Compact generation and Grothendieck duality

We saw last semester that compact generation of a triangulated category was a very useful property. Here we establish that is compactly generated for a quasi-projective schemes over an affine base. In addition, we provide some useful replacements for deriving common functors.

Lemma. Let be a -graded ring. Then for is a set of compact generators for .

Proof. (Expand to view)

One checks that Thus, if is right orthogonal to all it must be acyclic.

 

Proposition. If is finitely-generated in degree over , then is compactly generated.

Proof. (Expand to view)

Let . Recall that we [have](2022_01_11.md) a SOD Given by and . Thus, This vanishes for all if and only if is acyclic. As we see that is acyclic also.

 

Lemma. If is compactly-generated and is an open subscheme, then so is .

Proof. (Expand to view)

Let denote the inclusion. Then the counit is an isomorphism. Thus, if is a set of compact generators for , then is for .

 

Corollary. Let be quasi-projective over an affine base. Then any compact object is quasi-isomorphic to a bounded complex of locally-free sheaves.

Proof. (Expand to view)

We have seen that form a set of compact generators. As we saw [last semester](https://738.f21.matthewrobertballard.com/notes/2021_10_05/), the compact objects in are summands of iterated cones over maps to objects from the set of compact generators. Such a complex is bounded and has locally-free components.

 

We saw last time that is the Serre functor on . In fact, this is a consequence of a more general fact known as Grothendieck duality. One formulation of Grothendieck duality is that for a map of schemes possesses a right adjoint to . This is not automatic in full generality. But it is true with mild conditions on and .

For the projection \pi : \mathbb{P}^n_k \to \operatorname{Spec} k} we have

Originally proof of the existence of a right adjoint was arduous. However, [Neeman](https://www.ams.org/journals/jams/1996-9-01/S0894-0347-96-00174- 9/S0894-0347-96-00174-9.pdf) demonstrated how Brown Representability can be used to give a clean proof.

Thanks to Brown Representability, we know that for to have a right adjoint is equivalent to the natural map being an isomorphism for all sums. This question is local on so we reduce to asking -- when does taking cohomology commute with direct sums?

One useful fact: formation of the Čech complex for an affine cover where all intersections are affine commutes with direct sums.

Proposition. For any map between quasi-projective schemes over an affine base, there is a right adjoint to .

Proof. (Expand to view)

From the discussion above, we reduce to the checking the case where is affine. Here we are computing global sections over a quasi-affine scheme which can be done using a Čech complex.

 

Remark. Neeman proves the existence of whenver and are quasi-compact and quasi-separated.

Deriving tensor and sheaf Hom

We have implicitly used both the tensor product and sheaf Hom of -modules.

Recall that is locally-free if each has a neighborhood with Since exactness can be checked locally, preserves exactness of complexes of -modules if is locally-free. Thus, if we can replace any complex with one whose components are locally-free, we can use these adapted objects to derive .

Proposition. For any quasi-projective schemes and any chain complex of quasi-coherent sheaves, there is complex with locally-free components and a quasi-isomorphism .

Proof. (Expand to view)

It suffices to prove this for projective . Any complex of quasi-coherent sheaves on is of the form for some complex of graded modules . There are enough K-projectives in K(\operatarname{GrMod} R). Indeed, the twists of suffice. Thus, we get a -projective and a quasi-isomorphism . Applying gives the result.

 

Note that despite the notation we cannot find enough K-projective complexes in .

Using this result, we take for some such .

Similarly, we can derive by replacing by a complex of locally-frees or by a K-injective. The choices are naturally quasi-isomorphic so we denote either by