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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.)

EDIT. Here is a sketch of the proof of the valence formula.

(1) Prove the case of level $SL_2(\mathbf{Z})$ via contour integration as in Serre's book.

(2) Let $\Gamma$ be a general finite-index subgroup, and write $d = [PSL_2(\mathbf{Z}): \bar\Gamma]$. Let $f \in M_k(\Gamma)$, and let $F \in M_{dk}(SL_2(\mathbf{Z}))$ be the product of the translates $f \mid_k \gamma$ as $\gamma$ runs over coset representatives for $\bar\Gamma \backslash PSL_2(\mathbf{Z})$. Note that $F \ne 0$, so the $SL_2(\mathbf{Z})$ valence formula applies to $F$, and we have $$ \sum_{w \in PSL_2(\mathbf{Z}) \backslash \mathcal{H}} \frac{v_w(F)}{n_{PSL_2(\mathbf{Z})}(w)} + v_{PSL_2(\mathbf{Z}) \cdot \infty}(F) = \frac{kd}{12}.$$

(3) Now show that the left-hand side of this last formula is equal to $$ \sum_{z \in \Gamma \backslash \mathcal{H}} \frac{v_z(f)}{n_{\Gamma}(z)} + \sum_{c \in C(\Gamma)} v_{c}(f)$$ by grouping together terms for all those $z$'s mapping to a given $w \in PSL_2 \backslash \mathcal{H}$ (and similarly for the cusps) -- this is a nice little computation with the orbit-stabiliser theorem.

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Hi, I know this post is old, but I am interested in proving the valence formula stated above. How does multiplying $f$ by its translates reduce to the case of $SL_2(\mathbb{Z})$? Thanks. – GiantTortoise1729 Mar 27 at 17:39
I added a sketch of the proof. – David Loeffler Mar 28 at 8:10
Thank you very much! – GiantTortoise1729 Mar 29 at 1:01

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