TITLE: Quotient of a smooth curve by a finite group and differentials
QUESTION [11 upvotes]: Let $X$ be a proper smooth connected curve over an algebraically closed field $k$ of characteristic $0$, and suppose that $X$ is equipped with a $k$-linear action of a finite group $G$. It makes sense to form the quotient curve $Y := X/G$, and $Y$ is $k$-smooth because it is normal. Is it true that the pullback of differentials gives the identification
$$H^0(Y, \Omega^1_{Y/k}) = H^0(X, \Omega^1_{X/k})^G?$$
If so, how does one prove this?

REPLY [8 votes]: Counterexample in the non-tame case.  This is not what was asked by the OP.  However, since it comes up in answers and comments above, I thought I would write down one example to illustrate the problem in the non-tame case.  
Let $k$ be a field of characteristic $p$.  Let $U$ be $\text{Spec}\ k[x][1/(x^{p-1}-1)]$, i.e., the open affine $D(x^{p-1}-1)$ in $\mathbb{A}^1_k$.  In other words, remove the closed points corresponding to elements of $\mathbb{F}_p^\times$.  Let $f:U\to U$ be the morphism of $k$-schemes with $f^*(x) = x/(1+x)$.  This is an automorphism of order $p$.  The quotient is $V = \text{Spec}\ k[y] \cong \mathbb{A}^1_k$ with quotient morphism $q:U\to V$, $q^*y = x^p/(1-x^{p-1})$.  This is a finite morphism that extends to a finite étale morphism $\mathbb{P}^1_k\setminus \{0\}\to \mathbb{P}^1_k\setminus \{0\}$ (this is one of those examples proving that $\mathbb{A}^1_k$ is not algebraically simply connected in characteristic $p$).
Consider the algebraic $1$-form on $U$, $$\alpha = \frac{x^{p-2}}{1-x^{p-1}} dx.$$  It is straightforward to compute that $f^*\alpha$ equals $\alpha$.  Yet $\alpha$ cannot be of the form $\phi^*\beta$ for any algebraic $1$-form $\beta$ on $V$.  Indeed, the module of algebraic $1$-forms on $V$ is generated by $dy$.  By direct computation, $$q^*dy = \frac{-x^{2p-2}}{(1-x^{p-1})^2}dx,$$ so that every form $q^*\beta$ vanishes to order at least $2p-2$ at $x=0$.