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ynawics on loKenrz Mampolds

ynaancs on lonemz Mamplds

SCOT Adams University of Minnesota

USA

V f e World Scientific WH

New Jersey • London •'Sim Singapore • Hong Kong

Published by World Scientific Publishing Co. Pte. Ltd. P O Box 128, Farrer Road, Singapore 912805 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

DYNAMICS ON LORENTZ MANIFOLDS Copyright © 2001 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 981-02-4382-0

Printed in Singapore by World Scientific Printers

To IOJIHH and AjieKceft

Preface

We begin with a quote from a biography of Felix Klein taken from the site http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Klein.html on 8 November 1999: Klein's synthesis of geometry as the study of the properties of a space that are invariant under a given group of transformations, known as the Erlanger Programme (1872), profoundly influenced mathematical development . . . The Erlanger Programme gave a unified approach to geometry which is now the standard accepted view. Transformations play a major role in modern mathematics and Klein showed how the essential properties of a given geometry could be represented by the group of transformations that preserve those properties. In other words, central to Klein's Erlanger program is the idea of studying geometry through symmetry. This is the main theme of this book, applied specifically to a class of geometric objects called Lorentz manifolds.

Contents

Preface

vii

Chapter 1 Introduction, History and Outline 1.1 Lorentz manifolds and relativity . . 1.2 Symmetries of Lorentz manifolds 1.3 Outline of succeeding chapters 1.4 Notation 1.5 Acknowledgements

1 1 5 7 8 9

Chapter 2 Basic Results and Definitions 2.1 Some set-theoretic notions 2.2 Some group-theoretic notions 2.3 Some topological notions 2.4 Some notions from linear algebra 2.5 Matrix concentration lemmas 2.6 First results on expansive sequences 2.7 Topological groups 2.8 Discrete groups 2.9 Proper actions 2.10 Bilinear and quadratic forms 2.11 Root systems 2.12 Minkowski forms - basic definitions

11 11 14 15 18 22 23 25 29 31 32 35 39

Chapter 3 Basic Differential Topology 3.1 Some differential topological notions

41 41

ix

x

3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Contents

Inheritability of continuity and smoothness to leanike submanifolds Definition of prefoliation and foliation Preliminary results to the Frobenius Theorem Uniqueness in the Frobenius Theorem Passage from local to global in the Frobenius Theorem The Frobenius Theorem Potential submersions Lorentz metrics - basic definitions

Chapter 4 Basic Lie Theoretic Results 4.1 Some Lie theoretic definitions and notation 4.2 Dynamical consequences of the Frobenius Theorem 4.3 exp, Ad and ad 4.4 The Lie group Lie algebra correspondence 4.5 Some facts about Lie subgroups 4.6 The Lie algebra oi[A,B] 4.7 Lie groups and Lie algebras from bilinear and quadratic forms 4.8 Abelian Lie groups 4.9 Miscellaneous results 4.10 Generalities on semisimple groups and algebras 4.11 Real Jordan decomposition 4.12 Consequences of results on real Jordan decomposition 4.13 Generalities on algebraic groups 4.14 Generalities on nilpotent groups and algebras 4.15 Generalities on the nilradical 4.16 Relationships between representation theories

47 49 51 52 54 55 57 62 65 65 70 75 80 83 90 91 93 96 112 128 131 134 138 141 149

Chapter 5 More Lie Theory 151 5.1 Connection-preserving diffeomorphisms form a Lie group . . . . 151 5.2 The isometry group of a pseudoRiemannian manifold is a Lie group 153 5.3 More results on expansive sequences 154 5.4 Lie groups densely embedded in other Lie groups 156 5.5 Generalities on the Levi decomposition 160 5.6 Large normalizers and centralizers 163 5.7 Representation theory 166

Contents

xi

Chapter 6 Minkowski Linear Algebra 175 6.1 Notations for important elements and Lie subalgebras of so(Qa) 175 6.2 Linear algebra of Minkowski vector spaces 176 6.3 Basic calculations 184 6.4 Embeddings of Lorentz Lie algebras 192 Chapter 7 Basic Dynamical Results 7.1 Kowalsky's Lemma 7.2 Higher jets of vector fields and metrics - notation 7.3 Matrix realizations of jets and calculus on jets 7.4 Miscellaneous results 7.5 A basic collection of rigidity results 7.6 Strongly lightlike and nontimelike vectors 7.7 Basic results on degenerate orbits 7.8 More on strongly lightlike and nontimelike vectors 7.9 Nonproperness and cocompact subgroups 7.10 Kowalsky subsets 7.11 Types of chaotic actions 7.12 Induction of actions: Definition 7.13 Induction of actions: Basic results 7.14 Riemannian dynamics

197 197 198 202 206 221 224 225 227 235 236 238 249 250 251

Chapter 8 Examples of Actions on Compact Lorentz Manifolds 261 8.1 The isometry group of a compact quotient of SL2(M) 261 8.2 Twisted Heisenberg groups 262 8.3 SL 2 (E), twisted Heisenberg and closure 268 Chapter 9 Examples of Nonproper Actions 9.1 Restriction-Induction 9.2 Examples constructed using a quadratic form in the Lie algebra 9.3 A nilpotent Lie group without nonproper Lorentz dynamics . . 9.4 Groups with SO(n, 1) or SO(n,2) as a local direct factor . . . . 9.5 Groups with a normal subgroup isomorphic to K

271 272 276 278 280 284

Chapter 10 Semisimple Groups Admitting a Nonproper Action 285 10.1 Locally free actions of SL3(K) 285

xii

10.2 10.3 10.4 10.5

Contents

Locally faithful actions of SL3(R) Locally faithful actions of Sp3(K) C R 6 x 6 Locally faithful actions of SU(2,1) Kowalsky's theorems

287 288 290 291

Chapter 11 Groups with Action on a Compact Lorentz Manifold 293 11.1 Local freeness 293 11.2 Nilpotent Lie groups 297 11.3 Solvable nonnilpotent Lie groups with Abelian nilradical . . . . 298 11.4 Solvable nonnilpotent Lie groups with nonAbelian nilradical preliminaries 300 11.5 Solvable nonnilpotent Lie groups with nonAbelian nilradical final results 306 11.6 Semisimple Lie groups 310 11.7 The general case 311 Chapter 12 The Isometry Group of a Compact Lorentz Manifold 315 12.1 Oseledec splitting 316 12.2 Preliminaries 317 12.3 A candidate for the new Killing field 319 12.4 Final classification result 326 Chapter 13 Highly Symmetric Compact Lorentz Manifolds 327 13.1 SL2(K)-actions on compact Lorentz manifolds 327 13.2 Actions of twisted Heisenberg groups on compact Lorentz manifolds - a sketch 331 Chapter 14 Locally Free Orbit Nonproper Lorentz Actions 335 14.1 Preliminary result to degeneration of Adg^j 336 14.2 Degeneration of Ad0<7j 337 14.3 Locally free actions of simply connected Lie groups 341 14.4 Compact subgroups not in the radical 345 14.5 Sequences with no large Kowalsky subsets 347 14.6 Compact connected stabilizers 350 Chapter 15 Orbit Nonproper Lorentz Actions 15.1 Preliminaries

357 357

Contents

15.2 15.3 15.4 15.5 15.6 15.7

xiii

A strongly vanishing element A strongly lightlike element Nilradical is free Nilradical is not free, but its center is free, Part I Nilradical is not free, but its center is free, Part II The main result

Appendix A The Borel Density Theorem A.l Notation and terminology A.2 Preliminaries A.3 The Borel Density Theorem Appendix B

363 366 367 372 375 377 381 381 382 383

Tameness of Algebraic Actions in Characteristic

Zero

387

Bibliography

391

Index

395

Chapter 1

Introduction, History and Outline

We begin with a very elementary motiviation of Lorentz geometry, to be followed by a more technical description of the contents of this book.

1.1

Lorentz manifolds and relativity

To understand the importance of Lorentz manifolds in physics, assume that the speed of light is 1 and imagine, at time t = — 1, a two-dimensional universe which is empty except for the following: (1) A particle called "he", located at (0,0) and not moving. (2) A particle called "she", located at (—1/2,0) and moving at velocity 1/2 toward him. (3) For each point on the circle x2 + y2 = 1, a particle of light moving at speed 1 toward him. In this male-centric universe, he perceives that all things are moving toward him and that he is located at the center of all the action. This proceeds until the great romantic moment t — 0, when he and she come together and are bathed in light from all sides. Then, alas, our two main players separate as do the particles of light. All objects in this universe then move away from him for the rest of time. So, for example, at time t = 8, he is still located at (0,0) and she is at (4,0). The light is located at the locus of points (x,y) satisfying x2 + y2 = 8 2 , i.e., the light forms a circle of radius 8 about the origin. To complete the picture, one should also imagine that all of these particles existed infinitely far in the past. So, for example, at time t = —2, she l

2

Introduction,

History and

Outline

would be located at (—1,0). Also, at t = - 2 , the light would form a circle x2 + y2 = 2 2 twice as large as at time t = — 1. To see everything at once, we can imagine a three-dimensional "spacetime", with axes labeled x, y and i. Our original universe is obtained by slicing our space-time with the plane t = — 1. In this slice, we see a circle of light located at points {(x,y, - 1 ) | x2 + y2 = 1}, plotted, say, in yellow. We see him at (0,0, —1), plotted, say, in blue. We see her at (—1/2,0, —1), plotted, say, in pink. At time t = 0, the yellow, blue and pink are all plotted at (0,0,0). At time t = 1, the yellow forms the circle {(x,y, 1) | x2 + y2 = 1}, the blue is plotted at (0,0,1) and the pink is plotted at (1/2,0,1). At time t = 2, the yellow forms the circle {(x,y,2) \x2 + y2 = 4}, the blue is plotted at (0,0,2) and the pink is plotted at (1,0,2). Et cetera. The set of yellow points is called the "light cone". It is the cone {(x,y,t) | x2 + y2 = t2}. The set of blue points is called his "lifeline". It is the i-axis. The set of pink points, her lifeline, is the line {(t/2,0, i) 1i € E}. Note that there is one particular particle of light which is chasing her toward him. This is the particle with lifeline {(t,0,t) 1i 6 R}. Let us call this particle LQ. Changing perspective, let us try to imagine what his happening from her point of view. She perceives herself at the center of the universe, so, from this point of view, her lifeline should be the i-axis. According to 19th century physics, from her point of view, he is traveling toward her with velocity —1/2; she would say that his lifeline is {(—i/2,0,i) \t € E}. What of the light cone? Let us focus for a moment on the light particle LQ that is chasing her toward him. According to 19th century ideas, from her point of view, she should see L0 as moving a velocity of 1/2 toward her. That is, from her point of view, the particle L0 should have lifeline {(i/2,0,i) | i € E}. More generally, under a 19th century understanding of physics, if he perceives an event which he plots at (x,y,t), then she will plot it at (a; — (t/2),y,t); in particular, she would plot the light cone as {(x-(t/2),y,t)\x2 + y2 = t2}. This need to constantly switch between perspectives of various participants is highly invidious, and we propose to resolve all this by postulating the existence of a single space-time M, on which all events occur, with no particular coordinates attached to it. What he does is to choose a preferred

Lorentz manifolds and

relativity

3

diffeomorphism $ : M ->• E 3 . To describe an event at a point m € M, he would say that it occurred at $(m). Similarly, she prefers her own diffeomorphism (or "coordinates") * : M -> R 3 , and she would say that this same event occurred at \t(m). Yet another observer might choose yet another coordiatization of the "real" space-time M. These coordinate systems (like $ and * ) are sometimes called "inertial reference frames". Changing coordinates from his to hers is given by the map F := \t o $ - 1 :M3 ->M 3 , and, in the preceding paragraph, we asserted that, under pre-relativistic mechanics, it would be reasonable to have F(x,y,t) = (x — (t/2),y,t). Here is the catch: Experimental evidence indicates that this is simply not correct. She does not perceive L0 as moving at speed 1/2. In fact, amazingly, the empirical evidence is that she perceives all the particles of light as moving at speed 1. Her plot of the light cone is exactly the same as his; it is £ := {(x,y,t)\x2 + y2 = t2}. Then F(£) = £, i.e., F preserves £. In particular, the formula F(x,y,t) — (x — (t/2),y,t) simply cannot be true, so 19th century physics fails to describe accurately our universe. In our modern understanding of the "overlap" map F, we know more than just that F(C) — £: Let Q : I 3 -> E be the quadratic form defined by Q(x,y,t) — x2 + y2 — t2, so that the light cone £ is the 0-level set of Q, i.e., £ = <5 -1 (0). Then the basic axiom of special relativity is that F preserves all the level sets of Q, not just the 0-level set £. That is, Q o F = Q. In other words, F preserves Q. The collection of linear transformations which preserve Q is called the orthogonal group of Q, and is denoted by O(Q), so the basic axiom asserts that F € O(Q). We now set this aside for a moment. The 18th and 19th centuries saw the beginnings of what is now called Riemannian geometry. In its simplest form, one imagines a smooth surface X C E 3 to which particles are constrained to move. Under this constraint, minimizing travel distance between two points p, q € X is a basic problem in the calculus of variations. For any tangent vector v to X, let \v\ denote the length of v. For any path 7 : [0,1] -> X from p to q, define the length of 7 by £(7) = / 0 |-y(t) | E given by the formula gx(v) = \v\2. Abbreviating gx as g for readability, we define the length of 7 to be £(7) := / 0 (fKiM)) 1 / 2 dt; we wish to find 7 from p to q minimizing length among all competing paths. If the surface X is a flat plane in E 3 , then, as usual, the distance minimizing curve is a straight line. However, if

4

Introduction,

History and Outline

the surface is curved, then the answer is likely to be more involved. Note that we do not really require the ambient space R3 containing X. All one needs is a Riemannian manifold, i.e., a manifold X together with a system of positive definite quadratic forms {gx : TXX —• R}xex, one on each tangent space to X. These forms should vary smoothly in x. Again define L(-y) := J 0 (giJit)))1^2 dt, and we look to minimize L over paths connecting a pair of points. It is important to note that there is not a unique parametric path that minimizes distance, even in a fiat plane P in R3. Given two points p,q £ P, the path 70 : [0,1] —> P defined by 70(t) — p(l — t) + qt minimizes the distance, but so does the path 71 : [0,1] -»• P defined by 71 (t) = 70(t 2 ). This is because L is invariant under reparameterization. Given a Riemannian manifold X and two points p, q € X, a better quantity to study is the energy of a path 7 : [0,1] -¥ X defined by E(j) :— J0 g(j(t)) dt. It turns out that energy minimizing paths are the same as constant speed length minimizing paths. (See §12 of [Mi73].) The positive definiteness of the quadratic forms ensures that the square root in the definition of L makes sense. However, for manifolds carrying other kinds of forms, one could still propose either to minimize E or to minimize L over paths for which R3, g corresponds to a pseudoRiemannian structure j $ on M . Since F preserves g, we see that g$ = g^. That is, while different observers will give different coordinates to the same point in M, they all agree on a particular

Symmetries

of Lorentz

manifolds

5

pseudoRiemannian metric on M. Let d > 2 be an integer and let V be a real d-dimensional vector space. If Q : V -> R is a quadratic form, then we say that Q is Minkowski if there is an isomorphism / : Rd —> V such that, for all (x\,..., xj) € E d , we have Q(f(xi,...,Xd)) — x\ + • • • + x\_x — xd. If g is a pseudoRiemannian metric on a manifold X, then we say that g is Lorentz if, for all a; € X, the quadratic form gx on TXX is Minkowski. With this terminology, special relativity, then, is the study of a flat Lorentz manifold, i.e., a vector space on which a single Minkowski form is used to create a pseudoRiemannian metric by identifying V with each of its tangent spaces. In our example, we described a space-time with three dimensions, but it is more typical, within special relativity, to analyze fourdimensional space-times. By contrast, general relativity considers a more general class of Lorentz manifolds, those whose metric comes from a certain system of differential equations. Once this metric is determined, the equations of motion of particles are denned by minimizing energy, in analogy with Riemannian geometry. The collection of symmetries of a pseudoRiemannian manifold (i.e., the collection of self-diffeomorphisms preserving the metric) forms a group called its isometry group. Typically, the kinds of Lorentz manifolds appearing in general relativity are four-dimensional and have compact isometry groups. Their isometry groups therefore act in an uncomplicated manner (e.g., with closed orbits), from a dynamical point of view. In this work, we set aside the motivations of special and general relativity, and focus on Lorentz manifolds with large, complicated isometry groups, and in any dimension.

1.2

Symmetries of Lorentz manifolds

In any kind of dynamics of groups, it is basic to determine the collection of groups that have "interesting" actions. As the definition of "interesting" is varied, one obtains many different areas within dynamical systems. Working within a particular area, once the list of groups is complete, a second (and usually harder) problem is to figure, for each one, the collection of interesting actions that it admits. If this can be completed, then one can reasonably claim to finished an area of dynamical systems.

6

Introduction,

History and

Outline

It has been known since early work of Cartan that the isometry group of any "structure of finite type" (a class of structures which includes Lorentz metrics) has a unique manifold structure that makes it into a Lie group. The basic structure theory of Lie groups asserts that any connected Lie group decomposes into two parts: its "semisimple" part and its "solvable" part. (See the first two paragraphs of §4.9, p. 96.) So, in tackling a problem in Lie theory, it is quite common to break the problem into semisimple and solvable parts. Semisimple groups break up into simple groups. (See Corollary 4.10.3, p. 113.) The best understood of the solvable groups are the nilpotent groups. Thus one typically tries to do a nilpotent form of the problem and a "simple" form (which, terminology notwithstanding, is often very complicated). If successful, one is emboldened to study solvable and semisimple problems. Finally, one may aim to solve the general problem. The basic goal of much of this book is to determine the collection of Lie groups admitting an interesting action on a Lorentz manifold. In different places we define "interesting" in different ways. We sometimes focus on the semisimple form of the problem, in which we wish to find all semisimple Lie groups with an interesting action. In other places we focus on simple groups, nilpotent groups and solvable groups. What might one require for an action to be interesting? Since manifolds are smooth objects, it is natural to require that the action be smooth. For actions on Lorentz manifolds, another basic requirement is that the action be by isometries, i.e., that each element act as an isometry of the Lorentz manifold. We will say that an action of a Lie group on a Lorentz manifold is isometric if it is smooth and by isometries. An isometric action of a Lie group G on a Lorentz manifold M is smooth iff the action is given by a smooth homomorphism from G to the isometry group Isom(M) of M. Any group can act trivially on any manifold, and such an action is probably not very interesting. By definition, the kernel of an action is the collection of all group elements which fix all points of the space; it is a normal subgroup. So, to be interesting, one might require that an action be faithful, which is to say that its kernel is trivial. Any non-faithful action descends to a faithful action of the group mod the kernel, and so the original action is simply lifted from a faithful action. A slightly less stringent requirement is that the action be locally faithful, which is to say that its kernel is discrete. If a Lie group G acts locally faithfully on a set M and if G ->• G is a covering homomorphism, then the lifted action of G on M is again locally faithful. So local faithfulness

Outline of succeeding chapters

7

is preserved on passing to covers; faithfulness is not. The Lie algebra of a Lie group is the same as the Lie algebra of any of its covering groups, so, from a Lie theoretic point of view, it is not uncommon to think of a group as being essentially unchanged by passing to a covering. Extending this philosophy, one might want to lift actions to coverings without considering the lifted action to be essentially different from the original action. From this point of view, local faithfulness is a reasonable requirement. The basic requirements that an action be isometric and locally faithful appear nearly everywhere in this book. However, every Lie group acts faithfully on itself by left-translation, preserving any left-invariant metric. This action has only one orbit, so the orbit structure is clearly as uncomplicated as can be. In the remainder of the book, we try out various additional dynamical requirements on our actions. The general philosophy of much of what follows is that, while any Lie group can act faithfully and isometrically on a Lorentz manifold, if one imposes even small additional requirements, then the list of groups is dramatically reduced.

1.3

Outline of succeeding chapters

In this book, we assume, on the part of the reader, a good basic understanding of measure theory, topology, and differential topology. A typical first-year graduate course in these subjects will cover the needed material. We require basic facts from Riemannian geometry and Lie theory, but we develop most of the needed definitions, and we liberally cite results from standard texts so that a student who is not versed in these areas should find the arguments complete. We should point out that, while we have made an effort to prove or give careful references for all of the results herein, it is nevertheless true that we rely on a significant level of mathematical maturity, especially in Chapter 15. In Chapter 2, we collect together the miscellany of elementary definitions and results that we will be using. Chapter 3 concerns basic results in differential topology, i.e., basic facts about smooth manifolds. In Chapters 4 and 5, we set up the facts from Lie Theory that we will need. In Chapter 4, one can find an introduction to many basic topics in the theory of real Lie groups, although, to keep the exposition within reasonable bounds, we do not start from scratch, and one will find a number of references to basic texts. We sometimes reprove some well-know facts if we feel that the

8

Introduction,

History and Outline

exposition in the literature can readily be improved, or is not available. For the most part, Chapter 4 develops results that are generally of interest in Lie Theory, whereas Chapter 5 collects results that are of special interest in this book. Basic information on the structure of Minkowski vector spaces and their isometry groups can be found in Chapter 6. Chapters 2, 3, 4 and 6, as well as Appendices A and B, contain material that is well-known. Some of the material in Chapter 5 may also be described as well-known. In Chapter 8, we give a collection of examples of actions of Lie groups on compact Lorentz manifolds. Eventually, in Chpaters 11 and 12, we show that the collection of Lie groups analyzed in Chapter 8 is, in some sense, exhaustive. In Chapter 13, following M. Gromov and A. Zeghib, we show that, for some of these groups, we can describe their possible actions on compact Lorentz manifolds. In Chapter 9, we consider noncompact Lorentz manifolds. We give examples of groups admitting actions on such manifolds that exhibit some level of complicated orbit structure. In Chapters 10, 14 and 15, we show that the list of groups analyzed in Chapter 9 is, again, exhaustive. Chapter 10 focuses on semisimple groups acting on Lorentz manifolds, and describes work originally due to N. Kowalsky. Chapter 14 focuses on free actions and actions with compact stabilizers. Chapter 15 studies general isometric actions of general simply connected Lie groups. The problem of proving completeness of the various lists of groups obtained requires some new basic dynamical techniques, which are found in Chapter 7, which may be thought of as a kind of "toolbox" to be used in proving the results of Chapters 10-15. Appendix A contains an elementary proof of the Borel Density Theorem. Appendix B contains an elementary proof that algebraic actions are smooth.

1.4

Notation

There is a significant amount of notation in this book, but, to aid the reader, we have tried to reference it in the index, under "ZZZ Notation". Some notation is introduced and used repeatedly in many different chapters, and the reader would be well-advised to become familiar with it as quickly as possible. This includes Qd, B, I/>B, ^>{d\ ft(d), N(d]', V{d\ d (d m m 1 1 Nl \ E\f, e \ XB, Xg , X% , OMV, OUT *, MV, AfV *, vt -» v, Sm

Acknowledgements

9

and g|h. All these can be found under "ZZZ Notation" in the index. Throughout this book, if a Lie group is denoted by a capital Latin letter (perhaps with some adornments like asterisk and/or subscript), then the Lie algebra of that Lie group will always be denoted by the corresponding small fraktur character (with the same adornments). So, for example, if H£ is some Lie group, then its Lie algebra is denoted \j%. 1.5

Acknowledgements

This book was written with the support of the National Science Foundation (DMS-9703480 and DMS-0072165). I also acknowledge with gratitude the support of my department, which generously gave me a paid one-semester leave to provide time for the writing of this book. D. Witte has answered innumerable questions over the years, and, without this help, the results appearing here might be far less sharp. I also thank my wife, IOJIHH, who provided constant support and who tolerated, with good humor, my occasional weekend absences to work on this manuscript.

Chapter 2

Basic Results and Definitions

Throughout this text, Z denotes the set of integers, Q denotes the set of rational numbers E denotes the set of real numbers and C denotes the set of complex numbers. Let M+ := (0, oo) be the set of positive real numbers. If S and T are sets, then S\T := {s e S\s $T}. A Borel space is a set with a a-algebra. The sets in the u-algebra are referred to as Borel sets. A Borel measure on a Borel space (X, B) is a countably additive, [0, oo] such that n(X) ^ 0. If X and Y are Borel spaces, then a map / : X ->• Y is Borel if, for all Borel subsets B C Y, we have that f~x(B) is a Borel subset of X. If X is a topological space, then, unless otherwise specified, X will be given the smallest cr-algebra containing its topology, so that X is also a Borel space.

2.1

Some set-theoretic notions

Let X and Y be sets and let / : X ->• Y be a function. A subset A C I i s called a fiber of / if, for some y € Y, we have A = f~x (y). A subset AC. X will be said to be /-saturated if, for some B C Y, we have f~1(B) = A. If / : X -4 Y is surjective, then a cross-section to / is a map g : Y -> X such that, for all y £ Y, we have f(g(y)) — y-

L E M M A 2.1.1 Let X and Y be sets and let f : X ->Y be a function. Let A,B C X. Assume that A is f-saturated. Thenf(AnB) = (f (A))n{f (B)). 11

Basic Results and

12

Definitions

Proof. We have f(AC\B) C f(A) and f(Ar\B)Cf(B), so it suffices to show that (f(A)) n (/(B)) C / ( 4 n B). Choose ,4' C y such that f~1(A') = A. Let ye (f(A))n(f(B)). We wish to show that y £ f(AnB). Choose b £ B such that f(b) = ?/. We have

f(b) = yef(A) so b £ / _ 1 ( ^ ' ) = ^y

So

>

since

f(f-1(A'))CA',

=

& £ B , we get 6 6 A n B. It follows that

= f(b)ef(AnB).

a

Let X be a set, let v, w £ X and let y be a, collection of subsets of X . For any integer n > 1, we define a 3^-chain of length n from v to w to be an (n + l)-tuple (XQ, • • •, xn) £ Xn+1 such that xo — v, such that xn = w and such that for all integers k £ [l,n], there exists 7 £ ^ such that both :Efc_i € y and Xk € y . We say that v and u; are 3^-chain connected if, for some integer n > 1, there exists a ^-chain of length n from u t o w . Let X be a set and let y be a collection of subsets of X. We say that X is [y-connected if any two points of X are y-chain connected. Let R be an equivalence relation on a set X. For all x € X, we let R(x) := {y e X \ (x,y) £ R} denote the equivalence class of x. For all subsets X0 C X, we define R(X0) := [J R(x). x£X0

Let X be a set and let S C X x X. Then the equivalence relation on X generated by S is the smallest equivalence relation on X containing S. Let 5* := {(y,x) | (z,2/) € S} and let £> := {{x,x)\x £ X}. If R is the equivalence relation generated by S, then (a;, y) £ R iff there exists an integer n > 1 and there exist zo, • • •, zn £ X such that • zo = x; • zn = y; and • for all integers i £ [1,n], we have (ZJ_I,Zi) £ SU Sl ISD. Let X be a set and let S C X x X. For all a; € X, let S(x) :={y£X\

(x, y) £ S},

S\x)

:= {y £ X \ (y, x) £ 5}

and E(x) := (S(x)) U (Sl(x)) U {x}. Let £ := {£(a;) | a; £ X } . Let i? be the equivalence relation on X generated by S. Then, for all x,y £ X, we have: (a;, y) £ R iff a; and y are £-chain-connected. Let R and 5 be two equivalence relations on a set X. Then we say that R has countable index in S if every equivalence class of 5 is a countable

Some set-theoretic

notions

13

union of equivalence classes of R. In particular, if R has countable index in S, then Jf? is a sub-equivalence relation of S, i.e., R C S. L E M M A 2.1.2 Let X be a set and let Xi be a sequence of subsets of X. Assume thatX = ( J Xt. For all i, let Ri be an equivalence relation on Xi. i

Assume, for all i,j, for all x £ Xi, that there is a countable subset C C Xj such that (Ri(x)) D Xj C Rj(C). Let R be the equivalence relation on X generated by\\ Ri. Then, for all i, Ri has countable index in R\Xi. i

Proof. For all i, let Si be the set of equivalence classes of Ri and let Si := £i x {i}. Let S := [j Si. For all 5 = (E,i) £ S, for all j , let Aj(S) := {F £ £j | EHF ^ 0} C £j; then A'j{S) is countable. For all S £ S, for all j , let Aj(S) := ( ^ ( 5 ) ) x {j} C Sj C S; then A,(S) is countable. For all S e S, let A(S) := ( J Aj(S) C 5; then A(S) is countable. For all 3

So C <S, let yl*(<So) := M ^4(5). Then, for all countable So C <S, we have se5 0 that .4*(<So) is countable. From the above definitions, we have: for all i, for all T = (F,j) 6 <S, for all y 6 -F n X i ; we have {Ri(y),i) G «4(T). For all S € 5 , let Ci(S) := ^1(5); then Ci(5) is countable. We make the following recursive definition: For all S £ S, for all integers n > 1, C n +i(5) = ,4*(C n (S)). By induction, we see, for all S £ S, for all integers oo

n > 1, that Cn{S) is countable. For all S eS,

let C(5) := ( J Cn(S); then n=l

C(5) is countable. Fix i and fix a; € Xj. We wish to show that there is a countable subset BCSi such that (R\Xi)(x) = \J B. Let B := {B e £t\B C (R\Xi)(x)}. Then, as Ri C R\Xi, we conclude that (R\Xi)(x)

= M B. We wish to BeB

show that B is countable. Let E := Ri(x) £ £u S := (E,i) £ Si and V := {D ££i\(D,i) eC(S)}. Then C(S) is countable, so T> is countable, so it suffices to show that B C T>. Fix B £ B. We wish to show that B £ V. We have B £ B C £{, so B ^ 0. Fix y 6 5 . Then y S S € £",, so B = Ri(y). As B £ B, we get B C (i?|Xi)(x). Then y G (R\Xi)(x) C i?(a;).

14

Basic Results and

Let £ := \^J £j.

Definitions

Since R is generated by M Rj and since y £ R(x), it

i

j

follows that x and y are f-chain connected. Choose an integer m > 1 and an £ -chain (zo,...,zm) of length m from x to y. For all integers fc £ [l,m], choose Ek £ £ such that Zk-i,Zk £ i?*. For all integers A; € [l,m], choose jk such that .E^ 6 £jk and then define Sk •= (Ek,jk) e Sjh. We have x 6 £ n ^ , so E0 £ >IJ 0 (5), so 5 0 £ -4j 0 (5). For all integers k £ [l,m], we have Zk-i £ £ * - i nEfc, so Ek £ -/^(Sfc-i), so Sk £ Ajh (Sk-i). Let n := m + 1. Then 5 m € C„(5). Let j := j m , F := Em and T := (F, j) = Sm. Then y = zm £ £Jm = F. Moreover, T £ C„(5). so A(T) C ^*(C„(5)). As y £ B n F , we see that B e A\(T), so (B,t) € A ( T ) C A{T). Then (B,i) 6 A{T) C ^*(C n (5)) = C„+i(S) C C(5). So, by definition of V, we conclude that B £ V.

2.2

•

Some group-theoretic notions

In any group G, the identity element of G will be denoted \Q and the center of G will be denoted by Z(G). A group G is centerfree if Z(G) = {1G}For any group G, for any H,K C G, we denote by [H, K] to be the subgroup of G generated by {/ifc/i_1fc_1 \h £ H,k £ K}. Let G be a group. If S C G, then we define C G (S) := {
NG(S) := {g £ G \ gSg'1

= S}.

For all g £ G, we define CG(g) := CG({g}) • Let G be a group and let S, T C G. We say that 5 normalizes T if, for all s £ S, we have sT = Ts. We say that 5 and T centralize each other if, for all s £ S, for all* £ T, we have st — ts. Alternatively, in this situation, we will say that S centralizes T or that T centralizes S. If G is a group, then Aut(G) denotes the automorphism group of G. If G is a group and N is a normal subgroup of G, then we define a homomorphism Intjy : G -¥ Aut(iV) by (IntAr(<7))n = gng~l. Unless otherwise specified, by "action" we always mean left action. If a group G acts on a set X, then the set of orbits of this action is denoted G\X. For a right action, the set of orbits is denoted X/G.

Some topological notions

15

Let a group G act on a set X . For all x € X, StabGO*;) '•— {g & G \ gx = x} denotes the stabilizer in G of x. The G-action on X is free if, for all a; € A", we have Stabc(a;) = {1G}- The kernel of the G-action on X is | ) StabG(z). The action is faithful if its kernel is { 1 G } . xex If a group G acts on two sets X and Y and if / : X -> Y is a function, then we say that / is G-equivariant if, for all g € G, for all a; € X , we have /(gz) = g(f(x)). 2.3

Some topological notions

Following [SS78], we adopt the point-set topological convention that normal implies regular implies Hausdorff. We will use lcsc as an abbreviation for the phrase "locally compact, second countable and Hausdorff". Note that, with our conventions, "lcsc" implies Hausdorff. Let S be a subset of a topological space X . We say that S is p r e c o m pact in X if the closure in X of S is compact. We leave it as an exercise for the reader to prove: S is closed in X iff, for all a; € X , there exists an open neighborhood N of x in X such that S n N is closed in the subspace N of X. We say that 5 is locally closed in X if, for all s € S, there exists an open neighborhood N of s in X such that 5 n iV is closed in the subspace N of X. Note that S is locally closed in X iff there exist subsets U, C C X such that U is open in X , such that G is closed in X and such that 5 = U n G. Let X and Y be topological spaces and let / : X —» Y be a function. We say that / : X —¥ Y is open if, for any open subset U of X , we have that f(U) is open in Y. We say that / : X —> Y is closed if, for any closed subset G of X , we have that / ( G ) is closed in Y. If xi is a sequence in an lcsc topological space X, then "xi —> oo in X" means "xi leaves compact subsets of X " . That is, x% -*• oo in X iff for all compact sets X C X, the set {i | X{ € X } is finite. Let X and Y be lcsc topological spaces and let / : X —> Y be continuous. Then / is said to be proper if, for every compact K C Y, we have that f~x(K) is compact. Equivalently, / : X —• Y is closed and every fiber of / is compact. A map is nonproper if it is not proper. The map / : X —> Y

16

Basic Results and

Definitions

is nonproper iff there is a sequence xt in X such that x, -» oo in X, but such that f(xt) is convergent in Y. A topological space is Polish if it is separable and admits a complete metric compatible with its topology. By basic arguments in point-set topology, any lcsc topological space is Polish (see, e.g., i) =J> iv) of Theorem 5.3, p. 29 of [Kec95]). A topological space X is called a Baire space if, for any sequence Xi, X2,... of closed subsets of X, if X = Xi U X2 U • • •, then there exists an nonempty open subset U of X and there exists an integer i > 1 such that U C Xi. In other words, X cannot be written as a countable union of closed sets each of which has empty interior in X. The Baire category theorem asserts that any locally compact Hausdorff topological space is Baire and that any Polish space is Baire. A topological space is discrete if all of its subsets are open. A topological space X is zero-dimensional if, for every x € X, for every open neighborhood U of a; in X, there is an open neighborhood V of x in X such that V is closed in X and such that V C U. A topological space X is totally disconnected if, for any subspace S of X, if S is connected, then S has at most one element. A topological space X is totally pathdisconnected if every path in X is constant, i.e., if, for every continuous 7 : M. -> X, there exists x £ X such that j(R) = {x}. A discrete topological space is zero-dimensional. A zero-dimensional topological space is totally disconnected. A totally disconnected topological space is totally path-disconnected. There is a connected, totally path-disconnected subspace of the Euclidean plane E 2 (see the space denoted " F " under Example 128 on p. 145 of [SS78]). A subset of a topological space is discrete (resp. zero-dimensional, resp. totally disconnected, resp. totally path-disconnected) if, when endowed with its inherited topology, it is discrete (resp. zero-dimensional, resp. totally disconnected, resp. totally path-disconnected). Note that, under this definition, {1/n | n > 1} is a discrete subset of E, so a discrete subset may not be closed. However, a discrete subset of a T\ topological space is always locally closed. By Example 75 on p. 93 of [SS78], there is an infinite countable Hausdorff topological space which is connected. However, in non-pathological situations, countable topological spaces are highly disconnected. For example, since E cannot be written as a countable union of proper closed subsets, it follows that any countable Ti topological space is totally pathdisconnected. Thus, any countable subset of a 7\ topological space is totally path-disconnected. Moreover, by Lemma 2.3.1, p. 17 below, any countable

Some topological

notions

17

regular topological space is zero-dimensional. Consequently, any countable subset of a regular topological space is zero-dimensional. L E M M A 2.3.1 Any countable regular topological space is metrizable and zero-dimensional. Proof. Let X be a regular topological space. Assume that X is countable as a set. We wish to show that X is metrizable and zero-dimensional. Since X is Lindelof, by Lemma 1 of Chapter 4 on p. 113 of [Kel55], X is normal. Then, by Theorem 16 of Chapter 4 on p. 125 of [Kel55], X is metrizable. Let x 6 X and let Y be a closed subset of X. Assume that x £ Y. We wish to show that there is a subset C of X which is both open and closed in X such that x € C C X\Y. By Lemma 4 of Chapter 4 on p. 115 of [Kel55], choose a continuous function / : X ->• E such that f(X) C [0,1], such that f(x) = 0 and such that f(Y) = {1}. Since X is countable, f(X) is countable. Choose t e (0,1) such that t £ f(X). Let C := /"^([O.i]). Then C is closed in X, x e C and C D Y = 0. Moreover, as C = / _ 1 ( ( - 1 , * ) ) , we see that C is open in X. • L E M M A 2.3.2 Let X be a topological space. Let y be a collection of subsets of X. Assume, for all Y £ y, that Y is connected in the relative topology inherited from X. Assume that X is y-connected. Then X is connected. Proof. Let V, W be nonempty open sets in X such that V f~l W = 0 and such that V U W = X. We aim for a contradiction. Choose v € V and w 6 W. Choose an integer n > 1 and choose an (n + l)-tuple (z0,..., zn) £ Xn+1 such that ZQ = v, such that zn = w and such that for all integers k € [l,n], there exists Y 6 y such that both Zk-i 6 y and Zk € V. Choose an integer i G [l,n] such that Zj_i € V and such that ^ € W. Choose Y ey such that Zi_i,Zj G V. Give F the relative topology inherited from X. Then Y is connected. Let A :=V C\Y and let £? := W n V. Then A and B are open in Y. Moreover, A D B = 0 and A U B = 7 . Finally, as Zj_i € A and Zi £ J3, we see that A ^% ^ B. Then F is not connected, contradiction. • The next result is a partial converse to Lemma 2.3.2, p. 17. L E M M A 2.3.3 Let X be a connected topological space and let U be an open cover of X. Then X is U-connected.

Basic Results and

18

Definitions

Proof. For all x,y E X, write x ~ y if x and y are W-chain connected. Then ~ is an equivalence relation on X. For all x 6 X, let Vx be the equivalence class of x, so T4 = {«/ £ X | x ~ y}. We wish to show, for all a; € X, that K = X. For all a; 6 X, by definition of ~ , we see that there is a subset Ux C U such that Vx = M [/. In particular, for all x 6 X, we see that Vx is open. J7€M»

Since {Vx | a; 6 X } is a partition of the connected topological space X by nonempty open sets, we conclude, for all x 6 X, that Vx — X. D 2.4

Some notions from linear algebra

By vector space, we will always mean a finite-dimensional real vector space, unless otherwise specified. By complex vector space we will always mean a finite-dimensional complex vector space, unless otherwise specified. If V is a vector space, then V* denotes the dual vector space consisting of all linear transformations V -» R. For any real vector space V, by Vc we denote the complex vector space C ® R V; yc is the complexification of V. For any complex vector space V, by VR we denote the underlying real vector space V, after forgetting the complex structure; VR is the reallocation of V. By construction, for any complex vector space V, as sets, we have VR = V. Let V be a complex vector space and let J : VR —> VR be defined by J{v) — \f^\v. Let V be the complex vector space whose underlying real vector space is the same as that of V, but on which multiplication by y/— 1 is defined by \/—Tu = — (J(v)). Then scalar multiplication in V is defined by (a + &\/—l)v = av — b(J(v)). The complex vector space V is the conjugate of V. We define VR := (V) R ; by construction, as sets, we have VR = V. For any vector space V, for all v, w € V, we define v±iw

:=

(1 ® v) ± {y/^l

w)

€

VC.

For any vector space V, the map (x, y)

H>

x + iy

:

V ©V

->•

(V C ) R

is an isomorphism of real vector spaces, so V©V = ( V C ) R . For any complex vector space V, the identity map x H+ X : VR -> VR is an isomorphism of

Some notions from linear algebra

19

real vector spaces, so VR = VR. For any complex vector space V, the map (x,y)

^

(x + y/=ly)

- i(y + y/=lx)

:

V©V

is an isomorphism of complex vector spaces, so V © V = vector space V, the map v + iw

M-

v — iw

:

Vc

—>•

-> ( V R ) C (VR)C.

For any

Vc

is an isomorphism of complex vector spaces, so Vc = Vc. All of the isomorphisms described in this paragraph are natural. If Vi is a sequence in a vector space V and if v € V, then we will say that Vi converges in direction to v and write Vi —>• v if v ^ 0 and there exists a sequence ti in E such that • U -> 0 in R; and • tiVi —> v in V. Let V be a vector space and let T; be a sequence in End(V). An element t) e V i s said to be Kowalsky for T; if there is a sequence Vi in V such that Vi —> 0 in V and such that Ti(vi) —• v in V. We say that a subset S C V i s Kowalsky for T; if all of its elements are Kowalsky for T;. Note that, if we denote by VQ the set of all v e V such that v is Kowalsky for Ti, then VQ is a vector subspace of V. For all integers m, n > 1, let Wnxn denote the set of all m x n matrices with real entries. For all integers d > 1, for all integers i,j £ [l,d], let i5>.' denote the d x d matrix with a one in the (i, j) entry and zeroes elsewhere. For all integers d > 1, let e[',..., ey denote the standard basis of E d x l . Let V be a vector space, let d := dim(V) and let B be an ordered basis of V. Let ipB : E d ~> V be the vector space isomorphism which carries the standard ordered basis of Rd to B. Let 4>B '• R d x l -> V denote the vector space isomorphism which carries the standard ordered basis of E d x l to B. For any v € V, we denote (p^iv) by [V]BLet V and V be vector spaces and let T : V -> V be a linear transformation. Let d := dim(V) and let d' := dim(V). Let B be an ordered basis of V and let £ ' be an ordered basis of V. Then [T]Jf G E d ' x d is defined by: for all v € E d x l , we have ( ^ o T o <j>B)(v) = [T\$v. We call [T]f the matrix of T with respect to $ and B' . Let V be a vector space and let T : V -t V be a linear transformation. Let B be an ordered basis of V. Then the matrix of T with respect to B, written [T] B , is defined by [T]B := [T]%.

20

Basic Results and Definitions

If V is a complex vector space and if T : V -» V is a complex linear map, then we say that T is complex diagonalizable if there is an ordered basis of V with respect to which the matrix of T is diagonal. If V is a vector space, then a linear map T : V —> V is real diagonalizable if there is an ordered basis of V with respect to which the matrix of T is diagonal. The complexification of T is denoted T c : Vc -» Vc. We say that T is complex diagonalizble if Tc : Vc -»• Vc is complex diagonalizable. We say that T : V -» V is elliptic if {exp(tX)} t6 R is precompact in GL(g). A linear transformation T is elliptic iff both T is complex diagonalizable and all characteristic roots of T are pure imaginary. If T : V -> V is an endomorphism of a vector space V, then there exists a unique triple D, E, N : V —> V of endomorphisms of V satisfying: • • • • •

D : V —> V is real diagonalizable; E : V -¥ V is elliptic; N : V -> V is nilpotent; [D, E] = [D, N] = [E, N] = 0; and T = D + E + N.

The equation T = D + E + N is then said to be a real Jordan decomposition on V. We call D the real diagonalizable part of T. We call E the elliptic part of T. We call N the nilpotent part of T. Let T : V —>• T^ be an endomorphism of a vector space V. Choose D,E,N e End(V) such that T = £> + .E + J V i s a real Jordan decomposition on V. Let S C End(F) be the M-linear span of T,T2,T3,.... Then

D,E,N

£S.

Let V be a vector space. A subset <S of End(V) is said to be real Jordan closed if, for every T € S, for every D,E,N € End(V), i f T = £> + jE;-|-A/' is a real Jordan decomposition on V, then D,E,N £ S. Let F b e a real vector space and let / : V -+ V be the identity map. A complex structure on V is a linear map J :V —> V such that J 2 = —I. If W is a vector subspace of a vector space V, then we define codim v (W0 := dim(V) - dim(W). L E M M A 2.4.1 Let V and V be vector spaces and let f,D,E,N :V -> V and f',D',E',N' : V —> V be linear transformations. Assume that both f — D + E + N and f = D' + E' + N' are real Jordan decompositions. Let T : V —> V be a linear transformation. Assume that r o / = / ' o r . Then 1 T o D = D o T and r o E = E' o r and r o N = N' o r .

Some notions from linear algebra

21

The preceding lemma is basic and is left as an exercise for the interested reader, with the following hint: If r is either surjective or injective, then Lemma 2.4.1, p. 20 can be argued from uniqueness of the real Jordan decomposition. A general r may be written as the composition of a surjective map followed by an injective map, so the general case can be seen to follow from the surjective and injective cases. L E M M A 2.4.2 Let V be a vector space and let S : V ->• V be complex diagonalizable. Assume that every characteristic root of S : V —> V is contained in E U v^TK. For all A 6 M, let V\ := {v G V \ S(v) = Xv}. Let VD '•= / J ^A- Let T : V —> V be the real diagonalizable part of S. Let AeR\{0}

VE be an S-invariant vector subspace ofV. Assume that VD + VE = V and that VD l~l VE = {0}. Then T\VD = S\VD and T\VE = 0. Proof. We have S(VD) C VD and S(VE) C VB. The transformation S\VE : VE —> VE is complex diagonalizable and all of its characteristic roots are contained in y/^lR. Then S\VE '• VE —> VE is elliptic. The transformation S\VD -VD~>VD is real diagonalizable. Define D : V -> V by D\VD = S\VD and D\VE = 0. Define E : V -¥ V by E\VD = 0 and E\VE = S\VE- Then S = D + E + 0isa, real Jordan decomposition on V. Then, since T is the real diagonalizable part of S, we get T = D. Then T\VD = D\VD = S\VD and T\VE = D\VE = 0. • The next result appears as Lemma 4.4 in [A99b]. L E M M A 2.4.3 Let k>2

be an integer and let

V = V0 2 Vi 2 V2 D • • • D Vk = {0} be a chain of vector subspaces of V. Assume, for all i £ { 0 , . . . , k}, that T(Vi) C Vi. Let V. := (F 0 /Vi) © • • • © (Vi-i/Vi). Let T„ : K -*• V, 6e tte mop induced 6y T : V -^ V. Then dim(T,(K)) < dim(T(V)). Proof. It suffices to prove this in the case where k = 2. Let V := V/Vi and let 7r : V ->• V be the canonical map. Identifying V\ with Vi/{0}, we have V, =V@Vi. Then d = dim(V") = dim(V„). Let K denote the kernel of T : V —> V and let L denote the kernel of T. : V, -> V*. Let 7? := 7r(if). Let d K := dim(JiT), let dL := dim(L) and let n := (dim(ii')) + (dim(.K" n Vi)). The surjective homomorphism

22

Basic Results and

Definitions

•K\K : K -> K~ has kernel K C\Vi,so dK = n. We have ~K © (if n Vi) C I,, so n < d i . Then da = n < d^. Then dim(T,(K)) = d-dL BiAB'1 : End(V) -> End(V). Then Y : V -» V is nilpotent. Proof. Let Yi be a sequence in End(V) such that Y* —> 0 in End(V) and such that BiYiB^1 -> Y. Let d := dim(V). We may assume that V = Md. Let K := SO(d) and let A+ := { o i ^ j f + • • • + adE$ | oi > • • • > a d > 0}. Since we have GL(V) = KA+K, we may assume, for all i, that Bi G A+. Then V is strictly upper triangular, and, in particular, is nilpotent. D 2.5

Matrix concentration lemmas

Let d > 1 be an integer and let J := { 1 , . . . , d}. For all j G J, let ej := e), '. For all j , A; € J, let Ejk := E<.f. Let e j , . . . , e ^ be the basis of (M d x l )* dual to e i , . . . , e d , so that, for all j,Z € J, we have e*(ej) = fy. Let {-Ej*m}j,me.J be the basis of (E d x d )* dual to {.Ejjjj.kej, so that, for all j,k,l,m G J, we have Efm{Ejk) = SjiSkm. For all E C J, a matrix u € Rdxl is said to be c o n c e n t r a t e d on E if, for all j G <7\E, we have e*Av) = 0. For all II C J x J, a matrix M G K d x d is said to be c o n c e n t r a t e d on II if, for all (j, k) G (J x J ) \ n , we have £<-A(Af) = 0. L E M M A 2.5.1 Let D := EEu H 1- REdd be the set of diagonal dxd real matrices. Let Ai be a sequence in D. Let v G Rdxl be Kowalsky for Rdxl -> W*xl. Let E denote the set of j G J such that {Ejj(Ai)}i is unbounded in R. Then v is concentrated on E. Proof. Choose a sequence vt in R d x l such that Vi -> 0 and such that AiVi -»• v in M d x l . Then, by matrix multiplication, for all j G J \ E , we have e^AiUi) ->• 0, so e*^(v) = 0 . D L E M M A 2.5.2 Let D* ~ {aiEn the set of invertible diagonal dxd

+ ••• + adEdd \ au... ,ad G K\{0}} be real matrices. Let Ai be a sequence

First results on expansive sequences

23

in D*. Let M £ K d x l be Kowalsky for X h+ AiXA'1 : Rdxl ->• R d x l . Lei LT denote the set of (j,k) £ J x J such that {(Ejj(Ai))/(E%k(Ai))}i is unbounded in R. Then M is concentrated on II. Proof. Choose a sequence M» in ffidxl such that Mt ->• 0 and such that AiMiA^1 -> M in R d x l . Then, by matrix multiplication, we see, for all 0', k)e(Jx J ) \ n , that E^AiMiAr1) -> 0, so £* fc (M) = 0. D 2.6

First results on expansive sequences

Let W be a vector space and let V» be a sequence of vector subspaces of W. Let Vi be a sequence in W. Then we say that v» is a sequence in V; or that i>j is in V{ if, for all i, we have t>j € V;. Let W be a vector space and let Tj be a sequence in End(W). Let Vi be a sequence of vector subspaces of W. Then we say that Ti is expansive on Vi if, for every sequence Vi in Vi, we have: if {Ti(vi)}i is precompact in W, then u, -» 0 in W. Note that we do not assume, for any i, that Ti{Vi) C Vj. Note that, if W is a vector subspace of W, if V C W and if, for all i, we have Ti(W) C W , then Tj is expansive on V* iff Ti\W is expansive on Vi. In particular, if G is a Lie group, if H is a Lie subgroup of G if 5i is a sequence in H and if Vj is a sequence of vector subspaces of h, then Adg(ffj) is expansive on V, iff Ad[,((;t) is expansive on V*. Consequently, we may simply say that Adgi is expansive on Vi, without ambiguity. Similarly, if G is a Lie group, if N is a normal Lie subgroup of G, if Vj is a sequence of vector subspaces of g/n and if gt is a sequence in G, then we say that Adgt is expansive on Vi if Adg/n(gi) is expansive on Vi and, again, there is no worry of ambiguity. The connection between expansiveness and Kowalsky subspaces is: L E M M A 2.6.1 Let W be a vector space. LetTi be a sequence in End(TV). Let Ui be a sequence of vector subspaces ofW. Let V be a vector subspace ofW. Let k := dim(V). Assume, for all i, that dim(T;({/;)) = k. Assume that Ti is expansive on Ui- Assume that Ti(Ui) —> V in the topological space of k-dimensional vector subspaces of W. Then V is Kowalsky for T{. Proof. Let v £ V. We wish to show that v is Kowalsky for Tj. Choose a sequence u^ in Ui such that Ti{ui) —>• v. Because Ti is expansive on U, we see that Ui -> 0 in U. Thus v is Kowalsky for Tj. •

24

Basic Results and Definitions

L E M M A 2.6.2 Let W be a vector space and let Ti be a sequence in End(W) Let Xi and yi be precompact sequences in W. Let p,q 6 W. Assume that both Ti{xi) —^ p and Ti(yi) —*• q in W. Assume that p and q are linearly independent in W. Then Tt is expansive onffircj+ Ryt. Proof. Let Vi be a sequence in Vi :— Mxi + Ryi and assume, for a contradiction, that {Ti(vi)} is precompact in W, but that Vi •/> 0 in W. Passing to a subsequence, we may assume that Vi is bounded away from 0 in W. Since p and q are linearly independent, passing to a tail, we may assume, for all i, that Xi and yt are linearly independent. For all i, choose (o,i,bi) £ E 2 such that Vi = diXi + biyi. Since Vi is bounded away from 0 in W and since {xi}i and {?/;}« are both precompact in W, it follows that (ai,bi) is bounded away from (0,0) in E 2 . For all i, let p\ := Ti(xt) and let q\ :=Ti(yi). Choose sequences s; and ti in E such that Sj —• 0 and ti -» 0 in E and such that Sip[ —> p and Uq[^ q in W. Since p ^ 0 ^ q, by passing to a tail, we may assume, for all i, that Si ^ 0 7^ t». For all i, we have Ti(vi) = aip[ + biq\. For all i, let p; := Sip\, let gi := Uql, let a* := aj/s, and let ft := &*/*»; then Ti(vt) = aiPi + fcqi. Consequently, {ctiPi+/3iqi}i is precompact in W. Moreover, we havep; —>• p and qi -» q in W. Since (aj,6j) is bounded away from (0,0) in E 2 and since st —t 0 and ii —>• 0 in E, it follows that (aj,/3j) —• oo in K2. Choose a sequence Mj in E such that Ui —>• 0 in E and such that {ui(a;i,/3i)}i is precompact in E2 \{(0,0)}. Passing to a subsequence, assume that Ui(ati, ft) is convergent in E 2 \{(0,0)}. Let («',/?') := lim Wi(ai,ft); then (a',ft) € E 2 \{(0,0)}. i—>oo

Since u, —> 0 in E and {aiPi + Piqi}i is precompact in W, we conclude that Ui{ctiPi + Piqi) —> 0 in V. On the other hand, Ui(aiPi + /3iqi) -> a'p + /3'g in V. Then a'p + ft'q' = 0, so, by linear independence of p and q, we have (a', ft) = (0,0), contradiction. • Let W be a vector space and let T, be a sequence in End(W). Let V be a sequence of vector subspace of W. Then we say that Ti is expansive on V if, for every sequence v^ in V, we have: if {Ti(vi)}i is precompact in W, then «j —> 0 in V. That is, Tj is expansive on the constant sequence V, V,V,.... Note that we do not assume, for any i, that Tj(V) C V. COROLLARY 2.6.3 Let W be a vector space and let Ti be a sequence in End(W). Let x,y,p,q G W. Assume that both Ti(x) -*• p and Ti(y) —v q

Topological groups

in W. Assume that p and q are linearly independent in W. expansive on Kx +My.

25

Then Ti is

L E M M A 2.6.4 Let W be a vector space and let Ti be a sequence in End(W). LetV andW be a vector subspaces ofW. Assume VnW = {0}. Let Vi be a sequence of vector subspacse of V. Let W := W/W. Let n : W -> W be the canonical linear map. Let Ti be a sequence in End(W). Assume, for all i, that 7roTj = T,o7r. Assume that Ti is expansive on 7r(V;). Then Ti is expansive on Vi. Proof. Let Vi be a sequence in V; and assume that Ti(vi) is precompact in V. We wish to show that Vi —> 0 in V. Let V := 7r(V). For all i, let V := ir(Vi). For all i, let vi := ir(yi). Then for all i, we have 7r(Ti(uj)) = Ti{vt). So, since TJ(UJ) is precompact in W, it follows that Tt{vi) is precompact in W. So, as Ti is expansive on Vi, we see that v\ -* 0 in W. Then 7r(uf) ->• 0 in V. AsVnW = {0}, it follows that the map v •-»• 7r(u) : V —> V is a vector space isomorphism. So, as 7T(DJ) —> 0 in V, we conclude that Vi —> 0 in V. D 2.7

Topological groups

By topological group, we will always mean a Hausdorff topological group. Let G be a topological group. The complement of an open subgroup of G is a union of cosets and is therefore open. Consequently, any open subgroup of G is closed. Consequently, if G is connected, then no proper subgroup of G can be open. Consequently, if G is connected, if S C G and if S contains a nonempty open subset of G, then 5 generates G, i.e., no proper subgroup of G contains S. If G is a topological group, then G° denotes the connected component of G containing 1Q. If J? is a closed subgroup of a topological group G, then H° denotes the connected component of H (in the relative topology inherited from G) containing la- If G is a topological group, then we define Z°(G) := (Z(G))°. Let G be a topological group. For S CG, let C%(S) := (CG(S))° and AT° (S) := (NG(S))°. For g G G, let C°G(g) •= (CG(
26

Basic Results and Definitions

(resp. right-invariant) if, for all Borel S C G, for all g € G, we have fi(gS) = fi(S) (resp. n(Sg) = l*(S)). A left-invariant (resp. right-invariant) positive Borel measure on G is called a left Haar measure (resp. right Haar measure) on G. Let G be an lcsc group. Let A denote either the set of all left Haar measures on G or the set of all right Haar measures on G. Then A ^ 0. Moreover, if A, p, € A, then there exists a > 0 such that A = ap,. Moreover, for all A € A, A is positive on all nonempty open subsets of G and finite on all compact subsets of G. Moreover, if, for some A 6 A, we have \(G) < oo, then G is compact. Conversely, if G is compact, then, for all A 6 A, we have A(G) < oo. Let G be a topological group acting on a set X. Then (Stabc(a;)) 0 is called the connected stabilizer in G of x, and is denoted Stab^(a;). The G-action on X action is locally free if, for all x € X, Stabc(x) is discrete in G. The action is locally faithful if its kernel is discrete in G. Let G be a topological group acting on a topological space X. Then the G-action on X is continuous if the map (g, x) t-> gx : G x X —> X is continuous. L E M M A 2.7.1 Let G be an lcsc topological group and let G act continuously on topological spaces X and Y. Assume that Y is Baire. Assume that the G-action on Y is transitive. Let f : X —> Y be a continuous G-equivariant map. Then f : X —>Y is open. Proof. For all x £ X, let hx : G —>• X be defined by hx(g) = gx; then hx is continuous. For all x € X, let px :— f o hx : G -¥ Y. For any U C X, for any a; e I , we have PxiK1^)) = f(U fl (Gx)). So, for any U C X, we have f(U) = M Pxih^iU)). We wish to show, for any open subset U of X, that f(U) is open in Y. It therefore suffices to show, for all x € X, that px : G -» Y is open. Fix x £ X and fix an open subset W of G. We wish to show that px(W) is open in Y. Let p := px : G -> Y. For any subset A C Y, let A° denote the interior in Y of A and let A denote the closure in Y of A. Fix w € W. We wish to show that p{w) € {p(W))°. Let W := w~lW. Then W is an open neighborhood of \Q in G. Let y0 := P(1G)- Then y0 = / ( ^ ) Since wy0 — p(w) and since w(p(W')) — p(W), it suffices to show that
(P(W'))°.

27

Topological groups

For all A C G, let A denote the closure in G of A. Choose an open neighborhood V of la in G such that V is compact and such that V V C W. Choose a sequence
i

so, by transitivity of G on Y, we conclude that MffiVyo = Y. i

Since Y is Baire, choose io such that (gi0Vyo) ^ 0. So, since we have ffj^1 {OioVyo) = {Vyo) J w e conclude that (Vy0) ^ 0. Since V is compact, it follows that Vyo is compact, so Vyo is closed in y . So, as Vyo 2 V2/o> we conclude that Vy0 D Vy~o~. Then (Vy0)° 2 (Vy^)° ^ 0. Choose x £ (Vy0)°. Then x £ Vyo, so choose v £ V such that x = i;?/o. Then 2/o = v~xx e w-MCVyo)"] = (v-Wyo)0 For all g £G,we 2/o € (p(W))°.

have p(g) =

S(P(1G))

C ( V ^ F i / o ) 0 C (W"i/ 0 )°.

