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?


You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.

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.

  • $\begingroup$ 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! $\endgroup$ – Marty Aug 22 at 4:09
  • 2
    $\begingroup$ 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. $\endgroup$ – Peter McNamara Aug 22 at 4:20
  • $\begingroup$ 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? $\endgroup$ – Marty Aug 22 at 5:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Aurélien Djament Aug 22 at 5:58
  • $\begingroup$ @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. $\endgroup$ – Peter McNamara Aug 22 at 6:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.