Differential Manifolds and Theoretical Physics (Pure and Applied Mathematics, Vol 116)

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Author: W. D. Curtis

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Differential Manifolds and Theoretical Physics

This is a volume in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks

Editors: SAMUEL EILENBERG A N D HYMAN BASS A list of recent titles in this series appears at the end of this volume.

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Differential Manifolds and Theoretical Physics W. D. CURTIS F. R. MILLER Department of Mathematics Kansas State University Manhattan, Kansas

1985

ACADEMIC PRESS, INC. (Harcourt Brace Jovanovich. Publishers)

Orlando San Diego N e w York London Toronto Montreal Sydney Tokyo

COPYRIGHTO 1985, BY A C A D E M I C P R E S S , INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE A N D RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHER.

ACADEMIC PRESS, INC. Orlando,Florida 32887

United Kingdom Edition published by ACADEMIC PRESS I N C . (LONDON) LTD 24/28 Oval Road, London NWI 7DX

Library of Congress Cataloging in Publication Data Curtis W. D. Differential manifolds and theoretical physics. Bibliography: p. Includes index. 1. Geometry, Differential. 2. Mechanics. 3. Field theor (Physics) 4. Differentiable man if ol ds. I. Mirler, F. R. 11. Title. QC20.7.052C87 1985 530.1'5636 84-16861 ISBN 0-12-200230-X (alk. paper)

PRINTED IN THE UNITED S T A T E S OF A M t R l C A

85868788

9 8 7 6 5 4 3 2 I

To our wives Beverly and Helen

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Contents

xv

Preface

Chapter 1. Introduction Mathematical Models for Physical Systems

1

Chapter 2. Classical Mechanics Mechanics of Many-Particle Systems Lagrangian and Hamiltonian Formulation Mechanical Systems with Constraints Exercises

5 7 11 14

Chapter 3. Introduction to Differential Manifolds Differential Calculus in Several Variables The Concept of a Differential Manifold Submanifolds Tangent Vectors Smooth Maps of Manifolds Differentials of Functions Exercises

Chapter 4.

16 23 26 28 33 35 36

Differential Equations on Manifolds Vector Fields and Integral Curves Local Existence and Uniqueness Theory The Global Flow of a Vector Field Complete Vector Fields Exercises

vi i

39 40 53 55 57

...

CONTENTS

Vlll

Chapter 5. The Tangent and Cotangent Bundles The Topology and Manifold Structure of the Tangent Bundle The Cotangent Space and the Cotangent Bundle The Canonical I-Form on T*X Exercises

61 66 68 70

Chapter 6. Covariant 2-Tensors and Metric Structures Covariant Tensors of Degree 2 The Index of a Metric Riemannian and Lorentzian Metrics Behavior Under Mappings Induced Metrics on Submanifolds Raising and Lowering Indices The Gradient of a Function Partitions of Unity Existence of Metrics on a Differential Manifold Topology and Critical Points of a Function Exercises

Chapter 7.

72 74 74 77 79 83 84 84 87

90 92

Lagrangian and Hamiltonian Mechanics for Holonomic Systems Introduction The Total Force Mapping Forces of Constraint Lagrange’s Equations Conservative Forces The Legendre Transformation Conservation of Energy Hamilton’s Equations ~-FOIITIS Exterior Derivative Canonical 2-Form on T*X The Mappings # and b Hamiltonian and Lagrangian Vector Fields Time-Dependent Systems Exercises

94 95 96 99 99 103 105 106

110 112 114

114 115 121 124

Chapter 8. Tensors Tensors on a Vector Space Tensor Fields on Manifolds

127 129

CONTENTS

IX

The Lie Derivative The Bracket of Vector Fields Vector Fields as Differential Operators Exercises

Chapter 9.

Differential Forms Exterior Forms on a Vector Space Orientation of Vector Spaces Volume Element of a Metric Differential Forms on a Manifold Orientation of Manifolds Orientation of Hypersurfaces Interior Product Exterior Derivative Poincare Lemma De Rham Cohomology Groups Manifolds with Boundary Induced Orientation Hodge *-Duality Divergence and Laplacian Operators Calculations in Three-Dimensional Euclidean Space Calculations in Minkowski Spacetime Geometrical Aspects of Differential Forms Smooth Vector Bundles Vector Subbundles Kernel of a Differential Form Integrable Subbundles and the Frobenius Theorem Integral Manifolds Maximal Integral Manifolds Inaccessibility Theorem Nonintegrable Subbundles Vector-Valued Differential Forms Exercises

Chapter 10.

132 135 137 138

141 146 149 150 151 154 156 156 161 161 162 163 165 168 168 170 171 172 172 173 176 184 185 187 188 189 191

Integration of Differential Forms The Integral of a Differential Form Stokes’s Theorem Transformation Properties of Integrals w-Divergence of a Vector Field Other Versions of Stokes’s Theorem Integration of Functions The Classical Integral Theorems Exercises

196 199 20 1 203 204 207 208 210

CONTENTS

X

Chapter 11. The Special Theory of Relativity Basic Concepts and Relativity Groups Relativistic Law of Velocity Addition Relativity of Simultaneity Relativistic Length Contraction Relativistic Time Dilation The Invariant Spacetime Interval The Proper Lorentz Group and the Poincare Group The Spacetime Manifold of Special Relativity Relativistic Time Units Accelerated Motion-A Space Odyssey Energy and Momentum Relativistic Correction to Newtonian Mechanics Conservation of Energy and Momentum Mass and Energy Changes in Rest Mass Summary Exercises

213 220 222 222 223 223 224 225 227 229 233 234 235 236 236 237 237

Chapter 12. Electromagnetic Theory The Lorentz Force Law and the Faraday Tensor The 4-Current Doppler Effect Maxwell’s Equations The Electromagnetic Plane Wave The 4-Potential Existence of Scalar and Vector Potentials in R3 Exercises

239 243 245 246 248 250 25 1 253

Chapter 13. The Mechanics of Rigid Body Motion Hamiltonian Systems and Equivalent Models The Rigid Body O(3) and SO(3) Space and Body Representations The Geometry of Rigid Body Motion Left-Invariant 1-Form Symmetry Group Adjoint Representation Momentum Mapping Coadjoint Representation Space Motions with Specified Momentum Coadjoint Orbits and Body Motions

255 256 256 259 26 I 263 264 264 265 266 266 267

xi

CONTENTS

Special Properties of SO(3) Stationary Rotations Classical Interpretation-Inertial Tensor, Principal Axes Stability of Stationary Rotations Poinsot Construction Euler Equations Phase Plane Analysis of Stability Exercises

Chapter 14.

27 1 274 274 277 280 282 283 284

Lie Groups Lie Groups and Their Lie Algebras Exponential Mapping Canonical Coordinates Subgroups and Homomorphisms Adjoint Representation Invariant Forms Coset Spaces and Actions Exercises

286 289 289 290 29 I 292 293 296

Chapter 15. Geometrical Models Geometrical Mechanical Systems Liouville’s Theorem Variational Principles Forces Fixed Energy Systems Configuration Projections Lorentz Force Law Pseudomechanical Systems Restriction Mappings Rigid Body and Torque Gauge Group Actions Moving Frames and Goedesic Motion Basic Theorem Local (Lemma 15.36) Basic Theorem Global (Theorem 15.39) Principal Bundle Model Using a Special Frame The Souriau Equations Structure of the Lie Algebra of the Lorentz Group Construction of a Gauge Invariant 2-Form Curvature Form The Souriau GMS Appendix: Conservation Laws Exercises

297 298 300 30 1 304 305 306 306 307 308 310 31 1 314 316 319 32 1 322 322 327 328 329 332

xii

CONTENTS

Chapter 16. Principal Bundles and Connections; Gauge Fields and Classical Particles Principal Bundles Reduction of the Structural Group Connections on Principal Bundles Horizontal Lifts of Vectors Curvature Form and Integrability Theorem Horizontal Lifts of Curves Associated Bundles Parallel Transport Gauge Fields and Classical Particles Natural 2-Form on Coadjoint Orbits Pseudomechanical System for Particles in a Gauge Field Sternberg’s Theorem Geometrical-Mechanical System for Particles in a Gauge Field Affine Group Model Exercises

335 337 337 338 339 34 I 34 1 342 342 343 345 346 347 349 35 1

Chapter 17. Quantum Effects, Line Bundles, and Holonomy Groups Quantum Effects Probability Amplitudes Probability Amplitude Phase Factors DeBroglie and Feynman Phase Factors and 1-Forms COW and Bohm-Aharanov Experiments Complex Line Bundles and Holonomy Groups Integral Condition for Curvature Form Bundle Description of Phase Factor Calculation Remarks-Geometric Quantization Holonomy and Curvature for General Lie Groups Exercises

354 355 355 356 356 358 360 362 365 366 367 368

Chapter 18. Physical Laws for the Gauge Fields Gauss’s Law in Electromagnetic Theory Charge Conservation Curvature and Bundle-Valued Differential Forms Covariant Exterior Derivative Covariant Derivative of Sections and Parallel Transport The Group of Gauge Transformations

37 1 372 373 375 376 377

...

CONTENTS

Xlll

The Killing Form The Source Equation and Currents for Gauge Fields Exercises

379 379 382

Bibliography

387

Index

389

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__

Preface

There has always been a fundamental relationship between the concepts of geometry and the study of the physical world. The tremendous achievements of this century in both mathematics and physics, together with active current research efforts, indicate that this affair will continue. We will not attempt a comprehensive description of all of the recent developments-this would require many volumes-but will present the concepts of modern differential geometry in the study of classical mechanics, field theory, and simple quantum effects. In doing so, we start with Newton, Lagrange, and Hamilton and continue through Riemann, Maxwell, Einstein, Cartan, Chern, Yang, Souriau, and Sternberg. The idea of invariance is an essential ingredient and we introduce gauge invariance, bundles, and connections. This book has been successfully used for introductory one-year courses in the theory of differential manifolds and their relationship to various areas of physics. We begin by assuming modest mathematical preparation (advanced calculus, linear algebra, and some point-set topology) and maintain a high level of rigor throughout. The text carefully relates basic mathematical and physical concepts in much detail, but also includes material of a more advanced nature, bringing the reader close to the frontiers of current research. Some of the exercises present important enhancements and extensions of the theory. The following paragraphs give a detailed outline of the book. In the first two chapters, we introduce some basic ideas of classical mechanics of systems of particles and some ideas about the description of physical systems. We discuss the evolution of a physical system as a curve in a space ofstates. The evolution of the system is governed by a vector field on state space which specifies how the state changes in time. In the very simple cases, the state space can often be identified as an open set in some Euclidean space. When more complex cases are considered, we find this is no longer true. The presence of constraints, for example, xv

xvi

PREFACE

results in state spaces that are not open sets. In all cases, however, we find that the spaces which arise look locally like open sets in some Euclidean space, that is, they admit local coordinate systems. These examples motivate the introduction of differential manifolds as spaces on which the concepts of tangent vector to a curve and of vector field are meaningful. In particle mechanics the state space is naturally seen to correspond to the space of tangent vectors on conjiguration space, thus motivating introduction of the tangent bundle of a differential manifold. In Chapter 3, after some preliminaries on multivariable calculus, we introduce the basic ideas about differential manifolds, charts, atlases, submanifolds, tangent vectors, and smooth maps of manifolds. Then in Chapter 4 we introduce the notion of vector field and prove the fundamental result that a smooth vector field has a smooth flow. This result is one of the cornerstones on which the theory of manifolds rests. In Chapter 5 we describe the construction of the tangent and cotangent bundles of a differential manifold. These will serve as the state space and phase space for certain mechanical systems to be considered later. In Chapter 6 we describe general metric structures on manifolds. We discuss questions of existence for Riemannian and Lorentzian metrics. Raising and lowering indices is discussed here. Chapter 7 is a critical chapter. We use the mathematical formalism introduced in Chapters 3-6 to give an extremely elegant description of the mechanics of particle systems (begun in Chapter 2). We obtain the beautiful geometrical result that for a holonomic system subject only to forces of constraint the trajectory of the system is always a geodesic of configuration space for the kinetic energy metric. We introduce 2-forms, in particular the fundamental canonical 2-form on the cotangent bundle of a differential manifold. We establish the basic connection between the Lagrangian and Hamiltonian formulations of mechanics by means of the Legendre transformation. On state space and phase space we construct vector fields whose integral curves satisfy Lagrange’s and Hamilton’s equations, respectively. This relates back to the ideas of Chapter 1. At this point we have developed numerous important ideas about differential manifolds and have seen how naturally they relate to basic concepts in classical mechanics. Our purpose has not been to develop techniques for actually solving particular problems in mechanics. Rather we hope to have shown how the geometrical ideas in manifold theory can elucidate the structure of classical mechanics. Also, we believe that the physcial ideas have served as concrete motivation for introduction of much of the manifold theory. In subsequent chapters we shall try to maintain this interplay between the geometrical ideas of manifold theory and the physical ideas which arise in various branches of theoretical physics. The

PREFACE

xvii

geometrical notions play a basic role in conceptual understanding of physical ideas. And further, the geometry can play an important role from the computational point of view as well, by focusing attention on those quantities that really are basic. That is, the geometry points the way to the correct formulation of a physical problem and suggests which computations should be made. Conversely, the study of various physical notions continually motivates the study of particular mathematical ideas. In Chapters 8 and 9 we introduce further aspects of manifold theory. In Chapter 8 we introduce tensors of arbitrary degree and explore the fundamental Lie derivative in considerable detail. Chapter 9 is one of the most important chapters in the book. Most of the later work depends on the ideas contained here. We give a detailed development of Cartan’s calculus of differential forms. Orientation of manifolds is discussed in detail. We discuss the Hodge *-operator for a metric of arbitrary index and use it to define the classical differential operators, e.g., divergence, curl, Laplacian. We then discuss the fundamental Frobenius integrability theorem. This is the second major theorem of the book and is basic for subsequent work. We prove the inaccessibility theorem for integrable subbundles. In Chapter 10 we investigate integration on manifolds. We define integrals of n-forms over oriented n-manifolds. We relate the behavior of the integral of a form under motions generated by a vector field with the Lie derivative of the form with respect to the vector field. We prove the general Stokes formula relating the integral of an ( n - 1)-form over the boundary of an n-manifold to the integral of the exterior derivative of the form over the manifold, and we also give a proof for cylinders. Then we show how the classical integral theorems of Green, Gauss, and Stokes follow from the general theorem. In Chapters 11 and 12 we study the basic features of Einstein’s special theory of relativity and Maxwell’s equations in relativistic form. The structure of the spacetime of special relativity provides a beautiful example of the manifold theory we have developed. Spacetime is the set of all “events.” On spacetime there is a privileged class of charts which are the global coordinate systems known as Lorentz frames. These correspond to the physical notion of a Cartesian coordinate system rigidly attached to an inertial reference body. We show that the coordinate transformations relating these charts are the usual Lorentz transformations. These charts then form an atlas and hence determine a differential structure on spacetime. We then see that the 4-vectors of special relativity are just the tangent vectors for the differential structure determined by the Lorentz frames. We proceed to develop many of the standard topics of special relativity.

xviii

PREFACE

In Chapter 12 we develop Maxwell’s equations in terms of differential forms. We establish existence of the electromagnetic field tensor using the physical properties of the Lorentz force law. We then give some properties of this tensor and some examples. In Chapter 13 we study the mechanics of a rigid body moving in 3dimensional space with one point fixed and no external forces. We show that the state space for this system (it is a system with holonomic constraints) can be identified with T S 0 ( 3 ) , the tangent bundle of the group S 0 ( 3 ) , and that the kinetic energy metric is invariant under left translation. We use this chapter as a first example of concepts that will be developed later such as Lie groups and Lie algebras, adjoint and coadjoint representations, and symmetry groups and conservation laws. We relate these constructions to the usual angular momentum, inertial tensor, and principal axes, and to the Euler equations. We give a proof of the stability theorem (in both space and body coordinates) for rotations around the major and minor principal axes. In Chapter 14 we develop the properties of Lie groups, some of which were encountered in Chapter 13 for SO(3). These include the associated Lie algebra, subgroups, coset spaces, invariant differential forms, and invariant measures. We also study orbit spaces and vector fields obtained when Lie groups act as transformation groups. In Chapter 15 we further geometrize the study of mechanical systems by considering the trajectories (in phase or state space) as families of curves tangent to the kernel of a 2-form. We first show how this is related to Hamiltonian mechanics and obtain a proof of Liouville’s theorem on the invariance of phase space measure. We show the relationship to variational principles-locally if the 2-form is closed, globally if the 2-form is exact. Second, we see that the 2-form contains the classical forces and can describe nonconservative forces (for example, the rigid body of Chapter 13 subject to an external torque). Finally, we see that this geometrical approach in which the explicit time parameter is lost is well suited to the description of relativistic mechanics and interactions involving gauge fields. We define pseudo-mechanical systems which can incorporate gauge invariance with phase space, and we construct an invariant 2-form. We then obtain the particle phase (or state)-motions from equivariant restriction mappings. We use moving frame methods in these constructions leading to the concept of principal bundle. We show how the 1-form defining the metric connection is a factor of the 2-form defining the pseudo-mechanical system which gives geodesic motion. We also see that the restricted systems can be described by specifying a 2-form on a principal bundle. We give a development of the equations of Souriau which govern the evolution of the states of a

PREFACE

xix

charged particle with an internal spin in the presence of an electromagnetic and gravitational field. We end the chapter with an appendix on conservation laws. In Chapter 16 we study the structures of principal bundles, associated bundles, connections, and parallel transport. We consider homogeneous spaces and general affine connections in detail to present the work of Sternberg and co-workers in describing the interaction of classical particles with gauge fields. We describe how this relates to Chapter 15. In Chapter 17 we describe the concept of holonomy groups and show how complex line bundles are determined by closed 2-forms subject to a quantization condition. We explain the use of these concepts in describing quantum interference effects. The Bohm-Aharanov effect and the Colella, Overhauser, and Werner (COW) experiment are examples. In Chapter 18 the physical laws for the gauge fields are developed. Using the principle that a gauge transformation generates a divergence free current and a generalization of Gauss’s law, the Yang-Mills equations for the gauge fields are explained. The frame dependent nature of the charge densities is discussed. We use 0 to denote the empty set and Ito indicate the end of a proof.

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I Introduction

MATHEMATICAL MODELS FOR PHYSICAL SYSTEMS When we wish to describe a physical system in a “mathematical” way we try to construct some sort of mathematical structure which, in some sense, “represents” those aspects of the system which are of interest to us. This structure is then a “mathematical model” of the physical system. In this chapter we discuss some basic ideas concerning models for particle motion in classical mechanics. We will see that we are led to consider mathematical “spaces” on which the notion of “tangent vector” is meaningful. We will see how “physical laws” can be incorporated into the model as additional structures on these spaces. EXAMPLE 1.I Consider a single particle, of mass m, free to move along a straight line. We are not interested, at present, in the particular nature of the particle but, rather, we want to follow its motion. Thus, at each instant of time, we are interested only in the position and velocity of the particle. If x is the instantaneous position of the particle and u is its instantaneous velocity, we say the ordered pair (x,u ) is the instantaneous state of the system. The set of all possible states, called the state space, can thus be represented mathematically as the xu plane RZ. Suppose the particle is acted upon by a force field

F(x) =

~

k > 0.

kx,

(1.1)

Let the state of the particle at time t = to be (x,,, u,,). What will be the state at other times? To answer this we need a physical law, that is, a law of motion. That law is, of course, Newton’s second law, which says mx”

=

F

(1.2)

or, in the present case, x“ = -(k/m)x. 1

(1.3)

2

1 . INTRODUCTION

Note that the law (1.3)does not tell us directly what the state will be at other times. Rather, it tells how the state changes at each point. Thus (1.3) is equivalent to XI

( 1.4)

u‘ = - k x / m .

= u,

If we define a v e c t o r j e l d on R2 by (a, b) E R2

@(u, b) = (b, - ka/m),

(1.5)

then we see that, if the state of the system at time t is c(t) = (x(t),

VW),

then c‘(t) = @(c(t)).

A curve c satisfying (1.6) is said to be an integral curve of the vector field @. Thus, the physical law (1.2)determines a structure on the state space, namely, the vector field @. Given the state ( x , , v,) at t = t o , the vector field CD determines uniquely the state at all other times. That is, there is a unique curve, called the trajectory of the system with the given initial state, c: R

--f

R2,

such that c(to) = (x,, u,)

and

c’(t) = @(c(t))

for all t E R .

The curve c is to be found by “integrating the vector field,” which means by solving the system of differential equations (1.4) subject to the condition x(to) = x,, u(t,) = u,. This can be explicitly done for the simple example at hand. In fact the reader can easily check that we have c(t) = ( A cos(wt

+ 6),

-

A o sin(ot + 6)),

where w = (k/m)”2

A

= (xi

+(u~/o~))”~,

and

6={

-arctan(u,/ox,) -arctan(vo/wxo)

ot, - ot, -

+n

if x, > 0, if x, < 0.

The system considered in Example 1.1 is a time-independent mechanical system. If the force field (1.1) were replaced by a force depending on position, velocity, and time, F

= F(t, X , V )

(1.7)

MATHEMATICAL MODELS FOR PHYSICAL SYSTEMS

3

then @ would be replaced by @(t,X, U) = (u, 112- F ( f , X, u)).

(1.8)

Because of the explicit time dependence, we say this system is a timedependent mechanical system. Here, @ is not a vector field on state space R 2 but rather, a time-dependent vector field. The mathematical theory of vector fields and time-dependent vector fields is developed in Chapter 4. EXAMPLE 1.2 A mass m is fixed on the end of a rigid rod of negligible mass having length 1. One end of the rod is fixed at a point P in space so that the mass can move about P subject to the condition that it always be a distance 1 from P . Figure 1.1 depicts the situation. Figure l . l a shows the physical system, while Fig. l . l b shows the sphere M , of radius 1, centered at P , which is the space of all possible positions for m. This sphere is called the configuration space of the system. Suppose we are only interested in the motion of the particle. Then we take, as the state of the particle, the pair of x2, x3), u = (u', u2, u3), where x is three-dimensional vectors (x,u), x = (XI, the position vector of m and v is the velocity vector of m (with respect to some Cartesian coordinate system as shown). Since the mass must stay on the sphere M , we see u must be tangent to M . Thus our state space S does not consist of all pairs of 3-vectors but, instead, we have

S

=

{(x, u)lx E M and u is tangent to M at x).

configuration space M

mass m FIGURE 1.1

4

1 . INTRODUCTION

Although S is not a Euclidean space, nor an open set in one, we shall see that S is a space on which notions such as tangent vector, vector field, and time-dependent vector field have meaning. If we have a force field then the force field will determine a vector field on the state space S, as it did in Example 1.1. However, the construction of this vector field is not so simple as in Example 1.1. First, because S is not just an open set in Euclidean space; secondly, because, if we specify an external force field (e.g., a gravitational field or an electromagnetic field, if rn has a charge), we also must consider the force of constraint exerted by the rod to keep the mass on the sphere. This latter force is not known in advance. These difficulties will be overcome when we develop Lagrangian and Hamiltonian mechanics on manifolds in Chapter 7. The preceding examples exhibit the following general features: There is a state space S that represents those aspects of the system which are to be studied. Given a state so at time t o , the state of the system at other times t is determined by the physical laws which apply to the system. More precisely, the physical laws determine, for each state so and time t o , containing to, and (a) an open interval I(so.ro), s such that 4 ( s o , r o ) ( t O ) = so. (b) a map 4 S 0 . t " i I ( S " . t O ) -+

We interpret &,to,(t) as being the state of the system at time t if the state at time to is so. Frequently, we have = (- 00, co) but this is not always so. The curve in (b) is called the trajectory of the system determined by the condition that, at t = t o , the state is so. When we say that the physical laws determine the trajectories of the system, we do not mean that the trajectories are necessarily explicitly calculable. Rather, we mean we can prove that, given so and t o , the laws determine an and a mapping satisfying (a) and (b). As already disinterval I(so,ro) cussed, the physical law makes its appearance as a mathematical structure, e.g., a vector field or time-dependent vector field, which says what the tangent vector to the trajectory must be at each (s, t). Thus we need, as a state space, a space on which the notion of tangent vector to a curve makes sense. Such a space is called a differential manifold and the study of differential manifolds is a central theme of this book. We shall see that the manifolds which arise in the construction of models for physical systems have additional structures on them. These structures, which include various types of metrics, tensor fields, and differential forms, will be discussed in detail in the coming chapters. The significance of the dialogue between C. N. Yang and S. S. Chern will become clear to the reader.

2 C I assica I Mec ha n ics

As discussed in Chapter 1, we want to develop Lagrangian mechanics and generalized coordinates (i.e., manifold theory). The purpose of this chapter is to introduce the reader to some of the basic ingredients of Lagrangian and Hamiltonian mechanics for the special case when configuration space is an open set M in Euclidean space R" and state space is M x R". We will generalize these constructions to the manifold setting later. For now, we also restrict attention to time-independent systems. We begin with Newton's law of motion and show how to rewrite it so as to get Lagrange's equations. We then introduce phase space, defined to be M x (R")*,and obtain Hamilton's equations. We obtain vector fields on state and phase space whose integral curves are the trajectories of the system. We see that, in the Hamiltonian form, the vector field is obtained in a very simple way from the energy function. The trajectories of the Lagrangian and Hamiltonian systems are related by the Legendre transformation. These constructions easily generalize to the manifold setting of Chapter 7.

MECHANICS OF MANY-PARTICLE SYSTEMS Consider a system of n particles moving in space. Conjiguration space is a subset M c R3". A state is a pair (4, u) E M x R3", q = (q', . . . ,q3")rep. . . , u3") representing veresenting position of the n particles and u = (d, locities. If the masses of the particles are m , , . . . ,m,, define M i , i = 1, . . . , 3n, by M 3 i - 2 = M3i-1= M 3 i = mi.

This is just convenient notation.

5

6

2. CLASSICAL MECHANICS

EXAMPLE 2.1 (Two-body problem) Consider a system of two particles of masses m , and m, moving freely in space. If q , = (ql,q2, q3), q, = (q4, q 5 , q6) are positions, then we define configuration space to be

M

=

{(ql, q 2 )E R~ x

R ~ I ~ z, q,).

The requirement q1 # q2 means that our model does not describe collisions. Given a position q E M the velocities are arbitrary so our state space is S

=

M x R6.

Note that M is open in R6. For a given (4, u) E S the trajectory &,") may not be defined on all of R since, if the initial state is such that the particles will collide at some later time, the trajectory will be undefined after the collision time. EXAMPLE 2.2 (Rigid body motion) We will look at this important example in more detail later. We haven particles n 2 4.Let q i = (q3'-', q3').The particles are constrained by the requirement that there be constants cij such that

lqi - qjl = cij

for i , j

=

1,. . . , n.

The configuration space M is a subset of R3".Here M is not open. We shall see later that M is a diflerential submanifold of dimension 6 in R3". The state space will not be all of M x R3" but rather a subset called the tangent bundle of M and denoted T M . Now suppose M is open in R3" and S = M x R3". Let F = ( F ' , . . . , F3") be the total force. Generally we assume forces depend on position and velocity so

F': M x R3" + R ,

1 = 1,. . . , 3 n .

Note we are assuming our forces are time independent. If (q'(t))is a position as a function of t then Newton's second law says Mi

d'q'

= F'(q(t),~ ( t ) ) ,

u(t) = dq/dt.

(2.1)

Suppose we assume our forces are conservative. Then there is a function, called the potential energy function, V:M+R

such that F'(q, u) = -(dV/aq')(q), i = 1, . . . , 3n. Thus, in a conservative system, F depends on q but not on u. If q(t) is a path followed by the system in

7

LAGRANGIAN AND HAMILTONIAN FORMULATION

configuration space and u(t) = (dq/dr)(t),then (q(t),u(r)) is a trajectory in state space. The trajectories are solutions of dqi/dt = vi dd/dt = - l/Mi

I = 1, . . . , 3n

If we define a vector field on state space, (D: M x R3" -+ R3" x R3n,

by

then trajectories of the system, i.e., solutions of (2.2) are exactly integral curves of 0. This 0 is called the infinitesimal generator for the system.

LAGRANGIAN AND HAM ILTO N IAN FORM ULATION DEFINITION 2.3

Let T :M x R3"

--f

R be given by

3n

iMi(~i)2.

T(4, U) = i= I

T is called the kinetic energy function. Let

K:

M x R3" --i M be projection.

DEFINITION 2.4

The Lagrangian of the system is the map

L: M x R3" + R

defined by L

=

T - ( V o x).

Now dL/8ui = Mid, dLl84' = - i-JV/dqi.If (q(t),u(t))is a trajectory, then

(_ 8L) dt doi

-

=

("

do' (q(t),~ ( t ) ) )= Mi -. dt doi dt

So Newton's law reads

Equations (2.4), which in the present context simply represent rewriting Newton's law, are called Lagran ye's equations.

8

2. CLASSICAL MECHANICS

Later, when we discuss mechanics in the presence of constraints, we will see that similar equations still hold but that they are no longer simply a rewriting of Newton's law. Now, still assuming M is open, we turn to an alternative formulation of mechanics, the Hamiltonian formulation. We have

M = configuration space, S = M x R3" = state space. We define phase space to be

S* = M x (R3")*.

(2.5)

Let ( e l , . . . , e3,J be the standard basis for R3",( e l , . . . , e3")the corresponding dual basis in (R3")*. Define a map

9: M x R3" -+ M x (R3")* 9( 4 ,

dei)

= (4,

7

Misiei).

If we write only the coordinates, 9 has the form 9(4',. . . , 4 -I n , u 1 , . . . , u3") = (4', . . . ,q3", M,u', . . . , M3,,u3"). It is traditional to write points in S* as

( d , . . , q3n>P I , . . .

7

P3").

Note that the p i are just momentum components. We see that 2' has an inverse given by

9l(4, P ) = ( 4 , P l l M I , . . . > P3fl/M3J

(2.7)

Now the transformation 9, which is called the Legendre transformation, seems rather trivial since it is just multiplication of various coordinates by ' can also be viewed in the following way which certain scalars. However, 2 will be generalized later. We have an inner product 9: R3" x R3" -+ R defined by g(u, w ) =

3n

C i=

. .

M~u'w'.

1

This defines an isomorphism

6: R3" + (R3")*

by

[ ~ ( u ) ] ( w=)g(u, w).

9

LAG RANG IAN AND HAM I LTON IAN FORM U LATlON

We then have Y'(q, u ) = (q , G ( U ) ) .

The proof of (2.9) is left as an exercise. Note also that ?L/du' = M i d so that we may rewrite V as

The point here is that these last two descriptions of V will generalize to mechanics on manifolds, to be covered later, whereas the simple idea of scalar multiplication on certain coordinates will not determine a well-defined (independent of coordinates) mapping in the context of manifolds. Now, as before, a trajcctory of our system is a curve in state space, c: I -+S, satisfying certain differential equations. We shall refer to Y c as a trajectory in phase space. The image of Y' c will be called an orbit in phase space. DEFINITION 2.5 An integral on the phase space of a mechanical system is a function 4: M x (R3")*+ R that is constant on each orbit. An integral is often called a first integral in the literature. PROPOSITION 2.6 4: M x (R3")*+ R is an integral if and only if for each trajectory in phase space, c ( r ) = (q(t),p ( t ) ) , we have

where V is the potential energy function. PROOF: Left as an exercise.

1

Suppose U is an open set in a finite-dimensional vector space E . A irector jield on U is just a map f : U + E . An itzteyral curve off is a map c: 1 + U such that c'(t) = ,f(c(t)) for all t E 1. (To differentiate a curve in a vector space pick a basis and then differentiate the components of the curve.) Thus, a vector field on M x (R3")* is a mapping <: M x (R3")*-+ R3" x (R3")*.

If we express ( as t(q, P ) = (A'(% PI, . . . A""(% PI, BI(4, P P ' 3

+ . . . + B3,(4, p)e3"),

then the condition that a curve c(t) = (q'(t),. . . , q""(t),pl(t)e'

+ . . . + p3,(t)e3")

10

2. CLASSICAL MECHANICS

be an integral curve o f t is that (2.10) The trajectories in S = M x R3" uniquely determine the infinitesimal generator @ [see Eq. (2.3)]. There is a unique vector field @*: M x (R3")*+ R3" x (R3")*

(2.11)

such that its integral curves are exactly the phase space trajectories. This means c is a state space trajectory if and only if 9 0 c is an integral curve of @*. We now obtain an explicit expression for this vector field. Let c ( t ) = (4(t),u ( t ) ) be a trajectory in S . Then

Let 9 c(0) = (4,p). Then @*(4,p ) = (9c)'(O). Now 0

0

(9c)'(t)= ($(t), . . . , q3"(t),M l d l ( t ) e l + . . . + M3,zj3"(t)e3") 0

(1,

xi

So we have @*(4,p ) = (pi/Mi)ei, -(aV/d4')(4)ei).This vector field can be obtained from a function H on phase space, as we now show. DEFINITION 2.7 The Harniltonian of the system is the map H : M x (R3")*+ R defined by

Notice that aH/aqi = dv/aqi

(Xi

and

aH/ap,

= pi/Mi.

xi

Thus we see that @* = (dH/Jpi)ei, -(dH/dq')e'). Note that H is the total energy of the system. The following proposition expresses the law of conservation of energy. PROPOSITION 2.8

H is an integral of the system.

11

MECHANICAL SYSTEMS WITH CONSTRAINTS

PROOF: By Proposition 2.6 we must show

But a H / a q i = avlaqi,

So this does hold.

qi =

pi/Mi,

aH/ap, = p i / M i .

I Trajectories t + (q(t),p ( t ) ) in phase space are those

PROPOSITION 2.9

curves satisfying

dq'ldt

=

aH/api

dpJdt

and

=

-dH/?q'

(Hamilton's equations). PROOF: Trajectories are integral curves of

dq'ldt

= pJMi = aH/&

and

a* and so satisfy

dpi/dt = -avjdq'

=

-dH/aq'.

I

EXAMPLE 2.1 0 Refer to Example 1.1. This is a conservative system with V ( q ) = q k q 2 . The kinetic energy function is T(q, u) = $mu'. The Lagrangian is L(q, u) = +mu2 - +kq2. The single Lagrange equation is

+

(d/dt)(mu) k4

=0

or,

mq

+ kq = 0

which is, of course, just Newton's law. The Hamiltonian is H(4, p ) = ( p 2 / 2 m )+ $k42. We see the Legendre transformation is given by Y(q,4 = k, mu).

MECHANICAL SYSTEMS WITH CONSTRAINTS We wish to generalize the preceding work to include systems subject to constraints. For an n-particle system under constraint, configuration space will again be a subset M of R3", but it will not generally be open. Also state space will not be all of M x R"". This is because if (q, u ) is a possible state there must be a curve E(c) = (4(t),4 ( t ) )in state space such that q(0)= q, q(0)= u. But q(t) is a curve in M so u must be tangent to M at q. We shall, in fact, assume that any (q, u) in M x R3" for which u is tangent to M at 4 is a possible state. This means that, due to the constraints, motion is restricted to M , but any path in M is kinematically possible, i.e., does not violate the constraints. Such systems are called holonomic systems. Given an arbitrary subset M c R N there may not be any nontrivial tangent vectors at points of M .

12

2. CLASSICAL MECHANICS

FIGURE 2.1

EXAMPLE 2.11

Take M c R 2 as in Fig. 2.1.

At p1 and p2 no nonzero tangent vectors exist. At other points of M there would be a well-defined one-dimensional space of tangent vectors. We shall only consider systems for which M is a special type of subset, a diflerential submanifold of dimension d in R3".Then, at each q E M , we will have a d-dimensional vector space of tangent vectors. This space, denoted T,M, is called the tangent space to M at q. We can then form a new space T M defined by TM =

u

(2.12)

TqM,

qsM

which will be the state space of our system. EXAMPLE 2.12 Consider a mass m moving in the plane under constraints which confine the particle to the unit circle. Thus

M

= {(x, y ) l x 2

+ y 2 = l}.

M is a differential submanifold of R 2 . A tangent vector at q E M is any vector in R 2 which is perpendicular to q. Thus S = { ( q ,U ) E R 2 x R 2 1141 = 1, q 10). EXAMPLE 2.13 (Atwood's machine) We have two masses m , and m2 connected by a (massless) string and arranged over a frictionless pulley as shown in Fig. 2.2. Let (x', y ' ) , (x'. y 2 ) be the coordinates of the two masses. Then

y'

+ y 2 = l',

where

I'

=

1 -nu,

and

x1 = -a ,

x2 = a .

MECHANICAL SYSTEMS WITH CONSTRAINTS

13

FIGURE 2.2

So M = { ( - a , y , a, I' - y)10 < y < I'}. M is thus an open segment on the line in R4 given by

L={(-LI,~,u,I'--)I~ER}. M is a differential submanifold of dimension 1 in R4. The presence of constraints introduces certain complications into the problem of solving a mechanics problem. For example: (a) The coordinates x l , . . . , . x are ~ ~not independent but are related by the constraints. Thus the equations

Mi? = F ' ( x 1 , . . . , X3n, 2 , .. . , P) are not independent. (b) The forces of constraint are not in general known until the motion of the system has been found. This is because the forces of constraint are known only by their effect on the system. Thus, in Example 2.12, the forces of constraint are whatever is necessary to keep the particle on the circle. This depends on the velocity of the particle and hence is not known a priori. This problem can be handled by introducing the constraining forces as additional unknowns, but it does represent a complication in the problem. We will formulate the laws of mechanics in such a way as to avoid these problems. When our configuration space is a manifold, problem (a) can be eliminated by using, instead of the usual 3n Cartesian coordinates, the socalled generalized coordinates. Given a point rn E M there is an open set U

14

2. CLASSICAL MECHANICS

in M and a homeomorphism 4: U + 4 ( U ) c Rd, where 4 ( U ) is open in Rd. If ( q l , . . . , qd)= 4, then points u E U are specified by giving the coordinates ql(u), . . . qd(u).The paths of particles in U correspond to paths in + ( U ) and so we can write down equations in $(U), which correspond to the equations governing the behavior of the paths in U which the system might follow. In other words, we express the problem in terms of local coordinates. If we do this correctly, we can arrange that, in terms of the generalized coordinates, the forces of constraint do not appear at all. If we do this, then problem (b) is solved. Also, in view of the general philosophy that physical theory should not be dependent on a particular choice of coordinates, we shall try to formulate things in a way which is clearly coordinate independent. In order to achieve this, we introduce the theory of differential manifolds and the various geometrical objects associated with them. Once this is done, we can take the various objects we have already introduced, e.g., forces, potentials, kinetic energy functions, Lagrangians, Hamiltonians, integrals, and see how they appear in this new framework.

EX ER C IS ES 2.1 Verify Eq. (2.9). 2.2 Prove Proposition 2.6. 2.3 In Example 2.10 write Hamilton's equation and verify that H is an integral for the system. 2.4 Find the force of constraint when the system of Example 2.12 is subject to an external force field F(x, y ) = ( - kx, 0). (See Exercise 2.6.) 2.5 Suppose that M c RNis open,f : M x (RN)*+ R and g: M x (RN)*+ R are smooth functions. Define the Poisson bracket off and g by

where (ql, . . . , q N , p l ,. . . ,p N ) are the coordinates of (xqiei,xpiei).If H : M x (RN)*-+ R is the Hamiltonian function of a mechanical system (see Definition 2.7), then a function 4 is an integral of the system if and only if { H , 4 } = 0. 2.6 Suppose a particle of mass m is moving in 3-space subject to an applied force Fa. Also suppose that the particle is constrained to move on the surface f ( x , y, z ) = 0 by a force of constraint FC, which is perpendicular

15

EXERCISES

to the surface so that FC = ,iV,f. From calculus we know that the acceleration vector can be written in the form a = (d2s/dt2)T+ k(ds/dt)’N, where T is the unit tangent vector to the path, N = dT/ds, and k is the curvature of the path. Thus we have mu = m(d2s/dt2)T km(ds/dt)2 N = Fa + FC. Assuming, for convenience, that 1 (Vf 1 1 = 1 we obtain FC = 2 Vf, where i. = km (dsjdt)’ N . Vf - Fa . Vf.

+

3 Introduction to Differential Manifolds

DIFFERENTIAL CALCULUS IN SEVERAL VARIABLES Let U be open in R",f:U -+ R. We say that f is Ck on U if all partial derivatives of order at most k exist and are continuous on U . We say f is C" on U i f f is Ck on U for all k 2 1. Iff is continuous on U we say f is Co on U . Iff: U R", we say f is Ck if each component f is Ck. +.

THEOREM 3.1 Let f': U +. R" be C', where f = ( j ' ,. . . ,f").For each xo E U there is a unique linear transformation Axo: R" + R" such that

Moreover, with respect to the standard bases of R" and R", the matrix of lxo is ((df i/~xj)(xo)). PROOF: Let (el, . . . , en),(el,

. . . , ek) be the standard bases in R", R". Define

We must verify

Let xo = C1=lxbei, x = CrYlxiei. So

16

17

DIFFERENTIAL CALCULUS IN SEVERAL VARIABLES

So, by the triangle inequality it is enough to show, for each i, f'(x)

-

fi(xo) -

x-X"

(3fi

Cn

j = 1 SX'

1

(xo)(xj - xi)

= 0.

Let 1

yo = b o , .

Y o = xo,

y,

+ (x'

= xo

yi = yi-

1

y,

- X6)V1,

+ (xi

-

y,-

x:Jei,

1

= (XI,

=

. . x",,

xi,. . . , x;)

( x ,. . . , x n - l , X t ) ,

yn = (x 1 , . . . , x").

cj"=

Now f i ( x ) - f i ( x o )= I fi(~l,i ) .f"( the Mean Value Theorem says

for some

Qj E

and y j - y j - ,

jlj-

= ( x j - x&)ej, so

(0, 1). So

Given E > 0, choose 6 > 0 so that if Iz - xoI < 6 then I(dfi/dxj)(z) (afi/dxJ)(xo)l < 8. Then, if Ix - xoI < 6 we have "

n

.

j= 1

I nElx

-

xoJ.

For uniqueness, suppose y were another map satisfying the condition stated in the theorem. Then 0

=

lim i-0

I.f(xo

+ te,)

-

f ( x o ) - ty(ej)(

Ill

18

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

DEFINITION 3.2 The unique linear map in Theorem 3.1 is called the derivative off at xo and is denoted by Df (xo). THEOREM 3.3 (Chain rule) Let U be open in R", V open in R", f :U R", g: V + RP C'-maps with f ( U ) c V . Then g 0 f :U -+ RP is C' and

D(g f ) ( x o ) = Dg(f(x0))

O

O

-+

(xo E U ) .

Df(X0)

For a proof see [30]. Let I c R be an open interval and c:I -+R be a C'-map. Also, let f :U 4 R be C ' , where U is open in R" and c ( l ) c U . Then D ( f 0 c)(t)* 1 = Df(c(t))Dc(t) . 1. If (f c)'(t) E R is the derivative as defined in calculus books, then D ( f c)(t). s = (f c)'(t)s. Now REMARKS: (a)

0

0

0

This is just what the Chain Rule says. The matrix of Df(c(t))is the row vector ((af/dx')(c(t), . . . , (df/ax")(c(t)))and the matrix of Dc(t) is the column vector (i'(t), . . . , i ( ~ )so) ~ the, product is the 1 x 1 matrix whose entry is 1 (af/axj)(c(t))ij(t). (b) For f :R" -+ R", g: R" + R P the (i,j)-entry of D(g 0 f ) ( x ) is (a(g 0 f ) i / axj)(x). Thus we have

In classical notation, if y = f ( x ) and z

= g(y), then

Suppose U c R" and V c R" are open, f :U -+ V and g: V -+ U are Ck-maps which are inverse to one another. Let x E U , y = f(x) E V . Then and

D g ( y ) o D f ( x ) = Id,,

It follows immediately that n

D f ( x ) D g ( y ) = IdRm. 0

= m.

DEFINITION 3.4 I f f : U -+ V is a Ck-map having a Ck-inverse, then f is called a Ck-diffeomorphism.

19

DIFFERENTIAL CALCULUS IN SEVERAL VARIABLES

We observed above that, iff: U + V is a Ck-diffeomorphism, then Df(x) is a linear isomorphism for x E U . The following fundamental result is a partial converse. A proof may be found in [30]. Let k 2 1. THEOREM 3.5 (Inverse function theorem) Let f :U + R" be a Ck-map, c R", such that Df(xo): R" + R" is a linear isomorphism. Then there

xo E U

are open sets U , , V, containing xo,f(xo), respectively, such that f I U , is a Ck-diffeomorphism of U , onto V,.

We now develop some consequences of the inverse function theorem which will be important in the later development of manifold theory. PROPOSITION 3.6 Let f : U --t R n + k be C' where U is open in R". Assume xo E U and that Df(xo):R" + R n f k is a monomorphism. Then there is a neighborhood V of f(xo) and a C'-diffeomorphism 4: V -+4 ( V ) c R n f k such that, in some neighborhood of xo, the composite 4 0 f is defined and given by q5 of(x1,. . . , x") = (xi,.. . , x", 0,.. . ,0). '

-

PROOF: Let U =

{(x', . . . ,x", y ' , . . . , yk) E R n f k l xE U}.The matrix Df(X0) = ( ( ~ f ' / W ( x o ) )

has n linearly independent columns so it also has some set of n linearly independent rows. We assume rows 1, . . . , n are independent (if not, reorder the coordinates in R"+k). Define ,f 0 --t R n f kby

?(XI,. . . , x", y', . . . , yk) = (f'(X),f2(X), . . . f"(X),f"+ '(4 + y', . . . ,f"+k(X) + y k ) 3

so that

Df(X0,O) = I 0

1

This matrix is nonsingular. Thus, we can find a neighborhood W of (xo,0) in Rn+kand a neighborhood V of f(xo) such that f W + V is a Ck-diffeomorphism. Let 4 =f"-': V + Wand W, = {x E R")(x,0)E W}. W, isaneighborhood of xo in R" and f maps W, into V so 4 f is defined on W, and 4 f(x) = 4(f"(x,0))= (x,0)as asserted. 0

0

20

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

PROPOSITION 3.7 Let U be open in R"+k,f : U + R" a C-map, xo E U , and assume Df(xo):R"+k+ R" is surjective. Then there is an open neighborhood V of xo and a Cr-diffeomorphism 4: V -+ $( V ) c R n + ksuch that

f

0

4-'(x1,. . . , x " + ~=) (x', . . . , x")

for (x', . . . , x " + ~E) $(V).

PROOF: The matrix ((df'/axj)(x,)) has linearly independent rows and so, some set of n columns is linearly independent. Suppose, for convenience, that columns 1,. . . , n are independent. Define f U -+ R"+k by

Ax', ...

9

X",

1

Y .* 9

9

Y k ) = ( f ' ( x ,Y ) , * . *

9

f "(X,Y ) , Y', . . ,Yk). *

0

This matrix is clearly nonsingular so there is a neighborhood V of xo such that f maps V diffeomorphically onto an open set W. Let 4: V + W be the restriction of f" to V. If (x, y ) E & V ) we must show f 0 $-'(x, y ) = x. But f 0 $- ' ( x ,y ) = (x, y). So if 4- ' ( x ,y ) = (u, u), then f(u, u) = ( f (u, u), u) = (x, y ) so that f(u, u) = x. Thus the result is proved. I Thus far we have discussed calculus for mappings between R" and R"'. It is convenient to extend this to mappings between finite-dimensional real vector spaces E and F. We do this using the fact that such spaces are isomorphic to R" and R", where n is the dimension of E and rn the dimension of F , and we define derivatives of a mapping in a manner independent of the choice of such isomorphisms. For an alternate approach using norms see [181. Thus, let E be a finite-dimensional vector space of dimension n. A choice of a basis for E gives a linear isomorphism a : R" + E . If al:R" + E comes from the choice of a second basis then a-' a, : R" + R" is a linear isomorphism and, hence, a homeomorphism. Thus, every finite-dimensional vector space has a unique topology determined by its isomorphisms with R". 0

21

DIFFERENTIAL CALCULUS IN SEVERAL VARIABLES

Suppose that F is a vector space of dimension m and p: R" + F is a linear isomorphism. Let f :U + F where U is open in E. We will say that f is C' if p-' o f o a : a - ' ( U ) + R" is C' for all choices of a and fl. If 8-l o f o a is C' for one choice of a, fl then it is C' for all choices, for if a l , p1 is another choice, then p;' a .f' a , = p;' P o p-' f a 0 a - ' a1 f a, a - 1 a1 is C'. which is C , since each of p;' fi, fl-' For xo E U , we have a linear mapping fl o D(p-' f a ) ( a P 1 ( x 0 ) )a - ' from E to F . This linear mapping is independent of the choices of a and p. Consider two choices a, jand a ' , fll. Then we want 0

0

1,

c)

0

0

p D(p-' 0

0

,f

0

M ) ( ~ ' ( X ~ 0) )LY'=

'

0

0

0

PI D(p1' 0

f

0

0

0

C I ~ ) ( C ( ; ' ( X ~ ) )0

a;'.

But p;' o f o a1 = (B; p) (b-' '-> f cl a ) (a-' a) so, by the chain rule and the fact that the derivative of a linear map is that linear map, we see D(p;'

0

0

f al)(a; '(x~))= ([j; 0

0

' 8)

Thus, multiplying on the left by desired equation.

'J

0

D(p-'

0

0

f

0

CI)(CL-'(X~))0

p1 and on the right by

(U' 0 al).

a ; ' , we get the

DEFINITION 3.8 The linear mapping p D(B-' f a ) ( a - ' ( x , ) ) ~ a - ' is called the derivative of f a t xo and is denoted by Dj(x,). As shown above, this is independent of the choice of a and P. (1

0

0

For vector-valued functions of a real variable it is convenient to regard the derivative at a point as a vector in the range space. This is done as follows: given c: I + E, I an open interval, we have Dc(t) E L(R, E). Now L(R, E ) N E via A+ A(1) so we define c'(t) = Dc(t)(l). If (el, . . . , en) is a basis for E and c(t) = cJ(t)ej(summation on j, see below), then the representative 2 I + R is c^ = (c', . . . , c"). Then the representative of c'(t) is (d'(t), . . . , d"(t)) so c'(t) = i.j(t)ej. So again, to differentiate a curve in a vector space pick a basis and differentiate the components of the curve for the basis. The resulting vector is independent of the choice of basis. In the above paragraph, cj(r)ej really means cj(t)ej. Henceforth, a repeated index appearing once as a subscript and once as a superscript is to be summed over its range of values. For instance, if k is restricted by 1I k In, then Ti,bkmeans xi=, T,,hk.Exceptions will be noted explicitly. We can extend the preceding ideas to define higher derivatives and partial derivatives along a subspace. We shall state things in the context of R" and R" but, as above, everything can be done using arbitrary finite-dimensional vector spaces. Suppose f :V + R" is C" where V c R" is open. For x E V we have D f ( x ) E L(R",R") and, hence, a map D f : I/ .-+ L(R",R"). Let ( e l , . . . , en)and (e;, . . . , e;) be the standard bases in R" and R". Define A{ E L(R",R") for

cj"=l

22

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

1 Iis m, 1 I j I n, by

,

A?e. I' = e!I

(no sum o n j ) ,

Ije, = 0

for k # j.

It is an easy exercise to show that L(R",R") and that we have

{Ail 1

I i I m, 1 Ij 5 n } is a basis for

D f ( x ) = (df'/dxj)14. NOTE: According to the summation convention, i is summed from 1 to m and j is summed from 1 to n, in the preceding formula.

It follows that Df is C" so we can define D'f

=

D ( D f ) : I/-+ L(R", L(R", R")).

There is a natural isomorphism (see Exercise 3.10)

L(R",L(R",R")) + L'(R", R"), where L'(R", R") is the space of bilinear mappings from R" x R" into R". Thus, we may regard D z f ( x ) as being in L2(R",R"). The next result shows that, in fact, D z f ( x ) lies in L:(R", R"), the space of symmetric bilinear maps. PROPOSITION 3.9

For any h, k E R" we have

PROOF: Fix k E R" and define, for this k, g ( x ) = Df(x)k. If ev: L(R",R") + R" is given by ev(A) = Ak, then g = ev O f . Then Dg(x) = D(eu)(Df(x))0 D(Df)(x).But ev is linear so D(eu)(Df(x))= ev so Dg(x)h = ev 0 D(Df)(x)h= (D(Df)(x)h)k.Now observe that g ( x ) = (df/dx') (x)k' so 0

Dg(x)h = ag (x)hj = (x)k'hj, 2x1 d X j axi ~

which completes the proof.

I

Similarly, we can define DPf = D ( D p - ',f).There is an isomorphism L(R", L p - '(R",R")) -+ LP(R",R"),

where LP(R",R") is the space of p-multilinear maps from R" x . . . x R" into R" (See Exercise 3.10). Thus DPfcan be viewed as a mapping into the space LP(R",R") and, as in the case p = 2 above, it actually takes values in the subspace L:(R", R") of symmetric multilinear maps.

23

THE CONCEPT OF A DIFFERENTIAL MANIFOLD

DEFINITION 3.10 Suppose V c R", U c R" are open sets and F : V x U -+ R' is of class C'. For (x,y ) E V x U we define partial derivative linear mappings D , F ( x , y ) :R" -+ R' and D,F(x, y ) :R" -+ R' by D , F ( x , y)h = D F ( x , y)(h, 0) and D,F(x, y)k = DF(.x, y)(O, k). Note that DF(x, y)(h, k ) = D , F ( x , y)h + D,F(x, y)k. Now D , F : V x U L(R", R') so we can form D,D,F and D f F . Similarly for D,D,F and DZF (if F is C'). -+

THEOREM 3.1 1 (Implicit function theorem) Suppose V c R", U c R" are open sets and F : V x U R" is of class C . Let (x,, yo) E V x U and zo = F(x,, yo). Suppose that D,F(x,, y o ) :R" -+ R" is a linear isomorphism. Then there is an open set V,, xu E V, c V , such that there is a unique mapping g : V, -+ U satisfying F(x, g ( x ) )= zo for all x in V, and g(x,) = y o . Furthermore, g is of class c' and -+

D g ( x ) = -D,F(x, g(x))-l 0 D,F(x, g(x))

for x E V,.

PROOF: Define H : V x U -+ R" x R" by H(x, y ) = ( x , F ( x , y)). Then DH(x,,y,)(h, k ) = (h, D F ( x o , y,)(h, k ) ) . Thus, 0 = D H ( x o , yo)@, k ) gives h = 0 and 0 = D F ( x o , y,)(O, k ) = D , F ( x , , y,)k so that k = 0 also, by our hypothesis. Thus, D H ( x o , yo) is a linear isomorphism and by the inverse function theorem there is a C'-mapping G : V , x W, -+ V x U so that xo E V, c r! zo E W,, and H(G(x, z ) ) = (x, z ) for (x, z ) E Vo x W,. Clearly, G(x, z ) = (x, G o ( x , z ) ) and F(x, Go(x, z ) ) = z for (x, z ) E V, x W,. Take g(x) = G,(x, zo). Finally, F ( x , g ( x ) ) = zo Vx E V, gives, by differentiating, D , F ( x , g ( x ) ) D,F(x, dx)) O D g ( x ) = 0 so Dg(x) = -D,F(x, g(x))- " D,F(x, g W ) . I

+

THE CONCEPT OF A DIFFERENTIAL MANIFOLD It is possible to define the concept of a continuous function f : U Rk, where U c R", but the concept of continuity can be studied in a more general setting. Thus, we realize what is needed is a topological structure. The general study of topological spaces and continuous maps between them has resulted in the discovery of many new areas in which the notion of continuity is important. Similarly one can define the notion of a C' function f :U -+ Rk, U open in R". But this limited environment does not allow the concept of differentiability, of approximating nonlinear objects by linear ones, to fully express itself. Given a "smooth surface" in R 3 , one can speak of a differentiable function on that surface. Thus "submanifolds" of euclidean spaces provide structures, on which the idea of differentiability makes sense, which are more general -+

24

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

than open sets. The general notion of a diflerential manifold provides a general framework for the concept of differentiable function. Differential manifolds or, simply, manifolds, are a fundamental class of objects whose applications are only beginning to be appreciated. We shall introduce these objects and study the various geometrical structures which live on them. We shall see that various types of mechanical systems and, indeed, the laws of mechanics themselves, can be formulated in a very elegant manner in terms of geometric objects on differential manifolds. CONVENTION:From now on, unless the contrary is stated, all maps are assumed C". "Smooth" is a synonym for C".

Let X be a topological space. DEFINITION 3.12 A chart on X is a pair ( U , 4), where U is open in X and 4 : U -+ @ ( U )c R" is a homeomorphism of U onto an open subset of R". We say charts ( U , 4), ( V , $) are C"-related if either

(a) U n V = @

or

(b) U n V # @

and the maps

4

0

I+-

': $(U n V )-+ 4 ( U n V )

and

Ic/ 4- ':4 ( U n V )-+ $(U 0

n V)

are C". An atlas OM X is a family of charts {(Ui, 4Jli E A}, where { U i l i E A} is an open cover of X and every pair of charts is C"-related. DEFINITION 3.13 A diflerential structure on X is an atlas R which is maximal in the sense that, if ( U , 4) is a chart on X which is C"-related to every chart of R, then ( U , 4) E R. PROPOSITION 3.14 Let @ be an atlas on X . Then there is a unique differential structure R on X such that @ c R. PROOF: Let R be the set of all charts on X which are C"-related to every chart in 0.Then, of course, (D c R and if R is an atlas, it is clearly maximal and hence, is a differential structure. Uniqueness is then immediate, for if R' were another differential structure with the desired properties, then @ c R' c R by definition of R. But then maximality of R'forces R' = R. So we need to take ( U , d), (V, $) in R and show 4 ' : $(U n V )+ 4 ( U n V ) is C". Pick x E $(U n V ) . We show 4 0 I + -is' C" in a neighborhood of x . Choose (Ui, 4i)E @ such that $-'(x) E Ui.Now 0

4i

0

$ - ' : $ ( U i n V ) + R"

is C"

25

THE CONCEPT OF A DIFFERENTIAL MANIFOLD

and

+u4;1:4i(UnUi)-+R"

kC".

Choose an open neighborhood W of ,\ contained in $(U n V n U J . On W (bl $ - ' s o 4 $-'isC"onW,asdesired. I wehave4 I,-'=* 4;' DEFINITION 3.15 A diflerentiul n-rnun@fd is a pair ( X , R) where X is a second-countable Hausdorff space and R is a differential structure on X such that each chart in R takes values in R". REMARK: When we speak of a differential manifold X we mean we have some specific differential structure in mind. When we speak of a chart on X , ( U , 4),it is assumed ( U , 4)is in the particular differential structure. Note that to specify a differential manifold we must specify a second-countable Hausdorff space X and an atlas on X . Then the differential structure is uniquely determined by Proposition 3.14.

EXAMPLES 3.16 (a) An open set in R", say X , is a differential nmanifold with atlas { ( X , lx)). (b) Let Y be a differential n-manifold, X an open subset of Y . If R is the differential structure on Y , define a ditrerential structure RlX on X by RIX = { ( U ,4) E RI u c X ) . (c) Letf': U + R be C", U open in R". Let

x = {(XI,.

. . , x " , J ( s ' , . . . , x"))l(x', . . . , x") E U } .

X is an n-manifold, an atlas being { ( X ,$)}, where * ( X 1 , . . . , X", X f l + 1 ) = (x', . . . , X'I). (d) Let

. . . , x'"

(XI,

' ) E R"+

'

4: X

-+ U

is given by

nf I

We construct a Ccc-atlason S" consisting of two charts, the charts being stereographic projection from the north and south poles, respectively. Thus let

u , = s" - {(O,O, Then,

4 + :U +

-+

. . . ,-

u- = S"

-

{(O, 0,. . . , + 1)).

R" is defined by

4+(.u',. . . ,.x

n+

so that

l)},

I

) =(I

+ .x"+')-l

(XI,.

. . , x")

26

4 -; U -

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

R" is defined by

and

We see that

This is C"; in fact, 4 + 0 41' is just inversion in the unit sphere. This defines a differential structure on S" which is called the standard differential structure on S". (We shall show later this is the same differential structure that S" gets as a submanifold of R"".)

SUBMANIFOLDS Given Y c , X , X a differential manifold, there is, in general, no natural way to give Y a differential structure. However, if Y is sufficiently "nice," then there will be an induced differential structure on Y. Such "nice" subsets are called submanijolds. DEFINITION 3.17 Let X be an n-manifold, Y c X . We say Y is a k-dimensional submanifold if, for each y E Y, there is a chart ( U , 4) on X such that y E U , 4(y) = 0 E R", and such that

4(U n Y ) = 4 ( U ) n Rk x 0 =

{ .E 4(u)lzk+1= Z k + 2 = . . . = z" = 0).

INTUITIVE REMARK: A manifold is a space which looks locally like an open set in a vector space. A submanifold looks locally like a subspace of a vector space. That is, we can choose a chart at each y E Y which "straightens out Y locally." See Figure 3.1.

y

If Y is a submanifold of X , Y gets a differential structure as follows. Choose Y. Let ( U , 4) be a chart on X with y E U such that

E

4(U n Y ) = &(V) n Rk x 0. We call such a chart a submanifold chart for Y . Let 4 ( U ) = { ( X I , . . . ,x")l ( X I , . . . ,Xk, 0,. . . ,0) E + ( U ) ) . h_

27

SUBMANI FOLDS

FIGURE 3.1

Then

4 defines, in an obvious manner, a homeomorphism 6:u n Y --+ J@) c ~ k .

We also refer to ( U n Y ,$) as a submunifold chart for Y . PROPOSITION 3.18 The collection of all submanifold charts on Y forms an atlas and hence, defines a differential structure on Y.

The proof is left as an exercise. THEOREM 3.19 Let f :R" --f Rk be C", uo E Rk. Suppose f - ' ( u 0 ) # and, for all x E f - '(u0), D f ( x ) :R" -+ Rk is surjective. Then f - ' ( u 0 ) is an (n - k)dimensional submanifold of R". PROOF: Choose xo ~ f - '(u"). We must construct a submanifold chart at xo. By Proposition 3.7, there is a neighborhood U of xo and a diffeomorphism 4: U 4 ( U ) c R" such that, for z E 4 ( U ) , f 4-'(z) = (z', . . . , zk).Let $(xo) = zo = (zh, . . . , z;). Then ,f(xo) = f 4- ' ( z 0 ) = ( z h , . . . , z",. Let x E U . Then x ~ f - ' ( u ~if. )4(x) = z where zi= zb for 1 5 i k. So 4 ( U n f - ' ( u , ) ) = . 4 ( U ) n { z E R"(z' = zb, 1 i 5 k ) . If 8:R"-+ R" is given by 8(z) = (zkC1- z:+', . . . , Z" - z;, Z' - z;, . . . , zk - z:) then II/ = 8 4: U --+ 8 ( 4 ( U ) ) is a chart on R", $(xo) = 0 and $(U n f-'(uo)) = $ ( U ) n { z I z " - ~ + '- . . . z" = O}. This proves the result. I -+

0

0

0

EXAMPLES 3.20 (a) Let f : R"" -+ R be defined by f ( x ) = Then f is C" and f - ' ( O ) = S". If .x ~ f - ' ( 0 ) ,then

. . . , -(x) ax"

)

= (2x0,.. . ,2x")

# ( 0 , . . . ,O)

( x i ) 2- 1.

28

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

since ~ ~ = , ( x=~1.) Then ~ D f ( x ) :R"" -+ R is onto, so S" is a submanifold of Rnf1.We shall see later that the submanifold differential structure on S" is the same as the differential structure defined in Example 3.16. (b) Letf(x, y) = xz - y 2 ; f - ' ( 0 ) is not a submanifold sincex2 - y2 = 0 is not locally homeomorphic to R' near (0,O).This does not contradict Theorem 3.19 since (aflax)(O,0) = 0 = (dflay)(O, 0). (c) Let F , G: R 3 -+ R be such that whenever x E R 3 and F(x) = 0 = G(x), we have V F ( x ) x VG(x) # 0. Then M = (x E R 3 I F(x) = G(x) = 0} is either empty or a submanifold of dimension 1. Proof of this assertion is part of Exercise 3.2. (d) Let f : U + R be C", U open in R". The graph off is G, = {(XI, . . . , x " , f ( x ' , . . . , x"))~(x',. . . , x")

E

U}.

This is a submanifold of dimension n in R"", as one sees using the function F(x1,. . . , X"+l) = x n + l -f(x',

defined on 0 Exercise 3.2.

=

{(XI,.. . , x

" +

')1(x1, . . . , x")

E

. . . , x")

U } . Verification is part of

TANGENT VECTORS We originally began discussing manifolds as objects on which the idea of tangent vector makes sense. We now go to the details. Let X be an n-manifold, I an open interval in R , C: I + X a curve which we assume to be continuous. DEFINITION 3.21 We say that c is a Cm-curue if, for each chart ( U , 4 ) on X with c ( I ) n U # @, 4 c is a C"-map. 0

4 0 c is defined on c- ( U ) ,which is open, so it makes sense to require c to be C".

NOTE:

4

0

Suppose c(to)= xo E X . We want to define the concept of the velocity ), vector c'(to). For a curve in R" having the form c(t) = ( ~ ' ( t ) , . . . , ~ " ( t ) we want to have c'(to)= (c?(fo), . . . , C"(to)). For our curve c in X , choose a chart ( U , 4) at xo and let 4 0 c = (c', . . . , c"). Our vector c'(to) is going to be an object which, in each chart about xo, is represented by an n-tuple of numbers. The representative n-tuple for ( U , 4) is (?'(to), . . . , ?'(to)). Now suppose ( V , $) is another chart at xo, with $ n c = (c?, . . . ,C"). Then the (V, Ic/) representative of c'(to) is required to be (i'(to), t2(to),. . . , ;"(to)). But, in a neighborhood of t o , we have c= 4-l 4 c, $ 0

$ 0

0

cl

29

TANGENT VECTORS

so we have = D($ c '

4-')(4

0

to))

(3.1)

Thus, the representatives of c'(t,) in the two charts are related by the derivative of the coordinate transformation. If we write 4 = (x', . . . , x"), $ = (f', . . . ,in) and write 4 - l as 0

2'

=

X'(x1,

. . . , x"),

2"

= .f"(X',.

. . , x"),

then D($ 4-')(4 c(to))= [d.?/(7.d], evaluated at (x;, . . . ,x): Then (3.1) takes the form 0

0

= 4(xo).

(3.2)

? ( t o ) = (iiXi/C?xJ) Cj(to)

4 c(to) = (x;, . . . ,x:)). Thus, if we want to define tangent vectors so that the velocity vector of a curve has its usual form in R", then the coordinate representatives of the vector for different coordinate systems will be related by a certain transformation law, namely, (3.1) or (3.2). This motivates the following definition. ( j summed from 1 to n and partials evaluated at

DEFINITION 3.22

0

Let xo E X and

@,.<) = { ( U , $ ) l ( U , 4) is a chart and xo E U } .

A tangent vector at xo, say v, is a function u:J1%,, -+ R" such if (v', . . . , v") and V', . . . , 17")are the values of u on ( U , d), ( V , $), then

or, in coordinate form as above, Gi = (df'/dxj)

UJ

( , j summed from 1 to n).

(3.4)

REMARK: Equations (3.3), (3.4) are called the contravariant transformation law and a tangent vector v at xo is often called a contravariant vector at xo.

NOTATION:Given a tangent vector v at x and a chart ( U , 4) at x we have u((U,4 ) )E R". This is the ( U , 4)-representative of v. We shall shorten this

30

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

cumbersome notation by referring to the $-representative and writing v+ rather than u( ( U , 4)). DEFINITION 3.23 The tangent space at X E Xis the set of all tangent vectors at x and is denoted T,X.

Define addition and scalar multiplication on T'X as follows: let u, w E T,X, c E R, ( U , $) a chart at x . Then define (u

+ w)+= u+ + w g

and

(CU)= ~

(3.5)

cv+.

If ( V , $) is another chart at x, then we have (u

+w

)= ~ uJI

+ wJI

and

(c& = C

We must show (3.5) and (3.6) are related properly by (3.3) so that u cu are actually tangent vectors. We have UJI

=

w

O

$-l)($(x))u+

Hence UJI

so u

+ w E T,X

+ WJI =

and

w"

WJI =

N$

$-1)(4(x))(u+

O

(3.6)

U ~ .

+ w and

$-1)(4(x))w+.

+ w+)

and, similarly, cu E T'X.

PROPOSITION 3.24 T,X is a vector space. If dx$: T,X = u g , then d x $ is a linear isomorphism.

4

R" is defined by

d&(u)

We leave the proof as an exercise. Thus, we have d,.. is the map which takes a vector u and gives its $-representative. Given ( U , 4) a chart at x and z E R", there is a unique u E T,X such that d,$(u) = z. Thus, to specify a vector it is enough to specify its $-representative for any particular chart ( U , 4). We have a basis for T,X given by ( d X $ ) - l e k ,k = 1,. . . , n, where ( e l , . .. , e n ) is the standard basis in R". The vector ek in R" is characterized by the fact that the directional derivative of a function f in the direction ek is just d f / d x k . Therefore, we shall adopt the following notation for the above coordinate basis vectors. DEFINITION 3.25 Let ( U , 4)be a chart at x, 4 = (xl,. . . ,x"). The vector (d/dxk)Ixis defined to be ( d X $ ) - ' ( e k ) . If ( V , $) is another chart, $ = (y', . . . , y"), and if (d/dy')I,, i = 1,. . . , n, are the corresponding vectors, we have

It follows that (d/dx')l,, 1 5 i 5 n, is a basis for T,X and, if u+ = (19,. . . , u"), then u = u'(d/dx')l, (sum on i).

31

TANGENT VECTORS

EXAMPLE 3.26 Let X = R 2 . Consider the chart 4(x, y) = (x, y). Define another chart 9: U + R 2 , where U = {(x,y)I y # 0 or x 3 O}. We will require JI$ ( U ) = {( 8)lv I,> 0, - n < 8 < n} and define + ( U ) U by t+!-'(r, 8) = (rcos 8,rsin 6). Let u be a vector at q E X with uJI = (v,, ve) and v+ = (ox, vy). Then

-

- r sin 8

cos H that is, V, =

vY

u, cos 8 - r sin 6 vo,

= U,

sin 8 + r cos 8 u0,

Also

a

-=

ar

cos 0 -

a + sin 8-a

ax

aY

and

a

-=

a8

- r sin 8-

2

ax

+ r cos 8-.a

aY

On R" we usually identify tangent vectors with their I,,-representatives. Then alax and slay become the vectors e l and e,, and alar and d/a8 are the vectors associated with polar coordinates as shown in Figure 3.2. Our original motivation for the definition of tangent vector was our desire to talk about the velocity vector of a C"-curve on a manifold.

ar

FIGURE 3.2

32

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

Definition 3.27 Let c: I -+ X be a C"-curve, c(to)= xo. The velocity vector of c at t = to is the vector c'(t,,) E T,,X satisfying dxo4(c'(to))= (d, c)'(to)for each chart ( U , d,) at xo. (Here d, 0 c maps into R" so (4 c)' (to) is the usual componentwise derivative.) 0

NOTE: Let

7

d, c = (c', . . . , c"), i.e., the ci are the +coordinates of c. Then 0

(4 " c)'(to)= (?'(to), . . . W o ) ) 3

so

Thus, given d,, we have the basis (a/dx')l,,, 1 I iI n, and to find the components of c'(to) just differentiate the d,-components of c. Also, if ( U , d, = (x', . . . x")) is a chart on X and (xh,. . . , x:) E d,(U), then, for t in some neighborhood of 0, we may define c(t) = 4 - ' ( x h 7 . . . , x i t , . . . , x:). This is a parametrization of the jth-coordinate curve and we see that c'(0) = (d/dxj)I,,. Now consider xo E R" and v E T,,X. Let (v', . . . , v") be the l,.-representative of v. In discussions of vectors in R" in which a vector is simply defined as an n-tuple it is this identification which is being made. For example, let c: I -+ R" be a curve. With respect to the chart (R", l,,,), c has its standard Cartesian components (c'(t),. . . , c"(t))so the 1,"-representative of c'(to)is the usual (?'(to), . . . ,?"(to)). Consider X a k-dimensional submanifold of R", xo E X . We wish to identify T,,X as a linear subspace of R". Now any v E T,,R" can be realized as c'(0) for some C"-curve c in R" with c(0) = xo. Similarly, any v E T,,X can be realized as c'(0) for a C"-curve c in X . A map T,,X -+ T,,R" is defined as follows. Let v E T,,X. Write v = c'(O), c a C"-curve in X . Now c is also a Coo-curvein R" so c has a velocity vector c'(0) E T,,R". The map is zi E T,,X + ~ ' ( 0E )T,,R". We claim this is a welldefined injective linear map. To see this, let ( U , 4) be a submanifold chart at xo, so that d,(xo) = 0 and #J(U n X ) = d,(U) n Rk x (0).If 4 = ( y ' , . . . , y"), then $ = ( y ' , . . . , y k ) gives a chart $: U n X $(U n X) for X . Let c be as c = (c', . . . , ck), above and let d, c = (el, . . . , c"). Then

+

-+

0

and also c'(0) E T,,R" is given by

6

0

33

SMOOTH MAPS OF MANIFOLDS

But y' c = 0 for i 2 k + 1 so C'(0) = 0 for i 2 k we have defined can be described as follows: If 0

+ 1. So we see that the map

then the image of v is

or, in terms of the components, (v1, . . . , u k ) + (l", . . . , vk, 0, . . . , 0). This shows our map is independent of c and that it is injective and linear. Note that if we identify Tx,,R"with R" via (R", lRn)then the image of Tx,,X is precisely: {(v', . . . , t i " ) E R" I there is a C"'-curve in X , c(0) = xo, c = (c', . . . , c") and i'(0)= u*, i = 1, . . . , n} This latter set is usually taken as the space of tangent vectors to X at xo, so we have established the connection between the usual advanced calculus notion of tangent space and the general manifold formalism. When dealing with submanifolds of R", it is often convenient to make use of the above identification and we shall feel free to do so.

SMOOTH MAPS OF MANIFOLDS Let f': X

+

Y be a continuous map between two differential manifolds.

DEFINITION 3.28 If xo E X we say f is C" at xo if, for each choice of charts ( U , $), ( V , t,b) at xo,f(xo), respectively, the composite o f 0 $ - I , which is defined near 4(xo), is C" on a neighborhood of $(xo). REMARK: If there exists charts ( U , $), ( V , $) at xo,f(x0) such that $ f' c 4 - l is C" on a neighborhood of &x0) then f is C" at xo.See Exercise 3.7. 0

DEFINITION 3.29

If f ' : X + Y is C" at each x E X we say that f is C"

on X . REMARK: f : X + Y is C" on X if for each pair of charts ( U , $), ( V , $) on X , Y the composite $ f c 4-l is C" on its domain, 4 ( f - ' ( V ) ) . 0

REMARKS 3.30 (a) If U , V are open in R", R" then f':U has its usual meaning.

+

V being C"

34

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

(b) If c: 1 + X is a curve in X we can use the identity chart on I , so C" in the sense of Definition 3.29 agrees with C" as defined in Definition 3.21. (c) f : X -+ R is C" iff foreachchart(U, $) on X , f + - ' : $ ( U ) + R i s C". 0

DEFINITION 3.31

A C"-difleomorphism f : X

-+

Y is a C" map having a

C"-inverse. EXAMPLES 3.32 (a) Let X be R with the differential structure determined by the chart (R, lR).Let Y be R with the differential structure determined by the chart (R, +) where $(x) = x 3 . Then f : X + Y by f ( x ) = x113 is a C "-diffeomorphism. (b) g: X + X , g(x) = x113 is not C". (c) h:X --* X , h(x) = x 3 is a C"-homeomorphism but h is not a diffeomorphism. (d) The identity map from any manifold to itself is always a C"-diffeomorphism. However, $ : X -+ Y , +(x) = x , is not a diffeomorphism.

We will now see that a smooth mapping between manifolds induces a linear mapping between tangent spaces, generalizing the notion of derivative as given in Theorem 3.1. Let f : X Y be C", f ( x o ) = yo; let u E TxoX.Choose a curve c in X such that c'(0) = v. Then f c is a smooth curve in Y so, since f c(0) = f ( x o ) = yo, we have (f c)'(O)E TyoY.We claim c'(0)+ (f c)'(O) defines a linear transformation TxoX+ TyoY. Choose charts ( U , $ = (x', . . . ,x")), (V, $ = (y', . . . , y")) about xo, y , and arrange to have f ( U ) c I/. Let c have $ o c = (cl, . . . ,c"). Thus, c'(0) = xb, ?(O) = vi. Let f 0 = ( f ' , . . . ,f"), so that 0 .f o c = ( f ' ( c l ( t ) ,. . . , c"(t)),. . . ,f"(c'(t), . . . , c"(t)).Then(f 0 c)'(O) has $-representative -+

0

0

0

0

+-'

$ 0

(g

(XA,

. . . , X",)cj(O),

. . . , df" ~

(XA,

. . . , X",)cj(O)

axj

)

Thus, in terms of local representatives, we see our map is

(v', . . . , 0") -+

(af'

-

axj

VJ,. aj-2 V J. , . . . , af"

axJ

vj)

axj

so that the local representative of v is mapped by the matrix ((8y/8xj)($(x0))). DEFINITION 3.33 Letf:X + Y be C", withf(x,) = yo. The linear map described above is called the tangent off at xo, and is denoted by

T x , f :TX,X + Ty,Y.

The above calculation shows that the matrix of Tx,f with respect to the bases (J/dxj)lxo,1 I j I n, and (a/dyi)lyo,1 I i I m, is [(dfi/dxj)(+(xo))]. The following is a global version of the chain rule:

35

DIFFERENTIALS 0 F FUN CTl ONS

THEOREM 3.34 Letf:X --t Y and g: Y -+ Z be C". Then (a) g 0 f is C". (b) If xo E X and yo =f(xo), then

PROOF: (a) Left as an exercise. (b) Let u E T,,X and let c be a smooth curve with c'(0) = u. Then T J g f ) u = ((g f ) o c)'(O) = (g o (fo c))'(O) = T,,g(f O c)'(O) = T y o ( g ) ( ~ x o f ( u ) )I . D

PROPOSITION 3.35 (Inverse function theorem) Let f : X + Y be C", f ( x o ) = yo and assume T,,f: T,,X -+ T,,Y is an isomorphism. Then there exist open sets W,, W, about x o , yo such that f maps W, to W, and f: W, + W, is a diffeomorphism. PROOF: Choose charts ( U , 4), (V, $) about xo, yo. Let

$ 0

f

0

4-l

=

(f',. . . ,f"). Then ( ( l f / d x j ) ( 4 ( x o ) ) )is nonsingular so there exist neighborhoods, 0, ? of +(xo), $ ( y o ) such that $ f 4-l maps 0 onto ? diffeomorphically. Let W, = 4- '(O), W, = $ - I ( f).Then the the local representative of 0

f using (WI, 4 I Wlh (W2, $1

0

W,) is

$..f'+p:O

4

p.

Since this is C" with a C'"-inverse so is f :W, -+ W,.

I

For a curve c: I -+ X there is, for t o E I , T,,c: T J -+ T,,,,,X and also &(to) E T,(,,,X. We claim c'(to)= T f o c ( ( d t o l ~ ) ~ 'Now ( l ) ) . (d,,l~)-'(l)is that vector in T,,I whose 1,-representative is 1. If d: I -+ I is the curve d ( t ) = t then &(to) is that same vector. So T f o c ( ( d z ~ l l , ) ~= l ( T,,c(d'(t,)) l)) = (c 0 d)'(to) = ~ ' ( t , ) . So, if we identify the 1-dimensional vector space T,,I with R via dfolr,we could write c'(to)= Ttoc(l).

Now consider f : X + R, x o E X . We have T,,X .+ Tf(,,,R. We identify TS(,,)R with R using the identity chart and we get a map

DIFFERENTIALS OF FUNCTIONS DEFINITION 3.36 If f : X R the diflerential o f f at x o is df(xo)E L(T,,X, R). The differential off, denoted by df, is the map which, to xo E X , assigns df(xo). Given a chart 4 = ( x ' , . . . , x") on X , the maps xi: U -+ R are C" so we may define dxi(xo)E L(T,,X, R ) . -+

36

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

PROPOSITION 3.37 Let f : X + R be C", x o E c be a smooth curve in X with c'(0) = 0. Then

df(X0)U = df(xO)u= ( ( d f ( x o )

I

(See Definition 3.27).

R)

(f

Txof)(u)

O

X, and let u E TxoX.Let

C)'(O).

c)'(o)= (

=( d f ( x o ) R ) ( f

R

f c)'(o)

PROPOSITION 3.38 (dx'(xo), . . . ,dx"(xo))is the basis dual to

PROOF: Let cj(t)= 4 - ' ( x & . . . , x&

+ t , . . . ,x:).

Then, by Proposition

3.37,

Let f :R 3 + R be given by f ( x , y, z ) = ( x ~ + ~ ~ + z ~ ) ~ ~ ~ + C" on R 3 - (03. We have

EXAMPLE 3.39

x

+ 3y. f i s

af (1,0,0) = $(x2 + y2

-

ax

+ z y 2 x + lI(,,,,,,

=4

and

af

-

dY

(1,0,0)

= 3,

af

-

aZ

(1, 0,O) = 0

+

df(1, 0,O) = 4 d x 3 d y . Using spherical coordinates, (r, 6,4) we havef(r, 8, u) = r3 + r cos 4 sin 8 + 3r sin 4 sin 6. Our point (x, y, z) = (1,0,0) is now (r, 6, 4) = (1, 7c/2,0). Then

SO

af/ar

=

so df(1, z/2,0) = 4dr

3

+ 1 = 4,

ay-/ae = 0,

af/a$

=

3,

+ 3d4.

EXERCISES 3.1 Prove Proposition 3.18. 3.2 Work out the details of Examples 3.20(c) and (d). 3.3 (a) If X is a manifold, Y c X , then Y is a submanifold iff for every open set U c X we have Y n U = $3or Y n U is a submanifold of u.

37

EXERCISES

uicl

(b) Let Y c U i , U i open in X . If, for each i, Y n U i is a submanifold of U i then Y is a submanifold of X . (c) Let ( U , 4) be a chart on X , Y c U . Then Y is a submanifold of U iff $ ( Y ) is a submanifold of $(U). 3.4 Suppose ( ( U i ,$i))isA is a family of charts and c: I + X is a continuous U i and such that 4i c is C" for all i. Then c curve with c(1) c UieA is a CaJ-curve. 0

3.5

Prove Proposition 3.24.

3.6 Let X be a k-submanifold of R". (a) Let ( V , $) be a chart in the differential structure of X . Then $ - ' : $ ( V ) - + R".Showthat $-'isaC"-mapfrom$(V') c Rkinto R". (b) Let I) = (y', . . . , y k ) .Show that under the identification discussed above we have ( d l d y j )I x" = (d$ - '/dyj)($(xo)).Here I,-' is viewed as a function of y', . . . , y k with values in R" so (alC/-'/dyj)($(x,)) is in R". (c) The above identifications are still valid where k = n in which case X is an open set in R". Then the only identification which is not trivial is Tx,R" R". Consider the case n = 2. We have polar coordinates $ = (r, 8) on U = {(x, y ) E R21y # 0 or x > o}. Calculate ajar, 8/88 as points in R 2 . (d) Let X c R3 be the surface z = x 2 - y2. Given ( x o , yo, x i - y i ) E X find a basis for T,X, p = ( x o , y o , x i - y;), representing elements of T,X as elements of R3. 3.1

Prove the validity of the remark following Definition 3.28.

3.8 Prove the validity of the remark following Definition 3.29. 3.9 Show that: (a) composites of smooth mappings between manifolds are smooth; (b) if Y is a submanifold of X and i : Y + X is inclusion, then i is C". 3.10 Let E and F be finite-dimensional vector spaces. Show that: (a) L(E, L(E, F ) ) 2: L2(E,F); (b) L(E, L"-'(E, F ) ) N L"(E, F), where L"(E,F ) is the space of n-multilinear mappings from E into F . ( X I ,. . . , x") is a chart near x o and f : X + R is C", then the components of cif(xo) with respect to the basis (dxl(xo), . . . , dx"(x,)) are

3.11 If

38

3. INTRODUCTION TO DIFFERENTIAL MANIFOLDS

3.12 Let f be given in spherical coordinates byf(r, O,$) = r tan 8. Consider the point (r, O,b) = (1, 7c/4,0). Find constants A , B, C such that

df(l,71/4,0)= A d x

+ B dy + C dz.

3.13 Let Y be a submanifold of X . A mapf: Y -+ 2 is C" if and only if, for every Y E Y , there is a neighborhood W of y in X and a C"-map F: W + 2 such that F(w)= j ( w ) , for w E W n Y. 3.14 (Lagrange multiplier method) Suppose that V c R" is an open set and we have smooth functions f : V + R, gl: V -+ R, . . . , gk: V -+ R such that, 0 = yl(x) = g2(x) = . . = gk(X) implies that rank (dgi/dxj) = k. Let M = { X E R"IO = g l ( x ) = . . . = gk(X)}. Then (a) M is a submanifold of R" and, at each X E M, T,M = n{ker dg,(x)I 1 s i s k } . Suppose that x o E M and W is an open set in M containing x o such that f ( x ) If ( x o ) for each x in W . (b) Use a coordinate chart ( U , $ = u 1 , u 2 , . . . , u " - ~ )on M with x o E U c W to show that (by beginning calculus) df(xo) I T,,M =0. The fact that there are numbers (the Lagrange multipliers) Al, A,, . . . ,Ak such that df(xo) = Cr= Aidgi(xo) now follows from (a), (b) and the following basic fact from linear algebra: (c) suppose E is a vector space and a, pl, p,, . . . ,p k are each in E*. If n{ker pill I i I k } c ker a then there are numbers A l , &, . . . ,1, such that a = &pi. '

4 D iffere nt i a I Equations on Manifolds

The theory of flows and integral curves of vector fields on manifolds is one of the fundamental building blocks of which manifold theory is constructed. The theory developed in this chapter will be indispensable for our later work. The development given follows closely that given by Lang [ 191. We begin with the local existence and uniqueness theory for systems of ordinary differential equations. Then we discuss the corresponding results on manifolds. We discuss the flow of a vector field and obtain the basic results on smoothness of the flow of a smooth vector field.

VECTOR FIELDS A N D INTEGRAL CURVES A vector field on a manifold X is an assignment of a vector to each x E X so that the vectors "vary smoothly with x." DEFINITION 4.1 A C"-uectorjeld 5 on Xis an assignment, to each x E X , of a tangent vector ((x) E T,X, such that the following smoothness condition holds: If ( U , 4) is a chart on X and (+: + ( U ) -+ R" is defined by (+(z)= dx4(<(+-'(z))), where 4(x) = z , then <+ is C".

The map tS is the local representative of 5 for the chart ( U , 4) or, simply, the +-representative of 5. Suppose we define functions 5': U -+ R by

<(x) = <'(x)(ajax')l,,

x

E

u,

or simply

5 = 5' a p x i . . , 5") 4-l so &, is C"

on

U.

Then <+ = ( ( I , . if and only if each 5' is. The only difference between < I , . . . ,4" and the components of tS is that one is defined on U , the other on $(U). 0

39

40

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

EXAMPLE 4.2 (a) Let X be open in R". Since we identify vectors at points of R" with points in R" we see that a Cm-vectorfield on X corresponds simply to a C"-map 5: X -+ R". (b) Consider S2. We identify Tx,S2with the set of all v E R 3 with xo I v. Thus a Cm-vector field on S2 is simply a C"-map 5:s' -+ R 3 with t(x) Ix, Vx E S2 (check that the smoothness part of Definition 4.1 will hold). A specific example is

t(x, y, 4 = (xz,yz, z 2 - 1). DEFINITION 4.3 An integral curue of ( with initial condition xo E X is a Cm-curve c: I + X , I an open interval about 0, with c(0) = xo and for t E I , c'(t) = t(c(t)).

REMARK 4.4 The condition that c be an integral curve of 5, when expressed in terms of local coordinates, is the condition that a certain autonomous first-order system of differential equations should hold. We proceed to show this.

Assume c'(t) = <(c(t)) and let (c', . . . , c"), ( ( I , . . . , t")be the local representative of c, 5. Then (i.'(t),

. . . ,i."(t)) = (tycyt), . . . , c"(t)), . . . , t"(c'(t),. . . , c"(t)))

so (c', . . . , c") is a solution of the autonomous first-order system

i.'

1.

= <1(c', . . . , c")

(4.1)

c'(0) = xb.

i."

= ("(c',

. . . ,c")

We wish to investigate existence and uniqueness of integral curves. We first prove some basic local results about systems of differential equations. Then we use these results to obtain global results on manifolds.

LOCAL EXISTENCE AND UNIQUENESS THEORY Let E be a finite-dimensional vector space with norm

11 11.

Suppose that

f:[a, b] -+ E is continuous. DEFINITION 4.5

[a, b] + R. Define

Choose a basis e , , . . . , e n for E . Let f=f'ej, = (j: f j ) e j .

s: f

f j :

41

LOCAL EXISTENCE AND UNIQUENESS THEORY

Thus, to integrate a vector-valued function, we pick a basis, represent f via components for the chosen basis and integrate componentwise. It is a simple exercise to see that the result does not depend on the basis we chose.

PROOF: For each integer m let

cJ=l

where tk = a + (k/m)(b- a) and S, = S i e j . From elementary calculus we know that lim,+m S i = je f j . Thus lim,,,+mS , = j: f . But

DEFINITION 4.7 Let f : J x U -+ E be C p where J is an open interval about 0, U open in E . We call f'a CP-time-dependentvector field on U .

If x, E U , an integral curve off at xo is a map a: 1 0 E I c J s.t. (a) 4 0 ) = x, (b) a'(t) = f ( t , a(t))for t

E

+

U , I an open interval,

I.

DEFINITION 4.8 Let f : J x U A local ,pow for f at x, is a map

4: I"

--f

E be a C p time-dependent vector field.

x U"

+

u

such that (a) (b) (c) tegral

I , c J is an open interval about 0, U , c U is an open set containing xo, if x E U o and 4x:I, + U is defined by 4,(t) curve off at x.

=

&(t,x), then

4xis an in-

DEFINITION 4.9 f : J x U + E satisfies a Lipschitz condition on U uniformly on J if there is a K > 0 such that Ilf(t7x) - f ( t , y)ll

I Kllx - yll

for

x, y

E

U, t

EJ.

42

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

LEMMA 4.10 Suppose f :J x U + E is such that D,f:J x U -+ L(E, E ) is continuous. Then, for each xo E U , there is an open neighborhood U , of x, and an open interval J , about 0 such that f :Jo x U o -+ E is Lipschitz on U , uniformly on J,.

The proof is left as an exercise. The following proposition makes use of the contraction mapping principle (see [26]). Suppose that 11 11 is a norm on E and J is an open interval in R. Let B(J, E ) be the space of bounded continuous functions from J into E; B(J, E ) becomes a complete metric space with the metric d(f7 9) = SUP Ilf(t)- s(t)II. tsJ

Let Jb = (- b, b) for b > 0, B(x, a) the open ball of radius a about x and B(x, a) the closed ball of radius a about x. Let U be open in E , J an open interval about 0, U , 0 < a < 1, such that B(xo, 2a) c U . Assume f : J x U -+ E is continuous and satisfies PROPOSITION 4.1 1

x,

E

(a) If(t, x)( I L for t E J , x E U , where L 2 1. (b) f is Lipschitz on U uniformly on J with Lipschitz constant K 2 1. Then, if 0 < b < a / L K , f has a unique local flow a: J b x B ( x 0 , a)

u.

PROOF: The theorem asserts that for x E B(x,, a), there is a unique integral curve o f f , defined on J b , with initial condition x. Let C be the set of all continuous maps c: J , B(x,, 2 4 , c(0) = x. -+

i f B ( J b , E ) is the space of bounded continuous functions J b + E with the sup norm, then C is a closed subset so C is a complete metric space. Define S : C + C by

43

LOCAL EXISTENCE AND UNIQUENESS THEORY

But bK < a / L < 1 so S is a contraction map. Hence, there is a unique continuous map c: Jb+ B ( x o , 2a) such that c(t) = x

+ !O' f ( s , c(s)) ds

for t E J , .

Suppose C: J b + U is an integral curve off at x. We claim t(Jb) c B(xo, 2a), in which case the uniqueness of the fixed point of S implies P = c. But P(t) - x = f ( s , P(s)) ds, which implies

yo

so IlC(t)

-

xo(I I (IC(t)-

This proves the proposition.

