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£ S-. ',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 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 , . . . , ^ ) . 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. •
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',
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
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
^:= 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
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 :=
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 =
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),
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
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
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
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
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 -+ (
62
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L E M M A 3.8.7 Let M be a manifold and let T be a proper foliation of M. Letp := dim(J'). Let cj): M —> M/J7 be the canonical map. Then
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
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
Basic Lie Theoretic
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).
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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
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
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Basic Lie Theoretic
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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
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
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(R(w)) n W = {w}. Assume, for a contradiction that this fails, i.e., that there are two sequences Wi and w[ in W such that (1) Wi -> 0 in W and w\ -> 0 in W; (2) for all i, (wi,w'i) € i?; and (3) for all i, we have Wi ^ w;J. For all i, let m* := ^>_1(0,«;;) and let m^ := V' _1 (0,w^). Then we have mi —> m and mj —> m. For all i, since (u>j,u4) € i?, it follows that (rriijm'j) € T\ then Gm; = T{rrii) = J^(mJ) = GmJ. For all i, choose gi € G such that pimj = mj. The G-action on M is proper, so, after passing to a subsequence, we may assume that gi is convergent in G. Let g^ := lim gi. Since, for all i, we have gitrii = m'i, it follows that 1GChoose a connected open neighborhood Go of 1 Q in G and an open neighborhood M 0 of m in M such that GoM 0 C V. Passing to a tail, we may assume, for all i, that gt € Go- Passing to a further tail, we may assume, for all i, that rrii e Mo. Then, for all i, we have mj = gimi € GoMo C V. Let 7r : Rd —> R9 be projection onto the last g coordinates, defined by Tr(ti,... ,td) = {tp+i,... ,td). Let p := n o ip •. V -^ W. Then, for all i, p{rrii) = Wi and /9(mJ) = w^. For all v € V, we have p((Gv) n V) = R(p(v)); since R(p(v)) is countable, it follows that U(p(i;)) is totally disconnected in W. So, since Go is connected, for all m 6 Mo, the map g H-> p(gm) : Go —> R(p(v)) is constant. It follows, for all m e Mo, for all g E Go, that p(gm) = p(m). Then, for all i, we have u;J = /9(m'f) = p(girrii) = p(m$) = Wi, contradicting (3) above. End of proof of Claim 1. Choose W as in Claim 1. If necessary, replacing W by a smaller open neighborhood of 0 in W, we may assume that W is diffeomorphic to Rq. Let /? : W -> W be a diffeomorphism. By Claim 1, we have P.(R\W) = P*(T^\W). Then P,(R\W) = T«. Define a : W x W -> Rd by a(a;,u/) = (a;,£(«;)). Let £/ := ^(W x W). Let 0 := a o V> : *7 -> Kd. Then 0»(^|C7) = a.(T? x (R\W)) = T* x so <j>*{F\U) =
TPxT«.
{fi.{R\W)\ D
Dynamical consequences of the Frobenius
Theorem
73
For any Lie algebra g, for any manifold M, a g-action on M is a Lie algebra homomorphism from g to the infinite-dimensional Lie algebra of vector fields on M. For any Lie algebra g, a g-manifold is a manifold together with a g-action on M. If G is a Lie group and M is a G-manifold, then differentiating the action turns M into a g-manifold. Let Q be a Lie algebra and let M be a manifold. Let V denote the infinite-dimensional Lie algebra of vector fields on M and let / : g —> V be a g-action on M. Then we say that / is locally faithful if / is injective. We say that / is locally free if, for every X e fl\{0}, for every m e M, f(X) does not vanish at m. Let a Lie group G act smoothly on a manifold M. Let / be the g-action on M obtained by differentiating the G-action on M. Then the G-action on M is locally faithful iff / is locally faithful. Moreover, the G-action on M is locally free iff / is locally free. Let g be a Lie algebra and let M be a g-manifold. A g-manifold M is locally faithful (resp. locally free) if the g-action on M is locally faithful (resp. locally free). Let M be a manifold and let V denote the infinite-dimensional Lie algebra of vector fields on M. Let g be a Lie algebra and let / : g -» V be a g-action on M. Let U be an open subset of M. Let W denote the infinite-dimensional Lie algebra of vector fields on U. Then the g-action on U inherited from / is X i-> (f(X))\U : g -> W. For any Lie group G, for any manifold T, let aa(T) be the G-manifold G xT, with G-action given by g(g',t) — (gg',t). Differentiating the action turns aa(T) into a g-manifold. For any Lie group G, for any nonempty open subset U of G, for any manifold T, let /3G{U,T) denote the g-manifold U xT, with g-action inherited from the g-action on ao{T) = G x T. The next two results give local descriptions of locally free g-manifolds and of free, proper G-manifolds. They follow from Lemma 4.2.1, p. 71 and Lemma 4.2.2, p. 71. Their proofs are left as exercises. L E M M A 4.2.3 Let g be a Lie algebra. Let M be a locally free g-manifold. Let mo £ M. Then there exists an open neighborhood Mo of mo in M, there exists an open neighborhood U of I Q in G and there exists a manifold T such that Mo is g-manifold isomorphic to PG{U,T). L E M M A 4.2.4 Let G be a Lie group. Let M be a free, proper G-manifold. Let mo € M. Then there exists a G-invariant open neighborhood Mo of mo
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Basic Lie Theoretic
Results
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 —> 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 =
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 =
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 ' := xg. Then U',M/ G r e G . Moreover, vg = t/g' and wg = w'g' and i> = 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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4.4
Basic Lie Theoretic
Results
The Lie group Lie algebra correspondence
A subgroup of a Lie group G is a Lie subgroup if it is a leaflike submanifold of G. The leaflike manifold structure on the subgroup is referred to here as its Lie manifold structure. The resulting leaflike topology on the subgroup is called the Lie topology; it may be finer than the relative topology. If G is a Lie group and H is a Lie subgroup of G, then h is a Lie subalgebra of g. The collection of Lie subgroups of a Lie group is closed under a variety of basic set-theoretic and group-theoretic operations: Let G be a Lie group. In Corollary 4.5.3, p. 85 below, we see that the subgroup of G generated by any collection of connected Lie subgroups of G is again a connected Lie subgroup; its Lie algebra is generated by the Lie algebras of the groups in the collection. In Corollary 4.5.9, p. 87 below, we see that the subgroup of G generated by a countable collection of Lie subgroups of G is again a Lie subgroup; its Lie algebra is not necessarily generated by the Lie algebras of the groups in the collection. According to Corollary 4.5.6, p. 86 below, if A and B are Lie subgroup of a Lie group G, then [A, B] is again a Lie subgroup; however, by Example 4.6.5, p. 91 below, the Lie algebra of [A, B] need not be [o, b]. By Lemma 4.5.13, p. 89 below, the intersection of an arbitrary collection of Lie subgroups of G is again a Lie subgroup of G; its Lie algebra is the intersection of the Lie algebras of the groups in the collection. If H is another Lie group and if / : G -> H is a Lie group homomorphism, then, according to Corollary 4.5.7, p. 86 below, the image f(G) is a Lie subgroup of H; its Lie algebra is (df)(g). It is possible for a Lie subgroup to be connected in the inherited topology but not connected in the Lie topology: For example, let T := R/Z, let G := T 2 be the two-torus, choose X G g such that H0 := {exp(tX)} <£ R is dense in G, let g € G\Ho and let H be the subgroup of G generated by {g} U Ho . Then Ho is a connected Lie subgroup of G and H is a countable union of cosets of Ho- If a topological space contains a dense connected subset, then it is connected, so H is connected in the relative topology inherited from G. On the other hand, in the Lie topology, the cosets of Ho in H are all open, so, as Ho C H, we see that H is not connected. By a connected Lie subgroup of a Lie group, we mean a Lie subgroup which is connected in the Lie topology. By Lemma 4.4.4, p. 82, if G is a Lie group, then if •-> h is a one-to-one correspondence between connected Lie subgroups of G and Lie subalgebras of g. Under this correspondence,
The Lie group Lie algebra correspondence
81
