Faithfully representable analytic groups

by

Andy R. Magid

University of Oklahoma

In this paper, analytic group means a connected complex Lie group.

If G is an analytic group and p : G ~ GLn¢ , where p(g) = [Pij(g)] ,

is an analytic representation, the functions Pij : G ~ ~ are called the

coordinate functions of p . The set R(G) of all coordinate functions

of all analytic representations of G forms a complex algebra, under

pointwise operations, called the algebra of representative functions on

G . G acts on R(G) in the following way : if f ~ R(G) and x ~ G ,

x • f : G ~ C given by (x • f)(y) = f(yx) and f • x : G ~ ~ given

by (f • x)(y) = f(xy) are in R(G) . For x in G , let L x : G ~ G

be given by Lx(Y) = xy and R x : G ~ G be given by Rx(Y) = yx . Then

x • f = fR x and f • x = fL x

The algebra of representative functions on a group with a faithful

representation has the following description, due to Hochschild and Mostow

[3, Section 3] :

Theorem. Let G be an analytic group with a faithful representation.

Then R(G) = A[Q~ where Q = exp(Hom(G,~)) and

i) A is a finitely generated subalgebra of R(G)

2) Ax = A for all x in G

3)- G ~ alg~(A,~)(= Max(A)) by x ~ (evaluation at x) is bijective.

385

Conversely, if A satisfies i), 2), 3), then R(G) = A[Q] .

We note that if A satisfies i), 2), 3) of the theorem, then (G,A)

is an affine algebraic variety over ~ such that L : G ~ G is a mor- x

phism for all x in G ; we say that (G,A) is a left algebraic group.

The above theorem shows that finding left algebraic group structures on G

is equivalent to determing R(G) . If (G,A) is a left algebraic group strut

ture, the core of the structure, C(G) , is the set of all x in G

such that R is a morphism: C(G) = {x I xA = A} . It turns out that x

C(G) is an algebraic group [4, Cor. 1.5, p. 1047] and that this alge-

braic group determines A [5, Thm. 2.3, p. 174].

Groups with a faithful representation have an intrinsic characteri-

zation [3, p. 113]; we will show here how, from this characterization, a

representation can be constructed which yields a left algebraic group

structure.

For our purposes, "algebraic group" will mean "affine complex alge-

braic group." Thus algebraic groups always possess faithful representa-

tions. Embedding an analytic group in an algebraic group, therefore, pro-

duces a faithful representation. We show that if G is an analytic group

with a faithful representation, then G is a normal analytic subgroup of

an algebraic group G' such that G' is a semi-direct product of G and

an algebraic torus T ; then ~[G'] T is the coordinate ring of a left alge-

braic group structure on G .

We also use the following notations and conventions: a torus always

means a multiplicative algebraic torus (product of GLIC) . If G is a

group and x in G , l(x) is the inner antomorphism given by x . A

reductive group is an analytic group with a faithful representation such

that every representation is completely reducible. A reductive group

is algebraic, and in fact the image of a reductive group under any

386

representation is Zariski-closed.

Definition. An analytic group G is an FR group if G admits a faith-

ful finite dimensional analytic representation; i.e. if there is an injec-

tire analytic homomorphism G ~ GL ~ for some n . n

FR groups have the following intrinsic characterization, due to Hochschild

and Mostow:

Theorem. The analytic group G is FR if and only if G is a semi-direct

product of a closed , solvable , simply connected normal subgroup K and

a reductive subgroup P [3, p. 113].

Proofs of the "only if" assertion can be found in [2, Thm. 4.2, p. 86]

(from a representation-theoretic point of view) and in [6, Thm. i0, p. 880]

(from a group-theoretic point of view). This paper presents a new proof of

the "if" assertion, which explains, among other things, why an FR group

carries a left algebraic group structure.

It should be further mentioned that Hochschild and Mostow have a

slightly stronger characterization of FR groups: a simply connected sol-

vable normal subgroup L of an analytic group H is called a nucleus if

H/L is reductive, and they show that there exists a reductive subgroup

Q of H such that H is the semi-direct product of L and Q [2,

Thm. 3.6, p. 95]. Their proof uses the fact that a reductive group is

the complexification of a compact real Lie group. This paper also con-

tains a new proof of this, which avoids the use of compact real forms.

We now fix the following notation: the analytic group G is a semi-

direct product of the simply connected closed solvable normal subgroup K

387

and the reductive subgroup P .

Lemma: Lie(K) is a sum (not necessarily direct) of Lie subalgebras N

and C , where N is a nilpotent ideal of Lie(G) , C is a Cartan sub-

algebra of Lie(K) , and [Lie(P),C] = 0 [3, Lemma 2.1, p. 113].

