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Let $A,B$ be subfactors of a II$_1$ factor $M$ with $A*B\simeq M$. That is, $A$ and $B$ are freely independent with respect to the trace and $M\simeq A\vee B$. We'll call $B$ a free complement for $A$ in $M$.

My first question is simple (to state): if a free complement exists is it necessarily unique? My intuition says the answer should be ''no'', but I'm not aware of any simple examples?

Moving from the general to the specific, I'm really interested in the following example of Voiculescu. Let $H$ be a real, infinite dimensional Hilbert space and $\Phi(H_{\mathbb{C}})=\mathbb{C}\Omega\oplus\bigoplus_{n=1}^\infty{H^{\otimes n}_{\mathbb{C}}}$ the full Fock space over its complexification. If $\ell(f)$ is the left creation operator defined by $\ell(f)\Omega=f$, $\ell(f)\xi=f\otimes\xi$ when $\xi\perp\Omega$ and $\mathrm{h}$ is a subspace of $H$ then consider the von Neumann algebra $$L(\mathrm{h})=\{s(f):~f\in\mathrm{h}\}''\qquad\hbox{where}\qquad s(f)=\frac{\ell(f)+\ell(f)^*}{2}.$$ $L(H)$ is isomorphic to the free group factor $L(F_\infty)$ acting standardly with cyclic trace vector $\Omega$. If $\mathrm{h}\oplus\mathrm{k}=H$ then $L(\mathrm{h})*L(\mathrm{k})\simeq L(H)$. The second question is a special case of the first: is $L(\mathrm{k})$ the only free complement to $L(\mathrm{h})$ in $L(H)$?

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up vote 5 down vote accepted

If you want uniqueness up to equality, then the answer is no since if $B$ is a free complement for $A$ in $M$ then so is $u B u^*$ for any unitary $u \in A$. This is the case also in the specific example you give above.

If you want uniqueness up to isomorphism then this is more subtle since showing that II$_1$ factors are (non)isomorphic is usually quite difficult. I can't think of an example for this situation in general, but I do have an example if you assume that the free group factors $L\mathbb F_2$ and $L\mathbb F_3$ are isomorphic.

Indeed, Dykema and Radulescu showed in ( that if the free group factors are isomorphic then for any II$_1$ factors $M$ and $N$ we have that $M * N \cong M * N * L\mathbb F_\infty$. In particular, this would hold if $M$ were a II$_1$ factor with property (T), and $N = R$ were the hyperfinite II$_1$ factor. But in this case by a result of Ioana, Popa, and myself ( there would then also exist an isomorphism $\theta: M * R \to M * R * L\mathbb F_\infty$ such that $\theta(M) = M$. This would give a situation where $R$ and $R * L\mathbb F_\infty$ are both free complements of $M$, but $R * L\mathbb F_\infty$ is not hyperfinite and hence not isomorphic to $R$.

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Thanks, that's cleared up some false preconceptions! It seems free group factors give away nothing for free! – Ollie Margetts Oct 11 '11 at 11:04

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