Suppose we have a full exceptional collection (F1,...,Fn) of coherent sheaves on a smooth projective variety X. The number n of sheaves in this collection is equal to the rank of the Grothendieck group K0(X).

Is there any relation between n and the rank of the Picard group Pic(X) or the dimension of X?

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    $\begingroup$ If $X = \mathbb{P}^n,$ then its Picard group has rank one for any n, but a full exceptional sequence consists of n+1 line bundles. So I doubt there is any relation in general between the two. $\endgroup$ Commented May 26, 2010 at 13:00
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    $\begingroup$ One simple relation is $n \ge \dim X + 2\rho(X) - 1$ (if $\dim X \ge 3$), where $\rho(X)$ is the rank of the Picard group. $\endgroup$
    – Sasha
    Commented May 26, 2010 at 19:19

1 Answer 1


As a far as I know, there is no obvious relation between the number of objects in a full exceptional collection and the dimension of $X$, unless you impose extra conditions on the full exceptional collection, as in Bondal and Polishchuk's 'Homological properties of associative algebras', where they prove that when the collection is 'geometric', then ${\rm rank}(K_{0})(X)={\rm dim} X + 1$. Their hypotheses apply for example to $\mathbb{P}^{n}$, where the standard full exceptional collection is $\mathcal{O},\mathcal{O}(1), ..., \mathcal{O}(n)$.

But for examples like $\mathbb{P}^{1} \times \mathbb{P}^{1}$ these hypotheses do not apply, since the standard full exceptional collection is $\mathcal{O}, \mathcal{O}(1,0), \mathcal{O}(0,1), \mathcal{O}(1,1)$, so ${\rm rank}(K_{0}(X))={\rm dim} X + 2$.

In general, a variety $X$ will not admit a full exceptional collection of objects in its derived category of coherent sheaves, for instance because $K_{0}(X)$ is usually not free of finite rank. And some varieties cannot carry any exceptional objects at all. For instance, if $X$ is smooth projective with trivial canonical bundle, then if you checked that $Hom(E,E)$ were one dimensional as needed for $E$ to be exceptional, Serre duality would imply that $Ext^{n}(E,E)$ is also one dimensional.

To me it seems more natural to ask for a nice 'compact generator' of the derived category of a variety. This is a perfect complex $E$ on $X$ (one locally quasi-isomorphic to a finite length complex of vector bundles) that generates the category $Perf(X)$ of perfect complexes in finitely many steps for each object or $D_{QCoh}(X)$, the unbounded derived category of quasi-coherent sheaves, in infinitely many steps. This is a generalization of the sum $E$ of the objects in a full exceptional collection, with no extra conditions imposed on the $Ext$ algebra of $E$.

Such compact generators exist in great generality, due to work of Bondal and van den Bergh. Given such a compact generator and letting $A=RHom(E,E)$, the derived endomorphism dg algebra of $E$, we get equivalences $RHom(E,?): D_{QCoh}(X) \rightarrow D(A)$ and $RHom(E,?): Perf(X) \rightarrow Perf(A)$, just as we would if $E$ were the sum of objects in a full exceptional collection.

One could ask if you can find concrete examples of such compact generators, and indeed you can. If $X$ is projective of dimension $d$, let $L$ be an ample, globally generated line bundle. Then it is easy to show using a standard criterion for compact generation that $E=\mathcal{O}\oplus L \oplus \cdots \oplus L^{d}$ is a compact generator of the derived category of $X$.

So one can always say that if $X$ is projective, then the minimal number of line bundles (not necessarily exceptional) needed to generate the derived category of $X$ is bounded by ${\rm dim} X + 1$.

I would also guess that fewer line bundles will not suffice, but I'm not sure.


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