I learned the following fact from Bruhat and Tits's paper "Homomorphismes “abstraits” de groupes algebriques simples" Section 3.18 that

Let $k$ be a local field. Suppose that a $k$-group $H$ acts $k$-rationaly on a $k$-variety $M$ and $x$ is an element of $M(k)$ such that the map $h\mapsto hx$ for $h\in H$ of the group $H$ onto the orbit $Hx$ is separable . Then $H(k)x$ is locally closed in $M(k)$ with respect to metric Hausdorff topology.

Separable means its differential is surjective. I think any morphism is separable in characteristic 0. I have seen some examples for which the action is not separable, but all orbit are locally closed with respect to metric Hausdorff topology.

Q1: Is there a weaker (than separable) condition to ensure that all orbits are locally closed?

Q2: Do we know any examples for which there exists a non-locally closed orbit with respect to metric Hausdorff topology in positive characteristic case? It will be more interesting to have such action, which is also linear.

Thanks to user81562, I think we have the following:

If $F$ is a non-archimedean local field and $X$ algebraic variety defined over $F$. If $G$ is a linear algebraic $F$-group and $G\times X \rightarrow X$ is $F$-rational action. Then the action is constructive see Thm A on page 57 and in particular, all its orbits are locally closed (see proposition 6.8 c in http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf).


1 Answer 1


The orbits are always locally closed for the Hausdorff topology, even in positive characteristic. This follows from the appendix in the paper of Bernstein and Zelevinskii, Representations of the group GL(n,F), where F is a local non-Archimedean field. Uspehi Mat. Nauk 31 (1976), no. 3(189), 5–70.


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