+ IIx - xoll < a + a = 2a.

I

DEFINITION 4.12 Given f : J x U -+ E , E > 0, I c J open intervals, we say that a C'-map 4: I -, U is an &-approximatesolution if, for all t in I, we have ((4'(t)- f ( t >4(Nl(5 8. LEMMA 4.13 (Gronwall's inequality) Let j , g: [a, b) -+ R be continuous and nonnegative. Suppose A 2 0 and, for t E [a, b), we have f ( t ) I A f ( s ) g ( s )ds. Then f ( t ) I A exp[Ji y(s) ds] for t E [a, b). In particular, if A = 0, then f ( t ) = 0 for t E [a, b).

+

1;

PROOF: First do the case A > 0. Let h(t) = A + j: f(s)g(s)ds. Then h(t) > 0 and h'(t) = f ( t ) g ( t )I h(t)g(t). Thus h'/h 5 g so In h(t) - In A I j: g(s) ds. Therefore, h(t) I A exp[ji g(s) ds] so, since f ( t ) I h(t), we have the lemma for A > 0. In case A = 0 we have

f ( t )I

+ j'f ( s ) g ( s )ds

so f ( t ) I ~exp[f, g(s) ds] for all

E

for

E

> 0,

> 0 and hence, f ( t ) I 0 as desired. I

PROPOSITION 4.14 Let f : J x U -+ E be Lipschitz on U uniformly on J with Lipschitz constant K . For i = 1 , 2 let ci > 0 and let 4i:l-+ U be an E,-approximate solution. If t, to E I and E = E~ + e 2 , then

Il4l(t)- 42(t)lJ 5

Il4Ato) - 42(to)ll

eKlf-*ol+ ( E / W eK'f-zo'.

PROOF: First consider the case t 2 t o . We have

IldXt) - f ( t , 4i(t))llI

Ei,

t

E 1,

i = 1,2.

44

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

By Gronwall's inequality,

K This is the desired result in case t 2 t o . For the case t I t o , define the functions -

f ( - J ) x U-tE,

&, 32:- I

f(s,x)=

-

f(-s,x),

u,

3i(s) = cPi( -s). Then, f is uniformly Lipschitz with constant K and 6, is an &,-approximate solution. Now - t o I - t and each lies in - I , so J J i l ( - t ) - 82(-t)ll

+

I (I181(-to) - 4 4 - t o > ( l

+ (E/W

eK(lo-')

or

(I41(t)-

+

I (I141(to) - 42(t0)(1 ( E / K eKlt-tol. )

I

PROPOSITION 4.1 5 Let f : J x U + E be continuous, with Ilf(t, x)ll s L on J x U and let f satisfy a Lipschitz condition, uniformly on J , with con-

45

LOCAL EXISTENCE AND UNIQUENESS THEORY

stant K . Let x, E U . Then there is an open interval J , about 0, J , c I , and an open set U,, x, E U , c U , such that (a) f has a unique local flow 4: J , x U , -+ U , (b) q5 is continuous and satisfies a Lipschitz condition. PROOF: By Proposition 4.1 1, J , , U , exist so that (a) holds. Also, if t, s E J,, x, y E U , we claim, for J , small enough, we can prove

( ( 4 kx) - 44%YIII

5 eKllx - Yll

+ Llt - SI.

(4.2)

It is all right to shrink J , , for by Proposition (4.11), (a) will not be affected. So assume J , c ( - 1, 1). Now

Il4(4 x) - 46, Y)ll I Il4k x) - 4 4 Y)ll + Il4k Y ) - 4% Y)(l and II$(t, x) - 4(t,y)ll I IIx - yl(eK by Proposition 4.14. But since llfll I L we have Il$(t, y) - 4(s, y)ll I Llt - .TI, so (4.2) is proved, which proves the proposition. 1 PROPOSITION 4.1 6 Let f : J x U + E be Lipschitz uniformly on J . Let I c J be an open interval, t o E I , I $ ~ 42: , I + U integral curves of f with 41(to) = 4Jto). Then 41 = 4 2 . PROOF: Take E = 0 in Proposition 4.14. I PROPOSITION 4.17 Let f : J x U -+ E be Lipschitz uniformly on J = (a, b). Let 0 E I = (u,, b,) c J and let a: I -+ U be an integral curve such that a has no extension to an integral curve on any larger interval. Assume exis-

tence of (a) E > 0 such that a(b, - c, b,) c U , (b) B > 0 such that I f ( t , a(t))l I B for t E (b, - E, bo). Then bo = b (a similar result holds for left end points). PROOF: Let a(0) = x,. Then for t a(t) = x,

+

E

I we have

1;

f ( s , a(s))ds.

So, for t , , t2 in (b, - E , b,) we get

Hence, lim,,,, a(t) = u exists and, by (a), u E U . Suppose b, < b. We claim there is an open interval I‘ about b, and an integral curve 8:I‘ -+ U s.t. p(b,) = u. For let f ( J - b,) x U + E be defined by y(t,x) = f ( t b,, x). Let I” c J - b, be an open interval about 0, I” -+ U an integral curve of

+

46

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

?with p(0)= u. Then we define (a) I' = I" + b,, (b) PO) = k t - bo). Now, P(b,) = &O) = u and p(t)= F(t - b,) =f(t - b,, F(t - b,)) = f ( t , P(t)). Consider a, P on I' n (b, - E , b,). Each is an integral curve off. P is defined on some interval (b, - E ~ b,, + E J where (b, - E ' , b,) c I' n (b, - E, b,). On this interval define

: :{

bo - ~1 < t < bo, b, I t < b, E ' ;

YO) =

+

limt+bo-a(t) = u = P(b,) so y is continuous. If we show Y(t)= u

+

Jb', f(s, Yb))

ds,

(4.4)

then y is an integral curve off, so we have extended a to a larger interval. On b, - E' < t < b,, (4.4) reads a(t) =

+

oJ:

f(s, ~ ( s ) )ds.

Since a'(s) =f(s, a(s)) on t < s < b, and lims+bo-a(s) = u, (4.4) follows, for t < b,, from the fundamental theorem of calculus. For bo < t < bo e l , (4.4) says P(t) = 2) + yoP'(s) ds, which is true for the same reason. I

+

EXAMPLE 4.18 Define f:(- 1, 1) x R + R by f(t, x) = x2/(t - 1). Consider the initial value problem

dx/dt

= f(t,

x),

x(0) = -2. By separating variables, we find the solution to be x = - l/(lnJt- 11

+ +). + 1) for, as t + 1

This solution is valid only on (- co, -e-'/' - e-1/2from the left, lnlt - 11 + -$ and x + - 03. How does this compare with Proposition 4.17? If a(t, x) satisfies D,a(t, x) = f ( t , a(t, x)), a(0, x) = x,f is C' and a is C', then D2a is a solution of the linear differential equation

444 x) = D2fk a ( t , x ) ) satisfying

qo, x) = 1,.

w,x)

47

LOCAL EXISTENCE A N D UNIQUENESS THEORY

We will prove that a is C’ with respect to x by showing that the solution A(t, x) is D2a. We first prove a theorem about linear differential equations (see also Exercise 4.12). PROPOSITION 4.19 Let J be an open interval about 0, E and F vector spaces, V an open set in F . Let

g : J x V + L(E, E )

be continuous. Then there is a unique A: J x V -+ L(E, E ) such that (a) D,E.(t, x) = g(t, x) 0 A(t, x) for ( t , x) E J x I/ (b) i(0, x) = 1, for x E I/. Furthermore, this A is continuous. PROOF:

Fix x E I/. We show there is a unique A,: J

(a) AX(0) = I,? (b) 23)= g,(t)

0

-+

L(E, E ) such that

A&), where gAt) = g(t, 4.

Define f :J x L(E, E ) -+ L(E. E ) by f ( t , w) = g,(t) w. Let J’ be an open interval contained in J with J’ compact and J’ c J . Then, on J’ x L(E, E), 1 is Lipschitz uniformly on J’, because 0

Ilf(4 4 - f k

411

Ilsx(t)lI

IIU - 011

and g, is bounded on J’ (since J” is compact). If we have A: J’ -+ L ( E , E ) such that A(0)= l,, X ( t ) = f ( t , A(t)), then A is unique by Proposition 4.16. We claim there is a maximal interval J , about 0, J , c J , such that there is a A : J , -+ L(E, E ) which is an integral curve off with A(0) = 1,. For take J , = { t E 513 a J’ about 0 as above and a i defined on J’}.J , is the union of all open intervals J’ about 0 such that J’ is compact, J’ c J and 3:J’ -+ L(E, E ) with A(0)= l,, A‘ = f ( t , A). We claim J , = J . Let J = (a, b),J , = (a,, b,) and suppose b , < b. We show this leads to a contradiction, so b , = b (and a similar argument shows a, = a). Choose finite numbers a,, b, such that a, < a, < 0, b, < b2 < b. Consider f on ( a 2 ,b2) x L(E, E). Proposition 4.17 applies to ] ” : ( a 2b,) , -+ L(E, E). By maximality of J , , A cannot be extended to an integral curve on any larger subinterval of ( a 2 ,b,). Let E > 0 be such that b , - E > 0. If we show (a) W , - E, b,) = W , EL (b) 3B > 0 such that [If@, A(t))ll I B, t

E (b,

- 8, b,),

then, by Proposition 4.17, b , = b 2 , a contradiction. Now (a) is clearly true. Consider (b). Now (If(t, A(t))II I Ilg,(t)ll IIA(t)ll. But g,(t) is continuous on

48

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

[0, b,], so let C > 0 be such that IIgx(t)ll I C for 0 I t I b,. Then note that

i(r)

=

I

+ ji y,(s)

3'

A(s)ds,

t

E

[0, b,),

which gives

IlW)ll 5 1 + c J;

IlA(s)ll ds

So, by Gronwall's inequality, for

Il;l(t)ll I ect I eCb2

t

E

[0, bl).

So, for t E (b, - E , b,), Ilf(t, E,(t))llI CeCb2= B. Thus we have proved J , = J . We have shown that for each x E V 3 a unique A x : J + L(E, E ) with A,(O) = 1, and &(t) = f ( t , A,(t)) for t E J . This defines A: J x V L(E, E ) satisfying (a) and (b) and it remains only to prove continuity of A. Fix ( t o ,x,) E J x V . Let I be a compact subinterval of J having 0 and t o in its interior. Choose C > 0 so that Ill(t, xo)ll I C for t E I . Choose an open neighborhood V , of x, I/, c V , such that there is a K > 0 with

-

for all ( t , x) E I x V,.

Ilg(t, x)ll I K

For

(r, x) E I

11%

x V, we have

x) -

W,, .xo)ll s I l i ( r , x)

-

i ( r , xo)ll + 1(W,xo) - W,, xo)ll.

The term IlL(t, xo) - i(ro, xo)ll can be made small by making t near t o , so we need to see that IlA(t, x) - i ( t , xo)ll can be made small by making t near to and x near x,. We have

JIW(t,xo) - d t , 4

O

A(4 xo)l(

IpAG xo) - y(t, xo) " 4 4 X0)ll + Ils(4xo) 5 Cllg(4 x)

-

-

Ilnk X0)ll

g(4 4 11

s(t,x0)II.

Given E > 0 there is, by a simple compactness argument, an open neighborhood V, of xo, V, c V, such that for ( t , x) E I x V,,

Ils(L x) -

s(t3

X0)ll

5 E/C.

For such (t,x) we see that

-

1p14t,x,)

-

y(t, x)

('

3 4 , xo)(lI E.

So, t A(t, x,) is an &-approximate solution of the differential equation of which the map t + A(t, x) is a solution. Therefore, by Proposition 4.14, there is a constant K , such that

Il44 x) -

w,x0)II I

EK1.

49

LOCAL EXISTENCE AND UNIQUENESS THEORY

We can take eK[lenplh of I ]

K, =

K

We have used the fact that 1(0,x) = 1, COROLLARY 4.20

Let f :E

+E

= A(0, x,).

I

be linear. There is a unique a: R x E

+E

such that (a) a is continuous, (b) Dla(t, x) = f ( a ( t ,x)) for t E R and x E E , (c) a(0, x) = x for x E E, (d) if a,(x) = a(t, x), then ( a J f E Ris a 1-parameter group of linear transformations of E, that is, No =

1,.

a,, t 1 2 = atl "at,.

PROOF: Define g: R x E + L ( E , E ) by g(t, x) = j . Let 1:R x E L(E, E ) be as in Proposition 4.19. Define a(t, x) = A(t, x)x; a is continuous, (b) holds and a(0, x) = A(0, x)x = l,(x) = x. By Proposition 4.16, a is clearly unique if (b) and (c) hold. Only (d) remains. If two integral curves R E agree at one point they are equal everywhere, by Proposition 4.16. We claim d(t, x) does not depend on x. To see this, show A(t, x,)y = d(t, x,)y for all t , xl,x,, y. Let q5i(t)= L(t, xJy, for xi and y fixed. Then, for i = 1,2,4;(0) = i(0, xi)y = y. .X;))Y = (s(t,xi) * 4 4 xJ)y = f ( W ,XJY) = Also, 4Xt) = D , ( W , XJY) = (Dl .f(q5i(t)).Here, as in the verification of (b), one uses the chain rule as follows: Let F : R + L ( E , E ) be defined by F(t)= A(t, x i ) and let G: L(E, E ) + E be defined by G ( A ) = Ay. Then 4 J t ) = G F ( t ) so &(t) = DG(F(t))F'(t)= F'(t)y = D,A(t, xi)y.We conclude 4 , = ( p 2 , so 2(t, x) does not depend on x and hence, a(t, x) = lb(f, 0)x. Then clearly, a, = l ( t , 0) is linear. Finally, we verify the group property. We want a ( t , t,, x) = a(t,, a(t,, x)). Fix t , and show, for all t , q5(t) = a(t t,, x) and $ ( t ) = a(t, a(t,, x)) are equal. Now, (p(0)= a(t,, x) = $(O), so if we show both 4 and $ are integral curves, we are done. Clearly $ is an integral curve and -+

-+

w>

+

+

= D,a(t

so

+ r 2 , x) = f'(a(r+ r2,x)) = f ( W ) ) ,

4 is an integral curve. I

THEOREM 4.21 Let J be an open interval about 0, U open in E , f : J x U -+ E a time dependent CP-vector field, p 2 1. Let x, E U . There is an open subinterval J , containing 0 and an open set U,, x, E U , c U such that

(a) f has a unique local flow a : J , x U , + U , (b) a is Cp,

50

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

(c) For ( t , x) E J o x U , we have D1D244 X) = D J ( t , a(t, x)) D 2 4 t , X) and D2a(0, X) = 1,. PROOF: By Proposition 4.15 we can choose J , , U o s.t. there is a unique flow a: J , x U o + U which is continuous and satisfies a Lipschitz condition. Define 0

9: J 1 x Uo --t L(E, E )

g(t, X) = D2f(t, a(t, x)).

by

By shrinking J , , U , , if necessary, we may assume g is continuous and bounded on J , x U , . Let J , be an open interval about 0 with J, c J1. Corresponding to g:J, x U,

+ L(E, E ) ,

there is a unique 1:J, x U , -+ L(E, E ) for which (a) D , W , x) = g(t, x) (b) l(0, X) = 1,.

0

W,x)

We now claim D2a:J o x U o + L(E, E ) exists and, in fact, D,a(t, x) = l ( t , x) for (t, x) E J , x U , . Since, by Proposition 4.19,lis continuous, we will have shown D2a is continuous. Now D,a(t, x) = f ( t , a(t, x)) on J , x U , so D,a is continuous on J o x U , . Once we show D2cx = A, we conclude (a) c1 is C' on J o x U o and (b) statement (c) of the present theorem holds. We verify the claim as follows:

Fix x E U,. For h small and t E J , we have O(t, h)

defined by

+ h) - a(t, x) and D,e(t, h) = ~ , a ( tx, + h) D,a(t, X) = f ( t , a(t, x + h ) ) O(t, h) = a(t, x

-

-

f ( t , a(t, XI).

Thus we have p1m

h) - g ( 4 x ) W , h)ll

4,x + 4 )- f ( t , a@,XI) - D 2 f ( t , 44 x))% 411 5 IlW?h,ll supllD2f(t9Y) - D2f(4 a(t, X))ll? = Ilf(4

+

where the sup is over all y on the line segment joining a(t, x) and ~ ( tx, h). By the Lipschitz condition satisfied by a, there is a K > 0 s.t. IlO(t, h)ll 5 Kllhll, so

p,O ( 4 h)

-

g(h M t , h)ll 5 KIJhllS U P IID2f(t2Y) - D Z f k Y

46 x))((.

51

LOCAL EXISTENCE AND UNIQUENESS THEORY

As stated above, by Proposition 4.15 we can assume that Ila(t,, x l ) a(t,, xz)ll I Alt, - t,l Bllx, - xzll on J , x U,, where A > 1 and B > 1. For each t E 7, c J , , the continuity of D 2 f at (t, a(t, x)) gives an open interval I , c J , and 6, > 0 so that s E I , and llz - a(t, x)ll < 6, implies that IID,f(x, z ) - Dzf(t, a(t, x))ll < 4 3 k . Let V, = I , n {sl(s - tl < 6J2A). Use and take 6 = min hti/2B. the compactness of J,, to get J, c Now suppose llhll < 6. If t E J,, then s E for some i. Thus

+

uy= V i

Vi

and so that which implies

Hence, and which gives

so that

This shows that, if

then lim $(h) = 0. h+O

Now IlD,O(t, h) - g(t, x)O(t, h)l(Illhl($(h)for t E J,, so O is a (Ihll$(h)-approximate solution of the differential equation satisfied by t -,A(t, x)h, that is, D,(A(t, x)h) = g(t, x ) 0 A(t, x)h. Note that x is fixed here, so the differential equation we are talking about at the moment is y:J, x E

+

E,

y(t, U) = g(t, X ) U .

52

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

Since O(0, h) = h = L(0, x)h, we conclude, by Proposition 4.14, there is a constant C,,such that p(t,

h) -

w,x,q

for

5 C,llhll$(h)

t E Jo.

Note in Proposition 4.14 the constant does not depend on which Eapproximate solutions we are considering, so C, is independent of h. Since limb+, $(h) = 0, we have proved D2a(t,x) = L(t, x), as asserted. We have also shown

4 ) D,a(t,

DID244 x) = D2fk

x)

O

and D2a(0,X)

=

1.,

Now that we have established a is C' and that Eqs. (c) hold, we proceed to establish by induction, that a is actually C p i f f is Cp. Suppose, inductively, that the theorem is true for p - 1 and that f is Cp.Then, given x, we get a J , and a U , and a unique flow a : J , x U , + U such that (c) holds and with a being CP-I.To show a is C p we show D,a and D,cc are C p - ' . By the chain rule, we conclude that D,cr(t, x) = f ( t , a(t, x)) is C p - ' , by the inductive hypothesis. To show D2a is C p - ' on J' x U',J' a subinterval of J , containing 0, U' open, with x, E U' c U , we show D 2 a is "part" of the flow of a C p - ' vector field. Define F: J , x U , x L(E, E ) + E x L(E, E ) by F(t, x, w ) = (0,D 2 f ( t ,

XI)

O

w).

Clearly F is C p -'. Define p: J , x U , x L(E, E ) + E x L(E, E ) by

At, x, w ) = (x, D,a(t,x)

w).

Now we have P(0, x, w ) = (x, D 2 4 0 , X ) w ) = (x, w)

and DIAL, x

W ) = (0, D1D2a(t, X)

= (0, D Z f ( 4 =

0

W)

@, $1

F(t, X, D2a(t, X)

O

0

D,a(t, 4 w ) F(t, p ( t , X, w)). O

W)=

So p is a flow for F , so p is Cp-' on some J' x U' x W', J', U' as above, W' a neighborhood of 1, in L(E, E ) . So, on J' x U', the map (t, x) + 7r2 p(t, x, lE)= D2a(t,x) is C P - ' . I 0

We now translate our results to manifolds.

53

THE GLOBAL FLOW OF A VECTOR FIELD

THE GLOBAL FLOW OF A VECTOR FIELD PROPOSITION 4.22 Let ( be a vector field on X , x o E X . Then there is an interval I about 0 and an integral curve of (, c: I + X , with c(0) = x o . Further, if c^: f + X also satisfies the above conditions then c = t on I n f. PROOF: Choose a chart ( U , 4) about xo and consider the equation dxldt = (&x). The existence results already proved show that this differential equation has a solution c": I -+ &U), E(0)= 4 ( x o ) .Then 4 - l c": I -+ X is an integral curve of ( with 4- c F(0)= xo.The uniqueness results for differential equations imply that, if c, t are integral curves of ( with c(0) = c^(O), then c = c^ on some neighborhood of 0. To prove c = 2 on I n f, let 0

A

=

{r E I

n r^lc(t) = t ( t ) } .

Clearly A is closed in I n f (Here we use the assumption that our manifold X is a Hausdorff space.) If t o E A, let d(t) = c(t + to), J(t) = t(t to). Then d, d^ are integral curves of ( defined near 0 and d(0) = J(O), so d = d^ on some neighborhood of 0. Hence c = t on some neighborhood of t o . Thus A is open so A = I A f. 1

+

Given xo E X , let

9 = { I 11 is an open interval about 0 and there is an integral curve c: I + X with c(0) = xo}. Let I,,

=

UIE9I ; I,,

is an open interval about 0.

PROPOSITION 4.23 There is an integral curve of 4, with initial value xo, defined on I,, and I,, is the largest interval on which such an integral curve exists. PROOF: Define c,,: I,,

X as follows. Given t E I,, choose I E 9 with -+ X with c(0) = xo, so define c,,(t) = c(t). In other words, cAolI= c. By Proposition 4.22, cx0 is well defined and is clearly an integral curve at xo. Also it is clear that I,, is maximal as stated. 1 +

t E 1. Then there is an integral curve c: I

DEFINITION 4.24 We call c,,, the maximal integral curtie at xo and call 1," the maximal interval of existence at xo. REMARKS: (a) If 5 is a vector field on X , . x 0 e X , we sometimes speak of "the integral curve through xO" meaning the maximal integral curve ex": l x , +

x.

(b) In Definition 4.1, the notion of a C"-vector field on a manifold was introduced. It is clear that, by replacing ''00" by a nonnegative integer p ,

54

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

we can define the notion of a CP-vector field. It is equally clear that a vector field is C" if and only if it is C p for all p 2 0.

<

Given a vector field of class Cp, p 2 1, define t E I,}. 9(<) is called the domain of the flow of <. The (global)flow of is the map DEFINITION 4.25

9(<) c R x X by 9(<) = {(t, x)l

<

a : 9 ( < )+ X

where a,: I ,

-+

defined by a(t, x)

X is the maximal integral curve of

= a,(t),

< at x.

<

THEOREM 4.26 If is a CP-vector field, p 2 1, on X , then 9(<) is open in R x X and the flow of <,a : 9 ( < ) + X , i s Cp. PROOF: Fix xo E X . Let I:, c I,, consist of all t E I,, for which there is a b > 0 and an open set U about xo such that (t - b, t b) x U c 9(<) and c1 is CP on this product. is clearly an open subset of I,, and Theorem 4.21 implies that 0 E If we show I:, is closed in I,,, then we conclude ZZ0 = I,,, in which case our theorem will be proved. So let so E I,, lie in the closure of I:,. There is a neighborhood I/ of a(so, xo) and an a > 0 such that there is a unique CP-map p: ( - a, a) x V -+ X with b(0, x) for x E V and s.t. Bx:( - a , a) + X is an integral curve of <. This follows by Theorem 4.21. we can choose t , E I:, such that Since so E

+

so

E(t, -

a, t ,

+ a).

We can find b > 0 and a neighborhood of xo, say U , such that a is C p on ( t , - b, t , + b) x U . Further, if we choose t, close enough to so and b, U small enough, we can arrange for a to map ( t , - b, t , b) x U into I/. For t E (ti - a, t , a), x E U , define

+

+

4@,x) = B(t - t , , @,,XI), so that 4 is a Cp-map and 4(tl,x) = a(t,, x). For fixed x E U , t + 4(t,x) is an integral curve so 4 extends a to (tl - a, t , + a) x U . Since so E (tl - a, t , + a), we have shown so E 1:" as desired. I DEFINITION 4.27

9t(<)

COROLLARY 4.28

5 and t E R let = (x E Xl(4 x) E W ) ) .

For a vector field

9,(<) is open in X for all t E R.

PROOF: Obvious from Theorem 4.26.

as

I

PROPOSIXION 4.29 ~ ~ : 9 +~ X( , given ( ) by a,(x) = a(t, x), is as smooth is; ~ ~ ( 9=~9J<) ( < ) )and ~ - ~ : 9 .-+-5B,(<) ~(< is ) the inverse of a,.

<

55

COMPLETE VECTOR FIELDS

The proof of this and the next two results are left as exercises. PROPOSITION 4.30

If x

E

gn,(5) and a,(x) E gS((), then x E gf+s(() and

a,(ur(x)1 = us + f(X).

COMPLETE VECTOR FIELDS A vector field

DEFINITION 4.31

5 on X

is complete if g(() =R x X.

COROLLARY 4.32 The flow a: R x X + X of a complete C"-vector field is a 1-parameter group of diffeomorphisms. PROPOSITION 4.33 Let 4 be a vector field on X , xo E X , I,, = (a, b) with b < co. If A c X is compact, then there is E > 0 such that, for r E (b - E , b), u,,(t) 4 A . PROOF: Suppose not. Then there is a sequence ( t , ) in I,, converging to b, such that axo(tn)E A for all n. Since A is compact, some subsequence converges. By a change of notation we may suppose uxo(t,)+ y , E A . Now 9(5) is open, so there is a 6 > 0 for which there is a neighborhood U of y o with (-6,6) x U c 9(() Choose . IZ so large that u,,(t,) E U and b - t , < 6. Then f l : ( t , - 6, t, + 6) + X , given by B(t) = a(t - t,, uxo(t,)),is an integral curve with P(t,) = u,,(t,). So P extends ax,,to t, 6 > b, contradicting maximality o f b , b). I

+

COROLLARY 4.34

If

5 has compact

support, then ( is complete.

is a compact set K c X such that 5(x) = 0 if x 4 K . Now choose x E X and show I , = R . If x 4 K , then the maximal integral curve at x is c,(t) = x for t E R, so I , = R in that case. If x E K , then cx(Ix)c K , since integral curves outside K are all constant. But then Proposition 4.33 implies I,= R . PROOF: There

COROLLARY 4.35

A smooth vector field

i on a compact manifold is

complete. PROOF: 5 automatically has compact support. EXAMPLE 4.36

diffeomorphism

I

Let x o , y 0 E R", Ilroll, llyoII both <1. There is a C"-

4:R" + R" such that

(a) 4(x,) = Y o , (b) 4(x) = x if llxll 2 1.

56

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

FIGURE 4.1

To see this let u,

= yo

-

x,.

Choose a C"'-function f : R" .+ R such that

.f(x) = 0 for llxll 2 1 and f ( x ) = 1 for llxll I 6 , where 6 is chosen so that IIxoll, llyoll < 6 < 1 (see Exercise 4.13).

Define a Coo-vectorfield on R" by t(x) = f ( x ) v o (see Fig. 4.1) so that 5 has compact support in B(0; 1). Then 4 has a flow a: R x R" + R" and a,(x) = x for IlxI1 2 1, since t(x) = 0 for such x. Let c(t) = x, tv, for t in [0, 11. Then c'(t) = u, and t ( c ( t ) )= f(c(t))v, = u,, because IIc(t)ll < 6 for 0 I t I 1. Also c( 1) = y o , so if we consider the diffeomorphism a,: R" + R", we conclude that aI(xo)= y o . Therefore, take 4 = a,.

+

REMARK 4.37 Let f': J x U + E be C p , U open in E. Let J , be an open subinterval of J containing 0, U , an open subset of U and suppose there is a flow 01: J , x U , .+ U. Theorem 4.21 shows that, given x E U,, a is C p on J , x U , for some open interval J1containing 0 and some open set U , containing x. We now show that a is C p on all of J , x U , . To see this we use Theorem 4.26 applied to the vector field j? J x U + R x E , f(t, x) = (l,f(t, x)). Let &:53(,f)-+ J x U be the global flow off. 9 ( f )c R x J x U and we claim that, if ( t l ,x,) E J , x U , , then ( t l ,0, x J E ~ ( and ~ ) &(t,, 0, x,) = ( t ,cc(t,, xl)). If so, then clearly a is C p . Fix x, E U,. Let c : J , .+ J x U by ~ ( t=) ( t , a(t, xi)).

40) = (O,a(O, x 1 ) and

= (0, x 1)

57

EXERCISES

So B,,,,,

is defined at least on J , , i.e., ( t l , 0, xl) E 9 ( f )and oi(t,, 0, x,) = asserted. We can carry this further yet. Suppose f : J x V x U + E is Cp, V open in F , U open in E . We can view .f as a tinze-dependent uectorjield depending on parameters in I/. Given u, E V , xo E U , a local flow for f is a map a: J o x V , x U , + U for which a(0, u, x) = x and D l a ( t , u, x) = f ( t , u, ~ ( tu,, x)) for all t E J,, u E V , , and x in U , (where we assume J , is an open interval about 0, V, and U , are open, u , E V, c V , and x, E U , c U ) . We say c( is a local ,pow about (uo, xo).We leave it as an exercise (see Exercise 4.1 1) to show that a is C p . c ( t J = ( t l , a ( t , , xl)) as

EXERClS ES 4.1 Show for f : U

F C', U open in E , x, E U we have

-+

Ilf(x) - .1'(-%) - Df(xo)(x - xo)l( =

lim 4.2

IIx - XOII

-0

Ilx-XOII

o.

1: f does not depend on choice of basis.

4.3 (a) I f f : J x U

+

E is C" and

c(

is an integral curve o f f ) then

c(

is

c p + I.

(b) Let a: I -+ U be continuous, 0 E I c J . Then a is an integral curve at .xo if and only if, for t E I ,

4.4 Prove Lemma 4.10. 4.5 Let f :R 2 + R 2 be given by f(-x, Y) = (2x

-

y , 3y).

(a) Find the flow off, a : R x R Z + R2. (b) Define 2: R + L(R2,R 2 )by &(t)u = a(t, u) for t E R, u E R 2 . Show that 2(t s) = &(t)&(s)for t , s E R and show &'(O) = f (regarding ,f as a matrix).

+

4.6 Let t ,

E

I,, and let c,,(t,) I,,

=

I,,

-

t,

4.7 Prove Proposition 4.29. 4.8 Prove Proposition 4.30. 4.9 Prove Corollary 4.32.

=

y o . Then

and

cJt)

= c,,,(t

+ to).

58

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

4.10 Let X be a connected C" - manifold. Let Diff"(X) be the group of C"-diffeomorphisms of X onto X . Clearly Diffm(X)acts on X as a transformation group. Use the result of Example 4.36 to show that Diffm(X)acts transitively on X, that is, given x and y in X there is a diffeomorphism f such that f(x) = y. Would this be true if X were not connected? If m 2 2, is Diff"(x) rn-transitive, that is, given distinct points x l , . . . ,x, in X and distinct points y,, . . . , y, in X, does there exist f E Diff"(X) such that f(xi)= yi for i = 1, . . . ,m? 4.11 Refer for notation to the discussion in the last paragraph of this chapter. Our object is to prove c1 is C p . Define f J x V x U + R x F x E by f(t, u, x) = (l,O,f(t, q x ) ) . Let 4 : 9 ( J ) - + J x V x U be the flow o f f . Show: If ( t , , u , , x , ) E J , x V, x U,, then ( t 1 , 0 , u l , x l ) ~ 9 ( f ) and

a@,, 0, u1, x1) = ( t l , V 1 r a(t1, U l , Xl)). 4.12 Modify the proof of Proposition 4.19 to show the following: Given J c R an open interval containing 0, V a topological space, and continuous mappings G : J x V-+L(F,F), 4: V + F . Then there is a unique I : J x V -+ F so that (a) D,d(t, x) = G(t, x)(A(t,x))V(t, x) E J x V (b) I(0, X) = ~ ( x ) V XE V . Furthermore, this

A is continuous.

4.13 This exercise results in construction of a C" function f :R" -,R such that f(x) = 0 for 1x1 2 1 and f ( x ) = 1 for 1x1 5 6 , where 0 < 6 < 1 is given.

(a) Let

Show, by induction, that h: R -+ R is C". (b) Let g: R -+ R be defined by g(x) = h(x

Show g is C", g (- 1, 1).

=0

+ l)h(l

-

x).

on ( - G O , - 1 3 and on [l,

GO), and

g > 0 on

'

(c) Let A = j k g(x) dx and define $(x) = A - s" g(t) dt. Show 4 = 0on(-~0,-1],0<~<1on(-1,1)and~=1on[1,co). (d) Given 6 with 0 < 6 < 1, define f :R" -, R by f(x) = 4(a1xI2 b). Show that a, b can be chosen so that f has the properties stated at the beginning of this exercise.

+

59

EXERCISES

4.14 Denote by GL(n, R) the group of all nonsingular n x n real matrices. As usual we may regard GL(n, R ) as consisting of matrices or of invertible linear transformations of R". Let L(R",R") be the vector space of all n x n real matrices or, alternatively, all linear transformations of R". (a) Show GL(n, R ) is open in L(R", R"). (Hint: Consider det: L(R", R") -+ R.) (b) A smooth 1-parameter subgroup of GL(n, R ) is a smooth map y: R -+ GL(n, R) such that y(s t ) = y(s)A(t) for all s, t E R . Given f E L(R",R"), show that its flow CI: R x R" -+ R" defines a smooth 1-parameter subgroup 2: R 4 GL(n, R ) such that %(O) = f. Suppose y is a smooth I-parameter subgroup of GL(n, R ) such (c) that y' (0) = f E L(R",R").We want to conclude only one such y exists. To do this define a(t, x) = y(t)x and show CI is the flow of the vector field f:R" -+ R". (d) Conclude that y + y' (0) defines a one-to-one correspondence between the set of smooth 1-parameter subgroups of GL(n, R ) and L(R",R").

+

4.15 Given A E L(R", R") can we give an explicit description of the corresponding 1-parameter subgroup of GL(n, R)? a:, where A is the Define a norm on L(R",R") by IIA11' = matrix (aij). Show that this is indeed a norm and that IIAB(I I IlAll IIBII. If is a sequence in L(R", R") and A E L(R",R"),then the series C:= A, converges to A , written I = A, if IimN+mIICt= A, - ,411 = 0. The series converges absolutely if Ckm,l IlAkll converges. Show that, if a series converges absolutely, then it converges to some element of L(R", R"). If A E L(R",R") define exp(A) by exp(A) = (Ak/k!).Show that the series is absolutely convergent. We are aiming at a proof that the 1-parameter subgroup of GL(n, R ) corresponding to A E L(R", R") is given by y ( t ) = exp(tA). Bk = B, with both series Show that if C:==lA, = A and ck = AB, where ck = being absolutely convergent, then

c?=,,

AjBkPj. Show that if A, B

E

L(R",R") satisfy A B

exp(A Conclude exp((t

+ $4)

=

BA, then

+ B ) = exp(A) exp(B).

= exp(tA) exp(sA).

60

4. DIFFERENTIAL EQUATIONS ON MANIFOLDS

Show (d/dt)exp(tA) = exp(tA)A. [Hint: k-'(exp((t + h)A) exp(tA)) = k-'(exp(kA) - Z)exp(tA) because of (f).] Conclude y ( t ) = exp(tA) defines the smooth 1-parameter subgroup of GL(n, R ) with y'(0) = A . (h) Show det(exp(A)) = exp(trace(A)). (i) Let A be skew-symmetric, i.e., A' = - A . Show that exp(A) is an orthogonal matrix.

(8)

5 The Tangent and Cotangent Bundles

THE TOPOLOGY AND MANIFOLD STRUCTURE OF THE TANGENT BUNDLE First, we define the tangent bundle of X as the set

TX

=

u

TxX,

x e x

together with the topology and differential structure which we will now describe. Given a chart ( U , 4) on X , we have, for x E U , the linear map

d x 4 : T,X

-+

R".

Let T : T X X be canonical projection, i.e., T ( U ) Now, for a chart as above, we define -+

T$:z-'(U)

+

+ ( U ) x R"

T4(v)= (4(x),d,d(u))

=x

if u E T,X.

by for

u E T,X.

(5.1)

It is clear that T 4 is a bijection. DEFINITION 5.1 The standurd topology on T X is defined by requiring W c T X to be open if and only if, for each chart ( U , 4) on X , Td(W n T - ' ( U ) )is open in R" x R". PROPOSITION 5.2 If ( U , $ ) is a chart on X , then T - ' ( U ) is open in T X and T4:T - '( U ) + 4(U ) x R" is a homeomorphism. PROOF: We first show that, for ( U , d), ( V , $) charts, T$(t-'(U n V ) is

open and

T@ (T$)-': T $ ( t - ' ( U n V ) )4 T ~ ( T - ' ( U n V)) 0

is a homeomorphism. 61

62

5. TANGENT AND COTANGENT BUNDLES

First, T$(z-'(U n V ) ) = $(U n V ) x R", which is open. Also,

T 4 ( T $ ) - ' ( x ' , . . . ,x", u',. . . , u") 0

, . . . ) x " ) , dY' -(x)d

,...)

az3

azj

where 4 = ( y ' , . . . ,y"), z = (z', . . . ,z"). This is clearly a continuous map; indeed it is C". The inverse is clearly also C". Now, to show z - ' ( U ) is open we show T$(z-'(U) n z - ' ( V ) ) is open. But this set is T$(z-'(U n V ) ) = $(U n V ) x R", which is clearly open. If W is open in z - ' ( U ) , then, by definition of the topology, Td(W) is open so T 4 is an open map. If Z c 4 ( U ) x R" is open, we must show T$(z-'(V)n ( T 4 ) - ' ( Z ) )is open. But this set is equal to T $ (T+)-l(Z n 4 ( U n V ) x R") 0

so it is open, since T $ (T4)-' is a homeomorphism. 0

We have noticed during the above proof that T$ 0 (T4)-' is C". Then {(z-'(U), T4)I(U,4) is a chart on X }

is a C"-atlas. Charts in this atlas are called natural charts on T X , and they define the standard diferential structure on TX. Thus, TX is a C"-manifold of dimension 2n once we show it has the required topological properties, that is, TX is Hausdorff and second-countable. We leave this as an exercise. Given a C"-vector field and a chart ( U , 4 = ( x ' , . . . ,x")) on X we defined, in Definition 4.1, the local representative t4:q5(U) + R" by t4(z) = d&(<(x)) if x = 4- '(z). This means the following diagram commutes:

<

From this we conclude that the smoothness condition for vector fields is exactly the condition that < : X + TX

be a C"-map.

63

TOPOLOGY AND MANIFOLD STRUCTURE

So a Cm-vector field is characterized as a Cw-cross section of the tangent bundle. LEMMA 5.3

T:

TX

-+

X is C'"

PROOF: Let ( U , 4) be a chart on X and use the resulting natural chart ( T - ' ( U ) , T 4 ) on T X . Then

4 which proves

0

T

T

0

(Td)-'(X', . . . , X",

is C".

Z',

. . . ,Z") = ( X I , . . . , X"),

I

For x E X , T,X is an n-dimensional vector space so T,X z R" and one might think T X would be topologically the same as X x R". This condition is not generally true. Locally, however, it is true. Let ( U , 4) be a chart on X . Then the following diagram commutes:

If there exists a C"-diffeomorphism

h: T X

-+

X x R"

such that n 1 h = T and such that, for each x E X , hlTx,: T,X -+ {x} x R" is a linear isomorphism, then we say X has trivial tangent bundle. In general, no such h exists. But locally there are such maps; we see 0

h

=

(4-l x

1) T & r - ' ( U ) J

-+ U

x R"

satisfies the conditions. So generally, although TX is not trivial, it is locally trivial. Let p : E -+ B be a C"-map between two manifolds. Suppose for b E B the set p-'(b), called the j b e r over b, is an n-dimensional real vector space. Suppose for each b E B there is an open set U about b and a C"-map h : p - l ( U ) -+ U x R", such that (a) 711 h = P, (b) for b E B hlP-,(b): p - l ( b ) + {b} x R" is a linear isomorphism. Then we say p : E + B is an n-dimensional, Cm-vector bundle over B. Thus T : T X + X is an n-dimensional vector bundle over X (where n = dim(X)). Note that, in general, n need not be equal to the dimension of B. The notion of vector bundle is fundamental in the development of manifolds and differential geometry. The map nl:X x R" X is a vector bundle, --f

64

5. TANGENT AND COTANGENT BUNDLES

a rather uninteresting one, called the trivial vector bundle. A vector-valued function f : X + R" can be viewed as a cross section of the trivial bundle viaf(x) = (x, f ( x ) ) . In a similar way, a vector field 5 on X is a cross section of the tangent bundle. Since the tangent bundle may not be trivial it may not be possible to view 5 as the "graph" of a vector-valued function globally, but this can be done locally over chart domains. DEFINITION 5.4 A C"-map f : X + Y is an imbedding if,

(a) f i s 1-1, (b) T,f': T,X + Tf,,,Y is 1-1 for x E X , (c) f is a homeomorphism of X onto f ( X ) . PROPOSITION 5.5 Let f : X + Y be an imbedding. Then f ( X ) is a submanifold of Y and f :X + f ( X ) is a diffeomorphism.

PROOF: Pick xo E X and let f ( x o ) = y o . Choose charts (U1,4'), (Vl, t+h1) about xo,yo such that f ( U , ) c V , . Let f ; = t,h1 of. 4;':4,(Ut)+$,(V1). Let dim X = n and dim Y = n r. Now D f l ( 4 , ( x 0 ) ) :R" -+ R"" is injective so by Proposition 3.6 there is an open set W , c IcI1(V1) and a diffeomorphism 8,: W , + O,(W,)c R"+*such that, in a neighborhood N , of 4'(x0), 8, ofl(z', . . . , z") = (z', . . . , z", 0,. . . ,O). Let P" = $;'(w,), U = 4;'(ivl). Let 4: U --t 4 ( U ) c R" be given by 4 = 4 t I N , , $: V' -+ $ ( V )c R"" by $ = 0, o $,. Then of 4 - ' : $ ( U ) + $(V') is given by ( z ' , . . . , z") + (z', . . . , z", 0,. . . ,O). Now $ ( f ( U ) ) = 4 ( U ) x 0 c R" x 0. Since f ( U ) is open in f ( X ) , we can shrink V', if necessary, to get f(U) = f ( X ) n V'. Let W be open in R"+'such that W n R" x 0 = $ ( f ( U ) ) .Let V = $-'($(V') n W). Then yo E V n f ( X ) and we claim

+

0

$( V n f ( X ) )= $ ( V ) n R" x 0.

To show this note V c V' so $(V n f ( X ) ) c $(V' n f ( X ) )= $ ( f ( U ) ) c $(V) n R" x 0. For the reverse inclusion note that z E $ ( V ) n R" x 0 so z E $(I") n W n R" x 0 and hence z E $(V)n $ ( f ( U ) ) . Thus, z E $(V 'n f ( U ) )= $(V' n S m )so

z

E

$W n f ( X ) )n $ ( V ) = $(V'

f-l

f W ) n V ) = $(V

f(X)).

f-l

Thus, ( V , $) is a submanifold chart for f ( X ) at y o . So we get a chart for f ( X ) , namely (V nf ( X ) ,$), $ the first n components of $. Now ( U , 4) is a chart on X , V n f ( X ) = f ( U ) , ( f ( U ) ,$) is a chart on f ( X ) and

$ f 4- ' ( z )= z, 0

0

so f is a diffeomorphism of X onto f ( X ) .

I

65

TOPOLOGY AND MANIFOLD STRUCTURE

PROPOSITION 5.6

Iff: X

+

Y is an imbedding, then so is Tf: T X

-+

TY.

PROOF: There exist charts ( U , +), ( V ,$) about xo, y o , f ( x o ) = y o such that f ( U ) c Vand such that for ( z ' , . . . , z") E ( U ) ,$ 0 f 0 4 - ' ( z ' , . . . ,z") = (z', . . . , z", 0,. . . , 0) and f ( U ) = f ( X ) n V . Then T$ Tf (T$)-':+(U) x R" -+ $( V ) x R"+*is given by (z', . . . ,z", w l , . . . , w") + (z', . . . , z", 0,. . . , 0, w l , . . . , w", 0,. . . , 0). It follows that Tj. maps TX homeomorphically onto T f ( T X ) .The form of Tfin the charts given above shows T,(Tf) is injective V u E T X . Thus Tf is an imbedding as asserted. I 0

0

Thus, if X is a submanifold of Y, then T X may be regarded as a subset of T Y . In particular, if X is a submanifold of R", then T X may be regarded as a subset of X x R", namely, TX

{(x, v) E X x R"lu is tangent to X at x}.

The preceding results show that, if TX is represented as a subset of X x R", the correct topology and differential structure result. EXAMPLE 5.7

TS' is represented as a subset of R 2 x R 2 = C x @: TS'

z:TS'

=

{ ( z , w)E S' x C I Re(Zw) = 0},

+ S'

is given by

z(z, w ) = z.

Define h: TS' + S' x R by h(z, w ) = ( z , wZ/i); h is a C"-map, h-'(z, I+) = ( z , izl). For fixed z, h maps T,S' + R linearly and the following commutes: TS'-S'

h

x R

so S' has trivial tangent bundle. Thus, there exist nowhere vanishing vector fields on S', since any C" f : S' + R defines a C" section of n,:S' x R + S' and hence, via the isomorphisms, a C"-vector field. EXAMPLE 5.8 Let TS2 = ( ( 2 , w ) E S2 x R 3 1 z I w } with z: T S 2 + S 2 given by z(z, w) = z. Let f :S2 + R 3 be C" with f ( z ) Iz V z E S2. Then 5: S2 -+ TS2, given by r(z) = (z, f ( z ) ) ,is a Ca'-vector field. It is not possible to trivialize TS2 since, if we could, then we could also construct vector fields on S2 having no zeros. But it is a well-known fact that every continuous vector field on S2 has a zero (see [22]; also Exercise 10.8).

66

5. TANGENT AND COTANGENT BUNDLES

THE COTANGENT SPACE AND THE COTANGENT BUNDLE x

Given a differential n-manifold, we have defined the tangent space at X , denoted T,X. Now we define

E

DEFINITION 5.9 The cotangent space of X at x is defined to be the dual of T,X and is denoted T Z X . Thus,

T,*X

= L(T,X,

R).

DEFINITION 5.10 The cotangent bundle T * X of X is defined to be T*X = T,*X, together with the differential structure described below. We have a natural projection

uxsx

z*: T * X +

x

given by z*(M,) = x

if a,€ T Z X .

Let x E X, ( U , 4) a chart about x. We want to show that we can set up, using this chart, a linear isomorphism (d,+)*: T,*X + (R")*. Consider a linear transformation I : E + F. We can define a linear transformation A*: F* -+E* by 1*(a) = a A. In general, 1:E + F does not give rise to any map E* + F* but, of course, if 1 is an isomorphism, we can define A*: E* + F* by A* = ( I - ')*. Since we do have a linear isomorphism dx4:T,X 4R", we thus have a linear isomorphism 0

(d&)*: T,*X

-+

(R")*.

We have seen in Proposition 3.38 that if 4 = (XI, . . . ,x"), then (dx', . . . ,dx") gives a basis for T,*X which is the dual basis corresponding to (a/ax', . . . , a/dx"). Let (el,. . . , e n ) be the standard basis for R", ( e l , . . . ,e") the dual basis. Then [Definition 3.251

(d,4)(8/8xi)= e,, Given a

E

(d,4),(dxi) = e'.

T,*X, let a = aidx'. Then (d,+),a

(5.2)

= ate'.

DEFINITION 5.1 1 a l , . . . ,a, are called the components of a for the coordinate system ( U , 4). CLASSICAL INTERPRETATION: We see that, given a chart ( U , 4) at x we have, for a E T:X, an n-tuple of numbers (a1,. . . , a,,). Suppose ( U , 4 = (x', . . . ,x"))

67

COTANGENT SPACE AND COTANGENT BUNDLE

and ( V , $

= (y',

. . . , y")) are charts at x and

(dx+)*(~~) =

. . ., an),

(dx$)*(~)

= ( P I , . . .,

PJ

Here we should really write (dX+)*(xx) = Miei but we use the convention that a point in (R")*is written as a row of its components for the basis e l , . . . , e". We want to compute (dX$)*(dx4); '(aiei); [(dxlCI)*

0

(dx4); '(@iei)]ej= [(dx4), '(Uie') d x K ' ] e j 0

=

[aiei d x 4 0 d X $ - ' ] e j 0

. axk

= ctie'[@

aXk

e k ] = cxi

ax'

el(ek)= ai -. dY' aYJ

So (dX+)*(dx4); ' ( q e ' ) = ai(dxi/ayj)ej.Since (dX$)*(ax)= Bjei we conclude

pj = (dx'/dyj)ai

(5.3)

(couariant transformation law). Classically, one spoke of a couariant vector at x as consisting of an assignment, to each chart about x , of an n-tuple such that the transformation law (5.3) holds for each pair of charts. Now, in a manner analogous to what we did for the tangent bundle, we make T*X into a C"-manifold of dimension 2n, n = dim X . The projection z*: T * X -+ X will be shown to be C". As before, a local triviality condition will be shown, so that z*: T*X X is another n-dimensional vector bundle over X . Given a chart ( U ,4) on X and given x E U , we have (dx4)*:T,*X -+ (R")*. We thus have a map -+

(T+),:z*-' ( U )-+ 4 ( U ) x (R")*,

(5.4)

defined by

(T4)*(4= (4 for

CI E

.t*(a), (dx4)*(4)

TZX.

Clearly ( T 4 ) , is a bijection. If r: (R")*-+ R" is the isomorphism e' --t e i , then we get a bijection (@)*:z * - ' ( U ) -,4 ( U ) x R" by (@), = (1 x r) (Td)*. 0

DEFINITION 5.12 The standard topology on T * X is defined by the requirements that W be open in T*X if and only if, for each chart ( U , 4) on X , the set (%)*(W n z * - ' ( U ) ) is open in 4 ( U ) x R". PROPOSITION 5.13 With respect to the standard topology on T * X , if ( U , 4) is a chart on X , then z * - ' ( U ) is open and (%)*: z*-l(U) -+ 4 ( U ) x R" is a homeomorphism.

68

5. TANGENT AND COTANGENT BUNDLES

The proof of this we leave as an exercise. See the proof of Proposition 5.2. In analogy to what was done for T X , we let the standard diflerential sturcture on T * X be defined by the atlas

{(z*-'(U), (%)*)l(U,

4) is a chart on X } .

(5.5)

The charts of the atlas in (5.5) are called natural charts for T*X PROPOSITION 5.14

z*: T * X + X is C".

PROOF: Given a chart ( U , 4) on X and the corresponding natural chart on T*X we must show that the composite

4 is C". But

4

[ J

T*

which is clearly C".

-

(T$)*l(xl,.

0

T*

0

(%);I

. . , x", u l , . . . , u") = (XI,. . . , x")

I

DEFINITION 5.1 5 A diferential l-forrn on X is a C"-map w : X + T * X such that T* 0 w = 1,. According to this definition, w ( x ) E TZX for each x E X . If w is a 1-form and ( U , 4) is a chart on X then we can define the local representative of w,

a+:4(U)

+

(R")*,

(5.6)

= Z. by w&z) = (dX4)*(4x)), where Thus w+ is the map making the following diagram commute.

T* - '( U )

=cp(

U ) x (R")*

I

(l.W+)

+

>4(U)

THE CANONICAL 1 - F O R M ON T*X Let ( U , 4 = ( q l , . . . ,q")) be a chart on X . If w E T Z X , where x is in U then we can write w = pi&'. Thus we have a map z*-'(U)

+ &U)

x R"

defined by

and this map is precisely (%)*.

w + (q',. . . ,q", p i , .

. . , p,) (5.7)

69

THE CANONICAL 1 -FORM ON T ' X

Now T*X is a manifold, so we may speak of differential 1-forms on T * X . Such a form then, is a map a: T*X + T*(T*X)such that ~ ( wE) T:(T*X)

for each w E T* X .

Let x E X , w E T,*X, and u E T,,(T*X). We have we have

T*:

T*X

-+

X and hence

T,T*: To,(T * X )+ T,X. Define O(o):T,( T * X ) -+ R by

8(0)(4 = 47-J*(41;

(5.8)

0 is a 1-form on T * X , called the cunonicul l-forrn. Note that there is no analogous construction on T X ; 8 represents a very special feature which T*X has. To check 0 is C", use a natural chart. PROPOSITION 5.1 6 P1>.

8 = p i dq', in the coordinate system

( 4 l , . . . , 4",

. > PJ '

PROOF: We will prove this by showing that if o has coordinates (q', p i ) ,

then 0(o)(d/84')= pi 0 ( o ) ( d / 8 p , )= 0

l
In terms of the chart (4i,pi) on T*X and the chart (4') on X , the map given by s*(qi,p j ) = (qi).

Thus, T,T* has matrix 1 0

)"'

0 . . . 0. ... 0 0 ... 0 . . ... 0

.

...

.

1 0 ... 0

so T,T*(d/dqi)

= d/d4i E

T,T*(d/dpi)

=0E

T,X

and

If o has coordinates ($, pi), then &x)

T,x.

= (4') and

o = p i d q i . Thus,

e(o)(a/ay') = w(d/dq') = pi

T*

is

5. TANGENT AND COTANGENT BUNDLES

70 and

o(w)(a/ap')= 0. REMARK: This canonical form 8 will appear as a crucial element in the later development of Hamiltonian mechanics.

EXER CIS ES 5.1 Show that Definition 5.1 defines a topology. 5.2

Show that TX and T*X with the topologies which we defined are Hausdorff and second countable.

5.3 Show that Definition 5.12 defines a topology. 5.4

Prove Proposition 5.13.

5.5

Show that the collection of charts given by (5.5) is a C"-atlas on T * X .

5.6

If w is a differential 1-form on X , and ( U , 4), 4 = (ul,. . . , u") is a chart, then we have unique functions w l , . . . ,w, so that w = wi du' on U . Show that the condition that w be C" is equivalent to the condition that wl,. . . , w, be C" for each chart.

5.7

(Linear map bundle) Let p: E -+ B and q: F dimension rn and n, respectively. Define

U E , F) =

u

-,B be vector bundles of

L(Ex,Fx).

X C B

Define n:L(E, F ) -+ B in the obvious way. (a) Show that for each b in B there is an open set U containing b and diffeomorphisms 4: p-'(U) -+ U x R", $: q - ' ( U ) U x R" with properties as required in the definition of vector bundle given in the text. (b) Given U as above, define -+

&n-'(U) by 8 ( I ) = (x, $x

o

I 4; 0

-+

l)

U x L(R", R") if n ( I )= x.

Since L(R'", R") R"", such 0 give L(E, F ) the structure of a vector bundle. Define the topology and differential structure on L(E, F ) by requiring each map 0, constructed as above be a diffeomorphism. Show that this determines a unique topology, which is second countable and Hausdorff, and a unique differential structure. Show that n:L(E, F ) B is then a smooth rnn-dimensional vector bundle. -+

71

EXERCISES

5.8

In Exercise 5.7, take 4 : F -+ B to be the trivial 1-dimensional vector bundle, that is, F = B x R and 4(b, t ) = b. Then the linear map bundle n:L(E, F ) + B is called the dual of p : E + B and is denoted by p*: E* -+ B. Show that the dual of the tangent bundle of M is the cotangent bundle of M .

5.9 (Bundle of bilinear maps) Continue notation as in Exercise 5.7. Define L2(E,F ; R ) =

u

L2(E,, F,; R )

XEB

and define rc: L2(E,F; R ) + B in the obvious way. Given in Exercise (5.7),define [: n- '(V)

+

and t,h as

U x L2(R", R"; R)

by [(a)= (x,a (4, x t,hx)-') if n(a)= x. Use these mappings to define a topology and a differential structure on L2(E,F; R ) so that n: L2(E,F; R ) -+ B becomes a smooth mn-dimensional vector bundle. For further examples of vector bundles constructed from T M and T * M by methods similar to those used above, see Exercise 8.17. 0

6 Covariant 2-Tensors and Metric Structures

We have defined the notion of tangent vector and cotangent vector on a manifold. These objects require only the existence of the differential structure for their existence. However, we have, as yet, no way of defining the length of a vector or of saying when two vectors are orthogonal. In order to define these notions we need an additional structure, a metric structure. This consists of a bilinear form on each tangent space satisfying an appropriate smoothness condition. We first consider some algebraic results about bilinear forms on a vector space.

COVARIANT TENSORS OF DEGREE 2 Let V be an n-dimensional real vector space. DEFINITION 6.1 T 2 ( V )is the set of all bilinear maps V x V + R. These maps are called couariant tensors of degree two on V . Let el, . . . ,en be a basis for V . The numbers

gij = g(ei, e;),

are called the components of g w = w'ej, then

E

i, j

=

1, . . . , n,

T,(V) for the basis el, . . . , en. If u . .

g(u, w ) = g ( u i e i ,w;ej) = u'wjg(ei, ej) = giju'wJ

so g is completely determined by the g i i . Let a,p E V*. Define the tensor product a 0p E T2(V)

by

0B(v, w ) = a(u)B(w).

72

=

u'e,,

73

COVARIANT TENSORS OF DEGREE 2

PROPOSITION 6.2 With notation as above, the elements ei 0eJ, i , j = 1,. . . , n, give a basis for T J V ) and if g E T 2 ( V )has components gij for the basis e , , . . . , en, then (/ = gije' 0d .

PROOF: If g E T 2 ( V )is given by y = ai,$

0ej, then . .

g(ek, ef) = aijei8ej(ek,et) = aijei(ek)ej(e,)= a,&&

= ukf

so xkl = g k p . This proves linear independence. Given g, to show g we let u = ukek, w = weee.Then g(V, w ) = gk(VkWC SO gijei8 &(u, w ) = gijei 8 d ( u k e k w' , e = UkW'yijs;sj

as desired.

L)

=

= gijei 8eJ,

vkwfgijei0e-'(e,, e,)

= gij"iwj,

I

DEFINITION 6.3

g E T 2 ( V ) is a metric if g is symmetric and non-

degenerate. REMARK 6.4

The inverse of the matrix (gij) is denoted

(gij).

LEMMA 6.5 Let g be a metric on V . Then there is a basis e l , . . . , e, for V such that y(ei, ej) = 0 for i # , j and g(ei, ei) = k 1 for each i (such a basis is said to be orthonormal for 9). PROOF: We prove this by induction on n = dim V . For n = 1 the result is clear. Assume the result for n and prove it for the case dim V = n + 1. We first claim there is a vector u1 E V such that g(ul, u I ) # 0. For if g(u, u) = 0 Vu, then for any u, w E V we have

0 = g(u

+ w, u + w ) = y( u , u ) + 2g(u, w ) + g(w, w ) = 2g(u, w).

Thus, g(u, w) = 0 so g = 0, contrary to the nondegeneracy requirement. Let W = { w E V l g ( v , , w ) = 0). W is a linear subspace of V and is of dimension n. This is because iff': V + R is given by f ( u ) = g(u, u l ) then W = f - '(0). But f is not identically 0 since g is nondegenerate. Now g is nondegenerate on W , for if w 1 E Wand g(wI, w ) = 0 Vwl E W , then since y(wl, u l ) = 0, we have g(wl, u ) = 0 Vu E V , contrary to nondegeneracy on V . Then, by induction, there is a basis e , , . . . , en for W , which is orthonormal in the above sense. Since g(ul, u l ) # 0 we may, if needed, multiply by a scalar to arrange g(ul, u l ) = k 1. Then e l , . . . , e,,, 11, is a basis for V and satisfies the requirements of the lemma. I

74

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

THE INDEX OF A METRIC DEFINITION 6.6 The index of a metric g is the number of vectors in an orthonormal basis for which g(ei, ei) = - 1. PROPOSITION 6.7 The index of g is independent of the choice of orthonormal basis and is, in fact, the dimension of the highest dimension subspace of I/ on which g is negative definite. PROOF: Clearly q is negative definite on the trivial subspace (0) so there is a subspace W (not necessarily unique) of largest dimension on which g is negative definite. Let e l , . . . , en be an orthonormal basis, ordered so that g(ei,ei) = - 1 for 1 5 i I k, q(ei,ei) = 1 for k 1 5 i I n. Let E be the subspace spanned by el, . . . ,ek. We claim dim E = dim W , which will prove uiei and observe that our proposition. Let 0 # u E E. Then write u =

+

x

+

k

k

g(u, u) =

g(ei, ej)uiuJ= -

i,j=l

C (ui)’ i= 1

< 0.

So q is negative definite on E and hence, by definition of W , dim E 5 dim W . uiei and set A(u) = Define A: W -,E as follows. Given u E W write v = viei; E, is well defined, linear, and clearly takes values in E . If we show A is 1-1, we can conclude dim W I dim E, as desired. Suppose u E W and A(u) = 0. If we write u = xl=luiei, then ui = 0 for 1 I i I k, so u = cl=k+ viei. Then g(u, u) = x y = k + ( u ~ which )~ is >O, contrary to the fact that g is negative definite on W, unless u = 0. This shows u = 0, as desired. 4

cZ=,

RIEMANNIAN AND LORENTZIAN METRICS DEFINITION 6.8 A metric on V is called Riemannian if its index is 0. A metric having index n - 1 is called a Lorentzian metric. DEFINITION 6.9 If g is a Lorentz metric on V , then a Lorentz frame is a basis e l , . . . , en which is orthonormal and which is so ordered that g ( e , , el) = 1, g(ei,ei) = - 1 for i = 2,. . . , n. Thus, for an orthonormal basis, a Riemannian metric has matrix of components

rl

o

...

01

75

RIEMANNIAN AND LORENTZIAN METRICS

and for a Lorentz frame a Lorentz metric has matrix

DEFINITION 6.10

Let g be a Lorentz metric on V. A vector

UE

V is

called (a) timelike if g(u, u) > 0, (b) null or lightlike if g(u, u) = 0, (c) spacelike if g(u, u) < 0. If W is an (n - 1)-dimensional subspace of V , then V' is the 1-dimensional subspace of all normal vectors for W. We classify W as follows: (a') W is timelike if each normal vector is spacelike, (b') W is null if each normal vector is null, (c') W is spacelike if each normal vector is timelike. REMARK: If g is a metric on V , dim I/ = n, and W c V is a k-dimensional subspace, then

wL=

(0 E

Vlg(u,w ) = 0 v w E W }

is an (n - k)-dimensional subspace. Furthermore, (W')'

=

W.

PROPOSITION 6.11 Let W be an (n - 1)-dimensional subspace of V , g a Lorentz metric on V. Then g, restricted to W , is

(a) negative definite if W is spacelike, (b) Lorentzian if W is timelike, (c) degenerate if W is null. We leave the proof as an exercise. DEFINITION 6.12 A C"-metric y on a manifold X is an assignment, to each x E X , of a metric g(x) on TxX such that the following smoothness condition holds: Let ( U , 4 = (x', . . . , x")) be a chart and define gi; U + R by giAx) = g(x)((d/dxi)(,, (d/dxj)lx).Then g i j should be C" for all i and j . A manifold with a C"-metric is called a pseudo-Riemannian manifold.

REMARK: ((d/dx1)Ix,. . . , (d/dx")l,) is a basis for T x X and (dx'(x), . . . , dx"(x)) give the dual basis. Then (giAx)) is just the matrix of components of

76

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

g(x) for the basis ((a/dx')l,). Thus, on U we have

9 = gij dx'

0dx'.

Sometimes we will find it convenient to consider g i j as defined on # ( U ) rather than U . That is, we may want to consider gij 0 4-l. We shall denote both these simply as gij, it being clear from the context whether we are considering the functions as being defined on U or on &U). So a metric is C" iff the components of g in every chart are C". DEFINITION 6.1 3 A Riemannian metric on X is a C" metric g such that g(x) is Riemannian for each x E X . A Lorentz metric on X is a C" metric g such that g(x) is Lorentzian for each x E X . More generally if, for each x, w ( x ) E T,(T,X), we say w is a covariant, degree 2, tensorfield on X . Given a chart ( U , 4 = ( X I ,. . . ,x")), we can write

w = wij dx'

dxj,

wij: U

+ R;

w is a C" tensor field if wij is C" for all charts.

Classical Interpretation Let w be a covariant, degree 2, tensor field. Let ( U ,( x ' , . . . , x")), (0, (Z',. . . , X")) be charts. Write w = wij dx'

0dxj

on U ,

= Wij di'

0dXj

on U .

w

Now we ask: What is the relationship between wij and

aij?

PROPOSITION 6.14 a x k axe w.. = w -?

1~

8x1 8x1

ke.

PROOF:

a

a

Classically, a covariant 2-tensor w is an assignment to each chart, of a matrix of functions so that the transformation law, given in Proposition 6.14, holds. Let Y be an n-dimensional manifold with a Lorentz metric, X an (n - 1)dimensional submanifold. Then if i:X + Y is inclusion, T,i: T,X -,T,Y idenifies T,X as a subspace of T,Y.

77

BEHAVIOR UNDER MAPPINGS

DEFINITION 6.15 X is a spucelike hypersurface in Y if for all x , T J is a spacelike subspace of T,Y. Similarly, we can speak of timelike hypersurfaces and null hypersurjaces. DEFINITION 6.16 Given 1-forms a,/I on X , we define a covariant 20 fl by (a 0 p)(x) = a(.u) 0 p(x).

tensor a

NOTATION:The set of C" vector fields on X is Fh(X).The set of C" 1-forms on X is Fy(X). The set of C" 2-tensors on X is F;(X).

BEHAVIOR UNDER MAPPINGS Given a C" map f :X -+ Y and a vector field 5 on X , there is, in general, no vector field on Y which can be said to correspond to 5 under f . Suppose there is a vector field '1 on Y such that the following diagram commutes:

TX 7 7T Y

Then we say 4 and '1 are f-related. EXAMPLE: (a) Let f : R 2 -+ R be defined by f ( x , y ) = y . As we have seen, vector fields on R 2 and R can be identified with mappings 5: R 2 -,R2 and '1: R + R. Define 4 and '1 by ( ( x , y ) = ( y x , y - 1) and ~ ( z = ) z - 1. Then 4 and '1 are f-related as follows:

'1 O f ( x , Y ) = '1( Y ) = Y

-

1,

Tf05(~,y)=Tf(y~,y-l)=y-l. (b) With f as above, let 4: R 2 + R 2 be given by t ( x , y ) = ( y , xy). Then there is no vector field on R which is f-related to 4. The proof of this is Exercise 6.6. DEFINITION 6.17

define f,(

E

(a) Let , f : X + Y be a diffeomorphism. If

4 E Fh(X)

FA(Y) by

f * < =T f . ( c . f - ' . Then f*( is f-related to 4; f,S is called the push-forward of 4 by f . (b) I f f is a diffeomorphism and '1 E F h ( Y ) , then the pull-back of '1 by f , denoted f * q , is given by f*'1 = ( T f ) - ' '1 o f . 0

Note that f*q

= (f-'),q.

78

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

REMARK 6.1 8 We see that vector fields cannot generally be "transported" from one manifold to another via a smooth map. One major advantage of using covariant objects, e.g., l-forms, covariant 2-tensors, is that they do have nice mappings properties.

DEFINITION 6.1 9 Let f : X -+ Y be a C"-map, w E F y ( Y ) . The pull-back of w by f , denoted f * w , is in Y y ( X ) and is defined by

for u E T x X . ( f *w)(x)u = w(f ( x ) ) T , f ( v ) If a E F i ( Y ) define f * a E Fi(X)by, for u, w E T,X, 0

( f *@)(X)(U,

w) = .(f(x))(Txf(u),

TAW)).

PROPOSITION 6.20 f * w , f * a are C" tensor fields on X PROOF: We give the proof for f * a . Pick xo E X . Let U be a chart domain in X , 4 = ( x ' , . . . , x") being the chart map, where x,, E U . Let ( V ,Ic/) be a chart in Y , f ( x o ) E V , Ic/ = ( y ' , . . . ,y"). We may assume, by shrinking U if necessary, that f ( U ) c V. We need only compute the components off * a for the chart ( U , 4) and show they are C". Let a = aij dy' 0 dyj. Here we use Latin indices in Y , Greek in X , so Latin indices sum from 1 to rn, Greek from 1 to n. Let f * a = a,, dx" 0 dx'. Then

Let t,b

o f o

4-l = ( f ' , . . . f " ) .

This is clearly C".

1

Then

INDUCED METRICS ON SUBMANIFOLDS

PROPOSITION 6.21

If w

=

wi dy' then f'*w = (df'/dx') wi dx'. If

79 R =

x i j dy' 0 dyj then

The proof follows from the above calculation. Suppose X is a submanifold of Y, i: X + Y inclusion. If w is a covariant tensor field on Y of degree 1 or 2 (a degree 1 covariant tensor is just a 1-form), then i*w is a tensor field on X . Recall that, if i is as above, then Ti is a diffeomorphism of T X onto a submanifold of T Y . We can thus identify tangent vectors on X with certain tangent vectors on Y. Now if w is a 1-form then for u, E T,X we have i*w(x)(u,) = w(x)(T,i(uJ). Thus, via the isomorphism T,i: T,X

4

T,i( T,Y) c T,Y,

i*w(x) is just w ( x ) restricted to T,i(T,X). Similarly, for a 2-tensor a on Y, we may regard i*a as being a restricted to T,i( T,X) x T,i( T,X).

INDUCED METRICS ON SUBMANIFOLDS PROPOSITION 6.22 If y is a Reimannian metric on Y , then i*g is a Reimannian metric on X . PROOF: We certainly get a 2-tensor on X and the only question is whether it is symmetric and positive definitc. Symmetry is clear from the definition of pull-back. Let u # 0 be in T,X. Then i*g(x)(u,u) = g(x)(T,i(u), T,i(u)). But T,i is injective so T,i(u) # 0 so, since ,I/ is positive definite, we see i*g is also. PROPOSITION 6.23 If X is an ( n - 1)-dimensional submanifold of the nmanifold Y and if g is a Lorentz metric on Y, then

(a) If X is a spacelike-hypersurface, then (- i*y) is a Riemannian metric on X . (b) If X is a timelike-hypersurface, then i*g is a Lorentz metric on X . (c) If X is a null-hypersurface, then i*g is a degenerate covariant 2-tensor. PROOF: (a) We need only show that on each T,X the tensor i*y is negative definite. But T,X is a spacelike hyperplane in T,X so Proposition

80

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

6.11 says g restricted to TxX is indeed negative definite. (b) and (c) follow similarly. I PROPOSITION 6.24 Iff:

X

f*(a

Y is C" and a, b, E F y ( Y ) , then

-+

0 P) = f*. 0 f*P.

We leave the proof of this as an exercise. PROPOSITION 6.25

If f :X

-+

Y is C", a E F y ( Y ) , h: Y --+ R, C", then

f * ( h a ) = (h f ) f * a . 0

PROOF: Of course (ha)(y)= h(y)a(y). Also we might denote h o f by f * ( h ) and call it the pull-back of h by f . Then the formula is f * ( h a ) = f * ( h ) f * ( u ) .The rest of the proof is now left as an exercise. I PROPOSITION 6.26

Let f : X

+

Y and h: Y

--f

R be C". Then f * d h =

d(h f)= d(f*h). 0

PROOF: Let x E X , u E

T,X. Choose a curve c in X with c'(0)

(f*dh)(x)(u)= d h ( f ( x ) ) ( T x f ( u ) ) = d h ( f ( x ) ) ( ( f C)'(O)) = ( h 0 f c)'(O), d(f*h)(x)(u)= ( ( f * h ) c)'(O) = ( h

= u.

Then

O

0

0

0

f o

c)'(O).

We thus have the proposition. We could also give an alternate proof, based on local coordinates, as follows: Choose coordinates x p on X , y' on Y. Then we have

dh = - ah dy dY'

'

SO

afi ah f * d h =--dx'

ax'

ayl

and

Let X be a submanifold of Y, i : X -+ Y inclusion. If h: Y --+ R, then h 0 i = hl,. Also i* dh is just dh restricted to T X , under our usual identification. Thus, Proposition 6.26 gives

i* dh = d(h1,). This will be an important computational aid.

(6.1)

81

INDUCED METRICS ON SUBMANIFOLDS

EXAMPLE 6.27 g = d x 0 d x + d y 0 d y + dz 0 dz is the standard metric on R 3 . Let X = { ( x ,y , z)E R 3 ( x 2+ y’ + z 2 = r’). X is a submanifold of R 3 . A chart on X is given by

x

=r

sin 0 cos 4,

0 < 0 < n,

y

=r

sin 0 sin 4,

0 < 4 < 2n,

z = r cos 0, and we have i*y

i*dx

=

i* d x

0 i* d s

= d(xl,),

+ i* dy 0 i* d y + i* dz @ i*

dz,

etc.

Now, on X , (0,4)is a chart and x

=r

sin 8 cos 4, so

i* d x = r cos 0 cos 4 d0 - r sin 0 sin 4 d 4 .

Similarly, i* d y

=

rcos 0 sin 4 d8

+ r sin 8cos 4 d 4 and i* dz = -rsin

0d0.

so i*g

0 cos 4 d8 - r sin 0 sin 4 d 4 ) 0 ( r cos 0 cos 4 d0 r sin 0 sin 4 d 4 ) + ( r cos 8 sin 4 d8 + r sin 8 cos 4 d 4 ) 0 ( r cos 0 sin 4 d0 + r sin 0 cos $J d 4 ) + r2 sin’ 0 d0 @ d8 = (r’ cos’ 0 cos’ 4 + r’ cos2 0 sin’ 4 + r2 sin’ 8) d6 0 d 0 + (r’ sin’ 0 sin2 4 + r’ sin2 0 cos’ 4) d 4 0 d 4 = r2 d8 @ d 0 + r2 sin’ 8 d& 0 d 4 .

= ( r cos -

Thus, the Riemannian metric on S:

r’(d0 0 d8

=

{ ( x , y , z ) E R31x’

+ y 2 + z2 = r’> is

+ sin’ 8 @ 0 d4).

Let a, fi E S y ( X ) . We have defined the tensor product now define a new product.

c1@

fi E Y : ( X ) . We

DEFINITION 6.28 The symmetric product of a and fi is denoted ctfi and is defined by ( x B = ;(a 0 fi fi 0 a). Clearly (x 0 p # p 0 a in general, but ctP = pa always. Note (xu = ct 0 a and we will often write (x’ for (xu.

+

Let g be a metric and express g locally as g = g,,,, dx’ Q dx”.

82

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

Since g,,”

= gvpwe

see

Thus, the local form of a metric is often written g = g,, dx’ dx’. EXAMPLES 6.29

(a) The metric on R 3 is dx2

+ dy2 + dz2.

(b) The metric on SF induced from the metric on R 3 is r2 do2 + r2 sin2 8 d4’. (c) The following example comes from special relativity: Let X c R4 be X = { ( t , x, y, z ) l t 2 - x2 - y2 - z2 = 0, t > O}. We have a chart on X , ( 4 8, 4), given by t = t,

x = t sin 8 cos 4,

y = t sin 8 sin 4,

z =t

cos 8.

Let the Lorentz metric on R4 be 9 = dt2 - dX2 - dy2 - dz2.

Then dt = dt,

+ t cos 8 cos 4 d8 - t sin 8 sin 4 d4, dy = sin 8 sin 4 dt + t cos 8 sin 4 do + t sin 8 cos 4 d4,

dx = sin 8 cos 4 dt

dz = cos 8 dt

-

t

sin 8 do.

Then i*g = dt2 - dt2 - t 2 do2 - t 2 sin2 8 dcp2 = -t2(d8’ + sin2 Q d4’). This is a degenerate metric as it should be. If p = (t, x, y, z) E X , then (t, x, y, z) also represents a normal vector at p. For suppose (a, p, y, 6) is tangent to X at p . Then let (t(s), x(s), y(s), z(s)) be a curve in X with 0 = s -+ p and having velocity at s = 0 equal to (a, g, y,6). Then 2ti

-

2 x i - 2 y j - 2zi = 0,

and ta - x g - yy - z6 = 0

83

RAISING AND LOWERING INDICES

so (a, p, y, 6) I ( t , x , y, z ) as asserted. Since ( t , x , y, z ) is a normal vector which is null, we see X is a null hypersurface so that the induced metric on X should be degenerate.

RAISING AND LOWERING INDICES Let g be a metric on X . For x isomorphism

E

X the bilinear form gx on T,X induces an

G,: T,X

-+

TZX

defined by G,(W

= gx(u, w).

We say G;'(a) is obtained from a by raising indices, while G,(u) is obtained from u by lowering indices. COMPONENT FORM: If u = ui aldx', then G,(u) = g i j d dx' because

while gijd dx'

($)

= gijuj6i = g k j u j .

DEFINITION 6.30 If (u') are the components of u E T , X , let ui = g i j u j . So we have G,(u' a/dxi) = uidx' or, in terms of components, (u') + (ui).

If w E TCX, let w

= w idx'.

Then if

(g'j)

= (gij)-',

we claim

G; ' ( w ) = g'jwj8/dx'.

To see this, apply G, to both sides, so that w

=

Gx(g"wjd/dx')

is the equation to be proved. But G,(g'jwj (?/ax') = gkig'jwj dXk

and

skisi'

so gkig'jwj dXk = 6iWi A X k

=

wj d d ,

as desired. For ( w i )given wi = gijwj. Then C;

ywidxi) = wj a / a x j

=6i

84

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

or, in terms of components, (wi)

-+

NOTE: Given

(4.

(d),to get (ui)just use

and, given (wi), get (wi)by

THE GRADIENT OF A FUNCTION DEFINITION 6.31 Given f : X -+ R C", we have the vector field grad f , defined to be the result of raising indices applied to df. The vector field grad f is called the gradient off. EXAMPLE 6.32 Let (xi) be Cartesian coordinates in R". Let (gij) be the resulting metric components. Then, (gij) is the identity matrix, so ui = ui. Thus, in terms of components, raising and lowering indices looks like the identity. This is why it is correct to define the gradient to have components given by (afpx');

is correct in Cartesian coordinates. NOTE: According to the definition of gradient we have

grad f = gij afpxj d/axi.

PARTITIONS OF UNITY Let f: R -+R be given by ,-'Ix2,

LEMMA 6.33

f is C" on R.

x > 0, x so.

(6.3)

85

PARTITIONS OF UNITY

We leave the proof of this as an exercise. Let 0 I a < b. Set g(x) = f(x - a ) f ( b - x). Then we see g is C", g ( - co,a) u (b, a), while g > 0 on (a, h). Now let A = J?"u,g(x) dx and then define h: R --+ R by

=0

on

The functionsf, g, and h are pictured in Fig. 6.1. Note that h is C", h(x) = 0 for x I a, 0 < h(x) < 1 on (u, b) and h(x) = 1 for b I x. PROPOSITION 6.34 Let { U , ) [be an open cover of X . Then there is an atlas { ( Q ,$ , ) } k such that the covering { Vkfkis a locally finite refinement of { U i } i and such that &(Vk) = B(0; 3 ) and { W, = $kl(B(O; is an open cover of X . PROOF: By local compactness and second countability we can find a sequence U:, UT, . . . covering X and refining { Uiji such that DT is compact.

FIGURE 6.1

86

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

We now construct a sequence A , , A , , A , , . . . of compact sets such that (a) U k A k = X , (b) Ai c Int(A,+,) for each i.

Let A = 0:. Suppose we have constructed Ai.Letj be the smallest integer for which j 2 i and Ai c V : v . . . u U?.

,.

,

Then let A i + = 0;v . . . v 07 u Or+ Clearly Ai c Int(Ai+,) and Ai+ 2 =X. U: u . . * u U?+ Clearlyj + co as i + oc), so Let A , = @ = A - , . For i 2 1 the set Ai - Int(A,-,) is compact and contained in the open set Int(Ai+,) - Ai_,.For x E Ai - Int(AiP1)choose a chart (V,, 4,) such that

Ui

(a) Vx c Int(Ai.1) - A i - 2 , (b) 4 X W 2 = B(0; 3 1 3 (c) Each V, lies in some U i . Let W, = 4L1(B(0; 1)). Choose finitely many V:S such that the W: cover Ai - Int(Ai- Say @i = {(V,,, 4,,), . . . ,(V,", 4x,,)}.Let @ = ai. Write 6& = {(h, ( b k ) } k . Clearly { W k ) k covers X . By (c), { l / k } k refines {U,),. We need Only show { % } k is locally finite. Let xo E X . Let i 2 1 be the least positive integer for which x o E A i . Then

Ui

xo E Int(A,+,) - A i - 2 . But then only those V;; from 4Y1 v . . * u % i + 3 can intersect the open set Int(Ai+,) - A i - , . Since there are only finitely many V;; in %, u . . . u % i + 3 , we conclude local finiteness as desired. I DEFINITION 6.35 A manifold X admits C"-partitions of unity if, given any locally finite open cover { U i } i ,there exists a family of Cm-maps I); X [0,1] such that supp(I),) c U i and ICli(x)= 1 for all x in X .

xi

THEOREM 6.36 Every manifold X admits C"-partitions of unity. PROOF: Let { U i } i be a locally finite open cover. Let { ( v k , &))k be as in Proposition 6.34. For each k there is a C"-map h k : X + [O, 11 such that h k ( X ) = 1 for x E w k = bL'(B(0; 1)) and h k ( X ) = 0 outside &'(B(Q 2)). To construct h k we need only construct a C"-map H : B(0; 3) + [O, 11 such that

H(z) = 1

for IzI I 1

EXISTENCE OF METRICS ON A DIFFERENTIAL MANIFOLD

87

and H(z)= 0

for

IzI 2 2.

Then h, = H 4, on V, and hk = 0 on X - V,. But we have shown there is hI 1, h = 0 on (-00, I] and h = 1 on a C"-map h : R + R such that 0 I [4,a). Then set 0

H(z)= 1

-

h(lz12).

So we see hk exists. Now for each k choose some i = o(k) such that vk c U a ( k ) . Then let, for $ i is C' and supp($,) c U i (Why?). Now define $ i= each i, $ i = xo(k)=i_hk; Note Cj t+bjnever vanishes, since the W: cover X , so is C" and {t,hi} is the desired partition of unity. I

Ji/Ejqj.

-

EXISTENCE OF METRICS ON A DIFFERENTIAL MANIFOLD THEOREM 6.37

Every manifold X has a Riemannian metric.

PROOF: Cover X by charts ( U i ,4i)so that the cover { U i } iis locally finite ~ .n each U i we have a Rieand has a subordinate partition of unity { I X ~ } O , (x', . . . , x"). Let gi be such mannian metric, e.g., (dx')' + . . . + ( d ~ " )q5i~ = a metric on Ui. Then gigi is a symmetric covariant 2-tensor on U i which vanishes outside a compact subset of U i . Thus, we may define aigi to be zero outside U i and we have a tensor on all of X . Then define

B

=

C Nisi; I

g is a symmetric 2-tensor on X and is C" because, near any point of X , the sum is a finite sum, by local finiteness. If u # 0 is in T,X, then g(x)(u,

0)

=

Ci ai(x)gi(x)(u,u) > 0,

since no term is negative while, at least one must be positive (namely, choose i for which x E a,:'((O, 13)). This shows g is a Riemannian metric on X . I DEFINITION 6.38 A C"-Jield of lines on X is a choice, for each x E X , of a one-dimensional subspace L ( x ) c T,X. The following smoothness condition is required: For each xo E X there is an open neighborhood U of xo and a C"-vector field 5 defined on U , such that for x E U , the line L(x)is the span of ((x), that is, L(x) = (((x)) = {l<(x)1.2E R ) .

88

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

REMARK:

A C"-field of lines is also called a C", one-dimensional distribu-

tion on X . THEOREM 6.39

A manifold X admits a Lorentz metric iff it admits a C"-

field of lines.

PROOF: Suppose X admits a C"-field of lines. We construct a Lorentz metric on X as follows. By Proposition 6.30, let g be a Riemannian metric on X . Let x E X . Let ux E T x X be a unit vector for g with ( u , ) giving the line at x . Then u, is unique up to sign. Define a Lorentz metric on X by defining g^(x)(ux,wx) = 29(ux, ux)g(ux,wx) - d u x ,

W J

Now g^(x)is well defined independent of the choice of ux. Clearly g*(x)is symmetric and bilinear. Let e l , . . . ,en be an orthonormal basis for g(x) such that e l = ux. Then s*(x)(e,,e l ) = 2 d e 1 , e , ) g ( e , , e l ) - d e 1 , e l ) = 1, g^(x)(ei,ei) = 2g(e,, e i ) g ( e l ,ei) - g(ei, ei) =

-

1

s*(x)(ei,ej) = &(el, ei)g(e,, ej) - g(ei, ej) - 07

i = 2, . . . , n, i#j.

So e l , . . . en is also orthonormal for g^(x),with el timelike, e 2 , . . . en spacelike. For smoothness of g, choose a neighborhood U of any prescribed point of X such that (a) There is a Cm-vector field 4 on U such that ( t ( x ) ) gives the field of lines on U and c(x) is a unit vector, (b) U is a chart domain, 4 = (x', . . . , x") a chart map. Then we have

so the components of

6 in this chart are g^. . = 2gik5kg. (8 18

-

g..

This shows that s* is a C"-metric, as asserted. Now suppose g^ is a C"-Lorentz metric on X and construct a C" field of lines. Choose a Riemannian metric g on X . We have maps G: T,X

-+

T,*X,

6:T,X

+ T,*X,

G(u)= Y ( U , -),

G(u)= g^(u, -);

EXISTENCE OF METRICS ON A DIFFERENTIAL MANIFOLD

89

G-' G: T,X -+ T,X is a linear isomorphism which depends in a C" way on x. If (xi)is a coordinate system, gii, Gij components of g, yI, ( g i J )= ( g i j ) -l , then, with respect to the basis i3/?xi, the map G has matrix ( g i j i j k ) , that is, 0

0

e

G- ' G ( d / d x k )= g i j i j k ( d / d x i ) .

e

We claim G-' has a unique positive eigenvalue for each x and that the corresponding eigenspace has dimension 1. To see this, choose a basis e l , . . . , en for T,X such that 0

S k i , e j )=

0, 1,

i #j, i =j .

For this basis the matrix of G 6 is the same as that of 6 for this basis. By linear algebra, however, we may choose e l , . . . , en orthonormal for g so that 4 is diagonal for e l , . . . , en. For such a basis the matrix of G-' c 6 is diagonal and has 1 positive eigenvalue and n - 1 negative ones. Thus the eigenspace for the positive eigenvalue is one-dimensional, as asserted. But if a matrix A has components which are C" functions of x and if, for each x, A(x) has a unique positive eigenvalue and a corresponding onedimensional eigenspace, then there is a scalar function of class C", L(x), and a C" vector function p(x) such that i(x) is the unique positive eigenvalue of A ( x ) and p(x) is an eigenvector for L(x). This smooth choice of eigenvector gives a vector field which is defined locally and spans our line in T,X, the line being the eigenspace of the positive eigenvalue. I Which manifolds admit smooth fields of lines? We briefly mention the ideas involved in answering this question. If there is a smooth vector field 5 which is nonzero at every point of X then clearly this defines a smooth field of lines. For this there is the following result. THEOREM 6.40 (a) A connected manifold admits a smooth nowhere vanishing vector field if and only if its Euler number x ( X ) is zero, x ( X ) = ( - 1)4dim H,(X), where n = dim X . If X is noncompact then x ( X ) = 0. If the manifold X is simply connected (its fundamental homotopy group is zero), then a smooth field of lines determines a nowhere vanishing vector field. Thus, (b) A connected, simply connected manifold X admits a smooth field of lines if and only if x ( X ) = 0. For a discussion of these results, see Steenrod's book [31]. Also see the paper of Markus [21].

CJ=o

COROLLARY 6.41

S 2 has no Lorentz metric.

90

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

PROOF: Ho(S2;R ) E R, H , ( S 2 ; R ) = 0,

1=2#O.

H2(S2;R ) E R so x(S2) = 1 - 0 +

[

TOPOLOGY AND CRITICAL POINTS OF A FUNCTION EXAMPLE 6.42 We present an interesting application of gradients and flows. Let X be a manifold,f: X .+ R C" such that for a < b the setf-'([a, b]) is compact. Suppose a, < b , and f - ' ( [ a , , b,]) contains no critical points off, i.e., if x ~ f - ' ( [ a , ,b,]), then df(x) # 0. Then we assert: f - ' ( ( - c o , a , ] ) is diffeomorphic to f - '(( - co,b,]), that is, there is a diffeomorphism fl: X .+ X such that fl(f-Y(- a , a l l ) ) = f - Y - 00, b 1 3 . Now to prove this, let g be a Riemannian metric on X . We will first show there is an interval (a, b), a < a , < b , < b such that (gradf)(x) # 0 for x ~ f - ' ( ( a ,b)). If no such interval exists, then for every E > 0, there is an x E ( a , - E, b , E ) such that grad f ( x ) = 0. Since f - l ( [ a , - 1, b , + 11) is compact, if we choose x, ~ f - ' ( ( a ,- n-', b , + PI-')) with (grad f)(x,) = 0, we may assume x , + x ~ E Xand we see x o ~ f - ' ( [ a l , b l ] ) . Therefore, (grad f)(xo) = limn+"(grad f)(x,) = 0, contrary to hypothesis. Now let h: R -+ R be C" with h = 1 on [a,, b,] and h = 0 outside (a, b), h(x) 2 0 Vx E R . Define a vector field 5 on X by

+

If x $ f - ' ( ( a , b)), then h(f(x)) = 0 so [(x) = 0, while for x €f-l([al, b,]),

The vector field 5 has compact support so let a:R x X + X

be its flow. Let

8:X

.+

X be defined by

B(x) = E(b1 - a 1, x). Claim: /lmaps f - ' ( ( - co,a,]) onto f - ' ( ( tegral curve of through xo E X . Then

(f so f

0

ax,,is a

O

-

co,b J ) . First let axo be an in-

%J'(t) = g(grad f ( ~ x o ( t&)x)o,( t ) ) )

= h(f(@X,(t))),

nondecreasing function of t. Then, in particular, f ( f l ( x ) )2 f(x).

91

TOPOLOGY AND CRITICAL POINTS OF A FUNCTION

Let f ( x ) I a,. Then f ( B ( x ) )= f M b 1

bi -ai

-

a l l ) = f ( x ) + Jo I a,

h(f(%is))) ds

+ b, - a , = b,,

which gives B ( f - ' ( ( - . o ? a I l ) ) C f - Y - m , bll). For the reverse inequality let ,f(x) I b , . Let x' = P-'(x). We must show f(x') I a,. If not, then a , < f ( x ' ) I h,. Also, for 0 I t I b , - a,,f(ccx, ( t ) )I b , so bl 2 f ( x )= f ( B ( x ' ) ) = f'(x')

+

lo hi

-01

h(f(a,,

(4))ds > a,

+ b,

-

a,

= bi,

a contradiction. Figure 6.2 illustrates the meaning of this result.

FIGURE 6.2

92

6. COVARIANT 2-TENSORS AND METRIC STRUCTURES

EXERCl S ES 6.1 If e l , . . . , en is a basis for V with dual basis e l , . . . ,en and if a fl = fl j ej , what are the components of a 0 B?

= a$,

6.2 Let g be a metric on V and e l , e 2 , . . . , en a basis for V. Then the matrix (gij) is symmetric and nonsingular. 6.3 Prove the validity of the remark following Definition 6.10. 6.4 Prove Proposition 6.1 1. 6.5 Show that

CI

0 /I is a C"-tensor field, if a and fl are l-forms on X .

6.6 Prove part (d) of Example 6.17. 6.7 Let f : X + Y be C" and suppose related and f is onto.

5 E Yh(X)

and q

E Yh(Y) are

f-

(a) Let xo E X and c: I -+ X an integral curve of ( with c(0) = xo. Then f 0 c is an integral curve of q with initial condition f ( x o ) . (b) Suppose 5 is complete. Then so is q. (c) Given an example where q is complete but 5 is not. (d) Assume 5 (and hence v] also) is complete. Let a:R x X X, fl:R x Y -+ Y -+

be the flows of

5, v ] .

For t E R we then have at: x

-+

x,

/It: Y + Y

and the following diagram commutes: X S . Y ~

x-Y

Ipc

6.8 Prove Proposition 6.24. 6.9 Finish the proof of Proposition 6.25. 6.10 Consider spherical coordinates (r, 0 , 4 )in R 3 , with the usual Euclidean metric g = d x 2 + dy2 + d z 2 . Compute the matrix of components (gij) and its inverse (gij). Given f : R 3 -+ R what are the components of grad f in spherical coordinates? 6.11 Consider R4 with the Lorentz metric g Given f : R4 -+ R calculate grad f .

