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Given any unital non-commutative subalgebra $\mathcal M$ of $L(\mathbb F_2)$ is it true that $\mathcal M' \bigcap L(\mathbb F_2)^\mathcal U = \mathbb C I$ for any free ultrafilter $\mathcal U$?

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  • $\begingroup$ Hi Chris. It probably doesn't make any difference, but is M supposed to be self-adjoint as well? $\endgroup$
    – Yemon Choi
    Jul 30, 2015 at 17:29

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Hah, should have gone down the hall at UVa first. The following negative answer was given to me by Scott Atkinson:

The hyperfinite II$_1$ factor $\mathcal R$ is the minimal object among II$_1$ factors. As such, there is a copy of $\mathcal R$ sitting as a subalgebra of $L(\mathbb F_2)$. Thus by an abuse of notation, $$\mathbb C I \neq \mathcal R' \cap \mathcal R^\mathcal U \subset \mathcal R' \cap L(\mathbb F_2)^\mathcal U$$

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    $\begingroup$ This is basically the only case where you have a negative answer. Ozawa showed that if $\mathcal M$ is a II$_1$ subfactor of $L(\mathbb F_2)$, then either $\mathcal M \cong \mathcal R$ or else $\mathcal M' \cap L(\mathbb F_2)^{\mathcal U}$ is a direct sum of matrix algebras: ams.org/mathscinet-getitem?mr=2079600 $\endgroup$ Jul 30, 2015 at 22:41

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