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Let $M$ be a subset of $B(\mathcal{H})$ (the space of bounded linear operators) such that $M'$ is a von Neumann algebra.

As we know if $M$ is invariant under involution, then $M'$ is a von Neumann algebra. My question is about the converse of it. Is $M$ invariant under $*$-operation, if $M'$ is a von Neumann algebra?

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  • $\begingroup$ What about $M=\{\lambda I\}$, where $\lambda\in\mathbb C\setminus\mathbb R$? $\endgroup$ – Aweygan Feb 26 '18 at 5:11
  • $\begingroup$ @Aweygan, Since $M=\{\lambda I\}_{\lambda\in \Bbb C - \Bbb R}$ is invariant under involution, $M'$ is a von Neumann algebra. $\endgroup$ – saeed Feb 26 '18 at 5:15
  • $\begingroup$ No, not $M=\{\lambda I:\lambda\in\mathbb C\setminus\mathbb R\}$, I mean fix some $\lambda\in\mathbb C\setminus\mathbb R$, and put $M=\{\lambda I\}$. In this case $M'=B(\mathcal H)$, but $M$ is not closed under involution $\endgroup$ – Aweygan Feb 26 '18 at 5:17
  • $\begingroup$ @Aweygan But it is not my question. I need to prove or give a counterexample for the convers $\endgroup$ – saeed Feb 26 '18 at 5:17
  • $\begingroup$ @Aweygan Thanks. it is a nice example. Would you please give me a non proper example? $\endgroup$ – saeed Feb 26 '18 at 5:19
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Fix some $T\in B(\mathcal H)$ that is normal but not self-adjoint, and put $M=\{T\}$. Then $M'$ is a von Neumann algebra, but $M$ is not self-adjoint.

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