Timeline for The quotients of double cosets $P_\theta \backslash P_\theta w P_\Omega$ are algebraic varieties over $k$
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| Sep 1, 2023 at 13:54 | vote | accept | D_S | ||
| Aug 19, 2023 at 18:40 | answer | added | LSpice | timeline score: 2 | |
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| Dec 12, 2018 at 23:01 | history | edited | Michael Hardy | CC BY-SA 4.0 |
deleted 23 characters in body; edited title
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| Dec 12, 2018 at 16:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 14, 2018 at 15:38 | comment | added | LSpice | (In fact, I see that you have already discussed this analogue of the big cell at another question.) | |
| Nov 14, 2018 at 15:37 | comment | added | LSpice | (original) In particular, it can be used to compute dimensions. In fact, it can maybe be leveraged to understand the structure of the general quotient in a way reminiscent of the proof of existence of a reductive group with a given root datum; I don't know. Anyway, I will think further about your question about existence of rational points. | |
| Nov 14, 2018 at 15:36 | comment | added | LSpice | (Commenting here rather than on my answer because it's wrong and I'll eventually delete it.) As you mention, the direct-product decomposition Casselman claims on p. 12 can fail, but nonetheless it is true that the natural map $\prod N_\alpha \to P_\Theta\backslash P_\Theta w P_\Omega$ is an embedding with open image. (This is some sort of general-parabolic analogue of the 'big cell'.) (rest) | |
| Nov 12, 2018 at 15:40 | comment | added | LSpice | The choice of representative only conjugates the embedding $w^{-1}P_\theta w \cap P_\Omega \to P_\Omega$, hence cannot change parabolic-ness. I agree with you that the fixed-point theorem doesn't seem to say anything about cohomology; I was conflating it with the statement that $(G/P)(k) = G(k)/P(k)$, the latter of which may (I'm not sure) most easily be proven using a Bruhat-type decomposition and the fact about lifting of rational points that is the subject of your question! | |
| Nov 10, 2018 at 21:38 | comment | added | D_S | By the way, how can you get the injectivity of $H^1(k,P) \rightarrow H^1(k,G)$ from the Borel fixed point theorem? | |
| Nov 10, 2018 at 21:07 | comment | added | D_S | Maybe it is not true for general $w$? But Casselman does make a canonical choice of representative $w_0$ for $W_{\theta} w W_{\Omega}$, namely the unique one of minimal length. Maybe in this case $w_0^{-1}P_{\theta}w_0 \cap P_{\Omega}$ is parabolic. | |
| Nov 9, 2018 at 17:18 | comment | added | LSpice | I think that it's a consequence of Borel's fixed-point theorem that the map $\mathrm H^1(k, P) \to \mathrm H^1(k, G)$ always has trivial fibres for $P$ a parabolic subgroup of $G$. However, $w^{-1}P_\theta w \cap P_\Omega$ isn't necessarily a parabolic subgroup of $P_\Omega$, so I can't see how to use that here. | |
| Nov 9, 2018 at 17:15 | comment | added | LSpice | Proposition 6.6 of Borel discusses general quotients of this shape. As usual for abstract groups, one may identify the quotient with $w^{-1}P_\theta w \cap P_\Omega\backslash P_\Omega$, and similarly for the quotient on the level of rational points. Therefore, the question becomes whether $\mathrm H^1(k, w^{-1}P_\theta w \cap P_\Omega) \to \mathrm H^1(k, P_\Omega)$ has trivial fibres. | |
| Nov 9, 2018 at 16:55 | history | asked | D_S | CC BY-SA 4.0 |