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?

3$\begingroup$ Yes, the pullback of a form from the quotient is invariant, so you need only check that an invariant form is the pullback of a form. Given an invariant form on $X$, one can define a form locally on $Y$ away from the ramification locus by defining it as the value on one of the covering sheets. It is easy to check that this form extends holomorphically to all of $Y$. For instance, using RiemannHurwitz (or really just the local description of holomorphic maps $z\mapsto z^n$) to verify that the extension is holomorphic. $\endgroup$– Philip EngelCommented Dec 24, 2015 at 8:09

$\begingroup$ What does it mean for the action of $G$ on a curve to be "$k$linear"? Or do you just mean that the action is defined over $k$? $\endgroup$– Noam D. ElkiesCommented Dec 25, 2015 at 4:24

$\begingroup$ @NoamD.Elkies: I just mean that the action is "defined over $k$" as you say, i.e., $X$ is a $k$scheme, $G$ is also a $k$scheme (a "constant" one), there is a $k$scheme morphism $G \times_k X \rightarrow X$ such that blah blah blah, etc. $\endgroup$– Lisa S.Commented Dec 25, 2015 at 18:34
3 Answers
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, $\mathcal O_X (\sum_{x\in X} (e_x1) x)$, where $e_x$ is the ramification index at $x$. So we get an isomorphism of $G$equivariant invertible sheaves
$$ \Omega^1_{X/k} \simeq \pi^* \Omega^1_{Y/k} \otimes \mathcal O_X (\sum_{x\in X} (e_x1) x)$$
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 pushfroward 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
$$K_X = \pi^* K_Y +\sum_{x\in X} (e_x1) x $$
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 :
$$H^0(X,\Omega_{X/k}^1)^G= H^0(Y,\Omega_{Y/k}^1)$$
Counterexample in the nontame 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 nontame case.
Let $k$ be a field of characteristic $p$. Let $U$ be $\text{Spec}\ k[x][1/(x^{p1}1)]$, i.e., the open affine $D(x^{p1}1)$ 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 $q:U\to V$, $q^*y = x^p/(1x^{p1})$. 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 $U$, $$\alpha = \frac{x^{p2}}{1x^{p1}} dx.$$ 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, $$q^*dy = \frac{x^{2p2}}{(1x^{p1})^2}dx,$$ so that every form $q^*\beta$ vanishes to order at least $2p2$ at $x=0$.

1$\begingroup$ Thanks. In your last formula, shouldn't the numerator be $x^{2p 2}$ instead of $x^p$. $\endgroup$– Lisa S.Commented Dec 24, 2015 at 19:09

$\begingroup$ @LisaS. Thank you for catching the mistake. I have now corrected the exponent. $\endgroup$ Commented Dec 24, 2015 at 20:20

$\begingroup$ @jasonstarr How do you conclude that the ring of invariants is $k[\frac{x^{p}}{1x^{p1}}]$? I see that $\frac{x^{p}}{1x^{p1}}$ is the product of the orbit of $x$ under the $\mathbb{Z}/p$action, namely $\{x,\frac{x}{1+x},\dotsc,\frac{x}{1+(p1)x}\}$. $\endgroup$ Commented Jan 10, 2017 at 21:22

1$\begingroup$ OK here's one argument but there's probably an easier way. Set $y := \frac{x^{p}}{1x^{p1}}$. Since $k[x,\frac{1}{1x^{p1}}]$ is integral over $k[y]$ (namely $x^{p}+yx^{p1}=y$ and $\frac{1}{1x^{p1}} = x^{p1}+yx^{p2}+1$) and $k[y]$ is normal, the inclusion $k[y] \subseteq k[x,\frac{1}{1x^{p1}}] \cap k(y)$ is an equality. Moreover $(k(x))^{\mathbb{Z}/(p)} = k(y)$. $\endgroup$ Commented Jan 10, 2017 at 22:11

