# Why can I divide an affine variety by the action of the general linear group?

Let $G\subseteq\mathrm{Gl}_n(\mathbb{C})$ be a subgroup of the general linear group and assume that $\rho:G\to\mathrm{Gl}(V)$ is a representation. Understand the complex vector space $V$ as an affine algebraic variety. Then, it appears to be well-known that the quotient $V/G$ has the structure of an algebraic variety such that the quotient map $\pi:V\to V/G$ is a morphism of varieties. However, I cannot find a proof for this statement. There is an abundance of proofs for the case where $G$ is finite, using the Reynolds operator and corollaries of Hilbert's basis theorem, but I would like to see a proof in the general case. Thanks very much in advance!

Edit: I forgot to mention that I assume $G$ to be reductive and the action on $V$ is regular. You are free to assume even more about $G$ if that allows you to provide a comprehensible reference for a proof.

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–  Misha Apr 17 '12 at 17:15
You're not going to figure this out without rolling up your sleeves and reading a bit of GIT, or an equivalent like Dolgachev's book. For instance you write "the quotient $V/G$" as if it's obvious what this means. For finite $G$ it's obvious, but in general you need to talk about the so-called semi-stable locus for the action and the notion of a "good geometric quotient". If you have digested these things then you will understand Angelo's answer below, too. –  Dan Petersen Apr 17 '12 at 18:46
In fact, I am asking for conditions that I do not have to talk about (semi-)stability. I can very much understand $V/G$ as a set, and now I am asking when it is possible to understand it as $\mathrm{Spec}(A^G)$, where $A$ is the coordinate ring of $V$ and $A^G$ are the functions left invariant by $G$. –  Jesko Hüttenhain Apr 17 '12 at 21:25

If $G$ is reductive, try looking at Fogarty, Kirwan, Mumford, Geometric Invariant Theory, p. 27

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While I was hoping for a reference other than Mumford's book, this is a reference, so I am accepting the answer. –  Jesko Hüttenhain Apr 18 '12 at 7:07
Here is another reference: Theorem 1.24 in Michel Brion's "Introduction to actions of algebraic groups" [Les cours du CIRM 1 (2010), p.1-22], available online at ccirm.cedram.org:80/ccirm-bin/item?id=CCIRM_2010__1_1_1_0 –  Bart Van Steirteghem Apr 18 '12 at 13:22
That looks very good, thanks a lot! –  Jesko Hüttenhain Apr 18 '12 at 17:37

If by $V/G$ you mean the space of orbits, this is not true. Consider $\mathbb C^*$ acting on the affine space $\mathbb A^1$ by multiplication; the space of orbits has two points, but the only variety with two points is disconnected, while $\mathbb A^1$ is connected.

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$\mathbb C^\ast$ is the canonical example of a reductive group... –  Dan Petersen Apr 17 '12 at 18:21
Good point, my comment made no sense. –  Jesko Hüttenhain Apr 17 '12 at 21:19

To amplify what others have pointed out, it doesn't make sense to get an affine variety here as a "quotient" unless all orbits of G are closed (a condition not met even by the natural action of the general linear group) and the variety itself has regular functions given by a finitely generated algebra: the fixed points of G on the algebra of polynomials in n variables if $n=\dim V$. At the least you need G to be a Zariski-closed subgroup of the general linear group, but the natural condition for finite generation of polynomial invariants is typically satisfied in characteristic 0 by focusing just on reductive algebraic groups.

All of this goes back to classical invariant theory and Hilbert's 14th problem, but by now it has become part of the broader study of GIT. Thanks to the proof by Haboush of Mumford's Conjecture, reductive groups over an algebraically closed field of any characteristic yield finitely generated algebras here, but more is needed to get a genuine quotient.

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