What can be said about a complex curve $C$, if its jacobian $J(C)$ has the maximal Picard number?

It is natural to expect that for a general curve of given genus its Jacobian has Picard rank 1 (isn't it?). What are the basic references for this fact?

Is it true, that there exists only finite number of curves of given genus, wich Jacobians are of the maximal Picard rank? If it is, then what is the reference?

  • 1
    $\begingroup$ A reference for the fact that the Jacobian of a very general curve has Picard number one should be found somewhere in: Birkenhake & Lange - Complex abelian varieties. A direct reference is also: Koizumi: The ring of algebraic correspondences on a generic curve of genus g. For your other question, there are some facts about curves with Jacobians having maximal Picartd number in: Beauville - Some surfaces with maximal Picard number. $\endgroup$ – Bernie Jul 18 '17 at 9:26

According to Math. Res. Lett. 7 (2000), no. 1, 123–132 (https://arxiv.org/abs/math/9909052) , if $g\ge 2$ is an integer and $f(x)$ is a complex polynomial of degree $2g+1$ without multiple roots, has coefficients in a (sub)field $K$, is irreducible over $K$ and its Galois group over $K$ is the full symmetric group $\mathbf{S}_{2g+1}$ then the jacobian $J(C_f)$ of the genus $g$ hyperelliptic curve $$C_f: y^2=f(x)$$ has no nontrivial endomorphisms, i.e., its endomorphism ring $\mathrm{End}(J(C_{f}))$ coincides with the ring of integers $\mathbb{Z}$; in particular, its Picard number is $1$.

For example, let $t_1, \dots t_{2g+1}$ be complex numbers that are algebraically independent over the field $\mathbb{Q}$ of rational numbers. Let us consider the field $L=\mathbb{Q}(t_1, \dots, t_{2g+1})$ generated by all $t_i$'s and let $K$ be its subfield of symmetric functions in $t_1, \dots t_{2g+1}$. Then $$f_t(x):=\prod_{i=1}^{2g+1}(x-t_i) \in K[x]$$ and the Galois group of $f(x)$ over $K$ is $\mathbf{S}_{2g+1}$. This implies that $\mathrm{End}(J(C_{f_t}))=\mathbb{Z}$ and the Picard number of $J(C_{f_t})$ is $1$.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.