Dynamics on Lorentz Manifolds

£ 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.

60

Basic Differential

Topology

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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Topology

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

66

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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

72

Basic Lie Theoretic

Results

(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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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

Some facts about Lie subgroups

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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Results

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 •

Some facts about Lie subgroups

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
Lie groups and Lie algebras from bilinear and quadratic forms

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.

•

Generalities on semisimple groups and algebras

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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121

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]. •

Generalities

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125

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. •

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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.

Real Jordan decomposition

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

130

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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. •

132

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Results

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

Basic Lie Theoretic

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.

136

Basic Lie Theoretic

Results

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}.

Generalities

on algebraic groups

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. •

138

4.14

Basic Lie Theoretic

Results

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. •

Generalities

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.

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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.

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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

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• 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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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

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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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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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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. •

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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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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].

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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.

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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

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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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More Lie Theory

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.

174

More Lie Theory

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). •

178

Minkowski Linear Algebra

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

182

Minkowski Linear Algebra

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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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

186

Minkowski Linear Algebra

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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Minkowski Linear Algebra

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 ^

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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.

204

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.

Matrix realizations of jets and calculus on jets

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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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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Results

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 7.3.1, p. 204, we have: Xm - 0 iff X%m - 0. The implications X%m = O ^ X

C 8

=0

and

XB = X% = • X% = 0

follow from the definitions.

•

The following is based on Lemma 5.2, p. 216 of [A00c]. L E M M A 7.4.13 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let mo £ M and let W G stob B (mo). Let B be an ordered Qd-basis of TmoM. Let S C C denote the set of characteristic roots of a,dW : g ->• g. Then: (1) If {exp[£(Wg m )]} t€ R is not precompact in SO°(Qd), then the homomorphism 11-> exp(tW) : R —> G is proper and injective. Now assume, additionally, that M is connected and that the G-action on M is locally faithful. Then all of the following are true: (2) For all A € E, if W%m 6 XH^, then &AW : g -» g is complex diagonalizable and S C {—A, A} U ^/^TE. (3) If WjQm e P ( d ) , then the real diagonalizable part of ad W : g -> g is zero and S C \/^T]]L ls (4) If {Ads(exp(tW))}teR not precompact in GL(g), then the homom morphism 11-> exp[t(WB )] : M. -> SO (Qd) is proper and injective. (5) IfW£m e AfW, then S = {0} and {adW)3g = {0}.

Miscellaneous results

213

Proof. Proof of (1): Let H := Stab G (m 0 ). Let 1 : H -»• GL(T m o M) be the isotropy representation. By Lemma 7.4.7, p. 209, (C11B)(W) = W\. We conclude, for all t £ R, that lB(exp(£W)) = exp(tWi). So, because {exp(Wi)}t e R is not precompact in SO°(Qd), we see that {exp(tW)}teR is not precompact in G. Then, by Lemma 4.9.10, p. 101, the homomorphism t H4 exp(tW) : R -> G is proper and injective. .End of proof of (1). We now assume that M is connected and that the G-action on M is locally faithful. Define Wi := W | r o . Define W* : E d x l -> R d x l by W,(u) = Wiw and define W* : E d x d -> E d x d by W*(M) = WXM - MWX. We define W : E d x l © E d x d -> E d x l © E d x d by W(Af,v) = (W(Af),W,(i;)). Let S* be the set of characteristic roots of W* : E d x l - i E d x l . Let 5* be the set of characteristic roots of W* : E d x d -4 E d x d . Let 5 be the set of characteristic roots of W : Rdxl © E d x d -> E d x l © R d x d . Then 5 = S.US*. Let r : g -> E d x l 8 E d x d be defined by r ( X ) = ( X g m , X £ m ) . By Lemma 7.4.11, p. 211, we see that T is injective. Moreover, for all X € g, we have r((ad W)A") = W(r(X)). It follows that SCS. Proof of (2): By assumption, Wx 6 \U(d\ so W. and W* are both diagonalizable over C, so W is diagonalizable over C Then, by injectivity of r, we see that &dW : Q —> g is diagonalizable over C, as well. Since Wi e XH^d\ we calculate that S . U 5 * C {-A, A} U sf^lR. So, since S C 5 = 5* U S*, we are done. £nd of proof of (2). Proof of (3): Since Wi € W^d\ the real diagonalizable part of the map v \-¥ W\v : Rdxl —» R d x l is zero. Consequently, the real diagonalizable parts of W* and of W* are both zero. Then, by injectivity of r, the real diagonalizable part of ad, W : g —> g is zero, as well. Since Wi € AP ( d ) , we calculate that 5* U 5* C - / ^ l E . So, since we have S C 5 = 5* U 5*, we are done. Enrf of proof of (3). Proof of (4): Assume, for a contradiction, that this is false. Then, by Lemma 4.9.10, p. 101, {exp(Wi)}t 6 R i s precompact in SO°(<3d). Then W* and W* are both elliptic, so W is elliptic. Then, by injectivity of T, we see that a.dW : g -4 g is elliptic, as well. Then {exp(t(ad g W))} t£ R is precompact in GL(g). For all t € E, exp(i(ad B (W))) = A-dg(exp(tW)). Then {Ad B (exp(tW))}t e R is precompact in GL(g), contradiction. End of proof of (4). Proof of (5): Since 0 is the only characteristic root of a nilpotent linear transformation, it suffices to show that (a,dW)3g = {0}. Fix Y € g and

214

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define Z := (ad W)3Y. Then

Results

We wish to show that Z = 0.

Z§m = (W*Y{Ygm) Z%m = (W*)3(Ygm)

= (Wi) 3 (Y^ m ) = (adWi)3(y^m).

and

We have Wx £ N(d). Since T £ M^d\ by (1), (2) and (3) of Lemma 6.3.15, p. 190, we see that (Wi) 3 (]R g is real diagonalizable and S C {-A, 0, A}. We will not need this fact. The converse of (3) of Lemma 7.4.13, p. 212 is untrue: We define W := E$ - E$. Let G := {exp(tW)\t € R} act on M := R 3 x l by matrix multiplication. Give M the flat Lorentz metric corresponding to the Minkowski quadratic form Q3. Let m :— 0 and let B be the standard basis of the vector space TmM = T 0 R 3 x l . Then Wmo = 0 and Wgm = E[f - E£>. Let / := E[f + E$ + E^f. For all t € B, we compute that exp[t(W§m)] = I + tW - (t2/2)E[f. It follows that the m homomorphism t i-> exp[£(Wg )] : M. —> SO°(Q3) is proper and injective. However, since G is Abelian, we see that {Adg(exp(tW))}teu is trivial, hence precompact in GL(fl). The converse of (4) of Lemma 7.4.13, p. 212 is also untrue: We define X := E$ - E£}. Let G := R act on M := R 4 x l by t.v = [exp(tX)]v. Let W := (d/dt)t=ot € ToR = g. Give M the flat Lorentz metric corresponding to the Minkowski quadratic form Q4. Let m := 0 and let B be the standard basis of TmM = T 0 R 4 x l . Then Wmo = 0 and W%m = E$ - E$. Since G = R, it follows, for all Y € £l\{0}, that t i-> exp(iF) : R -> G is proper and injective. In particular, t >-> exp(tW) : R —> G is proper and injective. However, since W^m is antisymmetric, we see that {exp[i(Wg m )]} t€ R is precompact in SO°(Qd). m

L E M M A 7.4.14 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let d := dim(M). Let mo € M. Let W € 5tab B (mo). Suppose &&W : g —> g is not elliptic. Then there is an ordered Qd-basis B ofTmoM such that W£m € R+ ri{d) UV{d).

Miscellaneous results

215

Proof. Let C be an ordered Q^-basis of TmoM. Since adg(W) is not elliptic, {Ad8(exp(£V7))}teR is not precompact in GL(g). Then, by (4) of Lemma 7.4.13, p. 212, v H-> W£mv : Rdxl -> K d x l is not elliptic. Then, by Lemma 6.3.13, p. 189, choose x G SO(Qd) such that we have (Ada;)(W c Lm ) G R+ «(<*) U V^. Let E := (e^ d ) ,... ,e{dd)) be the standard ordered basis of E d x l . Let B be the image of £ under the linear transformation v H4 fcix^v) : Rdxl - • TmoM. Then, for all v G K d x l , we have (f>B(xv) = c{v). Let / : M -t M be the identity transformation defined by f(m) = m. Then [(eZ/)mo]£ = z. Then, by (3) of Lemma 7.3.7, p. 205, we have x{W^m)x~x = Wgm. Then, W%m = x{W^m)x~1 = (Adx)(W£m), m

sow£ eR+nWuvw.

D

L E M M A 7.4.15 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo G M and let W G stab 0 (mo). Assume that a,dW : a —> g is nonzero and nilpotent. Let d := dim(M). Then there exists an ordered Qj-basis B ofTmoM such

thatW£m e p M .

Proof. A nonzero nilpotent linear transformation cannot be complex diagonalizable, so ad W : 0 —• n is not complex diagonalizable, and, in particular, is not elliptic. By Lemma 7.4.14, p. 214, choose an ordered Q^-basis B of TmoM such that W%m G 1+ ft(d) U V{d). As ad W : a ->• 0 is not complex diagonalizable, by (2) of Lemma 7.4.13, p. 212, we get W%m £ WH{d). Then W%m G (R+ USd) U V^)\{RU^) C V{d). D L E M M A 7.4.16 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo G M and let H := StabQ(mo). Let d :— dim(M). Let B be an ordered Qd-basis of TmoM. Let X £ t) and assume that X%m G V(d). Let T : g -» g be the nilpotent part of adX : g -» g. Then, for all P G 0, we have (T(P))%™ =N?\P§m) and (T(P))%™ = [Af^d),P^m]. Proof. Define a linear map S' : Rdxl -> l d x l by S'(v) = X%mv. Let T : Rdxl -> Rdxl be the nilpotent part of 5 ' . Since X%m G V^d), it follows, for all veRdxl, that T'[v) = N^d)v. Define S" : Rdxd -> Rdxd by m S"(V) = [X% ,V]. Let T " : Rdxd -> Rdxd be the nilpotent part S". Since d) XLm e 7? (d) ) i t f o l l o w S ) f o r a l l v G Rdxd^ t h a t T»(y) = [M^ ,V]. Let S := adX : g -> 0. Define /i : g -> K d x l by p(X) = X%m. Define

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v : g -*• Rdxd

Results

by v(X) = X%m. For all P e 0, we have

((adX)P)g T O = ( X | m ) P £ m

and

((adX)P)§m = [ X | m , P ^ m ] .

That is, fx o S = S' o /j, and 1/ o S = S" °v. So, by Lemma 2.4.1, p. 20, we have / i o T = T ' o / j a n d i / o T = r o ( / . Then, for all P <E 0, we have both CT(P))g m = MT(P)) = T'(»(P))

= T'{P$m)

=

N^d){P§m)

and (T(P))%™ = i/(T(P)) = T"(i/(P)) = T"(P%m) = [7V2(d), P £ m ] .

•

COROLLARY 7.4.17 LeiC? 6e a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. We define d := dim(M) and fix an ordered Qd-basis B of TmoM. Let mo € M. Define H := S t a b ^ m o ) . Let X G F) and assume that X§m £ V^. Assume that a d X : 0 —• 0 is nilpotent. Then, for all P £ g, we have both ((adX)P)° m =M^d){P§m) and ( ( a d X ) P ) § m = [7V2(d), P | m ] . Proof. Let T :— ad B (X). Since a d X : 0 —> g is nilpotent, it follows that T : 0 —)• 0 is the nilpotent part of a d X : 0 —• 0. The result then follows from Lemma 7.4.16, p. 215. D L E M M A 7.4.18 Let G be a connected Lie group acting isometrically on a Lorentz manifold M. Let mo £ M and let H := StaboOno). Define d :— dim(M). Let B be an ordered Qd-basis ofTmoM. Let X 6 0 and assume that X%m e AfW U V^d). Then there exists T' € tfW such that, for all P € 0, we have {T{P))%m = T'{P§m)

and

(T(P))%m = (ad T ' ) ( P B " " ) .

Proof. UX%m € V{d\ then, by Lemma 7.4.16, p. 215, setting V := W2(d), we are done. We therefore assume that X%m $ V^. Then X%m € N{d) and T = a d X : 0 -> 0. Let V := X%m. For all P e 0, [X, P]cBm = (X%m)P§m and the result follows.

and

[X, P]LBm = [Xkm, P£ r o ], •

L E M M A 7.4.19 Let G be a Lie group acting smoothly on a manifold M. Let Qi be a sequence in G, let X, Y € 0 and let m,m' £ M. Let ti be a sequence in E and assume that ti[(Adgi)X] —> Y. Assume that Xm = 0 and that gim -¥ m'. Then Ym< = 0.

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217

Proof. For all i, let Xt := (Ad gi)X and let m, := <7,m; then (Xi)m. — 0. We have rrn -)• m' and tjXj -> y . For all i, we have (i».2f»)mi = 0, so, letting i —> oo, we get Ymi = 0 . D The following is a rephrasing of Lemma 6.5, p. 248 of [AS97a]. L E M M A 7.4.20 Let G be a Lie group acting smoothly on a compact manifold M. Let mo G M, let W, X, Y € g and let k > 0 be an integer. Assume that Xmo = 0, that (a.dW)kX = Y and that (a,dW)Y = 0. Then there exists m' € M such that Ymi = 0. Proof. We may assume that Y ^ 0. If k = 0, then X = Y, and we are done with m' — mo. We therefore assume that k > 1. By compactness of M, choose a sequence Si in R such that Sj —>• +oo and such that (exp(siW))mo is convergent in M. For all i, let gi :— exp(siW). Let TO' := lim g,mo. For all integers j > 0, let Xj := (adW^) J X. By oo

Taylor's formula, for all i, (Adg^X

= y^(l/j\)s{Xj.

Since Xk = Y and

j=o

since 0 = Xk+i = Xk+2 = • • •, we conclude that (Adgi)X then follows from Lemma 7.4.19, p. 216.

—^ Y". The result •

The following is Lemma 6.7 of [AS97a]. L E M M A 7.4.21 Let N be a connected nilpotent Lie group acting smoothly on a compact manifold M. Suppose the action of Z°(N) on M is locally free. Then the action of N on M is locally free, as well. Proof. Let n o , n i , . . . be the descending central series of n. Assume, for a contradiction, that there are P e n\{0} and m 6 M such that Pm = 0. Let S denote the set of integers s > 0 such that, for some P € n s \{0}, for some m G M, we have Pm — 0. Let t := max S. Choose X G ti t \{0} and mo G M such that Xmo = 0. By local freeness of Z°(N) on M, we see that X £ 3(n). Choose W € n such that [W,X] ^ 0. By nilpotence of n, choose an integer fc > 1 such that (adW) k X £ 0 and such that (&dW)k+1X = 0. Let Y :— (a,dW)kX. Since k > 1, we see that Y G [n,n t ] = n t + i . By Lemma 7.4.20, p. 217, choose m' G M such that Ym> = 0. Then t + 1 G S, so t + 1 < max S = t, contradiction. • L E M M A 7.4.22 Let G be a Lie group acting smoothly on a compact manifold M. Let N be the nilradical of G°. Assume that g has no simple direct

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summand. Assume that the action of Z°(N) on M is locally free. then G-action on M is locally free, as well.

Then

Proof. Replacing G by G°, we may assume that G is connected. Fix X G 0\{O}, m 0 G M. Say, for a contradiction, that Xmo = 0. By Lemma 4.15.7, p. 144, we have c0(n) = j(n). By Lemma 7.4.21, p. 217, the TV-action on M is locally free. Then X £ n, so, since cB(n) C n, we conclude that X $. cB(n), so choose W € n such that [W, X] ^ 0. Since W G n and since n is an ideal of g, we conclude, for all integers j > 1, that (ad W)jX G n. As W G n, it follows that a,dW : g -> g is nilpotent. Choose an integer k > 1 such that Y := (ad W)*^" ^ 0 and such that (adW)Y = 0. Since k > 1, we get 7 6 n. By Lemma 7.4.20, p. 217, choose m' G M such that ym< = 0. Since the iV-action on M is locally free, and since Y G n\{0}, we conclude that Ymi ^ 0, contradiction. • Let a group G act on a Borel space X. Then we say that the action is by Borel automorphisms if, for all g € G, the map x H-> gx : X —> X is Borel. The next result was first observed by R. Zimmer and C. Moore. P R O P O S I T I O N 7.4.23 Let G be a connected noncompact simple Lie group acting by Borel automorphisms of a standard Borel space M preserving a finite Borel measure [i. Then, for fi-a.e. m £ M, either m is a G-fixpoint or Stab<3(m) is discrete. Proof. Let P denote the collection of all vector subspaces of g. Define / : M -> P by f(m) = stabg(m). For all g G G, for all m G M, we have f(gm) = (Adg)(f(m)). Let / / := f*(n). Since fj, is G-invariant, fi' is (Ad G)-invariant. Let Vo denote the collection of all ideals of g. Let G' := Ad B (G). Then VQ is the collection of all G'-invariant vector subspaces of g. Moreover, G' is a noncompact simple connected Lie subgroup of GL(fl). By Lemma 4.13.4, p. 137 (with U replaced by g, with G replaced by G' and with \x replaced by / / ) , /i' is supported on VoConsequently, for ju-a.e. m G M, stab 0 (m) is an ideal of g. Then, by simplicity of g, for fi-a.e. m G M, we see either that stabc(w) = {0} or that siaba(m) = g\ this implies either that Stabc(m) is discrete or that Stab G (m) = G. D Let an lcsc group G act on a measure space M and let Go C G. Let S C M be a subset of positive measure. We say that S is recurrent for

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219

the Go-action on M if, for a.e. s £ S, there exists a sequence gt in Go such that gi -> oo in G and such that, for all i, we have g^s £ S. The following result is called the Poincare Recurrence Lemma. L E M M A 7.4.24 Let G be a noncompact Icsc topological group. Let G act by measure-preserving transformations of a finite measure space M. Let GQ C G and assume that GQ is not precompact in G. Then any subset of positive measure in M is recurrent for the Go-action on M. Proof. Let 5 C M be a subset of positive measure. Let Ki, K2,... be a sequence of compact subsets of G such that Ki C K2 C • • • and such that G = |^J Kn. For all n, let Sn denote the set of all s £ S such that n

{
is null. Fix n and

n

assume, for a contradiction, that the set 5„ has measure e > 0. Recursively choose a sequence gi,g2,. • .'mG0 such that, for any integer * > 2, we have gt $ (giKn) U • • • U (gi-iKn). For all j , k, we have: if j > k, then g^gj ^ Kn, so ((g^g^S^HS = 0, so (sjS n ) n (gkS) = 0, so, as 5„ C 5 , we get ( ^ 5 „ ) n (fffc5„) = 0. Then { p A l i is an infinite collection of disjoint subsets of M, all with measure e > 0, contradicting the assumption that M has finite measure. • Let X and M be manifolds. Then a mapping F : X -» M is a local diffeomorphism if, for all a; G X , there exists a neighborhood X' of a; in X and a neighborhood M ' of f(x) in M such that F(X') = M ' and such that F\X' : X ' -> M ' is a diffeomorphism. L E M M A 7.4.25 Lei a Lie jro«p G aci on a connected manifold M. Assume that the G-action on M is smooth and locally free. Let T be a foliation on M and assume that T is G-invariant. Assume, for all m 6 M, that {TmT) +gm = TmM and that (T m J") n flm = {0}. Let Y be a leaf of T. Then (g,y) i-> gy : G x Y -> M is a covering map. Proof. Let C denote the set of leaves of T. For any L £ £, define • G x L -»• M by $ L ( 5 , Z ) = gl. Let $ := $ y . We wish to show that * : G x 7 - > M i s a covering map. For any L £ C, for any z £ G x L, (d$L)z : TZ(G x i ) - > T^L^M is an isomorphism of vector spaces. Then, by the Inverse Function Theorem, for all L £ C, the map $z, : G x L —> M is a local diffeomorphism, and, in $L

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particular, it is an open map; consequently, GL = $ i ( G x L) is an open subset of M. Then U := {GL \ L G £ } is an open cover of M . Moreover, as T is G-invariant, we see that U is a partition of M. So, since M is connected, we conclude, for all L E C, that GL = M, i.e., that $ j , : G x I -> M is surjective. In particular, \P : G x Y —> M is surjective. Fix mo G M. We wish to show that there is a neighborhood Mo of mo in M such that * | ( * _ 1 ( M 0 ) ) : \ t _ 1 ( M 0 ) ->• M 0 is a trivial covering map. Since f : G x 7 -> M is surjective, choose go G G and 5(0 £ ^ such that mo = *&(go,yo) = goVo- Let X := goY and let Z := G x X. Define / : Z -> G x y by / ( 5 , x ) - {ggo^^x). Let $ := $ x : Z -»• M. Then / is a diffeomorphism and $ o / = $. It therefore suffices to show that there is a neighborhood M 0 of m 0 in M such that $ | ( $ _ 1 ( M 0 ) ) : $ - 1 ( M 0 ) -»• M 0 is a trivial covering map. We have m 0 = goVo G goY — X. Let zo := (lG,m 0 ) G G x X = Z. Since $ = $ x = G x X - ^ M i s a local diffeomorphism and since $(zo) = "^o, choose a neighborhood Z, of ZQ in Z and a neighborhood M* of mo in M such that $(Z«) = M* and such that $|Z» : Z, -> M» is a diffeomorphism. We have (lG,mo) = ZQ G Z* C Z = G x X , so choose a neighborhood Gi of 1G in G and a neighborhood XQ of mo in X such that Gi x Xo C Z*. Then $|(Gi x X 0 ) : Gi x X 0 -» M is injective. Choose a neighborhood Go of 1G in G such that G0~1Go C Gi. Let Z 0 := Go x X0 and let Mo := $(Z 0 ). Then Z 0 is an open neighborhood of zo in Z, Mo is an open neighborhood of mo in M and $|Zo : Zo —• Mo is a diffeomorphism. We wish to show that $ | ( $ _ 1 ( M 0 ) ) : $ _ 1 ( M 0 ) -»• M 0 is a trivial covering map. Let T := $ _ 1 (mo). For all 7 G T, choose <77 G G and z 7 G X such that 7 = (<77,a;7). For all 7 € T, let G 7 := Go5 7 , let X 7 := p 7 " 1 X 0 and let Z 7 := G 7 x X 7 ; then Z 7 is a neighborhood of 7 in Z. For all 7 G T, let h-y : Z 7 -» Z 0 be defined by hy(g,x) = (gg^jg^x); t h e n $ | Z 7 — ($|Z 0 )o/i 7 . So, for all 7 G T, since both /i 7 : Z 7 —> Zo and $|Zo : Zo —> Mo are diffeomorphisms, we conclude that $|Z 7 : Z 7 ->• Mo is a diffeomorphism. It therefore suffices to show:

(1) *~1(Af0)= (J Z7;and 7er (2) for all 7, (5 G T, if ZynZs?

0, then 7 = <5.

Proo/ 0/ (1): For all 7 G Y, we have $(Z 7 ) = M 0 , so Z 7 C fc-^Mo). Fix z G $ _ 1 ( M 0 ) . We wish to show, for some 7 G T, that z G Z 7 .

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221

Choose g G G and x £ X such that z = {g,x). We have gx = $(g,x) = $(2) € Mo = *(Z 0 ) = $(G 0 x X 0 ), so choose h £ Go and 1/ 6 I 0 such that gx = $(h,y) — hy. We have x,y £ X, so, since g_1hy = a;, we conclude that (g~lhX) f l l ^ f l . Then, since J" is G-invariant, g~1hX = X. So, as m 0 € X, we have g~1hmo £ X . Let 7 := (h~1g,g~lhmo). Then 7 G T. We wish to show that 2 £ if7. Since (p 7 ,a; 7 ) = 7 = (h"1g,g~1hmo), we see that g 7 = /i _ 1 g. Then g = hg^ G Go3 7 = G 7

and

a; = s _1 /i«/ = ff7 X2/ € ff^Xo = X 7 ,

so 2 = (5, a;) G G 7 x X 7 = Z 7 . £rad of proof of (1). Proof of (2): Assume that z £ Z^H Zg. We wish to show that 7 = 8. Choose g G G and x 6 l such that 2 = (g, x). Then (3, x) = z G Z 7 n Zs = (G 7 x X 7 ) n (Ga x Xj). Then g G G 7 D Gg and a; G X 7 D X$- Let o := gg~x and 6 := gsj" 1 . As g G G 7 = Go<77, we get o G Go- As g E Gg = Gogs, we get b G Go. We have b"1a G G ^ G o C Gi. As a; G X 7 = g~ 1 Xo, we get p := gyx G X 0 . As x G X<5 = fi^Xo, we get g := g^x G X 0 . Then $(6 _ 1 a,P) = b~lap = ^ g ~ * p = gsx = q - $ ( l G ) g ) , so, as $|(Gi x X 0 ) is injective, we conclude that (& -1 a,p) = ( l e g ) , SO b _ 1 a = 1G. Then 5,55"1 = 6 _ 1 a = 1 G , so g1 = gg. We have (p 7 ,a; 7 ) = 7 G T = $ _ 1 ( m 0 ) and (5,5,xg) = 5 G T = $ _ 1 ( m o ) . Then #7a;7 = $(g 7 ,a: 7 ) = m 0 = $(gg,xg) = 5,51,5. So, as p 7 = 35, we get a;7 = xg Then 7 = (g 7 ,a; 7 ) = (gs, xg) = 5. End of proof of (2). • 7.5

A basic collection of rigidity results

The results in this section are all similar to Lemma 7.4.10, p. 211. They assert that if information is known on an open subset of a connected geometric manifold, then, in certain situations, that information propagates to the rest of the manifold. L E M M A 7.5.1 Let M be a connected manifold. Let V be a connection on M. Let f : M —> M and g : M —• M be diffeomorphisms which preserve V. Assume that f and g agree to order 1 at some point of M. Then f = g.

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Proof. Let C be the set of all m € M such that / and g agree to order 1 at m. Then C is a closed subset of M. By assumption, ( 7 ^ 0 . So, as M is connected, it suffices to show that C is an open subset of M. Fix m 0 € C. We will show that, for some open neighborhood MQ of mo in M, Mo C C. Choose an open neighborhood U of 0 in TmoM and an open neighborhood M 0 of m 0 in M such that exp mo is defined on U, such that exp mo (E/) = Mo and such that e := exp mo : U -> Mo is a diffeomorphism. We wish to show, for all m € M 0 , that / ( m ) = g(m). Let m' := f(m0). Since m 0 £ C, we have m! — g(mo). Let <> / := (4f) mo : T m o M -)• T m -M

and

t/> := (dg) mo : T m o M -> T ^ M .

Since m 0 £ C, we have <j> = ip. Let J7' := {U) and let M ' := f(M0). Then exp m , is defined on U' and exp m ,([/') = M ' and e' := exp m , : U' -» M ' is a diffeomorphism. By naturality of the exponential map, for all m £ Mo, we have / ( m ) = e'((/>(e_1(m))) and (e _1 (m))); so, since <> / = ip, we get / ( m ) = g(m). D L E M M A 7.5.2 on M. Let f : preserve V. Let H({m e M | / ( m )

Lei M be a connected manifold. Let V be a connection M —> M and g : M -> M be diffeomorphisms which fi be a measure on M in the smooth measure class. If = g(m)}) > 0, then f =g.

Proof. Let J : M -> M be the identity map, defined by I(m) = m. Let h := / _ 1 o g. Then /i({m £ M | /i(m) = m}) > 0. We wish to show that h = I. For all m £ M , let Um be a neighborhood of 0 on TmM and let Vm be a neighborhood of m in M such that exp m is defined on J7m, such that expm(Um) = Vm and such that em := exp m : Um —>• Vm is a diffeomorphism. Since M is Lindelof, let C be a countable subset of M such that [ J Vc = M. Choose c £ C such that n{{m £ Vc | /i(m) = m}) > 0. cec Let := (dh)c : TCM -> T C M. Let [/ := Uc n <^_1(^c)- Let V := e(t/). Let e := ec|f7 : [/ —> V. Then e is a diffeomorphism. Let u := e*(fi\V). Then v is in the smooth measure class of U. By naturality of the exponential map, for all u £ £/, e((«)) = h(e(u)). Then i/({u £ U\4>{u) = u}) > 0. Let S := {x £ T c M|^(a;) = x}. Then i/(5 fl U) > 0. So, since 5 is a vector subspace of TCM, since 17 is an open neighborhood of 0 in TCM and since v is in the smooth measure class on

A basic collection of rigidity

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223

U, we conclude that S — TCM. Then, for all x G TCM, we have (x) = x. Then, for all x 6 U, we have h(e(x)) = e((f>(x)) = e(x). Then, for all m G e([/), we have h(m) = m. So, since e(U) = V, we see that h\V = I\V. So, since c € V and since V is open in M, it follows that h and / agree to order 1 at c. Then, by Lemma 7.5.1, p. 221, we see that h — I. • L E M M A 7.5.3 Let W and X be two Killing vector fields on a connected pseudoRiemannian manifold (M,g). Let U be a nonempty open subset of M and assume, for all m 6 U that Wm and Xm are linearly dependent. Assume, for some mi € M, that Wmi ^ 0. Then, for some c G E, we have X = cW. Proof. By Lemma 7.4.10, p. 211, fix ui € U, such that WUl ^ 0. Choose c £ B such that XUl = cWUl. Replacing X by X — cW, we may assume that XUl = 0. We wish to show that X = 0 on M. Let d := dim(M). Let Q G Q F ( E d x l ) have the same signature as the tangent spaces of M . Let B be an ordered Q-basis of TUlM. Since XUl = 0, we get X§m = 0. Let Y be the linearization of X at « i . By Lemma 7.4.6, p. 209, we have X%m = Yg1. Then, by Lemma 7.4.8, p. 210, X%m has rank < 1. By Lemma 7.4.1, p. 206, Xjjm G so(Q). Then, by Lemma 4.7.1, p. 93, X%m = 0. Then, by Lemma 7.4.11, p. 211, X = 0 on M. U L E M M A 7.5.4 Let X and Y be Killing vector fields on a connected Lorentz manifold M. Then there is a dense open subset M\ of M such that, for all me At!, we have dim(EX m + WYm) = dim(RX + MY). Proof. Let d := dim(EX + MY). Then d G {0,1,2}. For all m G M , let dm := dim(KX m + MYm). For all m G M , we have dm < d. Let M0 :- {m e M \dm < d-1} and let Mi := {m G M \ dm = d}. Then Mi = M\MQ. Since m \-t dm : M —> E is lower semicontinuous, it follows that Mo is a closed subset of M , so Mi is an open subset of M. We wish to show that M x is dense in M . If d — 0, then X — 0 = Y, so Mi = M, and we are done. We therefore assume that d G {1,2}. Assume, for a contradiction, that Mi is not dense in M . Let U be a nonempty open subset of M such that U n Mi = 0. Then U C M 0 . If d = 1, then there exists a Killing vector field Z on M such that EX + WY = EZ, and, by Lemma 7.4.10, p. 211, we see, for some u G U, that Zu 7^ 0, so dim(EZ u ) = 1, so du — 1 = d, so u $ Mo, contradicting U C M 0 . Then d = 2.

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Then, for some m\ G M, we have Ymi ^ 0. For all u G U, we have u G Mo, so du < d — 1, i.e., dim(RX'u + KKU) < 1; then Xu and Yu are linearly dependent. Then, by Lemma 7.5.3, p. 223 (with W replaced by Y), we have X G WY, so d < 1, contradiction. •

7.6

Strongly lightlike and nontimelike vectors

Let G be a Lie group acting isometrically on a Lorentz manifold M. Let d := dim(M). For all integers j G [l,d], let ej :— e^ , so e\,... ,e<j is the standard basis of E d x l . Let m G M. Note that, for any X G 0, X m is lightlike iff there is an ordered Qd-basis B of T m M such that X§m G Mei. Moreover, for any X G 0, X m is not timelike iff there is an ordered Qd-basis B of T m M such that X%m G l e i H 1- Ked-i. These two observations partially motivate the following series of definitions. For any X G 0, for any ordered Qd-basis B of TmM: • We say that X is strongly lightlike with respect to B if both Xgm G l e i and X | m G .AAd); and • We say that X is strongly vanishing with respect to B if both Xgm = 0 and X%m G AA^); and • We say that X is strongly nontimelike with respect to B if both X%m G Mei + • • • + Ee d _i and X £ m G !W ( d ) + A/"(d). For any X G 0, we say X is strongly lightlike at m (resp. strongly vanishing at m, resp. strongly nontimelike at m) if there is an ordered <3,2-basis B of TmM such that X is strongly lightlike with respect to B (resp. strongly vanishing with respect to B, resp. strongly nontimelike with respect to B). For any S C g , for any ordered Qd-basis B of TmM, we say that S is strongly lightlike with respect to B (resp. strongly vanishing with respect to B, resp. strongly nontimelike with respect to B) if, for all X G S, we have that X is strongly lightlike with respect to B (resp. strongly vanishing with respect to B, resp. strongly nontimelike with respect to B). For any 5 C g, we say that S is strongly lightlike at m (resp. strongly vanishing at m, resp. strongly nontimelike at m) if there is an ordered <2d-basis B of TmM such that S is strongly lightlike with respect to B (resp. strongly vanishing with respect to B, resp. strongly nontimelike with respect to B).

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orbits

225

L E M M A 7.6.1 Let a Lie group G act locally faithfully and isometrically on a connected Lorentz manifold M. Let X € 0\{O} and let m € M. Assume that X is strongly vanishing at m. Then (adX) 3 g = {0} and 11-> exp(tX) : E —> G is proper and injective. Proof. Let d := dim(M). Let B be an ordered Qd-basis of TmM such that X is strongly vanishing with respect to B. We have X^m = 0 and = {0}. XLm € ^ ( d ) B y ( 5 ) o f L e m m a 7 4 1 3 j p - 212, we have {sAXfg m m Since X ^ 0 and X% = 0, by Lemma 7.4.11, p. 211, we X% ^ 0. Then {exp[t(X| m )]} t e K is not precompact in SO°(<2,i). Then X%m = 0, so we have X € stabs(m). By (1) of Lemma 7.4.13, p. 212, we see that the homomorphism t H-> exp(tX) : E —> G is proper and injective. • L E M M A 7.6.2 Let a Lie group G act locally faithfully and isometrically on a connected Lorentz manifold M. Let A,B € g. Let m € M and let B be an ordered Qd-basis of TmM. Assume that A and B are strongly lightlike with respect to B. Then [A, B] = 0. Proof. Let d := dim(M) For all integers i € [l,d], let ej := ef'\ then e i , , . . , ed is the standard basis of E d x l . Since A and B are both strongly lightlike with respect to B, we conclude that A%m,B%m e Eei and that m A^nBLm = 0. £ ^ ( d ) C h o o s e ( S ; i ) € R 2 \ { ( 0 ) 0 ) } s u c n that (sA + tB)% By, if necessary, interchanging A with B and s with t, we may assume that * 7^ 0. Let C :=sA + tB. Then C§m = 0 and Cjfm e A/"(d). We have [C,A]%m = C%m(A%m) € Af^ei = {0}. Moreover, we have [C,A]%m = [CJ}m,A*sm] 6 [M**>,tfW] = {0}. So, by Lemma 7.4.11, p. 211, we get [C,A] = 0. Then 0 = [A,C] - [A,sA + tB] = t[A,B]. So, since t ^ 0, we conclude that [A, B] = 0. •

7.7

Basic results on degenerate orbits

Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let d := dim(M). Let mo 6 M. Assume that gmo is a degenerate vector subspace of the Minkowski space TmoM. L E M M A 7.7.1 Let B be an ordered Qd-basis ofTmoM. If some element of g\{0} is strongly lightlike with respect to B, then the kernel of Qd \g%m is Me[d).

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Results

Proof. For all integers i G [l,d], let a := e\> G R d x l . Let X G fl\{0} be strongly lightlike with respect to B. Then X%m G l e i and X%m G AfW. Since gmo is degenerate, it follows that Qd \ g%m is degenerate. Then, since e\ is <3\{0}. Let 5 := l e i + • • • + l e d - i C l d x l . CZaim i : g g m C S. Proof: Fix Y G g. We wish to show that Y§m G 5 . By (1) and (2) of Lemma 6.3.15, p. 190, we have (X%m)2Ygm G lej. So, as (X%m)2Y$m = ((adX) 2 Y)g m G g%m and as ex 0 g g m , we conclude that (X%m)2Y$m = 0. Then, by (7) of Lemma 6.3.15, p. 190, we see that Y§m G S. End of proof of Claim 1. Since ei 0 flgm, g%m n (Kei) = {0}. Let L := {x E Rdxl | T m o M. Then, for all P G 0, we have / ( P £ m ) = /([P m o ] B ) = P m o . We let / ' := fo, : l d x l -> T m o M . Then, for all P G 0, we have f'(Pgm) = /'([PmolflO = J W Define 1 := ( J ' ) o / G 0 ( Q ) . Let i : M ->• M be the identity map. Then ff d

[Wmoll' = 9Let Q be the Minkowski quadratic form on TmoM. Let K C TmoM be the kernel of Q|g m . By Lemma 7.7.1, p. 225, we have / -1 (.fir) = Kei and ( / ' ) - 1 W = Kei. Then #(Rei) = Kei.

More on strongly lightlike and nontimelike vectors

227

Let B G SBF(R d x l ) be the polarization of Qd G QF(M d x l ). For all S C R d x l , let Sx := {v G l d x l |Va G S,B(v,s) = 0}. Calculation shows that JV is exactly the set of all H G so(Qd) such that all three of the following conditions hold: H(Rdxl)

C (Kei)- 1 ,

//((Mei)- 1 ) C Me1;

# ( B e i ) = {0}.

Ass(Rei) = E e 1 } it follows that ^(Eex)-1-) = (lei)- 1 . Then, for all H G J\f, we have g~lHg 6 N. That is, s _ W g C Af. Since y is strongly lightlike with respect to B', we conclude both that Y§m e Eei and that Yg,m G A/". By (3) of Lemma 7.3.7, p. 205 (with m and m' replaced by mo and with / : M -> M replaced by i : M —> M), we have both g{Y§m) = y ^ m and ^ ( y i ' " 1 ) ^ - 1 = Y£,m. Then ^ B C m = P _ 1 ( i ^ m ) G 5 _ 1 (Kei) = Kei and Y£m = < / _ 1 ( ^ m ) P € respect to B.

ff-1^

C N. Then y is strongly lightlike with O

L E M M A 7.7.3 Let S be the set of all X € g such that X is strongly lightlike at mo. Then all of the following are true: (1) S is a vector subspace of g; (2) S is strongly lightlike at mo; and (3) for all X,Y &S, we have [X,Y] = 0. Proof. We may assume that S ^ {0}. Fix W G ^ { 0 } . Choose an ordered Qd-basis B of TmoM such that W is strongly lightlike with respect to B. By Lemma 7.7.2, p. 226, S is the set of all X G 0 such that X is strongly lightlike with respect to B. That is, S = {X G 0 | X%m G Me[d) and X%m G AA d) }. Consequently, S is a vector subspace of 0, proving (1). Moreover, S is strongly lightlike with respect to B and, in particular, is strongly lightlike at mo, proving (2). Part (3) follows from Lemma 7.6.2, p. 225. • 7.8

M o r e on strongly lightlike a n d nontimelike vectors

L E M M A 7.8.1 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo G M. As-

228

Basic Dynamical

Results

sume that gmo is not Minkowski. Let H := StabG(mo). Let d := dim(M). Let B be an ordered Qd-basis ofTmoM. If\)%m n (Af^ \j-pW) ^ {0}, then g g m C Me[d) + • • • + Me^i • Proof. Let X £ f) and assume that X%m £ (A/"(d) U "P (d ))\{0}. Assume, moreover, that g^m £ Me[ ' + • • • + Med\. We aim for a contradiction. Let T : g -» g be the nilpotent part of a d X : 0 -» g. Let T" be as in Lemma 7.4.18, p. 216. Then, for all P £ g, we have (T(P))g m = T'(P§m)

and

( T ( P ) ) § m = (adT'XPj" 71 ).

Fix Y £ 0 such that Y§m $ Me[d) +••• + Medd\. Let Z := T 2 ( F ) . Then Z%m = ( T ' ) 2 ^ " 1 ) , so, by (7) of Lemma 6.3.15, p. 190, we see that Z%m ^ 0. Moreover, by (1) and (2) of Lemma 6.3.15, p. 190, we have Z§m £ B e ^ . Thus e[d) £ 0 g m % Me{d) +••• + ^ 4 - 1 • Then, by Lemma 6.2.16, p. 182, we conclude that 0 g m is a Minkowski vector subspace of (Rdxl, Qj), contradiction. • L E M M A 7.8.2 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo £ M and assume, for some X € 0\{O}, that X is strongly vanishing at mo. Then Stab G (mo) is noncompact. Proof. By Lemma 7.6.1, p. 225, t <-)• exp(tX) : R ->• G is proper, so {exp(tX)}t € R is not precompact in G. So, since {exp(tX)}teR C Stab G (m), we conclude that Stab G (m) is noncompact. • L E M M A 7.8.3 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let d :— dim(M). Let m £ M. Let B be an ordered Qd-basis of TmM. Let S be a vector subspace of g and assume that S is strongly lightlike with respect to B. Then: (1) There exists a vector subspace S' C S such that dim(5/5') < 1 and such that S' is strongly vanishing with respect to B. In particular, ifdim(S) > 2, then there exists X £ <S'\{0} such that X is strongly vanishing with respect to B. (2) If M is connected, if the G-action on M is locally faithful and if dim(S) > 2, then Stab G (m) is noncompact.

More on strongly lightlike and nontimelike

vectors

229

Proof. Let N := N^. Let J := { 1 , . . . , d}. For all t 6 I, let e< := ef } . F o r a l H . j G J, let Etj = E$?. Define $ : S -> E d x l by $(X) = X g m . Then, since S is strongly lightlike with respect to B, we have $(S) C Ke[ d) , so dim($(5)) < 1. Let 5 ' := ker($). Then dim(S/5') < 1. Proof of (1): Fix X E S'. We wish to show that X is strongly vanishing with respect to B. Since S is strongly lightlike with respect to B and since X £ S' C S, X is strongly lightlike with respect to B. Since X & S' — ker($), it follows that $ ( X ) = 0, i.e., that X%™ = 0. Then X is strongly vanishing with respect to B. End of proof of (1). Begin proof of (2): We have dim (5') > (dim (5)) — 1 > 1, so choose X € S'\{0}. Then X is strongly vanishing at m, so, by Lemma 7.8.2, p. 228, we are done. End of proof of (2). • L E M M A 7.8.4 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let d := dim(M). Let m E M. Let B be an ordered Qd-basis of TmM. Let H := Stab^(m). Let X € f) and Y € 0. Let T : g -> 0 be a linear transformation. Assume that at least one of the following is true: (a) X%m G J^(d) and T = a d X : 0 -> Q; or (b) X^m G V{d) and T is the nilpotent part of a d X : 0 -*• 0. Then all of the following are true: (1) IfY is strongly nontimelike with respect to B, then T(Y) is strongly lightlike with respect to B. (2) If Qm is not Minkowski, then T2(Y) is strongly vanishing with respect to B. (3) The vector T(Y) is strongly nontimelike with respect to B, and the vector T2(Y) is strongly lightlike with respect to B. Now assume, additionally, that M is connected and that the G-action on M is locally faithful. Then both of the following are true: (4) If X 7^ 0 and if Y is not strongly nontimelike with respect to B, then T2(Y) ^ 0 and T2(Y) is strongly lightlike with respect to B. (5) IfY is strongly lightlike with respect to B, then T(Y) = 0. Proof.

By Lemma 7.4.1, p. 206, we have Y$m G so(Qd).

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Results

Let T" be as in Lemma 7.4.18, p. 216. Then, for all P G g, we have {T{P))%m = T'(Pgm)

and

(T(P))%m =

(&dT')(P£m).

Proof of (1): Since Y is strongly nontimelike with respect to B, we have Y§m G ffie[d) + • • • + K e ^ i and Y$m G ! f t ( d ) + A/"(d). Then, by Claim A and by (2) and (5) of Lemma 6.3.15, p. 190, we have {T{Y))%m G Me[d) and {T{Y))^m G A/"(d). Thus T ( y ) is strongly lightlike with respect to B. End of proof of (1). Proof of (2): Replacing G by G°, we may assume that G is connected. Let M be the connected component of M containing m,. Then replacing M by Mo, we may assume that M is connected. Let K be the kernel of the action of G on M. Replacing G by G/K, we may assume that the G-action on M is faithful. If X = 0, T = 0, so T 2 ( F ) = 0, and we are done. We therefore assume that X ^ 0. We have h£ m D V^ ^ {0}, so, by Lemma 7.8.1, p. 227, we conclude that Y§m G Me[d) + • • • + B e ^ x . Then, by Claim A and by (2) and (3) of Lemma 6.3.15, p. 190, ( T 2 ( y ) ) g m = 0, while, by Claim A and by (4) and (5) of Lemma 6.3.15, p. 190, {T2{Y))^m G tfW. End of proof of (2). Proof of (3): By Claim A and by (1) and (4) of Lemma 6.3.15, p. 190, we have {T{Y))%m G Me[d) +••• + K e ^ i and (T(Y))%m G WH^ +AfW. Then T(Y) is strongly nontimelike with respect to B. By Claim A and by (1) and (2) of Lemma 6.3.15, p. 190, we have (T 2 (F))g TO G Me[d). By Claim A and by (4) and (5) of Lemma 6.3.15, p. 190, we have (T2(Y))%m G Af{d}. Thus T2(Y) is strongly lightlike with respect to B. End of proof of (3). Proof of (4): By (3), T2(Y) is strongly lightlike with respect to B. Since X # 0 and X%m = 0, by Lemma 7.4.11, p. 211, we X§m ^ 0. Then, as V G {N^d),X^m}, we see that T ^ 0. Since Y is not strongly nontimelike with respect to B, it follows either that P§m <£ Ee[d) +••• + H e ^ or that P%m £ WH^ +N{d). By Claim A and by (7) and (8) of Lemma 6.3.15, p. 190, either we have ( T 2 ( F ) ) g m ^ 0 or we have {T2{Y))^m ^ 0. In either case, T2{Y) ^ 0. End of proof of (4). Proof of (5): By Claim A and by (3) and (6) of Lemma 6.3.15, p. 190, we have ( r ( F ) ) g m = 0 and (T(Y))%m = 0. Then, by Lemma 7.4.11, p. 211, we have T(Y) = 0. End of proof of (5). • L E M M A 7.8.5 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let d :— dim(M). For all integers i G [l,d], let e* := e\ .

More on strongly lightlike and nontimelike vectors

231

Let m £ M. Let B be an ordered Qd-basis ofTmM. Let X, Y £ g. Assume X is strongly vanishing with respect to B. Then the following are all true: (1) If Qm is degenerate, then [X, Y]^m € l e i • (2) If M is connected, if the G-action on M is locally faithful, t / I ^ O andifY§m g Hex + • • • + Red-!, then ( ( a d X ) 2 r ) g m £ ( l e ^ - f O } . m (3) If Yg € Hei + • • • + l e d _ i , then [X, Y)%m £ Hei. Proof. Let T := X§m. Since X is strongly vanishing with respect to B, we have X%m - 0 and T € N(d). For all P £ g, we have [X,P]%m = T ( P £ m )

and

[A",P]gm = ( a d T ) ( P £ m ) .

Proof of (1): Replacing G by G°, we may assume that G is connected. Let M be the connected component of M containing m. Then, replacing M by M 0 , we may assume that M is connected. Let K be the kernel of the action of G on M. Replacing G by G/K, we may assume that the G-action on M is faithful. If X = 0, then [X, Y]%m = 0 € Kei, and we are done. We therefore assume that X ^ 0. By Lemma 7.8.1, p. 227, we have g g m C Eej + • • • + E e ^ ! . Then, by (2) of Lemma 6.3.15, p. 190, we are done. End of proof of (1). Proof of (2): By (1) and (2) of Lemma 6.3.15, p. 190, we have that ( ( a d X ) 2 r ) g m £ Bei. Since X ^ 0 and X%m = 0, by Lemma 7.4.11, p. 211, we X%m ^ 0. By (7) of Lemma 6.3.15, p. 190, ( ( a d X ) 2 F ) g m / 0. End of proof of (2). Proof of (3): This follows from (2) of Lemma 6.3.15, p. 190. End of proof of (3). • L E M M A 7.8.6 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let d :— dim(M). Let m £ M and let B be an ordered Qd-basis ofTmM. Let X, Y £ g. Then all of the following are true: (1) If X is strongly lightlike with respect to B and if Y is strongly nontimelike with respect to B, then [X, Y]gm £ Me[ . (2) If X and Y are both strongly nontimelike with respect to B, then [X,Y)%meRe[d) +--- + l e ^ 1 . (3) Assume that X is strongly lightlike with respect to B. Then we have

[X,F]gro€le^)+--- + Ee^1.

232

Proof.

Basic Dynamical

Results

Proof of (1): We have X%mEMe(d\

XimeA/-(d);

Y£m e WHS*> + Af{d),

Y§m £ Me[d) +••• +

So, since [X,Y]%m = {X%m)(Y§m)

- (Y£m)(X§m)

Redd\.

and since

{AfW) (Re[d) + • • • + K e ^ ) C Me[d) ( R H ^ +AfW)(me[d)) C !e< d ) ,

and

we are done. £ W of proof of (1). Proof of (2): We have X | m € RH^ + N(d), Y£m £ HW(d) + AA^,

x £ m 6 Re
So, since [X,Y]%m = {X%m){Y§m)

- {Y%m){X%m)

and since

{WH{d) + Mid))(Me[d) +••• + Me{dd\) C Me[d) +••• + K e j ^ , we are done. End of proof of (2). Proof of (3): By Lemma 7.4.1, p. 206, Y£m £ so(Qd). We have

x^meN^d\ m

Y£

x%meMe[d),

£ so(Qd),

So, since [X, Y\%m = (X%m)(Ygm) (tf(d))(Rdxlj

Q d)

Y§m £ Rdxl. - (Ygm){X§m)

Me[d) +••• + Medd\

and since and

(so(Q d ))(Re( ) C l e ^ + • • • + B e ^ ! , we are done. End of proof of (3).

•

L E M M A 7.8.7 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let d := dim(M). Let m £ M and let B be an ordered Qd-basis of TmM. Let X,Y £ Q. Assume that 1 ^ 0 , that Xmo = 0, that X%m £ H^ and that [X,Y] = Y. Then Y is strongly lightlike with respect to B. Proof. Since Xmo = 0, X%m = 0. Since X ^ 0 and X%m = 0, by Lemma 7.4.11, p. 211, X%m £ 0. We have (X%m)Ygm = [X, F ] g m = Y§m. So, as X%m £ HW\{0}, it follows that Y§m £ Me[d).

More on strongly lightlike and nontimelike

233

vectors

By Lemma 7.4.1, p. 206, we have Y$m € so(Qd). We have (a,dX%m)Y£m

= [X%m,Y£m]

= [X,Y]%m =

Y£m.

So, as X G ft(d)\{0} and Y£m G so(Qd), it follows that Y ^ m 6 Af^l

D

L E M M A 7.8.8 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let X, Y G g. Let d := dim(M). Let m G M. Let B be an ordered Qd-basis of TmM. Assume that I / O , that X is strongly vanishing with respect to B and that Y is not strongly nontimelike with respect to B. Let Z :— [X, [X, Y]]. Then Z ^ O and Z is strongly lightlike with respect to B. Moreover, either (1) Z is strongly vanishing with respect to B; or (2) both WYm +MZm and gm are Minkowski vector subspaces

ofTmM.

Proof. Since X is strongly vanishing with respect to B, it follows that m X§ = 0 and that X%m G AfW. We have Z%m = {X%m)2Y§m and Z%m = (adX%m)2Y£m. Let T := a d X : g -> g. Then (a) of Lemma 7.8.4, p. 229 holds. By (4) of Lemma 7.8.4, p. 229, we see that Z ^ 0 and that Z is strongly lightlike with respect to B. Case 1: Y§m G Me[d) +••• + K e ^ . Proof in Case 1: By (2) and (3) of Lemma 6.3.15, p. 190, we have Zgm — 0. Since Y is not strongly nontimelike with respect to B and since Y^m G Me\ ' + ••• + Med\, we conclude that Y£m g WH^ +Md). By Lemma 7.4.1, p. 206, we have y B Lm G so(Q d ). Then, by (4) and (5) of Lemma 6.3.15, p. 190, Z%m e Af^. Then Z is strongly vanishing with respect to B. End of Case 1. Case 2: Y§m £ Me[d) + • • • + B e j ^ . Proof in Case 2: By (1) and (2) of Lemma 6.3.15, p. 190, we have Z%m G Me[d). By (7) of Lemma 6.3.15, p. 190, we have Z§m ^ 0. Then e[d) G RZ%m. Then e[d) G WYgm + RZ%m £ Me[d) +••• + R e ^ , so, by Lemma 6.2.16, p. 182, WY£m+RZ$m is a Minkowski vector subspace of ( E d x l , Qd). Then WYm + KZ m is a Minkowski vector subspace of TmM. So, since WYm + RZm C g m C TmM, by Lemma 6.2.13, p. 181, gm is a Minkowski vector subspace of TmM. End of Case 2. • L E M M A 7.8.9 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let d := dim(M). Let m G M

234

Basic Dynamical Results

and let B be an ordered Qd-basis ofTmM. Let H := Stab G (m). Assume, for some X £ s\{0}, that X is strongly vanishing with respect to B. Let n be an ideal of g and assume that nm is not a Minkowski vector subspace of the Minkowski vector space TmM. Assume that fjnn = {0}. Then n is strongly nontimelike with respect to B. Proof. Let Y £ n and assume, for a contradiction, that Y is not strongly nontimelike with respect to B. Let Z := ( a d X ) 2 y . We have Z £ (adX) 2 n C n. By Lemma 7.8.8, p. 233, Z ^ 0 and Z is strongly lightlike with respect to B. As n (~l h = {0} and 0 ^ Z £ n, we conclude that Z £ h. Then (1) of Lemma 7.8.8, p. 233 fails, so (2) of Lemma 7.8.8, p. 233 is true. Then WYm + MZm is a Minkowski vector subspace of TmM. So, since RY^, + I&Zm C n m , by Lemma 6.2.13, p. 181, we see that n m is a Minkowski vector subspace of TmM, contradiction. • L E M M A 7.8.10 LetG be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let X,Y,Z £ g. Let m £ M and let B be an ordered Qd-basis ofTmM. Assume that X is strongly vanishing with respect to B. Assume that (&dX)Y = Z and that (adX)Z = 0. Then Y is strongly nontimelike with respect to B and Z is strongly lightlike with respect to B. Proof. If X = 0, then the result follows, so we assume that X ^ Let T := a d X : g ->• g. Since T2(Y) = 0, we conclude, from (4) Lemma 7.8.4, p. 229, that Y is strongly nontimelike with respect to Then, as Z = T(Y), by (1) of Lemma 7.8.4, p. 229, we see that Z strongly lightlike with respect to B.

