Polynomials and Representations XXX

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 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 continuousanalyticrationalpolynomial 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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