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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 $\tau (2)=-24$.

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

For a weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(z)|$, with $y=\mathrm{Im}z$, is unchanged by the modular group: $\mathrm{Im}(\gamma z)=y/|cz+d{|}^2$ and $|f(\gamma z)|=|cz+d{|}^k|f(z)|$ cancel exactly.

Every orbit meets the standard fundamental domain of the modular group $\mathcal{F}=\{z:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, where $y\ge \sqrt{3}/2$. The Fourier expansion of a modular form at the cusp has $f(q)=qh(q)$, with $h$ holomorphic near zero, since $f$ is a cusp form. Thus $|f(x+iy)|=O(e^{-2\pi y})$ uniformly in $x$ as $y\to \infty$, and $y^{k/2}|f|$ tends to zero there. On the remaining compact part of $\mathcal{F}$ it is bounded by continuity. Invariance therefore gives

For every $y>0$, Fourier coefficient extraction givesFor $n\ge 1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the Fourier coefficient bound for a cusp formThe assertion concerns positive indices; the constant coefficient vanishes because $f$ is a cusp form.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and $G=G{L}_2(\mathbb{Q}{)}^{+}$. The double-coset Hecke algebra consists of $\Gamma$-bi-invariant complex functions on $G$ supported on finitely many double cosets, with convolutionEach double coset has finitely many orbits under left multiplication by $\Gamma$, by rational conjugation of finite-index modular subgroups. Thus the sum is finite and independent of representatives. The characteristic functions of the double cosets form its basis, and its right action on invariant modular forms is $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$, using the determinant-normalized slash operator.

For a positive integer $n$, let ${\mathcal{M}}_n$ be all integral two-by-two matrices of positive determinant $n$. It is $\Gamma$-bi-invariant. Define $T(n)$ to be its indicator function, equivalently the sum of the distinct double cosets it contains, each with coefficient one. This is the convention consistent with the requested formula. For composite $n$, it need not be the single double coset of $\mathrm{diag}(1,n)$: for instance $2I\in {\mathcal{M}}_4$ cannot lie in that double coset, since multiplying by unimodular integral matrices preserves the greatest common divisor of the entries.

Prove all the needed subgroup facts directly. For a finite-index subgroup $L\subseteq {\mathbb{Z}}^2$, take $a$ to be the least positive first coordinate appearing in $L$ and $d$ the least positive second coordinate on its intersection with the second axis. Euclidean division then shows that the first-coordinate projection is $a\mathbb{Z}$ and $L\cap (\{0\}\times \mathbb{Z})=\{0\}\times d\mathbb{Z}$. Positivity follows, for example, because the finite quotient group kills a nonzero multiple of each coordinate vector. Choose $(a,b)\in L$ and reduce $b$ modulo $d$ to $0\le b<d$. Every vector of $L$ has first coordinate a multiple of $a$, and subtracting that multiple of $(a,b)$ leaves a multiple of $(0,d)$. Therefore these two vectors form a basis of $L$. The parameters $a,d,b$ are unique. Reducing the first coordinate modulo $a$ and then the second modulo $d$ gives precisely $ad$ quotient representatives, so $[{\mathbb{Z}}^2:L]=ad$.

Apply this row Hermite normal form in rank two to the row lattice of an integral matrix $A$ with $\det A=n$. Its row lattice contains $n{\mathbb{Z}}^2$, since $\mathrm{adj}(A)A=nI$, so it has finite index. Its two rows and the displayed two rows are bases of the same row lattice. The two inverse change-of-basis matrices have integer entries, so their determinants are integers whose product is one. Thus the change-of-basis matrix has determinant $\pm 1$; since both orientations are positive, its determinant is one. Thus each orbit under left multiplication by $\Gamma$ has exactly one of the determinant-n matrix representatives for Hecke operatorsThere are ${\sum }_{d\mid n}d$ such representatives, proving finiteness as well as the formula. The subgroup argument is a proof of the relevant Hermite normal form, not an invocation of an unproved lattice classification.

Consequently the normalized Hecke operator isIt preserves $M_k(\Gamma )$: right multiplication by $\Gamma$ permutes the left-multiplication orbits in ${\mathcal{M}}_n$, giving invariance, and cusp holomorphy under rational slash operators gives the holomorphy of each term at every cusp.

