# External tensor product of irreducible representations is not irreducible?

I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false.

Dubious claim: If $$(\pi, V)$$ and $$(\rho, W)$$ are irreducible representations of two groups $$G$$ and $$H$$, respectively, then the "external" tensor product $$\pi \boxtimes \rho$$ is an irreducible representation of $$G \times H$$.

Of course this is true and well-known in the usual cases, e.g., when $$G$$ and $$H$$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar.

Is there a nice counterexample for complex representations of some infinite groups? Published somewhere?

Let E/F be a field extension. Let $$G=H=E^\times$$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $$E\otimes_F E\to E$$ is a submodule.
• I think this works. So in the context of complex representations, let $G = H = {\mathbb C}(T)^\times$ acting on the complex vector space $V = {\mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here! – Marty Aug 22 at 4:09
• Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G \times H}(\pi \boxtimes \rho$ is ${\mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here? – Marty Aug 22 at 5:17