The Missing Direction DifEerential Geometry on Heisenberg … · 2020. 4. 7. · described by Heisenberg manifolds- The research tools used in the thesis involve techniques of Dserential

The Missing Direction and DifEerential Geometry

on Heisenberg Manifolds

Ovidiu C;ilin

A thesis submitted in conformity with the requirements for the degree of Ph-D.

Graduate Department of Mathematics University of Toronto

@ Copyright by Ovidiu Calin 2000

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The Missing Direction and DWerential Geometry on

Heisenberg manifolds Ph.D* 2000

Ovidiu C a b Department of Mathematics

University of Toronto

Abstract

The purpose of this thesis is to investigate the conjugate points along the

sub-Riemannian geodesics and to compute the Carnot-Caratheodory dis

tance in the case of a step 4 sub-Laplacian Ax = +(x: + X;) where Xi =

a, + 4x2 1 xI2& , X2 = a., - 4zl [z12& The conjugate points are generating

the t-axis and there are an infinite sub-Riemannian geodesics of different

lengths which join the origin and any k e d point on the t-axis.

In the second part of the thesis we constnict a Riemannian metric which

extends the subFüemannian metric on the 3-dimensional Heisenberg mani-

folds on R3 and prove that the manifold is Einstein along the horizontal

distribution, Le. the Ricci tensor is proportional with the metric on the hori-

zontal vector fields. Finally we estimate the occurrence of the h t conjugate

points on these geodesics on the Heisenberg group and on the bdimensional

Heisenberg manifoIds in generaL

Acknowledgments

First and foremost, 1 would like to thank my thesis supervisor Professor Peter Greiner for his patient guidance, care and encouragement during my graduate studies.

1 t h d professor Bernard Gaveau of PUnFversite Pierre et Marie Cinie (Paris VI) for many helpful discussions and suggestiom. 1 thaak professor Richard Be& of Yale Universiity who carefiily read my thesis and made valuable sugestions-

1 also want to thank to the staff of the Department of Mathematics at University of Toronto in particular to Ida Bulat and Marie Bachtis who have assisted me greatly during my years at the Universiity.

Finally, I gratefdly acknowledge the financial support of the Department of Mathematics, Toronto Open Fellowship, OGS and OGSST.

Contents

1 Introduction 1

1.1 Sub-Riemannian, Contact and Hormander , , , , , . . . 1

1-1.1 Sub-Riemdan manifolds . - . , , . , . - . . , . - 1

1.1.2 Contact manifoIds , . . , - - - . , . - , . , , . , , . . 1

1.1-4 Horizontal objects . . . . . - . . . . . . . . . - . . . . 3

1.2 Thesub-Riemannianmetric . - - . - . . . - - , - . . - . - . 5

1-2.1 Construction . . . . - - - . - . , - - . . . - . - - - - 5

1.2.2 Sub-Riemannian metnc on the Heisenberg group . - . . 6

1.3 Sub-Riemannian geodesics as horizontal curves . , , . , . . . - 8

1-3.1 Heisenberg group case . . . . , . . , , . . . . . .. . , 9

1.3.2 Hhnander manifolds case - . - . . . - , , . , . , . , - 9

1.3.3 Sub-Riemannian geodesics as constant speed curves . . 10

1.4 The H d t o n i a n associate to X-distribution , . , . - . - - . 12

1.5 The Horizontal connection , - - , . - . - - . , . . . . , - . , - 13

. . . . . . . . . . . . . . . . . . . 1.5.1 Horizontal divergence 15

. . . . . . . . . . . . . . . . . 1.5.2 The Energy of a function 16

. . . . . . . . 1.5.3 Carnot-Carathéodory distance and action 17

. . . . . . . . . . 1.5.4 Geodesics via Hnmiftonian formalism 19

Extrinsic Approach 21

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Curvature 21

. . . . . . . . . . . . 2.1.1 Cartan curvature of a distribution 21

. . . . . . . . . . . . . . . . . . . 2.1.2 R-sectional curvature 22

. . . . . . . . . . . . . . . . . 2.1.3 (1,l )-Tensor of cufvature 23

. . . . . . . . . . . . . . 2.1.4 Scaling factors and curvature 25

. . . . . . . . . . . . . . . . . . . 2.1.5 0 as a hc t ion of K: 25

. . . . . . . . . . . . . 2.1.6 Cufvature on Heisenberg group 27

. . . . . . . . . . . 2.2 The Equation of sub-Riemannian geodesics 28

. . . . . . . . . . . . . . . . . . . . 2.2.1 Pickingthemetrie 28

. . . . . . . . . . . . . . . . . 2.2.2 The horizontal constraint 29

. . . . . . . . . . . . . . . . . 2.2.3 The variational problem 30

2.2.4 The energy and the length . . . . . . . . . . . . . . . . 34

. . . . . . . . . . 2.2.5 Physical signiscance of the cufvature 36

. . . . . . . . . . . . . . . . . . . . 2.3 The skew-symmetric mode1 38

. . . . . . 2.3.1 The curvature on the skew-symmetric mode1 39

2.3.2 The solution of the Euler-Lagrange equations in the . . . . . . . . . . . . . . . . . . . skew-symmetric case 41

2.3.3 Cufvature integral invariant for the skew-symmetric case 43

. . . . . . . . . . . . . 2.4 Missing direction for contact manifolds 45

. . . . . 2.5 A more general model than the skew-symmetric case 49

. . . . . . 2.5.1 A geometncd interpretation for coordinate t 52

3 Conjugate points and Jacobi vector fields 55

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Definitions 55

. . . . . . . . . . . . . . . 3.2 Curvature dong Jacobi vector fields 56

. . . . . . . . . . . 3.3 Exponential map on Homander manifolds 57

. . . . . . . 3.3.1 Ekponential map on the Heisenberg group 61

3.4 The rotational symmetry of the Euler-Lagrange equations . . 64

. . . . . . . . . . . . . . 3.4.1 The Euler-Lagrange equations 64

. . . . . . . . . . . . . . . . . . 3.4.2 The 1-connection form 65

. . . . . . . . . . . . . . . . 3.4.3 The rotational symmetry 67

. . . . . . . . . . . . . . . . . . 3.4.4 The conjugate points 68

4 The Main Result 71

. . . . . . . . . . . . . . . . . . . . 4.1 The Study of a step 4 case 71

. . . . . . . . 4.1.1 The Euler-Lagrange system of equations 71

. . . . . . . . . . . . 4.1.2 Solutions which &art from ongin 73

. . . . . . . . 4.1.3 An explicit solution in polar coordinates 77

. . . . . . . . . . . 4.1.4 The t-component dong the solution 80

. . . . . . . . . . . . . . . . 4.1.5 Conjugate points to origin 81

4-1.6 The length of the geodesics between the origin and (0.0. t) . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.1.7 Ekplicit solutions which join O to (O. O. t) . . . . . . . . 89

4.1.8 Solutions which start outside the origin . . . . . . . . . 91

. . . . . . . . . . . . . . . . . . . 4.1.9 Case sgn(k) = sgn(0) 92

4-1-10 Case sp(k) # ssgn(0) . k # O . . . . . . . . . . . . . 95

4.2 Conjugate points for a step k + 2 case. keven . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.2.1 Euler Lagrange equations . . . . . . . . . . . . . . . . 98

4.2.2 Solutions which &art fiom ongin . . . . . . . . . . .. 99

4.2.3 An explicit solution in polar coordinates . . . . . . . . 103

4.2.4 The number of loops . . . . . . . . . . . . . . . . . . . 105

4.2.5 Particular Cases . . . . . . . . . . . . . . . . . . . . . . 107

4.2.6 The t-component dong the solution . . . . . . . . . . . 109

4.2.7 Conjugate points to origin . . . . . . . . . . . . . . . . 110

. . . . . . . . . . . . . . . . . . . . . . . 4.3 The second variation 112

4.3.1 The second variation of the action on Heisenberg group 113

4.3.2 The accessory Vanational problem for S(4) . . . . . . . 114 . . . . . . . . . . . . . . 4.3.3 Getthg the Jacobi vector field 117

. . . 4.3.4 The second variation for the skew-symmetric case 119

4.3.5 The second variation for 3-nim Heisenberg . . . . . . . . . . . . . . . . . . . . . . . . . m d o l d s 120

5 Horizontal objects as Mt of Riemannian objects 122

5.1 A family of metrics which apprmimate the Carnot-Carathéodory metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

. . . . . . . . . . . . . . 5.1.1 The construction of the family 124

. . . . . . . . . - * . . . . . 5.1.2 An orthonormai triplet .. 126

. . . . . . . . . . . . . . . . . . . 5.1.3 The bicharocteristics 126

. . . . . . . . . . . . . . . . . . . . . 5.2 A few important tensors 127

. . . . . . . . . . . . . . . . . 5.2.1 The Christoffi Symbols 127

. . . . . . . . . . 5.2.2 The Riemannian Tensor of Curvature 128

. . . . . . . . . . . . . . . . . . . . . 5.2.3 The Ricci Tensor 129

. . . . . . . . . . . . . . . . . . . . . 5.2.4 The Ricci Scdar 129

6 Horizontal Einstein space property 130

6.1 Horizontal Einstein space property on the Heisenberg group . 130

. . . . . . . . 6.1.1 An analogous of the Gaussian cunmture 132

. . . . . . . . . . . . . . . 6-2 Estimation of the Conjugate Points 135

. . . . . . . . . . . . . . . 6.3 Properties of Levi-Civita connection 138

. . . . . . . . . . . . 6.4 Carnot-Carathéodory distance as a Iimit 144

. . . . . 6.5 Horizontal Einstein property on Honnander manifolds 147

. . . . . . . . . 6.5.1 Construction of the Riemannian metric 147

. . . . . . . . . . . . . . . . . . . . 6.5.2 Christoffel SymboLo 148

. . . . . . . . . . . . . . . . . . . . . . . . 6.5.3 Ricci Tensor 149

. . . . . . . . . . . . . . . . . . 6.5.4 The Einstein Property 150

6.5.5 An Estimation of the ocmence of the fkt conjugate point . . . . . . . . . . . . . . . . . . . . . . 154

Appendix A ......................................................... -155

Appendix C ......................................................... -162

Appendix D .................................~....................... -163

Appendix E .......................................................... 164

References ........................................................... 165

List of Figures

Figure 1: ~ h e sped and the cwature ~ ( 4 ) ................... -37 Figure 2: The BI&-tube ........................................ -44 Figure 3: The geometrical interpretation for nîj(r) ............... -54 Figure 4: The subRiemanni'an geodesic on the Heisenberg group . -63 Figure 5: The potential V(r) and r,, ........................... -74 Figure 6: A rotation of w = 7r/6 ................................. -76 Figure 7: After a rotation of r / 3 ................................ -78 Figure 8: The x-projection of the solution ........................ 80 Figure 9: The shapes of the soIution for m = 1,2, 3 .............. -90 Figure 10: The solution in the phase plane ....................... 94 Figure 11: The x-projection of the solution ....................... 94 Figure 12: The solution in the phase plane ...................... -96 Figure 13: The x-projection of the solution ...................... -97 Figure 14: 2k + 1 sector areas .................................. -106

........ Figure 15: The trajectory in the Heisenberg case (k = 0) -107 Figure 16: The trajectory for the case k = 2 .................... -108 Figure 17: The trajectory for the case k = 4 .................... -108

List of Appendices

Appendk A* . The solution of Euler-Lagrange system on Hi ...... 155 ................ Appendix Br The lengths of the geodesics on Hl 160

Appendix C: The harmonic oscillator, another example ......... -162 Appendix D: The physicai interpretation for fl and w ........... -163 Appendix E: EIliptic hctions .................................. -164

List of Notations

the field of real numbers a horizontai non-integrable distribution of dùn 2n in RZnC' 2n linear independent vector fields which span the horizontal distribution 'f. the set of vector fields on the manifold M the tangent space at p of the manifold M the set of smooth functions on the manifold M the Heisenberg-Laplacian d&ed as f (X: + -- + Xh) the horizontal connection the boundary of the domain U the 3-dimensional Heisenberg group the Lie derivative in V direction the dvectional curvature 1-connection form suc. that k m w = 3d 2-form deked as dw the orthogonal group of matrices of dimension n the action dong a solution of Euler-Lagrange equations the Hamiltonian associated to the fundion f the Carnot-Caratheodory distance between A and B a Riemannian metric which approximates the subfiiemannian metric Christoffel symbois the Levi-Civita connection associated to hA the Hessian of the fimction f the standard cornplex structure of the plane a plane rotation of angle a the order of conjugacy of the conjugate points O and P the expondal map at p the Ricci tensor on vector the fields U and V the Ricci scalar the length of the c u v e c the mal inner product on IL" the vectorial product on the exterior product for differentiai forms the 3-dimensional Heisenberg group

Preview

The subject of the thesis is situateci at the intersection of three important fields of the modem Mathematics: Partial DEkrentiaI Equatios, Differential Geometry and Calculus of Variations. It is related to the ideas of Be& R., Gaveau B. and Greiner P. who have developed the theory of Green hctions for subelliptic operators on Heisenberg manifolds (see [B Gl , [BGGl] , [BGGZ] , [BGG31, [BGG41). The thesis is concerned with a geometrïd approach of Heisenberg manifolds pursuing ideas fiom papes ([St], [Tl, N, [Gro], etc) or introducing new ones, regarding notions of curvature, connections, second d a t i o n , Jacobi vector fields and conjugate points. The main objective is to understand how quantum behavior of particles is described by Heisenberg manifolds- The research tools used in the thesis involve techniques of Dserential Geome- try (like the the- of connections, contact manifolds, tensors and curvature), Equations, Lagrangian and H d t o n i a n formalism. Specific chapter descriptions, with the emphaasis of the original results, are as follows.

Chapter 1: Starting with 2n vector fields on R2"+l one defines the Hormander structure, the subRiemaanian metric and coIlStNcts a Hamiltonian as the principal symbol of the sub-Laplcian. Using the Legendre transform to the above Hamiitonian, it is associated to a Lagrangian- This consists of an energy term and a constraint term. The solutions of the Euler-Lagrange equations are horizontal and constant speed cunres- The horizontal connection is d&ed as well as other horizontal objects which wil l be used in the next chapter, like energy of a hction, action and Carnot-Carathéodory distance-

Chapter 2: This chapter is regaxding an ez t r i~ ic approach, namely an a p proach which makes use of objects defined using the 1-form w or 0 = ch. Dinérent sorts of curvature are definecl. One of the above curvatures (the directional one) is usehl when one considers the variational problem and

writes the Euler-Lagrange equations. If a s p d direction for the missing direction is considered (the similar direction for &) the Euler-Lagrange equations take a very elegant invariant form. Something simiIar, but on fibrations and without using the crimature and the migsug direction can be found in Fr021 - The fùst variation is computed and Euler-Lagrange equations are found. Written invariantly (namely v4d = -x(&) this equation may be interpreted as Newton or Lorenz equation. Rom the geometric point of view, it equates the acceleration to curvature- Unlike in RiemaMian Geometry, where the geodesics aze defined as zero accekration m e s , on Heisenberg manifolds the gwdesics are bent by a force due to the nonintegrabiliw of the horizontal distribution- Physical interpretations for the theorems are provided- The solutions of Euler-Lagrange equations are interpreted as tra- jectories of charged particles in magnetic fields. The nonzero curvature in the formula v ~ & = -AK(&) means that the particle is accelerated all the time. The 2-fonn of curvatture S1 has the significance of the magnetic field, and di2 = O is the Maxwell equation. The directional curvatture in Euler- Lagrange equations plays the roIe of Lorenz force whkh bends the geodesics- Particular cases of Heisenberg manifolds are examined, solving the Euler- Lagrange equation explicitly and showhg geometrical interpretations for the Miiable t (as a trace of the product between the curvature matrix and the projections area matrix) .

Chapter 3: The chapter deals wi th the relation between Jacobi vector fields and exponential map. Similar results hold in Riemannian geometry. The above results are applied on the Heisenberg group. A study of the occurrence of conjugate points along the t-zuàs is done.

Chapter 4: This chapter contains the study of a step 4 operator- The Euler- Lagrange equations are solved and it is proved that the conjugate points are generating the t-axis (Theorem 4.1). A family of geodesics of difkent lengths is obtained between the origin and any fixed point on the t-axis (Theorem 4.2). In the case of the solutions which do not start from the origin a qua& tative study is done, Some of the above results are generalized to the k + 2 step case (k even).

