# Cohomologies with line bundle vs. coherent coefficients

I recently learned in a lecture that the derived category of a smooth variety is generated/spanned by (complexes of) locally free sheaves. (Unfortunately I haven't been able to find a more precise statement.)

Motivated by this, I was wondering if in certain cases, knowledge of cohomologies with line bundle coefficients implies knowledge of the cohomologies with arbitrary coherent coefficients.

For example, suppose that $X$ is a smooth quasi-projective variety (over $\mathbb C$) such that $H^i (X, \mathcal L) = 0$ for $i > 0$ and all line bundles $\mathcal L$; and suppose that all vector bundles are decomposable (i.e. as direct sums of line bundles). In particular, all locally free sheaves have no cohomology in positive degree.

Does this imply that $H^i (X, \mathcal F) = 0$ for $i > 0$ and all coherent sheaves on $X$?

• If $X$ is projective and every line bundle has vanishing higher cohomology, then $X$ must be a bunch of points by Serre duality. I suspect something like this could be true also in the quasi-projective case. – Daniele A Jun 16 '16 at 18:54
• @DanieleA But Serre duality is only true for projective varieties. I think that if X is the complement of an ample divisor of a projective variety, then X is affine and all higher cohomologies vanish. But of course X is not a bunch of points. – Earthliŋ Jun 17 '16 at 6:11
• Oh yes, sorry I was being stupid. – Daniele A Jun 22 '16 at 7:27

Yes, and for this the abelian version of the theorem on the derived category suffices: Since any coherent sheaf has a finite resolution by locally free sheaves, iterated use of the long exact cohomology sequence together with the assumed acyclicity of locally free sheaves shows that all coherent sheaves are acyclic.

Edit I am sorry that my first answer was too short to be comprehensible. From your post, I had - probably mistakenly - thought that you had actually proved the theorem about the derived category that you quoted, and assumed that it would have been the 'usual' way via proving an 'abelian version' of it first. Anyway, let me give details now.

There is an important (absolute) property of schemes, the resolution property or the existence of enough locall free sheaves:

Definition: A Noetherian scheme $X$ is said to have enough locally free sheaves if any coherent sheaf ${\mathscr G}$ on $X$ admits an epimorphism ${\mathscr F}\twoheadrightarrow {\mathscr G}$ from a locally free sheaf.

Example 1: An affine scheme has enough locally frees as these correspond to projective modules.

Example 2: Any quasi-projective scheme over a Noetherian ring has this property. Interestingly for your problem, you can even choose ${\mathscr F}$ to be a sum of line bundles: see Corollary 5.18 in Hartshorne.

If a scheme has enough locally free sheaves, you may iterate the choice of locally free covers and see that any coherent sheaf admits a (in general unbounded) resolution by locally free sheaves.

Now, suppose your scheme $X$ is regular of finite Krull dimension $n$ (a smooth variety over a field would do, for example) and that $\ldots\to {\mathscr F}_n\to {\mathscr F}_{n-1}\to\ldots\to{\mathscr F}_0\to {\mathscr G}\to 0$ is a locally free resolution of the coherent sheaf ${\mathscr G}$. Then the kernel ${\mathscr K} := \text{ker}\left({\mathscr F}_n\to {\mathscr F}_{n-1}\right)$ is a locally free sheaf, too: Since this is a local property, it suffices to check that all stalks are free, and this follows since the local rings ${\mathscr O}_{X,x}$ are regular local of dimension $\leq n$, hence of gobal dimension $\leq n$. Hence, truncating the resolution after step $n$ yields a finite resolution of ${\mathscr G}$ by locally free sheaves.

Now, if you have any cohomological functor ${\mathsf T}^{\ast}$ on an abelian category (i.e. a family of functors equipped with connecting homomorphisms turning short exact sequences into long exact sequences - derived functors like sheaf cohomology, for example), then the category of objects ${\mathscr G}$ such that ${\mathsf T}^i(X)=0$ for $i>0$ - the ${\mathsf T}$-acyclic objects - is closed under cokernels. Iterating, it follows that if you have a finite resolution $0\to {\mathscr F}_n\to\ldots\to{\mathscr F}_0\to {\mathscr G}\to 0$ such that all ${\mathscr F}$ are ${\mathsf T}$-acyclic, then ${\mathscr G}$ is ${\mathsf T}$-acyclic, too.

Applying this to your situation, you indeed get the vanishing of all cohomology of line bundles suffices to show the vanishing of all cohomology (and hence your variety is already affine, by Serre's criterion).

• Looking at the definition of "coherent" sheaf, I think that such finite resolutions by locally free sheaves only exist locally. (Is this different in the derived category? What do you mean by "abelian version of the theorem on the derived category"?) Is this what you are saying with the "iterated use of the long exact cohomlogy sequence", that to pass from local to global, there can be no obstructions ever, since at each step we are dealing with locally free sheaves (and in principle with their non-trivial extensions, which don't exist)? – Earthliŋ Jun 16 '16 at 11:35
• Would you be able to direct me to some references? – Earthliŋ Jun 16 '16 at 11:36
• @Earthliŋ Concerning what part precisely? – Hanno Jun 16 '16 at 11:49
• @Earthliŋ Sorry for my stupid question, I was in a hurry and missed your longer first comment. I will answer later when I have time. – Hanno Jun 16 '16 at 15:45
• Oh, that's not a big thing, on a Noetherian scheme any coherent sheaf on an open subscheme is isomorphic to the restriction of a coherent sheaf on the whole scheme. So the result on projective schemes implies the quasi-projective case. – Hanno Jun 17 '16 at 10:17