0. of B. is D

L E M M A 7.8.11 Let G be a Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let V be a vector subspace of g and assume that dim(T^) > 2. Assume, for all X £ V\{0}, that there exist Y,Z £ g\{0} such that (adX)Y = Z,

(a.dX)Z = 0

and

(adZ)V^{0}.

Then, for all m £ M, V is not strongly lightlike at m. Proof. Let m £ M, let B be an ordered Qd-basis of TmM and assume, for a contradiction, that V is strongly lightlike with respect to B.

Nonproperness

and cocompact

subgroups

235

By (1) of Lemma 7.8.3, p. 228, choose X G V\{0} such that X is strongly vanishing with respect to B. By assumption, choose Y,Z £ g such that ( a d X ) F = Z, such that (adX)Z - 0 and such that (a.dZ)V ^ {0}. By Lemma 7.8.10, p. 234, Z is strongly lightlike with respect to B. Then, by Lemma 7.6.2, p. 225, (adZ)T^ = {0}, contradiction. •

7.9

Nonproperness and cocompact s u b g r o u p s

L E M M A 7.9.1 Let G be an Icsc topological group acting continuously on an Icsc topological space X. Let H be a subset of G and let K be a compact subset of G. Assume G = KHK. Assume that the G-action on X is nonproper. Then there exists a convergent sequence X{ in X and a sequence hi in H such that hi —> oo in G and such that hiXi is convergent in X. Proof. Choose a sequence gi in G and a precompact sequence x\ in X such that gi —> oo in G and such that gix\ is precompact in X. For all i, choose fcj,Zj € K and hi € H such that gi — kihilf, then hi = k^ g^J . Since gi —> oo in G and since ki and U are precompact in G, it follows that hi —> oo in G. For all i, let xt := lix\. Then Xi is a precompact sequence. For all i, we have hiXi — l~1(gix'i). So, since U is precompact in G and since gix\ is precompact in X, it follows that hiXi is precompact in X. Passing to a subsequence, Xi and h both convergent in X. • Let an Icsc group G act continuously on an Icsc topological space X. For all x € X, the map x >->• gx : G -> X will be called the orbit map based at x. We say that the action is orbit proper if for all x 6 X, the orbit map g i-> gx : G -t X is proper. An action is orbit nonproper if it is not orbit proper. Equivalently, there is a sequence gt in G and there is some x 6 X such that gi -> oo in G and such that gix converges in X. Any proper action is orbit proper, i.e., orbit nonproperness is stronger than nonproperness, but the converse fails (Example 7.11.7, p. 243). Moreover, any continuous action of a compact second countable topological group is proper and therefore orbit proper. L E M M A 7.9.2 Let G be an Icsc topological group acting continuously on an Icsc topological space X. Let H be a subset of G and let K be a compact subset of G. Assume that G = KB.. Assume that the G-action on X is orbit nonproper. Then there exist XQ G X and a sequence hi in H such that

236

Basic Dynamical Results

hi —)• oo in G and such that hiXo is convergent in G. Proof. Choose XQ € X and a sequence oo in G and such that gixo is convergent in X. Let y := lim giXQ. For all i, choose i—yco

ki £ K and hi £ H such that pj = fci/ij. Passing to a subsequence, choose k 6 K such that &, —* k. Then /ijZo — k^giXo —»• fc_1j/. Moreover, for all i, we have /ij = k~xgi, so, as A;* is convergent and as gi —> oo in G, we conclude that hi —> oo in G. • L E M M A 7.9.3 Let G be an Icsc topological group acting continuously on an Icsc topological space X. Let H be a cocompact closed subgroup of G. Assume that the G-action on X is nonproper (resp. orbit nonproper). Then the H-action on X is nonproper (resp. orbit nonproper). Proof. Choose a compact subset K C G such that G = KB.. Then G = KHK. Use Lemma 7.9.1, p. 235 for nonproper and Lemma 7.9.2, p. 235 for orbit nonproper. • 7.10

Kowalsky subsets

L E M M A 7.10.1 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let m' € M. Let gi be a sequence in G and let rrii be a precompact sequence in M. Assume that giTUi —> m'. Let S C g. Assume that S is Kowalsky for AdB (<;;). Then S is strongly lightlike at m'. Proof. Let d := dim(M). Let A+ := {exp(W^)\t > 0} C SO 0 (Q d ). By Lemma 4.10.38, p. 127, fix a compact subgroup K\ of SO°(<5d) such that SO 0 (Q d ) = KiA+Kx. Let F be a finite subset of 0(Qd) such that F • SO°(Qd) - 0(Qd)- Let K := FKt. Then K is compact and we have 0(Qd) = KA+K. For all i, let Ci be an ordered Qd-basis of TmiM and assume that {C;}; is precompact in the frame bundle of M. For all i, let mj := M be denned by fi(m) = gim and we define hi := (dfi)mi

• TmiM

-> Tm>.M. Then, for all i, we have {hife\ e

0(Qd);

pi

choose ki, k\ € K and a, G A+ such that [/ij]c! = k[a,iki. For all i, let Bi be the image under v i-t <j>Ci{k~lv) : E d x l -> TmiM of the standard ordered

Kowalsky

237

subsets

basis of E d x l . Then, for all i, for all v £ E d x l , we have Ci {k~lv) = <j)Bi (v). For all i, let B\ be the image under v K> <j>c(k'iV) : M d x l -> T m ; M of the standard ordered basis of E d x l . Then, for all i, for all v £ R d x l , we have &{(*{«) = &»•(«). So, for all t, [^]g = (*{) -1 • N § • Kl = ai e ^ + Moreover, for all i, Z3j is an ordered <2d-basis of TmiM and #< is an ordered Qd-basis of Tm'.M. Since {Ci}i and {C,-}, are both precompact in the frame bundle of M, and since K is compact, it follows that {Bi}i and {B[}i are both precompact in the frame bundle of M. Passing to a subsequence, let B' be an ordered Q^-basis of Tm<M such that B\-) B'. We wish to show that S is strongly lightlike with respect to B'. For all i, for all X £ 0, we have ( / J ) * ( X M ) = ((Adgi)X)M', so, as o-i = [hi]jg. = [(d/j) mi ] B i, by (3) of Lemma 7.3.7, p. 205, we have
= ((Ad 5 i )X)g/"

and

^(X^a"

1

= ((Ad5i)*)£f-

So, since 5 is Kowalsky for Adg(<7;), it follows that (1) S%,m is Kowalsky for D 4 d j D : E d x l -4 R d x l ; and (2) S£,m is Kowalsky for V H> OjVo,-1 : Wxd -»• E d x d . For all i, we have a* € A+. So, by (1) and by Lemma 2.5.1, p. 22, we see that every element of Sg,m is concentrated on {1}. That is, we have S g m C Ee[d). It remains to show that S£,ro C Af(d\ Let J := { 1 , . . . ,d}. Let II := ({1} x J) U ( J x {<*}). For all i, we have Oj 6 ^4 + . So, by (2) and by Lemma 2.5.2, p. 22, we see that every element of S|, m is concentrated on II. By Lemma 7.4.1, p. 206, 5 | , m C so(Qd). So, since Af^ is exactly the collection of elements M £ so(Qd) such that M is concentrated on II, we get S£,m C M{d). D Since strongly lightlike implies lightlike, Lemma 7.10.1, p. 236 implies Lemma 7.1.1, p. 197. We view Lemma 7.10.1, p. 236 as a 1-jet generalization of Lemma 7.1.1, p. 197. One may also form higher jet generalizations; see Lemma 8.1 of [A99b]. L E M M A 7.10.2 Let G be a connected Lie group and let a be a real split torus of g. Let H £ a\{0}, let Hi be a sequence in a and let U be a sequence in (0,oo). Assume that ti -> 0 in R and that tiHi —> H in a. Let $ be the set of roots of a on g. For all (f> € $ , let g-rootspace of a on g. Let $+ := {<j> £ $\{H) > 0}. Let W := 0 g^. Then W is Kowalsky for Ad e (exp(i?i)).

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Proof. For all i, let ai := exp(Hi). Let $ G $+ and let X G g^. We wish to show that X is Kowalsky for Ad B (a;). For all i, let Si := e^Hi\ For all i, we have (Adoj)X = SiX, so (Adaj)(X/si) = X. We have U • <j>(Hi) = 4>{UHi) -> <j)(H) > 0, so since ti is a positive sequence and ij —> 0, we see that (j>(Hi) —> +oo. Then Sj —> +oo, so X/si —• 0. So, as (Ado,)(X/sj) —> X, we are done. •

7.11

Types of chaotic actions

By Lemma 7.11.12, p. 246 below, an action of an lcsc group G on an lcsc topological space X is orbit proper iff both every G-orbit is closed in X and every point of X has compact stabilizer in G. Consequently, if the action of G on X has an orbit that is not closed in X then it is orbit nonproper. Moreover, if the stabilizer of any point of X is noncompact in G, then the action is orbit nonproper. Moreover, any action of a noncompact lcsc group on a compact topological space is orbit nonproper. For these reasons, we think of orbit nonproperness as a very weak dynamical condition. In this section, we consider some other (more stringent) dynamical conditions that indicate complicated orbit structure. We also examine how these various dynamical conditions interact. Let a group G act on a measure space M. Let G act on L2(M) by (gf)(m) = / ( g _ 1 m ) . Let 1 G L 2 (M) be the constant function denned by l(m) = 1. Let (•, •) be the Hilbert inner product in L2(M). Define 2 V := {/ G L (M) | (/, 1) = 0}. Then we say that the G-action on M is mixing if, for all / , / ' G V, we have (gf, / ' ) -» 0, as g —> oo in G. If the G-action on M is mixing and if H is a noncompact closed subgroup of G, then the if-action on M is mixing as well. Moreover, if the G-action on M is mixing, then the G-action on M is ergodic. For an introduction to mixing actions, see, e.g., §6 of [AOl]. A Borel space Y is said to be countably separated if there exists a countable collection X of Borel subsets of Y which separates points, i.e., which satisfies the property: For all y,y' £ Y, if y ^ y', then there exists B € X such that y e B and y' ^ B. Let G be a group acting on a Borel space X. Let G\X denote the set of G-orbits in X and let ir : X —> G\X be the quotient map. Then G\X is a Borel space under the rule that a subset B C G\X is Borel in G\X iff n'^B) is Borel in X. We say that the G-action on X is tame if G\X is countably separated. Otherwise, we

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say that the action is nontame. A Borel space X, with cr-algebra B, is said to be standard if there is a Polish topology r on X such that r generates B i.e., such that B is the smallest cr-algebra containing r . By the Baire category theorem, any Polish space is Baire, and it follows that any countable Polish space is discrete. Consequently, a countable standard Borel space has discrete cr-algebra, meaning that every subset is in the cr-algebra. Moreover, an uncountable standard Borel space is isomorphic to M with the a-algebra generated by its usual topology. (See Remark (i) on p. 451 of [Ku66] or Theorem 15.6, p. 90 of [Kec95].) Any standard Borel space is countably separated. Let a group G act on a topological space X. We say that the G-action on X is tame if it is tame with respect to the cr-algebra generated by the topology of X. Otherwise, we say that the action is nontame. Assuming G is lcsc and that the G-action on X is continuous, by Lemma 7.11.6, p. 242 below, the G-action on X is nontame iff there is some x € X such that the G-orbit of x "nontrivially limits on itself" in the following sense: There exists a sequence x and such that the image of gi in G/(Stab%(x)) leaves compact subsets of G/(Stabg(a;)). Let a group G act on a set S and let s G S. Let H := Stabo(s) and let it : G —• G/H be the canonical map. Then the factored orbit map based at s is the injective map n(g) H> gs : G/H —> S. Its image is the orbit Gs of s. Let G be a Polish topological group acting on a standard Borel space B. Assume that the action map (g,b) i-»- gb : G x B -> B is Borel, i.e., that the action is Borel. According to Theorem 1.1, p. 335 of [BK93], there is a Polish topology on B which generates the Borel structure on B and with respect to which the G-action on B is continuous. It follows, for all b £ B, that Stabc(b) is closed in G. Moreover, with this topology on B, for all b € B, if we set H := Stabc(&), then the orbit Gb is the image of the Polish space G/H under the factored orbit map gH t-> gb : G/H -> B based at b. Consequently, we see, from Theorem 15.1, p. 89 of [Kec95], that every G-orbit in B is Borel. (Compare these statements with [Zi84, Theorem 2.1.19, p. 16] and [Zi84, Corollary 2.1.20, p. 16], in which G is assumed to be lcsc.) If G is a Lie group, then, for all b e B, Stabc(b), being closed, is a Lie subgroup, and we let stobc(&) denote the Lie subalgebra of g corresponding to Stabc(b). The next lemma implies that ergodicity (with respect to a smooth measure class) implies topological transitivity. Thus, by Corollary 7.11.14,

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p. 247, if a smooth action is properly ergodic (with respect to the smooth measure class), it follows that it is nontame and orbit nonproper, giving further evidence of the weakness of orbit nonproperness, as a dynamical condition. L E M M A 7.11.1 Let M be a second countable topological space let G be a group acting on M. Assume, for allfirG G, that m H-> gm : M -¥ M is continuous. Let \i be a measure on M such that every nonempty open set has positive fi-measure. Assume that the G-action on (M, //) is qmp and ergodic. Then, for a.e. m G M, we have that Gm is dense in M. Proof. Let B be a countable base for M such that 0 ^ B. It suffices to show, for a.e. m £ M, for all U G B, that (Gm) f~l U ^ 0. Since B is countable, it suffices to show, for all U G B, for a.e. m 6 M, that (Gm) C\U # 0. Fix U G B. For all m G M, we have: (Gm) n U ^ 0 iff m G Gt/. It therefore suffices to show that GU is conull in M. By ergodicity, since GU is G-invariant, it follows that GU is either null or conull in M. Since U is a nonempty open set, it follows that U has positive measure, so GU is not null in M. D L E M M A 7.11.2 Let A be a dense subset of a Hausdorff topological space So and give A the relative topology inherited from SQ. Assume that A is locally compact. Then A is open in SQ. Proof. Fix a G A. We wish to prove that there is an open neighborhood U of a in So such that U C A. For any A C So, let A denote the closure in So of A. Since A is locally compact, choose an open neighborhood U of a in 5o and choose a compact subset K of A such that ADU C K. As A is dense in So, it follows that U C AJYU, SOU CK = KCA. • L E M M A 7.11.3 Let A be a subset of a Hausdorff topological space S and give A the relative topology inherited from S. Assume that A is locally compact. Then A is locally closed in S. Proof. Let S 0 be the closure in S of A. Give S 0 the relative topology inherited from S. Then A is dense in So- Then, by Lemma 7.11.2, p. 240, we see that A is open in So- Choose an open subset U of S such that A — U fl So- Since U is open in S and since So is closed in S, we see that A is locally closed in S. •

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There is a converse for Lemma 7.11.3, p. 240: Any locally closed subspace of a locally compact topological space is again locally compact. So the terms "locally closed" and "locally compact" are interchangeable, when applied to subsets of a locally compact topological space. By Theorem 3.11, p. 17 of [Kec95], a subspace of a Polish space is Polish iff it is a, Gs. Any closed subset of a Polish space is a G j ; consequently, any locally closed subset of a Polish space is a Gs- Then any locally closed subset of a Polish space is again Polish, in its relative topology. L E M M A 7.11.4 Let G be an Icsc topological group acting continuously on a Polish space S. Let s £ S. Give the orbit Gs of s the relative topology inherited from S, and give G/(Stabc(s)) the quotient topology. Then Gs is locally closed in S iff the factored orbit map map G/(Stabc(s)) —>• Gs based at s is a homeomorphism. Proof. We let -K : G —t G/(Stabo(s)) be the canonical map and we let p : G/(Stabc(s)) —• Gs be the factored orbit map based at s. If p is a homeomorphism, then, since G/(Stabc(s)) is locally compact, we see that Gs is locally compact, so, by Lemma 7.11.3, p. 240, Gs is locally closed in S, proving "if". Assume that Gs is locally closed in S. Then Gs is Polish and therefore Baire. The map p is a continuous bijection, so, since G acts transitively on Gs, Lemma 2.7.1, p. 26 shows that p is a homeomorphism. • The following result is due to J. Glimm [G6l] and E. G. Effros [E65]. L E M M A 7.11.5 Let G be an Icsc topological group acting continuously on a Polish space S. For all s £ S, give the orbit Gs of s the relative topology inherited from S and give G/(Stabc(s)) the quotient topology. Then the following are equivalent. (1) For all s € S, Gs is locally closed in S. (2) The G-action on S is tame. (3) For all s € S, the factored orbit map G/(Stabc(s)) —> Gs based at s is a homeomorphism. Proof. This is a slight rephrasing of Theorem 2.1.14, p. 12 of [Zi84], taking into account that, in [Zi84], the term "smooth" is used for "tame". Note that (1) <^=>- (3) follows from Lemma 7.11.4, p. 241. •

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In Lemma 7.11.5, p. 241, if we only assume that G is Polish (rather than the stronger lcsc), then we have (1) = > (3) ==> (2), but both (2) =>• (3) and (3) => (1) are false. Let an lcsc topological group G act on a Polish space 5. Let so € 5. Assume that Go := Stabc(so) is noncompact. Choose a sequence g\ in Go such that gl -> oo in G. If e, is any sequence in G, if e, -> 1G and if we define a sequence gt in G by gi = e ^ , then we will have gtSo = €ig[so — CiSo —>• SoThus the existence of a noncompact stabilizer allows us to prove that the action is orbit nonproper using a sequence g, which, while not necessarily contained in a noncompact stabilizer, is nevertheless "almost" contained in a noncompact stabilizer in the sense that its image in G/(Stabo(s)) is precompact. According to the next lemma, nontameness is equivalent to being able to set up gtso —• so without being forced to use a sequence gi that is almost contained in the stabilizer of so. L E M M A 7.11.6 Let G be an lcsc topological group acting continuously on a Polish space S. For all s G S, let ps : G —• G/(Stabc(s)) be the canonical map. Then the G-action on S is nontame iff there exist s € S and a sequence gi in G such that giS —• s and such that ps{gi) -> oo in G/(Stab G (s)). Proof. Proof of "only if": By (3) =>• (2) of Lemma 7.11.5, p. 241, choose s € S such that the factored orbit map p : G/(Stabc(s)) —> Gs is not a homeomorphism. Then, since p : G/(Stabc(s)) —>• Gs is a continuous bijection, choose a sequence hi in G and h' € G such that ps(hi) -ft ps(h') and p(ps{hi)) ->• p(ps(h')). For all g e G, we have p{ps(g)) = gs, so his = p(ps(hi)) ->• p(ps(h')) = h's. For all i, let gi := (h')~lhi. Then

P.(9i) = (hT'iPsihi))

A (hT'iPsih1))

=p.(lG)

and p;s = (h')~1hiS -> (ft') - 1 /i's = s. We wish to show that ps(<7i) ->• oo in G/(Stabc(s)). Assume, for a contradiction, that, after passing to a subsequence, ps{gi) is convergent in G/(Stabc(s)). Since ps is a surjective open map, choose a convergent sequence Xi in G such that, for all i, we have p3(xi) = p3{gi)- For all i, choose yt e Stabo(s) such that gi = Xiyi. Let x := lim Xi. Then 5iS = XiyiS

= XiS - > i s .

So, as <7;s —> s, we conclude that xs = s, i.e., that x 6 Stabc(s). Then

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we get ps(gi) = ps(xiVi) = ps{xi) -» p3{x) = PS(1G), contradiction. End of proof of "only if". Proof of "if": Let s G S and let gi be a sequence in G. Assume that giS —> s and that ps(gi) —> oo in G/(Stabo(s)). Then the factored orbit map ps(g) i-> gs : G/(Stabc(s)) -> Gs based at s is not a homeomorphism, so, by (2) =$• (3) of Lemma 7.11.5, p. 241, the G-action on S is nontame. End of proof of "if". • Let W be a vector space. Then (v,w) H- (d/dt)t=0(v

+

tw):WxW-*TW

is a bijection, and its inverse TW —tWxW will be called the standard trivialization of TW over W. Recall that nonproper is weaker than orbit nonproper. That is, any orbit nonproper action is nonproper. The converse, however, is untrue: E X A M P L E 7.11.7 There is a smooth R-action on a manifold which is nonproper, but orbit proper. Proof. Let M := E 2 \{(0,0)}. Let * : TE 2 - > l 2 x l 2 be the standard trivialization of TE 2 over E 2 . Let $ := * | T M : TM -4- M x E 2 . Let p : M x E 2 —>• E 2 denote projection onto the last two coordinates and let (j> := p o $ : TM ->• E 2 .

Let / : M -»• (0,oo) be a smooth function such that f{x,y) -> +oo, as (x, i/) -> (0,0). Let M :- {(x, y, f(x, y)) | ( i , j ) e M } C M x l b e the graph of / . Let -K : M —> M be the projection map Tr(x,y,z) = (x,y). Then TT : M —> M is a diffeomorphism. Let g be the Riemannian metric on M C E 3 inherited from a fiat Riemannian metric on E 3 . Let g :— 7r»((Vm) £ (0, oo) x {0}. Then V is the "right-pointing" unit vector field on (M, g). The vector field V on M is the image under IT of a vector field V on M. Since V is a complete vector field on M, we conclude that V is a complete vector field on M. For all t € E, for all m G M, we let tm denote the image of m under the time t flow of V. This defines a smooth, orbit proper E-action on M. For all y € ( - 2 , 2 ) \ { 0 } , the set E x {y} is an orbit of this E-action. Moreover, (-oo,0) x {0} and (0,+oo) x {0} are both orbits, as well. We

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wish to show that the E-action on M is nonproper. For all integers i > 1, let m ; := (—1,1/i), let mj := (1,1/i) and choose U € K such that ^ m , = mj. Let moo := ( _ 1 , 0 ) and let m'^ := (1,0). We wish to show that U -» oo in E. Suppose, for a contradiction, otherwise. After a subsequence, choose too € E such that U -> too. Then we have tooTUoo = m'^. Then moo and mj^ are in the same orbit. However, moo € (—oo,0) x {0}, while m'^ 6 (0,+oo) x {0}, so moo and m ^ are in different orbits, a contradiction. • Example 7.11.7, p. 243 gives a method for producing many examples of smooth, nonproper, orbit proper actions. A specific and well-known example is as follows: Let E act on M := E 2 \{(0,0)} by t.(x,y) = (2tx,2~ty). For any m €E M we have that t.m —» oo in M, as |i| —> oo so the action is orbit proper. On the other hand, we have (2~™, 1) —• (0,1) in M , as n -» +00, but n . ( 2 _ n , l ) = (l,2~ n ) -> (1,0) in M, as n ->• +oo, showing that the action is nonproper. L E M M A 7.11.8 Let a Lie group G act continuously on an Icsc topological space X. Let g, be a sequence in G and let xo, x' £ X. Assume that gi -> oo in G and that giXo —> x' in X. Assume that StabG(x') = {1G}- Then there exists an open G-invariant open neighborhood XQ of x1 in X such that the G-action on Xo is locally free and orbit nonproper. Proof. Let X0 := {x € X | stabB(a;) = {0}}. Claim 1: Xo is open in X. Proof of Claim 1: Let yi be a convergent sequence in X and let y := lim yi. Assume, for all i, that j/j ^ X0. We i—>oo

wish to show that y £ XQ. For all i, siabg(yi) ^ {0}; choose Wi € (stob g (i/i))\{0}. Passing to a subsequence, choose a sequence Sj in E and choose W 6 0\{O} such that SiWi -> W. For all i, we have {siWi)yi = 0. Taking the limit as i -> oo, we have Wy = 0. Then O^W e siabe{y), so siabs{y) ^ {0}, so y £ X0. End of proof of Claim 1. Since Stabg(a:') = {1G}> we see that x' € Xo- By Claim 1, Xo is open in X . By definition, Xo is G-invariant. Then Xo is an G-invariant open neighborhood of x' in X . Moreover, by definition of X 0 , the G-action on Xo is locally free. Since giXo -+ x' and since Xo is an open neighborhood of x' in X , choose io such that gi0xo £ XQ. By G-invariance of XQ, we conclude that

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xo € XQ and, a fortiori, that, for all i, we have g^xo € Xo- Then the G-action on Xo is orbit nonproper. • For any metric space M, let Isom(M) denote the set of isometries of M, i.e., the set of distance preserving homeomorphisms of M. Give Isom(M) the compact-open topology, i.e., the topology of uniform convergence on compact sets. If M is locally compact (resp. compact), then, by Ascoli's Theorem (see, e.g., Theorem 7.17, p. 233 of [Kel55]), Isom(M) is locally compact (resp. compact). A metric space (M,d) is proper if, for every m € M, the function d(m, •) : M —»• R is proper. In any metric space, compact sets are closed and bounded. A metric space M is proper iff every closed, bounded set in M is compact. A proper metric space is locally compact. On any manifold M, a Riemannian metric g determines a distance function d: MxM —¥ [0, oo) such that, for all m, m' € M, the distance d(m, m1) is the infimum of the Riemannian lengths of smooth paths from m to m!. Then Isom(M, g) = Isom(M, d). Moreover, by the Hopf-Rinow Theorem, (M, g) is complete iff (M, d) is proper. L E M M A 7.11.9 Let H be the diffeomorphism group of a compact manifold M. Let T\ be the topology of C°° convergence on H. Let g be a Riemannian metric on M. Let G := Isom(M,g). Then ( G , n | G ) is compact. Proof. Let TO be the compact-open topology on H. As described above, let d be a distance function on M such that G = Isom(M, d). Then, by Ascoli's Theorem (see, e.g., Theorem 7.17, p. 233 of [Kel55]), ( G , T 0 | G ) is compact. By Lemma 5.2.1, p. 154, we have TQ\G = TI\G. • L E M M A 7.11.10 LetM be a proper, second countable metric space. Then the action of Isom(M) on M is proper. Proof. Let fc be a sequence in Isom(M) and let m, be a sequence in M. Assume that both {m,} and {/i(mj)}j are precompact in M. We wish to show that fi has a convergent subsequence in Isom(M). Let d denote the metric on M. For all m £ M, since {d(m,m,i)}i is bounded, it follows that {d(fi(m), fi(mi))}i is bounded, which, by precompactness in M of {/i(mj)} and by properness of (M,d), implies that {fi(m)}i is precompact in M .

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Let F denote the closure in Isom(M) of {fi}i C Isom(M). Since we have F C Isom(M), it follows that F is equicontinuous. So, by Ascoli's theorem (see, e.g., Theorem 7.17, p. 233 of [Kel55]), we see that F is compact. Passing to a subsequence, we may assume that ft converges uniformly on compact sets. Let /oo := lim fi. i—>oo

By a similar argument, we may, after passing again to a subsequence, assume that f^1 converges uniformly on compact sets. Let Poo := lim/j" . For all i, since fi and f^1 are inverses, it follows that /*> and g^ are inverses. Then /oo G Isom(M) and fi -»• /oo in Isom(M). • E X A M P L E 7.11.11 There is an Icsc metric space X such that the action of Isom(X) on X is orbit nonproper. Proof. For all x,y G E, let d0(x,y) — min{|x - y|, 2}. Let X be a set such that E C I and X\R has exactly one element. Let x0 G X\R, so that X = t U {XQ}. Define a metric d on X by

(

d0(x,y), 0,

1, Let G := Isom(X,d). For any function ,+ , . _ f f(x), \x0,

ifz,j/GE; if x — y = XQ] and otherwise. / : E -> E, define / + : X - • X by if a; G E; and iix = x0.

For all t £ R, define ft : E -> M by ft(x) = x + t and define gt : E ->• E by gt{x) = -x + t. Then G = {/t+ 11 G E}U {g? | i G E} and G is noncompact. So, as Stabc(zo) = G, we see that the action of G on X has a noncompact stabilizer, and is therefore orbit nonproper. • L E M M A 7.11.12 Let G be an Icsc topological group acting on an Icsc topological space X. Then the G-action on X is orbit proper iff for all x G X, we have both that Gx is closed in X and that Stab(j(:r) is compact. Proof. For all x € X, let fx : G -¥ X be the orbit map based at x, defined by fx(g) = gx; then the image fx(G) of fx is the orbit Gx. For all x G X, the map fx:G—*X has compact fibers iff Stabo(a;) is compact. The G-action on X is orbit proper iff, for all x G X, the map fx:G-}X is proper. A continuous map between Icsc topological spaces is proper iff it has closed image and compact fibers. We conclude that the G-action on

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X is orbit proper iff, for all x € X, both Stabc(a;) is compact and Gx is closed in X. • L E M M A 7.11.13 Let an Icsc topological group G act continuously on an Icsc topological space X. If the G-action on X is topologically transitive and nontransitive, then it is nontame and orbit nonproper. Proof. Fix mo G M such that Gmo is dense in M. By nontransitivity, Gmo ^ M. Then Gmo is not locally closed in M; in particular, Gmo is not closed in M. By Lemma 7.11.12, p. 246, the G-action on X is orbit nonproper. By (2) =$• (1) of Lemma 7.11.5, p. 241, the G-action on X is nontame. • C O R O L L A R Y 7.11.14 Let an Icsc topological group G act continuously on an Icsc topological space X. Let n be a Borel probability measure on X such that, for any nonempty open U C X, we have fi(U) > 0. Assume that the G-action on (X,fi) is qmp and properly ergodic. Then the G-action on X is nontame and orbit nonproper. Proof. By Lemma 7.11.1, p. 240, the G-action on X is topologically transitive. By definition of properly ergodic, the G-action on X is nontransitive. The result then follows from Lemma 7.11.13, p. 247. D L E M M A 7.11.15 Let an Icsc group act continuously on an Icsc topological space. Assume that the action is orbit proper. Then the action is tame and every stabilizer is compact. Proof. By Lemma 7.11.12, p. 246, every orbit is closed and every stabilizer is compact. Then, since closed implies locally closed, we conclude, from (1) = > (2) of Lemma 7.11.5, p. 241, that the action is tame. • The converse of Lemma 7.11.15, p. 247 is not true (see Example 7.11.17, p. 248, below), but we do have the following partial converse: L E M M A 7.11.16 Let G be an Icsc topological group acting continuously on an Icsc topological space X. For all x, let the orbit Gx have the relative topology inherited from X. If the G-action is tame with compact stabilizers, then, for all x G X, Gx is Icsc and the map g t-> gx : G —> Gx is proper. Proof. Fix x € X. Let H := Stabo(a;). Since any locally closed subset of an Icsc topological space is again Icsc, by (2) =£• (1) of Lemma 7.11.5,

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p. 241, we see that Gx is lcsc. Let <j>: G —¥ X be the orbit map based at x, defined by {g) = gx. It remains to show that (f>: G —> Gx is proper. Since the G-action on X has compact stabilizers, we conclude that H is compact. Let TT : G -> G/H be the canonical map. Then IT : G -t G/H is proper. Let ip : G/H —> X he the factored orbit map based at x, Then 1p 0 7T = (f>.

Then ip(G/H) = <j)(G) = Gx. Since the G-action on X is tame, it follows, from (2) =>• (3) of Lemma 7.11.5, p. 241, that tp : G/H ->• Gx is a homeomorphism. Therefore if> : G/H —> Gx is proper. Since both 7r: G —> G/H and ip : G/H —¥ Gx are proper, we are done. • Lemma 7.11.16, p. 247 only asserts that x H> gx : G -> Gx is proper; it may happen that x i-+ px : G —> X is not proper. In fact, we have: E X A M P L E 7.11.17 There is a smooth, tame, free, orbit nonproper action of E on a manifold. Proof. Let d\, 82 he the standard framing of R 2 . Fix a smooth function a : R -> M such that all of the following hold: • • • • • •

for all r € for all r £ for all r € o(l) = 0; for all r € for all r £

E, we have a(—r) = a(r); [0,1/2], we have a(r) = 1; (1/2,1), we have 0 < a(r) < 1; (1,3/2), we have - 1 < a(r) < 0; and [3/2,00), we have a(r) = —1.

Define smooth vector fields P and Q on E 2 by: At any z = (x, y) € E 2 , let Pz ••= -y(di)z +x(d2)z and Qz := [ ( a ( x 2 + j / 2 ) ) - a ; - ( 5 1 ) z ] + [(a(z 2 + y 2 ) ) • y • ( & ) , ] . Let 5 := E 2 \ { ( - 1 , 0 ) } and let X := E 2 \ { ( 0 , - 1 ) } . Let V : S -»• X be the diffeomorphism (fractional linear transformation) defined by ,, , ^ • w -

/ ^

2;/ 1 - a? - y a \ + 1)2+2,2- ( x + 1 ) 2 + 2 / 2 ; -

Then V(0,0) = (0,1), ^(1,0) = (0,0), V(0,1) = (1,0), ^ ( 0 , - 1 ) = ( - 1 , 0 ) . Let C := {(a;,0)|a; € E} and let T := {{x,y)\x2 + y2 = 1}. Then V>(T\{(-1,0)}) = C. Let So := 5\{(0,0)} and let X0 := X\{(0,1)}. Let V0 he the image of (P + Q)\SQ under ip\So : So -¥ XQ, SO VQ is a nowhere

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vanishing vector field on Xo- Let g denote the standard flat Riemannian metric on R 2 . Let V := Vo/y/g(Vo,Vo) be the normalization of VQ to a unit vector field on Xo- The flow of V defines a smooth R-action on Xo. Then

• V is a complete vector field on Xo; • C is an orbit of the flow of V; and • for every x £ Xo\C, the closure in Xo of Mx is (Ma;) U C.

We wish to show that the R-action on Xo is tame, free and orbit nonproper. Every R-orbit in Xo is locally closed in Xo; thus, by (1) =>• (2) of Lemma 7.11.5, p. 241, we conclude that the R-action on X 0 is tame. Moreover, V is a unit vector field on Xo, so V is nowhere-vanishing on Xo, so the R-action on Xo is free. No orbit in XQ\C is closed in X 0 , so, by Lemma 7.11.12, p. 246, the R-action on X 0 is orbit nonproper. •

7.12

Induction of actions: Definition

Let P be a subgroup of a group G. There is a basic process called induction (or suspension), which takes an action of P on a set M and produces an action of G on a set denoted G xp M: Define an action of P on G x M by p.(g,m) = (gp~l,pm). Define G Xp M := P\(G x M). The G-action on G x M defined by g'(g,m) = (g'g,m) factors to a G-action on G Xp M. If G is a Lie group, if P is a closed subgroup of G and if M is a manifold on which P acts smoothly, then the P-action on G x M is proper. So, from Lemma 4.2.6, p. 74 (with G replaced by P and M replaced by G x M) and by Corollary 3.8.3, p. 58, there is a unique manifold structure on G Xp M such that the canonical map G x M —> G Xp M is a submersion. We give GxpM this manifold structure. Then the G-action on G xp M is smooth. Let G be a Lie group and let P be a closed subgroup of G. Then the content of the remarks above is that induction M i - » G x p M is a functor from P-manifolds to G-manifolds. If G is a complicated Lie group, then it can be difficult to produce actions of G. So induction is a useful process in this setting, because one can hope to look for an uncomplicated (e.g., Abelian) closed subgroup P in G, produce many examples of P-actions, then induce them to G.

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Results

Induction of actions: Basic results

Let P be a closed Lie subgroup of a Lie group G. Let q : G -»• GjP be the canonical map. For all g G G, let [g] := q(g). Let P act smoothly on a manifold M. Let M' := G xP M be the G-manifold induced from the P-manifold M. Let 7r : M -> Af' be the canonical map. For all g € G, for all m G M, let [<7,m] := it(g,m). The following omnibus lemma records how the dynamical properties of the P-action on M and of the G-action on M' are related. L E M M A 7.13.1 All of the following are true: (1) If P acts nonproperly on M, then P acts nonproperly on M', so G also acts nonproperly on Af'. (2) If P acts orbit nonproperly on M, then P acts orbit nonproperly on M', so G also acts orbit nonproperly on M'. (3) If P acts properly ergodically on M (with respect to the smooth measure class), then G acts properly ergodically on M' (again, with respect to the smooth measure class). (4) There is a G-equivariant surjective submersion M' —> GjP. (5) Let A := {Stab P (m) | m 6 M } . Let A' := {Stab G (m') | m! <E AT}. Then A' = {gSg'11 S € A,g € G}. (6) If P is normal in G, then, for all ml G M', Stabc(m') C P . (7) If the P-action on M is free (resp. locally free, resp. faithful, resp. locally faithful), then the G-action on M' is free (resp. locally free, resp. faithful, resp. locally faithful). (8) Assume that P-action on M extends to a smooth G-action on M. Let G act on (G/P) x M by: g(x,m) = (gx,gm). Then there is a G-equivariant diffeomorphism between M' and (G/P) x M. (9) If P acts minimally on M, then G acts minimally on M'. Proof. For all g G G, let [g] denote the image of g under the canonical map G -»• G/P. For all g G G and m G M, let [g, m] denote the image of (g, m) € G x M in M' = G xP M. Define i : M -> M' by t(m) = [lo,m]. Then i is P-equivariant and continuous. So, if the P-action on M is nonproper (resp. orbit nonproper), then the P-action on M' is nonproper (resp. orbit nonproper), proving (1) and (2). The map S t-» t _ 1 ( 5 ) is a one-to-one correspondence between G-invariant subsets of M'

and

P-invariant subsets of M;

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its inverse is S H4 G(t(S)). Under this correspondence, negligible G-invariant subsets of M' correspond to negligible P-invariant subsets of M, which proves (3). Under this correspondence, closed G-invariant subsets of M ' correspond to closed P-invariant subsets of M, so (9) follows from the fact that an action is nonminimal iff there is a proper, nonempty, closed, invariant set. The map [g,m] i-4 [g] : M' -4 G/P is well-defined and satisfies the conditions of (4). For all g € G, for all m e M, we have StabcQff, m]) — g(Stabp(m))g~1; this implies (5). Then (6) follows immediately from (5), as does (7). The map [g,m] 1-4 ([g],g.m) : G XpM -4 (G/P) x M is a well-defined G-equivariant diffeomorphism, proving (8). • The following records in more detail how the map G Xp M -4 G/P of (4) of Lemma 7.13.1, p. 250 is constructed, and how to identify M with the fiber in G Xp M over [IQ]- Its proof is left as an exercise for the reader. L E M M A 7.13.2 Let M0 := g - 1 ( [ M ) - Define maps p : M' -4 G/P and L: M -t M' by p([g,m]) — [g] and i(m) = [ l c m ] . Then • p : M' —t G/P is a surjective submersion; • Mo is a P-invariant closed submanifold of M'; • L{M) = M 0 ; and • L : M —¥ M0 is an P-equivariant diffeomorphism. By Lemma 4.9.6, p. 99, q : G —> G/P is a principal P-bundle; in the language of bundle theory, [g,m] 1-4 [g] : G Xp M -4 G/P is the bundle associated to q by the P-action on M. Since left-translation defines an action of G by principal bundle automorphisms of q : G —> G/P, it follows that G acts by automorphisms of the associated bundle G Xp M —t G/P.

7.14

Riemannian dynamics

By adapting the argument of Lemma 7.10.1, p. 236, we can study Riemannian dynamics: L E M M A 7.14.1 Let G be an Ad-proper Lie group. Let G act locally faithfully and isometrically on a Riemannian manifold M. Then the action of G on M is proper.

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Proof. Let gi be a sequence in G, let m, and m\ be sequences in M and let W o c i n ^ € M. Suppose, for all i, that ginii = m^. Suppose that gi ->• oo in G. Suppose that both mj -»• moo and m!i -»• m ^ in M. We wish to obtain a contradiction. Since G is Ad-proper, we have Adg(gi) -+ oo in GL(g). By Lemma 4.9.34, p. 110, we choose l £ g such that at least one of the following holds: • {(Adgi)X}i is not precompact in g; or • {(Adg^~1)X}i is not precompact in g. By, if necessary, interchanging each rrii with m^ and replacing each gi by s j - 1 , we may assume that {(Ad<7t)X}j is not precompact in g. Passing to a subsequence, choose a sequence £j in E and choose X'^ £ fl\{0}, such that both U ->• 0 in R and ^ [ ( A d ^ X ] -*• X ^ in JJ. For each i, choose orthonormal bases Bi and B\ of TmiM and Tm'. M, respectively. Passing to a subsequence, choose orthonormal bases Boo and B'^ of Tm<x>M and TTO/ M, respectively, such that, in the frame bundle of M, we have both Bi -> Boo and B\ -> i ? ^ . Let d := dim(M). For all i, define / s : M -> M by /»(»") = St^n and let * , := [(#t)m«]?! £ 0(d). For all i, let X[ := UftAdgjX]. Then X[ -> X ^ . For all », we have (X]) M = U[(fi)t((Xi)M)}, so, by m (3) of Lemma 7.3.7, p. 205, we have both *i[*i(Xg )] = (X^g/™ and

umxkr^T1] = (^')i:m-

Since 0(d) is compact, {^i}i is precompact in GLd(K). Then, since Xg -> X£™ in E d x l , we see that {*i(Xg m )}i is precompact in K d x l . As U -4 0, this hnplies that (X.Og,™ = ii[*i(X*£ m )] -> 0. Thus ( X ^ ) g £ = 0. Since Xg™ -> Xg™ and since {\tj}i is precompact in GLd(R), we conclude that {* i (A'| m )* J ~ 1 }j is precompact in Rdxd. As U ->• 0, this implies that (X t ')§, m = * i [ * i ( ^ - ) * r - 1 ] -* 0. Thus ( X ^ = 0. As ( X ^ ) g m = 0 a n d ( X ^ ) £ m = 0, by Lemma 7.4.11, p. 211, we have X'oo = 0, contradicting the assumption that X ^ € fl\{0}. D m

The next three results are entirely Lie theoretic and not dynamical. They are known, but the use of dynamics to prove them is novel, so far as I know. L E M M A 7.14.2 Let G be an Ad-proper Lie group. Let H be a Lie group. Let p : G -> H be an injective homomorphism. Then p(G) is closed in H.

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Proof. Assume, for a contradiction, that p(G) is not closed in H. By Lemma 5.4.3, p. 158, let G act minimally, properly ergodically, freely and isometrically on a Riemannian manifold M. Fix m e M . Then Gm is dense in M. Since the action is properly ergodic, it is nontransitive. Fix m' £ M\(Gm). Choose gi in G such that gitn —> m'. Then, since ml ^ Gm, it follows that gi -> oo in G. Then the G-action on M is orbit nonproper, and therefore nonproper. This contradicts Lemma 7.14.1, p. 251. • C O R O L L A R Y 7.14.3 Let G be a connected, semisimple Lie group with finite center, let H be a Lie group and let f : G -> H be a homomorphism. Then f{G) is closed in H. Proof. Let K := ker(/). If K = G, then / ( G ) = {1 G }, and we are done, so we assume that K ^ G. Then, by Lemma 4.10.23, p. 122, GjK is a connected, semisimple Lie group with finite center. So, replacing G by GjK, by Corollary 4.9.3, p. 97, we may assume that / is injective. By Lemma 4.13.3, p. 135 (with p : L -)• GL(V) replaced by Ad : G ->• GL(fl)), we see that Ad B (G) is closed in GL(g). The center of G is finite, hence compact. So, by Lemma 5.4.1, p. 156, G is Ad-proper. Therefore, the result follows from Lemma 7.14.2, p. 252. • C O R O L L A R Y 7.14.4 Let H be a Lie group and let G be a connected Lie subgroup of H. If G is semisimple, then G is closed in H. L E M M A 7.14.5 LetG be a connected, simple Lie group with finite center, let H be a Lie group and let f : G —> H be a homomorphism. Assume that f(G) ^ {ljj}. Then f : G -> H is proper. Proof. Let K := ker(/). Since / ( G ) ^ {Iff}, we see that K ^ H. So, since G is simple and if is a normal closed subgroup of G, we conclude that K is discrete. Then, by (1) of Lemma 4.9.16, p. 103, K C Z(G). Then K is finite and therefore compact. By Lemma 7.14.3, p. 253, / ( G ) is closed in H. Then, by Lemma 2.7.7, p. 29, / : G -¥ H is proper. D L E M M A 7.14.6 LetG be a connected Lie group, letN be a normal closed subgroup ofG and let Lo be a semisimple connected Lie subgroup ofG. Then both of the following are true: (1) If Z(LQ) is finite, then LQN is a closed subgroup of G.

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(2) If LQ is a normal subgroup of a semisimple Levi factor of G and if N contains the nilradical of G, then LQN is a normal subgroup of G. Proof. Proof of (1): Let n : G -t G/N be the canonical Lie group homomorphism. By Corollary 7.14.3, p. 253 (with / : G -> H replaced by n\L0 : L0 -* G/N), 7r(Lo) is a closed subgroup of G/N. Then, since LQN = Tr~1(ir(Lo)), we see that LQN is a closed subgroup of G. End of proof of (1). Proof of (2): Let L be a semisimple Levi factor of G and assume that L0 is a normal subgroup of L. By Corollary 4.5.3, p. 85, the Lie algebra of LoN isfo+ n . By Corollary 4.6.3, p. 91, it suffices to show that [fo + n , g] C fo +n. Let r be the solvable radical of g. Then g = l + t , so [fo,fl]C [lo, I] + [lo, t]. Since l0 is an ideal of I, we get [l0,I] C fo. By (2) of Lemma 4.15.5, p. 143, we have [fo,r] C n. Then [fo,g] C fo + n. Since n is an ideal of g, we get [n, g] C n. Then [fo + n, g] C [fo, g] + [n, g] C fo + n + n = fo + n. End of proof of (2). • L E M M A 7.14.7 Let G be a connected semisimple Lie group. Then G is noncompact iff there is a closed subgroup G' of G such that g' is Lie algebra isomorphic to sh($). Proof. Proof of "if": Since g = sfo(IR), we see that g is noncompact. Then, by Lemma 4.10.32, p. 124, G' is noncompact. Then G is noncompact. End of proof of "if". Proof of "only if": By Lemma 4.10.32, p. 124, g is noncompact. By (1) = $ . (2) of Lemma 4.12.2, p. 132, choose N € g\{0} such that the map ad./V : g -¥ g is nilpotent. By Lemma 4.10.33, p. 125, choose a Lie subalgebra g' of g such that g' = sl2(IR). Let G' be the connected Lie subgroup of G corresponding to g'. Then G' is simple. By Corollary 7.14.4, p. 253, G' is closed in G. End of proof of "only if". • L E M M A 7.14.8 Let L be a connected simple Lie group. Then there is a one-dimensional connected Lie subgroup C of L such that, for any vector space V, for any nontrivial representation p : L —> GL(V), p(C) is Lie group isomorphic to the additive circle M/Z. Proof. Any one-dimensional, connected, compact Lie group is isomorphic to M/Z.

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Case 1: L is noncompact. Proof in Case 1: Let V := SL2(E) and let C := SO(2) C V. By Lemma 7.14.7, p. 254, by, if necessary, replacing L by a subgroup, we may assume that I is Lie algebra isomorphic to V. Let c be the Lie subalgebra of I corresponding to c' C ['. Let C be the connected Lie subgroup of L corresponding to c. Then C is one-dimensional. Let V be a vector space and let p : L —> GL(y) be a nontrivial representation. Let Co := p{C). We wish to show that Co is one-dimensional, connected and compact. Since C is connected, Co is connected, as well. Since p is nontrivial and L is simple, it follows that p is faithful. Then C and Co are Lie group isomorphic, so, since C is one-dimensional, we conclude that Co is onedimensional. It remains to prove that Co is compact. Let Lo := p(L). Since Adi<(C) is compact, it follows that Ad[(C) is compact. Then Ad[0(Co) is compact. By Lemma 4.13.3, p. 135, we know that LQ is an almost algebraic subgroup of SL(V). Let Z0 := Z(L0). Then, by Lemma 4.13.1, p. 135, Z0 is an almost algebraic subgroup of SL(V). Since Lo is semisimple, it follows that ZQ is discrete. Then, Lemma 4.13.2, p. 135 (with G replaced by Zo), we see that Z0 is finite. By Lemma 4.13.3, p. 135, Ad[ 0 (L 0 ) is a closed subgroup of SL(Io). Let r := Ad : L0 -> SL([ 0 ). Then T(C0) = Ad [ o (C 0 ), so T(CO) is a compact subgroup of SL(l 0 ). As r has finite kernel and closed image, by Lemma 2.7.7, p. 29, r is proper. So, since T _ 1 ( T ( C O ) ) = ZQCO and since T(CO) is compact, we see that ZQCQ is a compact subgroup of Lo- By Lemma 4.5.12, p. 88, (ZoCo)0 — Co- Then Co is closed in Z 0 Co, and is therefore compact. End of Case 1. Case 2: L is compact. Proof in Case 2: Let T be a maximal torus in L. By Theorem 1.6, p. 159 of [BtD85], T is nontrivial. Let C be a closed subgroup of T such that C = R/Z. Let V be a vector space and let p : L -> GL(V) be a nontrivial representation. We wish to show that p(C) is one-dimensional, connected and compact. Since C is connected, p(C) is connected, as well. Since p is nontrivial and L is simple, it follows that p is faithful. Then C and p(C) are Lie group isomorphic, so, as C is one-dimensional, we conclude that p(C) is one-dimensional. Since C is compact and p is continuous, it follows that p(C) is compact. End of Case 2. • L E M M A 7.14.9 Let Go be a connected Lie group, let Lo be a semisimple

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Levi factor of Go and let Ro be the solvable radical of Go- Assume that Lo is simple with finite center. Let G\ be a Lie group and let (f> : Go -> Gi be a homomorphism. Assume that oo in Lo- Then {hri) —> oo in G\. Proof. Replacing G\ by the closure in Gi of (j>(Go), we may assume that <j)(Go) is dense in Gi. Let R\ be the closure in Gi of (Ro). Then, since <j>{Go) is dense in Gj and since Ro is normal in G, we conclude that i?i is normal in G\. Let •n : Gi -> G\ /Ri be the canonical homomorphism. Since Ro is solvable, it follows that (f>(Ro) is solvable, so Ri is solvable. For all i, since 0(r*) £ i?i, we get 7r(0(rj)) = l G l / f l l . Since L0 is simple and since (L0) ^ {1GI}> we see that (j>(L0) is simple. Then \L0) : Lo - • G1/R1 and if := ker(/). Then K° is a normal connected closed subgroup of Lo and K = L0 D (-1(i?i)) 7^ L 0 , so, by simplicity of Lo, we conclude that K° = { 1 L 0 } . Then JFtT is a totally pathdisconnected subgroup of Lo- Thus, by (1) of Lemma 4.9.16, p. 103, since K is normal in Lo, we have K C Z(Lo). In particular, K is finite. By Corollary 7.14.3, p. 253, / ( L 0 ) is closed in G i / i ? i . It follows, by Lemma 2.7.7, p. 29, that / : LQ -> G1/R1 is proper. Then, as /, -> 00 in L 0 , we see that /(£,) —• 00 in Gi/R\. That is, n(<j)(li)) —> 00 in G1/R1. For all i, we have 7r(^(rj)) = l G ] / f l l , so n((li)) = •n^ihri)). Then 7r(0(Z»rj)) ->• 00 in G1/R1, so {Uri) -» 00 in Gi. D L E M M A 7.14.10 Lei G be a connected Lie group and assume either that Ad g (G) is noi closed in GL(g) or that Z(G) is noncompact. Then there exists a free, orbit nonproper, isometric action of G on a connected complete Riemannian manifold. Proof. If Z{G) is noncompact, then by Lemma 5.4.4, p. 159 and by Lemma 5.4.3, p. 158, there is an orbit nonproper, isometric action of G on a Riemannian manifold. We therefore assume that Z{G) is compact. Then Ad B (G) is not closed in GL(g). By Lemma 5.4.5, p. 159, choose a closed subgroup Go of G such that Go is isomorphic to the additive group M and such that Ad 0 (Go) is not a closed subgroup of GL(g). Then, because Go = K, choose X 6 g[(g) such that {exp(tX)} t€ R = Ad B (Go). Then {exp(tX)}tem is not closed in GL(g), so the map t (-»• exp(iX) : R -> GL(JJ) is not proper. Then,

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by Lemma 4.9.10, p. 101, {exp(tX)}t£R is precompact in GL(g). That is, Adg(Go) is precompact in GL(g). We therefore choose Q € QF(fl) such that Q is positive definite and (AdGo)-invariant. Fix g0 £ GOVOGO} a n d l e t C be the cyclic subgroup generated by go. Since Go — R, we see that C is infinite and is cocompact in Go- The form Q is invariant under Ad 0 (G). So, by Lemma 4.3.5, p. 79, there is a G-invariant Riemannian metric on G/C. Let H :— (GL(fl))0. Give H a left-invariant Riemannian metric. Let G act on H by g.h — (Adg(g))h. This action is isometric, and the stabilizer of any point is Z(G). Because C is an infinite cyclic group, it follows that C has no nontrivial, finite subgroups. Since Go — R, it follows that C is discrete in Go- Then, as Go is closed in G, we conclude that C is discrete in G. So because Z(G) is compact, we see that C D (Z(G)) is finite, and therefore trivial. This shows that the action of C on H is free. Since C is cocompact in Go, let if be a compact subset of Go such that CK — GQ. Then Adg(ii') is a compact subset of Ad B (G) and we have (Adg(C))(Adg(K)) = Ad 8 (Go). So, since Ad 8 (Go) is not closed in H, it follows, from Lemma 2.7.8, p. 29, that Ad B (C) is also not closed in H. Consequently, the action of C on H has nonclosed orbits. Then, by Lemma 7.11.12, p. 246, the G-action on H is orbit nonproper. By (2) and (7) of Lemma 7.13.1, p. 250, induction preserves orbit nonproperness and freeness. So, the induced action of G on G Xc H, is free and orbit nonproper. By (8) of Lemma 7.13.1, p. 250, we see that there is a G-equivariant diffeomorphism between GXQH and (G/C) x H. The Riemannian metrics on G/C and on H are both homogeneous and are therefore complete. Since the action of G on G/C preserves a complete Riemannian metric, as does the action of G on H, the G-action on G x p H also preserves a complete Riemannian metric. Since G and H are connected, it follows that G x c H is connected, as well. • L E M M A 7.14.11 LetG be a connected Lie group. Assume that Ad 0 (G) is not closed in GL(g). Then there is a connected Lie group H and an infective homomorphism f : G —¥ H such that f(G) ^ H and such that / ( G ) is dense in H. In particular, /(G) is not closed in H. Proof. By Lemma 7.14.10, p. 256, choose a connected complete Riemannian manifold M and a free, orbit nonproper, isometric action of G on M. Let H := Isom°(M). Let / : G -> H be the Lie group homomorphism

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Basic Dynamical

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giving the action of G on M. Since the action is free, it follows that the action is faithful, which implies that / : G —> H is injective. The action of G on M is orbit nonproper and, by Lemma 7.11.10, p. 245, the .H-action on M is orbit proper. Choose a compact subset K of M such that Go := {g € G | gK D K ^ 0} is not precompact in G. Let H0 := {h 6 H \ hK D K ^ 0}. Then H0 is compact and we have f~1(H0) = Go. We conclude that the map / : G -> H is nonproper. Because / : G -> H is injective, it follows that ker(/) is trivial, hence compact. Then, by Lemma 2.7.7, p. 29, /(G) is not closed in H. Let H be the closure in H of f(G). Then /(G) is dense in H. Since /(G) is not closed in H, it follows that /(G) ^ H. • Let •A/T' (resp. OAfV) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), locally faithful, isometric action on a connected Lorentz manifold. Let NT1* (resp. OAfV1*) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), locally free, isometric action on a connected Lorentz manifold. Let J\fV* (resp. ONV1) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), free, isometric action on a connected Lorentz manifold. Let NVn (resp. OMV-R) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), locally faithful, isometric action on a connected Riemannian manifold. Let ATP^ (resp. OAfV-j[) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), locally free, isometric action on a connected Riemannian manifold. Let NVK (resp. OHV^) be the collection of connected Lie groups admitting a nonproper (resp. an orbit nonproper), free, isometric action on a connected Riemannian manifold. If G acts nonproperly and isometrically on a Riemannian manifold M, and if M ' is any Lorentz manifold, then G acts nonproperly and isometrically on the Lorentz manifold M x M' by the rule g(m,m') = (gm,m'). Consequently, NVn C HV and OMVn Q OMV. Similarly, we have NVl-l C AfVlf and OAfv!£ C OAfVlf. Similarly, we have MV^ C AfVf and OATV& C OMVs. Since orbit nonproper implies nonproper, we have OMV C MV,

OAfVlf C AfVlf,

OAfVnQAfVn,

ONVllcNV1^,

OAfVf C AfVf,

ONV^cNV^.

