′i in RBU(n, d) is replaced by ti then we

obtain the rigidity matrix R(Kn, p) of the framework (Kn, p).

With the help of Definition 3.2.1 we can now also give the formal definition

of our symmetry-adapted notion of generic for a set R(G,S,Φ).

Definition 3.2.2 Let G be a graph with V (G) = v1, v2, . . . , vn, Kn be the

complete graph with V (Kn) = V (G), S be a symmetry group in dimension

d, Φ be a map from S to Aut(G), and BU be a basis of U =⋂

x∈S Lx,Φ.

A map p : V (G) → Rd is said to be (S, Φ, BU)-generic if the following

holds: If the determinant of any submatrix of R(Kn, p) is equal to zero, then

the determinant of the corresponding submatrix of RBU(n, d) is (identically)

zero.

The map p is said to be (S, Φ)-generic if p is (S, Φ,BU)-generic for some

basis BU of U .

A framework (G, p) ∈ R(G,S,Φ) is (S, Φ, BU)-generic if p is an

(S, Φ, BU)-generic map, and (G, p) is (S, Φ)-generic if (G, p) is (S, Φ, BU)-

generic for some basis BU of U .

58

Theorem 3.2.1 Let G be a graph, S be a symmetry group, and Φ be a

map from S to Aut(G). If (G, p) ∈ R(G,S,Φ) is (S, Φ)-generic, then (G, p) is

(S, Φ, BU)-generic for every basis BU of U =⋂

x∈S Lx,Φ.

Proof. Suppose S is a symmetry group in dimension d and the vertex set

of G is V (G) = v1, v2, . . . , vn. Let (G, p) ∈ R(G,S,Φ) be (S, Φ)-generic, say

(G, p) is (S, Φ, BU)-generic, where BU = u1, . . . , uk is a basis of U . Let

B∗U = u∗1, . . . , u∗k be another basis of U . Then we need to show that (G, p)

is (S, Φ,B∗U)-generic. Let

p1

p2

...

pn

= t1u1 + . . . + tkuk = t∗1u∗1 + . . . + t∗ku

∗k ,

where ti, t∗i ∈ R for all i = 1, . . . , k. Then there exists an invertible matrix of

real numbers (sij) such that

t1 = s11t∗1 + . . . + s1kt

∗k

......

...

tk = sk1t∗1 + . . . + skkt

∗k .

(3.1)

Let RBU(n, d) be the symmetry-adapted indeterminate rigidity matrix cor-

responding to BU with variables t′1, . . . , t′k, and RB∗

U(n, d) be the symmetry-

adapted indeterminate rigidity matrix corresponding to B∗U with variables

t∗′

1 , . . . , t∗′

k . Then note that if for i = 1, . . . , k, we replace the variable t′i in

RBU(n, d) analogously to (3.1) by

t′i = si1t∗′1 + . . . + sikt

∗′k , (3.2)

59

then we obtain the matrix RB∗U(n, d).

If each t∗′

i in RB∗U(n, d) is replaced by t∗i , then, by Remark 3.2.1, we

obtain the rigidity matrix R(Kn, p). Consider the determinant of a submatrix

of R(Kn, p) which is equal to zero. The determinant of the corresponding

submatrix of RBU(n, d) is a polynomial in t′1, . . . , t

′k, say

∑a(a1,...,ak)t

′a11 · . . . · t′ak

k , where a(a1,...,ak) ∈ R. (3.3)

Since (G, p) is (S, Φ,BU)-generic, the polynomial in (3.3) is the zero polyno-

mial. If in (3.3) we replace the variables t′i as in (3.2), then we again obtain

the zero polynomial. On the other hand, this polynomial is the determi-

nant of the corresponding submatrix of RB∗U(n, d). This says that (G, p) is

(S, Φ, B∗U)-generic and the proof is complete. ¤

Note that it follows directly from Definition 3.2.2 that the set of (S, Φ)-

generic realizations of a graph G is an open dense subset of the set R(G,S,Φ).

Moreover, as we will show next, the infinitesimal rigidity properties are

the same for all (S, Φ)-generic realizations of G.

Lemma 3.2.2 Let G be a graph with V (G) = v1, v2, . . . , vn, S be a sym-

metry group in dimension d, and Φ be a map from S to Aut(G). If for some

framework (G, p) ∈ R(G,S,Φ), the points p1, . . . , pn span an affine subspace of

Rd of dimension k, then for any (S, Φ)-generic realization (G, q) of G, the

points q1, . . . , qn span an affine subspace of Rd of dimension at least k.

Proof. Let (G, p) ∈ R(G,S,Φ) be a framework for which the points p1, . . . , pn

span an affine subspace of Rd of dimension k. Then there are k + 1 affinely

independent points among p1, . . . , pn, say wlog p1, . . . , pk+1. Let A be the

60

k × d matrix defined by

A =

(p1 − p2)1 (p1 − p2)2 . . . (p1 − p2)d

(p1 − p3)1 (p1 − p3)2 . . . (p1 − p3)d

...... . . .

...

(p1 − pk+1)1 (p1 − pk+1)2 . . . (p1 − pk+1)d

.

Then the rows of A are linearly independent and hence there exists a k ×k submatrix B of A whose determinant is non-zero. Fix a basis BU of

U =⋂

x∈S Lx,Φ and let RBU(n, d) be the symmetry-adapted indeterminate

rigidity matrix for R(G,S,Φ) corresponding to BU . Then the determinant of

the submatrix B′ of RBU(n, d) which corresponds to B is not identically

zero.

Now, let (G, q) be an (S, Φ)-generic realization of G and suppose the

points q1, . . . , qn span an affine subspace of Rd of dimension m < k. Then

the matrix A which is obtained from A by replacing each (pi)j by (qi)j has a

non-trivial row dependency, which says that the determinant of every k × k

submatrix of A is equal to zero. This contradicts the fact that (G, q) is

(S, Φ)-generic and that the determinant of B′ is not identically zero. ¤

Theorem 3.2.3 Let G be a graph, S be a symmetry group, and Φ be a map

from S to Aut(G) such that R(G,S,Φ) 6= ∅. The following are equivalent.

(i) There exists a framework (G, p) ∈ R(G,S,Φ) that is infinitesimally rigid

(independent, isostatic);

(ii) every (S, Φ)-generic realization of G is infinitesimally rigid (indepen-

dent, isostatic).

61

Proof. Suppose S is a symmetry group in dimension d. Let (G, p) ∈ R(G,S,Φ)

be infinitesimally rigid and let (G, q) be an (S, Φ)-generic realization of G.

Suppose first that |V (G)| ≥ d. Then, by Remark 2.2.3, the points p(v),

v ∈ V (G), span an affine subspace of Rd of dimension at least d−1. Therefore,

the infinitesimal rigid motions arising from d translations and(

d2

)rotations

of Rd form a basis for the space of infinitesimal rigid motions of (G, p) (see

[3, 33] for details), and hence we have

rank(R(G, p)

)= d|V (G)| −

(d + 1

2

).

By the definition of (S, Φ)-generic,

rank(R(G, q)

) ≥ rank(R(G, p)

).

By Lemma 3.2.2, the points q(v), v ∈ V (G), also span an affine subspace

of Rd of dimension at least d− 1, which says that nullity(R(G, q)

) ≥ (d+12

).

Therefore,

rank(R(G, q)

) ≤ d|V (G)| −(

d + 1

2

).

It follows that

rank(R(G, q)

)= d|V (G)| −

(d + 1

2

),

and hence (G, q) is infinitesimally rigid.

Suppose now that |V (G)| ≤ d − 1. Then the dimension of the space of

infinitesimal rigid motions of (G, p) is strictly smaller than(

d+12

)(see again

[3, 33] for details). Therefore, we have nullity(R(G, p)

)<

(d+12

), and hence

rank(R(G, p)

)> d|V (G)|−(

d+12

). It follows from Theorem 2.2.2 that G is a

complete graph and the points p(v), v ∈ V (G), are affinely independent. By

62

Lemma 3.2.2, the points q(v), v ∈ V (G), must also be affinely independent,

and hence (G, q) is infinitesimally rigid.

If (G, p) is independent, then it follows from the definition of (S, Φ)-

generic that (G, q) is also independent. Therefore, if (G, p) is isostatic, so is

(G, q). ¤

So, being infinitesimally rigid (independent, isostatic) is an (S, Φ)-generic

property. This gives rise to

Definition 3.2.3 Let G be a graph, S be a symmetry group, and Φ be a

map from S to Aut(G). G is said to be (S, Φ)-generically infinitesimally rigid

(independent, isostatic) if all realizations of G which are (S, Φ)-generic are

infinitesimally rigid (independent, isostatic).

Examples 3.1.1 and 3.1.2 show that a graph G which is (S, Φ)-generically

isostatic is not necessarily (S, Ψ)-generically isostatic, where Φ and Ψ are

two distinct maps from S to Aut(G).

In Example 3.1.1, (Cs, Φa)-generic realizations in R(K3,3,Cs,Φa) are isostatic,

whereas all the realizations in R(K3,3,Cs,Φb) are infinitesimally flexible, because

the joints of any realization in R(K3,3,Cs,Φb) lie on a conic section, as we al-

ready observed in the beginning of this section.

In Example 3.1.2, the graph Gtp is (C2, Ψa)-generically isostatic, but none

of the realizations in R(Gtp,C2,Ψb) is isostatic. This follows from the pure condi-

tion for Gtp, which says that a 2-dimensional realization of Gtp is not isostatic

if and only if the triangles p1 p2 p3 and p4 p5 p6 are perspective from a line

[71]. Equivalently, by Desargues Theorem, a 2-dimensional realization of Gtp

is not isostatic if and only if the triangles p1 p2 p3 and p4 p5 p6 are perspective

63

from a point or at least one of those triangles is degenerate.

For an example in 3-space, consider the complete graph K4 with V (K4) =

v1, v2, v3, v4, a symmetry group Cs = Id, s in dimension 3, and the maps

Υa and Υb from Cs to Aut(K4), where Υa maps Id to the identity auto-

morphism id of K4 and s to (v1 v2)(v3)(v4), and Υb maps both Id and s to

id. Then K4 is (Cs, Υa)-generically isostatic, but all realizations in R(K4,Cs,Υb)

are infinitesimally flexible, because all the joints of a realization in R(K4,Cs,Υb)

must lie in the mirror plane corresponding to s and are therefore coplanar.

.

..

.. ....

.p1 .p2

.p3

.p4

.(a)

...

..

..

..

.p1

.p2

.p3

.p4

.(b)

Figure 3.5: A 3-dimensional realization of (K4, Cs) of type Υa (a) and of type

Υb (b).

Remark 3.2.2 Let G be a graph, S be a symmetry group in dimension d,

and Φ : S → Aut(G) be a homomorphism. Frameworks in the set R(G,S,Φ),

particularly (S, Φ)-generic realizations of G, can then be visualized in a very

intuitive way via the following approach.

The map of S × V (G) onto V (G) that sends (x, v) to Φ(x)(v) defines a

group action on V (G) and the orbits Sv = Φ(x)(v)|x ∈ S form a partition

64

of V (G). Let v1, . . . , vr be a subset of V (G) obtained by choosing one

representative from each of these orbits, and recall from Definition 2.3.2 that

for every x ∈ S, Fx denotes the symmetry element corresponding to x. If for

i = 1, . . . , r, we define

F (vi) =⋂

x∈S with Φ(x)(vi)=vi

Fx,

then for every framework (G, p) ∈ R(G,S,Φ) and for every i ∈ 1, . . . , r, the

point p(vi) must be contained in the subspace F (vi) of Rd.

Note that the positions of all joints of a framework (G, p) ∈ R(G,S,Φ)

are uniquely determined by the positions p(v1), . . . , p(vr) of the joints(v1, p(v1)

), . . . ,

(vr, p(vr)

)and the symmetry constraints imposed by S and

Φ. In other words, we may construct frameworks in R(G,S,Φ) by first choosing

a point p(vi) ∈ F (vi) for each i = 1, . . . , r and then letting S and Φ determine

the positions of the remaining joints. In particular, note that we obtain an

(S, Φ)-generic framework (G, p) in this way for almost all choices of points

p(vi) that satisfy p(vi) ∈ F (vi) for i = 1, . . . , r.

Consider, for example, the set R(K4,Cs,Υa) an element of which is shown in

Figure 3.5 (a). The orbits for the group action from Cs× V (K4) onto V (K4)

are given by v1, v2, v3, and v4. If (K4, p) is a framework in R(K4,Cs,Υa),

then both p3 and p4 must be contained in the mirror plane Fs of s, because

F (v3) = F (v4) = FId ∩ Fs = Fs. Furthermore, since v1 and v2 are vertices

of the same orbit, the position of the point p2 is uniquely determined by the

position of p1 and the symmetry constraints imposed by Cs and Υa. Since

F (v1) = FId = R3, the point p1 may be chosen to be any point in R3; how-

ever, if p1 lies in the mirror plane of s, then p1 = p2, in which case (K4, p) is

not a framework.

65

We will return to this approach in Chapter 6.

We conclude this section by giving a few more interesting properties of

(S, Φ)-generic frameworks.

Theorem 3.2.4 Let G be a graph, S be a symmetry group, and Φ be a map

from S to Aut(G). Further, let (G, p) ∈ R(G,S,Φ), S ′ be a subgroup of S, and

Φ′ = Φ|S′. If (G, p) is (S ′, Φ′)-generic, then (G, p) is also (S, Φ)-generic.

Proof. Suppose S is a symmetry group in dimension d and G is a

graph with n vertices. Let (G, p) ∈ R(G,S,Φ) be (S ′, Φ′)-generic. We fix

a basis BU = u1, . . . , uk of U =⋂

x∈S Lx,Φ and extend it to a basis

BU ′ = u1, . . . , uk, uk+1, . . . , um of U ′ =⋂

x∈S′ Lx,Φ′ . Consider a subma-

trix of the rigidity matrix R(Kn, p) whose determinant is equal to zero. We

need to show that the determinant ∆ of the corresponding submatrix of

the symmetry-adapted indeterminate rigidity matrix RBU(n, d) is identically

zero.

Let ∆′ =∑

a(a1,...,am)t′a11 · . . . · t′am

m , where a(a1,...,am) ∈ R, be the deter-

minant of the corresponding submatrix of RBU′ (n, d). Then ∆′ is the zero

polynomial since (G, p) is (S ′, Φ′)-generic. But note that ∆ is a polynomial

that is obtained from ∆′ by deleting all those terms in ∆′ that have one or

more variables in t′k+1, . . . , t′m. Thus, ∆ is also the zero polynomial. ¤

The converse of Theorem 3.2.4 does not hold, as the following example

shows.

Example 3.2.1 The realization (K3,3, p) in Figure 3.6 is (C2v, Φ)-generic,

where C2v = Id, C2, sh, sv is a symmetry group in dimension 2 and Φ :

66

C2v → Aut(K3,3) is defined by

Φ(Id) = id

Φ(C2) = (v1 v6)(v2 v5)(v3 v4)

Φ(sh) = (v1 v5)(v2 v6)(v3 v4)

Φ(sv) = (v1 v2)(v5 v6)(v3)(v4).

However, (K3,3, p) is not (Cs, Φa)-generic, where Cs is the subgroup of C2v

generated by sv and Φa = Φ|Cs is the map we defined in Example 3.1.1.

.

..p5

..p3

. .p6

..p1 . .p2

..p4

.sv

.sh

Figure 3.6: A realization of K3,3 that is (C2v, Φ)-generic, but not (Cs, Φa)-

generic, where Cs is the subgroup of C2v generated by sv and Φa = Φ|Cs.

Corollary 3.2.5 Let G be a graph, S be a symmetry group, and Φ be a map

from S to Aut(G). If (G, p) ∈ R(G,S,Φ) is generic (in the sense of Definition

2.2.16) then (G, p) is also (S, Φ)-generic.

Proof. Suppose S is a symmetry group in dimension d and G is a graph with

n vertices. Let (G, p) ∈ R(G,S,Φ) be generic. Then Φ maps the symmetry op-

eration Id ∈ S to the identity automorphism id of G, for otherwise the map

q of every realization (G, q) in R(G,S,Φ) is non-injective, contradicting the fact

67

that (G, p) ∈ R(G,S,Φ) is generic. So, Φ|C1 = I, where I : C1 → Aut(G) maps

Id to id, and we have⋂

x∈C1 Lx,I = LId,I = Rdn.

Now, observe that the indeterminate rigidity matrix R(n, d) is equal

to the symmetry-adapted indeterminate rigidity matrix RBRdn(n, d), where

BRdn is the canonical basis of Rdn. Therefore, (G, p) is generic if and only if

(G, p) is (C1, I)-generic.

The result now follows immediately from Theorem 3.2.4. ¤

The converse of Corollary 3.2.5 is of course false. A (Cs, Φb)-generic real-

ization of K3,3, for example, where Cs and Φb are as in Example 3.1.1, has

all of its joints on a conic section and is therefore not generic (in fact, such

a framework is even non-regular).

3.3 Of what types Φ can a framework be?

Let (G, p) be a d-dimensional framework with point group symmetry P .

Then (G, p) ∈ R(G,S) for every subgroup S of P .

Fix a subgroup S of P . Then it follows from Theorem 3.1.1 that there

exists a map Φ : S → Aut(G) such that (G, p) ∈ R(G,S,Φ). The following

examples show that it is possible for (G, p) ∈ R(G,S) to be of more than

just one such type Φ. Note that each of these examples is a non-injective

realization.

Example 3.3.1 Let Gt be the graph of two triangles sharing an edge and

C2 = Id, C2 be the half-turn symmetry group in dimension 2. Figure 3.7

(a) shows a realization (Gt, p) of (Gt, C2) that is of type Θa as well as Θb,

68

where Θa : C2 → Aut(Gt) is defined by

Θa(Id) = id

Θa(C2) = (v1 v2)(v3)(v4),

and Θb : C2 → Aut(Gt) is defined by

Θb(Id) = id

Θb(C2) = (v1 v2)(v3 v4).

...p1

..p2

. .

.(a)

.p3 = p4

.

..

.. ....

.p1 .p2

.p3

.p4 = p5..

.(b)

Figure 3.7: A realization of (Gt, C2) of type Θa and Θb (a) and a realization

of (Gbp, Cs) of type Ξa and Ξb (b).

Example 3.3.2 Consider the graph Gbp of a triangular bipyramid and a

symmetry group Cs = Id, s in dimension 3. The framework (Gbp, p) in

Figure 3.7 (b) is a realization of (Gbp, Cs) that is of type Ξa as well as Ξb,

where Ξa : Cs → Aut(Gbp) and Ξb : Cs → Aut(Gbp) are defined as in Example

3.1.3.

69

Since for a given framework (G, p) in a set of the form R(G,S), the spec-

ification of a type Φ : S → Aut(G) plays a key role in a symmetry-based

rigidity analysis of (G, p), it is natural to ask how we can find all the types

Φ of (G, p), how these types are related to each other, and under what con-

ditions (G, p) is of a unique type.

The following definition is essential to answer all of these questions.

Definition 3.3.1 Let (G, p) be a framework. Then we denote Aut(G, p) to

be the set of all α ∈ Aut(G) which satisfy p(v) = p(α(v)


Given a framework (G, p) and an automorphism α ∈ Aut(G, p), it is easy

to see that only vertices of G that have the same image under p can possibly

belong to the same permutation cycle of α. In particular, for every frame-

work (G, p) with an injective map p, we have Aut(G, p) = id, as we will

see in the proof of Corollary 3.3.3.

For the framework (Gt, p) in Figure 3.7 (a), we have Aut(Gt, p) =

id, (v3 v4)(v1)(v2) and for the framework (Gbp, p) in Figure 3.7 (b), we have

Aut(Gbp, p) = id, (v4 v5)(v1)(v2)(v3).Clearly, Aut(G, p) is a subgroup of Aut(G).

Theorem 3.3.1 Let G be a graph, S be a symmetry group, and Φ be a

map from S to Aut(G). Further, let (G, p) ∈ R(G,S,Φ) and x ∈ S. Then

Φ(x)Aut(G, p) = Aut(G, p)Φ(x), and an automorphism α of G satisfies

x(p(v)

)= p

(α(v)

)for all v ∈ V (G) if and only if α is an element of

Φ(x)Aut(G, p).

Proof. First, we show that Φ(x)Aut(G, p) = Aut(G, p)Φ(x). Since the

cosets Φ(x)Aut(G, p) and Aut(G, p)Φ(x) have the same cardinality, it suffices

70

to show that Φ(x)Aut(G, p) ⊆ Aut(G, p)Φ(x). Let α ∈ Φ(x)Aut(G, p), say

α = Φ(x) β, where β ∈ Aut(G, p). Then for v ∈ V (G), we have

x(p(v)

)= x

(p(β(v)

))= p

(Φ(x)

(β(v)

))= p

(α(v)

).

Since we also have x(p(v)

)= p

(Φ(x)(v)

), it follows that p

(α(v)

)=

p(Φ(x)(v)

)for all v ∈ V (G). Therefore,

p(α (

Φ(x))−1

(v))

= p(v) for all v ∈ V (G),

and hence α (Φ(x))−1 ∈ Aut(G, p). Thus, α ∈ Aut(G, p)Φ(x).

Now, α ∈ Aut(G) satisfies

x(p(v)

)= p

(α(v)

)for all v ∈ V (G)

if and only if

p(α(v)

)= p

(Φ(x)(v)


if and only if

p(α (

Φ(x))−1

(v))

= p(v) for all v ∈ V (G)

if and only if

α (Φ(x)

)−1 ∈ Aut(G, p)

if and only if

α ∈ Aut(G, p)Φ(x) = Φ(x)Aut(G, p). ¤

Corollary 3.3.2 Let G be a graph, S be a symmetry group, Φ be a map from

S to Aut(G), and (G, p) ∈ R(G,S,Φ). Then for every Ψ : S → Aut(G) distinct

from Φ, we have (G, p) /∈ R(G,S,Ψ) if and only if Aut(G, p) = id.

71

Proof. It follows directly from Theorem 3.3.1 that Aut(G, p) = id if and

only if for every x ∈ S, the automorphism Φ(x) is the only automorphism of

G that satisfies x(p(v)

)= p

(Φ(x)(v)

)for all v ∈ V (G). ¤

Corollary 3.3.2 asserts that the type Φ : S → Aut(G) of a framework

(G, p) ∈ R(G,S) is unique if and only if Aut(G, p) only contains the identity

automorphism of G. In particular, we have the following result.

Corollary 3.3.3 Let G be a graph, S be a symmetry group, and Φ be a map

from S to Aut(G). If the map p of a framework (G, p) ∈ R(G,S,Φ) is injective,

then (G, p) /∈ R(G,S,Ψ) for every Ψ : S → Aut(G) distinct from Φ.

Proof. Let α be an element of Aut(G, p). Then we have p(v) = p(α(v)

)

for all v ∈ V (G), and since p is injective it follows that v = α(v) for all

v ∈ V (G). Thus, α is the identity automorphism of G and the result follows

from Corollary 3.3.2. ¤

The following examples show that the converse of Corollary 3.3.3 does

not hold, that is, a framework (G, p) ∈ R(G,S) that is of a unique type

Φ : S → Aut(G) can possibly have a non-injective map p.

Example 3.3.3 The framework (G, p) in Figure 3.8 (a) is a non-injective

realization of (G, C2) (since p5 = p6) with Aut(G, p) = id. So, (G, p) ∈R(G,C2) is of the unique type Φ : C2 → Aut(G), where Φ(Id) = id and

Φ(C2) = (v1 v2)(v3 v4)(v5 v6).

Example 3.3.4 The framework (G, p) in Figure 3.8 (b) is a non-injective

realization of (G, C3) (since p4 = p5 = p6) with Aut(G, p) = id. So,

72

(G, p) ∈ R(G,C3) is of the unique type Φ : C3 → Aut(G), where Φ is the

homomorphism defined by Φ(C3) = (v1 v2 v3)(v4 v5 v6).

.

..p6

..p5

..p3 . .p4

..p1

. .p2

.(a)

...p5..p6

..p4

..p3

..p1 . .p2

.(b)

Figure 3.8: Non-injective realizations with Aut(G, p) = id.

Remark 3.3.1 Let (G, p) ∈ R(G,S,Φ) be a framework with Aut(G, p) = idand let (G, q) ∈ R(G,S,Φ) be an (S, Φ)-generic framework. It follows immedi-

ately from the definition of (S, Φ)-generic (Definition 3.2.2) that two joints

(vi, qi) and (vj, qj) of (G, q) can only satisfy qi = qj if pi = pj. This says that

(G, q) also satisfies Aut(G, q) = id. Therefore, by Corollary 3.3.2, being of

a unique type is an (S, Φ)-generic property.

Remark 3.3.2 If a framework (G, p) ∈ R(G,S) is of distinct types Φ1, . . . Φk,

where k ≥ 2, then (G, p) is not (S, Φt)-generic for some t ∈ 1, . . . , k, as the

following argument shows.

Suppose to the contrary that (G, p) is (S, Φi)-generic for all i = 1, . . . , k

and let l ∈ 1, . . . , k. Since Aut(G, p) 6= id, there exist vertices v 6= w

of G such that p(v) = p(w) and α(v) = w for some α ∈ Aut(G, p). Since

(G, p) is (S, Φl)-generic, there must exist non-trivial symmetry operations

x, y ∈ S such that Φl(x)(v) = v and Φl(y)(w) = w, and the symmetry

73

elements corresponding to x and y must be the origin 0 = p(v) = p(w). If

for each x ∈ S with Φl(x)(v) = v, we replace Φl(x) by α Φl(x), then we

obtain a map Φt, t 6= l, with the property that for all x ∈ S, Φt(x)(v) 6= v.

Thus, (G, p) is not (S, Φt)-generic, a contradiction.

As an example, consider the framework (Gt, p) in Figure 3.7 (a). (Gt, p)

is (C2, Θa)-generic, but not (C2, Θb)-generic, because p3 = p4 and Θb(v3) = v4

(see Example 3.3.1).

The framework in Figure 3.7 (b) is a realization of (Gbp, Cs) of type Ξa

and Ξb which is neither (Cs, Ξa)-generic nor (Cs, Ξb)-generic, because p4 = p5

(see Example 3.3.2).

3.4 When is a type Φ of a framework a ho-

momorphism?

We will see in the next chapter that in order to use techniques from

group representation theory to analyze the rigidity properties of a symmetric

framework (G, p) ∈ R(G,S,Φ), we need Φ to be a homomorphism. In this

section, we therefore investigate the natural question of whether a type Φ :

S → Aut(G) of a given framework (G, p) ∈ R(G,S) is in fact a homomorphism

(rather than just a map).

Theorem 3.4.1 Let S be a symmetry group and (G, p) be a framework in

R(G,S) with Aut(G, p) = id. Then the unique map Φ : S → Aut(G) for

which (G, p) ∈ R(G,S,Φ) is a homomorphism.

74

Proof. Let x and y be any two elements of S. Then Φ(y) Φ(x) ∈ Aut(G)

satisfies

(y x

)(p(v)

)= y

(p(Φ(x)(v)

))= p

((Φ(y) Φ(x)

)(v)


and, by Corollary 3.3.2, Φ(y)Φ(x) is the only automorphism of G with this

property. Thus, Φ(y x) = Φ(y) Φ(x). ¤

In particular, it follows from Corollary 3.3.3 and Theorem 3.4.1 that if

the map p of (G, p) ∈ R(G,S) is injective, then the unique type Φ of (G, p) is

a group homomorphism.

Theorem 3.4.2 Let S be a symmetry group, Φ : S → Aut(G) be a map,

and (G, p) be a framework in R(G,S,Φ).

(i) If Φ is a homomorphism, then Φ(S) is a subgroup of Aut(G);

(ii) if Φ(S) is a subgroup of Aut(G) and Φ(x) = Φ(y) whenever Φ(y) ∈Φ(x)Aut(G, p), then Φ is a homomorphism.

Proof. (i) It is a standard result in algebra that the homomorphic image of

a group is again a group.

(ii) Let x and y be any two elements of S. By the same argument as in

the proof of Theorem 3.4.1, we have

(y x

)(p(v)

)= p

((Φ(y) Φ(x)

)(v)


It follows from Theorem 3.3.1 that Φ(y x) ∈ (Φ(y) Φ(x)

)Aut(G, p).

By assumption, Φ(S) contains at most one element of each of the cosets

of Aut(G, p). Since Φ(S) is a group, the element of the coset(Φ(y)

75

Φ(x))Aut(G, p) that lies in Φ(S) must be Φ(y) Φ(x). It follows that

Φ(y x) = Φ(y) Φ(x) and the proof is complete. ¤

For a framework (G, p) ∈ R(G,S) with Aut(G, p) 6= id, there does not

necessarily exist any homomorphism Φ : S → Aut(G) for which (G, p) ∈R(G,S,Φ), as the following examples illustrate.

.

. . . .

.

.

....

.

.

.

.

.

.

.v2 .v3

.v4.v1

.(a)

.

.

.

.

.p2, p4

.p1, p3

.(b)

Figure 3.9: A graph G (a) and a realization (G, p) ∈ R(G,Cs) (b) for which

there does not exist a homomorphism Φ : Cs → Aut(G) so that (G, p) is of

type Φ.

Example 3.4.1 Consider the graph G and the 2-dimensional realization

(G, p) of G shown in Figure 3.9 (a) and (b), respectively. Let s be the

reflection whose mirror line is shown in Figure 3.9 (b). All vertices of G

that are illustrated with the same color in Figure 3.9 have the same image

under p. Observe that the ‘14-turn-automorphism’ σ of G that permutes

the vertices v1, v2, v3, and v4 according to the cycle (v1 v2 v3 v4) satisfies

s(p(vi)

)= p

(σ(vi)

)for all vi ∈ V (G). Thus, s is a symmetry operation

of (G, p), and hence (G, p) is an element of R(G,Cs), where Cs = Id, s.Note that Aut(G, p) = id, σ2. Therefore, by Theorem 3.3.1, id and σ

76

are the two automorphisms of G that can turn Id ∈ Cs into a symmetry oper-

ation of (G, p). Similarly, either one of the elements of σAut(G, p) = σ, σ3can turn s ∈ Cs into a symmetry operation of (G, p). It now follows from The-

orem 3.4.2 (i) that there does not exist any homomorphism Φ : Cs → Aut(G)

such that (G, p) ∈ R(G,Cs) is of type Φ, because we cannot choose two ele-

ments, one from each of the cosets Aut(G, p) and σAut(G, p), that form a

subgroup of Aut(G).

.

..v1

..v2

..v3

..v4 .

.v5

..v6

..v7

. .v8

..v9

.(a)

.

. .

.p1, p4, p7

.p2, p5, p8 .p3, p6, p9

.(b)

Figure 3.10: A graph G (a) and a realization (G, p) ∈ R(G,C3) (b) for which

there does not exist a homomorphism Φ : C3 → Aut(G) so that (G, p) is of

type Φ.

Example 3.4.2 Consider the graph G and the 2-dimensional realization

(G, p) of G shown in Figure 3.10 (a) and (b), respectively. As in the previous

example, all vertices of G that are illustrated with the same color in Figure

3.10 have the same image under p. Note that (G, p) is an element of R(G,C3),

where C3 = Id, C3, C23 is a symmetry group in dimension 2, because the

automorphism γ = (v1 v2 . . . v9) of G satisfies C3

(p(vi)

)= p

(γ(vi)

)for all

vi ∈ V (G) and γ2 satisfies C23

(p(vi)

)= p

(γ2(vi)

)for all vi ∈ V (G).

We have Aut(G, p) = id, γ3, γ6, and hence γAut(G, p) = γ, γ4, γ7 and

77

γ2Aut(G, p) = γ2, γ5, γ8. Since C3 ∈ C3 has order 3 and each element in

γAut(G, p) has order 9 it follows that there does not exist any homomorphism

Φ : C3 → Aut(G) such that (G, p) is of type Φ.

Note that Examples 3.4.1 and 3.4.2 can easily be extended to obtain

further examples of frameworks (G, p) and symmetry groups S with the

property that there exists no homomorphism Φ : S → Aut(G) for which

(G, p) ∈ R(G,S,Φ).

78

Chapter 4

Using group representation

theory to analyze symmetric

frameworks

It is a common method in engineering, physics, and chemistry to apply

techniques from group representation theory to the analysis of symmetric

structures (see, for example, [26, 27, 34, 35, 44, 45]). In particular, some

recent papers have used these techniques to gain insight into the rigidity

properties of symmetric frameworks consisting of rigid bars and flexible joints

[15, 25, 43, 44, 53].

One of the fundamental observations resulting from this approach for

studying the rigidity of symmetric frameworks is due to R. Kangwai and S.

Guest ([44]): given a symmetric framework (G, p) and a non-trivial subgroup

S of its point group, there are techniques to block-diagonalize the rigidity

matrix of (G, p) into submatrix blocks in such a way that each block corre-

79

sponds to an irreducible representation of S. A number of interesting and

useful results concerning the rigidity of symmetric frameworks are based on

this block-diagonalization of the rigidity matrix [15, 25, 43]. However, since

the main focus of the work in [44], as well as in [15], [25], and [43], lies

on applications in engineering and chemistry, many of these results are not

presented with a mathematically precise formulation nor with a complete

mathematical verification.

In this chapter, we establish several major results. First, in Section 4.1,

we use the mathematical foundation we established in the previous chapter

to give a complete proof for the fact that the rigidity matrix of a symmetric

framework can be block-diagonalized in the way described above. Fundamen-

tal to this proof are our mathematically explicit definitions for the ‘external’

and ‘internal’ representation which were introduced in [25] and [44] only by

means of an example, and Lemma 4.1.1 which establishes the key connection

between these two representations.

Secondly, in Section 4.2, we apply the results of Section 4.1 to give a de-

tailed mathematical proof for the symmetry-extended version of Maxwell’s

rule given in [25]. This rule provides further necessary conditions (in ad-

dition to Maxwell’s original condition given in Theorem 2.2.7) for a sym-

metric framework to be isostatic. While the symmetry-extended version of

Maxwell’s rule, as formulated in [25], is only applicable to 2- or 3-dimensional

frameworks with injective configurations, we establish a more general result

in Section 4.2, namely a rule that can be applied to both injective and non-

injective realizations in all dimensions. The proof of this result is based on

Theorem 4.2.2 which in turn relies on the fact that the rigidity matrix of a

80

symmetric framework can be block-diagonalized as described in Section 4.1.

An alternate approach to proving the symmetry-extended version of

Maxwell’s rule in [25], as well as various generalizations of this rule to other

types of geometric constraint systems, is presented in [53] (see also Chapter

7).

In order to apply the symmetry-extended version of Maxwell’s rule to a

given framework (G, p), it is necessary to determine the dimensions of the

subspaces of infinitesimal rigid motions of (G, p) that are invariant under

the external representation. While in [25], the question of how to find the

dimensions of these subspaces is only briefly addressed and not answered

completely from a mathematical point of view (in particular, for all frame-

works in dimensions higher than 3, this question is not addressed at all), in

Section 4.2, we describe in detail how to determine the dimensions of these

subspaces for an arbitrary-dimensional framework.

The results of Sections 4.1 and 4.2 will also be presented in the paper

[56].

Since in [25] and [44], the rigidity properties of a symmetric framework

are studied from both the kinematic and static point of view simultaneously,

we develop the corresponding mathematical theory in this chapter in the

same manner.

In Section 4.3, we use the symmetry-extended version of Maxwell’s rule

to show that a symmetric isostatic framework in 2D or 3D must obey some

very simply stated restrictions on the number of structural elements that

are ‘fixed’ by various symmetry operations of the framework. In particular,

it turns out that a 2-dimensional isostatic framework must belong to one

81

of only six point groups. For 3-dimensional isostatic frameworks, all point

groups are possible. However, there still exist restrictions on the placement

of structural components. While analogous restrictions on the number of

‘fixed’ structural components can be established for symmetric frameworks

in an arbitrary dimension using the results of Section 4.2, we focus our atten-

tion on frameworks in dimensions 2 and 3, since they are of special interest

for current applications.

Most of the results in Section 4.3 appeared in the joint paper [15]. The

derivations also appeared there, and Sections 4.1 and 4.2 now provide a proof

that these methods are correct.

Finally, in Section 4.4 we use the results of Sections 4.1 and 4.2 to estab-

lish necessary conditions for a symmetric framework to be independent or

infinitesimally rigid.

4.1 Block-diagonalization of the rigidity ma-

trix

4.1.1 Basic definitions in group representation theory

We need the following notions from group representation theory.

Definition 4.1.1 Let S be a group and V be an n-dimensional vector space

over the field F . A linear representation of S with representation space V

is a group homomorphism H from S to GL(V ), where GL(V ) denotes the

group of all automorphisms of V . The dimension n of V is called the degree

of H.

82

Two linear representations H1 : S → GL(V1) and H2 : S → GL(V2) are

said to be equivalent if there exists an isomorphism h : V1 → V2 such that

h H1(x) h−1 = H2(x) for all x ∈ S.

Definition 4.1.2 Let S be a group, V be a vector space over the field F

and H : S → GL(V ) be a linear representation of S. A subspace U of V is

said to be H-invariant (or simply invariant if H is clear from the context)

if H(x)(U) ⊆ U for all x ∈ S. H is called irreducible if V and 0 are the

only H-invariant subspaces of V .

Note that the property of irreducibility depends on the field F . Since

we only consider frameworks in the real vector space Rd, the representation

space of any linear representation in this thesis is assumed to be a real vector

space.

In the examples throughout this thesis we use the Mulliken symbols (see

Appendix A or [6, 19, 37]) to denote the irreducible representations of a given

group. This is one of the standard notations in group representation theory

and its applications.

Definition 4.1.3 A linear representation H : S → GL(V ) is said to be

unitary with respect to a given inner product 〈v, w〉 if

〈H(x)(v), H(x)(w)〉 = 〈v, w〉 for all v, w ∈ V and all x ∈ S.

Remark 4.1.1 A unitary representation has the property that the orthog-

onal complement of an invariant subspace is again invariant [60].

Definition 4.1.4 Let H : S → GL(V ) be a linear representation of a group

S and let U be an invariant subspace of V . If for all x ∈ S, we restrict the

83

automorphism H(x) of V to the subspace U , then we obtain a new linear

representation H(U) of S with representation space U . H(U) is said to be a

subrepresentation of H.

Definition 4.1.5 Let H1 : S → GL(V1) and H2 : S → GL(V2) be two

linear representations of a group S. Then H1 ⊕ H2 : S → GL(V1 ⊕ V2)

is the representation of S which sends x ∈ S to H1 ⊕ H2(x), where H1 ⊕H2(x)

((v1, v2)

)=

(H1(x)(v1), H2(x)(v2)

)for all v1 ∈ V1 and v2 ∈ V2.

Definition 4.1.6 Let S be a group and F be a field. A matrix representation

of S is a homomorphism H from S to GL(n, F ), where GL(n, F ) denotes the

group of all invertible n× n matrices with entries in F .

Two matrix representations H1 : S → GL(n, F ) and H2 : S → GL(n, F )

are said to be equivalent if there exists an invertible matrix M such that

MH1(x)M−1 = H2(x) for all x ∈ S, in which case we write H1 w H2.

Let S be a group, V be an n-dimensional vector space over the field F ,

and H : S → GL(V ) be a linear representation of S. Given a basis B of

V , we may associate a matrix representation HB : S → GL(n, F ) to H by

defining HB(x) to be the matrix that represents the automorphism H(x)

with respect to the basis B for all x ∈ S. HB is then said to correspond

to H with respect to B. Note that two matrix representations H1 and H2

correspond to equivalent linear representations if and only if H1 w H2.

84

4.1.2 The internal and external representation

Given a graph G, a symmetry group S, and a homomorphism Φ : S →Aut(G), we define two particular matrix representations of S, the external

and the internal representation, both of which depend on G and Φ. These

two representations play the key role in a symmetry-based rigidity analysis

of a framework (G, p) ∈ R(G,S,Φ). Note that our definitions of these represen-

tations are mathematically explicit definitions of the external and internal

representation introduced in [25] and [44].

Definition 4.1.7 Let G be a graph with V (G) = v1, v2, . . . , vn and

E(G) = e1, e2, . . . , em, S be a symmetry group in dimension d, and Φ

be a homomorphism from S to Aut(G). For x ∈ S, let Mx denote the or-

thogonal d× d matrix which represents x with respect to the canonical basis

of Rd.

The external representation of S (with respect to G and Φ) is the matrix

representation He : S → GL(dn,R) that sends x ∈ S to the matrix He(x)

which is obtained from the transpose of the n×n permutation matrix corre-

sponding to Φ(x) (with respect to the enumeration V (G) = v1, v2, . . . , vn)by replacing each 1 with the matrix Mx and each 0 with a d× d zero-matrix.

The internal representation of S (with respect to G and Φ) is the matrix

representation Hi : S → GL(m,R) that sends x ∈ S to the transpose of the

permutation matrix corresponding to the permutation of E(G) (with respect

to the enumeration E(G) = e1, e2, . . . , em) which is induced by Φ(x).

Remark 4.1.2 It is easy to verify that both the external representation He

and the internal representation Hi of S (with respect to G and Φ) are in fact

85

matrix representations of the group S, provided that Φ is a homomorphism.

If, however, Φ is not a homomorphism, then He and Hi are also not homo-

morphisms, in which case neither He nor Hi is a matrix representation of the

group S.

Example 4.1.1 To illustrate the previous definition, let K3 be the com-

plete graph with V (K3) = v1, v2, v3 and E(K3) = e1, e2, e3, where

e1 = v1, v2, e2 = v1, v3 and e3 = v2, v3. Further, let Cs = Id, sbe the symmetry group in dimension 2 with

MId =

1 0

0 1

and Ms =

−1 0

0 1

,

and let Φ : Cs → Aut(K3) be the homomorphism defined by Φ(s) =

(v1 v2)(v3). Then we have

He(Id) =

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

, He(s) =

0 0 −1 0 0 0

0 0 0 1 0 0

−1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0 −1 0

0 0 0 0 0 1

,

Hi(Id) =

1 0 0

0 1 0

0 0 1

, Hi(s) =

1 0 0

0 0 1

0 1 0

.

For further examples, see [44] or [45].

86

...p1

..p2

..p3

.e1

.e3.e2

Figure 4.1: A framework (K3, p) ∈ R(K3,Cs,Φ).

4.1.3 The block-diagonalization

In this section, we use the mathematically explicit definitions of the ex-

ternal and internal representation from the previous section to prove that the

rigidity matrix of a symmetric framework can be transformed into a block-

diagonalized form. Basic to this proof is Lemma 4.1.1 which discloses the

essential mathematical connection between the external and internal repre-

sentation.

Recall from Section 2.2 that in the study of infinitesimal rigidity, we

consider the equation

R(G, p)u = z,

where R(G, p) is the rigidity matrix of a framework (G, p), u ∈ Rd|V (G)| is a

column vector that represents an assignment of d-dimensional displacement

vectors to the joints of (G, p), and z ∈ R|E(G)| is the column vector that

represents the distortions in the bars of (G, p) that are induced by u. The

component of z that corresponds to the edge vi, vj of G is also known as

the strain induced on the bar (vi, pi), (vj, pj) by u.

87

Similarly, in the study of static rigidity, we consider the equation

R(G, p)T ω = l,

where the column vector ω ∈ R|E(G)| is a stress of (G, p) and the column

vector l ∈ Rd|V (G)| is the load on (G, p) which is resolved by ω.

Now, suppose (G, p) is a symmetric framework in the set R(G,S,Φ), where S

is a symmetry group in dimension d and Φ : S → Aut(G) is a homomorphism.

Then, using the notation of Definition 4.1.7, and assuming that the ith row

of the rigidity matrix R(G, p) of (G, p) corresponds to the edge ei of G,

we have the following fundamental property of the external and internal

representation of S (with respect to G and Φ).

Lemma 4.1.1 Let G be a graph, S be a symmetry group, Φ be a homomor-

phism from S to Aut(G), and p ∈ ⋂x∈S Lx,Φ.

(i) If R(G, p)u = z, then for all x ∈ S, we have R(G, p)He(x)u = Hi(x)z;

(ii) if R(G, p)T ω = l, then for all x ∈ S, we have R(G, p)T Hi(x)ω =

He(x)l.

Proof. (i) Suppose R(G, p)u = z. Fix x ∈ S and let Mx be the orthogonal

matrix representing x with respect to the canonical basis of Rd. Also, let

Φ(x)(vi) = vk and Φ(x)(vj) = vl, and let ef = vi, vj and eh = vk, vl.Then, since p ∈ ⋂

x∈S Lx,Φ, we have

Mxpi = pk and Mxpj = pl.

By the definition of Hi(x), we have

(Hi(x)z

)h

= (z)f .

88

...pi

..pj

. .Mxpi = pk

. .Mxpj = pl

.ef .eh

.Mx

.Mx

.

.

.(z)h

.uk

.ul

.Hi(x)

.He(x)

.He(x)

.

.

.(z)f

.Mxui

.Mxuj

Figure 4.2: Illustration of the proof of Lemma 4.1.1 (i).

Similarly, it follows from the definition of He(x) that if u ∈ Rdn is replaced

by He(x)u, then uk ∈ Rd is replaced by Mxui and ul ∈ Rd by Mxuj. By the

definition of R(G, p), we have

(R(G, p)u

)h

= (z)h = (pk − pl) · uk + (pl − pk) · ul.

Therefore,

(R(G, p)He(x)u

)h

= (pk − pl) ·Mxui + (pl − pk) ·Mxuj

=(Mxpi −Mxpj

) ·Mxui +(Mxpj −Mxpi

) ·Mxuj

=(Mx(pi − pj)

) ·Mxui +(Mx(pj − pi)

) ·Mxuj

= (pi − pj) · ui + (pj − pi) · uj

= (z)f .

The penultimate equality sign is valid because the canonical inner product

on Rd is invariant under the orthogonal transformation x ∈ S. This proves

(i).

(ii) Suppose R(G, p)T ω = l. Fix x ∈ S and let Φ(x)(vi) = vk. Then,

since p ∈ ⋂x∈S Lx,Φ, we have

Mxpi = pk.

Let vi1 , vi2 , . . . , vij be the vertices in V (G) that are adjacent to vi, and let

eft = vi, vit for t = 1, 2, . . . , j. Further, choose an enumeration of the j

89

...pi

..pj

..Mxpi.= pk

..Mxpj.= pl

.(z)f .(z)h

.ui

.uj

.uk

.ul

.x

..pi

..pj

..pk

. .pl

.(z)h .(z)f

.Mxuk.Mxui

.Mxul

.Mxuj

.x

Figure 4.3: Illustration of the proof of Lemma 4.1.1 (i) in the case where x

is a reflection.

vertices that are adjacent to vk in such a way that

Mxpit = pkt ,

and let eht = vk, vkt for t = 1, 2, . . . , j. For the vertex vk, the equation

R(G, p)T ω = l yields the vector-equation

(pk − pk1)(ω)h1 + . . . + (pk − pkj)(ω)hj

= lk. (4.1)

...pi . .Mxpi = pk

..pi1

..pij

.ef1

.efj

..Mxpi1 = pk1

..Mxpij = pkj

.eh1

.ehj

.Mx

.Mx

.Mx

..lk

.(ω)h1

.(ω)hj

. .Mxli

.(ω)f1

.(ω)fj

.He(x)

.Hi(x)

.Hi(x)

Figure 4.4: Illustration of the proof of Lemma 4.1.1 (ii).

If l ∈ Rdn is replaced by He(x)l, then on the right-hand side of equation

(4.1), lk ∈ Rd is replaced by Mxli and if ω is replaced by Hi(x)ω, then the

90

left-hand side of equation (4.1) is replaced by

(pk − pk1)(ω)f1 + . . . + (pk − pkj)(ω)fj

=(Mxpi −Mxpi1

)(ω)f1 + . . . +

(Mxpi −Mxpij

)(ω)fj

= Mx

((pi − pi1)(ω)f1 + . . . + (pi − pij)(ω)fj

)

= Mxli.

This completes the proof. ¤

In the following, we again let G be a graph with V (G) = v1, v2, . . . , vnand E(G) = e1, e2, . . . , em, S be a symmetry group in dimension d, and Φ

be a homomorphism from S to Aut(G).

Let He be the external and Hi be the internal representation of S (with

respect to G and Φ). Then we let H ′e : S → GL(Rdn) be the linear represen-

tation of S that sends x ∈ S to the automorphism H ′e(x) which is represented

by the matrix He(x) with respect to the canonical basis of the R-vector space

Rdn. Similarly, we let H ′i : S → GL(Rm) be the linear representation of S

that sends x ∈ S to the automorphism H ′i(x) which is represented by the

matrix Hi(x) with respect to the canonical basis of the R-vector space Rm.

So, the external representation He corresponds to the linear representation

H ′e with respect to the canonical basis of Rdn and the internal representation

Hi corresponds to the linear representation H ′i with respect to the canonical

basis of Rm.

