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Let $S$ be a smooth projective surface with an ample divisor $X\subset S$. Consider the moduli stack of vector bundles $F$ on $S$ such that

1) $c_1(F)=0$

2) $c_2(F)=n$

3) The restriction of $F$ to $X$ is fixed (say, isomorphic to a bundle $M$).

$\mathbf{Questions:}$

1) Is it true that for any $n$ and $M$ the above stack is of finite type?

2) Is it true, that for any given $M$ the above stack is empty for $n$ sufficiently small?

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1 Answer 1

Take Q 2). Assume that we are in characteristic zero, that $F$ has rank $2$ and that $X$ is smooth. If $n<0$ then $F$ is Bogomolov unstable: there is a short exact sequence $$0\to O(A)\to F\to I_Z.O(-A)\to 0$$ with $A^2>0$ and $A.X>0$ and $I_Z$ the ideal sheaf of a $0$-dimensional subscheme $Z$. In fact $c_2(F)=\deg Z-A^2$, so $A^2\ge -n$. Suppose that we can make $n$ arbitrarily small (i.e., large and negative). Then (index theorem) $A.X$ is arbitrarily large (and positive). On the other hand, restrict the inclusion $O(A)\to F$ to $X$; this gives a non-zero homomorphism $O_X(A)\to M$, which is impossible for fixed $M$.

So yes, your stack is empty in this case (rank $2$, char. $0$). In higher rank Bogomolov gives a destabilizing flag and maybe an elaboration of the argument above will then work. I don't know about char. $p$.

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Where did you use char=0? By the way, I am also not sure what you mean by $O(A)$ - do you mean that you have chosen an embedding into ${\mathbb P}^n$? –  Alexander Braverman Sep 14 '11 at 22:54
    
No I have not chosen $O(A)$, it is given by Bogomolov's theorem, which is this. Suppose that $S$ is a smooth projective surface over a field of char. zero and $F$ a rank $2$ vector bundle on $S$ with $c_1(F)^2>4c_2(F)$. Then there is a short exact sequence $$0\to O(A)\to F\to I_Z O(B)\to 0$$ where $O(A),\ O(B)$ are invertible sheaves on $S$ and $I_Z$ is the ideal sheaf of a $0$-dimensional subscheme $Z$ of $S$ such that $(A-B)^2>0$ and $(A-B).X>0$ for all ample $X$ on $S$. There is a version for bundles of higher rank and there are counter-examples in char. $p$. –  inkspot Sep 15 '11 at 10:53
    
Thank you very much! –  Alexander Braverman Sep 15 '11 at 13:22
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