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

Let $X$ be a smooth projective curve over a field $k$ and let $L$ be a line bundle on $X$. I will denote by $S$ the total space of $L$ -- this is a smooth surface over $k$ containing $X$ (as the zero section). Let $S^0$ be the complement to the zero section. Let also $G$ be some reductive group (you can assume that $G$ is $GL(n)$ or $SL(n)$ if you wish). I need some finiteness results about moduli spaces of $G$-bundles on $S$ and $S^0$. Specifically, I would like to prove (or disprove) the following statements:

1) Assume that $deg(L)<0$. Consider all $G$-bundles $F$ on $S$ whose restriction to $X$ is fixed (i.e. it is isomorphic to a fixed $G$-bundle on $X$). Is it true that when $k$ is finite, the number of isomorphism classes of such bundles is finite? For general $k$, is it true that this (non-algebraic) stack can be covered by a scheme of finite type over $k$? I think that it is easy to prove this when we replace $S$ by the formal neighbourhood of $X$ in $S$ (by deformation theory), but is it true for $S$ itself?

2) Consider now $G$-bundles on $S^0$ and assume that $deg(L)\neq 0$ (note that $S^0$ doesn't change when we replace $L$ by $L^{-1}$, so you may as well assume that $deg(L)<0$). In the case when $k$ is finite I would like to prove the number of isomorphism classes of such bundles is finite. This is easy if $G=GL(1)$ (i.e. when we talk about line bundles); also it is not difficult when $X={\mathbb P}^1$ (in this case the number of isomorphism classes is finite for any $k$ -- not necessarily finite). For general $k$ there should probably be a statement that the (non-algebraic) stack $Bun_G(S^0)$ can be covered by a scheme of finite type over $k$ (but again, I just need the statement for the finite field).

share|improve this question

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.