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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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<
a1 ::::; a2 ::::; ... ::::; a p
<
a p +1 = ... = an-
Then every point (0, t) is connected to the origin by an infinite number of geodesics. Every point (x, t) with x" = (X2p+1, X2p+2,"" X2n) -=1= is connected to the origin by a finite number of geodesics. On the other hand, there are points (x, t), with x -=1= 0, but x" = 0, which are connected to the origin by an infinite number of geodesics. If (x, t), x -=1= 0, is connected to the origin by an infinite number of geodesics, then the infinity of the number of geodesics connecting (x, t) to (0,0) is "smaller" than the infinity of the number of geodesics connecting (0, t) to (0,0).
°
The technique of complex Hamiltonian mechanics is introduced in chapter 5. It consists in defining a quantum Hamiltonian and introducing a complex modified action. The latter is used in obtaining the lengths of real geodesics and in finding a geometric formula for the fundamental solution of sub-elliptic operators.
Chapter 6 is dedicated to applications in Quantum Mechanics and Electromagnetism. The main application is the fundamental solution and the Schrodinger kernel (propagator) for the Heisenberg operator by means of complex Hamiltonian formalism. Acknowledgments: This work owes much to the generous help and encouragement of many people. First, we would like to thank Professors Richard Beals, Bernard Gaveau, Stere Ianus, Adam Koranyi, Duong H. P hong , Yum-Tong Siu, and Elias M. Stein for their advice and help. Heartfelt thanks to the Mathematics Departments at Eastern Michigan University, Georgetown University and the University of Toronto for providing excellent research environments for us. We also owe Sandy Becker and Benjamin Jellema many thanks for technical support and help with technology.
PREFACE
ix
Last but not least, we would like to express our gratitude to International Press, especially Professor Shing-Tung Yau, series editor in making this endeavor possible. The Authors
lCh 6
CHAPTER 1
Geometric Mechanics on the Heisenberg Group The Heisenberg group and its sub-Laplacian are at the cross-roads of many domains of analysis and geometry: nilpotent Lie groups theory, hypoeUiptic second order partial differential equations, strongly pseudoconvex domains in complex analysis, probability theory of degenerate diffusion process, subRiemannian geometry, control theory and semiclassical analysis of quantum mechanics, see e.g., [13], [16], [30], [31]' and [44]. Here we give a detailed discussion of the behavior of the subRiemannian geodesics on the Heisenberg group, which is the paradigm of the theory. Research on the subRiemannian geometry induced by the sub-Laplacian and its analytic consequences has been studied extensively in the past ten years (see e.g., [3], [13], [4], [19], [18] and [26]). To the best of our knowledge, the first time these geodesics have been computed by Koranyi, see [39], [38], and [40]. In this chapter we shall provide a systematic method to handle problems along this direction. The Heisenberg group is the prototype for the subRiemannian manifolds of step 2, which makes this case very special in the class of subRiemannian manifolds. There are a few ways to introduce the Heisenberg group. In the first section we shall show that all of these are equivalent.
1.1. Definitions for the Heisenberg group
Let G be a noncommutative group. If h, kEG are two elements, define the commutator of hand k by [h,k] = hkh-1k- 1 = hk(kh)-l. If [h,k] = e, we say that hand k commute (e denotes the unit element of G). The set of elements which commute with all other elements is called the center of the group Z(G) = {g E G; [g, k] = e, Vk E G}. If K <J G is a subgroup of G then let [K, G] be the group generated by all commutators [k, g] with k E K and 9 E G. When K = G, then [G, G] is called the commutator subgroup of G. 1.1. Let G be a group. Define the sequence of groups (f n(G))n>l by fo(G) = G, f n+1(G) = [f n(G), G]. G is called nilpotent if there is an n E-N such that r n(G) = e. The smallest integer n with the above property is called the class of nilpotence of G. DEFINITION
The subset of M3(JR) given by (1.1 )
G~
{O r 0;
a, b,' E
R}
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
2
defines a noncommutative group with the usual matrix multiplication. Consider the matrices bl b3 al a3 ) 1 b2 (1.2) 1 ~2 A= , 1 Then
AB~U
(1.3)
(1.4)
B~U
U°
A-'
~
U°
al
+ bl
a3+
1
a2
°
1
-al
ala2 - a3
1
-a2
1
),
°
b3+ a l b2
+ b2
B-'~
~ [G,G] ~ ([A,B];A,B
E
G)
)
U°
-b l 1
and hence the commutator subgroup is
r,(G)
)
bl b2 -b 2 1
! ~);k
~ { (~
E
-
b3
)
1R}.
Let
Then AC
=
(~ ~ c~ k ) ~CA 001
and therefore [A,C] = AC(AC)-l = h. Hence r2(G) = [rl(G),G] = h = e, and the group G is nilpotent of class 2. G is called the Heisenberg group with 3 parameters. The nilpotence class measures the noncommutativity of the group. In the following we shall associate with this group a noncommutative geometry of step 2. This geometry will have the Heisenberg principle built in. The bijection ¢ : IR3
--t
M 3 (IR),
1(x"x"t)
~ (~
r ~2
)
induces a noncommutative group law structure on IR3 (1.5)
(Xl, X2, t) 0 (X~, X;, t')
=
(Xl
+ X~, X2 + X;, t + t' + X1X;).
The zero element is e = (0,0,0) and the inverse of (Xl,X2, t) is (-Xl,
-X2,X1X2
-t).
IR3 together with the group law (1.5) will be called the nonsymmetric 3-dimensional Heisenberg group. This group can be regarded also as a Lie group. The left
1.1. DEFINITIONS FOR THE HEISENBERG GROUP
3
translation La : G ----7 G, Lag = ag, Vg EGis an analytic diffeomorphism with inverse L;;l = La-I. A vector field X on G is called left invariant if
Va,g E G. The set of all left invariant vector fields form the Lie algebra of G, denoted by L(G). The Lie algebra of G has the same dimension as G and it is isomorphic to the tangent space TeG. We shall use this result in the following proposition in order to compute a basis for the Lie algebra of the Heisenberg group. 1.2. The vector fields
PROPOSITION
X = 8Xl ,
(1.6)
Y
= 8X2 + X18t,
T = 8t
are left invariant with respect to the Lie group law (1.5) on PROOF.
]R3.
Consider the notation X3 = t. In this case the left translation is
L(al,a2,a3)(X1, X2, X3) = (a1
+ Xl, a2 + X2, a3 + X3 + a1 x 2)'
Let X be a left invariant vector field. Then Va = (a1,a2,a3) E G, Xa In local coordinates Xa = Li X~8Xi' The components are given by
=
(La)*X e.
(1. 7) where Xi is the i-th coordinate and Xe is the value of the vector field X at the origin. Let b = (b 1, b2, b3) E G. Then
(Xl (X2 (X3
0 0 0
La)(b) La)(b) La)(b)
x1(Lab) = x1(ab) = a1 x2(Lab) = x2(ab) = a2 x3(Lab) = x3(ab) = a3
+ b1 = x1(a) + x1(b), + b2 = x2(a) + x2(b), + b3 + a1b2 = x3(a) + x3(b) + x1(a)x2(b).
Dropping b, yields
X1(a)+X1, x2(a) + X2, x3(a) + X3 + x1(a)x2. Substituting in equation (1.7) and using Xe
= e8Xl + e8X2 + e8X3 , yields
X~
Xe(X1(a)+XI)=e,
X~
Xe(X2(a)
X~
Xe(X3(a) +X3 +x1(a)x2) =e+X1(a)e.
+ X2)
=
e,
Hence, the left invariant vector field X depends on the parameters ~i
X
(e
e8Xl + e8X2 + + x1e)8x3 e8Xl + e(8X2 + X18x3) + e8X3 eX+ey+eT,
and the Lie algebra is generated by the linear independent vector fields X, Y and T. 0 In the following we shall show that the nonsymmetric model can always be reduced to a symmetric model, using a coordinate transformation.
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
4
PROPOSITION
1.3. Under the change of coordinates
the vector fields are transformed into (1.8)
PROOF.
The proof follows from the following relationships at
4an
a X2 a Xl
aY2 a Yl
-
2YI aT' 2y2a T.
o Making YI = Xl, Y2 = X2, 'T = -t in (1.8) we consider the vector fields Xl = a Xl + 2x 2a t , X 2 = a X2 - 2x l a t , X3 = at on lR 3 = lR; x lR t . We are interested in a Lie group law on lR 3 such that Xl, X 2 and X3 are left invariant. This shall be done using the Campbell-Hausdorff formula. The constants of structure are denoted by and are defined by
cfj
3
[Xi, X j ] =
I>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),
1.2. THE HORIZONTAL DISTRIBUTION
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
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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) = (<1>(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>(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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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,
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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),
1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP
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
1. GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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<
(1.74)
(m+ !)1f 2
-Xm
< _1_. m1f
PROOF. As J.L is an odd function, it suffices to show that it is a monotone increasing diffeomorphism of the interval (0,1f) onto (0, +00). We note that sin x x cos x vanishes at x = 0 and it is increasing on (0, 1f). Then
!J.LI(X)
2
= sin x
- xcosx
sin3 x
= { = 1/3, x = 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
1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP
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
<-.
< -1, there is a
~ between x and y such that
f(x) - f(y) = !,(~)(x - y) < -(x - y). Hence x - y < f(y) - f(x). Choosing x
f(
7r) m7r+ -2
= m7r +
~,
cos(m7r + ~)
= sin(m7r +
~)
y=
Xm
and using
°
= ,
and (1. 76) yields 1
0< (m+ -2)7r -
Xm
1 m7r
< cotx m <-.
o
1.
32
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
14
12-
-2
12
Figure 1.3: Critical points of J.L are solutions of tan x
= x.
The number of geodesics which join the origin with an arbitrary given point is givenin the following theorems, which can be found in [3]. THEOREM 1.52. There are finitely many geodesics that join the origin to (x, t) if and only if x =f. O. These geodesics are parametrized by the solutions ( of (1. 77)
It I
IIxl1 2
= p,(().
There is exactly one such geodesic if and only if (1. 78)
where Xl is the first critical point of p,. The number of geodesics increase without bound as Itl/llxl1 2 - t 00. Let (1 < (2 < ... < (N be the solutions of {I. 77). The square of the length of the geodesic associated with the solution (m is
(1.79)
s;, = (si~(mfllxI12.
PROOF. The enumeration of geodesics follows from Lemma 1.51. The line Itl/llxl1 2 intersects the graph of p, finitely many times. There is only one intersection if and only if inequality (1. 78) holds. See the graph of p, in Figure 1.2. As the geodesic was parametrized by arc length, its length is given by the value of the parameter sf. Dividing the square of the equation (1.68) by equation (1.69) yields the square of the length
Y
=
S2f = (~)2R2 sinCPf . CPf satisfies equation (1.71). For each CPf = -(m we obtain the length described by
formula (1.79).
0
1.6.
LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP
33
The natural dilations of the Heisenberg Laplacian operator
~ (OXl + 2x 2 0t)2 + ~ (OX2 _ 2XI at) 2 are (Xl, x2, t)
>. > 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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
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,
1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP
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.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
PROOF. Using
rmax = then
rm(
J2!2 = IN = J~~ =J
It I sin(
t(
= -2r~ax
1 <1>
<1>0
It I
sin 2 (
1<1>-<1>0
sin 2 udu.
0
Hence
tm(
_It_I (sin(2(
o
COROLLARY 1.57. The projection of the m-th geodesic on the x-plane is a circle with the radius (1.84) and area
It I
O"m=-·
4m
We note that r m and tm depend on
I;
THEOREM 1.58. The total curvature of a closed curve in is an integer.
]R3
is 2nx, where X
It is known that X represents the number of knots of the curve c. In the case of a plane curve, X = 1. This result was proved independently by Fechnel. For these proofs the reader may consult Millman and Parker [46J.
We shall show a similar property for the geodesics between the origin and points on the t-axis. We shall prove the following: PROPOSITION 1.59. The total curvature of a geodesic between the origin and the point (0,0, t) is 2nm, where m = 1,2, ... is an integer which gives the number of rotations of the geodesic. PROOF. Consider the curve c : [0, sfJ -+ ]R3 joining (0,0,0) and (0,0, t), parametrized by the arc length. Then Ic(u)1 = 1, and by Proposition 1.38, the curvature is k(u) = Ic(u)1 = 4101. Hence, the total curvature is
1 S
!
k(u) du = 4101sf.
1.6. LAGRANGIAN FORMALISM ON THE HEISENBERG GROUP
37
From the boundary value problem
r
-4B2r
r(O) it follows that 2Bs f 1,2,...
0, r(sf)
=0
= m7r. Therefore, the total curvature is equal to 27rm, m = 0
Geodesics between any two arbitrary points. Given two points Pl(Xl, Yl, h) and P2(X2, Y2, t 2), we shall study the connectivity by geodesics in the cases (Xl, yd = (X2, Y2) and (Xl, Yl) =f. (X2, Y2). Making a left translation by (-Xl, -Yl, -tl), the point PI will become the origin and the point P2 will have the coordinates
(X2 - Xl, Y2 - Yl, t2 - tl - 2(YlX2 - XlY2)). By Proposition 1.46 the left translation of a geodesic is a geodesic of the same length. Then theorems 1.52 and 1.53 become: THEOREM 1.60. Given the points Pl(Xl,Yl,tl) and P2(X2,Y2,t 2), there are finitely many geodesics between PI and P2 if and only if (Xl, yd =f. (X2, Y2)' These geodesics are parametrized by the solutions ( of
It2 - tl - 2(YlX2 - xlY2)1 (X2 - xd 2 + (Y2 - Yl)2
(1.85)
= (.-t(().
There is exactly one such geodesic if and only if (1.86)
where Cl is the first critical point of (.-t. The number of geodesics increase without bound as It2 - h - 2(YlX2 - xlY2)1 [(X2 - xd 2 + (Y2 - yd2] -+ 00. Let (1 < (2 < ... < (N be the solutions of (1.85). The square of the length associated with the solution (m is
( .(r::
)2[(X2 - xd 2 + (Y2 - Yl)2]
Sln"m
l/((m) (It2 - h - 2(YlX2 - xlY2)1
+ [(X2
- Xl)2
+ (Y2
- Yd 2]),
where l/(x)
=
X
.
