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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?

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1 Answer 1

up vote 10 down vote accepted

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 $N_1' \cap (N_1 * N_2) = \mathcal Z(N_1) = \mathbb C$.

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I should remark that the same proof shows that for any non-principal ultrafilter $\omega$ on $\mathbb N$, the asymptotic commutant $N_1' \cap (N_1 * N_2)^\omega$ is trivial if and only if $N_1$ is a non-amenable ${\rm II}_1$ factor. (A ${\rm II}_1$ factor is amenable if and only if the coarse bimodule contains almost central vectors.) –  Jesse Peterson Aug 4 '12 at 22:25
Great answer. Thanks! –  Ollie Margetts Aug 4 '12 at 23:54
@JessePeterson: If $N_i$ are ${\rm II}_1$ factors, is it true that $N_1 \subset B(L^2(N_1 * N_2, \tau))$ and its commutant in $B(L^2(N_1 * N_2, \tau))$, are also ${\rm II}_1$ factors ? –  Sébastien Palcoux Jul 31 '14 at 17:44
@SébastienPalcoux: If $N \subset M$ is an inclusion of ${\rm II}_1$ factors, then $N' \cap \mathcal B(L^2(M))$ is anti-isomorphic to the basic construction $\langle M, N \rangle = (JNJ)' \cap \mathcal B(L^2(M))$. This is always a semi-finite factor, and is finite if and only if $N$ is a finite index subfactor of $M$. In the case of free products, $N \subset N * B$ is finite index only in the case $B = \mathbb C$. –  Jesse Peterson Aug 1 '14 at 17:37
Typo: $\langle M ,N \rangle$ should be replaced by $\langle M ,e_N \rangle$. –  Sébastien Palcoux Aug 6 '14 at 4:23

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