Let $S$ be a $K3$ surface. Is it true that any sheaf on $S$ with zero Chern classes is isomorphic to $\mathcal{O}_S^{\oplus n}$ for some $n$? If not, do you have any counterexample?
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1$\begingroup$ Are you interested in all sheaves or just coherent ones? It is just that I have only ever seen the definition of Chern classes for coherent sheaves. $\endgroup$– Daniel LoughranDec 6, 2011 at 14:54
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$\begingroup$ I mean coherent sheaves $\endgroup$– ginevra86Dec 6, 2011 at 15:25
1 Answer
The answer is no. Here is a counterexample.
Take an ample divisor $L$ on $S$ and let $Z \subset S$ be a zero-dimensional subscheme of length $\ell(Z)=L^2$.
Now consider the coherent sheaf $$\mathscr{F}=\mathscr{O}_S(-L) \oplus \mathscr{O}_S(L) \otimes \mathscr{I}_Z.$$
Straightforward computations show that $$c_t(\mathscr{O}_S(-L))=1-Lt, \quad c_t(\mathscr{O}_S(L)\otimes \mathscr{I}_Z)=1+Lt + \ell(Z) t^2,$$
hence $c_t(\mathscr{F})=1$.
So $\mathscr{F}$ has zero Chern classes, but it is not isomorphic $\mathscr{O}_S^{\oplus 2}$ because it is not locally free.
Note that this construction holds for any smooth projective surface $S$, in fact the assumption $K3$ is not used here.