+ sm2 X
X2 -
.
smxcosx
Theorem 1.55 becomes: THEOREM 1.61. Given the points PI (Xl, Yl, tl) and P2(X2, Y2, t2), with Xl and Yl = Y2, the geodesics that join PI and P2 have lengths
s;, = m7rlt2 -
hi.
=
X2
1.
38
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
1.7. Carnot-Caratheodory distance Recall that a curve c(s)
= (X1(S), X2(S), t(s))
i-
2X2X1
is horizontal if c(s) E
Hc(s),
i.e.,
+ 2X1X2 = O.
By Proposition 1.11, any two points P and Q can be joined by a horizontal curve. We have shown that the curve can be considered smooth. Hence the set {c; c(O)
= P,c(l) = Q,c horizontal curve} =I- 0.
The length of a horizontal curve c is
C(c)
=
fa1 )g(c(s),c(s))ds,
where 9 : H x H --+ :F is the subRiemannian metric. The Carnot-Caratheodory distance is defined as de : JR3 x JR3 --+ [0,00), (1.87)
dc(P, Q)
=
inf{C(c); c E S}.
One may verify that de verifies the distance axioms: (i) dc(P, P) = 0, (ii) dc(P, Q) = d(Q, P), (iii) dc(P, R) ::; dc(P, Q) + d(Q, R). The main result of this section is to show that the Carnot-Caratheodory distance de is complete, i.e., any Cauchy sequence with respect to de is convergent. We shall prove this using the equivalence between the geodesic completeness and the completeness as a metric space. DEFINITION 1.62. If for any point P, any geodesic c(t) emanating from P is defined for all t E JR, the geometry is called geodesically complete. The following theorem can be found in Strichartz [50], [51]: THEOREM 1.63. Let M be a connected step 2 subRiemannian manifold. (a) If M is complete, then any two points can be joined by a geodesic. (b) If there exists a point P such that every geodesic from P can be indefinitely extended, then M is complete. (c) Every nonconstant geodesic is locally a unique length minimizing curve. (d) Every length minimizing curve is a geodesic. It is known that geodesics starting from the origin on the Heisenberg group are infinitely extendable. Using (b) of Theorem 1.63 we obtain: THEOREM 1.64. The Carnot-Caratheodory metric de is complete.
Computing the Carnot-Caratheodory distance. In the following we shall describe two ways to obtain the Carnot-Caratheodory distance. First method: From (c) and (d) of Theorem 1.63 we obtain a way to compute the CarnotCaratheodory distance as
dc(P, Q) = {C(c), where c is the shortest geodesic joinning P and Q}.
1.7. CARNOT-CARATHEODORY DISTANCE
Using Theorem 1.53 with m to (x, t) is
=
39
1, the Carnot-Caratheodory distance from the origin
dc(O, (x, t)) = lI((l)(ltl + IlxI12). In particular, if the points are on the same vertical line, the Carnot-Caratheodory distance squared is proportional to the Euclidian distance de ((x, t), (x, t,))2
=
7rlt' - tl = 7rd E ((x, t), (x, t')).
Second method: Another way to obtain the Carnot-Caratheodory distance is using the complex action. We shall discuss this issue in more detail later in the Complex Hamiltonian Mechanics chapter, (ch 5). Consider the modified complex action function
f(x, t, T) = Tg(X, t, T)
=
-iTt + T coth(2T) Ilx11 2,
where
= -it+ fu7{(X,~)-H}dS
g(X,t,T)
-it + coth(2T)llxI1 2
=
is the complex action. Like the classical action, the complex action 9 satisfies the Hamilton-Jacobi equation, see [3]
g
+ H (x,aax ) = O.
ag aT
Using Theorem 1.66 of [3], there is a unique critical point with respect to T of the modified complex action function f in the strip {IIm(T)1 < 7r/2} given by Tc(X, t) = iBc(x, t), where Be is the solution of
t
=
f.L(2TB)llxI1 2
in this interval. At the critical point
1 f(x, t, Tc) = "2dc(O, (x, t)). This works only in the case x
=1=
o.
Application to the fundamental solution. We plan to investigate whether we can use the subRiemannian geometry to construct the fundamental solution of the operator ~x
xl + x:; (aXl +2x2atf
+ (aX2 -2x 1at f
a;l + a;2 + 4at (x2aXl -
x1aX2)
+ 4(xr + x~)8;'
It is expected that the subRiemannian distance (Carnot-Caratheodory distance) will play a role similar to the one the Euclidian distance plays for the Laplacian, i. e., the fundamental solution is the inverse of the Carnot-Caratheodory distance
C K(x, t; 0, 0) = dc(x, t) We need the following lemma.
C
(lxl4
+ t2)1/2'
40
1.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
LEMMA 1.65. For any functions f, hE C 2(JR 3 ), we have (1.88)
Ax(fh)
2 f Ax h + hAf + 2 L Xi (f)Xi(h)
Ax(f2)
2 2fAxf + 2 LXi (f)2
i=l
(1.89)
i=l
(1.90)
221 L(Xd)2 - pAxf.
AX(y)
PROOF. For i
= 1,2 we
X'f(fh)
J3
.=1
have
Xi (Xi (fh)) =Xi(fXd+hXd)
+ Xi ( h Xd) X'fh + h x'f f + 2Xi (f)X i (h).
Xi (f Xih) f
Summing in the above relation with i = 1,2 we obtain relation (1.88). Making f = h we obtain the second relation. By Lemma 1.65, we have (1.91)
0 = Ax (fy) = fAx
(y) + yAxf + 2
t
Xi (f) Xi
(y) .
• =1
Using
Xi(fy) =0 we get
Xi
(y) = - )2 Xi (f)
and substituting in (1.91) yields (1.92)
Ax (
21 f1) = J32", LXi(f) - pAxf.
o LEMMA 1.66. If d is the Carnot-Caratheodory distance from the origin, dc(x, t) = then
v'lxl4 + t 2,
PROOF. From X1(d 2) = 2dX1(d) we get
(1.93)
Xl (d)
= 2~X1 (d 2).
The right side of (1.93) can be computed explicitly
+ 2X10d((xi + x~)2 + t 2) 41xl 2x2 + 4X1t. (OX2
1.7.
CARNOT-CARATHEODORY DISTANCE
41
Hence (1.94) Similarly,
(1.95) Summing the squares of (1.93) and (1.95) yields
Xl (d)2
~ ((lxI2 x2 + Xl t)2 + (lxl 2 xl -
+ X 2(d)2
X2t)2)
~ (lxI4(xi + X~) + t 2(xi + X~)) 41x1 2 . D
LEMMA 1.67. We have
PROOF. Using the formula for Xl(d 2) from the previous proof
(OX2 + 2XlOt)(4(xi + X~)X2 4((xi + x~) + 2x~ + 2xi)
+ 4xlt)
121x1 2 . Similarly,
X;(d 2) =
121x1 2 .
Summing the squares of the last two relations yields the result.
D
LEMMA 1.68. We have
PROOF. Substitute the formulas of Lemma 1.66 and Lemma 1.67 into the relation
D
PROPOSITION 1.69. For (x, t)
i-
(0,0), we have
42
1.
GEOMETRIC MECHANICS ON THE HEISENBERG GROUP
PROOF. Choosing f = d in the third formula of Lemma 1.65 and using the relations given by Lemma 1.67 and Lemma 1.68 yields
o One may show that in the sense of distributions
~xG)
= 15(0,0)'
For details see Folland [25]. 1.8. Exercises EXERCISE 1.1. Let 15>.
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¢.
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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.
2.2. THE HORIZONTAL DISTRIBUTION
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)
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
= (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.
GEOMETRIC ANALYSIS OF STEP
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) <
00.
= 0 and the system
becomes
~ = r¢(¢ + 160r 2) { ¢ = -40r 2
(2.27)
o= constant.
In particular, when 0 = 0, the system (2.27) becomes T= 0, Hence res) = reo) sand ¢(s)
¢=O,
0=0.
= ¢(O) =constant. The i(s) = -r4(s)¢(s) = 0,
t-component satisfies
and hence
t(s) = teO) = O. We arrive at the following result: PROPOSITION 2.11. Given a point P(x, 0), there is a unique geodesic between the origin and P. It is a straight line in the plane {(x, 0); x E 1R2 } of length IxIFrom now on, unless otherwise stated, we shall assume that 0 < O. Substituting the second equation of (2.27) into the first one, we obtain an equation for res) (2.28)
Let (2.29)
T = -480 2 r5.
2.5. SOLUTIONS WHICH START FROM THE ORIGIN
57
be the potential energy. Then the equation (2.28) can be written as the Newton's equation
r=
(2.30)
-V'(r),
where V'(r) = dV(r)/dr. Let r = p in the equation (2.28). Then
r=
dp = dp dr = p dp = -48(Pr 5 ds dr ds dr
Integrating in dp pdr
(2.31 )
=
-48(Pr 5
yields 2
~ = -8(Pr 6 + E 2
or (2.32)
where E is a constant, which denotes the total energy of the system. Using
and the initial condition r(O) = 0 yields
Hence
~r2 + V(r)
(2.33)
=
E
is the law of conservation of energy. The total energy E is equal to the value of the constant Hamiltonian Ho along the solutions
dH ds
8H·
8H.
8H·
8H·
8x ~ + Btt + 7i[~ + Te() .
-~x
.....
+ x~ -
()t
8H
+ t() + -8s
=
8H
+ as 1 2 0 = } H = Ho = -Jx(O)J = E. 2
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
58
r max
Figure 2.2: The potential V(r) and rmax In the following we shall discuss the components l' and ¢ of the geodesics.
The radius 1'(s) The radius 1'(s) starts at 0 and increases (r > 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
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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)
2.5. SOLUTIONS WHICH START FROM THE ORIGIN
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)),
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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
'
2.5. SOLUTIONS WHICH START FROM THE ORIGIN
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<1> sin4 / 3 (3(u -
t(¢) - t(¢o)
1
¢o)) du
<1>0
4 __ r 4 3 max
3 (<1>-<1>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
GEOMETRIC ANALYSIS OF STEP
4
CASE
PROOF. The proof is based on the construction of such a geodesic.
(i) case u < We choose
°°
e<
and require T
= u.
From equation (2.61) one finds
4 rmax
(2.64)
where Q = 1l1(7r/3). The desired geodesic will be given by
(d;l)
r(¢)
1/4
sin 1 / 3 (3(¢ - ¢o),
u r(ci>-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
GEOMETRIC ANALYSIS OF STEP
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¢
H 2/31- m7r sin4 __34 (_0) /3u du 8()2
(HO
0
4m )2/317r . 4/3 -()2 sm u duo 3 8 0
-
= Vb/2. K
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
68
Hence
_ m(HO)2/3 3 (}2 Q,
(2.76)
t -
J
where Q = o71: sin4 / 3 u duo We used
t
4
= 2
H5
Q m 3 K4'
Therefore,
(2.77)
m
= 1,2,3 ...
The physical interpretation of the above relation is the fact that the energy is discrete and depends on It I. The geometrical interpretation is that the geodesics joining the origin 0 to
P(O, 0, t) have discrete lengths. Indeed, let 'Y : [0,1] -> 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
K4
(2.80)
(dd 4
The constant Q follows. LEMMA
= 4Q Itl·
= Jo71: sin4 / 3 u du can be expressed in terms of gamma functions as
2.18. Q = ~ f(1/6) "fff. 4 f(2/3)
2.7. EXPLICIT SOLUTIONS CONNECTING THE ORIGIN TO PROOF. Making the substitution t
(0,0, t)
69
= sin u yields
r
l r~ r~ Q = io sin 4/ 3 udu = 2 io 2 sin 4/ 3 udu = 2 io t 4/ 3(1 - t2)-1/2 dt.
With the substitution t 2 = x the integral becomes a beta function, which can be expressed in terms of gamma functions Q
=
tx
1 / 6 (1
_
X)-1/2
io
Using f(1/2) = )(71"), r(7/6) =
dx = B(7/6 1/2) = f(7/6) f(1/2). ' r(7/6 + 1/2)
i r(1/6) and r(7/6 + 1/2) =
~f(2/3), we obtain
Q = i f (1/6)f(1/2) = ~ r(1/6) y'1i ~ f(2/3)
4 f(2/3)
.
o 2.7. Explicit solutions connecting the origin to (0,0, t) In the previous section we have shown that there are infinitely many geodesics joining the origin with points on the t-axis. In this section we shall find explicit formulae for these geodesics. From (2.74) and (2.76) we may compute rmax as a function of t
24Hg
4
(2Ho)2
rmax = 3 m4K4
= 12 m4K4
31tl = mQ'
It follows that for a given t, the maximal radius rmax is quantized
31tl
4
(2.81 )
rmax
mQ'
m=1,2,3 ...
As m increases, rmax decreases. Using Proposition 2.13 one obtains that the solution in polar coordinates, which joins 0 to (0,0, t) is given by (2.82)
m=1,2,3 ...
The t-component is computed by integration
t m (¢)
=
{4> -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.
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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).
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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}.
2.8. SOLUTIONS WHICH START OUTSIDE THE ORIGIN
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
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
f max
Figure 2.8: The x-projection of the solution
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.
2.8. SOLUTIONS WHICH START OUTSIDE THE ORIGIN
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
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
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,
2.8. SOLUTIONS WHICH START OUTSIDE THE ORIGIN
77
see Figure 2.10.
Figure 2.11: The x-projection of the solution
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
GEOMETRIC ANALYSIS OF STEP
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)-<1>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)-<1>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<4 p4«n+"2)Q,
where
Q=
1 r(1/6)
4 r(2/3) ..;;r ;: :;: 1.82.
Then the equation (2.105) has not more than 2n + 1 solutions and not less than 2n - solutions. All solutions belong to the interval (n + ~)7r ) .
