Mathieu Dutour Sikiric- Perfect forms and perfect Delaunay polytopes

8/3/2019 Mathieu Dutour Sikiric- Perfect forms and perfect Delaunay polytopes

1/55

Perfect forms and perfect Delaunay polytopes

Mathieu Dutour SikiricRudjer Boskovic Institute, Croatia

November 25, 2011


2/55

I. Lattices, packings

and coverings


3/55

Lattice packings

A lattice L Rn is a set of the form L = Zv1 + + Zvn.

A packing is a family of balls Bn(xi, r), i I of the sameradius r and center xi such that their interiors are disjoint.

If L is a lattice, the lattice packing is the packing defined bytaking the maximal value of > 0 such that L + Bn(0, ) is apacking.


4/55

Density of lattice packings

Take the lattice packing defined by a lattice L:

(L)

The packing density has the expression

(L) =

(L)n vol(Bn(0, 1))

det L with (L) =

1

2 minvL{0} ||v||,

vol(Bn(0, 1)) the volume of the unit ball Bn(0, 1) and det Lthe volume of an unit cell.


5/55

Empty sphere and Delaunay polytopes

Definition: A sphere S(c, r) of center c and radius r in an

n-dimensional lattice L is said to be an empty sphere if:(i) v c r for all v L,

(ii) the set S(c, r) L contains n + 1 affinely independent points.

Definition: A Delaunay polytope P in a lattice L is apolytope, whose vertex-set is L S(c, r).

c

r

Delaunay polytopes define a tesselation of the Euclidean spaceRn


6/55

Lattice covering

For a lattice L we define the covering radius (L) to be thesmallest r such that the family of balls v + B

n(0, r) for v

L

cover Rn.

The covering density has the expression

(L) =

(L)n vol(Bn(0, 1))

det(L) 1

with (L) being the largest radius of Delaunay polytopes

The only general method for computing (L) is to computeall Delaunay polytopes of L.


7/55

II. Gram matrix

formalism


8/55

Gram matrix and lattices

Denote by Sn the vector space of real symmetric n n

matrices andSn

>0 the convex cone of real symmetric positivedefinite n n matrices.

Take a basis (v1, . . . , vn) of a lattice L and associate to it theGram matrix Gv = (vi, vj)1i,jn Sn>0.

Example: take the hexagonal lattice generated by v1 = (1, 0)

and v2 =

12 ,

32

2v

v1

Gv =12

2 11 2


9/55

Isometric lattices

Take a basis (v1, . . . , vn) of a lattice L withvi = (vi,1, . . . , vi,n) R

n and write the matrix

V =

v1,1 . . . vn,1...

. . ....

v1,n . . . vn,n

and Gv = VT V.

The matrix Gv is defined byn(n+1)

2 variables as opposed to n2

for the basis V.

If M Sn>0, then there exists V such that M = VT V (Gram

Schmidt orthonormalization) If M = VT1 V1 = V

T2 V2, then V1 = OV2 with O

T O = In(i.e. O corresponds to an isometry ofRn).

Also if L is a lattice ofRn with basis v and u an isometry of

R

n

, then Gv = Gu(v).


10/55

Arithmetic minimum

The arithmetic minimum of A Sn>0 is

min(A) = minxZn{0}

xTAx

The minimal vector set of A Sn>0 is

Min(A) =

x Zn | xTAx = min(A)

Both min(A) and Min(A) can be computed using someprograms (for example sv by Vallentin)

The matrix Ahex =

2 11 2

has

Min(Ahex) = {(1, 0), (0, 1), (1, 1)}.


