Timeline for Irreducible quotient of $U\otimes V$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2013 at 6:54 | comment | added | Tobias Kildetoft | Given the new formulation, I would say this is not a research level question. The tensor product is a direct sum of simple modules, and the highest weights of those are given by the Littlewood-Richardson rule. Also, the formulation is a bit strange, as a finite dimensional highest weight module is the same as a finite dimensional simple module. | |
| Aug 7, 2013 at 19:23 | comment | added | Jim Humphreys | P.S. Given the new formulation, complete reducibility makes it easy to control the highest weight occurring in such a tensor product. So there seems to be no real problem here. | |
| Aug 7, 2013 at 19:01 | history | edited | Yilan Tan | CC BY-SA 3.0 |
More details
|
| Aug 7, 2013 at 16:08 | comment | added | Jim Humphreys | The formulation is a bit loose and symbols undefined, as comments indicate. In any case, you should specify what kind of field you are working over and whether any of the modules is assumed to be finite dimensional. (And some motivation for the question might help.) | |
| Aug 7, 2013 at 13:49 | history | edited | Yilan Tan | CC BY-SA 3.0 |
More details.
|
| Aug 7, 2013 at 9:50 | answer | added | Alexander Premet | timeline score: 1 | |
| Aug 7, 2013 at 7:04 | comment | added | Tobias Kildetoft | Why will that tensor product have a unique irreducible quotient? | |
| Aug 7, 2013 at 6:37 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
added missing dollar signs for math; fixed mistake introduced by my previous edit
|
| Aug 7, 2013 at 6:18 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
removed deprecated tag 'algebra'; replaced tags with more appropriate ones; added question marks to question
|
| Aug 7, 2013 at 4:59 | review | First posts | |||
| Aug 7, 2013 at 6:03 | |||||
| Aug 7, 2013 at 4:41 | history | asked | Yilan Tan | CC BY-SA 3.0 |