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The logarithmic derivative of the modular discriminant gives $E_2={\Delta }^{\prime }/(2\pi i\Delta )$. Differentiating its weight-twelve transformation yields the anomalous term $6c(c\tau +d)/(\pi i)$. Subtracting the displayed inverse-height term cancels that anomaly. The result transforms with weight two under the modular group but is not holomorphic, so it is not a holomorphic modular form.

For a finite-index subgroup of the modular group, the quotient of the complex upper half-plane becomes a compact Riemann surface after adjoining its cusp classes. At an elliptic fixed point use the quotient coordinate for the finite stabilizer; at a cusp use its exponential local parameter. Weight-zero modular forms descend to holomorphic functions on this compact surface, and hence are constant by the maximum modulus principle.

For a rational matrix $\gamma$ of positive determinant, use the positive real power of $\det \gamma$ in the displayed slash operator for modular forms. It satisfies $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$ by the automorphy factor identity and multiplicativity of the determinant. It agrees with the usual operator on the modular group. The determinant factor is essential for the stated normalization of the Hecke operators.

An elliptic point of a modular curve is the image of a point of the complex upper half-plane with a nontrivial stabilizer subgroup in the effective modular subgroup. For the full modular group the stabilizer subgroup orders are two and three, represented by $i$ and $e^{2\pi i/3}$. A local disk coordinate $u$ upstairs descends as $u^e$ when the stabilizer subgroup order is $e$. The central matrices $\pm I$ are factored out before counting this order.

For a finite-index subgroup of the modular group, take the union of translates of the standard fundamental domain of the modular group over left 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.

The subgroupis a congruence subgroup of the modular group.

Let $P_w(E_4,E_6)$ be weighted homogeneous of modular weight $w$, and suppose ${\sum }_w{P}_w(z)=0$ as a holomorphic function. Apply the modular group matrices $\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At a fixed point $z$ in the complex upper half-plane, this gives ${\sum }_w(nz+1{)}^w{P}_w(z)=0$ for every integer $n$. A polynomial with infinitely many roots vanishes identically, so every $P_w(z)=0$. Within one modular weight, the monomials reduce to powers of $E_4^3/E_6^2$, after taking out a common factor. This ratio is nonconstant, as its Fourier expansion begins $1+1728q+\cdots$, so its powers are linearly independent. This proves algebraic independence even when the generators are regarded as analytic functions rather than elements of a formal graded ring.

A function on the complex upper half-plane is modular invariant if it is unchanged by the fractional-linear action of the modular group. This does not imply holomorphicity: a nonholomorphic Eisenstein series is modular invariant, whereas a modular function additionally satisfies meromorphicity and cusp conditions.

The closed standard fundamental domain of the modular group isIts vertical sides are paired by $T:z\mapsto z+1$, and its circular sides by $S:z\mapsto -1/z$. It is a fundamental region with boundary identifications. To obtain exactly one representative of each orbit, use $-1/2<x\le 1/2$ and, on $|z|=1$, retain the half with $x\ge 0$. We will include both vertices when discussing the closure.

To prove existence of a representative, fix $z\in \mathbb{H}$. Among primitive integer pairs $(c,d)$ choose one minimizing $|cz+d|$. The minimum exists: below any fixed bound, $|c|y\le |cz+d|$ bounds $c$, and then the real part bounds $d$, so only finitely many pairs occur. Extend this primitive pair to the bottom row of a determinant-one integer matrix $\gamma$. Sincethis makes the imaginary part maximal over the orbit. Translate by an integer to put the real part in $[-1/2,1/2]$. If the resulting point had modulus below one, inversion by $S$ would increase its imaginary part, a contradiction. Thus every orbit meets $\mathcal{F}$.

For uniqueness in the interior, note that $y\ge \sqrt{3}/2$ throughout $\mathcal{F}$ and that $|cz+d|\ge 1$ for every primitive pair. If $|c|\ge 2$, then $|cz+d|\ge |c|y\ge \sqrt{3}>1$. For $|c|=1$, the minimum over integer $d$ is attained at $d=0$ or, at an endpoint, at a neighboring integer: $|z|\ge 1$ and $|z\pm 1{|}^2=|z{|}^2\pm 2x+1\ge 1$. All larger $|d|$ give a strict inequality. If $c=0$, primitivity gives $d=\pm 1$.