From group representation theory we know that every finite group has, up

to equivalency, only finitely many irreducible linear representations and that

every linear representation of such a group can be written uniquely, up to

equivalency of the direct summands, as a direct sum of the irreducible linear

91

representations of this group [42, 60]. So, let S have r pairwise non-equivalent

irreducible linear representations I1, I2, . . . , Ir and let

H ′e = λ1I1 ⊕ . . .⊕ λrIr, where λ1, . . . , λr ∈ N ∪ 0. (4.2)

For each t = 1, . . . , r, there exist λt subspaces(V

(It)e

)1, . . . ,

(V

(It)e

)λt

of the

R-vector space Rdn which correspond to the λt direct summands in (4.2), so

that

Rdn = V (I1)e ⊕ . . .⊕ V (Ir)

e , (4.3)

where

V (It)e =

(V (It)

e

)1⊕ . . .⊕ (

V (It)e

)λt

. (4.4)

Let(B

(It)e

)1, . . . ,

(B

(It)e

)λt

be bases of the subspaces in (4.4). Then

B(It)e =

(B(It)

e

)1∪ . . . ∪ (

B(It)e

)λt

is a basis of V(It)e and

Be = B(I1)e ∪ . . . ∪B(Ir)

e (4.5)

is a basis of the R-vector space Rdn.

Consider now the matrix representation He that corresponds to the linear

representation H ′e with respect to the basis Be. For x ∈ S, we have

He(x) = T−1e He(x)Te,

where the ith column of Te is the coordinate vector of the ith basis vector

of Be relative to the canonical basis, that is, Te is the matrix of the basis

transformation from the canonical basis of the R-vector space Rdn to the

basis Be. The column vectors of He(x) are the coordinates of the images of

92

the basis vectors in Be under H ′e(x) relative to the basis Be. So, for each

x ∈ S, the matrix He(x) has the same block form, namely

He(x) =

(A

(I1)e

)1(x)

. . . 0(A

(I1)e

)λ1

(x)

. . .(A

(Ir)e

)1(x)

0. . .

(A

(Ir)e

)λr

(x)

.

The block-matrix(A

(It)e

)j(x) represents the restriction of the linear trans-

formation H ′e(x) to the subspace

(V

(It)e

)j

with respect to the basis(B

(It)e

)j.

Since for a given t, each of the subspaces(V

(It)e

)j, j = 1, . . . , λt, corresponds

to the same irreducible linear representation It, we can choose the bases of

the subspaces(V

(It)e

)j

in such a way that

(A(It)

e

)1(x) = . . . =

(A(It)

e

)λt

(x) =: A(It)e (x).

In the following we assume that the basis Be is chosen in this way.

The above observations about the linear representation H ′e of S can be

transferred analogously to the linear representation H ′i of S. Let the direct

sum decomposition of H ′i be given by

H ′i = µ1I1 ⊕ . . .⊕ µrIr, where µ1, . . . , µr ∈ N ∪ 0. (4.6)

For each t = 1, . . . , r, there exist µt subspaces(V

(It)i

)1, . . . ,

(V

(It)i

)µt

of the

R-vector space Rm which correspond to the µt direct summands in (4.6), so

that

Rm = V(I1)i ⊕ . . .⊕ V

(Ir)i , (4.7)

93

where

V(It)i =

(V

(It)i

)1⊕ . . .⊕ (

V(It)i

)µt

. (4.8)

Let(B

(It)i

)1, . . . ,

(B

(It)i

)µt

be bases of the subspaces in (4.8). Then

B(It)i =

(B

(It)i

)1∪ . . . ∪ (

B(It)i

)µt

is a basis of V(It)i and

Bi = B(I1)i ∪ . . . ∪B

(Ir)i

is a basis of the R-vector space Rm.

Consider now the matrix representation Hi that corresponds to the linear

representation H ′i with respect to the basis Bi. Let Ti be the matrix of the

basis transformation from the canonical basis of the R-vector space Rm to

the basis Bi. Then for x ∈ S, we have

Hi(x) = T−1i Hi(x)Ti.

So, the matrix Hi(x) has the same block form for each x ∈ S, namely

Hi(x) =

(A

(I1)i

)1(x)

. . . 0(A

(I1)i

)µ1

(x)

. . .(A

(Ir)i

)1(x)

0. . .

(A

(Ir)i

)µr

(x)

,

and for each t = 1, 2, . . . , r, we can choose the bases of the subspaces(V

(It)i

)j

in such a way that

(A

(It)i

)1(x) = . . . =

(A

(It)i

)µt

(x) =: A(It)i (x) = A(It)

e (x).

94

In the following we assume that Bi is chosen in this way.

Definition 4.1.8 With the notation above, we say that a vector v ∈ Rdn

is symmetric with respect to the irreducible linear representation It of S if

v ∈ V(It)e . Similarly, we say that a vector w ∈ Rm is symmetric with respect

to the irreducible linear representation It of S if w ∈ V(It)i .

We are now in the position to state the fundamental theorem for analyzing

the rigidity properties of a symmetric framework using group representation

theory.

Theorem 4.1.2 Let G be a graph, S be a symmetry group with pairwise non-

equivalent irreducible linear representations I1, . . . , Ir, Φ be a homomorphism

from S to Aut(G), and p ∈ ⋂x∈S Lx,Φ.

(i) If R(G, p)u = z and u is symmetric with respect to It, then z is also

symmetric with respect to It;

(ii) if R(G, p)T ω = l and ω is symmetric with respect to It, then l is also

symmetric with respect to It.

Proof. (i) Suppose S is a symmetry group in dimension d and G is a graph

with n vertices. Let u ∈ (V

(It)e

)j. By the direct sum decomposition of V

(It)e

in (4.4), the result follows if we can show that z = R(G, p)u ∈ V(It)i . By

the decomposition of R|E(G)| into direct summands in (4.8), z has a unique

decomposition of the form

z =r∑

α=1

µα∑

β=1

zα,β, where zα,β ∈(V

(Iα)i

)β.

95

We now interpret R(G, p) : Rdn → R|E(G)| as a linear transformation and

for given m and k, we define the projection map Rm,k corresponding to

R(G, p)|(V

(It)e

)j

by

Rm,k :

(V

(It)e

)j→ (

V(Im)i

)k

u 7→ zm,k

.

We need to show that for all m 6= t, Rm,k is the zero map. So, let m 6= t.

Clearly, Rm,k is a linear transformation.

The image of Rm,k is an H ′i-invariant subspace of

(V

(Im)i

)k, as the fol-

lowing argument shows. Fix x ∈ S and let z′ be in the image of Rm,k, say

z′ = Rm,k(u′). Then, by assumption, H ′

e(x)(u′) ∈ (V

(It)e

)j

and, by Lemma

4.1.1 (i), H ′i(x)(z′) is the image of H ′

e(x)(u′) under Rm,k.

Since Im is an irreducible linear representation of S,(V

(Im)i

)k

and 0are the only H ′

i-invariant subspaces of(V

(Im)i

)k. If the image of Rm,k is the

null-space, then we are done, otherwise Rm,k is surjective.

Next, we show that the kernel of Rm,k is an H ′e-invariant subspace of

(V

(It)e

)j. Fix x ∈ S and let u′ be in the kernel of Rm,k, that is, Rm,k(u

′) = 0.

Then, again by Lemma 4.1.1 (i), the image of H ′e(x)(u′) under Rm,k is

H ′i(x)(0) = 0, and hence H ′

e(x)(u′) is also in the kernel of Rm,k.

Since It is an irreducible linear representation of S, we either have

ker (Rm,k) =(V

(It)e

)j, in which case we are done, or ker (Rm,k) = 0,

in which case Rm,k is injective.

So, assume Rm,k is bijective. Let the matrix that represents Rm,k with

respect to the bases(B

(It)e

)jand

(B

(Im)i

)k

be denoted by Rm,k. Then Rm,k is

an invertible matrix. Let u be the coordinate vector of an element in(V

(It)e

)j

relative to the basis(B

(It)e

)j

and let z be the coordinate vector of the image

96

of u under Rm,k relative to the basis(B

(Im)i

)k. Then, by Lemma 4.1.1 (i),

for any x ∈ S, we have

Rm,k

(A(It)

e

)j(x)u =

(A

(Im)i

)k(x)z =

(A

(Im)i

)k(x)Rm,ku,

and hence also

Rm,k

(A(It)

e

)j(x) =

(A

(Im)i

)k(x)Rm,k.

Therefore,

Rm,k

(A(It)

e

)j(x)R−1

m,k =(A

(Im)i

)k(x) =

(A(Im)

e

)k(x) for all x ∈ S,

which says that It and Im are equivalent representations, a contradiction.

This completes the proof of part (i).

With the help of Lemma 4.1.1 (ii), part (ii) can be proved completely

analogously to part (i). ¤

Theorem 4.1.2 (i) says that if u ∈ Rdn is an assignment of displacement

vectors to the joints of a framework (G, p) ∈ R(G,S,Φ) and u is symmetric

with respect to It, then the strains induced on the bars of (G, p) by u must

also be symmetric with respect to It. Similarly, Theorem 4.1.2 (ii) says that

if ω is a resolution of an equilibrium load l on (G, p) ∈ R(G,S,Φ) and ω is

symmetric with respect to It, then l must also be symmetric with respect to

It.

An immediate consequence of Theorem 4.1.2 is that the matrices R(G, p)

and R(G, p)T can be block-diagonalized in such a way that the original rigid-

ity problems R(G, p)u = z and R(G, p)T ω = l are decomposed into subprob-

lems, where each subproblem considers, respectively, the relationship between

vectors u and z and vectors ω and l that are symmetric with respect to the

same irreducible linear representation It. This is specified in

97

Corollary 4.1.3 Let G be a graph, S be a symmetry group with pair-

wise non-equivalent irreducible linear representations I1, . . . , Ir, Φ be a ho-

momorphism from S to Aut(G), and p ∈ ⋂x∈S Lx,Φ. Then the matrices

T−1i R(G, p)Te and T−1

e R(G, p)T Ti are block-diagonalized in such a way that

there exists (at most) one submatrix block for each irreducible linear repre-

sentation It of S.

Proof. Suppose R(G, p)u = z, and let u be the coordinate vector of u rela-

tive to the basis Be and z be the coordinate vector of z relative to the basis Bi.

Further, let R(G, p) be the matrix that represents the linear transformation

R(G, p) with respect to the bases Be and Bi, that is,

R(G, p) = T−1i R(G, p)Te.

Then, by changing coordinates relative to the canonical bases of Rdn and Rm

into coordinates relative to the bases Be and Bi, the equation

R(G, p)u = z

is converted into the equation

R(G, p)u = z.

By Theorem 4.1.2 (i), the matrix R(G, p) is block-diagonalized in such a way

that there exists (at most) one submatrix block for each irreducible linear

representation It of S and the submatrix block corresponding to It is a matrix

of the size dim(V

(It)i

) × dim(V

(It)e

). In particular, a submatrix block can

possibly be an ‘empty matrix’ which has rows but no columns or alternatively

columns but no rows.

98

Similarly, if we denote ω to be the coordinate vector of ω relative to the

basis Bi, l to be the coordinate vector of l relative to the basis Be, and

R(G, p)T = T−1e R(G, p)T Ti,

then we may carry out the same changes of coordinates as above to convert

the equation

R(G, p)T ω = l

into the equation

R(G, p)T ω = l.

By Theorem 4.1.2 (ii), the matrix R(G, p)T is again block-diagonalized in

such a way that there exists (at most) one block for each It. ¤

Remark 4.1.3 Note that the matrix R(G, p)T is equal to the transpose of

the matrix R(G, p) if and only if both of the matrices Te and Ti are orthogonal

matrices (i.e., T−1e = T T

e and T−1i = T T

i ) if and only if both Be and Bi

are orthonormal bases. Since the external and internal representation are

both unitary representations (for all x ∈ S, He(x) and Hi(x) are orthogonal

matrices), the invariant subspaces in (4.3) and (4.7) are mutually orthogonal

(see [24, 60], for example). Thus, Be and Bi can always be chosen to be

orthonormal.

Example 4.1.2 Let K3, Cs = Id, s, and Φ be as in Example 4.1.1 and

consider the framework (K3, p) ∈ R(K3,Cs,Φ) shown in Figures 4.1 and 4.5,

where

p1 =

−1

0

, p2 =

1

0

, and p3 =

0

2

.

99

The rigidity matrix of (K3, p) is given by

R(K3, p) =

(p1 − p2)1 (p1 − p2)2 (p2 − p1)1 (p2 − p1)2 0 0

(p1 − p3)1 (p1 − p3)2 0 0 (p3 − p1)1 (p3 − p1)2

0 0 (p2 − p3)1 (p2 − p3)2 (p3 − p2)1 (p3 − p2)2

=

−2 0 2 0 0 0

−1 −2 0 0 1 2

0 0 1 −2 −1 2

.

The symmetry group Cs has two non-equivalent irreducible linear repre-

sentations both of which are of degree 1. In the Mulliken notation, they are

denoted by A′ and A′′. A′ maps both Id and s to the identity transforma-

tion, whereas A′′ maps Id to the identity transformation and s to the linear

transformation A′′(s) which is defined by A′′(s)(x) = −x for all x ∈ R. We

have

R6 = V (A′)e ⊕ V (A′′)

e

and

R3 = V(A′)i ⊕ V

(A′′)i .

It is easy to see that the elements of the subspace V(A′)e of R6 are of the form

u1

u2

−u1

u2

0

u3

, where u1, u2, u3 ∈ R,

100

(see Figure 4.5 (a)), so that an orthonormal basis B(A′)e of V

(A′)e is given by

B(A′)e =

1√2

0

− 1√2

0

0

0

,

0

1√2

0

1√2

0

0

,

0

0

0

0

0

1

.

Similarly, the elements of the subspace V(A′′)e of R6 are of the form

u1

u2

u1

−u2

u3

0

, where u1, u2, u3 ∈ R,

(see Figure 4.5 (b)), so that an orthonormal basis B(A′′)e of V

(A′′)e is given by

B(A′′)e =

1√2

0

1√2

0

0

0

,

0

1√2

0

− 1√2

0

0

,

0

0

0

0

1

0

.

Orthonormal bases B(A′)i and B

(A′′)i for the subspaces V

(A′)i and V

(A′′)i of R3

101

can be found analogously (see Figure 4.5 (c), (d)). We let

B(A′)i =

1

0

0

,

0

1√2

1√2

.

and

B(A′′)i =

0

1√2

− 1√2

.

Therefore, we have

...p1

..p2

..p3

.e1

.e3.e2

.

(u1

u2

).

(−u1

u2

)

.

(0u3

)

.(a)

..p1

..p2

..p3

.e1

.e3.e2

.z2

.z1

.z2

.(c)

...p1

..p2

. .p3

.e1

.e3.e2

.

(u1

u2

)

.

(u1

−u2

)

.

(u3

0

)

.(b)

..p1

..p2

..p3

.e1

.e3.e2

.z1

.0

.−z1

.(d)

Figure 4.5: (a, b) Vectors of the H ′e-invariant subspaces V

(A′)e (a) and V

(A′′)e

(b) of R6; (c, d) vectors of the H ′i-invariant subspaces V

(A′)i (c) and V

(A′′)i

(d) of R3.

102

Te =

1√2

0 0 1√2

0 0

0 1√2

0 0 1√2

0

− 1√2

0 0 1√2

0 0

0 1√2

0 0 − 1√2

0

0 0 0 0 0 10 0 1 0 0 0

and

Ti =

1 0 00 1√

21√2

0 1√2− 1√

2

.

Thus,

R(K3, p) = T−1i R(K3, p)Te =

−2√

2 0 0 0 0 0

−1 −2 2√

2 0 0 0

0 0 0 −1 −2√

2

and

R(K3, p)T = T−1e R(K3, p)T Ti =

−2√

2 −1 00 −2 0

0 2√

2 00 0 −10 0 −2

0 0√

2

.

Remark 4.1.4 In the previous example, we were able to find the invariant

subspaces V(A′)e , V

(A′′)e of R6 and V

(A′)i , V

(A′′)i of R3 by inspection because Cs is

a small symmetry group with only two elements. This is of course generally

not possible. There are, however, some standard methods and algorithms

for finding the symmetry adapted bases Be and Bi for any given symmetry

group. Good sources for these methods are [24, 50], for example.

As we will see in Section 4.2, knowledge of only the sizes of the subma-

trix blocks that appear in the block-diagonalized rigidity matrices of a given

symmetric framework allows us to gain significant insight into the rigidity

properties of the framework. Since, with the aid of character theory, the

103

sizes of these submatrix blocks can be determined very easily without ex-

plicitly finding the bases Be and Bi, there exist a number of applications of

Corollary 4.1.3 (such as the symmetry-extended version of Maxwell’s rule we

will discuss in the following sections) that do not require finding the block-

diagonalized rigidity matrices explicitly.

Remark 4.1.5 The matrices R(G, p)TR(G, p) and R(G, p)R(G, p)T are also

of interest in some areas of rigidity theory [17, 44]. In structural engi-

neering, these matrices are called the stiffness matrix and the flexibility

matrix, respectively. It follows immediately from Corollary 4.1.3 that if

p ∈ ⋂x∈S Lx,Φ, then these matrices can also be block-diagonalized in such a

way that there exists (at most) one block for each irreducible representation

It of S. In fact, it is easy to see that the matrices T−1e R(G, p)TR(G, p)Te

and T−1i R(G, p)R(G, p)T Ti have the desired block-form.

In the following sections of this chapter, as well as in Chapter 6, we will

discuss some interesting applications of the fact that the rigidity matrix of

a symmetric framework can be block-diagonalized in the way described in

Corollary 4.1.3.

4.2 A symmetry-extended version of

Maxwell’s rule

Recall from Section 2.2.5 that Maxwell’s rule (Theorem 2.2.7) gives a

necessary condition for a d-dimensional framework (G, p) to be isostatic. If

104

(G, p) is a 2- or 3-dimensional symmetric framework with an injective con-

figuration, then the symmetry-extended version of Maxwell’s rule given in

[25] provides further necessary conditions for (G, p) to be isostatic. Though

the rule in [25] is a useful tool for engineers and chemists to analyze the

rigidity properties of symmetric structures in 2D and 3D, it is unsatisfactory

from a mathematical point of view since it cannot be applied to frameworks

in dimensions higher than 3 and since a complete mathematical proof of

this result has not been provided. In the following sections, we aim to give a

mathematical proof not only for the rule in [25], but also for an extended rule

that can be applied to a symmetric framework with a possibly non-injective

configuration in an arbitrary dimension.

In this section, we first develop all the necessary mathematical back-

ground that was omitted in [25]. This background consists of three major

parts. First, we show that the subspaces R and T of all rotational and trans-

lational infinitesimal rigid motions of a given symmetric framework (G, p)

are invariant under the external representation H ′e (see Lemma 4.2.1). This

allows us to define subrepresentations of H ′e for the subspaces R and T . We

then prove that the block-diagonalized form of the rigidity matrix of (G, p)

gives rise to additional necessary conditions for (G, p) to be isostatic (see

Theorem 4.2.2). The symmetry-extended version of Maxwell’s rule is based

on these conditions. Finally, we describe in detail how to determine the

dimensions of the H ′e-invariant subspaces of R and T . This is essential in

applying the symmetry-extended version of Maxwell’s rule to a given sym-

metric framework.

Using some basic techniques from character theory, all the results of this

105

section combined will allow us to formulate the symmetry-extended version

of Maxwell’s rule given in [25] (as well as its extension to higher dimensions)

as a mathematical theorem in Section 4.2.2.

For the remainder of this chapter, we will continue to use the notation of

the previous section.

4.2.1 The necessary conditions

As before, we let G be a graph, S be a symmetry group in dimension

d with pairwise non-equivalent irreducible linear representations I1, . . . , Ir,

Φ be a homomorphism from S to Aut(G), and (G, p) be a framework in

R(G,S,Φ).

In this section, we make the additional assumption that the points p(v),

v ∈ V (G), span all of Rd.

Recall from the previous section that we have the decomposition

Rdn = V (I1)e ⊕ . . .⊕ V (Ir)

e (4.9)

with

V (It)e =

(V (It)

e

)1⊕ . . .⊕ (

V (It)e

)λt

(4.10)

of Rdn into H ′e-invariant subspaces.

While the scalars λt (as well as the subspaces that appear as direct sum-

mands in (4.9)) are uniquely determined in this decomposition, the subspaces

that appear as direct summands in (4.10) are not [60]. In order to derive the

symmetry-extended version of Maxwell’s rule, the subspaces in (4.10) shall

now be chosen appropriately.

Since the points p(v), v ∈ V (G), span all of Rd, the subspace N =

106

ker(R(Kn, p)

)of Rdn, where Kn is the complete graph on V (G), is the

space consisting of all infinitesimal rigid motions of (G, p). This space can

be written as the direct sum

N = T ⊕R,

where T is the space of all translational and R is the space of all rotational

infinitesimal rigid motions of (G, p). More precisely, a basis of T is given by

Tj| j = 1, . . . , d, where for j = 1, . . . , d, Tj : V (G) → Rd is the map that

sends each v ∈ V (G) to the jth canonical basis vector ej of Rd, and a basis of

R is given by Rij| 1 ≤ i < j ≤ d, where for 1 ≤ i < j ≤ d, Rij : V (G) → Rd

is the map defined by Rij(vk) = (pk)iej − (pk)jei for all k = 1, . . . , n [81].

Each of the maps Tj and Rij is of course identified with a vector in Rdn (by

using the order on V (G)).

Note that in the context of static rigidity, T is the space of all translational

loads and R is the space of all rotational loads on (G, p).

Using the notation of the previous paragraph we have the following result.

Lemma 4.2.1 For every dimension d, the subspaces T , R, and N of Rdn

are H ′e-invariant.

Proof. Fix a dimension d. We show first that N = ker(R(Kn, p)

)is

H ′e-invariant. Since p ∈ ⋂

x∈S Lx,Φ, it follows from Lemma 4.1.1 that if

R(Kn, p)u = z, then for all x ∈ S, we have

R(Kn, p)He(x)u = Hi(x)z, (4.11)

where Hi is the internal representation of S with respect to Kn and Φ. Let

u ∈ N , i.e., R(Kn, p)u = 0. Then for any x ∈ S, we have

Hi(x)R(Kn, p)u = Hi(x)0 = 0.

107

By (4.11), we have Hi(x)R(Kn, p)u = R(Kn, p)He(x)u, and hence

R(Kn, p)He(x)u = 0.

Thus, for all x ∈ S, He(x)u ∈ ker(R(Kn, p)

), which says that N is H ′

e-

invariant.

Next, we show that T is also H ′e-invariant. Let x ∈ S and let, as usual, Mx

denote the orthogonal matrix that represents x with respect to the canonical

basis of Rd. Then for j = 1, . . . , d, we have

He(x)Tj =

Mxej

...

Mxej

= (Mx)1jT1 + . . . + (Mx)djTd.

It follows that T is H ′e-invariant.

It remains to show that R is H ′e-invariant. Since for all x ∈ S, He(x)

is an orthogonal matrix, H ′e is a unitary representation (with respect to the

canonical inner product on Rdn). Therefore, the subrepresentation H′(N)e of

H ′e with representation space N is also unitary (with respect to the inner

product obtained by restricting the canonical inner product on Rdn to N).

So, by Remark 4.1.1, it suffices to show that R is the orthogonal complement

of T in N .

Let t be any element of T and r be any element of R. Then

t =

w

...

w

for some w ∈ Rd

108

and

r =

V p1

...

V pn

for some skew-symmetric matrix V .

Since the point∑n

i=1 pi must be fixed by every symmetry operation x ∈ S,

we may wlog define an origin so that∑n

i=1 pi = 0. Then the inner product

of t and r is given by

t · r =n∑

i=1

wT V pi

= wT V

n∑i=1

pi = 0.

This gives the result. ¤

Since, by Lemma 4.2.1, N is an H ′e-invariant subspace of Rdn, it fol-

lows from Maschke’s Theorem (see [42, 51, 60], for example) that N has an

H ′e-invariant complement Q in Rdn. We may therefore form the subrepresen-

tation H′(Q)e of H ′

e with representation space Q. Since H′(Q)e is a direct sum

of irreducible linear representations of S, say

H ′(Q)e = κ1I1 ⊕ . . .⊕ κrIr, where κ1, . . . , κr ∈ N ∪ 0, (4.12)

we obtain, analogously to (4.10), a decomposition of Q of the form

Q = V(I1)Q ⊕ . . .⊕ V

(Ir)Q ,

where

V(It)Q =

(V

(It)Q

)1⊕ . . .⊕ (

V(It)Q

)κt

. (4.13)

Similarly, since both T and R are also H ′e-invariant subspaces of Rdn, we

may form the subrepresentations H′(T )e and H

′(R)e of H ′

e with respective rep-

resentation spaces T and R. This gives rise to a decomposition of T of the

109

form

T = V(I1)T ⊕ . . .⊕ V

(Ir)T ,

where

V(It)T =

(V

(It)T

)1⊕ . . .⊕ (

V(It)T

)θt

,

and to a decomposition of R of the form

R = V(I1)R ⊕ . . .⊕ V

(Ir)R ,

where

V(It)R =

(V

(It)R

)1⊕ . . .⊕ (

V(It)R

)ρt

.

We can now choose the decomposition in (4.10) in such a way that

V (It)e = V

(It)Q ⊕ V

(It)T ⊕ V

(It)R . (4.14)

In the following we assume that the subspaces(V

(It)e

)jare chosen in this way.

We are now in the position to derive the necessary conditions for

(G, p) ∈ R(G,S,Φ) to be isostatic upon which the symmetry-extended version

of Maxwell’s rule is based.

Theorem 4.2.2 Let G be a graph, S be a symmetry group in dimension d

with pairwise non-equivalent irreducible linear representations I1, . . . , Ir, and

Φ : S → Aut(G) be a homomorphism. If (G, p) is an isostatic framework

in R(G,S,Φ) with the property that the points p(v), v ∈ V (G), span all of Rd,

then for t = 1, 2, . . . , r, we have

dim(V

(It)Q

)= dim

(V

(It)i

). (4.15)

110

Proof. Suppose first that dim(V

(It)Q

)> dim

(V

(It)i

)for some t. In this

case we give two separate arguments to show that (G, p) is not isostatic, one

that is based on infinitesimal rigidity and another one that is based on static

rigidity. This will later allow us to obtain information about both kinematic

and static rigidity properties of symmetric frameworks with the symmetry-

extended version of Maxwell’s rule.

Since dim(V

(It)Q

)> dim

(V

(It)i

), it follows from Corollary 4.1.3 that there

exists an element u 6= 0 in V(It)Q that lies in the kernel of the linear trans-

formation which is represented by the matrix R(G, p) with respect to the

bases Be and Bi. In other words, u is an infinitesimal flex of (G, p) (which is

symmetric with respect to It), and hence (G, p) is not isostatic.

Alternatively, it follows from Corollary 4.1.3 that there exists an element

l in V(It)Q that does not lie in the image of the linear transformation which

is represented by the matrix R(G, p)T with respect to the bases Be and Bi.

This says that l is an unresolvable equilibrium load on (G, p) (which is sym-

metric with respect to It), so that we may again conclude that (G, p) is not

isostatic.

Suppose now that dim(V

(It)Q

)< dim

(V

(It)i

)for some t. Then, analo-

gously as above, there exists an element ω 6= 0 in V(It)i that lies in the kernel

of the linear transformation which is represented by the matrix R(G, p)T with

respect to the bases Be and Bi. This says that ω is a non-zero self-stress of

(G, p) (which is symmetric with respect to It). So, it again follows that (G, p)

is not isostatic. ¤

Example 4.2.1 Recall from Example 4.1.2 that for the framework (K3, p) ∈

111

R(K3,Cs,Φ) shown in Figure 4.6, we have

dim(V (A′)

e

)= 3

dim(V

(A′)i

)= 2

dim(V (A′′)

e

)= 3

dim(V

(A′′)i

)= 1.

It is easy to see that the 2-dimensional space T of all translational infinites-

imal rigid motions of (K3, p) can be written as the direct sum

T = V(A′)T ⊕ V

(A′′)T ,

where V(A′)T is the space of dimension 1 generated by the infinitesimal rigid

motion shown in Figure 4.6 (a), and V(A′′)T is the space of dimension 1 gen-

erated by the infinitesimal rigid motion shown in Figure 4.6 (b). Moreover,

the 1-dimensional space R of rotational infinitesimal rigid motions of (K3, p)

is clearly generated by the infinitesimal rigid motion shown in Figure 4.6 (c),

so that R = V(A′′)R and dim

(V

(A′)R

)= 0. It follows from equation (4.14) that

...p1

..p2

..p3

.(a)

...p1

..p2

..p3

.(b)

...p1

..p2

..p3

.(c)

Figure 4.6: (a) A basis for the subspace V(A′)T ; (b) a basis for the subspace

V(A′′)T ; (c) a basis for the subspace R = V

(A′′)R .

112

dim(V

(A′)Q

)= dim

(V (A′)

e

)− dim(V

(A′)T

)− dim(V

(A′)R

)= dim

(V

(A′)i

)= 2

and

dim(V

(A′′)Q

)= dim

(V (A′′)

e

)−dim(V

(A′′)T

)−dim(V

(A′′)R

)= dim

(V

(A′′)i

)= 1,

so that the conditions (4.15) in Theorem 4.2.2 are satisfied for the isostatic

framework (K3, p).

In general, finding the dimensions of the subspaces V(It)Q and V

(It)i by

inspection is not as easy as in the previous example. In the following, we

therefore describe a systematic method, based on techniques from character

theory, for determining the dimensions of these subspaces, so that we can

apply Theorem 4.2.2 to a symmetric framework with an arbitrary point group

in any dimension. We begin by introducing the necessary vocabulary.

Definition 4.2.1 Let A = (aij) be an n× n square matrix. The trace of A

is defined to be Tr(A) =∑n

i=1 aii.

It is an important and well-known fact that the trace of a matrix is

invariant under a similarity transformation [19, 37]. This gives rise to

Definition 4.2.2 Let H : S → GL(V ) be a linear representation of a group

S, B be a basis of V , and HB be the matrix representation that corresponds to

H with respect to B. The character χ(H) of H is the function χ(H) : S → R

that sends x ∈ S to Tr(HB(x)

).

For a fixed enumeration x1, . . . , xk of the elements of the group S, we

will frequently also refer to the vector(Tr

(HB(x1)

), . . . , T r

(HB(xk)

))as

the character of H.

113

In the following we need some well-known results from character theory

which we summarize in

Theorem 4.2.3 [19, 37, 42, 60] Let S be a group with r pairwise non-

equivalent irreducible linear representations I1, . . . , Ir and let H : S →GL(V ) be a linear representation of S with H = α1I1 ⊕ . . . ⊕ αrIr, where

αt ∈ N ∪ 0 for all t = 1, . . . , r.

(i) If H = H1 ⊕H2 for some linear representations H1 and H2 of S, then

χ(H) = χ(H1) + χ(H2);

(ii) χ(H) can be written uniquely as a linear combination of the characters

χ(I1), . . . , χ(Ir) as

χ(H) = α1χ(I1) + . . . + αrχ(Ir);

(iii) For every t = 1, . . . , r, we have

αt =1

‖χ(It)‖2

(χ(H) · χ(It)

).

We first explain how we can determine the dimensions of the subspaces

V(It)i for all t = 1, . . . , r.

It follows from the direct sum decomposition of H ′i in (4.6) that for t =

1, . . . , r, the dimension of V(It)i is the degree of It multiplied by µt. Since

the degree of each irreducible linear representation It can be read off from

the character tables in Appendix A (or, if a more complete list of character

tables is required, from the tables in [2, 4, 37]), we only need to determine

the values of the µt. This can easily be done by means of the formula given in

Theorem 4.2.3 (iii), because the characters of the irreducible representations

114

It can simply be read off from the above-mentioned character tables and

the character of H ′i can be found by setting up the internal representation

matrices Hi(x), x ∈ S.

Finding the dimensions of the subspaces V(It)Q for all t = 1, . . . , r requires

a little more work. It follows from (4.14) that for t = 1, . . . , r, we have

dim(V

(It)Q

)= dim

(V (It)

e

)− dim(V

(It)T

)− dim(V

(It)R

).

The dimensions of the subspaces V(It)e can be determined in the analogous

way as the dimensions of the subspaces V(It)i : for t = 1, . . . , r, the dimension

of the subspace V(It)e is equal to the degree of It multiplied by λt. Note that

the values of the λt in (4.10) can again easily be computed with the help of

Theorem 4.2.3 (iii) since the character of H ′e can be found by setting up the

external representation matrices He(x), x ∈ S.

For t = 1, . . . , r, the dimension of the subspace V(It)T is the degree of It

multiplied by θt and the dimension of the subspace V(It)R is the degree of It

multiplied by ρt. So, in order to determine the dimensions of the subspaces

V(It)T and V

(It)R with the formula in Theorem 4.2.3 (iii), it only remains to

determine the characters χ(H′(T )e ) and χ(H

′(R)e ).

We first show how to compute the character χ(H′(T )e ). It follows directly

from the proof of Lemma 4.2.1 that if S is a symmetry group in dimension

d and x ∈ S, then the matrix that represents the linear transformation

H′(T )e (x) with respect to the basis T1, . . . , Td is the orthogonal matrix Mx

that represents x with respect to the canonical basis of Rd. This says that

for a fixed enumeration x1, . . . , xk of the elements of S, we have

χ(H ′(T )e ) =

(Tr(Mx1), . . . , T r(Mxk

)).

115

For example, if S is a symmetry group in dimension 2, then the compo-

nent of χ(H′(T )e ) that corresponds to the identity Id ∈ S is equal to 2, each

component of χ(H′(T )e ) that corresponds to a rotational symmetry operation

Cm ∈ S is equal to 2 cos(

2πm

), and each component of χ(H

′(T )e ) that corre-

sponds to a reflection s ∈ S is equal to 0 (see also Table 4.1 in Section 4.3.2).

For a symmetry group S in dimension 3, the explicit values of the com-

ponents of χ(H′(T )e ) are summarized in Table 4.2 in Section 4.3.3.

The character χ(H′(R)e ) can be determined similarly. As an example, we

compute the character χ(H′(R)e ) in the case where S is a symmetry group in

dimension 2. Every element of S is then either the identity Id, a rotation

Cm about the origin by an angle of 2πm

, or a reflection s in a line through the

origin. Note that R is a one-dimensional subspace of R2n a basis of which is

given by R12. Let Cm be a rotational symmetry operation of (G, p) with

MCm =

cos

(2πm

) − sin(

2πm

)

sin(

2πm

)cos

(2πm

)

.

Then, by using the definition of the external representation He of S (with

respect to G and Φ) and the fact that (G, p) ∈ R(G,S,Φ), it is easy to verify

that

He(Cm)R12 = R12.

Similarly, if s is a reflectional symmetry operation of (G, p) with

Ms =

cos (θ) sin (θ)

sin (θ) − cos (θ)

,

then it is again easy to verify that

He(s)R12 = −R12.

116

It follows that the matrices which represent the linear transformations

H′(R)e (Id), H

′(R)e (Cm), and H

′(R)e (s) with respect to the basis R12 are the

1 × 1 matrices (i.e., scalars) 1, 1, and −1, respectively. Therefore, if d = 2,

the character χ(H′(R)e ) is the vector defined as follows: each component of

χ(H′(R)e ) that corresponds to the identity Id ∈ S or a rotational symmetry

operation Cm ∈ S is equal to 1, and each component of χ(H′(R)e ) that cor-

responds to a reflection s ∈ S is equal to −1 (see also Table 4.1 in Section

4.3.2).

Note that analogous calculations as above can easily be carried out for

any symmetry group in dimension d > 2 as well. For a symmetry group S in

dimension 3, the values of the components of χ(H′(R)e ) are again summarized

in Table 4.2 in Section 4.3.3.

Example 4.2.2 Let us apply the methods described above to the framework

(K3, p) ∈ R(K3,Cs,Φ) from Example 4.2.1. From the representation matrices in

Example 4.1.1 we immediately deduce that χ(H ′e) = (6, 0) and χ(H ′

i) = (3, 1).

Therefore, if we let

H ′e = λ1A

′ ⊕ λ2A′′

H ′i = µ1A

′ ⊕ µ2A′′,

then, by the formula in Theorem 4.2.3 (iii), we have

λ1 =1

2

(6 · 1 + 0 · 1) = 3

λ2 =1

2

(6 · 1 + 0 · (−1)

)= 3

µ1 =1

2

(3 · 1 + 1 · 1) = 2

µ2 =1

2

(3 · 1 + 1 · (−1)

)= 1.

117

Further, for the characters χ(H′(T )e ) and χ(H

′(R)e ), we have, as shown above,

χ(H′(T )e ) = (2, 0) and χ(H

′(R)e ) = (1,−1). So, if we let

H ′(T )e = θ1A

′ ⊕ θ2A′′

H ′(R)e = ρ1A

′ ⊕ ρ2A′′,

then, again by the formula in Theorem 4.2.3 (iii), we obtain θ1 = 1, θ2 = 1,

ρ1 = 0, and ρ2 = 1. Since both A′ and A′′ are linear representations of degree

1, it follows that

dim(V

(A′)Q

)= dim

(V (A′)

e

)− dim(V

(A′)T

)− dim(V

(A′)R

)

= 3− 1− 0

= 2

dim(V

(A′′)Q

)= dim

(V (A′′)

e

)− dim(V

(A′′)T

)− dim(V

(A′′)R

)

= 3− 1− 1

= 1

and

dim(V

(A′)i

)= 2

dim(V

(A′′)i

)= 1.

4.2.2 The rule

Using the mathematical background established in the previous section,

we can now prove a symmetry-extended version of Maxwell’s rule that can

be applied to both injective and non-injective symmetric realizations in any

dimension. Note that for dimensions 2 and 3, Theorem 4.2.4 is a rigorous

118

mathematical formulation of the rule given in [25].

The condition (4.16) in Theorem 4.2.4 is obtained by combining all of

the conditions in (4.15) into a single equation using some basic techniques

from character theory. This enables us to check the conditions in (4.15) with

very little computational effort, so that the symmetry-extended version of

Maxwell’s rule is in the same spirit as Maxwell’s original rule in the sense

that it only requires a simple count of joints and bars that are ‘fixed’ by

various symmetry operations.

From now on we will simplify our notation of the previous section by

denoting the characters χ(H ′e), χ(H ′

i), χ(H′(Q)e ), χ(H

′(T )e ), and χ(H

′(R)e ) by

Xe, Xi, XQ, XT , and XR, respectively.

Theorem 4.2.4 (Symmetry-extended version of Maxwell’s rule)

Let G be a graph, S be a symmetry group in dimension d with pairwise non-

equivalent irreducible linear representations I1, . . . , Ir, and Φ : S → Aut(G)

be a homomorphism. If (G, p) is an isostatic framework in R(G,S,Φ) with the

property that the points p(v), v ∈ V (G), span all of Rd, then we have

XQ = Xi. (4.16)

Proof. Suppose XQ 6= Xi. Then, by Theorem 4.2.3 (ii) and equations (4.6)

and (4.12), we have

κ1χ(I1) + . . . + κrχ(Ir) 6= µ1χ(I1) + . . . + µrχ(Ir),

which implies that κt 6= µt for some t ∈ 1, . . . , r. Therefore, dim(V

(It)Q

) 6=dim

(V

(It)i

). The result now follows from Theorem 4.2.2. ¤

119

So, by checking the condition (4.16) in Theorem 4.2.4, we implicitly check

all the conditions in (4.15). Since we have

H ′e = H ′(Q)

e ⊕H ′(T )e ⊕H ′(R)

e ,

it follows from Theorem 4.2.3 (i) that

XQ = Xe −XT −XR.

Note that we have already shown how to compute each of the above char-

acters in the previous section. In fact, for dimensions 2 and 3, the characters

XT and XR can be read off from Tables 4.1 and 4.2 in Section 4.3. So, in

order to check condition (4.16) for d = 2 or d = 3, it only remains to compute

the characters Xe and Xi.

So far, our method of determining Xe and Xi has been to set up the

external and internal representation matrices He(x) and Hi(x) for all x ∈ S,

and then to determine the traces of these matrices. In the following, we

generalize the method of P. Fowler and S. Guest presented in [25] to deter-

mine the characters Xe and Xi without explicitly finding the external and

internal representation of S. This will simplify significantly the calculations

required to apply the symmetry-extended version of Maxwell’s rule to a given

framework.

Recall from Section 3.3 that for a framework (G, p) ∈ R(G,S), there exists

a unique map Φ : S → Aut(G) such that (G, p) ∈ R(G,S,Φ) if and only if

Aut(G, p) = id.

Definition 4.2.3 Let G be a graph with V (G) = v1, . . . , vn, S be a sym-

metry group, Φ be a map from S to Aut(G), (G, p) be a framework in R(G,S,Φ),

120

and x ∈ S. A joint (vi, pi) of (G, p) is said to be fixed by x with respect to Φ

(or simply fixed by x if Aut(G, p) = id) if Φ(x)(vi) = vi.

Similarly, a bar (vi, pi), (vj, pj) of (G, p) is said to be fixed by x with

respect to Φ (or simply fixed by x if Aut(G, p) = id) if Φ(x)(vi, vj

)=

vi, vj.The number of joints of (G, p) that are fixed by x (with respect to Φ) is

denoted by jΦ(x) and the number of bars of (G, p) that are fixed by x (with

respect to Φ) is denoted by bΦ(x).

Recall from Definition 4.1.7 that for x ∈ S, the external representation

matrix He(x) is obtained from the transpose of the permutation matrix cor-

responding to Φ(x) by replacing each 1 with the d× d orthogonal matrix Mx

and each 0 with a d× d zero-matrix. Note that the transpose of the permu-

tation matrix corresponding to Φ(x) has a 1 in the diagonal if and only if

the corresponding joint of (G, p) is fixed by x with respect to Φ. Therefore,

a joint of (G, p) can make a contribution to the trace of He(x) only if it is

fixed by x with respect to Φ. So, for a fixed enumeration x1, . . . , xk of the

elements of S, we have

Xe =(Tr

(He(x1)

), . . . , T r

(He(xk)

))

=(jΦ(x1)Tr(Mx1), . . . , jΦ(xk)Tr(Mxk

))

= XJ ×XT ,

where XJ = (jΦ(x1), . . . , jΦ(xk)) and × denotes componentwise multiplication.

Similarly, for x ∈ S, the internal representation matrix Hi(x) has a 1 in

the diagonal if and only if the corresponding bar of (G, p) is fixed by x with

121

respect to Φ. Thus,

Xi = (bΦ(x1), . . . , bΦ(xk)).

So, condition (4.16) in the symmetry-extended version of Maxwell’s rule

can be written as

XJ ×XT −XT −XR = Xi, (4.17)

and each of the characters in (4.17) can be determined with very little compu-

tational effort. In the following we refer to (4.17) as the symmetry-extended

Maxwell’s equation.

Example 4.2.3 The symmetry-extended version of Maxwell’s rule, applied

to the framework (K3, p) ∈ R(K3,Cs,Φ) from Example 4.2.2, yields the counts

XJ = (jΦ(Id), jΦ(s)) = (3, 1)

XT = (2, 0)

XR = (1,−1)

XQ = XJ ×XT −XT −XR = (3, 1)

Xi = (bΦ(Id), bΦ(s)) = (3, 1).

Thus, condition (4.16) in Theorem 4.2.4 is satisfied for the isostatic frame-

work (K3, p).

Remark 4.2.1 Suppose the symmetry-extended version of Maxwell’s rule

detects that a framework (G, p) ∈ R(G,S,Φ) is not isostatic. Then we may use

Theorem 4.2.3 (iii) and the proof of Theorem 4.2.2 to obtain information

on the symmetry properties of self-stresses of (G, p), infinitesimal flexes of

(G, p), and unresolvable equilibrium loads on (G, p) in the following way.

122

Suppose for the framework (G, p), we have XQ 6= Xi. Using the formula

in Theorem 4.2.3 (iii), we may then determine the values of the κt and µt

in (4.12) and (4.6) for all t = 1, . . . , r. By the proof of Theorem 4.2.4, there

must exist t ∈ 1, . . . , r such that κt 6= µt.

It follows from the proof of Theorem 4.2.2 that if κt > µt, say κt − µt =

k > 0, then there exist k linearly independent infinitesimal flexes of (G, p)

which are symmetric with respect to It, as well as k unresolvable equilibrium

loads on (G, p) which are symmetric with respect to It.

Similarly, if κt < µt, say µt − κt = k > 0, then there exist k linearly

independent self-stresses of (G, p) which are symmetric with respect to It.

4.2.3 Example and further remarks

To illustrate how the symmetry-extended version of Maxwell’s rule can

give a significantly improved insight into the rigidity properties of a sym-

metric framework in comparison to Maxwell’s original rule, we consider the

framework (K3,3, p) ∈ R(K3,3,C2v ,Φ) from Example 3.2.1.

The symmetry group C2v = Id, C2, sh, sv has four non-equivalent irre-

ducible linear representations A1, A2, B1, and B2 each of which is of degree

C2v Id C2 sh sv

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

123

1. The characters of these representations are shown in the table above (see

also Appendix A).

We have

XJ = (jΦ(Id), jΦ(C2), jΦ(sh), jΦ(sv)) = (6, 0, 0, 2)

XT = (2,−2, 0, 0)

XR = (1, 1,−1,−1)

XQ = XJ ×XT −XT −XR = (9, 1, 1, 1)

Xi = (bΦ(Id), bΦ(C2), bΦ(sh), bΦ(sv)) = (9, 3, 3, 1).

Since XQ 6= Xi, we may conclude that (K3,3, p) is not isostatic. Note that

Maxwell’s original rule would not have detected this because |E(K3,3)| =

2|V (K3,3)| − 3 = 9.

With the help of the formula in Theorem 4.2.3 (iii) we obtain

XQ = 3A1 + 2A2 + 2B1 + 2B2 and

Xi = 4A1 + 2A2 + 2B1 + B2,

which implies that (K3,3, p) has a non-zero self-stress which is symmetric with

respect to A1 and an infinitesimal flex (as well as an unresolvable equilibrium

load) which is symmetric with respect to B2 (see Figure 4.7).

Remark 4.2.2 Given a framework (G, p) ∈ R(G,S), we need to specify a

type Φ : S → Aut(G) in order to apply the symmetry-extended version of

Maxwell’s rule (Theorem 4.2.4) to (G, p) and S, because Φ determines the

characters Xe and Xi. Of course, we also need to make sure that Φ is a

homomorphism, for otherwise the external and internal representation (with

respect to G and Φ) are not matrix representations of S (see Remark 4.1.2).

124

.

..p5

..p3

. .p6

..p1 . .p2

..p4

.sv

.sh

.(a)

.

..p5 .

.p3

..p6

..p1 . .p2

..p4

.sv

.sh

.(b)

.

..p5 .

.p3

..p6

..p1 ..p2

..p4

.sv

.sh

.(c)

Figure 4.7: (a) An infinitesimal flex of (K3,3, p) ∈ R(K3,3,C2v ,Φ) which is sym-

metric with respect to B2 (the displacement vector at each joint of (K3,3, p)

remains unchanged under Id and sv and is reversed under C2 and sh). (b)

An unresolvable equilibrium load on (K3,3, p) which is symmetric with respect

to B2. (c) A self-stress of (K3,3, p) which is symmetric with respect to A1 (the

tensions and compressions in the bars of (K3,3, p) remain unchanged under

all symmetry operations in C2v).

125

Recall that if Aut(G, p) = id, then, by Corollary 3.3.2, (G, p) ∈ R(G,S)

is of a unique type Φ and, by Theorem 3.4.1, Φ is a homomorphism, so that

the external and internal representation are uniquely determined in this case

and Theorem 4.2.4 can be applied to (G, p) and S in a unique way. By Corol-

lary 3.3.3, this is in particular the case if (G, p) is a framework whose map

p is injective. In the following section, we will see that if p is injective, then

the characters Xe and Xi can be found in a particularly easy way (without

determining the type Φ) by simply examining the geometric positions of the

joints and bars of (G, p).

Since in [25] only injective realizations in R2 and R3 are considered, The-

orem 4.2.4 includes the symmetrized version of Maxwell’s rule given in [25]

as a special case.

If Aut(G, p) 6= id, then, by Theorem 3.3.1, (G, p) ∈ R(G,S) is not of

a unique type, and hence we may apply Theorem 4.2.4 to (G, p) and S by

using any homomorphism Φ for which (G, p) ∈ R(G,S,Φ).

Note that Examples 3.4.1 and 3.4.2 show that there exist frameworks

(G, p) ∈ R(G,S) for which we cannot apply the symmetry-extended version

of Maxwell’s rule to (G, p) and S at all, because there does not exist any

homomorphism Φ so that (G, p) ∈ R(G,S) is of type Φ.

Remark 4.2.3 Let G be a graph, S be a symmetry group in dimension d,

and Φ : S → Aut(G) be a homomorphism, so that the set R(G,S,Φ) contains a

framework (G, p) with the property that the points p(v), v ∈ V (G), span all

of Rd. Then it follows from Lemma 3.2.2 and Theorem 3.2.3 that the con-

dition (4.16) in the symmetry-extended version of Maxwell’s rule (Theorem

4.2.4) is also a necessary condition for G to be (S, Φ)-generically isostatic.

126

In particular, if the condition (4.16) does not hold (i.e., if XQ 6= Xi) then we

may conclude that every framework in the set R(G,S,Φ) is not isostatic.

