MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be an affine algebraic variety over an algebraically closed field $k$ of characteristic zero. Let $G$ be a reductive algebraic group acting on $X$. In this setting, there exists a categorical quotient variety $X / G$. Are there nice conditions (involving $X$ and/or $G$) that imply that $\mathrm{dim}(X / G) = \mathrm{dim}(X) - \mathrm{dim}(G)$? As pointed out in the comments, at a minimum one would have to require the action to be faithful.

share|cite|improve this question
Certainly at a minimum the action ought to be faithful. Possibly you also want it to be free (otherwise consider the action of $\text{GL}_n$ on $\mathbb{G}_a^n$). – Qiaochu Yuan Jul 24 '12 at 17:28
I'm also bothered by the looseness of the formulation (as well as the reason for requiring characteristic 0). What precise notion of "geometric quotient variety" is invoked here? Must a "reductive" group be connected? (For a finite group acting on an affine variety, there is a nice theory which leads to equality of dimensions here, with $\dim G =0$ as an algebraic group.) Also, the question seems open-ended, with an indefinite number of correct answers. The literature on such group actions is wide-ranging and involves GIT too. – Jim Humphreys Jul 24 '12 at 20:27
I believe "geometric quotient" should be replaced by "good quotient" in this question as the natural quotient, given by taking the spectrum of invariant functions, is not always geometric in GIT's sense. – Olivier Benoist Jul 24 '12 at 20:44
Yes Oliver I believe you are probably right. I am not an expert on this stuff. Forgive me for the naive question regarding your terminology: is "good quotient" the same thing as "categorical quotient"? – Phillip Williams Jul 25 '12 at 16:40
Yes Phillip, "categorical quotient" would do the job ! Being a "good quotient" is stronger (roughly, one requires the properties that hold for quotients of affine varieties by a reductive group to hold for all good quotients). If a good quotient exists, then it is the categorical quotient. To be a "geometric quotient" is even stronger : for such quotients, the geometric points of the quotient should be in one-to-one correspondence with the orbits of geometric points of $X$. – Olivier Benoist Jul 25 '12 at 19:36
up vote 5 down vote accepted

A sufficient condition is that there should exist a closed orbit of maximal dimension (i.e. of dimension $\dim(G)$). Indeed, when it is the case, the stable locus is nonempty, and the image of the stable locus in $X/G$ is an open subset of $X/G$ of dimension $\dim(X)-\dim(G)$. This happens for instance if $G$ acts properly on $X$ (in this case all orbits are closed of maximal dimension).

It will not be possible to give a nice necessary and sufficient condition, because the property $\dim(X/G)=\dim(X)-\dim(G)$ is not a very natural one in this context. For example, it does not imply that the action is faithful : take $G=\mathbb{G}_m\times\mathbb{G}_m$ acting on $X=\mathbb{A}^2$ by $(s,t)\cdot(x,y)=(sx,sy)$.

share|cite|improve this answer
This criterion will certainly be useful; thank you. Can you give me a reference for the statements in your first paragraph? – Phillip Williams Jul 25 '12 at 16:41
The reference I know is Mumford's GIT. More precisely : apply the theory of Chap.1, par.4 to $X$ and to the trivially $G$-linearized line bundle $\mathcal{O}_X$. First, considering the section $1\in H^0(X,\mathcal{O}_X)^G$, you see that all points of $X$ are semistable. The hypothesis about the closed orbit of maximal dimension shows that the stable locus is nonempty, by Amplification 1.11. Then you look at the categorical quotient provided by Thm. 1.10. Restricted to the stable locus, the quotient is geometric. For such a quotient, the dimensions add up (see the discussion before Prop 0.2). – Olivier Benoist Jul 25 '12 at 19:55
OK, thanks! I had been looking at Mumford but this directs my reading a lot more. – Phillip Williams Jul 26 '12 at 15:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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