Timeline for Orbits of unipotent groups over local fields are closed?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 7, 2018 at 17:03 | comment | added | S. carmeli | I hope im right here. The map $H/(H_x)^0 \to H/H_x$ is a Galois cover with Galois group $H_x / (H_x)^0$. I might use characteristic 0 again here, my intuition is mostly p-adic, but I hope not :-) Anyway, the group $H_x / (H_x)^0$ is finite, and finite Galois cover is always proper. Given this, using the cohomological triviality of $(H_x)^0$ we get that the $k$-orbit in $H / (H_x)^0(k)$ is closed, so its image by the proper projection to $H / (H_x)(k)$ is closed as well. But this is exactly the $H(k)$ orbit (I use here the fact that taking $k$-points of a finite Galois cover gives proper map.) | |
| Mar 7, 2018 at 16:55 | comment | added | D_S | Can you explain a little more? Why is this map proper, and how can this be used to show that $H^1(H_x)$ is trivial from the fact that $H^1(H_x^{\circ})$ is trivial? | |
| Mar 7, 2018 at 7:57 | comment | added | S. carmeli | The connectivity is not a big deal. The map $(H/(H_x)^0)(k) \to (H/H_x)(k)$ is proper, so you can always reduce such a claim to the case where the stabilizer is connected, I think. | |
| Mar 7, 2018 at 2:54 | comment | added | D_S | "every unipotent group has a filtration with associated graded consist of several copies of $\mathbb G_a$..." What about when the unipotent group is not connected? In that case, it is not split | |
| Mar 6, 2018 at 16:40 | comment | added | Laurent Moret-Bailly | It is still true in char. $p>0$ that the orbits of $H(k)$ are closed in $X(k)$ for the valuation topology. First, the algebraic orbit (scheme-theoretic image of the orbit map) is a closed subscheme $Z$ of $X$ by Rosenlicht, so you may replace $X$ by $Z\cong H/H_x$. The projection $H\to X$ is then an $H_x$-torsor, and the result follows from Theorem 1.2 in my paper with O. Gabber and P. Gille (Algebraic Geometry 5 (2014) 573–612). This works over any henselian valued field $k$ whose completion is a separable extension of $k$. | |
| Mar 6, 2018 at 7:25 | history | edited | S. carmeli | CC BY-SA 3.0 |
added 56 characters in body
|
| Mar 6, 2018 at 7:25 | comment | added | S. carmeli | true, this works only for characteristic 0. thanks. | |
| Mar 6, 2018 at 7:12 | comment | added | nfdc23 | The method of proof in this answer is incorrect when ${\rm{char}}(k)=p>0$ since the scheme-theoretic stabilizer $H_x$ is merely a unipotent group scheme and not necessarily smooth nor connected nor (even if smooth and connected) $k$-split. In particular, it has no reason to have a filtration with successive quotient $\mathbf{G}_a$, and over imperfect fields of characteristic $p$ (including local function fields) the degree-1 cohomology of a form of $\mathbf{G}_a$ can be nontrivial and even infinite. | |
| Mar 6, 2018 at 6:57 | comment | added | S. carmeli | more or less, but you need to be careful here. The statement that the orbits of $H(k)$ are classified by the cohomology is stringer than the exactness of the corresponding long sequence of cohomology, as a sequence of pointed sets. The action of $H(k)$ on $(H(\bar{k}).x)(k)$ is important. | |
| Mar 6, 2018 at 6:46 | vote | accept | D_S | ||
| Mar 6, 2018 at 6:40 | comment | added | D_S | Thank you. Does the cohomology set $H^1(k,H_x)$ arise from looking at the exact sequence of pointed sets $1 \rightarrow H_x(\overline{k}) \rightarrow H(\overline{k}) \rightarrow H(\overline{k}).x \rightarrow 1$? | |
| Mar 6, 2018 at 6:33 | history | answered | S. carmeli | CC BY-SA 3.0 |