Riemannian dynamics

259

Since free implies locally free implies locally faithful, we conclude that OAfV* C OAfVlf C OAfV,

AfVf C AfVlf C AfV,

Let MV£^ denote the collection of connected Lie groups G such that G admits a free, minimal, properly ergodic, isometric action on a connected Riemannian manifold. By Corollary 7.11.14, p. 247, MVS^ C OAfV^. Each of the ^-subscripted collections {MVn, NVl-l, N^n, OAfVn, OMVll, OhfVSn, MV£h) of groups described above contains MV£^ and is contained in AfVn- Thus (4) •£=>• (5) of the following result asserts that these ^-subscripted collections are all the same. In other words, there is a well-defined collection of groups each of which admits a complicated Riemannian action, and the exact collection does not change as the technical definition of "complicated" varies over a wide range of possibilities. It would be of interest to know whether some similar statement could be made about Lorentz manifolds. However, we will eventually show that SO 0 (2,3) E ONV\NVlf, so the range of dynamical conditions will have to be restricted somewhat from the Riemannian case. L E M M A 7.14.12 Let G be a connected Lie group. Then the following are equivalent: (1) G is Ad-nonproper. (2) Either Z{G) is noncompact or Ad B (G) is not closed in GL(g). (3) There is a connected Lie group H and an injective homomorphism f : G -> H such that f(G) ^ H and such that f(G) is dense in H. (4) GeMVSi. (5) GeMVnProof. Since Z(G) is the kernel of AdB : G ->• GL(g), by Lemma 2.7.7, p. 29 (with / replaced by Ad g : G -> GL(JJ)), we see that (1) and (2) are equivalent. By Lemma 5.4.4, p. 159 and Lemma 7.14.11, p. 257, we see that (2) =*• (3). By Lemma 7.14.2, p. 252, we have (3) =*> (1). Proof of (3) =$• (4): Let 7 be a left-invariant Riemannian metric on H. Let G act on (if, 7) by g.h = f(g)h. By Lemma 5.4.2, p. 157, This action is free, minimal and properly ergodic. By construction, this action is isometric. End of proof of (3) =>• (4). Since MV£^ C OMVfn C NVfn C NV-R, we get (4) = • (5). Lemma 7.14.1, p. 251 shows that (5) => (1). •

260

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Results

The equivalence of (2) and (3) of Lemma 7.14.12, p. 259 is well-known, and can be argued by purely Lie theoretic means (as opposed to invoking dynamical arguments, as we have done). See Theorem 2.9, p. 640 of [Po94]. We have MV£^ C OMV^ C OAfVf. Therefore, by (1) = • (4) of Lemma 7.14.12, p. 259, we see that any Ad-nonproper connected Lie group admits a free, orbit nonproper, isometric action on a connected Lorentz manifold. However, we can do better: L E M M A 7.14.13 Let G be a connected Ad-nonproper Lie group. Then there exists a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action of G on a connected Lorentz manifold. In particular, we have G € OAfVf. Proof. Using (2) =J> (3) of Lemma 7.14.12, p. 259, choose H and / as in (3) of Lemma 7.14.12, p. 259. By Lemma 4.5.7, p. 86, G0 := f(G) is a connected Lie subgroup of H and / : G -> Go is a surjective homomorphism. Since / is injective and surjective, by Lemma 4.8.2, p. 94, / : G -> Go is an isomorphism of Lie groups. Then Go is a connected Ad-nonproper Lie group; it follows that dim(Go) > 1. Since Go C H, we see, from Lemma 4.4.5, p. 83, that dim(Go) < dim(if). Then dim(il) > 2. Let 7 be a left-invariant Lorentz metric on H. Let G act on (H,j) by g.h = f{g)h. By construction, this action is isometric. By Lemma 5.4.2, p. 157, it is also minimal, properly ergodic and free. So, 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. D

Chapter 8

Examples of Actions on Compact Lorentz Manifolds

8.1

The isometry group of a compact quotient of SL2(K)

Let G := PSL 2 (R). Let p : SL 2 (E) -*• G be the canonical homomorphism. Let e := 1 G = p(E[f + E$). Let K : g x g -> K be the Killing form of G. Then — n has signature (2,1) and is therefore Minkowski. Let 7 be the unique left-invariant Lorentz metric 7 on G such that j e — — K. By Lemma 4.3.4, p. 78, we see that 7 is right-invariant. By [Bor63], every semisimple Lie group admits a cocompact discrete subgroup. Let T be a cocompact discrete subgroup of G and let M := G/T. Because 7 is right-invariant and because the canonical map ir : G —> M is a covering map, there is a unique Lorentz metric 70 on M such that TT*(7O) = 7 . As 7 is left-invariant, it follows that 70 is invariant under the translation action of G on M. For all g € G, let fg : M —> M be defined by fg(m) = gm. Let Hi := {fg \g e G} be the image of the Lie group homomorphism g \-¥ fg : G -> Isom°(M, 70). Note that g i-> fg : G —> Hi is a surjective Lie group homomorphism with discrete kernel and is therefore a covering homomorphism. Then G is locally isomorphic to Hi. L E M M A 8.1.1 We have Isom°(M,7 0 ) = Hi. In particular, G is locally isomorphic to the connected isometry group of a compact Lorentz manifold. Proof. Let H := {/ £ Isom(G,7) | /(e) = e}. Let I : G -> G be the identity. Let x :— p(E[2 + ^21 )• ^et a : G -> G be defined by a(g) = xga; and let 6 : G -> G be defined by b(g) = g denote the differential at e of h. 261

262

Examples of Actions on Compact Lorentz

Manifolds

Define D : H -> 0 ( K ) by D(h) = dh. We have D(T) = Adg(G). Also, for all x 6 0 ( K ) , we have [D(S)] n [X(SO°(K))] ^ 0. That is, every coset in 0 ( K ) of S O ° ( K ) has nonempty intersection with D(S). So, since we have D(T) = Adfl(G) = SO°(K)> we conclude that D(ST) = 0 ( K ) . AS if preserves the Levi-Civita connection of 7, by Lemma 7.4.11, p. 211, D:H -> 0 ( K ) is injective. So, since £>(ST) = 0 ( K ) , ST = if. Let mo := 7r(e). Let H0 :— {/ S Isom(M, 70) | / ( m 0 ) = mo}. Then, by transitivity of G on M, we get Isom(M,70) — ffoifi. By Lemma 4.5.12, p. 88, we wish to show that (ffoifi )/ffi is countable. It suffices to show that Ho is countable. As T is discrete, we have NG(T) = C%(T). By the Borel Density Theorem (Corollary A.3.2, p. 384), we have CG(T) = CG(G) = Z(G). Then CG(T) is trivial, so N^(T) is trivial, so NG{T) is countable. Let T := {h G TI /i(r) = T}. Then the map g >-> Int(g) : NG(T) -> T' is surjective. Then T' is countable. Since S is finite, (ST)/T is finite, so, by Lemma 4.5.12, p. 88, we have (ST)° = T°. Then, as T is connected, we get (ST)° = T. Let if' := {h € ff|/i(r) = T}. Since if' is a closed subgroup of Isom(6 ! , 7), it follows that H' is a Lie subgroup of a Lie group, so, by Proposition 4.5.5, p. 85, H'/[(H')°] is countable. We have (H')° C H° = (ST)° = T. Then (H')° C J J ' n r = T'. Because T' is countable, because (if')° C T and because H'/[(H')°] is countable, we conclude that if' is countable. Since % : G —>• M is a Riemannian covering map and since 7r(e) = mo, it follows, from covering space theory, that there is an injective map H0 -> H'. Then if0 is countable. •

8.2

Twisted Heisenberg groups

Let M+ := (0,00) and let Q + := Q n K+. Let Vi := E 2 . Define ui : Vj. xVi -> E by ui((x,y), (x',y')) = xy'-x'y. Then (V^wi) is, up to isomorphism, the only two-dimensional symplectic vector space. For any 6 € K, let i?^ : T^ -> Vi be the rotation defined by Re(x,y) = ( (cos#)a; - (smd)y , (sin#)a; + (cos6)y ). Fix an integer k > 2. Let 14 := (Vi)* and define w^ : Vjt x Vj. ->• E by

w*((«i, ••-,««!), («!,-•-,«*)) = 5 3 u > i ( « i X ) .

Twisted Heisenberg

263

groups

Then (Vk,uik) is, up to isomorphism, the only (2fc)-dimensional symplectic vector space. Fix an integer k > 1 and let d := k + 2. We define I := { 1 , . . . ,d} and I2 := I x I. Let Si := {(1,i) \i G / } , let S2 := {(«',d)\i € 1} and let S3 := {(», i) | * e / } • Let S := Si U 5 2 U S3 C J 2 . Let Gfc be the subgroup of SLd(E) consisting of all matrices ^ J Cy ^jj *>i such that, for all (i,j) G I 2 \ S , we have cy = 0 and such that, for all i € I, we have c;i = 1. Let B be the ordered basis ^12

>• • • i - c ' l , d - l »

^dl

> • • -i&d.d-l'

-^ld

of (jfc. Then B satisfies the Heisenberg relations. Let £)Ufc be the Heisenberg Lie algebra associated to (14,iJk)- Let t i , . . . , i* : Vi -> (Vi)fc = Vk be the coordinate inclusion maps. Let ei,e2 be the standard ordered basis of E 2 . For all i G {1,2}, for all integers j G [1,A;], let e^ :— tj(ej) 6 Vk- Let a : Vk -» Vfc © R == gWfc and /? : E -»• Vfc © E = gUk be the coordinate inclusions. Let B'k be the ordered basis a(en),...,a(eu),

a ( e 2 i ) , . . . ,a(e2fc),

£(1)

of flWfc. Then SjJ. satisfies the Heisenberg relations. Let £& : gk - • flwj. be the vector space isomorphism sending # to $£.. Then Sfc : Qk —> Quk is a Lie algebra isomorphism. Fix an integer k > 1 for this paragraph. Fix A i , . . . , A* € R+ for this paragraph. Let A := ( A i , . . . , A*) G K+. Let R act on Vk by the rule t.(vi,...,vk)

=

(R\lt(vi),...,R\kt{vk))-

This action preserves the symplectic structure w&, and so induces an action of E by Lie algebra automorphisms of gUh. Then, by the isomorphism Sfc, this action induces an action of E by Lie algebra automorphisms of Qk- By monodromy, this, in turn, induces an action of E by Lie group automorphisms of Gk- We define H\ := E K GkThese groups H\ are not distinct: If a > 0, if k > 1 is an integer and if A e l J . , then (t, g) t-> (at, g) : Ha\ ->• H\ is a Lie group isomorphism. A Lie algebra f) will be said to be a twisted Heisenberg Lie algebra if, for some integer k > 1 and some A G E+, we have that I) is Lie algebra isomorphic to h*. A Lie algebra I) will be said to be a rationally twisted Heisenberg Lie algebra if, for some integer k > 1 and some A G Q+,

264

Examples of Actions on Compact Lorentz

Manifolds

we have that f) is Lie algebra isomorphic to h.\. A connected Lie group is twisted Heisenberg if its Lie algebra is. A connected Lie group is rationally twisted Heisenberg if its Lie algebra is. L E M M A 8.2.1 Let g be a Lie algebra. Let k > 1 be an integer. Let n be a Lie subalgebra of g and let (Xi,..., Xk, Y\,..., Yk, Z) is an ordered basis of n satisfying the Heisenberg relations. Assume that dim(fl/n) = 1. Let S :=RX1+---+RXk+MYi+---+WYk. LetW G fl\n. Let A i , . . . , Xk € E+. Assume, for all integers i G [1, k], that [W,Xi] = AjYi and [W, YJ] = -AjXj. Then all of the following hold: (1) g is a twisted Heisenberg Lie algebra; (2) {adW)SCS; and (3) if {exp(t&ds(W))}ten Q GL(5) is compact, then g is a rationally twisted Heisenberg Lie algebra. Proof. Let A := (Ai,...,Afc) G E+. Let Gk and H\ be as above. Let a : R ->• RK gk — i)x and r : gk -» KK gk = h^ be the coordinate inclusions. Let W :— o-(l) G t)\. Let B'k be the ordered basis of gk defined above. Then B'k satisfies the Hausdorff relations. Choose X[,..., X'k,Y{,..., Yk\ Z' G gk such that B'k = (X[,... ,x\,Y(,... ,Yk1,'Z'). For all integers i G [I,*], let XI := T(X<) and let Y( := r ( F / ) . Let Z' := T(Z'). For all integers i G [1, k], we have \W',Xj[\ = XiYf and [W',YJ] = -\X[. Let / : g -> hA be the linear map defined by • for all integers i G [1, k], f(X{) • f(W) = W and f(Z) = Z'.

= X[ and f(Yi) = Y>; and

Then / : g -»• hA is a Lie algebra isomorphism, proving (1). By assumption, for all integers i G [l,k], we have [W,Xj] = AjYf and that [W,Yi\ = -XiXi. Then (a,d W)S C S, proving (2). It remains to prove (3). Let G0 :— {exp(t&ds(W))}t^s. C GL(5). Assume that Go is compact. We wish to show, for some u G H+, that, for all integers i G [1, k], we have u\j G Z. Let I : S -¥ S be the identity map; then / = 1G 0 . For all t G R, let <j>t •= exp(tads{W)) : S -» 5; then ^ t G G0- Since E is noncompact, E is not Lie group isomorphic to Go- So, by Lemma 2.7.4, p. 28, the kernel of the homomphism t H-> (j>t : E -> Go cannot be trivial. Choose u G M+ such that 02iru = ^- For a l i i G E, for all integers i G [1, A], we calculate 0t(XO - (cos^Ai))^ + (sinftAi))^.

Twisted Heisenberg groups

265

So, for all integers i G [1, k], since Xi and Yt are linearly independent and since fonuiXi) = I(Xi) = Xi, we conclude both that cos(27ruAj) = 1 and that sin(27rwAj) = 0; it follows that uA, £ Z. D L E M M A 8.2.2 Let k > 1 be an integer. Then Aut(gfc) acts transitively on (3(0*))\{O). Proof. Let K := { 1 , . . . , k}. Choose a basis X\,..., Xk, Y i , . . . , Yk, Z of gk such that RZ = $(gk), such that, for all i G K, we have [Xi, Yi] = Z and such that, for all i, j 6 K, we have [Xi,Xj] —0 = [Yi, Y}]. Fix Z' e (3(flfc))\{0}. We wish to show that there is an automorphism ct: Qk —> Qk such that a(Z) = Z'. As 3(0*;) = RZ, choose s e K\{0} such that Z' = sZ. Define a linear map a : gk -> 9k by: • for all i G K, a(Xi) = sXi and a{Yi) = Yi; and • a(Z) = sZ. Then a e Aut(gfc) and a(Z) = sZ = Z'.

D

L E M M A 8.2.3 Let k > 1 be an integer and let To be a discrete subgroup of Z(Gk). Then there is a cocompact discrete subgroup V of Gk such that

r 0 cr. Proof. Since Z(Gk) is Lie group isomorphic toffi,and since every discrete subgroup of E is cyclic, it follows that To is cyclic. If necessary, we may replace r 0 by a larger discrete subgroup of Z(Gk) and assume that r 0 is nontrivial. Choose z £ Fo\{lG fc } such that To = {z n } n ezLet d := k + 2. Let A := Gk n (Zdxd). Then A is a cocompact discrete subgroup of G. Moreover, A n (Z(Gk)) is nontrivial. So, since Z(Gk) is Lie group isomorphic to E and since every discrete subgroup of E is cyclic, choose x e An(Z(Gjt)) such that An(Z(G*)) = { z " } n G Z . Since An(Z(G*)) is nontrivial, it follows that x ^ lok- By Lemma 8.2.2, p. 265, choose an automorphism p : Gk —• Gk such that p{x) = z. Let T :— p(A). • L E M M A 8.2.4 Let k > 1 be an integer and let \ € Q £ . Let I^ 6e a discrete subgroup of Z{H\). Then there is a cocompact discrete subgroup V of H\ such that Ti C V. Proof. Let H := Hx, let e:=\H = ( 0 , 1 G J . and let G := {0} K Gk. Let 7r : H —• i f / G be the canonical homomorphism. By Lemma 2.8.1, p. 30

266

Examples of Actions on Compact Lorentz

Manifolds

and Lemma 2.7.2, p. 27, we wish to show that there is a cocompact discrete subgroup r of H such that •Tier; • r n G is a cocompact discrete subgroup of G; and • 7r(r) is a cocompact discrete subgroup of H/G. Let KQ be the kernel of the K-action on Gk- Since A 6 Q^, it follows that KQ is a nontrivial cyclic subgroup of R. Then KQ is a discrete cocompact subgroup of E. Let K := K0 x {lGk}. We compute that Z(H) - K • Z(G). Then Z(G) = (Z(H)) D G. By assumption, Ti is a discrete subgroup of Z(H). By Lemma 4.9.21, p. 105, Ti is a finitely generated Abelian group. Let I^ := Ti D G. Then Ti = [Ti n (Z(H))] n G = r x n [(Z(H)) n G] = r x n (Z(G)). Since there is an injective homomorphism Ti/T^ ->• if/G and since i J / G is Lie group isomorphic to M, we conclude that Ti/F^ is torsion-free. So, by the theory of finitely generated Abelian groups, choose a subgroup A of Ti such that A r i = Ti and A n r j = {e}. We have A n G C Tj n G = r x . Then A 0 G C A n Ti = {e}. Moreover, Tj = A r i C A r 0 . By Lemma 8.2.3, p. 265, let To be a cocompact discrete subgroup of G such that Ti C r 0 . Then A n T 0 C A n G = {e}. Since A C Ti C Z(H), it follows that To and A centralize each other. Then II := ToA is a subgroup of H. As T 0 C G, it follows that ( r 0 A ) n G = r 0 ( A n G ) . So, since n = T 0 A and since A fl G = {e}, we get II n G = To- Then II fl G is a cocompact discrete subgroup of G. We have 7r(r 0 A) = TT(A) C 7r(ri) C v{Z(H)) = n(K • Z(G)) = n(K). So, since TT(K) is a discrete subgroup of H/G, we conclude that 7r(r 0 A) is discrete in H/G as well. Case 1: II C G. Proof in Case 1: Let T := KU. Since II C G, we have both (KU) n G = (K n G)n and n = n n G. By definition of K and G, we get K n G = {e}. Then r n G = (KU) n G = (K n G)U = U = U n G. Then T n G is a cocompact discrete subgroup of G. Since II = T 0 A C G = ker(?r), we get ?r(r) = n(KU) = n(K). Then 7r(r) is a cocompact discrete subgroup of H/G. End of proof in Case 1. Case 2: UgG. Proof in Case 2: Let T := U. Then

rnG = nnG = (r0A) n G, so r fl G is a cocompact discrete subgroup of G.

267

Twisted Heisenberg groups

A nontrivial discrete subgroup of R is cocompact. As II %. G = ker(7r), we have 7r(II) 7^ {7r(e)}. So, since 7r(II) is a discrete subgroup of H/G and since H/G is Lie group isomorphic to M, we conclude that 7r(II) is a cocompact discrete subgroup of H/G. That is, 7r(r) is a cocompact discrete subgroup of H/G. End of proof in Case 2. • L E M M A 8.2.5 Any twisted Heisenberg Lie algebra admits an aA-invariant Minkowski form. Proof. Fix an integer k > 1 and fix A := ( A i , . . . , A*,) G R+. We wish to show that \)\ admits an ad-invariant Minkowski form. Let K := {1,...,A;}. Let Xi,...

,Xk,

Yi,...,Yfc,

Z

be a basis for g satisfying the Heisenberg relations. For alH G K, define X[ := (l/VK)Xi and Y( := (l/VK)Yi. Let S := {X[,.. .,X'k,Y{,... ,yfc'} and T := {Z, W}. Define a symmetric bilinear form B : h^ x \)\ —> E such that S is 5-orthonormal, such that S is B-orthogonal to T and such that B(W,W)=0

= B(Z,Z),

B(W,Z)

= l.

This form has signature (2k + 1,1), and so is a Minkowski form. Computation shows, for all Q, U, V G S U T, that B([Q, U], V) + B(U, [Q, V}) = 0. Since S U T spans h*, it follows that B is (adh.\)-i nva ri an 1;• COROLLARY 8.2.6 A connected rationally twisted Heisenberg Lie group admits a locally free, isometric action on a compact Lorentz manifold. Proof. Let H be a rationally twisted Heisenberg Lie group. Choose k > 1 and A G M^. such that the universal cover of H is Lie group isomorphic to H\. Let p : H\ -> H be a covering homomorphism. Let Ti := ker(p). Then Ti is a discrete normal subgroup of H\, so, by (1) of Lemma 4.9.16, p, 103, we see that Tj C Z(H\). By Lemma 4.9.2, p. 96, H is Lie group isomorphic to H\/Ti. We therefore wish to show that H\/Ti admits a locally free, isometric action on a compact Lorentz manifold. Equivalently, we wish to show that if A admits a locally free, isometric action on a compact Lorentz manifold such that T\ is contained in the kernel of the action. By Lemma 8.2.4, p. 265, let T be a cocompact discrete subgroup of H\ such that Ti C T. Define M := H\/T. Let -K : H\ -> M be the canonical map. Since T is discrete, it follows that the standard transitive action of

268

Examples of Actions on Compact Lorentz

Manifolds

H\ on M is locally free. For all h € H, since Tj C Z(HX), it follows that Ti = hTih-1, so Ti C hTh-1 = StabHx(n(h)). Then Ti is in the kernel of the action of H\ on M. By Lemma 8.2.5, p. 267, let Q be an ad-invariant Minkowski form on h. Let e := 1HX- Let 7 be the left-invariant Lorentz metric on H\ such that 7e = <5- By Lemma 4.3.4, p. 78, 7 is right-invariant, and so there is a unique Lorentz metric g on M such that 7r* (g) = 7. The left-invariance of 7 implies that 3 is .H^-invariant. • We do not know if "locally free" can be replaced by "free" in Corollary 8.2.6, p. 267. 8.3

SL2(K), twisted Heisenberg and closure

Let Q denote the class of connected Lie groups which admit a locally faithful, isometric action on a compact connected Lorentz manifold. L E M M A 8.3.1 For any a connected Lie group G, the following all hold: (1) (2) (3) (4) (5) (6) (7)

If g is If Q is If G € IfG£G If G € If G £ If g is

Lie algebra isomorphic to sl2(E), then G & Ga rationally twisted Heisenberg Lie algebra, then G € Q. G and H is a connected Lie subgroup of G, then H £ Q. and K is a compact connected Lie group, then Gx K 6 Q. G and G is a covering group of G, then G € GG and A is an Abelian connected Lie group, then G x A € GHeisenberg, then G EG-

Proof. Result (1) comes from Lemma 8.1.1, p. 261. Result (2) comes from Corollary 8.2.6, p. 267. Results (3) and (5) come from the definition of G- For result (4), if G acts on M, then G x K acts on M x K by the rule (g,k)(m,k') — (gm,kk'). If M carries a G-invariant Lorentz metric, then, by giving K a left-invariant Riemannian metric, we see that M x K carries a (G x i?)-invariant Lorentz metric. Proof of (6): Let if be a compact connected Lie group such that A is a covering group of K. Then, by (4), G x K € G- So, since G x A is a covering group of G x K, by (5), G x A € G- End of proof of (6). Proof of (7): Let G' be a twisted Heisenberg Lie group. Then Q is isomorphic to a Lie subalgebra of Q'. By (2), G' € G- Since G is isomorphic to a connected Lie subgroup of G", by (3), G € G- End of proof of (7). •

SL2(M), twisted Heisenberg and closure

269

Let Go denote the collection of all rationally twisted Heisenberg groups. One of our goals (Theorem 11.7.3, p. 313) is to prove that Q is exactly equal to the closure of Go U {PSL2(K)} under the following operations: • • • •

passage to connected subgroups; passage to covering groups; products by compact, Abelian connected Lie groups; and products by compact, semisimple connected Lie groups.

By Lemma 4.8.3, p. 95, for any compact, Abelian connected Lie group A, there is an integer n > 0 such that A is Lie group isomorphic to ( E / Z ) n . Moreover, there is a complete classification (up to local isomorphism) of compact, semisimple connected Lie groups; see, for example, §V.5, pp. 209216 of [BtD85]. By an "effective" description of a class of groups, we mean one which is sufficiently structural in nature that, given any reasonable presentation of a Lie group, one may quickly determine whether or not it is the given class. So the content of the preceding paragraph is that Theorem 11.7.3, p. 313 will effectively describe G-

Chapter 9

Examples of Nonproper Actions

In this chapter, we begin the investigation of actions of Lie groups on (possibly) noncompact Lorentz manifolds. An initial observation is that any connected Lie group G admits a free isometric action on a connected Lorentz manifold: Fix any Minkowski form Q on g := T\GG and, for each g e G, let 7 9 be the image of Q under the differential at 1Q of the translation map x *-+ gx : G —> G. Then 7 := {lg}g€G is a left-invariant Lorentz metric. Then one simply lets G act on (G, 7) through left-translation. So, to make an interesting classification problem, one must require some dynamical conditions. One of the key points of Chapters 10, 11, 14 and 15 is that even under very mild dynamical hypotheses (nonproper and orbit nonproper, defined below), the list of groups that may act becomes dramatically restricted. We devote a substantial portion of these three chapters to describing effectively the collections ONV1* and OMV defined in §7.14, p. 251. Let <S denote the collection of simple Lie groups with finite center. In Chapter 8 (in Theorem 10.5.2, p. 292 and Theorem 10.5.1, p. 291), we will give an account of [Ko96], which describes exactly SnAfV1* and SC\J\fV. We will prove, among other things, that

SnAfV

= SnONV

and

SnAfPlf

= Sn

ONVlf.

In Chapters 12 and 13 (see Theorem 14.3.2, p. 342 and Theorem 15.7.4, p. 379), we determine effectively the collection of simply connected Lie groups in OAfVlf and OMV. We do not, at this point, have an exact description of OMV or of OMVli. 271

272

Examples of Nonproper

Actions

The universal cover of any element of OJ\fV (resp. ONV1*) is again in OAfP (resp. OAfV1*), so we do have an exact description, up to local isomorphism, of the groups in OMV (resp. ONV1*). However, it is possible for two connected Lie groups to be locally isomorphic, and for one to be in OAfVlf, while the other is not even in MV. (See Example 9.3.2, p. 280.) We do not know whether NV\OAfV = 0. We do not know whether AfVlf\OAfVlf = 0.

9.1

Restriction-Induction

Let G be a Lie group and let N be a closed subgroup. The point of much of Lemma 7.13.1, p. 250 is that induction of actions from N to G is a functor which preserves many dynamical properties. However, when one is examining iV-actions which preserve a geometric structure {e.g., a Lorentz metric), then the induced G-actions will typically not have an invariant geometric structure of the same type. So, in geometry-preserving dynamics, one might be led to the conclusion that induction is not so useful. The intent of this section is to describe two situations in which induced actions do, in fact, continue to preserve geometric structure. L E M M A 9.1.1 Let G be a Lie group acting isometrically on a Lorentz manifold M. Let N be a normal subgroup of G. Then the G-action on G XN M preserves a Lorentz metric. Proof. By (8) of Lemma 7.13.1, p. 250, it suffices to show that the diagonal G-action on (G/N) x M preserves a Lorentz metric. Since N is normal in G, we see that G/N has the structure of a Lie group, and therefore admits a left-invariant Riemannian metric. The product of the Riemannian metric on G/N with the Lorentz metric on M is a Lorentz metric on (G/N) x M and is G-invariant. D If G and N are connected, then Lemma 9.1.1, p. 272 follows from Lemma 9.1.3, p. 273, below (with H := G). COROLLARY 9.1.2 Let G be a connected Lie group acting isometrically on a Lorentz manifold M. Let G' and G* be a normal closed subgroups of G. Assume that the G'-action on M is orbit nonproper. Assume, for all m € M, that S t a b ^ m ) is compact and that Stabc(m) C G*. Let M' := G XQI M. Then all of the following are true.

Restriction-Induction

(1) (2) (3) (4)

The The For For

273

G' -action on M' is orbit nonproper. G-action on M' preserves a Lorentz metric. all m' € M', Stab G (m') is compact. all m' eM', Stab G (m') C G ' n G , .

Proof. Conclusion (1) follows from (2) of Lemma 7.13.1, p. 250. Conclusion (2) follows from Lemma 9.1.1, p. 272. Since the G-action on M has compact stabilizers, and since G' is a closed subgroup of G, we see that the G'-action on M has compact stabilizers. Then Conclusion (3) follows from (5) of Lemma 7.13.1, p. 250. For all m £ M, we have Stab G /(m) C G' nG„, so, since G' n G» is a normal closed subgroup of G, Conclusion (4) also follows from (5) of Lemma 7.13.1, p. 250. • For any vector space V, let BF(V) denote the vector space of bilinear forms on V. For all , ip 6 V* let B^, € BF(V) be defined by the equation B^(v,w) = (0(«)) • C 0 M ) . The map <j>®xj)^>- BH : V* ®RV* -»• BF(V) is then an isomorphism of vector spaces, giving a natural identification between V ® R V* and BF(V). Let M be a manifold and let V -> M be a vector bundle over M. For each m € M, the fiber of V over m will be denoted Vm. For each m € M, let Xm Q Vn®nV£ denote the subset of V £ ® R V £ = BF(V m ) corresponding to MinkSBF(V m ) C BF(V m ). Then X := ( J Xm C F* ® V* is a (smooth) fiber subbundle of the bundle V* ® V* -> M. A Lorentz bundle metric on V is a smoothly varying system of Minkowski quadratic forms, one on each fiber of V. More precisely, it is a smooth section of the fiber bundle X —• M. A Riemannian bundle metric on V is defined similarly, except that the quadratic forms are required to be positive definite rather than Minkowski. Note that a Riemannian (resp. Lorentz) metric on a manifold is a Riemannian (resp. Lorentz) bundle metric on its tangent bundle. If V is a vector bundle over a manifold M and if W is a vector subbundle of V, then a vector bundle complement to W in V is a vector subbundle W of V such that, for all m € M, we have both Wm + W'm = Vm and

wmnwu = {0}, L E M M A 9.1.3 Let N be a connected normal closed subgroup of two connected Lie groups G and H. Let H act on a manifold M. Assume that M admits an N-invariant Lorentz metric. Assume that Ad n (G) C Adn(-H"). Then the G-action on G x # M preserves a Lorentz metric.

274

Examples of Nonproper

Actions

Proof. For all g £ G, let [g] denote the image of g under the canonical map G —> G/N. For all g £ G and m £ M , let [g, m] denote the image of (g, m) £ G x M in M ' := G xN M. Define n : M' -> G/N by n([g,m]) = [g]. Let i : M -*• M ' be defined by t(m) = [l G ,m]. Let M 0 := 7r - 1 ([l G ]). Then, by Lemma 7.13.2, p. 251, • Mo is an TV-invariant closed submanifold of M; • i(M) = Mo; and • i: M -> M 0 is an A^-equivariant diffeomorphism. Let V := (TM')|M 0 , so V -*• M 0 is a vector bundle. Let T denote the set of smooth sections of V -»• Mo- The action of G on M ' differentiates to an action of G on TM'. Then V C W is AMnvariant. Define an Abaction on T by (n.7)(m) = n(rf(n~1m)). We wish to show that the G-action on TM' preserves a Lorentz bundle metric. Since M ' = GM0, it follows that TM' — GV. Moreover, for all g,g' £ G\N, we have: [g] = [g'] iff gM0 n g'M0 ? 0 iff gV n p'V ^ 0. It therefore suffices to show that the TV-action on V preserves a Lorentz bundle metric. Since the Abaction on TM preserves a Lorentz bundle metric, it follows that there exists an AT-invariant Lorentz bundle metric on TMo- It therefore suffices to show that there is an AT-invariant vector bundle complement C to TMo in V such that the vector bundle C —> Mo admits an AMnvariant Riemannian bundle metric. For X € g, we define X+ := —XM>- Then, for all m € M', we have X~i = [d/dt]t=o[(exp(tX))m]. For X 6 g, let X+ := X+\M0, so X+ € T. For all n£N, for all X € 0, we_have ((Adn)X)+ = n.(X+). For each Y £ h, we define F x := —YM- Then, for all m G M , we have Ym = [d/dt]t==o[(exp(tY))m]. Recall that i : M ->• Mo is an AT-equivariant diffeomorphism. For each Y £ h, let F x := t * ( y x ) , so F x is a section of T M 0 -> M 0 , so y x € T. For all n £ N, for all Y £ h, we have ((Adn)F)x = n . ( y x ) . For all P £ n C g n h, we have P + = Px. Let A; := (dim(G)) - (dim(Ar)). Then k = (dim(M')) - (dim(M)). Let Xi,..., Xk be a collection of linearly independent elements of g such that RXi -\ 1- RXk is a vector space complement to n in 0. Then the span S of X*,..., X£ is a vector bundle complement to TMo m V. Let := Ad n : G -> Aut(n) and V := Ad n : H -»• Aut(n). Then d(/> = ad n : 0 - • 5er(n) and dt/' = ad n : ^ ->• 5et(n). By assumption, 4>(G) C V ( ^ ) , so ( # ) ( 0 ) C ( # ) ( h ) . Let I := {1, . . . , * } . For each i £ I,

Restriction-Induction

275

choose Yi £ t) such that (dij))(Yi) - {d(j))(Xi). Then, for all i G / , for all t G E, we have i[>(exp(tYi)) = (^(exp(iXj)). For all i G I, let At := X/" - Y* G T. Let C denote the span of Ai,...,AkFor all i, Y* is a section of TM0 ->• M 0 . So, since 5 is a vector bundle complement to TM0 in V, it follows that C is a vector bundle complement to TMQ in V, as well. Moreover, since X\,..., .X^ is a framing of S, we conclude that A±,..., Ak is a framing of C. Give C the unique Riemannian bundle metric 7 such that the framing Ai,...,Ak is orthonomal. We wish to show both that C is iV-invariant and that 7 is iV-invariant. It suffices to show, for all i £ I, that Ai is iV-invariant. Let iVo be an open neighborhood of ljv such that N0 C exp(n). Then, since N is connected, it follows that N0 generates N. Fix i G I and n £ No. We wish to show that n.Ai = Ai. For all t G K, set /i t := exp^Fj) and gt := exp(tXj). For all t € R, Mh)

= i>{exp{tYi)) = 4>(exp(tXi)) = <j>(gt).

So, for all t G R, for all Z G n, we have (Ad ht)Z = (Adgt)Z. Differentiating this with respect to t and then setting t = 0, we see, for all Z G n, that fTi, Z] = [Xu Z], which gives (ad Z)Yt = (ad Z ) ^ . Then, for Z G n, y , (ad Z)nYj _ y > (ad Z)"Xj ^ n! ~ ^ n! n=l

n=l

SO

[(Ad(expZ))y i ] - Yi = [(Ad(expZ))Xi] -

Xt.

Since n G iV0 C exp(n), fix Z 0 G n such that n = exp(Z 0 ). Then, replacing Z by Zo above, we get ((Adn)Yi) -Yi = ((Adn)Xi) For all P G n, we have P+ = Px. [((Adn)Yi) - Y^ so (n.(Y*))

- Xt G n.

Then = [((Adn)Xi)

-

Xi}+,

- Y* = (n.(X+)) - X+. Then (n.Ai) -M

so n.Ai = At.

= [n.(X+ - Y*)} - (X+ -Y*)=

0, D

276

Examples of Nonproper

Actions

In Lemma 9.1.3, p. 273, if we drop the requirements that G, H and N are connected and replace "Ad„(G) C Ad„(iJ)" by "Intjv(G°) C IntN(H0)", then the resulting statement (which is Corollary 4.4 of [A99a]) is true. Note that this generalizes Lemma 9.1.3, p. 273, because, if G, H and N are connected, then "Ad n (G) C Ad„(.ff)" and "IntAr(G0) C Intjv(^ 0 )" axe equivalent statements. We do not need this generalization here, and its proof is slightly harder. COROLLARY 9.1.4 Let G be a connected Lie group and let H be a connected closed subgroup ofG. Let H act isometrically on a Lorentz manifold M. Assume, for some ideal h' of g, that I) + h' — g and that h D h' = {0}. Then the induced G-action on G x # M preserves a Lorentz metric. Proof. We have [h,h'] C h H h' = {0}, so ad,,(h') = {0}. Then we have adf,(g) = ad(,(f) + h') = ad(,(h). Then Adf,(C?) = Adt,(H), so, by Lemma 9.1.3, p. 273 (with N := H), we are done. •

9.2

Examples constructed using a quadratic form in the Lie algebra

The following is Lemma 7.4 of [A99b]. It gives an algebraic condition which implies that a group will have some interesting Lorentz dynamics. L E M M A 9.2.1 Let G be a connected Lie group and let Q € MinkQF(g). Let G act on MinkQF(g) via the Adjoint representation. Assume that Stabg(<5) is noncompact. Then there exists a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action of G on a connected Lorentz manifold. Proof. Let T be an infinite cyclic subgroup of StabQ(Q) such that T is discrete in G. Let S1 denote the circle, as a Riemannian manifold. Let T act by irrational rotation on S1. The T-action on S1 is then free, minimal, properly ergodic and isometric. Let M := G xrS1. By Conclusions (3), (7) and (9) of Lemma 7.13.1, p. 250, the G-action on M is free, minimal and properly ergodic. By Corollary 7.11.14, p. 247, the action is then nontame and orbit nonproper. It remains to show that the G-action on M preserves a Lorentz metric. Let e := IG- Let h! be the left-invariant Lorentz metric on G such that h'e = Q. Then, by Lemma 4.3.4, p. 78, h' is right-invariant, as well. Let h"

Examples constructed using a quadratic form in the Lie algebra

277

be the Riemannian metric on S1. Then h" is T-invariant. As h! is Lorentz and h" is Riemannian, it follows that h' x h" is a Lorentz metric on GxS1. Let 7r : G x S1 -* M be the canonical map. Let T act on G x S1 by -y(g,s) = (ff7 _1 ,7s), so that M = T\(G x S1). Then h! x h" is invariant under this action of Y on G x S1, so there is a unique Lorentz metric h on M such that 7r*(/i) = h' X h". The left G-invariance of /i' implies that h is G-invariant. • L E M M A 9.2.2 Let G be a connected Lie group and let N denote the nilradical of G. Let Ni be a connected closed subgroup of Z{N). Assume that JVi is normal in G and that N\ is simply connected. Let Q € QF(ni) be positive definite or Minkowski. Assume that Ad n i (G) C CO°(Q). Then G admits a locally faithful, isometric action on a Lorentz manifold such that a noncompact connected closed subgroup of N± stabilizes a point. In particular, G G ONV • Proof. Let d := dim(tii). Let n := d + 2. For all x e M d x l , let x € E n x l be the matrix such that • the (1,1) and (n, 1) entries of x are both 0; and • for all integers j € [2,d + 1], the (j, 1) entry of x is equal to the (j — 1,1) entry of x. For all x 6 R d x l , let Bx be the nxn

matrix such that

• the nth column of Bx is x; • the first row of Bx is the transpose of — x; and • all other entries of Bx are 0. Choose i £ {0,1} such that the signature of Q is (d — i,i). If i = 0, then define Q' £ Q F ( R d x l ) by Q(aie[d) + • + ade{dd)) = a\ + • • • + a\. If i = l, then let Q' := Qd £ Q F ( K d x l ) . Fix a vector space isomorphism p : ni ->• E d x l under which Q £ QF(ni) corresponds to Q' € QF(IR d x l ). Define Q 6 Q F ( R n x l ) by the rule: For all s . t e R , for all x € E d ,

Q(s4 n ) + £ + te^) = 2st + Q'{x). Then, as the signature of Q' is (d — i,i), it follows that the signature of Q is (d-i + l,i + l). For v £ ni, let Cv := exp(B p(v) ) e E n x n . Let V0 := {G„ | v G m } . Since both N\ and VQ are connected, simply connected, d-dimensional, Abelian

278

Examples of Nonproper Actions

Lie groups, it follows, from Lemma 4.8.3, p. 95, that they are isomorphic. We may therefore assume that Ni = VoFor a > 0, for all g 6 SO°(<3')> let Aag be the n x n matrix such that • • • •

the (1,1) entry of Aag is a; the (n, n) entry of Aag is 1/a; the middle d x d block of Aag is g; and all other entries are 0.

Let H := {AagCv\a > 0,g € SO°(Q),w G m } . Then H C SO°(Q). Moreover, iVi is a normal subgroup of G and of H. If i = 0, then • letM:=KBXl; • give M the flat Lorentz metric corresponding to Q; and • and let mo := 0 € M. If z = 1, then . let M : = { i £ M" x l |Q{x) = - 1 } ; • give M the Lorentz metric inherited from the flat signature (d, 2) metric on E " x l corresponding to Q; and • let m 0 := 4 " } - e ^ G M. Matrix multiplication gives a locally faithful, isometric action of H on M and computation shows that S t a b ^ (mo) is noncompact. We have Ad n i (G) C CO°(Q) = Ad n i (iT). Then, by Lemma 9.1.3, p. 273, the G-action on M' := G Xjy, M preserves a Lorentz metric. By (7) of Lemma 7.13.1, p. 250, the G-action on M ' is locally faithful. By (5) of Lemma 7.13.1, p. 250, for some m' € M', Stab/v^m') = StabjVi(mo); in particular, Stab^r^m') is noncompact. Then, by Lemma 7.11.12, p. 246, the G-action on M' is orbit nonproper, so G € OAfV. D

9.3

A nilpotent Lie group without nonproper Lorentz dynamics

In Example 9.3.1, p. 279 below, we establish the existence of a connected two-step nilpotent Lie group TV which does not admit a nonproper, locally faithful, isometric action on a connected Lorentz manifold. As a corollary, we see, in Example 9.3.2, p. 280 below, that that if two connected Lie

A nilpotent Lie group without nonproper Lorentz dynamics

279

groups are locally isomorphic, and if one of them admits a dynamically complicated smooth action preserving a Lorentz metric, it does not follow that the other does. The following is Example 7.1 of [A99a]. E X A M P L E 9.3.1 There is a connected two-step nilpotent Lie group N with compact center such that N admits no locally faithful, nonproper, isometric action on a connected Lorentz manifold. Proof.

Let n be the 8-dimensional Lie algebra with basis

X,X',X",

Y,Y',Y",

Z,Z'

such that X", Y", Z, Z' € j(n), such that [X,Y] = Z,

[X',Y'] = Z',

[X,X'] = X",

[Y,Y'] = Y",

and such that [X,Y'] = 0, [X',Y] = 0. By Corollary 4.9.19, p. 104, let N be a connected Lie group with Lie algebra n such that Z(N) is compact. Since n is two-step nilpotent, it follows that N is as well. Let M be a connected Lorentz manifold and assume that iV acts nonproperly and isometrically on M. We aim for a contradiction. Choose convergent sequences mi and mj in M and choose a sequence rij in AT such that ni —> co in N and such that, for all i, we have rijmj = mj. Let moo := hm rrii and m'^ :— lim m'j. Let H'^ := Stab^ r (m' 00 ). »—>oo

i—>oo

Let N0 := Z(N). Then iVo is compact and {X",Y",Z,Z'} is a basis of n 0 . Let q : N -> N/N0 be the canonical Lie group homomorphism. Define / : E 4 -> N by / ( a , b, c, d) = exp(aX + bX' + cY + dY'). Then g o / : I 4 -> N/No is an isomorphism of connected, simply connected Abelian Lie groups. In particular, q o f : M4 -> N/N0 is surjective, so N = (/(E 4 ))iVo. Choose a sequence z, in A'o and a sequence (aj,6j,c,,di) in R4 such that, for all i, we have ni = (f(ai,bi,a,di))zi. Then, for all i, niZ^1 = / ( a j , bi, Ci, di) = exp(ajX + fyX' + ciY + diY'), so Ad n (n i 2rr 1 ) = Ad„(exp(ajX + biX' + ciY + diY')). z~l e N0 = Z(N), so Ad n (z t - 1 ) is trivial, so Ad n (ni) = AdniniZ-1)

For all i, we have

= exp(ad„(aiX + b^X' + ctY + diY1)).

280

Examples of Nonproper

Actions

Then, for all i, we compute (Ad m)X = X-

hX" -CiZ

and

(Ad ni)X'

= X' + aiX"

- dtZ'

and

(Ad m)Y' = Y' + btZ' + aY".

and (Ad m)Y = Y + aiZ-

diY"

Since n* -> oo in N, since {z~l}i C N0 and since N0 is compact, it follows that ntz^1 —> oo in N. For all i, niZ^1 = /(ai,6j,c,,dj), so (a,i,bi,Ci,di) —> oo in 1R4. Passing to a subsequence, we have either that (bi, Ci) -> oo in R2 or that (a*, dj) -> oo in M2. We shall assume the former; the proof in the latter case is similar. Passing to a subsequence, choose (6, c) € R2 such that (bi,Ci) -*• (6, c) in E 2 . Then (Adrc^X -* -bX" - cZ and (Adn«)r' -»• bZ' + cY". Let S := E ( - 6 X " - cZ) + R(bZ' + c y " ) . Then S is Kowalsky for Adn(r»i). By Lemma 7.10.1, p. 236, 5 is strongly lightlike at m'. So, by (1) of Lemma 7.8.3, p. 228, choose W € 5\{0} such that W is strongly vanishing at mj^. Then, by Lemma 7.6.1, p. 225, t t-> exp(tW) : R -> G is proper, so {exp(tW)}t£R is not precompact in G. However, W € no and JV0 is compact, so we have a contradiction. • E X A M P L E 9.3.2 There exist connected Lie groups G and H such that g is Lie algebra isomorphic to h, such that G £ hlV and such that H admits a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action on a connected Lorentz manifold. Proof. Let JV be as in Example 9.3.1, p. 279. Let G := N. By Example 9.3.1, p. 279, G $. MV. Let H be the universal covering group of G. Since h is a nilpotent Lie algebra, we see, from Lemma 4.14.3, p. 138, that 3(h) 7^ {0}. So, as H is connected, it follows that Z(H) is nontrivial. So, as H is connected and simply connected, it follows, from Lemma 4.14.6, p. 140, that Z(H) is noncompact. By Lemma 5.4.1, p. 156, H is Ad-nonproper. Then, by Lemma 7.14.13, p. 260, we are done. •

9.4

Groups with SO(n, 1) or SO(n, 2) as a local direct factor

If a Lie group H acts isometrically on a Lorentz manifold M, and if H' is another Lie group, then the direct product H x H' acts on M x if' by

Groups with SO(n, 1) or SO(n,2) as a local direct factor

281

the formula (g,g').(m,x') — (gm,g'x'). Because H' carries a left-invariant Riemannian metric, this action preserves a Lorentz metric. Moreover, if the if-action on M is nonproper (resp. orbit nonproper), then the action of H x H' on M x H' is nonproper (resp. orbit nonproper). Thus certain dynamical assumptions on a direct factor of a Lie group G will imply similar dynamical conclusions about G itself. By contrast, at the Lie algebra level, by Example 9.3.2, p. 280, it is possible for a connected Lie group G $ NV to be locally isomorphic to a connected Lie group H € ONV* • So, if G and H are connected Lie groups, if h | g and if H G ONV?, we cannot conclude that G € MV. Thus, relatively strong dynamical assumptions on a local direct factor of a group G do not even imply weak dyamical conclusions for G. Nevertheless, we see below (in Lemma 9.4.1, p. 281 and Corollary 9.4.2, p. 281 and Lemma 9.4.5, p. 283), that knowledge that a connected Lie group G has a specific local direct summand sometimes implies that G admits nontrivial Lorentz dynamics. L E M M A 9.4.1 LetG be a connected Lie group and suppose that sl 2 (IR)|0. Then G admits a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action on a connected Lorentz manifold. Consequently, GeOMVf. Proof. Choose ideals () and h' of g such that g = h + h', such that hnh' = {0} and such that f) = sl2(M). Let H be the connected Lie subgroup of G corresponding to h. By Lemma 4.10.32, p. 124, H is noncompact. By Corollary 7.14.4, p. 253 (with G and H interchanged), H is closed in G. Let F be the Killing form of f) and let F' be any positive definite quadratic form on h'. Let Q := F © F'. Then H C S := Stab G (Q). Thus 5 contains a noncompact closed subgroup, and is therefore noncompact. The result then follows from Lemma 9.2.1, p. 276. • COROLLARY 9.4.2 Let G be a connected Lie group and suppose either that that so(2,1) |g or that so(2,2) \g. Then G admits a free, minimal, properly ergodic, nontame, orbit nonproper, isometric action on a connected Lorentz manifold. Consequently, G 6 ONV*. Proof. By Coincidence X.6.4(i), p. 519 of [He78], so(2,1) S s( 2 (R). By Coincidence X.6.4(x), p. 520 of [He78], so(2,2) S s!2(IR) ®s[ 2 (lR). Thus Corollary 9.4.2, p. 281 follows from Lemma 9.4.1, p. 281. •

282

Examples of Nonproper

Actions

L E M M A 9.4.3 Let n > 3 be an integer. Let H be a connected Lie group and assume that f) is Lie algebra isomorphic to so(n, 1). Then there exist a connected Lorentz manifold M, a point mo € M and a locally faithful isometric action of H on M such that Stab#(mo) is noncompact. Proof. Let d := n + 1. Let M := Rdxl with the flat Lorentz metric corresponding to Qd. Let m 0 := 0 £ Rdxl = M. Let H0 :- SO°(Qd). Then HQ has a natural linear action on M; in particular mo is a fixpoint. Since HQ has trivial center, by Lemma 4.9.29, p. 108, there is a covering map 7r : H -> H0. Pulling back the action of HQ on M, we have a locally faithful action of H on M such that mo is a fixpoint. Then StabH(mo) = H and, by Lemma 4.10.32, p. 124, H is noncompact. • L E M M A 9.4.4 Let n> 3 be an integer. Let H be a connected Lie group and assume that \) is Lie algebra isomorphic to so(n, 2). Then there exist a connected Lorentz manifold M, a point mo £ M and a locally faithful isometric action of H on M such that Stab#(mo) is noncompact. Proof. Let Q be a signature (n,2) quadratic form on V :— Rn+2. Let M' := {v G V | Q(v) — - 1 } . Then M' is connected and M' inherits a Lorentz metric from the flat (n, 2)-pseudoRiemannian structure on V corresponding to Q. Let (•, •) be the polarization of Q. For all v € ^ \ { 0 } , let Rv : V -» V be the reflection defined by Rv(w) = w — 2((v,w)/(v,v))v; then Rv € 0(Q). Let Go := SO°(Q). Then Go acts faithfully and isometrically on M'. For all m,m' € M', there exists v eV such that Rv(m) — ml. Thus 0(Q) acts transitively on M'. Then, by Lemma 2.7.4, p. 28, for all m € M', we have that g i-> pm : O(Q) -> M ' is an open map. So, since Go is open in O(Q), we conclude, for all m £ M', that Gom is an open subset of M'. Since Go has finite index in O(Q), it follows that M' has only finitely many Go-orbits, all of which are open. So, as M' is connected, we see that Go acts transitively on M'. Moreover, for all m 6 M', StabG(rci) is noncompact. Let / be the (n + 2) x (n + 2) identity matrix. If n is even, then Z(Go) = { / , - / } . If n is odd, then .Z^Go) = {/}. In either case, the action of Z{GQ) on M' is free, so there is a unique manifold structure on M := M'/(Z(Go)) such that the canonical map M' -¥ M is a smooth covering map. Let -K : M' -> M be the canonical map. Let 7' denote the Lorentz metric on M'. Since 7' is (^(Go))-invariant, let 7 be the unique Lorentz metric on M such that 7r*(7) = 7'. Let G 0 := G Q / ( Z ( G O ) ) .

