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Let $M$ be an infinite dimensional non-type $I$ factor, given $\xi$ in $\mathcal{H}$, does there exist a not identify operator $x$ in $M$ such that $x\xi=\xi$, I have tried with taking projection $P_{\xi}:\mathcal{H}\rightarrow [M'\xi]$, this works unless $P_{\xi}\neq I$, but how to tackle the case when $P_{\xi}=I$.

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No, there does not necessarily exist such an $x$. For example, if $M$ is a $II_1$ factor with trace $\tau$, $\mathcal{H} = L^2(M,\tau)$ and $\xi = 1$ (the identity of $M$, seen in $L^2(M,\tau)$), then $x\xi=\xi$ if and only if $x=1$.

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  • $\begingroup$ That means cyclic vectors for $M'$ is creating the problems $\endgroup$
    – user136400
    Commented Apr 30, 2019 at 7:57

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