TITLE: Presenting the Hyperfinite II_1 Factor
QUESTION [5 upvotes]: 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)?

REPLY [8 votes]: 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.