Let $X$ be an Abelian variety over a field $k$; $L$ line bundle on $X$. I would like to calculate the cohomology of the Mumford line bundle $\Lambda(L)=m^*L\otimes p_1^*L^{-1}\otimes p_2^*L^{-1}$; where $m$ is the group law on $X$ and $p_i$ are the standard projections $X\times X \to X$.

I know that $R^np_{2,*}\Lambda(L)=0$ if $n\neq g=:dimX$ and $R^gp_{2,*}\Lambda(L)=i_*O_{K(L)}$ where $K(L)$ is the kernel of the map $\phi_L:X\to \widehat{X}$ and $i$ its inclusion into X. Let us suppose $K(L)$ finite.

I want to use Leray spectral sequence and what I´ve got so far is that, since the stable page is everywhere 0 but in the g-th row, $H^n(X\times X,\Lambda(L))=0$ for each $n<g$ and $H^n(X\times X,\Lambda(L))=H^n(X,i_*O_{K(L)})=H^n(K(L),O_{K(L)})$ and hence it equals 0 for each $n>0$ since $dimK(L)=0$.

The questions are the following:

1)Since this is the first time I do calculation with spectral sequences, is what I´ve written before correct?

2)Is it true that $dim H^g(X\times X, \Lambda(L))=deg\phi_L$? if the answer is yes, why?


This is a good opportunity to learn about derived categories:$$R\Gamma(X \times X, \Lambda(L)) = R\Gamma_X \circ Rp_{2,*} \Lambda(L) = R\Gamma_X \circ Ri_* \mathcal{O}_{K(L)} = \Gamma(\mathcal{O}_{K(L)}),$$placed in degree zero.

What is written seems correct but not super obvious.

To compute $R^i p_{2, *} \Lambda(L)$, observe that by cohomology and base change, these are supported along $K(L)$ and hence Artinian modules. Then try to use the arguments from Mumford's acyclicity lemma to conclude. I am pretty sure this works at least if $\phi_L$ is separable.

At any rate, once we have $R_{p_{2, *}} \Lambda(L) = i_* \mathcal{O}_{k(\Lambda)}[-n]$, then by a trivial spectral sequence argument, $H^{n + k}(\Lambda(L)) = H^k(\mathcal{O}_{k(\Lambda)})$, and so $H^k(\Lambda(L)) = 0$ for $k \neq n$. This is not what is written in $(2)$ above.

  • $\begingroup$ That is because I've written it wrong, what I meant was $H^n$ where n is the dimension of the variety. However, what do you mean by $k(\Lambda)$? $\endgroup$ – Symòn Jul 4 '16 at 18:29

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.