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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.

Take the even primitive Dirichlet character ${\chi }_{12}$ whose values at $1,5,7,11$ modulo twelve are $1,-1,-1,1$. Its Gauss sum of a Dirichlet character is $\sqrt{12}$. The displayed series $A$ has $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$ and $A(\tau +1)=e^{\pi i/12}A(\tau )$. Thus $A^{24}$ is the normalized weight-twelve modular discriminant. It has no interior zeros by the valence formula for the modular group. Its logarithmic derivative equals $E_2/24$, so integration gives the Dedekind eta function product. The odd character modulo four similarly produces a series $B$ with $B^8=A^{24}$ and leading term $q^{1/8}$; consequently $B=A^3$.

The ratio of modular discriminants has weight zero and level ${\Gamma }_0(2)$. It has one simple pole at infinity and one simple zero at the zero modular cusp, with no zeros or poles in the half-plane. It is a spherical coordinate by the single-pole criterion for a spherical coordinate. The Klein j-invariant is $(t+256{)}^3/t^2$, with $t(e^{2\pi i/3})=-256$.

Put $S_j={\sum }_{n\ge 1}{\sigma }_j(n)q^n$. Expanding $E_4=1+240S_3$ and $E_6=1-504S_5$ gives $\Delta =(5S_3+7S_5)/12+100S_3^2+8000S_3^3-147S_5^2$. For every integer $d$, $5d^3+7d^5$ is divisible by twelve, by separate congruences modulo three and four. Consequently every coefficient of the modular discriminant is integral, without using a product expansion.

Choose $4a_j+6b_j=k-12j$ with nonnegative exponents, for every $j$ with $k-12j\ge 0$ and $k-12j\ne 2$. The Eisenstein series and the modular discriminant give modular forms with integer coefficients and distinct first powers of $q$. The valence formula for the modular group shows that these form a basis. Successively subtracting their leading terms proves that integral coefficients up to the greatest such $j$ force all coefficients to be integral. This is more precise than merely spanning by arbitrary monomials in $E_4,E_6$.

The modular discriminant has a simple zero at infinity and no zero in the upper half-plane, by the valence formula for the modular group. Multiplication by it identifies weight-$k$ modular forms with weight-$(k+12)$ cusp forms. Conversely dividing a cusp form by it preserves interior holomorphy and removes one order of vanishing at infinity.

The three nonzero half-period classes make the odd derivative of the Weierstrass elliptic function vanish. Their values are distinct: two equal values would force at least four zeros for a function with only one double pole. Thus they are three distinct roots of $4X^3-g_{2}X-g_3$, whose monic discriminant gives the displayed factor. This lattice discriminant differs by $(2\pi {)}^{12}$ from the normalized modular discriminant for the lattice $\mathbb{Z}+\mathbb{Z}\tau$.

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 product defining the modular discriminant converges locally uniformly for $|q|<1$ and has integral Fourier coefficients. Its logarithmic derivative isTherefore $D\Delta =E_2\Delta$. Comparing the coefficient of $q^n$ gives $n\tau (n)=\tau (n)-24{\sum }_{j=1}^{n-1}{\sigma }_1(j)\tau (n-j)$, orThis also fixes the initial value $\tau (1)=1$.

For the Ramanujan tau congruence modulo five, first justify the link between this product and $E_4,E_6$. Question 1 constructs ${\Delta }_0=(E_4^3-E_6^2)/1728\in S_{12}$ with first coefficient one. Its Serre derivative is in $S_{14}$, which vanishes by the valence formula for the modular group: a nonzero such cusp form would contribute at least $1+1/2>14/12$, since it vanishes at infinity and at $i$. Hence $D{\Delta }_0=E_2{\Delta }_0$. The product and ${\Delta }_0$ have the same differential equation and leading coefficient. Equivalently their quotient has derivative zero near $q=0$ and limit one; the identity theorem givesAlso $D_6{E}_6\in M_8=\mathbb{C}{E}_4^2$. Its constant coefficient is $-1/2$, so the Serre derivative gives $2D{E}_6=E_2{E}_6-E_4^2$.

