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.

2$\begingroup$ @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... $\endgroup$– Daniel LittFeb 8, 2021 at 23:29

1$\begingroup$ 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)... $\endgroup$– Daniel LittFeb 8, 2021 at 23:46

1$\begingroup$ mathoverflow.net/questions/380487/…  here you find a similar question. $\endgroup$– user122276Feb 9, 2021 at 9:30
2 Answers
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 Ktheory of surfaces can thus be generated by locally free sheaves of rank $1$ and $2$, and analogous statements hold in higher dimension.


1$\begingroup$ I'm probably missing something silly  why is the sentence "If the rank 2 bundle đť’¬ were in the subgroup of K0 generated by line bundles, then its Ktheory class would therefore have to be 2[det(đť’¬)]" true? $\endgroup$ Feb 10, 2021 at 16:07

$\begingroup$ @DanielLitt Thanks for pointing this out; I think it is a mistake, which should be fixed now. $\endgroup$– ssxFeb 10, 2021 at 18:17


1$\begingroup$ 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. $\endgroup$ 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).

1$\begingroup$ To be clear, that's Lemma 2.9 of Chapter 12. $\endgroup$ Feb 9, 2021 at 0:23

$\begingroup$ @DanielLitt: Thanks for the counterexample. Are there some known examples where the question has a positive answer? I am interested in dimension at least $3$. $\endgroup$ Feb 9, 2021 at 0:42

$\begingroup$ @user45397 Well, itâ€™s true for projective spaces! $\endgroup$ Feb 9, 2021 at 0:45

2$\begingroup$ @user45397: It is also true for toric varieties. $\endgroup$– SashaFeb 9, 2021 at 7:56