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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.

For cusps $\alpha ,\beta \in {\mathbb{P}}^1(\mathbb{Q})$, the modular symbol $\{\alpha ,\beta \}$ is the oriented path class in $H_1(X(\Gamma ),C;\mathbb{Z})$, where $C$ is the finite cusp set. Reversing paths changes the sign, concatenating paths gives $\{\alpha ,\beta \}+\{\beta ,\gamma \}=\{\alpha ,\gamma \}$, and the group identifies translated paths. A continued fraction expansion subdivides any rational-endpoint path into determinant-one edges. Representatives of the finitely many cosets therefore supply Manin symbols $[g]=\{g0,g\infty \}$. With inversion $S$ and translation $T$, reversing an edge and taking the boundary of an ideal triangle giveThese relations give a finite rational linear algebra presentation. Integral computation additionally removes the elliptic torsion in this naive presentation; the actual relative homology group of the compact surface is torsion-free.

The boundary of $\{\alpha ,\beta \}$ is $[\beta ]-[\alpha ]$ in $H_0(C)$. Its kernel is $H_1(X;\mathbb{Q})$ by the long exact sequence of a pair. A Hecke operator acts by summing the images of each path under its finite determinant-$n$ correspondence. Rational endpoints are again rational, so continued fractions reduce each image back to Manin symbols. Finite row reduction produces rational matrices for the relative symbols and for the boundary kernel. Integration of a weight-two cusp form against these absolute cycles is compatible with the Hecke action. The resulting decompositionexplains the relation between these matrices and cusp forms. Strictly, a rational matrix on absolute homology is not automatically a rational matrix on the holomorphic summand in an arbitrary complex basis. For subgroups preserved by real conjugation, the conjugation eigenspaces provide rational models of the cusp-form representation. More generally one should specify a compatible rational model or use the doubled homological representation. This distinction matters in interpreting the essay's rationality claim for an arbitrary subgroup; the algorithm on homology is always rational.

For the supplied subgroup, use $P=(1736)(254)$ for translation and $Q=(17)(26)(34)$ for inversion, following the PDF's naming. Translation has two cycles, so there are two cusps of widths four and three. Inversion fixes just coset five, giving one elliptic orbit of order two. The order-three product acts with cycles $(1)(253)(467)$ in one of the two equivalent composition conventions, giving one elliptic orbit of order three. Since $Q^2=1$, the central $-I$ acts trivially and the effective index is seven. The genus formula for a modular curve givesThe long exact sequence of a pair now givesOne can see this directly from the symbols. Over $\mathbb{Q}$, $Q$ gives $m_7=-m_1$, $m_6=-m_2$, $m_4=-m_3$ and $m_5=0$. The fixed order-three coset gives $m_1=0$, and the other triangle relations give $m_3=-m_2$. Thus $(m_1,\dots ,m_7)=(0,t,-t,t,0,-t,0)$. After removing elliptic torsion the integral group is free of rank one, agreeing with the surface calculation. Since $g=0$, the absolute boundary kernel and $S_2(\Gamma )$ both vanish: the cusp-form Hecke matrices for this particular example are the empty $0\times 0$ matrices.

The map $X(\Gamma )\to X(\Gamma (1))\cong {\mathbb{P}}^1$ has degree $d$, the effective projective index. Ramification occurs only over the two elliptic points and the modular cusp. Over the order-two elliptic point, $r_2$ points have index one and the other $(d-r_2)/2$ points have index two, contributing $(d-r_2)/2$ to the ramification divisor. The analogous order-three contribution is $2(d-r_3)/3$.

The local degree at a modular cusp is its effective cusp width. The cusp widths sum to $d$, so the modular cusp contribution is $d-r_{\infty }$. The Riemann-Hurwitz formula, $2g(X)-2=d(2g(Y)-2)+{\sum }_P(e_P-1)$, now yieldsTherefore the genus formula for a modular curve isUsing the index in ${\mathrm{SL}}_2$ without accounting for its center would give the wrong degree when $-I\notin \Gamma$.

The genus formula for a modular curve iswhere $\mu =[PS{L}_2(\mathbb{Z}):\overline{\Gamma }]$, $e_2,e_3$ count elliptic orbits of orders two and three, and $c$ is the number of cusps.

For ${\Gamma }_0(3)$,There are two cusps, represented by infinity and zero. The congruence $x^2+1\equiv 0(\mathrm{mod}3)$ has no solution, so there is no elliptic orbit of order two. The congruence $x^2+x+1\equiv 0(\mathrm{mod}3)$ has the single solution $x=1$, giving one elliptic orbit of order three. Thereforewhich is the Modular curve X0 3 calculation.

For $p\ge 5$, the Gamma 1 congruence subgroup has $(p-1)/2$ modular cusps of cusp width one and $(p-1)/2$ of cusp width $p$. Primitive columns modulo $p$, taken up to sign, are acted on by $(a,c)\mapsto (a+bc,c)$. The classes with $c=0$ have cusp width one, the others cusp width $p$. There are no effective elliptic stabilizers. The genus formula for a modular curve gives genus $(p-5)(p-7)/24$.

For $N\ge 3$, the principal congruence subgroup has projective index $\mu =\frac{1}{2}{N}^3{\prod }_{p\mid N}(1-p^{-2})$, no effective elliptic stabilizers, and all cusp widths equal to $N$. Hence the number of cusps is $\mu /N$, and the genus formula for a modular curve gives the displayed formula. The exceptional levels one and two both have genus zero; the full genus-zero list is $N=1,2,3,4,5$. Level six has genus one and every larger level has positive genus.

For a torsion-free effective modular group with genuine, untwisted modular cusp periods, a meromorphic weight-$k$ form has integer local orders. The tensor differential $f^{12}(d\tau {)}^{6k}$ has interior order $12\mathrm{ord}f$ and modular cusp order $12\mathrm{ord}f-6k$. Its canonical degree and the genus formula for a modular curve yield the displayed valence formula, with $d$ the projective index. Regular modular cusp periods are important in odd weights.