Geometric Analysis on the Heisenberg Group and Its Generalizations (Ams Ip Studies in Advanced Mathematics)

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Author: Ovidiu Calin; Der-Chen Chang; Peter Greiner

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AMSjlP

Studies in Advanced Mathematics S.-T. Vau, Series Editor

American Mathematical Society . International Press

AMSjlP

Studies in Advanced Mathematics Volume 40

Geollletric Analysis on the Heisenberg Group and Its Generalizations Ovidiu Calin Der-Chen Chang Peter Greiner

American Mathematical Society

•

International Press

Shing-Tung Yau, General Editor

2000 Mathematics Subject Classification. Primary 53C17, 53C22, 35H20; Secondary 46E25, 20C20.

Library of Congress Cataloging-in-Publication Data Calin, Ovidiu. Geometric analysis on the Heisenberg group and its generalizations j Ovidiu Calin, Der-Chen Chang, Peter Greiner. p. cm. - (AMSjIP studies in advanced mathematics, ISSN 1089-3288 j v. 40) Includes bibliographical references and index. ISBN-13: 978-0-8218-4319-2 (alk. paper) ISBN-lO: 0-8218-4319-2 (alk. paper) 1. Nitpotent "tie grottt>s. I. Chang, Der-Chen E. II. Greiner, P. C. (Peter Charles), 1938III. Series. 2007060760

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© 2007 by

the American Mathematical Society and International Press. All rights reserved. The American Mathematical Society and International Press retain all rights except those granted to the United States Government. Printed in the United States of America.

@)

The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at http://vww . ams. org/ Visit the International Press home page at URL: http://vww.intlpress . com/ 10987654321

12 11 10 09 08 07

This book is dedicated to Professor Yum-Tong Siu for his teaching, friendship and encouragement.

Contents Preface

Vll

Chapter 1. Geometric Mechanics on the Heisenberg Group 1.1. Definitions for the Heisenberg group 1.2. The horizontal distribution 1.3. Horizontal connectivity theorem 1.4. Hamiltonian formalism on the Heisenberg group 1.5. The connection form 1.6. Lagrangian formalism on the Heisenberg group 1.7. Carnot-Caratheodory distance 1.8. Exercises

1 1 7 10 13

18 23 38 42

Chapter 2. Geometric Analysis of Step 4 Case 2.1. Elliptic functions 2.2. The horizontal distribution 2.3. SubRiemannian geodesics 2.4. Lagrangian formalism 2.5. Solutions which start from the origin 2.6. The length of the geodesics between the origin and (0,0, t) 2.7. Explicit solutions connecting the origin to (0,0, t) 2.8. Solutions which start outside the origin 2.9. Geodesics between the origin and points away from the t-axis 2.10. Geodesic completeness in step 4 case 2.11. Exercises

45 45

Chapter 3. The Geometric Analysis of Step 2(k + 1) Case 3.1. The horizontal distribution 3.2. Horizontal connectivity 3.3. SubRiemannian geodesics 3.4. Euler-Lagrange system in polar coordinates 3.5. Geodesics starting at the origin 3.6. Carnot-Caratheodory distances from the origin 3.7. Particular cases 3.8. Conjugate points to the origin 3.9. The use of the hypergeometric function 3.10. Geodesics starting outside the origin 3.11. Geodesic completeness 3.12. Global connectivity by geodesics 3.13. Kaehler metrics 3.14. The classical action

85 85 86

v

47 51 53 56 64

69 71

78 82 82

88 91 91 97 100 105 106

110 113

115 117

118

CONTENTS

vi

3.15.

Exercises

122

Chapter 4. Geometry on Higher Dimensional Heisenberg Groups 4.1. The Heisenberg group Hn 4.2. Hamiltonian mechanics on Hn 4.3. The classical action 4.4. The horizontal distribution 4.5. The Carnot-Caratheodory distance 4.6. The shortest geodesic 4.7. Caustics 4.8. Exercises

125 125 126 129 129 130 136 140 141

Chapter 5. Complex Hamiltonian Mechanics 5.1. The harmonic oscillator and the Heisenberg group 5.2. The quantum Hamiltonian 5.3. Complex subRiemannian geodesics 5.4. Complex subRiemannian geodesics on the Heisenberg group 5.5. Complex connectivity 5.6. The complex action 5.7. Hamilton-Jacobi equation 5.8. Geodesic completeness 5.9. Solving Hamilton-Jacobi equation 5.10. Theorem of Jacobi and applications 5.11. The lengths of the real geodesics 5.12. Modified complex action function on Hn 5.13. A geometric formula for the fundamental solution 5.14. The volume element on the Heisenberg group 5.15. Exercises

145 145 147 147 148 153 162 166 169

Chapter 6. Quantum Mechanics on the Heisenberg group 6.1. Linear Harmonic Oscillator 6.2. The operators Z and Z and the energy quantification 6.3. The ground-state 'l/Jo and the states 'l/Jn 6.4. The operator aT - Z 6.5. Heisenberg derivative 6.6. The fundamental solution for D.H 6.7. The fundamental solution for D.>. = D.H - ~A[Xl' X 2] 6.8. An application to the eigenfunctions of D.H 6.9. The Schrodinger kernel 6.10. The heat kernel on HI 6.11. Maxwell's equations 6.12. Quantization of energy 6.13. Kepler-type laws 6.14. Exercises Bibliography

199 199 201 203 206 211 212 213 215 216 222 228 233 235 236 239

Bibliography

239

Index

241

171 174 177 184 188 190 194

Preface This book deals with a detailed analysis of the Heisenberg group and its extensions to superior steps, and their applications to Physics. The approach is using the complex and real Hamiltonian and Lagrangian formalism developed in the subRiemannian manifolds context. The exposition is enriched with chapter exercises which facilitate the reader's understanding. An overview for the Reader: This work is a text for a course or seminars designed for graduate students interested in the most recent developments in the subRiemannian manifolds and sub-elliptic operators theory. It is useful for both pure and applied mathematicians and theoretical physicists working in the quantum mechanics area.

Scientific Outline: This book deals with the study of subRiemannian manifolds, which are manifolds with the Heisenberg principle built in. This brings the hope that Heisenberg manifolds (step 2 subRiemannain manifolds) will playa role in Quantum Mechanics in the future, similar to the role played by the Riemannian manifolds in Classical Mechanics. Some people also speculate that superior steps subRiemannian manifolds may playa similar role in Quantum Field Theory. The subRiemannian geodesics behavior, which is very different than in the Riemannian case, plays an important role in finding heat kernels and propagators for the Schr6dinger equation, as well as in finding fundamental solutions for sub elliptic operators using a very ingenious geometric method involving complex modified action. One of the novelties of this book is to introduce the complex Hamiltonian mechanics techniques and apply them in describing the fundamental solutions and heat propagators. The fact that the propagator depends on each sub Riemannian distance is an expected fact in Quantum Mechanics. Our method is quite different and new. The fact that each path has a contribution to the propagator led Feynman to introduce the path integrals. In this work we can manage without them, just working in the proper geometric framework. For macro objects, like airplanes, planets or bullets, only the classical path is important and their motion is described by a unique geodesic in the context of Riemannian geometry. The situation for the sub-atomic particles is completely different and this book indicates the correct framework to study their motion, which is the subRiemannian geometry. The main issue which is the base of this interpretation is the fact that local and global in subRiemannian context are the same. Hence there are points arbitrary close which can be joined by infinitely many geodesics with distinct lengths and which are used in the asymptotic expansion of the heat kernel on the Heisenberg group. vii

PREFACE

viii

The first chapter contains a detailed description of the geometry of the Heisenberg group which is the prototype of the step 2 subRiemannian manifolds. It shows that the Carnot-Caratheodory distance plays a very important role in finding the fundamental solution of the Heisenberg operator which retrieves the well-known Folland's formula. The second chapter deals with a step 4 subRiemannian manifold and its geometry. The tools used to describe the geodesics involve complicated and explicit computations involving elliptic functions. Chapter 3 deals with the geometry of a step 2(k + 1) subRiemannian manifolds. As in chapter 2, the closed form solutions require the use of hypergeometric functions. We provide also a qualitative approach and we manage to prove global connectivity and geodesic completeness. Chapter 4 is concerned with the non-isotropic multi-dimensional Heisenberg group. In this case, the group law will be n

(x, t)

0

(x', t') = (x

+ x', t + t' + 2

L

aj [X2jX;j_1 -

X2j-1X~jl).

j=l

In the isotropic case a1 = a2 = ... = an, the results of Chapter 1 carryover with no change, except that each family of geodesics from the point (0, t) to the origin is parametrized by the (2n - I)-sphere. However, on a general non-isotropic Heisenberg group, the analysis is more laborious. More precisely, suppose that

0
tXk. k=l

From

[Xl,

X 2] = -4at and

[Xl,

at]

= [X 2, at] = 0, the constants of structure are C3l2 --

j

-

4,

= 1,2,3.

If X = (XI,X2,X3) and Y = (YI,Y2,Y3), a locally Lie group structure is given by the Campbell-Hausdorff formula:

(1.9)

(x

0

Y)i

. Xj Yk + 121 "L-- xp Yj Xs Cpjk Cks. + ... = Xi + Yi + 2"1" L-- cjk j,k

k,s,p,j

In our case, we obtain a globally defined group structure

(X oyh (x oYh

(x

0

yh

because the term xp Yj Xs C;j

+ YI, X2 + Y2, 1 X3 + Y3 - 2" . 4(XIY2 -

Xl

cL =

X2yt),

O. We arrived at the following result.

PROPOSITION 1.4. The vector fields Xl = a Xl + 2X2at, X 2 left invariant with respect to the following Lie group law on lR 3

(1.10)

= a X2

-

2x l a t are

1.1. DEFINITIONS FOR THE HEISENBERG GROUP

5

The Lie group Hl = (R,3, 0) is called the symmetric three dimensional Heisenberg group. The unit element is e = (0,0,0) and the inverse is (Xl,X2,t)-1 = (-Xl, -X2, -t). We shall study the subRiemannian geometry associated with this model. The geometry has the Heisenberg uncertainty principle (1.11)

built in. This brings the hope, that Heisenberg manifolds (step 2 subRiemannian manifolds) will play a role for Quantum Mechanics in the future, similar to the role played by the Riemannian manifolds for Classical Mechanics. (see e.g. [2], [1], [14]). In Quantum Mechanics, the states of a quantum particle (position, momentum) are described by differential operators. It is known that two states, which cannot be measured simultaneously correspond to operators, which do not commute (see [45]). For instance, if x and p are the position and the momentum of a particle, then one cannot measure them simultaneously and hence, we write [x, p] -I 0. The state of the particle is measured using a radiation beam sent towards the particle. The radiation is reflected partially back. Using the variation of frequency between the sent and reflected beams, the Doppler-Fizeau formula will provide the speed of the particle. This method will provide accurate results if the radiation will not significantly change the speed of the particle i. e., its kinetic energy K = mv 2 /2. This means the radiation has a low energy and hence, a low frequency, because Eradiation = hv. Therefore the wave length of the radiation). = l/v will be large. In this case the position of the particle cannot be measured accurately (see [32]). In order to measure the position accurately, the radiation wave length has to be as small as possible. In this case the frequency v is large as will be the energy Eradiation' This will change the kinetic energy of the particle and hence its velocity. Hence, one cannot measure accurately both the position and the speed of the particle. The Heisenberg uncertainty principle, fundamental in the study of quantum particles, can be found also in other examples at the large scale structure. Let's assume that you are watching high-street traffic from an airplane. You will notice the position of the cars but you cannot say too much about their speed. They look like they are not moving at all. A policeman on the road will see the picture completely differently. For him, the speed of the cars will make more sense than their position. The latter is changing too fast to be noticed accurately. The Heisenberg group is also closely linked with theory of functions in several complex variables. Let us start with the unit disc {w E C: Iwl ~ I}. Consider the Cayley transform:

(1.12)

w

i-z i+z

=-.

°

It is known that the transformation (1.12) maps the real line Im(z) = together with the point "00" into the unit circle Iwl = 1, and the upper half plane {z = X + iy E C : Im(z) > o} into the disc Iwl < 1. From the point of view of Fourier analysis, there are two groups of holomorphic self-mappings on it which are of importance to us. The first one is the translation group, z = X + iy ----t (x + h) + iy,

6

1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP

h E JR, which is isomorphic to R This group acts transitively on each level set {z = x + iy E JR~ : y = c, c ?: O}, which is responsible for the fact that many important operators, e.g., Hilbert transform and Cauchy operator, are convolution operators. The second one is the dilation group, Z -> 8z, 8> 0, which determines the homogeneity of the kernels of these operators. The analogues of these groups are equally significant. A complex analogue of the upper-half plane JR~ is the Siegel upper-half space

Its boundary is the "paraboloid"

The domain U is biholomorphically equivalent with the unit ball

via the generalized Cayley transform:

2izI + Z2'

WI=--

i

i -

W2

Z2

= -.--. 2

+ Z2

In this correspondence the boundary aU, together with the "point at 00" maps onto the unit sphere in ([:2. It is known that the Heisenberg group HI gives the translations of the domain U. The action of an element (z, t) E HI on the domain U is given by (1.13) Since 16 surface"

+ zII2 -

IZII2 = Im{ i(26 . ZI Im(6) -

+ IZII2)}, 161 2

this mapping preserves each "level

= constant.

It follows that the mapping (1.13) is simply transitive on the boundary aU: the boundary aU therefore may be identified as the orbit of the "origin" (0,0) under the action of the group HI:

The second group consists of dilations. The action on the domain U is given by 8> O. It is easy to see that these are automorphisms of the group structure of HI' Summarizing our discussion, the group HI is isomorphic to the boundary aU. Similar to analysis in JR2, we may use the Heisenberg group and the Siegel upper-half space to "approximate" a bounded strongly pseudoconvex domain in ([:n+ 1 . This phenomena indeed happened in many situations, see Folland-Stein [24] , Phong-Stein [49], [48], and Greiner-Stein [28].

1.2. THE HORIZONTAL DISTRIBUTION

7

1.2. The horizontal distribution Unlike on Riemannian manifolds, where one may measure the velocity and distances in all directions, on Heisenberg manifolds there are directions where we cannot say anything using direct methods. On the Heisenberg group, an important role is played by the distribution H generated by the linearly independent vector fields Xl and X 2:

x

---?

Hx

=

spanx{XI , X 2 },

called the horizontal distribution. As [Xl, X2J = -40t rj H, the horizontal distribution H is not involutive, and hence, by Frobenius theorem, it is not integrable, i.e., there is no surface locally tangent to H. A vector field V on }R3 is called horizontal if and only if Vx E H x , \:Ix. A curve c : [0, 1J ---? }R3 is called horizontal if the velocity vector c( s) is a horizontal vector field along c( s). Horizontality is a constraint on the velocities and it is called a non-holonomic constraint.

In this chapter we shall construct many horizontal objects, i.e., geometrical objects which can be constructed directly from the horizontal distribution and the subRiemannian metric defined on it. The main goal is to recover the external structure of the space, such as the missing direction Ot by means of horizontal objects. PROPOSITION

1.5. A curve c = (XI,X2, t) is horizontal if and only if

(1.14) PROOF.

i

= 2(XIX2 - XIX2).

The velocity vector can be written as

C

XIOX1 + X20X2 + tOt Xl (OXl + 2X20t) - 2XlX20t

+ 2XIX20t + iOt XIX I + X2X2 + (i + 2XIX2 - 2X2XI)Ot. +X2(OX2 - 2XIOt)

Hence, C E H iff the coefficient of Ot vanishes i.e., (1.14) holds. COROLLARY

1.6. A curve c =

(Xl,

o

X2, t) is horizontal if and only if

(1.15)

In the following we shall give a geometrical interpretation for the t component of a horizontal curve. This will be used in the proof of the connectivity theorem later. Using the polar coordinates Xl = r cos cp, X2 = rsincp, equation (1.14) becomes

t

2(XIX2 - XIX2) -2r2¢(sin2 ¢ + cos 2 cp) -2r 2 ¢.

In differential notation (1.16) Let r = r (cp) be the equation in polar coordinates ofthe projection of the horizontal curve on the x-plane. The area of an infinitesimal triangle with vertices at the origin,

8

1.

GEOMETRIC MECHANICS ON THE HEISENBERG GROUP

(r(¢), ¢) and (r(¢ + d¢), ¢ + d¢) is ~r(¢)r(¢ + d¢)d¢ ~ ~r2(¢)d¢. Integrating, we obtain the area swept by the vectorial radius between the initial angle ¢o and ¢, see Figure 1.1.

B

A

Figure 1.1 : The area swept by the vectorial radius between two points in the plane.

Taking the derivative, we obtain dA d¢

= ~r2(A.) 2

'1',

or (1.17) Dividing the equations (1.16) and (1.17) yields dt dA = -4,

(1.18)

which says that the t-component is roughly the area swept by the vectorial radius on the x-plane, up to a multiplication factor. The negative sign in (1.18) shows that when t is increasing, the rotation in the x-plane is done clock-wise. The equation (1.18) is valid only if t is not constant. The areal velocity is defined as dA

0=

Ts.

Areal velocity also appears in Kepler's laws: all of the planets have plane trajectories, which are ellipses with the sun in one of the focuses. The areal velocity o is constant along the motion. THEOREM 1.7. A curve c in]R3 is horizontal if and only if the rate of change of the t-component is equal to 40, i.e., i = 40. PROOF.

C = (Xl,

X2, t) is a horizontal curve iff

i = 2(XIX2 which in polar coordinates is

XIX2),


9

and hence

o The characterization of horizontal curves with t constant is given in the following result. PROPOSITION 1.8. A smooth curve c(s) is horizontal with t(s) = t constant if and only if c(s) = (as, bs, t), with a, bE IR, a2 + b2 =I O. PROOF. If c(s) is horizontal with t constant, the equation i = -2r2¢ yields ¢ =constant. Hence the projection on the x-space is a line which passes through

the origin. If c(s) = (as,bs,t), with t constant, then

2(X1X2 - X1X2) = 2(abs - abs) = 0 =

i

o

and hence the horizontality condition (1.14) holds.

The following proposition shows that the left translation of a horizontal curve is a horizontal curve. PROPOSITION 1.9. If c(s) is a horizontal curve, then c(s) = Lac(s) is a horizontal curve, for any a E H l .

(1.19)

a2 a3

C2 C3

+ C2 ==} C2 = C2 + C3 - 2(a1c2 - a2ct) ==} Z:3

= C3 -

2(a1c2 - a2c1).

Using that c is horizontal, equation (1.14) yields C3

C3 - 2(a1c2 - a2ct) 2(C1C2 - C1C2) - 2(a1c2 - a2c1)

2(c1(a2

+ C2) -

c2(a1

+ C1))

2 (Z:1 C2 - Z:2(1)'

o

From equation (1.14) it follows that c(s) is horizontal. COROLLARY

(1.20) PROOF.

1.10. The velocity of the horizontal curve c(s)

Z:(s)

=

(LaLC(S) = C1(S)X11c(s)

= Lac(s) is

+ C2(S)X2Ic (s)'

As c(s) is horizontal, from equations (1.15) and (1.19) we have Z:( s)

+ Z:2 (s )X2Ic(s) C1 (s )X11c(s) + C2 (s )X2Ic(s) c1(s)(LaLX11c(s) + c2(s)(La)*X2Ic(s) (LaL (C1(S)X11c(s) + C2(S)X2Ic(s))

Z:1 (s)X 1Ic(s)

(LaLc(s).

o


10

1.3. Horizontal connectivity theorem On the Heisenberg group H 1 , the vector fields Xl, X 2 and [X 1 ,X2 ] generate the tangent space of]R3 at every point. Such a subRiemannian manifold is called step 2. In the case of a general sub Riemannian manifold, the number of brackets needed to generate all directions +1 is called the step of the manifold. The higher the step, the more noncommutative and the harder it will be to study the geometry. The step 1 corresponds to Riemannain geometry, which is the commutative case. The step condition was used independently by Chow [22] and Hormander [33] to study connectivity and hypoellipticity, respectively. Using Hormander's theorem, the Heisenberg Laplacian !::.H = ~(Xr + xi) is hypoelliptic, i.e., !::.H U = f E

Coo

==} U

E

Coo.

In the following we shall prove Chow's connectivity theorem (see [22]) in the particular case of the Heisenberg group H 1 .

1.11. Any two points in Hl can be joined by a piece-wise horizontal curve, i. e., a curve tangent to the horizontal distribution. PROPOSITION

PROOF. Let P and Q be two points in ]R3. Let tp and tQ be the t-coordinates of P and Q. We distinguish between the following two cases: case (i): tp -=I- tQ Consider the number 0: = tp - tQ -=I- o. Let P1 and Q1 be the projections on the x-plane of the points P and Q. Consider in the x-plane a curve : [0,1] - t ]R2 which joins P1 and Q1, such that the area situated between the graph of and the segments OP1 and OQ1 is equal to 0:/4. The area will be considered positive in the case of a counter clock-wise rotation of the curve between P 1 and Q1, see Figure 1.2. If (s) = (xds), X2(S)), then consider the function

(1.21 )

We claim that ¢ : [0,1] ----t ]R3 defined as ¢(s) = ((s), t(s)) is a horizontal curve joining P and Q. Differentiating in (1.21) we obtain the horizontality condition i(s) = 2(X2(S)X1(S) - X1(S)X2(S)) and hence, ¢(s) is a horizontal curve. We shall check that ¢ joins the points P and Q.

¢(O) = ((0), t(O)) = (x(P1 ), tp) = P. Using t(O) = tp, integrating between 0 and 1 in equation (1.18) yields

t(l)

t(O) - 4(A(I) - A(O)) t(O) - 0: = tp - (tp - tQ) tQ,

and hence, ¢(1) = ((1), t(I)) = (X(Q1), tQ) = Q.

1.3. HORIZONTAL CONNECTIVITY THEOREM

11

O~~______~_________ X~2_

P,

Figure 1.2: The projection of a horizontal curve.

case (ii): tp = tQ = t Let R = (0,0, t). From Proposition 1.8 the curves Ci : [0,1] CI(S)

-+

]R3

= (SXI(P),SX2(P),t),

C2(S) = (SXI(Q), SX2(Q), t) are horizontal and join the points R with P and R with Q, respectively. Then the piece-wise defined curve

¢( s) = { is horizontal with ¢(O)

CI (1

- 2s), c2(2s -1),

~:::; S :::; ~ 2":::; S :::; 1

o

= cI(l) = P and ¢(1) = c2(1) = Q.

The piece-wise condition in the above proposition can be dropped. The proof can be modified such that any two given points can be connected by a horizontal smooth curve. In order to do that, we should take advantage of the group law. Given the points P(XI,YI,td and Q(X2,Y2,t2), a translation to the left by (-Xl, -YI, -td will transform them into 0(0,0,0) and S(x', y', t'), with x'

= X2 - Xl,

Y' = Y2 - YI,

t' = t2 -

tl -

2(YIX2 - XIY2).

If t' -I 0, applying case (i) of the previous proposition, we get a smooth horizontal curve c(s) joining 0 and S. If t' = 0, then c( s) = (sx', sy', 0) is horizontal and joins 0 and S, see Proposition 1.8. Translating to the left by (Xl, YI, h), the points 0 and S are sent into P and Q, respectively. Applying Proposition 1.9, the curve c(s) = (CI(S),C2(S),C3(S)) is sent into the horizontal smooth curve between P and Q

(1.22) (Xl, YI, td 0 c( S)

=

(Xl

+ cd S), YI + C2( S), tl + C3( S) -

2(XI C2 (s) -

YI CI (s)).

12

1.


COROLLARY 1.12. By a left translation, the t-axis c( s)

=

(0,0, s) is transformed

into

o

PROOF. It is an obvious consequence of equation (1.22).

In the following we shall find an explicit smooth horizontal curve between the origin and a given point P( Xl> X2, t).

= (X1(S),X2(S), t(s)),

LEMMA 1.13. For any a,b,a,(3 E lR, the curve c(s)

as + bs 2 as 2 + (3s3

Xl (s ) X2(S)

2 3 2 5 --aas - a(3s 4 - -b(3s 3 5 is horizontal and passes through the origin. t(s)

PROOF. A computation shows that

X1X2 - X1X2

(a

+ 2bs)(as 2 + (3s3) 2

=

3

- (as

+ bs 2)(2aa + 3(3s2)

1.

4

-aas - 2a(3s - b(3s =-t 2 and hence the curve is horizontal. For s = 0 the curve passes through the origin.

0

We shall show that given Xl> X2, t, we can chose the constants a, b, a, (3 such that the curve given by Lemma 1.13 passes through the point P(XI, X2, t) for s = 1. This means that a, b, a and (3 are solutions of the following system

a+b a+(3

2

2

--aa - a(3 - -b(3. 3 5

t

By computation 15t

= -lOa( a + (3) - 5a(3 - 6b(3 -lOax2 - (5a + 6b)(3 = -10ax2 - (5Xl + b)(3 -lOaa - 15a(3 - 6b(3

-10(xl - b)X2 - 5Xl(3 - b(3 = -10xlx2

+ lObx2

- (5Xl

+ b)(3.

Hence 15t + lOxlx2 - 10bx2

Choose b such that 5Xl

+ b -I- O. (3

=

+ (5Xl + b)(3 =

O.

Then

-15t - lOxlx2 + lObx 2 ' 5Xl + b

and

a = X2 - (3,

a = Xl - b.

Hence we have constructed a family of horizontal curves between the origin and P, which depends on the parameter b.

1.4. HAMILTONIAN FORMALISM ON THE HEISENBERG GROUP

13

1.4. Hamiltonian formalism on the Heisenberg group The Heisenberg group is a good environment to apply the Hamiltonian formalism. Consider the Hamiltonian H : T*IR(x,t) ---+ IR given by (1.23) which is the principal symbol of the Heisenberg Laplacian (1.24)

ax!

+ 2X20t, X 2 = OX2 - 2X10t. In Quantum Mechanics, the procedure where Xl = of obtaining the operator (1.24) from the Hamiltonian (1.23) is called quantization. It is natural to consider the Hamiltonian system

(1.25)

X

oH/o~

i

oH/o(} -oH/ox -oH/ot.

~ iJ The solutions c(s)

=

(x(s),t(s),~(s),(}(s)) of the system (1.25) are called

bicharacteristics. DEFINITION 1.14. Given two points P(xo, to), Q(xl,td E 1R3 , a geodesic between P and Q is the projection on the (x, t)-space of a bicharacteristic c : [0, T] ---+ 1R 3 , which satisfies the boundary conditions:

(x(O), t(O)) = (xo, to),

(1.26)

The most basic questions we shall answer in this chapter are: • Given any two points, can we join them by a geodesic? • How many geodesics are between any two given points? We shall start by proving the following result. PROPOSITION

1.15. Any geodesic is a horizontal curve.

PROOF. Let c( s) system (1.25)

=

(Xl (s),

X2 (s), t( s)) be a geodesic. From the Hamiltonian

and then

i

oH o(} 2X2(6

+ 2X2(}) -

2Xl(~2 - 2Xl(})

2X2Xl - 2X1X2, which is the horizontality condition (1.14). Hence, any geodesic is a horizontal curve. 0

14


Solving the Hamiltonian system. We shall solve the Hamiltonian system explicitly. We start with the observation that H does not depend on t. Then . 8H ()=--=O 8t and hence, () = constant. The equations

.

8H

= 86'

Xl

become (1.27) Differentiating, yields

Xl = ~l

(1.28) Using ~

X2 = ~2

+ 2X2(),

2Xl().

-

= -8Hj8x and the system (1.27) we obtain

(1.29) (1.30) From systems (1.28) and (1.29)

Xl = 4()X2,

(1.31)

X2 = -4Bxl

with constant B. The system (1.31) can be written as

X(s) = 4()Jx(s) ,

(1.32) where

(~1 ~)

J=

and X = (Xl Xl)' The equation (1.32) describes the projection of the geodesic on the x-space. We shall show that this is a circle. With the substitution y(s)

(1.33)

= x(s) equation (1.32) becomes y(s) = 4()Jy(s),

with the solution

y(s) = e40JS y(0). Therefore x(s) = e40J8 y(0). Integrating and using that J and e40Js commute, yields x(s)

1

e40Ju y(0) du

x(O)

+

x(O)

+ ~J-l e40J8y(0)lu=s

8

4()

1 x(O) - _Je 40JS y(0) 4()

(1.34) where K

u=o 1 + -.J-ly(O) 4()

e40Js K

= -Jy(0)j(4B) and C

+ C, = x(O) + K.

We need the following result.

1.4.

HAMILTONIAN FORMALISM ON THE HEISENBERG GROUP

(~1 ~)

LEMMA 1.16. If J =

15

then e4IJJs = R 41Js, where Rex denotes the

rotation by angle a in the x-plane.

PROOF. A computation yields 00

e

4IJJs

L n=O

(4Bs)n

00

L

- I

r

n!

00

=

L k=O

I

(4Bs)4k+2

k=O

(4k

+ 2)!

(4Bs )4k

00

L

- J

k=O

(4k

(4IJS)4k (4IJs)4k+2 (4k)f' - (4k+2)! ~ _ (4IJs)4k+1 (4IJs)4k+3 k=O (4k+l)! + (4k+3)! cos(4Bs)

+ 3)!

(4IJS)4k+1 _ (4IJS)4k+3) (4k+l)! (4k+3)! (4IJs)4k _ (4IJs)4k+2 (4k)! (4k+2)!

00

- sin( 4Bs)

(4Bs )4k+l

L (4k + I)! k=O

(4Bs)4k+3

"

(

00

+J

(4k)!

Sin(4Bs)) cos( 4Bs)

R 41Js '

o Using Lemma 1.4, the equation (1.34) becomes (1.35)

x(s) = R4IJsK

As Ix(s) - CI = IR4IJsKI C with the radius

= IKI = constant,

+ C.

x(s) will describe a circle centered at

IKI = I-JY(O) I = ly(O)1 = Ix(O))1 4B 41BI 41BI'

(1.36)

PROPOSITION 1.17. Consider a geodesic which joins the points P(xo, to) and Q(Xl' td, with to :f. t l · (i) The projection of the geodesic on the x-space is a circle or a piece of a circle with end points Xo and Xl. (ii) If the projection is one complete circle, with Xo = Xl, let (J' be its area. Then (J' =

Ih -4 tol .

PROOF. (i) comes from the solution of the Hamiltonian system discussed above. (ii) By Proposition 1.15, any geodesic is horizontal. From equation (1.18), the area of the projection on the x-plane and the t-component of a horizontal curve are related by 4dA

=

-dt.

Integrating, yields 4(J'

11'

4 dA = -

11'

dt

to - h, where x(r)

o

= Xl, t(r) = h.

COROLLARY 1.18. The radius of the projection circle is R

= jl\--",tol .

16

1.


If we let () ----7 0 in (1.36), the radius of the circle WI ----7 00. This corresponds to a circle with infinite radius, which is a line. This case is covered by the next result. PROPOSITION 1.19. If () a line.

PROOF. If ()

=

0, the projection of the geodesic onto the x-space is

= 0 the equation (1.32) yields x = 0, which corresponds to a 0

li~.

Proposition 1.15 claims that geodesics are horizontal curves. We shall show that the converse is false. PROPOSITION 1.20. There are horizontal curves which are not geodesics.

PROOF. Consider c(s) = (s2/2,s,s3/3). The curve is horizontal, because the horizontality condition holds

i(s) =

S2

=

2(s2 - s2/2) = 2(:hx2 - X1 X2)'

On the other hand, the system (1.31) becomes 4() to a contradiction. The t-component Using the Hamiltonian equation

i(s)

i = 8H/8(),

=

1 and 0

= -4()s,

and (1.36) yields

2(X2(S)X1(S) - X1(S)X2(S)) 2(x(s), J(x(s))) 2(e 40Js K + C, J(4()Je 40Js K)) 2(e40Js K, _4()e 40Js K) + 2(C, _4()e 40Js K) -8()IKI 2 - 8()(C, e40Js K).

Integrating we obtain

t(s)

=

J(

-8()IKI2 - 8()(C, e40Js

As

pc,

:s

e40Js K) =

4()pT JC, e40Js K)

K)) ds. pC, 4()Je 40Js K) =

4()(C, e40Js K)

then

Hence

t(s) where C 1

= -8()IKI 2s -

= t(O) + 2PC, K).

2PC, e40Js K)

+ C1

which leads 0

1.4. HAMILTONIAN FORMALISM ON THE HEISENBERG GROUP

17

HXI

The conservation of energy. Let E = + x~) be the kinetic energy. One may show that the Hamiltonian is equal to E along the geodesics, and hence E is a first integral for the Hamiltonian system. In the following proposition we shall give a direct proof. PROPOSITION

PROOF.

1.21. The kinetic energy is preserved along the geodesics.

Using equation (1.32) dE

d s

-d

ds

xi + x~ 2

=

...

..

X1 X1 + X2 X2

(x, x) = 4(} (x,.J x) = O.

o In the following we shall show that the projection of the solution is a circle, using the conservation of energy. Let 2E = R2 with R > 0 constant. For conservative systems with X1(s)2 + X2(S)2 = R2, there is a function a = a(s) such that

(1.37)

X1(S) = Rcosa(s),

X2(S) = Rsina(s).

Then

X1(S)

=

-Rsina(s) a(s),

and the system (1.31)

Xl X2

4(}X2 -4(}X1

becomes sina(s)(4(} + a)

0

cosa(s)(4(} + a)

O.

Adding the squares, yields a( s)

=

4(}. Integrating, we obtain

a(s)

=

-4(}s - ao.

Hence equations (1.37) can be integrated

Xl(S)

R18eos(-4(}s-ao)ds

~ sin(4(}s + ao) + Xl(O). X2(S)

R18sin(-4(}s-ao)ds

R

4(} cos(4(}s + ao) where (0) = _ Rsinao 4(}' We obtain the solution

Xl

x(s)

+ X2(0),

_ Rcos ao X2 (0) - 4(} .

~( sin(4(}s + ao), cos(4(}s + a o)) + x(O),

which is the parametric equation of a circle of radius ~ centered at x(O).

18

1.

GEOMETRIC MECHANICS ON THE HEISENBERG GROUP 1.5. The connection form

Let x ----> 1t x = spanx {Xl, X 2} be the horizontal distribution on JR.3. A connection I-form is a non-vanishing form w E T*JR.3 such that kerxw = 'Hx. The form w is unique up to a multiplicative factor. In this chapter we shall choose the standard I-form with the property w(ot} = 1, which is

w = dt - 2(X2dx1 - X1dx2).

(1.38)

DEFINITION 1.22. The curvature 2-form of the distribution 'H is defined as

n : 1t x 'H ----> F(R 3 ) n(U, V) = dw(U, V).

(1.39)

In our case n = 4dx1 1\ dX2. If the horizontal distribution 'H belongs to the intrinsic subRiemannian geometry, the form n describes the extrinsic geometry of the Heisenberg group. In general, the 2-form n describes the non-integrability of the horizontal distribution. DEFINITION 1.23. The pair (JR. 3,w) is called a contact manifold if w 1\ vanishes.

n never

In our case w 1\ n = 4dt 1\ dX1 1\ dX2 and hence, the Heisenberg group becomes a contact manifold. The following theorem shows that, locally, all contact manifolds are the same as the Heisenberg group, see Cart an [21]. THEOREM 1.24. (Darboux) Each point p of a contact manifold admits a local coordinate system t, Xl, X2 in a neighborhood U such that w = dt- 2(X2dxl -X1dx2).

1.5.0.1. The osculator plane. Let c( s) = (Xl (s), X2 (s), t( s)) be a curve. The osculator plane at c( s) is defined by span {c( s), c( s) }. PROPOSITION 1.25. Let c(s) be a curve. Then the curve c(s) is horizontal if and only if the osculator plane at c( s) is the horizontal plane 'He( s), for any s. PROOF. If span{c(s),c(s)} = 1te(s) , then c(s) E 1te(s). Hence, the curve is horizontal. If the curve c( s) is horizontal, c( s) E 'Hc( s)' It suffices to show c( s) E 'He( s)' From the horizontality condition (1.14), (1.40) Differentiating in equation (1.40) yields

l

=

2X2X1

+ 2X2Xl -

2X1X2 - 2X1X2

(1.41 ) The acceleration vector along c( s) is C

X10x! + X2oX2 + lOt Xl (OX! + 2X2ot) - 2X2X1ot

+ 2X1X2ot + lOt + (t" - 2X2X1 + 2X2XI)at

+X2(oX2 - 2X10t) X1X1 X1X1

+ X2X2 + X2 X 2,

since we used equation (1.41). Hence, c E 'He and the osculator plane is horizontal.

o

1.5.

THE CONNECTION FORM

19

COROLLARY 1.26. For a horizontal curve c (1.42) (1.43)

DEFINITION 1.27. Let J : H ----> H be defined by J(Xr) = -X2' J(X 2) = Xl. J is called the complex structure of the horizontal plane. We shall use J in order to write the equations for the geodesics on the Heisenberg group. The following result shows that the geodesics satisfy a Newton type equation. The left side is the acceleration, while the right side is the force, which keeps the distribution bent. As before, 0 is a constant. PROPOSITION 1.28. A curve c is a geodesic on the Heisenberg group if and only if (i) c is a horizontal curve and (ii) c satisfies

c = 40Jc.

(1.44)

PROOF. If c(s) is a geodesic, by Proposition 1.15, c(s) is horizontal. Using Corollary 1.26 and the system (1.31), C

X1X1 + X2X2 40X2Xl - 40X I X 2

40X2J(X2) + 40x 1 J(X 1 ) 40J(X 1 X 1 + X2 X 2) 40J(c). We shall now prove the converse: if (i) and (ii) hold, then c is a geodesic. We shall use the definition 1.14. The horizontality condition (i) can be written as i = 8H/OO, which is the Hamiltonian equation for t. Using a similar computation as in the first part, equation (1.44) written in components becomes the system (1.31). Let Xl(S) and X2 (s) be solutions of this system. Define the following curve in the cotangent space

"((s)

= (XI(S),X2(S),t(s),6(s),6(s),O),

where ~l

= Xl - 2X2(S)e,

6 = X2 + 2x10,

with 0 constant. Then "((s) satisfies the bicharacteristics system (1.25) for the Hamiltonian (1.23). Then the projection on the (x, t)-space is a geodesic and hence c( s) is a geodesic. 0

20


The subRiemannian metric. DEFINITION 1.29. A non-degenerate, positive definite bilinear form gx : 1ix x 1ix ......, F(JR.) at any point x E JR.3, is called a subRiemannian metric.

We shall consider the subRiemannian metric in which the vector fields Xl, X 2 are orthonormal. The subRiemannian metric will become a Kaehler metric on the horizontal distribution, as we shall explain later. The following definitions can be found, for instance, in Kobayashy and Nomizu, see [37]. DEFINITION 1.30. A Hermitian metric on a real vector space V with a complex structure J is a non-degenerate, positive definite inner product h such that

h(JX, JY) = heX, Y),

for X, Y E V.

We associate with each Hermitian metric of a vector space V a skew-symmetric bilinear form on V. DEFINITION

1.31. The fundamental 2-form is defined by

(X, Y) = heX, JY),

for all vector fields X and Y.

A Hermitian metric on a vector space V with a complex structure J is called a Kaehler metric if its fundamental 2-form is closed. The relationship with the subRiemannian metric is given in the following proposition. PROPOSITION 1.32. The subRiemannian metric 9 in which {X 1 ,X2 } are orthonormal is a Kaehler metric on 1i x , for any x E JR.3. The fundamental 2-form satisfies 4 = O. Hence

O(U, V) = 4g(U, JV),

for all horizontal vectors U and V.

PROOF. Consider V = 1t x . We shall show first that 9 is a Hermitian metric. Let U = u1Xl + U 2X 2 and V = VI Xl + V 2X 2 be two horizontal vector fields. Using JX 1 = -X2 and JX 2 = Xl, yields

JU = _U l X 2 + u2Xl,

Using the orthonormality of Xl and X 2 we obtain

g(JU, JV)

g(U 2Xl - U l X 2, V2 Xl - V1 X 2) U 1V1 + U 2V2 g(U l Xl + u2X 2, VI Xl + V2 X 2) g(U, V),

1.5. THE CONNECTION FORM

21

and hence 9 is invariant by J. The 2-form 0 is closed because it is exact 0 = dw.

O(u l Xl + U 2 X 2, VI Xl + V2 X 2) (U l V2 - U 2V l )O(X l ,X2) (U l V2 - U 2Vl)(Xdw(X2)) - X 2(w(Xd) - W([Xl' X 2])) 4(U l V2 - U 2V l )w(Ot) 4(U 1 V2 _ U2Vl)

O(U, V)

4g(U l Xl + U 2X 2, V 2X 1 4g(U, JV).

VI X 2)

-

o Using the skew-symmetry of 0 we obtain: COROLLARY

1.33.

g(U, JU) = 0,

for any horizontal vector U.

The geometrical interpretation of 0 is given below. PROPOSITION 1.34. Let 7r : JR.(x,t) horizontal plane onto the x-plane

->

JR.;

be the projection which sends the

Then, for any horizontal vectors U and V, the area of the parallelogram generated by 7r*(U) and 7r*(V) is equal to IO(U, V)I/4. PROOF.

We have

O(U, V)

=

(

)

4 dXl /\ dX2 (U, V)

41 U(xd U(X2)

dXl (U)

dXl (V)

= 4 dX2(U) dX2(V) 1

V(Xl) 1= 41 U 1 VI 1 V(X2) U2 V 2 .

Using the interpretation of the determinant as an area, the proof is complete. PROPOSITION

1.35. Let c(s) be a geodesic curve. Then

eo(u, c) = g(U, c) PROOF.

0

for any horizontal vector U.

Using the Kaehler property of the metric 9 and the geodesics equation

(1.44)

eo(u, c) = 4eg(U, Jc) = g(U, c).

o

°

If e = 0, then g(U, c) = for any horizontal vector U and then c = 0. Hence Cl(S) = C2(S) = 0. We obtain the following result: COROLLARY

1.36. Let c(s) be a geodesic for which the momentum

Then

c(s) = (as with a, ao, b, bo and to constants.

