Today’s main goal is the Leray theorem (though at the end I have to ask a question):

Theorem 1 Let {\mathcal{F}} be a sheaf on {X}, and {\mathfrak{U} = \{ U_i, i \in I\}} an open cover of {X}. Suppose\displaystyle H^n( U_{i_1} \cap \dots \cap U_{i_k}, \mathcal{F}|_{ U_{i_1} \cap \dots \cap U_{i_k}}) = 0

for all {k}-tuples {i_1, \dots , i_k \in I}, and all {n>0}. Then the canonical morphism\displaystyle H^n( \mathfrak{U}, \mathcal{F}) \rightarrow H^n( X, \mathcal{F})

is an isomorphism for all {n}

 

This seems rather useless, because the theorem presupposes the vanishing of (regular) cohomology on the covering. However, in many cases it turns out to be helpful. If {X} is a separated scheme, {U_i} an open affine cover of {X}, and {\mathcal{F}} quasi-coherent, it applies. The reason is that each of the intersections { U_{i_1} \cap \dots \cap U_{i_k}} are all affine by separatedness, so {\mathcal{F}} has no cohomology on them by a basic property of quasi-coherent sheaves. This gives a practical way of computing sheaf cohomology in algebraic geometry. Hartshorne uses it to compute the cohomology of line bundles on projective space.

Another instance arises when {\mathcal{O}} is the sheaf of holomorphic functions over some Riemann surface {X}. In this case {\{U_i\}} is a covering of charts. It is a theorem (which I will eventually prove) that for any open subset of {\mathbb{C}} (which any intersection of the {U_i}‘s is isomorphic to), the sheaf {\mathcal{O}} has trivial cohomology.

With this, let’s now prove the result.

Proof of the Leray theorem

We’re going to establish this by induction on {n}. For {n=0}, we know it already. Given {\mathcal{F}}, pick an injection {\mathcal{F} \rightarrow \mathcal{G}} where {\mathcal{G}} is flasque (e.g. injective) and take the cokernel {\mathcal{H}}:

\displaystyle 0 \rightarrow \mathcal{F} \rightarrow \mathcal{G} \rightarrow \mathcal{H} \rightarrow 0.

Lemma 2 There is a functorial long exact sequence\displaystyle \dots \rightarrow H^n(\mathfrak{U},\mathcal{H}) \rightarrow H^{n+1}(\mathfrak{U},\mathcal{F}) \rightarrow H^{n+1}(\mathfrak{U},\mathcal{G}) \rightarrow H^{n+1}(\mathfrak{U},\mathcal{H}) \rightarrow \dots .

 

This follows from the fact that there is an exact sequence of Cech complexes of sheaves as defined previously:

\displaystyle 0 \rightarrow \mathcal{C}( \mathfrak{U} , \mathcal{F}) \rightarrow \mathcal{C}( \mathfrak{U} , \mathcal{G}) \rightarrow \mathcal{C}( \mathfrak{U} , \mathcal{H}) \rightarrow 0.

which leads to a short exact sequence of abelian groups of global sections, since

\displaystyle \mathcal{C}( \mathfrak{U} , \mathcal{F})

has trivial (regular) cohomology by assumption. In general, however, it is not true that Cech cohomology is a {\delta}-functor, i.e. sends short exact sequences to long exact ones.

We have a big commutative (see below) diagram

Now for any {n} (possibly {n=0}!), the rightmost terms are both zeros by flasqueness, since {\mathcal{H}} has trivial cohomology by the long exact sequence. The left-most and second-to-left most maps are isomorphisms by the inductive hypothesis, so in

the left two arrows are isomorphisms. Thus {H^{n+1}( \mathfrak{U}, \mathcal{F}) \rightarrow H^{n+1}( X, \mathcal{F})} is an isomorphism too, proving the Leray theorem.

What I still haven’t been able to figure out here is why the diagram is commutative. In particular, how do we know that the morphisms between Cech cohomology and derived functor cohomology commute with the boundary maps?

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