Remark 4.2.4 There exist a number of further classical counting rules, sim-

ilar to Maxwell’s rule, that can predict the rigidity and flexibility properties

of various other types of structures, such as pinned frameworks (i.e., frame-

works that have some of their joints firmly anchored to the ground), body-

bar frameworks, and body-hinge frameworks, for example. For each of these

rules, symmetry extensions can be derived using techniques from group rep-

resentation theory (see, for example, [25, 34, 36, 53, 56]). We will discuss

some of these rules in Chapter 7.

4.3 Restrictions on the number of fixed joints

and bars of symmetric isostatic frame-

works

In this section, we show that the necessary conditions given in the

symmetry-extended version of Maxwell’s rule for a 2- or 3-dimensional sym-

metric framework (G, p) ∈ R(G,S,Φ) to be isostatic are equivalent to some

very simply stated restrictions on the number of joints and bars of (G, p)

that are fixed by various symmetry operations in S.

The basic results in this section are from the joint paper [15].

127

4.3.1 Fixed versus geometrically unshifted

We begin by summarizing some simple observations regarding the geo-

metric position of a joint or a bar of a framework (G, p) ∈ R(G,S,Φ) that is

fixed by an element in S (with respect to Φ).

Definition 4.3.1 Let x be a symmetry operation of a framework (G, p).

Then a joint(v, p(v)

)of (G, p) is said to be geometrically unshifted by x if

x(p(v)

)= p(v) or equivalently, if p(v) is contained in the symmetry element

Fx corresponding to x.

Similarly, a bar(

v, p(v)),(w, p(w)

)of (G, p) is said to be geometri-

cally unshifted by x if x(p(v), p(w)) = p(v), p(w) or equivalently, if the

undirected line segment p(v)p(w) is equal to the undirected line segment

x(p(v)

)x(p(w)

).

..

.

.(a)

. .

.

.

.s

.(b)

.C2

Figure 4.8: Geometrically unshifted bars in dimension 2: (a) a bar that is

geometrically unshifted by a half-turn C2; (b) possible placement of a bar that

is geometrically unshifted by a reflection s.

Theorem 4.3.1 Let G be a graph, S be a symmetry group, Φ : S → Aut(G)

be a map, (G, p) be a framework in R(G,S,Φ), and x be an element in S.

(i) If a joint j of (G, p) is fixed by x (with respect to Φ), then j is geomet-

rically unshifted by x;

128

(ii) if a bar b of (G, p) is fixed by x (with respect to Φ), then b is geometri-

cally unshifted by x;

(iii) if p is injective and a joint j of (G, p) is geometrically unshifted by x,

then j is fixed by x;

(iv) if p is injective and a bar b of (G, p) is geometrically unshifted by x,

then b is fixed by x.

Proof. (i) Let j =(v, p(v)

)be a joint of (G, p) that is fixed by x (with

respect to Φ). Then we have x(p(v)

)= p

(Φ(x)(v)

)= p(v), which says that

j is geometrically unshifted by x.

(ii) Let b =(

v, p(v)),(w, p(w)

)be a bar of (G, p) that is fixed by x

with respect to Φ. Then we have x(p(v), p(w)) =

x(p(v)

), x

(p(w)

)=

p(Φ(x)(v)

), p

(Φ(x)(w)

)= p(v), p(w), which says that b is geometrically

unshifted by x.

(iii) Let j =(v, p(v)

)be geometrically unshifted by x. Then we have

p(v) = x(p(v)

)= p

(Φ(x)(v)

). Since p is injective, it follows that v = Φ(x)(v).

Thus, j is fixed by x.

(iv) Let b =(

v, p(v)),(w, p(w)

)be geometrically unshifted by x.

Then we have p(v), p(w) = x(p(v), p(w)) =

x(p(v)

), x

(p(w)

)=

p(Φ(x)(v)

), p

(Φ(x)(w)

). Since p is injective, it follows that v, w =

Φ(x)(v), Φ(x)(w) = Φ(x)(v, w). Thus, b is fixed by x. ¤

It follows from Theorem 4.3.1 that if the map p of a framework (G, p) ∈R(G,S,Φ) is injective, then a joint or a bar of (G, p) is geometrically unshifted

by x ∈ S if and only if it is fixed by x. Therefore, if (G, p) is a framework

whose map p is injective, then the determination of whether a joint j or a

129

.

.. ...(b)

..

..

.Cm

.(a)

Figure 4.9: Possible placement of a bar that is geometrically unshifted by: (a)

any rotation Cm, m ≥ 2 (in dimension 3); (b) a half-turn C2 (in dimension

3) alone.

...

..

.(a)

..

..

.s.(b)

Figure 4.10: Possible placement of a bar that is geometrically unshifted by

a reflection s (in dimension 3): (a) lying in the reflection plane; (b) lying

perpendicular to the reflection plane.

130

.

..

...Sm

.(a)

.

..

...i = S2

.(b)

Figure 4.11: Possible placement of a bar that is geometrically unshifted by:

(a) any improper rotation Sm, m ≥ 2 (in dimension 3); (b) an inversion

i = S2 (in dimension 3) alone.

bar b of (G, p) is fixed by x ∈ S only requires a simple examination of the

geometric positions of j and b with respect to the symmetry element corre-

sponding to x. Figures 4.8, 4.9, 4.10 and 4.11 show the possible geometric

positions of a bar that is geometrically unshifted by the relevant symmetry

operations in dimensions 2 and 3.

If the map p of (G, p) ∈ R(G,S,Φ) is not injective, then a joint or a bar

that is geometrically unshifted by x ∈ S is not necessarily fixed by x (with

respect to Φ).

For example, the joints (v3, p3) and (v4, p4) of the framework (Gt, p) ∈R(Gt,C2,Θb) in Example 3.3.1 are both geometrically unshifted by C2, since

both p3 and p4 lie on the center of rotation of C2, but they are not fixed

by C2 with respect to Θb, since Θb(C2)(v3) = v4. Similarly, the bar(v3, p3), (v4, p4)

of the framework (Gbp, p) ∈ R(Gbp,Cs,Ξb) in Example 3.3.2

is geometrically unshifted by s, but not fixed by s with respect to Ξb, since

Ξb(v3, v4) = v3, v5 6= v3, v4.So, if the map p of a framework (G, p) ∈ R(G,S,Φ) is not injective, then

131

we can only find the joints and bars of (G, p) that are fixed by x ∈ S (with

respect to Φ) by considering the graph automorphism Φ(x).

4.3.2 Isostatic frameworks in dimension 2

Suppose (G, p) is an isostatic framework in R(G,S,Φ), where S is a sym-

metry group in dimension 2, Φ : S → Aut(G) is a homomorphism, and the

points p(v), v ∈ V (G), span all of R2.

Since S is a symmetry group in dimension 2, every element of S is of one

of the following three types: the identity Id, a rotation Cm, where m ≥ 2,

or a reflection s. This allows us to calculate the (2-dimensional) symmetry-

extended Maxwell’s equation (4.17) for (G, p) componentwise, as shown in

Table 4.1.

In this table we distinguish a half-turn C2 from a rotation Cm, where

m > 2, because there may exist bars of (G, p) that are fixed by a half-turn,

but there cannot be any bars of (G, p) that are fixed by a rotation Cm, where

m > 2.

By Table 4.1, the symmetry-extended Maxwell’s equation for the isostatic

framework (G, p) ∈ R(G,S,Φ) reduces to the following four equations:

Id: |E(G)| = 2|V (G)| − 3 (4.18)

C2: 2jΦ(C2) + bΦ(C2) = 1 (4.19)

Cm, m > 2: (jΦ(Cm) − 1) cos

(2π

m

)=

1

2(4.20)

s: bΦ(s) = 1, (4.21)

132

Id C2 Cm, m > 2 s

XJ |V (G)| jΦ(C2) jΦ(Cm) jΦ(s)

XT 2 −2 2 cos(

2πm

)0

XR 1 1 1 −1

XQ 2|V (G)| − 3 −2jΦ(C2) + 1 2(jΦ(Cm) − 1) cos(

2πm

)− 1 1

Xi |E(G)| bΦ(C2) 0 bΦ(s)

Table 4.1: Calculations of characters in the 2-dimensional symmetry-

extended Maxwell’s equation.

where a given equation applies when the corresponding symmetry operation

is present in S.

Some observations arising from this set of equations are:

(i) Since every symmetry group contains the identity Id, equation (4.18)

holds and simply restates the condition in Maxwell’s original rule (The-

orem 2.2.7).

(ii) If S contains a half-turn C2, then it follows from (4.19) and the fact that

both jΦ(C2) and bΦ(C2) must be non-negative integers that jΦ(C2) = 0

and bΦ(C2) = 1. In particular, since all bars of (G, p) (except the one

that is fixed by C2 with respect to Φ) and all joints of (G, p) occur in

pairs, it follows that |V (G)| is even and |E(G)| is odd.

(iii) If S contains a rotation Cm, m > 2, then it follows from (4.20) that

either jΦ(Cm) = 0 and m = 3 or jΦ(Cm) = 2 and m = 6. However,

if jΦ(Cm) = 2 and m = 6, then S also contains a half-turn C2 = C36

with jΦ(C2) = 2, contradicting (4.19). Therefore, S cannot contain a

133

rotational symmetry operation Cm with m > 3 and when either C2 or

C3 is present in S, then jΦ(C2) = 0 and jΦ(C3) = 0. Note that if S

contains a rotation C3, then all joints and bars of (G, p) occur in sets

of 3.

(iv) Finally, equation (4.21) says that if S contains a reflection s, then we

must have bΦ(s) = 1. However, (4.21) does not impose a restriction on

the number of joints that are fixed by s (with respect to Φ).

An immediate consequence of these observations is that the point group

of (G, p) must be one of the following six symmetry groups in dimension 2:

C1, C2, C3, Cs, C2v or C3v. Figure 5.48 depicts examples of small 2-dimensional

isostatic frameworks for each of these symmetry groups.

Group by group, the above results can be summarized as follows.

Theorem 4.3.2 Let G be a graph, S be a symmetry group in dimension 2,

Φ : S → Aut(G) be a homomorphism, and (G, p) be an isostatic framework

in R(G,S,Φ) with the property that the points p(v), v ∈ V (G), span all of R2.

(i) If S = C1, then |E(G)| = 2|V (G)| − 3;

(ii) if S = C2, then |E(G)| = 2|V (G)| − 3, jΦ(C2) = 0 and bΦ(C2) = 1;

(iii) if S = C3, then |E(G)| = 2|V (G)| − 3 and jΦ(C3) = 0;

(iv) if S = Cs, then |E(G)| = 2|V (G)| − 3 and bΦ(s) = 1;

(v) if S = C2v, then |E(G)| = 2|V (G)|−3, jΦ(C2) = 0 and bΦ(C2) = bΦ(s) = 1

for all reflections s ∈ C2v;

134

.. .

.

.(a)

. .

..

.(b)

. .

.

..

.

.(c)

.

.

.. .

. .

.

.(d.i)

.(d.ii)

. .

.

.

.(e)

...

.

..

.

.

. .

.(f.ii)

.(f.i)

. .

.

Figure 4.12: Examples, for each of the possible point groups, of small 2-

dimensional isostatic frameworks: (a) C1; (b) C2; (c) C3; (d) Cs; (e) C2v; (f)

C3v. For each of Cs and C3v, two examples are given, where in (i) each mirror

has a bar centered at and perpendicular to the mirror line, whereas in (ii)

each mirror has a bar that lies in the mirror line. For C2v, the bar lying at

the center of C2 must lie in one mirror line and perpendicular to the other.

135

(vi) if S = C3v, then |E(G)| = 2|V (G)| − 3, jΦ(C3) = 0 and bΦ(s) = 1 for all

reflections s ∈ C3v;

(vii) S must be one of the above symmetry groups.

Remark 4.3.1 By the above results, an isostatic framework in the plane

cannot possess any rotational symmetry operation Cm with m > 3. These

kinds of restrictions for isostatic symmetric frameworks in the plane become

intuitively plausible if one looks at some simple examples of frameworks with

m-fold rotational symmetries.

Suppose, for instance, that (G, p) is a 2-dimensional framework with point

group Cm, where m is a multiple of 4. Then |E(G)| must be an even number,

because every bar of (G, p) belongs to a ‘symmetry orbit’ of size m or m2.

Since 2|V (G)| − 3 is an odd number, however, it follows that the Maxwell

count |E(G)| = 2|V (G)| − 3 cannot be attained by (G, p).

Similarly, if m > 3 is odd, then every bar of (G, p) belongs to a ‘symmetry

orbit’ of size m. So, (G, p) can only satisfy the Maxwell count if at least one

of the joints of (G, p) is fixed by the m-fold rotation Cm, and we have already

seen in the previous sections that this kind of restriction has the potential

for affecting the rigidity properties of (G, p).

Symmetric subgraphs

Let S, Φ, and (G, p) ∈ R(G,S,Φ) be as in Theorem 4.3.2. Then, besides

the conditions for G, S, and Φ given in Theorem 4.3.2, the Laman con-

ditions for all nontrivial subgraphs of G are also necessary conditions for

(G, p) to be isostatic. Further, if H is a subgraph of G with the prop-

136

erty that |E(H)| = 2|V (H)| − 3 and (H, p|V (H)) is a symmetric framework,

say (H, p|V (H)) ∈ R(G,S′,Φ′), then the corresponding conditions in Theorem

4.3.2 for S ′ and Φ′ are clearly also necessary conditions for (G, p) to be

isostatic (provided that the points p(v), v ∈ V (H), span all of R2 and

Φ′ : S ′ → Aut(H) is a homomorphism).

However, if S ′ is a subgroup of S and Φ′ : S ′ → Aut(H) is defined by

Φ′(x) = Φ(x)|V (H), then these conditions are implicitly contained in the over-

all conditions for S and Φ. This is obvious for the conditions concerning the

number of joints that are fixed by any symmetry operation, since this number

is either 0 or unrestricted, and for the conditions concerning the number of

bars that are fixed by a rotation Cm, where m > 2, since this number is 0.

If x is a half-turn or a reflection, then we must have bΦ′(x) = 1 for (G, p)

to be isostatic. But if bΦ(x) = 1, then we have bΦ′(x) ≤ 1, and bΦ′(x) cannot

be zero, for otherwise |E(H)| is an even number, contradicting the count

|E(H)| = 2|V (H)| − 3. Thus, bΦ(x) = 1 implies bΦ′(x) = 1.

In particular, it follows that while the Laman conditions, as well as the

conditions in Theorem 4.3.2 concerning G, S, and Φ, are necessary con-

ditions for the graph G to be (S, Φ)-generically isostatic (see also Remark

4.2.3), there are no additional necessary conditions concerning ‘symmetric

subgraphs’ for G to be (S, Φ)-generically isostatic.

In Chapter 5, we consider whether the conditions in Theorem 4.3.2 for

G, S, and Φ, together with the Laman conditions, are also sufficient for G

to be (S, Φ)-generically isostatic.

137

4.3.3 Isostatic frameworks in dimension 3

Suppose that (G, p) is an isostatic framework in R(G,S,Φ), where S is a

symmetry group in dimension 3, Φ : S → Aut(G) is a homomorphism, and

the points p(v), v ∈ V (G), span all of R3.

Recall from Section 2.3 that since S is a symmetry group in dimension 3,

every element of S is of one of the following types: the identity Id, a rotation

Cm, where m ≥ 2, a reflection s, or an improper rotation Sm. This gives rise

to the componentwise calculations for the 3-dimensional symmetry-extended

Maxwell’s equation (4.17) for (G, p) shown in Table 4.2.

In this table we again distinguish a half-turn C2 from a rotation Cm,

where m > 2, and an inversion i = S2 from an improper rotation Sm, where

m > 2.

By Table 4.2, the symmetry-extended Maxwell’s equation for the isostatic

framework (G, p) ∈ R(G,S,Φ) reduces to the following six equations:

Id: |E(G)| = 3|V (G)| − 6 (4.22)

C2: jΦ(C2) + bΦ(C2) = 2 (4.23)

Cm, m > 2: (jΦ(Cm) − 2)

(2 cos

(2π

m

)+ 1

)= bΦ(Cm) (4.24)

s: jΦ(s) = bΦ(s) (4.25)

i: 3jΦ(i) + bΦ(i) = 0 (4.26)

Sm,m > 2: jΦ(Sm)

(2 cos

(2π

m

)− 1

)= bΦ(Sm), (4.27)

138

Id

C2

Cm

,m

>2

si

Sm

,m

>2

XJ

|V(G

)|j Φ

(C2)

j Φ(C

m)

j Φ(s

)j Φ

(i)

j Φ(S

m)

XT

3−1

2co

s( 2

π m

) +1

1−3

2co

s( 2

π m

) −1

XR

3−1

2co

s( 2

π m

) +1

−13

−2co

s( 2

π m

) +1

XQ

3|V(G

)|−

6−j

Φ(C

2)+

2( 2

cos( 2

π m

) +1) (j

Φ(C

m)−

2)j Φ

(s)−3

j Φ(i

)

( 2co

s( 2

π m

) −1) j Φ

(Sm

)

Xi

|E(G

)|b Φ

(C2)

b Φ(C

m)

b Φ(s

)b Φ

(i)

b Φ(S

m)

Tab

le4.

2:C

alcu

lation

sof

char

acte

rsin

the

3-di

men

sion

alsy

mm

etry

-ext

ende

dM

axwel

l’s

equat

ion.

139

where a given equation applies when the corresponding symmetry operation

is present in S.

Some observations arising from this set of equations are:

(i) By equation (4.22), (G, p) must satisfy the condition in Maxwell’s orig-

inal rule (Theorem 2.2.7).

(ii) If S contains a half-turn C2, then equation (4.23) holds. The solutions

of (4.23) are

(jΦ(C2), bΦ(C2)) = (2, 0), (1, 1), (0, 2).

A bar(

v, p(v)),(w, p(w)

)of (G, p) contributes to bΦ(C2) if and only

if Φ(C2) either fixes both v and w or interchanges v and w. However, if

(G, p) has a fixed bar of the first kind (i.e., a bar whose corresponding

joints are both fixed by C2 with respect to Φ), then this bar contributes

2 to jΦ(C2) and 1 to bΦ(C2), contradicting (4.23). Thus, the vertices

corresponding to the joints of any bar that is fixed by C2 with respect

to Φ must be images of each other under Φ(C2). This says in particular

that all bars included in bΦ(C2) must lie perpendicular to the C2 axis.

(iii) If S contains a rotation Cm, m > 2, then equation (4.24) holds. The

non-negative integer solution jΦ(Cm) = 2 and bΦ(Cm) = 0 is possible for

all m. For m > 2, the factor 2 cos(

2πm

)+ 1 is rational at m = 3, 4, 6,

but generates a further distinct solution only for m = 3:

m = 3:

0(jΦ(C3) − 2) = bΦ(C3),

and hence bΦ(C3) = 0, but jΦ(C3) is unrestricted.

140

m = 4:

jΦ(C4) − 2 = bΦ(C4).

If S contains C4, then S also contains C2 = C24 . Therefore, we must

have jΦ(C4) = jΦ(C2) = 2 and bΦ(C4) = bΦ(C2) = 0.

m = 6:

2(jΦ(C6) − 2) = bΦ(C6).

If S contains C6, then S also contains C2 = C36 and C3 = C2

6 , and

hence we must have jΦ(C6) = jΦ(C3) = jΦ(C2) = 2 and bΦ(C6) = bΦ(C3) =

bΦ(C2) = 0.

Thus, bΦ(Cm) must be 0 for all m > 2 and only in the case m = 3 may

jΦ(Cm) depart from 2.

(iv) Suppose S has the group I of all rotational symmetries of a regular

icosahedron as a subgroup. Then it follows from (iii) that for every

C5 ∈ S, we have jΦ(C5) = 2. Thus, all the ‘natural’ vertices of the

regular icosahedron must be present (as joints) in (G, p).

Similarly, if S has the group O of all rotational symmetries of a regular

octahedron as a subgroup, then for every C4 ∈ S, we have jΦ(C4) = 2,

and hence all the ‘natural’ vertices of the regular octahedron must be

present (as joints) in (G, p).

(v) If S contains a reflection s, then equation (4.25) says that the number

of joints of (G, p) that are fixed by s with respect to Φ is equal to the

number of bars of (G, p) that are fixed by s with respect to Φ.

(vi) If S contains an inversion i, then it follows from (4.26) that (G, p) has

neither a joint nor a bar that is fixed by i with respect to Φ. If p is

141

injective, this says that there is no joint and no bar that is located at

the center of the inversion i.

(vii) If S contains an improper rotation Sm, m > 2, then equation (4.27)

holds. The non-negative integer solution jΦ(Cm) = 0 and bΦ(Cm) = 0 is

possible for all m. For m > 2, the factor 2 cos(

2πm

) − 1 is rational at

m = 3, 4, 6, but generates no further solutions:

m = 3:

−2jΦ(S3) = bΦ(S3),

and hence jΦ(S3) = bΦ(S3) = 0.

m = 4:

−jΦ(S4) = bΦ(S4),

and hence jΦ(S4) = bΦ(S4) = 0.

m = 6:

0jΦ(S6) = bΦ(S6),

and hence bΦ(S6) = 0. But if S contains S6, then S also contains i = S36 ,

so that jΦ(S6) = 0.

Symmetry operation by symmetry operation, the above results can be

summarized as follows.

Theorem 4.3.3 Let G be a graph, S be a symmetry group in dimension 3,

Φ : S → Aut(G) be a homomorphism, and (G, p) be an isostatic framework

in R(G,S,Φ) with the property that the points p(v), v ∈ V (G), span all of R3.

Then

142

(i) |E(G)| = 2|V (G)| − 3;

(ii) if C2 ∈ S, then (jΦ(C2), bΦ(C2)) = (2, 0), (1, 1), (0, 2);

(iii) if C3 ∈ S, then bΦ(C3) = 0;

(iv) if Cm ∈ S, where m > 3, then jΦ(Cm) = 2 and bΦ(Cm) = 0;

(v) if s ∈ S, then jΦ(s) = bΦ(s);

(vi) if i ∈ S, then jΦ(i) = bΦ(i) = 0;

(vii) if Sm ∈ S, where m > 2, then jΦ(Sm) = bΦ(Sm) = 0.

In contrast to the 2-dimensional case, the conditions we derived from the

3-dimensional symmetry-extended Maxwell’s equation do not exclude any

point group. For every symmetry group S in dimension 3, we can construct

a fully triangulated convex polyhedron that has S as its point group and

is isostatic by the Theorem of Cauchy and Dehn [10, 22, 76]. One possible

approach to construct such frameworks is to begin with the regular triangu-

lated convex polyhedra (the tetrahedron, octahedron and icosahedron), and

to expand them by using operations of truncation and capping. In fact, for

every symmetry group S in dimension 3, an infinite family of isostatic frame-

works with point group symmetry S can be created in this way.

For example, to generate isostatic frameworks with only the rotational

symmetries of a given triangulated polyhedron, we can ‘cap’ each face with

a twisted octahedron, consistent with the rotational symmetries of the un-

derlying polyhedron. The resultant polyhedron will then be an isostatic

framework with the rotational symmetries of the underlying polyhedron, but

143

none of the reflectional symmetries. An example of the capping of a regular

octahedron is shown in Figure 4.13.

(a) (b)

Figure 4.13: A regular octahedron (a), and a convex polyhedron (b) generated

by capping every face of the original octahedron with a twisted octahedron.

The polyhedron in (b) has the rotational but not the reflectional symmetries of

the polyhedron in (a). If a framework is constructed from either polyhedron

by placing bars along edges, and joints at vertices, the framework will be

isostatic.

..(a) .(b)

One interesting possibility arises from consideration of symmetry groups

that contain 3-fold rotational symmetry operations. Equation (4.24) allows

an unlimited number of joints, though not bars, that are fixed by a 3-fold

rotation. Thus, if we start with an isostatic framework and add joints sym-

metrically along a 3-fold axis using vertex 3-additions (see Definition 2.2.18),

the resultant frameworks will remain isostatic as long as each of the new

joints is added so that it is not coplanar with the three joints it is linked

to. So, for instance, we can ‘cap’ every face of an icosahedron to give the

compound icosahedron-plus-dodecahedron (the second stellation of the icosa-

144

hedron), as illustrated in Figure 4.14, and this process can be continued ad

infinitum adding a pile of ‘hats’ consisting of a new joint, linked to all three

joints of an original icosahedral face (see Figure 4.15).

(a) (b)

Figure 4.14: An icosahedron (a), and the second stellation of the icosahe-

dron (b). If a framework is constructed from either polyhedron by placing

bars along edges, and joints at vertices, the framework will be isostatic. The

framework (b) could be constructed from the framework (a) by ‘capping’ each

face of the original icosahedron.

..(a) .(b)

Similar constructions starting from octahedral and tetrahedral symmetric

isostatic frameworks can be envisaged.

Symmetric subgraphs

Let S be a symmetry group in dimension 3, Φ : S → Aut(G) be a

homomorphism, and (G, p) be a framework in R(G,S,Φ) with the property that

the points p(v), v ∈ V (G), span all of R3. Then, besides the conditions in

Theorem 4.3.3 for the symmetry operations in S (with Φ as the underlying

type), the conditions in Theorem 2.2.8 for all nontrivial subgraphs of G

are also necessary conditions for (G, p) to be isostatic. Further, if H is a

145

...

..

..

..

..

..

Figure 4.15: A series of ‘hats’ added symmetrically along a 3-fold axis of an

isostatic framework leaves the framework isostatic.

subgraph of G with the property that |E(H)| = 3|V (H)| − 6 and (H, p|V (H))

is a symmetric framework, say (H, p|V (H)) ∈ R(G,S′,Φ′), then the conditions in

Theorem 4.3.3 for the symmetry operations in S ′ (with Φ′ as the underlying

type) are clearly also necessary conditions for (G, p) to be isostatic (provided

that the points p(v), v ∈ V (H), span all of R3 and Φ′ : S ′ → Aut(H) is a

homomorphism).

Suppose S ′ is a subgroup of S and Φ′ : S ′ → Aut(H) is defined by

Φ′(x) = Φ(x)|V (H). Then, unlike the plane, some of these conditions (for

example, the ones for reflections and half-turns in S ′) are not covered by the

conditions for the symmetry operations in S (with Φ as the underlying type).

This is demonstrated by the examples shown in Figures 4.16 and 4.17.

The conditions for a 3-fold rotation C3 ∈ S ′ (i.e., bΦ′(C3) = 0 and jΦ′(C3)

is unrestricted) and for any improper rotation Sm ∈ S ′ with m ≥ 2 (i.e.,

jΦ′(Sm) = bΦ′(Sm) = 0) clearly follow from the conditions for S and Φ.

If Cm is a rotation with m > 3, then the conditions for Cm and Φ are

bΦ(Cm) = 0 and jΦ(Cm) = 2. Thus, bΦ′(Cm) = 0 and jΦ′(Cm) ≤ 2. Since we

also have |E(H)| = 3|V (H)| − 6, it is easy to see that if jΦ′(Cm) < 2, then

m must be equal to 6. But a 6-fold rotation C6 implies the presence of the

146

...

..

....

..

.. ...(a)

...

..

....

..

.. ..

.. ..

.(b)

...

..

....

..

.. ..

..

..

....

..

.. ..

.. ..

.(c)

Figure 4.16: 3-dimensional frameworks with mirror symmetry satisfying

Maxwell’s original rule. (a) A framework which is not isostatic, since

bΦa(s) > jΦa(s); (b) a framework which is not isostatic, since bΦb(s) < jΦb(s);

(c) a framework which satisfies bΦc(s) = jΦc(s), but is not isostatic, because

it contains the frameworks depicted in (a) and (b) (Φa, Φb, Φc are uniquely

determined by the injective realizations).

147

.

.. .. ....

.. ......

.(a)

.

.. .. ....

.. ......

.(b)

.

.. .. ....

.. ......

.. .. ....

.. ......

.(c)

Figure 4.17: 3-dimensional frameworks with half-turn symmetry satisfying

Maxwell’s original rule. (a) An isostatic framework; (b) a framework which

is not isostatic, since bΦb(C2) = jΦb(C2) = 0; (c) a framework which satisfies

jΦc(C2) = 0 and bΦc(C2) = 2, but is not isostatic, because it contains the non-

isostatic framework depicted in (b) (Φb and Φc are uniquely determined by

the injective realizations).

half-turn C36 . So, in order to check all the conditions for S ′ and Φ′ which are

not covered by the conditions for S and Φ, it suffices to check the conditions

for reflections and half-turns in S ′ (with Φ′ as underlying type).

We will get back to this observation in Section 5.5.2 of the next chapter,

where we discuss necessary and sufficient conditions for a graph to be (S, Φ)-

generically isostatic.

148

4.3.4 A remark on non-injective realizations

Let G be a graph, S be a symmetry group in dimension 2 or 3 and

Φ : S → Aut(G) be a homomorphism. If (G, p) ∈ R(G,S,Φ) is an (S, Φ)-

generic framework whose map p is not injective, then (G, p) is not isostatic,

as the following argument shows.

By Remark 3.3.2, there exist two joints(v, p(v)

)and

(w, p(w)

)of (G, p)

such that Φ(x)(v) = v and Φ(y)(w) = w, where x and y are two (not nec-

essarily distinct) symmetry operations in S whose corresponding symmetry

elements are the origin. So if S is a symmetry group in dimension 2, then x

and y must be rotations, and if S is a symmetry group in dimension 3, then

x and y must be improper rotations. Therefore, in dimension 2, we have

jΦ(Cm) > 0 for some rotation Cm ∈ S and in dimension 3, we have jΦ(Sm) > 0

for some improper rotation Sm ∈ S. By Theorems 4.3.2 and 4.3.3, it follows

that (G, p) is not isostatic.

Note that if a framework (G, p) ∈ R(G,S,Φ) is a non-injective realization

of G that is not (S, Φ)-generic, then (G, p) can possibly be isostatic.

For example, the non-injective and non-(C2, Φ)-generic realization of G in

Example 3.3.3 is clearly isostatic (see also Figure 3.8). Similarly, the non-

injective and non-(C3, Φ)-generic realization of G in Example 3.3.4 is also

isostatic.

For an example in 3-space, consider the non-injective realization (Gbp, p)

given in Example 3.3.2 (see also Figure 3.7 (b)). (Gbp, p) is clearly isostatic

and neither (Cs, Ξa)-generic nor (Cs, Ξb)-generic.

149

4.4 Necessary conditions for independence

and infinitesimal rigidity

Now that we have established necessary conditions for a symmetric frame-

work (G, p) ∈ R(G,S,Φ) to be isostatic, it is natural to ask whether we can also

find similar necessary conditions for (G, p) to be independent or infinitesi-

mally rigid. This question begs to be investigated particularly for frameworks

in dimension 2, since we have seen in Section 4.3.2 that there are only six

possible point groups that allow isostatic frameworks in the plane, whereas

an independent framework, as well as an infinitesimally rigid framework, can

clearly be constructed for any point group.

Let G be a graph, S be a symmetry group in dimension d, Φ : S → Aut(G)

be a homomorphism, and (G, p) ∈ R(G,S,Φ) be a framework with the property

that the points p(v), v ∈ V (G), span all of Rd.

Consider the characters XQ =∑r

t=1 κtχ(It) and Xi =∑r

t=1 µtχ(It),

where XQ, Xi, and χ(It), t = 1, . . . , r, are defined as in Section 4.2.2. It

follows immediately from the results of Section 4.2 that for (G, p) to be inde-

pendent, we need to have κt ≥ µt for each t = 1, . . . , r. Similarly, for (G, p) to

be infinitesimally rigid, we need to have κt ≤ µt for each t = 1, . . . , r. So, by

expressing each of the coefficients κt and µt in terms of |V (G)|, |E(G)|, jΦ(x),

and bΦ(x), where x ∈ S, using the formula in Theorem 4.2.3 (iii), we may

obtain necessary conditions for (G, p) to be independent and infinitesimally

rigid, respectively.

For example, for the symmetry group C2 in dimension 2, we obtain the

following result.

150

Theorem 4.4.1 Let G be a graph, C2 = Id, C2 be the half-turn symmetry

group in dimension 2, Φ : C2 → Aut(G) be a homomorphism, and (G, p) ∈R(G,C2,Φ) be a framework with the property that the points p(v), v ∈ V (G),

span all of R2.

(i) If (G, p) is independent, then |E(G)| ≤ min2|V (G)| − 2 − (2jΦ(C2) +

bΦ(C2)), 2|V (G)| − 4 + 2jΦ(C2) + bΦ(C2)

;

(ii) if (G, p) is infinitesimally rigid, then |E(G)| ≥ max2|V (G)| − 2 −

(2jΦ(C2) + bΦ(C2)), 2|V (G)| − 4 + 2jΦ(C2) + bΦ(C2)

.

Proof. By Table 4.1, we have

XQ = XJ ×XT −XT −XR = (2|V (G)| − 3,−2jΦ(C2) + 1)

Xi = (|E(G)|, bΦ(C2)).

Since C2 has the two pairwise non-equivalent irreducible representations A

and B, we also have

XQ = κ1χ(A) + κ2χ(B)

Xi = µ1χ(A) + µ2χ(B).

Using the formula in Theorem 4.2.3 (iii) and the fact that A and B have the

characters χ(A) = (1, 1) and χ(B) = (1,−1), we obtain

κ1 = |V (G)| − jΦ(C2) − 1

κ2 = |V (G)|+ jΦ(C2) − 2

µ1 =1

2

(|E(G)|+ bΦ(C2)

)

µ2 =1

2

(|E(G)| − bΦ(C2)

).

151

By the results of Section 4.2, we have that if (G, p) is independent, then

κt ≥ µt for t = 1, 2, and hence

|E(G)| ≤ 2|V (G)| − 2− (2jΦ(C2) + bΦ(C2))

|E(G)| ≤ 2|V (G)| − 4 + 2jΦ(C2) + bΦ(C2).

Similarly, if (G, p) is infinitesimally rigid, then κt ≤ µt for t = 1, 2, and hence

|E(G)| ≥ 2|V (G)| − 2− (2jΦ(C2) + bΦ(C2))

|E(G)| ≥ 2|V (G)| − 4 + 2jΦ(C2) + bΦ(C2).

This gives the result. ¤

.. .

..

.(a)

.. .

..

.(b)

.. .

...

.(c)

.. .

..

..

.(d)

.. .

..

.(e)

.. .

...

.(f)

Figure 4.18: Independent frameworks in R(G,C2,Φ) with jΦ(C2) = bΦ(C2) = 0

(a), jΦ(C2) = 0, bΦ(C2) = 2 (b), and jΦ(C2) = 1, bΦ(C2) = 0 (c). Infinitesimally

rigid frameworks in R(G,C2,Φ) with jΦ(C2) = bΦ(C2) = 0 (d), jΦ(C2) = 0, bΦ(C2) =

2 (e), and jΦ(C2) = 1, bΦ(C2) = 0 (f).

Similar results can of course easily be established for an arbitrary sym-

metry group S in dimension d. If S has r pairwise non-equivalent irreducible

representations, then we obtain r necessary conditions (namely a system of

r linear inequalities) for the given framework (G, p) ∈ R(G,S,Φ) to be inde-

pendent and infinitesimally rigid respectively using the above method. The

structures of these types of systems of linear inequalities, however, are still

to be examined.

152

Sufficient conditions for a graph to be (S, Φ)-generically independent or

(S, Φ)-generically infinitesimally rigid will be discussed in Section 5.5.3.

153

Chapter 5

Necessary and sufficient

conditions for a graph to be

(S, Φ)-generically isostatic

In this chapter, we establish symmetric versions of some of the key results

in generic rigidity theory, such as Laman’s Theorem ([33, 46]), Henneberg’s

Theorem ([40, 33]), and Crapo’s Theorem ([20, 33, 67]) (see also Section

2.2.4).

These symmetrized results provide us with various techniques to give a

‘certificate’ that a given graph is (S, Φ)-generically isostatic. Furthermore,

they enable us to generate all (S, Φ)-generically isostatic graphs by means

of an inductive construction sequence. With each of the main results of this

chapter, we also lay the foundation to design algorithms that decide whether

a given graph is (S, Φ)-generically isostatic.

The statements in part (ii) of the Theorems 5.2.1, 5.3.1, and 5.4.1 were

154

conjectured in [15]. The extensions to the Henneberg-type inductions and

tree partitions are new, as are all the proofs. A summary of these results will

be presented in [57].

Note that the proofs of the sufficiency results do not depend on the group

representation theory used in Chapter 4 to establish the necessary counts.

5.1 Preliminary remarks and results

Let G be a graph with |V (G)| ≥ d+1, S be a symmetry group in dimen-

sion d, and Φ : S → Aut(G) be a homomorphism. If G is (S, Φ)-generically

isostatic, then G must satisfy the conditions identified in Theorem 2.2.8 and,

by the results of the previous chapter (and by Remark 2.2.3), the symmetry-

extended Maxwell’s equation for S and Φ must also be satisfied. In particular,

if S is a symmetry group in dimension 2, then G must satisfy the Laman con-

ditions (see Theorem 2.2.9), and the conditions identified in Theorem 4.3.2

for S and Φ must also hold.

Note that if S is the trivial group C1 in dimension 2, then the homo-

morphism Φ must send Id ∈ C1 to the identity automorphism of G, and

all of these conditions combined reduce to the Laman conditions which, by

Laman’s Theorem, are not only necessary, but also sufficient for G to be

(C1, Φ)-generically isostatic (i.e., generically 2-isostatic).

It was conjectured in [15] that Laman’s Theorem can be extended to each

of the non-trivial symmetry groups in dimension 2 that allow isostatic frame-

works, i.e., to each of the groups C2, C3, Cs, C2v, and C3v.

In this chapter, we verify this conjecture for the symmetry groups

155

S = C2, C3, Cs. That is, for each of these groups, we show that the con-

ditions identified in Theorem 4.3.2 for S and Φ, together with the Laman

conditions, are necessary and sufficient for G to be (S, Φ)-generically iso-

static. For the groups C2v and C3v, we restate (and, in the case of C3v, also

extend) the corresponding conjectures given in [15].

Recall from Section 2.2.5 that Hennebergs’s Theorem (Theorem 2.2.12)

and Crapo’s Theorem (Theorem 2.2.14) also provide characterizations of

generically 2-isostatic graphs. We show that these results also have sym-

metric analogs for each of the symmetry groups C2, C3, and Cs. More pre-

cisely, we define symmetrized inductive construction techniques as well as

symmetrized 3Tree2 partitions for S = C2, C3, Cs with the aid of which we

will be able to derive conditions that are again necessary and sufficient for

G to be (S, Φ)-generically isostatic.

In our initial efforts to establish characterizations of (S, Φ)-generically

isostatic graphs we focused our attention on the Laman-type conjectures

given in [15]. Our initial attempts to prove these conjectures, however, were

unsuccessful, so that these problems remained open for quite some time. This

was partly due to the fact that we mistakenly considered the symmetry group

Cs to be the easiest starting point for our investigation of these conjectures.

The proof for the symmetric version of Laman’s Theorem for Cs, however,

turns out to be considerably more complex than the ones for the ‘rotational’

symmetry groups C2 and C3. In fact, among all of the symmetry groups

C2, C3, Cs, C2v, and C3v, the group C3 allows the easiest and most natural proof

for a symmetric version of Laman’s Theorem (as well as for a symmetric

version of Crapo’s Theorem). Although this initially came as somewhat of a

156

surprise, we can now identify some clear indications for this.

For example, Crapo’s Theorem uses partitions of the edges of G into

three edge-disjoint trees, so that it is most natural to extend this result to

the cyclic group C3 of order three. Moreover, the condition jΦ(C3) = 0 implies

that for any subgraph H of G with full C3 symmetry we must have that

both |V (H)| and |E(H)| are multiples of three, so that H cannot satisfy the

count |E(H)| = 2|V (H)| − 4 or |E(H)| = 2|V (H)| − 5. This turns out to be

extremely useful in the proof of the Laman-type result for C3.

The difficulties that arise if one wants to establish characterizations of

(S, Φ)-generically isostatic graphs for the symmetry groups S = C2v and

S = C3v in dimension 2, as well as for any symmetry group S in dimension

3, are briefly discussed in Section 5.5.

In the same section, we present some algorithms that decide whether a

given graph G is (S, Φ)-generically isostatic.

We begin with the following two Lemmas that we will be using frequently

in the proofs throughout this chapter.

Lemma 5.1.1 Let G be a graph with |V (G)| ≥ 3 that satisfies the Laman

conditions. Then

(i) G has a vertex of valence 2 or 3;

(ii) if G has no vertex of valence 2, then G has at least six vertices of

valence 3.

Proof. (i) The average valence in G is

2|E(G)||V (G)| =

2(2|V (G)| − 3)

|V (G)| = 4− 6

|V (G)| < 4.

157

Since G satisfies the Laman conditions and |V (G)| ≥ 3, it is easy to see that

G has no vertex of valence 0 or 1.

(ii) Suppose G has no vertex of valence 2 and k vertices of valence 3,

where k < 6. Then the average valence in G is at least

3k + 4(|V (G)| − k)

|V (G)| = 4− k

|V (G)| > 4− 6

|V (G)|

contradicting (i). ¤

Lemma 5.1.2 Let G be a graph that satisfies the Laman conditions and

let v be a vertex of G with NG(v) = v1, v2, v3. Further, let α ∈ Aut(G)

and(v α(v) . . . αn(v)

)be the permutation cycle of α containing v. If

v, α(v), . . . , αn(v) is an independent set of vertices in G, then

(i) there exists i, j ⊆ 1, 2, 3 such that for every subgraph H ′ of

G′ = G − v, α(v), . . . , αn(v) with vi, vj ∈ V (H ′), we have |E(H ′)| ≤2|V (H ′)| − 4;

(ii) if i, j ⊆ 1, 2, 3 is the only pair for which (i) holds, then

αk(vi), αk(vj) 6= αm(vi), α

m(vj) for all 0 ≤ k < m ≤ n, and

G′ +αt(vi), α

t(vj)| t = 0, 1, . . . , n

satisfies the Laman conditions.

Proof. (i) It follows from Theorems 2.2.9 and 2.2.11 that there exists

in, jn ⊆ 1, 2, 3 such that Gn = G−αn(v)+αn(vin), αn(vjn) satis-

fies the Laman conditions. By the same argument, there exists in−1, jn−1 ⊆1, 2, 3 such that Gn−1 = Gn − αn−1(v) +

αn−1(vin−1), αn−1(vjn−1)

satisfies the Laman conditions. Continuing in this fashion, we arrive at a

graph G0 with V (G0) = V (G) \ v, α(v), . . . , αn(v) = V (G′) and E(G0) =

158

E(G′) ∪ αn(vin), αn(vjn), . . . , vi0 , vj0

that satisfies the Laman condi-

tions. Therefore, every subgraph H of G0−vi0 , vj0

with vi0 , vj0 ∈ V (H)

satisfies |E(H)| ≤ 2|V (H)| − 4. Since V (G′) = V(G0 −

vi0 , vj0)

and

E(G′) ⊆ E(G0 −

vi0 , vj0)

, it follows that every subgraph H ′ of G′ with

vi0 , vj0 ∈ V (H ′) satisfies |E(H ′)| ≤ 2|V (H ′)| − 4.

.

..v

. ..

.

.αn(v)...

..α(v) .

..

.G

.

. .....

.

..

.G0

.

. .....

.

..

.G′

Figure 5.1: Illustration of the proof of Lemma 5.1.2.

(ii) Wlog we suppose that i, j = 1, 2 is the only pair in 1, 2, 3 for

which (i) holds. Then there exists a subgraph H1 of G′ with v1, v3 ∈ V (H1)

satisfying |E(H1)| = 2|V (H1)| − 3 and a subgraph H2 of G′ with v2, v3 ∈V (H2) satisfying |E(H2)| = 2|V (H2)| − 3. Since G′ is invariant under α

(recall Definition 2.1.10), αk(H1) and αk(H2) are also subgraphs of G′ for all

1 ≤ k ≤ n. Moreover, for all 0 ≤ k ≤ n, we have

αk(v1), αk(v3) ∈ V

(αk(H1)

)

|E(αk(H1)

)| = 2|V (αk(H1)

)| − 3

and

αk(v2), αk(v3) ∈ V

(αk(H2)

)

|E(αk(H2)

)| = 2|V (αk(H2)

)| − 3.

159

By Theorems 2.2.9 and 2.2.11, there exists in, jn ⊆ 1, 2, 3 such that

Gn = G − αn(v) +αn(vin), αn(vjn) satisfies the Laman conditions.

Likewise, for all 0 ≤ k ≤ n − 1, there exists ik, jk ⊆ 1, 2, 3 such that

Gk = Gk+1 − αk(v)+αk(vik), α

k(vjk) satisfies the Laman conditions.

Since for all 0 ≤ k ≤ n, we have G′ ⊆ Gk, and hence αk(H1), αk(H2) ⊆ Gk,

we must have ik, jk = 1, 2 for all k. In particular, αk(v1), αk(v2) 6=

αm(v1), αm(v2) for all 0 ≤ k < m ≤ n and G0 = G′+

αt(v1), αt(v2)| t =

0, 1, . . . , n

satisfies the Laman conditions. ¤

For each of the groups C2, C3, Cs, we will prove a symmetrized version of

Crapo’s Theorem by using an approach that is in the style of Tay’s proof

(see [67]) of Crapo’s original result. This requires the notion of a ‘frame’,

i.e., a generalized notion of a framework that allows joints to be located at

the same point in space, even if their corresponding vertices are adjacent.

Definition 5.1.1 Let G be a graph with V (G) = v1, v2, . . . , vn. A frame

in R2 is a triple (G, p, q), where p : V (G) → R2 and q : E(G) → R2 \ 0are maps with the property that for all vi, vj ∈ E(G) there exists a scalar

λij ∈ R (which is possibly zero) such that p(vi)− p(vj) = λijq(vi, vj).

Definition 5.1.2 The generalized rigidity matrix of a frame (G, p, q) in R2

is the matrix

R(G, p, q) =

...

0 . . . 0 q(vi, vj) 0 . . . 0 −q(vi, vj) 0 . . . 0

...

,

i.e., for each edge vi, vj ∈ E(G), R(G, p, q) has the row with(q(vi, vj)

)1

and(q(vi, vj)

)2

in the columns 2i − 1 and 2i, −(q(vi, vj)

)1

and

160

−(q(vi, vj)

)2

in the columns 2(j − 1) and 2j, and 0 elsewhere.

We say that (G, p, q) is independent if R(G, p, q) has linearly independent

rows.

Remark 5.1.1 If (G, p, q) is a frame with the property that p(vi) 6= p(vj)

whenever vi, vj ∈ E(G), then we obtain the rigidity matrix of the frame-

work (G, p) by multiplying each row of R(G, p, q) by its corresponding scalar

λij. Therefore, if (G, p, q) is independent, so is (G, p).

Lemma 5.1.3 Let (G, p, q) be an independent frame in R2 and let pt :

V (G) → R[t] × R[t] and qt : E(G) → R[t] × R[t] be such that (G, pa, qa)

is a frame in R2 for every a ∈ R. If (G, pa, qa) = (G, p, q) for a = 0, then

(G, pa, qa) is an independent frame in R2 for almost all a ∈ R.

Proof. Note that the rows of R(G, pt, qt) are linearly dependent (over the

quotient field of R[t]) if and only if the determinants of all the |E(G)|×|E(G)|submatrices of R(G, pt, qt) are identically zero. These determinants are poly-

nomials in t. Thus, the set of all a ∈ R with the property that R(G, pa, qa)

has a non-trivial row dependency is a variety F whose complement, if non-

empty, is a dense open set. Since a = 0 is in the complement of F we can

conclude that for almost all a, (G, pa, qa) is independent. ¤

Each time Lemma 5.1.3 is applied in this chapter, the polynomials in

R(G, pt, qt) are linear polynomials in t.

Since the characterizations of (S, Φ)-generically isostatic graphs can be

given in the most natural way if S = C3, followed by S = C2 and S = Cs, the

sections in this chapter are arranged according to this order.

161

5.2 Characterizations of (C3, Φ)-generically

isostatic graphs

5.2.1 Symmetrized Henneberg moves and 3Tree2 par-

titions for C3

We need the following inductive construction techniques to obtain a sym-

metrized Henneberg’s Theorem for C3.

Definition 5.2.1 Let G be a graph, C3 = Id, C3, C23 be a sym-

metry group in dimension 2, and Φ : C3 → Aut(G) be a homomor-

phism. Let v1, v2 be two distinct vertices of G and v, w, z /∈ V (G).

Then the graph G with V (G) = V (G) ∪ v, w, z and E(G) =

E(G) ∪ v, v1, v, v2, w, Φ(C3)(v1), w, Φ(C3)(v2), z, Φ(C23)(v1),

z, Φ(C23)(v2)

is called a (C3, Φ) vertex addition (by (v, w, z)) of G.

.

.γ(v1)

.γ(v2)

..

.

. .

.

.γ2(v1).γ2(v2)

.v1

.v2

.

.γ(v1)

.γ(v2)

..

.

. .

.

.

. .

.γ2(v1).γ2(v2)

.v1

.v2

.z

.v .w

Figure 5.2: A (C3, Φ) vertex addition of a graph G, where Φ(C3) = γ and

Φ(C23) = γ2.

Definition 5.2.2 Let G be a graph, C3 = Id, C3, C23 be a symmetry

group in dimension 2, and Φ : C3 → Aut(G) be a homomorphism.

