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A Dirichlet character $\psi$ modulo $N$ specifying the displayed transformation law is the nebentypus of a modular form. The action of $-I$ forces $\psi (-1)=(-1{)}^k$ for a nonzero form. It describes a character eigenspace of the diamond action rather than an additional analytic condition at modular cusps.

A simultaneous eigenfunction of all normalized level-one Hecke operators in positive even weight, with nonzero constant Fourier coefficient, is a multiple of the normalized Eisenstein series. The constant coefficient forces its $T_n$ eigenvalue to be ${\sigma }_{k-1}(n)$, and $a_1(T_{n}f)=a_n(f)$ forces $a_n(f)=a_1(f){\sigma }_{k-1}(n)$. Subtracting the matching constant multiple of the Eisenstein series leaves a cusp form. Unless every coefficient vanishes, its prime-index coefficients grow like $p^{k-1}$, contradicting the Fourier coefficient bound for a cusp form $O(p^{k/2})$ when $k\ge 4$. Weight two has no nonzero modular forms, by vanishing of weight-two level-one modular forms.

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 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.

Put $q=e^{2\pi i\tau }$ and ${\sigma }_j(n)={\sum }_{d\mid n}{d}^j$. Separate the $m=0$ terms in the lattice Eisenstein series. Differentiating the cosecant partial-fraction identity $2k-2$ times gives, for $\Im z>0$,The initial identity is ${\sum }_n(z+n{)}^{-2}={\pi }^2{\csc }^2(\pi z)=-4{\pi }^2{\sum }_{r\ge 1}r{e}^{2\pi irz}$; differentiation accounts for the factorial and the sign. Pair $m$ and $-m$ and collect the products $mr$. Absolute convergence of the lattice sum for $k\ge 2$, and local uniform convergence of the resulting power series, justify these operations. Thus the requested Fourier expansions areHere $E_{2k}=G_{2k}/(2\zeta (2k))$, and the Bernoulli numbers have convention $B_2=1/6$. The second formula uses Euler's even-value identity for the Riemann zeta function, $2\zeta (2k)=(-1{)}^{k+1}{B}_{2k}(2\pi {)}^{2k}/(2k)!$. In particular $E_4=1+240\sum {\sigma }_3(n)q^n$ and $E_6=1-504\sum {\sigma }_5(n)q^n$.

We need the rational, rather than merely complex, version of the polynomial ring of level-one modular forms. Recall how the valence formula for the modular group supplies its proof. For a nonzero weight-$w$ modular form, with $\rho =e^{2\pi i/3}$,Integrate $f^{\prime }/f$ around a truncated standard fundamental domain of the modular group, indenting around boundary zeros. The paired vertical sides cancel by translation; pair the two halves of the circular edge using $f(-1/\tau )={\tau }^{w}f(\tau )$. The logarithmic derivative of ${\tau }^w$ supplies $w/12$, while the top edge supplies $v_{\infty }(f)$. The half-disc and third-disc at the two elliptic points account for the fractions $1/2,1/3$. The argument principle then gives the formula. All orders in it are nonnegative for a holomorphic modular form. It follows that negative weights have no forms, $M_0=\mathbb{C}$, and $M_2=0$: a weight-two form vanishes at $i$, contributing at least $1/2>2/12$.

Define ${\Delta }_0=(E_4^3-E_6^2)/1728$. It is a weight-twelve cusp form and its expansion starts with $q$. The valence formula for the modular group forces its order at infinity to be exactly one and leaves no interior zeros. Thus division by ${\Delta }_0$ identifies $S_w$ with $M_{w-12}$. Every nonnegative even $w\ne 2$ has a monomial $E_4^a{E}_6^b$ of weight $w$: solve $2a+3b=w/2$, taking $b=0$ when $w/2$ is even and $b=1$ when it is odd and at least three. Subtract the constant coefficient of a modular form times such a monomial and divide by ${\Delta }_0$. Induction on weight proves generation by $E_4,E_6$.

If the original Fourier coefficients are rational, every step preserves rationality: both generators and ${\Delta }_0$ have rational expansions, and formal division by ${\Delta }_0=q+O(q^2)$ does too. This proves the rational structure of level-one modular forms. Write $E_{2k}=\sum r_{a,b}{E}_4^a{E}_6^b$ with $r_{a,b}\in \mathbb{Q}$ and $4a+6b=2k$. Rescaling givesThe Bernoulli number formula above makes each zeta ratio rational: its powers of $\pi$ cancel precisely because the weights agree. This proves the required rational coefficients for the unnormalized lattice Eisenstein sums.