Groups with SO(n,l) or SO(n,2) as a local direct factor

283

Then G'0 acts transitively and isometrically on M, and, for all m' G M, S t a b c (m') is noncompact. Because G'0 is simple and because the action of G'Q on M is nontrivial, it follows that the kernel of this action is discrete, i.e., that this action is locally faithful. By Corollary 4.9.17, p. 104, G'0 is centerfree. Then, by Lemma 4.9.29, p. 108, there is a covering map H -> G'0. Pull back the action of G'0 on M to an H-action on M. The resulting action is again locally faithful and isometric and has noncompact stabilizers. • By Corollary 9.4.2, p. 281, if so(2,1) 10 or if so(2,2) | g, then we have G e OMVf. Let n > 3 be an integer. In Theorem 10.5.1, p. 291, we will show that SO°(n, 1) $ OAfVlf and that SO°(n,2) £ OMVlf. However: L E M M A 9.4.5 Let n > 3 be an integer. Let G be a connected Lie group and suppose either that so(n, 1) | g or that so(n,2)|g. Then G admits a locally faithful, isometric action on a connected Lorentz manifold such that the stabilizer of some point is noncompact. Consequently, G 6 OAfV. Proof.

Choose ideals () and h' of g such that

• fl = h + h'; • rj n h' = {0}; and • either f) S so(n, 1) or h = so(ra,2). Let H be the connected Lie subgroup of G corresponding to h. By Corollary 7.14.4, p. 253 (with G and H interchanged), H is closed in G. By Lemma 9.4.3, p. 282 and Lemma 9.4.4, p. 282, let M be a connected Lorentz manifold, letTOO€ M and let H act on M isometrically so that Stabij(mo) is noncompact. By Corollary 9.1.4, p. 276, the induced action of G on M' :— GXHM preserves a Lorentz metric. By (7) of Lemma 7.13.1, p. 250, the G-action on M' is locally faithful. By (5) of Lemma 7.13.1, p. 250, there is some m' € M ' such that Stabc(m') = Stabff(m 0 ). In particular, Stabc(m') is noncompact. Then, by Lemma 7.11.12, p. 246, the G-action on M ' is orbit nonproper. Then G € OMV. • L E M M A 9.4.6 Let n > 2 be an integer. Let G be a connected Lie group and suppose either that so(n, 1) | g or that so(n, 2) | g. Then G £ OMV. Proof. We have OMVj C OMV, so, if n = 2, then, by Corollary 9.4.2, p. 281, we are done. If n > 3, then we are done by Lemma 9.4.5, p. 283. •

284

9.5

Examples of Nonproper

Actions

Groups with a normal subgroup isomorphic to M

L E M M A 9.5.1 Let G be a connected Lie group and suppose that G contains a normal subgroup which is isomorphic to the additive Lie group R. Then G admits a free, minimal, properly ergodic, nontame, orbit nonproper isometric action on a Lorentz manifold. In particular, G 6 OAfV*. Proof. Let R+ denote the multiplicative group of additive reals and let E+ act on R by multiplication. Let H :=R+ K R. Let N := {1} K E. Then Aut°(n) = Adn(H). We may assume that N is a normal subgroup of G. Then Ad n (G) C Aut°(n) = Adn(H). By Corollary 7.11.14, p. 247, properly ergodic implies nontame and orbit nonproper. So, by (3), (7) and (9) of Lemma 7.13.1, p. 250 and by Lemma 9.1.3, p. 273, it suffices to find an action of if on a manifold such that the restriction of this action to N is free, is minimal, is properly ergodic and preserves a Lorentz metric. Let S := SL 2 (E). Define a homomorphism / : H -¥ S by

/(M))

y/a

by/a

0

1/y/a

Then / : H —> S is injective and proper, and f(N) is unipotent. By [Bor63], every semisimple Lie group admits a cocompact discrete subgroup. Let T be a cocompact discrete subgroup of 5. By Theorem 1.12, p. 22 of [R72], r has no nontrivial unipotent elements. Let M := S/T. Let S act on M via the standard transitive action. Pull this action back along / : H -»• S to an action of H on M. Let M ' := G xN M. By [Ve75], the AT-action on M is minimal. By Moore's Theorem (see Theorem 2.2.19, p. 22 of [Zi84] or Theorem IV.12, p. 118 of [EN89] or Theorem 1, p. 183 of [A98]), we see that the 5-action on M is mixing with respect to Haar measure on M. So, as / : H -4 S is a proper homomorphism and as iV is a noncompact subgroup of H, we see that the iV-action on M is mixing, as well. Then the AT-action on M is properly ergodic. Since every element of f(N) is unipotent, while T has no nontrivial unipotent elements, it follows, for all s € S, that (f(N)) C\ sTs-1 = {Is}Therefore, the restriction to N of the .ff-action on M is free. •

Chapter 10

Semisimple Groups Admitting a Nonproper Action

In this chapter, we give an account by examples of the work of N. Kowalsky in [Ko96].

10.1

Locally free actions of SL3 (ffi.)

Let G :— SLs(K). In this section, we demonstrate that any locally free, isometric action of G on a connected Lorentz manifold is proper. Since we focus on such a specific group, we can give a concrete proof, but the reader should bear in mind that the result generalizes; see Theorem 10.5.1, p. 291. Let S := SBF(g). Let V denote the set of vector subspaces of g. For F G S, let 1F := {V G V|F(V,V) = {0}} be the set of all F-isotropic subspaces of Q. For all F G S, let dp := max{dim(V) | V G Ip} be the maximum dimension of an F-isotropic vector subspace of V. We define Si := {F € <S I d,F < 2}. Then <Si is an open subset of the vector space S. The Adjoint representation of G on g induces a continuous action of G on S defined by the rule: For all F G <S, for all g G G, we define g.F G <S by (g.F)(X,Y) = F ( ( A d 5 ~ 1 ) X , (Adg'^Y). Then Si is G-invariant. Give Si the topology inherited from the vector space topology on S. Then the G-action on Si is continuous. L E M M A 10.1.1 The Adjoint action of G on Si is proper. Proof. Assume, for a contradiction, that the G-action on Si is nonproper. For all j,k G {1,2,3}, let Ejk := E$. For all A,fi,v e t , we define 285

286

Semisimple

Groups Admitting

a Nonproper

Action

diag(A,v,fi) := XEu + fj,E22 + vE33. Let A+ := {diag(A,A*,z/) | A > fi > v > 0, Xfiu = 1} C G and let K := SO(3) C G. Then G = KA+K. By Lemma 7.9.1, p. 235, we may choose a sequence Oj in A+ and a convergent sequence F, in <Si such that Fl := ai.Fi is convergent in <Si and such that a, —> oo in G. Let Foo := lim Fi G Si and let F^ := lim F/ € Si. i—• oo

i—>oo

For all i, choose Xi,fii,Vi G R such that a* = diag(A,,/Lti,i/i). Then, for all i, we have Aj > yu» > i/j > 0 and Xi^Vi = 1. Since a* —• oo in A + , after passing to a subsequence, we have either that Aj/^j —> +oo or that /ij/i'i —¥ +oo. We assume the former; the proof in the latter case is similar. For all i, let on := Xi/fn and let ft := Xi/vi. Then QJJ -» +oo. Moreover, for all i, we have \n > Vi, so ft = a^fii/vi) > a.i. Then ft —• +oo. Let X := E12 and Y := Ei3. Then X , y G g. For all i, we have ( A d a " 1 ) * = ( l / a 4 ) X and (Ada'^Y = (l/ft)Y, so FUX,Y)

=

{l/ai)(l//}i)(Fi(X,Y)).

Letting i -> oo, we get F^(X,Y) = 0 • 0 • (Foo(X,Y)) = 0. Similar arguments show that F^(X,X) = 0 and that F^(Y,Y) = 0. Then I X + WY G 1FL, so dF> > 2. Then F^ $ Si, contradiction. D T H E O R E M 10.1.2 LetG act isometrically on a connected Lorentz manifold M. Assume that the G-action on M is locally free. Then the G-action on M is proper. Proof. For all m € M, let (•, • ) m be the Minkowski symmetric bilinear form on TmM, and define Bm G SBF( fl ) by Bm(X,Y) = {Xm,Ym)m. For all m G M, since (•, • ) m is Minkowski and X >-t Xm : g ->• TmM is injective, it follows, from Corollary 6.2.3, p. 176, that Bm G <Si. The map m •-> Bm : M -> Si is G-equivariant and continuous. By Lemma 10.1.1, p. 285, the G-action on <5>i is proper. Then the G-action on M is properD T H E O R E M 10.1.3 Let G act isometrically on a connected Lorentz manifold M. Let Mo be a compact subset of M. Let gi be a sequence in G and let mi be a sequence in Mo- Assume that gi —> oo in G. Assume, for all i, that girrii G MQ. Then, for some m G MQ, we have that Stab G (m) ^ {1G}Proof.

Say, for contradiction, that for all m G M 0 , Stab^(m) = {1G}-

Locally faithful actions

o/SL3(K)

287

For all m G M , let (•, • ) m be the Minkowski symmetric bilinear form on TmM. For all m G M, define B m G SBF( 0 ) by 5 r o ( X , F ) = (Xm, Ym)m. For all m G M, since X i-» X m : g -4 T m M is injective, it follows, from Corollary 6.2.3, p. 176, that Bm G Si. The map m \-¥ 2?m : M —> <Si is G-equivariant. So, for all i, we have 9i-Bmi = Bgimi. So, since J3 m ; and # 5 i m i are both precompact in Si, while gi —> 00 in G, we conclude that the G-action on Si is nonproper, contradicting Lemma 10.1.1, p. 285. • COROLLARY 10.1.4 Let G act isometrically on a connected Lorentz manifold M of finite Lorentz volume. Then the action of G on M is trivial. Proof. Let /x be the measure on M corresponding to the Lorentz volume. Then \i is in the smooth measure class on M. Assume, for a contradiction, that the G-action on M is nontrivial. Then, by Lemma 7.5.2, p. 222, fi({m E M | Stab G (m) = G}) = 0. Then, by Proposition 7.4.23, p. 218, Mx := {m e M \ S t a b ^ m ) = {la}} is a [iconull subset of M . Using Theorem 17.11, p. 107 of [Kec95], let M0 be a compact subset of M such that MQ C M I and such that ^(M 0 ) > 0. By Lemma 7.4.24, p. 219, Mo is recurrent for the G-action on M . Then, by Theorem 10.1.3, p. 286, for some m G M0, Stabo(m) ^ { I G } , so m £ Mi. Since Mo C Mi, we have a contradiction. •

10.2

Locally faithful actions of SL 3 (E)

Let G := SL 3 (E). In this section, we improve on Theorem 10.1.2, p. 286 and show that G has no nontrivial, nonproper Lorentz actions. This is due to N. Kowalsky; see Theorem 5.2, p. 630 of [Ko96]. T H E O R E M 10.2.1 Any nontrivial, isometric action ofG on a connected Lorentz manifold is proper. Proof. Assume, for a contradiction, that G acts nontrivially, nonproperly and isometrically on a connected Lorentz manifold M. Since G is simple, it follows that the kernel of the action of G on M is discrete. That is, the G-action on M is locally faithful. For all j,k G {1,2,3}, let Ejk := E$. For all s,t,u G E, we define diag(s,t,u) := sEu + tE22 + uE33.

288

Semisimple

Groups Admitting

a Nonproper

Action

We define A := {diag(s,t,u) | s > 0,i > 0,u > 0, stu — 1} and we also define A+ := {diag(s, t, u) | s > t > u > 0,stu = 1}. Let K := 5 0 ( 3 ) . Then G = KA+K. So, by Lemma 7.9.1, p. 235, we may choose a sequence a, in A+ and a convergent sequence rtii in M such that aiirii is convergent and such that Oj —> oo in G. For all i, let mj := ajmj. Let moo : = lun mi i—>oo

and let m^, := lim mj. i—+oo

For all «, choose Hi £ a such that exp(-ffj) = Oj. Then ilj —>• co in a. Passing to a tail, we may assume, for all i, that Hi ^ 0. Choose a sequence ti in (0, oo) such that ijiZ; is precompact in o\{0}. Then ti -» 0 in E. Passing to a subsequence, choose ifoo G o\{0} s u c n t fla * *«-^« "^ -^oo in aFor all i, choose Pi,qt,ri G E such that Hi = diag(pi,<7j,r;). Choose p, q, r € E such that if^o = diag(p, g, r). As {if G o| exp(H) € A + } is a closed cone containing the sequence Hi, it follows that exp(£foo) € A+. Then p > q > r. Since ifoo 6 o C g = sl 3 (E), we see that p + q + r = 0. So, since Hoo ^ 0, we see either that p ^ q or that q ^ r. We will assume that q 7^ r, noting that the proof in the other case is similar. So, under our assumption, p>q>r. Let V := MEis + EE23. Let QK. denote the set of all elements of 0 which are Kowalsky for Ad 8 (a;). By Lemma 7.10.2, p. 237, V C 5*;. Claim A: For all X € F \ { 0 } , there exist Y,Z € fl\{0} such that (&dX)Y = Z, such that (adX)Z = 0 and such that (adZ)V ^ {0}. Proof of Claim A: Let s := M.(En - E22) + ^E12 + ME21 and let S be the connected Lie subgroup of G corresponding to s C Q. Then the action of Ad 5 on T^\{0} is transitive, so we may assume that X = E13. Let Y := £32 and Z := (adX)Y. Then Z = E12. Then {&dX)Z = 0. Moreover, (adZ)JS23 = Si3 ^ 0, so ( a d Z ) F ^ {0}. End of proof of Claim A. By Claim A and Lemma 7.8.11, p. 234, On the other hand, a s V C g/c, by Lemma 7.10.1, p. 236, V is strongly lightlike at m'^, contradiction. •

10.3

Locally faithful actions of Sp 3 (E) C E 6 x 6

Let V := E 6 x l . For all integers i G [1,6], let a := e\6); then e i , . . . ,e 6 be the standard basis of V. Let u : V x V -> E be defined by w(

aiei H

ha 6 e 6

,

Mi H

1- b6ee

)

= a\bQ + a2&5 + 0364 — 0463 — a5&2 — a6&i

Locally faithful actions o/Sp 3 (l) C l 6 x 6

289

Let G := {g € SL6(M) | Vv,w G V,u(gv,gw) = u(v,w)}. Then G = Sp 3 (E) has real rank = 3. In this section, we show that G has no nontrivial, nonproper Lorentz actions. This is originally due to N. Kowalsky; see Theorem 4.1, p. 625 of [Ko96]. T H E O R E M 10.3.1 Any nontrivial, isometric action ofG on a connected Lorentz manifold is proper. Proof. Assume, for a contradiction, that G acts nonproperly and isometrically on a connected Lorentz manifold M. Since G is a simple Lie group and since the G-action on M is nontrivial, we see that the kernel of this action is discrete, so the G-action on M is locally faithful. For all integers j,k G [1,6], let Ejk := E$. For all s,t,u € E, let diag(s, t, u) := sEu + tE22 + uE33 - uEi4 - tE55 - s £ 6 6 . Let o := {diag(s,i,u) \s,t,u G E}. Let A := exp(a). Let $ be the set of roots of a on g. For all <j> £ $, let $$ denote the (/>-rootspace of a on 0. Define a, /?, 7 € a* by a(diag(s,i,u)) = 2u, /?(diag(s, t, u)) =t — u, 7(diag(s, t, u)) = s - t. Then A := {a, /?, 7} is a base of the root system $. Let o + := { T e a | a ( r ) > 0 , / ? ( T ) > 0 , 7 ( T ) > 0 } . Let A+ :— exp(a + ). By Lemma 4.10.38, p. 127, choose a compact subgroup K of G such that G = KA+K. By Lemma 7.9.1, p. 235, choose a sequence a, in A+ and a convergent sequence m; in M such that Ojmj is convergent. For all i, let m'- := cumi. Let moo '•— lim "ii and let m' := lim m^. For all i, choose Hi G a + such that exp(Hi) = a*. Then Hi ->• 00 in o. Passing to a tail, we may assume, for all i, that i?j ^ 0. Let ij be a sequence in (0,oo) such that tiHt is precompact in a\{0}. Then ti -> 0 in E. Passing to a subsequence, choose Hoo G a\{0} such that tiHi -> i?oo in a\{0}. Choose a,b,c€R such that H^ = diag(a,b,c). Let jjjc be the set of all elements of g which are Kowalsky for Ad B (aj). Since, for all i, we have Hi e a+, we conclude that H^ G o + . That is, we have a(ifoo) > 0 and /?(i?oo) > 0 and 7(-Hr00) > 0. So, since a,/?,7 is a

290

Semisimple Groups Admitting a Nonproper Action

basis of o* and since # « , ^ 0, it follows that at least one of the following must hold: a(-ffoo) > 0 or /?(i?oo) > 0 or 7(#oo) > 0. Case 1: ^{HJ) > 0. Proof: Let P := Ei2-E56 and let Q := E13-Ei6. Then P G fl7 and Q € gp+y. Let V := MP + MQ. By Lemma 7.10.2, p. 237, we conclude that V C g*:C/ai'm A: For any X € V\{0}, there exist Y, Z £ fl\{0} such that ( a d X ) y = Z, ( a d X ) Z = 0 and (adZ)V ^ {0}. Proo/ o/ CTatm A: Let s : = M(i?23 ~ E45) + 9.(E32 — £54) + ^-(E22 — E33 + E44 — E55).

Let S be the connected Lie subgroup of G corresponding to s C g. Then the action of Ad 5 on V\{0} is transitive, so we may assume X — P. Let Y := E31 - EM and Z := (adX)Y. Then Z = E54 - E32. Then (ad Z)Q = E12 - E56 ^ 0, so (ad Z)V ^ {0}. End of Proof of Claim A. By Claim A and Lemma 7.8.11, p. 234, we see that V is not strongly lightlike at m'^ On the other hand, as V C JJJC, by Lemma 7.10.1, p. 236, V is strongly lightlike at m'^, contradiction. End of Case 1. Case 2: /?(.ffoo) > 0. Proof: This case is similar to Case 1, except that we interchange fj and 7. End of Case 2. Case 3: a(i/oo) > 0. Proof: Let P x := E3i, let Qi := E24 + E35 and let Ri := Eu + E36. Then Pi £ ga, Qx £ g Q+|3 and i?i € fla+;S+7. By Lemma 7.10.2, p. 237, Pi,Qi,Ri € fl/c. By Lemma 7.10.1, p. 236, choose an ordered Qd-basis B of Tm>^M such that gic is strongly lightlike with respect to B. By (1) of Lemma 7.8.3, p. 228, choose (a,/3) 6 K 2 \{(0,0)} such that X := aPi + PQi is strongly vanishing with respect to B. Let Y := £41 + #63 and Z := ( a d X ) F . Then Z = a ( £ 3 i - ^64) + P(E2i - E65)

and

(adX)Z

= 0.

By Lemma 7.8.10, p. 234, we see that Z is strongly lightlike with respect to B. Then, by Lemma 7.6.2, p. 225, we see that (ad Z)R\ = 0. However, (adZ)R1 = 2aE3i + fi(E24 + E35) ^ 0, contradiction. End of Case 3. D 10.4

Locally faithful actions of S U ( 2 , 1 )

Let V := C 3 * 1 . Let ex,e2,e3 be the standard complex basis of V. Let Q : V -> C be defined by Q{z\ei + z2e2 + z3e3) = 1z\zi + \z2\2. Let G := {g 6 SL3(C) | Vu € V, Q(gv) = Q(v)}; then G is Lie group isomorphic to SU(2,1). In this section, we show that G has no nontrivial, nonproper

Kowalsky's

theorems

291

Lorentz actions. This is originally due to N. Kowalsky; see Theorem 5.1, p. 629 of [Ko96]. T H E O R E M 10.4.1 Any nontrivial, isometric action ofG on a connected Lorentz manifold is proper. Proof. Assume, for a contradiction, that G acts nonproperly and isometrically on a connected Lorentz manifold M. For all j,k 6 {1,2,3}, we define Ejk := E$. For all s £ 1, we define diag(s) := sEu — SE33. Let 0 := {diag(s) | s € E}

and

o + := {diag(s) | s > 0}.

We define A+ := exp(o + ). By Lemma 4.10.38, p. 127, choose a compact subgroup K of G such that G = KA+K. By Lemma 7.9.1, p. 235, choose a sequence a, in A+ and a convergent sequence rrii in M such that aiirii is convergent. For all i, let rn'i := a;mj. Let rrioo := lim mi and let i->oo

m^j := lim m\. i—+00

For all i, choose Hi £ o + such that exp(ifj) = a*. Then Hi —> 00 in a. .Passing to a tail, we may assume, for all i, that Hi ^ 0. Let U be a sequence in (0,00) such that tiHi is precompact in o\{0}. Then U —> 0 in E. Passing to a subsequence, choose Hoo G a\{0} such that UHi —• Hoo in o\{0}. Choose p e M\{0} such that H^ = diag(p). Since {H e o| exp(i?) G A+} is a closed cone containing the sequence Hi, exp(i?oo) £ A+. Then p > 0. Let X := E12-E23 E g and Y := v / -T-Ei2+\/-T-E23 e 0- Let QK. denote the set of all elements of g that are Kowalsky for Ad B (aj). By Lemma 7.10.2, p. 237, X,Y E fljc- By Lemma 7.10.1, p. 236, gK is strongly lightlike at ml. Then, by Lemma 7.6.2, p. 225, it follows that [X, Y] = 0. However, matrix multiplication shows that [X, Y] = 2\/:::T.Ei3 ^ 0, contradiction. • 10.5

Kowalsky's t h e o r e m s

Let G be a noncompact connected simple Lie group. Let G act isometrically on a connected Lorentz manifold M. By using the techniques of §10.1, p. 285, one may prove the following result. For details, see Theorem 2.1, p. 615 of [Ko96].

292

Semisimple

Groups Admitting

a Nonproper

Action

T H E O R E M 10.5.1 Assume that Z(G) is finite. Assume that the action of G on M is locally free and nonproper. Then g is Lie algebra isomorphic tofi[2(E). By using the techniques of §10.2, p. 287, §10.3, p. 288 and §10.4, p. 290 one may give a proof of the following result, which is the main theorem (Theorem 5.1, p. 629) of [Ko96]. For details of this alternate proof, see Theorem 7.1 of [AOObj. T H E O R E M 10.5.2 Assume that Z(G) is finite. Assume that the action ofG on M is locally faithful and nonproper. Then there is an integer n>2 such that g is Lie algebra isomorphic either to so(n, 1) or to so(n + 1,2). Finally, using the techniques of §10.1, p. 285, we may generalize Corollary 10.1.4, p. 287, and prove: T H E O R E M 10.5.3 Suppose M has finite Lorentz volume and g is not Lie algebra isomorphic to skiM). Then the G-action on M is trivial.

Chapter 11

Groups with Action on a Compact Lorentz Manifold

Throughout this chapter, let Q denote the class of connected Lie groups admitting a locally faithful, isometric action on a compact connected Lorentz manifold. We consider the problem of classifying the collection Q of simply connected groups in Q. Our approach to this problem follows [AS97a]. In [Ze95], through different methods, A. Zeghib also obtains the same classification of Q. His result has the advantage that it also says something about elements of Q that are not simply connected; however it stops slightly short of an effective description of Q.

11.1

Local freeness

The next result asserts that any nontrivial lightlike flow on a Lorentz manifold is locally free. It is a special case of Lemma 3.2, p. 251 of [BEM88]. L E M M A 11.1.1 Let X be a Killing vector field on a connected Lorentz manifold M. Assume, for all m € M, that Xm is lightlike. Assume, for some mi € M, that Xmi ^ 0. Then, for all m € M, we have Xm ^ 0. Proof. Fix mo € M, and assume, for a contradiction, that Xmo = 0. Let V := TmoM and let B e SBF(y) denote the Minkowski symmetric bilinear form on V. Then (V, B) is a Minkowski vector space. Let Y be the linearization of X at mo; then Y is a linear vector field on V. For all v € V, let LV : V -» T„V be the vector space isomorphism defined by LV(W) = (d/dt)t=0(v + tw). For all v e V, let YJ := i " 1 ^ ) G V. Since Y is linear, it follows that v t-¥ YJ : V ->• V is a linear map. In particular, for 293

294

Groups with Action on a Compact Lorentz Manifold

all s G M, for all v € V, we have Ys'„ = s F j . By Lemma 7.4.5, p. 208, for all v eV,we have B(v, y„') = 0. Choose a convex open neighborhood U of 0 in V on which the exponential map exp mo is defined, such that U' := expTOo (U) is open in M and such that e :— exp mo \U : U —• U' is a diffeomorphism. Since X does not vanish everywhere on M and since X is a Killing vector field, it follows, from Lemma 7.4.10, p. 211, that {m G M | Xm ^ 0} is dense in M. By Lemma 7.4.4, p. 207, for all u 6 U, we have (de)(Yu) — Xe(uy Then {u G U \ Yu ^ 0} is dense in U. Fix u0 € U such that Y„0 ^ 0 and such that B(uo,uo) < 0. Let I := {t € R\ tu0 € f/}. For all t € J\{0}, we have (l/t)(YtU0)

= (l/t)(LtUo(YjUo)

= ituo{{llt){YL0))

For all t e I, we have (de)(Yi„0) = Xe(tUoy (de)( ttuo (y t ; o )) = (de)((l/t)YtU0)

=

HuoKJ.

Then, for all t € A { 0 } ,

= {l/t)Xe(tUo)

G T e ( t t I o ) M.

By assumption, X is everywhere lightlike, so we see, for all t G A { 0 } , that (de)(itu0(Yu0)) is lightlike. The set of lightlike vectors in TM is closed, so, letting t -> 0, we see that (de)(io(Y^0)) e TmoM is lightlike. For all v G T0(TmoM), we have (de)(i0(v)) - v. We conclude that Y^0 € TmoM is lightlike. That is, B{Y^,Y^0) = 0. So, since B(u0,Y^0) = 0 and since B(uo,uo) < 0, it follows, from Lemma 6.2.1, p. 176 (with ^o replaced by uo), that Y^0 = 0. Then YUo = tUo(Y^0) = 0, contradiction. • L E M M A 11.1.2 Let G be a Lie group acting locally faithfully on a connected Lorentz manifold M. Assume, for all X G g, for all m G M, that Xm is lightlike. Then dim(g) < 1 and the action of G on M is locally free. Proof. Claim 1: The G-action on M is locally free. Proof of Claim 1: Fix X 0 G fl\{0} and let X := ( X 0 ) M - Then X is a Killing vector field on M and we wish to show, for all m G M, that X m ^ 0. The G-action on M is locally faithful, so, for some mi G M, X m i ^ 0. So, by Lemma 11.1.1, p. 293, we are done. End of proof of Claim 1. Claim 2: dim(g) < 1. Proof of Claim 2: Fix mo G M. Since gmo is an isotropic vector subspace of TmM, it follows, from Corollary 6.2.3, p. 176, that dim(g mo ) < 1. By Claim 1, the map P H> Pmo : g -* gmo is injective. Then dim(g) < 1. End of proof of Claim 2. •

Local freeness

295

L E M M A 11.1.3 Let G be a connected Lie group acting isometrically on a compact pseudoRiemannian manifold M. Let mo G M. Let (•, •) be the nondegenerate symmetric bilinear form on TmoM. Define B £ SBF(g) by B(P, Q) — (Pmo,Qmo)Let X £ g and assume that the map adX : g —> g is nilpotent. Then B is (&dX)-invariant. Proof. For all m € M, let (•, • ) m denote the nondegenerate symmetric bilinear form on TmM. Then (•, •) = (•, -)m0- F ° r all m 6 M define Bm G SBF(g) by Bm{P,Q) = (Pm,Qm)m. Then B = Bmo. The map m \-t Bm : M -» SBF(g) is continuous, so, by compactness of M, we see that {Bm}mGM is a compact subset of the vector space SBF(g). For all i e B , let gt := exp(iX). Fix Y, Z G g. We wish to show, for all t G E, that B{{Adgt)Y, {Adgt)Z) = B(Y, Z). Define V : K -»• R by the equation ip(t) = B((Adgt)Y, (Adgt)Z). For all t € E, let mt := ^ 1 m 0 . Then V'(t) = Bmt(Y,Z). By compactness of {B m } m € M C SBF(fl), V(I^) is compact. By Taylor's formula, for all t, for all W G fl, ( A d ^ ) W = E £ i ( 1 / * 0 ( ( A d X ) W ) . For all W G g, by nilpotence of a d X : g —> g, we see, for all sufficiently large i, that (Ad Xyw = 0, and so £ i-+ (Adg t )V7 : E —> n is a polynomial map. So, as B : 0 x g —> E is bilinear, we conclude that ip : E —>• E is a polynomial map. So, by compactness of ip(R), ip : E -> E is constant. Then, for all t € E, we have B((Adgt)Y, (Adgt)Z) = i/>(t) = ip(0) = B(Y, Z). D L E M M A 11.1.4 LetG be a connected semisimple Lie group with no compact factors. Let G act isometrically on a compact pseudoRiemannian manifold M. Let mo G M. Let (•, •) be the nondegenerate symmetric bilinear form on TmoM. Define B G SBF(0) by B(X,Y) = (Xmo,Ymo). Then B is (Ad G)-invariant. Proof. Let 0o denote the set of nilpotent elements of 0. For all X G 0o, since a d X : g -t g is nilpotent, it follows, from Lemma 11.1.3, p. 295, that B is (adX)-invariant. Since G has no compact factors, it follows, from Lemma 4.12.4, p. 133, that 0o generates 0. Then B is (ad0)-invariant. So, as G is connected, the result follows. • L E M M A 11.1.5 Let G be a connected Lie group acting isometrically on a compact connected Lorentz manifold M. Let N be the nilradical of G. Assume that Z°(N) acts locally faithfully on M. Assume that g admits no simple direct summand. Assume that n has no nonzero Abelian direct summand. Then dim(3(n)) = 1 and the G-action on M is locally free.

296

Groups with Action on a Compact Lorentz

Manifold

Proof. For all m G M, let (•, • ) m denote the Minkowski symmetric bilinear form on TmM. For all m G M define Bm G SBF(g) by the equation Bm(P,Q) = {Pm,Qm)m- By Lemma 11.1.3, p. 295, we see, for all m G M, that Bm is (adn)-invariant. Since n has no nonzero Abelian direct summand, it follows from Lemma 4.9.38, p. I l l , that j(n) C [n, n]. Thus, for all m€ M, Bm(i(n),i(n)) C B m (j(n), [n,n]) = Bm(fo(n),n],n), so, since [3(n),n] = {0}, we get B m (j(n),3(n)) = {0}. Then, by Lemma 11.1.2, p. 294, dim(3(n)) < 1 and the action of Z°(N) on M is locally free. So, by Lemma 7.4.22, p. 217, the action of G on M is locally free, as well. Since g admits no simple direct summand, we conclude, from Corollary 4.10.3, p. 113, that g is not semisimple. Then, by Lemma 4.15.14, p. 148, n ^ {0}. Then, by Lemma 4.14.3, p. 138, we have j(n) ^ {0}, so, since dim(3(n)) < 1. we conclude that dim(3(n)) = 1. • COROLLARY 11.1.6 LetG be a connected Lie group acting locally faithfully and isometrically on a compact Lorentz manifold M. Assume that g is Heisenberg. Then the G-action on M is locally free. Let UQ denote the set of all upper triangular 2 x 2 real matrices. Let Go := (Uo n (SL2(M)))°. Then Go is sometimes called the ax + b group, because it isomorphic to the group {x *-¥ ax + b : R —• R \ a > 0, b € M] of affine transformations of M. We have g0 = U0 f~l G W ) L E M M A 11.1.7 Let f) :— sI2(M). Let G be a connected Lie group such that g is Lie algebra isomorphic either to go or to h. Let G act locally faithfully and isometrically on a compact Lorentz manifold M. Then the G-action on M is locally free. Proof.

Choose gi € {go, h} and an isomorphism a : gi -> g. Let T0:=(J

^ J e O

and

X0 := ( °Q J ) e h.

Then g 0 = ! T 0 +RX0 and [T0, X0] = 2X 0 . Let T := a(T0) and X := a(X0). Claim 1: For all m G M, Xm ^ 0. Proof of Claim 1: By Lemma 11.1.1, p. 293, it suffices to show, for all m € M, that Xm is lightlike. Fix m G M . We wish to show that Bm(X,X) - 0. By Lemma 11.1.3, p. 295, the form Bm is (adX)-invariant. We have Bm(2X,X) = Bm([T,X],X) = Bm(T,[X,X]) = Bm(T,0) = 0, so we get Bm(X,X) = 0. End of proof of Claim 1.

Nilpotent Lie groups

297

Assume, for a contradiction, that Y € fl\{0}, mo € M and Ymo = 0. Since a d X : h —» h is nilpotent, choose an integer i > 0 such that Z := (aAXyY ^ 0 and such that ( a d X ) Z = 0. By Lemma 7.4.20, p. 217, choose mi € M such that Zmi = 0. Calculation shows that C(,(X0) = MX0. So, since ( a d X ) Z = 0, we conclude that Z € MX. Since 0 / 2 6 1 1 and since Zmi = 0, we see that Xmi = 0, contradicting Claim 1. • 11.2

Nilpotent Lie g r o u p s

Let Gtf denote the collection of nilpotent Lie groups in Q. Theorem 11.2.1, p. 297 below is originally due to Gromov (see the "Example" in §5.3.E5 of [Gr88]), and it classifies the simply connected Lie groups in GtfT H E O R E M 11.2.1 Let "HA denote the collection of connected nilpotent Lie groups N such that there exist Lie algebras f) and a such that I) is Heisenberg or {0}, such that a is Abelian and such that n is Lie algebra isomorphic to f) © a. Then: (1) for all N 6 HA, if N is simply connected, then N S Qj^; and (2) for all N € QM, we have N e HA. Proof. Proof of (1): Since N 6 %A and since N is simply connected, choose connected Lie groups H and A such that I) is either Heisenberg or {0}, such that o is Abelian and such that N is Lie group isomophic to H x A. If h = {0}, then N is a connected Abelian Lie group; in this case, by (6) of Lemma 8.3.1, p. 268, we are done. We therefore assume that f) is Heisenberg. By (7) of Lemma 8.3.1, p. 268, H e Q. Then, by (6) of Lemma 8.3.1, p. 268, H x A € GM, so N e GM- End of proof of (1). Proof of (2): Let A denote the collection of all Abelian Lie algebras a such that a | n . Let d :— max{dim(a) |o € A}. Choose a E A such that dim(a) = d. Choose a Lie algebra no such that o © n 0 ^ n. By Ado's Theorem (Theorem 3.17.8, p. 237 of [Va74]), let N' be a connected Lie group such that n' is Lie algebra isomorphic to no- Then a © n' = n. Replacing N' by its universal covering group, we may assume that N' is simply connected. Since N' is a covering group of a connected Lie subgroup of N, we see, from (3) and (5) of Lemma 8.3.1, p. 268, that N' € Q. Then N' £ Ghf- Replacing N by N', we may assume N is simply connected and that n has no nonzero Abelian direct summand. We may assume that N is

298

Groups with Action on a Compact Lorentz

Manifold

nontrivial, i.e., that n ^ {0}. We will show that n is Heisenberg. Let N act locally faithfully and isometrically on a compact connected Lorentz manifold M. For all m € M, let (•, - ) m denote the Minkowksi symmetric bilinear form on TmM. For all m e M, define Bm e SBF(n) by Bm(X,Y) = (Xm,Ym)m. Fix m0 € M. Let B := Bmo. Then, by Lemma 11.1.3, p. 295, B is (adn)-invariant. Let t := ker(J3). Since n admits no simple direct summand, by Lemma 11.1.5, p. 295, we see that dim(3(n)) = 1 and that the iV-action on M is locally free. Then V >-> Vmo : n -»• TmoM is injective, so, by Lemma 6.2.11, p. 180 and by (1) and (2) of Lemma 6.2.6, p. 178, exactly one of the following is true: (1) B is Minkowski; (2) B is positive definite; or (3) B is positive semidefinite and dim(t) = 1. By Lemma 6.2.4, p. 176, we see that (1) and (2) are false, so (3) is true. Then t ^ {0}. Since B is (adn)-invariant, we conclude that 6 is an ideal of n. Then, by Lemma 4.14.3, p. 138, we see that (3(n))nt ^ {0}. So, since dim(3(n)) = 1 = dim(t), we conclude that a(n) = 6. By Theorem 3.6.2, p. 196 of [Va74], we see that Z(N) is closed in N. Let 7? := N/(Z(N)). Then n = n/(j(n)) = n/t. Let j) : n -» n be the natural map. Let B denote the unique symmetric bilinear form on n such that B = p*(B). Then B e SBF(n) is nondegenerate. By (3), B is also positive semidefinite. Then, by Lemma 2.10.1, p. 34, we see that B is positive definite. Since B is (adn)-invariant, B is (adn)-invariant. Then, by Lemma 6.2.4, p. 176 (with n replaced by n and B replaced by B), n is Abelian. Then [n, n] C t. Since dim(6) = 1, dim([n, n]) < 1. So, as n is nonAbelian, dim([n, n]) = 1. So, by Lemma 4.14.5, p. 140, n is Heisenberg. End of proof of (2). • We do not know whether the assumption that N is simply connected is necessary in (1) of Theorem 11.2.1, p. 297.

11.3

Solvable nonnilpotent Lie groups with Abelian nilradical

Let UQ denote the set of all upper triangular 2 x 2 real matrices. Let Go ••= (UQ n SL2(IR))0 be the ax + b group. Then there is a basis {V0, Z0} of 0o such that [VQ, Z0] = Z0.

Solvable nonnilpotent Lie groups with Abelian nilradical

299

P R O P O S I T I O N 11.3.1 Let G be a solvable connected Lie group acting locally faithfully and isometrically on a compact connected Lorentz manifold M. Assume that G is not nilpotent. Assume that the nilradical N of G is Abelian. Then (1) dim(n) = (dim(g)) — 1; and (2) there is an Abelian Lie algebra a such that g is Lie algebra isomorphic to go © 0Proof. Let gx := [g,n]. Since g is solvable, by (2) of Lemma 4.15.5, p. 143, [g,g] C n, so [g, [g,g]] C g x . Since g is not nilpotent, [g, [g,g]] ^ {0}. Then gi ^ {0}. Since n is an ideal of g, we have gi C n. Then we have [fli.n] C [n,n] - {0}. Then 8 l C c,(n). For all m £ M, let (•, • ) m denote the Minkowksi symmetric bilinear form on TmM. For all m € M, define Bm 6 SBF(g) by the formula Bm(X, Y) = (Xm,Ym)m. By Lemma 11.1.3, p. 295, we see, for all m G M, that Bm is (ad n)-invariant; consequently, we have # m ([g,n],c B (n)) = 5 m (g,[n,c 0 (n)]). So, since [g,n] = gi and since [n,cB(n)] = {0}, we see, for all m £ M, that •B m (gi,c fl (n)) = {0}; then 5 m ( g i , g i ) C B m (gi,c 0 (n)) = {0}. Let Gi be the connected Lie subgroup of G corresponding to gi. By Lemma 11.1.2, p. 294 (with G replaced by Gi), dim(gi) < 1. As gi ^ {0}, fix Z € gi\{0}. Then gi = MZ, dim(gi) = 1 and Z € gi C n. Since gi is an ideal of g, it follows that (adZ)g C gi. Let 6 denote the kernel of the map a d Z : g —> gi. Since dim(g!) = 1, it follows that the codimension in g of fi is < 1. We have 6 = ts{Z) = cB(]RZ) = c B (g!). So, since gi is an ideal of g, we conclude that t is an ideal of g. Let to := [6> [6,6]]- Then t0 C [g, [g, g]] C g : . Then [t, to] C [t, gi] = [c 0 (gi),gi] = {0}. Thus 6 is a nilpotent ideal of g, so t C n. Then the codimension in g of n is < 1. Since g is not nilpotent, we see that g ^ n, so dim(n) = (dim(g)) — 1. Choose U G g\n. Then g = WJ + n. We have [g, [g, [g,g]]] C [g,gi]. So, since g is not nilpotent, we conclude that [g,gi] ^ {0}. We have [g,gi] = [RU + n,RZ]. Since Z € n and since n is Abelian, we conclude that [n, RZ] = {0}. Then {0} # [g,fli] C R[U, Z] + [n, MZ] = R[U, Z\.

300

Groups with Action on a Compact Lorentz

Manifold

Then [U, Z] ^ 0. Moreover, gi is an ideal of g, so [g,gi] C glm Then [U,Z] G [fl,fli] C 0 1 = MZ. Choose A € M\{0} such that [[/,Z] = XZ. Let V := (1/X)U. Then [V, Z] = Z and g = B£7 + n = RV + n. Moreover, g[ := RV' + RZ is a Lie subalgebra of g and is isomorphic to goWe have (adV)n C [g,n] = gi = MZ. Let a be the kernel of the linear transformation adV : n -» MZ. Then o has codimension < 1 in n. Moreover, Z e n , and, as [V, Z] = Z, we see that Z ^ a. Then n = a + MZ and dim(o) = (dim(n)) - 1. Then g = RV+n = a+MV+MZ = a + g'1 and dim(a) = (dim(g)) — 2. By definition of a, we have [o, RV] = {0}. Also, [a,MZ] C [n,n] = {0}. Then [o,gi] = [a,RV + RZ] = {0}. Since n is Abelian and since o C n, we conclude that [a, a] = {0}. Then the map (X, Y) i-> X+Y : o©g'j -> g is a Lie algebra isomorphism. So, since g[ = g0, we see that g = a © go• 11.4

Solvable nonnilpotent Lie groups with nonAbelian nilradical — preliminaries

Let G be a connected solvable Lie group with non Abelian nilradical N. Let n' := [n, n]. Let ia :— n/n' and let n : n —> n be the canonical Lie algebra homomorphism. Assume that G is not nilpotent. Let G act locally faithfully and isometrically on a compact connected Lorentz manifold M. For all m G M, let (•, • ) m be the Minkowski symmetric bilinear form on TmM and define Bm G SBF(g) by Bm{X,Y) = (Xm,Ym)m. Define B E SBF(g) by B(X,Y)

= f

Bm(X,Y)dm.

Then B is (AdG)-invariant.

JM

For all S Cg, let S1- := {Y G g | B{Y,S) = {0}} denote the ^-orthogonal complement in g of S. For all X G g, let X1- :— {X}1-. L E M M A 11.4.1 We have dim(n') = 1 and n' C 3 (n). Proof.

By Theorem 11.2.1, p. 297, choose Lie algebras h and a such that

• h is either Heisenberg or {0}; • o is Abelian; and • n is Lie algebra isomorphic to h © a. Since [h © o, h © o] = [h, h] © {0} C [h, h] © g = ^(h © o), we get [n, n] C 3 (n). That is, we get n' C j(n). Since h is Heisenberg, we have dim([h,h]) = 1.

Solvable nonnilpotent Lie groups with nonAbelian nilradical - preliminaries

301

Then dim([Jj 8 a, f) © a]) = dim([h, h] 0 {0}) = 1, so dim([n,n]) = 1. That is, dim(n') = 1. • L E M M A 11.4.2 For all meM,

we have Bm(n,n')

= {0}.

Proof. Fix W € n and m € M . We wish to show that Bm(W, n') = 0. Since n is nonAbelian, choose X,Y £ n such that [X, Y] ^ 0. By Lemma 11.4.1, p. 300, we have dim(n') = 1. Then, as [X, Y] € n', we get n' = R[X, Y]. It therefore suffices to show that Bm(W, [X, Y]) = 0. By Lemma 11.1.3, p. 295, we see that Bm is (adn)-invariant. Then B([X,Y],Y) = B(X,[Y,Y}) and B(W,[X,Y]) = B([W,X],Y). Because n' = R[X,Y] and Bm([X,Y],Y) = Bm(X,[Y,Y]) = Bm(X,0) = 0, we conclude that Bm(n',Y) = {0}. Then Bm(W,[X,Y}) = Bm([W,X],Y) € Bm(n',Y) = {0}. D L E M M A 11.4.3 For all meM,

for all Z € n'\{0}, we have Zm ^ 0.

Proof. Fix Z 6 n'\{0}. By local faithfulness, for some mi e M , we have Zmi y£ 0. By Lemma 11.4.2, p. 301, for all m e M , Zm is lightlike. Then, by Lemma 11.1.1, p. 293, for all m £ M, Zm ^ 0. • L E M M A 11.4.4 For all m S M, the form Bm\(n x n) is positive semidefinite. Proof. Let mo E M. We wish to show that nTOo is a positive semidefinite vector subspace of the Minkowski vector space TmoM. Since n is nonAbelian, n' ^ {0}. Fix Z £ n'\{0}. By Lemma 11.4.3, p. 301, we see that Zmo ^ 0. By Lemma 11.4.2, p. 301, we see that Bmo{n,Z) = {0}, so ( n m o , Z m o ) m o = {0}, so n m o is a degenerate vector subspace of TmoM. So, from (2) of Lemma 6.2.6, p. 178, we see that n m o is positive semidefinite. • L E M M A 11.4.5 We have n' = {P £ n| B(P,P)

= 0}.

Proof. By Lemma 11.4.2, p. 301, for all P € n', for all m £ M , we have Bm(P,P) = 0. So, for all P € n', we have B(P,P) = 0 . Let P £ n and assume that B(P, P) = 0. We wish to show that P € n'. By Lemma 11.4.4, p. 301, we see, for all m € M, that Bm(P,P) > 0. So, since B{P,P) = 0, it follows, for a.e. m £ M, that Bm(P,P) = 0.

302

Groups with Action on a Compact Lorentz

Manifold

Then, by continuity o f m 4 Bm(P,P) : M -¥ M, we see, for all m G M, that Bm(P,P) = 0 . By Lemma 11.4.1, p. 300, {0} ^ n' C j(n). Choose I , F £ n such that [X,Y] ^ 0. Let Z := [ X , r ] . Then 0 ^ £ G n' C 3 (n). By Lemma 11.4.3, p. 301, for all m € M, we have Zm ^ 0. By Lemma 11.4.2, p. 301, for all m G M , we have Bm(P,Z) = 0 and Bm(Z,Z) = 0, so Bm(P,P) = Bm(P, Z) = Bm(Z, Z) = 0, so EPm + RZm is an isotropic vector subspace of TmM. Then, by Corollary 6.2.3, p. 176, for all me M, d i m ( ! P m + MZm) < 1, so, since Zm ^ 0, Pm G RZm. Fix mi G M. Choose a G l such that Pmi = aZmi. Then (P — OLZ)M is a Killing vector field on M which is everywhere lightlike and which vanishes at m i . By Lemma 11.1.1, p. 293, (P —
is positive semidef-

Proof. It follows, from Lemma 11.4.4, p. 301, that B\(n x n) is positive semidefinite. Then, by Lemma 2.10.1, p. 34, ker(£|(n x n)) = {P Gn\B(P,P)

= 0}.

So, by Lemma 11.4.5, p. 301, ker(B|(n x n)) = n'.

•

L E M M A 11.4.7 The group Ad„(G) is precompact in GL(n). Proof. By Lemma 11.4.6, p. 302, ker(5|(n x n)) = n'. Let F be the unique symmetric bilinear form on n such that B|(n x n) = ir*(F). Then F is nondegenerate. Since -B|(n x n) is invariant under Ad n (G), it follows that F is invariant under Ad w (G). That is, Ad„(G) C 0(F). By Lemma 11.4.6, p. 302, B\(n x n) is positive semidefinite, so F is positive semidefinite as well. So, since F is nondegenerate, by Lemma 2.10.1, p. 34, we see that F is positive definite. Then O(F) is compact. So, since Adn(G) C 0 ( F ) , we are done. • L E M M A 11.4.8 We have n' C 3 ( f l ). Proof. We wish to show that adn-(fl) = {0}. Let H0 :— Ad n -(G). Let / : n' -t n' be the identity map. It suffices to show that H0 = {I}.

Solvable nonnilpotent Lie groups with nonAbelian nilradical - preliminaries

303

By Lemma 11.4.1, p. 300, n' C 3(n). Then there exists a unique bilinear map w : n x n -¥ n such that, for all X,Y £ n, ui(ir(X),Tr(Y)) = [X,Y]. Note that cj(n x tt) C n' and that w is (AdG)-invariant. Since n is nonAbelian, fix A, B € n such that [A, B] ^ {0}. Let A := 7r(A),

B := ir(B),

C := [A,B] E n'\{0}.

Then C = CJ(A,B). By Lemma 11.4.7, p. 302, both (AdG)2~ and (AdG)B are precompact in n. So, as H0C = (AdG)C C w([(AdG)A] x [(AdG)B]), we conclude that HQC is precompact in n'. By Lemma 11.4.1, p. 300, dim(n') = 1. It follows, for every nontrivial subgroup H of GL(n'), for every v € n'\{0}, that Hv is not precompact in n'. So, since H0C is precompact in n' and C £ n'\{0}, H0 = {I}. D L E M M A 11.4.9 We have (n')"1 = n and dim(g/n) = 1. Proof. Let V := n'. Since both B and V are (AdG)-invariant, we conclude that V1- is (Ad G)-invariant. Then V1- is an ideal of g. We have B(n,[n, V-1-]) = B([n,n], V x ) = B(n',V)

= B(V,VX)

= {0}.

Then [n, V-1] C n x . Moreover, n is an ideal of g, so [n, V x ] C n. Then we have [ n , y x ] C n x n n = ker(B|(n x n)). By Lemma 11.4.6, p. 302, we have ker(B|(n x n)) = n'. Consequently, [n, V"-1-] C n'. By Lemma 11.4.8, p. 302, [n',g] = {0}. Since G is solvable, g is solvable, so, by (2) of Lemma 4.15.5, p. 143, we have [0, g] C n. Then [[[Fx,v^v^v1-] c [UelV-U] c [[n,yx],0] c [n',0] = {0}. Then V1- is a nilpotent ideal of g, so V1- C n. By Lemma 11.4.1, p. 300, we have dim(n') = 1, So, since V = n' and since dim^/V- 1 ) = dim(V), it follows that dim(g/l^- L ) = 1. Then, as V1- C n, we get dim(g/n) < 1. By assumption, g is not nilpotent, so g ^ n. Then dim(g/n) = 1. Since V1- C n, and since dim(fl/y- L ) = 1 = dim(g/n), we conclude that V1- — n. That is, ( n ' ) x = n. • L E M M A 11.4.10 The form B is Minkowski. Proof. By Lemma 11.4.9, p. 303, dim(fl/n) = 1. By Lemma 11.4.6, p. 302 and Lemma 11.4.1, p. 300, B|(n x n) is positive semidefinite with one-dimesional kernel. Then, by Lemma 6.2.17, p. 183, it suffices to show that B is nondegenerate. Let P € ker(S). We wish to show that P = 0.

304

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Manifold

We have P G ker(B) = § n g i C g i C (n')- 1 , so, by Lemma 11.4.9, p. 303, we get P e n . Then P G ker(B|(n x n)), so, by Lemma 11.4.6, p. 302, we conclude that P G n'. By Lemma 11.4.9, p. 303, we have (n')- 1 = n ^ g. By contrast, (RP)1- = g, so MP ^ n'. On the other hand, by Lemma 11.4.1, p. 300, we have dim(n') = 1 and P G n'. Then P = 0. • L E M M A 11.4.11 Assume thatg has no nonzero Abelian direct summand. Then n is Heisenberg. Moreover, there exist W G fl\n and a vector subspace S C n such that (ad W)S C S, S + n' = n and S D n' = {0}. Proof. By Lemma 11.4.1, p. 300, dim(n') = 1. Choose Z G n'\{0}. Then n' = RZ. By Lemma 11.4.10, p. 303, B is Minkowski. By Lemma 11.4.2, p. 301, B(Z,Z) = 0. Choose Wi e g such that a := B(Wi,Z) # 0. Then W1 i RZ. Let b := B{Wx,Wi). Let s := 6/(2a). Let W0 := W± - sZ. Then W0 £ RZ. We have B(W0,W0)

= b-2sa

=0

and

B(W0,Z)

= a - s • 0 = a ^ 0.

