Polynomials and Representations XXX

Representations of GL_n and U_n

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 $GL_n \mathbb{C}.$ We fix the following notations:

$G$ denotes $GL_n\mathbb{C}$ for some fixed $n$ ;
$K$ is the subgroup $U_n \subset GL_n(\mathbb{C})$ of unitary matrices; this group is compact.

Now, a representation

$\rho: G = GL_n\mathbb{C}\to GL_N\mathbb{C}$

is said to be continuous / analytic / rational / polynomial if it is so in each entry of $GL_N\mathbb{C}.$ For example:

$G \to \mathbb{C}^*, \ g \mapsto \det(g)^{m}$ is rational for each integer n but polynomial only for non-negative values of m.
$G \to \mathbb{C}^*,\ g \mapsto |\det(g)|$ is continuous but not analytic.

Note that the above definition makes sense for G since it is an open subset of $\mathbb{C}^{n^2}$ , the space of all $n\times n$ matrices.

Definition. For a coordinate-free variant, we say that the representation $\rho : G = GL_n\mathbb{C} \to GL(V)$ is continuous / analytic / rational / polynomial if:

for any $v\in V$ and $\alpha \in V^\vee$ , the function $G\to \mathbb{C},$ $g\mapsto \alpha(\rho(g)(v))$ 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.

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Main Theorem

We begin with a basic result in complex analysis.

Theorem. Let $f(z_1, \ldots, z_n)$ be an analytic function on $\mathbb{C}^n.$ If $f=0$ on $\mathbb{R}^n$ , then $f=0.$

Proof

When n=0, $\mathbb{R}^n = \mathbb{C}^n$ has only one point so this is trivial. Suppose n>0 and $f\ne 0.$ Treat f as a function in $z_n$ and we get:

$f(z_1, \ldots, z_n) = z_n^d g(z_1, \ldots, z_{n-1}) + z_n^{d+1} h(z_1, \ldots, z_n)$

where $g\ne 0$ and h are analytic. Suppose $f=0$ on $\mathbb{R}^n.$ Then $f z_n^{-d} = g + z_n h$ is analytic and on $\mathbb{R}^n - (\mathbb{R}^{n-1} \times \{0\})$ it is zero; since this set is dense in $\mathbb{R}^n,$ we have $g + z_n h=0$ on $\mathbb{R}^n.$ Thus $g=0$ on $\mathbb{R}^{n-1} \times \{0\}$ . 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 $f : G\to\mathbb{C}$ is an analytic function and $f|_K = 0$ , then $f=0$ .

Proof

Let $h : M_{n\times n}(\mathbb{C}) \cong \mathbb{C}^{n^2} \to \mathbb{C}$ be defined by $h(A) := f(\exp(2\pi i A))$ , which is analytic. We have:

$A \in M_{n\times n}(\mathbb{R})\implies \exp(2\pi i A) \in U_n= K$

so h=0 when restricted to $M_{n\times n}(\mathbb{R})$ . By the above theorem, h=0. Note that any diagonalizable matrix can be written as $\exp(A)$ for some matrix A, so f(B) = 0 if B is diagonalizable. Since the set of diagonalizable matrices in $GL_n\mathbb{C}$ is dense, f = 0. ♦

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Consequences

From the main lemma we get, in turn:

Corollary 1. Let $V, W$ be finite-dimensional analytic representations of G. Then:

$\text{Hom}_K(V, W) = \text{Hom}_G(V, W).$

Proof

A linear map $f:V\to W$ is G-equivariant if and only if for each $g\in G$ , $\rho_W(g)\circ f = f\circ \rho_V(g)$ . This is equivalent to:

$\lambda \in \text{Hom}_\mathbb{C}(V, W)^\vee, g\in G \implies\lambda(\rho_W(g) \circ f - f\circ \rho_V(g))= 0.$

Since $f, \rho_V(g), \rho_W(g), \lambda$ are all analytic, by the main lemma, this condition remains equivalent when we replace G by K, in which case the condition says that $f\in \text{Hom}_K(V, W).$ ♦

Corollary 2. Let $V, W$ be finite-dimensional analytic G-modules. They are isomorphic as G-modules ⇔ they are isomorphic as K-modules.

Proof

Indeed, saying $V\cong W$ (as G– or K-modules) just means there is an invertible $f\in \text{Hom}_G(V, W)$ or $\text{Hom}_K(V, W)$ . Apply corollary 1. ♦

Corollary 3. Let V be a finite-dimensional analytic G-module. A subspace $W\subseteq V$ 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 $g\in G$ , $\rho_V(g)(W) \subseteq W$ ; this is equivalent to:

if $w\in W$ and $\lambda \in V^\vee$ satisfies $\lambda(W) =0$ , then the function $G\to \mathbb{C}, g \mapsto \lambda(\rho_V(g)(w))$ vanishes.

By corollary 1, this condition is equivalent when we replace G by K, but now the condition says $\rho_V(g)(W)\subseteq W$ for all $g\in K.$ ♦

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:

$V = W_1^{\oplus m_1} \oplus W_2^{\oplus m_2} \oplus \ldots \oplus W_r^{\oplus m_r},$

where the $W_i$ are pairwise non-isomorphic K-irreps. By corollary 3, each $W_i$ is a G-irrep; the $W_i$ 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.

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