Let $f:X\to Y$ be a continuous map of complex manifolds.

Consider $H^p(Y;R^qf_*V)$, where $V$ is a vector bundle on $X$, and $R^qf_*V$ are the higher direct images. (This is the $E_2^{p,q}$ term in the Leray spectral sequence associated to $f$ for the vector bundle $V$.)

Is it the case that $H^p(Y;R^qf_*V)=0$ for $q>\text{dim}_{\mathbb{C}}\,X-\text{dim}_{\mathbb{C}}\,Y$?

I recall seeing this stated somewhere, but I have been unable to find this again, and I haven't been able to prove this statement to myself.

  • 1
    $\begingroup$ Which dimension are you using? $\endgroup$
    – Ben
    Oct 19, 2016 at 16:51
  • $\begingroup$ @Ben: Thanks for the comment! I was being sloppy as I really had in mind complex manifolds, but then tried to make the question more general. I've now specified that $X$ and $Y$ are complex manifolds, and I've specified that I mean the complex dimension. $\endgroup$
    – diracula
    Oct 19, 2016 at 17:07
  • $\begingroup$ I see; what you've seen most probably is that if $f\colon X\to Y$ is a proper holomorphic map, then $R^if_* V = 0$ for all vector bundles $V$ and all $i>$maximal fibre dimension. This happens to be $\dim(X)-\dim(Y)$ if $f$ is surjective. Interested in this? $\endgroup$
    – Ben
    Oct 19, 2016 at 17:29
  • $\begingroup$ I was curious to see a counterexample to the prior question about topological spaces (with $\dim(X)-\dim(Y)$ replaced by the maximal fibre dimension) where $\dim(X)$ is the smallest number such that $H^i(X,F) = 0$ for all $i>n$ and all sheaves of abelian groups $F$. $\endgroup$
    – Ben
    Oct 19, 2016 at 17:34
  • $\begingroup$ @Ben The result you mention for a proper holomorphic map, and/or any related results, would be very useful and interesting. Could you please direct me with a reference for this or related results? If you provide this as an answer, I would be inclined to accept it, though if you prefer I can leave off accepting for a little while, so that someone may feel more of an incentive to give an answer that contains also some information on the sorts of counterexamples you are curious to see for the question you mention. $\endgroup$
    – diracula
    Oct 19, 2016 at 21:54

1 Answer 1


Here is a heavy machinery proof for the vanishing $R^if_*V$. It uses the notion of coherent $\mathcal{O}_X$-modules and complex spaces, since we have to work on the fibres which may not be manifolds. I'll try to formulate the answer in such a way that knowledge of coherent modules and complex spaces is not necessary in order to follow the logic as long as you trust me in everything I'm attaching big names to. (Just pretend coherent module means vector bundle and complex space means complex manifold.)

Theorem (Andreotti-Grauert) — For any complex space $X$ and each coherent $\mathcal{O}_X$-module $F$ the sheaf cohomology group $H^i(X,F)$ vanishes as soon as $i>\dim(X)$. (See, e.g., Corollary 4.15 in Demailly's Complex Analytic and Differential Geometry.)

Note that this is precisely the claim in case $Y$ is just a point.

CorollaryLet $f\colon X\to Y$ be a proper holomorphic map between reduced complex spaces and let $d$ be the maximal dimension the fibres of $f$ attain. Then $R^if_*(F) = 0$ for all coherent $\mathcal{O}_X$-modules $F$ and all $i>d$. More precisely, for each $y\in Y$, $(R^if_*(F))_y = 0$ for all $i>\dim(f^{-1}(y))$.

