Let $X$ be a complex projective manifold and $G$ a finite group acting on $X$, then is the quotient manifold $Y=X/G$ still a projective manifold?
This question comes from a paper of Cao Junyan《On the approximation of Kähler manifolds by algebraic varieties》p403-p404, which talk about the finite covering $\hat X$ of $X$, I will explain it as follows:
Let $X$ be a compact Kähler manifold with Hermitian semipositive anticanonical bundle, by the structure theorem (cf[DPS96]), the universal cover $\tilde X$ can be splited as $\tilde X=\mathbb C^q\times Y\times P$, $Y$ denotes a product of Calabi-Yau manifolds and $P$ denotes a projective manifold, and there exist a normal subgroup $\Gamma$ of $\pi_1(X)$ such that $G=\pi_1(X)/\Gamma$ is a finite group, let $\hat X=\tilde X/\Gamma$, obviously we have $X=\hat X/G$, then there exist a smooth fibration $\pi:\hat X\rightarrow F$ where $F:=(\mathbb C^q\times Y)/\Gamma$ is a Ricci-flat compact manifold. I know we can always deform a Ricci-flat manifold to a projective manifold, so a projective deformation of $F$ may induce a projective deformation of $\hat X$, if the quotient map of $G$ may keep the projectivity, it may also induce a projective deformation of $X$ which is exactly what I need, so my question is:
Is the quotient of $\hat X$ by a finite group $G$ acting on $\hat X$ still projective?
Can anyone help me to figure out whether it is true? Thanks in advance!
Best Answer
There has been an answer in the comments, but the OP seems to prefer a more analytical language. Here is the canonical textbook reference:
As was noticed in the comments, in the given situation, the quotient map $f\colon \hat X\to X$ is a finite unramified cover. Thus, a corollary of the Kodaira Embedding Theorem in Griffiths and Harris' Principles of Algebraic Geometry, p. 192, applies. It states that for such a finite unramified cover, the base is projective if and only if so is the covering space.
The proof of the implication we need more or less goes like this. By the Kodaira Embedding Theorem, it is sufficient that there be a closed, positive $(1,1)$-form whose cohomology class is rational. If $\hat X$ is projective, then there exists such a $(1,1)$-form $\omega$ on $\hat X$. Since $f$ is a finite unbranched cover, a suitable coordinate chart $U$ around $p\in X$ is covered by a bunch of charts $\coprod_{i=1}^n V_i=f^{-1}U$ and a $(1,1)$-form $\omega'$ is defined as the trace: $\omega'|_U=\sum_{i=1}^n \omega|_{V_i}$ (through obvious identifications). The proof is concluded by the observation that this defines a closed, positive $(1,1)$-form on $X$ whose cohomology class is rational.