+ ao, bs + bo, to),

e vanishes.

22

1.


PROOF. The fact that Cl(S) = as+ao and C2(S) show that t(s) is constant. We have

t(s)

= bs+bo is obvious.

We shall

2(X2Xl -X1X2) 2(C2 Cl - C1(2)

2((bs + bo)a - (as

+ ao)b)

0,

o

i. e., t is constant.

The following proposition deals with the metric properties of the velocity and acceleration. PROPOSITION 1.37. Let c(s) be a geodesic. Then (i) The velocity c and the acceleration c vector fields are perpendicular in the subRiemannian metric. (ii) The magnitude of c in the subRiemannian metric is constant along the geodesic. (iii) The magnitude of cin the subRiemannian metric is constant along the geodesic. PROOF. (i) For U = C, Proposition 1.35 yields

g(c, c) = 8n(c, c) = o. (ii) Differentiating and using (i) yields :sg(c, c) = g(c, c)

+ g(c, c) = 0,

i.e. g(c(s), c(s)) is constant. (iii) The vector field c is horizontal, and using equation (1.44) and (ii) we have g(c, c) = g(48Jc, 48Jc) = 1682g(JC, JC) = 168 2g(c, c) = constant.

o In the classical theory of three dimensional curves, the curvature becomes along a curve c(s) is a function defined by k(s) = 11'(s)l, where T(s) = c(s)/lc(s)1 is the unit tangent vector. If s is the arc length parameter, Ic(s)1 = 1 and the curvature becomes k(s) = Ic(s)l. PROPOSITION 1.38. The curvature of a geodesic curve is constant, k( s)

= 4181.

PROOF. Consider the geodesic c( s) parametrized by the arc length. Then

k(S)2 = Icl 2 = g(c, c) = 168 2g(c, c) = 1682 .

o The curvature of a geodesic depends on the momentum 8. As we shall show later, 8 is a Lagrange multiplier which describes the number of rotations of the geodesics around the t-axis. Hence, we shall be able to prove a Gauss-Bonett type theorem for the geodesic curves. The following proposition provides the complex structure of the horizontal distribution in function of n and the basic horizontal vector fields.

1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP PROPOSITION

23

1.39. For any horizontal vector field U, we have

J(U) =

~ (D(Xl , U)Xl + D(X2' U)X2).

As both sides are linear, it suffices to check the relation only for the basic vector fields Xl and X 2. Using D(Xl' X 2) = -D(X2' Xl) = 4, and D(Xi' Xi) = 0, yields PROOF.

JXl = -X2 = JX2 = Xl =

~D(X2,XdX2 = ~(D(Xl,XdXl +D(X2,XdX2)

~D(Xl,X2)Xl = ~(D(Xl,X2)X1 +D(X2 ,X2)X2).

o 1.6. Lagrangian formalism on the Heisenberg group In this section we shall find the subRiemannian geodesics and characterize their lengths. The horizontality condition is a constraint on velocities, i.e., they are nonholonomic. The Lagrangian which describes the geodesics has a non-holonomic constraint. This constraint can be expressed using the I-form w. PROPOSITION

1.40. If ¢(s) is a horizontal curve, then

l

(1.45)

PROOF.

As ¢*w is a I-form on

JR,

w=O.

then ¢*w and ds are proportional

¢*w(s)

h(s) ds,

=

where the proportionality function h(s) is

h(s) = ¢(S)w(:J. Let ¢( s) be defined on [0, 1]. Then

l

w =

11

¢*w =

11

h(s) ds =

11 (:J) 11 w(¢*

because

ds =

11

¢(s)(w)

(:J

ds

w(¢(s)) = 0,

¢ E ?t.

0

We shall associate a Lagrangian L ; TJR 3 -+ JR with the Hamiltonian (1.23). This will be done using a formal Legendre transform in (x, i), see [14]. In the case of a quadratic Hamiltonian, the Lagrangian is given by the maximal distance between the hyperplane < ~,x > +ei and the convex surface given by the Hamiltonian in JR6;

L(x, t, x, i)

max(6xl UJ

+ ~2X2 + ei - H(~, e, x, t))

maxF(x, t, x, i,~, e). fjJ

24

1.


In our case the Hamiltonian is degenerate, so that the equivalence between the Lagrangian and the Hamiltonian formalism of Classical Mechanics might not hold. We shall still proceed formally, and set 8F 8~

(1.46)

8F 8()

= 0,

=0

as in Classical Mechanics, where the partial derivatives vanish when the maximum is reached. The equations (1.46) can be written as

.

(1.48)

8H

.

Xi = 8~i'

(1.47) Xl =

6 + 2X2(),

X2 =

8H

t =

6-

7iB =

0,

i=

2XI(),

2X2XI - 2XIX2.

Using relations (1.48), we formally define the Lagrangian by

. + 6X2 . + ()t. - "21(6 + 2X2() )2 -"21(6 -

L(x, t, X, i)

6XI

(Xl - 2X2())X1

2X1()

)2

+ (X2 + 2XI())X2 + ()i - ~(xi + X~)

.2 + x.2) ()(i (Xl 2 +

-

2X2 X. I

~(xi + X~) + ()(i -

.) -"21(.2 .2) + 2xlx2 Xl + x2

2X2XI

+ 2XIX2).

Using the I-connection form w, we write the Lagrangian as L(e, e)

(1.49)

= ~(xi + x~) + ()w(e),

where e = (Xl, X2, t). Let the action integral be (1.50)

S(e, T)

=

r

Jo

L(e, e) ds

=

r ~(xi + x~) 2

Jo

ds

+ ()

jw. c

The action has two parts: the kinetic energy and the non-holonomic constraint. In this case the constant () is a Lagrange multiplier. The curves e, which are critical points for the action S(e, T), satisfy the Euler-Lagrange equation

d (8L)

(1.51 )

8L

ds 8e = 8e· The partial equivalence between the Lagrangian and the Hamiltonian formalism is given in the following result.

PROPOSITION 1.41. A solution of the Euler-Lagrange equation (1.51) is a geodesic if and only if it is a horizontal curve. PROOF. As any geodesic is a horizontal curve, it suffices to show that a horizontal solution of (1.51) is a geodesic. If e = (Xl, X2, t), the equation (1.51) becomes on components Xl

= 4()X2,

X2

=

-4()X1,

Proposition 1.28 completes the proof.

B= 0 ==} () = constant.

o

1.6.

LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP

25

REMARK 1.42. The Euler-Lagrange equation (1.51) does not imply horizontality of solutions, which corresponds to the Hamiltonian equation i = 8H/8B. Hence the Hamiltonian and the Euler-Lagrange systems are not equivalent. This behavior is specific to subRiemannian geometry and has no analog in the Riemannian case.

Lagrangian symmetries. The symmetry of the Lagrangian influences the symmetry of the solutions of the Euler-Lagrange system and hence the geodesics. The study of the Lagrangian symmetries makes easier the understanding of geodesic behavior.

PROPOSITION 1.43. The Lagrangian (1.49) is left invariant with respect to the Heisenberg Lie group structure, i.e., L(c,~) = L(c, c), where c = La(c), for any a E H 1.

c = (C1' C2, (3) ===} C1 = C1 ===} C2 = C2

PROOF. If c = (C1, C2, C3), C1

a1

C2

a2 a3

C3

+ C1 + C2 + C3 -

2(a1 c2 - a2 c t}

and a

===}

= (a1, a2, a3), then

~3 = C3 - 2(a1c2 - a2 ( 1)

Then the kinetic energy is left invariant 1

2

2

1,2

·2

2"(C1 +(2) = 2"(c1 +(2). The horizontal constraint is also invariant C3 - 2(a1c2 - a2cd - 2(a2 C3 - 2C2C1

+ c2h + 2(a1 + Ct}C2

+ 2C1C2·

o

Hence, the Lagrangian is preserved by left translations.

COROLLARY 1.44. The solutions of the Euler-Lagrange equation (1.51) are left invariant by the translations on H 1 . In the following we shall use Noether's theorem approach to find first integrals of motion for the Lagrangian (1.49)

L(x, t, X, i)

=

~(xi + x~) + B(i -

2X2X1

+ 2X1X2).

The following theorem can be found in Arnold [2]. For its generalizations on manifolds one may consult [14]. THEOREM 1.45. (Noether) To every one-parameter group of diffeomorphisms (hs)s of the coordinate space M of a Lagrangian system which preserves the Lagrangian, corresponds a first integral of the Euler-Lagrange equations of motion J : TM -.IR given by (1.52)

J(

.) = (8L dhs(q)

q,q

>l" uq

dS

I )' 8=0

where (,) denotes the standard inner product on 1R3.

26

1.


In the case of the Heisenberg group, M

~~ = (:~ , :~ , ~~) =

= 1R3, q = (x, t) and

(Xl - 2Bx2, X2

+ 2BxI, B) .

Consider the following three independent one-parameter groups of diffeomorphisms

hs(x, t) = La(s)(x, t) = (al(s)

+ Xl, a2(s) + X2, a3(s) + t -

2(al(s)X2 - a2(s)xJ)),

with a(s) E {(O,O,s),(O,s,O),(s,O,O)}. The associated vector field is

dhs(q)

_ { (0,0,1),

~Is=o-

(0,1,2xI)' (1,0,-2x2)'

for a(s) :: (0,0, s) fora(s)-(O,s,O) for a(s) = (s,O,O).

Hence, formula (1.52) provides the following three functional independent first integrals

h

B = constant,

12

X2 + 4BxI = constant, Xl - 4Bx2 = constant.

h

Differentiating, we obtain the Euler-Lagrange system

Xl = 4Bx2 { X2 = -4BxI B = constant, which is equivalent to (1.51). The rotational symmetry of the Lagrangian will provide a non obvious first integral. The Lagrangian is invariant by the one-parameter group of rotations hs(x, t) = (Rsx, t). Using Rs(x) = eJsx, we have = Je Js and hence, the vector field generated by the rotation is

1f1:

dhs(q) I - dS

8=0

=

(Jx,O) = (X2, -Xl, 0).

The first integral associated to the rotation vector field is the kinetic momentum with respect to the t-axis I

(Xl - 2Bx2)X2 + (X2 + 2BxJ) ( -Xl) (XIX2 - X2XI) - 2Blx1 2.

i = 2XIX2 - 2X2Xl, we get the first 21 = i - 4Blxl 2 = constant.

Using the horizontality condition (1.53)

integral

PROPOSITION 1.46. If c is a geodesic, for any a E HI we have (i) the left translation c = Lac is also a geodesic, (ii) the geodesics c and c have the same length. PROOF. (i) Let c be a geodesic and let c = Lac. From Proposition 1.41 the curve c is horizontal and solves the equation (1.51). From Corollary 1.44 and Proposition 1.9, the curve c is a solution of the Euler-Lagrange equation (1.51) and it is horizontal. Using Proposition 1.41, we find that c is a geodesic. (ii) From Corollary 1.10

1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP

and hence

Icl 2 = I"W.

27

Proposition 1.37, (ii) yields

£(c) =

11 lei

lei

ds =

Ii:I =

=

11 Ii:I

ds = £(c). D

Connectivity by geodesics. Consider a geodesic joining the points P and Q. By a left translation, the point P can be transformed into (0,0,0). By Proposition 1.46, the geodesic is transformed into another geodesic of the same length, which starts at the origin. Hence, it makes sense to study the metric properties and the connectivity only for geodesics starting from the origin.

Because of the Lagrangian rotational symmetry, it is useful to use polar coordinates Xl = r cos ¢, X2 = r sin ¢. The Lagrangian becomes (1.54) The symmetries for the Lagrangian will provide symmetries for the geodesics. The Lagrangian (1.54) is invariant under the symmetry S: (r, ¢, t; ()) --; (r, -¢, -t; -()).

(1.55)

If (r(s), ¢(s), t(s)) is a geodesic corresponding to (), then (r(s), -¢(s), -t(s)) is a geodesic corresponding to -(). Consequently, whatever statement is made for the geodesics joining the origin with the point (r(s), ¢(s), t(s)), when () > 0, it can also be made for the geodesics between the origin and (r(s), -¢(s), -t(s)), when () < O. This allows us to do the analysis only for the case () > 0 and () = o.

1.6.0.2. Euler-Lagrange equations. A computation shows

(1.57)

(1.58)

8L .. 8r = r¢(¢ + 4()),

d 8L --=r ds 81'

(1.56) 8L

-. = r

8¢

2·

¢+ 2()r

8L

2

8¢

,

8L _ () 8t - ,

= 0,

8L

at =0,

and hence r(s), ¢(s) and () satisfy the Euler-Lagrange system (1.59)

f=r¢(¢+4()) { r2(¢ + 2()) = C(constant) () = ()o (constant).

If a geodesic starts at the origin, then r(O) = 0 and hence C = O. The second equation of (1.59) yields ¢ = -2(). With the assumption () > 0, the argument angle ¢ will rotate clock-wise. The Euler-Lagrange system becomes f

(1.60)

{

= -4()2r

¢ = -2() () = ()o (constant).

28

1.

When ()

= 0,


the system (1.60) becomes

(1.61)

~:~

{

() = O.

PROPOSITION 1.47. Given a point P(x, 0), there is a unique geodesic between the origin and P. It is a straight line in the plane {t = O} of length lxi, and it is obtained for () = O. PROOF. In the case () = 0, the Hamiltonian is H(~) = ~~r + ~~~. From the Hamiltonian equation t = 8H / 8() = 0, and hence t( s) is constant. Since t(O) = 0, it follows that t(s) = 0 and the solution belongs to the x-plane. Using the equation f = 0, it follows the solution is a straight line. 0 From now on, unless otherwise stated, we shall assume () > O. From the system (1.60) we can arrive at a first integral of energy. PROPOSITION 1.48. (i) The function J(r, t, ()) = f2 +4()2r2 is a first integral for the system (1.60). (ii) ~J(r,t,()) is equal to the energy of the system, i.e., J = xi +x~. PROOF. (i) Differentiating

:s J(r, t, ()) (ii) In polar coordinates, xi proof.

Let c(s)

=

+ x~ =

= (XI(S),X2(S),t(s))

2f(f

+ 4()2r)

+ r2¢2.

f2

=

O.

Using

¢

-2() completes the

o

be a geodesic. Since c(s) is a horizontal curve,

C = XIXI

+ X2X2

Let g denote the subRiemannian metric in which Xl, X 2 are orthonormal. Then

Ic(s)l; =

g(c(s), c(s)) = xi(s)

+ x~(s).

If s is the arc length along the geodesic, the curve c( s) becomes unit speed

xi(s)

+ x~(s) = 1.

Proposition 1.48 yields (1.62) Separating the variables, we have

dr

-r.;=~;:==;;: =

VI -

4()2r2

±ds '

and integrating, we obtain

1 . 2() arcsm(2()r(s)) = ±s, which yields (1.63)

1

.

r(s) = ±2() sm(2()s),


29

where we consider

+

sign for 2mr ::; 2(}s ::; (2n + 1)7r, and

-

sign for (2n

+ 1)7r ::; 2(}s ::;

(2n

+ 2)7r.

This yields a circle of diameter 1/(2(}) which passes through the origin. LEMMA

1.49. If ct>(0)

= ct>o,

then

2 r(ct»2 = (21(}) sin2(ct> - ct>o),

(1.64)

= 4~2 sin2(ct> - ct>o~ -

t(ct» - t(ct>o)

(1.65)

2(ct> - ct>o).

PROOF. From the second equation ofthe system (1.60) we obtain ct>(s) -ct>(0) = -2(}s. Substituting in (1.63) yields (1.64). One of the Hamiltonian equations yields

. oR. ae = 2(XIX2 -

t =

.

2'

XIX2) = -2r ct>.

Integrating between ct>o and ct> and using equation (1.64) we have t ( ct» - t ( ct>o )

=

- 2

r" r2 (ct>) dct>

J",o

-~l"'-"'o 4(}2

a

= - 2 ( I(}) 2

2

.

r" sin2(ct> - ct>o) dct>

J",o

2 d __I_sin2(ct>-ct>o)-2(ct>-ct>o) sm u u - 4(}2 2 .

o 1.6.0.3. Boundary conditions. We are interested in connecting the origin with a point P(x, t) by a geodesic c : [0, sfl --t JR3. Consider the following boundary conditions:

(1.66) (1.67)

X(O) = 0,

Ilx(sf)11

t(O) = 0,

ct>(0) = ct>o,

t(sf) = t,

= R,

ct>(sf) = ct>f'

Because of the rotational invariance around the t-axis, we may choose ct>o = O. From i = -2r2¢ and ¢ = -2(}, we get i = 4(}r 2 > O. Hence t(s) is increasing and if t(O) = 0, then t(sf) > O. LEMMA

1.50. The following relations among the boundary conditions take place:

=

(1.68)

ct>f

(1.69)

(S in ct>f)2

-2(}Sf, =

4(}2R2,

(1. 70)

I sin(2ct>f) - 2ct>f t = 4(}2 2 '

(1.71)

R2 = -J.L(ct>f) = J.L(2(}sf),

t

where

(1. 72)

J.L(x)

=

x -.-2- -

sm x

cotx.

30


PROOF. Integrating in the second equation of the system (1.60) and using the boundary conditions yields (1.68). Equations (1.64) and (1.65) together with the boundary conditions (1.66)-(1.67) yield equations (1.69) and (1.70). Eliminating 40 2 from equations (1.69) and (1.70) yields (1.71). D

The behavior of the function J.L given by (1. 72) is very important in understanding the subRiemannian geometry of the Heisenberg group. The function J.L was first used by Beals, Gaveau and Greiner in [3]. The graph of J.L for x 2 0 is given in Figure 1.2.

60 50 40 30 20 10 0

~

U

:

10

5

15

x

20

Figure 1.2: The graph of J.L(x) for x 2: O. LEMMA 1.51. J.L is a monotone increasing diffeomorphism of the interval (-1f, 1f) onto R On each interval (m1f, (m + 1)1f), m = 1,2, ... , J.L has a unique critical point x m . On this interval J.L decreases strictly from +00 to J.L(x m ) and then increases strictly from J.L(x m ) to +00. Moreover

(1. 73) 0
1/3, x E (0,1f).

The first identity holds as an application of the l'Hospital rule:

. sin x - x cos x 1Im--~-x-o sin3 x

. cos x - cos x + x sin x 1Im--------~-----x-o 3 sin2 x 1. x 1 - hm -.-- =-. 3 x-o smx 3


31

The second inequality holds because 1 II ( ) -J-L x 2

=

X

°

+ 2x COS 2 X - 3 cos x sin x sin 4 x

> .

The numerator vanishes at x = 0, and its derivative is 4sinx(sinx - xcosx) > 0, x E (0,7r). Therefore J-L is a diffeomorphism of the interval (0,7r) onto (0,00). In the interval (m7r, (m + 1)7r) the function J-L approaches +00 at the endpoints. In order to find the critical points, we set

!J-L'(x) 2

=

sinx - xcosx sin3 x

=

1- xcotx sin4 x

0.

=

Hence the critical point Xm is the solution of the equation x = tan x on the interval (m7r, (m + 1)7r). Note that

x+7r - cot (x + 7r) sm (x+7r) x -.,,---- - cot(x + 7r) sin 2 (x + 7r) . 2

J-L(x)

7r sin x

+-2

7r

+ -'-2-' sm x

so the successive minimum values increase by more than 7r. From Figure 1.3 we have (1. 75)

Using

Xm =

tanx m , yields

cotx m = -

(1. 76)

1

Xm

Let f(x) = cot x. As

f' (x)

= - sin\ x

1 m7r

O.

(>'XI, >'X2, >.2t),

---+

We are looking for a formula for the length of geodesics, different than (1.79), such that sf is homogeneous of degree 1 with respect to the above dilations. This is done in the next result. THEOREM 1.53. Consider the geodesics joining the origin with the point (x, t), x i- O. Let (1, ... ,(N be the solutions of equation (1.77). Then the square of the lengths is given by (1.80)

where

v(x)

=

x2

.

+ sm2 x

x

.

- smxcosx

PROOF. Equation (1.77) yields It I + IIxl1 2 = JL((m)llxI1 2 + IIxl1 2 = (1

+ JL((m))llxI1 2,

and hence IIxl1 2 = 1 +

~((m) (It I + IlxI12).

U sing equation (1. 79) and the definition of JL given in (1. 72)

s;n

Ci~(mfllxI12 = Ci~f(m . 1 + ~((m)) (It I + Ilx112) . 2(

Sin

m

+ (m

~!.

Sin

2(

m

co

t(

m

(It I + Ilx112)

= v((m) (It I + IlxI12).

o

For the graph of v see Figure 1.4.

x

PROPOSITION 1.54. The projection of the geodesics joining the origin and (x, t), Ilxll

i- 0 are circles or arcs of circle with diameters of at least Ilxll > x . I sin(ml - II II

(1.81)

PROOF. From the first equation of Lemma 1.49, the diameter of the circle is 1/2(). Using relation (1.69) yields

1

R

2()

I sin 4>fl'

Replacing R by Ilxll and 4>f by (m, we arrive at relation (1.81).

o

34

1.


10 8 6

4-

2

o

4

2

x

8

6

10

Figure 1.4: The graph of v(x).

1.6.0.4. Geodesics between the origin and points on the t-axis. We have seen that the number of geodesics increases without bound as Itl/llxl1 2 ----> 00. The following theorem deals with the limit case Ilxll = 0. This result was found by Gaveau (see [26J and [3]). THEOREM 1.55. The geodesics that join the origin to a point (0,0, t) have lengths Sl, S2 ..• , where

s~ = m7l'ltl.

(1.82)

For each length

Sm,

the geodesics of that length are parametrized by the circle

PROOF. We shall treat the problem as the limit case Ilxll Then the solutions (m ----> m7l'. Relation (1.80) becomes

s~

---->

§1.

°

of Theorem 1.53.

= v(m7l')ltl,

with the coefficient given by

v(m7l') =

m 27l'2 . = m7l'. 2 m7l' + sm m7l' - smm7l' cosm7l' .

D The above proof is using Theorem 1.53. In the following we shall provide a direct proof. The approach will use polar coordinates and the fact that the solution is given by r(¢) = rmaxl sin(¢ - ¢o)l. Consider the solution "( parameterized by [0,1]' with the end points "((0) = (0,0,0) and "((1) = (0,0, t). Using that hi is constant along the solution, we have the identity in Cauchy's inequality

{I

£("() =

10

{1

li'(s)1 ds

=

(10

1/2

dS)

{I

(10

1/2 li'(sW)

= v2E,


35

where E is the constant value of the energy along the solution

E_~1'12_~('2+ '2) - 2' - 2 Xl x2' 1.6.0.5. Quantization of (). () is constant along the solution and it is an eigenvalue for the following boundary value problem

r

-4()2r

r(O)

0, r(1)

= O.

The solution is r(s) = Asin(2()s). The boundary conditions yield 2()

(1.83)

= m7r,

with m integer. Each value of () will determine a certain length.

= t. One has -2() and then cP(s) = cPo - 2()s, where cPo = cP(O) and cPI = cP(1) = cPo - 2() = cPo - m7r. One may integrate between cPo and cPI = cPo - m7r in the equation 1.6.0.6. The energy and t. Consider the boundary condition t(1)

¢= i

= -2r2¢ .

-2

r~ax fo- m1r sin2 u du = 2 r~ax fo m1r sin2 u du

m1r 2 -u 1 - -1. 2 r max( SIn 2u) I 2 4 0 From the conservation of energy

2 2 = r maxm7r = 2()rmax'

~(r2 + r2¢2) = ~(r2 + 4r2()2) = E. 2

When r = r~ax' then r

2

= O. Hence E

2

rmax

and using t

= 2()r~ax'

2()2'

one obtains

E t= O. The lengths are (£m)2 PROPOSITION

= 2E = 2()t = m7rt,

m

= 1,2,3 ...

1.56. The equations of the geodesics starting at the origin are

JIt I

m7r

sin(cP - cPo)

_It_I (Sin(2(cP 2m7r

cPo)) - 2(cP - cPo)), m;:::: 1.

36

1.


PROOF. Using

rmax = then

rm(.

given by 15>. (x, t) = (..\x, ..\2t).

:]R3 ---->]R3

i) Show that (b>')>'EIR form a group, called the one-parameter group of dilations. ii) Show that (15).)* Xl = ..\-1 Xl, (15).)* X 2 = ..\-1 X 2 , where Xl, X 2 are the Heisenberg vector fields. iii) The length of a horizontal curve is multiplied by 1..\1 under the action of 15>.. iv) Show that dc(b>.p,b>.q) = 1..\ldc(p,q), for any two points p,q E ]R3. v) Prove the following estimation

~ (Ixl + IW/2) :s dc(O, (x, t)) :S 4(lxl + IW/2). EXERCISE 1.2. Define the unit ball centered at p as

B(p, 1) = {(x, t) where dc(x, t) =

E ]R3; dc(x,

t) :S I},

(lxl 4 + t 2)1/2.

i) Draw the unit ball centered at the origin. ii) Find the volume of the unit ball. iii) How is the volume of the unit ball changing under left translations? iv) How is the volume of the unit ball changing under dilations? v) Draw the unit ball centered at a point (xo, to) in the cases a)

Xo = 0,

°

b)

Xo -10.

> such that GI[-E,E] x [-E,E] x [_E2,E2] C B(O,E) c G2[-E,E] x [-E,E] x [_E2,E2].

vi) Find two constants GI , G2

vii) Show that Val(B(p, E)) = E4 Val(B(O, 1)),

where Val denotes the usual volume defined by the Lebesgue measure.

1.8.

EXERCISE 1.3. (see [20]) Let

o is in the interior of I and c(O) =

EXERCISES

C = (CI' C2, C3) :

43

I

----+

HI be a Cl-curve, such that

g. Show that if c~(O)

such that

B(go, E) C

U

i- 0, then there exists E > 0

Hc(s),

sEl

where B(go, E) = {g E H : dc(gC;lg) ::; E}, and Hp denotes the horizontal plane through the point p. EXERCISE 1.4. Find the expression for the exponential map of the Lie group HI as subgroup of GL(3,JR).

EXERCISE 1.5. Show that on every vertical line in JR3 the Carnot-Caratheodory metric is locally equivalent to JEuclidean metric.

CHAPTER 2

Geometric Analysis of Step 4 Case In this chapter we shall study the behavior of subRiemannian geodesics on a certain step 4 subRiemannian manifold defined on JR.3 = JR.; x JR. t by the vector fields

(2.1) where

Xl

= OXl + 4X21x120t,

Ixl 2 = xI + x~.

X 2 = OX2 - 4X11x12ot,

Since [Xl, X 2 ]

= -16IxI 2 0t,

[Xl, [Xl, X 2]] = -32x10t, [Xl, [Xl, [X 1,X2]]]

= -32ot ,

it follows that the hypoelliptic operator ~x = ~ (Xl + xD is step 4 along the t-axis and step 2 elsewhere. The geometry of this subRiemannian model can be found in Calin's Ph.D. thesis, see [12]. We shall compute the length of the subRiemannian geodesics between the origin and any point on the t-axis. We can see from the calculation that the t-axis is the conjugate locus of this geometry. We shall also characterize the number of sub Riemannian geodesics between the origin and any other point. Finally, we shall state the connectivity and completeness theorems for this model, see [17]. The study of the above step 4 model requires the use of elliptic functions. The Heisenberg group needs only elementary trigonometric functions, while steps greater than 4 require hypergeometric functions. We shall provide in the following the definitions of the elliptic functions used in the next sections. For a detailed description the reader may consult Lawden [42]. 2.1. Elliptic functions The integral

(W

Z =

dt J(l _ t 2 )(1 _ k 2 t 2 )'

Jo

Ikl < 1

is called an elliptic integral of the first kind. The integral exists if w is real and Iwl < 1. Using the substitution t = sin lJ and w = sin ¢, yields Z

= fa \11 _

:~ sin lJ 2

If k = 0, then z = sin -1 w or W = sin z. Similarly, the above integral is denoted by sn- 1 (w; k), where k =f. O. The number k is called the modulus. Thus

z = sn- 1 w = (OW

In

The function w

dt

J(l - t 2 )(1 - k 2 t 2 )

= sn z is called a Jacobian elliptic function. 45

2. GEOMETRIC ANALYSIS OF STEP 4 CASE

46

Similar to the trigonometric functions, it is convenient to define other elliptic functions

=

cnz

V1 -

sn 2 z,

V1 -

dn z =

k2 sn 2 z.

A few properties of these functions are sn(O) = 0,

cn(O) = 1,

sn( -z) = sn(z),

dn(O)

= 1,

cn( -z) = cn(z),

d d d dz snz = cnzdn z, dz cn z = -sn z dn z, dz dn z = -k 2sn z cn z,

-1 :s; cn z :s; 1,

-1:S; sn z :s; 1,

O:S; dn z :s; 1. , \, ,

"\

.;

J\

.\

\

\ I.

\ \ ,I 6"

sn(z,k)

! I

cn(z,k)

B

10

12

dn(z,k)

Figure 2.1: The graphs of functions sn(z, k), cn(z, k) and dn(z, k) for k = 0.3 and 0.7

Let K = K(k) =

rl

dt

Jo V(1 - t2)(1 -

k 2t 2)

= ('/2

Jo

d()

-v7'i=-=k2=s=in=;:;2=()

be the complete Jacobi integral. Then, as real functions, the elliptic functions sn and cn are periodic functions of principal period 4K. Unlike in the step 2 case, where the vector fields Xl and X 2 are left invariant with respect to the Heisenberg group law, in this case we have: PROPOSITION 2.1. There is no Lie group on and X 2 given by (2.1).

]R3

underlying the vector fields

Xl

PROOF. Assume there is a Lie group law on ]R3 such that Xl and X 2 are left invariant. Then [X I, X 2 ] = -161x1 2at is also left invariant. The vector fields [Xl, X 2 ] and are left invariant and proportional. A result of Lie group theory says that the proportionality function has to be a constant, i.e., Ixl 2 =constant, contradiction. 0

at

This will make the study of the step 4 model more difficult than the Heisenberg group, where any point could be translated to the origin.

2.2.

THE HORIZONTAL DISTRIBUTION

47

2.2. The horizontal distribution Like in the Heisenberg group case, the horizontal distribution is defined by

H :x

--*

Hx = span{XI' Xd,

where Xl, X 2 are the vector fields given by (2.1). PROPOSITION

2.2. A curve c = (Xl, X2, t) is horizontal if and only if

(2.2)

t = 41x12(XIX2 - XIX2).

The velocity vector can be written as

PROOF.

C

+ X20X2 + tat Xl (OXl + 4X21x120t) -

4XIX21xI20t

+X2(OX2 - 4Xllxl 2ot}

+ 4xd21xI20t + tat

XIOX1

XIXI Hence,

cE H

+ X2X2 + (t + 4XllxI2X2 -

4X2IxI2XI}Ot.

iff the coefficient of at vanishes, i.e. (2.2) holds.

o

Therefore a horizontal curve c = (Xl, X2, t) can be written as

(2.3) COROLLARY 2.3. The vector field [Xl, X 2 ] is not horizontal, and hence H is not an integrable distribution. PROOF.

(2.3).

[Xl, X 2 ] is proportional to at and it cannot be written in the form 0

The horizontality and non-integrability of H can be described also by the 1connection form

(2.4) A curve c = (XI,X2,t) is horizontal if and only if w(c) = O. Hence the horizontal distribution H = kerw. The non-integrability of H can be expressed also by the condition w /\ dw i= 0, called the Frobenius non-integrability condition. Indeed, w /\ dw

[dt - 41x12(X2dxI - Xldx2)]/\ 161xI2dxI /\ dX2 161xI 2dt /\ dXI /\ dX2 i= 0, for Ixi i= O.

Therefore the model is not a contact manifold along the t -axis. In polar coordinates Xl = r cos cP, X2 = r sin cP, equation (2.2) yields t 4(XIX2 - XIX2)lxI 2 -4r4¢(sin2 cP + cos 2 cP) -4r 4 ¢.

In differential notation

(2.5)

dt = -4r 4 d¢,

and hence the I-connection form becomes w = dt

+ 4r 4 d¢.


48

Horizontal connectivity. As the vector fields (2.1) and their iterated brackets span the tangent space at each point, by Chow's theorem (see [5]), any two points can be joined by a piece-wise smooth horizontal curve. However, in this section we shall prove the horizontal connectivity by a direct method. THEOREM 2.4. 1) Given a point P(XI, X2, t) E ]R3, there is a smooth horizontal curve connecting the origin and the point P. 2) Given two points P, Q E ]R3, there is a piece-wise smooth horizontal curve connecting P and Q.

PROOF. 1) In order to prove the theorem, we shall construct an explicit solution. We shall consider three cases: X2 i=- 0, Xl i=- 0 and Xl = X2 = O. In the case X2 i=- 0, set

xds) X2(S) t(s)

as 2 +bs, as 2, 4 -"7ba(a2

+ a 2)s7 -

4 4 "3b2aas6 - 5"b 3as 5 ,

s

E

[0,1]

where a, b and a are constants, which will be found later from the boundary conditions t(l) = t. The curve starts at the origin Xl(O)

Xl(S)X2(S) - Xl(S)X2(S) 4(Xl(S)2 + X2(s)2)

= X2(0) = t(O) = O. Using -abs 2, 4(a 2 + a 2)s4

=

+ 8abs 3 + 4b2s 2,

one may show by computation

i(s) = 4(Xl(S)2

+ x2(sf) (Xl(S)X2(S)

- Xl(S)X2(S)),

i. e., the curve is horizontal. Next, we shall find the constants a, a, b. For s = 1 we have

a+b a t 4

Set

t=

X2, 1 2 2 12 13 --ba(a + a ) - -b aa - -b a. 735

105t / 4. The above system yields

t

-15ba(a 2 + a 2) - 35b2aa - 21b3a -15bx2(a2 + X22) - 35b2x2a - 21b3x2 -15bx2((XI - b)2 + X22) - 35b2X2(XI - b) - 21b3x2 -x2(b3 + 5xl b2 + 15IxI 2b).

The cubic equation in b

has at least a real solution h. Choose a

= Xl - h.


In the case

Xl

49

=I- 0, we consider the horizontal curve o.S2,

as 2 + bs, 4

4

4

-bo.(a 2 + 0.2)S7 + -b 20.as 6 + -b3 0.s 5 . 7 3 5 In a similar way we arrive at the following cubic equation b3 + 5X2b2

t

+ 151xI 2 b = -. Xl

In the case Xl = X2 = 0, we need to find a horizontal curve joining the origin and the point (0,0, t). We shall assume t < 0. Set

XI(S) X2(S) t(s)

rcos(ns), rsin(ns), ts, s E [0,1]'

with the parameter r to be found from the horizontality condition

i(s) where ¢

= ns.

Then t(s)

=

-4Ix(s)1 4 J>(s)

= -4r4n, r

For t

= -4r4n,

and hence

= (~;r/4.

> 0, we consider XI(S) X2(S) t(s)

rcos( -ns), rsin(-ns), ts, s E [0,1].

2) Let c and 'Y be curves joining the origin with the points P and Q, respectively. Then 'Y 0 c l is a piece-wise smooth curve joining P and Q. 0 Recently Gromov [29] proved that Chow's piecewise smooth integral curves may be replaced by smooth integral curves. We shall adapt our direct proof of Chow's theorem to obtain global connectivity. It is interesting to note that if x~ =I- Xl or x~ =I- X2, i. e., the points are not on the same vertical line, the horizontal curve has quadratic x-components. THEOREM 2.5. Given two points P(x~,x~,tO), Q(XI,X2,t) E smooth horizontal curve connecting P and Q.

PROOF. We shall consider the cases: X~ =I- Xl, X~ =I- X2 and x~ In the case x~ =I- X2, choose

JR.3,

there is a

= Xl, X~ = X2.

as 2 + bs + x~, o.s2 + XO 2' where a, band

0.

are constants, which will be found from the boundary conditions XI(O)=X~,

xI(l)=XI,

= to, t(l) = t. t(O)


50

A computation shows

Xl(S)X2(S) - Xl(S)X2(S)

b(xg - X2)S2

+ 2(XgXl - bxg -

X~X2)S

+ bxg,

(4(Xl - X~ - b)2 + 4(X2 - xg)2)S4

4(Xl (s)2 + X2(S)2)

+8b(x l - X~ - b)S3

+ (8X~(XI

- X~ - b)

+4b2 + 8Xg(X2 - xg)) S2 +8bsx~

+ 4(x~)2 + 4(xg)2.

Integrating in the horizontality condition (2.2) yields

t(S)

11

to

+

to

+ ~ (4(x l - x~ - b? + 4(X2 - xg)2) (bxg - bX2)s7

4(Xl(S)2

+ X2(S)2) (Xl(S)X 2(s) - Xl (S)X2(S)) ds

°

°

°

1 4 + ( '3b(xl - Xl - b)(bx2 - bX2) + 6(4(xl - Xl - b) 2

+4(X2 - xg)2)(2xgXl - 2bxg - 2X~X2))S6

+G(8X~(XI - x~ -

+~b(Xl - x~ -

+ 4b 2 + 8xg(X2 - xg))(bxg - bX2)

b)(2xgxl - 2bxg -

+~(4(Xl - x~ -

+(2bx~(bxg -

b)

2X~X2)

b)2 + 4(X2 - xg)2)bxg)s5

bX2)

+ ~(8X~(XI

-

x~ -

b)

+4b2 + 8Xg(X2 - xg))(2Xgxl - 2bxg - 2X~X2) +2b2(Xl -

x~ -

+~x~b(2xgxl -

b)xg)S4

+ G(4(x~)2 + 4(xg)2)(bxg - bX2)

2bxg -

2X~X2)

+~ (8X~(XI - x~ -

b) + 4b 2 + 8xg(X2 - xg) )bxg) s3

+G(4(x~)2 + 4(xg)2)(2xgxl -

2bxg -

2X~X2) + 4b2xgxns2

+ (4(x~)2 + 4(xg)2) bxgs. For

S

= 1 the boundary conditions for the x-component provide

(2.6)

a

(2.7) (2.8)

t

Xl - x~ - b, X2 - x 2 , t(l ),

°

where t(l) can be found substituting

(2.9)

S

=

1 in the above formula. We obtain

t = P(b),

2.3. SUBRIEMANNIAN GEODESICS

51

where P(b) is the following polynomial in b

If X2 =I- x~, the above polynomial is of degree 3, and hence the equation (2.9) has at least a real solution b. Using the boundary conditions (2.6) we get a and D:. The other two cases are similar and are left to the reader. 0

2.3. SubRiemannian geodesics Consider the Hamiltonian H defined as the principal symbol of the operator

~x =

!(Xr+Xn

DEFINITION 2.6. The subRiemannian geodesics between the origin 0 and the point P(x, t) are the projections on the (x, t)-plane of the solutions of the Hamiltonian system j; {

(2.11)

=

aHja~

i = aHlaB ~

= -aHjax

iJ = -aHjat with the boundary conditions (2.12)

x(O) = t(O) = 0, x(l) = x, t(l) = t.

PROPOSITION 2.7. Any subRiemannian geodesic is a horizontal curve. There are horizontal curves, which are not subRiemannian geodesics.

52


= (Xl (s), X2(S), t(s))

PROOF. Let c(s) Hamiton's equations

8H

be a subRiemannian geodesic. Using

2

-8 = 6 + 4X21xI (),

6

8H

2

86 = 6 - 4Xlixi (),

~~

i

(6 + 4X2IxI2(}) 4X21xI2 -

=

(6 - 4XllxI2(}) 4XllxI2

41x12(XlX2 - x2xd,

which is the horizontality condition (2.2). For the second part of the proof, we need to prove first the following result LEMMA

2.8. If c(s) = (Xl (s), X2(S), t(s)) is a subRiemannian geodesic, then ~l

-8(}XlXlX2

Xl

16(}lxI2x2, constant.

() PROOF.

+ 4(}lx12x2 + 8(}xi X2,

The Hamiltonian equations yield 8H

iJ

6

-7ft

= 0

-

=

88H

Xl

===} ()

= constant,

-[Xl 4(}8x1 (x2IxI 2) + X2( -4(})8x1 (xllxI 2)]

-4(}XlX2 2Xl + 4(}X2(lxI 2 + 2xi) -8(}XlXlX2 + 4(}lx12X2 + 8(}xix2.

Differentiating Xl

= 6 + 4X2(xI + x~)(},

we get

~l + 4X21x12(} + 4X2(2xlXl + 2X2X2)(} -8(}XlXlX2 + 4(}lx12X2 + 8(}xix2

Xl

+4X2IxI2(} + 8XlX2Xl(} + 8X~X2(} 8(}lx12X2 + 8(}(xi + X~)X2 161x12x 2.

o We shall show that the horizontal curves constructed in the proof of Theorem

2.4 do not verify the second equation of Lemma 2.8. First, we note ()

-=1=

o.

If ()

= 0,

then the Hamiltonian becomes 1

2

H = 2(~1

2

+ ~2)

with the Hamiltonian system ~i

8H

-- = 0

8X i

8H 8~i = ~i

===}

===}

~i =

constant,

Xi(S) = ~iS,

2.4. LAGRANGIAN FORMALISM

53

which is in contradiction with the formulas provided by the proof of Theorem 2.4. In the case x2(1) i= 0, these are (2.13)

s E [0,1].

Using a computation shows that the relation Xl nomial identity

= 160lxI2x2 becomes the following poly-

(320a(a 2 + a 2))s5 As the coefficients vanish,

+ 640abas 4 + 320ab2s 3 and 0 i= 0, we have aba = 0,

2a

=

0,

Vs E [0,1].

a = 0.

