2
$\begingroup$

$\DeclareMathOperator{\Pic}{Pic}$ Let $f: M \to B$ be a smooth morphism of smooth varieties over $\mathbb C$ such that the natural map $\mathcal O_B \to f_* \mathcal O_M$ is an isomorphism (i.e. $f$ is proper with connected fibers). I know that the following holds[1, Lemma 2.7]:

If $L \in \Pic(M)$ is such that for all $b \in B$, the restriction $L_b = L|_{f^{-1}(b)}$ is trivial, then $$L \cong f^* L'$$ for some line bundle $L' \in \Pic(B)$.

Does a similar statement hold for algebraic equivalence? So assume $L_b$ is algebraically equivalent to zero for all $b$, which means that $$0 = c_1(L_b) \in H^2(f^{-1}(b), \mathbb C).$$ Is it true that $L$ is algebraically equivalent to $f^* L'$ for some $L' \in \Pic(B)$?

Equivalently, using the cited fact above, we may ask if there is a line bundle $L'' \in \Pic^0(M)$ such that $L_b \cong L''_b$ for all $b \in B$. Then $L \otimes L''^\vee \cong f^* L'$ for some $L' \in \Pic(B)$, since $L \otimes L''^\vee$ is trivial on the fibers of $f$.


[1] Kleiman, The Picard Scheme

$\endgroup$

1 Answer 1

2
$\begingroup$

No. A counterexample is given by the Poincaré line bundle on the product $A \times A^\vee$ for any abelian variety $A$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .