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May 12, 2019 at 15:38 comment added Nik Weaver Any weak* closed ideal of a von Neumann algebra $M$ has the form $pM$ for some central projection $p$. The quotient $M/pM$ is isomorphic to $(1-p)M$. Thus the image of $M$ under any normal (i.e., weak* continuous) $*$-homeomorphism is a von Neumann algebra.
May 12, 2019 at 12:13 comment added Matthias Ludewig I know that there is a unique map to every von-Neumann completion; but how do you see it is surjective?
May 12, 2019 at 5:51 comment added Nik Weaver No way! Every von Neumann completion is a quotient of the universal enveloping algebra. It's as far from being a factor as you can get.
May 12, 2019 at 4:00 history asked Matthias Ludewig CC BY-SA 4.0