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Let $X$ be a projective nonsingular variety (integral Noetherian scheme of finite type that is proper over a field $k=\overline{k}$ such that $\Omega^1_X$ is locally free). Suppose one knows the singular cohomology groups

$$H^i(X,\mathbb{Z})$$ for all $i$, or even the cohomology ring $$H^*(X,\mathbb{Z})$$

This tells us what the sheaf cohomology is for the constant sheaf $\underline{\mathbb{Z}}$ is:

$$H^i(X,\underline{\mathbb{Z}})$$ since it's isomorphic to $H^i(X,\mathbb{Z})$ as above. My question is: Can this be used to say anything about the cohomology of a coherent sheaf of modules on $X$? For instance:

  • Does $\underline{\mathbb{Z}}$ correspond to any quasi-coherent sheaf on $X$?
  • Does there exist a coherent sheaf $\tilde{M}$ on $X$ such that we can we use $H^*(X,\mathbb{Z})$ to say anything about the sheaf cohomology $H^i(X, \tilde{M})$? I would suspect not, since singular cohomology considers $X$ only as a topological space, and not as a ringed space. But it's not even clear to me how to generalize singular cohomology within sheaf cohomology of a quasi-coherent sheaf. I am assuming that quasi-coherent sheaves are $\Gamma$-acyclic, so we wouldn't need to resolve them. I suppose a separate question I am having here is: are locally free sheaves $\Gamma$-acyclic in general?
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    $\begingroup$ The sheaf cohomology for constant sheaves on irreducible spaces is basically trivial: see here. $\endgroup$
    – Zhen Lin
    Commented Jan 1, 2023 at 2:31
  • $\begingroup$ So the singular cohomology of any variety (for me, varieties are irreducible) vanishes in cohomological degree greater than or equal to 1. But if we take singular homology with coefficients in a field $k$, by (usual) Poincaré duality for a compact, orientable manifold, won't we have $H_0(X, k)$ and [after applying duality, if there is no torsion] $H_n(X,k)$ are non-zero [or, if we like, both $H^0(X,\underline{k})$ and $H^n(X,\underline{k})$ are non-zero]. How does this not contradict cohomology of the constant sheaf $\underline{k}$ vanishing in cohomological degrees $\geq 1$? $\endgroup$
    – locally trivial
    Commented Jan 1, 2023 at 17:55
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    $\begingroup$ If you want to think about varieties as compact orientable manifolds you should not be talking about arbitrary algebraically closed fields $k$, and you should specify what topology you are asking about. $\endgroup$
    – Zhen Lin
    Commented Jan 1, 2023 at 22:19
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    $\begingroup$ To be more explicit: For a complex variety, two sensible cohomology theories you can consider are a) the sheaf cohomology (in the Zariski topology) of quasi-coherent sheaves and b) singular cohomology, which can be identified with the sheaf cohomology (in the complex topology) of the constant sheaf. You seem to be trying to mix the two, considering the (Zariski) sheaf cohomology of the constant sheaf - as Zhen Lin mentioned, this only gives you trivial cohomology groups. However, there is a more indirect link given by the Hodge-de Rham spectral decomposition. $\endgroup$
    – dhy
    Commented Jan 2, 2023 at 4:38

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