162

Let v1, v2, v3 be three distinct vertices of G such that v1, v2 ∈ E(G)

and not both of v1 and v2 are fixed by Φ(C3) and let v, w, z /∈ V (G).

Then the graph G with V (G) = V (G) ∪ v, w, z and E(G) =(E(G) \ v1, v2, Φ(C3)(v1), Φ(C3)(v2), Φ(C2

3)(v1), Φ(C23)(v2)

) ∪v, vi| i = 1, 2, 3

∪ w, Φ(C3)(vi)| i = 1, 2, 3 ∪

z, Φ(C23)(vi)| i = 1, 2, 3

is called a (C3, Φ) edge split (on

(v1, v2, Φ(C3)(v1), Φ(C3)(v2), Φ(C23)(v1), Φ(C2

3)(v2)); (v, w, z)) of

G.

.

..

.

. .

.

.

. .

.γ2(v1).γ2(v2)

.v1

.v2 .γ(v1)

.γ(v2).γ2(v3)

.v3 .γ(v3)

.

..

.

. .

.

.

. .

.

. .

.γ2(v1).γ2(v2)

.v1

.v2 .γ(v1)

.γ(v2).γ2(v3)

.v3 .γ(v3)

.z

.v .w

Figure 5.3: A (C3, Φ) edge split of a graph G, where Φ(C3) = γ and Φ(C23) =

γ2.

Definition 5.2.3 Let G be a graph, C3 = Id, C3, C23 be a symmetry

group in dimension 2, and Φ : C3 → Aut(G) be a homomorphism. Let

v0 be a vertex of G that is not fixed by Φ(C3) and let v, w, z /∈ V (G).

Then the graph G with V (G) = V (G) ∪ v, w, z and E(G) = E(G) ∪v, w, w, z, z, v, v, v0, w, Φ(C3)(v0), z, Φ(C2

3)(v0)

is called a

(C3, Φ) ∆ extension (by (v, w, z)) of G.

163

...

.

.γ(v0).γ2(v0)

.v0

...

.

.

. .

.γ(v0).γ2(v0)

.v0

.z

.v .w

Figure 5.4: A (C3, Φ) ∆ extension of a graph G, where Φ(C3) = γ and

Φ(C23) = γ2.

Remark 5.2.1 Each of the constructions in Definitions 5.2.1, 5.2.2, and

5.2.3 has the property that if the graph G satisfies the Laman conditions,

then so does G. This follows from Theorems 2.2.9 and 2.2.12 and the fact

that we can obtain a (C3, Φ) vertex addition of G by a sequence of three

vertex 2-additions, a (C3, Φ) edge split of G by a sequence of three edge 2-

splits, and a (C3, Φ) ∆ extension of G by a vertex 2-addition followed by two

edge 2-splits.

In order to extend Crapo’s Theorem to C3 we need the following sym-

metrized definition of a 3Tree2 partition.

Definition 5.2.4 Let G be a graph, C3 = Id, C3, C23 be a symmetry group

in dimension 2, and Φ : C3 → Aut(G) be a homomorphism. A (C3, Φ) 3Tree2

partition of G is a 3Tree2 partition E(T0), E(T1), E(T2) of G such that

Φ(C3)(Ti) = Ti+1 for i = 0, 1, 2, where the indices are added modulo 3.

164

.

.

. .

.γ2(v)

.v .γ(v)

.

.

.

. .

.

...w

.γ(w)

.γ2(w)

.

.γ2(v)

.v .γ(v)

Figure 5.5: (C3, Φ) 3Tree2 partitions of graphs, where Φ(C3) = γ and

Φ(C23) = γ2.

5.2.2 The main result for C3

Theorem 5.2.1 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23 be a

symmetry group in dimension 2, and Φ : C3 → Aut(G) be a homomorphism.

The following are equivalent:

(i) R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically isostatic;

(ii) |E(G)| = 2|V (G)| − 3, |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G with

|V (H)| ≥ 2 (Laman conditions), and jΦ(C3) = 0;

(iii) there exists a (C3, Φ) construction sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

such that

(a) Gi+1 is a (C3, Φi) vertex addition, a (C3, Φi) edge split, or a (C3, Φi)

∆ extension of Gi with V (Gi+1) = V (Gi) ∪ vi+1, wi+1, zi+1 for

all i = 0, 1, . . . , k − 1;

165

(b) Φ0 : C3 → Aut(K3) is a non-trivial homomorphism and for

all i = 0, 1, . . . , k − 1, Φi+1 : C3 → Aut(Gi+1) is the homo-

morphism defined by Φi+1(x)|V (Gi) = Φi(x) for all x ∈ C3 and

Φi+1(C3)|vi+1,wi+1,zi+1 = (vi+1 wi+1 zi+1);

(iv) G has a proper (C3, Φ) 3Tree2 partition.

We break the proof of this result up into four Lemmas.

Lemma 5.2.2 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23 be a


If R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically isostatic, then G satisfies the

Laman conditions and we have jΦ(C3) = 0.

Proof. The result follows from Laman’s Theorem (Theorem 2.2.9), Theorem

4.3.2, and Remark 2.2.3. ¤



If G satisfies the Laman conditions and we also have jΦ(C3) = 0, then there

exists a (C3, Φ) construction sequence for G.

Proof. We employ induction on |V (G)|. Note first that if for a graph G,

there exists a homomorphism Φ : C3 → Aut(G) such that jΦ(C3) = 0, then

|V (G)| ≡ 0 (mod 3). The only graph with three vertices that satisfies the

Laman conditions is the graph K3 and if Φ : C3 → Aut(K3) is a homomor-

phism such that jΦ(C3) = 0, then Φ is clearly a non-trivial homomorphism.

This proves the base case.

166

So we let n > 3 and we assume that the result holds for all graphs with

n or fewer than n vertices.

Let G be a graph with |V (G)| = n + 3 that satisfies the Laman condi-

tions and suppose jΦ(C3) = 0 for a homomorphism Φ : C3 → Aut(G). In the

following, we denote Φ(C3) by γ and Φ(C23) by γ2. By Lemma 5.1.1, G has

a vertex of valence 2 or 3.

We assume first that G has a vertex v of valence 2, say NG(v) = v1, v2.Note that v, γ(v) and γ2(v) are three distinct vertices of G, because jγ = 0.

Suppose two of these vertices are adjacent, wlog v, γ(v) ∈ E(G). Then

γ(v), γ2(v), γ2(v), v ∈ E(G), because γ ∈ Aut(G). Let G′ = G −v, γ(v), γ2(v). Then

|E(G′)| = |E(G)| − 3 = 2|V (G)| − 6 = 2|V (G′)|.

Since |V (G)| ≥ 6, we have |V (G′)| ≥ 3, and hence G′ violates the Laman

conditions, a contradiction.

Therefore, v, γ(v), γ2(v) is an independent subset of V (G), which says

that the six edges v, vi, γ(v), γ(vi), γ2(v), γ2(vi), i = 1, 2, are all pair-

wise distinct. Thus,

|E(G′)| = |E(G)| − 6 = 2|V (G)| − 9 = 2|V (G′)| − 3.

Also, for H ⊆ G′ with |V (H)| ≥ 2, we have H ⊆ G, and hence

|E(H)| ≤ 2|V (H)| − 3.

Therefore, G′ satisfies the Laman conditions.

Let Φ′ : C3 → Aut(G′) be the homomorphism with Φ′(x) = Φ(x)|V (G′) for

all x ∈ C3. Then we have jΦ′(C3) = 0, and hence, by the induction hypothesis,

167

there exists a sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G′, Φ′)

satisfying the conditions in Theorem 5.2.1 (iii). Since G is a (C3, Φ′) vertex

addition of G′ with V (G) = V (G′) ∪ v, γ(v), γ2(v),

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G′, Φ′), (G, Φ)

is a sequence with the desired properties.

Suppose now that G has a vertex v of valence 3, say NG(v) = v1, v2, v3,and no vertex of valence 2. Note that v, γ(v) and γ2(v) are again three

distinct vertices of G, as are vi, γ(vi) and γ2(vi) for each i = 1, 2, 3, because

jγ = 0. We need to consider the following three cases (see also Figure 5.6):

Case 1: v, γ(v), γ2(v) is an independent subset of V (G) and all three

of these vertices share a common neighbor, say wlog v1. Since

γ ∈ Aut(G), this says that each of v, γ(v) and γ2(v) has the same

neighbors, namely v1, γ(v1) and γ2(v1).

Case 2: v, γ(v), γ2(v) is an independent subset of V (G) and there is no

vertex in G that is adjacent to all three of these vertices.

Case 3: v, γ(v), γ2(v) is not independent in G and hence

v, γ(v), γ(v), γ2(v), γ2(v), v ∈ E(G).

Case 1: By Lemma 5.1.2 (i), there exists a pair of vertices in

v1, γ(v1), γ2(v1), say wlog v1, γ(v1), such that for every subgraph H

of G′ = G − v, γ(v), γ2(v) with v1, γ(v1) ∈ V (H), we have |E(H)| ≤

168

. ..

.

.

. .

.γ2(v)

.v .γ(v)

.

.. .

.... .

.

.

. .

.γ2(v)

.v .γ(v)

.

.. .

..

.

.

. .

.γ2(v)

.v .γ(v)

. ..

.

.

. .

.γ2(v)

.v .γ(v)

. .(Case 1) .(Case 2) .(Case 3)

Figure 5.6: If a graph G satisfies the conditions in Theorem 5.2.1 (ii) and

has a vertex v of valence 3, then G is a graph of one of the types depicted

above.

2|V (H)|− 4. Further, since G′ is invariant under γ (recall Definition 2.1.10),

every subgraph H of G′ with γ(v1), γ2(v1) ∈ V (H) or γ2(v1), v1 ∈ V (H) also

satisfies |E(H)| ≤ 2|V (H)| − 4.

Suppose there exists a subgraph H of G′ with v1, γ(v1), γ2(v1) ∈ V (H)

and |E(H)| ≥ 2|V (H)|−5. Then the subgraph H of G with V (H) = V (H)∪v, γ(v), γ2(v) and E(H) = E(H) ∪ z, v1, z, γ(v1), z, γ2(v1)| z =

v, γ(v), γ2(v)

satisfies

|E(H)| = |E(H)|+ 9 ≥ 2|V (H)|+ 4 = 2|V (H)| − 2,

contradicting the fact that G satisfies the Laman conditions. So, every sub-

graph H of G′ with v1, γ(v1), γ2(v1) ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 6.

Now, let G = G′ +v1, γ(v1), γ(v1), γ

2(v1), γ2(v1), v1. Then we

have

|E(G)| = |E(G′)|+ 3 = |E(G)| − 6 = 2|V (G)| − 9 = 2|V (G)| − 3,

169

and hence G satisfies the Laman conditions.

Further, if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ C3, then Φ(x) ∈Aut(G) for all x ∈ C3 and Φ : C3 → Aut(G) is a homomorphism. Since

we also have jΦ(C3) = 0, it follows from the induction hypothesis that there

exists a sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

satisfying the conditions in Theorem 5.2.1 (iii). Since G is a (C3, Φ) edge

split of G with V (G) = V (G) ∪ v, γ(v), γ2(v),

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case 2: By Lemma 5.1.2 (i), there exists i, j ⊆ 1, 2, 3, wlog

i, j = 1, 2, such that for every subgraph H of G′ = G− v, γ(v), γ2(v)with vi, vj ∈ V (H) we have |E(H)| ≤ 2|V (H)|−4. Since G′ is invariant under

γ, every subgraph H of G′ with γ(v1), γ(v2) ∈ V (H) or γ2(v1), γ2(v2) ∈ V (H)

also satisfies |E(H)| ≤ 2|V (H)| − 4. Note that v1, v2, γ(v1), γ(v2) and

γ2(v1), γ2(v2) are three distinct pairs of vertices (though not edges, by the

above counts) of G, as the following argument shows.

Suppose v1, v2 = γ(v1), γ(v2). Then v1 = γ(v2) and v2 = γ(v1), be-

cause G satisfies jγ = 0. Therefore, γ(v1) = γ2(v2), and hence v2 = γ2(v2),

contradicting jγ = 0. Similarly, γ(v1), γ(v2) 6= γ2(v1), γ2(v2) and

v1, v2 6= γ2(v1), γ2(v2).

We claim that G = G′ +v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2)

satis-

fies the Laman conditions. We clearly have

|E(G)| = |E(G′)|+ 3 = |E(G)| − 6 = 2|V (G)| − 9 = 2|V (G)| − 3.

170

Suppose there exists a subgraph H of G′ with v1, v2, γ(v1), γ(v2) ∈ V (H)

and |E(H)| = 2|V (H)| − 4. Then there also exists γ(H) ⊆ G′ with

γ(v1), γ(v2), γ2(v1), γ

2(v2) ∈ V(γ(H)

)and |E(

γ(H))| = 2|V (

γ(H))| − 4, as

well as γ2(H) ⊆ G′ with γ2(v1), γ2(v2), v1, v2 ∈ V

(γ2(H)

)and |E(

γ2(H))| =

2|V (γ2(H)

)|−4, because G′ is invariant under γ. Let H ′ = H ∪γ(H). Then

|E(H ′)| = |E(H)|+ |E(γ(H)

)| − |E(H ∩ γ(H)

)|

≥ 2|V (H)| − 4 + 2|V (γ(H)

)| − 4− (2|V (H ∩ γ(H)

)| − 4)

= 2|V (H ′)| − 4,

because H ∩ γ(H) is a subgraph of G′ with γ(v1), γ(v2) ∈ V(H ∩ γ(H)

).

Since H ′ is also a subgraph of G′ with γ(v1), γ(v2) ∈ V (H ′), it follows that

|E(H ′)| = 2|V (H ′)| − 4.

Similarly, it can be shown that H ′′ = H ′ ∪ γ2(H) satisfies

|E(H ′′)| = 2|V (H ′′)| − 4,

because H ′ ∩ γ2(H) is a subgraph of G′ with v1, v2 ∈ V(H ′ ∩ γ2(H)

). How-

ever, H ′′ is invariant under γ and satisfies jγ|V (H′′) = 0, so that |V (H ′′)| ≡ 0

(mod 3) and |E(H ′′)| ≡ 0 (mod 3), contradicting the count |E(H ′′)| =

2|V (H ′′)| − 4.

Therefore, every subgraph H of G′ with v1, v2, γ(v1), γ(v2) ∈ V (H),

γ(v1), γ(v2), γ2(v1), γ

2(v2) ∈ V (H), or γ2(v1), γ2(v2), v1, v2 ∈ V (H) satisfies

|E(H)| ≤ 2|V (H)| − 5.

It is now only left to show that for every subgraph H of G′ with

v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2) ∈ V (H), we have |E(H)| ≤ 2|V (H)| −6. Suppose to the contrary that there exists a subgraph H of G′ with

171

v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2) ∈ V (H) and |E(H)| = 2|V (H)| − 5. Then

there also exist γ(H) ⊆ G′ and γ2(H) ⊆ G′ with the same properties, because

G′ is invariant under γ. Let H ′ = H ∪ γ(H). Then

|E(H ′)| = |E(H)|+ |E(γ(H)

)| − |E(H ∩ γ(H)

)|

≥ 2|V (H)| − 5 + 2|V (γ(H)

)| − 5− (2|V (H ∩ γ(H)

)| − 5)

= 2|V (H ′)| − 5,

because H∩γ(H) is a subgraph of G′ with v1, v2, γ(v1), γ(v2) ∈ V(H∩γ(H)

).

Since H ′ is also a subgraph of G′ with v1, v2, γ(v1), γ(v2) ∈ V (H ′), it follows

that

|E(H ′)| = 2|V (H ′)| − 5.

Similarly, it can be shown that H ′′ = H ′ ∪ γ2(H) satisfies

|E(H ′′)| = 2|V (H ′′)| − 5,

because H ′ ∩ γ2(H) is a subgraph of G′ with v1, v2, γ(v1), γ(v2) ∈ V(H ′ ∩

γ2(H)). However, H ′′ is invariant under γ and we have jγ|V (H′′) = 0, so that

|V (H ′′)| ≡ 0 (mod 3) and |E(H ′′)| ≡ 0 (mod 3), contradicting the count

|E(H ′′)| = 2|V (H ′′)| − 5.

Thus, G = G′+v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2)

indeed satisfies

the Laman conditions.

Further, if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ C3, then Φ(x) ∈Aut(G) for all x ∈ C3 and Φ : C3 → Aut(G) is a homomorphism. Since

we also have jΦ(C3) = 0, it follows from the induction hypothesis that there

exists a sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

172


split of G with V (G) = V (G) ∪ v, γ(v), γ2(v),

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case 3: Note that G′ = G− v, γ(v), γ2(v) satisfies

|E(G′)| = |E(G)| − 6 = 2|V (G)| − 9 = 2|V (G′)| − 3.


|E(H)| ≤ 2|V (H)| − 3,

so that G′ satisfies the Laman conditions.

If we define Φ′ by Φ′(x) = Φ(x)|V (G′) for all x ∈ C3, then Φ′(x) ∈ Aut(G′)

for all x ∈ C3 and Φ′ : C3 → Aut(G′) is a homomorphism. Since we also

have jΦ′(C3) = 0, it follows from the induction hypothesis that there exists a

sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G′, Φ′)

satisfying the conditions in Theorem 5.2.1 (iii). Since G is a (C3, Φ′) ∆

extension of G′ with V (G) = V (G′) ∪ v, γ(v), γ2(v),

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G′, Φ′), (G, Φ)

is a sequence with the desired properties. ¤



If there exists a (C3, Φ) construction sequence for G, then G has a proper

(C3, Φ) 3Tree2 partition.

173

Proof. We proceed by induction on |V (G)|. Let V (K3) = v1, v2, v3and wlog let Φ : C3 → Aut(K3) be the homomorphism defined by

Φ(C3) = (v1 v2 v3). Then K3 has the proper (C3, Φ) 3Tree2 partition

E(T0), E(T1), E(T2), where T0 = 〈v1, v2〉, T1 = 〈v2, v3〉 and T2 =

〈v3, v1〉.Assume, then, that the result holds for all graphs with n or fewer than n

vertices, where n > 3.

Let G be a graph with |V (G)| = n + 3 and let Φ : C3 → Aut(G) be a

homomorphism such that there exists a (C3, Φ) construction sequence

(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

satisfying the conditions in Theorem 5.2.1 (iii). By Remark 5.2.1, G satisfies

the Laman conditions, and hence, by Remark 2.2.5, any 3Tree2 partition of

G must be proper. Therefore, it suffices to show that G has some (C3, Φ)

3Tree2 partition. We let Φ(C3) = γ and Φ(C23) = γ2.

By the induction hypothesis, Gk−1 has a (C3, Φk−1) 3Tree2 partitionE

(T

(k−1)0

), E

(T

(k−1)1

), E

(T

(k−1)2

). In the following, we compute the in-

dices i of the trees T(k−1)i modulo 3.

Suppose first that G is a (C3, Φk−1) vertex addition by (v w z) of Gk−1 with

NG(v) = v1, v2, where w = γ(v) and z = γ2(v). Since Φk−1(C3) = γ|V (Gk−1)

we have NG(w) = γ(v1), γ(v2) and NG(z) = γ2(v1), γ2(v2). Note that

both v1 and v2 belong to exactly two of the trees T(k−1)i . Therefore, there

exists l ∈ 0, 1, 2 such that v1 ∈ V(T

(k−1)l

)and v2 ∈ V

(T

(k−1)l+1

). It follows

that γ(v1) ∈ V(T

(k−1)l+1

), γ2(v1), γ(v2) ∈ V

(T

(k−1)l+2

)and γ2(v2) ∈ V

(T

(k−1)l

).

174

So, if we define T(k)l to be the tree with

V(T

(k)l

)= V

(T

(k−1)l

) ∪ v, z

E(T

(k)l

)= E

(T

(k−1)l

) ∪ v, v1, z, γ2(v2),

T(k)l+1 to be the tree with

V(T

(k)l+1

)= V

(T

(k−1)l+1

) ∪ v, w

E(T

(k)l+1

)= E

(T

(k−1)l+1

) ∪ v, v2, w, γ(v1),

and T(k)l+2 to be the tree with

V(T

(k)l+2

)= V

(T

(k−1)l+2

) ∪ w, z

E(T

(k)l+2

)= E

(T

(k−1)l+2

) ∪ w, γ(v2), z, γ2(v1),

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (C3, Φ) 3Tree2 partition of G.

.

.γ(v1)

.γ(v2)

..

.

. .

.

.γ2(v1).γ2(v2)

.v1

.v2

.

.γ(v1)

.γ(v2)

..

.

. .

.

.

. .

.γ2(v1).γ2(v2)

.v1

.v2

.z

.v .w

Figure 5.7: Construction of a (C3, Φ) 3Tree2 partition of G in the case where

G is a (C3, Φk−1) vertex addition of Gk−1.

Suppose next that G is a (C3, Φk−1) edge split on

(v1, v2, γ(v1), γ(v2), γ2(v1), γ2(v2)); (v, w, z) of Gk−1 with E(Gk) =

(E(Gk−1) \ v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2)) ∪ v, vi| i =

1, 2, 3 ∪ w, γ(vi)| i = 1, 2, 3

∪ z, γ2(vi)| i = 1, 2, 3, where w = γ(v)

175

and z = γ2(v). Wlog we may assume that v1, v2 ∈ E(T

(k−1)0

). Then

γ(v1), γ(v2) ∈ E(T

(k−1)1

)and γ2(v1), γ

2(v2) ∈ E(T

(k−1)2

). Note that

v3 belongs to a tree T(k−1)l , where l 6= 0. Suppose v3 ∈ T

(k−1)1 . Then

γ(v3) ∈ T(k−1)2 and γ2(v3) ∈ T

(k−1)0 . So, if we define T

(k)0 to be the tree with

.

..

.

. .

.

.

. .

.γ2(v1).γ2(v2)

.v1

.v2 .γ(v1)

.γ(v2).γ2(v3)

.v3 .γ(v3)

.

..

.

. .

.

.

. .

.

. .

.γ2(v1).γ2(v2)

.v1

.v2 .γ(v1)

.γ(v2).γ2(v3)

.v3 .γ(v3)

.z

.v .w


G is a (C3, Φk−1) edge split of Gk−1.

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, z

E(T

(k)0

)=

(E

(T

(k−1)0

) \ v1, v2) ∪ v, v1, v, v2, z, γ2(v3)

,

T(k)1 to be the tree with

V(T

(k)1

)= V

(T

(k−1)1

) ∪ w, v

E(T

(k)1

)=

(E

(T

(k−1)1

) \ γ(v1), γ(v2))

∪w, γ(v1), w, γ(v2), v, v3,

and T(k)2 to be the tree with

V(T

(k)2

)= V

(T

(k−1)2

) ∪ z, w

E(T

(k)2

)=

(E

(T

(k−1)2

) \ γ2(v1), γ2(v2)

)

∪z, γ2(v1), z, γ2(v2), w, γ(v3),

176

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (C3, Φ) 3Tree2 partition of G. If v3 ∈

T(k−1)2 , then we obtain a (C3, Φ) 3Tree2 partition of G in an analogous manner.

Finally, suppose that G is a (C3, Φk−1) ∆ extension by (v w z) of Gk−1 with

E(G) = E(Gk−1) ∪v, w, w, z, z, v, v, v0, w, γ(v0), z, γ2(v0)

,

where w = γ(v) and z = γ2(v). Wlog we may assume that v0 ∈ V(T

(k−1)0

).

Then γ(v0) ∈ V(T

(k−1)1

)and γ2(v0) ∈ V

(T

(k−1)2

). So, if we define T

(k)0 to be

...

.

.γ(v0).γ2(v0)

.v0

...

.

.

. .

.γ(v0).γ2(v0)

.v0

.z

.v .w


G is a (C3, Φk−1) ∆ extension of Gk−1.

the tree with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)= E

(T

(k−1)0

) ∪ v, v0, v, w,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ w, z

E(T

(k)1

)= E

(T

(k−1)1

) ∪ w, γ(v0), w, z,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v, z

E(T

(k)2

)= E

(T

(k−1)2

) ∪ z, γ2(v0), z, v,

177

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (C3, Φ) 3Tree2 partition of G. ¤



If G has a proper (C3, Φ) 3Tree2 partition, then R(G,C3,Φ) 6= ∅ and G is

(C3, Φ)-generically isostatic.

Proof. Suppose G has a proper (C3, Φ) 3Tree2 partition

E(T0), E(T1), E(T2). By Theorem 3.2.3, it suffices to find some framework

(G, p) ∈ R(G,C3,Φ) that is isostatic. Since G has a 3Tree2 partition, G satisfies

the count |E(G)| = 2|V (G)| − 3 (see Remark 2.2.5), and hence, by Theorem

2.2.5, it suffices to find a map p : V (G) → R2 such that (G, p) ∈ R(G,C3,Φ) is

independent. In the following, we again denote Φ(C3) by γ and Φ(C23) by

γ2.

Let e0 = (0, 0), e1 = (1, 0), and e2 = (12,√

32

). Also, for i = 0, 1, 2, let Vi

be the set of vertices of G that are not in V (Ti), and let (G, p, q) be the

frame with p : V (G) → R2 and q : E(G) → R2 defined by

p(v) = ei if v ∈ Vi

q(b) =

e2 − e1 = (−12,√

32

) if b ∈ E(T0)

e0 − e2 = (−12,−

√3

2) if b ∈ E(T1)

e1 − e0 = (1, 0) if b ∈ E(T2)

.

We claim that the generalized rigidity matrix R(G, p, q) has linearly in-

dependent rows. To see this, we first rearrange the columns of R(G, p, q) in

such a way that we obtain the matrix R′(G, p, q) which has the (2i − 1)st

column of R(G, p, q) in its ith column and the (2i)th column of R(G, p, q) in

178

.

.T1

.T2

.T0

.

. .

.V2

.V0 .V1

.e2

.e0 .e1

Figure 5.10: The frame (G, p, q).

its (|V (G)|+ i)th column for i = 1, 2, . . . , |V (G)|. Let Fb denote the row vec-

tor of R′(G, p, q) that corresponds to the edge b ∈ E(G). We then rearrange

the rows of R′(G, p, q) in such a way that we obtain the matrix R′′(G, p, q)

which has the vectors Fb with b ∈ E(T0) in the rows 1, 2, . . . , |E(T0)|, the

vectors Fb with b ∈ E(T1) in the following |E(T1)| rows, and the vectors Fb

with b ∈ E(T2) in the last |E(T2)| rows. So R′′(G, p, q) is a matrix of the

form

−12

12

√3

2−√

32

......

−12

12

√3

2−√

32

−12

12

−√

32

√3

2...

...

−12

12

−√

32

√3

2

1 −1... 0

1 −1

.

Clearly, R(G, p, q) has a row dependency if and only if R′′(G, p, q) does.

Suppose R′′(G, p, q) has a row dependency of the form

∑

b∈E(G)

αbFb = 0,

179

where αb 6= 0 for some b ∈ E(T2). Then, since T2 is a tree, we have

∑

b∈E(T2)

αbFb 6= 0.

Thus, there exists a vertex vs ∈ V (T2), s ∈ 1, 2, . . . , |V (G)|, such that

∑

b∈E(T2)

αb(Fb)s = C 6= 0.

Since vs ∈ V (T2), vs belongs to either T0 or T1, say wlog vs ∈ V (T1) and

vs /∈ V (T0). Therefore, (Fb)s = 0 and (Fb)|V (G)|+s = 0 for all b ∈ E(T0) and

∑

b∈E(T1)

αb(Fb)s = −C.

This says that

∑

b∈E(T1)

αb(Fb)|V (G)|+s =∑

b∈E(G)

αb(Fb)|V (G)|+s = −√

3

2C 6= 0,

a contradiction. Therefore, if∑

b∈E(G) αbFb = 0 is a row dependency of

R′′(G, p, q), then αb = 0 for all b ∈ E(T2).

So, it is now only left to show that the matrix R(G, p, q) which is obtained

from R′′(G, p, q) by deleting those rows of R′′(G, p, q) that correspond to the

edges of T2 has linearly independent rows. This can be done by multiplying

R(G, p, q) by appropriate matrices of basis transformations and then using

arguments analogous to those above. So, as claimed, the frame (G, p, q) is

independent.

Now, if (G, p) is not a framework, then we need to symmetrically pull

apart those joints of (G, p, q) that have the same location ei in R2 and whose

vertices are adjacent. So, wlog suppose |V0| ≥ 2. Then we also have |V0| =

|V1| = |V2| ≥ 2, because E(T0), E(T1), E(T2) is a (C3, Φ) 3Tree2 partition

180

of G. Since E(T0), E(T1), E(T2) is proper, one of 〈V0〉 ∩ Ti, i = 1, 2, say

wlog 〈V0〉∩T2, is not connected, and hence 〈V1〉∩T0 and 〈V2〉∩T1 are also not

connected. Let A be the set of vertices in one of the components of 〈V0〉 ∩T2

and γ(A) and γ2(A) be the vertex sets of the corresponding components of

〈V1〉 ∩ T0 and 〈V2〉 ∩ T1, respectively. For t ∈ R, we define pt : V (G) → R2

and qt : E(G) → R2 by

pt(v) =

(−12t,−

√3

2t) if v ∈ A

(1 + t, 0) if v ∈ γ(A)(

12(1− t),

√3

2(1 + t)

)if v ∈ γ2(A)

p(v) otherwise

qt(b) =

(1 + 12t,√

32

t) if b ∈ EA,V1\γ(A)

(1 + 32t,√

32

t) if b ∈ EA,γ(A)

(−12− t,

√3

2) if b ∈ Eγ(A),V2\γ2(A)

(− 12− 3

2t,√

32

(1 + t))

if b ∈ Eγ(A),γ2(A)

(− 12(1− t),−

√3

2(1 + t)

)if b ∈ Eγ2(A),V0\A

(−12,−

√3

2−√3t) if b ∈ Eγ2(A),A

q(b) otherwise

,

where for disjoint sets X,Y ∈ V (G), EX,Y denotes the set of edges of G

incident with a vertex in X and a vertex in Y . Then (G, pt, qt) = (G, p, q) if

t = 0. Therefore, by Lemma 5.1.3, there exists a t0 ∈ R, t0 6= 0, such that

the frame (G, pt0 , qt0) is independent. This process can be continued until we

obtain an independent frame (G, p, q) with p(u) 6= p(v) for all u, v ∈ E(G).

Then, by Remark 5.1.1, (G, p) is an independent framework and the right

translation of (G, p) yields an independent framework in the set R(G,C3,Φ). ¤

181

.

.

. .

.V2 \ γ2(A)

.V0 \ A

.V1 \ γ(A)

.e2

.e0

.e1

.A

.γ(A)

.γ2(A)

.T0

.T1

.T2

.

.

.

.

. .

Figure 5.11: The frame (G, pt, qt).

Lemmas 5.2.2, 5.2.3, 5.2.4, and 5.2.5 provide a complete proof for Theo-

rem 5.2.1.

Next, we show that there also exists a direct geometric proof for the fact

that condition (iii) implies condition (i) in Theorem 5.2.1, i.e., that the exis-

tence of a (C3, Φ) construction sequence for G implies that R(G,C3,Φ) 6= ∅ and

that G is (C3, Φ)-generically isostatic.

Note that the basic geometric arguments used in the proofs of Lemmas

5.2.6, 5.2.7, and 5.2.8 can easily be generalized for other symmetry groups

S as well. With these generalized geometric techniques we can construct

classes of (S, Φ)-generically isostatic graphs for additional symmetry groups

S. These techniques also allow us to prove (or at least conjecture) character-

izations of (S, Φ)-generically isostatic graphs in situations where symmetric

tree partitions are too complex. Moreover, they provide significant results

for (S, Φ)-generically independent graphs.


symmetry group in dimension 2, and Φ : C3 → Aut(G) be a homomorphism

so that R(G,C3,Φ) 6= ∅ and jΦ(C3) = 0. Let G be (C3, Φ)-generically isostatic,

182

G be a (C3, Φ) vertex addition (by (v, w, z)) of G and Φ : C3 → Aut(G) be

the homomorphism with Φ(x)|V (G) = Φ(x) for all x ∈ C3 and Φ(C3)|v,w,z =

(v w z). Then R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically isostatic.

Proof. As usual, we denote Φ(C3) by γ and Φ(C23) by γ2. Let E(G) =

E(G) ∪ v, v1, v, v2, w, γ(v1), w, γ(v2), z, γ2(v1), z, γ2(v2)

(see

also Figures 5.2 and 5.12). Since R(G,C3,Φ) 6= ∅, jγ = 0, and G is (C3, Φ)-

generically isostatic, there exists an isostatic framework (G, p) ∈ R(G,C3,Φ)

with p(v1) 6= p(v2). Since C3 is an isometry, it follows that p(γ(v1)) 6=p(γ(v2)) and p(γ2(v1)) 6= p(γ2(v2)).

Now, let p : V (G) → R2 be such that p|V (G) = p and p(v) is not collinear

with p(v1) and p(v2). Further, let p(w) = C3(p(v)) and p(z) = C23(p(v)).

Then (G, p) ∈ R(G,C3,Φ). Moreover, since p(v), p(v1), and p(v2) are not

collinear and C3 is an isometry of R2, it follows that p(w), p(γ(v1)), p(γ(v2))

are not collinear and p(z), p(γ2(v1)), p(γ2(v2)) are also not collinear. Thus,

we may repeatedly apply the proof of Proposition 3.1 in [68] to show that

the framework (G, p) is isostatic (see also Lemma 2.1.3 in [81], for example).

It now follows from Theorem 3.2.3 that G is (C3, Φ)-generically isostatic. ¤

...

.. .

.

.(G, p)

...

.. .

.

.

. .

.p(z)

.p(v) .p(w)

.(G, p)


183

Lemma 5.2.7 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23

be a symmetry group in dimension 2, and Φ : C3 → Aut(G)

be a homomorphism so that R(G,C3,Φ) 6= ∅ and jΦ(C3) = 0.

Let G be (C3, Φ)-generically isostatic, G be a (C3, Φ) edge-split (on

(v1, v2, Φ(C3)(v1), Φ(C3)(v2), Φ(C23)(v1), Φ(C2

3)(v2)); (v, w, z))) of G

and Φ : C3 → Aut(G) be the homomorphism with Φ(x)|V (G) = Φ(x) for

all x ∈ C3 and Φ(C3)|v,w,z = (v w z). Then R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-


Proof. We again denote Φ(C3) by γ and Φ(C23) by γ2. Let E(G) =

(E(G) \ v1, v2, γ(v1), γ(v2), γ2(v1), γ

2(v2)) ∪ v, vi| i = 1, 2, 3

∪w, γ(vi)| i = 1, 2, 3

∪ z, γ2(vi)| i = 1, 2, 3

(see also Figures 5.4 and

5.13). Since R(G,C3,Φ) 6= ∅, jγ = 0, and G is (C3, Φ)-generically isostatic, there

exists an isostatic framework (G, p) ∈ R(G,C3,Φ) so that p(v1), p(v2), p(v3) are

three non-collinear points. (In fact, if (v1 v2 v3) is a permutation cycle of γ,

then the points p(v1), p(v2), p(v3) are three distinct points of an equilateral

triangle in R2 for any (C3, Φ)-generic framework (G, p). Otherwise, at least

one of the vertices v1, v2, v3 lies in a permutation cycle of γ that does not

contain the other two vertices, so that the points p(v1), p(v2), and p(v3) must

still be non-collinear for any (C3, Φ)-generic framework (G, p).) Since C3

is an isometry, it follows that the points p(γ(v1)), p(γ(v2)), p(γ(v3)) are not

collinear and the points p(γ2(v1)), p(γ2(v2)), p(γ2(v3)) are also not collinear.

Now, let p : V (G) → R2 be such that p|V (G) = p and p(v) is the midpoint

of p(v1) and p(v2). Further, let p(w) = C3(p(v)) and p(z) = C23(p(v)). Then

p(w) is the midpoint of p(γ(v1)) and p(γ(v2)), and p(z) is the midpoint of

p(γ2(v1)) and p(γ2(v2)). Thus, we may repeatedly apply the proof of Propo-

184

.

..

.

. .

..

. .

.(G, p)

.

..

.

. .

..

. .

.

. .

.p(z)

.p(v) .p(w)

.(G, p)


sition 3.2 in [68] (or the proof of Theorem 2.2.2 in [81]) to show that the

framework (G, p) is isostatic. So, by Theorem 3.2.3, G is (C3, Φ)-generically

isostatic. ¤



so that R(G,C3,Φ) 6= ∅. Let G be (C3, Φ)-generically isostatic, G be a (C3, Φ)

∆ extension (by (v, w, z)) of G and Φ : C3 → Aut(G) be the homomorphism

with Φ(x)|V (G) = Φ(x) for all x ∈ C3 and Φ(C3)|v,w,z = (v w z). Then

R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically isostatic.

Proof. We again denote Φ(C3) by γ and Φ(C23) by γ2. Let E(G) =

E(G) ∪ v, w, w, z, z, v, v, v0, w, γ(v0), z, γ2(v0)

(see also Fig-

ures 5.4 and 5.14). Since R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically isostatic,

there exists an isostatic framework (G, p) ∈ R(G,C3,Φ) so that the points

p(v0), p(γ(v0)), p(γ2(v0)) are pairwise distinct. By the symmetry of (G, p),

the points p(v0), p(γ(v0)), p(γ2(v0)) are the points of an equilateral triangle

in R2 (see Figure 5.14 (a)).

Let a be the point in R2 that is obtained by reflecting p(v0) in the line that

goes through p(γ(v0)) and p(γ2(v0)). Then the points a, p(γ(v0)), p(γ2(v0))

185

are not collinear, so that the framework (G′, p′), where V (G′) = V (G)∪z,E(G′) = E(G) ∪ z, γ(v0), z, γ2(v0)

, and p′ : V (G′) → R2 is defined by

p′|V (G) = p and p′(z) = a, is isostatic by the proof of Proposition 3.1 in [68]

(see also Figure 5.14 (b)).

Now, let G′′ be an edge 2-split of G′ (on z, γ(v0); w) so that E(G′′) =

E(G′) \ z, γ(v0) ∪ w, z, w, γ(v0), w, v0

, and let p′′ : V (G′′) →

R2 be defined by p′′|V (G′) = p′ and p′′(w) = C23

(p′(z)

). Then

p′′(z), p′′(γ(v0)), p′′(w) are three distinct collinear points in R2, so that

(G′′, p′′) is isostatic by the proof of Theorem 2.2.2 in [81] (see also Figure

5.14 (c)).

Now, note that G is an edge 2-split of G′′ (on w, v0; v). Let p : V (G) → R2

be defined by p|V (G′′) = p′′ and p(v) = C3

(p′′(z)

). Then (G, p) ∈ R(G,C3,Φ),

and since p(v), p(v0), p(w) are three distinct collinear points in R2, it follows

again from the proof of Theorem 2.2.2 in [81] that (G, p) is isostatic (see also

Figure 5.14 (d)). By Theorem 3.2.3, G is (C3, Φ)-generically isostatic. ¤

. ..

.

.(G, p)

.(a)

. ..

.

..p′(z)

.(G′, p′)

.(b)

. ..

.

.

.

.p′′(z)

.p′′(w)

.(G′′, p′′)

.(c)

. ..

.. .

..p(z)

.p(v) .p(w)

.(G, p)

.(d)


Remark 5.2.2 Note that it follows from the results in [71], for ex-

ample, that if a framework (G, p) ∈ R(G,C3,Φ) is isostatic and G

is a (C3, Φ) ∆ extension (by (v, w, z)) of G with E(G) = E(G) ∪

186

v, w, w, z, z, v, v, v0, w, γ(v0), z, γ2(v0), then a framework

(G, p) with p|V (G) = p is isostatic if and only if the three points p(v), p(w),

and p(z) are chosen in such a way that the three lines p(v)p(v0), p(w)p(γ(v0)),

and p(v)p(γ2(v0)) do not meet in a point.

Theorem 5.2.9 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23 be a


If there exists a (C3, Φ) construction sequence for G, then R(G,C3,Φ) 6= ∅ and

G is (C3, Φ)-generically isostatic.

Proof. We proceed by induction on |V (G)|. If G = K3 and Φ : C3 →Aut(K3) is a non-trivial homomorphism, then we can clearly construct a

non-degenerate (and hence isostatic) equilateral triangle (K3, p) that lies in

the set R(K3,C3,Φ). This proves the base case.

Assume, then, that the result holds for all graphs with n or fewer than n




(K3, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ).

By the induction hypothesis, R(Gk−1,C3,Φk−1) 6= ∅ and Gk−1 is (C3, Φk−1)-

generically isostatic. Further, by the definition of the (C3, Φ) construction

sequence, we have jΦk−1(C3) = 0. So, if G is a (C3, Φk−1) vertex addition of

Gk−1, then the result follows from Lemma 5.2.6; if G is a (C3, Φk−1) edge split

of Gk−1, then the result follows from Lemma 5.2.7, and if G is a (C3, Φk−1)

∆ extension of Gk−1, then the result follows from Lemma 5.2.8. ¤

187

5.3 Characterizations of (C2, Φ)-generically

isostatic graphs

We will find that the symmetric versions of Laman’s Theorem for the

symmetry groups C2 and Cs cannot be proved as straightforwardly as for the

group C3. In the following, we demonstrate one of the additional difficulties

that arises in proving the Laman-type results for C2 and Cs by means of some

simple examples.

Let G be a graph that satisfies the conditions of Case 1 or Case 2 in

the proof of Lemma 5.2.3, i.e., G satisfies the Laman conditions, jγ = 0,

and G has a vertex v with NG(v) = v1, v2, v3 so that v, γ(v), γ2(v)is an independent subset of V (G). Then it follows from this proof that

if there exists a pair i, j ⊆ 1, 2, 3 such that every subgraph H of

G′ = G−v, γ(v), γ2(v) with vi, vj ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 4,

then G = G′ +vi, vj, γ(vi), γ(vj), γ2(vi), γ

2(vj)

satisfies the Laman

conditions.

For the symmetry groups C2 and Cs, the situation can possibly be more

complicated, as the following examples illustrate.

Let G be the underlying graph of the frameworks in Figure 5.15 (a) and

(b), C2 = Id, C2 and Cs = Id, s be symmetry groups in dimension 2,

and Φ : C2 → Aut(G) and Ψ : Cs → Aut(G) be the homomorphisms that

map C2 and s to the automorphism (v w)(v1 v6)(v2 v7)(v3 v8)(v4 v9)(v5 v10)

of G, respectively. Observe that G satisfies the Laman conditions and

that the conditions in Theorem 4.3.2 also hold for C2 and Φ, as well as

for Cs and Ψ. Moreover, every subgraph H of G′ = G − v, w with

188

.

..p1

..p2

..p3.

.p4

..p5

..p6

..p7

. .p8

..p9

..p10

..p(v)

.

.p(w)

.(a)

.

..p1

..p2

..p3

..p4

..p5

..p6

..p7

..p8

..p9

..p10

..p(v) . .p(w)

.(b)

Figure 5.15: A realization of (G, C2) of type Φ (a) and a realization of (G, Cs)

of type Ψ (b).

v1, v2 ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 4. However, G = G′ +v1, v2, v6, v7

does not satisfy the Laman conditions, since for the sub-

graph H = 〈v1, v2, v4, v5, v6, v7, v9, v10〉 of G, we have |E(H)| = 2|V (H)|−2.


titions for C2

We need the following inductive construction techniques to obtain a sym-

metrized Henneberg’s Theorem for C2.

.

.

..

..v1

.v2.γ(v1)

.γ(v2) .

.

..

.

. .

.v1

.v2 .γ(v1)

.γ(v2).v .w

Figure 5.16: A (C2, Φ) vertex addition of a graph G, where Φ(C2) = γ.

Definition 5.3.1 Let G be a graph, C2 = Id, C2 be the half-turn

symmetry group in dimension 2, and Φ : C2 → Aut(G) be a homo-

189

morphism. Let v1, v2 be two distinct vertices of G and v, w /∈ V (G).

Then the graph G with V (G) = V (G) ∪ v, w and E(G) = E(G) ∪v, v1, v, v2, w, Φ(C2)(v1), w, Φ(C2)(v2)

is called a (C2, Φ) vertex ad-

dition (by (v, w)) of G.

Definition 5.3.2 Let G be a graph, C2 = Id, C2 be the half-turn symme-

try group in dimension 2, and Φ : C2 → Aut(G) be a homomorphism. Let

v1, v2, v3 be three distinct vertices of G such that v1, v2 ∈ E(G) and v1, v2is not fixed by Φ(C2) and let v, w /∈ V (G). Then the graph G with V (G) =

V (G) ∪ v, w and E(G) =(E(G) \ v1, v2, Φ(C2)(v1), Φ(C2)(v2)

) ∪v, vi| i = 1, 2, 3

∪ w, Φ(C2)(vi)| i = 1, 2, 3

is called a (C2, Φ) edge

split (on (v1, v2, Φ(C2)(v1), Φ(C2)(v2)); (v, w)) of G.

.

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3)

.

.

.

.

.

. .

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3).v .w

.

.

.

.

.

.. .

Figure 5.17: A (C2, Φ) edge split of a graph G, where Φ(C2) = γ.

Remark 5.3.1 Each of the constructions in Definitions 5.3.1 and 5.3.2 has

the property that if the graph G satisfies the Laman conditions, then so does

G. This follows from Theorems 2.2.9 and 2.2.12 and the fact that we can

obtain a (C2, Φ) vertex addition of G by a sequence of two vertex 2-additions,

and a (C2, Φ) edge split of G by a sequence of two edge 2-splits.

In order to extend Crapo’s Theorem to C2 we need the following sym-

metrized definition of a 3Tree2 partition.

190

.

..v1

..v2

...γ(v1).γ(v2)

.

...v1

..v2

.. .γ(v1)

...v3 .γ(v2)

.γ(v3).

Figure 5.18: (C2, Φ) 3Tree2 partitions of graphs, where Φ(C2) = γ. The edges

in black color represent edges of the invariant trees.

Definition 5.3.3 Let G be a graph, C2 = Id, C2 be the half-turn sym-

metry group in dimension 2, and Φ : C2 → Aut(G) be a homomorphism. A

(C2, Φ) 3Tree2 partition of G is a 3Tree2 partition E(T0), E(T1), E(T2) of

G such that Φ(C2)(T1) = T2 and Φ(C2)(T0) = T0. The tree T0 is called the

invariant tree of E(T0), E(T1), E(T2).

5.3.2 The main result for C2

Theorem 5.3.1 Let G be a graph with |V (G)| ≥ 2, C2 = Id, C2 be the

half-turn symmetry group in dimension 2, and Φ : C2 → Aut(G) be a homo-

morphism. The following are equivalent:

(i) R(G,C2,Φ) 6= ∅ and G is (C2, Φ)-generically isostatic;


|V (H)| ≥ 2 (Laman conditions), jΦ(C2) = 0, and bΦ(C2) = 1;

(iii) there exists a (C2, Φ) construction sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

such that

191

(a) Gi+1 is a (C2, Φi) vertex addition or a (C2, Φi) edge split of Gi with

V (Gi+1) = V (Gi) ∪ vi+1, wi+1 for all i = 0, 1, . . . , k − 1;

(b) Φ0 : C2 → Aut(K2) is a non-trivial homomorphism and for all

i = 0, 1, . . . , k − 1, Φi+1 : C2 → Aut(Gi+1) is the homomorphism

defined by Φi+1(C2)|V (Gi) = Φi(C2) and Φi+1(C2)|vi+1,wi+1 =

(vi+1 wi+1);

(iv) G has a proper (C2, Φ) 3Tree2 partition whose invariant tree is a span-

ning tree of G.


Lemma 5.3.2 Let G be a graph with |V (G)| ≥ 2, C2 = Id, C2 be the

half-turn symmetry group in dimension 2, and Φ : C2 → Aut(G) be a ho-

momorphism. If R(G,C2,Φ) 6= ∅ and G is (C2, Φ)-generically isostatic, then G

satisfies the Laman conditions and we have jΦ(C2) = 0 and bΦ(C2) = 1.

Proof. The result is trivial if |V (G)| = 2, and it follows from Laman’s

Theorem (Theorem 2.2.9), Theorem 4.3.2, and Remark 2.2.3 if |V (G)| > 2.

¤



morphism. If G satisfies the Laman conditions and we also have jΦ(C2) = 0

and bΦ(C2) = 1, then there exists a (C2, Φ) construction sequence for G.

Proof. We employ induction on |V (G)|. Note first that if for a graph G,

there exists a homomorphism Φ : C2 → Aut(G) such that jΦ(C2) = 0, then

192

|V (G)| ≡ 0 (mod 2). The only graph with two vertices that satisfies the

Laman conditions is the graph K2 and if Φ : C2 → Aut(K2) is a homomor-

phism such that jΦ(C2) = 0 and bΦ(C2) = 1, then Φ is clearly a non-trivial

homomorphism. This proves the base case.



Let G be a graph with |V (G)| = n+2 that satisfies the Laman conditions

and suppose jΦ(C2) = 0 and bΦ(C2) = 1 for a homomorphism Φ : C2 → Aut(G).

In the following, we denote Φ(C2) by γ. By Lemma 5.1.1, G has a vertex of

valence 2 or 3.

