Let $k$ be a field, $X= \text{Spec}\,A$ be an affine scheme, with $A$ a finitely generated $k$-algebra. $G=\text{Spec}\,R$ is a linearly reductive group acting rationally on A, i.e. every element of $A$ is contained in a finite-dimensional $G$-invariant linear subspace of $A$. By Nagata's theorem, $A^G$ is a finitely generated $k$-algebra. We have the affine GIT quotient $X \rightarrow X/G := \text{Spec}\,A^G$ induced by the inclusion $A^G \rightarrow A$ of $k$-algebras.

Question. Is the affine GIT quotient, viewed as a map of the underlying topological spaces, necessarily an open map? It does not need to be an open immersion of schemes. If not, any counterexample?


2 Answers 2


Categorical quotients are in general very far away from being open. In fact, it is a theorem of Chevalley (I think) that a morphism $\pi:X\to Y$ onto a normal variety $Y$ is open if and only it is equidimensional. Unless $G$ is finite, this is a very rare condition for quotient morphisms.

The idea behind this is the following: Assume that the fiber $X_y$ has bigger dimension than the generic fiber dimension $\dim X-\dim Y$. Then there is a curve $C\subseteq Y$ passing through $y$ such that the fiber dimension jumps over $C$. Let $Z$ be the closure of $\pi^{-1}(C)\setminus X_y$ in $X$. Then for dimension reasons $X_y\not\subseteq Z$. Consider the open subset $U:=X\setminus Z$ of $Z$. Then $\pi(Z)=(Y\setminus C)\cup\{y\}$ is only constrictible but not open.

More formally, the openness of a morphism is expressed by the so-called going-down property of Cohen-Seidenberg.

Let's try this out for the quotient of $G=\mathbf G_m$ acting on $\mathbf A^3$ by $(tx,ty,t^{-1}z)$. Then $$ \pi:\mathbf A^3\to\mathbf A^2:(x,y,z)\mapsto (u,v):=(xz,yz) $$ is the quotient morphism mapping the plane $\{z=0\}$ to $(0,0)$. Take $C=\{u=0\}$. Then $\pi^{-1}(C)=\{x=0\}\cup\{z=0\}$ and the image of $U=\mathbf A^3\setminus\{x=0\}$ is $$\Bigl(\mathbf A^2\setminus\{u=0\}\Bigr)\cup\{(0,0)\},$$ which is not open.

PS: Quotient maps have some properties not shared by other morphism:

  • Images of closed $G$-stable subsets are closed.
  • A subset of $Y$ is open iff its preimage in $X$ is open.
  • The morphism is surjective.

These properties hold universally, i.e., even after base change.

  • $\begingroup$ This is a great answer! Thanks for sorting it out. I was uneasy about my "proof" and am glad to see I was wrong (it's the best way to learn). $\endgroup$ Jun 13, 2016 at 13:27


The original answer below is wrong. The mistake is that when restricting the open set $U$ to $U\cap X^{ps}$, the image of the GIT projection (call the map $\pi$) of $U$ may not equal the image of the quotient map $p:X^{ps}\to X^{ps}/G$. Certainly, $p(U\cap X^{ps})\subset \pi(U)$ but the point is that $U$ can contain some non-polystable points that get removed when intersecting and can change the topology of the image. This in fact was my original intuition with my attempted counter-example, but the example I choose was in fact open. So instead of trying another example, I decided to challenge my "gut" by trying to prove myself wrong and surprisingly came up with a "proof".

Anyway, Friedrich Knop's answer is right. I feel like I should delete this "answer," but on the other hand, sometimes failed attempts are instructive to others so I am not sure I will. I added a remark at the end that might be useful to the OP since the OP expressed interest in understanding the strong topology of $X//G$.

Failed Counter Example:

Any open orbit maps to a point, so generally the GIT quotient is not an open map (see comments for the mistake).

Failed Proof of Openness:

We work over $\mathbb{C}$. Take an open set $U\subset X$ then $U\cap X^{ps}$ is open in $X^{ps}$ (with respect to the relative topology) where $X^{ps}$ is the set of polystable points (points with closed orbits). Therefore $U\cap X^{ps}$ maps to an open set in $X^{ps}/G$ since $p:X^{ps}\to X^{ps}/G$ is an open map (this follows from the definition of the quotient topology and the fact that $G$ acts by homeomorphisms). But that set is equal to the image of $U$ under the GIT projection (this step is the mistake). Hence it is open in $X//G$ since $X//G\cong X^{ps}/G$ (see for example Theorem 2.1 here).

Weak Correction:

Whenever $p(U\cap X^{ps})\supset\pi(U)$ for all open $U,$ then $\pi$ is an open map in the strong topology. This follows from the above failed proof, since it fills the gap with an assumption.


On the other hand, if one wants to understand the space $X//G$ in the strong topology one can replace $\pi:X\to X//G$ by $p:X^{ps}\to X^{ps}/G$. The latter is open while we now know the former might not be, but as a space in the strong topology $X//G$ remains homeomorphic to $X^{ps}/G.$ More still, as per Proposition 3.4 here, the usual quotient $X/G$ is homotopic to $X//G$.

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    $\begingroup$ Sean, I am confused. The orbit of $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$ is not open. Indeed, it is contained in the subvariety $\mathrm{Tr} = 2$. $\endgroup$ Jun 12, 2016 at 23:35
  • $\begingroup$ I don't mean it is open in $\mathrm{SL}(2,\mathbb{C})$, it is open in the Zariski closure of that orbit. The closure contains exactly two suborbits, the other being the identity (which is a point). $\endgroup$ Jun 12, 2016 at 23:37
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    $\begingroup$ Okay, but how is that a counter-example to the map being open? $\endgroup$ Jun 12, 2016 at 23:38
  • $\begingroup$ Let $X$ be the orbit closure. The GIT quotient is a point. The open subspace maps to that point. So the open set maps to a ....wait, it is clopen. Ooops. Thanks for the helpful comment :) $\endgroup$ Jun 13, 2016 at 13:28

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