Let $X$ be an (algebraic) K3 surface, then we have $H^{2,0}(X)=\langle \omega\_X\rangle$, where $\omega\_X$ is the period. Suppose $G=\langle g\rangle$ is a finite group acting on $X$ and $g$ as an automorphism of $X$ doesn't fix $\omega\_X$(e.g. $g$ is purely non-symplectic), then why is $h^{2,0}(X/G)=0$?


What's in my head is
$$
H^k(X,\mathbb{Q})^G\cong H^k(X/G,\mathbb{Q}). 
$$ by the Leray-Cartan-Serre spectral sequence. But it requires free action and it's of coefficient $\mathbb Q$. Do we have
$$
H^{i,j}(X)^G\cong H^{i,j}(X/G) 
$$ in general?


Also, I am not sure if we always have Hodge structure on normal varieties in general. Here $X/G$ is a normal variety, I guess we can define $H^{i,j}(X/G):=H^j(X/G,\omega\_{X/G}^i)$? I know the canonical divisor is well-defined for normal varieties, so for my case that $X$ is K3, $H^{2,0}(X/G)$ is well defined but is $H^j(X/G,\omega\_{X/G}^i)$ well-defined in general?



The Hodge number $h^{2,0}$ of (finite) quotient variety of a K3 surface
Your variety $X/G$ is an orbifold; on
a singular variety, the Hodge decomposition
does not work, but on an orbifold, it works
just as well. Then $G$ acts on $H^\*(X,{\Bbb Q})$, and
$H^\*(X/G,{\Bbb Q})$ is the space of $G$-invariants
(this is more or less a definition of
$H^\*(X/G)$ for an orbifold).
Similarly, $H^{p,q}(X/G)$ is the space
of $G$-invariants in $H^{p,q}(X)$.
Then $H^{2,0}(X/G)= H^{2,0}(X)^G=0$ because your $G$-action
is not holomorphically symplectic.


To use this argument, you would probably
need to compare the orbifold cohomology
with the usual cohomology of $X/G$.
This is not very hard to do, see
<https://www.pnas.org/content/42/6/359>
I. Satake,
"ON A GENERALIZATION OF THE NOTION OF MANIFOLD",
PNAS June 1, 1956 42 (6) 359-363.


