3
$\begingroup$

Let $Y$ be an algebraic curve of genus $g \geq 1$ defined over a number field $K$. If $Y$ has a $K$-point, then one can define the Abel-Jacobi map which embeds $Y$ into its Jacobian variety $\text{Jac}(Y)$. Torelli's theorem says that, conversely, $\text{Jac}(Y)$ essentially determines the curve $Y$ (this is an exact statement over an algebraically closed field).

The issue is that many (perhaps most... a theorem of Bhargava states concretely that most, in terms of natural density, of hyperelliptic curves of large genus defined over $\mathbb{Q}$ do not have any $\mathbb{Q}$-points) algebraic curves of genus $g \geq 1$ do not have $K$-rational points at all. Nevertheless, their Jacobian can still be defined over $K$.

For example, many genus one curves given by $C_f: z^2 = f(x,y)$, for $f$ a binary quartic form defined over $\mathbb{Q}$, do not have rational points. Indeed, even when $C_f$ is everywhere locally soluble, it may still fail to have $\mathbb{Q}$-rational points. Nevertheless it is clear that its Jacobian $\text{Jac}(C_f)$ is an elliptic curve defined over $\mathbb{Q}$, hence has at least one rational point. Moreover every elliptic curve is isomorphic to its Jacobian, so every Jacobian abelian variety of genus 1 is the Jacobian of a genus one curve with a rational point.

My question is: do there exist Jacobian abelian varieties $A$ defined over a number field $K$, of dimension $g > 1$, such that for all algebraic curves $Y$ defined over $K$ with $\text{Jac}(Y)$ is isomorphic to $A$ over $K$, $Y$ has no $K$-rational point?

$\endgroup$
  • $\begingroup$ If you can find a curve $Y/K$ with the following property: for every $Y'/K$ such that $Y$ and $Y'$ are isomorphic over $\bar K$, $Y'(K)=\emptyset$. Then $Jac(Y)$ is such an example. $\endgroup$ – Roman Fedorov Dec 14 '18 at 14:13
  • $\begingroup$ @RomanFedorov What about a higher genus curve $Y$ over $\mathbb{Q}$ with no non-trivial automorphisms and $Y(\mathbb{Q}) = \emptyset$? Such a curve has no twists. (Does this really answer the question though? It seems to me that there could still be a curve $Y'$ with $Y'(\mathbb{Q})\neq \emptyset$ such that $Jac(Y') \cong Jac(Y)$. This isomorphism won't respect the theta divisor.) $\endgroup$ – Ariyan Javanpeykar Dec 23 '18 at 23:26
  • $\begingroup$ @Stanley Yao Xiao: you say "this is an exact statement over an algebraically closed field" -- this is not unless you work with polarized abelian varieties. From the above discussion it seems to follow that there exists a polarized Jacobian abelian variety $A/K$ such that whenever $Jac(Y)$ is isomorphic to $A$ as polarized abelian variety, $Y$ has no rational points. $\endgroup$ – Roman Fedorov Jan 8 at 13:39
  • $\begingroup$ @AriyanJavanpeykar: Of course, you are right. $\endgroup$ – Roman Fedorov Jan 8 at 13:39

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.