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?



Quotient of a smooth curve by a finite group and differentials
Yes, the formula holds, even when the action is not free. Here is the principle of a proof for the case of a tame action (which covers the characteristic $0$ case). Denote by $\pi:X\to Y=X/G$ the quotient morphism.


First consider the exact sequence of $G$-sheaves on $X$


$$ 0 \to \pi^\* \Omega^1\_{Y/k} \to \Omega^1\_{X/k} \to \Omega^1\_{X/Y} \to 0$$


Twisting by $(\Omega^1\_{X/k})^{\vee}$ gives


$$ 0 \to \pi^\* \Omega^1\_{Y/k} \otimes (\Omega^1\_{X/k})^{\vee} \to \mathcal O\_X \to \Omega^1\_{X/Y} \otimes (\Omega^1\_{X/k})^{\vee} \to 0$$


In other words, $\pi^\* \Omega^1\_{Y/k} \otimes (\Omega^1\_{X/k})^{\vee}$ identifies with the ideal sheaf of the ramification locus which is, since the action is tame, $\mathcal O\_X (-\sum\_{x\in X} (e\_x-1) x)$, where $e\_x$ is the ramification index at $x$. So we get an isomorphism of $G$-equivariant invertible sheaves


$$ \Omega^1\_{X/k} \simeq \pi^\* \Omega^1\_{Y/k} \otimes \mathcal O\_X (\sum\_{x\in X} (e\_x-1) x)$$


Now consider any $G$-invariant divisor $D$ on $X$. By a local analysis it is easy to convince oneself that 


$$\pi\_\*^G (\mathcal O\_X (D) )\simeq \mathcal O\_Y(\left[\frac{\pi\_\* D}{\# G}\right])$$


where $\pi\_\*^G$ is the functor push-froward and take the invariants, $[\delta]$ is the integral part of the divisor with rational coefficients $\delta = \frac{\pi\_\* D}{\# G}$, taken coefficient by coefficient.


If $K\_X$ (resp. $K\_Y$) is the canonical divisor of $X$ (resp. of $Y$) then we have


$$K\_X = \pi^\* K\_Y +\sum\_{x\in X} (e\_x-1) x $$


so 


$$\left[\frac{\pi\_\* K\_X}{\# G}\right] = K\_Y +\sum\_{y\in Y} [1-\frac{1}{e\_y}]y $$


hence $[\frac{\pi\_\* K\_X}{\# G}]= K\_Y$ and finally $\pi\_\*^G (\Omega\_{X/k}^1) \simeq \Omega\_{Y/k}^1$. Taking global sections on $Y$ gives finally :


$$H^0(X,\Omega\_{X/k}^1)^G= H^0(Y,\Omega\_{Y/k}^1)$$


