# Generating the derived category with line bundles

The following lemma is useful and well-known:

LEMMA If $L^{\pm 1}$ is ample on proper scheme over a field $k$, then some number of powers $\mathcal{O},L,...,L^{m}$ generate the unbounded derived category of quasi-coherent sheaves $D(X)$ (or split generate the subcategory of perfect complexes).

QUESTION: What about a converse? Suppose that I know some number of powers of $L$ generate $D(X)$. Then can I conclude that $L^{\pm 1}$ is ample?

The best I can do so far is see that the restriction of $L$ to any integral curve $C$ in $X$ has non-zero degree. (Since by adjunction $\mathcal{O},L,...,L^{m}$ generates $D(C)$, but if $L$ had degree $0$ on $C$, there would be something orthogonal $\mathcal{O},L,...,L^{m}$, for instance a generic line bundle of degree $g-1$ having no cohomology.)

Something I don't know yet: does the degree of $L$ must have the same sign on all curves? This would be useful for numerical tests of ampleness.

Note: I think that one doesn't need properness in the above lemma, but I am willing to assume it to get a converse. It makes life easier when restricting to closed subschemes.

Note 2: When saying a collection of objects generates a triangulated category with all coproducts, like $D(X)$, one usually means that you take the smallest triangulated subcategory closed under all coproducts and containing the the collection. Once you have all coproducts, then idempotents automatically split, by a standard argument called, I think, the Eilenberg swindle. If you are working with a smaller triangulated category having only finite coproducts, like perfect complexes on a scheme, then the smallest triangulated subcategory containing a collection might not be 'thick', in the sense that some idempotents might not split, so in this case one usually adds in the missing summands. To emphasize this, some people speak of 'split generation'.

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I think you mean to say that these powers split-generate the derived category (otherwise, all varieties would have finite rank $K_0$ which fails already for curves). – Mohammed Abouzaid Aug 17 '10 at 14:51
Can you give a reference for the lemma? Thanks! – Fei YE Aug 17 '10 at 15:29
One reference would be arXiv:0804.1163, Theorem 4. If you are willing to assume $X$ proper, then you can give a shorter argument by replacing $L$ with some power that is both ample and globally generated, use this to get a finite morphism to some projective space, and then pull-back the usual generator $\mathcal{O}\oplus \mathcal{O}(1)\oplus \cdots \mathcal{O}(n)$ from there. – Chris Brav Aug 17 '10 at 19:34
What are your assumptions on $X$? At the very least you will need irreducible... – Arend Bayer Aug 18 '10 at 18:10
I am happy to assume $X$ is irreducible and even integral, but why do you think the former is necessary? – Chris Brav Aug 19 '10 at 7:13