= dt2

-

d x 2 - d y 2 - dz2.

EXERCISES

93

6.12 Prove Lemma 6.33. 6.13 Observe that the proof of Lemma 6.5 shows that if g is a metric on a vector space I/ and y(u, u) = 1, then there is an orthonormal basis for I/ containing u. Suppose 61 is a Lorentz metric; let u, w be nonzero vectors in V with g(u, w) = 0. (a) Show that if u is timelike, then w is spacelike. (b) Show that if u is null then w is either null or spacelike and, if w is null, then w is a multiple of u. (c) If u is spacelike, can anything be said about w?

7 Lagra ng i a n and Hamiltonian Mechanics for Holonomic Systems

INTRODUCTION Consider again a system of n particles of masses m , , . . . , m,. In Chapter 2 we considered a state of the system to be a point (x, u) E R3" x R3". As before, we introduce M i , 1 Ii I 3n, defined by M 3 j = M3j-1 = M3j-2

= mj.

In view of the diffeomorphism TI: T R 3 n + R3" x R3",

I the identity on R3", we may view such a state as a point in TR3". If ( X I , . . . ,x3") are the standard coordinates on R3", we define a metric as follows: DEFINITION 7.1

The kinetic energy metric o f the system is 3n

T

(Mi(dxi)2).

= i= 1

NOTE: If u E TXR3"is the state of the system, then &T,(u,u) is the total kinetic energy of the system. If the configuration space M is an open set in R3",then the state space is TM z M x R3" and the kinetic energy metric is a Riemannian metric on M . DEFINITION 7.2 The kinetic energy function is the map T TM - + R given by T(u)= iTx(u,u) (see Definition 2.3).

94

95

THE TOTAL FORCE MAPPING

Now recall the case of a system subject to holonomic constraints. In this case configuration space will no longer be open, but rather, will be a submanifold of dimension d in R3". We think of d as the number of degrees of freedom of the system. The assumption that the constraints are holonomic will be taken to mean that the system position is confined to M but any velocity tangent to M is a possible state. So, for a holonomic system, we have a configuration space M and T M is the state space. The kinetic energy metric restricts (i.e., pulls back) to a Riemannian metric, also denoted T, on M . Thus, so far, our mathematical model of a system of particles subject to holonomic constraints consists of (a) a C"-manifold M the conjguration space; (b) the tangent bundle T M , the state space; (c) a Riemannian metric T on M , the kinetic energy metric. Now we consider a holonomic system having configuration space M , a ddimensional submanifold of R3". We have standard coordinates xl,. . . , x3" on R3". For each particle there is a force acting on the particle. The total force on the system has 3n components F , , . . . , F3". We consider the forces as being functions of position and velocity. Thus, at present, we are not considering time-dependent forces. Thus, we have

Fi:T M -+ R.

THE TOTAL FORCE MAPPING DEFINITION 7.3 The total f o r w is the mapping

F: T M

+ T*R3"

given by

F(v) = Fk(V) dXk(.X)

for

We see that the following diagram commutes: TM

where i is the inclusion.

T*R3"

V E

T,M.

96

7. HOLONOMIC SYSTEMS

DISCUSSION 7.4 Although the total force looks like a differential 1-form it is actually not. For instance, a 1-form on T M should take values in T * ( T M ) rather than T*R3".If i: M + R3" is inclusion, we will want to "pull F down to M by i," that is, form i*F. Since we have only discussed pull-backs of 1-forms an explanation is needed. We simply define i*F in the obvious way; namely, (i*F)(v)E T Z M , for u E T x M , is given by (i*F)(u)(w)= F(u)(T,i(w)).Then it is clear that

i*F(u) = Fk(u)(i*~ x ~ ) ( x ) .

Note i*F is a map T M + T * M and i*F is fiber preserving, i.e., if v E T , M , then (i*F)(v)E T Z M . If the force F is not velocity dependent, then we may regard F as a map F: M + T*R3" and then i * F M + T*M is actually a 1-form on M . This will be true, for example, in the important case of conservative forces.

FORCES OF CONSTRAINT Now in an actual system, we may split the total force into a sum F = Fa F", where Fa is the total applied force and F" is the force of constraint. We have the following definition describing constraint forces.

+

DEFINITION 7.5 A force F": T M + T*R3" is a force of constraint if, for u E T,M, F'(v) vanishes on T x M c TxR3",that is, i*FC= 0.

REMARK: The "constraints" encountered in actual systems satisfy this requirement. In Goldstein's Classical Mechanics the condition is expressed by saying that "the virtual work of the forces of constraint is zero" (page 17). We shall show that the laws of motion can be formulated in such a way that the constraint forces do not appear in the equations. It should be noted that Goldstein's "virtual work. . . is zero" does not simply mean that the force does no work on the system as the system pursues its course. Consider a particle of mass WI and electric charge e moving in a magnetic field B. The force is F = e(v x B), where v is the instantaneous velocity of the particle. The work done by the force is clearly 0 but F is not a force of constraint since F(v)(w)# 0 for some w and, since M = R3, every w is tangent to M . Note that, in our notation, the magnetic force is a map

F: R 3 x R 3 -+ R 3 x (R3)*

defined by F(x, V ) = (x, (u2B3 - u3B2)dx'

+ (dB2

- u2B') d X 3 ) .

+ (u3B'

-

u'B3) dX2

97

FORCES OF CONSTRAINT

Note that, whenever configuration space is open in R3n,any force of constraint must be zero [according to Definition 7.51. According to Newton’s law, if c = (c‘, . . . , c3“)is the path in R3“followed by the system, then

MiZi = Fi(c, C)

1 I i 5 3n.

( i not summed),

(7.1)

REMARK: Newton’s laws, as written above, do not fit in with index conventions. Suppose we ask what Mi?’ really is. Recall our kinetic energy M,(dx’)’. We may express T in usual tensor notation as metric T =

c:!

T

=

T i jdx‘ dx’,

where

Now ( C i ) represents the acceleration vector in Cartesian coordinates (XI,.. . ,x3“),so (TjjEj)gives the components of the 1-form obtained from the acceleration by lowering the index. Thus, Newton’s second law may be written

T IJ. 2 = F j ,

i

=

1,. . . ,3n

(sum onj).

(7.2)

For each t ,

Tijc(t))cj( t )dx’(c(1 ) ) = Fi(c’(t)) dx’(c(t)).

(7.3)

Both sides lie in Tf;,,R3”.We have the map i*: T$,,R3”-+ Tf;,,M.

Operating on both sides of (7.3) with i* (see Discussion 7.4) gives

Tkji?i* d X k = Fki* d X k .

(7.4)

Let ( U , 4 = (ql, . . . ,q d ) )be a chart on M near c(t). Let us express (7.4) in terms of this chart. Let 4 c = c- = (-’, c . . . , (“1). Now i* dxk(c(t))= (dxk/Sq”). dq”(c(t)),where dxk/dq’ is evaluated at c‘(t).Thus, from (7.4) we get 0

Ti,Zj(dxi/dqp)dq”

=

F:(dx’/aq”)dq’.

(7.5)

We therefore obtain

T,’E~axi/aq = F~ axi/aqF’, p

=

1, . . . ,d .

(7.6)

Now F i dxi/a@‘are the components of i*F in terms of the basis dq’, . . . ,dqd. If we decompose F as F = Fa + Fc, then i*FC = 0 by Definition 7.5, so

98 i*F

7. HOLONOMIC SYSTEMS = i*Fa.Thus

(7.6) becomes T

Let Q,

= F;

~

aXi/aq, ~ C ~ = F; axi/aq’.

(7.7)

axi/aq” so that Q, dqp = i*Fa.Then we have T ~ a X~i / acg =~ Q,.

Now

so we get

But T i j is constant, so

where the quantities T,, = Tijaxi/aqpdxj/aq”are the components of T on M in the (q’, . . .,qd)-coordinates.Corresponding to the coordinates (ql,. . .,qd) on M , we have the natural chart (q’, . . . , qd, v’,. . . vd) on T M . In terms of this, the kinetic energy function T : TM + R is given by T(q,v ) = $T,,(q)v’u’. So we have

(7.10) Thus, (aT/dv’)(E,i)= TJc)? and

99

CONSERVATIVE FORCES

and, therefore,

LAG RANGE‘S EQUATIONS We see (7.8) becomes p = 1, . . . , d.

-Q,,

(7.11)

In mechanics books, Eqs. (7.1 1) are called Lagrange’s equations; Q, are called the generalized force components. The reason for the adjective “generalized” is that mechanics books tend to refer to Cartesian coordinates as coordinates and refer to other coordinates, such as ql, . . . qd, as “generalized coordinates.” The Q, are the components of our force in the coordinate system q l , . . . qd.

CONSERVATIVE FORCES Suppose Fa: T M + T*R3”is such that there is an open set U C“-map V :U + R such that Fa(o)= - d V ( x )

for x

E

M,

V E

3

M and a

T,M.

Such a force is said to be conseroatiue. A conservative force does not depend on velocity. NOTE: Since i*Fa = Q, dq’, we see that a system is conservative if there is a V :M + R with Q , = - aV/dqp. For conservative systems, we call V :M + R the potential function. DEFINITION 7.6 For a conservative, holonomic system, the Lagrangian

L: T M

--*

R is defined by

L= T-(VOT). Thus

qv)= T ( u ) - V ( x ) .If we use coordinates (q”, up) on dL

dT

c3L

avp

avfi’

aqv

- --

-

dT

dV

aqN

aqp’

T M , we see

100

7. HOLONOMIC SYSTEMS

Thus Lagrange's equations (7.11) take the form p = 1,..., d.

=0,

(7.12)

EXAMPLE 7.7 (Simple pendulum) Choose (x, y , z ) coordinates in R3 so that the z axis points toward the reader out of the page and the x and y axes are as shown in Fig. 7.1. The mass is constrained to move on the circle M = {(x, y , 0) x2 + y2 = t2}.The total applied force Fa:T M -+ T*R3 is given by

I

Fa(v)= my dx

for u a vector at (x, y , 0).

The kinetic energy metric on R 3 is T

= m(dx2

+ dy2 + dz').

Consider the coordinate system 0 on M (usual polar angle),

x

=P

cos 8,

y

=P

sin 8,

z

= 0.

The applied force, when pulled down to M is Qe d0

= mg

d(L cos d ) = - m g l sin 0 dd,

so Q B= -gm( sin 8. The kinetic energy function is T(6,d) = 4me2d2. This is seen as follows: First note we are using a notation common in physics, wherein if (ql, . . . ,qd) are coordinates on M , then ( q l , . . . , qd, q l , . . . , q d ) are the coordinates of

FlGURE 7.1

101

CONSERVATIVE FORCES

the corresponding natural chart on T M . So (8,d) is the natural chart corresponding to the chart 8 on M . Now in R3, T = m(dx2 + dy2 + dz2). On M , T

+ (i* dy)' + (i* dz)') = m ( [ d ( t cos 8)12 + [ d ( t sin €41' + [do]') = m(d2 sin2 8 do2 + d 2 cos2 8 do2) = me2 de2. = m((i* dx)'

Therefore, r(6,O) = +T

(B qae, d ape) = +md2e2

as asserted. The Lagrange equation is then - ('T) -

dt

I38

aT - Qe, 138

which becomes

m&

-

0 = -my[ sin 8

le + g sin 8 = 0.

or

This is the usual equation of motion for the pendulum. In practice one does not go through all these steps. One first introduces the coordinate 8. Then writing the kinetic energy function in terms of 6, 0 is done by simply expressing kinetic energy in these variables; linear velocity = Pd,

T

= $mt29'.

Then note the force is - d V , where V ( x , y, z ) = -mgx. Express V using 8, i.e., V = - m g t cos 6. Then L = 3mP2d2 m g l cos 8,

+

aLla6

= m128

and

aLJa6 = -mgd sin 8,

so (d/dt)(aL/a8)- (aL/aO)= 0 becomes m t 2 8 + m g l sin 8 = 0

or

te + g sin 6 = 0.

EXAMPLE 7.8 (Atwood's machine) We return to Example 2.13 using the notation adopted there. We consider only x and y coordinates, since the motion is clearly planar. We have the force

F: R4 x R4 + R4 x (R4)*, defined by F(x', y', x 2 , y 2 , i' jl, , 2,j 2 )= ( x ' , y ' , x 2 , y 2 , 0, m,g - z, 0, m2g - z), where z is the tension in the string. We now claim:

FC(x1, y', x 2 , y2, 2,j ' , i2, j 2 ) = ( X l , y', x2,y2, 0,

-?,

0,

-7)

is a force of constraint. Use the coordinate y on M . A vector in TM has vector part (0, A, 0, -A), so when it is acted upon by -T dy' - z dy2 we get -z(A) -+A) = 0. So

102

7. HOLONOMIC SYSTEMS

the definition of force of constraint is satisfied. The applied force arises from the potential V ( x l ,yl, x2, y 2 ) = -m,gy’ - m2gy2 or, V ( y )= -mlgy - m,g. (k‘ - nu - y). We have T = fil ((dx’)’ + (dy1)2)+ m2((dx2)2+ (dy2)2)so, since on M we have x2 = a,

y’ = y ,

x1 = -a,

y2 = e -na

-

y,

we get T

= m,

dy2 + m2 dy2

= (ml

+ m2)dy2

and thus, T(Y,3) = +(m1 + m2)j2.

Thus the Lagrangian is L ( Y ,A = +(ml + m2)Y

+ m,gy + m2g(e- nu - y )

and hence dL/dj = (m,

so (m, + m2)y - (ml - m,)g

+ m,)j,

dL/dy

= (m,- rn,)g,

= 0. Thus

Y=(

m1- m2 )4. m1 + m2

This equation is easily solved for y as a function of t .

Motion Subject Only to Forces of Constraint EXAMPLE 7.9 (Force-free motion with constraints) Suppose a system has configuration space M and the only forces are constraints. We then have

L = T : T M -+ R

and, if (ql, . . . , qd)are coordinates on M, we have T(q”7 4”)

= ts,v(q)4”4v.

This gives

aT/aq = g p v q v ,

103

THE LEGENDRE TRANSFORMATION

so the Lagrange equations become

Interchange

0 and v in this last equation

to get (7.14)

Add eqs. (7.13) and (7.14) and divide by 2 to get (7.15) We introduce the notations. (7.16)

q"= 9 w v >PI.

(7.17)

Then (7.15) becomes gpvqv

+

[BY, ,1]4%j"

(7.18)

= 0.

Multiply by 9'" and sum on p to get the set of equations q"

+ rFv@q" = 0,

CI =

1,

. . . ,d .

(7.19)

Equations (7.19) have as solution curves those curves which are geodesics on M for the Riemannian metric T = gpv dq" dq' on M . Geodesics are important objects in Riemunniun geometry and will be studied in more detail later. are called the ChristofSel symbols of the j r s t and The symbols [pv, p], r$" second kinds, respectively. We summarize the above results by saying: For a system of particles moving under the influence of holonomic constraints, but with no applied forces, each trajectory is a geodesic of configuration space with respect to the kinetic energy metric.

TH E LEG ENDR E TRANSFO R MATlON Let M be a manifold, g a metric on M . The Legendre transformation 9: 7'M sponding to g is given by DEFINITION 7.10

L?(v)(w)= gx(u, w )

for v, w E T,M.

+

T * M corre-

104 b;p

7. HOLONOMIC SYSTEMS

is a bundle isomorphism, i.e., the following diagram commutes,

TM

5T*M

and 9 gives a linear isomorphism T,M PROPOSITION 7.1 1

+ T,*M for

each x E M.

Y is a C"-diffeomorphism.

PROOF: Let ( V ,d, = ( q ' , . . . , q " ) ) be a chart on M . We get charts (T-'(U), Td,),(z*-'(U), (@),) on TM, T*M, respectively. We must show

(G),F c-

0

(Td,)-'

is C" with a Cm-inverse. We have

and

so 9'o (Td,)-'(q, q) = qigiXq)dqj(d,-'(q)).Then note that (5)*(digij(q) dq'(4- l ( 4 ) ) )= (q, gij(q)qi),

so (%)*o 9 o (Td,)-'(q,4) = (4, gij(q)qi).This is clearly C" with C" inverse given by (qi, Pi)

+

(qi, d'(q)Pj).

I

Recall if g is the kinetic energy metric of a particle system, we have the kinetic energy function

T :T M and the Lagrangian L = T L(q, 4 ) = $g,,@'q"

T(q,4) = &py(q)4p4v

R, -

V z. So we have

-

0

V(q)

and

dL/dqp = gpvqv.

Thus, the local representative of 9 is 2 ( 4 , 4) = (4,(dL/%?) ( 4 , 4 ) ) . See Exercise 7.12 for an invariant description of equation (7.20).

(7.20)

105

CONSERVATION OF ENERGY

CONSERVATION

OF ENERGY

Consider a system with configuration manifold M , kinetic energy function T : TM + R , potential I/: M + R, and Lagrangian L = T - I/. 5. DEFINITION 7.1 2

v

L

The energy function is E: TM

-+

R , given by E

=

T

+

5.

THEOREM 7.13 (Conservation of energy) If c is a trajectory of the system in T M , then E c is constant. 0

PROOF: We work locally to show (dldt) (E(c(t)))= 0. We first claim

E(q, 4.)

= (aL/aq')q' - L(q,

4).

(7.21)

This is seen as follows:

aL/di'

= dT/d~j',

774,4) = +gij4'q'

so that aT/aLj' = gij$; thus (aL/dq')q'= gij4jq'= 2T(q, q), so

(aLla4')ci' - U q , 4) = 2T(q, 4) - (T(q,4) - V q ) ) = T(q,4) + V q ) = E(q, 4). so

The first and third terms add up to zero, since Lagrange's equations hold, and the second and fourth terms cancel, so the proof is complete. The quantity (dL/&j') (4, 4)$, which appeared in the preceding proof, is the coordinate representation of a function on TM as follows: Define a function A: TM

+R

by A(v) = U ( u ) u . We see that

We also have E=A-L.

which is clear from the local equation E(q, 4) = (dL/dLj')4' - L ,

derived earlier.

106

7. HOLONOMIC SYSTEMS

HAMILTON'S EQUATIONS Consider a system with configuration space M . W e have shown that the equations of motion are the Lagrange equations, that is, if c is a trajectory of the system in state space T M , then, in local coordinates, c satisfies Lagrange's T M -+ T * M , we can conequations. Given the Legendre transformation 9: sider phase space trajectories, i.e., curves 9 c where c is a trajectory in state space. We want to know what differential equations are satisfied by these phase space trajectories. 0

DEFINITION 7.14 The Hamiltmian a ( 9 - l(a))- L ( 9 - '(a)). LEMMA 7.15 H = E PROOF: E = A

-

H:T*M

-+

R is defined by H ( a ) =

9-'.

0

L so we must show

A(.rl(a))

= +rl(a)).

But A(u) = ~ ( u ) SO u

A ( Y - y a ) )= 9(9-l(a))(9-'(Lx))= a(9-ya)).

I

Now consider H : T * M -+ R . Given coordinates (q', . . . , qd) on M , we get coordinates ( q l , . . . ,qd, q', . . . , qd) on T M and ( q l , . . . , qd , p l , . . . , p d ) on T*M. The Legendre transformation is described by pi = (dL/aq')(q,4). The Hamiltonian is given in local coordinates by H ( q , P ) = pi$

-

L(q, 4).

+

Now dH = (aH/dqi)dq' ( d H / d p , ) d p , in general, but we also have d H = pi dq' + 4, d p , - dL. Now really q i = Lji 9-', L = L 0 9-', so we are interested in 0

d(q' o

9 - l )

= (2-')* dq'

and

d(L 9-') = (2-')* dL. 0

Then

=

($

o

9-l)

dqi + p , ( Y - l ) * dq'.

107

HAMILTON'S EQUATIONS

Therefore dH

= pi(9-')*

dqi + (4'

0

9 - l )

dp,

-

d(L

0

9 - l )

Thus we have

Let c(t) = (ql(t),. . . , qd(t),ql(t), . . . ,d d ( t ) )be a state space trajectory. Then Lagrange's equations hold (we assume a conservative system), L dt( g adi ( q , d ) ) - : ( q ,84 q)=O,

i = 1 ,..., d.

Now 9 0 c is a trajectory in phase space. Let this be written (ql(l), . . * qd(t), 7

...

9

Pd(t)).

Then (8H/dpi)(2 c(t)) = q'(t), by Eqs. (7.22), and 0

We summarize as follows: THEOREM 7.16 Let (ql(t), . . . , qd(t),pl(t), . . . , p d ( t ) ) be a phase space trajectory in a natural cotangent bundle chart. Then (Hamilton's equations)

(7.23) EXAMPLE 7.1 7 (Central force motion-inverse square law) Consider a mass m moving in space under the influence of a force field generated by the potential

V:R3 - ( 0 ) + R,

V ( X )=

-

k/lxI,

k > 0.

There are no constraints, so configuration space is M = R 3 - {0}, and (ql, q2, q 3 )are the usual Cartesian coordinates. We identify T M with M x R3 with coordinates (ql, q2, q3, q l , q2, Q3)

= (494).

108

7. HOLONOMIC SYSTEMS

Also, we identify T * M with M x R 3 with coordinates

GI1>qZ. q3,Ply Pz > P 3 ) = (4, PI, We have the Lagrangian

m,4)

=

fm14I2 - WI).

TM The Legendre transformation 2: Y(q,4) = (4, m4)

and

-+

T * M is given by

2- l(4, P) = (4, P / 4 .

The Hamiltonian is H(q, P) = (lPI2/W + U q ) .

Simply writing down Lagrange's or Hamilton's equations is not very illuminating. Consider GI, G,, G,: T * M -,R defined by Gl(q7

P) = q 2 p 3 - q 3 p 2 ,

G2(q,P) = q3pl - q1p3,

(7.24)

G3(q, P) = q1PZ - q2Pl.

Recall that an integral, or constant of the motion, of a system is a map 4: T * M + R which is constant on each phase space trajectory (cf. Definition 2.5). We claim the functions in (7.24)are integrals. In Chapter 15 we will see how these quantities result from symmetries of the Hamiltonian system. We check the claim for G,, letting (q'(t),p j t ) ) be a trajectory. We have d dt

- Gl(q, P) = 4'6, - q3d2

+ qZp3- ri3pz = q2p3 - q3P2

where, in the second equality, we used p i = mq'. But aH/aq'= aV/aqi = kqi/)q13,so we get

The quantities G I , Gz, G, are familiar as the components of the angular momentum vector of the system and we have proved conservation of angular momentum (for this system). Consider now motion beginning with initial condition (qh, q i , 0, 46, &, 0).

109

HAMILTON‘S EQUATIONS

Then we will always have (see Exercise 7.2).

q3(t)= 0 = d 3 ( t ) = p 3 ( t ) Consider cylindrical coordinates, q’

=

r cos 8,

q2 = r sin 8, 43

= z.

4’ 4’

= cos

We have

=

43 =

sin ~i + r cos 86,

i.

+

Then L(r, 8, z, i, 6, i) = &(i’ r’@ that the Lagrange equations are

d -

dt

d -

dt

~i- r sin 04,

+ 2’) + (

(mi) = mrg2 -

k / J m ) . One easily sees

kr (r’

+z~)~/’’

(mr2@= 0,

d - kr ( m i )= dt ( r 2 + z2)3/2

-

We know z

=0 =

i, so the equations reduce to

mi: = mro2 - ( k / r 2 ) ,

(d/dt)(mr26)= 0.

(7.25)

We will not solve these equations here, but refer the reader to classical mechanics texts, e.g., Goldstein. The Legendre transformation is given by

Y ( r , 8, z , i, 0, i) = (r, 0, z , mi, mr’8, mi),

y -‘(r,6, z , p r , p e , P A

= ( r , 0, z ,

p,./m, peImr2,p,/m).

The Hamiltonian is

=

5

k i (p; + + Pi) 2m Jm-

110

7. HOLONOMIC SYSTEMS

and Hamilton’s equations are

If, due to initial conditions, z and i are 0, we get mi

= p,,

mr2e = pe, 0 = p,,

p.r = - -Pe-2

k

mr3

r2’

fie = 0, p, = 0.

We conclude d dt

- (rnr’e) = 0,

k d (mi) = Pe2 - dt mr3 r2‘

-

~

(7.26)

Note Eqs. (7.25) and (7.26) are the same. We now want to consider the equations of mechanics from the point of view of flows of vector fields. We shall show that there exist vector fields on T M and on T * M whose integral curves are precisely those curves which, when written in natural coordinates, satisfy Lagrange’s equations and Hamilton’s equations, respectively. We will see that the construction of these vector fields is accomplished in a very natural way by means of a canonical 2form on T * M . This canonical 2-form represents what is called a symplectic structure on T * M . First some general facts about skew-symmetric tensors will be discussed.

2-FORMS DEFINITION 7.18 A skew symmetric 2-tensor on a vector space V, is a bilinear map o:V x V + R such that o ( u l , u2) = - o ( u 2 , vl) for all u l , v2 E V . We often call a skew symmetric 2-tensor on V simply a 2-form on V . The set of all 2-forms on V is a vector space, denoted A2(V).

111

2-FORMS

DEFINITION 7.1 9

The antisymmetry operator .d:T 2 ( V )-+ A2(V) is def-

ined by .do(v,,v 2 )

= t(W(U1, " 2 ) - w(v2, UJ).

We say .dois the skew-symmetric part of if o is skew-symmetric. DEFINITION 7.20 CI A

The wedge product

(to.

A

p = 2,d(C!0p).

Note that .do= o if and only

: V* x V* -, A2(V) is defined by

Let e l , . . . , e n be a basis of V with dual basis e l , . . . , e n . PROPOSITION 7.21

{ e i A ejI

1I i < j 5 n ) is a basis for A2(V), so

PROOF: Write Q E A J V ) as (11 = wijei 0ej. Then

o = .do= wii.d(ei8 e') = (w$)e'

Now

(toij = w ( e , , ej) = - w ( e j , e i ) = --toii

A

ej

so that

This proves our set spans For indcpendence, note that if then, for k < e we have

0=

alje' A

XI

<

a,,el

A

eJ = 0,

eJ(ek,e,)

I<.!

=

C a,,(e' 0eJ

- P'

0e')(ek,e,.)

I

= ak/.

SUMMATION CONVENTION On a n n-dimensional space we have the con-

ven tion that

1,

We now agree oIIJIeLA el = < o,,e' A eJ. Thus, bars around the indices means sum only over pairs (i, j ) for which i < j .

112

7. HOLONOMIC SYSTEMS

The following come from the calculations in the proof of Proposition 7.21: (a) If w = wijei 0 eJ, then w = o l i j l e Ai ej. (b) If, o = o l i j l e Ai ej and if w i j = - w j i for i > j and wii = 0, then w wijei eJ.

o

EXAMPLE 7.22

tl13e1A e3

=

In R3, any 2-form is uniquely expressible as tl12e1A e2 +

+ t123e2A e3.

A differential 2,form on a manifold X is a tensor field F i ( X ) such that o(x) is skew-symmetric for every x E X. The set of differential 2-forms on X is denoted by A2(X). We sometimes refer simply to a 2-form on X . Let ( U , 4 = (q', . . . , 4")) be a chart on X. Then there exist uniquely defined functions w i j such that DEFINITION 7.23

w

E

w = wij dq'

0dqj

on U

and we also have o = olijl dq'

A

dqJ

on U .

We often consider a function of class C", fi X 4R to be a differential O-form. We then use the notation Ao(X) as an alternate notation for Cm(X,R).

EXTE R I0R D E R IVATIVE Also we may write Al(X) rather than Fy(X). Then we have the map d: Ao(X)-+ A,(X).

We now define an operation d: A,(X) + A2(X)

called exterior derivative. Eventually we shall define differential forms of arbitrary k , denoted A,(X). Then we will have an exterior derivative operator d: A,(X) A,, l(X). At present we have only the case k = 0, so we now define d for k = 1. -+

PROPOSITION 7.24 Let ( U , 4 = ( q ' , . . . , q " ) ) be a chart on X . There is a unique map d u : A l ( U )+ A2(U) such that

(a) d,(a + B) = d u ( 4 + d d B ) for a, B E Al(U), (b) d a ( f j l ) = d f A a fd,u for f~ Ao(U), tl E Al(U), (c) d,(df) = 0 for all f E Ao(U ) .

+

113

EXTER 10R D ER IVATIVE

PROOF: Let a E Al(U). Write a = aidq', where the ai are uniquely determined smooth functions. Define d,(a) = daj A dqi. We must verify (a)-(c). We leave (a) to the reader. For (b), let a = xi 4';then fa = fai dq', so

&(fa)

+ a, df)A dq' = fdC$ A dq' + 4 f A a; dq' = f d a + d j ' A a.

=

d(.fai) A dq' = (.f'da,

For (c),

since d2flaqJ aqi is symmetric in i , j while d 4 ' ~dq' is skew-symmetric. This proves existence. But if any map d, satisfies (a)-(c), then d,(ai d 4 ' ) = dcl; A d4'

so uniqueness is proved. THEOREM 7.25

+ Mjd,(d4') = dai

A

d4',

I

There is a unique map d: A,(X) -+ A,(X)

such that if a E A,(X) and ( U , 4) is a chart on X , then dal,

= dU(aIu).

PROOF: To prove uniqueness, let x E X . Then, if ( U , 4) is a chart at x, (da)(x)= d,(al ,)(x). For existence we must show that if ( U , $), ( V , +) are charts at x, then

I

d,(a W x ) = d " ( E I V ) ( . 4 .

Let

4 = (x', . . . , xn),

= ( J ) ' , . . . , y").

Then if a

= ai dx' = Jj dy',

du(alu)(X) = da,(.x)A d X ' ( X )

and d,(al,)(u)

= dBj(x) A

~Y'(x).

Now on U n V we have two coordinate systems (xi),( y j ) .Now d,, " a ( x ) = da,(x) A dx'(x), but also d,, "a(.x) = dp,(x) A dyj(x). These two expressions for the uniquely determined form d,, " a must be equal, so dai(x)A d.xi(x) = dp,(x) A dyj(x), as desired. PROPOSITION 7.26

d: A,(X) + A2(X) satisfies

(a) d(a + p) = da + dp for 2, /j E Al(X), (b) d(fa)= d f ~a + f da for .f E A,,(X), a E Al(X), (4 Wf) = 0, f E AO(X).

114

7. HOLONOMIC SYSTEMS

CANONICAL 2-FORM ON T * X In Eq. (5.8) we defined the canonical 1-form 0 E A1(T*X).We now have DEFINITION 7.27

The canonical 2-forrn R on T*X is defined by

=

-do. PROPOSITION 7.28

then R

If (ql, . . . , q”, p l , . . . ,p,) is a natural chart on T * X ,

= dq’ A d p , .

The proof is left as an exercise.

THE MAPPINGS b AND # Let o:V x V - t R be a bilinear mapping. If u E V , let ub E V* be defined by Ub(U’) =

w(u, u’).

If the bilinear mapping w is nondegenerate, then the map u -+ u b is an isomorphism and the inverse V* .+ V is denoted c( + a*. We note that if g: V x V-+ R is a metric, then the corresponding b and # maps are precisely the operations of lowering and raising indices with the metric. LEMMA 7.29 Let e , , . ... , en,u l , . . . , u, be a basis for the 2n-dimensional e’ A u’. space V. Let el, . . . , en, ul, . . . ,un be the dual basis. Let w = Then

(a) w is a nondegenerate 2-form, (b) e j b= uJ, ujb = -&, j = 1, . . . n. 9

PROOF: The matrix of w for the given basis is

This is true because o ( e , ,e j ) = 0 = o ( u i , u j )

for all i, j,

115

HAMILTONIAN AND LAGRANGIAN VECTOR FIELDS

This matrix is nonsingular so (a) holds. To prove (b), write ejbas ejb= rkek S k u k . Then r k = ejb(ek) =

+

w(ej,ek) = 0,

so ejb= uj. Similarly, one sees ujb

= -ej.

We leave the details to the reader.

I Now let T * X be the cotangent bundle of a manifold X . We have the canonical 2-form R on T * X . For z E T * X , we have the map b: T , ( T * X ) -, T T ( T * X ) . The form R is nondegenerate by Lemma 7.29 since, if (qi, p j ) are natural coordinates, we have

i2 = dq' A dp,.

(7.27)

Thus there is an inverse map for b, #: TT(T * X ) + T,( T*X).

Given a vector field

5 on T*X, we have a (tb)(Z)

(7.28)

1-form on T * X defined by

= (((4)b.

(7.29)

Given a 1-form w on T * X , we get a vector field o ' on T*X by defining d ( Z )

(7.30)

= (w(z))'.

HAMILTONIAN AND LAGRANGIAN VECTOR FIELDS DEFINITION 7.30 Let 5 be a vector field on T * X . We say 5 is a Hamiltonian uectorJield if there is a C"'-map H: T * X + R such that

( = (dH)'.

We say H is a Hamiltonian for 5. Note that, if (qi, pi)are natural coordinates on T * X , then, by Lemma 7.29, we have (a/aqi)b = dpi >

(dq')'

= - (ajdp,),

(a/api)b = - d d ,

(dp,)'

=

(7.31)

(a/&').

THEOREM 7.31 Let 5 be a Hamiltonian vector field on T * X with Hamiltonian H . Let (qi,pi) be natural coordinates on T * X . Suppose c(t) =

116

7. HOLONOMIC SYSTEMS

(qi(t),p i ( t ) ) represents an integral curve of

5 in these coordinates. Then

PROOF: The theorem asserts that the local representative of 5 in the given coordinates is

In other words, we are asserting that aH a ( d H ) =-----. api a41

a a q api

aH

THEOREM 7.32 Let M be the configuration space of a conservative, holonomic system. Let H: T * M -+ R be the Hamiltonian of the system. Then the phase space trajectories of the system are the integral curves of the Hamiltonian vector field (dH)? PROOF:

Immediate from Theorems 7.16 and 7.31.

I

REMARK 7.33 The only place in the proof of Theorem 7.31 where we used the assumption that our coordinates (qi, p i ) were natural coordinates was where we wrote = dq'

A

dp,.

If (Qi, Pi) is an arbitrary chart on T * X , then R is a linear combination of dQ' A dQj, dQi A dP,, and dPi A dP,. If we happen to have a coordinate system (Qi, P i ) for which R = dQi A d P i , then just as in the proof of Theorem 7.31, we conclude the integral curves (Qi(t),P j ( t ) ) satisfy Qi = aH/aPi,

Pi = -aH/aQi.

That is, Hamilton's equations hold in the (Qi, Pj) coordinates. This motivates the following definition.

117

HAMILTONIAN AND LAGRANGIAN VECTOR FIELDS

DEFINITION 7.34 A chart (Qi, P j ) on T * X is a canonical chart and Q’, P j are canonical coordinates if

R

=

dQ’ A dP,.

The differential equations

..

Q‘

Pi= -aH/aQ‘,

= 8H/ilPi,

(7.32)

which hold in any canonical coordinate system, are often knownas Hamilton’s canonical equations. There are techniques for constructing canonical coordinates in a given problem so as to make the problem simple. For more information on these methods see [ 2 ] . EXAMPLE 7.35 Consider the system of Example 7.7. We have the coordinate 8 on M and corresponding charts (0, (0, ps) on T M , T * M . O u r Lagrangian, Legendre transformation, and Hamiltonian are given by

e),

L = + m t 2 d 2+ my[ cos 8, H

= ($/2mt2) - mg/

~ ( 0 , s=) (0, mt’d),

cos d.

The canonical 2-form on T * M is given by

R

=

dOAdp,

and we have dH

= mg/

sin 8 d0

+ (pe/m/’)dp,.

The #-operation is defined by [see Eq. (2.31)] dP

= -?/?pH,

dpi

=

8/80,

so

a +Ps a ( d H ) $= -my/ sin 8 ?p, me2 80’ Suppose (e(t),ps(t)) is a phase space trajectory, i.e., an integral curve of (dH)’. Then

4 = pe/mP2,

po = - m g t sin 6.

Of course these are precisely Hamilton’s equations, as required. Now (dH)*E Y h ( T * M ) can be pulled back to T M , by 9, to give 9 * ( d H ) # )E F A ( T M ) . We compute

118

7. HOLONOMIC SYSTEMS

noting that p e = rnt28, so

a - i a ap, me2 ad'

- - ~-

a

a

a8

a8

- ---

so that Z*((dH)')

9 .

= -- sin

e

8

a + 8 -.a a0 a0

If (8(t), e ( t ) ) is a state space trajectory, then d8/dt

=

8,

d8ldt = -(g/d) sin 8.

These are, of course, Lagrange's equations, as obtained in Example 7.7, since the above system of two equations is equivalent to the single second-order equation

8 + ( g / t ) sin 8 = 0. Now we return to our formulation of mechanics on state space. We construct a 2-form on T X and show that the state space trajectories are exactly the integral curves of (dE)', where E: T X -+ R is the total energy function. Suppose L: T X -+ R is the Lagrangian of a conservative holonomic T X + T * X be the Legendre transformation of the system. system. Let 9: DEFINITION 7.36

The Lagrangian 2-form corresponding to L is

QL

=

9*R. We leave the proof that QL is non degenerate as Exercise 7.7. We thus have b and # operations, b: FA( T i ) + F :( T X ) DEFINITION 7.37

with inverse #: F(: T X ) .+ FA(T X ) .

The vector field (dE)# is called the Lagrangian vector

jield for L.

We shall use subscript coordinates to indicate partial derivatives, e.g., L,i = a q a g i , L~~~~ = a2L/aqi a g j . LEMMA 7.38

In a natural chart (q', 4') we have 52,

= L,,,,

dq'

A

dq'

+ L4i4jdq' A dq'.

The proof is left as an exercise. Now our energy function E has differential dE

=

E,i dq'

+ E,,

dq'.

We want to compute (dE)'. If we write (dE)'

= c('(d/dqi)

+ fl(d/dq'),

then

119

HAMILTONIAN A N D LAGRANGIAN VECTOR FIELDS

there is a (2n x 2n)-matrix such that

We want to determine this matrix. Let A = (L4z.ij);A is invertible and, in fact, A = (gij), since L = i g i j q i $ - V(q). Let B = (LqIqj). LEMMA 7.39 The matrix of b is

This says (a) (d/dqJ)b= (L,pqt - L q t q J ) dq' dq'. (b) (d/dq')b = -LqLq,

+

dg',

Lqt4J

The proof is left as an exercise. The matrix inverse of

1 -:1

------+----

is

[

0

i I

A-' A - ' ( B ' - B)A-'

------+-------------

-A-'

I.

We leave it as Exercise 7.10 to show that the above two matrices are indeed inverse to one another. (Compare this with the formula for inverting a (2 x 2)-matrix,

[:,-[='-I:

1

d

.I

-b

.)

So the components of (dE)# for the basis dldq', d/dq' are given as the column

120

7. HOLONOMIC SYSTEMS

This says =

[ T I ]

A

Eqn

which proves the lemma. Now we compute fl. These are the entries in the column

+ A-'(B'-B)A-' Thus, by Lemma 7.40, we see Lqkgjp'

But recall that L,k$

= gkj.

=

-E y k

+ (L@ik

-

L$kdj)qJ.

Then we get

pi = -gikEqk + gik(LqJqk- L,kqj)4'.

(7.33)

Therefore we have (dE)'

4' 7+ ( -gikEqk l3

=

a9

+

. a

yik(L,jqk -

L4kqj)q')@.

Let t = (2,d') be an integral curve of (dE)'. Then d ( t ) = t'(t)and

ci = -gik(?))E,k(4 + gik(4(L,,k

-

L4kYj)?'.

This gives L4r4' . c'

=

+

-(2Lqr4, - Lqr) (Lqj,. - L,,,,)t'

or L 4'4' . ?i

-

L,,

+ LgrqJi.J = 0.

(7.34)

121

TIME- DEPEN DENT SYSTEMS

This can be rewritten as

(7.35) Thus, we may summarize as follows: In a natural coordinate system, the differential equations describing the integral curves of (dE)’ are the first-order equations resulting from Lagrange’s equations by the standard trick of “doubling the number of variables.” We also see: The Lagrangian vector field (dE)‘ is the iizfinitesiinul generutor of the dynamical group of the conservative holonomic system having E : T X + R as total energy function.

TIME-DEPENDENT SYSTEMS Mechanical systems involving time-dependent forces or dissipative forces (producing time-dependent energies) can sometimes be described using a Lagrangian function which depends explicitly on time. Thus we consider

L:R x T X + R . The Lagrange equations corresponding to this Lagrangian are

(7.36) where (qk,q k )is a natural chart on TX and (q’)k= dqk/dt, 1 5 k I n = dim X . The evolution of such a system can also be described by the flow of a vector field on T * X obtained using the symplectic structure of T*X, just as in the time-independent case. However, now the Hamiltonian will depend on time and hence, the vector field will be a time-dependent vector field. For each t we define (a) (b) (c) (d)

.Yr:T X + T * X , A,: T X + R , E,: T X + R , H,: T*X -+ R ,

by holding t fixed in L and proceeding as before. Thus, in local coordinates,

.IP,(q,4 ) ( 4 = (;L/(%k)(f,4, 4)uk, At(q, 4) = y’r(q> Ci)(Y,

4 1 9

E,(q, 4 ) = A , ( % 4) - L(f, q>4, ffr(y, P ) == Et(9’; ‘ ( q ,P ) ) .

122

7. HOLONOMIC SYSTEMS

We assume, as before, that 2, is a diffeomorphism for each t. The phase flow, i.e., the motion of the system in phase space, is constructed as before, using the canonical 2-form Q on T * X and the time-dependent Hamiltonian H ( t , q, p ) = H,(q, p ) . DEFINITION 7.41

The Hamiltonian vector field X , on T * X is defined by

THEOREM 7.42 If q = q(t) is a curve in X which satisfies the Lagrange equations (7.36) and (q(t),p ( t ) ) = gt(q(t),q'(t)),then (q(t),p ( t ) ) is an integral curve of X , on T * X . PROOF: The proof is obtained just as in Theorem 7.32, by replacing H by H , and 2' by Yt.We arrive at

and

When we describe the evolution by a vector field on T X , the situation changes more drastically due to: (a) time-dependent 2-forms QL,, = YTSZ, (b) the appearance of the dissipative vector field Z,, given by Zr(q, 4) = T y ; '((d/dt)(yt(q,

4))).

Because we have a time-dependent 2-form Q,,t, we have a timedependent sharp operation which we denote #*. Then we have DEFINITION 7.43 The time-dependent Lagrangian vector field X , on T X is defined by X,(t, 5 ) = (dE,)3f(t). Then, for each fixed t, we have

T 9 t ( X , ( t , q 7 4)) = X,(t, 2 ' t ( 4 , 4 ) ) . DEFINITION 7.44

Define a time-dependent vector field X , on T X by 4, 4) = x E ( t , q? 4.1

-

zt(q,

4).

THEOREM 7.45 If q = q(t) satisfies the Lagrange equations (7.36), then (q(t),q ( t ) )is an integral curve of the time-dependent vector field X , = X , - Z i on T X .

123

TIME- D EPENDENT SYSTEMS

PROOF: Define

4:R

x T X +. R x T * X by @(t,q, 4 ) = ( t , Ip,(q, q ) ) and let

dt) = 4k q ( 0 7 4(t))Then Now

L'

is an integral curve of X ,

W t , 4>4)(1,4 6 )

= (1,

+ (i3/&)

on R x T * X by Theorem 7.42.

il + d YLq,4 ) )

T Y t ( q , 4)(4",

and

EXAMPLE 7.46 Suppose that a conservative mechanical system is described by L(q, q). We will see in Chapter 15 that the form R, is preserved under the motion. Suppose additional work is done on the system resulting in FTR, = f(t)R, where F,: T X + T X gives the trajectories on T X , f ( 0 ) = 1 and f ( t ) > 0. We then take e(r,q, q ) = f(t)L(q,4), which gives E"(t,q, 4) = f ( t ) E ( q ,4). We also get Ri, = f(t)R, and, since E = f ( t ) E , the Lagrangian vector field XE(r, q, 4) = X,(q, q ) is the original time-independent field. The dissipative vector field is given by Z't(4>4) = ( f " / f P ) ) T Y - ' ( $ ( 4 , 4 ) ) ,

where Y is the Legendre transformation determined by L(q, 4 ) and

{a% 4) + fY(4,4))

a q , 4) =

in

T,,,.,,(T*X).

t=n

Thus f l * ( t , q, q) = X J q , 4) ( f ' ( t ) / f ( r ) ) T Y - ' ( $ ( q4)). , z(t,q, q) = ea'($q2 - f q 2 ) leads to -

For

example,

m,

494) = X d q , 4) = 4(VW - q ( a / W ,

Z,(q, 4) which gives d q / d t

=

=

kj(?/?q),

q, d q / d t

=

-q

8 , = q(a/aq) - ( q + kj)(a/?4), j.4, so ( d 2 q / d t 2 )+ L(dq/dt) + q = 0.

-

EXAMPLE 7.47 Suppose that a conservative mechanical system is described by L(q, q), so the equations are (d/dt)(dL/&jk)= dL/dqk, and external

1 24

7. HOLONOMIC SYSTEMS

+

forces are imposed to give us (d/dt)(dL/dqk)= (dL/dqk) Fk(t) as dynamical equations. Then take L(t,q, q) = L ( t , q, 4) Fk(t)qk. In this case 9 , ( q , 4) = 3 ( q , 4.) which implies that 2, = 0 and RL, = R,. Then E(t, q, q) = E(q, q) F,(t)q", so that X E is time dependent and its flow determines the dynamics because 2, = 0. For example, L(t, q, q ) = f ( q 2 - q 2 )+ f ( t ) q gives 3 , ( q , q ) = 4, SZet = d q A d 4 , 2, = 0,

+

j77L.2 2q

+ t q 2 - f(t)q,

x, = 4(d/%)

- (4 - f ( t ) ) ( d / W

So the dynamics is given by dq/dt = q

and

dq/dt = -q

+f(t)

or simply, (d2q/dt2)+ 4 = .f(t). COMMENT: In Chapter 15 we will describe time-dependent systems in terms of structures on the manifold R x X . Our analysis will, in fact, also include systems which cannot be described using Lagrangian functions.

EXE R CI S ES 7.1 See the section on conservative forces in the text. Let i: M inclusion. Then

i*F"(u,) Show that if F"(u,)

=

E

T,*M

so i*F": T M

+ R3" be

+ T*M.

-dV(x), then i*F"(v,) = - d ( VIM)(x).

7.2 This exercise refers to Example 7.17. Show that if n is the plane in R 3 determined by the initial position and velocity, then the particle always moves in n. (Use vector notation, r = (q', q2, q3), v = ($, q2, q3), and show n is the plane determined by r x v.) 7.3 Show that if a, p are 1-forms on the vector space V , then a A p = a 0 B - B O a. 7.4 Let ei, 1 I i 1 4 , be the standard basis for R4, ei,1 I i I 4, the dual basis. If w : R4 x R4 -+ R is given by w((a', a', a 3 , a4), (b', b2, b3, b4)) = u3b2- u2b3 a2b4 - a4b2, express w as a linear combination of the basis forms e' A 8,i < j .

+

7.5 Prove Proposition 7.28.

125

EXERCISES

7.6 If X is connected and H , , H , are Hamiltonians for the vector field on T * X , then HI and H , differ by a constant. 7.7 Prove that R,, as defined in Definition 7.36, is nondegenerate. 7.8 Prove Lemma 7.38. 7.9 Prove Lemma 7.39. 7.10 Show that if A , B, and C are n x n matrices with A and C invertible, then

7.11 Let SZ be the canonical 2-form on T * X (see Definition 7.27). For functions f:T * X + R and g: T * X + R define the Poisson bracket { f,g } : T * X -+ R by { f , g } = SZ(dfr,dyff). If H is a Hamiltonian function for a vector field ( on T * X show that f is constant along the integral curves of ( if and only if { H , f } = 0 (see also Exercise 2.5). 7.12 Let M be a manifold and L : T M + R be a smooth mapping. For u E T m M define LF(u): T m M + R by P ( u ) ( w )= (d/dt)l,=,L(u + tw), using the fact that T,M is a vector space so that u + tw is well defined. We thus obtain 2':T M + T * M , which is called the Legendre transjormation determined by L . When L = T - V z as considered in the text, LF is a diffeomorphism. However, show that if ( M , g ) is a then ,9 is not a Riemannian manifold and we take L(u) = ,/= diffeomorphism. 0

7.13 Let M be the configuration space of a holonomic system, H : T * M + R the Hamiltonian function, and X , the Hamiltonian vector field. If S: M x R + R is a smooth function, then we have

where t is the coordinate on R , and we have d,S: M x R

+ T*M

defined by holding t fixed. Define a time dependent vector field X , on M by X , = Tz* X , c- d,S. Show that if 0

(7.37)

126

7. HOLONOMIC SYSTEMS

on M x R, then every integral curve (m(t),t ) of X , produces a Hamiltonian trajectory d,S(m(t), t). Equation (7.37) is called the HamiltonJacobi equation. If F = 51 + dH A dt on T * M x R where R is the canonical 2-form on T * M and g: M x R --t T * M x R by g(m, t ) = (d,S(m, t), t), show that the Hamilton-Jacobi equation implies that g*F = 0. Show that the preceding constructions work for time-dependent Hamiltonian functions H: T * M x R -+ R.

8 Tensors

TENSORS O N A VECTOR SPACE Let E be an n-dimensional real vector space. DEFINITION 8.1

A tensor of type (:) on E is a multilinear map t:E x

... x E

Gi2

x E* x . . . x E * + R ;

z z

r is the cooariant degree oft, s is the contravuriunt degree oft, r s is the total degree oft.

+

The set of all tensors of type (:) is a vector space denoted T;(E).We agree that T:(E) = R. EXAMPLES 8.2

(a) There is a canonical isomorphism e: E -+ E**, defined by [e(u)](a) = c( E E*. By definition, we have T k ( E ) = E**. Whenever it is convenient, we can make use of the canonical isomorphism and identify TA(E) with E. (b) A 1-form on E is a member of E* which, in the present notation, is

a(u) for u E E,

TW).

(c) Second-degree covariant tensors, discussed previously, are elements of T;(E).

127

128

8.TENSORS

REMARK: If a E T;;(E),p E T"(E), y E T"(E), then (a 0B) 0y = a 0(P 07).

Because of this, we simply write a 0 j0y. THEOREM 8.4 Let e l , . . . , e, be a basis for E with dual basis e l , . . . , e".

Then the set

{ e i l Q . . . O e i ~ ~ e j , O . . . O e<ji ~k I< ln , 1 5 j l < n } is a basis for T;(E).Thus dim T"E) NOTE: In general a 0p # a 0p = p

=

nr+S.

p 0a. However, if

a E T;(E), j E T",E), then

0a E T"E).

If a E T"E) and e l , . . . ,en is a basis of E, then the components of a for the given basis are the numbers . + . , .r = a(eil,. . . , ei,,ejl, . . . ,e'"). a ! Il 'I. '.. PROPOSITION 8.5 a = a{:::.'teilQ . . . @ eir Q e j l O .. . @ ej,, where the summation convention is in force.

We leave this proof as an exercise. EXAMPLE 8.6 (Kronecker delta) There is a canonical isomorphism Ti@) E L(E, E ) given by a + 4,Oi(u) being that vector in E such that, if w E E*, then w[Oi(u)] = a(u, w). Let { e , } be a basis for E. We have components

aj

= a(ej,ei).

On the other hand, Oi E L(E, E ) has a matrix claim = a;. To see this we must show that

Oifi

Oij such that 2(ej) = 2jei. We

ei(B(ej))= a(ej,ei).

But this is true by definition of 8. So when we have a = ajej Q e i , we have also defined a linear map E -+E whose matrix is the matrix of components (a)). We define the Kronecker delta 6 E T : ( E ) to be the tensor corresponding to the identity map in L(E, E). Given 1 I a I r, 1 I b I s there is a linear map C:: T;(E).+ T;: : ( E ) called contraction on the "a"-covariant index and "6''-contravariant index. Rather than give a general definition of contraction, which involves notational difficulties, we consider some special cases. Consider C:: Ti(E)+ T:(E). Let t E Ti(E). Fix u , E E, a , E E*. There is a bilinear map z : E x E* + R , (u, a) -+ t(u, u l , a,, a). This corresponds to a linear map z*: E + E .

129

TENSOR FIELDS ON MANIFOLDS

We define (Cft)(u,,a l ) = trace(?); Cft is well defined. We claim (C:t)(u,,a , ) is a bilinear function of (u,, al). We verify directly that C: t(u, + u',, al)= Ctt(ul, a ; ) + C:t(u;, al). Let

+ u;, a ; , a),

T;(U, a ) = t(l', u1 t2(u,

a ) = r(u, u,,

T3(U,

a ) = r(u, u;, a,, a).

+

@I,

a),

+ +

+

Then z1 = t 2 t,, so ?, = t 2 ?,. Therefore, trace(?,) = trace(?,) trace(?,). This shows Cff(c, (11, a , ) = Cft(u,, a,) Cit(u;, a l ) . Let e l , . . . , e n be a basis for E, e', . . . , e n the dual basis. We have C:t(u,, a , ) = t ( e i ,u,, a l , ei), where the summation convention is used. For the bilinear map t:(u, a ) -+ r(u, I ] , , a , , a) corresponds to a linear map ? whose trace is the sum of its diagonal elements. Then trace(?) = ei(?(ei))= t(ei, ei), so ~ f t ,(a l~) = z ( e i ,e') = t ( e i ,u a ,, ei).

+

,

COMPONENT DESCRIPTION:C f t E T : ( E ) , so Cft has components (C21 t)ji

-

-

Cft(ej, e') = t(e,, ej, ei, ek) = ti: ( k is summed). Similarly, if t E T:(E), then C:t(al, a23 ~

3 =)

t(ei, a,,

a29

ei,

that is, 3 i j k - ijrk

(C,r) - t r .

TENSOR FIELDS ON MANIFOLDS DEFINITION 8.7 A tensorjeld of type (i)on a manifold X is an assignment to each x E X of a tensor T(x)E Ts(T,X). We require the following smoothness condition: Let ( U , 4 = (XI,. . . ,x")) be a chart. Define T j ; ;:;, U+Rby

We require these component functions be C"' for every choice of coordinates. The set of all (;)-tensor fields on X is denoted F ; ( X ) . Consider a tensor field of type (i).Let T

=

0

Ttj dx' 0 dx' 0__ ?.uk

in (XI,. . . ,x")-coordinates,

130

8. TENSORS

T

=

a

Tfj d i i 0df' 0a.wk

in (TI, . . . , ?')-coordinates.

How are the T!j related to the Tfj? We have

For a tensor field of type

(s)

we have (see Exercise 8.6)

REMARK 8.8 In classical differential geometry, an (:)-tensor field was an assignment, to each coordinate system, of component functions Ti::::{; such that the transformation law (8.2) holds. Multilinear mappings were never mentioned. But the whole reason for requiring the transformation law (8.2) is as follows: Let (ufl,), . . . , (of,,), (ail'),. . . , (a?)) be r contravariant and s covariant vectors. We want the expression

T{; ;;{ ,'

,. . .

yir

a(,') . . .

(r)

~1

@.is

(S)

to be a scalar, i.e., independent of coordinates. That is exactly what the transformation law (8.2) guarantees. Suppose T is an (:)-tensor field. If ui E Ft(X),d E F y ( X ) for 1 I i < r, 1I j I s, we have T(u,,. . . , t i r , a', . . . , 2): X -+R, given by x

.+

T(x)(ul(x), . . . , ur(x),a'(x), . . . , as(x)).

-

Thus a tensor field T E F ; ( X ) defines a map, also denoted T , T:Y h ( X ) x . . . x Y h ( X ) x Y y ( X ) x . . . x Y y ( X ) -+ F ( X ) v r factors

s factors

where P ( X ) = C"(X, R). This map is multilinear over F ( X ) , i.e., it is additive in each variable separately and satisfies T ( u , ,. . . , f ~ , ., . . , u,, a ' , . . . , CC')= f T ( u , , . . . , u,, a', . . . , a')

for f

E

P(X).

131

TENSOR FIELDS ON MANIFOLDS

-

THEOREM 8.9 If r-factors

s-factor5

b

T:Y A ( X ) x . . .

x $Y(X) x

x

. . . x Y ? ( X )4 F ( X )

is multilinear over F(X), then T arises from a unique (;)-tensor field as above. PROOF To simplify the notation, we give the proof in the case r So we have T : Y A ( X )x Y A ( X ) x Y Y ( X ) + F(X).

=

2, s

=

1.

Let xo E X , u l , u 2 E T,,X, a E T$X. There exist vector fields t l , t2on X and a 1-form q on X with tl(x0) = u l , t2(x)= t i 2 , q(x0)= a. Define T(XO)(U,? L i z ,

co = T(51>52.v ) ( x o ) .

We need to show T(x,)(u,, u2, a ) does not depend on tl, t2,v. For example, if we replace 4 , by another vector field 5, with f,(xo) = u l , we must show

TtP,, 5 2 , v)(xo) = T(51, 5 2 , v ) ( x o )

or, T(5l - F l ? 5 2 , v)(xo) = 0.

So we show if 5(xo) = 0, then T(5,t 2 ,q)(xo)= 0. Let ( U , (ql, . . . , 4")) be a chart about xo. Let y:X + R be C" with g = 0 on X - C, C c U compact, and 61 = 1 on some neighborhood of xo, say I/. Let 5 = l1a/dq'. Then y2( = y t ' g (7/?4' = g('O,, where the vector field 8, = g d/dql is defined on all of X . Then s2T(5,5 2 , v ) = T(q'5, 5 2 ,

v ) = YS1T(Q,, 5 2 , v).

FIGURE 8.1

132

8.TENSORS

Now g(xo) = 1, (‘(x0) = 0 for all i so we get

T(t, 5 2 , v M x O )

=

T ( g 2 t >5 2 7 v ) ( x ~= ) (gt’)(Xo)T(ei,5 2 9 v ) ( x O )

= 0-

Similarly, T(<,,1 2 , q)(xo) is independent of the choice of t2,q. Thus T(x,)(v,, u 2 , a) is defined. We leave as exercises the proofs that the tensor giving rise to T is unique and that the tensor field just defined satisfies the smoothness condition. EXAMPLE 8.10 Fh(X) is a module over the ring F(X). Hom(FA(X), F ( X ) ) is the set of F(X)-linear maps FA(X) -,F(X). By Theorem 8.9, Hom(FA(X), F ( X ) ) E Fy(X). Similarly, Hom(Fy(X), F ( X ) ) 2 YA(X).

THE LIE DERIVATIVE Let X and Y be vector fields on M . We define an operation which provides a measure of the rate of change of Y in the direction of X. DEFINITION 8.11 The Lie derivatiue L,Y is the vector field defined by L,Y(p) = (d/dt)I,=,[(a:Y)(p)],where (a,),is the flow of X and a: = (ap,)* as in Example 6.17; that is, (a: Y ) ( p )= Tcc - t ( Y(ctt(p))).

Let a: 9(X)-+ A4 be the flow of X. Given p E M , there is DISCUSSION: an open neighborhood U of p and an E > 0 such that ( - E , E ) x U c 9(X). Then for It1 < E we have a,: U .+ a,(U), a diffeomorphism. So a : Y ( p ) T,M ~ and t -+ a:Y(p) is a Cm-curve in T,M whose derivative at t = 0 is LxY(p). More generally, DEFINITION 8.1 2 If X E F h ( M ) and T E Fz(M), the Lie derivative of T with respect to X is

LxT(P) = ( 4 4It :=.0

VP).

So L,T(p) is obtained as the derivative of a curve in T;(T,M). Here (g*T)(p)(v,,. . . , v,, wl,. . . , U S ) =

T(a,(p))(Ta,(u,),. . . , Ta,(v,), w1 0 Ta-,, . . . , os0 Ta-,).

We now give some useful formulas for computing Lie derivatives. Further results are in the exercises (see Exercise 8.9). CONVENTION 8.13 It is convenient to use the “comma notation” for partial derivatives. According to this, we write aj’/axJas l jand a2Ti/axkaxe

133

THE LIE DERIVATIVE

as T:,lk.Similarly, if f is a function of certain variables including t , we might write f,, for df/dt. We now have the basic formulas for Lie derivative of a function, a vector field, and a 1-form.

PROOF: Let M : 9 ( X ) + M be the flow of X . Given p E U , choose E > 0 and an open set W c U about p such that ( - E , E ) x W c 9 ( X ) and a((-&, E ) x W )c U . We write a on W x ( - E , F.) as a(t, q) = (a'(t, q), . . . , ~ " ( 4)). t , We write a,(q) = a(t, 4). We also note that ai,,(O,q) = X'(q). Now consider (a).

-bf(P)= (~/dt)Jt=o(MjEf)(P) = (Wt)l,=,f(a(t, PI) = .fi(p)~i.t(O~ P) = (.fiXi)(P).

so c ( ' , ~ ~ (pO) ,= 0

and

ai.AO, p ) = bij.

Thus,

(L,Y(p))' =

- X',,(p)Y'(p)

+ b"Yj,J(p)X~(p)

134

8.TENSORS

or

PROPOSITION 8.15

Let X

E

Fh(M). Then:

L,[Y + Z ] = L,Y + L,Z, for Y , Z E FA(M), Lx(ol w 2 ) = L,w, L,w,, for wl, w 2 E Fy(M), L,( f Y ) = fL,Y + (L, f ) Y , for f E 9(M), Y EF : ( M ) , L,( fw) = fLxw + (L,f ) w , for f~ B ( M ) , w E F:(M), L , [ w ( Y ) ] = L,w(Y) + w(L,Y), for Y E F;(M), w E Fy(M). We leave this proof as an exercise. (a) (b) (c) (d) (e)

+

+

Moving Frame Description Let e l , . . . ,e, be a basis for T,M. For It1 < E we have a diffeomorphism a,. Let ei(t)= Tpat(ei). Since Tpat:T p M + Tar(p,M is an isomorphism,

(el(&. . . , e,(t)) is a basis for Tat(p)M. So if c: (- E, E ) + M is the integral curve of X given by c(t) = at@), then el(t),, , . , e,(t) is a basis for T,,,,M. The functions el(t), . . . , e,(t) form what is called a moving frame along c.

Suppose Y is a vector field and Y(c(t))= Yi(t)ei(t). Then, a : ~ ( p= ) ( ~ , a , ) - ' ~ ( c (=t )( )~ ~ a , ) - ~ ( Y ' ( t )= e , Yi(t)ei, (t))

so (d/dt)l,=,cr:Y(p)= Yi(0)ei.We have proved

135

THE BRACKET OF VECTOR FIELDS

PROPOSITION 8.16

LxY(p) = Yi(0)ei.

If c is the integral curve of X through p and v E T p M , then the vector T,cc,(v) E T,,,,M is said to be obtained from v by Lie transport along c.

THE BRACKET OF VECTOR FIELDS The special case of Lie derivative, L x Y , is worth looking at in more detail. DEFINITION 8.17 If X , Y E $,!,(M), the Lie bracket is [ X , Y ] E T , ! , ( M ) defined by [ X , Y ] = L,Y. PROPOSITION 8.1 8

(a) (b) (c) (d)

The bracket satisfies

[ X , X I = 0, [ X , Y ] = - [Y , X I , [ X , aY b Z ] = a [ X , Y ] b [ X , Z ] for a, b E R, [ X , [ Y , Z ] ] + [ Z , [ X , Y l l + [Y, [z,X I ] = 0.

+

+

PROOF: The local formula for [ X , Y ] is

[ X , Y ] i = Y’<,X’ - x’,jY’ Using this, the various parts of the proposition are easily checked. PROPOSITION 8.1 9

If X , Y E Y , ! , ( M )f’, , g

I

E F ( M ) , then

C f X , g Y 1 = f g [ X , Y I + f ( L x g ) Y - g(L,f’)X. PROOF: Write everything out in local coordinates. I REMARK 8.20 Part (d) of Proposition 8.18 is called the Jacobi identity. Note that it can be written as

LX[Y, 21 = [LXY, 21 + [ Y , LXZ]. Thus, we may say that the operation L x is a derivation with respect to bracket. DEFINITION 8.21 A Lie alyebru over the reals is a real vector space E together with a bilinear map . : E x E -+ E such that

(a) x . x = 0 for x E E, (b) x . y = - y . x for x,y E E , (c) x . ( y . z ) z .(x. y ) y . ( z . x) = 0 for x, y , z

+

+

E

E.

136

8. TENSORS

REMARK: Usually in dealing with Lie algebras the multiplication is written [x, y ] rather than x . y. PROPOSITION 8.22

Y h ( M ) is a Lie algebra with respect to the Lie

bracket. PROOF: This is exactly what Proposition 8.18 says.

I

REMARK 8.23 F A ( M ) is an infinite-dimensional Lie algebra. We shall see further examples of Lie algebras of finite dimension when we consider the theory of Lie groups. LEMMA 8.24 (d/dt)I,=,,a:Y(p) = T,~JL,Y(a,,(p)))or, equivalently,

Tp@,,((d/dt)l, =,,a:Y(P))= - b Y ( E t 0 ( P ) ) .

We leave the proof as an exercise. LEMMA 8.25 Let X , Y be vector fields with flows LY, p. Let p E~ > 0 such that for It1 < E ~ (sI , < z2 we have

E

M . Assume

there exist

dt, B(s, P I ) = B(s, 4 t , PI). Then [ X , Y ] ( p )= 0. PROOF: Differentiate the assumed equation with respect to s at s = 0. We

get TP%(Y(P))= Y ( a 4 ) .

So a:Y(p)

= (T,,cc,)-'Y(a,(p)) =

Y ( p )and hence,

[ X , Y X P ) = L,Y(P)

= (d/dt)l,=,Y(d = 0.

I

LEMMA 8.26 Assume [ X , Y ] = 0 in some open set U . Suppose E ' ,

0, p

E

U are such that both

44 P(s, P I ) are defined and lie in U , for It/ < E 4

t

3 B(s3

PROOF: Let t be fixed with It1

P(s, @(t> P))

and ~ Is1 ,

<E

~ Then .

P I ) = P(s, 44 PI).

< E , . Let

E~

>

137

VECTOR FIELDS AS DIFFERENTIAL OPERATORS

c2 is an integral curve of Y with initial value cc(t,p). We claim c1 is too. That will prove our result. c;(s) = 7',~(s,p)c(,(Y(p(s,p))). We claim this equals Y ( 4 4 B(s, PI)). We prove

Y(B(s,P I )

= (7',,,,,,,%-

Y Y ( 4 4 P(s, P ) ) ) ) .

(*I

(d/dt)(~p,,*,,~,)- Y ( d L P(s, P ) ) ) ) = ( T p ( s , p )'(L*Y(a(t, w P(s, P I ) ) ) = 0

since L,Y = [ X , Y ] = 0. So the right side of (*) is independent oft. The left equals the right at t = 0. I

VECTOR FIELDS AS DIFFERENTIAL OPERATORS Given X E Fh(M) and f' E F(M), we can form a new scalar d f ( X )E F(M). If we fix X , this defines an operator, also denoted by X ,

X : F ( M )+ 3 ( M )

by

X ( j )= df(X).

As is often done with linear operators, we sometimes write simply X f . Since X f ( m ) = (dW01, = o f ( c ( t ) ) , where c is any curve with c'(0) = X(m), we see that X f is the directional derivative off in the direction of X . In case X = i?/dx' is a coordinate vector field in some chart, 4 = (XI,. . . , Y"), we get

again justifying the (d/dx')-notation. If X and Y are vector fields, consider X , Y and [ X , Y] as operators. Then we have THEOREM 8.27

[X, Y ] f

=

X ( Y f )- Y ( X f ) .

PROOF: Choose coordinates and write X = X id/Jx', Y = Y' c?/2xi. Then

Yf

= df( Y ) = f j Y J ,so

that

X ( Y f ) = ( f j Y j )kXk = f j k Y J X k+ .fjYJ,,Xk. Interchanging X and Y subtracting, we get

X ( Y f ) - Y ( X f ) = Y',,XYj - X ' , Y y j . But by Proposition 8.14, we see

[ X , Y ] j '= f j ( Y',kXk- X',,Yk), which completes the proof.

I

138

8.TENSORS

REMARK: Theorem 8.27 shows that, in terms of operators, bracket is the usual commutator.

EXER C IS ES 8.1 Show the validity of the remark following Definition (8.3). 8.2 Prove Proposition (8.5). 8.3 How are the components of a @ /? related to those of a and /?? How about a + /?? If c is constant, how are the components of ca related to those of a?

8.4 Find the components of the Kronecker delta 6 for any basis {ei} of E . Show that if t E T:(E) and the components of t are the same in every basis, then t = a6 for some a E R . 8.5 (a) (b) (c) (d)

If t = tjej@ ei, find Ctt. Find Cis where 6 is the Kronecker delta. If t = tfFei @ ej @ ek @ e, @ em, find an expression for CfCft. If u l , . . . ,u, E E , a l , . . . ,a, E E*, then t = a, 0 . . . 0a, 0u1 0 . . . @ us E T;(E).If 1 I iI r, 1 I j s s, show that

ai(uj)al * * . @ &,@. . . @ a, @ el @ . . . @ Zj 8 . . . @ e, (where over a term means to delete that term). For example,

Cit

=

"^"

C:(a @ u) = a(u).

(s)

has components T{::.':j;,Tt::::!;in two coor8.6 If a tensor of type dinate systems (xi),(Xi), then

8.7 Show that if the assignment of components to each coordinate system is made in such a way that (8.2) holds, then the components are the components of an (;)-tensor field.

8.8 Prove proposition 8.15.

8.9 Let X be a vector field on a manifold M and p be an element of M . Then: (a) If e l , . . . , e, is a basis for T,M and c is the integral curve of X through p , we have the moving frame ei(t), . . . ,e,(t) as in the text. Then { e i l ( t )@ . . . @ eir(t)

ej,(t)

. . . @ ejJt)l 1 I i, I n, I

I j, I n}

139

EXERCISES

is a basis for Ts(T,,,,M). If T E T"M) let T(c(t))= Ti,!;.';!;(t)eil(t) 0 . . * 0ej,(t).

Then show that

(b) If T E FS(M), T' e Fs:(M), then L,(T 0 T') = (L,T) 0 T' + T 0L,T'. A special case is when s = r = 0, i.e., T = f E F ( M ) . Then T 0 T' is just f T ' and L,(,fT') = (L,f)T' + fL,(T'); (c) Show that Lx6 = 0, where 6 E F : ( M ) is the Kronecker delta tensor; (d) Let TEF;(M), X I , . . . ,X , E F;(M). Then T ( X , , . . . , X,)E F(M)and L,[T(Xl,. .

. ?

Xr)] = ( L X T ) ( X l , . ., X r ) r

+ 1 T ( X , , . . . , L x X i , . . . ,X,); i= 1

(e) L , commutes with contractions, i.e., if T E Fs(M), 1 I a I r, 1 I b I s, then L,(c:T) = C:L,T.

8.10 Prove Lemma 8.24. 8.11 Suppose X , Y E Fh(M) with [ X , Y ] = 0 on M . Suppose a, P are the flows and that we have t, s E R, p E M for which P(S7

PI)

and

P(s,

44 P))

exist. Need they be equal? 8.12 With notation as in the previous exercise suppose P(s, a(t, p ) ) exists. Is it necessarily true that a(t, P(s, p ) ) exists?

8.13 Let X be the vector field on R Z defined by X ( a , b) = (-b, a). At (1,O) E R 2 we have the basis el = (1,0), e2 = (0,1) for T(l,o)R2.Let c(t) be the integral curve of X with c(0)= (1,O). Calculate the moving frame (el(t),e 2 ( t ) )along c obtained from e l , e2 by Lie transport. If Y(a,b) = (a + b, - l), compute L,Y(l, 0). If Z(a, b) = (a, b), compute L,Z. 8.14 Suppose X is a vector field on S 2 , given in spherical coordinates by x(0,4) = 4 8/80. Let Y(0,4) = (ape) + C$(d/8C$).Compute L,Y(0, 4).

8.15 Let

(XI,.

. . ,x") be a coordinate system on a manifold [d/dxi,a/axq

=0

for

1 I i, j I n.

M . Show that

140

8. TENSORS

8.16 (a) Suppose that f :M + N is a diffeomorphism and that X and Yare vector fields on M. If (V, 4) is a coordinate chart on N then ( j - ' (V ) , 4 o f ) is a coordinate chart on M . Use this fact to give a simple proof that [f,X,f,Y] = f,([X, Y]). (b) Suppose that f :M + N is a smooth mapping, that X, and X, are vector fields on M , and that Yl and Y, are vector fields on N satisfying

Tmf(Xi(rn))= X ( f ( r n ) ) Show that T,f(CX,, XZl(4) = CYI,

Vm E M *

Yzl(f(rn)).

8.17 (Tensor bundles) Let M be a smooth manifold. Define T;W) =

u

CUrnM)

meM

and define the projection

z;:T"M)

+M

in the obvious way. Let ( U , 4) be a chart on M . Then, for each rn in U we have the linear isomorphism dm4:T,M + R". Define (T4);:(z;)-' ( U )

+

4(U) x T m " )

by (T4)Xt)= (&Xt)), t o ((drn4)-' x . * . x ( d n d ) - ' x ( d m 4 ) * x . . . x (d&)*).

Since T;(R")2 ' Rk ( k = dr+')),we may use the collection of all such mappings as charts in a differential structure. Show that the topology for which all maps (T4); are homeomorphisms is a uniquely determined, second countable Hausdorff topology. Then show that these charts form a smooth atlas and hence determine a differential structure on T"(M). Show that, with this differential structure, r;: T ; ( M )+ M is a smooth vector bundle. Show that a smooth tensor field of type 0 on M is precisely a smooth cross section of this bundle (which is called the bundle of tensors of type on M ) .

(s)

9 Differentia I Forms

EXTERIOR FORMS ON A VECTOR SPACE Let E be an n-dimensional real vector space. An exterior form of degree p on E (or briefly a p-form on E ) is a tensor w E T;(E) which is skew-symmetric, i.e., interchanging any pair of variables changes the sign of the form. DEFINITION 9.1

NOTATION:Denote by A,(E) the vector space of all p-forms on E . Ao(E) is defined to be R . Thus A,(E) is a linear subspace of T:(E). If A: E + F is linear, then A*w = w (Ax . . . xA) is in A J E ) when w is in A,(F). Let s k be the set of all permutations on the set {1,2,. . . , k } : If o E s k define 6 : E k + Ek, where Ek = E x . . . x E (k-factors), by 8 ( u l , . . . , uk) = 0

...

(uu(l)?

7

uu(k)).

LEMMA 9.2

= z"

0

$.

. . , Uk)=Z"(w1,.. . , wk) where W i = U u ( i ) ; z"(W1,. . . , w,)= . . . , w,(k)).so N'qj) = U u ( , ( j ) ) and so fo8(U,,. . . , u k ) = ( T o z ( u l , . . . , u k ) . I

PROOF:

(W,(i),

If o, z E S k , then

Z"O$(u,,.

DEFINITION 9.3

If T E T f ( E ) ,o E s k , define oT E T@) by oT

=

T 0 8.

From Lemma 9.2 it immediately follows that o(rT)= (a " z)T. DEFINITION 9.4

For T E T f ( E )define d T

(9.1) E Ak(E)by

(antisymmetrization operator). Recall that

sgn(o) =

1, - 1,

o is an even permutation, o is an odd permutation. 141

142

9. DIFFERENTIAL FORMS

PROPOSITION 9.5

T E T f ( E )is in &(E) if and only if, for all a E S,, aT = sgn(a)T.

Proof is left as an exercise. PROPOSITION 9.6 (a) If T E T f ( E )then d T E I\,@), (b) If T E Ak(E)then d T = T , (c) If T E T f ( E )then d ( d T ) = d T . PROOF: (a) Let

7

E

sk.

Then

As a runs over Sk so does 7 a, so 0

1 sgn(z

0

O)(T

0

a)T =

arSr

Hence z d T (b) d T

= sgn(z)dT as = (l/k!)

1sgn(p)pT. PEsk

asserted.

c sgn(a)aT

= (l/k!)

c sgn(a) sgn(a)T

a€Sk

UfSk

= (l/k!) k!T = T.

(c) is immediate from (a) and (b) DEFINITION 9.7

Let

(1) E

&(E),

v] E

I AAE). Define 0 A v]

E Ak+!(E)

by

LEMMA 9.8 (a) A : A,(E) x A,(E) + I\k+/(E) is bilinear. (b) If i: E +F is linear and c1 E I\k(F),j? E A#), then A*(a A j?) = A*ct A A*@. If 0 E Ak(E) and v] E &(E), then 0 A v] = (- l)kdv]A w. (C) PROOF: (a) Left as an exercise. (b) We need (k t)!d ( A * c t 0 i*j?) = i*(ca.sk+d sgn(o)a(cc O j?)). But

+

143

EXTERIOR FORMS ON A VECTOR SPACE

while (k + t)!d(A*a0

A*/$

c

=

sgn(o)o(l*a 0 A*@,

U€Sk+/

so (b) holds. (c) Let p E sk+fbe given by

t. We see easily sgn(p) = (- l ) k Land q

So q A w = (- l ) k ' A~ q as asserted.

@ o = p(m @

q). Therefore

I

PROPOSITION 9.9 Let w E T:(E), 0 q ) = d ( o0 d q ) .

y~ E

T;(E). Then &'((do) 0 q)=

d(O

PROOF: d o = (l/k!)

For

k

OEESk,

let

curSk sgn(a)(oo)so we have

oESk+,

act as o on ( 1 , . . . , k } while leaving k

+ t fixed. Let H c sk+fconsist of all such o. Then

Then we have

For each a E H ,

+ 1,. . . ,

144

9. DIFFERENTIAL FORMS

So summing over c1 E H gives k ! d ( o Q yl) and hence, multiplying by Ilk! leaves d ( o Q yl), as desired. The proof of the second equation is similar. I PROPOSITION 9.1 0

( k + d + m)! 4 k!&!m!

( C O A ~ ) = A ~ O A ( ~ A = ~ )

0

0 4 0 Q,

where o,yl, 8 are forms of degree k, d , m. PROOF:

+ + 0 0) + ( k + d + m)!( k + d)! 4 d B ( W 0 I?) 0 6 ) ( k + L)!rn! k!d! ( k + L + m)! d ( m 0 q 0 0). ( k d m)! ( k &)!m! d ( ( w A yl)

( o A ~ ) A=~

k!L!m!

Similar calculation gives o A (yl A 0). PROPOSITION 9.1 1 o1

A ' .

I

Let a',. . . , orE Al(E). Then

. AW'

= r!d(o'

0 . . . Q or)

The proof is left as an exercise. THEOREM 9.12 Let e l , . . . , e, be a basis for E with dual basis e l , . . . , e". Then {e'l A . . . A eikJ 1 I d, < i, < . . . < i, I n>is a basis for Ak(E).There= (:). fore, dim PROOF: Let o E Ak(E).We write

o = oil . . ',eil0. . . @ eik. ,

Then

Now in this sum we do not have only terms with i , < * . * < i,. But, if i,, . . . , i, are not all different, then eil A . . . A elk = 0. Thus we may sum only over

EXTERIOR FORMS

145

ON A VECTOR SPACE

i,, . . . , i, different. But then, by changing the order of the factors, ascending order may be achieved, so that our proposed basis does indeed span Ak(E). Suppose we have a l i l ...ir,eilA . . . A eik= 0 (recall that this means sum over i, < . . . < ik). Choose a particular set ,jl, . . . , j , with 1 Ij , < . . . < j , < n. Then we have 0 = a l i l ..ik,eil . A . . . A eik(cjl,. . . ,ejk).Since j , < . . . < j k there is only one choice of i,, . . . , i, which gives a nonzero value, namely i, = jl, . . . , ik = j,. But ejl A . . . A ejk(cjl, . . . , ejk)= 1, so the sum above reduces to just a j , .j , . This proves linear independence. [ ,,

PROPOSITION 9.13 Let w ' , . . . , o k ~ A 1 ( E )Then . { w ' , . . . , w k } is a linearly independent set if and only if o1A . . . A wk # 0. PROOF: If { w ' , . . . , w k } is linearly independent, choose a', . . . ,an-, E A , ( E ) such that wl,. . . ,uk,a ' , . . . , is a basis. Then w' A . . . A wk is a member of the basis for Ak(E) given in Theorem 9.12 so w' A . . . A cok # 0. Conversely, if { w ' , . . . , w k } is a dependent set, then one vector in the set is a linear combination of the others. Suppose, without loss of generality, that 0 ' = @2(02

+

+ akak.

' ' '

Then w'

A..

. A wk = (a2w2

+ . . . + akwk)

A

= M2W2 A m 2 A W 3 A

' ' '

A Wk

+

+ (*I2 Lo3 + a k W k A 0 2 A . . . A w =k 0. M4W4 A

A

. . A wk

w2 A . A

A

'

.

'

U3W3 A 0 'A' A Wk

+

'

' A Uk

' ' '

[

NOTE: dim A,(E) = 1 since (i) = 1. A basis for A,(E) is {el A . . . ~ e " } , where e l , . . . , en is any basis for E*. DEFINITION 9.14

ul,.

A k-form w on E is decomposable if there exist w = v' A . . . A uk.

. . , vk E E* such that

REMARK: Exercise 9.5 says if dim E decomposable.

=

n, then every ( n - 1)-form is

PROPOSITION 9.1 5 Let o E A,(E), u l , . . . , u, a basis for E . Let wj = afiui.Then w(w,,. . . , w,) = det(aj)w(u,, . . . , u,). PROOF: Let u l , . . . , v" be the dual basis for u l , , . . , u,. There is a i. E R such that w = la1A . . . A u". Now u1 A . . . A u"(ul, . . . , 0,) = 1 so w(u,, . . . ,

146

u,)

= I.

9. DIFFERENTIAL FORMS

NOW w(W1,.

. . , W,)

= IU'

A.

. . A U n ( W 1 , . . . , W,)

1 (sgn

=A

0),$1)a$2)

. . . ax(,) = I det(aj).

assv

I

The fact that dim A,(E) = 1 is quite important as it lies at the heart of the concept of determinant. DEFINITION 9.1 6 Let I : E unique number a such that

3

E be linear. If w E A,(E) is not 0, there is a I*w

We dejine det(I) to be this

= am.

cl.

NOTE: Ifcw is any other n-form then A*(cw) = cA*w = cclw = acw, so det(I) does not depend on the choice of w. Let el, . . . , en be a basis for E . If I: E + E is linear, then A has matrix, for the basis e l , . . . ,en, (aj), where A(ej) = a:ei.

Then we have PROPOSITION 9.17

det(I) = det(a:).

PROOF: Let 0 # w E A,@). Then

w(Ie,, . . . , Ie,)

= det(I)w(e,,

. . . , en).

But Proposition 9.15 shows w(Ae,, . . . , Ie,) = det(aj)w(e,, . . . ,en). Since w(el, . . . , en) # 0, our result is proved.

1

ORIENTATION OF VECTOR SPACES Let dim E

=

n in what follows.

DEFINITION 9.18 Let ( u l , . . . , u,), ( w l , . . . , w,) be ordered bases for E . We say these bases are similarly oriented if wj = a;uiwith det(a;) > 0. If the two bases are similarly oriented, we write ( u l , . . . , u,) (w,,. . . , w,,).

-

We leave it as an exercise (Exercise 9.7) to show that relation on the set of all ordered bases of E .

- is an equivalence

147

ORIENTATION OF VECTOR SPACES

FIGURE 9.1

DEFINITION 9.19 An orientation of E is an equivalence class of ordered bases. An oriented vector space is a vector space together with a choice of orientation. REMARKS 9.20 If p is an orientation on E, the pair ( E , p ) is an oriented vector space. If ( u l , . . . , u,) is a basis in p , we say ( v l , . . . ,v,) is positively oriented (with respect to p). Note that each vector space has exactly two orientations.

I nt u itive Discussion In Fig. 9.1, we see that ( e l ,e,) ( u ~ u,), , since ( e l ,e,) gives the counterclockwise orientation to the plane while (ul, u,) gives the clockwise. Suppose we want to continuously change ( e l ,e,) into (vl, u,), maintaining a basis at each stage. Intuitively we cannot do this because, in order to change from counterclockwise sense of rotation to clockwise sense, at some stage the first vector must pass through the line given by the second in order to “get to the other side.” This idea is captured by our definition in terms of determinants. For let +

+ afe,, u2 = aie, + ate,,

u1 = a i e ,

a: a:

< 0, a: < 0, < 0, a t > 0.

So det(af) = < 0. Now if our deformation is given by (fl(~),f2(~)), 0 I7 5 1, A(0) = e i , A(1) = ui let

L{T)= a;{T)ei.

148

9. DIFFERENTIAL FORMS

Then

Fo

1 0

(a;@)) =

(a$)) = (txj).

Now det(a;(z)) must be 0 for some z E (0, 1) since it is positive for z = 0, negative for z = 1. So there is a z where fl(z) and f2(z) are dependent, i.e., lie along the same line. On the other hand, if (ul, u2) ( w l , w2), then according to Exercise 9.8 there is a continuous matrix function (a;(z)) with (a;(O))= Identity and a:{l)ui = wj. Thus, if fj(z) = aj(z)ui, then (fl(z),f2(z)) is a basis for R 2 and J(0)= u i , A(1) = w i . Note that what Exercise 9.8 really proves is that GL'(2, R ) is path-connected, where GL'(2, R ) is the group of all invertible (2 x 2) real matrices which have positive determinant. More generally, we can argue that GL+(n,R ) is path-connected, so that the result generalizes to arbitrary dimensions. Thus, to orient a vector space, we must say which bases (ordered bases, of course) are positively oriented. We now give an alternative formulation in terms of n-forms.

-

DEFINITION 9.21

If o,v] are nonzero n-forms on E, write o

Iv] for some I > 0.

-

v]

if o =

This divides the nonzero n-forms into two equivalence classes. Let (ul,. . . , u,), ( w l , . . . , w,) be bases with duals ( u l , . . . , u"), (w', . . . , w"). LEMMA9.22

(ul , . . . , un)-(wl

, . . . , w,) if and only if

~ A . . . A U " -

w l A . . . A Wn.

Assume wj = a$, with det(aj) > 0. Suppose w1 A . . . A w" = show A > O . We have W ~ A . . . A W ~ ( W ~ , . . . , W ,1,) = Iu' A . . .A un(wl,. . . ,w,) = 1 det(a$d A . . . A U " ( U ~ ,. . .,u,) = I det(a$. Hence we see I det(a$ = 1, so A > 0. Conversely, assuming A > 0, the above calculation shows det(txj) > 0. I PROOF:

I ~ A . * . A u " We .

This shows that specification of an equivalence class of bases determines uniquely an equivalence class of n-forms and conversely. Note that, given an equivalence class of n-forms, the equivalence class of bases is as follows: Pick an n-form o in the given equivalence class. A basis ( u l , . . . u,) is in the corresponding equivalence class if and only if o ( u l , . . . ,u,) > 0. If ( E , p) is an oriented vector space and ( u l , . . . , u,) is a positively oriented basis, an n-form u is called positiuely oriented if u(ul,. . . ,u,) > 0. DEFINITION 9.23 The standard orientation of R" is that determined by (el, . . . , en), the standard basis, or, equivalently, by el A * . . A en.

149

V O L U M E ELEMENT OF A METRIC

VOLUME ELEMENT OF A METRIC Let ( E , p) be an oriented vector space; p determines an equivalence class of nonzero n-forms. Suppose y: E x E + R is a metric on E . We claim g picks out a specific positively oriented n-form. Let (el,. . . , en)be a positively oriented orthonormal basis. Such bases do exist. Define w = el

A'.

.Aen.

w is a positively oriented n-form. Suppose (ul,

. . . , u,) is another positively

oriented orthonormal basis. We claim u1

. A U"

= e A . . .A

en,

Let u j = aiei. The matrices A = (.y(uj, u k ) )and B = (g(ej, ek))are equal, since both bases are orthonormal, (see remark below) d u j , uli) = aj&Mei, e,)

or, if C = (aj), A = C'BC.

Taking determinants we see (det(C))2= 1, so det(C) = f 1. But det(aj) > 0 so det(a;) = 1. Hence we see

u"

u l A .A. .

= det(cc'Je' A .

. . A en = e1 A . . . A en

as desired. REMARK: Actually, the statement A = B is not quite correct. Both matrices are diagonal with f l diagonal entries and both have the same number of 1's. But their placement on the diagonal may be different. However, det A = det B # 0.

+

DEFINITION 9.24 The volume element of the metric with respect to the given orientation is e1 A . . . A en, where (el, . . . , en)is any positively oriented orthonormal basis. Let ( u l , . . . , u,) be any positively oriented basis (not assumed to be orthonormal). Write u, = &;e,. Then y(v,, u k ) = a;a$g(e,, el) or, in matrix form, A = C'BC,

where A = (g(vj, vJ), B = (g(e,, eJ), C = (a:). Then det(A) = det(B)(det(C))' so = det(C), where we have used the facts, det(B) = f 1, det(C) > 0. Let g denote det(yJk)= det(y(u,, ok)). Thus, given a basis, g represents the determinant of the matrix of components of the metric. This notation may

Jm

150

9. DIFFERENTIAL FORMS

seem confusing but it is commonly used and is, in fact, convenient in calculations. The above calculations then give

4 = det(c$). Now by Proposition 9.15, we see that, if u l , . . . , u, is any basis with dual then det(clj)u' A . . . A u" = e' A . . . A en. For basis u', . . . , u", and if uj = let Iu' A . . A U" = e' A . * A en. Then 9

9

1 = iv' since e'

A.

. . A u"(ul, . . . ,u,)

A.

= el A .

. . A e"(ul, . . , , u,)

= det(olj)

. . A e"(e,, . . . , en) = 1. We have shown

PROPOSITION 9.25 Let ( u l , . . . ,u,) be a positively oriented basis for E where E has a metric g. Then the metric volume element is A . . . A u", where g = det(g(u,, uj)). If (u,, . . . , u,) is not positively oriented, then ( u 2 , u l , u 3 , . . . , u,) is, so the metric volume element is

mu'

mu2

A Y' A U 3 A ' . ' A U" =

-mu'

A.

. ' A 0".

Here note that, although the matrix of g on ( u 2 , u l , u 3 , , . . , u,) is different from the matrix on (ul, u z , . . . ,u,), the two have the same determinant. So we may state PROPOSITION 9.26 If E is an oriented vector space with a metric, then the metric volume element is f ~ U ' A . ' ' A U " ,

where u , , . . . , u, is any basis, g = det(g(u,, u j ) ) , and positively oriented, "-"otherwise. NOTES 9.27

"+"

(a) For a Riemannian metric, g > 0, so

is for ( u l , . . . , u,)

m is replaced by

(b) For a Lorentz metric, g < 0, so we write the Lorentz volume form as &U'

A.

. . A U".

DIFFERENTIAL FORMS ON A MANIFOLD On a manifold X , all of the preceeding algebraic constructions can be carried out at each point x E X , taking E = T,X. Thus a differential form of degree p on X , call it o,is an assignment w(x) E Ap(TxX)for each x E X . Suppose that ( V , 4) is a coordinate chart on X with coordinates x', . . . ,x".

