TITLE: Is there a characterization of free groups in terms of the unitary dual?
QUESTION [13 upvotes]: 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?

REPLY [4 votes]: 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 $\tilde{G}\to \tilde{F_S}$.