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This question concerns alternative characterizations of free group factors. The ping pong lemma is a well-known criteria for the freeness of a group. I've often wondered if there is a ping pong like criterion that can be used to determine if a given type $II_{1}$ factor is a free group factor, e.g. a ping pong-like criterion for the action of the factor on some Hilbert space.

Question: Is there a ping pong lemma analogue for group von Neumann algebras?

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I'm not even aware of any criterion at the purely vN level which characterizes the class of free group factors within the class of all group vN algebras -- but there are people reading MO who would no much more about this than I do. (Of course that class might consist of only one vN alg up to isomorphism, which is one reason I'm slightly pessimistic.) –  Yemon Choi Jan 31 '11 at 18:25
Indeed, Yemon, I think this is the case. One other such criterion has recently been proposed by Popa: whether the "free flip" can be connected to the identity. (This is even stronger in that it goes beyond characterizing the free group factors within group von Neumann algebras...) –  Jon Bannon Jan 31 '11 at 18:45
.."this" above being that no such abstract criterion is yet available. I would also be very surprised if my question were accessible. The idea is that if there is a decomposition of the Hilbert space on which the von Neumann algebra acts (I'm thinking something like Voiculescu's "free product" of Hilbert spaces) then one could in some setting deduce that the acting factor must be an interpolated free group factor. (This is kind of crazy, but who knows?) –  Jon Bannon Jan 31 '11 at 18:51
Also, we should emphasize that this question has little to do with whether free group factors are isomorphic. It asks only whether a factor is a free group factor...so somehow may avoid the issue Yemon is concerned with. –  Jon Bannon Jan 31 '11 at 21:25
Well, isn't this (vaguely) the kind of problem which motivated Murray and von Neumann and others to look at Property P, Property Gamma, and others? Maybe someone who knows about l^2-Betti numbers can tell us if something along those lines might be useful... –  Yemon Choi Feb 1 '11 at 3:27
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1 Answer

up vote 11 down vote accepted

I am guessing that the answer is "yes" if you interpret the question in the following way. Let $A_i$ be some subalgebras of a von Neumann algebra $(M,\tau)$ and assume that there are mutually orthogonal Hilbert subspaces $H_i$ of $H=L^2(M)$ so that for all $i$, $x (H\ominus H_i) \subset H_i$ whenever $x\in A_i$ with $\tau(x)=0$.

Let us also assume that $1 \perp \oplus_i H_i$ (probably this is not necessary).

Then if $y = x_1 \dots x_n$ with $x_j \in A_{i(j)}$, $i(1)\neq i(2)$, $i(2)\neq i(3)$, etc. and $\tau (x_j) = 0$, we have: $x_n 1 \in H_{i(n)}$ since $1\in H\ominus H_i$; $x_{n-1} x_n 1 \in H_{i(n-1)}$ since $x_n 1 \in H_{i(n)} \subset H\ominus H_{i(n-1)}$ (because $i(n)\neq i(n-1)$ and so $H(i(n))\perp H(i(n-1))$; $x_{n-2} x_{n-1} x_n 1 \in H_{i(n-2)}$ since $x_{n-1} x_n 1 \in H_{i(n-1)}\subset H\ominus H_{i(n-2)}$, etc. Thus We get that $x_1\dots x_n 1 \in H_{i(1)} \perp 1$, so that $\tau(y)=0$. It follows that $A_1,\dots,A_n$ are freely independent.

(Conversely, if $M$ is generated by $A_1,\dots,A_n$ and they are free inside of $M$, then $L^2(M) = \mathbb{C}1 \oplus \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$, where $L^2_0(A_j) = \{1\}^\perp \cap L^2(A_j)$. Then you can take $H_j = \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots; j_1= j} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$ and then $H_j$ are orthogonal and $H\ominus H_j$ is taken to $H_j$ by any $x\in A_j $ with $\tau(x)=0$).

If you now make some assumption (e.g. that $A_j$ are finite-dimensional, abelian or hyperfinite) then it follows from Ken Dykema's results (see e.g. his paper on Interpolated free group factors in Duke Math J.) that the von Neumann algebra they generate inside of $M$ is an interpolated free group factor. This is similar to the assumption you have put on the group (since the subgroup generated by a single element in the ping-pong lemma is necessarily abelian).

On the other hand, you raise the much bigger question of whether there exists some criterion that singles out free group factors -- just as the various functional-analytical criteria were shown by Connes to be equivalent to hyperfiniteness. Unfortunately, not much in known in this direction (note that a similar question exists on the ergodic equivalence side of things: is there a functional-analytic way of recognizing treeable actions? Or Bernoulli actions of free groups?)

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Thank you for the answer. This is precisely the sort of interpretation I thought may be possible. –  Jon Bannon Feb 1 '11 at 12:58
Especially, I like the comments regarding ergodic equivalence in the last paragraph. –  Jon Bannon Feb 3 '11 at 22:11
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