151

ORIENTATION OF MANIFOLDS

Thenon v w = wlil ,...,i?l dxii A . . . A dxiP.We will say that w is a smooth form if the component functions w l i , , . i,l are smooth for all coordinate charts (V, 4). The set of all smooth forms on X of degree p will be denoted by Ap(X).Suppose that 5 is a vector field on X and o E A p ( X ) .Since w E F:(X), the Lie derivative L p , as defined in Chapter 8, is an element of F : ( X ) . But it is a simple exercise to show that, since w E A p ( X )c F:(X), L p E A p ( X ) . Now, iff: X + Y is a smooth mapping and w E AP(Y)>then we define the pullback f *w E A p ( X )by ,,,

f*o(x)(u,, . . . up) = (Nf(X))(TXf'(Ul), . . . > T x f ( u p ) ) . 2

Recall that we have studied pullbacks in Chapter 6 for the cases p = 1 , 2 [see (6.19), (6.20), (6.22)-(6.24)]. Of course, Ao(Y) is the ring of smooth real valued functions on Y and f * ( g ) = g .f. 0

ORIENTATION OF MANIFOLDS Let X be an n-manifold. For x E X let px be an orientation for T,X. Thus px is an equivalence class of bases. We want the orientations to "vary continuously with x" in some sense. DEFINITION 9.28 An orientation p of X is a choice of orientation px for T x X ,for each x E X , such that for each xo E X there is a neighborhood U of x,, and continuous vector fields t l , . . . , 5, on U such that for x E U , (tl(x), . . . , <,,(x))E p x . A manifold for which such an orientation can be constructed is said to be orientable. DEFINITION 9.29 Let p be an orientation of X . A chart ( U , 4 x")) is called positively oriented if, for each p E U ,

= (x',

...,

(dldx'l,, . . . , a/ax"lp) E P p . PROPOSITION 9.30 An n-manifold X is orientable if and only if X has an atlas R = { ( U i ,4i)}isuch that, if d i= (xl,. . . ,x")and d j = ( y l , . . . , y"), then det(dyi/dxi) > 0 at all points in the domain of 4j 0 4,:'. PROOF: If X is orientable, then one easily sees that X has an atlas consisting of positively oriented charts for some orientation p. If 4i = (x', . . . , x"), 4 j = ( y l > .. . , y") are two positively oriented charts for p, we claim det(dy'/dxj) > 0. Now d/dxj = (dyi/dxj)(d/dyi). But both bases (d/dxi(,), (d/dyil,) are in px for x E U i n U j . So the matrix relating them must have positive determinant as desired. Conversely, suppose an atlas 0 exists with the stated properties. If x E X , choose ( U i ,di)E R with x E U i .Let px be determined by (a/dxilx,. . . ,a/dx"l,).

152

9. DIFFERENTIAL FORMS

-

If ( U j ,q5j = ( y ' , . . . , y")) is another such chart, then ( 3 / d y ' l x , . . . , d/dy"l,) (d/dx'(,, . . . , 3/3x"l,), since det(dy'/dx') > 0. The vector fields (d/dx', . . . , d/dx"), for charts in R, satisfy the continuity requirement of Definition 9.28.

I An atlas with the properties of Proposition 9.30 is called an oriented atlas. Such an atlas uniquely determines an orientation of X . Thus, specification of an orientation of X is equivalent to specification of an oriented atlas. Now we give another way of specifying orientation in terms of diferential forms. Let p be an orientation of X . Let R = { ( U i ,4i)}ibe a positively oriented atlas. Fix i and let 4' = (x', . . . , x"). Then mi = dx' A . . . A dx" is a posbe a C" partition of unity subordinate itively oriented n-form on U i . Let to the cover (Ui)i. Then aimi is an n-form on all of X vanishing (by definition) outside of U i . Let o = criwi; a is a C" n-form. If ( u l , . . . ,u,) E p,, then aioi(u,, . . . , u,) 2 0 for all i and strict inequality holds for some i, so ~ ( u , ., . . , u,) > 0. Thus we see o is positively oriented for p . In particular, w(x) is never the zero n-form for any x E X . Conversely, suppose o is a nowhere zero C" n-form on X . We can define an orientation of X as follows: For x E X let p , be the orientation of T,X defined by the n-form m(x). For xo E X let ( U , 4 = (x', . . . ,x")) be a chart about x, with U connected. Consider (3/dx', . . . ,djdx"). For x E U consider o(x)(d/dx'l,, . . . , d/dx'l,). This function is always positive or always negative on U . Thus, either (dldx', d/dx2,. . . , d/3xn) or (d/dx2, d/3x1,. . . , djdx") give the local vector fields defined near xo as required in Definition 9.28. If a l ,o2are two nowhere vanishing n-forms we say they are orientation equivalent if there is a positive Cm function ,f:X + R such that col = fa2.The above arguments then prove

El

THEOREM 9.31 There is a 1-1 correspondence between orientations of X and orientation equivalence classes of nowhere vanishing n-forms on X . In particular, X is orientable if and only if there is a C" n-form on X which vanishes nowhere. EXAMPLE 9.32 Consider the atlas R discussed in Example 3.12(d). We have

=

{(U,,

4+),( K ,4 - ) } on

S" as

and

"[

3x' (x')2

X'

I

+ . . . + (.x")2

1x126; - x'(2xj) =

1x14

.

We must consider the determinant of the matrix having this last expression

153

ORIENTATION OF MANIFOLDS

as (i,j)-entry. Consider x

= (1,0, . . . , 0) so

the matrix is

p:

;].

So the determinant at x = ( I , 0, . . . , 0) is - 1. Since the domain of $ + $1' is R" - 0, which is connected, we conclude the determinant is everywhere negative. Thus, this atlas is not oriented. But if $ - is modified by interchanging two of its coordinate functions the resulting chart, together with ( U + ,$+), gives an oriented atlas. Thus, S" is orientable. The orientation defined by the atlas just described is called the standard orientation of S". The following proposition provides a simple method for showing that a given manifold is not orientable. 0

PROPOSITION 9.33 Suppose X has charts ( U , $), ( V , $) with U , V connected. Assume X is orientable. Then det(D($ $ - ' ) ) cannot change sign on $(U n V ) . 0

PROOF: Let p be an orientation of X . Now p defines an orientation on U and on V. Then, since U is connected, $ is either positively oriented, i.e., $ = (x', . . . , x") and (i?/dxllx,. . . , d/dx"l,) E p, for x E U , or (a/dx'I,, . . . , d/Jxnl,) 6 px for all x E U . Similarly for V . If (d/dx'l,, . . . , d/dxnl,) 6 p.yfor all x E U , call ( U , #I = (x', . . . , x")) negatively oriented. Now if ( V , rc/ = ( y ' , . . . , y")), then d/dxj = (LJy'/~xj)(d/?x'). If both charts are positively oriented or both negatively oriented, then they are similarly oriented so det(ay'/Jx') > 0. If one is positive, one negative, then det(dy'/dxj) < 0 on $(U n V ) .So in any case, det(D(rc/ $ - I ) ) = det(Jy'/?.u.') does not change sign on U n V . 0

-

(The Mobius strip-A nonorientable manifold) Let R 2 I - 1 < y < 1). Define an equivalence relation on Z by (x, y) (x 2, - y ) for all x E R , - 1 < y < 1. Let X = Z / - . Let n : Z + X be the canonical projection. Let U , = {n(x,y ) = [x, y ] I - 1 < x < l}, EXAMPLE 9.34

Z

-

= {(x, y ) E

+

u, = {[x, y]lO

< x < 2).

n:(0, 2) x n:( - 1, 1) x (- 1, 1) + U , is a homeomorphism with inverse ( - 1, 1) + U , is a homeomorphism with inverse &. ( U , , $ 1 ) and ( U , , 4,) are charts covering X . U , n U 2 consists of two disjoint open sets, namely,

n((O,1) x ( - 1, 1)) LJ n ( ( l , 2 ) x (- 1, 1)). We have$,

o$;':((O,

l ) u ( l , 2 ) ) x (-1, 1 ) + R 2 , $ 1 ~$;'(x,y)=(x,y)for

154

9. DIFFERENTIAL FORMS

O < x < l , - k y < l , 4 1 0 ~ ; ' ( ~ 7 y ) = ( ~ - 2 , - y ) f 1o
ri L

minant

+ 1. On (1, 2) x

(- 1, l), the Jacobian matrix is

which has deter-

[i-:I,

which has

determinant - 1. By Proposition 9.33 it follows that the Mobius strip X is a nonorientable C"-manifold.

-

EXAMPLE 9.35 (The real projective plane) Define P 2 ( R )= R 3 - {O}/-, where p Ap for p E R 3 - 0, A # 0, in R. Let n:R 3 - (0)+ P2(R) be the projection. Define three charts on P,(R) as follows. (Let [ x , y , z] = n(x, y , z)):

so

d2 4;' 0

is defined and C" on its domain, which is ((s,01s f 0).

Similarly the other coordinate transitions are C". We leave it as an exercise to show that P,(R) is a nonorientable manifold.

0 R IENTATIO N 0 F HYPER S U R FACES Suppose M is an (n - 1)-dimensional submanifold of R". We often call M a hypersurface in R". Suppose M is oriented. Then consider the set of normal vectors to M at p E M . This is a one-dimensional subspace of T p R n and so contains exactly two unit vectors. We single out one of these as follows:

155

ORIENTATION OF HYPERSURFACES

DEFINITION 9.36 The positive unit normal to M at p is that unit normal, N , , such that, if (ul, . . . , u,- 1) is a positively oriented basis for TPM c TpR", then ( N p ,ul, . . . , v,is a positively oriented basis for T,R".

FIGURE 9.2

The proof that Definition 9.36 is unambiguous is Exercise 9.14. The outward unit normal on S2 at P is the positive unit normal. This follows from Exercise 9.13, since ajar is that normal, provided you know (6, 4) is a positively oriented chart on S 2 . Suppose we specify a smooth unit normal field to M . This means we have a map f : M + R" such that f is C" and f ( p ) is a unit normal to M at p for all p E M (here we are regarding T,R" as R"). This specifies uniquely an orientation for M . We shall define this orientation by requiring (ul, . . . , u,- 1) be positively oriented if and only if ( f ( p ) , u l , . . . , u,- 1) is a positively oriented basis for TpR". EXAMPLE 9.37

L E M M A 9.38 The class of bases prescribed above is in fact an orientation of T,M and defines an orientation for M . PROOF: Let ( w l , . . . , w,above sense. Then

( u l , . . . , un-l) be positively oriented in the

where det(a:) > 0. But (a;) has the form

1:

0

O '.

156

9. DIFFERENTIAL FORMS

(pf) will

have a positive determinant. This shows ( u l , . . . , v n - J and . . , w n P 1 )are similarly oriented as desired. Let p E M . Choose coordinates (x', . . . , x"-') on M near p so that, at p , (a/ax', . . . , d/ax"-') is

so

(wl,.

positively oriented. We want to show this basis is positively oriented in a neighborhood of p . But if

where ( y ' , . . . , y") are Cartesian coordinates on R", then dy' A . . . A dy"(f(z), 8/ax1I=,. . . ,3/8x"-'I,) > 0 for z near p , by continuity. This shows that we do have an orientation for M . I Note that we have shown that specifyinga smooth unit normal on M is equivalent to specifying an orientation of M . On an oriented Riemannian manifold we have seen that there is a uniquely defined n-form, the Riemannian volume element. In case M is an oriented hypersurface in R" we have an explicit description of the Riemannian volume element on M .

INTERIOR PRODUCT DEFINITION 9.39 Let o be a k-form on T,M, let u E T,M. The interior product of u and o is ivo,an (n - 1)-form on T,M, given by

ivw(ul,. . . , 0,-

1)

= w(u, u l ,

. . . , 0,- ').

PROPOSITION 9.40 Let N be the positive unit normal on an oriented hypersurface M in R". Then the Riemannian volume element on M is ow= i,(dxl A . . . A dx"). PROOF: Let p E M and let ( u l , . . . , u,- ') be a positively oriented orthonormal basis for T,M. Then ( N p ,u l , . . . , u,- ') is a positively oriented orthonormal basis for T,R". Then

. . A dxn(ul,. . . , u,- 1) = dx' A . . . A dx"(N,, u l , . . . , u,- = 1. This shows i, dx' A . . . A dx" is the Riemannian volume element on M . I i,dx'

A.

EXTER I0R D E R IVATIVE PROPOSITION 9.41 Let U be an open set on a manifold M such that U is the domain of some chart. There is a unique family of maps d,: Ak(U)+ Ak+' ( U ) ,k 2 0, such that

(a) (b) (c) (d)

iff

E

Ao(U) = Cm(U,R ) , then d,f

d,(f + Y) if

CI E

d,(d,cc)

=

d,f+

is the usual differential off;

d"g;

A,(U), fl E A,(U), then d,(cr A p) = (d,a) = 0 for every cr E A,(U), k 2 0.

A

+

/l (- l)%A

(&a);

157

EXT E R I0R DER IVATIVE

PROOF: We first prove uniqueness. If d, satisfies the above properties and a E I\k(U), then

a is a sum of terms,

with f E C " ( U , R ) , i , < . . . < i ,.

fdxi'r\...r\dxik

Then d,( f dxil A . . . A dx'") = f P dx' A dx" A . . . A dxik, so we see d,cc is uniquely determined. For existence, let a = c l l i , . , i k l dx" A . . . A dxB and define d,a = a l i l ,. i.k l , L dxl A dxil A . . . A dxik. Note that we are defining the operation d, in terms of a particular coordinate system but, if we show (a)-(d) hold, then it follows that the operation is independent of coordinates because of uniqueness. Now (a) is clear, as is (b). Note that if i,, . . . , ik are all different but are not necessarily in ascending order, then we still have ,

d,( f dxil A . . . A dx'") = J j d x j ~dxil A . . ' A dxik.

Because supposej,, . . . , j , are i,, . . . , i, rearranged in ascending order. Then fdx" A . . . A dxik= (- 1)"f dxj' A . . . A A dxJk,where rn is the number of interchanges needed to effect the reordering. So

d,( f dXi' A . . . A dXik)= d,( ( - 1)"f dXj' A . . . A dXjk) = (- l ) m f t d x ' ~dx" A . . . A dxjk = J t dxeA dxilA . . . A dxik. Now we prove (c). Write LI = a l i l ...i k l

dxil A . . . A dxil,

=

Plj,.. . j,l

dxj' A . . . A dxL

so that CIA

b = c l l i l . . . i k l Blj,. . .,il

dxilA . . . A dxikA dxjl A . . . A dxj'.

so d,(a

A

a

j)= axp [clli,. . . iklbljl,, .j,l] dxp A dx"

A.

. . A dxikA dxjl A . . . A dxj'

-

A . . . A dxj' dxPA dxil A . . . A dxj/ = ( - l)kali,. . .iklPljl. . _jfl.p dx" A . . . A dxikA ~ X P A dxJ1A . . . A dxj' (dux)A P = ( - l)kCcA dub + dua A p.

- ali,.. . i k & ? l. j. j ,l l ., p dxP A dx''

+ PIj,.. .jlpli,,

,,ikl.p

+

This proves (c). To prove (d), it is enough to show d,(d,a) = 0, where a is of the form f dxilA . . . A dxik.But d,r = , f e dxe A dxi' A . . . A dxik so d,(d,a)

= Jl,

by equality of mixed partials.

dx'

A

I

dx'

A

dxil A . . . A dxik= 0

158

9. DIFFERENTIAL FORMS

Note that if V is open in U , then d&l")

= (dual/".

We now globalize this. PROPOSITION 9.42 On any manifold M there is a unique family of maps d:Ak(M)+ Ak+ ,(M) such that:

(a) Iff E A,(M), df is the usual differential off, (b) d(a P) = da dp, d(CI A b) = d E A fl (- l)kCr A dp if CI E I \ k ( h f ) , (C) (d) dda = 0.

+

+

+

PROOF: Let a E I\k(M). Then if p E M we must define da(p). Let p where ( U ,(x', . . . , x")) is a chart on M . Define

E

U,

4 P ) = 4A@Iu)(P). If ( V , ( y ' , . . . , y")) is another chart about p , then choose an open set W with p E W c U n I/. Then d,(al,)(p) = d&al,)(p) = dv(alv)(p), so that da(p) is well defined. Since (da)l, = d,(al,) we see that da is a smooth ( k 1)-form as desired. So we have a map d: Ak(M)4 A,+ ,(M) as desired. We must verify (a)-(d).

+

(a) For f E A,,(M), (df)u = d u ( f l U )= differential o f f on U . Since M is covered by charts, (a) holds. (b) Clear. (c) On any chart domain U we have ( 4 a A P))Iu = du(alu A

PIU)

+ (- l)kaIuA du(PIu) = (da)lu A Plu + (- l)kaluA (dP)Iu = du(aILJ A P I U = (da A

P + (-

l ) kA ~Lip)[,.

(d) Similar. We now prove uniqueness. Suppose a E Ak(M),p E M . We want to write down a formula for da(p). First we argue: If a = ,8 in a neighborhood of p , then da(p) = dP(p). For suppose a = /lon a neighborhood I/ of p . Let k be a C" function on M which satisfies

Then ha

= kb

k

=

k

=0

1

nearp, outside V .

on M , so we get d ( W ( p ) = dk(p) A

+ k(p) A

=ddp).

159

EXT E R I0R D E R IVATlVE

and d(hP)(P) = dB(P) so da(p) = dP(p). Now, given

and express

CI

CI

and p we can choose coordinates around p

locally as a = a l i , ...i k l dx"

A

. . . A dxik.

We can choose functions A i l . , , i kX, i which are C" on M and agree with a i l . .. i k , x'

near p . Then M =

Ali, . . . ikI dXi1 A . . . A dXik

near p. So da(p) = d ( A l i , . . i k l ) ( pA) dX"(p) A . . . A d X i k ( p ) = d(ali1. . . i k l ) ( p ) ~ d x i l ( p .) ~~d. x . '"(p).

This proves uniqueness.

I

PROPOSITION 9.43 (d commutes with pullbacks) Let f : X smooth map, w a k-form on Y. Then f * d o = d ( f * o ) .

+

Y be a

PROOF: The case k = 0 is Proposition 6.26. Now locally o can be writ-

ten as o = w l i l . . i k l d x i l A . . . A dxik.

Then ,f*o = (olil.. . i k l L f ) f * dx" A . . . ~ , fdxik * - oli, , . . i k l 0 f d(x" f ) A . . . A d(xik 0 f ) 0

SO

d f * o = d ( o l i ,. . irl 0 f ) A d(xil f )A . . . A d(xik f ) . Now dw = do+, . . . i k ldx" A . . . A dxik so f * d o = f * dmlj,. . . j k l A f * dxil A . . . A f * dxik.

Since f * dx'j

0

,

= d ( x ' ~o f )

and f * d o i , .. . i k

= d ( o i l. . . ik

0

f ) , we have the result.

I A subset U of R" is said to be starlike with respect to p if, for x in U , the closed line segment from p to x is in U . PROPOSITION 9.44 Let U be an open set in R", starlike with respect to the origin. Let w be a k-form on U such that d o = 0. Then there is a (k - 1)form M on U with da = w.

160

9. DIFFERENTIAL FORMS

PROOF: We define a linear map H : A,(U) + At- l(U) for all 8 2 1 such that, if o is any [-form, then

+ H(do).

o = d(H(o))

Then if d o = 0 we get o = d ( H ( o ) ) ,so that we may take CI = H(o). Let o = f dxil A . . . A dxif (with no assumption about the ordering of the indices). We define e

Ho(x)=

1 (-1Y-I u= 1

(1:

1

A

t t - ' f ( t x ) dt xi, dxil A . . . A dxiaA . . .A dxi'.

If the factors are permuted by 6,so f is replaced by (sgn a ) f , then the form H o ( x ) is unchanged, so that H is well defined. Then e

e

x

Now d o = H do(x) =

xj dxj

A

(j: teJj(tx)d t ) xi- dx' dxil A .

A

-. d x i l A ' . ' A dx'.

A.

. . A dxie.

. . A dxit, so that we have

(j: t y j ( t x )d t ) x j d x i l e + (- 1)" (j:

A.

. . A dxie

t'lj(tx) dt)

-.

xi= dxj A d x i l A . . . A dxla A . . . A dxi'.

a= 1

Therefore dHo

+ H do = /

(1: + (1:

=

)

t'- ' f ( t x ) dt dxil A .

. . A dxiC

t Y j ( t x )d t ) xj d x i l A .

. . A dxiC

(j: ( / t t - ' f ( t x ) + t f f j ( t x ) x j )dt

= ( t y ( t x ) l h ) dxil A .

1

d x i l A . . . A dxi'

. . A dxie = f ( x ) d x i l A .

. . A dxi' = ~ ( x ) . a

161

DE RHAM COHOMOLOGY GROUPS

REMARK: For more insight into the mapping H see Exercise 10.8. The requirement that U be starlike from 0 can be relaxed considerably using Proposition 9.43. Let U be diffeomorphic to a starlike region and let w be a closed k-form on U . Let f: U + V be a diffeomorphism with V starlike from 0. Then ( f - ' ) * w is a closed k-form on V since d((f-')*w) = ( f - ' ) * d o = 0. So write ( f - ' ) * o = dp. Then f * f l E Ak- l(U) and d(f*b) = f * dp = f * ( f - ' ) * c o = w so w = da where a = f*P. Thus the actual requirement of starlikeness is unnecessary. However, some requirement on U is necessary. For example, if 0-

x2

-y

+

thenon U = R2 - {(0,0 ) }we havedw For if there were, then

j:n w(c(t))c'(t)dt =

X

dx+-

x2

y2

=

+y

dY,

2

0. Thereisno a: U

(a c)(t)dt 0

=

rR

(U

+

0

R withda

= w.

c)' dt

= a(c(270) - a(c(0))= 0,

which a direct calculation shows to be false. Here c(r) = (cos t , sin t),

POINCARE L E M M A PROPOSITION 9.45 (Poincare lemma) Let w be a k-form on M with dw = 0. Then for p E M there exists a neighborhood U of p and a (k - 1)form a on U such that da = w on U . PROOF: Immediate from Proposition 9.44.

I

DE R H A M COHOMOLOGY GROUPS Since d2 = 0 it follows there is a quotient vector space

and H o ( M ) = ker(d:A,(M) + A,(M)).Note that H o ( M )E R' where c is the number of connected components of M . A form w for which dw = 0 is said to be closed, and if w = da, then (0is said to be exact. This quotient is called the qth D e Rham cohomology group of M . The result of Proposition 9.44 is that on a starlike subset of R" we have H q ( M ) = 0 for all q > 0.

162

9. DIFFERENTIAL FORMS

MANIFOLDS WITH BOUNDARY Let H" = { x E R"lx" 2 O } . H" is called the upper half-space in R". We denote by dH"the set dH"= { X E R " ~ x "= O } . (9.3) Let U be open in H", f:U -+ R" a map. We say f is C" on U if, for each p E U , there is an open set V in R" containing p and a C" map F : V R", such that P = f on U n V. Thus f is C" if, near each point, f extends to a C" map on some open set in R". It is thus possible to speak of the derivative Df(p), even for points of U n dH".Ordinary rules of calculus, e.g., the chain rule, hold as usual. We want to discuss manifolds with boundary. We consider, for this purpose, charts which have as images open sets in H". If X is a topological space, let U be open in X and 4: U --t 0 be a homeomorphism of U onto an open set 0 in H". We consider such to be charts on X . An n-dimensional manifold with boundary is a second countable Hausdorff space which is covered by charts in the more general sense just discussed. We say charts ( U , $), ( V , $) are C"-related if $ 4-l and 4 0 $- are C" on their respective domains. Just as before, we may consider a C"-atlas or a diferential structure (see Definition 3.13). --f

0

DEFINITION 9.46 Let X be an n-dimensional manifold with boundary. The boundary of X , denoted dX,is given by dX = { p E XI there is a chart ( U , 4) in the differential structure of X such that 4 ( U )is open in H" and 4 ( p ) E dH"}. PROPOSITION 9.47 Let p open in H" and $ ( p ) E dH".

E

ax,( V , $) a chart with p in V . Then $(V) is

PROOF: Let ( U , 4) be as in Definition 9.46. Then both $ 4-l and 4 $ - I are C". If the proposition were false, then, by restricting 4 if necessary, we may assume $(V) is open in R". Now 4 $-' is defined on an open set in R" and D(4 $-')($(p)) is a linear isomorphism. By the Inverse Function Theorem, there is an open set W in R" such that im(4 I,!-') 3 W 3 &). But this cannot be, since 4 ( U ) c H" and 4 ( p ) E dH". [ 0

0

0

0

0

DEFINITION 9.48 If X is a manifold with boundary, the interior of X , denoted Int X , is given by Int X = X If p E Int X and ( U , 4) is a chart about p , then either 4 ( U ) is open in R" or, if 4 ( U ) is open in H", then

ax.

4(P)4 dH"All of our basic constructions on manifolds go through with no change for manifolds with boundary. We have, at each p E X , an n-dimensional tangent space T p X . This is true even if p E ax.We have the tangent bundle, co-

163

INDUCED 0 R I ENTATION

tangent bundle, Riemannian metrics, differential forms, orientations, and volume elements, just as before. The theory of integral curves of vector fields does not go through so well. At boundary points there may be no integral curve at all through the point.

+

EXAMPLE 9.49 Let X = {(x,y) E R 2 Ix2 y 2 5 1). Consider the vector field d/dx on X. What is the integral curve passing through (0, l)?

At points of Int X there is, of course, no problem. L E M M A 9.50 Let X be an n-manifold with boundary. Then dX has a natural structure of an (n - 1)-dimensional manifold without boundary. PROOF: Let A? be the set of all charts ( U , $) on X such that $(U )is open in H " a n d U n d X # $ . L e t Q , , = {(Unax,$I,,,,)l(U,$)~.~}.Note$I,,,, maps U n dX onto an open set in dH" = R"-'. Since, for any two charts ( U , $), ( V ,9) in d,$ 0 9-l and 9 $ - I are C", it follows that Qax defines a natural differential structure on dX. I 0

Let x E ax.Then we may regard T,(dX) as an (n - 1)-dimensional linear subspace of T,X, namely, all vectors in T,X which can be obtained as c'(O), where c: ( - E , E ) -+ X is a Cm-curvewith c(t) E dX for --E < t < E. Suppose g is a Riemannian metric on X . Then g defines a Riemannian metric on dX.There are exactly two unit vectors in T,X, x E a x , which are perpendicular to T,(dX). We distinguish one of these as follows. Let ( U , 6) be a chart about x with U n dX # 4, let u E T,X and let $ = (x', . . . , x"). Then write u = ui d/axi. Now, if u is a normal to a x , then U" # 0. We say u is inward pointing if u" > 0 and outward pointing if un < 0. This depends only on u not being tangent to a x ; being a normal is not needed.

INDUCED OR I ENTATION Let X be an oriented n-manifold with boundary dX. Then the orientation of X induces a natural orientution on dX as follows:

ax

DEFINITION 9.51 The induced orientation on is defined by requiring ( u l , . . . , u,_ 1) E T,(dX) be positively oriented if and only if (u, u l , . . . , u n _ 1)

is a positively oriented basis for T,X whenever u E TpX is outward pointing (see Fig. 9.3). PROPOSITION 9.52

T,(dX), for p E ax.

This definition gives a well-defined orientation at

PROOF: We first argue that, if u, u' are both outward pointing vectors at We p , then if (u, u l , . . . , u n P l ) is positively oriented so is (u', u l , . . . , u,-

164

9. DIFFERENTIAL FORMS

1

0 Pvectors outward pointing

ax, w i t h induced orientation

FIGURE 9.3

have

m i 011

U VI 9

I

1J

a > 0 since both u and u' are outward pointing. Hence the matrix has positive determinant as asserted. Now we argue that, if u is outward pointing and ( u l , . . . ,u ~ - ~ ) , (v;, . . . , V L - ~ ) are bases for T,(dX), then they are similarly oriented if and only if (u, u l , . . . , u,- I ) and (u, u;, . . . , ub- are similarly oriented bases for T p X .We have

The determinant of the matrix is equal to that of the submatrix (a:) so the result is proved. 1 PROPOSITION 9.53 Let X be an oriented, n-dimensional, Riemannian manifold with Riemannian volume element w x . Let N be the outward unit normal field on ax.Then the Riemannian volume element on dX is

wax = hv4wx).

Proof is left as Exercise 9.35.

165

HODGE "-DUALITY

HODGE *-DUALITY Let E be an n-dimensional real vector space which is oriented. Let g be a metric on E , i.e., a nondegenerate symmetric bilinear form. Let E be the volume element on E corresponding to the orientation and the metric (see Definition 9.24). Let 4: E + E* be the isomorphism defined by the metric

4.

4MW)= The map

(9.4)

4 (which is actually the b-operator for y) induces a map

4*: Tf(E)-+ T k ( E ) defined by 4*a(B', . . . , p)= a((flfl1,. . . , 4-'pk).

(9.5)

This is an isomorphism; in terms of components, 4* raises all the indices. Let a E Ak(E).Then 4 * a 0 E lies in Ti(E).Let C: Ti(E)-+ Tf-k(E)be the k-fold contraction contracting the ith contravariant with the ith covariant n). index for 1 I i I k (we are assuming k I

CI

DEFINITION 9.54 The Hodge *-operator *:Ak(E)-+ An-k(E)is given by * a = (l/k!) C(4,a @ E).

-+

Let (el,. . . , en) be a basis for E . Let a = a l i l . . ikleil A . . . A eik,E = E

__

~ ~ ,inleil . A

. . . A e'n; ail_ _ . i k , E

~ . ,. . j,

are the tensor components of a, E. Then 4 * a has components g i i j i . . . gikjkaj,, . j k so * a has components

=

,

(*a)jl...jn-k = (l/K!)d""' - .(i

So if a

= a l i l ., .ikleil A.

I

. . . ikl

kEil..

.ikjl...

E i l . . . i k j l . . .jn

k .

. . A eik,then

* a = alil,,.ikl E

~ . . i~k l j ,. . . .jn-klejl A.

. . A ejn-l..

(9.6)

EXAMPLE 9.55 (a) Let E = R 3 with the usual metric and orientation. Let (el, e2, e3) be the standard basis. We claim

*e 1 -- e 2 ~ e 3 , e1 = 6:e' so 4*e'

= hiei

*el

=

*e2 = e3 Ael,

*e3 --e1Ae2.

(we have identified TA(E) = E** with E). Thus,

(e')'EiljkleJA ek = c123e2A e3 = e2 A e3.

166

9. DIFFERENTIAL FORMS

Similarly, e2

= Sf.'

e3 = 63ej (b) Let E

= R4

so

*e

A e' = ~ 2 1 3 e 'A - 82~jlkIlek j

so *e3 = dj3EjlkllekA e'

= E312e' A

e3 = e3 A e', e2 = e'

A

e2.

with the usual orientation and the Lorentz metric g =

eo @ eo - e' @ e' - e2 0 e2 - e3 0 e3,where (eo, e l , e 2 , e3)is the standard

basis. The volume element is E = eo A e1 A e2 A e3. Verify the equations (a) (b) (c) (d) (e) (f)

*eo = el A e2 A e3, *el = eo A e2 A e3, *e2 = - eo A e l A e3, *e3 = e0 A e1 A e2, *(el A e3) = -eo A e2, *(eoA el A e3)= - e 2 .

Here is the pattern: Suppose a is a wedge of some e"s. The dual is the wedge of the others. The order of the others is such that the wedge in c1 followed by that in * a gives E. Then put a minus on * a each time one ei appears in a with i = 1, 2, or 3. Thus *(el A e 2 )= (- l)'eO A e3

since e' A e2 A e0 A e3 = E. If a, fl are k-forms then 4 * a

E

Tk,(E)so we have the complete contraction

C(4,a

0 B) E R .

(9.7)

Define the inner product of a and /? by (a, P) = (l/k!)C(4*a 0 PI.

(9.8)

Then, in terms of components, (a, p) = alii ...i k l g 1, 1 " ' l k.'

Clearly (a, p) is symmetric and bilinear in a and PROPOSITION 9.56

p.

For any k-forms a, p we have aA

PROOF:

(9.9)

*fl

= (a,

a)&.

167

HODGE *-DUALITY

Then we have

aA*fl

=

alil,.,iklfi~j~~~~j~l cjI . . .jklsl

. . . S n - klei i

A

. . .A

,SI

A.

. . A esn-k.

In this sum we always have

i,<...
j,<.'.<j,,

s,<'~'<s,-,.

The only nonzero terms are those for which i, = j t , 1 I /I k, since { i l , . . . , i k } n {sl,. . . ,s,-,} = @ a n d { j , , . . . ,j,} n {sl,. . . , s,-,} = @.So

= (a, fib

since, for each choice of i , < . . . < i,, the sum in brackets has exactly one nonzero term which is E . 1 COROLLARY 9.57 a A * a = (a, a)&for any k-form a.

Consider an oriented vector space with metric g and positively oriented orthonormal bases e l , . . . , en. Let i , < . . . < i, and let a = eil A . . . A eik. Then the only nonzero component of a with indices in ascending order is ail. . . i k = 1. Let m be the number of indices among i,,. . .. , ik . for which g(ei,,ei,) = - 1. Then = (- l)m. So (a, a) = a l j , ,.,jkl aJ1"'Jk= (- l)m. Let L,, . . . ,en-, be such that ( i , , . . . , i,, e l , .. . , en-,)is an even permutation of (1, . . . ,n). Then aA((-

so * a = (-1)metlA..

l)meel A ' . ' A een-k)= (-

= (a, a)&.

.Aeen-k

Suppose g(ei,ei) = - 1 for exactly s values of i ( g has index s). Let a = . . . A e'k. Then * a = (- l)meelA . . . A & n - k as above. Then * * a = (- 1)"( - l)m'eilA . . . A eik for appropriate m'. Now (il, . . . , i,, L , , . . . , t , -,) and ( f 1 , . . . , en-,,i,, . . . , ik) are related by a permutation of parity k(n - k). To (- I),'"-,) we must multiply (- l)m+m,where hi is the number of v such that g(ety,e,,) = - 1. But m + rti = s so e i ~A

* * a = (-

l)k(n-k)+Sa.

168 (-

9. DIFFERENTIAL FORMS

PROPOSITION 9.58

If the metric is of index s, then, on k-forms,

** =

l)k(n-k)+s

Suppose dim E = 3 and g is positive definite. Then so * * a = tl for any form a. O n the other hand, suppose dim E = 4 and g is a Lorentz metric. Then we easily show (- l ) p ( 4 - p ) + 3 = ( - 1 ) P + so, if tl is a p-form, then **a = (- 1)P' 'a. REMARK 9.59

** = (-

iy(3-g)

= +1

DIVERGENCE AND LAPLACIAN OPERATORS We have now covered the basic algebraic properties of the *-operator. Going to the manifold setting we now have for 0 I k I n.

* : A k ( M+ ) I\a-k(M)

(9.10)

Let M be an n-manifold, with metric g of index s, and assume M is oriented, with volume element E. DEFINITION 9.60

The divergence operator

6 : A k ( M ) - + A k - , ( M ) , k = 0 , 1 , . . . , n isdefinedby

6a = (- l ) n ( k + l ) + s +*d*a l

for a E A k ( M ) .

REMARKS: (a) For tl E Ao(M), 6a = 0 E Ap1(M) = (0). This is actually a matter of definition since the above definition does not apply when k = 0. (b) It appears 6 depends on the choice of orientation for M. But changing orientation changes the sign of E and hence, changes the sign of *. Since * appears twice in the definition of 6, 6 is not changed. (c) See Exercise 10.4 for an indication of the origin of the sign in the definition of 6tl.

DEFINITION 9.61

The Laplace-Beltrami operator on M is

A = 6d + d6: Ak(M)-i A,(M),

k 2 0.

We consider two important special cases.

CALCULATIONS IN THREE-DIMENSIONAL EUCLIDEAN SPACE We use Cartesian coordinates (XI, x2, x3), the usual metric and orientation, so that E = dx' A dx2 A dx3. Since the matrix (yij) is the identity matrix, raising and lowering indices does not change the value of the components, . ' i.e., v' = u i .

169

CALCULATlO NS I N TH R EE- DI M ENSI0NAL EUCLI DEAN SPACE

The Gradient of a Scalar (9.1 1)

So take f

E

A0(R3),compute d f , and then raise the index to get grad f

=

Vf.

The Divergence of a Vector Field Let 5 = vieibe a vector field. Lower the index to get f = ui dx'; * f is a 2-form so d * f is a 3-form and * d * t is a 0-form. 6( = - * d * f is thus a 0-form. We claim I

I

div

5 = viei,f = v i dx'

5 = -6g;

so *f = v1 d x 2 A d x 3 + u2 d x 3 A dx'

(9.12)

+ v3 dx'

A

dx2 and

therefore,

So div ( = *d*f = - 6 f , as asserted.

The Curl of a Vector Field Let 5 = u'e,, = vi dx'; then * d f E A , ( R 3 )so, raising the index, we get a vector field (*df)#.We have curl

5 = (*d@

(9.13)

The verification of (9.13) is left to the reader.

The Laplacian Thus, we have obtained the three classical differential operators of vector analysis. For CI E A0(R3),ACI= d 6a + 6 da = 6 da = -div(grad a) so we have: On functions, A is the negative of the usual Laplacian, Af

=

-vy.

(9.14)

Area 2-Forms Let

5 be a vector field, 5 = 5' d/dx' and ?=tidx'. Then *[= 5'EiljkIdxj A dxk.

DEFINITION 9.62 The area 2-forrns d 2 C i ,i = 1,2,3, are defined by d 2 C i = d x j dxk ~ ( = *dx'). Note that * f = 5' d 2 C i .

170

9. DIFFERENTIAL FORMS

Let M be an oriented surface in R 3 . Let N be the positive unit normal on M , E~ the volume element on M . PROPOSITION 9.63 Let p

E

M and

5 E T,R3. Then,

on T,M, we have

(5, N ) z M = ti d2Ci = *F. PROOF: Let (ul, u2) be a positively oriented orthonormal basis for T P M and write 5 = aN + bu, + cu2. Then ( 5 , N)EM(ul, u2) = a, 5' d2Xi(u1,u2) = E(& u l , u2) = a, where we have used i , ~ = *v" (see Exercise 9.37). Since the 2-forms (t,N ) E~ and tid 2Ci agree on the pair (ul, u2) they agree on T,M. 1

CALCU LATl ONS IN MIN KOWS KI SPAC ETI M E We consider R4 with standard coordinates (xo, x', x 2 , x 3 ) , Loreqtzian metric g = ( d ~ ' )-~(dx')2 - (dx2)2- ( d ~ ~and ) ~standard , orientation, for which E = d x o A dx' A d x 2 A d x 3 .

Gradient of a Scalar If f: R4 -,R is C", grad f = (df)# = (the result of raising the index on d f ) = g p y v .We see that df = J" dx" so that (9.15)

The Divergence of a Vector Field Let 5 = uiei be a vector field, f = ui dx' the corresponding 1-form. Note that we are now working with the Lorentz metric g given above so that we no longer have ui = u'. In fact we have uo = uo and ui = -ui for 1 Ii I3. Now, according to Definition 9.60, we have

-

S t = *d*r. We compute

@ as

5 = u0 d x o - U' * f = uo dx'

dx'

- u2

d x 2 - u3 dx3,

d x 2 A d x 3 - u' dxo A d x 2 A d x 3 + u2 dxo A dx' A d x 3 + u3 d x o A d x 2 A d x l ,

so d * f

= uPO&

+ u,'~E+

A

U:~E

+

and we get

U:~E= U ~ ~ E

@ = -div 5.

(9.16)

171

GEOMETRICAL ASPECTS OF DIFFERENTIAL FORMS

The d'Alembertian 0 Iff is a scalar we define

of= div(grad f). We see of= -6

f = J o o - 111 - 1 2 2 - J 3 3 .

df so

(9.17)

Volume 3-Forms Let

5 = va a/axa be a vector field. Lower the index to get 4 = u, dx". Then

*f is a 3-form,

*f = V " E , , ~ , ~ , , ~dx" A dxa A dxY. DEFINITION 9.64 The volume 3-forms d3C,, p = 0, 1, 2,3, are defined by d3XP = E~~~~~~ dx" A dxBA dxY(= g,,y*(dxv)).So we have

*r

= 5"

d3X,.

(9.18)

Let M c R4 be a spacelike hypersurface which is oriented (see Definition 6.15). Then the normal vectors on M are timelike vectors. There is a unique unit normal N defined by the orientation (see Definition 9.36). Let cM be the volume form on M . PROPOSITION 9.65

Let p

E

M and

4 E T,,R". Then, on T,M, we have

( 4 , N ) c M = ("d3Ca= *f. PROOF: Similar to the proof of Proposition 9.63.

Suppose M is a 3-manifold in Minkowski spacetime. In general we cannot define a unit normal (e.g., at those points where normal vectors are null). But given a vector field 4 on M , we may always consider 5"d3Ca.

which is just

*?.

This allows us to define "flux integrals" without referring to normal components.

GEOMETRICAL ASPECTS OF DIFFERENTIAL FORMS Differential forms can be used to specify submanifolds of a differential manifold by determining the tangent space at every point. More generally, smooth subbundles of the tangent bundle can be described using differential forms.

172

9. DIFFERENTIAL FORMS

SMOOTH VECTOR BUNDLES DEFINITION 9.66 Let M be a differential manifold. A smooth vector bundle of dimension k over M is a differential manifold B and a C"-map p : B -+ M such that

(1) for each m E M the set p-'(rn) has the structure of a real k-dimensional vector space; (2) for each m E M , there is an open set U containing m and a diffeomorphism h : p - ' ( U ) -+ U x Rk such that (a) n , 0 h = p , where n,(u, u) = u, (b) for each x E U , h \ p - l ( x ) : p - l ( x-+) ( X I x Rk is a linear isomorphism.

A pair ( U , h) as described in (2) is called a local uector bundle trivialization of p : B + M . REMARK 9.67 We saw in Chapter 5 that the tangent and cotangent bundles are smooth vector bundles. For the tangent bundle z: T M + M , there is a special local vector bundle trivialization ( U , h,) associated with a chart ( U , 4) on M . If we write 4 = ( X I , . . . ,x"), then h, is given by

h i yrn, u) = ui d/dx'(, for rn E U , u E R". Similarly, for the cotangent bundle we have (V, h$), where h,*- '(m,U) = uidx'(m).

From these, we can construct local trivializations for more general vector bundles over M . Let us define (see Exercise 8.17) Ts(M)=

u

TXTrnM),

meM

Qp(M)=

u

Ap(TmM)*

msM

In a manner analogous to that given for T M and T*M we can give the structure of a smooth vector bundle to each of T:(M) + M and Rp(M) + M . We leave the details of these constructions to the exercises.

VECTOR SU BBUNDLES DEFINITION 9.68 Suppose p : B + M is a smooth vector bundle over M . A smooth vector subbundle of B of dimension 'L is a submanifold C c B such that:

(1) For each rn E M , C, = p - ' ( m ) n C is a vector subspace of B , = p-'(m), of dimension t. (2) p i C: C + M is a smooth vector bundle.

KERNEL

173

OF A DIFFERENTIAL FORM

LEMMA 9.69 Suppose p : B + M is a smooth vector bundle of dimension k and C c B is a smooth vector subbundle of dimension t. Then for each m E M there is a local vector bundle trivialization of B, ( U , h), such that m E U and

h ( p - ' ( U ) n C) = U x R' x [O) c U x Re x Rk-'. Such an ( U , h) is called a C-suhhundle local triuialization of B. PROOF: Let m E M . Choose local trivializations ( U , , h,) for B and ( U , , h,) for C such that m E U , n U , . Then we have

I

p - l ( U , n U,) n C

h2

I ( U , n U,) x Re.

Write h , h; '(x, v) = (x, @(x)(u)), where cD(x) E L(RG,Re x Rk-' ) and @ is smooth. Now dim(cD(m)(Re))= so there is a linear isomorphism a: R' x Rk-' + Re x Rk-'such that IT, a @(m):R' + R' is an isomorphism, where n,:Re x Rk-e -+ Re is projection. Define Y(x) = IT^ a @(x) for x E U , n U , . Then, since Y ( m )is an isomorphism, it follows that there is a neighborhood W of m, contained in U n U , , for which Y(x) is invertible for all x E W . Now for each x E W we have 0

0

0

0

Re

Rk-8

= im(a '

0

0

cD( u)) 0((0) x Rk-').

Let a2(x)be projection onto the second summand of this decomposition. We leave it as an exercise to show u,: W -+ L(R' x Rk-', Rk-e)is smooth (see Exercise 9.22). Now for x E W , define v(x):Re x Rk-e + R' x Rk-' by v ( x ) = (nl, u2(x)0 a ) and h : p - ' ( W )+ W x (Re x Rk-e) by h(5) = (Id x v(p(()))(h,([)). It is easily checked that ( W ,h) has the desired properties. I

KERNEL OF A DIFFERENTIAL FORM DEFINITION 9.70 Let E be a vector space and a a k-form on E . We define the kernel of a by

ker(a) = { u E Ela(v, u 2 , . . . , u k ) = O If k

=

for all u 2 , . . . , uk E E } .

1 we define ker(a) = { u E Ela(u) = O } .

REMARK: Let a define a linear transformation

I,: E

+ A k p,(E)

by

I,(u) = iva,

174

9. DIFFERENTIAL FORMS

that is, Then clearly ker(a) = ker(Ia). LEMMA 9.71 Let E be a vector space of dimension n and a', . . . , ak a set of 1-forms such that a' A . . . A txk # 0. Then, if a = a' A . . . A ak, we get

ker(tx) =

n

ker(a')

lsjsk

and ker(a) has dimension n - k. PROOF: By Proposition 9.13 we can choose a basis e l , . . . , en for E so that & = aj for 1 < j < k. Then

ker(a) = { u E E l u is in the span of e k + l ,.. . ,en} =

0

ker(aj).

l s j s k

I

It is not always true that the kernel of a k-form on an n-dimensional space has dimension n - k. We can assert this in general only for decomposable forms. In fact, if a is a 2-form, then dim(ker(a)) = n - 2 if and only if a is decomposable (see Exercise 9.23). THEOREM 9.72 Let M be a differential manifold of dimension n, z: T M -+ M the tangent bundle and B c T M a smooth subbundle of dimension k. For each m E M there is an open set U containing m and n - k smooth 1-forms a,, . . . , an-k,on U such that z - '(x) n B = ker(a, A . . . A an-&)) for all x E U . PROOF: The result is immediate from Lemma 9.69 on the existence of B-subbundle trivializations of T M . I

Suppose that a is a smooth k-form on M such that there is an integer t for which dim ker(a(x)) = t for all x E M . We dejine ker(a) by ker(a) =

u ker(a(m)). mEM

We want to show that ker(a) is a smooth subbundle of T M . This fact is a special case of the following theorem. THEOREM 9.73 Let p : B M and q: C M be smooth vector bundles over M of dimensions k and t,respectively. Suppose that 4 : B + C is a smooth mapping such that -+

-+

(a) for each m E M 4(B,) c C , and 4: B, -+ C, is linear, (b) there is an integer r such that, for all m E M dim ker(6)Bm)= r. Then ker

4 = UmEM ker(4 I B,) is a smooth subbundle of B.

175

KERNEL OF A DIFFERENTIAL FORM

PROOF: For each m E M , ker($IB,) is a vector subspace of B,. We will

show that ker(q5) is a submanifold and a subbundle of B. Let m E M . Choose local trivializations ( U , , h , ) and ( U , , h,) for Band C such that m E U , n U , . We have the commutative diagram

p-l(u, n u,) A

( u , n u,) x (R'

q-I(u,n u,) A

(u,n u,) x

4

I

x

~ k - r )

( ~ e - ~ + xr ~ k - r ) .

We may assume h,(ker(41Bm)) = { m } x (R' x (0)) and h,(im($IB,)) (m} x ((0) x Rk-'). Write h, $ h;'(x, 5) = (x, l(x)5), where

=

0

I:U , n U ,

-+

L(Rr x R k - * ,

x Rk-')

is smooth. Let a,:R'-ki' x Rk-' -+ R k - * be a projection. Then (1, A(m): (0) x Rk-' -+ (0) x R k P ris an isomorphism. Thus there is an open set W containing m with W c U , n U , and 0, I(x)((O}x Rk-' an isomorphism for all x E W. Then 0

'

0

Rr

Rk-r -

ker(a,

0

l(x)) 0((0) x Rk-')

for all x E W . Let p(x) be the projection onto the second factor of this decomposition. Then p: W -+ L(Rk,R k - ' ) and p is smooth (see Exercise 9.22). Note that ker(a, A(x)) = ker(A(x)) so that I - p(x) is the projection onto ker(A(x)) of the above decomposition. If 71, and 71, are the projections of R' x R k - * , define v: W -+ L(Rk,R k ) by v(x) = (nl,p ( x ) ) . Then v(x) is an isomorphism for x E Wand v(x)(ker(A(x)))is precisely R' x (0). Finally, taking h(5) = (1 x v(p(5))) h , and restricting W to lie in a coordinate chart on M about m, gives a submanifold chart and a subbundle trivialization for ker(4) c B. I 0

0

COROLLARY 9.74 If a is a smooth k-form on M and dim(ker(a(m)))= G for all m E M , then ker(a) is a smooth subbundle of T M . PROOF: Define I,: T M -+ Rk- , ( M ) by 1,1 T,M following Definition 9.70). Then ker(a) = ker(la).

=

(see the remark

I

There are other natural ways of specifying a k-dimensional subbundle of a vector bundle p : B -+ M . Basically, we must specify, for each x E M , a k-dimensional linear subspace of B,, say C,, such that these subspaces vary smoothly with x. This smoothness condition can be given as in Theorem 9.73 or Corollary 9.74 or we can specify that, for each point m, there must be k C" local sections of B which give a basis for C, for x near m (see Exercise 9.24).

176

9. DIFFERENTIAL FORMS

INTEGRABLE SUBBUNDLES AND THE FROBENIUS THEOREM DEFINITION 9.75 Let C be a smooth k-dimensional subbundle of T M , dim(M) = n. We say C is an integrable subbundle if, for each m E M, there is a chart ( U , 4) on M about m such that, for each x E U , we have

TX$(C,) = {4(x,> x ( R k x ( 0 ) )= {$(x))

x

R",

That is, there is a natural tangent bundle chart which is also a local Csubbundle trivialization (see Fig. 9.4). We noted above that if C is a subbundle of TM, then, for any m E M, there is an open set U about m and k C" vector fields X I , . . . , xk on U such that, at each p E U , { X , ( p ) , . . . , X,(p)) gives a basis for C,. We refer to such vector fields as forming a local basisfor C on U . Then an equivalent definition of integrable subbundle is DEFINITION 9.75' Let C be a smooth k-dimensional subbundle of TM, dim(M) = n. We say C is an integrable subbundle if, for each m E M, there is a chart ( U , 4 = (XI,.. . ,x")) on M about m such that a/ax', . . . , a/dxk are a local basis for C on U . If C c TM is a k-dimensional subbundle, then given m E M it is always possible to find a k-dimensional submanifold Z of M such that m E Z and T,Z = C,. But can we always find a submanifold Z passing through m such that T,Z = Cx for eoery x in Z? We shall see that the answer in general is no! However, if C is integrable, then the answer is yes. To see this, pick m and choose ( U , 4) as in Definition 9.75. If $(m) = h, then take

z= I $ - ' ( { y

E

4(V)ly' = hi

for i

=k

+ 1, . . . , n}).

This Z has the desired properties. The situation is depicted in Fig. 9.4. Suppose dim(M) = n, a is a nowhere vanishing 1-form on M and C = ker(a). Corollary 9.4 implies that C is a smooth ( n - 1)-dimensional subbundle of TM. It follows from Definition 9.75 that if C is integrable, then there is a collection ( U i , f i , gi)i such that (Ui)iis an open cover for M and That is, l/gi is an integratingfuctor for c1 on a = g& (i not summed) on Ui. Ui. Conversely, suppose there is such a collection ( U i ,fi, gi)i. If m E M, then m E U i for some i. Proposition 3.7 implies that there is a coordinate chart ( W , I) = (x', x2, . . . , x")) such that m E W c U i and f = x1 on W. Thus C = ker(dx') on W so that the surfaces x1 = constant are tangent to C. This shows that C is integrable. We have shown that C is integrable if and only if there exists a system of local integrating factors for a.

177

INTEGRABLE SUBBUNDLES AND THE FROBENIUS THEORFM

An important special case, which is actually the case shown in Fig. 9.4, is when we have a one-dimensional subbundle C of T M . As noted above (also see Lemma 9.69), we have, for each m E M , a neighborhood U of m on which there is a smooth vector field X such that ( X ( p ) >is a basis for C , for each p E U . The following result says that, locally, such a vector field can be straightened out by a chart so that the corresponding vector field in Euclidean space is one of the coordinate vector fields. What this shows is that any one-dimensional subbundle is integrable. For subbundles of dimension greater than 1, integrability cannot be assured unless extra conditions are imposed. The fundamental result, which we prove shortly, is the Frobenius integrability theorem, which gives necessary and sufficient conditions for a subbundle to be integrable. THEOREM 9.76 (Straightening out theorem) Let X be a C" vector field on M , rn E M , and assume X(m) # 0. Then there is a chart ( U , 6 = (xl,. , . ,x")) on M with 4(m) = 0 such that, on U , X = d/dx'.

PROOF: Let P : g ( X )+ M be the flow of X . Choose an open set W about m and an E > 0 such that (-c, c) x W c 9(X).Then choose an (n - 1)dimensional submanifold N containing m and contained in W such that X(rn) q! T,N. Let ( V , I//) be a submanifold chart on N with m E I/ and $(rn) = 0; let $ ( V ) =

c R"-'. Define

O:(-E,

x

V+

W

by H(t, X) = P(t, $-'(x)). T,N, i it is easy Clearly 0 is C" and O(0,O) = m. Using the fact that X ( m )+ to show that qo,,,e: T ( ~ . , , x~ RR"- 1) T,W E)

--+

is a linear isomorphism. Thus, by the inverse function theorem (cf. Theorem 3.354, there is an open set 0 about the origin in R" (= R x R " - l ) and an open set U about m such that H : 0 + U is a diffeomorphism. Take 4 = 8 and we have our chart ( U , 4). To see X = (?/ax1note for (ql, q 2 , . . . , q " ) E 0 we have

'

P(4'

+ t , $ - 'h2, . '

'

,471

=O

= X(P(ql, I / / - ' ( 4 2 > .. . , q " ) ) ) = X(O(q',

. . ., 4")).

I

We now turn to the Frobenius integrability theorem, one of the most important results in differential manifold theory. The proof we give is a recent one due to Chern and Wolfson [4].

178

9. DIFFERENTIAL FORMS

FIGURE 9.4

THEOREM 9.77 (Frobenius integrability theorem) Let M be an nmanifold and let C be a smooth k-dimensional subbundle of T M . Then C is integrable if and only if the following two equivalent conditions hold:

(a) If U c M is open and X , Y are vector fields on U taking values in C (ie., X ( p ) and Y ( p )are in C, for all p E U ) , then [ X , Y ] takes values in C. (b) If X , , . . . , X , are a local basis for C on some open set U , then there exist C" functions Cij:U -+ R, for i, j , r = 1,. . . , k, such that [Xi, X j ] = CijXr. PROOF: We leave as Exercise 9.25 the proof that (a) and (b) are equivalent. We first prove that, if C is integrable, then (a) holds. In this proof we shall not use the summation convention since different sums may extend over different ranges of the index. Let C be integrable and suppose U , X , and Y are as in (a). Given p E U , choose a chart ( V , q5 = ( X I , .. . , x")) with p E V c U such that a/dx', . . . , d/dxk give a local basis for C on V . We may express X and Y as

[xi='

Then [ X , Y ] = I:= (b',iai-, ~ ' , ~ ba/axr ~ ) ] by Theorem 8.14. Thus [ X , Y ] lies back in C , as asserted.

INTEGRABLE SUBBUNDLES AND THE FROBENIUS THEOREM

179

Now we consider the converse, which is the more important (and more difficult) part of the theorem. We assume (b) holds and show that C is integrable. We proceed by induction on k. For k = 1 the conclusion is immediate from Theorem 9.76. Here the hypothesis (b) is not used since, in this case, it is automatically satisfied. Now suppose k 2 2 and suppose the result has been proved for subbundles of dimension I k - 1. Choose p E M and let X , , . . . ,x k be a local basis for C in a neighborhood of p . By Theorem 9.76 we may choose coordinates ( y ' , . . . , y") about p so that, in a neighborhood of P , Xk

(9.19)

= a/ayk.

Define vector fields X i for j = 1,. . . , k - 1 by

xi = xj

-

dyk(Xj)Xk.

(9.20)

Then, using (9.19) and (9.20) we see for j

dyk(XJ)= 0

=

1,. . . , k - 1.

(9.21)

x;-,,

X , form a local basis for C so we may, by The vector fields X i , . . . , (b), express each [ X i , X J ] in terms of this basis as [ X i , X J ] = d:jX',

+ . . . + d!;

'XL-

,+ d i j X k .

(9.22)

Because of (9.21), if we apply dyk to both sides of (9.22), we get dij = 0 (here we used dyk([Xi,X i ] ) = 0 which follows from (9.21) and Theorem 8.27). Now, in a neighborhood of p , define a ( k - 1)-dimensional subbundle c to be the span of X i , . . . , X i - , . By (9.22) with dij = 0, we see that (b) holds for so that, by the inductive assumption, c is integrable. Therefore there is a coordinate system (z', . . . ,z") about p for which d/az', . . . , a/dzk-' gives a local basis for c. Since a/dzj is expressible in terms of X i , . . . , X i - we see

e

dyk(d/8zj)= 0

for j

=

1,. . . , k - 1.

(9.23)

Now djaz', . . . , a / a z k - ' , d/dyk is a local basis for C near p . If we write [ d / d z j , d/dyk] in terms of this local basis, then, applying dyk, we see the coefficient of d/dyk is zero so that we have (9.24) Now express a/dyk in terms of d/dzi, i

=

a yG.

d

-=

dyk

1,. . . , n,

fl=l

(9.25)

180

9. DIFFERENTIAL FORMS

Then we get (see Theorem 8.14) (9.26) Comparing (9.24) and (9.26) we conclude ap/azj= 0

for j

=

1,. . . ,k

-

1; p

=k,.

. . ,n.

r"

aJaz". Therefore tk,. . . , are functions of zk, . . . , z" only. Let zk = Then ajaz', . . . , a/azk-', Z , is a local basis for C since a/&', . . . , a/azk-', a/8yk is a local basis and (9.25) holds. Suppose we perform a diffeomorphism affecting only the coordinates zk,. . . , z". That is, we replace (z', . . . , z") by ( z l , . . . , z k - ' , wk, . . . , w"), where the WJ depend only on (zk, . . . ,z"). Then a/&', i = 1,. . . ,k - 1 are unaffected, while a / a d depend only on a/azk,. . . , a/dz". Thus, by Theorem 9.76, we may choose this transformation so that

zk

(9.27)

= a/aWk,

since zk depends only on the coordinates (zk,. . . , z"). In the coordinate systern (z', . . . , zk-' , wk, . . . , w") a local basis for C is then given by a/azl, . . . ,a/azk- I ,

which proves C is integrable.

a/awk,

1

We have seen that smooth subbundles of 71M can be described locally (and in some important cases, globally) in terms of differential forms. The question of integrability for subbundles is closely related to the operation of exterior differentiation of differential forms. To see this we first give some useful formulas relating exterior derivatives, exterior products, and Lie derivatives. THEOREM 9.78 Let M be a differential manifold, V , V,, V,, . . . smooth vector fields on M , a, a,, a,,. . . smooth 1-forms on M and o a smooth k-form on M . Then

c,"::

O!AW(V,, V,,. . . , & + l ) = (-1y-'a(5)W(Vl,.. ., (2) i,(a A o)= (i,a)o - a A i,o, cI1 A . . A L,aj A ' ' ' A E P , (3) L,(a, A C!, A . . . A U p ) = (4) L, d o = dL,W, (5) L,o = di,o + i, do, (6) da(V19 = vi(a(b))- b(a(Vi)) - ~ ( [ V I , V2IL (7) L o ( V 1 , . . ., v,) = ~ , ( o ( V , , .. ., v,)) - EL14 v 1 , .. . , [ V , V ] , . . . , &I,

(1)

cj"=

'

4,. . . , &+l),

181

INTEGRABLE SUBBUNDLES AND THE FROBENIUS THEOREM

(8) do(V1, . . . , % + 1 )

=

C42: ( - l ) i - l y ( o ( V l , . . . , t,.. . , % + I ) ) + C l < i < i < k + l ( - l)i+j w ( [ l / ; ,Vj], Vl,

PROOF: ( 1 )

aAW(1/1,

a A o = (k

' . . , t , .. . , 5,.. . , % + I ) .

+ l),d(cw0w) so that 1

. . . , & + I ) = -

k!

sgn(a)a(Vu(l,)w(T/,(2), ...> ~

E

s

~

+

&(k+l)).

I

Let si+ = {c E sk+ I a ( 1 ) = j } for j = 1, . . . , k + 1. Suppose a E si+1 . The vectors ( v ~ ( .~,). ,, VU(k+1))are a permutation, t, of ( v ~.,. ., V;, . . . , %+J. The sign of the permutation z is ( - 1)'- sgn(a). To see this, start with (K, V,, . . . , . . , % + J and do f interchanges to achieve the ordering (q,Vu(2)r... , Then do r interchanges to get the ordering (Vl, . . . , V,+ Then sgn(z) = ( - I)', sgn(o) = ( - 1)' and we see ( - 1)'+' = ( - 1 ) j - l so we get sgn(z) = (-1y-.' sgn(0). So

6,.

1

k+l

c (-1)j-1a(y)o(V1

k+ 1

= i

=2

Comparing with the above calculation gives (2).

5

) . . . ) ) . . . )G + l ) .

182

9. DIFFERENTIAL FORMS

(3) Let (F,) be the flow of V . Then, for rn E M , we have

Thus the derivative is computed in the fixed vector space Ap(TmM). Now the map T:M x . . . x T:M + Ap(TmM)given by (pl,. . . ,p,) + p1A . . ' A P p , is multilinear, so we get

(4) The case when w is a smooth 0-form, i.e., a function, is obtained by a local coordinate calculation of both sides (see Exercise 9.26). The general case follows if we prove the case o = f d 4 , A . . . A d4,, f , 4 1 , . . ,4, smooth functions. We know dL,(d$, A * * * A d4,) = 0 by (3) and the present result for 0-forms. Thus (using Exercise 8.9) dL,(fd4,

A.

. . A d 4 J + fL,(d$, A . . . A db,)) = (dLvf) A d41 A . . . A d4, + (Lvf) d(d4, A . . A @), + d f ~L,(d4, A . . .A d4,) + f d L v ( d 4 ,A . . . A d 4 J

. . A d4,) = d(L,f(d$,

A.

'

= L , dfA

= L , d(

fd4

A ' .

A.

. r \ d 4 , f dfA Ly(d41A ' .

.Add,)

. . A d+,,).

This proves (4). (5) For o = f , a 0-form, L,f= df(l/), ivf = 0 (by definition) and i, df= df(V), so the result holds for 0-forms. For higher degree forms we work locally and first prove the result for w = dxil A . . . A dxir where (x', , , . , x") is a local coordinate system on M . Repeated application of (2) and writing V = V' i3/dxi gives i,(dxii

A.

. . A d x ' ~= )

r

1 (j= 1

Then

? . 1y-1Vij dxi1 A . . . A . dxlj A . . . A dxiv.

183

INTEGRABLE SUBBUNDLES AND THE FROBENIUS THEOREM

Of course, iv d o

L,w = L,(dxil

=0

A.

since dw

. . A dxir)=

= 0.

Now

r

dx"

A

. . . A L , dxiJA

. . . A dx"

j= 1

r

=

dxi1 A . .

. A ~ VA ~ . . .JA dx'.

j= 1

This proves the result for o = dx" a = dxil A . . . A dx'.. We have

. . A dx'.. Finally we prove it for o = fa,

+ ( L v f )=~, f ( i , da + diva) + (iv d f ) E , d(fa)= i,(df~ a) + i v f da = (i, df)a df i,a + i v f da

L,(fa)

i,

A.

= fL,a

-

A

and d(i,fa)

=

d(fi,a)

= df A

i,a

+ f' diva.

Comparing the right sides of the last three equations gives the result. (6) Work locally with = V $ d/8xi,j = 1,2 and a = aidx'. We get (a) da(Vl, V2)= dajAdxi(V{d/dx', Vk, 8/8xk)= a,,jV{V\ - U ' , ~ V $ V ; , (b) vi(a(V2)) = I/,(atVi)= (ajVi),jV{, (c) V2(a(Vi)) = (aiV\),jVi? (d) .([V1, V 2 ] ) = [V,, V2]'ui = u i V2i . j V { - u ~ V ~ , ~ V $ . Comparing the right sides of (a), (b), (c) and (d) gives (6). (7) This is just the statement that Lie derivative commutes with contraction and was given as an exercise in Chapter 8 (see Exercise 8.9(f)).

(8) do(V1,. . . , b+1)= (iv, du)(V2,.. . b+i) 1

=

L,p(l.;, . . .

1

K + d - 4 i " 1 4 ( V 2 , . . . K+A. I

The desired result for k = 1 is just (6). Inductively, assume the result for k - 1. Then the term d(ivlo)(V2,.. . , V,,,) can be evaluated since i,,w is a ( k - 1)-form. Use (7) to express the term L V l o ( V 2 ,. .. , b+l). The details are left as Exercise 9.27. This completes the proof of the theorem. I THEOREM 9.79 Let o be a smooth k-form on the n-manifold M and suppose, for all m E M ,

dim(ker(w(m)))= d.

184

9. DIFFERENTIAL FORMS

Then (1) ker(w) is a smooth vector subbundle of TM, (2) If V,, V, are vector fields with values in ker(o), then iv,iv, d o = i[V1,V*]~,

(3) ker(o) is integrable if and only if i v l i v , dw are vector fields having values in ker(w).

=0

whenever V, and V2

PROOF: (1) ker(o) is a subbundle by Corollary 9.74.

(2) This is immediate from Part (8) of Theorem 9.77. (3) Clear from (2) and the Frobenius theorem. I The following corollary is an important special case. COROLLARY 9.80 Suppose a is a smooth 1-form on M, dim(M) = n. Suppose a(m) # 0 for all m E M. Then

(1) ker(a) is a smooth vector subbundle of dimension n - 1, (2) If V, and V2 are vector fields with values in ker(a), then da(V,, V,) - .([Vl > V21), (3) ker(cc) is integrable if and only if dalkerca) = 0.

=

THEOREM 9.81 Let B c TM be a smooth vector subbundle. B is integrable if and only if for every pair of B-valued vector fields and every 1-form M with B c ker M we have da(Vl, V2)= 0.

Proof is left as Exercise 9.28.

INTEGRAL MANIFOLDS Let M be an n-manifold, B c T M a smooth integrable vector subbundle of dimension k. For each x E M we have a k-dimensional subspace B, c T,M. DEFINITION 9.82 An integral manijold of B is a pair (N,f), where N is a connected k-dimensional manifold and f : N -+ M is a smooth one-to-one mapping such that for every x E N we have

T,f(T,N)

=

Bf(,)

If, in addition, f maps N homeomorphically onto f(N), then f(N) is a k-dimensional submanifold of M. In many of the cases we want to consider this will not be true so that, in general, an integral manifold need not be a submanifold of M . Of course, if N happens to be a k-dimensional submanifold of M such that, for x E N, T,N = B,, then (N, i ) is an integral manifold of B, where i: N + M is inclusion.

185

MAXIMAL INTEGRAL MANIFOLDS

EXAMPLES 9.83 (a) Let B = ((.u, y , A, -n)lx, y , jbE R ) . Then viewing B as a subspace of TR2 = R Z x R Z we see B is a subbundle. Note B = ker(a), where a = d x dy, so B is a subbundle. Since dcc = 0, it follows from Corollary 9.79 that B is integrable. Given ( p , q ) E R 2 there is an integral manifold N containing ( p , 4). Here we may take N = { ( p t, q - t ) l t E R } and f = inclusion. (b) Take M = R 3 - ( 0 ) and let B c T M be defined as

+

+

B

=

ker(a),

+

where a = x dx y dy - z dz. B is a two-dimensional vector subbundle of T M . Let N be the submanifold defined by x2

+ y2

-

C a constant.

z 2 = C,

Then N is an integral manifold of B. For we see T(,,Y,Z,Nc R3 is given as T(,,y,z,N= ( ( 1 1 ,

U , MI)I2XU

= {(u, 11,

lY)lC((X,

+2

y~ ~ Z W= 0 } y , z)(u, u, w ) = 0)

= ker(a(.t, y , z)).

Again du

= 0 so

B is integrable.

MAXIMAL INTEGRAL MANIFOLDS We now wish to show that an integrable subbundle has a maximal integral manifold through every point. Thus, let B be a k-dimensional integrable subbundle of T M . DEFINITION 9.84 (a) The piecewise smooth curve y : Z + M , I an interval, is tangent to B if y'(t) E By(,,for all t E I. (b) For m E M we denote by W ( m )the set of all x E M for which there is a piecewise smooth curve y : [u, h] + M , tangent to B, with ?(a)= m, y(h) = x. We shall define a differential structure on W(m)in such a way that (W(m),i ) is an integral manifold of B. DEFINITION 9.85 A B-slice of M is a pair (S, $) where S c M , $:S and the following condition holds:

+ Rk

there is a chart ( U , 4) on M such that

R~I

lxil < c, i = 1, . . . , n} for some E > 0, (a) 4 ( ~=){ x E (b) Tm4(Bm) = {4(m)}x (Rk x 10)) for all m E U , (c) there are constants a k + . . . , a,, such that S = $-'({x for all i, x j = ajfor k 1 ~j I r z } ) and II/ = 41s.

+

(S, $) is said to be obtained from ( U , 4).

E

R"I lxil < E

186

9. DIFFERENTIAL FORMS

THEOREM 9.86 Let m E M be fixed. Let O(m) be defined by O(m) = {$-'(V)I there is a B-slice (S, $) such that S c W(m),V is open in Rk and I/ c $(S)}. Let t(m)be the topology for W(m)generated by O(m).Let d ( m ) = { ( S , $)l(S, $) is a B-slice and S c W(m)}.Then, with topology r(m),d ( m )is a C"-atlas for W(rn).

Proof is left as Exercise 9.29. Provide W(m)with the C"-differential structure determined by d ( m ) .We have required, as part of our definition of differential manifold, that the topology be second countable. Thus we need to show that t(m) is a second countable topology on W(m).This is a nontrivial fact and is established in the following theorem. THEOREM 9.87 With the topology t(m)the space W(m)is connected and is second countable. PROOF: Let x E W(m).There is a piecewise smooth curve y: [a, b] + M which is tangent to B, with y(a) = m, and y(b) = x. Clearly y([a, b]) c W(m). We claim y: [a, b] + W(m)is continuous, i.e., continuous with respect to the topology t(m) on W(m).Let c E [a, b] and m, = y(c). Choose a chart ( U , 4) on M such that $(U) = {x E R"I1xil< E } , Tp4(B,) = {&I)} x (Rk x (0)) for all p E U and 4(mo)= 0. Then 4 y: J -+ R" (where J is a connected open subset of [a, b] containing c), 4 y is smooth and 4 y(c) = 0. Since (4 y)' ( t )= q(t+$(y'(t)),which lies in { 4 ( y ( t ) ) } x ( R k x {0}), we see that, if 0

0

S

=~ - ' { x ~< ~x E, i ~ =

1,.

0

0

. . , n, xj = 0,j = k + 1,. . . , n},

then y ( J ) c S. Since ( S , 4 IS) is a chart for W(m)we see y: [a, b] + W(m)is continuous at c so, since c was arbitrary, y is continuous viewed as a map into W(m).Thus W(m)is connected. The proof that W(m)is second countable is given in [5, pp. 96-98]. Our W(m)is a submanifold of M according to the terminology of [5] (not necessarily according to our terminology, however). Lemmas 1-4 in [ 5 ] are detailed and clear and hold for C" as well as analytic manifolds. I Thus, we now have a smooth manifold W(m)and the inclusion i: W(m)-+ M. By construction the pair (W(m),i) is an integral manifold of B containing m and is called the maximal integral manifold of B through m. REMARK 9.88 W(m) is a connected differential manifold. However, it may not be a submanifold of M. Let S' be the unit circle in the complex numbers. Take M = S' x S'. We may represent TM as

TM

=

{(z,ilz, w, iiw)lz, w E S',

A E R).

187

INACCESSIBILITY THEOREM

Take X ( z , w ) = (z, iz, w, ivw), where v is a small irrational number. Let

z , ) E M, A E R } c T M . B = {AX(Z,w ) ~ ( W Then B is an integrable subbundle of T M and each W(m)is dense in M . The proofs of these statements are left as Exercise 9.30.

INACC ESS IB I LlTY TH EO R EM DEFINITION 9.89 If x, y are points in M we say y is B-accessible from x, written xBy, if y E W(x).

x B y defines an equivalence relation on M , (b) If S, and S, are B-slices of M and x,By, for some x1 E S, and y, LEMMA 9.90

(a)

E S,,

then xBy for all x E S,, y E S,. Proof is Exercise 9.31. We see that if S is a B-slice and m E M then either S c W(m) or S n W (m)= 0. In the former case, we can speak of the slice S being accessible from m. THEOREM 9.91 (Inaccessibility theorem) Let m E M . Then,in any open set containing m there exist points which are not B-accessible from m. More precisely, let ( U , 4) be a chart about rn such that

(a) 4 ( ~=) {xllx'l < E, i = 1 , . . . , H I , (b) Tp4(Bp)= { 4 ( P ) } x (Rk x (01) for all P E U , (4 4(m) = 0. Then there exist at most countably many B-slices obtained from ( U , 4) which are B-accessible from m (note we are assuming B integrable). PROOF: Let S be a B-slice obtained from the chart ( U , 4) such that S is a B-accessible from m. Then S c W(m)and, by construction of the topology on W(m),S is open in W(rn).If there were uncountably many such slices, then W(m) would contain a uncountable collection of disjoint open sets, contradicting second countability. 1

If we denote the set of B-accessible equivalence classes by M/B, then M/B

=

{W(m)lmE M )

and we have the canonical projection IT: M .+ MJB

by

.(m)

=

W(m).

188

9. DIFFERENTIAL FORMS

NONl NTEGRABLE SU BBUN DLES Let B be a nonintegrable k-dimensional vector subbundle of T M . The definition of the set W(m) still makes sense, even though B is not integrable. However, without the assumption of integrability we cannot give W(m)the structure of a k-dimensional manifold as we did before. Also the inaccessability theorem no longer holds. We present the following example.

EXAMPLE 9.92 Let (x,y, 8,4) be coordinates on R4 and define a subset B of TR4 by the requirement that each of the following l-forms vanish on B: cl1 =

Thus, B

= ker(cr')

dx - sin 8 d4,

c12 =

dy

+ cos 8 d 4 .

n ker(c12). By Lemma 9.71 we see

B

= ker(w),

where w

= cll A ~1,.

Thus, B is a two-dimensional subbundle of TR4. Now we have

o=dx~dy+sinedyr\d4+cosBdxr\d4. By Theorem 9.79, B is integrable if and only if i,,i,, dw = 0 whenever V,, V, are vectors in ker(w). Now it is easily checked that ker(w) is spanned by the vector fields

By Theorem 9.78(c) we see . . E ~ , dw Z ~=,d 4

+ cos 8 dy

-

sin 6 dx # 0,

so that B is nonintegrable. We claim if m E R4 then W(m)= R4. That is, any two points are accessible from one another. It is enough to show any point (xo, y o , 8,, &) is accessible from (0,0, 0, 0). For this, move along appropriate integral curves of V , , V,. Details are left as Exercise 9.32. REMARK 9.93 The subbundle B of the preceding example actually arises in mechanics. If we refer to [12, p. 151, we see that our subbundle B is the state space for a mechanical system which consists of a disk of radius 1 which rolls without slipping in the xy plane. The fact that all points are accessible from each other can be seen from the physical picture presented on p. 15 of [12]. The disk can be rolled to any desired point in the plane and then we can arrange any desired values of 8, 4. The desired value of 8

189

VECTOR-VALUED DIFFERENTIAL FORMS

is obtained by simply rotating the disk about the vertical axis through its center. Then the desired value of 4 is obtained by rolling the disk about a circle returning to the original x , y , 0 but with 4 changed by an amount depending on the radius of the circle. This is an example of a mechanical system subject to nonholonomic constraints. The configuration space is R4 but the no-slip condition causes state space not to be all of TR4 but rather, the subbundle B. If B were integrable, we could consider the system with configuration space restricted to one of the maximal integral manifolds W(m). The restricted system is holonomic in the sense that the state space is the complete tangent bundle of the configuration manifold. In the example just given this cannot be done, since B is not integrable.

VECTOR-VALU ED DIFFER ENTlAL FORMS In many situations, including some discussed later in this book, there arise differential forms whose values lie in some vector space other than the real numbers. Much of the discussion closely parallels what we have previously done but, as there are some new features, we give a brief discussion here. Let E be a real vector space of finite dimension, E* its dual. DEFINITION 9.94 A smooth E-iiulued k-form on M is an assignment to each m E M of an alternating multilinear map of degree k, a(m), from TmM into E such that the following smoothness condition holds: For every j- E E*, (Acr)(m)(ul,.. . , u k ) = A[a(m)(ul,.. . , u,)] defines a smooth real valued k-form 1-a on M . REMARKS 9.95 (a) The E-valued form a can be expressed in terms of components by choosing a basis for E. Then the smoothness criterion in Definition 9.94 is equivalent to smoothness of all components. (b) We could have constructed a smooth vector bundle flk(M;E ) = fl,(M) @3 E = (&(TmM) @ E). Then an E-valued k-form is a smooth cross-section of this bundle.

UmGM

Choose a basis e l , . . . , e, for E and write R-valued k-form. DEFINITION 9.96

dcr, is the ( k

M = Miei

where each cii is an

The exterior derivatiue of the k-form c!

+ 1)-form defined by

da = (da’)e,.

=

criei,denoted

190

9. DIFFERENTIAL FORMS

It is a simple matter to check that da does not depend on which basis is used in E. The notions of tensor product and wedge product parallel those previously given except that a bilinear “pairing” is used to “multiply” function values. DEFINITION 9.97

Suppose v: E x F

Let a be an E-valued k-form and p an F-valued [-form. is a bilinear mapping. Define

+G

(a 0, p ) ( m ) ( u l , .. . 9

Ok+l)

= v [ a ( m ) ( U l ?. . *

3

Uk)? p(m)(uk+

1,.

..

9

uk+t)].

Also, define

Note that S is the antisymmetry operator and is defined exactly as before. The fact that the values are vectors makes no difference here. REMARK: In applications, v is often some very natural pairing and we often omit writing it as a subscript. THEOREM 9.98 Suppose e l , . . . , e, is a basis for E, fl, . . . ,f,is a basis for F and v: E x F -,G is a pairing. Then we have

(a) If a = aiei, p = Bjfj and gij = v(ei, f j ) , then cc A,P = aiA p j g i j , (b) d(a A, p) = da A, p (- l ) k aA, dp where k = deg(a).

+

PROOF:

(a) a A,P(U1, . . . > uk,

Uk+ 1 ,

.. .> Uk+t)

EXERCISES

191

EXERCISES 9.1 Prove Proposition 9.5. 9.2 Prove Lemma 9.8(a). 9.3 Prove Proposition 9.11. 9.4 If w E A,(E), where dim(E) = n, then there exist u l , . . . , u" in E* for which w = u1 A . . . A 0". 9.5 (a) If dim(E) = n, show that every ( n - 1)-form on E is decomposable. (b) Let (d,u2, u 3 , u") be a basis for (R4)*. Show that u 1 r \ u 2 u3 A u4 = o is not decomposable.

+

9.6 Let dim(E) = II. (a) If w E A,(E) is nonzero and e l , . . . ,en is a basis for E, then o ( e l , . . . , en)# 0. (b) If y, I.: E + E are linear show, using Definition 9.16, that det(y 3.) = det(y) det(2). 0

9.7 Prove that -, as defined in Definition 9.18, is an equivalence relation on the set of all ordered bases of E . 9.8 Let ( u l , u2), ( w l ,w2) be similarly oriented bases for R2. Then there is a matrix of continuous functions (af(z)),0 I z I 1, det(aj(z)) > 0 V t such that (ai(0))= I and ai(l)ui = w j . 9.9 (a) Let p, p' be two orientations of X . Suppose for some xo E X , pxo= p i o . Then p x = pk. for all x in the component of X containing xo. (b) If X is orientable, then X is connected if and only if X has precisely two orientations. 9.10 Suppose X has charts ( U , 9) and ( V , i+b) with X connected. Then X is orientable.

=

U v V and U n V

9.1 1 (a) In Example 9.35 show that each U iis open and each homeomorphically onto R2. (b) Show P2(R)is not orientable.

9imaps U i

9.12 (Higher-dimensional real projective spaces) Define P 3 ( R )= (R" - O)/-, where p Ap for p E R4 - 0,A E R - 0. Let n: R4 - 0 + P 3 ( R )be projection. Define charts, this time four of them, by analogy with Example 9.35. Show that P 3 ( R )is orientable. What about P,(R) for arbitrary n?

-

9.13 Consider the standard spherical coordinates on R3, (Y, 8, 4). r > 0, 0 < 8 < n, 0 < 4 < 2n.

192

9. DIFFERENTIAL FORMS

(a) Show that this is a positively oriented chart. (b) Calculate the standard Riemannian metric on R 3 in the spherical coordinate system. (c) Calculate the Riemannian volume element in spherical coordinates.

9.14 If ( u l , . . . , u " - ~ )is a positively oriented basis for T,M and (Np, 019. ..>

un-1)

is a positively oriented basis for T,R", then if (u;, . . . , V L - ~ ) is any other positively oriented basis for T,M, show that (N,, u ; , . . . , is positively oriented. This shows the definition of positive unit normal is unambiguous.

9.15 (a) Let e, = (d/ax')l,, 1 I i I n. Let n

n.

j,:dx'A...r\dx'/\...r\dx"

w = i= 1

be an ( n - 1)-form where, as usual, the "hat" over dx' indicates omission of dx' from the wedge product. Then show J ( p ) = 4p)(el. .. .

,4, . . . , en).

(b) Let N = n'e,. Show that

(c) In the case M

=

S2, N

i, dx A dy A dz

= (x,y , z )

=

so

x dy A dz - y dx A dz

+ z dx A dy.

Does this agree with the formula given in Exercise 9.13? That is, take the formula given by that exercise for the Riemannian volume element and compute the interior product with N = a/&. Then compare with the above result.

9.16 (Components of p-forms) Given (bjl.. j k ) , 1 I ji I n, we introduce the corresponding skew-symmetrized quantities b,, . .j,, by ,

,

(a) Show that (bj,.. .jk) is skew-symmetric in j , , . . . , J k if and only if bj, . .jk = brjl .j k l . (b) A tensor, given as A = A i l . , i,dx" 0 . . . 0 d x ' ~ is , a p-form if . = A,',.. and only if A i l . .i,, ,

,,

,

193

EXERCl S ES

(c) Let A

=

A i l . ,. i,

dx" A . . . A dxiP. Show that

-

A , i , . . ip]

= ( l/p!)Ail... .ig.

(d) Prove that if A is a p-form, with components chart, then the coordinates of d A in that chart are

in some

(dA)ii...ip+l= (P + l ) ( - l ) P A I i , . . . i p , i p + I ~ .

9.17 For each of the following l-forms, see if it is closed and, if so, find a O-form whose exterior derivative is the given form. (a) w (b) w

+

= xy dx (4x' - y ) dy = ex cos y dx ex sin y

+

dy

9.18 If X is a manifold with boundary, we have defined the notations of Show that the inward and outward pointing vector at a point p E definition does not depend on the choice of chart.

ax.

9.19 If X = {(x,y)lx2 + y 2 5 1 ) and ( r , 0) is the usual polar coordinate system, then, if we use the usual Riemannian metric in the plane, show that a/& is the outward pointing unit normal at each point of d X .

9.20 (a) In Minkowski space we use the usual time coordinate xo = t , but replace the spatial coordinates (x', x2,x3) by (r, t3,$). Find the volume 3-forms d3C,, d'C,,

d 3 C , , d3C,.

(b) Let f be a C"-function on R 4 . Express Af in the coordinates ( 4 r> 0, 4). 9.2 1

In a manner analogous to that used for the tangent and cotangent bundle, put a topology and differential structure on T:(M) and R,(M) so that both become smooth vector bundles over M .

9.22 Let W be open in R",j W + L( R', R' x Rk-' ) smooth. Let f = (f1, f 2 ) , where fi, f 2 take values in L(R', R')), LIR', Rk-'), respectively. Assume fi(x) is an isomorphism for all .x E W . Then for each x E W we have R' x Rk-'

=

im(f(x)) @ ( ( 0 )x Rk-').

Let 0 2 :W -+ L(Re x Rk-', Rk-') be given by projection on the second summand above. Prove that o2 is smooth. 9.23 Prove that if a is a 2-form on the n-dimensional vector space E , then dim(ker(a)) = n - 2 if and only if a is decomposable. 9.24 Suppose p: B + M is a smooth vector bundle and that, for each rn E M , C, is a k-dimensional linear subspace of B , = p-'(rn). Suppose for

194

9. DIFFERENTIAL FORMS

each m E M there is a neighborhood U of m and k Cm-vectorfields on U , X,, . . . , X,, such that X , ( p ) , . . . , X,(p) is a basis for C, for each p E U . Then C is a smooth k-dimensional vector subbundle of B. 9.25 Prove that conditions (a) and (b) of Theorem 9.77 are equivalent. 9.26 By working in a local coordinate system, prove that, iff is a 0-form, V a vector field, then L , df = dL,f: 9.27 Complete the proof of part (8) in Theorem 9.78. 9.28 Prove Theorem 9.81. 9.29 Prove Theorem 9.86. As part of the proof show that O(m)does in fact generate a topology. This means if Y,, Y2 E O(m)and p E Y, n Y2,then there is a Y, E U(m)with p E Y3 c Y, n Y,. 9.30 Verify the comments in Remark 9.88. Proceed as follows: Define p : R 2 + M by p(s, t) = (e2nis,e'""). Given (so, t o ) E R 2 let L(so, to) = {(s, t ) I t - to = v(s - so)}. Show p : L(so,to) + M gives an integral manifold p(so, to).Since p is locally a diffeomorphism it can be used to construct appropriate charts showing integrability. Alternately, p could be used to locally represent B as the kernel of a 1-form and then use Corollary 9.80. For the denseness property, first argue that p(L(so,t o ) ) is dense in A4 for all (so, to). Since each W(m)contains p(L(so,t o ) ) if p(s,, t o ) E W(m),this shows each W(m)is dense. In fact, if p(so, to)E W(m),then W(m)= P(L(S0, t o ) ) . 9.31 Prove Lemma 9.90. 9.32 In Example 9.92, prove that W((0,0, 0,O)) = R4. 9.33 Consider H" as an oriented manifold with the standard orientation. View dH" as R"-' in the obvious way. How does the induced orientation on R"-' compare with the natural orientation? 9.34 Consider the standard orientation of S". How does this compare with the orientation obtained by viewing S" as the boundary of the ( n + 1)-ball B"+l = (X E

R n f l 11x1 I l }

the ball having the standard orientation from R""? 9.35 Prove Theorem 9.53. 9.36 Let E be an oriented vector space, g a metric on E, E the metric volume form in An(E).Show that if a E R is a 0-form, then * a = a&.

195

EXERCISES

9.37 Let E be an oriented real vector space with metric g and metric volume element E. Let LY E A,(E) and let 2' = a # , that is, v results from r by raising the index. Show that * a = i,s. 9.38 Let M be the sphere of radius ro > 0 about the origin. We have spherical coordinates (r, 0 , 4 ) in R 3 . Find the three area 2-forms d2C,, d2C,, d2C,. Find expressions for the restrictions of these forms to M in the coordinate system (0, 4) on M .

+

9.39 Continuing the preceding exercise, let 4 = Se(i3/a0) t4((a/84)be a vector field on M . Calculate Sf as a function of (0, 4). Iff: M + R is a scalar, calculate Af as a function of (0, 4). 9.40 Suppose that M is a manifold and T M = B 0 C, where B and C are smooth integrable subbundles. Show that for each rn, E M there is a chart ( U , 4 = xi, . . . , x") with rn, E U such that

for each m E U . 9.41 Suppose that B c T M is a smooth subbundle and that X is a complete vector field on M with flow ( 4 J r c RSuppose . also that if Y is any B-valued vector field on M , then L,Y is B-valued. Show that, for = B,t(m). every t E R and m E M , Tm4r(Bm) 9.42 In a manner similar to the construction in Exercise 8.17, give the set Q,(x) =

u

A,(T,X)

xex

the structure of a smooth vector bundle over X , and show that a differential p-form on X is just a smooth section of this bundle. Show that this bundle is a smooth vector subbundle of the bundle of covariant tensors of degree p defined in Exercise 8.17.

I0 Integration of Differential Forms

In this chapter we consider integration on smooth manifolds. We shall define the integral of a continuous n-form with compact support (this is defined below) over an oriented n-manifold. We shall see that this allows us to integrate a continuous function with compact support over an oriented Riemannian manifold. This generalizes the notion of surface integral, familiar from calculus. The main result of the chapter is Stokes’s theorem (Theorem 10.6). This theorem is of fundamental importance in manifold theory and has as consequences the three classical integral theorems, usually associated with the names of Green, Gauss, and Stokes.

THE INTEGRAL OF A DIFFERENTIAL FORM We begin by recalling some properties of smooth partitions of unity on a smooth n-manifold M . From Theorems 6.34 and 6.36 we know that, given there is a locally finite atlas ((V,, $ j ) ) j , J with any atlas on M , ( ( U i ,4i))ia,, each 6 contained in some U i (the cover (y)jEJ refines the cover (Ui)isl)and a family of P - m a p s (pj)jaJ,p j : M + [0, 11, such that the support of p j lies p j ( x ) = 1 for all x E M . Recall that, because in 5 for eachj and such that of the local finiteness of (V;.),iEJ, each m E M has an open neighborhood W on which all but finitely many p j vanish. This implies that, if K c M is compact, then there is a finite subset J’ c J such that if j E J - J‘, then p j ( x ) vanishes on K . In this chapter, when we refer to a partition of unity ( p j ) j s J , we shall assume it is related to a locally finite atlas as above. In analogy to the familiar definition for functions, we have

cj

DEFINITION 10.1 If o is a differential form on M , the support of o is the closure in M of the set of x E M for which w(x)# 0. This set is denoted supp(w). If supp(w) is compact, we say o has compact support. 196

197

THE INTEGRAL OF A DIFFERENTIAL FORM

sLI

We let p denote Lebesgue measure in R" so that integrals ,f d p , for U c R", should be interpreted as Lebesgue integrals. However, in most cases to be considered, f will be continuous with compact support so Riemann integrals will suffice. A basic result from analysis that we will need is the change of variable formula, which we state. THEOREM 10.2 Let U be open in R", f:U + V a diffeomorphism onto the open set V in R". If g E L ' ( V ) , then (g -f)Idet(Df)I is in L'(U) and

lv

9 dP = Ju (9 f)ldet(Df)l dP.

A proof can be found in [18].

Henceforth, we assume M is a smooth n-manifold (possibly with boundary) and assume M is oriented; thus we have chosen a particular orientation v for M . Let o be a continuous n-form with compact support on M . We want to define the integral of o over M , denoted jM

w.

We first define the integral in case there is a positively oriented chart ( U , 4) with supp(w) c U . Indeed, in that case, we may write o = h d.u'

A.

. . A dx"

on U ,

where 4 = (x', . . . , x") and observe that supp(h) = supp(w) is a compact set in U . Since h: U -+ R is continuous, we may define (10.1)

To see that this is well-defined, we prove LEMMA 10.3 Let ( U , 4) and ( V , $) be positively oriented charts with supp(o) c U n V , 4 = (x', . . . , x") and $ = (y', . . . ,y"). Let o = h dx' A * . ' A dx" on U and o = k dy' A . . ' A dy" on V . Then

PROOF: Since supp(h) c U n V and supp(k) c U n V , we may replace n V ) and $(V ) by $(U n V ) in the equation to be proved. On U n V we have

b( U ) by 4( U

h dx' Let F

= II/

0

4- ': $(U

A..

. A dun = k dy'

A'.

. Ad?/".

(1 0.2)

n V ) + $(U n V ) . Then, by Proposition 9.15 and

198

10. INTEGRATION OF DIFFERENTIAL FORMS

Eq. (10.2), we get

h

0

4-l

= (k 0

4-l) det(DF).

Thus on 4 ( U n V ) , we have h

0

4-l = ((k I+-') F ) det(DF). 0

0

(10.3)

Then Theorems 10.2 and 10.3 give, using det(DF) > 0 (recall that both charts are positively oriented),

which proves the lemma. [ Suppose (PJi.1 is a partition of unity and that w and ( U , 4) are as above. Then p i o is a continuous n-form having compact support in U for each i. Since supp(w) is compact, we see that p i o is identically zero if i is not in a certain finite subset I' c I . Then

Thus, from (10.1) it easily follows that

Now drop the assumption that supp(o) lies in a single chart. Let (Pi)iEI be any smooth partition of unity (recall that we always assume that our piu partitions of unity are subordinate to some locally finite atlas). Then is already well defined, by (10.l), so we may define

sM

If supp(w) lies in a chart, then we have, in (10.1) and (10.5), two definitions of

IM o.However, by (10.4), the two agree. It remains to show that, if (qj)jeJis another partition of unity, then

But, for each i,

199

STOKES'S THEOREM

so that we have

where only finitely many terms are nonzero. Similarly,

Thus,

j M w is well defined for any (0 having compact

support.