We assume first that G has a vertex v of valence 2, say NG(v) = v1, v2.Then γ(v) 6= v since jγ = 0. Also, γ(v) 6= v1, v2, for otherwise, say wlog

γ(v) = v1, the graph G′ = G− v, γ(v) satisfies

|E(G′)| = |E(G)| − 3 = 2|V (G)| − 6 = 2|V (G′)| − 2,

contradicting the fact that G satisfies the Laman conditions, since |V (G′)| ≥2.

Thus, the edges v, v1, v, v2, γ(v), γ(v1), γ(v), γ(v2) are pairwise

distinct. Therefore,

|E(G′)| = |E(G)| − 4 = 2|V (G)| − 7 = 2|V (G′)| − 3.


|E(H)| ≤ 2|V (H)| − 3,


Let Φ′ : C2 → Aut(G′) be the homomorphism with Φ′(x) = Φ(x)|V (G′) for

193

all x ∈ C2. Then we have jΦ′(C2) = 0 and bΦ′(C2) = 1, because none of the

edges we removed was fixed by γ. Thus, by the induction hypothesis, there

exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G′, Φ′)

satisfying the conditions in Theorem 5.3.1 (iii). Since G is a (C2, Φ′) vertex

addition of G′ with V (G) = V (G′) ∪ v, γ(v),

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G′, Φ′), (G, Φ)


Suppose now that G has a vertex of valence 3 and no vertex of valence 2.

Then, by Lemma 5.1.1, G has at least six vertices of valence 3. Therefore,

since bγ = 1, there exists a vertex v ∈ V (G) with valG(v) = 3, say NG(v) =

v1, v2, v3, and v, γ(v) /∈ E(G). Since jγ = 0, we have γ(vi) 6= vi for all

i = 1, 2, 3, and hence we only need to consider the following two cases (see

also Figure 5.19):

Case 1: vs = γ(vt) for some s, t ⊆ 1, 2, 3. Wlog we assume v1 = γ(v2).

Then we also have v2 = γ(v1).

Case 2: The six vertices vi, γ(vi), i = 1, 2, 3, are all pairwise distinct.

Case 1: Since γ(v1, v2) = v1, v2, it follows from Lemma 5.1.2 (i)

and (ii) that there exists i, j ⊆ 1, 2, 3 with i, j 6= 1, 2, say wlog

i, j = 1, 3, such that for every subgraph H of G′ = G − v, γ(v) with

vi, vj ∈ V (H), we have |E(H)| ≤ 2|V (H)| − 4. Since G′ is invariant under

γ, every subgraph H of G′ with γ(v1), γ(v3) ∈ V (H) also satisfies |E(H)| ≤2|V (H)| − 4.

194

.

.v1 = γ(v2)

.v3

.v2 = γ(v1)

.γ(v3).v .γ(v). .

.

.

.

.

.(Case 1)

.

.v1

.v3

.v2

.γ(v1)

.γ(v3)

.γ(v2).v .γ(v)

.

.

.

.

.

.. .

.(Case 2)


has a vertex v of valence 3, then G is a graph of one of the types depicted

above.

Note that v1, v3 and γ(v1), γ(v3) are two distinct pairs of vertices

(though not edges, by the above counts), for otherwise we have γ(v1) = v3

(since jγ = 0), and hence v3 = v2, a contradiction.

We claim that G = G′ +v1, v3, γ(v1), γ(v3)

satisfies the Laman

conditions. We clearly have

|E(G)| = |E(G′)|+ 2 = |E(G)| − 4 = 2|V (G)| − 7 = 2|V (G)| − 3.

Suppose there exists a subgraph H of G′ with v1, v3, γ(v1), γ(v3) ∈ V (H)

and |E(H)| = 2|V (H)| − 4. Then the subgraph H of G′ with V (H) =

V (H)∪v, γ(v) and E(H) = E(H)∪v, vi| i = 1, 2, 3∪γ(v), γ(vi)| i =

1, 2, 3

satisfies

|E(H)| = |E(H)|+ 6 = 2|V (H)|+ 2 = 2|V (H)| − 2,

contradicting the fact that G satisfies the Laman conditions.

Therefore, every subgraph H of G′ with v1, v3, γ(v1), γ(v3) ∈ V (H) satis-

fies |E(H)| ≤ 2|V (H)| − 5.

Thus, as claimed, the graph G = G′ +v1, v3, γ(v1), γ(v3)

satisfies

195

the Laman conditions.

Further, if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ C2, then

Φ(x) ∈ Aut(G) for all x ∈ C2 and Φ : C2 → Aut(G) is a homomorphism.

Since we also have jΦ(C2) = 0 and bΦ(C2) = 1, it follows from the induction

hypothesis that there exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)


split of G with V (G) = V (G) ∪ v, γ(v),

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case 2: By Lemma 5.1.2 (i), there exists i, j ⊆ 1, 2, 3 such that

for every subgraph H of G′ = G − v, γ(v) with vi, vj ∈ V (H), we have

|E(H)| ≤ 2|V (H)| − 4. Suppose first that wlog i, j = 1, 2 is the only

pair in 1, 2, 3 with this property. Then, by Lemma 5.1.2 (ii), G = G′ +v1, v2, γ(v1), γ(v2)

satisfies the Laman conditions.

Further, if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ C2 then Φ(x) ∈Aut(G) for all x ∈ C2 and Φ : C2 → Aut(G) is a homomorphism. Since we

also have jΦ(C2) = 0 and bΦ(C2) = 1 it follows from the induction hypothesis

that there exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)

196


Suppose now that there exist two distinct pairs in 1, 2, 3, say wlog

1, 2 and 1, 3, such that every subgraph H of G′ with v1, v2 ∈ V (H) or

v1, v3 ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 4. Then every subgraph H of

G′ with γ(v1), γ(v2) ∈ V (H) or γ(v1), γ(v3) ∈ V (H) also satisfies |E(H)| ≤2|V (H)| − 4, because G′ is invariant under γ.

Suppose there exists a subgraph H of G′ with vi, γ(vi) ∈ V (H) for all i =

1, 2, 3 and |E(H)| = 2|V (H)| − 4. Then the subgraph H of G with V (H) =

V (H)∪v, γ(v) and E(H) = E(H)∪v, vi| i = 1, 2, 3∪γ(v), γ(vi)| i =

1, 2, 3

satisfies

|E(H)| = |E(H)|+ 6 = 2|V (H)|+ 2 = 2|V (H)| − 2,


Thus, every subgraph H of G′ with vi, γ(vi) ∈ V (H) for all i = 1, 2, 3

satisfies the count |E(H)| ≤ 2|V (H)| − 5.

Now, suppose there exist subgraphs H1 and H2 of G′ with

v1, v2, γ(v1), γ(v2) ∈ V (H1) and v1, v3, γ(v1), γ(v3) ∈ V (H2) satisfying

|E(Hi)| = 2|V (Hi)| − 4 for i = 1, 2. Then there also exist γ(H1) ⊆ G′ and

γ(H2) ⊆ G′ with v1, v2, γ(v1), γ(v2) ∈ V(γ(H1)

)and v1, v3, γ(v1), γ(v3) ∈

V(γ(H2)

)satisfying |E(

γ(Hi))| = 2|V (

γ(Hi))| − 4 for i = 1, 2. Let

H ′i = Hi ∪ γ(Hi) for i = 1, 2. Then

|E(H ′1)| = |E(H1)|+ |E(

γ(H1))| − |E(

H1 ∩ γ(H1))|

≥ 2|V (H1)| − 4 + 2|V (γ(H1)

)| − 4− (2|V (H1 ∩ γ(H1)

)| − 4)

= 2|V (H ′1)| − 4,

197

because H1 ∩ γ(H1) is a subgraph of G′ with v1, v2 ∈ V(H1 ∩ γ(H1)

). Since

H ′1 is also a subgraph of G′ with v1, v2 ∈ V (H ′

1), it follows that

|E(H ′1)| = 2|V (H ′

1)| − 4.

Similarly,

|E(H ′2)| = 2|V (H ′

2)| − 4.

So, both H ′1 and H ′

2 have an even number of edges. Moreover, both of these

graphs are invariant under γ, which says that neither E(H ′1) nor E(H ′

2)

contains the edge e of G that is fixed by γ.

Note that H ′1 ∩ H ′

2 is a subgraph of G with v1, γ(v1) ∈ V (H ′1 ∩ H ′

2).

Therefore, we have

|E(H ′1 ∩H ′

2)| ≤ 2|V (H ′1 ∩H ′

2)| − 3,

because G satisfies the Laman conditions. Since H ′1 ∩ H ′

2 is also invariant

under γ and E(H ′1∩H ′

2) does not contain the edge e, |E(H ′1∩H ′

2)| is an even

number. The above upper bound for |E(H ′1 ∩H ′

2)| can therefore be lowered

to

|E(H ′1 ∩H ′

2)| ≤ 2|V (H ′1 ∩H ′

2)| − 4.

Thus, H ′ = H ′1 ∪H ′

2 satisfies

|E(H ′)| = |E(H ′1)|+ |E(H ′

2)| − |E(H ′1 ∩H ′

2)|

≥ 2|V (H ′1)| − 4 + 2|V (H ′

2))| − 4− (2|V (H ′

1 ∩H ′2)| − 4)

= 2|V (H ′)| − 4.

This is a contradiction, because H ′ is a subgraph of G′ with vi, γ(vi) ∈ V (H ′)

for all i = 1, 2, 3.

198

So, for i, j = 1, 2 or i, j = 1, 3, say wlog i, j = 1, 2, we

have that every subgraph H of G′ with vi, vj, γ(vi), γ(vj) ∈ V (H) satisfies

|E(H)| = 2|V (H)| − 5.

Thus, G = G′ +v1, v2, γ(v1), γ(v2)

satisfies the Laman conditions

and if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ C2, then Φ(x) ∈ Aut(G)

for all x ∈ C2 and Φ : C2 → Aut(G) is a homomorphism. Since we also have

jΦ(C2) = 0 and bΦ(C2) = 1, it follows from the induction hypothesis that there

exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)




morphism. If there exists a (C2, Φ) construction sequence for G, then G has

a proper (C2, Φ) 3Tree2 partition whose invariant tree is a spanning tree of

G.

Proof. We proceed by induction on |V (G)|. Let V (K2) = v1, v2 and let

Φ : C2 → K2 be the homomorphism defined by Φ(C2) = (v1 v2). Then K2

has the proper (C2, Φ) 3Tree2 partition E(T0), E(T1), E(T2), where T0 =

〈v1, v2〉, T1 = 〈v1〉, and T2 = 〈v2〉. Clearly, T0 is a spanning tree of

K2. This proves the base case.

199





(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



G must be proper. Therefore, it suffices to show that G has some (C2, Φ)

3Tree2 partition whose invariant tree is a spanning tree of G. In the follow-

ing, we denote Φ(C2) by γ.

By the induction hypothesis, Gk−1 has a (C2, Φk−1) 3Tree2 partitionE

(T

(k−1)0

), E

(T

(k−1)1

), E

(T

(k−1)2

)whose invariant tree T

(k−1)0 is a spanning

tree of Gk−1.

Suppose first that G is a (C2, Φk−1) vertex addition by (v w) of Gk−1

with NG(v) = v1, v2. Since Φk−1(C2) = γ|V (Gk−1), we have NG(w) =

γ(v1), γ(v2). Note that v1, v2, γ(v1), γ(v2) ∈ V(T

(k−1)0

), because T

(k−1)0 is a

spanning tree of Gk−1. Also, v2 belongs to either T(k−1)1 or T

(k−1)2 , say wlog

v2 ∈ V(T

(k−1)1

). Then γ(v2) ∈ V

(T

(k−1)2

). So, if we define T

(k)0 to be the tree

with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)= E

(T

(k−1)0

) ∪ v, v1, w, γ(v1),


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v2,

200


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ w, γ(v2),

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (C2, Φ) 3Tree2 partition of G whose

invariant tree T(k)0 is a spanning tree of G.

.

.

..

..v1

.v2.γ(v1)

.γ(v2) .

.

..

.

. .

.v1

.v2 .γ(v1)

.γ(v2).v .w


G is a (C2, Φk−1) vertex addition of Gk−1. The edges in black color represent

edges of the invariant tree T(k)0 .

Suppose next that G is a (C2, Φk−1) edge split on

(v1, v2, γ(v1), γ(v2)); (v, w) of Gk−1 with E(Gk) =(E(Gk−1) \

v1, v2, γ(v1), γ(v2)) ∪ v, vi| i = 1, 2, 3

∪ w, γ(vi)| i = 1, 2, 3.

First, we assume that v1, v2 ∈ E(T

(k−1)0

), and hence γ(v1), γ(v2) ∈

E(T

(k−1)0

). Note that v3 belongs to either T

(k−1)1 or T

(k−1)2 , say wlog

v3 ∈ V(T

(k−1)1

). Then γ(v3) ∈ V

(T

(k−1)2

). So if we define T

(k)0 to be the tree

with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)=

(E

(T

(k−1)0

) \ v1, v2, γ(v1), γ(v2))

∪v, v1, v, v2, w, γ(v1), w, γ(v2),

201


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v3,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ w, γ(v3),

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


invariant tree T(k)0 is a spanning tree of G.

.

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3)

.

.

.

.

.

. .

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3).v .w

.

.

.

.

.

.. .

.

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3)

.

.

.

.

.

. .

.v1

.v2

.v3

.γ(v1)

.γ(v2)

.γ(v3).v .w

.

.

.

.

.

.. .


G is a (C2, Φk−1) edge split of Gk−1. The edges in black color represent edges

of the invariant trees.

Assume now that v1, v2 /∈ E(T

(k−1)0

). Then wlog v1, v2 ∈ E

(T

(k−1)1

)

and γ(v1), γ(v2) ∈ E(T

(k−1)2

). In this case we define T

(k)0 to be the tree

202

with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)= E

(T

(k−1)0

) ∪ v, v3, w, γ(v3),


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)=

(E

(T

(k−1)1

) \ v1, v2) ∪ v, v1, v, v2

,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)=

(E

(T

(k−1)2

) \ γ(v1), γ(v2))

∪w, γ(v1), w, γ(v2).

ThenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


invariant tree T(k)0 is a spanning tree of G. ¤



morphism. If G has a proper (C2, Φ) 3Tree2 partition whose invariant tree is

a spanning tree of G, then R(G,C2,Φ) 6= ∅ and G is (C2, Φ)-generically isostatic.

Proof. Suppose G has a proper (C2, Φ) 3Tree2 partition

E(T0), E(T1), E(T2) whose invariant tree T0 is a spanning tree of G.

By Theorem 3.2.3, it suffices to find some framework (G, p) ∈ R(G,C2,Φ)

that is isostatic. Since G has a 3Tree2 partition, G satisfies the count

|E(G)| = 2|V (G)|− 3, and hence, by Theorem 2.2.5, it suffices to find a map

203

p : V (G) → R2 such that (G, p) ∈ R(G,C2,Φ) is independent. In the following,

we again denote Φ(C2) by γ.

Let Vi be the set of vertices of G that are not in V (Ti) for i = 0, 1, 2.

Then V0 = ∅ since T0 is a spanning tree of G. Let e1 = (0, 0) and e2 = (0, 1)

and let (G, p, q) be the frame with p : V (G) → R2 and q : E(G) → R2

defined by


q(b) =

(0, 1) if b ∈ E(T0)

(−1, 0) if b ∈ E(T1)

(1, 0) if b ∈ E(T2)

.

.

.T1

.T2.V1

.V2

.e1

.e2

.T0

.

.

Figure 5.22: The frame (G, p, q).



such a way that we obtain the matrix R′(G, p, q) which has the (2i − 1)st

column of R(G, p, q) in its ith column and the (2i)th column of R(G, p, q) in

its (|V (G)|+ i)th column for i = 1, 2, . . . , |V (G)|. Let Fb denote the row vec-

tor of R′(G, p, q) that corresponds to the edge b ∈ E(G). We then rearrange

the rows of R′(G, p, q) in such a way that we obtain the matrix R′′(G, p, q)

which has the vectors Fb with b ∈ E(T0) in the rows 1, 2, . . . , |E(T0)|, the

204

vectors Fb with b ∈ E(T1) in the following |E(T1)| rows, and the vectors Fb

with b ∈ E(T2) in the last |E(T2)| rows. So R′′(G, p, q) is of the form

1 −1

0...

1 −1

−1 1... 0

−1 1

1 −1... 0

1 −1

.



∑

b∈E(G)

αbFb = 0,

where αb 6= 0 for some b ∈ E(T0). Since T0 is a tree, it follows that

∑

b∈E(T0)

αbFb 6= 0.

Thus, there exists a vertex vr ∈ V (T0), r ∈ 1, 2, . . . , |V (G)|, such that

∑

b∈E(T0)

αb(Fb)|V (G)|+r = C 6= 0,

and hence∑

b∈E(G)

αb(Fb)|V (G)|+r = C 6= 0,

205

a contradiction.

So, suppose∑

b∈E(T1)∪E(T2)

αbFb = 0,

where αb 6= 0 for some b ∈ E(T1) ∪ E(T2), say wlog b ∈ E(T1). Since T1 is a

tree, we have∑

b∈E(T1)

αbFb 6= 0,

and hence there exists a vertex vs ∈ V (T1), s ∈ 1, 2, . . . , |V (G)|, such that

∑

b∈E(T1)

αb(Fb)s = D 6= 0.

Then∑

b∈E(T1)∪E(T2)

αb(Fb)s = D 6= 0,

because the trees T1 and T2 have disjoint vertex sets. This is again a contra-

diction, and hence the frame (G, p, q) is indeed independent.

Now, if (G, p) is not a framework, then we need to symmetrically pull

apart those joints of (G, p, q) that have the same location ei in R2 and whose

vertices are adjacent. So, wlog suppose |V1| ≥ 2. Then, since G has the

(C2, Φ) 3Tree2 partition E(T0), E(T1), E(T2), we have γ(V1) = V2, and

hence |V1| = |V2| ≥ 2. Since E(T0), E(T1), E(T2) is proper, one of 〈V1〉∩Ti,

i = 0, 2, is not connected. Note that T2 ⊆ 〈V1〉, and hence 〈V1〉 ∩ T2 is con-

nected. Thus, 〈V1〉 ∩ T0 is not connected. Therefore, 〈V2〉 ∩ T0 is also not

connected. Let A be the set of vertices in one of the components of 〈V1〉 ∩T0

and γ(A) be the set of vertices in the corresponding component of 〈V2〉 ∩ T0.

206

For t ∈ R, we define pt : V (G) → R2 and qt : E(G) → R2 by

pt(v) =

(t, 0) if v ∈ A

(−t, 1) if v ∈ γ(A)

p(v) otherwise

qt(b) =

(−t, 1) if b ∈ EA,V2\γ(A)

(−2t, 1) if b ∈ EA,γ(A)

(−t, 1) if b ∈ Eγ(A),V1\A

q(b) otherwise

,


incident with a vertex in X and a vertex in Y . Then (G, pt, qt) = (G, p, q)

.

.T1

.T2.V1 \ A

.V2 \ γ(A)

.A

.γ(A)

.e1

.e2

.T0

.

.

.

.


if t = 0. Therefore, by Lemma 5.1.3, there exists a t0 ∈ R, t0 6= 0, such that

the frame (G, pt0 , qt0) is independent.

If (G, pt0) is still not a framework, then V1 \ A or A, say wlog V1 \ A,

contains at least two vertices that are adjacent in G, as does V2 \γ(A). Since

E(T0), E(T1), E(T2) is proper, one of 〈V1\A〉∩Ti, i = 0, 2 is not connected.

If 〈V1 \ A〉 ∩ T0 is not connected, then 〈V2 \ γ(A)〉 ∩ T0 is also not con-

nected. Let B and γ(B) be the vertex sets of components of 〈V1 \A〉∩T0 and

V2 \ γ(A)〉 ∩ T0, respectively. Then we can pull apart the vertices of B from

207

(V1 \A) \B and the vertices of γ(B) from (V2 \ γ(A)) \ γ(B) in an analogous

way as before in order to obtain a new independent frame.

If 〈V1 \A〉 ∩T2 and 〈V2 \ γ(A)〉 ∩T1 are not connected, then we let B and

γ(B) be the vertex sets of components of 〈V1\A〉∩T2 and 〈V2\γ(A)〉∩T1, re-

spectively. In this case, we may pull apart the vertices of B from (V1\A)\B in

direction of the vector (0,−1) and the vertices of γ(B) from (V2\γ(A))\γ(B)

in direction of the vector (0, 1) to obtain a new independent frame.

This process can be continued until we obtain an independent frame

(G, p, q) with p(u) 6= p(v) for all u, v ∈ E(G). Then, by Remark 5.1.1,

(G, p) is an independent framework and the right translation of (G, p) yields

an independent framework in the set R(G,C2,Φ). ¤


rem 5.3.1

Remark 5.3.2 Note that the geometric proofs of Lemmas 5.2.6 and 5.2.7

can easily be adapted to also give a direct geometric proof that condition (iii)

implies condition (i) in Theorem 5.3.1, i.e., that the existence of a (C2, Φ)

construction sequence for G implies that R(G,C2,Φ) 6= ∅ and that G is (C2, Φ)-


Remark 5.3.3 Let G be a graph with |V (G)| ≥ 3, C2 = Id, C2 be the

half-turn symmetry group in dimension 2, and Φ : C2 → Aut(G) be a ho-

momorphism. If G is (C2, Φ)-generically isostatic, then we can modify the

construction in the proof of Lemma 5.3.4 to obtain proper (C2, Φ) 3Tree2

partitions of G whose invariant trees are not spanning. In particular, it can

be shown that if G is (C2, Φ)-generically isostatic, then there must exist a

208

proper (C2, Φ) 3Tree2 partition of G whose invariant tree is just a single edge

of G. However, the existence of a proper (C2, Φ) 3Tree2 partition of G whose

invariant tree is not spanning is not sufficient for G to be (C2, Φ)-generically

isostatic. This is because a vertex of G that does not belong to the invariant

tree of such a (C2, Φ) 3Tree2 partition can possibly be fixed by Φ(C2), and

hence jΦ(C2) may not be zero.

For example, consider the complete graph K3 with V (K3) = v1, v2, v3and let Φ be the homomorphism from the symmetry group C2 to Aut(K3)

defined by Φ(C2) = (v1 v2)(v3). Then K3 has the proper (C2, Φ) 3Tree2 par-

tition E(T0), E(T1), E(T2), where T0 = 〈v1, v2〉, T1 = 〈v2, v3〉, and

T2 = 〈v1, v3〉. Since v3 is fixed by Φ(C2), K3 is not (C2, Φ)-generically iso-

static. In fact, every realization in the set R(K3,C2,Φ) is a degenerate triangle.

If however G has a proper (C2, Φ) 3Tree2 partition (whose invariant tree

is not necessarily a spanning tree of G) and we also impose the condition that

jΦ(C2) = 0, then it is quite easy to show that we must also have bΦ(C2) = 1.

In other words, the two conditions that G has any proper (C2, Φ) 3Tree2 par-

tition and jΦ(C2) = 0 are sufficient for G to be (C2, Φ)-generically isostatic.

209

5.4 Characterizations of (Cs, Φ)–generically

isostatic graphs


titions for Cs

We need the following symmetrized inductive construction techniques to

obtain a symmetrized Henneberg’s Theorem for Cs.

Definition 5.4.1 Let G be a graph, Cs = Id, s be a symmetry group in

dimension 2, and Φ : Cs → Aut(G) be a homomorphism. Let v0 be a vertex

of G that is not fixed by Φ(s) and v /∈ V (G). Then the graph G with

.. ..v0 .σ(v0)

.. .

.

.v0 .σ(v0)

.v

Figure 5.24: A (Cs, Φ) single vertex addition of a graph G, where Φ(s) = σ.

V (G) = V (G) ∪ v and E(G) = E(G) ∪ v, v0, v, Φ(s)(v0)

is called a

(Cs, Φ) single vertex addition (by (v)) of G.

Definition 5.4.2 Let G be a graph, Cs = Id, s be a symmetry group

in dimension 2, and Φ : Cs → Aut(G) be a homomorphism. Let v1, v2, v3

be three distinct vertices of G such that v1, v2 ∈ E(G), σ(v1) = v2, and

σ(v3) = v3. Further, let v /∈ V (G). Then the graph G with V (G) = V (G) ∪

210

v and E(G) =(E(G)\v1, v2

)∪v, vi| i = 1, 2, 3

is called a (Cs, Φ)

single edge split (on v1, v2; v) of G.

.

.v1 .v2 = σ(v1)

.v3 = σ(v3)

. .

. .

.v1 .v2 = σ(v1)

.v3 = σ(v3)

. .

.

..v

Figure 5.25: A (Cs, Φ) single edge split of a graph G, where Φ(s) = σ.


dimension 2, and Φ : Cs → Aut(G) be a homomorphism. Let v1, v2 be two dis-

tinct vertices of G and v, w /∈ V (G). Then the graph G with V (G) = V (G)∪v, w and E(G) = E(G) ∪ v, v1, v, v2, w, Φ(s)(v1), w, Φ(s)(v2)

is

called a (Cs, Φ) double vertex addition (by (v, w)) of G.

..v2 .σ(v2)

.v1 .σ(v1). .

. .

..v2 .σ(v2)

.v1 .σ(v1). .

. .

. ..v .w

Figure 5.26: A (Cs, Φ) double vertex addition of a graph G, where Φ(s) = σ.


dimension 2, and Φ : Cs → Aut(G) be a homomorphism. Let v1, v2, v3 be

three distinct vertices of G such that v1, v2 ∈ E(G) and v1, v2 is not fixed

by Φ(s). Further, let v, w /∈ V (G). Then the graph G with V (G) = V (G) ∪

211

v, w and E(G) =(E(G) \ v1, v2, Φ(s)(v1), Φ(s)(v2)

) ∪ v, vi| i =

1, 2, 3 ∪ w, Φ(s)(vi)| i = 1, 2, 3

is called a (Cs, Φ) double edge split (on

(v1, v2, Φ(s)(v1), Φ(s)(v2)); (v, w)) of G.

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

. ..v .w

Figure 5.27: A (Cs, Φ) double edge split of a graph G, where Φ(s) = σ.


dimension 2, and Φ : Cs → Aut(G) be a homomorphism. Let v1, v2, v3, v4 be

four distinct vertices of G with v1, v2, v3, v4 ∈ E(G) and Φ(s)(v1, v2) =

v3, v4. Further, let v /∈ V (G). Then the graph G with V (G) = V (G)∪vand E(G) =

(E(G) \ v1, v2, v3, v4

)∪ v, vi| i ∈ 1, 2, 3, 4

is called

a (Cs, Φ) X-replacement (by (v)) of G.

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

..v

Figure 5.28: A (Cs, Φ) X-replacement of a graph G, where Φ(s) = σ.

Remark 5.4.1 Each of the constructions in Definitions 5.4.1, 5.4.2, 5.4.3,

5.4.4, and 5.4.5 has the property that if the graph G satisfies the Laman

212

conditions, then so does G. This follows from Theorems 2.2.9, 2.2.12, 2.2.13,

and the fact that we can obtain a (Cs, Φ) double vertex addition of G by a

sequence of two vertex 2-additions and a (Cs, Φ) double edge split of G by a

sequence of two edge 2-splits.

In order to extend Crapo’s Theorem to Cs we need the following sym-

metrized definitions of a 3Tree2 partition.


dimension 2, and Φ : Cs → Aut(G) be a homomorphism. A (Cs, Φ) 3Tree2

⊥ partition of G is a 3Tree2 partition E(T0), E(T1), E(T2) of G such that

Φ(s)(T1) = T2 and Φ(s)(T0) = T0. The tree T0 is called the invariant tree of

E(T0), E(T1), E(T2).

Remark 5.4.2 Let E(T0), E(T1), E(T2) be a (Cs, Φ) 3Tree2 ⊥ parti-

tion of a graph G. Then the vertex set of the invariant tree T0 of

E(T0), E(T1), E(T2) does not contain a vertex v ∈ V (G) with Φ(s)(v) = v,

for otherwise v ∈ V (T1) implies v ∈ V (T2) and vice versa, contradicting the

fact that v only belongs to exactly two of the trees Ti. Therefore, it is easy

to see that E(T0) must contain an edge e = v, w of G with Φ(s)(v) = w.

Let G be a graph and Φ : Cs → Aut(G) be a homomorphism such that

E(G) contains an edge e = v, w with Φ(s)(v) = v and Φ(s)(w) = w. Then

it follows immediately from the previous remark that G cannot have a (Cs, Φ)

3Tree2 ⊥ partition. However, G may have a symmetric 3Tree2 partition of

the following kind:

213


dimension 2, and Φ : Cs → Aut(G) be a homomorphism such that there

exists an edge e = v, w ∈ E(G) with Φ(s)(v) = v and Φ(s)(w) = w. A

(Cs, Φ) 3Tree2 ‖ partition of G is a 3Tree2 partition E(T0), E(T1), E(T2)of G such that e ∈ E(T1), Φ(s)(T1 − v) = T2 and Φ(s)(T0) = T0. The tree

T0 is called the invariant tree of E(T0), E(T1), E(T2).

.. .

.

.

.v1 .σ(v1)

.v3 = σ(v3)

.v2 = σ(v2)

.T1 .T2.T0

.(a)

.

.. .

. ..

.v2 = σ(v2)

.v4 .σ(v4)

.v3 .σ(v3)

.v1 = σ(v1)

.T1 .T2

.T0

.(b)

Figure 5.29: A (Cs, Φ) 3Tree2 ⊥ partition of a graph (a) and a (Cs, Φ) 3Tree2

‖ partition of a graph (b), where Φ(s) = σ.

Remark 5.4.3 Let E(T0), E(T1), E(T2) be a (Cs, Φ) 3Tree2 ‖ partition of

a graph G. Since T2 is a tree, so is Φ(s)(T2) = T1−v, and hence valT1(v) =

1. Also, v ∈ V (T0), for otherwise we have v ∈ V (Ti) for i = 1, 2, which

contradicts the facts that Φ(s)(T1−v) = T2 and that v /∈ V(Φ(s)(T1−v)

)

since v is fixed by Φ(s). Moreover, there does not exist a vertex x ∈ V (G)

with x 6= v, x ∈ V (T0), and Φ(s)(x) = x, for otherwise x ∈ V (T1) implies

x ∈ V (T2) and vice versa, contradicting the fact that x only belongs to

exactly two of the trees Ti.

214

Remark 5.4.4 Let G be a graph and Φ : Cs → Aut(G) be a homomorphism

such that E(G) contains an edge e = v, w with Φ(s)(v) = w. Then G

cannot have a (Cs, Φ) 3Tree2 ‖ partition E(T0), E(T1), E(T2), for otherwise

e ∈ E(T0) and, by Remark 5.4.3, there also exists a vertex in V (T0) that is

fixed by Φ(s), which implies that there must exist a cycle in T0.

5.4.2 The main result for Cs

Theorem 5.4.1 Let G be a graph with |V (G)| ≥ 2, Cs = Id, s be a sym-

metry group in dimension 2, and Φ : Cs → Aut(G) be a homomorphism. The


(i) R(G,Cs,Φ) 6= ∅ and G is (Cs, Φ)-generically isostatic;


|V (H)| ≥ 2 (Laman conditions), and bΦ(s) = 1;

(iii) there exists a (Cs, Φ) construction sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

such that

(a) Gi+1 is a (Cs, Φi) single or double vertex addition, a (Cs, Φi) sin-

gle or double edge split, or a (Cs, Φi) X-replacement of Gi with

V (Gi+1) = V (Gi) ∪ vi+1 or V (Gi+1) = V (Gi) ∪ vi+1, wi+1 for

all i = 0, 1, . . . , k − 1;

(b) Φ0 : Cs → Aut(K2) is a homomorphism and for all i =

0, 1, . . . , k − 1, Φi+1 : Cs → Aut(Gi+1) is the homomorphism de-

fined by Φi+1(s)|V (Gi) = Φi(s) and Φi+1(s)(vi+1) = vi+1 whenever

215

V (Gi+1) = V (Gi) ∪ vi+1 and Φi+1(s)|vi+1,wi+1 = (vi+1 wi+1)

whenever V (Gi+1) = V (Gi) ∪ vi+1, wi+1;

(iv) G has a proper (Cs, Φ) 3Tree2 ⊥ partition or a proper (Cs, Φ) 3Tree2 ‖partition.


Lemma 5.4.2 Let G be a graph with |V (G)| ≥ 2, Cs = Id, s be a sym-

metry group in dimension 2, and Φ : Cs → Aut(G) be a homomorphism.

If R(G,Cs,Φ) 6= ∅ and G is (Cs, Φ)-generically isostatic, then G satisfies the

Laman conditions and we have bΦ(s) = 1.

Proof. The result is trivial if |V (G)| = 2, and it follows from Laman’s

Theorem (Theorem 2.2.9), Theorem 4.3.2, and Remark 2.2.3 if |V (G)| > 2.

¤

Lemma 5.4.3 Let G be a graph with |V (G)| ≥ 2, Cs = Id, s be a symmetry

group in dimension 2, and Φ : Cs → Aut(G) be a homomorphism. If G

satisfies the Laman conditions and we also have bΦ(s) = 1, then there exists

a (Cs, Φ) construction sequence for G.

Proof. We employ induction on |V (G)|. The only graph with two vertices

that satisfies the Laman conditions is the graph K2, and hence the result

trivially holds for |V (G)| = 2. This proves the base case.



Let G be a graph with |V (G)| = n+1 that satisfies the Laman conditions

216

and suppose bΦ(s) = 1 for a homomorphism Φ : Cs → Aut(G). In the

following, we denote Φ(s) by σ. By Lemma 5.1.1, G has a vertex of valence

2 or 3.

Case A: G has a vertex v of valence 2, say NG(v) = v1, v2.

Case A.1: Suppose v is fixed by σ. Then σ(v1) = v2, because bσ = 1.

So, G′ = G − v clearly satisfies the Laman conditions and if we define

Φ′ : Cs → Aut(G′) to be the homomorphism with Φ′(x) = Φ(x)|V (G′) for all

x ∈ Cs, then we have bΦ′(s) = 1, and hence, by the induction hypothesis,


(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G′, Φ′)

satisfying the conditions in Theorem 5.4.1 (iii). Since G is a (Cs, Φ′) single

vertex addition of G′ with V (G) = V (G′) ∪ v,

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G′, Φ′), (G, Φ)


Case A.2: Suppose v 6= σ(v). Then σ(v) 6= v1, v2, for otherwise the

graph G′ = G− v, σ(v) satisfies

|E(G′)| = |E(G)| − 3 = 2|V (G)| − 6 = 2|V (G′)| − 2,


Thus, the edges v, v1, v, v2, σ(v), σ(v1), σ(v), σ(v2) are pairwise

distinct. Therefore,

|E(G′)| = |E(G)| − 4 = 2|V (G)| − 7 = 2|V (G′)| − 3

217

and for H ⊆ G′ with |V (H)| ≥ 2, we have H ⊆ G, and hence

|E(H)| ≤ 2|V (H)| − 3,


Let Φ′ : Cs → Aut(G′) be the homomorphism with Φ′(x) = Φ(x)|V (G′) for

all x ∈ Cs. Then we have bΦ′(s) = 1, and hence, by the induction hypothesis,


(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G′, Φ′)

satisfying the conditions in Theorem 5.4.1 (iii). Since G is a (Cs, Φ′) double

vertex addition of G′ with V (G) = V (G′) ∪ v, σ(v),

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G′, Φ′), (G, Φ)


Case B: G has a vertex v of valence 3, say NG(v) = v1, v2, v3, and no

vertex of valence 2.

Case B.1: Suppose σ(v) = v. Then wlog σ(v1) = v1 and σ(v2) = v3,

because bσ = 1. So, the edge v, v1 of G is fixed by σ and bσ = 1 implies

that v2, v3 /∈ E(G).

We claim that the graph G = G − v +v2, v3

satisfies the Laman

conditions. Clearly, we have

|E(G)| = 2|V (G)| − 3.

Let G′ = G−v and suppose there exists a subgraph H of G′ with v2, v3 ∈V (H) and |E(H)| = 2|V (H)| − 3. Since G′ is invariant under σ, σ(H)

is also a subgraph of G′ and we have v2, v3 ∈ V(σ(H)

)and |E(

σ(H))| =

218

2|V (σ(H)

)| − 3. Note that H ∩ σ(H) is a subgraph of G with v2, v3 ∈V (H ∩ σ(H)), and hence |E(

H ∩ σ(H)) ≤ 2|V (

H ∩ σ(H))| − 3. Since

H ∩ σ(H) is invariant under σ and E(H ∩ σ(H)

)does not contain the edge

v, v1, |E(H ∩ σ(H)

)| is an even number. Thus, we have |E(H ∩ σ(H)

)| ≤2|V (

H ∩ σ(H))| − 4. It follows that the graph H ′ = H ∪ σ(H) satisfies

|E(H ′)| = |E(H)|+ |E(σ(H)

)| − |E(H ∩ σ(H)

)|

≥ 2|V (H)| − 3 + 2|V (σ(H)

)| − 3− (2|V (H ∩ σ(H)

)| − 4)

= 2|V (H ′)| − 2,

contradicting the fact that G satisfies the Laman conditions. So, as claimed,

the graph G satisfies the Laman conditions.

If we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs, then Φ(x) ∈ Aut(G)

for all x ∈ Cs and Φ : Cs → Aut(G) is a homomorphism. Since v2, v3 is the

only edge that is fixed by Φ, we also have bΦ(s) = 1. So, by the induction

hypothesis, there exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

satisfying the conditions in Theorem 5.4.1 (iii). Since G is a (Cs, Φ) single

edge split of G with V (G) = V (G) ∪ v,

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case B.2: Suppose v is not fixed by σ. By Lemma 5.1.1, G has at

least six vertices of valence 3. So, since bσ = 1, we may assume wlog that

v, σ(v) /∈ E(G).

Let G′ = G − v, σ(v) and suppose there exists a subgraph H of G′

219

with vi, σ(vi) ∈ V (H) for all i = 1, 2, 3 and |E(H)| ≥ 2|V (H)| − 4. Then

the subgraph H of G with V (H) = V (H) ∪ v, σ(v) and E(H) = E(H) ∪v, vi| i = 1, 2, 3

∪ σ(v), σ(vi)| i = 1, 2, 3

satisfies

|E(H)| = |E(H)|+ 6 ≥ 2|V (H)|+ 2 = 2|V (H)| − 2,


Thus, every subgraph H of G′ with vi, σ(vi) ∈ V (H) for all i = 1, 2, 3

satisfies the count |E(H)| ≤ 2|V (H)|−5. In the following, we will frequently

make use of this fact.

Case B.2.1: Suppose that for every pair i, j ⊆ 1, 2, 3, we have

σ(vi, vj) 6= vi, vj. Then we need to consider the following two subcases

(see also Figure 5.30):

Case B.2.1a: The vertices vi, σ(vi), i = 1, 2, 3, are all pairwise distinct.

Case B.2.1b: One of the vi, say wlog vi = v1, is fixed by σ and the

vertices v2, v3, σ(v2), σ(v3) are pairwise distinct.

Case B.2.1a: By Lemma 5.1.2 (i), there exists i, j ⊆ 1, 2, 3 such

that for every subgraph H of G′ = G−v, σ(v) with vi, vj ∈ V (H), we have

|E(H)| ≤ 2|V (H)| − 4.

Suppose first that wlog i, j = 1, 2 is the only pair in 1, 2, 3 with

this property. Then, by Lemma 5.1.2 (ii), G = G′+v1, v2, σ(v1), σ(v2)

satisfies the Laman conditions and if we define Φ by Φ(x) = Φ(x)|V (G) for

all x ∈ Cs, then Φ(x) ∈ Aut(G) for all x ∈ Cs and Φ : Cs → Aut(G) is a

homomorphism. Since we also have bΦ(s) = 1, it follows from the induction

hypothesis that there exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

220

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

. ..v .σ(v)

.(Case B.2.1a)

.

.v1 = σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

.

. .

. .

. ..v .σ(v)

.(Case B.2.1b)


has a vertex v with NG(v) = v1, v2, v3 such that σ(vi, vj) 6= vi, vj for

all i, j ⊆ 1, 2, 3, then G is a graph of one of the types depicted above.

satisfying the conditions in Theorem 5.4.1 (iii). Since G is a (Cs, Φ) double

edge split of G with V (G) = V (G) ∪ v, σ(v),

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


So, suppose there exist two distinct pairs in 1, 2, 3, say wlog 1, 2and 1, 3, such that every subgraph H of G′ with v1, v2 ∈ V (H) or

v1, v3 ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 4. Then every subgraph

H of G′ with σ(v1), σ(v2) ∈ V (H) or σ(v1), σ(v3) ∈ V (H) also satisfies

|E(H)| ≤ 2|V (H)| − 4, because G′ is invariant under σ.

Suppose there exist subgraphs H1 and H2 of G′ with v1, v2, σ(v1), σ(v2) ∈V (H1) and v1, v3, σ(v1), σ(v3) ∈ V (H2) satisfying |E(Hi)| = 2|V (Hi)| − 4 for

i = 1, 2. Then there also exist subgraphs σ(H1) ⊆ G′ and σ(H2) ⊆ G′ with

v1, v2, σ(v1), σ(v2) ∈ V(σ(H1)

)and v1, v3, σ(v1), σ(v3) ∈ V

(σ(H2)

)satisfying

|E(σ(Hi)

)| = 2|V (σ(Hi)

)| − 4 for i = 1, 2. Let H ′i = Hi ∪ σ(Hi) for i = 1, 2.

221

Then

|E(H ′1)| = |E(H1)|+ |E(

σ(H1))| − |E(

H1 ∩ σ(H1))|

≥ 2|V (H1)| − 4 + 2|V (σ(H1)

)| − 4− (2|V (H1 ∩ σ(H1)

)| − 4)

= 2|V (H ′1)| − 4,

because H1 ∩ σ(H1) is a subgraph of G′ with v1, v2 ∈ V(H1 ∩ σ(H1)

). Since


1) it follows that

|E(H ′1)| = 2|V (H ′

1)| − 4.

Similarly, we have

|E(H ′2)| = 2|V (H ′

2)| − 4.

So, both H ′1 and H ′

2 have an even number of edges. Moreover, both of these

graphs are invariant under σ, which says that neither E(H ′1) nor E(H ′

2)

contains the edge e of G that is fixed by σ. Note that H ′1 ∩H ′

2 is a subgraph

of G with v1, σ(v1) ∈ V (H ′1 ∩H ′

2), and hence satisfies the count

|E(H ′1 ∩H ′

2)| ≤ 2|V (H ′1 ∩H ′

2)| − 3,

because G satisfies the Laman conditions. Since H ′1 ∩ H ′

2 is also invariant

under σ and E(H ′1∩H ′

2) does not contain the edge e, |E(H ′1∩H ′

2)| is an even

number, and hence the above upper bound for |E(H ′1 ∩H ′

2)| can be lowered

to

|E(H ′1 ∩H ′

2)| ≤ 2|V (H ′1 ∩H ′

2)| − 4.

Thus, H ′ = H ′1 ∪H ′

2 satisfies

|E(H ′)| = |E(H ′1)|+ |E(H ′

2)| − |E(H ′1 ∩H ′

2)|

≥ 2|V (H ′1)| − 4 + 2|V (H ′

2)| − 4− (2|V (H ′1 ∩H ′

2)| − 4)

= 2|V (H ′)| − 4.

222

This is a contradiction, because H ′ is a subgraph of G′ with vi, σ(vi) ∈ V (H ′)

for all i = 1, 2, 3.


have that every subgraph H of G′ with vi, vj, σ(vi), σ(vj) ∈ V (H) satisfies

|E(H)| ≤ 2|V (H)| − 5.

Thus, G = G′ +v1, v2, σ(v1), σ(v2)


and if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs, then Φ(x) ∈ Aut(G)

for all x ∈ Cs and Φ : Cs → Aut(G) is a homomorphism. Since we also

have bΦ(s) = 1, it follows from the induction hypothesis that there exists a

sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case B.2.1b: By Lemma 5.1.2 (i), there exists i, j ⊆ 1, 2, 3 such

that for every subgraph H of G′ = G−v, σ(v) with vi, vj ∈ V (H) we have

|E(H)| ≤ 2|V (H)|−4. If wlog i, j = 1, 2 is the only pair in 1, 2, 3 with

this property, then, by Lemma 5.1.2 (ii), G = G′ +v1, v2, σ(v1), σ(v2)

satisfies the Laman conditions and we obtain a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)

with the desired properties in the same way as in Case B.2.1a.

So, suppose there exist two distinct pairs i, j and q, r in 1, 2, 3 such

223

that every subgraph H of G′ with vi, vj ∈ V (H) or vq, vr ∈ V (H) satisfies

|E(H)| ≤ 2|V (H)| − 4. If one of the pairs i, j or q, r is the pair 2, 3,then the proof of Case B.2.1a also applies.

So, suppose that every subgraph H of G′ with v1, v2 ∈ V (H) or v1, v3 ∈V (H) satisfies |E(H)| ≤ 2|V (H)| − 4 and that there exists a subgraph A

of G′ with v2, v3 ∈ V (A) and |E(A)| = 2|V (A)| − 3. Then every subgraph

H of G′ with v1, σ(v2) ∈ V (H) or v1, σ(v3) ∈ V (H) also satisfies |E(H)| ≤2|V (H)| − 4, because σ(v1) = v1 and G′ is invariant under σ.

Suppose there exist subgraphs H1 and H2 of G′ with v1, v2, σ(v2) ∈ V (H1)

and v1, v3, σ(v3) ∈ V (H2) satisfying |E(Hi)| = 2|V (Hi)|−4 for i = 1, 2. Then

there also exist σ(H1) ⊆ G′ and σ(H2) ⊆ G′ with v1, v2, σ(v2) ∈ V(σ(H1)

)

and v1, v3, σ(v3) ∈ V(σ(H2)

)satisfying |E(

σ(Hi))| = 2|V (

σ(Hi))| − 4 for

i = 1, 2. Let H ′i = Hi ∪ σ(Hi) for i = 1, 2. Then

|E(H ′1)| = |E(H1)|+ |E(

σ(H1))| − |E(

H1 ∩ σ(H1))|

≥ 2|V (H1)| − 4 + 2|V (σ(H1)

)| − 4− (2|V (H1 ∩ σ(H1)

)| − 4)

= 2|V (H ′1)| − 4,

because H1 ∩ σ(H1) is a subgraph of G′ with v1, v2 ∈ V(H1 ∩ σ(H1)

). Since


1) it follows that

|E(H ′1)| = 2|V (H ′

1)| − 4.

Similarly, we have

|E(H ′2)| = 2|V (H ′

2)| − 4.

Let H ′ = H ′1 ∪ H ′

2. If V (H ′1 ∩ H ′

2) contains at least two vertices, then we

can derive a contradiction in the same way as in Case B.2.1a. So, suppose

224

V (H ′1 ∩H ′

2) = v1. Then

|E(H ′)| = |E(H ′1)|+ |E(H ′

2)| − |E(H ′1 ∩H ′

2)|

= 2|V (H ′1)| − 4 + 2|V (H ′

2)| − 4− (2|V (H ′1 ∩H ′

2)| − 2)

= 2|V (H ′)| − 6.

Since G′ is invariant under σ, σ(A) is a subgraph of G′ with σ(v2), σ(v3) ∈V

(σ(A)

)and |E(

σ(A))| = 2|V (

σ(A))| − 3.

Suppose |V (A∩ σ(A)

)| ≥ 2. Then |E(A∩ σ(A)

)| ≤ 2|V (A∩ σ(A)

)| − 3,

because G satisfies the Laman conditions, and hence A′ = A ∪ σ(A) is a

subgraph of G′ with v2, v3, σ(v2), σ(v3) ∈ V (A′) satisfying

|E(A′)| = |E(A)|+ |E(σ(A)

)| − |E(A ∩ σ(A)

)|

≥ 2|V (A)| − 3 + 2|V (σ(A)

)| − 3− (2|V (A ∩ σ(A)

)| − 3)

= 2|V (A′)| − 3.

It follows that |E(A′)| = 2|V (A′)| − 3, because A′ ⊆ G. Since H1 ∩ A′ is a

subgraph of G with at least two vertices, namely v2, σ(v2) ∈ V (H1 ∩A′), we

have

|E(H1 ∪ A′)| = |E(H1)|+ |E(A′)| − |E(H1 ∩ A′)|

≥ 2|V (H1)| − 4 + 2|V (A′)| − 3− (2|V (H1 ∩ A′)| − 3)

= 2|V (H1 ∪ A′)| − 4.

This is a contradiction, because H1 ∪ A′ is a subgraph of G′ with vi, σ(vi) ∈V (H1 ∪ A′) for all i = 1, 2, 3

So, suppose every subgraph A of G′ with v2, v3 ∈ V (A) and |E(A)| =

2|V (A)| − 3 satisfies |V (A ∩ σ(A)

)| ≤ 1. Let Amin be a subgraph of G′ that

225

satisfies v2, v3 ∈ V (Amin) and |E(Amin)| = 2|V (Amin)|−3 and has the small-

est number of edges among all such subgraphs of G′. Note that v1 /∈ V (Amin),

for otherwise v1, v2 ∈ V (Amin), and hence |E(Amin)| ≤ 2|V (Amin)|−4. Also,

Amin is connected as we see as follows.

