# Grothendieck group generated by classes of invertible sheaves

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.

• @TimCampion: I'm not sure I understand your hint -- won't such resolutions be of infinite length in general? It seems to me one needs to allow more general vector bundles to get everything... Feb 8, 2021 at 23:29
• I think K3 surfaces should be a counterexample to the original question (writing up something now). In general for smooth projective things K_0 is indeed generated by vector bundles, but it's definitely not so obvious (and it's false without smoothness)... Feb 8, 2021 at 23:46
• mathoverflow.net/questions/380487/… - here you find a similar question.
– user122276
Feb 9, 2021 at 9:30

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.

• Thanks, that helps. Feb 9, 2021 at 2:36
• I'm probably missing something silly -- why is the sentence "If the rank 2 bundle 𝒬 were in the subgroup of K0 generated by line bundles, then its K-theory class would therefore have to be 2[det(𝒬)]" true? Feb 10, 2021 at 16:07
• @DanielLitt Thanks for pointing this out; I think it is a mistake, which should be fixed now.
– ssx
Feb 10, 2021 at 18:17
• @SWS: Thanks, looks good now! Feb 10, 2021 at 22:08
• Sort of fun to note that if you want to show that line bundles don't span $(K_0)_\mathbb{Q}$, you end up using that the square root of $2$ isn't rational. Feb 10, 2021 at 22:11

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).

• To be clear, that's Lemma 2.9 of Chapter 12. Feb 9, 2021 at 0:23
• @DanielLitt: Thanks for the counter-example. Are there some known examples where the question has a positive answer? I am interested in dimension at least $3$. Feb 9, 2021 at 0:42
• @user45397 Well, it’s true for projective spaces! Feb 9, 2021 at 0:45
• @user45397: It is also true for toric varieties. Feb 9, 2021 at 7:56