Quotient of a smooth curve by a finite group and differentials

Question

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?

Answer 1

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 :

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

Answer 2

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 $x=0$.

Answer 3

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 quasi-projective $X$ 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.)