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Let $\mathscr{F}^{\bullet}$ be a complex of sheaves of abelian groups on a scheme $X$, e.g., $\Omega_X^0\to\Omega_X^1\to\Omega_X^2\to...$. In my mind there are three ways of definition the cohomology.

  1. Just take the global section, it forms a chain complex of abelian groups. Then we can just compute the usual cohomology.

  2. We can define the cohomology sheaves $\mathscr{H}^i:=\mathrm{ker}(d^i)/\mathrm{im}(d^{i-1})$ as a quotient sheaf. Then take its global section.

  3. Find an injective resolution of $\mathscr{F}^{\bullet}\to I^\bullet$. Then take global section on $I^\bullet$ and compute the cohomology of these chain of abelian groups.

Are they actually the same? Why do we take the the last one to define the de Rham cohomology?

CO2
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1 Answers1

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No, these are very badly not the same. Consider a sheaf $\mathcal{F}$ and the complex which has $\mathcal{F}$ in degree zero and is trivial elsewhere. Then procedures 1 and 2 produce the same wrong answer: global sections of $\mathcal{F}$ in degree zero, always. But we can find $X$ and $\mathcal{F}$ so that a single sheaf has interesting cohomology by the third procedure: take $X=\Bbb P^1_\Bbb C$ and $\mathcal{F}=\mathcal{O}(-2)$, for instance.

The reason we want to take hypercohomology (your third way) is that it makes de Rham cohomology useful: for a smooth complex variety $X$, we get that $\Bbb H(X,\Omega_X^\bullet) \cong H^i(X^{an},\Bbb C)$, where the first term is the de Rham cohomology on a scheme and the second is the de Rham cohomology of the complex manifold $X^{an}$. This MO post lays out more reasons.

KReiser
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  • Thank you! Is it true that when $X$ is affine variety over $k$. Then the first two definitions are actually correct? – CO2 Dec 09 '20 at 11:00
  • Without going from sheaves of abelian groups to sheaves of $\mathcal{O}X$-modules, the same counterexample works because $\Bbb P^1\Bbb C\cong\Bbb A^1_\Bbb C$ as topological spaces. You'll need at least quasi-coherent sheaves of $\mathcal{O}_X$-modules, and then the result should hold. – KReiser Dec 09 '20 at 20:44
  • Last question: where do we actually take the derived objects? In sheaf of abelian groups or $O_X$-modules? For cohomology of one sheaf, they are the same. So I hope they are the same as well here? – CO2 Dec 13 '20 at 23:05
  • What do you mean "take the derived objects"? Replacing the complex of sheaves $\mathcal{F}^\bullet$ with an injective resolution of sheaves $I^\bullet$ is a quasi-isomorphism and the only part of the calculation of hypercohomology that happens inside the derived category. – KReiser Dec 13 '20 at 23:22
  • Say, $\Omega_{X/S}$ is a sheaf of $O_X$-module but also an abelian sheaf. So now, I have two choices: one is to take an injective resolution in the cateogry of sheaves of $O_X$-module another is to take in sheaf of abelian groups. – CO2 Dec 13 '20 at 23:29
  • In sufficiently nice situations ($X$ noetherian, for instance), it does not matter and you get the same answer. In cases where it does matter and the answers could be different, it is the author's responsibility to indicate which choice they are making. – KReiser Dec 13 '20 at 23:37