Polynomials and Representations XXVIII

Starting from this article, we will look at representations of $G := GL_n\mathbb{C}$ . Now, $GL_n\mathbb{C}$ itself is extremely complicated so we will only focus on representations of particular types. Generally, for any topological group G, we want:

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

to be a continuous homomorphism of groups.

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Continuous Representations of Topological Groups

Let G be a topological group.

Our main objects of study are continuous representations $\rho: G \to GL_N\mathbb{C}.$ These correspond to $\mathbb{C}[G]$ -modules (also called G-modules) V of finite complex dimension, with an additional condition for continuity:

for any $v\in V$ and $\alpha \in V^\vee$ (dual), the map $G\to \mathbb{C}$ given by $g\mapsto \alpha(g(v))$ is continuous.

[Note the above condition condition is independent of coordinates and does not require a choice of basis. When we replace G with $GL_n\mathbb{C}$ later, we can also replace the word continuous with “analytic” to obtain a narrower class of representations, but let’s not get ahead of ourselves.]

A function $G\to\mathbb{C}$ of the form $g\mapsto \alpha(g(v))$ as above is called a matrix coefficient. E.g. if we fix a basis for V which gives us $G \to GL_N\mathbb{C},$ picking entry (i, j) in the matrix gives us a matrix coefficient $G\to\mathbb{C}.$

Henceforth, all representations of topological groups are assumed to be continuous and finite-dimensional.

Constructing New Representations

These constructions are identical to the case for representations of finite groups.

Given any complex vector space V, the trivial action takes gv = v for any $g\in G, v\in V$ ; it is clearly continuous.

Given a G-module V, a vector space is said to be G–invariant if Clearly, the resulting is continuous.
- We say that V is irreducible if $V\ne 0$ and it has no non-trivial G-invariant submodules. An irreducible representation V is often called irrep for short.

If $W\subseteq V$ is G-invariant, then V/W is a G-module; clearly it is continuous.

If V is a G-module, the dual $V^\vee$ is also continuous, where G acts via

$G\times V^\vee \to V^\vee, \qquad (g, \alpha) \mapsto \alpha\circ g^{-1} \in V^\vee.$

Given G-modules V, W, the action of G on $V\oplus W$ and $V \otimes_{\mathbb C} W$ are continuous as well.

Finally we have $\text{Hom}_{\mathbb C}(V, W) \cong V^\vee \otimes_{\mathbb C} W$ naturally. Using the above constructions, we get a continuous G-representation for this space via:

$g\in G, f:V\to W \mapsto \rho_W(g)\circ f \circ \rho_V(g^{-1}) : V\to W.$

How do we know that the above constructions are all continuous?

Here’s one example. To show that $V\otimes_{\mathbb{C}} W$ is a continuous representation, pick a basis $e_i$ (resp. $f_j$ ) of V (resp. W) so that $e_i \otimes f_j$ form a basis of $V\otimes W.$ The matrix for $g: V\otimes W \to V\otimes W$ is obtained from the matrices of $g:V\to V$ and $g:W\to W$ by multiplying entries; thus the map $G\to \mathbb{C}$ taking g to a fixed entry of the matrix for $g:V\otimes W\to V\otimes W$ is continuous.

Finally, a linear map of G-modules $f:V\to W$ is said to be G–equivariant if

$g\in G, v\in V \implies f(gv) = g\cdot f(v).$

Note that this is equivalent to saying G acts trivially on $f\in \text{Hom}_{\mathbb C}(V, W).$

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Compact Groups

From now till the end of the next article, G is assumed to be compact and Hausdorff; this case will be instrumental to the general theory even though $GL_n\mathbb{C}$ is not compact. Here are some common examples of compact topological groups in representation theory:

$\begin{aligned} U_n &= \{ M \in GL_n \mathbb{C} : M \overline{M}^t = I\},\\ SU_n &= \{ M \in GL_n\mathbb{C} : M\overline{M}^t = I, \det(M) = 1\},\\ O_n &=\{ M\in GL_n\mathbb{R} : M M^t = I\}, \\ SO_n &= \{M \in GL_n\mathbb{R} : MM^t = I, \det(M) = 1\}, \\ T &= \{ z \in \mathbb{C} : |z| = 1\}.\end{aligned}$

These groups are compact because they are closed and bounded subsets of $\mathbb{C}^{n\times n}.$ E.g. the condition for $M \in U_n$ can be written out as a sequence of quadratic equations in $\text{Re}(z_{ij})$ and $\text{Im}(z_{ij})$ for $1\le i,j\le n.$ Thus the set $U_n$ is closed; it is bounded since the condition gives:

