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?



modular forms, invertible sheaves, and quotients
You have to be a bit careful what "vanish" means in this context. For $k = 2 \bmod 4$, at an elliptic point of order 4, "vanishing as a section of the sheaf" and "vanishing as a function on the upper half-plane" aren't the same thing; it's easy to check that $E\_4$ is a local basis of the sections of $\mathcal{G}\_k$ in a neighbourhood of the elliptic point. So there is no contradiction with invertible sheaves of large enough degree having no base points -- a base point is a point where every *global* section fails to be a basis of the *local* sections, which is a different phenomenon.


The sheaves $\mathcal{G}\_k$ are always invertible except in the trivial case of odd $k$ and $-1 \in \Gamma$. But they're still morally the wrong objects in the presence of elliptic points; one should really work with $X(\Gamma)$ as a stack, as you are well aware.


