Timeline for Prove that a Verma module is projective only if its highest weight is dominant?
Current License: CC BY-SA 3.0
11 events
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| Oct 30, 2015 at 12:32 | comment | added | Jim Humphreys | @Tobias: Yes, it's always a problem with placement of exercises. As usual my excuse is lack of opportunity to work through the material in the book with actual students, so the feedback comes later. | |
| Oct 30, 2015 at 9:31 | comment | added | Tobias Kildetoft | @JimHumphreys Ahh, that explains why I could not make my hint into a full solution (I wrote the original hint before I had a closer look at what was presented before the exercise). | |
| Oct 29, 2015 at 14:58 | comment | added | Zhihua Chang | @JimHumphreys Thank you very much! I will continue reading the book. | |
| Oct 29, 2015 at 14:52 | comment | added | Jim Humphreys | @Zhihua: Maybe I can prevent some confusion by pointing to the revision list for my book, where this misplaced exercise is rewritten. It's much easier to prove in the stronger form after section 4.7. See the current list of revisions at the AMS page ams.org/publications/authors/books/postpub/gsm-94 | |
| Oct 29, 2015 at 10:14 | comment | added | Tobias Kildetoft | Hmm, I was thinking of using Proposition 1.4, but you need some integrality for that. | |
| Oct 29, 2015 at 10:09 | comment | added | Zhihua Chang | @TobiasKildetoft Indeed, this is my question. I do not know which $L(\mu)$ does appear as a composition factor of $M(\lambda)$. Is it possible that $M(\lambda)$ is irreducible when $\lambda$ is dominant? | |
| Oct 29, 2015 at 9:46 | comment | added | Tobias Kildetoft | Not $M(\lambda)$ but $M(\mu)$. Think of what submodules are in Verma modules. | |
| Oct 29, 2015 at 9:44 | comment | added | Zhihua Chang | @TobiasKildetoft Thanks for the comments. I know that if $L(\mu)$ is a composition factor of $M(\lambda)$, then $\mu\leq \lambda$. But why $M(\lambda)$ has a composition factor $L(\mu)$ with $\mu\neq\lambda$? | |
| Oct 29, 2015 at 9:28 | comment | added | Tobias Kildetoft | Hint: If $\lambda$ is not dominant, then there is some $\mu\neq \lambda$ such that $L(\lambda)$ is a composition factor of $M(\mu)$. Now apply BGG reciprocity. | |
| Oct 29, 2015 at 9:26 | review | Close votes | |||
| Oct 29, 2015 at 16:50 | |||||
| Oct 29, 2015 at 7:31 | history | asked | Zhihua Chang | CC BY-SA 3.0 |