= ffj/o- Then p(W)

= W'j/o- Thus •

A subgroup B of an lcsc topological group G is said to be cocompact in G if there is a compact subset K of G such that G = BK. Equivalently, there is a compact subset K of G such that G — KB. If if is a closed subgroup of an lcsc topological group G, then B is cocompact in G iff G/H is compact in the quotient topology. L E M M A 2.7.2 Let B be an lcsc topological group and let G be a normal closed subgroup of B. Let IT : B —> BjG be the canonical homomorphism. Let T be a closed subgroup of B. Assume thatTtlG is cocompact in G and that 7r(r) is cocompact in B/G. Then T is cocompact in B. Proof. Let e := l j j . Let K be a compact subset of G such that we have K(T n G) = G. Then G C KT. Let L be a compact subset of B/G such that L • 7r(r) — B/G. Let L be a compact'subset of B such that L C ir(L). Then L r G = n^^Lr)) = ^ ( Z - T r C r ) ) = ^(B/G) = B. Since G is normal in if, we have G r = TG. Then H = LYG = LGT C L ( i i T ) r = LAT C if, so B = LKT. So, since LK is a compact subset of B, it follows that T is cocompact in if. D L E M M A 2.7.3 Let G be a lcsc topological group. Let A and B be closed subgroups of G. Assume that AC\ B = {1G} and that AB = G. Assume

28

Basic Results and

Definitions

that A normalizes B. Let AQ be a cocompact closed subgroup of A and let BQ be a cocompact closed subgroup of B. Assume that AQ normalizes BQ. Then AQBQ is a cocompact closed subgroup of G. Proof. The multiplication map F : A \x B —> G defined by F(a, b) = oh is a continuous bijective homomorphism. By Lemma 2.7.1, p. 26, the map F : A K B -> G is open. Then F : A tx B -» G is an isomorphism of lcsc topological groups. Since A0 is closed in A and since BQ is closed in B, it follows that A0 x B0 is closed in AxB, and therefore that ^o-Bo = F(A0 * B0) is closed in G = F(A K B). It remains to show that AQB0 is cocompact in G. Let Co := AoB0. Choose a compact subset K of G such that BQK = B. Then C0K = A0B0K = A0B. Then KC0 — A0B. Choose a compact subset K' of G such that K'A0 = A. Then K'K(AoBo) Then

AQBQ

= K'KCo = K'A0B

= AB = G.

is cocompact in G.

L E M M A 2.7.4 Let G and H be lcsc topological groups. be a continuous surjective homomorphism. Let K denote and let n : G —> G/K be the canonical homomorphism. isomorphism f : G/K —tHof lcsc topological groups such

O Let f : G -* H the kernel of f Then there is an that f on = f.

Proof. By abstract group theory, fix an isomorphism / ' : G/K —> H of abstract groups such that f on = f. Since / : G -> H is surjective, we see that / ' : G/K —> H is surjective. Since / : G —> H is continuous, we see that / ' : G/K -> H is continuous. It remains to show that / ' : G/K -> H is open. Let G act on H by g • h — (f(g))h. By surjectivity of / : G -> H, this action is transitive. So, as / ' : G/K —> H is G-equivariant, we see, from Lemma 2.7.1, p. 26, that / ' : G/K -> H is open. • COROLLARY 2.7.5 Let G and H be lcsc topological groups. Then any continuous surjective homomorphism G —> H is open. L E M M A 2.7.6 Let G and H be lcsc topological groups and let f : G -> H be a continuous, bijective homomorphism. Then f is an isomorphism of topological groups.

Discrete groups

29

Proof. Let K := ker(/). Then, as / is injective, we have K = {1G}- Let •n and / ' be as in Lemma 2.7.4, p. 28. Then 7r and / ' are both isomorphisms, so, as / = / ' o 7r, we conclude that / is an isomorphism, as well. D L E M M A 2.7.7 Let G and H be Icsc topological groups and let f : G -» H be a continuous homomorphism. Then f : G —>• H is proper iff both ker(/) is compact and f(G) is closed in H. Proof. "Only if" follows from the fact that / is proper iff both / is closed and every fiber of / is compact. Assume that K := ker(/) is compact and that Ho := f(G) is closed in H. We wish to show that / : G —> H is proper. Since Ho is a closed subgroup of H, the inclusion Ho —>• H is proper. Let n : G —> GjK be the canonical homomorphism. By Lemma 2.7.4, p. 28, choose an isomorphism / ' : GjK -> H0 of topological groups such that / = / ' o 7r. Since / ' and TT are both proper, it follows that / : G —> Ho is as well. So, since the inclusion Ho -» H is also proper, it follows that / : G -> H is proper. • L E M M A 2.7.8 Let G be an Icsc topological group, let K be a compact subset ofG and let A be a closed subset ofG. Then AK and KA are closed subsets of G. Proof. The maps (a, k) \-t ak : A x K -» G and (k, a) *-* ka : K x A -> G are both proper, and so have closed image. • 2.8

Discrete groups

Any subspace of a second countable topological space is again second countable, and a discrete, second countable topological space is countable. Thus, a discrete subset of a second countable topological space is countable. By the Baire Category Theorem, any countable Polish topological space is discrete. By Theorem 3.11, p. 17 of [Kec95], a Gs subspace of a Polish space is again Polish; then any countable, Gs subset of a Polish topological space is discrete. In particular, a countable, closed subgroup of a Polish topological group is discrete. A locally closed subgroup of a topological group is closed, so a discrete subgroup of a topological group is closed. If G is a Polish topological group and if T is a subgroup of G, the T is discrete in G iff r is a countable closed subgroup of G.

30

Basic Results and

Definitions

If G is a topological group and if V is a subgroup of G, then T is discrete in G iff 1 Q is not an element of the closure in G of T \ { 1 G } iff there exists an open neighborhood U of 1Q in G such that U (~\ V = {1G}L E M M A 2.8.1 Let H be a topological group and let G be a closed subgroup of H. Let 7r : H —> H/G be the canonical map. Let T be a subgroup of H. Assume that TC\G is discrete in G and that TT(T) is discrete in H/G. Then T is discrete in H. Proof. Let e := l j j . Let U be an open neighborhood of n(e) in H/G such that U n 7r(r) = {n(e)}. Then ( T T " 1 ^ ) ) D F = F n G, Let V be an open neighborhood of e in if such that V n ( r n G ) = {e}. Let W := (n'1 (U))nV. Then W nT CW n(TnG) = {e}. Thus F is discrete in H. O A subgroup r of an lcsc group G is said to be a lattice in G if both T is discrete in G and the transitive G-action on G/T preserves a finite Borel measure. L E M M A 2.8.2 Let H be a closed subgroup of an lcsc topological group G. Then the following are equivalent: (1) (2) (3) (4) (5)

H is discrete. H is a zero-dimensional topological space. H is totally disconnected. H is totally path-disconnected. H° = {1G}.

Proof. The implications (1) =>• (2) = > (3) =>• (4) follow from the definitions. Since H° is a connected Lie group and is therefore path-connected, we see that (4) =>• (5). We assume that H° = {1 G } and we wish to show that H is a discrete subset of G. Let T denote the discrete topology on H and let \x be the relative topology on H inherited from G. Then the identity map (H,T) -> (H,fj.) is a bijective continuous homomorphism of lcsc groups, so, by Lemma 2.7.6, p. 28, we see that it is an isomorphism of topological groups. That is, the discrete topology on H agrees with the relative topology. That is, H is a discrete subset of G. D

Proper actions

2.9

31

Proper actions

Let G be an lcsc topological group acting on an lcsc topological space X. The G-action on X is proper if (g,x) H-> (a;,gx) : GxX —> X xX is proper. Equivalently, for any compact subset K C X, {g £ G \ gK n K ^ 0} is a compact subset of G. When G is a discrete group, many writers prefer the term properly discontinuous instead of proper. The G-action on X is nonproper if it is not proper. Equivalently, there are sequences gi in G and x, in X such that gi —> oo in G, such that a;, converges in X and such that giXi converges in X.

L E M M A 2.9.1 Let G be an lcsc topological group acting properly on an lcsc topological space X. Then the quotient topology on G\X is Hausdorff.

Proof. Fix x, y £ X. Assume that Gx ^ Gy. We wish to show that there are, in X, G-invariant open neighborhoods UQ and Vo of x and y, respectively, such that Uo n Vo = 0. It suffices to show that there are, in X, open neighborhoods U and V of x and y, respectively, such that (GU) n (GV) = 0. Let V\ be a precompact open neighborhood of y in X. By properness of the action, choose a compact subset C CG such that, for all g £ G\C, we have gx £Vi. For all S C X, let S denote the closure in X of S. Choose an open neighborhood V of y such that V C Vi\(Ga;). Then (Gx) n V = 0. Choose an open neighborhood U\ of x in X such that U\ is compact. By properness of the action, choose a compact subset K CG such that, for all g £ G\K, for all z £ V, we have gz £ U\. As (Gx) n V = 0, it follows that x £ GV. In particular, x £ KV. Since K is compact and since V is closed in X, it follows that KV is closed in X. Let U := E/i\(ifV). Then U is an open neighborhood of x. Moreover, U C C/i and [7 n (2fF) = 0, so, by definition of K, we get U n (GV) = 0. Then C7 n (GV) = 0. Then (GU) n (GV) = 0. • Let G be an lcsc topological group acting properly on an lcsc topological space X. Because the canonical map X -* G\X is open, by Lemma 2.9.1, p. 31, it follows that the quotient topology on G\X is lcsc. (Recall that, under our definitions, lcsc implies Hausdorff.)

32

2.10

Basic Results and Definitions

Bilinear and quadratic forms

Let F b e a vector space and let B : V x V -> E be a function. We say that the map B is a bilinear form on V if, for all v € V, w t-> B(v, w) : V -> E

and

w

H-> B(IU,

u) : V ->• E

are both linear transformations. We say that B is symmetric if, for all v,w G V, B(v,w) — B(w,v). We say that B is antisymmetric or alternating if, for all v,w G V, we have B(v,w) = —B{w,v). If B is a bilinear form, then there exist unique symmetric and antisymmetric bilinear forms S and A such that B = S + A. The kernel of a symmetric bilinear form B : V x V —> E is ker(B) := {v € V | Vw G V, £(u,u;) = 0}. We say that B is degenerate if ker(B) ^ {0}. We say that B is nondegenerate otherwise. If {vi,... ,Vd} is a basis of V, then B is degenerate iff the matrix [B(vi,Vj)] has determinant = 0. Let V be a vector space. If B : V x V -4 E and B ' : V x V - • K are bilinear forms, then we say that B and B' are equivalent if there is a vector space isomorphism f :V ->V such that, for all v, w € V, we have B'{f(v),f(w)) — B(v,w). Let d :— dim(V). There is a nondegenerate antisymmetric bilinear form on V iff d is even. If d is even, then any two nondegenerate antisymmetric bilinear forms on V are equivalent. A symmetric bilinear form B : V x V —> E is positive definite if, for all v G ^ \ { 0 } , we have B(v,v) > 0. It is negative definite if, for all v £ V\{0}, we have B(v,v) < 0. It is positive semidefinite if, for all v G V, we have B(v,v) > 0. It is negative semidefinite if, for all v &V, we have B(v, v) < 0. It is definite if it is either positive definite or negative definite. It is indefinite if it is not definite. An inner product space is a vector space V together with a positive definite symmetric bilinear form B :V xV -> R. Let V be a vector space and let B : V x V -> E be a symmetric bilinear form. A vector v G V is isotropic with respect to B if B(v,v) = 0. A subset VQ C V is isotropic with respect to B if, for all v,w G Vo, we have B(v,w) = 0. A vector subspace VQ of V is positive definite (resp. negative definite, resp. positive semidefinite, resp. negative semidefinite, resp. definite, resp. indefinite, resp. degenerate, resp. nondegenerate)

Bilinear and quadratic

forms

33

with respect to B if B\(VQ X Vo) is positive definite (resp. negative definite, resp. positive semidefinite, resp. negative semidefinite, resp. definite, resp. indefinite, resp. degenerate, resp. nondegenerate). Let V denote the collection of all positive definite subspaces of V. Let M denote the collection of all negative definite subspaces of V. Let p := max{dim(HA) | W € V} and let q := max{dim(W) | W E N}- Then (p, q) is the signature of B. Let V be a vector space and let B : V x V -> E be a symmetric bilinear form of signature (p, q). Then dim(ker(B)) = (dim(V)) — (p + q). In particular, B is nondegenerate iff p + q = dim(y). Let d > 1 be an integer, let V and V be d-dimensional vector spaces and let B : V x V -> E and B' : V x V -» E be symmetric bilinear forms. Let (p, q) be the signature of B and let (p1, q') be the signature of B'. Then B and B' are equivalent iff (p,q) = (p',q')Let V be a vector space and let d := dim(F). Let I := { 1 , . . . , d} and let I2 := I x I. Let A := {(i,i) \i 6 / } C J 2 . For any nondegenerate symmetric bilinear form i? : V x V —> E, there is a basis {vi,..., v^} of V such that, for all i € / , we have B(vi,Vi) € { + 1 , - 1 } and such that, for all (i, j) 6 I 2 \ A , we have B(v,,Vj) = 0. That is, [B(vi,Vj)] is a diagonal matrix and every diagonal entry is either + 1 or — 1. If p is the number of diagonal entries equal to + 1 and q is the number of diagonal entries equal to —1, then B has signature (p, q). Let V be a vector space. A function Q : V —> E is said to be a quadratic form if there exists a bilinear form B :VxV —> R such that, for all v 6 E d , Q(v) = B(v,v). For any quadratic form Q on a vector space V, there exists a unique symmetric bilinear form B on V such that, for all v £ V, Q(v) = B(v,v). The symmetric bilinear form B is said to be the polarization of Q. The signature of a quadratic form is, by definition, the signature of its polarization. A quadratic form is positive definite (resp. negative definite, resp. positive semidefinite, resp. negative semidefinite, resp. definite, resp. indefinite, resp. degenerate, resp. nondegenerate) if its polarization is. We define the kernel of Q to be the kernel of the polarization of Q; it is denoted ker(Q). A vector v € V is isotropic with respect to Q if Q(v) — 0. A subset VQ C V is isotropic with respect to Q if it is isotropic with respect to the polarization of Q. A vector subspace Vo of V is positive definite (resp. negative definite, resp. positive semidefinite, resp. negative semidefinite, resp. definite, resp. indefinite, resp. de-

34

Basic Results and

Definitions

generate, resp. n o n d e g e n e r a t e ) with respect to Q if <2|Vb is positive definite (resp. negative definite, resp. positive semidefinite, resp. negative semidefinite, resp. definite, resp. indefinite, resp. degenerate, resp. nondegenerate). Two quadratic forms are equivalent if their polarizations are. Let V and V be vector spaces and let Q : V -¥ E and Q' : V -»• E be quadratic forms. Then Q is equivalent to Q' iff there is a vector space isomorphism / : V -> V such that, for all v € V, we have Q'(f(v)) = Q{v). A function Q : Rd —> E is a quadratic form iff Q is a homogeneous polynomial of degree two, i.e., iff there is asystem {cij}i
In this case, the polarization B : M.d x E d -* M. of Q is given by the rule: For all x = (xu...,xd)e Rd, for all y = (ylt... ,yd) € W, B{x,y) = \

/ j ^ijy^iVj l
T

XjUi)

+

y ^ ckkXkVk

.*=!

Let V be a vector space, let d := dim(F) and let Q : V -> R be a quadratic form. Then <5 is degenerate iff there exists a linear transformation L : V -> E d _ 1 and a quadratic form F : IT*-1 -> E such that Q = F o L. Let V be a vector space. Then SBF(F) denotes the vector space of all symmetric bilinear forms on V, and QF(V) denotes the vector space of all quadratic forms on V. Let V be a vector space and let B £ SBF(V). A linear transformation T : V —¥ V is said to be symmetric with respect to B (resp. antisymmetric with respect to B) if, for all v, w 6 V, we have B(T(v),w) — B(v,T(w)) (resp. B(T(v),w) = -B(v,T(w))). If T : V -» V is symmetric (resp. anti-symmetric) with respect to B, then the finite-dimensional spectral theorem asserts that T is real diaongalizable (resp. elliptic). Let V be a vector space and let B € SBF(V). For any X,Y CV, we define B(X,Y) := B(X xY) = {B(x,y)\x £ X,y e Y}. For any X CV, for any v eV,we define B(X,v) := S(X,{u}) and B(u,X) := B({v},X), so, by symmetry, B(X, v) = B(w,X). L E M M A 2.10.1 Let V be a vector space and let Q e QF(V). that Q is positive semidefinite. Then ker(Q) = Q _ 1 (0). Proof.

Let B £ SBF(V^) be the polarization of Q.

Assume

By definition of

Root

systems

35

ker(Q), we have ker(Q) C <2 _1 (0). Fix v e Q _ 1 ( 0 )- We wish to show that v E ker(Q). Fix w £ V and let o := B(v,w). We wish to show that a = 0. Assume, for a contradiction, that o ^ O . We have B(v,v) = Q(u) = 0. Let b := B(w,w). Let * := -(b + l)/(2a). Then Q(tu + to) = B(t« + WJ, tu + w) = 0 + 2ta + 6 = - 1 < 0, contradicting the assumption that Q is positive semidefinite. D C O R O L L A R Y 2.10.2 Let V be a vector space and let Q 6 QF(V) be nondegenerate. Then (1) Q is positive definite iff Q is positive semidefinite; and (2) Q is negative definite iff Q is negative semidefinite. Proof. For both (1) and (2), "only if" follows from the definitions. Proof of "if" in (1): Since Q is nondegenerate, ker(Q) = {0}, so, by Lemma 2.10.1, p. 34, we have Q - 1 (0) = {0}, so Q is definite. Then, as Q is definite and positive semidefinite, we see that Q is positive definite. End of proof of "if" in (1). Proof of "if" in (2): Since — Q is nondegenerate and positive semidefinite, by "if" of (1) of Corollary 2.10.2, p. 35, we see that — Q is positive definite, and so Q is negative definite. End of proof of "if" in (2). • The following is a consequence of the Chain Rule. L E M M A 2.10.3 Let V be a vector space. For all v e V, let iv : V -> TVV be defined by iv(w) — (d/dt)t=o(v + tw). Let 6 > 0. Let I := (—6,6). Let a,b : I —> V be smooth. Let x := a(0) and y := 6(0). Let x' := (d/dt)t=0(a(t))

e TXV

and

y' := (d/dt)t=0(b(t))

€ TyV.

Let F € SBF(V). Define f : I -> E by f(t) = F(a{t), b{t)). Let f : I -> R denote the derivative of f. Then /'(0) = F(i~l(x'),y) + F(x, i~1 (?/'))• 2.11

Root systems

Let (V, B) be an inner product space. For all v €V, V1 := {w e V\B(v,w)

define

=0}.

For all v € ^ \ { 0 } , define rv : V -> V be the reflection in vx defined by rv(w) =w- 2[(B(v,w))/(B(v,v))]v; then

36

Basic Results and

Definitions

• rv(v) = —v; and • for all w € vx, we have rv(w) — w. If $ C V, then we say that V is a root system in (V, B) if: • $ is finite, $ spans V, 0 ^ $; and • for all (j),ip £ ^, there exists n £ Z such that r^ip) = ip + n € $. We say that $ is reduced if, for all £ $, we have (K^) n $ = {4>, -(f)}. Note that what we call a "reduced root system" is, in [Hu72], simply called a "root system". Let $ 0 := {€$ | Vt e (0, l),t<j> $ $ } . If $ is a root system, then $o is a reduced root system, and the root system results of [Hu72] may then be applied to $o- Because of this, one sees that many of the results of [Hu72] extend to not necessarily reduced root systems. Let $ be a root system in (V, B), let r > 0 be an integer and let 8i,..., 8r € $. Let A := {<$i,..., Sr} C $ and let 5+

:=

{xifii + • • • + xr5r \x\,...

,xr £ Zfl [0, oo)},

S-

:=

{xiSi + • • • + xrSr \xi,...

,xr € Zfl (—oo,0]}.

We say that A is a base of $ if both of the following hold: • A is a basis of the vector space V; and • $ C 5+US'-. By (i) of Theorem X.3.6, p. 458 of [Hu72], every root system admits a base. If A is a base of $ and if € $ , then we say that

£ S-.

Let V be an inner product space, let $ be a root system in V and let A be a base of <J>. Let 6,6' £ A. We say that 8 is adjacent to 5' in A if 5 + 6' £ $. The graph whose vertices are elements of A and whose edges are defined by adjacency is called the simplified Dynkin diagram of A. In fact, every base of a root system has, associated to it, a "Dynkin diagram", which is a graph whose vertices are the roots in the base and some of whose edges are directed. (See §11.2 of [Hu72].) The Dynkin diagrams associated to two bases of one root system are isomorphic. Two vertices 8,5' in the Dynkin diagram of A are adjacent iff 8 + 8' £ $ , so the simplified Dynkin diagram is formed from the Dynkin diagram by, first, making all edges undirected and, then, allowing only one edge to connect a pair of points. That is, to obtain the simplified Dynkin diagram, we

Root systems

37

eliminate all the arrows from the Dynkin diagram and replace any doubled or tripled edge by a single edge. Let V be an inner product space, let $ be a root system in V and let A be a base of $ . If k > 1 is an integer, then a path of length A; in A is a {k + l)-tuple (So,...,5k) £ Ak+l such that, for all integers i G [1,k], we have that 5t-i is adjacent to <5* in A. A subset A 0 C A is connected if, for any 8,6' € A 0 , there is an integer fc > 1 and a path (So, ..., 8 k) of length k in A such that 8Q — 8 and such that 8k = 8'. We say that a root system is irreducible if it admits a connected base. Since the Dynkin diagrams of bases of a root system are all isomorphic to one another, we see that any base of an irreducible root system is connected. The following lemma is a special case of (and, in fact, is equivalent to) the general assertion that, in any root system, if a root string lies on a line not passing through 0, then it is unbroken. L E M M A 2.11.1 Let V be an inner product space and let $ be a root system in V. Let 0, ' € $ and assume that 4> and (/>' are linearly independent. Let k > 0 be an integer and assume that (j) + k(f>' € <J>. Then <j> + (/>' € $ . Proof. Let (•, •) be the positive definite symmetric bilinear form on V. Assume, for a contradiction, that (f> + (t>' £ $• Then, as 4>,<j)' € $, it follows, from Lemma 9.4, p. 45 of [Hu72], that ((p',) > 0. Let J denote the set of integers j > 0 such that (j> + j<j>' € $ . Then k € J. Let m := min J . As cj> + 4>' ^ * . w e g e t m > 1. Let V := ' + Tn(j>. Then <j>,ij) € $, while ip - 4> £ $, so, by Lemma 9.4, p. 45 of [Hu72], we see that (ft,^) < 0. However, (0,ip) = (',4>) > (',<£') > 0, contradiction. Q If X is a graph, then a spanning tree in X is a subgraph which is a tree and whose vertex set is the vertex set of X. Any connected graph admits a spanning tree. A pending vertex in a graph is a vertex which is adjacent to exactly one other vertex. Any finite tree has a pending vertex. L E M M A 2.11.2 Let V be an inner product space, let $ be a root system in V, let A be a base of $ and let A 0 C A be connected. Then ^ J 8 € 3>. <5eA0 Proof. (J. Stembridge.) The proof is by induction on dim(V). Let V0 be the E-span of A 0 and let $ 0 ~ * fl V0. If A 0 C A, then dim(V0) < dim(V), and we are done by applying the induction hypothesis

38

Basic Results and Definitions

to the base Ao of the root system $o in Vo. We therefore assume that A = A 0 . Then A is connected. Let D be the simplified Dynkin diagram of A, so D is a connected finite graph. Let T be a spanning tree in D. Any finite tree has a pending vertex; let Si be a pending vertex in T. Let A' := A\{#i}, let V be the B-span of A' and let $':= n V. Then A' is a base of the root system $ ' in V. Moreover, by construction, A' is connected. Let <j> := 2_] $• By applying <5£A'

the induction hypothesis to the base A' of the root system $ ' in V, we conclude that (/>€$'. We wish to show that <j> + Si € $. Let B be the positive definite symmetric bilinear form on V. Claim 1: For some 6 G A', we have B{5,5\) ^ 0. Proof of Claim 1: Assume, for a contradiction, that B(S,Si) = 0. Let S := ESi and T := {v eV\ B(v,Si) = 0}. Let X := A n 5 and let Y := A n T. Then A = X U Y. Let W be the subgroup of GL(V) generated by {rg\S € A}. We have A C S U T, so W preserves both S and T. By parts (c) and (d) of Theorem 10.3 on p. 51 of [Hu72], we have $ = WA. Then $ = (WX)U(WY). So, since W preserves S and T, we get $ C SUT. For all S G X, for all 5' 6 Y, we have S E S\{0} and 8' £ T\{0}, so S + S' £ S U T; as S U T = $, we conclude that J is not adjacent to S'. So, since A = X U V, we see that A is not connected, contradiction. End of proof of Claim 1. For all S 6 A', by Lemma 10.1, p. 47 of [Hu72], we have B{5,5{) < 0. Then, by Claim 1, we get B(,Si) < 0. Let k := -2{B(4>,Si))/(B{Si,Si)). Then k > 0 and r ^ {(f) = <j> + kSi. As $ is a root system, we see that k G Z and that r ^ ^ ) G $. Then + kSi G $. Since A is a basis of V, we see that and Si are linearly independent. Then, by Lemma 2.11.1, p. 37, we have <j> + Si G 3>. • L E M M A 2.11.3 Let (V,B) be an inner product space and let $ be a root system in (V,B). Let v0 G V and let H := {v G V\B(v,v0) > 0}. Then there is a base A of $ such that A C H. Proof. If v0 — 0, then H = V and we may let A be any base of $. We therefore assume that VQ ^ 0. For all w G V, let

Hw~{v£V\B(v,w)

>0},

H'w:={veV\B(v,w)

> 0}

Minkowski forms - basic

definitions

39

and let $+(io) := $ n H'w. Then HVo = H. For all w e V, we define X ^ - { D G F I B{V,W) = 0}. Let 5 := | J <j) . For all w £ V \ 5 , we have $ n w;-1 = 0, so $ + (u;) = $ l~l Hw. For all <^ e $, the interior in V of x is empty; therefore, the interior in V of S is empty. Let $ i := $ D (V\if„ 0 ). For all (j> € <J>i, we have <> / ^ i?„ 0 , so choose an open neighborhood 11$ of VQ such that, for all u € Ify, we have £ Hu. Let

^:= n ^v Then £/ is an open neighborhood of vo in V. Moreover, for all u € U, we have $ i C V\if u , so V \ * i D Hu. As the interior of in V of 5 is empty, we see that U £ S. Choose vx G U\S. Then # W l C V \ $ i . Then $C\HVl C $ n ( V \ $ i ) = $ \ $ i = $ n ff„0. Since ui € Vr\Sr, we get $+( W l ) = $r\HVl. By Theorem' 10.1, p. 48 of [Hu72], choose a base A of $ such that A C $+(vi). So, since $+(vi) = $ n HVl C $ n # „ 0 , it follows that AC.$r\HV0CHV0=H. D

2.12

Minkowski forms — basic definitions

For all integers d > 2, let Qd : Rd -> R be the quadratic form denned by Qd(xi, • . . , Xd) = 2x\Xd + x\ + • • • + xd_x. A quadratic form Q on a vector space V will be said to be Minkowski if there is some d > 2 and an isomorphism V <—> Rd such that Q corresponds to Qd. A nondegenerate quadratic form is Minkowski iff, for some integer d > 2, it has signature (d — 1,1). A symmetric bilinear form on a vector space is Minkowski if it is the polarization of a Minkowski quadratic form. A Minkowski vector space is a vector space together with a Minkowski quadratic form. Let F be the quadratic form on Rd given by F(x\,..., xd) = x\-\ \- xd_t — xd. Then the isomorphism Rd -»• Rd given by ( x i , . . . , xd) \-¥ (x2,...,

xd-i, xi + (xd/2), xi -

(id/2))

carries the form Qd to F, and so the definition here of a Minkowski form agrees with the definition given in §1.1. Let V be a vector space. Then MinkQF(F) denotes the collection of all Minkowski quadratic forms on V, while MinkSBF(y) denotes the set of Minkowski symmetric bilinear forms on V. Let (V, Q) be a Minkowski vector space and let v E V. We say that v

40

Basic Results and

Definitions

is timelike (resp. spacelike, resp. lightlike) if Q(v) < 0 (resp. Q(v) > 0, resp. Q(v) = 0). Note that a vector is isotropic with respect to Q iff it is lightlike.

Chapter 3

Basic Differential Topology

3.1

Some differential topological notions

By smooth, we will always mean C°°. By manifold, we will always mean smooth manifold without boundary. A manifold is, by assumption, Hausdorff, finite dimensional, locally compact and second countable. For any manifold M, the tangent bundle of M will be denoted TM and, for each m 6 M, the tangent space to M will be denoted TmM. Vector fields are, by definition, smooth. Let M and iV be manifolds, let / : M -»• N be a diffeomorphism and let V be a vector field on M. Then the vector field f*{V) on M is defined

by (/.00)„ = (df)(vf-Hn)). By vector bundle, we will always mean real vector bundle, unless otherwise stated. Fiber bundles and vector bundles are always assumed smooth. A section of a fiber bundle or of a vector bundle is, by definition, smooth. Let M be a manifold and let E be a vector bundle over M. We define p := rank(£). A framing of E is a collection Xi,...,Xp of sections of E such that, for all m £ M, (Xi)m,..., (Xp)m is a basis of Em. A framing of TM will be called a tangent framing of M Let M be a manifold and let / : M -> M be a diffeomorphism. For any vector field V on M, we say that / preserves V if f*(V) = V. For any framing T of TM, we say that / preserves T if it preserves each of the vector fields in T. Let X and Y be manifolds and let / : X -> Y be smooth. We say that / is a local diffeomorphism if, for all x € X, there is a neighborhood X0 41

42

Basic Differential Topology

of x in X and a neighborhood Y0 of /(a;) in Y such that f(Xo) — Y0 and such that /|Xo : XQ —> Y0 is a diffeomorphism. A trivial covering map is a trivial fiber bundle with discrete fiber. A covering map is a fiber bundle with discrete fiber. Any covering map is a local diffeomorphism. Let M be a manifold and let E be a vector bundle over M. The rank of E, denoted rank(£J), is the vector space dimension of the fibers of E. Let c and d be integers satisfying 1 < c < d. Let M be a d-dimensional manifold and let S C M. Then S is a c-dimensional submanifold of M if, for all s £ S, there exists an open neighborhood U of s in M and there exists a diffeomorphism <j> : U -» Rd such that ^ ( 5 f l [ / ) = l c x { 0 } d _ c . We will say that 5 is a c-dimensional leaflike submanifold if, for all s 6 S, there exists an open neighborhood U of s in M, there exists a diffeomorphism <j> : U -> E d and there exists a countable subset T C M d_c such that 4>{S n £/) = Rc x T. Let M be a manifold. Then any O-dimensional leaflike submanifold of M is a O-dimensional subset of the toplogical space M. Moreover, any O-dimensional submanifold of M is discrete in M. (Recall, that, in our terminology, discrete does not imply closed.) Let M be a manifold and let d := dim(M). A subset of M is a submanifold (resp. leaflike submanifold) if, for some integer c € [l,d], it is a c-dimensional submanifold (resp. c-dimensional leaflike submanifold) of M. Any submanifold is a leaflike submanifold. For any leaflike submanifold L of M, L is a submanifold of M iff L is locally closed in M iff the leaflike topology on L is the same as the relative topology on L inherited from M. In particular, any closed leaflike submanifold of a M is a closed submanifold of M. If M 0 :—R2 is the Euclidean plane and if So is a carefully drawn figure eight in Mo, then there are two distinct manifold structures on So under which the inclusion So —>• Mo is an immersion (see the figures at the bottom of p. 25 in §1.31 of [Wa83]). By contrast, for any leaflike submanifold S of a manifold M, there is a unique manifold structure on S such that the inclusion map S —¥ M is an immersion; we call this the leaflike manifold structure on S from M . The topology of this manifold structure is called the leaflike topology on S from M; it may be finer than the relative topology on S inherited from M. However, if S is a submanifold of M, then the leaflike topology on S from M agrees with the relative topology

Some differential topological notions

43

on S inherited from M . If V is a vector field on a manifold M, and if {(f>t} denotes the local flow of V, then, for any tensor a on M, the Lie derivative of a along V is written Cya and is defined by Cya — (d/dt)tz=o{<j>ta)- If V a n d W are vector fields on a manifold M, then we define [V, W] := CyW. L E M M A 3.1.1 Let V and W be vector fields on a manifold M and let f : M -> E be a smooth function. Then [V, W]f = VWf - WVf. Proof. By the product rule, we have Cv(Wf) = (CvW)f + W{Cvf)So, since Cv{Wf) = VWf, since CVW = [V,W] and since Cvf = Vf, we get VWf = [V, W)f + WVf, and the result follows. • By Lemma 3.1.1, p. 43, if V and W are vector fields on a manifold M, then, for any / € C°°(M), we have [V, W]f — —[W, V]f; since a vector field is determined by its effect on functions, we conclude that [V, W] = —[W, V]. Similar arguments prove that if V, W and X are vector fields on a manifold, then \V + W,X] = [V,X] + [W,X] and [V,W + X] = [V,W] + [V,X]. Finally, similar arguments prove that if V and W are vector fields on a manifold and if a e E,then [aV, W] = a[V, W] and [V, aW] = a[V, W]. Thus, [ •, • ] is E-bilinear and antisymmetric. For any three vector fields V, W and X on a manifold M, a short computation shows, for all / € C°°(M), that [V, [W,X]]f = [[V, W],X]f + [W, [V,X]]f; it follows that [V, [W, X}] = [[V, W],X} + [W, [V, X}]. This equation is a form of the Jacobi identity. Let X and Y be manifolds. Let / , : X —> Y be a sequence of smooth maps. We say that fi satisfies the convergence-divergence dichotomy if one of the following two (mutually exclusive) conditions holds: • for all x 6 X, fa(x) is not a convergent sequence in Y; or • fi : X —> Y converges in C°° on compact sets. The next lemma asserts that the convergence-divergence dichotomy is a local condition. L E M M A 3.1.2 Let X and Y be manifolds and let fi : X ^ Y be a sequence of smooth maps. Assume that X is connected. Assume, for all m € M, that there is an open neighborhood U of m in M such that the sequence fi\U : U —> Y satsifies the convergence-divergence dichotomy. Then f satisfies the convergence-divergence dichotomy.

44

Basic Differential

Topology

Proof. Fix x0 £ X and assume that fi(xo) is convergent in Y. We wish to show that fi : X —> Y converges in C°° on compact sets. Let X' be the set of all x £ X such that, for some open neighborhood V of x in X, the sequence fi\V : V —> X ' is convergent in C°° on compact sets. Since C°° convergence on compacta is a local property, it suffices to show that X' = X. Fix xi eX. We wish to show that xi £ X'. Let U be an open cover of X such that, for all U £ U, U ^ 0 and the sequence fi\U : U -» Y satisfies the convergence-divergence dichotomy. By Lemma 2.3.3, p. 17, xo and zi are W-chain connected. Fix an integer n > 1 and a t/-chain ( z o , . . . , zn) from xo to x\. Then zo G X' and we wish to show that zn £ X'. Fix j £ { l , . . . , n } and assume that Zj-i £ X ' . We wish to show that Zj £ X'. Choose U £ U such that ZJ-I,ZJ 6 U. Since Zj-i £ Y, it follows that fi(zj-i) is convergent in Y. Then, as fi\U : U —> Y satisfies the convergence-divergence dichotomy, it follows that fi\U : U —> V is convergent in C°° on compact sets. So, as Zj £ U, we conclude that Zj £ X'. D L E M M A 3.1.3 Let X be a manifold and let gi : X —>• X 6e a sequence of diffeomorphisms of X. Let J- be a tangent framing of X and assume, for all i, that gi preserves T. Then gi has the convergence-divergence dichotomy. Proof. Let d := dim(X). Let V\,..., Vd be the vector fields defining T. Let J := { 1 , . . . , d}. For all j £ J, let X has the convergence-divergence dichotomy. For all 6 > 0, let Is := (-6, S) and let Us := if C Rd. For all j £ J, for all 6 > 0, let Ci(j) be the set of all a; € X such that, for all t £ Is, (t>j(x) is defined; then Cs(j) is closed in X. For all j £ J, for all 8 > 0, for any closed A C X , let C^Q', ^4) be the set of all x £ Cj(5) such that, for all t £ Ig, we have $(:r) € A; then C$(j, A) is closed in X . We recursively define, for all ji,.. .jk £ J, for all 6 > 0, for all integers k > 2, CdO'i,... .j*) := Cs(ji,Cs(J2, • • -Jk))] then C ^ j i , . . .,jk) is closed in X . For all integers fc > 1, for all j i , . . . , jk £ J, we have \^J Cg(ji,... ,jk) = X. <5>o For all integers k > 1, for all ji,...,jk € J, for all ( u i , . . . ,u*;) £ Ik, let

G!=«--^:^

i*)^-

Some differential topological notions

45

For all 6 > 0, let X's := Cs{d, d - 1 , . . . , 1) and let X4:=C4(l,2,...,d-2,d-l,d,d,d-l,d-2,...,2,l). Choose So > 0 such that x0 € Xj 0 . For 8 £ (0,So), define ips : Us —*• -X" by ^ ( s i , . . . , S d ) = ^.'.iSl(a;o),

and let Va := ips(Us)- By the Inverse Fucntion Theorem, choose p < So such that Vp is open in X and such that ipp : Up —¥ Vp is a diffeomorphism. Then i 6 Xj 0 C X p . Let [7 := Up, let V := Vp and let x ~ Vv • U -> V. Fix « o £ V and assume that <7i(«o) is convergent in X . We wish to show that gi\V : V —>• X converges in C°° on compact sets. Choose (uii,... ,Wd) € U such that vo = d.d.'1 Wl(%o)- Since zo S X p , we conclude that v0 £ X'. For all i, let xt := —IUJ. Then ^.V.^Cuo) = ^oSince vo G X'p, we see, for all i, that gi(vo) G X1,. Note that X£ is closed in X and that (fix}.'.'dXd : X'p -» X is continuous. So, as fft(^o) is convergent in X , we conclude that x\.'dXd(gi(vo)) is convergent in X . For all i, we have 9i{x0) = 9i((fi}::dXd(vo)) = 4>V':dXd(9i(vo))- Then gi(x0) is convergent in X . Let x' := lim gi{xo). i—too

Since xo £ Xp, we conclude, for all i, that Pi(xo) € Xp. So, since X p is closed in X , it follows that x' G X p . For all i, define the map Xi • U ->• X by Xi(*i, • • •, sd) = ^f.'.i' 1 (ffi(^o)). Define the map x' : C7 -> X by x'( s i> • • • >sd) = ^df.'.'i*1^')- Since <7j(a:o) —• x' in X, it follows that Xi —> x' i n C°° o n compact sets. For all i, for all s = ( s i , . . . , s<j) £ [7, we have

Xi(s) = 4>dlT(9i{zo)) = 9i{.4>d*T&*)) = 3<(lM«))For all i, for all u G £/, we have Xi( u ) = Si(Vv(u))- Then, for all u e V, we have Xi(x -1 (t>)) = 9iiv). That is, &|V = Xi° X" 1 - T h e n 5«l^ -> x' ° X" 1 in C°° on compact sets. • Let V and W be vector spaces and let T : V -> W be a linear transformation. Then the rank of T is dim(T(V)). Let M and JV be manifolds and let / : M -t N be smooth. We say that / is of constant rank if there is some integer r > 0 such that, for all m e M, the rank of (d/) m : TmM -> Tf^N is = r. In this case, the integer r is called the rank of / .

46

Basic Differential

Topology

L E M M A 3.1.4 Let M and N be manifolds and let 4> : M ->• N be a smooth map of constant rank. Let C C N be countable. Then <j>~~x (C) is a leaflike submanifold of M. Proof. Fix m G M. We wish to show that there is an open neighborhood U of m in M such that (<^ -1 (C)) fl U is a leaflike submanifold of U. Let n := • E« be defined by TT(XI,. .. ,Xd) = (xi,...,xq). Let t : E g —• E e be defined by t(a;i,... ,x g ) = (xi,...,xq,0,...,0). By Theorem 1.15.5, p. 86 of [He78], choose open neighborhoods U of m in M and V of n in iV and choose diffeomorphisms a : U -»• E d and /? : E e -»• F such that ^(J7) = V and such that is countable and we have ad^iC^nU) =a((<j>\U)-1(CnV)) = n-1{D) = W x D. D L E M M A 3.1.5 Let M be a manifold and let L and L' be leaflike submanifolds of M. Let d := dim(M) and let r 6 [0,d] be an integer. Assume, for all m € LFi L', that dim((T m L) n (TmL')) = r. Then L D L' is a r-dimensional leaflike submanifold of M. Proof. Fix m € LC\L'. We wish to show, for some open neighborhood U of m in M, that L (1 L' fl U is an r-dimensional leaflike submanifold of U. Let p := dim(L) and q := d — p. Choose • an open neighborhood V of m in M; • a countable subset S C I ' ; and • a diffeomorphism * : V ->• Rd such that * ( L n V J ^ f f x S . Define TT : E d -^ W by 7r(a;i,...,a; d ) = ( i p + i , . . . , ^ ) .

Let ip := 7r o * : V -> E«. Then £ n V = V - 1 ^ ) - Moreover, for all i; G L D V, T„L is the kernel of (dtp)v : TVV -> T ^ I R ' . Let p' := dim(L') and q' := d — p'. Choose • an open neighborhood P" of m in M; • a countable subset S' C 1 ' ; and • a diffeomorphism * ' : V ->• E d

Inheritability

of continuity

such that * ' ( £ ' n V ) = f '

and smoothness

to leaflike submanifolds

47

x S'. Define TT' : Erf - • W' by

•n'(xi,...,xd)

=

(xp,+i,...,xd).

Let V' := TT' o ¥ ' : V -+ R?'. Then L ' n V = (V»') -1 (S')- Moreover, for all u e i ' f l V, we have ker((d^')») = TVL'. Let 17 := V n V . We wish to show that LCiL' C\U is & leaflike submanifold of U. We have m eU. Let k := q + q', let AT := Rk and define : U -¥ N by (w) = {ip(w),il;'(w)). Let C := 5 * S" C AT. Then C is countable. We have LDL' nU = ~l(C). For all u E U, the kernel of the linear map (d<j>)u : TUU ->• T^^N is equal to (ker(dV>)u) n (ker(dt/>')u)- Thus, for u £ U, ker((d)u is equal to d — r. In particular, we see that <j> has constant rank. Therefore, by Lemma 3.1.4, p. 45, since LF\L'C\U — (f)"1 (C), we see that L n L' H (7 is a leaflike submanifold of U. D 3.2

Inheritability of continuity and smoothness to leaflike submanifolds

Let M be a manifold and let X be a leaflike submanifold of X. Give X the leaflike manifold structure from M. L E M M A 3.2.1 Let Y be a locally topological space and let f : Y —> M be a function. Assume that f(Y) C X. Then f :Y —>• X is continuous iff f :Y —*• M is continuous. Proof. "Only if" follows from continuity of the inclusion X —> M, so we need only prove "if". Let d := dim(M), p := dim(X) and q := d — p. We may assume that M = f x 1 ' and that there is a countable subset C C f such that X = W x C. Since continuity is a local condition and since Y is locally connected, we may assume that Y is connected. Then choose c € C such that f{Y) C W x {c}. Let -K : M -» X be defined by n(s,t) = (s,c). Then / : Y -> X is the composite of the continuous maps / : Y —> M and -K : M —> X, and so is continuous. • Since the map n of the proof of Lemma 3.2.1, p. 47 is also smooth, the same argument shows:

48

Basic Differential

Topology

L E M M A 3.2.2 Let Y be a manifold. Let f : Y ->• M be a function. Assume f(Y) C X. Then f :Y -> X is smooth iff f :Y ->• M is smooth. L E M M A 3.2.3 Let M be a manifold and let m £ M. Let L leaflike submanifolds of M. Assume that m £ L C V. Let component of L containing m, in the leaflike topology on L. the component of L' containing m, in the leaflike topology on Lo Q L0.

and L' be LQ be the Let L'Q be L'. Then

Proof. The inclusion map L —> M is continuous, so, by Lemma 3.2.1, p. 47, the inclusion map i : L -» L' is continuous. Then t(L 0 ) is connected in the relative topology inherited from L' and m £ L(L0), SO t(L0) C L'0. Since t: L —> L' is inclusion, we have i(Lo) = LQ. Then L0 C L'0. • L E M M A 3.2.4 Let L be a leaflike submanifold of a manifold M. Then L is path-connected in its leaflike topology iff L is path-connected in its relative topology inherited from M. Proof. Let a denote the leaflike topology on L and let r denote the relative topology on L inherited from M. The inclusion map (L, a) —> M is continuous, so the identity bijection (L, a) —> (L, r ) is continuous. So, if (L, a) is path-connected, then (L, r ) is, as well. Now assume that (L, r ) is path-connected. Fix 1,1' € L. We wish to show that there is a continuous map 7 : [0,1] —> (L, a) such that 7(0) = / and 7(1) = /'. Since (L,r) is path-connected, let 7 : [0,1] —> (L,T) be a continuous map such that 7(0) = / and 7(1) = I'. Then 7 : [0,1] —> M is continuous, so, by Lemma 3.2.1, p. 47, 7 : [0,1] —> (L,a) is continuous. • By Lemma 3.2.4, p. 48, for a leaflike submanifold, there is no abiguity in asking whether it is path-connected. By contrast, a leaflike submanifold may be connected in the inherited topology and, at the same time, disconnected in the leaflike topology: Let T 2 := E2 / Z 2 be the two-torus, as an additive Lie group. Let n : M2 —> T 2 be the canonical homomorphism. Let a £ E\Q. Let L := {(t,at)\t £ R} C R 2 . Let x := (0,a/2). Then 2 2 x £ R \(Z + I?) and 2x £ L + Z . Let L0 := ir(L) C T 2 . Let x := n(x). Then x £ T 2 \ L 0 and 2x £ L0 and let L := L0 U (x + L0). Then, as T 2 is connected and as L is dense in T 2 , we conclude that L is connected in the relative topology, inherited from T 2 . However, by Lemma 3.4.1, p. 51 below, in the leaflike topology on L, both LQ and x + LQ are open, so, since LQ fl (x + LQ) = 0, we see that L is disconnected.

Definition

3.3

of prefoliation

and

foliation

49

Definition of prefoliation and foliation

For all integers p > 0, let T ' C f x W be the transitive equivalence relation on W i.e., the equivalence relation with one equivalence class, defined by Tp := W? xW. For all integers q > 0, let T09 C W x W be the trivial equivalence relation on W i.e., the equivalence relation defined by TQ := {{x,x) \x E K ? }; each equivalence class of TQ has only one element. If X and Y are sets, if R C X x X is an equivalence relation on X and if S C Y x Y is an equivalence relation on Y, then we define an equivalence relation R x S C (X x Y) x (X x Y) on X x Y by R x S := {((x,y), (x',y')) | (x,x') € fl and {y,y') € 5 } . Then the (R x S)-equivalence class of an element (x, y) € X x Y is the product of the .R-equivalence class of x by the 5-equivalence class of y. Let R C X x X be an equivalence relation on a set X. We say that R is countable if each of its equivalence classes is countable. If XQ C X, then we define R\X0 C X0 x XQ to be the restriction of R to Xo defined by R\X0 := R n (X 0 x X 0 ); it is an equivalence relation on XQ. Let X and V be sets and let : X —> V be a bijective function. Let i? be an equivalence relation on X. Then we define *(i?) to be the equivalence relation on Y defined by <^*(i?) = {((j)(x),: M —> Rd and there is a countable equivalence relation R on W such that » {T) =TP x R. Equivalently, there is a diffeomorphism 4>: M -¥ Rd such that Tp x Tj has countable index in <j)*{T). We say that T is ap-dimensional prefoliation on M if, for all m G M, there exists an open neighborhood M0 of m in M such that J-\MQ is a trivial p-dimensional prefoliation on M. An equivalence class L of a prefoliation is a leaflike submanifold; we give it the leaflike manifold structure, i.e., the unique manifold structure making the inclusion map L —> M an immersion. An equivalence class of a prefoliation, together with its leaflike manifold structure, is called a leaf of the prefoliation. If T is a p-dimensional prefoliation on M, then we define TT := I ) TL, where the union is taken over all equivalence classes L

L of !F; then TT is a p-dimensional vector subbundle of TM. We say that

50

Basic Differential

Topology

J7 is a p-dimensional foliation on M if T is a p-dimensional prefoliation on M such that every leaf is connected in its leaflike topology. We say that T is a prefoliation on M (resp. foliation on M) if there is some integer p £ [0, d] such that T is a p-dimensional prefoliation (resp. foliation) on M; in this case, we define dim(JT) := p. Let M be a manifold and let d := dim(M). Let p £ [0,d] be an integer and let q := d—p. In this book, we will say that an equivalence relation T on M is a proper p-dimensional prefoliation on M if, for all m £ M, there is an open neighborhood U of m in M and a diffeomorphism (j> : U —> Rd such that
Preliminary results to the Frobenius Theorem

3.4

51

Preliminary results to the Frobenius Theorem

L E M M A 3.4.1 Let T be a prefoliation on a manifold M and let C be a countable collection of leaves of T. Let LQ := M L. Then L§ is a leaflike Lee submanifold of M. Moreover, the leaflike manifold structure v on LQ is equal to the disjoint union, over all L € C, of the leaflike manifold structure vi, on L. Consequently, for all L € C, L is open in {LQ,V). Consequently, for all L e C, for all I 6 L, we have Ti(L, VL) = Ti(L0, v). Proof. It follows from the definitions of prefoliation and of leaflike submanifold that L0 is a leaflike submanifold on M. Let r denote the disjoint union, over all L € C, of the leaflike manifold structure on L. Then the inclusion map (LO,T) -> M is an immersion, so, by definition of leaflike manifold structure, r is the leaflike manifold structure on LQ. • L E M M A 3.4.2 Let M be a manifold and let T and T' be equivalence relations on M. Assume that T is a prefoliation on M and that T has countable index in T'. Then T' is a prefoliation on M and TT = TT'. Proof. Let d := dim(M), p := dim(J") and q := d — p. We may assume that M = W x W and that TQ := T? x T0» has countable index in T. Then T§ has countable index in T', so T' is a prefoliation. Moreover, by Lemma 3.4.1, p. 51, for all m € M, we have Tm(T{m)) = Tm{P{m)). That is, for all m € M, we have TmT = TmP. Then TT = TT'. • L E M M A 3.4.3 Let M be a manifold and let A be a vector subbundle ofTM. Let Xi,...,Xp be a framing of A and assume, for all integers i,3 £ [1>PL that [Xj,Xj] is in A. Then A is involutive. Proof. Let A and B be vector fields in A. We wish to show that the commutator vector field [A, B] is in A. Choose fi,...,fpE C°°{M) such that A = frXi + 1- fpXp. Choose 9i,--.,gP € C^iM) such that B = giXi + ••• + gpXp. Then we have , [A,B] = £i[fi-Xi,grXj],BO

[A,B]=\Ytfi-gr[Xi,Xi\\

+

52

Basic Differential

J2Si • (xi9j) • * ; i,i

}

-

Topology

£ > • (Xjfi) • Xi \ hi

so [A,B] is in A.

•

Let M be a manifold and let A be a vector subbundle of the tangent bundle of M. We say that a framing Xi,... ,XP of A is commuting if, for all integers i, j € [l,p], we have [X;,X,-] = 0. The next result is left as an exercise for the reader. L E M M A 3.4.4 Let M be a manifold, let k > 1 be an integer and let Xi,... ,Xk be vector fields on M. Assume, for all integers i,j € [l,k], that [Xi,Xj] = 0. Assume, for all m € M, that (Xi)m,..., (Xk)m are linearly independent vectors in TmM. Let d :— dim(M) and let di,...,dd be the standard framing of Rd. For all 6 > 0, let Is := (-6, S) and Vs := if C Rd. Then, for all m 0 € M, there exist • an open neighborhood MQ of mo in M; • 6 > 0; and • a diffeomorphism <j>: Mo —> Vs such that, for all integers i £ [l,fc], we have *(Xi\Mo) = 9i|Va3.5

Uniqueness in the Frobenius Theorem

L E M M A 3.5.1 Let Y be a manifold and let M be a manifold. Let T be a prefoliation on M. Let (j) : Y —> M be a smooth map. Assume that (d(j))(TY) C TT. Then there is a countable subset C C M such that Y C T{C). Proof. Let d := dim(M). Let p := dim(!F). Let q := d — p. We may assume that M = Rd and that To := Tp x T0? has countable index in T. Let 7r : Rd ->• W be projection onto the last q coordinates, defined by 7r(*i,..., tj) = (tp+i ,...,td). Then, for all w £ TFo, we have (dir)(w) — 0. By Lemma 3.4.2, p. 51, we have TT0 = TT. Let V := TT O : Y -> W. For all v 6 TY, we have (d(j>)(v) £TT = TT0, so (di/>)(v) = (d7r)((d0)(«)) = 0. Then ip is constant on each connected component of Y, which implies that C := tp(Y) is countable. Since Y C V _ 1 (C) = W xC = T0{C) C ^ ( C ) , we are done. D

Uniqueness in the Frobenius Theorem

53

Given a set M with a manifold structure A, for any open subset U of (M, A), let A|?7 denote the relative manifold structure on U inherited from M. Let M be a set, let 7 be a countable set and, for all i e I, let Mi be a subset of M. Assume, for all i, j £ I, that M; n Mj ^ 0 =$> i = j , i.e., that {Mi}i£i is a pairwise disjoint collection of subsets of M. Let d > 0 be an integer. For all i, let Aj be a d-dimensional manifold structure on M». Then there is a unique manifold structure A on M such that, for all i, we have: Mj is open in (M, A) and A|M» = A;. We say that A is the disjoint union of A;, as i ranges over I. L E M M A 3.5.2 Let Y be a connected manifold and let M be a manifold. Let J7 be a prefoliation on M. Let <j> :Y —i M be a smooth map. Assume that (d<j>){TY) C TP. Then there is a leaf L of P such that 4>(Y) C L. Proof. By Lemma 3.5.1, p. 52, choose a countable subset C C M such that (Y) C P(C). Let C := {?{c)\c € C}. By Lemma 3.4.1, p. 51, LQ := F{C) is a leaflike submanifold of M and, for all L € C, L is an open subset of Lo- By Lemma 3.2.1, p. 47, the map : Y —> LQ is continuous. Then {(/>_1(L) | L G C} is a partition of Y into open subsets. So, since Y is connected, choose L £ C such that 0 - 1 ( L ) = Y. Then <£(Y) Q L. D L E M M A 3.5.3 Let Y be a connected manifold and let M be a manifold. Let T be a prefoliation on M. Let cj> : Y —• M be a smooth map. Assume, for all y 6 7 , that {d4>){TyY) — T^^T. Then there is a leaf L of T such that 4>(Y) is an open subset of L. Proof. By Lemma 3.5.2, p. 53, choose a leaf L of T such that (j>(Y) C L. By Lemma 3.2.2, p. 48, <j> : Y -t L is smooth. For all y e Y, we have (d(f>)y(TyY) = THy)F, so (d<j>)y(TyY) = THy)L. That is, 0 : Y -> L is a submersion. Then, by the Implicit Function Theorem, we conclude that <^>: Y —> L is open. In particular, <j>(Y) is an open subset of L. • L E M M A 3.5.4 Let T and T' be foliations on a manifold M and assume that TT C TT'. Then T C T'. Proof. By Lemma 3.5.2, p. 53, we see that each leaf of T is contained in a leaf of P'. That is, PC P. • P R O P O S I T I O N 3.5.5 Let P and P' be foliations on a manifold M and assume that TP - TP. Then P - P .

54

Basic Differential

Proof. 3.6

Topology

By Lemma 3.5.4, p. 53, T C P and T' C T.

U

Passage from local t o global in the Frobenius Theorem

Let M be a manifold and let U be a countable open cover of M . Let A be a vector subbundle of TM. For all U € H, let Fez be a prefoliation on U and assume that TTu = A|J7. Let T be the equivalence relation on M generated by I ) Tuueu L E M M A 3.6.1 The equivalence relation T is a prefoliation on M and TT = A. Proof. For all V,W € U, for all « G 7 , by Lemma 3.5.1, p. 52 (with T replaced by Tw and with the map <j>:Y —>• M replaced by the inclusion map (Fv(w)) n W —> TV), there exists a countable subset C CW such that {Tv{v))r\W C FW(C). By Lemma 2.1.2, p. 13, we see, for all U eU, that Tu has countable index in T\U. Then, by Lemma 3.4.2, p. 51, we see, for all U E U, both that T\U is a prefoliation on U and that T{F\U) = TFu, whence T(F\U) = A|t7. So, since U is an open cover of M, we see that T is a prefoliation on M and that TT = A. • L E M M A 3.6.2 Assume, for allU £ U, thatTu J7 is a foliation on M and TJ- = A.

is a foliation on U. Then

Proof. By Lemma 3.6.1, p. 54, we see that T is a prefoliation on M and that TT = A. Fix m € M and let X := T(m). Let ^ 0 denote the leaflike topology on X from M . We wish to show that (X, fio) is connected. For all U&U, let ya := {Tu{x) \ x £ U n X } . Let y := ( J 3^c/- Then X is 3^-connected. By Lemma 2.3.2, p. 17, we wish to show, for all Y € y, that Y is connected in the relative topology inherited from (X, /xo). Fix Y 6 y. Wish to show that (Y,no\Y) is connected. Choose U £ U and z e C/flX such that Y = Fu(x). Let i/o be the leaflike topology on Y from U. Let / : V -> M be the inclusion map. Then / : (Y, u0) ->• M is continuous and / ( F ) C X , so, by Lemma 3.2.1, p. 47, we see that / : (Y, vo) -¥ (X,no) is continuous. Since T\j is a foliation, it follows that (Y, UQ) is connected, so, by continuity of / : (Y,vo) -> (X,no), we conclude that (f(Y),fj,o\(f(Y))) is

The Frobenius

Theorem

connected. Since / : Y -> M is the inclusion map, we have f(Y) Then (y,/io|y) is connected.

3.7

55

= Y. D

The Frobenius Theorem

L E M M A 3.7.1 Let M be a manifold and let A be a vector subbundle of the tangent bundle TM. Suppose that A is locally integrable, i.e., suppose, for all m 6 M, that there exists an open neighborhood MQ ofm in M such that A|Mo is integrable. Then A is integrable. Proof. Since M is Lindelof, let U be a countable open cover of M such that, for all U £U, A\U is integrable. For all U £ U, let Tu be a foliation on U such that A\U = TTu. Let T be the equivalence relation generated by M TJJ. Then, by Lemma 3.6.2, p. 54, T is a foliation on M and V&A

A = TT. Then A is integrable.

•

Let M be a manifold and let A and A' be two vector subbundles of TM. We say that A' is complementary to A in M if, for all m € M, we have A m n A'm = {0} and A m + A'm = TmM. Note that this is equivalent to saying that A' is a vector bundle complement to A in TM. L E M M A 3.7.2 Let M be a manifold. Let A be a vector subbundle Let p be the rank of A. Let mo € M. Then there are

ofTM.

• an open neighborhood Mo of m in M; • a vector subbundle A' ofTM^; and • a framing X\,..., Xp of A|Mo such that • for all integers i,j € [l,p], [-2Q,X;] is in A'; and • A' is complementary to A | M Q in MQ. Proof. Let d := dim(M). We may assume M = Rd and m 0 = 0 € Rd. Let di,...,dd be the standard framing of E d . For all integers i € [i-,d], we define vi := (di)o £ TQM. By a change of basis, we may assume that Ao = Mvi + • • • + Rvp. Let T be the vector subbundle of TM spanned by the vector fields dp+\,... ,ddWe have To n Ao = {0}. Let Mo be an open neighborhood of 0 in Rd such that, for all m e MQ, we have T m n Am = {0}. By dimension count,

56

Basic Differential

Topology

for all m G M0, for all integers i G [l,p], we have Tm + Am = TmM0Then r | M 0 is complementary to A|M 0 in M 0 . For all integers j G [l,d], let 3j := 3j|Mo. For all integers i G [l,p], choose a section Wi of T\Mo and a section Xi of A|Mo such that d[ = Wi + Xi. For all integers i G [l,p], d

c h o o s e / ^ + 1 , . . . , / J G C ° ° ( M o ) such that VKi= ] T

/«3£.

fc=p+i

Let A' := T|Mo be the vector subbundle of TM0 spanned by the vector fields d'p+1,... ,d'd. Fix integers i, j G [l,p]- We wish to show that [X^X,] is in A'. We have [d'^d'j] = 0, so [d'^d'j] is in A'. We have [d'i, Wj] = (d'4+1)d'p+1

+ ••• + (d'iti)d'd,

so [dl, Wj] is in A'. Similarly, [d'j, Wi] is in A'. Finally, we compute that

[wi,wj}= J2 E k=p+l

l=p+l

\(flMfhd'i) - ((fttd'ifiM) L

so [Wh Wj] is in A'. We have Xi=&i-

Wi and Xj = d'j - Wj, so

[xh Xj] = [di d'j] - [dl, Wj] + [d'j,Wi] + [wu Wj]. We conclude that [Xi,Xj] is in A'. L E M M A 3.7.3 Let M be a manifold. Let A be a vector subbundle If A admits a commuting framing, then A is integrable.

