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 $x$. 



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


