Timeline for Algebraically simple Banach algebras
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2011 at 21:08 | comment | added | Yemon Choi | It may be worth remarking that a notorious open question in Banach algebras asks: does there exist an infinite-dimensional commutative Banach algebra (necessarily non-unital and radical) which is algebraically simple? The question has been open since the 1970s (I think). | |
| Dec 4, 2011 at 16:48 | comment | added | Bill Johnson | Right, Sellapan. There are many Banach algebras that have a largest (necessarily closed) ideal which you can mod out to get an example. | |
| Dec 4, 2011 at 16:03 | answer | added | Alain Valette | timeline score: 9 | |
| Dec 4, 2011 at 13:56 | history | edited | Sellapan Nathan | CC BY-SA 3.0 |
edit
|
| Dec 4, 2011 at 13:53 | comment | added | Sellapan Nathan | Perhaps the Calkin algebrea $B(H)/K(H)$ is a good candidate. | |
| Dec 4, 2011 at 13:43 | comment | added | Sergei Akbarov | Besides this, do you mean algebras with identity? | |
| Dec 4, 2011 at 13:41 | comment | added | Sergei Akbarov | Yes, the ideal $F(H)$ of all finite-dimensional operators in $K(H)$ is non-trivial. But as far as I understand, if you don't claim that $J$ is closed, then $K(H)$ is not a counterexample for you. I.e. $K(H)$ is not algebraically simple... | |
| Dec 4, 2011 at 13:34 | comment | added | Johannes Hahn | @Sergei: I think, that's exactly the point: is this algebraically simple. In other words: Are there non-closed two-sided ideals other than {0} and the whole algebra?? | |
| Dec 4, 2011 at 13:33 | comment | added | Sergei Akbarov | Will the algebra $K(H)$ of compact operators on a Hilbet space satisfy you? It has not non-trivial (closed) two-sided ideals. | |
| Dec 4, 2011 at 13:20 | history | edited | Sellapan Nathan |
cat
|
|
| Dec 4, 2011 at 13:09 | history | asked | Sellapan Nathan | CC BY-SA 3.0 |