I have a couple of questions regarding the descent theory of line bundles on abelian varieties under isogenies in positive characteristic.

Let $X$ be an abelian variety and $L\in Pic(X)$ a line bundle. Denote as usual by $K(L)=\ker \varphi_L$ the kernel of the isogeny $$\varphi_L:X\rightarrow \hat{X}=Pic^0(X), \quad x\mapsto \varphi_L(x)=t_x^{\ast}L\otimes L^{\vee}$$

Over $\mathbb{C}$, $K(L)_0$ (connected component containing 0) is a subtorus, so there is a natural map $p:X\to \bar{X}=X/K(L)_0$ and the Appell-Humbert description of line bundles $L=L(H,\chi)$ provides simple conditions for $L$ to descend to the quotient $\bar{X}$ via $p$, namely for the existence of $\bar{L}\in Pic(\bar{X})$ such that $L=p^{\ast}(\bar{L})$.

Lemma 3.3.2 in Lange-Birkenhake's book shows that one such $\bar{L}$ exists iff $L_{|K(L)_0}$ is trivial. They also show that if $\bar{L}$ exists, then $h^0(L)=h^0(\bar{L})$: clearly $h^0(\bar{L})\leq h^0(L)$, and if the inequality were strict, then some section of $L$ would restrict to a non-trivial section of $L_{|K(L)_0}$, which contradicts the first statement.

Using this, in Lemma 3.3.3 they compute $h^0(L)$ in terms of the Appell-Humbert data $(H,\chi)$: in particular they show that if $L_{|K(L)_0}$ is non-trivial, then $h^0(L)=0$.

In positive characteristic, since $K(L)$ is a closed subgroup scheme of $X$, we have a (fppf) quotient $p:X\rightarrow \bar{X}=X/K(L)$, and the theory of Theta groups provides conditions for a line bundle to descend under an isogeny: one needs the commutator pairing $e^L:K(L)\times K(L)\rightarrow G_m$ to be trivial ($\equiv 1$).

My questions are:

  1. Is the dimension of the space of global sections also preserved upon descending (namely if $\bar{L}\in Pic(\bar{X})$ is such that $L=p^{\ast}\bar{L}$, is it true that $h^0(L)=h^0(\bar{L})$ in positive characteristic?

  2. Is $e^L\equiv 1$ equivalent to $L_{|K(L)}$ being trivial in positive characteristic?

Thanks in advance for any insight.

  • 1
    $\begingroup$ I assume that in positive characteristic you still take the quotient modulo $K(L)_0$. If this is not reduced then what you ask for is not true; for example, $L$ could be ample with $K(L)_0$ non-trivial and then sections will not be preserved. $\endgroup$ – ulrich Oct 12 '13 at 4:39

Your Answer

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

Browse other questions tagged or ask your own question.