• ofTM.

Proof. Let d := dim(M) and let p be the rank of A. Let Xi,..., Xp be a commuting framing of A. Fix m G M. By Lemma 3.7.1, p. 55, we wish to show that there is an open neighborhood Mo of m in M such that A|Mo is integrable. For all integers i,j G [l,p], we have [Xj,X,] = 0. Let d\,...,dd, Vs, Mo and <j> be as in Lemma 3.4.4, p. 52. Let To := (Tp x T°)\VS. Then TFo = ( # ) ( A | M 0 ) . Let Tx := (^- 1 )*(^ 0 )- Then TTi = A|M 0 . Since T0 is a foliation, T\ is a foliation, and so A|Mo is integrable. D L E M M A 3.7.4 Let M be a manifold and let J7 be a prefoliation on M. Then TT is involutive. Proof. Let d := dim(M), let p := dim(J r ) and let q := d — p. We may assume that M = Rd and that To := Tp x T° has countable index in T.

Potential submersions

57

Then, by Lemma 3.4.2, p. 51, we have TT = TTQ. Let X and Y be vector fields in M. Assume that X and Y are in TT. We wish to show that [X, Y] is in TT. Let 7r : Rd -> E ? be projection onto the last q coordinates, defined by ?r(ii,..., td) = (tp+i,..., td). For all smooth / : E« ->• M, let 717 : E d -» E be the smooth function defined by 7T/ = fon. For any vector field V on M, all three of the following are equivalent: • V is in TT; • V is in TTo; and • for all smooth / : W -» E, we have V717 = 0. Let / : W ->• E be smooth. We wish to show that [X, Y}iTf = 0 By assumption, X and Y are in TT, so XiTf = 0 and Ynf •• 0. Then, by Lemma 3.1.1, p. 43, [X,Y]nf = X F T T / - YXnf = 0. D We can now state and prove the Frobenius Theorem: T H E O R E M 3.7.5 Let M be a manifold and let A be a vector subbundle of the tangent bundle TM. Then A is involutive iff A is integrable. Proof. Since "if" follows from Lemma 3.7.4, p. 56, we need only prove "only if". Fix m 6 M. By Lemma 3.7.1, p. 55, we wish to show, for some open neighborhood MQ of m in M, that A|Mo is integrable. Let p be the rank of A. Choose Mo, A' and X\,... ,XP as in Lemma 3.7.2, p. 55. By Lemma 3.7.3, p. 56, it suffices to show that Xi,..., Xp is commuting. Fix integers i, j 6 [l,p]- We wish to show that [Xi,X,-] = 0 on Mo. By Lemma 3.7.2, p. 55, we have that [Xj,X,] is in A'. On the other hand, by involutivity of A, we have that [Xi,Xj] is in A|M 0 . So, as A' is complementary to A|Mo in M 0 , we conclude that [Xi, Xj] — 0 on M 0 . • Theorem 3.7.5, p. 57 combined with Proposition 3.5.5, p. 53 shows that, for any manifold M, the map T \-> TT is a one-to-one-correspondence between foliations on M and involutive vector subbundles of TM. 3.8

Potential submersions

Let X and Y be manifolds and let / : X -> Y be smooth. If x G X and if y :— f(x), then we say that / is submersive at x (resp. immersive at x) if the differential df : TXX -> TyY is surjective (resp. injective). We say that

58

Basic Differential

Topology

f is a submersion (resp. immersion) if / is submersive (resp. immersive) at every point of X. L E M M A 3.8.1 Let X and Y be manifolds and let f : X -» Y be smooth. Let X' be a manifold and let a : X —» X' be a surjective submersion. Let Y' be a manifold and let ft : Y —»• Y' be a submersion. Let f':X'—> Y' be a function. Assume that f o a = f3 o f. Then f : X' —> Y' is smooth. Proof.

This follows from the Implicit Function Theorem.

•

COROLLARY 3.8.2 Let X, Y and Y' be manifolds and let f : X -> Y and f : X —¥ Y' be surjective submersions. Let : Y —»• Y' be a bijective function such that o / = / ' . Then is a diffeomorphism. COROLLARY 3.8.3 Let X be a manifold, let Y be a set and let the function f : X —t Y be surjective. Then there is at most one manifold structure XonY such that f : X —> (Y, A) is a submersion. Let X be a manifold, let Y be a set, let / : X —> Y be a function and let Y0 := f(X). We will say that / is a potential submersion if there is a manifold structure on YQ with respect to which / : X —> YQ is a submersion. By Corollary 3.8.3, p. 58, a potential submersion f : X -> Y induces a manifold structure on YQ. Let X be a topological space, let Y be a set and let / : X —> Y be a function. We will say that / is potentially open if, for every open subset U of X, the set f~1(f(U)) is open in X. Equivalently, after giving Y the quotient topology induced by / , the map / : X —> Y is open. Let X be a manifold, let Y be a set and let / : X —> Y be a function. We will say that / is a local potential submersion if, for every x € X, there is an open neighborhood U of x in X such that f\U : U —> Y is a potential submersion. Let A" be a topological space, let Y be a set and let / : X —> Y be a function. We will say that / is locally potentially open if, for every x £ X, there is an open neighborhood U of x in X such that f\U:U—*Y is potentially open. If X and Y are topological spaces and if / : X -¥ Y is surjective, continuous and open, then the topology on Y is the quotient topology defined by / ; it follows that / is potentially open. By the Implicit Function

Potential submersions

59

Theorem, any surjective submersion is open. Therefore, any potential submersion is potentially open. Consequently, any local potential submersion is locally potentially open. For maps with connected domain and fibers, locally potentially open is the same as open: L E M M A 3.8.4 Let X be a connected topological space, let Y be a set and let f : X —• Y have connected fibers. Assume that f is locally potentially open. Then f : X —¥Y is potentially open. Proof. Let V be an open subset of X. We wish to show that f~1{f(V)) _1 is open in X. Fix xo € / ( / 0 0 ) - We wish to show that there is an open neighborhood Xo of xo in X such that Xo C / - 1 ( / ( V ) ) . Let 2/0 := f(x0). Let F := / ^ ( l / o ) . Then 2/o = /(xo) € / ( / " W ) ) ) =

f(V),

so choose Do £ V such that yo = f(vo). Then vQ € f~1{yo) — F. Let U be an open cover of X such that, for every U 6 U, the mapping f\U : U -» y is potentially open. Let W := { F f~l (7117 G W}; then W is an open cover of the connected topological space F. So, by Lemma 2.3.3, p. 17, v and x are W-chain connected. We may therefore choose an integer n > 1 and a W-chain ( z 0 , . . . , z„) of length n from Do to x 0 - F ° r all integers k E [l,n], choose Uk £ U such that z/t_i,Zfc € FP\Uk. For all integers fc £ [0,n], let fk := f\Uk', then fk : 17^ -4 Y is potentially open. Let Vb := V and make a recursive definition: For integers k £ [l,n], let V/t := Vfc-i U [/fc_1(/fc(17fc n Vfc_i))]. By induction, for all integers k € [0,n], • Vjt is open in X; • f(Vk) = f(V); and • zk € Vfc. Let Xo := Vn. Then Xo is open in X. Moreover, zo = zn € V^ = Xo, so X 0 is an open neighborhood of x0 in X. Also, as /(Xo) = f(Vn) = f(V), wegetXoCrH/OO). • For maps with connected domain and fibers, locally potentially submersive is the same as submersive: L E M M A 3.8.5 Let X be a connected manifold, let Y be a set and let f : X —> Y have connected fibers. Assume that f is a local potential submersion. Then f : X —>Y is a potential submersion.

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Proof. Let U be an open cover of X by sets U such that f\U:U-*Y is a potential submersion. By the Implicit Function Theorem, for every x £ X, there is an open neighborhood V of x in X such that f\V : V -> Y is potentially open. Then, by Lemma 3.8.4, p. 59, / : X —• Y is potentially open. Let TQ denote the quotient topology on Y induced by / . Then / : X —> (Y,To) is an open map. Consequently, for every open subset W of X, the quotient topology on f(W) induced by f\W agrees with the relative topology on f(W) inherited from (Y,TQ). For all U £U, let Xu denote the manifold structure on f(U) such that f\U : U —>• (/(£/), Aj/) is a submersion; then, by the Implicit Function Theorem, f\U : U —> (f(U),Xu) is open. For all U £U, since the mapping f\U : U -» (f(U),Xu) is surjective, continuous and open, it follows that the topology of Xu is the quotient topology on f(U) induced by f\U, and so agrees with the relative topology on f(U) inherited from ToIt suffices to show that there is a manifold structure A on f(X) such that, for all U £ U, we have Xv = A|(/(17)). Fix U,U' £ U and fix y £ (/([/)) fl(/([/')). We wish to show that there is a T0-open neighborhood W of y in Y such that W C (/([/)) n (/(£/')) and XV\W = Xv, \W. By Lemma 2.3.3, p. 17 (with X replaced by f~l{y) and with U replaced by {U n {f~x{y)) | U € U}), choose an integer k > 1 and U0,...,Uk € U such that • Uo = U and Uk = U'; and • for all integers i £ [1, k], we have Ut-i r\UiC\ (/ - 1 (2/)) ¥" 0For each integer i £ [1,fc], let Vj := Ui-idUi and W^ := /(Vj). By openness of / : X —• (Y, ro), we see, for all integers i € [1,A;], that Wi is a ro-open neighborhood of y. Let W := Wi fl • • • D W*. We wish to show that Xu\W = \u'\W. For each integer i £ [l,fc], /|C/»_i : C/»_i -> (/(i7»_i), At/i_1) is a submersion, so the mapping /|Vf : Vf —> (Wi, At/^JWj) is a submersion. For each integer i £ [0,k], f\Ui : Ui —> (fiU^jX^) is a submersion, so f\Vi : Vi -> (Wf,At/JWi) is a submersion. For each integer i £ [l,k], as the maps f\Vt : VJ -)• (Wi.A^.JWi) and f\Vt : Vi -> (Wi,A^|Wi) are both surjective submersions, we see, from Corollary 3.8.3, p. 58, that A^_, \Wt = XVi \Wi-, then A ^ . , \W = XVi \W. Then XV\W = XUo\W = ••• = XUk\W = XV,\W. D Both Lemma 3.8.5, p. 59 and Lemma 3.8.4, p. 59 fail without the assumption that the fibers of / are connected: Let X := ( 0 , 3 ) C I and let

Potential submersions

61

Y := (0,2) C K. Define / : X -> Y by

{

x,

ifxG(0,2);

1, if i = 2; x-2, if a; € ( 2 , 3 ) . Then /|(0,2) : (0,2) -»• y and /|(1,3) : (1,3) -> Y are both bijective and therefore are potential submersions. Consequently, / : X -> Y is a local potential submersion. Consequently, / : X -¥ Y is potentially open. However, / - 1 ( / ( l / 2 , 3 / 2 ) ) = (1/2,3/2) U {2} U (5/2,3), so / : X -> Y is not potentially open. Then / : X —» Y is not a potential submersion. Finally, a potentially open map which is a local potential submersion need not be a potential submersion: Let T := R/Z and let 7r : R ->• T be the canonical homomorphism. Let I := [—1/2,1/2], let J := [—1/3,1/3] and let K := [—1/4,1/4]. Let a : I —> I be a homeomorphism such that, for all a; £ if, we have a(x) — x3 and such that, for all x £ I\J, we have a(x) = x. Let g : T -» T be the unique function satisfying g o IT = -K O a; then g : T —> T is a homeomorphism. Let ft : T -> T be the unique function satisfying: for all x € R, ft(7r(a;)) = 7r(2a;). Then ft : T -> T is 2-to-l. Let / := ft o g : T —> T; then / is 2-to-l. Then / is potentially open. For every x £ T, there is an open neighborhood U of x such that f\U : U -+ T is injective; then / is a local potential submersion. We leave it as an exercise to the reader to show that / is not a potential submersion. Let X be a manifold, let Y be a set and let / : X -> Y be a potential submersion. Then, by the Implicit Function Theorem, there is an integer p > 0 such that every fiber of / is a p-dimensional closed submanifold of X. We call p the fiber dimension of / . L E M M A 3.8.6 Let p > 0 be an integer. Let M be a manifold, let Y be a set and let <> / : M -+ Y be a function. Let C be the set of connected components of M. Assume, for every X € C, that <j>\X : X —> <j>(X) is a potential submersion with fiber dimension p. Assume, for all X,Y € C, that, ifX^Y, then (f{X)) n (f(Y)) = 0. Then : M -> Y is a potential submersion with fiber dimension p. Proof. Let X be the set of connected components of M. For all X £ X, let Ax be the manifold structure on <j>(X) such that <j>\X : X -+ ((X),Xx) is a submersion. Let d := dim(M) and let q := d — p. Then, for all X £ X, we have dim(: M -> (Y, A) is a submersion with fiber dimension p. •

62

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L E M M A 3.8.7 Let M be a manifold and let T be a proper foliation of M. Letp := dim(J'). Let cj): M —> M/J7 be the canonical map. Then is a potential submersion with fiber dimension p. Proof. Let Y := M/J7. Let X be a connected component of M. Let / := Y. By Lemma 3.8.6, p. 61, it suffices to show that / : X —> Y is a potential submersion with fiber dimension p. Every fiber of / is a leaf of !F, and is therefore connected. By definition of "proper foliation", <j>: M —• Y is a local potential submersion, so / : X —> Y is a local potential submersion. By Lemma 3.8.5, p. 59, / : X —> Y is a potential submersion. Every fiber of / : X —• Y is a leaf of F, and so is a p-dimensional closed submanifold of X. Consequently, f : X -> Y has fiber dimension p. •

3.9

Lorentz metrics — basic definitions

Let M be a manifold. A smoothly varying system of quadratic forms, one on each tangent space of M, is called a quadratic differential on M. Let d := dim(M), let G := GL d (R) and let FM denote the frame bundle of M. Identifying Rd with Rdxl, matrix multiplciation defines an action of G on E d , which then induces an action of G on QF(E d ). More precisely, a quadratic differential on M is a smooth section of the associated vector bundle FM xG QF(E d ) ->• M. Let M be a manifold. Let d := dim(M). Let Q be a quadratic differential on M. Then Q is said to be nondegenerate if it is nondegenerate at every point of M. In this situation, the signature of Q is constant on each connected component of M. If Q is nondegenerate and has constant signature, then we say that Q is a pseudoRiemannian metric on M. If Q is nondegenerate with signature (d, 0) at every point of M, then Q is said to be a Riemannian metric on M. If d > 2 and if Q is nondegenerate with signature (d — 1,1) at every point of M, then Q is said to be a Lorentz metric on M. A pseudoRiemannian manifold (resp. Riemannian manifold, resp. Lorentz manifold) is a manifold together with a pseudoRiemannian metric (resp. Riemannian metric, resp. Lorentz metric). A vector field X on a Lorentz manifold M is said to be everywhere lightlike (resp. everywhere spacelike, resp. everywhere timelike) if,

Lorentz metrics - basic

definitions

63

for all m G M, Xm is a lightlike (resp. spacelike, resp. timelike) vector in the Minkowski space TmM. For any pseudoRiemannian manifold M, Isom(M) denotes the group of isometries of M. We call Isom(M) the isometry group of M. Let V be a vector space and let d := dim(V). A map a : AdV —> [0, oo) is said to be a volume density element on V if there exists an isomorphism w : AdV ->• R such that, for all x G AdV, we have a(x) = |w(a;)|. Let M be a manifold and let d := dim(M). Let V := Ad(TM). A volume density on M is a smooth map a : V —• [0, oo), such that, for all m £ M, a\Vm is a, volume density element on TmM. Let V be a vector space, let d := dim(V), let F be a quadratic form on V, and let Q be a quadratic form on E d x l . An ordered Q-basis of (V, F) is an ordered basis B of V such that Q o B = F . Let (V, Q) be a Minkowski vector space and define d := dim(V). Let £ := ei A • • • A e^ € Ad(IR V, we let /» : A d (E d ) ->• Ad(V) be the induced isomorphism. For any ordered Qd-basis B of (V,Q), let £g := (V"B)*(0- Because every element of O(Qd) has determinant either + 1 or — 1, it follows that there is a unique volume density element v on V such that, for every ordered Q^-basis B of V, we have V(£B) = 1Let M be a Lorentz manifold. By the remarks of the preceding paragraph, for each m G M, we have a natural volume density element on TmM. These elements vary smoothly from tangent space to tangent space, so they give rise to a well-defined volume density, called the Lorentz volume density of M. This volume density induces a cr-finite measure fi on M; if /x(M) is finite, then we say that M has finite Lorentz volume. Let (M, g) and (N, h) be pseudoRiemannian manifolds. Let / : M -> N be smooth. We say that / is isometric if, for all v G TM, we have h((df)(v)) = g(v).

Chapter 4

Basic Lie Theoretic Results

4.1

Some Lie theoretic definitions and notation

Let k be either E o r C . A Lie group over k is a A;-manifold with a group structure, such that multiplication and inversion are smooth maps over k. A Lie algebra over k is a A;-vector space Q, together with an antisymmetric bilinear map (X, Y) H->- [X, Y] : g x g —> g satisfying the Jacobi identity. (See [Va74], §2.2, pp. 46-47.) According to the remarks following Lemma 3.1.1, p. 43, if M is a manifold, then the collection of vector fields on M, under the operation of [ •, • ], forms an infinite-dimensional Lie algebra over K. In this book, except where otherwise stated, Lie groups and Lie algebras will be assumed to be finite-dimensional. Moreover, they will be assumed to be over E, unless otherwise specified. An action of a Lie group G on a manifold M will be said to be s m o o t h if the action map (g, m) H4- gm : G x M —> M is smooth. If 0 is a complex Lie algebra, then 0R denotes the underlying real Lie algebra of g, forgetting the complex structure. Similarly, if G is a complex Lie group, then G R denotes the underlying real Lie group of G. For any Lie algebra g, define ad : g —> gi(g) by (a.dX)Y = [X, Y]. When the Lie algebra g is unclear, we use ad 0 instead of ad. For any Lie algebra g, ker(ad 0 ) = 3(0). Therefore, if 3(0) = {0}, then ad : g -> g((0) is injective. Let 0 be a Lie algebra. For all s C 0, we define c0(s)

:=

{ X € 0 | ( a d X ) s = {O}},

n„(«)

:=

{X<E0|(adX)sCs}.

For all X 6 0, we define cg(X)

:= cg({X}) 65

and ng(X)

:= n B (EX). If

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Results

a

> & Q 9, then we say that a normalizes b if, for all X 6 a, for all F e b , we have [X, F] € b. If o, b C g, then we say that a and b centralize each other if, for all X 6 a, for all F € b, we have [X, Y] = 0. Alternatively, in this situation, we will say that o centralizes b or that b centralizes o. For any Lie algebra g, for any h,6 C g, by [h,fi], we denote the Lie subalgebra of g generated by {[X, Y] | X e h, Y 6 6}. Let g be a Lie algebra. The derived series of g is the sequence J>o, 5i, • • • of ideals of g recursively defined by £>o := 9 and fy+i := [fy, 5,]. We say that g is solvable if there is some i such that fy = {0}. A connected Lie group is solvable if its Lie algebra is solvable. By Corollary 4.6.4, p. 91 below, a connected Lie group is solvable iff it is solvable as an abstract group. The descending central series of g is the sequence Co, Ci,... of ideals of g recursively defined by c0 := g and Cj+i := [g, C;]. For any integer n > 1, we say that g is n-step nilpotent if c n _i ^ {0} and cn = {0}. We say that g is nilpotent if there is some integer n > 1 such that c„ = {0}. A connected Lie group is nilpotent if its Lie algebra is nilpotent. By Corollary 4.6.3, p. 91 below, a connected Lie group is nilpotent iff it is nilpotent as an abstract group. A Lie algebra g is Abelian if [g, g] = {0}. A Lie algebra is simple if it is nonAbelian and its only ideals are {0} and g. A Lie algebra is semisimple if it is nonzero and admits no nonzero Abelian ideal. A Lie algebra is semisimple iff it admits no nonzero solvable ideal. If V is a vector space and if J : V —> V is a complex structure, then

gl(V,J)

~

si(v,j) :=

{X egl{V)\XJ=

JX},

{Xegi(v,J)\x,JXesi(v)}.

For any vector space V, let GL(F) denote the Lie group of invertible automorphisms of V, let gi(V) denote the Lie algebra of endomorphisms of V and let SL(V) := {g € GL(V) | det(s) = 1}. Then the Lie algebra of SL(V) is given by sl(V) = {X € gl(V) \ tr(X) = 0}. For any complex vector space V, let GL(V) denote the complex Lie group of invertible complex automorphisms of V, let gl(V) denote the complex Lie algebra of complex endomorphisms of V and let SL(V) := {g £ GL(V) | det(g) = 1}. Then al(V) := {X <Efll(V)| tr(X) = 0}.

Some Lie theoretic definitions

and notation

67

For all k € {ffi, C} for all integers n > 1, we define GLn(fc) := {g e knXn \ det(g) ? 0} SLn(fc) := {g £ knxn | d e % ) = 1}. Let 0 be a Lie algebra. A real g-module is a vector space V, together with a representation p : g ->• fl[(V). For all X € g, for all v € V, we typically denote (p(X))(i>) by Xv. A complex g-module is a complex vector space V, together with a homomorphism 0 -> (£)I(V))R of real Lie algebras. As before, we typically denote (p(X))(v) by Xv. Let 0 be a complex Lie algebra. A 0-module is a complex vector space V, together with a homomorphism g -> g((V) of complex Lie algebras. Again, (p(X))v is denoted Xv. Let G be a Lie group. A real G-module is a vector space V together with a Lie group homomorphism p : G —)• GL(V). For all g £ G, for all v € V, we typically denote (p(g))v by gv. A complex G-module is a complex vector space V together with a homomorphism p-.G-t (GL(V))R of real Lie groups. Again, we denote (p(g))v by gv. Let G be a complex Lie group. A G-module is a complex vector space V together with a homomorphism p : G —» GL(V) of complex Lie groups. Again, we denote (p(g))v by gv. Let G be a Lie group. For any vector space V, for any representation p : G -> GL(V), the derivative dp : 0 -» 0t(V) is a representation of 0. For any complex vector space V, for any representation p : G —• ( G L ( V ) ) R , the derivative dp : 0 —> ({JI(V))R is a representation of g. Thus any real (resp. complex) G-module gives rise to a real (resp. complex) 0-module. Similarly, for any complex Lie group G, a G-module gives rise to a 0-module. Let V be a real vector space. For any Lie subgroup G of GL(V), we consider V to be a real G-module via the inclusion G -> GL(V). For any Lie subalgebra 0 of gl(V), we consider V to be a real 0-module via the inclusion representation 0 —• Ql(V). Let V be a complex vector space. For any Lie subgroup G of ( G L ( V ) ) R , we consider V to be a complex G-module via the inclusion representation G -> ( G L ( V ) ) R . For any Lie subalgebra 0 of (BI(V))R, we consider V to be a complex g-module via the inclusion representation g -t (gl(V))R. For any complex Lie subgroup of GL(F), we consider V to be a G-module via the inclusion representation G -¥ GL(V). For any complex Lie subalgebra 0 of gl(V), we consider V to be a g-module via the inclusion g -¥ gl(y).

68

Basic Lie Theoretic

Results

Let A be either a Lie group or a Lie algebra; A may be real or complex. Let V be an A-module; if A is real, then V may be real or complex. A vector subspace W of V is A-invariant if AW C W. An A-submodule of V is an ^4-invariant vector subspace of V. Let A be either a Lie group or a Lie algebra; A may be real or complex. Let X and Y be A-modules; if A is real, then X and Y may either be both real or both complex. A linear transformation T : X -> Y is A-equivariant if, for all x £ X, for all a £ A, we have T(ax) = a(T(x)). Let A be either a Lie group or a Lie algebra; A may be real or complex. Let V be an ^-module; if A is real, then V may be real or complex. Let W be a vector subspace of V. Let I: W -> W be the identity map. We say that W is A-irreducible if both W is A-invariant and the only ^-invariant vector subspaces of W are {0} and W. We say that W has only scalar ^-intertwining if, both W is A-invariant and, for any vl-equivariant map T : W —> W, we have T 6 MI. Otherwise, we say that W has nonscalar A-intertwining. Note that, if W has only scalar A-intertwining, then W is g-irreducible. Let G be a Lie group; G may be real or complex. Let V be a G-module; if G is real, then V may be real or complex. Let W be a G-submodule of V. We say that W is G-trivial if, for all g € G, for all w e W, we have gu; = w. Otherwise, we say that W is G-nontrivial. Let g be a Lie algebra; g may be real or complex. Let V be a 0-module; if fl is real, then V may be real or complex. Let W b e a 0-submodule of V. We say that W is fl-trivial if, for all gW = {0}. Otherwise, we say that W is fl-nontrivial. Let g be a real Lie algebra. For any real g-module V, Vc is a complex g-module, in which the representation is denned by X(z v) = z ® Xv. For any complex g-module V, VR is a real g-module; recalling that, as sets, VR = V, the action of g on VR is the same as the action of g on V. For any complex g-module V, V is a real g-module; recalling that, as sets, V = V, the action of g on V is the same as the action of g on V. For any a real g-module V, following §2.4, p. 18, V © V is naturally isomorphic to (V" C )R, as real g-modules. Also following §2.4, p. 18, for any complex g-module V, VR is naturally isomorphic to VR, as real g-modules. Also, following §2.4, p. 18, for any complex g-module V, V © V is naturally isomorphic to (VR) C as complex g-modules. Also, following §2.4, p. 18, for any real g-modules V, Vc is naturally isomorphic to Vc. In the category of real and complex G-modules, we may also define Vc,

Some Lie theoretic definitions and notation

69

VR and V. Again, V © V £ {VC)R, VR = VR and V © V £ (V R ) C ._ Let g be a Lie algebra. Then, for any The complex g-module V is called the conjugate of V. As with modules over Lie algebras, if V is a real Gmodule, then ( V C ) R is naturally isomorphic to V © V and if V is a complex G-module, then (VR) C is naturally isomorphic to V © V. As with modules over Lie algebras, if V is a real G-module, then Vc is naturally isomorphic toyc. Let G be a group and let V be a real G-module. Let S :— MI. Let T denote the set of G-equivariant linear transformations V —> V. We say that V h a s only scalar intertwining if T = S. Otherwise, we say that V has nonscalar intertwining. Let G be a Lie group and let V be a real G-module. A bilinear form B : V x V ->• R is said to be G-invariant if, for all v, w € V, for all g 6 G, we have B(gv,gw) = B(v,w). We say that Q € QF(V) is G-invariant if, for all v &V, for all g € G, we have Q(gv) = Q(v). Let g be a Lie algebra and let V be a real fj-module. We say that B e SBF(V) is g-invariant if, for all v,w € V, for all X € fl, we have B{Xv,w) + B{v,Xw) = 0. We say that Q E QF(V) is g-invariant the polarization of Q is g-invariant. Let g be a Lie algebra and let B € SBF(fl). For X G 0, we say that B is (ad X)-invariant provided, for all Y,ZeQ, we have B([X,Y],Z)

+ B(Y,[X,Z})

= 0.

We say that B is (ad 0)-invariant or ad-invariant provided, for all X € 0, B is (ad X)-invariant. Let 0 be a Lie algebra. We say that Q € QF(g) is ad-invariant or (ad 0)-invariant if its polarization is. A vector subspace a of a Lie algebra 0 is said to be a r e a l split t o r u s of 0 if adg(o) := {adX : 0 -> g \ X € a} is simultaneously real diagonalizable, i.e., if there is a basis B of 0 such that, for all X £ a, for all B € B, we have (a,dX)B e MB. Equivalently, a is Abelian and, for all X € a, the linear transformation a d X : 0 —> 0 is real diagonalizable. Let g be a semisimple Lie algebra. We say that a Lie subalgebra of 0 is a m a x i m a l r e a l split t o r u s of 0 if it is maximal (under inclusion) among the collection of real split tori in 0. By Lemma 4.10.13, p. 117 below, for any two maximal real split tori a and b of g, there is a Lie algebra automorphism / : 0 —> 0 such that /(a) = b. In particular, the maximal real split tori all

70

Basic Lie Theoretic

Results

have the same dimension, and this dimension is called the real rank of g. Assume that a is a real split torus of a semisimple Lie algebra g. For all cf> G o*, define 0 0 := {X G g \ MA G o, {&&A)X = (<j){A))X}. For all G a*, we say that is a root of a on g if both (f> 7^ 0 and 0^ ^ {0}. For any root cj) of o on g, we call 0^, the -rootspace of o on 0. Let g be a Lie algebra, let V be a real g-module and let J : V —• V be a complex structure on V. Then we say that J is g-invariant if, for all X € g, for all v € V, we have X Ju — J X u . Let G be a Lie group, let V be a real G-module and let J : V —> V be a complex structure on V. Then we say that J is G-invariant if, for all g £ G, for all w G V, we have gJv = Jgu. If g and \) are Lie algebras then we say that 0 is a direct summand of f) and we write g\\) if there exists a Lie algebra 0' such that f) is Lie algebra isomorphic to 0 © 0'. If p : G —> H is a homomorphism of Lie groups that is also a covering map of manifolds then we say that p is a covering homomorphism. If G and H are Lie groups, then we say that G is a covering group of H if there exists a covering homomorphism G -> H. By covering space theory, for any connected Lie group G, one may lift multiplication G x G -* G and inversion G —> G to the universal covering space G of G. With these maps G x G —t G and G —> G, the manifold G becomes a Lie group called the universal covering group of G. The universal covering map G —> G is then a covering homomorphism. For any Lie group G, a G-manifold is a manifold M together with a smooth action of G on M. A G-manifold is faithful (resp. free, resp. proper) if the G-action on M is faithful (resp. free, resp. proper). Let a group G act on a pseudoRiemannian manifold M. We say that the action is by isometries if, for all g G G, the map m H-* gm : M -> M is an isometry of M. Now assume that G is a Lie group. Then we say that the action of G on M manifold is isometric if it is smooth and by isometries.

4.2

Dynamical consequences of the Frobenius Theorem

Let M be a manifold and let G be a Lie group acting smoothly on M . We define T := {(m,m') G M\Gm = Gm'} C M x M so that T is the

Dynamical consequences of the Probenius Theorem

71

equivalence relation on M whose equivalence classes are the G-orbits in M. That is, for all m £ M, we have T(m) = Gm. L E M M A 4.2.1 Assume that the G-action on M is locally free. Then T is a prefoliation. Moreover, if G is connected, then J- is a foliation. Proof. Let p := dim(G). Let V be the infinite-dimensional Lie algebra of vector fields on M. Let X\,...,XP be a basis of g. Define a vector subbundle A of TM by A m = gm. Then (XI)M,•••, (XP)M is a framing of A. Since {XM \ X £ g} is a Lie subalgebra of V, it follows, for all integers i,j £ [l,p], that [{Xi)M,{Yi)M] is in A. Then, by Lemma 3.4.3, p. 51, A is involutive. So, by Theorem 3.7.5, p. 57, let To be a foliation on M such that TTo = A. By Lemma 3.5.3, p. 53, for all m £ M, G°m is an open subset of To(m). Then, for all leaves L of To, {G°1}I^L is a partition of L into connected open sets; so, since L is connected, we conclude, for all I £ L, that G°l = L. Then, for all m £ M, we have G°m = T0(m). Let S be a countable subset of G such that G°S = G. For all m £ M, T(m) = Gm = G°Sm = T0(Sm). For all m £ M, Sm is countable. Then TQ has countable index in T. Then, by Lemma 3.4.2, p. 51, we see that T is a prefoliation. If G is connected, then, for all m £ M, T(m) = Gm = G°m = To(m); in this case, T = To, so T is a foliation. • L E M M A 4.2.2 Assume that the G-action on M is free and proper. Then T is a proper prefoliation. Moreover, if G is connected, then T is a proper foliation. Proof. Let d := dim(M), let p := dim(G) and let q := d - p. By Lemma 4.2.1, p. 71, we see that T is a prefoliation. Also, by Lemma 4.2.1, p. 71, if G is connected, then T is a foliation. It therefore suffices to show that T is proper. Fix m £ M. We wish to show that there are an open neighborhood U of m in M and a diffeomorphism : U —> Md such that 4>*(T\U)=TPxT0J. Since T is a prefoliation, choose an open neighborhood V of m in M, a countable equivalence relation R on W and a diffeomorphism ip : V —> Rd such that tp*(T\V) = TP xR. We may assume that ip(m) - 0 £ Rd. Claim 1: There is an open neighborhood W of 0 in W such that R\W = T Q | W . Proof of Claim 1: We wish to show that there is an open neighborhood W of 0 in W such that, for all w £ W, we have

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(R(w)) n W = {w}. Assume, for a contradiction that this fails, i.e., that there are two sequences Wi and w[ in W such that (1) Wi -> 0 in W and w\ -> 0 in W; (2) for all i, (wi,w'i) € i?; and (3) for all i, we have Wi ^ w;J. For all i, let m* := ^>_1(0,«;;) and let m^ := V' _1 (0,w^). Then we have mi —> m and mj —> m. For all i, since (u>j,u4) € i?, it follows that (rriijm'j) € T\ then Gm; = T{rrii) = J^(mJ) = GmJ. For all i, choose gi € G such that pimj = mj. The G-action on M is proper, so, after passing to a subsequence, we may assume that gi is convergent in G. Let g^ := lim gi. Since, for all i, we have gitrii = m'i, it follows that 1GChoose a connected open neighborhood Go of 1 Q in G and an open neighborhood M 0 of m in M such that GoM 0 C V. Passing to a tail, we may assume, for all i, that gt € Go- Passing to a further tail, we may assume, for all i, that rrii e Mo. Then, for all i, we have mj = gimi € GoMo C V. Let 7r : Rd —> R9 be projection onto the last g coordinates, defined by Tr(ti,... ,td) = {tp+i,... ,td). Let p := n o ip •. V -^ W. Then, for all i, p{rrii) = Wi and /9(mJ) = w^. For all v € V, we have p((Gv) n V) = R(p(v)); since R(p(v)) is countable, it follows that U(p(i;)) is totally disconnected in W. So, since Go is connected, for all m 6 Mo, the map g H-> p(gm) : Go —> R(p(v)) is constant. It follows, for all m e Mo, for all g E Go, that p(gm) = p(m). Then, for all i, we have u;J = /9(m'f) = p(girrii) = p(m$) = Wi, contradicting (3) above. End of proof of Claim 1. Choose W as in Claim 1. If necessary, replacing W by a smaller open neighborhood of 0 in W, we may assume that W is diffeomorphic to Rq. Let /? : W -> W be a diffeomorphism. By Claim 1, we have P.(R\W) = P*(T^\W). Then P,(R\W) = T«. Define a : W x W -> Rd by a(a;,u/) = (a;,£(«;)). Let £/ := ^(W x W). Let 0 := a o V> : *7 -> Kd. Then 0»(^|C7) = a.(T? x (R\W)) = T* x so <j>*{F\U) =

TPxT«.

{fi.{R\W)\ D

Dynamical consequences of the Frobenius

Theorem

73

For any Lie algebra g, for any manifold M, a g-action on M is a Lie algebra homomorphism from g to the infinite-dimensional Lie algebra of vector fields on M. For any Lie algebra g, a g-manifold is a manifold together with a g-action on M. If G is a Lie group and M is a G-manifold, then differentiating the action turns M into a g-manifold. Let Q be a Lie algebra and let M be a manifold. Let V denote the infinite-dimensional Lie algebra of vector fields on M and let / : g —> V be a g-action on M. Then we say that / is locally faithful if / is injective. We say that / is locally free if, for every X e fl\{0}, for every m e M, f(X) does not vanish at m. Let a Lie group G act smoothly on a manifold M. Let / be the g-action on M obtained by differentiating the G-action on M. Then the G-action on M is locally faithful iff / is locally faithful. Moreover, the G-action on M is locally free iff / is locally free. Let g be a Lie algebra and let M be a g-manifold. A g-manifold M is locally faithful (resp. locally free) if the g-action on M is locally faithful (resp. locally free). Let M be a manifold and let V denote the infinite-dimensional Lie algebra of vector fields on M. Let g be a Lie algebra and let / : g -» V be a g-action on M. Let U be an open subset of M. Let W denote the infinite-dimensional Lie algebra of vector fields on U. Then the g-action on U inherited from / is X i-> (f(X))\U : g -> W. For any Lie group G, for any manifold T, let aa(T) be the G-manifold G xT, with G-action given by g(g',t) — (gg',t). Differentiating the action turns aa(T) into a g-manifold. For any Lie group G, for any nonempty open subset U of G, for any manifold T, let /3G{U,T) denote the g-manifold U xT, with g-action inherited from the g-action on ao{T) = G x T. The next two results give local descriptions of locally free g-manifolds and of free, proper G-manifolds. They follow from Lemma 4.2.1, p. 71 and Lemma 4.2.2, p. 71. Their proofs are left as exercises. L E M M A 4.2.3 Let g be a Lie algebra. Let M be a locally free g-manifold. Let mo £ M. Then there exists an open neighborhood Mo of mo in M, there exists an open neighborhood U of I Q in G and there exists a manifold T such that Mo is g-manifold isomorphic to PG{U,T). L E M M A 4.2.4 Let G be a Lie group. Let M be a free, proper G-manifold. Let mo € M. Then there exists a G-invariant open neighborhood Mo of mo

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in M and there exists a manifold T such that MQ is G-manifold isomorphic to aG(T). L E M M A 4.2.5 Let M be a manifold and let g be a finite-dimensional Lie algebra of vector fields on M. For all m G M, let Am :— {Xm | X G g}. Let k := dim(g) and assume that dim(A m o ) = k. Let E be a vector subbundle ofTM. Assume that E is 0-invariant, i.e., assume for any X G g, for any vector field Y in E, that [X, Y] is in E. Let mo € M. Let w G Emo. Then there is an open neighborhood Mo of mo in M and there is a vector field W on Mo such that Wmo = w, such that W is in E\Mo and such that, for all X E g, we have [X, W] = 0 on M0. Proof. Since dim(A m o ) = k, it follows that there is an open neighborhood M' of mo in M such that, for all m G M', we have dim(A m ) > k. On the other hand, for all m G M, we have dim(A m ) < dim(g) = k. So, for all m G M', we have dim(A m ) = k. Replacing M by M', we may assume, for all m G M, that dim(A m ) = k. By Ado's Theorem (Theorem 3.17.8, p. 237 of [Va74]), let G0 be a Lie group such that g0 is Lie algebra isomorphic to g. Let / : g0 -> g be a Lie algebra isomorphism. Then / is a go-action on M. For all m G M, X H-» (f(X))m : go -> A m is an isomorphism, so the 0O-action on M is locally free. So, by Lemma 4.2.3, p. 73, choose an open neighborhood U of 1G 0 in Go and choose a manifold T such that M is g-manifold isomorphic to 0GO(U,T). We may then assume that M = /3a0(U,T). We have m0eM

= /3Go(U,T) =

UxT,

so choose no G U and to € T such that m0 = (uo,to). Let M := ao0(T), so M = Go x T is a Go-manifold containing M. Let W be any vector field on M such that W^0 = w and such that W is in E. Differentiating the Go-action on M, we obtain a Go-action on TM. Let N be a convex open neighborhood of 0 in g0 such that V := exp(iV) is open in Go and such that VUQ C U. Then, by g0-invariance of E, for all m G {uo} x T, for all e G Em, for all v G V, we have ve G E. Define a vector field W on M by W(9it) = 9uo1^rL0,t)Then W is G 0 -invariant, so, for all X G go, we have [Xg, W] - 0. For all m G {u0} x T, we have Wm = W'm. Then Wmo = W^o = w. Let M 0 := (VUQ) x T and let W := W\M0. D

exp, Ad and ad

75

L E M M A 4.2.6 Let G be a Lie group acting on a manifold M. Assume that the G-action on M is smooth, free and proper. Then the canonical map M -> G\M is a potential submersion. Proof. Let G0 := G/G°. Then G0 is discrete. Let a : G -¥• G0 be the canonical Lie group homomorphism. Let M0 := G°\M and let ir: M ->• M 0 be the canonical map. Let Go act on MQ by the rule {a{g)){-K{m)) = Tr(gm). Let p : Mo -»• Go\Mo be the canonical map. Let r : M ->• G\M be the canonical map. We wish to show that r is a potential submersion. Let T := {(m,m')\G°m = G°m'}. Then MjT = M 0 . Since G acts freely and properly on M, it follows that the G° -action on M is free and proper, as well. Then J7 is a, proper foliation, by Lemma 4.2.2, p. 71. Then, by Lemma 3.8.7, p. 62 we see that ir : M -¥ Mo is a potential submersion. Give Mo the unique manifold structure with respect to which •K : M —> M0 is a submersion. The G-action on M is smooth, free and proper, so the Go-action on Mo is smooth, free and proper. Since the Go-action on MQ is proper (i.e., properly discontinuous) and free, there is a manifold structure on Go\Mo such that the canonical map p : Mo -¥ GO\MQ is a smooth covering map. Any smooth covering map is a submersion, so p : Mo —>• Go\Mo is a potential submersion. Give Go\M 0 the unique manifold structure such that p is a submersion. Define a map /3 : G0\M0 -+ G\M by /?(p(7r(m))) = T(m). Then 0 is a bijection and is therefore a potential sumbersion. Give G\M the unique manifold structure such that /? is a submersion. We wish to show that r : M -¥ G\M is a submersion. Since /?, p and n are all submersions and since r = (3opoir : M —> G\M, we are done. •

4.3

e x p , A d and ad

Let G be a Lie group. Let G act on G from the left and right by left and right translation. Differentiating these two actions, we arrive at left and right actions of G on TG. Let e := 1Q and let g := TeG. Define Ad : G -+ GL(g) by (Adg)X = gXg~l. When the Lie group G is unclear, we use AdG or Ad 0 instead of Ad. Note, for all g £ G, that Ad g : g —¥ g is the differential at e of the map Int G defined by

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(Intg)x = gxg 1. The kernel of Ad : G -> GL(g) is CG(G°). So, if G connected, then the kernel of Ad : G ->• GL(g) is Z(G). So, if G is connected and centerfree, then Ad : G —> GL(g) is injective. Let S C G. For any V C g, we say that S normalizes V if, for all s G 5, we have (Ad s)V C V. For any connected Lie subgroup if of G, 5 normalizes H iff 5 normalizes h. Let H be a Lie subgroup of G. Let it : g - • g/f) be the canonical map. For all h € if, we have (Ad /i)f) C h, so there is a unique map A : g/f} ->• g/f) such that, for all X G g, 7r((Ad/i)X) = A(7r(X)). This map A is denoted AdB/(,(/i). When it will not cause confusion, A will be denoted Adh. Let T denote the set of vector fields on G. Let G act on T from the left and right by the rules: (gV)x = g(Vg-ix) and (Vg)x = (Vxg-i)g. For any V G T, we say that V is left-invariant if, for all g G G, we have f : R -»• G be the integral curve of XL satisfying (f>^ (0) = e. Similarly, let <j>^ : E —> G be the integral curve of XR satisfying (j)^(Q) = e; In Lemma 4.3.2, p. 77 below, we show, for all X G g, that &L = 'R- F o r a n ^ £ 8. w e define exp(Z) := *(1) = ^ ( 1 ) . In cases where exp is unclear, the map exp : g -> G is denoted by exp G Care must be taken not to confuse the Lie-theoretic exponential map just defined with the exponential map resulting from a connection on a manifold. By construction, (c?(exp))e : g -> g is the identity map, so, by the Inverse Function Theorem, we see that exp(g) contains an open neighborhood of 1G in G, and it follows that exp(g) generates G, i.e., that no proper subgroup of G contains exp(g). However, the exponential map exp : g ^ G may not be surjective. For X £ g, the one-parameter subgroup of X in G is {exp(tX)}teuLet G* := GL(g). Then, for all X G g, we have exp G (X) G G, we have adg(X) G g* and we have exp G ,(ad B (X)) = Ad 0 (exp G (X)) G G*. Let B G SBF(g). For g G G, we say that B is (Ad (^-invariant pro-

77

exp, Ad and ad

vided, for all Y,Z £ g, we have

B((Adg)Y,(Adg)Z)

= B(Y,Z).

We say B is (Ad G)-invariant or Ad-invariant provided, for all g G G, B is (Ad<7)-invariant. We say Q G QF(fl) is Ad-invariant or (Ad G)-invariant if its polarization is. L E M M A 4.3.1 Let X,Y G g. Then [X,Y] =

-[XR,YR]e.

Proof. Define f :G-+G by f(g) = g~1. For all P e g , because / * ( P L ) is right-invariant and because we have (f*(PL))e = (df)(P) = -P, we see that f*(PL) = -PR. So U{XL) = -XR and U{YL) = -YR and f.{[X,Y)L) = -[X,Y]R. We have [X,Y]L = [X L ,y L ], so /*([X,y] L ) = [ / . ( X L ) , / . ( y L ) ] . It follows that - [ X , y ] f l = [-X f l , - Y R ] , so [ X , y ] f l = -[XR,YR]. Evaluating at e gives the result. • L E M M A 4.3.2 Let X G g. Then £ = <£*. Proo/. For all t G M, let Vt : G -> G denote the time t flow of X f l . Then, for all a; G G, for all i,u G M, we have ipt(ipu(x)) = i/'t+u^)- For all t G K, let p t := Vt(e)- Then g0 = e and (d/dt)t=0gt = X. For this paragraph, fix i, u G I t We have <7t<7„ = (ipt(e))9u- Since X R is right-invariant, it follows, for all g,x G G, that (tpt(x))g = ipt(xg). In particular, we have (ipt(e))9u = ipt(9u)- Then Ptffu = (il>t(e))gu = ipt(9u) = tpt(.ipu(e)) = ipt+u(e) = gt+uFor this paragraph, fix t G E. For all s £ 1 , we have 9t9s = 9t+s = 9s+t = 9s9t, so gt9s9Tl = 9s- Then 9tXgrl

=

(9t9s9t s=0

) =

d_ ds

gs = x. 8=0

By right-invariance of XR, for all g,x € G, we get (XR)xg = [(XR)x]g. In particular, (X f l ) S t = [(Xij)e]fft, so (XR)gt = Xgt. Similarly, by leftinvariance of XL, we get (XL)gt = gt[{Xi)e], so {XL)gt = 9tX. Since X = gtXgt1, we get X fft = {gtXg^)gt = gtX. Then (XR)gt = (XL)gt.

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As 11-> gt is an integral curve of XR, we get (d/dt)gt = (Xn)gt. Then (d/dt)gt — {Xi,)gt. Then t >->• gt is an integral curve of XL, as well. Then, for all t £ R, we have (/>%(t) = gt = X(gt).

D

L E M M A 4.3.3 Let G be a Lie group. Then the differential at e := 1Q of Ad : G -» GL(g) is ad : g -* fll(fl). Proof. Let a : fl ->• gl(g) be the differential at e of Ad : G -> GL(g). Fix X,Y eg. We wish to show that (a(X))Y = (adX)Y. For all t 6 R, let Vt : G -> G denote the time i flow of Xx,. For all t £ 1, let gt := V'tCe)- Then gQ = e and (d/dt)t=ogt = -X\ Then (d/dt)t=o[(Ad5t)y] = (a(X))Y. Moreover, by definition, we have (adX)Y = [X,Y] = [X L ,Y L ] e . We therefore wish to prove that (d/dt) t=o[( Ad gt)Y] = [.XL, Yx,]e. By definition of [ •, • ] and Lie derivative, we have [XL,YL]

= CXL(YL)

=

(d/dt)t=0[rt(YL)].

We therefore wish to show that (d/dt)t=0[(Ad gt)Y] = (d/dt)t=0[(ipt(YL))e]. Fix t e l . We will show that (Ad gt)Y = (tpl(YL))e. For all g € G, left-invariance gives (Y^g = g(Yi)e — gY, so {YLg-%

= {YL)gg-1

= gYg'1 = (Ad g)Y.

In particular, we have (Ad<7t)Y = (YiJg^1)e. By left-invariance of XL, for all x,y 6 G, we have ipt(xy) — x(ipt(y))In particular, for all x G G, we have ipt(xe) = x(ipt(e)), so ipt(x) = xgt. Then, for all y e G, we have tptiygi1) = yfl^ft = 2/, so, applying Vt_1 to this equation, we get j/p ( _1 = ij)^l(y). Differentiating this with respect to j/ywe see, for all t; € TG, that t/'t (w) = vg^1. Consequently, for any vector field V on G, we have rpt(V) = Vg^1. In particular, ipl{YL) = YLg^1. Then ( A d 5 t ) y = ( F L S * - 1 ^ = ffl(YL))e. • L E M M A 4.3.4 Lei 7 6e o left-invariant quadratic differential on G and assume that 7 e € SBF(g) is Ad-mi/anan£. JTien 7 is right-invariant. Proof. Let x,g £ G and fix vectors i>,u; € TXG. We wish to prove that l(vg,wg) =i(v,w). Let t/ := vx_1, let u/ := wx-1 and let = v'x and iu = w'x. We wish to

exp, Ad and ad

79

show that ^(v'g'jtv'g1) = ^(v'x,w'x). It therefore suffices to show, for all h£G, that j{v'h,w'h) = j{v',w'). Fix h e G . Let v" := h~lv'h and u>" := h^w'h. Then u" = (Adft- 1 )*;' and to" = (Ad/i - 1 )™'. Moreover, v",w" € T e G. Then, by Ad-invariance of 7 e , it follows that 7(1/', to") = j(v',w'). Since 7 is left-invariant, we get j(v'h,w'h) —j(h~1v'h,h~1w'h). Then 7 (t//i, w'h) = 7(1;", to") = 7(1;', to'). D Let G be a Lie group and let if be a closed subgroup of G. Let H act on G by h.g = gh~x. Then, by Lemma 4.2.6, p. 74, the quotient map 7r : G -> G/J? is a potential submersion. We will endow G/if with the unique manifold structure making n a submersion. Then, by Lemma 3.8.1, p. 58, the standard action by translation of G on G/H is smooth. Let x := 7T(1G). Then H fixes x and so, by differentiation, acts on TX(G/H). The Adjoint representation of H on g preserves h, giving rise to the "Adjoint representation" of H on g/h. There is a natural .ff-equivariant vector space isomorphism identifying TX(G/H) with g/h. If the closed subgroup H is normal in G, then G/H is a Lie group. We leave the following as an exercise to the reader. L E M M A 4.3.5 Let H be a closed subgroup of G and let Q € QF(g/h) be (Ad H) -invariant. Then there is a unique G-invariant quadratic differential 7 on G/H such that 7 e = Q. If Q is positive definite, then 7 is a Riemannian metric. If Q is Minkowski, then 7 is a Lorentz metric. In this book, the Lie algebra of a Lie group is, by definition, its tangent space at the identity, with Lie algbra structure defined, as in §4.3, p. 75, via extension to /e/t-invariant vector fields (i.e., by [X, Y] := [XL,YL]S, as above). By Lemma 4.3.1, p. 77, it would be slightly different to change convention and use right-invariant vector fields. Throughout this book, if a Lie group is denoted by a capital Latin letter (perhaps with some adornments like asterisk and/or subscript), then the Lie algebra of that Lie group will always be denoted by the corresponding small fraktur character (with the same adornments). So, for example, if H£ is some Lie group, then its Lie algebra is denoted h j . For any Lie group G, the Lie algebras of G and of G° are the same. If AT is a normal closed subgroup of G, then n is an ideal of g and the Lie algebra of G/N is naturally isomorphic to the quotient Lie algebra g/n.

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Basic Lie Theoretic

Results

The Lie group Lie algebra correspondence

A subgroup of a Lie group G is a Lie subgroup if it is a leaflike submanifold of G. The leaflike manifold structure on the subgroup is referred to here as its Lie manifold structure. The resulting leaflike topology on the subgroup is called the Lie topology; it may be finer than the relative topology. If G is a Lie group and H is a Lie subgroup of G, then h is a Lie subalgebra of g. The collection of Lie subgroups of a Lie group is closed under a variety of basic set-theoretic and group-theoretic operations: Let G be a Lie group. In Corollary 4.5.3, p. 85 below, we see that the subgroup of G generated by any collection of connected Lie subgroups of G is again a connected Lie subgroup; its Lie algebra is generated by the Lie algebras of the groups in the collection. In Corollary 4.5.9, p. 87 below, we see that the subgroup of G generated by a countable collection of Lie subgroups of G is again a Lie subgroup; its Lie algebra is not necessarily generated by the Lie algebras of the groups in the collection. According to Corollary 4.5.6, p. 86 below, if A and B are Lie subgroup of a Lie group G, then [A, B] is again a Lie subgroup; however, by Example 4.6.5, p. 91 below, the Lie algebra of [A, B] need not be [o, b]. By Lemma 4.5.13, p. 89 below, the intersection of an arbitrary collection of Lie subgroups of G is again a Lie subgroup of G; its Lie algebra is the intersection of the Lie algebras of the groups in the collection. If H is another Lie group and if / : G -> H is a Lie group homomorphism, then, according to Corollary 4.5.7, p. 86 below, the image f(G) is a Lie subgroup of H; its Lie algebra is (df)(g). It is possible for a Lie subgroup to be connected in the inherited topology but not connected in the Lie topology: For example, let T := R/Z, let G := T 2 be the two-torus, choose X G g such that H0 := {exp(tX)} <£ R is dense in G, let g € G\Ho and let H be the subgroup of G generated by {g} U Ho . Then Ho is a connected Lie subgroup of G and H is a countable union of cosets of Ho- If a topological space contains a dense connected subset, then it is connected, so H is connected in the relative topology inherited from G. On the other hand, in the Lie topology, the cosets of Ho in H are all open, so, as Ho C H, we see that H is not connected. By a connected Lie subgroup of a Lie group, we mean a Lie subgroup which is connected in the Lie topology. By Lemma 4.4.4, p. 82, if G is a Lie group, then if •-> h is a one-to-one correspondence between connected Lie subgroups of G and Lie subalgebras of g. Under this correspondence,

The Lie group Lie algebra correspondence

81

normal connected Lie subgroups of G correspond to ideals of g. Let H be a closed subgroup of a Lie group G. Then, by Theorem 2.12.6, p. 99 of [Va74], H is a Lie subgroup of G. In particular, if is a closed submanifold of the manifold G. Moreover, the Lie topology u o i i G agrees with the relative topology r inherited from H, because the identity map (G, a) —> (G, r) is a continuous, bijective homomorphism, and is therefore, by Lemma 2.7.6, p. 28, a homeomorphism. If H is a Lie subgroup of a Lie group G, then H° denotes the connected component of H, in the Lie topology, containing I Q . Two connected Lie groups are locally isomorphic if they have isomorphic Lie algebras. Prom the Scholium after Theorem 3, p. 49 of [C46], we see that two connected Lie groups are locally isomorphic iff they have isomorphic universal covering groups. Let G be a Lie group and let T be a foliation on G. We say that T is left-invariant (resp. right-invariant) if, for all x, x' G G, for all g G G, we have: if (x,x') G T, then (gx,gx') G T (resp. (xg,x'g) G T). Let a Lie group G act smoothly on a manifold M. Let J b e a foliation on M. We say that T is G-invariant if, for all m,m' G M, for all g € G, we have: if (m, m') G T, then (gm, gm') G T. (That is, the action of G on M sends leaves to leaves; however, it does not necessarily send a point in a leaf to a point in the same leaf.) Let G act on G from the left and right by left and right translation. Differentiating these two actions, we arrive at left and right actions of G on TG. Let e := 1 G , so g = TeG. A subset A C TG is left-invariant (resp. right-invariant) if GA C A (resp. AG C A). For any S C g, there is a unique left-invariant A C TG such that A n g = g. In particular, for any vector subspace V of g, there is a unique left-invariant distribution A C TG such that A e = V. L E M M A 4.4.1 Let G be a Lie group and let J- be a left-invariant foliation on G. Let L := T{e). Then L is a connected Lie subgroup of G. Proof. Since L is a leaflike submanifold of G and since L is connected in the leaflike topology, it suffices to show that L is a subgroup of G. Fix a, b G L. We wish to show that ab~1 G L. By G-invariance of !F, we see that a b - 1 L is a leaf of T. Moreover a = (a6 _1 )6 G ab~xL. So, since a G (ab^L) n L, we get ab^L - L. So ab'1 = ab~le G ab^L = L. •

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L E M M A 4.4.2 Let G be a Lie group. Let A be a left-invariant vector subbundle ofTG. Let H be a connected Lie subgroup of G and assume that Ae = f). Let J7 be a foliation on G and assume that TJ- = A. Then T is left-invariant and H = !F(e). Proof. Claim 1: T is left-invariant. Proof of Claim 1: Let g € G and define / : G -»• G by f(x) = gx. Let T\ := f*{T). We wish to prove that T\ = T. By left-invariance of A, we have /*(A) = A. Then we have TT\ = f*{TT) = /»(A) = A = TT, so, by Proposition 3.5.5, p. 53, we have T\ = T. End of proof of Claim 1. By Lemma 3.5.3, p. 53 (with <j) :Y —> M replaced by the inclusion map H -> G), choose a leaf L of T such that H is an open subset of L. Since e € H C L, we see that L = .F(e). Then, by Lemma 4.4.1, p. 81, L is a connected Lie subgroup of G. As H is an open subgroup of the connected Lie group L, we conclude that H = L. Then if = .F(e). • L E M M A 4.4.3 Let G be a Lie group. Let f) be a Lie subalgebra of g. Let A be the left-invariant vector subbundle of TG such that A e = h. Then A is involutive. Proof. For all X £ g, let XL denote the left-invariant vector field on G such that (XL)e = X. Let p := dim(H). Let X\,..., Xp be a basis of h. Then X{,..., Xp is a framing of A. By definition of the Lie algebra structure on g = TeG, for all integers i, j € [l,p], we have [X^Xf] = [XuXj]L, so [X^Xf] is in A. Then, by Lemma 3.4.3, p. 51, we see that A is involutive. • L E M M A 4.4.4 Let G be a Lie group and let h0 be a Lie subalgebra of g. Then there exists a unique connected Lie subgroup HofG such that \) = \)Q. Proof. Let A be the unique left-invariant vector subbundle of TG such that A e = h 0 . By Lemma 4.4.3, p. 82, A is involutive. So, by Theorem 3.7.5, p. 57, choose a foliation T of G such that TT = A. Let H :— T(e). By Lemma 4.4.1, p. 81, H is a connected Lie subgroup of G. Moreover, f) = TeH = TeT — A e = l)o, proving existence. For any connected Lie subgroup H' of G, if h' = h 0 , then, by Lemma 4.4.2, p. 81, we have H' — T{e) = H, proving uniqueness. • Let G be a Lie group. For any connected Lie subgroup H of G, H is generated by exp(h). Thus, if H and H' are connected Lie subgroups of G

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83

and if f) = h', then H = H'. So, by Lemma 4.4.4, p. 82, the map H H-> f) is a one-to-one correspondence between connected Lie subgroups of G and Lie subalgebras of g. In the notation of Lemma 4.4.4, p. 82, we say that H is the connected Lie subgroup of G corresponding to hoLet G be a Lie group. For any Lie subalgebra I) of JJ, the connected Lie subgroup corresponding to I) is exactly the subgroup of G generated by exp(h). For any connected Lie subgroup H of G, we have: H is normal in G iff f) is an ideal of g. Let G and H be Lie groups and assume that G is connected and simply connected. From the Scholium after Theorem 3, p. 49 of [C46], one sees that / i-> df is a one-to-one correspondence between Lie group homomorphisms G —> H and Lie algebra homomorphisms g —• h. Under this correspondence, / is surjective iff df is surjective. Moreover, / has discrete kernel iff df is injective. Moreover, for any Lie subgroup Ho of H, we have: f(G) C Ho iff (#)(fl) C fo. L E M M A 4.4.5 Let G be a Lie group, let A be a Lie subgroup of G and let B be a connected Lie subgroup of G. Assume that A C. B and that dim(A) > dim(B). Then A = B. Proof. Since a C b and since dim(a) = dim(b), we get o = b. Then, by Lemma 4.4.4, p. 82, A0 = B. Then A0 C A C B = A0, so A = B. • A connected Lie group is simple if its Lie algebra is simple. Under this definition, a simple Lie group can have nontrivial center, so it may not be simple as an abstract group. However, in (1) of Lemma 4.10.22, p. 121 below, we argue that if G is simple and centerfree, then G is simple as an abstract group. A connected Lie group is semisimple if its Lie algebra is semisimple. The real rank of a connected semisimple Lie group is, by definition, the real rank of its Lie algebra.

