Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a group G acts on a variety X and a quotient exists, that is, we have a variety Y and a regular map $\pi : X \rightarrow Y$ so that any regular map $\varphi :X \rightarrow Z$ to another variety Z factors through $\pi$ if and only if $\varphi (p) = \varphi (g(p)) \forall p \in X, g \in G$.

I'm trying to prove that the points of Y correspond to orbits of G on X, i.e.

$\pi (p) = \pi (q) \iff \exists g \in G: g(p) = g(q) $

However, I am stuck. The only triviality I was able to show is that, assuming $\pi (p) = \pi (q)$, we'd have $\pi (g_1(p)) = \pi (g_2(q)) \forall g_1, g_2 \in G$. I guess it boils down to choosing the right variety Z and then make use of the fact that Y is a quotient, but I don't know how. I'd be grateful for any hints.

EDIT: I just realized I have a bad typo in this post. $g(p) = g(q)$ should be $g(p) = q$ , sorry!!

share|improve this question
add comment

3 Answers

Here's another perspective: suppose $G$ acts on $X$ and that a variety $Y$ exists which parameterizes the orbits of $G$ on $X$, with a regular morphism $\pi: X \rightarrow Y$ such that $\pi(x') = \pi(x) \Leftrightarrow x' = g \cdot x$ for some $g \in G$. Then, since any point $y \in Y$ is closed in the Zariski topology and $\pi$ is a continuous map, $\pi^{-1}(y)$ -- an orbit of $G$ in $X$ -- must be closed.

So if $G$ acts on $X$ with non-closed orbits, it cannot have a quotient in the category of algebraic varieties whose points corresond to orbits on $X$. In QiL's example, the orbit $k^*$ is not closed in $\mathbb{A}^1_k$.

share|improve this answer
add comment

By construction, $g(p)=q$ implies that $\pi(p)=\pi(q)$. But the converse is false in general. For example, suppose $k$ is an algebraically closed field, and let $G=k^*$ act on $X=\mathbb A^1_k$ by $(\lambda, z)\mapsto \lambda z$. Then the quotient $X\to Y$ is just the structural morphism $X\to Y=\mathrm{Spec}(k)$ (see below). So $\pi$ is constant. But $0, 1\in \mathbb A^1_k$ are not in the same orbit.

Computation of $X/G$. Let $f: X\to Z$ be any $G$ invariant morphism. Then $f(k^*)$ is one point because $G$ acts transitively on $k^*\subset X$. By continuity of $f$, $f$ is constant. So $f$ factors through the structural morphism $\rho$ of $X$. Therefore $X\to Y$ is equal to $\rho$.

If $G$ is a finite group acting on a quasi-projective variety $X$, then $G$ act transitively in the fibers of $X\to Y$. See Mumford, Abelian varieties, p 55.

share|improve this answer
Thanks! I lost access to my guest account so I can't comment/check your answer, that cleared some things up for me. However, see my original question, I included a bad typo when I copied the formulas. I was able to prove "$\leftarrow$" which like you said follows by construction, but I'm stuck with the other implication. –  Michi89 Nov 13 '11 at 10:34
OK this was a small typo that I also copied :). Does the counterexample above convinces you ? –  user18119 Nov 13 '11 at 14:34
add comment

Sadly I still can't comment because I lost the other account.

So the counterexample holds? It does convince me, however I am confused because I got the idea that the converse should be true from p. 123 in Harris' "Algebraic Geometry, A First Course".

He mentions that this "factorization property" I described in the beginning of my question is stronger than:

There exists $\pi: X \rightarrow Y $ surjective so that: $\pi (p) = \pi (q) \iff \exists g \in G: g(p) = q$

So in case I didn't misunderstand the text, the converse should be true. Your construction is presented one page later, but as a counterexample to the statement that a quotient always exists: This supposedly does not hold here because "[...] there does not exist a surjective morphism from $\mathbb{A}^1$ onto a variety with two points".

I am an absolute beginner in the field of AG. Are there maybe some definitions of quotients which are not equivalent or did I just get something terribly wrong?

share|improve this answer
The definition you presented is a so called "categorical quotient", and is weaker than the definition of "geometric quotient" as in Harris (weaker in the sense that a geometric quotient is a categorical quotient, but not in inverse). When $G$ is finite and $X$ is quasi-projective, then both definitions coincide. –  user18119 Nov 13 '11 at 17:10
By categorical quotient you mean the definition that any regular map to another variety Z factors through $\pi$ iff it is G-invariant? This is the definition Harris gives, or at least what should be stronger than the characterization only through $\pi$ without another variety Z. I just need to sort the terminology in my head, maybe then I'll get some understanding of this concept :) –  Michi89 Nov 13 '11 at 17:40
I checked with the book, Harris requires first that $G$ acts transitively at the fibers of $X\to Y$ (p.123, line -8), then a few lines below he adds the factorization property. –  user18119 Nov 13 '11 at 17:46
Yes, that's the way I understood it; But doesn't he take the factorization property as the more sensible definition of a quotient? ("We want to require something a little stronger, namely, a quotient should be..") The word "stronger" let me believe that from this, it should follow that G acts transitively on the fibers. Sorry if I'm a nuisance, but I don't get it yet :( –  Michi89 Nov 13 '11 at 17:55
It really has to be understood as "a quotient should further be...". It must satisfy the transitivity condition, and a minimality property (factorization). Hope this helps. –  user18119 Nov 13 '11 at 21:40
show 1 more comment

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.