# Quotient of a variety and orbits

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!!

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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 $\mathbb{A}^1_k$.

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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.

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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

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?

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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