 Solution

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.