Tensor Product of Factors of type $II_{1}$

Question

Let $\mathcal{R}$ be a type $II_{1}$ factor. Is it true that $\mathcal{R}\otimes\mathcal{R}$ and $M_{2}(\mathcal{R})$ are still type $II_{1}$ factors ? If so, any reference ? (I am asking this question because if this is the case, then I'd conclude the existence of isomorphism between $\mathcal{R}$ - the hyperfinite factor of type $II_{1}$ - and $\mathcal{R}\otimes\mathcal{R}$ and $M_{2}(\mathcal{R})$ , which is quite useful to fully understand a proof of a result I'm reading online )... Thanks you !

Answer 1

The symbol $\mathcal{R}$ is most unfortunate because it is reserved for the hyperfinite ${\rm II}_1$ factor...

Anyhow, let $\mathcal{M}$ be a ${\rm II}_1$ factor. Then indeed $\mathcal{M}\otimes \mathcal{M}$ and $M_2(\mathcal{M})$ are ${\rm II}_1$ factors too. More generally, suppose that $\mathcal{M}$ and $\mathcal{N}$ are ${\rm II}_1$ factors with (unique) normal traces $\tau$ and $\sigma$, respectively. Then $\mathcal{M}\otimes \mathcal{N}$ is a ${\rm II}_1$ factor too.

Indeed, $\mathcal{M}\otimes \mathcal{N}$ has trivial centre (easy computation). It admits also a normal trace, simply take $\tau\otimes \sigma$.

This in particular applies to $\mathcal{N} = M_2(\mathbb{C})$, in which case $\mathcal{M}\otimes \mathcal{N}\cong M_2(\mathcal{M})$.