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Mar 27, 2020 at 13:34 vote accept Math-Phys-Cat Group
Mar 27, 2020 at 13:10 answer added Nik Weaver timeline score: 2
Mar 27, 2020 at 12:24 comment added Math-Phys-Cat Group I changed again my question. I forgot the commuting requirement. Now $\operatorname{Nil\mathcal{H}}$ is just the vector subspace of bounded operators such that $T^2=0$. This is enough for my problem.
Mar 27, 2020 at 12:20 history edited Math-Phys-Cat Group CC BY-SA 4.0
Fixed grammar and fixed ambiguities on the statement
Mar 26, 2020 at 22:26 comment added Nik Weaver Thank you, but I'm still confused. How do I tell whether $T$ belongs to Nil$(\mathcal{H})$? It has to commute with everything else in Nil$(\mathcal{H})$. So I already need to know what is in Nil$(\mathcal{H})$.
Mar 26, 2020 at 22:02 comment added Math-Phys-Cat Group And I tried to clarify the meaning of "subalgebra of those nilpotent bounded operators which mutually commute" editing my question. Actually, it was bad written. Sorry for this and thanks for the comment.
Mar 26, 2020 at 22:00 history edited Math-Phys-Cat Group CC BY-SA 4.0
added 76 characters in body
Mar 26, 2020 at 21:55 comment added Math-Phys-Cat Group PI = Polynomial Identity.
Mar 26, 2020 at 19:40 comment added Nik Weaver What are "PI's" in the first sentence, and can you clarify what you mean by "subalgebra of those nilpotent bounded operators which mutually commute"?
Mar 26, 2020 at 17:42 history edited YCor CC BY-SA 4.0
added tag, removed capitals
Mar 26, 2020 at 17:37 comment added Math-Phys-Cat Group Thank you for this remark.
Mar 26, 2020 at 17:33 comment added Tomasz Kania All infinite-dimensional separable Hilbert spaces are isometric hence they have the same algebras of bounded operators. As you you look at algebraically defined entities, the question does not depend on the particular incarnation of $H$.
Mar 26, 2020 at 17:30 review First posts
Mar 26, 2020 at 17:49
Mar 26, 2020 at 17:26 history asked Math-Phys-Cat Group CC BY-SA 4.0