4.5

Some facts about Lie subgroups

L E M M A 4.5.1 Let G be a Lie group and let A and B be Lie subgroups ofG. Then AC\B is a Lie subgroup ofG. The Lie algebra of AC\B is o n b . Proof.

Let r := dim(o n b). Differentiating the left-translation action

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of G on G, we get a left action of G on TG. For all c G A n B , we have dim((T c ^) n (T C S)) = dim((co) n (cb)) = dim(c(on b)), so dim((TcA) n (T C JB)) = dim(o n b) = r. So, since A and B are leaflike submanifolds of G, we conclude, from Lemma 3.1.5, p. 46, that A n -B is a leaflike submanifold of G. So, as Ar\B is also a subgroup of G, we see that A ("1.0 is a Lie subgroup of B. The proof that the Lie algebra of A n B is o fl b is left as an exercise. • We will improve on Lemma 4.5.1, p. 83 in Lemma 4.5.13, p. 89. The intersection of two connected Lie subgroups of a Lie group is necessarily (by Lemma 4.5.1, p. 83) a Lie subgroup, but is not necessarily a connected Lie subgroup. Let X be a topological space. Then we say that X is arcwise conn e c t e d if, for any x, x' € S, there exists a continuous, injective map / : [0,1] ->• X such that /([0,1]) C 5, such that /(0) = x and such that / ( l ) = x'. Note that path-connected means almost the same thing; in the definition of path-connected, however, the map / is not required to be injective. If X is arcwise-connected, then X is path-connected. If X is path-connected and Hausdorff, then X is arcwise-connected. Recall that a Lie subgroup of a Lie group is said to be connected if it is connected in the Lie topology. For any locally path-connected topological space X, for any subset S C X, we have: S is a component of X iff S is a path-component of X. Any manifold is locally path-connected, so it is connected iff it is path-connected. In particular, a Lie group is connected iff it is path-connected. Consequently, for any Lie group G, for any Lie subgroup H of G, H is connected (in the Lie topology) iff H is pathconnected. This raises the question of whether a path-connected subgroup of a Lie group is necessarily a connected Lie subgroup. The following result, due to H. Yamabe, answers this affirmatively. T H E O R E M 4.5.2 Let G be a Lie group and let H be a subgroup of G. Then H is a connected Lie subgroup of G iff H is path-connected. Proof. "Only path-connected. path-connected, a connected Lie

if" follows from the fact that any connected manifold Conversely, since H is Hausdorff, it follows that, if H then H is arcwise-connected, in which case, by [Y50], H subgroup of G.

is is is •

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85

We remark that, for the applications we need in this book, it is sufficient to prove that, for any Lie group G, for any subgroup H of G, if any two points of H may be connected by a path that is contained in H and is piecewise C 1 in G, then H is a connected Lie subgroup of G. This result is weaker than Theorem 4.5.2, p. 84, but is easier to prove. (See Theorem 1.6*, pp. 43-44 of [IT91].) COROLLARY 4.5.3 Let G be a connected Lie group and let S be a collection of connected Lie subgroups ofG. Let H be the subgroup ofG generated by (J S. ThenH is a connected Lie subgroup of G and the Lie algebra f) of H is generated by I ) s. ses Proof.

For all S G S, S is path-connected. So, since 1Q G P | S, it

follows that M S is path-connected. Then H is path-connected, so, by Theorem 4.5.2, p. 84, we see that if is a connected Lie subgroup of G. Let h* be the Lie subalgebra of Q generated by I ) s. For all S G S, we ses have S C H, so s C h. Then h* C h. It remains to show that f) C h». Let H* be the connected Lie subgroup of G corresponding to h«. For all S € S, we have s C h*, so, as S is connected, we see that S C Ht. Then H CH,. Then!) C h„. • L E M M A 4.5.4 Let H be a Lie subgroup of a Lie group G and define T C G x G by T := {(a;, y) € G x G \ xH = yH}. Then J7 is a prefoliation on G. Moreover, if H is connected, then T is a foliation. Proof. Let H act on G by h.g = gh~x. This action is free. We have T = {(x,y) G G\H.x = H.y}. Then, by Lemma 4.2.1, p. 71 (with G replaced by H and with M replaced by G), the result follows. • P R O P O S I T I O N 4.5.5 Let G be a Lie group and let H be a subgroup of G. Then H is a Lie subgroup of G iff H has countably many pathcomponents, in the relative topology inherited from G. Proof. Proof of "only if": Let r be the Lie topology on H. Let i : H -t G be the inclusion map. Then i : (H,r) —• G is continuous. Since H is a

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Lie subgroup of G, we see that (H, T) is second-countable and locally pathconnected and therefore has countably many path-components. Then t(H) has countably many path-components in the relative topology inherited from G. Since t(H) = H, we are done. End of proof of "only if". Proof of "if": Let H* be the path-component of H containing 1Q. Then H/H* is countable. By Theorem 4.5.2, p. 84, we see that H* is a connected Lie subgroup of G. Let T :— {(x,y) | xH* — yH*}. By Lemma 4.5.4, p. 85, we see that T is a foliation on G. Let C be the collection of all leaves L of T such that L C H. Every leaf of T is a left-coset of H* in G, so C C H/H*, so C is countable. So, since H = M L, by Lemma 3.4.1, p. 51, we conclude L€C

that H is a leaflike siibmanifold of G. So, as H is also a subgroup of G, we see that if is a Lie subgroup of G. End of proof of "if". • The next result is due to Preudenthal ([Pr41, Anwendung 9.1, p. 1055]). COROLLARY 4.5.6 Let G be a Lie group and let A,B C G. If A and B are both Lie subgroups (resp. connected Lie subgroups) of G, then [A, B] is a Lie subgroup (resp. a connected Lie subgroup) of G. Proof. Let S := {a6a _ 1 6 _ 1 1 a G A, b € B}. Then S generates [A,B], so, by Corollary 4.5.8, p. 87 (resp. Theorem 4.5.2, p. 84), it suffices to show that S has only countably many path components (resp. is path-connected). Since A and B are Lie subgroups (resp. connected Lie subgroups) of G, by Lemma 4.5.5, p. 85 (resp. Theorem 4.5.2, p. 84), it follows that A and B both have only countably many path components (resp. are pathconnected). Then AxB has only countably many path components (resp. is path-connected). So, since the map (a,b) i-t aba~1b~1 : A x B —> S is continuous and surjective, we see that S has only countably many pathcomponents (resp. is path-connected). • Let A and B be connected Lie subgroups of a Lie group G. Recall, from Corollary 4.5.6, p. 86, that [A, B] is a Lie subgroup of G. It appears to be a common mistake, appearing in the published literature, to think that the Lie algebra of [A, B] is [a, b]. This is often true (Corollary 4.6.3, p. 91), but not always (Example 4.6.5, p. 91). COROLLARY 4.5.7 Let G and H be Lie groups and let f : G -> H be a Lie group homomorphism. Let HQ :— f(G). Then HQ is a Lie subgroup of H, (HO)0 = f(G°) and h0 = (df)(g). Moreover, f : G -> H0 is a

Some facts about Lie subgroups

87

surjective Lie group homomorphism. Moreover, if G is connected, then Ho is a connected Lie subgroup of H. Proof. Since G has countably many path-components, Ho does as well. Then, by Proposition 4.5.5, p. 85, H0 is a Lie subgroup of H. By Lemma 3.2.2, p. 48, / : G -> H0 is smooth. Then / : G -> H0 is a surjective Lie group homomorphism. The remaining statements are left as exercises for the reader. • Let G be a Lie group. By Corollary 4.5.7, p. 86, Ad a (G) is a Lie subgroup of GL(JJ). If G is connected, then the kernel of Ad : G -> GL(g) is equal to Z(G), so, by Lemma 4.9.2, p. 96, we see that Adg(G) is Lie group isomorphic to G/(Z(G)). C O R O L L A R Y 4.5.8 Let G be a Lie group and let S C G. Assume that S has countably many path-components, in the relative topology inherited from G. Then the subgroup of G generated by S is a Lie subgroup ofG. C O R O L L A R Y 4.5.9 Let G be a Lie group and letS be a countable collection of Lie subgroups ofG. Let H be the subgroup ofG generated by I ) S. ses Then H is a Lie subgroup of G. Proof. For all 5 € <S, by Lemma 4.5.5, p. 85, we see that S has only countably many path components. So, as S is countable, we conclude that X := M S has only countably many path components. Then, by ses Corollary 4.5.8, p. 87 (with S replaced by X), we see that H is a Lie subgroup of G. • We remark that, in Corollary 4.5.9, p. 87, it is not necessarily true that f) is generated by (^J s. For example, let G := SL2(E) and let ses

-=(--)• H(v)HLet J denote the 2 x 2 identity matrix and let T := {/, g0}. Then N and T are Lie subgroups of G whose union generates G. However, t = {0}; it follows that n U t = n, s o n U t generates only n.

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L E M M A 4.5.10 Let G and H be Lie groups and let HQ be a Lie subgroup of H. Let f : G —»• H be a homomorphism of Lie groups. Then f~1(Ho) is a Lie subgroup of G whose Lie algebra is ( (g,f(g)) : G -> G x H. Then, by Corollary 4.5.7, p. 86, T is a Lie subgroup oiGxH. Then, by Lemma 4.5.1, p. 83, E := (G x H0)nT is a Lie subgroup oiGxH. Let -K : G x H -»• G be projection onto the first coordinate. Then p := 7r|H : S -> G is a Lie group homomorphism. We have / _1 (.ffo) = p(H). So, by Corollary 4.5.7, p. 86, / _ 1 ( H ) is a Lie subgroup of G. We leave it as an exercise for the reader to show that the Lie algebra of f~1(E) is (d/) -1 (f)o). ^ For any additive Abelian group, for all integers k > 1, for all x € G, let kx denote the fc-fold sum of x. An additive Abelian group G is divisible, if, for all g £ G, for all integers A; > 1, there exists x £ G such that kx = g. L E M M A 4.5.11 Let G be a connected Lie group. Then G admits no proper subgroup of finite index and G admits no proper Lie subgroup of countable index. Proof. Let if be a proper subgroup of G. We wish to show that G/H is infinite and that, if H is Lie, then G/H is uncountable. Since exp(g) generates G, choose X € g such that exp(X) ^ H. Define / : K -» G by /(*) = exp(tX). Let A :- / _ 1 ( # ) - Then A is a proper subgroup of R and there exists an injective map R/^4 -> G/H. Since R is divisible, since divisibility passes to quotient groups and since no nontrivial finite Abelian group is divisible, it follows that R admits no proper subgroup of finite index. Thus, R/A is infinite. It follows that G/H is infinite. Now assume that if is a Lie subgroup of G. We wish to show that the G/H is uncountable. By Lemma 4.5.10, p. 88, A is a Lie subgroup of R. By Proposition 4.5.5, p. 85, A/A° is countable. Since ^4° is a proper Lie subgroup of R, we conclude that a is a proper Lie subalgebra of R, so o = {0}. Then ^4° is trivial, so, as ^4/^4° is countable, A is countable. Then R/A is uncountable. Then G/H is uncountable. • L E M M A 4.5.12 Let G be a Lie group, let A be a Lie subgroup of G and let B be a subgroup of G. If A C B and if B/A is countable, then B is a Lie subgroup of G and A0 = B° and a= b.

Some facts about Lie subgroups

89

Proof. By Proposition 4.5.5, p. 85, we see that A has countably many path-components. Then B has countably many path-components, so, by Proposition 4.5.5, p. 85 again, B is a Lie subgroup of G. Since B/A and A/A° are both countable, it follows that B/A° is countable. Then B°/A° is countable. Then, by Lemma 4.5.11, p. 88, A0 = B°. Then a = b. • L E M M A 4.5.13 Let G be a Lie group. Let S be a collection of Lie subgroups of G. Then H := O S is a Lie subgroup of G and h = | ) s. S€S

S€S

Proof. Claim 1: if is a Lie subgroup of G. Proof of Claim 1: For any subgroup A of G, let A* denote the path-component of 1Q in A, in the relative topology on A inherited from G. For any subgroup A of G, by Theorem 4.5.2, p. 84, we see that A* is a connected Lie subgroup of G and we define 8A := dim(A*). For any T C S, let AT := f] T, let A*T := (AT)* and let dT := SAT. T€T

Then H = As. Let d :— mm{dj^}, where T ranges over all finite subsets of S. Choose a finite To C S such that d = d^0. For all S €. S, let Ts •— Fo U {S}; then, as f s 3 ^ o , we see that Aj:s C A?0, so A*-ps C Aj? . On the other hand, by minimality of d, for S 6 «S, we see that d, so d^s > djr0, so dim(Aj7s) > dim(A$r0); so, since A^ C A*To, by Lemma 4.4.5, p. 83, we get A*-ps = Apo. For all S £ <S, we have 5 S F s , so Ajrs C 5; so, since ^>s = ^ - o ' w e Set A k £ 5 -

Then A

>o £ f l

5

= ^

so A

?o ^ A*s-

As

<S D Fo, it follows that A 5 C A F 0 , so A£ C A^ o . Then A*s = A*To. By Lemma 4.5.1, p. 83, A?0 is a Lie subgroup of G. We have A*^ = A*s = H* C H = As C Ajr0. By Proposition 4.5.5, p. 85, we see that Aj:0/Ajr is countable. Then, because A^o C H C ytjr0, it follows that H/A*^ is countable. Then, by Lemma 4.5.12, p. 88, if is a Lie subgroup of G. End of proof of Claim 1. Claim 2: t) = f] s. Proof of Claim 2: Let h, := f] s. For all S € <S, ses ses we have if C S, so h C s. Then h C h*. We wish to show that h* C h. Let if, be the connected Lie subgroup of G corresponding to ()*. For all 5 € S, we have h, C s, so if, C 5 ° . Then if, C f] S° C f] S = H. ses ses Then h, C h. End of proof of Claim 2. •

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Basic Lie Theoretic

Results

The Lie algebra of [A, B]

The results in this section were pointed out to me by D. Witte, and the original draft of this section was written by him. It is quite possible that the results herein have been observed by others. If S is a subset of a Lie algebra g, then we will denote by (5) the Lie subalgebra of g generated by S. If S is a subset of a group G, then we will denote by (S) the subgroup of G generated by S. NOTATION 4.6.1 We make the following recursive definition: For all integers n > 3, for any Lie algebra g, for all W\, • •., Wn € g, we define [W1,...,Wn]:=[\Wi,...,Wn-1],Wn]. Suppose a and b are Lie subalgebras of a Lie algebra g. Let T(a, b) be the smallest ideal of (aUb) containing [a, b]. More concretely, by the Jacobi identity, we see that T(o, b) is the vector subspace of g spanned by {[Y,Z,X1,...,Xn]

\ Y e a, Z e b, Xi e aUb, n =

0,1,2,3,...}.

The correct characterization of the Lie algebra of [A, B] is: L E M M A 4.6.2 Let A and B be connected Lie subgroups of a Lie group G and let H := [A, B]. Then H is a connected Lie subgroup of G and t) —

r(o,b). Proof. By Corollary 4.5.6, p. 86, H is a connected Lie subgroup of G. We wish to show that T(a, b) = h. (C) By Thm. 2.1(iii), p. 18 of [Go80], we see that [A,B] is normal in {A U B), so f) is an ideal of (o U b). So, since [o, b] C h, we get T(o, b) C f). (D) We may assume that G = (AU B). Then, since o and b normalize T(a, b), it follows that A and B normalize T(a, b). Then G normalizes T(a, b). Thus T(o, b) is an ideal of g. Let N be the connected Lie subgroup of G corresponding to T(o, b). Then n = T(a, b). Let G :- G/N. Then g :— g/n. Let a : G —• G be the canonical homomorphism. We have [dcr(o), (da)(b)] = (do-)([a, b]) C (do-)(T(a, b)) = (d
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91

COROLLARY 4.6.3 Let A and B be connected Lie subgroups of a Lie group G and let H := [A, B]. Assume that A normalizes B and B normalizes A. Then H is a connected Lie subgroup of G and t) = [a, b]. Proof. By Lemma 4.6.2, p. 90, if is a connected Lie subgroup of G and f) = r ( o , b). Since a normalizes b and vice versa, we conclude that [a, b] is an ideal of (o U b), which gives T(o, b) = [a, b]. • COROLLARY 4.6.4 Let G be a connected Lie group. Then [G, G] is a connected Lie subgroup of G and the Lie algebra of[G,G] is [g,g]. Proof.

This is immediate from Corollary 4.6.3, p. 91.

•

E X A M P L E 4.6.5 Let G be a connected Lie group and let A and B be connected Lie subgroups of G. Then it does not necessarily follow that the Lie algebra of [A, B] is [a, b]. Proof. Let G := SL(2,E). Let A := {aE[f + (l/a)E^ \a > 0}. Let B := SO(2). Let H := [A, B]. As BAB = G, it follows that (A U B) = G, so, by Corollary 4.5.3, p. 85, we have (a U b) = g. So, as g is simple, by Lemma 4.6.2, p. 90, we have f) = g. Then [A, B] = H = G. However, since dim(o) = 1 = dim(b), it follows that dim([a, b]) = 1. Then [a, b] ^ g. •

4.7

Lie groups and Lie algebras from bilinear and quadratic forms

Let Q be a quadratic form on a vector space V. We define

0(g)

:= {FeGL(V)\FoQ

= Q},

SO(Q)

:=

{F £ 0(Q) | det(F) = 1},

SO°(Q)

:=

(SO(Q))°.

Each of these is a closed submanifold of GL(V), and is a Lie group under the resulting manifold structure. The Lie algebra of SO(<2) is, by our convention, denoted so(Q); it is equal to the Lie algebra of SO°(<5). Let J : V —> V denote the identity transformation, and define: S

:=

{ A / | A £ K } CGL(V),

S+

:=

{A/|A>0}CGL(F),

5*

:=

{A/|A^0}CGL(V).

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We define CO(Q) := (5,)(0(Q)). Then C0(<2) is a closed Lie subgroup of GL(V). We define CO°(Q) := (CO(Q))°. Then CO°(Q) = (5+)(S0°(Q)) and co(Q) = S + so(Q). Let d > 1 be an integer. Let B be the standard ordered basis of Rd. The map T i-» [T] B : End(E d ) -» l d x d is a vector space isomorphism and allows us to identify Rdxd with End(E rf ). Under this identification, for any quadratic form Q £ QF(M d ), we may identify so(Q) as a subset of M d x d . Similarly, using the standard ordered basis of Rdxl for any Qo € Q F ( R d x l ) , we identify so(Q 0 ) as a subset of Rdxd. If Q £ QF(R d ) and Q0 e QF(E d><1 ) correspond under the standard isomorphism Rd •<—> Rdxl, then we have so(Q)=so(Q0)CRdxd. If Q and Q' are equivalent quadratic forms, then 0(Q) = O(Q') and SO(Q) S SO(Q') and SO°(Q) S SO°(Q') and so(Q) S so(Q')- K
CO(B) := CO(Q), SO°(B) := SO°(Q).

We leave it as an exercise for the reader to show, for all X e so(B), for all u,io € V, that B(Xv,w) + B(v,Xw) = 0. In particular, for all X € so(-B), for all v 6 V, we have B(X«,t;) = 0. We compute

Let 3 be an integer. Then so(Qd) is the collection of matrices a w 0

-V*

B —w1

0 u —a

such that a e R, such that u,«; £ R(d~2)xl and such that B £ £(d-2)x(d-2) satisfies B* = —B. To remember this representation of so(d— 1,1), we refer to the three blocks 0, B and 0 as the "blocks on the backwards diagonal".

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93

With this, one notices that each block is the negative transpose of the block which is located across from it on the other side of the backwards diagonal. This forces the (1, d) and (d, 1) entries to be equal to 0. There is a common confusion: One must remember that v1 and w* refer to the transposes of v and w in the usual sense, not in any backward sense. Thus, for example, the (1,2) entry of a matrix in so(Qd) is the negative of the (2, d) entry of the same matrix. Moreover, B is antisymmetric across its main diagonal, not across its backwards diagonal. So, for example, the (2,3) entry is the negative of the (3,2) entry. Also, for example, if i £ { 2 , . . . , d — 1}, then the (i,i) entry is = 0. If P,<1 > 1 a r e integers, if p + q — d and if Q e QF(E d ) is given by Q(wi,... ,wp,xi,... ,xq) — w2 + • • • + w2 — x\ — • • • — x2q, then we define 0(p,q) := 0 ( g ) , SO(p,q) := SO(Q) and SO°(p,g) := SO°(Q). Let d > 1 be an integer. Define Q e QF(R d ) by Q(xi,...,xd)

=x\ +

Vx2d.

We define SO(d) := SO(Q) and SO°(d) := SO°(<2). L E M M A 4.7.1 Let V be a vector space and let B 6 SBF(F) be nondegenerate. Let X € o(B) and assume that dim(X(V)) < 1. Then X = 0. Proof. Assume, for a contradiction, that X ^ 0. Then dim(X(V)) = 1. Choose v0 6 V\{0} such that X{V) = Rv0. Let W :— {v € V\B(VQ,V) = 0}. Then W is a proper subspace of V. So, since K :— ker(X) is also a proper subspace of V, choose v\ € V\(JK" U W). We have X(v{) € X{V) = Rv0. Choose a € M such that X(vi) - av0. Since v\ ^ K, we get X(vi) ^ 0, so a ^ 0. Since X € o(B), we get B(X(vi),vi) = 0. Then a-B(v0,vi) = Biavo^!) = B(X(vi),vi) = 0. So, since a ^ 0, we get B(vo,vi) = 0, so vi G W, contradiction. D

4.8

Abelian Lie groups

For any additive Abelian group, for all integers k > 1, for all x 6 G, let kx denote the fc-fold sum of x. This gives G the structure of a Z-module. So, since any Z-module has an underlying additive group structure, we identify additive Abelian groups with Z-modules.

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L E M M A 4.8.1 Let V be a vector space and let D be a discrete additive subgroup of V. Let d := dim(y). Then there are an integer m € [0,d] and a vector space isomorphism f : V -> Rd such that f(D) = Z m x { 0 } d - m . Proof. We may assume that D / {0}. Let wi,... ,wm be a maximal linearly independent subset of D. Let W := Rwi + • • • + Rwm and let Do := Zwi + • • • + 7Lwm. By maximality of m, we have D C W. As w\,..., wm E D, we get DQ C D. Since the vectors u>i,..., wm are linearly independent, we have • W/Do is compact in the quotient topology; • the abstract groups Do and Z m are isomorphic; and • dim(W^) = m. Because D is a discrete subset of W and because DQ C D, we conclude that D/D0 is a discrete subset of W/D0. Then, by compactness of W/D0, we see that D/D0 is finite. We identify additive Abelian groups with Z-modules. Since D0 = Z m and since Do has finite index in D, we see that D is finitely generated. Moreover, since V, being a vector space, is torsion-free and since D C V, we conclude that D is torsion-free. Then D is a finitely generated, torsionfree Abelian group, so choose an integer n > 1 such that D = TLn. Let F := D/D0; then F is finite. We have an exact sequence {0} - • Z m -> Z n -> F -» {0} of Z-modules. Since JF is finite, we get F ®z Q = {0}By tensoring the exact sequence with the flat Z-module Q, we obtain an exact sequence {0} -> Q m -> Q" ->• {0} -4 {0} of Q-modules. Then Q"1 is isomorphic to Q™, as Q-modules, i.e., as Q-vector spaces. Then we have m = dim Q (Q m ) = dim Q (Q n ) = n. Then D S Z n = Z m . Choose vi>---,vm € D such that Z^i + h Zvm = D. We have wx,..., wm £ D C Rvt H h Kj; m , so W C Rvx H 1- Rvm. Since dim(W) = m, we conclude that vi,... ,vm are linearly independent. Extend vi,..., vm to a basis vi,..., Vd of V. Let e i , . . . , e Rd by f(vi) = e^. Since D = Z«i + • • • + Zu m , it follows that /(£>) = Zei + • • • + Ze m . That is, f(D) =Zmx {0}d-m. O

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L E M M A 4.8.2 Let G and H be Lie groups and let f : G -» H be a bijective continuous homomorphism. Then f is an isomorphism of Lie groups. Proof.

This is Theorem 2.11.3, p. 93 of [Va74].

•

L E M M A 4.8.3 Let G be an Abelian Lie group. Let T := R / Z . Then there is a discrete Abelian Lie group F, and there are integers m,n>0 such that G is Lie group isomorphic toT x T m x W1. Proof. We may assume that G is an additive group. By Corollary 2.13.3, p. 105 of [Va74], the exponential map exp : g -> G° is a covering homomorphism. Let A denote the kernel of this homomorphism. Then G° is Lie group isomorphic to the additive group g/A. Define d := dim(G) and, by Lemma 4.8.1, p. 93, choose an integer m 6 [0, d] and a vector space isomorphism / : g -> E" such that /(A) = Z m x {0} d ~ m . Let n:=d-m. Then G ° S T m x l n . Then G° is divisible. So, by Proposition 3.16, p. 158 of [J80], we see that G° is injective as a Z-module. Then, by Proposition 3.17, p. 160 of [J80], choose a Z-submodule T of G such that (x, y) t-t x + y : T © G° -» G is a Z-module isomorphism. Let 7T : G ->• G/G° be the canonical map. Then 7r|r : F ->• G/G° is an isomorphism of abstract groups. So, as G/G° is countable, we see that r is countable. Give F the discrete (zero-dimensional) manifold structure. Then (x, y) i-» x + y : T x G° -> G is a continuous map which is also an isomorphism of abstract groups. It follows, from Lemma 4.8.2, p. 94, that (x, y) 1-4 x + y : F x G° —• G is an isomorphism of Lie groups. Then G^TxG0 ^TxTm xl". • L E M M A 4.8.4 Let V and W be vector spaces. Let f : V -> W be a homomorphism of Lie groups. Then f :V —> W is a linear transformation. Proof. By assumption, / is additive, i.e., for all v,v' G V, we have / ( " + V) = (/(«)) + (/(«')). Then (/(0)) + (/(0)) = / ( 0 + 0) = / ( 0 ) , so /(0) = (/(0)) + (/(0)) - (/(0)) = (/(0)) - (/(0)) = 0. Since / is additive, it remains to show, for all v € V, for all r £ M., that f(rv) = r{f(y)). Claim 1: For all fc € Z, for all w € V, we have f{kw) = k(f(w)). Proof of Claim 1: If fc = 0, then /(fcw) = /(0) = 0 = k(f(w)) and if fc > 0, then we are done by additivity. We therefore assume that fc < 0. Let I := — fc.

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Then, by additivity, f(lw) = l(f(w)) and f(kw) + f(lw) = /(O) = 0. Then f(kw) = -(f(lw)) = -l(f(w)) = k(f(w)). End of proof of Claim 1. Fix v € V. We wish to show, for all r € K, that f(rv) — r(f(v)). Since Q is dense in 1R, it suffices to show, for all q 6 Q, that f(qv) = q(f{v)). Fix q € Q. Choose an integer n > 1 such that nq € Z. Let m := nq. By Claim 1, f(n(v/n)) = n(f(v/n)) and f(m(v/n)) = m(f(v/n)). We have n(f(v/n)) = f{n(v/n)) = f(v), so f(v/n) = (l/n)(f(v)). Then f((m/n)v) = f(m(v/n)) = m(f(v/n)) = m ( l / n ) ( / ( « ) ) = «(/(«)). Therefore /(^u) = f((m/n)v) = q(f(v)). D 4.9

Miscellaneous results

Let g be a Lie algebra. The solvable radical of g is the unique maximal solvable ideal of g, with respect to set theoretic inclusion. Let t denote the solvable radical of a Lie algebra g. A Lie subalgebra [ of g is called a semisimple Levi factor of 0 if • I is semisimple or zero; • l n t = {0}; and • l + r = 0. Let G be a connected Lie group. The solvable radical of G is the connected Lie subgroup of G corresponding to the solvable radical of 0. It is a solvable normal subgroup of G. It is maximal among connected solvable normal subgroups of G. It is closed in G. A connected Lie subgroup of G is a semisimple Levi factor of G if its Lie algebra is a semisimple Levi factor of 0. By Theorem 3.14.1, p. 225 of [Va74], any Lie algebra has a semisimple Levi factor so any connected Lie group has a semisimple Levi factor. The next result is left as an exercise for the reader. L E M M A 4.9.1 Let g be a Lie algebra. Let I be a semisimple subalgebra of 0 and let r be a solvable ideal of g. Assume that I + t — 0. Then I is a semisimple Levi factor of g and r is the solvable radical of g. We have a smooth analogue of Lemma 2.7.4, p. 28: L E M M A 4.9.2 Let G and H be Lie groups and let f : G -*• H be a continuous surjective homomorphism. Let K denote the kernel of f and let

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7r: G -¥ G/K be the canonical Lie group homomorphism. Then there is an isomorphism f : G/K -> H of Lie groups such that f OTT = f. Proof. Choose / ' as in Lemma 2.7.4, p. 28. By Lemma 4.8.2, p. 94, we see that / ' is an isomorphism of Lie groups. • COROLLARY 4.9.3 Let G and H be Lie groups and let f : G -> H be a continuous homomorphism. Let K be the kernel of f. Let n : G —> G/K be the canonical homomorphism. Then there is an injective homomorphism f : G/K -> H of Lie groups such that f on = f. Proof. Let HQ := f(G). By Corollary 4.5.7, p. 86, Ho is a Lie subgroup of G and / : G —> Ho is a surjective Lie group homomorphism. Then, by Lemma 4.9.2, p. 96 (with / : G ->• H replaced by / : G ->• H0), let / ' : G/K —>• Ho be a Lie group isomorphism such that f'on = / . Then / ' : G/K —> H is an injective Lie group homomorphism. • COROLLARY 4.9.4 Let G, A and B be Lie groups and let p : G -> A and q : G —> B be Lie group homomorphisms. Assume that p(G) = A and that ker(p) C ker(g). Then there is a Lie group homomorphism f : A —» B such that f op = q. Proof. Let K := ker(p) and let a : G -> G/K be the canonical homomorphism. By Lemma 4.9.2, p. 96, let p' : G/K -> A be a Lie group isomorphism such that p' OCT = p. Let L := ker(<7) and let T : G -> G/L be the canonical homomorphism. By Lemma 4.9.2, p. 96, let q' : G/K —> B be a Lie group isomorphism such that q'or = q. Since K = ker(p) C ker(g) = L, let / ' : G/K —> G/L be the canonical homomorphism. Then f o a = T. Let f :=q' o f o (p1)'1 :A->B. Then fop = q. D L E M M A 4.9.5 Let G and H be Lie groups and let f : G -t H be a Lie group homomorphism. Assume that f(G) = H and that ker(/) is discrete in G. Then f : G —• H is a covering homomorphism. Proof. Let K := ker(/). Let n : G -> G/K be the canonical homomorphism. Choose / ' : G/K -> H as in Lemma 4.9.2, p. 96; then / ' : G/K -> H is a Lie group isomorphism. Since K is discrete, it follows that 7r : G —>• G/K is a smooth covering homomorphism. Then, because / = / ' o 7r, we see that / : G —> H is a covering homomorphism. •

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Let (x, h) \-¥ x.h : X x H —»• X be a right action on a set X by a group H. The action on X by H is said to be simply transitive if it is both transitive and free; equivalently, for all x, x' G X, there is a unique h £ H such that x.h = x'. For H a Lie group, a principal pre-H-bundle over M consists of • • • •

two manifolds P and M; a Lie group if; a smooth right action on P by if; and a smooth function n : P —> M

such that, for all m € M, the fiber 7r _1 (m) is invariant by H and simply transitive by if. We will typically omit any notation for the action and say simply that ir : P -> M is a principal pre-if-bundle. If U is an open subset of M, then V := 7r _1 ([/) is invariant by H and ir\V : V -> U is then a principal pre-if-bundle. Let M be a manifold and let H be a Lie group. Let -K : P —• M and 7r' : P ' —• M be two principal pre-if-bundles over M. A map / : P —> P ' is said to be a principal pre-if-bundle isomorphism over M if all three of the following hold: • / : P -» P' is a difFeomorphism; • 7r' O / = 7r; and • / is equivariant by if, i.e., Vp € P,V7i € if, f(p-h) =

(f(p)).h.

Let M be a manifold and let if be a Lie group. The standard trivial principal pre-if-bundle over M is the projection map M x if —• M together with the right translation action in the second coordinate, given by (m, h).h' — (m, /i/i'). A principal pre-H-bundle over M is trivial if it is isomorphic to the standard trivial principal pre-if-bundle over M. Let M be a manifold and let if be a Lie group. Let 7r : P —> M be a principal pre-ff-bundle over M. For any m € M, we say that 7r is locally trivial near m if there is an open neighborhood U of m in M such that 7r|(7r-1(J7)) : 7T_1(C7) -> U is a trivial principal pre-if-bundle over U. We say that n is locally trivial if it is locally trivial near each point of M. Let if be a Lie group. A principal if-bundle is a locally trivial principal pre-if-bundle. In the preceding five paragraphs, replacing "manifold" by "topological space", "Lie group" by "topological group", "smooth" by "continuous" and

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"diffeomorphism" by "homeomorphism", one arrives at the definition of a continuous principal ff-bundle. Let H be a Lie group. Let £ : P -»• M be a principal il-bundle. Let H act smoothly on a manifold X. Then the action of H on P x X given by h(p, x) = (ph_1, hx) is smooth, free and proper. Let Px^X := H\(P x X) and let TT : P x X -* P x # X be the canonical map. Then, by Lemma 4.2.6, p. 74, we see that n is a potential submersion. We give P XJJ X the manifold structure such that n is a submersion. Define q : P x # X -> M by g(7r(p,a;)) = £(p). Then g : P x # X -> M is a fiber bundle with fiber X. We call q : P xH X ->• M the bundle with fiber X associated to 7r. L E M M A 4.9.6 Let G be a Lie group and let H be a closed subgroup of G. Then the canonical map G —> G/H is a principal H-bundle with respect to the right action on G by H given by right translation: g.h = gh. The preceding result is left as an exercise. Note, however, that the assumption that G and H are Lie groups is essential. Otherwise the canonical map G —> G/H need not even be a continuous principal if-bundle; H. Becker provided the following elementary example: Let T be the circle in the complex plane, given by T := {z € C : \z\ — 1}; it is a multiplicative Lie group. Let § := {+1, - 1 } C T. Let G := T z with the product topology and with coordinate-by-coordinate multiplication. Then G is a compact metrizable Abelian topological group and H := S z is a closed subgroup of G. Assume, for a contradiction, that p : G —> G/H is a continuous principal H-bundle. Identifying G/H with (T/S) z , then p : T z -> ( T / S ) z is the Zfold product of the canonical homomorphism q : T —> T / S . Let U := TT Uj jez be a basic open neighborhood of the identity in G/H = ( T / S ) z and let s : U —> G be a continuous cross-section to p|(p _ 1 (^))- Choose k € Z such that Uk = T / S . Let i : Uk -> TT Uj be the fcth coordinate inclusion. Let u : T z ->• T be the fcth coordinate projection. Let « : ( T / S ) z -> T/S be the fcth coordinate projection. The composite v o i is equal to the identity on C/fc = T/S. The composite p o s is equal to the identity on U. The composite q o u is equal to the composite » o p o n T z = G. Therefore qo u o s o i = v o p o s o i = i ; o j is the identity on Uk = T / S . Consequently, u o s o i : T/S -4- T is a continuous cross-section of the canonical map q : T -> T / S . However, by covering space theory, such a continuous cross-section does not exist, contradiction.

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L E M M A 4.9.7 Let K be a connected compact Lie group, let K be a covering group of K. Assume that K is connected. Then there is a connected compact subgroup C of K and an Abelian connected Lie subgroup B of K such that K = BC = CB. Proof. By Theorem V.8.1, p. 233 of [BtD85], fix a connected, simply connected compact Lie group L, a connected Abelian Lie group A and a covering homomorphism LxA—^K. Replacing A by its universal covering group, we may assume that A is simply connected. Then, by covering space theory, there is a covering homomorphism q : L x A —> K. Let B := q(A) and let C := q(L). • L E M M A 4.9.8 Let G be a connected semisimple Lie group. Then there exists a compact subgroup C of G and two Abelian connected Lie subgroups A and B of G such that G = CBABC. Proof. By Theorem IX. 1.1, p. 402 of [He78], choose connected Lie subgroups K and A of G such that K is a covering group of a connected compact Lie group, such that A is Abelian and such that G = KAK. By Lemma 4.9.7, p. 100, choose a connected compact subgroup C of K and an Abelian connected Lie subgroup B of K such that K = BC = CB. Then G = KAK = CBABC. D If C is a collection of Lie groups, then a Lie group G is said to be an extension of groups in C if there exists an integer n > 1 and closed subgroups Go,. •. ,Gn of G such that { I G } = G I C ••• C Gn = G, and such that, for all integers i G [l,n], we have both: • C?i_i is a normal closed subgroup of Gf, and • Gi/Gi-i is Lie group isomorphic to an element of C. L E M M A 4.9.9 Let G be a connected Lie group. For all W € Q, define Sw •= {exp(iW)}t e K Q G. Then there are an integer k > 1 and Xi,...,Xk Gg such that G = Sxx • • • Sxk • Proof. If G is compact, then this follows from the fact that the canonical coordinates of the second kind (p. 89 of [Va74]) cover an open neighborhood U of the identity, and from the fact that (by compactness) there exists an integer n > 1 such that {ui • • • un \ u\,..., un £ U} — G.

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If dim(G) = 1, then g is Abelian, so G is Abelian. So, if dim(G) = 1, then, by Lemma 4.8.3, p. 95, either G = E or G = E/Z; in either case, we may set k = 1 and let X\ be any nonzero element of g. If G is semisimple, then, choose C, A and B as in Lemma 4.9.8, p. 100 and use the fact that the result is true for compact and Abelian groups. An extension of connected Lie groups satisfying the lemma is again a group satisfying the lemma. Any connected Lie group is an extension of a connected semisimple Lie group and a connected solvable Lie group. Any connected solvable Lie group is an extension of connected Abelian Lie groups. By Lemma 4.8.3, p. 95, we see that any connected Abelian Lie group is an extension of connected one-dimensional Lie groups. The lemma follows. • L E M M A 4.9.10 Let G be a Lie group and let X € g. Then exactly one of the following is true: (1) 1i-> exp(tX) : E —• G is a proper, injective homomorphism; or (2) {exp(tX)}t£R is precompact in G. Proof. Define / : 1 -> G by f{t) = exp(iX). Assume /(R) is not precompact in G. We wish to show that / : R —> G is proper and injective. Replacing G by the closure of / ( E ) , we may assume that /(R) is dense in G. Then G is a connected Abelian Lie group. Let T := E / Z . By Lemma 4.8.3, p. 95, choose integers m , n > 0 such that G is Lie group isomorphic to T m x R". We may assume that G = Tm x E n . Since /(E) is not precompact in G, it follows that G is noncompact. So, as G = T m x R n , we conclude that n > 1. Let p : G -> E" be projection onto the second factor. By Lemma 4.8.4, p. 95, p o f is a linear transformation. Them p o / : E —>• Era is linear and has dense image, and is therefore surjective. It follows that n < 1. So, as n > 1, we get n = 1. Then p o / : R -» W1 is an isomorphism of vector spaces, and is therefore proper. It follows that / : E -> G is proper. By Lemma 2.7.7, p. 29, ker(/) is compact. As E admits no nontrivial compact subgroup, ker(/) is trivial, so / : E -> G is injective. • L E M M A 4.9.11 Let Go be an Abelian connected Lie subgroup of a Lie group G. Assume that Go is Lie group isomorphic to the additive group E and assume that Go is not closed in G. Let Go denote the closure of Go in G. Then, for some integer n > 2, Go is Lie group isomorphic to T m .

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Proof.

Because Go is nontrivial and is not closed in G, we conclude that Then, by Lemma 4.4.5, p. 83, 0 < dim(G 0 ) < dim (Go"). Then dim(Go) > 2. Since Go — E, we see that Go is connected and Abelian; it follows that Go is connected and Abelian. So, by Lemma 4.8.3, p. 95, it suffices to show that Go is compact. Fix X e flo\{0}. Since G 0 = E, we see that {exp(tX)}t€R = G0. So, since Go is not closed in G, it follows, from Lemma 2.7.7, p. 29, that the map t H-^ exp(tX) : R —> G is not a proper homomorphism. Then, by Lemma 4.9.10, p. 101, Go is precompact in G, i.e., Go is compact. • IGSGOCGO".

L E M M A 4.9.12 Let G be a noncompact connected Lie group. Then there exists X G g such that t M- exp(iX) : E —>• G is a proper homomorphism. Proof. For all W e g, let Sw •= {exp(tW)}t€R C G. By Lemma 4.9.9, p. 100, choose an integer k > 1 and X\,...,Xj. 6 Q such that we have G = Sxx • • • Sxk- As G is noncompact, choose an integer i G [1,/J] such that Sxi is not precompact in G. Let X :— Xi. Then {exp(tX)}teu is not precompact in G. By Lemma 4.9.10, p. 101, we are done. • Let g be a Lie algebra. A linear map / : g —¥ g is a derivation if, for all X,Y eg, we have f([X,Y]) = [f(X),Y] + [X,f(Y)]. The Jacobi identity asserts, for all X £ g, that a d X : g —> g is a derivation. For any Lie algebra g, let Oer(g) denote the Lie subalgebra of gl(g) consisting of all derivations of 0. Then adB(fj) C 5et(jj). L E M M A 4.9.13 Let g be a Lie algebra, let \) := Der(fl) C gl(g) and let go := ad 0 (0) Q h. Then flo is an ideal ofi) and, for all T 6 c^(0o), we have T(0) £ 3(0). Proof. For all X E 0, for all T G f), we have [T,adB(X)] = ads(T(X)). Thus [h,g0] = [l),ad0(0)] C ad 8 (0) = 0o- Then 0 o is an ideal of h. For all T e c„(0O) a n d X G 0, we get ad B (T(X)) = [T,ad 0 (X)] G [cb(flo),flo] = {0}, soT(X)Gker(ad0)=3(0). D The next result is the real Jordan analogue of Corollary 3.1.14, p. 160 of [Va74]. L E M M A 4.9.14 Let g be a Lie algebra. Then Dcr(0) is real Jordan closed in End(0).

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Proof. Let 6 £ Det(g) and let D,E,N £ End(g). Assume that the equation 5 = D + E + N is a real Jordan decomposition on g. We wish to show that D,E,N S5et(g). Let V be the vector space of bilinear maps g x g —> g. Let uo € V be defined by t ) 0 ( X , r ) = [X,Y]. For all T e End(g), let T £ End(V) be defined by (T'(v))(X,Y) = (T(v(X,Y))) - (v(T(X),Y)) (v(X,T(Y))); then T : g -+ g is a derivation iff T"(i>0) = 0. Moreover, for all T € End(g), if T : g -t g is real diagonalizable (resp. elliptic, resp. nilpotent), then T' : V —• V is real diagonalizable (resp. elliptic, resp. nilpotent). Moreover, for all T,U £ End(g), we have [T, U]' = [T, U']. Then, as 6 = D + E + N is a real Jordan decomposition on g, it follows that 8' = D' + E' + N' is a real Jordan decomposition on V. As 5 € Oet(g), we get <5'(^o) = 0. So, since D', E', N' are elements of the real linear span of 5', (S1)2, (S')3,..., we see that D'(v0) = E'{v0) = N'(v0) = 0. Then D,E,N £ Oer(g). D The following is Lemma 7.14 of [A99b]. L E M M A 4.9.15 Let g be a Lie algebra and let 5 : g —> g be a derivation. Assume that 6 : g —• g is a complex diagonalizable linear transformation. Let f) be the Lie subalgebra of g generated by 6(g). Then I) is an ideal of g. Proof. Let t := <5_1(0) denote the kernel of 6 : g —> g. For all X £ t, for all Y £ g, we have [5(X),Y] = [0,Y] = 0, so, as S is a derivation, we get 5{[X,Y\) = [6{X),Y] + [X,5(Y)] = [X,5(Y)], so [X,5{Y)] = S([X,Y}) £ (5(g). Then, for all X£t, (adX)(<5(g)) C <5(g). Let i := 5(g). Then, for all X 6 6, we have (adX)i C i. So, since I) is generated by i, we conclude, for all X £ t, that (adJf)f) C h. That is, 6 C n B (h). We also have i C f ) C n , ( i ) ) . Because S : g -> g is complex diagonalizable, we get g = (ker((5))+(<5(g)). Then g = 6 + i C n B (h). That is, f) is an ideal of g. • L E M M A 4.9.16 Let G be a path-connected topological group and let A be a normal, totally path-disconnected subgroup of G. Let w : G —• G/A be the canonical homomorphism. Then (1) A C Z(G); and (2) Z(G/A)=v(Z(G)). Proof.

Proof of (1): Let S £ A, g £ G. We wish to show that dg = gd.

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Choose a continuous path c : R —> G such that c(0) = \Q and such that c(l) = g. Define 7 : R -> G by 7(4) = [(c(*)) -1 ] •<$• [c(t)]. Since A is normal in G, we have 7(E) C A. So, since A is totally path-disconnected, we have that 7 is constant. So, since 7(0) = 8, we get 7(1) = 8. Then g~l5g = 8, so 8g — g8. End of proof of (1). Proof of (2): As n(Z(G)) C Z(TT(G)) = Z(G/A), we wish to prove that Z(G/A) C TT(Z(G)). Fix 3 E G, and assume that n(g) G Z(G/A). We wish to show that ir(g) & TT(Z(G)). It suffices to show that g € Z(G). Fix x £ G. We wish to show that ya; = xg, i.e., that x~lgx = g. Let c : R —>• G be a continuous path such that c(0) = 1Q and such that c(l) = a;. Define 7 : R ->• G by j(t) = [(c(i))" 1 ] • g • [c(t)]. Since 7r(p) e Z(G/A), it follows, for all t € M, that ir(j(t)) = n{g). Then, for all t 6 R, we have 7(f) € 7r_1(7r(g)). Since 7r-1(7r(<7)) = gA, we conclude that 7r_1(7r(flr)) is totally path-disconnected. Then 7 is constant. So, since 7(0) = g, we get 7(1) = g, i.e., x~1gx = g. End of proof of (2). • C O R O L L A R Y 4.9.17 Let G be a connected Lie group and assume that Z(G) is discrete. Then G/(Z(G)) is centerfree C O R O L L A R Y 4.9.18 LetG be a path-connected topological group, let H be a topological group and let f : G —> H be a continuous homomorphism. Let A := ker(/) and assume that A is totally path-disconnected in G. Then (1) A C Z(G); and (2) Z(f(G)) = f(Z(G)). Proof. Let n : G -> G/A be the canonical Lie group homomorphism. Let / ' : G/A -> H be an injective homomorphism such that / ' o 7r = / . By Lemma 4.9.16, p. 103, we have A C Z(G) and Z(G/A) = n(Z(G)). Then f{Z{G)) - f'(n(Z(G))) = f'{Z{G/A)). Let H' := f{G). Since the homomorphism / ' : G/A -> H' is bijective, f'(Z(G/A)) = Z(H'). Then f{Z{G)) = Z(ff') = Z(/(G)). D C O R O L L A R Y 4.9.19 Lei go be a Lie algebra. Then there exists a connected Lie group G with compact center such that g is Lie algebra isomorphic to goProof. By Theorem 3.17.8, p. 237 of [Va74], let G' be a connected Lie group such that g' = flo- By Lemma 4.8.3, p. 95, choose a discrete Abelian group r , choose integers m,n > 0 and choose a Lie group isomorphism

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<> / : Z(G') -> T x T m x E n . Let A := ^(T x {0} x Z n ) . Then A is a discrete subgroup of Z(G'), and, therefore, of G. Let G := G ' / A . Then g = g' = g 0 . Let 7T: G' -»• G be the canonical Lie group homomorphism. By (2) of Lemma 4.9.16, p. 103 (with G replaced by G'), we have Z{G) = n(Z(G')), so Z(G) = ( r x T m x E n ) / ( r x {0} x Z") S T m x T n , so Z(G) is compact.

D

L E M M A 4.9.20 Any connected Lie group is homotopy equivalent to a finite CW-complex. Proof. Let G be a connected Lie group. By §1.8, p. 6 of [R72] (or Theorem 6, p. 530 of [149]), G has a compact subgroup K to which it is homotopy equivalent. Since K is a closed subgroup of G, it follows that if is a closed submanifold of G. Then G is homotopy equivalent to a compact manifold. Any compact manifold is homotopy equivalent to a finite CW-complex. (See, for example, Theorems 1.2 and 4.1 in Chapter 6 of [H76], on p. 147 and on p. 166, respectively.) • L E M M A 4.9.21 Let G be a connected Lie group and let D be a discrete subgroup of Z(G). Then D is finitely generated. Proof. Let G be the universal covering group of G and let D be the preimage of D under the universal covering homomorphism G ^> G. Then D is a discrete normal subgroup of G. By (1) of Lemma 4.9.16, p. 103, we see that D C Z(G). By Lemma 4.9.20, p. 105, G/D has the homotopy type of a finite CW-complex. Consequently, ni(G/D) is finitely generated. As D is isomorphic to ni(G/D), D is finitely generated. So, as there is a surjective homomorphism D —> D, D is finitely generated, as well. • L E M M A 4.9.22 If G is a connected Lie group, then [Z(G)]/[Z°(G)] finitely generated.

is

Proof. As [Z(G)]/[Z°(G)] is a discrete central subgroup of G/[Z°(G)], the result follows from Lemma 4.9.21, p. 105. • The following two lemmas are versions of Schur's Lemma.

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L E M M A 4.9.23 Letg be a Lie algebra and let V and W be real g-modules. Assume that V is g-irreducible. Let f : V —> W be a g-equivariant linear map. Let Wo := f(V) and assume that Wo ^ {0}. Then f : V -» Wo is an isomorphism of real g-modules. Moreover, if W is g-irreducible, then f :V —> W is an isomorphism of real g-modules. Proof. Since f(V) = Wo, we need only show that / : V —> W is injective. Since Wo ^ {0}, it follows that ker(/) C V. Moreover, by g-equi variance, ker(/) is a g-invariant vector subspace of V. Then, by g-irreducibility of V, we see that ker(/) = {0}, so / : V -»• W is injective. Then / : V ->• Wo is an isomorphism of real g-modules. Now assume that W is g-irreducible. Then, as Wo ^ {0}, we see that Wo = W. Then / : V —¥ W is an isomorphism of real g-modules. • L E M M A 4.9.24 Let g be a Lie algebra and let V be a real g-module. Let J :V —> V be a g-invariant complex structure on V. Let T € flI(V, J ) . Assume that no proper nonzero vector subspace ofV is both g-invariant and J-invariant. Assume that T : V —> V is g-equivariant. Let I : V -> V be the identity transformation. Then there exist a, b £ E such that T — aI-\-bJ. Proof. Let V be the complex g-module defined by VR = V, and with complex structure denned by y/^lv = Jv. Then V is an irreducible complex g-module and T : V —• V is a g-equivariant complex linear transformation. By Schur's Lemma (see, e.g., Lemma 6.1, p. 26 of [Hu72]), choose c 6 C such that T = cl. Choose a, b £ E such that c = a + byf^l. Then T = al + by/^11 = al + bJ. • L E M M A 4.9.25 Let g be a Lie algebra and let V be an irreducible real gmodule. Let W be a g-submodule ofV@V. Assume that {0} ^ W ^ V@V. Then W is isomorphic to V, as a g-module. Proof. Define p, q : V ® V -> V by p(v, v') = v and q(v, v') = v'. Either p(W) ? {0} or q(W) ? {0}. We will here assume that p(W) ^ {0}; the proof in the other case is similar. By irreducibility of V, we see that p(W) = V, i.e., that the map p\W : W -> V is surjective. It therefore suffices to show that p\W is injective. Assume, for a contradiction, that ker(p|W) ^ {0}. We have {0} ^ kei(p\W) C ker(p) = {0} ®V. Then, because {0} ®V is g-irreducible, ker(p|W) = {0}©V. Then {0}®V = ker(p\W) C W. Define

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t : V -» V © V by i(w) = (w,0). Let V0 := r^W). As { 0 } ® F C W, it follows that W = V0 © V. Then {0} ^ p{W) = V0, so, by irreducibility of V, we have V0 = V. Then W = V © V, contradiction. • L E M M A 4.9.26 Let g be a Lie algebra and let V be an irreducible real g-module. Then V has nonscalar g-intertwining iffV admits a Q-invariant complex structure. Proof. Let I : V —> V be the identity map. Let T denote the set of g-equivariant linear transformations V —¥ V. Let S :— MI. If there is a g-invariant complex structure J on V, then J £ F\S, so T C S, proving "if". For "only if", assume that TC <S. We wish to show, for some J 6 J7, that J 2 = -I. Choose T € F\S. Then T : V -> V is g-equivariant, but T £ RZ. Let V := V <8>R C denote the complexification of V. Let Tc

: V ->• V be

the complexification of T. Let I : V —• V be the identity map. Then Tc £ CI and T : V -> V is g-equivariant. Then, by Schur's Lemma (see, e.g., Lemma 6.1, p. 26 of [Hu72]), we see that V is not irreducible. Let W be a nonzero proper irreducible complex g-submodule of V. The real g-modules VR and V © V are isomorphic. So, since {0} ^ Wfa C VR, we conclude, from Lemma 4.9.25, p. 106, that Wfo = V. Let J : V -> V correspond to «; i-> \f^lw : W —> W under a g-equivariant vector space isomorphism WR -> V. Then J 2 = -I and J e T. D Our interest here is in modules over Lie groups and Lie algebras; however, we point out that Lemma 4.9.26, p. 107 is true for any finite-dimensional real module V over an abstract group G, with much the same proof. An lcsc topological group G is compactly generated if there is a compact subset of G which generates G, i.e., which is not contained in any proper subgroup of G. L E M M A 4.9.27 Let G be a Lie group. Then G is compactly generated iff G/G° is finitely generated. Proof. A discrete group is compactly generated iff it is finitely generated. If G is compactly generated, then G/G° is compactly generated and discrete, hence finitely generated. Now assume that G/G° is finitely generated. We wish to show that G is compactly generated. Let 7r : G —> G/G° be the canonical homomorphism. Let F be a finite subset of G such that 1Q E F and such that ir(F) generates G/G°. Let U

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be a precompact open neighborhood of 1Q in G°. Then U generates G°. Let C be the closure of U and let K — FC. Then K is compact. Let Go be the subgroup of G generated by K. We wish to show that Go = G. Let H := 7r(G0). Since 1G € F, we see that C C K, so U C K, so G° C G 0 . Then G 0 = T T - 1 ^ ) . AS X generates G 0 , it follows that TT(K) generates H. As 7r(iir) = ^(i*1), we see that G/G° is generated by n(K). Then tf = G/G°. Then G 0 = T r " 1 ^ ) = T T - ^ G / G 0 ) = G. D L E M M A 4.9.28 Lei A fee an Abelian Lie group and assume that A is compactly generated and noncompact. Then A contains an infinite discrete cyclic subgroup. Proof. By Lemma 4.8.3, p. 95, choose a discrete Abelian group T and choose integers m, n > 0 such that A is Lie group isomorphic to T x T m x W1. If n > 1, then K™ contains an infinite discrete cyclic subgroup, and we are done. We may therefore assume that n = 0. Since A is compactly generated, it follows, from Lemma 4.9.27, p. 107, that r is finitely generated. Since A is noncompact, since n = 0 and since T m is compact, it follows that T is infinite. Any infinite, finitely generated Abelian group contains an infinite cyclic subgroup, so T contains an infinite cyclic subgroup. • L E M M A 4.9.29 Let G be a connected Lie group and let Go be a connected centerfree Lie group. Assume that g is Lie algebra isomorphic to 0o- Then there is a covering homomorphism G —> GoProof. Let G be the universal covering group of G and let Go be the universal covering group of Go- Let n : G —> G and 7To : Go —> Go be the universal covering homomorphisms. Since g = go, it follows that G is Lie group isomorphic to Go- Let / : G -> Go be a Lie group isomorphism. Let p := 7T0 o / : G -> Go. Then p(Z(G)) C Z{G0) = { W , so Z(G) C ker(p). By (1) of Lemma 4.9.16, p. 103, ker(7r) C Z{G) Then ker(Tr) C ker(p). It follows, from Corollary 4.9.4, p. 97, that there is a Lie group homomorphism / : G —> Go such that p—fo-K. Since 7To : Go —> Go is a covering map and since / : G —> Go is an isomorphism, we conclude that p : G -> Go is a covering map. Then ker(p) is countable. We have ker(/) C 7r(ker(p)). Then ker(/) is a countable closed subgroup of G, so ker(/) is discrete in G.

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Since TTQ : Go -> Go is surjective and since / : G -» Go is an isomorphism, we conclude that p : G -» Go is surjective. Then f : G -t G0 is surjective. So, since / : G —• Go has discrete kernel, it follows, from Lemma 4.9.5, p. 97, that / : G ->• Go is a covering homomorphism. • L E M M A 4.9.30 Let G be a Lie group, let N be a normal subgroup of G and let N' be a normal connected Lie subgroup ofG. Assume that NC\N' is totally disconnected. Then [N,N'] = {1G}Proof. LetaeN and define / : N' -> G by f(b) = aba^b'1. We wish to show that f(N') = {1 G }. As N and N' are normal subgroups of G, we have [N, W] C N C\ N', so f(N') C NnN'. So, since N' is connected, since NnN' is totally disconnected, since / is continuous, we conclude that / : N' —• G is constant. So, since / ( l j y ) = 1G, we have /(iV') = {1 G }• L E M M A 4.9.31 Let G be a Lie group and let gi be a sequence in G. Let X € g be Kowalsky for Adg(gi). Then a d X : g —¥ g is nilpotent. Proof. Choose a sequence Xi -> 0 in g such that (Ad gi)Xi -> X. Let V := 0. Let Y := a d X e End(V). For all i, let 5* := A d 9 i € GL(V) and let y := adXj € End(V); we then have ad B ((Adc/i)Xj) = BiYiB^1, so BiYiBr1 -> y . Since X* -> 0 in g, it follows that y -> 0 in End(V). The result follows by Lemma 2.4.4, p. 22. • L E M M A 4.9.32 Let V be a vector space and let gi be a sequence in GL(V). Assume that {gi} is not precompact in GL(V). Then either {gi} is not precompact in End(V) or {g^1} is not precompact in End(F). Proof. Assume, for a contradiction, that { oo in GL(V). Passing again to a subsequence, we may assume that gi and g^1 are both convergent in End(y). Let a := limgj and let b := lim^r 1 . Let I : V —> V be the identity transformation. Then ab = \im gtg^~ = I, so a € GL(V). Then gi —t a in GL(V), contradicting the fact that gi —> oo in GL(V). D L E M M A 4.9.33 Let V bea vector space and letgi be a sequence in End(V) Assume that {gi}i is not precompact in End(V). Then there exists v E V such that {gi(v)}i is not precompact in V.

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Proof. Let d := dim(V). We may assume that V = Rdxl and that gi is a sequence in Rdxd such that {gi}, is not precompact in Rdxd. We wish to show, for some v € Rdxl, that {gtv}i is not precompact in Rdxl. For all integers i 6 [1, d], let e% := e\ ', so that e i , . . . , e^ is the standard basis of R d x l . For all g € Rdxd, for all integers j 6 [l,
True by Lemma 4.9.32, p. 109 and Lemma 4.9.33, p. 109.