1
(0,
2. GEOMETRIC ANALYSIS OF STEP 4 CASE
80
(ii) If there is an n E N* such that 3 t 1 4 p4=(n+"2)Q, there are exactly 2n+ 1 nonzero solutions for the equation (2.105), and they belong to (0, (n
+ ~)11"].
The largest solution is (n
+ ~)11".
(iii) If there is an n E N* such that 3 t 1 4 p4>(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.
GEOMETRIC ANALYSIS OF STEP
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].
3. THE GEOMETRIC ANALYSIS OF STEP
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
3. THE GEOMETRIC ANALYSIS OF STEP
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 <,,2
2k
EJR
= EJ6 '
2B Xl IX12k -_ EJR.
EJ6
From the horizontality condition
.Xl ) t. = 211 X 2k (.XlX2 - X2
=
EJR 80'
We shall show in the following that .
EJR
6=--· EJXl Using a computation from the previous proof, EJR + 2Bx21xI2k VXl On the other hand, the definition of 6 yields -~ = -4BkxlX2XllxI2k-2
(3.7)
~l
:8
+ 4BkxrX2IxI2k-2.
(Xl - 2Bx21x12k)
4B(k + 1)x21x1 2k - 2B
:8
(x2IxI2k)
+ 4Bx21xI2k - 2Bx21xI2k - 2Bkx2IxI2k-2(2xlXl + 2X2X2) 4Bkx21xI2k + 4BlxI2kX2 - 2Bx21xI2k - 4BkxlX2Xllxl2k-2 - 4Bkx~lxI2k-2X2 -4BkxlX2XllxI2k-2 + 2Bx21xI2k + 4Bkx2 (lxI 2k - 2(xr + X~) - IxI2k-2X~) -4BkxlX2XllxI2k-2 + 2Bx21xI2k + 4BkxrX2IxI2k-2. 4Bkx21xI2k
Using (3.7) we get .
EJR
6=--,
EJXl and hence 6 satisfies the Hamiltonian system. By direct computation, EJR VX2
~
=
-4BkxlX2X2IxI2k-2
+ 2Bxllxl2k + 4Bkx~XllxI2k-2.
Differentiating in the definition of 6, and using ];2 = -4B(k + 1)lxI 2k xl, yields
6
d~ (X2 + 2Bxllxl2k) d
+ 1)l x l2k Xl + 2B d8 (xllxI 2k ) -4Bklxl 2k Xl - 4Blxl 2k Xl + 2Bxllxl2k + 2BkxllxI2k-2(2xlXl + 2X2X2) -4BklxI2k-2xl(Xr + x~) + 4Bkxixllxl2k-2 - 2Blxl 2k Xl + 4BkxlX2X21x12k-2 -4B(k
4BkxlX2X21x12k-2 - 2Blxl2kxl - 4Bkx~XllxI2k-2 EJR EJX2 '
3.5. GEODESICS STARTING AT THE ORIGIN
and hence
6
satisfies the Hamiltonian equation.
91
o
3.4. Euler-Lagrange system in polar coordinates In polar coordinates
Xl
L=
(3.8)
= r cos ¢,
= r sin ¢, the Lagrangian becomes
X2
~(r2 + r 2J?) + e(i + 2r 2k+ 2¢). 2
The momentum e is constant because iJ a Lagrange multiplier.
=
-8H/8t
=
0, and can be considered as
The symmetries for the Lagrangian (3.8) will provide symmetries for the geodesics. The Lagrangian (3.8) is invariant under the symmetry
S : (r, ¢, t; e)
(3.9)
---7
(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 and the point (r(s), ¢(s), t(s )), when e > 0, can be also made for the geodesics between the origin and (r(s), -¢(s), -t(s)), when e < O. This allows us to do the analysis only for the case > O. Euler-Lagrange equations. A computation shows
e
8L .. 2k 8r =r¢(¢+4(k+1)er ),
(3.10)
d 8L .. ds 8r = r,
(3.11)
8~ = r2j, + 2er 2(k+1) 8¢
~
8L _
(3.12)
e
=
0,
8L
at =0,
8t - ,
and hence r (s), ¢( s) and
(3.13)
8L 8¢
,
e satisfy the Euler-Lagrange system {
r=
r¢(¢ + 4(k + 1)er2k) r2(¢ + 2er 2k ) = C(constant)
e = eo (constant).
3.5. Geodesics starting at the origin If the geodesic starts at the origin, then r(O) = 0 and hence C = O. The second equation of (3.13) yields ¢ = -2er2k. With the assumption e > 0, the argument angle ¢ rotates clock-wise.
When
(3.14)
e = 0, the system (3.13) becomes
{~:~ e
= o.
92
3. THE GEOMETRIC ANALYSIS OF STEP
2(k + 1)
CASE
PROPOSITION 3.9. Given a point P(x,O), 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 0 = O.
Hi
PROOF. In the case 0 = 0, the Hamiltonian is H(~) = + ~~~. From the Hamiltonian equation i = aH/aO = 0, t(s) is constant, and as t(O) = 0, it follows that t(s) = 0 and the solution belongs to the x-plane. Using the equation r = 0, it follows the solution is a straight line. 0
From now on, unless otherwise stated, we shall assume 0 > O.
Solving the Hamiltonian system for 0 #- o. As dH/ds = (aH/a~)~+(aH/aO)iJ+ (a H / ax ) x + (a H / at )i + a H / as = 0, the Hamiltonian H is preserved along the solutions. From the Hamiltonian equations
+ 2X21xl2kO, 6 - 2X11x12kO,
Xl = aH/a6 = 6 X2 = aH/a6 = and the Hamiltonian (3.5) becomes
(3.15)
H =
1(.2.2) 1 2 Xl +x 2 = 2'
where we considered s the arc length parameter along the solution. In polar coordinates, xi + x~ = r2 + r2 ¢2 . Using ¢ = - 20r 2k , yields
(3.16)
r2
+ 40 2r 2(2k+l)
= 1.
When r = 0 we get r = r max , with
(3.17)
rmax
=
(201)$ .
We have the following simultaneous system of equations
{
(3.18)
dr / ds = ±-I1 - 40 2r 2(2k+l) d¢/ds = -20r2k.
The" +" sign corresponds to the case when the radius r(s) increases between 0 and rmax. Similarly, the" -" sign is considered for the case when r(s) decreases between rmax and O. In the following we denote by CPo the initial argument. Taking the" + " sign, dividing the equations in (3.18) and integrating, we obtain the angle swept by the vectorial radius between r(O) = 0 and r = rmax w The substitution v w
=
= 20r 2k + 1 =
l
rmax
-20r 2k dr. -II - 40 2 r 2 (2k+1)
yields
1 2k + 1
11 v'f=V2 dv
0
-~
1
- 2k + 1 (arcsin(l) - arcsin(O))
=
2(2k + 1)'
It is interesting to note that the angle w is independent of O. The negative sign shows the rotation is done clock-wise.
GEODESICS STARTING AT THE ORIGIN
3.5.
93
Taking the" -" sign, dividing the equations in (3.18) and integrating, we obtain the angle swept by the vectorial radius between r = rmax and r(O) = 0
w
=
1
0
vI -
Tmax
I
2()r2k dr4()2r2(2k+l) -
T ,,wX
-
VI -
0
2()r 2k 4()2 r 2(2k+1)
-7r
w = 2(2k + 1)" Hence we have the following proposition: PROPOSITION 3.10. The projection of the solution on the x-space passes again through the origin after it sweeps an angle equal to w + iii = 2k";1. THEOREM 3.11. For ¢o - ¢ E [0, given by (3.19)
r(¢)
=
2k~1],
the solution in polar coordinates is
1 ) 2k~1 sm • 1 ( ( 2() 2k + 1 (2k
+ 1)(¢0 -
¢) ) .
PROOF. We have two cases to investigate: Case ¢o - ¢ E [0, 2(2;+1)]: The radius r increases between r(O) = 0 and rmax. Considering the positive sign and dividing equations (3.18) we obtain d¢ _2()r 2k
VI -
dr
4()2r2(2k+1)
Integrating between 0 and r rI.. 'I-' -
rI..
'1-'0
=-
I
T
0
2()r2k
1
arcsin(2()r 2k +1 ). VI _ 4()2r2(2k+1) = ___ 2k + 1
Taking the sine on both sides yields (3.19).
Case ¢o - ¢ E [2(2;+1)' (2k~1)]: The radius r decreases between rmax and dividing equations (3.18) d¢ dr
Integrating between ¢o the right side, yields
2(2;+1)
VI -
¢ - ¢o
Considering the negative sign and
2()r 2k 4()2 r 2(2k+1)
and ¢ on the left side and between rmax and r on
7r
(3.20)
o.
+ 2(2k + 1)
=
1T Tmax
2()r 2k
VI _ 4()2 r 2(2k+1)·
Using the decomposition
the right side of (3.20) becomes
(T 2(2k + 1) + Jo VI 7r
7r
2(2k + 1)
1
2()r 2k MPr 2(2k+1) . (
+ 2k + 1 arcsm
2()r
2k+1)
.
3.
94
THE GEOMETRIC ANALYSIS OF STEP 2{k + 1) CASE
Then (3.20) is equivalent with the following sequence of identities: ¢
+
¢
7l'
_1_ arcsin(2/Jr2k+1) 2k + 1 2/Jr2k+1
¢o) )
2/Jr2k+l ,
2k + 1 sin ((2k + 1)(¢ - ¢o) + 7l') -
0
+ 1) (¢ -
- sin ((2k
o
which leads to (3.19). COROLLARY 3.12. For any angle ¢ and number pEN,
r(¢)2p
=
1 ) 2;~1 . ~ ( 2/J sm 2k +1 ((2k
+ 1)(¢0 -
¢)).
{;:::t7l'] , then
PROOF. Similarly, one may show that if ¢o - ¢ E [27::;1'
r(¢)
1 ) 2k~1 sm • 2k1+ 1 ( (2k + 1)(¢0 - ¢) ) . = (_l)m ( 2/J
Raising to an even power eliminates the minus sign.
o Consider the boundary conditions
r(O) = 0, ¢(O) = ¢o, t(O) = 0, r(s) = R, ¢(s) = ¢1, t(s) = t. Using the rotational symmetry, we may assume ¢o = O. In the following theorem the parameter s represents the arc length. THEOREM 3.13. The following relations take place among the boundary conditions:
1
{2k+l)
(3.21 )
o
dv 2k sin 2k +1 (v)
1
= -
(2k
+ 1) (2/J) 2k+1 s,
(3.22)
2 (1) 2~~!:) 1{2k+1)
(3.23)
t = --- -
(3.24)
R2{k+l)
2k + 1 2/J t
0
= Jik( -¢1),
for t > 0, R
=1=
'
t > 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 {
d¢
sin2~~' ((2k + 1)¢)
io
=
-(20) 2k~' s.
Changing the variable, (2k+l)
o
dv 2k sin 2k +I (v)
1
=
-(2k + 1)(20) 2k+l s,
which is equation (3.21). Replacing ¢ by ¢l, r by R, and p by k + 1 in Corollary 3.12, yields equation (3.22). From the Hamiltonian equation
As ¢ = -20r 2k < 0, t(s) is increasing from t(O) = 0 to t > O. Making p = k Corollary 3.12 and integrating between ¢o = 0 and ¢l, yields
t = -2 ( -
1 ) 2~~!i)
+ 1 in
1 1(2k+l)
20
which is equivalent to (3.23). Dividing equations (3.23) and (3.22) yields equation (3.24).
o
The function f.Lk' The behavior of the function f.Lk given by (3.25) is very important in understanding the subRiemannian geometry defined by the vector fields (3.1). For the sake of simplicity we shall denote in the following f.Lk by f.L. The function f.L(x) is odd. It suffices to study it just for x > 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
the geodesics of that length are parametrized by the circle
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(
=
r~ax sin ~ (3(
After m clock-wise loops, the angle is
m1T
3'
Integrating between
i=
-4r4~,
and using the boundary conditions t(<po) = 0, t(
=
1 1
<1>0- ";"
-4r~ax
4
sin 3 (3(
<1>0
4 4 -rmax
3
m7r
0
4mr 4 2 3 max 4m
4
Trmax B
,.1 Sln 3
1 7r
0
/
2
4m 4 vdv = -rmax 3
1 7r
0
4m 4 , 4 3 vdv=-r sm
3 max
,.13 sm vdv
11 0
1 x 61 (I-x)-?:dx
(76' 21) = TrmaxQ, 2m 4
Hence 4 rmax
31tl 2mQ' D
Heisenberg group case. When k (3.41)
a a + 2X2-;:), UX1 Ut
Xl = ~
= 0 the vector fields
a aX2
X2 = -
-
a at
2X1-,
define a step 2 model. They are left invariant vector fields with respect to the I-dimensional Heisenberg group HI = (JR3, 0), with the group law
(x, t)
0
(x', t') = (x
+ x', t + t' + 2X2X~
-
2X1X~),
In this case, we may recover Theorems 1.52 and 1.53 and hence Theorems 1.36 and 1.41 in [1], We state the result as the following theorem:
3.
102
THE GEOMETRIC ANALYSIS OF STEP
2(k
+ 1) CASE
THEOREM 3.22. There are finitely many geodesics that join the origin to (x, t) if and only if x 1= O. These geodesics are parametrized by the solutions 'Ij; of:
It I
(3.42)
IIxll 2 = p,('Ij;).
There is exactly one such geodesic if and only if:
It I < p,(x1)lIxIl 2 . The number of geodesics increases without bound as 11~1112 --+ 00. If 0 ~ 'lj;1 < ... < 'lj;N are the solutions of equation (3.42), there are N geodesics joining the origin and their lengths are given by
s~
(3.43)
= V('Ij;m)
(It I +
IIx1l2) ,
where
x2
V(x) =
(3.44)
x
..