STOKES'S THEOREM We begin with some preliminary results, which will lead to the general form of Stokes's theorem. In this section, assume M is an oriented n-manifold with boundary M. Let w be a smooth ( n - 1)-form on M . Recall that the interior of M is defined by Int(M) = M - dM and that (ul, u 2 , . . . , in T,(dM) is positively oriented if ( N , ul, . . . , u,is positively oriented in T,M, where N in T,M is outward pointing. LEMMA 10.4 Let ( U , 4) be an oriented chart on M with U n dM empty and $ ( U ) a bounded open set in R". Suppose o is an ( n - 1)-form on M having compact support in U . Then

PROOF: Let

4 = ( X I , . . . , x") and

let

(11

be given on U by

Then

so that we get (10.6) Choose a cube C, = {x E R"1 lxil 5 u for i = 1, . . . , n} containing $(U) in its interior. If g is any smooth function with compact support in $(U), then -2d57p = O ,

+(a)c7x'

for j = 1 , . . . , n.

200

10. INTEGRATION OF DIFFERENTIAL FORMS

This is because the integral is equal to

which is seen to be zero by treating it as an iterated integral, integrating with respect to xj first, and using the fact that g vanishes on the boundary of the cube. Since the right side of (10.6) is the sum of such integrals, we have the lemma. LEMMA 10.5 Let ( U , 4) be an oriented chart on M with U n i3M = r! a nonempty set. Assume that 4 ( U ) is a bounded open set in H". Let o be an ( n - 1)-form on M having compact support in U . Then

PROOF: Express o on U as 0

(- l)"Ei

=

dx'

A . . . A z i A . .

. Adx".

i= 1

We see it is enough to prove the result in case 0 = (-

for some j. Let ,f = ,f

1)j-Y dx' A , .

. . A dx"

4- ',so that f has compact support in 4 ( U ) .We have

so that

If j = 1, . . . , n - 1 this integral is zero since we can integrate first in the xjdirection and use the fact that f vanishes outside a compact set in $(U). For j = n we get

Now consider J a M o;we see ( V , 6)is a chart on ciM, where &in) = ( ~ ' ( m .) ., . , x"-' ( m ) ) . The orientation of the chart ( V , 8) is ( - 1)" times the induced orientation which we use on d M . The form o on dM has compact support in V . If j = 1,. . . , n - 1, then clearly w is 0 on V , since dx" = 0 on V . Thus,

201

TRANSFORMATION PROPERTIES OF INTEGRALS

the lemma holds for j < n. Thus, consider = (~

1y -

d.yl . . .

If

because of the orientation of the chart 8

I)',-

= (-

J'M

I(

I(.*-', ...,

-

-J$w which gives the lemma in this case. -

(I/, +),

1.

we see

6-

f'

I)

dp

0) dx' . . . d x " - ' ,

.Y"-',

I

We can now prove our main result. THEOREM 10.6 (Stokes's theorem) Let A4 be an oriented n-manifold with boundary aA4. Let w be a smooth ( n - 1)-form on M having compact support. Then

IM

dw =

j<,M

0.

PROOF: Cover M by positively oriented charts ( ( U j ,$ j ) ) j E J , with each set $i(Uj) a bounded open set in H". Let (pi)ierbe a subordinate partition of unity. There is a finite set I' c 1 such that if i E 1 - l', then supp(p,) n supp(w) = @. We claim, for each i E I',

(1 0.7)

This follows from Lemmas 10.4 and 10.5. But (I)=

c

pi",

i E I '

so summing (10.7) for i

E

l', we get the desired result.

I

TRANSFORMATION PROPERTIES OF INTEGRALS THEOREM 10.7 Let ( M i ,vi) be oriented n-manifolds for i = 1, 2 and f : M l + M , an orientation preserving diffeomorphism. If (0 is an ti-form having compact support on M , , then JM,

.1'*0 =

The proof is left as Exercise 10.5.

JM2

0.

10. INTEGRATION OF DIFFERENTIAL FORMS

202

EXAMPLE 10.8 Let M = {XI - 1 I x I l} with the standard orientation induced from R . If w = x 2 d x and f :M + M is defined by f ( x ) = - x , then

LEMMA 10.9 Let

5 be a vector field with flow (4Jtand

w be a differ-

ential form. Then (a) $:(i,O) = q4:4; (b) ( d / 4 1 t = r o ( 4 : 4 = 4,*,(L,w); = o whenever both sides are defined, if and only if Leo = 0. (c) PROOF: First note that

5 commutes with

its own flow, that is,

The last equality follows, since

41*,Lrw= 4; d i p

+ 4t*,isdw = dis4,*,w+ i, d$,*,w = L,&w.

I

THEOREM 10.10 Suppose that M is an oriented n-dimensional manifold, ( is a complete vector field on M , and w is an n-form on M such that L p = 0. Let (4Jtbe the flow of 5 and suppose that N c M is a compact n-dimensional submanifold (with boundary). Then

jNw j,,

for t 2 0. o PROOF: Since each 4ccan be continuously deformed to the identity it is orientation preserving. Thus jgf(N) w = J N q5:w = J N w by Lemma 10.9. I =

w-DIVERGENCE OF A VECTOR FIELD

203

a-DIVERGENCE OF A VECTOR FIELD If w represents a volume element for M, we see that L p = 0 implies that the flow of 5 is volume preserving. This motivates the following definition. DEFINITION 10.11 Let M be an n-dimensional manifold and w a nowhere vanishing n-form on M. If 5 is any vector field on M we define the a-divergence of [ by the equation

L p

=

(div, 5)o.

(10.8)

PROPOSITION 10.12 Let A4 be an oriented Riemannian manifold, ( a vector field on M , and E the Riemannian volume element. If denotes the form obtained from [ by lowering the indices then div, 5 = -@'= *d*r.

r

PROOF: Choose a chart (x") and write E

=

dx'

A.

. . A dx"

We have

LCc= [d(&)(5)] dx'

+ & dx'

A

A.

. . A dx"

L; dx2 A . .

. A

-t &(Lr

dx"

dx') A dx2 A . . . A dx"

+ . . . + & dx'

A ' . . A

L, dx"

so (10.9)

204

10. INTEGRATION OF DIFFERENTIAL FORMS

so

Taking

* again and comparing with (10.9) gives the result. I

OTHER VERSIONS OF STOKES'S THEOREM We now consider some alternate versions of Stokes's theorem. Theorem 10.6 does not apply when the space is an n-dimensional rectangle

K

=

{(x', . . . , x")la' I xi I b' for 1 I i I n),

since.K is not a manifold with boundary (due to the edges and corners). We obtain a version of Stokes's theorem in this case. If (ul, . . . , 0), is an ordered basis, then we denote by [v,, . . . , vn] the orientation containing this basis. Thus, if ( el ,. . . , en) is the standard basis for R", then v = [ e l , . . . ,en] is the standard orientation for R". If K is the n-rectangle defined above, we define the various faces of K by

for j

=

Kf

=

{(x', . . . , bj, . . . , x")la' I xi I b' for i # j } ,

Kj:

=

{(x', . . . , d , . . . ,x")la'i xi I b' for i Z j } ,

1 , . . . , n. Let w be an (n - 1)-form on K , by which we mean

c n

W =

Widxlr\...r\~ir\...r\dx",

i= 1

where each wi is a continuous function on K . We define the integral of w over Kf and Kj: by w = ([K,

+

n.

1)j-l

Scj

wj(x', . . . ,bj, . . . , x")dx' . . . dx' . . . dx"

and JK, -

w = (- 1)' Jcj

COAX', . . . ,uj, . . . , x") dx'

n.

. . . dxJ. . . dx",

where Cj = {(x', . . . , i j , . . . , x")la' I xi I b' for i # j } . If o is C' on K , we define

OTHER VERSIONS OF STOKES'S THEOREM

205

and we define

What we have done in the preceding discussion is to proceed just as we would have if everything really were smooth. Since they are not, we just define things in analogy to how they work in the smooth case. Thus, the factor (- 1 ) j - I in the integral of o over K t is there because the map 4:Kf Cj, given by 4(x1, . . . , b', . . . , x") = (x", . . . , 6',. . . ,x"), is (the analog of) a chart and its "orientation" is ( - l)j- times the orientation of Kj'. Similarly, we see why the integral of do and dto itself are defined as they are. Now the boundary of K consists of all the faces, K : and KJ:, so we integrate over K by summing the integrals over these faces. Thus, ---f

'

Then, with the above definitions, we have THEOREM 10.13 (Stokes's theorem for a rectangle) If o is a C' (n - 1)form on the n-rectangle K , then

We leave the proof as Exercise 10.6. EXAMPLE 10.14 Consider the case n = 2, as shown in Fig. 10.1. The various faces and their orientations are shown in the figure. Writing o = P(x', xz) dx' Q(x', xz) dx', we get

+

where C is the boundary, oriented counterclockwise. Thus, Theorem 10.13 reduces to Green's theorem on the rectangle in this case. We now obtain Stokes's theorem for a generalized cylinder in R". Suppose M is a compact (n - 1)-dimensional submanifold with boundary in R". Suppose X is a smooth vector field in neighborhood of M , such that X is not tangent to M at any point of M (in particular, X is nowhere zero on M ) . Let (4Jtbe the flow of X . For sufficiently small positive a, the map ( t , ~ ) - + 4 ~ (defines x) a diffeomorphism of a neighborhood of [O,a] x M onto an open set in R". Let fi be the image of [0, a] x M , that is,

fi = u { 4 t ( M ) 1 0I tI a}.

206

10. INTEGRATION OF DIFFERENTIAL FORMS

FIGURE 10.1

The situation is pictured in Fig. 10.2. If a is an n-form defined near A,then, taking ( x i )to be the standard coordinates of R", we write c( = h dx' A . . . A dx" and then define JG

M =

JG h dp.

The assumptions made above imply that M is an orientable hypersurface for T,M is posand we orient M by requiring that the basis (ul, . . . ,u,itively oriented if and only if (X(rn),u l , . . . , u,,-~) is a positively oriented basis for T,R", and orient each $,(M) so that 4, is orientation preserving. Now ah?f = M u 4,(M) u S where S = U { 4 , ( d M ) ( OI t I u } . Suppose that m E $,(dM) and N is an outward pointing vector to 4,(dM). Then (w',. . . , w,-J is positively oriented for Tm(4,(dM))if ( N , w l , . . . , w,-J is positively

FIGURE 10.2

207

INTEGRATION OF FUNCTIONS

oriented for ‘Tm(4t(M)),which is true if ( X , N , w,, . . . , w , - ~ )is positively oriented for T,R”. But we may assume that N is also an outward pointing vector for S at m, so that ( - X , w , , w z , . . . , w,)is the boundary orientation on S. Notice that the boundary orientation on M (the bottom of A?) is the negative of that defined above while the boundary orientation on +,(M) (the top of A?) is the orientation we have defined. With these comments, we get THEOREM 10.15 Let o be a smooth (n - 1)-form defined on a neighborhood of @ and let M, @,(M)and S have their boundary orientations. Then we have Ji? do = J M

+

L M )0

+

Is

0.

PROOF: Using the orientation we have defined on the sets 4 r ( M ) ,we get

JI,(M)

Lxt’l

j

= 8 < I M )ixdw

Integration of the above equation from t rewriting,

+ Jd,(M) &xu). =

0 to t

=u

gives, after some

INTEGRATION OF FUNCTIONS Let M be an oriented n-manifold with metric y. Let metric volume element.

E

be the associated

DEFINITION 10.16 Let f’: M + R be continuous with compact support. Then the integral of f ‘ over M is denoted by

208

10. INTEGRATION OF DIFFERENTIAL FORMS

and is defined to be

One can use standard techniques of analysis (e.g., the Reisz representation theorem, see [18]) to show that the above definition leads to a unique positive Bore1 measure on M such that the integral of a continuous function f with compact support with respect to this measure equals the integral of the differential form f'. This explains the above notation. The familiar surface integrals of calculus are integrals of the type we are considering (see Exercise 10.1).

THE CLASSICAL INTEGRAL THEOREMS Let X be an oriented, n-dimensional, Riemannian manifold with boundary ax,with E , t the Riemannian volume elements on X and ax,respectively. Let N be the outward unit normal on ax. LEMMA 10.17

Let x E ax,w E TZX. Then on Tx(i?X)we have *w

=

4 N X ) t X .

PROOF: Let (e2,. . . , en) be a positively oriented orthonormal basis for Tx(dX). Then ( N x ,e 2 , . . . ,en) is a positively oriented orthonormal basis for T x X . Let (o',. . . , o n be ) the corresponding dual basis. We must show that on T X ( d X )we have, for eachj, *wj = oj(Nx)2x.

The right side of this equation is S{ w 2 A . . . A o".The left side is (- 1)j. A & A . * . A O" which, when restricted to Tx(8X),is 0 if j # 1 and O1 0 2 A . . . A O " = t i f j = 1. I THEOREM 10.18 (Gauss's divergence theorem) Let 4 be a smooth vector field with compact support on X . Then

Jx div(0 4% =

jax ( 4 , N ) 4%

have jx div(<)dpL,=jx(div(5))E= j X - ( f i & = -jx *(@) - jx *(- * d * a = j x d * f = fax = j d X = Sax (4, N ) 4%.I PROOF: We

m)t

t

=

The Divergence Theorem in R3 Let X be a compact, oriented 3-manifold with boundary in R 3 . If vector field on X and N is the outward unit normal on ax,then

IX W4)dv jax(5, N ) dS, =

4 is a (10.10)

209

THE CLASSICAL INTEGRAL THEOREMS

where dV = d p & ,dS = dpi. If dinates, then we have

(.XI,

x', x3) are the standard Cartesian coor-

jxdiv 4 d V jx ti,'dx' =

We have the area 2-forms d 2 C i , i

= 1,

2, 3, and (10.10) can be rewritten as

dx' r l x Z d.u3 =

Jx

dx2 dx3.

j?x5'd2Zi.

(10.11)

The Divergence Theorem in Spacetime Let X be an oriented 4-manifold with boundary in spacetime R4, ( a vector field with compact support on X . Then

Ix

div 5 dp,

= =

Ix

(div ( ) E

lx*(div <)

Recall that *f = ("d3C, in a Lorentz frame. Writing, as is often done, d4V rather than d p E ,we see (10.12) We view (10.12) as the spacetime version of the divergence theorem

The Classical Stokes's Theorem Let X be a compact, oriented 2-manifold with boundary in R 3 . Let ( , ) be the standard Riemannian metric on R 3 . Let 5 be a smooth vector field on an open set U containing X . Let N be the positive unit normal on X . Let d A be the Riemannian volume element on X and let dS be the Riemannian volume element on d X . S X is an oriented l-manifold so, at each x E S X , there is a unique unit vector z, which gives a positively oriented orthonormal basis for T,(?X). We call z the positive unit tangent on FX. THEOREM 10.19

PROOF: Let

(Stokes's theorem)

(XI, x2, x3)

Under the conditions stated above,

be standard coordinates on R3. Let 5

=

5'?/?x',

4 = (, dx'. We first claim that on T ( d X ) , 5 = ((,z) dS. For each side is a

21 0

10. INTEGRATION OF DIFFERENTIAL FORMS

1-form on T,(dX), for each x, so apply both sides to T,; &rx) = (t,, r,) and dS](t,) = (tX,tx) since, by definition, the Riemannian volume element gives 1 when applied to a positively oriented orthonormal basis. We next show on T X we have

[(t,, T,)

df Now curl we get

5 is related

=

(curl 5, N ) dA.

to * d f by raising indices. Applying Proposition 9.63

* ( * d f )= (curl Since

5, N ) dA.

** = 1 on forms in R 3 , we have the desired equation. Now

Green's Theorem in R 2 Let X be a compact 2-manifold with boundary in R2. Let o = P dx

+

Q d y be a 1-form on X . Then Stokes's formula says

sx do

= /ax

or, explicitly,

The integral on the left is

while the integral on the right is the line integral of the vector field ( P , Q) along the curve ax,traversed in the direction of the positive unit tangent, i.e., in the counterclockwise direction.

EXERCISES 10.1 Let U be open in R 2 , $: U R 3 a smooth parametrization of the 2manifold X c R 3 . This means $ mapping U onto X is a diffeomorphism. Then ':X -+ U is a chart on X and we choose X to be oriented in such a way that I,-'is a positively oriented chart. Denote points in U by (u, 0). Let r , = d/du, rv = 3/80 be the coordinate base vector fields on X . Let E be the metric volume element on X . --f

21 1

EXERCISES

Let Iru x r,l be the magnitude of the cross product ru x rv. Then 0 x r,,l E L ' ( U , p) and forf E L ' ( X , p&)we have

f E L ' ( X , pe)iff(.f $)lr,,

lxf&& J" f ( $ h

o))lr, x r,I du do.

=

IY,

How is x rvl related to the quantity the Riemannian volume element?

& which usually appears in

10.2 (a) Let X be a compact, oriented, n-manifold, o a positively oriented n-form. Assume X has no boundary. Then o cannot be exact. (b) Show that the above conclusion may be false if X is not compact. (Hint: Let a = ( - 1)J-l.d dx' A ' . . A & ? A , . . A dx" and compute da.) 10.3 Fill in the details of the proof of Theorem 10.13. 10.4 Suppose M is a compact oriented n-dimensional pseudo-Riemannian manifold without boundary. Define an inner product on k-forms by ( a , p ) = Jw a A *p. Show that (da, p ) = ( a , Sp).

10.5 Prove Theorem 10.7. To do this, let ( P ~ ) ~be , a partition of unity on M,. Then, if qj = p j f , we see that ( q j ) j E is a partition of unity on M , . Now the result follows easily from the basic definition of the integrals involved. 0

J

10.6 Prove Theorem 10.13 by treating the integral of do as a sum of terms, each of which can be handled as an iterated integral. 10.7 Suppose that M is a compact oriented n-dimensional Riemannian manifold without boundary. For each k 2 0 we have the differential operators

d:A k ( M )

+

h:Ak(M)+A,-,(M)

(take A-,(M) = (0)),

+

by A = d6 6d. Furthermore, we have an inner product on each &(M) given by ( a , p ) = a A *p. For each k 2 0 let xk(M)= { a E A,(M)IAa = 0) (the harmonic kzfbrms). Show that (use Exercise 10.4):

A: Ak( M ) + I\k( M )

SiCr

(a) ( a , a) 2 0 and ( a , a ) = 0 implies a = 0; ) { a E Ak(M)Ida= 0 and 6a = 0} since AN = 0 gives 0 = (b) X k ( M = (Au, a ) = . ' .; , and xk(M) are (c) the vector subspaces d(Ak- * ( M ) )6(Ak+l(kf)) orthogonal in Ak(M); @ a(&+ , ( M ) )@ % k ( M ) C & ( M ) as an (d) We thus have d ( I \ k - ,(kf)) orthogonal internal direct sum. The fact that this is actually

21 2

10. INTEGRATION OF DIFFERENTIAL FORMS

equality is the Hodge decomposition theorem, and follows from the fact that CI E XCk(M)'implies CI = Am for some w E &(M) (see ~91). (e) the kth De Rham cohomology group Hk(M)is isomorphic with X k ( M )the , space of harmonic k-forms. Since H k ( M )is determined by the topology of M , this relates the kernel of the differential operator A to the topology of M .

10.8 (Boothby's proof of the hairy ball theorem [ 3 ] )

Let M be a manifold without boundary and N = [0, 13 x M = { ( t , m)10 I t I 1, m E M } . Then N is a manifold with boundary. Define 9;:Ak+, ( N ) + /\k(M) by Y w ( m )= sh [i,w(t, rn)l TmM]dt, where ( ( t , m) = (d/ds)l,=,(t + s, m). Show that d 9 w + 9 dw = w1 - w o , where o,= jTw, with j,: M + N by jt(m) = (t,m). Take M = S2 c R 3 . Suppose there is a smooth vector field X. S2 --t R 3 such that X(m)E TmS2and IIX(m)ll = 1 for each m in S2. Define H : N = I x S2 + S2 by H(t, m) = (sin nt)X(rn)+ (cos m)m. Thus H(0, m) = m and H(1, m) = - m . Let R = i*(x' dx2 A dx3 - x 2 dx' A dx3 + x3 dx' A dx2), where i: S2 + R 3 . Then dR = 0, but R = 3 dx' A dx2 A dx3 # 0, where B = { u E R3111uII I 1). Thus, there is no 1-form /? on S2 for which R = dp. Let A(m) = H ( 1, rn) = - m = A"(m),where A:R3 + R3 by = -x. Since det A = - 1, A" is orientation reversing. But takes the outward normal to Sz at m to that at -rn so that T,S2 must be orientation reversing. Thus A*R = -R. Let w = H*(R). Now

ssz

SB

a(,)

a

j:o(rn)=R( -m)(TH(l, m)T,,,j,)=O(Am)TA= A*R(m)= -Q(m) and j,*w(m)= R(m)(TH(O,m)'i"j,(m))

= R(m).

By (b), since dw = 0, d ( 9 w )= jTw - jgw = - R - R = -2R, which is a contradiction to (d). Show that this proof extends to S2"c R2"+'. For S2"-' c R2" = { ( x ' ,y', x2,y 2 , . . . , x", yn)Ixi,y' in R ) we have that X(x', yi)= Cy= (xi (dldy') - yi(d/dx'))gives a nowhere vanishing tangent vector field to S2"-'. If M c R" is open and starlike with respect to the origin, let F: I x M + M by F(t, x) = tx. Define X : A f ( M )+ L I - ~ ( Mby ) X ( w ) = 9(F*w).Show that this 2 is exactly the mapping used in the proof of Proposition 9.44.

The Special Theory of Re I a t ivity

BASIC CONCEPTS AND RELATIVITY GROUPS The physicist defines spacetime as the set of all events. This corresponds to the definition of a differential manifold as an abstract topological space. The existence of coordinate charts allows us to label points by parameters. In physics we want to record the occurrence of events using clocks and measuring rods, such as (saying) “the particle was at a certain position at a certain time.” To do this, one uses a reference frame, which defines time and spatial position. Associated with each reference frame is a four-dimensional Cartesian coordinate system. Different reference frames produce different labels for the same event and different relationships between events. The group (or pseudogroup), each element of which corresponds the label of an event in some reference frame to the label of the same event in another reference frame is called the relativity group of the theory. Since specific reference frames are constructed in terms of physical properties, the relativity group is obtained from physical considerations (a purely mathematical derivation must be based on some fundamental assumptions about the nature of our observation of the physical world). By a physical process we mean an ordered sequence of events in spacetime. Of special interest to us is a physical process which corresponds to successive observations of a particle. The classical continuum assumption is that we can consider such processes to be a continuous curve which is called the world line of the particle. Since the study of such curves is the subject of classical mechanics, spacetime becomes a smooth manifold (see the discussion in Chapter 1) and the relativity group must consist of smooth transformations. Thus spacetime is a four-dimensional smooth manifold M . On this manifold there is a collection of smooth curves F , each of which represents the world line of a particle. The choice of a basic unit of measurement is given by the proper time axiom: For each y E F , there is a differential 1-form 8, defined on y c M , such 8, represents the proper time elapsed that if y 1 c y, then O,, = O,ly, and 21 3

214

11. THE SPECIAL THEORY OF RELATIVITY

during yl. The motivation for this axiom is the idea of a simple basic process (an atomic clock). However, we must point out that it is here joined with the continuum assumption. REMARK 11.I The proper time axiom is satisfied in general relativity theory where there is a Lorentzian metric dz2 defined on M so that Or = P I y V y E F. This fixes the clocks in a reference frame 92 as follows: If W is a reference frame with Cartesian coordinates ( t , x, y , z ) and y = { ( t ,x, y , z)lto I t I t l } for fixed (x,y, z), then J, 0, = t , - t o .

In the special theory of relativity one assumes that there is a collection 9 of reference frames with the following property: If y E 9 represents the world line of a particle subject to no external force That is, in 9, the particle and 9 E 4 then y appears as a straight line in 9. is observed to move with constant velocity. Each element of 9 is called an inertial reference frame. An equivalent way of looking at this is the following: If 9 can be specified physically then the concept of force is defined. The elements of 9 are assumed to define global charts for M . We want to determine the relativity group G of 9.Since each group element must correspond straight lines to straight lines, G is a subgroup of the affine group of R4. We now derive the subgroup of G which are correspondences under which the origins of the reference frames represent the same event. In this derivation (which follows that of [25]) no use is made of the proper time assumption. It is a result that proper time can be described using a metric in special relativity. Consider two inertial rejerence frames 9,W:

9:Spatial origin 0, Cartesian coordinates (x, y , z), time coordinate t . 82: Spatial origin I", Cartesian coordinates (2,j , Z),time coordinate t. Assume that at t = t = 0 the origins coincide and the coordinate directions are the same, while generally, W is undergoing uniform translatory motion with velocity u in the direction of the x-axis. Thus the location 6 is given by x = ut, y = z = 0. We wish to find the functions which describe the transformation between these two coordinate systems. These equations have the form t = AEt + A ~ +X A:y A!z,

X = A,$

+ A:x

+ + Aiy + A ~ z ,

+ A:x + A ; y + A:z, Z = Ait + A:x + A:y + Aiz,

j

= Ait

(11.1)

21 5

BASIC CONCEPTS AND RELATIVITY GROUPS

where the matrix (At) must depend only on the relative velocity u. We proceed to identify the various matrix elements. STEP1:

The axes coincide at t

=

I

=

0 so if t

= 0, x = 0, z = 0, then

X = 0,

Z = 0 and hence we get

If t

= 0,

x = 0, y

0 = Aiy,

y arbitrary, so A;

= 0;

0 = Aiy,

y arbitrary, so A;

= 0.

=0

then X = 0, j = 0 and thus

0 =A~z,

z arbitrary, so A:

=

0 = Aiz,

z arbitrary, so A:

= 0.

0;

Similarly, A: = 0, A: = 0. We have therefore reduced (1 1.1 ) to

I = A:t

+ A ~ +x A:y + A;Z, (11.2)

STEP2: The origin has coordinates ( t , ut, 0, 0) in 9. The origin 0 has coordinates (I, 0, 0,O) in 8.From (1 1.4) we get

0 = AAt

+ Aiut,

SO

A:

+ A:.

= 0;

0 = A;t,

so

A;

=

0;

0 = A&

so

A;

=

0.

Similarly 0 has coordinates ( t , 0, 0,O) in ."R while 0 has coordinates (f, -uf, 0, 0) in B!. From (11.2) we get - u t = Aht;

The first and fourth of these give

-A:u,

0 = A;t;

Ah

=

0 = Ait;

Ah

+ A ~ =u 0

t = A:t.

A:

= A:.

which combines with

to give

21 6

11. THE SPECIAL THEORY OF RELATIVITY

So we have reduced (1 1.2) to

(11.3)

Consider an event which has coordinates (0, 0, y, 0) in 9.In B the coordinates are (t,0, j , 0). By homogeneity I must be 0, since otherwise t would depend on distance from the x-axis, violating homogeneity. So t =0; but then 0 = A:y

Similarly A:

=0

A:

SO

= 0.

and our transformation becomes

+A~x, X = -I\O,ut + A:x, t = A:t

(1 1.4)

A;Y> A:z.

'j -

Z=

Now by isotropy, A; = A: and both are positive since.the axes in 9, B have the same positive directions at t = 0. Similarly, A: > 0 since at t = 0 we have X = Agx. Thus (1 1.4) becomes

(11.5)

By the principle of relativity, the equation expressing ( t , x, y, z ) in terms of

(f,2,j,5)is obtained by interchanging barred and unbarred coordinates and replacing u by -u. Thus, t = A:(

x

=

Y= Z=

-A:(

-

u)t

+ A:(

- u)( -

-

u)X,

u)I + A:(

-

u)X,

(11.6)

A 3 - u)Y, A:(

- u)Z.

21 7

BASIC CONCEPTS A N D RELATIVITY GROUPS

If (11.5) is inverted directly we get

(11.7)

Comparing coefficients in (1 1.6) and (1 1.7) we get (11.8)

(11.9) (11.10) Now A:( - u) must equal A:(u), since the factor by which transverse lengths change cannot depend on whether the motion is to the right or to the left. So Ai(u) = 1 by (1 1.10).Now A:(u) is the factor by which time scales change and hence, A: is an even function of u. By (1 l.9),A: is an odd function of u. Our transformation thus becomes

t = A:t

+ AYx

(A: > O),

z= z . This form depends only on the principle of special relativity [25], homogeneity of space, and isotropy of space. Suppose we assume absolute time, i = t. Then A: = 0, A: = 1 so we get t = t,

x=x j

=

y,

z = z.

-

ut,

(11.12)

21 8

11. THE SPECIAL THEORY OF RELATIVITY

This is the Galilean transformation. It is based on the relativity principle, spatial homogeneity, spatial isotropy, and the explicit assumption of absolute time. These assumptions are in conflict with a fundamental experimental fact: the velocity of light is the same in all inertial frames. Let a flash of light occur at t = 0 at the origin 0. A spherical shell of light spreads out in all directions with velocity c. A point (t, x,y, z ) on this shell satisfies x 2 + y2 + z 2 = c2t2. The corresponding events in 32 have coordinates (f, 2,j , 2). The Galilean transformation, -

t

=

t,

x = x - ut, j = Y,

z= z , implies (X + + j 2 + 2’ = c21z. This is not a sphere about the origin 8, so the light is not spreading out with the same velocity in all directions in the frame 32. The assumption of absolute time is just that, an assumption, based on “common sense.” The approach taken by Einstein was to assume, instead of absolute time, the law of constancy of velocity of light. This leads to a different determination of A: and A:. We now proceed to this. REMARK 11.2

Let us assume (11.11) and the postulate of constancy of the velocity of light and determine A: and A:. Consider an event ( t , 0, ct, 0) which measures the position of a light signal sent along the y-axis, beginning at t = 0 at 0. The coordinates of this event in iZ! are, by ( l l . l l ) ,

t = Att, X = A:(

-

ut),

j = ct,

z = 0. But this light signal travels with velocity c, starting at the origin at l = 0, so its distance from 8 after t seconds is cl. Hence 22

+ y 2 + 52

= c2p,

which gives

(A:)2U2t2+ C2t2 = C2(A:)2t2,

21 9

BASIC CONCEPTS A N D RELATIVITY GROUPS

so that ( Q 2 u 2 + c2 = c ~ ( A : ) ~ . Solving for A: and using A: > 0 gives (11.13)

Now consider sending a signal of light along the positive x-axis beginning at 0 at t = 0. After t seconds the pulse is described in 9 by (t, ct, 0,O) and i n . g by(t,ct,O,O). But b y ( l l . l l ) ,

+

(a) t = A:L Ayct; (b) c t = Ag(ct - ut). Multiply (a) by c and subtract from (b) to get

0

=

AyC2t

+ @t

or, canceling t,

0 = Ayc2

+ A$.

We then conclude

Thus, we obtain the Lorentz transjormation equations -

t= 2=

t

(I

-

(ux/c2)

- (u2/c2))”2

x

-

ut

(1 - (u2/c2))”2’

’ (11.14)

Y’Y,

z= z. Henceforth, we shall use x l , x 2 , x3 rather than x, y , z and xo = ct rather than t. Also, let fi = u/c, y = 1/(1 - fi2)’’2. We then can rewrite (11.14) as

.uo = y(x0 - fix’), 2’ = y(x1 - fiXO), 22

- x 2, -

.f3= x3.

(11.15)

220

11. THE SPECIAL THEORY OF RELATIVITY

In matrix form, these equations become

(11.16)

The transformation (I 1.15) which transforms from an inertial coordinate system (x”) to another inertial coordinate system (2”)moving with velocity u in the x’-direction is called a velocity boost along the x’-axis with velocity u. It is clear that similar equations can be obtained for velocity boosts along the x2- and x3-axes. Formulas for velocity boosts along the coordinate axes are (boost in x’-direction)

(boost in x2-direction)

(boost in x3-direction)

RELATIVISTIC LAW OF VELOCITY ADDITION Let (2”)have velocity u2 in the 2’-direction and let (2”)have velocity u1 in the x’-direction. What is the velocity of (2”) relative to (x”)? We shall derive Einstein’s formula for this which says that the “common sense” answer, u1 u 2 , is incorrect. The origin of (2”) is given in (%”) by

+

(t,u2i,0, 0).

22 1

RELATIVISTIC LAW OF VELOCITY ADDITION

The corresponding events, in (x”)-coordinates, are obtained by inverting (1 1.16), which gives

Y

[

PY 0 0- ci

Brr 0 0

u2T

0 0 1 0

0 0

0 0

0 1

The velocity in (xu)is then (pyci

+ u,yt)c

-

u1

+ u2

yci + Pyu2t 1 + (u1u2/c2) Thus we have the velocity of (2’)measured by (xu).In terms of p we can write

i= ( p + j7/(1 + PF). < 1 we will have Note that as long as IPI < 1 and Define the velocity parameter a by

P = tanh CI. If B = tanh 5, then tanh(tl + a)=

-

1.

(P).

a = tanh-

Then

Ijl<

tanh a + tanh i? 1 + tanh a tanh i?’

+

= tanh(a E ) and hence Cr = tl + 5, that is, a is additive. Now tanh a = and y = (1 - f12)-1’2. Also, cosh’ a - sinh2 tl = 1 so 1 - tanh’ CI = l/cosh2 a and hence, cosh tl = ( 1 - tanh’ R ) - ” ~ . Therefore,

So,

B

cosh a

sinh tl = By.

and

=y

Thus we may say: If (2”)moves in the .u’-direction with velocity u, then tanh a = B

cash

tl

=

= u/c,

1 (1

~

P2)’/2

The matrix form of the boost from cosh a -sinhtl 0 0

1

-

(XI‘)

(1 - ( u 2 / c 2 ) 1 / 2 ’

to (2”)becomes

-sinh a cosha 0 0

(11.17)

222

11. THE SPECIAL THEORY OF RELATIVITY

RELATIVITY OF SIMULTANEITY Consider two reference frames (x”), (X”), with (2”)obtained by a velocity boost along the x’-axis. Then

+ pyx’, = pyx0 + yx’,

xo = yxo x’

x2 = 2 2 , x3 = 23.

Suppose two events a, b are simultaneous in (2’). Let 2”= 2‘ - gfl. Then do = 0 and we get d o = pyd’.

So if a, b have different %‘-coordinates, we conclude do # 0, i.e., a and b are not simultaneous as viewed from (x”). This is called the relativity of simultaneity.

R ELATlVl STI C LENGTH CONTRACTION Consider a rod of length e stretching from (0,0, 0) to (/,O, 0) in (Z1,X2, X3)coordinates. In (?)-coordinates the ends of the rod are given by ( c t = Xo, O,O, 0), ( c t = To, e, 0, 0). The ends of the rod in (x”) are

+ pyt,

ct = yct,

ct = yct

x1 = pyct,

x’ = pyct+ ye,

x2 = 0,

x2 = 0,

x3 = 0,

x3 = 0.

Suppose we observe the ends of the rod at some time to in the reference frame (x”). Then the left end is given by (ct,, flct,, O,O), the right end by (ct,, j c t , yC(1 - p2), 0, 0), so the length of rod in the coordinate system (x”) is

+

yt(1 - p’)

= t(1 -

p y .

Thus the length in (xp) is shorter than the length in (2”)by the factor (1 - j2)’/’. Thus, judging from (x”), the moving rod has a length which is (1 - p2)1/2times the length of the rod in its rest frame. This phenomenon is historically known as the Lorentz-Fitzgerald contraction.

223

THE INVARIANT SPACETIME INTERVAL

RE LATlVl STI C TI M E D I LATlO N Consider a reference frame (x”)and consider an observer moving with velocity u down the positive x’-axis. The observer measures time with his own clock. How do his time intervals compare with those measured in (x”)? Let (2”)be a reference frame moving with our observer. Consider two events measured in (X”), (cf,

7

o,o, O),

(cf2 2

o,o, 0).

The time difference measured by the observer, i.e., measured in (2”)is fz - tl . What is the time separation of these events measured in (x”)? We have ct, = ycl,

+ pyo = yct,,

Ct,

= yct,,

t, - El is shorter than t , - t,. Thus, we see that the observer’s clock appears to be running slower by a factor y-l = (1 - p2)1’2when recorded in the (.xp) frame. This is known as relativistic time dilation. We remark that the clocks of the (x”) frame would appear to be running slower to the observer (see Exercise 11.1). so t 2 - t , = y(t2 - t,); the time interval

THE INVARIANT SPACETIME INTERVAL We see that temporal separations and spatial separations may be different when viewed from different reference frames. But let a, b be events, d = b - a. View the separation in our usual two frames (x”),(2‘) Then . do

=

] d o + pyd’,

d’

=

pydo + yd’,

d 2 = 22,

d 3 = 23, from which we readily calculate that (do),

-

(d’),

-

( d 2 ) 2- ( d 3 ) , = (ao),- (d’), - (a’),

-

(d3),.

This quantity, which is independent of reference frame, is called the spacetime interval between a and b. The fact that the spacetime interval is independent of the choice of Lorentz frame will allow us to define a natural Lorentz metric on spacetime (see Definition 11.6).

224

11. THE SPECIAL THEORY OF RELATIVITY

THE PROPER LORENTZ GROUP A N D THE POINCARE GROUP Consider linear transformations of R4 which preserve the Lorentz metric g =( d ~ ’) ~( d ~ ’-) (dx2)2 ~ -( d ~ ~ ) ~ .

(11.18)

Any matrix

(11.19)

where (ui)is a (3 x 3)-rotation matrix, is such a transformation, The velocity boosts along the coordinate axes, such as cosh c( sinha 0 0 sinh M cosha 0 0

(11.20)

also have this property, as we showed above. DEFINITION 11.3 The proper Lorentz group Y i is the group of linear isomorphisms of R4 generated by transformations of the types (1 1.19) and (11.20).

REMARK 11.4 We have seen that if (xp), (9) are two Lorentz frames whose origins x” = 0, X p = 0 coincide then the transformation expressing 2” in terms of xfl does in fact lie in 9J. (Recall we assumed all our Lorentz frames had their spatial coordinates oriented similarly.) DEFINITION 11.5 The PoincarP group is the subgroup

generated by

9 of GL(4),

91and the group of spacetime translations.

Every 4 E 9 is uniquely expressible as 4 a translation. Consider a linear transformation of R4,

which preserves the metric g . be the matrix of g, i.e.,

=A

4: R4 .+ R4, Let 4 be given

0

t, where

A

E

9 1 and z is

by the matrix (a:). Let (gJ

225

SPACETIME MANIFOLD OF SPECIAL RELATIVITY

# v, p=v=o,

0, 1, -1,

Then gap = a:a;y,,

so that, taking 1 = a$aal;g,,

= (a:)’

Hence (a:)’ 2 1, so a: 2 1 or a: equation Yagr

p=v>l.

CI =

fi

= 0,

we obtain (a;)’

(a:)2.

-

(a;)’

-

1. Taking determinants in the matrix

=

-

-

a:a;g,,

gives det(ga8)= [det(at)]’ det(g,,). Since det(gab)= - 1, we conclude det(u:) = f 1. It can be shown that 9:is precisely the set of those linear transformations preserving g for which (a) a: 2 1; (b) det(at) =

+ 1.

If we write 2” = atx” and consider points with arbitrary xo and xi = 0 for j = 1, 2, 3, then

xo = a:xo. Thus, a: 2 1 implies that the transformation preserves time sense. Such a transformation is called orthochronous. This is the reason for the arrow in 9 i . The “+” indicates that the transformation preserves spatial orientations. When we refer to Lorentz transformations, we mean transformations in 9;.

THE SPACETIME MANIFOLD OF SPECIAL RELATIVITY As we have discussed, spacetime is the set M of all events. Each inertial reference frame, together with its system of Cartesian coordinates and clocks, gives a chart on M which we shall refer to as a Lorentzjrame. If (xu)is the system of four coordinates associated to a certain Lorentz frame, we shall often refer to “the Lorentz frame (x”).” If (x”) and (2”)are two Lorentz frames, we have seen that the coordinate transformation is of the form

+ b, where b = (b”)is a point in R4 and A E 91. x 4 2 = AX

226

11. THE SPECIAL THEORY OF RELATIVITY

DEFINITION 11.6 Let g be the metric on M whose (x”) representation is ( ~ x O ) ~ ( d ~ ’) ~( d ~-~( d) ~~ ~The ) ~metric . g is a well-defined Lorentz metric on M . We call g the spacetime metric.

Thus, M is a four-dimensional Lorentz manifold which is diffeomorphic to R4.At each point p E M we have a four-dimensional vector space T p M of tangent vectors to M at p . A convenient convention used in relativity is to refer to v E T P M as a 4-vector. This is convenient, since we also want to refer to 3-vectors. Given a Lorentz frame (x”), a 3-vector is a 4-vector of the form (0, a’, u2, u3), that is, a vector which is purely spatial in the given reference frame. But note that this same vector, when viewed in a different Lorentz frame, may have a nonzero time component, so there is no welldefined subset of T p M consisting of 3-vectors. When we speak of a 3-vector we must have a definite Lorentz frame in mind. Consider a particle moving in space. The set of events which make up the history of this particle, defines a curve in spacetime, called the world line of the particle. Given a Lorentz frame (x”), the world line appears as (ct, x’(t), x2(t),x3(t)). We shall use Greek indices p, v, a,. . . as indices on 4vectors and tensors. Latin indices i, j, k, . . . are used as 3-vector indices. So Greek indices take values 0, 1, 2, 3 while Latin indices take values 1, 2, 3. If the world line is (ct, x’(t), x2(t),x3(t)), then (il(t), i2(t), i3(t)) is the usual 3-velocity of the particle, as measured in the given reference frame. We now want to define a four-dimensional velocity vector for this world line. Choose a Lorentz frame (x”) and write xp = x”(t). We cannot define the 4-velocity to be (?(t)) because this involves parametrizing by t and differentiation with respect to t, while t is not independent of the reference frame. What is needed is an invariant parametrization of the curve. We now obtain this parameter, called proper time. Choose a Lorentz frame (x”) and write the world line as (x”(t)). Then xo = ct and (xi@))gives the spatial position in the given reference frame, as a function of t. The 3-vector (ii(t)) satisfies (i’(t))2 + (i2(t))2+ (i3(t))2< c2. Thus the 4-vector having (xp)-components (i”(t)) is a timelike vector, g(dx/dt, dxldt) > 0. Pick to and dejine z(t)

=

c

s‘ /= dt ’ dt to

dt;

(1 1.21)

z(t) is called the proper time elapsed on the world line from to to t. We note that z depends both on to and on the particular world line under

consideration and that dz

1 c

- = - (g(i, i ) ) ’ ’ 2

dt

> 0.

227

RELATIVISTIC TIME UNITS

Let us reparametrize the world line by z. Writing x(z), we see

Now if we choose a different to, z will be affected only by an additive constant. If we use a different Lorentz frame we again get a parametrization of the world line by a parameter f, and

+

It is then easily verified that .7 = z k, for some constant k. Thus we have, for the world line of any particle, a natural parameter z defined up to an additive constant, with the 4-vector dx/dz satisfying dx dx

“(di.di) = c DEFINITION 11.7 The 4-vector dx/dz is called the 4-uelocity of the par-

ticle. So every particle has, at each point on its world line, a 4-velocity vector dx/dz = u satisfying g(u, u) = c2. If we work in a Lorentz frame (xp),and if u is the magnitude of the particle’s 3-velocity, then 1 dt C - -2 2 1/2‘ dz (2- u2)”2 (1 - (0 /c 1 So u = 1/(1 - ( U ~ / C ~ ) ” ~ ) ( Cul,, v2, c 3 ) , where (v’, u 2 , u 3 ) are the 3-velocity components. Thus,

(4 velocity components) u’ =

U‘

(1

-

UZ/C2)’/2

for i

=

1, 2, 3.

RELATIVISTIC TIME UNITS We shall measure spatial distances in meters as is usual. But now, we shall also measure time in meters.

228

11. THE SPECIAL THEORY OF RELATIVITY

DEFINITION 11.8 A meter of time is the time in which a light signal trav-

els 1 meter (m). Since light travels 3.0 x lo8 m/s, we see that 1 m of time =

1

3.0 x lo8

=

3.3 x 10-9 s.

The result of using meters of time is that velocities become dimensionless and c = 1. The speed of any material particle is 0 5 u < 1 in any Lorentz frame. The equations derived up to now remain valid if, wherever c appears we replace it by 1. Note that, since x” = ct, xo is in meters of time. Consider the Lorentz frame (x”).Consider a world line (x”)given by xB = x”(t). Then xo(t) = t and (dx’ldt, dx2/dt,dx3/dt) is the 3-velocity of the particle in the given reference frame. Let v(t) be the ordinary magnitude of this 3-velocity. Consider an interval TI I t I T,. Partition this interval TI = to < t , < . . . < t, = T,. Assume these subintervals are so short that the 3-velocity is essentially constant on each one. Consider a subinterval [ t i t i ] . There is a Lorentz frame (2”)such that, during the interval t i - < t < ti we have the particle at rest in (2”).According to a clock in (2’) the time elapsed as t goes from t i - l to ti is approximately where ti lies in [ t i - l , t i ] . Since the particle is at rest in (X”),this elapsed time is the time which elapses on a clock carried by the particle. Thus, as t goes from TI to T,, the elapsed time, as measured by the particle’s clock, is given approximately by

the approximation becoming better as the lengths of all the subintervals go to 0. Thus, the exact amount of time which passes on the particle’s clock as t goes from T , to T , is dt.

(11.22)

If y is the spacetime metric and dxo dx’ dx2 dx3

then (11.22) is (11.23)

ACCELERATED MOTION-A

SPACE ODYSSEY

229

Thus we have the important fact: The proper time between two events on the world line of a particle is the elapsed time as measured by a clock traveling with the particle.

ACCELERATED MOTION-A

SPACE ODYSSEY

Let z -+x(z) be the world line of a particle, parametrized by proper time. We have defined the 4-velocity of the particle to be u = dx/dt.

We now want to define the 4-acceleration,

for each z. We define a(z) by giving its components in an arbitrary Lorentz frame (x”). If x(z) = (x”(z))in the Lorentz frame (x”) then we define a”(z) = d2x/dt2.If (9) is a Lorentz frame the components (iP(z)) are related to (a”(7))as the components of a vector should be, since the Lorentz transformations are linear. The vector a(z) is thus determined by giving its components in a Lorentz frame. Choose a Lorentz frame (x’). If u, w E TPM we have g(u, w) = U’W’,

the inner product of u and w. If u is the 4-velocity, then 1

(1 1.24)

u’a, = 0.

(1 1.25)

UQU’ =

and hence

Thus, 4-velocity and 4-acceleration are always perpendicular. Now, at any point on the world line (xp(z))we can construct a Lorentz frame in which the particle is instantaneously at rest. We call such a frame an instantaneous rest frame of the particle. In the rest frame at z = zo we have U’(5’) = (1, 0, 0, 0).

Now aw(zo)= (a’, a’, a’, a 3 ) so ( 1 1.25) implies a’ = a’u, = 0.

Thus, the 4-acceleration is (0, a’, a’, a3). The acceleration at any time is a’ = d2xp’/dz2,

230

11. THE SPECIAL THEORY OF RELATIVITY

so we have dx” dx’ dt dz dt dz’ d2xp- dx” d 2 t

dz2

d2x” dt

dt dz2+

dt2 (z)

(11.26)

‘

If we measure these components in a frame in which the particle is instantaneously at rest, then dx’ldt

= 0,

i

=

1,2,3;

dtJdz

=

1.

(11.27)

So we have, from (1 1.26) and (1 1.27): In the instantaneous rest frame of the particle, the spatial components of 4-acceleration are d2x‘/dz2= d2xiJdt2.

(11.28)

In the instantaneous rest frame of the particle, the 4-acceleration is (0, a) where a is the ordinary 3-acceleration in that reference frame. If la1 = a,then u”ap = - a 2 . Thus, the frame independent scalar uPupcan be interpreted as a measurable quantity by noting (- apu,,)’12is the magnitude of the acceleration measured by the observer riding with the particle. A rocket leaves the earth on a mission to explore deep space. It travels in a straight line away from earth. The acceleration of the rocket is used to create an artificial gravity equal to that on earth, i.e., 1 g. Thus, the rocket accelerates so that, in the instantaneous rest frame of the rocket the observer feels an acceleration of magnitude 1 g. Therefore, we have a”ap = - g 2 .

The rocket accelerates with the constant acceleration 1 g. as above. Note this does not necessarily mean that the 4-acceleration is constant. The rocket accelerates away from earth for 10 years. Then the rocket reverses its thrust so as to give a 1-g deceleration for 10 years. At the end of 20 years (these time intervals are measured by clocks on the rocket) the rocket is once again at rest with respect to earth. Then the process is reversed for the return to earth which takes another 20 years of rocket time. The rocket arrives back at earth having traveled for 40 years according to the rocket passengers. We consider the following questions: (a) How much time passed, as measured by clocks on earth, during the rocket’s absence? (b) How far from earth did the rocket go, as measured from the frame of the earth?

ACCELERATED MOTION-A

231

SPACE ODYSSEY

Let ( t ,x, y , z ) be the Lorentz frame of the earth (we ignore the fact that the earth is not quite an inertial frame). The rocket starts at the origin at t = 0 and moves in the positive x-direction. Then all yz-coordinates are always 0. In our calculations, t will be in meters so we will have to convert years to meters to get numerical answers. The following equations hold. Let t be elapsed time as measured by the rocket clock, so that t is proper time along the world line of the rocket. Then the following conditions hold: t=O

dt/dz

=

u0

dx/dt

=

u1

when when

x=O

(u"2

z = 0, z

=0,

u1 = O when z uo = 1 when t

= 0,

- (u')2 =

= 0,

1,

uouo - u'al = 0,

(a0)2 - ( d ) 2

=

-92.

Now (aO)2

= (a')2 - g2 = (uOuO/u~)2- 92,

so 0 2

(a

1 (u112 - ( a 0 2( u0 )2 -- -9

2

1 2

(u)

5

which gives -(a0)2 = -gZ(u')2.

Then we get u0 = gul, since a', u1 2 0 and a' = uouo/u1 = uOgu'/u' = guo.

so duO/dt= qu',

~'(0)= 1,

du'/dr

~ ' ( 0= ) 0.

= gu0,

(1 1.29)

The initial value problem (11.29) has the solution

u0(z) = cash gt ~ ' (= t ) sinh

gt.

Then t

= (l/g) sinh g T ,

x

=

(l/g) cosh gz

-

(l/g).

(1 1.30)

232

11. THE SPECIAL THEORY

OF RELATIVITY

FIGURE 11.1

If we view this motion in the tx-plane, then we see the world line of the rocket is (part of) a hyperbola as shown in Fig. 11.1. Note that the world line is asymptotic to the line x = t - (l/g) as t --* co. Suppose that, at t = 0, a light beam is fired after the rocket, beginning at x = - l/g. Then the conclusion is that, as long as the rocket continues its 1-g acceleration, the light beam will not overtake the rocket. See Exercise 11.2 for further discussion of this space journey problem. For obvious reasons, motion subject to constant acceleration as above is known, in relativity theory, as hyperbolic motion. We now answer question (a) and (b) posed earlier. ANSWERTO (a): Elapsed time is four times the elapsed time during the first 10-year segment of the trip. We see that 10 years = 9.46 x 10l6 m, since we have 9.46 x 1015m/year,

g = 9.8 m/s' = 1.09 x

(l/g) sinh gz

= 9.17 x

m-';

lOI5 sinh 10.3 = 1.36 x lo2' m.

so t = 1.36 x l0''m

=

1.44 x lo4 year.

Total elapsed time for the trip is 57,600 years! ANSWERTO (b): If we compute x when z tance from earth is twice as far;

=

10 years the maximum dis-

x = 9.17 x 1 0 ' 5 ( ~ ~ 10.3 ~ h- 1) =

1.36 x lo2' m

=

14,376 light-years.

So the rocket gets 28,800 light-years from earth!

233

ENERGY AND MOMENTUM

ENERGY AND M O M E N T U M Let (ct, x’(t), x2(t),x3(t))be the world line of a particle of mass m > 0. We introduce proper time z as parameter and write the world line as ( ~ ” ( 7 ) ) . Then the 4-velocity is (u”) = (dxP/dt). DEFINITION 11.9 The 4-momentum is (p”) = (muP).Thus

po

= myc

(where y

=

(1 - u ~ / c ~ ) - ’ / ~ ) .

(11.31)

pi = mu’y

The energy of the particle, as measured in the Lorentz

DEFINITION 11.I 0

frame (x”),is E

= poc.

In relativistic units (c = 1) we have PO = my, pi = mu’?,

E

= po.

(11.32)

Note that the 4-momentum satisfies (conventional units)

p’p,, = m2c2,

(1 1.33a)

(relativistic units)

p”,,

(1 1.33b)

= m2.

Consider the case of a photon. The photon has zero rest mass. Also it travels at the speed of light so that, if we define proper time t along its world line as in (1 1.21), then z = 0 for all r. Thus we can not reparametrize by proper time to define 4-velocity as we did previously. Indeed we shall not define the 4-velocity of a photon at all (nor does anyone else define such a vector). However, we d o define a 4-momentum vector ( p ” ) for a photon. To accomplish this, let ( x p )be a Lorentz frame and let u be a 3-vector (in the given Lorentz frame) which is a unit vector in the direction of motion of the photon. We need to introduce certain physical quantities which are

h h

Planck’s constant, =

h/21~,

o

circular frequency of the photon,

v

frequency of the photon,

i

wavelength:

k

= 2z/i

wave number.

Then the energy of the photon is taken to be E

=

h\j

( = h o = hc/i).

(11.34)

234

11. THE SPECIAL THEORY OF RELATIVITY

The momentum 3-vector (pi)is defined to be

(Pi)= (E/c)u and the ordinary 3-vector length of this vector is the magnitude of the 3momentum of the photon. The wave 3-vector (k') is defined by (k') = ku. We define the wave 4-vector to be

(k')

= (o/c,

k', k 2 , k3),

where (k', k 2 , k 3 ) are the components of the wave 3-vector. Finally, we define the Cmomentum vector of the photon to be (p') = h(k').

(11.35)

Note that in this case, as in the case of particles of positive mass, we have the relationship between energy and the 0-component of 4-momentum, E

= pot,

E

= PO.

or, in relativistic units,

In the case of positive rest-mass particles we have (11.33). The analogous result in the case of zero rest mass would be that

'P'P

= 0,

that is, the 4-momentum of a photon is a null vector. This is true and we refer to Exercise 11.3 for the details.

R E LATlVi STI C CO R R ECTl0 N TO NEWTONIAN MECHANICS The question arises as to the correction to Newton's laws in a given inertial frame W E 9,in the special theory of relativity. The correct result is obtained by simply replacing the usual 3-momentum in W by the relativistic 3-momentum (m dx'ldz). Thus, we get (d/dt)(relativistic 3-momentum) = force Of course, this force is a 3-vector dependent upon W. in 9.

235

CONSERVATION OF ENERGY A N D M O M E N T U M

Similarly, it is the total relativistic 3-momentum that is conserved in an isolated system. Finally, what about the time rate of change of the other component E of dpldz in W?We claim that dEldt = F v, which is the classical idea of energy and work. Note that it is the change in energy which is due to mechanical effects. These claims have been verified by experiment-for example, in particle accelerators.

-

CONSERVATION OF ENERGY AND M O M E N T U M Suppose we have a collection of particles and the sum of their 3-momenta (mdx'ldz) gives a total 3-momentum (Pi). Some of these particles may be photons, neutrinos, some positive rest-mass particles. These particles interact in some way and then we add up the 3-momenta of the particles after interaction. Call this total ( P i ) .The law of conservation of momentum says ( P i ) = ( P i ) .This is supposed to be true in every Lorentz frame. Now suppose E l , E , are the total energies of all the particles before and after the interaction, in a given Lorentz frame. We claim that it follows from the conservation of momentum that El = E,. To see this, note (El, P1),( E , , P,) are 4-vectors and, in every Lorentz frame, PI = P,. The following result then proves E l = E,. LEMMA 11.I 1 Let (b') be a 4-vector with (bl, b2, b 3 ) = 0 in every Lorentz frame. Then bo = 0 so (b') is the zero vector.

PROOF: Let (6') be the components of b in a frame related to (xp) by a boost in the xl-direction. Then

6'

= ybo

-

Pyb',

-

b2 = b2, The second equation says 0 = -bybo

-

b'

=

-Pybo

+ ybl,

b3 = b3.

+ 0, so bo = 0. I

REMARK 11.I 2 Similarly, if energy conservation is assumed, then conservation of 3-momentum follows. Thus we see that actually total 4-momentum is conserved. We state this as the

FUNDAMENTAL LAWOF NATURE: In any physical interaction the total 4momentum of the particles involved is the same before the interaction as after.

236

11. THE SPECIAL THEORY OF RELATIVITY

MASS AND ENERGY The energy of a particle of mass m is E = poc = mc2y, where y = (I - p2)-lI2 and /3 = u/c with u the ordinary spatial speed of the particle (in the given Lorentz frame). If the particle is at rest in the Lorentz frame, then we see E = mc2. Thus, a particle at rest has, as a consequence of its mass, a certain amount of energy given by the above well-known formula. In our discussions, m has always denoted the so-called rest mass of the particle, a scalar quantity having the same value in all Lorentz frames. Sometimes a relativistic muss Z is defined by f i = my. Then, the energy of the particle is always related to Z by E = Zic2. The difference between the energy of a particle measured in a Lorentz frame (XI’)and the rest energy mc2 of that particle is called the relativistic kinetic energy and is given by E , = E - mc2. See Exercise 11.5 for the relations with the Newtonian concept of kinetic energy. The law of conservation of 4-momentum says the total energy of a collection of particles does not change if they interact. However, the amount of rest energy (energy due to rest mass) and the amount of kinetic energy need not to preserved. The following simple example illustrates this.

CHANGES IN REST MASS Energy is conserved in interactions, but there is no law which says the total rest mass of the particles must be the same before and after an interaction. EXAMPLE 11.I 3 Suppose two objects, each of rest mass m, move toward each other and collide, forming a single object as depicted in Fig. 11.2. Specifically, let the 4-momenta of the particles before the collision be (2m, 2m($/2)i), (2m, -2m($/2)i), where i is the unit vector in the x-direc/ ~2 to arrive at the above tion. We are using v = &2 so y = (1 - ~ * ) - l = 4-momenta. After collision the 4-momentum of the resulting particle must be

(4m, 090, O),

237

EXERCISES

FIGURE 11.2

so the rest mass of the particle formed will be 4m, rather than 2m as might have been expected. The kinetic energy of the initial particles is converted into rest mass of the resultant particle.

SUMMARY The spacetime of special relativity is a four-dimensional Lorentzian manifold. The set which forms spacetime is defined as the set of all “events,” where “event” is left as a primitive notion. The topology and differential structure are determined by introducing the notion of Lorentz frames on spacetime, which provide charts. The Lorentz (or, more generally, Poincare) transformations enter as the coordinate transformations relating pairs of Lorentz frames. We see that the 4-vectors of special relativity are precisely the tangent vectors corresponding to this differential structure. We note that threedimensional vectors, such as ordinary velocity, are not well-defined objects on the spacetime manifold; rather they are tied to choices of coordinate system. We define 4-momentum and note how deep consequences result from the fact that energy and 3-momentum form a 4-vector; for example, the (unforeseen) fact that conservation of energy and conservation of (relativistic) 3-momentum actually imply one another.

EXERClS ES 11.1 Refer to the discussion in the text of relativistic time dilation. Show that, if events a, and u2 are the successive ticks of one of the clocks in the (x”)frame, then the time difference between these events, as measured in the frame (?), is shorter by the factor (1 - fi2)-1/2than the time interval as measured in (xu). This shows the two reference frames obtain the same results when measuring each other’s clock rates, as required by the principle of relativity. 11.2 (a) Refer to the discussion of the space odyssey in the text. Suppose a rocket is to travel in the x-direction with constant 1-g acceleration. Let a light beam be fired after the rocket starting, at t = 0,

238

11. THE SPECIAL THEORY OF RELATIVITY

some distance behind the rocket. How many miles behind the rocket must the light beam start, in order that it never catch the rocket? (b) In the space journey in the text, replace each 10-year segment by a 20-year segment. How far from earth does the rocket get? If the rocket leaves in the year 2000 A.D., what year will it be when the rocket returns to earth? 11.3 Show that the 4-momentum of a photon is a null vector by verifying that p”pp = h2(W2

- k2)

=0

11.4 Verify Remark 11.12 by proving: If (bp)is a 4-vector such that bo = 0 in every Lorentz frame, then b’, b2 and b3 must also vanish in every Lorentz frame and hence (b’) = 0. 11.5 Let u be the ordinary spatial speed of a particle of mass m as measured in a Lorentz frame (x’). Show that the energy E can be expressed as a sum of terms equal to (a) the rest energy mc2, (b) the classical kinetic energy +mu2, (c) terms proportional to ( u / c )which, ~ for Newtonian speeds u << c, are negligible. 11.6 Let (x’) be a Lorentz frame. Now introduce a new chart (t,r, 8,4) by

xo = c,

x1 = r cos 4 sin 8,

x2 = r sin 4 sin 8,

x3 = r cos 8.

Express the spacetime metric g in terms of this new chart. 11.7 Let %? be a class of particles, any two of which have the same rest mass. For instance the class of electrons is such a class. Use the law of conservation of 4-momentum to show that a particle P E %? cannot split into a photon and another particle Q E W (note that we are assuming no other particles or fields are involved so that the 4-momentum of the initial particle P equals the sum of the 4-momenta of Q and the photon). 11.8 The previous exercise says than an isolated electron cannot emit a photon. Note that isolated atoms can and do emit and absorb photons. Why does the result of the previous exercise not hold for, say, the class of oxygen atoms?

72 Electromagnetic Theory

THE LORENTZ FORCE LAW AND THE FARADAY TENSOR Let (x@)be a Lorentz frame and let E, B be electric and magnetic fields as measured in (x"). We shall begin by using SI units (which is based on the MKS system [19]) later shifting to relativistic units. Let a charge q move in space. The force of the electric and magnetic fields on the particle is given by the Lorentz force law as

F = q(E

+ v x B),

(12.1)

where Y is the particle 3-velocity. Let p be the relativistic 3-momentum of the particle. Then

dpldt

= q(E

+ v x B).

(12.2)

Multiply by dtldr to get

dp/dr = q((uo/c)E + u x B);

(12.3)

ui= dx'/dr = vi dtldz, uo = c dtldr, since we are working with coordinates (x' = ct, XI,x2, x3). NOWE = p 0 c SO dpo 1 d E 1 - -- = - q E * v dt c dt c

~

and hence

dpO = !!E . ". d.r

c

Thus the 4-vector dpldr is expressible as

dpldz

= ((q/c)E * U, (quo/c)E

239

+ qu x B).

(12.4)

240

12. ELECTROMAGNETIC THEORY

The point is that dpldr is a linear function of the 4-velocity (up).It follows there is, at each event m, a linear map F : T,M + T,M, such that

dpldr

(12.5)

= qF(u).

NOTE: We are defining F , by giving its value on an arbitrary 4-velocity vector u. Since u is not an arbitrary 4-vector at m but, rather, is a future pointing timelike vector with upup = c 2 , one might think F , is not fully defined. But there are "enough" such vectors as the following result states. As usual, g is the standard Lorentz metric on R4. LEMMA 12.1 Let 4, $ : R4 + R4 be linear transformation with 4 ( w ) = $ ( w ) whenever w is a vector with g ( w , w ) = c2. Then 4 = I).

The simple proof is left as an exercise. In a given frame ( x p )we have F = (Fg). Equation (12.5) becomes

dp"/dr

= qFtfuv

(12.6)

(relatiuistic form of the Lorentz-force law). From Eq. (12.6) we determine the matrix of F to be

E'lc E3/c

E2/c B3 0

0 B2

-B'

-i:/c] (12.7)

B' 0

'

We may raise or lower indices using the spacetime metric g to get F,,

= gpaFt

and

F""

= g""F;.

We find

(12.8)

and

(F'")

=

YO -E'/c E'lc 0 E2/c B3 E3/c - B 2

I

-E2/c -B3 0

-B1

B'

0

1

-E3/c1 B2

(12.9)

REMARK: The matrices (12.7), (12.8), (12.9) are given in the coordinates (xo = ct, x', x2,x3). If, instead of this, we use (xo = t, xl, x2, x3), then the

24 1

THE LORENTZ FORCE LAW AND THE FARADAY TENSOR

matrix of g becomes 0

(.Z

0 0 0

Then uo = d t / d t and p o

=

-1

0 0

-

0 0 0 0 1 0 0 -1.

E / c 2 , so we get EZ/c2 E3/cZ B3 - B2 0 B' B' 0 -

B3

B'

=(F3,

= (FJ,

( 12.1 0)

(12.11)

0

and -E2/c2

E'/c2 E2/c2

0 B3

-

B3 0

-E3/c2.

B2 -

B'

= (F"").

( 12.12)

0

B' If we use relativistic units we get the matrices

( 12.1 3 )

( 12.1 4)

E' E2 LE3

0 B3 -B2

-B3 0

B'

-B' 0

= (F"').

1

( 12.15)

242

12. ELECTROMAGNETIC THEORY

CONVENTION 12.2 For the remainder of this chapter we shall use relativistic units (see Remark 12.4). Note that (F,") is a skew-symmetric tensor. This tensor F is called the Faraday tensor or the electromagnetic jield tensor. Recall we may write

F = FPV dxfl 0 dx'

= iF,,

dx"

A

dx" = FlavldxP A dx".

If we use relativistic units and label the coordinates (t, x, y , z), we can write

+

+

+

+

+

+

F = E,dt A d x E, dt A d y E , dt A d z B , d y Adx By d x Adz + B, dz A dy. = E x dt A dx E, dt A dy E , dt A dz - B, d y A dz - . . - B , d z A d x - B,dxr\dy. EXAMPLE 12.3 (Electric and magnetic field of a moving charge) Consider a charge q at rest in a Lorentz frame (l,X, j , 5) which is moving with velocity p along the positive x-axis of the Lorentz frame ( t , x, y, z). Let q lie at X = j = Z = 0. Then in the (f, 2,j , 2) frame we have

E,

= kilr3,

E,

=

kyp3,

-

E, = kqr3,

B-X

(Y = (22 + y2 + 52)1'2).

-

-

= B-Y = B.Z = 0,

Thus the matrix of F in the Lorentz frame (i,X,p, F) is 0 -kZ/F3 (Fl,v)

= -kj//r3

[-k5/Y3

kX/-3 r 0 0 0 0

0

0

We want to compute the electric and magnetic field in the ( t , x, y , z ) frame so we must compute the matrix (FJ of F in ( t , x, y , 2)-coordinates. Write X" = A;x'. Then F,,

If A

= (A;),

= AiA(Fgp.

(12.16)

then (12.16) has the matrix form

V,")

= ~Y&)A.

The transformation matrix for the transformation from (t, x, y , z) to (i,2,j , 5) is

243

THE 4-CURRENT

so the desired matrix is

x

0

So E Now

= (k/F3)(X, y j ,

75) and B

= ( k / f 3 ) ( 0, ByZ,

x = -pyt + yx, f 3=

y j

[y2(x

j = y,

- pt)2

+ yz +

yz

B y j ) in (2,j, 5)-coordinates.

z= z, 2213’2,

So we get, in the ( t , x,y , z) frame,

The electric field E is radial from the location of the particle, but its magnitude is not the same in all directions. The magnetic field B is always perpendicular to the x-axis and circles around the x-axis. In fact, if v = (B, 0,O) is the 3-velocity of the charge, then

B=vxE.

THE 4-CURRENT Suppose (2‘) is related to (x”)by a boost down the x-axis as usual. Suppose in (2”)there is no current but there is a uniform tube of charge along the x-axis having cross sectional area 1 and linear charge density p. Let p be the velocity of (2”)with respect to (x”).An observer in the (x”) frame judges there to be both a charge density and a current density along the x-axis. Consider a piece of the tube of charge of length G, at rest in (X”). This piece is moving in the x-direction with velocity /3 in (x”). Let p be the charge density in (x”). Both observers agree on how much charge is in the piece so

p 7G = p G / y , since &/y

is the length of our piece as measured in (2’). This shows that P

= PY.

(12.17)

244

12. ELECTROMAGNETIC THEORY

FIGURE 12.1

What current density does (x”) see? Consider a cross section of the tube of charge at x = xo. How much charge crosses this section between t = to and t = to + At? This situation is depicted in Fig. 12.1. Since the charges are moving to the right with velocity P, the charges passing xo from to to to A t are those in the shaded tube of length P At. The amount of charge in this tube is, by (12.17),

+

pP At = pyp At.

Now this means the current density in the x-direction is Jx

= PYP.

(12.18)

If we form the 4-tuple ( p , J,, J,, J,) and similarly in (2”)form we see (J”) = (PY, PYP, O,O),

(J?

(p,J,, J,, J,)

( P , o,o, 0).

=

:: :];:I

Thus these components transform like components of a 4-vector, that is,

[i]=[L 0

0

0

1 0

J2.

0

0 1

J3

Note we are using relativistic units. There is a 4-vector, the current density 4-vector or, briefly, the 4-current whose components in any Lorentz frame are (J”) = (p, J’, J 2 , J 3 )

(Ccurrent in relativistic units). If SI units are used, the vector has components (J”) = (pc, J’, J 2 , 5 3 )

(4-current in SI units). REMARK 12.4 If we use (xo = ct, x i ) , then all coordinates xfl have the same units. Then Jacobians (a2”/axv)are dimensionless.A 4-vector must have the same units in all components. Otherwise, when applying Lorentz trans-

245

DOPPLER EFFECT

formations, we add apples and oranges. Similarly for more general tensors. (F’J has as entries, components of the vectors c - ’ E and B, which have SI units c-’E

(volt/m)(sec/m),

c,

B tf tesla

= volt-sec/m2.

Such arguments, based on units, allow one to correctly put in c when going from relativistic units to SI units. For instance, in relativistic units, the 4current is given by (J’) = ( p , J). In SI units we claim that (J’) = (pc, J). To see this, note that the units of p and J are coulombs/m3 and coulombs/s-m2, respectively. Thus if we replace p by p c , all components will have the same units.

DOPPLER EFFECT A “source” moves along the x-axis with velocity u > 0 and emits light with frequency f,. An observer at the origin of the ( t , x, y, z ) frame sees this light and judges it to have frequency ,fo. CASE1 : The source is moving toward the observer. Let (t,2,j,5)be the , 0, 0), (t2,0, 0,O) be the beginning and rest frame of the source. Let ( t l 0, end of a period, so that T , = iz - f,. The coordinates of these events in the (t,x, y, z ) frame are (ytl, Byt,, 0, 0), (yf2, byi2,0,O). Here fl < f 2 < 0. The times when the events are observed at the origin are yl, - &tl, yt2 - pyt2. Therefore the observed frequency is

so the observed frequency is greater than the frequency judged by the source. The light is “blue shifted.” CASE2: The source moves away from the observer. Using notation as above we see that the times f l , I , must satisfy 0 < tl < 12.The times of observation are

so

The observed frequency is less than the frequency as judged by the source. The light is “red shifted.”

246

12. ELECTROMAGNETIC THEORY

MAXWELL’S EQUATIO NS We use relativistic units and let (x”) = ( t , x , y , z ) be a Lorentz frame. We have an electromagnetic field given by (F’J and a 4-current (J”). We have the free space values c0, p o where copo = c-’, c0

=

8.85 x lo-’’

cou12-sec2 kg-m3

in MKS units or 7.96 x 105

cou12 kg-m

~

in relativistic units. Maxwell’s equations now read, using c

=

1,

(a) V x E = -dB/dt, (b) V - B = 0, (c) v x B = (aE/at) + p o ~ , (4 v * E = (1/&o)P = POP.

(12.19)

Now we have

+

+

F = E x dt A dx E , dt A d y E , dt A dz - B , d y ~ d z - B , d z ~ d x -B , d x A d y .

We have then

Thus the Maxwell equations (a), (b) combine in the single equation (1 2.20)

dF = 0.

Now consider Maxwell equations (c), (d). If p, J are 0, these are

v x B = aE/at,

v .E = 0.

These are just (a), (b) if B is replaced by E and E is replaced by - B . Let

247

MAXWELL'S EQUAT10 N S

( M J be obtained from ( F J by replacing B by E, E by - B so that

ro

-B, 0 E, -E,

B,

(MiI") =

B, LB,

-B, -E, 0 Ex

-B,] -Ex 0

1

If ( M @ Jcan be shown to be the matrix of components of a tensor M , then

JE,

dE,

-~ - - -

")

d x A dy Adz.

The Maxwell equations (c), (d) require comparing the four quantities in parentheses with the components (pOJ'). To do this we need a 3-form, so we lower the index on J p and form the dual.

pgJ

dt - J , d x - J , d y - J , d z ) =po(pdxr\dyAdz+ J,dyAdtr\dz + J,dtAdxAdz + J , d x A dt A d y ) = - p o J , dt A d x A dy + poJ, d z A d t A d x - p o J , d y A d z A dt + pop d x A d y A dz.

= pg(p

The Maxwell equations (c), (d) become dM = -p*J0

7

(12.21)

* d M = -pO**J = -poJ. Note that M = * F , so M is indeed a tensor. Then we have *d*F = - p o J . We can write this as SO

6F

=

-poJ.

Thus Maxwell's equations are

dF

=

0,

In the source free case, i.e., in case J

dF=0,

6F

=

= 0,

-poJ.

(12.22)

we see Maxwell's equations are

6F=0.

(12.23)

248

12. ELECTROMAGNETIC THEORY

In terms of components, Maxwells equations are

F,,,,,,]

FBV.,= -/-@',

= 0,

where F[pv,m]

= 4[Fpv,a

- Fpn,v

+ Fvn,,,

-

Fvp.a

+

FEW,,

-

F,v,pI

is the skew-symmetrization of F,,,,.

THE ELECTROMAGNETIC PLANE WAVE Let F,, = A,, exp(ik"x,), (A,,) = A , a constant 2-form and k = (P) a constant vector. In order that the sourceless Maxwell equations hold we find (1 2.24)

(12.25) Now we first note: LEMMA 12.5 The 4-vector k must be null. PROOF: By (12.24) we have

+ A,,k,, + A,,k,

A,,,k,

= 0.

Contracting with k" gives

+ A,,kVP + A,,k,k"

A,,k,ka

= 0,

where the last two terms on the left vanish by (12.25). So, since A,, # 0 we see k,k" = 0. I Now note that (12.24) says kr\A=O. PROPOSITION 12.6

such that A

=kA

e.

If k A A

= 0,

k a nonzero vector, then there is an

e

0 2 ,0 3 ,with coo = k. Write PROOF: Choose a basis for 1-forms, oo,ol,

A

=

+ AO2woA o2+ A03w0 A o3 + ~ , , w ' o2+ A 1 3 ~ 'A w 2 + A 2 3 u 2A

Ao,wo A o' A

03.

Then kAA

=ooAA=A

+

, 2 0 0 ~ ~ 'A ~, , w~o A2f B ' A W 3

+ A2,w0AW2AW3.

249

THE ELECTROMAGNETIC PLANE WAVE

We conclude A , ,

=

A = A,,u'

0, A13 A

U'

= 0,

A,,

= 0.

+ A,,u'

So

AU '

+ A,,?'

A

m3 = k A d,

where

+ A,,02 + AO3w3. I

d = A,,u'

Thus, if (12.24), (12.25) are satisfied by A , k, then by Proposition 12.7,

A p , = kpCv - [,k, and so AM' = k"!' - Pk'. Also, using Lemma 12.6, we find 0 = A"k, k'Cvk, - Pk'k, = tvk,kp, which shows /"k,,

=

= 0.

We summarize as follows. PROPOSITION 12.7 Let A be a nonzero 2-form and k be a nonzero vector. Then Eqs. (12.24) and (12.25) hold iff the following are true:

(a) k is null, (b) there is a vector P with A (c) P k , = 0.

=

k A P,

We see what the significance of k is, namely, the propagation vector. What about L? Write k = i(1, n), n a unit vector giving the direction of propagation of the wave. Then write P = (lo, 1). Now P is not unique, since adding to a given P a multiple of k does not affect (b) or (c), so there is a unique choice of / for which Eo = 0. Thus, take P = (0, I) and we have

So (P)= (0, -Eo) and B, = n x E,, where E(x) = E, exp(ikpx,) and B(x) = B, exp(ik'x,). We see from P k , = 0, that

-

-

E n = 0 = B n. Also, from the above equations we see E . B = 0,

IE( = (BI,

E, x B,

=

nlE12

250

12. ELECTROMAGNETIC THEORY

TH E 4- POTENTIAL Since d F = 0 there exists, at least locally, a 1-form A with d A = F . This A is called a 4-potential for F. Now we can replace A by A + d f . This freedom of choice in the potential is called gauge freedom. We have 6 A + 6 df

=

6A

+ Af.

Given A, we choose f with Af = - 6 A . Then A’ = A + d f satisfies 6A‘ = 0. The condition 6A’ = 0 is called the Lorentz gauge condition. Now the other Maxwell equation is 6F

=

-pLgJ.

This becomes 6 d A = - p o J . In the presence of the Lorentz gauge condition, 6 A = 0, we get 6 dA = 6 d A + d 6 A = AA, so Maxwell’s equations have the form 6 A = 0,

AA =

-POJ.

(12.26)

Recall i f f is a scalar on spacetime, we have

of Also if A is a 1-form A

=

=foo

-1 1 1 -f 2 2

-1 3 3 .

A, dx’, then

6 A = -A’ . Recall of = -Af where A is the Laplace-Beltrami operator of the Minkowski metric. If A = A , dx” is a 1-form, ( x p ) a Lorentz frame, then 9,

AA

= (AA,)

dx”

or, equivalently,

6dA

= (Ap,p),v dx” - (0 A,) h”.

Thus Maxwell’s equations in the form (12.26) can be written A,,”

= 0,

O A ” = POJ,

(12.27)

(Maxwell’s equations in a Lorentz frame with Lorentz gauge condition). EXAMPLE 12.8 (4-Potential of a moving charge) Consider two Lorentz moving along the x’-axis with velocity p. We frames, (x”)and (2”)with (2’) continue to use relativistic units. Then 2” = Agx”, where the matrix is as in Example 12.3. Suppose we have a charge q at rest at the origin of the (2’) frame. Then, in (%”)-coordinates,the 4-potential is given by

(A,) = (q/47Te0J,0, 0, 0).

251

EXISTENCE OF SCALAR AND VECTOR POTENTIALS IN R 3

This does satisfy the Lorentz gauge condition and, referring to Example 12.3, we see d A is the correct electromagnetic field tensor. We want the components of A in the (x") frame. We have A,

=

AfA,

(since a?/ax" = At, this is just the usual transformation law for components of a covariant vector). Thus, A,

= ?A,,

A1

=

Writing it out explicitly, using

A,

=0 =

-PYA,, (l,x, y,

A2

= 0,

A3

= 0.

z ) rather than (xo,xl, x2, x3), we get

A,.

EXISTENCE OF SCALAR AND VECTOR POTENTIALS I N R3 We have defined (see Equation (9.2)) the De Rham Cohomology group W ( X )for any manifold X , as the space of closed q-forms modulo the space of exact q-forms. The Poincare lemma says that if X is diffeomorphic to a starlike subset of R", then H 4 ( X )= 0 for q > 0. The famous De Rham theorem says that if X is an oriented n-manifold, then the DeRham cohomology vector space H 4 ( X ) is isomorphic to the singular cohomology group H 4 ( X ; R). Generally H 4 ( X ; R) E H , ( X ; R)*. These results lead to statements on the existence of potentials. If (A,) = (#, A) one calls 4 a scalar potential and A a vector potential. SCALARPOTENTIALS: Consider a vector field E on a region U in R3. If E has a potential then E = Vb for some scalar 4, in which case, curl E = curl(grad 4) = 0. Conversely if curl E = 0 can we be sure E = V 4 for some 4. The condition which guarantees this is that U be simply connected as we now show. Let = o be the 1-form corresponding to E. Then curl E = *do= 0. Since U is simply connected, we have 0 = zI(U) H , ( U ; Z ) ,so H,(U, R ) = 0 and hence, H1(U,R ) = 0. Now *do= 0, so d o = 0 and hence, w = d 4 for some 0-form 4. Then E = grad 4. So existence of scalar potentials is a question of whether closed 1-forms are exact and this involves H'(U).

e

VECTORPOTENTIALS Suppose B is a vector field in a region U in R3

and div B

= 0 in

U . Does B have a vector potential, i.e., is there a vector field

252

12. ELECTROMAGNETIC THEORY

A on U with curl A = B? It is often believed that, if U is simply connected, this will be so, but we shall show this is not so. Let B = w. Then 0 = div B = *d*w and thus d * o = 0. Since *ais a closed 2-form we want to ask whether *w = da for some 1-form a. If so, then w = **o= *da, so B = curl A where A = a. Thus, to guarantee the existence of vector potentials we need H 2 ( U )= 0. This is not the condition of simple connectivity. We consider two specific examples. EXAMPLE 12.9 (Magnetic field of a line current) Let U = R3 ((0, 0, z)lz E R } . Let B = (p01/271r)e0,in cylindrical coordinates, so that B is the magnetic field due to a current I running in the positive z-direction along the z-axis (see Fig. 12.2). We have H ' ( U ) # 0, H 2 ( U )= 0. B thus has a vector potential, indeed B = V x A where A = -(p01/271)ln r e , . But B has no scalar potential, even though V x B = 0. For if B = V$, then the line integral of B around a closed curve will be zero. But consider a curve which links the z-axis, such as

y(8) = (cos 8, sin 8, 0)

for 0 I 8 I 271.

We have

EXAMPLE 12.1 0 (Magnetic monopole field) Let B = (k/r2)e,, using spherical coordinates, and U = R3 - ((0, 0, O)}. B has a scalar potential since H I ( U ) = 0 (since U is simply connected). Indeed, $ = - k/r is a scalar potential. But H 2 ( U )# 0 and we argue B has no vector potential, even though div B = 0 in U :B = ( k / r 2 )dr so div B = *d* ((k/r2)dr) = *d(k sin 8 dO A d d ) = 0, which verifies div B = 0. For B to have a vector potential, * ( ( k / r 2 dr) ) = k sin 0 dkJA dd must be exact. But, restricted to S 2 = {(x, y , z) I x2 y2 z2 =

+ +

B

FIGURE 12.2

253

EXERCISES

l}, * ( ( k / r 2 dr) ) = kc, where E is the volume element on S 2 . So k sin 6 d6 A d+ cannot be da for a 1-form a, for if it were, then

0=

Is.

da

=

Js2

k sin 0 dO A d 4

=

k4n # 0.

So we have a divergence free vector field on a simply connected region, having no vector potential on that region.

EXERCl SES 12.1 (Scalar invariants) There are three obvious scalars that can be constructed from an electromagnetic field tensor F , namely, (a) +FB"F@", (b) $F""*F,,, (c) * F P v * F p , . Show that

~ F P " F I= , , IBI2 - IE12,

-

'FP"*FP,, 2 = E B, $*F'I'"*F

11V

= lEl2 -

\BIZ.

12.2 Suppose in a Lorentz frame ( x p ) there is a nonzero electric f d d E, and B = 0. Is it possible to transform to another Lorentz frame (2") so that E = 0, B # 0 in (.XI')? 12.3 Let (A,) = (4, A). One calls 4 a scalar potential, A a vector potential. Interpret the condition d A = F in terms of 4, A, E, B. What does the Lorentz gauge condition say? 12.4 (Gauss's law) The equation dF = - p o J can be written as d*F = - p g J . Let (t,x , y , z ) be a Lorentz frame (recall that we are using relativistic units), S' = { ( t ,x, y, z ) ( t = t o , x 2 + y' + z2 = l}, and B = { ( t ,x , y, z ) l t = t o , x 2 + y2 + z 2 5 l}. Show that

= Jud*F

- JYi

-

E , dx A d y

+ E , d x A d z - E , d y A dz.

254

12. ELECTROMAGNETIC THEORY

Now, working on the spacelike hypersurface t = to with the Euclidean metric dx2 + dy2 + dz2, we have * e = E , d x r \ d y - E E , d x r \ d z + E,dyAdz,

so that, using Proposition 9.63, we get p o I B p d x r , d y ~ d z SS =*E.ndA.

*e

Note that the dual is computed with respect to the Euclidean metric on the hypersurface. Of course, we get the same result for any two-dimensional surface S bounding a three-dimensional region B in a t = to slice. 12.5 Let F = k A teikxbe an electromagnetic plane wave as described in the text; we can assume that L = (0,t').Then the 1-form -iCeikx is a 4potential for F. Find a 4-potential for * F = * ( k r \ t ) e i k Xof the form - imeikx where m = (0,m).Here kx = k"x,. 12.6 Let E be the volume 4-form on the spacetime of special relativity and F be a 2-form. Show that (a) F A* F = V2 " " F " V E , (b) F A F = -'*F'"F E 2 PV ' (C) * F A * F = -LFPvF 2 PVE. 12.7 Let F be the electromagnetic field tensor. Show that FaYF,,,,

+ ~ g a r F P Y F=P y+$(FaYFvp+ *Fav*Fvp) + B2)

=i +(E2

L

Bx E

(B x E)T - EiEj - BiBj + f h i j ( E 2

where 1

-1 -1

-1

1

+ B2)

73 The Mechanics of Rigid Body Motion

HAMILTONIAN SYSTEMS AND EQUIVALENT MODELS In our previous discussions of classical mechanics we have made use of a configuration manifold X , the state space T X and the phase space T * X . Given the Lagrangian function L : TX + R we constructed the Legendre transformation 2:T X + T * X and used it to define the Hamiltonian function H: T * X + R. On T * X there is a naturally defined 1-form 0, the canonical 1-form, and there is the canonical 2-form R = -do. Using the #-mapping defined by R we constructed the Hamiltonian vector field X , = ( d H ) $on T*X. The integral curves of the vector field X , were the phase space trajectories of the system, that is, they are the curves satisfying Hamilton’s equations. A corresponding picture was obtained on T X . We defined the Lagrangian 2-form R, = 2 * R and then used the resulting #-map to define the Lagrangian vector field X , = (dE)‘ on T X . The integral curves of X , were shown to be the curves satisfying Lagrange’s equations. We now begin to take a more general point of view according to which a model of a physical system involves curves in some appropriate space, the curves giving the evolution of the system. It will be seen that it is too restrictive to require that the space always be TX or T * X . In order to treat the motion of a rigid body we shall use the notion of a Hamiltonian system. DEFINITION 13.1 A Hamiltonian system 2 is a triple ( M , w, H), where M is a smooth manifold, o is a closed, nondegenerate 2-form on M , and H is a smooth real-valued function on M . We refer to M as the phase space, (0 the symplectic form, and H the Hamiltoniunfunction of the Hamiltonian system. Given a Hamiltonian system 2, there is a #-map which converts 1-forms on M into vector fields on M . If CI E Ty(M),we write the corresponding vector field as as. 255

256

13. MECHANICS OF RIGID BODY MOTION

The vector field X , = (dH)' is called the Hamiltonian vector .field. The integral curves of X , are the trajectories of the Hamiltonian system 2. Thus for any initial state m, E M the Hamiltonian system prescribes a unique evolution beginning at m, . Let 2 = ( M , o,H ) and 2' = ( M ' , a', H')be Hamiltonian systems. Suppose f :M -+ M' is a diffeomorphism such that .f *o'= o and H' f = H. We then say the Hamiltonian systems X and X' are isomorphic and the mapping f is called an isomorphism of Hamiltonian systems. If # and #' correspond to 2 and X ' , then it is left as an easy exercise to show ( T f ) # = #' (Tf),. Also f * X , = X,,, and so a curve c in M is a trajectory of 2 if and only iff 0 c is a trajectory of Y ' . If X models some physical system then we regard an isomorphic Hamiltonian system 2'as an equivalent model for the physical system. The notion of equivalent model will be important for our discussion of rigid body motion. 0

0

0

THE RIGID BODY We shall consider the motion of a rigid body having one point heldfixed. We consider the rigid body to be a system of N + 1 particles of masses m,, . . . ,m,. Let m, always be located at the origin of an inertial Cartesian coordinate system which is regarded as fixed in space. These Cartesian axes are usually referred to as space axes. The assumption that our system is a rigid body is expressed by requiring that there exist constants cij > 0, 0 5 i, j 5 N , i # j , such that, if R,, . . . , R , are the position vectors of m,, . . . , m N , then IRi - Rjl = c i j . Here is the usual Euclidean norm of v and for v, w E R3 we let ( u , w ) be the usual inner product in R 3 . We shall assume N 2 3 and we assume that not all the masses lie in a single plane. We choose some configuration of the system as a reference configuration and denote by to= 0, t1,.. . , 5, the position vectors for the system in this reference position. Since not all the masses lie in a plane, we see that some three of tl,.. . , lN are linearly independent and we may assume the masses are labeled so that tl, t2,t3are linearly independent. Two important groups of matrices will be of interest.

11

O ( 3 ) AND SO(3) DEFINITION 13.2 The orthogonal group 0(3) consists of all linear maps A: R3 -+ R 3 such that for all x, y in R 3 , ( A x , Ay) = ( x , y). The rotation group SO(3) consists of all matrices A in O(3) such that det(A) = 1.

257

O ( 3 ) AND SO(3)

THEOREM 13.3 Let R , , R , , . . . , RN be the position vectors of an arbitrary configuration of the rigid body. Then there is a unique A E 0(3) such that

At; = Ri, PROOF:

{t,,t2,t3} is

i = 0 , . . . ,N .

a basis for R 3 so the equations A t i = R , , i =

1,2,3, define a unique linear transformation of R 3 . We must show that this transformation is orthogonal and that it satisfies A t i = R j for i = 4,. . . , N . To prove orthogonality we first show ( A t i , A t j ) = (ti,t j )for i,.j = 1, 2, 3. This holds because 2 ( ~ < i~ ,

t j =) I ~ t i 1+ ~IAtjl’

- IAti - A<jI2 IRiJz+ IRjI2 - IRi - Rilz = + ci”o ci”. =

~

(ti1’+ 1CjI2 - lti - t j I 2 = Vt;, tj).

