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?

  • $\begingroup$ 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

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