Hi, Is it true that all irreducible unitary representations of a residually finite group are finite dimensional?
Actually I suspect that it is not, but cannot find any example.
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2$\begingroup$ How about the free group with two generators? Every representation of every 2-generated group is a representation of $F_2$. $\endgroup$– user6976Commented May 14, 2011 at 12:02
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3$\begingroup$ Mark is right. Here is for example a nice result by Michael Cowling and Tim Steger: Let $G$ be a non-compact simple Lie group, and let $\pi$ be a unitary irreducible representation of $G$; assume that $\pi$ is not in the discrete series of $G$ (you exclude countably many rep's in a continuum). Then the restriction of $\pi$ to any lattice $\Gamma$ in $G$, is an irreducible representation of $\Gamma$. $\endgroup$– Alain ValetteCommented May 14, 2011 at 13:24
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$\begingroup$ To clarify: when you talk about an irreducible representation $\pi:G \to {\rm End}(V)$, do you mean that there is no proper $G$-invariant subspace other than zero, or do you mean that there is no proper, closed $G$-invariant subspace other than zero $\endgroup$– Yemon ChoiCommented May 15, 2011 at 3:57
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$\begingroup$ I believe the standard definition is no closed $G$-invariant subspaces. $\endgroup$– Benjamin HayesCommented May 23, 2011 at 7:31
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As Mark Sapir pointed out, this is of course false. There are tons of infinite dimensional irreducible representations of residually finite groups. In fact, I think that any finitely generated group for which the answer is yes, is virtually abelian.
However, you could also ask whether all irreducible representations of a given residually finite group are weakly contained in finite dimensional representations. This is true for $\mathbb F_2$ but for example not for $SL(3,\mathbb Z)$. For the group $\mathbb F_2 \times \mathbb F_2$, this problem is open and equivalent to Connes Embedding problem.
answered May 21, 2011 at 16:26
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2$\begingroup$ @ Andreas: Correct, the assumption in the OP ("every unitary irrep is finite dimension") implies that the group is type I, so by Thoma's theorem the group is virtually abelian. Your sentence "You could ask...", ending with a (in)famous open question, is a bit tough (:-). $\endgroup$ Commented May 21, 2011 at 17:06
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1$\begingroup$ Rats! I was going to say that! $\endgroup$ Commented May 21, 2011 at 17:09