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For an integer $k\ge 2$ and $\mathrm{Im}z>0$, differentiating the partial-fraction expansion of $\pi \cot \pi z$ givesBoth differentiated series converge locally uniformly. This kernel computes the Fourier expansion of a modular form for holomorphic Eisenstein series.

For even nonnegative weight, the displayed dimension follows from multiplication by the modular discriminant and the constant-coefficient map. Every nonnegative even weight except two has a product $E_4^a{E}_6^b$ of Eisenstein series with constant term one, so $\dim M_k=1+\dim M_{k-12}$ for even $k\ge 4$. The initial dimensions are one in weight zero and zero in weight two; negative and odd weights have no nonzero level-one modular forms.

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 $E_4,E_6$.

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.

Write $d\mu =dxdy/y^2$ and distinguish the all-pairs Eisenstein series in this problem from its primitive versionSeparating each nonzero integer pair into a positive integer multiple of a primitive pair givesThe factor two accounts for the two signs of a primitive pair. It must not be omitted in the Rankin–Selberg unfolding.

Because $f\overline{g}{y}^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\le x<1$, $y>0$:The $x$ integral of the product Fourier expansions is ${\sum }_{n\ge 1}a(n)\overline{b(n)}{e}^{-4\pi ny}$. The Gamma function integral then givesThese steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and modular cusp decay show the square-coefficient series converges for $\mathrm{Re}s>k$; the Cauchy-Schwarz inequality gives the same absolute-convergence range for the mixed series.

Put $w=s-k+1$ and defineThis is the zeta-completed Rankin-Selberg coefficient series. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.

The exponential decay of cusp forms at infinity dominates the polynomial growth in $y$ of the Eisenstein series and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb{C}$, with possible simple poles only at $s=k-1,k$, and its functional equation isIndeed $w\mapsto 1-w$ corresponds to $s\mapsto 2k-1-s$. If $⟨f,g⟩=0$, both pole residues vanish and the completion is entire.

The normalization of the poles can also be made explicit. The constant term of the given Eisenstein series isThe second term has residue one at $w=1$, since the pole of $\zeta (2w-1)$ has residue $1/2$; the functional equation gives residue minus one at $w=0$. Hence the residues of $\mathcal{R}$ at $k$ and $k-1$ are respectively $⟨f,g⟩$ and $-⟨f,g⟩$, the Petersson inner product.

It is important to separate this completed statement from the raw $F$. Inverting the completion gives meromorphic continuation of $F$, but zeros of $\zeta (2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow 2k-1-s$. At the two displayed completion poles, for a nontrivial modular cusp modular weight,The latter is finite because the pole of $\Gamma (w)$ cancels that of the integral at $w=0$, using $\zeta (0)=-1/2$. In particular, taking $f=g\ne 0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal{R}$, not the raw series with all completion factors suppressed.

Finally suppose the two $L$-series have Euler products. With first coefficients normalized to one, their coefficients are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Thereforeinitially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized coefficients inside it.

For the usual degree-two Hecke products of Question 3, take roots ${\alpha }_p,{\beta }_p$ with sum $a(p)$ and product $p^{k-1}$, and ${\gamma }_p,{\delta }_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the Euler factor of a coefficientwise product of Hecke eigenforms:Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta (2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and use the determinant-normalized slash operatorThe automorphy factor identity proves $(f{|}_k\alpha ){|}_k\beta =f{|}_k(\alpha \beta )$, and positive scalar matrices act trivially. Let ${\mathcal{M}}_n$ be all integral matrices of determinant $n>0$. It is stable under left and right multiplication by $\Gamma$. Its left orbits have unique representativesIndeed Bézout's identity reduces the first column to $(a,0)$, the determinant fixes $d=n/a$, and a row shear reduces $b$ modulo $d$. These are the determinant-n matrix representatives for Hecke operators. For composite $n$, ${\mathcal{M}}_n$ can contain more than one double coset; keeping every integral matrix, including nonprimitive ones, is essential.

Define the Hecke operator byChanging a representative on the left leaves its summand unchanged. Right multiplication by $\Gamma$ permutes left orbits of ${\mathcal{M}}_n$, so the result satisfies the weight-$k$ transformation law. Each summand is holomorphic. Its expansion at infinity is bounded, and vanishes there when $f$ is a cusp form; the group has just one cusp class. Therefore $T_n$ preserves both $M_k$ and $S_k$, and $T_1$ is the identity.

