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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?

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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 $\mathbf A^2\setminus\{u=0\}\cup\{(0,0)\}$ 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.

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  • $\begingroup$ 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). $\endgroup$ – Sean Lawton Jun 13 '16 at 13:27
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Preamble:

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.

Remark:

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 $\mathrm{Tr} = 2$. $\endgroup$ – David E Speyer Jun 12 '16 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). $\endgroup$ – Sean Lawton Jun 12 '16 at 23:37
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    $\begingroup$ Okay, but how is that a counter-example to the map being open? $\endgroup$ – David E Speyer Jun 12 '16 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 :) $\endgroup$ – Sean Lawton Jun 13 '16 at 13:28

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