$\begingroup$ @MinseonShin. I do not actually remember the argument I had in mind, but your argument sounds great. $\endgroup$ Commented Jan 11, 2017 at 0:07
The answer is positive if the action is free: in that case, $f \colon X \to Y$ is a $G$torsor. This means that the natural map \begin{align*} G \times X &\to X \times_Y X \\ (g,x) &\mapsto (gx, x) \end{align*} is an isomorphism. Recall that $Z \mapsto \Omega_{Z/k}$ is a sheaf on the étale site of $Y$ (this uses that étale maps have no differentials, so for $g \colon Z \to Y$ étale we have $\Omega_{Z/k} = g^*\Omega_{Y/k}$, and the latter is a sheaf by fpqc descent).
Now the sheaf condition on the étale covering $f \colon X \to Y$ of $Y$ reads $$0 \to H^0(Y,\Omega_{Y/k}) \to H^0(X, \Omega_{X/k}) \to \prod_{g \in G} H^0(X,\Omega_{X/k}), $$ where the second map is $\omega \mapsto (\omega  g (\omega))_{g \in G}$. Thus, the global $1$forms on $Y$ are exactly the $G$invariant $1$forms on $X$. $\square$
I don't know whether the result is still true if the action has fixed points, i.e. if the map $X \to Y$ is ramified.
(This is probably not very useful, as most interesting group actions on curves I can think of are not free. Similarly, most morphisms of curves are ramified.)
(However, what I say above is very general, i.e. applies to all smooth quasiprojective $X$ and all $\Omega^i$. In particular, you can apply it to the open part where the map $f \colon X \to Y$ is unramified, and then try to extend to a global $1$form, cf. Philip Engel's comment.)

1$\begingroup$ In the tame case, the extension over the ramification locus follows from the proof of Proposition 3.2 of a draft of my article with de Jong, Cubic Fourfolds and Spaces of Rational Curves, math.stonybrook.edu/~jstarr/papers/newpair6.pdf In the published version, the argument was replaced by a (bad) citation. $\endgroup$ Commented Dec 24, 2015 at 11:36

4$\begingroup$ In the spirit of Engel's comment but being more "algebraic", it suffices to prove the map $\Omega_Y\rightarrow f_{\ast}(\Omega_X)^G$ of sheaves is an isomorphism. The map of completed stalks at $y$ is $\widehat{\Omega}^1_{\widehat{O}_y/k}\rightarrow (\widehat{\Omega}^1_{\widehat{O}_x/k})^{G_x}$ with $f(x)=y$ and stabilizer $G_x$ at $x$. By tameness $\widehat{O}_y=k[\![t]\!]$ and $\widehat{O}_x=k[\![T]\!]$ where $T^e=t$ and $G_x=\mu_e$ acts on $T$ via scaling. But ${\rm{d}}T/T={\rm{d}}t/t$ and the target consists of 1forms $f(T){\rm{d}}T/T$ for $f \in Tk[\![T]\!]^{\mu_e}=tk[\![t]\!]$. QED $\endgroup$– nfdc23Commented Dec 24, 2015 at 13:49

$\begingroup$ @nfdc23. The argument that Johan and I gave is similar to what you write (we also pass to completions), but it is formulated in a way that applies to arbitrary finite extensions where the ramification indices (e.g., orders of stabilizer groups) are prime to the characteristic. Perhaps that case reduces (via base change) to the case of a group quotient. $\endgroup$ Commented Dec 24, 2015 at 14:13

$\begingroup$ @JasonStarr: Ah, OK. I hadn't looked at the link you give (which I now see treats a far more general setting). Well, since the case of curves in the question can be settled in just a few lines by this sort of tameness reasoning, hopefully for the OP it is instructive to see both the short version and the broader context of the longer version. $\endgroup$– nfdc23Commented Dec 24, 2015 at 16:07

2$\begingroup$ @nfdc23. I know that the OP does not ask this, but I am writing an answer giving the (standard) nontame example where this fails. $\endgroup$ Commented Dec 24, 2015 at 16:08