Timeline for Adelic description of moduli of $G$-bundles on a curve
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2012 at 2:33 | comment | added | user27056 | Dear RK: By Steinberg's theorem (Serre's conjecture), any connected smooth linear algebraic group over a perfect field of cohomological dimension $\le 1$ has vanishing degree-1 Galois cohomology. By Tsen, this applies to the function field of your Riemann surface (viewed as an algebraic curve, say), so your principal bundle extends across the missing point by "gluing" it to the trivial bundle over a Zariski-open neighborhood of the puncture. Now comes the overkill reference in English: apply Theorem 3 (also see Remark 2(b)!!) in the short 1995 paper by Drinfeld and Simpson in MRL #2. :) | |
| Nov 17, 2012 at 1:31 | comment | added | user27056 | Dear Justin: You have the good fortune to be a graduate student at Harvard, so all you need to do is to talk to people in your department. (I don't know any references; I figured it out for myself by thinking carefully. Beauville-Lazslo is helpful to "algebraize" from the completed data.) Anyway, one should only consider triviality at the generic point (not Zariski-locally), and in practice it is only reasonable to consider $G$-torsors for the etale topology (this is why Serre "invented" it...). So basically, this stuff is "only" good for the simply connected case (or variants like GL$_n$). | |
| Nov 16, 2012 at 23:54 | comment | added | Raju | On a related note, xbnv: do you know of a non-German reference for the fact that a semi-simple principle G-bundle on a once punctured Riemann surface is trivial? | |
| Nov 16, 2012 at 19:35 | comment | added | Will Sawin | (as I did in my answer) | |
| Nov 16, 2012 at 19:34 | comment | added | Will Sawin | $PGL_n$ is reductive, so that might not be enough. If you just take "$G$-bundle" to mean "$G$-bundle locally trivial in the Zariski topology" then you're fine. | |
| Nov 16, 2012 at 19:29 | comment | added | Justin Campbell | @xbnv: I was not aware that this was so subtle! Do you have a reference for these statements? I mostly care about the cases where $k$ is finite or algebraically closed and $G$ is reductive. | |
| Nov 16, 2012 at 19:25 | comment | added | user27056 | This answer overlooks crucial hypotheses without which there is no such bijection (only an injection): one has to know that all $G$-bundles are trivial at the generic point and that all $G$-bundles over finite extensions of k are trivial (such as for finite $k$ and connected $G$). Those conditions are where the "almost everywhere integral" aspect of adelic points is logically relevant for surjectivity. So things are good for GL$_n$ but certainly not PGL$_n$, for example. And likewise for simply connected semisimple $G$ and finite $k$ one is in good shape by Harder's theorem, etc | |
| Nov 16, 2012 at 19:08 | comment | added | Justin Campbell | Thanks! This is all very nice. I see the shape of it now, and I should work out the details for myself... | |
| Nov 16, 2012 at 19:06 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 219 characters in body
|
| Nov 16, 2012 at 19:01 | comment | added | Will Sawin | In the world of categories, quotienting by a group action where $g(x)=y$ can be equivalently seen as adding a morphism, $g$, from $x$ to $y$. When you take a quotient by an action that's not free, like in a double coset, this gives the objects some automorphisms. This is usually the right thing to do for a moduli problem - it usually finds the correct automorphism groups for your objects. This allows us to find the category structure - it's exactly the category induced as a group quotient. Then we say that the trivial bundle corresponds to the identity in $G(\mathbb A_K)$. | |
| Nov 16, 2012 at 18:57 | comment | added | Will Sawin | Well I mistyped some things, which might be part of the problem. I also don't think this is the best proof of the bijection - it's just the first one I thought of. The idea is that, since principal $G$-bundles form a category, we should first understand the categorical structure on $G(\mathbb A_K)$. This categorical structure comes naturally from the double coset structure. Then we use the fact that sections are the same thing as maps from the trivial bundle. Then I guess a better way to say it is that the sections form a locally trivial sheaf. | |
| Nov 16, 2012 at 18:48 | history | edited | Will Sawin | CC BY-SA 3.0 |
deleted 3 characters in body
|
| Nov 16, 2012 at 18:44 | comment | added | Justin Campbell | Thanks for the explanation. I have to admit that I find this very hard to follow. The first construction is very interesting, but I still don't see how to get from an adelic point to local sections of the corresponding $G$-bundle. I can't understand the other direction at all... This is probably my fault, but could you perhaps go through the constructions a bit more slowly? | |
| Nov 16, 2012 at 18:39 | comment | added | Will Sawin | The level structure on a principal $G$-bundle is, for each local ring $R$ of $X$, a fixed section of the bundle pulled back to $\hat{R})$, or, equivalently, a compatible family of sections of the bundle pulled back to $R/m$, $R/m^2$, etc. This is clear from the categorical way of looking at it - just literally pull back to that ring! | |
| Nov 16, 2012 at 17:44 | history | answered | Will Sawin | CC BY-SA 3.0 |