•

L E M M A 4.9.35 Let V be a vector space. Let gi be a sequence in SL(V). Assume that {gi}i is not precompact in SL(y). Then, for some v £ V, we have that {gi(v)}i is not precompact in V. Proof. Since SL(V) is closed in End(V) and since {gt}i is not precompact in SL(V), it follows that {gt}i is not precompact in End(V). The result then follows from Lemma 4.9.33, p. 109. • L E M M A 4.9.36 Let G be a connected Lie group, let G be the universal covering group of G and assume that Z(G) is connected. Then Z(G) is connected. Moreover, G is simply connected iff Z{G) is simply connected. Proof. Let n : G -> G be the universal covering homomorphism. Let A := ker(Tr). By (1) and (2) of Corollary 4.9.18, p. 104, we see that A C Z(G) and that Z(G) = TT(Z(G)). SO, since Z{G) is connected, it follows that Z(G) is connected, as well. By Theorem 3.18.2, p. 238 of [Va74], if G is simply connected, then any normal connected Lie subgroup of G is simply connected, so, in particular, Z(G) is simply connected. Now assume that Z(G) is simply connected. We wish to show that A = {1^}. Letp := n\(Z(G)). Thenp : Z(G) -»• Z(G) is surjective and ker(p) = A. So, as A is a discrete subgroup of Z(G), we see, from Lemma 4.9.5, p. 97,

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that p is a covering homomorphism. So, as Z(G) is simply connected, we see that p : Z(G) -> Z(G) is an isomorphism. Then A = ker(p) = {lg}- • Let X be a topological space and let A be a collection of continuous maps X —> X. Let a, j3 : X —> X be continuous maps. Then we say that a is homotopic to /3 through A if there exists a continuous map i f : I x [ 0 , l ] - > I such that H( •, 0) = a, such that H( •, 1) = /?, such that, for all t G [0,1], we have H{ •, t) € A. L E M M A 4.9.37 Let G be a connected Lie group and let T be a closed, normal subgroup of G. Let d :— dim(T). Assume that T is Lie group isomorphic to Rd/Zd. Then T C Z°(G). Proof. Since T is connected, it suffices to show that T C Z(G). Fix g E G. We wish to show, for all t G T, that gtg~x = t. Define <j) : T -> T by (j)(t) — gtg~l. Let / : T -> T be the identity map defined by I(t) = t. We wish to show that <j> = I. The map : T -> T is an automorphism of the Lie group T. Moreover, by path-connectedness of G, <j> is homotopic to I through Aut(T). Since T ^ E d / Z d , let F : Rd ->• T be a covering homomorphism such that ker(F) = Z d . Let ,4 be the set of all additive Lie group automorphisms a : Rd -*• E d such that a{1d) - Zd. Let «4o be the set of all additive Lie group automorphisms a : Rd -¥ Rd such that, for all z € Z d , we have a(z) — z. Let / : E d ->• E d be the identity map defined by I(r) = r. By covering space theory, let tp : Rd -+ Rd be the (unique) continuous map satisfying both <j> o F = F ° ip and V(0) = 0. Since cj> : T —> T is a Lie group automorphism, it follows (from covering space theory) that \j) € A. Moreover, since (j> is homotopic to J through Aut(T), it follows (from covering space theory) that ip is homotopic to I through A. So, since Zd is discrete and since I G Ao, we conclude that ip £ AoBy Lemma 4.8.4, p. 95, we see that tp : Rd ->• Rd is a linear map. The only linear map in Ao is / . Then ip = I, so <j)°F — Fotp = FoI = F. Then, by surjectivity of F : Rd -¥ T, we conclude that cj> = I. O If V is a vector space and if W, W are vector subspaces of V, then we say that W is a vector space c o m p l e m e n t in V to W if both W(l W = {0} and W + W = V. L E M M A 4.9.38 Leifl 6e a Lie algebra. Theng admits no nonzero Abelian direct summand iff i(g) C [g,g\.

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Proof. Let 3 := 3(0) and let g' := [0, g\. If g has a nonzero Abelian direct summand, then there are ideals 0 and b of g such that o is Abelian, such that o ^ {0}, such that oflb = {0} and such that g = a+ b; then a C 3 and 0 2 fl'i so 3 2 fl'i proving "if". Assume that g admits no nonzero Abelian direct summand. We wish to show that 3 C g'. Let 30 := g' fl 3. Let 6 be a vector space complement in 3 to 30. Then 1 + 30 = 3 and 6 n 30 = {0}. Since 6 C 3, we conclude that £ = £ 0 3 . Then

en f l ' = !n3n0' = en( 0 'n3) = en3o = {o}. Let c be a vector space complement in g to 6 such that g' C c. Then c (~l t = {0} and c + 1 — g. Since [g, c] C g' C c, we see that c is an ideal of 0. Since t C 3 = 3(0), we see that [0,6] = {0} C 6, so 6 is an ideal of 0. Then (X, Y) \-t X + Y : c © 6 —• 0 is a Lie algebra isomorphism, so 0 is Lie algebra isomorphic to c © t. In particular, 61 0. Since 6 is Abelian, and since 0 admits no nonzero Abelian direct summands, we conclude that 6 = {0}. Then 3 = 6 + 30 = 3o = 0' H 3 C 0'. • L E M M A 4.9.39 Let fo be a semisimple Lie subalgebra of a Lie algebra g. Then there exists a semisimple Levi factor I of g such that to C (. Proof.

4.10

This follows from Corollary 3.14.3 of [Va74].

•

Generalities on semisimple groups and algebras

Let 0 be a Lie algebra. For all 1 , 7 6 9, let fxY '• a d X o adY : g ->• 0, so that, for all Z £ g, we have fXy{Z) = [X,[Y,Z]]. The Killing form of 0 is the ad-invariant symmetric bilinear form n : g x g -> R defined by K(X, Y) = tr(/jcy). Then K, is (ad0)-invariant. Let 0 be a semisimple Lie algebra, let 0 be a maximal real split torus of g and let $ C o* be the set of roots of 0 on 0. Let K be the Killing form on 0. Then K|(O X O) is positive definite. Define $ : 0 —> o* by ($(X))(y) = K(X,Y). Then $ : a -» a* is a vector space isomorphism. Define F : a* x a* -> K by F{a,P) = K ^ - ^ a ) , * - 1 ^ ) ) . Then the map F : a* x a* -> K is a positive definite symmetric bilinear form and $ is a root system in (a*,F); see Example (ii) before Lemma X.3.2, p. 456 of [Hu72]. We will denote F by RKF( 0 ,o) € SBF(a*). (Here, "RKF" stands for "restricted Killing form".) Note that $ need not be a reduced

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root system (see, e.g., Exercise VII.5, p. 347 of [He78]). The semisimple Lie algebra 0 is simple iff $ is irreducible. L E M M A 4.10.1 Let g be a Lie algebra. Then g is semisimple iff its Killing form is nondegenerate. Proof. This is Theorem 3.9.2, p. 210 of [Va74], although what we call the Killing form is, in [Va74], called the "Cartan-Killing form". • L E M M A 4.10.2 If g is a semisimple ideal of a Lie algebra h, then there exists an ideal g' of f) such that g + g' = h, such that 0 H g' = {0} and such that [g,g'] = {0}. Proof. Let n denote the Killing form of h, so that K|(0 X 0) is the Killing form of 0, which, by Lemma 4.10.1, p. 113, is nondegenerate. Let 0' := {X 6 I) | K(X,S) = {0}}. For all X e h, (adX)0 C 0, so, by (ad h)-invariance of K, we get (adX)g' C 0'. That is, g' is an ideal of h. Since /c|(g x g) is nondegenerate, we get gflg' — {0} and 0 + 0' = h. Then, as 0 and g' are both ideals in h, we have [0, g'] C 0 D 0' = {0}. • C O R O L L A R Y 4.10.3 Let I) be a semisimple Lie algebra. Then there exists an integer k > 1 and there exist simple Lie algebras l\,..., Ifc such that \] is Lie algebra isomorphic 10 Ii ©•••©[*. Proof. The proof is by induction on dim(h). If h is simple, then the result follows (with fc = 1 and [1 = h), so let 0 be a proper nonzero ideal of h. Choose 0' as in Lemma 4.10.2, p. 113. Then h is Lie algebra isomorphic to 0 © g'. By induction, both 0 and 0' are isomorphic to direct sums of simple Lie algebras, so f) is, as well. • For any nonAbelian Lie algebra 0, 3(0) is a proper ideal of 0. So, for any simple Lie algebra s, we have 3(0) = {0}. Then, by Corollary 4.10.3, p. 113, we conclude, for any semisimple Lie algebra I, that 3(Q = {0}. Consequently, for any semisimple Lie group G, Z(G) is discrete in G. C O R O L L A R Y 4.10.4 Let I) be a semisimple Lie algebra. Then there exists an integer k > 1 and there exist simple ideals h,- • • ,h °fty such that the addition map (Xi,... ,Xk) •->• X\ + • • • + Xk : (1 © • • • © h -> h is an isomorphism of Lie algebras.

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Proof. By Corollary 4.10.3, p. 113, we may assume that f) is a direct sum of simple Lie algebras. Choose an integer k > 1 and simple Lie algebras S i , . . . ,Sk such that I) = Si0- • -©5*. For all integers j G [1,fc],let LJ : Sj -»• f) be the jth coordinate inclusion and let lj := tj(Sj). • COROLLARY 4.10.5 Let g be a semisimple Lie subalgebra of a Lie algebra h. Then nt,(g) = 0 + (Cf,(fl)). Proof. Replacing h by ni,(0), we may assume that g is an ideal of h, and we wish to show that I) = g + (CI,(JJ)). Choose g' as in Lemma 4.10.2, p. 113. Then, as [0,0'] = {0}, we get fl'Cc6(ji). Thenh = 0 + 0 ' C 0 + (c„(0))Ch, soh = 0 + (cft(0)). • L E M M A 4.10.6 Let g be a semisimple Lie algebra. Then [g,g] = g. Proof. By Corollary 4.10.3, p. 113, we may assume that 0 is simple. Then [0,0] is a nonzero ideal of 0, so, by simplicity, we get [0,0] — 0. • Let 0 be a Lie algebra and let V be a real 0-module. We say that V is completely reducible if, for every 0-invariant vector subspace W of V, there exists a ©-invariant vector subspace W of V such that W + W = V and such that W D W = {0}. The next result is sometimes called Weyl's Theorem. L E M M A 4.10.7 Let g be a semisimple Lie algebra. Then any real gmodule is completely reducible. Proof. See Theorem 3.13.1, p. 222 of [Va74] or Theorem 8 of §111.7 on p. 79 of [J62]. • L E M M A 4.10.8 Let g be a simple Lie algebra. Then there exist a vector space V and a faithful irreducible representation p : g —> gl(V). Proof.

Let V := g and let p :— ad : 0 -»

fll(V).

L E M M A 4.10.9 Let g be a semisimple Lie algebra. Then there exist a vector space V and a faithful representation p : g -> fll(V). Proof. By Corollary 4.10.3, p. 113, we may assume that 0 is simple. The result then follows from Lemma 4.10.8, p. 114. D

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In fact, Ado's Theorem asserts that Lemma 4.10.9, p. 114 is true for any Lie algebra not just semisimple Lie algebras. (See Theorem 3.17.7, p. 237 of [Va74].) Similarly, Lemma 4.10.10, p. 115 below can be generalized. (See Theorem 3.17.8, p. 237 of [Va74].) L E M M A 4.10.10 If G is a connected semisimple Lie group, then there exists an integer d > 1 and a connected Lie subgroup G' of GLd(R) such that G is locally isomorphic to G'. Proof. Choose V and p as in Lemma 4.10.9, p. 114. Let d := dim(V). Fixing a basis of V, we may assume that V = Rd. Let G' be the connected Lie subgroup of GLd(R) corresponding to the Lie subalgebra p(g) of gld(R). Then g is Lie algebra isomorphic to g'. D For any Lie algebra g, let Aut(g) denote the closed subgroup of GL(g) consisting of all Lie algebra automorphisms of g; let Aut°(g) := (Aut(g)) 0 . Then Aut°(g), being closed in Aut(g), is closed in GL(g). Let g be a semisimple Lie algebra and let / : g —• g be the identity transformation. Let 6 : g —• g be a map. We say that 6 is an involution of g if both 6 G Aut(g) and 0 o 0 = I. Note that, in our terminology, the identity I: g —r g is an involution. Let 6 be an involution of a Lie algebra g. Let K : g x g -> M. be the Killing form of g. Define B : g x g -> R by B(X,Y) = K{X,6(Y)). Then, for all X, Y € g, (ad B (X)) o (ad 9 (0(y))) = 0o (a,dB(6(X))) ° (ad B (F)) ofl" 1 , so, taking the trace of both sides, we get B(X, Y) = B(Y,X). Then we have B G SBF(g). We say that 6 is a Cartan involution of g if B is negative definite. By the paragraph preceding Theorem III.7.2, p. 183 of [He78], together with Proposition 111.7.4(h), p. 184 of [He78], any semisimple Lie algebra admits a Cartan involution. L E M M A 4.10.11 Let g be a semisimple Lie algebra and let 0 : g —> g be a Cartan involution. Let X G g. Then both of the following are true: (1) If 6{X) = —X, then a d X : g -> g is real diagonalizable. (2) If 6(X) — X, then sAX : g -> g is elliptic. Proof. Let K be the Killing form of g. Define a symmetric bilinear form B : g x g -> K by B(Y,Z) = K(Y,0(Z)). Then, by definition of "Cartan involution", B is negative definite. If 6{X) — —X, then, for all Y, Z G g,

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we have B((a,dX)Y,Z) = B(Y, (aAX)Z); that is, aAX : g -> g is symmetric with respect to B. If 6(X) = X, then, for all Y, Z E g, we have B((adX)Y, Z) = -B(Y, (adX)Z); that is, adX : g -»• g is anti-symmetric with respect to B. By the Spectral Theorem, a linear transformation which is symmetric (resp. anti-symmetric) with respect to a definite symmetric bilinear form is real diagonalizable (resp. elliptic). The results follow. • Let n > 2 be an integer and let go := sl n (R). Then the map 6Q : go —> 0o defined by 6Q{X) = ~X* is a Cartan involution. Note that, in this case, {X G go | 0o(X) = —X} is the subset of go consisting of symmetric matrices. For this reason we will make the following definition: If g is a semisimple Lie algebra and if S C g, then S is s y m m e t r i z a b l e if there is a Cartan involution 6 : g -> g such that, for all X € S, we have 9(X) = —X. Let g be a semisimple Lie algebra. A Lie subalgebra h of g is a complex split torus of g if ad B (h) is simultaneously complex diagonalizable; equivalently, both [h, h] = {0} and, for every X € h, ad X : g ->• g is complex diagonalizable. A Lie subalgebra h of g is a maximal complex split torus of g if it is maximal (with respect to inclusion) among complex split tori of g. If g is a Lie algebra, then g c denotes the complexification of g; it is a complex Lie algebra. Let g be a semisimple Lie algebra and let h be a Lie subalgebra of g. Following [He78] (see the definition in the second paragraph of §IX.4 on p. 418), we say that h is a Cartan subalgebra of g if both • h c is a maximal Abelian subalgebra of g c ; and • for all X E h c , a d X : g c - • g c is diagonalizable over C. We leave it as an exercise to the reader to verify that h is a Cartan subalgebra of g iff h is a maximal complex split torus of g. L E M M A 4.10.12 Let g be a semisimple Lie algebra and let a be a vector subspace of g. Then a is a real split torus of g iff [a, a] — {0} and a is symmetrizable. Proof. Proof of "only if": By the definition of "real split torus", we see that o is Abelian. It therefore suffices to show that o is symmetrizable. Since a is a real split torus, o is a complex split torus. Let h be a maximal complex split torus of g such that o C h. From Corollary IX.4.2,

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p. 419 of [He78], we know that f) is invariant under some Cartan involution 6 of fl. Let £ := {X G g 16{X) = X} and let p := {X G g 16{X) = -X}. We wish to show, that o C p. Let X G a. We wish to show that X G p. Since i) is ^-invariant, we have f) = (hnp) + (hnt). Choose D G hnp and £ G h n ! such that X = D + E. By Lemma 4.10.11, p. 115, ad£> : g -»• g is real diagonalizable and ad.E : g -> g is elliptic. As a d X : g -» g and ad£) : g —> g are both real diagonalizable, and as [X, D] G [h, h] = {0}, we see that ad(X — D) : g —> g is real diagonalizable. However, X — D = E and a.d.E : g ->• g is elliptic. We conclude that adg(E) = 0. So, since 3(g) = {0}, we get E = 0. Then X = I » G h n p C p . End of proof of "only if". Proof of "if": Choose a Cartan involution 6 : g —> g such that, for all X G a, we have 9{X) = -X. As a is Abelian, it follows that adB(o) is an Abelian Lie subalgebra of gl(o). Moreover, by (1) of Lemma 4.10.11, p. 115, we see, for all X G 0, that a d X : g —> g is real diagonalizable. A commuting collection of real diagonalizable linear transformations is simultaneously real diagonalizable, so o is a real split torus. End of proof of "if". • Let G := SL2(IR), let 1) be the set of diagonal matrices in g = s^CK) and let h' := so(2) C g. Then f) and h' are both maximal complex split tori in g; however, for all g G G, we have (Adg)f) ^ h'. On the other hand, note the complexifications of F) and h' are conjugate by an element of SLa(C). The possibility of nonconjugate maximal tori does not occur for maximal (complex) split tori in complex semisimple Lie algebras. Moreover, it does not occur for maximal real split tori in (real) semisimple Lie algebras: For any Lie algebra g, let Inn(g) denote the connected Lie subgroup of GL(g) group corresponding to the Lie subalgebra ad 8 (g) of gl(g). Then Inn(g) is generated by {exp(ad B (X)) | X G g}. For any connected Lie group G, we have Ad B (G) = Inn(g). In particular, we see that, if G and G' are connected Lie groups and if g = g', then Ad B (G) = Ad g /(G'). In other words, the Adjoint image in GL(g) of a connected Lie group G is completely determined by its Lie algebra g. Consequently, the main content of the next lemma is that if G is a connected semisimple Lie group then any two maximal real split tori of g are conjugate by an element of Ad fl (G). L E M M A 4.10.13 Let g be a semisimple Lie algbra and let 0 and b both be maximal real split tori of g. Then there exists f G Inn(g) such that f(a) = b.

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Proof. For every Cartan involution 6 of g, let 6# := {X G g 16(X) = X} and let pe := {X € 0 | 0(X) = -X}. By Lemma 4.10.12, p. 116 choose Cartan involutions (j) and xj) of g such that a is a maximal Abelian vector subspace of p^, such that b is a maximal Abelian vector subspace of p^. By Theorem III.7.2, p. 183 of [He78], we may assume that <j> = ip. The result then follows from Lemma V.6.3 (ii), p. 247 of [He78]. D L E M M A 4.10.14 Let I be a Lie algebra, let k > 1 be an integer and let hi • • • i h be simple Lie subalgebras of I. Assume that (Xu...,Xk)i->X1

+ --- +

Xk:l1®---®lk-*l

is an isomorphism of Lie algebras. Let t C I. Then 6 is an ideal of I iff there exists a subset S C { 1 , . . . , k} such that t = > J Ij. ies Proof. The "if" part follows from the fact that a sum of ideals is an ideal. It remains to prove "only if". Let ip : h © • • • © h —> I be the Lie algebra isomorphism defined by tp(Xi,..., Xk) = Xi H + Xk- For all integers i € [1, k], let pt : I -> k be -1 the composition of ip with the ith coordinate projection. Let S be the collection of integers i G [l,k] such that k C 6. Let S' be the set of integers i G [1, Jfe] such that Pi{t) ^ {0}. Then S C S' and ^ Ij C t C ^ T [;. We wish to show that S = S'. Let i £ S'. We wish to show that i £ S. We have Pi (6) = {0} and we wish to prove that [; C 6. Claim 1: [k,i\ C [j. Proof of Claim 1: For every integer j G [l,/c]\{i}, we have pj([li, I]) = \pj(U),Pj(l)] = [{0}, I,] = {0}. The result follows. End of proof of Claim 1. Choose X e t such that Pi(X) ^ 0. Since i(U) = {0}, choose Y G U such that \pi(X),Y] ^ 0. By Claim 1, we get \pi{X),Y] G k. Since Y G [», we have pt(Y) = Y. We have [X,Y] G [I, Ij] C Ii, so [X,Y] =Pi([X,Y]). Then [X,Y] = \pi{X),Pi{Y)] = b i W , > 1 e U\{0}. Moreover, [X, Y] G [6, [] C {. Then t n I* ^ {0}. So, as t D l* is an ideal of I; and as [j is simple, we get 6 n k — k- Then U = t H U C t, so i £ S. D COROLLARY 4.10.15 If I is a semisimple Lie algebra and if 6 is a nonzero ideal of I, then I, as a Lie algebra, is semisimple. Proof.

True by Corollary 4.10.4, p. 113 and Lemma 4.10.14, p. 118.

•

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119

COROLLARY 4.10.16 Let I be a semisimple Lie algebra, Let f) be a Lie algebra and let f : I —> f) be a nonzero Lie algebra homomorphism. Then /([) is a semisimple Lie subalgebra of\). Proof.

True by Corollary 4.10.4, p. 113 and Lemma 4.10.14, p. 118.

•

COROLLARY 4.10.17 Let I be a semisimple Lie algebra. Let t be an ideal of I. Let to be an ideal of 6. Then to is an ideal of I. Proof. By Corollary 4.10.4, p. 113, choose an integer k > 1 and simple ideals ii,.. .,lk in I such that (Xu...,Xk)

h->

Xx + --- + Xk

:

d © • • • © lk

->

[

is a Lie algebra isomorphism. By Lemma 4.10.14, p. 118, choose a subset S C {1,...,A;} such that 6 = ^ U- By Lemma 4.10.14, p. 118 again, tes choose S0 C S such that to = Y^ U- For all i € So, we know that k is an ideal of I. It follows that to is an ideal of I.

•

L E M M A 4.10.18 Letg andt) be Lie algebras. Let f : g —> () be a Lie algebra homomorphism. Lett be the solvable radical of g. Let I be a semisimple Levi factor of g. Let Y 6 I and Z € t. Assume that f(Y + Z) — 0. Then both f{Y) = 0 and f(Z) = 0. Proof. Since I is semisimple or zero, it follows, from Corollary 4.10.16, p. 119, that /(I) is semisimple or zero. So, since ( / ( 0 ) n ( / ( r ) ) is a solvable ideal of /([), we see that (/([)) n (/(t)) = {0}. We have f(Y) e /(I) and f(Y) = f(-Z) G /(c). Then f(Y) € (/([)) n (/(t)) = {0}. Consequently, f(Z) = (f(Y + Z)) - (f(Y)) = 0 . • L E M M A 4.10.19 Let g be a Lie algebra, let r be the solvable radical of g and let I be a semisimple Levi factor of g. Let t be an ideal of g. Then tf)t is the solvable radical of the Lie algebra t and t fl I is a semisimple Levi factor of the Lie algebra t. Proof. Let t' := t D r and let [' := t D [. Then t' is a solvable ideal of t. Since I' is an ideal of I, by Corollary 4.10.15, p. 118, V is semisimple or zero. So, by Lemma 4.9.1, p. 96, it suffices to show that r' + V = t. We have v' + l'Ct + t = t. Fix X e t. We wish to show that X <E r' + V. Choose

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Y £ t and Z e I such that X = Y + Z. We wish to show that Y ex' and Z &{'. Let f) := g/6 and let / : g -» h be the canonical homomorphism. We have X e 6, so / ( X ) = 0, so by Lemma 4.10.18, p. 119, we get f(Y) = 0 and f(Z) = 0. Then Y, Z £ I. Then Y € 6 f~l r = 6' and Z £ t D 1 = ['. D L E M M A 4.10.20 LeiG be a connected semisimple Lie group and assume that G is simply connected (resp. centerfree). Then there exist an integer k > 1 and connected, simply connected (resp. centerfree), simple Lie groups Si,..., Sk such that G is Lie group isomorphic to Si x • • • x Sk • Proof. Case 1: G is simply connected. Proof in Case 1: By Corollary 4.10.4, p. 113, choose an integer k > 1 and simple ideals Si,.. .Sk of g such that the map (Xi,..., Xk) *-> Xi-\ \-Xk : fii©- • -©Bfc -> Q is a Lie algebra isomorphism. Let Si,..., Sk be the connected Lie subgroups of G corresponding to S i , . . . , Sk, respectively. Then the map / : Si x • • • x Sk —> G defined by / ( s i , . . . ,Sk) = si • • • Sk is a surjective homomorphism with discrete kernel. It is therefore a covering homomorphism. So, since G is simply connected, / is a homeomorphism. Then, by Lemma 4.8.2, p. 94, / is an isomorphism of Lie groups. Since G is simply connected and since G is homeomorphic to Si x • • • x Sk, we conclude, for all integers i 6 [l,k], that Si is simply connected. End of proof in Case 1. Case 2: G is centerfree. Proof in Case 2: Let G be the universal covering group of G. Let n : G —> G be the universal covering map. By (2) of Corollary 4.9.18, p. 104, Z(ir(G)) -j(Z(G)), so, since 7r(G) = G and since Z(G) = { 1 G } , w e conclude that Z(G) C ker(7r). On the other hand, by (1) of Lemma 4.9.16, p. 103, we have ker(7r) C Z(G). Then ker(?r) = Z(G). Then, by Lemma 4.9.2, p. 96, G is Lie group isomorphic to G/(Z(G)). By Case 1, choose an integer k > 1 and connected simple Lie groups Si,..., Sk such that G is Lie group isomorphic to Si x • • • x Sk • For all integers i_€ [1, k], let Zi := Z(§i) and let S{ := Si/Zi. Since Z(G) = Zx x • • • x Zk, G/(Z(G)) is Lie group isomorphic to 5i x • • • x SkFor any integer i G [l,k], 5, is simple, so Zi is discrete; then Si is simple. Since G S Si x • • • x Sk, Z(G) = (Z(Si)) x • • • x (Z(Sk)). So, since G is centerfree, for any integer i £ [l,k], Si is centerfree. For any integer i € [l,k], Si is connected, so Si is connected. End of proof in Case 2. D

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L E M M A 4.10.21 Let G be a connected, simply connected semisimple Lie group and let N be a normal connected Lie subgroup ofG. Then N is closed in G and G/N is simply connected. Proof. By Lemma 4.10.20, p. 120, choose an integer k > 1 and connected, simply connected, simple Lie groups S i , . . . , Sfc such that G is isomorphic to Si x • • • x Sfc. We may assume that G = Si x • • • x Sfc. By Lemma 4.10.14, p. 118, after, if necessary, reordering, we may assume, for some integer j € [l,k], that N = Si x ••• x Sj. Then N is closed in G. Moreover G/N S Sj+i x • • • x Sfc, so G/N is simply connected. • L E M M A 4.10.22 Let G be a connected semisimple Lie group. Let N be a normal subgroup of G. Then (1) ifG is simple and Z(G) — {IG}, then N = {1G} or N = G; (2) N is closed in G; (3) there exists a normal connected Lie subgroup N' of G such that NC\N' is discrete in G and such that: G = NN1, [N, N'] = { I G } , n + n' = 0, n n n' = {0} and [n, n'] = {0}; (4) Z{N) C Z(G); and (5) if N CG, then G/N is semisimple. Proof. Proof of (1): Assume N ^ { I G } - We wish to show that N = G. Fix no 6 iV\{l<3}. Let S denote the subgroup of G generated by {gnog~l \g € G}. Then S C N. Moreover, S is path-connected, so, by Theorem 4.5.2, p. 84, we see that S is a connected Lie subgroup of G. Since S ^ { I G } , we see that s ^ {0}. Since S is a normal Lie subgroup of G, we see that s is an ideal of g. Then, since g is simple, we conclude that s = 0. Consequently, S = G. Then G = S C N C G, so N = G. End of proof of (1). Proof of (2): Let G be the universal covering group of G. Let IT : G —> G be the covering homomorphism. We wish to show that G\N is open in G. By Corollary 2.7.5, p. 28, •K is open, so it suffices to show that TT~1(G\N) is open in G. As T T - 1 ( G \ A ' ' ) = ^ ( ^ ( i V ) ) , it suffices to show that 7r_1(iV) is closed. Replacing G by G and N by TT"1(N), we may assume that G is simply connected. Let N* be the path-component of I G in N. By Theorem 4.5.2, p. 84, N* is a connected Lie subgroup of G. Then, by Lemma 4.10.21, p. 120, we see that N* is a closed subgroup of G and that G/N* is simply connected.

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Replacing G by G/N* and N by N/N*, we may assume that N is totally path-disconnected in G. Then, by Lemma 4.9.16, p. 103, N C Z(G). Then, as Z(G) is discrete in G, we see that N is discrete in G. So, as N is a discrete subgroup of G, N is closed in G. End of proof of (2). Proof of (3): By (2), we know that N is closed in G, and is therefore a Lie subgroup of G. By Lemma 4.10.2, p. 113, choose an ideal n' of g such that n + n ' = g, such that nfln' = {0} and such that [n, n'] = {0}. Let N' be the connected Lie subgroup of G corresponding to n'. Then G — NN' and NC\N' is discrete. Then, by Lemma 4.9.30, p. 109, we have [N,N'] = {1G}End of proof of (3). Proof of (4): Fix N' as in (3). Then N and N' centralize each other and G = NN'. Then Z(N) C Z{NN') = Z(G). End of proof of (4). Proof of (5): Since N C G, we get n C g . Then, by Corollary 4.10.16, p. 119, g/n is semisimple. Then G/N is semisimple. End of proof of (5).D L E M M A 4.10.23 Let G be a connected semisimple Lie group with finite center and let N be a normal closed subgroup of G. Assume that G ^ N. Then G/N is a semisimple Lie group with finite center. Proof. By (5) of Lemma 4.10.22, p. 121, G/N is semisimple. It remains to show that Z(G/N) is finite. Choose N' as in (3) of Lemma 4.10.22, p. 121. Let A := N (1N'. Then A is discrete in G. By (4) of Lemma 4.10.22, p. 121 (with N replaced by TV'), we see that Z{N') C Z(G), so Z(N') is finite. Let p : G -> G/N be the canonical homomorphism. Then p\N' : N' -+ G/N is surjective and ker(p|iV') = A, so G/N is isomorphic (as an abstract group) to N'/A. It therefore suffices to show that Z(N'/A) is finite. Let n : N' —> N'/A be the canonical homomorphism. By (2) of Lemma 4.9.16, p. 103 (with G replaced by N'), Z(N'/A) = n(Z(N')). So, since Z(N') is finite, it follows that Z(N'/A) is finite. • L E M M A 4.10.24 Let G be a connected semisimple Lie group and let N be a normal closed subgroup of G. Then N C (Z(G))N°. Proof. Fix x £ N. We wish to show that x 6 (Z(G))N°. Let TV' be as in (3) of Lemma 4.10.22, p. 121. Then NnN' is a discrete normal subgroup of G, so, by (1) of Lemma 4.9.16, p. 103, NnN' C Z{G). Since n + n' = g, it follows that d7r|n' : n' —> g/n is surjective. Then 7r|iV' : N' -> G/N° is surjective. Then G = N'N°. Choose a £ N' and

Generalities on semisimple groups and algebras

b e N° such that x = ab. We have a = xb~l € N{NQ)~l

aeNnN'CZ(G).Thenx

123

C N.

Then

= ab£{Z(G))N°.

D

COROLLARY 4.10.25 Lei G be a connected semisimple Lie group and let N be a normal closed subgroup of G. If Z{G) is finite, then N/N° is finite. Moreover, if G is centerfree, then N is connected. L E M M A 4.10.26 Let I be a semisimple Lie algebra and let 6 be an ideal of I. Then there exists an ideal t' of I such that t D t' = {0}, such that [I, t'] = {0} and such that 6 + t' = I. Proof. By Corollary 4.10.15, p. 118, we see that 6 is semisimple. Therefore, by Lemma 4.10.2, p. 113, we are done. • L E M M A 4.10.27 For any semisimple Lie algebra g, 0er(g) = ad B (g). Proof. Let I) := 0et(g) C g[(g) be the Lie subalgebra of derivations of g. Let flo := ad B (g) C h. We wish to show that h = g0Since g is semisimple, we see that 3(g) = {0}. Then ad : g ->• f) is injective, so go is Lie algebra isomorphic to g, and is therefore semisimple. So, by Corollary 4.10.5, p. 114, we have n(,(g0) = go + (c(,(flo))- By Lemma 4.9.13, p. 102, for all T 6 C(,(g0), we have T(g) C 3(g) = {0}, so T = 0. Then C(,(fl0) = {0}. By Lemma 4.9.13, p. 102, g0 is an ideal of h, i.e., \) = n(,(g 0 ). Then f) = ti(,(fl0) = flo + (cf,(flo)) = flo + {0} = flo• L E M M A 4.10.28 Let g be a semisimple Lie algebra. Then ad g (g) is real Jordan closed. Proof. By Lemma 4.10.27, p. 123, we have Der(g) = ad g (g). So, by Lemma 4.9.14, p. 102, we are done. • L E M M A 4.10.29 Let G be a connected semisimple Lie group. Then we have Ad B (G) = Aut (fl). In particular, Ad B (G) is closed in GL(g). Proof. By Lemma 4.10.27, p. 123, we have ad g (g) = 0et(g). So, since the Lie algebra of Ad 0 (G) is adfl(g) and since the Lie algebra of Aut°(g) is 0er(g), we conclude that Ad B (G) = Aut°(g). Since Aut°(g) is closed in GL(g), we are done. •

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Let g be a semisimple Lie algebra. We say that g is compact if K is negative definite. Otherwise we say that g is noncompact. Using Lemma 4.10.32, p. 124 below, one sees that a connected semisimple Lie group is compact iff its Lie algebra is compact. So, by Corollary 4.9.19, p. 104, one sees that a semisimple Lie algebra is compact iff it is isomorphic to the Lie algebra of a compact semisimple Lie group. We say that a connected semisimple Lie algebra g has no compact factors if every nonzero ideal of g is noncompact. We say that a semisimple Lie group has no compact factors if its Lie algebra has no compact factors. L E M M A 4.10.30 Let I be a semisimple Lie algebra. Then there exist semisimple ideals 6 and m of I such that all of the following are true: (1) t is compact; (2) m has no compact factors; (3) (X,Y) i-> X + Y : Z(Bm —¥ I is a Lie algebra isomorphism. Proof. Let 6 be maximal (with respect to set-theoretic inclusion) among compact ideals of I. By Lemma 4.10.2, p. 113, choose an ideal m of I such that trim = {0} and such that 6 + m = I. Then (X,Y) >->• X+ Y : 6 0 m -> [ is a Lie algebra isomorphism. By Corollary 4.10.15, p. 118, 6 and m are both semisimple. It remains to show that m has no compact factors. Assume, for a contradiction, that m' is a nonzero compact ideal of m. By Corollary 4.10.17, p. 119, m' is an ideal of I. Then 6+m' is a compact ideal of [ and 6 C 6 + m', contradicting maximality of 6. • L E M M A 4.10.31 Let t be a compact semisimple Lie algebra. Let X € 6. Assume that a d X : 6 —¥ 6 is nilpotent. Then X = 0. Proof. Let K : t x t -> E denote the Killing form on t. Then K is negative definite. Since a d X o a d X : 6 ->• t is nilpotent, it follows that K(X,X) = 0. Then, as K is negative definite, we get X = 0. • L E M M A 4.10.32 Let G be a connected semisimple Lie group. Then G is compact iff g is compact. Proof. By Lemma 4.10.1, p. 113, the Killing form K of g is nondegenerate. Then, by (2) of Corollary 2.10.2, p. 35, K is negative definite iff K is negative semidefinite. The result then follows from (1) of Proposition 3.39, p. 75 of [CE75]. •

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Let g be a Lie algebra and let X,Y,T G g. We say that X,Y,T standard s ^ M ) basis of g if X, Y, T is a basis for 0 satisfying [X, Y] = T,

[T, X] = 2X,

is a

[T, Y] = -2Y.

The following fact is sometimes called the Jacobson-Morozov Lemma. L E M M A 4.10.33 Let g be a semisimple Lie algebra and let X S fl\{0}. Assume that a d X : g —> g is nilpotent. Then there exist Y,T £ g such that X,Y,T is a standard s^(K) basis of some Lie subalgebra of g. Proof.

This is Theorem LX.7.4, p. 432 of [He78].

•

The following is a slight modification of Lemma 6.9, p. 470 of [AOOa]. The proof we present is modified from the last paragraph (on p. 617) of the proof of Theorem 2.1 of [Ko96]. L E M M A 4.10.34 Let I be a semisimple Lie algebra and assume that the real rank of I is — 1. Assume that I not Lie algebra isomorphic to s ^ W Let a be a maximal real split torus of I and let <j> £ a* be a root of a on I. Let V be the -rootspace of a on I. Then dim(V) > 2. Proof. Since I ^t s{%{W) and since the real rank of [is = 1, using Section 3 and 4, on pp. 518-543 in Appendix C of [Kn96], together with the Coincidences on pp. 519-520 in §X.6.4 of [He78], we see that exactly one of the following is true: • • • •

for some integer n > 3, I — so(n, 1); or for some integer n > 2, I = su(n, 1); or for some integer n > 2, I = sp(n, 1); or t is isomorphic to the Lie algebra denoted f4(_2o) i n [He78] and denoted "F II" on p. 542 of [Kn96].

In each of these cases, we can examine the rootspaces and see directly that they have dimension > 2. • L E M M A 4.10.35 Let I be a simple Lie algebra. Assume that I ^ B12(1R). Let a be a maximal real split torus of I. Let X € a\{0}. Let $ be the set of roots a on I. For each a e $ , let la be the a-rootspace of I. Define $+:={a6$|a(X)

> 0}. Then $ + ^ 0 and dim I ^

la j > 2.

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Proof. Let F := RKF([,o) € SBF(o*). Then $ is a root system in (a*,F). Since X € o\{0}, we see that o ^ {0}, so dim(a) > 1. That is, the real rank of I is > 1. By Lemma 2.11.3, p. 38, let A be a base for $ such that, for all a 6 A, we have a(X) > 0. Since A spans a*, and since X ^ 0, choose ao € A such that a0(X) ^ 0. By choice of A, we have a0(X) > 0. We conclude that a0(X) > 0. That is, a 0 € $ + . Then 4>+ ^ 0. Let 1+ := Y^ la. We wish to show that dim([+) > 2. If dim(a) = 1, i.e., if the real rank of [ is = 1, then, by Lemma 4.10.34, p. 125, we have dim([ ao ) > 2, so, since lao C [+, we are done. We therefore assume that dim(a) > 2. Since [ is simple, it follows that $ is irreducible, so A is connected. Let A £ A be adjacent to ao in A. Let 70 := «o + /?o- Then 70 € $ . By choice of A, we have Po(X) > 0. So, since a0(X) > 0, it follows that 70(X) > 0. Then 70 € $ + . Then lao ^ {0} and (7o f {0}, so dim([ ao + t7o) > 2. So, since l ao + I7o C [ + , we are done. • L E M M A 4.10.36 Letg andi) be Lie algebras, let f : g —> f) be a surjective Lie algebra homomorphism, and assume that () is semisimple. Then there exists a Lie subalgebra go C g such that f\go '• 0o —> f) is a Lie algebra isomorphism. Proof. Let r denote the solvable radical of g. Then / ( t ) is a solvable ideal of h, so, by semisimplicity of h, we see that /(r) = {0}. Let [ be a semisimple Levi factor of JJ. Then /(I) = /(I + r) = /(g) = h. Let 6 denote the kernel of / | I : I ->• h. By Lemma 4.10.26, p. 123, let go be an ideal of I such that g0 fl t = {0} and such that g0 +t = I. Then /Iflo : go —> f) is a Lie algebra isomorphism. • L E M M A 4.10.37 Let G be a connected semisimple Lie group and fix a Cartan involution 6 : g -> g. Let 6 :— {Y € g 16{Y) = Y} and let K be the connected Lie subgroup ofG corresponding to i. Then Adg(K) is a compact subgroup o/GL(g). Proof. Since Ad g (G) is Lie group isomorphic to G/(Z(G)), and since Z(G) is a discrete subgroup of G, it follows, from Corollary 4.9.17, p. 104, that Ad 0 (G) has trivial center. The result then follows from (i) of Theorem VI.1.1 on p. 252 of [He78], with G replaced by Ad„(G) and K replaced byAdgtJO. •

Generalities

on semisimple

groups and algebras

127

L E M M A 4.10.38 Let G be a semisimple Lie group and let a be a maximal real split torus of g. Let $ C o* be the set of roots of a on Q. Let A be a base of $ . Let a+ := {T € a | \/S € A, S(T) > 0}. Let A+ := exp(a+) C G. Assume that Z(G) is finite. Then there is a compact subgroup KofG such that G = KA+K. Proof. By Lemma 4.10.12, p. 116, fix a Cartan involution 9 : g -» g such that, for all X € o, 0(X) = - X . Let t := { F e g 10(F) = F } . Let A" be the connected Lie subgroup of G corresponding to t. By Lemma 4.10.37, p. 126, we see that C := Adg(K) is a compact subgroup of GL(g). As Z(G) is finite, the map AdB|.ftf : K —> C has finite kernel. Then K is compact. By Theorem IX.1.1, p. 402 of [He78], G = KA+K. • L E M M A 4.10.39 Let V be a vector space and let g be a semisimple Lie subalgbra ofgl(V). Then g C si(V). Moreover, if J is a Q-invariant complex structure on V, then g C sl(V, J). Proof.

By Lemma 4.10.6, p. 114, we have g = [g,g]. Then gcl0i(V),fli(V)]c«[(y).

Now let J be an invariant complex structure on g. For all X 6 g, we have X 6 g C g[(V, J) and we have X,JX EgC sl{V), so X <E sl{V, J). D L E M M A 4.10.40 Let V be a vector space and let g be a semisimple Lie subalgbra of f) := sl(V). Assume that V has only scalar g-intertwining. Then ri(,(g) = g. Proof. By Corollary 4.10.5, p. 114, it suffices to show that C[,(g) = {0}. Let T £ C(,(g). We wish to show that T = 0. Let d := dim(V). Let I : V —• V be the identity transformation. Since V has only scalar g-intertwining, choose a € E such that T = a/. Then 0 = tr(T) = da, so a = 0. Then T = al = 0. • L E M M A 4.10.41 Le£ V be a vector space, let J be a complex structure on V and let g be a semisimple Lie subalgbra of f) := sl(V, J). Assume that V is g-irreducible. Then ti|,(g) = g. Proof. By Corollary 4.10.5, p. 114, it suffices to show that C(,(g) = {0}. Let T 6 C(,(g). We wish to show that T = 0.

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Let d := dim(V). Let I: V —> V be the identity transformation. Since V is 0-irreducible, by Lemma 4.9.24, p. 106, choose a, b £ M. such that T = al + bJ. Since J is a complex structure, we get t r ( J ) = 0. Then 0 = tr(T) = da and 0 = tr(JT) = -db, so a = b = 0, so T = 0. • 4.11

Real Jordan decomposition

L E M M A 4.11.1 Let V be a vector space. Let g be a Lie subalgebra of gl(V). Let X,D,E,N € g and assume that X = D + E + N is a real Jordan decomposition on V. Then &ds(X) = (adB(£>)) + (adB(.E)) + (adg(N)) is a real Jordan decomposition on g. Proof. For all A,B € {D,E,N}, since [A,B] = 0, it follows, from the Jacobi identity, that [adB(A),adB(.B)] = 0. Since D : V ->• V is real diagonalizable, it follows that ad D : 0l(V) —> gi(V) is real diagonalizable. So, as g is invariant under ad£> : fll(F) -> fll(V), we conclude that the restriction ad£> : g —> g is real diagonalizable. Similar reasoning shows that ad-E : g —> g is elliptic and that ad N : g —» g is nilpotent. • L E M M A 4.11.2 Let V be a vector space. Let g be a Lie subalgebra ofgliV). Assume that g is real Jordan closed. Assume that%{g) = {0}. Let X,D,E,N e g. Assume thatads(X) = (ad„(£>)) + (adB(.E)) + (ad„(iV)) is a real Jordan decomposition on g. Then X — D + E + N is a real Jordan decomposition on V. Proof. Since %(g) = {0}, it follows that ad : g -> gl(g) is injective. Since g is real Jordan closed, choose D', E', N' 6 fl such that X = D' + E' + N' is a real Jordan decomposition on V. By Lemma 4.11.1, p. 128, &dg{X) = (ad B (D')) + (adfl(£?')) + (ad 0 (iV')) is a real Jordan decomposition on g. Then, by uniqueness of real Jordan decomposition, we conclude that &dB(D) = adB(£>')> that &dg(E) = adg(E') and that adB(iV) = adB(JV'). So, by injectivity of ad : 0 —> gl(g), we get D = D', E = E' and N = N'. Then X = D + E + N is a real Jordan decomposition on V. • L E M M A 4.11.3 Let V be a vector space. Then sI(V) is real Jordan closed. Moreover, for any complex structure J on V, we have that sI(V, J ) is real Jordan closed.

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129

Proof. Since elliptic and nilpotent transformations have trace = 0, we see that sl(V) is real Jordan closed. Let J be a complex structure on V. Let X G sl(V, J) and D,E,N e gl{V). Assume that X = D + E + N is a real Jordan decomposition. We wish to show that D,E,N G s[(V, J). Let S C End(V) be the M-linear span of X,X2,X3,.... Then we have D,E,N G 5 . So, since XJ = JX, we get DJ = JD, EJ = JE and NJ = JN. That is, D,E,N G gl(V,J). We wish to show that

D,E, N, JD,JE,JN

esl{V).

Because X = D + E + N is a real Jordan decomposition and because D,E,N Gfll(V,J ) , it follows that JX - JE + JD + JN is a real Jordan decomposition. So, since sl(V) is real Jordan closed and since we have X, JX G gi(V), we are done. • L E M M A 4.11.4 Let V be a vector space and let g and f) be Lie subalgebras of gl(V). Assume that g C f) and that ti(,(g) = g. Assume that f) is real Jordan closed. Then g is real Jordan closed, as well. Proof. Choose D,E,N G gl(V) such that X = D + E + N is a real Jordan decomposition on V. We wish to show that D,E,N G 0. Since f) is real Jordan closed, we get D,E,N £ h. For all P 6 h, let P := a d P : f) -* h. By Lemma 4.11.1, p. 128 (with g replaced by h), we see that X = D + E + N is a real Jordan decomposition on h. As X G g, we see that X(g) C g. Let 5 C End(g) be the IR-linear span of X, X2, X3,.... Then D,E,N G S. So, since X(g) C g, we conclude that D(g) C fl, that E(g) C g and that iV(fl) C g. Then D,E,N e it(,(fl) = fl.D L E M M A 4.11.5 Lei V be a vector space and let g be a Lie subalgebra of gl(V). Assume that g is simple and that V is g-irreducible. Then g is real Jordan closed. Proof. Case 1: V has only scalar g-intertwining. Proof in Case 1: Let I) := sl(V). By Lemma 4.11.3, p. 128, we know that I) is real Jordan closed. By Lemma 4.10.40, p. 127, ti(,(fl) = g. Then, by Lemma 4.11.4, p. 129, we see that g is real Jordan closed. End of proof in Case 1. Case 2: V has nonscalar g-intertwining. Proof in Case 2: By using Lemma 4.9.26, p. 107, choose a g-invariant complex structure J on g. Let f) := sl(V,J). By Lemma 4.11.3, p. 128, we know that 1} is real Jordan closed. By Lemma 4.10.41, p. 127, ni,(g) = g. Then, by Lemma 4.11.4, p. 129, we see that g is real Jordan closed. End of proof in Case 2. D

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L E M M A 4.11.6 Letg be a semisimple Lie algebra, letV be a vector space and let p : g —• gi(V) be a representation. Let X, D,E,N 6 g. Assume that adg(X) = (ade(.D)) + (adB(.E)) + (adB(7V)) is a real Jordan decomposition on g. Then p(X) — (p(D)) + (p(E)) + (p(N)) is a real Jordan decomposition on V.

Proof. By Corollary 4.10.3, p. 113, we may assume that g is simple. By Lemma 4.10.7, p. 114, we may assume that p is irreducible. We may assume that p(g) ^ {0}. Then ker(p) ^ g, so, as g is simple, we have ker(p) = {0}, so p is injective. Then, replacing 0 by p(g), we may assume that g is a Lie subalgebra of gi(V) and that p : g —>• gi(V) is the inclusion map. By Lemma 4.11.5, p. 129, we see that g is real Jordan closed. Since g is semisimple, we have • 3 ( 0 ) = {0}. The result then follows from Lemma 4.11.2, p. 128. C O R O L L A R Y 4.11.7 Letg be a semisimple Lie algebra, letV be a vector space and let p : g -»• gl(V) be a representation. Let X € 0. Assume that a d X : 0 —»• 0 is real diagonalizable (resp. elliptic, resp. nilpotent). Then p(X) : V —> V is real diagonalizable (resp. elliptic, resp. nilpotent).

Proof. Assume that &dX : g -»• 0 is real diagonalizable. It follows that adg(X) = (ad B (X)) + 0 + 0 is a real Jordan decomposition on g. Then, by Lemma 4.11.6, p. 130, we see that p(X) = (p(X)) + 0 + 0 is a real Jordan decomposition on V. Consequently, p(X) : V —> V is real diaongalizable. A similar argument works for "elliptic" and "nilpotent". D

L E M M A 4.11.8 Let V be a vector space and let g be a semisimple Lie subalgebra of gi(V). Then g is real Jordan closed.

Proof. Let X £ 0 be given. We wish to show, for some D,E,N € 0, that X = D + E + N is a real Jordan decomposition on V. By Lemma 4.10.28, p. 123, fix D,E,N eg such that ad fl (X) = (adg(£»)) + (adfl(-E)) + ( a d g ( ^ ) ) is a real Jordan decomposition on 0. Let p : g -> 0l(V) be the inclusion map. Then, by Lemma 4.11.6, p. 130, we see that X = D + E + N is a real Jordan decomposition o n V . •

Consequences of results on real Jordan decomposition

4.12

131

Consequences of results on real Jordan d e c o m p o s i t i o n

L E M M A 4.12.1 Let g be a Lie algebra and let X,T e Q. Let X G E\{0} and assume that (adT)X — XX. Then a d X : g -> g is nilpotent. Proof. Let S := a d T : g ->• g and let W := a d X : g - • g. We wish to show that W : g -+ g is nilpotent. Let f) := gl(g). Then 5,T G J). Let J : h -> f) be the identity transformation. For all P G h, define P ' := a d P : f) -> h. Then, for all P,Q € f), we have P'(Q) = [P, Q] = PQ - QP. By the Jacobi identity, we have [ad 0 (T),ad g (X)] = ad B ([T, X]). So, since [T,X] = (adT)X = AX, this gives [S,W] = XW. Then we have S'(W) = [S,W] = XW. Let D : g ->• g be the real diagonalizable part of 5 : g -¥ g. By Lemma 4.11.1, p. 128, D' : \) -> f) is the real diagonalizable part of S" : f) —»• I). Then D' — A/ is the real diagonalizable part of S" — A J, so D'-XI is an element of the M-linear span of S'-XI, (S'-XI)2, (S'-XI)3,.... So, since (S" - XI)(W) = 0, we conclude that (£>' - XI)(W) = 0. Then [D,W] = D'(W)=XW. For all n G M, let g,, := { F G fll-D(^) = fiY}. For all /i G E, since [D,W] = XW, it follows that W(fl„) C g A+/1 . Then, for all / i f R , for all integers i > 0, we have W'g^ C 0JA+/ILet S :— {/j, € M|gM ^ {0}} be the set of eigenvalues of D : g ->• g. Since D : g -> g is real diagonalizable, we see that g = Y j g^. Then, for i*es l all integers i > 0, we have W (g) C Y^ g»A+^- Since 5 is finite and A ^ 0, ij.es

choose an integer i0 > 0 such that, for all g is nilpotent. D In fact, Lemma 4.12.1, p. 131 can be improved: Letting Jo : g —• g be the identity transformation, if there exist an integer m > 0 and A G K\{0} such that [(adT) - A7o]nX = 0, then a d X : g -»• g is nilpotent. The proof is essentially the same. Lemma 4.12.1, p. 131 has quick a dynamical proof: Proof. Let Si be a sequence in M such that SjA —• +00. For all i, define gi := exp(siT); then (Adg^X = eSiXX, so (Adgi)(e~SiXX) = X. Then X is Kowalsky for Adg(gi). Therefore, by Lemma 4.9.31, p. 109, we conclude that a d X : g -> g is nilpotent. •

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L E M M A 4.12.2 Let g be a semisimple Lie algebra. Then the following are equivalent: (1) (2) (3) (4)

g is noncompact. For some D G fl\{0}, a d D : g —>• g is real diagonalizable. For some N e fl\{0}, ad AT: g -»• g is nilpotent. The real rank of g is > 1.

Proof. Let K : g x g ->• R denote the Killing form of g. Proof of (1) =>• (2): Since g is noncompact, we conclude that K is not negative definite. By Lemma 4.10.1, p. 113, K is nondegenerate. Then K is not negative semidefinite. Choose X G g such that K(X, X) > 0. By Lemma 4.10.28, p. 123, choose D,E,N e g such that ad fl (X) = (adB(£>)) + (ad B (£)) + (ad0(7V)) is a real Jordan decomposition on g. We wish to show that D ^ 0. Assume, for a contradiction, that D = 0. As (ad N) o (ad N) : g —> g and (ad N) o (ad E) : g —> g are both nilpotent linear transformations, we conclude that K(N, N) = K(E, N) = 0. Then we have K(X, X) = K(E,E). AS &&E : g -> g is elliptic, it follows that all characteristic roots of ad E : g -t g are pure imaginary, which implies that all characteristic roots of the map (ad E) o (ad E) : g —> g are nonpositive real numbers. Then K(E, E) < 0. Then 0 < K{X, X) = K(E, E) < 0, contradiction. End of proof of (1) => (2). Proof of (2) =$• (1): Let 5 be the set of characteristic roots of the linear transformation a d D : g —¥ g. Then S C E, so, for all s € S, we have s 2 > 0. So, as K(D, D) is a linear combination of elements of {s2 \ s € S}, with positive integer coefficients, we conclude that K(D,D) > 0. Then K is not negative definite, i.e., g is not compact. End of proof of (2) =$• (1). Proof of (2) =>• (3): Since 3(0) = {0}, we see that ad : g ->• g[(g) is injective. Then, as D ^ 0, it follows that ad£) : g -¥ g is nonzero. Then, since adD : g -> g is real diagonalizable, choose N G 0\{O} and A £ M\{0} such that (ad D)N = AN. Then, by Lemma 4.12.1, p. 131, we see that adiV : g -¥ g is nilpotent. End of proof of (2) =$• (3). Proof of (3) ==• (2): By Lemma 4.10.33, p. 125, choose Y,D G g such that N,Y,D is a standard s ^ W basis of some Lie subalgebra go of g. Since [D, N] = N ^ 0, we see that fl^O. Moroever, adD : g0 -> g0 is real diagonalizable. So, by Corollary 4.11.7, p. 130 (with p : g —>fll(V)replaced

Consequences of results on real Jordan

decomposition

133

by ad :flo-> 0t(fl))i w e s e e t n a t a d D : 0 -> 0 is real diagonalizable. End of proof of (3) = > (2). Proof of (2) 4=$- (4): This follows from the definition of "real rank". End of proof of (2) 4=^ (4). • L E M M A 4.12.3 Let g be a noncompact semisimple Lie algebra. some Lie subalgebra of g is isomorphic tofife(K).

Then

Proof. By (1) =i> (3) of Lemma 4.12.2, p. 132, choose N e g such that adJV : 0 —> g is nonzero and nilpotent. Then, by Lemma 4.10.33, p. 125, we are done. • L E M M A 4.12.4 Let g be a semisimple Lie algebra with no compact factors. Let flo denote the set of all X £ g such that a d X : 0 ->• 0 is nilpotent. Then 0o generates g, i.e., no proper Lie subalgebra of g contains goProof.

Let 0i be the Lie subalgebra of 0 generated by 0o- Assume, for a

contradiction, that 0i 7^ 0. Since 0O is (ad B (0))-invariant, it follows that 01 is an ideal of 0. By Lemma 4.10.2, p. 113, choose an ideal g[ of 0 such that 01 D g[ = {0} and such that 0! + g[ = g. Then g[ 7^ {0}. Since 0 has no compact factors, we see that g[ is noncompact. Then, by (1) =>. (3) of Lemma 4.12.2, p. 132, choose JV € 0i\{O} such that the map ad JV : 0i ->• 0i is nilpotent. By Corollary 4.11.7, p. 130 (with p : g -+ 0l(V) replaced by ad : 0^ —> g[(g)), we see that adJV : 0 —> g is nilpotent. Then JV e 0o- Then 0 ^ N £ g0 n g[ C 0i n g[ = {0}, contradiction. • Let d > 1 be an integer. Let x\,... ,Xd : Rd —> R be the coordinate projections. Let / : E d ->• E be a function. We say that / is a monomial of degree 0 if / is constant. For any integer k > 1, we say that / is a monomial of degree k if, for some integers ii,...,ik € [l)<^j we have f = Xit • • -Xik. For any integer k > 0, we say that / is a homogeneous polynomial of degree k if / is a finite E-linear combination of monomials of degree k. We say that / is a monomial if there exists an integer k > 0 such that / is a monomial of degree k. We say that / is a polynomial if / is a finite E-linear combination of monomials. Let d > 0 be an integer, let 7 b e a d-dimensional vector space and let / : V —• E be a function. We say / is a polynomial if, for some vector space isomorphism / i : E d - ^ F , / o / i : E d - ^ E i s a polynomial.

134

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Results

Let V and W be vector spaces and let : V —>• W be a function. We say that (f> is polynomial if, for every linear map L : W —• R, we have that L o (ft ; V —• R is a polynomial. For any vector space V, for any polynomial / : R —> V, we have: / : R -> V is proper iff / is constant. L E M M A 4.12.5 Lei G be a semisimple Lie group with no compact factors. Let X € 0\{O}. Then there exists a sequence gi in G such that (Ad gi)X -¥ oo in g. Proof. Let go be the set of W G 0 such that ad W : g —>• g is nilpotent. By Lemma 4.12.4, p. 133, we see that g is generated by go. So, since X $ {0} = 3(0), choose y £ 0o such that (a,dY)X 7^ 0. Since Y € g0 and since (ad Y)X ^ 0, choose an integer k > 1 such that (ad Y)kX ^ 0 and such that (a,dY)k+1X = 0. For all integers j > 0, let Xj := (adY^X. Then 0 = Xk+i = Xk+2 = • •; while Xk ^ 0. For all t € R, we have (Ad(exp(tY)))X - XQ + tXx + (t2/2\)X2 + ••• + (tk/kl)Xk. Then the map t H-» (Ad(exp(iF)))X :ffi-> 0 is a nonconstant polynomial, and is therefore nonproper. Let ti be some sequence in R such that \ti\ —¥ +00. For all i, define g^ := exp(ijF). •

4.13

Generalities on algebraic groups

Let V be a vector space, let S C V and let V denote the set of all polynomials V -4 R. Let I := {/ € V \ f(S) = {0}}. The Zariski closure in V

of S is f|/ _ 1 (°)/ex Let V be a vector space. A subset of V which is equal to its own Zariski closure in V is said to be Zariski closed in V. Let C be the collection of subsets of V which are Zariski closed in V. If S CV and if S is the Zariski closure in V of S, then S £ C. For all C 6 C, C is closed in the ordinary vector space topology on V. The collection C is closed under finite union and arbitrary intersection. Let V be a vector space. A subset S of GL(V) is Zariski closed in.GL(V) if it is the intersection of GL(V) with a Zariski closed subset of End(V). A subgroup of GL(F) is algebraic if it is Zariski closed in GL(F). Any algebraic subgroup of GL(V) is closed in GL(V), and is therefore a Lie subgroup of GL(V). A subgroup of SL(F) is algebraic iff it is Zariski

Generalities

on algebraic groups

135

closed in End(V). As before, the collection of subsets of GL(V) which are Zariski closed in GL(V) is closed under both finite union and arbitrary intersection. For any S C GL(V), The Zariski closure in GL(T^) of S is the intersection of all subsets of GL(F) which both are Zariski closed in GL(V) and which contain S. Let V be vector space and let G be a subgroup of GL(V). In this book, we say that G is almost algebraic if there is an algebraic subgroup H of GL(y) such that H° C G C H. Any almost algebraic subgroup of GL(F) is closed in GL(V). A finite union or arbitrary intersection of algebraic (resp. almost algebraic) subgroups of GL(V) is again algebraic (resp. almost algebraic). If 0 is a Lie algebra, then Aut(jj) is an algebraic subgroup of GL(g), so Aut°(g) is an almost algebraic subgroup of GL(g). L E M M A 4.13.1 Let V be a vector space and let G be an algebraic (resp. almost algebraic) subgroup of GL(V). Then Z(G) is an algebraic (resp. almost algebraic) subgroup of GL(V), as well. Proof. For all g e GL(V), let Cg := {x € GL(V)\gx = xg}; then Cg is algebraic. Then C :— f ] Cg is algebraic, so Z{G) — C D G is algebraic g€G

(resp. almost algebraic).

