Orbits of unipotent groups over local fields are closed?

Let $H$ be a connected, unipotent linear algebraic group defined over a local field $k$. Let $H \times_k X \rightarrow X$ be an action of $H$ on an irreducible, affine $k$-variety $X$ which is defined over $k$. Then the orbits of the group action $H(\overline{k}) \times X(\overline{k}) \rightarrow X(\overline{k})$ are Zariski-closed in $X(\overline{k})$.

What about the orbits of the group action $H(k) \times X(k) \rightarrow X(k)$? By the general theory of analytic manifolds, the orbits are locally closed subvarieties of $X(k)$. Are these orbits always closed, as in the algebraically closed case?

• Can you further explain how this relates to Kostant-Rosenlicht (1961, Thm 2); Birkes (1971, Thm 12.1); Borel (1991, Prop. 4.10)? – Francois Ziegler Mar 6 '18 at 8:16
• Sorry, I am not sure what you are asking. Those theorems just say that the orbits of $X(\overline{k})$ under a unipotent group are Zariski closed, right? – D_S Mar 6 '18 at 16:54
• Just curious and unsure how algebraic closure enters, in these refs. – Francois Ziegler Mar 6 '18 at 18:17
• If $V \subseteq \mathbb A_k^n$ is an affine scheme of finite type over a field $k$, then $V(\overline{k})$ is an affine variety in the classical sense, i.e. the set of zeroes in $\mathbb A_k^n(\overline{k}) = \overline{k}^n$ of some set of polynomials. In Borel and the other references, they are working with classical varieties. So when they talk about a variety $V$, their $V$ is really our $V(\overline{k})$. – D_S Mar 6 '18 at 19:30
• I see; thank you! – Francois Ziegler Mar 7 '18 at 12:23

Yes. For a unipotent group $H$ the orbits of $H(k)$ in $X(k)$ coinside with the intersections of the form $H(\bar{k}).x \cap X(k)$, hence even Zarizki closed. In general, the orbits of $H(k)$ on sets of the form $H(\bar{k}).x \cap X(k)$ are classified by the first cohomology $H^1(k,H_x)$ for $H_x$ the stabilizer of $x$ in $H$. But Galois cohomology of unipotent groups vanish. This is a generalization of the fact that $H^1(k,\mathbb{G}_a) = \{1\}$ and in fact follows from this, since every unipotent group has a filtration with associated graded consist of several copies of $\mathbb{G}_a$, so by induction and some long exact sequences it follows.
• Thank you. Does the cohomology set $H^1(k,H_x)$ arise from looking at the exact sequence of pointed sets $1 \rightarrow H_x(\overline{k}) \rightarrow H(\overline{k}) \rightarrow H(\overline{k}).x \rightarrow 1$? – D_S Mar 6 '18 at 6:40
• 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. – S. carmeli Mar 6 '18 at 6:57
• 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. – nfdc23 Mar 6 '18 at 7:12
• 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 $k$. – Laurent Moret-Bailly Mar 6 '18 at 16:40