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Let $X,Y$ be smooth projective varieties over $\mathbb C$ where $X$ is the universal cover of $Y$. Assume that the fundamental group of $Y$ is finite and has order $d$. Then we want to show that $\chi (X)=d \chi(Y)$ where $\chi(\cdot)$ means Euler characteristic with respect to the sheaf of regular functions.

Any ideas here? I guess it might be an easy application, but I don't see the connection between the chern class and the Todd class of $X$ and $Y$ in order to apply the Riemann Roch.

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Crossposted from math.SE: – Zev Chonoles Mar 10 '13 at 6:39
The question is purely topological, so HRR is not needed here. Just choose a triangulation of $Y$ and the induced triangulation of $X$. – Sasha Mar 10 '13 at 8:25
The characteristic classes are of the tangent bundles, and the tangent bundle pulls back to the tangent bundle, along etale maps. – Allen Knutson Mar 10 '13 at 12:45
@Sasha : the question seems to be about the "holomorphic" Euler characteristic (Euler characteristic of the sheaf of regular functions) and not the topological Euler characteristic. – user25309 Mar 10 '13 at 17:22
Let $\pi:X\to Y$ be the given morphism. You may compute that $\chi(X,{\cal O}_X)=\chi(Y,\pi_*({\cal O}_X))$. Now using either the fact that $\pi_*({\cal O}_X)$ is flat as a $C^\infty$-bundle or Grothendieck-Riemann-Roch together with Allen Knutson's remark, you conclude that ${\rm ch}(\pi_*({\cal O}_X))=d={\rm deg}(\pi)$. Now applying Hirzebruch-Riemann-Roch, you get that $\chi(Y,\pi_*({\cal O}_X))=\int_Y{\rm ch}(\pi_*({\cal O}_X)){\rm Td}(X)=d\cdot \int_Y{\rm Td}(X)=d\chi(X,{\cal O}_X)$. – Damian Rössler Mar 10 '13 at 23:08

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