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 $x\_{\xi}(\xi)=\xi$?



On existence of certain operators in von Neumann algebra
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. 