normal connected Lie subgroups of G correspond to ideals of g. Let H be a closed subgroup of a Lie group G. Then, by Theorem 2.12.6, p. 99 of [Va74], H is a Lie subgroup of G. In particular, if is a closed submanifold of the manifold G. Moreover, the Lie topology u o i i G agrees with the relative topology r inherited from H, because the identity map (G, a) —> (G, r) is a continuous, bijective homomorphism, and is therefore, by Lemma 2.7.6, p. 28, a homeomorphism. If H is a Lie subgroup of a Lie group G, then H° denotes the connected component of H, in the Lie topology, containing I Q . Two connected Lie groups are locally isomorphic if they have isomorphic Lie algebras. Prom the Scholium after Theorem 3, p. 49 of [C46], we see that two connected Lie groups are locally isomorphic iff they have isomorphic universal covering groups. Let G be a Lie group and let T be a foliation on G. We say that T is left-invariant (resp. right-invariant) if, for all x, x' G G, for all g G G, we have: if (x,x') G T, then (gx,gx') G T (resp. (xg,x'g) G T). Let a Lie group G act smoothly on a manifold M. Let J b e a foliation on M. We say that T is G-invariant if, for all m,m' G M, for all g € G, we have: if (m, m') G T, then (gm, gm') G T. (That is, the action of G on M sends leaves to leaves; however, it does not necessarily send a point in a leaf to a point in the same leaf.) Let G act on G from the left and right by left and right translation. Differentiating these two actions, we arrive at left and right actions of G on TG. Let e := 1 G , so g = TeG. A subset A C TG is left-invariant (resp. right-invariant) if GA C A (resp. AG C A). For any S C g, there is a unique left-invariant A C TG such that A n g = g. In particular, for any vector subspace V of g, there is a unique left-invariant distribution A C TG such that A e = V. L E M M A 4.4.1 Let G be a Lie group and let J- be a left-invariant foliation on G. Let L := T{e). Then L is a connected Lie subgroup of G. Proof. Since L is a leaflike submanifold of G and since L is connected in the leaflike topology, it suffices to show that L is a subgroup of G. Fix a, b G L. We wish to show that ab~1 G L. By G-invariance of !F, we see that a b - 1 L is a leaf of T. Moreover a = (a6 _1 )6 G ab~xL. So, since a G (ab^L) n L, we get ab^L - L. So ab'1 = ab~le G ab^L = L. •
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L E M M A 4.4.2 Let G be a Lie group. Let A be a left-invariant vector subbundle ofTG. Let H be a connected Lie subgroup of G and assume that Ae = f). Let J7 be a foliation on G and assume that TJ- = A. Then T is left-invariant and H = !F(e). Proof. Claim 1: T is left-invariant. Proof of Claim 1: Let g € G and define / : G -»• G by f(x) = gx. Let T\ := f*{T). We wish to prove that T\ = T. By left-invariance of A, we have /*(A) = A. Then we have TT\ = f*{TT) = /»(A) = A = TT, so, by Proposition 3.5.5, p. 53, we have T\ = T. End of proof of Claim 1. By Lemma 3.5.3, p. 53 (with <j) :Y —> M replaced by the inclusion map H -> G), choose a leaf L of T such that H is an open subset of L. Since e € H C L, we see that L = .F(e). Then, by Lemma 4.4.1, p. 81, L is a connected Lie subgroup of G. As H is an open subgroup of the connected Lie group L, we conclude that H = L. Then if = .F(e). • L E M M A 4.4.3 Let G be a Lie group. Let f) be a Lie subalgebra of g. Let A be the left-invariant vector subbundle of TG such that A e = h. Then A is involutive. Proof. For all X £ g, let XL denote the left-invariant vector field on G such that (XL)e = X. Let p := dim(H). Let X\,..., Xp be a basis of h. Then X{,..., Xp is a framing of A. By definition of the Lie algebra structure on g = TeG, for all integers i, j € [l,p], we have [X^Xf] = [XuXj]L, so [X^Xf] is in A. Then, by Lemma 3.4.3, p. 51, we see that A is involutive. • L E M M A 4.4.4 Let G be a Lie group and let h0 be a Lie subalgebra of g. Then there exists a unique connected Lie subgroup HofG such that \) = \)Q. Proof. Let A be the unique left-invariant vector subbundle of TG such that A e = h 0 . By Lemma 4.4.3, p. 82, A is involutive. So, by Theorem 3.7.5, p. 57, choose a foliation T of G such that TT = A. Let H :— T(e). By Lemma 4.4.1, p. 81, H is a connected Lie subgroup of G. Moreover, f) = TeH = TeT — A e = l)o, proving existence. For any connected Lie subgroup H' of G, if h' = h 0 , then, by Lemma 4.4.2, p. 81, we have H' — T{e) = H, proving uniqueness. • Let G be a Lie group. For any connected Lie subgroup H of G, H is generated by exp(h). Thus, if H and H' are connected Lie subgroups of G
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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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
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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.
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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. •
90
4.6
Basic Lie Theoretic
Results
The Lie algebra of [A, B]
The results in this section were pointed out to me by D. Witte, and the original draft of this section was written by him. It is quite possible that the results herein have been observed by others. If S is a subset of a Lie algebra g, then we will denote by (5) the Lie subalgebra of g generated by S. If S is a subset of a group G, then we will denote by (S) the subgroup of G generated by S. NOTATION 4.6.1 We make the following recursive definition: For all integers n > 3, for any Lie algebra g, for all W\, • •., Wn € g, we define [W1,...,Wn]:=[\Wi,...,Wn-1],Wn]. Suppose a and b are Lie subalgebras of a Lie algebra g. Let T(a, b) be the smallest ideal of (aUb) containing [a, b]. More concretely, by the Jacobi identity, we see that T(o, b) is the vector subspace of g spanned by {[Y,Z,X1,...,Xn]
\ Y e a, Z e b, Xi e aUb, n =
0,1,2,3,...}.
The correct characterization of the Lie algebra of [A, B] is: L E M M A 4.6.2 Let A and B be connected Lie subgroups of a Lie group G and let H := [A, B]. Then H is a connected Lie subgroup of G and t) —
r(o,b). Proof. By Corollary 4.5.6, p. 86, H is a connected Lie subgroup of G. We wish to show that T(a, b) = h. (C) By Thm. 2.1(iii), p. 18 of [Go80], we see that [A,B] is normal in {A U B), so f) is an ideal of (o U b). So, since [o, b] C h, we get T(o, b) C f). (D) We may assume that G = (AU B). Then, since o and b normalize T(a, b), it follows that A and B normalize T(a, b). Then G normalizes T(a, b). Thus T(o, b) is an ideal of g. Let N be the connected Lie subgroup of G corresponding to T(o, b). Then n = T(a, b). Let G :- G/N. Then g :— g/n. Let a : G —• G be the canonical homomorphism. We have [dcr(o), (da)(b)] = (do-)([a, b]) C (do-)(T(a, b)) = (d
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91
COROLLARY 4.6.3 Let A and B be connected Lie subgroups of a Lie group G and let H := [A, B]. Assume that A normalizes B and B normalizes A. Then H is a connected Lie subgroup of G and t) = [a, b]. Proof. By Lemma 4.6.2, p. 90, if is a connected Lie subgroup of G and f) = r ( o , b). Since a normalizes b and vice versa, we conclude that [a, b] is an ideal of (o U b), which gives T(o, b) = [a, b]. • COROLLARY 4.6.4 Let G be a connected Lie group. Then [G, G] is a connected Lie subgroup of G and the Lie algebra of[G,G] is [g,g]. Proof.
This is immediate from Corollary 4.6.3, p. 91.
•
E X A M P L E 4.6.5 Let G be a connected Lie group and let A and B be connected Lie subgroups of G. Then it does not necessarily follow that the Lie algebra of [A, B] is [a, b]. Proof. Let G := SL(2,E). Let A := {aE[f + (l/a)E^ \a > 0}. Let B := SO(2). Let H := [A, B]. As BAB = G, it follows that (A U B) = G, so, by Corollary 4.5.3, p. 85, we have (a U b) = g. So, as g is simple, by Lemma 4.6.2, p. 90, we have f) = g. Then [A, B] = H = G. However, since dim(o) = 1 = dim(b), it follows that dim([a, b]) = 1. Then [a, b] ^ g. •
4.7
Lie groups and Lie algebras from bilinear and quadratic forms
Let Q be a quadratic form on a vector space V. We define
0(g)
:= {FeGL(V)\FoQ
= Q},
SO(Q)
:=
{F £ 0(Q) | det(F) = 1},
SO°(Q)
:=
(SO(Q))°.
Each of these is a closed submanifold of GL(V), and is a Lie group under the resulting manifold structure. The Lie algebra of SO(<2) is, by our convention, denoted so(Q); it is equal to the Lie algebra of SO°(<5). Let J : V —> V denote the identity transformation, and define: S
:=
{ A / | A £ K } CGL(V),
S+
:=
{A/|A>0}CGL(F),
5*
:=
{A/|A^0}CGL(V).
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Basic Lie Theoretic Results
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 i s nondegenerate, then the index in the group O(Q) of the subgroup SO(<3) is = 2. If Q is nondegenerate and indefinite, then the index in SO(<2) of SO°(Q) is also = 2. If Q is definite, then SO(Q) = SO°(Q), i.e., then SO(<5) is connected. Let V be a vector space, let Q 6 QF(V) and let B be the polarization of Q. Then we define 0 ( B ) := 0 ( g ) , SO(B) := SO(Q),
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".
Abelian Lie groups
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
Abelian Lie groups
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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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Basic Lie Theoretic
Results
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
Miscellaneous
results
97
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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Basic Lie Theoretic
Results
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
Miscellaneous results
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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].
