Let $M\subset B(\mathcal{H})$ be an infinite dimensional vN algebra in standard form. Fix $\xi\neq 0 \in \mathcal{H}$, does there exist $M\ni x_{\xi}\neq I$ such that $x_{\xi}(\xi)=\xi$?
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This is false. Consider, for example, the case of M being the von Neumann algebra of bounded complex-valued functions on an infinite countable set I. It acts on the Hilbert space of square-summable functions on I, which is the standard form of M. Take any ξ∈H that is everywhere nonvanishing (e.g., n↦1/n). Then any x∈M such that xξ=ξ must satisfy x=1.
Much more generally, for any M that admits a faithful finite trace τ, the element ξ=τ^{1/2} is an element in the standard form of M and is a counterexample to your claim: if xξ=ξ, then (x−1)ξ=0, so rsupp(x−1)lsupp(ξ)=0, but lsupp(ξ)=lsupp(τ^{1/2})=1, hence rsupp(x−1)=0, i.e., x−1=0 and x=1.
answered Sep 23, 2019 at 19:06
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$\begingroup$ @Dimitri, If I assume $M$ is the hyperfinite $\mathrm{II_{1}}$ factor, then still false!! $\endgroup$– sibaniCommented Sep 23, 2019 at 19:09
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1$\begingroup$ $I$ must be countable for this to work, or else there is no nowhere vanishing $\xi$. $\endgroup$ Commented Sep 23, 2019 at 19:35
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2$\begingroup$ @Desperatemathematician: For a II_1 factor M you can construct a counterexample by taking a faithful finite trace τ on M, then ξ=τ^{1/2} is an element in the standard form of M that is a counterexample to your claim: if xξ=ξ, then (x−1)ξ=0, so rsupp(x−1)lsupp(ξ)=0, but lsupp(ξ)=lsupp(τ^{1/2})=1, hence rsupp(x−1)=0, i.e., x−1=0 and x=1. $\endgroup$ Commented Sep 23, 2019 at 22:41