Let C = exPK(C) and N = exPK(N) . Then N and C are closed,

simply connected analytic subgroups of K , with N normal in G and

hence K , and the elements of C and P commute. In addition, we can

(canonically) regard C and N as unipotant algebraic groups. Let s :

C +Aut(N) be given by s(c) = l(c) I N . Let K I be the (analytic)

semi-direct product Nx C ; when convenient, we regard N and C as s

subgroups of K I . There is an analytic homomorphism f : K I ~ K given

by f(n,c) = nc . Let L be its kernel and let L be the connected c

component of the identity in L . Then KI/L c ~ K has the discrete group

L/L c as kernel; since K is simply connected L/L c is trivial so L

is connected. Actually, L = {(x,x -I) I x 6 N Q C} , so L is also nil-

potent.

Next, we look at the image s(C) of C under s . Aut(N) = Aut(Lie(N))

is an algebraic group and hence the Zariski closure s(C) of s(C) in Aut(N)

is an algebraic group, which is nilpotent since C is nilpotent. Thus ~ =

UxT where U is the unipotent radical of s(C) and T is a torus. Let p :

s(C) ~ U and q : s(C) + T be the projections. As noted above, we can regard

as a unipotent algebraic group. Let D be the algebraic group CxT .

Now r = ps : C + U is an analytic homomorphism of unipotent algebraic

groups, and hence algebraic. Let t : D ~Ant(N) be given by t(c,x) =

(r(c)x) . Since t is just rxid T followed by the inclusion of UxT

into Aut(N) , t is algebraic. We can thus form the semi-direct product

K 2 = NxtD and K 2 is an algebraic group.

388

We will now embed K I in K 2 . There is an analytic map g : C ~ D

which sends c to (c,qs(c)) . Let h : K I + K 2 be given by h(n,c) =

(n,g(c)) . Then h is an injective analytic homomorphism by construction.

Since D = g(C)T , K 2 = h(KI)T . Moreover, it is clear that T N h(K I) =

{e} • We claim that h(K I) is Zariski-dense in K 2 . It suffices to show

that g(C) is Zariski-dense in D . Let g(C) be the Zariski-closure of

g(C) in D . Since C is nilpotent, so is g(C) , so g(C) = VxS

where V is the unipotent radical of g(C) and S is the unique maxi-

mal torus of g(C) . The projection from g(C) to C is surjective and

hence so is the projection from g(C) to C . Thus V projects onto C .

The projection from g(C) to T has qs(C) as image, and hence the

image of g(C) in T is Zariski-dense. Thus g(C) projects onto T

and hence S projects onto T . It follows that g(C) = D .

Now h(K I) is Zariski-dense in K 2 , so (K2,K 2) = (h(Kl) , h(Kl))

h(Kl) and h(K I) is normal in K 2 . It follows that the algebraic group

K 2 is an analytic semi-direct product of h(K I) and T , and h : K I ~ K 2

is an analytic embedding of K I in an algebraic group with Zariski-dense

image.

We now need to examine h(L) (recall that L is the kernel of

K I ~ K). First, we look at s(N N C) . If we identify Aut(N) with

Aut(Lie(N)) , then s(~) = Ad(x) for x in C . Since N is nilpotent,

if x ( C N N , Ad(x) is unipotent, and it follows that s(N N C) ~ U ,

so qs(x) = e if x ( C n N . Thus g(C N N) : (C N N)x{e} . Now L =

{ (x,x-l) I x E C n N} so if y = (x,x -I) ( L , h(y) = (x,x-l,e) . Hence

h(L) is contained in the unipotent radical of K 2 and it follows that

h(L) is Zariski-closed in K 2 (recall that L is simply connected).

Since L is normal in K 1 and h(Kl) is Zariski-dense in K 2 , we also

389

have that h(L) is normal in K 2 . We can thus form the algebraic group

quotient K' = K2/h(L) , and we have an analytic embedding 1 : K ~ K'

induced from h . Moreover, £(K) is Zariski-dense in the algebraic group

K' and K' is the semi-direct product of l(K) and a torus T' which is

the image of T in K'

Finally, we need to embed G in an algebraic group. This will be

done by extending the action of P on K to an action on K' and then

forming semi-direct products. We begin by looking at the action of P

on N in G : we have a homomorphism v : P ~ Aut(N) given by v(p) =

l(p) I N . Since P is reductive, v(P) is an algebraic subgroup of

Aut(N) [6, Prop. 5, p. 878]. As noted above, P and C commute in G ,

so v(P) and s(C) commute in Aut(N) , and hence v(P) and the Zariski-

closure UxT of s(C) commute in Aut(N) . In other words, the actions

of P and D on N commute. As an algebraic variety, K 2 = NxD . For

p in P , K 2 ~ K 2 by (n,d) ~ (pnp-l,d) is a morphism of varieties, and

since P and D commute on N , this is an algebraic group automorphism

of K 2 . Thus w~ can form the semi-direct product algebraic group K2xP =

G 2 . Since P commutes with N n C , L remains normal in G 2 , and we

have the algebraic group quotient G' = G2/L ; and G' is a semi-direct

product of K' and P . Let m : G ~ G' be l on K and the identity

on P . Then m is an analytic embedding of G in an algebraic group.

