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?



Orbits of unipotent groups over local fields are closed?
Edit: This answer is valid only in characteristic 0!


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. 