•

L E M M A 4.13.2 Let V be a vector space and let G be an almost algebraic subgroup o/GL(V). Then G° has finite index in G. In particular, if G is discrete, then G is finite. Proof. As G is almost algebraic, let H be an algebraic subgroup of GL(V) such that H° C G C H. Then, by Lemma 3.2.3, p. 48, G° = H°. By [Wh.57], H° has finite index in H. Then G° has finite index in G. • The following is Lemma 4.4, p. 463 of [AOOa]. L E M M A 4.13.3 Let L be a connected, semisimple Lie group, let V be a vector space and let p : L —• GL(V) be representation. Then p(L) C SL(V) and p(L) is almost algebraic, so p(L) is a closed subgroup of SL(V). Proof. Let J : V —¥ V be the identity transformation. By Lemma 4.10.6, p. 114 and Corollary 4.6.3, p. 91, we have L = [L,L]. Then p(L) = p([L,L]) C [GL(V),GL(V)] C SL(V). Let L0 := p(L). It remains to show that L0 is almost algebraic.

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Case 1: V has only scalar (/o(L))-intertwining. Proof in Case 1: Let G:=SL(V). Then G is an algebraic subgroup of GL(F). By Lemma 4.10.40, p. 127, we have lo = TIB(IO), SO L0 = NQ(10)- SO, as NQ(10) is algebraic, it follows that LQ is almost algebraic. End of proof in Case 1. Case 2: V is (/9(jL))-irreducible and has nonscalar (p(Z/))-intertwining. Proof in Case 2: By Lemma 4.9.26, p. 107, choose a complex structure J : V -»• V which is (p(L))-invariant. Let G := SL(V,J). Then G is an algebraic subgroup of GL(V). By Lemma 4.10.41, p. 127, we have l0 = nB([o), so LQ = N^(lo). So, as No(k>) is algebraic, it follows that L0 is almost algebraic. End of proof in Case 2. We now consider the general case. By Lemma 4.10.7, p. 114, choose an integer k > 1 and choose /9-invariant vector subspaces V\,..., Vjt of V such that (vi,..., Vk) i-> v\ 4- • - • + Vk : V\ (B • • • © 14 -> V is an isomorphism of vector spaces. Let K := { 1 , . . . ,k}. For all i £ K, let G; := GL(Vi) and define p{ : L -»• G» by pi(Z) = (p(0)|Vi. For all i € K, let L* := ^ ( L ) . By Case 1 and Case 2, we see, for all i & K, that Li is an almost algebraic subgroup of SL(Vj). Let t : Gi x • • • x Gfc —> GL(V) be the standard injection; so, for all (9i, • • • ,9k) e Gi x • • • x Gk, for all i £ K, we have (i(g1}..

.,9k))\Vi

= gt.

Let P := Li x • • • x Lk- For all i 6 K, let 7Tj : P -> X, be the ith coordinate projection. Let (J := i(P). Then L CQ. Since, for all i £ K, we know that Li is a connected and almost algebraic subgroup of GL(Vi), it follows that Q is a connected and almost algebraic subgroup of GL(T^). Then NQ(1) is an almost algebraic subgroup of GL(F). Let L0 := t _ 1 ( £ ) . For all i G K, the composition of t _ 1 : L -) LQ with 7Tj|Z/o '• LQ -* Li is equal to pi : L —> Li, and is therefore surjective. We have cp(Io) C Ci^ti) © • • • © cik(lk). For all i £ if, we have that [j is simple, so ^ ( I ; ) = 3(1;) = {0}. It follows that CP(IQ) = {0}. By Corollary 4.10.5, p. 114, we have [0 + cp(Io) = n p (I 0 ). Then [0 = n p (Io). Then L0 = N%(lo). Then L = t(L0) = L(N°P(L0)) = JVg([). So, as JVg(I) is an almost algebraic subgroup of GL(V), we are done. • Let A denote the class of all commutative E-algebras with multiplicative unit. For all A £ A, let 1^ denote the multiplicative unit of A. Let U be a vector space. For any A £ A, let UA denote the A-module A ®R U, and let det^ : End^(!7 A ) -)• A be the determinant map over A. Let SLy be the affine algebraic group such that, for all A £ A, the yl-points of S L a are given by SLV(A) := {T £ EndA(UA) | det^(T) = 1A}.

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137

Let U be a vector space and let Q be an E-algebraic subgroup of SL[/. By Whitney's Theorem (Theorem 1.2.1, p. 23 of [Va74]), £(R) is a closed submanifold of SL[/(E). We have SLt/(E) = SL([/); give 5(E) the unique manifold structure such that the inclusion SL[/(R) —> SL(?7) is an immersion. Then ^(E) is a Lie group. L E M M A 4.13.4 Let U be a vector space. Let G be a connected semisimple subgroup of GL(U) with no compact factors. Let V denote the topological space of all vector subspaces of U. Let VQ denote the collection of all G-invariant vector subspaces of U. Let [i be any G-invariant probability measure on V. Then /i(Vo) = 1, i.e., fi is supported on VoProof. By Lemma 4.13.3, p. 135, there is an E-algebraic subgroup of SLf/ such that (5(E)) 0 = G. Replacing Q by the Zariski closure in Q of G, we may assume that G is Zariski dense in Q. There is an E-algebraic variety V on which Q acts E-algebraically such that V is the real points of V and such that the action of G on V is the real points of the action of Q on V. Then, by Theorem A.3.1, p. 384, it suffices to show that G is generated by the union of its split algebraic 1-psgs. Let go denote the set of all X € g such that a d X : g -)• g is nilpotent. By Lemma 4.12.4, p. 133, flo generates g. It therefore suffices to show, for every X € 0o> that {exp(iX)} t£ H is a split algebraic 1-psg of G, or, equivalently, of GL(U). Fix X € floLet p : g -> gl{U) be the inclusion, so that, for all X e g, we have p(X) = X 6 End(C/). By Corollary 4.11.7, p. 130, we see that X : U ->• U is nilpotent. Choose an integer n > 0 such that Xn = 0. Then, for all n

t € E, exp(iX) = ^2(tk/k\)Xk.

Then t H-> exp(tX) : E -*• End(£7) is a

fc=0

polynomial, so {exp(£X)} t6 R is a split algebraic 1-psg of G. O L E M M A 4.13.5 Let G be a noncompact semisimple Lie group with finite center. Let gi be a sequence in G and assume that gi —> oo in G. Then there exists X € g such that (Ad<7;)X —» oo in g. Proof. Since G is semisimple, we see, from Lemma 4.13.3, p. 135, both that Ad B (G) C SL(JJ) and that Ad B (G) is closed in SL(fl). Moreover, the kernel of Ad : G —> SL(g) is equal to Z(G), and is therefore finite. Then, by Lemma 2.7.7, p. 29, we see that Ad : G -* SL(JJ) is proper. Then Ad0(• oo in SL(g). So, by Lemma 4.9.35, p. 110 (with V replaced by g and gi replaced by Ad 0 (gj)), we are done. •

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Generalities on nilpotent groups and algebras

L E M M A 4.14.1 Let n be a nilpotent Lie algebra, let X £ n and let F C n . Assume that V C (adX)V. Then V = {0}. Proof. Assume, for a contradiction, that V ^ {0}. Let n 0 , n i , . . . be the descending central series of n. Choose an integer i > 0 such that ^ C t i ; and V £ ni+1. Then V C (adX)V C [n,n«] = n i + i , which gives a contradiction. D L E M M A 4.14.2 Let n be a nilpotent Lie algebra and let X £ n. Then n„(X) = c n (X). Proof. Prom the definitions, we see that c n (X) C n n (X). Let Y £ n n (X). We wish to show that Y £ c„(X). We have [X,Y] £ EX. Assume, for a contradiction, that [X,Y] ^ 0. Then EX = ( a d r ) ( E X ) , so, by Lemma 4.14.1, p. 138 (with X replaced by Y and V replaced by EX), we see that EX = {0}. Then X = 0, so [X, Y] = 0, contradiction. D L E M M A 4.14.3 Let n be a nilpotent Lie algebra. Then, for any nonzero ideal m of n, we have m fl (}(n)) ^ {0}. In particular, if n ^ {0}, then 3(n) ? {0}. Proof. Let mo := m. For integers i > 0, recursively define rri;+i := [n, m]. Since m is an ideal of n, we see, for all integers i > 0, that mi C m. Moreover, by nilpotence of n, for any sufficiently large integer i, we have m = i {0}- Choose j such that rrij ^ {0} and such that nij+i = {0}. Then [n,nij] = mj + i = {0}, so nv,- C 3(n). So, as {0} 7^ m^ C m, we conclude that m n ( j ( n ) ) ^ {0}. D L E M M A 4.14.4 Let n be a nilpotent Lie algebra. Let m be an ideal of n. Assume that m ^ {0}. Then [n, m] C m. Proof. Since m is an ideal of n, we have [n, m] C m. Assume, for a contradiction, that [n, m] = m. Let no,tti,... be the descending central series of n. Choose an integer k > 0 such that raCnj and m £ n^+i. Then m = [n, m] C [n, n/t] = n* + i, contradiction. •

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on nilpotent groups and algebras

139

In some sense, the collection of 2-step nilpotent Lie algebras is the least complicated after the Abelian ones and we next describe how they may be classified: Let V and W be vector spaces and let B : V x V —> W be antisymmetric and bilinear. We define a Lie algebra structure on QB '•= V © W by declaring that {0} © W is central in g, and by, for each v, v' £ V, setting [(v,0), (v',0)] = (0,B(v,v')) e {0} © W C fl. Then gB is 2-step nilpotent. Now suppose that g is a 2-step nilpotent Lie algebra. Set W :— 3(0). Since [0, [0,0]] = {0}, it follows that [0,0] C W. Let V be a vector space complement in 0 to W. Define B : V x V -» W by B(v,v') = [v,v']. Then 0 = 0 B . Note that B is nondegenerate. Thus understanding 2step nilpotent Lie algebras is equivalent to understanding antisymmetric nondegenerate vector-valued bilinear forms. Let (V, B) be a symplectic vector space. This means that V is a vector space and that B : F x y - > l i s a symplectic form on V, i.e., B is an antisymmetric, nondegenerate real-valued bilinear form on V. Then 0B is called the Heisenberg Lie algebra of (V, JB). A Lie algebra will be said to be Heisenberg if it is isomorphic to the Heisenberg Lie algebra of some symplectic vector space. The collection of 2-step nilpotent Lie algebras with one-dimensional center is exactly the collection of Heisenberg Lie algebras. These Lie algebras may be considered the least complicated of all the 2-step nilpotent Lie algebras. A connected Lie group is Heisenberg if its Lie algebra is. Let k > 1 be an integer and let 0 be a Lie algebra. Assume that dirn(g) = 2k + 1. Let B

=

(

Xu...,Xk,

Yu...,Yk,

Z

)

be an ordered basis of 0. Then we say that B satisfies the Heisenberg relations if all of the following hold:

.

mZ=i(g);

• for all i € K, we have [Xi, Yi\ — Z; • for all i, j 6 K, if i ^ j , then we have [Xi, Yj] = 0; and • for all i,j € K, we have [Xi,Xj] = 0 = [Yi, Y,-]. A Lie algebra is Heisenberg iff it admits an ordered basis satisfying the Heisenberg relations. An odd-dimensional vector space does not carry a symplectic form. If (V, B) and (V',B') are symplectic vector spaces and if dim(^) = dim(V),

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then there is an isomorphism V —> V which carries B to B', and so the Heisenberg Lie algebras of (V, B) and of (V',B') are isomorphic to one another. That is, for each even integer n > 2, there is, up to isomorphism, exactly one symplectic vector space of dimension n, and as a result there is, up to Lie algebra isomorphism, exactly one Heisenberg Lie algebra of dimension n + 1. L E M M A 4.14.5 Letn be a nilpotent Lie algebra. Assume dim([n, n]) = 1. Assume that n admits no nonzero Abelian direct summand. Then n is Heisenberg. Proof. Let 3 := 3(n). We wish to show that dim(3) = 1 and that n is 2-step nilpotent. Let n' := [n,n]. By Lemma 4.9.38, p. I l l , we have 3 C n'. It follows that dim(3) < dim(n') = 1. By Lemma 4.14.3, p. 138, 3 ^ {0}. Then dim(3) = 1. Since dim(n') = 1, we see that n' ^ {0}, so, by Lemma 4.14.4, p. 138, [n,n'] C n ' . Then dim([n,n']) < dim(n') = 1. Then [n,n'] = {0}. We conclude that n is 2-step nilpotent. • L E M M A 4.14.6 Let N be a connected, simply connected nilpotent Lie group and let K be a compact subgroup of N. Then K is trivial, i.e., K = {1N}. Proof. Assume, for a contradiction that K By Theorem 3.6.2, p. 196 of [Va74], the feomorphism. Fix X € 6\{0}. Since X ^ pact in n. Then, since exp : n —> N is a that {exp(nX)}n€z is not precompact in N. {exp(nX)} n 6 z C K, contradiction.

^ {ljv}map exp : n -> N is a dif0, {nX}„ e z is not precomdiffeomorphism, we conclude However, K is compact and •

L E M M A 4.14.7 LetG be a Lie group. If Z°(G) is simply connected, then any connected Abelian normal closed subgroup of G is simply connected. Proof. Let A be a connected Abelian normal closed subgroup of G, and assume, for a contradiction, that A is not simply connected. Let T := R/Z. By Lemma 4.8.3, p. 95, choose integers m,n > 0 such that A is Lie group isomorphic to T m x W1. We may therefore assume that A = T m x R™. Since A is not simply connected, we see that m > 1. Let

Generalities

on the nilradical

141

T := T m x {0}. Then, by Lemma 4.14.6, p. 140, we know that T £ Z°(G). So, by Lemma 4.9.37, p. I l l , we see that T is not a normal subgroup of G. We therefore fix g0 E G such that goTg^1 ^ T. Define / : A -*• A by f(9) = 90990*. Then f(T)?T. Let 7r: E m x W1 —¥ A be the canonical map. By covering space theory, let / : E m x E n ->• E m x E" be the Lie group homomorphism satisfying 7T o / = / o 7T. Then /(kerfr)) C ker(Tr). That is, / ( Z m x {0}) C Z m x {0}. By Lemma 4.8.4, p. 95, the map / : E m x E n -> E™ x R n is linear. Then we have / ( E m x {0}) C E m x {0}. Then f(T) C T. Since / : A -» A is a diffeomorphism, we have dim(/(T)) = dim(T). Then f(T) = T, contradiction. • L E M M A 4.14.8 Let N be a connected nilpotent Lie group. Then Z(N) is connected. Moreover, N is simply connected iff Z{N) is simply connected. Proof. Let N denote the univesal covering group of N. By Corollary 3.6.4, p. 198 of [Va74], Z(N) is connected. The result then follows from Lemma 4.9.36, p. 110. • 4.15

Generalities on the nilradical

The nilradical of a Lie algebra g is the unique maximal nilpotent ideal of fl, where maximal is with respect to set-theoretic inclusion. The nilradical of a Lie group G is the unique maximal connected nilpotent normal subgroup of G; its Lie algebra is the nilradical of g. L E M M A 4.15.1 Let G be a connected, simply connected Lie group. Then the nilradical of G is simply connected. Proof. In fact, by Theorem 3.18.2, p. 238 of [Va74], any normal connected Lie subgroup of G is simply connected. (Note: What we here call a connected Lie subgroup is, in [Va74], called an "analytic" subgroup.) • The following is a slight variant of Lemma 14.4, p. 482 of [A00a]. L E M M A 4.15.2 Let g be a Lie algebra and let n be the nilradical of g. Let X,Y £ n. Assume that the codimension in n of n„(X) is < 1. Then ng(X)Cng([X,Y}).

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Proof. Let Xi := [X,Y]. Then Xx £ (adX)n. Fix W € n B (X). We wish to show that W S n B (Xi), i.e., that [W,XX] 6 MXl. If Xi = 0, then [W, Xi] = 0 € RXi, and we are done. We therefore assume that Xi ^ 0. Because n is nilpotent, it follows, from Lemma 4.14.2, p. 138, that the kernel of a d X : n —> n is nn(X). Then a d X : n —> n has kernel of codimension < 1 in n, so dim((adX)n) < 1. As 0 ^ Xi G (adX)n, we conclude that (adX)n = RXi. As W e n B (X), we get [W,X] € MX, so [[Ty,X],F] e E [ X , r ] = KXi. Since F e n and n is an ideal of g, we conclude that [W,y] e n. Then [X, [W, Y]] e (adX)n = EXi. By the Jacobi identity, we have [W,Xx) = [W, [X,Y}} = [[W,X],Y] + [X, [W, Y]}, so [W,XX] 6 KXi + RXi = RXi.

D

Let V be a vector space. Let J : V —>• V be the identity. A linear transformation T : V —> V is said to be unipotent if there exists an integer n > 1 such that (T - I)n = 0. A subgroup G of Gh(V) is said to be unipotent if every element of G is unipotent: The following is Lemma 4.1, p. 213 of [AOOc]. L E M M A 4.15.3 Let G be a connected Lie group, let N be the nilradical of G and let L be a connected semisimple Lie subgroup of G. Then LN is closed in G. Proof. (D. Witte.) Let L*, N*, and H* be the Zariski closures in GL(fl) of Ad B (I/), AdB(iV), and Adg(ZJV), respectively. Because Adg(L) is connected and semisimple, by Lemma 4.13.3, p. 135, we have Ad g (L) = (L*)°. By Corollary 1.5.3.7, p. 47 of [Bou75], we know that AdB(A^) is unipotent. Then, by Lemma 3.20 of [Wi95], we conclude that Adg(N) = N*. Therefore, by Lemma 3.17 of [Wi95], we get Adg(LN) = (H*)°. This implies that Ad 0 (LiV) is closed in GL(g), so its inverse image I under Ad 0 : G -4 GL(g) is closed in G. We have I = LN(Z(G)), so, from the fact that Z°(G) C N, we see that LN = 1°. Therefore LN is closed. • L E M M A 4.15.4 Let G be a connected Lie group. Assume that the nilradical of G is simply connected. Let Go be a connected Lie subgroup ofG. Assume that go *s o one-dimensional ideal of g. Then Go is a normal subgroup of G and Go is Lie group isomorphic to the additive Lie group R.

Generalities on the nilradical

143

Proof. Let N be the nilradical of G. Then N is simply connected. Since go is an ideal of Q, it follows that Go is a normal subgroup of G. Any one-dimensional connected Lie group is isomorphic either to K or to M/Z. So, since dim(Go) = dim(jj0) = 1, we see that Go is Abelian. Since Go is a normal Abelian connected Lie subgroup of G, it follows that Go C N. So, by Lemma 4.14.6, p. 140, Go is noncompact, so Go is not isomorphic to R/Z. Then G 0 is isomorphic to E. • L E M M A 4.15.5 Let g be a Lie algebra, let r denote the solvable radical of g and let n denote the nilradical of g. Then: (1) for all X € r, we have: ad X : t -> r is nilpotent iff X £ n; and (2) [ 0 , r ] C n . Proof.

This is a p a r t of Theorem 3.8.3(iii), p. 206 of [Va74].

•

L E M M A 4.15.6 Let g be a Lie algebra, let n be the nilradical of g, let I be a semisimple Levi factor of g and let lo be an ideal of I. Assume that [lo,n] = {0}. Then • • • •

[lo,r] = {0}; lo is an ideal of g; for some ideal jji of g, both lo f~l 0i = {0} and 0 = lo + 01/ and lolfl-

Proof. By (2) of Lemma 4.15.5, p. 143, we have [lo,r] C n, so we get (adlo)(r/n) = {0}. By assumption, (adlo)n = {0}. By Lemma 4.10.7, p. 114, the adjoint representation of lo on r is isomorphic to the direct sum of the adjoint representations of lo on n and on r/n. Then (adlo)r = {0}, i.e., [lo,r] = {0}. By Lemma 4.10.26, p. 123, choose an ideal li of I such that lo D li = {0} and such that I = l0 + li- Then [lo, li] C [Q n li = {0}, so Po,8] C [Mo] + [lo.li] + [ M C lo + {0} + {0} = l0. That is, lo is an ideal of 0. Let 0i := li + r . We have [li, li] C l! and [r, li] = [li,r] C r and [t,r] C r and [r, l0] C t and [li, l0] = {0}. Then fei.fl] = [ti + t,fl] = [li + 1 , l 0 + li +1] C li + r. That is, 0i is an ideal of 0.

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As l0 and gi are both ideals of g, as g = (o + gi and as to PI gi = {0}, we conclude that the addition map (X, Y) \-¥ X + Y : To © gi —> g is a Lie algebra isomorphism. Then lo | 0D L E M M A 4.15.7 Let g be a Lie algebra with nilradical n. Assume that g admits no simple direct summand. Then c8(n) = 3(n). Proof. Let r be the solvable radical of g. Since j(n) C c 0 (n), we assume that X e (c B (n))\(3(n)) and aim for a contradiction. Claim 1: c t (n) = 3(n). Proof of Claim 1: Let c := c t (n) and let 3 := 3(n). Then [c,n] = {0} and c n n = 3. By (2) of Lemma 4.15.5, p. 143, we get [g,r] C n. Then [c, [c, c]] C [c, [g,t]] C [c, n] = {0}. Then c is a nilpotent ideal of g, so c C n. Then cc(n) = c = cfln = 3 = 3(11). End of proof of Claim 1. Let ( be a semisimple Levi factor of g. Choose Y £ [ and Z G r such that X = Y+Z. By Claim 1, we see that c t (n) = a(n). T h e n X <E (c B (n))\(c t (n)), so X £ t, so Y ^ 0. By Lemma 4.10.18, p. 119, as ad„(X) = 0, we see that ad n (F) = 0 and that ad n (Z) — 0. Let t := C[(n). Then t is the kernel of ad n : ( -> gl(n). Then 0 7^ Y e 6. As t is an ideal of I and as I is semisimple, by Corollary 4.10.15, p. 118, 6 is semisimple. Then, by Corollary 4.10.4, p. 113, choose a simple ideal [1 of t. By Corollary 4.10.17, p. 119, li is an ideal of [. We have [li,n] C [6,n] = [ci(n),n] = {0}. Then, by Lemma 4.15.6, p. 143, (11 g, so g admits a simple direct summand, contradiction. • L E M M A 4.15.8 Let g be a semisimple Lie group, letn denote the nilradical of g and let I denote a semisimple Levi factor of g. Then [g,g] C [ + n. Proof. Let r denote the solvable radical of g. By (2) of Lemma 4.15.5, p. 143, we have [g, r] C n, so [[, r] = [r, [] C n and [r, t] C n. Then, since g = I + r , w e g e t [ g , 0 ] = [l + r,r + t]C[[,l] + [[,r] + [t,r] + [ t , r ] C [ + n. • The following is Lemma 6.3, p. 466 of [A00a]. L E M M A 4.15.9 Let G be a connected Lie group with simply connected nilradical. Let R be the solvable radical of G. Let P be a connected normal closed subgroup of G. Then there exists a cocompact connected closed subgroup P\ of P such that Pi is normal in G and such that Pi
Generalities

on the nilradical

145

follows that [G, P] is a normal subgroup of G, as well. Let M be the closure in G of [G,P]. Then M is normal in G and [G,P] C Af C P . Since [P,P] C M , it follows that P / M is Abelian. Since P is connected, P/M is connected, as well. Let T := E/Z. By Lemma 4.8.3, p. 95, choose integers m, n > 0 and choose an isomorphism (f>: P/M -¥ T m x Rn. Let Pi be the subgroup of P such that M C Px and <£(Pi/M) = {0} x E n . Then P / P i = T m , so P / P i is compact. Moreover, because [G,Pi]C[G,P]CMCPi, it follows that Pi is normal in G. Suppose if is a connected compact subgroup of Pi fi R. We wish to show that K — {1G}Let 7r : Pi -> P i / M be the canonical homomorphism. Because P\jM is Lie group isomorphic to {0} x W1, it follows that Pi / M admits no nontrivial compact subgroups, so ir(K) is trivial, so if C M. Let S be the solvable radical of M. Because (MflP) 0 is a solvable normal connected Lie subgroup of M, we must have (M n R)° C S. Then, as if is connected and as K C M n R, we have K C S. Let L be a semisimple Levi factor of G and let N be the nilradical of G. By Lemma 4.15.8, p. 144, we have [G,G] C LAT, so [G,P] C LN. By Lemma 4.15.3, p. 142, LN is closed in G. It follows that M C LA^. Since M is normal in G, it follows that S is normal in G. Then 5 is a connected solvable normal closed subgroup of LN, so S is contained in the solvable radical of LN. That is, S C N. Then K CS C N,so, by Lemma 4.14.6, p. 140, K = {1 G }. • Let g be a Lie algebra and let V be a real g-module. Then we say that V is trivial (or g-trivial) if, for every X £ g, for every v 6 V, we have Xw = 0. Otherwise, we say that V is nontrivial (or g-nontrivial). Let 0 be a Lie algebra and let V and W be real g-modules. We say that V and W are isomorphic (or isomorphic as real g-modules) if there exists a g-equivariant vector space isomorphism V —¥ W. The next result is a variant of Lemma 6.8, p. 468 of [AOOa]. It states that, under certain circumstances, if a Lie subalgebra has nontrivial adjoint representation on the nilradical, then it also has nontrivial adjoint representation on the center of the nilradical. L E M M A 4.15.10 Let g be a Lie algebra with nilradical n. Let fo be a Lie subalgebra of g and assume that the adjoint representation of IQ on g

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is completely reducible. Let no 2 n i =? n 2 5 • • • be a chain of ideals in g such that n = no and such that, for some j , we have n,- = {0}. Assume, for all j , that [n,rij] C rij+i. Assume, for all j , that the adjoint representation of [o on rij/rij+i either is trivial or has only scalar intertwining. Assume that g admits no one-dimensional ideal. Assume that [to,n] ^ {0}. Then [lo,i(n)} * {0}. Proof. Assume, for a contradiction, that [fo>3(n)] = {0}. Let g° := c8(Io). Let n° := c n (l 0 ) = 0° n n. Since [lo,n] ^ {0}, it follows that n° C n. Let Jx denote the collection of integers j > 0 such that the adjoint representation of [o on xij/rij+i is nontrivial. Choose an integer r > 0 such that both n r £ n° and n r + i C n°. Fix an (ad lo) -irreducible vector subspace W\ C n r such that W\ <£ n°. Then, as n r + i C n°, we get Wi % n r +i, so the image in n r / n r + i of W\ is nonzero and hence, by Lemma 4.9.23, p. 106, is isomorphic to W\ as a real (ad (0)-module. Also, W\ <£ n°, so Wi is (ad[o)-nontrivial. Then the adjoint representation of lo on n r / n r + i is nontrivial, i.e., r £ J\. Fix some X £n° for this paragraph. We have (adX)Wi

C [n,n r ] C n r + i C n°,

so (ad X)W\ is trivial as a real (ad Io)-module. Since W\ is (ad lo)-nontrivial, (&&X)Wi is not isomorphic to W\ as a real (adlo)-module. Then, by Lemma 4.9.23, p. 106, we have (adX)Wi = {0}. As X € n° is arbitrary, we get \Wi,n°] = {0}. By assumption, we have [IO,3(TI)] = {0}, so j(n) C n°. On the other hand, Wi £ n°. Then Wi £ a(n). That is, [Wx,n] ^ {0}. Choose an integer s > 0 such that [Wi,n s ] ^ {0} and such that [Wi,n s + i] = {0}. Let W2 be a nonzero (ad lo)-irreducible vector subspace of n s such that [Wi, W2] ^ {0}. Then, since [Wi,n°] = {0}, it follows that W2 £ n°. Thus the adjoint representation of lo on W2 is nontrivial. Since [WUW2] ± {0}, while [W 1 ; n s + i] = {0}, we get W2 £ n f + i . So the image in n s /n s +i of W2 is nonzero and hence, by Lemma 4.9.23, p. 106, is isomorphic to W2 as a real (ad Io)-module. Therefore the adjoint representation of lo on n s / n s + i is nontrivial, i.e., s € J\. Fix some X 6 n r + i for this paragraph. We have (adX)W2C[nr+1,fl]Cnr+1Cn°, so (adX)W 2 is trivial as area! (ad lo)-module. Since W2 is (ad lo)-nontrivial,

Generalities on the nilradical

147

we conclude that (adX)W 2 is not isomorphic to W2 as a real (ad [o)-module. Then, by Lemma 4.9.23, p. 106, (&dX)W2 = {0}. As X G n r + i is arbitrary, [nr+1,W2] = {0}. As [WUW2] 5* {0}, fix X1eW1,X2e W2 such that [XUX2] ^ 0. Let Z := [Xi,X2]. Since dim(KZ) = 1 and since g admits no one-dimensional ideal, we conclude that [0,KZ] 2 ^ - Fix an (adlo)-irreducible vector subspace W0 in g such that [W0,1&Z] 2 ^ZWe have Z = [XX,X2] G [Wi,VK2] Q [n r ,n] C n r + i C n°. Since [Wo,KZ] g KZ, we conclude that (&dZ)W0 ^ {0}. By Lemma 4.9.23, p. 106, this shows (adZ)Wo is isomorphic to Wo as a real (adlo)-module. However, (adZ)Wo C [n r+ i,fl] C n r + i C n°, which shows that (adZ)Wo is trivial as a real (ad(o)-module. We conclude that Wo is trivial as a real (ad Io)-module. That is, W0 C g°. Let qi := n r / n r + i and let q2 := n s / n s + i . Since [Wo,KZ] 2! K£> choose Y 6 Wo such that [Y,Z] <£ RZ. Since Y eW0Cg°, [Y,lo] = {0}. Then ad q i (Y) and ad qi (Io) centralize each other. Similarly, ad q 2 (y) and ad q2 (lo) centralize each other. Because the adjoint representations of fo on qi and on q2 both have only scalar intertwining, it follows that both ad Y : qi -> qi and a d F : q2 —> q2 are scalar maps. In particular, (a,dY)Xi € RXi + n r + i and (ad Y)X 2 € MX2 + n s + i . Choose ci,c 2 € R, Pi G n r + i and P2 G n s + i such that both (adY)Xi = CiXi + Pi and (adY)X2 = c2X2 + P2. We have (adY)Z = (ady)[Xi,X 2 ], so, by the Jacobi identity, (adY)Z

= [ciXi + PUX2] + [Xi,c2X2 + P2].

We have [Pi,X2] G [nr+i,W2] = {0} and [XUP2] G [Wuns+i] = {0}. Then [Y, Z] = (adY)Z = (ci + c2)[Xi,X2] = (ci + c2)Z G MZ, contradiction. D The next two results are Lemma 6.1, p. 465 and Corollary 6.2, p. 466 of [A00a]. L E M M A 4.15.11 Let G be a connected Lie group. Then the nilradical of G is simply connected iff Z° (G) is simply connected. Proof. Let JV denote the nilradical of G. By Theorem 3.6.2, p. 196 of [Va74], any connected subgroup of a connected, simply connected, nilpotent Lie group is again simply connected. So, as Z°(G) C JV, if JV is simply connected, then Z°(G) is simply connected. Assume, for a contradiction, that Z°(G) is simply connected and that JV is not simply connected.

148

Basic Lie Theoretic Results

By Lemma 4.14.8, p. 141, Z(N) is connected, but not simply connected, contradicting Lemma 4.14.7, p. 140. • COROLLARY 4.15.12 Let G be a connected Lie group and assume that g admits no one-dimensional ideal. Then Z(G) is discrete and the nilradical of G is simply connected. Proof. Since any vector subspace of 3(0) is an ideal of 0, we conclude that 3(g) = {0}, i.e., that Z(G) is discrete. Then, by Lemma 4.15.11, p. 147, we see that the nilradical of G is simply connected. • L E M M A 4.15.13 Let g be a Lie algebra, let n be a nilpotent ideal of g and let V be a vector subspace of g. Assume that n + V = g. Assume, for all P G ^ \ { 0 } , that a d P : g —)• g is not nilpotent. Then n PI V — {0} and n is the nilradical of g. Proof. For all P G n, since n is a nilpotent ideal of g, it follows that a d P : g —> g is nilpotent. By contrast, for all P G ^ \ { 0 } , a d P : g —• g is not nilpotent. Then nfl (V\{0}) = 0. Then nl~l V = {0}. Let no be the nilradical of g. Then, since n is a nilpotent ideal of 0, we see that n C no- Let X G no- We wish to show that X G n. Choose Y G n and Z G V such that X - Y + Z. Then Z = X - Y, so, since X G no and since Y G n C no, we conclude that Z G no- Then, since no is a nilpotent ideal of g, it follows that ad Z : g —> g is nilpotent. So, since Z G V, we conclude that Z = 0. Then X = Y G n. • L E M M A 4.15.14 Let g be a Lie algebra, let r be the solvable radical of g and let n be the nilradical of g. Assume that g ^ {0}. Then g is semisimple

iffv = {0}iffn = {0}. Proof. If g is semisimple, then 0 admits no solvable ideal, so r = {0}. Since n C r, we see that, if r = {0}, then n = {0}. Assume that n = {0}. We wish to show that g is semisimple. By (2) of Lemma 4.15.5, p. 143, [Q,V] C n = {0}. Then [r,t] = {0} and [g, t] C r, i.e., r is an Abelian ideal of g. In particular, r is a nilpotent ideal of 0. Then r C n = {0}. Then g is semisimple. •

Relationships

4.16

between representation

theories

149

Relationships between representation theories

The results in this section are not needed for this book. However, they basic to Lie theory, so we include them for the sake of completeness. There are many different kinds of representation theory: We have real and complex modules over real Lie groups and over real Lie algebras. Moreover, we have modules over complex Lie groups and over complex Lie algebras. The basic content of this section is that, for semisimple real or complex Lie algebras and for semisimple connected real or complex Lie groups, the problem of classifying irreducible modules reduces to two tasks: • Classify irreducible modules over complex semisimple Lie algebras. • For each real semisimple Lie algebra g, compute the action of the complex conjugation involution V •-> V on the class of irreducible complex modules over g. The process through which solutions to these two problems handle other kinds of modules is sketched below. We leave proofs as exercises to the interested reader. By Lemma 4.10.7, p. 114, for any (real) semisimple Lie algebra g, the problem of classifying all real 0-modules is reduces to the problem of classifying irreducible real fj-modules. We remark that Weyl's Theorem is a universal phenomenon for semisimple Lie theory. In particular, Lemma 4.10.7, p. 114 is also valid for • • • • •

complex modules over (real) semisimple Lie algebras; real modules over connected (real) semisimple Lie groups; complex modules over connected (real) semisimple Lie groups; modules over complex semisimple Lie algebras; and modules over connected complex semisimple Lie groups.

Let A be a complex Lie group or a complex Lie algebra. Let 1Z% denote the collection of all isomorphism classes of irreducible A-modules. Let A be a (real) Lie group or a (real) Lie algebra. Let 7 ^ denote the collection of all isomorphism classes of irreducible real ^-modules. Let 1Z% denote the collection of all isomorphism classes of irreducible complex A-modules. Define aA:1l%^ 11% by aA{V) = V. Let K% := HcAjaA\ element V G 1Z% satisfies one of the following two conditions: • there exists V eTZ% such that V = V and V = {V}; or

an

150

Basic Lie Theoretic

Results

• there exists V € ~R% such that V ^ V and V = {V, V}. 1. Modules over groups can be calculated from modules over algebras. Let G be either a connected (real) Lie group or a connected complex Lie group. If G is real, let F € {E,C}. If G is complex, let F := C. Let G be the universal cover of G. Let is : G —• G be the universal covering homomorphism. Let K := ker(7r), so if is isomorphic to the fundamental group 7i"i(G). For any G-module V, give V a G-module structure by, for all g £ G, for all ?; € V, defining gv := (7r(<7))r;. This gives an injection a:1l£ -* 71^. For any V € 7l£, we have: V € a(K%.) iff V is ^-trivial. G

G

The Lie algebra of G is isomorphic to $j, so differentiation gives a map yS : ft£ -> K%. From the Scholium after Theorem 3, p. 49 of [C46], one G

"

sees that /? is a bijection. Thus, via ^ o a , w e may identify 7£Q as a subset of TZg and reduce the calculation of TZQ to the calculation of that subset, or, equivalently, to the calculation of a(7^g). 2. Complex modules over real algebras can be calculated from modules over complex algebras. Let g be a (real) Lie algebra. Let f) := gc. For any complex g-module V, there is a unique extension of the {(-representation on V to a representation of g c on V. This gives a map 7 : 71^ -> Ttfj. The map 7 is bijective, so we may reduce calculation of 7£g to calculation of 71? • 3. Real modules can be calculated from complex modules and complex conjugation. Let A be a (real) Lie algebra or a (real) Lie group. For any V 6 T^A, one of the following two mutually exclusive possibilities occurs: (1) Vc is irreducible, whence Vc € 72.^. (2) for some V € 71% we have Vc = V ® V. We define S : 71^ -> 7*5 by the rules: • in Case (1), define S(V) := {Vc_}; and . in Case (2), define 8(V) := {V, V}. We remark that, in Case (2), it may happen that V = V, in which case we have S(V) = {V}. The map 6 : 72* —> 72^ is bijective, so we may reduce calculation of 72^ to calculation of 71% and of cr^ : 72^ -> 72^.

Chapter 5

More Lie Theory

5.1

Connection-preserving diffeomorphisms form a Lie group

Let M be a manifold, let V be a smooth connection on M and let FM denote the frame bundle of M. Let G be the collection of diffeomorphisms of M which preserve V. Let To denote the compact-open topology on G. Let T\ denote the topology on G of C°° convergence on compact subsets of M. Differentiating the action of G on M, we obtain an action of G on FM. The action of (G,TI) on FM is smooth. L E M M A 5.1.1 Let f € FM and let gi be a sequence in G. Assume that gif is convergent in FM. Then gi is convergent in T\. Proof. Let X := FM and let -K : X -> M be the frame bundle map. Let d := dim(M). Let L ~ GLd(M). Let n := dim(i). Then -K : X -»• M is a principal L-bundle. For all x € X, let Vx C TXX be the kernel of (d-K)x : TXX —• T^^M. Then V is a vector subbundle of TX. For all Y e I, for all x € X, let Yx := (d/dt)t=0(x.(exp(tY))). Let V 1 , . . . ,Yn be a basis of I. For all x € X, for all integers j € [l,n], let ysJ : = ( F i ) x ; then (T^ 1 ,... ,Y?) is an ordered basis of Vx. The connection A determines a vector subbundle H of TX such that, for all x EX, we have both Hx+Vx = TXX and HXDVX = {0}. Let m S M and let x = (a;i,..., xj) be an ordered basis of TmM. Then x e X. Define a := (d-?r)x\Hx. Then a : i l x -> T m M is a vector space isomorphism. For all integers k 6 [l,d], let Zx :— a"1(xk). Then (Zl,..., Z*) is an ordered basis of Hx. Then (y x \ . . . , Yf, Z\,..., Z*) is an ordered basis of TxX. 151

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Theory

Then (Y1,..., Yn, Z1,..., Zd) is a framing of X and, by construction, it is preserved by the action of G on X. So, by Lemma 3.1.3, p. 44, we see that the sequence x i-> gtx : X -)• X satisfies the convergence-divergence dichotomy. So, since gif is convergent in X, x H->- gix : X —> X converges in C°° on compact sets. Then gi : M —• M converges in n . D L E M M A 5.1.2 Let f 6 FM, let gi be a sequence in G and let g^ e G. Assume that g, —• g^ in TQ. Then gtf —• g^f in FM. Proof. Let ir : FM - > M b e the frame bundle map and let m := ir(f). Let m\ := gim. Let m'^ := g^m. Then mj -¥ m'^. If x e M, if V is an open neighborhood of x in M and if U is an open neighborhood of 0 in TXM, then we will say U is good for V at x if all of the following hold: • the exponential map is denned on U; • exp(£7) = V; and • exp \U : U —> V is a diffeomorphism. For all x € M, if V is an open neighborhood of x in M, then we will say that V is good at x if there is an open neighborhood of 0 in TXM which is good for V at x. Choose an open neighborhood V of m'^ in M such that V is good at m'oo- Replacing V by a smaller open neighborhood and passing to a tail, we may assume, for all i, that V is good at mj. For all i, let U[ be an open neighborhood of 0 in Tm>.M which is good for V at m\. Let U^ be an open neighborhood of 0 in Tmi^M which is good for V at m'^. For all i, we define e\ := exp \U!; then ej : U- -> V is a diffeomorphism. We also define e ^ := exp {U^; then e ^ : £7^, ->• V is a diffeomorphism. For all S C M, let 5 denote the closure in M of 5. Let V be a precompact open neighborhood of m in M such that V is good at m and such that gooV ^ y'- Since #, -> p ^ in ro, by passing to a tail, we may assume, for all i, that giV C V'. Let [7 be the open neighborhood of 0 in TmM which is good for V at m. Let e := exp \U\ then e : £7 -> V is a diffeomorphism. Let C be the collection of continuous maps U -> T M . By naturality of the exponential map, for all i, we have that (dgi)m :TmM -4- Tm>.M agrees on U with (e^) _1 o gt o e : U -> U[. Similarly, (c^oo)m : TmM -> T ^ M agrees on £7 with ( e ^ ) - 1 o g M o e : [/ -> E//^. So, since g, converges to g in TQ it follows that (dgi)m\U converges to (dg^mlU in the compact-open

The isometry group of a pseudoRiemannian manifold is a Lie group

topology on C. Then, for all u € U, we have {dgi)m(u) in TM. So, since (dgi)m is a sequence of linear maps, and of U is TmM, we conclude, for all w € TmM, that (dgi)m(w) in TM. Then, for all x € FmM, we have &£ ->• (jooZ in FM. we have gif -> goo/ in F M .

153

->• (d(/oo)m(u) since the span —> (dgao)m(w) In particular, •

L E M M A 5.1.3 We have r 0 = n . Proof. Let #i be a sequence in G and let g £ G. If 3 in T 0 . We wish to show that gi -^ g va. T\. Fix / G FM. Since <7i —• 3 in T 0 , by Lemma 5.1.2, p. 152, we have 9if -> gf- So, by Lemma 5.1.1, p. 151, we see that gi is convergent in TI. Choose g' 6 G such that #j -> g' in T\. Then gi -* g and gi -> g' in TQ , so 5 = 9'- Then ^ ->
T h e isometry group of a pseudoRiemannian manifold is a Lie group

Let M be a pseudoRiemannian manifold and let FM denote the frame bundle of M . Let V be the Levi-Civita connection of M. Let G be the

154

More Lie

Theory

collection of diffeomorphisms of M which preserve V. Let To denote the compact-open topology on G. Let T\ denote the topology on G of C°° convergence on compact subsets of M. Let G' := Isom(M). Let T'0 denote the compact-open topology on G'. Let T[ denote the topology on G' of C°° convergence on compact subsets of M. Differentiating the action of G on M, we obtain an action of G on FM. The action of (G, n ) on FM is continuous. L E M M A 5.2.1 We have

T^=T[.

Proof. By Lemma 5.1.3, p. 153, we see that r 0 = TJ. SO, since TQ is the relative topology on G' inherited from To and since r{ is the relative topology on G' inherited from T\ , we conclude that TQ = T[ . • By Lemma 5.1.4, p. 153, (G,TI) is locally compact. Then, since G' is Ti-closed in G, we conclude that (G',T[) is locally compact. Then, by Lemma 5.2.1, p. 154, (G',TQ) is locally compact. According to Theorem 2 in §5.2 on p. 209 of [MZ64], there is a unique Lie group structure on G' compatible with TQ. We give G' this Lie group structure. Then, by the Theorem in §5.2 on p. 212-213 of [MZ64], the action of G' on M is smooth. Following this discussion, for any pseudoRiemannian manifold N, we give Isom(iV) the unique Lie group structure compatible with the compactopen topology on N. We define Isom0(N) := (Isom(N))0. We call Isom 0 (N) the connected isometry group of N. Let H be a connected Lie group acting isometrically on M. Define a homomorphism / : H —> G' by (f(h))(m) = hm. Then, by continuity of the iJ-action on M, it follows that / : H —> (G',TQ) is continuous. Then, by Theorem 2.11.2, p. 93 of [Va74], / : H -» G' is smooth. That is, an isometric action of a Lie group on a pseudoRiemannian manifold defines a smooth map to the isometry group of the manifold.

5.3

More results on expansive sequences

Let G be a nontrivial connected Lie group. For every normal closed subgroup iV of G, let ITN • G -¥ G/N be the canonical homomorphism and if gi is a sequence in G, then we say that gi diverges mod normal subgroups in G if, for every normal connected closed subgroup ./V of G, we have: if

More results on expansive

sequences

155

N C G, then we have itN{gi) ->• oo in G/N. L E M M A 5.3.1 Let G be a connected semisimple Lie group with finite center and no compact factors. Assume that g is not Lie algebra isomorphic to sl2(IK). Let gi be a sequence in G. Assume that gi diverges mod normal subgroups in G. Then there is a sequnce V, of two-dimensional vector subspaces of g such that a subsequence of Ad gi is expansive on ViProof. Case 1: G is not simple. Proof in Case 1: Let o be an ideal of g such that {0} ^ o ^ g . By Lemma 4.10.26, p. 123, choose an ideal b of g such that o + b = g, such that o + b = g and such that [a, b] = {0}. Since {0} ^ a ^ g, it follows that g ^ b ^ {0}. Let A (resp. B) be the connected Lie subgroup of G corresponding to o (resp. b). Then A ^ { 1 G } i1 B, a fl b = {0} and G = AB. Moreover, A and B centralize each other. Then Z(A) C Z(G) and Z(B) C Z{G). So, since Z(G) is finite, it follows that both Z(A) and Z(B) are finite. For all i, choose a, € A and 6; G B such that gi = aibi. Since gi diverges mod normal subgroups in G, it follows that TTA(gi) —> oo in G/A and that nsigi) -> oo in G/B. For all i, we have itA(.9i) = ^A{bi) and •Ksigi) = 7i"s(aj)- It follows that bi —> oo in B and that a^ -» oo in A. Then, by Lemma 4.13.5, p. 137, choose X £ a and Y G b such that (Ada,i)X -t oo in a and such that (Adbi)Y -> oo in b. Passing to a subsequence, choose X' £ o and Y' eb such that (Adaj)X -1- X' in o and ( A d 6 i ) r - ^ y ' i n b. Then, by Corollary 2.6.3, p. 24, Adft is expansive on RX + MY. For all i, let Vi := MX + MY. End of proof in Case 1. Case 2: G is simple. Proof in Case 2: By Lemma 4.10.38, p. 127, choose Lie subgroups A and K of G such that o is a maximal real split torus in g such that K is a compact subgroup of G and such that G = KAK. For all i, choose ki,k € K and a* € A such that 0} and we define g + := V^ ga. Then, by Lemma 4.10.35, p. 125, we see that $ + ^ 0 and that dim(g + ) > 2. Choose X, y G g + such that X and Y are linearly independent. Then (Adoj)X -»• X and ( A d a ^ F ->• Y. Thus, by Corollary 2.6.3, p. 24, we see

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More Lie Theory

that Adoj is expansive o n K X + M F . For all i, let V, := ^(RX End of proof in Case 2.

+WY). O

L E M M A 5.3.2 Let Go be a connected Lie group. Let R be the solvable radical of Go- Assume that Z{GQ/R) is finite. Assume that g0/v is not Lie algebra isomorphic to 5(2 (1R). Let no • Go —> Go/R be the canonical homomorphism. Let gi be a sequence in Go- Assume that ito(gi) diverges mod normal subgroups in GQ/R. Then there is a sequence Vi of two-dimensional vector subspace of go such that Ad gi is expansive on Vi. Proof. Let Go := Go/R. For all i, let ~gl : = ^oC^i)- By Lemma 5.3.1, p. 155, choose a sequence Wi of two-dimensional vector subspaces of go such that Ad gi is expansive on W». Let V be a vector subspace of go such that V + r = go and such that VC\t= {0}. L e t p := (d7r0)\V; t h e n p : V ->• g 0 / t is an isomorphism of vector spaces. For all i, let Vi := p~1(Wi). Since Ad(7To(<7i)) is expansive on (d-ir0)(Vi), by Lemma 2.6.4, p. 25, we see that Ad<7f is expansive on Vi. O 5.4

Lie groups densely embedded in other Lie groups

If G is a Lie group, then we will say that G is Ad-proper if the Adjoint representation Ad : G -> GL(g) is a proper map. If G is not Ad-proper, then we say that G is Ad-nonproper. Equivalently, there is a sequence gt in G such that gi —> oo in G and such that AdB( GL(g). The result then follows from Lemma 2.7.7, p. 29. • An action on a topological space is minimal if every orbit is dense and topologically transitive if some orbit is dense. Then minimality is stronger than topological transitivity and, for nontransitive actions, we see that topological transitivity implies orbit nonproperness. Below, we give some other standard dynamical conditions and indicate their relationship with properness and orbit properness. A measure space is a Borel space with a Borel measure. A subset of a measure space is null if it is contained in a Borel set of measure zero. A

Lie groups densely embedded in other Lie groups

157

subset of a measure space is conull if its complement is null. A subset 5 of a measure space space is measurable if there is a Borel subset B such that both S\B and B\S are null. Let G be a topological group acting on a measure space M. The action is said to be Borel if the action map G x M -»• M is Borel. The G-action on M is said to be quasi-measure preserving or qmp if both • it is Borel; and • for all g £ G, for all null sets Z C M, we have that gZ is null. We say that the G-action on M is essentially transitive if both • it is qmp; and • there is a conull orbit, i.e., there is some m £ M such that is null.

M\(Gm)

The G-action on M is ergodic if both • it is qmp; and • every G-invariant measurable set is null or conull. We say that the G-action on M is properly ergodic if it is both ergodic and not essentially transitive. The null algebra of a measure on a Borel space is the collection of all null sets. We say that two measures on a Borel space are equivalent if they have the same null algebra. The class of a measure is its equivalence class, i.e., the collection of all measures that have the same null algebra. Quasi-measure preservation, essential transitivity, ergodicity and proper ergodicity all depend only on the measure class of a measure. L E M M A 5.4.2 Let G be a topological group and let H be an lose topological group. Let i : G —> H be an injective homomorphism. Assume that i(G) is dense in H. Let ft be a left Haar measure on H. Let G act on H by the rule g.h — (b(g))h. Then the G-action on H is minimal and free. Moreover, the G-action on (H, /x) is ergodic. Moreover, if t(G) ^ H, then the G-action on (H, n) is properly ergodic. Proof. The G-orbits in H are the right cosets of L(G) in H. Since t(G) is dense in H, it follows that the right cosets of t(G) are all dense in H. Therefore the G-action on H is minimal. Since L : G —> H is injective, we see, for all g £ G, for all h £ H, that g.h = h =$• g = 1Q- Therefore

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Theory

the G-action on H is free. Finally, the G-action on (H,n) is ergodic by Lemma 2.2.3, p. 20 of [Zi84]. Assume that u(G) ^ H and assume, for a contradiction, that the G-action on (H, fi) is essentially transitive. Choose ho € H such that G.ho is conull. Since t(G) ^ H, it follows that (i(G))h0 ^ H. That is, G.h0 ^ H. Choose h'0 € H\(G.h0). Then (G.h0) n (G.h'o) = 0. Define / : H -¥ H by /(/i) = hholh'0. Then /*(;«) is a left Haar measure on H, so choose A > 0 such that / * ( / J ) = A/z. Then, for any conull subset S of (H,n), f(S) is conull in (H,f*(n)). Then /(G./i 0 ) is conull in (if, A//), and, therefore, in (H,fj,). Since both G./io and f(G.ho) are conull in (H,fi), it follows that (G./i0) D (f(G.h0)) ± 0. However, f(G.h0) = G.h'0 and (G./io) n (G.h'0) = 0, contradiction. • Let d > 1 be an integer. If M is a smooth d-dimensional manifold and S C M, then we say that S is negligible if, for every open set U C M, for every diffeomorphism c : {/ -> E d , we have that c(S D U) is a set of Lebesgue measure zero in E d . There is a unique measure class on M whose null algebra is the collection of negligible subsets of M; this measure class is called the smooth measure class. On a Lie group, Haar measure is in the smooth measure class. For any manifold M, for any nonempty open subset U of M, for any measure fi in the smooth measure class of M, we have fi(U) > 0. For actions of groups on manifolds, unless otherwise specified, the action is said to be ergodic (resp. properly ergodic) if it is ergodic (resp. properly ergodic) with respect to the smooth measure class. L E M M A 5.4.3 Let G be a Lie group. Assume that there are a Lie group H and an injective homomorphism f : G —> H such that f(G) is not closed in H. Then there exists a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action of G on a Riemannian manifold. Proof. Let i ? be the closure in H of f(G). Then f(G) C H. By Corollary 4.5.7, p. 86, /(G) is a Lie subgroup of H. Let G act on H by the rule g.h — (f(g))h. By Lemma 5.4.2, p. 157, this G-action on H is free, minimal and properly ergodic. Since minimal implies topologically transitive and since properly ergodic implies nontransitive, it follows, from Lemma 7.11.13, p. 247, that the G-action on H is nontame and orbit nonproper. Any Lie group admits a left-invariant Riemannian metric. Any left-invariant Riemannian metric on H is G-invariant. •

Lie groups densely embedded in other Lie groups

159

L E M M A 5.4.4 Let G be a connected Lie group such that Z(G) is noncompact. Then there exists a connected Lie group H and an infective homomorphism f : G -4 H such that f(G) C H and such that f(G) is dense in H. In particular, f(G) is not closed in H. Proof. Let T := K/Z. By Lemma 4.9.22, p. 105, [Z{G)]/[Z°(G)] is finitely generated. So, by Lemma 4.9.27, p. 107, Z(G) is compactly generated. By Lemma 4.9.28, p. 108, let S be an infinite, discrete cyclic subgroup of Z(G). Let e : S -4 T be an injective homomorphism to the circle group. Let T := e{S). Then T is not closed in T. Let N := {(s,e(s))\s € S}. Then AT is a closed, normal subgroup of G x T. Let H := (G x T)/iV. Let p : G x T —> H be the canonical homomorphism. Define a Lie group homomorphism / : G -4 H by f(g) = p(g, IT). We have / ( G ) = p(Gx{lj}) so p _ 1 ( / ( G ) ) = (G x {1T})N = G x T. Then / ( G ) = p{G x V). So, since G x T is dense in G x T, it follows that / ( G ) is dense in H. We have P _ 1 ( / ( G ) ) = G x T ^ G x T = p~l{H), so / ( G ) C if. It remains to show that f : G —> H is injective. Let g € G and assume that /(g) = 1#. We wish to show that g = 1QWe have p(g, IT) = /(• exp(tX) : R -> G is both proper and injective. By Lemma 4.9.10, p. 101, it suffices to show that Sx is not precompact in G. Assume, for a contradiction, that the closure K in G of Sx is compact. Then f(K) is a compact subgroup of H and f(SX) C /(AT). Then we have G* C / ( i f ) C / ( G ) , contradiction. •

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5.5

More Lie Theory

Generalities on the Levi decomposition

L E M M A 5.5.1 Let g be a Lie algebra, let I a semisimple Levi factor of g and let t be the solvable radical of g. Let f : g —¥ t) be a surjective Lie algebra homomorphism. Then /(I) is a semisimple Levi factor of F) and /(r) is the solvable radical o/h. Proof. As /(I) is a semisimple Lie subalgebra of F), as / ( r ) is a solvable ideal of F) and as (/([)) + (/(t)) = h, we are done by Lemma 4.9.1, p. 96.D The following is Lemma 7.7 of [A99b]. L E M M A 5.5.2 Letg be a Lie algebra, letv be the solvable radical of g and let I be a semisimple Levi factor ofg. Let X\ e I, Pi € r. Set Wi := Xi +Pi. Then ad Wi : g —> g is nilpotent iff both ad Xi : g —> g is nilpotent and ad Pi : g —»• g is nilpotent. Proof. Let n be the nilradical of 0. Proof of "only if": Let g := g/r. Then the map adWi : $j -» g is nilpotent. Let IT : g —> g be the canonical Lie algebra homomorphism. Then p := ir\l: I —¥ g is a Lie algebra isomorphism. We have po (adi(Xi)) o p " 1 = adj(Xi) = adg(Wi). Therefore, for all integers n > 0, wehavepo[(ad[(X 1 )) n ]op _1 = (ad^(Wi))n. Then a d X i : I - • Us nilpotent. Then, by Corollary 4.11.7, p. 130 (with g replaced by I, with V replaced by g and with p replaced by ad : I —> gl(g)), we conclude that a d X i : g —> g is nilpotent. It remains to show that the map ad P : g —• g is nilpotent. Let tj := r and recursively define tj + i := [n, tj]. This yields a chain r i 5 *2 2 r 3 5 " " of ideals in 0 such that, for all sufficiently large i, we have tj = {0}. For all integers i > 1, let Sj := tj/tj+i. It suffices to show, for all integers j > 1, that ad Pi : Sj ->• Sj is nilpotent. Fix j . Since adWi : 0 -> 0 and a d X i : 0 -> g are both nilpotent, it follows that adWi : Sj —¥ Sj and a d X i : Sj —> Sj are both nilpotent. By (2) of Lemma 4.15.5, p. 143, we see that [Xi,Pi] € n, so [Wi,X1] = [X1+PUX1]

= [P1,X1]€

n,

which implies that (ad[Wi,Xi])sj = {0}, so ([adWi,adXi])Sj = {0}. Then adWi : Sj -¥ Sj and a d X i : Sj -¥ Sj are commuting nilpotent

Generalities on the Levi decomposition

161

endomorphisms. So, as Pi = W\ - X\, it follows that ad Pi : Sj -> Sj is also nilpotent. End of proof of "only if". Proof of "if": By (1) of Lemma 4.15.5, p. 143, Pi £ n. Let gi := g and recursively define gj+i := [n,g,]. This yields a chain gi 2 02 2 03 2 • • • of ideals in g such that, for all sufficiently large i, we have gi = {0}. For all integers i > 1, define q* :— gi/gi+i- It suffices to show, for all integers j > 1, that ad Wi : qj -> qj is nilpotent. Fix j . Since Pi e n, it follows that (adPi)qj = {0}. Then adWi : qj -> qj is equal to a d X i : qj -> qj. As a d X i : g -» g is nilpotent, we see that a d X i : qj —> qj is nilpotent, so ad W\ : qj —• qj is nilpotent. • COROLLARY 5.5.3 Let g be a Lie algebra, let r be the solvable radical of g, let n be the nilradical of g, and let I be a semisimple Levi factor of g. Let Xi € I, Pi € r. Set Wx := Xi+Px. Then ad Wi : g -» g is nilpotent iff both ad Xi : g -» g is nilpotent and Pi G n. Proof. This follows by combining (1) of Lemma 4.15.5, p. 143 with Lemma 5.5.2, p. 160. • L E M M A 5.5.4 Let g be a Lie algebra, let n be the nilradical of g and let I be a semisimple Levi factor of g. Let X € g. Assume that ad X : g —> g is nilpotent. Then there exists an ideal to of I such that X Gfo+ n and such that lo has no compact factors. Proof. Let r be the solvable radical of g. Let n : g —> g/t be the canonical Lie algebra homomorphism. Then ir\{ : I -> g/v is a Lie algebra isomorphism. Choose Y € I such that ir(Y) = n{X). Since ad A" : g ->• g is nilpotent, a d X : g/x -> g/r is nilpotent, so adY : I -> I is nilpotent. Choose 6 and m as in Lemma 4.10.30, p. 124. Let a : 6©m -»• [ be a Lie algebra isomorphism. Choose (P, Q) 6 t 0 m such that a(P, Q) = Y. Then ad P : 6 —• 6 is nilpotent and ad Q : m -» m is nilpotent. By Lemma 4.10.30, p. 124, we see that P = 0. Let lo := a ( { 0 } © m ) . Then [0 is an ideal of l and fo has no compact factors. We have ir{X) = n(Y), so X € Y + t. Let Z := X - Y. By Corollary 5.5.3, p. 161 (with W\ replaced by X, Xi replaced by Y and Pi replaced by Z), Z e n. Since Y = a(P,Q) = a(0,Q), we conclude that Y e lo- Then X = Y + Zelo + n. • The following is similar to Lemma 4.8, p. 463 of [A00a].

