## What is a closed orbit that is not separable? (ANSWERED)

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What is an example of an action of a linearly reductive group variety acting on an affine variety with the property that there exists a closed orbit that is not separable?

To be more precisely, let's work over a fixed algebraically closed field $k$. Suppose that we are given an affine variety $X$ and a group variety $G$ acting on $X$. Given a closed point $x \in X$, we define the orbit $\operatorname{O}(x)$ to be the image of the map $G \to X$ given by $g \mapsto g x$ . We say that the orbit is separable if the natural map $$G \to \operatorname{O}(x)$$ given by $g \mapsto gx$ is separable.

This question is only interesting in characteristic $p>0$. In this case, the condition that $G$ is linearly reductive is very strong: it implies $G$ is the product of a multiplicative torus and a finite group of order prime-to-$p$.

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This happens for every $G$ of positive dimension, since all one needs is a non-smooth subgroup scheme (such as kernel of Frobenius). Indeed, if $H$ is a closed $k$-subgroup scheme of $G$ then let $X = G/H$ equipped with the natural left $G$-action (so $X$ is smooth, since $G$ is smooth, regardless of how "bad" $H$ may be). Then the orbit map through $x = 1$ is the natural surjection $G \rightarrow X$ which is not separable precisely when $H$ is not smooth. The simplest example is $G = {\rm{GL}}_1$ acting on $X = G$ via $t.x = t^p x$, whose orbit map through $x = 1$ is $t \mapsto t^p$. – BCnrd Sep 12 2010 at 5:39
Thanks! The formulation in terms of the stabilizer really clarifies things! – jlk Sep 12 2010 at 6:57
Can I suggest that @BCnrd leave his answer as an answer, so that the question can be marked as "answered" rather than have it expressed so in the title? I know that BCnrd doesn't like to leave answers, so another option is for OP to leave copy the answer into the answers (and mark the answer as CW if you want to make sure you don't pick up points for someone else's answer). I would do so, but I wanted to give BCnrd a chance. – Theo Johnson-Freyd Sep 13 2010 at 2:37
Believe me, Theo, you are far from the first to have suggested that. – Ben Webster♦ Sep 13 2010 at 3:56
@BCnrd, Ben Webster submitted your answer, and I have accepted it so the problem will be marked "Answered." Please let me know if you prefer that I not do this. – jlk Sep 13 2010 at 4:28

## 1 Answer

BCnrd writes: "This happens for every G of positive dimension, since all one needs is a non-smooth subgroup scheme (such as kernel of Frobenius). Indeed, if H is a closed k-subgroup scheme of G then let X=G/H equipped with the natural left G-action (so X is smooth, since G is smooth, regardless of how "bad" H may be). Then the orbit map through x=1 is the natural surjection G→X which is not separable precisely when H is not smooth. The simplest example is G=GL(1) acting on X=G via $t\cdot x=t^px$, whose orbit map through $x=1$ is $t\mapsto t^p$."

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Here is your obligatory +1. – S. Carnahan♦ Sep 13 2010 at 4:09
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