2
$\begingroup$

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 $x$.

$\endgroup$
0
$\begingroup$

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 $\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.