11/55

Reexpression of previous definitions

Take a lattice L = Zv1 + + Zvn. If x L,

x = x1v1 + + xnvn with xi Z

we associate to it the column vector X =

x1...

xn

We get ||x||2 = XTGvX and

det L =

det Gv and (L) =1

2

min(Gv)

(L)

For Ahex = 2 11 2

, det Ahex = 3 and min(Ahex) = 2


12/55

Changing basis

Ifv and v are two basis of a lattice L then V = VP with

P GLn(Z

). This implies

Gv = VTV = (VP)TVP = PT{VTV}P = PTGvP

If A, B Sn>0, they are called arithmetically equivalent if there

is at least one P GLn(Z

) such that

A = PTBP

Lattices up to isometric equivalence correspond to Sn>0 up to

arithmetic equivalence. In practice, Plesken/Souvignier wrote a program isom for

testing arithmetic equivalence and a program autom forcomputing automorphism group of lattices.All such programs take Gram matrices as input.


13/55

An example of equivalence

Take the hexagonal lattice and two basis in it.

v1

v2

v1 = (1, 0) and v2 =

12 ,

3

2

2v

v1

v1 =

52 ,

3

2

and v2 = (1, 0)

One has v1 = 2v1 + v2, v2 = v1 and P = 2 1

1 0

Gv =

1 1212 1

and Gv =

7 52

52 1

= PTGvP


14/55

Root lattices Let us take the lattice

An =

x Zn+1

s.t.

n+1i=1

xi = 0

If we take the basis vi = ei+1 ei then we get the Grammatrix A = (aij)1i,jn with ai,i = 2, ai,i+1 = ai+1,i = 1 andai,j = 0 otherwise.

Let us take the lattice

Dn =

x Zn s.t.

ni=1

xi 0 (mod 2)

For the basis v1 = e1 + e2, v2 = e1 e2, vi = ei ei

1 we get

Gv =

2 0 1 0 . . . 00 2 1 0 . . . 01 1 2 1 . . . 0

0 0 1 2. . . 0

..

. 0

. . .. . .

. . .10 . . . . . . . . . 1 2


15/55

III. Perfect and

eutactic forms


16/55

Hermite function

If A Sn>0 then the arithmetic minimum is

min(A) = minxZn{0} xT

Ax

and the set of minimal vectors is

Min(A) = x Zn : xTAx = min(A)

The Hermite function on the space Sn>0 is

(A) =min(A)

(det A)1/n

The density of the lattice packing L associated to A is

(L) =

(A)nvol(Bn(0, 1))

2n

Finding lattice packings with highest packing density is the

same as maximizing the Hermite function.


17/55

Perfect forms

A form A is extreme if there is a neighborhood V of A in Sn>0

such that

If B V with B = A then (B) < (A)

A matrix A Sn>0 is perfect (Korkine & Zolotarev, 1873) if

the equation

B Sn and xTBx = min(A) for all x Min(A)

implies B = A.

Theorem: (Korkine & Zolotarev, 1873) If a form is extremethen it is perfect.

Perfect forms are rational forms.

If A is perfect then (A)n is rational.

f f


18/55

A perfect form

Ahex = 2 1

1 2

corresponds to the lattice:

v1

v2

If B =

a b

b c

satisfies to xTBx = min(Ahex) for

x Min(Ahex) = {(1, 0), (0, 1), (1, 1)}, then:

a = 2, b = 2 and a 2c + b = 2

which implies B = Ahex. Ahex is perfect.

A f f


19/55

A non-perfect form

Asqr =

1 00 1

has Min(Asqr) = {(0, 1), (1, 0)}.

See below lattices LB, Lsqr associated to matricesB, Asqr S

2>0 with Min(B) = Min(Asqr):

v1

v2

v1

v2

E i f


20/55

Eutactic forms

A form A Sn>0

is eutactic (Voronoi, 1908) if there existv > 0 such that

A1 =

vMin(A)vvv

T

Theorem: (Voronoi, 1908) A form A is extreme if and only ifit is perfect and eutactic.

Theorem: (Ash, 1977)

(i) If A is not an eutactic form then it is topologically ordinary

point for (ii) If A is an eutactic form then it is a critical but topologicallynon-degenerate point for .

(ii) is a topological Morse function.