Chapter 5: This chapter contaias a construction of a 1-parameter family of Riemannian metrics which extend the subRiemannian metric on the Heisen- berg group. Christoffel symbols, Riemannian tensor of curvature, Ricci tensor and Rlcci scdar are computed for this metric -

Chapter 6: Here it is stated that the Ricci tensor is proportional with the metric dong the horizontd distrr'bution (Einstein prop-) and provide bounds for occurrence of the conjugate points dong the geodesics in the metric introduced in chapter 5. There are also some results regarding an approximation of the Carnot-Carathéodory distance using Riemannian distances. A study of the Levi-Civita connection and sectional cutvatures on the Heisenberg group is done.

Chapter 1

Introduction

1.1 Sub-Riemannian, Contact and Hormander

structures

1.1.1 Sub-Riemannian manifolds

A subRiem;uinian manifold is often defined as a m d o l d M of dimension n together with a distribution 'H of m planes (m 9 n) and a Riemannian metric g on X. One can mite it as a triplet (M, Yi, 9). The notation for 3C : x -, 3C, stands for the horizontal distribution and a vector field X E 3C, Le. X, E 7C, is called horizontal vector field. A study of sub-Riemannian manifolds was done in [BC], [BG], [Be], [Gro] , [Be], [Mo], [St] , [Tl etc.

1.1.2 Contact manifolds

As a special case of sub-Riemannian manifo1às one may define the contact manifolds. A contact structuw on M, where dim M=2n+l, is a distribution 31 : x + 3C, of rank 2n which is non-integrable.

A contact structure on M is dehed by a 1-form w which satisfies the Robe- nius nonintegrab'ity condition

The horizontal distribution is given by 'fL = k m W. When 272 i- 1 = 3 Ekobenius condition is equivalent to # O (see [Ar]) which means in fact that the Cartan curvature is nonzero. This condition is also quivalent to nonintegrabilily of the distribution 3C.

1 1.3 H6rrnander manifolds

A manifold M of odd dimension 2n + 1 is cded a Hormander manifold if there are dehed (Iocally) 2n lin- independent vector fields {XI, X2, --Xh) on M such that the foUowing HOrmander condition holds: (*) The vector fields {XI, X2, -.Xh) Lie-genemte the tangent space of M ut every point. By Lie-generute we mean that Xi together with their iterated brackets [Xi, Xi], [Xi, [Xjl X,,]].-. span the tangent space at each point of M. Hormander proved (see[Hq) that if (*) holds, then the operator Ax = '(x: 2 + ... + Xi) is hypoelliptic, nameIg Axu = f E Cm * u E Cœ. If Xi, X2, ..-X2* together with the f b t brackets generate the tangent space at every point (as in the case of the Heisenberg group) then Ax is said step 2 operator. In generd, if the maximal number of brackets needed to generate the tangent space at every point is k - 1, the operator Ax is said to be step k. When step k = 2 the Hormander manifold is d e d Heisenberg manifold. Condition (*) is known in sub-Riemannian geometry also as Chow's condition, after the name of the mathematician who proved (see[Ch]) the con- nectivity theorem for a sub-Riemannian case which is the analogue of the Hopf-Rinow's theorem from Riemannian geometry. The theorem States that

Theorem 1.1 (Chow) If condition (*) hokh ut every point on a connectai manifold M, then any two points can be joined by a horizontal curve, Le. a eurue tangent to the horizontal distribution 3C.

Here the Hhander manifolds will be studied for the case M = P -

1.1.4 Horizontal objects

A horizontal objeet (HO) is defined as a mathematical object which can be constructed directly fiom the horizontal distribution 31 and the metric g on R. In the case of a m;mifoId, when just the vector fields XI, -., Xzn are given, a HO depends just on the vector fieIds Xi and on the mehic g, which makes Xi's orthonormal. If we lïve on the horizontal distribution 3C we may know nothing whatsoever about the extemal stmctures of the space. We can measure the lengths of the horizontd vectors in the metrîc g, on 3L, and measure the distances using the Carnot-Carathéodory distance. W we can know are HO and the geometry built up is using HO. One goal is to recover the missing direction (MD) by means of HO geometry- For instance the X-Lsplacian Ax = f xzl X? is a HO. Another HO is the X-gradient of a function

Using the g metric on 3L one can define

and construct another HO which is the Hamiltonian or the energy of f

There is a theorem which states that H(V f) = O ifE f is a constant function. If one defines the energy of f on U as

then &(f) is not a HO because it depends on dv which involves the Iocd coordinates 21, .-, Xzn+l- However, a r d t states that the crît id points of

satisfy the equation

(1-7) Axf = O

so that f is a HO- If one considers

it tum out that the Green's function is also a HO- The bicharacteristics (~ (s ) , as)) are also HO because the Hamiltonian is defhed as the principal -bol of Ax as

where (, ) denotes the Euclidian inner product. The x-projection of the bicharacteristics, which are solutions of Euler-Lagrange equations are also HO and the action

will also be a HO. This is related to the Carnot-Carathéodory distance dc which is another important HO. Another HO dehed in the thesis is the horizontal connection D : 3C x3C -i 3C

which is useful when one wrïtes the Euler-Lagrange equations in an invariant fonn on the Heisenberg group. Rom the connection D one can be consfruct other HO as torsion, X-trace of D which is the horizontal divergence- The horizontal divergence of solutions flow $(s) is also a HO and

1.2 The sub-Riemannian metric

1.2.1 Construction

The sub-Riemannian metric appears in the literature under a varie@ of names: singular Riemannian metrics ([Hal), Carnot-Carathéodory metrics ([Be], [Gr], [Pa]), sub-Riemamian metrics ([Stl) and nonholonomic Riema- nnian metrïcs ([Ve])-

cf v E &, then v = CE, viXi(x). As Xi, X2, ..Xk are linearly independent and the hinctions vc are unique. Dehe the length of the v e r v as

By polarkation g, can be d&ed as g, : 31, x ?l= L, R,

In the case v 6 Hz we define

Following an idea in [Be] one can Say that if a, : R2- -+ 31, is defined as

then let p, = 0;' and define

In this met& one c m verify that the vector fields Xi's are orthonormal

Definition 1.1 The positive defmtte quadmtic fotm g, d e w on 3C, is called the subRiemIinriian metEic associated m*th the vector field system xi, --1 x2,- A curve 4 : [a, b] -+ M îP mlled a horizontal curve i f &s) E ?f4(+ Vs 's [a, b]. Onci the met* g is defned, one can nteasure the length- if the horizontal curues as

and the Carnot-Cbathéodory distance as

which is a HO (horizontal objeet).

1.2.2 Sub-Riemannian metric on the Heisenberg group

Heisenberg group constittutes a paradigm for the theory, The 3-dimemional Heisenberg group Hl may be realized as Ra x R = {(x, t)) endowed with the group law

The vector fields

(1.22) xi = a,, + ZX& , xz = a,, -2x1at , T = a,

are left invariant and generate the Lie algebra of &. The distribution defined by XI, X2 is not integrable, as

As XI , Xa, -& span the tangent space to @, the Hormander condition is satisfied in step 2 case. The Heisenberg Laplacian on is the left-invasirnt operator

(1 -24)

and it is hypoelliptic. The principal symbol for AH is given by the formula

It is denoted with H to suggest that it wil l be considered as a Hamütonian. In fact H is a quadratic form on T*@, with (2, t) E #Z3 and (cl O) E T(S,,~ l@. The next step is to associate to the Hamiltonian H a Lagrangian L : TR3 +

It consists in considering the convex d a c e given by the Hamiltonian and computing the m e a l distance fkom the hyperplane (c, x) + + to thk sur-

It is supposed that the supremum is reached. In this case the derivatives are zero (1.27)

The &st of the above equations can also be wrîtten as

which are in fact Hamilton's equations. Computing in (1.27) one obtains

The second equation in (1.28) can be written as

Using (1.29) it tunis out that dong the solutions #(s) = (xl (s) , x2 (s) , t (s) ) we have (1.30) t(s) - 222 (s) xi (9) + xl ( s ) x ~ (s) = O

As will be shown, formula (1.30) is a necessary and sacient condition for 4 to be horizontal, The Lagrangian becomes

Using (1.30) one obtains that dong the solutions of Euler-Lagrange equatiom (and in general dong horizontal m e s )

which is in fact a halt of the squared Iength of the tangent &s) in the sub- Riemannian metric- The sub-Riemannian metrîc can be viewed as

Observation In general the Lagrangian will take the form

1.3 Sub-Riemannian geodesics as horizontal cmves

The Hamiltonian is the principal symbol of Ax

The projections of the solutions of the Hamilton's system

%(O) = O 1 x(1) = x

on the x-plane are c d e d sub-Riemannian geodesics.

1.3.1 Heisenberg group case

In the case of the Heisenberg group, one of the Hamilton's equation states that dong the bicharacteristics

(1.34) f (s) = ~x~(s)x&) - 221(~)*2(~)

On the other hand we have the following result

Proposition 1.2 A curve O(s) = (zi(s), q (s) , t(s)) is ho~mntal i# the relation (1.34) takes place. In particular the sub-Riemannian geodesics on the Heisenberg p u p are horizontal curves.

Proofi Indeed, one can write

)(s) = (ii&), k2 (s) , @)) = *&)x~ + =2 (s)& + E(s)& =

= a1(s)xi + x2(s)x2 + (t(s) - ZZ~(S)X~(S) + ~X~(S)X~(S) )&

Hence ks) is horizontal iff

Le. (1.34) takes place.

The next subsection generalizes this result.

1.3.2 Hormander maaifolds case

Let XI, Xz, .., X2= be the vector fields which d&e the horizontal distribution. If 4(s) = (3C1(8), -.x2,,(s), t (s) , f i (s) , c2 (s) , e (~ ) ) is the solution for the bicharacteristics system, then the projection on the (x, t)-hmerpIane is the solution for the Euler-Lagrange system of equations.

Proposition 1.3 The sub-Riemannian geodesics are horizontal Cumes.

Proof: The associate Hamiltonian is

and Hamilton's equations are

wbich can also be written as

and hence

wbich becomes

1.3.3 Sub-Riemannian geodesics as constant speed curves

If 4 is a horizontal curve, &s) = C ii(s)Xi, then the length of # in the

Proposition 1.4 Along the sub-Riemannian geodesics the length of the speed, measumd in the sub-Riemannian metrie, i s constant-

Proofi As we have seen, the subRiemannian geodesïcs are horizontal m e s , so that, if #(s) = (xi@), . . Z~~+~ (S ) ) is a solution,

The Hamilton's eauations

for the Hamiltonian 2n

where

and the H d t o n i a n dong the solution can be wrïtten as

On the 0 t h hand the Hamiltonian H does not depend explicitly on the parameter s, so that

i-e. H is constant almg solutions. If we take in consideration that the length of the tangent d>(s) in the subRiemannian metric is exactly Jx:(s) + .. f xh(s), it tums out that the speed &s) has a constant length. rn

1.4 The Harniltonian associate to %distribution

We have seen that the Hamiltonian associate to 2n linear independent vector fields XI, Xz, . . - , X2,, is

where ( , ) denotes the usual inner product on Hn+l. It seems that the H d t o n i a n depends on the vector fields XLs. In fact it depends just on the distribution 7C and the Carnot-Carathbdory metric on it which d e s the vector fields XLs orthonormal. Indeed, let {&, .., &) be an orthonormai frame in 3C. Then at each point x exists a matrix 4 E O(%) such that Xk = AkjEj . Then

So that one can talk about the Kamiltonian associate to the distribution X and Carnot-Carathéodory metric g on 'fL dehed as

where (Ej) , is any orthonormal trame in ?&. The system of bicharacteristics is

The sub-Riemannian geodesic flow is

Hence the solution 9 depends on the horizontal distribution 3L and the sub- Riemannian metric g.

1.5 The Horizontal connection

Defmition 1.2 Let z -t 3C, be the horuontal distdmtion and {&, .., E2n) an orthonormal bas& in ?&:

where ( , } denotes in th& case the sub-Riemannian metRc on RZnf'. Dejine the horizontal connection as

The following result shows that the definition is correct.

Proposition 1.5 The definition of the connection D does not depend on the CEk).

Proof: Pi& another orthonormal base É = {Ex). Then exïsts a matrix A E O(2n) such that É = AE. Then the following computation takes place at a point z

and the coaoection D does not depend on the basis,

- In paaicular one can choose Ek = Xk, k = 1,2n, provided that ( , ) is a mefxic such that

Therefore

(1.40)

where V and W are horkontal vector fields and {XI, .., Xh) are the given vector fields which span 3L.

Proposition 1.6 D id a lànear metric connedion.

Proof= D is IR-linear in both azguments and F(BZ2n+1)-linear in the fi& argument. One has to verify Leibnitz d e of differentiation

where we used

For the second part one has

2 n 2 n 2n

C(Wv, xi)) (Xc, W ) + ~ ( V , xi) U(W, Xi) = U ( ~ ( W , xi) (xi, w)) = i=l i=1 ï=1

and hence D is a metric connection. m

1.5.1 Horizontal divergence

Deanition 1.3 Let Z be a horizontal vector field. One defines the horizontal divergence of Z as the trace of the horizontal connection

where the Dace .ds taken over W.

Proposition 1.7 Using the fields Xk, the horizontal diveryence can k ex-

Proposition 1.8 If Vx f = C, X'(f )Xk M the X-gmdient off, then

Proofi This is just a computation which is using (1.43)

1.5.2 The Energy of afunction

Definit ions

Let p + 3Lp = spanp{Xl, .., X2*) = 'Wp be a non integrable distribution on R2"*'. If 4 is a function, the X-gradient of 4 is defined as

e l

The energy density of 4 is de- as

and the energy on a bounded set 0

The invariance of V x f and H(V$)

The vector fields XI, .X2, are orthonormal in the Carnot-Carathéodory metric. If &, .-& îs amther orthonormal frame, then Ei = x, A-&&, where

so that the X-gradient depends just on the distribution 3C and the metrïc on it, In a similar manner, H(Vq5) does not depend on the orthonormal base used

H(V#) = O i- f - f . # = constant

Proof: Suppose that q5 is not constant. Then there exists a point p such that (Vg), # O where V is the gradient written in a metric on Pf? Let c = q5@) and Sc = q5-l(c) be the hyperdace through p. As Xi(@) = O it turns out that 4 is constant dong the integral curves of XI, so that the integral cuves through p will stiU be in S, and thus XI E X(S,), i.e. is a vector field tangent to the d a c e Sc, at least in a neighborhood of p. Similarly we get Xa, Xs, ..X, E X(S,). It tums out that the distribution {Xi, -&} is (locally) integrable with Sc integral hypersurface, which is a contradiction. Hence q5 is a constant function. rn

1.5.3 Carnot-Carathéodory distance and action

The relation between the action and the Carnot-Carathéodory distance is given in the following result:

Proposition 1.10 If S(T) is the action @en by a minimizerfrom O to z(r) then 1

Proofi The action is

where #(s) = (zl(s), .., x ~ ~ ( s ) , t(s)) satisfies Euler-Lagrange equations. As

(1 -5 1) br d m da 5 ([(c 5: (s) )ds) II2 (Ir O d ~ ) II2

As along the bicharacteristics the energy is constant, it turns out that in the above relation there is an equalitsf and thus

Fom where we get the desired result.

1.5.4 Geodesics via Hamiltonian formahm

In this section M will be a clifferenfiable manifold endowed with a Rlema- nnian metric g. Let # : I -+ (M, g) be a curve and consider the Hamiltonian

The goal of this section is to show that the solutions q5(s) for Hamilton's system

(1.54)

are geodesics.

hence (1.56) pk = #&k

In the following we'll compute 8gij/Bxk- As gi*g, = 6:, one obtauis

Multiplying by g'j and sum over s we get

Differentiate in (1.56) and get

Substitute (1.56), (1.57), (1.58) in (1.55) and obtain

Making in the right haad side c = b, d = r one obtains

6b9Lb + r r b d b # r = 0

Multiplying by gks and using that gkb gl' = 6; and rZb = gks rrbk one obtains

(1-59) & +rib #r = O

which is the equation of geodesic in local coordinates and which is quivalent to the invariant equation v&# = O, where V is the Levi-Civita connection wîth respect to the metric g-

Chapter 2

Extrinsic Approach

2.1.1 Cartan curvature of a distribution

Let x -t M = sp,(Xi, -.Xm) be a horizontal distribution on R ~ " ~ ' .

Definition 2.1 The connection 1-form M a fonn w E T*R~"+' such that

w $ 0

(2-1) k e r , w = K

The connection 1-form is unique up to a multiplicative factor.

Proposition 2.1 If s + is a non-integrable distribution, then, locally, the connection 1-form is not closed (dw # 0).

