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Sep 24, 2019 at 19:36 history edited Dmitri Pavlov CC BY-SA 4.0
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Sep 23, 2019 at 22:41 comment added Dmitri Pavlov @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.
Sep 23, 2019 at 19:35 comment added Robert Furber $I$ must be countable for this to work, or else there is no nowhere vanishing $\xi$.
Sep 23, 2019 at 19:09 comment added sibani @Dimitri, If I assume $M$ is the hyperfinite $\mathrm{II_{1}}$ factor, then still false!!
Sep 23, 2019 at 19:08 vote accept sibani
Sep 23, 2019 at 19:06 history answered Dmitri Pavlov CC BY-SA 4.0