If two points in $\mathcal{F}$ are equivalent, applying this inequality to the transformation and its inverse shows their imaginary parts are equal. For two interior points equality forces $c=0,d=\pm 1$, hence an integer translation. Their real parts lie in an interval of length one, so the translation is zero. The only matrices acting trivially are $\pm I$. On the boundary, the equality cases give precisely the stated vertical and circular identifications. This establishes the fundamental-domain claim, including its boundary convention.

An elliptic point of a modular curve means a fixed point of a nonidentity element of the effective modular group $PS{L}_2(\mathbb{Z})$; the central matrix $-I$ is not counted, since it acts trivially everywhere. The equality cases above show that the only such points in the closed region areTheir stabilizer subgroups in $S{L}_2(\mathbb{Z})$ arewhereIndeed $S^2=-I$, $(ST{)}^3=(TS{)}^3=-I$, and their fractional-linear fixed-point equations give the listed points. To see that these are the full stabilizer subgroups, equality in $|cz+d|\ge 1$ leaves only bottom rows $(0,\pm 1),(\pm 1,0)$, with the extra neighboring rows at the vertices. The fixed-point equation then fixes the top row. This yields exactly four matrices at $i$, six at each vertex, and just $\{\pm I\}$ elsewhere. These are the elliptic stabilizers of the modular group. After passing to $PS{L}_2(\mathbb{Z})$ their effective orders are two and three. The two vertices represent the same elliptic orbit.

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.

Use the determinant-normalized slash operator and let ${\mathcal{M}}_n$ be all integral matrices of positive determinant $n$. Their left $S{L}_2(\mathbb{Z})$ cosets have the unique determinant-n matrix representatives for Hecke operatorsInteger row operations reduce the first column to $(a,0{)}^T$, and a row shear then reduces $b$ modulo $d$. DefineChanging representatives does not affect the sum, while right multiplication by the modular group permutes the cosets. Hence $T_{n}f$ has the same modular transformation law. Rational positive-determinant slash transforms are holomorphic on $\mathbb{H}$ and preserve modular cusp vanishing: choose an integral matrix carrying infinity to their rational image modular cusp, after which the remaining matrix is upper triangular. Its positive dilation sends the original decaying modular cusp expansion to another decaying expansion. Thus these are well-defined Hecke operators on $S_k$.

For the Fourier expansion, insert $f(z)={\sum }_{m\ge 1}a(m)e^{2\pi imz}$. The sum over $b$ vanishes unless $d\mid m$, in which case it is $d$. Writing $m=d\ell$, the resulting exponential is $q^{a\ell }$, and its factor is $n^{k-1}{d}^{1-k}=a^{k-1}$. ThereforeIn particular, the first coefficient of $T_{n}f$ is $a(n;f)$.

Assume $f\ne 0$ is a simultaneous Hecke eigenform. The first-coefficient identity gives $a(n)=\lambda (n)a(1)$. If $a(1)=0$, all coefficients vanish, contradicting $f\ne 0$. After dividing by $a(1)$, we may therefore suppose $a(1)=1$, and then $\lambda (n)=a(n)$.

We derive the relations needed for the Euler product. For coprime $m,n$, applying the coefficient formula twice splits the divisors into their disjoint prime supports and yields $T_m{T}_n=T_{mn}$. For one prime define formal Fourier operators $U_p(\sum a(r)q^r)=\sum a(pr)q^r$ and $V_{p}f(z)=f(pz)$. The coefficient formula givesMultiplying this finite sum by $T_p=U_p+p^{k-1}{V}_p$ shows thatThese auxiliary operators need not separately preserve level one; the identities concern their action on Fourier series. Applied to the eigenform, they giveThe prime-power generating function consequently satisfiesMultiplicativity and unique prime factorization now prove the Euler product of a Hecke eigenform:For the original unnormalized form, the right side is multiplied by $a(1)$ and $a(p)$ in the factors is replaced by $\lambda (p)=a(p)/a(1)$. The zero form has $L=0$ and is excluded from the nonzero-eigenform product assertion.