Proof: The contrapositive is: If w 2s closed then the horizontal distribution às in$egmble-

If w is closed, it is exact: there is a fiinction f such that, locally w = df- Then w(X<) = O can be written as df (Xc) = O or Xi(f) = O, V i = 1,2n. This means that f is constant along the integrd curves of Xl, -.X2m. It follows that the hyperdace C = {f = concastant} is a integrable hypersurface for the distribution ')C-

In our case the horizontal distribution is always non-integrable so that w is not closed-

Definition 2.2 The curvature 2-fom of the dishibution is defined as 0:3Cx3C+3 (2-2) n(x, Y) = &(x, Y)

The 2-fom R is also d e d Cartan curuature and measures the degree of nonintegrab%@ of the horizontal distribution.

Let r be a 2-plane. The sectional (Cartan) cuntatu~ along ?r is defined as

where {X, Y) is any orthonormal system in r. The above definition is correct in the sense that if {x, Y) is another orthonormal system in n, with the same orientation as {X, Y), then

Indeed, if

as (c d) is an orthogonal matrll.

The sectional cufvatwe is defined for every plane .rr not necessary horizontal. In this case the metric in which the kames are orthonormal is a metric which extends the sub-Riesnamian met&-

2.1.3 (1,l)-Tensor of curvature

Definition 2.3 Let K: : 3L + X be defined as

called the (1,l)-tensor of nimature, w k {a, -.E2=) îp an orthonormal fkme in the Carnot-Camthéodory metrz'cC

K mesures the curvature dong one direction U. It is indeed a tensor because

The definition of K: does not depend on the orthonormal fiame chosen. In- deed, there is an orthogonal matrix A E O(2n) SU& that Ek = AkjXj- SO that

2n

K(U) = C R(U, AkjXj)AkjXj = l-1

Defhîtion 2.4 If $(s) is a horizontal cume we defie the cumatu+e vector field along 4 as ~(4 ) .

Proposition 2.2 In the contact case manifold the curvatum vector field along any horizontal cvrve 2s nontero euerywhere

Proofi If q5 Is a horizontal curve, then

Suppose that ~ ( 4 ) = O at a certain point p. Then xj&fi2(~j, &), = O, for every k = 1,2n. If = a(&, Xi), then (njk@), = 0 which is a linear homogeneous system with a nonzero solution 6,. Et follows that det (Ogj), = O, i-e. !2 is a degenerate form- This is a contradiction. (0 is a closed, non-degenerate form, Le. a simplectic one). Hence ~ ( 6 ) # O everywhere. I

De5ition 2.5 The scalar (Cartan) curvatuie of a horizontal curue $(s) is the length of the m a t u r e vector field, measured in the subRiernannian metric

Hence 2n

The curvature vector field and the scaiar (Cartan) curvature can be defined for every curve, not just for the horizontal ones. If c(s) is an arbitrary m e , then define the c m t u r e vector field for c(s) as K(@), where @ = proj&

2.1.4 Scaling factors and curvature

As we have seen, the horizontal distribution is given by W = ker w, where w is the 1 -fom of comection- IIi fact w is not unique, If is a nonzero smooth fuaction, then the 1-form 3 = a w gives a h the distribution H as ker G = R- We Say that the 1-con.necfiecfion fotm is unique up to a multiplicative factor. The cufyature vector field dong a horizontal curve #(s) with respect to w is

and with respect to iS is

As 4 and Xk are horizontal, w(#) = w(&) = O, so that the above determinant is equd to zero. We arrive to the following result

Proposition 2.3 The cumatvre is defined up to a multiplicative factor, namely, if W = aw then (2.8) R(4) = an(& for any horizontal curve 4.

The relationship between the 2-form Q and the directional curvature K is given in the following result

Proposition 2.4 If g 6 the Carnot-Camthéodory m&cY then for every U , W € X

(2-9) Q(v, W ) = -dv, qw))

= n(w; u) = -n(u, w) which proves the proposition.

CoroIlary 2.6 For every U, W E 'U the following relation tukes place

(2.11) w([U, WI) = du, wv)

Corollary 2.7 For euery horizontul vector field U E 3L we have

(2.12) QJ)) = 0

namely K(U) is normal to U in the Carnot-Cumthéodory rnetric.

Proof: Make U = W in the Proposition 2.4 and use that Q(U, U) = O

2.1.6 Curvature on Heisenberg group

Curvature of the distribution

Consider on Pf' the foilowing two vector fields

which span the horizontal distribution- The 1-form of connection is

The curvature 2-form is

The curvatuce dong the pair of vectors (Xi, X2) is

Fkom the geometrical point of view !R(X, Y) means the area of the projected parallelogram (X, Y) on the (x l , x2)-plane. If ?r : @ + H Z ~ ~ ~ , ) is the cane nicd projection (which sen& the horizontal distribution ont0 E&,) then $(X, Y) is the area of the parallelogram generated by ?r,(X) and %(Y).

The constant cuwature of sub-Riemannian gwdesics

It is possible to show that the scalar Cartan curvature along every horizontal unit speed curve on Hl is constant. In particular, the scalar cuirvature along the solutions of Euler-Lagrange system (sub-Riemannian geodesics) will be constant-

The cumature R is described by a main the {XI, X2) system of vectors

Let c(s) = (xl(s) , z2 (s) , t ( s ) ) be a horizontal cmve, namely f (s) = xl(s) XI + x2 (s)X2- Then the curvature vector is

where is the comp1ex structure of the plane- A computation shows

Hence the curvature is constant and equal to 4 dong every horizontal unit speed curve.

2.2 The Equation of sub-Riemannian geodesics

In this paragraph an equation satisfied by the sub-Riemannian geodesics will be established- It consists of equating two terms: the acceleration and the curvature.

2.2.1 Picking the metric

The horizontal (nonintegrable) distribution H is generated by the vector fields {XI, X2, ... Xan}. There is a one-form w such that ker(w) = R. We assume that there is locaily a unique direction T (detennined up to a multiplication by a non-vanishing fuction) such that, if C2 = dw , then n(U, T) = 0, for a l l vector fields U. In fact it is enough to assume that U is a horizontal vector field. A Riemannian metic which makes {Xi, ..Xh, T) orthonormal such that the length of T is \/X is chasen. Denoting this metric by hA one may wrïte

In order to emphasize the length of T sometimes the notation TA will be used.

2.2.2 The horizontal constraint

Let {XI, .., X2n) be linearly independent vector fields on R2n+1 which define a non-integrable distribution 3L. The problem of finding subRiemannian geodesics becomes the following constraint problem:

PROBLEM: Minimze the length of a curue #(s) between two poànts, subject to the nonholonomic construânt

(2.18) &s) E s ~ n ( X 1 , - - y X2n) =

I£ w is the 1-form of connection, Le. ker w = X, the following result hoIds.

Proposition 2.8 If # is a horkontal curve, then

Proof= Let 6 : [O, 11 -t lUh*'. As f w is a 1-form on R, then @w and ds will be proportional

~ * w ( s ) = h(s) ds

The hc t ion h(s) is given by

Hence

h ( ~ ) d~ = JI1 m*(w>(.l(;) ds =

2.2.3 The variational problem

Let A and B be two fixed points and

be the space of aIl smooth m e s from A to B parametrized by [O, 11 on a difkrential manifoId A'K The energy with respect to the metric hA is

(2.21) E A : h + R ,

The tangent space to SZAB at y0 is

(2.22) { W ; W is a vector field dong 7 0 with W(0) = W(1) = 0)

and the tangentid map of Ex is defined between

If 70 is a minimum on the space for EA(r), then (EA) , , = 0. If one is interested in hding the c m e s which joui A and B that have the smallest energy among all the horizontal c w e s between A and B it is necessary to search for a minimum of Ex on the subspace

Consider the minimum of Ex on the set of zeros of

Hence one will search the following functional

where p is a Lagrange multiplier, w is a 1-form such that w(Xi) = O, or 3L = ker W . For instance in the case of the Heisenberg group Xi = 8,+2z1&,

X2 =a,, -2z& and w = d t + 2 ~ ~ & ~ -2x2dx1 and& =4dzl h d ~ = O . If 7 is a minimimi for EA + pG, then

which means that

Fbding (EA),,~ (W) This is present in almost ail books of Differential Geometry (see for instance [KN] or [Sp]) but will be induded here for a continuiw of ideas and d o d ~ of notations- Let ~.(s) = ~ ( s ) + EW(S) a evariation of y with fixed ends. Then

where we used that VA is the Levi-Civita co~mection. As W and j commute and using that VA is a metric connection, the above integral becomes

1

= hA(w4) 1; -b hA(WJ7$4)

As W(0) = W(1) = 0, the first term in the above relation is zero. Hence the e s t variation for the energy is

.. - ~

Considering the same variation

one obtaïns

1 1

= 4 ~ W W (i.(s>> = J/ ww i) Hence the tangentid application of G at 7 is

This formula is present for instance in MO^]. The right hand side does not depend on the metric h? This d l be worked witil something as hA( . , W) wiU be emphasized- If {Xi, Tj, ) is an orthonormal system in hA, where the horizontal distribution is X = span(Xl, ..Xz,), then define KA : X x X as

As TA was chosen such that for every vector field V one has Q(V, TA) = 0, the last tenn does not depend on X and

The relationship between R, hA and K: is given in the foiiowing result.

Proposition 2.9 For every vector field V

Proofi As V is decomposed in

then, using the orthonormalitg of the vector fields in the metric hA one has the computation

h A ( ~ ( # ) , V) = h A ( ~ X ( 6 ) , V) =

Hence, the nrst variation for G becomes

becomes

1' hA(-va) + pK($), W) = 0 , V W E X

Hence the Euler-Lagrange equations may be written as

(2.36) v;i. = n(9)

where p is a constant (Lagrange multiplier).

As the nght hand side does not depend on A, it turns out that also the left hand side does not depend on A- The constant p has the physical interpretation of electrical charge of a unit mass particle in a constant magnetic field given by a closed 2-fonn 0 = dw (whidi has the ker pointing to 4 direction). The acceleration is ~ $ 6 and the Lorenz force is /rlC(j) -

Theorem 2.10 The Euler-lAgmnge equation for the jhctional (2.26) is

v;+ = pK(?)

Proposition 2.11 If y ts a solution of (2.36) then +y has a constant speed.

because is a skew-symmetric form and TA is a vector field with the property a(?, TA) = o. w

2.2.4 The energy and the length

As the length of a c m e does not depend on the parameterization, one may consider the curve unit speed parametrized. h m the Holder inequaliw given in (1.51) one has (2.37) WJ) 5 dm

where # ïs defined on [O, 71. For a unit speed paxameterization the above inequality becomes an identity and to minimbe I ( # ) it is enough to minim- imize the energy E(q5). The problem becomes:

PROBLEM 1: iKinfm2'ze the energy of a mit speed cume cvrue between two points, subjeCt to the eonstmiizt &s) E ?d#(sl, Vs E [O, T] - The problem can be fonnulated in temm of Lagrange multipliers (see for instance [Mo]):

PROBLEM 2: Find the rninzinizers # for the finctional

whem p 2s a Lagrange multiplier-

The energy is defined as

where g is a Riemannian metric chosen such that it extends the sub-Riern;innim metric- To construct such a metric one considers a non-horizontal direction V and define g such that

Horizontal acceleration and V-connection

Let 4 be a solution for the Euler-Lagrange equation v4$ = p ~ ( $ ) - As v*# is horizontal,

On the other hand, the horizontal co~mxtïon D is given by

Hence

(2.41)

It is proved in chapter 6 that for the Heisenberg group case the right hand side is zero.

2.2.5 Physical sisnificance of the curvature

The geodesics are bent by the curvature in order to remain in the horizontal distribution all the time. Indeed, it was shown that if a horizontal curve 6 is a sub-Riemannian geodesic (solution of Euler-Lagrange system) then the following equation takes place

where V stands for the Levi-Civita connection in a certain metric which extends the subRiemannian metric. This can be recognized as Newton's

equation. As the le& hand side is an aeceleration with respect to V, the right hand term which is the curvature dong the minimizer, should have the signiûcance of a force. In Riemannian geometry the geodesîcs are given by v~) = O whüe in this case the perturbation is due to the cunrature of the distribution 7C. One can show that the above Newton's equation in 3 dimensions becomes Lorenz's equation

where m is the mas, e is the electnc charge, B is the length of the magnetic field and 3 is the complex structure z(v1, y) = (-uz, vl). The right side term is called the Lorenz force and is perpendicular to the magnetic field which is t-orientecl- The charged particle travels dong the x-plane which is perpendicular to the magnetic field. Supposing that the trajectory is unit speed, -5B stands for the curvature of the x-trajectory. AsAsnimùig that the magnetic field is constant, the x- trajectory wiU be a piece of a &de. This explains why on the Heisenberg group we get the scalar (Cartan) curvature constant. As ~ ( 6 ) stands for the Lorenz force (which is normal to the trajectory of the particle ), a missing direction is given by the magnetic field which is normal to the plane generated by the trajectory and Lorenz force (see figure

1).

Figure 1: The speed 6 and the curvature ~ ( 6 )

2.3 The skew-symmetric model

This case corresponds to the 2n + 1 dimensional Heisenberg group H,. The 2n vector fields are dehed as

where

(2 -45)

or B = 2Az where A is a skew-symmetric (At = -A) non-sinpliil: matrix- This model was studied in PGGl], [BGG2], [BGG3], [BGG4]. It is included here to exemplify the above theory on a concrete case-

Proposition 2.12 The 1-form of connection in the skew-synmetric case is g2ven by

Proof= As X = a. + B(z)& a computation gives

w(X) = (dt - B(x)&)(X) = (dt - B ( ~ ) d z ) ( & f B(x)&) =

= dt(& + B(x)&) - B(x) dx(& + B(x)&) = O + B(x) - B(x) + O = O

So that ker w = 3C. rn

Proposition 2.13 The curvature 2 - f o m matri& in the skew-sgmmetric case

Hence

(2 -48)

This gives a geometrid interpretation for the constants ai,- that appear in the fields Xi = + 2ajkx& as a sectional (Cartan) ctuvature of the horizontal distribution in (Xi, Xj)-dir~ti0n. Hence

1

2.3.1 The curvature on the skew-symmetric mode1

As the 1-connection form associated to the horizontal vector fields

and the 2 - m t u r e form is

it follows that the curvature along any horizontal curve 4 is

one obtains

Proposition 2.14 The curvature along any horizontal curue # in the skew-

Due to the antisymmetry of the matrix A = (aij) one obtains

Proposition 2.15 For any horizontal curve in the skew-symnaetric case the curvature K(&) is nomal to the tmjectory of 4.

In fact this result was proved in a more general case, but it is included here to show its significance in the skew-symmetric case.

2.3.2 The solution of the Euler-Lagrange equations in the skew-symmetric case

The Lagrangian in this case is

where it was used that At = -A- Then

Hence the Eulediagrange equations become 5 - 28Ax = 2BAx , or

where 8 is a constant which in this section will be assumed positive. If q5 = (x, t), then the nght hand side means nothing else than the curvature dong the curve q5

2 n

and the equation may be written as

which is a particdar form of the g e n d equation v&$ = p ~ ( # ) - In order to solve equation (2.54) the substitution x = u should be used. The equation becornes (2.57) 'Li = 4ûAu

A solution is (2 -58) u(s) = e*Asu(~)

As u(0) = *(O), denoting v = x(0) , the initial speed ,

As the matnx A is skuw-symmetnc

e 4 e ~ ( e 4 8 ~ ) ~ = e 4 e ~ e(BAT~ - - e ~ ~ ( A M * ) S = eo =

which means that e4@& E O(2n). Using this one obtains

which means that the length of the speed vector is constant dong the trajectory. In 0th- words and in another contact the Hamiltonian is presemed. Going back to equation (2.54) it is observeci that the solution has the fonn

H the solution starts from the origin, z(0) = 0, the constant C is given by

and the solution becomes

As e-4eAs is a power series of A, it follows that it commutes with A-' and the solution becomes

The above solution lies on a 2n-dimensional sphere centered in and with radius R = &IA-'vl, as the following computation shows

So that x(s ) is a spherical curve. Hence

Proposition 2.16 The solution of (2.54) which starts at the o r i ' wtlr the initial speed v , in the skew-symmetric case, G a gmat c i d e on the sphem S2"(+A-'v , &lA-'vl) given by (2.64)

The solutions of the EuterLagrange system have constant cumature-

Proposition 2.17 If x(s) LP a solution for the Euler-Lagmnge system of equations, then (2.65) In(+)) 1 = 41Avl

where x(0) = O and x(0) = v.

