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Given a smooth projective curve $C$ such that a group $G$ acts on it, how exactly is the quotient curve ${X/G}$ defined? I cannot seem to find any resources covering explicitly it's definition. I am aware of the fact that there are various categorical equivalences between geometrical objects (varieties) and algebraic objects (function fields) and maybe these are at work here, but I'm uncertain

CoffeeBean
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  • There is a set of properties a quotient $X/G$ should have if it exists (it should be "universal" in a certain sense), and then one tries to construct a scheme $\pi: X \rightarrow Y$ with "these" properties. In many cases such a quotient does not exist. – hm2020 Jan 05 '23 at 14:09
  • The projective line $C:=\mathbb{P}^1$ may be constructed as a quotient $U/\mathbb{G}_m$ where $\mathbb{G}_m$ is the group-scheme of units. Any invertible sheaf on $C$ may be constructed using a character of $\mathbb{G}_m$ https://math.stackexchange.com/questions/4599310/a-description-of-line-bundles-on-projective-spaces-mathcalo-mathbbpn/4601133#4601133 – hm2020 Jan 05 '23 at 14:11
  • By this I mean the following: There is an algebraic vector bundle of rank one $\pi: L(d) \rightarrow C$ with sheaf of sections $\mathcal{O}(d)$. The vector bundle $L(d)$ is constructed using a character of $\mathbb{G}_m$. – hm2020 Jan 05 '23 at 14:27
  • If you have a group $G$ acting on a set $S$ it follows the "quotient" $S/G$ is by definition the set of equivalence classes, and if $S$ has a topology you get an induced topology on $S/G$ such that the natural map $\pi: S \rightarrow S/G$ is continuous. If you want to do this for schemes there are many problems involved. You want a scheme $\pi:X \rightarrow X/G$ where the "points" of $X/G$ correspond 1-1 to the orbits of $X$ under $G$. To construct such quotients is a non-trivial problem in algebraic geometry (called "geometric invariant theory"). – hm2020 Jan 05 '23 at 15:50
  • @hm2020 thank you very much for your comments. I have to admit I'm not familiar with scheme theory, the extent of my knowledge is essentially just basic Algebraic Geometry – CoffeeBean Jan 07 '23 at 13:05

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One easy way to define a quotient of a curve $C$ under the action of the finite group $G$ is as the curve with function field $K(C)^G$, the $G$-invariants of the function field $K(C)$ under the action of $G$. This works because the categories of regular curves over a base field $k$ with nonconstant morphisms and finitely generated field extensions of $k$ with transcendence degree one are canonically equivalent. To see that $K(X)^G$ is again a function field, note that $[K(X)^G:K(X)]$ is a finite Galois extension of degree $|G|$ - so $K(X)^G$ is again of transcendence degree one. For the claim that $K(X)^G$ is again finitely generated, see here on MO.

There are other ways: if $G$ is finite, then we can cover $X$ by $G$-invariant affine opens (take an affine open neighborhood $U$ of $x\in X$ and consider $\bigcap_{g\in G} gU$ which is open since $G$ is finite and affine as $X$ is separated) and compute the quotient on each of these affine pieces, then glue back together, using that for an affine scheme $\operatorname{Spec} R$, the quotient by $G$ is $\operatorname{Spec} R^G$.

KReiser
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  • I'm not sure why you got downvoted. Just a question, though: how do we know also that ${K(X)^G}$ is also an algebraic function field of transcendence degree $1$? Thank you – CoffeeBean Jan 06 '23 at 15:02
  • @CoffeeBean I've added a little material to the answer. – KReiser Jan 07 '23 at 03:54
  • Thank you so much! This is perfect. I was a little confused how you knew that ${K(X) / K(X)^G}$ would be a Galois extension, but I found out on the Wikipedia page that an equivalent formulation of a field extension ${E/F}$ being Galois is that $F$ is the fixed field of $F$ under some finite subgroup ${G \leq \text{Aut}(E)}$. Thank you! – CoffeeBean Jan 07 '23 at 13:13