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## Derivation of von Neumann algebra which is zero on MASA

Are there any example of $II_1$-factor $M$ with maximal abelian von Neumann subalgebra $A$ and non-zero derivation $\delta:M\rightarrow B(H)$ such that $\delta(a)=0$ for every $a\in A$?

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Let $M= L \mathbb F_2$ and $H = \ell^2 \mathbb F_2$, where $\mathbb F_2 = \langle a,b \rangle$ is the free group on two generators and $B(H)$ is a bimodule via the left and right multiplication with the left-regular representation $\lambda \colon L \mathbb F_2 \to B(\ell^2 \mathbb F_2)$.

Define $\delta(x) = [x,\lambda(a)]$. Then, $\lambda(a^{\pm1})'' \subset L \mathbb F_2$ is a MASA (as can be shown) and $\delta$ vanishes on it. However, $\delta(\lambda(b)) \neq 0$ so that $\delta$ does not vanish on $L \mathbb F_2$.

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 Thanks, Andreas! In fact, if $\delta$ takes value in compact operators then it is $0$ on the whole $M$. Thanks for the fast clarification! – Kate Juschenko Jan 13 2011 at 14:21 What is the precise theorem? Does this hold for $H = L^2(M,\tau)$? – Andreas Thom Jan 13 2011 at 14:25 It is the following: $\delta: M \rightarrow K(H)$ is a derivation of $II_1$-factor and A - MASA, if $\delta|_{A}=0$ then $\delta=0$. (it is contained in the proofs of Popa, JFA 1987). In particular it holds for $H=L^2(M, \tau)$. – Kate Juschenko Jan 13 2011 at 14:39 Thanks for the reference. – Andreas Thom Jan 13 2011 at 14:42

Sorry to be naive but why can't you just represent $M$ on a separable Hilbert space and take the inner derivation induced by any non-central element in the masa? This maps into $M\subset B(H)$, vanishes on the masa, but is non-trivial on $M$ because the element is non-central.

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