Suppose to the contrary that Amin = A1∪A2, where V (A1)∩V (A2) = ∅.Clearly, one of A1 or A2 has at least two vertices. If wlog |V (A1)| = 1 and

|V (A2)| ≥ 2, then

|E(Amin)| = |E(A1)|+|E(A2)| ≤ 2|V (A1)|−2+2|V (A2)|−3 = 2|V (Amin)|−5

and if |V (A1)|, |V (A2)| ≥ 2, then

|E(Amin)| = |E(A1)|+|E(A2)| ≤ 2|V (A1)|−3+2|V (A2)|−3 = 2|V (Amin)|−6.

In both cases, we have a contradiction to the fact that |E(Amin)| =

2|V (Amin)| − 3. So, Amin is indeed connected.

Now, consider H ′ ∪ Amin. We have

|E(H ′ ∪ Amin)| = |E(H ′)|+ |E(Amin)| − |E(H ′ ∩ Amin)|.

Note that v2, v3 ∈ V (H ′ ∩ Amin) so that

|E(H ′ ∩ Amin)| ≤ 2|V (H ′ ∩ Amin)| − 3.

We claim that |E(H ′ ∩ Amin)| < |E(Amin)|.Since Amin is connected, there exists a v2 − v3 path P in Amin and P

does not contain an edge incident with the vertex v1, because v1 /∈ V (Amin).

Let E(P ) denote the set of edges of P . Since V (H ′1 ∩ H ′

2) = v1, we have

v2 ∈ V (H ′1), v2 /∈ V (H ′

2), v3 ∈ V (H ′2), v3 /∈ V (H ′

1), and every v2 − v3 path in

H ′ = H ′1 ∪H ′

2 must contain an edge incident with v1. Thus, E(P ) * E(H ′).

226

So, as claimed, |E(H ′ ∩ Amin)| < |E(Amin)|.By the minimality of |E(Amin)|, we can conclude that

|E(H ′ ∩ Amin)| ≤ 2|V (H ′ ∩ Amin)| − 4.

Thus,

|E(H ′ ∪ Amin)| = |E(H ′)|+ |E(Amin)| − |E(H ′ ∩ Amin)|

≥ 2|V (H ′)| − 6 + 2|V (Amin)| − 3− (2|V (H ′ ∩ Amin)| − 4)

= 2|V (H ′ ∪ Amin)| − 5.

Now, consider (H ′ ∪ Amin) ∪ σ(Amin). We have

|E((H ′ ∪ Amin) ∪ σ(Amin)

)| = |E(H ′ ∪ Amin)|+ |E(σ(Amin)

)|

−|E((H ′ ∪ Amin) ∩ σ(Amin)

)|.

Note that σ(v2), σ(v3) ∈ V((H ′ ∪ Amin) ∩ σ(Amin)

)so that

|E((H ′ ∪ Amin) ∩ σ(Amin)

)| ≤ 2|V ((H ′ ∪ Amin) ∩ σ(Amin)

)| − 3.

By the definition of Amin, σ(Amin) is a subgraph of G′ with σ(v2), σ(v3) ∈V

(σ(Amin)

)and |E(

σ(Amin))| = 2|V (

σ(Amin))| − 3 and σ(Amin) has the

smallest number of edges among all such subgraphs of G′. We claim that


)| < |E(σ(Amin)

)|.Since Amin is a connected subgraph of G′, so is σ(Amin). Therefore, there

exists a σ(v2)−σ(v3) path P in σ(Amin) and E(P ) does not contain an edge

incident with the vertex v1, because v1 /∈ V(σ(Amin)

). Also, by assumption,

|V (Amin ∩ σ(Amin)

)| ≤ 1, which says that E(P ) does not contain an edge

of Amin. Since V (H ′1 ∩H ′

2) = v1, we have σ(v2) ∈ V (H ′1), σ(v2) /∈ V (H ′

2),

σ(v3) ∈ V (H ′2), σ(v3) /∈ V (H ′

1), and every σ(v2)− σ(v3) path in H ′ ∪Amin =

227

(H ′1 ∪H ′

2) ∪Amin must contain an edge incident with v1 or an edge of Amin.

Thus, E(P ) * E(H ′ ∪Amin). So, as claimed, |E((H ′ ∪Amin) ∩ σ(Amin)

)| <|E(

σ(Amin))|.

By the minimality of |E(σ(Amin)

)|, we can conclude that


)| ≤ 2|V ((H ′ ∪ Amin) ∩ σ(Amin)

)| − 4.

Thus,

|E((H ′ ∪ Amin) ∪ σ(Amin)

)| = |E(H ′ ∪ Amin)|+ |E(σ(Amin)

)|

−|E((H ′ ∪ Amin) ∩ σ(Amin)

)|

≥ 2|V (H ′ ∪ Amin)| − 5 + 2|V (σ(Amin)

)| − 3

−(2|V ((H ′ ∪ Amin) ∩ σ(Amin)

)| − 4)

= 2|V ((H ′ ∪ Amin) ∪ σ(Amin)

)| − 4.

This is a contradiction, since (H ′ ∪Amin)∪ σ(Amin) is a subgraph of G′ with

vi, σ(vi) ∈ V((H ′ ∪ Amin) ∪ σ(Amin)

)for all i = 1, 2, 3.


have that every subgraph H of G′ with vi, vj, σ(vi), σ(vj) ∈ V (H) satisfies

|E(H)| ≤ 2|V (H)| − 5. Thus, G = G′ +v1, v2, σ(v1), σ(v2)

satisfies

the Laman conditions and if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs,

then Φ(x) ∈ Aut(G) for all x ∈ Cs and Φ : Cs → Aut(G) is a homomorphism.

Since we also have bΦ(s) = 1, it follows from the induction hypothesis that


(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

228



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case B.2.2: Suppose there exists exactly one pair i, j in 1, 2, 3such that σ(vi, vj) = vi, vj. Then we need to consider the following two

subcases (see also Figure 5.31):

Case B.2.2a: Exactly two of the vertices vi, i = 1, 2, 3, say wlog v1 and

v2, are fixed by σ.

Case B.2.2b: There exists a pair i, j ⊆ 1, 2, 3 such that σ(vi) = vj.

Wlog we assume σ(v1) = v2.

Case B.2.2a: Since σ(v1, v2) = v1, v2, it follows from Lemma 5.1.2

(i) and (ii) that there exists i, j ⊆ 1, 2, 3 with i, j 6= 1, 2 such that

for every subgraph H of G′ = G − v, σ(v) with vi, vj ∈ V (H), we have

|E(H)| ≤ 2|V (H)| − 4.

If for every subgraph H of G′ with v1, v3 ∈ V (H) or v2, v3 ∈ V (H), we

have |E(H)| ≤ 2|V (H)| − 4, then the proof of Case B.2.1a applies.

So, suppose wlog that every subgraph H of G′ with v1, v3 ∈ V (H) satisfies

|E(H)| ≤ 2|V (H)|− 4 and that there exists a subgraph A of G′ with v2, v3 ∈V (A) and |E(A)| = 2|V (A)|−3. Since G′ is invariant under σ and vi = σ(vi)

for i = 1, 2, every subgraph H of G′ with v1, σ(v3) ∈ V (H) also satisfies

|E(H)| ≤ 2|V (H)|−4 and σ(A) is a subgraph of G′ with v2, σ(v3) ∈ V(σ(A)

)

and |E(σ(A)

)| = 2|V (σ(A)

)| − 3.

229

.

.v1 = σ(v1)

.v2 = σ(v2)

.v3 .σ(v3)

.

.

. .

. ..v .σ(v)

.(Case B.2.2a)

.

.v1 = σ(v2) .v2 = σ(v1)

.v3 .σ(v3)

. .

. .

. ..v .σ(v) .

.v1 = σ(v2) .v2 = σ(v1)

.v3 = σ(v3)

. .

.

. ..v .σ(v)

..(Case B.2.2b)


has a vertex v with NG(v) = v1, v2, v3 such that σ(vi, vj) = vi, vj for

exactly one pair i, j ⊆ 1, 2, 3, then G is a graph of one of the types

depicted above.

We claim that the graph G = G′ +v1, v3, v1, σ(v3)

satisfies the

Laman conditions. Clearly,

|E(G)| = 2|V (G)| − 3.

Suppose there exists a subgraph H of G′ with v1, v3, σ(v3) ∈ V (H) that

satisfies |E(H)| = 2|V (H)| − 4. Then σ(H) is also a subgraph of G′ with

v1, v3, σ(v3) ∈ V(σ(H)

)that satisfies |E(

σ(H))| = 2|V (

σ(H))| − 4. Let

230

H ′ = H ∪ σ(H). Then

|E(H ′)| = |E(H)|+ |E(σ(H)

)| − |E(H ∩ σ(H)

)|

≥ 2|V (H)| − 4 + 2|V (σ(H)

)| − 4− (2|V (H ∩ σ(H)

)| − 4)

= 2|V (H ′)| − 4,

because H ∩ σ(H) is a subgraph of G′ with v1, v3 ∈ V(H ∩ σ(H)

). Since H ′

is also a subgraph of G′ with v1, v3 ∈ V (H ′) it follows that

|E(H ′)| = 2|V (H ′)| − 4.

So, since H ′ has an even number of edges and is invariant under σ, E(H ′)

does not contain the edge e that is fixed by σ.

Let A′ = A ∪ σ(A). Then H ′ ∩ A′ is a subgraph of G with v3, σ(v3) ∈V (H ′ ∩ A′), and hence satisfies the count

|E(H ′ ∩ A′)| ≤ 2|V (H ′ ∩ A′)| − 3.

Since H ′ ∩A′ is invariant under σ and E(H ′ ∩A′) does not contain the edge

e, |E(H ′ ∩ A′)| is an even number. Therefore, the above upper bound for

|E(H ′ ∩ A′)| can be lowered to

|E(H ′ ∩ A′)| ≤ 2|V (H ′ ∩ A′)| − 4.

Note that if V(A∩ σ(A)

)= v2, then |E(

A∩ σ(A))| = 2|V (

A∩ σ(A))| − 2

and if |V (A)∩σ(A)| ≥ 2, then |E(A∩σ(A)

)| ≤ 2|V (A∩σ(A)

)|−3, because

A ∩ σ(A) is a subgraph of G. Therefore,

|E(A ∩ σ(A)

)| ≤ 2|V (A ∩ σ(A)

)| − 2,

231

and hence

|E(A′)| = |E(A)|+ |E(σ(A)

)| − |E(A ∩ σ(A)

)|

≥ 2|V (A)| − 3 + 2|V (σ(A)

)| − 3− (2|V (A ∩ σ(A)

)| − 2)

= 2|V (A′)| − 4,

Thus,

|E(H ′ ∪ A′)| = |E(H ′)|+ |E(A′)| − |E(H ′ ∩ A′)|

≥ 2|V (H ′)| − 4 + 2|V (A′)| − 4− (2|V (H ′ ∩ A′)| − 4)

= 2|V (H ′ ∪ A′)| − 4.

This is a contradiction, since H ′ ∪ A′ is a subgraph of G′ with vi, σ(vi) ∈V (H ′ ∪ A′) for all i = 1, 2, 3.

Thus, G = G′ +v1, v3, σ(v1), σ(v3)


and if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs, then Φ(x) ∈ Aut(G)

for all x ∈ Cs and Φ : Cs → Aut(G) is a homomorphism. Since we also

have bΦ(s) = 1, it follows from the induction hypothesis that there exists a

sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case B.2.2b: By Lemma 5.1.2, for i, j = 1, 3 or i, j = 2, 3, say

wlog i, j = 1, 3, we have that every subgraph H of G′ = G − v, σ(v)

232

with vi, vj ∈ V (H) satisfies |E(H)| ≤ 2|V (H)| − 4. Since G′ is invariant

under σ, every subgraph H of G′ with σ(v1), σ(v3) ∈ V (H) also satisfies

|E(H)| = 2|V (H)| − 4. Moreover, if there exists a subgraph H of G′ with

v1, v3, σ(v1), σ(v3) ∈ V (H), then vi, σ(vi) ∈ V (H) for all i = 1, 2, 3, and hence

|E(H)| ≤ 2|V (H)| − 5.

Therefore, G = G′ +v1, v3, σ(v1), σ(v3)

satisfies the Laman con-

ditions and if we define Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs, then

Φ(x) ∈ Aut(G) for all x ∈ Cs and Φ : Cs → Aut(G) is a homomorphism.

Since we also have bΦ(s) = 1, it follows from the induction hypothesis that


(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)


Case B.2.3: Finally, suppose G has no vertex of valence two, no vertex

of valence three that is fixed by σ, and every 3-valent vertex v of G (except

possibly the two vertices that are incident with the edge e ∈ E(G) that is

fixed by σ) has the property that σ(u) = u for all u ∈ NG(v).

Let T denote the set of 3-valent vertices of G. Then |T | = 2k, k ∈ N,

and, by Lemma 5.1.1, k ≥ 3. Also, let e = y, z be the edge of G that is

fixed by σ.

We claim that G has a vertex v with NG(v) = v1, v2, v3, σ(vi) = vi

for all i = 1, 2, 3, and valG(vi) = 4 for some i ∈ 1, 2, 3. Suppose to the

233

contrary that there does not exist such a vertex v in V (G).

We suppose first that y, z /∈ T . Then for every v ∈ T , we have σ(u) = u

for all u ∈ NG(v). Let N =⋃

v∈T NG(v) and suppose |N | = m ≤ k + 1. The

subgraph 〈T ∪ N〉 of G induced by T ∪ N satisfies |V (〈T ∪ N〉)| = 2k + m

and |E(〈T ∪N〉)| ≥ 6k. Thus,

2|V (〈T ∪N〉)| − 3 = 2(2k + m)− 3 ≤ 6k − 1 < |E(〈T ∪N〉)|,

which is a contradiction to the fact that G satisfies the Laman conditions.

Therefore, |N | ≥ k + 2. By assumption, every vertex in N has valence at

least 5 in G. Since at most two vertices in N can possibly be incident with

e, it follows that at least k vertices in N must have valence at least 6 in G.

Therefore, the average valence in G is at least

2k · 3 + 2 · 5 + k · 6 +(|V (G)| − (2k + k + 2)

) · 4|V (G)| = 4 +

2

|V (G)| ,

which contradicts the fact that the average valence in G is 4 − 6|V (G)| (see

Lemma 5.1.1).

Suppose now that y or z is a vertex in T . Then σ(y) = z and both y

and z are in T . Let T ′ = T \ y, z and let N ′ =⋃

v∈T ′ NG(v). Suppose

|N ′| = m ≤ k. The subgraph 〈T ′ ∪ N ′〉 of G induced by T ′ ∪ N ′ satisfies

|V (〈T ′∪N ′〉)| = 2k−2+m and |E(〈T ′∪N ′〉)| = (2k−2) ·3 = 6k−6. Thus,

2|V (〈T ′ ∪N ′〉)| − 3 = 2(2k − 2 + m)− 3 ≤ 6k − 7 < |E(〈T ′ ∪N ′〉)|,

which is a contradiction to the fact that G satisfies the Laman conditions.

Therefore, |N ′| ≥ k + 1. By assumption, every vertex in N ′ has valence at

least 5 in G, and since y, z /∈ N ′, every vertex in N ′ must even have valence

at least 6 in G. Therefore, the average valence in G is at least

2k · 3 + (k + 1) · 6 +(|V (G)| − (2k + k + 1)

) · 4|V (G)| = 4 +

2

|V (G)| ,

234

which again contradicts the fact that the average valence in G is 4 − 6|V (G)|

(see Lemma 5.1.1).

So, as claimed, there exists a vertex v ∈ V (G) with NG(v) = v1, v2, v3,σ(vi) = vi for all i = 1, 2, 3, and valG(vi) = 4 for some i ∈ 1, 2, 3, say wlog

valG(v1) = 4 with NG(v1) = v, σ(v), w, σ(w).

.

.v3 = σ(v3)

.v1 = σ(v1)

.v2 = σ(v2)

.σ(w) .w

.v .σ(v)

.

.

.. .

. .

Figure 5.32: If a graph G satisfies the conditions in Theorem 5.4.1 (ii), has no

vertex of valence two, no vertex of valence three that is fixed by σ, and every

3-valent vertex v of G (except possibly the vertices that are incident with the

edge that is fixed by σ) has the property that σ(u) = u for all u ∈ NG(v), then

there exists v ∈ V (G) with NG(v) = v1, v2, v3, σ(vi) = vi for all i = 1, 2, 3,

and valG(vi) = 4 for some i ∈ 1, 2, 3.

Let G′ = G − v1. We claim that G = G′ +v, w, σ(v), σ(w)

satisfies the Laman conditions. We have

|E(G)| = |E(G)| − 2 = 2|V (G)| − 5 = 2|V (G)| − 3.

Suppose there exists a subgraph H of G′ with v, w, σ(v), σ(w) ∈ V (G′) that

satisfies |E(H)| ≥ 2|V (H)| − 4. Then the subgraph H of G with V (H) =

V (H) ∪ v1 and E(H) = E(H) ∪ v1, v, v1, σ(v), v1, w, v1, σ(w)

satisfies

|E(H)| = |E(H)|+ 4 ≥ 2|V (H)| = 2|V (H)| − 2,

235


Thus, every subgraph H of G′ with v, w, σ(v), σ(w) ∈ V (H) satisfies the

count |E(H)| ≤ 2|V (H)| − 5.

Suppose there exists a subgraph H of G′ with v, w ∈ V (H) that satisfies

|E(H)| = 2|V (H)| − 3. Then |V (H)| ≥ 3 since v, w /∈ E(H). Since

G′ is invariant under σ, σ(H) is also a subgraph of G′ and σ(H) satisfies

σ(v), σ(w) ∈ V(σ(H)

)and |E(

σ(H))| = 2|E(

σ(H))|−3. Let H ′ = H∪σ(H).

Then

|E(H ′)| = |E(H)|+ |E(σ(H)

)| − |E(H ∩ σ(H)

)|.

Suppose first that E(H ∩ σ(H)

)= ∅. Then v2, v3 /∈ V (H). Thus, v is an

isolated vertex in H, and hence

|E(H − v)| = |E(H)| = 2|V (H)| − 3 = 2|V (H − v)| − 1.

This contradicts the fact that G satisfies the Laman conditions, because

|V (H − v)| ≥ 2.

Suppose now that |V (H ∩ σ(H)

)| ≥ 1. If |V (H ∩ σ(H)

)| = 1, then

|E(H ∩ σ(H)

)| = 2|V (H ∩ σ(H)

)| − 2, and if |V (H ∩ σ(H)

)| ≥ 2, then

|E(H ∩ σ(H)

)| ≤ 2|V (H ∩ σ(H)

)| − 3, because H ∩ σ(H) is a subgraph of

G. Thus,

|E(H ∩ σ(H)

)| ≤ 2|V (H ∩ σ(H)

)| − 2,

and hence

|E(H ′)| = |E(H)|+ |E(σ(H)

)| − |E(H ∩ σ(H)

)|

≥ 2|V (H)| − 3 + 2|V (σ(H)

)| − 3− (2|V (H ∩ σ(H)

)| − 2)

= 2|V (H ′)| − 4.

236

This is a contradiction, because v, w, σ(v), σ(w) ∈ V (H ′).

It follows that every subgraph H of G′ with v, w ∈ V (H) or σ(v), σ(w) ∈V (H) satisfies |E(H)| ≤ 2|V (H)| − 4.

Therefore, as claimed, G satisfies the Laman conditions and if we define

Φ by Φ(x) = Φ(x)|V (G) for all x ∈ Cs, then Φ(x) ∈ Aut(G) for all x ∈ Cs and

Φ : Cs → Aut(G) is a homomorphism. Since we also have bΦ(s) = 1, it follows

from the induction hypothesis that there exists a sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

satisfying the conditions in Theorem 5.4.1 (iii). Since G is a (Cs, Φ) X-

replacement of G with V (G) = V (G) ∪ v1,

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (G, Φ), (G, Φ)



group in dimension 2, and Φ : Cs → Aut(G) be a homomorphism. If there

exists a (Cs, Φ) construction sequence for G, then G has a proper (Cs, Φ)

3Tree2 ⊥ partition or a proper (Cs, Φ) 3Tree2 ‖ partition.

Proof. We proceed by induction on |V (G)|. Let V (K2) = v1, v2 and

let Φ : Cs → K2 be the homomorphism defined by Φ(s) = (v1 v2). Then

K2 has the proper (Cs, Φ) 3Tree2 ⊥ partition E(T0), E(T1), E(T2), where

T0 = 〈v1, v2〉, T1 = 〈v1〉 and T2 = 〈v2〉. Let Ψ : Cs → K2 be the

homomorphism defined by Ψ(s) = id. Then K2 has the proper (Cs, Ψ) 3Tree2

‖ partition E(T0), E(T1), E(T2), where T0 = 〈v1〉, T1 = 〈v1, v2〉 and

T2 = 〈v2〉. This proves the base case.

237



Let G be a graph with |V (G)| = n + 1 and let Φ : Cs → Aut(G) be a

homomorphism such that there exists a (Cs, Φ) construction sequence

(K2, Φ0) = (G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)



G must be proper. Therefore, it suffices to show that G has some (Cs, Φ) ⊥or (Cs, Φ) ‖ 3Tree2 partition. In the following, we denote Φ(s) by σ.

By the induction hypothesis, Gk−1 has a (Cs, Φk−1) ⊥ or (Cs, Φk−1) ‖3Tree2 partition

E

(T

(k−1)0

), E

(T

(k−1)1

), E

(T

(k−1)2

).

Case 1: Suppose G is a (Cs, Φk−1) single vertex addition by v of Gk−1

with NG(v) = v0, σ(v0). Note that v0 is a vertex of T(k−1)1 or T

(k−1)2 . Wlog,

we assume v0 ∈ V(T

(k−1)1

). Then σ(v0) ∈ V

(T

(k−1)2

). So, if we define

T(k)0 = T

(k−1)0 ,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v0,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v

E(T

(k)2

)= E

(T

(k−1)2

) ∪ v, σ(v0),

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition

of G.

238

.. ..v0 .σ(v0)

.. .

.

.v0 .σ(v0)

.v

Figure 5.33: Construction of a (Cs, Φ) ⊥ or (Cs, Φk−1) ‖ 3Tree2 partition of

G in the case where G is a (Cs, Φk−1) single vertex addition of Gk−1.

Case 2: Suppose G is a (Cs, Φk−1) single edge split on v1, v2; v of Gk−1

with NG(v) = v1, v2, v3. Then Φk−1(s)(v1) = σ(v1) = v2 and Φk−1(s)(v3) =

σ(v3) = v3. By Remark 5.4.4,E

(T

(k−1)0

), E

(T

(k−1)1

), E

(T

(k−1)2

)must be a

(Cs, Φk−1) ⊥ 3Tree2 partition of Gk−1. Clearly, v1, v2 ∈ E(T

(k−1)0

)and, by

Remark 5.4.2, v3 /∈ V(T

(k−1)0

). Therefore, v3 ∈ V

(T

(k−1)1

). So, if we define


V(T

(k)0

)= V

(T

(k−1)0

) ∪ v

E(T

(k)0

)=

(E

(T

(k−1)0

) \ v1, v2) ∪ v, v1, v, v2

,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v3,

and

T(k)2 = T

(k−1)2 ,

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ‖ 3Tree2 partition of G.

Case 3: Suppose G is a (Cs, Φk−1) double vertex addition by (v, w) of

Gk−1 with NG(v) = v1, v2. Since Φk−1(s) = σ|V (Gk−1), we have NG(w) =

239

.

.v1 .v2 = σ(v1)

.v3 = σ(v3)

. .

. .

.v1 .v2 = σ(v1)

.v3 = σ(v3)

. .

.

..v

Figure 5.34: Construction of a (Cs, Φ) ‖ 3Tree2 partition of G in the case

where G is a (Cs, Φk−1) single edge split of Gk−1. The edges in black color

represent edges of the invariant trees.

σ(v1), σ(v2).

Case 3.1: If v1, v2 /∈ V(T

(k−1)0

), then v1, v2, σ(v1), σ(v2) ∈ V

(T

(k−1)i

)for

i = 1, 2. In this case, we define

T(k)0 = T

(k−1)0 ,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v, w

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v1, w, σ(v2),


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v, w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ v, v2, w, σ(v1).

ThenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


of G.

Case 3.2: If v1 ∈ V(T

(k−1)0

)and v2 /∈ V

(T

(k−1)0

), then σ(v1) ∈ V

(T

(k−1)0

)

and v2, σ(v2) ∈ V(T

(k−1)i

)for i = 1, 2. So, if we define T

(k)0 to be the tree

240

..v2 .σ(v2)

.v1 .σ(v1). .

. .

..v2 .σ(v2)

.v1 .σ(v1). .

. .

. ..v .w

Figure 5.35: Construction of a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition of

G in the case where G is a (Cs, Φk−1) double vertex addition of Gk−1 and

v1, v2 /∈ V(T

(k−1)0

).

with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)= E

(T

(k−1)0

) ∪ v, v1, w, σ(v1),


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v2,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ w, σ(v2),

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


of G.

Case 3.3: If both v1 and v2 are vertices of T(k−1)0 and vi 6= σ(vi) for

i = 1, 2, then wlog v2 ∈ V(T

(k−1)1

), and hence σ(v2) ∈ V

(T

(k−1)2

), so that

the previous construction in Case 3.2 can be used to obtain a (Cs, Φ) ⊥ or

(Cs, Φ) ‖ 3Tree2 partition of G.

241

Case 3.4: If both v1 and v2 are vertices of T(k−1)0 and vi = σ(vi) for

some i ∈ 1, 2, say wlog v1 = σ(v1), then, by Remarks 5.4.2 and 5.4.3,E

(T

(k−1)0

), E

(T

(k−1)1

), E

(T

(k−1)2

)is a (Cs, Φk−1) ‖ 3Tree2 partition of G

and v2 6= σ(v2). Suppose wlog that v2 ∈ V(T

(k−1)1

). Then σ(v2) ∈ V

(T

(k−1)2

)

and the construction in Case 3.2 can again be used to obtain a (Cs, Φ) ‖3Tree2 partition of G.

..v2 .σ(v2)

.v1 .σ(v1). .

. .

..v2 .σ(v2)

.v1 .σ(v1). .

. .

. ..v .w

Figure 5.36: Construction of a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition of G in

the case where G is a (Cs, Φk−1) double vertex addition of Gk−1 and at least

one of v1 or v2 is a vertex of T(k−1)0 . The edges in black color represent edges

of the invariant tree.

Case 4: Suppose G is a (Cs, Φk−1) double edge split on

(v1, v2, σ(v1), σ(v2)); (v, w) of Gk−1 with E(Gk) =(E(Gk−1) \

v1, v2, σ(v1), σ(v2)) ∪ v, vi| i = 1, 2, 3

∪ w, σ(vi)| i = 1, 2, 3.

Case 4.1: Suppose v1, v2 ∈ E(T

(k−1)0

). Then we also have

σ(v1), σ(v2) ∈ E(T

(k−1)0

).

Case 4.1a: If v3 /∈ V(T

(k−1)0

), then v3, σ(v3) ∈ V

(T

(k−1)i

)for i = 1, 2. In

this case we define T(k)0 to be the tree with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)=

(E

(T

(k−1)0

) \ v1, v2, σ(v1), σ(v2))

∪v, v1, v, v2, w, σ(v1), w, σ(v2),

242


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v3,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ w, σ(v3).

ThenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


of G.

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

. ..v .w

Figure 5.37: Construction of a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition

of G in the case where G is a (Cs, Φk−1) double edge split of Gk−1,

v1, v2, σ(v1), σ(v2) ∈ E(T

(k−1)0

)and either v3 /∈ V

(T

(k−1)0

)or v3 ∈

V(T

(k−1)0

)and σ(v3) 6= v3. The edges in black color represent edges of the

invariant trees.

Case 4.1b: If v1, v2, σ(v1), σ(v2) ∈ E(T

(k−1)0

)and v3 ∈ V

(T

(k−1)0

)

with σ(v3) 6= v3, then wlog we have v3 ∈ V(T

(k−1)1

), and hence σ(v3) ∈

V(T

(k−1)2

), so that the previous construction in Case 4.1a can again be used

to obtain a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2 partition of G.

Case 4.1c: Suppose v1, v2, σ(v1), σ(v2) ∈ E(T

(k−1)0

)and v3 ∈

V(T

(k−1)0

)with σ(v3) = v3. Then, by Remarks 5.4.2 and 5.4.3,

243

E

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ‖ 3Tree2 partition of G and σ(vi) 6=

vi for i = 1, 2. Since T(k−1)0 is connected, we have that for i = 1 or i = 2, there

exists a v3 − vi path in T(k−1)0 that does not contain the edge v1, v2, say

wlog P = v3, e1, . . . , em, v2 is a v3− v2 path in T(k−1)0 not containing the edge

v1, v2. Then σ(P ) is a v3 − σ(v2) path in T(k−1)0 not containing the edge

σ(v1), σ(v2) and P and σ(P ) do not share a common vertex other than v3,

for otherwise there exists a cycle in T(k−1)0 . Assume wlog that v2 ∈ V

(T

(k)1

),

and hence σ(v2) ∈ V(T

(k)2

). Then we define T

(k)0 to be the graph with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)=

(E

(T

(k−1)0

) \ v1, v2, σ(v1), σ(v2))

∪v, v1, v, v3, w, σ(v1), w, v3,

T(k)1 to be the graph with

V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v2,

and T(k)2 to be the graph with

V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)= E

(T

(k−1)2

) ∪ w, σ(v2).

Clearly, the graphs T(k)1 and T

(k)2 are trees and T

(k)0 is connected. If there

exists a cycle C in T(k)0 , then C must contain at least one of the edges v, v3

or w, v3, for otherwise C does not contain any edge incident with v or w

and there exists a cycle in T(k−1)0 .

Suppose first that C contains only one of the two edges v, v3 and

244

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 = σ(v3)

. .

. .

.

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 = σ(v3)

. .

. .

.

. ..v .w


where G is a (Cs, Φk−1) double edge split of Gk−1, v1, v2, σ(v1), σ(v2) ∈E

(T

(k−1)0

), v3 ∈ V

(T

(k−1)0

)and σ(v3) = v3. The edges in black color represent

edges of the invariant trees.

w, v3, say wlog C contains v, v3, but not w, v3. Then C contains

the edge v, v1, but not w, σ(v1). Thus, there exists a v3 − v1 path P ′

in T(k−1)0 that does not contain the edge v1, v2. This is a contradiction,

because v3, e1, . . . , em, v2, v1, v2, v1 is also a v3 − v1 path in T(k−1)0 distinct

from P ′.

So, suppose C contains both edges v, v3 and w, v3. Then C also

contains the edges v, v1 and w, σ(v1). Thus, there exists a v1 − σ(v1)

path P ′′ in T(k−1)0 that does not contain the edges v1, v2 and σ(v1), σ(v2).

But v1, v1, v2, v2, em, . . . , e1, v3, σ(e1), . . . , σ(em), σ(v2), σ(v1), σ(v2), σ(v1)

is also a v1 − σ(v1) path in T(k−1)0 distinct from P ′′.

Thus, T(k)0 is a tree and

E

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ‖ 3Tree2

partition of G.

Case 4.2: Suppose v1, v2 /∈ E(T

(k−1)0

), say wlog v1, v2 ∈ E

(T

(k−1)1

).

Then we also have σ(v1), σ(v2) ∈ E(T

(k−1)2

).

245

Case 4.2a: If v3 ∈ V(T

(k−1)0

), then σ(v3) ∈ V

(T

(k−1)0

). In this case we

define T(k)0 to be the tree with

V(T

(k)0

)= V

(T

(k−1)0

) ∪ v, w

E(T

(k)0

)= E

(T

(k−1)0

) ∪ v, v3, w, σ(v3),


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)=

(E

(T

(k−1)1

) \ v1, v2) ∪ v, v1, v, v2

,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ w

E(T

(k)2

)=

(E

(T

(k−1)2

) \ σ(v1), σ(v2))

∪w, σ(v1), w, σ(v2).

ThenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


of G.

Case 4.2b: If v3 /∈ V(T

(k−1)0

), then v3 ∈ V

(T

(k−1)i

)for i = 1, 2, and we

define

T(k)0 = T

(k−1)0 ,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v, w

E(T

(k)1

)=

(E

(T

(k−1)1

) \ v1, v2)

∪v, v1, v, v2, w, σ(v3),

246

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

. ..v .w

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

.

.v1 .σ(v1)

.v2 .σ(v2)

.v3 .σ(v3)

. .

. .

. .

. ..v .w


where G is a (Cs, Φk−1) double edge split of Gk−1, v1, v2 ∈ E(T

(k−1)1

)and

σ(v1), σ(v2) ∈ E(T

(k−1)2

). The edges in black color represent edges of the

invariant tree.


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v, w

E(T

(k)2

)=

(E

(T

(k−1)2

) \ σ(v1), σ(v2))

∪ w, σ(v1), w, σ(v2), v, v3.

ThenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is again a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2

partition of G.

Case 5: Finally, suppose that G is a (Cs, Φk−1) X-replacement by v of

Gk−1 with E(G) =(E(Gk−1)\

v1, v2, v3, v4)∪v, vi| i ∈ 1, 2, 3, 4

.

Then Φk−1(s)(v1, v2) = v3, v4. Wlog we assume Φk−1(s)(v1) = σ(v1) =

v3 and Φk−1(s)(v2) = σ(v2) = v4.

247

Case 5.1: Suppose v1, v2 /∈ E(T

(k−1)0

), say wlog v1, v2 ∈ E

(T

(k−1)1

).

Then v3, v4 ∈ E(T

(k−1)2

). So, if we define

T(k)0 = T

(k−1)0 ,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v

E(T

(k)1

)=

(E

(T

(k−1)1

) \ v1, v2) ∪ v, v1, v, v2

,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v

E(T

(k)2

)=

(E

(T

(k−1)2

) \ v3, v4) ∪ v, v3, v, v4

,

thenE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2


of G.

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

..v


where G is a (Cs, Φk−1) X-replacement of Gk−1, v1, v2 ∈ E(T

(k−1)1

)and

v3, v4 ∈ E(T

(k−1)2

).

Case 5.2: Suppose v1, v2, v3, v4 ∈ E(T

(k−1)0

). Since T

(k−1)0 is a tree

and Φk−1(s)(T

(k−1)0

)= T

(k−1)0 , there either exists a v1 − v3 path that does

not contain the vertices v2 and v4 or a v2− v4 path that does not contain the

vertices v1 and v3 in T(k−1)0 . Suppose wlog that P is a v2 − v4 path in T

(k−1)0

248

that does not contain the vertices v1 and v3. Wlog we may also assume that

v2 ∈ V(T

(k−1)2

), and hence v4 ∈ V

(T

(k−1)1

). If all the vertices and edges of

P , as well as the edges v1, v2 and v3, v4, are deleted from T(k−1)0 , then

the resulting subgraph of T(k−1)0 has at least two components, namely the

components A with v1 ∈ V (A) and σ(A) = B with v3 ∈ V (B).

Case 5.2.1: Suppose V (A) = v1. Then we also have V (B) = v3.

Case 5.2.1a: If v1 ∈ V(T

(k−1)1

), then v3 ∈ V

(T

(k−1)2

). In this case we


V(T

(k)0

)= V

(T

(k−1)0

) \ v1, v3

E(T

(k)0

)= E

(T

(k−1)0

) \ v1, v2, v3, v4,


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v, v3

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v3, v, v4,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v, v1

E(T

(k)2

)= E

(T

(k−1)2

) ∪ v, v1, v, v2.

Case 5.2.1b: If v1 ∈ V(T

(k−1)2

), then v3 ∈ V

(T

(k−1)1

). In this case we


V(T

(k)0

)= V

(T

(k−1)0

) \ v1, v3

E(T

(k)0

)= E

(T

(k−1)0

) \ v1, v2, v3, v4,

249


V(T

(k)1

)= V

(T

(k−1)1

) ∪ v, v1

E(T

(k)1

)= E

(T

(k−1)1

) ∪ v, v1, v, v4,


V(T

(k)2

)= V

(T

(k−1)2

) ∪ v, v3

E(T

(k)2

)= E

(T

(k−1)2

) ∪ v, v2, v, v3.

In both cases,E

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ⊥ or (Cs, Φ) ‖ 3Tree2

partition of G.

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

..v

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

.

.v1 .v3 = σ(v1)

.v4 = σ(v2) .v2

. .

. .

..v


where G is a (Cs, Φk−1) X-replacement of Gk−1 and v1, v2, v3, v4 ∈E

(T

(k−1)0

). The edges in black color represent edges of the invariant tree.

Case 5.2.2: Finally, suppose |V (A)| = |V (B)| = m ≥ 2. Then we first

carry out the same construction as in Case 5.2.1. Subsequently, we delete all

the edges of A and B from E(T

(k)0

), one edge from both A and B at a time,

and add them to either E(T

(k)1

)or E

(T

(k)2

)in the following way.

250

Let A be the subgraph of A that only contains the single vertex v1 and

let B be the subgraph of B that only contains the single vertex σ(v1) = v3.

Let v1, z be an edge of A. Then v3, σ(z) is an edge of B. By the

construction in Case 5.2.1, v1, v3 ∈ V(T

(k)i

)for i = 1, 2. Also, σ(z) 6= z and

z, σ(z) ∈ V(T

(k)0

), which says that either z ∈ V

(T

(k)1

)and σ(z) ∈ V

(T

(k)2

)

or z ∈ V(T

(k)2

)and σ(z) ∈ V

(T

(k)1

).

We now delete the edges v1, z and v3, σ(z) from E(T

(k)0

)and if z ∈

V(T

(k)1

), then we add v1, z to E

(T

(k)2

)and v3, σ(z) to E

(T

(k)1

), and

if z ∈ V(T

(k)2

), then we add v1, z to E

(T

(k)1

)and v3, σ(z) to E

(T

(k)2

).

Subsequently, we add the vertex z to V (A), the vertex σ(z) to V (B), the edge

v1, z to E(A), and the edge v3, σ(z) to E(B). If we then have A = A,

then B = B andE

(T

(k)0

), E

(T

(k)1

), E

(T

(k)2

)is a (Cs, Φ) ⊥ or (Cs, Φ) ‖

3Tree2 partition of G.

Otherwise, there exists an edge x, y in E(A) \ E(A) with x ∈ V (A)

and y ∈ V (A) \ V (A), and hence there also exists the edge σ(x), σ(y) in

E(B) \ E(B) with σ(x) ∈ V (B) and σ(y) ∈ V (B) \ V (B). Note that since

x ∈ V (A) and σ(x) ∈ V (B), we have x, σ(x) ∈ V(T

(k)i

)for i = 1, 2. So, we

can repeat the above construction step for the edges x, y and σ(x), σ(y).This process can be continued until A = A and B = B. ¤


group in dimension 2, and Φ : Cs → Aut(G) be a homomorphism. If G has a

proper (Cs, Φ) 3Tree2 ⊥ partition or a proper (Cs, Φ) 3Tree2 ‖ partition, then

R(G,Cs,Φ) 6= ∅ and G is (Cs, Φ)-generically isostatic.

Proof. Case 1: Suppose G has a proper (Cs, Φ) ‖ 3Tree2 partition

E(T0), E(T1), E(T2). In the following, we again denote Φ(s) by σ. There

251

exists an edge e = w, z ∈ E(T1) such that σ(w) = w and σ(z) = z and,

by Remark 5.4.3, valT1(w) = 1, w ∈ E(T0), and no other vertex of G that is

fixed by σ is a vertex of T0.

Since G has a 3Tree2 partition, G satisfies the count |E(G)| = 2|V (G)|−3.

Therefore, by Theorems 2.2.5 and 3.2.3, it suffices to find some framework

(G, p) ∈ R(G,Cs,Φ) that is independent.

Let Vi be the set of vertices of G that are not in V (Ti) for i = 0, 1, 2 and

let (G, p, q) be the frame with p : V (G) → R2 and q : E(G) → R2 defined by

p(v) =

(0, 1) if v ∈ V0

(−1, 0) if v ∈ V1

(1, 0) if v ∈ V2 \ v(0, 0) if v = w

q(b) =

(1, 0) if b ∈ EV1,w or b ∈ EV2\w,w

(2, 0) if b ∈ EV1,V2\w

(−1, 1) if b ∈ E(T1) \w, z

(1, 1) if b ∈ E(T2)

(0, 1) if b = w, z

,


incident with a vertex in X and a vertex in Y .



such a way that we obtain the matrix R′(G, p, q) which has the (2i−1)st col-

umn of R(G, p, q) in its ith column and the (2i)th column of R(G, p, q) in its

(|V (G)|+ i)th column for i = 1, 2, . . . , |V (G)|. Let Fb denote the row vector

of R′(G, p, q) that corresponds to the edge b ∈ E(G). We then rearrange the

252

..w

.(−1, 0) .(1, 0).(0, 0)

.(0, 1)

.T1.T2

.V1 .V2 \ w

.V0

.T0

. .

.

.

Figure 5.42: The frame (G, p, q) in Case 1 of the proof of Lemma 5.4.5.

rows of R′(G, p, q) in such a way that we obtain the matrix R′′(G, p, q) which

has the vectors Fb with b ∈ E(T0) in the rows 1, 2, . . . , |E(T0)|, the vectors

Fb with b ∈ E(T1) \w, z in the following |E(T1)| − 1 rows, the vector

Fw,z in the next row, and the vectors Fb with b ∈ E(T2) in the last |E(T2)|rows. So R′′(G, p, q) is of the form

1 −1

... 0

2 −2

−1 1 1 −1

......

−1 1 1 −1

0 1 −1

1 −1 1 −1

......

1 −1 1 −1

.



∑

b∈E(G)

αbFb = 0,

253

where αb 6= 0 for some b ∈ E(T0). Since T0 is a tree, it follows that

∑

b∈E(T0)

αbFb 6= 0.

Thus, there exists a vertex vs ∈ V (T0), s ∈ 1, 2, . . . , |V (G)|, such that

∑

b∈E(T0)

αb(Fb)s = C 6= 0.

Since vs ∈ V (T0), vs belongs to either T1 or T2.

Suppose first that vs ∈ V (T2) and vs /∈ V (T1). Then (Fb)s = 0 and

(Fb)|V (G)|+s = 0 for all b ∈ E(T1) and we have

∑

b∈E(T2)

αb(Fb)s = −C.

This says that

∑

b∈E(T2)

αb(Fb)|V (G)|+s =∑

b∈E(G)

αb(Fb)|V (G)|+s = −C 6= 0,

a contradiction.

So, suppose that vs ∈ V (T1) and vs /∈ V (T2). Then (Fb)s = 0 and

(Fb)|V (G)|+s = 0 for all b ∈ E(T2) and we have

∑

b∈E(T1)

αb(Fb)s = −C.

Note that vs 6= w, because valT1(w) = 1 and (Fw,z)s = 0 for all s =

1, 2, . . . , |V (G)|. Also, vs 6= z, since z /∈ V (T0). Therefore,

∑

b∈E(T1)

αb(Fb)|V (G)|+s =∑

b∈E(G)

αb(Fb)|V (G)|+s = C 6= 0,

which is again a contradiction. So, if∑

b∈E(G) αbFb = 0 is a row dependency

of R′′(G, p, q), then αb = 0 for all b ∈ E(T0).

254

It is now only left to show that the matrix R(G, p, q) which is obtained

from R′′(G, p, q) by deleting those rows of R′′(G, p, q) that correspond to the

edges of T0 has linearly independent rows. Clearly, R(G, p, q) has linearly in-

dependent rows if and only if the matrix R(G, p, q) has linearly independent

rows, where R(G, p, q) is obtained by deleting the row Fw,z from R(G, p, q).

In order to show that R(G, p, q) has linearly independent rows we may mul-

tiply R(G, p, q) by appropriate matrices of basis transformation and then use

arguments analogous to above. So, as claimed, the frame (G, p, q) is inde-

pendent.

Now, if (G, p) is not a framework, then we need to symmetrically

pull apart those joints of (G, p, q) that have the same location in R2 and

whose vertices are adjacent. So suppose |V1| ≥ 2. Then it follows that

|V1| = |V2 \w| ≥ 2, because σ(V1) = V2 \w. Since E(T0), E(T1), E(T2)is proper, one of 〈V1〉 ∩ Ti, i = 0, 2, is not connected.

Suppose first that 〈V1〉 ∩ T0 is not connected. Then 〈V2 \ w〉 ∩ T0 is

..w.(−1, 0) .(1, 0).(0, 0)

.(0, 1)

.T1.T2

.V1 \ A .(V2 \ w) \ σ(A)

.A .σ(A)

.V0

.T0

. .

.

.

. .

. .

Figure 5.43: The frame (G, pt, qt) in the case where 〈V1〉∩T0 is not connected.

also not connected. Let A be the set of vertices in one of the components of

〈V1〉 ∩ T0 and σ(A) be the set of vertices in the corresponding component of

255

〈V2 \ w〉 ∩T0. For t ∈ R, we define pt : V (G) → R2 and qt : E(G) → R2 by

pt(v) =

(−1− t,−t) if v ∈ A

(1 + t,−t) if v ∈ σ(A)

p(v) otherwise

qt(b) =

(1 + t, t) if b ∈ EA,w

(2 + t, t) if b ∈ EA,(V2\w)\σ(A)

(1 + t,−t) if b ∈ Eσ(A),w

(2 + t,−t) if b ∈ Eσ(A),V1\A

q(b) otherwise

.

Suppose now that 〈V1〉 ∩ T2 is not connected. Then 〈V2 \ w〉 ∩ T1 is

also not connected. Let B and σ(B) be the vertex sets of components of

〈V1〉 ∩ T2 and 〈V2 \ w〉 ∩ T1, respectively. In this case, for t ∈ R, we define

pt : V (G) → R2 and qt : E(G) → R2 by

..w.(−1, 0) .(1, 0).(0, 0)

.(0, 1)

.V1 \B .(V2 \ w) \ σ(B)

.V0

.T0

. .

.

.. ...

.T2 .T1

.

. .

.B .σ(B)

Figure 5.44: The frame (G, pt, qt) in the case where 〈V1〉∩T2 is not connected.

pt(v) =

(−1− t, 0) if v ∈ B

(1 + t, 0) if v ∈ σ(B)

p(v) otherwise

256

qt(b) =

(1 + t, 1) if b ∈ EB,V0

(−1− t, 1) if b ∈ Eσ(B),V0

q(b) otherwise

.

In both cases, we have (G, pt, qt) = (G, p, q) if t = 0. Therefore, by

Lemma 5.1.3, there exists a t0 ∈ R, t0 6= 0, such that the frame (G, pt0 , qt0)

is independent. This process can be continued until we obtain an indepen-

dent frame (G, p, q) which has the property that if p(u) = p(v) for some

u, v ∈ E(G), then u, v ∈ V0.

Suppose (G, p) is still not a framework. Then |V0| ≥ 2 and since

E(T0), E(T1), E(T2) is proper, 〈V0〉 ∩ T1 or 〈V0〉 ∩ T2 is not connected.

In fact, since σ(〈V0〉 ∩ T1) = 〈V0〉 ∩ T2, both 〈V0〉 ∩ T1 and 〈V0〉 ∩ T2 are

not connected. Let A be the set of vertices in one of the components of

〈V0〉 ∩ T2 and σ(A) be the set of vertices in the corresponding component

of 〈V0〉 ∩ T1. We denote A ∩ σ(A) by D and A ∪ σ(A) by F . Clearly,

ED,V0\F = ∅, EA\D,V0\F ⊆ E(T1) and Eσ(A)\D,V0\F ⊆ E(T2). Further, we have

EA\D,σ(A)\D = ∅ as the following argument shows.

Suppose to the contrary that there exists x, y ∈ E(G) with x ∈ A \D

and y ∈ σ(A) \ D. Then x, y ∈ E(T1) or x, y ∈ E(T2), say wlog

x, y ∈ E(T2). Since x, y ∈ E(〈V0〉), it follows that x, y ∈ E(〈V0〉∩T2).

Therefore, since x ∈ A, y must also be a vertex of A, contradicting the fact

that y ∈ σ(A) \D.

Finally, note that EA\D,D ⊆ E(T2), because if x, y ∈ E(T1), where

x ∈ A \D and y ∈ D, then we must have x ∈ σ(A), contradicting x ∈ A \D.

Similarly, we have Eσ(A)\D,D ⊆ E(T1).

257

So, for t ∈ R, we define pt : V (G) → R2 and qt : E(G) → R2 by

pt(v) =

(−t, 1 + t) if v ∈ A \D

(t, 1 + t) if v ∈ σ(A) \D

(0, 1 + 2t) if v ∈ D

p(v) otherwise

qt(b) =

q(b) + (t, t) if b ∈ EV2\w,σ(A)\D

q(b) + (−t, t) if b ∈ EV1,A\D

q(b) + (0, 2t) if b ∈ EV2\w,D

q(b) + (0, 2t) if b ∈ EV1,D

q(b) otherwise

.

Then (G, pt, qt) = (G, p, q) if t = 0. Therefore, by Lemma 5.1.3, there exists

..w

.(−1, 0) .(1, 0).(0, 0)

.(0, 1)

.V1 .V2 \ w

.V0 \ F

.T0

. .