Proofi Using the fonnulas for the cu~vature developed in the previous section,

Hence IEC(x(s)l = 4[eaAS~vl = 41Avl

2.3.3 Curvature integral invariant for the skew-symmetric case

The 2-form of curvature is

The associated mat& is of tank 2n:

Obviously, an eigenvector correspondïng to the eigenvalue X = O is

The vector & is the curl vector field for the 1-form w which defines the W- distribution. Let a be the tube in &direction (the tube generated by the $-integral curves) and 71, y2 two cuives which generate the same tube (see figure 2). Then

Figure 2: The &tube hdeed, as Cl(%, 2) = O, V Z and 2 is tangent to the tube a, then

On the other hand, applying Stokes, one obtains

Rom (2.68) and (2.69) one obtains (2.67). Using Stokes again, relation (2.67) becornes

where Dl and 4 are two sections through the tube, with aDi = ri. Thus the curvature 2-fonn is an integral invariant-

Proposition 2.18 in &difection JD 52 ts consemed.

In this case is also a missing direction ( w ( g ) # O) and it follows that flD Q is preserved dong a missing direction.

The case of the Heisenberg group

In this case the missing direction is V = & and R = 4dxl /\ dz2 and

which is constant for every section D in the tube.

2.4 Missing direction for contact manifolds

Let XI, be the vector fields which span the horizontal distribution. In local coordinates

2n+l Proposition 2.19 Let 4 ( s ) = (xl(s), .., x2=(s), t ( s ) ) be a curue in QZ - Then q5 2s horàzontal 28

linearly dependent:

An expansion &er the Iast row gives

where Dk are the quotients of the deternVnants gïven in the proposition. 4

Associate the 1-form of connection

As in the skew-symmetric case

Proposition 2.20 i s a hotirontal cuive,

Observation is the volume generated by the projections of vectors Xl(z), .., Xh(x)

on the hyperplane (zl, z 2 , .., x ~ ) .

The 2-form of curvature 0 = & in the general case is

Assume that w A (dw)" # O. Rom the contact stnictures theory, Q measures the noriintegrabilty of the horizontal distribution 3L. As w = dt-&(z, t )hk defines the nonintegrable disfxïbution 7L, applying the Darboux theorem, it follows that there exists a locai system of coordinates in which w can be written in the normal fonn (see [Ar]):

In these normal coodinates the 2-forni 0 becomes

The matrix Cl , is

The rank is maximal, equal to 2n. It is obvious that the vector at is a eigenvector correspondhg to the zero eigendue. Hence in the nomal coordinates a, is the c d direction for w:

This can also be WIitfen as a(&, 2) = O for every vector field 2, namely, the curvature vanishes on every %plane which contains Ot. One arrives at the following result

Theorem 2.21 There is a vector field V svch that the curvature on each %plane which contaim V vanishes. firtherrnore, V is a missing direction and the direction with the above property ik unique.

Proofi Edtence: Choose the vector field V such that in the normal coordinates V = a,. Then (2 -81) Q(K 2) = i2ijv'Zi = O

RijVC = O- V is a missing direction because

Uhiqueness: As the ma& f irj is skew-symmetricY det Qj = O. Then there is at Ieast an eigenvector V such that the eigenvaiue is O- As the rank is maJEimum (equal to 2 4 , the direction V is unique with this property, up to a scaling factor-

The above result gives a way to define a missing direction using the curvature (in the Cartan sense). In general the following result takes place

Proposition 2.22 Let a be a tube of integnrl curves of the above misshg direction V and Dl, D2, two arbihry sections in the tube. Then

Proof= As V is a tangent to the tube 0, 1, = O. U h g Stokes it results

Using Stokes theorem again we obtain the desired result.

In the following proposition Lvw denotes the Lie derivative of w with respect to V.

Proposition 2.23 If q5 is any closed m e then

whem V O the mLPsing direction defmed by Q(V, W) = O for dl W E X(R~*$-').

Proofi Using Cartan decomposition and Stoke's Theorem (see PI), one obtaiiis

2.5 A more general model than the skew- symmetric case

A generalization for the model on R3

Deno t ing (2.87)

one can wrïte - (2 -88) &=a., +&(x)&, Vk=l ,2n

The Hamiltonian associated to the Heisenberg Laplacian

Proposition 2.24 A curue c(s) = (xi (s) , -.z%, t ( s ) ) is horizontal iff

and C(s) is horizon& ifE the coefficient of at is zero.

Proposition 2.25 The LagmngMn associated to the above Harnittonian b

xi = ci + Bj(z)B, one obtains L just in t e r m ~ of X, x and i:

Proposition 2.26 The solution of the Euler-Lagmnge equations are horizontal curves.

Proof: The solution for Euler-Lagrange equations is the projection on R2nf l of the solution of Hamilton system of equations (bicharactenstics) £rom x nL2"+I. One of the equations give us

As the other equafions provide

it turas out that 2n

j=l

and in the virtue of the Proposition 2.25, one obtains the solution of the Euler-Lagrange equations horizontal curve.

Corollary 2.27 Along the solution the Lagmngian is constant.

Proof: As the solution is horizontai, it Is obvious that

and use the fact that the energy H is preserved dong the solution.

Corollary 2.28 Along the solutions

2.5.1 A geometrical interpretation for coordinat e t

Consider on R2*+' 2n vector fields: Z n

where a,k = -Ukj are red number~. Using the previous notation

where A = (aij)ij a d x = (xk)k- Using Corollary 2.29 one obtains along the horizontal cwes that

is the area swept by the vectorial radius from O to x(r) in the (x j , xk)-plane along the projection of the solution of Euler-Lagrange system in this plane,

a computation gives

Hence (2.103) ~ ( T ) = ~ ( O ) - ~ ~ U ~ ~ ~ ~ ~ ( T ) = ~ ( O ) + ~ ~ U ~ ~ U ~ ~ ( T )

the skew-symmeetric matrix of the areas associatecl to the projection of the solution in (x j9 xk)-planes (se figure 3). Hence (2.104) t (r) = t(0) + 2 Trace A m ( r ) ( ) and the variable t dong the sohtion can be expressed as a trace of a certain product of matrices, each of them with a geometrical sï@cance. As Proposition 2.14 states that 4 . = a, then

Figure 3: The geometrical interpretation for a,@)

Chapter 3

Conjugate points and Jacobi vect or fields

3.1 Definitions

This chapter de& with the Jacobi vector fieIds along solutions of Euler- Lagrange equations in terms of an exponential map. Some similaritles between sub Riemannian geometry and Riemannian geometry may be observed here.

Let 7 : [O, r ] + PL" be a solution of Euler-Lagrange equation and c : [O, r ] x (-E, E ) -t IL" a smooth E-variation of the mbimïzer 7, such that c(s, O) = ~ ( s ) and s + c(s,u) (u fixed) is a solution of Euler-Lagrange equation-

DeWtion 3.1 The vector field of variation

is called Jacobi vector field along 7 associated to the above variation- A = ~ ( 0 ) and B = y(r) use wnjugate points along the minirnizer 7 if there

is a Jacobi vector field atong 7 such that J(0) = J(r) = O and J(s) # O for euery s E (O, r).

3.2 Curvature along Jacobi vector fields

Proposition 3.1 Let c(s) be a sub-Riemannian geodesic which starts ut the ongin and let P lie the first wnjugate point witlr O dong c(s) . Denote b y V ( s ) a Jacobi uector field along c(s) and S(s) as the action between O and c(s). Then we have

(3-2) l1 ~(v(S))(S(S)) d.9 = 0

where P = c(1) and X: i s the cumature in V-difection.

Proof= Let c. = FE(c) be a smooth variation of c, such that for every E > O, cc is a sub- Riemannian geodesic, As c, is a horizontal curve and w is the 1-connection

Hence /LW = O

C

where V is the Jacobi vector field associated to the variation. As V is zero at the end points of c,

where ivw = w(V) is a function which is zero at the ends of c. Using Cartan decomposition (see m)

and the fact the above integral is zero, one obtains

which can &O be Wfitten as

Using that E = fiXi, then

where it was used the known fact that Ei(s) = Xj(S)- Hence

l1 K(V) (S) ds = O

3.3 Exponential map on Hormander manifolds

The exponential map is defined as ezp, : 'H, + BP,

where 7. is the minimiler which starts at p = y@) with the initial speed v = *(O). The vector v is horizontal at p. As a first observation, for every X f I?Z

Indeed, as s + rAv(s) and s + r,(As) are solutions for the same Euler- Lagrange system of equations with the same initid conditions, %(s) = y,(Xs) and chooaing s = 1, the desireci result is obtained- The difEerential of the exponential map is d&ed between the following tangent spaces

(3-6) (d e&, : Z(%) -r TIPL"

where x E 'fl, and x' = ezp,(x) (see dÏagrgram 1). In this section a formula will be given for the differentid of exponential fwiction in ter- of the Jacobi vector fields which is an andogous of the same formula fkom Riemamian geometry (see for instance [ON]). Let s -t x + SV be a d a t i o n in 3L, of the horizontal vector z, where v E ?4, is an arbitrary vector field- For each vector in the variation there is a sub Riemannian T,+, with the initid speed x f su which &arts at p. Define c(t, s) = y.+&) where c : [O, l] x (-E, E) + R? Then y. (t) = c(t, O) and c(t, s) = expp(t(z + su)). The Jacobi vector field is

The value of the Jacobi field in t = 1 is given by the following computation

where $(s) = x + su and $(O) = v, G(0) = x. Hence results the following nice formula which relates the exp map and Jacobi field

In a similar manner one may obtain the value of the Jacobi vector field for every t

The vector v may be found as in the foUowing

Hence v = &O), namely v denotes the initial speed of J. A necessaq and d c i e n t condition to have conjugate points is stated in the next r d t .

Proposition 3.2 Let y : [OJ] -f E" a sub-Riemannian geodesic on a E6mander manifold Then ~ ( r ) ia wn*gote to ~ ( 0 ) a h g 7 iff (d =&(O)

has a nontràvial kernel,

Proof= Indeed, as J( t ) = (d exp)t+(o) ( ~ J ( o ) ) , ~ ( 0 ) and y@) are coqjugate J(T) = O or (d ~ X ~ ) ~ , ( T J ( O ) ) = O, which means that TJ(O) E Null(d e ~ p ) ~ ( ~ ~ . rn

Let P = y(r) conjugate with ?(O) = O along 7. Then there is a Jacobi vector field J along y such that J(0) = J(r) = O. Denote by JOp the set of all Jacobi vector fieIds dong -y wbich vanish at O and P.

Proposition 3.3 Jop Zs a vector space with dim Jop 5 n - 2.

Proof: Let JI, Jz E &P. Then, if vl = J(O) and v* = &(O), JI and J2 are given by

where Js is the Jacobi vectorfidd with J(0) = J1(0) + Jz(0) = O and ~ ~ ( 0 ) = VI + v2- AS J3(r) = JI@) + J2(7) = O, it tums out that J3 E 30p- In a similar manner, if X is a real number, and J E TOP it may be written that

X J(t) = (d e ~ p ) ~ ( o ~ (th) = j(t)

It is easy to check that J E gop- Hence zoP is a vector space.

diagmm 1 The matrix associatecl to the linear function (d exp), is in Mn,(a-i) (R). Hence the rank (d exp), 5 n - 1. On the other hand, as exp cannot be constant, then rank (d exp), 2 1. As (d e ~ p ) ~ ( ~ ) has a nontnvial kernel, it turns out that (3.10) 1 5 rank(d e ~ p ) , ( ~ ) 5 n - 2

Hence (3.11)

Deenition 3.2 If P = ~ ( r ) is conjugate to O = ~ ( 0 ) along 7, then ordconj(0, P) =dim (zop) is called the o d e r of conjugacy of P and O along 7-

The order of wnjugacy indicates how many linearly independent variations with fixed ends at O and P along the sub-Riemannian exists.

Corollary 3.4 If O and P are two configate points along a solution of Euler-Lagmnge equatiow on a Hômander manifold, then

Corollary 3.5 P Xs conjugate ~ 6 t h O along y iff

3.3.1 Exponential map on the Heisenberg group

In this subsection it wiii be shown how the previous theory works for the Heisenberg group. In this case Xl = a, + 2z2& X2 = - 2x&. The horizontal minimizer which starts at the origin with the initial speed v and 0 = 1 is denoted by c%& (s) = (xl(s), x2 (s) , t(s)) , where (see Appendix A)

1 (3.14) q(s) = -(vl sin 4s - v2 COS 4s + v2)

4

then one can mite 1 1

and the exponential map is

As stated before, ~ ( r ) is conjugate with O = 7(O) alongy(0) ïfF d exp is degenerate, or

(3.18)

As n = 3, it was obse~ed that the rank is qua1 to 1 in the conjugate points. So that, in conjugate points, any 2 x 2 minor must be zero. In particdar,

which means that

Hence, it is observed that T = 7r/2 as the srnaIlest positive solution. It tunis out that the first conjugate point to O is P = 7(7r/2). Hence

And the order of conjugacy is

In the following will be shown that the conjugate point &(7r/2) to O = &(O) along q5,, belongs to the t-axïs. Indeed,

Hence the first conjugate point to the origin is (O, O, T lv12) ( s e figure 4).

Figure 4: The subRiemannian geodesic on the Heisenberg group

Between two consecutive conjugate points the velocity of the minimir:er describes a complete rotation of angle 271. Indeed,

and hence ?t #d,) = (VI, Y, 0) =

which is the initial speed v = &(O). Hence, under the hypothesis 8 = 1 the conjugate points to origin occur for r = mr/2. In the Appendix B, where r is aisumeci fixed (r = 1) and #(L) is conjugate to O then the possible values of 8 are 8, = m7r/2, m nonzero integer .

3.4 The rotational symrnetry of the Euler- Lagrange equations

R e c d that if XI = a, + A1(x)at, X2 = 8+2 - A2 (z)& are two vectOr fields on R3 which span the horizontal distribution 31 = span{Xl, X2), then the H d t o n i a n associateci to the above vector fieIds is

and the Lagrangian obtained from H using the Legendre transform is

where 8 is constant because

3.4.1 The Euler-Lagrange equations

In the foIIowUig the Euler-Lagrange equations will be presented and the solutions will be characterized using the symrnetries of 12 and W .

A computation shows that

aL aAl , OAz , - = -8-x1 + e-x2 as2 a22 as2

Then the Euler-Lagrange equation

becomes (3.26)

And (3.27)

becomes (3.28)

Thus the Euler-Lagrange system of equations is obtained

9 = constant

3.4.2 The 1-connection form

As the Lagragian is

where #(s) = (z(s), t (s) ) , it tums out that the l-connection form is

(3.31) w = dt - Al(x) hl + A2(z) dz2

Indeed, this can be checked by direct computation that

o(X1) = w(X2) = O

Hence ker w = X- The curvature 2-form is

Proposition 3.6 The distribution 3L i s not integmble iff

The functions Al and A2 have to be picked such that the above relation is satisfied-

Lemma 3.7 C2 is invariant under rotations umund the t-azis if

is invannant under the sante type of rotations.

Proofi Let

As dxl A is preserved by rotations, will be preserved iff the coefncient of A &2 is the same. rn

Lemma 3.8 0 is invariant under rotations amund t-sais iff [XI, X2] is also invariant.

By computation aA2 [x,,x*~ = -(% ax2 + -)& dx1

and use Lemma 3.7-

3.4.3 The rotational symmetry

Denote O(z) the coefficient of &cl A ka in the expression of R

@(s) is a nonzero smooth hction. Then the Euler-Lagrange equations may be written as

If Z = (4). z = ( 2 : ) and J (E) = ( ) is the compla structure of -u

the plane, the above system may be &O wrïtten as

Suppose that 8(x) is in-ant under rotations Le. @(&z) = O(x). If xf = &x then x' = Rax and .7(xf) = g(&x) = R,J(x)). The systern

which gives I = $@(z)x- Hence the foilowing is obtained

Lemma 3.9 The Euler-Cagrcrnge quatiow a m invariant by nifations provided that O(%) i s invariant by rotations.

3.4.4 The conjugate points

In [St] it is proved that in step 2 case, on any e b d around the origin there is at least a conjugate point with the ongin. In general step case it is not known if this is still true. In the following it will be assumed that there are conjugate points with the ongin. Moreover, it WU be supposed that f2 is invariant under rotations around the t-ruris. Then, if x(s) = (xl(s), q(s)) is a solution for the Euler- Lagrange system, then every rotation of x(s) will also be a solution

(3.35) Raz (s) = (cos a z&) - sin a z2 (s) , sin a zl (s) + cos u z2 (s))

for the system. Let P be the first conjugate point with O dong 4(s) = (x(s ) , t(s))- Supp- that P = #(T). Let #a = (x,(s), ta(s)) be a smooth &-variation for q5(s), where

As w = dt + Al(x)dzi - A2(x)dx2 is invafiant under rotations, Al(x)di l - A*(X)&~ will be the same. Hence

which means that t,(s) - t ( s ) = k =constant. As aLl the curves &art nom origin, t,(O) - t(0) = O, then k = O. It tunis out that t,(s) = t ( s ) , Le. the t-component remaïns at the same height d m g rotation. If P = #(T) is conjugate point with O = #(O), then ais0 P = &(T)- If P would not be on the t-axis, then &(T) will describe a circle parallel with the x-plane, a contradiction. The foUowing result has been obtained

Proposition 3.10 If w 6s invariant by rofat~ons (of x-plane), then We conjugate points are on the t---

Proof: Let W be the vector field of rotation- If w is preserved by W, then Lww = O. Hence

O = dLww = Lw& = LwQ

i.e. R is preserved by W. Using Lemma 3.9 Q(x) will be invariant under rotations and apply the above consideratiom.

