Let $K$ be a local field, $G$ a (connected) reductive $K$-group, and $P \le G$ a parabolic subgroup. Is the map $G(K) \rightarrow (G/P)(K)$ necessarily surjective, and, if so, then why?
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5$\begingroup$ @abx: Nope, not so: consider $\mathrm{SL}_n(K) \rightarrow \mathrm{PGL}_n(K)$ (for $K$ of characteristic $0$) for one of many counterexamples. $\endgroup$– Question MarkCommented Oct 18, 2014 at 6:12
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1$\begingroup$ @abx: For a more geometric example consider a smooth projective geometrically connected curve $X_0$ over the finite residue field $k$ of $K$ such that $X_0(k)$ is empty, and let $X$ be a proper flat lift of $X_0$ over $O_K$. The generic fiber $X_K$ is smooth, proper, and geometrically connected curve over $K$ with no $K$-points. $\endgroup$– user27920Commented Oct 18, 2014 at 12:34
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2$\begingroup$ Even more explicit, how about the map $C\to \text{Spec}(\mathbb{R})$ where $C$ is the smooth projective conic defined by $x^2+y^2+z^2=0$? $\endgroup$– Daniel LittCommented Oct 18, 2014 at 21:09
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2$\begingroup$ @DanielLitt, there is probably some sort of personality test inherent in the question of whether your example or QuestionMark's is the first one to come to mind. :-) $\endgroup$– LSpiceCommented Jan 13, 2015 at 20:41
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1$\begingroup$ @LSpice: That's the case for maps which are quotients by free group actions (as in QuestionMark's example) but the situation is rather more complicated in general. The sense in which my example is "cohomological" is that the curve in question exhibits a Severi-Brauer curve whose associated Brauer class (in $H^2(K, \mu_2)$) is non-trivial. $\endgroup$– Daniel LittCommented Jan 15, 2015 at 7:32
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1 Answer
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The map G(K) to (G/P)(K) is surjective over any field K. Here is a link to an explanation by Brian Conrad. http://math.stanford.edu/~conrad/249CS13Page/handouts/parsurj.pdf
answered Oct 18, 2014 at 9:53