.

.

.

. .

.D

.A \D .σ(A) \D

.T2 .T1

.

. .

. .

.


a t0 ∈ R, t0 6= 0, such that the frame (G, pt0 , qt0) is independent.

Now, if |A \ D| ≥ 2, then |σ(A) \ D| = |A \ D| ≥ 2. Since

E(T0), E(T1), E(T2) is proper, 〈A \ D〉 ∩ T1 or 〈A \ D〉 ∩ T2 is not con-

nected, say wlog 〈A \ D〉 ∩ T2 is not connected. Then 〈σ(A) \ D〉 ∩ T1 is

258

also not connected. Let B be the set of vertices in one of the components of

〈A \D〉 ∩T2 and σ(B) be the set of vertices in the corresponding component

of 〈σ(A) \ D〉 ∩ T1. Then, by using arguments analogous to above, we can

pull apart the vertices of B from (A \D) \ B in the direction of the vector

(−t, t) and the vertices of σ(B) from (σ(A) \ D) \ σ(B) in the direction of

the vector (t, t) in order to obtain a new independent frame.

This process can be continued until we obtain an independent frame

(G, p, q) with p(u) 6= p(v) for all u, v ∈ E(G). Then, by Remark 5.1.1,

(G, p) is an independent framework and, if necessary, an appropriate rota-

tion of (G, p) about the origin yields an independent framework in the set

R(G,Cs,Φ).

Case 2: Suppose G has a proper (Cs, Φ) ⊥ 3Tree2 partition

E(T0), E(T1), E(T2). Let Vi be the set of vertices of G that are not in

V (Ti) for i = 0, 1, 2 and let e0 = (0, 1), e1 = (−1, 0), and e2 = (1, 0). We let

(G, p, q) be the frame with p : V (G) → R2 and q : E(G) → R2 defined by


q(b) =

(2, 0) if b ∈ E(T0)

(−1, 1) if b ∈ E(T1)

(1, 1) if b ∈ E(T2)

.

The proof that (G, p, q) is independent and that we can construct an

independent framework (G, p) ∈ R(G,Cs,Φ) is analogous to the proof of Case

1. ¤


rem 5.4.1.

259

..e1 .e2

.e0

.T1.T2

.V1 .V2

.V0

.T0

. .

.

Figure 5.46: The frame (G, p, q) in Case 2 of the proof of Lemma 5.4.5.

Remark 5.4.5 By generalizing the geometric proofs of Lemmas 5.2.6 and

5.2.7, and by using the fact that the framework (G, p) which is obtained

from an isostatic framework (G, p) by performing an X-replacement on G

and placing the new vertex in G at the point of intersection of the two bars

that were removed from (G, p) is again isostatic (see the proof of Proposition

3.9 in [68], for example), it is straightforward to also give a direct geometric

proof that condition (iii) implies condition (i) in Theorem 5.4.1, i.e., that the

existence of a (Cs, Φ) construction sequence for G implies that R(G,Cs,Φ) 6= ∅and that G is (Cs, Φ)-generically isostatic.

Remark 5.4.6 Theorem 5.4.1 still holds if we omit (Cs, Φi) single edge splits

in condition (iii). However, all the other inductive construction techniques,

including the (Cs, Φi) X-replacement, are necessary to characterize all (Cs, Φ)-

generically isostatic graphs in terms of an inductive construction sequence.

260

.

.

.

.

.

.

.

.

.

.

. .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Figure 5.47: Two frameworks whose underlying graphs satisfy the conditions

of Case B.2.3 in the proof of Lemma 5.4.3 with respect to the types Φ that

are uniquely determined by the injective realizations. Any symmetrized Hen-

neberg’s sequence for any of these two graphs needs to include a (Cs, Φi)

X-replacement.

5.5 Conjectures, algorithms, and further re-

marks

5.5.1 Dimension 2

For the symmetry groups C2v and C3v in dimension 2, it was conjectured in

[15] that the following Laman-type characterizations of (C2v, Φ)- and (C3v, Φ)-

generically isostatic graphs hold.

Conjecture 5.5.1 Let G be a graph with |V (G)| ≥ 2, C2v be a symmetry

group in dimension 2, and Φ : C2v → Aut(G) be a homomorphism. Then

R(G,C2v ,Φ) 6= ∅ and G is (C2v, Φ)-generically isostatic if and only if

(i) |E(G)| = 2|V (G)| − 3 and |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G with

|V (H)| ≥ 2 (Laman conditions);

261

(ii) jΦ(C2) = 0 and bΦ(C2) = bΦ(s) = 1 for both reflections s ∈ C2v.



R(G,C3v ,Φ) 6= ∅ and G is (C3v, Φ)-generically isostatic if and only if

(i) |E(G)| = 2|V (G)| − 3 and |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G with


(ii) jΦ(C3) = 0 and bΦ(s) = 1 for all reflections s ∈ C3v.

If one wants to prove these conjectures in the analogous way as the sym-

metrized Laman’s Theorems for C3, C2, and Cs, one has to consider two basic

cases: first, the case where the given graph G has a vertex of valence 2, and

secondly, the case where G has a vertex of valence 3 and no vertex of valence

2.

For each of the groups C2v and C3v, the first case can be treated in a

straightforward fashion by using appropriate symmetrized versions of a ver-

tex 2-addition.

We have seen in the previous section that for vertices of valence 3, the

presence of a single reflection s in the symmetry group S gives rise to a

large number of subcases that need to be treated separately, where each

subcase corresponds to a particular allocation of the 3-valent vertex and its

three neighbors to the permutation cycles of the graph automorphism Φ(s).

Since the symmetry groups C2v and C3v contain more than just one reflection

(namely two and three, respectively), the number of subcases that need to

be considered for these groups is even larger than it was in the case of Cs.

262

So, while we suspect that the above conjectures can be proven in this way,

the number of cases that need to be treated in these proofs is unreasonably

large for the scope of this thesis.

We conjecture that (C3v, Φ)-generically isostatic graphs can also be char-

acterized by means of symmetrized 3Tree2 partitions. We need the following

definitions.

Definition 5.5.1 Let G be a graph, C3v = 〈C3, s〉 be a symmetry group in

dimension 2, and Φ : C3v → Aut(G) be a homomorphism. A (C3v, Φ) 3Tree2

⊥ partition of G is a 3Tree2 partition E(T0), E(T1), E(T2) of G such that

(i) Φ(C3)(Ti) = Ti+1 for i = 0, 1, 2, where the indices are added modulo 3;

(ii) there exists i ∈ 0, 1, 2 such that Φ(s)(Ti) = Ti and Φ(s)(Ti+1) = Ti+2.

.

.γ2(v)

.v .γ(v) = σ(v)

.γ2(w)

.w

.γ(w).

. .

.

.

..T0

.T2

.T1

.(a)

.

.γ2(x)

.x .γ(x) = σ(x)

.γ2(w)

.w

.γ(w)

.γ2(v)

.v

.γ(v)

.

. .

.

.

..

.

..T0

.T1.T2

.(b)

Figure 5.48: A (C3v, Φ) 3Tree2 ⊥ partition of a graph (a) and a (C3v, Φ)

3Tree2 ‖ partition of a graph (b), where Φ(C3) = γ and Φ(s) = σ.

Definition 5.5.2 Let G be a graph, C3v = 〈C3, s〉 be a symmetry group in

dimension 2, and Φ : C3v → Aut(G) be a homomorphism such that there

263

exists an edge e = v, w ∈ E(G) with Φ(s)(v) = v and Φ(s)(w) = w. A

(C3v, Φ) 3Tree2 ‖ partition of G is a 3Tree2 partition E(T0), E(T1), E(T2)of G such that

(i) Φ(C3)(Ti) = Ti+1 for i = 0, 1, 2, where the indices are added modulo 3;

(ii) e ∈ E(T1), Φ(s)(T1 − v) = T2 − Φ(C3)(v) and Φ(s)(T0 −Φ(C2

3)(v)) = T0 − Φ(C23)(v).



R(G,C3v ,Φ) 6= ∅ and G is (C3v, Φ)-generically isostatic if and only if G has a

proper (C3v, Φ) 3Tree2 ⊥ partition or a proper (C3v, Φ) 3Tree2 ‖ partition.

Due to the structure of the group C2v, there does not seem to exist an

analogous characterization of (C2v, Φ)-generically isostatic graphs in terms of

symmetrized 3Tree2 partitions.

Algorithms

An immediate consequence of the symmetrized Laman’s Theorems for C3,

C2, and Cs (and the above conjectures for C2v and C3v) is that there is (would

be) a polynomial time algorithm to determine whether a given graph G is

(S, Φ)-generically isostatic. In fact, although the Laman conditions involve

an exponential number of subgraphs of G, there are several algorithms that

determine whether they hold in c|V (G)||E(G)| steps, where c is a constant.

The pebble game ([39]) is an example for such an algorithm. The additional

264

symmetry conditions for the number of fixed structural elements can trivially

be checked in constant time, from the graph automorphisms.

5.5.2 Dimension 3

There is no known characterization of generically 3-isostatic graphs,

although we have the necessary conditions identified in Theorem 2.2.8:

|E(G)| = 3|V (G)| − 6 and |E(H)| ≤ 3|V (H)| − 6 for all subgraphs H of

G with |V (H)| ≥ 3. Recall from Section 2.2.5, however, that there are

a number of inductive construction techniques which are known to preserve

the generic rigidity properties of a graph. Suppose we are given such a generi-

cally 3-isostatic graph G with a construction sequence. Further, suppose S is

a symmetry group in dimension 3 and Φ : S → Aut(G) is a homomorphism.

Then the graph G can only be (S, Φ)-generically isostatic if G satisfies all the

conditions given in Theorem 4.3.3 for the symmetry operations in S (with Φ

as the underlying type) as well as the corresponding conditions for all sub-

graphs H with the full count |E(H)| = 3|V (H)|−6 and either reflectional or

half-turn symmetry. One may ask whether all of these conditions combined

are also sufficient for G to be (S, Φ)-generically isostatic. The following ex-

ample shows that this is in general not the case.

Consider the (Cs, Φ)-generic realization (K4,6, p) of the complete bipar-

tite graph K4,6 with partite sets X = v1, v2, v3, v4, and Y = v5, . . . , v10shown in Figure 5.49. The graph K4,6 is generically 3-isostatic, and for the

reflection s ∈ Cs, we have jΦ(s) = bΦ(s) = 0, so that all the symmetry con-

ditions given in Section 4.3.3 are satisfied. However, the mirror symmetry

forces the four points p1, p2, p3, p4 corresponding to the vertices in X to be

265

.

....

....

..

....

..

.....p1

.p5

.p6

.p7

.p8

.p9

.p10

.p2

.p3

.p4

Figure 5.49: A (Cs, Φ)-generic realization of the complete bipartite graph K4,6.

coplanar, so that it follows from the results in [71] or [75] that (K4,6, p) is

infinitesimally flexible.

Further examples and a more detailed investigation of how ‘flatness’

caused by symmetry gives rise to additional necessary conditions for a graph

to be (S, Φ)-generically isostatic, where S is a symmetry group in dimension

3, will be presented in [59]. This builds on [70].

What these ‘failures from flatness’ show is that symmetry in dimension

d > 2 induces extra conditions for a graph G to be (S, Φ)-generically isostatic

beyond those of

(a) G being generically d-isostatic and

(b) the symmetry-extended Maxwell’s rules for G and all subgraphs of G

with the full Maxwell count.

We conjecture that flatness is the only additional concern, and that it can be

made into a finite set of added combinatorial conditions, for each symmetry

group.

266

5.5.3 Independence and infinitesimal rigidity

In Section 4.4, we described how to derive necessary conditions for a d-

dimensional symmetric framework (G, p) ∈ R(G,S,Φ) to be independent and

infinitesimally rigid, respectively, provided that the type Φ : S → Aut(G) of

(G, p) is a group homomorphism and that the points p(v), v ∈ V (G), span

all of Rd.

Given a graph G, it follows immediately from Lemma 3.2.2 and Theorem

3.2.3 that if there exists a framework (G, p) ∈ R(G,S,Φ) so that p(v), v ∈ V (G),

span all of Rd, then the conditions derived in Section 4.4 are also necessary

conditions for G to be (S, Φ)-generically independent and (S, Φ)-generically

infinitesimally rigid, respectively.

In this section, we investigate whether sufficient conditions for a graph

to be (S, Φ)-generically independent or (S, Φ)-generically infinitesimally rigid

can also be established.

In our exploration of these questions we restrict our attention to the

symmetry group C3 in dimension 2, since for this group we have the most

natural characterizations of (S, Φ)-generically isostatic graphs, as we have

seen in the previous sections. Similar explorations, however, can of course

also be carried out for other symmetry groups, with analogous corollaries for

C2 and Cs, and analogous conjectures for these groups and others.

Using the techniques described in Section 4.4, we obtain the following

result.

Theorem 5.5.4 Let G be a graph, C3 = Id, C3, C23 be a symmetry group

in dimension 2, and Φ : C3 → Aut(G) be a homomorphism, so that the set

267

R(G,C3,Φ) contains a framework (G, p) with the property that the points p(v),

v ∈ V (G), span all of R2.

(i) If G is (C3, Φ)-generically independent, then |E(G)| ≤ 2|V (G)| − 3 −2jΦ(C3);

(ii) if G is (C3, Φ)-generically infinitesimally rigid, then |E(G)| ≥ 2|V (G)|−3 + jΦ(C3).

Independence

We suppose first that jΦ(C3) = 0. In this case, we claim that for each of

the characterizations of (C3, Φ)-generically isostatic graphs given in Theorem

5.2.1, there also exists an analogous characterization of (C3, Φ)-generically

independent graphs. We need the following definitions.


in dimension 2, Φ : C3 → Aut(G) be a homomorphism, and v0 ∈ V (G). Then

(a) the graph G with V (G) = V (G)∪v, w, z and E(G) = E(G) is called

a (C3, Φ) partial vertex addition of order 0 (by (v, w, z)) of G .

(b) the graph G with V (G) = V (G) ∪ v, w, z and E(G) = E(G) ∪v, v0, w, Φ(C3)(v0), z, Φ(C2

3)(v0)

is called a (C3, Φ) partial ver-

tex addition of order 1 (by (v, w, z)) of G.

(c) the graph G with V (G) = V (G) ∪ v, w, z and E(G) = E(G) ∪v, w, w, z, z, v is called a (C3, Φ) ∆ addition (by (v, w, z)) of

G.

268

. ..

. .

.z

.v .w

.

.. . .

.. .

.

. .

.z

.v .w

..(a) .(b) .

. .

.

. .

.z

.v .w

..(c) .

Figure 5.50: A (C3, Φ) partial vertex addition of order 0 of a graph G (a), a

(C3, Φ) partial vertex addition of order 1 of a graph G (b), and a (C3, Φ) ∆

addition of a graph G (c).


in dimension 2, and Φ : C3 → Aut(G) be a homomorphism. A (C3, Φ)

3Forest2 partition of G is a partition of E(G) into the edge sets of three edge

disjoint forests F0, F1, F2 such that each vertex of G belongs to exactly two of

the forests and Φ(C3)(Fi) = Fi+1 for i = 0, 1, 2, where the indices are added

modulo 3.

A 3Forest2 partition is called proper if no non-trivial subtrees of distinct

forests Fi have the same span.

The proof of Theorem 5.2.1 can be extended directly to a proof of the

following result.

Corollary 5.5.5 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23 be a


so that jΦ(C3) = 0. The following are equivalent:

269

(i) R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically independent;

(ii) |E(G)| = 2|V (G)| − 3 and |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G with


(iii) there exists a sequence

(G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

such that

(a) G0 = K3 or G0 is the graph with three vertices and no edges, Gi+1

is a (C3, Φi) partial vertex addition of order 0 or 1, a (C3, Φi) vertex

addition, a (C3, Φi) edge split, a (C3, Φi) ∆ addition, or a (C3, Φi)

∆ extension of Gi with V (Gi+1) = V (Gi) ∪ vi+1, wi+1, zi+1 for

all i = 0, 1, . . . , k − 1;

(b) Φ0 : C3 → Aut(G0) is a non-trivial homomorphism and for

all i = 0, 1, . . . , k − 1, Φi+1 : C3 → Aut(Gi+1) is the homo-

morphism defined by Φi+1(x)|V (Gi) = Φi(x) for all x ∈ C3 and

Φi+1(C3)|vi+1,wi+1,zi+1 = (vi+1 wi+1 zi+1);

(iv) G has a proper (C3, Φ) 3Forest2 partition.

Corollary 5.5.6 Let G be a graph with |V (G)| ≥ 3, C3 = Id, C3, C23 be a


so that jΦ(C3) = 0. Then G is (C3, Φ)-generically independent if and only if G

is a subgraph of a (C3, Φ′)-generically isostatic graph G′ with V (G) = V (G′),

where Φ′ : C3 → Aut(G′) is defined by Φ′(x) = Φ(x) for all x ∈ C3.

270

Proof. If G is a subgraph of a (C3, Φ′)-generically isostatic graph G′ with

V (G) = V (G′), where Φ′ : C3 → Aut(G′) is defined by Φ′(x) = Φ(x) for all

x ∈ C3, then G is clearly (C3, Φ)-generically independent.

The converse follows immediately from Corollary 5.5.5 (iii). ¤

Whenever jΦ(C3) 6= 0, the analogous result to Corollary 5.5.6 does not

hold. In fact, given any independent framework (G, p) ∈ R(G,C3,Φ), where

jΦ(C3) 6= 0, there does not exist any isostatic framework (G′, p′) ∈ R(G′,C3,Φ′)

so that G ⊆ G′, V (G) = V (G′), and Φ′ : C3 → Aut(G′) is defined by

Φ′(x) = Φ(x) for all x ∈ C3, because an isostatic framework in R(G′,C3,Φ′)

must satisfy jΦ′(C3) = 0.

..

.

. ...

.

. .

Figure 5.51: Independent frameworks in R(G,C3,Φ) with jΦ(C3) = 1. These

frameworks cannot be contained in an isostatic framework that has the same

joints and also C3 symmetry.

Note also that if jΦ(C3) 6= 0, then a characterization of (C3, Φ)-generically

independent graphs in terms of symmetric 3Forest2 partitions, analogous to

the one given in Corollary 5.5.5, does not exist, because a vertex that is fixed

by Φ(C3) can only belong to either none or all three of the forests of a (C3, Φ)

3Forest2 partition. However, we conjecture that for jΦ(C3) 6= 0, the following

characterizations of (C3, Φ)-generically independent graphs hold.

Conjecture 5.5.7 Let G be a graph, C3 = Id, C3, C23 be a symmetry group

in dimension 2, and Φ : C3 → Aut(G) be a homomorphism, so that jΦ(C3) 6= 0

271

and the set R(G,C3,Φ) contains a framework (G, p) with the property that the

points p(v), v ∈ V (G), span all of R2. The following are equivalent:

(i) R(G,C3,Φ) 6= ∅ and G is (C3, Φ)-generically independent;

(ii) |E(G)| ≤ 2|V (G)| − 3− 2jΦ(C3), |E(H)| ≤ 2|V (H)| − 3 for all H ⊆ G

with |V (H)| ≥ 2, and for every subgraph H of G that is invariant

under Φ(C3), contains exactly m vertices that are fixed by Φ(C3), where

0 ≤ m ≤ jΦ(C3), and satisfies |V (H)| ≥ m + 3, we have |E(H)| ≤2|V (H)| − 3− 2m;

(iii) there exists a sequence

(G0, Φ0), (G1, Φ1), . . . , (Gk, Φk) = (G, Φ)

such that

(a) G0 is the graph with jΦ(C3) vertices and no edges, Gi+1 is a (C3, Φi)

partial vertex addition of order 0 or 1, a (C3, Φi) vertex addition,

a (C3, Φi) edge split, a (C3, Φi) ∆ addition, or a (C3, Φi) ∆ ex-

tension of Gi with V (Gi+1) = V (Gi) ∪ vi+1, wi+1, zi+1 for all

i = 0, 1, . . . , k − 1;

(b) Φ0 : C3 → Aut(G0) is the homomorphism that sends each x ∈ C3

to the identity automorphism and for all i = 0, 1, . . . , k− 1, Φi+1 :

C3 → Aut(Gi+1) is the homomorphism defined by Φi+1(x)|V (Gi) =

Φi(x) for all x ∈ C3 and Φi+1(C3)|vi+1,wi+1,zi+1 = (vi+1 wi+1 zi+1).

272

Infinitesimal rigidity

Suppose first that for a graph G, we have jΦ(C3) = 0. If G has a (C3, Φ′)-

generically isostatic subgraph G′ with V (G) = V (G′), where Φ′ : C3 →Aut(G′) is defined by Φ′(x) = Φ(x) for all x ∈ C3, then G is clearly (C3, Φ)-

generically infinitesimally rigid. We conjecture that the converse of this result

also holds and that it can be proven using symmetric 3Forest2 partitions of

graphs.

Whenever jΦ(C3) 6= 0, however, the analogous result does not hold. In fact,

given any infinitesimally rigid framework (G, p) ∈ R(G,C3,Φ), where jΦ(C3) 6= 0,

there does not exist any isostatic framework (G′, p′) ∈ R(G′,C3,Φ′) so that

G′ ⊆ G, V (G) = V (G′), and Φ′ : C3 → Aut(G′) is defined by Φ′(x) = Φ(x)

for all x ∈ C3, because an isostatic framework in R(G′,C3,Φ′) must satisfy

jΦ′(C3) = 0.

..

.

. ...

.

. .

..

.

Figure 5.52: Infinitesimally rigid frameworks in R(G,C3,Φ) with jΦ(C3) = 1.

These frameworks cannot contain an isostatic framework that has the same

joints and also C3 symmetry.

In order to obtain characterizations of (C3, Φ)-generically infinitesimally

rigid graphs in cases where jΦ(C3) 6= 0, one needs to apply methods beyond

those developed in this thesis. One might ask, for example, whether the

Lovasz-Yemini matroid partition algorithm (see [48]) can be appropriately

symmetrized. These questions, however, have not yet been explored.

273

Chapter 6

Symmetry as a sufficient

condition for a flex

Recall from Section 2.2.2 that if a framework (G, p) is infinitesimally rigid,

then (G, p) is also rigid. The converse of this result does not hold. However,

in the end of Section 2.2.5 we summarized some results which assert that

under certain conditions, the existence of an infinitesimal flex of (G, p) also

implies the existence of a flex of (G, p). In fact, it follows from Theorem

2.2.15 and Corollary 2.2.16 that for ‘almost all’ realizations of the graph G,

rigidity and infinitesimal rigidity are equivalent. If (G, p) is a symmetric

framework with a non-trivial point group, however, then the joints of (G, p)

typically lie in special, non-generic (and frequently also non-regular) posi-

tions, so that these results of Section 2.2.5 cannot be applied to (G, p).

In this chapter, we show that if (G, p) is a symmetric framework in the

set R(G,S,Φ), where S is a non-trivial symmetry group and Φ : S → Aut(G)

is a homomorphism, then the symmetry of (G, p) can be exploited to obtain

274

sufficient conditions for the existence of a flex of (G, p). We establish these

conditions by symmetrizing the methods in [3] and by using the fact that the

rigidity matrix of (G, p) can be block-diagonalized as described in Section

4.1.3. As a corollary of these results, one obtains the Proposition 1 stated

(but not proven) in [43].

In Section 6.3, we use these symmetry-adapted results to prove the exis-

tence of a flex for a number of interesting and famous examples of symmetric

frameworks in both 2D and 3D.

The structures analyzed in [35] and [64] can also be proven to be flexible

using the methods of this chapter.

The development of a symmetry-based rigidity analysis not only yields

additional sufficient conditions for the existence of a flex of a given frame-

work, but it also enables us to obtain valuable information about the sym-

metry properties of a detected flex. In particular, using the results of this

chapter, we can determine whether a framework (G, p) in R(G,S,Φ) possesses

a ‘symmetry-preserving’ flex, i.e., a flex which moves the joints of (G, p) on

differentiable displacement paths in such a way that all the resulting frame-

works remain in the set R(G,S,Φ).

Each of the frameworks in Section 6.3 which we prove to be flexible with

the results of this chapter possesses such a symmetry-preserving flex, and it

is precisely this kind of flex that the new methods detect in each case. While

detection of these flexes is not new, the verification of the flexes uses this

new approach, and is much simpler than previous methods. New flexes can

also be detected, and some will be presented in [58]. See also Chapter 7 for

further comments.

275

6.1 Alternate definitions of rigidity via the

edge function

We begin by introducing the edge function of a graph G and then refor-

mulating some of the basic definitions in rigidity theory stated in Section 2.2

in terms of this function. The introduction of symmetric versions of these

‘reformulated’ definitions will allow us to examine the rigidity properties of

symmetric frameworks in Sections 6.2 and 6.3.

Definition 6.1.1 Let G be a graph with V (G) = v1, v2, . . . , vn. For a fixed

ordering of the edges of G, we define the edge function fG : Rdn → R|E(G)| by

fG

(p(v1), . . . , p(vn)

)=

(. . . , ‖p(vi)− p(vj)‖2, . . .

),

where vi, vj ∈ E(G) and p(vi) ∈ Rd for all i = 1, . . . , n.

If (G, p) is a d-dimensional framework with n vertices, then f−1G

(fG(p)

)

is the set of all configurations q of n points in Rd with the property that

corresponding bars of the frameworks (G, p) and (G, q) have the same length.

In particular, we clearly have f−1Kn

(fKn(p)

) ⊆ f−1G

(fG(p)

), where Kn is the

complete graph on V (G).

The definitions of a motion, a flex, and a rigid motion of (G, p) (see

Definitions 2.2.3 and 2.2.4) can be rewritten in terms of the edge function of

G as follows.

Definition 6.1.2 Let G be a graph with n vertices and let (G, p) be a frame-

work in Rd. A motion of (G, p) is a differentiable path x : [0, 1] → Rdn such

276

that x(0) = p and x(t) ∈ f−1G

(fG(p)

)for all t ∈ [0, 1].

A motion x of (G, p) is a rigid motion if x(t) ∈ f−1Kn

(fKn(p)

)for all

t ∈ [0, 1] and a flex of (G, p) if x(t) /∈ f−1Kn

(fKn(p)

)for all t ∈ (0, 1].

The next result gives some alternate definitions of a flexible framework

all of which are equivalent to Definition 2.2.5, as shown in [3].

Theorem 6.1.1 [3] Let (G, p) be a framework in Rd with n vertices. The


(i) (G, p) is flexible;

(ii) there exists a flex x : [0, 1] → Rdn of (G, p);

(iii) there exists a motion x : [0, 1] → Rdn of (G, p) such that x(t) /∈f−1

Kn

(fKn(p)

)for some t ∈ (0, 1];

(iv) for every neighborhood Np of p ∈ Rdn, we have f−1Kn

(fKn(p)

) ∩ Np &

f−1G

(fG(p)

) ∩Np.

Remark 6.1.1 In Definition 6.1.2, we may replace the term ‘differentiable

path’ by the terms ‘continuous path’ or ‘analytic path’. The fact that all of

these definitions are equivalent is a consequence of some basic results from

algebraic geometry [3, 54, 83].

Let fG : Rdn → R|E(G)| be the edge function of a graph G. Then it is

an easy but important observation that the rigidity matrix of a framework

(G, p) is (up to a constant) the Jacobian matrix dfG(p) of fG, evaluated at

the point p ∈ Rdn.

277

Recall from Section 2.2.5 that a point p ∈ Rdn is said to be a regular

point of a graph G if there exists a neighborhood Np of p in Rdn so that

rank(R(G, p)

) ≥ rank(R(G, q)

)for all q ∈ Np. This definition may now

also be reformulated in terms of the edge function fG as follows.

Definition 6.1.3 A point p ∈ Rdn is a regular point of a graph G if

there exists a neighborhood Np of p in Rdn so that rank(dfG(p)

)=

max rank(dfG(q)

)| q ∈ Np.

6.2 Detection of symmetric flexes

Our main goal in this section is to find sufficient conditions for the exis-

tence of a ‘symmetry-preserving’ flex of a symmetric framework. As we will

see, the existence of a flex that preserves some, but not all of the symmetries

of a given framework can be predicted in an analogous way.

Let G be a graph with V (G) = v1, . . . , vn, S be a symmetry group in

dimension d with r pairwise non-equivalent irreducible linear representations

I1, . . . , Ir, Φ : S → Aut(G) be a homomorphism, and (G, p) be a framework

in R(G,S,Φ). In this chapter, I1 will always denote the trivial irreducible linear

representation of S, i.e., I1 denotes the linear representation of degree one

with the property that I1(x) is the identity transformation for all x ∈ S.

Recall from Section 4.1.3 that V(I1)e denotes the H ′

e-invariant subspace of

Rdn which corresponds to I1. So, p ∈ V(I1)e if and only if H ′

e(x)(p) = p for all

x ∈ S.

Further, recall from Section 3.2 that if (G, p) is a framework in R(G,S,Φ),

278

then p ∈ Rdn is an element of the subspace U =⋂

x∈S Lx,Φ of Rdn, where

Lx,Φ = ker(M(x) −PΦ(x)

)for all x ∈ S.

Note that it follows immediately from the definitions of U and V(I1)e that

U = V(I1)e , because p ∈ U if and only if M(x)p = PΦ(x)p for all x ∈ S if and

only if (PΦ(x))TM(x)p = p for all x ∈ S if and only if He(x)p = p for all

x ∈ S if and only if p ∈ V(I1)e .

So, since we are interested in flexes of (G, p) ∈ R(G,S,Φ) that preserve the

symmetry of (G, p), we need to restrict the edge functions fG of G and fKn of

Kn to the subspace V(I1)e of Rdn. In the following, we let fG : V

(I1)e → R|E(G)|

denote the restriction of fG to V(I1)e , and fKn : V

(I1)e → R(n

2) denote the

restriction of fKn to V(I1)e . The Jacobian matrices of fG and fKn , evaluated

at a point p ∈ V(I1)e , are denoted by dfG(p) and dfKn(p), respectively.

Definition 6.2.1 An element p ∈ V(I1)e is said to be a regular point of G in

V(I1)e if there exists a neighborhood Np of p in V

(I1)e so that rank

(dfG(p)

)=

max rank(dfG(q)

)| q ∈ Np. A regular point of Kn in V(I1)e is defined

analogously.

Definition 6.2.2 An (S, Φ)-symmetry-preserving flex of a framework

(G, p) ∈ R(G,S,Φ) is a differentiable path x : [0, 1] → V(I1)e such that x(0) = p

and x(t) ∈ f−1G

(fG(p)

) \ f−1Kn

(fKn(p)

)for all t ∈ (0, 1].

Lemma 6.2.1 Let G be a graph, S be a symmetry group, Φ : S →Aut(G) be a homomorphism, and (G, p) be a framework in R(G,S,Φ). If p

is a regular point of G in V(I1)e , then there exists a neighborhood Np of

p in V(I1)e such that f−1

G

(fG(p)

) ∩ Np is a smooth manifold of dimension

dim(V

(I1)e

)− rank(dfG(p)

).

279

Proof. The result follows immediately from Proposition 2 (and subsequent

remark) in [3]. ¤

Theorem 6.2.2 Let G be a graph with n vertices, S be a symmetry group,

Φ : S → Aut(G) be a homomorphism, and (G, p) be a framework in R(G,S,Φ).

If p is a regular point of G in V(I1)e and also a regular point of Kn in V

(I1)e ,

then

(i) rank(dfG(p)

)= rank

(dfKn(p)

)if and only if (G, p) has no (S, Φ)-

symmetry-preserving flex;

(ii) rank(dfG(p)

)< rank

(dfKn(p)

)if and only if (G, p) has an (S, Φ)-

symmetry-preserving flex.

Proof. Since p is a regular point of both G and Kn in V(I1)e , it follows from

Lemma 6.2.1 that there exist neighborhoods Np and N ′p of p in V

(I1)e so that

f−1G

(fG(p)

)∩Np is a manifold of dimension dim(V

(I1)e

)− rank(dfG(p)

)and

f−1Kn

(fKn(p)

) ∩N ′p is a manifold of dimension dim

(V

(I1)e

)− rank(dfKn(p)

).

Since f−1Kn

(fKn(p)

) ∩ N ′′p is a submanifold of f−1

G

(fG(p)

) ∩ N ′′p , where N ′′

p =

Np ∩N ′p, it follows that

rank(dfG(p)

) ≤ rank(dfKn(p)

).

Clearly, rank(dfG(p)

)= rank

(dfKn(p)

)if and only if there exists a neigh-

borhood N∗p of p in V

(I1)e such that f−1

Kn

(fKn(p)

) ∩ N∗p = f−1

G

(fG(p)

) ∩ N∗p .

Therefore, if rank(dfG(p)

)= rank

(dfKn(p)

), then there does not exist an

(S, Φ)-symmetry-preserving flex of (G, p).

If rank(dfG(p)

)< rank

(dfKn(p)

), then every neighborhood of p in V

(I1)e

280

contains elements of f−1G

(fG(p)

) \ f−1Kn

(fKn(p)

), and hence, by the proof of

Proposition 1 in [3] (and references therein), there exists an (S, Φ)-symmetry-

preserving flex of (G, p). This completes the proof. ¤

In order to make further use of Theorem 6.2.2, we need the following

fundamental observations.

Recall from Section 4.1.3 that with respect to the bases Be and Bi, the

rigidity matrix of a framework (G, p) in R(G,S,Φ) has the block form

R(G, p) =

R1(G, p) 0

. . .

0 Rr(G, p)

, (6.1)

where for t = 1, . . . , r, the block Rt(G, p) corresponds to the irreducible

linear representation It of S, and the size of the block Rt(G, p) depends

on the dimensions of the subspaces V(It)e of Rdn and V

(It)i of R|E(G)|. (In

particular, the block Rt(G, p) is an empty matrix if and only if both of the

coefficients λt and µt in equations (4.2) and (4.6) are equal to zero.)

Since with respect to the bases Be and Bi, the Jacobian matrix of fG,

evaluated at p, is (up to a constant) the matrix R(G, p), it follows that with

respect to the bases B(I1)e and Bi, the Jacobian matrix of fG, evaluated at

the point p ∈ V(I1)e , is (up to a constant) the matrix

R1(G, p)

0

...

0

.

Thus, we have

rank(R1(G, p)

)= rank

(dfG(p)

). (6.2)

281

Furthermore, note that if Kn is the complete graph on the vertex set

V (G), then with respect to the bases Be and Bi, where Bi is an appropriate

extension of the basis Bi, the rigidity matrix of (Kn, p) has a block form

analogous to the one of R(G, p) in (6.1), namely

R(Kn, p) =

R1(Kn, p) 0

. . .

0 Rr(Kn, p)

.

Clearly, Rt(G, p) is a submatrix of Rt(Kn, p) for all t = 1, . . . , r. Moreover,

analogously to (6.2), we have

rank(R1(Kn, p)

)= rank

(dfKn(p)

). (6.3)

Recall from Definition 4.1.8 that if u ∈ V(I1)e , then u is said to be sym-

metric with respect to I1. So, if we think of the vector u ∈ Rdn as a set

of displacement vectors with one vector at each joint of (G, p) ∈ R(G,S,Φ),

then u is symmetric with respect to I1 if and only if all of the displacement

vectors remain unchanged under all symmetry operations in S. A vector

u ∈ Rdn that is symmetric with respect to I1 can therefore also be termed

fully (S, Φ)-symmetric [35, 43] (see also Figure 6.1).

Theorem 6.2.3 Let G be a graph, S be a symmetry group in dimension d,

Φ : S → Aut(G) be a homomorphism, and (G, p) be a framework in R(G,S,Φ)

with the property that the points p(v), v ∈ V (G), span all of Rd. If p is a

regular point of G in V(I1)e and also a regular point of Kn in V

(I1)e and there

exists a fully (S, Φ)-symmetric infinitesimal flex of (G, p), then there also

exists an (S, Φ)-symmetry-preserving flex of (G, p).

282

.. .

.

.(a)

.

...

.

.

.

.(b)

.

. .

.

. .

.

.(c)

Figure 6.1: Fully (S, Φ)-symmetric infinitesimal motions of frameworks: (a)

a fully (Cs, Φ)-symmetric infinitesimal rigid motion of (K3, p) ∈ R(K3,Cs,Φ);

(b) a fully (Cs, Φ)-symmetric infinitesimal flex of (K3,3, p) ∈ R(K3,3,Cs,Φ); (c) a

fully (C3, Φ)-symmetric infinitesimal flex of (Gtp, p) ∈ R(Gtp,C3,Φ). Since each

of the above frameworks is an injective realization, the type Φ is uniquely

determined in each case.

Proof. Let Kn be the complete graph on V (G). Since (G, p) has a fully

(S, Φ)-symmetric infinitesimal flex and the points p(v), v ∈ V (G), span all

of Rd, we have nullity(R1(G, p)

)> nullity

(R1(Kn, p)

), and hence

rank(R1(G, p)

)< rank

(R1(Kn, p)

). (6.4)

Since, by (6.2), we have rank(R1(G, p)

)= rank

(dfG(p)

)and, by (6.3), we

have rank(R1(Kn, p)

)= rank

(dfKn(p)

), it follows from (6.4) that

rank(dfG(p)

)< rank

(dfKn(p)

).

The result now follows from Theorem 6.2.2. ¤

The above results concerning the subspace V(I1)e of Rdn may be transferred

analogously to the affine subspaces of Rdn of the form p + V(It)e , where t 6= 1.

283

More precisely, if we define a point q ∈ p + V(It)e to be a regular point of

a graph G in p + V(It)e if there exists a neighborhood Nq of q in p + V

(It)e so

that rank(dfG(q)

)= max rank

(dfG(q′)

)| q′ ∈ Nq, where fG denotes the

restriction of the edge function fG to p+V(It)e , then the following results can

be proved completely analogously to the Theorems 6.2.2 and 6.2.3.

Theorem 6.2.4 Let G be a graph with n vertices, S be a symmetry group,

Φ : S → Aut(G) be a homomorphism, and (G, p) be a framework in R(G,S,Φ).

If p is a regular point of G in p + V(It)e and also a regular point of Kn in

p + V(It)e , then

(i) rank(dfG(p)

)= rank

(dfKn(p)

)if and only if (G, p) does not have a

flex x with x(t) ∈ p + V(It)e for all t ∈ [0, 1];

(ii) rank(dfG(p)

)< rank

(dfKn(p)

)if and only if (G, p) has a flex x with

x(t) ∈ p + V(It)e for all t ∈ [0, 1].

Theorem 6.2.5 Let G be a graph, S be a symmetry group in dimension d,


with the property that the points p(v), v ∈ V (G), span all of Rd. If p is a

regular point of G in p+V(It)e and also a regular point of Kn in p+V

(It)e and

there exists an infinitesimal flex u of (G, p) with u ∈ V(It)e , then there also

exists a flex x of (G, p) with x(t) ∈ p + V(It)e for all t ∈ [0, 1].

Note that if we define ker (It) = x ∈ S| It(x) = id, where id is the

identity transformation, then ker (It) is a normal subgroup of S (see [42],

for example). Therefore, Theorems 6.2.4 and 6.2.5 provide us with sufficient

284

conditions for the existence of a flex of (G, p) that preserves the sub-symmetry

of (G, p) given by ker (It) and Φ|ker (It).

An important property of the subspace V(I1)e which does not hold for

the affine subspaces p + V(It)e , where t 6= 1, is that, by Corollary 4.1.3, for

every q ∈ V(I1)e , the rigidity matrix R(G, q) has the same block structure

as the rigidity matrix R(G, p). Thus, p ∈ V(I1)e is a regular point of G in

V(I1)e if and only if there exists a neighborhood Np of p in V

(I1)e so that

rank(R1(G, p)

) ≥ rank(R1(G, q)

)for all q ∈ Np.

Similarly, p ∈ V(I1)e is a regular point of Kn in V

(I1)e if and only if

there exists a neighborhood Np of p in V(I1)e so that rank

(R1(Kn, p)

) ≥rank

(R1(Kn, q)

)for all q ∈ Np.

The fact that regular points of G and Kn in V(I1)e can be characterized in

this way is essential to proving all the remaining results of this section. These

results will turn out to be very useful for practical applications of Theorem

6.2.3, as we will see in the final section of this chapter.

Theorem 6.2.6 Let G be a graph with n vertices, S be a symmetry group

in dimension d, Φ : S → Aut(G) be a homomorphism, and (G, p) be a

framework in R(G,S,Φ). If the points p(v), v ∈ V (G), span all of Rd, then p

is a regular point of Kn in V(I1)e .

Proof. Since the points p(v), v ∈ V (G), span all of Rd, there exists a

neighborhood Np of p in V(I1)e so that for all q ∈ Np, the points q(v), v ∈

V (G), also span all of Rd. It follows from the results of Section 4.2.1 that for

all q ∈ Np, the dimension of the subspace of Rdn consisting of all fully (S, Φ)-

symmetric infinitesimal rigid motions of (G, p) is equal to the dimension of

285

the subspace of Rdn consisting of all fully (S, Φ)-symmetric infinitesimal rigid

motions of (G, q). Therefore, we have rank(R1(Kn, p)

)= rank

(R1(Kn, q)

)

or equivalently, by (6.3), rank(dfKn(p)

)= rank

(dfKn(q)

)for all q ∈ Np.

Thus, p is a regular point of Kn in V(I1)e . ¤

By Theorem 6.2.6, the condition that p is a regular point of Kn in V(I1)e

may be omitted in Theorem 6.2.3.

Theorem 6.2.7 Let G be a graph, S be a symmetry group, Φ : S → Aut(G)

be a homomorphism, and (G, p) be a framework in R(G,S,Φ). If p is (S, Φ)-

generic, then p is a regular point of G in V(I1)e .

Proof. Suppose G is a graph with n vertices and S is a symmetry group

in dimension d with r pairwise non-equivalent irreducible representations

I1, . . . , Ir. Fix a basis BU = u1, . . . , uk of U = V(I1)e =

⋂x∈S Lx,Φ and let

p = t1u1 + . . . + tkuk. Then the symmetry-adapted indeterminate rigidity

matrix RBU(n, d) for R(G,S,Φ) (corresponding to BU) is a matrix in the vari-

ables t′1, . . . , t′k. More precisely, the entries of RBU

(n, d) are elements of the

quotient field of the integral domain R[t′1, . . . , t′k]. Over this field we can again

do linear algebra. We let R(G)BU

(n, d) denote the submatrix of RBU(n, d) that

corresponds to the submatrix R(G, p) of R(Kn, p), i.e., R(G)BU

(n, d) is obtained

from RBU(n, d) by deleting those rows that do not correspond to edges of G.

If we replace each variable t′i in R(G)BU

(n, d) with ti, then, by Remark 3.2.1, we

obtain the rigidity matrix R(G, p). Therefore,

rank(R(G, p)

) ≤ rank(R

(G)BU

(n, d)).

Since (G, p) is (S, Φ)-generic, we also have

rank(R(G, p)

) ≥ rank(R

(G)BU

(n, d)),

286

and hence

rank(R(G, p)

)= rank

(R

(G)BU

(n, d)). (6.5)

Now, let Te be the matrix of the basis transformation from the canonical basis

of the R-vector space Rdn to the basis Be, and let Ti be the matrix of the basis

transformation from the canonical basis of the R-vector space R|E(G)| to the

basis Bi, so that the matrix R(G, p) = T−1i R(G, p)Te is block-diagonalized as

in (6.1). Then, by Corollary 4.1.3, the matrix R(G)BU

(n, d) = T−1i R

(G)BU

(n, d)Te

has the same block form as R(G, p). For t = 1, . . . , r, let R(G)t (n, d) denote

the block of R(G)BU

(n, d) that corresponds to the block Rt(G, p) of R(G, p).

Since the rank of a matrix is invariant under a basis transformation, and

since the rank of a block-diagonalized matrix is equal to the sum of the

ranks of its blocks, it follows from equation (6.5) thatr∑

t=1

rank(Rt(G, p)

)= rank

(R(G, p)

)

= rank(R(G, p)

)

= rank(R

(G)BU

(n, d))

= rank(R

(G)BU

(n, d))

=r∑

t=1

rank(R

(G)t (n, d)

).

Since we clearly have rank(Rt(G, p)

) ≤ rank(R

(G)t (n, d)

)for each t, it fol-

lows that rank(Rt(G, p)

)= rank

(R

(G)t (n, d)

)for each t. This gives the

result. ¤

Corollary 6.2.8 Let G be a graph, S be a symmetry group in dimension d,


with the property that the points p(v), v ∈ V (G), span all of Rd. If (G, p)

is (S, Φ)-generic and (G, p) has a fully (S, Φ)-symmetric infinitesimal flex,

then there also exists an (S, Φ)-symmetry-preserving flex of (G, p).

287

Proof. The result follows immediately from Theorems 6.2.3, 6.2.6, and 6.2.7.

¤

In Section 6.3, we will use Corollary 6.2.8 to prove the existence of an

(S, Φ)-symmetry-preserving flex for a variety of symmetric frameworks.

Note that it follows from our discussion in Chapter 3 that the frameworks

in Figures 6.1 (b) and (c) are not (S, Φ)-generic, so that Corollary 6.2.8

cannot be applied to these frameworks. In fact, it can be verified that none of

the frameworks in Figure 6.1 possesses any flex, let alone an (S, Φ)-symmetry-

preserving flex.

Note that Corollary 6.2.8 is a symmetrized version of Corollary 2.2.16 in

Section 2.2.5. Next, we show that a symmetrized version of Corollary 2.2.17

can also be obtained from the previous results.

Corollary 6.2.9 Let G be a graph, S be a symmetry group in dimension d,


with the property that the points p(v), v ∈ V (G), span all of Rd. If the

block R1(G, p) of the block-diagonalized rigidity matrix R(G, p) has linearly

independent rows and (G, p) has a fully (S, Φ)-symmetric infinitesimal flex,

then there also exists an (S, Φ)-symmetry-preserving flex of (G, p).

Proof. Since the block R1(G, p) has linearly independent rows, p is a regular

point of G in V(I1)e . The result now follows from Theorems 6.2.3 and 6.2.6.

¤

Corollary 6.2.9 confirms the observation made by R. Kangwai and S.

Guest in [43]. Note that the condition that the block matrix R1(G, p) has

linearly independent rows is equivalent to the condition that the framework

288

(G, p) has no fully (S, Φ)-symmetric non-zero self-stress, i.e., a non-zero self-

stress in the subspace V(I1)i of R|E(G)|. In particular, it follows that if (G, p) is

independent (i.e., (G, p) does not possess any non-zero self-stress) and there

exists a fully (S, Φ)-symmetric infinitesimal flex of (G, p), then there also

exists an (S, Φ)-symmetry-preserving flex of (G, p).

In order to apply Corollary 6.2.9 to a given framework (G, p), we need

to compute the rank of the submatrix block R1(G, p). This can be done by

finding the block-diagonalized rigidity matrix R(G, p) with the methods and

algorithms described in [24, 50], for example.

The rank of the submatrix block R1(G, p) can also be determined directly

by finding the rank of an appropriate ‘orbit rigidity matrix’ whose columns

and rows correspond to a set of representatives for the orbits of the group

action from S × V (G) to V (G) that sends (x, v) to Φ(x)(v) (see Remark

3.2.2) and a set of representatives for the orbits of the group action from

S × E(G) to E(G) that sends (x, e) to Φ(x)(e), respectively. The kernel of

this matrix is the space of fully (S, Φ)-symmetric infinitesimal motions and

the cokernel of this matrix is the space of fully (S, Φ)-symmetric self-stresses.

Further details will be presented in [58].

6.3 Examples of flexible frameworks

We now consider a number of examples of flexible symmetric frameworks

in dimensions 2 and 3. For each of these frameworks, we use the results of

the previous section to prove the existence of an (S, Φ)-symmetry-preserving

flex.

289

6.3.1 Examples in 2D

We begin with a very simple example of an independent framework in R2

with point group Cs that possesses a symmetry-preserving flex.

Example 6.3.1 Let K2,2 be the complete bipartite graph with partite sets

v1, v2 and v3, v4, Cs be a symmetry group in dimension 2, and Φ : Cs →Aut(K2,2) be the homomorphism defined by

Φ(Id) = id

Φ(s) = (v1)(v2)(v3 v4).

The framework (K2,2, p) ∈ R(K2,2,Cs,Φ) shown in Figure 6.2 is clearly inde-

...p1

..p2

..p3 . .p4

Figure 6.2: A fully (Cs, Φ)-symmetric infinitesimal flex of the independent

framework (K2,2, p).

pendent and the symmetry-extended version of Maxwell’s rule applied to

(K2,2, p), Cs, and Φ gives the counts

XQ = (4, 2)× (2, 0)− (2, 0)− (1,−1) = (5, 1) = 3A′ + 2A′′

Xi = (4, 0) = 2A′ + 2A′′.

So, we have dim(V

(A′)Q

)= 3 > 2 = dim

(V

(A′)i

), and hence, by the results of

290

Section 4.2.2, (K2,2, p) has a fully (Cs, Φ)-symmetric infinitesimal flex. Thus,

by Corollary 6.2.9, (K2,2, p) also has a (Cs, Φ)-symmetry-preserving flex.

