Subfactor of hyperfinite one

35 Views Asked by At

Is there a strict subalgebra of the hyperfinite $II_1$ factor that is separable and type $II_1$ factor?

1

There are 1 best solutions below

0
On

Denote the hyperfinite $\mathrm{II}_1$ factor by $R$. Then $R$ is isomorphic to $R \bar{\otimes} R$. So if you fix an isomorphism $\pi: R \bar{\otimes} R \rightarrow R$, then $\pi(R \otimes \mathbb{C})$ would satisfy your requirement.