E l f f t f


21/55

Examples of perfect forms

The root lattice are all perfect:

Name Min |Min| det | Aut |An ei ej 2n(n + 1) n + 1 2(n + 1)!Dn ei ej 4n(n 1) 4 2

nn!E6 complex 72 3 103680E7 complex 126 2 2903040

E8 complex 240 1 696729600 Another remarkable lattice is the Leech lattice of dimension

24. Every vector v has v2 4 and det Leech = 1. There are 196280 shortest vectors (maximal number in

dimension 24) Its automorphism group quotiented by Id24 is the sporadic

simple group Co0 It plays a significant role in modular form theory and

Lorentzian lattice theory.

K lt l tti ki d it i i ti


22/55

Known results on lattice packing density maximization

dim. Nr. of perfect forms Absolute maximumof realized by

2 1 (Lagrange) A23 1 (Gauss) A34 2 (Korkine & Zolotarev) D45 3 (Korkine & Zolotarev) D56 7 (Barnes) E6 (Blichfeldt)

7 33 (Jaquet) E7 (Blichfeldt)8 10916 (DSV) E8 (Blichfeldt)9 500000 9?

24 ? Leech (Cohn & Kumar)

Remarks

The enumeration of perfect forms is done with the Voronoialgorithm.

The solution in dimension 24 was obtained by different

methods.


23/55

IV. Ryshkov cone

and the Voronoi algorithm

The Ryshkov cone


24/55

The Ryshkov cone

The Ryshkov cone Rn is defined as

Rn =

A Sn s.t. xTAx 1 for all x Zn {0}

The cone is invariant under the action of GLn(Z).

The cone is locally polyhedral, i.e. for a given A Rnx Zn s.t. xTAx = 1

is finite

Vertices of Rn correspond to perfect forms. For a form A we define the local cone

Loc(A) =

Q Sn s.t. xTQx 0 for x Min(A)

The Voronoi algorithm


25/55

The Voronoi algorithm

Find a perfect form (say An), insert it to the list L as undone.

Iterate For every undone perfect form A in L, compute the local coneLoc(A) and then its extreme rays.

For every extreme ray r of Loc(A) realize the flipping, i.e.compute the adjacent perfect form A = A + r.

If A is not equivalent to a form in L, then we insert it into Las undone.

Finish when all perfect domains have been treated.

The subalgorithms are:

Find the extreme rays of the local cone Loc(A) (use cdd or lrsor any other program)

For any extreme ray r of Loc(A) find the adjacent perfectform A in the Ryshkov cone Rn

Test equivalence of perfect forms using autom

Flipping on an edge I


26/55

Flipping on an edge I

Min(Ahex) = {(1, 0), (0, 1), (1, 1)}

with

Ahex =

1 1/2

1/2 1

and D =

0 1

1 0

v1

v2

Ahex

Flipping on an edge II


27/55

Flipping on an edge II

Min(B) = {(1, 0), (0, 1)}

with

B =

1 1/4

1/4 1

= Ahex + D/4

v1

v2

A

B

hex

Flipping on an edge III


28/55

Flipping on an edge III

Min(Asqr) = {(1, 0), (0, 1)}

with

Asqr =

1 00 1

= Ahex + D/2

v1

v2

Ahex

Asqr

Flipping on an edge IV


29/55

Flipping on an edge IV

Min(Ahex) = {(1, 0), (0, 1), (1, 1)}

with

Ahex =

1 1/2

1/2 1

= Ahex + D

v1

v2

Ahex

Ahex

The Ryshkov cone R2


30/55

The Ryshkov cone R2

+ (1,1)

+ (1,2)

+ (1,0)

+ (2,1)

+ (2,1)

+ (1,1)

+ (0,1)

+ (1,2)

1/2

1 1/2

1

1/21

1/2 1

3

3/2

1

3

3/2

3/2

1

3/2

Well rounded forms and retract


31/55

Well rounded forms and retract

A form Q is said to be well rounded if it admits vectors v1,. . . , vn such that

(v1, . . . , vn) form a basis ofRn

v1, . . . , vn are shortest vectors. Q[v1] = = Q[vn].

