While thinking about something else, I noticed recently the following result, which is certainly not new:
Let $latex G$ be a compact topological group [ADDITIONAL ASSUMPTION pointed out by Y. Choi: connected, Lie group], and let $latex \rho$ be a finite-dimensional irreducible unitary continuous representation of $latex G$ on a vector space $latex V$. Then the natural representation $latex \pi$ of $latex G$ on $latex \mathrm{End}(V)$ decomposes as a direct sum of one-dimensional characters if and only if $latex \rho$ is of dimension $latex 1$.
One direction is clear: if $latex \rho$ has dimension one, then $latex \pi$ is simply the trivial one-dimensional representation. For the converse, here is an argument with character theory.
As a first step, note that if $latex \rho$ (of dimension $latex d\geq 1$, say) has this property, then in fact $latex \pi$ decomposes as a direct sum of distinct one-dimensional characters: indeed, the multiplicity of a character $latex \chi$ in $latex \pi$ is the same as
$latex n_{\chi}=\int_{G}\chi(x)\mathrm{Tr}(\pi(g))dg,$
where $latex dg$ is the probability Haar measure on $latex G$, and since
$latex \mathrm{Tr}(\pi(g))=|\mathrm{Tr}(\rho(g))|^2,$
we get
$latex n_{\chi}\leq \int_{G}\mathrm{Tr}(\pi(g))dg=1$
by the orthogonality relations of characters. (Algebraically, this is just an application of Schur’s lemma).
Thus if we decompose $latex \pi$ into irreducible representations, we get
$latex \pi=\bigoplus_{1\leq i\leq d^2} \chi_i,$
where the $latex \chi_i$ are distinct one-dimensional characters. We then know by orthogonality that
$latex d^2=\int_{G} |\mathrm{Tr}(\pi(g))|^2 dg=\int_{G} |\mathrm{Tr}(\rho(g))|^4 dg.$
Now the last-integral is bounded by
$latex \int_{G} |\mathrm{Tr}(\rho(g))|^4 dg\leq \mathrm{Max}_{g}|\mathrm{Tr}(\rho(g))|^2 \times \int_G|\mathrm{Tr}(\rho(g))|^2dg\leq d^2,$
(since $latex |\mathrm{Tr}(\rho(g))|\leq d$). Comparing, this means that there must be equality throughout in this estimate, which in turn implies that $latex |\mathrm{Tr}(\rho(g))|=d$ for all $latex g\in G$. Since $latex \rho(g)$ is unitary of size $latex d$, this implies that $latex \rho(g)$ is scalar for all $latex g$, 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 $latex \rho$ is nicely reminiscent of the Larsen alternative (see Section 6.3 of my notes on representation theory, for instance…)
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)
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…)
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…
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.