Characters

Definition. The character of a continuous G-module V is defined as:

$\chi_V : G\to\mathbb{C}, \quad g \mapsto \text{tr}(g: V\to V).$

This is a continuous map since it is an example of a matrix coefficient.

Clearly $\chi_V(hgh^{-1}) = \chi_V(g)$ for any $g,h\in G$. The following are quite easy to show:

\displaystyle \begin{aligned}\chi_{V\oplus W}(g) &= \chi_V(g) + \chi_W(g); \\ \chi_{V\otimes_{\mathbb C} W}(g) &= \chi_V(g) \chi_W(g); \\ W\subseteq V\implies \chi_V(g) &= \chi_W(g) + \chi_{V/W}(g); \\ \chi_{V^\vee}(g) &= \chi_V(g^{-1}) = \overline{\chi_V(g)}.\end{aligned}

The last equality, that $\chi_V(g^{-1})$ is the complex conjugate of $\chi_V(g)$, follows from the following:

Lemma 1. For $g\in G$, the linear map $g:V\to V$ has all eigenvalues on the unit circle T.

Proof

Suppose the eigenvalues are $\lambda_1, \ldots, \lambda_N$ with the right multiplicity. Then $g^m$ has trace $\chi_V(g^m)= \sum_i \lambda_i^m.$ But $\chi_V : G\to \mathbb{C}$ is continuous; since G is compact the image is bounded. It remains to show: if $\sum_i \lambda_i^m$ is bounded for all integers m we must have all $|\lambda_i| = 1$.

Suppose $|\lambda_1| \ne 1$; replacing g by its inverse, we assume $|\lambda_1| > 1$ and $|\lambda_1| \ge |\lambda_2| \ge \ldots.$ If $|\lambda_1| > |\lambda_2|$, $|\lambda_1|^m$ increases much faster than the other $|\lambda_i|^m$ and we get a contradiction. On the other hand, if $|\lambda_1| = \ldots = |\lambda_k|$, write $\lambda_j = Re^{i\theta_j}$ for $1\le j \le k$ where R>1 and $\theta_j$ are real. Consider the continuous homomorphism

$\phi:\mathbb{Z} \mapsto (\mathbb{C}^*)^k, \ m \mapsto (e^{im\theta_1}, \ldots, e^{im\theta_k}).$

Its image lies in the torus group $T_k$ which is compact. Hence, there are arbitrarily large m such that $\text{Re}(e^{im\theta_1}), \ldots, \text{Re}(e^{im\theta_k}) > \frac 1 2$ (see exercise). So the real part of $\lambda_1^m + \ldots + \lambda_k^m$ exceeds $\frac 1 2 R^m$ and increases much faster than the remaining $|\lambda_i|^m.$ ♦

Exercise

Prove that if $f:\mathbb{Z} \to G$ is a continuous homomorphism to a compact Hausdorff group, then $f^{-1}(U)$ is infinite for any open neighbourhood U of e.

Lemma 2. For $g\in G$, the linear map $g:V\to V$ is diagonalizable.

Proof

Write $g:V\to V$ in block Jordan form. Suppose the matrix is not diagonal, say entry (1, 2) is 1 with corresponding eigenvalue $\lambda$. The matrix for $g^n : V\to V$ has entry (1, 2) equal to $n \lambda^{n-1}$ which is unbounded since $|\lambda| = 1.$ However, G is compact so taking entry (1, 2) of the representation $G\to GL(V)$ has bounded image, which is a contradiction. ♦

Orthogonality of Characters

The space of G-invariant elements of a G-module V is denoted by:

$V^G := \{v\in V: gv = v \, \forall\, g\in G\}.$

We have:

Proposition. The dimension of $V^G$ is:

$\displaystyle \int_{x\in G} \chi_V(x) dx.$

Proof

Take $\pi: V\to V$ via $v\mapsto \int_{x\in G} xv\cdot dx.$

Left-invariance then implies that the image of π lies in $V^G.$ Also, if $v\in V^G$ then $\pi(v) = v.$ Putting these two facts together gives us $\pi^2 = \pi$ so  π is the projection map onto $V^G.$ Hence:

$\displaystyle \dim V^G = \text{tr}(\pi) = \int_{x\in G} \text{tr}(x:V\to V)dx = \int_{x\in G} \chi_V(x)\cdot dx.$

as desired. ♦

In particular, replacing V by $\text{Hom}_{\mathbb C}(V, W)$, note that:

$\text{Hom}_{\mathbb C}(V, W)^G = \text{Hom}_{\mathbb C[G]}(V, W).$

So the dimension of this space is the integral of:

$\chi_{\text{Hom}(V, W)}(g) = \chi_{V^\vee \otimes W}(g) = \overline{\chi_V(g)} \chi_W(g).$

In other words:

$\dim \text{Hom}_{\mathbb{C}[G]}(V, W) = \int_{x\in G} \overline{\chi_V(x)} \chi_W(x)\cdot dx$

which we denote by $\left.$ Schur’s lemma then gives us:

Orthonormality of Irreducible Characters.