+ sm2 x -
smxcosx
,
and (3.45)
p,(x) =
x
cotx. sm x PROOF. It comes from Theorem 3.17. With the substitution k = 0 in equation (3.77) ('/) . 2 d . p, ('Ij;) _ 2 Jo sm v v _ 'Ij; - sm'lj;cos'lj; _ _ 'Ij;_ _ cot'lj; o sin2'1j; sin2'1j; - sin 2 '1j; , -.-2- -
which is equation (1.26) of [1]. Substituting k = 0 in (3.29) yields 'lj;2 v('Ij;) = ----...,,--(1 + p,o( 'Ij;)) sin2 'Ij;
'lj;2 sin2 '1j; + 1>-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,
the geodesics of that length are parametrized by the circle
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
3.8. CONJUGATE POINTS TO THE ORIGIN
lO5
Fact 1: If a loop belongs to the sector area m then the following loop will belong to the sector area m + 2k + 2. As the second loop belongs to the sector area 2k + 3, using Fact 1, the third will belong to (2k + 3) + (2k + 2) = 4k + 2 + 3 = 3 sector area. Fact 2: In general, the 2£ + 1 loop lies into the 2£ + 1 sector area. Indeed, using Fact 1 for 2£ times, the loop 2£ + 1 belongs to the class of
1 + (2k
+ 2) + ... + (2k + 2) = 1 + 2£(2k + 2) = 1 + 2£ + (4k + 2), which is the class 2£ + 1. From Fact 2 each odd loop lies in the corresponding odd number class area sector. Hence each odd area sector contains a loop. We will show that the even sector areas remain empty. This will come from the fact that the even loops lie also in the odd sector areas. Indeed, the second loop lies into 2k + 3 sector area, which is an odd number. In general, using Fact 1, the 2£ loop belongs to the class of 1 + (2£ - 1)(2k + 2) = 1 + 2(2£ - l)(k + 1), which is odd. As there are 4k + 2 = 2(2k + 1) sectors only half are odd. No loops belong to even class sector areas. Hence there only 2k + 1 loops. This completes the proof of the proposition. Using Proposition 3.25, N(O) = l. In this case, the single loop is in fact a circle which starts and closes at the origin. The equation in polar coordinates is r(¢) = rmax sin(¢ - ¢o), where ¢ - ¢o E [0,7l']. If k = 1, then N(l) = 3. There are 3 loops with an angle of 120 degrees between the symmetry axes. This corresponds to the step 4 case. The equation of the first loop is r(¢) = rmax sinl/3(3(¢ - ¢o)), where ¢ - ¢o E [0,7l'/3]. If k = 2, then N(2) = 5. The equation of the first loop is r( ¢) = rmax sin l / 5 (5(¢ - ¢o)), where ¢ - ¢o E [0, 7l' /5]. See Figures 3.1 - 3.4. 0 3.8. Conjugate points to the origin
Consider a geodesic (Xl, X2, t) which starts at the origin. In polar coordinates, one has r(¢o) = and t(¢o) = 0. When ¢l = ¢o - 7l'/(2k + 1), then r(¢l) = 0, which means that the x-component is zero. The t-component is given by Theorem 3.13, equation (3.23)
°
t(¢o -7l'/(2k
+ 1))
2 (1) 2~~!~) 1(2k+l)4>1 sm.
--- 2k + 1 2()
(v) dv
0
2 (1) 2~~!~) 17r sin
--- 2k + 1 2()
2(k+1) 2kH
2(k+1) 2k+1
(v) dv
=T
0
does not depend on the initial argument ¢o. On the other hand, for each ¢o there is a geodesic which starts with the initial argument ¢(O) = ¢o. A variation of ¢o corresponds to a variation of the geodesic. The point (0,0, T) remains fixed during the variation, and hence it is a conjugate point to the origin. The following theorem shows that the converse is also true. THEOREM 3.26. Given a point P(O, 0, T) on the t-axis, T # 0, there is a geodesic 'Y starting at the origin, such that P is the first conjugate point to the origin along 'Y.
3. THE GEOMETRIC ANALYSIS OF STEP
106
PROOF.
sider
T
2(k + I)
CASE
The proof is based on an explicit construction of the geodesic. ConT = T, we have
< 0 and hence, () < O. From the equation
-T(2k+l)(17r. 2(k+l) sm 2k+l (v) dv 2 0
(3.47)
)-1
-T(2k + 1) Q
Given ¢o, using Theorem 3.11, one may construct the projection on the x-plane of the geodesic 1) r(¢) = ( 2()
(3.48)
2k'+'
•
1
(
sm 2 k+ 1 (2k
+ 1)(¢o -
)
¢) .
The t-component can be found by integrating the horizontal condition
t(¢)
=
-2
('I> 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
<7(S)
1
-----r=;=:==~~ =
Jv(1 - v 2k +1 )
o
107
2(20) 2k+l S,
where (1( s) = (40 2) 2k'f-l and we considered 0 > O. The above integral can be written in terms of Barnes's extended hypergeometric function (see Erdelyi [23])
(3.50) r(8)
io
dv _ JV(I-v2k+l) -
Using 2J(1(s)
(3.51 )
=
1 ][
2~F([~ (1(s
1]
2'2(2k+l) , 1+ 2(2k+l) , (1(s)
2k+1)
.
2 (20) 2k'f- 1 r(s), the equation (3.50) becomes
1] 2k+ 1 1 ] [ r(s)F ( [ 2'2(2k+l)' 1+ 2(2k+l) , (1(s)
1) =s.
The function
F([~, 2(2k1+ 1)]' [1 + 2(2k1+ 1)]'z)
z ---.
is increasing on the domain [0,1]. See Figure 3.6.
1.12 1.1
1.08 1.06· 1.04 1.02 1~~~~~~~__~~~~
o
0.2
0.4
0.6
0.8
z
Figure 3.6: The function z ---. F(z). The minimum and maximum values are
F([~, 2(2k\ 1)]' [1 + 2(2k1+ 1)],0) =
1,
F([~, 2(2k1+ 1)]' [1 + 2(2k1+ 1)],1) =
F.
Choose r(s) such that (1(s) = 1, and hence the hypergeometric function reaches the maximum in relation (3.51)
r(s) =
C10)
2k'f-l ,
which is the same as r max , see formula (3.17). Let Sl be the arc length parameter for which r(sr)
(3.52)
rmax F =
S1·
= rmax.
Then (3.51) yields
3. THE GEOMETRIC ANALYSIS OF STEP
108
2(k + 1)
CASE
At 1'( Sl) the trajectory starts to bounce back, and at s = 2Sl it reaches the t-axis again, i.e., r(2sl) = o. Hence, the shortest geodesic will have the length
1\
(3.53)
=
2Sl
=
2rmax F.
If the geodesic winds m times around the t-axis, joining (0,0,0) and (0,0, t), the length is (3.54) The momentum () depends on the boundary condition t. This relationship will be found below. A Hamiltonian equation yields
i = He = -2r 2k +2J. We shall consider the following boundary conditions for the argument angles ¢o
= ¢(O),
Let () > 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 )
3. THE GEOMETRIC ANALYSIS OF STEP
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
3. THE GEOMETRIC ANALYSIS OF STEP
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 <
y(to) = Yo
on a right maximal interval J. Then either J and ly(t)l----; 00 as t ----; 8.
=
[to, to +a] or J
=
[to, 8),8:::; to +a,
The Hamiltonian system of equations can be written in the form (3.70). Consider y E ]R2n, y = (x,p), with X,p E ]Rn. Let H(y) be the Hamiltonian function, and set "YH(y)
=
8H) ( 8H 8x' 8p ,
and
Then
Hamiltonian system (3.71)
{
x(s) p(s)
= 8H/8p = -8H/8x
114
3. THE GEOMETRIC ANALYSIS OF STEP 2(k
+ 1)
CASE
can be written as
iJ(s) = f(y(s)),
(3.72)
where f(y) = .T'V H(y). Since f is independent of s, we may rewrite Lemma 3.33 as COROLLARY 3.34. Let f(y) be an arbitrary continuous function for y E and let y( s) be a solution for
iJ(s) = f(y(s)), on a right maximal interval I. Then either I as s ---+ 5.
]R2n
y(O) = Yo
= [0, (0) or I = [0,5) with ly(s)1
---+ 00
In other words, if the solution y( s) doesn't blow up for a finite value of the parameter s, then the maximal right interval is [0,(0). We are working in coordinates r, ¢, t with the corresponding momenta p, TJ, (). Consider a solution for the Hamiltonian system
y(s)
=
(r(s), ¢(s), t(s);p(s), TJ(s), ()(s)).
We shall show that ly(s)1 cannot blow up for finite values of s. This is equivalent to showing that each component of y( s) has this property. The momentum () is bounded because it is constant. Recall the Lagrangian in polar coordinates is
(3.73)
L
= ~(r2 + r 2 J}) + ()i + 2()r 2k +2 ¢.
Equation (3.73) yields i] = oLlo¢ = O. This implies that TJ(s) = TJ is a constant and hence bounded. The momentum p is given by p = oLlar = r. Using the conservation of energy, yields
Ip(s)1
Ir(s)1 ~
=
m.
As E is constant along the solutions, Ip( s) 1 is bounded. Here we used the standard definition of the momentum
oL
p
= oq'
and the Euler-Lagrange equation
. p
=
oL oq'
Next, we deal with the bounds for r(s). We have seen that for E > 0 there are exactly two positive roots rmin, rmax of the equation
V(r)
= E,
and
0< rmin ~ r(s) ~ rmax , so r(s) is bounded. The projection of the trajectory in the x-plane is contained in the ring defined by the circles of radii rmin and rmax. See the Figures 2.8 and 2.1l. In order to deal with the bounds for the ¢ component, we use (3.74)
¢(s) = {
rP(s) -
2()r2k(s),
-2()r 2k ,
if r(O) if r(O)
=I 0, = O.
3.12.
GLOBAL CONNECTIVITY BY GEODESICS
115
The inequality
0:::; rmin :::; r(s) :::; rmax yields the upper bound
14>(s)1 :::; Co, VS, with
(3.75)
if r(O) =f 0, if r(O) = O.
Co = {
Thus for all s
1¢>(s)l:::; I¢>(s) - ¢>(O) I + I¢>(O)I :::; Coisl
+ I¢>(O)I,
so lim 1¢>(s)1 :::; C8 + I¢>ol,
s->/i
and 1¢>(s)1 cannot blow up for finite 8. We still need to show that t(s) doesn't blow up for finite s. From the Hamiltonian equation
i(s) =
~~
=
-2r2(k+ll(s)4>(s).
Using the boundedness of 14>1,
It( s) I <
It(s) - t(O)1
+ It(O)1
:::; max li(u)llsl uE[O,s]
+ It(O)1
< 2r~~:lll4>(s)1 + It(O)1 :::; 2Cor;~~:1)Isl + It(O)1 Clisl
+ It(O)I·
Hence t(s) cannot blow up for a finite value of s. Using Corollary 3.34 the bicharacteristic solution y(s) = (r(s), ¢>(s), t(s),p(s), 'I](s), 8(s)) is defined on [0, +00). Using a similar argument to the left we get y( s) defined on the entire real line. In particular, the projection on the (r, ¢»-plane has the same property. 3.12. Global connectivity by geodesics Consider the subRiemannian geometry induced by the vector fields (3.1). In this section we shall prove the following global connectivity result: THEOREM 3.35. Given any two points A, B E 1R 3 , there is a geodesic joining A
and B. It is known that geodesics are locally length minimizing, see Hamenstadt [30], Lee
and Markus [43]' Strichartz [50], [51], Belai'che [7]. However, there are examples of length minimizing curves in steps greater than 2, which are not geodesics, see Liu and Sussmann [44], Sussmann [52]. DEFINITION 3.36. The distribution H generated by the vector fields Xl and X 2 given by (3.1) is called the horizontal distribution. A curve c is called horizontal if C E H. The Riemannian metric g defined on H with respect to which Xl and X 2 are orthonormal is called the subRiemannian metric.
116
3.
THE GEOMETRIC ANALYSIS OF STEP
2{k
+ 1) CASE
By Chow's theorem, [22], arbitrary points P and Q can be joined by a piece-wise horizontal curve. Using the subRiemannian metric, one may define the distance d(P, Q)
where length(c)
= inf{length(c); c horizontal, joining P and Q},
= II Jg(c,c).
DEFINITION 3.37. A subRiemannian manifold is called complete if it is complete as a metric space, i.e., the distance function d given by the subRiemannian metric is complete. The following theorem can be found in Strichartz [50], [51]: THEOREM 3.38. Let M be a connected step 2 sub Riemannian manifold. (a) If M is complete, then any two points can be joined by a geodesic. (b) If there exists a point P such that every geodesic from P can be indefinitely extended, then M is complete. (c) Every nonconstant geodesic is locally a unique length minimizing curve. (d) Every length minimizing curve is a geodesic.
Proof of Theorem 3.35 Let M = lR 3 \{(0, t); t E lR}. The subRiemannian model defined by the vector fields (3.1) is step 2 on M and step 2(k + 1) on {(a, t); t E lR}. Let A, B E lR 3 .
(i) A,B EM: From Theorem 3.32, any point of M has the property (b) of Theorem 3.38. Therefore M is complete. Using Theorem 3.38 (a), any two points in M can be joined by a minimizing geodesic. By property (d) of Theorem 3.38, the minimizing geodesic is a geodesic. Hence A and B can be connected by a normal geodesic. (ii) A E {(O,t);t E lR} and B E M: After a translation along the t-axis, Theorem 3.17 has the following consequence: COROLLARY 3.39. There are finitely many geodesics that join the point (0, to) to (x, t) if and only if x #- 0. These geodesics are parametrized by the solutions 1jJ of:
It - tol IxI 2{k+l)
(3.76)
=
J.L(7/J) ,
There is exactly one such geodesic if and only if:
where (3.77)
Cl
is the first critical point for J.L and _
2
J.L(x)-~k1
+
10(2k+l)x sin sin
2(H1) 2H1
~ 2 +1
(v) dv
(x)
Clearly, (ii) is an immediate consequence. (iii) A,B E {(O,t);t E lR}: This is the limit case, Ixl-> 0, of Corollary 3.39. Namely, (3.76) becomesJ.L(1jJ) Hence 7/J E 2:+1 Z. There are infinitely many geodesics between A and B.
= 00.
3.13.
