2
$\begingroup$

I hope my question is not too basic here.

I have a very specific setting and I am trying to not use "big theorems" to prove a well known fact algebraically. I hope someone can give me a hint.

Let $J/k$ be the Jacobian of a hyperelliptic curve $H$ of genus $2$ with one rational point at infinity $\infty$. Consider $\iota:H\to J$ given by the usual embedding $\iota(P)=[P-\infty]$. Let $\Theta:=\text{Im}(\iota)\in\text{Div}(J)$ be the theta divisor of $J$.

I need to use that $\deg\iota^*\Theta=2$, that is, $\iota^*\Theta=2\infty$. Poincare's formula helps as this is equivalent to calculate the self intersection $\Theta\bullet\Theta$, but I think a simpler argument could help using that $J$ is the jacobian of a hyperelliptic curve $H$ (which has a symmetric theta divisor isomorphic to $H$).

I want to calculate this special case by hand but I have some problems understanding the pullback of $\Theta$ under $\iota$. (In fact I will appreciate a definition of pullback of morphisms of varieties with different dimensions using Weil Divisors, a lot is documented, but just for curves or in general for line bundles)

I saw a proof in the book Diophantine geometry - An Introduction by Silverman and Hindry, theorem A.8.2.1 for general Jacobians and non-symmetric Theta divisors but I think in this genus $2$ case it can be simpler.

Also there is this question pullback of theta divisor Which is also very general.

I am working over $\overline{\mathbb{F}_q}$, and I am trying not to abuse of the $\mathbb{C}$ proofs and then using Lefschetz principle.

Improve this question
$\endgroup$
4
  • $\begingroup$ You may suppose that $\Theta=H$. Then ${\rm deg}(\iota^*(\Theta))=\Theta\cdot\Theta={\rm deg}(N)$, where $N$ is the normal bundle of $\Theta$ in $J$ (by the adjunction formula, see Prop. V.1.5 in Hartshorne). There is a canonical sequence $0\to N^\vee\to\Omega_{J}|_\Theta\to\Omega_\Theta\to 0$ so ${\rm deg}(N)={\rm deg}(\Omega_\Theta)=2\cdot 2-2=2$, since $\Omega_J$ is a trivial vector bundle. $\endgroup$
    – Damian Rössler
    Commented Mar 31, 2017 at 16:19
  • $\begingroup$ Dear Damian, thanks for the comment. I was trying to use in fact the adjuction formula. and what I got is that $\deg K_\Theta = \Theta\bullet\Theta + K_J\bullet \Theta$ which is equal to $2g-2=2$ by Riemann-Roch. Why in my case then it should happen that $K_J\bullet\Theta = 0$ ? $\endgroup$
    – Eduardo R. Duarte
    Commented Mar 31, 2017 at 17:20
  • $\begingroup$ Is this $0$ because the zero class in $J$ is the blowdown of the canonical divisor of $\text{Sym}^2(H)$ and $\text{Sym}^2(H)$ and $J$ are birational? $\endgroup$
    – Eduardo R. Duarte
    Commented Mar 31, 2017 at 17:23
  • $\begingroup$ $K_J \Theta=0$ simply because the canonical class of any abelian variety is trivial. $\endgroup$
    – Francesco Polizzi
    Commented Apr 1, 2017 at 7:49

0

Reset to default

You must log in to answer this question.