Corollary 3.11 If [XI , X2] ii invariant under rotations amund t-axis, then the conjugate points are on the t - d .

Example 3.1 (Heisenberg p u p ) The vectorfields are Xi = a,, f 22&, X2 = a, -2x& and the 1-connection form ïs

w = dt - 2 s 2 h l + 2x1dZ2

and [XI , Xzl = -4&

w and às [Xi, Xz] are invariant by rotations, so that the conjugate points with O dong the subRiernannian geodesics are on the t-axis,

Example 3.2 (4 step m e ) Take the vector fields Xl = a, + 4221~1~& X2 = a, - 4~11~1~&. The 1-connection form is

which is invan-ant by mtd2owY so that the w n . g a t e points, i$ ezist, are situated along t-axk

Example 3.3 (2p + 2 step cme) Take the vector fields XI = 8, + 4z2 (z (*pl&, X2 = 8, - k1 (x12Vt. Tlie 1-connection fonn ik

In the next chapters a study of the above cases wiiI be done. The first case is known (Heisenberg group case) and it is inchded in the Appendix A and B. For the second case wi l i be shown that the conjugate points are on the t-axis- The lengths of the geodesics between origin and these points are computed.

Chapter 4

The Main Result

4.1 The Study of a step 4 case

4.1.1 The Euler-Lagrange system of equations

Consider the vector fields

As [XI, X2] = -16 12 1 2&7 [XI , [Xi, &]] = - 3 2 d t and [XI, [Xi, [Xi, Xz]]] = -323 it huns out that the hypoelliptic operator Ax is step 4 (at origin). The associated Lagrangian wi l l be

As w = dt + 4 [ ~ 1 ~ ( r ~ d z ~ - x2dxi), then Q = & = 161~1~dx~ A dx2. It follows that this is not a contact case at x = O. Using polar coordinates xl = r cos # , ~2 = r sin one obtains

and the Lagrangian becomes

In the following the Euler-Lagrange system of equations wi l l be constructed for the above Lagangfan- A computation shows

d âL -- - ac - t and - = rd(6 + IW~')

ds % l3r

and the fkt Euler-Lagrange equation becomes

the second Euler-Lagrange equation is

which can be written as (4.9) r2(d, +@) = k where k is a constant-

Then it turns out that O =constant dong the solutions- r(s) and 4(s) wi. v e the Euler-Lagrange system

i.. = rd(4 + 160r2) t2($ + 48r2) = k

9 = constant

4.1.2 Solutions which s t a r t from origin

The solutions which &art fiom origin wil l be m n s i d d The system (4.11) will be considered with the fo110wing initial conditions

From the second equations one obtains k = O and the system (4.11) becomes

r = +(fj + l e ) q5 = -4er2

8 = constant

Substituthg the second equation in the first one it is obtained an equation for T (s) (4.14) T = -48o2r5

D e h e the potential energy as

Then the equation (4.14) can be written as a Newton's equation

where V' (r) = dV(r)/dr. Letting i = p the equation (4.14) can be written

It is obtained an quation in p and r

which can be solved by separation of the variables. Integrating

where C is a constant whkh denotes the total energy of the system. In fact C = 1;2 (0)/2 and using that

and the initial conditions, one gets C = )12(0) l2 and wilI be denoted by E. Hence (4.19) A i 2 + V(T) = E

2 is the law of cornefvation of energy. The total energy E is equal to the valne of the constant Hamiltonian Ho almg the solutions.

Figure 5: The potential V(r) and r,,

The radius r (s) The radius r(s) starts at O and increases (f > O) until i = O, when r = r-. After that r(s) decreases (f < O) to O. r,, can be found Ietting f = O in (4.19). Hence r,, is the positive solution of the equation (see the fig. 5)

For each level of energy E > O there is a comsponding r,, > O which is independent of the initial direction of the solution. The solution lies in the disk D(0, r-). The greater the energy E, the Iarger r- is. As E = 1 x(0) 1'12 depends just on the magnitude of the initial speed, r,, wiiI do the same-

One of the equations in (4.13) provides

As r(0) = O, $(O) = O. For 0 > 0, &O) c O, i-e. 4 is decreasing. For 0 < O, &O) > O, i-e. 4 is increasing. Relations (4.19) and (4.22) on be d e n as simultaneous equations

where ~ ( s ) was supposed to be inaeasing. A dlfferential equation connecting r and 4 but not containhg s can be obtained by division of the equations in (4.23)

d4 -Ur2 (4.24) -= dr J ~ E - 16O2r6

Integrating, the variation of the angle q5 is obtained

When r = r,, the above variation of q5 is denoted by w and is given by

Computing w Making the substitution

the integral (4.26) becomes

Hence the angle w swept by the vectorial radius between T = O and r = r,, is

(4.27) 6 if 8 c 0 W = { - ~ / 6 , if 8 > O

which agrees with the fact that the particle is moving dock-wise for 8 > O and counter-clodc-wise for 19 < O (see fig- 6).

Figure 6: A rotation of w = 7r/6 Between r = r,, and r = O the vectonal radius sweeps an angle 5. As r is decreasing, f < O and

dr (428) -=- * / 2 ~ - 1 6 B 2 ~ 6

ds Using again (4.22) and making the division one obtains

The angle 6 is given by

which is the same integral as that one given in (4.26). Hence

The solution passes through origin again after sweeps an angle equal to u + @ = 7r/3 (clock-Wise if û < O and counter-clodr-wise if 0 > O). This means that the angle between the tangents at origin is ~ / 3 . As this angle is commensurable with 27r, the solution will be periodic-

4.1.3 An explicit solution in polar coordinates

In order to avoid using sgn(B), one may assume B < O. Then 4 is increashg

whlch becomes

where r(s) 5 r-. Making the substitution

and using r,, given in (4.21) one may Wnte

Denoting q50 = #(O) and using that 9 < 0, the relation between 0 and r d l be given by

(4.34) 1 r3 4 - = - arcsin(J-) 3 *-

which is quivalent to

Case 4 E + ~ 1 % #(O) + 4 3 1 In this case T demeases kom r,, to O. Let M(r-, & + 1~16) denote the vertex and let A(r, 4) be a point as in the following figure, i.e. q5 E [#(O) + n/6, #(O) + n/3]. IL this region f < O and

As w - = - a r 2 ds

one obtoins by division

Figure 7: After a rotation of 4 3 Integrating between k + and q5 one obtauis

The right hand side integral can be decomposed as

and hence

The f i t integral is given in (4.26) and is e q d to 7r/6 (see &O (4.27)). The second integral appears in the right hand side of (4.32) with a negative sign and is given by the negative right hand side of (4-34)

which is the same as (4.35) but q5 E [i$o + r / 6 , & f 7t/3]. (see fig. 7). From (4.35) and (4.40) it turns out that

which means the particle starts at O with an asgument and bounces back with an éugument + 4 3 . Case $ E [$jo + +/3, & + 2~131 In this case the particle starts nom origin O with a new argument = 40 + ~ / 3 and arrives at O with an argument & + ~13. For 4 E [hl & + 7r/3] one may write using (4.35) and (4-40)

As 3(4 - #O) E [?r, 2 ~ 1 , s in sin negative here and the minus sign in front of the formula makes sense.

In general case the formula in polar coordinates r = r(4) can be WLitten as

The solution is periodic and the trajectory is closed. It contains 3 loops with an angle of 2r/3 between the symmetry axes (see fig- 8).

Figure 8: The x-projection of the solution

4.1.4 The t-component dong the solution

Suppose that (xi(s) , z2(s), t(s)) sathfies the Hamilton's system of equations given by the Hamiltonian

Using one of the Hamilton's equations

or in p o h coordinates (4-44) t = -4t4(5

Then t decreases as 4 inmeases (case 0 < 0). Assume # E [40, + 7r/3]. Using (4.42) and (4.44) one obtains

and does not depend on &. In other words T does not depend on the initial direction 40.

4.1.5 Conjugate points to origin

Consider a solution zi(s), ~ ~ ( 4 , t (s ) ) which starts at origin. In polar co- ( ordinates (4, r(q5)) one has r(qj0) = O and t(&) = O. When 4 = & + 7r/3 formula (4.42) provides r(& +7r/3) = O which means that the x-components at this point are zero. The t-component is given by (4.48)

It turns out that the point (O, 0,T) belongs to the solution and does not depend on the &- On the other hand, for each q50 there is a solution which starts with the initial argument 4(O) = A d a t i o n for corresponds to a rotation of the solution. The point (0, O, T) rem* fwed during the variation and belongs to all the solutions which are coming from the initial one by making & variable.

The foliowing result States that the conjugate points to origin dong the solutions is desaibing a missing direction which is in this case the t-axis.

Theorem 4.1 For every point P = (0, O, w ) on the t-axis, w # O, the= is a solution y which starts ut origin such that P is the Jirst point conrgate €0 O dong 7.

Proofi The proof is based on the coIlStruction of such a solution, (i) case w < O Choose 0 < O and require T = W . Rom relation (4-47) one nnds

where q = Q(7r/3). As r,, = ( E / S @ ~ ) ' / ~ it turns out that

Therefore choose a solution which starts at origih with an initial velocity of magnitude

The point P = (0, O, w ) belongs to this solution (fiom the construction of the solution). Performing a rotation in the x-plane of the initial veIocity, ail the solutions obtained will p a s through P (fiom the previous discussion) and P will be the fin& conjugate point with the ongin, (ii) case w > O In this case 8 > O and the rest of the discussion is similar with the case 8 < O.

4.1.6 The length of the geodesics between the origin (O, O, t)

It is already known from Theorem 4.1 that there are solutions which join ongin to any point on the t-axk. This section ammers two questions: 1. How rnany sub-Riemannian gwdesics of dgerent lengths (pammetrized by [O, 11) join the ohgin to (O, O, t), for a &ed t # O ? 2. What are these lengths?

The case of the Heisenberg case it is treated in the Appendix (B) and it is &O containeci in [BGG5I.

Theorem 4.2 (The Main Theorem) The geodegcs that join the origin to u point (O, O l t ) have Iengths dl, da, d3 ..., where

For euch length &, the gwdesics of that length are pammetfized by the circle

Proof: We need to find the Harniltonian path connecting the origin to (O, O, t ) . The path is considered to be paxametrized by [O, 11. The boundary conditions are ~(0) = r ( 1 ) = O , t ( 0 ) = O and t ( 1 ) = t. Using Euler-Lagrange equations in polar coordinates one obtains that the constant 0 v d e s the folIowing boundaq value problem

Writing A2 = 48B2 then (4.55) may be considered as

with the boundary conditions r ( 0 ) = r(1) = O. The conservation of energy leads to

1 1 -f2 + -p# = &, 2 6

where Ho is the constant d u e of the H d t o n i a n dong the soIutioas (denoted also by E in the previous section). Separate the variables

Then r is dehed by the relation

Making the substitution / A* \ 116

the above expression becomes

where

Making again the substitution u2 = t, one may wrïte

The Left haad side can be expressed in terms of eliiptic integrah. For this we need the foliowing result.

where b = 2 - fi, k = \/6/2 and &) = +: - 1. us+ ,

Using the above Lemma the relation (4.59) becomes

At the boundarïes s = 0, s = 1 we have r = O and hence a = O. Using the

expression ofg one has g(0) = Os = 2 - a = b and hence 4 1 - gF, = 0. It turas out that for s = O and s = 1 the le& hand side in (4.60) is zero. Then

s n ( z ~ / ~ ~ ~ / ~ ~ ( E ~ x ) ~ / ~ ) = 0

and hence

( m # O because Ha # O and û # O) where 4K is the period of sn ( , k) where k = Jb/2 and K is definecl by

From (4.61) one has

Using X = 4 4 6 the above f o d a becomes

and

Using (4.21) one obtaios

The lefi hand side of (4.62) is related to t as follows. As i = -4r44 or dt = -4r4 d@, integrating between and & - mr/3 one

Hence (4.63)

where Q = J'sin4I3 udu. We used 4 decreasing because 6 was assumed positive, accordhg to the formula 4 = -M. Piugging (4.62) in relation (4.63) one obtains

fFom where

(4.64)

The physicd interpretation of the above relation is the fact that the energy îs discrete and depends on (t( . The geometricd interpretation is that the geodesics joining the orïgin O to P(0, O, t) have discrete lengths. Fndeed, let 7 : [O, 11 + @ be a soIution between O identity in

l(r)

and hence (4.65)

and P . As 1j.l is constant in the sub-Riemannian metric, we have the Cauchy's inequality

Using (4.64) one obtauis that the lengths of the solution 7 are quantifid, namely, l(y) = where

These lengths do not depend on the angle &- When this angle is d e d the soIution is rotating around t-axk having the same Iength aiI the tirne- The constant Q is given In Lemma 4.4,

The shortest solution corresponds to m = 1. The distance between O and P(0, O, t) will be di where

D

Carnot-Carathéodory

The constant Q = sin4I3 u du can be expressed in terms of gamma functions as follows-

Making the substitution t = sinu one gets

With the substitution t2 = x the integral becomes a beta function which can be expressed in terms of gamma functions

Using r(1/2) = Ar), r(7/6) = ) l?(l/6) and r(7/6 + 1/2) = $r(2/3) one obtains

Q = ~ W / 6 ) ~ ( 1 / 2 ) =- 1 W / 6 ) fi $(2/3) 4 W / 3 )

Proof of the Lemma 4.3 We need to evaluate the integral

where f (x) = ~ ( 1 - x3) = -(x2 - Z) (1 + 2 f z2). Making the substitution

Pt+q x=- t t l

where -1-J3 -i+J3

P = 2 9 q = 2 f becomes equal to

the integral becomes

Using the -formula (see [Lad chapter3)

one obtainfi

4.1.7 Explicit solutions which join O to (O, O, t)

h m (4.61) and (4.63) one rnay compute r,, as a function of t

It tums out that r,, is quantized

As m increases, r- decreases. Using formula (4.42) one obtains that the solution in polar coordinates which joins O to (O, O, t) is given by

The t-component is computed by integration

Making the substitution v = 3(u - 4 0 ) one obtains

The above formula expresses r = r(&, but one may compute explicitIy the radius r = r(s) which is a solution for the boundary value problem

where X2 = 4802. This will be done in the following. Using the relation berneen the eliiptic function cn = 4- relation (4.60) becomes

Using the definition of g(a) one obtains

the above formula becomes

where k = 4 1 2 , b = 2 - f i , p = (-1- 4 ) / 2 . Figure 9: The shapes of the solution for m = 1,2,3

Above are sketched the fkst three geodesics and their projections on the x- plane (fig. The above

Theorem wha'ch join

(4.73)

(4.74)

The rad-

(4.75)

9). ARer that the pattern is periodic. resuits can be stated in a theorem

4.5 (Main Theorem) The equation of the sub-Riemannian geodesics the onHn to a @ed point (O, O, t) in polar coorduiates are

4.1.8 Solutions which star t outside the origin

The solutions with r(0) # O will be discussed. The system (4.11) will be considered with the init id condit ions

The second equation of (4.11) becornes

(4.77) $ = $ - a r z

where k # O. Substituthg in the first equation of (4.11) one obtains

k k P = + 16er2) = r (- ~2 - 4eP) (- r2 + 12eT2) e

one may write equation (4.78) as a Newton's Iaw


Writuig p = i the equation (4.78) becomes

where E is the constant of total energy.

Proposition 4.6 If r W a solution of Euler-Lagmnge system with initial conditions (4.76) then

In the fo110wïng a qualitative description of the solution will be done. The behavior of the solutions depends on the sign of k. There are two cases to be investigated sgn(k) = sgn(0) and sgn(k) # sgn(8). h each case wiil be drawn the graph of the potential energy V and the trajectory of the solution in the phase plane.

4.1.9 Case sgn(k) = sgn(9)

In this case V(r) has a zero and a global minimum at

and Y(+) = lim,+, V(r) = m.

Equilibrium points ro corresponds to a stable equilibrium point because

The corresponding q5 satisfies

and hence t$ is constant. This means that each point of the cirde C(0, ro) is an quilibrium point (constant solution). In each of these points the energy E -0 .

Limit circles When E > O there are exact two positive roots r d , r,, of the equation

such that (4.85)

The phase plane and the potential V are shown in figure 10.

