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Let $G$ be a simple algebraic group, $\mathcal{O}=\mathbb{C}[[t]], \mathcal{K}=\mathbb{C}((t))$ and let $\text{Gr}_G=G(\mathcal{K})/G(\mathcal{O})$ be the affine Grassmanian. My main question:

  • Why does $\text{Gr}_G$ parametrize the data of a $G$-bundle on $\mathbb{P}^1$, and a trivialization away from a point $x \in \mathbb{P}^1$?

There are 2 other related statements which I don't understand:

  • Why does $\text{Gr}'_G = G(\mathcal{K})/G(\mathbb{C}[t^{-1}])$ (the "thick" affine Grassmanian) parametrize the data of a $G$-bundle on $\mathbb{P}^1$, trivialized on the formal neighbourhood of $0$?
  • Why is $\text{Bun}_G(\mathbb{P}^1) = \text{Gr}'_G/G(\mathcal{O})$?
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When you write the symbol "$=$", do you mean an bijection on sets of complex points, or an isomorphism of sheaves of sets? Alexander Chervov's answer answers the point-set question, but the sheaf question requires a descent theorem (e.g., Beauville-Laszlo). –  S. Carnahan Jul 7 '13 at 1:49
    
Thanks! So how would you expand the answer below to answer the same question for sheaves of sets? –  Vinoth Jul 7 '13 at 8:35
    
@Scott what means sheaves of sets? –  Alexander Chervov Jul 7 '13 at 10:31
    
@AlexanderChervov The affine Grassmannian has a set of $S$-points for any scheme $S$, and this makes it a functor from schemes to sets. These sets satisfy descent along certain covers of $S$, making the affine Grassmannian a sheaf in most useful topologies. If you want to identify this sheaf with a moduli problem involving $G$-bundles, you have to work with $G$-bundles on all possible base changes of $\mathbb{P}^1$. –  S. Carnahan Jul 17 '13 at 4:57
    
@S.Carnahan Thank you ! –  Alexander Chervov Jul 17 '13 at 5:20

1 Answer 1

up vote 3 down vote accepted

I am not sure I understand your question correctly, it might be you asking something deeper. But let me write standard words (which you probably know). Let me consider the case of vector bundles, general case is similar.

Remark 1. If you have any manifold and two charts U_1 U_2 you can construct a bundle taking an a gluing map O^r(U_1)->O^r(U_2), which should be defined and invertible on the intersection of charts. Moreover we have a map $G(U_1) \backslash G(U_1 \cap U_2 )/ G(U_2)$ to the set if G-bundles. These more or less by definition. But saying it more explicitly we should say: take trivializations at charts $U_i$ and take gluing map as element of $G(U_1 \cap U_2)$. Different choices of trivializations are governed by $G(U_i)$.

Take $P^1$, point $x=0$ and chart around it: $U_1 = Spec k[[t]]$ (formal neighborhood of $x=0$), take another chart $U_2$ which is $P^1 \backslash x = Spec k[t^{-1}]$. Now the intersection of the two charts is $ Spec k((t))$.

Now if you apply the Remark 1 to $P^1$ and charts $U_1, U_2$ described above you will get that the map $ G[t^{-1}] \backslash G(k((t)))/ G[[t]] $ to $Bun_{P^1}$. This is what you are asked in your last question.

In your main question you ask about $G(k((t))/G(k[[t]])$. So element from $G(k((t)))$ defines a bundle, we kill all trivializations in first chart $U_1$ , but we have all trivializations at the second chart $U_2$ untouched, so we see that $G(k((t))/G(k[[t]])$ describes a bundle plus trivialization in $U_2$. That is what you want, similar about the second question.

Now why we get ALL bundles in that way. That is because any bundle on charts $U_i$ is trivial, since $k[[t]]$ is local ring and any module over local ring is trivial and $k[t^{-1}]$ are polynoms again any module is trivial since Euclid algorithm.

Actually similar words can be used to explain the adelic coset for vector bundles: Adelic description of moduli of G-bundles on a curve.

What is more interesting that the fact is true for any curve, not only $P^1$. This is because curve without point is actually affine manifold, and on affine manifold cohomology of coherent sheaves vanishes, so any bundle can be splitted into sum of 1-dimenstionals, and so reduce to the determinat of your bundles which is trivial since G is simple , but not just GL.

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