# Stabilisers of group actions

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.

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.

• 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?. Apr 19, 2015 at 15:18
• Unfortunately no, that would be quite interesting.
– abx
Apr 19, 2015 at 16:17
• 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.
– 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.

• 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? Apr 20, 2015 at 9:36
• @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 $\Gamma$. Apr 20, 2015 at 10:51