Proof. It suffices to prove that the completion $(R^if_*F)_y^\wedge = (R^if_*F)_y\otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y,y}^\wedge$ vanishes for all $i>d$ since the completion $\mathcal{O}^\wedge_{Y,y}$ of $\mathcal{O}_{Y,y}$ with respect to the maximal ideal is faithfully flat. By Grauert's comparison theorem (in the algebro-geometric world also known as the Theorem on formal functions), for each $y\in Y$, the completion of the stalk at $y$, $(R^qf_*F)_y^\wedge$, is isomorphic to a limit $$(R^qf_*F)^\wedge_y\cong \varprojlim\nolimits_k H^i\left(f^{-1}(y),F_{(k)}\right)$$ for certain coherent sheaves $F_{(k)}$, $k\in\mathbb{N}$, on the fibre $f^{-1}(y)$. Thus, if $i>\dim(f^{-1}(y))$, then $(R^qf_*F)_y = 0$. Hence, if $i>d = \max_{y\in Y}\{\dim(f^{-1}(y))\}$, then $R^qf_*F = 0$, as claimed.

Thanks to tracing's comment below, we can conclude:

CorollaryLet $f\colon X\to Y$ be a proper holomorphic map between complex spaces and let $d$ be the maximal dimension the fibres of $f$ attain. Then $H^p(R^qf_*(F)) = 0$ for all coherent $\mathcal{O}_X$-modules $F$, all $i>d$ and all $q$.

Next, note that we can't replace 'maximal fibre dimension' by $\dim(X)-\dim(Y)$. Let $Y$ be any smooth surface and let $f\colon X\to Y$ be the blow up of some point $y\in Y$. That's clearly a proper surjective map of complex manifolds and $\dim(X)-\dim(Y) = 0$. The sheaf $R^1f_*F$ is supported in $y$ and so $H^0(Y,R^1f_*F) = (R^1f_*F)_y$ for any coherent sheaf $F$ on $X$. Therefore, we only have to find a vector bundle on $Y$ such that $R^1f_*F \not=0$. Let's look at the line bundle $F := \mathcal{O}_Y(2E)$ where $E\subset Y$ is the exceptional divisor. We'll use Grauert's comparison theorem, so I have to tell you what the $F_{(k)}$ are: The ideal sheaf of $E$ in $Y$ is $\mathcal{O}_X(-E)$ and so $F_{(k)} = F\otimes_{\mathcal{O}_X}\mathcal{O}_X/\mathcal{O}_X(-kE)$. Note that $$\mathcal{O}_X(-kE)/\mathcal{O}_X(-(k+1)E) = \mathcal{O}_X/\mathcal{O}_X(-E)\otimes \mathcal{O}_X(-kE) = \mathcal{O}_E\otimes \mathcal{O}_X(-kE) = \mathcal{O}_E(k)$$ since the normal bundle of $E$ in $X$ is $\mathcal{O}_E(-1)$. For the same reason, $F_{(1)} = F\otimes \mathcal{O}_{E} = \mathcal{O}_{E}(-2)$ and we get each $F_{(k)}$ sitting in the exact sequence $$0\to \mathcal{O}_E(k-2) \to F_{(k+1)}\to F_{(k)}\to 0.$$ Since $H^1(E,\mathcal{O}_E(n))=0$ for all $n\geq -1$ this implies that $H^1(E,F_{(k+1)})\cong H^1(E,F_{(k)})$ for all $k\geq 1$; therefore, the limit in Grauert's comparison is constant: $$(R^1f_*F)^\wedge_y \cong \varprojlim\nolimits_k H^1\left(E,F_{(k)}\right) = H^1(\mathbb{P}^1,\mathcal{O}(-2))\not=0$$ and so $R^1f_*F\not=0$.

This also shows that $H^0(Y,R^qf_*V)$ need not vanish for $q>\dim(X)-\dim(Y)$. Right now, I don't have a counter-example for $p>0$.

  • 2
    $\begingroup$ You should be able to remove "reduced'' from the statement (for what its worth) since one could replace $F$ by the graded pieces of its $I$-adic filtration, where $I$ is the nil radical of $O_X$, and reduce to $X$ (and hence $Y$) being reduced. (Note that cohomology only depends on the underlying topological space, so once $X$ is reduced, we can replace $X \to Y$ by the induced map $X \to Y_{red}$, and so assume that $Y$ is reduced as well.) $\endgroup$
    – tracing
    Oct 20, 2016 at 20:27
  • $\begingroup$ Dear tracing, you're right, thank you. I added the statement referring to your comment. $\endgroup$
    – Ben
    Oct 20, 2016 at 21:19

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