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Generalities on semisimple groups and algebras
Let 0 be a Lie algebra. For all 1 , 7 6 9, let fxY '• a d X o adY : g ->• 0, so that, for all Z £ g, we have fXy{Z) = [X,[Y,Z]]. The Killing form of 0 is the ad-invariant symmetric bilinear form n : g x g -> R defined by K(X, Y) = tr(/jcy). Then K, is (ad0)-invariant. Let 0 be a semisimple Lie algebra, let 0 be a maximal real split torus of g and let $ C o* be the set of roots of 0 on 0. Let K be the Killing form on 0. Then K|(O X O) is positive definite. Define $ : 0 —> o* by ($(X))(y) = K(X,Y). Then $ : a -» a* is a vector space isomorphism. Define F : a* x a* -> K by F{a,P) = K ^ - ^ a ) , * - 1 ^ ) ) . Then the map F : a* x a* -> K is a positive definite symmetric bilinear form and $ is a root system in (a*,F); see Example (ii) before Lemma X.3.2, p. 456 of [Hu72]. We will denote F by RKF( 0 ,o) € SBF(a*). (Here, "RKF" stands for "restricted Killing form".) Note that $ need not be a reduced
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root system (see, e.g., Exercise VII.5, p. 347 of [He78]). The semisimple Lie algebra 0 is simple iff $ is irreducible. L E M M A 4.10.1 Let g be a Lie algebra. Then g is semisimple iff its Killing form is nondegenerate. Proof. This is Theorem 3.9.2, p. 210 of [Va74], although what we call the Killing form is, in [Va74], called the "Cartan-Killing form". • L E M M A 4.10.2 If g is a semisimple ideal of a Lie algebra h, then there exists an ideal g' of f) such that g + g' = h, such that 0 H g' = {0} and such that [g,g'] = {0}. Proof. Let n denote the Killing form of h, so that K|(0 X 0) is the Killing form of 0, which, by Lemma 4.10.1, p. 113, is nondegenerate. Let 0' := {X 6 I) | K(X,S) = {0}}. For all X e h, (adX)0 C 0, so, by (ad h)-invariance of K, we get (adX)g' C 0'. That is, g' is an ideal of h. Since /c|(g x g) is nondegenerate, we get gflg' — {0} and 0 + 0' = h. Then, as 0 and g' are both ideals in h, we have [0, g'] C 0 D 0' = {0}. • C O R O L L A R Y 4.10.3 Let I) be a semisimple Lie algebra. Then there exists an integer k > 1 and there exist simple Lie algebras l\,..., Ifc such that \] is Lie algebra isomorphic 10 Ii ©•••©[*. Proof. The proof is by induction on dim(h). If h is simple, then the result follows (with fc = 1 and [1 = h), so let 0 be a proper nonzero ideal of h. Choose 0' as in Lemma 4.10.2, p. 113. Then h is Lie algebra isomorphic to 0 © g'. By induction, both 0 and 0' are isomorphic to direct sums of simple Lie algebras, so f) is, as well. • For any nonAbelian Lie algebra 0, 3(0) is a proper ideal of 0. So, for any simple Lie algebra s, we have 3(0) = {0}. Then, by Corollary 4.10.3, p. 113, we conclude, for any semisimple Lie algebra I, that 3(Q = {0}. Consequently, for any semisimple Lie group G, Z(G) is discrete in G. C O R O L L A R Y 4.10.4 Let I) be a semisimple Lie algebra. Then there exists an integer k > 1 and there exist simple ideals h,- • • ,h °fty such that the addition map (Xi,... ,Xk) •->• X\ + • • • + Xk : (1 © • • • © h -> h is an isomorphism of Lie algebras.
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Proof. By Corollary 4.10.3, p. 113, we may assume that f) is a direct sum of simple Lie algebras. Choose an integer k > 1 and simple Lie algebras S i , . . . ,Sk such that I) = Si0- • -©5*. For all integers j G [1,fc],let LJ : Sj -»• f) be the jth coordinate inclusion and let lj := tj(Sj). • COROLLARY 4.10.5 Let g be a semisimple Lie subalgebra of a Lie algebra h. Then nt,(g) = 0 + (Cf,(fl)). Proof. Replacing h by ni,(0), we may assume that g is an ideal of h, and we wish to show that I) = g + (CI,(JJ)). Choose g' as in Lemma 4.10.2, p. 113. Then, as [0,0'] = {0}, we get fl'Cc6(ji). Thenh = 0 + 0 ' C 0 + (c„(0))Ch, soh = 0 + (cft(0)). • L E M M A 4.10.6 Let g be a semisimple Lie algebra. Then [g,g] = g. Proof. By Corollary 4.10.3, p. 113, we may assume that 0 is simple. Then [0,0] is a nonzero ideal of 0, so, by simplicity, we get [0,0] — 0. • Let 0 be a Lie algebra and let V be a real 0-module. We say that V is completely reducible if, for every 0-invariant vector subspace W of V, there exists a ©-invariant vector subspace W of V such that W + W = V and such that W D W = {0}. The next result is sometimes called Weyl's Theorem. L E M M A 4.10.7 Let g be a semisimple Lie algebra. Then any real gmodule is completely reducible. Proof. See Theorem 3.13.1, p. 222 of [Va74] or Theorem 8 of §111.7 on p. 79 of [J62]. • L E M M A 4.10.8 Let g be a simple Lie algebra. Then there exist a vector space V and a faithful irreducible representation p : g —> gl(V). Proof.
Let V := g and let p :— ad : 0 -»
fll(V).
L E M M A 4.10.9 Let g be a semisimple Lie algebra. Then there exist a vector space V and a faithful representation p : g -> fll(V). Proof. By Corollary 4.10.3, p. 113, we may assume that 0 is simple. The result then follows from Lemma 4.10.8, p. 114. D
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In fact, Ado's Theorem asserts that Lemma 4.10.9, p. 114 is true for any Lie algebra not just semisimple Lie algebras. (See Theorem 3.17.7, p. 237 of [Va74].) Similarly, Lemma 4.10.10, p. 115 below can be generalized. (See Theorem 3.17.8, p. 237 of [Va74].) L E M M A 4.10.10 If G is a connected semisimple Lie group, then there exists an integer d > 1 and a connected Lie subgroup G' of GLd(R) such that G is locally isomorphic to G'. Proof. Choose V and p as in Lemma 4.10.9, p. 114. Let d := dim(V). Fixing a basis of V, we may assume that V = Rd. Let G' be the connected Lie subgroup of GLd(R) corresponding to the Lie subalgebra p(g) of gld(R). Then g is Lie algebra isomorphic to g'. D For any Lie algebra g, let Aut(g) denote the closed subgroup of GL(g) consisting of all Lie algebra automorphisms of g; let Aut°(g) := (Aut(g)) 0 . Then Aut°(g), being closed in Aut(g), is closed in GL(g). Let g be a semisimple Lie algebra and let / : g —• g be the identity transformation. Let 6 : g —• g be a map. We say that 6 is an involution of g if both 6 G Aut(g) and 0 o 0 = I. Note that, in our terminology, the identity I: g —r g is an involution. Let 6 be an involution of a Lie algebra g. Let K : g x g -> M. be the Killing form of g. Define B : g x g -> R by B(X,Y) = K{X,6(Y)). Then, for all X, Y € g, (ad B (X)) o (ad 9 (0(y))) = 0o (a,dB(6(X))) ° (ad B (F)) ofl" 1 , so, taking the trace of both sides, we get B(X, Y) = B(Y,X). Then we have B G SBF(g). We say that 6 is a Cartan involution of g if B is negative definite. By the paragraph preceding Theorem III.7.2, p. 183 of [He78], together with Proposition 111.7.4(h), p. 184 of [He78], any semisimple Lie algebra admits a Cartan involution. L E M M A 4.10.11 Let g be a semisimple Lie algebra and let 0 : g —> g be a Cartan involution. Let X G g. Then both of the following are true: (1) If 6{X) = —X, then a d X : g -> g is real diagonalizable. (2) If 6(X) — X, then sAX : g -> g is elliptic. Proof. Let K be the Killing form of g. Define a symmetric bilinear form B : g x g -> K by B(Y,Z) = K(Y,0(Z)). Then, by definition of "Cartan involution", B is negative definite. If 6{X) — —X, then, for all Y, Z G g,
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we have B((a,dX)Y,Z) = B(Y, (aAX)Z); that is, aAX : g -> g is symmetric with respect to B. If 6(X) = X, then, for all Y, Z E g, we have B((adX)Y, Z) = -B(Y, (adX)Z); that is, adX : g -»• g is anti-symmetric with respect to B. By the Spectral Theorem, a linear transformation which is symmetric (resp. anti-symmetric) with respect to a definite symmetric bilinear form is real diagonalizable (resp. elliptic). The results follow. • Let n > 2 be an integer and let go := sl n (R). Then the map 6Q : go —> 0o defined by 6Q{X) = ~X* is a Cartan involution. Note that, in this case, {X G go | 0o(X) = —X} is the subset of go consisting of symmetric matrices. For this reason we will make the following definition: If g is a semisimple Lie algebra and if S C g, then S is s y m m e t r i z a b l e if there is a Cartan involution 6 : g -> g such that, for all X € S, we have 9(X) = —X. Let g be a semisimple Lie algebra. A Lie subalgebra h of g is a complex split torus of g if ad B (h) is simultaneously complex diagonalizable; equivalently, both [h, h] = {0} and, for every X € h, ad X : g ->• g is complex diagonalizable. A Lie subalgebra h of g is a maximal complex split torus of g if it is maximal (with respect to inclusion) among complex split tori of g. If g is a Lie algebra, then g c denotes the complexification of g; it is a complex Lie algebra. Let g be a semisimple Lie algebra and let h be a Lie subalgebra of g. Following [He78] (see the definition in the second paragraph of §IX.4 on p. 418), we say that h is a Cartan subalgebra of g if both • h c is a maximal Abelian subalgebra of g c ; and • for all X E h c , a d X : g c - • g c is diagonalizable over C. We leave it as an exercise to the reader to verify that h is a Cartan subalgebra of g iff h is a maximal complex split torus of g. L E M M A 4.10.12 Let g be a semisimple Lie algebra and let a be a vector subspace of g. Then a is a real split torus of g iff [a, a] — {0} and a is symmetrizable. Proof. Proof of "only if": By the definition of "real split torus", we see that o is Abelian. It therefore suffices to show that o is symmetrizable. Since a is a real split torus, o is a complex split torus. Let h be a maximal complex split torus of g such that o C h. From Corollary IX.4.2,
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p. 419 of [He78], we know that f) is invariant under some Cartan involution 6 of fl. Let £ := {X G g 16{X) = X} and let p := {X G g 16{X) = -X}. We wish to show, that o C p. Let X G a. We wish to show that X G p. Since i) is ^-invariant, we have f) = (hnp) + (hnt). Choose D G hnp and £ G h n ! such that X = D + E. By Lemma 4.10.11, p. 115, ad£> : g -»• g is real diagonalizable and ad.E : g -> g is elliptic. As a d X : g -» g and ad£) : g —> g are both real diagonalizable, and as [X, D] G [h, h] = {0}, we see that ad(X — D) : g —> g is real diagonalizable. However, X — D = E and a.d.E : g ->• g is elliptic. We conclude that adg(E) = 0. So, since 3(g) = {0}, we get E = 0. Then X = I » G h n p C p . End of proof of "only if". Proof of "if": Choose a Cartan involution 6 : g —> g such that, for all X G a, we have 9{X) = -X. As a is Abelian, it follows that adB(o) is an Abelian Lie subalgebra of gl(o). Moreover, by (1) of Lemma 4.10.11, p. 115, we see, for all X G 0, that a d X : g —> g is real diagonalizable. A commuting collection of real diagonalizable linear transformations is simultaneously real diagonalizable, so o is a real split torus. End of proof of "if". • Let G := SL2(IR), let 1) be the set of diagonal matrices in g = s^CK) and let h' := so(2) C g. Then f) and h' are both maximal complex split tori in g; however, for all g G G, we have (Adg)f) ^ h'. On the other hand, note the complexifications of F) and h' are conjugate by an element of SLa(C). The possibility of nonconjugate maximal tori does not occur for maximal (complex) split tori in complex semisimple Lie algebras. Moreover, it does not occur for maximal real split tori in (real) semisimple Lie algebras: For any Lie algebra g, let Inn(g) denote the connected Lie subgroup of GL(g) group corresponding to the Lie subalgebra ad 8 (g) of gl(g). Then Inn(g) is generated by {exp(ad B (X)) | X G g}. For any connected Lie group G, we have Ad B (G) = Inn(g). In particular, we see that, if G and G' are connected Lie groups and if g = g', then Ad B (G) = Ad g /(G'). In other words, the Adjoint image in GL(g) of a connected Lie group G is completely determined by its Lie algebra g. Consequently, the main content of the next lemma is that if G is a connected semisimple Lie group then any two maximal real split tori of g are conjugate by an element of Ad fl (G). L E M M A 4.10.13 Let g be a semisimple Lie algbra and let 0 and b both be maximal real split tori of g. Then there exists f G Inn(g) such that f(a) = b.
