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Let $k$ be a field, $X$ an algebraic variety, and $G$ a smooth algebraic group, acting on $X$ via $(g,x)\mapsto g\cdot x$.

Fixing $x$ in $X$ a $k$-point, there is a map $f_x:G\rightarrow X$ sending $g\mapsto g\cdot x$.

Now, assuming that $df_x:T_eG\rightarrow T_xX$ is surjective, how can one show that the orbit $G\cdot x$ is open in $X$?

NB: I expect that some restrictions on the field are necessary, but I'd like to keep them as minimal as possible!

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    $\begingroup$ What if the field has characteristic $p$, the variety $X$ is the multiplicative group, and the group is $\mu_p$? $\endgroup$ Oct 25 at 15:56
  • $\begingroup$ Is the derivative really surjective in this case? In any case, I've added a smoothness hypothesis! $\endgroup$ Oct 25 at 16:04
  • $\begingroup$ @ThibaultDécoppet, assuming that the action is multiplication and $x$ is the identity, the derivative in @‍JasonStarr's example is the inclusion of Lie algebras $\operatorname{Lie}(\mu_p) \to \operatorname{Lie}(\operatorname{GL}_1)$, which is an isomorphism precisely when $p$ is the characteristic. But this example does not qualify because $G = \mu_p$ is not smooth. $\endgroup$
    – LSpice
    Oct 25 at 16:40
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$\DeclareMathOperator\Im{Im}$I don't think that smoothness of $X$ is necessary, just the fact that it is geometrically integral (I assume your varieties are geometrically integral). (As noted by Jason Starr, smoothness of $G$ is necessary.)

Here is a possible argument, please let me know if there are any mistakes.

As noted by AlexIvanov, it suffices to prove openness after passing to the algebraic closure of $k$, so we can assume that $k$ is algebraically closed.

Let $Z$ be the scheme theoretic closure of $G$ in $X$. The set theoretic image $\Im(G)$ of $G$ is open inside $Z$ by Chevalley's theorem and translating using the group action. After replacing $X$ with the $G$-stable open subvariety $X \setminus (Z \setminus \Im(G))$, we can assume that the natural map $G \to Z$ is surjective, in other words the orbit is closed. What we need to show at this point is that $Z = X$. We will argue by contradiction. Assume $Z\subset X$ is a proper closed subset.

We know that the tangent space of $Z$ at the base point $x$ agrees with that of $X$ (at $x$) by assumption. Using the fact that $k$ is algebraically closed, we can translate to see that the tangent space of $Z$ at any of its closed points coincides with that of $X$. But note that since $G$ is reduced (because $G$ is smooth), $Z$ is reduced. Since $k$ is algebraically closed, $Z$ is smooth over an open subset. In other words, the dimension of the tangent space of $Z$ for all points in an open dense agrees with the Krull dimension of $Z$. If $Z \subset X$ is a proper closed subset, the dimension of $Z$ is strictly smaller than that of $X$. This is a contradiction, because the dimension of the tangent space of $X$ at any closed point is at least the Krull dimension of $X$.

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  • $\begingroup$ Where are you using the geometrically integral hypothesis? It looks to me like your are not using it! $\endgroup$ Oct 26 at 7:14
  • $\begingroup$ @ThibaultDécoppet It's used in the last part. First, since $X$ is geometrically integral, when "reducing" to $k$ algebraically closed, you still have that $X$ is integral. Then, in the last part, the statement that a proper closed subset (closed is missing) has smaller dimension than $X$ is false for disconnected $X$, but also for non-reduced $X$. $\endgroup$ Oct 26 at 8:33
  • $\begingroup$ Ok, right, but can't we just reason in irreducible components for the last step? As opposed to having to assume that $X$ is geometrically integral? $\endgroup$ Oct 26 at 10:08
  • $\begingroup$ @ThibaultDécoppet thank you for your comment, I think that you are right. Geometric irreducibility isn't needed as far as I can tell. We can first focus on the neutral component of $G$, which is irreducible, so the (scheme theoretic) orbit (over algebraic closure) will be irreducible. You can apply the reasoning above for each separate irreducible component containing it. This shows that the neutral component orbit is open. Then the rest of the connected components of the orbit (over algebraic closure) are translates, so open as well. $\endgroup$
    – afh
    Oct 26 at 12:03
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    $\begingroup$ @AriyanJavanpeykar Thanks, I added the "closed" in the last paragraph to clarify. $\endgroup$
    – afh
    Oct 26 at 12:19
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If $X$ is additionally assumed to be smooth, the orbit $G\cdot x$ is open in $X$ under the assumption that $T_eG \rightarrow T_xX$ is surjective.

Namely, the property "being an open subset" satisfies fpqc-descent, so we may (using the fpqc-cover $X \otimes_k k^\text{alg} \rightarrow X$) wlog assume that $k$ is algebraically closed. The condition imposed on tangent spaces implies that the map on tangent bundles $\mathcal{T}_G \rightarrow f_x^\ast\mathcal{T}_X$ induced by $f_x \colon G \rightarrow X$ is surjective in the fiber at $e$. This surjectivity extends to an open subset of $G$, i.e., there is some open $U\subseteq G$, such that $f_x|_U \colon U \rightarrow X$ induces surjections on tangent spaces at each point of $U$. By the last part of Alexander Bett's answer to Smooth morphism (algebraic geometry) vs. Submersion (differential geo) & Ehresman's Lemma (essentially, miracle flatness is used), $f_x|_U \colon U \rightarrow X$ is flat and as it is also of finite presentation (being a map between $k$-varieties), it follows that its image is open in $X$. Thus the $G$-orbit $G\cdot x$ contains an open subset of $X$. On the other side, $G\cdot x$ is a locally closed subset of $X$ (this holds for any action of a smooth algebraic group on a variety). Now, a locally closed subset of an irreducible variety (we may wlog assume that $X$ is irreducible) containing an open subset must be open itself.

I am not sure what happens for non-smooth $X$, but the only place where smoothness was used above is the application of miracle flatness. So, the smoothness hypothesis probably can be weakened.

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  • $\begingroup$ Very nice! Thanks a lot for your answer! Would you happen to know of a reference in the litterature? $\endgroup$ Oct 26 at 5:18
  • $\begingroup$ @AlexIvanov, oops, sorry for the typo I introduced in my edit. $\endgroup$
    – LSpice
    Oct 27 at 4:03

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