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Given a Hilbert space $\mathcal{H}$.

Consider a von Neumann algebra: $$M\subseteq\mathcal{B}(\mathcal{H}):\quad M=M''$$

Suppose a cyclic vector: $$\Omega\in\mathcal{H}:\quad\overline{\mathcal{M}\Omega}=\mathcal{H}$$

Regard the involution: $$S_0:\mathcal{M}\Omega\to\mathcal{H}:\quad S_0M\Omega:=M^*\Omega$$ This operator is closable!

But I didn't succeed checking this?
What if there's no cyclic vector?

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The arguments I know require that $\Omega$ is both cyclic and separating (i.e., also cyclic for the commutant).

When you only require that $\Omega$ is separating, I don't think that $S_0$ even makes sense. For instance, let $\mathcal M=B(\mathcal H)$, $\Omega=e_1$. Consider the (constant) sequence $E_{12}e_1=0$. Then $S_0E_{12}e_1=E_{21}e_1=e_2\ne0$, so $S_0$ is not even well-defined.

Now, if $\Omega$ is cyclic and separating, the map $S_0$ is well-defined, because if $M\Omega=0$, then $M=0$ by the separating property.

The key to the proof is to also consider the related map $F_0:\mathcal M'\Omega\ni N\Omega\longmapsto N^*\Omega\in \mathcal H$.

Note also that $S_0$ and $F_0$ are conjugate-linear (so some inner products below will look like they are the wrong way).

We have, for $M\in\mathcal M$ and $N\in\mathcal M'$, $$ \langle S_0M\Omega,N\Omega\rangle= \langle M^*\Omega,N\Omega\rangle=\langle N^*M^*\Omega,\Omega\rangle=\langle N^*\Omega,M\Omega\rangle. $$ This implies that $N\Omega$ is in the domain of $S_0^*$ and that $S_0^*N\Omega=N^*\Omega$. So $S_0^*$ is a (closed) extension of $F_0$. Then $S_0^*$ is densely defined, and it follows that $S_0$ is closable.

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  • $\begingroup$ Before, thanks for the nice answer!!! Now, why does it imply that the adjoint is densely defined? The domain of the related operator, i.e. the commutant, may be trivial, or? $\endgroup$ – C-Star-W-Star May 9 '16 at 13:33
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    $\begingroup$ You are welcome. As I mentioned somewhere in the answer, the key is that $\Omega$ is also separating, which implies it is cyclic for the commutant; so $\{N\Omega:\ N\in\mathcal M'\}$ is dense. $\endgroup$ – Martin Argerami May 9 '16 at 15:48
  • $\begingroup$ I saw it now - my mistake. But I couldn't check it yet by scratch. I will simply open a new thread on this. Thanks again alot!! :) $\endgroup$ – C-Star-W-Star May 10 '16 at 10:01

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