Hence the equation (2.13) becomes Xl(S) = bs, X2(S) = 0, which contradicts the condition x2(1) i= 0. In a similar way, one may show that the other horizontal curves provided by the Theorem 2.4 are not satisfying the Hamiltonian system of equations, and hence, they are not subRiemannian geodesics. 0 2.4. Lagrangian formalism We shall associate a Lagrangian with the Hamiltonian (2.10) using a formal Legendre transform. The Hamiltonian is not quadratic, so the Euler-Lagrange and the Hamiltonian equations won't be equivalent. In this section we shall investigate the relationship between the Lagrangian and Hamiltonian formalism. Using the Hamiltonian equations

Xl = 6 + 4X21xI 20 => 6 = Xl - 4X21X120, X2 = 6 - 4XllxI 20 => 6 = X2 + 4Xllx120, a formal application of the Legendre transform yields

L

6Xl

+ 6X2 + oi -

(Xl - 4X21x1 20)Xl

H

+ (X2 + 4Xllx1 20)X2 + oi - ~(xi + X~)

~(xi + X~) + O(i - 41x12(X1X2 -

x2xd).

If c = (Xl, X2, t), the Lagrangian becomes

(2.14)

1

L = 2(xi

+ x~) + Ow(c),

with w given by (2.4). A computation shows that the Euler-Lagrange system of equations

:s (;~) - ;~,

i = 1,2

for the Lagrangian (2.14) becomes

(2.15)

Xl = 160lxI2x2 { X2 = -1601x12xl 0= constant.

2.

54

GEOMETRIC ANALYSIS OF STEP

4

CASE

The following proposition will enable us to compute subRiemannian geodesics using the above system. The system (2.15) does not contain information about the t component. This will be recovered from the Hamiltonian, as the horizontality condition. PROPOSITION 2.9. A horizontal curve, which is a solution of the Euler-Lagrange system (2.15) is a subRiemannian geodesic. PROOF. Let (Xl, X2) be a solution of the system (2.15). Let

6

= 6 = (Xl, X2, t, 6, 6, 0)

We shall show that the horizontality condition

~

be defined by

Xl - 4X21xI 20 X2 + 4Xllx1 20. verifies the Hamiltonian system (2.11). From

.) Xl t. = 41 X12 (.X1X2 - X2

=

8H ae'

which is the second equation of (2.11). From the above construction of have

.

6,6, we

8H

X= 8[' As 0 = constant,

.

8H

O=O=7!it. We still need to show

.

8H

~i =

--8' Xi

i = 1,2.

Let i = 1. Taking the derivative with respect to the s variable in the definition of

6, one has

6

)O

Xl-4:s (X2I x I2

Xl - (4X21x120

+ 8X1X2X10 + 8X~X20)

160lxI2x2 - 4X21xI 20 - 8X1X2X10 - 8X~X20 80lxI2X2 + 80(xI + X~)X2 - 4X21xI 20 - 8X1X2X10 - 8X~X20 -8X1X2X10

+ 40lxI2X2 + 80XI x 2

+ 40X2(lx1 2 + 2xI) (x2IxI 2) + X2 (-40)8

-40X1X2 2Xl - [x 1408x1

x1

(xllxI 2 )]

8H 8Xl'

Hence

.

8H

.

8H

6=--· 8Xl A similar computation yields

6=--· 8X2 Hence (Xl, X2, t) is a subRiemannian geodesic. The converse of Proposition 2.9 is also true:

D

2.4. LAGRANGIAN FORMALISM

55

PROPOSITION 2.10. Let c = (Xl,x2,t) be a subRiemannian geodesic. Then c is a horizontal curve and (Xl, X2) satisfy the system (2.15).

PROOF. The horizontality part follows from Proposition 2.7. For the second part, Lemma 2.8 provides Xl = 16elx12x2, with e constant. In a similar way X2 = -16elxI2xl, which yields the system (2.15). o

In the next sections we shall characterize the subRiemannian geodesics using the Propositions 2.9 and 2.10. We shall solve the system (2.15) for different boundary conditions and construct the t-component of the subRiemannian geodesics using the horizontality condition. The system (2.15) can be written as

X = 16elxI 2Jx, where X = (Xl,X2), and J2 = -I, i.e., J is a 90° rotation of that for given boundary conditions

]R2.

We shall show

X(O) = Xo,

(2.16)

the constant e takes only discrete values. e plays the role of the eigenvalue for the system (2.15) with the boundary conditions (2.16). Because of Lagrangian rotational symmetry, we may rewrite the Lagrangian in terms of polar coordinates Xl = rcos¢ ,X2 = rsin¢. We have (2.17)

(2.18) and the Lagrangian becomes

(2.19) Since e is constant along the solutions of the Hamiltonian system, it can be considered as a Lagrange multiplier. The symmetries of the Lagrangian will provide symmetries for the geodesics. The Lagrangian L is invariant under the symmetry

S: (r, ¢, t; e)

---'>

(r, -¢, -t; -e).

If (r( s), ¢( s), t( s)) is a geodesic corresponding to e, then (r(s), -¢(s), -t(s)) is a geodesic corresponding to -e. Consequently, whatever statement is made for the geodesics joining the origin with the point (r(s), ¢(s), t(s)), when e > 0, it can be also made for the geodesics between the origin and the point (r(s), -¢(s), -t(s)), when < 0. The Euler-Lagrange system of equations can be constructed from the above Lagrangian. A computation shows

e

(2.20)

d aL..

aL

..

2

ds or = rand or = r¢(¢ + 16er ).

The first Euler-Lagrange equation becomes

r=

(2.21)

r¢(¢ + 16er 2 ).

As (2.22)

aL

a¢

2'

=

r ¢

+ 4er

4

and

aL

a¢

=

0,

56

2.


4

CASE

the second Euler-Lagrange equation is

:s

(r2¢ + 40r 4) = 0,

which can be written as

(2.23) where C is a constant.

As

8~ = 0 8L 8t and 8t = 0,

(2.24)

then 0 = constant along the solutions. Hence res) and ¢(s) will verify the following Euler-Lagrange system

T =.r¢(¢ + 160r 2) { r2(¢ + 40r2) = C

(2.25)

o= constant.

The system (2.25) is the polar coordinates version of the system (2.15), and hence Propositions 2.9 and 2.10 can be applied.

2.5. Solutions which start from the origin The solutions which start from the origin will be studied. The system (2.25) will be considered with the following initial conditions

(2.26)

reO) = 0,

¢(O) = ¢o,

reo)

i- 0,

From the second equation of (2.25) one obtains C

¢(O)
0) until r = 0, when l' = 1'max. After that 1'(s) decreases (r < 0) to O. 1'max can be found making r = 0 in (2.33). Hence 1'max is the positive solution of the following equation (see Figure 2.2):

V(1') = E,

(2.34) which is (2.35)

For each level of energy E > 0 there is a corresponding 1'max > 0, which is independent of the initial direction of the solution. The solution (1'(s), ¢(s)) lies in the disk D(O,1'max). The greater the energy E is, the larger rmax. As the energy E = I±(OW /2 depends just on the magnitude of the initial speed, 1'max will do the same.

The angle ¢ The second equation of (2.27) provides

¢=

(2.36)

-481'2.

As 1'(0) = 0, then ¢(O) = O. We know that for 8> 0, ¢(O) < 0, i.e., ¢ is decreasing, and for 8 < 0, ¢(O) > 0, i.e., ¢ is increasing. The first two equations of system (2.27) can be written as simultaneous equations

J 2E - 1682 1'6 and -d¢ = -481' 2 ds ds' where 1'( s) was supposed to be increasing. A differential equation connecting ¢ but not containing s can be obtained dividing the equations in (2.37) (2.37)

(2.38)

-dr =

d¢ d1'

-481'2 J2E - 168 21'6

l'

and

2.5. SOLUTIONS WHICH START FROM THE ORIGIN

59

Integrating, we obtain the variation of the angle ¢ r(5)

¢(s) - ¢(O) = lo

(2.39)

-40x 2 V2E _ 160 2 x 6 dx.

r max the above variation of ¢ is denoted by wand is given by Trnax -40X2 (2.40) W= dx. o V2E - 1602 x 6 Computing w: Making the substitution

When r

=

l

the integral (2.40) yields

r1 (~)2 / 6 (~f/6 u 2du

-40 w

V2E -0

V1- u 6

lo

r

1

Wllo

Ivdv

h

-~sgn(O) ( arcsin(l) PROPOSITION

and r (2.41)

=

arcsin(O)) = -sgn(O)i'

2.12. The angle w swept by the vectorial radius between r = 0

rmax is

w={ 7f/6, -7f /6,

if 0 < 0 if 0> o.

This agrees with the fact that the particle is moving clock-wise for 0 > 0 and counter-clock-wise for 0 < 0, see Figure 2.3.

o

Figure 2.3: A counter clock-wise rotation of angle w

Between r = rmax and r decreasing, r < 0, and

=

0 the vectorial radius sweeps an angle

(2.42) Using again (2.37) and making the division, one obtains (2.43)

= 7r /6.

d¢

40r2

dr

V2E - 1602 r 6

w.

As r is


60

The angle ill is given by

l

1

rmax 40r2 -40r 2 dr = dr rmax v2E - 160 2r 6 0 v2E - 160 2r 6 ' which is the same integral as that one given in (2.40). Hence

ill

=

0

(2.44)

= w = -sgn(O)i'

ill

The curve (r (s), 4>( s)) passes through the origin again after sweeping an angle equal to w + ill = 7r /3 (clock-wise if 0 < 0 and counter-clock-wise if 0 > 0). This means that the angle between the tangents at the origin is 'lj; = 7r /3. Since 'lj; /27r is a rational number, the solution will be periodic.

Explicit solution in polar coordinates. In order to avoid using sgn(O), we shall assume 0 < O. Then 4> is increasing and the trajectory is counter-clock-wise. We shall consider a few cases:

• Assume 4> E [4>(0),4>(0) + 7r/6] and r E [0, rmax] The radius r is increasing and d4> dr

(2.45)

-40r2 V2E - 160 2r 6 '

which becomes x 2 dx

rCs)

(2.46)

4>(s) - 4>(0) = -40

10

v2E _ 160 2x 6 '

where r(s) ::::: rmax. Making the substitution

_ (160 2 ) 2E

(2.47)

1/6

U -

x,

and using rmax given in (2.35), one may write

4>(s)-4>(O)

_ ~ ( 2E ) 2/6 ( 2E ) 1/6 rCs)/rmax V2E 1602 160 2 10

o ( arCSIn . (r(s)3) - -Illl -3-

-

3u r max Let 4>0 = 4>(0) be the initial argument. Using 0 and r will be given by

(2.48)

u 2 du u6

vI -

. (0)) . arCSIn

< 0, the relationship between 4>

) 4> - 4>0 = -1 arcsin ( - r3 33 rmax

,

which is equivalent to

(2.49)

r(4)) = rmax sin 1/ 3 (3(4> - 4>0))'

• Assume 4> E [4>(0) + 7r /6,4>(0) + 7r /3] The radius r decreases from rmax to O. Let M(r max , 4>0 + 7r /6) be the vertex and let A(r, 4» be a point as in Figure 2.4, i. e., 4> E [4>(0) + 7r /6,4>(0) + 7r /3]. On this interval r < 0, and

(2.50)


61

Using

d¢ = -48r 2 ds ' by division of the last two expressions, we obtain d¢ 48r 2 (2.51 ) dr .../2E - 168 2 r6

Figure 2.4: After a rotation of 7r /3

Integrating between ¢o

+i

¢ - (¢o

(2.52)

and ¢, we have

1

48x 2 dx v'2E _ 1682

r

7r

+ '6)

=

Trnax

6' X

The right hand side integral can be decomposed as

1 10 l l l r

Trnax

r

r

+

=

=-

0

Tmax

",ax

+

0

r

0 '

and hence

7r

¢ - (¢o

(",ax

+ '6)

= - Jo

48x 2 dx ( 4 8 x 2 dx v'2E - 1682x6 + Jo .../2E - 1682x6

The first integral is given in (2.40) and it is equal to 7r/6 by Proposition 2.12. The second integral appears in the right hand side of (2.46) with a negative sign and it is given by the negative right hand side of (2.48) ( 48x 2 dx 1 r3 (2.53) Jo .../2E _ 1682x6 = -'3 arcsin(r~ax)' Hence (2.52) becomes

¢ - (¢o

7r

+ -) 6

r3 arcsin(-3-) 6 3 r max 1 r3 - arcsin(-3-) 3 r max 7r

1

= - - -

{:}

7r - - (¢ - ¢o) 3

{:}

sin (7r - 3(¢ _ ¢o)) =

{:}

r

=

;3 rmax

= rmax sin 1/ 3 (3(¢ - ¢o)),


62

which agrees with (2.49). We note that 7r (2.54) r(¢o + "3) = r(¢o) = 0, which means the solution starts at 0 with an argument ¢o and bounces back with an argument ¢o + 7r /3 . • Assume ¢ E [¢o + 7r /3, ¢o + 27r /3] The solution starts from the origin 0 with an argument ¢o = ¢o + 7r /3, and arrives at 0 with an argument ¢o + 7r /3. For ¢ E [¢o, ¢o + 7r /3] one may write using (2.49)

r( ¢)

rmax sinl/3(3( ¢ - ¢o)) rmax sin l / 3(3¢ - 3¢o + 7r) -rmax sin l / 3(3(¢ - ¢o)).

We note that sine function is negative for 3(¢ - ¢o) E [7r,27rJ, and the minus sign in front of the formula makes sense. In the general case the formula in polar coordinates r = r( ¢) can be written as (2.55) The solution in the (Xl, x2)-plane is periodic. It contains 3 loops with an angle of 27r /3 between the symmetry axes, see Figure 2.5.

Figure 2.5: The x-projection of the solution

As a conclusion, one may state the following result: PROPOSITION

(2.56)

2.13. The solution in polar coordinates satisfies

r(¢)2 = r;'ax sin 2 / 3(3(¢ - ¢o),

where 2

rmax E being the constant of energy.

1

="2

(E) ()2

1/3

'


63

2.5.0.8. The t-component along the solution. Suppose that (xt{s), X2(S), t(s)) is a geodesic, i. e., it satisfies the Hamiltonian system of equations given by the Hamiltonian (2.57)

One of the Hamiltonian equation is

.

(2.58)

t =

8H 80

=

41xl 2 (X2 Xl -

4'

X1X2) = -4r ¢.

If () < 0 then ¢ increases, and hence t decreases. Using Proposition 2.13, one obtains

dt

= -4r4 d¢ =

-4r~ax sin 4 / 3 (3(¢ - ¢o)) d¢.

Integrating, we have

-4r~ax 1 sin4 / 3 (3(u -

t(¢) - t(¢o)

1

¢o)) du

0

4 __ r 4 3 max

3 (-0)

0

sin 4 / 3 v dv

.

Set (2.59)

w(u)

r

3U

= io sin4/ 3 vdv.

Then (2.60)

t(¢) - t(¢O) =

4 -"34 r max W(¢ -

¢O).

Let (2.61 ) Then (2.62)

t(¢o

+ 7r/3) -

t(¢O)

=T

does not depend on the initial angle ¢o.

Conjugate points to the origin. Consider a subRiemannian geodesic (Xl (s), X2(S), t(s)) which starts at the origin. In polar coordinates r(¢o) = 0 and t(¢o) = o. When ¢ = ¢o + 7r /3 we have r( ¢o + 7r /3) = 0, which means that the x-component is zero. The only non-vanishing component is the t-component (2.63)

t(¢o

+ 7r/3)

= T.

It follows that the point (0,0, T) belongs to the geodesic and does not depend on the initial condition ¢o. A variation of ¢o corresponds to a rotation of the geodesic. The point (0,0, T) remains fixed during this variation and belongs to all geodesics which are obtained from the initial one by the variation of the initial angle ¢o. The following result states that the conjugate points to the origin along the geodesics are the t-axis. THEOREM 2.14. Given the point P = (0,0, u) on the t-axis, u i- 0, there is a geodesic" which starts at the origin, such that P is the first point conjugate to the origin along ,.

2.

64


4

CASE

PROOF. The proof is based on the construction of such a geodesic.

(i) case u < We choose

°°

e
-ci>o)

t(¢)

= QJo

sin 4 / 3 vdv.

It is easy to see that the point P = (0,0, u) belongs to the above geodesic. Performing a rotation in the x-plane of the initial velocity, all the geodesics obtained will pass through P, and hence P will be the first conjugate point. (ii) case w > In this case > and the rest of the proof is similar with the case < 0. 0

°°

e

e

2.6. The length of the geodesics between the origin and (0,0, t) It is already known from Theorem 2.14 that there are solutions which join the origin to any point on the t-axis. This section answers two questions: 1. How many subRiemannian geodesics of different lengths (parametrized by [0,1]) join the origin to (0,0, t), for a given t f o? 2. What are these lengths?

The length of a horizontal curve is measured with respect to the subRiemannian metric g : H x H ~ F such that g( Xi, X j ) = ISij , with the vector fields Xi given by (2.1). It is known that if c = (Xl, X2, t) is a horizontal curve of parameter s E [0,1]' c is horizontal if and only if The length of e is

le(s)lg = Jg(e(s),e(s)) = JX1(S)2 +X2(s)2, and the length of c is

£(c) =

11

le(s)lgds =

11

jxi+x§ds.

PROPOSITION 2.15. If c: [0, 1] ~ 1R 3 is a subRiemannian geodesic, then is constant, and £(c) = lelg.

le(s)lg

PROOF. If c is a subRiemannian geodesic, the Hamiltonian is preserved along the geodesic

X1(S)2

+ X2(s)2

---=-'----'------='-'--'2

= H0

(constant) ,

and hence

le(s)lg = J2Ho. The second part of the conclusion is obvious.

o

2.6. THE LENGTH OF THE GEODESICS BETWEEN THE ORIGIN AND (0,0, t)

65

In the following we shall find explicit lengths of the subRiemannian geodesics between the origin and points situated on the t-axis, see [12]. The following lemma will be useful. LEMMA

2.16.

1

0"(r)2

o

where b = 2 -

PROOF.

dt

----r:=;=:==.~ Jt(l - t 3 ) -

V3, k = v'b/2,

-

1

31 / 4

and g(a)

=

sn

-1

[J

~v'3

0"2+-2-

g(a)2]

1 - -- k

b2 '

-

,

l.

We need to evaluate the integral

r2

where f(x) = x(l - x 3 ) = _(x 2 Making the substitution

-

dx

Jo

Jf(x)'

x)(l

+ x + x 2 ). pt + q t +1 '

X=--

with p=

f

-12

V3

-1

q=

+ V3 2

becomes

- (t: 1)4 ((p2 - p)t 2 + q2 - q) ((p2 + p + 1)t 2 + (q2 + q + 1))

f

1 (3

- (t + 1)4 ("2 + 1

v'33)t 2 + ("23- v'3) (3 2 3) 3) "2t +"2

3 3

r,;

2

(t+1)4"2("2+v3)(t -b

2) ( 2

)

t +1,

= 7 - 4V3 and b = 2 - V3. Let

where b2

g (a)

=

V31+v3 2 a +-2-

-

1.

Then the integral becomes

r2 ~ _

r

b

2

1

dt

Jo J f(x) - J3 + 2V3 Jg(O") J(b 2 - t 2)(t 2 + 1) Using the formula (see Lawden, [9], chapter 3)

l

b

1

1

J(a 2 +t 2)(b2 -t 2)