Proposition 4.7 The solution of the Euler-Lupnge system (4.11) in the case sgn(k) = sgn(B) lies into the cimdar mum W (O, r,,,in, T-) .

Proof: As T reaches r, or r,, when i = O, then the equation of conservation of energy (4.82) shows that r,i,, , r,, are solutions for the equation V(r) = E and the solution T has the property

Figure 10: The solution in the phase plane

Existence of loops The equation 6 = O has the positive root

Hence 6 = O ïff r = ro . When the solution intersects the circle of equilibrium points C(0, rO) the sign of 6 changes, i.e. the trajectory is bouncing back, making loops in the crown W (O, r,, r-) ( s e fig. 11).


The width of the crown The width of the crown w = r,, - r e is a function of energy

Using the properties of the function V(r) one obtains the following (i) w(E) is an inaeasing function of energy E (ii) w(0) = O, case in which rmin = ro = r,,

The t-cornponent As = -4r4# and 6 = O on the cirde C(0, ro) then t will do the same. i(s) changes the sign when crosses the above cirde and t ( s ) reaches the maxima and minima on the same circle-

4.1.10 Case sgn(k) # sgn(0) , k # O

which means that 4 increases (demeases) for 0 < O (6 > 0) and hence there are no loops like in the fitst case. The potential energy is positive

and has a minimum at 114

(4.87) = (s) Indeed, Vr(r) = 4802r5 - 80kt - $ = $(4802r8 - 8ekr4 - ka) and

where u = r4. The quadratic equation in u has two roots

k k u1 = - and 7 4 = --

48 128

As just the second one is positive, rl will be chosen such that

The equilibrium solution The equilibrium solution in this case is the cùcle of radius rl. The corresponding q5 can be obtained fiom

and it is equd to

The circle is traversed dock-wise if k > O and counter-dock-Wise if k < O.

The bounds for r(s) There are two positive roots p,i, and p,, for the equation

V(T) = E , where E > V(rl)

They correspond to the points where 1; = O in the conservation law equation (4.82). Figure 12: The solution in the phase plane

In the phase plane the solution is rotating around the stable eqdibrim point (ri, O) between p , , , ~ ~ and p-


The trajectory in the (zi, x2)-plane belongs to the crown W (O, pmk, p-). The width of the crown inmeases as the total energy E increases, Le- the initiai speed increases (E = $ Ix(0) 12). The t-component As in this case 15 # O and i = -4r4tj it tums out that t ( s ) only increases (in the case 8 > O) or only demeases (in the case B c O).

4.2 Conjugate points for a step k + 2 case, k even

4.2.1 Euler Lagrange equations

Consider the vector fields on R3, where k = 2p is an even integer

For k = O one obtains the Heisenberg group case and for k = 2 one gets the previous step 4 example. The Hamiltonian H is defined as the principal symbol of the X-Laplacian

Using the Legendre transform one obtains the Lagrangian

Using polar coordinates zl = r cos 4 , 2 2 = r sin q5 one obtains

0 is a constant as the Hamilton's equation & = -BH/ût = O shows. En the following the Euler-Lagrange system of equations will be constructed for the above Lagrangian. A computation shows

d aL aL -- = i: and - = r#($ + 2(k + 2 ) h k ) ds ô+ &

and the fkst Euler-Lagrange equation becomes

aL dL - = r2$ + 20#+~ and - = O a# w


which can be written as (4.96) r2(4 + 2erk) = K

where K is a constant-

it turns out that 6 =constant dong the solutions. r ( s ) and #(s) will verify the Euler-Lagrange system

r = T&# + 2(k + 2 ) e y k ) r2($ +2erk) = K

9 = constant

4.2.2 Solutions which s t a r t fkom origin

The solutions which start fiom ongin will be considered. The system (498) will be considered with the following initial conditions

E'rom the second equations one obtains K = O and the system (4.98) becomes

Substituting the second equôtion in the first one it is obtained an equation in which q5 does not appear

Define the potentid energy as

Then the equation (4.101) can be wrïtten as a Newton's equation

where V1(r) = dV(z)/dr. Letting i = p the equation (4.101) can be wrïtten

It is obtained an equation in p and r

which can be solved by separation of the variables. Integrating

where C is a constant which denotes the total energy of the system. In fact C = ~ ~ ( 0 ) / 2 and using that

and the initial conditions, one gets C = iIx(0) I2 and will be denoted by E. Hence q

is the law of conservation of energy.

The radius r(s) The radius r(s) starts at O and increases (i > O) until P = O, when r = r-. After that t ( s ) decreases (F < O) to O. r,, can be found letting r' = O in (4.106). Hence r,, is the positive solution of the equation (see the above picture)

which is (4.108)

For each level of energy E > O there is a corresponding r,, > O which is independent of the initial direction of the solution. The solution Iies in the disk D(O, r-). The greater the energy E, the larger r,, is. As E = lx(0) 1'12 depends just on the magnitude of the initial speed, r,, wiII do the

The angle @ One of the equations in (4.100) provides

As r(0) = O? $(O) = 0. For B > 0, 4(0) < O, i-e. 4 is decreasing. For 8 < 0, $(O) > O, Le. 4 is increasing. Relations (4.106) and (4.109) can be written as simultaneous equatiom

where r ( s ) was supposed to be increasing. A differential equation connecting r and 9 but not containhg s can be obtained by division of the equations in (4.110)

Integrating, the variation of the angle q5 is obtained

When r = r,, the above variation of 4 is denoted by w and is given by

Computing w Making the substitution

the integral (4.113) becornes

Proposition 4.8 The angle w swept by the vectorial mdiw between r = O and T =r- .is

r 2(k+*l) ' w = { - 2 - f e c o

(4.114) -- 2(k+l) ' Z f o > o

(which agrees with the fact that the particte is moving dock-urise for 0 > O and counter-dock-wbe for 0 < 0).

Between T = T,, and r = O the vectorial radius sweeps an angle 6. As r is decreasing, f < O then

Using again (4.109) and making the division one obtains

The angle 6 is given by

which is the same integral as that one given in (4.113). Hence

The solution passes through ongin again after sweeps an angle equal to w + 6 = &-FI (clock-Wise if 8 < O and counter-dock-wise if B > O). This means that the angle between the tangents at ongin is &. As this angle is commensurable with 27r, the solution will be closed and periodic.

4.2.3 An explicit solution in polar coordinates

In order to avoid using sgn(B), one may assume 8 < O. Then 4 is increasing and the trajectory is followed conter-clock-Wise. lkmme 4 E [#(O), #(O) + *l and 7- E [O, G-l

The radius r is increasing and

where r(s) ( r,, and denote 4 = 4(s), h = 4(0). Making. the substitution

and using r,, given in (4.108) one may write

Using that 8 < 0, the relation between O and r will be given by

Integrating between + & and 4 one obtains

As the right hand side integral can be decomposed as

one obt;iins

28xk & + lr 4 2 ~ - 4&1222@+l)

The first integral is given in (4.1 13) and it is equal to & (see ais0 (4.114))- The second integral appears in the right hand side of (4.119) with the opposite sign and using (4.121) it is equal to

Hence (4.123) becomes

(4- 124) T = Tm= sin* ((k + 1)(# - b)) which agrees with (4.122). As a conclusion one may state the following

Proposition 4.9 For t$ - E [O, &], the solution b giuen in polar wor- dinates by (4.125) r = r-sin* ((k + 1)(# - &))

This piece of solution wil l be called the first Zoop. The trajectory closes periodidy after a certain number of loops.

4.2.4 The number of loops

For 5 4 5 ( + & the solution (r(#), 4) describes a loop. The angle at the root of the loop (the angle between the tangent lines at origin) is r / ( k + 1). As this is commensurabIe with 27r, it turns out that after a h i t e number of loops the trajectory becornes closed and periodic. The number of total distinct loops depends on the integer k and is denoted by N(k).

Proposition 4.10 The total numbers of loops is given by

(4.126) N(k) = 2k + 1

Proof= The angle of a loop is & = which means there are 2(k + 1) such aagIes around origin. This angles divide the plane in 2(k + 1) sector areas as below (fig. 14).

Figure 14: 2k + 1 sector areas

All the sector areas between + 2n7r and & + & + 2nr are denoted by î (the class of equivaence of the fîrst sector area). The sector areas between

+ & + 272" and & + & + 2n7r are denoted by 2 (the class of quivalence of the second sector area). The equivalence classes are modulo 2(k +- 1) (Le. - j + 2(k + 1 ) ~ = 7, where j E {l, 2,3, . . . ,2(k + 1))). Each loop belongs to a certain sector area, for instance the first loop (which &arts with & argument) fies into the sector area î. The second loop continues the first loop such that they have common tangent at O. As B < O (our initial assumption) the particle is moving counter-clock- - - wise in the sector area k + 1 + 2 = k +- 3. Indeed, an increment with k + 1 means to go to the c k s which corresponds to the sector situated with n forward. This happens for any loop. FACT 1: If a loop belongs to the sector area i. then the following loop w i ' - belong to the sector area r + k -f- 2. As the second loop belongs to the sector area k + 3, using FACT 1, the third - wiu belong to (k + 3) + (k + 2) = 2(k + 1) + 3 = 3 sector area. FACT 2: In general, the 2s + 1 loop lies into the 2s + 1 sector area- Indeed, using FACT 1 for 2s times, the loop 2s + 1 belongs to the class of

l + ( k + 2 ) + ...+( k+2) = 1+2s(k+2) =1f 2(k+l)s+2s - which is the class 2s + 1. Rom FACT 2 each odd loop lies in the correspondhg odd number class area

sector. Hence each odd area sector contains a loop. We will show that the even sector areas remain emptp. This wi l l corne nom the f x t that the even loops lie &O in the odd sector areas- - - Indeed, the second loop lies into k + 3 = 2p + 3 sector area which is an odd number- In general, using FACT 1, the 2s + 2 loop belongs to the class of

which is odd. As there are 2k + 1 = 2(2p + 1) sectors only half are odd. No loops belong to even class sector areas- Hence there only 2p t 1 loops. rn

4.2.5 Particular Cases

1. When k = O, then N(0) = 1. The single loop is in fact a circle which starts and closes at origin. The equation in polar coordinates is

(4.127) r(4) = t;Mt - 40)

where q5 - 4o E [O, n]. This corresponds to the Heisenberg case (see fig. 15)

Figure 15: The trajectory in the Heisenberg case

2. When k = 2, then N(2) = 3. There are 3 loops with an angle of 120 degrees between the symmetry axes. This corresponds to the step 4 case (at origin) studied before (see fig. 16). The equation of the first loop is

Figure 16: The trajectory for case k = 2

3. When k = 4, then N(2) = 5 (see fig. 17). The equation of the fùst loop is

(4.129) r (4) = r- ~in'/~(5(q5 - &))

where 6 - E [01~/5].

Figure 17: The trajectory for the case k = 4

4.2.6 The t-component dong the solution

Suppose that (zl (s) x2 (3) , t (s) ) satisfies the Hamilton's system of equations given by the Hamiltonian (4.90). Using one of the Hamilton's equations one has

or in polar coordinates (4.130) k+2 ' i=-2r q5

Then t decreases as 4 increases (case 0 < 0). Assume C$ E & + r/(k + l)]. Then one obtains

Proposition 4.11 For qj E [400, t,z50 + f 1

and hence t(#) --(#O) depends on the di#mence 4 - &.

D e h e 7r

(4.132) T = t ( # o f -) - t ( h )

Using Prop. 4.11 T will not depend on the initial argument #O

4.2.7 Coqjugate points to origin

Consider a solution (zl(s), z2 (s) , t (s)) which &arts a t origin. In polar coordinates (4, ~(4)) one has T(&) = O and t(&) = 0. When q5 = & + r/(k f 1)

then r (& +?r/(k + 1)) = O which means that the x-cornponents a t this point are zero. The t-component is given by (4.132)

It turns out that the point (0, 0, T) belongs to the solution and does not depend on the &- On the 0th- hand, for each & there is a solution which starts with the initial argument #(O) = A variation for corresponds to a rotation of the solution. The point (O, O, T) remains fixed during the variation and belongs to all the solutions which are coming fiom the initial one by making & variable.

The following result states that the conjugate points to origin along the solutions is generating a missing direction which is in this case the t-axis-

Theorem 4.12 For every point P = (O, O, w) on the t-axis, w # O, the= is a solution y which starts ut on-ph such that P i3 the first point conjugate ta O along 7-

Proof: The proof is based on the construction of such a solution. (i) case w < O Choose 0 < O and require T = W. From relation (4.133) one fin&

where q = $: s i n e v du. Then choose a solution which starts at origin with an initial velocity of magnitude where

with r,, @en in (4.135). The point P = (0, 0, w ) belongs to this solution (kom the construction of the solution). Performing a rotation in the x-plane of the initial velocity, ail the solutions obtained wi l l p a s through P (from the previous discussion) and P will be the fist conjugate point with the origin. (ii) case w > O h this case 8 > O and the rest of the discussion is similar with the case 8 < 0- H

4.3 The second variation

Up to now we have dealt with the Lagrangian of the form

whete w is the 1-form which defines the horizontal distribution 3C and h is a Riemannian rnetric which extends the subRiem;Lnnian rnetric- We were concerned about the mtical points of the functïonal

which are solutions of the Euler-Lagrange equation

where (4.139)

is the kst variation along q5 in the direction of q. We also were concemed about minnnizers of the action S@). If we denote

where satisfies Euler-Lagrange equation, then u = O is a critical point of f , namely f '(O) = O. If the solution 4 m h h i z e s S(+), a necessary condition is P(0) 2 O- This can be seen better from the Taylors formula, where f'(0) = O

In order to study this we need to define the second variation of S along 4 in direction q

(4.140)

The above ideas and the foilowing computation may &O be found in [Jq. Assuming that L is smooth enough (in our case it always be )

Writing in the abbreviate notation

where t# is the solution of the Euler-Lagrange equation. In the following the second variation and an extra-variational problem (accessory vanational pmblem) will be applied for the Heisenberg group in order to compute the Jacobi equations and Jacobi fields. Just for cornparison reasons a similar approach for the harmonie oscillator is included in the Appendix C-

4.3.1 The second variation of the action on Heisenberg WOUP

I£ in the formula for the Lagrmgian

Al(x) = 2x1 and A2(x) = 2x1 are chosen then it will be asriveci at the Lagrangian for the Heisenberg group

where 6 is a constant and q = (xl, 22, t) - A simple computation provides that

obtains that the second variation of S on the Heisenberg group is

4.3.2 The accessory variational problem for S(4)

In [JL] it is dehed the accessory variational problem. The minimizers of this new action are the Jacobi vector fields corresponding to the initial La- grangian. More geometricallly, the idea is explained in [Sp]. As has been stated eiulier, the subject of interest is about a positive second variation. Hence, if the minimum value of the action

is positive for a.ll 7 vector fields dong the solution tional problem consists in finding the minimum for

q5. The accessory varia-

where (4.147) *(% 4 = Q M + zLt&+ &lm

It is proved in [JL] that this minimizer is a Jacobi vector field. This Jacobi vector field is a solution of Euler-Lagrange equations with respect to the Lagrangian 8 ( q , f j ) . Our interest lies on nonzero solutions I ) such that g(0) = q(r) = O and q(s) # 0, for s E (0 ,~) . As it is known that if q is a Jacobi vector field then c q is still a Jacobi

vector field, V n E R\{O>, namdy a Jacobi vector field is unique up to a multiplication by scalars. I f q #O, let

(4-148)

Then r j = $q is a Jacobi vector field with

and any nonzero Jacobi field is equivaient with such a ''unit norm'' Jacobi vector field-

The constraints and the Lagrange multiplier

A Jacobi vector field q dong # : [O, T] + M minimizes the accessory problem

(4.150) q + 62s(#)9

where # is a solution for Euler-Lagrange equations, namey q5 satisfies

(4.15 1) m#J)il= O , Vl l

In order to avoid nonzero Jacobi vector fields dong 4, add the constraint

and deal with the following variational problem

where the constant p is a Lagrange multiplier and

Consider F itself as a new Lagrangian in q and i. E q is a minimum for J,' F(q) , then r ] will satisfy Euler-Lagrange equations with Lagrangian F, which are cailed Jacobi equations

Jacobi fields and conjugate points on the Heisenberg group

The Lagrandan for the accessory variational problem is obtained using the second variation formula (4.154)

As a F / q 3 = O, from (d/ds)(aF/âj3) = ( i 3 F / b ) we obtain aF/ûq3 = O. As = 2pm and p # O it tunui out that l)s = O and the Jacobi vector field will have the third component zem, namely q = (m, ')r, 0)- This agrees with the fact that the variation of the solutions on the Heisenberg group is a rotation around the t-m. As

and as

then the first Euler-Lagrange equation is

aF -= aF w2

2&-4Brh and -= &!2

48711 + 2pm


Hence the nonzero components m, of the Jacobi field satisfies the system

The system can also be written as

where q = (a, %) and = -Id. The above system is rotational invariant because if R, is a rotation in x-plane, RJ = J& and = &q safidies the same system .-

fi - UJ($) = pq

4.3.3 Getting the Jacobi vector field

From the definition of the Jacobi vector field, q should be nonzero between O and r and q(0) = q(r) = O. One may look for a solution of the following for m

R (4.161) rl(s) = f (4 sin(;s)

where f is a unknown smooth vector field which has no zeros on [O, r]. Then

7r 7r 7r i(s) = f (s) sin(-s) + f (s) - cos(-s)

T r T

The coefficients of cos and sin should be equd

The second equation may be written as

(4.163) f (s) - 20Jf (s ) = O

with f (s) # O. The solution is (4.164) f (4 = e 2û J s c

As e28 Js orthogonal m a e ( d& ezeJ. = e*Wr - Js = eO = l), then f (4 is a rotation applied to the vector C. The constraint condition is

It can be written as

Eience C should be a unit vector and the Jacobi vector fieid is

which means a rotation in (zl, q) - plane applied to the unit vector C. If the conjugate point 4(r) is not on the t-axk, this rotation will rotate 4(r) around a cïrcle. The single way when +(r) is h e d is when the circle has radius zero, nameiy #(r) is on the the t-axïs. As the order of conjugacy is 1 (because the dim =3) the dimension of the space of the Jacobi vector fields which vanish at O and T is 1, namely only one representative Jacobi vector field, the others being multiples of it.