162

More Lie Theory

L E M M A 5.5.5 Let G be a connected Lie group and let LQ be a connected semisimple Lie subgroup of G. Then there is a finite index subgroup of Z(LQ) which is a subset of Z(G). Consequently, if Z(G) is finite, then Z(LQ) is finite, as well. Proof. By Lemma 4.13.3, p. 135, we know that Ad 0 (L o ) is an almost algebraic subgroup of SL(g). Let ZQ := Z(Adg(L0)). Then Z0 is an almost algebraic subgroup of SL(g). Since Ad 0 (X o ) is semisimple, it follows that ZQ is discrete. Then, by Lemma 4.13.2, p. 135 (with G replaced by Zo), we see that ZQ is finite. So, since Ad B (Z(L 0 )) C Zo, we conclude that there is a finite index subgroup of Z(LQ) which is contained in the kernel of Ad 8 : G -> GL(g), i.e., which is contained in Z(G). O L E M M A 5.5.6 Let G be an Ad-proper connected Lie group. Assume that g admits no one-dimensional ideal. Then Z(G) is finite. Moreover, for any semisimple connected Lie subgroup LQ of G, we have that Z(Lo) is finite. Proof. Since G admits no one-dimensional ideal, by Corollary 4.15.12, p. 148, Z(G) is discrete. Since G is Ad-proper, it follows, from Lemma 2.7.7, p. 29, the kernel of Ad : G —> GL(g) is compact. That is, Z{G) is compact. Then Z(G) is compact and discrete, hence finite. By Lemma 5.5.5, p. 162, for any semisimple connected Lie subgroup LQ of G, Z{LQ) is finite. • L E M M A 5.5.7 Let g be a Lie algebra, let I be a semisimple Levi factor of g and let t be the solvable radical of g. Let XQ € I, let P G t and let X := X0 + P. Assume that a d X : g —> g is nilpotent. Let k > 0 be an integer. Assume that (adX) fc n = {0}. Then (adX 0 )*n = {0}. Proof. Let n be the nilradical of g. By Corollary 5.5.3, p. 161, we have that a d X 0 : 0 -> 0 is nilpotent and that F e n . We may assume that X0 ^ 0. By Lemma 4.10.33, p. 125, choose Y0,T0 € I such that X0,Y0,T0 is a standard s ^ R ) basis of a Lie subalgebra s of 0. Let no, tii, • • • be the descending central series of n. Since n is nilpotent, fix an integer r > 0 such that n r = {0}. For all integers i > 1, we define mj := rii-i/ni; then (adX)*mj = {0}. For all integers i > 1, we have ad m i (P) € ad m i (n) = {0}, so ad m ; (X) = ad m ,(X 0 ). Then, for all integers i > 1, (adXo^m; = (&dX)kmi = {0}. By Lemma 4.10.7, p. 114, the adjoint representation of s on n is completely T

reducible, so, as (ad s)-modules, n is isomorphic to ffi m*.

Large normalize™ and centralize™

Then(adX o ) f c n = {0}.

163

•

L E M M A 5.5.8 Letg be a Lie algebra, letso be a simple Lie subalgbra ofg and let n denote the nilradical of g. Let XQ € So\{0}, let P € n and define X :— X0 + P. Assume that (adXo)n ^ {0} and dim((adX)n) < 1. Then there is exactly one vector subspace ofn which is both (ad Bo) -nontrivial and (ad So)-irreducible. Proof. Since (adXo)n ^ {0}, the adjoint representation of So on n is nontrivial. By Lemma 4.10.7, p. 114, the adjoint representation of So on n is completely reducible. Existence follows. Assume for a contradiction, that V and V" are distinct (adso)-nontrivial, (adso)-irreducible vector subspaces of n. By irreducibility, we have V D V" = {0}. By simplicity of So, since X0 € So\{0} and since ady(so) ^ {0} ^ ady//(s 0 ), we conclude that adv-(Xo) ^ {0} # adv»(JC 0 ), »•«., that (ad Jf0)V" ^ {0} ^ (adX 0 )V". So, since (adX0)V C V since (adX 0 )V" C V" and since V n V" = {0}, we see that dim([(adX 0 )V"] + [(adX 0 )V"]) > 2- Then dim((adX 0 )n) > 2. Let Vo, V i , . . . be the descending central series of n. As n is nilpotent, fix an integer k > 0 such that Vk = {0}. Let V„ := (Vb/Vi) @ • • • © (V*_i/Vk). By Lemma 2.4.3, p. 21, we get dim((adX)V.) < dim((adX)n). By assumption, dim((adX)n) < 1. Then dim((adX)V*) < 1. By Lemma 4.10.7, p. 114, the adjoint representation of So on n is completely reducible, and is therefore isomorphic to the adjoint representation of s 0 on V*. Then we have dim((adX 0 )n) = dim((adX 0 )V*)- Since P e n , adv.(P) = 0, soady.(Xo) = ady.(X), sodim((adX 0 )V,) = dim((adX)Vi). Then dim((adX 0 )n) = dim((adX 0 )K) = dim((adX)V,) < 1. However, we already established that dim((adX 0 )n) > 2, contradiction. D

5.6

Large normalizers and centralizers

The following is a variant of Lemma 14.2, p. 480 of [A00a]. L E M M A 5.6.1 Let s be a semisimple Lie algebra and let X € s\{0}. Assume that a d X : s —> s is nilpotent and that the codimension in s of cs(X) is < 2. Then there is an ideal Si of s such that Si is isomorphic to 0(2(1^) and such that X 6 S\. Moreover, codim B (c s (X)) = 2.

164

More Lie Theory

Proof. Let s 0 := el2(R). By Lemma 4.10.33, p. 125, choose Y,T G s such that X, Y, T is a standard s^iR) basis of a Lie subalgebra Si of s. Then Si is Lie algebra isomorphic to SQ. By Lemma 4.10.7, p. 114, choose an (adsi)-invariant vector subspace V of s such that V D Si = {0} and such that V + Si = s. Let 6 be the kernel of ad : si —> fll(V). The codimension in s of c s (X) is < 2 so dim((adX)s) < 2. As X,Y,T is an sf2(K) basis of Si, it follows that dim((adX)si) = 2. Then dim((adX)s) = 2 and dim((adX)V) = 0. As ( a d X ) F = {0}, we conclude that X G I So, as X ^ 0 and as Si is simple, it follows that t = s±. Then V C c s (si). Then we have n B (si) D sx + (c s (si)) D Si + V = s. Then Si is an ideal of s. Since dim((adX)s) = 2, the codimension in s of cs(X) is = 2. • L E M M A 5.6.2 Let g 6e o semisimple Lie algebra and let X G fl\{0}. 77ien the codimension in g of c0 (X) is>2. Proof. Let go : = Cg(-^Q- Let n be the codimension in g of go- Assume, for a contradiction, that n < 2. Since 3(3) = {0}, n > 0. Then dim(fl/g0) = n = 1 and (adX)g0 = {0}. For all P G 0, let P ' := ad 0 (P) G End(fl). By Lemma 4.10.28, p. 123 (with V replaced by g and g replaced by ad 0 (g)), choose D, E, N G g such that the equation X' = D' + E' + N' is a real Jordan decomposition on g. Since 3(g) = {0}, we see that ad : g -> gl(g) is injective. Consequently, as ad B (X) = X' = D' + E' + N' = &dg(D + E + N), it follows that X = D + E + N. Let S C End(g) denote the E-linear span of X', (X1)2,.... Then N' G 5 , so ker(N') D ker(X'). That is, cg(N) D c B (X). Then codim0(c0(Af)) < codim 0 (c 0 (X)) = n = 1, so, by Lemma 5.6.1, p. 163, we have N = 0. Then &dX : g -> g is complex diagonalizable. Since dim(g/go) = 1, it follows that the unique characteristic root of a d X : g/g 0 -»• g/go is real. So, since 0 is the only characteristic root of the zero map a d X : g 0 —> go, we see that every characteristic root of ad X : g - • g is real. Then ad X : g -> g is real diagonalizable. Let 0 be a maximal real split torus of g such that X G a. Let $ be the set of roots of a on g. For all G $, let g^ denote the 0-rootspace of a on g. Choose V G $ such that V W # 0. Let B := RKF(g,a) G SBF(a*). Then $ is a root system in (a*,B). Then — tp G $. Since ip,—ip G <J>, it follows that g,/, 5^ {0} 7^ g_^,. Choose Y G 0 v \ W a n d Z e fl-t/>\{0}- F o r all (s,t) G K 2 \{(0,0)}, we have [X, Y + tZ} = (s- (
MX))

• Z) ? 0.

Large normalize™ and

centralizers

165

Then ( I F + WZ) f~l (cg(X)) = {0}, and y and Z are linearly independent. So codim B (c 0 (X)) > 2, contradiction. • The following is Lemma 14.3, p. 481 of [AOOa]. L E M M A 5.6.3 Let g be a Lie algebra. Assume that s^M) is not a direct summand of g. Let X € g and assume that a d X : g -> g is nilpotent. Assume that the codimension in g o/n B (X) is < 1. Let n denote the nilradical ofg. ThenX € n. Proof. Let r be the solvable radical of g and let I be a semsimple Levi factor of g. Choose Y £ I and Z G r such that X = Y + Z. By Corollary 5.5.3, p. 161, we see that a d y : g —> g is nilpotent and that Z € n. It suffices to show that Y = 0. Assume, for a contradiction, that Y ^ 0. As a d X : g —> g is nilpotent, choose an integer n > 0 such that (adX) n jj = {0}. Let p : g —> g/v be the canonical Lie algebra homomorphism. Let q := p\{ : I -> g/r; then q is a Lie algebra isomorphism. We define IT := q~l op : g —» I. Then 7r is a Lie algebra homomorphism and ir(X) = Y, so, for all Peg, 7r((adX)P) = (adY)(ir(P)). Then (ady)"I = n((a,dX)ng) = {0}, so a d y : ( -»• ( is nilpotent. Moreover, since 7r(X) = Y and since 7r : g —> I is surjective, we conclude that codim[(n[(y)) < codim g (n g (X)). Then codimi(ni(y)) < 1. So, since ( a d y ) ( m ( y ) ) C i y , we get dim((ady)l) < 2. So, since ci(Y) is the kernel of a d y : I —• [, we get codhri[(c[(y)) < 2. Then, by Lemma 5.6.1, p. 163, choose an ideal lo of I such that IQ is isomorphic to s b (R) and such that y £ lo- Moreover, by Lemma 5.6.1, p. 163, the codimension in I of C[(Y) is — 2. That is, the kernel oi adY : I -> I has codimension = 2 in [. Consequently, we have dim((ady)() = 2. Since p\[ : I -¥ fl/r is injective, we get dim(p((ady)l)) = dim((ady)[). Moreover, since p : g —>• I is a Lie algebra homomorphism, and since we have p{X) = Y = p{Y), we get p((adX)t) = p((ady)[). Then dim(p((ad Jf)l)) = dim(p((ady)0) = dim((ady)I) = 2. Fix a two-dimensional subspace V of (adX)I such that y n (ker(p)) = {0}. As ker(p) = r, we get V D r = {0}. Since (adX)(n 0 (X)) C MX and since codim B (n B (X)) < 1, we conclude that dim((adX)fl) < 2. Since F n ( ( a d X ) r ) C Vnv = {0} and dim(^) = 2, we conclude that dim(F + ((adX)r)) = 2 + [dim((adX)r)]. Then 2 + [dim((adX)r)] = dim(y + ((adX)r)) < dim((adX)fl) < 2.

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Theory

Then (adX)t = {0}, i.e., ad c (X) = 0 , so, by Lemma 4.10.18, p. 119 (with / replaced by ad : g -» fll(r))> w e have ad r (F) = 0. Then 7 ^ 0 and Y is in the kernel of ad : fo -> 9KX)-- So, by simplicity of lo, we conclude that adc(lo) = {0}, i.e., that [I0,t] = {0}. Then, by Lemma 4.15.6, p. 143, we have I0 10- Then sl2(K) | g, contradiction. • The next lemma is equivalent to Corollary 14.6, p. 483 of [A00a]. L E M M A 5.6.4 Let g be a Lie algebra and let n be the nilradical of g. Let X s n\(3(n)) and assume that the codimension in g ofng(X) is < 1. Then g admits a one-dimensional ideal. Proof. Since X £ 3(n), we conclude, from Lemma 4.14.2, p. 138, that nn(X) 7^ n. Then, as n fl (ng(X)) = nn(X), we see that ng(X) ^ g. Then codim 0 (n 0 (X)) ^ 0. So, since codim B (n 0 (X)) < 1, we conclude that codim B (n B (X)) = 1. We have nn(X) = nf"l (n B (X)). Therefore, codimn(nn(-X")) < 1. Since X 6 n\(3(n)), we get cn(X) ^ n. So, as n is nilpotent, by Lemma 4.14.2, p. 138, n n (X) ^ n. Thus the codimension in n of nn(X) is = 1. Choose Y E n\(n n (X)). Then Y e g\(n0(X)), so g = (n B (X)) + (WY). Let Xi := X and recursively define Xi+i :— [Xi, Y], for i — 1,2,3, Because n is nilpotent, choose an integer fco > 0 such that Xk0 — 0. Let / denote the set of all integers i > 0 such that n 0 (X;) = ng(X). Since Xi = X, it follows that l e i . Since g ^ ng(X), it follows that k0 £ I. Choose an integer k > 0 such that l,...,k £ I, while k + 1 fi I. Then ng(X) = n„(Xi) = • • • = ng(Xk) ? ng(Xk+1). So nn(Xk)

= n n (ng(Xk))

= n n (n B (X)) =

nn(X),

so the codimension in n of nn(Xk) is = 1. Moreover, Y fi n n (X) = nn(Xk), so [Xfc,F] $ MXk, i.e., Xk+i £ RXk. In particular, we have Xk+i ^ 0. By Lemma 4.15.2, p. 141, we have ng(Xk) C ng(Xk+i). So, since we have ng(X) = ng(Xk) ^ ng(Xk+1), we get n 0 (X) C ng(Xk+1). So, since the codimension in g of n 0 (X) is = 1, we conclude that nB(Xfc+i) = g, so MXk+i is an ideal of 0. As Xk+1 ^ 0, we have dim(ILX"fc+i) = 1. • 5.7

Representation theory

The next result is Lemma 7.11 of [A99b].

Representation

theory

167

L E M M A 5.7.1 Let Go be a connected Lie group and let No be the nilradical of Go. Let LQ be a semisimple Levi factor of Go • Let V be a vector space and let p : Go —• GL(V) be a representation. Let I : V —> V be the identity map. Assume that p(No) = {/} and that V has nonscalar (p(Lo))-intertwining. Then V has nonscalar (p(Go))-intertwining. Proof. Let Ro be the solvable radical of GoCase 1: Assume, for all r 6 Ro, that p(r) € MI. Let T : V -*• V be a (/9(Lo))-equivariant linear transformation such that T £ MI. For all r € Ro, because p(r) G MI, it follows that T(p(r)) = (p(r))T. That is, T : V -> V is (jo(.Ro))-equivariant. Then T is (p(L0.Ro))-equivariant, so, as Go — LoRoi we are done. End of Case 1. Case 2: Assume, for some r0 € -Ro, that p(r0) $L MI. Let Gi := p(Go). By Lemma 4.5.7, p. 86, we see that G\ is a connected Lie subgroup of GL(F) and that p : Go —• G\ is a surjective Lie group homomorphism. Let i?i be the solvable radical of G\. By Lemma 5.5.1, p. 160, we have p(Ro) = R\. By (2) of Lemma 4.15.5, p. 143, we have [go, to] C n 0 . Then [fli,ti] = p([flo,r0]) C (dp)(no) = {0}. Then, by Corollary 4.6.3, p. 91, [Gl,R1] = {I},i.e.,R1CZ(G1). Then p(r0) € p(Ro) = Ri C Z(Gi) = Z(p(G 0 )). Then p(r 0 ) : F -»• V is (p(Go))-equivariant. So, as p(fo) £ M/, we are done. End of Case 2. D L E M M A 5.7.2 Lei 0 6e a Lie algebra. Let V be an irreducible real gmodule. Assume that V has only scalar g-intertwining. Let B, B' € SBF(g) both be g-invariant. Assume that B ^ 0. Then B' € MB. Proof. As B ^ 0, it follows that ker(J5) C g. So, as ker(B) is g-invariant, and as V is fl-irreducible, ker(B) = {0}, i.e., B is nondegenerate. Let F : V -» V* and F' : V -> F* be denned by (F(u))(«;) = B(u,iu) and (F'(v))(w) = B'(v,w). As B is nondegenerate, it follows that the map F : V -»• F* is a vector space isomorphism. Let T := F _ 1 o F ' : V -> V. Then T is 0-equivariant, so, as V has only scalar g-intertwining, choose e e l such that, for all v €V, T(V) = cv. We wish to show that B' = cB. For all v EV,vte have F'(v) = F(T(v)) = F(cv). Then, for all v, w £ V, B'(y,w) = (F'(v))(w) = (F(cv))(w) = B(cv,w) = c- (B(v,w)). • Let g be a Lie algebra. Note that if V is an irreducible g-module which has nonscalar intertwining, then it is possible for there to exist g-invariant B, B' £ SBF(V) such that B ^ 0 and such that B' $. MB: Let g consist of

168

More Lie Theory

all 6 x 6 real matrices

(

-

)

such that 1 , 7 6 so(3). Let E 6 x l be a real g-module under matrix multiplication. Let I denote the 3 x 3 identity matrix. Let B and B' be the symmetric bilinear forms on R 6 x l whose matrices (with respect to the standard basis of E 6 x l ) are, respectively,

0-°,) -

(;.')•

Then B and B' are both g-invariant. Moreover, B ^ 0 and B' £ RB. The situation is better for positive definite symmetric bilinear forms invariant under a representation of a compact Lie group: Let K be a compact Lie group and let V be a real if-module. Then there is a if-invariant positive definite symmetric bilinear form on V. Moreover, if V is .^-irreducible and if B,B' € SBF(V) are both positive definite and -ftT-invariant, then there exists a > 0 such that B' = aB. Finally, if V is if-irreducible and if B 6 SBF(y) is nonzero, if-invariant and positive semidefinite, then B is positive definite. These facts are well-known and will not be needed here, so we omit their proofs. L E M M A 5.7.3 Let V be a one-dimensional vector space, let L be a connected Lie group and let p : L —> GL(F) be a representation. If L — [L,L], then p is trivial. Proof. Since V is one-dimensional, GL(F) is Abelian, so p(L) is Abelian, i.e., [p{L),p(L)} is trivial. We have p(L) = p([L,L]) C [p(L),p(L)]. Then p(L) is trivial. • L E M M A 5.7.4 Let X,Y,T be a standard B[2(IR) basis of a Lie algebra g. Let V be a vector space and let p : g —t gl(V) be a representation. Then p(T) : V —> V has an even number of nonzero characteristic roots. Proof. Let S be the set of characteristic roots of p(T) : V -> V. It follows from the representation theory of s ^ W (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]), that, for all A £ 5, we have —A € S. Consequently, 5\{0} has an even number of elements. •

Representation

theory

169

Recall, from §2.12, p. 39, that Q3 is a quadratic form on R3 defined by Q3(x,y,z) = 2xz + y2. L E M M A 5.7.5 Let g be a Lie algebra which is Lie algebra isomorphic to 5b(R)- Let V be a three-dimensional vector space. Let p : g —> gi(V) be an irreducible representation. Then, for some F € MinkQF(V), p(g) = so(F). Proof. As Q3 has signature (2,1), it follows from Coincidence (i) on p. 591 in §X.6.4 of [He78], that s[ 2 (!) = so(Q3). We therefore assume that g = so(<33). The canonical representation of so(Q3) on R 3 x l is irreducible. Moreover, by the representation theory of s ^ W (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]) there is, up to isomorphism, a unique three-dimensional irreducible representation of g. We therefore assume that V = R 3 x l and that p : so(Q3) -> g[3(E) is the inclusion. Let F := Q3. • L E M M A 5.7.6 Let LQ be a connected Lie group. Assume fo is Lie algebra isomorphic to sl2(R). Let V be an irreducible real L^-module. Assume that dim(V) = 3. Let U be a sequence in LQ. Then, after replacing li by a subsequence, there exists a two-dimensional vector subspace VQ of V such that, for all v € Vo, there is a precompact sequence Vi in V such that liVi ->• V.

Proof. Let p : LQ —> GL(V) be the representation underlying the real Lo-module structure on V. By Lemma 5.7.5, p. 169, fix Q £ MinkQF(V) such that p(L0) = SO°(Q). Replacing LQ by p(Lo), we may assume that L0 = SO°(<2) and that p : LQ —> GL(V) is the inclusion map. There is an isomorphism V —»• E 3 x l which carries Q to Q3; we may therefore assume that V = E 3 X 1 and that L0 = SO°(<33). For all A > 0, we define diag(A) := \E[f + ( 1 / A ) ^ € L0. We define A+ := {diag(A) | A > 1}. By Lemma 4.10.38, p. 127, choose a compact subgroup K of L0 such that L0 = KA+K. For all i, choose ki,k[ € K and choose a, € A+ such that li = kiaik[. Let Vi := Ee^3) + Ee 2 3) C R 3 x l =V. Then, for all w e Vi, by matrix multiplication, we have that {a^w}^ is precompact in V. Passing to a subsequence, assume that fcj is convergent in K. Let fcoo := lim ki. Let Vo := fcooVi. Let v € V0. We wish to show that i—too

there is a precompact sequence Vi in V such that Uvi —> v. Let w := k^v. Then w G Vi. For all i, let v* := (fc?)_1af 1 io. Then liVi = kiW —>• kooW = v.

•

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Theory

L E M M A 5.7.7 Let X,Y,T be a standard s^iM) basis of a Lie algebra g. Let V be a two-dimensional irreducible real g-module. Then there exists a basis a,b ofV such that Xb = a, Ya = b,

Xa = 0, Yb = 0.

Proof. This follows from the representation theory of s^IR) (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]). D L E M M A 5.7.8 Let X,Y,T be a standard sfeOR) basis of a Lie algebra g. Let V be a three-dimensional irreducible real g-module. Then there exists a basis a, b,c ofV such that Xc = 2b, Xb = 2a, Ya = b, Yb = c,

Xa = 0, Yc = 0.

Proof. This follows from the representation theory of 0l2(K) (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]). • L E M M A 5.7.9 Let S be a connected Lie group with finite sume that s is Lie algebra isomorphic to sfe (K). Let V be an S-module. Assume that dim(V) > 4. Let Si be a sequence that Si —> oo in S. Then, after passing to a subsequence, precompact sequences Vi and Wi in V and there exist v, w £

center and asirreducible real in S. Assume there exist two V such that

• v and w are linearly independent in V; • SiVi —>• v; •

SiWi

and

—>• W.

Proof. Let cp : sl2(M) -> s be a Lie algebra isomorphism. We define T := E$-E$ 6 «b(R). Let T := <j>(T') and let A+ := {exp(rT) | r > 0}. By Lemma 4.10.38, p. 127, there exists a compact subgroup K of S such that S = KA+K. We may therefore assume, for all i, that s* 6 A+. For all i, choose r-j > 0 such that s; = exp(rjT'). Then r-j -)• +oo. By the representation theory of sfeOR) (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]), choose v,w G ^ \ { 0 } and choose A, fj, > 0 such that A 7^ n, such that T'v = Xv and such that T'w = fj,w. For all i, let vt := v and let Wi := w. Then s^i = (exp(riT'))v and

SiWi = (exp(riT'))w

= eXriv ->• v, = e^'w

-*• w.

Representation theory

171

Because v and w are nonzero and in different eigenspaces of T" : V —> V, it follows that v and w are linearly independent. • COROLLARY 5.7.10 Let S be a connected Lie group with finite center. Assume that s is Lie algebra isomorphic to sl^W- Let V be an irreducible real S-module. Let Sj be a sequence in S and assume that Sj —> oo in S. For all i, define Ti : V -> V by Tj(u) = stv. Assume that dim(V) > 4. Then, after passing to a subsequence, there is a sequence Vi of two-dimensional vector subspaces of V such that Ti is expansive on Vi. Proof.

True by Lemma 5.7.9, p. 170 and Lemma 2.6.2, p. 24.

•

L E M M A 5.7.11 Let g be a Lie algebra. Let U be a real Q-module and let V and W be g-invariant vector subspaces ofU. Assume that . V + W = U andVHW = {0}; • U is g-irreducible; and • V is g-trivial, i.e., gV = {0}. Let UQ be a g-invariant vector subspace of U. Then (1) Either V QU0 orU0CW. (2) IfUo is g-nontrivial and g-irreducible, then Uo = V. Proof. Proof of (1): The addition map / : V © W -¥ U denned by f(v, u>) = v + w is an isomorphism of real g-modules. Let 7ry : V © W —• V and TTW '• V © W —> W be the projection maps onto the first and second coordinates, respectively. Define p := ny ° f~l : U -> V and q :— TTW ° f~l '• U ->• W. Assume that V <£ U0. We wish to show that U0 C W, i.e., that p(U0) = {0}. Since V is g-irreducible, it suffices to show that p(Uo) ^ V. As Uo n V C V, by g-irreducibility of V, we conclude that UQC\V = {0}. Then q\Uo : Uo -*• W is injective. By assumption, W is g-trivial, i.e., gW = {0} C W. Then q(gU0) = g(q(U0)) C gW = {0}. So, by injectivity of q\U0, we get gU0 = {0}. Then, as p : U -t V is g-equivariant, we get g(p(Uo)) — {0}. On the other hand, as V is nontrivial, we get gV ^ {0}. As g(p(Uo)) = {0} ^ gV, we conclude that p(Uo) ^ V. End of proof of (1). Proof of (2): As Uo is g-nontrivial, whereas W is g-trivial, we conclude that Uo 2 W. By (1), we see that V C C/0. As V is g-nontrivial, it follows

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that V ^ {0}. So, since {0} ^ V C Uo and since UQ is g-irreducible, we get U0 = V. End of proof of (2). U L E M M A 5.7.12 Let $ be a Lie algebra and assume that g is Lie algebra isomorphic to s^C^)- Let V be an irreducible real g-module. Then V has only scalar g-intertwining. Proof. We may assume that g — s ^ K ) . Let T := E$ - E$. Let Y := E$. Let d := dim(V). By the representation theory of sfeW (e.i+i; (2) Yvd = 0; and (3) for all integers i € [l,d], we have Tvi — (d — 2i + l)vi. Let I : V -¥ V be the identity transformation. Let 5 : V ->• V be a linear transformation, and assume, for all P € g, for all t; € V, that 5(Pu) = P(S{v)). We wish to show that S eRI. For any integer i € [l,d], by (3), we have T(S(vi)) = (d-2i + l)-{S{vt)), so S(vi) e ffiwi; choose A; € E such that S(vi) ~ XiVi. We wish to show that Ai = • • • = Ad. Fix an integer i s [l,d — 1]. We will show that Aj = Aj+i. A s l < i < d — 1 and as v\,...,Vd is a basis of V, we get Vi+i ^ 0. We have A, • Yvi = Y(XiVi) = Y{S{vi)) = S(Yvi), so, by (1), this gives AjUj+i = S(vi+i). Then \iVi+i = Aj+ii>j+i, so Aj = Aj+i. D If G is a Lie group and V is a real G-module, then we say that V is isotypic if there is an irreducible real G-module U and an integer k > 1 such that V and ®kU are isomorphic real G-modules. L E M M A 5.7.13 Let Go be a connected Lie group and let V be an isotypic real Go-module. Assume that V has nonscalar intertwining. Let v' € V\{0}. Then there exists a Go-equivariant linear transformation T : V -> V such that T(v') £ W. Proof. For any complex structure J : V —> V, we have J(v') ^ M.v', so, by Lemma 4.9.26, p. 107, we are done if V is Go-irreducible. We therefore assume that V is Go-reducible. Then, as V is Go-isotypic, choose an irreducible real Go-module U and an integer / > 2 such that V is isomorphic to ®lU. We may assume that V = ®lU. Let L := { 1 , . . . , / } . For

Representation

theory

173

alH 6 L, let 7Tj : 6 U -> U be the ith coordinate projection map and let H : U -> @lU be the ith coordinate inclusion map. Choose j G L such that 7Tj(t/') y£ 0. Fix X; 6 L\{j}. Let T := ifc ° ^ : V - • V. Then T : V -> V is Go-equivariant. We wish to show that T(v') £ Rv'. Let WQ := T(v'), let a £ l and assume, for a contradiction, that wo = av'. We have wo = T(v') = tjfe(7Tj («')). Since j ^ k, TTJ o tk = 0. Then a • Ttj(v') = Ttj{wo) = nj(tk(^j(v'))) = 0. So, as 7Tj(i/) ^ 0, we get a = 0. Then tfc(7r,-(i/)) = u>o = aw' = 0. So, since ik : 17 ->• ©'{/ is injective, we conclude that nj(v') = 0, contradiction. • L E M M A 5.7.14 Let G be a connected Lie group and let Go be a Lie subgroup of G. Let C C G G ( G O ) . Assume that CGQ = G. £e£ V be an irreducible real G-module. Then V is Go-isotypic. Proof. Let U be a nonzero Go-irreducible vector subspace of V. For all c £ C, elf is Go-invariant and cU is real Go-module isomorphic to U. For any integer n > 1, for any 7 = ( c i , . . . , cn) 6 Cn, let C/7 := ( c i * 7 ) e - - - e ( c „ I / ) and let Ay : {77 —> V be defined by A 7 ( u i , . . . ,vn) = v\ + • • • 4- ?;„. For any integer n > 1, for any 7 e C™, U7 is real Go-module isomorphic to @nU and dim(f/ 7 ) = n[dim(C/)]. Let S be the set of all integers n > 1 such that, for some 7 € Gra, Ay : U7 —> V is injective. For all n 6 5, we have n[dim([/)] < dim(V), so S is bounded above. Moreover, 1 £ S. Let m := max(S). Choose 7 = ( c 1 ; . . . ,c m ) € Cm such that Ay : C/7 -» V is injective. Let V0 :— Ay(E/7) = ciU-\ \-cmU C V. The map Ay : l/ 7 ->• Vb is an isomorphism of real Go-modules, so, since Ay = ®mU, it suffices to show that Vo = V. Assume, for a contradiction, that Vo ¥" V. Since CU = CG0U = GU, by G-irreducibility of V, we see that CU spans V. Choose Co € G such that CQU £V0. Then, as cot/ is Go-irreducible and as Vo is Go-invariant, we get {CQU) f~l Vb = {0}. Then ( X 0 , . . . ,Xm)

: (coC/) 0 (ClU) © • • • © (t^tf) -> V

is injective, so m + 1 6 S. Then m = max(S) > m + 1, contradiction.

•

L E M M A 5.7.15 LetX,Y,T be a standards^W basis o/a Lie algebras. Let V be an irreducible reals-module. Define d := dim(V). Then, for some v e V, we have Xd~1v ^ 0.

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Proof. By the representation theory of s[2(IR.) (e.g., Theorem 12 of §111.8 on pp. 85-86 of [J62]), choose a basis v\,. ..,v
= [{d - 1)!]V ^ 0. Let v := vd.

D

L E M M A 5.7.16 Let V be a vector space and let H := GL(V). Let G be a connected compact subgroup of H and assume that V is G-irreducible. Then there exists a positive definite Q 6 QF(V) such that g C so(Q) and such that C|,(fl) C co(Q). Proof. As G is compact, let Qo G Q F ( ^ ) be G-invariant and positive definite. Then G C SO(<2o)- Let J : V -> V be the identity map. If V has only scalar G-intertwining, then CH{G) C (E\{0})J C C O ( Q O ) ; SO, in this case, we may set Q :— QoWe therefore assume that V has nonscalar G-intertwining. Then, by Lemma 4.9.26, p. 107, let J be a g-invariant complex structure on V. Let Q '•— Qo + (Qo ° J)- As Qo is positive definite, we conclude that Q is positive definite. Since Qo and J are both G-invariant, we see that Q is G-invariant, i.e., that G C 0((J). Then g C o(Q) = so(Q). Since J 2 = -I, we have Q o J = (QQ o J) + (Q0 o (—/)). Since Q0 is a quadratic form, we have Qo ° ( - / ) = Qo- Then Q o J = (Q 0 o J ) + Q 0 = Q. Then J e so(Q). By Lemma 4.9.24, p. 106, we have C(,(g) = {al + bJ \ a, b € R}. So, since / £ co(Q) and since J € so(Q) C co(Q), we see that C(,(g) C co(Q). •

Chapter 6

Minkowski Linear Algebra

6.1

Notations for important elements and Lie subalgebras of so(Qd)

In the following definitions, "£>", "£", "ft", "A/"" and "7>" stand for the words "diagonal", "elliptic", "hyperbolic", "nilpotent" and "parabolic", respectively. Let d > 2 be an integer. Let 7 := { l , . . . , d } . For all j,k € 7, define Ejk := E\f. Let V^ •- £ n - Edd € so(Qd) . Let S[d^ be the collection of all matrices ]T] ayEjj in Rdxd

such that

• for all i € {l,d}, for all j € 7, we have ay- = 0; and • for all i, j € 7, we have a^- = —aji. Then £
._

V(d)

+

g(d)

g

Note that, if d € {2,3}, then ^ s o ( Q d )

= {0}. Let

_

lid = 2, then we define S^d) := {0}, P ^ := 0, J V ^ := {0}. Assume, for the remainder of this paragraph, that d > 3. For j € 7 \ { l , d } , let Mj := Eij — Ejd € 50(Qd). Let £2 be the collection of all matrices Y,aHEij in M dxd such that • for all i € {1,2, d}, for all j € 7, we have ay- = 0; and • for all i, j E 7, we have ay = — a,-;. Then f £ ° C so(Q d ). Note that, if d G {3,4}, then £^d) = {0}. Let V™ := 7V2(d) + 4 d ) C so(Qd) . Let JV/W := iUV2(d) +• • • + MA^1)1 C so(Q d ). 175

176

6.2

Minkowski Linear Algebra

Linear algebra of Minkowski vector spaces

L E M M A 6.2.1 Let (V,Q) be a Minkowski vector space. Let B G SBF(V) be the polarization of Q. For all v G V, let v1- :— {w 6 V \ B(v,w) = 0}. Let Vo G V\{0} be timelike. Then Q\VQ- is positive definite. Proof. Let n := (dim(V)) — 1. Then the signature of Q is (n, 1) and dim(t>o") = n. Since VQ is timelike, it follows that Q\Mvo is nondegenerate. Then Mv0 H v£ = {0}. Then dim(fc 0 + v£) = 1 + n = dim(y), so Rv0 +VQ- = V. Then ker(Q\vQ-) C ker(Q) = {0}, so Q|«o" i s nondegenerate. Let (p,q) be the signature of Q|IRDO and let (p',q') be the signature of Q\VQ. Then (n, 1) = (p,g) + (p',q'). Since i>o is timelike, it follows that (P,q) = (0)1)- Then (p',q') = (n,0). So, since 0 such that n*; £ 3(n) and such that tijt+i C ^(n). Choose X e n/fc\(3(n)). Let V := (adX)n. Then V ^ {0}. Then { 0 } / F C [n*,n] = n fc+1 C j(n). Then (adX)V C (adX)( 3 (n)) = {0}. Then B(V,V) = B(V,{&dX)n) = B((adX)V,n) = B({0},n) = {0}. Since V ^ {0}, we conclude that B is not definite. Assume, for a contradiction, that B is Minkowski.

Linear algebra of Minkowski vector spaces

177

Fix Z e V\{0}. By Corollary 6.2.3, p. 176, dim(V) < 1, so V = MZ. Since Z € V = (adX)n, choose Yen such that Z = (adX)Y. Fix Wen such that B(W,Z) ^ 0. Since (adX)W € (adX)n = V = RZ, choose A € R such that (adX)W = AZ. Let a := B ( Z , y ) . We have 0 ^ B(W,Z) = B(W,(adX)Y) = -B((adX)W,Y) = -B(XZ,Y) = -Xa. Then a ± 0. However, a = B((adX)Y,Y) = -B(Y, (adX)Y) = -a, so a = 0, contradiction. • We remark that Lemma 6.2.4, p. 176 cannot be improved to assert that B must be nondegenerate: Let n be the two-step nilpotent Lie algebra with basis {A, B, C, X, Y, Z} such that X,Y,Ze }(n) and such that [A,B] = Z,

[B,C]=X,

[C,A]=Y.

Let Q e QF(n) be defined by Q(aA + bB + cC + xX + yY + zZ) = 2ax + 2by + 2cz. Let B be the polarization of Q. Then the symmetric bilinear form B is nondegenerate and (adn)-invariant. L E M M A 6.2.5 Let (V,Q) be a Minkowski vector space. Let B e SBF(V) be the polarization of Q. For all v e V, let vx := {w e V \ B(v,w) = 0}. Let VQ e V\{0} be lightlike. Then: (1) every vector in i^VR^o) is spacelike; and (2) Q\VQ is positive semidefinite. Proof. Proof of (1): Fix w € VQ. Assume Q(w) < 0. We wish to show that w e R«o • We have 0 7^ vo € w1, and Q(vo) = 0, so w1- is not positive definite. If w were timelike, then, by Lemma 6.2.1, p. 176 (with vo replaced by w), it would follow that w1- is positive definite, which it is not. Consequently, w is not timelike, i.e., Q(w) > 0. On the other hand, by assumption, Q(w) < 0, so we get Q(w) — 0. Then Rv0 + Rw is isotropic, so, by Corollary 6.2.3, p. 176, we see that dhT^Ruo + Rw) < 1, so, as «o 7^ 0, we get w £ Ri>o. End of proof of (1). Proof of (2): As every vector in Rvo is lightlike, it follows from (1) that every vector in VQ is either spacelike or lightlike. That is, VQ is positive semidefinite. End of proof of (2). •

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L E M M A 6.2.6 Let W be a degenerate vector subspace of a Minkowski vector space (V, Q). Then (1) (2) (3) (4)

dim(ker(Q|W)) = 1; Q\W is positive semidefinite; ker(<3|W) is equal to the set of lightlike vectors in W; and any nonzero lightlike vector in W spans kei(Q\W).

Proof. Proof of (1): By Corollary 6.2.3, p. 176, dim(ker(Q|WO) < 1Since W is degenerate, dim(ker(Q|W)) > 1. End of proof of (1). Proof of (2): Let B : V x V ->• IK be the polarization of Q. Choose v0 G (ker(Q|W))\{0}. For all v € V, let vL := {w £ V\B(v,w) = 0}. Then W C v£, so, by (2) of Lemma 6.2.5, p. 177, we see that Q\W is positive semidefinite. End of proof of (2). Proof of (3): This follows from (2) and from Lemma 2.10.1, p. 34. End of proof of (3). Proof of (4): This follows from (1) and (3). End of proof of (4). • If (V, Q) is a Minkowski vector space, then the light cone of (V, Q) is Q- (0) = {veV\Q(v)=0}. l

L E M M A 6.2.7 Let (V,Q) be a Minkowski vector space. Let C := Q _ 1 (0) be the light cone of (V,Q). Then the R-span of £ is = V. Proof. Let d := dim(V). We assume that V = Rd and that Q : V -+M. is defined by Q(xi,... ,xa) = — x\ +x\ -\ \-x2d. Let I := { 2 , . . . ,d}. Let d e i , . . . , ed be the standard basis of K . Let S := {ei + a | i e 1} U {ei - a | i £ / } . Then S C C and the 1-span of S is V. The result follows.

•

L E M M A 6.2.8 Let (V,Q) be a Minkowski vector space and let v,w € V be linearly independent vectors. Assume that Q(v) = 0 — Q(w). Then Rv + Rw is a Minkowski vector subspace of (V, Q). Proof. Let S := Rv + Rw. Let B : V x V -> R denote the polarization of Q. Let t := B(v,w). By Corollary 6.2.3, p. 176 we know that S is not isotropic, so, as B(v,v) = 0 = B(w,w), we must have t ^ 0. Let x := w/t. Then B(v,v) = 0 = B(x,x) and B(v,x) = 1.

Linear algebra of Minkowski vector spaces

179

Let B := {v,x). Then B is an ordered basis of 5 and, under the vector space isomorphism (J)B : K 2 x l -¥ S, Q2 corresponds to Q\S. We conclude that S is Minkowski. • L E M M A 6.2.9 Let V be a vector space and letT:V-tVbea diagonalizable linear transformation. For all X € M., let

complex

Vx := {v e VI T(v) = Xv}. Assume {X £ R\{0} | Vx ^ {0}} = {-1,1}. Assume that dim(Vi) = 1 and that dim(V_i) = 1. Then, for some Q e MinkQF(V), T e so(Q). Proof. Since T is complex diagonalizable, choose linear transformations D,E : V —> V such that DE = ED, such that D is real diagonalizable, such that E is elliptic and such that T = D + E. As both D and E are complex diagonalizable linear transformations, as T — D + E and as DE = ED, we conclude that T is complex diagonalizable and that D is the real diagonalizable part of T. We may assume that V = E d x l . Let W := V-\ + V\. Since T is complex diagonalizable, choose a T-invariant vector subspace W C E d x l such that W + W = V and such that W D W = {0}. After a change of basis, we may assume that V_i = M e ^ , that Vx = Ee^d) and that W = Ee^d) + ••• + K e ^ . By Lemma 2.4.2, p. 21, we see that D\W = T\W and that D\W = 0. Then £» = X>(d). We have S | W = (T - D)\W = 0. Since W is T-invariant and D-invariant, it follows that W is ^-invariant. So, since E : V -> V is elliptic, we see that i ? | W : W ->• W is elliptic. Then, after a change of basis to W, we may assume that E S £[ . Then T = D + E£ V^

+ £[d) = H^

C so(Q d ). Let Q := Q,,.

•

L E M M A 6.2.10 Let V be a vector space and let T,E,N : V ->• V be linear transformations. Assume that EN = NE, that E is elliptic, that N is nilpotent and that T = E + N. Assume that dim(N(V)) — 2 and that dim(JV2(V)) = 1. Then there exists Q G MinkQF(V) such that T G so(Q). Proof. Let W := N(V) and X := N2(V) C W. Then N\V : V ->• W is surjective. By assumption, we have dim(W) = 2 and dim(X) = 1, so the codimension in W of X is = 1. Let Y := N_1(X). Then the codimension in V of Y is = 1. Moreover, as N(W) = X, we get W CY. Since £iV = iVS, we see that E(W) C W, that S ( X ) C X and that E(Y) C r . Since J5 : y —> V is elliptic, we see that any .E-invariant vector subspace of V has

180

Minkowski Linear Algebra

an E-invariant vector space complement. Let Y' be an ^-invariant vector space complement in V to Y. Then dim(y') = 1, so, as E is elliptic, it follows that E\Y' = 0. Choose v 6 Y'\{0}. Then v £ Y and Ev = 0. Let •w := iVw and let a; := JVu;. Then w; € W and a; 6 X. Since w ^ y and since Y = N-1(X), we have Nv f X. That is, w $ X. As X is iV-invariant, as N : V —> V is nilpotent and as dim(X) = 1, we get N(X) = {0}, i.e., X C ker(iV). The map N\W : W -> X is surjective. So, since dim(T^) = 2 and dim(X) = 1, we conclude that dim(ker(JV|T¥)) = 1. So, since X C ker(N\W) and since dim(X) = 1, we get X = ket(N\W). Thus, since w G W\X, it follows that Nw ^ 0. That is, x ^ 0. Since x e -X^\{0} and since dim(X) = 1, we conclude that X = Mx. So, since w € PF\X and since dim(W) = 2, we conclude that W = Mw + Mx. Then, since v $Y and since W C Y, we see that {v, w, x} is a linearly independent subset of V. We have N(Mv + Rw) = Mw + Mx = W = N(V). It follows that Rv + E.w + (ker(JV)) = V. We have Ew = ENv = NEv = N(0) = 0 and Ex = ENw = NEw = JV(0) = 0. Let d := dim(V). Since X is an .E-invariant vector subspace of ker(iV), let X' be an .E-invariant vector space complement in ker(iV) to X. Since the linear transformation E\X' : X' —> X' is elliptic, let Bo := (*i,..., td^3) be an ordered basis of X' such that [ £ J | X ' ] B 0 is an antisymmetric matrix. Let B := (x,w,t\,... ,td-z,— v). Then B is an ordered basis of V. Moreover, [N]B = A/"2 and [E]B € £2 • We may therefore assume that V = E d x l , that N = A/"2(d) and that E € ££°. Then N,E € so(Q d ). Let Q:=Qd. Then T = E + N e so(Qd) = so(Q). D L E M M A 6.2.11 Let (V, Q) be a Minkowski vector space and let W be a nondegenerate vector subspace of (V, Q). Then either Q\W is positive definite or Q\W is Minkowski. Proof. Let (p, q) be the signature of Q\W. We wish to show that q<\. Let W1- be the orthogonal complement of W with respect to the polarization of Q. Let (r,s) be the signature of QIW^. Since Q\W is nondegenerate, it follows that W C\WL - {0} and that W + W1- = V. Then (p + r, q + s) is the signature of Q, so, as Q is Minkowski, we see that q + s = 1. So, since 0 < s, we get q < 1. • L E M M A 6.2.12 Let (V,Q) be a Minkowski vector space and let U be a vector subspace of V. Then either Q\U is Minkowski or Q\U is positive

Linear algebra of Minkowski vector spaces

181

semidefinite. Proof. By Lemma 6.2.11, p. 180, we see either that Q\U is positive definite, in which case we are done, or that Q\U is degenerate, in which case (2) of Lemma 6.2.6, p. 178 completes the proof. • L E M M A 6.2.13 Let (V, Q) be a Minkowski vector space and let W,X CV be vector subspaces. Assume that W C X and that Q\W is Minkowski. Then Q\X is Minkowski. Proof. Since Q\W is Minkowski, Q\W is not positive semidefinite. Then Q\X is not positive semidefinite, so, by Lemma 6.2.12, p. 180, Q\X is Minkowski. • L E M M A 6.2.14 Let (V, Q) be a Minkowski vector space. Let d := dim(V). LetX G so(Q), and assume thatX : V —> V is nonzero and nilpotent. Then there exists an ordered Qd-basis BofV such that [X]B = Proof. Let (•, •) denote the polarization of Q. Since X G so(Q), it follows, for all v,w G V, that {Xv,w) = —(v,Xw). In particular, for all v G V, we get (Xv,v) = 0. For any S C V, let S 1 := {t; £ V | V s € §,<«,«> = 0 } . Claim 1: X2 ^ 0. Proof of Claim 1: Suppose, for a contradiction, that X2 = 0. For all v G V, {Xv,Xv) = -(v,X2v) = 0. That is, X(V) is isotropic. So, by Corollary 6.2.3, p. 176, dim(X(V)) < 1. As X / 0, we get X(V) ^ {0}. Choose w € (X(V))\{0}. Then w is lightlike and X(V) = M.w. Choose a vector v £ V such that Xv ^ 0 and such that (v,w) / 0. We have Xv G X(V) = Mw, so choose a £ M. such that Xv = aw. Then 0 = (Xv,v) = a{w,v). So, since (w,v) ^ 0, we see that a = 0. Then Xv = aw = 0, contradiction. End of proof of Claim 1. Claim 2: X3 = 0. Proof of Claim 2: Choose an integer n > 0 such that Xn ^ 0 and such that Xn+1 — 0. Choose an integer m such that n / 2 < m < (n/2) + 1. Then 2m > n, so X2m = 0. Then, for all v £ V, we have (Xmv,Xmv) = ( - l ) m ( v , X 2 m i ; ) = 0. Thus Xm(V) is isotropic, so, by Corollary 6.2.3, p. 176, we have dim(X m (V)) < 1. Fix v0 € Xm{V) such that Xm(V) = Mv0. Then X«o € X m + 1 ( V ) C X m ( y ) = Rv0. Choose A G M such that Xv0 = A^o- Since X : V -> V is nilpotent, it follows that 0 is its only eigenvalue. Then A = 0, so XVQ = Xvo = 0. Then Xm+1(V) = X{Xm{V)) C X(Hwo) = {0}, i.e., X m + 1 = 0. So, since

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Xn ^ 0, we conclude that n < m + 1. Then, as m < (n/2) + 1, we get n < (n/2) + 2, which yields n < 4, so n < 3. Then m < (n/2) + 1 < 5/2, so m < 2, so m + 1 < 3. Then, because Xm+1 = 0, we have X3 = 0. End of proof of Claim 2. Let C := Q~l($) denote the light cone of (V,Q). By Lemma 6.2.7, p. 178, the E-span of C is = V. By Claim 1, we have X2 ^ 0. Fix x € L such that X2x ^ 0. Let y := Xx and let z := Xy. We have z = X2x ^ 0, so y 7^ 0. By Claim 2, we get X z = 0. As X is nilpotent, we see that {x,y,z} is linearly independent. Let S :— Kx + Ez. We have (z,z) — (Xy,z) = —(y,Xz) = 0 and (x,y) = (x,Xx) = 0. We have (y, z) = {y, Xy) = 0. Then x and z are lightlike and y £ (Mx + Rz)1-. Since x and z are linearly independent lightlike vectors in V, by Lemma 6.2.8, p. 178, we conclude that 5 is a Minkowski vector subspace of (V,Q). By Lemma 6.2.13, p. 181, we see that T := Ear + % + Ez is Minkowski, as well. Then T1- is positive definite. So, as X^1- is nilpotent, we conclude that XlT1- = 0. Let ui,...,Ud-z be an orthonormal basis x for T . Let a := Q{y). Since S is Minkowski, it follows that S1- is positive definite. So, since y 6 5 J -\{0}, we conclude that a > 0. Let XQ := x/y/a, y0 :— y/y/a and ZQ := z/\/a. Then j/o 6 S1-1 = (Exo + Ej/o)"1"- We have (a;o,a;o) = Qi^o) — (Q(^))/« = 0. Similarly, we have (z0,z0) = 0. Also, (x0,zo) = (x0,X2x0) = -(Xx0,XxQ) = -Q(Xx0) = -Q(yo) = - 1 - Let B := ( z 0 , y o , u i , . . . ,Ud-3, —xo)- Then B is an ordered Qd-basis of V and

[x]B=M^d).

a

L E M M A 6.2.15 Let U be a vector subspace of a Minkowski vector space (V, Q). Let B e SBF(V) be the polarization of Q. Let U' be a nonzero vector subspace of U. If Q(U') = {0} and if B(U', U) # {0}, then Q\U is Minkowski. Proof. Since U' ^ {0} and Q(U') = {0}, it follows that Q\U is not positive definite. So, by Lemma 6.2.11, p. 180, it suffices to show that Q\U is nondegenerate. Assume, for a contradiction, that Q\U is degenerate. Then, by (3) of Lemma 6.2.6, p. 178, we see that U' C ker(Q|*7). Then B(U',U) = {0}, contradiction. • L E M M A 6.2.16 Letd>2 be an integer. Let V := E d x l . For all integers i € [i-,d], let ei := e\ '. Let W be a vector subspace of V. Assume that

Linear algebra of Minkowski vector spaces

ei £ W and that R ^ ^ l e H

183

r Med-i • Then Qd\W is Minkowski.

Proof. Since e\ € W and since Qd(ei) = 0, we see that W is not positive definite. By Lemma 6.2.11, p. 180, W is either Minkowski or degenerate. Assume, for a contradiction, that W is degenerate. By (4) of Lemma 6.2.6, p. 178, we see that Mei is the kernel of <2d|W. Let B : V x V ->• E be the polarization of Qd. Let 5 := {v € V | B(ei,u) = 0}. Then W C S. On the other hand, computation yields S — Eei H hEed_i, so, by assumption, we have W <£ S, contradiction. • L E M M A 6.2.17 Let V be a vector space and let Q € QF(V) be nondegenerate. Let W be a vector subspace ofV and assume that dim(V/W) = 1. Assume that Q\W is positive semidefinite with one-dimensional kernel. Then Q is Minkowski. Proof. Let B € SBF(V) be the polarization of Q. For all A CV, let A^- := {a; e V | B{x, A) = {0}}. For all x£V, let x1- := {ar}-1. Choose w € W\{0} such that En; is the kernel of Q\W. Then we have dimiV/w1-) = dim(Ew) = 1 = dim(V/W). Then dim(W) = dim(«;-L). So, since W C w 1 , we conclude that W — w1. We have B(w,w) = 0. Choose v0 € V such that B(uo,^) = —1/2. For all t e E, we have 2?(uo + tw, w0 + tw) = (B(VQ, V0)) — t. Let t0 := B(v0,v0) and let v := —2(uo + fo^)- Then B(v,v) — 0 and B(?;,w;) = 1. Let S :— Rv + Eu>. Then Q|5 is Minkowski. It therefore suffices to show that QIS1- is positive definite. We have S1- C w-1 = W. Let ?/ £ 5-1and assume that B(y,y) = 0. We wish to show that y — 0. Since B(w,w) = B(y,w) = B(y,y) — 0, it follows that Mw + My is isotropic. We have w £ W and y e Sx Cwx =W,so Rw+Ry C W. Then, by Lemma 2.10.1, p. 34, we have Rw + Ry C ker(<5|W). By assumption, dim(ker(Q|W0) = 1. Then dim(Ew + Ey) < 1. So, since w ^ 0, we get y £ Rw. Choose A € E such that y = Xw. We wish to show that A = 0. We have y € S1- C v1, so B(v,y) = 0. Then, as B(v,w) = 1, we get X = X-B{v,w) = B(v,Xw) =B(v,y)=0. D L E M M A 6.2.18 Let (V,Q) be a Minkowski vector space and let W be a vector subspace of V. Let Wc := {w E W \ Q{w) = 0}. Assume that W is not Minkowski. Then Wc is a vector subspace of W of dimension < 1. Proof. By Lemma 6.2.11, p. 180, we see that Q\W is positive definite or degenerate. If Q\W is positive definite, then Wc = {0} and the result

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follows. We therefore assume that Q\W is degenerate. Then, by (3) of Lemma 6.2.6, p. 178, ker(<5|W) = Wc- In particular, Wc is an isotropic vector subspace of (V, Q). So, by Corollary 6.2.3, p. 176, dim(W) < 1. • Recall, from §2.12, p. 39, that Q3 is a quadratic form on E 3 denned by Q3(x,y,z) = 2xz + y2. L E M M A 6.2.19 Let V := E 3 x l . Let I : V -> V be the identity map. Let v 6 V\{0} and let K be a compact subgroup ofSO°(Q3). Assume that Q3(v) = 0 and that Kv C Rv. Then K = {/}. Proof. Let v0 := e£3). Let L := {v £ V\Q3(v) = 0}. Then we have v,vo 6 L. Since 0(Q3) acts transitively on L\{0}, choose a € 0(Q3) such that av = VQ. Let KQ := aKaT1. Since 0{Q3) normalizes SO°(<33), we have K0 C SO°(Q3). Moreover, K0 is compact. Also, K0v0 = (aKa'^iav)

= aKv C a{Rv) = R(av) = Rv0.

Let U := {g € SO°(Q3)\gv0 C Rv0} and N := {g € SO°{Q3)\gv0 = v0}. Then N is a closed normal subgroup of U and U is a closed subgroup of SO(<53). Computation shows that N £ E and that [//AT ^ E. So, since E has no nontrivial compact subgroup, we see that U has no nontrivial compact subgroup. Then, as Ko C U, we see that KQ = {I}- Then we have K = o _ 1 K 0 a = {/}. •

6.3

Basic calculations

Let d > 2 be an integer. Define T : Rdxd ->• E d x d by letting T(M) denote the transpose M* of M in E d x d . Let J := { l , . . . , d } . For all j € J, let ej := e{d). For all j,k £ J, let Ejfc := E$. Let P := £> := 7>(d>. Let ft := ft Tw : W —> W has at exactly one nonzero real characteristic root. Proof. Let V := E d x l . Choose A € E such that T € Aft. Since T # 0, we conclude that A ^ 0. Then the set of characteristic roots of T : V -> V

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185

is contained in {—A, A} U \/-TR. Moreover, the A-eigenspace of T : V ->• V is Kei, while its (—A)-eigenspace is RedLet S := Kei + Med. Then Qd\S is Minkowski. If T\W : W -> W had two nonzero real characteristic roots, then we would have S C W, which, by Lemma 6.2.13, p. 181, would imply that Qd|W is Minkowski. However, Qd\W is degenerate, so T\W : W -4 W cannot have two nonzero real characteristic roots. Now assume, for a contradiction, that T\W : W -» W has no nonzero real characteristic roots. Then the set of characteristic roots of T\W : W —> W is contained in y/-WL Let S' := Re2 H h Re d _i and let TT : V ->• V / S ' be the canonical map. Since T(5") C S", fix T : V/S' -> V / S ' such that T o TT = vr o T. Let JV ™ n(W). Since Qd|S' positive definite, while Qd\W is degenerate, we conclude that WgS'. Then W # {0}. _ Let (j, be a characteristic root of T|W : W —• W. Then /u is a characteristic root of T\W : W -*• W, so fi € -/-TR- In particular, /x 0 {-A, A}. Since the set of characteristic roots of T\S : S —> S is {—A, A}, since TT\S : S —• V/S' is a vector space isomorphism and since T o TT = -n o T, it follows that the set of characteristic roots of T : V/S' -> V/S' is {—A, A}, as well. Then fj, € {—A, A}, contradiction. • L E M M A 6.3.2 Assume that d > 3 Let V := R ( r f - 2 ) x l . Define a linear transformation F : V -* Rdxd by Ffe) = A/i+i- Define a linear transformation 4> : R(d-2)x(d-2) _> Rdxd hy ^E^~2)) = Ei+1J+1. Define an injective Lie group homomorphism $ : GLd-2(R) —> GLd(R) by $(
This is a computation.

[<j>{X),F(v)] = F(Xv). (Ad(*(g)))(F(v)) = F(gv). •

L E M M A 6.3.3 Assume that d > 3. Let X,Y £ A/"\{0}. Then we have both ( a d X ) ( T ( y ) ) ^ 0 and {aAX)2{T(Y)) # 0. Proof. Let V, F and $ be as in Lemma 6.3.2, p. 185. Choose v £ V such that F(u) = X. Choose p € SO(d - 2) and choose A 6 R\{0} such that gv = Aei. Let X' := (l/A)(Ad$(g))X and let Y' := \(Ad$(g))Y. Then, by (2) of Lemma 6.3.2, p. 185, we have X' = F(e[d~2)). That is, X' = A/2. Replacing X by X' and Y by Y', we may assume that

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X = M2- Let I := { 2 , . . . , d - l } . Then {(adX) 2 {T{Mi)) | i € 1} is a linearly independent subset of so{Qd)- We conclude that the linear transformation Z H-> (adX)2(T(Z)) : T{M) -> so(Qd) is injective. So, since Y ^ 0, we see that ( a d X ) 2 ( T ( F ) ) ^ 0. It follows that (adX)(T(Y)) ^ 0. D L E M M A 6.3.4 Let V 6e a vector space and Zet Q € MinkQF(V). Assume tftat d = dim(y). Let 0 := so(Q) and let X,T,Y 6 g. Assume that a d T : g —• g is reaZ diagonalizable, that 7 / 0 ^ X , t/iat [T, X] = X and that [T,Y] = -Y. Then [X,Y] ^ 0. Proof. We may assume that V = Wdxl and Q = Qd. Then ED is a maximal real split torus in g. By Lemma 4.10.13, p. 117, we may assume that T € WD. Choose A € R such that T = XD. Then the eigenvalues of a d T : 0 -> 0 are {—A,0,A}. So, since [T,X] = X and since X ^ 0, we conclude that 1 € {—A,0, A}, so A € {—1,1}. By, if necessary, replacing T by — T, replacing A by —A and interchanging X and Y, we may assume that A = 1. Then T = T>. Since [T,X] = X, X e Af. Since [T,Y] = -Y, Y € T{N). So, as X ^ 0 ^ Y, by Lemma 6.3.3, p. 185, we have [X, y ] ^ 0. • L E M M A 6.3.5 Let g :- so(Qd). d > 3, then c0(A/"2) = M + £2.