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Proof. For every Cartan involution 6 of g, let 6# := {X G g 16(X) = X} and let pe := {X € 0 | 0(X) = -X}. By Lemma 4.10.12, p. 116 choose Cartan involutions (j) and xj) of g such that a is a maximal Abelian vector subspace of p^, such that b is a maximal Abelian vector subspace of p^. By Theorem III.7.2, p. 183 of [He78], we may assume that <j> = ip. The result then follows from Lemma V.6.3 (ii), p. 247 of [He78]. D L E M M A 4.10.14 Let I be a Lie algebra, let k > 1 be an integer and let hi • • • i h be simple Lie subalgebras of I. Assume that (Xu...,Xk)i->X1
+ --- +
Xk:l1®---®lk-*l
is an isomorphism of Lie algebras. Let t C I. Then 6 is an ideal of I iff there exists a subset S C { 1 , . . . , k} such that t = > J Ij. ies Proof. The "if" part follows from the fact that a sum of ideals is an ideal. It remains to prove "only if". Let ip : h © • • • © h —> I be the Lie algebra isomorphism defined by tp(Xi,..., Xk) = Xi H + Xk- For all integers i € [1, k], let pt : I -> k be -1 the composition of ip with the ith coordinate projection. Let S be the collection of integers i G [l,k] such that k C 6. Let S' be the set of integers i G [1, Jfe] such that Pi{t) ^ {0}. Then S C S' and ^ Ij C t C ^ T [;. We wish to show that S = S'. Let i £ S'. We wish to show that i £ S. We have Pi (6) = {0} and we wish to prove that [; C 6. Claim 1: [k,i\ C [j. Proof of Claim 1: For every integer j G [l,/c]\{i}, we have pj([li, I]) = \pj(U),Pj(l)] = [{0}, I,] = {0}. The result follows. End of proof of Claim 1. Choose X e t such that Pi(X) ^ 0. Since i(U) = {0}, choose Y G U such that \pi(X),Y] ^ 0. By Claim 1, we get \pi{X),Y] G k. Since Y G [», we have pt(Y) = Y. We have [X,Y] G [I, Ij] C Ii, so [X,Y] =Pi([X,Y]). Then [X,Y] = \pi{X),Pi{Y)] = b i W , > 1 e U\{0}. Moreover, [X, Y] G [6, [] C {. Then t n I* ^ {0}. So, as t D l* is an ideal of I; and as [j is simple, we get 6 n k — k- Then U = t H U C t, so i £ S. D COROLLARY 4.10.15 If I is a semisimple Lie algebra and if 6 is a nonzero ideal of I, then I, as a Lie algebra, is semisimple. Proof.
True by Corollary 4.10.4, p. 113 and Lemma 4.10.14, p. 118.
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COROLLARY 4.10.16 Let I be a semisimple Lie algebra, Let f) be a Lie algebra and let f : I —> f) be a nonzero Lie algebra homomorphism. Then /([) is a semisimple Lie subalgebra of\). Proof.
True by Corollary 4.10.4, p. 113 and Lemma 4.10.14, p. 118.
•
COROLLARY 4.10.17 Let I be a semisimple Lie algebra. Let t be an ideal of I. Let to be an ideal of 6. Then to is an ideal of I. Proof. By Corollary 4.10.4, p. 113, choose an integer k > 1 and simple ideals ii,.. .,lk in I such that (Xu...,Xk)
h->
Xx + --- + Xk
:
d © • • • © lk
->
[
is a Lie algebra isomorphism. By Lemma 4.10.14, p. 118, choose a subset S C {1,...,A;} such that 6 = ^ U- By Lemma 4.10.14, p. 118 again, tes choose S0 C S such that to = Y^ U- For all i € So, we know that k is an ideal of I. It follows that to is an ideal of I.
•
L E M M A 4.10.18 Letg andt) be Lie algebras. Let f : g —> () be a Lie algebra homomorphism. Lett be the solvable radical of g. Let I be a semisimple Levi factor of g. Let Y 6 I and Z € t. Assume that f(Y + Z) — 0. Then both f{Y) = 0 and f(Z) = 0. Proof. Since I is semisimple or zero, it follows, from Corollary 4.10.16, p. 119, that /(I) is semisimple or zero. So, since ( / ( 0 ) n ( / ( r ) ) is a solvable ideal of /([), we see that (/([)) n (/(t)) = {0}. We have f(Y) e /(I) and f(Y) = f(-Z) G /(c). Then f(Y) € (/([)) n (/(t)) = {0}. Consequently, f(Z) = (f(Y + Z)) - (f(Y)) = 0 . • L E M M A 4.10.19 Let g be a Lie algebra, let r be the solvable radical of g and let I be a semisimple Levi factor of g. Let t be an ideal of g. Then tf)t is the solvable radical of the Lie algebra t and t fl I is a semisimple Levi factor of the Lie algebra t. Proof. Let t' := t D r and let [' := t D [. Then t' is a solvable ideal of t. Since I' is an ideal of I, by Corollary 4.10.15, p. 118, V is semisimple or zero. So, by Lemma 4.9.1, p. 96, it suffices to show that r' + V = t. We have v' + l'Ct + t = t. Fix X e t. We wish to show that X <E r' + V. Choose
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Y £ t and Z e I such that X = Y + Z. We wish to show that Y ex' and Z &{'. Let f) := g/6 and let / : g -» h be the canonical homomorphism. We have X e 6, so / ( X ) = 0, so by Lemma 4.10.18, p. 119, we get f(Y) = 0 and f(Z) = 0. Then Y, Z £ I. Then Y € 6 f~l r = 6' and Z £ t D 1 = ['. D L E M M A 4.10.20 LeiG be a connected semisimple Lie group and assume that G is simply connected (resp. centerfree). Then there exist an integer k > 1 and connected, simply connected (resp. centerfree), simple Lie groups Si,..., Sk such that G is Lie group isomorphic to Si x • • • x Sk • Proof. Case 1: G is simply connected. Proof in Case 1: By Corollary 4.10.4, p. 113, choose an integer k > 1 and simple ideals Si,.. .Sk of g such that the map (Xi,..., Xk) *-> Xi-\ \-Xk : fii©- • -©Bfc -> Q is a Lie algebra isomorphism. Let Si,..., Sk be the connected Lie subgroups of G corresponding to S i , . . . , Sk, respectively. Then the map / : Si x • • • x Sk —> G defined by / ( s i , . . . ,Sk) = si • • • Sk is a surjective homomorphism with discrete kernel. It is therefore a covering homomorphism. So, since G is simply connected, / is a homeomorphism. Then, by Lemma 4.8.2, p. 94, / is an isomorphism of Lie groups. Since G is simply connected and since G is homeomorphic to Si x • • • x Sk, we conclude, for all integers i 6 [l,k], that Si is simply connected. End of proof in Case 1. Case 2: G is centerfree. Proof in Case 2: Let G be the universal covering group of G. Let n : G —> G be the universal covering map. By (2) of Corollary 4.9.18, p. 104, Z(ir(G)) -j(Z(G)), so, since 7r(G) = G and since Z(G) = { 1 G } , w e conclude that Z(G) C ker(7r). On the other hand, by (1) of Lemma 4.9.16, p. 103, we have ker(7r) C Z(G). Then ker(?r) = Z(G). Then, by Lemma 4.9.2, p. 96, G is Lie group isomorphic to G/(Z(G)). By Case 1, choose an integer k > 1 and connected simple Lie groups Si,..., Sk such that G is Lie group isomorphic to Si x • • • x Sk • For all integers i_€ [1, k], let Zi := Z(§i) and let S{ := Si/Zi. Since Z(G) = Zx x • • • x Zk, G/(Z(G)) is Lie group isomorphic to 5i x • • • x SkFor any integer i G [l,k], 5, is simple, so Zi is discrete; then Si is simple. Since G S Si x • • • x Sk, Z(G) = (Z(Si)) x • • • x (Z(Sk)). So, since G is centerfree, for any integer i £ [l,k], Si is centerfree. For any integer i € [l,k], Si is connected, so Si is connected. End of proof in Case 2. D
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L E M M A 4.10.21 Let G be a connected, simply connected semisimple Lie group and let N be a normal connected Lie subgroup ofG. Then N is closed in G and G/N is simply connected. Proof. By Lemma 4.10.20, p. 120, choose an integer k > 1 and connected, simply connected, simple Lie groups S i , . . . , Sfc such that G is isomorphic to Si x • • • x Sfc. We may assume that G = Si x • • • x Sfc. By Lemma 4.10.14, p. 118, after, if necessary, reordering, we may assume, for some integer j € [l,k], that N = Si x ••• x Sj. Then N is closed in G. Moreover G/N S Sj+i x • • • x Sfc, so G/N is simply connected. • L E M M A 4.10.22 Let G be a connected semisimple Lie group. Let N be a normal subgroup of G. Then (1) ifG is simple and Z(G) — {IG}, then N = {1G} or N = G; (2) N is closed in G; (3) there exists a normal connected Lie subgroup N' of G such that NC\N' is discrete in G and such that: G = NN1, [N, N'] = { I G } , n + n' = 0, n n n' = {0} and [n, n'] = {0}; (4) Z{N) C Z(G); and (5) if N CG, then G/N is semisimple. Proof. Proof of (1): Assume N ^ { I G } - We wish to show that N = G. Fix no 6 iV\{l<3}. Let S denote the subgroup of G generated by {gnog~l \g € G}. Then S C N. Moreover, S is path-connected, so, by Theorem 4.5.2, p. 84, we see that S is a connected Lie subgroup of G. Since S ^ { I G } , we see that s ^ {0}. Since S is a normal Lie subgroup of G, we see that s is an ideal of g. Then, since g is simple, we conclude that s = 0. Consequently, S = G. Then G = S C N C G, so N = G. End of proof of (1). Proof of (2): Let G be the universal covering group of G. Let IT : G —> G be the covering homomorphism. We wish to show that G\N is open in G. By Corollary 2.7.5, p. 28, •K is open, so it suffices to show that TT~1(G\N) is open in G. As T T - 1 ( G \ A ' ' ) = ^ ( ^ ( i V ) ) , it suffices to show that 7r_1(iV) is closed. Replacing G by G and N by TT"1(N), we may assume that G is simply connected. Let N* be the path-component of I G in N. By Theorem 4.5.2, p. 84, N* is a connected Lie subgroup of G. Then, by Lemma 4.10.21, p. 120, we see that N* is a closed subgroup of G and that G/N* is simply connected.
