# E. Kowalski's blog

## Another exercise with characters

While thinking about something else, I noticed recently the following result, which is certainly not new:

Let $G$ be a compact topological group [ADDITIONAL ASSUMPTION pointed out by Y. Choi: connected, Lie group], and let $\rho$ be a finite-dimensional irreducible unitary continuous representation of $G$ on a vector space $V$. Then the natural representation $\pi$ of $G$ on $\mathrm{End}(V)$ decomposes as a direct sum of one-dimensional characters if and only if $\rho$ is of dimension $1$.

One direction is clear: if $\rho$ has dimension one, then $\pi$ is simply the trivial one-dimensional representation. For the converse, here is an argument with character theory.

As a first step, note that if $\rho$ (of dimension $d\geq 1$, say) has this property, then in fact $\pi$ decomposes as a direct sum of distinct one-dimensional characters: indeed, the multiplicity of a character $\chi$ in $\pi$ is the same as
$n_{\chi}=\int_{G}\chi(x)\mathrm{Tr}(\pi(g))dg,$
where $dg$ is the probability Haar measure on $G$, and since
$\mathrm{Tr}(\pi(g))=|\mathrm{Tr}(\rho(g))|^2,$
we get
$n_{\chi}\leq \int_{G}\mathrm{Tr}(\pi(g))dg=1$
by the orthogonality relations of characters. (Algebraically, this is just an application of Schur’s lemma).

Thus if we decompose $\pi$ into irreducible representations, we get
$\pi=\bigoplus_{1\leq i\leq d^2} \chi_i,$
where the $\chi_i$ are distinct one-dimensional characters. We then know by orthogonality that
$d^2=\int_{G} |\mathrm{Tr}(\pi(g))|^2 dg=\int_{G} |\mathrm{Tr}(\rho(g))|^4 dg.$

Now the last-integral is bounded by
$\int_{G} |\mathrm{Tr}(\rho(g))|^4 dg\leq \mathrm{Max}_{g}|\mathrm{Tr}(\rho(g))|^2 \times \int_G|\mathrm{Tr}(\rho(g))|^2dg\leq d^2,$
(since $|\mathrm{Tr}(\rho(g))|\leq d$). Comparing, this means that there must be equality throughout in this estimate, which in turn implies that $|\mathrm{Tr}(\rho(g))|=d$ for all $g\in G$. Since $\rho(g)$ is unitary of size $d$, this implies that $\rho(g)$ is scalar for all $g$, and since it is assumed to be irreducible, it is in fact one-dimensional.

I see two interesting points in this argument: (1) is there a purely algebraic proof of the last part? I haven’t thought very hard about this yet, but it would be nice to have one; (2) the appearance of the fourth moment of $\rho$ is nicely reminiscent of the Larsen alternative (see Section 6.3 of my notes on representation theory, for instance…)

March 22nd, 2013 at 10:19 am

Posted in Exercise,Mathematics

### 4 Responses to 'Another exercise with characters'

1. no doubt I am missing something, but why can’t rho of g sometimes have trace zero? (This calculation reminds me of one I had to do a while ago with L^1 norms of characters)

Yemon Choi

22 Mar 13 at 12:14     Reply

2. You’re right that I was a bit too fast — the argument goes through if the group is a connected Lie group (the function |Tr(rho(g))|^2 is continuous and the equality implies that its value is either 0 or d^2 at any point, and so it must be constant, and equal to d^2 since it is a non-zero representation), but it might not work otherwise. (I’ll look for counterexamples in finite groups…)

Kowalski

22 Mar 13 at 15:46     Reply

3. Indeed, if one takes a dihedral group of order 8, the unique 2-dimensional representation has |tr(rho)|^2 taking values 0 and 4, and it is the direct sum of the four one-dimensional characters! Thanks for catching this…

Kowalski

22 Mar 13 at 15:50     Reply

4. I think this is an algebraic version of what you wrote down in the last step, although I can’t see why it is completely parallel:

Let $\chi(g) = Tr (\rho(g))$. Now $\pi(g)$ acts on End(V). Consider a weighted average of these maps: $$\int_G |\chi(g)|^2 \pi(g) dg$$
This is an endomorphism of End(V) that projects onto the subspace fixed by all $|\chi(g)|^2 g$. Thus its trace is the dimension of the fixed subspace, which has dimension at most $d^2$.

On the other hand, the trace of this operator is also $\int_G |\chi(g)|^4 dg$, which is $d^2$ as you have argued.

So every endomorphism of End(V) is fixed by $|\chi(g)|^2 g$. This implies that for any $v \in V$,
$|\chi (g)|^2 g \cdot T(v) = T(gv)$
Take $g$ to be the identity, we see that $|\chi(1)|^2 = 1$, meaning that $\rho$ is dimension 1 to start with.

Soarer

23 Mar 13 at 9:12     Reply