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)?


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.

| cite | improve this answer | |

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.

  • $\begingroup$ 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 $\endgroup$ – Yemon Choi Apr 5 '11 at 19:13
  • $\begingroup$ Or the union of all the $S_n$ as $n\to \infty$? $\endgroup$ – David Roberts Sep 11 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.