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Replacing G by G/N* and N by N/N*, we may assume that N is totally path-disconnected in G. Then, by Lemma 4.9.16, p. 103, N C Z(G). Then, as Z(G) is discrete in G, we see that N is discrete in G. So, as N is a discrete subgroup of G, N is closed in G. End of proof of (2). Proof of (3): By (2), we know that N is closed in G, and is therefore a Lie subgroup of G. By Lemma 4.10.2, p. 113, choose an ideal n' of g such that n + n ' = g, such that nfln' = {0} and such that [n, n'] = {0}. Let N' be the connected Lie subgroup of G corresponding to n'. Then G — NN' and NC\N' is discrete. Then, by Lemma 4.9.30, p. 109, we have [N,N'] = {1G}End of proof of (3). Proof of (4): Fix N' as in (3). Then N and N' centralize each other and G = NN'. Then Z(N) C Z{NN') = Z(G). End of proof of (4). Proof of (5): Since N C G, we get n C g . Then, by Corollary 4.10.16, p. 119, g/n is semisimple. Then G/N is semisimple. End of proof of (5).D L E M M A 4.10.23 Let G be a connected semisimple Lie group with finite center and let N be a normal closed subgroup of G. Assume that G ^ N. Then G/N is a semisimple Lie group with finite center. Proof. By (5) of Lemma 4.10.22, p. 121, G/N is semisimple. It remains to show that Z(G/N) is finite. Choose N' as in (3) of Lemma 4.10.22, p. 121. Let A := N (1N'. Then A is discrete in G. By (4) of Lemma 4.10.22, p. 121 (with N replaced by TV'), we see that Z{N') C Z(G), so Z(N') is finite. Let p : G -> G/N be the canonical homomorphism. Then p\N' : N' -+ G/N is surjective and ker(p|iV') = A, so G/N is isomorphic (as an abstract group) to N'/A. It therefore suffices to show that Z(N'/A) is finite. Let n : N' —> N'/A be the canonical homomorphism. By (2) of Lemma 4.9.16, p. 103 (with G replaced by N'), Z(N'/A) = n(Z(N')). So, since Z(N') is finite, it follows that Z(N'/A) is finite. • L E M M A 4.10.24 Let G be a connected semisimple Lie group and let N be a normal closed subgroup of G. Then N C (Z(G))N°. Proof. Fix x £ N. We wish to show that x 6 (Z(G))N°. Let TV' be as in (3) of Lemma 4.10.22, p. 121. Then NnN' is a discrete normal subgroup of G, so, by (1) of Lemma 4.9.16, p. 103, NnN' C Z{G). Since n + n' = g, it follows that d7r|n' : n' —> g/n is surjective. Then 7r|iV' : N' -> G/N° is surjective. Then G = N'N°. Choose a £ N' and
Generalities on semisimple groups and algebras
b e N° such that x = ab. We have a = xb~l € N{NQ)~l
aeNnN'CZ(G).Thenx
123
C N.
Then
= ab£{Z(G))N°.
D
COROLLARY 4.10.25 Lei G be a connected semisimple Lie group and let N be a normal closed subgroup of G. If Z{G) is finite, then N/N° is finite. Moreover, if G is centerfree, then N is connected. L E M M A 4.10.26 Let I be a semisimple Lie algebra and let 6 be an ideal of I. Then there exists an ideal t' of I such that t D t' = {0}, such that [I, t'] = {0} and such that 6 + t' = I. Proof. By Corollary 4.10.15, p. 118, we see that 6 is semisimple. Therefore, by Lemma 4.10.2, p. 113, we are done. • L E M M A 4.10.27 For any semisimple Lie algebra g, 0er(g) = ad B (g). Proof. Let I) := 0et(g) C g[(g) be the Lie subalgebra of derivations of g. Let flo := ad B (g) C h. We wish to show that h = g0Since g is semisimple, we see that 3(g) = {0}. Then ad : g ->• f) is injective, so go is Lie algebra isomorphic to g, and is therefore semisimple. So, by Corollary 4.10.5, p. 114, we have n(,(g0) = go + (c(,(flo))- By Lemma 4.9.13, p. 102, for all T 6 C(,(g0), we have T(g) C 3(g) = {0}, so T = 0. Then C(,(fl0) = {0}. By Lemma 4.9.13, p. 102, g0 is an ideal of h, i.e., \) = n(,(g 0 ). Then f) = ti(,(fl0) = flo + (cf,(flo)) = flo + {0} = flo• L E M M A 4.10.28 Let g be a semisimple Lie algebra. Then ad g (g) is real Jordan closed. Proof. By Lemma 4.10.27, p. 123, we have Der(g) = ad g (g). So, by Lemma 4.9.14, p. 102, we are done. • L E M M A 4.10.29 Let G be a connected semisimple Lie group. Then we have Ad B (G) = Aut (fl). In particular, Ad B (G) is closed in GL(g). Proof. By Lemma 4.10.27, p. 123, we have ad g (g) = 0et(g). So, since the Lie algebra of Ad 0 (G) is adfl(g) and since the Lie algebra of Aut°(g) is 0er(g), we conclude that Ad B (G) = Aut°(g). Since Aut°(g) is closed in GL(g), we are done. •
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Let g be a semisimple Lie algebra. We say that g is compact if K is negative definite. Otherwise we say that g is noncompact. Using Lemma 4.10.32, p. 124 below, one sees that a connected semisimple Lie group is compact iff its Lie algebra is compact. So, by Corollary 4.9.19, p. 104, one sees that a semisimple Lie algebra is compact iff it is isomorphic to the Lie algebra of a compact semisimple Lie group. We say that a connected semisimple Lie algebra g has no compact factors if every nonzero ideal of g is noncompact. We say that a semisimple Lie group has no compact factors if its Lie algebra has no compact factors. L E M M A 4.10.30 Let I be a semisimple Lie algebra. Then there exist semisimple ideals 6 and m of I such that all of the following are true: (1) t is compact; (2) m has no compact factors; (3) (X,Y) i-> X + Y : Z(Bm —¥ I is a Lie algebra isomorphism. Proof. Let 6 be maximal (with respect to set-theoretic inclusion) among compact ideals of I. By Lemma 4.10.2, p. 113, choose an ideal m of I such that trim = {0} and such that 6 + m = I. Then (X,Y) >->• X+ Y : 6 0 m -> [ is a Lie algebra isomorphism. By Corollary 4.10.15, p. 118, 6 and m are both semisimple. It remains to show that m has no compact factors. Assume, for a contradiction, that m' is a nonzero compact ideal of m. By Corollary 4.10.17, p. 119, m' is an ideal of I. Then 6+m' is a compact ideal of [ and 6 C 6 + m', contradicting maximality of 6. • L E M M A 4.10.31 Let t be a compact semisimple Lie algebra. Let X € 6. Assume that a d X : 6 —¥ 6 is nilpotent. Then X = 0. Proof. Let K : t x t -> E denote the Killing form on t. Then K is negative definite. Since a d X o a d X : 6 ->• t is nilpotent, it follows that K(X,X) = 0. Then, as K is negative definite, we get X = 0. • L E M M A 4.10.32 Let G be a connected semisimple Lie group. Then G is compact iff g is compact. Proof. By Lemma 4.10.1, p. 113, the Killing form K of g is nondegenerate. Then, by (2) of Corollary 2.10.2, p. 35, K is negative definite iff K is negative semidefinite. The result then follows from (1) of Proposition 3.39, p. 75 of [CE75]. •
Generalities
on semisimple
groups and algebras
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
for some integer n > 3, I — so(n, 1); or for some integer n > 2, I = su(n, 1); or for some integer n > 2, I = sp(n, 1); or t is isomorphic to the Lie algebra denoted f4(_2o) i n [He78] and denoted "F II" on p. 542 of [Kn96].
