Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a pretty basic question, so I'd be happy with either standard references or with explanations. Also, there's a good chance I'm confused about some things in the statement of the question, and corrections of these would also be cool.

If G is a finite group acting on a (commutative) $ \mathbb{C} $-algebra A, then we can define $Spec(A)/G$ as $Spec(A^G)$ (where $A^G$ is the invariant sublagebra of A), and there is an algebraic map $Spec(A) \to Spec(A^G)$ whose "fibers" correspond to orbits of G on $Spec(A)$.

If L is a lattice in $ \mathbb{C} $ then $ \mathbb{C} / L$ as a topological space can be given the structure of a scheme (it's an elliptic curve). However, this scheme is not $Spec(\mathbb{C}[x]^L)$, since the only polynomials invariant under the action of a lattice are the constant polynomials. Is there a construction which replaces $Spec(\mathbb{C}[x]^L)$ in this case?

Now let's assume X is a smooth affine scheme and G is a countable group acting freely on X (in my case $G = SL_2(\mathbb{Z})$, but I would guess this isn't too important). Since X is smooth we can also view X as a complex manifold $X^{an}$, and then the quotient $X^{an}/G$ is a topological space. Can X/G naturally be given the structure of a scheme? If the answer is no in general, are there conditions on the data X, G that ensure the answer is yes?

share|improve this question
    
The study of quotients of schemes by group actions is called geometric invariant theory and the standard reference is Mumford's book. You usually end up with something called a stack but in some cases you can say more –  Frank Mar 11 '10 at 20:53
4  
You definitely won't end up with a scheme in general. For instance, take a lattice $L$ in $\mathbf{C}^2$. For some $L$, the quotient $\mathbb{C}^2/L$ will admit an algebraic structure and other times it won't. In general, the map to the analytic quotient $X\to X/G$ won't be algebraic even if $X/G$ is algebraizable (e.g. elliptic curves), so even when $X/G$ is algebraizable, there's not going to be a general approach to proving it, and so there probably aren't any general theorems. Any results will depend crucially on the specifics of $X$, $G$, and $X/G$ and will require some honest work. –  JBorger Mar 11 '10 at 21:15
3  
Somehow the problem with the C-->C/L example is that C is a variety, C/L is a variety, but the map C-->C/L isn't a morphism of varieties (as it's not locally defined by polynomial equations). Using the word "scheme" instead of variety doesn't help here either---it's not a map of schemes. The quotient really is going on in another category. As James says, there are instances (for example the Baily-Borel theorem) where certain quotients of symmetric spaces by arithmetic groups are known to be schemes, but you can't really expect a general theorem. –  Kevin Buzzard Mar 11 '10 at 21:21
2  
Here's another way of putting things which I thought of on the bike: The statement that $\mathbf{C}/L$ is a variety is an algebraic statement about an analytic object. The content is essentially that there are meromorphic functions on $\mathbf{C}$ (the Weierstrass p-function and its derivative) which are invariant under $L$, separate points, and satisfy a polynomial relation. The existence of such a set is not obvious at first. To prove an analytic quotient is algebraizable, you're essentially going to have to come up with something similar. And that will require some real work. –  JBorger Mar 11 '10 at 22:24
    
Thanks for these comments, they were definitely useful. –  Peter Samuelson Mar 15 '10 at 16:06
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.