Ja 2 +b2

~~~~~~dt=

x

sn

_1[Jb2 -x 2 b] b ' Ja 2 +b2 '

one obtains

o

2.

66


4

CASE

THEOREM 2.17. The geodesics that join the origin to a point (0,0, t) have lengths d1 , d 2 , d3 ... , where (2.65) where (2.66)

r dt - io J(l - t2)(1 - k 2t 2) ,

K _

1

Q = ~ f(1/6) ..;rr 4 r(2/3) 1r.

(2.67)

For each length d m , the geodesics of that length are parametrized by the circle

§1.

PROOF. We need to find the subRiemannian geodesic connecting the origin to (0,0, t). The geodesic is parametrized by [0,1]. The boundary conditions are r(O) = r(l) = 0, t(O) = 0 and t(l) = t. Using the Euler-Lagrange equations in polar coordinates, one obtains that the constant () verifies the following boundary value problem (2.68)

r(O) = r(l) = O.

(2.69)

Let ). 2 = 48()2. Then (2.68) becomes

(2.70) with the boundary conditions r(O) law of energy

= r(l) = O. Integrating, we get the conservation

1

1

+ _).2 r 6 = Ho 2 6 . Ho is the constant value of the Hamiltonian along the solutions (denoted also by E). Separate the variables and obtain _f2

dr o-

.J2H

~).2r6

= ds .

The radius r is defined by the integral equation

[ V2H d: p2 6~ o

x

6.

Making the substitution _ (~)1/6 uH x,

6

0

the above expression becomes

(2.71)

r*)

io

du

V1- u6

= ( 4H5).2)1/6 S 3

where

a(r)

=

).2 ) 1/6 ( 6H r.

o

'

2.6. THE LENGTH OF THE GEODESICS BETWEEN THE ORIGIN AND (0,0, t)

Substituting u 2

= t,

67

one may write

l

(2.72)

dt

C7 (r)2

-Jrt

o

1=-===;t3""')

::;=(

2 - -(2H ),)1/3 8 - 3 1/ 6 0 .

The left side can be expressed in terms of elliptic integrals. Using Lemma 2.16, relation (2.72) becomes

)1- g~)2 =

(2.73) For

8

=

[24/33 1/ 12 (Ho),)1/3 8

= 0, and g(O) = ~~+:fl = 2 - v'3 = b, and 0,

have 8 = 0 and

8

8

=

8n

1, we have r

=

hence a hence

,

kJ.

O. Using the expression of 9 we

VI - g~ot = O. It turns out that for

= 1 the left hand side in (2.73) is zero. Then 8n

(24/33 1/ 12 (Ho),)1/3)

= 0,

and hence

m=I,2,3 ...

(2.74)

-I 0 because Ho -10 and () -I O. 4K is the period of 8n (. , k) with k is defined by the complete Jacobi integral

m

From (2.74) one has

Ho),

=

m 3 K3 2.3 1 / 4 '

Using), = 4v'3 (), the above formula becomes

Ho()

=

C~3~4)3.

Then () is given by mK () = ( 2.3 1 / 4

)3

/

Ho,

and

Using (2.35) we have

4 rmax =

(2.75)

4

Ho 2/3 2 HJ (02) = 3 m4K4 ·

The left hand side of (2.75) is related to t as follows. As dt = -4r 4 d¢, integrating between ¢o and ¢o - m7r /3, we obtain

1

<po- m7r / 3

t

=

-4

r;'ax sin 4 / 3 (3(¢ - ¢o)) d¢

1R3 be a geodesic between o and P. As 11'1 is constant in the subRiemannian metric (see Proposition 2.15), we have identity in Cauchy's inequality

r

1

£b) =

r

Jo 11'(8)1 d8 = (Jo

1

1/2

d8)

r

1

(Jo 11'(8)1 2 d8)

1/2

=

yhHo,

and hence

(2.78) Using (2.77) we see that the lengths of the solution 'Yare quantified, namely, £b) d m where

(2.79)

m

=

= 1,2,3, ...

These lengths do not depend on the angle -4r~(u)du= -12Qltl (4>sin4/3(3(U-¢0))dU. m i4>o

i4>o

Making the substitution v = 3(u - ¢o), we obtain (2.83)

-41tl

(3(4>-4>o)

t m (¢) = mQ io

Formula (2.82) expresses r

sin4/3(v) dv , =

m=1,2,3 ...

r(¢), but we may compute explicitly the radius

r = r(s), which is a solution for the boundary value problem (2.84) where A2 = 480 2 . This will be done in the following.


70

t

,,

m=l

4

31tl

rmax=Q

,

,

/ t

r 4 = 31tl

m=2

max

2Q

t

m=3

,

,,

Figure 2.6: The shapes of the solution for m = 1,2,3.

= Jl - sn 2 , relation (2.73)

Using the relation between the elliptic functions en becomes

g(u) = en [24/3 31/12(H >,)1/3 s b

0,

/Ii] 2 .

Since

then

g(u)

=

/Ii

b en[2mKs, 2].

Using the definition of g(u), we have u2

-

J3

-ben(2mKs,1)+1

+p '

2.8. SOLUTIONS WHICH START OUTSIDE THE ORIGIN

where p

=

-1;V3. Using that

2=

(~)1/3

2=

r 6H 0 the above formula becomes 0'

(2.85)

r2 (s) = m

where

71

(~)1/3 6H 0

r

2=

2= 2 (~)1/3 H0 r 2(mQ)1/22 II r, 3t

~(~)1/2[p+ ~ ]] 2 mQ 1 + b cn 2mK s , k

k = v'b/2, b = 2 - /3,

p

=

, m= 1,2,3 ...

(-1- /3)/2.

The first three geodesics and their projections on the x-plane are sketched in Figure 2.6. After that the pattern is periodic. The above results can be stated in the following theorem: 2.19. The equation of the subRiemannian geodesics which join the origin to a fixed point (0,0, t) in polar coordinates are THEOREM

(2.86)

~~sin4/3(3(¢-¢o)),

r;,,(¢) =

m=1,2,3 ...

-4ItI13c-O) sin4/3(v) dv.

tm(¢) = - mQ 0

(2.87)

The radius r( s) is given by

1(

2

(2.88)

r m(s) =

'2

31tl ) 1/2 [

mQ

p

/3

]

+ 1 + b cn[2mKs , k] ,

wherem=1,2,3 ...

2.8. Solutions which start outside the origin In this case, r(O) =I- 0 and the system (2.25) will be considered with the initial conditions

r(O) =I- 0,

(2.89)

¢(O)

= ¢o·

The second equation of (2.25) becomes

~=

(2.90)

C _ 4Br2

r2

'

where C =I- O. Substituting in the first equation of (2.25), we obtain

(2.91)

r

r~(~ + 16Br 2) = r(~

r

'3

C2 r

+ 8BCr -

- 4Br2)

(~ + 12Br2) {:?

48B 2r 5 .

Let (2.92) be the potential function. Equation (2.91) can be written as the Newton's equation (2.93)

r = -V'(r).


72

A computation shows that

v (r) = ~ (40r: -

(2.94) Writing p =

r

C)

2.

the equation (2.91) becomes dV

dp

--

p-

dr p2

dr

2 where E is the constant of total energy.

{::}

-V(r)

+ E,

PROPOSITION 2.20. If r is a solution of the Euler-Lagrange system (2.25) with initial conditions (2.89), then the energy is conserved

(2.95)

~r2 + ~ (40r4 2

r

2

- C)2 = E.

We shall give a qualitative description of the solutions of (2.95). There are two cases: sgn( C) = sgn( 0) and sgn( C) =1= sgn( 0). In each case we shall draw the trajectory in the phase plane.

(1) Case sgn(C) = sgn(O). In this case V(r) has a root and a global minimum at

_ (C)1/4 ro - 40 '

(2.96) and limx---+o+ V(r)

= limx---+ oo V(r) = 00, see Figure 2.7

Equilibrium points ro corresponds to a stable equilibrium point because V'(ro) =

°

and V"(ro)

> O.

The corresponding ¢ satisfies

°

C _ 411 2 _ C - 40r3 2 uro 2 ro ro and hence ¢ is constant. This means that each point of the circle C(O, ro) is an equilibrium point (constant solution). At each of these points the energy E = O. j _

If' -

Tangent circles: When E> 0 there are exactly two positive roots rmin, rmax of the equation (2.97) such that (2.98)

V(r) = E,

°< rmin :s; ro :s; rmax·

The phase plane and the potential V are shown in Figure 2.7. PROPOSITION 2.21. The solution of the Euler-Lagrange system (2.25) in the casesgn(C) = sgn(O) lies into the circular ring W(O,rmin,r max ) = {x E ]R2;rmin:S; Ixl :s; rmax}.


73

PROOF. As l' reaches 1'min or 1'max for r = 0, then the equation of conservation of energy (2.95) shows that 1'min , 1'max are solutions of the equation V(r) = E, and the solution r has the property

o < 1'min ::; r

::; 1'max.

D

v V(r)

r r nun .

:r

; max

r

Figure 2.7: The solution in the phase plane

Existence of loops: The equation ¢ = 0 has the positive root

(2.99)

r

= (C)1/4 4()

.

Hence ¢ = 0 if and only if l' = 1'0. When the solution intersects the circle of equilibrium points 8(0, ro), the sign of ¢ changes, i.e., the trajectory is bouncing back, making loops in the ring W(O, rmin, 1'max) , see Figure 2.8.

74


f max


The width of the ring: The width of the ring w = 1'max - 1'min is a function of energy w = w(E). Using the properties of the function V(1'), one obtains the following: (i) w(E) is an increasing function of energy E, (ii) w(O) = 0, case in which 1'min = 1'0 = 1'max , (iii) limE->oo 1'min = 0, limE->oo 1'max = 00, (iv) 1'min '" -;as E ~ 00. rmax The asymptotical correspondence (iv) comes from the analysis of the equation

C ~ 481' 3 - - = ±v2E, l'

when E is large. We may have l' = l' min ~ 0 or l' = l' max ~ 00. When consider l' = 1'min, the leading factor is ~. When l' = 1' max , the leading factor is 481'~ax' T m 1.n See Figure 2.7.

The t-component: As i = -41'4¢ and ¢ = 0 on the circle 8(0,1'0)' then t will do the same. i(s) changes the sign when the solution x(s) crosses the above circle. The component t(s) reaches the maxima and minima on the cylinder 8(0,1'0) x [0, +00). See Figure 2.9.


75

X2

Figure 2.9: The geodesic and its projection in the case sgn(C) =I- sgn(B).

(2) Case sgn(C) =I- sgn(O), C =I- O. Here

.

C

2

=I- 0, r which means that ¢ increases (decreases) for 0 < 0 (0) 0), and hence there are no loops. In this case the potential energy is strictly positive ¢

V(r)

2" - 40r

=

~ (40r: -

=

and it has a minimum at

Cf > 0,

_ (-C)1/4 120 .

(2.100)

rl -

Indeed,

and the equation of critical points

V'(r)

=0

is equivalent with the following quadratic equation

480 2 u 2 where u

= r4.

-

80Cu - C 2

= 0,

The equation in u has two roots Ul

C = 40 and

As only the second one is positive,

rl

U2

C = -120·

will be chosen such that

C

4

h) = -120· The equilibrium solution: The equilibrium solution in this case is the circle of radius The corresponding ¢ can be obtained from 2

•

2

r 1 (¢ + 40rl)

= C,

rl

centered at the origin.

76


and it is equal to

¢(s) =

4C ,J-12B) ("3 -C s+¢o·

The circle is traversed clock-wise if C > 0 and counter-clock-wise if C < O.

The bounds for r(s): There are two positive roots

Pmin

V(r)

=E

and

Pmax

for the equation

where E ~ V(rl),

,

and 0< Pmin :::;

They correspond to the points where

rl :::;

Pmax·

r = 0 in the conservation equation (2.95).

v V(r)

E > V(rJ

r r

Figure 2.10: The solution in the phase plane

In the phase plane the solution is rotating around the stable equilibrium point (rl,O) such that Pm in :::;

r(s) :::;

Pmax,


77

see Figure 2.10.


The trajectory in the x-plane belongs to the ring W(O, Pmin, Pmax), see Figure 2.11. The width of the ring increases as the total energy E increases, i. e., the initial speed increases, where E = ~ 1:i:(OW.

The t-component: In this case i i= 0 and i = -4r4¢. Hence t(s) only increases (if decreases (if e < 0). See Figure 2.12.

Figure 2.12: The geodesic and its projection in the case sgn(C)

e > 0)

#

sgn(O).

or only

2.

78


4

CASE

2.9. Geodesics between the origin and points away from the t-axis

We have seen that if Ixl = 0 and t -=I- 0, there are infinitely many geodesics between the origin and P. In this section we shall study the case Ixl -=I- o. THEOREM 2.22. Let P(x, t) E 1R3 be a point such that x -=I- 0 and t ~ O. There are finitely many geodesics between the origin and P. This number increases unbounded as t/lxl4 --t 00.

PROOF. Let p and q> be the polar coordinates for the endpoint (Xl,X2). Consider a geodesic which joins the origin and P. The t-component is computed by integration between cPo and q> 4

4 t=--r 3 max

(2.101)

1

3 (4)-0)

0

sin 4 / 3 vdv .

Writing p = r(q» in Proposition 2.13, yields (2.102) Eliminating

(2.103)

rmax

from equations (2.101) and (2.102), we obtain

3 t - - sin 4 / 3 (3(q> - cPo)) = 4 p4

1

3 (4)-0)

0

sin 4 / 3 v dv.

Let (2.104)

It is easy to see that F(x) is an increasing, unbounded and odd function. Let 'l/J = 3(cPo - q». The relation (2.103) becomes (2.105) Since the left hand side is a periodic and bounded function of variable 'l/J, the above equation has finitely many solutions, see Figure 2.13. This means that for p -=I- 0 there are only a finite number of subRiemannian geodesics joining the origin 0 and the point P. This number increases unboundedly as the quotient t/ p4 increases. For t = 0 the equation (2.105) has only the solution 'l/J = 0, i.e., cPo = cP = q>. In this case t(s) = O. Hence, the only solution is the segment line OP. 0

A detailed analysis of the number of solutions of equation (2.105) is done in the following lemmas, see [15].

2.9. GEODESICS BETWEEN THE ORIGIN AND POINTS AWAY FROM THE t-AXIS

79

F('I')

o

rt/2

rt

rt+rtl2

Figure 2.13: The functions )..( 'lj;)

2rt+rt/2

2rt

= ~ 1< sin4 / 3 'lj;

3rt

and F( 'lj;).

Let ).('l/J) = ~~sin4/3'l/J. The functions )...('l/J) and P('l/J) start from the origin and their graphs are tangent at 'l/J = O. The following lemma shows that the graph of P('l/J) is below the graph of )...('l/J) on a small neighborhood. LEMMA

2.23. Let t

> 0 and p =I- O. There is an t > 0 such that

0< )...('l/J) < P('l/J),

for

0

< 'l/J < t.

Hence, there are no solutions for the equation (2.105) on the interval (0, t). PROOF. The derivatives of )... at 'l/J P, as the following computation shows

= 0 blow up faster than the derivatives of

P'('l/J)

sin 4/ 3 'l/J ~ P' (0)

P" ('l/J)

~ sin1/ 3 'l/J cos'l/J ~ P"(O) = 0,

PIII('l/J)

4 4 cos 2 'l/J -. 9 sin 2 / 3 'l/J

)...' (0)

0,

3

= 0,

~

pili (0)

=+00

'

)..." (0) = +00.

o 2.24. Let t > 0 and p =I- O. N* be an integer such that

LEMMA

(i) Let n

(2.106)

E

1 3 t 1 (n-"2)Q(n+"2)Q, then in the interval (0, (n

+ ~)11"]

there are 2n distinct solutions for the equation

(2.105). However, the equation (2.105) may have other solutions outside of the above interval.

(i) The function x --- sin4 / 3 x is positive and periodic with the principal period 11". Let n E N* be the smallest positive integer such that PROOF.

~~ < F(n11" + ~). 4 p4 2 Then the equation (2.105) has no solutions in the interval (n11" + Hence, if

F((n -1)11"

(2.107)

~,+oo).

+~) < ~~ < F(n11" +~) 2 4 p4 2 '

there are three cases:

(1) If the graph of F( 'l/J) is above the graph of )..( 'l/J) on the interval (n11", (n+ ~ )11"), there are exactly 2n - 1 nonzero solutions. See Figure 2.13. (2) If the graph of F('l/J) is tangent to the graph of )..('l/J) on the interval (mr, (n + ~)11"), there are 2n nonzero solutions. The largest solution is a double solution. See Figure 2.14(a). (3) If the graph of F('l/J) intersects the graph of )..('l/J) twice on the interval (n11", n + ~11"), there are exactly 2n + 1 nonzero solutions. See Figure 2.14(b) .

...11 4p4

...11

---------------

o

nIt

4p4

nIt+It/2

Figure 2.14(a)

o

nIt

nIt+It!2 Figure 2.14(b)

2.9. GEODESICS BETWEEN THE ORIGIN AND POINTS AWAY FROM THE t-AXIS

All solutions belong to the interval (0, (n that

+ ~)1l"). In

81

the following we shall show

(2.108)

Using the properties of the integral of a periodic function, rn7r+~

io

4

sin 3 x dx

(2n

r~

+ 1) io

(2n+1)~

4

sin 3 xdx

= (n+

~)Q.

Replacing n by n - 1, yields (2.109)

Substituting (2.108) in (2.107) yields (2.106). (ii) If there is an integer n E N* such that t F ( n1l"+ -1l") =3- 2 4 p4'

then

1

3 t

+ 2)Q = "4 p4' and there is one more solution, equal to n1l" + i. (iii) The condition l~ > (n + ~)Q = F(n1l" + i) below the line y = ~ on the interval [0, n1l" + i]. (n

implies that the graph of F is The graph of F intersects the graph of A right before and after each turning point such as 1l", ... , n1l". Hence there are n+n intersections, which correspond to 2n solutions in the interval [0, n1l"+1l" /2]. See Figure 2.13. 0 As a conclusion, one may state the following theorem. THEOREM 2.25. Let P(x, t) be a point ofJR3, away from the t-axis and x-plane. There are not less than 2n - 1 and not more than 2n + 1 subRiemannian geodesics between the origin and the point P, where the integer n is defined by

1 (n - 2)Q

(2.110)

3

It I

< "4lxl 4 ~

(n

1

+ 2)Q·

PROOF. The number of subRiemannian geodesics is equal to the solutions of the equation (2.105). Applying Lemma 2.24 leads to the desired conclusion. 0

The limit case Ixl ---- 0 One may see from the Figure 2.14 that the solutions of equation (2.105) become in this case

o < 'l/J1

=

'l/J2

= 1l"

< 'l/J3

=

'l/J4

= 21l"

< 'l/Js

=

'l/J6

= 31l"

< ... ,

and hence we have an infinite number of geodesics joining the origin and P(x, t).

82

2.


4

CASE

2.10. Geodesic completeness in step 4 case

The following definition holds on any subRiemannian manifold. DEFINITION 2.26. If for any point P, any geodesic c(t) emanating from P is defined for all t E JR, the model is called geodesic ally complete. Theorems 2.27 and 2.28 are proved in a more general case in the next chapter. The first theorem states the completeness of the model, while the second one states the global connectivity property. THEOREM 2.27. Given a point P = (x~, xg, to) E JR3, any subRiemannian geodesic associated with the model defined by the vector fields (2.1) emanating from P can be extended indefinitely, i. e., the model is geodesically complete. THEOREM 2.28. Given any two points P, Q E JR3, there is a geodesic joining P and Q. 2.11. Exercises

= sn(z, k), cnz = cn(z, k), dnz = dn(z, k). Prove that

EXERCISE 2.1. Let snz

d -snz

cnzdnz,

d -cnz

-snzdnz,

d -dnz

-k 2 snzcnz.

dz

dz

dz

EXERCISE 2.2. Show that the following Maclaurin power series expansions hold about z = 0 snz cnz

1( 1 1 + k 2 )z 3 + ,(1 + 14k 2 + k 4 )z 5 - ... , 3. 5. 1 2 1+ 4'1. (1 + 4k 2 )z 4 - ... , 2.

Z -

,

,Z

lim sn( u, k)

k-+l

EXERCISE 2.4. a) If 0

+, (4k + k 4.

1221244

1 - ,k z 2. EXERCISE 2.3. Show that dnz

= tanh u,

) z - ....

lim cn( u, k)

k-+l

= sech u.

< u < K, prove the following inequalities:

u 1 cnu < dnu < 1 < < --. snu cnu b) If 0 < u < ~ show that u 1 cos u < 1 < -.- < - - . smu cosu EXERCISE 2.5. Consider the equation y = ay3 -by, where a and b are constants such that ~ = b-1 > O. Show that y(x) = sn(x, v'b'=1) is a solution for the above equation.

2.11. EXERCISES EXERCISE

83

2.6. Find the first integral of energy for the equation

r(8) = _Kr2m+l(8) where K is a constant and m is an integer. EXERCISE

2.7. Let

(h : ]R3

---+]R3

be given by J),

=

().x, ).4t).

i) Show that J~Xl = ).-1 Xl,

J~X2 = ).-1 X 2 ,

where Xl and X 2 are defined in (2.1). ii) Prove the following estimates C l (lxl 4 + Itl)1/4 :::; dc((O, (x, t)) :::; C2 (l x 14 + ItI)1/4,

where C l and C2 are two universal constants and dc((O, (x, t)) is the CarnotCaratheodory distance between the origin and the point (x, t).

CHAPTER 3

The Geometric Analysis of Step 2(k + 1) Case In this chapter we study the subRiemannian geometry on the step 2(k subRiemannian manifold defined on 1R 3 = 1R~ x IR t by the vector fields Xl

(3.1)

= aX! + 2X21x12kOt,

+ 1)

X 2 = OX2 - 2Xllxl2kOt,

where IxI 2k = (xi + x~)k, kEN. We shall compute the length of the subRiemannian geodesics between the origin and any point on the t-axis, where the conjugate locus is. We shall also characterize the number of subRiemannian geodesics between the origin and any other point. Finally, we shall prove the connectivity and completeness theorems for this model. The subRiemannian lengths are computed in two ways: using polar coordinates and using hyper geometric functions. With a similar proof as in chapter 2, we have the following result: PROPOSITION 3.1. There is no Lie group on 1R3 underlying the vector fields Xl and X 2 given by (3.1).

3.1. The horizontal distribution Let H be the horizontal distribution x -Y H x = span{XI' X2}' A curve c such that the velocity C E H, is called horizontal curve. PROPOSITION 3.2. A curve c = (Xl, X2, t) is horizontal if and only if (3.2)

PROOF. The velocity vector can be written as C

XI0x! + X20X2 + iOt Xl (aX! + 2X21x12kOt) - 2XIX21xI2kOt +X2(OX2 - 2Xllxl2kOt) XIX I

+ 2XIX21xI2kOt + iOt

+ X2X2 + (i + 2XllxI2kx2 -

2X2IxI2kxI)Ot.

Hence, C E 1i if and only if the coefficient of at vanishes. D

Therefore, a horizontal curve c = (Xl, X2, t) can be written as (3.3) COROLLARY 3.3. H is not involutive, and hence H is not an integrable distribution. 85

86

3. THE GEOMETRIC ANALYSIS OF STEP

2(k + 1)

CASE

at

PROOF. [Xl, X 2 ] is proportional to and it cannot be written in the form (3.3). Therefore H is not involutive. By Frobenius theorem, it is not locally integrable.

o The I-connection form

w = dt - 2lxl2k(X2dxl - Xldx2) describes the horizontality and non-integrability. A curve c = (Xl, X2, t) is horizontal if and only if w(c) = O. The horizontal distribution H = kerw. The Frobenius non-integrability condition can be expressed as

w /\ dw

[dt - 2lxl2k(X2dxl - Xldx2)]/\ 4(k + 1)lxl2kdxI /\ dX2

= 4(k + 1)lxl2kdxI /\ dX2 /\ dt # 0,

for

Ixi # o.

Hence the model is not a contact manifold along the t -axis. In polar coordinates Xl = r cos ¢, X2 = r sin ¢, equation (3.2) yields

t and hence w

2(XlX2 - XIX2)lx1 2k _2r2kJy,

= dt + 2r 2k d¢. 3.2. Horizontal connectivity

The vector fields (3.1) satisfy Chow's condition, i.e., Xl , X 2 and their iterated Lie brackets span the tangent space of JR3 at each point. The number of brackets plus 1 is called the step of the model, which in this case is 2(k + 1). By Chow's theorem, (see [22]) any two points can be joined by a piece-wise smooth horizontal curve. In the following we shall prove horizontal connectivity by a direct method. The following theorem is the generalization to 2(k + 1) step of a similar theorem from chapter 2. THEOREM 3.4. 1) Given a point P(Xl, X2, t) E JR3, there is a smooth horizontal curve which joins the origin and the point P. 2) Given two points P, Q E JR3, there is a piece-wise smooth horizontal curve which joins P and Q.

PROOF. 1) An explicit solution will be constructed. We shall analyze the following three cases: X2 # 0, Xl # 0 and Xl = X2 = O. In the case X2 # 0, set

xds) X2(S)

as 2 +bs, as 2, s E [0,1].

The constants a, b and a will be found later from the boundary conditions

t(l) = t. Hence a

+b a

Xl ===}

X2.

a =

Xl -

b,

3.2. HORIZONTAL CONNECTIVITY

The curve starts at the origin Xl (0)

87

= X2 (0) = t(O) = 0. We have

From the horizontality condition

i(s)

2(XI(S)2

+ X2(s)2)k (X I (S)X2(S)

((as 2 + bs)2

+ a 2s 4)

k

- XI(S)X2(S))

(-2abs 2)

k

-2abs 2

L Ck(as 2 + bs)2r a 2(k-r) s4(k-r)

r=O k 2r -2ab C"k a2(k-r) C~raPb2r-p] sP+4k-2r+2. r=O p=o

L

[L

Integrating,

t(s)

= t(O) -2ab ~ =0

Substituting a

k

2r

r=O

p=o

L C"ka 2(k-r) [L C~raPb2r-p] sP+4k-2r+3 /(p + 4k -

= Xl - b, a = X2 and taking s = 1 in the above relation, yields

(X2)2(~-r)+1

(3.4)

2r + 3).

k

=

p

~ ~ C"kC~r ,(Xl - b~b2r-p~/(p + 4k -

2r + 3).

degree=2r+1 The expression on the right side is a polynomial of degree 2k + 1, and hence the equation (3.4) has at least a real root b. One may substitute b back to find a. Then, for any given Xl, X2 =1= 0, t, there are constants a, a and b such that the boundary conditions are satisfied. The case X I =1= is similar. In the case Xl = X2 = 0, we need to find a horizontal curve joining the origin and the point (0,0, t). We shall assume t < 0. Set

°

XI(S) X2(S) t(s)

rcos(7rs),

r sin(7rs), ts, s E [0,1],

with the parameter r to be found from the horizontality condition

i(s) = -2Ix(s)12k¢(s) = -2r2k7r, where ¢

= 7rS. Then t(s) = -2r 2k 7rs, and hence making t = 1 yields

r=(~t;f/2k. For t

> 0, we consider XI(S) X2(S) t(s)

rcos(-7rs), rsin( -7rs),

ts,

s E

[0,1].


88

2(k + 1)

CASE

2) Any two points can be connected by a horizontal curve which passes through the origin. D

It is interesting to note that if the point P( xl, X2, t) is not on the t-axis, the xcomponents of the horizontal curve are quadratic in s.

3.3. SubRiemannian geodesics The principal symbol of the sub-elliptic operator ~x =

HXf + Xi)

is given

by

(3.5)

H(~,O,x,t) = ~(6 +2X21x12kOf + ~(6 - 2X11xl2kOf

and it will be called Hamiltonian. DEFINITION 3.5. A subRiemannian geodesic connecting the points P(xo, to) and Q(xf, tf) is the projection on the (x, t)-space of a solution of the Hamiltonian system

i=Ho,

with the boundary conditions x(O)

= xo, t(O) = to, x(sf) = xf, t(sf) = tf.

As dH/ds = aH/as = 0, the Hamiltonian is conserved along the geodesics. In the following sections we cannot assume sf = 1 and H = ~ simultaneously. If we use the arc length parametrization, we take H = ~, but in this case sf is left free. If the parametrization used is not by arc length, we may fix sf = 1 and consider the constant of energy H as a parameter. Euler-Lagrange system. The Lagrangian associated with the Hamiltonian (3.5) is given formally by the Legendre transform L(x, y, t, X, y, i)

+ 6X2 + oi ~(xi + x~) + O(i -

6X1

H 21x12k(X2X1 - X1X2)).

The Euler-Lagrange system of equations d aL ds aX1

aL aX1 '

d aL ds aX2

aL aX2 '

d aL ds ai

aL at

becomes (3.6) PROPOSITION 3.6. Any subRiemannian geodesic is a horizontal curve. There are horizontal curves which are not sub Riemannian geodesics.

3.3.

SUBRIEMANNIAN GEODESICS

89

PROOF. Let c(s) = (XI(S),X2(S),t(s)) be a subRiemannian geodesic. Then, using the Hamiltonian equations, we have [)H

2k

[)6 = 6 + 2X21xl (), [)H

2k

[)~2

= 6 - 2Xllxl (),

~~

=

(~l + 2X2IxI2k()) 2X21xl2k -

(6 - 2XllxI2k()) 2Xllxl2k

21x12k(XIX2 - X2XI), which is the horizontality condition (3.2). For the second part of the proof, we may check that the horizontal curves constructed in the proof of Theorem 3.4 do not satisfy the system (3.6), and hence they are not subRiemannian geodesics.

o PROPOSITION 3.7. Any sub Riemannian geodesic satisfies the system (3.6). PROOF. From the Hamiltonian system . () =

[)H

-7ft

=

- ;~ = -[Xl 2()[)x

6

0 ===> ()

1

=

constant.

(x2IxI 2k ) +X2(-2())[)Xl(XllxI2k)]

-2()kXIX22xllxI2k-2 + 2()X2 (lxl2k + 2kxilxI 2k - 2) -4()kxIX2XllxI2k-2

Xl

+ 2()X21x1 2k + 4()kxix 2IxI 2k - 2.

d ( 6 + 2X21xl 2k () ) ds

~l + 2X21x12k() + 2X2()klxI2k-2(2xIXI + 2X2X2) -4()kxIX2XllxI2k-2 + 2()X21x1 2k + 4()kxix 2IxI 2k - 2 +2X2IxI2k() + 2X2()klxI2k-2(2xIXI + 2X2X2) 4()lx1 2k X2 + 4()kx2IxI2k-2(xi + x~) 4()(k + 1)lxI 2k x2'

Similar calculations will lead to

o THEOREM 3.8. A horizontal curve which satisfies the system (3.6) is a subRiemannian geodesic. PROOF. Let s ~ (xds), X2(S), t(s)) be a horizontal curve. Consider

6 6 ()

Xl - 2()X21x1 2k X2 + 2()Xllxl 2k constant.

90


We shall show that

(Xl,

2(k + 1)

CASE

X2, t, 6, 6, B) satisfies the Hamiltonian system. Indeed, EJR

0=80'

iJ Xl

6 + 2Bx21xI

. X2

c 0,

0,

with (3.25) PROOF. Using

(2k+1)x 2(k+l) _ 2 fo sin 2k+l (v) dv Jik(X) - - 2 k1 ~ + sin 2k+l (x)

¥s = -2/Jr2k and Corollary 3.12 with p = k, yields 2k ( - (2k + 1)¢) . -d¢ = -2/J ( -1 ) 2~~1 sin2k+1 ds

2/J

Separating d¢

1

---;:-:------ = -(2/J)2k+1 ds

sin2;~1 ((2k

+ 1)¢)

,

3.5. GEODESICS STARTING AT THE ORIGIN

95

and integrating { O. Furthermore, since limx--+o f.L(x) = 0, we shall define f.L(0) = O. Indeed, applying I'Hospital rule

lim f.L(x)

x--+o

lim _2_ . x--+o

2k

+1

2(k+l)

+ 1) sin 2k+l ((2k + l)x) 2~~1~) sin2k~' ((2k + l)x) cos((2k + l)x) (2k + 1) (2k

. 2sin2k~'((2k+1)x) hm =0. cos((2k + l)x)

x--+o

LEMMA

3.14. (i) f.L satisfies the ordinary differential equation f.L'(X)

=2-

2(k

+ 1) cot((2k + l)x)f.L(x).

(ii) If Xc is a critical point for f.L( x), then f.L(x c)

=

1 -k- tan((2k +1

+ l)xc).

3.

96

THE GEOMETRIC ANALYSIS OF STEP 2(k + 1) CASE

PROOF. (i) Differentiating in equation (3.25)

I/'(x)

=

t""

_2_. (2k 2k + 1

+ 1) sin sin

4(Z+1)

2 +1

4(k+1) 2k+1

((2k + 1)x)

((2k

+ 1)x) (2k+l)x

1

_ _2_2(k+1)sin2kTf((2k+1)x)cos((2k+1)x)Jo 2k + 1 4(k+1) sin 2k+1 ((2k + 1)x) 2[1- (k

sin

2(Z+1) 2·+1

(v)dv

+ 1) cot((2k + 1)x)J.l(x)]

(ii) It is a simple consequence of (i).

o The next result deals with the second derivative of J.l. LEMMA 3.15. If Xc

> 0 is a critical point for J.l, then J.l"(x c) > 4(2k + 1) > 0,

and hence all the positive critical points are local minima. PROOF. Differentiating in relation (i) of Lemma 3.14

J.l"(x) =

+ 1)(1 + cot 2((2k + 1)x))J.l(x)] + cot((2k + 1)x)(2 - 2(k + 1) cot((2k + 1)x)J.l(x)) 2(k + 1) [ - (2k + 1)J.l(x) + 2 cot((2k + 1)x) - (4k + 3) cot((2k + 1)x)J.l(x)]. 2(k + 1) [ - (2k

The relation (ii) of Lemma 3.14 yields

J.l"(xc) =

-

2(k

+ 1) [ -

2k + 1 k + 1 tan((2k + 1)xc)

(4k + 3) k + 1 cot 2((2k

+ 2cot((2k + 1)xc)

+ 1)xc) tan((2k + 1)xc) ]

+ 1) [tan((2k + 1)xc) + cot((2k + 1)xc)] 2: 4(2k + 1), used the inequality a + ~ 2: 2, for a > O. In the above case a = tan((2k + 1)xc) = (k + 1)J.l(x c) > O. 2(2k

where we

o LEMMA 3.16. J.lk is a monotone increasing diffeomorphism of the interval ( - 2k~1' 2k~1) onto lR. On each interval (27::1' (r;:~{7r), m = 1,2, ... , J.lk has a unique critical point Xm. On this interval J.lk decreases strictly from +00 to J.lk(X m) and then increases strictly from J.lk(Xm) to +00. Moreover (3.26)

with q > 0 independent of m. PROOF. As J.l is odd and J.l(2k~1' -) = +00, it suffices to show that J.l is increasing on (0, 2:+1). As J.l(0) = 0 and J.l(x) > 0 for X > 0, J.l is increasing on a right neighborhood (0,8), with 8 > 0 small enough. Let

I = {x E

7r

(0, -k--);J.l 2 +1

increasing on I}.

97

3.6. CARNOT-CARATHEODORY DISTANCES FROM THE ORIGIN

I =I- 0 because (0,8) c I. The set I is an interval, because otherwise, there are a < c < b with a, bEl and c tf- I. JL is increasing at a and decreasing at c and hence it has a maximum point in the interval (a, c), which is in contradiction with Lemma 3.15. Let 'Y = sup I. Suppose 'Y =I- 2k~I' There is an a such that 'Y < a < 2k~1 with JLI (O,'Y) increasing and JLI b,a) decreasing. Then there is a local maximum for JL on

b,a), contradiction. It follows that 'Y = 2k~1 and I = (0, 2k~1)' Hence JL is a monotone increasing diffeomorphism of the interval ( - 2k~I' 2k~l) onto R As JL( 27::;1' +)

= JL( (r;::tr ,-) = +00,

there is at least a critical point in the in-

(r;::t'r).

terval (27::;1' As all the critical points are local minima, the critical point to be unique (more than one critical point will imply the existence of a local has maximum, contradiction). We shall compute a lower bound for the difference between two successive minima. r(2k+l)x+1I". 2(k+1) ( ) d 2 _ . Jo __ sIn 2k+1 v v 2k + 1 2(k+1) sin 2k+1 ((2k + l)x + 1f)

1l" + J;2k+1)x+1I"

2

J0

2k + l '

2(k+1) sin 2k+1

~

sin

2k+1

((2k

q

((2k

+ l)x)

+JL(x),

+ l)x)

and hence

JL (x

+ 2k: 1)

- JL( x) > q > O.

The constant q is independent of m and it is evaluated below: q

2

--2k + 1 4

-2k + 1

111" sin 2(k+1) 2k+1 (v) dv =

11

4 ---

0

t

2(k+1) 2k+1

2k

+1

111" /2 sin 2(k+1) 2k+1 (v) dv 0

(1 - t 2 )-1/2 dt = -22k

0

+1

11

_1_

X2(2k+ 1) (1- X)-1/2

0

dx '

and hence q can be expressed in terms of beta functions as

q

=

2k

2

+1 .B

(4k 4k

+ 3 1) + 2' 2 > O.

The values of JL are shifted up by at least q when JL changes from (27::;1' to

(m+2)1I") ( (m+l)1I" 2k+I' ~.

(r;::i1l")

H ence the mInImum .. . WI'11 b e s h·ft I ed up b y t a leas t q, z.e.,

JL(xm+d - JL(xm) > q. Note that limm->oo JL(xm)

=

+00.

o

The function JL has a similar behavior for all k. The graph of the function JL in the step 2 case (k = 0) is given in Figure 1.2. 3.6. Carnot-Caratheodory distances from the origin We shall study the connectivity and the length of the geodesics starting from the origin. The complete picture is given in theorems 3.17 and 3.18.

3.

98

THE GEOMETRIC ANALYSIS OF STEP 2(k + 1) CASE

THEOREM 3.17. There are finitely many geodesics that join the origin to (x, t) if and only if x=/:. o. These geodesics are parametrized by the solutions 'l/J of:

It I IlxI12(k+1)

(3.27)

= p('l/J) ,

There is exactly one such geodesic if and only if:

It I < p(xt} Ilxll2(k+1), where Xl is the first critical point for p. The number of geodesics increases without I boun d as IlxI1 2It(k+1) - ; 00. If 0 :s: 'lj;1 < ... < 'l/J N are the solutions of equation (3.27), there are exactly N geodesics, and their lengths are given by

s~k+l)

(3.28)

=

I/('l/Jm) (It I + IlxI12(k+1)),

where [ fa(2k+1),p sin- 2k2k+1 (v) dv ] 2(k+1) (3.29)

I/('l/J)

=

+ 1)2(k+1) (1 + p('lj;)) sin

(k

3.0±U' 2k+1

+ 1)'lj;)

((2k

PROOF. The enumeration of geodesics follows from Lemma 3.16. In order to compute the lengths, multiply equation (3.22) with equation (3.21) raised to power 2(k + 1) to eliminate

e

(3.30)

(sm)

2(k+1) _ -

[r(2k+1),pm . (k) (_1_)2 +1 2(k+1) sm 2k 1 R

_..2l£.... (

v

)

d ] 2(k+1)

Jo 2k+1 V -"-"--'''------;:-2;.,-( k-:-+I"'")------'--'-------=-----

+

sin

((2k

2k+1

+ l)'l/Jm)

On the other hand, (3.24) yields

It I + R 2(k+1)

=

(1

+ p('l/Jm) )R2(k+1).

Then (3.31)

(Sm)2(k+1) =

1/( 'l/Jm)

(It I + R 2(k+1») =

1/('l/Jm)(l

+ p('l/Jm))R 2(k+1).

Choosing 1/ as in formula (3.29), the right side of the equations (3.30) and (3.31) are the same.

o

The graph of the function 1/ in the step 2 case (k = 0) is given in Figure 1.4. We note that the function 1/ is not increasing. But the values at 27:";1 are increasing as m increases. THEOREM 3.18. The geodesics that join the origin to a point (0, t) have lengths Sl, S2,"" where m )2k+1 M 2(k+1) It I (s )2(k+1) = ( - (3.32) m 2k + 1 Q'

with the constants M and Q expressed in terms of beta functions M

B( 4k 1+ 2'"21) '

Q

4k + (2k+1)q=2B ( 4k+2'"2'

3 1)

3.6. CARNOT-CARATHEODORY DISTANCES FROM THE ORIGIN

For each length PROOF.

Sm,

When


Ilxll

99

§1.

= 0, the equation

is replaced by J..L('IjJ) = +00. This equation has solutions 'ljJm = 2;:;1' m = 1,2, .... The function J..L has vertical asymptotes at these points. We obtain infinitely many geodesics joining the origin with (0, t) with lengths given by formula (3.28)

(s

(3.33)

m

)2(k+1) = v(~) Itl. 2k+ 1

In the following we shall compute v (27::;1) in terms of beta functions. Using formula (3.29) and the definition of the J..L function, after cancellations we get fm7r [ Jo

.

~( )

sm 2k + 1 v 2

d ] 2(k+1)

v

m7r

2(k+l) 2k+l

+ 1)2(k+1). 2k+l fo sin (v)dv (2m)2(k+1) [f07r/2 sin~ (v) dv] 2(k+1) (2k

7r

2(2k + 1)2k+l . m fo sin

= (~)2k+1.

(3.34)

2k

[ f7r /2

-2k (

sin2kTf v) dv

Jo

+1

2(k+1) 2k+l

7r

fo sin

2(k+l) 2k+l

]

(v) dv

2(k+l)

(v) dv

The bottom integral was found as a part of computation of q in Lemma 3.16

1 7r

o

sin

2(k+l) 2k+l

(v) dv

=

3 1)

(4k + 4k + 2' 2

B --- - .

With the substitutions v = arcsin u and t = u 2 , the top integral becomes {7r/2

10

-2k

sin 2k + 1 (v) dv -1 2

11

C

4k+l 4k + 2

0

1 1 ( -1- 1 ). (1-t)-2dt=-B 2 4k + 2' 2

Hence (3.34) yields

(

2m )2k+l 2k

(

+1

~)2k+l

2k+ 1

B

1

(

22k+2

~,~

)

B (4k+3 4k+2'

2(k+1)

l) 2

M2(k+1)

Q

Substituting in formula (3.33) we get the distances (3.32).

o

100

3.

THE GEOMETRIC ANALYSIS OF STEP

2(k

+ 1) CASE

3.7. Particular cases

=

Step 4 case. When k ( 3.35 )

Xl =

fj -fj Xl

1 the vector fields

fj + 2X2 IX 12 -fj' t

X2 =

fj 12 fj - f j - 2Xllx -fj' t

x2

define a step 4 model. There is no Lie group law on lR,3 underlying these vector fields. Theorems 3.17 and 3.18 in this case are: THEOREM 3.19. There are finitely many geodesics that join the origin to (x, t) if and only if x t- o. These geodesics are parametrized by the solutions 'I/J of:

It I IIxl14 = p('I/J) ,

(3.36)

There is exactly one such geodesic if and only if:

It I < p(xdllxI1 4 . The number of geodesics increases without bound as 1I~1114 ----* 00. If 0 ::::; 'l/Jl < ... < 'l/JN are the solutions of equation (3.36), there are exactly N geodesics, and their lengths are given by

(3.37) where f3x [ Jo

v(x) =

(3.38)

. - a( )d ]4 sm 3 v V 4'

16(1

+ p(x)) sina(3x)

and

() _ 2 _ p (x ) - PI X - 3

(3.39)

J03X

sin! (v) dv 4 • sina x

THEOREM 3.20. The geodesics that join the origin to a point (0, t) have lengths where

Sl, S2, ... ,

(3.40) with M = B ( ~, ~ ). For each length Sm, the geodesics of that length are parametrized by the circle §l .

PROOF. Using the properties of beta functions and taking k

M

B(~ ~) = r(~)r(~).

Q

2B(~ ~) = 2r(~)r(~) = ~ ra)r(~)

6'2 6' 2 M 2

Then

r(~)

r(~)

2

r(~)

= 1,

yields

3.7. PARTICULAR CASES

101

Substituting k = 1 in equation (3.32), yields

(3m)3M4 crltl = (m)3 3 . 2M3 Itl,

(sm)4 = which leads to (3.40).

D

The geodesic of length Sm intersects the t-axis between (0,0,0) and (0,0, t) m - 1 times. The projection of the geodesics on the x-plane consists of loops which return back to the origin after a clock-wise rotation of ~. The projection is tangent to a circle which narrows down as m increases. The radius of the circle is given by the following proposition: PROPOSITION

3.21. The m-th geodesic between the origin and (0, t) is tangent

(2~b) 1/4, where Q = 2B(i, ~).

to a cylinder centered on the t-axis with the radius PROOF.

The solution in polar coordinates satisfies

r( -S~~J ~os '" sin 2 'Ij;

'lj;2 2 sin 'Ij; + 'Ij; - sin 'Ij; cos 'Ij; ,

o

which is (3.44).

When k = 0, the constants M and Q can be computed explicitly. Hence we recover Theorem 1.55 if chapter 1. Since the proof is different, we state it as a new theorem as follows. THEOREM 3.23. The geodesics that join the origin to a point (0, t) have lengths S1, S2, ... , where (3.46) For each length

Sm,


PROOF. Substituting k = 0 in Theorem 3.18 yields

M

B(~ ~) = r(~)f(~) =

Q

2B 2'2

f(l)

2'2

(~

(y'7r)2 1

~) _ f(~)f(~) _

-2r(~+~)

= 7r '

~f(~)2 =

-2 f(2)

7r.

§1.

3.7. PARTICULAR CASES

103

Then equation (3.32) becomes

o Corollary 3.12 in this case becomes

with Tmax = 210. The projection on the x-plane is a circle of radius 1/(20). In this case the solution reaches the t-axis again after a clock-wise rotation (for 0 > 0) of angle 2w = 1r. The m-th geodesic between the origin and (0, t) intersects the t-axis m - 1 times (without counting endpoints). Because of the rotational symmetry, we may always choose the initial argument ¢o = O. After m loops the angle is ¢1 = m7r. The equation (3.23) yields

We arrive at: PROPOSITION

