Timeline for Representations are determined by characters : Groups and Lie algebras
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 14, 2020 at 9:02 | history | bounty ended | CommunityBot | ||
| S Jan 14, 2020 at 9:02 | history | notice removed | CommunityBot | ||
| S Jan 6, 2020 at 7:50 | history | bounty started | GA316 | ||
| S Jan 6, 2020 at 7:50 | history | notice added | GA316 | Draw attention | |
| Jan 3, 2020 at 16:52 | answer | added | Bugs Bunny | timeline score: 5 | |
| Dec 28, 2019 at 18:31 | comment | added | GA316 | @BenjaminSteinberg. Thanks. But I am interested in general Kac moody setup where algebras are infinite-dimensional mostly. | |
| Dec 28, 2019 at 15:37 | comment | added | Benjamin Steinberg | For finite dimensional algebras over algebraically closed fields, simple modules are characterized up to isomorphism by their characters. Probably algebraically closed is not needed. | |
| Dec 28, 2019 at 13:29 | comment | added | GA316 | @BertramArnold Such series exists for infinite-dimensional modules as well? because I remember for arbitrary highest weight modules of Kac-Moody algebras such series doesn't exist. Kindly explain to me more. thanks. | |
| Dec 28, 2019 at 10:43 | comment | added | Bertram Arnold | Characters are additive in short exact sequences, so the character can't tell the difference between a direct sum and a nontrivial extension. In particular, the character of any module can be written as a sum of characters of simple modules (namely those of its Jordan-Hölder composition series). | |
| Dec 28, 2019 at 8:02 | comment | added | GA316 | @JayTaylor please explain, why is it always possible to write a character (of a module over a Lie algebra) as a sum of characters of irreducibles? $\chi = \chi_1 + \cdots + \chi_n$? | |
| Dec 28, 2019 at 7:41 | comment | added | Jay Taylor | Assume a module $M$ is determined up to isomorphism by its character $\chi$. Write $\chi = \chi_1 + \cdots + \chi_n$ as a sum of irreducibles. Let $M_i$ be a simple module affording $\chi_i$. The direct sum $M_1 \oplus \cdots \oplus M_n$ affords $\chi$ and by assumption is isomorphic to $M$. Hence $M$ is a direct sum of simple modules, so semisimple. Complete reducibility holds. | |
| Dec 28, 2019 at 6:30 | history | asked | GA316 | CC BY-SA 4.0 |