KAEHLER METRICS
117
3.13. Kaehler metrics
With the definitions given in Chapter 1, we have the following generalization of Proposition 1.32 PROPOSITION 3.40. The subRiemannian metric 9 in which {Xl ,X2 } are orthonormal is a Kaehler metric on H. x , for any x E JR3. The fundamental 2-form cP satisfies 4(k + 1)lxl 2k cp = n. Hence
n(U, V) = 4(k + 1)lxI2kg(U, JV),
for all horizontal vectors U and V.
PROOF. Consider V = H. x . We shall show first that 9 is a Hermitian metric. Let U = U l Xl + U 2X 2 and V = VI Xl + V 2X 2 be two horizontal vector fields. Using JX l = -X2 and JX 2 = Xl, yields
JU = -U l X 2 + U 2X l ,
JV = -V l X 2 + V 2X l .
Using the orthonormality of Xl and X 2
g(u 2Xl - U l X 2, V2 Xl - VI X 2) UlVl + U 2V2 g(U l Xl + U2X 2, VI Xl + V2 X 2) g(U, V),
g(JU, JV)
and hence 9 is invariant by J. The 2-form
n(U, V)
n is closed because it is exact n =
dw.
n(U l Xl (U l V2 (U l V2 (U l V2 -
+ U 2X 2, VI Xl + V2 X 2) U 2V l )n(Xl , X 2) U 2V l )(Xl (w(X 2 )) - X 2 (W(X l )) - W([X l ,X2 ])) U 2V l )w(4(k + 1)lxl 2k 8d 4(k + 1)lxI2k(U l V2 - U 2Vl) 4(k + 1)lxI2kg(U l Xl + U 2X 2 , V2 Xl - VI X 2) 4(k + 1)lxI2kg(U, JV).
o The skew-symmetry of n yields: COROLLARY 3.41.
g(U, JU)
= 0,
for any horizontal vector U.
The following result was proved in Chapter 1 and it is true in general step: PROPOSITION 3.42. For any horizontal vector field U we have
J(U) =
~(n(Xl,U)Xl +n(X2,U)X2).
PROOF. As both sides are linear, it suffices to check the relation only for the basic vector fields. Using n(X l ,X2) = -n(X2,Xd = 4, yields
JXl = -X2 = JX 2 = Xl =
~n(X2,Xl)X2 = ~(n(Xl,Xl)Xl +n(X2,xdX2) ~n(Xl' X2)Xl = ~ (n(xl , X2)Xl + n(X2' X 2)X2 ).
o
3. THE GEOMETRIC ANALYSIS OF STEP
118
2(k + 1)
CASE
3.14. The classical action The case of step 2(k + 1). The Lagrangian associated with the vector fields
Xl = OX}
(3.78)
+ 2X21x12kOt,
X 2 = OX2 - 2Xllxl2kOt
is given in polar coordinates by
~(xi + x~) + O(i - 40r 4k +2) 2 E + oi - 40 2 r 4 k+ 2 .
L
Consider the action along the geodesics with the classical boundary conditions
x(O) = 0,
t(O)
=
0,
where r = Ixl. The classical action is given by
S(r, t, 0, r)
=
1T
L ds =
1T (E + oi - 40 2r 4k+2) ds
Er
+ Ot -
40 2
1 T
r 4k +2(s) ds
Er + Ot - 40 2 J, Hence
S = Er + Ot - 40 2 J,
(3.79) where (3.80)
We shall compute in the following the integral J. The energy conservation law xi + x~ = 2E, written in polar coordinates is f2 + 40 2 r 4k + 2
= 2E.
This is equivalent to
(
3.81
ds
)
=
dr V2E _ 402r4k+2 '
where r(O) = 0 and f(O) > O. Substituting (3.81) in relation (3.80), yields T r4k+2 4J = 4 dr. (3 82) . 0 V2E - 40 2r 4k +2
i
Substituting
_ (20 2 ) 4k~2 v-u E '
20 2 ) 4k~2 a-=r ( E
in relation (3.82), yields 4J
=
r
4u 4k +2 du io V2E - 402U 4k +2
4 =
v'2E io
_4_. ~. (~) 4k~2 f'Y v'2E 202 20 2 io Then (3.83)
r
u 4k+2 du
VI _2~2 u4k+2
v 4k + 2 dv VI - v 4k + 2 .
3.14. THE CLASSICAL ACTION
119
where I
(3.84)
= fa
v 4k +2 dv .
vI - v
Jo
4k + 2
In the following we shall compute the integral I. Integrating in (3.81) yields
Integrating by parts in (3.84) and using the above relation, one obtains an equation for I I
r
v 4k + 2 dv
Jo VI - v4 k+2
=
r
_1_ 2k + 1
r v.
Jo
(2k
+ l)v 4k +l
VI - v4k + 2
dv
viI - v 4k +21' dv J - -1 - [O-\iI - 0-4k+2 - a VI - v 4k + 2 dV] _1_ v [_ 2k + 1 o
+1
2k
0
1
- - - 02k 1[
+
-12k 1
+
+1
1- 0-4k+2
V
[V1 - 0-
_1_ [_ 2k
l l VI _ +l VI a
-
1- v 4k +2 dv v 4k +2 ]
0
0-4k+2
a
0
1 dv v 4k + 2
o-Vl _0-4k+2 + r(2k+1 Eke)
2kt,.1
-
l
0
a
v 4k +2
VI _ v 4k + 2
dv ]
I].
Then
and solving for I, yields (3.85) 1
. ) Usmg In the following we shall compute the first term in ( 3.85. we have
0- =
) r (202 If"
4k+2
'
3.14. THE CLASSICAL ACTION
119
where
1=
(3.84)
l
V4k+2 dv
u
VI -
o
v 4k +2 .
In the following we shall compute the integral I. Integrating in (3.81) yields
r
io
du V2E - 4(PU 4k +2
=7
r
{=}
io
du
VI - 40
2 U
2E
V2E7
=
4k+2
Integrating by parts in (3.84) and using the above relation, one obtains an equation for I
I
r
io
V4k +2 dv VI - v4k+2
-12k + 1
l
u
1
= 2k+1 io
+ l)v 4k +1
(2k V·
VI _ v4k+2
dv
v [-Vl- v 4k +2]' dv
0
_ _1_ 2k + 1
[a Vl _ a 4k+2 _ ior VI _ v4k+2 dV] u
- -1- [aVl - a 4k+2 2k + 1
l VI +l VI -
1 - v 4k +2 dv ] v 4k +2
u
0
-1- [- aVl - a 4k+2 2k + 1
u
0
_1_ [ _
2k + 1
r
1 dv v 4k +2
aVl _ a4k+2 + 7(2k+1 Eke) 2k~j-1
-
l
u
v 4k +2 dV] v 4k +2
VI -
0
I].
Then
and solving for I, yields (3.85)
__ aVl - a 4k+2 12(k+l)
_7_ .
+ k+l
((}Ek) 2k
2k't-l
. 1
. ) Usmg In the following we shall compute the first term in ( 3.85. we have
0"
= r (202 E )
4k+2
'
3. THE GEOMETRIC ANALYSIS OF STEP 2(k
120
+ 1)
CASE
Substituting in (3.85), relation (3.83) becomes
4J
=
Hence
_ Er
(3.86)
4J -
()2 (k
_
+ 1)
r
()2 (k
+ 1)
j E _ ()2 2
4k+2
r
.
Substituting (3.86) in equation (3.79) yields
S
=
Er + ()t -
()2
[Er ()2 (k + 1) -
r
()2 (k
+ 1) VIE '2 -
()2 4k+2]
r
Er- + -r- jE- - ()2 r4k+2. Er + ()t - k+1 k+1 2 Hence the classical action is given by
S=
(3.87)
()t
+ _1_ [kEr + rJ E k +1 2
r
()2 4k+2].
The action S = S(r, t, (), r, E) depends on the boundary conditions r
= r(r) = Ix(r)l,
t
= t(r),
the energy constant E, the time r and the momentum constant (). The Heisenberg group case. In the case k = 0, formula (3.87) becomes
(3.88) In this case the constant E can be explicitly computed in terms of r, rand (). Making k = in relation (3.81) yields
°
ds =
Integrating between
dr
J2E - 4()2r2
°
and r yields
r ----r=d=u== .1 2
io
V
1
49 2 2E
U
=
v'2E r.
3.14. THE CLASSICAL ACTION
PROPOSITION
121
> 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
3. THE GEOMETRIC ANALYSIS OF STEP
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.
4.4. THE HORIZONTAL DISTRIBUTION
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<
L ajJ..L(2aj7r/2an)r;,
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.
GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS
(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
<
0'2
< ... a < O'm.
Let Vk denote the eigenspace of A2 , which is associated with the eigenvalue -40'~. We set k
(4.46)
Wk =
U Vj,
R~ =
j=l
L
r;'
aj=ak
The following results deals with all the geodesics between (0,0) and a given point (x, t), x i:- 0, t > 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.
GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS
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
<
x2
-
a(x - sin x cos x) 2 ' sin a
a < x.
the smaller
4.6. THE SHORTEST GEODESIC
137
We set
1jJ(x) = 2X2 - a(~x - sin2x) 2sm 2 x
(4.56) and note that
1jJ'(x) = (x - a){l'(x) >
°
if a < x < 7r.
This proves (4.55). (ii) The only other possibility which needs to be considered is that (4.33) holds with m = 1, and there is a 0 E Crr/2a n , rr/2ak) such that n
(4.57)
It I = L
aj{l(2ajO)r;.
j=1
(4.58) This follows from (4.55), and hence the lemma is proved.
D
COROLLARY 4.18. Let x = 0, t =I- 0, and fix a pair of integers (j, m), j = 1,2, ... ,n, m = 1,2,3, .... There are infinitely many geodesics of the same length djm(O, t) between (0,0) and (0, t), where d
Jm
(0 )2 ,t
=
2mrrltl 2' aj
The Carnot-Caratheodory distance of (0, t) from the origin is 2
rrltl
2
d(O,t) =dn1 (0,t) = - . an
REMARK 4.19. We note that different pairs (j, m) may yield the same geodesic. The previous lemmas lead to the following result: THEOREM 4.20. Suppose x' =I- 0, x" (i) If
= 0.
n
(4.59)
It I < L
aj{l(2ajrr /2a n )r;,
)+1
then the Carnot-Caratheodory distance d(x, t) is given by n
(4.60)
d(x, t)2 = 2S(x, t, 1; 0) = 20 [It I +
L aj cot(2ajO)r;] , j=1
138
4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS
where B is the unique solution of n
It I =
L aj/l(2a B)r; j
j+1
in the interval (0, 1l" /2a n ). (ii) If (4.59) fails, then there are infinitely many shortest geodesics of the same length from (0,0) to (x, t), parametrized by the sphere
~II(II(O)II = ~[Itl- taj/l(2aj1l"/2an)r;], 2 2a n
(4.61)
j=1
where we set ( = ((', ("), (" = ((2p+l, ... , (2p). The length of these geodesics yield the Carnot-Caratheodory distance of (x, t) from the origin: 2S(s,t,I;1l"/2an )
d(x,t)2
n
21l" /2an [It I +
(4.62)
L aj cot(2aj1l" /2an )r;] . j=1
The distance d(x, t) is homogeneous of degree 1 with respect to the dilations
(x, t)
(4.63)
-*
(Ax, A2 t),
A> 0,
it is bounded above and below by multiples of Ixl function of (x, t), although it is not differentiable.
+ IW/ 2
and it
is
a continuous
°
PROOF. We note that it suffices to prove the result for t > 0, since for t < we simply switch to -1l"/2a n in our arguments. We also note that formulas (4.61) and (4.62) are justified by the convention (4.20), since rp+1 = ... = rn = 0. Lemma 4.12 proves the statement (4.62), and Lemmas 4.47, 4.15, 4.16 and 4.17 imply (4.61) and (4.62). To prove that d(x, t) is continuous on Hn we only need to show that
lim d(x', x", t)
(4.64)
X"-----I'O
= d(x',O,t).
We obtain the above limit from the part (ii) of the proof of Lemma 4.15, if we set k = 1. Homogeneity with respect to the dilations (4.63) is a consequence of the fact that Be(x, t) is homogeneous of degree 0. The quotient d(x, t)/(Ixl + IW/2) is homogeneous of degree and positive on Hn \{O}, so it is bounded above and below. Thus to finish the proof we are left with showing that d(x, t) is not differentiable. Let
°
n
(4.65)
Ep(x, t)
= It I -
L aj/l(2aj1l" /2an )r; j=1
denote the excess function defined on (4.66)
Hp
= {(x, t)
E H n,x
= (x',
On.
Lemmas 4.13 and 4.16 show that for points in Hp, the initial momentum ~(O) is not uniquely defined, and we obtain a multitude of bicharacteristics. We may refer to such points as caustic points. We need the following result: LEMMA 4.21. All those first derivatives of S(x, t, 1; Be) which are normal to Hp are discontinuous on {(x, t) E Hn; Ep(x, t) > 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
4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS
{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)
<
oo}
is the Bergman space. (For more detailed discussion of the Bergman kernel, readers may consult [41].) (i) Show that the Bergman kernel K(z, w) is conjugate symmetric, i.e., K(z, w) =
-=-::c;--'--'-;-
K(w, z). (ii) Prove that the Bergman kernel is uniquely determined by the properties that it is an element of A2(n) in z, is conjugate symmetric, and reproduces A2(n). (iii) Let n 1 and n 2 be two domains in en. Let f: n 1 -+ n 2 be biholomorphic. Show that det Jcf(z)Ko2 (f(z), f(w)) det Jcf(w) = K o , (z, w). Here det Jcf(z) is the complex Jacobian of the mapping f.
EXERCISE 4.8. Since A2(n) is a subspace ofthe Hilbert space L2(n), thus there exists a complete orthonormal basis {¢dkEN for A2(n). Let V be a compact subset of n. (i) Show that the series 00
L ¢k(Z)¢k(W) k=l converges uniformly on V x V to the Bergman kernel K (z, w). (ii) Let n = Bn be the unit ball in en. By the uniqueness of the power series expansion for an element of A2(Bn), the element zk = zr' z~n form a complete orthonormal system on Bn. Setting
Z;2 ...
