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It's well known that all hyperfinite $\mathrm{II}_1$ factors are isomorphic. I risk the wrath of MathOverflow elders to ask if a particular isomorph is easier than others to handle. In particular, is there a presentation for which it's obvious that the commutant is also the hyperfinite $\mathrm{II}_1$ factor (presented in the same way)?

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2 Answers 2

Another classical construction is to see the hyperfinite $\mathrm{II}_1$ factor as the infinite tensor product of the two by two matrices

$$ \mathcal{R}=\otimes_{n=1}^{\infty}{M_{2}(\mathbb{C})} $$

acting on its $L^{2}$ closure $L^{2}\Big(\otimes_{n=1}^{\infty}M_{2}(\mathbb{C})\Big)$ by right multiplication.

Then its commutant is given by the left multiplication.

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I suppose what you mean is that you want the commutant of the hyperfinite $II_{1}$ coming from a certain construction to have arisen from essentially the same construction.

The closest I can think of to this is the classic one: if you take the left group von Neumann algebra of the group of those permutations of $\mathbb{Z}$ having finite support, then the commutant will have arisen as the right group von Neumann algebra of this group.

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Or any discrete ICC amenable group, but I agree that your example is probably the easiest and most natural of such - as far as I know –  Yemon Choi Apr 5 '11 at 19:13

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