=

Thus A preserves inner products on the basis vectors and hence, is orthogonal. Now pick k 2 4 and show A t , = R,. In the calculation made above we saw that ( R i , R j ) = (ti,t j ) holds for all, i, j . Write Ck = x i t i (all sums from 1 to 3). Also write R, = PiRi. We want to show cli = P’ for i = 1, 2, 3, for then A t k = A d t i = & A t i = aiRj = BiRi = R,. Now if j is 1, 2 or 3, we see ( t k , t j )= cli(ti,t j ) and ( R , , R j ) = P i ( R i ,R j ) . Then also, ( t k ,t j ) = Pi(ti,t i )since ( R , , R,J = ( E k , tj)and ( R i , R j ) = (ti,t j ) .Since the(3 x 3)matrix ((ti,t j ) )is nonsingular we conclude cli = p. I THEOREM 13.4

The matrix A obtained in the preceding theorem actu-

ally lies in SO(3). PROOF: We must appeal to some physical reasoning. If R , , . . . , RN give some configuration of the system, it must be possible to move the body continuously in space so that at t = 0 it is in the reference configuration t,, . . . , tNwhile at t = 1 it is in configuration R , , . . . , R,. Thus, there are N continuous curves $,, . . . , t,hN in R 3 such that $i(0)= ti, i+hi(l)= R , and for each t E [0, 11, ( $ , ( t ) , . . . , $ N ( t ) )= ( A , < , ,. . . , At
258

13. MECHANICS OF RIGID BODY MOTION

Then B(t) = A,B(O) so IB(t)l = lAtl IB(0)land

rt: B(0) =

LEI

t2

"I, t;

so (B(O)(# 0. By continuity IB(t)l and lB(O)I have the same sign so lAtl = 1 for all y (We already know lAtl = & 1 by orthogonality.) Thus IAl = IA,] = 1, as asserted. I We have defined the general linear group GL(n, R ) in Exercise 4.14. GL(n, R ) is an open set in the vector space M , of real n x n matrices so that GL(n, R ) is a manifold of dimension n2. SO(3) is a subgroup of GL(3, R ) and will, in fact, be shown to be a submanifold of GL(3, R ) of dimension 3. This will be shown in the next chapter. This has the consequence that operations such as matrix multiplication and inversion are C" in SO(3). Since SO(3) is a submanifold of M , , we have TSO(3) identified as a subset of T M , z M , x M , (see Chapter 3). Recall that for A E S0(3), u E TAS0(3) we can find a curve A(t) in SO(3) with A(0) = A , A'(0) = u. Now A(t) is a matrix in M , so we can differentiate the entries of A(t) forming the matrix A(t). The identification of TSO(3) with a subset of M , x M , is given by u + ( A , A(0)).Under this identification TJO(3) is a subspace of M,, e being the identity in SO(3). We identify this subspace of M,. PROPOSITION 13.5 T,S0(3) is the set of all skew-symmetric matrices in M3

*

PROOF: If A(t) gives a curve in SO(3) with A(0) = e , then A ( t ) , 4 ( ~=) e~ so k(0)A(O)T+ A(0)A(O)T= 0 and hence, A(0)+ A(0)T= 0. This shows everything in TJO(3) is skew-symmetric. Conversely, let B be skew-symmetric. We saw in Exercises 4.14-4.15 that the curve A(t) = exp(tB) satisfies A(0)= e, k(0)= B. If we show exp(tB) E SO(3) for all t, then we are done. From Exercise 4.15, exp( - t B ) = exp(tB)- and it is easy to see that exp(tBT) = exp(tB)T. Since BT = - B we conclude exp(tB)-' = exp(tB)T, that is, exp(tB) E O(3). But det(exp(tB)) is a continuous function of t and for t = 0 gives the value 1 so exp(tB) has positive determinant; thus exp(tB) E SO(3). I

Henceforth in this chapter, the group SO(3) will be denoted by G and we will view T,G as the set of skew-symmetric (3 x 3)-matrices. We have the exponential map exp: T,G + G and y B ( t ) = exp(tB) defines the 1-parameter subgroups of G. Given a choice of reference configuration t l , .. . , we define a:G + R3N by @(A)= ( A t , , . . . ,A t N ) . Then TQ TG + R 3 N x R 3 N identifies TG with

rN

259

SPACE AND BODY REPRESENTATIONS

the state space in R3N x R 3 N .It can be shown that @ and T @ map their domains diffeomorphically onto the submanifolds @( c),T @ (TG),respectively. We are using the notation G for SO(3) to emphasize that much of the development does not depend on special properties of SO(3) but only on the fact that SO(3) is a group which is a differential manifold whose group operations are smooth. Such a group is called a Lie group. The following notational convention will be convenient. If A(t) is a curve in G then A'(t) = dA/dt is, as usual, an element of T,,,,G. On the other hand, k(t) is the matrix function obtained by differentiating each of the matrix elements. Thus, when we identify TG with a subset of M , x M , , we get dA/dt

= ( A ( t ) ,k(t)).

(13.1)

SPACE AND BODY REPRESENTATIONS The identification of T G as a subset of M , x M 3 depends on the fact that G is a group of (3 x 3)-matrices. We now consider two important representations of TG (which are valid for any Lie group). For A , E G, define LAo:G + G and RAo:G -+ G by LA,(A)= A,A, RAo(A)= A A , . These mappings are called, respectively, left and right translation by A , . Clearly = LAi1 and similarly for RAO.LA, and RA, are smooth so we can define

2: G x T,G

+ TG

by

il(A, B) = T,LA(B),

(13.2)

p: G x T,G + TG

by

p ( A , B) = T,R,(B).

(1 3.3)

If we represent TG by matrices, we have I ( A , B ) = ( A , AB) and p ( A , B) = ( A , BA). Considering the mapping @: G + R 3 N we get T @ p : G x T,G 0

TQ, p ( A , B,

=

( A t ,.,. .

9

+ R3N x

R3N, ..>

BAtN).

(13.4)

The right side of (13.4) is a state of the system in R3N x R 3 N .We would describe this state by saying that the instantaneous velocity is the injinitesimal rotation B applied to the instantaneous configuration ( A t , , . . . , A t N ) . The mapping T@ p is called the space representation of the state space. The matrices A, B which correspond to a given state under this mapping have direct interpretation in terms of the observed motion of the rigid body in physical space. On the other hand, the mapping T @ 1 is called the body representation of state space. This representation is useful in describing mathematical (global) properties of the motion of a rigid body, as we shall now see. 0

0

260

13. MECHANICS OF RIGID BODY MOTION

Recall from the general development of Lagrangian mechanics we have a kinetic energy metric T defined for states u = (xl,.. . , x N ,u l , . . . ,uN), w = ( x l , .. . , x N ,wl,. . . , wN) by T,(u, w) = I mi(ui, wi). Then the kinetic energy function was defined on state space by T(u)= $T,(U,u) for u as above (see Definitions 7.1 and 7.2). If we want to obtain the motion of the rigid body in the absence of external forces, we simply use T as the Lagrangian and proceed as usual in classical mechanics. We first use @ to pull the structures over to G. Define h = @*T to be the metric on G corresponding to the kinetic energy metric on configuration space under the mapping @. Thus

xyz

h(A)(P,4 = T(T,@(P),TA@(4), PROPOSITION 13.6

for all A,

E

B, 6 E TAG.

(13.5)

h is a left-invariant metric on G, that is, h = Lz,h

G.

PROOF: Pick A , , A E G, 8,6 E TAG.Then we want to show

Note we have used the fact that A , preserves the Euclidean inner product < >

>. I

Now the motions of the rigid body that we see in physical space correspond to curves A(t) in G such that the positions of the masses at time t are A ( t ) t , , . . . , A ( t ) t N .We called such curves in G space motions of the rigid body. THEOREM 13.7 The space motions of the rigid body subject to no external force are the geodesics of the left-invariant metric h on G. PROOF: As shown in Example 7.9, the trajectories of a holonomic system without external forces are the geodesics in configuration space for the kinetic energy metric. Under our mapping @ the group G corresponds to configuration space and the metric h corresponds to the kinetic energy metric. Thus the geodesics of our two metrics correspond under @ so the theorem holds. I

261

GEOMETRY OF RIGID BODY MOTION

THE GEOMETRY OF RIGID BODY MOTION We now present an outline of the constructions involved in our analysis. The required details and definitions are then given in the following pages. We encourage the readers to consult this outline while studying the rest of this chapter to avoid losing their way in a mass of details. (1) O n G = SO(3) there is a left invariant metric h such that if A(t) is a geodesic, then y ( t ) = ( A ( t ) t l , . . . , A ( f ) t Nis) a motion of the rigid body. The geodesics are called space motions of the body. (2) Using 1:G x T,G + TG, construct the Hamiltonian system

2

=

(G x T,G,

6, H )

whose Hamiltonian trajectories are those curves (A(t),B(t))= I-'(dA/dt), where A(t) is a space motion. In matrix form this gives B(t) = A(t)-'k(t). (3) G acts as a group of symmetries on 2 which leads to the momentum map p: G x T,G + (T,G)* given by p ( A , B)(C) = h,(B, A d ( K ' ) ( C ) )satisfying If A ( t ) is a space motion, then p(A(t),A ( t ) - ' k ( t ) )= p o is constant. po is called the conserved momentum of the trajectory.

(4) Let p o E (T,G)* be fixed. Suppose (A(& B(t))is a Hamiltonian trajectory having momentum po. The equations h,(B(t), Ad(A(t)- I)( )) = p o , B(t) = A(t)-'k(t) lead to two reductions of the differential equations on G x T,G. We use #:(T,G)* 4 T,G by h,(v*, B ) = v(B) and b = # - I . (a) Let Z,,(A) = TL,((Ad*(A-')(p,))#). Then A(t) is a geodesic with conserved momentum po iff A(t) is an integral curve of Z p 0 . (b) Consider p(t) = B(t)b = Ad*(A(t)-')(po) = p o . Ad(A(t)) in (T,G)*. For any C E T,G, p(t)(C)= p(t)([p(t)*,C ] ) so that p(t) is an integral curve of the vector field Y , on (T,)* given by Y,(v)(C) = v([v#, C ] ) . Such integral curves are called body motions. (5) Given a body motion p ( t ) there is a unique space motion with the property A(0) = e associated with it. For then po = p(0) and A(t) is determined by Z,, and initial condition A(0) = e. Such a space motion is called a special space motion. If A(t) is any space motion then A(t) = A(O)A"(t)where A(t) is a special space motion. (6) The vector field Y , is tangent to each Ad* orbit in (T,G)*. Thus once po is chosen we are reduced to X , the Ad* orbit through p o . Such an orbit carries a natural symplectic structure o by o(v)(tB(v), tc(v)) = v([B, C ] ) , where tB(v)( ) = v([B, ( )I) V B E T,G. Defining T(v)= $h,(v#, P),we get a Hamiltonian system ( X ,o,7')and Y , is the Hamiltonian vector field.

262

13. MECHANICS OF RIGID BODY MOTION

(7) Special properties of SO(3).We have a: R 3 + T,G by a(u)(w) = u x w. Then [cT(u),a(w)] = a(u x w ) and Ad(A)(au) = ~ ( A u )Define . Z (T,G)* + R 3 by ( 6 ( p ) , u ) = r*(p)(u) = p(au). Then 6(Ad*(A)p) = A6(p). (a) Define a symmetric positive definite matrix R by h,(ou, aw) = (Ru, w) Vu, w E R 3 , and let eigenvectors be u,, u,, u, with eigenvalues I,, I,, I,. Let S = R - ' . Then b = 6 - ' R K 1 , # = oS6. (b) X, = {p E (T,G)*l(a"(p),&)) = r 2 } is an Ad* orbit. Let TE = {P E (TeG)*I T(P)= E } .

Then C?(TE) = { u E R31(u, Su) = 2 E ) = {&Ui IX I ; ( U i ) 2 = 2 E } is an ellipsoid. Every body motion is contained in some T , n .X,, where E is the total energy and r is the total momentum. (c) p ( t ) is a body motion iff m(t) = 6 ( p ( t ) ) is an integral curve of Y(u)= u x Su. Thus, eigenvectors of S give trivial body motions which give stationary rotations of the rigid body. If m(t) = rui V t then A(t)ui = ui V t . (d) Let v ( t ) be a body motion and A(t) be the special space motion determined by v(t). Define W(t)= Ad(A(t))(v(r)$)

and

w(t) = a- '(W(t));

o(t)is

the usual angular velocity Vector (why?). (8) Interpretation: (a) The Kinetic energy metric h on G , when represented on G x ( R 3 x R 3 ) using p (1 x a): G x R 3 + TG, becomes 0

so I(A)(n,n) gives the moment of inertia of rigid body about n, n a unit vector. (b) I(A)(u, w ) = (R,u, w ) defines R , . Its eigenvectors u,(A) diagonalize I(A) and are the principal axes for ( A t , , . . . , AtN);its eigenvalues are the moments of inertia about these axes. (c) I(A)(u,w ) = I(e)(A-'u, K ' w ) gives R , = A R A - I . (Note that Re = R.) Thus u,(A) = Au,; these are called body axes. (d) 6p(A(t),A ( t ) - ' k ( t ) )= C m i ( A ( t ) t ix k(t)ti)= 6(po); m(t)= A(t)-'(C?p,,) gives the motion of the angular momentum vector as seen by an observer moving with the body. With these remarks in hand we now resume the main flow of the development. Since h is left-invariant it is useful to use the body representation 1:G x TeG -,TG in studying the geodesics of h. Let & , be the metric on the trivial

263

LEFT-INVARIANT 1-FORM

vector bundle G x T,G

+G

determined by h and 1.Thus, by definition,

@ ( A , B),( A , C ) )= h(TLA(B),TLA(C))= h,(B, C),

where the second equality follows from left-invariance of h. Using the metric h on G, we can construct, in the usual way, a Hamiltonian system (TG, Q h , h), where the function h: TG -+ R is the kinetic energy function for the metric h, h(v) = ih(u, v). Recall the metric h induces a Legendre transformation 9: TG + T*G, and Qh = 9 * R , where R is the canonical 2-form on T*G. The trajectories of this Hamiltonian system are the curves c’(t)in TG where c ( t ) is a geodesic of h in G. Now using I., we construct an equivalent model, an isomorphic Hamiltonian system on the manifold G x T,G. The symplectic form on G x T,G is j”*Rh= fi and H : G x T,G + R is H = h 1, where h: TG + R is the kinetic energy as above. We have the formula 0

(1 3.6)

H ( A , B) = +h,(B, B).

We consider the form fi in more detail. Recall on T*G there is the canonical 1-form 8 given by @a) = CI Tat*:Ta(T*G)+ R, for a E T*G. Let 8 = ;i*2?*8. Then -d8 = lb*9*( - do) = 1*9*(R) = i*Rh= fi. For the form we have the formula 0

e

&A, B ) = [ 6 p ]*(A,B)] 0 Tn,

where z: G x T,G

+G

(1 3.7)

is projection on the first factor. Now

= h(TeLA(B), [T A(A>B)](w) = Y(T,LA(B))(w)

w, = he(B,

TALA-l(W)).

LEFT-INVARIANT 1 -FORM DEFINITION 13.8 For A E G let O(A):TAG T,G be defined by 8(A) = TALA-,. 8 is called the cunonicul left-invariunt TeG-valued 1-form on G . Note that we then have --f

$(A, B) = h,(B, B(A) Tz( -)).

(13.8)

0

The blank is to be filled by a vector in T(A,B)(Gx T,G). We now have the Hamiltonian system X = (C x T,G, fi, H ) . This is isomorphic to the Hamiltonian system (T*G,R, h) under the mapping 3’i. 0

PROPOSITION 13.9 The trajectories of the Hamiltonian system X correspond under A: G x TeG + TG to the curves A’(t),where A(t) is a geodesic of the metric h on G.

PROOF: This is left as an easy exercise using previous results.

I

264

13. MECHANICS OF RIGID BODY MOTION

SYMMETRY GROUP We now exploit the fact that G acts on the Hamiltonian system 2 as a group of symmetries to obtain conservation laws which we shall see determine the trajectories of %. PROPOSITION 13.10 For A , E G define C,,: G x 7',G [,,(A, B) = (A,A, B). Then we have

+G

x T,G by

(a) the diagram

commutes, (b) L'fo@= @, (c) H L', = H . 0

The proof is left to the reader. = so (.,, is an isomorphism of X onto 8; A , + t,, It follows that defines a left action of G on 8as a group of isomorphisms. We call this the symmetry action of G on 2. Let B , E T,G. Define the vector field tBoon G x T,G by

{?,,a a,

(13.9)

(13.10)

ADJOl NT REPRESENTATION DEFINITION 13.11 The adjoint representation of G is the homomorphism Ad: G + Aut(T,G) given by

Ad(A)(B)= A B A - ' . This definition makes sense since G is a matrix group. An equivalent definition is Ad(A)(B)= TL,TR,- ,(I?). LEMMA 13.12

vB,(A, B) = h,(B, Ad(A-')(B,))

(13.11)

265

MOMENTUM MAPPING

exp(tB,)A

dr

=

TR,(B,).

*=,

MOMENTUM MAPPING DEFINITION 13.13 The momentum map associated with the symmetry

action of G on 2 is p: G x T,G

+

TZG,

defined by p(A, B)(C)= h,(B, Ad(A-')(C)). THEOREM 13.14 If ( A ( t ) ,B ( t ) )is a trajectory of the Hamiltonian system 2,then B(t) = A(t)-'k(t)and

d p ( A ( t ) ,B ( t ) )= 0 dt

-

(conseraation of momentum). PROOF: B(t) = A - l ( t ) k ( t ) , since ( A ( t ) ,B ( t ) )= X ' ( d A / d t ) by Proposition 13.9. Let X , = ( d H ) # , where # is with respect to a. We want to show dp(X,) = 0, which will prove the theorem, since the trajectories of 2 are the integral curves of X,. Since p takes values in TZG, we view p as a TZGvalued 1-form on G x T,G and hence, dp(X,) takes values in TZC. Let B, E T,G. We must show ( d p ( X I f ) ) ( B , )= 0. Now the following diagram commutes:

G x T,G-T,TG

by Lemma 13.12, where euB,(u)= cc(B,). Also, euBo is linear so that

d(eu,,,

7

p ) = euBo dp. 0

13. MECHANICS OF RIGID BODY MOTION

266

COADJOI NT REPRESENTATION There is another description of the momentum map. DEFINITION 13.1 5 The coadjoint representation of G on TZG is given by Ad*(A)(4) = 4 Ad(A-'). Note that, because of the A-' on the right side, the coadjoint representation gives a left action of G on TZG. Using the i4 and b mappings corresponding to the metric h, on T,G we then have 0

LEMMA 13.16 p(A, B) = Ad*(A)(B,) for all ( A , B) E G x T,G. The proof is left as an exercise.

SPACE MOTIONS WITH SPECIFIED MOMENTUM DEFINITION 13.17 Let A(t) be a space motion of the rigid body. The element po = p(A(t), A ( t ) - ' k ( t ) )in T,*G (which is independent oft) is called the conserved momentum of the motion. REMARK: po can be identified as the conserved total angular momentum of the system (see Eq. (13.30)).

We now show that if the conserved momentum pLois specified, the equations of motion reduce to a first-order differential equation. THEOREM 13.18 Let po E TZG. Suppose that A(t) is a curve in G which satisfies the differential equation

dA/dt

=

TLA((Ad*(A-')(po))').

(13.12)

Then p(A(t),A - '(t)k(t))= po for all t in the interval I on which A is defined and A(t) is a geodesic of h. PROOF: The differential equation gives

A-'(t)k(t) = (Ad*(A-'(t))po)'

267

COADJOINT ORBITS AND BODY MOTIONS

so that A d * ( A ( t ) ) [ ( A - ' ( t ) k ( t ) ) = , ] p o . But then, by Lemma 13.16, we have

p ( ~ ( t )A, - ' ( t ) k ( t ) )= p o Now let

A

for all t i n I .

be a geodesic of G such that

(a) ! ( t o ) = W O ) , (b) A'(t0) = T~,,,,,((Ad*(A(tO)- l)Po)Y? where to is some given point in I . By Theorem 13.14, vo = p(a(t), k ' ( t ) A " ( t ) ) is independent oft. Now vo = Ad*(A(t))((A-'a(t)),) by Lemma 13.16 so dA"/dt

=

T L J ((Ad*(A-')vo)').

Choose t = to. Put t = to to get v o = po. The uniqueness theorem for differential equations now shows A = A" locally, so A is a geodesic. I COROLLARY 13.19 A curve A ( t ) in G is a geodesic of h if and only if (d/dt)p(A(t),A - ' ( t ) k ( t ) )= 0 for all t .

The proof is left as an exercise. DEFINITION 13.20

For p o E T,*G the vector field Z,, on G is defined by Z,,(4

=

TL,((Ad*(A- l)Po)fl).

We then have COROLLARY 13.21 A curve A ( t )in G is a space motion of the rigid body with conserved momentum po if and only if A ( t ) is an integral curve of Zp,.

COADJOINT ORBITS AND BODY MOTIONS The preceding results, in particular the form of the vector field Z p 0 ,suggests the importance of the orbits in T,*G under the right representation of G given by A + Ad*(A-I). Of course, as point sets, these are the same as the orbits of the coadjoint representation. Let X be such an orbit. We will show that X can be given the structure of a Hamiltonian system in such a way that the Hamiltonian flow determines space motions of the body with conserved momenta in X . There is an operation on T,G called the bracket. This will be discussed in more general terms in the next chapter. For now, recalling that we are viewing T,G as the set of skew-symmetric 3 x 3 matrices, we define [B, C]

=

BC

-

CB.

It is easily checked that [B, C ] lies in T,G if B and C do.

(13.13)

268

13. MECHANICS OF RIGID BODY MOTION

DEFINITION 13.22 ad: T,G -+ L(T,G, T,G) is given by ad(B)C = [ B , C]. We have Ad: G -+ L(T,G, T,G) and, since L(T,G, T,G) is a vector space, we may review T , Ad as a map T , Ad: T,G -+ L(T,G, T,G). Then we have

PROPOSITION 13.23 T , Ad@) = ad(B). The proof is left as a straightforward exercise. Now we define, for each B E T,G, a vector field tBon TZG. We shall view this vector field as a map tB:TQG -+ TZG. DEFINITION 13.24 For B

E

T,G let

tBbe given by

t B ( v )= v ad(B). 0

LEMMA 13.25 t B ( v )= d/dt(,=,,[Ad*(exp(-tB))v]

Again, the proof is left as a simple exercise. LEMMA 13.26 If v E T,*G and X is the Ad*-orbit containing v, then every vector in T v X is of the form (,(v) for some B. Also, if tB(v) = tc(v), then v ( [ B - C, D]) = 0 for all D E T,G. PROOF: The first assertion will be proved in greater generality in the next chapter so we omit it here. If rB(v) = tc(v), then v([B, D]) = v([C, D]) for all D so v ( [ B - C,D]) = 0 for all D. I

DEFINITION 13.27 T:T,*G -+ R is defined by

T ( v )= ;h,(v', v'). If we define an inner product h^, on TZG by &?(/A 4 = he(@, v'),

then we have

T ( v )= ~ v ( v ' = ) f h , ( ~ ' ,v')

= $,(v,

v).

DEFINITION 13.28 Denote by Y, the vector field on TZG given by

YdV) = t"'(V). PROPOSITION 13.29 Y, is a smooth vector field and if X c T,*G is an orbit of the coadjoint representation, then, for v E X , Y,(v) is tangent to X . We leave the proof as an exercise.

269

COADJOINT ORBITS AND BODY MOTIONS

DEFINITION 13.30 Let a,p E T V X .Define o(v)(a,p) = v([B, A ] ) , where

A, B

E

T,G are such that CA(v) = a, <,(v)

LEMMA 13.31 PROOF: If

=

8.

o(v)(a,p) is a well-defined 2-form on X .

tx(v) = a, tB(v) = /1,

then for all D, E

E

T,G we have

v ( [ B - B, D ] ) = 0, v ( [ A - A, E l ) = 0.

But then

+

v([B, A ] ) - v([B,A ] ) = v([B - B, A] [B, A - A ] ) = v([B - B, A ] ) - v ( [ A - A, B])= 0.

This shows o is well defined. Clearly o ( v ) is bilinear and skew-symmetric. In general it will be shown later that o is closed and nondegenerate. Thus ( X ,w,T ) is a Hamiltonian system, (see Theorem 16.17). LEMMA 13.32 Let v E X , C

E

T,G. Then

dT(<,(v))= [ - v ad(v#)](C)= v([C, v']). 0

PROOF: Let

4(t)= Ad*(exp(- tC))v.Then by

Lemma 13.25 &O)

=

&-(v).

Then d ~ ( t c ( 0 )=) =

ltz0

~(4(t= ))

$(

+Le($(t), t=O

4(t))= i e ( 4 ( 0 ) 3 4(0))

h,(&O)#, v # ) = d ( O ) ( V # ) = v([C, v").

1

PROPOSITION 13.33 The Hamiltonian vector field of ( X ,o,T ) is given

by

yT\K-

PROOF: We must show that for v E X and

8 E T,,X

we have

W(V)(YT(V),8) = dT(v)(B).

Choose C E T,G so that p = tc(v). Then dT(v)())= dT(v)(t,(v)) = v([C, v*]). But w(v)(YT(v),B) = o(v)(<,a(v),&-(v)) = 4( [C, v']), so the proof is complete.

I Now let X be an orbit of the coadjoint representation and let po E X . Define OMo:G + X by O,,(A) = Ad*(A-')po

( 13.14)

27 0

13. MECHANICS OF RIGID BODY MOTION

Clearly O,, maps G onto X and it is easy to check that the following diagram commutes: G

'G

RAa

1

0wo.1

x

QNO

Ad'(Ao')+

x

We have LEMMA 13.34 tB(Op0(A)) = TO,, PROOF:

0

TL,(B).

By Lemma 13.25,

LEMMA 13.35

Y,(@,,(A)) = TO,,(Z,,(A)) for A in G .

PROOF: T@,,(z,,(A)) =

T@,,TLA([Ad*(A-')p0]')

=

= S , w o , A , * ( ~ p o ( 1~ = ) YT(qLo('4)).

TO,,TL,((O,,(A))*)

I

We now establish a correspondence between integral curves of YT in T,*G and certain space motions of the rigid body. If A ( t ) gives a space motion, then (A(O)t1,. . . , A(O)t,) gives the configuration of the system at t = 0. Thus, the condition A(0) = e is just that the configuration at t = 0 is the reference configuration chosen at the beginning of our discussion. The particular reference configuration chosen is, of course, of no importance except that it is necessary to choose one in order to set up the correspondence between configuration space and the group SO(3) = G. Having said that, we shall refer to a space motion A(t), satisfying A(0) = e, as a special space motion. If A ( t ) is a space motion and A , E G, then act) = A,A(t) is also a space motion. This is because a space motion of the rigid body is a geodesic of the left-invariant metric h (also see Exercise 13.9). Therefore, any space motion A(t) can be written as A ( t ) = AoA(t), where A , = A(0) and A(t) is a special space motion.

271

SPECIAL PROPERTIES OF SO(3)

FIGURE 13.1

Let v(t) be an integral curve of Y, in T f G . There is a unique geodesic A(t) in G such that THEOREM 13.36

(a) 4 0 ) = e ,

(b) oyco)(4t)) =v(a (c) dA/dt = Z,,(,,)(A(t)),so that v(0) is the conserved momentum of A(t). PROOF: Of course there is a unique geodesic A ( t ) satisfying (a) and (c). To see that (b) holds, note that 4(t)= Ov(o)(A(t))satisfies 4(0)= v(0)and d4ldt = To"&"(o)(4) = (by lAxnma 13.35) = y,(ov(o)(4) = YT(4(t)), so 4 ( t ) = v(t), as (b) asserts. I

DEFINITION 13.37 The trajectories of Y, in T,*G are called the body motions of the rigid body.

If A ( t ) is any space motion, then A ( t ) has a conserved momentum, p o E T f G , and it is an immediate consequence of Corollary 13.21 and Lemma 13.35 that v ( t ) = O,,(A(t)) is an integral curve of Y,, i.e., a body motion. If it happens that A(0) = e, then v(0) = p o . Thus, there is a one-to-one correspondence between special space motions and body motions. This correspondence is depicted in Fig. 13.1.

SPECIAL PROPERTIES OF S O ( 3 ) Much of what we have done up to now actually works for any Lie group. We now consider some results which depend on the specific nature of SO(3). LEMMA 13.38 There is an isomorphism

6:R 3 + T,G

such that:

(a) a(u)w = u x w (Here ~ ( u is) viewed as a matrix so a(u)w is the vector

272

13. MECHANICS OF RIGID BODY MOTION

obtained as the product of the skew-symmetric matrix o(u) with the column vector w.) (b) [o(u),4w)] = 4 0 x 4. (c) Ad(A)(o(u))= o(Au)for A in G. The proof is left as an exercise. It is easily checked that o is given by 0 o ( d ,u 2 , u 3 ) = [

0'

-u3

-;I; u2

( 13.15)

o is a linear isomorphism so there is a linear isomorphism o*:T,*G

+ (R3)*,

defined by (13.16)

o*(a)u = a(cT(u)).

We also have 6: T,*G-,R 3 , given by ( 13.17)

6(a) = (o*(a))',

where the #-map, #:(R3)*+ R 3 , corresponds to the standard inner product ( ) o n R3. NOW (c?(Ad*(A-')p), U ) = p(Ad(A)(o(u)))= p(a(Au))= (6(p),A u ) = (A-'c?(p), u ) for p E T,*G,u E R 3 . Therefore, 6(Ad*(A-')p)

= A-'i?(p)

for

A

E

SO(3).

( 13.18)

There is a symmetric, positive dejnite matrix R such that h,(o(u), o(w)) = (Ru, w)

for u, w E R 3 .

(13.19)

For a description of R in terms of the reference configuration, see Eq. (13.29). Since R is symmetric and positive definite, we have eigenvectors u,, u2, u, in R 3 , with corresponding eigenvalues I , , I , , I,, such that

(a) (ui, t i j ) = Sij for i, j (b) I,, I,, I , are positive.

=

1, 2, 3,

(1 3.20)

By Eq. (13.18), the orbit of p E T,*G under the action of Ad*(A-') is mapped by 6 onto a sphere in R 3 centered at the origin. Such a sphere is, of course, characterized by its radius. Let us introduce the notation

PROPOSITION 13.39 Each X,

tion.

is an orbit of the coadjoint representa-

273

SPECIAL PROPERTIES OF S O ( 3 )

PROOF: As noted above, if p € E r , then for any A E G we have o"(Ad*(A-')p) = A - ' Z ( p ) so Ad*(A-')p also lies in Xr since A is orthothen 6(p) and 6(p) have the same norm gonal. O n the other hand, if i E X,, in R 3 so there is an A E G such that 6 ( j ) = A-'a"(p). Then o"(Ad*(A-')p) = A-'o"(p) = G(p), so Ad*(A-')p = p, which completes the proof. I

By Proposition 13.29, YT gives a vector field on Er for each r > 0. By Proposition 13.33, Y, on X, is the Hamiltonian vector field corresponding then dT(I;)= dT(Y,(v)) = to T. Hence, if v(t) is an integral curve of Y, in Yr, w(v)(Y,(v), Y,(v)) = 0 whence T is constant along v. We introduce the notation, for E > 0, ZE = {/A E

T*,GIT ( p ) = E } .

(1 3.22)

Note that each body motion, that is, each integral curve of Y,, lies in Xr n zE for some r and E . LEMMA 13.40 Let S = R - ' . Then the b and !4 maps on T,*G are given

by (a) b = 6 - l o R o ,--I, (b) # = a o S o G .

(13.23)

PROOF: Clearly if (a) holds so does (b). To prove (a) let u, w E R 3 . We ) ~ 6-'(Ru) or 6 ( a ( ~ )= ~ )Ru. But want b a = 6-l R so we show ( c J ( v ) = ( ~ ( C J ( V ) ~w) ) , = [a*(a(u),)]w = a(u)b(a(w)) = h,(a(u), ~ ( w ) = ) (Ru, w). Since w is arbitrary, we are done. I 0

0

Thus, h e w , p? and we see that

= h,(oSW,

a m 4 ) = ( R S ( m , S 4 p ) ) = <4PL),S

6 ( t E= ) {U E R 3 ( ( u ,S U ) = 2E)

W >,

=

Since (ul, u2, u3) is an orthonormal basis in R 3 , C(zE) is an ellipsoid centered at the origin. Let v(t) be a body motion and let m(t) = 6(v(t)). The curve m(t) lies in the intersection of a sphere and an ellipsoid. For v(t) lies in Xr for some r as shown in Proposition 13.39. Thus (6(v(t)), Z(v(t))) = r2 so m(t) lies in the sphere of radius r centered at the origin. Also v(t) lies in some z E for all t , so m(t) = 6(v(t)) lies in the ellipsoid given in Eq. (13.24). Note that this ellipsoid may be described as the ellipsoid centered at the origin having J;?E12u2 and W , u 3 . principal axes We now obtain explicitly the vector field on R 3 which has as its integral curves the curves m(t) = 6(v(t)), where v(t) is a body motion.

mIu1,

274

13. MECHANICS OF RIGID BODY MOTION

STAT I 0 NARY R OTATl 0 N S DEFINITION 13.42

Let Y

=

{v E T*,GI6(v)is a nonzero eigenvector of S } .

Y is called the set of stationary rotations of the rigid body. THEOREM 13.43 Let v E Y. Then v(t) = v for all t gives a body motion of the rigid body. Let Go = { A E GIAd*(A-')v = v} and let A(t) be the special space motion determined by v(t) = v. Then Go is the group of rotations leaving C(v) fixed and t + A(t) is a group homomorphism of the real numbers onto Go with k(0)= v'. PROOF: Y,(v) = O"-'Y(6(v)) by Lemma 13.41, so that YT(v) = 6-'(6(v) x S8(v)) = F 1 ( 8 ( v ) x A8(v)) = 0, where A is the eigenvalue corresponding to the eigenvector 8(v). Thus v is a critical point of Y,, so v(t) = v for all t is an integral curve. Let A(t) be the corresponding special space motion given by Theorem 13.36. Since Coy(A(t))= v for all t(by Theorem 13.36) we see A(t) lies in Go for all t. Now for A E Go we have &(A) = TLA(v') so that A(t) = exp(tv') (see the discussion following Definition 4.23). Thus A(t) is a homomorphism with k(0)= 0'. Since K 1 6 ( u )= 6(Ad*(A-')(v)) (see Eq. 13.18)),it follows that A lies in G o if and only if A-'o"(v) = 6(v), that is, if and only if A is a rotation leaving 6(v) fixed. I

CLASS I CA L I NTER P R ETATI 0 NINERTIAL TENSOR, PRINCIPAL AXES Recall we have the kinetic energy metric Ton configuration space in R3N and it was pulled back to give a metric on G . This new metric was h = @*T.

CLASSICAL INTERPRETATION-INERTIAL

TENSOR, PRINCIPAL AXES

275

We have the mapping p

0

(1 x a):G x R 3 -+ TG,

so we have the metric I(A):R 3 x R 3 + R for each A E G by It is left as an exercise to show that

I ( A ) is called the inertia tensor of the rigid body for the configuration ( A t l , . . . , A t N ) .It is just the kinetic energy metric represented on G x R 3 by the mapping @ p 0 (1 x CJ). If n is a unit vector in R3, then 0

where di is the perpendicular distance from A t i to the axis determined by n. This is illustrated in Fig. 13.2. Thus, I(A)(n,n) gives the moment of inertia of the rigid body about the axis n. Because I(A) is an inner product on R 3 , there is a symmetric, positive definite matrix RA such that (13.28) I(A)(o,w ) = ( R # , w>. There is a basis u l , u 2 , u3 for R3 such that (a) ( u i , uj> = dij, (b) Each ui is an eigenvector of RA with positive eigenvalue Zi(A).

FIGURE 13.2

276

13. MECHANICS

OF RIGID BODY MOTION

The basis u l , u 2 , u3 determines a Cartesian coordinate system in R 3 and the axes so determined are called principal axes for the body in the configuration ( A t l , . . . ,A t N ) .The numbers Zi(A) are the moments of inertia about the principal axes and hence are called the principal moments of inertia. LEMMA 13.44 I ( A ) ( u , w ) = Z(e)(A-'v, A - ' w ) .

PRooF:Just compute both sides using Eq. (13.25), making use of the leftinvariance of h and using Lemma 13.38 (c). [ It follows that (R,u,

W) =

( R A - b , A-'w)

=

(ARA-'v,

W)

so R , = A R A - ' . Therefore, RA and R have the same eigenvalues and the eigenvectors of RA are Au, where u is an eigenvector of R . Thus principal axes for the configuration ( A t , , . . . , AtN) are obtained from those for (tl,.. . ,t N ) by rotation by A. If we fix principal axes for ( t l ,. . . , tN)by a basis (ul, v 2 , u3) of R3, then we determine a basis Vi(A)= Aui for each A and the inertia tensor has the matrix

1"

(Zij(A))= 0

Lo

OZ2 0O l 0

131

with respect to the basis V l ( A ) , V2(A),V,(A). The vector fields V l , V,, V , provide what are called body axes. They are a basis which rotates with the body in R 3 . Now the momentum mapping p: G x TeG + (T,G)* defined by (13.13) and (13.16) gives the total angular momentum of a motion ( A ( t ) t , , A ( t ) t 2 , .. . , A(t)tN),when composed with 6:(TeG)*-+ R 3 . We have 6p(A(t), A -

'A(t )) = 6(Ad*(A)(A-.'A),) = A(6 0 b(A-'A)) = A(t)(RC'(A-'A)),

using (13.18) and (13.23). By (13.25), I(e)(u,W ) = he(ou,

so that Re given by (13.28) is equal to R defined by (13.19). Thus, using (13.26),

Thus,

277

STABILITY OF STATIONARY ROTATIONS

the total angular momentum of the system at time t (note that this derivation is valid for any motion of the rigid body). If A ( t ) is a geodesic of G (motion with no external forces) and A(0) = e, we have the body motion v(t) = Ad*(A(t)-')(p,), where 6(po)is the constant angular momentum vector of the motion. Thus m(t) = G(v(t)) = A(l)-'(6po) gives the motion of the angular momentum vector as seen by an observer moving with the rigid body.

STABILITY OF STATIONARY ROTATIONS Let u l , u,, u3 be an orthonormal basis of eigenvectors for R, with eigenvalues I,, I,, I,. As shown in Theorem 13.43, for each i there is a oneparameter group of rotations A(r) such that the corresponding body motion is v(t) = v for all t , where v = 6- I(zii). These space motions are rotations of the rigid body about a fixed axis and we now consider the question of stability of these motions. Assume the rigid body is unsymmetrical, that is, I , , I,, and I, are distinct and assume I , < I, < 1,. We shall prove the well-known results that the stationary rotations about u , and u j are stable while the rotation about u, is unstable. The following result says that the stationary rotations about u1 and u, are stable with respect to small changes in the initial value v(0). Since v(0) is the conserved momentum of the motion (which is, of course, the total angular momentum of the system), we are considering stability with respect to small changes in the angular momentum of the system. THEOREM 13.45 (Body stable theorem) Assume I , < I, < I,. Let v, E Y be such that G(vo) = roul or 3(v0) = ~ 0 ~ Let 3 . E , = T(v,). If v(t) is a body motion with v(0) close to v,, then E = T(v) is close to E,, r = ((6(v(O)), 6 ( ~ ( 0 ) ) ) )is" ~close to ro and v(t) is near vo for all t. PROOF: Since ro = ((6(vo), 8 ( ~ , ) ) ) " ~and T is continuous, it is clear that if v(0) is near vo, then E is near E , and r is near ro . To show that v(t) remains near v o for all t, we argue geometrically. We con= rouj. Let T(vo)= E,. Now we have, for any p E TZG, sider the case 8(vo)

T ( p ) = #,(pP, p')

=~

( R o '(p'), ~ ~ ' ( p ' ) ) = t ( R K 'aS6(p), 0-'oS8(p))

Thus, for the stationary rotation Eo

I],,

= *(6(p), S6(p)).

we have

= +(6(~0), SG(\lo)> = + ( r 0 ~ 3 I, T ' T ~ U ~ = ) +($I,).

Thus, ro = (213E0)1'2,that is, the sphere ( u , u ) = rg and the ellipsoid given by (v, Sv) = 2E0 are tangent at the points f r , ~ , . Now consider the body

278

13. MECHANICS OF RIGID BODY MOTION

principal a x i s determined by I,, \3V I /

FIGURE 13.3

motion v(t) and let T(v(0))= E , (C(v(O)), C(v(0))) = r2. Then v(t) always lies in the intersection of the sphere ( u , u ) = r2 and the ellipsoid ( u , Su) = 2E. Now since 1;’ is the smallest eigenvalue of S, we get (u, S u ) 2 Z;’ ( u , u ) so 2E 2 1; l r 2 . Thus, (2EZ3)”’ is just slightly greater than r (or (2E13)1/2= r, but never do we get (2EZ,)’I2 < r). Thus C(v(t)) lies on the intersection of a sphere of radius r and an ellipsoid having largest principal half-axis (2EZ3)’/’ with 0 < (2EZ3)’/2- r << 1. All points on this intersection are very near rug which is near r 0 0 3 . Thus v(t) is near vo for all t. The situation is pictured in Fig. 13.3. Notice that the ellipsoid and the sphere intersect in a small closed curve all of whose points lie near (2E13)’i2U3. The situation for a(vo)= youl can be understood in a similar manner. This completes the proof. I To obtain results about the motion we see in physical space we must work with space motions. Thus, let v(t) be a body motion and let A(t) be the special space motion determined by v(t) (see Theorem 13.36). Define W(t)and o(t)by (13.31) LEMMA 13.46 Let notation be as above. Then:

(a) (b) (c) (d)

W(t)= k ( t ) A - ‘ ( t ) E T,G; if U(t)= Ad(A(t))(U,), then U ( t ) = [W(t),U(t)]; if u(t) = A(t)uo, then zi(t) = o(t)x u(t); o(t)= A(t)Sm(t),where m(t) = c?(v(t)).

279

STAB1LlTY 0 F STAT10 NARY ROTATIO N S

PROOF: A(t) satisfies dA/dt = TL,((Ad*(A- ')v(O))') = TL,((cOyco,(A))p) = TL,(v(t)#) = A(t)v(t)', where we used Theorem 13.36. Therefore v(t)' = A-'k, so Ad(A(t))(v(t)') = kA -', which proves (a). For (b) we start with U ( t )= Ad(A(t))Uo = A(t)U,A(t)- '. Differentiation gives

0 = AU&'

+ A U , -dtd( A - ' )

=

AUoA-' - A U o A - ' A A - '

=

[AA-', AUoA-']

=

[W, U ] ,

proving (b); (c) follows from (b), since zi(t) = A(t)uo = a-'(d/dt Ad(A)Uo), where o-'(Uo) = uo. So ti(t) = a-'([W, U ] )= a-'(W) x o - ' ( U ) = w x u. Finally, to prove (d) we write w(t) = o-'(Ad(A(t))(v(t)')) = A ( t ) o - '(v(t)#) = A(t)a-loSG(v(t)) = A(t)SG(v(t)) = A(t)Sm(t). 1 REMARK 13.47 Part (c) of Lemma 13.46 shows that w(t) is the usual angular velocity vector. LEMMA 13.48 Let v(t) be a body motion with associated special space motion A(t). Let m(t)= G(v(t)).Then m(t) = A(t)-'(m(0)). PROOF: Let

v

E R3

(G(v(t)), v)

be arbitrary. Then (G(v(0) Ad(A)), U) = ~ ( 0 ) Ad(A)(a(v)) = v(O)(o(Au)) = (o*v(O))(Au) = (G(v(O)), A u ) = (A-'G(v(O)), v) = (A-'m(O), v). I

=

0

0

LEMMA 13.49 Suppose m, is a nonzero eigenvector of S with eigenvalue a. Suppose (m(t)- m,( < E for all t. Then

(a) Iw(t) - am(O)l < (a+ IS~)E, (b) l 4 t ) - w(O)(< 2(a + 1SI)E, (c) lw(t) - am,l < (2a + ISl)E, for all t (IS1 is the operator norm of S). PROOF: The assumption Im(t) - rn,l < E gives ISm(t) - am,l < IS(&and lam, - am(t)I < a&, so (Sm(t)- am(t)J< (a IS\)&.Therefore, IA(t)Sm(t)aA(t)m(t)l < ( a + l S [ ) ~since JAl = 1. But by Lemmas 13.46 and 13.48 w(t) = A(t)Sm(t)and A(t)m(t)= m(0) so (a) holds. To prove (b) use (a) to get lw(t) w(0)I s Jw(t)- am(0)I + (w(0)- am(O)(< 2(a+ ISI)E. Finally, Iarn(0) - am,l < a& so, using (a), lo(t)- am,l < (2a IS()&. I

+

+

280

13. MECHANICS OF RIGID BODY MOTION

The next result is the expected one, that for a stationary rotation the angular velocity vector is constant. LEMMA 13.50 Let v

E

9. Then

w ( t ) = aG(v)

for all

t,

where S(6(v))=

CCG(v). PROOF: We have m(t) = G(v(t)) = G(v) for all t. So by Lemma o(t)= A(t)Sm(t)= A(t)SG(v) = A(t)crG(v). Therefore, by Lemma 13.46 h(t)= w(t) x w(t) = 0, so w(t) is constant. But w(0) = a6(v).

13.46, again,

THEOREM 13.51 (Space stable theorem) Suppose I , < I , < I , . Let v, E Y be such that G(v,) = rool or G(v,) = r,vj. Let w, be the constant angular velocity vector corresponding to stationary rotation v,. If A(t) is a space motion with angular velocity vector o ( t )and if w(0)is sufficiently close to w,, then w(t) will remain close to w, for all t. PROOF: By Lemma 13.46, o(t)= A(t)Sm(t), so w(0) = Sm(0). Thus, v(0) = 6-‘Rw(O) and v, = G-’Rw, so that v(0) is close to v,. Now apply Theorem 13.45 to conclude that v ( t ) is close to v, for all t. It follows that m(t) = c?(v(t)) is near m, = G(v,) for all t. By Lemma 13.49, we get Iw(t) - amO[ remains small for all t. Here m, is an eigenvector of S, M is the eigenvalue and w, = Sm, = am, so o(t)remains near w, for all t . I

POINSOT CONSTRUCTION The force-free motion in space of a rigid body can be described by considering the motion of a comoving ellipsoid, the so-called “inertial ellipsoid.” Corresponding to each such motion one can construct a fixed plane in space and the motion of the body satisfies the requirement that the inertial ellipsoid roll without slipping on this “invariable plane.” We now show how these ideas arise from our description of rigid body motion. Suppose the rigid body moves so that its conserved momentum is p, and its energy is E . Assume, for specificity, that at t = 0 the body is in the reference configuration and that the Cartesian coordinates have been chosen so that, at t = 0, the standard basis e l , e 2 , e 3 , defines principal axes. Then for any configuration determined by A E G, the axes A e , , Ae, , Ae, are principal axes for that configuration. DEFINITION 13.52 If A E G gives the configuration of the body, then the inertial ellipsoid for the configuration is

8E.A = { U E R ~ I I ( A ) (U) u ,= 2 E ) .

281

POINSOT CONSTRUCTION

Note that &LA

( 1 3.32)

= 48E.e)

If A(t) is the space motion under consideration, then the family of ellipsoids & E . A ( r ) move with the body, by Eq. (13.32). DEFINITION 13.53 The inuariahlr plane (for the given motion) is given by

ll = { U E R 3 I p o ( a ( ~= ))2E). Note that

n = { C' E R3 I (u.6(po))= 2E}.

(1 3.33)

THEOREM 13.54 (Poinsot construction) Let w ( t )be the angular velocity and ll is tangent to &E,A(t) at w(t). for our space motion. Then o(r)E n n 8b.A(t) The instantaneous velocity of the point of contact is zero. PROOF: First we show w(t) E

&t,n(t,.

We have

I(A(t))(@(t),d t ) ) = I ( A ( d) ( A ( t ) W t ) ,A(t)Sm(t)) = I(e)(Sm(t),Sm(t)) = h,(aSm(t), aSm(t)) = h,(aSG(v(t)),aS&(v(t))) = he(\!', = 2E 17')

Now we show o(t)E

n;

p o ( a ( N ) ) )= p o ( A ( f ) A -'(0) = h,(A - '( t)k(t),Ad(A = Iz,(A-lA, =

~

' (t))A(t)A

'( t ) )

APAAPA)

h,(A-'k, A - ' A )

=

h,(\~(t)',

\j(f)')

=

2E.

Note that we have made use of the important relationship between A(t)and v(t), A-'(t)k(t)= v(t)' (see the proof of Lemma 13.46). Thus Il is given by

n = w ( t ) + rr-'(ker

po)

for all t .

Suppose u ( ( ) is a curve in 8fi,A(r) with u(0)= w(t) (here we have fixed t). Then I(A(t))(u((),u ( ( ) ) = 2E for all 5. Thus 2 E = I(e)(A-'u(t), A - ' u ( ( ) ) = h,(aA-'u((), o A - ' u ( ( ) ) = ( R A - '(f)u((), A - ' ( t ) u ( ( ) ) so differentiating with respect to gives ( R A ( t ) - ' dv/dyT, A ( r ) - ' u ( t ) ) = 0, where we have used the fact that R is symmetric. Setting yT = 0 and using u(0) = w(t), we get

<

A(t)-

'

-

, RA(t)-'w(t)

282

13. MECHANICS OF RIGID BODY MOTION

So, letting d

= dv/dt15=o, we

have

0 = ( K ' ( t ) d , R A - ' ( t ) w ( t ) ) = h,(aA-'(t)d, a A ( t ) - ' o ( t ) ) = h,(aA(t)- 'd,

oA(t)-'A(t)Sf5(v(t)))= h,(oA(t)- 'd, oSc?(v(t))) = h,(aA(t)-'d, v(t)') = he(v(t)', Ad(A(t)- ')(ad)) = po(od).

'

Thus d E 6 (ker po), as desired. The fact that the instantaneous velocity of the point of contact is h(t)= w ( t ) x w ( t ) shows that the inertial ellipsoid does indeed roll without slipping on the invariable plane. I

EULER EQUATIONS These equations make their appearance in every treatment of the motion of a rigid body. It appears that they are differential equations for the angular velocity vector ~ ( t )But . actually, as usually written, they govern the components of o(t)with respect to body axes, that is, axes which rotate with the body. Consider the vector m(t)= c?(v(t)).By Lemma 13.41 we have m(t) = m(t) x Srn(t). Choose a basis vl, u 2 , v3 for R 3 which is orthonormal and such that ui is an eigenvector of R with eigenvalue Zi. Let vi(t) = A(t)vi. Thus the vi(t) give principal axes for the body at time t. Then let o(t)= ti(t)ui(t).Now o(t)= A(t)Srn(t)

so

c?=

Thus, if we write m(t) = mi(t)vi,then m'(t) = ti(t)Zi.The component form of the equation h(t)= m(t) x Sm(t) is d ( t ) = m2(t)rn3(t)(z;

'

-

1;

rh"t) = m'(t)m3(t)(z;'

-

ITI),

-

1;').

rh3((t)

= m'(t)m2(t)(I;

I),

(13.34)

We replace m'(t) by t i ( t ) I i to get (13.35)

283

PHASE PLANE ANALYSIS OF STABILITY

Equations (13.35) are, of course, Euler's equations for the case of no external torque. Thus, Euler's equations are essentially the differential equations which govern body motions.

PHASE PLANE ANALYSIS OF STABILITY Consider the ellipsoid € given by

8 = (U

E

R 3I( u , S U ) = 2E},

where E is a fixed positive number. The vector field Y, whose integral curves correspond under 5 to body motions, is tangent to 8.In fact, for u E 8 the vector Su is normal to & at u (see Exercise 13.13). Therefore, Y(u)= u x Su is tangent to 8.Y has critical points at the vectors u E & which are eigenvectors of S. These are precisely the points f(2EZi)112ui,i = 1,2,3. Each of these defines a stationary rotation. We have already seen that for i = 1 or 3 the critical points are stable. For i = 2 we now show the critical point is unstable. This is the famous result that rotation about the intermediate principal axis is unstable. The proof will consist simply of choosing local coordinates about the critical point and viewing the vector field in local coordinates, as an autonomous system in a neighborhood of the origin in R2, with a critical point at the origin. Now

Consider the critical point w 2 = (2EZ2)1/2u2.Coordinates (s, t ) in a neigh&,, where u1 = s, a 2 = borhood of w2 can be defined by (s, t ) + [Z2(2E- (s2/1,)- ( t 2 / Z 3 ) ) ] 1 / 2and , a3 = t. The coordinate vector fields on € for this coordinate system are

-=(,.*,o), a

'r=(o.+). a

- tz

a 1,

as

a2z3

Now Y(u)= u x

su =

+

( (lj a2u3

where u = ulul a2u2 basis (ul, u 2 , ug). So

-

li>.

a1a3(;

-

;),

ulu'(;

-

;)),

+ u3u3 and the components on the right are for the

;( <) ay + (<

Y(u) = a2a3

i

-

a

a1a2

-

l a K) 'r.

284

13. MECHANICS OF RIGID BODY MOTION

If we replace Y(u) by a-' Y(o), the integral curves are simply reparametrized with direction unchanged so we can study the phase portrait, near the origin, of

This is a linear system having matrix

The eigenvalues are the roots of

Since I, < I, < I, the roots are real and distinct so the critical point at the origin is a saddle point which is, of course, unstable. Note that the instability results from I , < I, < I,. If 1, were either the largest or smallest principal moment, then the eigenvalues would have been purely imaginary and the critical point would be a stable center type. This again verifies stability of stationary rotation about the long and short principal axis.

EXERCISES 13.1 If f :2 + X' is an isomorphism of Hamiltonian systems and #, #' correspond to the two systems, show that ( T f ) # = #' (Tf),. 0

0

13.2 T(G x T,G) z TG x (TeG x T,G). Since T G has been identified as a subset of M , x M , we have the identification T(G x TeG) c M , x M , x TeG x TeG. Show that under this identification

l B o ( A , B, = ( A , B O A , B, If, instead of identifying TG as a subset of M , x M , , we identify it with G x T,G using 1,show that SB,,(A,B) = ( A , Ad(A-')B,, B, 0). 13.3 Prove Proposition 13.9. 13.4 Prove Proposition 13.10. 13.5 Prove Corollary 13.19. 13.6 Prove Proposition 13.23.

EXERCISES

285

13.7 Prove Lemma 13.25. 13.8 Prove Proposition 13.29. 13.9 Show that if A(t) is a space motion with conserved momentum p,, then for A,,E G, the curve A(t) = A,A(t) is a space motion with conserved momentum Ad*(& ')p,. 13.10 Prove Lemma 13.38.

13.11 Prove Eq. 13.26. 13.12 Let A(t) be a space motion with conserved momentum p, E TZG. How is p , related to the usual notion of total angular momentum? 13.13 Let & = { u E R 3 I (u, S u ) = 2 E ) . Show that Sv is normal to & at u. To do this let u(t) be a curve in Q with u(0) = u. Then use d / d t ) , , , (u(t), Su(t)) = 0. 13.14 Suppose that p o E (T,G)*, A , E G, and Ad*(& ' ) p 0 = p,. If A ( t ) is a trajectory of Z,, and $ t ) = A,A(t), show that A"(t)is a trajectory of z,,and O,,(4t)) = O,,(A(t)). 13.15 Suppose that p ( t ) is a periodic trajectory of Y, with period z. Let A(t) be the special space motion determined by p(t) and H = A(z). Then A(t) = H"A(t - nz) for nz I tI ( n + 1)z. If Go = { A E G l O , , , ( A ) = p(O)},then Go is the group of rotations about Z(p(0))and, thus, is isomorphic to S'. A ( t ) is periodic or not according to whether { H " l n = 1, 2, 3,. . .} is finite or dense in Go. 13.16 Use the phase plane analysis around an unstable stationary rotation to show that there are nonperiodic body motions.

74 Lie Groups

LIE GROUPS AND THEIR LIE ALGEBRAS DEFINITION 14.1 A Lie group is a group G which has also been given the structure of a smooth manifold so that the group operations p: G x G + G by p(g, h) = gh and a: G G by a(g) = g-' are smooth. We will use the symbol e to denote the identity element of the group.

-

In Definition 14.1, G x G is given the natural product differential structure; if ( V l , 4,) and (V,, 4,) are charts on G, then (V, x V,, 41 x 4z) is a chart for G X G. It follows that, for (9, h) E G X G, T(,,,,G X G = T,G X T h G . This allows US to define T l ( g , h ) pT:g G--* T,hG and TZ(,,h)p:ThG + TghGby Tl(g,h)p(t)= T(g,h)p(t,O)

and

TZ(g,h)dc)

=

c).

T(g,h)p(o?

(14.1)

For each g E G we get diffeomorphisms

L,:G-G

by L,(h) = g h

and

R,:G-G

and

T,,R,

by R,(h)= hg.

Clearly, TgoL, =

T2(g,goP

= TW7,gdP.

(14.2)

The fact that the L, are diffeomorphisms allows us to construct an atlas for G from a single chart around e. In fact, we have PROPOSITION 14.2 Suppose that G is a topological group which is second countable and Hausdorff, V is an open set containing e, and 4: V 4 ( V ) c R" is a homeomorphism of V with an open set in R". Suppose there is an open set W, with e E W, c V so that p(W, x W,) c V , a(W,) c V and the mappings 4 p o (4-l x 4-,), 4 0 a0 4-l are smooth. Choose an open set W containing e so that W = a( W )and p( W x W ) c W, . For each g E G let W, = L,(W) and 4, = 4 L;'. Then {(W,, 4,)lg E G} is a smooth atlas which gives G the structure of a Lie group. 0

0

We leave the details of the proof to the reader. 286

287

LIE GROUPS AND THEIR LIE ALGEBRAS

DEFINITION 14.3 Let A E T,G. Define a vector field A on G by A(g)= T,L,(A). A is called the left-invariant vectorfield determined by A .

The reason for calling these vector fields "left-invariant'' is explained by the following lemma. LEMMA 14.4

(a) If g E G and A E T,G, then (L,)*A= A,

(b) If X E F i ( G ) and (L,),X

=

X for all g E G, then

X

(c) If g

E

= X(e),

G and A , B E T,G, then

(L,)*[A,B ] PROOF: First, note that Lgh = L,

TL,,

0

=

=

[A, B ] .

Lh gives us

TL, TL,. 0

(14.3)

NOW (L,)*A(h)= TL,(A(g-'h)) = TL, TLg-l 0 TLh(A)= TL,(A) = A(h), SO that (a) is proved. Suppose that (L,),X = X for each g E G. Then X ( g ) = ((L&X)(g)= TL,(X(e)),so (b) is proved. Finally, (c) follows from part (a) and Exercise 8.16. I 0

Lemma 14.4 shows that the set of left-invariant vector fields on G form a Lie subalgebra of the Lie algebra of all vector fields on G, and that this Lie algebra is isomorphic, as a vector space, with T,G. We now use this isomorphism to give T,G the structure of a Lie algebra. DEFINITION 14.5 The Lie algebra of the Lie group G is the vector space T,G provided with the bracket operation defined by

[A, B]

=

[A, B](e).

REMARK 14.6 Some authors define the Lie algebra of G to be the Lie algebra of all left-invariant vector fields on G. The isomorphism A + A establishes the natural correspondence between these definitions. EXAMPLE 14.7 Let G = GL(n)c L(R",R") 1: R"'. G is an open subset of R"' and thus has a natural manifold structure; p: G x G + G is the restriction of 12: L(R",R") x L(R", R") + L(R",R"), which is bilinear and thus smooth. Inversion in G can be seen to be smooth from the formula 9 - l = (l/det g ) i where the matrix for y* is obtained from that of g by replacing each element by its cofactor and taking the transpose. Doing calculus on G as an

288

14. LIE GROUPS

open set in L(R", R"), we have basic facts:

(0

means matrix multiplication) the following

(a) p(g, h) = g h and so, since L, and R , are linear, 0

DIP(97 h)( 1 = ( ) h? O

(b) p ( g , cx(g))

=

D2P(9, h)( 1 = 9 O (

e, where a(g) = g-', and thus

DlP(Y9 4 9 ) ) ( 1 + D2I*(97 4 9 ) ) Da(s)( O

by (a), ( ) g - l 0

1;

+g

o

1 = 0;

Da(g)( ) = 0 which gives Da(y)( )

=

-y-l

0

(14.4)

( ) 0 9-l;

(c) A , B E L(R",R") N T,(GL(n))correspond to A, B given by

A(g) = 9 A, 0

B(g) = g B; 0

therefore, DB(g)(A(g))- D A ( g ) ( B ( g ) )= ( g A ) B 0

[A, B]

=

0

-

(g 0 B ) 0 A, so

[A, B ] ( e ) = A B - B A. 0

0

So the bracket on T,(GL(n)),using left-invariant vector fields, is the usual commutator. We leave it as an exercise to show that, for a general Lie group, if we use right-invariant vector fields the induced bracket on T,G is the negative of the one of our construction. THEOREM 14.8 Let G be a Lie group and A E T,G. Then the leftinvariant vector field A is complete and, if y: R -+G is the integral curve of A with initial condition y(0) = e, then y is a one-parameter subgroup of G. Conversely, if y: R -+ G is a smooth mapping such that y ( t s) = y(t)y(s)V t , s E R and (dy/dt)(O) = A in T,G, then y is an integral curve of A.

+

PROOF: By Theorem 4.1 1 we get P:( - E , E ) -+ G such that b(0) = e and (dB/dt)(r)= A(P(t))= TLs(&4). Let y1 = /I(+&).Define p , : ( - f ~ ,$ E ) -+ G by Pl(t) = glb(t - 4~). Then

(dB,/dt)(t) = TL,,(d/I/dt)(t- + E ) ) = TL,,A(fl(t - f E ) )

44, = 4Pl(O), so Pl(t) is an integral curve of A. Since P1(&) = g1 = P(+E), we get that b1 = fi on ( - + E , E), so that /I1gives an extension of b to ( - E , $8). Clearly, this proce= 491P(t -

dure can be repeated to give y: R -+ G extending to all of R. Let go E G be given. Let P ( t ) = goy([). This gives the integral curve of A with initial condition go and completes the proof that A is complete. Taking go = y(s) also gives us that y(s t ) = y(s)y(t). Finally, suppose y: R + G is smooth and satis-

+

289

CANONICAL COORDINATES

+

fies y(s t ) = y(s)y(t). Let A = (dy/dt)(O). Applying d/dtl,=, gives (dy/dt)(s)= TL,,,,(A),so dy/dt = Aly, which completes the proof. I

EXPONENTIAL MAPPING By Theorem 14.8 we get a smooth mapping r:R x T,G G so that y ( t ) = r(t,A ) is the one-parameter subgroup of G which is the integral curve of A -+

with initial condition e. We define the exponential muppiny of G, exp: T e G + G , by exp(A) = r(1,A). Note that exp(0) = e. Let p(t) = y(st) = r(st, A ) . Then (dp/dt)(t)= s(dy/dt)(st)= sA(y(st))= sTTL,(,,,(A)= sA(b(t))and b(0) = y(0) = e. Thus b(t)= r(t,sA) so we get the useful facts: T(st, A ) = r(t,sA) and T(s, A ) = r(l,sA). Thus the one-parameter subgroup generated by A is given by y ( t ) = exp(tA), which shows that the exponential mapping takes straight lines to one-parameter subgroups. THEOREM 14.9 There is an open set 0 c TeG such that 0 E 0 and exp: c" + exp(O) is a diffeomorphism between open sets.

PROOF: We have exp: T,G -+ G so that, identifying T,(T,G) with TeG in the usual way, we obtain To exp: T,G -+ TeG by

The last equality follows from the fact that y ( t ) = exp(tA) is the one-parameter subgroup generated by A , as noted above. The result now follows from the inverse function theorem. I

CAN 0 N I CA L CO 0 R D INATES Thus (exp(L'),(exp 10)- ') can serve as a coordinate chart around e. These are called canonical coordinutes for G. Suppose that ( V , 4) is a coordinate chart for G , e E V , and 4(e) = 0 E R". Let m: W x W + 4( V ) , W c &( V )open and containing 0, be the group multiplication. Then (a) m(0, x) = x so D,m(O, x) = id, D,D,m(O, x) = 0, (b) m(x, 0) = x so D,m(x, 0) = id, D , D , m ( x , 0) = 0, (D,D,m(O, 0)v)w = (D,Dim(O, 0)w)u (c) = D2m(0,O)((O, 4, (u, O)), D2m(0,O)(X, Y ) ( X , Y) = 2(D,D,m(O, 0)x)y;

290

14. LIE GROUPS

thus, Taylor's theorem of advanced calculus gives m(x, Y ) = x

+ Y + D1D2m(O9O)(X)Y + R(x, Y),

where

Let us now assume that we are dealing with canonical coordinates. Then 2tx = m(tx, t x ) = 2tx t2(D,D2m(0,0)x)x R(tx, t x ) so

+

+ 0 = t2(DlD2m(0,0)x)x + R(tx, tx).

Dividing by t2 and letting t + 0 gives (D,D,m(O, 0)x)x = 0. It follows that (_D,D,m(O, 0)x)y = -(D,D,m(O, 0)y)x.If A , B E T,G, then A ( x ) = D2m(x, O)A, B = D,m(x, 0)B so that [A, B]

=

[A, B](0)= (D,D,m(O, 0)A)B - (D,D,m(O, 0)B)A = (2D,D2m(O,0)A)B.

(14.5)

Thus, in canonical coordinates, m(x, Y) = x + y

+ 4[x, Y ] + R(x, Y).

(14.6)

LEMMA 14.10 Let A, B E T,G, y ( t ) = exp tA, b(t) = exp tB, and x(t) = y(t)b(t)y(t)-'b(t)- Define the curve j ( r ) = x(&) for r 2 0. Then (dj/dr)(O)= [ A , BI. PROOF: Consider the function q(t) = m(m(tA, tB), m( - tA, - tB)). Use Eq. (14.6) to show that lim,+o tC2q(t) = [ A , B]. We leave the details as an exercise for the reader. I

SUBGROUPS AND HOMOMORPHISMS THEOREM 14.11 Let G be a Lie group and H a closed connected subgroup. Then H is a submanifold of G and is a Lie group.

PROOF: We outline the proof.

(a) Let 2? be the set of all vectors tangent to curves in H at e. We claim that if A E X , then exp s A E H Vs. We use canonical coordinates, exp: 0 + exp (0).Write (expl8)-' = log. We will also choose a norm 1 1 on T,G. Suppose A E %'. By reparametrizing the curve y ( t ) which determines A , we may assume that lIAI1 = 1. Let 7, = log(?(lln)). Then + 0 and ny, + A , since A = q,,

29 1

ADJOINT REPRESENTATION

(d/dt)(,=, log(y(t)).Since ((All= 1 we get that ~ , , / / ~+~A,.,Choose ~ ~ integers m, so that rnnllYnII + s. Then rn,?,, + sA, so exp(rn,?,) --t exp sA. But exp(rn,y,) = exp(V,)"" = y y n E H so, exp(sA) E H . (b) It now follows from Lemma 14.10 that 2 is a Lie subalgebra of T,G. (c) Write T,G = X @ %. Define 0:T,G G by @ ( A @ B) = exp A exp B. Show that there is an open set 0 , c 8 so that @: 0, -,a(&,)is a diffeomorphism between open sets. (d) We claim 3 an open set 0, c 0, about 0 such that exp(0, n 2)= 0(02 n 2)= @(02) n H . Suppose not. There is a sequence 7, = En 0 fin E 0 so that 7, -,0, a, # 0, and exp(E,) exp(P,,) E H. Thus 5, -,0 and exp(E,) E H . By choosing a subsequence, we can assume that ti,/ 1 a, 1 -,A E X , 1 A 1 = 1. Using the argument given in (a), we see that exp(sA) E H Vs, so A E 2.This is a contradiction. (e) From (d) it follows that (O,,0) is a submanifold chart for H around e. Apply L, for h E H to get submanifold charts everywhere. Let I/ = @(02 n H). Since H is connected it follows that H = V". Thus H is second countable and is a Lie group by Proposition 14.2. [ --f

u;=,

PROPOSITION 14.12 Let G, and G , be Lie groups and h : G , + G , a smooth group homomorphism. Then we have

(a) exp T,h (b) T,h: T,G, 0

=h --t

exp, T,G, is a Lie algebra homomorphism. 0

PROOF: (a) Let A E T,G,. Then y ( t ) = exp t A is a one-parameter subgroup of G I so h y ( t ) is a one-parameter subgroup of G. By Theorem 14.8, 0

h y ( t ) = exp(tB), 0

where B

dl

h y(t) = T,h(A).

=-

dt

(b) follows from (a) and Lemma 14.10.

0

f=O

I

ADJOl NT REPRESENTATION As we did in Chapter 13 for S0(3), we define the adjoint representation of a Lie group on its Lie algebra, Ad: G + L(T,G, T,G) by

Ad(g) = T,-,L, LEMMA 14.13

[A, B ] .

T , Ad: T,G

0

T,R,-

+ L(T,G,

, = T,R,_,

0

T,L,.

(14.7)

T,G) is given by T , Ad(A)(B) =

292

14. LIE GROUPS

INVARIANT FORMS DEFINITION 14.14 The canonical left-invariant T,G-valued l-forrn on G is

defined by

8(g) = TL,-

1.

We wish to calculate do. For 2 E (T,G)* we see 8, = 2 0 8 is a smooth l-form on G. If A , B E T,G, then O,(g)(A(g)) = 2(A) a constant. Thus, by Theorem 9.78(f),

dO,(g)(A, B) = -O,(g)([A, B ] ) =

- 4 [ A BI).

But dO,(g)(A, B) = A(dO(y)(A, B ) ) so, since 2 was arbitrary, we get

dO(g)(A, B) =

-

Q(g)([A,B ] ) = - [ A , B ] .

(14.8)

Let n = dimension of G. By a left-invariant n-form, we mean an n-form o such that L,*o = o Vg E G. We now construct all such. Choose a basis el, el., . . . , en for T,G and let e l , . . . , e" be the dual basis. Define l-forms 2' by Z' = ei 8. A simple calculation gives us L?Z' = Z' Vg E G, 1 I i I n. It follows that, if o = F' A Z2 A . . . A 2", then 0

L,*o = 0 and every left-invariant n-form on G is of the form rw for some number r. REMARK: Clearly w never vanishes and thus defines an orientation for G. Thus, if M c G is a compact n-manifold with boundary p U ( M ) = is defined. We then see that o = Jy L,*w = Jy w, so that pU(g(M))= pU(M),and we can construct, from o,a left-invariant measure. We see that

SS(M)

SM

293

COSET SPACES AND ACTIONS

it is unique up to multiplication by a constant. An analogous construction will give a right-invariant measure.

COSET SPACES A N D ACTIONS If H is a closed subgroup of a Lie group G, then by G / H we mean the set { g H l g E G}. Note that y l H = g2H if and only if g; ‘gl E H . Let n:G + G / H be projection. We define a topology on G / H by: a set 6 c G / H is open if and only if n-’(8) is open in G. Then n is an open mapping and, since H is closed, the topology on GIH is Hausdorff. We want to define a differential structure on G / H . We will use some constructions from the proof of Theorem 14.1 1. X = T,H is the Lie algebra of H . T,G = X 02 for some vector subspace X . 0:T,G -+ G by @ ( A 0 B) = exp A exp B. There is an open set 0 c T,G so that CD: 0 + @(P) is a diffeomorphism and exp(0 n A?) = @(F n A?)= @(0)n H . Now choose an open set V c O(0)so that

d V , V ) c @(0), @ - l ( V ) = w10 w,,

V - ’ = V,

e E 1/,

where W, and W, are open in A” and H , respectively. We may also assume that 0 = 8,08,. LEMMA 14.15

n 0 O: W, -+ n( V ) is a homeomorphism.

PROOF: Let g H E n(V),y E I/. Thus g = @(wl 0w2)= exp w1 exp w,. But exp w 2 E H , so n(exp wl)= n(g) and we get TC 0 @(wl) = n(g). Suppose that n @(wl)= z @(ul). Then exp(-u,) exp w l = h = exp u,, u2 E 0,. Thus, @(w,) = exp w1 = exp u1 exp u2 = @(u, 0u2) so w1 = u1 0u,. This gives u2 = 0 so w1 = u l . We have shown that n @I W, is a bijection. The continuity is clear. For any U c W, open, n @ ( U )= n @(U 0 W2),which is open, since @ 10 is a homeomorphism and n is an open mapping. I 0

0

0

0

0

Thus, the pair (n(V ) ,(n@ I wl)-’) can be used as a chart around n(e)in G / H . To get an atlas we move these charts around using left translations. Let go E G. The mapping Lsn:G -+ G factors through TC to give &,,: G / H + G/H by E,,,(gH) = gogH. Let us denote the chart around n(e),which we just construtted, by (P, I)).For each go E G, let Ps,, = I,,,,(V ) and I)~,, = I) I,,, I .

- -

0

THEOREM 14.16

{(c,,

I)so)lgoE G} forms a smooth atlas and thus de-

fines a differential structure for G / H . PROOF: Note that each L,, is a homeomorphism, so I)s,, is a homeomorphism, It only remains to check the smoothness of the coordinate transformations. We leave this to the reader. I

294

14. LIE GROUPS

The following theorem gives a useful identification of the tangent spaces of G / H . THEOREM 14.17 Let M = G / H , where H is a closed subgroup of the Lie group G . For each g E G, define 4,: G + M by 4g= 'II 0 R,. Then Te4,:TeG + T,,,,M and we have

(a) (b) (c) T,M

ker Te4, = Ad(g)(X), where 2 is the Lie algebra of H, if 'II(s1)= 4 9 ) then Ad(g1)W) = Ad(g)(%) and 4g= 4 9 0 for each m E M there is a natural isomorphism Om:T,G/Ad(g)(%) induced by TeiJgfor any g E 'II-'(~).

PROOF: First note that Ad(g)(A)= (d/dt)l,,,(g

+

exp t A g-'). Thus,

Te4g(AW)A) = (d/dt)lt=o4g(g~ X t PA g-') = (d/dt)lt=on(g~ X LA). P Now if A € 2 ,so that exp t A E H , we get 7t(g exp tA) = n(g), so that But T b ~ g T,+,(Ad(g)A) = (d/dt)l,=,n(g) = 0. Thus Ad(g)(%) c ker is onto so, by dimension arguments (using that Ad(g) is an isomorphism), Ad(g)(%) = ker T&,. Suppose g,H = gH. Then g1 = gh, so that 4g,(z) = zglH = zghH = zgH = +,(z). Thus, Ad(gl)(2) = ker T&,, = ker TeiJg= Ad(g)(2). This establishes (a) and (b) from which (c) follows.

I

We now consider smooth actions of Lie groups on manifolds. DEFINITION 14.18 Let G be a Lie group and M a differential manifold. By a smooth left action of G on M we mean a smooth mapping a: G x M + M so that a(gh, m) = a(g, a(h, rn)) and a(e, rn) = m, Vg, h, m (we often omit the a and simply write grn for a(g, m)).If A E TeG, we obtain a vector field on M by a(m) = (d/dt)l,=,(exp tA)(m).

a

NOTE:

a

a ( m ) = Tla(e, m)(A),so that A -+ is a linear mapping of TeG into

Sh(M). PROPOSITION 14.19

A

[a, s] = - [ A ,

B].

PROOF: We use canonical coordinates on G, local coordinates on M . Then B(z) = D,a(O, z)B, so DB(z)(&)) = D,D,a(O, z)(D,a(O, z)A)B. Also, 4x9 4 Y , 4) = a(m(x,Y ) , 4 gives D,a(x, 4 Y , 4) = D,a(m(x, Y ) , 4 O D,m(x, Y). Thus,

D,D,a(x, 4 Y , 4 ) DlU(Y, 2) = D:a(m(x, Y), z)D,m(x, Y ) O D,m(x, Y) +D,a(m(x, Y), 2) D,D,m(x, Y). O

O

295

COSET SPACES AND ACTIONS

Taking x

=

y

=0

gives

D2Dla(0, z ) D,a(O, z ) = D:a(O, z ) 0

+ D,a(O, z )

0

D,D,m(O, 0).

Let a : G x M + M be a left action of G on M. Ifm E M , we get 4,:G + M by 4,(g) = gm, and if H = { h E G ( h m = m ) , then we get that H is closed and the following diagram commutes:

GiH The orbit of G through m is defined by 0,

=

{gmlg E G).

4,:

G / H + M is smooth and T$, is PROPOSITION 14.20 (a) The map injective. (b) If is a homeomorphism onto 0, then 0, is a submanifold and is a diffeomorphism.

4,

6,

PROOF: 4,: G -+ M gives Te4n,: T,G + TM. In fact T&,(A) = A(m). Clearly A? c ker T&,, where 2 is the Lie algebra of H . Suppose A(,) = 0. Let y ( t ) = (exp tA)rn. For each g E G, define 8,: M + M by 8,(fi) = g k Now y(t) = a(exp t A , m), so that

dy/dt

=

T,a(exp tA, m)TL,,,,,(A).

Also a(exp tA, exp sA, m) = a(exp(t + s)A, rn) and hence (apply (d/ds)l,=,), T2a(exp t A , m)&n) = T,a(exp t A , m)TL,,, &I) so TO,,, ,,(a(m)) = (dy/dt)(t). Thus dy/dt = 0 so y ( t ) = m and A E 2. This gives ker TJ$, = 2. But then, since Ten:T,G -+ T,,,,(G/H) has kernel 2,we see T,,,)dm:T,,,,(G/H) -+ T,M has zero kernel. It now follows easily that Tnc,,6, is injective for all g E G. This proves (a) and (b) follows from Proposition 5.5. I

296

14. LIE GROUPS

EXERCISES 14.1 Let G be a Lie group and h: R G a continuous one-parameter subgroup. Use canonical coordinates to show that k is smooth around 0 and hence, smooth. --f

14.2 Let G be a Lie group. If B , , . . . ,B, is a basis for T,G, define $: T,G + G by II/(CtiBi)= exp(t'B,) exp(12B2). . . exp(t"B,); show that there is an open set 0 c T,G containing 0 such that $: 0 -+ $(O) is a diffeomorphism between open sets. Thus, (II/(O), (I) 10)-') is a coordinate chart for G around e (Hint: Compute Toe$.) 14.3 Let G and H be Lie groups and h: G H a continuous homomorphism. Use the results of Exercises 14.1 and 14.2 to show that h is smooth. --f

+ +

+

14.4 Verify formula (14.6) in the form rn(tA, tB) = t A t B (t2/2) [ A , B ] 0(t3) for the matrix group GL(n, R ) by using the power series for exp A and log(I C), where A and C are matrices.

+

14.5 Let G be a Lie group and g E G be fixed. Use Proposition 14.12 to show that [Ad(g)B, Ad(g)C] = Ad(g)([B, C ] ) VB, C E T,G. 14.6 Let o = 2' A 2' A . . . A F" be the left-invariant n-form on the n-dimensional Lie group G which we constructed in this chapter. Now suppose that G is compact and connected. Let A = o and (3 = A - o.For each k E G let o h : G -+ G by &(g) = kgk-'. Show that

iG

(a) ot(c5)is left invariant, (b) ot((3)= cxh(3 for some number ah, (c) all = JG a,,3 = Jc R t - I(&) = 1 V h E G, (d) R t - I(&) = 6 V h E G. Thus,

(3

is also right invariant.

14.7 Let H be a closed subgroup of the Lie group G and let G / H have the differential structure which was described in the text. Prove that the projection 71: G G/H admits smooth local sections, that is, if g E G, there is an open set W in GIH with n(g)E W , and a C" map II/: W + G such that 71 $(z) = z for all z E W . -+

0

14.8 Use the result of (14.7) to show that iff: G -+ M and F: G / H -+ M are mappings with F n = f , then F is smooth if and only iff is smooth. 0

14.9 With notation as in (14.7), let y: I -+ G/H be a smooth curve and a E I . Let g E G such that n(g)= y(a). Then there is a smooth curve jj:I -+ G such that y = 71 c> and Y(a) = y.

75 Geometrical Models

GEOMETRICAL MECHANICAL SYSTEMS N

Let ( M , w, H ) be a Hamiltonian system. We construct a new manifold = M x R and define a 2-form 6 on N by G=w+dHr\dt,

(15.1)

where t is the coordinate on R . Recall that, if 0 is a 2-form on a vector space V , then ker !2 = { u E VlQ(v, w)

= 0 for

ker !2 is called the kernel of SZ. We may represent vectors in T,,,,,N as u u E T,M, u E R.

all w E V } ;

+ u (a/&) or

as (u, a), where

+

LEMMA 15.1 ker G(m, t ) = ( u X , ( m ) a ( d / d t ) l u E R ) , where X , is the Hamiltonian vector field on M .

+ a d / d t , w + /j ii/i)t) = 0 for all w + fl d/dt,then o ( u , w) + w(u, w) = u dH(w) for all w and dH(u) = 0. Therefore, u-lu = X,,(m), so u + a/& does lie in the span of X,(m) + a/& as desired. It is easily checked that X,(m) + a/& does lie in PROOF: If G(u

p dH(u) - u d H ( w ) = 0 for all w, fl. Then

CI

ker d,so the lemma is proved.

I

Thus, ker 6 is one-dimensional at all points. Now let m(t) be a trajectory of ( M ,w , H ) . Then the curve in N , given by n(t) = (m(t),t), is tangent to ker & (for ri = (n3, 1) = (X,(m), 1) which is in ker 6).Conversely, let n be any curve in N which is nowhere stationary (ri # 0) and suppose that ri always lies in ker (5. If n(z) = (m(z),y ( r ) ) , then ri = (ni, j ) so, since ri E ker 6,we see m = iX,(m) and ri = j(X,(m), 1). Thus j is never 0, so we can choose a new parameter, say t, such that n ( ( ) = (m((),5). Then ri(0 = (ni(<),1) = (X,(m), 1) so m ( ( ) is an integral curve of X,. Thus any nowhere stationary curve in N which is everywhere tangent to ker 6 can be reparametrized to be of the form n(t) = (m(t),t ) , where m is a trajectory of ( M , o,H). 297

298

15. GEOMETRICAL MODELS

Consider equivalence classes of nowhere stationary curves in N , everywhere tangent to ker 6,where curves obtained by reparametrization from each other are equivalent. Then these equivalence classes are in one-one correspondence with the nonconstant trajectories of ( M , o,H). This motivates the introduction of the following concept. DEFINITION 15.2 A geometrical mechanical system (GMS) is a pair ( M , o),where u is a 2-form on M such that dim(ker w ) = 1 for all m E M . A trajectory of ( M , u) is a curve m: I + M for which m(t) is never 0 and h(t) lies in ker(o(m(t)))for all t in 1. If dw = 0, we say that the GMS is closed. If u = -du for some 1-form c1 we say that the GMS is exact. Consider again the GMS ( N , 6)of Lemma 15.1;

N

=

M x R, G =

+ dH A d t ,

where ( M , o,H ) is a Hamiltonian system. Let ?H(mr t ) = X,(m) + a/&, so that X,,(m, t ) E ker 6 ( m , t ) by Lemma 15.1. Then L z H 6 = dizH6 if, dw = 0. Thus, if the dimension of M is 2 4 we get that L*f, applied to, 6,6A 6,. . . , Gfl gives zero. These forms are called the invariants of Cartan. They are invariants in the sense that 4:(Gk) = Gk, where 4s(m,t ) = (t,bs(m,t), t s) is the flow of by Lemma 10.9. Also, suppose that K is any compact 2k-dimensional submanifold of N (possibly with boundary), and let k(s) =

+

x,,

SUK)

+

".

LEMMA 15.3 The quantity k(s) is independent of s; i.e., an invariant of

the flow. PROOF:

Using Lemma 10.9, we get that

LIOUVILLE'S THEOREM We will now see that this result includes the Liouville theorem on the invariance of phase space measure under the Hamiltonian flow. Note that our proof includes the case when the Hamiltonian function depends upon t . Suppose that dim K = 2n and K c M x (0). Let qS:M + M by $,(m) = *s(m, 0).

299

LIOUVILLE'S THEOREM

THEOREM 15.4 (Liouville's theorem) For all s, jGs(K)

= j K wn'

PROOF: We have that j$s(K)o"= f K @(on). But $:(on) I K = 4:(G") the result follows from Lemma 15.3. I

I K , so

REMARK 15.5 One can consider other 2n-forms p = po" such that p = J K p for all s. As above, the condition for this invariance is

Lf,(p6") = 0. Since L x , ~ " = 0, this condition becomes Lf,p of natural phase space coordinates (qi,pi), this is

= 0.

In terms

+

or (@/at) { p , H } = 0, where { p , H } is the Poisson bracket of functions on phase space (see Exercise 7.1 1).

iZ&

REMARK 15.6 If ( N , G)is any closed GMS and 2 is a vector field such that = 0, we get that

Lz6

=

i, d 6

+ dizG = 0,

so Gkis an invariant of the flow of 2 for k = 1 , 2 , . . . . Suppose now that our Hamiltonian system is obtained from Lagrangian mechanics, as described in Chapter 7; that is, M = T*Q. Then there is the canonical 1-form Q on M so that w = a = - dQ and thus, if we take g = 0 - H dt, we get G = -dg. LEMMA 15.7 Let z*: T*Q + Q be the natural projection, so that Tz*: T T *Q + TQ. Suppose that 5 = g + a/&, where E T,,,(T*Q) satisfies Tz*(() = 9;'(a).Then

Bia, t)(?) =

w,3;Y4).

PROOF: Choose natural coordinates and write ci as ( p , q). Then &a, t)r= O(cc)( - H(a, t ) = piqi - H(q, p , t), where (4, q) = Tt*(<). Thus

6@)

= (aL/aqi)qi-

~ ( q4,, t ) = q t , 4 , d ) .

I

COROLLARY 15.8 (a) &(R,(ci, t ) ) = L(t, 9 ; ' ( q , p ) ) for ci E T*Q, t E R . (b) Suppose that y: [a, b] + Q is a curve. Let c(t) = (Tt($(t)), t),a curve in T*Q x R. Then jc = j! (L(t,$ ( t ) )dt.

300

15. GEOMETRICAL MODELS

PROOF: (a) %,(a, t ) satisfies the condition of the previous lemma by Eq. (7.22). (b) Let a(t) = g(i(t)), so that c(t) = (a(t),t ) and take ?= E(t). Then, by ‘ ( a ( t ) ) )= L(t, y(t)). Lemma 15.7, O(i.(t)) = L(t,2’;

REMARK 15.9 Part (b) of Corollary 15.8 suggests that if a GMS (P, w ) is exact, that is, if there is a 1-form a so that w = -da, then its trajectories are obtained from a variational principal. We will now show that this is true.

VAR I AT10 NA L P R I NC I P LES We shall indicate the role of “variational integrals” on an exact GMS. Suppose a is a 1-form on a manifold M . If y: [a, b] -+M is Cm-curve, we define the variational integral corresponding to c1 by

Given y, we want to consider variations of y. A variation (y,) consists of curves Ye:

[a(&), b(41

+

M,

for E in some interval (- r, r), such that tI b(~)} (a) y ( ~ t, ) = y,(t) is a C“-map from { ( E , t)I - r < E < r, a(&)I into M ; (b) ( d y / d ~ ) (t~) , ( d y / d t ) ( ~~, ) C ( E E ker(a) ) when either (1) t = a(&)and C(E)= a‘(&)or, (2) t = b ( ~and ) C(E)= b’(e);

+

(4

Y o = Y.

DEFINITION 15.10 A solution of the variational problem defined by Eq. (15.2) is a curve y: [a, b] + M such that for all variations of y we have

THEOREM 15.11 Let ( M , w ) be an exact GMS, a a 1-form on M such that w = -da, and J the variational integral corresponding to a. If (y,) is any variation of a curve y in M , then

301

FORCES

where t ( t ) = (d/d&)I,=,y,(t).Therefore, y is a solution of the variational problem if and only if y ' ( t ) E ker(w(y(t)))for all t E [a, b]. The proof is left to the reader. We can describe the behavior of measures associated with a nonclosed GMS. Consider ( N , 15)where N = M x R, 6 ( m , t ) = w(m) a(m, t ) A dt, where o is a nondegenerate 2-form on M and a, = a(m, t ) is a 1-form on M for all t. Let Z be a vector field on N such that

+

(a) iz15 = 0, (b) dt(Z) = 1

+

Let 4,(m, t ) = ( ~ ) ~ (t), m t, s) give the flow of Z . Take GS:M $,(m, 0) and cc), = $,*a.

-+

M by $,(m)=

PROPOSITION 15.12 (a) (dids)(w,)= $;(das),

(b) (d/ds)(oS)= k ~ : ( d a ,A w k -'). PROOF: L,o = di,w = -di,(a A dt) = -d((iza) dt - a) = da - d(iza)A dt. Use Lemma 10.7 and restrict to M x (0).(b) follows from (a). I

FORCES The idea of a geometrical mechanical system can be used to treat problems involving nonconseroative .forces. Consider the usual form of Newton's law,

mi dx'ldt

( i not summed),

= pi

dpi/dt = F i ( x , p, t )

for

(15.3)

1I i 5 3n.

On (t,x , p)-space, equations may be written in differential form as dpi - Fi dt dx'

-

(pi/mi)dt

= 0, =0

(i not summed).

(15.4)

A solution of (15.4) is a curve t -+ (t, x(t),p(t))such that the differential forms in (15.4) vanish in directions tangent to the curve. Let (15.5)

302

15. GEOMETRICAL MODELS

6 can be rewritten as

xi

so that (3 is closed if and only if the 1-form a,(x, p ) = Fi(x, p , t ) dx' is closed for each t. For this to happen we must have that F iis independent of p and that the 1-form "(x) = Fi(x, t ) dx' is closed for each t . Thus, the force of an electromagnetic field acting on a charged particle leads to a GMS where the form is not closed when starting from Newton's law on 3space. We will see, however (Example 15.18), that, when written in relativistic form using the Faraday tensor (see Chapter 12),.the electromagnetic interaction can be described by a GMS where the 2-form is closed. Returning now to Newton's law, we know that a,(x) = F i ( x , t ) dx' is closed if there are local potentials, K ( x ) such that Fi(t, x) = -(dK/ax')(x). If the forces are conservative, we have F i= -(dV/dx')(x) for all x and, if we take H ( x , p ) = ((pJ2/2mi) V ( x ) , then 6 = dx' A dp, + d H A dt. Thus we have the GMS ( N , 6)corresponding to the Hamiltonian system ( M , wo,H), as discussed earlier. In many cases a GMS can be viewed as being obtained from a GMS ( N , Go), corresponding to a Hamiltonian system ( M , w o , H ) , by "perturbing" Go. For instance, the form 6 given in Eq. (15.5) is expressible as

xi

xi

xi

xi

+

6 =Go

xi

+

+ EAdt,

xi

(15.6)

xi

where 6,, = dx' A dpi d H A dt, a = - F idx', and H = $( ( ~ ~ ) ~ / r n ' ) . This has the following interesting interpretation. In the given coordinates, the GMS ( N , Go) corresponds to the Hamiltonian system ( M , w o , H ) , where wo = dx' A dp', H = (p')'/2rni. No forces act in these coordinates, since Hamilton's equations give 8' = - d H / d x i = 0. The forces are introduced by perturbing the GMS by forming G as in Eq. (15.6).

xi

xi

THEOREM 15.13 Let ( N , 6) be the GMS corresponding to the Hamiltonian system ( M , o,H ) . Let a, be the 1-form on M for each t E R and let 6 = (3 + a , A d t . Then

(a) dim(ker 6)= 1 at all points, so ( N , 6)is a new GMS; (b) Let 2 = X , + d / d t , 2 = X , + 2, where X , is a vector field on M for each t . Then Z E ker 6 in all cases, while 2 E ker 4 if and only if iXro = a, for all t. PROOF: We first prove (b); then (a) follows (recall that X , and X , are tangent to M and o is nondegenerate). The fact that 2 E ker (3 was proved =X , AZ, as Lemma 15.1. Now 6 = w + dH A dt + a, A d t . Consider

+

303

FORCES

t 2= Y, + p2. Then 6(ti>( 2 ) = ~ ( x+ ,AXH, X + PXH) + dH(Xt + AXH)p

+

+ p x , ) i a,(X, + A X H )-~ a,( + ~ X H ) L Y,) + A d H ( Y ) - /.diH ( X , ) d H ( x r ) p - dH( x ) A + l b x H ) P - a,(y pxH)A = w ( x , , T) a , ( x , + 2XH)P - ‘%,(k; pxH)A. - dH(

= w(X,,

+

+

+

+

+

Suppose 2 = X , + Z lies in ker 4.Then

+ xH)p - a t ( k ; + pxH)

+

w(xt,

=

for all p, Y,. Taking p = 0, we get w ( X , , Y,) = at(Y,)for all Y,, so ix,o= a,. Suppose, conversely, that iXlw= a,for all t . Taking A = 1, we get

cn(2,y + pz)

= w(x,>

X) +

+ xH)p

-

I:+ pxH),

Taking p = 0 gives d(2, y) = w ( X , , Y) - a,(Y,) = 0, since ix,w = a,.Taking y = 0, p = 1 gives &(Z,2)= a,(X, X , ) - a,(XH)= a,(X,) = 0 again, because ixtw= a,.Since any vector is y + pZ for some y, p, we are done. I A

+

Consider ( R 2 ,dx A d p , H ) ,

EXAMPLE (One-dimensional harmonic oscillator)

+ k 2 x 2 ) .The associated GMS is ( R 3 ,(3), where 6 = dx d p + ( p d p + k ’ x d x ) dt.

where H ( x , p ) = + ( p 2

A

A

Let us perturb

(3

by adding the term f ( x ) p dx A dt, so that

6 = d x A d p + ( p d p + k2x d x ) A d t

+ f ( x ) p dx

A

dt.

We compute the kernel of 6,using Theorem 15.13. We have

ker 6 is, as shown before, spanned by p-

a

ax

-

a

a

ap + -.at

k2x -

According to Theorem 15.13, we have ker 4 spanned by p

a --

ax

a k2x ap

+ atd + X , , -

where iXtw= f ( x ) pdx. It is easily seen that X , = - f ( x ) p d/dp, and we get a 2 a a 2=p - ( k x + f ( x ) p )- + -.

ax

ap

at

304

15. GEOMETRICAL MODELS

Any trajectory of the GMS (N, h)can be parametrized by t , i.e., as (x(t),p(t), t). Then @ = - k 2 X -f(x)p.

i= p ,

Thus, trajectories of the perturbed GMS are solutions of X

+f(x)i

+ k2x = 0.