Well rounded forms correspond to bounded faces of Rn.

Every form can be continuously deformed to a well roundedform and this defines a retracting homotopy of Rn onto apolyhedral complex WRn of dimension

n(n1)2 .

Every face of WRn has finite stabilizer, hence we can use it forcomputing the homology of GLn(Z) and other arithmeticgroups.

Actually, in term of dimension, we cannot do better: A. Pettet and J. Souto, Minimality of the well rounded retract,

Geometry and Topology, 12 (2008), 1543-1556.

References


32/55

References

G. Voronoi, Nouvelles applications des parametres continues a

la theorie des formes quadratiques 1: Sur quelques proprietesdes formes quadratiques positives parfaites, J. Reine Angew.Math 133 (1908) 97178.

M. Dutour Sikiric, A. Schurmann and F. Vallentin,Classification of eight dimensional perfect forms, Electron.

Res. Announc. Amer. Math. Soc. A. Schurmann, Computational geometry of positive definite

quadratic forms, University Lecture Notes, AMS.

J. Martinet, Perfect lattices in Euclidean spaces, Springer,

2003. S.S. Ryshkov, E.P. Baranovski, Classical methods in the

theory of lattice packings, Russian Math. Surveys 34 (1979)168, translation of Uspekhi Mat. Nauk 34 (1979) 363.


33/55

V. The lattice covering

problem

Empty sphere and Delaunay polytopes


34/55

p y p y p y p

A sphere S(c, r) of radius r and center c in an n-dimensional

latticeL

is said to be an empty sphere if:(i) v c r for all v L,

(ii) the set S(c, r) L contains n + 1 affinely independent points.

A Delaunay polytope P in a lattice L is a polytope, whosevertex-set is L S(c, r).

c

r

Equalities and inequalities


35/55

q q

Take M = Gv with v = (v1, . . . , vn) a basis of lattice L.

If V = (w1, . . . , wN) with wi Zn are the vertices of a

Delaunay polytope of empty sphere S(c, r) then:

||wi c|| = r i.e. wTi Mwi 2w

Ti Mc + c

TMc = r2

Substracting one obtains

{wTi Mwi wTj Mwj} 2{wTi wTj }Mc = 0

Inverting matrices, one obtains Mc = (M) with linear andso one gets linear equalities on M.

Similarly ||w c|| r translates into linear inequalities on M:

Take V = (v0, . . . , vn) a simplex (vi Zn), w Zn. If onewrites w =

ni=0 ivi with 1 =

ni=0 i, then one has

||w c|| r wTMw n

i=0

ivTi Mvi 0

Iso-Delaunay domains


36/55

y

Take a lattice L and select a basis v1, . . . , vn. We want to assign the Delaunay polytopes of a lattice.

Geometrically, this means that

1v

2v

2v

1v

are part of the same iso-Delaunay domain. An iso-Delaunay domain is the assignment of Delaunay

polytopes, so it is also the assignment of the Voronoi polytopeof the lattice.

Primitive iso-Delaunay If one takes a generic matrix M in Sn>0, then all its Delaunay

are simplices and so no linear equality are implied on M. Hence the corresponding iso-Delaunay domain is of dimension

n(n+1)2 , they are called primitive

Equivalence and enumeration


37/55

The group GLn(Z) acts on Sn>0 by arithmetic equivalence and

preserve the primitive iso-Delaunay domains.

Voronoi proved that after this action, there is a finite numberof primitive iso-Delaunay domains.

Bistellar flipping creates one iso-Delaunay from a giveniso-Delaunay domain and a facet of the domain. In dim. 2:

Enumerating primitive iso-Delaunay domains is doneclassicaly:

Find one primitive iso-Delaunay domain. Find the adjacent ones and reduce by arithmetic equivalence.