3.24. The m-th geodesic between the origin and (0, t) is tangent

to a cylinder which contains the t-axis and has the base diameter

(

~~ )

1/2

.

Note that in the step 2 case, the cylinder contains the t-axis, while in step 2(k + 1), k ::::: 1, the cylinder is centered at the t-axis. The projections on the x-space for three different values of m = 1,2,3 in the case of steps 2,4,6 and 8 are given in Figures 3.1 - 3.4.

-0.6 -0.8

Figure 3.1: x-projection in Heisenberg case.

Figure 3.2: x-projection in step 4 case.

104

3. THE GEOMETRIC ANALYSIS OF STEP 2(k

+ 1)

CASE

Figure 3.3: x-projection in step 6 case. Figure 3.4: x-projection in step 8 case.

The number of loops . The number of total distinct loops depends on the integer k and is denoted by N(k). We have the following result. PROPOSITION

3.25. The number of loops is given by

N(k) = 2k + 1. PROOF. The angle of a loop is 2k~1 = 4i~2' which means there are 4k + 2 such angles around the origin. The angles divide the plane in 4k + 2 sector areas as in Figure 3.5.

3 4

5

/

/. 1

4k+2

Figure 3.5: 4k + 2 sector areas. All the sector areas between r2k(¢) d¢. Jc/>o

o 3.9. The use of the hypergeometric function

In this section we shall compute the length of geodesics between (0,0,0) and (0,0, t), using the arc length parametrization, i. e., xi + x~ = 1, and hypergeometric functions. We recover the formula for the lengths obtained in the previous section. The Hamiltonian

H(~,(),x,t) = ~(6 + 2X2IxI2k())2 + ~(6 - 2XllxI2k())2, is preserved along the geodesics. Using the Hamiltonian equations

and considering the arc length parametrization, the Hamiltonian becomes H

1(.2

.2)

1

= 2 XI + x 2 = 2'

In polar coordinates xi + x~ = r2 + r2if;2 = r2 -2()r2k. The radius r(s) verifies the equation

r2

+ 4()2 r 2(2k+l)

+ 4()2 r2(2k+I),

where we used

if; =

= 1.

This can be written as

As the geodesic starts at the origin, r(O) sign in the right side. Integrating

res) Jo VI The substitution u

=

= O.

Then we shall consider the positive

dx 4()2x 2(2k+l)

= s.

x 2 yields

r

2

Jo

(s)

du

Ju(l _ 4()2 u 2k+l) = 2s.

3.9. THE USE OF THE HYPERGEOMETRIC FUNCTION

Let v = u(402)2k'f-l. Then du becomes

l

(3.49)

= (40 2 )-2k'f-l dv and the above integral equation dv

o. Then the angle ¢ moves clock-wise, i.e., decreasing. After one complete loop we have ¢(2Sl) = - 2k:1' and hence after m loops

mn ¢1 = m¢(2st} = - 2k + 1· Integrating between ¢o and ¢1, yields

Using the boundary conditions for t(s)

t(¢o)

=

t(¢l)

0,

=

t,

and the formula in polar coordinates for the solution given by Corollary 3.12, we have t

=

-2r~!2

1

¢1

sin((2k

+ 1)(¢ -

~

¢o)) 2k+l d¢.

¢o

Using the substitution v = (2k

=

t

+ 1)(¢ -

¢o) yields

2r2k+21m7r 2k+2 2mr 2k + 2 1 7r 2k+2 mQr 2k +2 sin(v)2k+l dv = max sin(v)2k+l dv = max 2k + 1 0 2k + 1 0 2k + 1 '

~

and hence

2k+2 _ (2k + l)t rmax mQ '

r 7r sin (v) ~ where Q = 2 Jo 2k+l dv. Substituting in the equation (3.54) we obtain the following result. THEOREM

3.27. The lengths of the geodesics joining the origin and the point

(0,0, t) are (3.55)

£;/;+2 = (2k

+ 1) m

2k+1 (2F)2k+2

Q

Itl.

3.9. THE USE OF THE HYPERGEOMETRIC FUNCTION

109

The Carnot-Caratheodory distance between (0,0,0) and (0,0, t) is £1 with

£ik+2 = (2k

(3.56)

+ 1) (2F~k+2Itl.

The above formulas are consistent with the formulas obtained in Theorem 3.18 if we take M = 2(2k + l)F. This provides a formula for F in terms of beta functions.

°

The constant F. In the case k = the constant F is related to the period of the sine function while in the case k = 1, it is related to the period of the elliptic function sn. Indeed, making o'(s) = 1 in equation (3.50), yields (3.57) For k

2F =

r JV(l dv- v2k+ 1

io

1)

= 0, in the step 2 case 2F

=

r dv = arcsin(2v _ 1)l io JV(l - v) 1

v

=1 v=O

= Jr,

i.e., half of the period of the sine function. For k = 1, in the step 4 case

2F -

r

1

- io

__ 2 K( J2 ) 31 / 4 2(1 + v'3)

dv

JV(l - v 3 )

-

,

where K(",) is the complete elliptic integral defined by

r

1

K(",)

=

io

dt

J(l - t2)(1 _ ",2t 2)

In the above case", = 2(1fv'3)' and K (

2(1

J2 v'3 ) + 3)

~ 1.598.

2F is proportional to 2K(",) which is half of the period of the elliptic function sn( . ,,,,). For the general step case, we shall write F as an infinite numerical series. PROPOSITION

3.28. We have 00

F-"~ - L... 2nn! n=O

where an PROOF.

series

=

TI;:~ (1 + 2p) 1 + 2n(n + 1)

.

The Barnes' extended hypergeometric function can be written as a

TIP r(n,+k) " t=O r(ni) k ([ ) [l 1 F n, d ,Z = L... k! q ~ Z k=O TIt=o r(d 00

i )


110

2(k + 1)

CASE

where [n] = [n1, ... ,n p ] and [d] = [d 1, ... ,dq ]. In our case p = 2 and q = 1. Then 1 1 1 d 1 =1+ 2(k+l) n2 = 2(k + 1)' Using r(z

+ 1) = F

zr(z), we have

=

F([~'2(k~I)],[1+2(k~1)],I)

f

~ r(~ + k) r(~ + k) r(~) r(2(k~1))

k=O k! 1 r(~

00

+ k)

~k! r(~)

+ ~) r(1 + 2(k~1) + k) r(1

1

·1+2k(k+l)"

Using

G+

k-

1) G+ k - 2) ... G+ k -

k

)rG)

I1;:~ (1 + 2p) r (~) 2k

2 '

yields the formula for ak.

o 3.10. Geodesics starting outside the origin

Recall Euler-Lagrange system of equations f = r¢(¢ + 4(k + l)Or2k) { r2(¢ + 20r 2k ) = C(constant)

(3.58)

o= 00 (constant).

The solutions with r(O) the initial conditions

-I- 0 will be discussed.

(3.59)

r(O)

The system (3.58) is considered with

-I- 0, (0) = 0·

The second equation of (3.58) becomes

¢= ~ -

20r2k. r Substituting in the first equation of (3.58), one obtains (3.60)

(3.61)

f

= C: + 4kOCr 2k - 1 - 40 2(2k + 1)r4k +1. r

Let (3.62)

Then, one may rewrite equation (3.61) as Newton's equation

(3.63)

r = -V'(r).

3.10. GEODESICS STARTING OUTSIDE THE ORIGIN

Substituting p

= r, equation

111

(3.61) becomes dp dV p---dr dr'

which yields

p2 2=-V(r)+E,

where E is the constant of the total energy. PROPOSITION 3.29. If r is a solution of the Euler-Lagrange system with initial conditions (3.59), then the energy is conserved:

1,2

(3.64)

-r 2

+ -21 (2Br

2k + 2 _C)2 r

= E.

We shall give a qualitative description of the solutions of (3.64). There are two cases: sgn(C) = sgn(B) and sgn(C) i= sgn(B). In each case we shall draw the graph of the potential energy V and the trajectory of the solution in the phase plane.

Case sgn(C) = sgn(B). In this case V(r) has a zero and a global minimum at

_ (C)

(3.65)

and limx-+o+ V(r)

ro -

1/2(k+l)

2B

'

= limx-+ oo V(r) = 00.

Equilibrium points: ro corresponds to a stable equilibrium point, V'(ro) = 0 and V"(ro) > O. The corresponding ¢ satisfies j _ 'f' -

2(k+1)

C _ 2B 2k _ C - 2Bro 2 ro 2 ro ro

=0

and hence ¢ is constant. This means that each point of the circle 8(0, ro) is an equilibrium point (constant solution). At each of these points the energy E = O.

Tangent circles: When E > 0, there are exactly two positive roots rmin and rmax of the equation (3.66)

V(r) = E,

such that (3.67)

o < rmin :::; ro :::; rmax.

PROPOSITION 3.30. The solution of the Euler-Lagrange system (3.58) in the case sgn(C) = sgn(B) lies in the circular ring W(O,rmin,r max ) = {x E ]R2jrmin:::;

Ixl :::; rmax}. PROOF. r is equal to rmin or rmax when r = O. The equation of conservation of energy (3.64) shows that rmin and rmax are solutions of the equation V(r) = E and the solution r has the property

o < rmin :::; r :::; rmax·

o

112


Existence of loops: The equation

¢=

(3.68)

r

2(k + 1)

CASE

0 has the positive root

=(C)~ 20 .

Hence ¢ = 0 if and only if r = roo When the solution intersects the circle of equilibrium points S(O, ro), the sign of ¢ changes, i.e., the trajectory is bouncing back, making loops in the ring W(O, rmin, r max ), see Figure 2.8.

The ring width: w = rmax - rmin is a function of energy w = w(E). Using properties of the function V (r), we find (i) w(E) is an increasing function of energy E; (ii) w(O) = 0 if and only if rmin = ro = rmax; (iii) limE-->oo rmin = 0, limE-->oo rmax = 00. The t-component: From the Hamiltonian equation .

t =

8H 8B

=

21xl 2k (X2 X1 -

X1 X2) = -2r

2(k+1) . ¢,

¢ = 0 on the circle S(O, ro) implies that £(s) changes sign and reaches the local maxima and minima when the geodesic crosses the cylinder S(O, ro) x R Case sgn(C)

# sgn(O), C # o. Here . C - 20r 2(k+1) ¢= 2 # 0,

r which means that ¢ increases (decreases) for 0 < 0 (0) 0) and hence there are no loops. In this case the potential energy is strictly positive 1 (20r 2k +2 - C)2 r > 0,

2

V(r) =

and has a minimum at (3.69)

r1 =

-c )~ ( 20(2k + 1) .

Indeed, V'(r) = 0 if and only if

402(2k + 1)r 4kH - 4kOCr 2k +2 - C 2 = O. Substituting u

=

r 2k +2 we obtain a quadratic equation in u 40 2(2k

with the roots

U1

+ 1 )u 2 -

C = 20

4kOCu - C 2 = 0,

and U2

C = - 20(2k + 1) .

As r1 is positive, we choose

(r )2(k+1) _ _ 1

-

C 20(2k + 1)

The equilibrium solution. The equilibrium solution is the circle of radius r1 centered at the origin. The corresponding ¢ can be obtained from

d(¢ + 20r~k)

= C,

3.11. GEODESIC COMPLETENESS

113

and it is equal to

cp(S)

- 2()r12k) + CPo·

( rrc

=

Bounds for r(s): There are two positive roots Pmin and Pmax for the equation

V(r) = E , and 0

< Pmin :::; r1

:::; Pmax.

In the phase plane the solution is rotating around the stable equilibrium point (r1,0) such that

Pm in :::; r(s) :::; Pmax· The projection ofthe geodesics in the x-plane is contained in the ring W(O, Pmin, Pmax). The width of the ring increases as the total energy E increases, where E = ~ Ix(o)j2. See Figure 2.11. The t-component: In this case i = _2r 2(k+1)¢ i= O. Hence t(s) increases if () > 0, decreases if () < 0, it is constant if () = O. 3.11. Geodesic completeness DEFINITION 3.31. If for any point P, any geodesic c(t) emanating from P is defined for all t E ]R, the geometry is called geodesic ally complete.

The main result of this section is the following theorem. THEOREM

3.32. The geometry induced by the vector fields (3.1) is geodesically

complete.

We note that geodesic completeness does not hold in general. A counter-example in the step 2 case is provided in Calin [4]. To prove Theorem 3.32 we need a few results regarding extendability of solutions of differential equations. The following result can be found for instance in Hartman [14]. 00,

LEMMA 3.33. Let f(t, y) be a continuous function on a strip to :::; t :::; to y E ]Rd arbitrary. Let y = y(t) denote a solution of

(3.70)

y' = f(t,y),

+a
0, there are finitely many solutions of the

3.43. Given E

equation (3.89) if and only if r PROOF.

=I O.

Equation (3.89) can be written as

~=

(3.90) Let '\(x)

2T2E

= (si~X)2.

(sin(21'h))2 20T

The graph of function '\(x) is given in the Figure 3.7.

0.04 0.03 0.02 0.01

15

x

20

25

30

Figure 3.7: The graph of >.(x).

If r > 0, the equation (3.90) has finitely many solutions, because limx-->oo '\(x) = O. 0

In the case r = 0, the equation (3.90) yields sin(20T) = 0 ===> 20T = k1r k E Z 20T ' and hence there are infinitely many solutions k1r 0= Ok = - . 2T

122


2(k + 1) CASE

The final computation of the action is given below. Using the equations (3.89) and (3.88)

S

J~ -

(}t

+r

(}t

+r

(}t

+ (}r2

(}t

+ (}r2 cot (2(}T).

(}2r2

(}2 2

_-,,--r__ _ (}2r2 sin 2 (2(}T)

1 - 1 sin 2 (2(}T)

Hence, the classical action in the Heisenberg group case is

S

(3.91 )

= (}[t + r2 cot(2(}T)],

where (} is a solution of the equation studied in chapter 1

t

(3.92)

2 = p,(2(}T), r

where p,(x)

=

x

cotx.

-.-2- -

sm x From equations (3.91) and (3.92) we have

S

= (} [t + r2 (

. ;(}T

sm (2(}T)

- p,(2(}T))]

(} (t + sm.2~Tr2 _ r2 P,(2(}T)) (2(}T) (} ( t+

2(}Tr2

sin 2 (2(}T)

-t

)

2(}2Tr2

sin 2 (2(}T )" We obtained the following result. PROPOSITION

3.44. The classical action on the Heisenberg group is given by 2(}2Tr2

S = -----;,,--:---:sin 2 (2(}T) , where (} is a solution of (3.92). For each solution (} of (3.92), the action S has infinitely many singularities for Tk = ~;, kEN. 3.15. Exercises EXERCISE

3.1. Let

= OXl + 2X21x12kOt, two vector fields with Ixl 2 = xI + x~. Xl

be

Show that

3.15. EXERCISES

123

form a basis of T(l,O) ffi T(O,l) of the surface ankH = {(Zl' Z2) E C 2 : Im(z2) = IZlI2(kH)}.

Here we use the coordinates Zl = Xl EXERCISE

+ iX2

and Re(z2) = t.

3.2. Show that the hypersurface an k+ l is equivalent to the "ellip-

soid" Ek+l = {(Wl,W2) E C 2 : IWll2(k+l)

+ IW212

=

l}

via the "generalized Caley transform":

3.3. Show that 2 2 2_ a a2 ( 21 X12k+2 -a a 2 + 4k ( + 1)1 X12k (X2a - X la- ) -a' a Xl+X2-a2+a2+4k+l) 2 a Xl a X2 Xl X2 t t and 2k 2 2 2 12k [X2' Xd = 4k(k + l)lxl - (Xl + X2) at + 4(k + l)lx at' EXERCISE

a

EXERCISE

F(f)(x, r)

a

3.4. We exploit the partial Fourier transform in the variable t

-

= f(x, r) =

Joo. e-zrtf(t)dt, -00

Show that F

( aatf ) (r) =

f(x, t)

=

Joo eztr. f(x, - r)-. dr -00

irf(r),

Consequently, (Llf)(r)

2 2 = -1 [ -a + -a

2 aXfax~

a a +4i(k+ 1)rlxI2k(X2- -XI-) aXl aX2

-4(k + Ifr2IxI4k+2]/

Here Ll

=

Xl + xi is the sub-Laplacian.

2n

CHAPTER 4

Geometry on Higher Dimensional Heisenberg Groups 4.1. The Heisenberg group Hn In this chapter we shall study the behavior of subRiemannian geodesics on higher dimensional Heisenberg groups, which are step 2 subRiemannian manifolds. n ) x lR t be endowed with the following non-commutative group Let lR 2n + l = lR 2Cx,y law n

(4.1)

(x, y, t)

(x', y', t') = (x

0

+ x', Y + y', t + t' + 2 L

aj (xjYj - Xjyj)),

j=l

where x = (Xl,oo.,X n ), x' = (x~,oo.,x~), Y = (Yl,oo·,Yn), Y' = (Yi, ... ,Y~) and aI, ... ,an are positive constants. The Lie group Hn = (lR 2n +l, 0) is called the 2n + 1 dimensional Heisenberg group. In this chapter we shall deal with the geometry induced by the following 2n vector fields

(4.2)

Xj

a

a

a a 1j = aYj - 2ajxj at'

= aXj + 2ajYj at'

j

= 1, ... , n.

The vector fields (4.2) are linearly independent, because the matrix coefficients

(In On

(4.3) has rank 2n, where

2ay ) -2ax

On In

( ),

() Yl

Xl

X=

y=

Xn

Yn

Since the vector fields Xl"'" X n , Yl

, ... ,

Yn and their first brackets

a

[X j ,1j] = -4aj at

span T(lR 2n + l ), then the induced geometry is step 2 at every point. Let (a, b, s) E Hn be arbitrary fixed. The left translation LCa,b,s) : Hn is defined as

LCa,b,s)(X, y, t) = (a, b, s) 0 (x, y, t). f on Hn is defined by

The left translation of function

(4.4)

(Lca,b,S)f)(x,y,t)

=

f((a,b,s) 125

0

(x,y,t)),

V(x,y,t) E Hn.

----t

Hn

126

4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS LEMMA

4.1. The vector fields Xi, Yi, i

Xj(L(a,b,s)f) "Yj (L(a,b,s)!) PROOF.

=

= 1, ... ,n are left invariant, i.e.

L(a,b,s)(Xjf), L(a,b,s) ("Yj f),

'V f function.

An explicit application of the chain rule yields

Xj(L(a,b,s)f)(X,y,t)

=

(O~

+2ajYj%t)f((a,b,s)o(x,y,t)) J

(O~ + 2ajYj %t)f( a + x, b + y, s + t + 2 Lai(bixi -

Yiai))

i

J

fXj ((a, b, s)

0

(x, y, t))

+2ajYjft((a, b, s) fXj ((a, b, s)

0

0

(x, y, t))

(x, y, t))

(x, y, t))

L(a,b,s) (fxj (x, y, t)

+ 2ajbj ft ((a, b, s) 0

+ 2aj (b j + Yj )ft (( a, b, s) 0

(x, y, t))

+ 2Yjft(x, y, t))

L(a,b,s) (Xjf)(x, y, t).

o

The proof of the second relation is similar.

It follows that we need to study the subRiemannian geodesics starting from the origin only. A left translation will provide geodesics starting at any other points.

4.2. Hamiltonian mechanics on Hn We shall relabel the vector fields (4.2) as follows (4.5)

X 2j -

1

a OX2J -1

a ut

a a X 2j = -;::;-- - 2ajX2j-1~' UX2j ut

= - - - + 2aJx2J~'

Setting

we have

A+AT A2

=

0,

= diag( -4ai, -4ai, -4a~, ... , -4a;').

The vector fields (4.5) may be denoted collectively by

a

X = ax

a

+ Ax at ,

and the group law (4.1) may be written in the form

(x, t)

0

(x', t/) = (x

+ x' , t + t' + (Ax, x) ) ,

where 2n

(x, y) = L XjYj· j=1 We consider the Hamiltonian mechanics associated to the symbol of -AH:

127

4.2. HAMILTONIAN MECHANICS ON Hn

H(x,~,

0) 1

2(~

+ OAx, ~ + OAx)

1

2((' (), where (

=

~

+ OAx.

(x(s),t(s),~(s),O(s))

(4.6)

In this notation the Hamiltonian equations for a curve take the form:

8H

x(s)

7i[

~(s)

8H -ax

i(s)

8t = (((s), Ax(s)),

O(s)

8H . 8t = 0, Le.,

((s),

=

=

OA((s),

8H

O(s) = 0(0),

where the dot denotes d/ds. We let the parameter s run from 0 to T > O. Because of the group invariance we shall study the paths relative to the origin and a point (x, t). The boundary conditions are (4.7)

x(O)

X(T)

= 0,

=

O(s) = O.

x,

The solutions (x(s), t(s)) of the Hamiltonian's system satisfying the boundary conditions (4.7) will be called subRiemannian geodesics. As usual, the Hamiltonian itself is constant along a curve satisfying (4.6):

1

H(x(s), ~(s), 0) == Ho == 2(((0), ((0)).

(4.8)

The following theorem provides the subRiemannian geodesics on H n , see Beals Gaveau and Greiner [3]. THEOREM

(4.9) x(s)

t - t( s)

(4.10)

4.2. The solution of (4.6) with the boundary conditions (4.7) is e(

s-r)IIA

sinh(sOA) X' sinh(TOA) ,

T - S ( (BA)2 X x) _ ~ ([sinh(2TBA) - sinh(2s0A)]OA x x) 20 sinh2(TOA)' 40 sinh2(TOA) , 2 n 2a 0 (T - s)" ) r2 L sin2(2a)'TO) ) )=1 _ ~ aj sin(4ajTB) - sin(4ajsO)r 2 L 2 ' sin (2ajTO) ) j=l 2

2

2

The value of the Hamiltonian H on this path is:

(4.11)

n 2a;02 2 x,x)=L 2 r·. 2 sinh (TBA) j=l sin (2ajTO) )

1

Ho=-(

(BA)2 2

2

rj =X 2 j-1 +X2j'

128

4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS PROOF.

The equations (4.6) imply that

+ BAx(s)

((s) = ((s)

= 2BA((s),

so one can solve for x(s) as a function of x, T and B, and then solve for t(s) as a function of x, t, T and B. We have

((s)

(4.12)

e2sllA ((0),

1 s

x(s)

= (2BA)-1{ e2sOA - I}((O)

(2BA)-1{((s) - ((O)},

(4.13)

t(s)

(4.14)

((r) dr

t(O)

+ iot

t(O)

+ OHo -

t(O)

+ OHo -

(((r), Ax(r)) dr = t(O)

Iiot (((r), ((r) + 2B

- ((0)) dr

sI t 2B io (x(r), ((0)) dr s

1 2B (x(s), ((0)).

Then

(4.15)

((0)=2BA(e

2rllA

-1)

-1

BA e -rliA x=sinh(TBA)x,

( ) _ (s_r)llAsinh(sBA) x s - e sinh( TBA) x.

We note that in the case B = 0 we have

((s) = ((0),

x(s) = ((O)s and t(s) = t(O).

Using that in any expression

(F(A)x, x) we may replace F by its even part

~[F(A) + F(At)] = ~[F(A) + F( -A)]. yields (4.16)

(x(s), ((0)) =

~(s~nh(2sBA) BAx, x). 2 smh2(TBA)

Using (4.17)

exp

(_~ ~) (_~~~~ ~~~), o

relations (4.8), (4.12) - (4.16) yield (4.9) - (4.11). REMARK 4.3. It is interesting to note that as (x, t) =I- (0,0). COROLLARY

T

=I- 0 implies that t(O) =I- 0 as long

4.4. We have n

t - t(O)

= L ajJ.1(2ajTB)r;, j=l

where we have set r; = X~j -1

+ X~j

and J.1( 'P) =

sin'l2 'P -

cot 'P.


129

4.3. The classical action The action integral associated to the Hamiltonian curve is

S(x, t, T; B)

=

for {(~(s), x( s)) + Bi(s) -

H(x(s), ~(s), B)} ds.

Because H is homogeneous of degree 2 with respect to

S=

{r

io

oH

oH

{(~'8[)+B8e-H}ds=

{r

io

(~,

B), we observe that

(2H-H)ds= THo,

where Ho is the constant value of the Hamiltonian along the solution. Using Theorem 4.2 we obtain various forms for the action. THEOREM

4.5. The action integral S(x, t, T; B) is given by:

S(x, t, T; B)

1

"2 (T( BA)2 [sinh( TBA)t2x, x) 1

[t - t(O)]B + 2(BAcoth(TBA)x, x)

n

[t - t(O)]B + L ajB cot(2ajTB)r;. j=l

4.4. The horizontal distribution We shall work with the skew-symmetric model

X

=

Ox

+ AXOt,

which means

2n j = 1, ... , 2n. Xj = OXj + 2 L ajkXkOt, k=l The horizontal distribution is defined by H = span{ Xl, ... , X 2n}. A curve is called horizontal if its velocity vector belongs to the horizontal distribution. Let c(s) = (Xl (s), X2 (s), ... ,X2n (s), t(s)) be a curve in 1l~2n+l. The velocity vector field is

(4.18)

2n (X1, ... ,X2n,i) = LXjOxj +iot j=l (x, ox) + iOt = (x, Ox + Axot - Axot) (x, X) - (x, Ax)ot + iOt.

+ iOt

We arrive at the following result: LEMMA

(4.19)

4.6. A curve c = (x, t) in lR. 2n + 1 is horizontal if and only if

i = (x, Ax).

130

4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS PROOF.

= '£XjXj

Since (X,X)

c=

E 1i, the equation (4.18) yields

(x, X)

+ (i -

(x, Ax) )at.

Since at t/:- 1i, it follows that c is horizontal if and only if the coefficient of at vanishes, i.e. equation (4.19) holds. 0 COROLLARY

4.7. Let w = dt - (dx, Ax) = dt -

L 2ajkXk dXj. j,k

Then w(at)

= 1, w(Xj ) = 0, "ij = 1, ... , 2n, i.e. 1i = kerw.

The I-form w is called the one-connection form. The following result provides a relation between the subRiemannian geodesics and the horizontal curves. PROPOSITION

4.8. i) Any subRiemannian geodesic is a horizontal curve.

ii) There are horizontal curves, which are not subRiemannian geodesics. PROOF. i) Let c(s) = (x(s), t(s)) be a subRiemannian geodesic. From the first equation of the Hamiltonian system (4.6) we have x( s) = «( s). Substituting in the third equation of (4.6) yields

i(s) = ((s),Ax(s)) = (x(s),Ax(s)), which is (4.19). Hence c(s)

ii) Set X2j-1(S) shall show that c( s) geodesic. We have

(x, Ax)

=

(x(s), t(s)) is horizontal.

s;

= s2/2, X2j(S) = S, j = 1, ... , nand t(s) = ,£7=1 aj. We = (Xl (s), ... , X2n (s), t( s)) is a horizontal curve, which is not a

2a1 (X2X1 - X1X2)

+ 2a2(X4x3 -

S2

S2

2a1(s2 - 2') +2a2(s2 - 2')

2'"

X3X4)

+ ... + 2an(X2nX2n-1 -

+ ... +2an (s2

X2n-1X2n)

s2

- 2')

.

n

s L....t aj = t(s),

j=l and Lemma 4.6 yields c = (x, t) horizontal curve. Since the Hamiltonian H

=

~JC(sW = ~ LX;(s) = ~(S2 + 1)

is not constant along the curve, the curve cannot be a subRiemannian geodesic.

0

4.5. The Carnot-Caratheodory distance We obtain classical Hamiltonian mechanics by replacing the boundary condition T = 1. Then necessarily

O(s) = 0 with the condition t(O) = 0, and setting n

t=

L ajJ.l(2ajO)r;' j=l

The behavior of the term on the right, as function of 0 depends on both rj and the distribution of the aj. In the isotropic case a1 = a2 = ... = an, the results

4.5.

THE CARNOT-CARATHEODORY DISTANCE

131

of Chapter 1 carryover with no change, except that each family of geodesics from (0,0) to (0, t) is parametrized by the (2n - I)-sphere. We shall adopt the following convention regarding the function J.l:

J.l( m1f)

(4.20)

=

m = 1,2,3, ... ,and 0 . 00 =

+00,

o.

In particular this means that n

L ajJ.l(2aj1f /2a n )rJ < +00

if and only if rj = 0 when aj = an.

j=1 In order to deal with the non-isotropic case, we assume that there is an index p with the property

and set

X = (x', x"),

XII

=

(x 2p+1, X2p+2, .. ·, X2n ) .

The following treatment of the non-isotropic case can be found in Beals, Gaveau and Greiner [3]. THEOREM 4.9. Suppose x" i= O. Then there are finitely many geodesics from (0,0) to (x, t). The geodesics are indexed by the solutions of n

It I =

(4.21)

L ajJ.l(2a/})rJ. j=1

In particular, the Carnot-Carathedory distance from (x, t) to the origin is d(x, t)2

(4.22)

where

(}e

=

2S(x,

Itl, 1; (}e),

is the unique solution on (4.21) in the interval [0, 1f /2a n ).

PROOF. The second part comes from the fact that the subRiemannian geodesics are traces of Hamiltonian paths, see Bismut [10] and Strichartz [50]. The associated action is one-half the square of the length. If x" i= 0, the right side of (4.21) is strictly increasing on [0,1f/2a n ), so there is an unique solution (}e of (4.22) in the interval [0, 1f /2a n ). In every other pole free interval the function on the right side of (4.21) decreases from +00 to a positive minimum at (}m, m = 1,2, ... , then it increases to +00. Furthermore the right side of (4.21) is bounded from below by a straight line through the origin with a positive slope, since this is true for the individual summands. Hence there are finitely many solutions of (4.21). D

REMARK 4.10. The minima ofthe right side of (4.21) do not necessarily increase with m. For instance in the case of H 2 , choosing 0 < a1 < a2 with a2 - a1 small enough, we can arrange that 2

2

j=1

j=1

L ajJ.l(2aj(}1)rJ > L ajJ.l(2aj(}2)rJ.

132

4.

GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS

REMARK 4.11. The lengths of the geodesics associated to a solution increases with O. We have done this in the case of HI in Chapter 1 and it can be done directly for the isotropic Hn, since the dependence in (4.21) is much simpler when the aj are identical. In the non-isotropic case the behavior depends crucially on the relative sizes of the aj and on the relative sizes of the rj. This case will be approached in Chapter 5 by means of the modified complex action f. In the following we shall study the geodesics when x" = O. We shall do a careful enumeration of all geodesics connecting (0,0) and (x, t). For this, we need a few preliminary results. LEMMA 4.12. Suppose x' 'I 0 , x" = 0 and t > O. Then there exist a finite number of geodesics connecting (0,0) and (x, t) which are indexed by the solutions of n

t = ~ ajp,(2a j O)rJ,

(4.23)

j=1

and their lengths increase with O. The square of the length of the geodesic associated with a solution of (4.23) is 2S(x, t, 1; 0). PROOF. If we replace an by the largest aj such that rj '10, then the lemma is an immediate consequence of Theorem 4.9 and Remark 4.11. If 01 < O2 ~ ... denote the solutions of (4.23), then 0 < 01 < 7r/2al' D Next we shall find the geodesics which are not associated with a solution 0 of (4.23). We assumed x' '10. Then t > 0 yields 0 > 0, and then (4.24) Therefore (4.12) and (4.17) imply (4.25)

[

cos( 4anO) - sin( 4a n O)

sin( 4anO)] [1 0] cos( 4a n O) = 0 1 '

so we have

2a n O = m7r,

(4.26) Assuming t(O)

m

= 1,2, ...

= 0, (4.14) yields Ot

(4.27)

=

~11((0)'112 + ~11((0)III12 - ~(x,,((O)').

Let (4.28)

(j)

= ((2j-l,(2j),

x(j) = (X2j-l,X2j),

j

= 1,2, ... ,n.

Then (4.29) Therefore 0 has to be a solution of 1 (4.30) Ot = 211 ((0)" 112

n

+ ~ aj Op,(2aj O)rJ. )=1

4.5.

THE CARNOT-CARATHEODORY DISTANCE

This can be turned around by saying that we may set t(O) choose ((0)" so that

t

~ 11((0)"11 2 = () [t -

(4.31 )

133

0 for a given () if we

=

aj {)J..L(2aj {))r; ].

j=1

Relation (4.14) yields the length dm(x, t) of such a geodesic: n

(4.32)

d:r, (x, t) = 2Ho = 2t{) + (x, ((0)) = 2{) [t + L

aj cot(2aj{))r;].

j=1 We have proved the following result: LEMMA 4.13. Given (x, t) E H n , x' -=I- 0, x" positive integer, such that

=0

and t -=I- 0, let m denote a

n

It I ~ L

(4.33)

ajJ..L(2ajm7r/2an)r;'

j=1 Then there are infinitely many geodesics of the same length dm(x, t),

between (0,0) and (x, t). They are parametrized by the sphere

(4.35)

~ 11((0)"112 = ~7r [It I n

t

ajJ..L(2ajm7r /2an)r;] ,

((0)"

E

1R 2n - 2P .

j=1

REMARK 4.14. We note that (4.33) implies that x" when rj -=I- O.

=

0 and that an -=I- ajm,

The Carnot-CaratModory distance d(x, t), i.e. the length of the shortest geodesic between (0,0) and (x, t), will turn out to be associated to a () in the interval (0, 7r /2a n ]. A geodesic associated to such a () always exists. There are two cases: (i) If we have n

(4.36)

t
0,

j=1 then the proof of Theorem 4.9 yields the existence of a geodesic with length d(x, t), where n

(4.37)

d(x, t)2 =

2{)

[t

+L

aj cot(2aj{))r;] ,

j=1 is associated with (), the unique solution of n

(4.38)

t=

L aj J..L(2aj {))r; j=1

in the interval (0, 7r /2a n ).

134

4.


(ii) If n

t ~

(4.39)

'L ajJ.L(2aj71'/2an)r; , j=l

then Lemma 4.13 yields a geodesic with length d(x, t), where

n

n

[t 'L aj cot(2aj71'/2an)r;]

2~ + 2a

d(x, t)2

j=l

71' [ ~ 2 ~ 2aj71'/2an rj2] 2 - t - ~ ajJ.L(2aj71'/2an)rj + ~ aj . 2 2an j=l j=l sm (2aj71' /2an) 11((0)"11 2 +

(4.40)

t

.(~aj71'/2an)2

r;'

j=l sm (2aj71' /2an)

In particular, d(x, t) always has the form (4.37) with the understanding that when (4.39) holds.

() = 71' /2a n

LEMMA 4.15. Let x' =f. 0, x" = 0, t > 0 and suppose that (4.33) holds for some positive integer m. Then

d(x, t) :::; dm(x, t),

(4.41 )

where d(x,t)

=

d1 (x,t).

PROOF. Let ()" E ((k - 1)71'/2an , k71'/2a n ) be the solution of (4.38) which increases as x" ---+ O. We shall prove that for x" small enough we have lim 2S(x',x",t,1;()") =dk(X',0,t)2,

(4.42) where k

X"~O

=

1 or k

= m.

(i) If k = 1 and (4.36) holds, then (4.42) follows from the behavior of the function J.L in the interval [0,71' /2a n ).

(ii) Otherwise we assume that (4.33) holds, and dk(x, t), k by (4.34). With x" =f. 0 we assume that ()" is a solution of

=

1 or m, is given

p

t - 'LajJ.L(2aj()")r; = anJ.L(2an()")llx"11 2

(4.43)

j=l in the interval ((k -1)71'/2a n , k71'/2a n ), which increases to k71'/2a n as x" lim an J.L(2a n()")llx"11 2 =

X"-->O

lim an X"-->O

2a n()"

sin 2(2a n ()")

---+

O. Then

Ilx"11 2

p

(4.44)

t - 'LajJ.L(2ajk71'/2an)r;'

j=l The square of the distance associated with ()" is given by (4.45)

2S(x' x" t l' ()") = 2 ~ (2aj()")2 r2 , '" ~ 2sin2(2a·()")) )=1)

+2

(2a n()")2 2sin2(2a n ()")

Ilx"11 2.

4.5. THE CARNOT-CARATHEODORY DISTANCE

135

Then (4.44) yields lim 2S(x' " x" t , l', e")

x" --+0

Therefore the distances ddx, t) and dm(x, t) with x" = 0, are limits of distances d(y, t), y" i:- 0, as y" ---4 0. When y" i:- 0, the distances are increasing with increasing see Remark 4.11. Hence this is also true for limits which implies (4.41). 0

e,

In the following we shall introduce more terminology and notation. Let -40'I, -40'~, ... , -40'~ denote the distinct eigenvalues of A2, where 0'1

0. LEMMA 4.16. Let x E W k , X rf. W k and a positive integer m, such that:

1•

Suppose there is an integerq, k < q::; n,

n

(4.47)

t 2:

L ajf.L(2ajm7fj2a )r;' q

j=l

Then there exist infinitely many geodesics of the same length, connecting (0,0) and (x, t). There is only one geodesic if (4.47) is an equality. This set of geodesics is parametrized by the sphere:

where C is an arbitrary positive integer, and the square of their length is given by n

(4.49)

2S(x, t, 1; m7f /2a q ) = 2 ;:7r

[It I + L

aj cot(2ajm7r /2a q )r;] .

j=l

q

Moreover, if m = m1 and m = m2 both satisfy (4.47), then ml < m2 implies

(4.50)

2S(x, t, 1; ml7r /2a q )

::;

2S(x, t, 1; m27r /2a q ).

136

4.


PROOF. It suffices to prove only (4.50). This follows from taking the limit of S(y, t, 1; 0) as y ~ x, where y has a nonzero projection into Vq and 0 is a solution of n

(4.51)

t =

I: ajf.L(2ajO)r;,

0 E ((m - 1)1T/2aq , m1T/2aq ),

j=1 which increase to m1T /2a q as y

~

o

x.

4.6. The shortest geodesic

Next we shall deal with the shortest geodesic between (0,0) and a given point (x, t), x =f. 0, t > O. Suppose x E W k , x¢:. Wk-1. Among the geodesic provided by Lemma 4.12 the shortest is associated with 0 E (0, 1T /2ak). Among the geodesics given by Lemmas 4.13 and 4.16 the shortest occurs when m = 1, and it is associated with 0, where 0 = 1T /2a q , q > k. Therefore the shortest geodesic is associated with a 0 E (0,1T/2ak). LEMMA 4.17. Suppose x =f. 0, x E W k , X ¢:. W k- 1. Then the shortest geodesic connecting (0,0) and (x, t) is associated with a 0 E (0,1T/2a n ], and its length, the Camot-Caratheodory distance of (x, t) from the origin, is given by d(x, t), where: n

(4.52)

d(x, t)2

= 20 [It I + I: aj cot(2ajO)r;]. j=1

PROOF. (i) Suppose It I > 0 is large, so that (4.33) holds with m = 1, and also (4.50) holds with m = 1 and with q > k. We need to show that

2S(x, t, 1; 1T /2a n ) < 2S(x, t, 1; 1T /2a q ), that is

1T I I ~ [ 2-t+L.,..

(2aj1T /2a n )2 ---f.L 2aj1T (2a1T /2)] a r·2 j=1 sin 2 (2aj1T /2a n ) 2a n J n J

2a n

(4.53)

I ~ < 2 ~ It+L.,.. 2a q

[ (2aj1T /2a q )2 _ 2aj1T f.L (2 a .1T /2 aq )] r 2 , 2 J J j=1 sin (2aj1T /2a q ) 2a q

see (4.40). Moving (21T/2a n )[t1 to the right side and replacing quantity

It I by

n

I: aj f.L(2aj1T /2a

q )r;.

j=1 Then (4.53) follows from

(4.54) Let a (4.55)

(2aj1T /2a n )2 sin 2 (2aj1T /2a n )

= 2aj1T /2a n and x = 2aj1T /2a q . Then (4.54) takes the form: a 2 - a(a - sinacosa) ------'--"2- - - - ' sin a

O}.

4.6. THE SHORTEST GEODESIC

139

PROOF. It suffices to prove the lemma for t > 0, which we shall assume in the rest of the proof. Let (x, t) E H p and let h denote one of the variables X2p+ 1, ... , X2n. Then S is given by

S(h)

S(X1, ... ,X2p,0, ... ,0,h,0, ... ,0,t,1,Oh)

~

t

(2a j Oh)2 r2 2 j=1 sin2(2ajOh) J

(4.67)

+ ~ (2a.i 0h)2 h2 2 sin2(2ajOh)

,

where p

(4.68)

t = L aj{l(2aj Oh)r; j=1

+ an{l(2ahOh)h2.

We shall calculate limh->o S'(h). Using

ip2) = ip{l , (ip) -·-2ip sm ip

d ( -d we obtain

S'(h) (4.69) Differentiating (4.68) with respect to h, yields (4.70)

0

=

..f:-. 2 , dOh 2 2 , dOh 2 ~ aj{l (2ajOh)Thrj + an{l(2anOh)h + an{l (2anOh)Thh . j=1

Substituting (4.70) into (4.69), we find

(4.71) Since h

S'(h) = 2anOh cot(2anOh)h. --t

0, Oh

--t 7r

lim [ 2a nOh h] h->O sin(2anOh)

/2a n , then (4.67) we have:

t

lim (2S(h) (2aj Oh)2 r2) h->O j=1 sin 2(2aj Oh) J 7r

[

2- t 2an

2] + ..f:-. ~ aj cot(2aj7r/2an )rj j=1

..f:-.

(2aj7r /2an)2 2 - ~ . 2 rj j=1 sm (2aj7r /2an )

p

(4.72)

~[t- Laj{l(2aj7r/2an)r;] = an

j=1

.!!...-Ep. an

Since Oh is symmetric in h, yields 1/2

(4.73)

lim dS(h) = h->O± dh

± ( 7r Ep )

=1=

0,

an

when Ep > O. Hence S'(h) is discontinuous at h

= O. This proves Lemma 4.21.

With Lemma 4.21 we have completed the proof of Theorem 4.20.

0 0

140

4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS

4.7. Caustics

Consider the set of caustics: (4.74)

Cp

= {(x,t)

E

Hp;Ep(x,t) > a}.

Let Sp(x, t) be the classical action on Hp: P

Sp(x, t) = (lpltl

(4.75)

+ L aj(lp cot(2aj(lp)r;,

(x, t) E H p ,

j=l

whenever (lp E (0, 7r /2a p) is a solution of p

It I = L aj/-t(2aj(lp)r;;

(4.76)

j=l

otherwise we extend (4.75) to all of Hp by continuity. Then (ii) in the proof of Lemma 4.17 implies that

Sp(x, t) > S(x, t),

(4.77)

if (x, t) E Hp and Ep(x, t) > O.

On the other hand we always have if (x, t) E Hp and Ep(x, t) < O.

Sp(x, t) = S(x, t),

(4.78)

Consequently,

Sp(x, t) 2: S(x, t),

(4.79)

(x, t)

E

Hp.

Suppose that

We set n

HPI = {(x, t)

(4.80)

E

Hn; L

r; = O},

j=PI +1 PI

(4.81)

Ep,(x,t) =

Itl- Laj/-t(2aj7r/2ap)r;,

(x,t) E H p"

j=l

and (4.82) Since Ep(x, t) > EpI (x, t) on HpI we obtain (4.83) The points of CPI are more caustic than the points of Cp \Cp" in the sense that they have more indeterminacy in their initial momenta, which leads to a larger set of bicharacteristics. Let SPI (x, t) be the classical (continuous) action on H PI : PI

(4.84)

SpI(X,t) =

(lPlltl + Laj(lPI cot(2aj(lp,)r;, j=l

(x,t) E H p"

EXERCISES

4.8.

whenever

()PI

141

E (0, 7r /2a p J is a solution of PI

It I =

(4.85)

'L,ajf..t(2aj()PI)r;; j=1

otherwise SPI is extended by continuity to all applies and we have

S(x, t) S Sp(x, t) S

(4.86)

Again the proof of Lemma 4.17

HpI

SPI

(x, t) S ... ,

where (x, t) belongs to the appropriate domain of definition of the corresponding actions. The sequence of inequalities (4.86) may be continued to So, the action on {(O,t);y E IR}. We refer to S(x,t) as the minimal action. Extending (4.83) we obtain a decreasing sequence of sets:

Cp :J CPI :J CP2 :J ... , whose intersection is the t-axis. 4.8. Exercises EXERCISE 4.1. Let n 2 1 and let Bn+!

c en+!

be the unit ball. Let n

Un+1 = {z E e n+1 :

Im(zn+1) >

'L, IZjI2}. j=1

= (Z1",.,Zn+1) with

Define F(w)

z. _ J -

Wj

1 + W n+1 '

j

.(1- W n +1) Zn+l = 2 (1 + W n+1) .

= 1, ... , n,

Verify that F maps Bn+! biholomorphically onto Un+1. The domain On+1 is called a Siegel upper half-space. EXERCISE 4.2. The set en x IR can be equipped with the group law

(z, t) . (w, s) = (z

+ w, t + s + 2Im(z· w)),

where z . w = 'L,?=1 ZjWj. Show that this binary operator makes en x IR into a noncommutative group which is the Heisenberg group Hn as we discussed in this chapter. EXERCISE 4.3. If (Z1"'" Zn, t + ilz'12) E 8Un+1, where z' = (Z1, ... , zn), identify the boundary 8Un+1 with the Heisenberg group Hn. Show that in this way 8Un + 1 becomes a group in a natural way. EXERCISE 4.4. If Z E Un+1, denote p(z) function on Un+! and

Z

(Z1"'" Zn, (Re(zn+1)

=

Im(zn+1) -

Iz'I2

be the "height"

+ ilz'12) + i(Im(zn+d -lz'12)

= (6, ... ,xn+1)+(0, ... ,0,ip(z)). If ((, t) = g E Hn, then define

go (Z1"'" zn+d = 9 0 (6, ... , X n+1)

+ (0, ... ,0, ip(z)),

where go (Z1, ... , zn+d is the group action of Hn on 8Un+!. In other words, each 9 E Hn induces an automorphism on Un+!, which preserves the level set

142


{z E Un + 1 : p(Z) = constant}. Denote this subgroup of the automorphism group AutUn+1 by N. i) To what subgroup of the automorphism group on Bn+1 does N c AutUn+1 correspond by way of the map F? ii) Compute the subgroup K of AutUn +1 consisting of elements that fix (0, ... ,0, i). To what subgroup of Aut Bn+1 does K correspond? EXERCISE 4.5. Let z = (Z1, ... , zn+d be a point in Un+1 and let E > the non-isotropic dilation as follows

o. Define

o",(z) = (EZ1, ... ,EZn,E2Zn+1). Show that 0", E AutUn + 1 and the set of all dilations {O"'} form a subgroup of AutUn+1. Denote this subgroup by A. To what subgroup of Aut Bn+1 does it correspond? EXERCISE

4.6. Show that AutUn +1

= K· A· N. See

[36].

EXERCISE 4.7. Let n c en. The Bergman kernel is the function K (z, w), z, wEn, which has the reproducing property

f(z) for all

f

E

=

10 K(z, w)f(w)dV(w),

A2(n), where

A2(n) = {fholomorphic onn:

10 If(zWdV(z)

=

Ilfll~2(O)

2: aj IZj 12}. k=l

Prove that the Bergman kernel for U~~l is given by

