2
$\begingroup$

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 $P_{\xi}=I$.

$\endgroup$
1
$\begingroup$

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 $x=1$.

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

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.