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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?

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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.

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  • $\begingroup$ 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

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