In each of these cases, we can examine the rootspaces and see directly that they have dimension > 2. • L E M M A 4.10.35 Let I be a simple Lie algebra. Assume that I ^ B12(1R). Let a be a maximal real split torus of I. Let X € a\{0}. Let $ be the set of roots a on I. For each a e $ , let la be the a-rootspace of I. Define $+:={a6$|a(X)
> 0}. Then $ + ^ 0 and dim I ^
la j > 2.
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Proof. Let F := RKF([,o) € SBF(o*). Then $ is a root system in (a*,F). Since X € o\{0}, we see that o ^ {0}, so dim(a) > 1. That is, the real rank of I is > 1. By Lemma 2.11.3, p. 38, let A be a base for $ such that, for all a 6 A, we have a(X) > 0. Since A spans a*, and since X ^ 0, choose ao € A such that a0(X) ^ 0. By choice of A, we have a0(X) > 0. We conclude that a0(X) > 0. That is, a 0 € $ + . Then 4>+ ^ 0. Let 1+ := Y^ la. We wish to show that dim([+) > 2. If dim(a) = 1, i.e., if the real rank of [ is = 1, then, by Lemma 4.10.34, p. 125, we have dim([ ao ) > 2, so, since lao C [+, we are done. We therefore assume that dim(a) > 2. Since [ is simple, it follows that $ is irreducible, so A is connected. Let A £ A be adjacent to ao in A. Let 70 := «o + /?o- Then 70 € $ . By choice of A, we have Po(X) > 0. So, since a0(X) > 0, it follows that 70(X) > 0. Then 70 € $ + . Then lao ^ {0} and (7o f {0}, so dim([ ao + t7o) > 2. So, since l ao + I7o C [ + , we are done. • L E M M A 4.10.36 Letg andi) be Lie algebras, let f : g —> f) be a surjective Lie algebra homomorphism, and assume that () is semisimple. Then there exists a Lie subalgebra go C g such that f\go '• 0o —> f) is a Lie algebra isomorphism. Proof. Let r denote the solvable radical of g. Then / ( t ) is a solvable ideal of h, so, by semisimplicity of h, we see that /(r) = {0}. Let [ be a semisimple Levi factor of JJ. Then /(I) = /(I + r) = /(g) = h. Let 6 denote the kernel of / | I : I ->• h. By Lemma 4.10.26, p. 123, let go be an ideal of I such that g0 fl t = {0} and such that g0 +t = I. Then /Iflo : go —> f) is a Lie algebra isomorphism. • L E M M A 4.10.37 Let G be a connected semisimple Lie group and fix a Cartan involution 6 : g -> g. Let 6 :— {Y € g 16{Y) = Y} and let K be the connected Lie subgroup ofG corresponding to i. Then Adg(K) is a compact subgroup o/GL(g). Proof. Since Ad g (G) is Lie group isomorphic to G/(Z(G)), and since Z(G) is a discrete subgroup of G, it follows, from Corollary 4.9.17, p. 104, that Ad 0 (G) has trivial center. The result then follows from (i) of Theorem VI.1.1 on p. 252 of [He78], with G replaced by Ad„(G) and K replaced byAdgtJO. •
Generalities
on semisimple
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127
L E M M A 4.10.38 Let G be a semisimple Lie group and let a be a maximal real split torus of g. Let $ C o* be the set of roots of a on Q. Let A be a base of $ . Let a+ := {T € a | \/S € A, S(T) > 0}. Let A+ := exp(a+) C G. Assume that Z(G) is finite. Then there is a compact subgroup KofG such that G = KA+K. Proof. By Lemma 4.10.12, p. 116, fix a Cartan involution 9 : g -» g such that, for all X € o, 0(X) = - X . Let t := { F e g 10(F) = F } . Let A" be the connected Lie subgroup of G corresponding to t. By Lemma 4.10.37, p. 126, we see that C := Adg(K) is a compact subgroup of GL(g). As Z(G) is finite, the map AdB|.ftf : K —> C has finite kernel. Then K is compact. By Theorem IX.1.1, p. 402 of [He78], G = KA+K. • L E M M A 4.10.39 Let V be a vector space and let g be a semisimple Lie subalgbra ofgl(V). Then g C si(V). Moreover, if J is a Q-invariant complex structure on V, then g C sl(V, J). Proof.
By Lemma 4.10.6, p. 114, we have g = [g,g]. Then gcl0i(V),fli(V)]c«[(y).
Now let J be an invariant complex structure on g. For all X 6 g, we have X 6 g C g[(V, J) and we have X,JX EgC sl{V), so X <E sl{V, J). D L E M M A 4.10.40 Let V be a vector space and let g be a semisimple Lie subalgbra of f) := sl(V). Assume that V has only scalar g-intertwining. Then ri(,(g) = g. Proof. By Corollary 4.10.5, p. 114, it suffices to show that C[,(g) = {0}. Let T £ C(,(g). We wish to show that T = 0. Let d := dim(V). Let I : V —• V be the identity transformation. Since V has only scalar g-intertwining, choose a € E such that T = a/. Then 0 = tr(T) = da, so a = 0. Then T = al = 0. • L E M M A 4.10.41 Le£ V be a vector space, let J be a complex structure on V and let g be a semisimple Lie subalgbra of f) := sl(V, J). Assume that V is g-irreducible. Then ti|,(g) = g. Proof. By Corollary 4.10.5, p. 114, it suffices to show that C(,(g) = {0}. Let T 6 C(,(g). We wish to show that T = 0.
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Let d := dim(V). Let I: V —> V be the identity transformation. Since V is 0-irreducible, by Lemma 4.9.24, p. 106, choose a, b £ M. such that T = al + bJ. Since J is a complex structure, we get t r ( J ) = 0. Then 0 = tr(T) = da and 0 = tr(JT) = -db, so a = b = 0, so T = 0. • 4.11
Real Jordan decomposition
L E M M A 4.11.1 Let V be a vector space. Let g be a Lie subalgebra of gl(V). Let X,D,E,N € g and assume that X = D + E + N is a real Jordan decomposition on V. Then &ds(X) = (adB(£>)) + (adB(.E)) + (adg(N)) is a real Jordan decomposition on g. Proof. For all A,B € {D,E,N}, since [A,B] = 0, it follows, from the Jacobi identity, that [adB(A),adB(.B)] = 0. Since D : V ->• V is real diagonalizable, it follows that ad D : 0l(V) —> gi(V) is real diagonalizable. So, as g is invariant under ad£> : fll(F) -> fll(V), we conclude that the restriction ad£> : g —> g is real diagonalizable. Similar reasoning shows that ad-E : g —> g is elliptic and that ad N : g —» g is nilpotent. • L E M M A 4.11.2 Let V be a vector space. Let g be a Lie subalgebra ofgliV). Assume that g is real Jordan closed. Assume that%{g) = {0}. Let X,D,E,N e g. Assume thatads(X) = (ad„(£>)) + (adB(.E)) + (ad„(iV)) is a real Jordan decomposition on g. Then X — D + E + N is a real Jordan decomposition on V. Proof. Since %(g) = {0}, it follows that ad : g -> gl(g) is injective. Since g is real Jordan closed, choose D', E', N' 6 fl such that X = D' + E' + N' is a real Jordan decomposition on V. By Lemma 4.11.1, p. 128, &dg{X) = (ad B (D')) + (adfl(£?')) + (ad 0 (iV')) is a real Jordan decomposition on g. Then, by uniqueness of real Jordan decomposition, we conclude that &dB(D) = adB(£>')> that &dg(E) = adg(E') and that adB(iV) = adB(JV'). So, by injectivity of ad : 0 —> gl(g), we get D = D', E = E' and N = N'. Then X = D + E + N is a real Jordan decomposition on V. • L E M M A 4.11.3 Let V be a vector space. Then sI(V) is real Jordan closed. Moreover, for any complex structure J on V, we have that sI(V, J ) is real Jordan closed.
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129
Proof. Since elliptic and nilpotent transformations have trace = 0, we see that sl(V) is real Jordan closed. Let J be a complex structure on V. Let X G sl(V, J) and D,E,N e gl{V). Assume that X = D + E + N is a real Jordan decomposition. We wish to show that D,E,N G s[(V, J). Let S C End(V) be the M-linear span of X,X2,X3,.... Then we have D,E,N G 5 . So, since XJ = JX, we get DJ = JD, EJ = JE and NJ = JN. That is, D,E,N G gl(V,J). We wish to show that
D,E, N, JD,JE,JN
esl{V).