Then cs(V) = WD + £x. Moreover, if

Proof. Let Oi := WD, a2 := N, o 3 := T{M) and a4 := E\. Let 7 := {1,2,3,4}. Then, for all i G I, (adl>)0i C a*. Moreover, the addition map (W, X,Y,Z)^W + X + Y + Z: ^ a» ->• 0 is a vector space isomorphism. It follows that cg(D) = y j c 0 i ( I ? ) . Computation shows that i€l

cai(V) = oi = WD, that c„2(X>) = {0}, that c03(£>) = {0} and that cai(D) = cx4 = £i- Then cB(V) = WD + Ex. Now assume d > 3. Let bx := r(7V),

b2 := WD+£U

b 3 := N,

b4 := {0}.

Let K := {1,2,3}. Then, for all k E K, (adA/^bfc = bfc+i- Moreover, the addition map (X, Y,Z)>-tX + Y + Z: ffi b* ->• g is a vector space isomorphism. It follows that cg(Af2) = / ^ ^ ( . A / ^ ) - By Lemma 6.3.3, p. 185,

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187

£61(^/2) = {0}. Computation shows Cb2(J^2) - £2 and (^(A^) = Af. Then CgCA/i) = AT + £2. • L E M M A 6.3.6 Let d > 3 6e an integer. Then there exists g e SO°(<2d) such that (Adg)V = -V. Proof.

Let X := Af2 and let F := T(X). Let 5

:= [exp(-\/2y)][exp(\/2X)][exp(-^y)].

Then computation shows that (Ad g)T> = —V.

O

L E M M A 6.3.7 Assume that d > 3. Let G = SO°(Qd) and let T e g . Assume that a d T : g -» g is real diagonalizable. Assume that there exists U e 0\{O} such that (adT)U € {-U,U}. Then there exists g € G such that (Ad g)T = V. Proof. Since WD is a maximal split torus of so(Qd), by Lemma 4.10.13, p. 117, we may assume that T S WD. Choose A € R such that T = XV. Then the set of real characteristic roots of a d T : g -> g is {-A,0,A}. Since [T,U] € {-U,U}, we conclude that {-1,1} n {-A,0, A} ^ 0. Then A € {1, - 1 } , so T € {—V,T>}. We may assume that T = —V, since we are otherwise done. By Lemma 6.3.6, p. 187, choose g € SO°(<5d) such that (Ad g)V = -V. Then, as T = -V, we get (Ad#)T = V. • COROLLARY 6.3.8 Assume that d>3. Let G be a connected Lie group and assume that g is Lie algebra isomorphic to so(Qd)- Let T,T' £ g. Assume that a d T : g —• g and a d T ' : g —> g are both real diagonalizable. Assume that there exist U,U' € 0\{O} such that (adT)[7 € { — U, U} and such that (adT')U' € {—U',U'}. Then there exists go S G such that (Ad 5 o )T = T'. Proof. We may assume G — SO(Qd)- By Lemma 6.3.7, p. 187, choose g,g' eG such that (Adg)T = V and (Adg')T' = V. Let 50 := ( s ' ) ~ V • L E M M A 6.3.9 Let G be a connected Lie group and assume that g is Lie algebra isomorphic to so(Qd). Let T € g and assume that a d T : g —> g is real diagonalizable. Let C := {g € G\ (Ad g)T = T } . Let n denote the (+l)-eigenspace of a d T : g -t g. Let i € { 0 , . . . ,dimn}. Then the Adjoint action of C on i-dimensional vector subspaces ofn is transitive.

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Proof. We may assume that G = SO(Qd). We may assume that n ^ {0}. Then G is nonAbelian, so d > 3. Then, by Lemma 6.3.7, p. 187, we may assume that T = V. Define V, F and $ as in Lemma 6.3.2, p. 185. We have n = F(V) and C ~D $(SO(d — 2)). So, since SO( 3. Let g be a Lie algebra which is isomorphic to so(Qd). Let T £ g. Assume that a d T : g —> g is real diagonalizable. Let n+ := {P £ 0 | (adT)X = X},

n_ := {P £ g | (adT)X = - X } .

Assume that n_ ^ {0}. Let Y £ n _ \ { 0 } . X £ n+ such that [X, Y] = T.

Then there exists a unique

Proof. We may assume g = so(Qd). As n_ ^ {0}, by Lemma 6.3.7, p. 187, we may assume T — V. Then n + = M and n_ = T{M). Define V, F and $ as in Lemma 6.3.2, p. 185. Choose v £ V\{0} such that T{F(v)) = Y. Choose g0 £ SO(d - 2) and A € E\{0} such that gov = Xe[ ~ '. Replacing Y by Y/X, we may assume that g0v = e[ ~ . Let Y' := ( A d # ( s o ) ) r . For all g e $(SO(d - 2)), for all W e so(Qd), we have T((Adg)W) = (Adg)(T(W)). By (2) of Lemma 6.3.2, p. 185, we have (Ad $(g0))(F(v)) = F(gov). Applying T to both sides of this, we conclude that Y' = T(F(g0v)). It follows that Y' = T(F(e[d~2))) = T(A4). So, replacing Y by Y', we may assume that Y = T(J^2)Let X := JV2. Then computation gives [X,Y] = T. For all X0 £ n+, if [X 0 , F] = T. then [X - X 0 , F] = T - T = 0, so, by Lemma 6.3.3, p. 185, we conclude that X — X0 — 0, i.e., that X = X0. • L E M M A 6.3.11 Assume that d> 3 and let g := so(Qd). LetD£so(Qd) and assume thatv i-> Dv : Rdxl —t Rdxl is nonzero and real diagonalizable. Then there exists g £ SO°(Qd) such that (Ad g)D £ 1+ V. Proof. As WD is a maximal real split torus in so(Qd), by Lemma 4.10.13, p. 117, we may assume that D £ RT>. Choose A £ E such that D = XV. Since D ^ 0, it follows that A ^ 0. If A > 0, then we are done, so we assume that A < 0. By Lemma 6.3.7, p. 187, choose g £ SQ(Qd) such that (Adg)(-V) = V. Then (Adg)D = -XV £ E+ V. •

Basic

189

calculations

L E M M A 6.3.12 Let g := so(Qd). Let N £ so{Qd) and assume that v H-> Nv : E d x l ->• E d x l is nilpotent. Then all of the following are true: (1) Ifd=2, thenN = 0. (2) If d = 3 and N ^ 0, tten i/iere exists # E SO(Qd) suc/i £/m£ ("3J If d > A and N ^ 0, i/ien tfiere exists g € SO0(Qd) such that (Adg)N = Af2. Proof. If d = 2, then we compute that so(<2, and (1) follows. We therefore assume that d > 3. By Lemma 4.10.33, p. 125, choose Y,T e so(Qd) such that N,Y,T is a standard s[ 2 (E) basis of some Lie subalgebra 0o ofso(<3<j). By Lemma 6.3.11, p. 188, we may assume D = V. Then, as [D, N] = N, we get N eN". Proof of (2): We have Af = EA/"2. Choose A 6 E such that N = XN2. Since N ^ 0, we see that A ^ 0. Let g := (l/A)Eu + E22 + XE33. Then g € SO(Q d ) and (Ad 5 )iV = gNg~l = N/X = A/"2. .End o/proo/ o/ (%). Proof of (3): Let V, F and $ be as in Lemma 6.3.2, p. 185. Choose v € V such that N = F(v). Since d - 2 > 2, choose a e SO(d - 2) and A > 0 such that av = Xe[d~2). Let x := $(o). Then a; 6 SO°(Qd). Let A/"' := (Adz)JV. Then, by (2) of Lemma 6.3.2, p. 185, we see that 1

N' = F(Xe[d~2)).

Then N' = XM2. Let

v

:= - B u + A

\d-l

+ A^dd. J'=2

Then y € SO°(Q d ) and (Ad^iV' = N'/X = N2. Let g := yz. Then we have (Adg)iV = (Ady)(Ada;)iV = (Ad y)N' = N2. End of proof of (3). • The preceding lemma could be used to give a more algebraic proof of Lemma 6.2.14, p. 181. The following "trichotomy" lemma asserts that any element of so(n, 1) is elliptic, (mixed) hyperbolic or (mixed) parabolic. L E M M A 6.3.13 Assume that d > 3. Let X € so(Qd)- Assume that E d x l -> E d x l is not elliptic. Then there exists g € SO(Qd) such that (Adg)X €M+HUV.

D 4 I D :

Proof. By Lemma 4.11.8, p. 130, choose D,E,N € so(Qd) such that the equation X = D + E + N is a, real Jordan decomposition on E d x l . Case 1: D •£ 0. Then, by Lemma 6.3.11, p. 188, we may assume that D = K+Z>. Since [D,E] = 0 and since [D,N] = 0, we conclude, from Lemma 6.3.5, p. 186, that E, N £ ED + £,x.

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Every element of WD + £\ is diagonalizable over C, so the only nilpotent element of WD + £\ is zero, so N = 0. Moreover, £\ is the set of elliptic elements of WD + £i, so E 6 £i. Then X = D + E + N = D + EeW+D

+ £1,

so X e 1+ H. End of Case 1. Case 2: D = 0. By assumption, X : Rdxl ->• Wdxl is not elliptic, so X ^ E. Then E + N = D + E + N = X^E, so N ^ 0. Then, by (2) and (3) of Lemma 6.3.12, p. 189, we may assume that N = J\f2- Then, as [N, E] = 0, we conclude, from Lemma 6.3.5, p. 186, that E e J\f+£2- Since £2 is the set of elliptic elements of N + £2, it follows that E G £2- Then X = D + E + N = E + Ne£2+N2C£2+Af = V. End of Case 2. D We remark that, in Lemma 6.3.13, p. 189, if d > 4, then we may choose a in SO°(Qd). Moreover, even if d = 3, as long as the real diagonalizable part of « 4 Xv : Rdxl -» Wdxl is nonzero, then we may choose g in SO°(<2d). The proofs of these refined statements are essentially the same as the proof found in Lemma 6.3.13, p. 189. Let g := E$ + E%>. Then g € 0(Q 2 ) and (Adg)D^ = -D^. Prom this we see that Lemma 6.3.13, p. 189 is even true for d = 2, provided we may choose g € 0(Qd)The following is Lemma 3.1 of [A99b]. L E M M A 6.3.14 Assume that d > 3. Let (V,Q) be a d-dimensional Minkowski vector space and let T £ so(Q). Assume that T : V —• V is not elliptic. Then there exists an ordered Qd-basis B of (V, Q) such that [T\Be(R+H)UV. Proof. Let BQ be an ordered <5d-basis of (V,Q). Let A := [T]B0. By Lemma 6.3.13, p. 189, choose g G SO(Qd) such that gAg'1 € (R+ U) U V. Let £ :— (e\ \...,ed'). Let B be the image of £ under the map v H- (f>Bo{g~lv) : Rdxl -> V; then, for all v € Rdxl, 3. Le£ T £ 7V\{0}. Then (1) T{Rdxl) c Rei + • • • + l e d - i ; (2) T(lfii + • • • + Me d _i) C Bei; (S)T(Se1) = {0};

Basic

(4) (5) (6) (7) (8)

calculations

191

(adT)(so(Qd))CWH+M; (adT)(WH + Af) C Af; (adT)AT = {0}; for all v G (R d x l )$Rei + • • • + l e d - i ) , we have T2(v) ^ 0; and for all X € (so(Qd))\(WH + Af), we have (adT) 2 (X) ^ 0.

Proof. Let V, F and be denned as in Lemma 6.3.2, p. 185. Choose v £V such that T = F(v). Choose g 6 SO(d - 2) and A 6 K\{0} such that gv = \e[d~2). Let V := (Ad(f>(g))T. By (2) of Lemma 6.3.2, p. 185, we see that (Ad(g))T = XAf2. The transformation 4>(g) :Rdxl -> Rdxl preserves Hex and l e i H h Kfid-i • The transformation Ad((p(g)) : so(Qd) -> so(Qd) preserves Af and H+Af. We may therefore replace T by (l/$(Ad((j)(g)))T, and assume that T = M2Then (1) - (6) are straightforward computations and (7) follows from the fact that T2 = -Eld. Proof of (8): Let X 6 so(Qd) and assume that (adT) 2 X = 0. We wish to show that X eWH+Af. We have so(Qd) = (T(Af)) + (WH) + TV, so choose A G T(Af), B G WH and C G Af such that A + B + C = X. We wish to show that A — 0. By (4) and (5) of Lemma 6.3.15, p. 190, (&dT)2(B + C) = 0. Then 0 = ( a d T ) 2 X = (zdT)2(A + B + C) = ( a d T ) 2 A Then, from Lemma 6.3.3, p. 185, we conclude that A = 0. End of proof of (8). • L E M M A 6.3.16 Let V be a vector space and let Q G MinkQF(V). Let T G (so(Q))\{0} be nilpotent. Then dim(T(V)) = 2, dim(T 2 (V)) = 1 and TZ(V) = {0}. Proof. Let d := dim(V). We may assume that V = K d x l and that Q = Qd- By Lemma 6.3.12, p. 189, we may assume that T = A/"2 . Then T(V) = B e ^ + l 4 d ) and T2(V) = K e ^ and T3(V) = {0}. D L E M M A 6.3.17 Let d > 2 be an integer. Let A G R\{0}. Let T G XH^. Let V C Rdxl be T-invariant. Assume that Qd\V is Minkowski. Then there exist v,w G y \ { 0 } such that Tv = —Xv and such that Tw = \w. Proof. Let T — D+E+0 nalizable part as T(V) C V,

D := XD^. Choose E G e[d) such that T = D + E. Then is a real Jordan decomposition in E d x l , so D is the real diagoof T. Then D is contained in the real span of T, T2, T3,..., so, we get D(V) C V. Let A := Ee[d), B := Ee2d) +••• + E e ^

192

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and C := Med>. Since D : R d x l -> E d x l is real diagonalizable, since A, B and C are the eigenspaces of D and since F is £>-invariant, choose vector subspaces A0 of A, Bo of B and Co of C such that V = AQ + Bo + CoSince Qd|(^4 + B) and <9d|(jB + C) are both degenerate, it follows, from Lemma 6.2.13, p. 181, that V & A + B and that V £ B + C. Then Co ^ {0} ^ A0. Fix « € C o \{0} and w € A o \{0}. D

6.4

E m b e d d i n g s of Lorentz Lie algebras

For all integers m,n > 1, if m < n, we define an injective E-linear map 4^ : E m x m -s- Rnxn by C O ^ ) = ^ n ) . This is the "upper left corner" embedding. L E M M A 6.4.1 Letp > 3 be an integer. Let q>0 be an integer and define F0 : R" -> E by F0(yi,..., 2/,) = y\ + • • • + y\. Define F : W x K? -> E by F(x,y) = <2P(x) + F0{y). Define 4> := (^+«|so(Q p ) : so(Qp) ->• 50(F). Let V> : so(<5p) —> so(F) be any nontrivial Lie algebra homomorphism. Then there exists g £ SO°(F) such that (Adg)(ip(so(Qp))) = 4>(so(Qp)). Proof. Let d := p+q. The signature of F0 is = (q, 0) and the signature of Qp i s = ( p - 1 , 1 ) . Then, the signature of F is = (g, 0) + ( p - l , 1) = (l,d—1), which is the signature of Qd, so so(F) is Lie algebra isomorphic to so(Qd). In particular, so(F) is semisimple with real rank = 1. Since so(Qp) is simple and tp is nontrivial, we have ker(^) = {0}, so V> : so(<3p) ->• so(F) is injective. Let T := V^. Let T' := 4>{T). Let a := I T . Let a' := KT". Then o is a maximal real split torus of so(Qp) and o' is a maximal real split torus of eo(F). By Corollary 4.11.7, p. 130, V>(<0 is a real split torus of so(F). We have dim(ip(a)) = dim(a) = 1, so, since the real rank of so(F) is = 1, it follows that tp(a) is a maximal real split torus of so(F). So, by conjugacy of maximal real split tori (Lemma 4.10.13, p. 117), we may assume that V»(o) = a'. Choose A £ E such that rp(T) = XT'. Since 1 is an eigenvalue of a d T : so(Qp) —> so(Qp), we conclude that 1 is an eigenvalue of a.d(ip(T)) : so(F) -> so(F). So, by Corollary 6.3.8, p. 187, we may assume that ip(T) = T'. Let n+ be the (+l)-eigenspace of a d T : so(Qp) ->• so(Qp) and let n_ be the (-l)-eigenspace of a d T : so(Qp) -> so(Qp). Similarly, let n^ be the

Embeddings of Lorentz Lie algebras

193

(+l)-eigenspace of adT" : so(F) ->• 00(F) and let n'_ be the (-l)-eigenspace ofadT':so(F)->so(F). We have (T) = T", so (n_) C n'_ and (n+) C n'+. Similarly, we have t/>(T) = T', so ^ ( n - ) C n'_ and V(n+) C n^.. Let C := {g € SO°(F) | (Adg)T' = T'}. Let i := dim(n + ). By Lemma 6.3.9, p. 187, the Adjoint action of C on n'+ is transitive on the set of i-dimensional vector subspaces of n'+. Then ^(n+) and {n+) are both i-dimensional vector subspaces of n' + . We may therefore assume that ip(n+) = (n+). We wish to show that ip(so(Qp)) = 4>(so(Qp)). Since dim(tp(so(Qp))) = dim(so(Q p )) = dim(4>(so(Qp))), it suffices to show that V(so(Q p )) C 4>(so(QP)). Since no proper Lie subalgebraof so(Qp) contains n+Un_, it follows that no proper Lie subalgebraof ip(so(Qp)) contains (i/>(n+))U(i/>(n_)). We have VKn+) = (n+) C (j>(so(Qp)), so it suffices to show that rp(n^) C (so(Qp)). Fix Y € n_. We wish to show that ip(Y) € (f>(so(Qp)). We may assume that 7 / 0 . By Lemma 6.3.10, p. 188, we choose X £ n+\{0} such that [X,Y] = T. Then we get i>(X) e V(n+) =
and MX),^00] =^(T)=T'. Choose X0 € n+ such that 4>(X0) — ip(X). Since X ^ 0, it follows that ip{X) / 0, so (T) = T'. Then

MX),(0(ro)) - WOO)] = [
= O.

We have <j)(Y0) € 0(n_) C n'_

and

^(y0)e^(n.)Cn'..

Then ((Y0)) - (ip(Y)) € n'_. Moreover, tp{X) 6 ip(n+) C n' + . Also, as X ^ 0, we have ip{X) £ 0. Let X' := ^ ( X ) . Then we have X ' € n ^ \ { 0 } . Let Y' := (0(KO)) - W-00)- Then r £ n'_. Then [X',Y'] = 0, so, by Lemma 6.3.4, p. 186 (with Q replaced by F and T replaced by T"), we see that Y' = 0, i.e., that <£(Y0) = ip(Y). Then ^(K) = 0(KO) € # n _ ) C 0(so(Q p )). D Let g be a Lie algebra, let d > 3 be an integer and let V be a real gmodule. Let 7" denote the dxd identity matrix. Assume that g is Lie algebra isomorphic to so(Qd). We say that V is standard if there are a Lie algebra isomorphism a : g —> so(Qd) and a vector space isomorphism /3 : V —> E d x l such that, for all X e fl, for all u e V, we have /?(Xv) = (a(X))(/3(v)). Computation shows {M £ R d x d |VW 6 a(g),MN = iVM} = R7. There-

194

Minkowski

Linear Algebra

fore V has only scalar g-intertwining and is a fortiori 0-irreducible. COROLLARY 6.4.2 Let Q be a Lie algebra and assume, for some n > 2, that g is Lie algebra isomorphic to so(n, 1). Let V be a nontrivial real g-module and let Q G MinkQF(F) be g-invariant. Then there exists an g-irreducible, g-nontrivial vector subspace V of V and a g-trivial vector subspace V" of V (1) dim(V') = n + 1; (2) the g-module V is standard, and therefore has only scalar intertwining, and therefore is irreducible; (3) V = V' + V" and V n V" = {0}; (4) Q\y is Minkowski; and (5) Q\V" is positive definite. Proof. Let d := dim(V). Let p := n + 1 and let q := d — p. We may assume that g = so(Qp). Since so(Qp) is Lie algebra isomorphic to a Lie subalgebra of so(Q), it follows that the real rank of so(Qp) is < the real rank of so(Q). That is, p - 1 < d - 1. Then q > 0. Let F and (/> be as in Lemma 6.4.1, p. 192. We may assume that V = Rd and that Q = F. Let V : 0 -> so(F) be defined by (^(^))u = -Xw. By Lemma 6.4.1, p. 192, we may assume that <j>(g) = ip(g). Let V := W x {0}« C Rd and let V" := {0} p x W C Md. D L E M M A 6.4.3 Letd > 3 be an integer. Letg be a Lie algebra and assume that g is Lie algebra isomorphic to so(Qd). Let T € 0. Let A,B £ g, let X,H>0 and assume that (adT)^4 = A^l and that (adT)-B = fiB. Then [A,B] = 0. Proof. We may assume that 0 = so{Qd) and that A ^ 0 ^ B. By Lemma 4.10.28, p. 123, choose D,E,N € g such that the equation adB (T) = ad g (D) + adB (E) + adB (N) is a real Jordan decomposition on on 0. Let / : 0 —> 0 be the identity transformation. Then the equation (ad 0 (T)) - XI = [(adB(Z?)) - XI] + ad 8 (£) + ad0(AT) is a real Jordan decomposition on 0, so (adB(L>)) - XI is in the IK-span of {[(ad B (T)) - XI]k \ke {1,2,3,...}}. Then A € ker[(ad B (T)) - XI] C ker[(ad B (D)) - XI],

Embeddings of Lorentz Lie algebras

195

so (adD)A — XA. Similarly, we have (&dD)B = \iB. Replacing T by D, we may therefore assume that ad T : g —> g is real diagonalizable. Replacing T by (1/A)T, we may assume A = 1. Then, by Lemma 6.3.7, p. 187, we may assume T = T>^. Then the set of eigenvalues of ad T : g —¥ g is {—1,0,1}, so, since (adT)jB = /j,B and fj, > 0, we conclude that \i = 1. Then (adT)B = B. As N^ is the (+l)-eigenspace of a d T : g -> g, it follows that A, Be M^. Then [A,B] e [TV'^.A/'W] = {0}. • L E M M A 6.4.4 Let n > 2 be an integer and assume that g is Lie algebra isomorphic to a Lie subalgebra of so(n, 1). Assume that g is semisimple and that g has no compact factors. Then there is an integer m € [2,n] such that g is Lie algebra isomorphic to so(m, 1). Proof. Assume, for a contradiction, that this is false. For all integers m > n, we have dim(so(m, 1)) > dim(so(n, 1)), so g is not Lie algebra isomorphic to so(m, 1). So, for any integer m > 2, we see that g is not Lie algebra isomorphic to so(m, 1). We may assume that g is a Lie subalgebra of so(n, 1). Then the real rank of g is < the real rank of so(n, 1), which is = 1. Since 0 is noncompact, we see, by Lemma 4.12.2, p. 132, that the real rank of g must be > 1. Then the real rank of g is — 1 and so g is simple. Fix T 6 fl\{0} such that a d T : g -> g is real diagonalizable. By Corollary 4.11.7, p. 130, a d T : so(n, 1) -4 so(n, 1) is real diagonalizable. For every A > 0, let Vx := {X € so(n,l) | (adT)X = XX}. For all A > 0, let gx := Vx n 0 = {X € g | (adT)X = XX}. Let V+ := ^ Vx. \>o By Lemma 6.4.3, p. 194, for all A,B G V+, we have [A,B] = 0. Let 0+ :=V+ H0 = ^

0A

"

A>0

Since 0 has real rank = 1 and since, for every integer m > 2, 0 is not Lie algebra isomorphic to so(m, 1), it follows, by the classification of rank one simple Lie algebras (Lemma 4.10.34, p. 125) and by computation, that there exist A, B £ 0+, such that [A,B] ^ 0. Since 0+ C V+, we have a contradiction. • COROLLARY 6.4.5 Let I be a noncompact simple Lie algebra and let V be a nontrivial real [-module. Assume that V admits an [-invariant Minkowski quadratic form. Then, for some integer m > 2, we have that I is Lie algebra isomorphic to so(m, 1).

196

Minkowski Linear Algebra

Proof. Let d := dim(V). Let p : [ —> fll(F) denote the representation of 0 on V. Let Q 6 MinkQF(V) be (-invariant. Then p(l) C so(Q). Since p is nontrivial and since I is simple, it follows that ker(p) = {0}, so p : I —> so(Q) is injective, so I is isomorphic to a Lie subalgebra of so(Q). In particular, so(Q) is nonAbelian, so d > 3. Let n := d— 1. Then n > 2 and so(<2) is Lie algebra isomorphic to so(n, 1). So, by Lemma 6.4.4, p. 195, we are done.D

Chapter 7

Basic Dynamical Results

Let a Lie group G act smoothly on a manifold M. For all X £ g, for all m e M, define Xm :— {d/dt)t=o{exp{—tX))m. For all l e g , define a := vector field XM on M by the rule (Xjvf)m Xm- Note that the minus sign appearing in the definition of Xm ensures that, for all X, Y € fl, we have [XM,YM] = [X, Y]M- That is, X i-t XM is a Lie algebra homomorphism from fl into the infinite-dimensional Lie algebra of vector fields on M. For all g 6 G, for all p € M, for all X € fl, {{Adg)X)gp € TgpM is the image of Xp G TPM under the differential at p of m <-¥ gm : M -> M.

7.1

Kowalsky's Lemma

The following lemma is basic to [Ko96]. Eventually, we will develop a higher jet analogue of it (Lemma 7.10.1, p. 236). L E M M A 7.1.1 Let a Lie group G act isometrically on a pseudoRiemannian manifold M. Let gi be a sequence in G, let mi be a precompact sequence in M and let ml £ M. Assume that gimi —> m' in M. Let X' € fl be Kowalsky for Ad B (pj). Then X'm, is a lightlike vector in the Minkowski space TmiM. Proof. We may assume that X' ^ 0. As X' is Kowalsky for Ad fl (pj), fix a sequence Xj in fl such that both Xi -* 0 and (Ad gijXi -+ X' in fl. For all m € M, let (•, • ) m denote the Minkowski symmetric bilinear form on TmM. For readability, if A, B 6 fl, and if m 6 M, then we will write {A, B)m to denote (Am,Bm)m. 197

198

Basic Dynamical

Results

For all i, let m\ := giirii and let X[ := (Ad• 0 in g and since m, is precompact in M, it follows that (Xj, Xi)mi -> 0 in E. For all i, {X[)mi, is the image of (Xj) m ; under the differential at m^ of the isometry m i->- gtm : M -»• M. It follows, for all i, that (X;,Xj) m v = (Xj',X t ') m '.. In particular, (X<,Xj)m>. -> 0 in 1, so, since X? -> X ' in g, (X',X')m> = O.D

7.2

H i g h e r j e t s of vector fields a n d m e t r i c s — n o t a t i o n

Recall that C denotes Lie derivative. Let d,r > 0 be integers. Let di,...,dd be the standard framing of Rd. d Let / , g : R -> W be smooth maps. We say that / a n d g agree t o o r d e r 0 at 0 if /(0) = g(0). Let k > 1 be an integer. We say that / a n d g agree t o o r d e r & at 0 if /(0) = g(0) and if, for all integers j € [l,k], for all integers pu ... ,pj e [1, d], we have (9 P1 • • • dPjf)(0) = (dPl • • • dPjg)(0). Let M and N be manifolds and let m € M. Let W denote the set of open neighborhoods of m in M. For any W € W, let C°°(T^,JV) denote the set of all smooth maps W —> N. An iV-valued local m a p n e a r m in M is an element of ( J C°°{W, N). The domain W of such a map will be denoted dom(/). If / and g are two AT-valued local maps near mm M, then we say that / a n d g agree near m if there is an open neighborhood U of m in (dom(/)) D (dom(gr)) such that f\U = g\U. This defines an equivalence relation on the set of all iV-valued local maps near mm M. An equivalence class of this equivalence relation is called a g e r m at m in M w i t h values in N. Let d := dim(M) and let r := dim(N). Let k > 0 be an integer. We say that / a n d g agree t o order k at m if there are • an open neighborhood U of m in (dom(/)) D (dom(p)); • an open neighborhood V of / ( m ) in N; and • diffeomorphisms <j>: Rd -»• U and ip : V -> W such that • • • •

: Rd -»• E r and t/> o (p|C7) o ^ : Rd ->• E r agree to order fc at 0.

Higher jets of vector fields and metrics - notation

199

This defines an equivalence relation on the set of all iV-valued local maps near m in M. An equivalence class of this equivalence relation is called a fc-jet at m in M with values in N. Let M and N be manifolds and let m € M. Let / and g be two N-valued local maps near m in M. Then / and g agree to order 0 at m iff / ( m ) = g{m). Moreover, / and g agree to order 1 at m iff both / ( m ) = g(m) and (df)m = (dg)m. Let M be a manifold and let £ be a fiber bundle over M. (In particular, E could be a tensor bundle, or any vector bundle.) A .E-valued local section near m in M is a smooth section of the restriction of E to some open neighborhood of m in M. Agreement near m (resp. agreement to order k at m) defines an equivalence relation on the set of S-valued local maps near m i n M . We restrict this equivalence relation to the set of Evalued local sections near m in M; an equivalence class of the restriction is called a germ of a section of E at m (resp. fc-jet of a section of E at m). Now specialize to the case where E = TM is the tangent bundle of M. A TM-valued local section near m in M is called a local vector field near m in M. A germ of a section of TM at m is called a germ of a vector field at m in M . A fc-jet of a section of TM at m is called a fc-jet of a vector field at m in M. Let X be a local vector field near m in M. Then the 0-jet of X at m is exactly determined by the value Xm of X at m. The conclusion of Lemma 7.1.1, p. 197 concerns X'm,, or, equivalently, the 0-jet of X'M at ml. We eventually wish to be able to analyze higher jets of X'M at m'. This requires that we develop a systematic notation and calculus for working with these jets. Let M be a d-manifold, let m € M, let X be a local vector field near m in M and let B be an ordered basis of TmM. Then Xm corresponds to an element of E d under the isomorphism ips : Rd —> TmM. This allows us to think of 0-jets concretely as elements of Rd. Now assume that M comes equipped with a smooth connection V and let U be an open neighborhood of 0 in TmM on which the exponential map exp : U -*• M is defined. Replacing U by a smaller open neighborhood if necessary, and defining V :— exp(f/), we may assume that exp : U —> V is a diffeomorphism. Let UQ := ipQl(U). Let e : Uo —>• V be the diffeomorphism defined by e(x) = exp(^e(x)). Let XB denote the germ at 0 of e*{X\V). Note that the value of XB depends on V; however we suppress this from the notation, because, in every situation we will encounter, the connection V will be

200

Basic Dynamical

Results

clear. In particular, on a pseudoRiemannian manifold, we always use the Levi-Civita connection. We will also need to analyze jets of tensors other than vector fields. Let r be a tensor bundle over M. If a is a r-valued local section near m, then we define as to be the germ at 0 of e*(• K such that A = /i<9i H + fddd. Let k > 0 be an integer. We say that A has degree k if f \ , . . . ,fa are all homogeneous polynomials of degree k. For example, A has degree 2 iff A is an R-linear combination of vector fields taken from the set {xiXjdk \ i,j, k 6 / } . Let A be a vector field on a d-dimensional vector space V. Then we say that A has degree k there is an isomorphism <j> : V -» Rd such that 4>*{A) has degree A;. If A has degree k, then, for any isomorphism : V -> R d , we have that <j>*(A) has degree k. We say that A is constant (resp. linear, resp. quadratic) if A has degree 0 (resp. 1, resp. 2). For all v € V, let tv : V —• TVV be the vector space isomorphism defined by t,v(w) = (d/dt)t=o(v + tw). Then A is constant iff there is a vector w £ V such that, for all v € V, we have Av = tv(w). Moreover, A is linear iff there is a linear map L : V —• V such that, for all v 6 V, we have Av = LV(L(V)). Let B be another vector field on V. If A has degree k and B has degree I, then [A, B] has degree k + 1 — 1. Let V be a vector space and let Q be the set of germs at 0 of vector fields on V. Let A € Q. Then we say that A has degree k if A is the germ at 0 of a vector field of degree k on V. We say that A is constant (resp. linear, resp. quadratic) if A has degree 0 (resp. 1, resp. 2). For all X, Y € Q, if X has degree k and Y has degree ^, then [X, Y] has degree

k+

e-i.

According to Taylor's theorem, for any A £ G, there is a unique sequence Ao, Ai, A2,... in Q such that • for all integers k > 0, Ak has degree k; and • for all integers k > 0, A and -A0 H 1- Ak agree to order k at 0. We call AQ the constant part of A and denote it by Ac. We call vli the

Higher jets of vector fields and metrics - notation

201

linear part of A and denote it by AL. We call Ai the quadratic part of A and denote it by A®. For all integers i > 0, we call the A, the degree i part of A. For all A, B 6 G, we have [A,Bf

= L

[A,B]

=

[A°,BL] C

Q

+ [AL,BC] L

L

[A ,B ] + [A ,B ] +

and [AQ,BC].

Let A, B € Q. For any integer i > 0, let A, denote the degree i part of A and let Bi denote the degree i part of B. Then, for all integers k > 0, k+1

the degree fc part of [^4, B] is V j [Aj, S/t + i_j]. If A is linear, then the above i=0

formulas give [A,B] C = [A,BC] and [A,B] L = [A,-BL]; in general, for all integers k > 0, the degree k part of [A,i?] is [A, Bk\- In particular, if A is linear, then [A,B]Q = [A,B®]. Let d := dim(M). Let G := GLd(K) and £ := QF(E d ). Let F M denote the frame bundle of M. Let .E0 := FM XQ £ • A E0-valued local section near m in M is called a local quadratic differential near m in. M. A germ of a section of EQ at m is called a germ of a quadratic differential at m in M. A fc-jet of a section of -Bo a t m is called a A;-jet of a quadratic differential at m in M. For all v € V, let t„ : V —• T„V be the standard identification defined by i„(u;) = (d/dt)t=o(v + tw). A quadratic differential Q on a vector space V is said to be flat if there is a quadratic form Q : V -> E such that, for all u e V, we have Q = Q o iv. In this case, Q is the flat quadratic differential on V corresponding to Q . Let Q be the germ at 0 of a quadratic differential on Rd. For every pair i,j 6 / , let Qij : Rd ->ffi.be defined by Qij(h,... ,td) = tttj and let Qij be the germ at 0 of the flat quadratic differential on Rd corresponding to Q^. For all integers k > 0, let Vk denote the collection of germs at 0 of homogeneous polynomials on Rd of degree k. We will say that Q has degree k if there are fy e Vk such that Q = ^ P / y Qjj. We say that Q is constant (resp. linear, resp. quadratic) if Q has degree 0 (resp. 1, resp. 2). We observe that a germ of a quadratic differential at 0 € V is constant iff it is the germ at 0 of a flat quadratic differential on V. Let V be a vector space. By Taylor's theorem, if Q is the germ at 0 of a quadratic differential on V, then there are unique germs Qo> Qi-, Q2, • • • such that

202

Basic Dynamical Results

• for all integers k > 0, Qk has degree k; and • for all integers k > 0, Q and Q 0 + I- Qfc agree to order k. We call Q 0 the constant part of Q and denote it by Qp'. We call Qi the linear part of Q and denote it by QL. We call Q2 the quadratic part of Q and denote it by Q®. For all integers i > 0, we call the Qi the degree i part of Q. Let A be the germ at 0 of a vector field on V. Let Q be the germ at 0 of a quadratic differential on V. If A has degree k and Q has degree £, then the Lie derivative £A(Q) of Q along A has degree fc + £ — 1. Moreover, (£AQ)C 0CAQ)L

= =

(^c(QL)) + (£^(Qc)) Q

(CAo(Q ))

L

+ (CA,(Q ))

and +

(CAQ(Qc)).

For all integers i , j > 0, let A, denote the degree i part of A and let Qj denote the degree j part of Q. Then, for all integers k > 0, the degree fc+i

k part of £ ^ Q is Y J ^ ^ (Q/t+i-i)- If A is linear, then the above formulas give (CA(Q))C =~CA(QC) and {CA{Q))L = CA(QL); in general, for all integers k > 0, the degree fc part of CA(Q) is / ^ ( Q * ) . In particular, if A is linear, then (£AQ)Q = CA{QQ). 7.3

Matrix realizations of jets and calculus on jets

Let d > 1 be an integer. Let I := { 1 , . . . , d}. Let 9 { , . . . , d'd be the standard framing of E d and let x[,..., x'd : E d —¥ E be the coordinate projections. For all i £ 7, let 9j to denote the germ at 0 of d[ and let Xi to denote the germ at 0 of x\. For all i € I, let e* := e^ , so e\,... ,e<j is the standard basis of E d x l . For all i,j e 7, let £ y := i?j- , so {i?^} is the standard basis of M dx ' i . Let ^ be the set of germs at 0 of vector fields on Rd. Let A 6 Q. Then there is a unique system of constants a i , . . . , a<j s E such that we have Ac = aid±-\ h addd- Moreover, there are unique constants 6i:7 e E such L that A = V j bijXidj. We define

A Cm

. = J^ ajej G E d x l

and

A i m := - ^

b^E^ € E d x d .

Matrix realizations of jets and calculus on jets

203

The superscript m stands for "matrix form". We do not have a convenient matrix form for the parts of A of degree > 2. Note that, in the formula defining ALm, there is a minus sign and that we use bji instead of bij. This is done to guarantee the following facts: For all A, B € Q, we have • . • •

if A is linear and B is constant, [A,B]Cm = (ALm)(BCm) <E M d x l ; Lm Lm Lm if A and B are both linear, then [A,B] = [A ,B ] e Rdxd; Cm Lm Cm Lm Cm dxl [A,B] = (A )(B ) - (B )(A ) e M ; and Cm if A is linear, then we have [A,B] = (ALm){BCm) e E d x l and [A,B]Lm = [ALm,BLm] 6 Rdxd.

We think of Xc and XCm as two concrete realizations of the 0-jet of X at 0. There is a standard isomorphism T0Rd <—• Rdxl. Under this isomorphism, XQ corresponds to XCm, so one sees directly that the information in the 0-jet of X at 0 is exactly the information in XCm. Similarly, we think of Xc + XL and (XCm,XLm) € Rdxl © Rdxd as two concrete realizations of the 1-jet of X at 0. Finally, we think of the sum Xc + XL + X® as a concrete realization of the 2-jet of X at 0. Let M be a d-dimensional manifold with a smooth connection and let X be a vector field on M. Let m 6 M and let B be an ordered basis of TmM. We then define X% := {XB)C, X% := (XB)L, X$ := {XB)Q', Xgm := (XB)Cm € E d x l , and X^m := (XB)Lm € Rdxd. Let a Lie group G act smoothly on a d-dimensional manifold M preserving a smooth connection. If X € g, if m £ M and if B is an ordered basis of TmM, then we define XB '•= {XM)B and X% := {XM)% and Xgm := {XM)%m G Kf"*1. Similarly, we define X ^ := ( X M ) | and X | m := ( X M ) | T e E d x d and X f := (XM)% Let a Lie group G act Smoothly on a manifold M. Let S C g and let m € M. We define Sm •— {Xm \X e fl}. Assume that the G-action on M preserves a smooth connection and let B be an ordered basis of TmM. Then we define SB := {XB\X eg}. We similarly define Sg, S£, 5 ^ , 5 g m and 5 g m . Note that, under these definitions gm = {Xm \ X G g} is the tangent space to the G-orbit at m. Wyoming; Some authors follow the convention that Gm denotes the stabilizer at m of G and that gm denotes the Lie algebra of Gm. We do not follow this, and instead use Stabc(»Ti) and stabg(m) for the stabilizer and its Lie algebra. Recall that, for any vector space V, for any v € V, [v]B := 4>~^l{v). The next three results are direct consequences of the definitions.

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Basic Dynamical Results

L E M M A 7.3.1 Let G be a Lie group acting smoothly on a manifold M preserving a smooth connection. Let d := dim(M). Let mo 6 M and let B be an ordered basis ofTmoM. Then [Pmo]B = P§m. L E M M A 7.3.2 Let d > 1 be an integer. Let A be the germ at 0 of a linear vector field on Rd. Let Q be a quadratic form on Rd. Let Q be the germ 0 of the flat quadratic differential on Rd corresponding to Q. Then CA(Q) = OiffALmeso(Q). L E M M A 7.3.3 Let M be a pseudoRiemannian manifold, let m G M and let d := dim(M). Let Q G QF(R d ) have the same signature as the tangent spaces of M. Let Q be the germ 0 of the flat quadratic differential on Rd corresponding to Q. Let B be an ordered Q-basis ofTmM. Then g$ = Q. L E M M A 7.3.4 Let Q be a germ at 0 of a quadratic differential on Rd. Assume that Q is linear. Then Q = V J Xj^Cg^Q)). Proof. For k,l e I, define Qkl G QF(Md) by Qki{h,.. .,td) - tktr, then let Qki be the germ at 0 of the flat quadratic differential on Rd corresponding to Qki- Choose {ajki}j,k,iei such that Q = ^ ajkiXjQki. Then, for j,k,l€l

all j 6 / , we have £a,.(Q) = Y^ o,jkiQki. The result follows. k,l€l

^

Let Go be the collection of germs at 0 of constant vector fields on Rd. The next lemma asserts that, in exponential normal coordinates, a metric is flat to first order. This is a well-known basic result in differential geometry. L E M M A 7.3.5 Let M be a d-dimensional manifold and let g be a pseudoRiemannian metric on M. Let m G M and let B be an ordered basis ofTmM. Thengfe = 0. Proof. Let d := dim(M). For all A G Go, let e(A) := A0 G T0Rd denote the value at 0 of A. Let h := gs- We wish to show that hL = 0. Fix j G / . Let U := dj. By Lemma 7.3.4, p. 204, it suffices to show that Cu{hL) — 0.. We have (Cu(h))c = Cu(hL), so Cu(h) and Cu(hL) agree at 0. We therefore wish to show that Cu(h) vanishes at 0. Let V, W G Go- We wish to show that (Cu(h))(V,W) = 0 on M.

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205

For all v G E d , let 7„ : E ->• E d be the curve 7(f) = tv. By definition of the exponential map, we see, for all v eRd, that the germ at 0 of 7,, is geodesic with respect to h. Let V be the Levi-Civita connection of h. Then, for all v £ E d , we have V-y„(7„) = 0. For all A € Go, there exists v £ E d such that 7„(0) = e(A), and we then have e{SIAA) = (V-y„ (>))(0) = 0. For all A, B 6 So, we have V^-B - V B A = [A, 5 ] = 0. Then, for all A,B eg0, 2 • t(VAB) = e((V A + B(A + B)) - (VAA) - (V B B)) = 0. We have U(h(V, W)) = h(S?uV, W) + h(V,

VuW).

So, as e(V[/V) = e(V[/X) = 0, we see that C7(/i(V,W)) vanishes at 0. Moreover, U, V, W € Q0, so we get [U, V] = 0 = [[/, W]. Then (£u(h))(V, W)

= =

([/(^(V, W))) - (/>([[/, VI, W)) - (h(V, [U, W})) U(h(V,W)).

Then (£u{h))(V, W) vanishes at 0. Since U, V and W are constant and since hL is linear, it follows that (£u(h))(V, W) is constant. Then (£u{h))(V, W) = 0 on M. • Matrix multiplication gives an action of the matrix GL<j(E) C E d x d on E d x l . Identifying E d x l with E d , we get an action of GL d (E) on E d . Differentiating, we obtain, for each integer k > 0, a natural action of GLd(E) on the set of germs at 0 of vector fields in E d , and on the set of fc-jets at 0 of vector fields in E d . Using these actions, we calculate: L E M M A 7.3.6 Let d > 1 be an integer and let A be a germ at 0 of a vector field on Rd. Let g £Rdxd and let B := gA. For all k, let Ak be the degree k part of A and let Bf. be the degree k part of B. Then (1) for all k, we have gAk = B^; (2) g(Ac) = Bc and g{AL) = BL and g(A^) = B®; and (3) g{ACm) = BCm and ff(Aim)g-1 = BLm. L E M M A 7.3.7 Let d > 1 be an integer. Let M be a d-manifold with a smooth connection and let f : M —± M be a connection-preserving diffeomorphism. Letm € M. Letm' := f(m). LetB be an ordered basis ofTmM and let B' be an ordered basis ofTm>M. Let g := [{df)m\% € GLd(E) be the matrix of (df)m : TmM —> Tm<M with respect to B and B'. Let X be a germ at m of a vector field on M and let X' := f*(X). Then:

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(1) g{XB)=X'B,; (2) g(X%) = {X% andg(X^) = (X% and g(X%) = (X')%; (3) g(Xg™) = (X%r andg^X^g-1 = (X%1».

and

Proof. Let := (df)m : TmM ->• T m «M. Fix an open neighborhood U of 0 in TmM such that exp m is denned on U. Then exp m< is defined on U' ~ (U) C TmM. Let e := exp m oxj)B : tp^l{U) ->• M,

e' := expTO, o^B, : tp^(U')

-» M.

By naturality of the exponential map, for u G £7, /(exp m (w)) = exp m , (cf>(u)). Moreover, since g = [<j>]^ , for all w G Rd, we have (ipB(w)) = ipB>(gw). Then, for all w G ip^iU), we have /(exPmW'BH)) = exp m ,((/i(^ B (u;))) = expm, (ipB, (gw)), so f(e(w)) = e'(gw). Conclusion (1) then follows from the definitions. Using Conclusion (1), Conclusions (2) and (3) follow from (2) and (3) of Lemma 7.3.6, p. 205, respectively. • 7.4

Miscellaneous results

L E M M A 7.4.1 Let a Lie group G act isometrically on a pseudoRiemannian manifold M. Let d := dim(M). Let Q G Q F ( R d x l ) . Assume that the signature of Q is the same as that of the tangent spaces of M. Let X E g and let me M. Let B be an ordered Q-basis ofTmM. Then Xfe™ G so{Q). Proof. Let g denote the Lorentz metric on M. We have 0 = CXB(gB), so 0 = {CXB{gB))C = {Cxc{g^)) + (CXL(g%)). By Lemma 7.3.5, p. 204, we have gfe = 0. Then 0 = £XL(g%). Let Q0 G QF(E d ) correspond to Q G QF(Md><1) under the standard isomorphism Rd -f—> E d x l . Then, under our identifications, we have so(Qo) = so(Q) C K d x d . Let Q denote the germ at 0 of the flat quadratic differential on Md corresponding to Q 0 . By Lemma 7.3.3, p. 204, we have g% = Q. Then 0 = CXL(Q), Lm which, by Lemma 7.3.2, p. 204, implies that (X%) G so(Q0). Then ( X | ) i m = X | m G so{Q0) = so(Q) C Rdxd. D L E M M A 7.4.2 Let a Lie group G act continuously on a topological space M. Then {m G M \stabe(m) = {0}} is a G-invariant open subset of M.

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Miscellaneous results

Proof. Let M 0 := {m G M | s t a b e ( m ) = {0}}. For all m G M , for all g G G, we have stab g (gm) = (Ad<7)(stabB(m)), so Mo is G-invariant. Let mi be a convergent sequence in M\MQ- Let m' := lim rrn. We wish to i—>oo

show that m' ^ Mo. For all i, choose Xi G (stab e (mi))\{0}. Let tt be a sequence in K such

that {ijXj}, is precompact in fl\{0}. Passing to a subsequence, choose X' G fl\{0} such that t{Xi -> X'. For all i, we have tjXj G stab 0 (mj). Then X' e stab B (m'). So, since X' ^ 0, we see that m' ^ M 0 . D L E M M A 7.4.3 Let a Lie group G act isometrically on a Lorentz manifold. Let Mo be the collection of all m G M such that gm is a Minkowski vector subspace ofTmM. Then Mo is a G-invariant open subset of M. Proof. The G-action on M is isometric, so Mo is G-invariant. Let m* be a sequence in M\Mo. Assume that m* is convergent in M and let m' := lim mi. We wish to show that m' $ MoFor all m G M , and let Q m be the Minkowski quadratic form on TmM. For all i, Qmi is not Minkowski, so, by Lemma 6.2.12, p. 180, Qmi must be positive semidefinite, so, for all X G 0, we have Qm{(Xmi) > 0. Then, for all X G 0, we have Qm>(Xm>) = lim Qmi{Xmi) > 0. Then gm> is positive i—>co

semidefinite, and is therefore not Minkowski. Then m' ^ Mo• Let M be a manifold, let V be a smooth connection on M and let X be a vector field on M . We say that X is V-Killing if, for all vector fields Y and Z on M, we have [X, VyZ] = (V [ x ,y]Z) + (Vy[X,Z]). Equivalent^, the local flow of X preserves V. Let M be a manifold. Let X be a vector field on M . Let m0 G M and assume that Xmo = 0. Let [/ be a precompact open neighborhood of mo in M . Let J be an open neighborhood of 0 in R such that, for a l i i G I , the time t flow of X is defined at every point of U. For all t G / , let ft : {/ -> M be the time £ flow of X on [/. Since X m o = 0, it follows, for all t G / , that ft{m0) = m 0 . For all t G I, let & := (dft)mo : TmoM -> TmoM be the differential at mo of ft. For all w G TmoM, let 1^ := (d/dt)t=o((t>t('w)). Since, for all t G / , the map 0 t : TmoM -»• TmoM is linear, it follows that y is a linear vector field on TmoM. The vector field V on T^,, M is called the linearization of X a t mo. L E M M A 7.4.4 Let M be a manifold with a smooth connection V. Let X be a V-Killing vector field on M. Let mo G M . Assume that Xmo = 0.

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Let Y be the linearization at mo of X. Let U be an open subset ofTmoM on which the exponential map of V is defined. Let e : U —> M be the exponential map. Then, for all u 6.U, we have (de)u(Yu) = Xe(uy Proof. We may assume that U is precompact in TmoM. Define I, ft and <j>t as above. Then, for all w 6 TmoM, we have Yw — {d/dt)t=o{(l>t{'w))By naturality of the exponential map, for all u € U, for all t e I, we have e((f>t(u)) = ft{e{u)). Computing (d/dt)t=o of both sides, we see, for all ueU,tha,t (de)u(Yu) = Xe{u). • Recall that a vector field V on a Lorentz manifold (M, g) is said to be Killing if Cy (g) = 0, or, equivalently, the local one-parameter flow of V preserves g. If V is the Levi-Civita connection of g, then any Killing vector field on (M, g) is V-Killing. L E M M A 7.4.5 Let (M, g) be a pseudoRiemannian manifold, let m € M, let V := TmM and let B := gm e SBF(V). Let X be a Killing vector field on (M,g). Assume that Xm = 0. Let Y be the linearization of X at m. For all v EV, let LV :V —> TVV be the vector space isomorphism defined by LV(W) = (d/dt)t=o{v + tw). Then, for allv € V, we have B{v,t~1{Yv)) = 0. Proof. Fix v eV. Let w :- i " 1 ^ ) . We wish to show that B(v,w) = 0. Let U be a precompact neighborhood of m in M. Choose 6 > 0 such that, for all t € J := (—6,5), the time t flow t(m) = m; let ipt '•= (d(j>t)m : V -> V. For all t £ I, let f(t) := -B(^) t (v),^(t;)). As X is a Killing vector field, it follows, for all t e I, that f(t) = B(v,v). Then /'(0) = 0. So, by Lemma 2.10.3, p. 35, B(w, v) + B(v, w) = 0. So, as B(v, w) = B(w, v), we get B(v, w) = 0. • Let d > 1 be an integer and let Y be the germ at 0 of a linear vector field Y' on Rd. Let x\,..., Xd : Rd —> R be the coordinate projections and let b\,...,dd be the standard framing of Rd. Let / := { 1 , . . . ,d}. For all i,j € I, let Eij := E\f. Choose 6 y G E such that Y' = ^ hj^idj. Then i,j£l

we define Ym := — \] i,j£l

bjiEij.

The superscript "m" stands for "matrix

form". Note, in the definition of Ym, the minus sign and the use of bji instead of 6»j. Since Y' is linear, it follows that Y' is complete. For all t € M, let 4>t : Rd -> Rd be the time t flow of Y'. Then i ^ ^, : 1 -> GLd(M) is a Lie group homomorphism. Let A := (d/dt)t=o(t) € flfd(E). Then

Miscellaneous

results

209

computation shows that, identifying gld(R) with Rdxd, we have A = — Ym. Let Z be the germ at 0 of another linear vector field on E d . Then we compute [Y, Z]m = YmZm - ZmYm = \Ym, Zm\. Let V be a vector space and let d := dim(V). Let B be an ordered basis of V. If Y is a linear vector field on V and if B is an ordered basis of V, then we define YB to be the the linear vector field on E d corresponding to Y under the vector space isomorphism xps '• Rd -> V and we define Y& to be germ at 0 of F B . Moreover, we define Yg := ( i s ) m . L E M M A 7.4.6 Let M be a manifold with connection V. Let X be a VKilling vector field on M. Let mo £ M and let B be an ordered basis of TmoM. Assume that Xmo — 0 and let Y be the linearization at mo of X. Then XB = YB and X%m = Yg. Proof. Let U be an open neighborhood of 0 in TmoM on which the exponential map of V is defined. By Lemma 7.4.4, p. 207, for all u £ U, we have (de)u(Yu) — Xe(uy Then, by definition of XB, (IPB)*(XB) is the germ at 0 of Y, so XB = YB- It then follows from the definitions that XLm_ym D Let G be a Lie group acting on a manifold M . Differentiating, we get an action of G on TM. Let mo € M and let H := StabG(m 0 ). Then the isotropy representation at mo is the representation I : H —> GL(T m o M) defined by (X{h))v = hv. Let H be a Lie group, let V be a vector space, let d := dim(y), let B be an ordered basis of V and let ~R : H ->• GL(V) be a representation. Then, by Us, we denote the representation TZB • H —> GLd(M) defined by KB(h) = V B 1 ° WW) ° *I>B- Then dJlB : f) -> flld(R) = Rdxd is the differential of TZB at 1#L E M M A 7.4.7 Let G be a Lie group acting locally faithfully on a manifold M preserving a connection. Let mo £ M and let B be an ordered basis ofTmoM. LetH := Stabo(m 0 ) and let! : H ->• GL(T m o M) be the isotropy representation at mo- Let X £ h. Then (dZe){X) = Xgm. Proof. For all t £ R, let gt := exp(-tX) and define / t ( m ) = gtm. Let y be the linearization at mo of XMp. 209, we have (XM)%m = ^B • That is, X£TO = Yg1. Let d := dim(M). By definition of XM, we see, for is the time t flow of XM- SO, by definition of the word

ft : M ->• M by By Lemma 7.4.6, all t e R , that ft "linearization", it

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follows that, for all t £ M, l(gt) : TmoM —^ TmoM is the time t flow of Y. Then, for all t £ R, lB(gt) : Kd -)• Rd is the time t flow of Yfc. Then (rfTB)(X) = -(d/dt)t=0(lB(9t)) = Yg> = X%m. D L E M M A 7.4.8 Let M be a manifold and let rn0 G M. Let W and X be vector fields on M. Assume that Wmo ^ 0 and that Xmo = 0. Assume, for all m € M, that Xm and Wm are linearly dependent vectors in TmM. Let Y denote the linearization of X at mo- Let B be an ordered basis ofTmoM. Let d := dim(M). Then Yg1 G Rdxd has rank < 1. Proof. We may assume that M is an open subset of Rd, that m 0 = 0 and that B is the standard ordered basis of T0Rd. Let M 0 := {m G M | Wm ^ 0}. Replacing M by Mo, we may assume, for all m € M, that Wm ^ 0. For all m e M, choose sm e R such that XTO = sTOWm. Define / : M ->• M by f(m) = sm. Then / : M ->• R is smooth and X = fW. Let /„ be the germ at 0 of / , let W* be the germ at 0 of W and let X* be the germ at 0 of X. Then Ym = X^m. We therefore wish to show that dim(X^ m (R d x 1 )) < 1. Since X0 = 0, we get X°m = 0. Since X = fW, we get X° = /.CTV,C. Then 0 = X f m = /(0) • Wf m . Since W0 ^ 0, we conclude that W?m ^ 0. Then /(0) = 0. Since X = / W , we get X»L = f^W^+f^W^. As /(0) - 0, we get / f = 0. Then X.L = tfW?, so, for all u 6 E d x l , X^ m r; G RW?m. Then X f " " ( E d x l ) C E ^ C m , so dim(X» Lm (]R dx1 )) < 1. D Much of the work in Chapters 14 and 15 is based on the idea that, for a vector field to be Killing with respect to a Lorentz metric, its various jets have components that satisfy many linear equations. For the vector field also to be in a complicated Lie algebra of vector fields, even more linear equations must be satisfied. Via tricks like Kowalsky's Lemma (Lemma 7.1.1, p. 197) and its higher jet generalization (Lemma 7.10.1, p. 236), dynamical requirements (like nonproperness) can impose even more linear conditions. If all the equations so obtained do not admit nontrivial solution, then the specified Lie algebra cannot be the Lie algebra of a Lie group acting locally faithfully with the given dynamical requirements. Of course, if one can immediately conclude that many components of the jets of vector fields are zero, then the linear problem simplifies. The next lemma asserts that, for a Killing vector field on a manifold with connection, if the components of the constant part of the vector field vanish, then so do all of the components except in the linear part.

Miscellaneous results

211

L E M M A 7.4.9 Let M be a manifold with a smooth connection V. Let X be a V -Killing vector field on M. Let mo £ M and let B be an ordered basis ofTmoM. Assume that X% = 0. Then XB=X%. Proof. Since X$ = 0, it follows that Xmo = 0. Let Y be the linearization at mo of X. By Lemma 7.4.6, p. 209, XB — YB- SO, since Y is linear, we conclude that YB is linear, so XB is linear. Then XB = XQ. O Let X be a vector field on a manifold M, let Z be the zero vector field on M, let m £ M and let k > 0 be an integer. We say that X vanishes to order k at m if X and Z agree to order k at m. L E M M A 7.4.10 Let M be a connected manifold with a smooth connection V and let X be a V-Killing vector field on M. Assume that X vanishes at every point of some nonempty open subset of M. Then X vanishes at every point of M. Proof. Let U be a nonempty open subset of M such that, for all u £ U, we have Xu = 0. Let A be the set of points m € M such that X vanishes to order 1 at m. We wish to show that A — M. Since U C A, we know that A ^ 0. So, since ^4 is closed in M and since M is connected, it suffices to show that A is open in M. Let a £ A. We wish to show that there is an open neighborhood V in M of a such that V C A. Let 6 be an ordered basis of TaM. Since a £ A, it follows that Xg = 0 and that X% = 0. Then, by Lemma 7.4.9, p. 211, we have XB = 0. Then the germ at a of X vanishes. Fix an open neighborhood V in M of a such that, for all v £ V, we have Xv = 0. Then V Q A. U Let G be a Lie group acting smoothly on a manifold M preserving a smooth connection. Let m € M, let B be an ordered basis of TmM and let X, Y £ g. It is a consequence of Lemma 7.4.9, p. 211 that if Xm = 0, then . . • • .

[X,Y]cB=\Xk,Y§]; [X,Y]k = [X*;,Y£]; [X,Yi% = [XB',Yg]; [ X , F ] g m = (X£ m )y J f m ; and [X,Y}%m = [X§m,Y£m].

These formulas will be used many times without comment in the sequel.

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L E M M A 7.4.11 LetG be a Lie group acting locally faithfully and smoothly on a connected manifold M preserving a smooth connection. Let X € 0, let m € M and let B be an ordered basis ofTmM. Assume that X%m = 0 and that X%m = 0. Then X = 0. Proof. We have X% = 0 and X§ = 0. Then, by Lemma 7.4.9, p. 211, we have XB = 0. Then XM vanishes on an open neighborhood of m in M. Then, by Lemma 7.4.10, p. 211, we see that XM — 0 on M. So, as the action of G on M is locally faithful, we conclude that X = 0. D L E M M A 7.4.12 Let G be a Lie group acting locally faithfully and smoothly on a manifold M preserving a smooth connection. Let X € g, let m £ M and let B be an ordered basis of TmM. Then Xm = 0 iff X%m — 0 iff Xg = 0iffXB = XL. Proof. By Lemma 7.4.9, p. 211, if X% = 0, then XB = Xfe. By Lemma 7.3.