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Most often than not, the sheaves appearing in algebraic geometry (with the Zariski topology) are $\mathcal{O}_X$-modules, instead of simple abelian sheaves.

Now, when dealing with topological spaces (for example, in Verdier duality) and in étale cohomology, it seems that abelian sheaves have the main role.

I wonder why. For example, Verdier duality works just fine as a statement about $\mathcal{O}_X$-modules (as in Spaltenstein's Resolution of unbounded complexes) and we recover the standard topological statements setting $\mathcal{O}_X=\underline{\mathbb{Z}}$. Even more, since ringed spaces always have fibered products (and the underlying topological spaces are the usual fibered products), base change also works well.

(I'm not sure about the étale case. I would love to know more about it.)

Is there some fundamental reason not to consider $\mathcal{O}_X$-modules in these cases?

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  • $\begingroup$ Not sure if this is helpful but because one works with abelian sheaves, pullback is always exact. If you work with $\mathcal{O}_X$-modules, you lose exactness. $\endgroup$
    – David Benjamin Lim
    Commented Sep 10, 2021 at 11:17
  • $\begingroup$ @DavidBenjaminLim I thought about this. Actually this isn't a real problem since the functor $fˆ*$ is exact if $f$ is flat and a morphism between ringed spaces with structure sheaves $\underline{\mathbb{Z}}$ is always flat. Similarly, a base change theorem for $\mathsf{R}f_!$, in the setting of ringed spaces, only holds for flat $f$. $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 11:19
  • $\begingroup$ @AUser for topological spaces $X$ we can consider the ringed space $(X,\underline{\mathbb{Z}})$ and then we recover abelian sheaves. Perhaps, as Wojowu said, étale cohomology theory is not powerful enough to handle such spaces but I'm pretty sure that Verdier duality holds just fine in this context, for example. Indeed abelian sheaves are useful, I never said otherwise :) $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 11:24
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    $\begingroup$ In addition to David E Speyer nice answer below, note that in algebraic geometry, you can study cohomology of coherent sheaf in Zariski topology, and when you go to etale topology cohomology of coherent sheaf remain the same, so you don't need to study them anymore, and you are also interested in cohomology with constant coefficient, and you want six functor formalism, and this forces you to study the constructible sheafs(which you can see as special A-module for your choice of $A=F_p,Z_p,Q_p$) $\endgroup$
    – ali
    Commented Sep 10, 2021 at 19:23

2 Answers 2

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We come to the question of what $\mathcal{O}_X$ should mean, when $X$ is a manifold or topological space. If $\mathcal{O}_X$ is smooth (or continuous) real valued functions, then we will always wind up studying $H^{\ast}(X, \mathbb{R})$. There is nothing wrong with this and, in fact, de Rham cohomology makes lots of use of sheaves of $\mathcal{O}_X$-modules, such as the sheaf of differential $k$-forms. But you might want to study $H^{\ast}(X, \mathbb{Z})$.

Alternatively, one could decide that $\mathcal{O}_X$ is the sheaf of locally constant $\mathbb{Z}$-valued functions. In this case, abelian sheaves are equivalent to $\mathcal{O}_X$-modules, so it is just a matter of terminology which one you say you are studying.

In algebraic geometry, the sheaf of locally constant $\mathbb{Z}$-valued functions is flasque in the Zarsiki topology (if your underlying space is irreducible), so you always wind up either working in the etale topology, or else studying some variant of de Rham cohomology (D-modules, crystals, etc). It is only in the topological world that you get to work with locally constant sheaves of abelian groups in the ordinary topology and learn something interesting.

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    $\begingroup$ Dear Prof. Speyer, that's basically what I thought. For topological spaces, both points of view seem to be very useful. Then why do people only consider one of them when studying Verdier duality, for example? Isn't the $fˆ!$ of a vector bundle interesting? $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 14:21
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    $\begingroup$ No, it isn't. For reasonable topological spaces (e.g. paracompact), the cohomology of vector bundles (or $\mathcal{O}_X$-modules) is trivial. $\endgroup$
    – abx
    Commented Sep 10, 2021 at 16:57
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    $\begingroup$ The fact that it is trivial is interesting, though. If you look through a book on de Rham cohomology, like Bott and Tu, you can make a game of spotting every time that a result follows from the fact that sheaf cohomology of vector bundles on paracompact manifolds vanishes. $\endgroup$
    – David E Speyer
    Commented Sep 10, 2021 at 17:00
  • $\begingroup$ @abx Interesting! Do you have a reference for that? Do the same holds for compact supported cohomology? $\endgroup$
    – Gabriel
    Commented Sep 11, 2021 at 9:49
  • $\begingroup$ @DavidESpeyer that seems very cool, do you have an example in mind of that? Thank you for your answer! $\endgroup$
    – Gabriel
    Commented Sep 11, 2021 at 9:50
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If you are using the cohomology of your sheaves to count points and the variety is modulo a prime then your point counts will be modulo that prime.

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  • $\begingroup$ Dear @ncfr, I'm not sure I understand what you mean. Could you explain it further? $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 9:51
  • $\begingroup$ I think @nfcr alludes to the Weil conjectures: to count the number of points on a variety $X$ defined over $\mathbb{F}_p$, you need a cohomology theory with values in a field of characteristic zero, while $\mathscr{O}_X$-modules give a cohomology with values in $\mathbb{F}_p$, hence a result modulo $p$. $\endgroup$
    – abx
    Commented Sep 10, 2021 at 10:26
  • $\begingroup$ Sure, but we can always take our variety $X$ and consider the ringed space $(X,\underline{\mathbb{Z}})$, can't we? My point is that we can develop the theory in the setting of $\mathcal{O}_X$-modules and then use the particular case of abelian sheaves, if needed. $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 10:46
  • $\begingroup$ The ringed space $(X,\underline{\mathbb Z})$ is not a scheme, and I don't think theory of etale cohomology can be developed in sufficient generality to deal with such a space. $\endgroup$
    – Wojowu
    Commented Sep 10, 2021 at 10:47
  • $\begingroup$ @Wojowu that's possible. What about the "topological case"? $\endgroup$
    – Gabriel
    Commented Sep 10, 2021 at 10:57

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