Because X = D + E + N is a real Jordan decomposition and because D,E,N Gfll(V,J ) , it follows that JX - JE + JD + JN is a real Jordan decomposition. So, since sl(V) is real Jordan closed and since we have X, JX G gi(V), we are done. • L E M M A 4.11.4 Let V be a vector space and let g and f) be Lie subalgebras of gl(V). Assume that g C f) and that ti(,(g) = g. Assume that f) is real Jordan closed. Then g is real Jordan closed, as well. Proof. Choose D,E,N G gl(V) such that X = D + E + N is a real Jordan decomposition on V. We wish to show that D,E,N G 0. Since f) is real Jordan closed, we get D,E,N £ h. For all P 6 h, let P := a d P : f) -* h. By Lemma 4.11.1, p. 128 (with g replaced by h), we see that X = D + E + N is a real Jordan decomposition on h. As X G g, we see that X(g) C g. Let 5 C End(g) be the IR-linear span of X, X2, X3,.... Then D,E,N G S. So, since X(g) C g, we conclude that D(g) C fl, that E(g) C g and that iV(fl) C g. Then D,E,N e it(,(fl) = fl.D L E M M A 4.11.5 Lei V be a vector space and let g be a Lie subalgebra of gl(V). Assume that g is simple and that V is g-irreducible. Then g is real Jordan closed. Proof. Case 1: V has only scalar g-intertwining. Proof in Case 1: Let I) := sl(V). By Lemma 4.11.3, p. 128, we know that I) is real Jordan closed. By Lemma 4.10.40, p. 127, ti(,(fl) = g. Then, by Lemma 4.11.4, p. 129, we see that g is real Jordan closed. End of proof in Case 1. Case 2: V has nonscalar g-intertwining. Proof in Case 2: By using Lemma 4.9.26, p. 107, choose a g-invariant complex structure J on g. Let f) := sl(V,J). By Lemma 4.11.3, p. 128, we know that 1} is real Jordan closed. By Lemma 4.10.41, p. 127, ni,(g) = g. Then, by Lemma 4.11.4, p. 129, we see that g is real Jordan closed. End of proof in Case 2. D
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L E M M A 4.11.6 Letg be a semisimple Lie algebra, letV be a vector space and let p : g —• gi(V) be a representation. Let X, D,E,N 6 g. Assume that adg(X) = (ade(.D)) + (adB(.E)) + (adB(7V)) is a real Jordan decomposition on g. Then p(X) — (p(D)) + (p(E)) + (p(N)) is a real Jordan decomposition on V.
Proof. By Corollary 4.10.3, p. 113, we may assume that g is simple. By Lemma 4.10.7, p. 114, we may assume that p is irreducible. We may assume that p(g) ^ {0}. Then ker(p) ^ g, so, as g is simple, we have ker(p) = {0}, so p is injective. Then, replacing 0 by p(g), we may assume that g is a Lie subalgebra of gi(V) and that p : g —>• gi(V) is the inclusion map. By Lemma 4.11.5, p. 129, we see that g is real Jordan closed. Since g is semisimple, we have • 3 ( 0 ) = {0}. The result then follows from Lemma 4.11.2, p. 128. C O R O L L A R Y 4.11.7 Letg be a semisimple Lie algebra, letV be a vector space and let p : g -»• gl(V) be a representation. Let X € 0. Assume that a d X : 0 —»• 0 is real diagonalizable (resp. elliptic, resp. nilpotent). Then p(X) : V —> V is real diagonalizable (resp. elliptic, resp. nilpotent).
Proof. Assume that &dX : g -»• 0 is real diagonalizable. It follows that adg(X) = (ad B (X)) + 0 + 0 is a real Jordan decomposition on g. Then, by Lemma 4.11.6, p. 130, we see that p(X) = (p(X)) + 0 + 0 is a real Jordan decomposition on V. Consequently, p(X) : V —> V is real diaongalizable. A similar argument works for "elliptic" and "nilpotent". D
L E M M A 4.11.8 Let V be a vector space and let g be a semisimple Lie subalgebra of gi(V). Then g is real Jordan closed.
Proof. Let X £ 0 be given. We wish to show, for some D,E,N € 0, that X = D + E + N is a real Jordan decomposition on V. By Lemma 4.10.28, p. 123, fix D,E,N eg such that ad fl (X) = (adg(£»)) + (adfl(-E)) + ( a d g ( ^ ) ) is a real Jordan decomposition on 0. Let p : g -> 0l(V) be the inclusion map. Then, by Lemma 4.11.6, p. 130, we see that X = D + E + N is a real Jordan decomposition o n V . •
Consequences of results on real Jordan decomposition
4.12
131
Consequences of results on real Jordan d e c o m p o s i t i o n
L E M M A 4.12.1 Let g be a Lie algebra and let X,T e Q. Let X G E\{0} and assume that (adT)X — XX. Then a d X : g -> g is nilpotent. Proof. Let S := a d T : g ->• g and let W := a d X : g - • g. We wish to show that W : g -+ g is nilpotent. Let f) := gl(g). Then 5,T G J). Let J : h -> f) be the identity transformation. For all P G h, define P ' := a d P : f) -> h. Then, for all P,Q € f), we have P'(Q) = [P, Q] = PQ - QP. By the Jacobi identity, we have [ad 0 (T),ad g (X)] = ad B ([T, X]). So, since [T,X] = (adT)X = AX, this gives [S,W] = XW. Then we have S'(W) = [S,W] = XW. Let D : g ->• g be the real diagonalizable part of 5 : g -¥ g. By Lemma 4.11.1, p. 128, D' : \) -> f) is the real diagonalizable part of S" : f) —»• I). Then D' — A/ is the real diagonalizable part of S" — A J, so D'-XI is an element of the M-linear span of S'-XI, (S'-XI)2, (S'-XI)3,.... So, since (S" - XI)(W) = 0, we conclude that (£>' - XI)(W) = 0. Then [D,W] = D'(W)=XW. For all n G M, let g,, := { F G fll-D(^) = fiY}. For all /i G E, since [D,W] = XW, it follows that W(fl„) C g A+/1 . Then, for all / i f R , for all integers i > 0, we have W'g^ C 0JA+/ILet S :— {/j, € M|gM ^ {0}} be the set of eigenvalues of D : g ->• g. Since D : g -> g is real diagonalizable, we see that g = Y j g^. Then, for i*es l all integers i > 0, we have W (g) C Y^ g»A+^- Since 5 is finite and A ^ 0, ij.es
choose an integer i0 > 0 such that, for all g is nilpotent. D In fact, Lemma 4.12.1, p. 131 can be improved: Letting Jo : g —• g be the identity transformation, if there exist an integer m > 0 and A G K\{0} such that [(adT) - A7o]nX = 0, then a d X : g -»• g is nilpotent. The proof is essentially the same. Lemma 4.12.1, p. 131 has quick a dynamical proof: Proof. Let Si be a sequence in M such that SjA —• +00. For all i, define gi := exp(siT); then (Adg^X = eSiXX, so (Adgi)(e~SiXX) = X. Then X is Kowalsky for Adg(gi). Therefore, by Lemma 4.9.31, p. 109, we conclude that a d X : g -> g is nilpotent. •
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L E M M A 4.12.2 Let g be a semisimple Lie algebra. Then the following are equivalent: (1) (2) (3) (4)
g is noncompact. For some D G fl\{0}, a d D : g —>• g is real diagonalizable. For some N e fl\{0}, ad AT: g -»• g is nilpotent. The real rank of g is > 1.
Proof. Let K : g x g ->• R denote the Killing form of g. Proof of (1) =>• (2): Since g is noncompact, we conclude that K is not negative definite. By Lemma 4.10.1, p. 113, K is nondegenerate. Then K is not negative semidefinite. Choose X G g such that K(X, X) > 0. By Lemma 4.10.28, p. 123, choose D,E,N e g such that ad fl (X) = (adB(£>)) + (ad B (£)) + (ad0(7V)) is a real Jordan decomposition on g. We wish to show that D ^ 0. Assume, for a contradiction, that D = 0. As (ad N) o (ad N) : g —> g and (ad N) o (ad E) : g —> g are both nilpotent linear transformations, we conclude that K(N, N) = K(E, N) = 0. Then we have K(X, X) = K(E,E). AS &&E : g -> g is elliptic, it follows that all characteristic roots of ad E : g -t g are pure imaginary, which implies that all characteristic roots of the map (ad E) o (ad E) : g —> g are nonpositive real numbers. Then K(E, E) < 0. Then 0 < K{X, X) = K(E, E) < 0, contradiction. End of proof of (1) => (2). Proof of (2) =$• (1): Let 5 be the set of characteristic roots of the linear transformation a d D : g —¥ g. Then S C E, so, for all s € S, we have s 2 > 0. So, as K(D, D) is a linear combination of elements of {s2 \ s € S}, with positive integer coefficients, we conclude that K(D,D) > 0. Then K is not negative definite, i.e., g is not compact. End of proof of (2) =$• (1). Proof of (2) =>• (3): Since 3(0) = {0}, we see that ad : g ->• g[(g) is injective. Then, as D ^ 0, it follows that ad£) : g -¥ g is nonzero. Then, since adD : g -> g is real diagonalizable, choose N G 0\{O} and A £ M\{0} such that (ad D)N = AN. Then, by Lemma 4.12.1, p. 131, we see that adiV : g -¥ g is nilpotent. End of proof of (2) =$• (3). Proof of (3) ==• (2): By Lemma 4.10.33, p. 125, choose Y,D G g such that N,Y,D is a standard s ^ W basis of some Lie subalgebra go of g. Since [D, N] = N ^ 0, we see that fl^O. Moroever, adD : g0 -> g0 is real diagonalizable. So, by Corollary 4.11.7, p. 130 (with p : g —>fll(V)replaced
Consequences of results on real Jordan
decomposition
133
by ad :flo-> 0t(fl))i w e s e e t n a t a d D : 0 -> 0 is real diagonalizable. End of proof of (3) = > (2). Proof of (2) 4=$- (4): This follows from the definition of "real rank". End of proof of (2) 4=^ (4). • L E M M A 4.12.3 Let g be a noncompact semisimple Lie algebra. some Lie subalgebra of g is isomorphic tofife(K).
