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Let $A \subset B \subset C$ be von Neumann algebras (and more specifically factors, if it helps) either all type II or all type III, acting on the same separable Hilbert space.

Call $A' \subset B$ the "unique complement" of $A$ in $B$, if $A'$ is the least sub-von-Neumann-algebra of $B$ such that the algebra generated by $A \cup A'$ is all of $B$.

Question 1: does $A$ always have a unique complement in $B$?

Question 2: when $A$ has unique complements $A'$ and $A''$, in $B$ and in $C$ respectively (with $B \subset C$), must we have $A' \subseteq A''$?

If anyone can point out examples/proofs I'd be very grateful!

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Remark: with von Neumann algebras, using the notation $A'$ for anything other than the commutant is highly unusual.

Let $R$ be the hyperfinite II$_1$-factor and take $B=R\oplus R\oplus R=R\otimes\mathbb C^3$, and $A=R\otimes I$. So we think $$ A=\{(a,a,a):\ a\in R\},\ \ B=\{(a,b,c):\ a,b,c\in R\}. $$ You can take a "complement" of $A$ to be $$ A_1=\{(\lambda I,b,c):\ \lambda\in\mathbb C,\ b,c\in R\}; $$ or $$A_2=\{(a,\lambda I, c):\ \lambda\in\mathbb C,\ a,c\in R\}; $$ so no uniqueness.

Factors do not improve things. For instance you could think of $B=L(\mathbb F_2)=W^*(a,b)$ where $a,b$ are the usual generators of the group. Take $A=W^*(a)$. Now you can take the other canonical masa $A_1=W^*(b)$ as a complement, but also the radial masa $A_2=W^*(a+b)$.

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  • $\begingroup$ Once again, many thanks ... I'm finally making progress on proving that vN algebras can be used for the purposes I'd had in mind ... but actually using them for this will require me to work through Kadison & Ringrose in a more thorough way than I've done so far! $\endgroup$ – Doug McLellan Mar 14 '16 at 17:17

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