The eigenvalue p

Substitute the formula found in (4.164) for f (s) in the first equation of (6.162) and obtain

and taking f (s) as a common factor

(-4e21 - (L)'I +8e21 - p l ) f (s) = O 7

As f (s) # O it tunüi out that the Lagrange multiplier can be obtained in the

These eigendues are discrete. If .r = 1 it is known that 0, = mr/2 (see Appendk B) and h e m

4.3.4 The second variation for the skew-symmetric case

The Lagranglan corresponding to the vector fields

where a j k = - a k j , is

If q = (x l , x2, . . ~ 2 ~ , t ) , a computation shows that Lqq =

where A = (aij),. Hence the second variation is

The accessory variational probkm is to minimize the fimctional

where p is a Lagrange multiplier. One of the Lagrange equations provides m,,+l= O which means that the Jacobi vector field has no component in the missing direction. The Jacobi equations in this case are

4.3.5 The second variation for 3-dim Heisenberg manifolds

Consider the Hormander manitold on R3 defhed by the vector fields

x1 = a, + Al (+t , X2 = a., - A2(Wt

The associatecl Lagrangian in this case is

A computation provides

where #(s) = (xl(s), x2(s), t (s ) ) is a solution of Euler-Lagrange equations. Denoting

(4.176)

one obtains

If the Hessian îs defined as Hess(f) = (&)y=i3 then the second variation

The Heisenberg case

In the Heisenberg case Kess(Ai) = Eess(A2) = 02,

-1 O and ( O 1) D(A) = (2 i2) = 2 5 . Hence the second variation becornes

(4.179)

Chapter 5

Horizontal objects as limit of Riemannian ob jects

In this chapter some of the horizontal objects (HO) considered previously WU be considered as a limit case of Riemannian objects- As it was defbed in the previous chaptrs, the sub-Riemannian metric is

Definition 5.1 A Riemannian metric h on M i s said to be compatible with a sub-Riemannian mefric (R, g) if the testriction g = hllr holdr.

A sub-Riemannian metric can be constructed fiom a given Riemannian one (see R MO^]). Pick a Riemannian metric 6 on M ( Bt2=+' in out case) such that Xi's are orthonormal, Denote V as the orthogonal distribution of 3t with respect to h:

and decompose the metric h as

whereg, :3t, x3t, + R andv* :V, xV, + R. Now construct a family of Riemamian metncs

The distribution x + V, is called the uertical dz'stribution with respect to the metric h. It is 1-dimensional and integrable. Let V E V , the vertical vector such that q(v V) = 1. Then {Xi, X2, -Xm, &v} is an orthonormal system with respect to hA. Construct the Laplacian -as a sum of squares

and associate a Hamiitonian HA as the principal symbol

where

(5-8)

is the sub-Riemannian Hamitonian or the principal symbol associate to the X-Laplacian . 2n

As X -t +ao then hA tends to a sub-Riemannian metric which is singular dong the vertical direction. The "geodesics" in this sub-Riemannian metric are x-projections of the bicharacteristics of the sub-Riemannian Hamiltonian H(x, C) and they are called s u b t i e m d a n geodesics.

The Riemannian geodesics in hA are x-projections of the bicharacteristics of the Riemannian Hamiltonian HA, As HA + H then the RiemannÏan geodesics will tend towards the subRiemannian geodesics. The Carnot-Carathéodory distance is also a pointWise M t of the Riema- nnian distances

where dx is the Riemannian distance in hA-metric-

In next paragraphs there will be constmcted an explicit 1-parameter famiIy of metrics which are compatible with the subRiemannian mekïc.

5.1 A family of metrics which approximate the Carnot-Carathéodory metric

5.1.1 The construction of the family

In this section wiU be constnicted a f d y of Riemannian metrics hX on the Heisenberg group Hl such that

- (5.11) hA(x i , Xi) = Sïj , where i , j = 1 ,2

(5.12) hA(&&)=X, h x ( x j 1 & ) = O

where the vector fields Xi are given by

Denoting the metric coefficients with respect to the system {a=,, a., , &) by -

(5.14) h$ = hA(azi,8.,) , where il j = 1,2 -

(5.15) h i = hA(a,,&) , h& =hX(&,&) , i = 1,2

then (5.11) can also be written as the following linear system of equations in which the unknowns are the metric coefficients

&+4z2 G3+442fht3 = 1 (5.16) g2 -4- h& + 4 ~ : h& =1

ht2 - 2~~ e3 + 2x2 h& - 4zix2 h& = 0. The relations (5.12) become

Using (5.17) the system (5.16) takes the simple fonn

from where one obtains the expressions for the metric coefficients

(5.19) h~l=1+4Ax~, h&=1+4Ax: &=-4Aztz2

As the coefficients h$ are symmetric in i and j the matrix of co&cients is

1+4Xz; -4Xz1z2 - 2 h 2 (5 -20) h ) = ( - 1 + 4Ax: 2Ax1 )

-2Xz* 2Xs1 X The determinant of this mafrix is

To compute the determinant one may multiply the 3-rd row by 2x2 aad add it to the nrst row and mdtipIy the 3 - d row by -2x1 and add it to the second row. If X # O , then the matrix (hk) can be inverted and the inverse is

1 O (5.22) (h$' = ( 0 1 )

2 q -211 f +41212 This matrix is not very complicated, a fsct which is good news for computing the Christoffel symboIs rjk as we will be noticed later on. - It may be observed that

which bas a zero determinant

5.1.2 An orthonormal triplet

Define the vector field which depends on X i

Then the triplet {Xi, Xa, TA) is an orthonormal system on (R3, hA) , for every X>O- The second order efiptic operator

has the following principal symbol

and

where . am

is the principal symbol for the Heisenberg Laplaciaa

5.1.3 The bicharacteristics


1 X: + xi + 2'' = a: +al, + (41.1' + + 4x2ax1 & - 4x1&&, =

and the Hamiltonian given in (5.25) can be wrïtten as

Denoting by q = (x, t) , p = (E, 8) and the coefficients of the inverse ma- by upper indices (h$)-' = hA", then the Hamiltonian becornes

It is known that the Hamilton's system

where the Harniltonian is given in the relation (5.27), provides q(s) as a geodesic in the Riemamian metric h& This r d t is proved in the section

1.5.4. The solutions (q(s), p(s)) = (x(s) , t(s), ~ ( s ) , 8 ) of the system (5.29) are called bàcharacteristics. Hence the fmt component of the bicharacteristics, namely q(s) = (x(s) , t ( s ) ) , where x(s) = (xl(s) , xz(s)), are geodesics in the metric h?

5.2 A few important tensors

5.2.1 The Christoffel Symbols

For a given Riemannian metric g the Christoffel symbols of the first kind are

rijk are not tensor coefficients, but they will be helpfid in c~nstntcting some tensors. With respect to the metric hA defined in (5.20) the nonzero Christoffel symbols of the first klnd are

The Chnstoffel spmbois of the second h d are defineci contracthg the symbols of the first kind (5.31) r;, = 9br,, The non-zero components in this case are

This symbols wiU be used in the future sections for computing the Riema- nnian tensor of curvature, the Ricci tensor, the sectionai curvature and the Levi-Civita connection.

5.2.2 The Riemannian Tensor of Curvature

The non-zero components of the 4covariant Riemann tensor of curvature are

2 2 Rlt12 = -12X + 16X (2, + x:) , RnI3 = ~X'X~

R1223 = 8 X 2 ~ 2 1 RI3l3 = 4A2 , RZS23 = a2

5.2.3 The RJcci Tensor

The Ricci tensor components given by contracting the Riemannian tensor with the metric g (5.32) &j = a jk lgH

The non-zero components of the Ricci tensor with respect to the metric hA are

5.2.4 The Eticci Scalar

The Ricci scalar is a function defined as the contraction of the Ricci tensor

Using the matrix of the inverse coefnaents given in (5.22), the above formulas for Ricci components and the formula (5.32) by direct computation one can arrive at (5 -34) R-8X

IR th& case the Ricci scalar is a constant.

Chapter 6

Horizontal Einstein space property

6.1 Horizontal Einstein space property on the Heisenberg group

Definition 6.1 A Riemannian space (M, g) is called Einstein space i f the Ricci tensor and the Riemannian metrac are proporttonal

where p is a mooth fulzction on M.

For the Heisenberg group something very similar to the Einstein space property will be proven. The following result states that the Heisenberg group endowed with the metrics hA can be considered as a skewed Einstein space.

Theorem 6.1 lf ~ g ) are the components of the Ricci tensor vith respect to the metnc hi;), then

(6-2) =

In the relation (6.2) X appears with a different signature in the left and right members. This k why it is calleci skewed Eirutein and not Einstein. Proofi From f o d a (5.20) flipping the sign one can obtain

Making some decompositions in the expressions for Ricci tensor components one can observe

RE) = 8 X ( l - 4 ~ ~ $ ) = ~ x G

Hence the relation (6.2) is verifid on components- It is also worthwhile to observe that the proportionality constant is just the Ricci scalar R. rn

Even if the Heisenberg group is not Einstein space in the usual manner, it is s t a Einstein dong the horizontal distribution. The following result will be used

Lemma 6.2 Let U be a horizontal vector field Then hA(U, U) does not depend on the value of A.

Proof: As U is a horizontal vector field, it may be written as

and hence it does not depend on the value of A-

In patticular for X # O

By polarkation and using (6.5) we get for every U,V horizontal vector fields that

2 h A ( c v ) =hA(lJ+V,U+ V)-hA(U,U) -~(')(v,v) =

= h(-A) (CI + V, u + V) - h(-*)(u, u) - hc-" (v, V) = 2 h(-A) (u, V)

Hence

(6-6) h A ( ~ ~ ) = h - A ( U , ~ ) y V U , V E ' H

Hence Theorem 6.1 becomes

Theorem 6.3 Along the horizontal distrtbution the Ricci tensor and the metric are proportional

The theorems presented in this section are of a great importance in computùig the sectional curvature along the horizontal distribution and foreseeing the occurrences of the conjugate points in hA-metric and Carnot-Carathéodory metric as weil.

6.1.1 An analogous of the Gaussian curvature

If Cu, V) is an orthonormal basis in a plane x c TpM, the sectional curvature at p along K is defined as

where R denotes the Riemamian tensor- In the case when M is a d a c e and the plane n = T,M and k, is called the Gawsian curuature at p. In our case the horizontal distribution is not integrable, so that one cannot associate a surface tangent to it and consider the Gaussian cmture, However one can stiU measure the curvature as the sectional curvature dong the horizontal distribution kw. Because {Xl, Xz) is an orthonond system in (X, hA),

The goal of this section is to compute explicitly for the Heisenberg group. However, another two sectional amratures must be computed beforehand.

Lemma 6.4 Let = span(Xl, T) and 1r2 = span(X2, T ) - Then

Proof= Using the expressions for the Riemannian tensor components, some dgebra shows

Proposition 6.5 The sectional m a t u r e dong the ho+ontal distribution with respect to the metric hA i s

Proofi Choosing in Theorem 6.3 the vector fields U = V = XI,

because XI has a unit length for all A. Using the definition of the Ricci tensor,

3

where CEi) is an orthonormal system in the metric hA. Denote Ei = Xi, - i = 1,2 and & =T, then

The first tenu in the right hand side is zero. Using the properties of the curvature, the value of the Ricci curvatwe along Xi found above may be written as

8 X = k x + k q

Using the value of k, provided by the previous lemma one obtains

This section wilL be finished by proving that the Ricci tensor is constant along the solutions of Euler-Lagrange equations.

Proposition 6.6 Let t# be a solution of Euler-Lagmnge equations svch that I$(o)~ = 1- Then (6.12) R@)(&), #(s)) = 8X

Proof: Using the Theorem 6.1 and Lemma 6.2 we have

= (&s), #(s)) = SA (x:(s) + 12:(s)) where 6(s ) = Zi(s) XI + x2(s)X2, because it is known that the soIutions of the Euler-Lagrange equatiom are horizontal cmes. It is also a known fact that the Hamiltonian is preserved dong the solutions, or, in other words, the tangent to the solution has constant length (constant speed). As the initial speed is qua1 to 1, it hnns out that x;(s) f z$(s) = 1 and hence

Our goal in the next section is to foresee the occurrence of the conjugate points along the geodesics in the hA- metric-

6.2 Estimation of the Conjugate Points

For each metric hA one has a Levi-Civita connection VA detkied by

where the Christoffel symbois r&(X) depend üneasly on X and are given in section 5.2.1. One may easily observe that

If X > O , then h(-A) is a Lorentz (++ -) metric with the diagonal form with respect to {XI, Xa, &) equal to

One may associate a Levi-Civita connection v(-~) to the Lorenz metric h(-A) in the usual manner (6.15) - - r;(-~)a~ Using (6.14) iit is observed that

which roughly speaking means that V A is an odd function in A. It is known that a ciirve # is a geodesic with respect to the metnc hA ifE

Using (6.16) one may also write

which meam that 4 is a geodesic also in the metnc h(-'). H e c e the following resdt has been obtained

Proposition 6.7 Cet X > O be @ed. Then 4 Is a geodesic in hA-metric iff 4 is a geodesic in h(-A) -metrie.

Another result regarding the Ricci tensor is the following

Proposition 6.8 If q5 : [O, q + Al3 = Hl is a unzt speed geodesic with respect to the metric hA with the horizontal initial velocity &O) E W, then the Ricci tensor is constant along it

Proof= Let # : [O, L] -t mM3 = H1 be a geodesic in hA-metnc, parametrized by arc length with $0) horizontal. Using Theorem 6.1

Using the above proposition it turns out that 4 is &O a geodesic in the h(-%netric. As the Iength of the speed îs constant along the geodesics,

As &O) is horizontal,

Hence (6.21) becomes hc-A) (&SI , &s)) = 1

and (6.20) may be -en as

This result is important because it wil l aid in finding an estimation for the £kst conjugate point dong the geodesics. This will be done in the next theorem,

Theorem 6.9 Let q5 : [O, LI -t (RI, h*) a geodesic parometrized b y arc length with horizontal initial speed. If L > ?r/(2\/X) then the= is a point r E (O, L) such that 4(r) is conjugate to #(O) along 4 and hence Q B not of minimal length.

Proof: Use Myers theorem (see [Sp] vol 4): Let M be an n-dimensional Riemannian manifold and # : [O, L] + M a geodesic pammetrized by arc length. Let r > O be a constant and suppose that

n - 1 m s ) , Q>(s)) 2 7 1 Qs

and that 4 has L > m. Then the= is a point T E (O, L) conjugate to O and q5 is not of minimal length. In order to get a proof for the theorem, choose n = 3 and r = 1 / ( 2 d ) and apply Proposition 6.8. rn

In other words, in the met& hA the first conjugate point to the orïgin is before lr/(2z/X) - The parameter X is related to the length of at

which means that the longer &, the closer to the origin the first conjugate point is. n o m the above theorem r e d t s that for every neighborhood V of p there is a X > O such that V contains a conjagate point dong any geodesic q5 with respect to the metric hA which starts at p.

