I am trying to understand the proof of the theorem of Artin-Grothendieck (theorem 3.1.13. in 'Positivity in algebraic geometry' by Lazarsfeld). Right in the beginning of the proof, he chooses a finite map $f: X \rightarrow \mathbb{C}^n.$ Here $X$ is some affine variety of dimension $n$, considered as complex analytic space. However smooth (i.e. a complex manifold) would totally suffice for me. He then claims that $$ H^p(X, \underline{\mathbb{R}}) = H^p(\mathbb{C}^n, f_*\underline{\mathbb{R}})$$ (he does it more generally for constructible sheaves but again this would totally suffice for me). I do not really understand why this shall be true. After all, $\underline{\mathbb{R}}$ is not a coherent sheaf (it is not even a sheaf of $\mathcal{O}_X$-modules), so we can not invoke GAGA + vanishing of higher direct images in the algebraic setting. Do you have any hints for me?

Moreover, I started to wonder if in this situation the push forward of the de Rahm complex on $X$ is still exact on $\mathbb{C}^n$ (of course this would yield the result above). If I try to picture it, then I think that the preimage of sufficiently small balls down stairs should still be contractible upstairs (or at least copies of such sets). However, sadly I do not have much knowledge of branched covers of manifolds and my intuiton already failed me several times in this proof.

Any help, hints or references would really be appreciated!

  • 2
    $\begingroup$ Hartshorne, chapter III, exercise 8.1. $\endgroup$
    – AG learner
    Jul 20, 2020 at 20:25

1 Answer 1


This may be "taking a sledgehammer to a mosquito," but the Leray spectral sequence for $f$ degenerates since the higher cohomology of the constant sheaf will vanish along the fibers (which are finite).

(More explanation if it helps -- you have a spectral sequence $$ H^p(\mathbb{C}^n, R^qf_*\mathbb{R}) \Rightarrow H^{p+q}(X, \mathbb{R}), $$ but for $q > 0$ the terms on the left vanish, so what you really have is an isomorphism $$ H^p(\mathbb{C}^n, f_*(\mathbb{R})) = H^p(X, \mathbb{R}). $$ As mentioned by @AGLearner in a comment, the appeal to the Leray spectral sequence is overkill -- given a map $f: X \to Y$ you can prove directly that the natural map $$ H^p(Y, f_*\mathcal{F}) \to H^p(X, \mathcal{F}). $$ is an isomorphism whenever the higher direct images of $\mathcal{F}$ vanish.

For deducing the vanishing of the higher direct images in this particular case, we just need that the fibers are discrete, so they have no higher singular cohomology (maybe we are appealing to something like Ehresmann's lemma implicitly here).

  • $\begingroup$ Thank you very much for your answer! I agree that this may be a bit overkill, but I am definitely willing to accept it. However, could you please elaborate a bit on what you mean? I only learned about the Leray spectral sequence quite recentely and do not quite see what you mean by this (I do know the basic definition and how to work with it directely, but do not know any kind of 'tricks'). Do you use one of the base change theorems to compute the $R^if_*$'s? $\endgroup$ Jul 21, 2020 at 8:28
  • $\begingroup$ Ok, then I think you missunderstood me. I am well aware of the fact that $R^if_*\mathcal{F} = 0 \Rightarrow H^p(\mathcal{F}) = H^p(f_*\mathcal{F})$ and of course as you pointed out, using the Leray spectral sequence in this case is definitely overkill. However, my question specifically was why $R^if_*\mathcal{F} = 0$. In the algebraic setting it is clear to me, but I do not know how to prove it in the analytic setting. $\endgroup$ Jul 21, 2020 at 14:48
  • $\begingroup$ @Photographer the fibers are zero dimensional and so they don't have higher cohomology. (the sheaf cohomology of $\mathbb{R}$ agrees with usual singular cohomology from topology) $\endgroup$
    – hunter
    Jul 21, 2020 at 16:02
  • $\begingroup$ Ah, now I see. I did not know that $(R^if_*\mathcal{F})_x$ really just is $H^i(f^{-1}(x), \mathcal{F}|_{f^{-1}(x)})$, but apparently if $f$ is topologically proper and the spaces reasonable this does hold. Thank you! $\endgroup$ Jul 21, 2020 at 20:53
  • $\begingroup$ @Photographer just seeing your last comment -- yeah I think there is something like Ehresmann's lemma that maybe I was eliding over. Basically we want that the pre-image of a sufficiently small neighborhood of $x$ might not deformation retracts onto the fiber over $x$. I really don't know the right conditions for that to hold, and on reading the original question it was exactly your question! So I'm not convinced I've really answered you. $\endgroup$
    – hunter
    Jul 29, 2020 at 17:10

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