These manipulations initially hold in an absolutely convergent right half-plane. For example, $y^{k/2}|f(z)|$ is bounded by modular invariance and modular cusp decay. The coefficient integral at $y=1/n$ gives $|a(n)|\le C{n}^{k/2}$, so $\mathrm{Re}s>k/2+1$ is sufficient. No analytic continuation is needed to justify the initial product.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

Let $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$ and $j=c\tau +d$. Differentiating the transformation law of a weight-$k$ modular form gives the derivative transformation of a weak modular formThe corresponding anomaly for $Dg$ has coefficient $\ell c/(2\pi i)$. Consequently these anomalies cancel in $\ell gDf-kfDg$, which transforms with weight $k+\ell +2$. The derivatives are holomorphic on the complex upper half-plane. In a Fourier expansion, $D({\sum }_{n\ge 0}{a}_n{q}^n)={\sum }_{n\ge 1}n{a}_n{q}^n$, so each term in the combination has zero constant coefficient and no negative powers. There is just one modular cusp up to the full modular group. ThereforeThis is, up to the chosen sign, the first Rankin-Cohen bracket.

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

Let $a_0$ be the constant Fourier coefficient and suppose first that $k>0$ is even and $M_k\ne 0$; thus $k\ge 4$. Put $\epsilon =i^k=(-1{)}^{k/2}$. The precise Mellin continuation of a noncuspidal modular form isIt is meromorphic on the whole plane, with possible simple poles only at $0,k$, of residues $-a_0,\epsilon a_0$, respectively. For a cusp form it is an entire function. A nonzero constant term makes both of these poles genuine.

We first justify the initial Mellin transform. Decompose $f=a_0{E}_k+h$ with $h\in S_k$. For a cusp form, $y^{k/2}|h(x+iy)|$ is invariant under the modular group and bounded on its fundamental domain, since it tends to zero at the cusp. Therefore it is bounded throughout the complex upper half-plane. Integrating the Fourier coefficient at height $y=1/n$ gives $|a_n(h)|\le C{e}^{2\pi }{n}^{k/2}$. The Eisenstein series coefficients satisfy ${\sigma }_{k-1}(n)\le \zeta (k-1)n^{k-1}$ by replacing divisors $d$ by $n/d$. Consequently $a_n(f)=O(n^{k-1})$, and absolute termwise integration is valid for $\Re s>k$:Indeed ${\int }_0^{\infty }{e}^{-2\pi ny}{y}^{s-1}dy=(2\pi n{)}^{-s}\Gamma (s)$, and the sum of the absolute integrals converges in that half-plane.

Write $I_f(s)={\int }_1^{\infty }(f(iy)-a_0)y^{s-1}dy$. Exponential cusp decay of the difference implies uniform convergence on compact subsets of the $s$-plane, including after any number of derivatives in $s$, so $I_f$ is entire. The modular transformation $f(i/y)=(iy{)}^{k}f(iy)$ givesOn the lower half of the Mellin integral, subtract the transformed constant explicitly:Substitute $y=1/t$ in the decaying term and integrate the two powers in the constant term. This yields, initially for $\Re s>k$,The right side supplies the claimed meromorphic continuation and its residues. Replacing $s$ by $k-s$ and multiplying by $\epsilon$, with ${\epsilon }^2=1$, leaves this expression unchanged, proving the functional equation with its correct sign. In particular the residue at $k$ is tied to the residue at zero by that equation; subtracting only $a_0$ at infinity and overlooking its transformed value at zero would miss these poles.

In weight zero every modular form is constant, so $L(f,s)=0$ and its completion is identically zero. The same holds for the zero form in odd, negative, or weight-two spaces. Thus the assertion covers all weights in the notation of the paper. There is no conjugation of $f$ in this level-one functional equation.

