A remark on the ultrapower algebra of the hyperfinite factor

Abstract

On page 43 in \cite{Po83} Sorin Popa asked whether the following property holds: \emph{If $\ddot{I}‰$ is a free ultrafilter on $\mathbb{N}$ and ${\mathcal{R}}_1\subseteq \mathcal{R}$ is an irreducible inclusion of hyperfinite II${}_1$ factors such that ${\mathcal{R}}^{\prime }\cap {\mathcal{R}}^{\ddot{I}}‰\subseteq {\mathcal{R}}^{\ddot{I}}{‰}_1$ does it follows that ${\mathcal{R}}_1=\mathcal{R}$?} In this short note we provide an affirmative answer to this question.

Minor corrections. Final version. To appear in Proceedings of AMS