K(

)

=

z, W

(n + I)! TI~=l ak 1fn+1(2Im(zn+1) _ L~=l ak zkWk)n+2'

For more details the reader may consult [9]. EXERCISE 4.10. Let A

=

{a{} be an m x n real matrix. Let

Ak = (aLa~, .. · ,ak) # 0 for 1 ::; k ::; n, so that {A 1, A 2,'" , An} C ]Rm is a set of n non-zero vectors. Write

A(z) =

(~a~IZkI2, ... ,~aklzkI2) E]Rm.

The quadratic manifold associated to A is the set L:A

= {(z, w)

E

en x em:

Im(w) = A(z)}.

Show that for m = 1 with Ak = {ad, ak > 0, L:A = BUn+1' Furthermore, show that when {A 1, ... ,A n } is the standard basis of ]Rn, then L:A is the Cartesian product of n copies of BU2 . For more details wee [47].

144


EXERCISE 4.11. We introduce coordinates on EA by identifying the point (z, t+ iA(z)) E EA with the point (z, t) E en x IR m, so that t = Re( w). When using these coordinates, we have the following identifications: 10 10 - - ...... - and - - ...... - - . OWj 2 otj OWj 2 otj Show that EA is a generic e R submanifold of en x em with real dimension 2n + m, e R dimension n, and m "missing directions" .

a

EXERCISE

a

4.12. Define real vector fields by

a X k = ~ + 2Xk+nAk' V't, UXk

X k+n

a

= - - - - 2XkAk' V't, OXk+n

Tj

=

a

;"It.

UJ

where V't = (a~l"'" a~m ) . Calculate the commutator structure of the vector fields Xk, k = 1, ... , 2n and T j , j = 1, ... , m and to conclude that EA is a eR submanifold of step 2. EXERCISE 4.13. Let (w,s) and (z,t) be two points in en x IRm. Define a product on en x IR m by setting

(4.87)

(w,s)· (z,t)

=

(w+z,s+t+2Im

[~AkWkZkl).

Prove that the product given in (4.87) defines a Lie group structure G A on en x IRm. Show also that the vector fields {X 1, ... , X 2n, T 1 , ... , T m} are left invariant and hence can be identified with the Lie algebra of the Lie group GA.

CHAPTER 5

Complex Hamiltonian Mechanics We recall that second order elliptic operators, essentially Laplace-Beltrami operators, induce a Riemannian geometry, and the attached inverses, i.e., fundamental solutions, heat kernels and wave kernels have analytic expressions in terms of the underlying geometric concepts, mainly in terms of the lengths of the appropriate geodesics. We strongly believe that the structure of the inverses attached to sub-elliptic operators should be understood just as precisely as the well known structure of the kernels attached to Laplace-Beltrami operators is understood. One may say that elliptic operators arise from classical mechanics and second order sub-elliptic operators have connections to quantum mechanics. In this chapter we shall describe the complex geodesics from the origin and we shall deal with the lengths of the real geodesics starting at the origin by means of complex Hamiltonian mechanics. An understanding of the dynamics in the complexified phase space is essential also for quantum mechanics, see [54].

5.1. The harmonic oscillator and the Heisenberg group We shall show that the harmonic oscillator leads to the Heisenberg operator by certain complex quantization. Consider a unit mass particle under the influence of force F(x) = x. Newton's equation is x = x. This is the equation which describes the dynamics of an inverse pendulum in an unstable equilibrium, for small angle x, see Figure 5.l. The potential energy is

U(x) = The Lagrangian L : TlR energy

-+

l

x

o

x2

F(u) du = --. 2

lR is the difference between the kinetic and the potential

L(x x) = K - U = ~X2 , 2

+ ~X2. 2

The momentum is p = ~i = x. The Hamiltonian associated with the above Lagrangian is obtained using the Legendre transform: H : T*lR -+ lR . .

212121212

H(x,p) =px-L(x,x) =p - 2 P - 2 x = 2 P - 2x. 145

5. COMPLEX HAMILTONIAN MECHANICS

146

Figure 5.1: The inverse pendulum problem.

Using ideas from Xavier [54], we consider the following complexification

(5.1) Hence H : T*C -+ C and

H(x,p)

Replacing

e = -i,

(5.2) Quantizing, i. e., making the changes PI -+ aXl' P2 -+ aX2 , e -+ where 1 2 1 2 ~x = "2 (aX1 - x2at) +"2 (aX2 + XI at}

at,

we get H -+

~x,

is the Heisenberg operator. REMARK 5.1. a) If we choose the complexification

P = PI + 2ix2 then we arrive at the operator

~x

121

2

= "2 (aX1 - 2X2at) +"2 (aX2 + 2Xlat} .

b) The expression e = -i was used in the literature for a long time without an explanation. We can see that this value for e is the natural one and it is related to quantum behaviors.

5.3. COMPLEX SUBRIEMANNIAN GEODESICS

147

5.2. The quantum Hamiltonian The previous assumption () = -i can be considered in the general case of the study of complex dynamics on the model defined by the vector fields

Xl = aX! + 2X2lxl2kat,

(5.3)

X2

= aX2 - 2Xllxl2kat.

In this case, the Hamiltonian becomes

l(

.

H(x,O='2 6-2~X2Ixl 2k)2

(5.4)

. 112k)2 , +'2l(6+2~XlX

and it will be called quantum Hamiltonian. DEFINITION 5.2. The PT-symmetry operations are defined as follows. Combined parity:

Time reversal:

T:

(Xl,X2,6,6)

~

(Xl,X2, -6, -6), i

~ -i.

A rigorous study of the one-dimensional complex Hamiltonian systems with PTsymmetry can be found in the recent work of Bender and coworkers [8]. We shall deal with a two-dimensional Hamiltonian in the variable x. The following result is obvious. PROPOSITION

5.3. The quantum Hamiltonian (5.4) is PT-symmetric.

This symmetry will influence the symmetry of the solutions of the Hamiltonian equations. Xavier and Aguiar [54] used complex solutions of the Hamiltonian system to study the semiclassical approximation of the coherent-state propagator in the case of a quartic potential. Consider complex solutions x(s) with Hamiltonian (5.4)

= (XdS),X2(S))

E

<e 2 for the Hamiltonian system

(5.5) with real boundary condition

x(O) PROPOSITION

(5.6) PROOF.

() =

-i.

= Xo E ~2,

x(l)

= x

E ~2.

5.4. The solutions of system (5.5) satisfy the system

2k { Xl = -4i(k + 1)lxl x2 2k X2 = 4i(k + 1)lxI Xl' It comes from the system (3.6) of chapter 3, with the substitution 0

5.3. Complex subRiemannian geodesics In the case of a complex subRiemannian geodesic we shall ask that the momentum () is complex and constant along the geodesic, with () = -i. To eliminate the over-determination of the Hamiltonian system, we shall assume non-standard boundary conditions, i.e., t(O) is not specified, since we fixed ().

148

5.

COMPLEX HAMILTONIAN MECHANICS

DEFINITION 5.5. A complex subRiemannian geodesic is the projection on the (x, t)-space of a solution of the Hamiltonian system x=H~,

~ = -Hx, with non-standard boundary conditions

x(O) = Xo,

() = ()o =

-i,

and Hamiltonian

H(x,

f

t,~, ()) = ~ (6 + 2()X21x1 2k + ~ (6 - 2()X11x1 2k

f·

Note the boundary conditions are real, i. e., Xo, x j E JR.2, t j E JR., while the geodesic components between the end-points might be complex. The following theorem is useful in future computations. THEOREM 5.6. A curve c( s) = (x( s), t( s)) is a complex subRiemannian geodesic (in the sense of Definition 5.5) if and only if the x-component satisfies the system (5.6) with boundary conditions

x(O)

=

Xo,

and the t-component is the unique solution of the equation

i = 21x12k(X1X2 -

(5.7)

X1X2),

with the final condition t(O) = tj. PROOF. "===>" Let c(s) = (x(s), t(s)) be a complex subRiemannian geodesic. Applying Proposition 5.4, the x-component satisfies the system (5.6). The Hamiltonian equation i = HI} yields the second part. " ~ " Is the same proof as in Theorem 2.4 of chapter 3, with () = -i. 0 Complex symmetries From the PT-symmetry ofthe quantum Hamiltonian (see Proposition 5.3), if c(s) = (x(s), t(s)) is a complex geodesic, then P(c(s)) = (-x(s), t(s)) and T(c(s)) = c(s) are also complex geodesics, where c(s) = (X1,X2,t) denotes the complex conjugation. Hence, we deal with quadruples of geodesics (±x, t), (±x, I). 5.4. Complex subRiemannian geodesics on the Heisenberg group The x-component We are interested in finding complex subRiemannian geodesics with real boundary conditions Xo = 0, Xj = x and tj = t on the Heisenberg group H 1 . To find them, we use Theorem 5.6. In this case, k = 0 and the x-component satisfies the complex system (5.8)

~1 = -.~ix2 X2 = 4ZX1, Xl = a1 + ib1, X2 = a2 + ib 2, with ai = ai (s) {

where and bi = bi (s) real functions. The system (5.8) can be decomposed in the following two real systems:

(5.9)

(1) {

~1 = 4~2

b2 = 4a1,

(II) {

~2 = -4~1

b1 = - 4a2,

5.4. COMPLEX SUBRIEMANNIAN GEODESICS ON THE HEISENBERG GROUP

149

with boundary conditions

(5.10)

i

= 1,2.

• Solving the system (1). Multiplying the first equation by 0,1, the second by h2 and subtracting, yields .. .

0,10,1 - b2b2 = 0 o Case

at - h~ =

2'2

===}

R2. There is a real function a = a( s) such that

a1(S) = Rcosha(s),

(5.11)

2

0,1 - b2 = ±R (constant). h2(S) = Rsinh a(s).

Differentiating and substituting in system (I), yields (5.12)

& sinh a

= 4 sinh a,

& cosh a

= 4 cosh a.

Then

&(s) = 4

(5.13)

===}

a(s) = 4s + aD,

see Exercise 5.1. Using (5.13), and integrating the relations (5.11), yields

= a1(0)+Rfo8COsh(4u+ao)dU

a1(s)

a1(0) +

~ (sinh(4s + aD) -

sinh aD ).

b2(0) + R fo8 sinh(4u + aD) du

b2(s)

b2(0) +

From the boundary conditions b2 (0) that aD verifies the equation

~ ( cosh(4s + aD) -

cosh aD ).

= b2 (1) = 0, it follows from the above relation

cosh(4 + aD) - cosh aD = 0,

(5.14)

which has the unique solution aD = -2, see Exercise 5.2. Using the boundary conditions a1(0) = 0, a1(1) = Xl and aD = -2, the expression for a1 (s) yields

Xl =

!!:.4 (sinh( 4 -

2) - sinh( -2))

===}

R=

~Xh1 2 .

~n

Hence, the solutions for the system (I) are

~lh (sinh(4s-2)+sinh2),

a1(s)

2sm 2

s E [0,1]. ~lh (cosh(4S - 2) - cosh2), 2sm 2 Other solutions can be obtained if in (5.11) we choose combinations of

b2 (s)

(5.15)

a1(S) = ±Rcosha(s),

h2 = ±Rsinha(s).

The corresponding solutions are as above but with sign combinations. o Case

at - h~ =

-

R2 . We shall show that this case is impossible. In this case

h2(S) = Rcosha(s),

ads) = Rsinha(s),

150


with a(s) = 4s + ao. Integrating, b2 (s)

=

1 s

b2 (0)

+R

cosh(4u + a) du

b2 (0)

+ ~ (sinh(4s + ao) -

sinhao).

From the boundary conditions b2 (0) = b2(1) = 0 we get sinh(4 + ao)

sinh(ao),

=

which is a contradiction, because sinh is one-to-one . • Solving the system (II). Multiplying the first equation by 0,2, the second by and subtracting, yields .. . 2·2 2 0,20,2 - blb l = 0 ===} 0,2 - bl = ±p (constant).

o Case O,~ -

br = p2.

There is a real function {3

bl

= (3( s) such that

(5.16)

Differentiating and substituting in system (II), yields {Jcosh{3

(5.17)

= -4cosh{3,

and then {3 = -4s + (30. Integrating in (5.16) and using a2(0) = bl (0) = 0 and {3 = -4s + {30, yields

-~ (sinh( -4s + (30) -

sinh{3o)

(5.18)

~ ( sinh( 4s -

(5.19)

-~ ( cosh( -4s + (30) - cosh{3o).

(30)

+ sinh (30 )

The boundary condition bl (1) = 0 yields cosh( -4 + (30) - cosh {30 = 0

===}

(30 = 2.

Substituting a2(1) = X2 in formula (5.18)

+ sinh (30) = f!.. sinh 2 4 2 Hence, the solutions of the system (II) are X2 = f!.. ( sinh( 4 - (30)

===}

P=

~Xh2

sm 2

.

+ sinh 2),

a2(s)

:2 (sinh( 4s - 2) 2sm h 2

bl(s)

-2s~:h2(cosh(4S-2)-cosh2),

SE[O,l].

Changing signs in (5.16)

o'2(S) = ±p cosh {3,

(5.20)

bl(s)

=

±p sinh {3,

leads to solutions ±a2(s) and ±bl (s), with a2 (s) and bl (s) provided above.

br

o Case O,~ = _p2. This case is impossible, see the proof in the previous case. The above discussion can be summarized into the following result:

5.4. COMPLEX SUBRIEMANNIAN GEODESICS ON THE HEISENBERG GROUP PROPOSITION

151

5.7. The solutions of system (5.8) with the boundary conditions

x(O) = 0,

x(1) = x

are ±x(s), with

x(s) = A(s)x ± iB(s)Jx,

(5.21 ) where A(s)

= sinh(4s - 2) + sinh 2, B(s) = cosh(4s - 2) - cosh 2 and

(5.22) PROOF.

s E [0,1],

J

[~1 ~],

=

x = 2si:h2·

If we choose all the possible sign combinations in

±x1A(s),

a2(s) = ±x2A(s),

±x1B(s),

b1(s) = -( ±x2B(s)),

we get

x(s) = ±(a(s) ± ib(s)) = ±(xA(s) ± iJxB(s)).

o This result is consistent with the PT-symmetries of the quantum Hamiltonian

f

H(x,~) = ~(6 -2iX2 + ~(6+2iX1)2. The t-component From the boundary condition t(1) = t E IR, while t(O) is not fixed. In this section we shall study the behavior of the t-component, and show that t(O) cannot be a real number unless x = O. Integrating in i = 2(X1X2 - X1X2), yields

t - t(s) = 211 (X1X2 - X1X2).

(5.23) Using the geodesics

xds) X2(S)

ads) a2(s)

+ ib 1(s) = x1A(s) - iX2B(s), + ib2(s) = x2A(s) + iX1B(s),

a computation provides

2(XIX2 - XIX2) = 8ilxI 2(1- cosh4s). Substituting in (5.23), yields

t - t(s)

8ilxl211 (1 - cosh(4u)) du sinh Sinh4S) 8z.1~12(( x 1 - s) - -4-- -4

2i Ixl We arrive at the following result.

2(

(sinh 4s - 4s) - (sinh 4 - 4) ) .

152

5.


PROPOSITION 5.8. The t-component is given by

t(s) = t - 2ilxl 2 ((Sinh4s - 4s) - (sinh 4 - 4)), where

x

~

X=---

2sinh2' Hence, for x =j:. 0, t(s) is never real for any s E [0,1). In particular,

t(O) = t -

ilxI2(~ smh 2

coth2).

PROOF. The function 'ljJ( s) = sinh 4s - 4s has the derivative 'ljJ' (s) = 4( cosh s 1) ~ 0, and hence 'ljJ is increasing and one-to-one. Imt(s) = -2IxI2('ljJ(S) -'ljJ(1)) > 0, for any 0 S; s < l.

~Ix~

t - 2ilxl 2(4 - sinh 4) = t -

t(O)

ilxI2(~ -

t-

smh 2

(2 - sinh 2 cosh 2)

coth2).

smh 2

o LEMMA 5.9. Let

x - - -cothx sinh2 x ' X

- - -cotx

JL(x)

sin2 x

'

be two odd functions. Then E(ix) = -iJL(x),

JL(ix) = -iE(X).

PROOF.

ix

f(ix)

2 - coth(ix) sinh (ix)

ix (. ) ix . .. )2 - -zcotx = . 2 +zcotx (2smx -sm x

-i(~ 2 sm x

JL( ix)

- cot x) = -iJL(x).

ix ix (. . )2 - cot(ix) = (' . h )2 - (-i cothx) smzx 2sm x .ix 2 -smh x

+ icothx = -i(~ smh

- cothx) = -if(X).

o Lemma 5.9 and Proposition 5.8 yield

t(O) = t - ilxI 2f(2) = t

+ IxI 2JL(2i),

which can be written as

t - t(O) Ixl 2

.

= JL( -22).

5.5.

If we consider () see chapter 1:

= -i,

COMPLEX CONNECTIVITY

153

we arrive at the same relation as in the case of real geodesics,

t

Ixl 2

=

J.1-(2()sf) ,

where t(O) = 0 and sf = 1. From this point of view, real and complex geodesics behave similarly. The symmetries for real geodesics induced by changing () into -(), became time reversal or conjugation T-symmetry for complex geodesics.

5.5. Complex connectivity Due to the non-standard boundary conditions, the usual definition of connectivity is not working in the case of complex subRiemannian geodesics. The difficulty arises from the fact that the equation i = 21x12k(XlX2 - XlX2) may become supradeterminate if the following two boundary conditions are required

t(O) = to,

t(l) = tf.

On the other hand, we have seen in the previous section that in the case x

=1= 0,

t(s) is real only for s = 1 and complex for 0 :::; s < 1. We have the following non-connectivity result on the Heisenberg group.

t

=1=

PROPOSITION 5.10. Given two distinct points P(x, t) and Q(O, to) in lR 3 , with to, there are no complex subRiemannian geodesics between Q and P.

PROOF. If x = 0, from Propositions 5.8 and 5.7, x(s) = 0, and t(s) = t, constant. P and Q cannot be connected because t =1= to. If x =1= 0, from Proposition 5.8, t(O) cannot be real, and hence t(O) = to E lR doesn't hold. D Hence, even in the step 2 case, the connectivity by complex geodesics fails. We need to give an appropriate definition for the connectivity in the complex case. For this we shall give up the t-component and we shall consider only the connectivity in the x-plane. DEFINITION 5.11. Let H be the quantum Hamiltonian (5.4). The complex connectivity property holds if given any two points P(xo) and Q(x f) in the xplane, there is a solution of the system (5.24)

x=H~,

satisfying the real boundary conditions x(O) = Xo, x(sf) = xf. In the following we shall prove the geodesic connectivity from the origin. In polar coordinates (r, 1», the system (5.25)

{ Xl = -4i(k + 1)lxl 2k x2 X2 = 4i(k + 1)lxl 2k xl

becomes (5.26)

4(k + 1)ir2k) r2(¢ - 2ir2k) = C(constant).

{ r = r¢(¢ -

154


As the solution starts from the origin, we have r(O) = O. ¢ = 2ir2k. Substituting in the first equation, yields

Hence C

o and

r . 2ir2k(2ir2k - 4(k + 1)ir2k)

r

-2ir2k+1(4k + 2)ir2k) 4(2k + 1)r 4k +l = -V'(r), where V (r) = - 2r4k+2. The following conservation of energy holds

~i'2 + V(r)

(5.27)

=

E,

where the constant of total energy may be complex. Sometimes, for some particular boundary conditions, the energy is real. Indeed, choose the boundary conditions (5.28)

x2(1)

= O.

Assume that c(s) = (Xl (s), X2(S)) is a solution of the Euler-Lagrange system (5.25) with the boundary conditions (5.28). Taking conjugation in the system (5.25), yields (5.29) which is the Euler-Lagrange system for ')'(s) = (xds), -X2(S)). Thus c(s) and ')'(s) satisfy the same system of differential equations and the same boundary conditions (5.28). Consequently, if they exist, for xl(1) small enough the solution is unique, i.e. Xl(S)

= Xl(S),

X2(S) = -X2(S) =? r2(s) E R

From (5.27) we get E E R Using the rotational symmetry of the Euler-Lagrange system we can always choose, using a rotation, x2(1) = O. Hence from now on we shall assume the energy E real. We have the following result: LEMMA 5.12. Given a real number R > 0, there is a real function r : [0,1] which is a solution for the boundary value problem

---*

JR.,

= 4(2k + 1)r 4k +l { rr(O) = 0, r(l) = R.

(5.30) The energy function E

= E(R) is increasing as a function of R.

PROOF. Assume r = r(s) is a real valued function. Then E will be real. The equation (5.27) becomes

i'2 - 4r 4k +2 = 2E dr

i' = J2E

+ 4r 4k + 2

res)

= ds

10

J2E + 4r 4k +2

du

= s.

J2E + 4u 4k +2

For s = 1, (5.31 )

rR

10

du

-V-;=2==E==+=4=u=T4k;=;+~2

-1

-

.

We shall show that there is an unique real number E > 0 which satisfies (5.31). Consider the function

r

R

T}(E) =

10

du

J2E

+ 4u4k+2'

5.5.

COMPLEX CONNECTIVITY

155

which is decreasing because

dry dE

(R du = - io J(2E + 4U4k+2)3 < 0.

Using lim ry(E)

0,

lim ry(E)

l

E/oo

0

E"-"O

R

du

1

IU=R

-2k-1 - - - - +00 2u + 4ku 2k u=o ,

and the continuity of ry, the equation ry(E) E(R), see Figure 5.2.

=

1 will have an unique solution E

=

ll(E)

o Figure 5.2: The uniqueness of E.

When R increases, in order to get the same value for the integral, E has to increase, too. With the previous E, let

r

du

w(r) = io J2E + 4U4k+2 As w'(r) > 0, w is increasing and hence it is one-to-one. As W(O) = 0, w(R) = 1, w is onto on [0,1]. Then the solution is defined by r(s) = w-1(s), s E [0,1].

o

The connectivity is given in the following result. THEOREM 5.13. Given a point P(x) in the plane, there is at least one complex geodesic in IR3 such that its projection onto the x-plane joins the origin and the point P. PROOF. Let R = Ixl and ¢f be the radius and argument of P, respectively, i.e. x = Re i 1>f. The numbers Rand ¢f are real. Let r(s) be given by the Lemma 5.12. Then the function ¢( s) is a solution of the Cauchy problem

which is ¢(s)

=

2i

JS r

2k (u)

du + ¢f'

The radius function r(s) is real but the argument ¢(s) is complex. Hence the xprojection of the complex geodesic will be complex with real end points, which are the origin and the point P(R, ¢f). 0

156

5.


An explicit computation of the radius r(s) and argument ¢(s) in general step is almost impossible. It can be done explicitly in the particular cases k = and k = 1, which correspond to step 2 and step 4 cases. We shall do this computation in the following sections.

°

The case k = steps:

°

(Heisenberg group case). We shall follow the next few

• Finding the argument function. In this case the argument is given by ¢( s) 2i(s - 1) + ¢t. • Finding the energy. The energy E is defined by

r

R

Jo

du V2E + 4u 2

=

l.

A computation of the integral yields

~ sinh- I

(VI u)

u=R

= 1 {=:? sinh- 1

(VI R)

= 2,

u=o

and hence the energy depends only on the boundary condition 2R2

(5.32)

E = sinh2 2·

We note the energy E is an increasing function of R. • Finding the radius r (s). The radius is defined by the integral equation ((s)

Jo

du = s V2E+4u 2

{=:?

sinh- 1

(

f2E2 r(s)) = 2s.

VB

Hence, using (5.32) r(s)

To conclude the case k

=

~2 SIn. h(2 s ) --

sinh(2s) R h . sin 2 0, we have the following result: =

THEOREM 5.14. Given a point P(x) in the plane, there is exactly one complex geodesic in ]R3, such that the projection onto the x-plane joins the origin and the point P.

= 1.

The case k

l

r (S)

o

The substitution v

The radius r(s) satisfies the integral equation du

~~===;=7 = V2E+4u 6

=

S {=:?

l

0

(2/ E)I/6 u yields a (S) d

l

--:::=v=

r (S)

du =

V1+j;u6

= 22/3 EI/3 S,

o VI + v 6 where a(s) = (2/E)I/6 r (s). Making v 2 = t, yields (5.33)

l

a2 (S)

o

dt

Jt(1

+ t3 )

= 25/ 3 El/3 s.

V2E s.

5.5. COMPLEX CONNECTIVITY

157

We need the following result: LEMMA

5.15. Let

g(a) =

2

a

Then

rr io

2

(s)

J3 +

b= 2+

fl=l - 1,

J3,

2

dt =3-1/4 sn- 1( !1_(g(a))2,k). Jt(l + t 3 ) V b

Let f(t) = t(l + t 3 ) = t(t + 1)(t 2 - t + 1) = (t 2 + t)(t 2 - t + 1). Making the substitution PROOF.

t=pu+ q u+ 1 '

(5.34)

+ p)u2 + [q(l + p) + p(q + l)]u + q(q + 1) (u + 1)4 2 (p2 _ P + 1)u + (2pq - p + 2 - q)u + (q2 - q + 1)

p(l

f(t(u))

+

(u+1)4

.

Choose p and q such that the coefficients of u vanish

q(1+p)+p(q+1) 0 2pq - p + 2 - q = O. The above system has the symmetric solution 1-

J3

1+J3

q - -2 -.

P=-2-' Then

[(~ -

f(t(u))

~

J3)u 2 + ~ + J3] (u + 1)4

1

[~u2 + ~]

(~_J3)[U2+3+2J3](U2+1)

2(u+1)4 2

=

3-2J3

~ (u~ 1)4

G- J3)

~

(J3-~)W-u2)(u2+1)

1

2(u+1)4

[u 2 - (2+ J3)2](U 2 + 1)

2

'

and hence

J7 =

!~ (J3 _~) J(b 2 - u 2)(u2 + 1).

V2

(u

2

+ 1)2

Using the substitution (5.34) and

-J3

dt = (u+1 )2 du,

J3

2

V~(J3-~)

J2J3-3'


158

yields dt

2

Jf(t)

J2J3 - 3

0"2

10

du

r(O")

J(b 2

Jb

lb V2J3 - 3 2

- U 2 )(U 2

+ 1)

du

g(O") J(b 2 - U 2 )(u 2

+ 1)'

The limits of integration of the above integral are computed as follows.

pu+q t = -u+l For t

=

=

u

q-t t-p

= --.

0, q

u For t

{=:::}

1 + J3

-- = - - p 1-

J3

=2+J3=b.

(J2, u

Following [42], p. 52,

In our case a = 1, x = g((J),

v'b 2

J _(=)

2 - x = bIb

2

'

J1+b2 = VI + (2 + V3)2 = V8 + 4V3 k

2

1

V2J3-3'

v'a +b 2

2

2V2 + V3 = 2Vb. b Vb v'a 2 +b2 2

2

1

1

V2J3 - 3 . 2Vb - Vb(2V3 - 3) 1

V(2 + J3)(2J3 -

=

3)

3- 1 / 4

.

Hence

o

159


• Quantification of the energy E Using Lemma 5.15, equation (5.33) yields

sn- 1 (

VI _(g(a;s))) k) 2,

{=?

(3 /42 E1/3 k) cn (3 /42 E k).

1- (g(a;s)))2

sn 2

1

5/ 3

S,

(g(a;s))) 2

2

1

5/ 3

1 / 3 S,

{=?

1/6 If s = 0, then a(O) = ( j ) r(O) = 0, and g(O) = 1 = 2 + J3 both left and right terms in the above equation are equal to 1. 1/6 ( )1/6 If s = 1, then a(l) = ( j ) r(l) = j R, and

../t!1 -

g(a(1))

Let

T

=

= b. Hence

=

E 1 / 3 , and

F(T)

~ 7+~v'3 [~v: 4=' -11'

PROPOSITION 5.16. The function F(T) has the following properties: (i) lim 7 ---+ 00 F(T) = 1, i.e. the line y = 1 is a horizontal asymptote for the graph of F; (ii) There is an unique TO> 0 such that F(TO) = 0; (iii) F is decreasing on (0, TO) and increasing on (TO, +(0). PROOF.

(i) We have

lim F(T) 7---+00

1 7 + 4J3

[2J3 -1]2 J3 - 1

1 In 2 f') (2 + y 3) 7 +4y3

=

_ 1 - 7 + 4J3

[J3+1]2 J3 - 1

l.

(ii) F(TO) = 0 if and only if 21 / 3 R2 J3 - 1 24 / 3 R2 - - + - - = v'3 {=? TO = - - . TO 2 J3 + 1 (iii) See Figure 5.3 below.

o


5.

160

0.6 0.5 0.4 0.3

0.2 0.1

o

2

4

6

8

10

12

14

16

18

20

tau

Figure 5.3: The graph of F(r).

THEOREM 5.17. Given R > 0, there is an infinite sequence of energies

o < E1 < E2 < E3 < ... < En < ... < +00, parametrized by the solutions r = E 1/ 3 of the equation

(5.35) PROOF. As F(r) / 1 and cn 2 oscillates between 0 and 1, the equation (5.35) has an infinite number of solutions. See Figure 5.4. 0

0.8 0.6

V

v

V

-- --

,..-

12

14

f-

--

,/

V

0.4

V 0.2

V '/

o

2

4

6

8

10

16

18

tau

Figure 5.4: The graphs of F(r) and cn2(31/425/3r,k).

REMARK 5.18. The energy E provided by Lemma 5.12 is E1 (the smallest one) . • Finding rn(S) There is a radius function r n (s) associated with each energy level En (n(S)

io

du J2En

+ 4u

= 6

s.


Let O"n(s)

=

(

in

)

1/6

161

rn(S). As before, an algebraic calculation provides

(g(a~(s))f

=

cn2(3 1/ 425/3E;/3 s,k)

and then (5.36) On the other hand

J3

2

(5.37)

J3-1

O"n(s) = 1 + g(O"n(s)) - - 2 -

and (5.38) Substituting (5.36) and (5.37) in (5.38), yields

(5.39)

r;(s) = (En)1/3 [ J3 2 1±bcn(31/425/3E~/3s,k)

_ J3

-1],

2

s E [0,1].

• Finding the argument ¢n (s) Let '"Yn = 31/425/3E~/3. Then we have

= ¢f + 2i

¢n(s)

JS r;(u) du

¢f

+ 2i( ~n

¢f

+ -2-

31 / 4 i

r/ [J3 JS 3

l'Yn "In

s

1±

dv

b~~bnu) -

+i

1 ± bcnv

(E ) 1/3

-?-

J32- 1 (s - 1)]

(J3 -

1)(1- s).

We shall show that the integral in the above expression can be computed explicitly using elliptic functions theory. Following [42], p. 93, for a 2 > 1

J+ 1

du acnu

_

__2_1-A(u, a ,k)+ a a-I ../(a2 - 1) v(a2 _ 1)(k2 + k/2a2) xtanh- 1 (

k2

+2 k/ 2 a 2 sdu, ) a -1

where

A(u,f3,k) = In our case a

ior

= ±b, and a 2 = (2 + J3)2 > f3 k

a

";a 2 -1

v"b = 2

±b ~

../2 + J3 2

=

1-

;~sn2 v.

l.

±

2 + J3 ~ ±0.8578, ../6+4J3

~ 0.9659.

162

5.


Following [42] p. 69, in the case 0 < (3 < k < 1 we have

A( u" (3 k)

[11

(3

8(u - F((), k))

= u + J(k 2 _ (32)(1 _ (32) 2 n 8(u + F((), k)) + u -

D(() k) ,

~UF((),k)],

where 8(u) = ()4(X), with x = 2Kxj-rr and 00

DO

2:

()4(z,q)=1+22:(-1)nqn2cos2nz=

(_1)nqn2e2inz,

n=-oo

n=l

and

F((),k) = sn- 1 ((3/k), E(u, k) = K =

1

7r/2

sin() = (3/k,

lou dn (v, k) dv,

E = E(K, k),

2

(1 - k 2 sin 2 ())-1/2 d() =

11

D((),k) = E(F((),k)),

1/

3 (1) (1) -

4 dt = --f - f J(l - t2)(1 - k 2t 2) 2ft 6

o 0 see [42] p. 86. To conclude the case k = 1, we have the following result:

3'

THEOREM 5.19. Given a point P(x) in the plane, there are infinitely many complex geodesics in ~3 such that their projections onto the x-plane join the origin and the point P. 5.6. The complex action Let (Xl (s), X2( s), t( s), 6 (s), 6( s), ()( s)), s E [0, r] be a solution for the Hamiltonian system given by the quantum Hamiltonian (5.4) with the non-standard boundary conditions (5.40) X1(0) =0, X2(0) =0, x1(r)=x1, x2(r)=x2, t(r)=t, ()=-i. DEFINITION 5.20. The complex action associated with the above solution is

(5.41)

g

= g(x, t, r) = -it + 10'1'

[(~, x) -

H(s)] ds.

One may write a formal relation between the complex action and the classical action S g = S - it(O), with

S where ()

= -i.

=

S(x, t, r, ())

=

loT [(~, x) + ()i -

H(s)] ds,

The Lagrangian is given by L

=

(~, x)

+ ()i -

H.

A computation (see chapter 3) yields

(5.42)

L

= ~ (xi + x~) + ()(i - 21x12k(X2X1 - X1X2))'

The Euler-Lagrange system is

5.6. THE COMPLEX ACTION

163

(5.43) with boundary conditions (5.40). Multiplying the first equation by Xl and the second by X2 and adding, yields XlXl + X2X2 = 0, i.e. 1('21 +X2 '2) = E , 2"X

where the energy constant E is real. Multiplying the first equation of the system (5.43) by X2 and the second equation by Xl, yields

XlX2 X2Xl Let r

=

Ixi.

40(k + 1)lxl2kx2x2 -40(k + 1)lxI2kx1Xl'

=

Subtracting, we have

40(k + 1)lxI2k(x2x2 + xlxd 20(k + 1)r2k(2x2x2 + 2XIXd 20(k + 1)r 2k (xi + xn 20(k + 1)r2k(r2} 20 ((r2)k+1)' and there is a constant

n,

called momentum, such that

XlX2 - X2Xl Since x(O)

n = O.

= 0, then

L

=

20r 2(k+1)

+ n.

Hence, the Lagrangian (5.42) becomes

E + O(t - 2r2k 20r 2k +2 ) E - it + 4r 4k +2 ,

= =

where we considered 0 = -i. The complex action becomes 9

=

S - it(O) = 11' Lds - it(O)

it + 4r4k +2 (s)) ds -

11' (E -

Er - it(r)

it(O)

+ 411' r 4k +2 (s) ds.

We need to work out the last term of the above relation. This will be done in the following result. LEMMA

5.21. Let r(s) = Jxi(s)

4

(5.44) PROOF.

i

+ x~(s)

and r(r) = r. Then

r 4k + 2 (s) ds = -r - JE o k+1 2 T

+ r 4k + 2 -

Let =

11' r 4k+2(s) ds.

ds =

dr J2E + 4r 4k +2 '

J From the proof of Lemma 5.12

-Er -. k+1

164


and substituting in J we have

1 r

J= 1

With v

= (2/ E) 4k+2 U

U4k+2

o v'2E + 4u 4k +2

duo

1

and a

= (2/ E) 4k+2 r,

4J =

Let

1 - - r 2k +1 > k+1 2 k+1' and hence (5.53)

r>

_1_ r2k+1 k+1

For k 2 1 the derivative of r is increasing faster than O(r2), and we shall show that r cannot be defined globally for any T E R

5.9. SOLVING HAMILTON-JACOBI EQUATION

171

Separating and integrating, (5.53) yields

J

dr r 2k + 1 r- 2k {:=> - -2k 1 {:=> 2k r

{:=>

> >

J

dr k+l 1 -(r-C) k+l 2k -(C-r) k+l k+l 2k(C-r)'

where C is the integration constant. Hence lim r 2k (r)

7'-C

and r(r) blows up for r

=

= +00,

C.

In the case k = 0, using Corollary 5.23 yields org

r

2rcoth(2r) ==* 2rcoth(2r),

which together with the initial solution r(O) = 0 has a global solution r( r) with r E [0,(0), see Exercise 5.3. We arrived at the following result. THEOREM 5.30. Let k ~ 1 be an integer. Then the complex geodesics starting at the origin cannot be extended indefinitely on the whole real line JR. This can be done only for the case k = 0, i.e. in the step 2 case.

With the general complex action formula developed in section 5.9 and similar computations, one may show that the complex geodesics starting outside of the origin cannot be extended indefinitely for the case k ~ 1.

5.9. Solving Hamilton-Jacobi equation In this section we shall work out the complex action reducing the HamiltonJacobi equation to an eiconal equation. This works out well because the Hamiltonian is conservative (since does not contain r explicitly).

• The Hamiltonian in polar coordinates In this paragraph we shall write the Hamiltonian (5.54) in polar coordinates. The associated Lagrangian is given by (5.47). In polar coordinates (cp, r) the Lagrangian is


172

Let Pcp, Pn Pt be the momenta associated with the coordinates ¢, r, t.

a~

Pcp

a¢

=

aL

r2A. + 20r2k+2===*A. = Pcp _ 20r2k 'f' 'f' r2 '

.

.

ar = r ===* r = Pr,

Pr

aL

- . = 0 ===* 0 = Pt· at

Pt

Legendre transform yields the Hamiltonian

pcp(p + Pr'r + Pti - L (r2(p + 20r 2k +2)(p + r2

H

+ oi

_~r2 _ ~r2(P2 - oi _ 20r 2k +2(p 2

2

~(r2 + r2(p2) 2

~p2 + ~ (PCP _ 2Ptr2k+l)2. 2

PROPOSITION

2

r

r

5.31. The Hamiltonian (5.54) can be expressed in polar coordi-

nates (¢, r) as (5.55) The Hamiltonian does not depend on the variables t and ¢. This yields the conservations

Pt Pcp

-Ht = 0 ===* Pt = 0 (constant), -Hcp = 0 ===* r2((p + 20r 2k ) = C (momentum constant).

• The eiconal equation We shall solve the following Hamilton-Jacobi equation in polar coordinates

as aT

(5.56)

+H

(

as as as)

¢, r, a¢' ar 'at

=

O.

We shall look for a solution S(¢, r, t, T) of (5.56) with separable variables

S(¢, r, t, T) = W(¢, r)

(5.57)

+ U(T) + V(t).

Substituting in the Hamilton-Jacobi equation (5.56) yields

U'(T)

+H

= O.

Using the constant value of the Hamiltonian along the solutions H = E we have

U(T) = -ET, where E is the energy constant. Consider E as a parameter. Substituting in (5.56) yields an eiconal equation for W

(5.58)

~ (aW)2 + ~ (~aw _ 2 av r2k+1)2 2

ar

2

r a¢

at

=

E.

173

5.9. SOLVING HAMILTON-JACOBI EQUATION

We shall consider the ansatz: W does not depend on ¢, i. e. oWj o¢

=

0, and

oVjot = (). Then

+ 0,

V(t) = ()t

where 0 is a constant. The equation (5.58) becomes a first order ODE

°o~ = ±J2E Taking ()

=

4()2r4k+2.

-i and the positive sign, we have

oW ~-----:-:---:or = V2E + 4r 4k +2.

= ro. Then W will depend on parameters E

We assume the geodesic starts at r(O) and ro

W(r, ro, E) =

l

r

J2E

for J2E + 4u4k+2 du _foro J2E + 4U4k+2 duo

+ 4u 4k+2 du =

o

,

,

v

,

, 'V

=It

=10

We shall compute each of the above integrals.

l

2E + 4u 4k +2 du o ..J2E + 4u 4k +2

II

r

2E

l

r

1

o ..J2E + 4u 4k +2 2E7+4J,

du +4

l 0

r

u 4k +2 du ..J2E + 4U 4k+2

with J defined in Lemma 5.21. Substituting the expression for J, yields

r

Replacing

7

by

70

. IE

E7

h = 2E7 + k + 1 V2" + r 4k +2 - k + 1 . = 0 and r by ro yields an expression for 10 10

=

ro k+1

E _ + r 04k +2 2

Summing up, yields

S(r, t, 7)

W(¢,r)

+ U(7) + V(t)

=

h(r) - Io(ro)

+ U(7) + V(t)

E7 2E7+ -r- JE - +r4k +2 - - - -rok+1 2 k+1 k+1 -it + _1_ (kE7 k+1

-E +r 4k +2 -E7+()t+O 2 0

IE + r4k+2 _ ro· rE + r + V2 V2 0

+ r·

4k 2 +

0) .

Let So = S(ro, 0, 0). Then the constant 0 = So. Taking So = 0 we arrive at the following result, obtained also by Beals, Gaveau and Greiner [4J in a different way. THEOREM

5.32. The complex action, which does not start at the origin, is given

by (5.59)

g(r, ro, t, E, 7) = -it + k: 1 (kE7

When ro = 0, we obtain Theorem 5.22.

+ rJ ~ + r 4k +2 -

ro

E2

+ r 04k +2 )

.

174

5.


5.10. Theorem of Jacobi and applications

In this section we shall show the non-uniqueness of the complex geodesics between any two given points for the case k 2:: 1. We provide the complete picture of the number of geodesics from the origin and from outside of the origin. In order to do this we need to take advantage of Jacobi's method of integration of the Hamiltonian system when a solution for the Hamilton-Jacobi equation is known. We start by recalling a few known definitions and results, see [27], vol. II, p.366. Consider a solution S(T, x, a) of the Hamilton-Jacobi equation

as aT

(5.60)

+ H(x, Sx)

=

0,

which depends on n parameters a = (a l , ... , an) E P. In our case n = 1 and a l = E the energy only, and x = r. Assume that S is defined for (T,X) E G. DEFINITION 5.33. A function S(T, x, a) is called a complete solution of the Hamilton-Jacobi equation (5.60) if we have (i) S E C2 (G x P) and det(Sxiaj) =f. 0 on G x P. (ii) For any a E P the function S( . , " a) is a solution of (5.60), i.e.

as(T,X,a) ( ( )) aT +Hx,SxT,x,a =0. We remark that a complete solution of (5.60) should not be mistaken for a general solution, see [27], vol. II, p.370. Consider the Hamiltonian system

p = -Hx(x,p).

(5.61)

Jacobi's method of obtaining a general solution x(T,a,b) and p(T,a,b) of the system (5.61) depending on 2n parameters a = (al, ... ,an ), b = (bl, ... ,bn ) by means of a complete solution S( T, x, a) of (5.60) consists of the following two steps: • Solve the n implicit equations with respect to x, obtaining a solution x = x(T,a,b) . • Construct the function p = p( T, a, b) defined by

p(T,a,b) = Sx(T,x(T,a,b),a). The formal statement of Jacobi's result is given below. THEOREM 5.34. {Theorem of Jacobi} Let S(T,x,a) be a complete solution of the Hamilton-Jacobi equation {5.60}, and suppose that x = x( T, a, b), p = p( T, a, b) are functions of class Cl satisfying the equations (5.62)

Sa(T,x(T,a,b),a) = -b,

p(T,a,b) = Sx(T,x(T,a,b),a).

Then x( " a, b), p( " a, b) is a solution of the Hamiltonian system {5.61} depending on 2n parameters a and b. PROOF. Differentiating with respect to a i in the Hamilton-Jacobi equation

(5.63)

Sr(T, x, a)

and inserting x

(5.64)

=

+ H(x, Sx(T, x, a)) =

0,

X(T, a, b), yields

Srai (T, x( T, a, b), a)

+ Hpk (X(T, a, b),p( T, a, b))Sxkai (T, x( T, a, b), a) =

O.

175

5.10. THEOREM OF JACOBI AND APPLICATIONS

Differentiating the first equation of (5.62) with respect to (5.65)

Sair( T, x( T, a, b), a)

T,

we obtain

+ Saixk (T, X(T, a, b), a))j;k = O.

Subtracting (5.64) and (5.65) yields

[j;k _ Hpk(x(T,a,b),a)]Sxkai(T,x(T,a,b),a) = O. The condition det Sxkai

=f. 0 implies j;k = Hpk(x(T,a,b),a),

which is the first equation of system (5.61). To derive the second equation of the Hamiltonian system (5.61), differentiate (5.63) with respect to xi and obtain

Srxi(T, x, a)

+ Hxi(T, x, Sx(T, x, a)) + Hpk (T, x, Sx(T, x, a))Sxkxi(T, x, a) =

which becomes after using the relations of

j;

0,

and P

(5.66) On the other hand, differentiating in the second relation of (5.62) with respect to and using (5.66), yields

T

Sxir( T, x( T, a, b), a) + Sxtxk (T, x, a)j;k -Hxi(T, X,p(T, a, b)).

o In our case a and b are I-dimensional. The complex action is a complete solution because

-(2k + I)r 4k +1 (E + 2r4k+2)3/2

{j2g

8E8r

= -

=f. 0,

for r > O.

We should solve

8g =-b 8E and get r

= r(T, E, b). _1_ [kT k +1

This leads to

+ !:.

1 _ ro 1 2 ../2E + 4r 4k +2 2 j2E + 4r~k+2

1= -b,

which can be rewritten as (5.67)

r = ro _ 2kT _ 2b( k ../2E + 4r 4k+2 j2E + 4r~k+2

+ 1),

with the solution r = r( T, E, b). The following result deals with the number of parameters of a solution which starts from the origin. PROPOSITION 5.35. Let k :::: 1. In the case when r(O) only on the energy parameter E. PROOF.

Making ro

= 0 in

= 0 the solution depends

the equation (5.67), yields

r

= -2kT - 2b(k + 1) . ../2E + 4r 4k+2 When T = 0, then r = r(O) = ro = 0, and hence b = 0 and b is not a parameter any more. Hence r = r(T, E) depends only on the energy parameter E.

(5.68)

o


176

Let

A(1')

l'

-

-----c==:::::==;:~

- ..j2E + 4r 4k +2

The function A(r) has the following properties (see Figure 5.5).

(i) A(r) ~ 0, with A(r) = 0 { = } l' = O. (ii) There is an unique critical point 1'1 > 0 for A(r). (iii) A(r) is increasing on (O,rd and decreasing on (1'1,00). (iv) The line y = 0 is a horizontal asymptote to +00 for A(r). (v) The maximum of A depends on the energy E and it is given by max A =

