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Let $A$ be a von Neumann algebra acting on a Hilbert space $H$. Write $\mathcal{B}(H)$ for the bounded linear operators on $H$. Suppose that $\rho:\mathcal{B}(H) \rightarrow \mathcal{B}(H)$ is an inner automorphism of $\mathcal{B}(H)$, that is $\rho(x) = uxu^{*}$ for some unitary operator $u$ on $H$. Suppose furthermore that $\rho$ leaves $A$ invariant, that is $\rho(A) = A$.

When is $\rho|_{A}:A \rightarrow A$ inner? In other words, when does there exist a unitary $u_{A} \in A$ such that $\rho|_{A}(a) = u_{A} a u_{A}^{*}$, for all $a \in A$.

I am not sure if it matters, but I am interested in the case that $A$ is a factor, and we may furthermore assume that $\rho$ also leaves $A'$ invariant.

Furthermore, we assume that the von Neumann algebra $A$ admits a cyclic and separating vector.

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    $\begingroup$ You have to be more specific. If the representation of $A$ on $H$ is standard then any automorphism is given by a conjugation by a unitary operator on $H$ that also preserves $A'$. So in general nothing can be said. $\endgroup$
    – Mateusz Wasilewski
    Commented Mar 20, 2018 at 17:14
  • $\begingroup$ @MateuszWasilewski I am not sure what It means for a representation to be standard. I do know that $A$ has a cyclic and separating vector in $H$. $\endgroup$
    – Peter
    Commented Mar 20, 2018 at 17:26
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    $\begingroup$ Oh, that should be sufficient. I have to think about the general case but if you are willing to assume that your von Neumann algebra is tracial and the cyclic and separating vector $\Omega$ defines the trace then it definitely works. The point is that if $\rho$ is an arbitrary automorphism of $A$, then it preserves the trace, by uniqueness of the trace. Then one can check that the map $x\Omega \mapsto \rho(x)\Omega$ extends to a unitary on $H$, which implements the automorphism. I was a bit to quick, so now I'll have to think about the general case (i.e. no trace). $\endgroup$
    – Mateusz Wasilewski
    Commented Mar 20, 2018 at 17:35
  • $\begingroup$ @MateuszWasilewski Great! I will add that to my question. However, I don't think the von Neumann algebra I have in mind is tracial. I don't quite understand your argument so far though. You claim $x \Omega \mapsto \rho(x) \Omega$ extends to a unitary, say $u$, on $H$, which implements the automorphism, then it's not clear to me that $u \in A$. Maybe my question was not quite clear, let me rephrase. $\endgroup$
    – Peter
    Commented Mar 20, 2018 at 17:44
  • $\begingroup$ What I meant is that in this case every automorphism is implemented by a unitary and therefore the assumption that it comes from an inner automorphism of $B(H)$ does not provide any further information. And in general it is a hard problem to determine whether an automorphism is inner. I think it would be better if you told us what algebra and what automorphism you have in mind. $\endgroup$
    – Mateusz Wasilewski
    Commented Mar 20, 2018 at 19:01

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Denote by $U(M)$ the group of unitary elements of a von Neumann algebra $M$.

$\rho|_A$ is inner iff $u\in U(A)U(A')$. Assume first that $u\in U(A)U(A')$, so there are $v\in U(A)$ and $w\in U(A')$ with $u=vw$. Then for $a\in A$ we have $$\rho(a)=uau^*=vwaw^*v^*=vav^*$$ On the other hand, if there exist $v\in U(A)$ such that $uau^*=\rho(a)=vav^*$, then $v^*ua=av^*u$, $a\in A$, whence $v^*u\in U(A')$ and $u=v(v^*u)\in U(A)U(A')$

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