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 $\mathcal U$?



A relative property gamma and $L(\mathbb F_2)$
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, $$\mathbb C I \neq \mathcal R' \cap \mathcal R^\mathcal U \subset \mathcal R' \cap L(\mathbb F\_2)^\mathcal U$$