(f,.)~

VE (2+k)

Figure 5.5: The graph of >"(1').

In the following we shall keep the energy E fixed. Making b = 0 in equation (5.68) yields

A(r) = -2kT, which has zero, one or at most two real solutions for negative values of T. (For positive values we need to take positive sign in W of the previous section). Hence we arrived at: PROPOSITION 5.36. Given the energy E and a point x, there are at most two solutions joining the origin and the point x.

We shall consider now the case 1'0 i- O. Given E, we shall find an interval where the parameter b belongs to. Equation (5.67) yields 0
jB cot(2aj B)r;

+ tB.

j=l

Then n

F'(B) = t - Lajp,(2aj B)r;' j=l

The lemma follows from the properties of the function LEMMA

5.45. For any (x, t) with x"

#0

p"

see chapter 1.

0

and t > 0, there is exactly one branch

of the set

rO = {T:

(5.103)

that goes to as T -+ 00. PROOF.

00

Imf(x,t,T)

= O,ReT > O,ImT > O}

in the quadrant Re T > 0, 1m T > O. On this branch Re f increases

Assume T =

S

+ iB,

with s, B > O. Then

I m ( T cot h) T

(5.104)

=

B sinh 2s - s sin 2B 2

2 sinh s + 2 sin 2 B

> 0,

since B sinh 2s - s sin 2B = B(sinh 2s - 2s)

+ s(2B - sin 2B) > O.

If B = ImT > 0 fixed, then (5.105)

lim Im(Tcoth)

= B,

uniformly for bounded B.

8-++00

When sin B = 0, then (5.106)

Im(B coth T)

B s

= B coth s 2: -.


186

We have the following estimation

:S

(1m

cos 0 sinO

Tcoth T)

+

-0 sinh2 S cos 2 0 0 cosh 2 S sin 2 S sin2 0 +-----------n----~~------(sinh2 S sin 2

+

0)2

2s cosh S sinh S cos 0 sin 0 + (sinh2 S + sin2 0)2 O(s+O) sinh2 S '

(5.107) as

S

---7

00.

We write n r2 1mf(x,t,T) = -st+1mL2ajTcoth(2ajT) ~.

(5.108)

)=1

Choose aj such that rj =1= o. Let 0 be large enough, such that sin(2aj O) = 0 and coth(2ajrj(0/2t)1/2) < 3/2. Using (5.105), (5.106) and (5.108), there will be at least one point T = So + iO, when 1m f(x, t, T) = O. Using (5.106) and (5.108) it follows that (5.109)

and (5.105), (5.108) and (5.109) imply that (5.110)

and hence So belongs to the interval 3

n

~2· JO ~ So ~ "2 ( L

(5.111)

v.d

k=l

t) 2

ak r

O.

From (5.107) and (5.111) we get the estimation (5.112)

a 1m Ln I-a S

j=l

r21 S +0 2ajTcoth(2ajT)-L ~ C. 2 ~ C 10e- C2VO , 2 smh (2a1s)

S> So,

where C, C 1 and C 2 are positive constants which depend only on the aj's and on (x, t). Then for suitable large 0 we have (5.113)

a t, s + zO) . < 0, aJ(x,

s > So,

and there is only one solution So of 1mf(x,t,s+iO) = O. It follows from (5.105) that no branch of f 0 can escape to 00 between two consecutive such lines 1m T = 0, with sin(2a j O) = O. Since 1m f = 0 on the imaginary T-axis, a branch can escape from the quadrant through the imaginary T-axis only at a critical point of f. By Lemma 5.44, there is no such escape for large O. This still leaves the question of a possible branching of fo between two such consecutive 1mT = 0 lines. Two such branches must join at the two So points, since they are unique. This implies the existence of a bounded region n on which 1m f is harmonic and non-constant and vanishes on an, which is a contradiction. Thus, for large ITI, there is exactly one branch of fo that goes to infinity within the quadrant.

5.12. MODIFIED COMPLEX ACTION FUNCTION ON Hn

187

In particular, f has no critical points on this single branch of roo Therefore, on this branch, Re f must increase or decrease, it cannot have a stationary point. Now, using cosh u sinh u - i sin v cos v sinh2 u + sin 2 v vcotv,

coth(u + iv) iv coth(iv)

implies that Re{ (u + iv) coth(u + iv) - iv coth(iv)} u cosh u sinh u + v sin v cos v ----,,------~--- - v cot v sinh2 u + sin 2 v

sinh2 u . 2 . 2 . (u coth u - v cot v). smh u + sm v It follows that when sin () = 0, we have

(5.114)

Re(TcothT) = scoths.

Relations (5.111) and (5.114) yield 2

n

(5.115)

Re f(x, t, So

+ i(})

= (}t

+L

2akSO coth(2akso) r; ,

k=1

which increases as () ~ 00. Consequently, Re f increases on the branch of r 0, which goes off to infinity, and we have derived Lemma 5.45. 0 PROPOSITION PROOF.

5.46. If f'(T)

= 0 and

Imf(T)

It follows from the next lemma.

= 0,

then T is imaginary.

o

LEMMA 5.47. Assume x" i= 0 and t > O. Let the critical points of f on the positive imaginary axis, counted according to multiplicity, be i(}l, ... , i(}2m+l, with

(5.116) Let

r

be the union of r 0 and the closed intervals

(5.117) Then r, oriented in the direction of increasing Re f, is a simple connected curve from 0 to 00. PROOF. The set r is a union of analytic arcs that meet only at critical points of f. At any critical point T i= 0 there will be at least one branch that is oriented toward T and one that is oriented away from T. The real part Re f increases along the branch of r that leads to 00. It follows that any point of r is contained in at least one maximal oriented contour rIC r. Moreover, r 1 must begin at 0 and eventually contain the branch that leads to 00. If r 2 is another such maximal oriented contour, either it coincides with r 1 or the union of the two contains a simple closed contour r 3. Suppose the latter; then r3 encloses a bounded region O. But then 1m f is harmonic and non-constant on 0 and vanishes on the boundary, which is a contradiction. Thus there is a unique such maximal contour, and it 0 coincides with r itself.


188

We still assume t > 0 and x"

=I o.

A solution of

n

It I = L

(5.118)

ajp,(2aj(})r;

j=l

corresponds to a critical point r = i(} of the modified complex action function f. The square of the length of the associated geodesic is 2f (i(}). Let (i(}k) be the critical points of f on the positive imaginary axis, numbered as in (5.116). Then (5.119) with strict inequality where the corresponding inequality in (5.116) is strict. This is the same as saying that the critical points occur in this order on the oriented curve r of Lemma 5.47. Any other ordering of the critical points along r would imply that r has self-intersections, which contradicts Lemma 5.47. COROLLARY 5.48. The lengths of the geodesic associated with a solution of (5.118) increases with ().

5.13. A geometric formula for the fundamental solution In this section the complex action will be used to find the fundamental solution

K(x, t; Xo, Yo) for the Heisenberg operator D..H = ~(Xf D..~,t) K(x, t, Xo, Yo)

=

6x -

+ X~), i.e.,

xo 6t- to,

is the Dirac distribution centered at ao and Xl = OXl + 2XlOt, X 2 = Due to the left invariance with respect to the Heisenberg translation it suffices to look for the fundamental solution only at the origin, because

where 6a -

OX2

ao

+ 2XlOt·

K(x,t;xo,to)

K((xo,to)-l 0H (x,t);O,o),

=

where 0H is the Heisenberg group law. One shall look for a fundamental solution in the form

K( x, t,. 0, 0) --

Joo E(x,(T)V(X, r) d ) -00

g x, t,

T,

T

see [3], [4]. In the above relation g = -it + r2 coth(2r) is the complex action and E = -oglor = sin~~~27) is the energy along the geodesics. The function v(x, r) is called the volume element and satisfies the transport equation (5.120)

where (5.121)

is called the first order transport operator. The following result shows that in fact T is the differentiation along the bicharacteristics. PROPOSITION

5.49. If f is a function and (x(s), t(s)) is a subRiemannian geo-

desic, then (5.122)

d

Tf = dJ(x(s), t(s), s)18=7'

5.13. A GEOMETRIC FORMULA FOR THE FUNDAMENTAL SOLUTION PROOF.

Using Xig

189

= Xi, i = 1,2, then the chain rule yields

d dJ(x(s), t(s), s)

d dJ(x(s), t(s), s)

=

(Xd) (Xl g)

of (.

+ at ! +

2

+ (X2f)(X2g)

. 2 .) X2X~ - X1 X 2,

of + as

=0

(Xd)(X 1g)

of + (X2f) (X2g) + as'

Evaluating the previous identity at the final value s COROLLARY

= T, we obtain

(5.122).

0

5.50. If E is the energy along the bicharacteristics,

TE=O. PROOF.

It comes from the fact that E is constant along the bicharacteristics.

o

The transport equation (5.120) follows from a formal computation.

Since

K(x, t; 0, 0) must satisfy ~HK(x,t;O,O) = 0,

(x, t)

i=

(0,0),

this formally leads to

o (5.123)

1: ~H(~V) 1

00

-00

dT

[~H(EV) _ (Xg, X(Ev))

9

g2

+ 2EvH('\7g) _ g3

where

(Xg, X(Ev)) = X 1 gX1 (Ev) and H('\7g)

+ X 2gX2(Ev)

= ~ [(X 1 g)2 + (X29)2]. The Hamilton-Jacobi equation

~~ + H('\7g) =

0

(~H9)EV] dT,

g2

190


yields

2EvH('\1g)

_ 2Evog/fh = Ev~~

or g2

g3

-~~(Ev) + ~(EV). g2

Assuming limr->±oo Ev / g2

or

or

g2

0, (5.123) yields

=

0=/00 [D.H(EV) _ T(Ev) + ~D.H9)EV]

dr 9 9 with T defined by (5.121). Corollary 5.50 and an integration by parts yield

(5.124)

-00

Under the assumption lim Tv r->±oo

+ (D.Hg)V

= 0

9

,

the identity (5.124) becomes

0= /

00

a

1

- (D.H(Ev) + -a (Tv + (D.H9)V))

dr r and hence it suffices that v satisfies equation (5.120). In the next section we shall solve the transport equation (5.120). -00

9

5.14. The volume element on the Heisenberg group

Considering

Xl = a computation shows

aX! + 2X20t,

Xlg

2Xl coth(2r) - 2ix2,

X 2g

= 2X2 coth(2r) + 2ixl,

Xfg = 2 coth(2r) , Xig = 2 coth(2r),

where 9

= (xi + x~) coth(2r) -

it.

Hence

D.xg = 2coth(2r). We need to solve the transport equation

D.x(Ev)

+ :r (Tv + (D.Xg)v)

=

O.

We are looking for a volume element v, which satisfies simultaneously both equations

D.x(Ev) Tv + (D.Xg)v

0

O.

5.14. THE VOLUME ELEMENT ON THE HEISENBERG GROUP

Consider the ansatz v

191

= A(r 2 )B(T). In the following we shall compute Tv. = 8X1 v = 2x1A'(r2)B(T). X 2v = 8 v = 2x2A'(r2)B(T). XlV

X2

Tv

+ (X 19)X1 V + (X 2 9)X2 V A(r2)B'(T) + 2(X1 coth(2T) - iX2)2x1A'(r2)B(T) +2(X2 coth(2T) + iX1)2x2A'(r2)B(T) A(r2)B'(T) + 4r2 coth(2T)A'(r2)B(T).

8r v

The equation Tv

+ 2coth(2T)v = 0

can be written as

A(r2)B'(T) + 4r2 coth(2T)A'(r2)B(T)

+ 2coth(2T)A(r2)B(T) =

O.

Dividing by A(r2)B(T) coth(2T) yields

B'(T) A'(r2) + 4r2 - - + 2 = O. B(T) coth(2T) A(r2) There is a separation constant C such that B'(T) Ccoth(2T) B(T) 2 A'(r2) -(C + 2). 4r A(r2) ---,----=-""-----

Integrating the above equations we obtain

B(T)

K1 sinh(2Tf/2

2

= K 2r

A(r)

-2(C+2) 4

,

where K 1 , K2 E IR are constants. Hence the volume element is

v=k

sinh(2T)c/2 r

~

,

4

with k constant. Then we choose the constant C such that tlx(Ev)

= 0, where

2r2 sinh(2T)c/2 Ev = SIn . h 2 (2 T ) . k r 2(C+2) 4 Obviously, tlxEv = 0 if Ev does not depend on r, i.e., when C = 2. In this case the volume element becomes _ k sinh (2T) v2' r

In the following we shall show that this is the only possible volume element, i. e., we cannot choose another value of C such that tlx(Ev) = O. Let m-2- 2(C+2) -

4'

We choose m such that the following condition is satisfied

tlx(rm) = O. The following Lemma yields only one good value, which is m

= 0, i.e., C = 2.


192

LEMMA

5.51. We have

.6. x (rm) = 2m 2 r2(m-l). PROOF.

(X?

A computation shows

+ X~)(rm)

4r2(m-2)m 2xi

=

+ 4r 2 (m-l)m -

+4r2(m-2)m2x~

_

4r 2(m-2)mxi

4r2(m-2)mx~

+ m2x~ - mx~]

4mr 2(m-1)

+ 4r 2(m-2) [m2xi -

4mr 2 (m-l)

+ 4r 2(m-2) [(m 2 - m)~]

mxi

+ 4r 2(m-2)m(m - 1)r2 4mr 2 (m-l)(1 + m - 1) = 4m 2r2m-2.

4mr 2(m-1)

o 5.14.0.9. The fundamental solution. In this section we shall compute explicitly the fundamental solution given by the integral

° -1 -

K( x, t,. 0, )

(5.125)

00

()

E(x, r)v(x, r) d

r,

9 x, t, r

-00

where E(x r) ,

=

21xl 2

9

sinh2(2r)'

=

(xi

+ x~) coth(2r) -

( ) _ k sinh (2r) ' v x, r I x 12

it,

K(x, t, 0, 0) may be thought of as the (action)-l summed on the characteristic variety over (0,0) with measure Ev. Changing the variable of integration from r to 9 yields ( E(x, r)v(x, r)dr g(x,r)

K(x, t; 0, 0)

JJR

{ ~ (_ ag )

(5.126)

JJR 9

dr

= _ {g+ v dg ,

Or

Jg _

where g±

= T->±OO lim 9 = ±(xi + x~)

- it.

In this notation we have

ei1r /2 v

where we chose k

=

1

2 -)17(g=+=_====;g):=;=(g=_=g_"'7) , = - -47r-

~. Then

K(x, t; 0, 0)

=

1 c_

dg V-,

9

where we made a branch cut

c=

(-00 - it, g-J

U

[g+, -it + (0),

9

5.14. THE VOLUME ELEMENT ON THE HEISENBERG GROUP

193

o

Figure 5.6: The "dumbbell" with waist at g±.

with upper and lower directed edges denoted by C+ and C_, respectively. See Figure 5.6. Integrals of v / 9 vanish on upper and lower semicircles as their radii increase without bound. Therefore we can integrate on a "dumbbell" with waist at g±. Inside this domain v / 9 has a simple pole at 9 = O. Hence i7r 2

v(g)

e / -1/2 --(g+) (-g_) -1/2 = Res I

= gOg

471'2

1 = --.

271'z

1

C_uC+

From the residue calculation, we know that

d9 _(27ri)-!j dg_ r Vdg- . JcrC+ ve vJc 9 -c 9 C_ 9 Hence, one has

and therefore we obtain K(x t· 0 0) , , ,

1 871'

1

= ------r,=-::::;=;===.=;:

Jl x l4 + t 2

dg v-. 9


194

One may show that in the distributions sense. This is the Folland-Stein formula on the Heisenberg group. Unfortunately this simple formula is just a lucky coincidence of too much symmetry, and on general non-isotropic Heisenberg groups the fundamental solutions are given in the form (5.126). The reason is that the fundamental solution must include all the distances, which necessitates the use of g, and the summation over all the distances means integration in the variable 7. REMARK 5.52. dc(x, t; 0, 0) = Jlxl4 from the origin, see chapter l.

+ t 2 is the Carnot-Caratheodory distance

5.15. Exercises EXERCISE

5.l. Let a(s) be a differentiable function such that asinha

Show that a(s) = 4s EXERCISE

+ ao,

= 4 sinh a,

acosha

where ao E IR is a constant.

5.2. Show that the equation cosh(4 + x) - cosh x

has the unique solution x EXERCISE

= 4 cosh a.

=0

= -2.

5.3. Consider the initial condition problem

r= i) Show that

2rcoth(27),

reO)

=

O.

J cothxdx = lnsinh(x) + C, for any x > O.

ii) Using the method of separation of variables, show that

r(7)

= Csinh(27)

where C is a constant. iii) Show that for any 0 :::; s :::; 7 we have _ ( ) sinh(2s) r (s ) - r 7 sm . h( 27 ). 5.4. i) In polar coordinates Xl = r cos ¢, X2 = r sin ¢, = Ox} + 2X21x12kOt, X 2 = OX2 - 2X11x12kOt, become

EXERCISE

fields Xl

sin¢ k. cos¢or - --04> + 2r2 +1 sm¢ot r . cos¢ k sm¢or + --04> - 2r2 +1 COS¢Ot. r ii) Show that

X 1g X 2g where 9 is the complex action.

cos¢org - 2ir 2k +1 sin¢ sin¢org + 2ir 2k +1 cos¢,

the vector

5.15.

EXERCISES

195

iii) Prove that

D.xg

=

1

2

2(X l

1

2

1

2

+ X 2 )g = 2(Org + -:;,Org).

EXERCISE 5.5. Let v = v(r, r), where r = Ixl and Xl, X 2 as in Exercise 5.4. i) Then v is rotational symmetric and translation invariant along the t-axis. ii) Show that

(Note that v and 9 satisfy the same equation!). EXERCISE 5.6. Find the functions A and B such that v(r, r) = A(r) satisfies the transport equation

ov ED.x v + (T - D.xg) or

+ B(r)

= 0,

following the steps: i) Show that

~(A"(r) + ~A'(r)), B'(r),

T(~~)

0;:.' + I)Xig) (XiB') = B"(r). i=l

ii) Choosing B such that B"(r)

= 0, show there is a constant G such that

-~B'(r) E

G,

=

r2 A"(r) + r A'(r) r 20;g + rOrg

G.

iii) Using the substitution r = eP , show that

r2 A" (r) + r A' (r) r 20;g + rOrg

o;A(e P ) =

o;g(e P ).

iv) Using ii) and iii) show that

o;A(e P ) = G o;g(e P ). v) Show that

A(r) B(r)

= Gg(r)+alnr+b, = Bo-GEr, Bo,G,a,bER

vi) Find the function v in both cases r(O) = 0 and r(O) = ro

-=1=

O.

196

5. COMPLEX HAMILTONIAN MECHANICS EXERCISE

5.7. The Heisenberg vector fields on HI in polar coordinates are

sin a", + 2r sin at, r cos X2 sin ar + - - a", - 2rcosat , r and the complex action is 9 = -it + r2 coth(2r). Find a solution of the transport equation cos ar

Xl

-

--

av

EAx v + (T - AXg) ar

= a(r)

under the form v

=0

+ b(r), using the following steps:

i) Show there is a separation constant C such that

r 2a"(r)

(5.127)

+ ra'(r) =

b"(r) + 2coth(2r)b' (r) + (coth2 r - l)C

=

C, O.

ii) Using the substitution r = eP show that the equation (5.127) has the general solution 1 a(r) = 2C(ln r)2 + Clln r + C2, iii) Let v

= coth(2r).

Show that Vi

iv) Let b(r)

= ~(coth(2r)).

(5.128) v) Let y

v 2 ).

Show that ~ satisfies the ODE

v2)~II(V) - V~I(V) - ~ = O.

(1 -

= ~'.

= 2(1 -

Show that y satisfies the linear equation _ C_1_ 1 - v2 Y - 4 1 - v2 '

I __ V_

y with the integrant factor J.L( v) =

v'1J2=1.

vi) Show that

v'1J2=1) v'1J2=1

y(v) = _ Cln(v + 4

vii) Show that COSh-l (coth 2r)

+

Co

J v2 -

, 1

Co ER

= In(cothr).

viii) Show that

J v'1J2=1

COSh-l (v) dv

ix) Show that

~(v) = - ~ (COSh-l (V))2 +

Co COSh-l (v) + C,

x) Show that

b(r) = -

~ (COSh-l (coth2r))2 +Cocosh-l(coth2r).

5.15. EXERCISES

197

xi) Show that the volume element is C v = a(r) + b(T) = gln(r 2 tanh T) In(r 2 coth T). EXERCISE 5.8. Let E(T) denote the energy along a complex geodesic c : [0, T] of a step 2 manifold. Prove that if E satisfies (i) E(T) is real,

~3

(ii) limlrl-->oo E( T) = 0, then the momentum () is pure imaginary.

~

CHAPTER 6

Quantum Mechanics on the Heisenberg group We have seen in the first section of chapter 5 how Heisenberg operator arose from the harmonic oscillator by a certain complex quantification. In this chapter we shall emphasize more relations between the quantum harmonic oscillator and the Heisenberg operator. It turns out that the quantification of energy property is identical for both models. Before starting this discussion we shall recall in the next section a few useful things about Quantum Mechanics.

6.1. Linear Harmonic Oscillator The potential energy of a material point of mass mo = 1 in an elastic force field IS

(6.1) where the constant w is the oscillator pulsation. Let E be the oscillator total energy. Classically, the particle will oscillate between Xl = -A and X2 = +A, where 2

E = ~A2. 2

From the quantic point of view, the harmonic oscillator is characterized by a density probability function 'l1 (x, t), which satisfies the one-dimensional Schrodinger equation (6.2)

f)1Ji ih-f)

h2 d2 d 2'l1

+ -2 t

x

= U(x)lJi,

with U(x) given by (6.1). The linear harmonic oscillator is one of the most important examples in elementary Quantum Mechanics. It describes vibrations in molecules and their counterparts in solids. The most eminent role of this oscillator is its linkage to the boson. The boson is one of the conceptual building blocks of microscopic physics, describing the models of the electromagnetic field, providing the basis of its quantization. It is named after the Indian physicist S. N. Bose, see

[11]. I t is a well-known fact that the energy of the linear harmonic oscillator is quantified by the relation

En =

(n + ~)hw,

n = 0,1,2, ...

where h is the Plank constant. Hence the harmonic oscillator, from the quantic point of view, may have only discrete values for the energy. The difference between 199

200

6. QUANTUM MECHANICS ON THE HEISENBERG GROUP

two successive levels of energy is always the same, equal to hw. The minimum energy the linear harmonic oscillator may have is Eo _ hw - 2 .

This is called the ground-state energy, see Figure 6.l. U(x)

.................................................. --- EZ hoo ........................................... El

hoo

x

o Figure 6.1: The energies of the linear harmonic oscillator are discrete.

The same quantification relation will be obtained in the Heisenberg case. The construction will be done by analogy with the elementary Quantum Mechanics construction, which uses the creation and anihilation operators. Schrodinger's equation (6.2) can be written as

oW

~

ihat = H(w), with the Hamiltonian operator

(6.3) where ~

h d

x=x

P = i dx' are the momentum and the position operators. The uncertainty Heisenberg relation is expressed as

(6.4) where IT is the identity operator. In the elementary Quantum Mechanics the following creation and anihilation operators are used

(6.5)

(6.6)

~ a

=

fw~

V2i x -

i ~ V2hw P,

6.2.

THE OPERATORS

Z

AND Z AND THE ENERGY QUANTIFICATION

They are obtained by a formal factorization of the operator h~ it. operators have the following properties (6.7)

201

The above

~+] [~a ,a

IT, 1 ~ 1 (6.8) -H+-IT hw 2 ' 1 ~ 1 -H--IT. (6.9) hw 2 One may find more about the linear harmonic oscillator in [45]. In the next section we shall find similar relations on the Heisenberg group.

6.2. The operators Z and

Z and the energy quantification

In this chapter we shall adopt the following notation for the Heisenberg vector fields (6.10)

Xl = OXl

+ 2hx20t,

X 2 = OX2 - 2hxIOt,

where h > 0 stands for the Plank's constant. The Heisenberg operator

(6.11)

D.H

=

1 2 "2Xl

1

2

+ "2X2

and the Hamiltonian operator (6.3) are both sum of squares of operators which do not commute. The relation [Xl, X 2 ] = -4hot is an analogue of the uncertainty relation (6.4). The vector fields Xl and X 2 correspond to some physical fields, which cannot be observed simultaneously. Let Z

(6.12)

(6.13) which correspond to the operators a;+ and a;- given in (6.5)-(6.6). The following proposition shows that Z and Z have similar properties to operators a;+ and a;given above. PROPOSITION 6.1. The operators Z and

Z verify

the following relations

2)

[Z,Z] ZZ

2hiot , 1 . "2D.H + zhot,

3)

ZZ

1 . "2D.H - zhot .

1)

PROOF. 1) It follows from 2) and 3) by subtraction. 2) We have 1 ZZ 4(X I + iX2 )(XI - iX 2 )

~(Xf + X? -

i[Xl' X 2])

. t· 41(Xl2 + X 22) + zho 2) It is obtained from the second one by conjugation.

o

6.

202

QUANTUM MECHANICS ON THE HEISENBERG GROUP

COROLLARY 6.2. The Heisenberg operator can be expressed as 2ZZ + 2ihot

= 2ZZ + [Z, ZJ, = 2ZZ - [Z, Z].

2ZZ - 2ihot

Consider the operator 1

2

ZZ + ZZ

(6.14)

1

2

"2(X1 + X 2 )

-

"2 iA [X1,X2]

+ A[Z, Z].

Consider a quantum system described by the wave function 1jJ : ]Ra ---> C, which satisfies the equation 1:1)..'1/) = O. The real number A will be called the energy of '1/'. Like in the case of the linear harmonic oscillator, we shall assume the existence of a ground state 1jJo, which is a solution of Z1jJo = o. We shall discuss the solutions of this equation in a later section. Here we are concerned only with the values of A. We shall find Ao such that l:1)..o1jJo = o. A() plays the role of the ground-state energy. Corollary 6.2 yields =0 -~

-

(2ZZ - 2ihot )1jJo = 2Z (Z1jJo) -2ihot 1jJo 1. "2 Z[X 1 , X 2]1jJo,

and hence Ao = + l. Z will be considered as creation operator. We can create excited states 1jJl, 1jJ2, 1jJ3, ... , where

1jJ1 = Z1jJo, 1jJn+l = Z1jJn = Zn1jJo. In this section we shall compute the energies An associated with the excited states 1jJn, i.e., such that l:1)..n1jJn = O. Using again Corollary 6.2 and the fact that commutes with Z and Z, we have

at

I:1 H (Z1jJo) = (2ZZ - 2ihot )Z1jJo Z(2ZZ)1jJ() - 2ihot 1jJ1 Z(2ZZ + 2ihot - 2ihot )1jJo - 2ihot 1jJ1 Z(I:1 H 1jJo) - 2ihot Z1jJo - 2ihot 1jJ1 Z(I:1 H 1jJo) - 4ihot 1jJ1 Z(2Z Z1jJo -2ihot 1jJo) - 4ihot 1jJ1 '-v-' =0

1

-

"2i[X1, X 2]Z1jJo

+ i[Xl' X 2 ]1jJ1

(1 + Di[X1, X ]1jJ1. 2

Hence I:1 H 1jJl =

and then Al

(6.15) Let 1jJn+l

=

(1 + ~)i[Xl' X ]1jJl,

I:1 H 1jJn = (n

=

2

3. We shall prove by induction that

+ ~)i[Xl' X 2]1jJn.

Z1jJn· A similar computation yields I:1 H 1jJn+l

= Z(I:1 H 1jJn) -

4ihot 1jJn+l.

6.3. THE GROUND-STATE 1/10 AND THE STATES .pH

203

Using the induction hypothesis (6.15) yields

z( (n + ~ )i[X j , X 1

2 ]1/In)

+ i[X j , X 2 ]1/In

-

(n

+ "2 )i[X j , X 2]Z1/In + i[Xl' X 2 ]1/In

(n

+ 1+

D

i[Xl' X 2 ]1/In+1.

Hence we have proved by induction that An n = 0, 1,2, ...

=

2n

+

1, where

~An 1/In

0, for

To conclude, we have the following comments. 1) The eigenvalue A/2 has only discrete values, equal to n + ~, which is the same quantification relation as in the linear harmonic oscillator case. 2) For the values A = 2n + 1 the operator ~A ceases to be hypoelliptic. A fundamental solution for ~A is constructed in [4]. The fundamental solution is extended meromorphically in A with obstructions at the simple poles ±A = 2n + l. 3) The negative sign of A is obtained if we start with a conjugate ground state 1P(J which satisfies Z1/l~ = 0, and define the the conjugate excited states 1/1~ = Z1/I~_1 = zn1/l~.

In a similar way, one may show that

~H1/I~ =

-(n + ~ )i[Xl' X2]1/I~,

n

= 0, 1,2, ...

i.e., ~-(2n+j)1/I~ = 0. 6.3. The ground-state 1/10 and the states 1/In The function

1/10 : 1R3 --+

C satisfies the ground-state equation

(6.16)

Z1/Io = 0.

Let 1/10 = U + iv, where U and v are real-valued functions. The following result is an analog of the Cauchy-Riemann equations. PROPOSITION

6.3.

1/10

= u+iv satisfies the ground-state equation

(6.16) if and

only if (6.17) PROOF.

2Z1/Io

(Xl - iX2 )1/io = (Xl - iX 2 )(u + iv)

+ iv) - iX2 (u + iv) Xju + iXjv - iX 2 u + X 2 v (Xju + X 2 v) + i(XIV - X2U). Xl(u

°

Then 21/10 = if and only if both the real and the imaginary components are zero, i.e., relation (6.17). 0


204

COROLLARY 6.4. Let 'l/Jo = u + iv be a solution of the ground-state equation (6.16). The following relations are equivalent i) 'l/Jo is constant, ii) v is constant, iii) u is constant. PROOF. We shall show that if v is constant, then 'l/Jo is constant. Using relation (6.17) we have X1u = X 2u = O. Hence, for any horizontal vector field W = aX I + bX2 , we have Wu = O. Let x i- O. We shall show that u(O) = u(x). From the connectivity theorem, there is a sub Riemannian geodesic joining the origin with the point x. Let c : [O,r] ......., ]R3 be the geodesic with c(O) = 0 and c(r) = x. The velocity vector c(s) = XI(S)X 1 + X2(S)X 2 is horizontal and hence c(s)u = 0, i.e., u is constant along the geodesic c(s) and hence u(O) = u(c(s)) = u(c(r)) = u(x). 0

The relationship between the states PROPOSITION 6.5. Let is the ground-state. Then 1) 'l/Jl = XI'I/JO = iX2'I/J0,

2) 3)

'l/JI = Z'l/Jl

'l/Jo

and

'l/Jl

is given by the following result.

Z'l/Jo be the first excited state, where

'l/Jl =

'l/Jo =

u + iv

2Zu = 2iZv,

= 6. H 'l/Jo.

PROOF.

1) Relation 2Z'l/Jo = (Xl - iX2)'l/Jo = 0 yields Xl'I/JO = iX2'I/J0. Then -

1

= Z'l/Jo = "2 (Xl + iX2)'l/Jo = XI'I/JO = iX2'I/J0·

'l/Jl

2) The equations (6.17) yield

Z'l/Jo = Xl'I/JO = X1(u + iv) = Xlu + iXlv -X2V + iXlv = i(Xl + iX2)v = 2iZv,

'l/JI

and

o Before solving the ground-state equation (6.16), we need a few results regarding the vector fields Z and Z. Let z (6.18)

= Xl + iX2

E C and define the operators

-

Oz

1

= "2(OXl

-

ioX2 )'

Oz

A straightforward computation shows (6.19) (6.20)

8z z = 0, ozz = 1,

1

= "2(OXl + ioX2 )'

6.3. THE GROUND-STATE Wo AND THE STATES Wn PROPOSITION

205

6.6. We have

(6.21) PROOF.

~(XI -

Z

iX2) =

~(aXl + 2hx2ad - ~(aX2 -

1

2(aX1 - iaX2 ) + h(X2

+ ixdat

=

-

az

2hxlat)

+ ihzat.

The second relation is obtained by conjugation from the first, because are conjugate operators.

8z

and

az

o Z and Z are the holomorphic and the anti-holomorphic tangent vector fields to the hypersurface where lDJ 2 (h) = {~ E

is the Siegel upper-half space in

([:2;

Im6 > h1612}

([:2.

has a group of biholomorphic affine transformations, which preserves lDJ 2 (h) and blDJ 2 (h). They are parametrized by (z, t) E ([: x R

([:2

(z,t)*~

(,

~~

= 6 + z, ~~ = 6 + t + ihlzl 2 + 2ih6z. This group acts simply transitively on the boundary, so blDJ 2 (h) can be identified with 1HI I (h) by matching a transformation (z, t) to the image of the origin

(z, t)

+---+

(z + t

+ ihlzI 2 ).

The group structure is that of the Heisenberg group lHIdh) with z = Xl + iz 2. In the following we shall construct a family of solutions for the equation (6.16). PROPOSITION

Z'¢o

=

2) Let

o.

Wi

=

PROOF.

6.7. 1) Let '¢O

= G(hlzl 2 + it) where G E CI(~) arbitrary. Then

F(hlzl 2 + it) where F

E CI(~)

arbitrary. Then Z'¢o =

o.

Using (6.19), a straightforward computation shows

Z'¢o

G'(hlzI 2 + it)Z(hlzI2 + it) G'(hlzI 2 + it)(8z + ihzat)(hlzl2 + it) G'(hlzI 2 + it)(h8 z lzI 2 - hz) = O.

Z'¢o

F'(hlzI 2 - it)(az - ihzad(hlzl2 - it) F'(hlzI 2 - it)(haz lzl 2 - hz) = O.

o The formulas for the excited states '¢n are given below.

206

6. QUANTUM MECHANICS ON THE HEISENBERG GROUP PROPOSITION

6.8. For any G

(6.22)

C1 (JR.), the function

= (2hzrG(n)(hlzI2 + it) = 0 with An = 2n + 1.

'l/Jn

satisfies the equation

~.xn '1/'11

We shall prove the relation by induction over n.

PROOF.

Z'l/Jo = ZG(hlzl 2 + it) G'(hlzl 2 + it)(oz - ihzot )(hlzl 2 + it) G'(hlzl 2 + it)(h'ozlzI2 + hz) 2hzG'(hlzl 2 + it),

'1/)1

and hence (6.22) is verified for n

'l/Jn+1

E

= 1. Assume (6.22) verified for some n ;::: 1. Then

Z'l/Jn = 2n h" (Z(zn)G(n l (hlzI 2 + it) 2nh n ( nzn-1 ~ G(n l (hlzI 2 + it)

+ z nZG(n l (hlzI 2 + it))

+ z nG(1I+1)(hlzI 2 + it)Z(hlzI2 + it))

=0

2nhnznG(n+1l(hlzI2 + ii)(hoz lzI 2 + hz) 2n+1hn+1zn+1G(n+1)(hlzI2 + it), which is formula (6.22) for n

+ 1.

o

An important space in Quantum Mechanics is the space of functions vanishing at infinity No = {f : JR.~ x JR. t -+ JR.; lim f(x, t) = O}. I(x,tll->oo

The function G should be chosen such that 'l/Jn E No, for any n ;::: For instance, if we choose G(u)

=

o.

~, then

1

cPo = hlzl 2 + it E No, and Proposition 6.8 yields

6.4. The operator 07" - Z In this section we shall investigate the non-commutativity relations aroused from the operators Z and Z. The calculus developed is a consequence of the Heisenberg non-commutativity relation (6.23) The 1-parameter group of diffeomorphisms associated to Z and the exponential function will be studied. LEMMA

6.9. For any integer n ;::: 1, we have

-n

-n-1

[Z, Z ] = nZ

-

--n-1

[Z, Z] = n[Z, Z]Z

.

THE OPERATOR aT - z

6.4.

207

PROOF. Relation (6.23) implies that [Z,1'] commutes with Z and 1'. This solves the second identity. For the first identity we shall use induction. The case -n -=T/-1 n = 1 is obvious. Assuming [Z, Z ] = nZ [Z, Z], we have

Z1'n1' _ 1'n+l Z

+ 1'n Z)1' - 1'n+1 Z [Z, 1"']1' + 1"1 Z1' - zn+l Z [Z, zn]1' + zn[Z, 1']

([Z, zn]

-n-1 - - -n nZ [Z,Z]Z+Z [Z,Z] nzn[Z, 1'] + 1'n[Z, 1'] (n + l)1'n[Z, 1'], D

which proves the induction step. THEOREM 6.10. Let f be an analytic function on lR. Then

[Z, f(1')] = !'(Z)[Z, 1'] = [Z,1'lf'(1'). PROOF. The function

f

can be written as a convergent power series

Lemma 6.9 yields

[Z,

[Z, f(1')]

L

L

f(n),(O)zn] n.

=

L

f(n\(o) [Z,1'n] n.

f(n),(O) nzn- 1[Z, 1'] = !,(1')[Z, 1']. n.

The second identity comes from the fact that [Z, 1'] commutes with the powers of Z and 1'. D Starting with a ground-state 'l/Jo, we have constructed the excited states 'l/Jn = 1'n'I/Jo. We can construct more general states of the form f(1')'l/Jo, with f analytic function. PROPOSITION 6.11. Let 'l/Jo be a solution of the ground-state equation Z'l/Jo = 0 and let f be an analytic function. 1) We have

Zf(Z)'l/Jo

!'(1')[Z, 1']'l/Jo

=

=

[Z,1']!'(1')'l/Jo.

2) In general, for any n E N, we have

PROOF. 1) We have

Zf(1') Using Z'l/Jo

=

=

[Z, f(1')]

+ f(1')Z.

0, Theorem 6.10 yields

Zf(1')'l/Jo

=

[Z,f(1')]'l/Jo

=

!,(1')[Z,1']'l/Jo.

208

6.


2) As [Z, Z] commutes with Z, using relation 1) and Theorem 6.10 yields

Z!,(Z)[Z, Z]'l/Jo = Z[Z, Z]!'(Z)'l/Jo [Z, Z] z!' (Z)'l/Jo = [Z, Z]2 !"(Z)'l/Jo !,,(Z)[Z, Z]2'l/JO.

Z2 J(Z)'l/Jo

o

By iteration we get formula 2). COROLLARY 6.12. We have

Zn J(Z)'l/Jo = J(n) (Z)( -2ihot )n'l/Jo = (-2ihot )n J(n) (Z)'l/Jo.

= zm'l/Jm

COROLLARY 6.13. Let 'l/Jm

be the m-th excited state. Then

zn'l/Jm = (m ~!n)!(-2ihodn'l/Jm-n'

0::; n::; m.

In particular,

zn'l/Jn PROOF. Choose J(Z)

= Zm

=

n!( -2ihot )n'l/Jo.

in Proposition 6.11 and get

znzm'l/Jo = m(m - 1) ... (m - n + 1) Z m-n( -2ihot )n'l/Jo

Zn'l/Jm

( m! ),(-2ihot )n zm -n'l/Jo. m - n . "-v--" ='Ij;m.-n

o Consider the operator eTZ with

T

E JR, defined by

(6.24) THEOREM 6.14. Let J be an analytic Junction and let v solution oj the Cauchy problem

Zu

OTU

UIT=O

=

J(Z)'l/Jo. Then the

on [0,00) xC

v,

is given by U

= J(Z + [Z, Z]T)'l/Jo.

PROOF. The solution can be written formally as u it using Proposition 6.11 U

=

eTZv

=

eTZ J(Z)'l/Jo

= ' " Tn ~n!

= eTZv. We shall compute

Zn J(Z)'l/Jo

n20

J(Z + T[Z, Z])'l/JO.

o

6.4.

THE OPERATOR aT -

209

Z

In the theory of the linear harmonic oscillator the excited states ('ljJn)n>O form a complete orthonormal system and the function v can be expanded always as v

=

L vn'ljJn = L vnzn'ljJo = f(Z)'ljJo. n~O

n~O

DEFINITION 6.15. PZ(7) = eTZv is called the I-parameter group of diffeomorphisms associated to the vector field Z.

Z belongs to the complexified Lie algebra of the Lie group H 1 . The above definition is motivated by the following properties. PROPOSITION 6.16. The following relations hold 1) pz(O) = v, 2) PZ(71 + 72) = PZ(T1) PZ(72), "171, T2 E~, 3) lTPZ(7) = Z. PROOF. Relations 1) and 3) follow from the Definition 6.15. We still need to show 2). Using Theorem 6.14

pZ(7deT2Z v

PZ(7d PZ(72)

= pZ(7deT2Z fCZ)'ljJo

pZ(7df(Z + 72[Z, Z])'ljJo eT1Z f(Z + 72[Z, Z])'ljJo

+ 72[Z,Z] + T1[Z, Z])'ljJo f(Z + (71 + 72)[Z, Z])'ljJo

f(Z

e(Tl+T2)Z f(Z)'ljJo = e(Tl+T2)Zv PZ(71

+ 72).

o The missing direction [Z, Z] plays an important role in the non-commutativity relations. The usual multiplication rule for exponentials ea+b = eae b does not work if a and b are replaced by non-commutative operators. This rules are present in Quantum Mechanics in many places and there are consequences of the uncertainty principle. For instance, we have

= eZ+[Z,zl'ljJo.

eZez'ljJo

This is obtained from the following result making s PROPOSITION 6.17. For any

7,

sE

eTZeSZ'ljJo

~

=7 =

1.

we have

= esZ+sT[Z,ZI'ljJo.

PROOF. Let f(Z) = e Sz and v = f(Z)'ljJo. Then

eTZeSZ'ljJo

= eTZv = f(Z + 7[Z, Z])'ljJo = esz+ST[Z,zl'ljJo.

o Because the operators e Z and eZ do not commute, we have another formula when their position is switched. THEOREM 6.18. For any (6.25)

0:,

f3 E C we have


210

PROOF.

For any

T

E

JR let

(6.26) We have (6.27) (6.28) because of Theorem 6.10 with fCZ) = eWz . Differentiating in relation (6.26) and using relations (6.27) and (6.28) yields

d~ T(T)

aZeTnZeT/1Z + eTaZ (JZe T/1Z (aZ + (JZ - (JZ)eWZeTJ3Z (aZ + (JZ)T(T)

+ eTaZ(JZe T/1Z - (JZemZeTJ3Z + eTaZ (JZe T{3z + ( - (JZe wz + emZ (JZ)e T{3Z

(aZ + (JZ)T(T)

+ [eTaZ, (JZ]e T{3z

(aZ + (JZ)T(T)

(aZ + (JZ)T(T) - (J[Z, eTaZ]e T{3Z

+ (JZ)T(T) (aZ + (JZ)T(T) (aZ

- Ta(J[Z, Z]e TaZ eT{3Z - Ta(J[Z, Z]T(T).

Hence T( T) satisfies the following linear ODE

d~T(T) + (Ta(J[Z,Z]- aZ -

(6.29)

(JZ)T(T) = O.

An integrant factor is e~T2a{3[Z.Zl-T(aZ+{3Z).

J.1(T) =

Multiplying by J.1, yields the exact equation d

dT (T(T)J.1(T)) = O. Hence

T(T) = CJ.1-1(T) = CeT(o:Z+{3Z)-~T2a{3[Z,Zl,

where C is a constant, Making

T

= 0 yields

C = T(O) = eo. aZ eO'{3Z = 1, and hence

T(T) Making

T

= 1 yields

=

eT(aZ+{3Z)-!T 2 a{3[Z,Zl.

relation (6.25).

o

Relation (6.25) written as

is an analog of the Campbell-Hausdorff formula from the Lie groups theory. This says that if X, Yare left invariant vector fields and a, (J E JR, then enX e{3Y = e nX +{3Y +!a{3[X,Yl+higher level brackets.

6.5. HEISENBERG DERIVATIVE

211

For the step 2 Lie groups there are no higher level brackets. We expect formula (6.25) to look more complicated in the case when there are many missing directions. 6.5. Heisenberg derivative In the elementary Quantum Mechanics the following Heisenberg differentiation rule holds dA

dt

where fi = - ~2 6.

+ U(x)

= ~[fi A] h

'

,

is the Hamiltonian operator and

A is an operator which

characterizes a real physical system. The operator is called conservative if ~~ = 0, i.e., [fi,A] = fiA - Afi = 0. If the operator A depends explicitly on the time t, the differentiation rule becomes

dA aA i ~ ~ dt = at + h[H, A]. This is a similar formula of the following relation from symplectic mechanics

df = af + dt at

L,

( af aH _ af aH) = af + H f aqi ap;

api aqi

at

{,},

where H is the Hamiltonian function and { , } is the Poisson bracket. The operator ~

l

~

{H,A}

=

~

~

h[H,A]

is called the quantum Poisson bracket. In the following we shall compute the quantum Poisson bracket and find the conservative operators of the type A = feZ), where f is an analytic function. The Hamiltonian operator in our case is ~

H

h'2

2

2

= -2(X 1 + X 2 ) =

2

-h 6. H .

PROPOSITION 6.19. If f is an analytic function, we have

[6. H , feZ)]

PROOF. Using 6. H

=

2Z1'CZ)[Z, Z] = 21'(Z)Z[Z,Z] = 2[Z, Z]1'(Z)Z.

2ZZ + [Z, Z], yields 6. H f(Z) - f(Z)6. H (2ZZ + [Z,Z])f(Z) - f(Z) (2ZZ + [Z,Z])

2ZZf(Z) + [Z,Z]f(Z) - 2f(Z)ZZ - f(Z)[Z,Z]

2( ZZf(Z) - f(Z)ZZ) = 2( Zf(Z)Z 2( Zf(Z) - f(Z)Z)Z = 2[Z, f(Z)]Z

f(Z)ZZ)

21'(Z)[Z, Z]Z = 2Z1'(Z)[Z, Z], where we have used that [Z, Z] commutes with Z and f(Z).

o

212


In our case

z

at =

-

2h [Z, Z].

This means that the bracket [Z, Z] corresponds to the time operator. An operator which does not depend explicitly on time in our case will not depend on [Z, Z]. The operator f(Z) does not depend explicitly on [Z, Z]. Hence f(Z) is conservative if and only if it commutes with D..H, i.e.

2Zj'(Z)[Z, Z] or f'(Z)

=

0,

= 0, i.e. f is a constant function. 6.6. The fundamental solution for D..H

The infinite matrix formalism in Quantum Mechanics was introduced in 1925 by W. K. Heisenberg. Using this idea we shall look for a fundamental solution in the form n,k~O

