Given a proper surjective morphism $f:X\rightarrow Y$ where $X$ and $Y$ are smooth projective varieties. The proper pushforward $f_!$ is the homomorphism that sends the class of a coherent sheaf $M$ in $K_0(X)$ to $\sum_i(-1)^i[R^if_*M]$. Furthermore assume $X$ and $Y$ are abelian varieties so they have trivial tangent bundles. Let $M$ be the structure sheaf $\mathcal{O}_X$. By the application of GRR it implies that higher chern classes of $f_![\mathcal{O}_X]$ are zero. Is it also possible $f_![\mathcal{O}_X]$ to have the generic rank equal to zero? In other words is it possible for $f_![\mathcal{O}_X]$ to be zero in $K_0(Y)\otimes \mathbb{Q}$? (I am interested in the case that $f$ is the covering from a Jacobian of a curve to the arbitrary abelian variety.)
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$\begingroup$ Of course, just take $f$ to be a composition $X \to \mathrm{Spec}(k) \to Y$. $\endgroup$– SashaCommented Apr 12, 2021 at 4:39
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$\begingroup$ Yes but in my case I meant a surjective covering between abelian varieties. $\endgroup$– user127776Commented Apr 12, 2021 at 6:23
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$\begingroup$ Was this assumption mentioned somewhere? If $f$ is a covering (flat finite morphism of positive degree) then the rank of $R^0f_*\mathcal{O}_X$ is positive (equal to the degree of the covering) and higher direct images vanish. $\endgroup$– SashaCommented Apr 12, 2021 at 6:44
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1$\begingroup$ @abx Which conclusion did you meant? $\endgroup$– user127776Commented Apr 12, 2021 at 8:31
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1$\begingroup$ By the GRR for $f$, the rank of ${\rm R}f_*[{\cal O}_X]=f_{!}[{\cal O}_X]$ is zero if ${\rm dim}(X)>{\rm dim}(Y)$ and is the generic degree of $f$ if ${\rm dim}(X)={\rm dim}(Y)$ (or did I misunderstand something?). $\endgroup$– Damian RösslerCommented Apr 12, 2021 at 9:00
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