## Is it known that “hyperfinite length” cannot distinguish free group factors?

Given a type $II_{1}$ factor $M$, Popa and Ge defined the hyperfinite length $l_{h}(M)$ of $M$ to be the minimum natural number $n$ such that there are hyperfinite subalgebras $R_{1}, R_{2},..., R_{n}$ such that $\overline{sp}R_{1} R_{2}... R_{n}=M$. Here the closure is in 2-norm with respect to the trace on $M$.

This was introduced in *On some decomposition properties for factors of type $\mathrm{II}_1$*, Liming Ge and Sorin Popa. Source: Duke Math. J. Volume 94, Number 1 (1998), 79-101.

Is it known that hyperfinite lenth cannot distinguish free group factors?

It is known that $l_{h}(M)=1$ iff. $M$ is hyperfinite, and $l_{h}(L(\mathbb{F}_{n}))\geq 3$ if $n \geq 3$. It is known that if $M$ is the tensor product of two type $II_1$ factors, then the hyperfinite length is less than or equal to 3.

I recall hearing somewhere that this idea is not going to pay off, but I don't know why. Perhaps the hyperfinite length of every finitely generated non-hyperfinite type $II_{1}$ factor is 3?

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I would expect that the hyperfinite length of $L(\mathbb F_n)$ is infinite for all $n \in \mathbb N$. At least this is, what the heuristics of $\ell^2$-Betti numbers would tell you. – Andreas Thom Nov 19 2010 at 13:47
what are know examples, where the hypefinite length can be computed? – Kate Juschenko Nov 19 2010 at 17:45
@Kate: As far as I know, we only have explicit computation of this number for the hyperfinite case. – Jon Bannon Nov 22 2010 at 19:15