Work now in the formal power series ring ${\mathbb{F}}_5[[q]]$. The integral expansions give $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$, $-24\equiv -504\equiv 1$, and $240\equiv 0$. The two identities reduce toThus $3\Delta \equiv -2D{E}_6\equiv 3D{E}_6$, and division by the nonzero element $3$ gives $\Delta \equiv D{E}_6$. Since the coefficient of $q^n$ in $D{E}_6$ is $-504n{\sigma }_5(n)$, we conclude

The preceding dimension of level-one modular forms gives $\dim M_{16}=2$ and $\dim S_{16}=\dim M_4=1$. Thus $E_4\Delta$, which begins with $q$, spans $S_{16}$. A normalized weight-sixteen Eisenstein series and $E_4^4$ both have constant term one, so their difference is a multiple of $E_4\Delta$.

There is a normalization error in the printed auxiliary expansion: for the usual normalized $E_{16}$, the coefficient multiplier is $16320/3617$, rather than $255/(8\cdot 3617)$. Indeed, the Fourier expansion of a normalized Eisenstein series gives $-32/B_{16}$, and the Bernoulli number $B_{16}=-3617/510$ gives $16320/3617$. The printed multiplier is smaller by a factor of $512$. With the correct normalization, the first Fourier coefficients yieldComparing all positive-index coefficients therefore gives the exact identityThe Fourier coefficients of $E_4^4$ are integers, as are those of $E_4\Delta$: $E_4$ has its integral divisor-sum expansion and the modular discriminant has the product $q{\prod }_{r\ge 1}(1-q^r{)}^{24}$. Also $3456000+16320=960\cdot 3617$, and $\gcd (16320,3617)=1$. Reducing the identity modulo $3617$ and cancelling the invertible multiplier proves the weight-sixteen Eisenstein congruenceThe desired congruence is valid; it is the auxiliary normalization in the PDF that requires correction.

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:

Logarithmically differentiate the convergent product of the modular discriminant. On compact subsets of $\mathbb{H}$, absolute convergence permits termwise differentiation and rearrangement of the resulting divisor sums:The division is valid because $\Delta$ has no interior zeros. Its weight-twelve inversion law is $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$. Taking a logarithmic derivative, including the derivative $1/{\tau }^2$ of $-1/\tau$, yieldsThus the Eisenstein series of weight two has the anomalous transformation

Let $y=\mathrm{Im}\tau$. Under inversion, $y$ becomes $y/|\tau {|}^2$. Using ${\tau }^2-|\tau {|}^2=2iy\tau$, the anomalous term cancels:The Fourier expansion gives $E_2(\tau +1)=E_2(\tau )$, and translation leaves $y$ unchanged. Since translation and inversion generate the projective modular group and weight two is unchanged by $-I$, the almost holomorphic weight-two Eisenstein series satisfiesIt is modular in the transformation sense, but is not a holomorphic modular form. Substituting $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$ into this law also gives the useful full-group anomaly

Write $\Gamma =S{L}_2(\mathbb{Z})$, $\mathbb{H}=\{\tau :\mathrm{Im}\tau >0\}$ and $q=e^{2\pi i\tau }$. The space $M_k$ consists of holomorphic functions on the complex upper half-plane satisfyingand holomorphic at the cusp: their Fourier expansion of a modular form is $f={\sum }_{n\ge 0}{a}_n{q}^n$. The subspace $S_k$ of cusp forms has $a_0=0$. Since all cusps are equivalent at level one, this condition treats every cusp. The element $-I$ forces odd-weight forms to vanish.

Use the permitted properties of the normalized Eisenstein series and modular discriminant:They have integer Fourier coefficients and weights $4,6,12$. For even $k\ge 0$, the assumed dimension formula gives $d=\lfloor k/12\rfloor +1$ except when $k\equiv 2(\mathrm{mod}12)$, when $d=\lfloor k/12\rfloor$. For $0\le j<d$, the remaining weight $k-12j$ is nonnegative and is not two. Choose $b_j\in \{0,1\}$ and $a_j\ge 0$ with $4a_j+6b_j=k-12j$. Every allowable remaining weight admits such a choice. The formsgive an integral triangular basis of level-one modular forms: their distinct first exponents prove independence, and their number $d$ proves that they span.

