Let $H=K\_1\oplus K\_2$ be infinite dimensional Hilbert spaces. Voiculescu's free Gaussian functor gives us free group factors $L(H)$, $L(K\_1)$, $L(K\_2)$ acting on the full Fock space $\Gamma(H)$ and, moreover, $L(H)=L(K\_1)\*L(K\_2)$.


But what is known about the relative commutant $L(K\_1)'\cap L(H)$? 


Certainly $L(K\_1)$ is non-amenable, so by Ozawa's result <http://arxiv.org/abs/math/0608451> the relative commutant must be atomic. But I don't know any more than that. In particular is it even nontrivial?



How well do we know relative commutants in $L(\mathbb{F}_\infty)$?
Given any two tracial von Neumann algebras $(N\_1, \tau\_1)$ and $(N\_2, \tau\_2)$ the $L^2$ space of the free product $(N\_1 \* N\_2, \tau)$ canonically decomposes as 
$$
L^2(N\_1 \* N\_2, \tau) = \mathbb C \oplus\_{n \in \mathbb N} \bigoplus\_{i\_1 \not= i\_2, i\_2 \not= i\_3, \ldots, i\_{n - 1} \not= i\_n } \overline {\otimes}\_{k = 1}^n L^2\_0(N\_{i\_k}, \tau\_{i\_k}),
$$
where $L^2\_0(N\_i, \tau\_i)$ is the orthogonal complement of the scalars.


From this decomposition it's not hard to see that as an $N\_1$ bimodule, $L^2(N\_1 \* N\_2, \tau)$ decomposes as a direct sum of one copy of the trivial bimodule $L^2(N\_1, \tau\_1)$ and copies of the coarse bimodule $L^2(N\_1 \overline \otimes N\_1, \tau\_1 \otimes \tau\_1)$. Specifically, if $\xi \in L^2(N\_1 \* N\_2, \tau)$ is a unit vector which is an ``elementary word'' starting and ending with vectors in $L^2\_0(N\_2, \tau\_2)$ then $x \xi y \mapsto x \otimes y$ extends to an isomorphism from the $N\_1$ bimodule generated by $\xi$ and $L^2(N\_1 \overline \otimes N\_1, \tau\_1 \otimes \tau\_1)$.


Thus $N\_1' \cap (N\_1 \* N\_2) \not= \mathcal Z(N\_1)$ if and only if the coarse $N\_1$ bimodule has non-zero central vectors, which, by identifying $L^2(N\_1 \overline \otimes N\_1, \tau\_1 \otimes \tau\_1)$ with Hilbert-Schmidt operators and then taking a spectral projection, is if and only if $L^2(N\_1, \tau\_1)$ has a finite dimensional $N\_1$-sub-bimodule, which is if and only if $N\_1$ has a non-trivial finite dimensional direct summand.


In the case you're looking at $N\_1$ is a ${\rm II}\_1$ factor and so $N\_1' \cap (N\_1 \* N\_2) = \mathcal Z(N\_1) = \mathbb C$.


