EDIT: Let $f\colon X\to Y$ be proper holomorphic submersive map of complex analytic manifolds. Let $\mathcal{F}$ be the sheaf of holomorphic sections of a holomorphic vector bundle over $X$. Assume that for some $i$ the function $y\mapsto dim H^i(X_y,\mathcal{F_y})$ is constant on $Y$. Then the Grauert theorem says that the sheaf $R^if_*\mathcal{F}$ is locally free of finite rank, and that for any point $y\in Y$ the canonical map from the fiber $$(R^if_*\mathcal{F})|_y\to H^i(X_y,\mathcal{F}_y)$$ is an isomorphism, where $X_y:=f^{-1}(y)$, $\mathcal{F}_y$ is the pull-back of $\mathcal{F}$ from $X$ to $X_y$.

Question. Whether the following generalization of the Grauert theorem holds? Under the above assumptions on $X,Y,\mathcal{F}, f$ and $i$, let us fix a point $y\in Y$ and a natural number $n$. Let $y^{(n)}$ be the $n$th infinitesimal neighborhood of $y$, $X_n$ the $n$th infinitesimal neighborhood of the fiber $X_y$, and $\mathcal{F}_n$ be the pull-back of $\mathcal{F}$ to $X_n$. Assume as previously that the function $z\mapsto dim H^i(X_z,\mathcal{F_z})$ is constant on $Y$. We have the canonical map $$R^if_*\mathcal{F}\otimes \mathcal{O}_{y^{(n)}}\to H^i(X_n,\mathcal{F}_n).$$ Is this map an isomorphism? If yes, is there a reference for this fact?

I think this is true in the context when all manifolds and morphisms are algebraic. I think it can be shown by the same method of Ch. III, $\S$ 12 of Hartshorne's "Algebraic Geometry" book. However I need the statement for complex analytic manifolds.

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    $\begingroup$ Your statement is wrong. The pushforward $f_*\mathcal{F}$ might be identically zero, yet $H^0(X_y,\mathcal{F}_y)$ nonzero for some points $y$. $\endgroup$ Jul 3, 2014 at 12:25
  • $\begingroup$ @JasonStarr: Thanks! I have corrected the statement. $\endgroup$
    – asv
    Jul 3, 2014 at 12:52
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    $\begingroup$ Not exactly what you want, but the only generalisation of the direct image theorem and its applications I know is due to Banica: 1972. I think the same version of the theorem is presented in the book by him and Stanasila, Algebraic Methods in the global theory of complex spaces. $\endgroup$
    – Ben
    Jul 4, 2014 at 7:36
  • $\begingroup$ @BenA.: Thanks for the useful reference. I think the book does contain the positive answer to my question. $\endgroup$
    – asv
    Jul 5, 2014 at 11:17

1 Answer 1


EDIT: I checked in the book "Algebraic Methods in the global theory of complex spaces" by Banica and Stanasila mentioned in a comment by Ben A. I think it does contain the positive answer to my question.

The statement is explicitly formulated on p. 119 of the book during the proof of Theorem 3.4. It is the claim that the map (**) is an isomorphism, where one should take $Y'=Y$ and $g=Id$.


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