Now take $f$ with integral coefficients. Subtract its constant coefficient times $F_0$. The remainder still has integral coefficients and has no constant term. Subtract its coefficient at $q$ times $F_1$, then its current coefficient at $q^2$ times $F_2$, and continue. Every subtraction uses an integer. After $d$ steps the remainder's first $d$ coefficients vanish; triangularity and the basis property force that remainder to be zero. ConsequentlyThis proves integral polynomial generation without dividing by $1728$. The relation $E_4^3-E_6^2=1728\Delta$ explains why the additional integral generator is useful, although over $\mathbb{C}$ two generators suffice. For $k=0$ the basis is the constant one; when $M_k=0$, including weight two and odd weights, the claim is vacuous.

Fix an even weight and use the determinant-normalized slash operator for modular formsIt is unchanged by positive scalar multiples of $A$ and satisfies $(f|A)|B=f|AB$. The determinant-n matrix representatives for Hecke operators areIndeed, integer row operations with determinant one send the first column of an integral determinant-$n$ matrix to $(a,0{)}^t$, with $a>0$; the second diagonal entry is $d=n/a>0$, and a row shear reduces $b$ modulo $d$. This gives unique representatives of the left modular-group orbits.

Define the Hecke operator byThe set of integral matrices of determinant $n$ is stable under right multiplication by the modular group. Right multiplication therefore permutes its left orbits, proving the modular transformation law for $T_{n}f$. Each summand is holomorphic in the upper half-plane. Its imaginary argument tends to infinity at the cusp, so the formula preserves cusp holomorphy and maps $M_k$ to itself and $S_k$ to itself.

Expand each summand in q. The finite sum over $b$ vanishes unless the Fourier index is divisible by $d$, when it equals $d$. The coefficient at $q^m$ then receives a contribution for each $a\mid \gcd (m,n)$, giving the Fourier coefficients of a composite-index Hecke operator formulaFor $m=0$ interpret the sum over all divisors of $n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$ and integral coefficients are preserved.

Here is a useful direct derivation of the Hecke multiplication relations. On formal Fourier series put $U_{p}f={\sum }_m{a}_{pm}{q}^m$ and $V_{p}f=f(q^p)$. They obey $U_p{V}_p=1$, although $V_p{U}_p$ is generally not the identity. The coefficient formula givesMultiplying by $T_p=U_p+p^{k-1}{V}_p$ and using $U_p{V}_p=1$ separates the terms with $j=0$ from the others and yieldsFor distinct primes, all the relevant shift operators commute. Thus $T_m{T}_n=T_{mn}$ when $\gcd (m,n)=1$, and each prime-power operator is a polynomial in its prime operator. This proves commutativity. Induction on the smaller of two prime exponents, using the displayed recurrence, gives $T_{p^r}{T}_{p^s}={\sum }_{j=0}^{\min (r,s)}{p}^{j(k-1)}{T}_{p^{r+s-2j}}$. Factoring general indices into primes then provesThe auxiliary shifts need not themselves preserve the level-one form space; they are used only to prove identities of coefficient sequences.

For cusp forms, the Petersson inner product isIt converges by exponential cusp decay, is positive-definite, and makes the Hecke operators self-adjoint. For the change-of-variable argument, the determinant-normalized slash action compensates exactly for the change in $y^k$, while hyperbolic area is invariant. Integrating a correspondence term in the opposite direction replaces its matrix by its inverse. The adjugate $A^{\#}=n{A}^{-1}$ is again an integral matrix of determinant $n$; it has the same slash action as $A^{-1}$, because scalar matrices act trivially. Summing over the two sets of representatives, equivalently over the common finite covering domains of the correspondence, interchanges $f$ and $g$ and proves $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. This explains Hecke operators are self-adjoint for the Petersson inner product with the present normalization.

