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

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

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

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

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

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

Apply the Serre derivative in weight twelve to the modular discriminant. The preceding result makes $D_{12}\Delta =D\Delta -E_2\Delta$ a weight-fourteen cusp form. The vanishing of weight-fourteen level-one cusp forms follows directly from the valence formula for the modular group: a nonzero such form has order at infinity at least one, and $f(i)=i^{14}f(i)=-f(i)$ forces order at $i$ at least one. Thus its weighted number of zeros would be at least $1+1/2>14/12$, a contradiction. Therefore $D\Delta =E_2\Delta$.

Compare the Fourier coefficients at positive indices using $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ and $E_2=1-24{\sum }_{r\ge 1}{\sigma }_1(r)q^r$. This gives $n\tau (n)=\tau (n)-24{\sum }_{r=1}^{n-1}{\sigma }_1(r)\tau (n-r)$. For the Ramanujan tau function, the concise recurrence isAt $n=1$ the sum is empty and both sides are zero; normalization supplies $\tau (1)=1$. For instance the next coefficient is $\tau (2)=-24$.

We use left multiplication by $\Gamma =S{L}_2(\mathbb{Z})$. For an integral matrix $A$ of determinant $n>0$, the Bezout identity gives a determinant-one row operation sending its first column to $(a,0{)}^t$, where $a$ is the positive gcd of that column. The resulting matrix is $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $d=n/a>0$. Adding a multiple of the second row to the first makes $0\le b<d$.

These representatives are unique. Left multiplication by a unimodular matrix preserves the gcd of the first column, so two representatives in one orbit have the same $a$ and hence $d$. A matrix taking $(a,0{)}^t$ to itself has the form $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$, so it changes $b$ by $td$; the prescribed range makes $b$ unique. Thus the determinant-n matrix representatives for Hecke operators are exactly ${\Pi }_n$. In the printed set, $0\le b<d$ already forces $d>0$, and $ad=n>0$ forces $a>0$.

For positive determinant extend the slash operator for modular forms byThe determinant factor makes this a right action: $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Right multiplication by $\delta \in \Gamma$ permutes the left cosets represented by ${\Pi }_n$, since it preserves the set of determinant-$n$ integral matrices. Hence $T_{n}f$ satisfies the modular transformation law. Its summands are holomorphic on the upper half-plane. For an upper-triangular representative,Substitute the Fourier expansion of a modular form. The sum over $b$ is zero unless the original index is divisible by $d$, in which case it equals $d$. Writing that index as $dr$ givesOnly nonnegative powers occur, so $T_{n}f$ is holomorphic at infinity, and all level-one cusps are equivalent to infinity. We have proved $T_{n}f\in M_k(\Gamma )$ and the Fourier coefficients of a composite-index Hecke operator formulaHere $\gcd (n,0)=n$, giving $a_0(T_{n}f)={\sum }_{a\mid n}{a}^{k-1}{a}_0(f)$. In particular cusp forms stay cusp forms. For nonzero level-one cusp forms the weight is an even integer at least twelve, so the powers $a^{k-1}$ are integers; the zero cusp space causes no exception. Thus every $T_n$ preserves integral Fourier coefficients, and closure under addition, multiplication and integer scalars provesThis is the integral Hecke algebra of level-one cusp forms.

We will also need commutativity. The coefficient formula directly givesFor the first identity, the divisors of $u$ and $v$ in the iterated coefficient sum combine uniquely into a divisor of $uv$. For the second, write a coefficient index as $p^t{r}_0$ with $p\nmid r_0$; the formula isApplying $T_p$ splits this into the two sums with shifts $t+1$ and $t-1$; their overlapping terms give exactly the stated recurrence. Since $T_1$ is the identity, induction expresses each $T_{p^r}$ as a polynomial in $T_p$, and coprime multiplicativity expresses every $T_n$ as a product of these polynomials. Distinct prime operators commute by the coprime identity. Hence all Hecke operators commute, and so do all elements of $\mathbb{T}$.

Let $\Lambda =S_k(\Gamma ,\mathbb{Z})$ and $m=\dim S_k(\Gamma )$. In the permitted basis $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, the modular discriminant begins $q+O(q^2)$ and the normalized Eisenstein series begin $1+O(q)$, soThe matrix of $a_1,\dots ,a_m$ on this integral basis is unitriangular and has determinant one. Integer elimination therefore gives an integral basis $f_1,\dots ,f_m$ with $a_i(f_j)={\delta }_{ij}$ for $1\le i,j\le m$. Thus $a_1,\dots ,a_m$ are a $\mathbb{Z}$-basis of the dual module ${\mathrm{Hom}}_{\mathbb{Z}}(\Lambda ,\mathbb{Z})$.

