modular forms, invertible sheaves, and quotients 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?
 A: 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.
A: I think my confusion has been cleared now, and finally understand what you've beeing trying to tell me (Sorry for being so slow!). If it's okay I will add what I understood also as an answer in case someone gets confused the same way as I did, please tell me if I write further nonsense.
My faith is now restored in the following statement:

Unless $-1 \in \Gamma$ and $k$ is odd, there exists an invertible sheaf $\mathcal{L}_k$ on $X(\Gamma)$ such that modular forms of weight $k$ for $\Gamma$ are in 1-to-1 correspondence with the global sections of $\mathcal{L}_k$. 

Suppose $\Gamma' \subset \Gamma$ is a small normal congruence subgroup, then there is a very nice invertible sheaf $\omega$ on $X(\Gamma')$ and modular forms of weight $k$ for $\Gamma'$ are global sections of $\omega^{\otimes k}$. Let $\pi : X(\Gamma') \rightarrow X(\Gamma)$ be the projection. In Diamond and Im, the sheaf $\mathcal{G}_k$ is constructed as $\pi_{\ast}(\omega^{\otimes k})^G$. I think it doesn't matter whether it is invertible or not (probably it is). It is coherent of generic rank 1 with possible torsion supported only at the elliptic points, so one can just define $\mathcal{L}_k$ to be the locally free part of $\mathcal{G}_k$.
My confusion came from this phenomenon: the identification of modular forms with global sections of $\mathcal{L}_k$ is unnatural. On $X(\Gamma')$, the global sections of $\omega^{\otimes k}$ are really modular forms, so $E_6$ as a section of $\omega^{\otimes 6}$ really vanishes at the orbit $[i]$. 
On the other hand, the global sections of $\mathcal{L}_k$ correspond to modular forms, but they're not really modular forms (not in the sense I had in mind, functions on the moduli space of elliptic curves). Of course, this is due to the fact that $\pi^{\ast} \mathcal{L}_k$ is in general not isomorphic to $\omega^{\otimes k}$. Thus on $X(SL_2(\mathbb{Z}))$, $E_6$ as a section of $\mathcal{L}_6$ does not vanish at $[i]$ (as David said, $E_6$ is the local basis). So the invertible sheaf $\mathcal{L}_6$ is actually trivial for having a nowhere vanishing section. (This explains why its space of global sections is 1-dimensional.)
Thanks eric and David for helping me out!
