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If there a group G acting on a variety V. The action is algebraic. What is the definition of algebro-geometric quotient of this action?

I hope you can give a very basic explanation.


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It is certainly possible to give the definition of a quotient of a variety by an algebraic group without mentioning algebraic stacks or spaces.

There are two distinctly different situations here, depending on whether the group is finite (and discrete) or a general algebraic group.

In the finite case things are a lot easier. Then you could take as your definition of a quotient that on an affine scheme given as the spectrum of a ring R, the quotient is the spectrum of the ring of invariants $R^G$. (This is not a good definition, but you could.) For general schemes you can try to glue together open affines to define a quotient globally, but this will not always produce a scheme (roughly speaking, you might have to identify too many points). When the original scheme is quasiprojective, this gluing procedure will give you a scheme (folklore), and in general it will always produce an algebraic space (Deligne).

In the more general case of algebraic groups one needs to be more careful and start distinguishing between different definitions of quotients, and start keeping track of whether your group is geometrically reductive or linearly reductive or neither. The quotient definition given by James Borger is called the categorical quotient. It is unfortunately not well behaved in the category of schemes. (Example: let $\mathbb{C}^\ast$ act on $\mathbb{C}^2$ by scaling. Then a categorical quotient exists in the category of schemes (!), however it is a single point.) The basic problem is that most properties that one wants from a quotient, like "the preimage of a point in X/G is a single G-orbit in X" will not hold unless one imposes extra conditions on top of the categorical one.

More well behaved notions are in particular good quotient and geometric quotient, but beware that there is here a morass of different definitions of quotients which can be more or less nice (there are things called weak quotients and semi-geometric quotients and probably more that I don't know, and these can be modified by adjectives like "uniform" or "universal" or "categorical", and there are nontrivial combinations of these properties like "good geometric" so these notions of quotients are not linearly ordered). There are also some differences in the literature between how these properties are defined. The point that makes finite groups so much easier is that for a finite group all these types of quotient will coincide anyway.

Just as an example let me give the definition of a geometric quotient $\pi : X \to X/G$:

  1. $\pi$ is G-equivariant;
  2. the geometric fibers of $\pi$ are the orbits of geometric points on $X$ (i.e. the property I mentioned earlier, but it is required only for points with coordinates in an algebraically closed field);
  3. $U \subset X/G$ is open if and only if $\pi^{-1}(U)$ is open;
  4. $(\pi_\ast O_X)^G = O_{X/G}$.

(Quoted from GIT)

The most important work in this area is the slightly intimidating GIT (Geometric Invariant Theory) by David Mumford. The main construction of the book shows that when the group is reductive and acts linearly on a projective variety $X$, there are canonically defined open dense subsets traditionally denotes $X^s$ and $X^{ss}$ such that the $X^s$ has a geometric quotient and $X^{ss}$ has a good quotient which is projective. When X is not projective one has to work with so called L-linear actions where L is a line bundle on X. This reduces to the case X projective by taking $L = O(1)$. GIT is written like a textbook so in principle you can start reading on page one, but there are friendlier introductions out there. I like Dolgachev's Lectures on Invariant Theory.

