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 scheme, $S$ a $K3$ surface and $F$ a flat family of coherent sheaves on $S$ parametrized by $X$. Let us assume that for every $x\in X$ $F_x$ is locally free, has fixed Chern classes and satisfies $h^1(F_x)=h^2(F_x)=0$. Does there exist a scheme $A$ together with a morphism $p:A\to X$ such that $p^{-1}(x)$ is isomorphic to $Aut(F_x)$ for every $x\in X$? (by $F_x$ I mean $F|_{X_x}$ with $X_x$ the fiber of the projection $X\times S\to X$ over the point $x$)

share|improve this question
    
Two quick comments: 1. the Chern class of Fx is automatically fixed; 2. the assumption h1=h2=0 is irrelevant, because you can always twist F by a line bundle to ensure this. –  t3suji Oct 9 '10 at 14:42
add comment

2 Answers

up vote 1 down vote accepted

There are standard ways of constructing this kind of objects, but I can't immediately think of a reference, so here it goes:

Let $p:Y\to X$ be a projective map (in your case, $Y=S\times X$), let $F$ and $G$ be coherent sheaves on $Y$ with $G$ flat over $X$. Then there is a scheme $H$ of finite type over $X$ whose fiber over $x$ is the space $Hom(F_x,G_x)$. (More concretely, $H$ represents certain natural functor.) In your case, $F=G$, and your $A$ is an open subscheme of $H$ consisting of invertible homomorphisms, i.e., automorphisms.

Sketch of construction of $H$: Write $F$ as the cokernel of a map $$d:O(-n_1)^{N_1}\to O(-n_2)^{N_2}$$ for $n_1,n_2\gg 0$. Morphisms from $O(-n)$ to $G$ correspond to sections of $p_*(G(n))$; by assumptions, it is a vector bundle if $n\gg0$. Let $G_n$ be the total space of this vector bundle. Then $d$ induces a map $$(G_{n_2})^{N_2}\to (G_{n_1})^{N_1},$$ and $H$ is the preimage of the zero section.

share|improve this answer
add comment

You most certainly need to assume that $S$ is proper (or something similar); even when $X$ is the spectrum of a field but $S$ is affine, say, you will not get what you want (unless you accept something silly like taking $\mathrm{Aut}(F)$ as the disjoint union of copies of $S$). Assuming some properness you are still probably in trouble unless you assume that $p_\ast F$ (where $p\colon X\times S\rightarrow X$ is the projection) is locally free and commutes with scalar extension. That there is some problem can be seen by letting $X=\mathrm{Spec}R$, $R$ a discrete valuation ring, $S=\\mathrm{Spec}\mathbb Z$ and $F$ the sheaf on $X\times S=X$ associated to $R\bigoplus R/m$, where $m$ is the maximal ideal. Hence we are looking for a group scheme over $X$ whose generic fibre is the multiplicative group and whose special fibre is $\mathrm{GL}_2$ and for which the specialisation of the multiplication group is the natural subgroup of $\mathrm{GL}_2$. Unless I am mistaken such a group scheme does exist (you don't require $A$ to be a group scheme and you can certainly construct such a scheme). It doesn't look very nice however and things become much worse if you let $X$ have higher dimension and then if $p$ is not an isomorphism you also have to fight with the fact that $p_*F$ may not commute with base change which makes things even worse.

Hence even though I don't have a specific counter example to the existence of your $A$ fulfilling your very weak conditions, if you do not assume that $S$ is proper and that $p_*F$ is locally free and commute with base change I doubt that you can get an answer that you can actually use.

share|improve this answer
    
Sorry, I think I have been really too generic in asking my question. My situation is the following: $S$ is a K3 surface, for every $x\in X$ $F_x$ is locally free, has fixed Chern classes and $H^1(S,F_x)=H^2(S,F_x)=0$. Are these conditions enough in order to conclude the existence of $A$? –  ginevra86 Oct 9 '10 at 10:21
    
Yes, in that case you do have commutation with base change so you are perfectly alright. –  Torsten Ekedahl Oct 9 '10 at 10:58
    
Thank you very much! Am I alright in the sense that $A$ always exist? Do you have any reference? –  ginevra86 Oct 9 '10 at 13:13
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.