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If $G$ is a countable discrete group, I'm curious if it is possible to decide whether $G$ is a free group only by looking at properties of $Rep(G)$, the collection of (equivalence classes of) strongly continuous unitary representations of G on separable Hilbert space.

If this question is too crude, consider the following refinement:

If $G$ is a countable discrete group, is it possible to decide whether $G$ is a free group only by looking at properties of $Rep(G)$ viewed as the unitary dual of $G$, the topological space of (unitary equivalence classes of) unitary representations of $G$ on separable Hilbert space, equipped with Fell's topology?

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Why is it ridiculous? It might well be possible to characterize $F_2$, say, by the space of representations of dimension 2 (homs into $SU(2)$). Why do you think this is not possible? – Mark Sapir Oct 24 '10 at 1:48
I have no mathematical reason to think it isn't possible. Only that if it were, it would probably be widely enough known that I would have heard about it. – Jon Bannon Oct 24 '10 at 2:17

I put it as an answer because it is too long for a comment. Perhaps you had in mind the following. There are groups with very "few" unitary representations. Perhaps if you take such a group $G$ and take the direct product $G\times F_2$, you get the same space of unitary representations as for $F_2$. For example, if you take $G$ to be a simple torsion infinite group (such exist by Novikov-Adian-Zelmanov). Then all finite dimensional unitary representations of $G$ are trivial. Now the space of finite dimensional unitary representations of $G\times F_2$ is the same as for $F_2$.

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Better yet take the free product $G*F$ whose representations on a space are pairs of representations of $G$ and $F$. Of course the question is about representations on a separable space which is quite different... – Torsten Ekedahl Oct 24 '10 at 4:21

There is a very strong form of residual finiteness that holds for free groups, property FD, introduced by Lubotzky and Shalom, which says that finite representations are dense in the unitary dual (with respect to the Fell topology). So property FD is certainly a necessary condition for the group to be free.

If $S\subset G$ is a generating set for $G$, then we may detect whether $S$ generates $G$ freely from the unitary dual. Let $F_S$ be the free group generated by $S$. Then the map $F_S\to G$ induces a bijection of unitary duals $\tilde{G}\to\tilde{F_S}$ if and only if $S$ generates $G$ freely. This is because free groups are residually finite, so if $w\in F_S$ were a relator $G$, then there would be a finite group $H$ and a homomorphism $\varphi:F_S\to H$ such that $\varphi(w)\neq 1$. So the finite unitary representation of $F_S$ corrsponding to $\varphi$ would not be in the image of $\tilde{G}\to \tilde{F_S}$.

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Does finite representation mean finite-dimensional or factoring over a finite quotient? Is $F_2 \times F_2$ known to be or not to be FD? – Andreas Thom Oct 24 '10 at 16:44
Yes, a finite representation in their sense is one whose image in $U(H)$ is a finite group. I think that the Fell topology is a functor that preserves Cartesian products, and clearly the finite representation functor has this property. So I suppose yes, $F_2 \times F_2$ is FD. – Greg Kuperberg Oct 24 '10 at 16:59
I guess what you say is true for free products, but a unitary representation of a cartesian product is given by a pair of commuting representations. I expect, that the classification of pairs of commuting representations up to conjugation is more subtle. – Andreas Thom Oct 24 '10 at 17:47
Oh, sorry, I was thinking of irreducible representations. – Greg Kuperberg Oct 25 '10 at 11:13

The equivalence classes of representation of the free group on two generators are in bijection with $(U(H) \times U(H))/U(H)$, i.e. pairs of unitaries up to simultaneous conjugation. I doubt that this, as a topological space characterizes the free group.

However, it is true that any group homomorphism between countable groups which induces an isomorphism on the space of unitary representations up to conjugation (on an infinite-dimensional separable Hilbert space) is an isomorphism of groups. One way to see this is to note that the induced map on maximal group $C^*$-algebras is monic and epic and hence an isomorphism (see this post).

As Mark showed, this obviously fails if the Hilbert space is too small.

Also, maybe my anwser to this question is helpful.

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