Timeline for Is the category of $Z^* $ algebra equivalent to the category of $C^*$ algebras
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2023 at 8:22 | comment | added | Ali Taghavi | So your answer is actually $A\mapsto A\otimes C_0(X)$. The tensor product of a C^* algebra and a Z^* algebra is again Z^* algebra. X being uncountable is not an approximately sigma compact. After all I am really curious about this category, the category of Z^* algebras. Please see my conversation with David Roberts about a precise structure for the category of C^* algebras(what kind of morphisms we are consideringt, etc). Any way it would be interesting that the two categories would be isomorphic in a reasonable sense | |
| Oct 8, 2023 at 21:28 | comment | added | Nik Weaver | Well, $C_0(X, A\oplus A) \sim C(X,A)$ because $X$ is in bijection with $X \coprod X$ (disjoint union of two copies of $X$). We need $X$ to be uncountable so that anything in $C_0(X,A)$ is zero off of a proper (because countable) subset of $X$, and therefore a zero divisor. | |
| Oct 8, 2023 at 11:11 | comment | added | Ali Taghavi | Do you mean that $C_0(X,A\oplus A)\sim C_0 (X\times X,A)$ hence isomorphic to $C_0(X,A)$ since $X$ homeomorphic to $X\times X$? Or you mean some things else? | |
| May 29, 2023 at 17:30 | comment | added | Ali Taghavi | Thank you and +1 very much for your answer | |
| May 24, 2023 at 12:40 | history | answered | Nik Weaver | CC BY-SA 4.0 |