Grothendieck group generated by classes of invertible sheaves

Question

Given a smooth, projective (complex) varieties $X$, is it true that the grothendieck group $K_0(X)$ of equivalence classes of coherent sheaves on $X$, is generated by clases of invertible sheaves i.e., any class of a coherent sheaf on $X$ can be written as a linear combination of classes of invertible sheaves on $X$? Any reference/idea will be most welcome.

Answer 1

Here is a different, perhaps more elementary example.

Consider the Grassmannian $\mathrm{Gr}(2,4)$ of lines in $\mathbf{P}^3$, and let $\mathscr{Q}$ be the universal quotient bundle. The line bundle $\mathrm{det}(\mathscr{Q})$ generates the Picard group of $\mathrm{Gr}(2,4)$. If $[\mathscr{Q}]$ were in the subgroup of $\mathrm{K}^0$ generated by line bundles, then we would have $[\mathscr{Q}]=\sum_i n_i [\mathrm{det}(\mathscr{Q})^{\otimes r_{i}}]$ for some integers $n_i, r_i$. Taking Chern classes implies $$c_2(\mathscr{Q})= m c_1(\mathscr{Q})^2$$ for some integer $m$, which cannot be true. Indeed, fix a point $p$ and a line $L$ in $\mathbf{P}^3$; the Chern class $c_1(\mathscr{Q})$ is geometrically represented by the Schubert cycle $\Sigma_1(L)=\{L' \ | \ L\cap L'\neq\emptyset\}$, while $c_2(\mathscr{Q})$ is represented by $\Sigma_2(p)=\{L' \ | \ p\in L'\}$. Then $$1=\int_{\mathrm{Gr}(2,4)} c_2(\mathscr{Q})^2\neq m^2\int_{\mathrm{Gr}(2,4)} c_1(\mathscr{Q})^4=2m^2.$$ The integral on the left is the number of lines through two distinct points in $\mathbf{P}^3$, while the integral on the right is the number of lines meeting four given (general) lines in $\mathbf{P}^3$.

I think only few varieties have the property which you want. It is certainly true for curves: if you have a locally free sheaf $\mathscr{E}$ on a curve, then you can find an exact sequence of locally free sheaves $$0\rightarrow\mathscr{E}'\rightarrow\mathscr{E}\rightarrow\mathscr{E}''\rightarrow 0,$$ where the ranks of $\mathscr{E}'$ and $\mathscr{E}''$ are strictly smaller than $\mathscr{E}$ (unless, of course, $\mathscr{E}$ is already of rank $1$). The same property holds if $\mathscr{E}$ is a locally free sheaf of rank $>2$ on a surface. Indeed, after twisting with an invertible sheaf, we may assume that $\mathscr{E}$ is globally generated; by a standard result (e.g. Exercise 8.2 in Chapter 2 of Hartshorne) we can then find a nonzero section $s:\mathscr{O}\rightarrow\mathscr{E}$ whose cokernel is locally free. The K-theory of surfaces can thus be generated by locally free sheaves of rank $1$ and $2$, and analogous statements hold in higher dimension.

Answer 2

The answer is "no" in general, even if one asks if the classes of invertible sheaves generated $K_0(X)$ as a ring; K3 surfaces provide a counterexample. For a K3 surface $X$ over $\mathbb{C}$, the subring of $$K_0(X)_\mathbb{Q}$$ has uncountable dimensional as a $\mathbb{Q}$-vector space; this follows, for example, from the fact that this vector space is isomorphic to the rational Chow ring of $X$ (which was proven to be very large by Mumford). But the subspace generated by line bundles has countable dimension (as the Picard group of $X$ is a finitely generated abelian group of rank at most 20).

EDIT: Here's a slightly different argument, with references. Consider any K3 surface $X/k$, with $k$ algebraically closed. Let $K=k(X)$ be the function field of $X$. Then by Lemma 2.9 here, the map $$K_0(X)_{\mathbb{Q}}\to K_0(X_K)_{\mathbb{Q}}$$ is injective; by Proposition 2.10, it is not surjective. But the span of the invertible sheaves is contained in its image (again by the discreteness of $\text{Pic}(X)$), so we're done. If you'd like an example over an algebraically closed field, you can extend to the algebraic closure of $K$ (which again works by Lemma 2.9 of the Huybrechts reference above).