This is very similar to the Voronoi algorithm for perfect forms.

The partition ofS2>0 R3 I


38/55

>0

If q(x, y) = ux2 + 2vxy + wy2 then q S2>0 if and only ifv

2 < uw and u > 0.

w

v

u

The partition ofS2>0 R3 II


39/55

We cut by the plane u + w = 1 and get a circle representation.

u

v

w

The partition ofS2>0 R3 III


40/55

Primitive iso-Delaunay domains in S2>0:

Optimization problem


41/55

The lattice covering problem is to find a lattice covering ofminimal density.

Thm. Given an iso-Delaunay domain LT, there exist a uniquelattice, which minimize the covering density over LT.

The effective lattice is obtained by solving a semidefiniteprogramming problem, so no exact solution, but approximate

solutions available at any precision. The local maxima that are found are defined by algebraic

integers. See for more details

A. Schurmann and F. Vallentin, Computational approaches tolattice packing and covering problems, Discrete &Compututational Geometry 35 (2006) 73116.

A. Schurmann, Computational geometry of positive definitequadratic forms, University Lecture Notes, AMS.

Known results on covering density minimization


42/55

dim. Best covering nr of iso-Delaunay

2 A2 (Kershner) 1 (Voronoi)3 A3 (Bambah) 1(Voronoi)4 A4 (Delone & Ryshkov) 3(Voronoi)5 A5 (Ryshkov & Baranovski) 222(Engel)6 L6 (conj. Vallentin)? ?7 L7 (conj. Schurmann & Vallentin)? ?

24 Leech (conj.)? ?

It turn out that the lattice of minimal covering density areunique for n 5

In general the best lattice coverings are expected to be

non-rational and with low symmetry.

But experimentations seemed to indicate that E6 is a localcovering maxima.


43/55

VI. Quadratic functions

and the Erdahl cone

The Erdahl cone


44/55

Denote by E2(n) the vector space of degree 2 polynomialfunctions on Rn. We write f E2(n) in the form

f(x) = af + bf x + Qf[x]

with af R, bf Rn and Qf a n n symmetric matrix

The Erdahl cone is defined as

Erdahl(n) = {f E2(n) such that f(x) 0 for x Zn}

It is a convex cone, which is non-polyhedral since defined byan infinity of inequalities.

The group acting on Erdahl(n) is AGLn(Z), i.e. the group ofaffine integral transformations

x b+ Px for b Zn and P GLn(Z)

Scalar product


45/55

Definition: If f, g E2(n), then:

f, g = afag + bf, bg + Qf, Qg

Definition: For v Zn, define evv(x) = (1 + v x)2.

We have

f, evv = f(v)

Thus finding the rays of Erdahl(n) is a dual descriptionproblem with an infinity of inequalities and infinite groupacting on it.

Definition: We also define

Erdahl>0(n) = {f Erdahl(n) : Qf positive definite}

Relation with Delaunay polytope


46/55

If D is a Delaunay polytope of a lattice L = Zv1 + + Zvnof empty sphere S(c, r) then we define the function

fD,v : Zn R

x = (x1, . . . , xn) n

i=1 xivi c2 r2

Clearly fD,v Erdahl>0(n).

On the other hand if f Erdahl(n) then there exists a latticeL of dimension k n, a Delaunay polytope D of L, a basisv of L and a function AGLn(Z) such that

f (x1, . . . , xn) = fD,v(x1, . . . , xk)

Thus the faces of Erdahl(n) correspond to the Delaunaypolytope of dimension at most n.

The perfection rank of a Delaunay polytope is the dimensionof the face it defines in Erdahl(n).

Perfect quadratic function


47/55

Definition: If f Erdahl(n) then

Z(f) = {v Zn : f(v) = 0}

andDom f =

vZ(f)

R+evv

We have f, Dom f = 0.

Definition: f Erdahl(n) is perfect if Dom f is of dimension

n+2

2 1 that is if the perfection rank is 1. This is equivalent to say that f defines an extreme ray in

Erdahl(n).

