Timeline for $K$-Theory of finite dimensional Banach algebras
Current License: CC BY-SA 3.0
15 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Feb 15, 2016 at 13:59 | vote | accept | Ali Taghavi | ||
| Feb 14, 2016 at 6:42 | answer | added | David Handelman | timeline score: 5 | |
| Jan 30, 2016 at 10:40 | comment | added | Ali Taghavi | @EricWofsey thanks for your comment. do you think that the space of strictly upper triangle matrices gives finite non trivial k_ group? | |
| Jan 30, 2016 at 10:17 | comment | added | Eric Wofsey | If you allow non-unital algebras, it is easy to find a finite-dimensional algebra $A$ such that $K_0(A)$ is trivial (for instance, let $A=\mathbb{C}$ with the zero multiplication; then over $\tilde{A}=\mathbb{C}[x]/(x^2)$ all projective modules are free). Whether $K_0(A)$ can be finite but nontrivial seems a lot harder. | |
| Jan 30, 2016 at 9:31 | comment | added | Ali Taghavi | @AlainValette Prof Valette, What is $K_{0}(A)$ where $A=$ is the space of strictly upper triangle matrices ? | |
| Jan 29, 2016 at 21:55 | comment | added | Alain Valette | @Fermando: Because, for non-unital algebras $K_0(A)$ is defined as $\ker[K_0(\tilde{A})\rightarrow K_0(k)=\mathbb{Z}]$, where $\tilde{A}$ is the unitization of $A$, so you don't deal with f.g. projective modules, but with differences of modules having the same dimension. | |
| Jan 29, 2016 at 16:27 | comment | added | Fernando Muro | @AlainValette Why so? | |
| Jan 29, 2016 at 16:11 | comment | added | Alain Valette | @Fernando: I think you are implicitly assuming that your algebra has a unit. | |
| Jan 29, 2016 at 13:49 | review | Close votes | |||
| Jan 29, 2016 at 20:57 | |||||
| Jan 29, 2016 at 13:44 | comment | added | Fernando Muro | I'm using that the algebra has finite dimension so that any f.g. projective module too, over the ground field. | |
| Jan 29, 2016 at 8:25 | comment | added | Ali Taghavi | @FernandoMuro where are you using finite dimensionality?As we know there are algebras with finite k group? | |
| Jan 29, 2016 at 7:41 | comment | added | Fernando Muro | No, not even removing the word Banach, taking dimension over the ground field defines a surjection onto an infinite cyclic group. | |
| Jan 29, 2016 at 7:36 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
deleted 15 characters in body
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| Jan 29, 2016 at 5:18 | history | asked | Ali Taghavi | CC BY-SA 3.0 |