Then
Proof. By (1) =i> (3) of Lemma 4.12.2, p. 132, choose N e g such that adJV : 0 —> g is nonzero and nilpotent. Then, by Lemma 4.10.33, p. 125, we are done. • L E M M A 4.12.4 Let g be a semisimple Lie algebra with no compact factors. Let flo denote the set of all X £ g such that a d X : 0 ->• 0 is nilpotent. Then 0o generates g, i.e., no proper Lie subalgebra of g contains goProof.
Let 0i be the Lie subalgebra of 0 generated by 0o- Assume, for a
contradiction, that 0i 7^ 0. Since 0O is (ad B (0))-invariant, it follows that 01 is an ideal of 0. By Lemma 4.10.2, p. 113, choose an ideal g[ of 0 such that 01 D g[ = {0} and such that 0! + g[ = g. Then g[ 7^ {0}. Since 0 has no compact factors, we see that g[ is noncompact. Then, by (1) =>. (3) of Lemma 4.12.2, p. 132, choose JV € 0i\{O} such that the map ad JV : 0i ->• 0i is nilpotent. By Corollary 4.11.7, p. 130 (with p : g -+ 0l(V) replaced by ad : 0^ —> g[(g)), we see that adJV : 0 —> g is nilpotent. Then JV e 0o- Then 0 ^ N £ g0 n g[ C 0i n g[ = {0}, contradiction. • Let d > 1 be an integer. Let x\,... ,Xd : Rd —> R be the coordinate projections. Let / : E d ->• E be a function. We say that / is a monomial of degree 0 if / is constant. For any integer k > 1, we say that / is a monomial of degree k if, for some integers ii,...,ik € [l)<^j we have f = Xit • • -Xik. For any integer k > 0, we say that / is a homogeneous polynomial of degree k if / is a finite E-linear combination of monomials of degree k. We say that / is a monomial if there exists an integer k > 0 such that / is a monomial of degree k. We say that / is a polynomial if / is a finite E-linear combination of monomials. Let d > 0 be an integer, let 7 b e a d-dimensional vector space and let / : V —• E be a function. We say / is a polynomial if, for some vector space isomorphism / i : E d - ^ F , / o / i : E d - ^ E i s a polynomial.
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Let V and W be vector spaces and let
4.13
Generalities on algebraic groups
Let V be a vector space, let S C V and let V denote the set of all polynomials V -4 R. Let I := {/ € V \ f(S) = {0}}. The Zariski closure in V
of S is f|/ _ 1 (°)/ex Let V be a vector space. A subset of V which is equal to its own Zariski closure in V is said to be Zariski closed in V. Let C be the collection of subsets of V which are Zariski closed in V. If S CV and if S is the Zariski closure in V of S, then S £ C. For all C 6 C, C is closed in the ordinary vector space topology on V. The collection C is closed under finite union and arbitrary intersection. Let V be a vector space. A subset S of GL(V) is Zariski closed in.GL(V) if it is the intersection of GL(V) with a Zariski closed subset of End(V). A subgroup of GL(F) is algebraic if it is Zariski closed in GL(F). Any algebraic subgroup of GL(V) is closed in GL(V), and is therefore a Lie subgroup of GL(V). A subgroup of SL(F) is algebraic iff it is Zariski
Generalities
on algebraic groups
135
closed in End(V). As before, the collection of subsets of GL(V) which are Zariski closed in GL(V) is closed under both finite union and arbitrary intersection. For any S C GL(V), The Zariski closure in GL(T^) of S is the intersection of all subsets of GL(F) which both are Zariski closed in GL(V) and which contain S. Let V be vector space and let G be a subgroup of GL(V). In this book, we say that G is almost algebraic if there is an algebraic subgroup H of GL(y) such that H° C G C H. Any almost algebraic subgroup of GL(F) is closed in GL(V). A finite union or arbitrary intersection of algebraic (resp. almost algebraic) subgroups of GL(V) is again algebraic (resp. almost algebraic). If 0 is a Lie algebra, then Aut(jj) is an algebraic subgroup of GL(g), so Aut°(g) is an almost algebraic subgroup of GL(g). L E M M A 4.13.1 Let V be a vector space and let G be an algebraic (resp. almost algebraic) subgroup of GL(V). Then Z(G) is an algebraic (resp. almost algebraic) subgroup of GL(V), as well. Proof. For all g e GL(V), let Cg := {x € GL(V)\gx = xg}; then Cg is algebraic. Then C :— f ] Cg is algebraic, so Z{G) — C D G is algebraic g€G
(resp. almost algebraic).
•
L E M M A 4.13.2 Let V be a vector space and let G be an almost algebraic subgroup o/GL(V). Then G° has finite index in G. In particular, if G is discrete, then G is finite. Proof. As G is almost algebraic, let H be an algebraic subgroup of GL(V) such that H° C G C H. Then, by Lemma 3.2.3, p. 48, G° = H°. By [Wh.57], H° has finite index in H. Then G° has finite index in G. • The following is Lemma 4.4, p. 463 of [AOOa]. L E M M A 4.13.3 Let L be a connected, semisimple Lie group, let V be a vector space and let p : L —• GL(V) be representation. Then p(L) C SL(V) and p(L) is almost algebraic, so p(L) is a closed subgroup of SL(V). Proof. Let J : V —¥ V be the identity transformation. By Lemma 4.10.6, p. 114 and Corollary 4.6.3, p. 91, we have L = [L,L]. Then p(L) = p([L,L]) C [GL(V),GL(V)] C SL(V). Let L0 := p(L). It remains to show that L0 is almost algebraic.
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Case 1: V has only scalar (/o(L))-intertwining. Proof in Case 1: Let G:=SL(V). Then G is an algebraic subgroup of GL(F). By Lemma 4.10.40, p. 127, we have lo = TIB(IO), SO L0 = NQ(10)- SO, as NQ(10) is algebraic, it follows that LQ is almost algebraic. End of proof in Case 1. Case 2: V is (/9(jL))-irreducible and has nonscalar (p(Z/))-intertwining. Proof in Case 2: By Lemma 4.9.26, p. 107, choose a complex structure J : V -»• V which is (p(L))-invariant. Let G := SL(V,J). Then G is an algebraic subgroup of GL(V). By Lemma 4.10.41, p. 127, we have l0 = nB([o), so LQ = N^(lo). So, as No(k>) is algebraic, it follows that L0 is almost algebraic. End of proof in Case 2. We now consider the general case. By Lemma 4.10.7, p. 114, choose an integer k > 1 and choose /9-invariant vector subspaces V\,..., Vjt of V such that (vi,..., Vk) i-> v\ 4- • - • + Vk : V\ (B • • • © 14 -> V is an isomorphism of vector spaces. Let K := { 1 , . . . ,k}. For all i £ K, let G; := GL(Vi) and define p{ : L -»• G» by pi(Z) = (p(0)|Vi. For all i € K, let L* := ^ ( L ) . By Case 1 and Case 2, we see, for all i & K, that Li is an almost algebraic subgroup of SL(Vj). Let t : Gi x • • • x Gfc —> GL(V) be the standard injection; so, for all (9i, • • • ,9k) e Gi x • • • x Gk, for all i £ K, we have (i(g1}..
.,9k))\Vi
= gt.
Let P := Li x • • • x Lk- For all i 6 K, let 7Tj : P -> X, be the ith coordinate projection. Let (J := i(P). Then L CQ. Since, for all i £ K, we know that Li is a connected and almost algebraic subgroup of GL(Vi), it follows that Q is a connected and almost algebraic subgroup of GL(T^). Then NQ(1) is an almost algebraic subgroup of GL(F). Let L0 := t _ 1 ( £ ) . For all i G K, the composition of t _ 1 : L -) LQ with 7Tj|Z/o '• LQ -* Li is equal to pi : L —> Li, and is therefore surjective. We have cp(Io) C Ci^ti) © • • • © cik(lk). For all i £ if, we have that [j is simple, so ^ ( I ; ) = 3(1;) = {0}. It follows that CP(IQ) = {0}. By Corollary 4.10.5, p. 114, we have [0 + cp(Io) = n p (I 0 ). Then [0 = n p (Io). Then L0 = N%(lo). Then L = t(L0) = L(N°P(L0)) = JVg([). So, as JVg(I) is an almost algebraic subgroup of GL(V), we are done. • Let A denote the class of all commutative E-algebras with multiplicative unit. For all A £ A, let 1^ denote the multiplicative unit of A. Let U be a vector space. For any A £ A, let UA denote the A-module A ®R U, and let det^ : End^(!7 A ) -)• A be the determinant map over A. Let SLy be the affine algebraic group such that, for all A £ A, the yl-points of S L a are given by SLV(A) := {T £ EndA(UA) | det^(T) = 1A}.
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'