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Let $f\colon X\to Y$ be a proper holomorphic map of complex analytic manifolds. Assume $f$ to be submersive for simplicity, but probably it is not important. Let $\mathcal{F}$ be a coherent sheaf on $X$; it may be assumed to be locally free for simplicity. For any point $y\in Y$ denote by $X_y$ the fiber $f^{-1}(y)$, and by $\mathcal{F}_y$ the pull-back of $\mathcal{F}$ to $X_y$. Fix an integer $i$ and define the function $\phi\colon Y\to \mathbb{Z}$ by $\phi(y):=\dim H^i(X_y,\mathcal{F}_y)$.

Question. Is it true that each level set $\phi^{-1}(m)$, $m\in \mathbb{Z}$, can be presented (at least locally) as a finite union of sets of the form $W_1\backslash W_2$, where $W_1,W_2$ are closed analytic sets?

A reference would be very helpful.

Remark. An algebraic version of this theorem is true. It follows from the algebraic version of the Grauert semi-continuity theorem which says that the function $\phi$ is upper semi-continuous. While the semi-continuity in Zariski topology does imply the positive answer to my question for algebraic varieties, the semi-continuity in analytic topology apparently does not.

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    $\begingroup$ Let $W_1 = \{\phi \ge m\} \subset Y$ and $W_2 = \{\phi \ge m+1\} \subset Y$. Grauert's theorem (valid without smoothness on $X$ or $Y$, with coherent $\mathscr{F}$ flat over $Y$) does state that $W_1$ and $W_2$ are analytic sets in $Y$. Why do you believe that his theorem gives a weaker conclusion than in the algebraic case? $\endgroup$ – user27920 Jul 10 '14 at 7:48
  • $\begingroup$ @user52824: Did I misunderstand the right definition of semi-continuity? Under the definition I know, the sets $W_1,W_2$ (as in your comment) are just closed, but I do not see why they should be analytic. (In the algebraic case the closedness is in the Zariski topology.) $\endgroup$ – orbits Jul 10 '14 at 8:37
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    $\begingroup$ In fact the semi-continuity theorem holds for the (analytic) Zariski topology, see for instance this note by Demailly. $\endgroup$ – abx Jul 10 '14 at 13:38

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