FIXED ENERGY SYSTEMS In certain important situations there is no natural choice of a global time coordinate. The following result says that surfaces of constant energy in a Hamiltonian system have a natural GMS structure. THEOREM 15.14 Let ( M , o,H ) be a Hamiltonian system, E a constant, and S, = { p E MIH(p) = E}. Assume dH never vanishes on S, so S, is a smooth codimension-one submanifold. Then (S,, o)is a GMS. PROOF: Let X , be the Hamiltonian vector field on M . Then dH(X,) = w(X,, X H )= 0, SO X , is tangent to S,. We now show that if X E T,SE, then o ( X , Y) = 0 for all Y E TpSEif and only if X is a multiple of X,(p). For if Y E T,S,, then w(X,(p), Y) = dH(p)Y = 0, which proves half of the assertion. For the converse, suppose w ( X , Y) = 0 for all Y E TpSE.Choose W E T,M with dH(p)W# 0. Then w(X,(p), W ) # 0. Let y = -o(W, X ) / w(l/t:X,) and Z = yX, X . Then if YE TpSE, we get w(Z, Y) = ~ o ( X , , Y) w ( X , Y) = 0, while o(Z, W ) = yo(X,, W ) o(X, W ) = 0 by choice of y. So w(Z, Y) = 0 for all Y E T p M so that nondegeneracy of o gives Z = 0. Hence X = -yX,, as asserted. I

+

+

+

A very nice example of the preceding theorem is the following application to pseudo-Riemannian manifolds. EXAMPLE 15.15 Let ( M , y) be a pseudo-Riemannian manifold and w = Q, the canonical 2-form on T*M. Let E # 0 be a constant, SE = { p E T * M ( g(p, p ) = E}. Then (S,, w ) is a GMS. The trajectories of this GMS cover (not necessarily affinely parametrized) geodesics on M . This last assertion follows from the fact that the trajectories are tangent to the Hamiltonian vector field X,, where the Hamiltonian is H ( p ) = g(p, p ) (see Example 7.9 and Exercise 15.8).

305

CONFIGURATION PROJECTIONS

CONFIGURATION PROJ ECTIONS In cases such as the preceding, we can no longer use Theorem 15.13 to consider the effect of perturbing the 2-form in a GMS. We introduce a new concept relevant to this. DEFINITION 15.16 Let ( M , o ) be a GMS. A smooth mapping onto n: M -r X is said to be a configuration projection if, for every 2-form 6 on X , ( M , w + n*6) is a GMS. REMARK: See Exercise 15.10

EXAMPLE 15.17 Let ( M , y) be a pseudo-Riemannian manifold. We have the GMS (S,, w ) as in Example 15.15. Then z*: S, + M is a configuration projection. To see this let 6 be a 2-form on M . First we argue that I5 = w + (r*)*S is nondegenerate on T,(T*M). This is easily seen by looking at the matrix of the form in a natural cotangent bundle chart. If (qi,pi) are coordinates, then o = dq' A dp,, (z*)*6 = fiij dq' A dq', so w (t*)*6 has matrix

+

which is clearly nonsingular. Now Y -r A( Y ) = (z*)*6(X,, Y ) is a linear functional on T,(T*M) for each p so there is a unique 2,such that

6(2,, Y ) = -A(Y).

(15.7)

Then we have 6 ( X H + 2,Y ) = w ( X H ,Y ) + (z*)*6(X,, Y ) + &(Z,Y ) = w(X,, Y), by Eq. (15.7). Thus, for any Y E T,T*M, we have shown d(X,

+ z,Y ) = W ( X H , Y ) .

(1 5.8)

From (15.8) we conclude first that 2,E T,S,; for this we need w(X,, 2,)= 0. But from Eqs. (15.7) and (15.8), cu(X,, 2,)= 6 ( X H+ Z,, 2,)= 6(X,, 2,) = -d(Z,, X,) = (r*)*6(X,, X,) = 0. Next, 2, X , is in ker d on S,, for if YE T,S,, then & ( X , + 2,Y ) = a,(X,,, Y) = 0. To prove that I5 restricted to T , S , has a one-dimensional kernel, we show that if S(z, Y ) = 0 for all Y E T,,S,, then 2 = A(X, + 2) for some A. This is true because Y+ B ( X , + Z , Y ) and Y-r G(2, Y ) are both linear functionals on T,(T*M) and both vanish on the codimension-one subspace T p S E .Therefore, there is a y such that y 6 ( X I f + 2, Y ) = d(Z, Y ) for all Y E T,T*M. So

+

306

15. GEOMETRICAL MODELS

+ Z ) , Y ) = 0 for all YE T,(T*M). Since ii, is nondegenerate on T,(T*M), we are done.

6(2- y ( X ,

LORENTZ FORCE LAW EXAMPLE 15.18 (Continue the preceding example.) T*: S, --+ M is a configuration projection. If M = R4 and g is the usual Lorentz metric of special relativity, then let

6 = -(q/2)F,, dx'

A

dx',

where ( F J are the components of the usual electromagnetic field tensor. Then, for m > 0, ( S m Z , o+ (z*)*6) is a closed GMS which gives the motion of a particle of charge q and mass m moving according to the Lorentz force law. The proofs of these facts are left as an exercise.

PSEUDOMECHANICAL SYSTEMS We now introduce a more general type of system which is useful in constructing geometrical mechanical systems to describe interactions. DEFINITION 15.19 A pseudomechanical system is a triple ( M , o,H ) , where o is a 2-form on M and H is a smooth, real-valued function on M such that

ker(o(m)) c ker(dH(m))

A trajectory of ( M , o,H ) is a curve y: I

--+

for all m in M . M such that

o(y(t))(j(t),-) = dH(y(t))

for t in I .

If ker(o) is an integrable subbundle, we say ( M , o,H ) is integrable. If d o = 0, we say that ( M , o,H ) is closed. REMARKS 15.20 (a) For any trajectory y(t), H(y(t)) is constant. This follows from ( d l d w w ) ) )= dH(Y(t))(j@)) = w(y(O)(j(t),W ) )= 0. (b) By Theorem 9.79, if d o = 0 and ker(w) has constant dimension, then ( M , w, H ) is integrable. If ( M , w, H ) is integrable, then there is a trajectory through every point. (c) If ker(o) # (0) on M , then there may be many trajectories through a point m E M . If dH(m) # 0 none of these trajectories is stationary at m. A system for which dH = 0 everywhere admits the constant trajectory at every point. If dH = 0 and dim ker w(m) = 1 Vm E M , then the nowhere stationary trajectories of ( M , w, H ) are exactly the trajectories of the GMS, ( M , o).

RESTRICTION MAPPINGS

307

NOTATION:We shall use PMS to mean pseudomechanical system. An IPMS is an integrable PMS.

R ESTR l CTl ON MAP PI NGS A PMS is often only an initial step in the geometrical description of a physical system. A simpler (or a related) PMS, and eventually a GMS, can be obtained using (or breaking) symmetries which occur in the system. Such constructions are determined by a restriction mapping, which we now describe. DEFINITION 15.21 Letf: M + N be a smooth mapping and rn E M . We say that f ( m ) is a proper value off i f f - '(f(rn)) is a submanifold of M . REMARK: By Theorem 3.19, we know that if T,fis surjective for each f(rn) is proper value o f f . A famous theorem of Sard

x ~ f - ' ( f ( r n ) )then ,

implies that iff is onto, then the set of points on N which are not proper values is a set of measure zero. See [32, pp. 45-50]. DEFINITION 15.22 A restriction mapping for an IPMS, ( M , w, H ) is a smooth map p: M + N , where N is a smooth manifold, so that for each trajectory rn: [a, b] + M there is a trajectory % [ a , b] + M satisfying

(a) G ( 4 = m(4, (b) Pu(Gi(t))= P ( r n ( 4 ) V t E [a, bl, (c) z ( f i ( t ) )= n(rn(t)),where n: M + M/ker(w) (see Lemma 9.90 and remark following Theorem 9.91). If n E image(p) is a proper value of p, the system (p- '(n),o l p - ' ( n ) , H lp- ' ( n ) ) is called a p-restricted system of ( M , w , H ) . This is a PMS (see the following lemma). LEMMA 15.23 Suppose p: M + N is a restriction mapping for the IPMS ( M , w, H ) and n E image(p) is a proper value of p. Then for each rn E p-'(n), ker(o Ip- '(n))(m)c ker dH(rn). PROOF: Let rn E p-'(n). If dH(rn) = 0, then the result is clear. Suppose dH(rn) # 0. Choose a trajectory rn: [0, 61 + M such that m(0) = m (Exercise 5.8). There is a trajectot-y K(t) such that G(0) =.m and p(G(t))= p(m) = n for 0I tI 6. Now w(m)(Gi(O), - ) = dH(rn(t)), so G(0) # 0 and, since G(0) E T,,, ( p - ' ( n ) ) , we see that ker(wIp-'(n))(rn) c ker(dH(m)). I REMARKS 15.24 As indicated above, restriction mappings occur in different contexts. In each case, however, the structure of the p-restricted

308

15. GEOMETRICAL MODELS

systems is important. Examples are: (a) If ( M , a,H ) is a Hamiltonian system, then ker(o(m)) = (0) for all m E M , so that k ( t ) = m(t) and thus, p gives conserved quantities. The construction of such a p from a Lie group of symmetries has been described in [l] and [27]. We did this in Chapter 13 for the rigid body and we will describe the p-restricted systems below. (b) If ( M , w, H ) is a Hamiltonian system such that dH(m) # 0 for all m E M , then H : M -+ R is a restriction mapping, each H(m) is a proper value, and the p-restricted systems determine geometrical mechanical systems which give the trajectories of the original system (see Theorem 15.14). (c) If ( M , o,H ) is such that d H = 0 everywhere, then every trajectory m: [a, b] + M is ker(o)-related to the stationary trajectory, fi(t) = m(a) for t E [a, b], and thus, any smooth mapping p: M + N is a restriction mapping. This will arise in the case of gauge symmetries. There, the initial dynamics is only that of symmetry, but a particle will determine a restriction mapping p: M + N so that the p-restricted systems are geometrical mechanical systems which determine the motion of the particle interacting with the gauge fields.

RIGID BODY AND TORQUE We now want to consider the rigid body in relation to our discussion of forces and restriction mappings. The model we used (in Chapter 13) for the rigid body was (G x T,G, w, H ) where G = S0(3), w = - dg, &A, B)(k,B) = h,(B, K'k) and H ( A , B ) = $he(& B). See Chapter 13 for details.

where p: G x T,G + T,*G, given by p ( A , B)C = (Ad*(A)B,)C, is the momentum mapping (see Chapter 13). We leave this proof and many other calculations in this discussion as exercises. The momentum mapping p: G x T,G + T,*G is a restriction map. For each p o E TZG, consider the system ( M p o ,W I M , ~ )where , M,,

= P-YPo) =

((4 B)JB= (Ad*(A-%oY}.

Let Go = { H E GIAd*(H-')pO = po}. If ( A , B) E M,, and H E Go, then &,(A, B) = ( H A , B) E M,, ,since Ad*(A - ' H - )po = Ad*(A- ') Ad*(H- ' ) p o =

'

309

RIGID BODY AND TORQUE

Ad*(A-')po

=

B,. Thus, we have a group action Go x M,,

--$

Mp,.

LEMMA 15.26 On MWowe have that

ker

( A , B ) = q,&G0(A, B)),

where Go(A,B) is the orbit of ( A , B ) under Go in M P o . The proof is left as an exercise. We know that each group G o is a circle group consisting of rotations about some fixed axis. Thus the restricted systems (M,,, oIM,,,, HIM,,) have a one-dimensional degeneracy along the orbits of Go. One can show that the quotient structures (MI,,,/Go,0, H), which are again Hamiltonian systems, are those giving the body motions as described in Chapter 13. Finally, we wunt to consider an external force (torque) acting on the rigid body. We take the G M S associated to the Hamiltonian system (G x T,G, 0 , H). Thus we form G x T,G x R , and use the 2-form cij = w + dH A dt. Let G, be a smooth time-dependent l-form on G. Perturb cij by adding on - G, A dt. Thus we have, on G x T,G x R, the 2-form Q = cij

-

G,~dt.

We have Q(A, B, t ) ( k ,B, $(A", B, f)= - d p ( k , @ ( A , B)(A"A-')+h,(B, A - ' k ) + he(& B ) T - he(& B)t - G,(A)kT G,(A)Ai, where we have used Lemma 15.25. Fix (A, B, t ) and (A, B, t') with t # 0 and suppose d(A, B, t)(k,B, t) (A", B, f) = 0 for all (A", B, f). Taking, successively, B = f = 0, A" = f = 0, A" = B = 0, gives -dp(A, B)(k,B)(A"A"-') G,(A)A"t= 0, (15.9)

+

+

h,(B, A -

'k)- h,(B, B)t = 0,

( 15.10)

h,(B,

B)f - G,(A)Af= 0.

(15.11)

Now p(A(t),B(t))C = h,(B(t), A (t)- ' C A ( t ) ) ,so

(g)

p(A, B)C = he(& A - l C A )

+ h,(B, - A - ' k A - ' C A ) + he@, A - l C A ) ,

which gives

Thus, dp(A, B)(k,E)(A"A-') = he(& A-'A") + he(& [ K I A " , K'k]), so that dp(A, B)(k,B)(kA-') = he(& A - l k ) . Therefore, (15.11) follows from (15.9) and (15.10). If we assume t= 1, then (15.10) gives B = A - ' A and (15.9)

310

15. GEOMETRICAL MODELS

becomes dp(A, B)(& B)(AA-') = G,(A)A

for all A.

(15.12)

Let F,(A) = G,(A)TR, be the force form pulled back to T,G using the space representation (see Chapter 13). Then (15.10) and (15.12) combine to give

d

- p ( ~ ( t )A, -

dt

' ( t ) k ( t ) )= F , ( A ( ~ ) ) for all t.

(15.13)

Thus, (15.13) is the equation of motion for a rigid body with torque (compare with Theorem 13.14).Note that (15.13) is just the statement that the rate of change of angular momentum equals the total external torque.

GAUGE GROUP ACTIONS We now introduce the concept of a gauge symmetry which was referred to in Remark 15.24(c). DEFINITION 15.27 By a gauge group action on a PMS, ( M , w, H), we mean a smooth left action of a Lie group G on M such mat

(a) each orbit is tangent to ker(w), (b) H is invariant under the action of G, (c) azw = o for all g E G, where a,: M + M is the action of g on M ; that is, w is a G-invariant form. LEMMA 15.28 Let ( M , w, H ) be a PMS with a gauge group action by G. If m: I + M is a trajectory and g is any smooth curve in G defined on I , then k(t)= g(t)m(t)is also a trajectory of ( M , o,H).

PROOF: Let a:G x M + M be the group action. Then f i ( t ) = a(g(t), m(t)), so that dfi/dt = T,a(g(t),m(t))g'(t) T,a(g(t),m(t))m'(t).But T,a(g(t), m(t)). g'(t) is tangent to an orbit and, thus, is in ker(o(fi(t))). Therefore, for any v E T,,,,M, we have w(&(t))(&'(t),v) =w(&(t))(T,a(g(t), m(t))m'(t),v)= o(m(t)). (m'(t),(Ta,~,,)-'u)=dH(m(t))(Ta,~,,)-'v=dH(fi(t))u. I

+

THEOREM 15.29 Let ( M , o,H ) be an IPMS with a gauge group action by a Lie group G . Suppose G acts on the left on the manifold F and assume each orbit is a submanifold of F. Suppose v: M + F is a smooth map such that

(a) v(gm)= gv(m)for all g E G, m E M ; (b) if m: I -+ M is a trajectory of ( M , o,H ) , then {v(m(t))lt E I } is contained in some orbit. Then v is a restriction mapping.

31 1

MOVING FRAMES AND GEODESIC MOTION

PROOF: Let m[a, b] --* M be a trajectory of ( M , o,H ) and u, We have the subgroup

G,,

=

(g E G l p ,

=

v(m(a)).

= u0>.

If Go, is the orbit of u,, then there is a commutative triangle

GIG

!I(,

where a&) = gu,, Oi, is the induced map on the quotient and p is the natural projection. Now Gu, is a submanifold of F and B,, is a diffeomorphism, so that t -+ Oi,;'v(m(t)) is a smooth curve in G/G,,. By Exercise 14.9 there is a smooth curve y: I -+ G such that y(t)uo = v(m(t))and y(a) = e, e being the identity element in G. Then f i ( t ) = y ( t ) - 'rn(t)is a trajectory by Lemma 15.28. Since y ( t ) can always be joined to e by a smooth curve and since G acts as a gauge group, we see n(m(t))= n(G(t)),where n: M -+ M/ker(w) is the projection. Finally v ( f i ( t ) )= v(y(t)-'rn(t)) = y(t)-'v(m(t))= u,. [

MOVING FRAMES A N D GEODESIC MOTION We now use the ideas developed previously to study, first the properties of geodesic motion on a pseudo-Riemannian manifold, and then more general particle motions. We have previously seen that geodesic motion can be described in terms of geometrical mechanical systems (Example 15.15). The development we now give obtains these systems as the final product of more basic physical-geometric objects, since it is our first example of a particle interacting with a gauge field. Recall that in our study of the rigid body we used an equivalent model in which the metric is "constant." For a general pseudo-Riemannian manifold this can be done only locally, using the notion of moving frames. Let ( M , g) be a pseudo-Riemannian manifold of dimension n. Let ( e l , . . . , e,) be the standard basis of R" and let h be the metric on R" for which h(ei, e j ) =

i

0 1

-1

Thus h is a metric of index n index of the metric g on M . For rn E M we consider Lfrn =

{g:R" + TrnM(a is

-

ifi#j, if i = j , 1 5 i 5 p , if i = j , p + l < i < n . p (see Definition 6.6). We take n-p to be the

a linear isomorphism and o*g(rn)= h } . (15.14)

31 2

15. GEOMETRICAL MODELS

Define Y =

u Ym,

(15.15)

msM

n : Y + M,

.(a)

if

=m

OE

9,.

(1 5.16 )

We now introduce a construction by which we make 9 into a smooth manifold in a special way. This is analogous to our construction of the tangent and cotangent bundles in Chapter 5. The result of our efforts will be our first example of what is known as a principal jiber bundle. LEMMA 15.30 For each mo E M there is an open set V containing m,

and a diffeomorphism

p: V x R" + TVsuch that

(a) for each mE I/, Pm:R" + TmV defined by isomorphism, (b) B;g(m) = h, for all m E V .

pm(t)= P(m, 5) is a

linear

PROOF: Construction of p is equivalent to construction of n (2"'-vector fields tl, . . . , t,, on V such that at each m e V we have

.I i

g(m)(ti(m),S j m ) ) =

0, 1, -1,

i #j, i =j , i=j,

The construction is outlined in Exercise 15.3.

l l i l p , p 1 I i I n.

+

I

A basis ( u l , . . . ,u,) for Tm0Msuch that

0,

g(mo)(ui,u j ) =

1,

-1,

i #j, i =j, l l i l p , i=j, p+l
is called an orthonormal frame at m o . Specification of an orthonormal frame is equivalent to specification of an isomorphism Om,: R" -+ TmoMsuch that B;,(s(mo)) = h. A local orthonormal frame about m, consists of n Cm-vector fields t l , .. . , t,, defined on a neighborhood V of mo, such that for each m in V , ( t l ( m ) ,. . . , t , ( m ) ) gives an orthonormal frame at m. As we have noted, this is equivalent to specification of a diffeomorphism p as in Lemma 15.30. We will often refer to ( V , B) as a local orthonormal frame. DEFINITION 15.31 O(n, h) consists of those transformations A in GL(n) which preserve h, that is, h(Au, A w ) = h(u, w) for all u, w in R". If ( V , j) is a local orthonormal frame, define fi V x O(n, h) + n - l ( V ) by P(m, A ) = p,

0

A.

31 3

MOVING FRAMES A N D GEODESIC MOTION

It is easily checked that manner.

fl maps

V x O(n, k ) onto n - ' ( V ) in a one-to-one

THEOREM 15.32 The family of "charts," { ( K 1 ( V ) , p - ' ) [ ( V ,p) is a local orthonormal frame} determines on 2? the structure of a smooth manifold such that (a) n : 2 --+ M is a smooth surjection with smooth local sections, (b) for each ( V , p), n - ' ( V ) is open in 2? and fi V x O(n, k ) - + x - ' ( V )is a diffeomorphism, (c) there is a smooth, free, right action of O(n, k ) on 2 by (CJ,A ) + CJ A . 0

PROOF: The proof, which is very similar to those given for the tangent

and cotangent bundles is left as Exercise 15.4.

I

DEFINITION 15.33 With the structures described in Theorem 15.32, (3, n, M ) is called the orthonormal ,frame bundle of ( M , g). DEFINITION 15.34 The canonical R"-valued 1-form 0 on 2 is defined by O(a) = CJ-'Ton for CJ E 3. 0

Note if CJ E n-'(m), then T,n maps T O Y to TmM.Then CJ-' maps TmM to Ton:T O Y+ R" is a well-defined linear map. R", so that CJ-' Now suppose CJ:V + Y is a smooth local section of Y over V . Note that CJ-', defined by (cJ-')(x) = (cJ(x))-':T,M + R", may be viewed as an R"valued differential l-form defined on V . We define 6: V x (R")* + T*V by 0

6(m, A)(v) = ~

Now if H : T * M

+R

C J -' ( m ) ~ ) .

( 15.17)

is defined by H(P) = hY(z*(p))(p,P),

then ( T * M ,ID, H ) is a Hamiltonian system and its trajectories cover geodesics in M . Here, as usual, o is the canonical 2-form on T*M. So, for V open in M , we have the Hamiltonian system (T*V, o,H). Given a section CJ as above, we use 6 to construct an equivalent model for (T*V,o,H). Now the canonical l-form a on T*M is given by ~ ( p=) p 0 Tpz*,where p E T,*M and T,z*: T p ( T * M -+ ) T,M is the tangent of the natural cotangent bundle projection z*:T * M + M . Since z* 0 6 = n l , n l : V x (R")*-+ V the which shows that 6, = 6*a is given first projection, we get Tz* Td = TTC,, by (15.18) 8,(m, A ) ( h , 1)= A(CJ(m)- l(h)). 0

By a slight abuse of notation, we write (15.18) as

O,(m, A) = 2 CJ-'(m). 0

(15.19)

314

15. GEOMETRICAL MODELS

Now if w,

= 6*w,

we then have W, =

-do,,

so that w,(m, A)((k,A), (fi,1))= I"a-'(m)h - h - ' ( m ) f i - A d(o-')(m)(m, hi). ( 1 5.20) We recall that 6'is an R"-valued 1-form, so that d(o-') is defined. These remarks are summarized in the following theorem. THEOREM 15.35 S,= (V x (R")*, w,, H) is a Hamiltonian system whose trajectories cover geodesics on V , where

(a) o,(m, A)(k,A), (fi, I ) )= Xo-'(m)m - Lo-'(rn)fi - A d(o-')(m)(m, G), (b) H(m, A) = #*(A, A), where h* is the metric on (R")* corresponding to h on R".

Let O(n, h) be the Lie algebra of O(n, h).

BASIC THEOREM LOCAL (L E MMA 15.36) LEMMA 15.36 For each local section valued 1-form 6, on V such that

on V there is a unique o(n, h)-

0

d(o- l)(m)(k, 6)= ( d u ( m ) f i ) ( 6'k) - (6,(m)m)(o- %). PROOF: We fix m in V. Then for m, hi in T,V we define

d ( 6 ' ) ( m ) ( k ,fi)b= T a p Y ( C 'riz)p(o-'hi)?dx". Here b is the operation of lowering indices in R" using the metric h. Taayis skew-symmetric in (B, y) since d(o- ' ) ( m )is a 2-form. ~ e f i n e6 a p y = + ( T a p 7 + pya + ~ y p a ) . NOW $Buy = + ( ~ p u y+ ~ u y p+ Tyap), so we see 6,,, is skew-symmetric in (a, fl). Thus we get a O(n, h)-valued 1-form 6,(m) on T,V by I

I

I

( 6 , ( m ) ~ ) u= 6;y(o-

liqup

a/axa.

The fact that &,(~-'k)~ lies in O(n, h) follows since in (a, B). Now, as one easily checks,

zapyis skew-symmetric

I

z a p y - day, =

Tapy,

so that we have d(o-')(m)(k, 6)= %,(o- ' m ) p ( f J - ' f i ) y = (6,(m)fi)o-'k

-

-

qp(o-'fi)y(o-'m)fl

(6,(m)m)o-'fi.

31 5

BASIC THEOREM GLOBAL (THEOREM 15.39)

This gives existence. For uniqueness, we note that any 6, with the required properties gives rise to a 8a,y satisfying

COROLLARY 15.37 For any a we have

o,(rn,i)((k,11, (fi,1))= k ' ( r n ) v i z

-

ia-'(rn)fi

+ 16, A a-'(rn)(ni,

fi).

We have thus described geodesic motion, using moving orthonormal frames, in terms of W,

+ A(6,Aa-').

= - & A 6 '

(1 5.21)

In Eq. (15.21), A is interpreted as the identity map on R" and d(a-') = - 6 , ~a-', as we have seen. In the model X u ,the Hamiltonian is independent of rn, so all forces are encoded in the form w,, just as in the case of the rigid-body model previously discussed. The term 16, A D - ' can be interpreted as representing forces in the given moving frame which change the form from its "inertial" value -dA A a-'. Note that o, = -dA A 6'if and only if d ( 0 - I ) = 0. We now show that the Hamiltonian systems X u associated to local orthonormal frames can be obtained from a single integrable pseudomechanical system with a gauge group action. Let A = 9 x (R")*.Define an action of O(n, h) on A by g ( 0 , A ) = (erg-', A g - ' ) . 0

Define & A+ T * M by

4(a, A ) = A

o

D-'.

PROPOSITION 15.38 (a) 4 induces a diffeomorphism $: A / O ( n , h) + T*M; (b) if CI is the canonical 1-form on T * M , then 4*a(o, A) = i O(a), where 9 is the canonical R"-valued 1-form on 2; (c) for g E O(n, h), let Ag:A + 4 by &(a, A) = (as-',Ag- '); then Af(4*a) = 4*a, that is, 4*a is invariant under the action of O(n, h); (d) if w = -d(d*a), then for each (a, A) in A, we have T(O(n,h)(o,A)) = ker ~ ( aA);, (e) taking R(o, A) = th*(A, A), we get an IPMS (A,o, with a gauge group action by O(n, h). 0

n)

PROOF: The proof is a routine calculation which we leave as Exercise

15.6. [

31 6

15. GEOMETRICAL MODELS

BASIC THEOREM GLOBAL (THEOREM 15.39) THEOREM 15.39 There exists a unique o(n, h)-valued 1-form 6 on 9

such that (a) 6, = a*6 for any smooth local section a, (b) 6(a)(tc(a)) = C for all C in O(n, h), where &-(a) = (d/dt)l,,,a exp(tC), (c) R,*6 = Ad(g- ')6, where R,: Y Y is defined by R,(o) = a . g, (d) T,Y = ker 6(0)0ker Tunfor all a E 9, (e) w = (- dA As)A 0, where w = - d($*cc) and A is interpreted as the @")*-valued 0-form on A given by (a, A) -,1. --f

+

m, = ~(0,). Choose a local section PROOF: We first define 6. Pick o, E 9, a: V --f 9 with a(mo) = go. Now a(V) is a submanifold of 9 and it is easy 0ker Tqon.So we define 6(0,) by to see T,,Y = Tu0(a(V)) 6(oo)(Trno4t)+ ~ c ( o o = ) ) du(mo)t

+ C.

(15.22)

Here we have used the fact that C -,&-(ao) gives an isomorphism o(n, h) -, ker T J K . We must show this is independent of the choice of a. Suppose 0': V' -+ 9 is another section with a'(m,) = a,. Then there is a smooth map $: V n V' -+ O(n, h) such that o'(m)= a(m)$(m)for all m E V n V'. Note that $(m,) = e, e being the identity in O(n, h). Let p: .9x O(n, h) -,9 be the action. Then p(p7 gh) = p ( p g , h). Therefore,

so, taking h = e, we get

where

31 7

BASIC THEOREM GLOBAL (THEOREM 15.39)

Thus, d(o'-') = -(I,-'

d$

+ I,/-'C ~ ~ $ ) A C J ' - ~ ,

so, by Lemma (15.36) we get 6,. = I,-'6,t+h + I/-'d+. Evaluating at mo, where $(mo)= e, we get T,p(ao, e ) = I

=

(1 5.26)

identity,

6 U b O ) = fi,(mo) + d*(mo). So if 6' is defined by Eq. (15.22), using CJ'instead of CJ, we get

~'(CJdTrno4t) + ~ C ( C J O )=) 6'(ao)(Trno'(t)- tccrn)(CJo)+ SC(CJO)) = 6,.(mo)< - C(rno) C

+

+ d*(mo)t - d*(mo)t + c = 6,(mo)S + c = J(CJO)(Trnoo(t)+ S ~ C J O ) ) . = 6,(mo)5

This concludes the proof that 6 is well defined. By the Eq. (15.22) defining 6, we immediately see that (a) and (b) hold. And (a) and (b) imply Eq. (15.22) so that 6 is uniquely determined. It now remains to be verified that (c), (d) and (e) hold. Consider (c); R;(CJo)(Trnoo(S) + SC(CJO))= 6(CJog)(TRgTrn,o(t)+ TR&SC(CJO))). If a'(m) = c(m)g,then T R g T r n o a=( ~Trnoo'(S). ) Also,

This proves (c).

31 8

15. GEOMETRICAL MODELS

Consider (d). If u lies in ker 6(oo) n ker Tu0n,then by (15.22) 6(o,)u = C = 0, where u = tc(o0). Thus u = 0. Now counting dimensions establishes (d) since, by (b), 6(c): T , 2 + O(n, k) is onto. Consider (e). Recall that 8 is defined by 8(o) = o-' Tun,where n:2 + M is the projection. We first show d8 = -6 A 8. Let o be a local section with o(mo)= go.Then o* d8 = -a*(6 A 8), because o* d8 = d(o*8) = d ( 0 - l ) while - o*(6 A 8) = -o*6 A o*8 = -6, A o-' and Lemma 15.36 then shows o* d8 = -o*(6 A 8). So 0

d6(4m))(Trno(u),Trn4w))= -(6 A @(o(m))(Trno(u), Trno(w)). It is easily argued that this implies d8 = -6 A 8. Then, since by Proposition 15.38(b)we have w = -d(q5*cc) = -d(I A 8),we get w = -dA A 8 - I A d8 = (- d l Ad) A 8 which proves (e). I

+

COROLLARY 15.40 Let o: V -+ 2 be a local orthonormal frame. Define 8: V x (R")*+ A'by 8(m,A) = (o(m), A). Then

(a) 6, = B*6, (b) W , = B*o, (c) f I o 8 = H . Thus, the Hamiltonian system 2, is obtained naturally from the IPMS (A,0,fI). DEFINITION 15.41 A connection on the frame bundle 2 is an o(n, k)valued l-form on 9 satisfying

(a) 6(4(tc(4) = C for C E o(n, h), (b) R,*6 = Ad(g-')S for g E O(n, h). Let o E 2, ( E T,,,,M. Then there is a unique

5h

in T , 2 such that

(1) Tun(5h)= 5, (2) t h E ker 6(o).

5, is called the horizontal lift of 4 at o. REMARK 15.42 The particular 6 constructed in Theorem 15.39 is uniquely determined by the fact that it is a connection and satisfies d8 = - 6 ~ 8 .To see that 6 is uniquely determined by these conditions, assume 6 is an O(n, k)-valued l-form on 2 which is a connection and which satisfies d8 = -6 A 8. For each local section o, define 6, to be o*6. We show d(o-') = -6, A o-', which shows 6, is uniquely determined. Since d8 = -6 A 8, we get o* d0 = -a*6 A a*8. Since a*@ = o- ', we see d(o- ') = - 6, A o- ', as asserted.

31 9

PRINCIPAL BUNDLE MODEL USING A SPECIAL FRAME

LEMMA 15.43 The projection n2:A

--*

(R")* is a restriction map.

PROOF: We show the hypotheses of Theorem 15.29 hold. O(n, h) acts on (R")* by (9, A) + A 9- '; n 2 ( g ( C , A)) = n21og-', Ag- ') = Ag-' = gn2(a, A). Now let rn: I + A be a trajectory of (A, o,H ) . Write m(t) as rn(t)= (a(t), A(t)). Then ( - d l Ad) A 0((6,A), (6,x)) = h*(A, 1)for all 6,1.Taking 6 = 0 gives 0

+

Xe(6) = h*@, 1)

for all 1,

so that H(6) = 2.

Taking

=0

gives &(6) = A(S(6)0(6)- 8(30(6)).

(15.27)

Choose 6 # 0 with 6(6)= 0. Since TOY= ker 6(a) 0 ker Ton,we can choose 6 with 6(6) = 0 and 0(6) an arbitrary point in R". Then Eq. (15.27) implies 1 = AS(&). Then it follows that (d/dt)h*(A, A) = 2h*(A, 1)= 2h*(A, l s ( 6 ) )= 0, since h*(A, A6(6))= - h*(A6(6),A) = - h*(A, Ad(6)). Thus 1lies in a level surface of h*. But any two vectors in the same h* level surface can be mapped to one another by an element of O(n, h). That is, all A(t) lie in one O(n, h)orbit. I We now consider the systems which are n,-restricted (see Definition 15.22 and Remark 15.24). Then For each A in (R")*,let A,= n;'(A), o,= olAn,and E , = g1A2. AA= Y x {A} E 9.E , is a constant and o,= A ~ 8A since dA = 0 on A,. Let G, = {g E O(n, h)lAg-' = A}. Then the standard right action of O(n, h) on A induces an action of G, on A2.

PRINCIPAL BUNDLE MODEL USING A SPECIAL FRAME THEOREM 15.44 For each A in (R")*,let Y 2= ( 9 , A d A 0, EL). Then 9, is a PMS with gauge action by G , . Furthermore,

(a) for each cs E 9 we have ker A6 A 0(o) = {(yo(l?)),, + <Ja)ly

E

R, C E T,G,},

',

(b) if 42:9 --t T*M is given by 4,(a) = A o 6 then 42induces an isomorphism between Y J G 2 and the GMS of energy E , in T*M.

320

15. GEOMETRICAL MODELS

+

PROOF: Let o E 2.Choose d E T U 2 and write 6 = d,, du, where dh E ker 6(o)and 6"E ker T,n. Suppose A6 A d(d, a") = 0 for all 6 E T u 9 .Then

lJ(d)e(a")- A6(a")O(6) = 0 for all a". Choose 5 so that a" E ker Tun.Then A6(6)8(&) = 0 for all such a". By (b) of Theorem 15.39, we conclude U(6)= yAb

for some y E R .

Now choose 6 to be in ker 6(o). Then 16(d)O(a")= 0 for all such 6. By (d) of Theorem 15.39, this implies U(d) = 0. Then it follows that 6(6)E T,G,. But e(6) = e(dh) SO dh = y(o(/2')), and 6(d) = 6(d,) SO that 6" = tC(O) for C = 6(d). This proves (a) and then (b) follows from Proposition 15.38. I Thus, the n,-restricted system determined by A E (R")* is isomorphic to the frame bundle 9 with the 2-form cod = A6 A 8 constructed from 6,8, and A. There is a "residual" gauge group action which, when factored out, gives the GMS which describes the motion of a particle with energy E , . We see that we can obtain the system Y, without the use of a Hamiltonian function H . In fact, 9,is determined by two actions of the group O(n, h): (a) the gauge group action on (4, o,Ed), (b) the standard action on (R")*. These actions determine the restriction map and lead to Y,. This is what o,G,) there is only the motion is referred to in (c) of Remark 15.24. On (A, of gauge symmetry, while in Y, we see ker o,contains a "residual" gauge action plus one extra dimension which determines the dynamics of the interaction. COROLLARY 15.45 For each A in (R")* define a vector field 2, on 2 by Z,(a) = (o(A'))h. Then if Ed > 0, 2, projects under +,:2 + T*M. to the Hamiltonian vector field on S,, c T * M . PROOF: This follows from the preceding theorem and the fact that M(Z,) = A(2) = 2E,. I REMARK 15.46 The integral curves of the vector fields Z , can be used to describe the motion of the tangent space along a geodesic under parallel with .(ao) = m(0) transport. Suppose m(t) is a geodesic in M . Choose oo E 9. and A E (R")*such that 2' = a; '(rh(0)). Let o(t)be the integral curve of Z , in 2 with a(0) = oo. Then o;'(rh(O)) =' A = O(d(t))= o(t)-'m(t). So if we define T(t):Tm,,,M -, Tm,,,M by T(t)= a(t)a;', we have k(t)= T(t)(m(O))." The transformations T(t)give parallel transport along the geodesic. The general definition of parallel transport determined by a connection, along any curve, will be given in Chapter 16.

321

THE SOURIAU EQUATIONS

THE SOURIAU EQUATIONS According to the general theory of relativity, the motion of a particle interacting with a gravitational field may be described as geodesic motion in a pseudo-Riemannian manifold of dimension four and index three (the spacetime manifold). We now obtain the equations of Souriau [28] for the motion of a charged particle with spin in a gravitational and electromagnetic field. This is an extension of the preceding discussion of geodesics, with h being the Minkowski metric on R4. In relativity theory, it has become standard to write the standard basis of R4 as e,, e l , e 2 , e3 so that k(e,, e,) = 1, h(e,, ei)= - 1, i = 1, 2, 3, and k(e,, e,,) = 0 for a # p. The group O(4, h) is called the Lorentz group and will, in the following discussion, be denoted by G. Also n : P + M is the Lorentz frame bundle for ( M , 9). We will use the components ha, and hap of the Minkowski metric where, as usual, ha,hDV= 8;. These will be used to raise or lower indices on tensors on R4. Let 9 be the Lie algebra of G. Either G or 9 can be viewed as matrices which operate on R4. That is, if A E G and T E 9, we have, for u E R4, Au = A(u"e,) = (AapuS)ea and Tu = T(u"e,)= (Ta,uB)e,. NOTE: When we write a (:)-tensor we offset the indices, writing T", rather than T;. This is done so that we can keep track of indices when raising or lowering is used. In the case of 8; there is no problem since if we lower or raise an index we get ha, or hap which are symmetric. LEMMA 15.47

Let A

= (A",) E

G. Then (A-')",

=

A,".

PROOF: Since A E G, we have h(Au, A w ) = h(u, w )for all u, w E

R4.Thus h,,

= k ATpApu, so that ha, = 8: = h A',hP"ABu= A,"Aou. This shows that the ? r! matrix C,where C",= A,", is the inverse of A , as desired. I

The action of G on R4 induces an action on (R4)*so that contraction is invariant. This action is the usual one given by (Ap)u = p(A- 'u).

To summarize, the actions of G are given as follows: let A

E

G, u E R4,

P E (R4)*; (Au)" = A",ufi,

( A p ) , = A,Bp,.

(15.28)

These actions induce actions on tensors of arbitrary type. For example, if L has components Lapy,then (ALyBy= AapApuAYrLPur. Such actions commute with contraction and with raising and lowering indices.

322

15. GEOMETRICAL MODELS

STRUCTURE OF THE LIE ALGEBRA OF THE LORENTZ GROUP Now G is a group of matrices, a subgroup of GL(4, R). G is determined by the requirement that it preserve the Minkowski metric h. Then the Lie algebra Y is naturally identified as a subalgebra of the Lie algebra of all 4 x 4 real matrices. This was discussed in Chapter 14. The condition that a matrix T be in Y is that, for all u, w E R4, h( Tv, W)

=

-

h(0, Tw).

(15.29)

In component form (15.29) becomes

+

ha8TapuPWa h,,VPTuSWB

= 0.

Then (To,+ T,,) = 0. In fact we have LEMMA 15.48 (T",) -+ (Tu,) is a vector space isomorphism @:Y -+

A2(R4). It satisfies @(ATA-') = A@(T)for A

E

G.

PROOF: (ATA-I)", = A",TP,(A-')", = A',TP,A," so that @ ( A T A - l ) u ,= A,,TP,A," = AaPT,,A,". But (A@(T)),,= AaPABaTpU.I

CONSTRUCTION OF A GAUGE INVARIANT 2-FORM REMARK 15.49 The isomorphism of Lemma (15.48) allows us to define, for S, T E A2(R4), ST, TS, and [S, TI = ST - TS. Let

N,

=

{ ( p , S) 10 # p

E

(R4)*, S E A2(R4), Spvpv= 0, and SpvSpv= s2>,

where s is a fixed number. For ( p , S ) E N,, A E G, A(p, S) = (Ap, AS) is again in N , , since the group action commutes with contractions. Thus, if A s= 2 x N , c 9 x (R4)* x A2(R4), we have a left action of G on Asby A(a, p , S) = (crA- ', Ap, AS). We now want to define a 2-form o on A,which will lead to the motions of a charged particle with spin in an electromagnetic and gravitational field, and is such that the action of G is a gauge action. We know that o should contain a term - d ( p e ) as already described for the case of zero charge and zero spin (here we are using p for the identity mapping on (R4)* instead of A). Also, recalling Example 15.18, there will be a term -qp( , ), where q is the charge of the particle, F = fi*(F), ii is given by fi(o,p , S) = n(a), and F is the 2-form on M describing the electromagnetic field (see Chapter 12). Now for (a,p , S) E As, 6(a) takes values in Y which is isomorphic to A2(R4) (Lemma 15.48), and S E A2(R4) N (A2(R4))* by This is the same contraction, so that S . 6 is a real-valued l-form on A,.

323

CONSTRUCTION OF A GAUGE INVARIANT 2-FORM

type of term as p .O. In fact, the use of the Poincare group (the Lorentz group together with the translation group of R4)would give p . 8 + S . 6 (see Chapter 16). We thus include a term -d(S . 6) in o.So far we have w = -d(pH) - qF( , ) - d(S .6). For this o,however, G is not a gauge action. We will show that there is a number CI such that

s)(cf,P, S)(6,6,S ) = - d ( m c f ,

Pw,p") - q & w , 6) - d(S . 6)(0, S)(cf, $(6, g) CIS.

o(~ P,,

+

[s,$1

(15.30)

defines a close 2-form on A,such that the action of G is a gauge action. First we need some preliminary lemmas. LEMMA 15.50 (a) Let o be a 2-form on a three-dimensional vector space E . Then o = CI A fl for some 1-forms CI and fi. (b) If w is a 2-form on a four-dimensional vector space E and there is a nonzero vector u E E so that i,w = 0, then o = C I Afl for 1-forms CI and 8. PROOF: (a) If w = 0 then o = 0 A fi. Suppose o # 0. Choose el, e2 so that w(e,, e2) = I. Choose f 3 so that {el, e 2 , f 3 } form a basis for E . Let e3 = f 3 - o(f3,e,)e1 + 4 f 3 , e,)e2. Then w(el, e3)

= o( e1,f3)

+ w(f33

el)w(el, e2)

=

and 4 e 2 , e3) = 4 e 2 , f 3 )

- w

3

,

e,)o(e,, el) = 0.

Let {el, e2, e 3 } be the dual basis to {el, e 2 , e 3 } .Then o = e l of (b) follows easily from (a) and we leave it as an exercise. COROLLARY 15.51 -$(SPY

A

e2. The proof

I

If (p, S ) E N , and Tap= - TBa,then S""TvpSPu=

Tp,)SY

PROOF: Since ( p , S) E N , , i,S = 0 and p # 0. Thus by Lemma 15.50 (b), Spy= L"Mv - L V M p .Writing out both sides of the equation gives the result. I

LEMMA 15.52 If (8, p", 9) E T ~ g , p . s )and A s Tab= - TDa,then FPTrw= (~/S~)T,,~S~~S~~~"~". PROOF: There is a curve (a(t), p(t), S ( t ) ) in A so that s" = (d/dt)l,=,S(t). By Corollary 15.51, we have that S7ySvpSp'= -s2Sp for each t, so that s2S.Tp(t)T,,, = S V ( t ) S y p ( t ) S P ~ ( tfor ) T ~each r t. Thus, s2FpTzp = ~yS,pSpBTpr ST"&SPWTwr + S"S,ps"PPTflr. But S~"Sv,Spo= -$SuDs".Ta,)S" by Corollary = 0 since SUBSu,= 2s2. Thus, the second term is zero. Anti15.51, and Sapgap symmetry shows that the first and third terms are the same, which gives the result.

+

324

15. GEOMETRICAL MODELS

PROOF: First note that for A, B, C in

[A,B]

A2(R4),

. c = ( [ A , B-yy)CaY= {A",B$

B",A$}C,Y = A,,Bfi,C"Y - B"PA,,C,Y = AapBPyCaY B?4,,CaY = A,,BByCaY ApyBB,Cay= 2A,,BflYCaY. -

+

+

Thus, [ [ T ,S], S ] . s" = 2 { 7 - 3 , S $ P - SadT~,S~),SfiySny} = 2T,dsd~S,YsbrY- 2(Sd"~aaySY~)T6p.

But the second term is zero by Corollary 15.51 and the fact that 15.52 now gives the result. [ Lemma 15.52 For a E 9 and [ E T U B we have O(a)t;, O(CT)t;, = 0.

t = th t h+ tu, t u ,where

s". S = 0

6(a)th 6(a)th = 0 and

5 E T u 9 ,v] E T u 9 we have that dS(o)((, q ) = LEMMA 15.54 For B E 9, d6(a)(th,q h ) - [ S ( 0 ) [ u , 6(cr)qul. PROOF: Let a. E 9 and rn, = .(ao). Choose a section B: V -, 9 so that mo E V and a(rn,) = 0,; a gives 0:V x G + 2' by 0(m,g) = o(rn)g an isomorphism. Let s'= @*6 and s^ = a*6. Then using the equation R1;6 = ad(A-')6 from Theorem 15.39, we obtain

+ A-~A",

&rn, A)(G,A) = A - ' ( & ~ ) G ) A

LLEMMA E M M A 15.55 15.55

For For AA,, B, B, CC in in A2(R4), A2(R4), [[AA,,BB]] .C .C = =A A .. [[BB,, C C].] .

325

CONSTRUCTION OF A GAUGE INVARIANT 2-FORM

PROOF: The calculation at the beginning of the proof of Corollary 15.53 shows that [ A , B] . C = 2AorpBByCay. Thus, [B, C ] . A = 2B,,CB,Aay = 2A,,B",CaY = +2A,,Ba,CYB = [ A , B] . C I

Now w, given by (15.30), can be written w(o, p , s)(6,p, S)(6,@,

S) = -PO(+

+ @q0)6 + p6(o)6e(o)6

p6(u)a(a)6 - Sh(o)6 Sd(a)6- s . d6(a)(cf,5) (15.31) - qF(a)(6,6)+ a s . [S, $1. -

+

For T E 9, < T ( a ) = ( d / d t ) l , = oo exp(tT) is a vector field on 2, and XT(a,

p , s, = ( ( d / d t ) l C = O exp tT)(a,p , s,

is a vector field on As.A simple calculation shows that XAa, p , S) =

(6,@,S) = (-tT(cr), -pT, [T, S ] ) . Since O(o)((,(o)) = 0 and 6(o)((T(a)) = T,

we get 6?6,$1 p , s)(xT(a, p , = pTO(o)6 0 (-p)TO(o)6 - 0 - [ T , S] . 6(c~)6- s". T

+ +

+ Sd6(o)((&),

6)

+ a s . [ [ T ,S ] , $1.

By Lemma 15.54, d6(a)(tT(o),6) = - [ T , 6(0)6] so we get w = - [ T , S ] . 6(0)6 - S . [ T , 6(0)6] - 5 * T a s . [[IT; S ] , q. Using Lemma 15.55 we get w = - S . T - as". [ [ T ,S], S ] so that, by Corollary 15.53, w = - 3 . T as2s". T = -(1 - as2)s". T. Thus, if we take c( = l/s2, we get then iXTm=

-

+

+

O V T E ~ . THEOREM 15.56 For each A E G, Lzw = w where L, is the action of A on A,. If we take a = l/s2 then for each (o,p , S ) E A, = 9 x N , we have that T(,,p,s,(G(o,P? S ) ) c ker 0. PROOF: The second part has been established by the preceeding calculations. Now L,(o, p , S ) = (OR,-I , p A - ', A S ) so that TL,(o, p , S)(6,p, 3) = (TR,-,d, pA-', AS). Invariance of the form w follows from the relations

T I T TRA-1 . = Tx,

~19 -+ M ,

O(aA-1) = Aa-lTn,

d(oA-')TR,-I( ) = A6(a)( )K1, dh(aA-')(TR,-i( ), TR,-i( )) = Ad 6(a)( , )K' ( A S ) . [AS, A

q = s . [S, q.

I

326

15. GEOMETRICAL MODELS

Thus o is invariant under the action of G and the G-orbits are tangent to ker o.We will now assume that our particles have positive mass so we restrict ourselves to A T = {(a,p , S ) E A s l p . p > 0). Let p , = e0 and So = sel A e2. LEMMA 15.57 A: = {(a, zp,A-', AS,)la E 9,A E Go, and z where Go is the connected component of the identity in G . PROOF: Clear.

ER+},

I

Writing p = zpoA-' and S = AS, gives 6 = ( i / z ) p - pT and S = [T, S ] , where T = A A - Thus a tangent vector at (a,p , S ) can be written as

'.

(6,Li, 9) = (6,(i/Z)P,0) + (0, -PT, [ T , S]) = (6+ 5Aah ( i l z ) ~ 0) , + (-tr(a), -PT,[T, S ] ) = ( 6 1 9 ( ~ / z ) P ,0) + X r . But we have seen that X , E ker w. Suppose that Y, = (6, ( i / z ) p ,0) + X , , E ker o(a, p , S). Then (6, ( i / z ) p , 0) E ker o.Let Y, = (6,(.?/z)p, 0) + X , , be any tangent vector. Then

0 = W V ' , Y2) = 4 6 , ( i / Z ) P , o w , ( W P . 0) = - (i/z)pe(a)6 + (qz)pe(a)6+ pqapqa)z - p s ( o ) ~ e (-o )s~. dqa)(8., 6 ) - qP(a)(6,6). Write 6 = 6 h + 5" (as in Lemma 15.54)and 6 = 6, + 6". Then we must have (if Y, is in ker w ) that

(l) p8(a)6h = O, (2) - p ( 6 ( a ) d " ) ( e ( o ) 6 h ) (3)

-(i/z)po(a)6h

+s

'

[d(a)6",6(0)6u]= 0 vcu,

+ p6(a)6"8(o)6h = s

'

d6(a) ( 6 h y 6 h ) f qP(:(a)( 8 h j Gh) V 6 h .

Here we have used Lemma 15.54.We will consider p = zp, and S = So. The group invariance gives the validity of our conclusions at all points in 4:. Write R = 6(o)17,,T = d(a)c?",v = B(o)bh.Equation (2)is p(Ru) = [T ,R] . S = - R . [ T , S ] VR. If pR = 0, then R . [ S , TI = 0. This shows that T = aieo A e' as so that T EA ~ p' + as. Now p S = 0 and [ S , S ] = 0, so that x.-s(a,p,S)=(5s(g),090). Thus Y1 =X,, +c~X-,+(~,+~,-~~S(O),(~/Z)P, 0). Since ax-,E ker o,the last term must be in ker w. Thus, we can assume T = aieo A e'. Then, using [el A e 2 ,eo A e'] = e0 A e2, [e' A e2,eo A e2] = -eO A el, [e' A e2, eo A e3] = 0 we get, using R = yieo A e' + R,, R, E A2(p'), that zyivi = 2sa1y2 - 2sa2y1,Vy,, y 2 , y 3 . Thus, v1 = -(2/z)cr2s,

xi

+

327

CURVATURE FORM

v2 = (2/z)a,s, and v3 = 0. Note that Eq. (1) implies uo = 0. Thus, 6, is given in terms of 6,. We now turn to Eq. (3). Define F: Y -+A2(R4) by F(a)(tl, t2)= F(z(o)). ( O < l ? a<2) and ( 4 1 , 4 2 ) = d6(a)(a(<1)h, a(<2)h), where a(
<

That is, ( - (i/z)p)( + ( p T ) t = p(o, S)(u, t), where p: A: + A2(R4). Now T= a g o A e', v = -(2/z) m 2 e l (212) scx,e2, as we have seen. Write p = pa pea^ ep so that p . S = 2sp12 and zp(u, 5 ) = -2a2splptS 2a1sp2p
+

-2

= -(2/Z)S(W10

+

- alP2012

a l z = -(2/z)sa,p1, = -(l/z)cx,p. s, a2z = -(2/z)s",pl2 a3z

= -(2/z)s(a2p13

= -(l/z)a,p.s, -

a1p23).

There are two cases: CASE1:

z2 # - p

*

S. In this a1 = a2 = 0, which leads to 6 = Oand i = 0, so

that Yl E T(G(o,p , s)). CASE2: zz = - p . S. In this case a , and in ker w in addition to T(G(o,p , S)).

ct2 are free, giving two dimensions

We summarize our results. THEOREM 15.58

Let

r: A: -+

R by T(a,p , S) = T(0, S) = - S . {(S. Q)(cT)

+ qF(o)}.Then ker w(a, p , S) = 7;u,p.s, (G(a, p , S)) except on the set p p p r =

r(a,S). On this set ker w(o,p,s) = 7;T;a.p,S) (G(o,p,S)) 0 Y ( o , p , S )where dimension "f(a,p , S) = 2.

CURVATURE FORM DEFINITION 15.59 The 92-valued 2-form D6(a) (6, 5 ) = d6(a) called the curvature 2-form on 9. LEMMA 15.60

D6 = 0.

( b h , (7h)

is

Let X = { i r TdiP1Gir ~ = O}. X is integrable if and only if

328

15. GEOMETRICAL MODELS

PROOF: Let D6 = 0. We will show that condition (1) of Theorem 9.77 is satisfied. Suppose 5 and q are vector fields on 2, which take values in X . For any A€%* we have (Ad)< = 0 = (16)~.Thus, by Theorem 9.78, d(A6) (5, q) = -(Ad) ([t,q ] ) . But d(A6) (5, q) = A dd(5, V ) = -A[X, S V ] =O. Thus, (Ah)([(,473) = 0 V A E 9*,which shows that S([& q ] ) = 0, so that [t, q ] takes values in 2.Conversely, suppose 2 is integrable. If 5, q are any vector fields, then 6 [ 5 h , qh] = 0. Thus, for I-€%*, A Ds(5, q ) = A d6(
REMARK 15.61 In the theory of general relativity, D6 is a measure of the gravitational field. (See [ 2 3 ] ) . Recall that R in Theorem 15.58 was constructed directly from D6.

We have seen that ker w is tangent to the orbit of G on A‘: except at points of a submanifold, 4;= { (a,p , S ) E 4;Ipppe = T(o, S)}. In [ 2 8 ] it is shown that the dynamics are determined by an equation of state, which we will write in the form pFpu = E +f(a,S), for some constant E.

THE SOURIAU G M S

+ f(o, S)}. Assuming

that E + from a restriction mapping v such that the v-restricted systems give the Souriau equations. We do this for any equation of state pPpe = E + f(a, S ) > 0, where f is invariant under the action of G. Let v: 4; + N , by Let .M:(f)

=

{(a,p , S) E 4: Ipepe = E

=a,we will obtain A : ( f )

f(o, S ) # T(o,S), so that 4:(f) n ATo

405 P, S ) =

+ fb,S), S ) .

Let p o = (1, 0, 0, 0), So = sel A ez and N s ( f )= v-’(p0, So). Then {(a,p , S) E A: Ip’pp = E +f(a, S)} = Go . N S ( f )Define . K : Y -,N S ( fby ) K(a) = (a,

JE+fo= e(a),so that

p o , so). For convenience we write K ( o ) = (a,e(o)po,So).That is p = e(a)poand S = So so that p = 0. Putting this into (1 5.31) gives

+

= de(a)&p0and

+

K*(w)(o)(&, 6) = -de(&)po8(6) de(6)po8(cf) epo6(u)8(6)- epob(6)B(cf) -so ‘ d6(8,d) - qF(r5, 6) = (epo)6A e(ir,8) - (dep,) A e(ir76) - qP(u,c) - S , . dS(6, 8). Recalling that d8

=

-6

A

8, we see that

K*(w) = -d(epo8) - qF - So d6,

(15.32)

329

APPENDIX: CONSERVATION LAWS

an amazingly simple formula. Of course there is a residual group invariance left, the subgroup of G which leaves p o and So fixed. Denote it by G(po,So). K is equivariant and e is invariant with respect to the action of G(po,So). The information determining the dynamics is contained in the forms 0 and 6 as before together with d6 and the function e: 2 .+ R, and the group action by G(po, So).Of course the orbits of G(p,,So) in 2 are tangent to ker K*(o). THEOREM 15.62 The system (Y,K*(w))/G(p,,,So)is a closed GMS which gives the trajectories of Souriau [28]. REMARK: In Chapter 16 we will obtain these trajectories as the motion of a particle in a gauge field as described by Sternberg (see Theorem 16.19 and the following remark).

AP PEN D IX: CO NS E RVATlO N LAWS Obtaining conservation laws from symmetries is one of the most important features of analytical mechanics. We describe these methods for a closed GMS ( P , w). There are three basic constructions: (a) Suppose that X is a vector field on P such that L,o 0 = i , do + di,o = di,w, so that i,w is closed.

= 0.

Then

PROPOSITION 15.63 Let [ixo] E H ' ( P ) be the De Rham cohomology class determined by i,w.

If [ixo]= 0, then there is a function K : P is constant along any trajectory.

+R

such that i,o

=

d K , and K

PROOF: Let p ( t ) be a trajectory. Then

( d / d t ) ( K ( p ( t ) ) )= d K ( d p / d t ) = i,o(dp/dt) since d p / d t is in the kernel of w.

= w ( X , d p l d t ) = 0,

I

(b) Suppose that o = -do, and that X is a vector field such that L,o 0. Then 0 = L , dd = dL,d, so that L,d is closed.

=

PROPOSITION 15.64 If [L,O] = 0 in H'(P), then there is a function G: P .+ R such that L,d = d G , and the function K = i,O - G is constant along any trajectory. PROOF: Since i,o = - i, dO = - L,d + di,d result follows from Proposition 15.63. I

(c) To obtain the next result we will need

=

-dG

+ di,d

=dK,

the

330

15. GEOMETRICAL MODELS

LEMMA 15.65 Let X , Y be vector fields on

P and

a a differential form.

Then (a) (b)

L[X,Yl" = L,L,a - LYL,@, i[x,yla = Lxiya - i y L , ~ .

PROOF: When a is a function (a) holds by Theorem 8.27. The proof now follows by induction on the degree of a, using that dL, = L,d and L,(a A p) = L,a A B C I A L,B. As to (b), it is trivial on functions. When c( is a 1-form, we have da(X, Y ) = X ( a ( Y ) )- Y(a(X))- a ( [ X , Y ] )(by Theorem 9.78); that is, iyi, da = Lx(iya)- iy(dixa)- i[x:yla,so we get i[x,yla= L,iya - iy(ix da di,)a, as desired. Again, use induction on degree of a. For a = a0 A B, where a. is a 1-form, using (Theorem 9.78)

+

+

L,yiy(CIo A p) = L,(iyao

A

= LxiyCro A

p-

/?f

A iyp)

iyao A

L,p - L , ~ oA iy/? - N o A L,I'yp.

Also, i,J%ao

A

B) = iy(Lxa0)A B + i Y b 0 A LXP) = iyL,ao A fi - L,ao A i y B + iyclo A Lxb -

A

iyLxp,

so that &XiY

- iYLX)@O

PROPOSITION 15.66 = 0 = Lyo. Then Thus, i[x,ylco = dK for

L,o

A

P) = (~[x,Ylao)A B - a0 A i[X,YlB = i[x,Y](ao A B). I

Suppose X and Yare vector fields on P such that L[x,ylw= 0 and, furthermore, [i[x,ylo]= 0 in H'(P). some function K , and K is constant along any tra-

jectory. PROOF: 0 = d i y o . Form pairs

( q ,K i ) such that

(a) P = Ui V, (b) iyo= dKi on

q. n 5 # 0. Then dK, = d K j on q n 5 so that L x K i = L,K,

Now suppose on vl. n 5, which shows that {L,K,), defines a global function K . Now i[x,y,w = L,iyw - iyL,o = L,iyo. Thus itx,ylo = L,(dKi) = d(L,K,) on q. Thus i t x , y l = ~ dK. I Suppose we have a smooth left action of a Lie group G on P (see Definition 14.18), and let f,:P --t P be defined byf,(p) = g p for all p in P. We say that G acts as a group of symmetries of (P,o) for each g E G.

DEFINITION 15.67

iff:@

=w

331

APPENDIX: CONSERVATION LAWS

For A E T,G, we obtain a vector field 2 on P by a ( p ) = (d/dt)J,=,(exp tA)(p). We leave it as Exercise 15.9 to show that LAO = 0. We will say that the Lie group G is simple if every element A of T,G can [ B i , C , ] . It then follows, by Propobe written as a sum of brackets, A = sition 14.19 that = [Bi,

a -xi

c,].

xi

THEOREM 15.68 Suppose that G acts as a group of symmetries on the closed GMS ( P , w), and that either

(a) P is simply connected, so that H ' ( P ) = 0, or (b) G is a simple Lie group. Then, for each A E T,G there is a function K,: P + R such that iAw and thus, K , is constant along each trajectory of (PI0). PROOF: This

appendix.

= dK,,

is a result of the previous results and discussion of this

I

If ( M , o,H ) is a Hamiltonian system and X is a vector field on M , such that L,w = 0 and L,H = 0, we can apply all of the above discussion to the GMS ( N , 6), where N = M x R and 6 = w + d H A dt as was described at the beginning of Chapter 15. Take r?(rn, t) = X(m) + 0 a/&. Then LgG = 0. In particular, if w = -do, L,H = 0, and L,8 = 0, then G = -d& where @ = 8 - H dt and Lg@= 0. Thus K = i f @ = i,8 is a conserved quantity. EXAMPLE 15.69 Consider motion in the q l , q,-plane with H ( q , , q,, p , ,

+

Jm.

+

p,) = (1/2m)(p: p i ) V(r), where r = This leads to di = dH/api = pi/m. There is rotational symmetry. H is invariant under

cos 6 sin 8 0 0

-sin 8 cos 8 0 0

0 0

Applying (d/d 8)ls=oto this transformation group leads to its generator

+

+

+

Clearly d H ( X ) = 0,O = p1 d q , p , dq, and L,B = - p z d q , p1 d ( - q , ) p 1 dq, p 2 d(qJ = 0. Thus, i,8 = - p l q 2 p 2 q l is a conserved quantity. This is the angular momentum about the origin. Of course, for a Hamilwe can consider vector fields tonian system ( M , w, H ) , when we form ( N , 6), on N which are more general than the consideration mentioned above. (2= X 0 a/at). This is especially true for time-independent Hamiltonians.

+

+

+

332

15. GEOMETRICAL MODELS

EXAMPLE 15.70 We saw in Example 7.46 that the damped system rnq + A q + k2q = 0 is obtained from the time-dependent Hamiltonian H(q, p , t ) = $(p2/rn)eCat+ $kZqzeat on T*R. We will find a conserved quantity using our methods. We have the forms

dj=dqAdp+dHAdt,

$=pdq-Hdt.

We want to find a vector field so that L$ j(a/ap) (a/&) this equation gives, using

+

di$ = d(ap - H ) = cc dp ig

d8

= -(a

d p - p dq

= 0.

Writing 3

= a(a/dq)

+

+ p da - dH,

+ d H ( 3 ) dt - dH),

that 0 = Lg8= p dcc

+ p dq + d H ( 3 ) dt.

We try d H ( 8 ) = 0, which leads directly to CI = -1q/2 and j = Ap/2. But then p dcc + j dq = -(A/2) p dq + (1p/2)dq = 0, as desired. Thus X = -(Aq/2)(d/dq) Ap/2(8/Bp) and K = ife" = -(A/2)pq - H is conserved. Writing t h s out, using 4 = dH/ap = (p/rn)e-", we get

+

+ (a/&)

-(eAt/2)(;lrnq4

+ rn(4)' + k'q')

= constant.

EXERCISES 15.1

Prove Lemma 15.25.

15.2

Prove Lemma 15.26.

15.3

Let ( M , g) be a pseudo-Riemannian manifold, and (V, 4) a coordinate system on M , rno E V. (a) Show that we may choose

4 = (xl, . . . ,x") so that

where index (g(rno))= n - p. (b) Let P(m) = span{d/ax', . . . ,a/axp), rn E V. With rn E M and 1' = 1, define

1

to obtain W c V such that g I P(m) is positive definite for rn E W.

333

EXERCISES

(c) Let Q(m)= P(m)’ c TmM.Show that TmM = P(m) 0 Q(m) and that q(m)I Q(m) is negative definite for m in some neighborhood of m,. (d) Let p(m):T,M -+Q(m) be the projection determined by the direct sum in (c). Take ti= d/dxi, 1 I iI p and C j = p(d/dxP+j), 1 < j I n - p. Apply the Gram-Schmidt orthonormalization procedure separately to (ti)and (Cj) to obtain a moving orthonormal frame. Note that our construction shows that the index of g is locally (e) constant . 15.4 Prove Theorem 15.32.

15.5 Establish the assertions in Example 15.18. 15.6 Prove Proposition 15.38. 15.7 Give two proofs of Theorem 15.11; one by direct calculation, the other using Stokes’s theorem. 15.8 Suppose that ( M , o,H ) is a PMS such that ker o is a smooth subbundle of TM. If g is a Riemannian metric on M , let B = (ker 0)’. Then there is a unique vector field Y taking values in B so that ( i y o ) I B= dHIB. Show that each integral curve of Y is a trajectory of ( M , 0,H). 15.9 Let a: G x P -+ P be a smooth left action of G on the GMS ( P , o), and let f,: P -+ P by f,(p) = a(g, p) for each g E G . Suppose that f,*o= o for each g E G. Show that LAW= 0, for each A E T,G by showing that 4(t,p) = a(exp tA, p) is the flow of 2. 15.10 Let ( M , o)be a GMS and 71: M -+ X be a smooth onto mapping. Suppose that there are subbundles B and C of TM such that (a) TM = ker o 0 B 0 C , B0 = w ~ C , (b) w ~ = (c) C c ker Tn. Show that 15.11

71

is a configuration projection.

Let Q be the configuration space of a mechanical system, L: TQ -+ R the Lagrangian function, and 9TQ -+ T*Q the Legendre transformation (which we assume is a diffeomorphism); we have E: TQ -+ R by E(u) = 9 ( u ) u - L(u) and H = E 9 - I . Suppose that G is a Lie group which acts smoothly on the left of Q. For g in G we have 4g: Q --t Q which gives T4g:TQ -+ TQ and T*4,: T*Q -+ T*Q by 0

(T*+,)(P)= P

O

Vi1.

334

15. GEOMETRICAL MODELS

Thus we obtain left actions of G on TQ and T*Q. Suppose that L T4, = L for each g in G. Show that, for each g in G, 0

(a) E o T4, = E (b) (T*$,) 0 9’= 9’ T$g (c) H 0 (T*4,) = H (d) If 8 is the canonical 1-form on T*Q, then (T*4,)*8 = 8 (e) If G acts on T*Q x R by (Dg(p, t ) = ((T*d,)p, t), then 0

@;(8 - H d t ) = 8 - Hdt Now, if A is in the Lie algebra of G, we obtain a vector field X on Q by

Show that Proposition (15.64) gives a conserved quantity K,: T*Q --+ R by K,@) = p ( X ( t * p ) ) .

16 Principal Bundles and Connections; Gauge Fields and Classical Particles

PRI NCI PAL BUNDLES In Chapter 14 we defined a smooth left action of a Lie group G on a manifold M to satisfy (gh)rn = g(hrn) and ern = rn. If '3 denotes the Lie alby A gebra of G, we obtained a linear mapping of '3 into where A(m) = d/dtl,=,(exp tA)(m).We saw that (Proposition 14.19) B] = - [ A , B]*. A smooth right action is a smooth mapping b: M x G + M such that b(rn, gh) = b(b(rn,y), h) and b(rn, e) = rn, or simply, m(gh) = (rng)h and me = m. Defining A*(m) = (d/dt)l,=,rn(exp tA), we get

[a,

[A*, B*]

=

[ A , B]*.

+a,

(16.1)

This follows from Proposition 14.9 because a(g, rn) = b(m, g- ') is a left action and $rn) = (d/dt)l,=,a(exp t A , rn) = (d/dt)l,=,b(y exp(-tA)) = -A*(rn), so A = - A * . Thus, [A*, B*] = [A,B] = - [ A , B ] = [ A , B ] * . DEFINITION 16.1 Let G be a Lie group and M a smooth manifold. A principal G-bundle over M is a smooth mapping n : P + M together with a right action of G on P such that

(a) uy = u for some u E P implies that g = e; (b) M is covered by open sets I/ for which there is a diffeomorphism 4:n - ' ( V ) + V x G such that 4 has the form 4(u) = (n(u),$ ( u ) ) and 4(ug) =

(44,$(4g). We say that P is the total space, M is the base space, and G is the structural group. REMARK 16.2 Condition (b) implies that each orbit of G in P is of the form n-'(rn) for some m E M , and by (a) and (b), each orbit is diffeomorphic 335

336

16. PRINCIPAL BUNDLES AND CONNECTIONS

to G. However, we cannot say that n: P -+ M is a bundle of groups since there is no unique way of specifying em,the identity at m. Notice that condition (b) could be replaced by condition (b'): Vm E M 3 an open set I/ containing m and a smooth mapping 6:V -,P so that n CT = id on M . 0

EXAMPLES 16.3

Let M be an n-manifold and G a Lie group.

(a) (Trivial bundle) Let n: M x G -+ M be the projection and let G act on the right of M x G by (m,g)g' = (m,gg'). This gives a principal G-bundle which is called the trivial principal G-bundle ouer M . (b) (Frame bundle) For m E M let Fm= {u:R" -+ TmMlu is a linear isomorphism}. Let

F M = IJ Fm. mcM

Let p:FM-+ M be defined by p(u) = m if u E Fm. The construction of the manifold structure on FM is very similar to that already used for the tangent and cotangent bundles. Let ( U , 4) be a chart on M . Define y4: p - '( U ) -+ U x GL(n, R )

by

Y ~ ( u=) MU),(L4) u). 0

Then y4 is bijection and the topology and differential structure on FMare defined by requiring these pairs (p-'(U), yb) to be charts. It is easily checked that everything works. The group GL(n,R ) acts on the right of FMby u A = u 0 A for u E Fm, A E GL(n, R). Note that y4(uA) = (p(u),(dm4) u A), so the requirements of Definition 16.1 are met. Thus, we have a principal GL(n, R)-bundle. (c) (Orthonormal frame bundle). If g is a pseudo-Riemannian metric on M then we have the orthonormal frame bundle (9, n,M ) with group O(n, h) as constructed in Chapter 15 (see Definition 15.33). We will denote 9 by Zg(M). (d) (Affine frame bundle). Let dm= (12R" -+ TmMIu(u)= uo(u) to, where uo E Fm and toE T m M ) ; let d M = and GA(n, R ) = { T: R" -+ R" I Tu = Au uo, A E GL(n, R), uo E R"). Using d,EF,x T,M tlm E M , there is a natural way of giving dMa differential structure. The action is u T = u T. dM is called the afJine frame bundle of M . (e) Write S3 = {(z1,z2)1z1,z2 E C with zlZl z2Z2 = l}. Define S3 x S' -+ S3 by ( ( z l ,z2), eie)-+ (zleiB,z2eie). Then S3/S1N PI(@)2: S2 and n:S 3 -+ S2 is a principal S'-bundle called the Hopfjbration. 0

+

ukEMdm,

0

+

0

+

337

CONNECTIONS ON PRINCIPAL BUNDLES

REDUCTION OF THE STRUCTURAL GROUP DEFINITION 16.4 Let x : P -+ M be a principal G-bundle and H a closed subgroup of G. A reduction of the structural group of P to H is a submanifold Q c P such that x:Q -, M is a principal H-bundle with the restricted action (thus QH = Q).

In Example 16.3 we have that .dMZJ cFM 2 .9,(M) give successive reducwith group GA(n, R ) to subgroups GL(n, R ) 2 O(n, h). Notice tions of d M there is a 1-1 correspondence between metrics g and reductions .9,(M) of F M once , the index of the metric is chosen.

CONNECTIONS O N PRINCIPAL BUNDLES In Chapter 15 we saw that a connection on the orthonormal frame bundle of a pseudo-Riemannian manifold was an essential part of obtaining the geodesic trajectories from a geometrical mechanical system. DEFINITION 16.5 Let x:P --t M be a principal G-bundle and % be the Lie algebra of G. For g E G we have R,: P -+ P by R,(u) = ug. By a connection on P we mean a %-valued differential 1-form a on P such that

(a) ct(u)(A*(u))= A for u E P, A E 9, (b) @(us)0 TR, = Ad(g-') a(u) for u E P, g (1

If x:P

-+

E

G.

M is a principal G-bundle, then we have Tx:TP -+ T M . For

UueP

u E P, define Vu= ker T u x c T,P. By Theorem 9.73, V = Vuis a smooth vector subbundle of T P . It is called the bundle of vertical vectors on P. Now suppose that a is a connection on P. For each u E P let 2, = ker a(u) and <# = .#u. Again Theorem 9.73 shows that is a smooth vector subbundle of T P . It is called the bundle of horizontal vectors on P. Note that 2 is determined by a, whereas ^Y- is independent of a. Of course,

uusp

to apply Theorem 9.73 we needed to know that both dim Vuand dim Xu are independent of u. But this follows directly from the definition of a principal bundle and of a connection. LEMMA 16.6 Let x:P -+M be a principal G-bundle and CI a connection on P. Suppose the dimension of M is n and the dimension of G is k. Then

(a) dimension of P is n + k, (b) for each u E P, V ,= {A*(u)IAE T,G} and dim Vu= k, (c) for each u E P, the dimension of a(u)(T,P) c T,G is k, so that dim Xu= n,

338

16. PRINCIPAL BUNDLES AND CONNECTIONS

(d) for each u E P, T,P = 2,0 V, so we write TP = 2 0 V , (e) for each u E P, TunI 2,:Xu--f T,,,,M is an isomorphism. PROOF: We will prove (b) and leave the rest as a simple exercise. Clearly {A*(u)lAE T,G} c V, and dim V, = dim(ker Tun)= (n k) - n = k. Suppose A E T,G and A*(u) = 0. Let y ( t ) = u exp tA. Then

+

dt

(u exp t A ) exp sA

Thus y ( t ) = u Vt, so exp t A = e V t and A

= A*(u exp = 0.

dim(A*(u)lA E T,G)

t A ) = A*(y(t)).

This shows that =

k

and we are done. Recall that A*(u) = T,b(u, e)(A),where b:P x G -+P is the right action. 1

HORIZONTAL LIFTS OF VECTORS If u E P and rn = n(u), define t,:T,M + Xu by t, = (T,nl&',)u E TJ4, the vector t,(u) E T,P is called the horizontal lift of u to u. LEMMA 16.7

'.

For

Let u E P and g E G. Then

(a) TR,(&) = X u g , = TR, 0 t,. (b) tug PROOF: Suppose

4 E T,P and

a(u)(() = 0. Then

a(us)(TR,(t))= Ad(g - ')(a(u)(O)= 0

so that TR,(%,) c Xu,. But T,R, is an isomorphism so we must have equality. (b) follows, since 'II R, = 'II gives TU,n TuR, = Tun. I 0

0

It is clear that the subbundle V c T P is integrable, for its integral submanifolds are the fibers K ' ( r n ) , rn E M . The horizontal subbundle 2 c TP is in general not integrable. This nonintegrability leads to the curvature 2form of the connection. For each u E P, let h,: T,P + 2,be the projection determined by T,P = Xu0 V,. In fact, h,

= t, 0

Tun,

(16.2)

so that Lemma 16.7 implies h,,

0

TR, = TR, h,. 0

(16.3)

339

CURVATURE FORM AND INTEGRABILITY THEOREM

CURVATURE FORM AND INTEG RABl LlTY THEOREM DEFINITION 16.8 Let a be a connection on the principal G-bundle n:P + G. Define a 9-valued 2-form Da on P by

Da(u)(t, 0= da(u)(h,t, hui). Da is called the curvature form of a. We have the equation of structure

W u ) ( t , i)= d'm(5,i) + [a(u)L4 U ) i I .

(16.4)

The derivation of this equation is the same as given in Lemma 15.54 for the connection we constructed on the orthonormal frame bundle of a pseudoRiemannian manifold. It was based on the following local representation: Let c: V + P , V open in M , be a section of 71: P + M (n c = id"). We define @: V x G + PI V by @(m,g) = a(m)g. Let a" = @*a and a^ = c*a. Then 0

by properties (a) and (b) of Definition 16.5. THEOREM 16.9 Let 71: P + G be a principal G-bundle, a a connection on P , and 2 c TP the horizontal subbundle determined by a. 2 is integrable if and only if Da = 0. PROOF: Let V c P be an open set and 5, ( ( u ) and [(u) are in 2, V u E V . Assuming that

ibe vector fields on V so that

Da = 0 we get, using Eq. (16.4), da((, [) = 0 on V. Using part (f) of Theorem 9.78, we get that a ( [ [ , i]) = 0 on I/, so that [(, takes values in X . By Theorem 9.77 %' is integrable. Note

[I

that while Theorem 9.78 was proved for real-valued differential forms, we can use it by considering 1 0 a V A E (T,G)*. Conversely, suppose that X is integrable. Let u E P and (, E X u ,[,, E 2,.Choose horizontal vector fields ( and [ defined on some open set containing u so that ((u) = to,[(u) = 5,. Since X is integrable, we get that a([(, [I) = 0 on V . Thus, for each A E (T,G)*, we have A da((, i)= d(A a)((, [) = -Act([(, [I) = 0 on V , so that A do!((,, to)= 0. Since A was arbitrary, da((,, to)= 0. That is, da(u) is 0 on horizontal vectors. By the definition of Da it follows that Da(u) = 0. I 0

REMARK: Thus we see that the definition of Da using the horizontal projection was so that the integrability of 2 is equivalent to the vanishing of Da.This is another important integrability theorem with many applications.

340

16. PRINCIPAL BUNDLES AND CONNECTIONS

We now investigate the effect on Da of the mappings R,. Thus consider g E G as fixed. By Eq. (16.4).

But R,*a = Ad(g-') 0 a, where Ad(g-') is a fixed linear transformation of T,G. Thus d(R,*a)= Ad(g-') 0 d ~ Also, . [Ad(q-')B, Ad(g-')C] = Ad(g-I)([& C]) from Exercise 14.5. Thus we get Wug)(TR,(S),TR,(i)) = Adk-

' ) ( W u K 0).

(16.6)

We have seen that a connection a on 7c: P + M defines a horizontal lifting of tangent vectors from M to P. More importantly, we have the following THEOREM 16.10 Let m(t) be a smooth curve in M for a I tI b and suppose x(u) = m(a). Then there is a unique smooth curve u(t) in P , for aI tI b, so that

(a) = u, (b) x(u(t)) = m(t) for a I t I b, (c) du/dt E iP,,([) for a I tI b. PROOF: We first work locally, choosing a section 0:V -,P with m(a)E V , a(m(a)) = u. As before, we have 0:V x G + PI V given by @(m,g) = o(m)g. We obtain u in the form u = 0 0 U,whereU(t) = (m(t),g(t)). Then, the condition o(u(t))(i(t)) = 0 becomes C(m, g)(rit, g) = 0, where ti = @*a.Then, by Eq. (16.5) we see g must satisfy

dgldt

= - TL,(t,(Ad(g(t)-' ) ) ( B ( t ) )=

TR,,t,(B(t)),

g ( 4 = e,

where B(t) = B(m(t))(dm/dt)and & = C * M . The local existence and uniqueness follows from the differential equation theorems in Chapter 4. We will assume that the curve m(t) is entirely contained in V , the domain of our section 6.The general case will follow from this since each curve can be broken up into finitely many such pieces. We claim 36 > 0 such that if g(t) is defined on [a, to] then it can be extended to a solution on [a, to + 61 n [a, b]. The result will then follow. Extend m(t) to (a - E, b + E). Define a vector field on G x ( a - E , b E ) by X ( g , r) = (TR,(B(r)), a/&). Since (e) x [a, b] is compact 36 > 0 3 (-6,d) x ((e) x [a, b ] ) c 9(X)(see Definition 4.25 and Theorem 4.26). Now suppose g(t) is defined on [a, to].Let (h(t), to t), h(0) = e, be the integral curve of X defined for (tl < 6. Define i ( t ) = h(t - to)g(to) for It - tol < 6. Then (dh/dt)(t)= TRh(t)(B(tO + t ) )so that (dG/dt)(t) = T R B ( t , ) ( T R h ( t - t , ) ( B ( t ) ) ) = T R B ( t ) ( B ( t ) ) , and i ( t o ) = &). This completes the proof. I

+

+

341

ASSOCIATED BUNDLES

HORIZONTAL LIFTS OF CURVES REMARK: Theorem 16.10 has important geometrical consequences. We have seen that principal bundles are often frame bundles. That is, each U E Pdefines a basis of tangent vectors at n(u) or a basis in the fiber of some other vector bundle over M . The curve u(t) then determines parallel frames along m(t). We say that the curve u(t), a I t I b, given by Theorem 16.10 is the horizontal lift of m(t) with initial condition u. LEMMA 16.11 Let m(t) be a smooth curve in M , a I t I b, n(ul) = n(uz) = m(a). If u ( t ) is the horizontal lift of m(t) with initial condition u1 and u2 = u,g, then w(t) = u(t)g is the horizontal lift of m(t) with initial condition u2.

PROOF: Use Lemma 16.7 and the equation dw/dt = TR,(du/dt).

I

ASSOCIATED BUNDLES Let n:P -+ M be a principal G-bundle over M . If a: G x F -+ F is a smooth left action of G on a manifold F , we construct the associated (to x by a ) bundle P x GF over M as follows: Define a left action of G on P x F by g(u,f) = ( u g - I , gf). Then define P xG F as follows: we write [u,f] = {(ug-', g f ) l g E G}, so that [ul,fl] = [uz,f2] if and only if there is g with u,=u,g-' and f 2 = g f l ; then P x , F = { [ u , f ] l u ~ P , f ~ F There ). is a natural projection nF:P xG F -+M by xF([u,f]) = n(u). Suppose CJ: V -+ P is a section of P. We obtain QO: V x F -+ nF1(V)by @,(m,f) = [o(m),f]. It is easy to see that is a bijection. Give P x F the topology so that each Q0 is a homeoa section of n} determorphism between open sets. Then ((71; ' ( V ) ,@;')lo mines a differential structure so that P x GF becomes a smooth manifold which is locally like V x F. THEOREM 16.12

We leave this proof as an exercise. An important case is when F is a vector space and a ( g , f ) = p(g)(f) with p(g) a linear isomorphism for each g. That is, p: G -+ GL(F) is a linear representation of G . Then n F : P xG F -+ M has the structure of a vector bundle over M , since each fiber nF'(m) carries a vector space structure by A,[u,fl] + A2[u,f2] = [u, A,fl + A2f2]. This is well defined because ~ , [ u g - ' , Sfll

+ 1b2[ug-',

sf21

=

[

w

1

= [ug-

3

4gf1 + A29f2l

', S(&fl +

M 2 ) 1

= [U7 W I

+ j'2f21.

342

16. PRINCIPAL BUNDLES AND CONNECTIONS

EXAMPLE 16.13 (a) Let FM be the frame bundle of M as defined in Example 16.3, G = GL(n, R ) and we take F = R" with the natural action of G. Then FM xG F N T M , an isomorphism given by [u, u] + u(u). (b) Let n: P + M be a principal G-bundle. Take F = G with the action of G on F by left translation. Then P x G G N P by [u, g ] 1: ug. (c) Let n: P + M be a principal G-bundle and Go c G a closed subgroup. G/G, = {gG,lg E G} is a smooth manifold (see Theorem 14.16) and there is a left action of G on G/Go by g(glGo) = (ggl)Go. Let 9 = P xG G/G,. There is a natural mapping $: P + P xGG/Go by $(u) = [u, Go]. Then $(ug) = $(u) if and only if [ug, Go] = [u, Go]. But [ug, Go] = [u, gG,], so $(ug) = $(u) if and only if g E Go. We thus write PIG, = P x G GIG,. If Q c P is a reduction of the group of P to Go then, Vrn E M , $(n-'(rn) n Q) is a set with exactly one element so that Q determines a global section x: M -+ P xG G/Go. Conversely, given x: M + P x G G/G,, taking Qm = $-'(x(rn)) and Q = Q, gives a reduction of the group of P to Go.

0,

PARALLEL TRANSPORT Let n: P -P M be a principal G-bundle and F = P x G F an associated bundle. We will denote n;'(m) by Fm, Now suppose that M is a connection on P . DEFINITION 16.14 The connection c( defines the process of paralEel transport in any associated bundle 9 along any piecewise smooth curve in M as follows: Let rn(t), a 5 t I 6, be a curve in M . Choose u E n-'(rn). Define T,:9m(a) + Fm(,) by T , ( [ u , f ] )= [ u ( t ) , f ] ,where u(t) is the horizontal lift of m(t) with initial condition u, as given by Theorem 16.10.

Of course, we must show that Tt is independent of the choice of u. Suppose w E n-'(m(a)). Then u = wg for some g E G, so that [ u , f ] = [ w g , f ] = [wg, g - ' g f ] = [w,g f ] so that, using w, T,([u,f]) = [w(t), g f ] , where w(t) is the horizontal lift of m(t) with initial condition w. But by Lemma 16.11, w(t) = u(t)g-', so we get T,([u,f]) = [u(t)g-', g f ] = [u(t),f], as before. REMARK 16.15 In Example 16.13 we saw that P is an associated bundle to n:P -+ M using F = G with the action of left translation. Thus parallel transport in P is defined. We leave it as an exercise to show that for u E n-'(m(a)),w ( t ) = T,(u)is the horizontal lift of m(t)with initial condition u.

GAUGE FIELDS AND CLASSICAL PARTICLES A connection a on a principal G-bundle n:P -+ M is what the physicist calls a gauge jield. If F is a space representing a set of states of some

NATURAL 2-FORM ON COADJOINT ORBITS

343

system which carries an action of G, then the associated bundle P x G F is a bundle of states over M . We have seen that a gauge field is exactly what is needed to relate states at different points along a curve in M through parallel transport. EXAMPLE 16.16 Let ( M , g) be a Lorentzian manifold, and R4 have the usual Lorentz metric and carry the action of the Lorentz group G. The momentum states for a particle of rest mass rn are given by

F

Then P x G F

'v

=

((p,)

~ 4 l ~ ,=p m 2 > .

Sm2.

NATURAL 2-FORM ON COADJOINT ORBITS Sternberg [33] has described classical particles which interact with a gauge field. Let 9 be the Lie algebra of G and 9* the dual space. We define the coadjoint action of G on 9*(as we did in Chapter 13 for SO(3)) by Ad*(g)(v) = v 0 Ad(g-'). For vo E 9* fixed, let Go = {g E GIAd*(g)(v,) = vo} and take F = G/Go. Then we have p: F -,V ,defined by AgGo) = Ad*(g)vo.

Also G acts on F by S(Y1Go) = (g91)Go.

There is a natural 2-form R defined on G/Go. We make use of Theorem 14.17. Let go be the Lie algebra of Go, and q: G -,G/Go the projection. If q(g) = f we take 4, = q R, and get T4q:9 -, Tf(G/Go)which factors to give an isomorphism @,-: 9/Ad(g)(g0)4 T,(G/G,), which is independent of the choice of g E q - ' ( f ) (Theorem 14.17). Suppose t, [ are tangent vectors to G/Go at f . Choose B, C E 9 so that T4,(B) = <,T4,(C) = t. We define 0

Q(f)(t,0 = - P ( f ) ( [ B , CI).

( 16.7)

We show this is well defined. Suppose T$,(B,) = T$,(B). Then B - B , = Ad(g)(A), where A E
using Exercise 14.5 and the fact that p ( f ) = vo 0 Ad(g- '). This shows that R(f) is well defined.

344

16. PRINCIPAL BUNDLES AND CONNECTIONS

THEOREM 16.17 R is a closed nondegenerate 2-form. Furthermore, if g E G, then t,*(R)= R, where L,(glGo) = (ggl)Go. PROOF: Suppose R(f)(5, [) = 0 V i E Tf(G/Co). Now 5 = T4,(B), so 0 = v,([Ad(g-')B, D])VD E 9. Write B , = Ad(g-')B. Then, using Lemma 14.13 again, we get (d/dt)l,=, Ad*(exp tB,)(vo) = 0. But if Z(v) = (d/dt)l,=, Ad*(exp tB,)(v), then P ( t ) = Ad*(exp tBl)(vo) is the integral curve of 2 with B(0) = vo. But Z(vo)= 0, hence exp t B , E Go V t , so that B , E g o , which gives B E Ad(g)(30) and shows that 5 = T4,(B) = 0 by Theorem 14.17. Thus Q is nondegenerate. 1

Now by Exercise 14.7, we have local sections of q: G -+ G/Go, around every point. Let B : V -+ G, V open in G/Go be one such. Then 8, = o*(8) is a 9 valued 1-form on V , where 8 is the left invariant 9-valued 1-form on G (see Definition 14.14). Let po = vo 0 0,, a 1-form on V. Then dp, = vo 0 do, = vo 0 o*(d8) and, using Eq. (14.8), we get dP~(f)(<,

=

v0(d0(g)(To(5)7

=

-vo(o(g)([A,

BI)),

where q(g) = f . A and B are left invariant vector fields so that A(g) = To((), B(g) = To([). Now A = TL,-,(Ta(l)), B = TL,-,(Ta([)), and dP,(f)(t,i) = - vo([A, BI). If A 1 = Ad(g)A, Bl = Ad(g)B, then since T4,(A1) = 5 and T+,(B,) = i.Thus, R = dp, on V so that dQ = 0 on V . Such V cover G/Go so dR = 0. Finally, the invariance of R under the left action of G follows from R(f)(5, [) = -vo([A1, B1]),where Tq(TL,A,) = 5 and Tq(TL,Bl) = i,d g ) = f . I We can now describe Sternberg's construction [33] for the motion of a particle in spacetime interacting with a gauge field defined by a connection on a principal G-bundle P -,M . We modify slightly the way in which T*M is handled. We saw in Chapter 15 that 4: 2 x (R")*-+ T*M induces a diffeomorphism Y x (R")*/O(n,h) -, T * M (Proposition 15.38). Thus, instead of forming a pullback of P (see Exercise 16.9) over T * M , we use the principal bundle over M given by

6:

L? 0 P = { ( B , U ) ~ E B L?,, u E P , for some rn E M } ,

with group O(n, h) x G acting by (a,M a , 9) = (w 4.

Let n: 3 -, M be the orthonormal frame bundle of the metric on spacetime M . Suppose that CI is a connection on P . We must choose vo E 3* (this is analogous to specifying the charge of a particle). We thus get p: G/Go -+ 9*

345

PSEUDOMECHANICAL SYSTEM

and the form R on G/G,, as we have just seen. There is a left action of O(n, h) x G on (R")* x G/G, by (a, g)(A, glGo) = ( A u p ' , (gg,)G,). Let ,A! = (2@ P ) x ((R")*x G/G,). We have the following differential forms on A: 0

(a) R pulled back from G/G,, which we will denote also by R, (b) p a pulled back from P x G/G, which we will still denote by p 0 a; on p x G/G,, P N ( U , f ) = P(.J') a(4, (c) w pulled back from 2 x (R")* which we will denote by w ; on 2 x (R")*,w = ( - d i ~ J )0A (see Theorem 15.39 and Definition 15.34). 0

O

cj

+

Define a 2-form on 4 by C

=w

-

d(p a) - R. 13

= 0, we see that d X = 0. Define H : A + R by Recall that h is a fixed metric on R", h* the induced metric on (R")*.The spacetime metric is obtained from h using 9.

Since dw

=

0 and dR

E(a,u, I., f )= +h*(A, A).

PSEUDOMECHANICAL SYSTEM FOR PARTICLES I N A GAUGE FIELD C, A) is an integrable PMS; THEOREM 16.18 (a) (A, (b) The action of O(n, h) x G on .&, given by (a, y)(o, u, A, f )= (oa-', u g - ' , I,u-', g f ) is a gauge group action; (c) Vz E A,ker C(z) is the tangent space to the O(n, h) x G orbit through Z.

