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For a finite-index subgroup of the modular group, take the union of translates of the standard fundamental domain of the modular group over right coset representatives. Boundary identifications are inherited from the subgroup action. The truncated region is compact, and its finitely many cusp ends have exponential parameters; this proves compactness of the compactified modular curve and boundedness of invariant cusp-form norms.

A modular function for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a meromorphic function on the complex upper half-plane that is invariant under $\Gamma$ and is meromorphic at every modular cusp:At a modular cusp of width $h$, meromorphicity means that the transformed function has a Laurent series in the local parameter $q_h=e^{2\pi i\tau /h}$ with only finitely many negative powers. Thus a modular function can have poles, in contrast to a modular form, which is required to be holomorphic on the half-plane and at the modular cusps. The Klein j-invariant is a fundamental example for the full modular group.

Some sources use the broader expression modular function of weight $k$ for a meromorphic function satisfyingwith the corresponding meromorphic cusp condition. This is also called a meromorphic modular form; the unqualified term modular function ordinarily means weight zero. Weight-zero modular functions descend to meromorphic functions on the compactified modular curve.^[ref]

Normalize the weight-four and weight-six Eisenstein series to have constant term one. Their Fourier expansions beginTheir absolutely convergent lattice sums transform by modular weights four and six under reindexing by the modular group, and their Fourier expansions prove holomorphy at its modular cusp. Thus they are elements of $M_4,M_6$.

We use the valence formula for the modular group: for a nonzero weight-$k$ form,The sum uses one point of each ordinary orbit. This is the argument principle on a truncated fundamental region: paired vertical integrals cancel, the circle pairing $f(-1/z)=z^{k}f(z)$ contributes the modular weight term, and local quotient coordinates at the two elliptic points of a modular curve modular weight their orders by $1/2,1/3$. All orders are nonnegative for a holomorphic modular form.

Define the modular discriminant byIt is a weight-twelve cusp form. Its order at infinity is one, exhausting the valence total $12/12=1$. Hence $\Delta$ has no zeros in $\mathbb{H}$. Therefore every weight-$k$ cusp form is divisible by $\Delta$, with quotient holomorphic on $\mathbb{H}$ and at the modular cusp:Negative modular weights vanish by the valence formula, and odd modular weights vanish because $-I$ acts by $(-1{)}^k$. Weight-zero forms are constant: they descend holomorphically to the compactified modular curve, and the maximum modulus principle applies. There are no weight-two forms: $f(i)=i^{2}f(i)=-f(i)$ forces a zero at $i$, whose valence contribution $1/2$ exceeds $2/12$.

Every nonnegative even modular weight other than two is $4a+6b$ with $a,b\ge 0$. If $k\equiv 0(\mathrm{mod}4)$, take $b=0$; if $k\equiv 2(\mathrm{mod}4)$ and $k\ge 6$, take $b=1$. For $f\in M_k$, choose such a monomial $h=E_4^a{E}_6^b$, whose constant term is one. Subtracting the constant term $c$ of $f$ gives $f-ch\in S_k$, soInduction on modular weight, together with $1728\Delta =E_4^3-E_6^2$, proves that every modular form is a polynomial in $E_4,E_6$.

For algebraic independence, first fix one modular weight $w$. All solutions of $4a+6b=w$ have the same residue of $a$ modulo three and the same parity of $b$. Consequently their monomials can be writtenfor suitable $a_0,b_0,L$. On an open set where the common factor and $E_6$ are nonzero, a linear relation would give a polynomial relation inThis is nonconstant. By the open mapping theorem, a polynomial vanishing on all its values must be the zero polynomial. Thus same-weight monomials are linearly independent.

Different modular weights cannot cancel either. Suppose ${\sum }_w{P}_w(E_4,E_6)=0$ with each $P_w$ weighted homogeneous. Transform by ${\gamma }_n=\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At any fixed $z\in \mathbb{H}$ this givesThe numbers $nz+1$ are infinitely many distinct values, so the polynomial in that number vanishes identically, and every $P_w(z)=0$. The same-weight independence then sets every coefficient to zero. This is the graded weights separate analytic polynomial relations argument. Hencewith algebraically independent generators. Multiplying the two Eisenstein series by their nonzero lattice-normalization constants does not alter either generation or independence.

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.

Adjoin ${\mathbb{P}}^1(\mathbb{Q})$ to the complex upper half-plane. Ordinary points retain their usual neighborhoods. At a rational boundary point $a=\sigma \infty$, a neighborhood basis consists of $a$ together with $\sigma \{z\in \mathbb{H}:\mathrm{Im}z>Y\}$, with $Y$ arbitrarily large. Different choices of $\sigma \in S{L}_2(\mathbb{Z})$ change it by a translation and sign, so give the same neighborhood system. Give $X(1)$ the quotient topology. The modular group acts properly discontinuously on $\mathbb{H}$; its modular cusp neighborhoods descend to punctured neighborhoods with their modular cusp point filled in. This quotient is Hausdorff: interior orbits can be separated by proper discontinuity, and a sufficiently high modular cusp neighborhood avoids any prescribed compact interior neighborhood.

Away from fixed points of the effective group $PS{L}_2(\mathbb{Z})$, a small disk is disjoint from its nonidentity translates, so its coordinate gives a holomorphic quotient coordinate. At an elliptic fixed point ${\tau }_0$, useIts effective stabilizer rotates $u$ by an $e$th root of unity, where $e=2$ at $i$ and $e=3$ at $\rho$. The quotient coordinate is $w=u^e$, not $u$. All rational boundary points lie in one modular cusp class. At infinity the stabilizer acts by $\tau \mapsto \tau +1$, andis a punctured-disk coordinate; adjoining $q=0$ fills in the modular cusp. These charts and their holomorphic transition maps define the complex structure of the compactified modular curve.

For compactness use the standard regionIts part with $\mathrm{Im}\tau \le Y$, $Y>1$, is compact: the imaginary part is at least $\sqrt{3}/2$. The quotient of its upper tail, with its two vertical sides identified and infinity added, is the closed disk $|q|\le e^{-2\pi Y}$. These two compact images cover $X(1)$, including the boundary-edge identifications. Thus $X(1)$ is a compact Riemann surface.

The $j$ function also extends over the modular cusp with one simple pole, since $j=q^{-1}+O(1)$. Its local expressions at elliptic points descend to holomorphic functions of $u^e$. By Question 1 it maps the remaining quotient bijectively onto $\mathbb{C}$, so its extension is a degree-one holomorphic map $X(1)\to {\mathbb{P}}^1$. A degree-one map of compact Riemann surfaces is biholomorphic: any local ramification would already give degree at least two. Therefore $X(1)$ is the Riemann sphere; this will also identify the base genus in Question 5.

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 central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.

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 $q{\prod }_{n\ge 1}(1-q^n{)}^2(1-q^{11n}{)}^2$.