# Jacobians of pointed curves

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?

• 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. – Roman Fedorov Dec 14 '18 at 14:13
• @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.) – Ariyan Javanpeykar Dec 23 '18 at 23:26
• @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. – Roman Fedorov Jan 8 at 13:39
• @AriyanJavanpeykar: Of course, you are right. – Roman Fedorov Jan 8 at 13:39