# Are horizontal divisors on abelian fibered hyperkähler manifolds proportional in $NS(X)$ up to vertical divisors?

Oguiso writes[1]

Theorem 1.1 Let $$f: X \to \mathbf P^n$$ be an abelian fibered HK [hyperkähler] manifold. Let $$K = \mathbf C(\mathbf P^n)$$ and let $$A_k$$ be the generic fiber of $$f$$. Then, $$\rho(A_K)= 1$$. Here $$\rho(A_K)$$ is the Picard number of $$A_K$$ over $$K$$.

It can happen that $$\rho(X_t) \geq 2$$ for all smooth closed fiber of $$f$$ ([...]). The statement is of arithmetical nature. Geometrically, it means that two horizontal divisors on $$X$$ are proportional in $$NS(X)$$ up to vertical divisors.

A divisor is called horizontal if it dominates $$\mathbf P^n$$, otherwise it is called vertical.

I'm struggling with the last sentence of Oguiso. Is that correct? Given two line bundles $$L_1, L_2$$ on $$X$$, the theorem tells us that they are proportional in $$NS(A_K)$$, so there exist $$n,m \in \mathbb Z$$ such that $$n L_1|_{A_K} + m L_2|_{A_K} = 0 \in NS(A_K).$$ I guess that Oguiso then alludes to an exact sequence of the kind $$Z^0(X \setminus U) \to NS(X) \to NS(U) \to 0,$$ where $$U = f^{-1}(V)$$ is the preimage of an open set $$V \subset \mathbf P^n$$, and $$Z^0(X\setminus U)$$ is the free group generated by the irreducible components of $$X \setminus U$$.

But I don't see a reason why $$nL_1 + mL_2$$ should vanish in $$NS(U)$$, for appropriate $$U$$. As far as I know, a line bundle being trivial in the Neron-Severi group of all fibers does not mean that it is trivial in the total Neron-Severi group.

Did I miss anything here?

[1] Keiji Oguiso, Picard number of the generic fiber of an abelian fibered hyperkähler manifold, 2009, arXiv:0803.1205

• This is simply false, already for K3 surfaces with an elliptic fibration. The quotient of $NS(X)$ by the subgroup generated by the vertical components + the zero section is the Mordell-Weil group of the generic fiber, it can very well be nontrivial.
– abx
Commented Aug 16, 2023 at 15:00
• @abx Thanks for pushing me to reading up on the Mordell-Weil group. From skimming through Huybrecht's book, this seems to work. If you write it as an answer I will be happy to accept it. It is just hard to believe that Oguiso didn't catch this, taking into consideration that he goes on to prove a Shioda-Tate formula in that situation. 🤷 Commented Aug 16, 2023 at 16:03
• I think the sentence by Oguiso that you quote is just a careless comment — the statement of the Theorem is correct. I will write my comment as an answer.
– abx
Commented Aug 16, 2023 at 19:33

## 1 Answer

This is simply false, already for K3 surfaces with an elliptic fibration. The quotient of $$NS(X)$$ by the subgroup generated by the vertical components + the zero section is the Mordell-Weil group of the generic fiber, it can very well be nontrivial. See for instance Elliptic surfaces by Schütt and Shioda, Adv.Stud. Pure Math. 60.