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For a torsion-free modular curve with regular modular cusps, let $f$ be a nonzero meromorphic weight-$k$ form and $C$ the reduced sum of modular cusp points. The local-order divisor of $f$, minus $C$, imposes exactly holomorphy in the interior and vanishing at every modular cusp on the product $f\phi$. Dividing any cusp form by $f$ gives the reverse identification with a Riemann-Roch space. The regular-cusp valence formula on a torsion-free modular curve and Riemann-Roch theorem compute the dimension when the resulting divisor has degree greater than the canonical degree.

The dimension assumption supplies a nonzero $h\in S_2({\Gamma }_0(11))$. The projective index of the Gamma 0 congruence subgroup is twelve: reduction modulo eleven identifies its cosets with the twelve points of ${\mathbb{P}}^1({\mathbb{F}}_{11})$. There are two modular cusps, by the preceding classification. There are no effective elliptic fixed points. An order-two fixed point would require a root of $x^2+1$ modulo eleven, which does not exist; an order-three fixed point would require a root of $x^2+x+1$, equivalent to a nontrivial cube root of unity in a group of order ten, which also does not exist.

The regular-cusp valence formula on a torsion-free modular curve therefore givesAll orders are nonnegative and both cusp orders are at least one. Thus $h$ has order exactly one at both modular cusps and has no zeros in the half-plane. One way to justify the degree count here is the genus formula for a modular curve: the same index, cusp and elliptic data give genus one. The invariant differential $h(\tau )d\tau$ has cusp order $\mathrm{ord}h-1$, because $d\tau$ is a nonzero constant times $d{q}_c/q_c$. Its total canonical degree $2g-2=0$ therefore makes the sum of the two cusp-adjusted orders and the interior orders zero, giving precisely the count above.

Let $H(\tau )=\Delta (\tau )\Delta (11\tau )$. By the product formula it has no zeros in the half-plane; by the cusp-order table it has order twelve at each cusp. Its weight is twenty-four. Therefore $H/h^{12}$ is a weight-zero modular function, with neither zeros nor poles anywhere on the compactified modular curve. It is constant by the maximum modulus principle on a compact surface. Normalize $h$ so its leading Fourier coefficient at infinity is one. Since $H=q^{12}+O(q^{13})$, that constant becomes one:This fixes the holomorphic twelfth root by its leading term and proves the weight-two discriminant root at level eleven is a nonzero member, hence a generator, of the given space. Taking an arbitrary pointwise twelfth root without this argument would not establish modularity: it could carry an unwanted character.

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.

The effective group ${\Gamma }_0(11)$ is torsion-free, has index twelve and has two regular cusps. If its weight-two cusp form space is one-dimensional, any nonzero member has order one at each cusp and no interior zeros by the regular-cusp valence formula on a torsion-free modular curve. Its twelfth power and $\Delta (\tau )\Delta (11\tau )$ have the same weight and divisor. Their ratio is a holomorphic function without poles on the compactified modular curve, hence a nonzero constant. Normalizing the leading Fourier coefficient to one proves the displayed formula, with the holomorphic root chosen to start with $q$. It equals $q{\prod }_{n\ge 1}(1-q^n{)}^2(1-q^{11n}{)}^2$.