Let $G$ be an algebraic group acting on an irreducible algebraic variety $X$ over an algebraically closed field $k$ of characteristic $0$.

Suppose there exists some point $x \in X$ whose stabiliser $G_x$ is trivial. Does there exist an open subset $U \subset X$ such that the stabiliser $G_u$ is trivial for all $u \in U$?

I'm open to the fact that the answer might be no in general, however I have an application in mind where my $X$ and $G$ are nice. If it helps, I can also assume that

  • $G$ is reductive.
  • $X$ is smooth.
  • For all $x \in X$, the stabiliser $G_x$ is finite and the orbit $Gx$ is closed.

2 Answers 2


Under some of your extra hypotheses, namely if $G$ is reductive and the orbit $Gx$ is closed, the answer is yes. This follows easily from Luna's slice theorem; see for instance these lectures, Proposition 5.7.

  • $\begingroup$ This is great, thanks! Out of interest, do you happen to know an example for which the answer to my question is no? e.g. for some non-reductive group $G$ which acts with non-closed orbits?. $\endgroup$ Apr 19, 2015 at 15:18
  • $\begingroup$ Unfortunately no, that would be quite interesting. $\endgroup$
    – abx
    Apr 19, 2015 at 16:17
  • $\begingroup$ Here is an example where $G$ is reductive and $X$ is smooth, but the answer is "no". Take $G = SL_2(\mathbb{C})$ acting on cubic forms in variables $x$, $y$. There is a $G$-stable open set of forms that vanish on three distinct lines with representative $u := xy(x-y)$ and $G_u \cong \mathbb{Z}/3$. The forms that have a double vanishing (e.g., $x^2 y$) have stabilizer $G_u \cong 1$. This example is typical for the case where $G$ is simple, $X$ is a linear representation, and a generic element of $X$ has finite stabilizer. $\endgroup$
    – Skip
    Dec 23, 2021 at 22:56

The answer abx gives is completely correct; I am just adding an example proving that some hypothesis is necessary. Let $k$ be a field of characteristic different from $2$. Let $G$ be the algebraic subgroup of $\textbf{GL}_{2,k}$ of the form $$ G =\left\{ \left[ \begin{array}{rr} 1 & b \\ 0 & \lambda \end{array} \right] \left\vert b\in \mathbf{G}_a,\ \lambda^2 = 1 \right. \right \}. $$ Denote by $U$ the connected component of the identity, which is isomorphic to $\mathbf{G}_a$ via the coordinate $b$. This is a normal closed subgroup of $G$ whose quotient is $\mu_2 = \{1,-1\}$. Denote by $U_{-1}$ the other connected component of $G$, which is isomorphic to $\mathbf{G}_a$ via the coordinate $b_{-1}$.

Let $\Delta$ denote $\mathbb{A}^1_k$ with coordinate $t$. Let $\Delta^*$ denote the basic open affine $D(t)$ inside $\Delta$. Let $\overline{U}$ denote $\mathbb{P}^1$ with homogeneous coordinates $[b_0,b_1]$; identify $U$ with the open subset $D_+(b_0)$ via $b=b_1/b_0$.

Inside $\Delta\times_k \overline{U}$, form the blowing up of the closed point $p$ where $t=0$ and where $b_0 = 0$. Next, on the exceptional divisor $E$ with homogeneous coordinates $[t,b_0/b_1]$, form the blowing up of the point $q$ with $[t,b_0/b_1] = [1,1]$, i.e., the zero scheme of $t(b_0/b_1)^{-1} - 1$. Denote the composite of these two blowings up by $$ \nu: \overline{X} \to \Delta \times_k \overline{U}. $$ Denote by $\overline{F}$ the exceptional divisor over $q$ with homogeneous coordinates $[t, t(b_0/t_1)^{-1} -1]$.
Denote by $\widetilde{U}_0$ the inverse image in $\overline{X}$ of $\{0\}\times U$. Denote by $H$ the strict transform in $\overline{X}$ of the "horizontal cross-section", $\Delta\times\{[0,1]\}$. Denote by $\widetilde{E}$ the strict transform of $E$ in $\overline{X}$. Finally, denote by $F$ the open affine complement in $\overline{F}$ of the intersection point $\overline{F}\cap \widetilde{E}$; this is isomorphic to $\mathbb{A}^1_k$ via the coordinate $$ b_F = t^{-1}(t(b_0/b_1)^{-1} -1). $$ Denote by $X$ the open complement in $\overline{X}$ of the Cartier divisor $\widetilde{E} + H$. This is a scheme over $\Delta$, and the fiber over $t=0$ is the disjoint union $\widetilde{U} = \text{Spec}(k[b])$ and $F=\text{Spec}(k[b_F])$.

Because the blowings up occur over points of $\Delta\times\{[0,1]\}$, the open immersion of the complement extends to an open immersion, $$ q_U: \Delta \times_k U \hookrightarrow X. $$ Next, consider the morphism, $$ q'_{-1}: \Delta^* \times_k U_{-1} \to \Delta \times_k \overline{U}, \ \ (t,b_{-1}) \mapsto (t,b_{-1} + t^{-1}). $$ This extends to a unique open immersion, $$ q_{-1}:\Delta\times_k U_{-1} \to X. $$ In fact, $\overline{X}$ is the minimal blowing up of $\Delta\times_k \overline{U}$ such that $q'_{-1}$ extends to an open immersion. This open immersion maps $\{0\}\times U_{-1}$ isomorphically to $F$ with identification of coordinates $b_{-1}\leftrightarrow b_F$.

The union of these two morphisms defines a morphism $$ q: \Delta\times_k G \to X. $$ This is a surjective, quasi-finite morphism of $\Delta$-schemes. In fact, this is a quotient of the group scheme $\Delta \times_k G$ over $\Delta$ by the closed subgroup scheme $\Gamma$ whose intersection with $\Delta\times_k U$ equals $\Delta \times_k \{0\}$ and whose intersection with $\Delta\times_k U_{-1}$ is $\{(t,b_{-1}) | tb_{-1} + 1 =0\}$.

There is an action of $G$ on $\Delta^*\times_k U$ by $$ b*(t,\beta) = (t,b+\beta), \ \ b_{-1}*(t,\beta) = (t,-b_{-1} - t^{-1}-\beta). $$ This extends to an action of $G$ on $X$. For $t\neq 0$, the stabilizer of $(t,0)$ is the fiber of $\Gamma$ over $t$. For $t=0$, the stabilizer in $\widetilde{U}$ of the point with $b=0$ is the identity subgroup.

In this example, the action of $G$ on $X$ is not closed. Equivalently, the morphism $q$ is not finite.

  • $\begingroup$ This seems pretty cool, however it is so technical I'm struggling to understand it. Is it possible to give an intuitive idea of where it comes from? Is this a standard example or your own example? Is it perhaps related to some kind of moduli problem or something in GIT? $\endgroup$ Apr 20, 2015 at 9:36
  • 2
    $\begingroup$ @DanielLoughran: That is literally the algebraic group of smallest dimension that admits a one-parameter family of closed subgroups (or rather, a closed subgroup scheme $\Gamma$ over the base $\Delta$) that specializes from $\mu_2$ to $\mu_1$. Now you just work out what is the quotient of $\Delta\times G$ by $\Gamma$. $\endgroup$ Apr 20, 2015 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.