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?

  • 1
    $\begingroup$ If $k$ is odd then the invertible sheaf $G_k$ will exist for sufficiently small $\Gamma$, for example $\Gamma(N)$ with $N>=3$ (no elliptic elements, no $-1$). But if you start trying to descend it to level 1, for example, you will definitely get the zero sheaf (in characteristic away from 2) because $-1$ acts as $-1$ which has no non-zero fixed points. So there's a trivial counterexample to the assertion that $G_k$ is invertible. $\endgroup$
    – eric
    Aug 9, 2014 at 12:19
  • $\begingroup$ Completely off-topic -- you need to be careful at cusps too. I am guessing that by $X(\Gamma)$ you mean the compatified curve, but then you have to be careful about defining $G_k$ at the cusps. For example if $k=1$ then there are problems with the "middle" cusp on $\Gamma_1(4)$ even though $\Gamma_1(4)$ doesn't contain any elements of finite order other than the identity. The problem is the irregular cusp, where $G_1$ doesn't behave too well. But this is a side issue. $\endgroup$
    – eric
    Aug 9, 2014 at 12:21
  • $\begingroup$ Finally, for your group theory question, if you're interested only in characteristic zero, then remember that a finite group acts semisimply in any situation where its order is invertible on the base, so everything is easy here. Are you interested in modular curves over finite fields or not? $\endgroup$
    – eric
    Aug 9, 2014 at 12:23
  • $\begingroup$ Dear eric, thank you for your reply. I should have quoted precisely the statement from Diamond and Im, which says "$\mathcal{G}_k$ is invertible unless $-1 \in \Gamma$ and $k$ is odd". So I was well aware of the case you mentioned. However, even when $k$ is even, I think the statement is still false, because of the argument I mentioned? $\endgroup$ Aug 9, 2014 at 14:35
  • $\begingroup$ I am interested in cases where there are no irregular cusps, only possibly elliptic curves. I am also interested in positive characteristic, but for now sticking to characteristic 0. I didn't understand your last comment, did you mean that $\pi_{\ast}^G \mathcal{L}$ will be invertible? Doesn't that contradict the situtation for modular forms? (since we said these sheaves are not necessarily invertible). $\endgroup$ Aug 9, 2014 at 14:39

2 Answers 2


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.

  • $\begingroup$ Dear David, thanks for your reply. if you check out Eyal Goren's book on Hilbert modular forms: books.google.dk/… He says on page 24 that these sheaves whose global sections are modular forms are generally not invertible. His argument is basically the same as mine. $\endgroup$ Aug 10, 2014 at 12:18
  • $\begingroup$ On the stack $[\Gamma]$ an invertible sheaf $\omega$ exists, I'm trying to look at the pushforward of $\omega^{\otimes k}$ to the coarse moduli curve $X(\Gamma)$, this should be the same as the sheaves $\mathcal{G}_k$. I'm wondering whether this pushforward is locally free on $X(\Gamma)$. I don't quite understand your second sentence. How is $E_4$ a local basis for $\mathcal{G}_k$ if $k \equiv 2 \pmod{4}$ ? $\endgroup$ Aug 10, 2014 at 12:19
  • $\begingroup$ Also, I thought that for any invertible sheaf on a curve the two things are equivalent: an invertible sheaf is generated by global sections if and only if at every point there is a global section not vanishing at that point. $\endgroup$ Aug 10, 2014 at 12:20
  • 1
    $\begingroup$ Can you exhibit a section of $\mathcal{G}_4$ on a neighbourhood of $[i] \in X(1)$ which is not a regular function times $E_4$? $\endgroup$ Aug 10, 2014 at 19:44
  • 1
    $\begingroup$ @Nadim: if we fix distinct points $P$ and $Q$ on projective 1-space $X$, and then we look at the sheaf of functions on $X$ that are holomorphic away from $P$ and $Q$, vanish at $P$ and are allowed a pole at $Q$, then that sheaf is isomorphic to the structure sheaf, and has no base points, but all the sections vanish at $P$ when considered as functions. Is this the point that's confusing you? $\endgroup$
    – eric
    Aug 11, 2014 at 9:42

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!


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