# Cover for Cech cohomology of the constant sheaf $\mathbb{Z}$ over $\mathbb{P}^n$

First of all consider $\mathbb{P}^n$ as a complex analytic manifold. In Griffiths and Harris's Principles of Algebraic geometry p.145 it is stated $$H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z},$$ that is, the second Cech cohomology group of $\mathbb{P}^n$ over the constant sheaf $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$.

I wanted to check this using the usual algebraic cover of open sets: $\mathcal{U} = \{ U_i \}_{0 \le i \le n}$, where $U_i = \{ x_i \neq 0 \}$, but already for $\mathbb{P}^1$ this fails, meaning that $\mathcal{U}$ is not fine enough for computing Cech cohomology.

Edit: As Rene remarked in the answer below, there is a theorem of Leray which states that given a sheaf $\mathcal{F}$ and a cover $\mathcal{U}= \{ U_i \}$ such that $$H^p(U_{i_0} \cap \dotsb \cap U_{i_k}, \mathcal{F}) = 0$$ for all finite intersections of $\mathcal{U}$ and all $p$, then $$H^p(\mathcal{U},\mathcal{F}) \rightarrow H^p(X, \mathcal{F})$$ is an isomorphism for all $p$.

This still gives rise to questions:

1. Is there a standard, or at least well known, cover of $\mathbb{P}^n$ satisfying this condition?
2. Is there any intuition on how to choose such covers?
3. Is there any other direct method of seeing the isomorphism $H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}$ ?
• They must mean something else than the sheaf cohomology because th constant sheaf is flabby and flabby sheaves have zero cohomology in positive dimensions. – Rene Schipperus Jan 24 '17 at 20:24
• @Rene: Lucas's question is about $\mathbb P^n$ endowed with its classical topology, as he states in the first sentence. In that topology $\mathbb Z$ is not flabby. – Georges Elencwajg Jan 24 '17 at 22:44
• @GeorgesElencwajg I dont quite understand are not constant sheaves always flabby ? Or is $\mathbb{Z}$ not constant ? – Rene Schipperus Jan 24 '17 at 23:00
• @GeorgesElencwajg Oh you mean you can have the union of two disjoint open sets. Duh, too much algebraic geometry. – Rene Schipperus Jan 24 '17 at 23:04
• Dear @Rene, yes you nailed it : disjoint open subsets are the problem. Hausdorff spaces are too difficult for us poor algebraic geometers :-) – Georges Elencwajg Jan 24 '17 at 23:11

$\text{}$1. Is there a standard, or at least well known, cover of $\mathbb{P}^n$ satisfying this condition?

Not an explicit one as far as I know.

$\text{}$2. Is there any intuition on how to choose such covers?

Any covering consisting of geodesically convex sets works.

$\text{}$3. Is there any other direct method of seeing the isomorphism $H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}$?

The simplest approach in my opinion is to use the cell complex structure of $\mathbb{P}^n$: it has exactly one cell in each even dimension $0$, $2$, $\ldots$ , $2n$. Then use the definition of homology for CW complexes, and apply it to this cell decomposition.

I hope that this will not be the only answer to this question, but I'd like the point out that the theorem of Leray gives sufficent conditions for $$H^n(\mathcal{U}, \mathcal{F})\rightarrow H^n(X,\mathcal{F})$$ to be an isomorphism for all $n$, namely that for all $n$

$$H^n(U_{i_0}\cap \cdots \cap U_{i_k},\mathcal{F})=0$$ for all finite intersection of open sets from the cover. This is not satisfied by the open covering given in the question.

• Indeed, thanks! But this also raises more questions: Is there a known standard cover for $\mathbb{P}^n$ which is acyclic in that sense? Is there any intuition in finding such covers? I'll include that on the question to make it more precise and complete. – Lucas Das Dores Jan 25 '17 at 12:09
• Yeah I was thinking about that. For $\mathbb{P}^1$ think of the tetrahedron. For higher dimensions I dont know, but with this topology and this sheaf, ( which by the way is not constant) the cohomology is the same as singular cohomology, so it becomes an algebraic topology question. Maybe start with simplicial decompositions. – Rene Schipperus Jan 25 '17 at 12:33