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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asked Jul 30, 2015 at 14:04
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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 ChoiCommented Jul 30, 2015 at 17:29
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1 Answer
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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, $$\mathbb C I \neq \mathcal R' \cap \mathcal R^\mathcal U \subset \mathcal R' \cap L(\mathbb F_2)^\mathcal U$$
answered Jul 30, 2015 at 18:10
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3$\begingroup$ 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 $\endgroup$ Commented Jul 30, 2015 at 22:41