Let W := Wo/a. Then B(W,W) = 0 and £ ( W , Z ) = 1. It follows that B\(RW +RZ) is Minkowski. Let S := {W, Z}±. Then B\(Sx S) is positive definite. Let o := 5n(j(n)). Then £?|(axa) is positive definite. As a C ^(n), we see that a is Abelian and that [n, a] = {0}. Since B(W, Z) ^ 0, it follows, from Lemma 11.4.2, p. 301, that W $. n. By Lemma 11.4.9, p. 303, dim(n) = (dim(fl)) - 1. Then WW + n = g. By Lemma 11.4.8, p. 302, we have Z G 3(0), so [W, Z] = 0. Then {W, Z} is (adP^)-invariant, so, by G-invariance of B, S is (ad W)-invariant. That is, (&dW)S C 5. Since j(n) is an ideal of 0, j(n) is (adW)-invariant. Then a is (ad W)-invariant. Moreover, [n, a] = {0}; in particular, we see that a is (adn)-invariant. Then, since RW + n = g, a is (ad0)-invariant. So, since B is G-invariant, we see that a1- is (adg)-invariant, as well. Then o and a-1 are both ideals of 0. Since B\(a x a) is positive definite, we see that aC\ax = {0}. Then o + o-1 = 0 . Then 0 is an Abelian direct summand of 0. By assumption, 0 has no nonzero Abelian direct summand, so 0 = {0}. Since RZ = n', by Lemma 11.4.9, p. 303, we have n = ZL. Therefore ± W nn = W-LnZ-L = {W,Z}1- = S. Then WL n (j(n)) = W1- n n n (3(n)) = S n (3(n)) = a = {0}. So, since W1- has codimension one in 0, we conclude that dim(3(n)) < 1. By assumption, the connected Lie group N is nonAbelian, so, in particular,

Solvable nonnilpotent Lie groups with nonAbelian nilradical - preliminaries

305

n ^ {0}. So, by Lemma 4.14.3, p. 138, 3 (dim(0)) - 1 = dim(n). So, since S + n' C n + n = n, we see that S + n' = n. D L E M M A 11.4.12 Assume thatg has no nonzero Abelian direct summand. Then, for all m € M, we have ker(B m |(n x n)) = n'. Proof. Let Q denote the Minkowski quadratic form on TmM. By Lemma 11.4.11, p. 304, n is Heisenberg, so, by Corollary 11.1.6, p. 296, the iV-action on M is locally free, s o ! i - > Xm : n —> TmM is injective. Then n'm ^ {0} and it suffices to show that ker(Q|n m ) = n ^ . By Lemma 11.4.2, p. 301, n' C ker(£? m |(n x n)), so n ^ C ker(<5|n m ). Then, since n ^ 7^ {0}, it follows that Q\nm is degenerate. So, by (1) of Lemma 6.2.6, p. 178, dim(ker(Q|n m )) = 1. So, as {0} ^ n ^ C ker(Q|n m ), we conclude that n'm = ker(<5|nm). D L E M M A 11.4.13 Assume thatg has no nonzero Abelian direct summand. Then, for all m 6 M, n = {X £ 0 | Bm{X,n') = {0}}. Proof. Fix m E M. By Lemma 11.1.3, p. 295, Bm is (adn)-invariant. For all ACg, let Av := {X € g\ Bm(X, A) = {0}}. Let V := n'. Assume, for a contradiction, that Vw / n. By Lemma 11.4.2, p. 301, we see that n C V v . By Lemma 11.4.9, p. 303, we have dim(0/n) = 1. Then, as n C V V C 0, we conclude that V V = 0. We have £ m ( n , [n,V v ]) = B m ([n,n],V v ) = Bm(V,Vw) = {0}. Then we have [n, V v ] C n v . Moreover, n is an ideal of 0, so [n, Vv] C n. Then we have [n,fl] = [n, Vv] C n v n n = ker(£ m |(n x n)). By Lemma 11.4.12, p. 305, we have ker(2? m |(n x n)) = n'. Consequently, [n,0] C n'. So, by Lemma 11.4.8, p. 302, [[n,0],fl] = {0}. Since G is solvable, by (2) of Lemma 4.15.5, p. 143, [0,0] C n. Then [[[0,0], 0L0] C [[n,g],0] = {0}. Since 0 is, by assumption, not nilpotent, we have a contradiction. •

306

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11.5

Manifold

Solvable nonnilpotent Lie groups with nonAbelian nilradical — final results

L E M M A 11.5.1 Let M be a compact Lorentz manifold and let A be a connected Lie subgroup of Isom(M). Assume that A is Abelian and that the A-action on M is locally free. Assume, for all m £ M, that o m is a Minkowski vector subspace ofTmM. Then A is precompact in Isom(M). Proof. Let 7 be the Lorentz metric on M. Let H denote the group of diffeomorphisms of M and let T\ denote the topology on H of C°°-convergence. Then Isom(M) is a closed subgroup of (H,Ti), so we wish to show that A is precompact in (H,TI). By Lemma 7.11.9, p. 245, it suffices to show that A preserves a Riemannian metric on M. Let A be the subbundle of TM defined by A := ( J am. Then, for all m£M

m 6 M, we have A m = o m . For all m e M, let fm : a -> A m be defined by fm(X) = Xm. Let Q be any positive definite quadratic form on o. Let g be the Riemannian bundle metric on A defined by g(Xm) = Q(X). Let A x := [J {v € TmM \ 7(11, A m ) = {0}} be the orthogonal complem£M

ment in TM to A with respect to 7. By assumption, for all m 6 M, o m is Minkowski, so, since o m = A m , we see that 7|A m is Minkowski, so 7 | A ^ is positive definite. Then 7IA-1- is a Riemannian bundle metric on A- 1 . Then h := g © (7IA-1-) is an vl-invariant Riemannian bundle metric on A + A- 1 . So, since A + A1- = TM, we see that h is an A-invariant Riemannian metric

on M.

•

L E M M A 11.5.2 Let M be a compact Lorentz manifold and let A and H be connected Lie subgroups of Isom(M). Let I : M ->• M be the identity map. Assume that • • • • • •

A is Abelian and A ^ {/}; A is precompact in Isom(M); H is nilpotent and H ^ {I}; f) has no nonzero Abelian direct summand; A normalizes H; and CA(H) is discrete.

Then dim(A) = 1.

Solvable nonnilpotent

Lie groups with nonAbelian

nilradical - final results

307

Proof. Let A denote the closure of A in Isom(M). Since A normalizes H, it follows that (AdA)h C h, so (AdA)ty C h. Then, by connectedness of H, we see that A normalizes H. For all X € o, for all t € R, we have exp(£ad|j(X)) = Ad(,(exp(iX)), so {exp(tadf,(X))} t € R C Ad<,(]4). So, since A is compact, we conclude, for all X € a, that {exp(tad(,(X))} t€ R is precompact in GL(h). So, for all X 6 o, a d X : f) -> (j is elliptic. Since C A ( H ) is discrete, it follows that c„(h) = {0}. Then, for all X € a\{0}, a d X : \] —> \) is nonzero. A nonzero elliptic linear transformation is not nilpotent, so, for all X 6 o\{0}, a d X : h —> h is not nilpotent. Let G :— AH. Then, by Corollary 4.5.3, p. 85, G is a connected Lie subgroup of Isom(M) and g = a + h . By Lemma 4.15.13, p. 148, we see both that () D o = {0} and that \) is the nilradical of g. Then, by Lemma 11.4.9, p. 303, dim(g/h) = 1. Since g = a + h , we have dim(a/h) = dim(o/(onh)). So, since oflf) = {0} and dim(g/h) = 1, we conclude that dim(o) = 1. Then dim(j4) = 1. • P R O P O S I T I O N 11.5.3 Let G be a solvable connected Lie group acting isometrically on a compact connected Lorentz manifold M. Assume that the nilradical of G is nonAbelian and that G is not nilpotent. Then there is an Abelian Lie algebra a and a rationally twisted Heisenberg Lie algebra h such that g is Lie algebra isomorphic to f) © o. Proof. Let V denote the collection of all Abelian Lie algebras p such that p | g. Let d := max{dim(p) | p e P } . Choose p £ V such that dim(p) = d. Choose a Lie algebra g' such that g' © p is Lie algebra isomorphic to n. Since the nilradical of g is nonAbelian, it follows that the nilradical of g' is nonAbelian. Since g is not nilpotent, it follows that g' is not nilpotent. By Ado's Theorem (Theorem 3.17.8, p. 237 of [Va74]), let G0 be a connected Lie group such that go is Lie algebra isomorphic to g'. Replacing G by Go, we may assume that g has no nonzero Abelian direct summand. Replacing G by its universal covering group, we may assume that G is simply connected. We wish to show that g is a rationally twisted Heisenberg Lie algebra. For all m 6 M, let (•, • ) m be the Minkowski symmetric bilinear form on TmM and define Bm 6 SBF(n) by Bm(X,Y) = (Xm,Ym)m. Define B G SBF(fl) by B(X,Y) 1

= [

Bm(X,Y)

dm. Then B is (AdG)-invariant.

JM

For all S C g, let S - := {X e g | B(X, S) = {0}} denote the B-orthogonal complement in g of S. For all X 6 n, let Xs- := {X}-1.

308

Groups with Action on a Compact Lorentz

Manifold

Let N be the nilradical of G. Let n' := [n, n]. By Lemma 11.4.11, p. 304, we see that n is Heisenberg. Choose W € fl\n and S C n as in Lemma 11.4.11, p. 304. Since n is Heisenberg, it follows that n admits no nonzero Abelian direct summand. Since G is solvable, it follows that g admits no simple direct summand. Then, by Lemma 11.1.5, p. 295, the G-action on M is locally free. By Lemma 11.4.6, p. 302, B|(nxn) is positive semidefinite with kernel n'. By Lemma 11.4.5, p. 301, we have n' = {X £ n\B(X,X) = 0}. Define B0 := B\(S x S). As 5 n n' = {0}, we see, for all X £ S\{0}, that B0(X,X) ^ 0. Moreover, Bo is positive semidefinite, i.e., for all X £ S, we have B0(X,X) > 0. Then, for all X £ S\{0}, we have B0(X,X) > 0. That is, Bo £ SBF(S) is positive definite. Let T := ad W : S -> S. Then, since B is (adW)-invariant, we conclude that T £ o(5o). By Lemma 11.4.9, p. 303, we have (n') 1 - = n, so, as W $. n, we see that B(W,n') ^ {0}. Choose Zen' such that B(W,Z) = 1. Then Z £ n'\{0}. Define <j>: K -> n' by (j>(t) = tZ. By Lemma 11.4.1, p. 300, we have dim(n') = 1. Then : E —> n' is a vector space isomorphism. Let CJ be the antisymmetric bilinear form on S defined by OJ(X,Y) = (p_1([X,Y]). Then, for all 1 , 7 6 5, we have [X, Y] = {OJ{X, Y))Z. Since n is Heisenberg, it follows that |(n) = n'. So, as SOn' = {0}, we get 5n(a(n)) = {0}. Then w is nondegenerate. For X,Y e S, as B is (adfl)-invariant, B([W,X],Y) = B(W,[X,Y]). So, as [X,Y] = (ui{X,Y))Z and as B(W,Z) = 1, B([W,X],Y) = u(X,Y). Then, for all X,Y e S, we have B0(T(X),Y) = B([W,X],Y) = u{X,Y). So, since u is nondegenerate, it follows that T : S -> S is invertible. Then, for any T-invariant vector subspace S' of S, T\S' : S' —> S' is invertible, so, since B0\(S' x 5") is positive definite, it follows that UJ\(S' X S') is nondegenerate. Since 5 admits a nondegenerate antisymmetric bilinear form, it follows that dim(S') is even. Let k := (dim(5))/2. Since T : S ->• S is invertible, since B0 is positive definite and since T € o(J50), choose T-invariant vector subspaces S i , . . . , S* of S such that Si 4- • • • + Sk = S and such that, for all integers i e [1, k], we have dim(Sj) = 2. Fix an integer i £ [1, k] for this paragraph. Let B, := J3|(S» x S,); then Bi is positive definite. Since T\Si G o(Bi), choose a basis X , y of Si and choose Aj > 0 such that T(X) - \{Y and such that T(Y) = -XiX. Since T\Si : Si -¥ Si is invertible, we conclude that Aj ^ 0. Then Aj > 0. Let

Solvable nonnilpotent

Hi := LJ(X,Y).

Lie groups with nonAbelian

nilradical - final results

309

We have Xi • Bi(Y,Y)

= Bi(\iY,Y)

= B0(T(X),Y)

=

Bi(T(X),Y)

= U(X, Y) = UJ(X, Y) = m.

So, since Bt is positive definite and since \ t > 0, we conclude that Hi > 0. Define Xi := X/y/ju and YJ := Y/y/JH. Then [W, Xi] = XiY

and

[W, Y{] = - A ^ .

Also, u{Xi,Yi) = {u(X,?))//* = 1. Then [Xi,Y$ = co{Xi,Yi)Z = Z. Then, as X € n' = 3(n), it follows that ( X i , . . . ,Xk, Yi,..., Yk,Z) is an ordered basis of n satisfying the Heisenberg relations. Moreover, we have S = RX"i + • • • + MXk + m\ + • • • + KYfc. Let A := { e x p ( W ) } t 6 R . By Lemma 11.4.9, p. 303, dim(g/n) = 1. So, since W $. n, we see taht g = MVF + n, so G = AN. By (3) of Lemma 8.2.1, p. 264, it suffices to show that Adg(A) is compact. Let c := WW + n'. By Lemma 11.4.8, p. 302, we have n' C 3(g). Then c is Abelian. Let C be the connected Lie subgroup of G corresponding to c. Then C is Abelian and A C C. Let / : G -> Isom(M) be defined by (/(p))(m) = gm; then / is a Lie group homomorphism. The G-action on M is locally faithful, so ker(/) is a discrete subgroup of G. Let A' be the closure in Isom(M) of f(A) and let C" be the closure in Isom(M) of f(C). Let N0 := f(N). By Corollary 4.5.7, p. 86, f(A), f(C) and iV0 are all connected Lie subgroups of Isom(M). By Lemma 11.4.13, p. 305, we see, for all m € M, that n ^ is an isotropic vector subspace of TmM and that Bm(Wm,n'm) ¥" {0}; so, since we have cTO = RWm + n'm, it follows, by Lemma 6.2.15, p. 182 (with U replaced by c m and U' replaced by n'm), that c m is a Minkowski vector subspace of TmM. Then, by Lemma 11.5.1, p. 306 (with A replaced by / ( C ) ) , we see that C" is compact. Then A' is a compact connected Abelian Lie group. Therefore, by Lemma 4.8.3, p. 95, there exists an integer n > 0 such that A' is Lie group isomorphic to (R/Z) n . Then, for any closed subgroup B of A', there is a closed subgroup B\ of A' such that BB\ — A' and such that BnBi is finite. Since A normalizes N, it follows that f(A) normalizes No- Then f(A) normalizes no, so A' normalizes no. So, as iVo is connected, we conclude that A' normalizes A^. Let B := CA'(NO)- Then B is a closed subgroup of A'. Choose a closed subgroup Bi of A' such that BB\ = A' and such

310

Groups with Action on a Compact Lorentz

Manifold

that B n Bi is finite. We have BC\Bi= CBl (N0). Then CBl (N0) is finite, hence discrete. Let I: M —• M be the identity map. Since G is not nilpotent and since G = AN, we conclude that [A,N] ^ {/}, so [A',N0] ^ {I}, so B ^ A1, so Bi 7^ {/}. Moreover, since G is not nilpotent, we conclude that G ^ A, so JV 7^ { 1 G } , so JV„ ^ {/}. By Lemma 11.5.2, p. 306 (with A replaced by B\ and with H replaced by No), we see that dim(.E?i) = 1. Since B centralizes No, Adno(B) is trivial. Then dim(Ad no (A')) = dim(Ad„ 0 (Si)) < dim(£i) = 1. Then Ad no (^4') has no proper dense connected Lie subgroup. Therefore we have Ad„ 0 (/(A)) = Ad n o (A'). Therefore, since A' is compact, we conclude that Ad no (/(-A)) is compact. Since / : N —> No is surjective with discrete kernel, we conclude that df : n —• no is an isomorphism of Lie algebras, and, in particular, of vector spaces. This isomorphism induces a Lie group isomorphism F : GL(n) ->• GL(n 0 ), and we have F(Ad„(A)) = Adno(f(A)). Then Adn(A) is compact. So, since the map a t-> a\S : Adn(A) -> Adg(A) is surjective and continuous, we see that Ads(^l) is compact. •

11.6

Semisimple Lie groups

P R O P O S I T I O N 11.6.1 Let G be a connected noncompact semisimple Lie group acting locally faithfully and isometrically on a compact connected Lorentz manifold M. Then there is a compact semisimple Lie algebra 6 such that g is isomorphic to s ^ W © CProof. Let o be a maximal real split torus of g and let $ be the set of roots of o on g. By Lemma 4.10.32, p. 124, g is noncompact. So, by Lemma 4.12.2, p. 132, we see that a ^ {0}. Let B := RKF(g, a) G SBF(o*) be the positive definite form on o* induced from the Killing form of g. Then $ is a root system in (a*,B). Let A be a base of * and let $ + be the set of roots that are positive with respect to A. For all ^ £ $, let g^ denote the 0-rootspace of o on g. Let n := ffi g^. Let A and N be the connected Lie subgroups of G corresponding to o and n, respectively. Then, by Corollary 4.5.3, p. 85, AN is a connected Lie subgroup of G and the Lie algebra of AN is o + n. Moreover, N is nilpotent and AN is solvable. For all X E a\{0}, a d X : n -¥ n is nonzero and real diagonalizable and

The general case

311

is therefore not nilpotent. Then, by Lemma 4.15.13, p. 148, o f l n = {0} and n is the nilradical of a + n. By (1) of Lemma 11.3.1, p. 298, if N is Abelian, then dim(a + n) = (dim(n)) + 1. By Lemma 11.4.9, p. 303, if N is nonAbelian, then dim(o + n) = (dim(n)) + 1. So, in either case, we have dim(a + n) = (dim(n)) + 1. Because o n n = {0}, it follows that dim(a + n) = (dim(o)) + (dim(n)) We conclude that dim(a) = 1. Let r be the rank of g. By definition of "real rank", we have r = dim(a). Then r = 1. Since g is noncompact, by Corollary 4.10.4, p. 113, let g' be a noncompact simple ideal of g. By Lemma 4.10.2, p. 113, choose an ideal 6 of g such that g' + t = g and such that g' n 6 = {0}. Then g is Lie algebra isomorphic to g' © t. By Corollary 4.10.15, p. 118, since f is an ideal of g, we see that 6 is semisimple. Let r' be the rank of g' and let s be the rank of 6. Then r' + s = r = 1. Since g' is noncompact, by (1) => (4) of Lemma 4.12.2, p. 132, it follows that r' > 1. Then r' = 1 and s = 0. By (1) = » (4) of Lemma 4.12.2, p. 132, since s = 0, we conclude that 6 is compact. Let G' be the connected Lie subgroup of G corresponding to g'. Since r' = 1, by (4) =$• (1) of Lemma 4.12.2, p. 132, we see that g' is noncompact, so, by Lemma 4.10.32, p. 124, G' is noncompact. Then, by Theorem 10.5.3, p. 292 (with G replaced b y G ' ) , 0 ' = sb(M). • 11.7

The general case

Let UQ denote the set of all upper triangular 2 x 2 real matrices. Let / be the 2 x 2 identity matrix. Since Z(GQ) = {I}, we see that G0 is Lie group isomorphic to Ad Bo (Go). Then, as Go admits no nontrivial compact subgroup, we conclude that AdBo(Go) admits no nontrivial compact subgroup. Let Go := (U0 D SL2(IR))0 be the ax + b group. Then Ad 0o (G o ) = Inn(g 0 ), so Inn(g 0 ) admits no nontrivial compact subgroup. Let G be a connected Lie group acting locally faithfully and isometrically on a compact connected Lorentz manifold M. Let R be the solvable radical of G and let L be a semisimple Levi factor of G. L E M M A 11.7.1 We have [[,t] = {0}. Proof. Assume, for a contradiction, that [I, t] ^ {0}. By Lemma 4.10.4, p. 113, fix a simple ideal l0 of [ such that we have r [foi ] 7^ {0}- Let LQ be the connected Lie subgroup of L corresponding

312

Groups with Action on a Compact Lorentz

Manifold

to lo- By Corollary 4.5.3, p. 85, L0R is a connected Lie subgroup of G. Replacing G by LQR, we may assume that L is a simple Lie group. Let N be the nilradical of G. Because [I, r] ^ {0}, by Lemma 4.15.6, p. 143, [l,n] ^ {0}. Let D denote the kernel of Ad : L -» GL(n). Since [l,n] ^ {0}, it follows that D j^ L. So, since L is simple and since D is a proper normal closed subgroup of L, we conclude that D is a discrete subgroup of L. By Lemma 7.14.8, p. 254, choose a one-dimensional connected Lie subgroup C of L such that, for any vector space V, for any nontrivial representation p : L —> GL(V), we have that p(C) is compact. We define H := CN. Then H is solvable. By Corollary 4.5.3, p. 85, H is a connected Lie subgroup of G. By definition of C, Ad(,(C) C GL(I)) is compact. Because Ad(,(C) is compact, we see, for all X 6 c, that a d X : t) —> f) is elliptic. Since D is discrete, for all X G c\{0}, we have (adX)n ^ {0}, so &dX : \) —> f) is nonzero. A nonzero elliptic linear transformation is never nilpotent, so by Lemma 4.15.13, p. 148, we see both that c n n = {0} and that n is the nilradical of f). Then H is not nilpotent. For any Abelian Lie algebra a, the Lie group Inn(go © o) is Lie group isomorphic to Inn (go), and therefore admits no nontrivial compact subgroup. On the other hand, Ad|,(C) C Ad^(H) = Inn(fj) and Ad(,(C) is compact and nontrivial. We conclude, for any Abelian Lie algebra a, that f) is not Lie algebra isomorphic to go © o. It follows, from (2) of Proposition 11.3.1, p. 298 (with G replaced by H), that iV is not Abelian. Let n' := [n,n]. Define B 6 SBF( 0 ) by B(X,Y)

= [ (Xm,Ym)mdm.

Then,

by constrcution, B is (Ad C?)-invariant and is therefore (adg)-invariant. By Lemma 11.4.1, p. 300, we have dim(n') = 1. Fix W € c\{0}. Then W $ n, so by Lemma 11.4.9, p. 303, we have B(W,v!) ^ {0}. Since l is simple, we have [[,[] = [. Then {0} ^ B(W,n') C B(l, n') = £([[, l],n') = B(l,[l,n']). Then[[,n']#{0}. Since N is a normal subgroup of G, we see that n is (Ad Gf)-invariant. It follows that n' is (AdG)-invariant. Since dim(n') = 1, we see that the map Ad : L —> GL(n') is a one-dimensional representation of the simple Lie group L. Since I = [I, fj, by Corollary 4.6.4, p. 91, L = [L, L], so, by Lemma 5.7.3, p. 168, Adn<(L) is trivial. Then [I,n'] = {0}, contradiction• L E M M A 11.7.2 If L is noncompact, then R is Abelian. Proof. By Lemma 4.10.32, p. 124, I is noncompact. It follows, from Lemma 4.12.3, p. 133, that I has a Lie subalgebra fo which is Lie algebra

The general case

313

isomorphic to s[ 2 (E). Let LQ be the connected Lie subgroup of L corresponding to lo- By Corollary 4.5.3, p. 85, L0R is a connected Lie subgroup of G. Replacing G by L^R, we may assume that I = sl2(K). Since t = sl2(IR), we choose a connected Lie subgroup 5 of L such that s = go- Let H := SR. Then H is a solvable subgroup of G. Moreover, by Corollary 4.5.3, p. 85, H is a connected Lie subgroup of G and h = s + r. By Lemma 11.7.1, p. 311, we see that [l,r] = {0}, so [s,r] = {0}. Then (P, Q) i-> P + Q : 5 © t ->• f) is a Lie algebra isomorphism. Since s is not nilpotent, we conclude that f) is not nilpotent. Let 0o := [go,go]- We have [s,s] © [r,r] S [h,h], so [s,s] | [h,h]. So, as [s, s] = 0o, and as dim(g 0 ) = 1, we conclude that [h, h] has a one-dimensional direct summand. Then fj is not Lie algebra isomorphic to the direct sum of a twisted Heisenberg Lie algebra and an Abelian Lie algebra. Therefore, by Proposition 11.5.3, p. 307, we see that the nilradical of f) is Abelian. Let t' := [r,r]. By (2) of Proposition 11.3.1, p. 298 (with G replaced by H), choose an Abelian Lie algebra a such that h = g 0 ffi a. We have s © r = h = 0o © o. So, since s = go, we get g 0 ffi r S 0O © o. So, since [go © r,0 O © t] 2 g 0 © r', while [g0 ffi a,g 0 ffi o] = g 0 , we conclude that 0o © *' - 0o- Then (dim(0 o )) + (dim(r/)) = dim(g 0 ), so dim(r') = 0, so r7 = {0}, so t is Abelian, so R is Abelian. • Let £ be the collection of Lie algebras consisting of (1) (2) (3) (4) (5)

the trivial group {1}. the Heisenberg Lie algebras; the rationally twisted Heisenberg Lie algebras; and 0O; and s[ 2 (E);

T H E O R E M 11.7.3 There exists q € C, there exists an Abelian Lie algebra o and there exists a compact semisimple Lie algebra t such that g is Lie algebra isomorphic to qffioffi6. Proof. By Lemma 11.7.1, p. 311, we see that [l,t] = {0}. Then the mapping (P, Q) i-> P+Q : [ffit -¥ g is a Lie algebra isomorphism, so g = Iffir. If L is noncompact, then, by Lemma 11.7.2, p. 312 and Proposition 11.6.1, p. 310, we are done. We therefore assume that L is compact. Then, by Lemma 4.10.32, p. 124, [ is a compact semisimple Lie algebra. Replacing G by R, we may assume that G is solvable. If G is nilpotent, then, by

314

Groups with Action on a Compact Lorentz

Manifold

Theorem 11.2.1, p. 297, we are done. We may therefore assume that G is not nilpotent. Let N be the nilradical of G. If N is Abelian, then, by (2) of Proposition 11.3.1, p. 298, we are done. On the other hand, if N is nonAbelian, then by Proposition 11.5.3, p. 307, we are done. • We have a converse of Theorem 11.7.3, p. 313: For any connected, simply connected Lie group G\, if there exist q £ C, an Abelian Lie algebra o and a compact, semisimple Lie algebra t such that 0 is Lie algebra isomorphic to q©a©t, then there is a locally faithful, isometric action on a compact connected Lorentz manifold; this is the content of the results in Chapter 8 (especially Lemma 8.3.1, p. 268). Moreover, by using Lemma 11.1.5, p. 295 together with Lemma 11.1.7, p. 296, we can see that, if G is a connected Lie group and if g € £,, then any locally faithful, isometric action of G on a compact connected Lorentz manifold is locally free. The remarks of the last paragraph, together with Theorem 11.7.3, p. 313 comprise Theorem 11.3, p. 260 of [AS97a].

Chapter 12

The Isometry Group of a Compact Lorentz Manifold

In Chapter 11, we classified, up to local isomorphism, the class Q of connected Lie groups admitting a locally faithful, isometric action on a compact connected Lorentz manifold. For any Lorentz manifold M, recall that Isom°(M) denotes the identity connected component of the Lie group Isom(M). For any Lorentz manifold M, let UCIsom°(M) be the universal covering group of Isom°(M). In this chapter, we wish to describe the class Q' of Lie groups G such that, for some compact connected Lorentz manifold M, G is Lie group isomorphic to UCIsom°(M). Note that Q' C Q. By (3) of Lemma 8.3.1, p. 268, Q is closed under passage to connected Lie subgroups. As we will see, this is not true of Q'. Let UQ denote the set of all upper triangular 2 x 2 real matrices. Let Go := (U0 n SL 2 (R))° be the ax + b group. By Lemma 8.1.1, p. 261, SL2(M) e Q' C g and so, by (3) of Lemma 8.3.1, p. 268, we have G0 € Q. In Theorem 1.2, p. 264 of [AS97b], one finds an effective classification of the simply connected Lie groups in Q'. The same result was obtained independently in [Ze98]. In our exposition here, we will not be so amitious. Since the hardest part of this classification is establishing that Go $ G', we will confine ourselves to this goal. Furthermore, while the proof in [AS97b] is complete, we will here omit some of the more tedious details. The interested reader may consult either [AS97b] or [Ze98]. Let Go act locally faithfully and isometrically on a compact connected Lorentz manifold M. For all m € M, let (•, • ) m denote the Minkowski symmetric bilinear form on TmM. Then (QO)M •= {XM \ X G flo} is a Lie algebra of Killing vector fields on M. The goal of this chapter, then, is to 315

The Isometry

316

Group of a Compact Lorentz

Manifold

prove that some Killing field on M is not contained in (BO)MDifferentiating the action of Go on M, we get an action of Go on the tangent bundle TM of M. This action then determines an action of Go on the set of vector fields on M by the rule: (gX)m — g(Xg-im). Let / denote the 2 x 2 identity matrix and let

Then I,A0,X0 [AQ,P]

^ -P.

€ flo- We calculate [A),X 0 ] = X0. For all s,t

€ E,

as := exp(sA0) € G 0 s

For all P € 0o\{O},

let

and

xt := exp(tX 0 ) 6 Go-

x

Then (Ada s )Xo = e Xo, and so asxtaj — xe*t- We also calculate, for all s , t £ i , that (Adxt)A0 = A0 — tX0. We will prove that there is a nonzero Killing field Y on M such that [(A0)M,Y] = -Y; then Y $. (Q0)M12.1

Oseledec splitting

Let S C t Assume, for some to that (t 0 , co) C S. Let / : S ->• (0, oo) be a function. For A 6 K, we say that / is of class £\ if, there exists t\ € K such that, for all t > ti, we have /(£) < eM*. We say that / has exponential decay if there is some fi < 0 such that / is of class f . We say that / is neutral if, for all A < 0, for all (i > 0, we have that / is both of class £\ and of class £fi. Let X be a compact manifold and let W be a, vector bundle over X. Let Cx • X -> W be the zero-section of W and let Zx := C,x{X) C W. Let S C t Assume, for some to S E, that (to, oo) C 5. Let 4> : S -> W be a function. We say that 0 has exponential decay (resp. <j> is neutral) if there exist • a function / : S -¥ (0, oo); • a compact subset C C W \ Z x ; and • a function ip : S —> C such that / has exponential decay (resp. / is neutral) and such that <j> = fip. Let C : M ->• T M be the zero section of T M and let Z := C(M). Let £ be the collection of all v e TM such that either veZorti-^ atv : E ->• T M

Preliminaries

317

has exponential decay. Let E° be the collection of all v G TM such that either v e Z or • both 11-¥- atv : E -> T M and t i-> a_ t i; : E -> TM are neutral. Finally, let E+ be the collection of all v G T M such that either v £ Z or i H> a,-tv : E —• TM has exponential decay. For all m G M, let E~ = E~ n T m M , let £ ^ = £ ° n T m M and let E+ = E+ n T m M ; then JE~, E^, and E+ are subspaces of TmM. We remark that, by the Oseledec splitting theorem (see [Ma89, Theorem V.2.1, p. 171] or [Pe77]), for a.e. m G M, (x,y,z) >-+ x + y + z : E^ © E^ © £ + -> T m M is an isomorphism of vector spaces. The following propostion appears as part of Lemma 15.1, p. 282 of [AS97b]. A substantial portion of [AS97b] is devoted to proving this result; however, the proof is quite tedious and can probably be substantially improved. For this reason, we have chosen here to omit the proof. P R O P O S I T I O N 12.1.1 The subsets E~, E°, E+ defined above are all three smooth vector subbundles of TM, and, for all m G M, the linear transformation (x,y,z) i-> x + y + z : E^ © E^ © i5+ —> TmM is an isomorphism of vector spaces. 12.2

Preliminaries

If X is a Lorentz manifold, then a vector subbundle V of TX will be said to be everywhere lightlike if, for all x G X, for all v G Vx, v is a lightlike vector in the Minkowski space TXX. L E M M A 12.2.1 The vector subbundles E+ and E~ ofTM erywhere lightlike.

are both ev-

Proof. We will consider E~\ the proof for E+ is similar. Denote the Lorentz metric on M by (•, •). Fix m £ M and v G E^. We wish to show that {v, v) = 0. For all s G E, (v,v) = (asv,asv). By compactness of M, choose a sequence s* in E such that s; —> +00 and such that Sjm is convergent in M. Let m' — lim Sim. By definition of E~, we get aSiv - • 0 G Tm<M. Then (v,v) — lim (aSiv,aSiv) = 0. •

318

The Isometry

Group of a Compact Lorentz

L E M M A 12.2.2 The vector field (X0)M Proof.

Manifold

is in E+.

For all s € E, we have esa-s((X0)M)

= es((Ada-s)X0)M

= es(e-sX0)M

It follows, for all m 6 M , that {esa-s((Xo)m)\s in TM, so (X0)m eE+.

=

(X0)M-

€ E} is precompact D

L E M M A 12.2.3 We have rank(S + ) = rank(.E~) = 1. Moreover, for all m e M, we have E+ = R((X0)m). Proof. By Lemma 12.2.1, p. 317, both E~ and E+ are everywhere lightlike, so, by Corollary 6.2.3, p. 176, we get rank(E~) < 1 and rank(2? + ) < 1. By Lemma 12.2.2, p. 318, for all m € M, we have (X0)m € E+. Then {XQ)M is everywhere lightlike. By Lemma 11.1.1, p. 293, for all m 6 M, we have (X0)m ^ 0. It follows, for all m € M, that E+ = R((X0)m). In particular, rank(i? + ) = 1. It remains to show that rank(2J~) > 1. Assume, for a contradiction, that rank(E~) = 0, i.e., that, for all m £ M, we have Em = {0}. Let d := dim(M). Let V :— Ad(TM); then V is a rank one vector bundle over M. The Lorentz metric on M induces a volume density a : V —>• [0, oo) on M. Let ((•, •)) denote the unique Riemannian bundle metric on V such that, for all v G V, we have {{v,v)) = (a(v))2 on M. Since the Lorentz metric on M is Go-invariant, it follows that ((•, •)) is Go-invariant, as well. Fix a point mo 6 M and fix a basis vi,...,Vd of TmoM such that vi € Emo, and such that V2,...,v& € E^lQ. Let v :— vi A • • • A vj, € Vmo. Then v^O, so ((v,v)) ^0. By compactness of M, choose a sequence Si in M. such that Si —> +oo and such that a_ Si mo is convergent in M. Let mi := lim o_ 8j mo. i—*oo

By definition of i ? + and £J°, s >-> a-sv\ : E —> T M has exponential decay, while, for all integers i € [2,d], s <-¥ a-svt : E -4 T M is neutral. It follows that s i-> a_gi> : E —• V has exponential decay. In particular, we have a-Siv - > 0 e Vmi. By Go-invariance of ((•, •)), for all s £ E, ((u,t;)) = ((a_ s w,o_ s u)). Then ((«,«)) = lim ((a-^v.a-gjt;)) = 0, contradiction. • i—¥oo

L E M M A 12.2.4 Denote the Lorentz metric on M by (•, •). Then, for all meM,we have (E^,E^)m =0= (E°m,Em)m.

A candidate for the new Killing field

319

Proof. Fix m e M . We will prove that (Em, Em)m = 0, the proof that 0 = (Em, J5+) m being similar. Fix v G Em and w G Em. We wish to show that (v,w)m = 0. The map s (->• asv : R -» TM is neutral, while s i-> a s w : R ->• T M has exponential decay. Then s •-)• (a s i;,a s «;) : E —• E has exponential decay. Then lim (asv,asw) = 0. For all s G R, we have (i>,w;) = (asv,asw), so s—>+oo

(i;,«;) =

12.3

lim (asv, asw) — 0.

Q

A candidate for the new Killing field

Let A := (A0)M and let X :— (X0)ML E M M A 12.3.1 For allm G M, there exists a unique vector Cm G TmM such that Cm G E^ and such that ((Xo)m,Cm) = 1. Moreover, the map m i-> Cm : M —> TM is smooth, so C is a (smooth) vector field on M. Proof. Fix m G M for this paragraph. Let a : TmM ->• E be defined by a(v) = ((X0)m,v). By Lemma 12.2.1, p. 317, both Em and Em are lightlike vector subspaces of TmM. By Lemma 12.2.3, p. 318, we see that rank(£+) = 1 = rank(JS-). By definition of E+ and E~,E+^E~. Then, by Corollary 6.2.3, p. 176, ( £ + , £ " ) # {0}. By Lemma 12.2.3, p. 318, we have E+ = E((X 0 ) m ). Then ((X0)m,Em) ^ {0}. So, since dim(B-) = 1, we conclude that a : TmM -> E is an isomorphism. Let Cm := a _ 1 ( l ) . Then ((X0)m,Cm) = 1, proving existence. For all v G Em, we have: ((X0)m,v)

= l

=>

a(v) = l

=»

v = a-1{l)

= Cm,

proving uniqueness. Smoothness of C follows from smoothness of X0, of E~ and of the Lorentz metric (•,•). • L E M M A 12.3.2 Then there exist vector fields H and Y on M such that [H,X] = X, [H,Y] = -Y, [X,Y] = H, such that [A,Y] = -Y, such that [A, H] = 0 and such that, for all m G M, the vectors Xm,Ym,Hm are linearly independent.

320

The Isometry

Group of a Compact Lorentz

Manifold

Proof. Since A and X are Killing vector fields on M, we conclude, for all vector fields S and T on M, that A(S, T) = ([A, S), T) + (S, [A, T]) and X(S,T) = ([X,S),T) + (S,[X,T}). Choose C as in Lemma 12.3.1, p. 319. Then, for all m £ M, we have ((Xo)raiCm) = 1) s o Cm 7^ 0. Since E~ is A-invariant and since C is in E~, it follows that [A,C] is in E~. As (X, C) = 1 on M, we get 0 = A(X,C) = ([A,X],C) + (X,[A,C]). Since [A0,X0] = X0, we have [A,X] = X. Then {X,[A,C]) = -([A,X],C) = - ( X , C ) = - 1 . Then, by uniqueness in Lemma 12.3.1, p. 319, we conclude that [A, C] = — C. By Lemma 12.2.3, p. 318, for all m 6 M, we have E+ = R({X0)m), i.e., £ + = RXm. Moreover, by Lemma 12.2.3, p. 318, for all m € M, we get dim(£J~) = 1, so, since 0 ^ Cm G Em, we conclude that i£~ = RCm. Let A' := [X, C]. By the Jacobi identity, we have [A,A'] = [[A,X},C} + [X,[A,C)}. Then [A, A'] = [X, C] + [X, -C] = 0. Then, for all s e B, asA' = A'. Then A' is in £ ° . By the Jacobi identity, [A,[X,A']] = p , X ] , A ' ] + [X,[A,4']]. Then we have [A,[X,A']] = [X,A'\ + [X,0] = [X,A']. Let W := [X,A']. Then [A,W] = W. Then, for all s € K, we have a s W = esW, so W is in E+. So, since, for all m 6 M, we have .E+ = MXm, we choose / € C°°(M) such that W = fX. Then / X = W = [A, W] = [A, fX) = (Af)X

+ f[A, X] = (Af)X

+ fX.

Then, as X is nowhere vanishing on M, we get Af = 0. By Lemma 12.2.4, p. 318, since A is in E° and C is in E~, (A, C) = 0. Then0 = X(A,C) = ([X,A],C) + (A,[X,C)), so (A,[X,C\) = ([A,X],C). Then (A, A') = (A, [X, C)) = ([A, X], C) = (X, C) = 1. Then A1 is nowhere vanishing on M. Moreover, A1 is in E°, so, by Lemma 12.2.4, p. 318, A' is everywhere Lorentz orthogonal both to E+ and to E~. For all m € M, we have that Em and Em are distinct lightlike lines in TmM, so the intersection of their orthocomplements is positive definite. Then A' is everywhere spacelike. Moreover, as A' is in E°, as X is in E+ and as C is in E~, we conclude, for all m € M, that .A^, Xm and C m are linearly independent.

A candidate for the new Killing field

321

By Lemma 12.2.4, p. 318, E° is Lorentz orthogonal to E . So, since A' is in E° and since C is in E~, we have {A',C) = 0. Then 0 = X(A',C)

= ([X,A'],C)

+

(A',[X,C\),

so ([X, A'], C) = -(A', [X, C]). We have / = f(X, C) = (/X, C) = (W, C), so, since W = [X, A'], we get / = ([X,A'],C)

= -(A',[X,C\)

=

~(A',A').

Since A' is everywhere spacelike, we conclude that (A1, A') is a nowhere vanishing function on M. Then / is a nowhere vanishing function on M. Since gC — [C, A'], the Jacobi identity gives [X,gC] = [[X,C},A'} + [C,[X,A']}. Then [X,gC] = [A1, A'} + [C, W] = 0 + [C, fX). (Xg)C + g[X, C] = (Cf)X

Then

+ f[C, X],

so (Xg)C + gA' = (Cf)X - fA'. Then ( / + g)A' - (Cf)X + {Xg)C = 0. Since A', X and C are everywhere linearly independent on M, we see that / + g = 0, that Cf = 0 and that Xg = 0. Since / is nowhere vanishing on M and since g = —/, we conclude that 3 is nowhere vanishing on M. Let H := (l/g)A' and let Y" := (l/g)C. Since Xg = 0 and since g is nowhere vanishing on M, we get X(l/g) = 0. Then [tf,X] - [(1/ 5 )A',X] = (l/ff)[A',X] = -(1/ 5 )[X,A'] =

-(l/g)W.

So, since g = —f and since W = / X , we get [if, X] = X . We have Xg = 0 and Cg = -Cf = 0. By Lemma 3.1.1, p. 43, we have [X, C]g = XCg - CXg. Then A'g = [X, C]g = 0. Then A'(l/g) = 0. Since Cg = 0, we get C{l/g) = 0. Then [tf,Y-] = [(l/g)A',(l/g)C]

= (1/ 5 2 )[A',C]

= - ( 1 / 9 2 ) [ C , A'] = -(l/ff 2 )(ffC) = -(l/g)C We have Ag = -Af

= 0, so A(l/g)

= -Y.

= 0. Then

[A,Y] = [A, (l/g)C] = ( 1 / S ) H , C ] = - ( 1 / S ) C = - K Then [X,Y] = [X, (1/ 5 )C] = (l/g)[X,C]

= (l/g)A'

= H.

322

The Isometry

Group of a Compact Lorentz

Manifold

Since A{l/g) = 0 and since [A, A'] = 0, we get [A,H] = [A, (1/g)A'] = (l/g)[A,A']

= 0.

Finally, since X, A' and C are everywhere linearly independent on M and since 1/g is nowhere vanishing on M, it follows that X, H and Y are everywhere linearly independent on M. D L E M M A 12.3.3 For all mGM, TM, as s -> +oo.

for all v G E~\{0},

e~a>2 a,-sv —> oo in

Proof. As [A 0 ,X 0 ] = Xo, we conclude that [A,X] = X . Choose Y as in Lemma 12.3.2, p. 319. By Lemma 12.3.2, p. 319, we have [A,Y] = -Y. Also, by Lemma 12.3.2, p. 319, we see that X and Y are everywhere linearly independent on M. In particular, X and Y are nowhere vanishing on M. In particular, Xm ^ 0 ^ Ym. As [A,Y] = -Y, we see, for all s G E, that asY = e~sY. Then Ym G E^. By Lemma 12.2.3, p. 318, we have dim(Em) = 1. We conclude that E~ = MYm. Define <j>: E -> M by 0(s) = a_ s m. Choose a G IR\{0} such that v — aYm. Then, for all s G M, we have e~s/2a-sv — e~s/2a-sYm = e _s / 2 e*y0( s ) = e s / 2 Y^( s ). So, as Y is nowhere vanishing on M, the result follows. D Let G act on C°°(M) by (s/)(m) = / ( s _ 1 m ) . L E M M A 12.3.4 i e t / G C°°(M), let k G M\{0} and assume, for all s £ l , that asf = eksf. Then f = 0 on M. Proof. Replacing / by | / | , we may assume that / > 0 on M. We define P := {m G M | f{m) ^ 0} and assume, for a contradiction, that P ^ 0 . Let \i denote the Lorentz measure on M. Then, since P is open and nonempty, n(P) > 0. For all integers n > 1, let Pn := {m e M \ f(m) > 1/n}. Then P — [J Pn. Choose an integer r > 1 such that n(Pr) > 0. We will n

assume that k > 0; the proof for k < 0 is similar. By Lemma 7.4.24, p. 219, choose x € Pr and choose a sequence Sj in (0,+co) such that Sj -> +oo and such that, for all i, we have aSix G P r We have e - * ^ -4 0 in K. For all i, we have aSix G P r , so f(aSix) > 1/k. However, we have f(aSix) = (a,-3if)(x) = e~kBif(x) -+ 0, contradiction. D

323

A candidate for the new Killing field

L E M M A 12.3.5 For some Killing field Y on M, [{A0)M,Y]

= -Y.

Proof. Let H and Y be as in Lemma 12.3.2, p. 319. Then [A, Y] = -Y. Let g denote the Lorentz metric on M. Let h :— CYg. We wish to show that Y is a Killing vector field on (M,g), i.e., that h = 0. We have [A,Y] = -Y, so, for all s £ R, asY = e~sY. Then Y is in E~. Claim 1: h{X,X) = h(X,Y) = 0. Proof of Claim 1: For every s £ K, a3[(CYg)(X,X)}

= (Ca,Yg)(asX,a3X)

= es •

(£Yg)(X,X),

so as[h(X,X)] = es-h(X,X). For all s £ R, we have

Then, by Lemma 12.3.4, p. 322, h(X,X)

a.[(CYg)(X,Y)]

= e~s •

= (Ca,Yg)(a8X,asY)

= 0.

((CYg)(X,Y)),

so as[h(X,Y)] = e~s • (h(X,Y)). Then, by Lemma 12.3.4, p. 322, we have h(X, Y) = 0. End of proof of Claim 1. From Claim 1, we have h(X, X) = h(X, Y) = 0. Also, by Lemma 12.2.4, p. 318, for any vector field V on M , if V is in E°, then h(V, Y) = 0. Then we have h{E+,E+) = {0}, h(E+,E~) = {0} and h(E° ®E~, E° ® E~) = {0}. It remains to show that h(E+,E°) = {0}. By Lemma 7.5.4, p. 223, let Mi be a dense open subset of M such that, for all m £ M i , we have dim(EH m + RAm) = dim(Rff + MA). For all m £ M , we have Hm, Am £ E° and Xm £ E+, so dim(EA m + MHm + RXm) = 1 + dim(MATO + Rff m ). So, for all m € Mi, dim(!L4 m + RHm + MXm) = 1 + dim(MA + RH), so dim(IM m +MHm+RXm) > dim(RA+RH+RX). On the other hand, for all m € M , we have dim(R4 m +MHm +MXm) < dim(RA + RH +RX). Then, for all m G Mi, we get dim(IL4 m + MHm + RXm) = dim(DL4 + M.H + RX). Since M x is dense in M, it suffices to show that h(E+,E°) — 0 on Mi. Let m 0 G Mi and let w £ E^0. We wish to show that /i(.E+ 0 ,w) = {0}. By Lemma 12.2.3, p. 318, we have RXmo = E^0. It therefore suffices to show that h(Xmo, w) = 0. Claim 2: E° ® E+ is X-invariant. Proof of Claim 2: Fix t0 £ R. We wish to show that xto(E° © E+) C E° ®E+. Let v £ E° © S + Choose v' G E _ and u" G E° ® E+ such that xt0w = v' + v". Assume, for a contradiction, that v' ^ 0.

324

The Isometry

Group of a Compact Lorentz

Manifold

For all s £ R, let ts := e 3t0. Note that t3 ->• 0 as s -> +00. For all s £ 1, we have a-sxta = Xt,a-S; then a-sv' + a-sv" = a-sxtov Since

D 6 £ ° ®

= zt„a_st>.

E+, we get e _ s / 2 a_ s w -» 0, as s -» +00. Then e~s'2Xt,a,-sv

= Xtse~s'2a,-Sv

—• 0,

as s —> +co. Also, since v" £ E°(BE+, we see that a_ s w" —>• 0, as s —» +00. Then e~ s / 2 a_ s t/' —»• 0, as s —> +00. Then, as s —>• +00, we have e - s / 2 a _ s 7 / = e-s/2xt,a-sv

- e-s/2a,-sv'

^ 0 - 0 = 0,

However, since v' G E~~ and since v' ^ 0, this contradicts Lemma 12.3.3, p. 322. that e~8/2a-sv' -¥ 00 in TM, a s s - > +00, contradiction. £/nrf of proof of Claim 2. By Claim 2, we see that E° + E+ is X-invariant. By definition of E~ and E° and B + , we see that E~, E° and E+ are all three A-invariant. Moreover, since [H, A] = 0, it follows that E~, E° and E+ are all three H-invariant. Combining this with Claim 2, we see that E° + E+ is invariant under every element of the Lie algebra EA + MH + MA. By Lemma 4.2.5, p. 74 (with 0 replaced by MA + MB + MX and E replaced by E° + E+), choose an open neighborhood Mo of mo in M and choose a vector field W defined on Mo such that: • Wmo - w; • Wis in (£° + £ + ) | M 0 ; a n d • [H, W] = 0 and [X, W] = 0 and [A, W] = 0. It suffices to show that h(X, W) = 0 on M0. As h = CYg, h(X,W) = Y(g(X,W)) - g([Y,X],W) g(X,[Y,W}) on M 0 . By Lemma 12.2.1, p. 317 and Lemma 12.2.4, p. 318, we have g(E+,E° + E+) = {0}. Therefore, since X is in E+ and since W is in (E° + E+)\M0, we get g(X, W) = 0 on M 0 . We therefore wish to show that g{[Y, X],W)+ g(X, [Y, W]) = 0 on M 0 . Since X is a Killing vector field on (M,g), we have X(g(Y, W)) = g([X, Y], W) + g(Y, [X, W]). So, as [X,W] = 0, we get -X(g(Y,W)) we get g(X,[Y,W]) = g([H,X],[Y,W]). -X(g(Y,W))+g([H,X],[Y,W]) = 0.

= g([Y,X],W). As [H,X] = X, We therefore wish to show that

A candidate for the new Killing field

325

Since X is a Killing vector field on (M,g), we have X(g{H, [Y, W])) = g([X, H], [Y, W)) + g(H, [X, [Y, W}]). The Jacobi identity gives [X, [Y, W]} = [[X, Y],W] + [Y, [X, W]]. So, since [X, W] = 0, since [X, Y] = H and since [H, W] = 0, we get [X,\Y,W]] = [H,W] + \Y,0] = 0. Then X(g(H, [Y, W])) = g([X, H], [Y, W}), or ~X(g(H,[Y,W]))

=

g([H,X},[Y,W]).

We therefore wish to show that -X(g(Y,W)) - X(g(H,[Y,W})) = 0. It therefore suffices to show that g(Y, W) + g(H, [Y, W]) = 0. By Lemma 12.2.3, p. 318, for all m e M, we have MXm = £ + , so choose a vector field W0 in E°\M0 and f £ C°°(MQ) such that W = W0 + fX on M 0 . Since [A, W] = 0 and [A, X] = X, we get [A, W0] + ((Af)+f)X = 0. Since A preserves E°, we see that [A, Wo] is in E°. On the other hand, [A,W0] = ~((Af) + f)X, so [A,Wo] is in E+. For all m e M, we have E°m n E+ = {0}. Then [A, W0] = 0 on M 0 . So, since [A,Y] = -Y, the Jacobi identity yields [A, [Y,W0]} = -[Y,W0]. For all s G E, we have as[Y,W] = e-»[Y,W]. Then [Y,W0] is in E~\M0. We have g{Y, W) = g(Y, W0) + f • (g(Y, X)). By Lemma 12.2.4, p. 318, so, since Y is in E~ and since W is in E°, we get g(Y, Wo) = 0. It follows that g(Y, W) = f • {g{Y,X)). We have g(H,[Y,W])

= g(H,[Y,W0 + fX})

= (g(ff,[Y,W 0 ])) + {(Yf)-(g(H,X)))

+ {f •

(g(H,[Y,X}))).

Since i? is in E° and [F,VF0] is in E~\Mo, by Lemma 12.2.4, p. 318, we get g(H, [Y, W0]) = 0. Since H is in E° and X is in E+, by Lemma 12.2.4, p. 318, we get g(H,X) = 0. Then g(H,[Y,W]) = f • (g(H,[Y,X})). So, as [X, Y] = H, we get g(H,[Y, W}) = -f • g(H,H). We therefore wish to show that [/ • {g(Y,X))] - [f • {g(H,H))] = 0. It therefore suffices to show that g(Y,X)=g{H,H). Since Y is in E~ and since H is in E°, by Lemma 12.2A, p. 318, we have g(Y,H) — 0. Since X is a Killing vector field on (M,g), we have X(g(Y,H)) = g([X,Y},H) + g(Y,[X,H]).

326

The Isometry

Then 0 = X(g(Y,H))

Group of a Compact Lorentz

= g([X,Y],H) g([X,Y],H)

Then g(H, H) = g([X, Y],H)= 12.4

=

Manifold

+ g(Y, [X,H]), so g(Y,[H,X\).

g(Y, [H, X}) = g(Y,X).

•

Final classification result

Let C be the collection of Lie algebras consisting of (1) (2) (3) (4)

*fe(R); the Heisenberg Lie algebras; the rationally twisted Heisenberg Lie algebras; and the trivial group {1}.

Notice that C is similar to the collection C denned just before Theorem 11.7.3, p. 313, except that the Lie algebra flo is omitted. The following is the main result of [AS97b] and of [Ze98]. T H E O R E M 12.4.1 Let G be a connected Lie group. Assume, for some compact connected Lorentz manifold, that G is Lie group isomorphic to UCIsom°(M). Then there exist

• an Abelian Lie algebra a; and • a compact, semisimple Lie algebra 6 such that g is Lie algebra isomorphic to g' © o © t. For simply connected Lie groups, the converse of Theorem 12.4.1, p. 326 is true. That is, if G is a connected, simply connected Lie group and if there exist • fl' € C; • an Abelian Lie algebra o; and • a compact, semisimple Lie algebra 6 such that fl = fl'ffia©6, then there is a compact connected Lorentz manifold M such that G S UCIsom°(M). For detailed proofs of the statements in this section, we refer the reader to [AS97b] and to [Ze98].