Ctk
=
r Iz lBn
k I2 dV(z),
4.8.
show that 4.8,
{Zk /
EXERCISES
143
y0k} is an orthonormal basis in A 2 (Bn). Thus by (i) of Exercise
(iii) Show that
(iv) Show that
21fn(k!)2 (n+k-1)!' 1f n k! (n+k)!' (v) Let z E Bn and 0 < r < 1. Denote e the point (1,0, ... ,0). Then show that
n! 1 KBn(z,re) = -1fn ( 1 - rZ1 ) n+1' (vi) Now applying part (iii) of Exercise 4.8, one may conclude that
n! 1 KBn(Z, w) = 1fn (1 _ z. w)n+1 where Z . W = L~=l ZkWk· EXERCISE 4.9. Let {a1, ... , an} be a set of n positive constants. Let
U~~l
n
= { z E
e n+1 :
2Im(zn+d > 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
4. GEOMETRY ON HIGHER DIMENSIONAL HEISENBERG GROUPS
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
5. COMPLEX HAMILTONIAN MECHANICS
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.
COMPLEX HAMILTONIAN MECHANICS
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
5. COMPLEX HAMILTONIAN MECHANICS
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.
COMPLEX HAMILTONIAN MECHANICS
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'
5. COMPLEX HAMILTONIAN MECHANICS
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
5.5. COMPLEX CONNECTIVITY
• 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
COMPLEX HAMILTONIAN MECHANICS
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.
5.5. COMPLEX CONNECTIVITY
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.
COMPLEX HAMILTONIAN MECHANICS
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
5. COMPLEX HAMILTONIAN MECHANICS
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 <7
V4k+2
dv. o v'1 + v 4k +2 We shall compute [ by cyclic integration. Before that we need to compute the integral 4k +2' It is known that [=
J; v'1::
du = v'2E + 4u 4k +2
{r
lo
T.
1
With the substitutions v = (2/ E) 4k+2 U and a = (2/ E) becomes
1
dv
<7
(5.45)
-r.:=====;n=;:=ii=
o v'1
+ v 4k +2
..!ct..!..
(2k
+ 1)[ = aV1 + a4k+2 -
r the above integral
k
= 2 2k + 1 E 2k+l
Hence
1
4k+2
T.
r v'1 + v +
- [.
2k + 1 2 ..!ct..!..
T)
lo
dv
4k 2
Solving for [ and using equation (5.45), yields [
=
1 2(k + 1)
(.f 1 + a ay
4k + 2 -
1 ----:-:-----:-aV1 + a 4k +2 2(k+1)
E2k~1
-
k E2k+l
T
--k----·
22k+l k+1
5.6.
THE COMPLEX ACTION
165
Then 2 2k'f.l E
4J
A\\ I ~
k
1
22k+1 E2k+l
Next we shall compute
2(k+1)
k+l
av'l + a 4k+2.
(~ rk~2 r
aVl + a 4k+2
Er
.1
ay 1 + a 4k +2 - - - .
J+ ~ 1
--r (-E2) 4k~2.j2 v'E
r 4k +2 =
Je
(~ rk~2
J
E 4k+2 --+r
2
(2-E )*' r
rV E + 2r 4k+2
J
E 4k+2 -+r
2
.
Then k ~ 22k+1E2k+1
4J
1 ( -2 ) ~v 2k+l r -E +r4k+2 - Er2(k+1) E 2 k+1
_r_. IE +r4k+2 _ Er . k+1V 2 k+1
o Lemma 5.21 is useful for computing the complex action. THEOREM 5.22. The complex action is given by g
=
-it
+ _1_ (kEr + r·
VIE2 + r4k+2) .
k+ 1
PROOF. A straightforward computation shows g
=
Er - it(r)
+ 4 foT r 4k +2(s)
. Er - zt(r)
+ -r- - + r 4k +2 - -Er-
-it
k+l
VE
k+l
2
+ _1_ (kEr + r·
VIE2 + r4k+2) '
k+1
where we used the boundary condition t(r) = t.
o The complex action from the origin is defined for any r > 0 for which r( r) makes sense. The complex action depends on the energy parameter E. Only in the step 2 case one may compute the energy E explicitly as a unique function of the boundary condition r = r( r). In the case k = 1 this is not possible because there are many energies E, which form an infinite increasing sequence. In the case when the complex action is not starting from the origin, there is the second parameter, the momentum n, see [4]. When the action starts from the origin, n vanishes. We shall deal more with this issue in section 5.10. In the step 2 case the energy E depends on r( r) = r as follows: r
r
io
du v'2E +
1
rJf
4u2 = 2" io
T
dv
1.
v'"f+V2 = 2" smh
-1 (
(2 )
VE r
.
5.
166
COMPLEX HAMILTONIAN MECHANICS
Solving for E yields (5.46) Using the above expression of the energy we can compute the complex action in the step 2 case without using the parameter E.
= 0, the complex action g = -it + r2 coth(2r).
COROLLARY 5.23. When k
PROOF. Making k
= 0 in Theorem g
=
is
5.22 yields
+ rJ ~ + r2.
-it
Using (5.46) we can work out the term under the square root.
E 2 - +r
r2
2
sinh2(2r)
r
+r
2
=
1 + sinh2(2r) 2 r sinh2(2r)
h2(2) 2 cosh2(2r) 2 2 = r cot r . sinh (2r)
Hence
g
-it + rJ
~ + r2 = -it + r2 coth(2r). o
5.7. Hamilton-Jacobi equation In this section we shall deal with the Hamilton-Jacobi equation for the complex action. We shall describe the complex geodesic flow as a horizontal gradient of the complex action. The following computations follow from classical ideas of Lagrangian and Hamiltonian formalism. The momentum () will be considered arbitrary, if otherwise stated. Recall the Hamiltonian
H(x, t,~, ()) =
~ (6 + 2()X2IxI2k) 2 + ~ (6 -
2()Xllxl 2k
f,
which is associated with the Lagrangian
(5.47)
L(x, t, X, i) =
~(xi + x~) + ()(i -
21x12k(XIX2 - XIX2))'
The variables Xl, X2, t are the coordinates and 6,6, () are the momenta. The following result provides a relation between the Lagrangian and the momenta. LEMMA 5.24. Along the solution (x, t,~, ()) of the Hamiltonian system, we have
() = aL ai'
. = aL
~.
aXi'
i=1,2.
PROOF. The first relation is obvious. Differentiating in the Lagrangian (5.47)
aL.
~ = Xl -
UXI
2()X21xl
2k
=
6,
where we used Hamiltonian equation
Xl = H6 = 6
+ 2()X2IxI 2k .
HAMILTON-JACOBI EQUATION
5.7.
A similar computation provides
6 =
167
BB!- • X2
o
The following lemma gives a relation between the momenta and the action. LEMMA 5.25. Along the solution (x, t,~, 0) of the Hamiltonian system we have og OXj og ot
(1)
(2)
~j,
j
= 1,2
0(= -i).
PROOF. (1) Using Euler-Lagrange equations and Lemma 5.24 og OXj
(2) Replace Xj by t in the calculation of (1) or just differentiate in the formula of 9 given by the Theorem 5.22.
o
THEOREM 5.26. The complex action given in Theorem 5.22 satisfies the HamiltonJacobi equation (5.48) PROOF. Consider the total derivative with respect to g=g(X,t,T)
T
of the complex action
dg dg. dg. og ---x·+-t+dT - dx j J dt OT'
(5.49) where x = dX(T)/dT and dg dT
(5.50)
i=
=
dt(T)/dT. On the other hand,
d dT
ior
L(x, t, (, 0) ds
= L(x, t,~, 0).
Substituting in (5.49) and using Lemma 5.25 yields og OT
dg _ ( og x dT OXj J
+ og i) ot
L(x, t, x, i) - (~jXj
+ Oi)
-((~jXj + Oi) - L(x, t,x, i)) = -H(x, t,~, 0) og Og) ( -Hx,t'ox'ot'
Hence (5.51 )
og OT
+H
(
og Og) x, t, ox' ot =
The Hamiltonian can be also written as
o.
5. COMPLEX HAMILTONIAN MECHANICS
168
og Og) H ( x,t, OX' ot
and the Hamilton-Jacobi equation (5.51) becomes (5.48).
o • Application to step 2 case We shall use the Hamilton-Jacobi equation to compute the energy in the step 2 case. Unfortunately, this is the only case when the computation can be performed explicitly. It is known that the action in this case is
g(7)
= -it+r 2 coth(27),
and it satisfies the Hamilton-Jacobi equation og
07 +H = O. As dH/d7 = 0, the Hamiltonian is constant along the solutions and its value will be the energy E = H. Then
E
- -og = - -0 07
o
07
(-.~t
_r2 07 coth(27) -2r
2
=
+ r 2 coth (27) ) -2r2(1- coth2(27))
sinh2 (27) - cosh 2(27) sinh2(27)
2
sinh (27)
•
• Application to complex geodesic flow By similarity with the definition of the gradient in the elliptic case, we have the following definition. DEFINITION
f
5.27. Let
f : [0,7]
-->
C be a function. The horizontal gradient of
is defined by
The following result provides the complex geodesic flow when the complex action is known. PROPOSITION
5.28. If c
=
(Xl, X2,
t) is a complex geodesic, then the velocity is
c='Vxg·
5.8. GEODESIC COMPLETENESS
169
PROOF. From the Hamiltonian equation i = He = 21x12k(XlX2 - XlX2). Following a computation made at the beginning of chapter 3
C
+ X20X2 + tat XlXl + X2X2 + (i - 21x12k(XlX2 XlXl + X2 X 2.
XlOXl
XlX2) )Ot
On the other hand, using Hamiltonian equations and Lemma 5.25 Xl
H6 =
6 + 2l1x21xl2k
og+ 2X2 IX12kOg OXI at Xlg· X2
Hf.2 =
6-
2l1XllxI2k
og _ 2XllxI2kOg OX2 at X 2 g.
Hence
o 5.8. Geodesic completeness DEFINITION 5.29. If for any point P in the x-plane, any complex geodesic c( s) = (x( s), t( s)) E ((:3 emanating from P is defined for all s E R, then we say the geometry is geodesic complete.
It suffice to show the geodesic completeness just for the x-component, because the t component will follow. In the following we shall study the geodesic completeness for complex geodesics. We shall investigate whether the solutions of system (5.6) are defined for all s E R We shall use the action 9 and the tools developed in the previous sections.
Considering the action 9 known, we shall investigate the system
In polar coordinates Xl
=
r cos , X2
or a",
=
r sin , we have
+ sin OX2 , -r sin OXl + r cos OX2'
cos OXl
Inverting, yields sin cos Or - - - a"" r . cos SIll Or + - - a",. r
170
5. COMPLEX HAMILTONIAN MECHANICS
In polar coordinates, the vector fields become
+ 2X21xl2kOt
Xl
OXl
X2
sin¢ cos ¢ a,. - - - 01> + 2r 2k +1 sin ¢ at. r OX2 - 2Xllxl2kOt . cos ¢ 2k I sm¢ a,. + - - 01> - 2r + cos¢ at. r
Relations 01>g = 0 and Otg = -i yield cos ¢ org - 2ir2k+1 sin ¢, sin¢ org + 2ir2k+l cos¢. Differentiating in the polar coordinates formula,
¢, rsin¢ + rcos¢ ¢.
Xl
r cos¢ - rsin¢
X2
Equating the last two sets of identities, we obtain cos ¢ org - 2ir2k+l sin ¢, sin ¢ org + 2ir 2k +1 cos ¢
r cos ¢ - r sin ¢ ¢ r sin ¢
+ r cos ¢ ¢
cos ¢( org - r)
(5.52) ,
+ sin ¢(r¢ -
2ir2k+1)
0,
Sin¢(org-r) -cos¢(r¢-2ir 2k +l )
0
( COS ¢ sin ¢
sin ¢ ) ( Or g - r - cos ¢ r¢ - 2ir2k+l '
v
det,tO
Hence
r ¢
org, 2ir2k.
Using the action g provided by Theorem 5.22,
_1_ [~J2E + 4r4k+2 + (4k + 2)r 4k +2 ] k+1 2 J2E + 4r 4k +2 1
1
1
- . -V2E + 4r 4k +2 > - - 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
<
r 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
5. COMPLEX HAMILTONIAN MECHANICS
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.
COMPLEX HAMILTONIAN MECHANICS
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
5. COMPLEX HAMILTONIAN MECHANICS
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<
1'0
V2E +
- 2kT - 2b(k + 1) < max A,
4rt k +2
which defines an interval for the parameter b by the double inequality (5.69)
A(ro) - max A - 2kT < 2b(k + 1) < A(ro) - 2kT.
5.11. THE LENGTHS OF THE REAL GEODESICS
177
PROPOSITION 5.37. Let k ~ 1. If ro =1= 0, the solution r = r(T, E, b) depends on two parameters E and b. Given E, the parameter b belongs to the interval defined by (5.69). Hence, for each given E, there are infinitely many solutions r(T, E, b) parametrized by b.
The parameter b is called momentum, see [4].
5.11. The lengths of the real geodesics The classical action from the origin is given by (5.70)
S is the integral of the Lagrangian along the bicharacteristic with boundary conditions:
(5.71)
x(O) = 0,
X(T) = x,
t(O)
=
0,
t(T) = t.
Here we fix to = t(O), so 0 cannot be chosen arbitrarily; it is a real constant along the path. In the case of the complex action, the boundary condition t(O) = 0 is replaced by 0 = -i. The complex action and the classical action are related by the formula 9 = S - it(O).
(5.72)
The Hamiltonian H is homogeneous of degree 2 with respect to (6,6,0), so
S(X, t, T) =
foT [(~, x) + oi -
H(s)] ds =
foT (2H -
H) ds = THo,
where Ho is the constant value of the Hamiltonian along the solution. The HamiltonJacobi equation may be rewritten as (5.73)
og +.:?:
OT
T
= 0,
so (5.74) DEFINITION
(5.75)
og OT
9
+ it(O) T
5.38. The modified complex action function is defined as
f(x, t, T) = Tg(X, t, T),
where g(x, t, T) is the complex action from the origin. We shall show that the critical points of f with respect to T play an important role in finding the lengths of the (real) subRiemannian geodesics between the origin and the point (x, t). This was showed by Beals, Gaveau and Greiner (see [3]) for the step 2 case, where the modified complex action function is
f(x, t, T) = -iTt + T coth(2T)(xi
+ x~).
