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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 '10 at 13:47
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@Jon, roughly speaking, a similar result in the equivalence relation setting is used in the proof of theorem 4.10 of the paper "Topological Properties of Full Groups" by J.Kittrell and T.Tsankov, and F. Maitre refined this process in his paper on generator problems. I guess when Ge and Popa introduced this "hyperfinite length", they also expect it to be related to the generator problem for Von Neumann algebras? –  Jiang Oct 29 '13 at 22:42
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@Jon, it is interesting that Ge and Shen has already defined the "generating length" for a type II_1 Von Neumann algebra as an analogy of the definition of cost in the equivalence relation setting, you can find it in page 368 of the book" Third International Congress of Chinese Mathematicians, 1st part". But the relation between hyperfinite length and generating length has not been mentioned in that paper. –  Jiang Oct 30 '13 at 13:19
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Indeed, Jiang, the initial expectation was that there would be a connection with the generator problem. –  Jon Bannon Oct 31 '13 at 2:12
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@Jon, roughly speaking, a result in equivalence relation setting is often easier to prove than its counterpart in Von Neumann algebras setting, but the result on the topological generators of the full group generated by the hyperfinite equivalence relation is proved rather later than its counterpart. This seems strange. –  Jiang Oct 31 '13 at 16:30
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