For a triangular representative the determinant-normalized slash operator is $f{|}_{k}A=n^{k/2}{d}^{-k}f((az+b)/d)$. Summing the Fourier expansion of a modular form over $b$ kills every index not divisible by $d$, by finite exponential orthogonality. ThusThe Fourier coefficients of a composite-index Hecke operator are thereforeIn particular $a_1(T_{n}f)=a_n(f)$. If $T_{n}f=\lambda f$, comparison of the $q$ coefficients gives

Finally suppose $a_0(f)\ne 0$ and $f$ is a simultaneous eigenfunction. Constant coefficients force ${\lambda }_n={\sigma }_{k-1}(n)$, so $a_n(f)=a_1(f){\sigma }_{k-1}(n)$ for all $n\ge 1$. Weight two cannot occur, by vanishing of weight-two level-one modular forms. For even $k\ge 4$, use the normalized Eisenstein series with its Fourier expansion of a normalized Eisenstein serieswhere $B_k$ is the Bernoulli number. Then $h=f-a_0(f)E_k$ is a cusp form with coefficients $(a_1(f)-a_0(f)c_k){\sigma }_{k-1}(n)$. The Fourier coefficient bound for a cusp form bounds these by $O(n^{k/2})$, but at arbitrarily large primes $p$ their magnitude is $|a_1-a_0{c}_k|(1+p^{k-1})$. Since $k-1>k/2$, the coefficient factor must vanish. All coefficients of $h$ then vanish, including its constant coefficient, so $h=0$ by its cusp expansion and the identity theorem. This proves the noncuspidal level-one Hecke eigenform characterization:

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

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.

Writeand suppose $p\mid s$. In the basis from part b, comparison of the constant term and the first $N$ nonconstant coefficients givesIndeed, $f_0$ has constant term one and no terms $q,ldots,q^N$, while $f_i$ has the sole term $q^i$ in that range.

SetEvery $f_i$ with $i\ge 1$ vanishes at infinity and is therefore a cusp form; moreover $f$ has integral coefficients. Comparing the coefficient of $q^n$ in the displayed identity givesReduction modulo $p$ kills the first term on the right. Since $p\nmid r$, cancellation of $r$ yieldsfor every $n\ge 1$, proving the Eisenstein congruence from a denominator prime.

Put $g=f{|}_k[{\alpha }_N]$. Termwise Mellin transform in the initial half-plane givesso $\Lambda (f,s)=N^{s/2}I(f,s)$.

From the definition of the slash action,or equivalentlySubstituting $y=1/(Nt)$ in the Mellin integral therefore givesMultiplying by $N^{s/2}$ and recognizing the completed function on the right yields

At infinity, both $f(iy)$ and $g(iy)$ decay exponentially because they are cusp forms. The transformation just used converts the behavior of $f(iy)$ near zero into the exponential decay of $g(i/(Ny))$ at infinity. Consequently $I(f,s)$ converges absolutely for every $s\in \mathbb{C}$ after splitting the integral at one and applying that substitution to the part near zero. It is locally uniformly convergent in $s$, hence entire, and agrees with the original Dirichlet series in its initial domain. This proves the analytic continuation and the stated functional equation, as summarized by the Mellin transform of a cusp-form L-function.

Put $\tau =x+iy$. The modular transformation law andshow that the invariant norm of a modular formis invariant under $\Gamma (1)$. On the region from part (a), it is bounded: it is continuous on every truncated region, while the cusp form condition makes it tend to zero as $y\to \infty$. Thusfor all $x\in \mathbb{R}$ and $y>0$.

The Fourier coefficient formula on one period givesand henceChoosing $y=1/n$ yieldsThis proves the Fourier coefficient bound for a cusp form.

Suppose for a contradiction that $b_0\ne 0$. The Eisenstein series in the question has constant term $2\zeta (k)$, sohas zero constant term and is therefore a level-one cusp form of weight $k$. Part (b) gives the coefficient bound $[q^n]h=O(n^{k/2})$. Since $b_n=O(n^{k/2})$ as well, the displayed Fourier expansion of $G_k$ would imply