The map $\alpha$ is well-defined by lattice preservation and is $\mathbb{Z}$-linear. Because $\alpha (T_j)=a_j$, it is surjective. For injectivity suppose $\alpha (T)=0$. For every $f\in \Lambda$ and every $n\ge 1$, commutativity givesThe constant term is also zero because $Tf$ is a cusp form. The identity theorem applied to its Fourier expansion of a modular form gives $Tf=0$. The integral basis spans the complex cusp space, so $T=0$ as an endomorphism. This proves the perfect integral Hecke pairing andIf $m=0$, both modules are zero and the asserted basis is empty.

Let $N+1=\dim M_k(S{L}_2(\mathbb{Z}))$. For each $0\le i\le N$, the dimension formula, equivalently the valence formula for the modular group, ensures thatfor some nonnegative integers $a_i,b_i$. DefineThe modular discriminant, $E_4$, and $E_6$ have integral Fourier coefficients and leading terms $q$, $1$, and $1$, respectively. HenceTheir distinct orders of vanishing make the $h_i$ linearly independent, so they form a basis.

Starting with $f_N=h_N$, define $f_i$ downwards by subtracting from $h_i$ the integral multiples of $f_{i+1},\dots ,f_N$ needed to kill the coefficients of $q^{i+1},\dots ,q^N$. This integer Gaussian elimination preserves all integral coefficients and givesThis is the integral echelon basis of level-one modular forms.

LetForthe matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 3b\\ c& d\end{array}\right)$ lies in $S{L}_2(\mathbb{Z})$ and satisfies $3\gamma \tau ={\gamma }^{\prime }(3\tau )$. The weight-twelve transformation law for the modular discriminant gives the same factor $(3c\tau +d{)}^{12}$ in numerator and denominator. Hence $f(\gamma \tau )=f(\tau )$, so $f$ is a weight-zero modular function of level ${\Gamma }_0(3)$.

The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$ givesThe Fricke involution exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.

Thus the morphism ${\varphi }_f:X_0(3)\to \hat{\mathbb{C}}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.

The dimension of level-one cusp forms is zero when $k$ is odd or $k<12$. For even $k\ge 12$ it isEquivalently, multiplication by the modular discriminant gives $S_k(\Gamma (1))=\Delta M_{k-12}(\Gamma (1))$, and the displayed formula follows from the standard dimension formula for $M_{k-12}(\Gamma (1))$.

Subtract the constant term of a modular form times a same-weight monomial in $E_4,E_6$. The remainder is a cusp form, hence divisible by the modular discriminant. Induction on weight and $1728\Delta =E_4^3-E_6^2$ express every form as a weighted homogeneous polynomial in these two Eisenstein series. The number of monomials of a given weight equals the dimension of level-one modular forms, so they form a basis and the graded ring has no polynomial relation.

For prime $p$, the two cusps of ${\Gamma }_0(p)$ have widths one and $p$. At infinity the modular discriminant product gives the first pair of orders. At zero use $S\tau =-1/\tau$ and the parameter $q_0=e^{2\pi i\tau /p}$. The weight-twelve transforms are $\Delta {|}_{12}S=\Delta (\tau )$ and $\Delta (p\tau ){|}_{12}S=p^{-12}\Delta (\tau /p)$, giving orders $p$ and one in $q_0$.

The modular discriminant satisfies $1728\Delta =E_4^3-E_6^2$, while the Serre derivative gives $2\Theta E_6=E_2{E}_6-E_4^2$. Their integral Fourier expansions have $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$. Reducing the two identities gives $\Delta \equiv \Theta E_6(\mathrm{mod}5)$. The coefficient of $q^n$ in $\Theta E_6$ is $-504n{\sigma }_5(n)\equiv n{\sigma }_5(n)$.

The Ramanujan tau function gives the Fourier coefficients at positive indices of the modular discriminant: $\Delta (z)={\sum }_{n\ge 1}\tau (n)q^n$, with $\tau (1)=1$. The Serre derivative and vanishing of weight-fourteen level-one cusp forms give $qd\Delta /dq=E_2\Delta$ and hence the convolution recurrence

There are no weight-four cusp forms at level two. A nonzero form would have a weight-twelve coset norm of a modular form with modular cusp order at least two, contradicting the simple modular cusp order and interior nonvanishing of the modular discriminant. The two independent displayed Eisenstein series span the weight-four space, since its cusp-constant map embeds it in a two-dimensional space.

The weight-sixteen cusp form space is spanned by $E_4\Delta$. Its leading coefficient and the Fourier expansion of a normalized Eisenstein series give $3617E_{16}=3617E_4^4-3456000E_4\Delta$. The coefficient multiplier $-3456000$ is congruent to $16320$ modulo $3617$, and is invertible there. Coefficient comparison gives the displayed modular congruence. Integrality of $E_4\Delta$ follows from the modular discriminant product.