Timeline for Maximal Submodule of a Verma Module
Current License: CC BY-SA 3.0
13 events
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| Oct 28, 2014 at 13:02 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 17, 2013 at 15:09 | comment | added | Jim Humphreys | @Bib: I'm not requiring $\mu$ to be minimal in such a set of weights, only asserting the existence of a subweight $\mu$ having "bad" $N(\mu)$. Locating such $\mu$ can be subtle. The difference between linkage and strong linkage of weights is crucial here. | |
| Dec 17, 2013 at 10:24 | comment | added | Bib | You mention a converse statement in your 4th paragraph : μ has to be a minimal element of { η / [M(η);L(ν)]>1 } or am i wrong ? Thanks | |
| Dec 9, 2013 at 9:44 | vote | accept | Bib | ||
| Dec 9, 2013 at 0:07 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 8, 2013 at 14:29 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 5, 2013 at 14:01 | comment | added | Jim Humphreys | @Vit: Here I only look at resolutions of simple highest weight modules (not necessarily finite dimensional) using direct sums of Verma modules. (See Remark 6.5 in my book for more detaile.) | |
| Dec 5, 2013 at 3:13 | comment | added | Vít Tuček | The answer in general of course depends on what exactly one means by "BGG-type resolution". The only result that goes in this direction that I know of is arxiv.org/abs/math/0604336 But they work in the category $\mathcal{O}$ attached to a parabolic subalgebra with abelian nilradical. | |
| Dec 4, 2013 at 18:45 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 4, 2013 at 17:05 | comment | added | Jim Humphreys | A side remark is that I've never seen usable conditions on a weight determining whether or not $L(\lambda)$ has a BGG-type resolution. There are obvious examples beyond the dominant integral case, but it doesn't seem possible to organize them nicely. Such a resolution would start with the maximal submodule of a Verma module and then try to proceed with Verma submodules. But usually the character formula for $L(\lambda)$ can't be as nice as the one given by Weyl-Kostant. | |
| Dec 4, 2013 at 14:58 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 4, 2013 at 14:06 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 4, 2013 at 13:57 | history | answered | Jim Humphreys | CC BY-SA 3.0 |