MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

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.

share|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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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