PROOF: We first note that a curve in Y 0 P is of the form (o(t),u(t)), where n,(cr(t))= n(u(t)).Thus Tn0(&) = Tn(i),so we will write a tangent vector to A as 5 = (6, u,, where fi = u h u, and ti, is determined by Tn(uh)= Tn,(&).Suppose that E ker C(z), z = (a,u, 1,f).Then C(z)(<,[) = 0 V[ = (6, ii,, 1,T). Suppose = (0, O , x , 0). Since 6 = 0 and ii, = 0 we can conclude ii = 0. The condition 0 = X(z)((, <) = A(O(6))vx

x,f),

<

+

<

gives us that d(6) = 0, which shows that dh = 0 and this gives u h = 0. Now by differentiating p(exp(tC)gG,)(B) and using Proposition 14.13 we get that 4 4 g G , ) ( f ) ( B ) = - p(q)([ C , B ] ) , where f = T4,(C). Thus

( 4 P 4 + Q)(u,gGo)(k f)(k7) O

=

-P(g){ [ C , a(u)u"] - [ D , a(u)zi]

+ [C, D ]

-

da(4 c)}

9

346

16. PRINCIPAL BUNDLES AND CONNECTIONS

T. Thus, %W))+ W W g + p(g){[C, .(u)ii]

where C and D satisfy T+,(C)

W ( Li)= -

= f , T4,(D) =

[D,a(u)ri] + [C, 03 - da(a, ii)}.

Take 6 = 0 = 6. Then 0 = p(g){[a(u)ti

+ C, D]}

VD E 9.

But in the proof of Theorem 16.17 we saw that this implies that C + a(u)tj E Ad(g)(g0). Thus we can take C = -a(u)tj. We now have C(z)((, C) = -A(@(&)) +Id(ir)O(r?) - p(g)(Da(u)(ti,6 ) ) using Eq. (16.4). But we have seen that ich = o so D C I ( U ) ( ~ii), = 0. We now have c(z)(<, () = -A@@)) + nd(&)e(q Since this is zero Vc? we get A = Id(&). In summary, ( = (6, k,, A, f )E ker C(z) implies (a) 6 , = 0 = c h f (b) = - T4,(a(uP), (c) A = A6(6).

f

Writing (d/dt)(,=o((r exp( - LA),u exp( - tB)7I exp( - tA), (exp tB)gGo) = (6,ti, irh = o = ti,, ~(a)ir = - A , ;Z = - I A , a(u)ti = -~ , f T=~ , ( B ) so that (a)-(c) above are exactly the conditions that (6,ti,, A,f) be tangent to the orbit of O(n, h) x G . This establishes (c). Since R is invariant under the group action we get ker X(z) = T,((O(n, h) x G)z) c ker dE?(z), which establishes (a). To obtain (b) we need to show the invariance of the form C. The invariance of R follows from Theorem 16.17. Then show the invariance of p CI and A 8 (recall w = -d(AO)). We leave the details as a simple exercise. I

f),we see that

0

0

0

Of course, the space &/O(n, h) x G is exactly the associated bundle B = (20 P ) x O(,,h) ((R")* x G/Go), the bundle of states obtained from (R")* x G/Go using dip 0 P . It follows from Theorem 16.18 that X induces a 2-form c on 9 so that C = (pr)*(c), where pr: & + 9 is the natural projection, and that 2 is both nondegenerate and closed. Clearly R induces a function on B which we will also denote by E?. We summarize our constructions.

STERNBERG'S THEOREM THEOREM 16.19 Let n:P + M be a principal G-bundle, where M is (part of) spacetime and thus carries an orthonormal frame bundle no:2 + M . Suppose that c1 is a connection on P (a gauge field). A choice of vo E 9* determines the bundle of states B for a particle interacting with the gauge

347

GEOMETRICAL-MECHANICAL SYSTEM

field c1. (9, c, A) is a Hamiltonian system which determines the evolution of the particle states. REMARK: In Theorem 16.19 the only property of the function

R(0,u, A,.f)= +*(A, A) which was needed for the construction was its invariance under the group action. Other invariant functions could be used (giving different trajectories, of course) and would reflect energy dependence on internal states. For example, to obtain the Souriau equations of Theorem 15.61 we would take R(a, ,?, S) = i h * ( I , A) - f ( o , S) (see Exercise 16.10).

GEOMETRICAL-MECHANICAL SYSTEM FOR PARTICLES IN A GAUGE FIELD

cE=elFE.

For E > O let F E = { s ~ F I R ( s ) = Eand } We know that is a GMS which determines the trajectories (see Theorem 15.14)(FE, with A = E. The map n3:4 -+ (R")*, given by n3(6,u, A,_sGo) = A is a restriction mapping, since each trajectory of the PMS (A, C,H ) has constant and thus n3 takes it into an O(n, h) orbit in (R")*. Let ,? E (R")* and E , = *h*(,?,A). Then A,= n;'(,?) = (9 0 P ) x {A} x G/G, N (9 0 P) x G/G, and we get the extension of Theorem 15.44. ,?6 A 0 - d(pc1) - Q, En). Then S , is a PMS For ,? E (R")* let S, = (AA, with a gauge action by G,. Furthermore, SJG, is isomorphic with the GMS (FE, Here G, = G, x G. But now we can do a further reduction. The = projection n4(cr, u, ,?,gG,) = gC, is a restriction map for S,. Let An,v, n'i '(Go)n A,.

cE)

eE).

THEOREM 16.20 For ,? E (R")* let S,,vo = (A,,vo, Ah A 0 - d(v,cc)). Then S,,vo is a PMS with a gauge action by G,. Furthermore, S,,vo/G,is isomor-

phic with the GMS (FE, &). Notice that, in terms of Theorem 16.20, the Hamiltonian function is reduced to a constant corresponding to a certain orbit. In this approach, the Hamiltonian is a labeling for the orbits corresponding to the rest mass of the particle. The constructions leading to Theorem 16.19 used the principal bundle 9 0 P with group O(n, h) x G. We have connections 6 and c1 on 9 and P, respectively. However, 2 was treated in a special way, since - d(prx) appears in C but there is no such term for 6. Instead we have w = --d(I* 0) in C, where 0 is the R"-valued l-form on 2 defined by 0(o) = 6' Trio. Such a In order to construct form 6' is special to a tangent frame bundle such as 9. 0

0

348

16. PRINCIPAL BUNDLES AND CONNECTIONS

the interaction entirely from connections it would be desirable to incorporate 6 into a connection on some principal bundle. This can be done by replacing linear frames by affine frames (e.g., enlarging the Lorentz group to the Poincare group). Thus, we replace 9 by a larger bundle 9 as follows: Let A(n, k ) be the semidirect product of O(n, k) with R". That is, we can write

with

[" '][b ;I=[; "y] 0 1 0 1

and, identifying u E R" with [;I,

[" 'lcu] ;'1, 0 1

1

=

+

so that [g f ] gives the transformation v + au 5. Of course, O(n, k ) c A(n, k ) as a subgroup so that O(n, k ) acts on A(n, h) through left translation. Let 9 = 9 x ~ ( ~A(n, , ~h)) be the associated bundle. For y o , r E A(n, k), defining [a, r]ro = [a, rro] gives 9'the structure of a principal A(n, h)-bundle. We thus have j : 9.+9 by j ( u ) = [u, id] and j ( 9 ) gives a reduction of the group of 9 to O(n, k). The Lie algebra of A(n, k ) is denoted by d ( n , k ) and can be represented by matrices [;f z], where A E O(n, k), u E R", and we see (16.8)

The adjoint representation of A(n, k ) on d ( n , k ) is given as follows. If

then A

u

A

v

+ uv

Ad(a)(A) -Ad(a)(A)' 0 0

=[ Thus

i'A5

+ a-'v 0

11.

(16.9)

349

AFFINE GROUP MODEL

Let A be a connection on 9'. Then we can write

where 6 is an o(n, h)-valued l-form on 9 and $ is an R"-valued 1-form on 9.Let

(a) 6, is a connection on 9, RB(4,) = a4, Va E O(n, h). (b) We leave the calculations as an exercise. Of course, A, determines A uniquely because of the A(n, h) action on 9. LEMMA 16.21

'

0

AFFINE GROUP MODEL DEFINITION 16.22 A connection A on B for which afine connection.

4,

=0

is called an

Now given 6, a connection on Y , define

which determines a unique affine connection A on 2'. Thus, there is a natural correspondence between connections on Y and affine connections on 9. Now ( d ( n ,h))* = O(n, h)* @ (R")*. Choose V, = ((0,A,), v o ) E d ( n , h)* x 9*.In the following we will write G for A(n, h) x G and

Go = {g We wish to calculate

so that

With

E

G1Ad*(J)(?,)

= go}.

on G/G,. Using Eq. (16.9), we see that

350

16. PRINCIPAL BUNDLES AND CONNECTIONS

where

so that U is the O(n, h) x G orbit of Vo. We then have

(2@ P ) x u c ( 9 0 P ) x G/G,. When we restrict Z = -d(P E ) - fi to (9'0P ) x U we get 0

Q(GoK(t1,l l h

since

t = 0,

v 1 = u2

= 0;

and G,

= W G O ) ( i l , 12),

(52,tJ) 0

&(a,u, gG,) = M ( o )

5 = 0, A = IZ0u-l, 4 = 8 on 2.We thus get THEOREM 16.23

Let

=

-d(p

o

E)

-

+p

0

a(u, gG,), since

fi on (9@P ) x G/Go. We have

embeddings I , J

2-k

A! = (2 @ P ) x (R")* x G/Go such that J * @ ) = Z*(X). Furthermore, if E FE,where pr: A + 9.

(9 0 P ) x G/Go = *h*

(Ao, Lo), then pr(image I ) =

REMARK 16.24 Theorem 16.23 shows that we can construct the GMS (FE, zE)entirely from a connection on the A(n, h) x G bundle 9 x P , the coadjoint representation of A(n, h) x G, and the reduction 9 ' x P of the group to O(n, h) x G. This construction is also due to Sternberg and co-workers [33, 341. We could take a more general Go; Go = ((So,Ao),vo) in d(n,h)* x 9*.This would give a spin-curvature coupling term.

351

EXERCISES

EXERCISES 16.1 Complete the proof of Lemma 16.6. 16.2 Give the proof of Theorem 16.12. 16.3 Give the proof of Theorem 16.16. 16.4 Suppose that a is a connection on the principal G-bundle 71: P + M and that Da is its curvature form. Show that R,*(Da) = Ad(9-l) 0 Da Vg E G. 16.5 Let S' = SO(2) = { z l z E @, Z Z = 1). We consider S' as a Lie group. Writing

gives two coordinate charts for S ' , say -7112 < 4 < 3x12 and 0 < 4 < 271. For each z E S' we have T,S' = {irzlr E R}. Thus T,S' = firlr E R } . Of course S' is abelian, so that the bracket is zero on TeS1and Ad(z) = Id V z E S'. (a) For either of the two coordinate systems, using ei4,show that d irzE T,S' is equal to r -. Thus, d l d 4 is the unit tangent vector d4 field pointing counterclockwise. (b) Show that id4 is canonical T,S'-valued 1-form on S' (recall that the Lie algebra has been identified with the set of purely imaginary complex numbers). 16.6 Let n: P + M be a principal S'-bundle and a a connection on P . Show that: (a) We can write a = ib, where fi is a 1-form on P . (b) Da = dcc = idp. (c) There exists a unique 2-form x on M such that TC*(X)= dp. It is defined by x(m)(u, w) = d p ( z ) (f,(u), L',(w)), where ~ ( z=) m and L', is horizontal lift. (d) If 0 : I/ + P is a section of 7c: P + M and = o*p, then x = dp. Thus dX = 0. (e) For M = R4 with the Minkowski metric, work through the construction of Theorem 16.19 in this case. Here Y R so get that 3* = R and the orbits of the coadjoint representation are points. Write v0(d/d4(,) = q. Show that the GMS ( S E Z E with ) E = m2 is the GMS (Sm2.w + (~*)*6)of Example 15.18.

p

-

352

16. PRINCIPAL BUNDLES AND CONNECTIONS

16.7 Derive Eq. (16.5). 16.8 Derive Eq. (16.8).

16.9 Suppose that n:P + M is a principal G-bundle over a manifold M and that f :Y + M is a smooth mapping. Define a set

f * P )= {(Y?4 l Y E

y and u E P f ( J

and mappings E : f * ( P )+ Y

by Z(y, u) = y

f l : f * ( P )4 P

by f"(y, u) = U.

We thus obtain a commutative diagram f * ( P )-5 P

Suppose that {(K, Cpi)liE I } is a system of local trivializations for P . That is, each K c M is open, M qli E Z}, and &i: n-l (q)+ q x G is a local trivialization. Define $i:iY1( f - ' ( K ) ) 4 f - l (F) x G by where 4i(u) = (nu,4i2u).Show that: Icli(y, u) = (y,

u{

(a) there is a unique differential structure onf*(P) so that each $i is a diffeomorphism between open sets; (b) defining (y, u)g = (y, ug) gives f * ( P ) the structure of a principal G-bundle; (c) if CI is a connection on P , then f;c(a)is a connection f * ( P ) . ii.:f*(P)+ Y is called the pullback of n:P

+M

under f .

16.10 In the constructions leading up to Theorem 16.19, we had

4= (2'@ P ) x ((R")*x G/Go), where 9 is the Lorentz frame bundle over spacetime M and n: P + M is a principal G-bundle with a connection. We also had defined the mapping p: GIG, + G*. (a) Take P = 2' with the connection 6 constructed in Chapter 15. Thus G is the Lorentz group. Recall that we have isomorphisms 3 N A2(R4) N %*.

For So in A, (R4), let Go = { A E GI Ad*(A)So = S o } so we have p: G/Go + 9?*. Then the form X is given by X = w - d(S .6) - Q.

353

EXERCISES

(b) Let A' = {(@,p,A,S)E cdlcr= u and AS = 0 ) (where the spin and momentum are given in the same Lorentz frame) and let

A s ={(o,A,S)I;1S=OandS.S=s2}, which was used in Chapter 15. If s2 = So . So, we have an imbedding of eA'in As, given by

1,S )

(0, 0,

+

( 0 , A S).

Show that C equals the form given in Eq. (15.30) (with F = 0) on A'. (c) To include that F term, we would take P = 2 63 P , , where 71: PI + M is a principal S1-bundle with a connection such that iqF is the curvature form on M (see Exercise 16.6). (d) Thus, (A1,X) is imbedded into ( A S , m )of Chapter 15. There is obtained using the diagonal map a group action of G on A', G + G x G, given by A + ( A , A ) . The map H(O,Eb,S) = +h*(A,

A) - f ( 0 , S )

is invariant under this action and factors to 9 = A 1 / G . Show 2) can be identified with the system (9, K * (a))/ that (5E, G(A0, So)of Chapter 15, where FEc 9 is given by l? = E .

77 Quantum Effects, Line Bundles, and Holonomy Groups

QUANTUM EFFECTS The quantum behavior of particles occurs in experiments in which particles exhibit a wavelike nature manifested by certain interference effects, a simple case of which we now describe. A beam of identical particles is split into two parts at a point S. Or we can consider S as a source point for beams of particles, all of which are reflected back from a barrier except for the beams which reach the points A and B, where there are slits in the barrier. The points A and B then act as secondary sources allowing particles to go off in all directions (see Fig. 17.1). We now place a detection screen at a distance d from the barrier (see Fig. 17.2). Let us coordinatize the screen by x. We are interested in the following question.

FIGURE 17.1

354

355

PROBABILITY AMPLITUDE PHASE FACTORS

PROBABI LlTY AMPLITUDES Of the total number of particles which reach the screen in a certain time interval, what fraction arrive at x? This fraction can be interpreted as the probability that a particular particle will arrive at x. A model which correctly describes this is based on three principles written down by Feynman [ 113: (a) There is a complex function @(x)so that lCD(x)lzis the probability that a particle will arrive at x. The complex number @(x)is called the probability amplitude for the recording of the arrival of the particle at x. (b) For each path p there is a probability amplitude CD, so that lCD,12 represents the probability that the particle traveled along p. If pi is the path li followed by ti,then @(x) = CD,, Thus, if an event can result from several different processes, you add the probability amplitudes to obtain the probability amplitude for the event. This is in contrast to classical probability theory in which you would add the probabilities of the possible processes which produce the event. (c) If the path consists of two parts such as p = A u t,then CD, = On@(.

+

PROBABILITY AMPLITUDE PHASE FACTORS Thus we obtain interference effects. If CD,,(x) A , ( ~ ) d @then ~(~),

(apl+

@,2)(@,,

+ 6,J

=

A:

=

Al(x)ei@l(x) and CD,(x)

=

+ 4 + 2A1’42 cos (41 - 4 2 )

+

so that \O(x)[’ varies between ( A , A,)’ and ( A , - A,), depending upon - $z. If 6 = 2nn, we get a maximum probability the phase diflerence 6 =

FIGURE 17.2

356

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

point and if S = (2n + 1)7c, we get a minimum probability point. Of course, A , and A , must be such that f @ ( x ) m d x= 1, but A , and A , have no effect on the locations along the screen of the max and min probability points. The location of these points relates to the properties of atomic particles and fields interacting with them. For example, suppose we know the locations of the max-min points for certain particles in an inertial frame subject to no force. When a field is turned on which interacts with the particles, the locations of the max-min points will shift. This is what we want to calculate. Thus we will not worry about A in @ = Ae'4 but will see how to determine the 4. We will refer to ei@as the probability amplitude phase factor.

DeBROGLIE AND FEYNMAN The idea of associating a wave disturbance with a particle motion was developed by Louis deBroglie in 1924. Later Richard Feynman gave us the principle that the probability amplitude phase factor of a path q(t) in the configuration space of a classical mechanical system with kinetic energy function T(q, q) and potential energy function U(q)is exp(i 1(T(q,4) - U(q))dt). The deBroglie wave for a particle of mass m in its rest frame is @(t,x,y, z ) = @oe-2"i", where v = mc2/h, c being the speed of light and h Planck's constant. We include the minus sign in the exponent to be in agreement with Feynman's prescription for particles moving with a velocity much smaller than c as we shall see. Using the Lorentz transformation (11.14) we see that the deBroglie wave of a particle of mass m in an inertial frame in which the particle is moving with velocity u parallel to the x-axis is @(t,x,y , z) = exp [-2niv{yt - y(u/c2)x}],

where y = l / , / l _ O . Now let p be any classical path of such a particle, so that p is given by t = t, x = xo + vt, t, < t < t g . Then @(B)= @ ( A ) . exp [ -2niv(t, - t , ) , / m ] , so that the phase change of @ from point A = (t,, xo + ut,) to B = (t,, xo + ut,) is S, = - 2 n v ~ , ~ ,where T, is the elapsed proper time along p from A to B. This is the phase change of the deBroglie wave along the classical path of the particle. Following Feynman, we shall take the probability amplitude phase factor of any future pointing timelike path p to be exp( - 2niv AT).

PHASE FACTORS AND 1 -FORMS Now let 6 be the curve in phase space obtained from using the relativistic energy momentum relations. Recall that

357

PHASE FACTORS AND 1 -FORMS

where v that

= dp/dt.

Since the metric is ds2 = c2 dt2 - dx2 - d y 2 - dz2, we get

If 8 is the canonical 1-form on phase space, then

Thus, j,- 8 = r n c 2 , / m ( A t ) ,so that 6,, = -(27c/h) culation is valid on any Lorentzian space-time.

Jp

8. In fact, this cal-

LEMMA 17.1 Let A4 be any four-dimensional manifold with a Lorentz metric g . If p(t) is a timelike curve in M and p(t) in T*M is defined by g(p(t))(md p / d q ) = fi(t), then, for a particle of rest mass m, the probability amplitude phase factor of p is exp( (- i/h) jo 0). If p is future pointing, 0 = mc2(A.rh where

so

is the elapsed proper time along p (h = (1/27c)h). REMARK 17.2 The lemma gives the phase factor if there are no external fields influencing the particles, otherwise it must be modified. We will see how to do this in the case of the electromagnetic field. Note that the gravitational field is included in the lemma since it is represented by replacing the space time of special relativity with a general Lorentzian manifold.

We now return to the situation of Fig. 17.2. With pi = ,Ii u Li, the elapsed time of p 1 equals the elapsed time p2 in our reference frame. Since we are assuming that the initial speeds are the same, we get that the proper time of 2, = the proper time of 1,. Now letting L, and L2 also denote the length of the paths, and writing v1 = t ,j A t , u2 = [ , / A t , where At is the elapsed time of path el and e,, the phase difference is

If L, = e,, then let u = u1 = u2 and we get 6 = 0 so 0 is a maximum probability point. Let us find the closest minimum probability point above 0

358

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

assuming that u << c and that d is much larger than the distance between A and B so that d, - d , is small. Using the binomial expansion, we get that 6 z (2.nmv/h)(t1- t,).Setting this equal to -n, we get that the minimum point closest to 0 from above occurs when t, - el = h/2rnu. Because of the small value of h, this value will be too small to be detected for particles that we are used to seeing in daily existence. It is only for atomic particles that h/2rnu is of detectable magnitude. REMARK 17.3 While the heuristic derivation of the formula was, via deBroglie, associated with the motion of a particle moving along a classical trajectory, we cannot consistently think in this way in dealing with quantum behavior. We have seen that the statement “the particle must go along path t,or path t,” would lead us to add the probabilities instead of the probability amplitudes. We cannot apply the principles of classical particle mechanics to curves and points internal to the experiment unless we include the measurement of the values of these quantities as part of the experiment. This inclusion may then change the outcome of the experiment. In addition, there is the Heisenberg uncertainty principle which states that (ApNAx) 2 h, where Ap is the uncertainty in the measurement of the momentum component in the xdirection, Ax is the uncertainty in the xposition, and h is Planck’s constant. For more details on these matters, see Feynman [ll].

COW AND BOHM-AHARANOV EXPERIMENTS We now consider a second interference effect (Fig. 17.3), where the elapsed times of all four paths are the same. We wish to calculate the phase difference between the paths SAD and SBD. Of course, if no fields interact with the particles, the phase difference is 0, since the proper times of the paths are equal. However, we assume there is a uniform gravitational field of strength

FIGURE 17.3

359

COW AND BOHM-AHARANOV EXPERIMENTS tl in

the negative z direction. The form of the proper time metric which represents such a field is dz2 =f(z) dt2 - (~(z)/c’) dz2 - (1/c2)dXz,

wheref(z) = 1 - (2az/c2) and g(z) = l/f(z). The derivation of this is given in [13]. Such an experiment has been carried out with neutrons by Colella, Overhauser, and Werner and is referred to as the COW experiment. For a discussion see [ 131. Let z1 be the height of BD and z2 the height of SA. Since the metric is invariant under translations in the x direction, the proper time of AD equals the proper time of BD and hence cancel in the phase difference. Now t(SA)

= Jf(z2)(At)’

{

=At I----

- ( 1/c2)(Ax)’

c2

cz

(where v = Ax/At)

so that

But (zz - z,)(Ax) 17.1,

= a,

the area enclosed by SADB, so we get, using Lemma

6 z (maa/hv)2n, which is the COW result. We consider one more example; that of Fig. 17.2, when the particles have a charge q and there is an electromagnetic field present. The formula for the probability amplitude phase factor must be modified as follows. In the presence of an electromagnetic field with a 4-potential A = A , dx”, the probability amplitude phase factor associated with p is defined as di’*)fl, where B=j6(-D-qA).

(17.1)

Here the coordinates are xo = ct, x1 = x, xz = y, x3 = z. Also A , = ($/c, -A), where - V 4 is the electric field if A = 0, and curl A is the magnetic field if 4 = 0 (see Eq. (12.8)).Of course, the 4-potential could be changed by adding df, where f is a scalar function, which would then change the phase

360

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

FIGURE 17.4

factor. But observable effects have to do with the phase difference, which is an integral around a closed curve and hence, is unaffected by df. Also, this prescription only works if there is a global 4-potential in the region of interest. This would not be the case, for example, in the field of a magnetic monopole (see Example 12.11). We will formulate a geometrical interpretation of the phase difference calculation which will not require a global 4-POten tial. The “correctness” of this definition in describing the phyiical world has been established by experiments such as those measuring the Bohm-Aharanov effect, in which a long solenoid carrying an electric current is placed as shown in Fig. 17.4. Here the result is very striking because the electromagnetic field (but not the 4-potential) is negligible along the particle paths. For a discussion see [lo, 351.

COMPLEX LINE BUNDLES AND HOLONOMY We will describe the phase difference calculations in terms of a principal S’-bundle associated with a certain GMS. At this point we ask the reader to work exercises (16.5) and (16.6). Suppose that n: P -+ M is a principal S’-bundle over M , and a is a connection on P. Now s’ acts naturally by complex multiplication on the complex plane C so we can form the associated bundle F = PxSlC.pr: PxSl@-+ M is called the complex line bundle associated with n:P -+ M . Defining

361

COMPLEX LINE BUNDLES AND HOLONOMY

z [ u , w ] = [u, z w ] gives multiplication of elements of

9 by complex numeiezw] = [u, zw]. In fact, P,is a onebers. Note that z[ue-ie, eiew] = dimensional complex vector space V m E M . If m(t), a I t I b, is a curve in M , then we have the parallel transport + Frn(,) (see Definition 16.14). operators T,:Fm(o) Suppose that o: V P is a section of P over the open set I/ c M , and that m(t) is in V for a I t I b. We have 0: V x S' -+ PI V by @(m,z ) =o(m)z, o? = @*a, & = o*a as discussed in Chapter 16. We have -+

&(m,z ) = d(m)

+ i d0

(where z = e"), by Eq. (16.5) and Exercise 16.5. As in Exercise 16.6 we write a = ip, where fl is an ordinary (real-valued) I-form on P, = (D*p = -%, a^ = o*fl = -id. Now a horizontal lift of m(t) is given by u(t) = (m(t),z(t)), aI tI b, where 0 = o?(m,z)(&, i)= d(m)& id and z(a) = zo is the initial condition. Therefore, i0 = - a*(m)&= - ifi(m)k, which gives = - fl(m)ni. Thus, z(t) = zo exp( - i I),where c, is m(s), a I sI t. Now we also get $:I/ x C -+ 91 I/ by $(m, w) = [o(m),w], so I,-' Tt $ can be written as

+

sCc

0

I+-' T, $(m(a), w) = 0

0

0

( (

m(t), exp ( - i P j j W ) -

(17.2)

Thus we have the basic formula for parallel transport in terms of a section o: V + P and I =o*(fl). For closed curves in M , we have the concept of holonomy operator, which we now describe for any principal bundle. DEFINITION 17.4 Let 7c:P-+ M be a principal G-bundle with connection a, m E M , and C , the set of all piecewise smooth curves y : [a, b] -+ M such that in = y(a) = y(b). For u E 7c- ' ( m )we define H i : C , + G as follows: Let yeC, and let y" be the horizontal lift of y with initial condition y"(a) = u. Now y"(b)E 7c- '(in),so there is a unique g E G satisfying f ( b ) = y"(a)g. Then we define H i ( y ) = y.

LEMMA 17.5 (a) The image Hz(C,) is a subgroup of G. (b) If u1 = ug, then H",'(y) = g-'(H:(y))g.

We leave the proof as an exercise. Hi(C,) is called the holonomy group of a at u. If 7c: P -+ M is a principal S'-bundle, with connection a = ip, we see, by Lemma 17.5, that Hi is independent of the choice of u E 7c-'(m), so we get H,: C , -+ S1. If F = P xs,C and y E C,, y ( t ) defined for a I tI b, then T,,:9, -+ F ,is given by multiplication, Tb(<)= H,(y)<, V< E 9,.

362

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

INTEGRAL CONDITION FOR CURVATURE FORM In Exercise 16.6 we saw that there is a unique closed 2-form x on M so that n * ( ~=) dfl. Let m E M and c:V+ P a section of P with m E V Suppose that y E C,, y(t), a I t I b, lies entirely within Vand that y is the boundary of an oriented compact 2-submanifold R c V . Now we saw that the horizontal lift u(t)= (y(t), z ( t ) ) with u(a) = (m, 1) is given by

Thus, u(b) = (m,exp(- i j, b), so that H,(y) = exp( - i 1, p). But by Stokes's theorem H,(y) = exp( - i SR 1)since x = d g by Exercise 16.6. PROPOSITION 17.6 Suppose that y E C , is the boundary of an oriented compact two-dimensional submanifold R c M . Then H,(y) = eie, where 8 = - j R x . PROOF: We have given the proof in the case where R is contained in the domain of a section of P. If not, break R up into finitely many pieces, each of which is obtained in the domain of a section. I COROLLARY 17.7 Suppose that K is an oriented two-dimensional sphere embedded as a submanifold of M . Then x = 2nn: for some integer n.

SK

PROOF: Fix m E K . For E small, let y,:[O, 11 -+ K be a parametrized circle of radius E with y,(O) = m = ~ ~ ( 1Then ) . y, divides K into two parts K , and D,, as shown in Fig. 17.5. Now assume that y, is parametrized so that its orientation is that induced from D,.Then the orientation is the negative of that induced from K , . By Proposition 17.7,

FIGURE 17.5

363

INTEGRAL CONDITION FOR CURVATURE FORM

x + 0 so we conclude that H,(y,) + 1. Thus, But, as E --t 0, for some integer n, which gives x = 2nn = 2nn, as asserted.

IK

SK, x + 2nn

Note that if the 2-sphere K is the boundary of a three-dimensional compact region B in M , then j K x = J B dx by Stokes’s theorem, so that jK x = 0 since d x = 0. Thus, if every 2-sphere in M is the boundary of a three-dimensional compact manifold in M , then Corollary 17.7 places no restriction on x. If Q: S 2 + A4 is any smooth mapping, then we get the pullback bundle 5 a*(P)+ S 2 , 6:Q*(P)+ P , and ic?*(p) gives a connection on this bundle (see Exercise 16.9). Applying Corollary 17.7 to o*(P) shows that j S 2 a*(x)= 2nn for some integer n. Suppose that M is a simply connected manifold and 6 is a closed 2-form on M such that js2~ * is6an integer for every smooth mapping Q: S2 + M . Then, there are collections THEOREM 17.8

where each r/; c M is open and di is a 1-form on where hiis a function on & n V,, such that (a) (b) (c) (d) (e)

v, and {Ljlr/I n 6 # a},

M = u{&li~Z}, dd, = 6 on di - h j = dfij on 5 n 5,

v,

f-. = -f..

11 JC ’ if V# n 5 n V, = $3then f i j V;n Q nV,.

+ fjk + f k ,

is an integral constant on

PROOF: We will indicate the ideas involved; the details require considerable knowledge of algebraic topology [16, 29, 311. Choose an open cover { V l i E l } for M such that each V#,qn5, n F n V, (if nonempty) is contractible to a point. For each i, there is a 1-form 6, on K, such that d6, = G.If n 5 # 0, then d(6, - S j ) = 0 on n 5, so there is a function gij on r/; n satisfying 6, - d j = dg,,. O n r/I n 5 n V,,d(gij g j k g k i ) = 0 so that g i j g j k gki = a i j k , a constant. The collection & = {aijkl& n Vj n V, # a} defines an element of f i 2 ( M ,R), the secondCech cohomology group with real coefficients. Now nn,(M)= 0 implies that n,(M) N H”,M, Z), the second singular homology group with integral coefficients. Define, for Q: S2 + M in n,(M), &(a) = j S z a*6. Then, since M is simply connected, C;, defines an element of the singular cohomology group with integral coefficients H ; ( M , Z ) . Finally, i: Z -+ R gives i,: H ; ( M , Z ) +

v

+

v

+

+

+

364

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

H?(M, R ) and there is an isomorphism 4: H?(M, R ) + kZ(M,R), which gives $(i,(G)) = 15.It follows that there is a collection of numbers {A,l V;- n 5# so that if we takeAj = gij + Aij, then (e) is satisfied. I

a}

REMARK: Suppose that 15 is a closed 2-form on a manifold M . Then [h] E H 2 ( M ) ,the second deRham cohomology group of M . If h satisfies the conclusion of Theorem 17.8, then we will write [I51 E H 2 ( M , 2)and say that [h] is an integral cohomology class.

We want to apply these ideas to the interference effects discussed at the beginning of this chapter. They all involve interactions which, for classical particles, are described by the GMS (Sm2,w + (z*)*6) of Example 15.18. We will see that the evolution of the probability amplitude phase factor will be given by parallel transport in a complex line bundle over a subset of S,, and that x cc (w + (r*)*6). There is an additional problem, however. The curves in Sm2 are lifts (using the relativistic energy momentum relations) of curves in spacetime and these lifts are not continuous, due to jumps in the momentum at certain points. Thus we must define a unique parallel transport along certain discontinuous curves. This is achieved by noting that we have e*: S,, + Q, where Q is a region of spacetime and i:x = 0, where iq:(e*)-'(q) + Smz is inclusion, for all q E Q. We will need the following lemma. LEMMA 17.9 Suppose that 7c:P + M is a principal S'-bundle, that M is connected and simply connected, and that a is a connection on P such that da = 0. Then, for each m E M , H,: C, + S' is given by H,(y) = 1, for each y E C,. Thus, if n is any other point in M , and we define T,,: F, + 8, by parallel transport along any piecewise, smooth curve from m to n, then T,, is independent of the choice of curve. Of course, 8 denotes the associated complex line bundle. PROOF: Fix m E M and uo E P,. Recall that, if y: [a, b] + M is an element of C, and u: [a, b] + P is the horizontal lift of y satisfying ~ ( a=) uo, then u(b) = u,H,(y). Now P , Z S' by uog -+ g so that H,(C,) E { u E P,lu is accessible from uo along the horizontal subbundle %}. Since M is simply connected, H,(C,) is a connected subgroup of S'. Thus H,(C,) = S' or H,(C,) = ( 1 ) . If H,(C,) = S', then every u E P , is %-accessible from uo. But da = 0 implies that % is integrable, so, since X is transverse to P,, we have a contradiction to the inaccessibility theorem. This establishes the first part of the conclusion. The fact that T,, is independent of the curve now follows directly. I

365

PARALLEL TRANSPORT DETERMINED BY A FOLIATION

BUNDLE DESCRIPTION OF PHASE FACTOR CALCU LATl ON We now present the bundle description of the phase factor calculation. We assume we have a GMS ( M , G) and T*: M -,Q such that dc3 = 0 and (PF1) [(1/2n)c3] E H 2 ( M ,Z ) , (PF2) M is simply connected, (PF3) if M , = (z*)-'(q) V q E Q, then $5 = 0, where i,: M , inclusion, (PF3) each M , is connected and simply connected.

-+

M is

We have seen that (PF1) is necessary for 6 to be obtained from the curvature form of a principal S'-bundle over M . In fact it is also sufficient. This follows by Theorem 17.8 and Exercise 17.7. Let n:P -,M be the principal S'-bundle and CI the connection on P constructed in Exercise 17.7, and let 9 be the associated complex line bundle.

Parallel Transport Determined By A Foliation DEFINITION 17.10 Suppose that y = (yl, y 2 , . . . , y k ) is a finite sequence of smooth curves in M so that if m i= initial point of y i is in M a i , ni = end point of y i is in M b i ,then bi = ai+ for i = 1,2, . . . ,k - 1. For any points rn E M a , and n E M,, we define Tmn(y):Fm + cFn by

Tmn = Tnkn

0

7'7,'

Tnk - j m k

0

Tyk

~

I o

. . '

0

TnZrn,

0

Ty, Tn,mz Tyj 0

0

0

Tmmt.

Here T Y Jdenotes parallel transport along y j and the T n J _ l m areJ given by Lemma 17.9 using property (PF2). Now assume that Q is a simply connected region in spacetime and that the structural group of the orthonormal frame bundle of Q has been reduced to the proper Lorentz group 9;(see Chapter 11). This means that spacetime is an oriented 4-manifold with a preferred time direction. We now define M

=

{ p E T*Qlpppp,= rn2 and p is future pointing},

so that M is an open connected subset of Sm2. Also, take 6 = -(l/h)(w + (z*)*6) where 6 = -qF as in Example 15.18. We assume that 6 is such that 6 satisfies (PFI) (see Remark 17.12). The interference effects we discussed related to a situation in spacetime similar to that described by Fig. 17.3. Let

366

17. QUANTLIM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

y, v be the lifts of SAD and SBD to M as described in Lemma 17.1. Then we get T,,(y): @, + @, and T,,,(v):F,+ F,, where m is a fixed element of M a , and n a fixed element of M,. Choose 5 E F,. Write = Tm,,(y)([) and Q2 = T,,,(v)(~). These depend upon t and do not determine unique complex numbers. However, there is a unique eigE S1 such that Q2 = eid@,,and this 6 is the phase difference which we calculated in the interference effects discussed at the beginning of this chapter. The validity of this claim follows from Lemma 17.1, Eq. (17.2),and Eq. (17.3).Note that 6 = - d B where B = -8 qA, if there is a 4-potential A defined on all of Q (the region of interest in spacetime). Now eigt= T,,,(y)-' T,,,(v)(t) so we get 0

THEOREM 17.1 1 The phase difference is 6, where g,(v u (-7)) = eid,g,,, being the extended holonomy operator, which is defined using the extended parallel transport operators of Definition 17.10, and is determined by the connection tl and the decomposition B = { M , I q E Q } .

R EM A R KS-G EO M ETR IC QUANTI ZATl 0 N REMARK 17.12 (1) The replacement tl -+ CI + in*(df) for f:M + R does not change the eidof Theorem 17.11. However, if n l ( M ) # 0, then different connections could produce different phase differences. (2) [(1/27r)6] E H 2 ( M ,2) is no restriction if H 2 ( M , R ) = 0. However, for an electron in the field of a magnetic monopole (see Example 12.11), H 2 ( M , R ) # 0 and we get the Dirac condition 2kq/h = integer. (3) We want to emphasize that our procedure depended not only upon the GMS ( M , G)but also upon (a) the method of lifting paths from Q to M , (b) the decomposition B = { M , I q E Q}. Actually, Q = M / B , so it is the decomposition of M by submanifolds, on each of which 153 restricts to be zero, that is essential. (4) The development we have given is related to the geometric quantization procedure of Konstant and Souriau. For each fixed time t , let M , be the time slice and o,= GIM,. One considers wave functions $, which are sections of F,, the complex line bundle determined by w,, and are parallel on each of the leaves of some foliation B,. Again 6 1K = 0 VK E 9,. Such a 9, is called a polarization of M,. One is interested in more than the simple interference effects we have discussed: for example, in obtaining the quantized energy levels of bound systems. The standard approach to quantum mechanics requires an inner product ($;, $,") between wave functions. This will be of the form $:$,". Thus, the product of two wave functions should be a differential form. This leads to a redefinition of wave functions as sections

sp

367

HOLONOMY AND CURVATURE FOR GENERAL LIFE GROUPS

of Ft tensor product with certain -$-forms (the square roots of differential forms). It is necessary (because of topological considerations) to generalize 9, to complex polarizations which are certain subbundles of the complexified tangent bundle T M ORC of M . This work is very intricate and the topic of recent and continuing research. For a nice discussion see [14, Chapter 51. ( 5 ) To carry through our development for a general GMS the construction of a polarization 9'is the difficult part. One can usually obtain a lift of curves in Q = M / 9 to M by considering classical trajectories. For example, gE)constructed in which decomposition should one use for the GMS (FE, Chapter 16 to describe a particle interacting with a general gauge field (suppose the group is G = S0(3))?

HOLONOMY A N D CURVATURE FOR GENERAL LIE GROUPS We now present generalizations of the preceding results about holonomy for principal S'-bundles to the case of principal G-bundles, for any Lie group G. Recall that if n:P + M is a principal G-bundle, M is a connection on P, we get H i : C, + G and H;(C,) is called the holonomy group of ct at u. Here u E n-'(rn). Let C: = {y E C,ly can be deformed continuously to m}. Hk(C2) c H;(C,) is called the restricted holonomy group of M at u. Let G, = H;(C,) and G: = H;(C;). GI] is a normal subgroup of G, and G,/GZ is countable.

LEMMA 17.13

PROOF: If yl, y 2 E C, we define y 2 y1 to be the curve obtained by first going around y1 and then around y 2 . Also, 7'; will be the curve obtained by traversing y1 in the opposite direction. Let g E G, and h E GI] be obtained v y can be contracted to y y which can be from curves y and v. The y contracted to a point. Thus g - l h g E GZ. Let n , ( M , rn) be the fundamental group of M with base point m. If y E n,(M), choose a representative y1 E C,. Then Hi(?,) E G,. If y 2 E C , is homotopic to yl, then y;' y 2 represents y-l 0 y = 1 in n , ( M , m).Thus H;(yl)-'H;(y2) E GZ, and we get H : nl(M, rn) G,/G:, a surjection. Since M is a second countable manifold, nl(M, m) is countable. I 0

'

(1

0

-'

0

0

--f

Now let Q, from u to

."I.

=

{."

E

PI there is a horizontal piecewise smooth curve y" in P

THEOREM 17.14 Suppose that G, is a closed subgroup of G, and that M is connected. Then Q. is a reduction of P to a principal subbundle with group G,.

368

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

PROOF: The assumption that G , is a closed subgroup of G (so that it is a submanifold and a Lie group by the proof of Theorem 14.11) makes this theorem a straightforward exercise. [

For a more general theorem which does not assume that G, is closed in G see [17]. From this reference we also obtain THEOREM 17.15 G , is a Lie group and G t is the connected component of the identity. Let 3 be the Lie algebra of G and 3, the Lie algebra of GI]. Then 9, is spanned by all elements of the form Da(ii)((, [) where u" E Q. and (, ( are horizontal tangent vectors at fi (recall that Da is the curvature form of the connection a). COROLLARY 17.16 (a) Suppose that M is connected, U E P, and G , = (e).Then there exists a global section 6:M -+ P , and thus that P is equivalent to M x G . (b) Suppose that M is connected and simply connected and that Da = 0. Then there exists a global section 6:M -+ P , so that P i s equivalent to M x G. PROOF: (a) Let m = n(u).For n E M choose a smooth curve y in M from rn to n. Let 7 be the horizontal lift of y with initial condition u. Define o(n)= endpoint of 7. Since G, = (e), this is independent of the choice of y. The smoothness of 6 can be seen using local trivializations. (b) Since M is simply connected, G, = GE by the proof of Lemma 17.13. But Dcc = 0 means that G t = (e) by Theorem 17.15. [ REMARK 17.17 In Theorem 17.11 a quantum phase difference was expressed in terms of a holonomy operator on a principal S'-bundle. The generalization to the nonabelian case is as follows: Let n:P -+ M be a principal G-bundle with a connection a, and F a set of states, carrying an action of G and a G-invariant inner product ( F can be a complex vector space with a complex-valued inner product). Form the bundle of states B = P x G F as described in Chapter 16. Let m E M , y E C,, and 4 E F,, It,is.a simple exercise to show that, if we choose u E P,, then the quantity (v, H:(y)v), where 4 = [u, u ] , is independent of the choice of u. It is this quantity which is eis in the previous discussion.

EXERCISES 17.1 Show that condition (PF2) implies that any two connections a, and a, on P such that da, = da, are related by a, = a, + in*(df)for some function f :M -+ R.

369

EXERCl S ES

17.2 Show that 2kqlh = integer for a electron in the field of a magnetic monopole. See Remark 17.12(2). 17.3 Answer the question raised in Remark 17.12(5) with G

= SO(3).

17.4 Supply the proof of Theorem 17.14. 17.5 Supply the proof of Lemma 17.5. 17.6 Let n:P --+ M be a principal S'-bundle and CI, = ibl, u2 = ip2 two connections on P. We have seen that there are closed 2-forms xI, x 2 on M so that "*(xi) = It/?,. Suppose that K is an oriented twodimensional sphere embedded as a submanifold of M . Show that JK x1 = S K x 2 .

17.7 Suppose that 6 is a closed 2-form on M such that [(1/2n)G] E H 2 ( M , Z ) is an integral cohomology class. Let {(v,, S,)}, {hj}be the collections described in the conclusion of Theorem 17.8. Define

C = ((i, m, z)li E I, m E and (i, m, z)

N

( j , m, ezffifl,f(m)z) if m E

(1) Show that

-

6 , and z E Sl)

6 n 5.

is an equivalence relation.

Writing [i, rn, z] = { ( j ,m, zl)l(i,m, z) {[i, m, z]li E I , m E 6 , z E Sl}, 71:

P -+ M

by

-

( j , m, z,)), we define P =

n([i,m, z ] ) = m,

and [i, m, z]z, = [i, m, zz,].

For each i E I , we have 4,: x S' Then q5J:1 4,: n x S' -+ 6 n (m,e2ffifij(m)z). Show that 0

-+

P by +,(m, z) = [ i , rn, z]. x S' by

4J:1

0

q5i(m, z) =

( 2 ) There is a unique differential structure on P such that each a diffeomorphism onto an open set of P.

4; is

(3) n:P -+M is a principal S'-bundle with the S'-action defined above.

Define a T,S' valued 1-form Eion 1/; x S1 by

E,(m, z)(ril,i) = 2ni4(m)riz + Ti. Show that (4) There is a unique T,S'-valued connection 1-form CI on P satisfying

370

17. QUANTUM EFFECTS, LINE BUNDLES, AND HOLONOMY GROUPS

(a) &+a = oZi Vi E I , (b) da = n*(i6). See also Exercise 17.8. 17.8 In the case that M = S2, the 2-sphere, and [(1/2n)6] E H 2 ( M , Z ) a direct construction of the desired principal S1-bundle can be given which is simpler than that of Exercise 17.7, because a system similar to {(K, Si)} can be explicitly specified. We can write S2 = D , u D,, where D , is the upper hemisphere, D, is the lower hemisphere and D, u D , N S'. In fact, assume we have a parametrization D, n D , = {eie10 5 6' 5 2n}. On each Di there is a l-form di such that d6, = (1/2n)6. {(D,, h1), (D,, a,)) plays the role of Si)} in Exercise 17.7 (we do not need that D , n D, is contractable in this construction). Define g : D , n D, + S' by

{(v,

ss,

Then y(0) = 1 and g(2n) = exp(i 6)= e2nin= 1 = g(0) so that g is well defined. Now let P = {[i, rn, z ] 1 i = 1,2, rn E Di, z E S'}, where [l, rn, z] = [2, rn, g(m)z] if rn E D , n D,; define an S1-action on P by [i, rn, z]zl = [i, rn, zz,], and define n:P + S2 by n([i,m, z ] ) = m. There is a unique differential structure on P such that c $ ~ :Di x S' + P by 4i(rn,z ) = [i, m, z ] gives diffeomorphisms between manifolds with boundary, and we obtain a principal S'-bundle. There is a unique connection a on P such that +;(a) = 2niSi + 6, where 9 is the canonical 1-form on S'.

78 Physical Laws for the Gauge Fields

GAUSS'S LAW IN ELECTROMAGNETIC THEORY Previously, we have used connections on principal bundles to determine particle trajectories and calculate probability amplitude phase factors. There is a reciprocity, however. Just as particles react to these fields the fields must react to particles and be restricted by physical laws. This is what we want to consider in this chapter. Let us first consider the case of electromagnetic theory. We have Maxwell's equations (12.22), which are of the form dF

= 0,

*d*F = tij,

where F is the 2-form which is the Faraday tensor. Now aF is the curvature form of a connection on some principal S'-bundle for certain constants a (which depend upon charges of particles that respond to the presence of F). This gives geometrical meaning to the equation dF = 0. The second equation is the source equation, specifying how charged sources determine the field F. Let us write this in integral form. First apply * to the equation to get d*F = ti*j. Now suppose that S is any closed two-dimensional surface in space time which is the boundary of a compact oriented 3-manifold B. Then *F = Se d*F = K je*j. If we take S to be 2-sphere in a constant time hyperplane in some Lorentz frame, then by Gauss's law *F = ti x (total charge inside of S) and thus the space component of *j represents a charge density. If we take S to be a cylinder such as that in Fig. 18.1, then we get Ampere's law modified with Maxwell's displacement current. The displacement current terms come from the integrals over the top and bottom of the cylinder.

ls

ss

371

372

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

FIGURE 18.1

CHARGE CONSERVATION Another basic fact follows from the equation d*j

= 0,

which is a consequence of d*F = .*j. Thus Sj = 0, which shows that the spatial components of j are not independent of the time component (charge density). Now for our spacelike 2-sphere S , consider S x [a, b], where [a, b] is a time interval. Then Je ( b ) *j - jB (a) *j Js [a,61 *j = Je ro,b] d*j = 0. This is charge Conservation, since Js [a,b] *j = -J: (Jsj n d A ) dt. That is, (charge inside of S at t = b) - (charge inside of S at t = a) = charge leaving through S . The equation d*j = 0 also allows us to define a total charge. If P is any spacelike hyperplane and Q = J p *j, then d*j = 0 and Stokes’s theorem shows that for any other spacelike hyperplane P‘, we get Q = jpf*j. Thus Q is independent of time and of Lorentz frame. We thus will consider the basic ingredients of electromagnetic theory to be

+

-

(a) a F = R is the curvature form of a connection, (b) Js (*R)(a-’) = K Je *j. (We will refer to this equation as the generalized Gauss law. As we have seen this encompasses Gauss’s law and the modified Ampere’s law). (c) d*j = 0, which is charge conservation.

373

CURVATURE AND BUNDLE-VALUED DIFFERENTIAL FORMS

CURVATURE AND BUNDLE-VALUED DIFFERENTIAL FORMS We wish to develop the analog of the preceding discussion for the general case of a connection on a principal G-bundle. We will develop analogs for the Faraday tensor (curvature form), the currents (sources), and the generalized Gauss law (link between geometry and the sources). First we develop some calculational machinery. Recall that if n:P -+ M is a principal S1-bundle and CI a connection on P , than a determines a closed 2-form on M (see Exercise 16.6). We obtain the analog of this for more general bundles. Let n: P -+ M be a principal G-bundle and cr: V + P be a section. If CI is a connection on P , then we have seen that

g(m, g)(% 8) = Ad(K1)(4m)k)+ TL,-i(g), where E = @*a, CD: V x G + P by @(m, g) = cr(m)g and a^ = o*a. From this it follows that the curvature form is given, on V , by

+

cD*(Da)(m,g)(rc1, g)(Ki, $) = Ad(g-')(da^(m)(i, 6) [.*(rn)k,i(m)Ki]). (18.1) () = Now for u E n-'(m) define a,,,a Cq-valued 2-form on T m M , by B,,(t7 Dcr(u)(f,((), f u ( ( ) )where , 9 is the Lie algebra of G and f , the horizontal lift as described in Chapter 16. It follows from Eq. (18.1) that

(18.2)

gug(t9 0 = w g - l)(gu(t, 0). Using the adjoint action of G on 9 we obtain the associated bundle

.d= P

XG9.

Since each Ad(g) is linear and [Ad(g)A, Ad(g)B] = Ad(g)([A, B]), d is a bundle of Lie algebras. LEMMA 18.1 Let m E M . Writing R(m)(t,() = [u, %,((, [)] for (, ( E T,M, u E .-'(in) defines an dm-valued 2-form R(m):T m M x T m M + d mIf. cr: V + P is a section we get $: V x 9 + d l V by $(m, A ) = [cr(m), A ] . Then $-'(R(m)((, ()) = (m, R,(m)((, [)) where R , is a 9-valued 2-form on V given by

RO(@(t>5) = d&(m)(t,i)+ [a^(m)t7a^(m)CI. Furthermore, R,

= a*(Da).

PROOF: This all follows from Eqs. (18.1) and (18.2).

I

374

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

LEMMA 18.2 Using the same notation as in Lemma 18.1 suppose that y:V+ G is smooth and (il: V - + P by al(m) = a(m)y(m). Then R,,(m) = Ad(y(m)- ')R,(m) Vm E V . PROOF:

R(m)(t, i)= [o(m),R,(m)(t,

01 and

R(m)(t, i)= [adm), R u m ( < ?01 = [o(m)y(m),R,,(mXt, 01 = [o(m),Ad(y(m))(R,,(m)(S,011. I The quantity R which we have constructed is our first example of a bundlevalued differential form. Previously we had considered vector-valued differential forms. We saw that we could take the exterior derivative of vector-valued forms componentwise (see the end of Chapter 9). But bundlevalued forms take values in different vector spaces at different points. However, if € is a vector bundle on M which is the associated bundle to the principal G-bundle z: P -+ M obtained from a linear representation of G on some vector space E, then &-valued differential forms on M correspond to certain E-valued differential forms on P in the way that R corresponds to Da. More precisely, suppose F(m):T,M x T,M x . . . x T,M -+ c?, is an €,-valued k-form on T , M . For u E n-'(m) define F(u)(tl, t 2 , .. . , t k ) E E by the condition F(m)(Tz(ti),Tn(
tk)].

7

It is clear that if m -+ F(m) is an &-valued differential form on M , then u -+ F(u) is an E-valued differential form on P. Furthermore,

(a) F(ug) 0 ( T R J k = g-'F(u), (b) F(u)(tl, t2,. . . , &) = 0 if T z ( t j )= 0 for some j.

(18.3)

Now suppose that (i: V + P is a section. We get $: V x E + € 1 V by $(m, u) = [a(rn), u ] . Define an E-valued k-form F , on V by $-'(F(m). ( X l , . . . , xk))= (m, F,(m)(X,, . . . , x k ) ) . We then get that F , = a*(P). This follows from $-'([o(m), F ( a ( m ) ) ( ~ o ( .~.,.),,T O ( X , ) ) ] )= F(m)(X,,. . . , xk)),since T z ( T a ( X j ) )= X i . Thus,

61([(i(m)> a*(P)(m)(Xl,. . . xk)]) >

= $-'(F(m)(Xl,.

..

9

xk))?

so (m, a*(F)(m)(xl, . . . > x k ) ) = $-'(F(m)(xlj . . . xk))> We summarize the preceding constructions with the following proposition: 9

PROPOSITION (1 8.3) (Trilogy). Let z: P -+ M be a principal G-bundle and € an associated vector bundle obtained from a linear representation of G on a vector space E . Then, a smooth €-valued differential k-form on M

375

COVARIANT DERIVATIVE OF SECTIONS

is determined by any one of the following three structures: (1) a smooth section F of the bundle L:(TM, 6);thus F(m):T,M x . . . x T,M -+ B, is an alternating multilinear mapping for each m E M , (2) a smooth E-valued k-form F on P satisfying Eq. (18.3), (3) a collection {( oi,F U i )I i E I } where (a) each r/: c M is open and M = (b) each ci: -+ P is a smooth section of P; (c) each FUiis a smooth E-valued k-form on r/:; (d) on r/: n V;. we have F,J(m) = y(m)- 'FUi(m),where ojm) = oi(m)dm).

<, <

uv;

Furthermore, these structures are related in the following three ways:

(4) [% F ( u ) ( < l , . . . >
-

COVAR IA N T EXTER I0R D E R IVATIV E We want to define the exterior derivative of the original €-valued k-form F. The exterior derivative dF" is an E-valued (k + 1)-form on P . In order to induce a form on M we need a form which vanishes if any one of its arguments is in ker Tn. Thus we have DEFINITION 18.4 Let a be a connection on P and w be any E-valued k-form on P . We define an E-valued (k + 1)-form on P by D"w(u)(
.

* 3

< k + 1)

= dw(u)(ku(
>

hu(
where k, is the projection onto the horizontal subspace determined by T,P = X u0 V , (see Eqs. (16.2) and (16.3)). Recall that if c:I/ .+ P is a section we have Ic/: I/ x E .+ € 1 I/ by @(m,u) = [ ~ ( m )u ,] , 0:I/ x G +. PI I/ by @(m,g) = c(m)g,Oi = e*u, and F , = o*F. Then, @*F(m, g ) ( 4 1 ?4 2 3 . . .

?

= g-'(Fu(m)(X1,. . .3 xk))?

where t j = ( X j ,Z j ) E T(,JI/ x G). It follows from Eq. (18.4), and the fact that d(fB)

= dffl

d(@*F)(m,9) = 9- d ~ , ( m ) qy)A 9-

(18.4)

+ fdp, that

l~,(m).

(1 8.5)

376

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

Here 0 is the canonical left invariant 1-form on G, O(g) = TL,-,. The wedge product is between vector-valued forms as described in Chapter 9. The pairing B x E -+ E is obtained from the representation of G on E . If this is given by p: G -+ GL(E), then ( A , v) + T,p(A)(v). Since horizontal projection at (m, g) takes (h,g) to (h,-(oi(m)m)g) we get that

Da(@*F)(m, g ) = g-'

0

(dF,(m)

+ & @ ) A F,(m))

0

(TZ)~".

(18.6)

Thus, by Proposition 18.3, we can make the following definition: DEFINITION 18.5

We define the couariant exterior derivative of F by

DaF(m)(Xl, - ..

9

Xk+l)

= [u? D a F ( u ) ( ~ l.,.~,'&+I)],

where n(u) = m and T n ( t j )= Xj. If we define D"F, by the condition

$-l(DuF(m)@l,. ..

7

Xk+l))

= (m, D"Fu(m)(Xl,. * . 9 xk+l)),

then D"F,(m)

= dF,(m)

+ &(m)A F,(m),

(18.7)

where oi = .*(a). PROPOSITION 18.6 Let R be the d-valued 2-form on M such that l? Daa (see Lemma 18.1). Then DaR = 0.

=

+

PROOF: We work on P . By Eq. (16.4) Daa = da C I Aa. Thus d(Daa)= da A a - M A da. Since a . h, = 0 we get that Da(Daa)= 0. I REMARK 18.7 Note that in general it is not true that Da(DaF)= 0. We will refer to the 2-form R of Proposition 18.6 as the d-valued curvature form of CI.

COVARIANT DERIVATIVE OF SECTIONS AND PARALLEL TRANSPORT A special case of Definition 18.5 is when F: M + 8 is a section of the vector bundle € (F(rn)E &,)considered as an &-valued 0-form. Then F:P + & is a function satisfying F(ug) = q-'F(u) and F(m) = [u, F(u)] Vu E n-'(m). DaF is then an 8-valued 1-form, called the covariant derivative of F , and Eq. (18.7) reads D"F,(m) = dFo(m) + .^(m)( ) ( F u ( 4 ) 7 where F,(rn) = P(o(m)). Here &(m)( )(F,(m)) = T,p(oi(rn)())(F,(m)), where p: G -+ GL(E) is the representation.

377

THE GROUP OF GAUGE TRANSFORMATIONS

In Chapter 16 we saw that a connection c( defines parallel transport in any associated bundle. For associated vector bundles we have THEOREM 18.8 Let m(t) be a smooth curve in M , a I t I b, and sup-+ be the parallel transport operpose F is a section of 6. Let T,:8m,a) ators defined in Chapter 16. Then T,(F(m(a)))= F(m(t))for all t if and only if D"F(rn(t))(dm/dt)= 0, a I t I b. PROOF: Let a: I/ -+ P be a section of P so that m(t)E I/ for a It < t , . Let u(t) be the horizontal lift of m(t) with u(a) = o(m(a)).Then u(t) = a(m(t))g(t) for g: [a, t l ] -+ G, and T,(F(m(a)))= [u(t),F,], where [a(m(a)), F,] = F(m(a)).

Thus $-'T,(F(m(a))) = ( m ( t ) , f ( r ) )where f ( t ) = g(t)(F,), so that d f / d t = g(t)(F,) = -&(rn)(riz)g(F,) = -&(m)m(,f(f))and f ( a ) = F,. Let t ( t ) = F,(rn(t)) so dd/dt = dF,(m(t))(riz(t))and / ( a ) = F a . Thus t ( t ) = f ( t ) iff DaF,(m(t)). (riz(t))= 0. Since the curve can be covered by finitely many sections such as a, the result follows. I

THE GROUP OF GAUGE TRANSFORMATIONS If n:P -+ M is a principal G-bundle, then the group of all diffeomorphisms, K : P + P such that (a) n K = n, (b) K(ug) = K(u)g ' d E~P, g E G, is called the group of gauge transformations and denoted by Gauge(P). Suppose that K E Gauge(P). For u E P we have that n ( K ( u ) )= n(u) and thus we get g E G so that K(u) = ug. Define k(u) = g. Thus we get k: P -+ G such that K(u)=uk(u). Now (ug)k(ug)=K(ug)= K(u)g= u(k(u)g) so gk(ug)= k(u)g or k(ug) = g-'k(u)g. Thus we have Gauge(P) 'v ( k : P -+ GI k(ug) = g - l k ( u ) g ) c Cm(P,G). Also, if K , k , , K , k , , then K , K , k , k , . If c: I/ + P is a section, then 0

-

-

Gauge(PIV)-C"(V,G)

by

0

-

Kttkttkoa.

Gauge(P) is an infinite-dimensional (Lie) group (see [6-81). Let d = P x 9 as previously discussed and T(M, d )be the smooth sections of d.T(M, d) plays the role of the Lie algebra of Gauge(P). We obtain exp: T(M, d ) Gauge(P) -+

as follows: Suppose A E T(M, d ) .There is A": P -+ 3 such that [u, A"(u)]= A(n(u))and A"(ug) = Ad(g- ')(A(u)).Define k(t):P --* G by k(t)(u)= exp(tA(u)). Then k(t) defines K ( t ) E Gauge(P) V t and we define exp(tA) = K ( t ) .

378

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

Suppose that & = P x E is an associated vector bundle coming from a linear representation p : G -+ GL(E).Then T,p: 9 + L(E, E). Now if A , E d,, we define A,: €, + €, by A,([u, u ] ) = [u, Bu], where A, = [u, B]. Thus A E T(M, d)defines A: 6 + & which is linear on the fibers. Similarly, K E Gauge(P) defines K: E -+ 8 by K([u,u ] ) = [u, k(u)u] where K c* k as previously described. Now suppose that a is a connection on P and K E Gauge(P). Then K*a is a %-valued 1-form on P . LEMMA 18.9 K*a is a connection on P. If (5: V + P is a section, a^ = c*a, p^ = c*(K*a), and & = k 0 (r where K H k, then

+

fl(m)(riz)= Ad(&(m)-')(oi(m)m) TLk(,)- IT&(rn)(m). PROOF: K

0

R,

= R,

K*a(:lt=;

0

K gives TK

exp t B )

0

T R, = T R , 0 TK . Now

= a(K(u))0

(d4It=,

u exp tB

TK -

1

K*a(ug) 0 T R, = a(K(ug))0 TK 0 TR, = a(K(u)g)0 TR, 0 TK = (Ad(g- ') 0 a(K(u))) T K = Ad(g-') 0 (a(K(u))0 T K ) = Ad(g- I ) 0 (K*a)(u). 0

Thus K*a is a connection. To obtain the formula for &m) we use @: V x G + PIV as we have done so many times previously. On V x G, K(m, g ) = (m,f(rn)g) so that TK(m,g)(m, g ) = (m, T&(rn)(lit)g fi(rn)g) and thus,

+

+

&(K(m,e ) ) 0 TK(m,g ) = &(rn, &(m))(m,Tf(m)m f(m)g) = Ad(f(m)-')(oi(m)ni) TLi(,)- ,(TC(rn)m f(m)g)

+ = Ad(&(m)-')(oi(m)m) + TLi(,)-

I

+ T&(rn)ri~ + g.

I

COROLLARY 18.10 Let R, = c*(D"a) and S , = a*(DK*"(K*a)).Then S,(rn) = Ad(,&(m)-')0 R,(m). Thus, if Daa = 0, then DK*"(K*a)= 0. PROOF: Using b(m)fi= ff-'(rn)(d(m)fi)&(m) + ff*O(m)fi,where O is the canonical %-valued left invariant 1-form on G, we get that dfl(m)(m,6)= [ k 1 T & ( 6 )Ad(&-')a^(m)]-[&-'T&(m), , Ad(&-')a^(6)]-[ff-'T&(m), k1 T&(6)]+ &-'&(u)(m, fi)&. Thus, @(rn)(m, 6)+ [p^(m)riz, p^(rn)fi] = &-'(da^(rn)(m, 6) [oi(m)h,oi(m)fi])L I

+

THE SOURCE EQUATION AND CURRENTS FOR GAUGE FIELDS

379

THE KILLING FORM Now the adjoint representation Ad: G + G L ( 9 )induces an action of 9 on itself which, by Lemma 14.13, is T,Ad(A)(B) = [ A , B ] . We define an inner product on 9 by K ( A , B ) = trace(T, Ad(A) 0 T,Ad(B)). This is called the Killing form on 9. PROPOSITION 18.11

(a) K ( A , [B, C]) = K ( B , [C, A ] ) = K(C, [ A , B ] ) . (b) If G is compact and $9 is semisimple, then K is negative definite. (4 K(Ad(S)(A),Ad(S)(B)) = W A , B ) vg E G. PROOF: We refer the reader to [15, Chapter 111 for the proof of (b); (a) and (c) are exercises. I REMARK 18.12 It follows from part (c) of the preceding proposition that K induces an inner product on each fiber of the bundle .d = P x G $9. We will often not use the symbol K but simply write A, . B , for A,, B, E d,.

THE SOURCE EQUATION AND CURRENTS FOR GAUGE FIELDS Suppose that c( is a connection on a principal G-bundle and R is the d-valued curvature form of Lemma 18.1. In the case of electromagnetism, we saw that it was (l/a)R = F which described the physical field. Similarly, in the nonabelian case, it is F = (l/g)R which corresponds to the Faraday tensor. In terms of local representatives as used in Lemma 18.1, if we write

P = (l/Y)a^

and

F,

=

(UM,,

then we obtain F , = d p + g[P, PI. Thus we see that g is a coupling constant affecting the self-interaction of the field p, and that the physics depends upon a geometrical object (the connection) and a coupling constant (for more on this see [24]). Now by Proposition 18.6, we have D"R = 0 from the fact that R comes from a, so the geometrical equation works the same way as in electromagnetic theory. What about Gauss's law? Suppose that we are on an oriented pseudoRiemannian manifold ( M , 9). Then the *-operator is defined on differential forms (see Chapter 9). This definition extends to vector-valued differential forms as follows: If w =

C oiei, i

380

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

then *o=

1i ( * o i ) e i .

But this also extends to bundle-valued differential forms because it is purely algebraic, performed at each point independently of other points. We can thus define the covariant divergence operator on bundle-value differential forms by = (- l ) n ( k + l ) + S + l * D a * o

if w is a bundle-valued k-form (see Definition 9.60). Note that this depends on both the metric on M and the connection a on P. Thus, the analog of the second Maxwell equation is 6"R = K j ,

(18.8)

where j is an d-valued l-form which represents an "external source." We want to understand this in terms of Gauss's law. But we cannot form js * R since * R takes values in different vector spaces at different points. However, suppose z E T(M, d). Then using the inner product on the fibers of d coming from the Killing form on we get that (*R) . T

=

* ( R . 7)

is an ordinary 2-form on M . DEFINITION 18.13 Let n:P -, M be a principal G-bundle over a region of spacetime, a a connection on P, and R the d-valued curvature form of a. If T E T(M, d)then the l-form xr = *d*(R . T ) is said to be the current induced by the infinitesimal gauge transformation z. These currents satisfy the generalized Gauss law for the field R in the form

(1 8.9)

for every three-dimensional submanifold B with boundary S and for every 7E q M , d). Clearly dx, = 0. The following gives a useful expression for xr. PROPOSITION 18.14

xr = 6"R T + R AD?, '

where R AD? is R . D"T with the spacetime index of D"T contracted with the second spacetime index of R .

381

THE SOURCE EQUATION AND CURRENTS FOR GAUGE FIELDS

PROOF: We establish the formula locally using a section 6:V + P, so that local representatives R,, T , are 92-valued. Then (writing R = R , and T = T,)

(xr), = dV(R,' . T ) = d,(R,')

+

.T

+ R,'

. 8,~.

Now DUV(R,')= dy(Rpv) [.^,,,R,'] and D",(T) = d , ~+ [d,,, z ] , where .^ = G*(CI) = dvdxv. Thus, (x& = Dav(R,') . T R,' . D a v ~ [d,, R,"] . T - R,' . [dy, T ] . But by part (a) of Proposition 18.11,

R,". [dv, T ]

so the last two terms cancel.

+

= d,, . [ T ,

R,']

= t . [R,',

d,]

I

Thus we see that the conserved current generated by an infinitesimal gauge transformation 's depends linearly on T and its derivatives. Such a current is naturally of the form xr = A . T B AD?. We have seen that to obtain Gauss's law (Eq. (18.9)) we must have B = R and A = 6"R. But, by considering Gauss's law for electromagnetism, we identify A with K j where j is the external source, and thus obtain Eq. (18.8). This equation then implies that 6xr = 0. Returning now to the integral Je *xr, where B is a spacelike 2ball, we see from xr = K j . t + R AD? that the charge density involves more than the external d-currentj. There is a term coming from the field R itself. To study this more closely we study the situation using a section g: I/ -, P. We say that 6 is a choice of a gauge on V . Now when we write R , j , T we will really mean R,, j,, z, which are 92-valued quantities. Also d = o*(a) as always, and

+

(x&l

But R," . [ d v , T ]

=z.

=

(W,) . T + RllV. (0 + [d'? T I ) .

[R,", .^,,I so we get that

(xr), = ( ~ j+, [RlI',d,,]) . T + R,'

. d,t.

But using the fixed gauge (r there is a special class of T , namely, those which are constant in the gauge 6. In fact, these are determined by elements of 92 and they are linear combinations of A A , , . . . , A , (a basis for 9).We will then get n-different charges inside of S,

Q~= JB

*(Kjp

+ [ R ~ ~ $,,I). v, A ~ ,

1s i I n.

where B is the interior of S. DEFINITION 18.15

The 92-current of a gauge

V,),

= "(j,),

6:I/ + P

is given by

+ [(R,),', 41,

wherej is the external current, R is the .&-valued curvature form of a, and 6 = rJ*(a).

382

18. PHYSICAL LAWS FOR THE

GAUGE FIELDS

We have shown that PROPOSITION 18.16 If z is an infinitesimal gauge transformation which is constant in the gauge (T, then the z-charge inside a spacelike 2-sphere S is given by Q, = f B * J , * z, = fs *R, . z., These charges are conserved.

REMARK 18.17 The gauge currents are gauge dependent and do not transform using Ad if the gauge is changed.

We have shown that the currents defined by Yang and Mills [36] using a fixed gauge (which in general exists only locally) can be obtained from globally defined currents which are generated by infinitesimal gauge transformations z E r(d)via Gauss's law for the 2-forms (*R) . z. The interest in gauge fields by the physicists came from quantum field theory, where the development of quantum gauge field theory represents the work of nearly thirty years. The following two considerations are basic to these developments. (a) The coupling constant g which we discussed is dimensionless in four dimensions. If we use units in which R = c = 1, then the physical fields have dimensions of inverse length. That is, [S,] = L - l . Since F,, = a,@, - a,B, + g[j,, p,] we get [F,,,] = L - 2 + gL-' so that g is dimensionless since both terms must have the same dimensions. In quantum field theory, one constructs expressions of the form g"M,. If [M,] = L-Nn,then the N , must be bounded above in order to have a well behaved theory. (b) The structure of the actions of the group Gauge(P) on the space of connections and on bundle valued differential forms. For a development of the elementary properties of these actions see Exercises (18.8)-( 18.15).

IF=,,

EXER C IS ES 18.1 Prove Proposition (18.3). 18.2 Derive Eq. (1 8.5). 18.3 Derive Eq. (18.6). 18.4 Prove parts (a) and (c) of Proposition 18.11. 18.5 Show that 6"R = K J , where R is the d-valued curvature form of a, implies 6"J = 0. 18.6 Let ( M , g) be a pseudo-Riemannian manifold of dimension n, and z: P + A4 be the principal G-bundle of orthonormal frames, where G is determined by index of g. Suppose that a is any connection on P , and that 8 is the R"-valued 1-form on P given by 8(u) = u-' . Tz.

383

EXERCISES

Then, (a) The R"-valued 2-form D"0 is called the torsion 2-forrn of the connection CI. (b) By Eq. (18.6) and Remark 15.42 the unique connection 6 constructed in Chapter 15 is determined by the requirement that its torsion form is zero. (c) Suppose that M = V c R" is an open set, g is the usual Euclidean metric on R", and e l , e 2 , . . . , e, is the standard basis for R". If 7c: V x O(n)+ V is the orthonormal frame bundle of ( V , g), we can specify a connection on P by defining a certain field 4 ( x ) of frames to be parallel. (i) If 4 ( x ) = ( e l , . . . , en) for each x E V , we get the usual flat torsion-free Euclidean connection. (ii) If A: V-+O(n)and we take 4 ( x )= ( A ( x ) e , , A(x)e,, . . .,A(x)e,), we get a connection which is flat, by Corollary (18.10), but unless A is constant, has nonzero torsion. 18.7 Use Proposition 18.3 to give a proof of Corollary 18.9. 18.8 Let 71: P -+ M be a principal G-bundle, a a connection on P , and 3 be the Lie algebra of G. (a) If B is any connection on P , then there is a unique d-valued 1-form 6 on M such that fi - a = & where 8 is the 9-valued 1-form on P determined by 6 (see Proposition 18.3). (b) If 6 is an d-valued 1-form on M , then c( + s" is a connection on P . Thus, the choice of a gives a correspondence; connections on P are identified with d-valued 1-forms on M . For each real number t , define K , E (c) Suppose that A E T ( M ,d). Gauge(P) by K , = exp(tA). By Lemma 18.9 K:a is a connection on P for each t. Let 6, be defined by K:CI - CI = 5,. Use Lemma 18.9 to show that

the covariant derivative of A using a. 18.9 Let n:P + M be a principal G-bundle and 8 be an associated vector bundle coming from a linear representation of G on a vector space E . If F is an €-valued k-form on M and F is the corresponding E-valued k-form on P , then for each K in Gauge(P), K*F defines an 6-valued k-form on M (use Proposition 18.3 and the equations K R , = R , 0 K , 71 K = n). Thus we have a representation of Gauge(P) on the bundle valued differential forms of any associated vector bundle. Show that K*F(u) = k(u)-'F(u) where k: P + G is determined by K ( u ) = uk(u). 0

0

384

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

18.10 Consider the action of Gauge(P) on the &-valued k-forms as described in Exercise 18.9. Let p: G -+ GL(E) denote the representation of G on E . For A E T(M, d ) ,we have A: P -+ 9 as described by Proposition 18.3. Let K , = exp(tA) so that K,(u) = u exp(tA(u)), and let F, = KFF. Show that F,(u) = - Tep(A(u))F(u). Recall that we have often written the right side of this equation simply as -A"(u)F(u), the T,p being understood. 18.11 Let

71:

P

-+

M be a principal G-bundle.

(a) If F is an &-valued p-form and G is an d-valued q-form on M , we obtain %-valued forms F and G on P. The Killing form h on 9 (which is an inner product) gives an ordinary ( p 4)form FA,,(? on P (see Definition 9.97). Use the invariance of h to show that this factors down to give a ( p q)-form FA,, G on M . (b) Suppose that M carries a pseudo-Riemannian structure. If F and G are ,d-valued k-forms, at least one of which has compact support, define ( F , G ) = fM F A *G. Show that, for any connection CI on P. we have

+

+

(D"F, G)

=

( F , 13%)

if F or G has compact support.

18.12 Let 71: P -+ M be a principal G-bundle over a Riemannian manifold M , A ( M , d ) be the d-valued 1-forms on M , and be the set of connections on P. Suppose CI is a connection on P.

(a) We have a bijection 2:A(M, d)-+ V given by Z(B) = c1 + B" as discussed in Exercise 18.8. (b) Let T o ( M , d ) be those elements of T ( M , d ) which have compact support and Gauge,(P) = exp(T,(M, d).If D: = D"I To(M, d ) ,we obtain mappings

r,,(M,d)

25.+A(M,&)

6"

r(M,d).

For A in To(M, d)let K , = exp(tA) and Z(B,) = KFa. Then, by Exercise 18.8, B, = D*,A,

385

EXERCISES

so we can identify the tangent space to the orbit of Gauge,(P) through a with image D;. (c) If B is in kernel 6", then ( B , D , A ) = 0 for each A in To(M,d ) . Thus, if G is compact and semisimple, kernel 6" n image D; = 0. 18.13 Let 71: P + M be a principal G-bundle, a be a connection on P , and K be in Gauge(P). If R is the .&'-valued curvature form of a and S is the d-valued curvature form of K*a, show that = K*I? (see Corollary 18.10 and Exercise 18.9). Thus S is obtained from R by the action of Gauge(P) on the ,&'-valued forms as described in Exercise 18.9. Referring to Lemma 18.9, use the fact that the a-representatives of K*a and K*I? can be considered to be the K u a-representatives of a and r? to show that

GK'"(KR)= K 6"R. Thus, (a) if 6"R = 0, then SK*"S = 0, so the Gauge(P) group action on the connections takes the set of solutions of the source free YangMills equation (Eq. (18.8) with j = 0) into itself; and (b) if 6"R = K j then dK*"S= ri(Kj). 18.14 Let x : P + M be a principal G-bundle over a pseudo-Riemannian manifold and a be a connection on P. We have 2:A(M, d)+ V as given in Exercise 18.12. Suppose that W is a vector bundle over M and that F:%?+ T(M, W ) is a differential operator. Then F Z: A(M, d)+ T(M, W ) is a differential operator. Define L:A(A4, .d)+ UM, W )by 0

dl

F L(C)(rn)= d t r=ij

and let T

=

0

Z(tC)(rn)

L D" on T ( M , d). 0

(a) If F(K*a) = F(a) for each K in Gauge(P), then image D" c S where S is the d kernel L. For example, take F(B) = valued curvature form of [j. Then L(C) = - 6 " R . C

-

6 ( R AC)

where R is the &'-valued curvature form of a. Thus L u D" = -6 x, where ~ ( r=) xr is the current operator of Definition 18.13. (b) Let W = a? and F Z = 6*. Show that L = 6" and T = 6" D". If M is Riemannian, Exercise 18.12 gives that kernel L n image D = (0). 0

C>

L'

386

18. PHYSICAL LAWS FOR THE GAUGE FIELDS

(c) If image T = image L and image D n kernel t = (0), then each B in A(M, d)can be written uniquely as B = D"A + C, where C is in ker L, by choosing A so that T(A)= L(B). Thus we obtain a tangent plane transverse to the action of Gauge(P). This plane is tangent to the set F(P) = constant, passing through tl.

18.15 Suppose M = R4 is the spacetime of special relativity. Let P = M x G and tl be a connection on P with d-valued curvature form R. For each ~ ( m=) (m,s(m))we have the o-current of Definition (18.15) J,. Define T , = J R 3 *( J,)

= total

%charge.

Suppose that al(m) = (m,s(m)g(m))and that g(m) = h for m outside of some large bounded set. Now R,,(m) = Ad(g(m)-')(R,(m)), so that T,, = lim r+

=

a,

lim r+m

S,, *(J,,) = lim j *R,, r-tm

Sr

Jsr Ad(g(m)- ')(*R,)

= Ad(h-

')( T,).

Thus the total charge is invariant under the action of Gauge,(P).

-

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388

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Sard, R. D. “Relativistic Mechanics.” Benjamin, New York. (1970). Simmons, G . F. “Introduction to Topology and Modern Analysis.” McGraw-Hill, New York. (1963). [27] Souriau, J. M. “Structure des System Dynamiques.” Dunod, Paris. (1970). [28] Souriau, J . M. Modele de particule a spin dans le champs electromagnetic et gravitationnel, Ann. Inst. H . Poincarh., XX, No. 4. (1974). [29] Spank, E. H. “Algebraic Topology.” McGraw Hill, New York. (1966). [30] Spivak, M. “Calculus on Manifolds. ”Benjamin, New York. (1965). [31] Steenrod, N. “The Topology of Fiber Bundles,” Princeton University Press, Princeton, New Jersey. (1951) 1321 Sternberg, S. “Lectures on Differential Geometry.” Prentice-Hall, Englewood Cliffs, New Jersey. 1964. [33] Sternberg, S. On the role of field theories in our physical conception of geometry. Lecture Notes in Math., No. 676. Springer-Verlag, Berlin. (1978). [34] Sternberg, S., and Ungar, T., Classical and prequantized mechanics without Lagrangians or Hamiltonians. Hadronic J., No. I . (1978). [35] Wu, T. T., and Yang, C. N. Concept of nonintegrable phase factors and global formulation of gauge fields. Phys. Reo. D,12, No. 12, 1975. [36] Yang, C. N., and Mills, R. L. Conservation of isotopic spin and isotopic gauge invariance. Phy. Reo. 96, No. 1 (1954).

[25] [26]

- Index

A Action of Gauge(P), 378, 383-386 Adjoint representation, 264, 291 Affine frame bundle, 336 Affine group model, 349-350 Ampere's law, 37 1-372 Antisymrnetry operator, 111, 141 Area 2-forms, 169 Associated bundle, 341 Atlas, 24 Atwood's machine, 12, 101

B Bilinear map bundle, 71 Bohm-Aharanov effect, 360 Boundary of a manifold, 162 Bracket of vector fields, 135 B-slice, 185 Bundle of horizontal vectors, 337 Bundle of linear maps, 70 Bundle of states, 343 Bundle of vertical vectors, 337 Bundle-valued differential form. 374-375

C Canonical coordinates, 289 Canonical left-invariant form, 263, 292 Canonical I-form, 68-69 Canonical 2-form, 114 Chain rule, 18, 35 Change of variable formula, 197 Charge conservation, 372, 382

Chart, 24 Choice of gauge, 38 1 Christoffel symbols, 103 C'-function, 16 Closed subgroup, 290 Coadjoint action, 343 Coadjoint representation, 266 Complex line bundle, 360-361 Configuration projection, 305, 333 Configuration space, 3, 5 , 95 Connection, 318, 337 Conservation of energy, 105, 235 Conservation laws, 329-332 Conservation of momentum, 235 Conservative forces, 99 Constraint, 11, 96 Continuum assumption, 213 Contravariant degree, 127 Coset spaces, 293-294, 296 Cotangent bundle, 66 Covariant degree, 127 derivative, 376 exterior derivative, 375-376 tensor, 72,127 vector, 67 COW experiment, 359 Critical points of a function, 90 Curl of a vector field, 169 Current defined by Yang and Mills, 382 determined by a gauge, 381 in electromagnetic theory, 243-245 induced by an infinitesimal gauge transformation, 380

389

INDEX Curvature form, 327, 329 &valued, 376 integral conditions for, 362-363

D D’Alembertian, 171 De Broglie, Louis, 356 De Broglie wave, 356 Decomposable form, 145 De Rham cohomology group, 161 Derivation, 135 Derivative as a linear map, 18, 21 Determinant, 146 Diffeomorphism, 18, 34 Differential form, 150-151 bundle-valued, 374 exterior derivative of, 156-159 integral of, 196-198 support of, 196 vector-valued, 189-190 Differential of a function, 35 Differential manifold, 25 Differential I-form, 68 Differential structure, 24 Differential 2-form, 112 Displacement current, 371 Divergence of a vector field, 169, 170, 203 Divergence operator, 168 on bundle-valued forms, 380 Doppler effect, 245

E Electromagnetic field tensor, 242 Electromagnetic plane wave, 248 Energy function, 105 Equation of state, 328 Equation of structure, 339 Equivalent model, 256 Euler equations, 282-283 Exponential mapping, 289 Exterior derivative, 112, 156-159 covariant, 375-376 Exterior form, 141

F Faraday tensor, 242 Feynman, Richard, 356

Field of lines, 89 First integral, 9 Fixed energy system, 304 Flat and sharp mappings, 114 Flow of a vector field global, 54 local, 42-52 Force free motion with constraints, 102-103 Forces, 95, 301-302 Forces of constraint, 4, 96 Four-acceleration, 229 Four-current, 243-245 Four-momentum, 233 Four-potential, 250 Four-vector, 226 Four-velocity, 227 Frobenius integrability theorem, 178 Function of class Ck,16 critical points of, 90 differential of, 35 energy, 105 gradient of, 84, 169, 170 integration, 207-208 potential, 99

G Galilean transformation, 21 8 Gauge field, 342, 371-386 Gauge group action on a PMS, 310 Gauge invariant 2-form, 322-325 Gauss’s divergence theorem, 208-209 Gauss’s law, 253, 371, 380 Geodesic, 103 Geodesic motion, 102-103, 304-305, 311-320 Geometrical mechanical system, 298 for particles in a gauge field, 347 Geometric quantization, 366-367 Global flow, 54 Gradient of a function, 84, 169, 170 Green’s theorem, 205, 210 Group of gauge transformations, 377-378

H Hairy ball theorem, 65, 212 Hamiltonian function, 10, 106, 255 system, 255-256

INDEX

391

vector field, 115, 122, 255-256 Hamilton-Jacobi equation, 126 Hamilton’s equations, 107, 117, 255 Harmonic differential form, 211 Harmonic oscillator, 303 Hodge decomposition theorem, 2 12 Hodge *-operator, 165- 168 Holonomic constraints, 95 system, 11, 95 Holonomy group, 361 operator, 361 and reduction of structural group, 367 restricted group, 367 Homomorphism continuous, 296 smooth, 291 Hopf fibration, 336 Horizontal lift, 3 18 of curves, 341 of vectors, 338 Hyperbolic motion, 232 Hypersurface null, spacelike and timelike, 77 orientation of, 154-155

1

Imbedding, 64 Implicit function theorem, 23 Index of a metric, 74 Inertial reference frame, 214 Inertia tensor of a rigid body, 275 Infinitesimal gauge transformation, 380 current induced by, 380 Infinitesimal generator, 7 Inaccessability theorem, 187 Integrable subbundle, 176 Integral cohomology class, 364 Integral curve, 2, 9 , 40 Integral of a differential form, 196-199 Integral of a function, 207-208 Integral manifold, 184 maximal, 185-186 Interior of a manifold with boundary, 162 Interior product, 156 Invariant measure, 292-293, 296 Invariant spacetime interval, 223 Invariant subbundle, 195

Inverse function theorem, 19

J Jacobi identity, 135

K Kernel of a differential form, 173 Killing form, 379 Kinetic energy function, 94 metric, 94 relativistic, 236 Kronecker delta, 128

L Lagrange multipliers, 38 Lagrange:s equations, 7, 99-100, 121, 255 Lagrangian function, 7, 99, 121, 255 time dependent, 121 two-form, 118, 255 vector field, 118 Laplace-Beltrami operator, 168 Left-invariant I-form, 263, 292 Left-invariant vector field, 287 Legendre transformation, 8, 103, 125, 255 Lie algebra, 135, 287 Lie bracket, 135 Lie derivative, 132 moving frame description of, 134-135 Lie group, 286 adjoint representation of, 29 1 canonical coordinates on, 289 canonical left-invariant I-form on, 292 closed subgroup of, 290-291 exponential mapping of, 289 homomorphism, 291, 296 invariant measures on, 292-293, 296 left action of, 294-295 Lie algebra of, 287 right action of, 335 Lightlike vector, 75 Linear map bundle, 70 Liouville’s theorem, 299 Lipschitz condition, 41 Lorentz-Fitzgerald contraction, 222

392

INDEX

Lorentz force law, 239, 306 Lorentz gauge condition, 250 Lorentz group, 224 Lorentz transformation, 2 19

M Magnetic field of a line current, 252 of a monopole, 252 of a moving point charge, 242 Manifold with boundary, 162 Maximal integral curve, 53 Maximal integral manifold, 186 Maxwell’s equations, 246-248 Meter of time, 228 Metric existence of, 87 index of, 74 left-invariant, 260 kinetic energy, 94 Lorentzian, 74, 76 Riemannian, 74, 76 volume element of, 149-150 Minkowski spacetime, 170- 171 Mobius strip, 153 Momentum mapping, 265 Moving frames, 311-320, 332-333

N Nonholonomic constraints, 189 Nonintegrable subbundles, 188 Null hypersurface, 77 subspace, 75 vector, 75

0 Orientation of a hypersurface, 154 induced on the boundary, 164 of a manifold, 151 of a vector space, 147 Orthochronous Lorentz transformation, 225 Orthogonal group, 256 Orthonormal frame, 312 Orthonormal frame bundle, 313, 336

P Parallel frames, 341 Parallel transport, 342, 376-377 determined by a foliation, 365 Partition of unity, 86 Phase factor calculations, 357, 359, 365-366, 368 Phase plane analysis, 283 Phase space, 5, 8, 106, 255 Physical process, 213 Poincare group, 224 Poincark lemma, 161 Poinsot construction, 280-281 Polarization, 366 Positively oriented basis, 147 Positive unit normal, 155 Potential function, 99 Principal axes, 276 Principal fiber bundle, 312, 335 Probability amplitude, 355 Probability amplitude phase factor, 356 Projective plane, 154 Projective spaces, 191 Proper Lorentz group, 224 Proper time axiom, 213 Proper value, 307 Pseudomechanical system, 306 closed, 306 integrable, 306 for particles in a gauge field, 345 Pseudo-Riemannian manifold, 75 Pullback bundle, 352 commutes with exterior derivative, 159 of a covariant tensor, 78 of a differential form, 151 of a vector field by a diffeomorphism, 77 Pushforward, 77

Q Quantum effects, 354 Quantum field theory, 382

R Reference frame, 2 13 Relativistic correction to Newtonian mechanics, 234

INDEX

393

energy, 233 4-momentum, 233 kinetic energy, 236 law of velocity addition, 220-221 length contraction, 222 mass, 236 time dilation, 223 time units, 227-228 3-momentum, 234 velocity. 227 Relativity group, 213 Rest frame, 229 Rest mass, 236 Restricted system, 307 Restriction mapping, 307 and particles, 308 Rigid body, 308-310 angular velocity vector of, 262, 279 body motion of, 267-268, 271 body representation of, 259 body stable theorem for, 277 conserved momentum of, 261, 266 Euler equations for, 282 inertial ellipsoid of, 280 inertia tensor of, 275 invariable plane of, 281 left invariant metric for, 260 moment of inertia of, 262, 276 outline of analysis of, 261-262 phase plane analysis for motion of, 283 Poinsot construction for, 280-281 principal axes of, 276 space motion, 260, 266-267 space representation of, 259 space stable theorem for, 280 special space motion, 261, 266-267 stationary rotation of, 274 Riemannian metric, 76 Rotation group, 256-258, 271 -274

S Sard’s theorem, 307 Scalar invariants in electromagnetism, 253 Sharp and flat mappings, 114 Simple pendulum, 100 Simultaneity in special relativity, 222 Skew symmetric tensor, 110, 141 Souriau equations, 32 1 Souriau GMS, 328-329, 352-353

Spacelike hypersurface, 77 subspace, 75 vector, 75 Space odyssey, 229-232 Spacetime, 2 I3 manifold, 225 metric, 226 Special relativity, 214 Standard topology on the cotangent bundle, 67 on the tangent bundle, 61 Starlike set, 159 State, I , 5 State space, I , 4,95, 255 Stationary rotation of a rigid body, 274 stability of, 277-280 Sternberg’s theorem, 346 Stokes’s theorem classical version, 209 for a generalized cylinder, 205-206 for a manifold with boundary, 201 for rectangles, 205 Straightening out theorem, 177 Structural group, 335 Subbundle of a vector bundle, 172 integrable, 176 Submanifold, 26 Submanifold chart, 26 Symmetric product, 81 Symmetry groups, 330-331, 333-334 Symplectic structure, 110

T Tangent bundle, 61 Tangent space, 30 Tangent vector, 29 Tensor, 127 Tensor bundles, 140 Tensor field, 129 Time dependent mechanical system, 121-124 vector field, 41 Timelike hypersurface, 77 subspace, 75 vector, 75 Topology and critical points, 90 Torque, 309-310

INDEX Torsion form, 383 Total charge for electromagnetism, 372 for gauge theory, 386 Total force mapping, 95 Trajectory, 2, 4 , 9 , 106, 255 Transformation law for contravariant vector components, 29 for covariant components, 67 for tensor components, 130 Trilogy, 374-375 Two-form on coadjoint orbits, 343-344

U Uncertainty principle, 358

v Variational principle, 300 Vector bundle, 63, 172 Vector field, 39 complete, 55 curl of, 169

as a differential operator, 137 divergence of, 169-170 global flow of, 54 Hamiltonian, 115, 255-256 integral curve of, 2, 40 Lagrangian, 118, 255 left invariant, 287 local flow of, 42-52 local representative of, 39 maximal integral curve of, 53 o-divergence of, 203 time dependent, 41 Vector subbundles, 172 Vector-valued differential form, 189-190 Velocity boost, 220 Velocity parameter, 22 I Velocity vector, 32 Virtual work, 96 Volume element, 149 Volume 3-forms. 171 W Wedge product, 111, 142, 190 World line, 213, 226

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