Chapter 13

Highly Symmetric Compact Lorentz Manifolds

13.1

SL 2 (K)-actions on compact Lorentz manifolds

Let G be a connected, simply connected Lie group and assume that g is Lie algebra isomorphic to s ^ K ) - Let K £ SBF(fl) denote the Killing form of g; then K is an (Ad G)-invariant Minkowski form. Let A denote the left-invariant Lorentz metric on G satisfying AiG = K. Since K is (AdG)invariant and since A is left-invariant, it follows, from Lemma 4.3.4, p. 78, that A is right-invariant, as well. Let G act locally faithfully and isometrically on a compact connected Lorentz manifold M. Let M be the universal cover of M. In this section, we expose a result of M. Gromov (see 5.4. A of [Gr88] or 0.13.A of [D'A88]) which asserts: T H E O R E M 13.1.1 There is a connected, simply connected manifold N, a Riemannian metric n on N, a smooth map h : N —>• (0, oo) and a diffeomorphism f : M —> G x N such that (1) for all m 6 M, for all g,g' € G, for all n € N, if f(m) = (g',n), then f(gm) = (gg',n); (2) ifh:GxN-> (0,oo) is defined by h(g,n) = h(n), and if'y :— Axr] denotes the product Lorentz metric onGxN, then f is an isometry of M and (N^-j). In other words, M is isometric, via a G-equivariant isometry, to a product of (G, A) with some Riemannian manifold (N^n), except that the product metric A x r/ might be multiplied by some real-valued, everywhere positive 327

328

Highly Symmetric

Compact Lorentz

Manifolds

function on N. The proof of Theorem 13.1.1, p. 327 appears at the end of this section. We already know (Theorem 10.5.3, p. 292) that, among connected simple Lie groups, the only groups admitting a nontrivial, isometric action on a compact connected Lorentz manifold are those locally isomorphic to SL2(M). Theorem 13.1.1, p. 327 essentially classifies the possible actions of such groups, up to covers. So the problem of understanding isometric actions of finite-center simple Lie groups on compact Lorentz manifolds is, for the most part, finished. For all m 6 M, let (-,-),„ be the symmetric bilinear form on TmM, and define Bm £ SBF(fl) by Bm(X,Y) = (Xm,Ym). L E M M A 13.1.2 For all m G M, there is cm > 0 such that Bm = cmK. Proof. Since g S s ^ M ) , K is Minkowski. By Lemma 11.1.4, p. 295, we know, for all m € M, that Bm is (Ad G)-invariant. As G is simple, it follows that the Adjoint representation of G on g is irreducible. Then, by Lemma 5.7.12, p. 172, the Adjoint representation of G on g has no nonscalar intertwining. Consequently, by Lemma 5.7.2, p. 167, for all m € M, there exists c m € R, that Bm = cmK. Fix m G M. We wish to show that cm > 0. By Lemma 11.1.7, p. 296, the action of G on M is locally free, so the map X \-t Xm : g ->• TmM is injective. Therefore dim(g m ) = dim(0) = dim(s[2(M)) = 3. So, by Corollary 6.2.3, p. 176, gm is not isotropic. Choose X € gm such that (Xm,Xm) ^ 0. Then cm • K(X,X) = Bm(X,X) = (Xm,Xm) ^ 0. It follows that c m 7^ 0. Assume, for a contradiction, that cm < 0. Let po denote the set of symmetric matrices in s ^ W - Then the restriction to po x Po of the Killing form of s ^ (R) is positive definite. Then g contains a two-dimensional vector subspace p that is positive-definite with respect to K, so p is negative definite with respect to Bm. Then p m is a two-dimensional, negative definite vector subspace of TmM, contradicting Lemma 6.2.2, p. 176. • By Lemma 11.1.7, p. 296, the action of G on M is locally free. Let A be the vector subbundle of TM defined by A m := gm. Then, by Lemma 13.1.2, p. 328, for all me M, A m is Minkowski. For all m € M, let Ai

:= {v e TmM | Mw 6 A m , (v, w)m = 0}.

SL.2(E) -actions

on compact Lorentz

manifolds

329

Then, for all m E M, Am is positive definite. Let A1- :— M

Am, so that

m£M

A1- is the vector subbundle of TM orthogonal to A. The left and right translation actions of G on G induce, by differentiation, left and right actions of G on TG. The action of G on M induces, by differentiation, an action of G on TM. Let G act on the set of vector fields on M by the rule: (gV)m g(Vg-im). L E M M A 13.1.3 Let m e M and let v € A m \ { 0 } . Then there exists a sequence gi in G such that giV —> oo in TM. Proof. Choose X e g such that Xm — v. Then X ^ 0. By Lemma 4.12.5, p. 134, choose a sequence gi in G such that (Adgt)X —> oo in g. Passing to a subsequence, assume, for a contradiction, that g^ is precompact in TM. For all i, let X{ := (Adgi)X and let mi := g^m. Then g^v = (Xi)mi. Passing to a subsequence, choose X' € fl\{0} and choose a sequence U in E such that ti —> 0 in E and such that UXi —> X' in g. Passing to a subsequence, choose m' 6 M such that mj —> m'. Let v' := (X')mi. Then ti{g%v) —> v'. So, since g^ is precompact in TM and ti —> 0, we conclude that v1 = 0. Then X' ^ 0 6 g and (X')m' = 0 e T m -M, contradicting Lemma 11.1.7, p. 296. • L E M M A 13.1.4 The distribution A1- is integrable. Proof. For any two vector fields X and Y in A x , we will denote by A(X, Y) and B(X, Y) the vector fields in A and in A x , respectively, such that [X, Y] = A(X, Y)+B(X, Y). Let q : TM - » M b e the tangent bundle map and define T := {(v,w) G A1- x A x | q(v) = q(w)}. For all / € C°°(M), for any two vector fields X and F i n A 1 , we have [X,fY] = f[X,Y] + (Xf)Y, so A([X,fY]) = f[X,Y]; a similar argument shows A([fX,Y}) = f[X,Y]. Because of this C°°-bilinearity of A, there is a smooth vector bundle map a : T —> A such that, for all vector fields X and Y in A1-, for all m e M, we have (^4(X,F)) m = a(Xm,Ym). Let X and Y" be vector fields in A-1. By Theorem 3.7.5, p. 57, we wish to show that [X,Y] is in A x . Let V := A(X,Y) and W := B(X,Y). Then, since [X, Y] = V + W and since W is in A-1-, it suffices to show that V = 0. Assume, for a contradiction that V ^ 0. Fix m0 E M such that V ^ ^ 0. By Lemma 13.1.3, p. 329, choose a sequence gt in G such that gi(Vmo) -¥ oo in T M . For all i, we define

330

Highly Symmetric Compact Lorentz Manifolds

mi := giino. For all i, we have giV = gi(A(X,Y)) = A(giX,giY); giiYmo) = (giV)mi = (A{giX,giY))mi. Then, for all i, we have 9i(Vmo) = a((giX)mi,(giY)mi)

=

then

a(gi(Xmo),gi(Ymo)).

So, since gi{Vmo) - • oo in TM, it follows that {gi(Xmo)}i U {gi(Ymo)}i is not precompact in TM. Define 7 : A-1 ->• M by j(v) = (v,v). For all g G G, for all v £ A 1 , we have j(gv) = y(v). For all K G K, let H^ := {w G A x |7(w) < K). By Lemma 13.1.2, p. 328, for all m G M, A m is Minkowski, so, for all m G M, Am is positive definite. That is, the restriction to A1- of the Lorentz metric on M is a Riemannian bundle metric on A-1. So, since M is compact, it follows, for all i f e i , that EK is a compact subset of A -1 . Let KQ := max{7(X m o ),7(y m o )}. Then, for all i, we have l(9i(Xmo))

= -y(Xmo) < KQ

and

7 ( ^ ( ^ 0 ) ) = l(Ym0)

so gi(Xmo) e EKo and gi(Ymo) G Ejf0. Since {gi(Xmo)}i not precompact in TM, we have a contradiction.

< K0,

U {si(y mo )}i is •

Proof of Theorem 13.1.1, p. 327: For all m G M, choose c m as in Lemma 13.1.2, p. 328. Since A is Lorentz, we see that Ax is Riemannian. By Lemma 13.1.4, p. 329, let T be a foliation of M such that TT = A- 1 . Let iVo be a leaf of T. Let 77 be the restriction to TNQ of the Lorentz metric on M. Since A is Lorentz, it follows that 77 is a Riemannian bundle metric on TN0. That is, 77 is a Riemannian metric on JV0. Define /o : G x N0 -> M by fo(g,n) — gn. Let TT : M —> M be the universal covering map. For all g £ G, for all X G 0 = Ti G G, we have Xg G TflG, so, for all n£N, for all Y G TniV, we have (Xg,Y) G TgG@TnN = T(g
G TnN0,

Then fi0 is a Lorentz metric on G x NQ. Claim 1: f0 : G x N0 -> M is isometric. Proof of Claim 1: Fix g0 G G and no G iVo- Let ZQ :— {go,no) G G x No and let m0 :— fo(zo) = <7o"oFix X G 0 and Y G TnoiV"o. Let W := (d/oW-X'So.lO- W e w i s h t o s h o w that p0((Xg0,Y),(Xgo,Y)) = (W,W)mo.

Actions of twisted Heisenberg groups on compact Lorentz manifolds - a sketch

We have (dfo)Zo(Xg0,0) = Xmo and that {dfo)Zo(0,Y) = g0Y. (dfo)Zo(Xgo, Y) — Xm0 + goY. We therefore wish to show that cnoK(X, X) + (Y, Y)no = (Xmo + g0Y, Xmo +

331

Then

g0Y)mo.

Since (X,Y) e ( A , A x ) = {0}, we get (Xmo +g0Y,Xmo

+ goY)mo = (Xmo,Xma)mo

+

{goY,g0Y)mo.

Since (•, •) is G-invariant, we conclude that (Y,Y)no = (goY,goY)mo. therefore wish to show that cnoK,(X,X) = (Xmo,Xmo)mo. By construction of c and B, we have cmoK[X,X)

= Bmo(X,X)

—

We

(Xmo,Xmo)mo.

It therefore suffices to show that c m o = c„ 0 . Fix P, Q 6 g such that K(P, Q) ^ 0. We wish to show that c m o • K(P, Q) = c„0 • K(P, Q). Let P' := {kdg)~xP and Q' := (Adg)~ l Q. Since mo = ffo^o, we have c m o • K ( P , g ) = Bmo(P,Q)

- 5„ 0 (P',Qf)

= cno •

K(P',Q').

Since K is (adg)-invariant, we conclude that n(P, Q) = K(P', Q'). It follows that c m o • (K(P, Q)) = c n o • (K(P, Q)). End of proof of Claim 1. By Lemma 7.4.25, p. 219, we see that /o : G x No —>• M is a covering map. By covering space theory, choose a covering map / i : M —> G x NQ such that fo ° fi = n- Let N denote the universal cover of iVo and let p : N —> No be the universal covering map. Define q : G x N —> G x No by q(g,n) = (g,p(n))- By covering space theory, choose a diifeomorphism / : M -¥ G x N such that q° f = fiThe required properties of / follow from the construction and Claim 1. End of proof of Theorem 13.1.1, p. 327. 13.2

Actions of twisted Heisenberg groups on compact Lorentz manifolds — a sketch

Let G be a twisted Heisenberg group and let N be the nilradical of G, so TV is a Heisenberg group. In this section, we sketch a result, due to A. Zeghib, which characterizes the possible actions of G on compact Lorentz manifolds. (See §1.6 of [Ze95].) Let Z := Z(G) act smoothly and locally freely on a manifold N. Let S denote the set of (Ad iV)-invariant Minkowski symmetric bilinear forms

332

Highly Symmetric

Compact Lorentz

Manifolds

on Q. Let Si denote the collection of all K G S such that, for any twodimensional vector subspace V of g, we have K(V, V) ^ {0}. (That is, <5i consists of those elements of S not admitting a two-dimensional isotropic vector subspace.) We leave it as an exercise to the reader to verify that iSi C MinkSBF(g) and that, for all K G SI, K is (AdG)-invariant. For all K G S, K is (AdG)-invariant, so, by Lemma 8.2.5, p. 267, we see that Si 7^ 0. For all K G <Si, let XK denote the left-invariant Lorentz metric such that (AK)iG = K. By Lemma 4.3.4, p. 78, we see that XK is right-invariant, as well. Fix, for this paragraph, a smooth, Z-invariant map h : N ->• Si and a Z-invariant Riemannian metric rj on N. We define a smooth G- invariant Lorentz metric 7° on G x N as by the rule: for all x — (g,n) E G x N, for all (v, w), (1/, w') G TgG © TnN = Tx(GxN), 72,„((«,«'), ( « > ' ) ) = Wn))(v,v')}

+ [V{w,w')].

Let N' := G x z J V . Let p : G x N ^ N' denote the canonical map. For all (g,n) G G x N, let [g,n] := p(ff,n) G N'. For all x = (g,n) e G x N, let fx : Z -> G x N be defined by /^(z) = (gz, z _ 1 n ) . For all x G G x AT, let Ex := (dfx)(i) C ^ ( G x JV); then 7° ]7) |S X is positive definite. For all x = {g,n) € GxN, let E^ CTx(Gx N) denote the orthogonal complement t o S , , in Tx{Gx N). Then, for all x G G x N, qx := (dp)|S^- : S^- -> Tp^N' is a vector space isomorphism. As S-1 is JV-invariant and 7°„|S J - is A^-invariant, there is a unique Lorentz metric 7^,, on N' such that (P*(7/I,TJ))|S J - = 7 ^ ^ - * - . The procedure defined above gives a method for producing a locally faithful, isometric action of G on a compact connected Lorentz manifold. Conversely, let G act locally faithfully and isometrically on a compact connected Lorentz manifold M. Let 7 denote the metric on M. Let M be the universal cover of M and let 7 be the lifted Lorentz metric on M. When we try to apply the argument of §13.1, p. 327 to a twisted Heisenberg group G, a new complication occurs because of the fact that these groups have one-dimensional center. Fix ZQ G (3(fl))\{0). Then H^o = 3(fl)- In the notation of Theorem 13.1.1, p. 327, for all m € M, we can prove that Am(Am x A m ) C R(Z0)m, but we cannot conclude that Am(Am x A m ) = {0}. So, in this situation, we cannot conclude that A1- is integrable. However, the argument can be modified to show that A x + M.(ZQ)M is integrable. Let AT be a leaf in the resulting foliation. Then

Actions of twisted Heisenberg groups on compact Lorentz manifolds - a sketch

333

N in (Z(G))-invariant. By Lemma 11.1.5, p. 295, we see that the action of G on M is locally free. For all m G M, let {•, • ) m be the symmetric bilinear form on TmM, and define Bm e SBF( 0 ) by Bm(X,Y) = (Xm,Ym). For all m 6 M, since <Si consists of Minkowski forms, we see that A m = 0m is Minkowski, so A m is positive definite. Then A1- is G-invariant, carries a G-invariant Riemannian metric, and is complementary in TN to ^(ZO)MGive R(Z0)M the Riemannian bundle metric such that (Z0)M has norm one everywhere, and combine this with the Riemannian bundle metric on A x . Then we arrive at a G-invariant Riemannian metric on N. Following the remainder of proof of Theorem 13.1.1, p. 327, we get: T H E O R E M 13.2.1 There is a connected, simply connected manifold N, a Riemannian metric r/ on N, an action of Z := Z(G) on N, a smooth Z-equivariant map h : N —> Si and a diffeomorphism f : M -> G Xz N such that (1) for all m € M, for all g,g' € G, for all n E N, if f(m) — [g',n], then f(gm) = [gg',n]; (2) / . ( 7 ) = 7 M This result appears as Theoreme 1.14 of [Ze95]. Since it is not difficult to classify the elements of Si, we see that Theorem 13.2.1, p. 333 provides a very clear picture of the possible actions of twisted Heisenberg groups on compact Lorentz manifolds. A similar result for actions of (untwisted) Heisenberg groups remains elusive.

Chapter 14

Locally Free Orbit Nonproper Lorentz Actions

In Chapters 8, 11, 12 and 13, we discussed symmetries of compact Lorentz manifolds. In this chapter, we go beyond compact manifolds, so we must take into account the fact that any Lie group will act on itself (by left translations) preserving any left-invariant Lorentz metric. Moreover, any Lie group G admits a left-invariant Lorentz metric: Let Q be a Minkowski quadratic form on T\GG, let G act on TG by the differential of left translation and, for all g E G, define j g G QF(TgG) by jg(v) = Q{g~xv). Then 7 is a left-invariant Lorentz metric on G. Thus, if we wish to describe the collection of connected Lie groups which admit a free, isometric action on a connected Lorentz manifold, then the answer is: All do. However, one of the key insights of N. Kowalsky in [Ko96] is that if one requires even such a weak dynamical condition as nonproperness, then the list of Lie groups admitting an isometric action on a (possibly noncompact) Lorentz manifold is quite restricted. In this chapter, pick up on this theme. Recall that ONV is the collection of connected Lie groups admitting a locally faithful, orbit nonproper, isometric action on a connected Lorentz manifold, and that ONV1* is the collection of connected Lie groups admitting a locally free, orbit nonproper, isometric action on a connected Lorentz manifold. In the present chapter, we aim to describe effectively the collection ONV1*. As is typical for Lie theoretic questions, one of our first goals should be to describe the collection of nilpotent Lie groups in OAfV1?; however even this turns out to be somewhat difficult: In Example 9.3.1, p. 279, we showed that it is possible to construct a connected nilpotent Lie group N with compact center such that N £ OMV. On the other hand, by (2) 335

336

Locally Free Orbit Nonproper Lorentz

Actions

of Theorem 14.3.2, p. 342 below, we will see that every connected, simply connected nilpotent Lie group is in ONVlf because any such group admits a normal closed subgroup isomorphic to the additive Lie group E. In fact, an effective determination of the collection of nilpotent Lie groups in ONV1* remains an open, though possibly tractable, question. Here, we will avoid this problem by only describing (in Theorem 14.3.2, p. 342) the subcollection of groups in ONV1* with simply connected nilradical. By Lemma 4.15.1, p. 141, any connected, simply connected Lie group has simply connected nilradical. So, as ONV1* is closed under passage to covering groups, we can claim to have an effective description of the simply connected Lie groups in ONV1*. Equivalently, we have an effective description of the Lie algebras of Lie groups in OAfV1?. Let G be a connected Lie group with simply connected nilradical. In Lemma 14.6.2, p. 353, we show that if G admits an isometric, locally faithful, orbit nonproper action on a connected Lorentz manifold such that every connected stabilizer is compact, then G € ONV1*. So, in Chapter 15, where we analyze OMV, we will begin with the fact that if G € {ONV)\{OMVlf) and if G has simply connected nilradical, then G admits an isometric, locally faithful action on a connected Lorentz manifold such that some connected stabilizer is noncompact. Recall that AfV is the collection of connected Lie groups admitting a locally faithful, nonproper, isometric action on a connected Lorentz manifold, and that AfV1* is the collection of connected Lie groups admitting a locally free, nonproper, isometric action on a connected Lorentz manifold. We expect that AfV = OAfV and that NVlf = OAfVlf, but, for now, both of these are open questions.

14.1

Preliminary result to degeneration of Adggi

Let / := [0, +00) be the topological space of nonnegative real numbers and let I := [0, +00] be the one-point compactification of [0, +00). Let / have the usual linear ordering <. Note, for all a G I, that a < +00. Let V be a vector space. Let d := dim(V). For all integers j € [1, d], let ej := efK For all integers j,ke[l,d], let Ejk := E$. Let A+ := {Ai-Bn + • • • + \dEdd I + 00 > Ai > • • • > \d > 0} C Idxd. Let A+ := {\iEu

+ ••• + \dEdd \ + 00 > Ax > • • • > \d > 0} C J d X d . Note

Degeneration

of Adggj

337

that A + is a compact topological space. Let <S denote the topological space of ordered bases of V. Let T; be a sequence in End(V). Let Bi and B\ be two sequences in <S. We say Ti is diagonally a d a p t e d t o Bi, B[, if all of the following hold: • for all i, we have [T\g\ € A+; • Bi and B\ are both convergent in <S; and • \T\Q. is convergent in A + . The following is based on Lemma 11.1, p. 477 of [AOOa]. L E M M A 14.1.1 Let Ti be a sequence in End(V). Then, after passing to a subsequence, there exist two sequences Bi and B[ in S such that Ti is diagonally adapted to Bi, B\. Proof. Let H := GL d (E). Then H acts on S by the rule: hB is the image under the vector space isomorphism v H-> (j>B(h~1v) : Rdxl -> V of the standard ordered basis ( e i , . . . , e<j) of E d x l ; then, for all v € Rdxl, we have <j>hB(v) = ^(h^v). Let d and C[ be two precompact sequences in S. For all i, we define fi := [Tifc\ = cl o Ti o <j)Ci. Let K := 0(d). Then Rdxd = KA+K, so choose sequences ki and U in K such that, for all i, we have kifiU € A+. For all fii,..., na € % let diag(/n,. ..,nd)

:= AiiBn + • • • + / ^ d e M d x d .

For all i, let Bi := l^d and let B\ :- kiC[. Then {Bi}i and {B^}i are both precompact in 5 . Moreover, for all i, [Tiltf. = kifiU, so pi] B " G A+. Because «S is compact, passing to a subsequence, we may assume that Bi and B[ are both convergent in <S. As A+ is compact, after passing to a subsequence, \Ti\B\ converges in A + . • 14.2

D e g e n e r a t i o n of

Adggi

This section is adapted from §17, pp. 495-498 of [AOOa]. Let p be a positive integer and let G be a p-dimensional connected Lie group. Assume that G is Ad-proper. Let M be a connected Lorentz manifold on which G acts isometrically. Assume that the action of G on M is locally faithful.

338

Locally Free Orbit Nonproper Lorentz

Actions

Let gt be a sequence in G. Let rtii and m\ be sequences in M. Assume, for all i, that giTUi = m^. Let moo, m ^ € M. Assume that rrn —> m^ in M, that m!i —> m'^ in M and that gi -¥ ooia G. Assume that Stab^rrioo) and Stab^mJ^) are both compact. Let / := [0, +00) be the topological space of nonnegative real numbers and let / := [0,+oo] be the one-point compactification of [0,+00). Let i" have the usual linear ordering <. Note, for all a E I, that a < +00. For all i, let bn,...,bip,b'il,...,Vip £ g. Assume both Si :=(fci, •••,&*)

and

B\ := (b'a,...

,b'ip)

are ordered bases of g. Let tool, • • •, &oop> &ooi' • • • > ^'oop Boo := (&001, • • •, boop)

and

e

fl- Assume both

B'^ := (6^1 > • • •. b'ooP)

are ordered bases of 0. Assume that Bi -¥ Boo and that B\ -¥ B'^. For all i, for all integers j 6 [l,p], let Ay > 0 and assume that (Adg%)bij = Ayfty. For all integers j € [l,p], let AQQJ 6 / and cLSSUIHG t licit \{j —^ AQQJ in I, as i -¥ 00. For all i, assume that A;i > • • • > Ajp. For all m € M, let (•, • ) m be the symmetric bilinear form on TmM, and define Bm G SBF( 0 ) by Bm(X,Y) = (Xm,Ym). For all fi1,..., np 6 K, let diag(/zi,..., fj,p) — HiE^ H 1- fj,pEpp'. Let A + := {diag(/ii,..., fj.p) \ + 00 > fii > • • • > fip > 0}. L E M M A 14.2.1 Assume that Aioo = +00. forAdg{gi). Then X € Wb'^ .

Let X € g be Kowalsky

Proof. Choose a sequence Xi in g such that Xi ~¥ 0 in 0 and such that (Adgi)Xi - • X in 0. We have ba -+ 6001, A»i -> +00, Xi->0,

(Adgi)Xi-*X

and

(Ad&X&a/Aa) -4 i ^ .

Then {X,b'^} is Kowalsky for AdB(<7j), so, by Lemma 7.10.1, p. 236, we see that {X,^^} is strongly lightlike at mj^. By (2) of Lemma 7.8.3, p. 228 and by the assumption that S t a b ^ m ^ ) is compact, we see that dim(RX + Mtooi) < 1. As b'^ ^ 0, the result follows. • L E M M A 14.2.2 Either Aooi = +00 or Aoop = 0. Proof. For all i, let Ft := [Ad 0 (&)]^ e GL P (R). As Ad : G -> GL(g) is proper and as gt -> 00 in G, we know that Ad0(<7j) -+ 00 in GL(0). So, as Bi

Degeneration o/Adgg;

339

and B\ are both convergent, we get Fj —> oo in GL P (R). For all i, we have Fi = diag(Aii,..., Xip). We conclude that Ft -t oo in A+. Then, for some integer j € [l,p], either AQOJ = +00 or AQOJ = 0. So, as Aooi > • • • > AooP, we are done. D L E M M A 14.2.3 Let j e [l,p] 6e on integer. If Xooi = +00 and

Proof.

AQOJ

^ 0,

We have A^Ay —• +00, we have [&a>&y] —• [&ooi,&ooj] and we

have (Adgi) ( \

1

\

) -> [&'ooi.&'ooj]>

as

* "^ +°°-

Then Lemma

p. 338 concludes the proof. COROLLARY 14.2.4 If X^

14.2.1, •

^ 0, then M1^

is an ideal of g.

Proof. By Lemma 14.2.2, p. 338, A^i = +00. Since AooP ^ 0 and since Xoop < • • • < Aooi, it follows, for all integers j € [l,p], that AQOJ 7^ 0. By Lemma 14.2.3, p. 339, for all integers j e [l,p], we have [ ^ 1 , 6 ^ ] £ IK&coi • As 5 ^ is an ordered basis of g, we conclude that Wool is an ideal of g. O L E M M A 14.2.5 We have A002 < +00. Proof. Suppose that A002 = +00. We aim for a contradiction. Since A^i > Aoo2 = +00, we see that A^i — +00, We have Xa —> +00, bi2 ->• &002 and (Adpj)(6j2/Ai2) ->• &£»2- So, by Lemma 14.2.1, p. 338, 6^2 6 Kftjjo!, which contradicts the fact that B'^ is a basis of g. O For all i, for integers j € [l,p], let Ay := l/Ai ) P +i_j, let fry := b'ip+l_j and let b'^ := 6j ] P + i_j. For integers j e [l,p], let Aooj := l/Aoo,p+i-j, let booj := b'^p+^j and let 6^- := b^p+i-j For all i, let & := 5 ~ \ let m; := m^ and let rh^ := mi. L E M M A 14.2.6 If X^i ^ +00, then Koop »'* on » 0.

340

Locally Free Orbit Nonproper Lorentz Actions

Proof. This proof is similar to the proof of Lemma 14.2.5, p. 339, except that we replace Ay with Ay, 6y with 6y, b'^ with 6y, AQOJ with Xooj, &ooj with booj, b'cej with b'^p gt with (ft, m; with Aj and m£ with m£. D L E M M A 14.2.8 Suppose thatg admits no one-dimensional ideal and that s b W is not a direct summand ofg. Let N denote the nilradical ofG. Then (1) -f-oo — AQOI > Aoo2 > • • • > AQQ^-I > Aoop = 0; (2)nB(b'ool) = Mb'ool+----rm'00^1; (3) the codimension in g ofn^b^) is = 1; (4) 6^i G j(n), so, in particular, n ^ {0}; (5) for all X € n\{0}, for all m G { f f i o o ^ ^ } , we have Xm ^ 0; (6) for all X e n ^ b ' ^ ) , we have Bm^b^^X) = 0; and (7) Bmi |(n x n) is positive semidefinite; its kernel isffi&^i• Proof. Proof of (1): By Lemma 14.2.6, p. 339, we have A ^ = +oo. By Corollary 14.2.4, p. 339, we have A^p = 0. By Lemma 14.2.5, p. 339, we have A,^ < +00, so Ao^ < Aooi- By Lemma 14.2.7, p. 339, we have Aoo,p-i > 0, so Aoo,p_i > Aoop. By construction, A ^ > • • • > Aoo)P_i. Proof of (2): Since g admits no one-dimensional ideal, it follows that n a(&ooi) 7^ 0' s 0 ^ suffices to show that

w>'00l+--- + m'00tP_1 c n , ( C ) . So fix an integer j € [l,p— 1]. We wish to show that [b'00l, b'^j] £ IRo^i. By (1) of Lemma 14.2.8, p. 340, we have Ao^p—1 > 0. Then A ^ > Aoo,p_i > 0. By (1) of Lemma 14.2.8, p. 340, we have A^i = +00. Then Lemma 14.2.3, p. 339 concludes the proof. Proof of (3): This follows from (2) of Lemma 14.2.8, p. 340. Proof of (4): By (1) of Lemma 14.2.8, p. 340, we have Aooi = +00. Then b'^ is Kowalsky for Adg(gi), so, by Lemma 4.9.31, p. 109, we see that ad b'^ : g -¥ g is nilpotent. By (3) of Lemma 14.2.8, p. 340, codim B (n 0 (6' ool )) = 1. By Lemma 5.6.3, p. 165, we have b'^ € n. By Lemma 5.6.4, p. 166, we get 0 ^ € j(n). Proof of (5): Since g admits no one-dimensional ideal, it follows, from Corollary 4.15.12, p. 148, that Z(G) is discrete and that N is simply connected. Since TV is a connected, simply connected, nilpotent Lie group, it follows, from Lemma 4.14.6, p. 140, that N admits no nontrivial, compact subgroup. Then, because S t a b ^ m ^ ) and S t a b ^ m ^ ) are both compact, Stab^moo) a n d Stab^(m c o ) are both trivial. The result follows.

Locally free actions of simply connected Lie groups

341

Proof of (6): Fix an integer j € [l,p - 1]. By (2) of Lemma 14.2.8, p. 340, it suffices to show that -Bm'^C&roi > ^'ooj) = 0By (1) of Lemma 14.2.8, p. 340, we have An Ay -> +oo. Moreover,

) and Bm,(b'n , fy) -»• ^ ( ^ ,

b'^).

For all i, we have (Adgi)bn = A;i&a and ( A d g , ) ^ = Aj 2 ^ 2 - This implies, for all i, that Bmi(bn , 6^-) = Bm<(Aii&a , A ^ ) , so ^ ( O i i , oi:j) =

——

.

Letting i -> +oo, we get B m i o ( 6 ^ , 6 ^ ) = 0. Proof of (7): Let Q be the Minkowski quadratic form on Tmi M. By (5) of Lemma 14.2.8, p. 340, it suffices to show that Q' := Q\nmi is positive semidefinite and that M(6^ol)m'oo = ker(<3'). By (4) of Lemma 14.2.8, p. 340, we see that n C cB(bJX)1), son C n 8 (6J )ol ). So, from (6) of Lemma 14.2.8, p. 340, we have B ^ ^ n i i ) = {0}. By (4) of Lemma 14.2.8, p. 340, we also see that b'^ 6 n. Then b'^ £ ker(B m ; J(n x n)). By (5) of Lemma 14.2.8, p. 340, we have (b'^U^ ? 0. Then 0 # (b1^)^ € ker(B m .J(n x n)) = ker(Q')Then Q' is degenerate. Then, by (1) and (2) of Lemma 6.2.6, p. 178, Q' is positive semidefinite and dim(ker(Q')) = 1. So, as 0 ^ (fc^i)™^ £ ker(<3'); we see that ker(<2') = ^ ( 6 ^ )m>ae. •

14.3

Locally free actions of simply connected Lie groups

Recall that ONV1* is the collection of connected Lie groups admitting an orbit nonproper, locally free, isometric action on a Lorentz manifold. In this section, in Theorem 14.3.2, p. 342, we effectively describe the collection of simply connected Lie groups in ONV1*. The only proof I know of the following lemma requires certain facts about algebraic groups and algebraic dynamics, which I do not wish to develop here. Consequently, the proof given below depends on some results in papers in the literature. It would be nice to find a proof that avoids the theory of algebraic groups. L E M M A 14.3.1 Let G be a connected Lie group and assume that the Adjoint action of G on MinkQF(g) is orbit nonproper. Then at least one of the following holds:

342

Locally Free Orbit Nonproper Lorentz

Actions

(1) G is Ad-nonproper; (2) G admits a normal closed subgroup isomorphic to E; or (3) s l 2 ( E ) | 0 . Proof. Assume, for a contradiction, that (1), (2) and (3) all fail. Let G be the Zariski closure of Ad B (G) in GL(g). By Lemma 19.3, p. 503 of [AOOa], we see that the Adjoint action of G on MinkQF(g) has compact connected stabilizers. Then, by Proposition 9.2, p. 475 of [AOOa] (with G replaced by G, with X replaced by SBF(g) and with U replaced by MinkQF(g)), we conclude that the G-action on MinkQF(g) is orbit proper. Because G is an Ad-proper Lie group, the map Ad : G —> G is proper. Then the Adjoint action of G on MinkQF(g) is orbit proper, contradiction. • By Lemma 4.15.1, p. 141, a simply connected Lie group has simply connected nilradical. Thus the following result effectively describes the class of simply connected groups in OMV1^. T H E O R E M 14.3.2 Let G be a connected Lie group with simply connected nilradical. Then G € OAfV1* iff at least one of the following holds: (1) G is Ad-nonproper; (2) G admits a normal subgroup isomorphic to E; or (3) sl 2 (E)| f l . Proof. By Lemma 7.14.13, p. 260, we have (1) = > G e ONV11. By Lemma 9.5.1, p. 284 we have (2) =*> G G OMVli'. By Lemma 9.4.1, p. 281, we have (3) =>• G € OMVls. Now assume that (1), (2) and (3) fail. Let G act locally faithfully and isometrically on a connected Lorentz manifold M. Assume, for a contradiction, that the G-action on M is orbit nonproper. Choose mo, m' £ M and choose a sequence oo in G and such that gtmo -> m'. Since G has simply connected nilradical and since G admits no normal subgroup isomorphic to E, it follows, from Lemma 4.15.4, p. 142, that g admits no one-dimensional ideal. Let p := dim(G). For all n\,..., fxp € E, let diag(/ii,..., fJ-p) :=

Mi^i? + • • • + lhtf$ •

By Lemma 14.1.1, p. 337, after passing to a subsequence, let Bi and B[ be sequences of ordered bases of g such that Ad gi : g —• g is diagonally adapted to Bi, B\. Let B^ :— lim Bi and let B'^ := lim B\. i—»oo

i—>oo

Locally free actions of simply connected Lie groups

343

Let J := [0, +00) be the topological space of nonnegative real numbers and let / := [0,+00] be the one-point compactification of [0,+00). Let / have the usual linear ordering <. Note, for all a € / , that a < +00. For all i, choose ba,..., bip, b'n,..., b'ip £ g such that Bi = (bn,..., bip) and such that B\ = (b'n,..., b'ip). Choose &001,•••, &ooP, &'ooi, •••,Voop <= 9 such that 8^ — (&ooi, • • •, &ooj>) and such that B'^ — (b1^,..., &^ p ). For all i, for all integers j € [l,p], choose Ay > 0 such that (Adgi)fry = Ayby. For all integers j 6 [l,p], let AQOJ be the limit in / of Ay, as i —• +00. For all i, let mi :— girriQ. For all m € M, let (•, -) m denote the Minkowski symmetric bilinear form on TmM. For all m 6 M, we define Bm G SBF(g) by Bm(X,Y) := ( X m , F m ) m . CZaim i : flm/ is a degenerate vector subspace of Tm'M. Proof of Claim 1: By (7) of Lemma 14.2.8, p. 340, B\{n x n) is degenerate and therefore not positive definite, so, by (5) of Lemma 14.2.8, p. 340, nm< is not positive definite. Then gmi is not positive definite, so, by Lemma 6.2.11, p. 180, gmi is either degenerate or Minkowski. Assume that gm- is a Minkowski vector subspace of TmiM. We aim for a contradiction. Let M 0 denote the set of all m e M such that gm is Minkowski. Then m' € M0 and Mo is a G-invariant subset of M. By Lemma 7.4.3, p. 207, M 0 is open in M. So, as gim0 -> m', for some i, we have gjm 0 € M 0 . Then, by G-invariance of Mo, we have mo £ Mo. Because gitno —• m', (Ad3j)B m o -> Bm', so {(Adpi)S m o }i is precompact in MinkSBF(fl). Consequently, the Adjoint action of G on MinkSBF(g) is nonproper, so the Adjoint action of G on MinkQF(g) is nonproper, so, by Lemma 14.3.1, p. 341, we have a contradiction. End of proof of Claim 1. For all i, let /j : TmoM -> TmiM be the differential at m' of the isometry m i->- gim : M -> M. Let d := dim(M). Let £ := (e^ , . . . ,e>d ') be the standard ordered basis of K d x l . For all v > 0, let D{y) := uE[f + E$

+ E$

+ • • • + E^d_2

+ E^ui_,

+ (l/u)EddJ

denote the diagonal dx d matrix with diagonal entries v, 1 , 1 , . . . , 1,1,1/v, from top left to bottom right. Let A+ := {D{v) \ v > 1}. By Lemma 4.10.38, p. 127, we let KQ be a compact subgroup of SO0(Qd) such that we have SO°(Q d ) = K0A+K0. Since SO°(Q d ) has finite index in 0(Qd), Let F be a finite subset of 0(Qd) such that (SO 0 (Q d ))F = 0(Qd). Let K := K0FUKQ. Then K is compact and we have 0(Qd) = KA+K. For all m € M, let Tm be the set of all ordered <2
344

Locally Free Orbit Nonproper Lorentz

Actions

For all m G M, let 0(<2d) act on Tm by the rule: gB is the image of £ under v i-> <J>B{.9~1V) : R d x l ->• T m M ; then, for all v 6 M d x l , we have gB(v) = <j>B{g~lv). Let J" := ( J ,F m . m

Let Cj and C,' be precompact sequences in T such that, for all i, we have both d 6 fmo and C\ G .F m i . For all i, choose ki,li G K and a, G A+ such that [file] = haik. For all i, let X>j := Ud and let X>< := krlC[. Then, for all i, we have £>i € Tmo, V\ G J^mi and [fi]x>. — a% £ ^ + - Moreover, {X>j}j and {X>i}i are both precompact in T. Passing to a subsequence, assume that T>i and T>\ are both convergent in T. Let V^ := limX>j and let D ^ :— limX^. For all i, choose /x, > 1 such that a, = D(fn). For all integers s € [l,d\, for all u e E d x l , we will let e*(v) denote the (s, l)-entry of v. For all integers s,t G [l,d], for all R G Rdxd, we will let E*t(R) denote the (s,i)-entry of R. For all i, for all integers j G [l,p], we define Sij := {bij)%™ we define and 8'^ := (byOg™. Then, for all i, we have A i i a ^ i = S'n. For all integers j G [l,p], let J ^ / := (ftocOg™ and let ^ := (^Og™By definition of A + , for all integers s,t G [l,d], if (s,t) 7^ (1,1), then 0 < E*t(ai) < 1. By (1) of Lemma 14.2.8, p. 340, we have 1/A a -> 0, so, for all integers s G [2,d], we have e*(Ajiaj(5ji) —• 0, i.e., that e*(<^x) —> 0. Then, for all integers s G [2, d], we have e * ^ ^ ) = 0. By local freeness, we have ( b ^ W ^ 0, so 8'^ ? 0. Then e ^ ) ? 0. A similar argument shows, for all integers s G [l,d — 1], that we have ^(Soop) = 0. Moreover, since (boop)mo ^ 0, we get $oop 7= 0, and we conclude that e^(<5oop) 7^ 0. Since 5'^ is concentrated on {1}, it follows that Qd-lightlike, we conclude that (&Jxji)m' is a lightlike vector in Tm'M. Then, by Claim 1 and by (3) of Lemma 6.2.6, p. 178, we see, for all v G flm', that Bm^b^^v) = 0. In particular, Bm'(b'ool,b'oop) = 0. Let F be the polarization of Qd- Then ^X^ooi,^oop) — 0- Then, since 8'^ is concentrated on {1}, we conclude that e*(<^ p ) = 0. For all i, we have (1/Ajp)aj<$ip = 8'ip. So, as a^ = D(ni), we conclude from matrix multiplication that {1 /\ip)(l/Hi)[e*d(8iP)] = ed(5'ip). Since e *d(Sip) ""*• e d(<W) ¥" 0 and since e*d(8'ip) -> e^((5^p) = 0, we conclude that (l/\iP)(l/ni) -> 0, as i ->• +00. For all i, for all integers j G [l,p], we let Ay := (fry)©™ a n ( ^ w e ^ et Aj- := (&y)p™. Then, for all i, we have (X^aiAna^1 = ^ x . Moreover, for

Compact subgroups not in the radical

345

all i, for all integers j G [l,p], by Lemma 7.4.1, p. 206, Ay, A^- G so(Qd)For all integers j G [l,p], let A ^ := (&ooj)£™ and let A ^ . := (b'^T. For all integers j G [l,p], by Lemma 7.4.1, p. 206, A ^ , A ' ^ G so(Qd). °° Let 5i:={(l,2),...,(l,d)}, 53:={(2,l),...,(d,l)},

S2:={(l,d),...,(d-l,d)}, S4:={(d,l),...,(d,d-1)}.

L e t S : = { l , . . . , d } x { l , . . . , d } . Since 1/Xn ->• 0, for all (s, t) € S\(SiUS 3 ), we have J E ^ A ^ ) = 0. Also, since (1/Aip)(l//Xj) ->• 0, by Lemma 2.5.2, p. 22, for all (s,t) G S3 U 54, we have ^ ( A ' ^ ) = 0. Let 7 := [b'^^b'^r. Then 7 = A ' ^ ^ , - A ^ p , so matrix multiplication shows, for all integers s G [2,d], that e*(7) = 0. Recall, for all integers s G [2,d], that effl^) = 0. Recall that effl^) ^ 0. Choose a G K such that such that a i ^ = 7. Let c := a b ' ^ - [ ^ l ^ ^ p ] - Then Cpl™ = o J ^ ! —7 = 0. Consequenctly cTO< = 0. By local freeness, this shc"ws that c = 0. That is, ab'^ = [b'^b'^}. Then [b'^b^] G Mb'^. By (2) of Lemma 14.2.8, p. 340, for all integers j G [l,p — 1], we have [&'ooi> b'ooj] € K&Jxji • Since 6 ^ , . . . , 6 ^ is a basis of g, we conclude that [b'oon o] Q KfeJxji > an d> moreover, that fc'^ ^ 0. Then IK^i *s a n ideal of g and dim(IR6^ol) = 1. Since g admits no one-dimensional ideal, we have the desired contradiction. End of proof of "only if". D

14.4

Compact subgroups not in the radical

Let Go be a connected Lie group. Let LQ be a semisimple Levi factor of Go- Let R be the solvable radical of Go- Assume that Z(LQ) is finite and that [0 is Lie algebra isomorphic to s ^ W - Let V be a vector space Let p : G0 -> GL(V) be a representation. For all g € Go, for all v G V, let gv := (p(g))v. Let I : V -»• V be the identity map. Assume V is (/9(Lo))-irreducible and (/9(L0))-nontrivial. Let K be a connected compact subgroup of Go. L E M M A 14.4.1 We have p(R) C MI. Proof. Let N be the nilradical of G 0 . Let V0 := {v G V \ Nv = {v}} be the set of JV-fixed vectors in V. By Engel's Theorem (Theorem 3.3, p. 12 of of [Hu72]), V0 ^ {0}. As N is normal in Go, V0 is (/>(Go))-invariant.

346

Locally Free Orbit Nonproper Lorentz Actions

Then Vo is (/9(Lo))-invariant, so, by (/9(Lo))-irreducibility of V, we see that Vo = V. That is, V is (p(JV))-trivial. That is, p(N) = {/}. By (2) of Lemma 4.15.5, p. 143, we have [G0,R] C N, so [L0,R] C N. Then [p(Lo),p(i?)] = {/}, so p(L0) centralizes p{R). As \Q is Lie algebra isomorphic to sfeOR), and as V is (p(L 0 ))-irreducible, we see, by Lemma 5.7.12, p. 172, that V has only scalar (/9(Lo))-intertwining. So, since p(L0) centralizes p(R), it follows that p(R) C RI. D R E M A R K 14.4.2 There exists a compact subgroup K' of L$ such that KR = K'R. Proof. Let n : Go —> Go/R be the canonical Lie group homomorphism. Let p := TT\L0 : L0 ->• G0/R. Let K' := P~1(-K{K)) . Since G0 = L0R, it follows that p : Lo —> Go/R is surjective. Then p(K') = n(K). Moreover, K' C L0, so TT(K') = p(K'). Then TT(K') = n(K). Then we have K'R = ir-\Tt{K')) = ir-^iriK)) - KR. We have ker(p) = Lo PI R. Then ker(p) is a proper normal closed subgroup of Lo, so, as Lo is simple, we see that ker(p) is a normal discrete subgroup of L0- Then, by (1) of Lemma 4.9.16, p. 103, ker(p) C Z(Lo)- So, since Z(Lo) is finite, it follows that ker(p) is finite. So, since p : L0 —> Go/R is surjective, by Lemma 2.7.7, p. 29, we see that p : L0 —> Go/R is proper. Then, as n(K) is compact, p~~l{ir(K)) is compact, i.e., K' is compact. • L E M M A 14.4.3 Assume that V is K-nontrivial, that vo e V\{0}. Then Kv0 £ Rv0.

that dim(V) = 2 and

Proof. Replacing Go by p(Go), we may assume that Go C GL(V). Since dim(V) = 2, we may assume that V = R 2 x l and that G0 C GL 2 (E). Let k := SO(2). Then, for all v £ V\{0}, we have Kv £ Rv. We have K?{I), so, since K is connected, by Lemma 4.4.5, p. 83, we get dim(.K') > 0. As K C GL 2 (E) is compact, choose g £ GL(F) such that gKg~l C K. As dim(ii') — 1, as dim(K) > 1 and as K is connected, by Lemma 4.4.5, p. 83, we conclude that gKg^1 = K. Then Kvo = g~lKgvo- Moreover, since gv0 ^ 0, we get K(gv0) £ R{gv0). Then Kv0 £ g^^gvo)) = H«o. • L E M M A 14.4.4 Assume that K £ R and that dim(V) = 3. Let h be a sequence in Lo- Let v' 6 V\{0}. Assume that v' is Kowalsky for p{h). Then Kv' £ Rv'.

Sequences with no large Kowalsky subsets

347

Proof. By Lemma 5.7.5, p. 169, choose F e MinkQF(V) such that p(L0) = SO°(F). Replacing G 0 by p(G0), assume that G 0 C GL(V) and that L0 = SO°(F). Identifying (V,F) with ( E 3 x l , Q 3 ) , assume that G 0 C GL 3 (R) and that L0 = SO 0 (Q 3 ). Assume, for a contradiction, that Kv' C Rv'. By Remark 14.4.2, p. 346, choose a compact subgroup K' of Lo such that KR = K'R. Then we have K'v' C K'Rv' = KRv'. By Lemma 14.4.1, p. 345, we have Rv' C Rv'. Then K'v' C KRv' C K(Rv') = R(Kv'). So, since Kv' C Rv', we get K'v' C Rv'. Let Vi be a sequence in V such that i>j -)• 0 and such that kvi ->• w'. For all i, we have Qz{Uvi) = Q3(vi). So, letting i —> oo, we get Qz{v') = 0. Then, by Lemma 6.2.19, p. 184 (with K replaced by K' and with v replaced by v'), we get K' = {/}. Then K C K'R = R, contradiction. •

14.5

Sequences with no large Kowalsky subsets

Let G be a connected Lie group. Let R be the solvable radical of G. Let L be a semisimple Levi factor of G. Let Lo be a connected normal closed subgroup of L. Let •K : G —> G/R be the canonical homomorphism. Assume that Z(L0) is finite. Let
Then Z{G0/R)

is finite and l0 is Lie

Proof. Let w : Go ->• Go/R be the canonical Lie group homomorphism. Let / := 7r|Lo : L0 -> Go/R- Because Go = LoR, it follows that / is surjective, so df : lo -*• 0oA i s surjective. Because Lo n R is discrete, it follows that / has discrete kernel, so df : (o -> floA ' s injective. Then d/ : (o -> So A *s a ^ ' e algebra isomorphism. By (2) of Corollary 4.9.18, p. 104, we have Z(f(L0)) = f(Z(L0)). Since / : Lo -» GQ/.R is surjective, we get Z(Go/R) — Z(f(L0)). Then

348

Z(G0/R)

Locally Free Orbit Nonproper Lorentz

= Z(f(L0)).

So, as Z(L0) is finite, Z(G0/R)

Actions

is finite.

•

L E M M A 14.5.2 The Lie algebra l0 is isomorphic to sl2(K). Proof. Assume, for a contradiction, that [0 ¥ sh(M). Let Go := L0R. Let w0 := 7r|G0. By Corollary 4.10.17, p. 119, any connected normal subgroup of GQ/R is a normal subgroup of G/R. So, since n(gi) converges mod normal subgroups in G/R, it follows that iro(9i) converges mod normal subgroups in GQ/R. By Lemma 14.5.1, p. 347, we see that Z{GQ/R) is finite and that lo is Lie algebra isomorphic to floAThen g0/x is not Lie algebra isomorphic to sfeOR). By Lemma 5.3.2, p. 156, choose a sequence Vj of two-dimensional vector subspaces of g such that Ad<7; is expansive on Vj. Passing to a subsequence, assume that (Ad<7,)Vi converges in the topological space of two-dimensional vector subspaces of 0. Let V := lim (Adgi)Vi. Then, by Lemma 2.6.1, p. 23, we conclude that i—*-oo

V is Kowalsky for Adg(gi), contradiction.

•

L E M M A 14.5.3 Let N be the nilradical of G. Let a and b be ideals of g such that a C b. Assume that [n, b] C a. Assume that the Adjoint representation of G on b/a is irreducible. Assume that the Adjoint representation of LQ on b/a is nontrivial. Then the Adjoint representation of LQ on b/a has only scalar intertwining. Proof. Let V := b/a. Assume, for a contradiction, that V has nonscalar (Ad Lo) -intertwining. Let I: V -»• V be the identity map. Let p := Ady : G -> GL(l^). Since [n, b] C o, it follows that p(N) = {I}. Let G 0 := L0R and Gx := GL(V). Let := p\Go : GQ -¥ G\. For all i, choose U € L0 and rt 6 R such that gt = lin. Since it(gi) diverges mod normal subgroups in G/R, we see that 7r(<7;) -> oo in G/R. For all i, we have 7r(rj) = IG/R, so ir(gi) = ir(h). Then 7r(Zj) ->• oo in G/R. Then k -¥ oo in G. By assumption, Z(Lo) is finite. By assumption, V is (Adio)-nontrivial. That is, p(L0) ^ {/}. By Lemma 14.5.2, p. 348, LQ is simple. Then, by Lemma 7.14.9, p. 255, we see that 4>(gi) —> oo in G\. That is, Adv(gi) —> oo in GL(F). By Lemma 4.9.34, p. 110, choose vo 6 V such that either {(Ad
Sequences with no large Kowalsky subsets

349

By Lemma 4.10.26, p. 123, choose a normal connected Lie subgroup L\ of L such that L0 centralizes L\ and such that L0Li — L. By (2) of Lemma 4.15.5, p. 143 we have [Li,R] C N. So, since p(N) = {/}, we conclude that p{L\) centralizes p(R). Since L\ centralizes LQ, it follows that p(Li) centralizes p(L0). Then p(L\) centralizes p(L0R). That is, p(Li) centralizes p{Go). So, since GQL\ — G and since V is (Ad G)-irreducible, by Lemma 5.7.14, p. 173 (with G replaced by p(G), with Go replaced by p(Go) and with C replaced by p{L\)), V is (AdGo)-isotypic. By Lemma 5.7.1, p. 167, V has nonscalar (Ad Go)-intertwining. By Lemma 5.7.13, p. 172, choose an (Ad Go)-equivariant linear transformation T : V -> V such that T(v') $ W. In particular, T(v') ^ 0. So, since (Adhi)vo —*• v' and since T is (AdGo)-equivariant, we conclude that (Ad/ii)(r(u 0 )) ->• T(v'). Then, since T(v') <£ W, it follows that i>i := T(vo) 0 M.v0. SO, since vo i1 0, we see that VQ and v\ are linearly independent vectors in V. Then dim(Ri;o + M^i) = 2. By Corollary 2.6.3, p. 24, we see that Ad hi is expansive on Mt>o + K^i • Let 7r : g —t g/a be the canonical Lie algebra homomorphism. Choose a vector subspace W of b such that W D o = {0} and TT(W) = Mv0 + Wfcvi. Then dim(VF) = dim(7r(VF)) = dim(Ri>0+Ki;i) = 2. By Lemma 2.6.4, p. 25, we see that Ad/ij is expansive on W. Passing to a subsequence, assume that (Ad hi)W converges in the topological space of two-dimensional vector subspaces of g. Let W' := lim (Ad hi)W. Then, by Lemma 2.6.1, p. 23, i—•oo

W is Kowalsky for Ad B (/i;). However, dim(W') = 2, contradiction.

•

L E M M A 14.5.4 Let N be the nilradical of G. If [lo,n] ^ {0} and if g admits no one-dimensional ideal, then [lo,3(n)] ^ {0}. Proof. For any ideal b of 0, we have that [n, b] is an ideal of g and that [n, b] C b. For any nonzero ideal b of g, let i?(b) denote collection of all ideals a of g such that [n, b] C a C b; then [n, b] € R(b), so R(b) ^ 0. Define #({0}) = {{0}}. For any ideal b of g, let M(b) denote the collection of all elements of R(b) that are maximal with respect to inclusion. Then, for any ideal b of g, for any a € M(b), the Adjoint representation of G on b/a is irreducible. Let no := n and recursively choose xij+i from M(rij). Then no 2 n i 2 n 2 ' ' "• Then, for all integers j > 0, we have n

j ¥" {0}

== -

*

dim(tij) > dim(n J + i).

