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.

  • 3
    $\begingroup$ 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 Mar 20 '18 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 Mar 20 '18 at 17:26
  • 1
    $\begingroup$ 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 Mar 20 '18 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 Mar 20 '18 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 Mar 20 '18 at 19:01

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 $u=v(v^*u)\in U(A)U(A')$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.