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

Let $\mathbb{H}$ be the upper half-plane and put $q=e^{2\pi i\tau }$. A weight-$k$ modular form for the modular group is a holomorphic function on $\mathbb{H}$ satisfyingwhose Fourier expansion at the modular cusp at infinity has only nonnegative powers of $q$. These functions form the vector space $M_k$. Its subspace $S_k$ consists of the cusp forms, whose constant Fourier coefficient is zero. For the full modular group every cusp is equivalent to infinity, so this is the complete cusp condition. The matrix $-I$ gives $M_k=0$ for odd $k$, and the valence formula for the modular group excludes nonzero forms of negative weight.

The normalized Eisenstein series start as $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$. Consequently the modular discriminantis a nonzero cusp form of weight twelve, with order one at infinity. Its contribution there already exhausts the valence formula, so it has no zero in $\mathbb{H}$. Multiplication by $\Delta$ takes $M_k$ injectively into $S_{k+12}$. Conversely, dividing any member of $S_{k+12}$ by $\Delta$ is holomorphic in $\mathbb{H}$, has weight $k$, and has no negative power at infinity because its numerator vanishes there. Thus multiplication by the modular discriminant gives

The valence formula for the modular group shows that a weight-zero form minus its constant term must vanish: otherwise it would be a cusp form with positive order at infinity and total zero order zero. Hence $\dim M_0=1$. A weight-two form satisfies $f(i)=i^{2}f(i)=-f(i)$, so any nonzero one would contribute at least $1/2$ to the valence formula, exceeding $2/12$. Therefore $M_2=0$.

Every even integer $k\ge 4$ has a representation $k=4a+6b$ with $a,b\ge 0$. Indeed, if $k/2$ is even, take $b=0$; if it is odd, take $b=1$ and $a=(k-6)/4$. The form $R_k=E_4^a{E}_6^b$ has constant term one. Taking the constant Fourier coefficient therefore maps $M_k$ onto $\mathbb{C}$, with kernel $S_k=\Delta M_{k-12}$. Starting with the negative-weight vanishing and the two initial dimensions givesand hence the dimension of level-one modular forms isThis also proves finite dimensionality, rather than presupposing it in the recurrence.

For the generation assertion, if $f\in M_k$ has constant coefficient $a_0$, subtract $a_0{R}_k$. The difference is a cusp form, and so equals $\Delta h$ with $h\in M_{k-12}$. Induction on weight expresses $h$ as a polynomial in $E_4,E_6$; substituting the polynomial expression for $\Delta$ does the same for $f$. Constants start the induction, while weight two contributes nothing. The result is a weighted homogeneous polynomial, assigning weights four and six to its variables. In fact the monomials $E_4^a{E}_6^b$ with $4a+6b=k$ number exactly the displayed dimension: $b$ runs from zero to $\lfloor k/6\rfloor$ with parity $k/2$. Their spanning property therefore makes them a basis in each weight. This identifies the polynomial ring of level-one modular forms:

Fix an even weight $k$ and write $\Gamma =S{L}_2(\mathbb{Z})$. Extend the slash operator for modular forms to positive-determinant rational matrices by the determinant-normalized slash operatorThe automorphy factor identity makes this a right action. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. Their left $\Gamma$-cosets have unique determinant-n matrix representatives for Hecke operatorsAn integral row operation obtained from Bézout's identity makes the first column $(a,0{)}^T$, and a row shear reduces the other top entry modulo $d$, proving this description. The usual Hecke operator isThis normalization has $T_1=1$. Right multiplication by $\Gamma$ permutes the left cosets in ${\mathcal{D}}_n$, so $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic in $\mathbb{H}$. Holomorphy at the cusp and preservation of cusp forms follow directly from the coefficient formula below; the full modular group has only one cusp orbit. Thus the operators act on both $M_k$ and $S_k$.

If $f={\sum }_{r\ge 0}{a}_r{q}^r$, summing over $b$ kills terms unless $d\mid r$. Writing $r=d\ell$ gives output exponent $a\ell$, so a term of exponent $m$ requires $a\mid m$ and has original index $r=dm/a=nm/a^2$. Its multiplier is $n^{k-1}{d}^{1-k}=a^{k-1}$. The Fourier coefficients of a composite-index Hecke operator are consequentlyFor $m=0$, this means $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$. For a prime number $p$,with a nonexistent fractional-index coefficient interpreted as zero.