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On a projective smooth variety $X$ over complex numbers (or rather compact Kahler) we have a specific set of sheaves, namely sheaves of holomorphic forms ${\mathcal \Omega}^p$ of various degrees. The cohomology $H^q(X, \Omega^p)$ of these sheaves "fit" together via Hodge decomposition into cohomology groups of our variety (manifold), which are of purely topological origin.

Assume we have a sheaf $\mathcal F$ on $X$ (say, coherent, although I don't really know how relevant this is).

Can we "fit" together groups $H^q(\Omega^p\otimes \mathcal F)$ "analogously" to Hodge decomposition (where we set $\mathcal F=O_X$) to get a purely topologically defined object (maybe originating now not from $X$ but from another variety)?

If sheaves $\Omega^p$ do not work for arbitrary such $\mathcal F$, can we find their respective analogs $\mathcal \Omega^p_{F}$ to fit the corresponding groups together as it is suggested above?

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2 Answers 2

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I think the situation you are looking for is when $\mathcal F$ is a local system. If $F$ is a locally constant sheaf on $X$, say of $\mathbf C$-vector spaces, then you can put $\mathcal F = F \otimes_\mathbf{C} \mathcal O$ to get a holomorphic vector bundle, which is canonically equipped with a flat connection $$ \nabla : \mathcal F \to \Omega^1 \otimes \mathcal F$$ such that $F = \mathrm{ker}(\nabla)$. Flatness means that when you extend $\nabla$ to a map $\Omega^d \otimes \mathcal F \to \Omega^{d+1}\otimes \mathcal F$ then $\nabla \circ \nabla = 0$, so you get a complex of sheaves $\Omega^\bullet \otimes \mathcal F$. The hypercohomology $\mathbb H^i(X,\Omega^\bullet \otimes \mathcal F)$ is isomorphic to the cohomology $H^i(X,F)$, which is a purely topological object ($F$ is just a complex representation of $\pi_1(X)$).

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  • $\begingroup$ Thank you, Dan, your answer clarifies the situation for locally constant sheaves. It's still interesting to know anything analogous about sheaves that are not of purely topological nature like locally constant ones. $\endgroup$
    – N B
    Commented Oct 9, 2013 at 15:14
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Suppose $X$ is compact. For simplicity, first assume that $\mathcal F$ is a holomorphic verctor bundle. Then we can organize the numbers $h^q=dimH^q(X, \mathcal O(\mathcal F))$ as follows: $$\chi(X,\mathcal F):=\sum_{q=0}^{dim X}(-1)^qh^q.$$ Then by Hirzebruch-Riemann-Roch theorem: $$\chi(X, \mathcal F)=\int_X ch(\mathcal F)td(X),$$ where $ch(\mathcal F)$ is the chern character of $\mathcal F$ and $td(X)$ is the Todd class of the tangent bundle of $X$, they are topological invariants of $\mathcal F$ and $TX$ respectively. Now if you want to consider $H^q(X, \Omega^p\otimes \mathcal F)$, you just fix $p$ and take sum of dimesions of the cohomology over $q$, then you get the above formula again, with $\mathcal F$ replaced by $\Omega^p\otimes \mathcal F$. For general cohorent analytic sheaf $\mathcal F$, it is well known that $\mathcal F$ can be resolved by a exact sequence of holomorphic vector bundles, then by using formal minus operation on vector bundles and splitting property of chern characters, you can do the same thing as above.

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  • $\begingroup$ This looks interesting, but it would be really enlightening to see if it is possible to combine all χ$(X, \Omega^p\otimes \mathcal F)$ on the left and respectively Chern classes in integrals on the right to something "without" $\mathcal F$. Thank you. $\endgroup$
    – N B
    Commented Oct 17, 2013 at 23:58
  • $\begingroup$ I don't think so. Since $\mathcal F$ is not an intrinsic object of $X$. So it is unreasonable to try to combine invariants of $\mathcal F$ to get a pure topological invariant of $X$. On the other hand, you can consider the category of all coherent sheaves on $X$, then you get some information of $X$, which is the content of K-theorey. In a word, you can not read out some topological invariants of $X$ just by considering a single eternal object $\mathcal F$. $\endgroup$
    – Entaou
    Commented Oct 18, 2013 at 10:39
  • $\begingroup$ Sorry, in the last sentence, "eternal" should be external. $\endgroup$
    – Entaou
    Commented Oct 18, 2013 at 10:47
  • $\begingroup$ But you can do so considering the sheaf of holomorphic functions and "convolutions" (tensor products) with sheaves of holomorphic forms. So the question was mainly about analogs of sheaves of differential forms, allowing to "average" any other sheaf (or a vector bundle) to a topological invariant. $\endgroup$
    – N B
    Commented Oct 18, 2013 at 16:26
  • $\begingroup$ And any other coherent sheaf is locally modelled using few copies of the sheaf of holomorphic functions. So this doesn't look that unreasonable. $\endgroup$
    – N B
    Commented Oct 18, 2013 at 16:29

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