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@Dan Peterson: I am sure you know this, but it might be helpful to emphasize that the semi-stable $X^{\text{ss}}$ locus and the stable locus $X^{\text{s}}$ depend on the choice of projective embedding (i.e., the linearization $L$). – jlk Feb 6 '11 at 18:12
what I am really interested is following: I have a group G, finitely presented. Consider R(G)=HOM(G,SL(2,C)), the space of all reps of G into SL(2,C). Then SL(2,C) acts on R(G) naturally, i.e. conjugation. So the usual quotient of this action is the space of orbits, which are all conjugacy classes of reps. Now the algebro-geometric quotient is not this. It is the character variety of G. Thanks. – Xuanting Cai Feb 6 '11 at 18:30
I know nothing about character varieties; maybe you would have better luck asking a question specifically about them. Anyway you are right that the set of all orbits in your set-up is in a natural way the set of isomorphism classes of representations of G in $SL(2,\mathbb{C})$. Whether or not the orbits correspond bijectively to points in the quotient is a more delicate issue. You will have to linearize the action and figure out which points are stable and semistable (or rather, someone else already has done this and you will have to read it :p). – Dan Petersen Feb 6 '11 at 19:28
Here I am assuming you are taking the "GIT quotient" which is the good quotient of $X^{ss}$ that I mentioned in my answer. It has the property that all the stable orbits correspond to actual points of the quotient (since the quotient of $X^s$ was supposed to be geometric) but some semistable orbits will be identified with each other. – Dan Petersen Feb 6 '11 at 19:32
Xuanting Cai, perhaps you should have explained your interest in the question. Dan's and the other answers are certainly relevant, but you might find the book by Lubotzky and Magid "Varieties of representations..." gives more specific answers. For the record the (closed) points of the character variety correspond isomorphism classes of semisimple representations, which are the stable points for the action. – Donu Arapura Feb 6 '11 at 20:11

The are several possible meanings. Which one it is would surely depend on the context.

The straight-up meaning is the one that works in any category. If $G$ acts on $X$ then a quotient is a universal object $X/G$ with a $G$-equivariant map $X\to X/G$, where $X/G$ has the trivial $G$ action. Here, 'universal' means that if $Y$ is any other such object, then there is a unique map $X/G\to Y$ commuting with the two maps from $X$. Then $X/G$ is unique up to unique isomorphism.

Unfortunately, if you're working in the category of schemes, such quotients sometimes don't exist. If you work in the slightly larger category of algebraic spaces (IMHO 'schemes done right'), they are more likely to exist, and there are even nice theorems (M Artin) to this effect. (There are also some not-so-nice theorems about scheme quotients in SGA 3.) You can also work in some big ambient topos, such as the category of sheaves of sets on the category of affine schemes, equipped with your favorite Grothendieck topology. In this big category, quotients always exist, and they have all the nice formal properties you could ask for (i.e. they are universal and effective, in the language of category theory).

As Steven Landsburg points out, you can also ask for the quotient in the stack-theoretic sense.

There are also various kinds of quotients that come up in geometric invariant theory, which are important if you're interested in projective algebraic geometry, but I'm embarrassed to admit that I never got to the bottom of what's going on there.

Finally, there might even be a theory of quotients in Weil's foundations for algebraic geometry, which I hear that some people working in algebraic groups still use.

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Dear James, how does the stack-theoretic quotient differ from the quotient of algebraic spaces or the sheaf-theoretic quotient (and do these two generally agree)? – Harry Gindi Feb 6 '11 at 12:21
@Harry Gindi: I believe the two notions are distinct in the case of the quotient of a point by the trivial action of a (non-trivial) group $G$. – jlk Feb 6 '11 at 18:04
In this case, the categorical quotient $X/G$ that James Borger describes is just the point. The stack-theoretic quotient is the classifying stack of $G$. In other words, an object over a scheme $T$ is a principal $G$-bundle $P \to T$. This is not equivalent to a $1$-category (as I am sure you know). – jlk Feb 6 '11 at 18:08
Harry, a stack is a 2-categorical object, whereas an algebraic space is a 1-categorical concept. In any situation where your action has nontrivial stabilizers (such as $G$ acting on a point, as jlk points out), the two will disagree, because the algebraic space quotient doesn't have enough n-categorical headroom to carry stabilizer information. The quotients in the categories of sheaves and algebraic spaces also typically disagree. (See Dan Peterson's example of $\mathbf{C}^*$ acting on $\mathbf{C}^2$.) They agree precisely when the algebraic space quotient is universally effective. – JBorger Feb 6 '11 at 21:37
Ah, thanks, gentlemen! – Harry Gindi Feb 6 '11 at 23:49

You are looking for the theory of stacks. The Wikipedia article will get you started. Or see the article by Tomas, which is quite readable (if you have the right background). See especially example 2.33 in that paper.

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