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A weight-$2n$ meromorphic modular form satisfies $f(\gamma z)=(cz+d{)}^{2n}f(z)$, while $d(\gamma z)=(cz+d{)}^{-2}dz$. The factors cancel in $f(z)(dz{)}^n$, giving a $\Gamma$-invariant meromorphic n-differential on the complex upper half-plane.

Away from elliptic points and cusps, descend this differential through local inverse branches of the quotient map. At an elliptic point of effective stabilizer order $e$, choose a coordinate $t$ in which the stabilizer acts by $t\mapsto \zeta t$, and use $w=t^e$ downstairs. If the lifted coefficient has order $r$, invariance forces $r+n$ to be divisible by $e$. The descended differential has integer orderIndeed $F(w)(dw{)}^n$ pulls back to $e^n{t}^{n(e-1)}F(t^e)(dt{)}^n$. At a modular cusp of width $h$, the coordinate $q_c=e^{2\pi iz/h}$ gives $dz=(h/(2\pi i))d{q}_c/q_c$. ThereforeBoth formulas provide meromorphic extensions over the omitted points, producing the unique differential with the prescribed pullback on the compactified modular curve.

For the character-valued question, write $X=X_0(N)$ and let $f$ be the chosen nonzero meromorphic form. The quotient of two forms with the same weight and Dirichlet character is a meromorphic function on $X$. At an interior point $P$ with effective elliptic order $e_P$ (one at ordinary points), let $r_P$ be the vanishing order of the lifted $f$. At a cusp let $v_c$ be its order in the width coordinate; this can be fractional because of the character multiplier. Multiplying by a meromorphic function $\phi$ changes these orders by $e_P{\mathrm{ord}}_P\phi$ and ${\mathrm{ord}}_c\phi$, respectively. Interior holomorphy requires a nonnegative order, while cusp vanishing requires a strictly positive order. The exact divisor is consequentlyOnly finitely many coefficients are nonzero. The order inequalities are equivalent to $\mathrm{div}(\phi )+D(f)\ge 0$. Division by $f$ proves the reverse inclusion, so the Riemann-Roch space givesThis extends the cusp-form divisor presentation to elliptic points and character multipliers.

Choose a nonzero meromorphic one-differential on $X$, and let $g$ be its weight-two pullback. For $f_t=f{g}^t$, the interior order is $r_P(f)+t{r}_P(g)$ and the cusp order is $v_c(f)+t{v}_c(g)$. The elliptic orders are one, two or three, so adding six to $t$ changes each interior floor by $6r_P(g)/e_P$, and each cusp ceiling by $6v_c(g)$. The valence formula for the modular group for $g$ says ${\sum }_P{r}_P(g)/e_P+{\sum }_c{v}_c(g)=d/6$, where $d=[S{L}_2(\mathbb{Z}):{\Gamma }_0(N)]$. ThusThe degrees tend to infinity. For sufficiently large $t$, the Riemann-Roch theorem therefore givesThe periodic formula obtained this way is an eventual formula. For small $t$, the exact answer includes the Riemann-Roch correction $\ell (K_X-D(f_t))$; it should not silently be discarded without the large-degree hypothesis.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.