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It should be a very simple question, but I couldn't find any reference concerning it.

Let $\mathcal{H}$ be a separable Hilbert space, $\mathcal{A}\subset\mathcal{B}(\mathcal{H})$ a von Neumann algebra and $\phi:\mathcal{A}\rightarrow\mathcal{A}$ an *-automorphism (linear map, preserving products and adjoints). The question is: does it exist a unitary operator $U\in\mathcal{B}(\mathcal{H})$ such $\phi(A)=UAU^*$ for all $A\in\mathcal{A}$.

I'm looking for a proof, a reference or a counterexample about the above question.

A remark about continuity: If I'm not confused, every *-automorphism (*-homomorphism indeed) between vN algebras is always continuous in the operator topology, but it may not be necessary continuous in the $\sigma$-weak topology. If the above statement is false in general, may it be true for $\sigma$-weak continuous *-automorphisms?

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If you had that property, it would mean that every automorphism of every von Neumann algebra extends to an automorphism of $B(H)$. That's very non true. For instance, let $p$ be a finite-projection, and consider the algebra $$ A=\{\alpha p+\beta(1-p):\ \alpha,\beta\in\mathbb C^2\}. $$ This algebra is of course $\mathbb C^2$ in disguise, and it has the automorphism $(\alpha,\beta)\longmapsto (\beta,\alpha)$. If you had your unitary, you would have $UpU^*=1-p$, impossible since $p$ is finite and $1-p$ is infinite. The automorphism is continuous in any reasonable topology.

Note that the example above even works in finite dimension, just by choosing the projection $p\in M_n(\mathbb C )$ with $\operatorname{Tr}(p)\ne n/2$.

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  • $\begingroup$ Completely clear. Thanks a lot! $\endgroup$ – Diego Jun 12 at 22:34
  • $\begingroup$ Please consider upvoting and not only my answer, but anything that you find useful on this site; that's how it works. $\endgroup$ – Martin Argerami Jun 13 at 3:28
  • $\begingroup$ Done! Thanks for point that out. $\endgroup$ – Diego Jun 13 at 13:30

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