350

Locally Free Orbit Nonproper Lorentz

Actions

Then, for some k, we have n^ = {0}. For all integers j > 0, a,- is an ideal of g and XVJ C n. Moreover, for all integers j > 0, the Adjoint representation of G on nj/nj+i is irreducible. Moreover, for all integers j > 0, [n, a,-] C tij+i. By Lemma 14.5.3, p. 348, we see, for all integers j > 0, that xij/xij+i either is (Ad I/o)-trivial or has only scalar (AdLo)-intertwining. By Lemma 4.10.7, p. 114, the adjoint representation of fo on g is completely reducible. Then, by Lemma 4.15.10, p. 145, [lo,3(n)] ^ {0}. •

14.6

Compact connected stabilizers

Let G be a connected Lie group. Assume that G is Ad-proper, that s ^ R ) is not a direct summand of g and that g admits no one-dimensional ideal. Let L be a semisimple Levi factor of G. Let R be the solvable radical of G. Let N be the nilradical of G. Let 7r : G -> G/R be the canonical Lie group homomorphism. Assume that G is not solvable. L E M M A 14.6.1 Let G act locally faithfully and isometrically on a connected Lorentz manifold M. Let LQ be a normal connected Lie subgroup of L. Let gi be a sequence in L$R and let mo,m' 6 M. Assume that gtmo —> ml in M. Assume that n(gi) diverges mod normal subgroups inG/R. Assume that Stab G (mo) and Stab G(m') are both compact. Assume that Stab^(m') C L0R. Then S t a b ^ m ' ) C R. Proof. If LQ = { 1 G } , then Stab^(m') = L0R = R, and we are done. We therefore assume that L0 ^ {1G}- By Lemma 5.5.6, p. 162, Z{LQ) is finite. Claim 1: Let hi be a subsequence either of <^ or of g^1. Then there is no two-dimensional vector subspace of g that is Kowalsky for Ad 0 (/ij). Proof of Claim 1: Let S be a two-dimensional vector subspace of g, let hi be a sequence in G and assume that S is Kowalsky for Ad B (/ij). If hi is a subsequence of gi, then, by Lemma 7.10.1, p. 236 (with rrii replaced by the constant sequence mo and with
Compact connected

stabilizers

351

By Lemma 14.5.2, p. 348, lo is Lie algebra isomorphic to sl2(]R). Then to is not a direct summand of Q. SO, by Lemma 4.15.6, p. 143, we have [I0,n] ^ {0}. Let U := j(n). By Lemma 14.5.4, p. 349, [l0,U] £ {0}. By (1) and (2) of Lemma 7.14.6, p. 253, we conclude that L0N is a normal closed subgroup of G. Then Adu(L0N) is a normal subgroup of Adu(G). As Adu(N) is trivial, we see that Adu(L0) = Adu(L0N). Then Adu(Lo) is a normal subgroup of Adjj(G). Let V be the sum of all (Ad Lo)-nontrivial, (Ad Lo)-irreducible vector subspaces of U. Then (AdG)V' C V. Let V be an (Ad G)-irreducible vector subspace of V. Then V is (Ad Lo)-nontrivial. By Lemma 4.10.6, p. 114, [L0,LQ] = L0. Then, by Lemma 5.7.3, p. 168, dim(V) > 2. For all i, choose Zj S LQ and ri £ R such that gi — 1^. Let I : V —¥ V be the identity map. Since V C U = 3(n), it follows that Ady(N) = {/} and that [n,V] = {0}. By (2) of Lemma 4.15.5, p. 143, [L0,R] C N. Then Ady(Lo) centralizes Adv(R)- By Lemma 14.5.3, p. 348 (with b replaced by V and o replaced by {0}), V has only scalar (Ad-Lo)-intertwining. Then V is (adLo)-hreducible and Ad^(i?) C MI. For all i, choose A; £ M such that Adv(rt) = Xil; since Ady(rj) is invertible, we conclude that A; ^ 0. For all i, for all v € V, we have (Adr^v = XiV. By (5) of Lemma 14.2.8, p. 340, for all m £ {m0,m'}, for all v £ n\{0}, we have vm ^ 0. Consequently, the linear maps X i-> Xmo : n -> TmoM and X H-> X m / : n -» T m -M are both injective. Let (•, •) be the Lorentz metric on M. For all m£M, define Bm € SBF(g) by Bm(X,Y) = (Xm,yro)m. Claim 2: Vm' is not positive definite. Proof of Claim 2: We define d := GL(V) and Go := L0R. Let := Ad : G 0 ->• Gi. Since we have 7r(/j) = 7r(gi) -> oo in G/.R, we get /j -> oo in L 0 - Since l0 = s[ 2 (K), i-o is simple. Then, by Lemma 7.14.9, p. 255, we have (• x in V. Then, by Lemma 7.1.1, p. 197, either Bmi(x,x) = 0 or Bmo(x,x) = 0. Assume, for a contradiction, that Vmi is positive definite. Then, by (5) of Lemma 14.2.8, p. 340, Bm,\(V x V) is positive definite. Then Bm>(x,x) > 0, so Bmo(x,x) = 0. Let L := Rx. Then we have Bmo(L,L) = {0}. For all t, let L; := (Adgi)L; then Bmi(U,U) = {0}. Passing to a subsequence, choose a line L' C V such that L» —>• L' in the

352

Locally Free Orbit Nonproper Lorentz

Actions

topological space of all lines in V. Then Bmi(L',L') = {0}. However, Bmi | (V x V) is positive definite, contradiction. End of proof of Claim 2. We have Vm' Q nm<. By (7) of Lemma 14.2.8, p. 340, nm< is positive semidefinite. Then Vm> is positive semidefinite, as well. Then, by Claim 2 and by (1) of Corollary 2.10.2, p. 35, it follows that Vm> is degenerate. Therefore, by (1) and (2) of Lemma 6.2.6, p. 178, Vmi is positive semidefinite with one-dimensional kernel. Choose w € V such that the kernel of Vmi is Rwm'. Let C be the set of all lightlike vectors in Vm>. Then, by Lemma 2.10.1, p. 34, C = Mwmi. Claim 3: If A, -> 0, then dim(y) < 3. Proof of Claim 3: Assume, for a contradiction, that dim(F) > 4. Then, by Corollary 5.7.10, p. 171 (with S replaced by L0 and s, replaced by IJ1), there is a sequence of twodimensional vector subspaces Ui of V on which A d / " 1 is expansive. Then, as |l/Aj| -)• +00, we see that Adg^1 is expansive on Ui, as well. Passing to a subsequence, assume that (Ad*?"1)**/* is convergent in the topological space of two-dimensional vector subspaces of 0. Let U' := lim (Ad*?"1)?/*. i—»oo

Then, by Lemma 2.6.1, p. 23, U' is Kowalsky for A d g ^ " 1 ) , contradicting Claim 1. End of proof of Claim 3. Claim 4: dim(V) < 3. Proof of Claim 4: By Claim 3, if A* —> 0, then we are done. So, passing to a subsequence, we may assume that inf |A*| > 0. i

Assume, for a contradiction, that dim(V) > 4. Then, by Corollary 5.7.10, p. 171, (with S replaced by L0 and Si replaced by k), there is a sequence of two-dimensional vector subspaces Ui of V on which Ad k is expansive. Then, since inf |A,| > 0, Adp* is expansive on U, as well. Passing to a i

subsequence, assume that (Ad gi)U is convergent in the topological space of two-dimensional vector subspaces of g. Let U' :— lim (Adgi)Ui. Then, i—>oo

by Lemma 2.6.1, p. 23, U' is Kowalsky for Ad0(*7j), contradicting Claim 1. End of proof of Claim 4Let Q be the Minkowski quadratic form on Tm>M. Let r : K -> SO°(<2) be the isotropy representation at m'. Let K := Stabg(m'). By assumption, we have K C LQR. Assume, for a contradiction, that K $.R. We have ker(Q|VTO<) = Wwm>. It follows that (T(K))wm> C Ru;m<. So, since ((Ad K)w)mi = {T{K))wmi, we see that ((Ad K)w)mi = Rwmi. So, by (5) of Lemma 14.2.8, p. 340, (Ad K)w C Rw. Then, by Lemma 14.4.3, p. 346, dim(V) ^ 2. So, by Claim 4, since dim(V) > 2, we get dim(V) = 3. Claim 5: The sequence A* of real numbers is bounded away from zero. Proof of Claim 5: Assume, for a contradiction, that, after passing to a

Compact connected stabilizers

353

subsequence, we have A; -> 0 in R. By Lemma 5.7.6, p. 169 (with U replaced by ljl), after replacing Z; by a subsequence, choose a two-dimensional vector subspace Vo of V such that, for all v £Vo, there is a precompact sequence Vi in V such that ( A d i ^ 1 ) ^ -» v, whence (Adg~1)(Xiv) -> v. Since A; -> 0 in R, we conclude that Vo is Kowalsky for Adg(g^1), contradicting Claim 1. End of proof of Claim 5. By Lemma 4.13.3, p. 135, we have Adv(L0) C SL(V). Since V is (Ad Lo)-nontrivial and since LQ is simple with finite center, by Lemma 7.14.5, p. 253, we see that Ad : LQ —> SL(V) is proper. Then, as U -> oo in LQ, we conclude that Adv(h) -> oo in SL(V). Then, by Lemma 4.9.35, p. 110, choose x £ V such that {(Adk)x}i is not precompact in V. Passing to a subsequence, choose x' £ V such that (AdZj)a; —»• x'. Then x' ^ 0 and x' is Kowalsky for Ady(Zj). By Lemma 14.4.4, p. 346, we have (AdK)x' g Rx'. Since (AdZj)a; -1- a;', by Claim 5, Xi(Adli)x ->• x'. For all i, (Adgi)x = (Adli)(Adri)x

-

Xi(Adh)x.

Then (Adgi)x —»• x', so a;' is Kowalsky for Adg(gi). Then, by Lemma 7.1.1, p. 197, x'm, is a lightlike vector in Vmi. Then x'm, £ C = Rwm>. Then, by (5) of Lemma 14.2.8, p. 340, x' 6 Mw. So, as x' ^ 0, we get Ex' = Rw. Then, as (AdK)w C Rw, we get (AdK)x' C E x ' , contradiction. • L E M M A 14.6.2 Let G act locally faithfully and isometrically on a connected Lorentz manifold M. Assume that the G-action on M is orbit nonproper. Then there exists some m £ M such that Stab G (m) is noncompact. Proof. Assume, for a contradiction, that, for all m £ M, the connected stabilizer Stabg(m) is compact. Since g admits no one-dimensional ideal, by Corollary 4.15.12, p. 148, we conclude that N is simply connected. By Lemma 4.15.9, p. 144 (with P replaced by G and Pi replaced by Gi), choose a cocompact connected normal closed subgroup G\ of G such that G\ D R admits no nontrivial connected compact subgroup. By Lemma 7.9.3, p. 236, we see that the d - a c t i o n on M is orbit nonproper. Then, by Corollary 9.1.2, p. 272 (with G' replaced by G\ and G* replaced by G), replacing M by G XQ1 M, we may assume, for all m £ M, that StabcOn) C G\. Let O be the set of all normal connected Lie subgroups L' of L such that L'R acts orbit nonproperly on M. Let L0 be an element of O that is minimal with respect to set-theoretic inclusion. Let Go := LQR. Then the

354

Locally Free Orbit Nonproper Lorentz

Actions

Go-action on M is orbit nonproper. By Lemma 5.5.6, p. 162, Z(LQ) is finite. Then, by (1) and (2) of Lemma 7.14.6, p. 253 (with N replaced by R), we see that Go is a normal closed subgroup of G. Then, by Corollary 9.1.2, p. 272 (with G' replaced by Go and G* replaced by Gi), replacing M by G XG 0 M, we may assume that the Go-action on M is orbit nonproper and that, for all m € M, we have Stabc(m) C G 0 f*1 G\. Let 7r : G —> G/R be the canonical homomorphism. Choose mo, m' € M and choose a sequence gi in Go such that gi —• oo in Go and such that gimo -> m'. Claim 1: G is not solvable and n(gi) diverges mod normal subgroups in GQ/R. Proof of Claim 1: Let Go := GQ/R. Let K be a proper normal connected closed subgroup of Go and let p : GQ —• Go/K be the canonical homomorphism. After passing to a subsequence, assume, for a contradiction, that {p(iT(gi))}i is precompact in Go/K. Let q := p o (7r|Go) : Go —> Go/K. Then {q(gi)}i is precompact in Go/K. Choose a precompact sequence Q in Go such that, for all i, we have lipi) = l(9i)- Passing to a subsequence, assume that c, is convergent in Go- Let c' := lim C*. Replacing gi by cjxgi and m' by (c') _ 1 m', we may i->oo

assume, for all i, that (KR)/(K°R), we conclude that (KR)/(K°R) is finite. Since, for all i, we have gi € KR, it follows that the action of KR on M is nonproper. Then, by Lemma 7.9.3, p. 236, K°R acts nonproperly on M , so K° e O, contradicting minimality of Lo- End of proof of Claim 1. As Go is a closed subgroup of G, we have gi -> oo in G. Moreover, Stab^(m') C Stab G (m') C G 0 n Gi C G 0 = L0R. By Claim 1 and Lemma 14.6.1, p. 350, we see that Stab^(m') C R. Then Stabg(m') C Gi H i?. Then S t a b ^ m ' ) is a connected compact subgroup of Gi n R, so, by choice of G\, we have StabJ^m') = { I G } Let Mo := {m e M | S t a b c ( m ) = { I G } } - By Lemma 7.4.2, p. 206, MQ is an G-invariant open neighborhood of m''. Let MQ be the connected component of MQ containing m'. Then MQ is a connected G-invariant open

Compact connected stabilizers

355

neighborhood of m'. Then, as g^mo -> m', we see that there exists j such that gjmo 6 MQ. Then, by G-invariance of MQ, we have m 0 £ Mg. Then, by G-invariance of M'0, for all i, we have pjmo € M'0. Replacing M by MQ, we may assume that the G-action on M is locally free. Then, by Theorem 14.3.2, p. 342, we have a contradiction. • Using Lemma 14.6.2, p. 353 (together with Lemma 4.15.4, p. 142), one may show that Theorem 14.3.2, p. 342 remains true even if "G £ ONV1*" is replaced by "G admits an orbit nonproper, isometric action on a connected Lorentz manifold with no noncompact connected stabilizers". This improves the "only if" part of Theorem 14.3.2, p. 342.

Chapter 15

Orbit Nonproper Lorentz Actions

Recall that ONV is the collection of connected Lie groups admitting a locally faithful, orbit nonproper, isometric action on a connected Lorentz manifold. This chapter is devoted to the proof of Theorem 15.7.4, p. 379 below, which effectively describes the collection of groups in ONV with simply connected nilradical. 15.1

Preliminaries

L E M M A 15.1.1 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo € M and let H := Stab G (mo). Let L be a semisimple Levi factor of G, let r be the solvable radical of g and let n be the nilradical of g. Assume that Z(L) is finite. Let XQ € !\{0} and let P € r. Assume that {exp(iJTo)}teK is not precompact in G. Let X := Xo + P and assume that X G h. Assume, for all integers n>2, that neither so(n, 1) nor so(n +1,2) is a direct summand ofg. Then (adX 0 )n ^ {0}. Proof. Assume, for a contradiction, that (adX 0 )n = {0}. By Coincidence X.6.4(i), p. 519 of [He78], so(2,l) 2 sl 2 (K), so we are done if s^IR) 10. Then sfeOR) is not a direct summand of g. By Corollary 4.10.4, p. 113, choose an integer k > 1 and choose simple ideals I i . . . , Ik oil such that the addition map a : Ii © • • • © I* —M defined by a(Yi,... ,Yk) = Y\ H (- Yk is an isomorphism of Lie algebras and such that, for all integers i € [l,k], the Lie algebra U is simple. For all integers i G [!,&], let 7Ti : li © • • • © It —> k be projection to the ith factor and let 357

358

Orbit Nonproper Lorentz

Actions

Pi '•= 7Tj " a - 1 : ( —>• k. As {exp(tXo)}teR is not precompact in G, choose an integer i0 £ [l,fc] such that {exp(i • \pi0(X0)])}teR is not precompact in G. Reordering, if necessary, assume i0 = 1. Then, by Lemma 4.9.10, p. 101, the Lie group homomorphism t \-> exp(i • [pi(X 0 )]) : M —> L\ is proper. Let 6 be the kernel of ad : I -> gt(n). Then 6 is an ideal of I, so, by Lemma 4.10.14, p. 118, choose a subset 5 C { l , . . . , i } such that t = ^ k. ies We have X0 6 6. So, since pi(X0) ^ 0, we see that 1 G S. Then l\ C 6. Then [li,n] = {0}. So, by Lemma 4.15.6, p. 143, we have h\g. Then [i is not Lie algebra isomorphic to sfeW- We wish to show, for some Y £ li\{0}, that Y is strongly vanishing at moLet L\ be the connected Lie subgroup of g corresponding to l\. Let ti be a sequence of real numbers such that U —> +oo. For all i, we define <7i := exp(fj • [p!(X 0 )]) and g[ := exp(tjXo). Then oo in L\. Since Z(Li) C Z(L) and since Z(L) is finite, we conclude that Z(Li) is finite. By Lemma 5.3.1, p. 155 (with G replaced by L\), choose a sequence Vi of two-dimensional vector subspaces of li such that Adg; is expansive on Vi. Passing to a subsequence, assume that Vi is convergent in the compact topological space of two-dimensional vector subspaces of li. Let V := lim Vi. i—>oo

Then, by Lemma 2.6.1, p. 23, V is Kowalsky for Ad^ (gt). Because li 10, we conclude, for all i, that (Ad g (^))|[i = Adit(n] 7^ {0}- Then 1'0 is noncompact. By Lemma 4.12.3, p. 133, let s 0 be a Lie subalgebra of [0 such that So is Lie algebra isomorphic to sl2(K). Since

Preliminaries

359

ad : 1'0 —> gi(n) is nontrivial, and hence (by simplicity of 1'0) faithful, we conclude that [so,n] ^ {0}. By Coincidence X.6.4(i), p. 519 of [He78], we have sl2(R) = so(2,1), so, by Corollary 6.4.2, p. 194 (with n replaced by 2, with g replaced by So, with V replaced by n and with Q replaced by F), choose a three-dimensional, (adso)-irreducible F-Minkowski vector subspace V of n and choose an (adso)-trivial vector subspace V" of n such that n = V + V" and such that V' n V" = {0}. Then ( a d s 0 ) F ' f {0} and V ^ {0}. Claim 1: [V',V"] = {0}. Proof of Claim 1: Fix Y G V". We wish to show that ( a d y ) V ' = {0}. Since [so, V"] = {0}, it follows that ad Y : V —> n is (adso)-equivariant. By Lemma 4.14.1, p. 138, (adY)V' ^ V, so, by (2) of Lemma 5.7.11, p. 171, (adY)V' cannot be both (adso)-nontrivial and (adSo)-irreducible. In particular, (adY)V' cannot be isomorphic to V as an (ads 0 )-module. So, by Lemma 4.9.23, p. 106, {adY)V = {0}. End of proof of Claim 1. By Lemma 4.14.1, p. 138, [V',V] 2 V. So, by (1) of Lemma 5.7.11, p. 171, [V',V] C V". Then (ads 0 )([V, V']) C (ads 0 )(V") = {0}. Let Xo,Y0, T0 be a standards [2 (K) basis of So- By Lemma 5.7.8, p. 170, choose a basis A, B, C of V such that (a,dX0)C = 2B, (adY0)A = B,

{adX0)B = 2A, (a,dY0)B = C,

(adX 0 )A = 0, (ady o )C = 0.

By the Jacobi identity, we have (adX 0 )([^4,C]) = 2[A, B]. Therefore [A,B] e (ads0)([V',V']) - {0}. By the Jacobi identity, we have both (adY 0 )([A,£]) = [A,C] and (adY0)([A,C]) = [B,C]. So, as [A,B] = 0, we get [A,C] = 0 and [B,C] = 0. So, since A,B,C is a basis of V, we conclude that [V'.V] = {0}. Let U := j(n). By Claim 1, we have [V, V"] = {0}. Then [V, n] = [V, V + V"} C [V, V] + [V, V"] = {0}, so V C 3(n) = U. Since ( a d s 0 ) y ^ {0} and s 0 C 1'0, we conclude that ( a d l 0 ) ^ ' ^ {0}- Then (adl 0 )C/ D (adI{,)V' # {0}. Moreover, since F | V is Minkowski and since V C [/, by Lemma 6.2.13, p. 181, F\U is Minkowski. By Corollary 6.4.5, p. 195, choose an integer n > 2 such that 1'0 = so(n, 1). So, by Corollary 6.4.2, p. 194 (with g replaced by l0, with V replaced by U, with Q replaced by F\U, with V replaced by V and with V" replaced by W), choose an (ad (0)-irreducible submodule V of U and an (ad [ 0 )-trivial

360

Orbit Nonproper Lorentz

Actions

submodule W of U such that (1) (2) (3) (4) (5)

dim(V) = n + 1; V has only scalar (adl 0 )-intertwining; U = V + W and V n W = {0}; F\V is Minkowski; and F\W is positive definite.

Since 1'0 is an ideal of lo> since to is an ideal of I and since I is semisimple, it follows, from Corollary 4.10.17, p. 119, that 1'0 is an ideal of I. By Lemma 4.10.26, p. 123, choose an ideal Ii in I such that l'Q + Ii = ( and 1'0 n (i = {0}. By (2) of Lemma 4.15.5, p. 143, we see that [r, 1'0] C n. So, as [Mo] C h n 1'0 = {0}, we get adu{[h +1,l0]) C adf/(n). Since C/ = 3(n), we have ad[/(n) = {0}. Then adt/([ti + r , 1'0]) = {0}. It follows that ad[/(li + r ) and adf/(lo) centralize each other. Then ad[/(Ii +1) preserves the sum V» of all (ad (o)-nontrivial, (ad [(^-irreducible vector subspaces of U. By (2) of Lemma 5.7.11, p. 171, we have V. = V. Then (ad(Ii + v))V C V. Let I : V —> V denote the identity map. By (4) above, we have F\V 6 MinkQF(^). We have ad„(lo) C so(F), so ad v (Io) C so(F\V), so ady([g) C so(F\V). Since a.du(h + t ) and adj/(lo) centralize each other, by (2) above, we conclude that ady(ti + r) C MI C co(F\V). So, since ady(Io) C so(F\V) C co(F|V) and since 1Q + Ii + r — g, we conclude that a d v ( 0 ) C co(F|F). D L E M M A 15.1.3 Let G be o connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let d := dim(M). Let XQ, YQ, TO be a standard s^lR) basis of a Lie subalgebra s of g. Let n be an ideal of g. Assume that nds — {0}. Let P € n. Let T\ := To + P. Let mo E M. Let H := S t a b ^ m o ) . Assume that I \ £ h. Then both of the following are true: (1) For some ordered Qd-basis B ofTmoM, (T x )^ m € ! + H{d). (2) Assume that n is nilpotent. Assume that I) D n = {0}. Then a nondegenerate vector subspace ofTmoM. Proof. Proof of (1): Let g := g/n. Let 7r: g —• g be the canonical homomorphism. Since ad To : s ->• s is real diagonalizable, by Corollary 4.11.7, p. 130 (with p : g -> gl(F) replaced by Ad : s -> flKfl)); it follows that ad T0 : fl -> 0 is real diagonalizable. Moreover, since adT 0 : s -> s is nonzero, and since ir\s : s —> g is an injective, (adTo)-equivariant map, we

Preliminaries

361

conclude that adT 0 : g -> g is nonzero. Since P € n, we conclude that adg-(To) = adg-(Ti). Then adTi : g -» g is nonzero and real diagonalizable, and therefore is not elliptic. We have IT O (ad B (Ti)) = (adg-(Ti)) o 7r, so, by Lemma 2.4.1, p. 20, we see that adTi : g -> g is not elliptic. By Lemma 7.4.14, p. 214, choose an ordered <3d-basis B of TmoM such that (Ti)^ m € 1+ n^ U"P (d) . Assume, for a contradiction, that (Ti)£ m € V{d). By (3) of Lemma 7.4.13, p. 212, we see that adTi : 0 -> 0 has no nonzero real characteristic roots. Then adTi : 0 ->• 0 has no nonzero real characteristic roots. However, since adTi : g -> g is a nonzero, real diagonalizable linear transformation, this is impossible. End of proof of (1). Proof of (2): Assume, for a contradiction, that n m o is degenerate. Choose B as in (1) of Lemma 15.1.3, p. 360. Then Qd\^%m is degenerate. We have n D s = {0}, so T0 ^ -P, so Ti ^ 0. Then, by Lemma 7.4.11, p. 211, we see that (Ti)§ m # 0. Let a : n g m -» n g m be defined by a(w) = {Ti)^mw. Then, by Lemma 6.3.1, p. 184, a : n g m ->• n g m has exactly one nonzero real characteristic root. Let Q be the Minkowski quadratic form on TmoM. Define a linear transformation / : n ->• n%m by f(X) = X§m. Then / o (ad„(Ti)) =aof. Since n n h = {0}, it follows that / : n -> ngTO is an isomorphism of vector spaces. We conclude that adTi : n —> n has exactly one nonzero real characteristic root. Let A be the unique nonzero real characteristic root of adTi : n -> n and let W := {X € n | (adTi)X = XX}. Let n o , n i , . . . be the descending central series of n. Choose an integer i > 1 such that njH W # {0} and ni+1nW = {0}. Let V := n i / n i + i . Then A is the only nonzero real characterstic root of adTi : V —> V. We have ady(P) € ady(n) = {0}, so ady(T 0 ) = ady(Ti). We conclude that A is the only nonzero real characterstic root of adT 0 : V ->• V. Let p :— Ad : s -¥ 0l(V). By Lemma 5.7.4, p. 168, there are an even number of nonzero real characteristic roots of p(To) : V —> V, i.e., of ad To : V -)• V. This gives a contradiction. End of proof of (2). D Let n be a nilpotent Lie algebra. Let n 0 , n i , . . . be the descending central series of n. Let P,Q £ n. We write P < Q (dcs n) if there is an integer i > 0 such that P, Q £ rtj and Q £ n i + i . We write P 0 such that P £ n i + 1 and Q 6 nj and Q ^ nj + i. Here "dcs" is an abbreviation for "descending central series". L E M M A 15.1.4 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo € M. Let

362

Orbit Nonproper Lorentz Actions

n be the nilradical of g. Let P G n\(3(n)) and assume that Pmo = 0. Then there exists Q G n\{0} such that Q < P (dcsn) and such that Q is strongly lightlike with respect to B. Proof. Choose Y G n such that [P,Y] ^ 0. Since P G n, it follows that a d P : g —> g is nilpotent. By Lemma 7.4.15, p. 215, choose an ordered basis B of TmoM such that P£m € "P (d) . Let T := a d P : g ->• g. Then T(Y) = [P, Y] ^ 0. By (3) of Lemma 7.8.4, p. 229, if Y is strongly nontimelike with respect to B, then, setting Q := T(Y), we are done. On the other hand, by (4) of Lemma 7.8.4, p. 229, if Y is not strongly nontimelike with respect to B, then, setting Q := T2(Y), we are done. • L E M M A 15.1.5 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo G M and let B be an ordered Qd-basis ofTmoM. Let H := Stabg(mo). Let gc denote the set of elements ofg which are strongly lightlike with respect to B. Let g^ denote the set of elements of g which are strongly nontimelike with respect to B. Let n be the nilradical of g. Assume that gc D n ^ {0} and that h n (3(n)) = {0}. Then \gc,9tf] C gcProof. By construction, gc is a vector subspace of g. Let X E gc and Y Gfljv"-We wish to show that [X,Y] G gcLet d :— dim(M). Let ei := e^ '. Let jio,tii,... be the descending central series of n. Choose an integer i0 > 0 such that gc n nj0 ^ {0} and such that gc n n i o + i = {0}. Fix Q €(gcn n i o )\{0}. Claim 1: For all P G n\{0}, if P < Q(dcsn), then P $ h. Proof of Claim. 1: Assume, for a contradiction, that P G h, i.e., that Pmo — 0. Since 0 ^ P G t) n n and since f) D (j(n)) = {0}, we get P G n\(a(n)). By Lemma 15.1.4, p. 361 (with Q replaced by Q'), choose Q' G n\{0} such that Q' < P (dcs n) and such that Q' is strongly lightlike with respect to B. Then 0 ^ Q' G gc n nj 0 +i, contradiction. End of proof of Claim 1. By Claim 1, since Q < Q(dcsn), we have Qmo ^ 0. So, since Q is strongly lightlike with respect to B, we have Q%m G (fei)\{0}. So, as X%m G Eei, choose a G E such that (X - aQ)%m = 0. Let X := X - aQ. Then X^m = 0. Moreover, as Q,X G gc, we have X G gc, i.e., X is strongly lightlike with respect to B. Then X is strongly vanishing with respect to B. By (1) of Lemma 7.8.6, p. 231 we conclude that [Q,Y]%m G Eei. So, since 0 ^ Q%m G Eei, choose /? G M such that {[Q, Y] - 0Q)%m = 0. Then

A strongly vanishing element

363

([Q,Y] - PQ)ma = 0, so, since [Q,Y] - 0Q < Q(dcsn), by Claim 1, we have [Q, Y]-/3Q = 0, so [Q, Y] = f3QeSc. Since X is strongly vanishing with respect to B and since Y is strongly nontimelike with respect to B, by (1) of Lemma 7.8.4, p. 229, we conclude that [X,Y] G Qc- Therefore, since we have [Q,Y] G gc, it follows that a[Q, Y] + [X, Y]eQc. Then [X, Y] = [oQ + X, Y] 6 D fl£. 15.2

A strongly vanishing element

For any ordered basis B = (vi,...,Vd) of any vector space V, we define switch(S) :— {vd,v2,V3,... ,Vd-2,Vd-i,vi), so that switch switches the first and last vectors of the basis. Let G be a connected Lie group acting isometrically on a connected Lorentz manifold M. Let mo G M and let H := Stabg(mo). Assume that H is noncompact. Assume that G is Ad-proper. Assume that g admits no one-dimensional ideal. Assume that sfeW is not a direct summand of g. Let N denote the nilradical of G. Assume that rj n (j(n)) = {0}. L E M M A 15.2.1 There exists X G fl\{0} such that X is strongly vanishing at mo • Proof. By Lemma 4.9.12, p. 102, choose XQ G rj such that the Lie group homomorphism 11-4 exp(tX) : R —> G is proper. Then, by Ad-properness of G, we see that {AdB(exp(£X0))}teR is not precompact in GL(g). Then adXo : fl —> Q is not elliptic. Let d :— dim(M). By Lemma 7.4.14, p. 214, fix an ordered Qd-basis B of TmoM such that either (X0)^m G 1+ U{d) or (X0)%m G 7>(d). Case 1: ( X 0 ) § m G K+W ( d ) . Proof in Case 1: We may assume, by multiplying X0 by a constant, that (X0)^m G % ( d ) . By (2) of Lemma 7.4.13, p. 212, we see that adXo : 0 —> 0 is complex diagonalizable and that the characteristic roots of adXo : 0 —> 0 are all contained in {1, — 1} U v^—TflL For all A G R, let Vx:={Y£g\ (&dX0)Y = AF}. Let VD := ^ ^ A C F . As a d X 0 : V -t V is complex diagonalizable, fix an (ad X0)-invariant vector subspace VE of V such that VD + VE = V and VJJ C\VE = {0}. We have VD = Pi + V-i, so 0 = V\ + VE + V_i. Moreover, the set of characteristic roots of a d X 0 : VE ->• VE is contained in ^/—TK- Then a d X 0 : VE ^ VB is elliptic. By Lemma 7.8.7, p. 232, for all W G Vi,

364

Orbit Nonproper Lorentz

Actions

W is strongly lightlike with respect to B. Also, by Lemma 7.8.7, p. 232 (with X replaced by —X and B replaced by switch(S)), for all W € V-i, W is strongly lightlike with respect to switch(S). So we may assume that dim(Vi) < 1 and dim(V_i) < 1; otherwise, using (1) of Lemma 7.8.3, p. 228, we would be done. Let T : g -¥ g be the real diagonalizable part of a d X 0 : g -» g. By the Jacobi identity, ad-Xo : g —> g is a derivation, so, by Lemma 4.9.14, p. 102, T : g —> g is a derivation. By Lemma 2.4.2, p. 21, we conclude that T(fl) = (adX 0 ). Then T(g) = V± - V-i = VD. Then, by Lemma 4.9.15, p. 103, the Lie subalgebra of 0o of g generated by VD is an ideal of g. Any one-dimensional vector subspace of a Lie algebra is a Lie subalgebra, so, if dim(Vb) = 1, then g0 = VD, contradicting the assumption that g admits no one-dimensional ideal. Moreover, if dim(Vo) = 0, then g — VE and adXo : fl —> 0 is elliptic, whence {Adg(exp(tXo))}teR is precompact in GL(g), contradiction. Thus dim(Vo) > 2. So, since dim(Vi) < 1, since dim(V_i) < 1 and since VD — V\ + V_i, we get dim(Vi) = 1 = dim(F_i). By Lemma 6.2.9, p. 179 (with T replaced by ad B (Xo)), there is an (ad Xo)-invariant Minkowski form on g. Then, by Lemma 9.2.1, p. 276, there exists a free, orbit nonproper, isometric action of G on a connected Lorentz manifold. Then, by Theorem 14.3.2, p. 342, we have a contradiction. End of Case 1. Case 2: ( X 0 ) § m € P ( d ) and gmo is Minkowski. Proof in Case 2: Let n '• S ~* fl/f) D e the canonical map. Let T : g —> g be the nilpotent part of a d X 0 : g -> 0. Since X0 € h, we have (adXo)f) C h. Let T" be the nilpotent part of a d X 0 : 0/1) -> 0/h. Then, by Lemma 2.4.1, p. 20, we have T o 7T = 7T o T. Let V := T'( 0 /h) and let W := T{V). Let V := T(g) and let W := T(V). As T' o 7T = 7T o T, we get TT(V) = V and n{W) = W. Let S := g%m C R d x l . As gmo is Minkowski, we see that Q := Qd\S is Minkowski. The kernel of the surjective map Y i-> Y§m : g ->• S is h, so there is an isomorphism i : g/\) -> S such that, for all Y € 0, we have t(7r(y)) = Y§m. Then, by Lemma 7.4.16, p. 215, for all Y G g/h, we have i{T'{Y)) = A/"2(d) • [i(Y)}. Since v >-> N^d)v : S -> S is an element of so(Q), we conclude that T" : g/l) —> g/t) preserves a Minkowski form. The kernel o f v ^ A/"2(<\ : Rdxl -> Rdxl is K := K e ^ + • • • + M e ^ i - Since Qdlif is positive definite, while Qd\S is Minkowski, we have S %. K. Then t; !->• A/2 u : 5 —>• S is nonzero. Then, since t: 0/f) -> S is an isomorphism, we conclude that T" : 0/I) —> 0/f) is nonzero.

A strongly vanishing element

365

The map T" : fl/f) ->• g/f) is nilpotent, nonzero and preserves a Minkowski form. Then, by Lemma 6.3.16, p. 191, dim(V') = 2 and dim(W') = 1. Consequently, we have dim(y) > 2 and dim(W) > 1. Suppose, for this paragraph, that dim(V) > 2. Then the canonical map V ->• V has nonzero kernel. That is, h n F ' ^ {0}. Choose Q 6 (hnV')\{0}We may assume that Q is not strongly lightlike with respect to B, since we are otherwise done. Choose P £ 0 such that T(P) = Q. Since Q is not strongly lightlike with respect to B, it follows, from (1) of Lemma 7.8.4, p. 229, that P is not strongly nontimelike with respect to B. So, by (4) of Lemma 7.8.4, p. 229, we see that T2(P) ^ 0 and that T2(P) is strongly lightlike with respect to B. Moreover, T2(P) = T{Q), and Q £ h, so T2(P) e T(h) C h. Then T2(P) is nonzero and is strongly vanishing with respect to B, and we are done. We therefore assume that dim(V) < 2. As we have already observed that dim(F) > 2, we get dim(V) = 2. Since T : g -> g is nilpotent, we have dim(T(V)) < dim(V). So, as W = T(V) and dim(V^) = 2, we get dim(W) < 2. However, dim(W) > 1, so we conclude that dim(W) = 1. By (3) of Lemma 7.4.13, p. 212, the real diagonalizable part of the map adXo : 0 —> g is zero. Then, by Lemma 6.2.10, p. 179, there is an (adXo)-invariant Minkowski form on g. Then, by Lemma 9.2.1, p. 276, there exists a free, orbit nonproper, isometric action of G on a connected Lorentz manifold, contradicting Theorem 14.3.2, p. 342. End of Case 2. Case 3: (X 0 )g m £ V(d) and gmo is not Minkowski. Proof in Case 3: Let fl£ denote the set of elements of g which are strongly lightlike with respect to B. Letfljv"denote the set of elements of g which are strongly nontimelike with respect to B. Let T : g -+ g be the nilpotent part of a d X 0 : g -> g. Then, by Lemma 4.9.14, p. 102, T : g ->• g is a derivation. For all P € g\gtf, by (2) and (4) of Lemma 7.8.4, p. 229, T2(P) is nonzero and strongly vanishing with respect to B, and we are done. We therefore assume that 0 = fljv"By (1) of Lemma 7.8.4, p. 229, we have T(gx) C gc. Therefore we have T(g) = T(gx) C 5 i . By (5) of Lemma 7.8.4, p. 229, T(gc) = {0}. If T(g) = {0}, then adXo is has precompact one-parameter subgroup in GL(g), a contradiction. So T(g) ^ {0}, so gc ^ {0}. If dim(0£) > 2, then, by (1) of Lemma 7.8.3, p. 228, we have nonzero strongly vanishing elements, and we are done. We therefore assume that dim(0 £ ) = 1. Then, as {0} £ T(g) C gc, we get T(g) = gc.

366

Orbit Nonproper Lorentz

Actions

Let 5 be the K-linear span of a,dB(X0), (ad g (X 0 )) 2 , (ads(X0))3, Since (adX 0 )r C t and since T 6 5, it follows that T(r) C r. Then T factors to a derivation T" : g/t -4- g/t. By Lemma 4.10.27, p. 123, we see, for some X' € g/t, that T' = a d X ' : g/t ->• g/t. Since dim(T(g)) = dim(g £ ) = 1, we conclude that dim(T"(g/t)) < 1. Then the centralizer in g/t of X' has codimension < 1. By Lemma 5.6.2, p. 164, no nonzero element of a semisimple Lie algebra has codimension < 1 centralizer, so X' = 0. Then T'ish) = {0}. Then T(g) C t. By Lemma 4.15.8, p. 144, we have [g,g] C l + n. Then (ad-X"0)g C [ + n. So, since T € 5, we get T(g) C [ + n. Then gc = T{g) C r f l ( [ + n) = n. By Lemma 15.1.5, p. 362, we have [g£,g.v] C gc. So, since dim(g,c) = 1 and since g = gjv", we see that g admits a one-dimensional ideal, contradiction. End of Case 3. •

15.3

A strongly lightlike element

Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Let mo € M and let if := Stabe(mo). Let d := dim(M). Let X e g\{0}. Let B be an ordered Qd-basis of TmoM. Assume that X is strongly vanishing with respect to B. Assume, for all integers n > 2, that neither so(n, 1) nor so(n + 1,2) is a direct summand of g. Assume that () n (j(n)) = {0}. Let n be the nilradical of g. L E M M A 15.3.1 There exists Yen such that Ymo ^ 0 and such that Y is strongly lightlike with respect to B. Proof. Let ny denote the collection of all W € n such that W is strongly vanishing with respect to B. Case 1: ny ^ {0}. Proof in Case 1: Let rto,ni,... be the descending central series of n. For all integers i > 0, let (ni)y be the set of all W £. rtj such that W is strongly vanishing with respect to B. Choose an integer IQ > 0 such that (n; 0 ) v ^ {0} and (n; 0 + 1 )v = {0}. Choose Wo € ( ( W J O ) V ) \ { 0 } . Then, by assumption, we get Wo ^ 3(n). Then, by Lemma 15.1.4, p. 361, choose Q G n such that Q < P(dcsn) and such that Q is strongly lightlike with respect to B. Since Q < P (dcs n), it follows that Q 6 rtj0+i, so Q is not strongly vanishing with respect to B. Thus, as Q is strongly lightlike with respect to B, we get Qmo ^ 0. End of Case 1.

Nilradical is free

367

Case 2: ny = {0}. Proof in Case 2: Let I be a semisimple Levi factor of g and let r be the solvable radical of g. Let xic denote the collection of all W £ n such that W is strongly lightlike with respect to B. From the definitions of strongly lightlike and strongly vanishing, for all W 6 n£\ny, we have Wmo ^ 0. So, since ny = {0}, it suffices to show that ri£ ^ {0}. Since ny = {0}, since X ^ 0 and since X is strongly vanishing with respect to B, we have X $ n. By (5) of Lemma 7.4.13, p. 212, a d X : g -*• g is nilpotent. So, since X £ n, by (1) of Lemma 4.15.5, p. 143, we see that X £ t. Choose l0 as in Lemma 5.5.4, p. 161. Then X € l0 + n and [0 is an ideal of I with no compact factors. Choose X0 £ lo and Pen such that X — X0 + P. Since X £ n, we conclude that X0 ^ 0. Since lo is semisimple, we have 3(^0) = {0}. Then ad[ 0 (X 0 ) ^ 0, so ad B (X 0 ) 7^ 0. As a d X : g -> g is nilpotent, by Lemma 5.5.2, p. 160, a d X 0 : g -> 0 is nilpotent. A nonzero nilpotent linear transformation cannot be elliptic, so {Ad0(exp(tXo))}tgK is not precompact in GL(g). Consequently, {exp(tXo)}te& 1S n ° t precompact in G. By Lemma 15.1.1, p. 357, (adX 0 )n ^ {0}. Then, by Lemma 4.10.18, p. 119 (with / replaced by ad„ : fl -> flKn))> w e hare (adX)n ^ {0}. Choose F e n such that (adX)Y ^ 0. Let Z := (a.dX)Y. Then we have 0 / 2 € [ g , n ] C n . If Z 6 rt£, then we are done, so we assume Z £ tt£. Then, by (1) of Lemma 7.8.4, p. 229, Y is not strongly nontimelike with respect to B. Then, by (4) of Lemma 7.8.4, p. 229, (ad X)2Y is nonzero and is strongly lightlike with respect to B. Moreover, ( a d X ) 2 F = ( a d X ) Z = [X,Z] e [fl,n] C n. Then (adX) 2 Y e n £ \ { 0 } . End of proof in Case 2. 15.4

D

Nilradical is free

Let G be a connected Lie group. Assume that G is Ad-proper and that 0 admits no one-dimensional ideal. Assume, for all integers n > 2, that neither so(n, 1) nor so(n + 1,2) is a direct summand of g. Let I be a semisimple Levi factor of g, let r be the solvable radical of g and let n be the nilradical of g. Let G act isometrically and locally faithfully on a connected Lorentz manifold M. Let d := dim(M). Let m 0 6 M and let H := S t a b ^ m o ) . Let X G fl\{0} and let y e n . Let B be an ordered Q^-basis of TmoM. Assume

368

Orbit Nonproper Lorentz

Actions

that X is strongly vanishing with respect to B and that Y is strongly lightlike with respect to B. Assume that Ymo ^ 0. Assume h fl n = {0}. For all integers i € [l,d], let e* := e\ '. L E M M A 15.4.1 The vector subspace n m o ofTmoM

is nondegenerate.

Proof. Assume, for a contradiction, that n m o is degenerate. Let fljv denote the collection of elements of fl that are strongly nontimelike with respect to B. Let gc denote the collection of elements of g that are strongly lightlike with respect to B. Let n/v" := nnjjjv"- Let ri£ := ndgcFor all S £ n, if Sgm € Kei, then there exists a eM. such that we have (S - aY)%m = 0, whence S - aY £ h, whence S-aYe1jr\n{0}, m whence S = aY € ti£. Thus, we see that n £ = {S £ n | S g £ Rei}. By (1) of Lemma 7.8.6, p. 231, [n£,ttAf]gm C Me!. Then [n £ ,njv] C n £ . By Lemma 7.8.9, p. 233, n = tw- Then [ti£,n] = [n£,njv] C n£. Then it£ is an ideal of n. Moreover, by (1) of Lemma 7.8.3, p. 228, a codimension one vector subspace of n£ is contained in hnn. So, as hfln = {0}, it follows that dim(ri£) < 1. We have Y € ti£, so ti£ ^ {0}. Then dim(n£) = 1. Since n£ is an ideal of n and since n is nilpotent, by Lemma 4.14.3, p. 138 (with m replaced by ti£), we see that ti£ D (3(tt)) ^ {0}. So, as dim(n£) = 1, we conclude that ri£ C j(n). Since X is strongly vanishing at mo, it follows, by (5) of Lemma 7.4.13, p. 212, that (adX) 3 g = {0}. Then a d X : fl -4 0 is nilpotent. Then, by Corollary 5.5.3, p. 161, choose X0 6 I and P € n such that a d X 0 : I -t I is nilpotent and such that X = XQ+P. Since O ^ X S f ) and since hnn = {0}, we conclude that X ^ n. So, since X = Xo + P € XQ + n, we conclude that Xo ^ 0. Then, by Lemma 4.10.33, p. 125, choose Y0,T0 € I such that X0,Y0,T0 is a standard sI2(M) basis of a Lie subalgebra so of I. For all integers n > 2, we know that neither so(n, 1) nor so(n + 1,2) is a direct summand of fl, so, by Lemma 15.1.1, p. 357, we see that (adXo)n ^ {0}. Then (ads 0 )n # {0}. We have n = tijv, so, by (1) of Lemma 7.8.4, p. 229, (adX)n C ti£. Then dim((adX)n) < 1. Since a d X : g -> fl is nilpotent, it follows that dim((adX) 2 n) < dim((adX)n). Then (adX) 2 n = {0}. By Lemma 5.5.8, p. 163, let V be the unique (adso)-irreducible, (adso)-nontrivial vector subspace of n. As V is (adso)-nontrivial, by Lemma 5.7.3, p. 168, we get dim(V') > 2. Since (adX) 2 n = {0}, it follows, from Lemma 5.5.7, p. 162, that (adX 0 ) 2 n = {0}. Then, by Lemma 5.7.15, p. 173, dim(V') < 2.

Nilradical is free

369

Then dim(V') = 2. By Lemma 5.7.15, p. 173, since dim(V) > 2, we have (ad-Xo)V' 7^ {0}- Choose u € V such that {&dX0)u ^ 0. We have (adXo)u + (adP)it = {a,dX)u e (adX)n C n £ C }(n). Claim 2: V C 3(11). Proof of Claim 2: Let no, r t i , . . . be the descending central series of n. Assume, for a contradiction, that V % 3(n). Choose i such that V C n» + (3(n)) and V % ttj + 1 + (3(n)). Let m := n i + i + (a(n)). Then V £ m. We have (adX 0 )u 6 (3(11)) - (adP)u. Then 0 ^ (adX 0 )u € (3(11)) — [n, r\i + 3(11)] C m — m = m. Also, we have (adX0)u e [s0,V] C V . Then 7 ' f l m / {0}. So, by (ads 0 )-irreducibility of V, we conclude that V C m, contradiction, ^nrf of proof of Claim 2. By Lemma 4.10.7, p. 114, let V" be an (adso)-invariant complement in n to V. Then V" is (ads 0 )-trivial. That is, [V",s0] = {0}. By Lemma 5.7.7, p. 170, choose a basis A, B of V such that (&dX0)B = A,

(adX 0 )A = 0,

( a d y 0 ) ^ = -B,

(a,dYo)B = 0.

Then (a,dT0)A = A and (adT 0 )B = -B. Recall that (adX)n C n £ . By Claim 2, we have A,B £ V C 3 (n). Then ( a d P ) B = 0. It follows that ( a d X ) P = (adX0)B = A. Then EA C (adX)n C nc, so, since dim(ri£) = 1, we conclude that EA = tt£. Fix P ' G V and P" € V" such that P = P ' + P " . Fix ot,P € E such that P ' = aA + )8B. We have ( a d X ) r 0 = [X0 + P'+ P " , Y0]. So, as • [-Xo) ^o] = ?o;

• [P', y 0 ] = [aA + 0B, Y0] = -aB; . [P",Y0] e[V",so] = {0},

and

we get (ad X)Yb = T0 - aB, so ( a d X ) 2 r 0 = [^0 + P ' + P " , T 0 - aB]. Therefore, since P' = aA + PB, since [P",To] € [F",s 0 ] = {0} and since [P",B] € [n,3(n)] = {0}, we get (adX) 2 Y 0 = [X0 + aA + pB,T0 - aB). We then calculate that (adX)2Y0 = -2X0 - 2aA + PB. Therefore, since X = X0 + aA + PB + P", we get 2X + ((ad X)2Y0) = 3PB + 2 P " . So, since 2X is strongly vanishing with respect to B and since, by (3) of Lemma 7.8.4, p. 229, (adX) 2 lo is strongly lightlike with respect to B, it follows that 3/?B + 2 P " G n £ = EA. Then, as A, B and P " are linearly independent, we get P — 0 and P " = 0. Then we have X — X0 + aA and 2X + ((adX) 2 F 0 ) = 0. Then (adX) 2 y 0 € h. So, since X is strongly vanishing with respect to B, we conclude, from (7) of Lemma 6.3.15, p. 190,

370

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that ( r 0 ) g m e ffiei + • • • + l e d _ ! . Then, by (2) of Lemma 6.3.15, p. 190, ((adX)F 0 )B m € Kei- Choose r 6 E such that {{adX)Y0) + TY e h. We have [X, Y0] = [X0 + ctA, Y0] = T0 - aB. Let P x := TY - aB. Then To + Pi = ((adX)Y0)+TY 6 f) and Pi 6 n. Then, by (2) of Lemma 15.1.3, p. 360, we see that n m o is nondegenerate, contradiction. • L E M M A 15.4.2 There are an (adg)-irreducible vector subspace V C j(n) and Q 6 MinkQF(V) such that dim(V) > 2 and ady(g) C co(<2). Proof. Since Ymo is nonzero and lightlike, we conclude that n m o is not positive definite. Then, by Lemma 6.2.11, p. 180 and by Lemma 15.4.1, p. 368, we see that n m o is Minkowski. As X is strongly vanishing at mo, it follows, from (5) of Lemma 7.4.13, p. 212, that (adX) 3 0 = {0}. Then a d X : 0 -> 0 is nilpotent. By Corollary 5.5.3, p. 161, choose Xo € [ and R e n such that X = XQ + R and such that ad Xo : 0 —• 0 is nilpotent. Since 0 ^ X € f) and since h n n = {0}, we conclude that X $. n. So, since X = Xo + R € Xo + n, we conclude that X 0 ^ 0. By Lemma 4.10.33, p. 125, choose Y0,T0 € I such that X0,Y0,T0 is a standard sfeW basis of a Lie subalgebra So of 1. Since n m o is Minkowski, it follows that Qd\n^m is Minkowski. On the other hand, Qd\(Mei + • • • + l e ^ - i ) is degenerate. So, by Lemma 6.2.13, p. 181, we have n g m £ Kei + • • • + Mea-i- So choose 5 € n such that S§m $. ffiei + • • • + Ke d _i. Choose 7 € ffi such that (Yo + jS)%m e Bei + • • • + Ee d _i. Let To* := (adX)(F 0 + iS). By (3) of Lemma 7.8.5, p. 230, we have (7o)e m € Rei • Choose 0 € R such that (T^ + /3F) m o = 0. We have T^+/3Y = [X0 + P , F 0 + 7-5] + 0Y- Let P := 7 ^ 0 , S] + [R, Y0 + 7 5 ] + /3Y. Then T^ + /3Y = T0 + P , so (T0 + P ) m o = 0. Moreover, since S, R, Y e n and since n is an ideal of 0, it follows that P e n . Let T\ := To + P . By (1) of Lemma 15.1.3, p. 360, choose an ordered Qd-basis C of TmoM such that (Ti)£ m e R+ H^. Choose A > 0 such that (Ti)£ m e AftW. Define p : 0 -> ffidxl, by p(X) = X ^ m . Since h n n = {0}, we see that p\n : n —• n £ m is an isomorphism of vector spaces. For all A £n, ((adTi)A)£ m = ( ( r i ) £ m ) • ( 4 g m ) . Moreover, by Lemma 7.4.1, p. 206, (Ti)£ m 6 so{Qd), s o v ^ (Ti)£ m t; : n g m -> n g m preserves Qd\n%m. So, because Qdln^" 1 is Minkowski and becausep\n : n -> n£?m is an isomorphism of vector space, we conclude that adTi : n -» n preserves a Minkowski form.

Nilradical is free

371

It follows, from Lemma 6.3.17, p. 191, that both A and —A are roots of the characteristic polynomial of ad Ti : n -> n. Let U and V be eigenvectors of adTi : n -¥ n with eigenvalues A and -A, respectively. Then, by Lemma 7.8.7, p. 232 (with X replaced by (1/\)U, BbyC and Y by U), we see that U is strongly lightlike with respect to C. Let C := switch(C). By Lemma 7.8.7, p. 232 (with X replaced by -(1/\)U, B by C and Y by V), we see that V is strongly lightlike with respect to C. Neither of these can vanish at mo, since we assume that I) n n = {0}. Then p(U) is a nonzero multiple of e\ and p(V) is a nonzero multiple of edThen ffiei + Med C p(n). Let V := V^. Then V(Rdxl) C p(n). Let 7r : g -> g/t be the canonical map. For all P G g, let P := 7r(P). For all S C g, let 5 := {P | P G 5 } . C7aim i ; (adT0)g" C ^. Proo/ o/ CTaim i : We have (Ti)^ m G A?^ d ), so the real diagonalizable part of (Ti)^ m : K d x l -»• E d x l is AP : M d x l -> M d x l . Let D : g -> Q be the real diagonalizable part of ad 0 (Ti) : g -> g. By Corollary 4.11.7, p. 130, adT 0 : g -> jj is real diagonalizable. Since adg-(P) G adg(n) C adg(t) = {0}, it follows that adj(Ti) = ad F (T 0 ). Then ad Ti : g -»• g is real diagonalizable. As 7r o (adg(Ti)) = (adg-(Ti)) o n, we see, from Lemma 2.4.1, p. 20, that •K o D = (adgCTi)) o 7T. That is, for all l e g , n(D(X)) = (adTi)(7r(X)). A similar argument shows, for all X G g, that p(D(X)) = (XD)(p(X)). Thenp(D(g)) = (\V)(p(g)) C XV(Rdxl) C p(n). Then D(g) C p " 1 ^ ) ) = n + (ker(p)) = n + h. Then (adTo)g = (adT0)(7r(g)) = (adri)(7r(g)) = 7r(D(g)) C 7r(n + h), so (adTo)g C (7r(n)) + (?r(h)) = {0} + h = h. End of proof of Claim 1. We have 2X 0 = (adT 0 )X 0 , so X 0 G (adT 0 )g. Because s0 n r = {0}, we see that X 0 ^ 0 G g. Then (adT 0 )g ^ {0}. Let go" be the Lie subalgebra of g generated by (adT 0 )g. Then, by Lemma 4.9.15, p. 103, go is an ideal of g. Since g is semisimple, by Corollary 4.10.15, p. 118, we see that g~S is semisimple. We have go" C h. By Lemma 4.10.36, p. 126, choose a Lie subalgebra g0 C [) such that 7r|go : go -* 0o" is a Lie algebra isomorphism. Then g 0 is semisimple, so, by Lemma 4.9.39, p. 112, choose a semisimple Levi factor lo oi g such that go C fo. Then, as go is an ideal of g, it follows that go is an ideal of loWe have P G n C r , s o P G t = {0}. Then XT = X 0 + R = X. Since X 0 G (adTo)g C go", choose Xi G go such that Xi = XQ. Choose R' G t

372

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Actions

such that Xi + R' = Xo + R- By Lemma 5.5.4, p. 161, let gi be an ideal of go with no compact factors such that Xi + R' € gi + r. Then X\ € gi + x. On the other hand, X\ € g0 C lo- Then X\ e (fli + t) n fo = fli- By Lemma 7.6.1, p. 225, we see that {exp(iX)} t£ R is not precompact in G. By Lemma 15.1.1, p. 357 (with X0 replaced by Xi and P replaced by R'), we have (adXi)n ^ {0}. Then (adfli)n ^ {0}. By Corollary 4.10.15, p. 118, gi is semisimple. By Corollary 4.10.4, p. 113, choose a simple ideal l\ of gi such that (adli)n ^ {0}. Let FQ be the Minkowski quadratic form on TmoM. Let F € QF(n) be the puUback of Fo along the map X i-> Xmo : n -> TmoM. Since h n n = {0}, we conclude that X i-)- X m o : n —> TmoM is injective. So, since n m o is Minkowski, we see that F € MinkQF(n). Then, by Lemma 15.1.2, p. 358, we are done. D 15.5

Nilradical is not free, but its center is free, Part I

Let G be a connected Lie group. Assume that G is Ad-proper and that g admits no one-dimensional ideal. Assume, for all n > 2, that neither so(n, 1) nor so(n + 1,2) is a direct summand of g. Let I be a semisimple Levi factor of g and let r be the solvable radical of g. Let N be the nilradical of G. Assume that N is simply connected. Let G act locally faithfully and isometrically on a connected Lorentz manifold M. Let m0 £ M and let H := Stabg(m 0 ). Let d := dim(M). For all integers i G [l,d], let e, := e\ '. Assume that \) n (a(n)) = {0}. Let P 6 (h n n)\{0}. Assume, for all Q 6 (1)0 n)\{0}, that P < <5(dcsn). Let B be an ordered Q d -basis of TmoM. Assume that P%m € P^h Let gc denote the collection of elements of g that are strongly lightlike with respect to B. Let g ^ denote the collection of elements of g that are strongly nontimelike with respect to B. Let n£ := Qc n n and njv := g^y n n. L E M M A 15.5.1 For all Q e g, we have both ((ad P)Q)%m = M^d){Q%m)

and ((adP)Q)%» = [N?>,Q%»\. Proof. Let T := a d P : g - • g. Since P 6 n, a d P : g -> g is nilpotent. The result then follows from Corollary 7.4.17, p. 216. • L E M M A 15.5.2 For allX € g, ifXgm 6 Kfii +• • •+Med-1, thenX In particular, the codimension in g of g^f is < 1.

