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Aug 22, 2019 at 6:31 comment added Marty @PeterMcNamara -- that sounds promising, thanks! I'm going to take a little while to digest the Jacobson density theorem, and will write to you if I can't put it together.
Aug 22, 2019 at 6:27 comment added Peter McNamara @Marty, you can use Jacobson density theorem on V and W to prove V\otimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
Aug 22, 2019 at 5:58 comment added Aurélien Djament If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
Aug 22, 2019 at 5:17 comment added Marty 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?
Aug 22, 2019 at 4:20 comment added Peter McNamara There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
Aug 22, 2019 at 4:09 comment added Marty 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!
Aug 22, 2019 at 4:06 vote accept Marty
Aug 22, 2019 at 0:55 history answered Peter McNamara CC BY-SA 4.0