 Solution

Fix an even weight and use the determinant-normalized slash operator for modular formsIt is unchanged by positive scalar multiples of $A$ and satisfies $(f|A)|B=f|AB$. The determinant-n matrix representatives for Hecke operators areIndeed, integer row operations with determinant one send the first column of an integral determinant-$n$ matrix to $(a,0{)}^t$, with $a>0$; the second diagonal entry is $d=n/a>0$, and a row shear reduces $b$ modulo $d$. This gives unique representatives of the left modular-group orbits.

Define the Hecke operator byThe set of integral matrices of determinant $n$ is stable under right multiplication by the modular group. Right multiplication therefore permutes its left orbits, proving the modular transformation law for $T_{n}f$. Each summand is holomorphic in the upper half-plane. Its imaginary argument tends to infinity at the cusp, so the formula preserves cusp holomorphy and maps $M_k$ to itself and $S_k$ to itself.

Expand each summand in q. The finite sum over $b$ vanishes unless the Fourier index is divisible by $d$, when it equals $d$. The coefficient at $q^m$ then receives a contribution for each $a\mid \gcd (m,n)$, giving the Fourier coefficients of a composite-index Hecke operator formulaFor $m=0$ interpret the sum over all divisors of $n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$ and integral coefficients are preserved.

Here is a useful direct derivation of the Hecke multiplication relations. On formal Fourier series put $U_{p}f={\sum }_m{a}_{pm}{q}^m$ and $V_{p}f=f(q^p)$. They obey $U_p{V}_p=1$, although $V_p{U}_p$ is generally not the identity. The coefficient formula givesMultiplying by $T_p=U_p+p^{k-1}{V}_p$ and using $U_p{V}_p=1$ separates the terms with $j=0$ from the others and yieldsFor distinct primes, all the relevant shift operators commute. Thus $T_m{T}_n=T_{mn}$ when $\gcd (m,n)=1$, and each prime-power operator is a polynomial in its prime operator. This proves commutativity. Induction on the smaller of two prime exponents, using the displayed recurrence, gives $T_{p^r}{T}_{p^s}={\sum }_{j=0}^{\min (r,s)}{p}^{j(k-1)}{T}_{p^{r+s-2j}}$. Factoring general indices into primes then provesThe auxiliary shifts need not themselves preserve the level-one form space; they are used only to prove identities of coefficient sequences.

For cusp forms, the Petersson inner product isIt converges by exponential cusp decay, is positive-definite, and makes the Hecke operators self-adjoint. For the change-of-variable argument, the determinant-normalized slash action compensates exactly for the change in $y^k$, while hyperbolic area is invariant. Integrating a correspondence term in the opposite direction replaces its matrix by its inverse. The adjugate $A^{\#}=n{A}^{-1}$ is again an integral matrix of determinant $n$; it has the same slash action as $A^{-1}$, because scalar matrices act trivially. Summing over the two sets of representatives, equivalently over the common finite covering domains of the correspondence, interchanges $f$ and $g$ and proves $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. This explains Hecke operators are self-adjoint for the Petersson inner product with the present normalization.

Commuting self-adjoint operators on the finite-dimensional cusp space have a common orthonormal eigenbasis: diagonalize one, restrict the others to its invariant eigenspaces and repeat. If $f$ is a nonzero common Hecke eigenform, $a_n(f)=a_1(T_{n}f)={\lambda }_n{a}_1(f)$. The first coefficient cannot vanish, since then every coefficient would vanish. Normalize it to one. The eigenvalue is then ${\lambda }_n=a_n(f)$, and the common eigenspace is one-dimensional because all its normalized Fourier coefficients are fixed. The eigenvalues are real by self-adjointness and are algebraic integers, since the Hecke operators act by integer matrices on the integral cusp-form lattice from Question 1. The integral echelon basis and the operator identity there also show that $T_1,\dots ,T_{d-1}$ are an integral basis of the integral Hecke algebra of level-one cusp forms; independence follows by evaluating their first-coefficient functionals on $g_1,\dots ,g_{d-1}$.

The eigenform's coefficient relations areThey give the Euler product of a Hecke eigenformin a sufficiently far right half-plane. The Mellin transform of a cusp-form L-function completes this to $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)={\int }_0^{\infty }f(iy)y^{s-1}dy$. Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential decay makes this an entire function of $s$, and the formula proves $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

On the noncuspidal line in positive even weights $k\ge 4$, the normalized Eisenstein series has eigenvalues ${\sigma }_{k-1}(n)$, as follows from the same divisor formula for its coefficients; $M_k=\mathbb{C}{E}_k\oplus S_k$. As a cuspidal example, $S_{12}$ is spanned by the normalized modular discriminant. Its coefficients $\tau (n)$ are the Hecke eigenvalues, and the recurrence gives $\tau (4)=\tau (2{)}^2-2^{11}=-1472$ from $\tau (2)=-24$. Thus the Hecke theory converts analytic forms into multiplicative arithmetic data.