Is an ultrapower of the hyperfinite factor still hyperfinite?

Question

Let $\mathcal{R}$ be the hyperfinite type $II_{1}$ factor and let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$.

Is it true that $\mathcal{R}^{\mathcal{U}}$ is never hyperfinite ? How can I see this ?

Thanks

I know that under Continuum Hypothesis, every $\mathcal{R}^{\mathcal{U}}$ is isomorphic. Also, every infinite dimensional subfactor of $\mathcal{R}$ is isomorphic to $\mathcal{R}$. Since $\mathbb{F}_{n}$ is sofic, $\mathcal{L}_{\mathbb{F}_{n}}$ can be embeddable into a suitable ultrapower $\mathcal{R}^{\mathcal{U}}$, but $\mathcal{L}_{\mathbb{F}_{n}}$ is never hyperfinite.

Can I argue, using the above lines, that under Continuum Hypothesis $\mathcal{R}^{\mathcal{U}}$ is never hyperfinite (if $\mathcal{U}$ is free, of course) ?

PS : I would like to know why was this question downvoted. Does it contain an obvious or childish mathematical mistake ? That is what is worrying me, not the reputation points. Thanks :)

Answer 1

As you say, an ultrapower of $R $ contains $L (\mathbb F_2) $, which is not hyperfinite, while every subfactor of the hyperfinite II$_1$ is hyperfinite.

Another way to see it is that $R $ is separable, while its ultrapowers aren't.

Answer 2

As we know the ultrapower of $\rm II_1$ factors is a $\rm II_1$ factor and also there is only one hyperfinite $\rm II_1$ factor up to isomorphism. So If $R^U$ is hyperfinite we must have $R\simeq R^U$, but it is impossible.