For a Fourier expansion $f={\sum }_{r\ge 0}{a}_r{q}^r$, the sum over $b$ vanishes unless $d\mid r$. Substitute $r=dt$ and then collect the coefficient of $q^m$ to obtain the Fourier coefficients of a composite-index Hecke operator:Here $\gcd (0,n)=n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0$; at $m=1$ it is $a_n$. This proves preservation of integral coefficients. At a prime,with $a_{m/p}=0$ if $p\nmid m$.

We can prove the entire Hecke multiplication relations directly, not just quote them. On formal power series put $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. Although these individual operators need not preserve level one, $U_p{V}_p=1$ and the coefficient formula givesMultiply on the left by $T_p=U_p+p^{k-1}{V}_p$. In the first part the $j=0$ term is $U_p^{r+1}$; every $j\ge 1$ reduces by $U_p{V}_p=1$ to $p^{j(k-1)}{V}_p^{j-1}{U}_p^{r-j}$. The second part, together with the $j=0$ term, is $T_{p^{r+1}}$. The remaining sum is $p^{k-1}{T}_{p^{r-1}}$. ThusFor coprime indices, the coefficient divisors split independently in the displayed formula, proving $T_m{T}_n=T_{mn}$ when $(m,n)=1$. Induction with the prime-power recurrence gives $T_{p^r}{T}_{p^t}={\sum }_{j=0}^{\min (r,t)}{p}^{j(k-1)}{T}_{p^{r+t-2j}}$. Combining the primes provesIn particular the modular Hecke algebra is commutative and is generated by the prime operators.

For cusp forms the Petersson inner product isCusp decay makes it finite and positive definite. The Hecke operators are self-adjoint for the Petersson inner product. To see the change-of-variables mechanism, fix a double coset $\Gamma \alpha \Gamma \subset {\mathcal{M}}_n$ and put $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. Representatives $\gamma$ of $H\backslash \Gamma$ give left representatives $\alpha \gamma$. Unfolding the sum in $⟨{\sum }_{\gamma }f{|}_k\alpha \gamma ,g⟩$ changes the integral over $\Gamma \backslash \mathbb{H}$ to the integral over $H\backslash \mathbb{H}$ of $(f{|}_k\alpha )\overline{g}$. The substitution $w=\alpha \tau$ preserves $dxdy/y^2$ and has $\Im w=(\det \alpha )y/|c\tau +d{|}^2$; these factors move the slash from $f$ to $g{|}_k{\alpha }^{-1}$. Refold over $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$ to get the reverse double coset. Since scalar matrices act trivially, ${\alpha }^{-1}$ has the same slash action as its adjugate ${\alpha }^{\ast }=n{\alpha }^{-1}$, an integral matrix of determinant $n$. The adjugate involution permutes all the double cosets in ${\mathcal{M}}_n$. The common real factor $n^{k/2-1}$ is unchanged, giving $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$.

Since $S_k$ is finite dimensional, commuting self-adjoint linear operators have a common orthonormal basis of Hecke eigenforms: diagonalize one operator, restrict the others to its eigenspaces, and repeat until these spaces cannot split further. If $f$ is a nonzero common Hecke eigenform with $T_{n}f={\lambda }_{n}f$, the coefficient at $q$ gives $a_n={\lambda }_n{a}_1$. Were $a_1=0$, every positive coefficient would vanish, contrary to $f\ne 0$ being cuspidal. Normalize $a_1=1$; then ${\lambda }_n=a_n$, and the Hecke multiplication relations yieldThe eigenvalues are real by self-adjointness. Their multiplicativity and prime-power recurrence give the Euler product of a Hecke eigenform, initially in an absolutely convergent half-plane:The Mellin transform in Question 5 supplies its analytic continuation and functional equation.

Finally, for positive even $k\ge 4$, $M_k=\mathbb{C}{E}_k\oplus S_k$, by subtracting the constant coefficient times the normalized Eisenstein series. The divisor identityfollows by multiplying the geometric sums at each prime; it and the constant-coefficient formula prove $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Thus $M_k$ has an Eisenstein series eigenline and a basis of cuspidal Hecke eigenforms. Odd and negative weights have no forms, $M_2=0$, and on $M_0=\mathbb{C}$ the operator is the scalar ${\sigma }_{-1}(n)$. These statements fix the normalization and describe the exceptional small weights as well.

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.

