I'm very confused about some contradicatory statements, and I hope someone can help me clarify this.

Let $\Gamma$ be a congruence subgroup. It is well known that modular forms of weight $k$ for $\Gamma$ can be constructed as global sections of a sheaf $\mathcal{G}_k$ on the modular curve $X(\Gamma)$. If $\Gamma$ contains neither -1 nor elliptic elements, then the sheaf $\mathcal{G}_k$ is actually invertible: there exists a universal elliptic curve $\mathcal{E}$ over $X(\Gamma)$, and $\mathcal{G}_k$ is the $k$th tensor power of the pullback of $\Omega_{\mathcal{E}/X(\Gamma)}$ via the zero section.

If $\Gamma$ contains elliptic elements, then there is no universal family above $X$. However, one can still construct the sheaf $\mathcal{G}_k$, this construction is described in Diamond and Im, Section 12.1. Take a normal congruence subgroup $\Gamma' \subset \Gamma$ not containing $-1$ or an elliptic element. In terms of modular curves, we have the following situation (if I understand correctly). If $G = \Gamma / \Gamma'$, then $X(\Gamma) = X(\Gamma')/G$ and we have a quotient map $\pi : X(\Gamma') \rightarrow X(\Gamma)$. Let $\mathcal{G}'_k$ be the invertible sheaf as described above. Then for any open $V \subset X(\Gamma)$, $G$ acts on $\mathcal{G'}(\pi^{-1}(V))$, so define $\mathcal{G}_k(V) = \mathcal{G}'_k(\pi^{-1}(V))^G$. This gives the sheaf we want.

The article of Diamond and Im says that this sheaf is invertible unless $-1 \in \Gamma$ and $k$ is odd. (without giving any justification). However I fail to see how this can be the case even for $k$ even. For example, say we construct this sheaf $\mathcal{G}_k$ for modular forms of level $1$ of weight $k \geq 6$ such that $k \equiv 2 \pmod{4}$. Suppose $\mathcal{G}_k$ is invertible. We know from the theory of invertible sheaves that if $\mathcal{G}_k$ is an invertible sheaf on a curve $X$ of genus $g$ such that $\deg \mathcal{G}_k \geq 2g$, then $\mathcal{G}_k$ has no base points. But in this case we find that any modular form of weight $k$ vanishes at the elliptic point $SL_2(\mathbb{Z})i$. A similar argument can be made with any group $\Gamma_0(N)$ that has elliptic elements.

**Question 1** So is this an error in the article? And if the sheaves $\mathcal{G}_k$ are not in general invertible, can we still say they are coherent?

**Question 2** Related to this question, I've been trying to look at the situation of a $G$-equivariant invertible sheaf on an affine curve (noetherian, integral, etc). Say the curve is $X = Spec A$, $G$ a finite group of automorphisms of $X$, and $\mathcal{L}$ a $G$-equivariant invertible sheaf on $X$. Let $Y = Spec(A^G)$ be the quotient curve, and consider the $G$-invariant pushforward $\mathcal{F} = \pi_{\ast}(\mathcal{L})^G$. If I understand things correctly, the algebraic description is then: we have a finitely generated $A$-module $M$, and an action of $G$ on $M$ such that:

$g(ax) = g(a)g(x)$ for all $a \in A$, $x \in M$, and $g \in G$.

Then the module $N = M^G$ is a finitely generated $A^G$-module. Now i'm asking, will $N$ be locally free? If the action of $G$ is free, then I know that it will be (as explained in Mumford, Abelian Varieties).

Suppose there is a prime $\mathfrak{q} \in A$ such that $g(\mathfrak{q}) = \mathfrak{q}$ for all $g \in G$, and let $\mathfrak{p} = \mathfrak{q}\cap A^G$ be the prime under it. Then I think $(M_{\mathfrak{q}})^G = N_{\mathfrak{p}}$, which means that $N$ is locally free hence also projective. This seems to say that $\pi_{\ast}(\mathcal{L})^G$ is invertible. So this is contradictory to what I'm asking about in Question 1.

I'm making a mistake (or several!) somewhere. Can someone point them out to me?