In the following two examples, we consider realizations of the complete

bipartite graph K4,4. This graph is generically rigid and ‘over-braced’ (i.e.,

|E(K4,4)| > 2|V (K4,4)| − 3) in dimension 2. However, we will show that

under certain symmetry conditions, 2-dimensional realizations of K4,4 not

only become infinitesimally flexible, but even flexible.

Example 6.3.2 Let K4,4 be the complete bipartite graph with partite sets

v1, v2, v3, v4 and v5, v6, v7, v8, C2v = Id, C2, sh, sv be a symmetry group

in dimension 2, and Φ : C2v → Aut(K4,4) be the homomorphism defined by

Φ(Id) = id

Φ(C2) = (v1 v3)(v2 v4)(v5 v7)(v6 v8)

Φ(sh) = (v1 v4)(v2 v3)(v5 v8)(v6 v7)

Φ(sv) = (v1 v2)(v3 v4)(v5 v6)(v7 v8).

Note that if we choose positions p1 and p5 in R2 for the vertices v1 and v5,

then the positions pi for all the remaining joints of a framework (K4,4, p) ∈R(K4,4,C2v ,Φ) are determined by the symmetry imposed by C2v and Φ. More-

over, for almost all placements of v1 and v5 off the two mirror lines corre-

sponding to sh and sv, we obtain a (C2v, Φ)-generic realization of K4,4 (see

also Remark 3.2.2). Such a (C2v, Φ)-generic realization (K4,4, p) is depicted

in Figure 6.3.

We have |E(K4,4)| = 16 and 2|V (K4,4)| − 3 = 13, so that Maxwell’s rule

detects three linearly independent self-stresses of (K4,4, p), but no infinites-

imal flex. Further, the symmetry-extended version of Maxwell’s rule gives

291

.

..p1 . .p2

. .p3..p4

..p5 . .p6

. .p7..p8

.sv

.sh

Figure 6.3: A fully (C2v, Φ)-symmetric infinitesimal flex of a (C2v, Φ)-generic

realization of K4,4.

the counts

XQ = (13, 1, 1, 1) = 4A1 + 3A2 + 3B1 + 3B2

Xi = (16, 0, 0, 0) = 4A1 + 4A2 + 4B1 + 4B2.

So, by the results of Section 4.2.2, we may conclude that (K4,4, p) has at least

one non-zero self-stress which is symmetric with respect to A2, at least one

non-zero self-stress which is symmetric with respect to B1, and at least one

non-zero self-stress which is symmetric with respect to B2 (see also Appendix

A). However, like Maxwell’s original rule, the symmetry-extended version of

Maxwell’s rule does not detect any infinitesimal flex of (K4,4, p). To detect

a fully (C2v, Φ)-symmetric infinitesimal flex of (K4,4, p) (as shown in Figure

6.3), we therefore need to analyze (K4,4, p) geometrically.

Since the intersection points of all three pairs of opposite sides of the

hexagon shown in Figure 6.4 (a) are collinear, it follows from the converse

of Pascal’s Theorem that the point p2 lies on the unique conic section deter-

mined by the five points p1, p3, p5, p6, p7. Similarly, it can be shown that the

remaining two points p4 and p8 must also lie on the same conic section (see

292

Figures 6.4 (b) and (c)).

.

..p5

..p1 . .p2

. .p3

..p6

..p7

.sv

.sh

.(a)

.

..p5

..p1

..p8

..p3

..p6

..p7

.

.

.sv

.sh

.(b)

.

..p5

..p6

..p2

..p8

..p4.

.p7

.sv

.sh

.(c)

Figure 6.4: Illustration of the proof that the joints of (K4,4, p) lie on a conic

section.

By the results in [75], it follows that (K4,4, p) possesses an infinitesimal

flex. In fact, it is shown in [75] that this infinitesimal flex assigns a velocity

vector that is directed outward from (and perpendicular to the tangent of)

the conic section containing the joints of (K4,4, p) to each joint correspond-

ing to one partite set of K4,4, and a velocity vector that is directed inward

from (and perpendicular to the tangent of) this conic section to each joint

corresponding to the other partite set of K4,4 (see also Figure 6.3).

Due to the symmetric positions of the joints of (K4,4, p), this infinitesimal

flex of (K4,4, p) is fully (C2v, Φ)-symmetric. Therefore, it follows from Corol-

lary 6.2.8 that (K4,4, p) also has a (C2v, Φ)-symmetry-preserving flex. This

flex is also known as ‘Bottema’s mechanism’ [8].

Remark 6.3.1 Note that some of the configurations that lie on the path of

293

the (C2v, Φ)-symmetry-preserving flex of (K4,4, p) are not (C2v, Φ)-generic.

For example, the symmetry-preserving flex of (K4,4, p) passes through a

configuration q with the property that the four points q1, q4, q5, q8, as well as

the four points q2, q3, q6, q7, are collinear (with each set of four points lying

on a line that is parallel to the mirror corresponding to sv). The framework

(K4,4, q) is therefore clearly not (C2v, Φ)-generic. However, by computing the

rank of R1(K4,4, q) and showing that it is equal to the rank of R1(K4,4, p),

where p is (C2v, Φ)-generic, the configuration q can be proven to be a regular

point of K4,4 in V(A1)e , where A1 is the trivial irreducible representation of

C2v. So, Theorem 6.2.3 can be used in this case to prove the existence of a

(C2v, Φ)-symmetry-preserving flex of (K4,4, q).

Note that the same reasoning can also be applied to any other non-

(C2v, Φ)-generic realization of K4,4 that is reached by the flex of (K4,4, p).

Example 6.3.3 Figure 6.5 shows another realization of the graph K4,4. This

framework is a (C2v, Ψ)-generic realization of K4,4, where C2v is the same

symmetry group as in Example 6.3.2 and Ψ : C2v → Aut(K4,4) is defined by

Ψ(Id) = id

Ψ(C2) = (v1 v4)(v2 v3)(v5 v8)(v6 v7)

Ψ(sh) = (v1)(v2)(v3)(v4)(v5 v8)(v6 v7)

Ψ(sv) = (v1 v4)(v2 v3)(v5)(v6)(v7)(v8).

The joints of the framework (K4,4, p) shown in Figure 6.5 again lie on a

conic section (namely the two mirrors), so that we may again conclude that

(K4,4, p) has an infinitesimal flex [75].

In fact, the infinitesimal flex in Figure 6.5 is clearly fully (C2v, Ψ)-

294

...p1

..p2 ..p3 ..p4

..p5

.

.

..p8

.p6

.p7

.sv

.sh

Figure 6.5: A fully (C2v, Ψ)-symmetric infinitesimal flex of a (C2v, Ψ)-generic


symmetric, so that it again follows from Corollary 6.2.8 that (K4,4, p) has

a (C2v, Ψ)-symmetry-preserving flex.

Note that this particular analysis extends to arbitrary K2m,2n realized

with these symmetries. In fact, all realizations of Km,n whose joints lie on

two perpendicular lines are flexible.

Remark 6.3.2 For the frameworks in Examples 6.3.2 and 6.3.3, the (C2v, Φ)-

generic and (C2v, Ψ)-generic failure to be rigid, respectively, is not predicted

by the combinatorial counts of Maxwell’s rule or the symmetry-extended

version of Maxwell’s rule. This demonstrates that these combinatorial counts

alone are not sufficient for rigidity.

6.3.2 Examples in 3D

In his famous paper from 1897, the French engineer R. Bricard proved

that if an octahedron in 3-space with no self-intersecting faces is realized as a

295

framework by placing bars along edges, and joints at vertices, then this frame-

work must be rigid [9]. Moreover, he showed that there exist three distinct

types of octahedra with self-intersecting faces whose realizations as frame-

works are flexible. Two of these three types of octahedra possess non-trivial

symmetries: Bricard octahedra of the first type have a half-turn symmetry

and Bricard octahedra of the second type have a mirror symmetry. In the

next example, we consider both of these types of symmetric Bricard octa-

hedra (as well as octahedra with dihedral symmetry) and use the results of

Section 6.2 to not only show that they are flexible, but also that they possess

a symmetry-preserving flex.

Various other treatments of the Bricard octahedra can be found in [5, 63],

for example. R. Connelly’s celebrated counterexample to Euler’s rigidity con-

jecture from 1776 (see [23]) is also based on a flexible Bricard octahedron (of

the first type) [11, 13].

Example 6.3.4 (a) Let G be the graph of the octahedron, C2 be a symmetry

group in dimension 3, and Φa : C2 → Aut(G) be the homomorphism defined

by

Φa(Id) = id

Φa(C2) = (v1 v3)(v2 v4)(v5 v6).

If we apply the symmetry-extended version of Maxwell’s rule to the (C2, Φ)-

generic framework (G, p) in Figure 6.10 (a), C2, and Φa, then we obtain the

counts

XQ = (6, 0)× (3,−1)− (3,−1)− (3,−1) = (12, 2) = 7A + 5B

Xi = (12, 0) = 6A + 6B.

296

So, from the coefficients of the trivial irreducible representation A we deduce

that dim(V

(A)Q

)> dim

(V

(A)i

). This says that (G, p) has a fully (C2, Φa)-

symmetric infinitesimal flex (see Section 4.2.2). It follows from Corollary

6.2.8 that (G, p) also has a (C2, Φa)-symmetry-preserving flex.

... ..

..

..

.. ..

.p2

.p3.p4.p1

.p5 .p6

.(a)

... ..

..

..

.. ..

.p2

.p4

.p3.p1

.p5 .p6

.(b)

.

..

....

..

....

.p2

.p4 .p3

.p1

.p5.p6

.(c)

Figure 6.6: Flexible octahedra: with point group C2 (a); with point group Cs

(b); with point group C2v (c).

Since the symmetry-extended version of Maxwell’s rule yields the same

counts for any (C2, Φa)-generic realization of G, any such realization has a

297

(C2, Φa)-symmetry-preserving flex.

(b) Similarly, if we apply the symmetry-extended version of Maxwell’s

rule to the framework in Figure 6.10 (b), the symmetry group Cs, and the

homomorphism Φb : Cs → Aut(G) defined by

Φb(Id) = id

Φb(s) = (v1 v3)(v2)(v4)(v5 v6),

then we obtain the counts

XQ = (6, 2)× (3, 1)− (3, 1)− (3,−1) = (12, 2) = 7A′ + 5A′′

Xi = (12, 0) = 6A′ + 6A′′.

Thus, from the coefficients of the trivial irreducible representation A′ we may

conclude that any (Cs, Φb)-generic realization of G, such as the realization

(G, p) in Figure 6.10 (b), has a (Cs, Φb)-symmetry-preserving flex.

(c) Finally, consider the (C2v, Φc)-generic framework (G, p) in Figure 6.10

(c), where Φc : C2v → Aut(G) is the unique type determined by the injective

realization of G. Although (G, p) is neither (C2, Φa)-generic nor (Cs, Φb)-

generic, we anticipate from the Examples 6.3.4 (a) and (b) that (G, p) pos-

sesses a flex that preserves both the C2 and the Cs symmetry defined in these

examples.

The symmetry-extended version of Maxwell’s rule applied to (G, p), C2v,

and Φc yields

XQ = (6, 0, 4, 2)× (3,−1, 1, 1)− (3,−1, 1, 1)− (3,−1,−1,−1)

= (12, 2, 4, 2) = 5A1 + 2A2 + 3B1 + 2B2

Xi = (12, 0, 4, 0) = 4A1 + 2A2 + 4B1 + 2B2.

298

Thus, by Corollary 6.2.8, the framework (G, p), as well as any other (C2v, Φc)-

generic realization of G, indeed possesses a (C2v, Φc)-symmetry-preserving

flex.

Remark 6.3.3 If G is the graph of the octahedron, Cs is a symmetry group

in dimension 3, and Φd : Cs → Aut(G) is defined by

Φd(Id) = id

Φd(s) = (v2 v4)(v1)(v3)(v5)(v6),

then G is (Cs, Φd)-generically isostatic. The framework (G, p) in Figure 6.7,

for example, is a realization of G in R(G,Cs,Φd) which is isostatic by Cauchy’s

Theorem [10, 22].

.

..

....

..

..

..

.p2

.p4

.p3

.p1

.p5

.p6

Figure 6.7: An isostatic octahedron in R(G,Cs,Φd).

Remark 6.3.4 The rigidity analysis of the symmetric octahedra given in

Examples 6.3.4 (a), (b), and (c) can directly be extended to an analysis of

symmetric frameworks that consist of an arbitrary 2n-gon and two ‘cone-

vertices’ that are linked to each of the joints of the 2n-gon. These kinds of

299

frameworks are also known as ‘double-suspensions’, and are studied in [12],

for example.

Example 6.3.5 (a) Consider the following construction. First, construct

a framework with mirror symmetry like the one shown in Figure 6.8 (a)

by taking a convex octahedron and placing bars along edges, and joints at

vertices. This framework is isostatic by the Theorem of Cauchy [10, 22].

Then attach a ring of eight triangles to this realization of the octahedron as

illustrated in Figure 6.8 (b), so that the resulting framework (G, p) has point

group Cs. If we apply the symmetry-extended version of Maxwell’s rule to

.

.p5 .p6

.p4

.p2

..

..

.. ..

.. ..

.p3.p1

.(a)

.

.p5 .p6

..

..

.. ..

.. ..

.p3.p1

.p2

.. ..

..

..

.p3

.p5

.p4

.p6

.p7 .p8

.p9

.p10

.(b)

Figure 6.8: A fully (Cs, Φa)-symmetric infinitesimal flex of the framework

(G, p) ∈ R(G,Cs,Φa).

(G, p), Cs, and the unique type Φa : Cs → Aut(G) determined by the injective

realization of (G, p), then we obtain the counts

XQ = (10, 4)× (3, 1)− (3, 1)− (3,−1) = (24, 4) = 14A′ + 10A′′

Xi = (24, 2) = 13A′ + 11A′′.

300

Thus, from the coefficients of the trivial irreducible representation A′ of Cs we

deduce that (G, p) has a fully (Cs, Φa)-symmetric infinitesimal flex. There-

fore, by Corollary 6.2.8, (Cs, Φa)-generic realizations of G also have a (Cs, Φa)-


(b) If we apply the symmetry-extended version of Maxwell’s rule to the

framework (G, p) in Figure 6.9, the symmetry group C2v, and the unique type

Φb : C2v → Aut(G) determined by the injective realization of (G, p), then we

obtain the counts

XQ = (10, 2, 4, 4)× (3,−1, 1, 1)− (3,−1, 1, 1)− (3,−1,−1,−1)

= (24, 0, 4, 4) = 8A1 + 4A2 + 6B1 + 6B2

Xi = (24, 0, 2, 2) = 7A1 + 5A2 + 6B1 + 6B2.

Therefore, any (C2v, Φb)-generic realization of G also possesses a (C2v, Φb)-


.

.p1

.p2

..

....

..

..

..

..

....

..

.p3

.p5

.p4

.p6

.p7

.p8.p9

.p10

Figure 6.9: A fully (C2v, Φb)-symmetric infinitesimal flex of the framework

(G, p) ∈ R(G,C2v ,Φb).

301

Remark 6.3.5 If we do not impose any symmetry constraints, then 3-

dimensional realizations of the graph G are in general rigid. In fact, G is

generically 3-isostatic, because it can be constructed from a triangle by a

sequence of vertex 3-additions and edge 3-splits.

Moreover, if Φc : C2 → Aut(G) is the homomorphism that maps the half-

turn in C2 to the automorphism (v1)(v2)(v3 v6)(v4 v5)(v7 v8)(v9 v10) of G, then

G is also (C2, Φc)-generically rigid.

If Φd : Cs → Aut(G) is the homomorphism that maps the reflection in

Cs to the automorphism (v1)(v2)(v4)(v5)(v3 v6)(v7 v10)(v8 v9) of G, then the

symmetry-extended version of Maxwell’s rule applied to a (Cs, Φd)-generic

realization (G, p), Cs, and Φd does not detect any fully (Cs, Φd)-symmetric

infinitesimal flex. However, it does detect an infinitesimal flex of (G, p) which

is symmetric with respect to the irreducible representation A′′ of Cs, so that

Theorem 6.2.5 can be used to further analyze the rigidity of (G, p).

Finally, let C4v = 〈C4, s〉 be the symmetry group that has the group C2v

defined in Example 6.3.5 (b) as a subgroup, and let Φe : C4v → Aut(G) be

the homomorphism defined by

Φe(C4) = (v1)(v2)(v3 v4 v6 v5)(v7 v9 v8 v10)

Φe(s) = (v1)(v2)(v7)(v8)(v3 v5)(v4 v6)(v9 v10).

Further, let (G, p) be a (C4v, Φe)-generic realization of G. Then (G, p) does

not possess any (C4v, Φe)-symmetry-preserving flex. However, (G, p) does

possess a flex which preserves the C2v symmetry defined in Example 6.3.5

(b). In fact, since the symmetry-extended version of Maxwell’s rule, applied

to (G, p), C2v, and Φb, detects a fully (C2v, Φb)-symmetric infinitesimal flex,

and since (G, p) can also be shown to be a regular point of G in V(A1)e , where

302

A1 is the trivial irreducible representation of C2v, the existence of this flex

follows from Theorem 6.2.3.

An alternative approach to detect this flex of (G, p) is to use Theorem

6.2.5 and the fact that the symmetry-extended version of Maxwell’s rule,

applied to (G, p), C4v, and Φe, detects an infinitesimal flex of (G, p) which

is symmetric with respect to the irreducible representation B1 of C4v (i.e., a

fully (C2v, Φb)-symmetric infinitesimal flex).

Note that a similar analysis can also be carried out to show the flexibility

of each of the structures examined in [35] and [64].

Remark 6.3.6 The construction in Example 6.3.5 can easily be generalized

as follows to create an infinite class of frameworks with point groups Cs or

C2v that possess a symmetry-preserving flex.

Instead of a convex octahedron, we may start with a convex ‘double-

suspension’ on a 2n-gon, where n > 2, i.e., a triangulated convex polyhedron

which is constructed from a 2n-gon by joining two new ‘cone-vertices’ to each

of the joints of the 2n-gon (see also Remark 6.3.4). This framework is iso-

static by the Theorem of Cauchy [10, 22]. If we then symmetrically attach

a ring of 4n triangles to this 2n-gon in the analogous way as we attached

the eight triangles to the points p3, p4, p5, p6 of the octahedral framework

in Figure 6.8, then, by using analogous arguments as in Example 6.3.5 (a),

we can prove that the resulting framework with point group Cs possesses a


It is easy to see that this kind of construction can also be used to ob-

tain frameworks with point group C2v that have a symmetry-preserving flex,

provided that n is even.

303

Example 6.3.6 Consider the complete bipartite graph K4,6 with partite

sets v1, v2, v3, v4 and v5, . . . , v10. This graph is generically 3-isostatic.

However, if Cs is a symmetry group in dimension 3 and Φ : Cs → Aut(K4,6)

is the homomorphism defined by

Φ(Id) = id

Φ(s) = (v1)(v2)(v3)(v4)(v5 v8)(v6 v9)(v7 v10),

then the symmetry-extended version of Maxwell’s rule applied to a (Cs, Φ)-

generic realization of K4,6, Cs, and Φ yields the counts

XQ = (10, 4)× (3, 1)− (3, 1)− (3,−1) = (24, 4) = 14A′ + 10A′′

Xi = (24, 0) = 12A′ + 12A′′.

Thus, any (Cs, Φ)-generic realization of K4,6 has a fully (Cs, Φ)-symmetric

...

......

.... ..

....

..

.p1

.p5

.p6

.p7

.p8

.p9

.p10

.p2 .p3.p4

Figure 6.10: A fully (Cs, Φ)-symmetric infinitesimal flex of a (Cs, Φ)-generic


infinitesimal flex. In fact, since dim(V

(A′)Q

)= 14 and dim

(V

(A′)i

)= 12,

any such framework must even possess a 2-dimensional space of fully (Cs, Φ)-

symmetric infinitesimal flexes. So, by Corollary 6.2.8, any (Cs, Φ)-generic

realization of K4,6 also has a (Cs, Φ)-symmetry-preserving flex.

304

Remark 6.3.7 The previous example can be generalized as follows to obtain

another infinite class of frameworks with mirror symmetry that possess a


It is shown in [75] that if a complete bipartite graph is realized in Rd so

that all the joints corresponding to one partite set of the graph lie on a quadric

surface within a hyperplane of Rd, then this realization is infinitesimally

flexible. So, let Km,2n be a complete bipartite graph with partite sets X and

Y of cardinality m and 2n, respectively, where 1 ≤ m ≤ 5 and n ≥ 1; further

let Cs = Id, s be a symmetry group in dimension 3, and let Φ : Cs →Aut(Km,2n) be a homomorphism so that Φ(s) fixes all the vertices in X and

maps each vertex vi ∈ Y to a vertex vj ∈ Y with j 6= i. Any framework in

the set R(Km,2n,Cs,Φ) is then infinitesimally flexible, because for any set of at

most five points in the plane there always exists a conic section containing

these points. In fact, it is easy to check that any such framework must have

a fully (Cs, Φ)-symmetric infinitesimal flex, so that, by Corollary 6.2.8, any

(Cs, Φ)-generic realization of Km,2n also has a (Cs, Φ)-symmetry-preserving

flex which moves all the joints corresponding to the vertices in X within the

mirror plane of s.

305

Chapter 7

Further work

In this thesis, we have developed a range of methods that allow us to

gain further insight into the rigidity and flexibility properties of symmetric

frameworks.

A discussion on how the results of Chapter 4 can be extended to a variety

of other geometric constraint systems, provided that they are presented in

matrix form (via their Jacobian, for example), is given by J.C. Owen and

S.C. Power in [53].

Since for each of the areas of rigidity (namely rigidity, infinitesimal rigid-

ity, static rigidity, and generic rigidity), we were able to derive symmetry-

adapted results for bar and joint frameworks in this thesis, we also do not

expect any major obstacles for developing symmetric versions of diverse re-

lated theories, such as the theory of parallel drawings ([81, 83]), scene analysis

([77, 79, 81, 83]), or the theory of tensegrity frameworks ([17, 54]), for exam-

ple. These possible fields of applications for our methods, however, have not

yet been carefully explored.

306

In the first section of this chapter, we describe a number of conjectures, as

well as some initial results, which indicate that the methods for analyzing bar

and joint frameworks derived in this thesis can also be applied to the rigidity

analysis of several related types of symmetric structures many of which have

important applications in fields such as engineering, chemistry, and biology.

Some additional promising directions for future work on problems relating

to rigidity and symmetry are presented in the rest of this chapter.

7.1 Rigidity of other types of symmetric

structures

7.1.1 Pinned frameworks

In mechanical and structural engineering, one is often interested in the

rigidity and flexibility properties of pinned frameworks, i.e., frameworks that

have some of their joints firmly anchored (‘pinned’) to the ground (see, for

example, [25, 44, 45, 53, 61, 62]). A formal definition of a pinned framework,

as well as some basic results concerning the rigidity of these structures can

be found in [61, 62], for example.

In [25], a symmetry-extended version of Maxwell’s rule is presented for

isostatic pinned frameworks. If we define an external representation (by tak-

ing into account only the unpinned joints) and an internal representation (by

taking into account only the unpinned bars, i.e., the bars that are incident

with at least one unpinned joint) in the same way as in Section 4.1.2, then we

can easily establish a mathematical proof for this rule by slightly modifying

307

the results of Section 4.2. In fact, since a pinned framework does not pos-

sess any infinitesimal rigid motions, a proof of this rule requires significantly

less work than the proof of the symmetry-extended version of Maxwell’s rule

given in Section 4.2.

An alternate proof for the symmetry-extended version of Maxwell’s rule

for pinned frameworks is presented by J.C. Owen and S.C. Power in [53]. In

that paper, the analysis of symmetric pinned frameworks is used as a starting

point to establish symmetry-extended counting rules for various geometric

constraint systems.

By appropriately adapting the analysis in Section 4.3, it is straightfor-

ward to show that the necessary conditions given in the symmetry-extended

version of Maxwell’s rule for a symmetric pinned framework (G, p) to be iso-

static are equivalent to some very simply stated restrictions on the number

of (unpinned) bars and joints of (G, p) that are fixed by various symmetry

operations of (G, p). While we have seen in Section 4.3.2 that there are only

six possible point groups that allow isostatic frameworks in the plane, it

..

.

..

.(a)

..

..

..

..

.

.(b)

Figure 7.1: Isostatic pinned frameworks in the plane: (a) with point group

C4; (b) with point group C4v.

turns out that an isostatic pinned framework can be constructed for any

308

point group in dimension 2.

We conjecture that the standard Laman-type conditions for a pinned

graph G (see [61], for example), together with the additional necessary con-

ditions concerning the number of fixed bars and joints, are also sufficient

for pinned 2-dimensional realizations of G which are as generic as possible

subject to the given symmetry conditions to be isostatic.

In particular, for the symmetry groups C2, C3, and Cs, we claim that the

proofs in Chapter 5 extend directly to proofs of the corresponding symmetric

versions of Laman’s Theorem for pinned frameworks.

7.1.2 Body-bar structures

Recall from Section 2.2.5 that no combinatorial characterization of gener-

ically d-rigid graphs has been found for dimensions d ≥ 3. Faced with these

difficulties, in contrast with the well developed theory in the plane, there has

recently been a careful study of a special class of frameworks, the class of

body-bar frameworks. These structures have a basically complete combina-

torial theory which exhibits all the key theorems and algorithms of the well

understood plane frameworks (see, for example, [66, 72, 81, 82]).

The underlying combinatorial structure for a body-bar framework in d-

space is a multigraph G which allows up to(

d+12

)edges between any pair of

‘vertices’ (bodies). This is motivated by the fact that the space of infinitesi-

mal motions of a full-dimensional rigid body in d-space (such as an isostatic

framework whose joints span all of Rd) has dimension(

d+12

). So, in order to

join two rigid bodies in Rd in such a way that the resulting structure is again

rigid, one needs(

d+12

)properly placed bars.

309

The configuration p of a d-dimensional body-bar framework (G, p) defines

the positions of all the end-points of the bars of (G, p) in Rd (i.e., the attach-

ment points of the bars on the bodies).

For more detailed definitions and further information on body-bar frame-

works, see [66, 72, 81, 82], for example.

The following key result has been proven in all dimensions (see also [82]).

Theorem 7.1.1 (Tay, 1980) [65, 66] Given a multigraph G, the following

are equivalent:

(i) For a generic body-bar configuration in Rd, p, the body-bar framework

(G, p) is isostatic;

(ii) |E(G)| = (d+12

)|V (G)|−(d+12

), and for all non-empty subsets V ′ of V (G)

that induce only the edges in E ′, we have |E ′| ≤ (d+12

)|V ′| − (d+12

);

(iii) E(G) is partitioned into(

d+12

)spanning trees.

There also exists a Henneberg-type characterization of generically d-

isostatic ‘body-bar multigraphs’ which is equivalent to the characterizations

given in Tay’s Theorem. We refer the reader to [29, 65] for details.

For a body-bar framework (G, p) that possesses non-trivial symmetries,

joint work with S. Guest and W. Whiteley shows that in addition to the con-

ditions in Theorem 7.1.1, there exist further necessary conditions for (G, p)

to be isostatic [36]. These are some very simply stated restrictions on the

number of bars and bodies that are fixed by various symmetry operations of

(G, p). While these extra conditions are analogous to the ones derived for bar

310

and joint frameworks in Section 4.3, the modified context holds the promise

of converting them into necessary and sufficient conditions for an arbitrary-

dimensional body-bar realization of G to be isostatic, provided that this re-

alization is as generic as possible subject to the given symmetry constraints.

An interesting example where the presence of certain symmetries forces

the realizations of a generically 3-isostatic ‘body-bar multigraph’ to be flex-

ible is illustrated in Figure 7.2. The depicted structures represent different

types of ‘Steward platforms’ which have an abundance of applications in

various areas of science and engineering (see [53], for example).

.

....

.... ..

..

.... ..

.. ....

.(a)

.

.... .. ....

..

....

.. ......

.(b)

Figure 7.2: 3-dimensional body-bar frameworks modeling different types of

‘Steward platforms’: the non-symmetric body-bar framework in (a) is iso-

static; the body-bar framework in (b) is flexible due to the presence of the

6-fold rotational symmetry, as predicted by the necessary counts derived in

[36].

Dimension 2

Although the main motivation for studying symmetric body-bar frame-

works is to establish combinatorial results for symmetric structures in dimen-

sions d ≥ 3, a discussion of symmetric body-bar frameworks in the plane is

311

also included in [36].

As one might anticipate from the analysis of symmetric isostatic pinned

frameworks (recall Section 7.1.1), there exists an isostatic body-bar frame-

work for any point group in dimension 2. However, restrictions on the number

of fixed bars and bodies still apply [36].

By modeling a body-bar framework as a framework (in the sense of Def-

inition 2.2.1) with isostatic bar and joint bodies of required symmetry and

applying the results for symmetric isostatic bar and joint bodies derived in

Chapter 5, it can be shown that the conditions in Theorem 7.1.1 (ii), to-

gether with the conditions in [36] concerning the number of fixed bars and

bodies, are also sufficient for a generic body-bar structure with point group

C2, C3, or Cs in dimension 2 to be isostatic.

For the remaining groups, the corresponding conjectures are stated in

[36]. In some cases, these could not be generalizations of the plane (bar and

joint) framework results, since there do not exist any isostatic plane frame-

works with point group Cm or Cmv, where m > 3 (recall Section 4.3.2).

All of these conjectures are promising directions for future work.

Dimension 3

Conjectures concerning combinatorial characterizations of 3-dimensional

symmetric body-bar frameworks in the spirit of Tay’s Theorem are also pre-

sented in [36]. We conjecture that the counts in Theorem 7.1.1 (ii) for the

multigraph G, together with the corresponding necessary conditions concern-

ing the number of fixed bars and bodies derived in [36], are also sufficient

for symmetric-generic body-bar realizations of G to be isostatic. Some ad-

312

ditional conjectures are also included in [36]. These are cast in terms of

symmetric tree partitions which are at the core of proofs for Tay’s Theorem

in all dimensions (Theorem 7.1.1), as well as for the proofs of combinato-

rial characterizations of symmetric bar and joint frameworks in dimension

2, as we have seen in Chapter 5. The fact that Tay’s Theorem can also be

proven using Henneberg-type constructions (see [29, 65]) gives another indi-

cation that the results in Chapter 5 can be extended to results for body-bar

frameworks in all dimensions.

7.1.3 Body-hinge and molecular structures

A body-hinge framework consists of a set of rigid bodies which are con-

nected, in pairs, along ‘hinges’. The bodies each move, preserving the con-

tacts along the hinges (see also Figure 7.3). These structures are of particular

interest in some important applications of rigidity [21, 69, 78, 81, 82].

Note that a body-hinge framework can be described as a special case of

a body-bar framework, because each hinge can implicitly be replaced by a

set of 5 independent bars, each of which intersects the hinge line [69, 81, 82].

Although the end-points of the bars of such a special type of body-bar frame-

work are forced to lie in non-generic positions, it is shown in [69] that body-

hinge realizations of a multigraph G with generic hinge assignments are in-

finitesimally rigid if and only if body-bar realizations of G with generic posi-

tions for the end-points of the bars are infinitesimally rigid. So, body-hinge

frameworks have the same efficient algorithms for testing generic rigidity as

body-bar frameworks [82].

Moreover, the Molecular Conjecture posed by T.-S. Tay and W. Whiteley

313

in 1984 proposes that the even more special class of body-bar frameworks

that arise in the models of molecular kinematics (i.e., the class of body-hinge

frameworks that have all hinges of each body concurrent in a point) also

have the same good combinatorial theory as general body-bar frameworks

[69, 82]. More precisely, the Molecular Conjecture states that generic body-

hinge realizations of a multigraph G are infinitesimally rigid if and only if

generic molecular body-hinge realizations of G are infinitesimally rigid, so

that, under generic conditions, the efficient counting algorithms for body-

bar frameworks also apply to molecular body-hinge frameworks.

Given certain symmetry constraints, we conjecture that, analogously to

the non-symmetric situation, the results (and conjectures) in [36] concerning

symmetric-generic body-bar frameworks also translate directly to symmetric-

generic body-hinge frameworks.

.

.(a)

.

.(b)

Figure 7.3: 3-dimensional body-hinge frameworks whose underlying multi-

graph is a hexagonal cycle: the non-symmetric body-hinge framework in (a)

is isostatic; the body-hinge framework in (b) is flexible due to the half-turn

symmetry, as predicted by the counts derived in [36].

We further conjecture that a symmetric version of the Molecular Con-

jecture holds, i.e., that under symmetric-generic conditions, body-bar frame-

314

works and molecular frameworks also possess the same rigidity properties.

We note that a number of biomolecules possess rotational symmetry, in-

cluding a number of virus shells which exhibit the symmetry of the rotational

icosahedral group. The potential for such applications, as well as for under-

standing human-built structures which are designed to have symmetry, is a

further motivation for giving explicit results for symmetric body-bar, body-

hinge, and molecular structures.

7.2 Coning, symmetry, and spherical frame-

works

Coning is a well recognized technique in rigidity theory which takes frame-

works in Rd to frameworks in Rd+1, preserving infinitesimal rigidity, indepen-

dence, and generic rigidity. We briefly recall the essential vocabulary and the

most basic result.

The cone graph of a graph G is the graph G ∗ v with V (G ∗ v) =

V (G)∪v, where v /∈ V (G), and E(G∗v) = E(G)∪v, w|w ∈ V (G).

If p∗ is a configuration for the cone graph G∗v so that no two joints corre-

sponding to adjacent vertices in G are collinear with p∗(v) or are parallel to

a hyperplane H of Rd+1, then we denote pH to be the projection of p∗ from

p∗(v) into H. Further, we call p∗ and pH a projection pair of configurations.

The following basic result is proven in [74].

Theorem 7.2.1 (Whiteley, 1983) Let G be a graph, G ∗ v be the cone

graph of G, and p∗ and pH be a projection pair of configurations. Then

315

(G ∗ v, p∗) is infinitesimally rigid (independent, isostatic) in Rd+1 if and

only if (G, pH) is infinitesimally rigid (independent, isostatic) in Rd.

Symmetric coning

Given a symmetric framework (G, pH) in H = Rd, we may place the

cone vertex v ‘symmetrically’ in Rd+1 to obtain a new (d + 1)-dimensional

framework (G ∗ v, p∗) which inherits the symmetry of (G, pH).

For example, if (G, pH) has point group Cs, then we may place the cone

vertex v off the mirror corresponding to the reflection in Cs, but on the

perpendicular extension of this mirror in Rd+1 to obtain a framework (G ∗v, p∗) which is symmetric with respect to this larger mirror (see also Figure

7.4).

.. ..

..H

.(G, pH)

... ....

..

..

.H

.(G ∗ v, p∗)

Figure 7.4: Illustration of ‘symmetric coning’: both the framework (G, pH)

in Rd and the coned framework (G ∗ v, p∗) in Rd+1 have mirror symmetry.

Similarly, if (G, pH) has point group Cm, then we may place the cone

vertex v off the old rotational axis, but on the perpendicular extension of

this axis in Rd+1 to construct a framework (G ∗ v, p∗) which is symmetric

316

with respect to this larger rotational axis.

Similar constructions can be envisaged for other symmetry groups as well.

We can identify several immediate uses for an exploration of ‘symmetric

coning’.

First, symmetric coning allows us to make predictions for the way the

symmetric Maxwell counts (i.e., the restrictions on the number of joints and

bars that are fixed by various symmetry operations) change between dimen-

sions. In particular, we can gain some initial insights into the expected

counting rules for 4-dimensional frameworks in this way (at least as a set of

basic examples).

Further, we can use symmetric coning as a tool for constructing frame-

works with symmetry-preserving flexes. We claim that if the symmetric

Maxwell counts detect a fully symmetric infinitesimal flex for the framework

(G, pH) in Rd, then they also detect a fully symmetric infinitesimal flex for

the symmetrically coned framework (G ∗ v, p∗) in Rd+1. So, with the help

of the results in Chapter 6, we can detect the flexibility of (d+1)-dimensional

realizations of G∗v (which are generic with the given symmetry) by apply-

ing the symmetry-extended version of Maxwell’s rule to the corresponding

d-dimensional framework (G, pH).

Moreover, in addition to predicting the flexibility of symmetric coned

frameworks combinatorially in this manner, we can use coning as a geomet-

ric tool for constructing new flexible frameworks.

Consider, for instance, the 2-dimensional realization of the graph K4,4

with point group C2v given in Example 6.3.2. By placing the cone vertex v

‘generically’ on the axis which constitutes the perpendicular extension of the

317

origin (i.e., the center of the half-turn C2 ∈ C2v) in R3, we may construct

a 3-dimensional framework (K4,4 ∗ v, p∗) which is generic with the point

group C2v in dimension 3. (Note that this requires the two sets of ‘symmetry

orbits‘ of vertices of K4,4 to be placed on distinct hyperplanes in R3). It can

be shown that the infinitesimal flex of (K4,4∗v, p∗) guaranteed by Theorem

7.2.1 again has the full C2v symmetry, so that it follows from Corollary 6.2.8

that (K4,4 ∗ v, p∗) also has a symmetry-preserving flex.

Note that the detection of this flex for (K4,4 ∗ v, p∗) requires a geo-

metric analysis, because the combinatorial counts of Maxwell’s rule and the

symmetry-extended version of Maxwell’s rule do not detect any infinitesimal

flexes for these over-braced frameworks.

We note that a d-dimensional symmetric framework (G, pH) can of course

also be coned in such a way that the resulting (d+1)-dimensional framework

(G∗v, p∗) possesses additional symmetries beyond the ones inherited from

(G, pH), and vice versa. In particular, it is a notable consequence of Theorem

7.2.1 and the results derived in this thesis that a symmetry-based rigidity

analysis of a symmetric framework (G, pH) in Rd can be used to gain insight

into the rigidity properties of a non-symmetric coned framework (G∗v, p∗)in Rd+1 with a symmetric projection.

Spherical frameworks

Suppose we cone a framework (G, pH) so that for the resulting frame-

work (G ∗ v, p∗), all the points p∗(w), w ∈ V (G), are equidistant (wlog

with distance 1) from the position p∗(v) of the cone vertex v. The frame-

work (G ∗ v, p∗) may then be interpreted geometrically and for a rigidity

318

analysis as a spherical framework lying on the unit sphere with center p∗(v).

Therefore, Theorem 7.2.1 can also be used to transfer rigidity results between

spherical and Euclidean space (see also [18, 74]).

In particular, the symmetry-extended version of Maxwell’s rule can be

applied to predict the infinitesimal rigidity of spherical frameworks that pos-

sess non-trivial symmetries.

Moreover, if a configuration p∗ is a regular point of the cone graph G∗vin V

(I1)e , and we project p∗ from p∗(v) onto the unit sphere, then the result-

ing configuration is again a regular point of G ∗ v in V(I1)e . Therefore,

by the results in Chapter 6, both the combinatorial and geometric meth-

ods for finding symmetry-preserving flexes in coned frameworks described in

the previous section implicitly also detect symmetry-preserving flexes for the

corresponding frameworks in spherical space.

The techniques described in this section, together with the results of

Chapter 4, might also help us to give a mathematically explicit explanation

for the symmetric version of the Danzerian counting rule for the spherical

circle packing problem presented in [28].

7.3 Symmetric global rigidity

A d-dimensional realization (G, p) of a graph G is said to be globally rigid

in Rd if all d-dimensional realizations of G with the same edge lengths as

(G, p) are congruent to (G, p). To determine whether a given framework is

globally rigid is a fundamental problem in distance geometry which also has

many important practical applications ranging from localization problems in

319

networks to the study of molecular stability [14, 18, 38, 41].

Studying the impact of symmetry on global rigidity appears to be an-

other very promising new direction in rigidity theory with a large number

of interesting open questions. Two separate fundamental problems can be

identified.

(1) Given a symmetric framework, one may ask whether there exists an-

other non-congruent framework with the same edge lengths that may

or may not be symmetric.

(2) Given a symmetric framework, one could examine whether there exists

another non-congruent framework with the same edge lengths and the

same symmetry.

.. .

.

.

.(a)

.. .

.

.

.(b)

.. .

.

.

.(c)

Figure 7.5: Frameworks with mirror symmetry which are not globally rigid

in R2. The framework (G, p) in (a) is also not ‘symmetric globally rigid’

in the sense of problem (2), since the framework in (b) is another non-

congruent realization of G in R2 with the same edge lengths and the same

mirror symmetry as (G, p); the framework in (c), however, is ‘symmetric

globally rigid’ within the set of all realizations of G in R2 with the same

mirror symmetry.

320

While none of these problems has been investigated carefully yet, the

second problem appears to be significantly easier than the first.

A natural starting point for the investigation of the second problem is

to find symmetric analogs to each of the necessary conditions for a generic

framework to be globally rigid given in Hendrickson’s Theorem (see [38] or

[14], for example). Note that for dimensions 1 and 2, the conditions in Hen-

drickson’s Theorem are also proven to be sufficient for generic global rigidity

[41].

Further results on global rigidity, including a necessary and sufficient con-

dition for a generic framework to be globally rigid in an arbitrary dimension,

can be found in [14, 31], for example.

321

Appendix A

Character tables of selected

point groups

In this appendix, we provide a list of character tables for some important

point groups. A more complete (and also more detailed) list can be found

in [6, 19, 37], for example. The irreducible linear representations in these

character tables are denoted by the Mulliken symbols. A nice description of

their meanings can be found in [19].

It shall suffice to mention here that the letter assigned to an irreducible

linear representation indicates the degree of the representation: all repre-

sentations of degree 1 are denoted by either A or B and all representations

of degree 2 are denoted by E. Further, all representations of degree 3 are

denoted by T , all representations of degree 4 by G and all representations of

degree 5 by H.

In a given matrix representation of a group S, the traces of all matrices

belonging to elements of S that lie in the same conjugacy class are identical

322

[6, 19, 37]. Therefore, in the character table of a group S, there exists exactly

one column for each conjugacy class of S. The number of elements in such

a conjugacy class is indicated by the coefficient in front of the element that

represents this conjugacy class in the character table.

Character tables for the groups C1, Cs and Ci

C1 Id

A 1

Cs Id s

A′ 1 1

A′′ 1 -1

Ci Id i

Ag 1 1

Au 1 -1

Character tables for Cm groups

C2 Id C2

A 1 1

B 1 -1

C3 Id C3 C23

A 1 1 1

E 2 -1 -1

C4 Id C4 C2 C34

A 1 1 1 1

B 1 -1 1 -1

E 2 0 -2 0

C5 Id C5 C25 C3

5 C45

A 1 1 1 1 1

E1 2 2 cos(

2π5

)2 cos

(4π5

)2 cos

(4π5

)2 cos

(2π5

)

E2 2 2 cos(

4π5

)2 cos

(2π5

)2 cos

(2π5

)2 cos

(4π5

)

323

C6 Id C6 C3 C2 C23 C5

6

A 1 1 1 1 1 1

B 1 -1 1 -1 1 -1

E1 2√

3 −√3 -2 −√3√

3

E2 2 −√3 −√3 2√

3√

3

Character tables for Cmv groups

C2v Id C2 sh sv

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

C3v Id 2C3 3sv

A1 1 1 1

A2 1 1 -1

E 2 -1 0

C4v Id 2C4 C2 2sv 2sd

A1 1 1 1 1 1

A2 1 1 1 -1 -1

B1 1 -1 1 1 -1

B2 1 -1 1 -1 1

E 2 0 -2 0 0

324

Character tables for Cmh groups

C2h Id C2 i sh

Ag 1 1 1 1

Bg 1 -1 1 -1

Au 1 1 -1 -1

Bu 1 -1 -1 1

C3h Id C3 C23 sh S3 S5

3

A1 1 1 1 1 1 1

E ′ 2 -1 -1 2 -1 -1

A′′ 1 1 1 -1 -1 -1

E ′′ 2 -1 -1 -2 1 1

Character tables for Dm groups

D2 Id C2(z) C2(y) C2(x)

A 1 1 1 1

B1 1 1 -1 -1

B2 1 -1 1 -1

B3 1 -1 -1 1

D3 Id 2C3 3C2

A1 1 1 1

A2 1 1 -1

E 2 -1 0

D4 Id 2C4 C2(= C24) 2C ′

2 2C ′′2

A1 1 1 1 1 1

A2 1 1 1 -1 -1

B1 1 -1 1 1 -1

B2 1 -1 1 -1 1

E 2 0 -2 0 0

325

Character tables for Dmh groups

D2h Id C2(z) C2(y) C2(x) i s(xy) s(xz) s(yz)

Ag 1 1 1 1 1 1 1 1

B1g 1 1 -1 -1 1 1 -1 -1

B2g 1 -1 1 -1 1 -1 1 -1

B3g 1 -1 -1 1 1 -1 -1 1

Au 1 1 1 1 -1 -1 -1 -1

B1u 1 1 -1 -1 -1 -1 1 1

B2u 1 -1 1 -1 -1 1 -1 1

B3u 1 -1 -1 1 -1 1 1 -1

D3h Id 2C3 3C2 sh 2S3 3sv

A′1 1 1 1 1 1 1

A′2 1 1 -1 1 1 -1

E ′ 2 -1 0 2 -1 0

A′′1 1 1 1 -1 -1 -1

A′′2 1 1 -1 -1 -1 1

E ′′ 2 -1 0 -2 1 0

326

Character tables for Dmd groups

D2d Id 2S4 C2 2C ′2 2sd

A1 1 1 1 1 1

A2 1 1 1 -1 -1

B1 1 -1 1 1 -1

B2 1 -1 1 -1 1

E 2 0 -2 0 0

D3d Id 2C3 3C2 i 2S6 3sd

A1g 1 1 1 1 1 1

A2g 1 1 -1 1 1 -1

Eg 2 -1 0 2 -1 0

A1u 1 1 1 -1 -1 -1

A2u 1 1 -1 -1 -1 1

Eu 2 -1 0 -2 1 0

Character tables for S2m groups

S4 Id S4 C2 S34

A 1 1 1 1

B 1 -1 1 -1

E 2 0 -2 0

S6 Id C3 C23 i S5

6 S6

Ag 1 1 1 1 1 1

Eg 2 -1 -1 2 -1 -1

Au 1 1 1 -1 -1 -1

Eu 2 -1 -1 -2 1 1

327

Character tables for the tetrahedral groups

T Id 4C3 4C23 3C2

A 1 1 1 1

E 2 -1 -1 2

T 3 0 0 1

Td Id 8C3 3C2 6S4 6sd

A1 1 1 1 1 1

A2 1 1 1 -1 -1

E 2 -1 2 0 0

T1 3 0 -1 1 -1

T2 3 0 -1 -1 1

Th Id 4C3 4C23 3C2 i 4S5

6 4S6 sh

Ag 1 1 1 1 1 1 1 1

Eg 2 -1 -1 2 2 -1 -1 2

Tg 3 0 0 -1 3 0 0 -1

Au 1 1 1 1 -1 -1 -1 -1

Eu 2 -1 -1 2 -2 1 1 -2

Tu 3 0 0 -1 -3 0 0 1

Character tables for the octahedral groups

O Id 8C3 3C2 6C ′2 6C4

A1 1 1 1 1 1

A2 1 1 1 -1 -1

E 2 -1 2 0 0

T1 3 0 -1 -1 1

T2 3 0 -1 1 -1

328

Oh Id 8C3 6C2 6C4 3C2(= 3C24) i 6S4 8S6 3sh 6sd

A1g 1 1 1 1 1 1 1 1 1 1

A2g 1 1 -1 -1 1 1 -1 1 1 -1

Eg 2 -1 0 0 2 2 0 -1 2 0

T1g 3 0 -1 1 -1 3 1 0 -1 -1

T2g 3 0 1 -1 -1 3 -1 0 -1 1

A1u 1 1 1 1 1 -1 -1 -1 -1 -1

A2u 1 1 -1 -1 1 -1 1 -1 -1 1

Eu 2 -1 0 0 2 -2 0 1 -2 0

T1u 3 0 -1 1 -1 -3 -1 0 1 1

T2u 3 0 1 -1 -1 -3 1 0 1 -1

Character tables for the icosahedral groups

I Id 12C5 12C25 20C3 15C2

A 1 1 1 1 1

T1 3 −2 cos(

4π5

) −2 cos(

2π5

)0 -1

T2 3 −2 cos(

2π5

) −2 cos(

4π5

)0 -1

G 4 - 1 -1 1 0

H 5 0 0 -1 1

329

Ih Id 12C5 12C25 20C3 15C2 i 12S10 12S3

10 20S6 15s

Ag 1 1 1 1 1 1 1 1 1 1

T1g 3 1+√

52

1−√52

0 -1 3 1−√52

1+√

52

0 -1

T2g 3 1−√52

1+√

52

0 -1 3 1+√

52

1−√52

0 -1

Gg 4 -1 -1 1 0 4 -1 -1 1 0

Hg 5 0 0 -1 1 5 0 0 -1 1

Au 1 1 1 1 1 -1 -1 -1 -1 -1

T1u 3 1+√

52

1−√52

0 -1 -3 −1−√52

−1+√

52

0 1

T2u 3 1−√52

1+√

52

0 -1 -3 −1+√

52

−1−√52

0 1

Gu 4 -1 -1 1 0 -4 1 1 -1 0

Hu 5 0 0 -1 1 -5 0 0 1 -1

330

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