A perfect quadratic function is rational.

Perfect Delaunay polytope


48/55

A Delaunay polytope P is perfect if vert P = Z(f) with f a

perfect quadratic function. It has at least

n+2

2

1 vertices.

Known results:dim. perfect Delaunay authors

1 [0, 1] in Z

2 3 4 5 (Deza, Laurent & Grishukhin)6 221 in E6 (Deza & Dutour)

7 321 in E7and ER7 in L(ER7)

8 27 (Dutour Sikiric & Rybnikov)9 100000 (Dutour Sikiric)

Voronoi algorithm on the Erdahl cone


49/55

From a given n-dimensional perfect Delaunay polytope Q of

form f we can define the local cone

Loc(f) = {g E2(n) s.t. g(x) 0 for x Z(f)}

The flipping algorithm finds the adjacent quadratic perfect

form g from a given perfect form f. The problem is Erdahl(n) is not locally polyhedral, i.e. the

rank of g can be lower than n.

The technique is to use a recursive algorithm for realizing the

enumeration. We start form [0, 1] Rn

1

and by subdivizionreach [0, 1]n (its local cone is the cut cone CUTn+1 occuringin combinatorial optimization).


50/55

VII. Covering maxima, pessima

and their characterization

Eutacticity

If f E d hl ( ) h d fi d h h


51/55

If f Erdahl>0(n) then define f and cf such that

f(x) = Qf[x cf] f

Then define

uf(x) = (1 + cf x)2 +

fn

Q1f [x]

Definition: f Erdahl>0(n) is eutactic if uf is in the relativeinterior of Dom f.

Definition: Take a Delaunay polytope P for a quadratic formQ of center cP and square radius P. P is called eutactic ifthere are v > 0 so that

1 =

vvertPv,

0 =

vvertPv(v cP),

Pn

Q1 = vvertPv(v cP)(v cP)

T.

Covering maxima


52/55

A given lattice L is called a covering maxima if for any latticeL near L we have (L) < (L).

Theorem: The following are equivalent: L is a covering maxima Every Delaunay polytope of maximal circumradius is perfect

and eutactic.

The following are perfect Delaunay polytope:

name # vertices # orbits Delaunay polytopesE6 27 1E7 56 2

ER7 35 4O10 160 6

BW16 512 4O23 94208 5

23 47104 709

Theorem: For any n 6 there exist one lattice L(DSn) whichis a covering maxima.There is only one perfect Delaunay polytope P(DSn) ofmaximal radius in L(DSn).

The infinite series

( )


53/55

For n even P(DSn) is defined as the lamination over Dn1 of one vertex the half cube 1

2Hn

1 the cross polytope CPn1

For n = 6, it is E6. For n odd as the lamination over Dn1 of

the cross polytope CPn1 the half cube

12 Hn1

the cross polytope CPn1

For n = 7, it is E7. Conjecture: The lattice DSn has the following properties:

L(DSn) has the maximum covering density among all covering

maxima Among all perfect Delaunay polytopes, P(DSn) has

maximum number of vertices

maximum volume

If true this would imply Minkovski conjecture.

Pessimum and Morse function property


54/55

For a lattice L let us denote Dcrit(L) the space of direction dof deformation of L such that increases in the direction d.

Definition: A lattice L is said to be a covering pessimum if thespace Dcrit is of measures 0.

Theorem: If a lattice L has all its Delaunay polytopes ofmaximum circumradius are eutactic and are not simplices then

Q is a pessimum.name # vertices # orbits Delaunay polytopesZn 2n 1

D4 8 1Dn (n 5) 2n1 2

E6 9 1

E

7 16 1E8 16 2K12 81 4

Theorem: The covering density function Q (Q) is atopological Morse function if and only if n 3.


55/55

THANK

YOU

Documents

Mathieu Dutour Sikiric- Perfect forms and perfect Delaunay polytopes