Let $f : X \rightarrow Y$ a projective morphism of quasi-projective algebraic varieties over $\mathbb{C}$. Assume that $X$ is smooth, that $Y$ is normal and that:

$$\textbf{R} f_* \mathcal{O}_X = \mathcal{O}_Y $$

(here $ \textbf{R} f_* $ denotes the derived push-forward of $f$). Let $E$ be a Cartier divisor on $X$ such that $E$ is numerically trivial on every fiber of $f$. Is there an integer $m$ such that $mE = f^* F$ for some Cartier divisor $F$ on $Y$ ?

In case $Y$ is a point, I think it is true, because $E$ numerically trivial implies that $mE$ is a deformation of $\mathcal{O}_X$ for some $m>>0$. One concludes with the equality: \begin{equation*} T_{\mathrm{Pic}^0(X), \mathcal{O}_X} = H^1(X,\mathcal{O}_X), \end{equation*} where $T_{\mathrm{Pic}^0(X), \mathcal{O}_X}$ is the tangent space to the neutral component of the Picard scheme at $\mathcal{O}_X$.

I guess that if $f$ is flat, the same argument should work with the "relative Picard scheme". However I am interested in the case where $f$ is not flat.

Many thanks in advance!

  • $\begingroup$ Related: is it true that the set of $y$ in $Y$ s.t. $E$ is trivial on $X_y$ is closed in $Y$? Probably not (as you said, the usual results i.e. "Seesaw theorem" require flatness) and maybe a counterexample to this would give a counterexample to your question? $\endgroup$ – Piotr Achinger Mar 29 '13 at 3:16
  • $\begingroup$ Related II: maybe we could ask first: "Suppose $f:X\to Y$ is a projective morphism (no assumptions on $Rf_* O_X$), $E$ is an invertible sheaf on $X$ which is trivial on every fiber. Does it follow that $E = f^* F$ for an invertible sheaf $F$ on $Y$?" $\endgroup$ – Piotr Achinger Mar 29 '13 at 3:18
  • $\begingroup$ @Piotr : Good questions! $\endgroup$ – Johan Mar 29 '13 at 9:57
  • 2
    $\begingroup$ @Piotr: for your first question: Let $E\subset X$ be the exceptional divisor of a blow up of a closed point $f:X\to Y$. Then the set $\{y\in Y\vert E|_{X_y}\text{ is trivial }\}=Y\setminus f(E)$ is open in $Y$. $\endgroup$ – Sándor Kovács Mar 29 '13 at 15:55


So the result is true and a proof has been nicely explained to me by Yoshinori Namikawa.

First, let's recall the following "classical" fact (whose proof can be found as lemma $7.7$ in the book Fourier-Mukai and Nahm transforms in Geometry and Mathematical Physics, for instance):

Lemma : Let $f : Y \rightarrow X$ be a proper morphism of varieties such that that

$$ \mathrm{R} f_* \mathcal{O}_X = \mathcal{O}_Y. $$

Let $E$ be a Cartier divisor on $Y$. Then $E$ is the pull back of a Cartier divisor on $X$ if and only if for all $x \in X$, there is a neighborhood $U$ of $x$ in $X$ such that $E$ restricted to $f^{-1}(U$) is trivial.

Let $x \in X$, and let $U$ be a contractible neighborhood of $x$ in $X$. Since $\mathrm{R} f_* \mathcal{O}_{f^{-1}(U)} = \mathcal{O}_U$, the exponential exact sequence shows that: $$ \mathrm{Pic}(f^{-1}(U)) = H^2(U,f^{-1}(U), \mathbb{Z}).$$

Let $F_x$ be the reduced scheme underlying $f^{-1}(x)$. As the open $U$ is contractible, we have $H^2(f^{-1}(U), \mathbb{Z}) = H^2(F_x, \mathbb{Z})$. So we only have to prove that $E$ is homologically trivial on $F_x$. Denote by $\mathrm{Pic}^h(F_x)$ (resp. $\mathrm{Pic}^{\tau}(F_x)$ the group of homologically trivial line bundles (resp. numerically trivial line bundles) on $F_x$. Since $F_x$ is reduced, a theorem of Matsusaka (see Kleiman : Towards a numerical theory of ampleness for a proof) tells us that the quotient group: $$ \mathrm{Pic}^{\tau}(F_x)/\mathrm{Pic}^h(F_x)$$ is of finite order. Hence some multiple of $E$ is homologically trivial on $F_x$, which conclues the proof.

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