Timeline for Weight spaces of representations of finite dimensional simple Lie algebras
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2019 at 6:28 | vote | accept | Ester | ||
| Jul 21, 2019 at 6:27 | comment | added | Ester | @Victor Thanks Sir. Your answer is quite helpful. | |
| Jul 20, 2019 at 23:14 | comment | added | Victor Protsak | @Ester I apologize for the gender mix-up. I have converted my comments into an answer and provided some additional details. This is as far as I am willing to go, and even a bit beyond that. Good luck with your project! | |
| Jul 20, 2019 at 23:08 | answer | added | Victor Protsak | timeline score: 3 | |
| Jul 20, 2019 at 17:01 | comment | added | Ester | Can you please provide a bit more detailed answer?Thank you. | |
| Jul 20, 2019 at 16:50 | comment | added | Victor Protsak | The last statement is a general property of multigraded rings, where the grading abelian group is the root lattice. I don't have Miller-Sturmfels book close at hand, but that will be a natural reference to check. | |
| Jul 20, 2019 at 16:37 | comment | added | Victor Protsak | Yes, we need finite generation of $U$ as a right $U_0$-module. The algebra $U$ is almost commutative (i.e. its associated graded algebra ${\rm gr\,}U$ is commutative and generated by degree 1 part) and the adjoint action of $\frak h$ is semisimple and preserves the filtration. So it is sufficient to prove the corresponding statement for the associated graded algebra: $S_{\lambda}$ is a finitely generated $S_0$-module. | |
| Jul 20, 2019 at 15:58 | comment | added | Ester | No problem at all! | |
| Jul 20, 2019 at 15:52 | history | edited | YCor | CC BY-SA 4.0 |
edited title
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| Jul 20, 2019 at 15:52 | comment | added | YCor | "Finite-dimensional simple Lie algebras" is not specific enough, I edited the title. | |
| Jul 20, 2019 at 14:21 | history | edited | Ester | CC BY-SA 4.0 |
edited title
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| Jul 20, 2019 at 13:00 | history | edited | Jim Humphreys |
edited tags
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| Jul 20, 2019 at 12:55 | answer | added | Jim Humphreys | timeline score: 1 | |
| Jul 20, 2019 at 7:11 | comment | added | Ester | I have been trying to prove that it is finitely generated for a number of days using the PBW theorem, but it has always evaded me. I know that proving your statement will give the desired result(though I needed the fact it is finitely generated as a right module). Can you please provide at least a sketch of the proof of your assertion here or in some other medium?Thank you. | |
| Jul 19, 2019 at 23:42 | comment | added | Victor Protsak | Yes, here is the last instance. The answer is affirmative and follows from the fact that $U_{\lambda}$ is a finitely generated left module over $U_0$, where the subscipts denote the grading of the universal enveloping algebra $U$ with respect to the adjoint action of the Cartan subalgebra. (There is too much notation for me to type a proper answer on a tablet.) | |
| Jul 19, 2019 at 18:20 | history | edited | YCor | CC BY-SA 4.0 |
added tag, removed capitals
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| Jul 19, 2019 at 18:19 | history | edited | Ester | CC BY-SA 4.0 |
deleted 1 character in body
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| Jul 19, 2019 at 18:15 | review | First posts | |||
| Jul 19, 2019 at 19:21 | |||||
| Jul 19, 2019 at 18:12 | history | asked | Ester | CC BY-SA 4.0 |