Let $k$ be a $p$-adic field, $G$ a connected reductive group over $k$ with minimal parabolic $P_0$ containing a maximal split torus $A_0$. Let $W = N_G(A_0)(k)/Z_G(A_0)(k)$ be the Weyl group, and $S \subset W$ the simple reflections from $P_0$.

For $\theta, \Omega \subset S$, we have the standard parabolic subgroups $P_\theta, P_\Omega$.

Each double coset $P_{\theta}w P_{\Omega}$, for $w \in W$ with a $k$-rational representative, is a locally closed subvariety of $G$, with $P_\theta wP_\Omega(k) = P_\theta(k)wP_\Omega(k)$.

How do we know that the quotient $P_\theta \backslash P_\theta wP_\Omega$ is an algebraic variety over $k$? This variety and its dimension are considered in Casselman's notes on representation theory, Chapter 6. I do not know the general theory of quotients of algebraic group actions which would make sense out of something like this.

Once this variety is made sense of, can we say that its $k$-rational points coincides with $P_\theta(k) \backslash P_\theta(k) wP_\Omega(k)$?

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    $\begingroup$ 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. $\endgroup$ – LSpice Nov 9 '18 at 17:15
  • $\begingroup$ 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. $\endgroup$ – LSpice Nov 9 '18 at 17:18
  • $\begingroup$ 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. $\endgroup$ – D_S Nov 10 '18 at 21:07
  • $\begingroup$ 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? $\endgroup$ – D_S Nov 10 '18 at 21:38
  • $\begingroup$ 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! $\endgroup$ – LSpice Nov 12 '18 at 15:40

Casselman's notes which you have cited, p. 12, show that the natural map $\prod N_\alpha \to P_\Theta\backslash P_\Theta w P_\Omega$, where the product runs over the obvious collection of roots, is an isomorphism of varieties. (In particular, you may regard $\prod N_\alpha$ as the quotient of $P_\Theta w P_\Omega$ by $P_\Theta$, thus explaining its structure as a variety.) Via the resulting $P_\Theta$-equivariant splitting $P_\Theta w P_\Omega \to \prod N_\alpha$, you can see that every rational point of the quotient lifts to a rational point of the product of root groups.

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    $\begingroup$ I think the result about $P_{\Theta}wP_{\Omega}$ being a direct product in that way is false (mathoverflow.net/questions/306675/…) although it is true when $\Theta$ and $\Omega$ are empty $\endgroup$ – D_S Nov 12 '18 at 15:43
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    $\begingroup$ In fact, I think it is true when either $\Theta$ or $\Omega$ is empty; but I see that you're right that it is not true in general. I will think further. $\endgroup$ – LSpice Nov 14 '18 at 17:10

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