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)?
2 Answers
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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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$\begingroup$ 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 $\endgroup$ Commented Apr 5, 2011 at 19:13
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$\begingroup$ Or the union of all the $S_n$ as $n\to \infty$? $\endgroup$– David Roberts ♦Commented Sep 11, 2020 at 0:44