The general results of this section were obtained by Calin, Chang and Greiner (see [17]). The geodesics starting from the origin are described by Theorem 3.17 of chapter 3:
178
5.
COMPLEX HAMILTONIAN MECHANICS
THEOREM 5.39. There are finitely many geodesics that join the origin to (x, t) if and only if x =1= O. These geodesics are parametrized by the solutions 'lj; of:
It I IxI 2(k+1) = J.L('lj;) ,
(5.76)
There is exactly one such geodesic if and only if:
It I < J.L( C1) IxI 2 (k+1), where C1 is the first critical point for J.L. The number of geodesics increases without bound as IXI2I(t~+1) -> 00. If 0 ::; 'lf1 < ... < 'lfN are the solutions of (5.76), then there are exactly N geodesics, and their lengths are given by
(5.77)
1'n
= 1, ... , N
where 2k ] 2(k+1) [ fa(2k+1).p sin- 2 k+l(v)dv
(5.78)
V('lj;)
=
(k+
1)2(k+1)
(1 +J.L('lj;)) sin
2(k+l) 2k+l
((2k+l)'lj;)
Differentiating (5.75), yields 8f = 8(7g) = 9
87
For t(O)
=1=
87
+ 7 8g
87
= 9 _ (g
+ it(O))
= -it(O).
0, relation (5.76) of Theorem 5.39 becomes
t - t(O)
(5.79)
= -J.L( ¢)
IxI 2 (k+1),
1> = 1> - 1>0,
and therefore (5.80)
Since
'*
=
-2Br 2k (see (3.18) of chapter 3), one has
(5.81)
¢=
-2B
1T r2k(s) ds.
Now (5.80) implies PROPOSITION 5.40. Let x action f(x, t, 7) if and only if
=1=
O.
7c
is a critical point for the modified complex
is a solution for the equation (5.82) Theorem 5.39 implies PROPOSITION 5.41. Let 71, ... ,7m be the critical points of the modified complex action f(x, t, 7). The numbers
5.11. THE LENGTHS OF THE REAL GEODESICS
179
satisfy the equation (5.83) and for each (j we have a geodesic connecting (x, t) to the origin. The length f!j of the geodesic parametrizied by (j is
f!~(k+l)
lI((j)(l t l + IlxI1 2(k+l») 1I(2i
foTj r2k(s)ds) (It I + IlxI12(k+l»).
Consequently, the knowledge of the critical points Tc of the modified complex action function gives the lengths of the geodesics. The rest of this section contains the calculation of Tj and the geodesic lengths for the step 2 and step 4 cases. Step 2 case: The Heisenberg group. Here k = O. In this case ¢j = 2iTj and Tj is purely imaginary. The lengths of the geodesics between the origin and (x, t), x -=I- 0, are given by (5.77): f!;
(5.84)
= 1I(2iTj) (It I + IlxI12).
Step 4 case. This is a lengthy calculation. We need to compute the integral
(5.85)
2i
foT r2(s) ds,
and we shall do that in terms of elliptic functions. Recall the Hamiltonian function H(~, (), x, t)
(5.86)
1 ( 6 + 2X21xl 2)2 =2 () + 21 ( 6 - 2Xllxl 2() )2 .
Using the Hamiltonian equations Xl = X2 =
6 + 2X2IxI2(), 6 - 2XllxI2(),
we have 1('2 .2) H = 2 Xl +X2
(5.87)
=
1 2'
where we have used the arc length parametrization. In polar coordinates Xl = rcos¢, x2 = rsin¢ we have = -2()r 2k and the conservation of energy (5.87) takes the form
#
(5.88) With ()
=
-i,
(5.89) and (5.90) Since r(O) = 0, then r( s) satisfies the integral equation (5.91)
r viI dx+
io
4x6
= ±s.
5. COMPLEX HAMILTONIAN MECHANICS
180
The substitutions x 2 = t, 4t 3 = u 3 and u = -v yield
liT
i v1 dx T
+ 4x 6
o
2
2
dt
Jt(l
0
+ 4t 3 ) du
JU(l
+ u3 ) dv
y'v(l - v3 ) Recall Lemma 2.16 of chapter 2: LEMMA
5.42.
i
a
dx
o Jx(1-x 3 )
.J3, k = Vb/2
where b = 2 -
__ 1 sn-1 (J1- h(a)2 k) -
b2
31 / 4
'
,
and h(a) =
.J3 .!.±..il -
a+
1
2
3
1.
Consequently,
i v1 dx T
+ 4x 6
o
and in view of (5.91),
V1 -
h(-4~~3r2)2'k)
(5.92)
sn- 1 (
From cn 2x - sn 2x
= 1 (see [42], p. 24 (2.1.4)),
==f24/331/4is.
(5.93) Again, by [42], p. 39 (2.6.12):
cn(iu, k) = nc(u, k') = where k'
(5.94) with
= V1 -
k 2 , so (5.93) becomes
bnc(24/33 1/ 4s,k')
k' = v'b/2 and b = 2 + .J3.
(5.95)
(1 k)' cn u, '
r 2 (S)_4- 1 / 3 -
Inverting
=
h(_4 1/ 3r2),
h,
we obtain
(1+.J3_ .J3 2 1 + bnc(24/3 3
1 / 4S,
k')
)
.
5.11. THE LENGTHS OF THE REAL GEODESICS
181
Thus (5.85) can be expressed as
r
2i Jo r2(s) ds
=
1+v'3.
~(2T)
_2 1/ 3
(5.96)
Va i
r
ds
Jo 1 + b nc(2 4 / 3 31/ 4 s, k') .
We are left with evaluating the second term on the right side of (5.96). With u one has
1
(T
i Jo
1
(5.97)
(iT
i Jo
= is,
du 1 + b nc( -i 24 / 3 3 1 / 4 S, k') du
1+bcn(2 4 / 3 31 / 4 s,k)'
since nc is an even function. Consequently, (5.98)
where
F(z)
1+ v'3
21/3
1 + v'3
31/ 4
~z -
Va
r
dv
Jo 1 + b cn(2 4 / 3 31 / 4 V, k) dv
(24/3 3 1/4 z
~z - -2- Jo
1 + b cn(v, k)'
Next we use [42], p. 93 (51):
with b2
< 1, where
Jo 1 + :~ sn2 v .
A(u, i(J, k) = {U
(5.99)
We recall that b= 2-
Va,
1 - b2
k
= Vb = J6 - V2 2
= 4Va - 6,
4'
k'=
vb 2
=
J6+V2
4'
b (J
= v'f=b2'
182
5. COMPLEX HAMILTONIAN MECHANICS
and b
(2 - J3)J2 (J3 -lhf2J3 - 3 \f2-J3.J2
(J3-1)·3 1/ 4
1 1 2 2J3 - 3
1 1-
-
b2
2 - J3
b
j2(2J3 - 3)
J2
2
(J3-1)3 1/ 4
2+J3
1 J3
1 2'
=--=-+-
1J2(J3-1) 4 \f2J3-3
J2(J3-1)
2J3
1 J2(J3-1) 4 31/4J2 _ J3
2 - J3 21/23 1/ 4 J2 - J3
1 2.3 1 / 4 '
1
J2( J3 - 1)
21/23 1/ 4
2
J3-1 2.3 1 / 4 ' Consequently,
[U dv _ (J3 io 1 + bcnv 3
~) A ( u, ~. J3 - 1 ) 2.31/4' k
+2
_
3
-1/4
tan
-1
(~) 2.31/4'
and
F(z) (5.100) To evaluate the second term more explicitly, we use the following elliptic functions:
p.50 (3.1.2) p.52 (3.2.2)
5.11. THE LENGTHS OF THE REAL GEODESICS
F(w, k') £(u, k) D(w, k')
sn-l(sinw, k')
1 1 J1U
=
1 w
(1 - k,2 sin 2 (})-1/2 d(),
p.51 (3.1.8) p.62 (3.4.25)
dn 2 (v, k) dv,
W
k,2 sin 2 (}d() =
£
£(k) = £(K,k) =
K
K(k) =
183
1 w
Jcos 2 ()
-
k 2 sin 2 (}d(),
p.63 (3.4.26)
111"1 2(1_ k 2 sin2 (})1/2d() = 1K dn 2vdv,
r l\l - k2
10
sin 2 (})-1/2 d() =
t
10
dt J(l - t 2)(1 - k 2t 2)'
see p.63 (3.5.4) and p.73 (3.8.1), where numbers refer to [42]. In particular,
A(u,i{3,k)
=
where 00
X
1 + 2 ~:)_1)nqn2 cos2nx cosh2ny, n=l 00
Y
= 22,)-1)nqn2 sin2nxsinh2ny, n=l
with q = exp( -7fK' / K), x = 7fu/2K and y = 7fF(w, k')/2K and sinw = {3/ see [42], p. 72 (3.7.47). (}4 stands for Jacobi's zeta function. In our case
J3-1 {3 = 2.3 1/ 4 '
k=V6-v'2 4
'
k'
= V6 + v'2 4
'
b
J {32 + k 2 ,
= 2 - /3.
An algebraic computation yields
We shall show the function A depends only on the Jacobi's epsilon function £(u, k). We have F(w, k') = sn- l (sinw, k') = sn- l (/3 - 1, k'). From [42], p. 41 (2.7.14) and p. 63 (3.4.27), amv = Jovdn-yd-y, and E(v,k') = D(am v, k'), with sn(v) = sin(am v). With w = am v, we have v = sn- l ( J3 -1, k'), and hence D(w, k') = £(sn-l(V3 - 1, k'), k').
5. COMPLEX HAMILTONIAN MECHANICS
184
If k
=
1{;1 = 0>-,1'12, then K' / K = v'3 see [42], p. q
86 (10). Hence
= e- v'3. 7r
Also 31 / 4 1 K = K(k) = 4y1r v'3 r(1/6)f(1/3), and K[2V3E -
Dividing by K2
= 16:v'3f2(1/6)f2(1/3), we get 7r
£,
K
(V3 + l)K] =~.
v'3 + 1
(
27r
)2
= 4v'3K2 + 2v'3 = r(1/6)f(1/3)
3 + v'3 + -6-'
Hence,
bV3u + 2.3 1 / 4 b[U{ E(am- 1 w, k') -
A(u, i{3, k)
+ with am- 1 w F(z)
=
[(f(1/6~;(1/3)
r
am- 1w
~ log ::~: ~ !~n,
+ 3 +6 v'3] am- 1w } +
= sn- 1( v'3 - 1, k'). Hence, the formula (5.100) becomes 1 ~/f\
-3
1/ 2
+21[U { E(am
T
-1' 27r )2 - 3-v'3] w, k) + [( f(1/6)f(1/3) 6 - am -I} w
+-i 1og 04(X - iY)] 2
2/ 3 z
04(X + iy)
43 14 - -1 tan -1 (Sd(2 / 3 / Z)) 2
2.3 1 / 4
'
with u = 24 / 3 31 / 4 z and am- 1w = sn- 1 (v'3-1,k'). Finally, we restate the last part of Proposition 5.41 for the step 4 case (k = 1). COROLLARY 5.43. Let Tj be the critical points of the modified complex action f(T) = Tg(T). Setting (j = F(iTj), the lengths of the geodesics between the origin and the point (x, t), Ixl f:. 0 are given by
cj = lI((j)(ltl + IlxI14). 5.12. Modified complex action function on Hn In this section we use the notions and notations introduced in chapter 4. Using the classical action for the vectors (4.5) given by the Theorem 4.5, n
S(x, t, T; 0) = [t - t(O)]O + L ajO cot(2ajTO)r;, j=l
where r; function
= X~j_1 + X~j' Let 9 = S + Ot(O) and define the modified complex action n
(5.101)
f(X,t,T) = -iTt + LajTcoth(2ajT)r;' j=l
5.12. MODIFIED COMPLEX ACTION FUNCTION ON Hn
185
Recall the notation x = (x ' ,x") ,
The following study of the modified complex action
f
can be found in [3].
LEMMA 5.44. For any (x, t) with x" # 0, and t 2: 0, the function f(T) f(x, t, T) has finitely many critical points on the imaginary axis. There is one critical point between the origin and the first pole of f on the positive imaginary axis, and it is a local minimum for f. There are either zero or two critical points (counting multiplicity) between each pair of poles on the positive imaginary axis; of such a pair of critical points the one nearer the origin is a local minimum and the other a local maximum for f.
PROOF.
Let n
(5.102)
F(B)
=
f(x, t, iB)
=
I>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: -.
5. COMPLEX HAMILTONIAN MECHANICS
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.
5. COMPLEX HAMILTONIAN MECHANICS
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
5. COMPLEX HAMILTONIAN MECHANICS
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.
5. COMPLEX HAMILTONIAN MECHANICS
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
5. COMPLEX HAMILTONIAN MECHANICS
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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.
QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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~,
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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 .
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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.
QUANTUM MECHANICS ON THE HEISENBERG GROUP
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
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
J,2 .
Therefore, we get the following bound for the modulus of
1
11,2 - l <
(m+~)1T
2"
0
Since have
U,
JR2+y2
_!JL
e
1sinh(R + iy)1
dy.
t, Y > 0, this implies that e- t;? ~ 1. Besides, 0 ~ y ~ (m + 1)7r, and we
Hence,
But
we have
1,2
z
.!JL
e'2u--dz ---., 0 sinh(z)
as R ---.,
+00.
as R ---.,
+00.
Similarly, one can show that
1,4
eit;? _Z_dz ---., 0
Now we need to evaluate
1,3
ill
Z
e 2u--dz sinh(z)
sinh(z)
1,3'
Here 1'3:
=
-
j-,3 e 2u--dz sinh(z)
-
j
[R, -R] ---.,
ill
R
C, 1'3(X) = X + i(m + 1)7r.