where M is an infinite transition probability matrix, i.e., Mkn = Mnk E [0,1]. In this way, we can take into account all possible transitions between the excited states and the conjugated states. Let A = hlzl 2 +it and Ii = hlzl 2 - it. We shall consider the following ground-states functions 1 * 1 'l/Jo = -a' 'l/Jo = Aa' A Proposition 6.8 yields

1) A~+n'

=

(-2hzto:(0:

+ 1) ... (0: + n -

'l/J~ =

(-2hzto:(0:

+ 1) ... (0: + n - l)_:+n'

'l/Jn

A

Assuming M diagonal matrix,

K

The operator D..H is homogeneous with respect to the dilations

(z, t) The term

Izl 2n IAI 2n

=

-+

(>.z, >.2t).

(

Izl2

h21z14 + t 2

)n

is not homogeneous with respect to the above dilations, unless n = O. Hence

Moo * K = IAI 2 a = Moo'l/Jo'l/Jo.

6.7. THE FUNDAMENTAL SOLUTION FOR fl.>.

= fl.H

- ~'>"[Xl. X2]

213

We shall determine a such that

D.HK=O,

for (z,t):r!=(O,O).

In the computation we shall use the following relations

ZA ZA

2hz,

=

0,

Z A=O, z::4 = 2hz.

Then Z7/J~

0,

Z7/Jo

-2ahz Ao+l ,

Z7/Jo = 0, Z7/J* = -2ahz o ::40+l'

and then

Using D.H

=

= n/.*

2ahlzl 2 -0+1' A ZZ + ZZ, yields Z(

n/.*)

z'/-'o

_

'/-'0

- _ 2ahlzl 2 Z(z7/Jo) = 7/Jo - Ao+l .

ZZ(7/Jo7/J~)

+ ZZ(7/Jo7/J~) = Z((Z7/Jo)7/J~) + Z((Z7/Jo)7/J~) z(-2ahZ ~) + Z(_l_ -2ahZ) Ao+l ::40

Ao ::40+1

-2ah (Z) -2ah-( z ) Ao+1 Z ::40 + ::40+1 Z Ao -2ah ( ) -2ah-( ) Ao+l Z z7/J~ + ::40+1 Z z7/Jo -2ah (7/J* _ 2ah 1Z 12 ) -2ah (7/J --.:. 2ah 1z 12 ) Ao+1 0 ::40+1 + ::40+1 0 Ao+1

(1

1

2 h 2ahlzl 2 - a Ao+1 A 0 - IAI2(0+l) + ::40+1Ao

-2ah

IAI 20

(~ _

-2ah . 2hlzl 2

IAI for a

=

)

~ _ 2ah 1Z 12 ) IAI2 +::4 IAI2

2ahlzl 2

A

-2ah (::4 + A _ 4ah 1z 12 ) IAI 2 0 A::4 AA 20

2ahlzl 2

- IAI2(0+1)

4ahlzl 2 = AA -

°

~. Hence the fundamental solution has the form

6.7. The fundamental solution for D..>.. = D.H - ~A[Xl,X2l In this case we shall look for a fundamental solution at the origin of the form

K(z, t; 0, 0) =

C7/Jo7/J~,


214

with

1 * 1 'l/Jo = An' 'l/Jo = Ai1 ' ~,\

is a linear combination of Z Z and Z Z ~,\

ZZ+ZZ+'\[Z,Z] ZZ+ZZ+'\ZZ-'\ZZ (1 + '\)ZZ + (1 - '\)ZZ.

A straightforward computation shows

__2_h_ ((1 + '\)a A a Ai1 A

+ (1- A)(3) + _Sa----'(3-'.I-'.zl..". 2h,. . ,. 2 A

A n+ 1 AM 1

h_i1 +1 (sa(3h Z I2 - 2[a + (3 + A(a - (3)]hlzI 2 Aa+lA 1

-2it[(3 - a - ,\((3 + a)]). Im~,\('l/Jo'l/J~)=O

===}

Re ~'\('l/Jo'l/J~) = 0

===}

(3-a=A((3+a), 4a(3 = a + (3 + A(a - (3) = (a

+ (3)(1 - A2).

The first relation of the above equations yields

and substituting in the second equation, yields

I+A 4a 21+A - - =a+a-1-,\

Multiplying by

1-,\

1 - A + 1 + '\(1 - A) _ A

I+A

1+'\

2(1 - A2)

1+,\ 1-,\ 2

(a-a 1+,\) -- . I-A

ntl+\) ' A f:. -1, yields 4a

Thus a =

+,\

and

(.l

fJ

=

1-,\ 1+,\ 2 1-,\

K(z, t; 0, 0)

=

1+,\ 2'

= 2(1

_ A).

Hence

1_,,_1+" CA-2-A 2

C (hlzl2 + it) 1;" (hlzl2 _ it) 1~" .

6.8. AN APPLICATION TO THE EIGENFUNCTIONS OF .0.H

215

6.8. An application to the eigenfunctions of 6. H

In this section se shall study the relation between the eigenfunctions of 6. H and the associated sub-elliptic Schrodinger equation. PROPOSITION

6.20. Let 'ljJ be an eigenfunction for 6. H , i.e., there is a constant

E E IR such that

6. H 'ljJ = E'ljJ.

1) Then

Z'ljJ satisfies the following Schrodinger equation

1]/ =

4ihot 1]/ + 6. H 1]/

2) Then

1]/* =

PROOF.

= EI]/.

Z'ljJ satisfies the equation

1) Using the commutativity between Z and Ot yields 6. H Z'ljJ = (2ZZ - 2ihot )Z'ljJ

6. H

(2ZZZ - 2ihot Z)'ljJ Z(2ZZ)'ljJ - 2ihot Z'ljJ Z(2ZZ + 2ihot Z(6. H

-

2ihot )'ljJ - 2ihot Z'ljJ

-

2ihot )'ljJ - 2ihot Z'ljJ

Z(E'ljJ - 2ihot 'ljJ) - 2ihot Z'ljJ EZ'ljJ - 2ihot Z'ljJ - 2ihot Z'ljJ EI]/ - 4ihot 1]/.

2) Comes from 1) by conjugation. D

Choosing E = 0, we have COROLLARY

6.21. If 6. H 'P

= 0 on U, then

cP

= Z'P satisfies

(6.30) on U. EXAMPLE

6.22. Choosing 1

'P

we have 6. H 'P

= Jh21z14 + t 2 '

= 0 on U = 1R 3 \{(0, O)}. cP =

Z

-hE

=

'P

satisfies equation (6.30) on U.

Then

(hlzl2

+ it)Jh 2lzl4 + t 2


216

6.9. The Schrodinger kernel Propagators. We shall find a Green function for the operator (6.31)

L = ihou

+ ~2 (Xr + Xi),

where h is the Plank's constant. The Green function P(x, t, U; xo, to, uo) is also referred to in Quantum Mechanics as a propagator. If W is a wave function, which satisfies the equation (6.32) with the initial condition Wo(x, t) time u > Uo is given by

w(x, t, u) =

= w(x, t, uo),

then the solution at any instant of

J

P(x, t, U; xo, to, uo)wo(xo, to) dxo dto·

In Quantum Mechanics the propagator is denoted by w(x, t, ulxo, to, uo). Equation (6.32) is a subelliptic Schrodinger equation with zero potential energy. It describes a free particle in the Heisenberg quantum framework. One may look for solutions of (6.32) of the type j

Substituting in (6.32) we arrive at ih1:Jj (x, t)'Pj(u) + h 2 t:. H 1:Jj (x, t) 'Pj(u)

= O.

Dividing by 1:Jj 'Pj yields ih'Pj(u) 'P(u)

+ h2 t:. H 1:J j (X,t)

=0.

1:Jj (x, t)

There is a separation constant E j , which has the significance of energy, such that ih'Pj(u)

E j 'P(u),

h 2 t:. H 1:J j

-Ej 1:J j .

Then 'Pj(u) = e- hi"E j

=

From the Classical Mechanics, Eu 2 we let fJ = Eju 2 • Then

u2

•

gu, where 9 is the action. Using this idea,

L 1:Jj (x, t)e- it . if·

W=

j

In the next section se shall show that the modified complex action f = 7g is replacing all fj's, while the sum is replaced by an integral. This is an important application of the fact that the modified complex action f can produce all subRiemannian distances dJ = 27 E j . This shows how useful complex Hamiltonian mechanics techniques are for describing the Schrodinger equation. We shall show that the formula for the Schrodinger's kernel at Xo 1S

1 )2 P(x,t,u)= ( - 2 'h 7rZ

U

;+00

e-

-00

where V (7) satisfies a certain transport equation.

if(x.t.r) hu

V(7)d7,

= 0, to = 0, Uo = 0

6.9. THE SCHRODINGER KERNEL

217

This shows that the propagator P depends on each subRiemannain distance d j . This is a well known fact of Quantum Mechanics stated by R. Feynman. He explained the double slit experiment assuming the fact that the electron takes all possible paths between any two points. Hence each path has a contribution to the propagator, which is defined as a path integral. For macro objects, like airplanes, planets or bullets, only the classical path is important and their motion is explained in the context of Riemannian geometry. The picture for the sub-atomic particles is completely different and the correct framework to study their motion is the subRiemannian geometry.

The Green function at the origin. A Green function P for the operator L is a distribution on ~;, x ~t with the following properties 0, for u > 0,

1) aup - AHP 2) lim P(x,t,u)

(6.33) (6.34)

8(x)8(t),

u"-,O

where 8 stands for the Dirac distribution. The following result will be useful. LEMMA

6.23. For any smooth function 'P we have

AHe'P = e'P(AH'P+ where

l\7x'P12 =

PROOF.

(Xl'P)2

+ (Xl'P)2

~1\7x'P12),

and AH = ~(Xf

+ X?).

We have Xje'P = e'P Xj'P and

XJe'P = Xj(e'P Xj'P) = e'P(Xj'P)2

+ e'PXJ'P.

Hence

D

The above lemma works for any vector fields Xl, X 2,···, X 2m . Let 9 = g(x, t, T) = -it+hcoth(2hT)lxI 2 be the complex action from the origin. 9 satisfies the Hamilton-Jacobi equation

The modified complex action (6.36)

1

ag

-aT + -1\7xgl 2

(6.35)

of OT

f = Tg =

-iTt

==}

= 0.

+ Thcoth(2hT)lxI 2

of

ag

= 9 + T aT

2

T aT

= f +T

satisfies

2ag aT'

Equations (6.35) and (6.36) yield (6.37)

l\7xfl 2 =

T

2

ag l\7xgl 2 = -2T 2 aT = 2(f -

Let 'P = ~. Then using (6.37) we have (6.38)

Of)

T aT .


218

As D.H'P

(6.39)

= h~ D.H I,

Lemma 6.23 yields

D.He'P = e'P ( -

~D.H 1 hu

_1_ (I _ /3 1 )).

aT

h2u2

Hence (6.40) and (6.41 )

. a ( -e'PV(T) 1 ) = --V(T) e'P ( -1 zhau

u2

u3

u

+ 2ih ) .

Relations (6.40) and (6.41) yield

(h 2 D.H - ihau )

(:2 e'PV(T))

Let V (T) satisfy the equation

dV T dT

+ (D.HI -

l)V(T)

=

O.

If we set V (T) = TV (T) then V satisfies

d(TV)

T~

+ ( D.H(Tg) -1 ) TV =

0,

which can be written as (6.42) and it is called the transport equation. A straightforward computation provides D.Hg = 2hcoth(2hT). Hence the transport equation (6.42) becomes

dv dT

+ 2h coth(2hT)v =

0,

and it can be solved by the method of separation. The solution is

V(T)

=

2hC sinh(2hT)'

6.9. THE SCHRODINGER KERNEL

= 1,

where C is a constant. If we employ V(O)

219

then 2hT

= TV(T) = sinh(2hT)

V(T)

Therefore 2

(h 6. H

-

(1

. 'P ) _ a (ih 'P 2hT2 ) 2hau ) u 2e V(T) - aT u3e sinh(2hT) .

alar vanishes at

As the expression under

(h 2 6. H -ihau )

J

OO

T

= ±oo, integrating yields

1

2" e'PV(T)dT=O, u

-00

\;fu>O.

Let

C P(x, t, u) = 2" u

JOO 2" 1 e'PV(T) dT, -00

U

where

r.p =

if(x,t,T) hu .

Hence the first relation of (6.33) is satisfied. To prove the second relation of (6.33) we note first that if 0 < 12h--y1 < 7f, then Re f(x, t, T + i--y sgnt) ~ tb)(lxI 4

+ t 2 )1/2

for some tb) > 0, see (2.7.7) of [5]. Consequently P can be put in the form P(x,t,u)

=

C j+oo+irsgnt

2"

u

if(x.t.. T)

e--"-u-V(T)dT.

-oo+irsgnt

This yields

)m

P(x, t, u) = 0, lim (dd u uniformly in (x, t) on compact subsets of 1R 3 \{0} for each fixed n, n = 0, 1,2, ... To derive 2) we need to compute u~O

r

1 ) dx dt. lim P(x, t, -R R~oo Jf[{

This was done in [5], see (2.8.13)-(2.8.15). In particular,

r

1 ) dx dt = (27fih)2C = 1, lim P(x, t, -R R~oo Jf[{

which yields the following result. THEOREM 6.24. The Schrodinger kernel, i.e., the kernel of e-ihut:. H at the origin is given by

1 P(x, t, u) where f function.

=

Tg

=

-iTt

= (27fhiu)2

j+oo -00

-if(x,t.T)

e

h"

+ hTcoth(2Th)(xI + x~)

2hT sinh(2hT) dT,

is the complex modified action

We have proved that P(x, t, u) is the kernel relative to the origin. It follows from the left invariance of 6. H that the full Schrodinger kernel is obtained by left translations. The Heisenberg convolution with P gives the solutions of the usual initial-value problem for the Schrodinger operator on the Heisenberg group Hi.

220


The Green function at any point. In order to get a closed form expression for the Schrodinger kernel, instead of using the Heisenberg convolution, we shall use direct computation. The computation and the propagator formula will be very similar to the one in the case when Xo = 0, to = O.

In the general case the complex action between the points (xo, yo) and (x, y) within time 7 is 9 = g(x, t, xo, to, 7)

=

-it

+ hcoth(2h7)(lxI 2 + Ixol2)

1 -2hlxllxol sinh(2hT)

(6.43) THEOREM

6.25. The complex action 9 given by (6.43) satisfies the Hamilton-

Jacobi equation

og

07 PROOF.

1

2

1

2

+ 2(X 1 g) + 2(X2 g) = O.

By direct computation. Let r = Ixl and ro = Ixol. We have

1 ) ro 2hxl ( coth(2h7) - - . h( h ' r sm 2 7

ro 1 ) 2hx2 ( coth(2h7) - -;: sinh(2h7 .

Therefore

221

6,9, THE SCHRODINGER KERNEL

Hence H(Vg)

On the other hand

ag

h(r

aT

2

2 a + rO)-aT coth(2hT) 2 (2

+ ro2)

-2 h r

1

sinh2(2hT)

( r2 . -22h2 smh (2hT) and hence

ag

aT

+ r6 -

=

a

1

2rroha- . h( h ) T sm 2 T cosh(2hT) + 4h 2 rro-.....,,---csinh2(2hT)

2rro cosh(2hT) )

-H(Vg). D

Following the same computation as in the proof of Theorem 6.24, the volume element v( T) satisfies

Using 2

X 1g

ro 1 ) = X 22 g = 2h ( coth(2hT) - -;: sinh(2hT) ,

yields D.Hg =

ro 1 2hcoth(2hT) - 2h-;: sinh(2hT)

The transport equation becomes

dv v In Ivi

( - 2h coth(2hT) 1

+ 2h ~ sinh~2hT)) dT ~

-2h 2h In Isinh(2hT)I

ro 1

+ 2h-;: 2h In Itanh(hT)I + C ~

C tanh(hTtO/r sinh(2hT) sinh(hTto/r 1 C --.--~~--~~ cosh(hT)ro/r 2smh(hT) cosh(hT) C sinh(hT) !:Q. -1 2 cosh(hT)~+l . r

We shall choose the constant C such that V(O)

->

1 for ro

->

!:Q.

V(T) = TV(T) = hT

sinh(hT) " -1 !:Q.' cosh(hT) ,. +1

The propagator's formula is given by the following theorem.

O. Hence C

= 2h and


222

THEOREM

6.26. The Schrodinger kernel is given by

P(x,t,xo,to,u) = (

27l'

h h'

)2

2U

J+oo e

-if(x,t,X().tp,T)

T

h"

-00

sinh(hT)~-l

cosh(hT)

c:ll. r

+1

dT,

where f = Tg = -iTt

+ Thcoth(2hT)(lxI 2 + Ixol2) -

2Thlxllxol .

ht

sm 2

h ) T

is the complex modified action function.

6.10. The heat kernel on HI

In this section v denotes the time variable and we shall consider the heat operator (6.44) The fundamental solution at the origin is the function P(x, t; v) defined on HI x lR+ such that the following relations are satisfied

lim P(y, t; v)

=

V ","0

b(y)b(t).

The heat kernel has been obtained in [6], [26], [34]. For a variational approach using Hamiltonian formalism, see [3]. In the following we shall use the results proved in the previous section. With the transformation v

= u/(ih),

Yj = xj/(ih)

the heat operator (6.44) becomes the Schrodinger's operator

. au zh

h 2 (2 +2 Xl + X 22) .

Then the Theorem 6.26 becomes THEOREM

(6.45)

6.27. The heat kernel at the origin is given by P(y t v) = -1, , (27l'v)2

Joo e-

f(y,t,r)/vV(T)

dT

-00

'

where f(y, t, T) = -iTt + hTcoth(2Th)(yi is the modified complex action and V(T) =

2hT

sinh(2hT) is the volume element.

+ y~),

6.10. THE HEAT KERNEL ON HI

223

The small time behavior when x =1= O. We shall use the steepest descends method in order to compute the small time asymptotics for the heat kernel. Consider the Laplace integral in the complex plane

F()")

=

i

V(z)eAh(Z) dz,

where the V(z), h(z) are holomorphic functions on a domain D, which contains the smooth piece-wise curve C. The following theorem will provide asymptotic expansions for F()") when the parameter).. ---4 00. If z = Z(T), tl ::; T ::; t2 is a parametrization of the curve C, we write

F()")

= It2 V(Z(T))z'(T)e A[¢(T)+i1)!(T)] dT, tl

where h = ¢ + i'IjJ, and ¢( T) = ¢(z( T)), 'IjJ(T) = ¢(z( T)). The following result can be found for instance in Iacob [35], vol. III, p. 83: THEOREM 6.28. Let C be a piece-wise smooth curve and the functions V(z) and h(z) satisfying 1) V(z), h(z) are holomorphic at each point ofC, 2) the maximum of the function ¢(z) = Re h(z) on the curve C is reached at a point Zo interior to C, 3) V(zo) =1= 0, h'(zo) = 0, hl/(zo) =1= 0, 4) W(z)leA¢(Z) dz < 00, for).. > O. Then for).. ---4 00 we have the following asymptotic expansion

Ie

= eAh(ZO) L an).. -n-~, 00

F()")

n=O

with the coefficients 00

_ v'27f(2n)! b an 2n, n.'2 2n

V(Z(T))Z'(T) =

L bnTn. n=O

The dominant term is

6.10.0.10. The small time asymptotics for the heat kernel. We shall assume

x

=1=

O. The contour for the integral (6.45) can be moved to the line ImT

P(y,t;u) =

= Bc:

r

e-f(y,t,T)/uV(T)dT. (21ru) JlmT=lJ c In order to apply Theorem 6.28, we shall let h = - f, ).. = l/u, and C = {ImT = Bc}. We know that the unique critical point of f with respect to T in the strip {IImTI < 1f/2h} is the point Tc = iBc, where Be satisfies (6.46)

_1_2

t

= hJL(2hO)IIYI12.

Then on the contour C = {1m T = Bc}, the function Re f has a strict minimum at Tc = iBc, and its value is d2(y, t)/2. We have that ¢ = Re h has a strict maximum at Zo = T c , which is in the interior of C.

224

6.


Since Zo is a strict maximum, h' (zo) V(zo) i= O. We have

v

Zo =

()

=

0 and h" (zo)

< O. It is obvious that

2hi()c sinh(2hi()e)

2hzo

sinh( 2hzo)

The condition 4) of the Theorem 6.28 is satisfied, because

i

IV(T)leA¢(T) dT

lmT=oc IV(T)e-f/ul dT

< e- d2 (x,t)/(2u) LIV(s + i()e) I ds < We also note that while the first derivative of derivative is

1" (iTe) = ~2 { I

.

uT T=1.()c

= . 24h2

sm (2h()e)

f

vanishes at

T

00.

= i()e,

the second

[1 - 2hBe cot(2hBe)]IIYI12.

Applying Theorem 6.28, we have

1 1 (27ru)2 F()") = (27ru)2

P(y, t; u)

-271' eAh(ZO) [V(zo) )..h"(zo)

+ 0(11 )..)]

27rU e-d2(x,t)/(2u) [V(iO ) + O(u)]. I"(i()c) e

(6.47) Using

27rU

~

.

1" (i()e) V (z()e)

=

sin 2hBe 27ru' 2h(1 - ahBe cot(2a()e) )l/211yll

2hBe sin 2hBe '

then (6.47) yields: THEOREM 6.29. Given a fixed point (x, t), x i= 0, let ()e denote the solution of (6.46) in the interval [0,71' 12h). Then the heat kernel has the following asymptotic expansion . ) _ _1_ -d(x,t)2/(2u) [()eV 27rU )] (1-2hBe cot(2hBe))l/21IYII +O(u ,

(

Py,t,u - (27ru)2 e

as U

""

O.

The small time behavior when x = O. In this section we shall assume h=l.

THEOREM 6.30. At points (0, t) with t > 0, we have the following expansion

1 00 P(O,t;u) = 4u 2 ~)_1)k+lke-e~(O,t)/2U k=l

as u --t 0+, where Ck(O, t) is the length of the k-th geodesic connecting the origin and the point (O,t). PROOF. As we know, the heat kernel for the operator 6. H has the form

P(x, t, u)

=

4 ;

71' u

21+

00

e-f(X,t,T)/UV(T)dT

-00

where

f(x, t, T)

= -iTt + Tcoth(2T)(xi + x~)

225

6.10. THE HEAT KERNEL ON HI

is the modified complex action and

V(T) = is the volume element. Since x 1 47T2U2

j +OO e

itT u

-00

= 0,

2T

sinh(2T)

we have to look at the integral

2T d sinh(2T) T

=

1

j+OO

87T2 U 2

-00

itT

T

e 2u sinh( T)

d T.

This integral can be evaluated explicitly by residue calculus. Let us consider the following contour integral first. See Figure 6.2.

Y,

R

-R

Figure 6.2: The paths 'Yi, i

Let 'Y2 : [0,

(m +

j "12

~)7T]

---->

= 1,4.

C and 'Y2(y) = R + iy.

Z d' eil3.. 2u - - - Z = Z sinh(z)

l

(m+~)7r

R

+ zy. dy. sinh(R + iy)

ei.J....[R+iy] 2u

0

But sinh(R + iy)

2 e R [ cos(y)

+ i sin(y)]

c

-

R [ cos(y)

- i sin(y)]

2 eR -

cos(y)

e- R

2

eR

+isin(y)

+ e- R 2

.

Hence, sinh( R + iy)

= cos(y)

sinh( R) + i sin(y) cosh( R).

It follows that

Isinh(R + iy)1 = Since R

Jcos 2(y) sinh2(R)

+ sin2(y)

COSh2(R)

= JSinh 2(R) + sin2(y).

> 0, we have

I sinh(R + iy) I ::::: I sinh(R) I = sinh(R) and hence

I sinh(R + iy)1 ::::: I sin(y)l·

226


J,2 .

Therefore, we get the following bound for the modulus of

1

11,2 - l
~n] 1JI dx J ,

I

where J = (jl < ... < jn-k) is the complementary multi-index to I ik), is called the Roge operator.

= (il < ...
2, the energy levels become more and more distant as m increases (see Figure 6.7)

o < E2 -

El < E3 - E2 < ... < Em+l - Em < ...

This is characterizing the energy spectrum for a charged particle in a static, nonconstant, rotational invariant magnetic field. Energy

Energy

E4~-----------------------

E"

E)~

E2~-----------------------

E 3- EJ = 1t ItI

E,

1E,-E J=

EJ

_______________________

~ ItI

EJ

~

_______________________

o Figure 6.6. Energy levels for k = 1.

Let fk(X)

= Ckx2-t,

Figure 6.7. Energy levels for k

=

2.

where the constant

C k = 2M ( (2k ~ l)1tl )

11k

Then

Em+l - Em = h(m + 1) - fk(m) = fU~m)' with m < ~m < m + 1. Hence, the difference between two consecutive energy levels is estimated as

(2 - ~)Ckml-t < Em+l -

Em
o.

1 W 2k _ 1

e > 0, cp is decreasing and t is increasing.

Then (6.61) becomes

(6.62) T is the period, i. e., the time needed for the particle to come back at the origin. Hence T1/2k = c·

rmax

is an analog of the third law of Kepler, with c = Qk/2k. For k one has r= - y7r.

T 1/2 _

= 1, the step 2 case,

r max ,

where rmax is the diameter of the circle in the x-plane. For k = 2, the step 4 case, one has 1

T 1/ 4

= (Vir(1/6))4 3 f(2/3)

.r max .

• The second law Let x(s) = (X1(S),X2(S)) be a curve in the plane. Consider polar coordinates Xl = r cos cp, X2 = rsincp, with r = Ixl. Let CPo = cp(O) and CP1 = cp(l). The vectorial --->

radius OP will sweep an area equal to

A

= 11 0

~r2 dcp. 2

The above formula can be also written in differential form dA =

See Figure 1.1.

~r2dCP.


236

The areal velocity is defined as

dA

a=-.

ds Let c = (Xl, X2, t) be a subRiemannian geodesic in step 2 case. Then the rate of change of the t-component is equal to 4a, i.e., i = 4a. For the proof, see Theorem 1.7.

6.14. Exercises EXERCISE 6.1. Let

~ Joo e-f(x,t,r)/uV(T) be the heat kernel (27ru)-00

P(x , t·, u) =

for D.H. i) Show that

r P(x,

J~3

t; u) dx dt

= 1.

ii) Prove that the function P is smooth on Hi x lim (dd

u->O

U

~

and satisfies

)m P(x, t; u) = 0,

uniformly in (x, t) in compact subsets on Hi "" 0, for each m = 1,2,3, .... EXERCISE 6.2. Find the full heat kernel

P(x, t, Xo, to, u) for D.H using

i) left translations;

ii) direct computation starting from the modified complex action function. EXERCISE 6.3. Show that the heat kernel

P(x, t, u) :::; C

e- d (x,t)2/ 2u { u2 min 1,

EXERCISE 6.4. Let f(X,t,T) plex action function. Show that

P(x, t, u) satisfies the estimate

Iu

Ilxl d(x, t)

},

(x, t, u) E Hi

= -irt + arcoth(2aT)llxI1 2 be

a2 { I . aT

X

~+.

the modified com-

2': 2 11 ax112, 3

r=dJc

where Oe is the solution of t

in the strip IRe 01
\) as >..

---t

=

[!;..eiCA-n2"-:!J:) +0(>..-3/2)

00.

J:

EXERCISE 6.8. Consider the function F(>..) = f(t)e AhCt ) dt, >.. E R Let M = maXtE[a,bjlh(t)1 < 00 and assume there is >"0 > 0 such that

lb

If(t)leAOh(t) dt < 00.

Show that

> >"0, C = constant. EXERCISE 6.9 (Watson). Let a > 0,(3 > -1 and g(t) be a infinitely differenv>..

tiable function on the interval [0, T]. The asymptotic expansion of the integral

G(>") is _

~

G(>") - a as >..

---t

=

00

foT e- At" t{3g(t) dt

L9 n=O

(n)

(0)

r ( ((3 + n + 1) / a) n!>..C{3+n+l)/a

'

00.

EXERCISE 6.10. i) Show that V(z)

V(z) =

rv

1 - ~h2z2

+ !~h2 + O(z6),

where

2hz sinh(2hz)

ii) Find the coefficient

a1

of the asymptotic expansion of the heat kernel.

Bibliography [1] R. Abraham and J. E. Marsden. Foundations of Mechanics. Addison Wesley, 1994. [2] V. 1. Arnold. Mathematical Models of Clasical Mechanics. Springer Verlag, 1989. [3] R. Beals, B. Gaveau, and P. C. Greiner. Hamilton-Jacobi theory and the heat kernel on Heisenberg groups. Number 79. J. Math. Pure Appl., 2000, pp. 633-689. [4] R. Beals, B. Gaveau, and P.C. Greiner. On a Geometric Formula for the Fundamental Solution of Subelliptic Laplacians. Number 181. Math. Nachr., 1996, pp. 81-163. [5] R. Beals, B. Gaveau, and P.C. Greiner. Complex Hamiltonian mechanics and parametrices for subelliptic Laplacians, I, II, III. Number 21. Bull. Sci. Math., 1997, pp. 1-36, 97-149, 195-259. [6] R. Beals and P.C. Greiner. Calculus on Heisenberg Manifolds. Ann. Math. Studies, Vol. 119, Princetown University Press, 1988. [7] A. Bellaiche. Sub-Riemannian Geometry. Progr. Math. 144, Birkhauser, Basel, 1996. [8] C.M. Bender, S. Boettcher, and P. N. Meisinger. J. Math. Phys., 1999. [9] C. Berenstein, D.C. Chang, and J. Tie. Laguerre Calculus and Its Applications on the Heisenberg Group, volume 22. AMS/IP, Studies in Advanced Math., 2001. [10] J. M. Bismut. Large Deviations and the Malliavin Calculus. Progr. in Math. 144, Birkhauser, Boston, 1984. [11] S. N. Bose. Plancks gesetz und lichquantentenhypothese, volume 26. Zeitschrift fur Physik, 1924. [12] O. Calin. Ph.D Thesis, University of Toronto. 2000. [13] O. Calin. Geodesics on a certain step 2 sub-Riemannian manifold. Number 22. Annals of Global Analysis and Geometry, 2002, pp. 317-339. [14] O. Calin and D. C. Chang. Geometric Mechanics on Riemannian Manifolds, Applications to Partial Differential Equations. Applied and Numerical Analysis, Birkhauser, 2004. [15] O. Calin and D. C. Chang. Geometric Mechanics on a step 4 subRiemannian manifold. Taiwanese Journal of Mathematics, June, 2005. [16] O. Calin, D. C. Chang, and P. C. Greiner. On a Step 2(k + 1) sub-Riemannian manifold, volume 14. Journal of Geometric Analysis, 2004, pp. 1-18. [17] O. Calin, D. C. Chang, and P. C. Greiner. Real and complex Hamiltonian mechanics on some SubRiemannian manifolds, volume 18. Asian Journal of Mathematics, 2004, pp. 137-160. [18] O. Calin, D. C. Chang, and P. C. Greiner. Geometric mechanics on the Heisenberg group. Bulletin of the Institute of Mathematics, Academia Sinica, 2005. [19] O. Calin and V. Mangione. Variational calculus on sub-Riemannian manifolds, volume 8. Balcan Journal of Geometry and Applications, 2003. [20] A. Calogelo, G. Carcano, and R. Pini. Optimization in Heisenberg group. to appear in Optimization, A journal of mathematical programming and operations research. [21] E. Cartan. Les groupes de transformations continus, injinis, simples. Ann. Ecole Norm. Sup. [22] W.L. Chow. Uber Systeme von linearen partiellen Differentialgleichungen erster Ordnung. Number 117. Math. Ann., 1939, pp. 98-105. [23] A. Erdelyi et al. Higher Transcendental Functions, volume I,IV. Bateman Manuscript project, McGraw-Hill, New York, 1955. [24] G. Folland and E.M. Stein. Estimates for the lh, Complex and Analysis on the Heisenberg Group. Number 27. Comm. Pure Appl. Math., 1974, pp. 429-522. [25] G.B. Folland. Introduction to PDEs, second edition. Princeton University Press, New Jersey, 1995. [26] B. Gaveau. Principe de moindre action, propagation de la chaleur et estimees sous-elliptiques sur certains groupes nilpotents, volume 139. Acta Math, 1977, pp. 95-153. 239

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[27J M. Giaquinta and S. Hildebrand. Calculus of Variations, I, II, volume 310. Springer, 1977. [28J P.C. Greiner and E.M. Stein. Estimates for the a-Neumann Problem, volume 19. Math. Notes Series, Princeton University Press, Princeton, New Jersey, 1977. [29J M. Gromov. Metric Structures for Riemannian and Non-Riemannian Spaces, with appendices by M. Katz, P. Pansu and S. Semmes, volume 152. Progr. Math., Birkkiiuser, Boston, MA, 1999. [30J U. Hamenstiidt. Some regularity theorems for Carnot-Caratheodory metrics, volume 32. J. Differential Geom., 1990, pp. 819-850. [31J P. Hartman. Ordinary Differential equations. Wiley, 1984. [32J S. Hawking. A Brief History of Time. Bantam, 1998. [33J L. Hormander. Hypoelliptic second order differential equations, volume 119. Acta Math., 1967. [34J A. Hulanicki. The distribution of energy in the Brownian motion in the Gaussian field and aanalytic-hypoellipticity of certain subelliptic operators on the Heisenberg group. Number 56. Studia Math., 1976, pp. 165-173. [35J C. Iacob, D. Homentcovschi, N. Marcov, and A. Nicolau. Matematici clasice si moderne. Ed. Tehnica, Bucuresti., 1983. [36J A. Kanpp. Lie groups, Lie algebras and cohomology. Math. Notes, Princetown University Press, New Jersey, 1988. [37J S. Kobayashi and K. Nomizu. Foundations of Differential Geometry, volume I-II. Wiley Interscience, 1996. [38J A. Koranyi. Geometric properties of Heisenberg groups. Advances in Math., 1985, pp. 28-38. [39J A. Koranyi. Geometric aspects of analysis on the Heisenberg group. Topics in Modern Harmonic Analysis, May-June 1982, pp. 209-258. [40J A. Koranyi and H. M. Riemann. Quasiconformal mappings in the Heisenberg group. Invent. Math., 1985, pp. 30g..338. [41J S. Krantz. Function of several complex variables. Wadsworth & Brooks/Cole, Pacific Grove, CA,1992. [42J D. F. Lawden. Elliptic FUnctions and Applications, volume 80. Applied Mathematical Sciences, Springer-Verlag New York Inc., 1989. [43J E. B. Lee and L. Markus. Foundations of Optimal Control Theory. Wiley, New York, 1968. [44J W. Liu and H. J. Sussmann. Shortest paths for sub-Riemannian metrics on rank two distributions, volume 118. Mem. Amer. Math. Soc., 1995. [45J A. Messiah. Quantum Mechanics. Dover Publications, 2000. [46J R. S. Millman and G. D. Parker. Elements of Differential Geometry. Prentice Hall, Inc., 1977. [47J A. Nagel, F. Ricci, and E.M. Stein. Singular Integrals with flag kernel and analysis on quadratic manifolds. Number 181. Journal of Functional Analysis, 2001, pp. 29-118. [48J D.H. Phong and E.M. Stein. Hilbert integrals, singular integrals, and Radon transforms II. Number 86. Inventiones Math., 1986, pp. 75-113. [49J D.H. Phong and E.M. Stein. Hilbert integrals, singular integrals, and Radon transforms 1. Number 157. Acta Math., 1986, pp. 99-157. [50J R. Strichartz. Subriemannian geometry. Number 24. J. Differential Geom., 1986, pp. 221-263. [51J R. Strichartz. Corrections to "Subriemannian geometry". Number 38. J. Differential Geom., 1989, pp. 595-596. [52J H. J. Sussmann. A cornucopia of four-dimensional abnormal sub-Riemannian minimizers in A. Bellaiche (ed.), Sub-Riemannian Geometry. Progr. Math. Birkkiiuser, Basel, 1996, pp. 341-364. [53J W. Thirring. A Course in Mathematical Physics, vol 1: Classical Dynamical Systems. Springer-Verlag, Berlin· Heidelberg· New York., 1978. [54J A. L. Jr. Xavier and M. A. M. de Aguiar. Semiclassical approximations to the coherent-state propagator for a particle in a box, volume 54. Physical Review A, 1996, pp. 1808-1819.

Index

action, 129, 169 integral, 24 affine transformations, 205 analytic function, 211 angle, 94 anihilation operator, 200 ansatz, 173, 191 arc length, 22, 36, 92 areal velocity, 8 argument function, 156 asymptotical correspondence, 74 asymptotics, 223

charged particles, 231 Chow's theorem, 10, 48, 86, 116 circular crown, 232 class of nilpotence, 1 classical action, 118, 177 Classical Mechanics, 5, 216 closed curve, 36 coherent-state propagator, 147 combined parity, 147 commutativity, 215 commutator subgroup, 2 complete orthonormal system, 209 completeness, 45, 113 complex action, 39, 163, 165, 175, 220 conjugation, 148 connectivity, 153 dynamics, 147 Hamiltonian, 147 Hamiltonian mechanics, 216 quantification, 199 solutions, 147 subRiemannian geodesic, 147, 148 symmetries, 148 complex action, 162 complex geodesics, 162 complexified Lie algebra, 209 complexified phase space, 145 components, 3 conjugate points, 63 conjugate points, 105 conjugated states, 212 connection form, 18 connectivity, 27, 45, 86 conservation equation, 76 of energy, 17, 57, 73, 111, 114, 179 contact manifold, 18 coordinate transformation, 3 creation operator, 200 critical point, 97 critical points, 75

Barnes' function, 109 beta function, 69, 97 bicharacteristic solution, 115 bicharacteristics, 13, 188 biholomorphic, 205 bilinear form, 20 boson, 199 boundary condition, 156 conditions, 13, 29, 87, 94, 101, 147, 149, 163, 177 value problem, 69 boundedness, 115 brackets, 86 Campbell-Hausdorff formula, 4, 210 Carnot-Caratheodory distance, 38, 130, 194 distances, 97 metric, 38 Cartan, 18 Cauchy problem, 208 Cauchy sequence, 38 Cauchy's inequality, 68, 233 theorem, 227 Cauchy-Riemann equations, 203 caustics, 140 Cayley transform, 5 chain rule, 189 241

242

curvature 2-form , 18 Darboux, 18 diffeomorphism, 3, 30, 97 dilations, 138 dimension, 3 Dirac distribution, 217 disk, 58 distribution, 7, 18 Doppler-Fizeau formula, 5 double slit experiment, 217 dumbbell, 193 eiconal equation, 171, 172 eigenfunctions, 215 eigenspace, 135 eigenvalue, 135 electromagnetic field, 199, 228 electron, 217 elliptic functions, 45 elliptic operators, 145 elliptic functions, 70, 161, 182 elliptic integral, 109 energy, 72, 156, 165, 189 quantification, 201 energy level, 160 energy parameter, 165 equations of motion, 25 equilibrium solution 75 112 Euler-Lagrange ' , equation, 24, 25 equations, 91, 167 system, 53, 55, 72, 88, 110, 111, 162 Euler-Lagrange equations, 27 exact equation, 210 excited states, 203, 205 explicit formulae, 69 solutions, 69 Fechnel,36 first integral, 17, 26 first order ODE, 173 Folland-Stein formula 194 frequency, 233 ' Frobenius, 47 Frobenius theorem, 7, 86 fundamental solution, 39, 145, 188, 192, 212, 213 gamma functions, 69 Gauss-Bonett, 22 geodesic, 13, 19, 24 geodesic completeness, 38, 82, 169 geodesic curves, 22 geodesic lengths, 179 geodesically complete, 38 geometric formula, 188

INDEX

global connectivity, 49, 82, 115 minimum,72 graph, 80, 98, 176 Green function, 216, 217, 220 ground-state, 204 energy, 200 equation, 207 ground-states functions, 212 Hormander's theorem, 10 Hamilton's equations, 92, 106 Hamilton-J acobi equation, 39 Hamilton-Jacobi equation, 166, 171, 174, 189 217,220 ' Hamiltonian, 54, 57, 88, 147, 172 equation, 16, 63 equations, 29, 53 formalism, 13, 166 operator, 200 system, 13, 17, 114, 147 system, 13 Hamiltonian, 23, 28 Hamiltonian equation, 28 Hamiltonian system, 51 Hamiton's equation, 52 Hankel's function, 237 harmonic oscillator, 145 heat kernels, 145 Heisenberg, 212 convolution, 220 derivative, 211 differentiation, 211 group, 1, 5, 7, 10, 13, 18, 19 26 38 45 , , 46, 101, 125, 179, 201, 205,' 23~ Laplacian, 10, 13, 33 Lie group structure, 25 manifolds, 5, 7 non-commutativity relation, 206 operator, 145, 188, 199 principle, 2 quantum framework, 216 translation, 188 Heisenberg group, 122, 219 Hermitian metric , 20 Hoge operator, 228 holomorphic, 223 horizontal connectivity, 48 curve, 9, 13, 16, 19, 49, 85, 88, 189 distribution, 7, 18, 47, 85 objects, 7 vectors, 21 horizontal connectivity, 10 horizontal curve, 23, 24, 38

INDEX

horizontality condition, 9, 16, 26, 87 hypergeometric functions, 45, 85, 106 hypersurface, 205 hypoelliptic, 10, 203 operator, 45 independent vector fields, 3 initial-value problem, 219 inner product, 20 inverse pendulum, 145 involutive, 7, 85 Jacobi's epsilon function, 183 method, 174 Kaehler metric, 20, 117 property, 21 Kepler's law, 8, 235 kinetic energy, 5, 17, 24, 25 Kobayashy, 20 I'Hospital rule, 30 Lagrange multiplier, 24, 91 Lagrangian, 25, 26, 91, 145, 231 formalism, 23, 53 symmetries, 25 system, 25 Laplace-Beltrami operators, 145 Laplacian, 39 Larmour orbits, 230 left invariant vector, 3 invariant vector field, 3 translation, 3, 12, 125 Legendre transform, 88 Legendre transform, 23, 145, 172 Legendre transform , 53 length of the geodesics, 64 Lie algebra, 3 brackets, 86 group, 2, 100, 209 group, 5 linear harmonic oscillator, 199 local maximum, 97 loops, 104 Lorentz equation, 231 magnetic vector field, 230 maximal right interval, 114 Maxwell's equations, 228 meromorphically, 203 method of separation, 218 metric space, 38 Milnor's theorem, 36 minimizing curve, 38

243

minimum energy, 200 missing direction, 209 modified complex action, 177, 217 momentum, 91, 165, 200 Newton's equation, 71, 110, 145 nil potence class, 2 Noether's theorem, 25 Nomizu,20 non-commutativity relations, 206 non-holonomic constraint, 7, 23, 24 non-integrability, 18 non commutativity , 2 nonsymmetric model, 3 orbit, 235 orthonormal, 20, 28 orthonormal frame, 230 orthonormality, 11 7 oscillator pulsation, 199 osculator plane, 18 parametrization, 88 particle, 5 path integral, 217 periodic function, 81 phase plane, 72, 73 physical fields, 201 Plank's constant, 201, 216, 233 Poisson bracket, 211 polar coordinates, 27, 60, 71, 93, 101, 171, 179 position operator, 200 potential energy, 145 projection, 17, 77 propagator, 216, 217, 220 quadratic equation, 75 quantification, 159 quantization, 199 quantum Hamiltonian, 148, 151, 162 quantum Hamiltonian, 147 Quantum Mechanics, 5, 145, 206 quartic potential, 147 reflected beams, 5 Riemannian geometry, 217 Riemannian manifolds, 5, 7 rotation, 61 rotational symmetry, 27, 55 Schriidinger equation, 199, 216 Schriidinger kernel, 216, 219 Schriidinger's equation, 200 sequence of energies, 160 Siegel upper-half space, 205 skew-symmetric, 129 skew-symmetry, 117

244

spherical symmetry, 231 stable equilibrium, 72 sub-atomic particles, 217 sub-elliptic operator, 88 operators, 145 Schrodinger equation, 215 sub-Riemannian geodesics, 45 subRiemannain distance, 217 subRiemannian distance, 39 geodesic, 63, 64, 88, 204 geodesics, 23, 45, 51, 53, 55, 71, 78, 81, 125, 126 geometry, 5, 18, 30, 39, 95, 115, 217 lengths, 85 manifold, 1, 10,38, 45 metric, 7, 20, 22, 28, 68, 115 model,116 subRiemannian metric, 233 symmetric solution, 157 symmetry, 55, 147 symplectic mechanics, 211 time reversal, 147 trajectory, 60, 112 translations, 25 transport equation, 188, 218 operator, 188 transport equation, 189 uncertainty principle, 5, 209 relation, 200 uncertainty principle, 5 volume element, 188, 190 Watson, 237 wave length, 5

INDEX

ISBN 978-0-8218-4319-2

9 780821 843192

Geometric Analysis on the Heisenberg Group and Its Generalizations (Ams Ip Studies in Advanced Mathematics)

Geometric Analysis on the Heisenberg Group and Its Generalizations (Ams Ip Studies in Advanced Mathematics)

Ip Studies in Advanced Mathematics)

Higher Franz-Reidemeister Torsion (Ams Ip Studies in Advanced Mathematics)

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Heat Kernel and Analysis on Manifolds (Ams Ip Studies in Advanced Mathematics)

Heat Kernel and Analysis on Manifolds (AMS IP Studies in Advanced Mathematics)

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Foundations of p-Adic Teichmuller Theory (Ams Ip Studies in Advanced Mathematics)

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A Course in Abstract Harmonic Analysis (Studies in Advanced Mathematics)

Geometric Relativity, Graduate Studies in Mathematics 201

Topics in geometric group theory

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Analysis (Graduate Studies in Mathematics)

Finite Group Theory (Graduate Studies in Mathematics)

Algebraic and Geometric Combinatorics (Mathematics Studies)

Harmonic Analysis of Mean Periodic Functions on Symmetric Spaces and the Heisenberg Group (Springer Monographs in Mathematics)

Geometric and cohomological methods in group theory

Complex Analysis (Cambridge Studies in Advanced Mathematics 107)

A Primer of Nonlinear Analysis (Cambridge Studies in Advanced Mathematics)

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Conformal differential geometry and its generalizations

A course on geometric group theory

Geometric Analysis on the Heisenberg Group and Its Generalizations (Ams Ip Studies in Advanced Mathematics)