G gjv.

Nilradical is not free, but its center is free, Part I

373

Proof. Assume, for a contradiction, that X £ gj/. Then, because X%m £ lex + • • • + Be d _i, we get Xjjm $ WH^ +M^dK By Lemma 7.4.1, p. 206, we have X%m G so(Qd). Let Y := ( a d P ) 2 X . Since P G n, we get Y € n and Y < P(dcsn). By Lemma 15.5.1, p. 372 and by (8) of Lemma 6.3.15, p. 190, we have that Y£m ^0,soY^0. By Lemma 15.5.1, p. 372 and by (2) and (3) of Lemma 6.3.15, p. 190, we have Ygm = 0. Then Y G h. Since F G (h n n)\{0}, we have P < Y (dcsn), contradiction. • L E M M A 15.5.3 The vector subspace gmo ofTmoM Proof.

is Minkowski.

By Lemma 15.5.1, p. 372, for all Q 6 g, we have

((ad P)Q)cBm = ^ 2 ( d ) (gg m ),

((adP)Q)£™ = [N?\QLBm]-

Case 1: QM C g. Proof in Case 1: By Lemma 15.5.2, p. 372, Choose X G 0 such that X§m £ ffiei + • • • + Med-x. Then, by (1), (2) and (7) of Lemma 6.3.15, p. 190, we have ( ( a d P ) 2 X ) g m G (Kei)\{0}. Then, by Lemma 6.2.16, p. 182, Qd\g%m is Minkowski. Then gma is Minkowski. End of proof in Case 1. Case 2: 0jv = 0. Proof in Case 2: By (1) of Lemma 7.8.4, p. 229, we have ( a d P ) 0 ^ C gc- Then, as gj^ = 0, we get (adP)g C gc. That is, (ad P)g is strongly lightlike with respect to B. Since O ^ P G h n n a n d h n (j(n)) = {0}, we get P G n\(a(n)). Choose Qo G n such that (adP)Q 0 ^ 0. Let Qi := (adP)Q 0 - Then Qi < P (dcsn). Then Qi $ h, so (Qi)mo ^ 0. Since Qi G (adP)0 = (adP)gj^, we see, by (1) of Lemma 7.8.4, p. 229 that Qi is strongly lightlike with respect to B. As 0 = 0AT, as 0 ^ Qi G nr = 0£ n n and as h n (j(n)) = {0}, by Lemma 15.1.5, p. 362, we conclude that gc is an ideal of g. So, because "£ = 0£ n n, we conclude that nc is an ideal of n. Since <3i G nc, we get nc i1 {0}. By Lemma 4.14.3, p. 138, we see that n£n(3(n)) ± {0}. Then g £ n( 3 (n)) ± {0}. If dim(g £ n(3(n))) > 2, then, by (1) of Lemma 7.8.3, p. 228, some element of (3(n))\{0} is strongly vanishing at mo, contradicting the assumption that I) n (3(n)) = {0}. On the other hand, if dim(g£ H (j(n))) = 1, then gc D (i(n)) is a one-dimensional ideal of g, contradiction. End of proof in Case 2. • L E M M A 15.5.4 The vector subspace nmo ofTmoM Proof.

is not Minkowski.

Assume, for a contradiction, that n m o is Minkowski.

374

Orbit Nonproper Lorentz

Actions

Then Qd\n%m is Minkowski. So, since Qd|(ffiei H h Kcd-i) is dem generate, by Lemma 6.2.13, p. 181, we have tVg £ Kei + • • • + Med-iChoose Wen such that W§m £ Eei + • • • 4- fed-i. Let Z := ( a d P ) 2 W . Then Zen. By (4) of Lemma 7.8.4, p. 229, we have 0 ^ Z e gc- Then {0}^fl£nn. By (2) of Lemma 7.8.6, p. 231, for all X,Y e QM, we have

[x,y]gmefei +--- + Ked_1, so, by Lemma 15.5.2, p. 372, we conclude that [X, Y]%m e Qtf. Then g// is a Lie subalgebra of g. Then n/y is a Lie subalgebra of n. By Lemma 15.1.5, p. 362, 0£ is an ideal of gjj. Then it£ is an ideal of njv- By Lemma 4.14.3, p. 138, a nonzero ideal of a nilpotent Lie algebra always meets the center, so nc n (3(iw)) ^ {0}. By (3) of Lemma 7.8.6, p. 231, we have [n, nc]%m € Kei + 1- Me d -i, so, by Lemma 15.5.2, p. 372, we get [n, nc] C gjj. Furthermore, we have K nc] C [n, g] C n. Then [n, nc] C g^ f) n = n/j-. Claim 1: f) D nc D {l{n^r)) = {0}. Proof of Claim 1: Assume, for a contradiction, that 0 ^ I 6 f) (1 ii£ fl foOw))- As 0 ^ I 6 I) and as f) n (j(n)) = {0}, we get X £ a(n). Choose Yen such that [X,Y] ^ 0. We have [X, n^f] Q [3(tw),tw] = {0}. Then Y £ n^r, so Y £ g_^. Since X e f) n nc, we see that X is strongly vanishing with respect to B. Then, by (4) of Lemma 7.8.5, p. 230, we have [X, [X, Y]] ^ 0. However, [X,[X,Y]]e[}(nM),[nc,n]], and [n, n^] C njv", so [X, [X, Y]] e [3(tw),n/v] = {0}, a contradiction. End of proof of Claim 1. By (1) of Lemma 7.8.3, p. 228, we conclude that the codimension in nc n (}(njv)) of f) fl nc D (3(njv)) is < 1. So, by Claim 1, we see that dim(n,c n (3(njv))) < 1. So, since nc n (3(ttjv)) ^ {0}, we conclude that dim(n/; n (3(tw))) = 1. Since gc is an ideal of g//, it follows that fljv normalizes gc- Since 0 normalizes n, it follows that g// normalizes n. Then 0jv normalizes 0£Dn. That is, 0JV normalizes n£. Since gx normalizes both g^r and n, it follows that 0jv" normalizes 0^/- n n. That is, 0jv normalizes njv"- We conclude that 0JV normalizes i(n_sf). Since gjy normalizes both nc and 3(n/v")> we conclude that 0jv normalizes nc D (3(iw)). Moreover, by Lemma 15.5.2, p. 372, we know that the codimension in 0

Nilradical is not free, but its center is free, Part II

375

of gjv is < 1. Then the normalizer in fl of ri£ PI (3(tw)) has codimension < 1 in g. So, as dim(ti£ PI (3(ttjv))) = 1 and as g admits no one-dimensional ideal, by Lemma 5.6.4, p. 166, we have ri£ D (3(njv)) C ^(n). Since Wgm $ Me1 + • • • + Me d -i, we have W $ g^. So, since the codimension in g of g^ is < 1, we have EW + gjv" = 0Since rt£ fl (j(iw)) C 3(11) and since I f £ n, we conclude that W centralizes nc n (3(tw)). So, since gx normalizes ric n (}(ntf)) and since KW + gx = g, we conclude that g normalizes nc n (3(tw))- That is, "£ H (3(njv")) is an ideal of g. However, dim(n£ n (3(n))) = 1 and g admits no one-dimensional ideal, contradiction. •

15.6

Nilradical is not free, but its center is free, Part I I

L E M M A 15.6.1 Let G be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Assume that g admits no one-dimensional ideal. Assume, for all integers n > 2, that neither so(n, 1) nor so(n + 1,2) is a direct summand of g. Let mo £ M. Let H := Stab^(mo). Assume that f) fl (3(n)) = {0} and that G is Ad-proper. Then f ) D n = {0}. Proof. Assume, for a contradiction, that h n n 7^ {0}. Let N be the nilradical of G. By Corollary 4.15.12, p. 148, since fl admits no one-dimensional ideal, we see that N is simply connected. For all integers i € \\,d\, let e; := e\ '. Let n o , n i , . . . be the descending central series of n. Choose an integer io > 0 such that h n nj0 7^ {0} and such that f) D n i o + 1 = {0}. Choose P £ (fj D n i o )\{0}. Then, for all Q € (f) n n)\{0}, we have P < Q. Since P 6 n, it follows that a d P : fl -» 0 is nilpotent. Let d := dim(M). By Lemma 7.4.15, p. 215, fix an ordered Qd-basis B of TmoM such that P%m € V^d\ By Corollary 7.4.17, p. 216, for all Q £ g, ( ( a d P ) Q ) g m = J \ # ° ( Q g m ) and ( ( a d P ) Q ) § m = [M?*,Q%»]. By Lemma 15.5.3, p. 373, gmo is Minkowski, so Qd|0g m is Minkowski. So, since Q<j|(IRei + • • • + Wed-i) is not Minkowski, we conclude, from Lemma 6.2.13, p. 181, that g%m £ Mex H + Ee d _i. m Choose A € g such that A% $. l e i + 1- Ked-i • Let B :- [P, A] and C := [P,B]. Then B, C € n. Moreover, we have B < P(dcsn) and C < P(dcsn). Moreover, by (3) of Lemma 7.8.4, p. 229, we see that B is strongly nontimelike with respect to B. Moreover, by (4) of Lemma 7.8.4,

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Orbit Nonproper Lorentz

Actions

p. 229, we see that C ^ 0 and that C is strongly lightlike with respect to B. By (5) of Lemma 7.8.4, p. 229, (ad P)C = 0. By choice of P , since C < P(dcsn), we conclude that C ^ h, i.e., that Cmo ^ 0. So, since C is strongly lightlike with respect to B, we have C g m <E (Eei)\{0). Then, by (3) of Lemma 6.3.15, p. 190, we see that i?g m $ Eei. So, since B is strongly nontimelike with respect to B, we have B%m € ( ] & > ! + . . . + Re«i_i)\(Eei). Claim 1: For all X £ n, HX§m £ Eei and X < C (dcsn), then X £ EC. Proof: Assume otherwise. Choose a £ E such that (X + a C ) g m = 0. Let Y := X + aC; then Y £ h. Since X and C are linearly independent, we get 7 ^ 0 . Moreover, V 6 f) fin. We have X < C (dcsn), so Y < C (dcsn). Since Y < C (dcsn) and C < P (dcsn), it follows that Y < P (dcsn). Then, by choice of P, we get Y fi h, contradiction. £Vid of proof of Claim 1. Let 0jv" denote the collection of X £ a such that X is strongly nontimelike with respect to B. Claim 2: n C g ^ . Proof: Suppose that X £ n and that X £ gjj. We aim for a contradiction. By Lemma 15.5.2, p. 372, we see that Xgm $ Kei + • • • + Ee d _i. Define Z := ( a d P ) 2 X . Then, by (1), (2) and (7) of Lemma 6.3.15, p. 190, we conclude that Z%m £ (Eei)\{0}. Then, by Lemma 6.2.16, p. 182, we see that Qd\n^m is Minkowski, so n m o is Minkowski, contradicting Lemma 15.5.4, p. 373. End of proof of Claim 2. Since QM is strongly nontimelike with respect to B and since C is strongly lightlike with respect to B, we conclude, from (1) of Lemma 7.8.6, p. 231, that ((adC)fljv-)em C Kei. Moreover, since C £ n, we see, for all Q £ (adC)gjv, that Q £ n and Q < C(dcsn). So, by Claim 1, we get (ad C)fljv" Q IRC. Then EC is an ideal of 0jv. Then, by Claim 2, we see that EC is an ideal of n. Then, by Lemma 4.14.3, p. 138, (EC) n (j(n)) # {0}. Then {0} C (EC) l~l (a(n)) C EC, so, because dim(EC) = 1, we conclude that (EC) n (a(n)) = EC, Then C £ EC = (EC) n (|(n)) C j(n). By the Jacobi idenity, (adP)[A,B] = [A,C] and (adP)[A,C] = [B,C]. Then (adP) 2 [A,B] = [P,C] £ [n,a(n)] = {0}. So, by (7) of Lemma 6.3.15, p. 190, [A,B]%m £ Eei + • • • + Merf_i. So, since [A,C] = (adP)[A,B], by (2) of Lemma 6.3.15, p. 190, we conclude that [A,C]%m € Eei. Moreover, as C 6 n, we get [A,C] £ n and [A,C] < C(dcsn). So, by Claim 1, we see that [A, C] £ EC. Since A%m ^ Eei -I + Med-i, we conclude that A $. 0JV"- By Lemma 15.5.2, p. 372, we see that the codimension in g of g^r is < 1. Then gjv + EA = g. So, since EC is an ideal of ajv" and since

The main result

377

[A, C] £ M.C, we see that EC is an ideal of g. However, by assumption, g admits no one-dimensional ideal, contradiction. •

15.7

The main result

Recall that ONV is the collection of all connected Lie groups admitting an orbit nonproper, locally faithful, isometric action on a connected Lorentz manifold. A Lie algebra g is reductive if there exist a semisimple Lie algebra I and an Abelian Lie algebra o such that g is Lie algebra isomorphic to I © a. A connected Lie group is reductive if its Lie algebra is reductive. Let TZTZT be the set of all pairs (Go, V) such that Go is a connected reductive Lie group and V is an irreducible real Go-module. (The letters 1Z71T are an abbreviation of "Reductive Representation Theory".) Let NCS be the set of all pairs (GQ,V) e TZTZT such that, for some locally faithful, isometric action of Go K V on a connected Lorentz manifold M, for some m 6 M, we have that Staby(m) is noncompact. (The letters NCS are an abbreviation of "Non-Compact Stabilizer".) Standard structure theory and representation theory allows for an effective description of TZTZT: One first describes all connected reductive Lie groups, and then, for each, analyzes its irreducible representation theory. In Theorem 15.7.1, p. 377, we consider the problem of describing the pairs in NCS. Given a good understanding of TZTZT, it is, in some sense, not surprising that we can identify, one by one, those pairs in TZTZT that are in NCS. However, a careful proof of the final result requires more representation theory than we have had space to develop here. We therefore refer the reader to [A99c], where the details of the proof are given. Let S denote the collection of connected Lie groups with simply connected nilradical. By Lemma 4.15.1, p. 141, <S includes all connected, simply connected Lie groups. In Theorem 15.7.3, p. 379, we show how the problem of effectively describing the groups in S D ONV can be reduced to the problem of finding those G £ S which admit an action on a Lorentz manifold for which the stabilizer of some point has noncompact intersection with the center of the nilradical of G. In Corollary 15.7.2, p. 378, we are able to reduce this problem to the problem of computing NCS, which is exactly the problem addressed in Theorem 15.7.1, p. 377. Our final effective description of S n ONV appears in Theorem 15.7.4, p. 379.

378

Orbit Nonproper Lorentz Actions

T H E O R E M 15.7.1 Let (G 0 , V) € WllT. For all g 6 Go, let Tg : V ->• V be defined by Tg(y) = gv. Then (G 0 , V) € NCS iff there exists Q e QF(V) such that both • Q is either positive definite or Minkowski; and • for all g 6 G 0 , we have Tg\V e CO°(Q). Proof. Proof of "only if": Suppose (G0,V) € NCS. Let G := G 0 <x V. Let L be a semisimple Levi factor of Go- Then L is a semisimple Levi factor of G, as well. We are done if (2) of Theorem 21.2 of [A99c] holds, so we assume that (1) of Theorem 21.2 of [A99c] holds. Fix a nonzero (adflo)-invariant vector subspace of VQ of V such that Adv0(L) is compact. Because V is Go-irreducible, we have V = V0. Then L\ := Ady(L) is compact. As I is semisimple or zero, by Corollary 4.10.16, p. 119, li is semisimple or zero. Let G\ := Adv(Go) and let Hi := G L ( ^ ) . By reductivity of Go, L= [Go, Go], so L is a normal subgroup of Go- Then Gi normalizes L\. Then gi C n(,,(Ii). Then, by Corollary 4.10.5, p. 114, we have 0i C \x + (c^ (h)). By Lemma 5.7.16, p. 174 (with G replaced by L\ and H replaced by Hi), choose a positive definite Q e QF(V) such that Ij C so(Q) and such that c b l (d) C co(Q). Then Ql C Ix + (c bl (Ii)) C co(Q). Then, for all $ € G 0 , Tg\V = Ady(p) € Gi C CO°(Q). Snrf o/proo/ o/ "on/y i/". Proo/ o/ "«/"•' This follows from Lemma 9.2.2, p. 277 (with G replaced by Go ix V and with JVi replaced by V). End of proof of "if". D COROLLARY 15.7.2 LetH be a connected Lie group acting locally faithfully and isometrically on a connected Lorentz manifold M. Assume that the nilradical N of H is simply connected. Assume that there is a noncompact closed subgroup of Z(N) which fixes a point of M. Then there exists an (AdH)-irreducible vector subspace V\ 0/3(11) and there exists Q € QF(Vi) such that both of the following hold: • Q is either positive definite or Minkowski; and . Ad V l Cff)CCO°(Q). Proof. Let Z := Z(N) and let V := 3 = j(n). Let H0 := Intz(H) and let Go := Ady(G). Since N is simply connected, it follows, from Theorem 3.6.2, p. 196 of [Va74], that the map exp : n —> N is a diffeomorphism. For all v € V, g £ G, we have exp((Ad 5)1;) = (Int <7)(exp(t>)). Let G := Ho * Z.

The main result

379

Define $ : G 0 K V -> G by $((Ad v (5),u)) = (Int z (0),exp(u)). Then $ is an isomorphism of abstract groups and $(V) = Z. Give G the Lie group structure such that $ is an isomorphism of Lie groups. Because Z is a normal subgroup both of G and of H, by Lemma 9.1.3, p. 273 (with N replaced by Z), the G-action on M' := G xz M preserves a Lorentz metric. By (7) of Lemma 7.13.1, p. 250, the action of G on M' is locally faithful. By assumption, for some m G M, S t a b ^ m ) is noncompact. So, by (5) of Lemma 7.13.1, p. 250, for some ml € M', Stabz(m') is noncompact. Then, because $ _ 1 : G -¥ Go tx. V is a Lie group isomorphism and because $ _ 1 ( Z ) = V, we see that (Go,V) € AfCS. The result therefore follows from Theorem 15.7.1, p. 377. • T H E O R E M 15.7.3 Let G € OMV. Assume that the nilradical N of G is simply connected. Then at least one of the following is true: (1) The group G is Ad-nonproper. (2) There is a locally faithful, isometric action of G on a connected Lorentz manifold such that a noncompact closed subgroup of Z{N) fixes a point. (3) The Lie algebra g admits a one-dimensional ideal. (4) For some integer n>2, either so(n, 1) | g or so(n + 1,2) | g. Proof. Say that (1), (2), (3) and (4) are false. We aim for a contradiction. By Coincidence X.6.4(i), p. 519 of [He78], we have s[ 2 (R) = so(2,l), so s^IR) is not a direct summand of g. By Lemma 14.6.2, p. 353 choose mo € M such that H :— Stabo(m 0 ) is noncompact. Since (2) is false, we know that Hf)(Z(N)) is compact, so, by Lemma 4.14.6, p. 140, we see that Hn(Z(N)) is trivial. Then hn(j(n)) = {0}. Choose X as in Lemma 15.2.1, p. 363. Choose Y as in Lemma 15.3.1, p. 366. Then, by Lemma 15.6.1, p. 375, f) D n = {0}. By Lemma 15.4.2, p. 370, fix an (adg)-irreducible subspace V C 3(11) and a Q £ MinkQF(V) such that ady(jj) C co(Q). Let ni := V. Then Ad n i (G) C CO°(Q). As m C j(n), we see that nx is an Abelian Lie subalgebra of g. Let iVi be the connected Lie subgroup of G corresponding to m . By Theorem 3.6.2, p. 196 of [Va74], JVX is simply connected and is closed in N. Moreover, since V is (adfl)-invariant, it follows that Ni is a normal subgroup of G. Then, by Lemma 9.2.2, p. 277, (2) holds, contradiction. • T H E O R E M 15.7.4 Let G be a connected Lie group with simply connected nilradical N. Then G € OMV iff at least one of the following is true:

380

Orbit Nonproper Lorentz

Actions

(1) The group G is Ad-nonproper. (2) There exists an (Adif)-irreducible vector subspace V\ of i{xi) and there exists Q G QF(Vi) such that Q is either positive definite or Minkowski and such that A d y ^ i / ) C CO°(<5). (3) For some integer n > 2, either so(n, 1) | g or so(n + 1,2) | g. Proof. Proof of "if": By Lemma 7.14.13, p. 260, (1) = • G € OMV. By Lemma 9.4.6, p. 283, ( 3 ) = ^ G e OMV. Assume that (2) holds. We wish to show that G € OMV. Let N\ be the connected Lie subgroup of G corresponding to V\. By Theorem 3.6.2, p. 196 of [Va74], iV"i is simply connected and Ni is closed in G. Because Vi is (Ad G)-invariant, it follows that iVi is normal in G. Then, by Lemma 9.2.2, p. 277, G € OMV. End of proof of "if". Proof of "only if": For every one-dimensional vector space Vi, for every Q 6 QF(Vi), we have (GL(Vi)) 0 = CO°(Q). So, for every one-dimensional ideal Vi of g, for every Q € QF(Vi), we have Ad Vl (G) C CO°(Q). Then (3) of Theorem 15.7.3, p. 379 implies (2) of Theorem 15.7.4, p. 379. By Corollary 15.7.2, p. 378 (with H replaced by G), we conclude that: (2) of Theorem 15.7.3, p. 379 implies (2) of Theorem 15.7.4, p. 379. By Theorem 15.7.3, p. 379, since G e OMV, we know that (1), (2), (3) or (4) of Theorem 15.7.3, p. 379 must hold. Since (1), (2), (3) and (4) of Theorem 15.7.3, p. 379 imply (1), (2), (2) and (3) of Theorem 15.7.4, p. 379, respectively, and we are done. End of proof of "only if". O Let G be a connected Lie group with nilradical TV. Let L be a semisimple Levi factor of G. Then one may effectively determine whether G satisfies (2) of Theorem 15.7.4, p. 379 by decomposing the Adjoint representation of L on j(n) into irreducible subrepresentations. For a version of Theorem 15.7.4, p. 379 that clarifies this remark, see Theorem 1.1 of [A99c].

Appendix A

The Borel Density Theorem

In this appendix, we give a simple proof of the Borel Density Theorem. I first heard the idea for the following proof from M. Burger in 1990. It was known to D. Witte in the mid-1980s.

A.l

Notation and terminology

If X is a locally compact topological space, then we write "a; —> oo in X" to mean "a; leaves compact subsets of X " . Let F b e a topological field, i.e., a field with a Hausdorff topology such that multiplication, reciprocation and addition are all continuous. We assume F is lcsc. We assume that 1/t —> 0, as t —• oo in F. (This rules out, for example, the possibility of F = Q, with the discrete topology.) For any variety V defined over F, give the F-points V(F) of V the Hausdorff topology coming from the topology on F. For all integers n > 1, let An denote affine n-space over F. Let Qa denote the additive group defined over F and let Qm denote the multiplicative group defined over F. Let Q be a connected algebraic group defined over F. Let G := Q{F). We will say that a subgroup H < G is a split algebraic one-parameter subgroup if there is a Go 6 {Ga, Qm} and there is a rational homomorphism h:Go-+g defined over F, such that H = h(G0(F)). If X is a topological space, if Fo C F is a compact subset of F and if / : F\F0 ->• X is a function, then we will say that / is decisive if either • f(t) —t oo in X, as t —)• oo in F; or 381

382

The Borel Density

Theorem

• there exists x £ X such that f{t) —> x, as t -> oo in F.

A. 2

Preliminaries

Note that if F = R and if fit) = sin(t), then / is not decisive. On the other hand, if / is rational function, i.e., a quotient of two polynomials with coefficients in F, then, because \/t —> 0, as t —> oo in F, it follows that / is decisive. This basic algebraic fact can be extended to algebraic parametric curves in any variety: L E M M A A.2.1 Let V be a variety defined over F. Let f be a rational map from A1 to V defined over F. Let FQ denote the set of points in F = A1 (F) where f is not defined. Then f : F\F0 -¥ V(F) is decisive. Proof. Let W Q A1 be the set of points where / is defined. Then W is an irreducible one-dimensional variety. Assume, for a contradiction, that / is not decisive. Then there exist v £ V(F) and a sequence ti in F such that • U -> oo in F, as i ->• oo; • f(U) -4 u, as i -> oo; and • /(*) -ft v, as t -l oo in F. Let V be an open neighborhood in V of v such that V is isomorphic to a closed subvariety of some affine space. Let Wo := / _ 1 ( V ) C W. Then Wo is a nonempty open subset of the one-dimensional variety W. It follows that U := W\W 0 is finite. Let Fi := U(F) C W(F) C A1 (F) = F and let FQ := F0U F\. Then FQ is finite, so, for all sufficiently large i, we have U^F^. Let / ' be the restriction of / to W 0 . Then / ' : F\F£ -> V'(F) is not decisive. For some integer n > 0, identify V with a closed subvariety of A n . For each integer i £ [l,n], let 7T; : A" -¥ A1 denote the ith coordinate projection map. Then, for some i, •Ki o / ' : F\F0 -4 F is not decisive. However, 7Tj o / ' is a rational function from A1 to A 1 , i.e., 7Tj o / ' is a quotient of two polynomials with coefficients in F. Since 1/i -> 0, as t —> oo in F, it follows that 7Tj O / ' : F\Fo —> F must be decisive, giving the desired contradiction. •

The Borel Density Theorem

383

L E M M A A.2.2 Suppose L is a noncompact, Icsc topological group acting continuously on an Icsc topological space X. Assume that the L-action on X is orbit proper. Then there is no L-invariant probability measure on X. Proof. Assume that /x is an L-invariant probability measure on X. We wish to obtain a contradiction. Choose a compact subset K C X such that n(K) > 1/2. Choose a sequence k in L such that k -» oo, as i ->• oo. For each A; € K, for all sufficiently large i, we have Uk ^ K. Choose io sufficiently large that <£K}> 1/2. Let K0 ~ {k € K\liok $ K}. lx{keK\liok Then fJ.(lioK0) - fj,(K0) > 1/2. On the other hand fj,(K) > 1/2 and li0Ko C\K — %. However /z is a probability measure, contradiction. • L E M M A A.2.3 Fix Qo £ {ga,Qm} and let G 0 := Go{F). Let V0 be a variety defined over F and let Qo act F-morphically on Vo- Assume that Vo := Vo(F) admits a probability measure fixed by Go- Then Vo has at least one point fixed by Go • Proof. We will assume that Go = Gm, so Go is the multiplicative group F* of the field F. If Q0 = Qa, then a similar proof will work. Since there is a Go-invariant probability measure on VQ, it follows from Lemma A.2.2, p. 382 that there exists a point v G Vo such that tv -/> oo in Vo as t —• oo in F*. Then either or both of the following must hold: (*) tv -ft oo in V0, as t ->• oo in F; or (**) tv -ft oo in Vo, as t -» 0 in F. We will assume (**), the proof in case (*) being similar. Then (^/t)v ft oo in Vo, as t ->• oo in F. By Lemma A.2.1, p. 382, the map t (->• (l/t)v : Go ->• V0 is decisive, and we conclude that there is a point vo € Vo such that (l/t)v —> vo, as t —>• oo in F. We will show that vo is a Go-fixpoint. Fix t0 S G 0 = F*. We wish to show that (l/to)^o = «o- As i -^ oo in F, (l/t0)[(l/t)v] -> (l/t0)vo. On the other hand, since t0t -^ oo in F , we get [l/(t0t)]v -»• v0- Then (l/io)^o = i>o, as desired. •

A.3

The Borel Density Theorem

Let S be the subgroup generated by the split algebraic one-parameter subgroups of G. Note that if G is semisimple with no compact factors, then S

384

The Borel Density

Theorem

is Zariski dense in Q. T H E O R E M A.3.1 (Borel Density Theorem.) Assume that S is Zariski dense in Q. Let V be a variety defined over F and let Q act F-morphically on V. Let V := V(F). Let \i be a probability measure on V and assume that /i is G-invariant. Let W denote the set of G-fixpoints in V. Then W 7^ 0 and [i is supported on W. Proof. By the ACC on closed irreducible subvarieties of a variety, we can choose an integer m > 0 and a collection H1,..., Hm of split algebraic oneparameter subgroups of G such that Q is the Zariski closure of the group H generated by H\ U • • • U Hm. Then any H-fixpoint in V is a (/-fixpoint, and is therefore a G-fixpoint. Therefore W is the set of .ff-fixpoints in V. For each integer i € [l,m], let W» denote the ifj-fixpoints in V. Then W = Wi n • • • n Wm. Assume, for some integer i0 € [1, m] that n is not supported on W, 0 . We wish to obtain a contradiction. Choose Qo £ {Qa, Qm} and a rational homomorphism h : Go —>• Q defined over F such that Hio = h(G0(F)). Let G 0 := Go(F). Pulling the (/-action on V back via h, we obtain an action of Go on V defined over F. Let Wo denote the F-subvariety of <7o-fixPomts in V and let WQ := Wo(F). Since Go is Zariski dense in Go, WQ is the set of Go-fixpoints in V. So, since h(Go) = Hi0, we get Wo = Wt0. Then fj, is not supported on WQ. Let V0 := V\W 0 and let V0 := V0(F). Then Go acts F-morphically on Vo and, passing to F-points, we obtain an action of Go on VQ. Since fi is not supported on Wo and since Vo = V\Wo, we conclude that fJ,(Vo) > 0. Then Vo has a Go-invariant probability measure given by normalizing the restriction of /* to Vo, i.e., given by (/x|Vo)/[/i(Vo)]From Lemma A.2.3, p. 383, we see that VQ must have a Go-fixpoint. Then Vo n Wo ^ 0, while Vo = V\Wo, giving the desired contradiction. • COROLLARY A.3.2 Assume that S is Zariski dense in G- Let T be any subgroup of G such that G/T admits a G-invariant probability measure. Then T is Zariski dense in GProof. Let H denote the Zariski closure in G of T. Since S is Zariski dense in G and since 5 C G, it follows that G is Zariski dense in G- It therefore suffices to show that G C %{F). Let H := H(F). We will show that H = G.. Since the G-action on G/H is transitive, it suffices to show that G/H has a G-fixpoint.

The Borel Density Theorem

385

Let V := G/H and let V := V{F). Since there is a G-equivariant Borel injection of G/H into V, we see from Theorem 2.4 that any G-invariant probability measure on G/H must be supported on G-flxpoints; it therefore suffices to show that G/H admits a G-invariant probability measure. However, T C H, so there is a G-equivariant map G/Y —> G/H. By assumption, G/T has a G-invariant probability measure; we conclude that G/H does as well. •

Appendix B

Tameness of Algebraic Actions in Characteristic Zero

Let fc be a local field (i.e., a nondiscrete, locally compact topological field) of characteristic 0. Let A" denote affine n-space over fc. Varieties are assumed to be irreducible by definition. If V is a fc-variety, then we denote the field of fc-rational functions on V by k(V). If V is an affine fc-variety, then we denote the ring of fc-regular functions on V by k[V]. The locally compact Hausdorff topology on fc induces a locally compact Hausdorff topology on the set fc-points of any fc-variety. If V is a fc-variety, then we denote the locally compact Hausdorff topological space of fc-points of V by V(k). T H E O R E M B.0.3 Let G be a k-group. Let V be a k-variety. Assume that G acts k-algebraically on V. Then the action ofG(k) on V(k) is tame. Proof. Fix v G V(fc). By (1) =>• (2) of Lemma 7.11.5, p. 241, we wish to show that (G(k))v C V(k) is locally closed in V(k). Replacing V by the fc-closure of Gv, we may assume that g H* gv.G -* V is dominant (Definition B.0.4, p. 388), hence generically submersive (Definition B.0.8, p. 389 and Theorem B.0.10, p. 390). Then, by G-equivariance, g i-t gv.G -* V is everywhere submersive (Definition B.0.8, p. 389). We will show that (G(k))v is open in T^(fc). Let Vi be the set of simple points of V. By standard algebraic geometry, Vi is fc-open and nonempty. Moreover, since G acts algebraically, it follows that Vi is G-invariant. Further, since Gv is fc-dense, Gv n V\ ^ 0. By G-invariance, Gv C V\, so we may replace V by V\ and assume that V is a smooth fc-variety. On passing to fc-points, g i-> gv. G(fc) —> V(k) is an everywhere submersive map of analytic fc-manifolds. So, by (3) of Lemma B.0.9, p. 389, its 387

388

Tameness of Algebraic Actions in Characteristic

Zero

image is an open set of V(k). That is, (G(k))v is open in V(k).

D

D E F I N I T I O N B.0.4 We say that a rational k-map V -> W is d o m i n a n t if its image in W is k-dense. We say that two dominant rational k-maps V —• W and V —• W are b i r a t i o n a l if there exist birational maps V <-¥ W and V •*-» W such that the composition V —> W —> W is equal to the composition V —> V —r W. R E M A R K B.0.5 Let V denote the category of k-varieties with dominant rational k-maps. Let T denote the category of finite type extension fields of k with k-algebra homomorphisms. Standard algebraic geometry shows that V i-> k(V): V —> T defines a contravariant equivalence of categories. If V is an affine k-variety, let Ty € k[V x A1] be the fc-regular function defined by projection onto A 1 . We identify k[V x A1] = A;[V][Ty], where Ty is transcendental over k[V]. D E F I N I T I O N B.0.6 Let V be an affine k-variety, f € k[V x A 1 ]. Assume that f € fc(V)[Ty] is monic irreducible and has degree > 1. Let Wf be the k-variety defined by the vanishing of f. (It is a closed k-subvariety of V x A1.) Let 7T/: Wf —> V denote the restriction of the projection V x A1 -> V. A rational k-map is simple if it is birational (in the sense of Definition B.0.4, P- 388) to some irf. A rational k-map is purely transcendental or p.t. if there exists a k-variety V and a nonnegative integer n such that <j> is birational (Definition B.0.4, P- 388) to the projection V x A" -> V. T H E O R E M B.0.7 Any dominant rational k-map is birational (in the sense of Definition B.0.4, p. 388) to a composition of a simple rational k-map followed by a p.t. rational k-map. Proof. By Remark B.0.5, p. 388, we can translate this theorem into algebraic terms: We wish to show that if K, L 6 T and K C L, then there exists an intermediate field K C M C L such that K C M is purely transcendental of finite transcendence degree and such that M C L i s simple (i.e., generated by a single element of L). The minimimum polynomial of a generator of L over M will then serve as the / of Definition B.0.6, p. 388. If K,L € T and K C L, then, by the existence of transcendence bases, there exists an intermediate field K C M C L such that K C M is purely

389

transcendental and such that M C L is algebraic and finite type, hence finite-dimensional. Since L is finite type over fc, it follows that M has finite transcendence degree over K. Since the characteristic of fc is 0, we have that M C l i s separable and finite-dimensional, hence simple. • D E F I N I T I O N B.0.8 A rational k-map :V -> W is generically submersive (resp. everywhere submersive,) if {v E V | d-Kv:TVv ->• TWf(v) contains a nonempty k-open subset ofV

is surjective}

(resp. is equal to V).

L E M M A B.0.9 The following all hold: (1) Generic submersiveness is an invariant of birational equivalence of rational maps (Definition B.0.4, p. 388). (2) Simple maps are generically submersive. (3) There is an analogous notion of everywhere submersiveness for maps between analytic k-manifolds. By the implicit function theorem, the image of an everywhere submersive map M —> M' between analytic k-manifolds is an open subset of M'. Proof. We will only prove (2). We use the notation developed in Definition B.0.6, p. 388 and in the text immediately preceding Definition B.0.6, p. 388. Let V be an affine fc-variety, / £ k[V x A 1 ]. Assume / € fc(V)[Ty] is irreducible monic and has degree > 1. By (1), we need only show that 717: Wf -> V is generically submersive. Let T := Ty, := 717, W := Wf. Choose a o , . . . , o n - i £ kty] such that f = Tn + an-iT"-1

+ ••• + a2T2 + aiT + a0.

Define / ' := nTn~l + (n - l ) o n _ i T " - 2 + • • • + 2a2T + a i . Then the set of points in W where dcj) is surjective on tangent spaces contains the set of points in V x A1 where / vanishes, but / ' does not. We are thus reduced to showing that the vanishing set of / ' does not contain the vanishing set of / . By the Nullstellensatz, this is equivalent to showing that there cannot exist both a g £ k[V x A1] and a positive integer n such that gf = (/')"• But / , / ' E fc(V)[T] are relatively prime (since the characteristic of fc is 0), and the result follows. • We now come to an algebraic analogue of Sard's theorem:

390

Tameness of Algebraic Actions in Characteristic

Zero

T H E O R E M B.0.10 A dominant rational k-map is generically submersive. Proof. Note that the composition of a dominant, generically submersive rational map followed by a generically submersive rational map is again generically submersive. So, since p.t. maps are dominant and generically submersive, the theorem follows from Theorem B.0.7, p. 388 and from (1) and (2) of Lemma B.0.9, p. 389. •

Bibliography

[A01] S. Adams: Containment does not imply Borel reducibility, to appear, Proceedings of the Hajnal Conference, ed. Simon Thomas, Dimacs series, AMS. [AOOa] S. Adams: Locally free actions on Lorentz manifolds, Geometric and Functional Analysis (GAFA) 10 (2000), 453-515. [AOOb] S. Adams: Dynamics of simple Lie groups on Lorentz manifolds, preprint, 2000. [AOOc] S. Adams: Orbit nonproper actions on Lorentz manifolds, Geometric and Functional Analysis (GAFA) 11 (2001), 201-243. [AOOd] S. Adams: Nontame Lorentz actions of scmisimplc Lie groups, preprint, 2000. [A99a] S. Adams: Induction of geometric actions, to appear, Geometriae Dedicata. [A99b] S. Adams: Transitive actions on Lorentz manifolds with noncompact stabilizer, preprint, 1999. [A99c] S. Adams: Orbit nonproper dynamics on Lorentz manifolds, preprint, 1999. [A98] S. Adams: Another proof of Moore's ergodicity theorem for SL(2, W), Contemporary Math. 215 (1998), 183-187. [AS97a] S. Adams and G. Stuck: The isometry group of a compact Lorentz manifold, I, Invent. Math. 129 (1997), no. 2, 239-261. [AS97b] S. Adams and G. Stuck: The isometry group of a compact Lorentz manifold, II, Invent. Math. 129 (1997), no. 2, 263-287. [B73] W. Baily: Introductory Lectures on Automorphic Forms, Princeton University Press, Princeton, NJ, 1973. [BK93] H. Becker and A. Kechris: Borel actions of Polish groups, Bull. Amer. Math. Soc. (N.S.) 28 (1993), no. 2, 334-341. [BEM88] J. Beem, P. Ehrlich and S. Markvorsen: Timelike isometrics and Killing fields, Geom. Dedicata 26 (1988), 247-258. [Bor63] A. Borel: Compact Clifford-Klein forms of symmetric spaces, Topology 2 (1963), 111-122. [Bou75] N. Bourbaki: Lie Groups and Lie Algebras, Part I, Addison-Wesley, 391

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Index

action of a Lie algebra on a manifold, 73 Ad-invariant quadratic form, 77 ad-invariant quadratic form, 69 Ad-nonproper, 156 Ad-proper, 156 additive, 95 agree near m, 198 agree to order 0 near 0, 198 agree to order k at 0, 198 agree to order k at m, 198 almost algebraic group, 135 anti-symmetric with respect to B, 34 antisymmetric bilinear form, 32 arcwise connected, 84 associated bundle, 99 ax+b group, 296

by isometries, action, 70 Cartan subalgebra, 116 centerfree, 14 centralize, 14, 66 chain, 12 chain connected, 12 class £", 316 class £x, 316 class (measure class), 157 closed mapping, 15 cocompact, 27 commuting framing, 52 compact semisimple Lie algebra, 124 compactly generated, 107 complementary distributions, 55 completely reducible, 114 complex diagonalizable, 20 complex module over a Lie algebra, 67 complex module over a Lie group, 67 complex split torus, 116 complex structure, 20 complex vector space, 18 complexification, 18 concentrated, 22 conjugate module over a Lie group, 69 conjugate vector space, 18 connected base of a root system, 37 connected isometry group, 154

Baire category theorem, 16 Baire space, 16 base of a root system, 36 basic axiom of special relativity, 3 bilinear form, 32 Borel action, 157, 239 Borel map, 11 Borel measure, 11 Borel sets, 11 Borel space, 11 by Borel automorphisms, action, 218 by isometries, 6 395

396 connected Lie subgroup, 80 connected Lie subgroup corresponding to a subalgebra, 83 connected stabilizer, 26 connected, y~, 12 constant, 201 constant germ of vector field, 200 constant part, 200, 202 constant rank smooth map, 45 constant vector field, 200 continuous action, 26 continuous principal bundle, 99 conull, 157 convergence in direction, 19 convergence-divergence dichotomy, 43 countable equivalence relation, 49 countable index subrelation, 12 covering group, 70 covering homomorphism, 70 covering map, 42 cross-section, 11 definite quadratic form, 33 definite subspace, 32, 33 definite symmetric bilinear form, 32 degenerate bilinear form, 32 degenerate quadratic form, 33 degenerate subspace, 32, 34 degree, 200, 201 degree i part, 201, 202 derivation, 102 derived series, 66 descending central series, 66 diagonally adapted, 337 direct summand, 70 discrete cr-algebra, 239 discrete subset, 16 discrete topological space, 16 disjoint union of manifold structures, 53 diverges mod normal subgroups, 154 divisible, 88 effective, 269

Index elliptic, 20 elliptic part, 20 energy, 4 equivalence relation generated by a relation, 12 equivalent (measures), 157 equivalent bilinear forms, 32 equivalent quadratic forms, 34 equivariant, 15, 68 ergodic, 157, 158 essentially transitive, 157 everywhere lightlike, 62 everywhere lightlike vector subbundle, 317 everywhere spacelike, 62 everywhere timelike, 62 expansive, 23, 24 exponential decay, 316 extension of Lie groups, 100 factored orbit map, 239 faithful, 6 faithful G-manifold, 70 faithful action, 15 fiber dimension, 61 fiber of a map, 11 finite Lorentz volume, 63 flat, 201 flat Lorentz manifold, 5 flat metric, 4 flat quadratic differential, 201 foliation, 50 framing of a vector bundle, 41 free G-manifold, 70 free action, 15 Frobenius Theorem, 57 generates, topology generates cr-algebra, 239 germ at m, 198 germ of a quadratic differential, 201 germ of a section, 199 germ of a vector field, 199

397

Index Haar measure, 26 Heisenberg Lie algebra, 139 Heisenberg Lie group, 139 Heisenberg relations, 139 homogeneous polynomial of degree A;, 133 homotopic through a family of maps, 111

kernel of a bilinear form, 32 kernel of a quadratic form, 33 kernel of an action, 6, 15 Killing, 208 Killing form, 112 Killing w.r.t. connection, 207 Kowalsky element, 19 Kowalsky subset, 19

immersion, 58 immersive at x, 57 in, 23 indefinite quadratic form, 33 indefinite subspace, 32, 33 indefinite symmetric bilinear form, 32 induction of actions, 249 inherited g-action, 73 inner product space, 32 invariant bilinear form, 69 invariant complex structure, 70 invariant foliation, 81 invariant quadratic form, 69 invariant symmetric bilinear form, 69, 76, 77 invariant vector subspace, 68 involution, 115 involutive vector subbundle, 50 irreducible root system, 37 irreducible vector subspace, 68 isometric, 6 isometric action, 70 isometric map, 63 isometry group, 5, 63 isomorphic modules, 145 isotropic subset, 176 isotropic vector, 32, 33 isotropy representation, 209 isotypic, 172

lattice, 30 lcsc, 15 leaf, 49 leaflike manifold structure, 42 leaflike submanifold, 42 leaflike topology, 42 left-invariant distribution, 81 left-invariant foliation on a Lie group, 81 left-invariant measure, 25 left-invariant vector field, 76 Lie algebra, 65 Lie algebra of a Lie group, 79 Lie derivative, 43 Lie group, 65 Lie manifold structure, 80 Lie subgroup, 80 Lie topology, 80 light cone of a Minkowski vector space, 178 lightlike, 40 linear, 201 linear germ of vector field, 200 linear part, 201, 202 linear vector field, 200 linearization of a vector field, 207 local diffeomorphism, 41 local map near m, 198 local potential submersion, 58 local quadratic differential, 201 local section near m, 199 local vector field, 199 locally closed, 15 locally faithful, 6 locally faithful g-action, 73

Jacobi identity, 43, 102 jet, 199 jet of a quadratic differential, 201 jet of a section, 199 jet of a vector field, 199

398 locally faithful g-manifold, 73 locally faithful action, 26 locally free fl-action, 73 locally free 0-manifold, 73 locally free action, 26 locally isomorphic, 81 locally potentially open map, 58 locally trivial principal pre-bundle, 98 localy diffeomorphism, 219 Lorentz, 5 Lorentz bundle metric, 273 Lorentz manifold, 62 Lorentz metric, 62 Lorentz volume density, 63 manifold, 41 manifold, 0-, 73 manifold, G-, 70 matrix of a linear endomorphism, 19 matrix of a linear transformation, 19 maximal complex split torus, 116 maximal real split torus, 69 measurable, 157 measure class, 157 measure space, 156 minimal, 156 Minkowski, 5, 39 Minkowski vector space, 39 module over a complex Lie algebra, 67 module over a complex Lie group, 67 monomial, 133 monomial of degree 0, 133 monomial of degree k, 133 negative definite quadratic form, 33 negative definite subspace, 32, 33 negative definite symmetric bilinear form, 32 negative semidefinite quadratic form, 33 negative semidefinite subspace, 32, 33 negative semidefinite symmetric bilinear form, 32

Index negligible, 158 neutral, 316 nilpotent Lie algebra, 66 nilpotent Lie algebra, n-step, 66 nilpotent Lie group, 66 nilpotent part, 20 nilradical of a Lie algebra, 141 nilradical of a Lie group, 141 no compact factors, 124 noncompact semisimple Lie algebra, 124 nondegenerate, 62 nondegenerate bilinear form, 32 nondegenerate quadratic form, 33 nondegenerate subspace, 32, 34 nonproper action, 31 nonproper foliation, 50 nonproper map, 15 nonscalar intertwining, 68, 69 nontame, 239 nontrivial submodule, 68 normalize, 66, 76 Notation: M+, 11 null, 156 null algebra, 157 one-parameter subgroup, 76 open mapping, 15 orbit map, 235 orbit nonproper action, 235 orbit proper, 235 ordered Q-basis, 63 orthogonal group, 3 path in a base of a root system, 37 pending vertex, 37 Poincare Recurrence Lemma, 219 polarization, 33 Polish topological space, 16 polynomial, 133 positive definite quadratic form, 33 positive definite subspace, 32, 33 positive definite symmetric bilinear form, 32

Index

399

positive root, 36 positive semidennite quadratic form, 33 positive semidennite subspace, 32, 33 positive semidennite symmetric bilinear form, 32 potential submersion, 58 potentially open map, 58 precompact, 15 prefoliation, 49, 50 preserves a tangent framing, 41 preserves a vector field, 41 principal bundle, 98 principal pre-bundle, 98 principal pre-bundle isomorphism, 98 product of equivalence relations, 49 proper G-manifold, 70 proper action, 31 proper foliation, 50 proper map, 15 properly discontinuous, 31 properly ergodic, 157, 158 pseudoRiemannian manifold, 4, 62 pseudoRiemannian metric, 4, 62

real Jordan closed, 20 real Jordan decomposition, 20 real module over a Lie algebra, 67 real module over a Lie group, 67 real rank of a Lie algebra, 70 real rank of a Lie group, 83 real split torus, 69 realification, 18 recurrent, 219 reductive Lie algebra, 377 reductive Lie group, 377 restriction of an equivalence relation, 49 Riemannian bundle metric, 273 Riemannian manifold, 4, 62 Riemannian metric, 62 right-invariant distribution, 81 right-invariant foliation on a Lie group, 81 right-invariant measure, 26 right-invariant vector field, 76 root, 70 root system, 36 rootspace, 70

qmp, 157 quadratic, 201 quadratic differential, 62 quadratic form, 33 quadratic germ of vector field, 200 quadratic part, 201, 202 quadratic vector field, 200 quasi-measure preserving, 157

saturated, / - , 11 scalar intertwining, 68, 69 semisimple Levi factor of a Lie algebra, 96 semisimple Levi factor of a Lie group, 96 semisimple Lie algebra, 66 semisimple Lie group, 83 separates points, 238 sequence of vectors in sequence of subspaces, 23 signature of a bilinear form, 33 signature of a quadratic form, 33 simple Lie algebra, 66 simple Lie group, 83 simply transitive, 98 smooth, 41 smooth action, 65 smooth measure class, 158

rank of a linear transformation, 45 rank of a smooth map, 45 rank of a vector bundle, 42 rational function, 382 rationally twisted Heisenberg Lie algebra, 263 rationally twisted Heisenberg Lie group, 264 real diagonalizable, 20 real diagonalizable part, 20

400

solvable Lie algebra, 66 solvable Lie group, 66 solvable radical of a Lie algebra, 96 solvable radical of a Lie group, 96 spacelike, 40 spanning tree, 37 spectral theorem, 34 split algebraic one-parameter subgroup, 381 stabilizer, 15 standard sl2(l&) basis, 125 standard Borel space, 239 standard module over S0(n, 1), 193 standard trivial principal pre-bundle, 98 standard trivialization, 243 strongly lightlike, 224 strongly nontimelike, 224 strongly vanishing, 224 sub-equivalence relation, 13 subbundle invariant under fl, 74 submanifold, 42 submersion, 58 submersive at x, 57 submodule, 68 switch, 363 symmetric bilinear form, 32 symmetric with respect to B, 34 symmetrizable subset, 116 symplectic form, 139 symplectic vector space, 139 tame, 238, 239 tangent framing of M, 41 timelike, 40 topological field, 381 topological group, 25 topologically transitive, 156 totally disconnected subset, 16 totally disconnected topological space, 16 totally path-disconnected subset, 16 totally path-disconnected topological space, 16

Index transitive equivalence relation, 49 trivial covering map, 42 trivial equivalence relation, 49 trivial module, 145 trivial prefoliation, 49 trivial principal pre-bundle, 98 trivial submodule, 68 twisted Heisenberg Lie algebra, 263 twisted Heisenberg Lie group, 264 unipotent group, 142 unipotent transformation, 142 universal covering group, 70 vanishes to order k at m, 211 vector bundle, 41 vector bundle complement, 273 vector space, 18 vector space complement, 111 volume density, 63 volume density element, 63 Weyl's Theorem, 114 Zariski closed, 134 Zariski closed in GL(V), 134 Zariski closure, 134 Zariski closure in GL(F), 135 zero-dimensional subset, 16 zero-dimensional topological space, 16 ZZZ Notation: 1 G , 14 ZZZ Notation: [h,{], 66 ZZZ Notation: [H,K], 14 ZZZ Notation: [T]B, 19 ZZZ Notation: [T\$, 19 ZZZ Notation: [v]B, 19 ZZZ Notation: [X,Y], 76 ZZZ Notation: Ad, 75 ZZZ Notation: ad, 65 ZZZ Notation: Ad 8 , 75 ZZZ Notation: ad„, 65 ZZZ Notation: Ad G , 75 ZZZ Notation: Ad 0 / [ ) (/i), 76 ZZZ Notation: aG(T), 73

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-dynamics on lonenrz Manifolds Within the general framework of the dynamics of "large" groups on geometric spaces, the focus is on the types of groups that can act in complicated ways on Lorentz manifolds, and on the structure of the resulting manifolds and actions. This particular area of dynamics is an active one, and not all the results are in their final form. However, at this point, a great deal can be said about the particular Lie groups that come up in this context. It is impressive that, even assuming very weak recurrence of the action, the list of possible groups is quite restricted. For the most complicated of these groups, one can also describe reasonably well the local structure of the actions that arise.

This advanced text is also appropriate to a course for mathematics graduate students w h o have completed their first year of study.

ISBN 981-02-4382-OI

www. worldscientific. com 4491 he