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Let $M_{k}(\Gamma)$ denote the space of weight $k$ modular forms for the congruence subgroup $\Gamma$. Are there any proofs of the finiteness of the dimension of $M_{k}(\Gamma)$ that don't rely on Riemann-Roch?

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Yes, one can prove finiteness of dimension without Riemann-Roch. One way is to prove the "valence formula" if $f$ is any nonzero modular form of weight $k$ and level $\Gamma$, then $$ \sum_{z \in \Gamma \backslash \mathcal{H}} \frac{v_z(f)}{n_\Gamma(z)} + \sum_{c \in C(\Gamma)} v_c(f) = \frac{k\, [PSL_2(\mathbf{Z}) : \overline{\Gamma}]}{12}$$ where $\overline{\Gamma} = \Gamma / (\Gamma \cap \{\pm 1\})$ is the image of $\Gamma$ in $PSL_2(\mathbf{Z})$, $n_\Gamma(z)$ is the order of the stabilizer of $z$ in $\overline{\Gamma}$, and $C(\Gamma)$ is the set of cusps of $\Gamma$. This is usually proved as an application of Riemann surface theory, but it can also be shown in a purely elementary way: by multiplying together $f$ and all of its translates by coset representatives for $SL_2(\mathbf{Z}) / \Gamma$ one reduces to the case where $\Gamma = SL_2(\mathbf{Z})$, and this can be proved by a contour integral around the boundary of the standard fundamental domain, cf. Serre's "A Course in Arithmetic".

With the valence formula in hand, it is clear that any modular form vanishing at $\infty$ to order $> \frac{k [PSL_2(\mathbf{Z}) : \overline{\Gamma}]}{12}$ is zero. So a modular form is uniquely determined by finitely many terms of its $q$-expansion at $\infty$, hence the space of modular forms is finite-dimensional.

(I used exactly this proof in a lecture course I taught last term.)

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