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:

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

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

The Hecke multiplication relations follow from these coefficients. For coprime indices, the two divisor choices combine uniquely and give $T_m{T}_n=T_{mn}$. For prime powers put $A=p^{k-1}$, and, on formal Fourier series, let $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_p(\sum a_m{q}^m)=\sum a_m{q}^{pm}$. These individual maps need not preserve level one, but $U_p{V}_p=1$, and the coefficient formula givesMultiplying on the left by $T_p=U_p+A{V}_p$ and cancelling $U_p{V}_p$ yieldsInduction gives $T_{p^u}{T}_{p^v}={\sum }_{j=0}^{\min (u,v)}{p}^{j(k-1)}{T}_{p^{u+v-2j}}$; combining distinct prime factors provesThe modular Hecke algebra is therefore commutative and generated by its prime-index operators. The identity $U_p{V}_p=1$, rather than the generally false reversed identity, is what proves the prime-power recurrence.

On $S_k$ use the positive definite Petersson inner productExponential cusp decay makes this integral finite. The Hecke operators are self-adjoint for the Petersson inner product. Here is the correspondence argument. For a positive-determinant matrix $\alpha$, set $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. The finite sum over the double coset $\Gamma \alpha \Gamma$ unfolds its pairing intoChanging variables $z=\alpha \tau$ preserves hyperbolic area, and $\mathrm{Im}(\alpha \tau )=\det (\alpha )y/|c\tau +d{|}^2$ shows that the determinant-normalized slash factors transfer $\alpha$ from $f$ to ${\alpha }^{-1}$ on $g$. The new quotient is by $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$; folding it back gives the operator for the reversed double coset. For $\alpha \in {\mathcal{D}}_n$, its adjugate $n{\alpha }^{-1}$ again lies in ${\mathcal{D}}_n$. Positive scalar matrices act trivially in the determinant normalization, so ${\alpha }^{-1}$ and its adjugate give the same action. The adjugate involution preserves the entire determinant-$n$ correspondence, and the external normalization $n^{k/2-1}$ is real. Thus

Finite-dimensional spectral theory now gives a simultaneous orthonormal eigenbasis on $S_k$, since the operators commute and are self-adjoint. For a common Hecke eigenform $f$ with eigenvalues ${\lambda }_n$, the first-coefficient formula gives $a_n(f)={\lambda }_n{a}_1(f)$. If $a_1(f)=0$, every coefficient vanishes and $f=0$, a contradiction. Rescale each eigenform to have $a_1=1$. The resulting basis is orthogonal, andAll these eigenvalues and coefficients are real. A fixed simultaneous eigenspace is one-dimensional, because its normalized Fourier coefficients are uniquely determined by the eigenvalues.

Applying the multiplication relations to an eigenform givesThe recurrence yields ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$. Multiplicativity then gives the Euler product of a Hecke eigenforminitially in a right half-plane, for example $\mathrm{Re}s>k/2+1$ using the Fourier coefficient bound for a cusp form. Modular inversion and the Mellin transform of a cusp-form L-function continue $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)$ to an entire function and give $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

There is also an integral structure. If $d=\dim S_k$, choose $F_j={\Delta }^j{R}_{k-12j}$, $1\le j\le d$, with $R_w$ a monomial in $E_4,E_6$ of weight $w$ and constant term one. These forms have integral Fourier coefficients and start with $q^j$; their first $d$ coefficients form a unitriangular matrix. They are a basis, and every cusp form with integral coefficients is an integral combination of them, by successive elimination of its first $d$ coefficients. The coefficient formula shows that $T_n$ preserves this lattice. Its characteristic polynomial is consequently monic with integer coefficients, proving that the Hecke eigenvalues are algebraic integers. Together with self-adjointness, this gives real algebraic integers without requiring any unproved bound for their sizes.

For even $k\ge 4$, the full space decomposes as $M_k=\mathbb{C}{E}_k\oplus S_k$. The normalized Eisenstein series is a simultaneous eigenform withIts coefficients are $-2k{\sigma }_{k-1}(m)/B_k$, and the elementary divisor identity ${\sigma }_{k-1}(m){\sigma }_{k-1}(n)={\sum }_{d\mid \gcd (m,n)}{d}^{k-1}{\sigma }_{k-1}(mn/d^2)$ verifies the eigenvalue formula, including the constant coefficient. Thus the cusp eigenbasis plus $E_k$ diagonalizes the operators on $M_k$. Weight two and odd weights have no forms; in weight zero the constant function has eigenvalue ${\sigma }_{-1}(n)$. As a concrete cuspidal example, $S_{12}=\mathbb{C}\Delta$, so $T_n\Delta =\tau (n)\Delta$ for its Fourier coefficients, and the recurrence gives $\tau (p^2)=\tau (p{)}^2-p^{11}$.

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.