Timeline for Relative weight lattice
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2018 at 17:12 | vote | accept | CommunityBot | ||
| Dec 13, 2018 at 19:02 | comment | added | LSpice | In fact I'm not quite sure how to make sense of $\check\Lambda_{Z_0(M)}$; according to the text it seems it should be a coweight lattice, but the coweight lattice of a torus is trivial. Anyway, tensoring with $\mathbb Q$ kills all my finite-index worries in the semisimple case; those are just issues with integer lattices, not rational vector spaces. | |
| Dec 13, 2018 at 18:58 | comment | added | LSpice | I don't see this statement on p. 8 of the linked paper. The closest that I see is the claim that $\smash{\check\Lambda}_{Z_0(M)}^{\mathbb Q} \to \smash{\check\Lambda}_{G, P}^{\mathbb Q}$ is an isomorphism. If that's the statement you mean, it is different in many ways from what you said. A probably minor point: the paper's definition of $\Lambda_{G, P}$ is different from yours. (It is a sublattice of $\Lambda_G$, not a quotient.) | |
| Dec 13, 2018 at 18:39 | comment | added | user100841 | this fact is true. I am not able to figure it out. This is written on page 8 of a paper of S.Schieder arxiv.org/abs/1212.6814 . This statement is also there in a paper of Dennis Gaitsgory. | |
| Dec 13, 2018 at 13:54 | comment | added | LSpice | OK, I think that I have fixed the issue. I apologise for rejecting your initial edit, by the way; I thought that you were just deleting my re-phrasing of the problem, and didn't realise that your goal was to harmonise notation. Now I can't undo my reject vote. I'll stick with $\mathrm X^*$, but I have added a note in order hopefully to avoid confusion. | |
| Dec 13, 2018 at 13:53 | history | edited | LSpice | CC BY-SA 4.0 |
Updated to fix my misunderstanding
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| Dec 13, 2018 at 12:31 | history | edited | LSpice | CC BY-SA 4.0 |
Updated to reflect notational inconsistency and misinterpretation of question
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| Dec 13, 2018 at 12:29 | comment | added | LSpice | Hmm, you are right; so I should have passed initially to the adjoint quotient of $G$, in which case, by the same argument, the problem becomes: if $G$ is adjoint, then does $M$ have connected centre? (Of course this destroys my counterexample.) I think that this is still false, but I will consider more. | |
| Dec 13, 2018 at 8:44 | comment | added | user100841 | Thanks for your answer but the weight lattice is not the same as the space of characters, to begin with. | |
| Dec 12, 2018 at 20:59 | review | Suggested edits | |||
| Dec 13, 2018 at 0:22 | |||||
| Dec 12, 2018 at 19:32 | history | answered | LSpice | CC BY-SA 4.0 |