Moreover, since l(K) is Zariski-dense in K' , m(G) is Zariski-dense

in G' Also, G' = K'P = /(K)TP = m(G)T , and the construction shows

G' to be a semi-direct product of m(G) and T , and that T and P

commute in G' .

We have thus established the following theorem:

390

Theorem i. Let G be an analytic group which is the semi-direct product

of a solvable, normal, closed, simply connected subgroup K and a reduc-

tive subgroup P . Then G is a Zariski-dense analytic subgroup of an

algebraic group G' such that G' is a semi-direct product of G and

a torus T , and T and P commute.

We recall how embeddings as in Theorem i yield left algebraic group

structures (see [6, Prop. 6, p. 878] for details): we have a hijection

G ~ G'/T , and G'/T is an affine algebraic variety. Since G'/T is

right cosets of T , for any y in G' , the map xT ~ yxT is a mor-

phism of G'/T , and hence, regarding G as an algebraic variety via

the above bijection, each L for x in G is a morphism. Moreover, x

if x E P , R× is a morphism of G since P and T commute. Thus P

is contained in the core of the left algebraic group structure.

Theorem I has a converse:

Theorem: Let G be a Zariski-dense analytic subgroup of the algebraic

group G' . Then there is a torus T in G' such that G' = GT (not

necessarily semi-direct product) and Lie(G') = Lie(G) @ Lie(T) (semi-

direct product) [6, Theorem 3, p. 878].

Further information about these matters is contained in [6] and [7].

We now turn to the problem of showing that an analytic group with a nucleus

is a semi-direct product of the nucleus and a reductive subgroup. We fix

the following notations: G is an analytic group and K is a closed,

solvable, simply-connected normal subgroup of G such that G/K is re-

ductive.

We begin with the case that G is solvable, so G/K is a torus. We

391

want to find a torus T in G such that T ~ G/K is an isomorphism. We

use induction on the dimension of K . Let K 0 be the last non-vanishing

turn in the derived series of K . K 0 is simply connected, closed, sol-

vable, and normal in G , and ~ = K/K 0 is a nucleus of ~ = G/K 0 . By

induction, there is a torus ~ in ~ such that T ~ G/K = G/K is an

isomorphism. Let G O he the inverse image of T in G . Then G0/K 0 =

so K 0 is a nucleus of G O . We note that K 0 is abelian. If T is

a torus in G O such that T ~ G0/K 0 is an isomorphism, then T ~ G/K is

an isomorphism, so we may replace G by G O and K by K 0 and assume

K is abelian, and hence a vector group. The action of G on K by con-

junction factors through G/K , and since G/K is a torus this means there

is a one-dimensional subgroup K 0 of K normal in G . By induction,

using arguments similar to the above, we may assume K 0 = K , i.e. that K

is one-dimensional. Then Aut(K) = ~* , and we have a homomorphism s :

G ~ Aut(K) where s(g) = I(g) I K . Let G 1 be the kernel of s and G c

be the connected component of the identity in

have a surjection G/K + G/G c Since

product of a torus and a compact group

tainted in the kernel of G/G 2 ~ G/G 1 .

group GI/G c , so A is trivial. Thus

G I . Then K ~ G c and we

G/K is a torus, G/G c is a direct

A . Since G/G I ~ C* , A is con-

But this kernel is the discrete

G/G is either trivial or is a c

one-dimensional torus. Thus the surjection G/K ~ G/G is split and c

G/K = G/GcX~ where ~ is a subtorus of G/K . Thus Gc/K is a torus

( i s o m o r p h i c t o ~ ) . S i n c e K i s c e n t r a l i n G , G i s a c t u a l l y a b e l i a n , c c

so G e = VxT I where V is a vector group and T is a torus. Let

be t h e c o n n e c t e d c o m p o n e n t o f e x p - l ( K ) i n ' L i e ( G ) . ~ i s i s o m o r p h i c t o c

K and Lie(G )/K is the universal covering of G /K . It follows that c c

G /K a n d G h a v e t h e same f u n d a m e n t a l g r o u p , a n d h e n c e t h a t c c

392

dim(Tl) = dim(Gc/K) : dim(G e) - i , so V is one-dimensional. The projec-

tion K ~ V is either then trivial or an isomornhism. If it's trivial,

Go/K = Vx(TI/K) , which is imnossible, since Gc/K is a torus. Thus

K + V is an isomorphism. If we identify T I and (~,)(n) then K =

{(v,~xP(~l,V),...,exp(~nV))} I v 6 V} for appropriate ~l,...,en in ~ .

