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Eduardo R. Duarte
Eduardo R. Duarte
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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]))=\infty$$\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 $\infty$.

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

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 $\kappa_4(\Phi_0([(x,y)-\infty]))=\infty$ 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 $\infty$.

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 $\infty$.

added 8 characters in body
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Eduardo R. Duarte
Eduardo R. Duarte
  • 795
  • 3
  • 9

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

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=0$$\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 $\kappa_4(\Phi_0([(x,y)-\infty]))=\infty$ 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 $\infty$.

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

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 $\Theta_n\cdot\Theta=0$ 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 $\kappa_4(\Phi_0([(x,y)-\infty]))=\infty$ 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 $\infty$.

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

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 $\kappa_4(\Phi_0([(x,y)-\infty]))=\infty$ 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 $\infty$.

Source Link
Eduardo R. Duarte
Eduardo R. Duarte
  • 795
  • 3
  • 9

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

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 $\Theta_n\cdot\Theta=0$ 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 $\kappa_4(\Phi_0([(x,y)-\infty]))=\infty$ 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 $\infty$.