Corollary 6.10 Let q5 : [O, L] + (ml, h*) a geodesic pazametrized by am length un'fh horizontal initial speed. If L > k r r l ( 2 d ) then thete are at least k wnjugate points r k E (O, L) such that #(rk) b wnjugate to #(O) dong #.

Proofi Apply inductively Theorem 6-9.

6.3 Properties of Levi-Civita connection

Denote by VA the Levi-Civita connection with respect to the metric hA de- h e d by (6.13). The following result states that the Levi-Civita connection with respect to &, &, , & is always a linear combination of XI and X2 and hence it is a horizontal vector field.

Theorem 6.11 For every X > O we have

(6 -24) v&$ = v$~$ = (22x1 + 51x2)

Proofi The identities between the connections corne out 6com the fact that the Levi- Civita connection is symmetric and

The second identities are just a computation using lin- algebra and the expressions for the Christoffel symbois given in the section 5.2.1.

v&, a== = ri2a, + ri2sZ2 + ri2& =

= -8Xxl& + O - ~ ~ X X ~ X Z ~ , = - ~ X X ~ X ~ -

As all F $ ~ = O , for k = 1,3, it tums out that

Vk& = O

The foilowing proposition shows that the vector fields Xl and X2 are geodesic vector fields.

Proposition 6.12 For e v e q X > 0,

(6.30) V$,X~ = O , V:,X~ = O

Proof: start by computing wing Proposition 6.11

v & ~ XI = viil (a, + 2x&) = v;& + 2z2v$ =l Bt =

= -8Xz2X2 + 2x2 2- = -4Xxz Xz and next we compute that

Using the above two relations and the previous theorem it is obtained that

v%~& = v8=,+2z,&xi = v8=,xlt2z2v;&xl =

= -4Ax2X2 + 2x2 2% = O

Observation As VA is symmetinc, the torsion i9 zero

v$,x* - vX2x1 = [X,, X2]

Using the above relations,

-2& - 28, = -4& = [XI, Xz]

a fact that can be verified directly using a direct computation. Another property of the Levi-Civita connection is given in the foIlowing theorem. This can &O be considerd as a definition for the D-connection as will be shown later on.

Theorem 6.13 The restntncfion of the eonnection VA on 3L x 3L does not depend on A.

Proofi Let U, V E 3C be two horizontal fields which have the decomposition

The coeiEcients Us, V' do not depend on A. When V&V is computed, one obtnins

= u(v') XI + V' v&xl + v2 v* + u(v2) X, =

(6.32) = u(v~) Xl f u(v2) X2 + V' V&X~ + Va V$X~

Before continuing the computation the third and the fourth term in the last relation must be computed, using the previous theorem about VxïXj- For the tbird term,

= u2 v*,xl = 2 u2at and for the fourth term,

= u1 vg1x2 = -2 u1 a, Hence relation (6.32) becornes

v$v = U(V1) Xl + U(V2) X2 + 2v1u2at

(6.33) = U(vl) Xi + U(v2) Xa + 2&t

As one may observe in relation (6.33) there is no A. Hence Vh,, is independent on X and the theorem is proved.

Denote by D g the horizontal component of V&V, namely

This is the honiontd connection defined and studied in section 1.5- The determinant has the geometrical sîgniscance of s parallelogram area generated by the vectors V and ET, denoted by area(U,V). Hence the following result îs obtained

Proposition 6.14 For alZ X > O we have the following decomposition

(6.35) V$V = DuV + 2 area(V, V)& , VU, V E R

Hence the horizontal comection D may be defineci to be the horizontal component of VA. This is independent of A.

Next result relates the accelerations in both connections

Corollasy 6.15 If q5 Zs a horizontal curve, then

(6.36) ~ $ 4 = 0,jd

Proof: As area(&,t&) = O, formula (6.35) gives the result.

CoroUary 6.16 If t$ ts a sub-Riemannian geodesic, then

(6.37) o*6 = PX($)

Proof: Use the above Corollary in the Euler-Lagrange equation.

Observations: 1. For a horizontal curve #, it does not matter in which connection the acceleration ïs taken,

2. Formula (6.35) can be considerd as a Gauss formula The horizontal part of VA k the connection on the distribution H and it is denoted by D and the vertical part P(U, V) = 2 area(U, V)& is Iike a second fundamental form. However, here things are dXerent because it is not a case of an integrable distribution- The above denominations are used in the integrable case.

6.4 Carnot-Carathéodory distance as a lirnit

Let P and Q be two points and denote the Riemannian distance measwd in the Riemannian metric hA by

where the length is

Remember that the Carnot-Carathéodory distance is measured as the Length along the solutions of Euler-Lagrange equatiom.

Proposition 8.17 If dc denotes the Carnot-Carathéodory distance,

(6.40) &(PI Q) = k~ ~A(P, 9) A+œ

Proof= The distance may be written as

where #A is the geodesic in hA metric between P and Q . When X + oo the geodesic tends to a solution of Euler-Lagrange equatiom ('csub- Riemannian geodesicl') which joins P and Q, denoted by &. If h, denotes

the Carnot-Carathéodory metric then it is observd that

As &, is horizontal, h&, &) does not depend on A, so that

The next result shows even more than that, namely the Carnot-Carathéodory distance is an increasing limit of the distances measmed in h*.

Proposition 6.18 If Xi < , then dx,(P, Q) 5 d ~ , (P, Q) .

Proofi Prove that X -t dA(P, Q) is increasing. As

then the derivative will be

where hl is hx,-, - and ho =

If = aXl + bXz + dr, as Xi, X2, are orthonormal in the metric hl, then

and the ccength in ho" is

Consider the m e q5 : [O, 11 + Al3 fixeci, differentiate and using (6.41),

We note that the derivative is zero i f i c = O, i-e- if the c m e is horizontal then the length measured in all hA are the same. The conclusion is that for a fixed #, if Al 5 & then

Taking infimiim over di q5 which join P and Q,

Observation On short, the above proof is using that

1$lS1 = a2 + b2 + c 2 ~ = 5 Q1+ d + c2& = [&$a

6.5 Horizontal Einstein property on Hormander manifolds

In this section it will be proven that the horizontal Einsteh property works even in a more general case of the three dimensional Ekmander manifold defined by the vector fields

First, construct a one parameter family of Riemannian metncs in which the above two vector fields are orthogonal.

6.5.1 Construction of the Riemannian metric

For any X > O define the vector field 4 = A&. It is desired to constnict a metic hX such that {XI, X2, TA) are orthonormal in h? First, the inverse of this mehic must be found, using the following method. Construct the Hamihonian as the principal syrnbol of the elliptic operator I(X: 2 + X," + Ti)

and let (hA)'i be the coefficients matrix of the quadratic form

where p = (cl, b, O ) - A computation shows that

The determinant is &t(hA)" = f. The inverse of the matrix (hA)" is denoted by h$ and is given by

This is the Riemannian metric in which {XI, X2, TA) is an orthonormal system. The proof is a simple verifkation

It is the same for the others. The geodesics in hA metrie are given by the q-projection of the solutions of the Hamilton's system with the Hadtonian (see section 1.5.4.)

where q = (x, t) and p = (& 8). When X -t m , HA + H , where H is the Hamiltonian associate to the operator i (X: + X;). The geodesics in hA wiU tend to the subRiemannian geodesics (solutions of the Euler-Lagrange equations) . The following paragraphs are dealing with Christoffel symbols a d Ricci curvature associated to the Riemannian metric hA.

6.5.2 Christoffel Symbols

The non-zero components of the Christoffel symbols of the first kind with respect to the metric hA are

and the non-zero components of the second kind Christoffel symbols are

rL= - M l (A; + A;)

1 1 r:2 = --X(A\ + A~)(A: - A:) + ZX(A; - A;) 2

1 r& = --AA2(AA; + A;) 2

6.5.3 Ricci Tensor

The Ricci teiisor with respect to the metric hA has the following non-zero components

and the Ricci scdar is X

(6 -47) R = -(Ai + A;)* 2

AU of the above components should have a X superscript but for the sake of simplicity it has been neglected.

6.5.4 The Einstein Property

The goal of this section is to write

(6.48) R;- = + fremainder

where

(6 -49)

The remainder measures the departue hom the Heisenberg group structure, as wi l l be observed.

Theorem 6-19 The following d a t i o n takes place

Proof: The proof is just a computation using the formulas for the Ricci t e m . Com- pute one component at a time

Hence Qll = 2A1Ay

Hence Qlz = -(AiAa + AlA2).

A ( 4 Hence R23 = R hp + *w and Qp = &. And for the last component

and hence Q33 = 0.

Observations 1- Qij depends just on the vector fields Xi and X2, namely Q, depends on the horizontal distribution 3C only. 2. In the Heisenberg group case A1(x2) = 2z2 and A2(z1) = 2z1 and hence A: = A! = O and Qv = O. It may be stated that Q meanies the departure fiom the Heisenberg structure. 3. QG is independent of X (this is the same for all hA). 4- If Qij = O then the Heisenberg structure is the Heisenberg group (up to some transformations). 5. The most important property of QG is given in the following result and states that Q is zero dong the horizontal distribution . This leads to the following Iemma

Lemma 6.20 If d(s) = (xl(s) , x2 ( s ) , t ( s ) ) , then

(6.52) Q(), 4) = 2 ( ~ ; b ~ ( s ) - A:& ( s ) ) @(s) - AIXI (s) + A& ( s ) )

Proofi Let 6 ( s ) = xl (s)Xl + X 2 ( s ) X2 + f (s)& Then a straightforward computation provides

Proposition 6.21 If 4 is a horizontal curve, €hem

Proof: It is known that 4 is horizontal ifE i - A& + A2x2 = O and hence, using the above lemma, Q(&& = 0. rn

Next result is the goal of this section. It states that dong the horizontal distribution the Homander manifold defined by the vector fields in (6.43) has the Einstein property.

Theorem 6.22 Let 4 be a horizontal curve. Then for any X # O

Proof: From the Theorem 6.19 and the Proposition 6.21 one obtains

As the length of a horizontal vector does not depend on metric hA, hA(& #) = h(-')(&, 6) and hence arrives at the desired resdt. rn

6.5.5 An Estimation of the occurrence of the fhst conjugate point

Proposition 6.23 If 4 LP a solution of Euler-Lagmnge equations with unit initial speed, then

\

Proofi As hA(& #) = 1 and

X R~ = -(Ai + A;)2

2 applying the Theorem 6.22 the result is immediate.

In the following a theorem will be given which generalizes the similar result fkom the Heisenberg group case.

Theorem 6.24 Let Xl = a, +Ai(z2)& X2 = a., -Aa(xt)& where Al and Aa are such that 2 E > O for i = 1,2. Then the first conjugate point to origin dong the (unàt speed) gwdesics in hA-metric happens before T/(&).

Using Myers theorem with r = l / (~ f i ) one obtains a conjugate point before TT- = ?t/(€,/X)- rn

Observations 1. The right hand side is a constant itf Al(x2) = k1q, Al(%=) = k2xl which can be reduced to a Heisenberg group case- 2. If Al and Aa are odd bctions, the derivatives A; and A& are even fwicti- ons and (A; + ïs a h even and bounded ftom below.

Appendix A

The solution of Euler-Lagrange system on Hl

Consider the vector fieIds:

X,=a , ,+Zs&, x* = a., - 2x14 and the Heisenberg Laplacian

The Hamiltonian is d e h d as the principal symbol for the above operator

From Hamilton's system of equations

Using Legendre transform the associate Lagrangian is:

L =&XI +cas2 +Bt - H ( c , ~ , t 9 ) .

PerfoIILUag the computations it is obtained

One way to approach the problem is to write the Lagrangian in polar coordinat es:

21 =rcosq'i, 22 =rSinq5,

where r = r ( s ) ,9 = q5(s), and

By direct computation

zl22 - alsa = r2$

The fkst expression is the energy and the second the areal velociW The Lagrangian becomes:

The Euler-Lagrange system is:

Computing term by term

and the b t Euler-Lagrange equation becomes:

For the second equation:

the second Euler-Lagrange equation becomes:

d -(t2(J + 28) ) = k (constant). ds

Considering B = 1, the Euler-Lagrange becomes

where k is a constant. As ~ ( 0 ) = O it turns out that k = O and hence # = -2 or

The first equation becomes -- T = - 4 ~

with the solution r(s) = ai cos 2s f sin 2s

As r(0) = O it turns out that r(s) = q sin2s Finding a2

where j.&) = f (s) cos $(s) - r(s) sin #(s)#(s)

and therefore

The length of v is I V [ = i (0 ) and the arg of v is 4(0). It turns out that

1 a2 = -14sin2~

2 and

#(s) = -2s + #(O)

Finding the t-component From one of the Hamilton equation

i = 2[28(~: + 1 2 ) + - xl&] Using the other Hamilton's equations

one obtahs t = 2(x1x2 - x1I2)

As in polar coordinates xiz2 - x1x2 = -r2Q,

Using the expressions for r ( s ) and &s)

In the following will be proved that the @st con*gate point to on& along the minimàzer is situated at t h e T = lr/2 away h m origin. hdeed, let u, be a variation of the Initial speed v. Then

The conjugate point corresponds for a r # O for which

Rom the third equation it tums out that

namely, the vari-ation must be a rotation. Ekom the second equation we pet &(O) = 4(O). Then v and v. are collinear. As they have the same length, there are the same, v = v,. Hence there is no variation, which is a contradiction- To avoid this we have to be situated in the particdar case r ( ~ ) = O where the angles #(O), &(O) play no role. Then sin2r = O and chosen the mallest T, we get T = r/2. In conclusion the fhst conjugate point is situated on the t-& at a disfance

from the origin. When the Length IV 1 is taken as a parameter, the conjugate point describes the t-axis.

Appendix B

The lengths of the geodesics on

Here will be computed the lengths of the solutions which join the origin and the point (O, O, t) in the case of the Heisenberg group. The approach wiU use polar coordinates and the fact that the solution is given by

The length and the Harniltonian Ho The solution 7 is considered parameterized by [O, 11 and using the fa& that 191 is constant along the solution, one has

O is constant along the solutions. It satisfies the following b o u n d q value problem

kom where 20 = mr, m integet. Each d u e of 0 will determinate a certain length. The Hamiltonian and t One of the boundary conditions is t(1) = t. Assuming B > O one has $ = -20 and 4 decreases. Denote A = t$(O). Using the equation I! = -2r24 one may integrate between and q50 - m?r

Using that

one obtains

The lengths lm The lengths are

The equations Using that

For computing t&) one integrattes dt = -2r2dq5 between A and 4

Hence

r, and t,,, depend on q50 but the corresponding length lm = J* does not*

Appendix C

The harmonic oscillator, another example

The Lagrangian for the harmonic d a t o r is given by

where m is the mass of the particle and X an elasticity constant. Then the second variation is

PS(#)? = br r n ( - ~ ~ r p ~ + @2) ds

In this case F is given by

where p is a multiplier. The Euler-Lagrange system for the Lagrangian F(q, i ) together with homogeneous boundary conditions is in fact the Jacobi system of equations

f i = -(A2 - p/m)q

V(0) = ~(7) = 0 Assuming A2 > p/m, the solution will be

R o m the boundary conditions one obtains ,/A= - ( p / m ) r = kr, k = l ,2 , 3 . . . and hence the Lagrange multiplier takes the values

The first multiplier is obtained for k = 1 and corresponds to the Jacobi vector field

7r q(s) = A sin(;s)

Appendix D

The physical interpretation for R and w

A magnetic field on p ( x , y, z) is given by a two form 0 whîch saMes the Maxwell's equations (6.56) dS1 = O

The case of conceni is when the magnetic field cornes fiom a potentid

(6.57) 0-dw

where w = wi d i i and

(6.58) Cl = n,dy ~ d z + Q , d z ~ d z + S l & ~ d y

The equation of motion for a particle with charge e and mass m is given by the Lorenz equation

The above equation is the iPngian

Euler-Lagrange equation for the folIowing La-

The associate Hamiltonian is

Example For w = dt - 22 dy + 2ydz then fl= 4& A dy and

with the Euler-Lagrange solutions circles in xy-plane, called Larmor orbits (see [Th]).

Appendix E

The intemal

is c d e d an ell5ptic i n t e p l of the fird kind The IntegraL exists ifw is real and such that Iwl c 1. Using the substitution t = sin0 and w = sin@ then

If k = O, then z = s i d w or w = sina. By analogy the above integral is denoted by sn-'(w; k) where k # O. Thus

The function w = sn z is called an elliptic function or a Jacobian elliptic jhction. By analogy with the trigonometnc functioas, it is convenient to define other elliptic functions

The properties of these functions are given in the following

then the principal period of sn z and cn s (as a real functions) is 4K.

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The Missing Direction DifEerential Geometry on Heisenberg … · 2020. 4. 7. · described by Heisenberg manifolds- The research tools used in the thesis involve techniques of Dserential