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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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    ...But here's the key point: taking the $k$th tensor power of$a$vector space -- as opposed to the exterior or symmetric power -- **imposes no relations**. Thus $(dz)^k$ does have$a$rigorous meaning: rigorously, it behaves as a formal $k$th power! By way of analogy, for any element $a$, we can consider $a^k$ in the free group generated by $a$, and of course no power of it will be trivial.2011-08-04

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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).