# Why does exactness of Cech complex not make Cech cohomology trivial?

In my class notes we proved that the (Cech?; I'm not sure what it's usually called) complex $$\underline{C}^\bullet((U_i), \mathcal{F})$$ (Hartshorne Section III.4) is an exact cochain complex, where $$(U_i)$$ is an open cover of $$X$$. Hartshorne (in the first definition of Section III.4) defines the $$p$$th Cech cohomology group of $$\mathcal{F}$$ with respect to the covering $$(U_i)$$ to be $$\check{H}^p((U_i), \mathcal{F}) = h^p(\underline{C}^\bullet((U_i), \mathcal{F})).$$ But wouldn't this lead to the Cech cohomology always being trivial? Having done examples in this area before, I've always ended up taking global sections of $$\underline{C}^\bullet((U_i), \mathcal{F})$$ and then taking the cohomology of the resulting complex. So from the examples I've done/what I've just said, I'd expect the Cech cohomology groups to be defined something like $$h^p(\Gamma (X , \underline{C}^\bullet((U_i), \mathcal{F}))$$, but they're not.

I think I'm missing something really obvious here. Thanks for any answers.

Edit: red_trumpet has answered my original question in the comments, though I'm still interested in Q1 and Q2 which I've written in the comments.

To expand on Q2, I now understand that in my class notes I have the sheafified Cech complex $$\mathscr{C}^\bullet((U_i), \mathcal{F})$$ which satisfies $$\Gamma(X, \mathscr{C}^\bullet((U_i), \mathcal{F})) = C^\bullet((U_i), \mathcal{F})$$ where $$C^\bullet((U_i), \mathcal{F})$$ is the global Cech complex (I've switched to Harthsorne's notation for this edit).

If $$X$$ is affine, then $$\Gamma(X, -)$$ is an exact functor. I presume in class that we proved $$\mathscr{C}^\bullet((U_i), \mathcal{F})$$ is exact, so by the identification above, $$C^\bullet((U_i), \mathcal{F})$$ too would be exact. Then putting extra conditions on $$X$$ such that Cech cohomology and normal sheaf cohomology agree, we get that $$H^p(X, \mathcal{F})$$ vanishes for $$p>0$$. Have we recovered a frequently stated "affine $$\implies$$ vanishing cohomology" result here?

• The Cech complex is not exact in general. Are you missing some assumptions ? (For instance $\mathcal{F}$ flasque ?) – Roland May 31 at 15:17
• $U_i$ are open affines but apart from that I don't think so; perhaps I'm thinking about the definition of the Cech complex the wrong way. – mathphys May 31 at 15:20
• Even with $U_i$ affines this is not true. Do you have a specific lemma in mind in Hartshorne Section III.4 or is this from your class notes ? – Roland May 31 at 15:21
• There are two versions of the Chech complex: A global one, $C^p(\mathfrak{U}, \mathscr{F})$ (p. 218), and a sheafified on, $\mathscr{C}^p(\mathfrak{U}, \mathscr{F})$ (p. 220). Taking global sections of the sheafified version yields the global Chech complex. The sheafified one is in fact exact save for the beginning (Hartshorne proves this in Lemma 4.2), but taking global sections destroys this property. – red_trumpet May 31 at 15:24
• @red_trumpet Yes, this makes sense, thank you. I suspect the one in my class notes is the sheafified version (the definitions actually match). I have two more quick questions (I've split them into two comments): Q1: firstly, the sheafified version (p. 220) is therefore the one that is an exact complex as I proved in class, correct? – mathphys May 31 at 15:38

I'll attempt to answer my own Q2. Let $$X$$ be a noetherian scheme with a finite open affine covering $$(U_i)$$ such that any finite intersection of the $$U_i$$ is affine $$(*)$$, and let $$\mathcal{F} \in \mathsf{QCoh}(X)$$. Then using the Leray spectral sequence and the fact that for an affine morphism of schemes $$f: X \rightarrow Y$$ and for $$\mathcal{F} \in \mathsf{QCoh}(X)$$ we have $$R^pf_* \mathcal{F} = 0$$ for all $$p>0$$, it can be shown that we have the canonical isomorphism $$\Gamma(X, \mathscr{C}^\bullet((U_i), \mathcal{F})) = C^\bullet((U_i), \mathcal{F}) \cong H^p(X, \mathcal{F}).$$ But then by what I've said at the end of my edit in the question, we indeed have for $$X$$ affine + noetherian + $$\mathcal{F} \in \mathsf{QCoh}(X)$$ + condition $$(*)$$ that $$H^p(X, \mathcal{F})=0$$ for all $$p >0$$, which is the result claimed at the beginning of section III.3, Hartshone p. 213, though actually Hartshorne does not state condition $$(*)$$; perhaps there's more work to be done if we drop it?