Commuting self-adjoint operators on the finite-dimensional cusp space have a common orthonormal eigenbasis: diagonalize one, restrict the others to its invariant eigenspaces and repeat. If $f$ is a nonzero common Hecke eigenform, $a_n(f)=a_1(T_{n}f)={\lambda }_n{a}_1(f)$. The first coefficient cannot vanish, since then every coefficient would vanish. Normalize it to one. The eigenvalue is then ${\lambda }_n=a_n(f)$, and the common eigenspace is one-dimensional because all its normalized Fourier coefficients are fixed. The eigenvalues are real by self-adjointness and are algebraic integers, since the Hecke operators act by integer matrices on the integral cusp-form lattice from Question 1. The integral echelon basis and the operator identity there also show that $T_1,\dots ,T_{d-1}$ are an integral basis of the integral Hecke algebra of level-one cusp forms; independence follows by evaluating their first-coefficient functionals on $g_1,\dots ,g_{d-1}$.

The eigenform's coefficient relations areThey give the Euler product of a Hecke eigenformin a sufficiently far right half-plane. The Mellin transform of a cusp-form L-function completes this to $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)={\int }_0^{\infty }f(iy)y^{s-1}dy$. Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential decay makes this an entire function of $s$, and the formula proves $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

On the noncuspidal line in positive even weights $k\ge 4$, the normalized Eisenstein series has eigenvalues ${\sigma }_{k-1}(n)$, as follows from the same divisor formula for its coefficients; $M_k=\mathbb{C}{E}_k\oplus S_k$. As a cuspidal example, $S_{12}$ is spanned by the normalized modular discriminant. Its coefficients $\tau (n)$ are the Hecke eigenvalues, and the recurrence gives $\tau (4)=\tau (2{)}^2-2^{11}=-1472$ from $\tau (2)=-24$. Thus the Hecke theory converts analytic forms into multiplicative arithmetic data.

The modular discriminant has the convergent productis a weight-twelve level-one cusp form, and has order one at infinity. Applying the degeneracy map for cusp forms with $M=1,N=D=p$ gives $\boxed{\Delta (p\tau )\in S_{12}({\Gamma }_0(p))}$. The original form $\Delta$ is also a cusp form on this subgroup.

There are precisely two modular cusps for prime level, represented by infinity and zero. To see this directly, write a rational cusp as $a/c$ in lowest terms. If $p\mid c$, completing $(a,c{)}^T$ to a determinant-one matrix gives a member of ${\Gamma }_0(p)$ taking infinity there. If $p\nmid c$, the equation $Ac-aC=1$ has a solution with $p\mid C$: adjust a Bézout solution modulo $p$, using the invertibility of $c$. The matrix $\left(\begin{array}{cc}A& a\\ C& c\end{array}\right)$ then lies in ${\Gamma }_0(p)$ and takes zero there. The two classes are distinct because their denominator divisibility is preserved by the group.

At infinity the cusp width is one, so $q$ is the local parameter. The products start with $q$ and $q^p$, giving orders one and $p$. At zero use $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Since $S{T}^h{S}^{-1}=\left(\begin{array}{cc}1& 0\\ -h& 1\end{array}\right)$, the cusp width is $p$ and the parameter is $q_0=e^{2\pi i\tau /p}$. The weight-twelve inversion law of the modular discriminant givesHere determinant one makes the printed and ordinary slash operators for modular forms agree. In terms of $q_0$, the leading terms are $q_0^p$ and $p^{-12}{q}_0$. Hence the prime-level cusp orders of discriminant degeneracy forms areThe scalar $p^{-12}$ changes no order of vanishing.

The infinite product converges locally uniformly in $\mathbb{H}$, because ${\sum }_{n\ge 1}|q{|}^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally uniform convergence, givesThe middle equality groups the absolutely convergent terms $n{q}^{nr}$ by $m=nr$. Thus $F^{\prime }/F=2\pi i{E}_2$.

Let $H(\tau )={\tau }^{-12}F(-1/\tau )/F(\tau )$. It is nonzero and holomorphic on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau )={\tau }^2{E}_2(\tau )+6\tau /(\pi i)$, its logarithmic derivative isSo $H$ is constant. At $\tau =i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau )={\tau }^{12}F(\tau )$. Also $F(\tau +1)=F(\tau )$ directly from $q=e^{2\pi i\tau }$. Since these two transformations generate $S{L}_2(\mathbb{Z})$, $F$ has the weight-twelve modular form transformation law.

At the unique modular cusp its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. HenceThis is the modular discriminant. The transformed argument contributes $6\tau /(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.

For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.