Some curves on the Jacobian of a genus $2$ curve and their image under certain maps (char $p$)

Question

I hope this question belongs here. The situation in this question is quite particular and specific.

I am trying to weak some of theory to measure the degree of some function on the Jacobian of a genus $2$ curve which is a rational function when restricted to certain curves inside the Jacobian.

I am currently using this degree for a bigger purpose, but there are "possible situations" where this rational function "may be constant". I have control of this when it it is constant $0$ or $\infty$. I claim this rational function when constant, it cannot be different from $0$ or $\infty$ when restricted to my curves.

Let $H/\mathbb{F}_q$ be a genus $2$ curve given by $y^2 = f(x)$ with $f$ of degree $5$ and $J$ its Jacobian with generic point $[(x_1,y_1)+(x_2,y_2)-2\infty]$.

Let $\Theta\subset J$ be the image of the map $\alpha:H\to J$ given by $(x,y)\mapsto [(x,y)-\infty]$. Consider $\Phi_n:=\phi+[n]\in \text{End}(J)$ ($q$Frobenius + $n$ map), and suppose that it is non-constant. Further, since $\Phi_n$ is non-constant, consider the curves $\Theta_n=\Phi_n(\Theta)\subset J$.

Let $\kappa:J\to \mathbb{P}^3$ be the $2:1$ map that realizes a surface $K$ birational to the Kummer surface via a basis of $\mathcal{L}(2\Theta)$ given by $\kappa_1:=1, \kappa_2:=x_1+x_2, \kappa_3=x_1x_2$ and $\kappa_4=\tfrac{F_0(x_1,x_2)-2y_1y_2}{(x_1-x_2)^2}$. This is a standard basis of symmetric-even functions in $J$ used in the literature to study the Kummer Surface associated to $J$ (e.g. Cassels and Flynn).

My claim is that if $\mathfrak{g}_n$ is the generic point of $\Theta_n$ and $\kappa_4(\mathfrak{g}_n)=c$ (is constant) then $c=0$ or $\infty$ which implies that $\Theta_n$ is a zero or a pole ($\Theta_n=\Theta$) of $\text{div}(\kappa_4)\in \text{Div}(J)$.

I believe in this claim since $\Theta_n$ is a curve and is symmetric with respect to $[-1]\in\text{Aut}(J)$. Generically, $\kappa_4|_{\Theta_n}$ is a rational function in $x$ since $\Theta_n$ is symmetric with respect to $[-1]^*$ (hyperelliptic involution extended to $J$) and its degree is well defined. This degree is the integer that I am using.

I have been studying the fibers $\kappa_4^{-1}(c)$ with $c\not\in\{0,\infty\}$ without success in order to prove that the generic point of the fiber cannot be of the form of $\Theta_n$. What I know about $\Theta_n$ is that is of course symmetric with respect to $[-1]\in\text{Aut}(J)$, smooth and irreducible ($\Theta$ is). Moreover, all the points of $J$ of the form $[2P-2\infty]$ are zeroes of $\kappa_4$, that is the "diagonal" in $J$ is in the divisor of zeros of $\kappa_4$. So, $\kappa_4|_{\Theta_n}$ being constant $c$ must contradict some geometrical feature of $\Theta_n$.

Motivating example.

Let $H:y^2 = x^5 - 5x$ over $\mathbb{F}_{49}$. we have that $\phi=-[7]\in \text{End}(J)$ therefore $\Phi_9=\phi+9=[2]$ is non-constant and $\kappa_4(\Phi_9(\Theta))=\kappa_4([2](\Theta))=0$.

I have not succeeded constructing an example where $\kappa_4(\Phi_n)(\Theta)=c\neq 0$, or I cannot justify that this cannot happen.

In case it is not possible to prove my claim, maybe it is easier to prove that there must be a translation of $\Theta_n$ by a $2$-Torsion point $w\in J$ such that $\kappa_4(\Theta_n^w)$ is non-constant. (Translating by a 2-Torsion preserves the symmetry with respect to $[-1]$ and therefore if $\mathfrak{g}_n^w$ is the generic point of $\Theta_n^w$ $\kappa_4(\mathfrak{g}_n^w)$ will be again a rational function in $x$.

Answer 1

Here I just check that when $\Theta_n$ is a curve in $J$, $\kappa_4$ cannot be a constant different from $0$ or $\infty$ when restricted to $\Theta_n$.

Let $H$ be a hyperelliptic curve of genus $2$ with a rational point $\infty$ defined over $\mathbb{F}_q$ with $(x,y)$ its generic point and consider its Jacobian $J$. Further, let $\Theta$ be the image of the curve in $J$ via the Abel-Jacobi map via $\infty$. Consider $\Phi_n:=\phi+[n]\in\text{End}(J)$ where $\phi$ is the $q$-Frobenius map. Consider $\Theta$ as an element of $\text{Div}(J)$ and take $\kappa_4\in\mathcal{L}(2\Theta)$ as explained in the question.

First note that $\Theta_n:=\Phi_n(\Theta)$ is symmetric with respect to $[-1]\in\text{Aut}(J)$. Suppose that $\Theta_n\subset J$ is a curve. It is easy to check using the symmetry with $[-1]$ that if $\Theta_n\not\in\text{Supp}(\text{div}(\kappa_4))$, we have that $\kappa_4(\Phi_n([(x,y)-\infty]))\in\mathbb{F}_q(x)$ (is a rational function). Therefore, since $\kappa_4\in\mathcal{L}(2\Theta)\subset\mathbb{F}_q(J)$ we have that $\deg\kappa_4(\Phi_n([(x,y)-\infty]))=\Theta\cdot\Theta_n$ (intersection number). If this function is $c\in\mathbb{F}_q^*\subset\mathbb{F}_q(x)$, then $\Theta_n\cdot\Theta=\deg\;c=0$ which is not possible since both curves share the $0$ point and also since it would imply that $\dim\;\Theta>\dim\Theta_n$.

So $\kappa_4$ when restricted to a the curve $\Theta_n$ cannot be constant unless its a zero or a pole of $\kappa_4$

The latter situations of $\Theta_n$ being a zero or a pole can happen in the case that $\phi=[n]$ (like in my motivating example), then $\Phi_{n+2}=[2]\in\text{End}(J)$ and therefore $\Theta_{n+2}=\Phi_{n+2}(\Theta)$ is the "diagonal" of $J$, hence $\kappa_4(\Phi_{n+2}([(x,y)-\infty]))=0$. Moreover for any hyperelliptic curve of genus $2$ we have that $\kappa_4(\Phi_0([(x,y)-\infty]))=\kappa_4([(x^q,y^q)-\infty])=\infty$ since $\Phi_0(\Theta)=\Theta$ is indeed a pole of $\kappa_4$.

Of course if $\Theta_n$ has dimension $0$ then it must be because $\Phi_n$ is the $0$ map and therefore when restricted to $\kappa_4$ it takes the value $\infty$.