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This is so far the most naive of my questions.

Weakly modular functions of weight $2k$ correspond to $k$-forms on $X(1)$, right? But $X(1)$ is a curve. So shouldn't there not be any $k$-forms for $k\geq 1$ or maybe $2$?

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The short answer is: wrong. You are confusing the operation of taking the $k$th tensor power of a line bundle with the $k$-fold wedge product of differential forms. It's probably best to think about it in terms of the "factor of automorphy": when you change from one coordinate system to another, for a weight $2$ form you get the usual chain rule factor, whereas for a weight $2k$ form you get its $k$th power. But in the above you can see that you are taking the power of the coefficient function, not the exterior power of the form. I think someone else will explain this better... – Pete L. Clark Aug 3 '11 at 21:15
(By the way, it's a perfectly good question...) – Pete L. Clark Aug 3 '11 at 21:18
And that's a perfectly good answer. Thanks! – Nicole Aug 3 '11 at 21:31
Here's a notational distinction I've encountered: if we have a wedge product of $k$ forms, we call it a "differential $k$-form" as usual. Corresponding to a weight $2k$ modular function, we instead have just a product of $k$ forms, and we call it a "$k$-fold differential form." (I believe this comes from "Algebraic Curves and Riemann Surfaces" by Rick Miranda.) – Jonas Kibelbek Aug 3 '11 at 22:25
Let me add to my first comment above. To understand the sections of a line bundle it is enough to understand the effect of a change of variables: "automorphy factor", so the first thing to say is what I said above about raising the automorphy factor to the $k$th power. But as others have said, it's a convenient formalism to write the sections as $f(z) (dz)^k$. If you want the $(dz)^k$ to be the actual $k$th power somewhere, the place for it to live is in the $k$th tensor power of the space of sections of the sheaf of differential $1$-forms.... – Pete L. Clark Aug 4 '11 at 4:02
up vote 3 down vote accepted

I don't know exactly how these objects work rigorously, but here's a formal explanation. We have expressions of the form $f(z) dz$ where $f$ is a meromorphic function on the upper half plane, and we'd like to think of these as meromorphic differential forms. Under an automorphism $g : \mathbb{H} \to \mathbb{H}$ they transform as

$$f(z) dz \mapsto f(gz) g' dz.$$

Writing $g = \frac{az + b}{cz + d} \in \text{PSL}_2(\mathbb{R})$ we get $g' = \frac{ad - bc}{(cz + d)^2} = \frac{1}{(cz + d)^2}$, so in particular a meromorphic modular form of weight $2$ is the same thing as one of these expressions.

More generally we can talk about expressions of the form $f(z) (dz)^k$ for $k > 1$, in a purely formal sense: we're not thinking about wedge products or anything. Anyway, these new fancy expressions transform as

$$f(z) dz \mapsto f(gz) g'^k (dz)^k$$

hence are the same thing as meromorphic modular forms of weight $2k$.

In rigorous language, the objects we're considering are "sections of tensor powers of the canonical bundle on $X(1)$"; the objects we've described are not differential forms in the usual sense (which are sections of exterior powers of a bundle).

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