It is well-known that if a reduced algebraic group $G$ acts on a separated reduced scheme $X$, and $G$ acts trivially on a dense open subscheme $U\subseteq X$, then the action is trivial.

If $X$ is non-reduced, the standard counterexample is $\def\AA{\mathbb A}$$\AA^1$ with an embedded point, with the group acting non-trivially on the embedded point. If $X$ is non-separated, the standard counterexample is $\AA^1$ with a doubled origin with $G$ swapping the two origins. My question is whether the separated hypothesis can be removed if $G$ is assumed to be connected.

Suppose $G$ is a connected group scheme acting on a reduced scheme $X$, and that $G$ acts trivially on a dense open subscheme $U\subseteq X$. Must the action be trivial?

For reference, the basic argument for the original version is that the graphs of the two morphisms $G\times X\to X$ (projection and action) are closed subschemes of $G\times X\times X$ (since the graphs are pullbacks of the diagonal of $X$, which is assumed separated) which share a common dense open (the image of $G\times U$). Since $G\times X\times X$ is reduced, this means that the two graphs agree, so the action and projection maps are the same.

For finite-characteristic people, I actually assume $G$ is reduced, but this is fine since the action on a reduced scheme must factor through the reduction of $G$ anyway.

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    $\begingroup$ "the action on a reduced scheme must factor through the reduction of G anyway" : what about $\alpha_p$ acting on $\mathbb{A}^1$ by translations (in char. $p$) ? $\endgroup$ – Matthieu Romagny Jul 4 '12 at 19:44
  • $\begingroup$ @Matthieu: Whoops! Embarrassingly, I got the universal property of reduction backwards; it's terminal among maps from reduced things, not initial among maps to reduced things. I'll edit to correct the error. $\endgroup$ – Anton Geraschenko Jul 5 '12 at 20:36
  • $\begingroup$ I guess you have swapped initial and terminal in your previous comment. $\endgroup$ – Fernando Muro Jul 6 '12 at 6:13
  • $\begingroup$ @Fernando: I think I got it right. If $Y$ is a reduced scheme, any morphism $Y\to X$ factors uniquely through $X_{\textrm{red}}$, so $X_{\textrm{red}}$ is the terminal object in the category of reduced schemes over $X$. $\endgroup$ – Anton Geraschenko Jul 6 '12 at 8:44

Let me work over an algebraically closed field $k$, let $X$ be a reduced $k$-scheme of finite type and let $G$ be a smooth connected $k$-group scheme acting on $X$, and acting trivially on an open dense subset $U$ of $X$. I will show that $G$ acts trivially on $X$.

I will only manipulate closed points. Since both $G$ and $X$ are reduced, it suffices to prove that for any $g\in G$ and $x\in X$, $gx=x$.

Let us fix $x\in X$. We may of course suppose that $x\notin U$. Let $V$ be an affine neighbourhood of $x$. By quasicompactness of $X$, the open subset $GV$ is covered by finitely many of the $gV$, say $GV=\cup_i g_iV$. Since $U$ is dense in $X$, it is possible to find a smooth curve $C$ and a morphism $C\to V$ whose image contains $x$ and intersects $U$. Up to removing points of $C$, we may suppose that $x$ is the only point of the image of $C$ not belonging to $U$.

Now, if $g\in G$, $gx\in g_i V$ for some $i$. Consider $g(C)$ (this is an abuse of notation for the composition of $C\to V$ and of the action of $g$) : its image contained in $g_iV$. It coincides generically with $g_i(C)$, hence, the morphisms $g(C)$ and $g_i(C)$ are equal by separation of $g_iV$ ($g_iV$ is even affine). This shows $gx=g_ix$. We have proved $Gx\subset ${$g_1x,\dots, g_rx$}. Since $G$ is connected, $Gx$ is also connected. This shows $Gx=${$x$} and finishes the proof.

Moreover, the hypothesis that $G$ is smooth may be removed as follows. First, we deal with the case where $X$ is separated and reduced. In this case, the arguments given by Anton in the question still work : the graphs of the action and of the projection are closed subschemes of $G\times X\times X$ isomorphic to $G\times X$ that coincide on an open dense subset $U$. Since $G\times X$ has no embedded point ($X$ is reduced and $G$ homogeneous), these two graphs need to coincide with the schematic closure of $U$ in $G\times X\times X$ and are thus equal.

Then we deal with the general case ($G$ connected, $X$ reduced) in the following way. Applying the result when the group is smooth connected and the space is reduced to $G^{red}$ acting on $X$ shows that $G^{red}$ acts trivially on $X$. In particular, $G^{red}$ stabilizes every affine open subset of $X$ ; this implies that $G$ also stabilizes every affine open subset of $X$. Applying the result when the group is connected and the space is separated and reduced to $G$ acting on any affine open subset of $X$ shows that $G$ acts trivially on any open subset of $X$, hence on $X$.

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  • $\begingroup$ Nice. I think we can avoid choosing a curve with the following tweak. The projection $p_1:\overline\Delta\to X$ from the closure of the diagonal has discrete fibers (since any scheme is locally separated). Since $G$ acts trivially on a dense open, $\overline\Delta$ is invariant under the action $g\cdot(x,y)=(x,gy)$ (I'm secretly using that scheme-theoretic closure commutes with flat base change). Since this action preserves the fibers of $p_1$, which are discrete, and $G$ is connected, it must be trivial. $\endgroup$ – Anton Geraschenko Jul 5 '12 at 21:27
  • $\begingroup$ Indeed ! I wonder whether the statement remains true when $X$ is a reduced (not necessarily locally separated) algebraic space ? $\endgroup$ – Olivier Benoist Jul 6 '12 at 8:54

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