Inheriting complex structure from a covering space (Griffiths and Harris)

Question

On page 16 of Griffiths and Harris' Principles of algebraic geometry, they write

In general, if $\pi: M\to N$ is a topological covering space and $N$ is a complex manifold, then $\pi$ gives $M$ the structure of a complex manifold as well; if $M$ is a complex manifold and the deck transformations of $M$ are holomorphic, then $N$ inherits the structure of a complex manifold from $M$.

My question is for the last statement does one need more assumptions on the covering $\pi: M\to N$? Or could anyone please explain to me how to construct the complex structure on $N$?

Answer 1

Let $X$ be a compact connected Riemann surface such that $\chi(X)$ is negative and divisible by $3$. Then $X$ admits an irregular topological covering $\pi: X\to Y$, where $Y$ is another Riemann surface. On the other hand, by the dimension count, a generic Riemann surface of hyperbolic type does not admit a nontrivial holomorphic covering over any surface. This is a counter example to the claim.

Edit. Regarding the dimension count. Let $X, Y$ be two compact Riemann surfaces of genera $g_X, g_Y$. Then every topolical covering map $p: X\to Y$ induces a holomorphic immersion $p^*: {\mathcal M}_{g_Y}\to {\mathcal M}_{g_X}$, where ${\mathcal M}$ denotes the moduli space. Here, $p^*$ sends a Riemann surface $Y\in {\mathcal M}_{g_Y}$ to the Riemann surface $X$ equipped with the pull-back of the complex structure on $Y$ via the covering $p$.

The (complex) dimensions of the respective moduli spaces are $3g_X -3 > 3g_Y-3$. Since there are only finitely many inequivalent covering maps $X\to Y$, it follows that a generic Riemann surface in ${\mathcal M}_{g_X}$ does not cover holomorphically any other Riemann surface.

A similar argument works for surfaces of finite type.

Next, it is a nice exercise in algebraic topology to prove the following: Every compact connected oriented $X$ surface of Euler characteristic which is not a power of $2$, and hence, divisible by some prime number $q\ge 3$, admits a nontrivial degree $q$ topological covering $p: X\to Y$ whose group of covering transformations is trivial. Now, if you equip $X$ with a randomly chosen complex structure, then holomorphicity of the group of covering transformations of such $p: X\to Y$ is an empty condition, while $X$ does not cover any surface of smaller genus holomorphically. Hence, the claim made in Griffiths and Harris is false. They simply forgot to require the covering to be regular.

Answer 2

Since $\pi$ is a covering, for each point $p\in N$ there is a neighborhood $U$ such that $$ \pi^{-1}(U) = \cup_\alpha V_\alpha $$ You can define a complex atlas on $N$ making $\left.\pi\right|_{V_\alpha}$ holomorphic.

If we suppose that the deck acts transitively and holomorphically on the local sheets of $\pi$, we have that the structure is well defined since every $V_\alpha$ will be biholomorphic to each other.