Introduction: The definition of "good categorical quotient" in geometric invariant theory (given below) seems fairly ad hoc to me, except that it looks very similar to the coequalizer of the action in the category of locally ringed spaces. I'm trying to understand this similarity.

Some background: Suppose we have a group action $G\times X \to X$ in the category $\bf Sch$ of schemes. The category $\bf LRS$ of locally ringed spaces is cocomplete (see propositiion 1.12 in A. Vezzani's thesis), so in particular, $G\times X \rightrightarrows X$ has a $\bf LRS$-coequalizer; call it $Q$. One can construct it as the coequalizer in the category of topological spaces equipped with the sheaf of $G$-invariant functions to make it a locally ringed space, as Vezzani's proposition explains more carefully.

If $G\times X \rightrightarrows X$ has a $\bf Sch$-coequalizer $Y$, then GIT calls $Y$ a "categorical quotient"; see, for example, R. Birkner's Introduction to GIT. Note that here we get a universal map $u: Q\to Y$ in $\bf LRS$.

$Y$ is called a "good categorical quotient" if it satisfies 3 properties, also found in Birkner's notes, which I prefer to re-express here in terms of the universal map $u$:

i) $u^\sharp : {\cal O}_Y \to u _* {\cal O}_Q$ is an isomorphism.

ii) $u: Q\to Y$ is a closed map (at the level of topological spaces).

iii) $u$ takes disjoint closed sets to disjoint closed sets.

(These together imply that $X\to Y$ is a categorical quotient, i.e. $\bf Sch$-coequalizer; see Mumford's GIT I.2 remark 6.)

All of these properties are clearly satisfied if $u$ is an isomorphism. That is, if the $\bf LRS$-coequalizer is a scheme, that scheme is a good categorical quotient.

Towards a converse, (i) implies that $u$ must be dominant, and hence by (ii) it is surjective. And (iii) implies it is injective on closed points, so it's looking a lot like an isomorphism. Can anything more be said?

1) When is a good categorical quotient an $\bf LRS$-coequalizer of the group action? I.e., are there nice and general conditions where $u$ must be an isomorphism?

2) If the answer is not always, what's a counterexample? (Answered by Anton)


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    $\begingroup$ The cocompletness of LRS is not due to Vezzani, it's already contained in Groupes algebriques (Demazure, Gabriel), 1970. I don't know the answer to your problem, but usually the notions concerning colimits of schemes are motivated by explicit applications, and not by abstract nonsense. I doubt that there is any connection. $\endgroup$ – Martin Brandenburg Mar 28 '11 at 10:19
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    $\begingroup$ I agree with Martin (and also don't know the answer). The goodness is a kind of wish list of properties that make the quotient useful in applications. Even if it (by chance) should be the same as the LRS quotient that would be more or less useless as there does not seem to be any way of really using that a scheme is the LRS quotient of a schematic equivalence relation. $\endgroup$ – Torsten Ekedahl Mar 28 '11 at 10:56
  • $\begingroup$ @ Martin, I prefer to give links to online references when possible. Thanks for the book recommendation. $\endgroup$ – Andrew Critch Mar 28 '11 at 16:56

The answer to the title question is NO. If $X\rightrightarrows Y\to Z$ is a coequalizer in $\bf LRS$, then the underlying topological space of $Z$ is the coequalizer of the underlying topological spaces (and the structure sheaf is the equalizer of pushforwards of the structure sheaves). On the other hand, good categorical quotients identify any two points whose orbit closures intersect. So one of my favorite good quotients, $\mathbb A^1/\mathbb G_m$, provides a counterexample. The good categorical quotient in schemes has a single point as its underlying topological space, but the quotient in $\bf LRS$ has two points.

I think the bad behavior on underlying topological spaces is probably the only problem. That is, a good categorical quotient is a coequalizer in $\bf LRS$ precisely when it is a geometric quotient (i.e. when we impose the additional condition that each fiber is a single orbit).

  • $\begingroup$ Thanks, Anton! But the relationship between (good) geometric quotients and categorical quotients is not clear to me... The LRS coequalizer Q, being a topological coequalizer, has as its topological fibres the "G orbits" in the sense that if there exists a (g,x,p) \in |GxX| (here p \in Spec(k(g)\otimes k(x)) mapping via the action to y \in X, then x and y map to the same point of Q, and conversely. But I don't see what this means for the geometric fibres ... can this be cleared up? $\endgroup$ – Andrew Critch Mar 28 '11 at 18:30
  • $\begingroup$ ...When X and G are finite type over an algebraically closed field, at least over closed points the topological fibres and geometric fibres are the same... but even for varieties, over non-closed points these may differ. For example when flipping a parabola about its axis: Say G is Z/2 acting on X=k[x,y]/(y=x^2) by flipping sign of the x coordinate, and X->Y is projection to the Y axis... the topological fibre over the generic point is 1 point, but a geomteric fibre over the generic point has 4 points... hmm... $\endgroup$ – Andrew Critch Mar 28 '11 at 18:54
  • $\begingroup$ Your question is only about the second paragraph, right? Say $q\in Q$ is a point of the LRS coequalizer, and $F=X\times_Q Spec(k(q))$ is the fiber (note: I think this is the topological fiber ; does the proof of Hartshorne, Ex. II.3.10 work in LRS?). Saying that $F$ is a single $G$-orbit is the same as saying that the map $G\times F\to F\times F$ (given by $(g,f)\mapsto (g\cdot f,f)$) is surjective. Surjectivity of this map can be checked on geometric points, so the topological fibers of $X\to Q$ are single $G$-orbits if and only if the geometric fibers are single $G$-orbits. $\endgroup$ – Anton Geraschenko Mar 28 '11 at 19:06
  • $\begingroup$ Whoops, I missed your second comment while composing the above. I think a geometric fiber over the generic point in your example only has 2 points, and that $G=\mathbb Z/2$ swaps them. $\endgroup$ – Anton Geraschenko Mar 28 '11 at 19:13
  • $\begingroup$ Bleh ... it's not true that a fiber is a single $G$-orbit topologically iff $G\times F\to F\times F$ is surjective. In the example of the parabola mapping to the line, the trivial group $G$ acts on the fibers. Over the generic point, the fiber $F$ is one point topologically, but $G\times F\to F\times F$ is the inclusion of one point into two points. Nevertheless, the direction I actually used is true: if $G\times F\to F\times F$ is surjective (which it is for a good geometric quotient), then $F$ is topologically a single $G$-orbit. $\endgroup$ – Anton Geraschenko Mar 28 '11 at 19:20

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