Finally $S_8=S_{10}=0$ because the quotient weights are negative, and $S_{14}=0$ because $M_2=0$. In each of these weights a form is determined by its constant coefficient. The indicated monomials have that coefficient equal to one, so

The required anomaly cancellation depends on the exact Eisenstein series of weight two lawHere is a way to establish it without assuming that $E_2$ is a modular form. The lattice definition of the completed nonholomorphic Eisenstein series in Question 6 is modular invariant. Its Poisson summation calculation, proved there, gives the finite Laurent coefficient at $s=1$ asThe residue is constant, so $C$ is also invariant. This calculation uses only convergent Gaussian integrals and the analytic infinite product, not modularity of that product. Using the Wirtinger derivative ${\partial }_{\tau }$, with ${\partial }_{\tau }y=1/(2i)$, differentiation of the locally uniformly convergent series givesDifferentiate $C(\gamma \tau )=C(\tau )$ to see that $E_2^{\ast }=E_2-3/(\pi y)$ transforms with weight two. Substituting $\Im (\gamma \tau )=y/|j{|}^2$ yieldssince $\overline{j}-j=-2icy$. This is the weight-two transformation from an invariant Eisenstein limit.

Combine this law with the derivative transformation of a weak modular form proved in part (i). The anomalous terms in $Df-(k/12)E_{2}f$ cancel exactly, leaving the weight-$k+2$ transformation law. Both functions are holomorphic on the complex upper half-plane, and their Fourier expansions contain no negative powers. Hence the Serre derivative satisfiesIts constant coefficient is $-k{a}_0(f)/12$; it need not be a cusp form when $f$ is not cuspidal.

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 $\mathbb{H}$ 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

Here being modular of weight $k$ means obeying the weight transformation law, or equivalently being invariant under the slash operator for modular forms:This transformation property alone does not impose holomorphy; a holomorphic modular form additionally satisfies interior and cusp holomorphy. That distinction is needed for the nonholomorphic completion in the next part.

Put $F(\tau )=f(N\tau )$. For $\gamma \in {\Gamma }_0(N)$, conjugation by $\mathrm{diag}(N,1)$ gives the integral determinant-one matrixSince the denominator in the latter action is $(c/N)(N\tau )+d=c\tau +d$, the full-group modularity of $f$ givesThus argument dilation lowers the group to the Gamma 0 congruence subgroup while retaining the weight. If $f$ is a holomorphic modular form, cusp holomorphy is preserved too: an integral matrix $\mathrm{diag}(N,1)\gamma$ of determinant $N$ factors as an integral determinant-one matrix times an upper triangular positive-determinant matrix. The corresponding slash translate is a constant times $f((a\tau +b)/d)$ with positive $a/d$, whose expansion at infinity has no negative exponents. This checks every rational cusp; for a cusp form its constant term remains zero. It is the level-one instance of the degeneracy map for cusp forms.

State the characteristic convention explicitly: for real $\alpha ,\beta$, define the theta function with characteristics byBinary subscripts ${\vartheta }_{ab}$ often denote this function with $(\alpha ,\beta )=(a/2,b/2)$, $a,b\in \{0,1\}$. On compact subsets of the $(z,\tau )$ domain, a summand and any derivative are bounded by a polynomial in $|n|$ times $e^{-c{n}^2+C|n|}$. Normal convergence therefore gives an entire function of $z$, holomorphic dependence on $\tau$, and legitimate termwise differentiation. In particular the heat equation for theta functions with characteristics isCharacteristics are defined modulo integers with a phase: $\theta [\alpha +1,\beta ]=\theta [\alpha ,\beta ]$ and $\theta [\alpha ,\beta +1]=e^{2\pi i\alpha }\theta [\alpha ,\beta ]$. Shifting the summation index proves the spatial quasi-periodsThese multipliers describe a degree-one line bundle on the complex torus with periods $1,\tau$, rather than an ordinary elliptic function. The logarithmic derivative is unchanged under one and decreases by $2\pi i$ under tau. Pairing opposite sides of a fundamental cell therefore makes its contour integral $2\pi i$. The argument principle counts exactly one zero in a cell, with multiplicity.

The characteristic function is a translate of the unshifted theta function:At $z=(1+\tau )/2$, the unshifted series cancels in pairs $n,-n-1$. Hence the zeros of theta functions with characteristics occur exactly and simply atReindexing also gives $\theta [\alpha ,\beta ](-z)=\theta [-\alpha ,-\beta ](z)$. For half-integer characteristics the parity is $e^{4\pi i\alpha \beta }$, so only $(1/2,1/2)$ is odd and vanishes at zero; the other three theta constants are nonzero.

The modular transformations of theta characteristics are obtained from Gaussian Poisson summation and elementary phase rearrangement. With the holomorphic branch of $\sqrt{-i\tau }$ positive at $\tau =i$,For the first formula, apply Poisson summation to $F(t)=e^{\pi i\tau (t+\alpha {)}^2+2\pi i(t+\alpha )(z+\beta )}$. Completing the square in its Fourier transform givesRearranging ${\sum }_{n}F(n)={\sum }_m\hat{F}(m)$ proves the stated inversion law. For translation by one, use $e^{\pi i{n}^2}=e^{\pi in}$ in the defining series; the remaining characteristic phase is precisely the displayed factor.