A remark on the ultrapower algebra of the hyperfinite factor
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mathcalomegahyperfinitesubseteqaffirmativealgebraanswerasked
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On page 43 in \cite{Po83} Sorin Popa asked whether the following property holds: \emph{If $\omega$ 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 $\mathcal R'\cap \mathcal R^\omega\subseteq \mathcal R^\omega_1$ does it follows that $\mathcal R_1=\mathcal R$?} In this short note we provide an affirmative answer to this question.
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