$|z_{i1}|^2 + |z_{i2}|^2 + \ldots + |z_{in}|^2 = 1, \quad\text{for } 1\le i\le n$

so each $|z_{ij}| \le 1.$

Haar Measure

To proceed, we will borrow a result from harmonic analysis:

Theorem. If G is a compact Hausdorff group, then there is a measure m on G (called the Haar measure) satisfying:

m(G) = 1;

m is “left-invariant”, i.e. for any continuous $f:G\to \mathbb{C}$ , we have:

$\displaystyle y\in G\implies \int_{x\in G} f(yx)\cdot dm(x) = \int_{x\in G} f(x)\cdot dm(x),$

The Haar measure is unique and in fact right-invariant, i.e. $\int_{x\in G} f(xy)\cdot dm(x) = \int_{x\in G} f(x) \cdot dm(x).$ We will implicitly assume the Haar measure is used when integrating over G.

Note

The Haar measure is usually defined for locally compact Hausdorff groups in general, in which case we have to modify the statements a little. In particular, left-invariant measures are not necessarily right-invariant.

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Semisimplicity of Continuous Representations

The following result is crucial.

Theorem. Let V be a G-module. If $W\subseteq V$ is a G-invariant subspace, then we can find a G-invariant subspace $W' \subseteq V$ such that $V = W\oplus W'.$

Proof

Let $\pi : V\to V$ be any projection map onto W, so $\pi^2 = \pi$ and $\text{im}(\pi) = W$ . Now define:

$\pi' : V\to V, \quad \pi'(v) := \int_{x\in G} x\pi(x^{-1}v)\cdot dx.$

[Explicitly, one may take an isomorphism $V\cong \mathbb{C}^N$ so each $v\in V$ is a tuple of complex numbers; the above integral can be carried out component-wise, as N integrals of continuous functions.]

We then have:

The image of $\pi'$ is in W, since $\pi(V)=W$ and $x(W)\subseteq W$ for $x\in G.$

If $v\in W$ then $x^{-1}v \in W\implies x\pi(x^{-1}v) = xx^{-1}v = v$ and $\pi'(v) = v.$

$\pi'^2 = \pi'$ : indeed if $v\in V$ then $\pi'(v) \in W$ by the first property and so $\pi'(\pi'(v)) = \pi'(v)$ by the second.

The map $\pi'$ is G-equivariant: for $y\in G$ we have $\pi'(yv) = y\pi'(v).$ Indeed by left-invariance we have: $\pi'(v) = \int_{x\in G} yx \pi(x^{-1}y^{-1}v) dx.$ Replacing v with yv gives us the desired outcome.

Hence, $\pi'$ is projection onto W and its kernel W’ satisfies $V = W \oplus W'.$ Since $\pi'$ is G-equivariant, W’ is a G-invariant subspace. ♦

Thus, every continuous representation of G can be written as a direct sum of irreps. Next, we have:

Schur’s Lemma. If $V, W$ are irreps of G, then:

$\text{Hom}_{\mathbb{C}[G]}(V, W) \cong \begin{cases} \mathbb{C}, \qquad &\text{if }V \cong W,\\ 0, \qquad &\text{otherwise.} \end{cases}$

The proof is identical to that of the usual Schur’s lemma.

Proof

Suppose $V\not\cong W$ and $f:V\to W$ is G-equivariant. Then $\text{ker}(f) \subseteq V$ and $\text{im}(f)\subseteq W$ are subrepresentations, so $\text{ker}(f) = 0, V$ and $\text{im}(f) = 0, W$ . By considering various cases we have either

$\text{ker}(f) = 0, \text{im}(f) = W$ , in which case f is an isomorphism or
$\text{ker}(f) = V, \text{im}(f) = 0$ , in which case f = 0.

It remains to show that $\text{Hom}_{\mathbb{C}[G]}(V,W) \cong \mathbb{C}$ when $V\cong W$ .

It suffices to show that any isomorphism $f:V\to V$ is a scalar: let $v\in V$ be an eigenvector of f with eigenvalue $\lambda\in\mathbb{C}$ so $f(v) = \lambda v.$ Then $f - \lambda\cdot I : V\to V$ is G-equivariant and its kernel is non-zero; hence $f - \lambda\cdot I = 0 \implies f = \lambda\cdot I.$ ♦

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