Thus G c = KxT I also and T I ~ T is an isomorphism. T is the kernel

of the additive characters of G and hence characteristic in G , and it c

follows that T I is normal, hence central, in G . Finally, let H be

the inverse image of G/G regarded as a subgroup of G/K in G . Then c

K g H and H/K is trivial or a one-dimensional torus. If H = K , G = G c

and we are done. If not, H is an extension of ~* by K . If H =

~*xK , let T 2 = C* ~ H ; then T 2 ~ H/K is an isomorphism. Otherwise

Lie(H) is the solvable non-abelian two-dimension el algebra, so the univer-

sal cover ~ of H is ~x ~ (semi-direct product) . The center of exD

is {O}x2~i~ and this contains the fundamental group of H . H is

not simply connected (since H/K isn't) so ~I(H) = {0}x2~im~ for some

m ~ 0 . Thus H = ~/~I(H) contains the torus T 2 = 0x~/~l(H) . T 2 N K

is the kernel of T 2 ~ H/K . This latter is a map of one-dimensional tori

so its kernel is either all of T 2 or finite. Thus T 2 N K = T 2 or

T 2 N K is finite. The former implies that T 2 ! K , which is impossible,

while the latter implies that T 2 N K is trivial since K has no finite

subgroup. Thus T 2 ~ H/K is an isomorphism. Now T I n T 2 ~ K N T I = {e} ,

and T I is normal in G . It follows that T = TIT 2 is a torus in G

and T ~ G/K is an isomorphism.

So far, we have been assuming G solvable. We now drop that assump-

tion. Let R be the radical of G . Then K ~ R and R/K = ~ is the

radical of G/K : ~ . Since G is reductive, T is a torus. Thus K

393

is a nucleus of the solvable group R , and by the above argument for the

solvable case there is a torus T in R such that T ~ ~ is an isomor-

phism. Let L = Lie(G) and ~ = Lie(T) . T acts on L via the adjoint

representation, and the projection L ~ ~ is a T -morphism. Since T

is a terus, the induced map L T ~ ~T is surjective, and since T is

central in ~ , ~T = ~ . Since ~ is reductive, there is a projection

~ [~,~] and [L,L] is semi-simple. Combining, we have a surjection

L T ~ [~,~] , and hence there is a subalgebra S of L T such that S

[L,L] is an isomorphism. Let S be the analytic subgroup of G with

Lie(S) : S . Then S is semi-simple and since S ! L T , S and T com-

mute. Under the projection G ~ G , S maps onto (G,G) inducing an iso-

morphism on Lie algebras. The kernel of S ~ (G,G) is thus central and

hence finite. Since the kernel is S N K and K has no finite subgroups,

the map is an isomorphism. Let P = TS . P is a subgroup of G since

S normalizes, in fact centralizes, T . Since G/R = G/T = (G,G) , S f]R

is finite. Let x = ts be in K N P where t ( T and s ( S . Since

T ~ R and K ~ R , s ( S N R , so s n = e for some n . Then x n = t n

is in K N T = {e} , so x n = e . Since K has no elements of finite order,

this means x = e . Thus K N P = {e} . Since P ~ ~ is by construction

surjective, P ~ ~ is an isomorphism.

We have thus shown:

Theorem 2. Let G be an analytic group and K a nucleus of G . Then

there is a reductive subgroup P of G such that G is the semi-direct

product of K and P .

The above proof relies on the fact that the center of a semi-simple

analytic group is finite. This can be established using complex groups

394

via the classification theorem of semi-simple complex Lie algebras and the

fact that all of these are algebraic, or by using compact real forms [i,

Thm. 2.1, p. 198]. Since the point of the above proof is to avoid compact

real forms, it would be nice to have an elementary direct proof of the

fact.

395

References

i.

2.

3.

4.

5.

6.

7.

G. Hochschild, The Structure of Lie Groups, Holden Day, San Francisco, 1965.

G. Hochschild and G. D. Mostow, "Representations and representative functions of Lie groups, III," Ann. Math. 70(1959), 85-100.

, "On the algebra of representative functions of an analytic group," Amer. J. Math. 83(1961), 111-136.

A. Magid, "Analytic left algebraic groups," Amer. J. Math. 99(1977), 1045-1059.

, "Analytic left algebraic groups, II," Trans. Amer. Math. Soc. 238(1978), 165-177.

, "Analytic subgroups of affine algebraic groups," Duke J. 44(1977), 875-882.

, "Analytic subgroups of affine algebraic groups," Pacific J. (to appear).

Andy R. Magid Mathematics Department University of Oklahoma Norman, OK 73019 USA