Holomorphic Euler characteristic of a finite cover

Question

Let $Y\to X$ be a finite cover of degree $k$ of compact complex manifolds. Is it true that the holomorphic Euler characteristic satisfies $$\chi(X,T^{1,0}X) k=\chi(Y,T^{1,0}Y)?$$ As it would be the case for the topological Euler characteristic.

Answer 1

Multiplicativity of the holomorphic Euler characteristic is a direct consequence of the Hirzebruch–Riemann–Roch theorem: For $E=T^{1,0}X$,

$$\chi(X, E)= \int_X ch(E) td(X)$$ where $ch(E)td(X)$ is a certain top-degree differential form on $X$. The point is that this form behaves nicely under pull-backs: If $f: Y\to X$ is a holomorphic covering map then $$ f^*(ch(E) td(X))= ch(f^* E) td(Y)= ch(T^{1,0}Y)td(Y) $$ This implies the multiplicativity property under holomorphic covering maps. Alternatively, if you do not like to use differential forms, you can use naturality of Chern classes.