If V, W are irreps of G, then

$\left< V, W\right> = \int_{x\in G}\overline{\chi_V(x)} \chi_W(x)\cdot dx= \begin{cases} 1, \ &\text{if } V\cong W, \\ 0 \ &\text{else.}\end{cases}$

Abelian Case

Suppose now G is compact Hausdorff and abelian.

Theorem. Each irreducible representation V of G is 1-dimensional.

Proof

First we show that every $g\in G$ acts as a scalar. By lemma 2 above, V decomposes as a direct sum of eigenspaces $W_\lambda$ for $g:V\to V.$ Each $W_\lambda$ is G-invariant since:

$g' \in G, v\in W_\lambda \implies g(g'v) = g'(gv) = g'(\lambda v) = \lambda (g'v) \implies g'v \in W_\lambda.$

Since V is irreducible, we have $W_\lambda = V$ so g acts as a scalar.

Since every element of G acts as a scalar, every vector subspace is G-invariant; so dim(V) = 1. ♦

Hence by lemma 1, every irrep of G is a continuous homomorphism $G\to T$, where $T\subset \mathbb{C}^*$ is the unit circle. The following special case is of interest.

Definition. The n-torus is the topological group $T^n \cong (\mathbb{R}/\mathbb{Z})^n.$

[In fact, any compact Hausdorff abelian G must be of the form (finite group) $\times T^n$ but we won’t be needing this fact.]

For i = 1,…,n, let $\chi_i : T^n \to T$ be the homomorphism taken by projecting to the i-th coordinate. More generally, for any $\mathbf m := (m_1, \ldots, m_n) \in \mathbb{Z}^n$, we define:

$\chi^{\mathbf m} := \chi_1^{m_1} \cdot \ldots \cdot \chi_n^{m_n} : T^n \to T.$

Thus it takes $\mathbf v =(v_1, \ldots, v_n)$ to $v_1^{m_1} v_2^{m_2} \ldots v_n^{m_n}.$

Theorem. Any irrep of $T^n$ is of the form $\chi^{\mathbf m}$ for some $\mathbf m \in \mathbb{Z}^n.$

Proof

Suppose n=1: for a continuous $\chi:T\to T$ write T as $\mathbb{R} / \mathbb{Z}$ and compose $\mathbb{R} \to \mathbb{R}/\mathbb{Z} \stackrel\chi\to \mathbb{R}/\mathbb{Z}.$ The subgroup $\mathbb{Z}$ lies in the kernel; since the kernel is a closed subgroup it is either $\mathbb{R}$ or $\frac 1 m \mathbb{Z}$ for some m>0. So $\chi$ is of the form $v\mapsto v^m$ for some $m\in\mathbb{Z}.$

For n>1 consider the homomorphism $\phi_i : T \to T^n$ for = 1,…,n, by taking $v\mapsto (1,\ldots, 1,v,1\ldots,1),$ where the i-th component is v. By the previous case, $\chi\circ\phi_i : T\to T$ is of the form $v\mapsto v^{m_i}$ for some $m_i\in\mathbb{Z}.$ Since $\mathbf v = \prod_i \phi_i\chi_i(\mathbf v)$ for all $\mathbf v\in T^n,$ we get:

$\chi(\mathbf v) = \prod_i \chi \phi_i\chi_i(\mathbf v) = \prod_i \chi_i(\mathbf v)^{m_i} = \chi^{\mathbf m}(\mathbf v).$ ♦

Corollary. A character of $T^n$ is a finite linear combination of $\chi^{\mathbf m}$ whose coefficients are positive integers.

Replacing $\chi_i$ with variables $x_i$, this gives us a polynomial in $x_1, \ldots, x_n, x_1^{-1},\ldots, x_n^{-1}$, also known as a Laurent polynomial in $x_1, \ldots, x_n.$ Hence we have a correspondence:

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