Representations of GLn and Un
Note: all representations of topological groups are assumed to be continuous and finite-dimensional.
Here, we will look at representations of the general linear group We fix the following notations:
denotes
for some fixed
;
is the subgroup
of unitary matrices; this group is compact.
Now, a representation
is said to be continuous / analytic / rational / polynomial if it is so in each entry of For example:
is rational for each integer n but polynomial only for non-negative values of m.
is continuous but not analytic.
Note that the above definition makes sense for G since it is an open subset of , the space of all
matrices.
Definition. For a coordinate-free variant, we say that the representation
is continuous / analytic / rational / polynomial if:
- for any
and
, the function
![]()
is so.
Our focus is in analytic representations of G. For K, we will only look at continuous representations since it does not have a natural analytic structure. It turns out there is a correspondence between these two classes of representations, so the former can be analyzed using our earlier tools.
Main Theorem
We begin with a basic result in complex analysis.
Theorem. Let
be an analytic function on
If
on
, then
Proof
When n=0, has only one point so this is trivial. Suppose n>0 and
Treat f as a function in
and we get:
where and h are analytic. Suppose
on
Then
is analytic and on
it is zero; since this set is dense in
we have
on
Thus
on
. By induction hypothesis, this gives g=0, a contradiction. ♦
From this, we obtain the following result, which is the main case of concern.
Main Lemma. If
is an analytic function and
, then
.
Proof
Let be defined by
, which is analytic. We have:
so h=0 when restricted to . By the above theorem, h=0. Note that any diagonalizable matrix can be written as
for some matrix A, so f(B) = 0 if B is diagonalizable. Since the set of diagonalizable matrices in
is dense, f = 0. ♦
Consequences
From the main lemma we get, in turn:
Corollary 1. Let
be finite-dimensional analytic representations of G. Then:
Proof
A linear map is G-equivariant if and only if for each
,
. This is equivalent to:
Since are all analytic, by the main lemma, this condition remains equivalent when we replace G by K, in which case the condition says that
♦
Corollary 2. Let
be finite-dimensional analytic G-modules. They are isomorphic as G-modules ⇔ they are isomorphic as K-modules.
Proof
Indeed, saying (as G– or K-modules) just means there is an invertible
or
. Apply corollary 1. ♦
Corollary 3. Let V be a finite-dimensional analytic G-module. A subspace
is G-invariant ⇔ it is K-invariant. Hence V is irreducible as a G-module ⇔ it is irreducible as a K-module.
Proof
W is G-invariant if and only if for all ,
; this is equivalent to:
- if
and
satisfies
, then the function
vanishes.
By corollary 1, this condition is equivalent when we replace G by K, but now the condition says for all
♦
Summary
Let V be a finite-dimensional analytic representation of G. Since K is compact, V as a K-module can be written as a direct sum of irreps:
where the are pairwise non-isomorphic K-irreps. By corollary 3, each
is a G-irrep; the
are also pairwise non-isomorphic G-modules by corollary 2.
Question
Given a K-module, can we necessarily lift it to an analytic G-module? We will explore this in the next article.