Z
e i--L. 2u [x+i(m+l)rr] 2
We shall split
X
+ i(m + 1)7r dx. + i(m + 1)7r]
sinh[x
-R
j R= j-1 + j1 + JR -R
-R
-1
1
When Ixl 2 1, we shall bound I sinh
[x + i(m + ~)7r] 121 sinh(x)I.
When Ixl ~ 1, we shall bound
Isinh [x + i(m + ~)7r] 12
1sin (m
+ ~)7r1
= 1cos(m7r)1 = 1.
Also, sinh(x) =
eX - e- x eX > 2 - 4
{o}
2e x - 2e- x > eX -
log 2
{o}
x> - 2- , -
6.10. THE HEAT KERNEL ON H,
227
e;. Therefore,
so, for x ::::: 1, we have sinh(x) :::::
So by Cauchy's theorem, we have
T dT+tj j ReituT sinh(T) -R
As R
---+
j=l
+00, we have
j +OO e
ii.. T
T
2".
-00
sinh(T)
dT
m
-27r 2
2) _1)kk(e-tu k=l
1f
)k
"fj
=27riI:Res(eituz-Zk=l
sinh(z)'
hi)
228
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
Now let m
-> 00
1 871' 2 U 2
j
and by Theorem 1.55, one has +OO ib:.
-00
e
2u
d
T
sinh(T)
~
271'
(ill;, sinh(z)' Z .) k71'~
871' 2 U 2 ~ Res e
=
T
2"
k=l
2 2 00 _71'_ " ( -l)k+lke871' 2 U 2 ~
k2,;,.'
k=l
-
1
00
2
" ( -l)k+lke-~.
4U2~ k=l
where Sk = £(0, t) is the length of the k-th geodesic connecting the origin and the point (0, t). In order to show that the series is convergent, we shall use the formula
-( with z = e-f,;1r, and 00
°< z <
L( _l)kk(e-f,;1r)k k=l
1 ~ k k )2=~(-1)kz l+z k=l
1. Then
1
e- f,;1r
(1 + e-f,;1r)2
-
ef,;1r(l
+ 2e-f,;1r + e- 2 f,;1r)
1
1
1
4cosh 2 (;~)' This completes the proof of the theorem.
D
6.11. Maxwell's equations
The first group of Maxwell's equations is laB
(6.48)
curl(E) = --;;7jt'
div(B) = 0,
where E : D x IR t -> 1R3 is the electric field, and B : D x IR t -> 1R3 is the magnetic field on a domain D C 1R3. If the electromagnetic field is static, then E and B are time independent. In this case (6.49)
curl (E) = 0,
div(B)
=
0.
The equations (6.48) and (6.49) are written in vectorial form. In the following we shall write them in differential form. For this we need to define the Roge operator. Let Ak (IR n) be the space of k- forms on IR n . DEFINITION
6.31. The operator
* : Ak(lRn) -> An- k (lR n ), defined by
*'f/ = L
sgn [>~n] 1JI dx J ,
I
where J = (jl < ... < jn-k) is the complementary multi-index to I ik), is called the Roge operator.
= (il < ... <
229
6.11. MAXWELL'S EQUATIONS
One may verify the following property: **r"/
Let dv
= dx 1 A ... A dxn
DEFINITION
=
(_I)k(n-k)1]k,
be the volume form on JRn.
6.32. If V is a vector field, define iv : Ak(JRn ) -; Ak-l(JRn) by (i V 1]k)(W1 , ... , Wk-d
= 1]k(V, WI"'"
W k-
1 ).
One may associate with each vector field V a I-form 1]~ defined by the equation
*1]~
= iv(dv).
A computation in local coordinates is provided below. If V = vector field, using the relation * * 1]1 = (-1 )n-l1]l, yields
1]~
Ei Via
(-It- 1 * (VI dx 2 A ... A dxn - V2 dx l A dx 3 A ... A dxn VI dx 1
DEFINITION
Xi
is the
± ... )
+ V2 dx 2 + ... + V n dxn.
6.33. The divergence of a vector field V is defined by the equation d(*1]~) = d(iv(dv)) := div(V) dv.
(6.50)
d( *1]~) and dv = dx l A dx 2 A dx 3 are 3-forms on JR3 and hence, they are
proportional. The proportionality function is div (V). divergence in local coordinates and get 3
divV
= I:
One may work out the
aV i
ax
i '
i=1
The following relations will be useful (*1]~ )(u, w)
(ivdv)(u, w) = dv(v, u, w) = det(v, u, w).
As *1]~ E A2(JR3), let (6.51) A computation yields
A2
= det(V, aX2 ' aX3 ) = VI, (*1]~) (aXIl aX3 ) = det(V, aXIl axJ = - v 2,
A3
(*1]~ ) (aXIl aX2 ) = det(V, aX! ,aX2 ) = V 3,
Al
*1]~ DEFINITION
(*1]~) (aX2l axJ
VI dx 2 A dx 3 + V2 dx 3 A dx l
+ V 3 dx l
A dx 2.
6.34. The 2-form associated with the magnetic vector field B
(BI, B2, B3) is defined by
Then
f1 = BI dx 2 A dx 3 Formula (6.50) becomes (6.52)
+ B2 dx 3 A dx l + B3 dx 1 A dx 2.
df1 = div(B) dv,
and hence, Maxwell's equation div(B)
= 0 can
df1=O.
be written in differential form as
=
230
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
The case of concern is when the magnetic field comes from a potential
n=dw, where W = 2:~=1 Wj dx j, and ker W = 'H is the horizontal distribution. The equation of motion for a particle with charge e, speed c and mass m is given by the Lorentz equation diJ dt
(6.53)
m-
=
e ---> -iJx B c '
where 13 = (n 1 , n 2 , fh) is the magnetic vector field. The equation (6.53) is the Euler-Lagrange equation for the following Lagrangian L(x, x) = ; xD: X{3b a {3
-
e:i;iWi(X) =
In the case of a plane curve with velocity 1/ B = (0,0, (x)), the right side of (6.53) becomes
--->
~mlxl2 =
ew(x).
(Xl, X2, 0) and magnetic field
(X)(X20Xl
-
(x).1'(XIOXl
X1 0xJ
+ X20X2)
(x).1'(1/),
while the left side of (6.53) is m(xI, X2, 0). For different values of we recover the models discussed in the previous sections . • Step 2 case: Heisenberg group Consider the vector fields
o
0
Xl = ~ +2X2!.'l' UXI ut which belong to kerw, where w = dt + 2(XI dX2 - X2 dxd. The magnetic field is given by the 2-form n = dw = 4dxI 1\ dX2' The corresponding Lagrangian is (6.54) The Euler-Lagrange equations for the Lagrangian (6.54) are (6.55) Let
d:~S)
(6.56)
= 40.1'¢(s),
with .1'2 = -I, which is the Lorentz equation (6.53) with mass m = 1, charge e = 0 and constant magnetic vector field 13 = (0,0,4). The projections on the (XI,X2)space are circles, called Larmour orbits (see [53]). These are the trajectories of charged particles in constant magnetic fields. The trajectory is a circle perpendicular on the magnetic field (see Figure 3.1). The velocity iJ = ¢( s), the Lorentz force ¢( s) and the magnetic field = (0,0,4) form a right oriented orthonormal frame. In the case of Heisenberg group (step 2) the magnetic field is constant and hence does not vanish. As a consequence, the trajectory of the particle will always be a
n
6.11. MAXWELL'S EQUATIONS
231
circle, even if it starts from outside of the origin (see Figure 6.3). This is true only for step 2 and it is not working for superior steps.
10
Figure 6.3: Larmour orbits outside the origin .
• Step 4 case Consider the vector fields Xl
8 +4X2 1x 128 = VX1 ~ l'l' vt
We have span{Xl' X 2} = kerw, where w = dt + 41x12(Xl dX2 - X2 dxI). The magnetic field is given by the 2-form n = dw = 16lxl2dxIAdx2' The corresponding Lagrangian is
(6.57) The Euler-Lagrange equations for the Lagrangian (6.57) are (6.58) Let ¢(s) = (Xl(S),X2(S)). Then (6.58) becomes
d¢(s) = ()¢(s) x ds
n,
which is the Lorentz equation (6.53) with mass m = 1, charge e = () and magnetic field oriented along the t-axis = (0,0, 16IxI2). This magnetic vector field vanishes at the origin and it has spherical symmetry. The trajectories for the charged particles depend whether the particle passes through the origin. Figure 3.2 represents the trajectory of a charged particle which passes through the origin. The trajectory is a union of loops of angle 1r /3 at the origin.
n
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
232
Figure 6.4: Trajectories outside of the origin.
If the trajectory is not passing through the origin, it looks like in Figure 6.4. The particle remains all the time between two circles. Using the results of section §3.1O, the particle will never leave the circular crown . • Step 2k case Consider the vector fields
xI
= ~ + 2kx IxI 2 (k-l) ~ aXI
2
at'
We have span{Xb X 2 } = kerw, where w = dt + 2klxI 2 (k-1)(XI dX2 - X2 dxd. The magnetic field is given by the 2-form n = dw = 4k2IxI2(k-l)dxI 1\ dX2. The corresponding Lagrangian is
(6.59) The Euler-Lagrange equations for the Lagrangian (6.59) are 2k X2, .. -- 4k2 ()I X1 . Xl
(6.60) Let ¢(s)
=
.. X2
. = - 4k2 ()I X12k Xl,
() = constant.
(XI(S),X2(S)). Then (6.60) becomes
d!~S)
= ()¢(s)
x
0,
which is the Lorentz equation (6.53) with mass m = 1, charge e = () and magnetic field oriented along the t-axis 0 = (0,0,4k2IxI2(k-I)). The magnetic field vanishes at the origin. Figures 3.2 and 6.3 show the trajectories through the origin in the cases k = 2 and k = 3. The trajectory is periodic and consists of loops with an angle of 2k~1 at the origin. The number of loops is N = 2k - 1. For k = 1 (step 2 case) there is only one loop, which is a circle (Larmour orbit), see Figure 6.2. For k = 2 (step 4 case) there are 3 loops, see Figure 3.2. In the cases k = 3 and k = 4 above, there are 5 and 7 loops, respectively (see Figure 3.3 and Figure 3.4). If the particle is not passing through the origin, the trajectory belongs to a
6.12. QUANTIZATION OF ENERGY
233
circular crown, like in Figure 6.4 or Figure 6.5.
Figure 6.5: Trajectories outside of the origin.
6.12. Quantization of energy The lengths are the same for all parametrizations. In this section we shall consider geodesics 'Y : [0,1] ---> ]R3 joining the origin and the point (0,0, t). As the Hamiltonian is preserved along the solutions, the velocity has constant length in the subRiemannian metric (in which Xl and X 2 are orthonormal), and hence we have identity in Cauchy's inequality
£("() = fo\Y(s) [ ds =
(fol
dS) 1/2 (fo1
h(S)[2
ds
r/
2
= V2E.
Hence the energies of a unit mass particle moving along the geodesic are given by Em
=
1 2 2£m,
m
=
1,2,3, ...
Theorem 3.27 yields (with k + 1 changed into k): THEOREM
6.35. The particle has discrete energies 0
<
E1
<
E2
< ... ,
given
by
For k
= 1,
and for k
= 2,
El is the ground-state energy, i. e., the lowest energy of the particle. All the other energy levels correspond to excited states. If the electron is in ground-state, then it can be excited up to the level of energy Em if it receives energy from a photon with energy hVm = Em - E l , where h is Plank's constant and Vm is the frequency of the photon. The difference Em+! - Em gives the distance between the energy levels. In the step 2 case, the energy levels are equidistant (see Figure 6.6) E2 - E1
=
E3 - E2
= ... =
E m+ l
-
Em
=
7r
2'[t[.
234
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
In the step 2 case (k = 1) the smallest energy is 7rltl/2. This corresponds to the ground-state energy hw /2 for a linear harmonic oscillator with pulsation w = 7rltl/ h, studied in this chapter, section §6.1 For step k > 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 <
(2 - ~)Ck(m + l)l-t .
• Case k = 2: We have 3
3
2C2'v/m < Em+l - Em < 2C2vm + 1,
with
!Ttl (y'7r r(
C2
=
V2
f;))
3r( ~)
3/2
~ 2.676vTt!.
Using 3C2 /2 ~ 4.014JitT, yields 4Vm1t1 < Em+l - Em,
m = 1,2,3, ...
235
6.13. KEPLER-TYPE LAWS
6.13. Kepler-type laws From Kepler's first law, all the planets have plane trajectory, which are ellipses with the sun in one of the focuses. The second Kepler's law says that the areal velocity 0: is constant along the motion. If T is the orbital period and r is the semi-major axis of the orbit, then the third Kepler's law says that T 2 / 3 and rare proportional. We shall work out an analogy between Kepler's laws and our models . • The third law
Integrating in
i = 8H/8e,
yields
2k 2k t(cp) - t(cpo) = - 2k _ 1r max w(cp - CPo), where r(2k-1)U
w(u)
=
Jo
2k
sin 2k -
1
vdv.
Let
(6.61) and
Qk If
2k
= 2k _
(7r) > 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
= 1<1>1 <1>0
~r2 dcp. 2
The above formula can be also written in differential form dA =
See Figure 1.1.
~r2dCP.
6. QUANTUM MECHANICS ON THE HEISENBERG GROUP
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 <
7r
/2a,
= a{.l(2aO)llxI12
and {.l(¢) =
SiS
EXERCISE 6.5. Prove that at points
P(x, t, u) = 4a1u 2
e-
¢ -
cot ¢.
(0, t), t
d (0.t)2 /2" { 1
i= 0, the heat kernel
+ 0 (e- d2 (0,t)/2U) },
as u -; 0+. EXERCISE 6.6. Find the small time behavior for the kernel of the Schrodinger
operator
= 0; x i= o.
a) in the case x b) in the case
6.14.
EXERCISES
237
EXERCISE 6.7. Consider Hankel's function of first type H (1)(>,)
1 = -:~7T
n
1
00
+i 11" eAsinhz-nz dz.
-00
Using Theorem 6.28 show that
HA1)(>\) 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.
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240
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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
