2
$\begingroup$

Recently in a paper we get the following result:

Let a discrete group $\Gamma$ act on a discrete abelian group $G$ by group automorphisms. Every irreducible unitary representation $\pi$ of $G\rtimes\Gamma$ on a Hilbert space $H$ satisfies $\dim{H_\Gamma}\leq 1$. Here $H_\Gamma$ is the space of $\Gamma$-invariant vectors in $H$.

We are not experts in representation theory.

My question is :

Does this result appear in the literature before?

$\endgroup$

1 Answer 1

0
$\begingroup$

I would think that this result would go all the way back to Frobenius. Anyway, the proof seems easy enough:

Let $V$ be the trivial $\Gamma$-module. We want to show that $\dim H_\Gamma=\dim\mathrm{Hom}_{\Gamma}(V,H)\leq 1$. To this end, note that by Frobenius reciprocity, $$ \mathrm{Hom}_{\Gamma}(V,H)\cong\mathrm{Hom}_{G\rtimes\Gamma}(\mathrm{Ind}_{\Gamma}^{G\rtimes\Gamma}V,H) $$ so it suffice to show that the right-hand side is at most 1-dimensional. As $H$ is irreducible, this amounts to showing that every irreducible summand of $\widehat{V}:=\mathrm{Ind}_{\Gamma}^{G\rtimes\Gamma}V$ appears with multiplicity 1 (by Schur's lemma).

Well, this is easy since $\widehat{V}\cong\mathbb{C}G$ as a $G$-module and $\mathbb{C}G$ decomposes as a direct sum of non-isomorphic $G$-modules. On the other hand, if $U\oplus U\leq\widehat{V}$ for some irreducible $G\rtimes\Gamma$-module $U$, then decomposing each $U$ as a $G$-module shows that there are multiple copies of the same irreducible $G$-module in $\widehat{V}$ upon restriction to $G$. This is a contradiction.

Hence, $\dim H_\Gamma\leq 1$.

$\endgroup$
4
  • $\begingroup$ Since these groups may be infinite, is Frobenius reciprocity still directly applicable? $\endgroup$
    – Yemon Choi
    Commented Jul 7, 2015 at 0:33
  • $\begingroup$ @YemonChoi I think it's okay. Isn't Fronenius reciprocity really a statement about adjointness of tensor product and hom. The proof I know is for modules over some rings S<R, though I don't recall the exact hypotheses. I will certainly double check. $\endgroup$
    – David Hill
    Commented Jul 7, 2015 at 0:52
  • $\begingroup$ It might be OK in this setting, I just occasionally get worried about induction versus co-induction. There are definitely problems for locally compact, non-discrete groups though $\endgroup$
    – Yemon Choi
    Commented Jul 7, 2015 at 2:28
  • $\begingroup$ @YemonChoi Frobenius reciprocity seems to be okay. The proof I know is fine for the situation. The only issue might be the equality between $\dim H_\Gamma$ and the hom space, but given we have an invariant inner product, that seems fine too. $\endgroup$
    – David Hill
    Commented Jul 7, 2015 at 3:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .