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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.

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.

The pairing $(T,f)\mapsto a_1(Tf)$ identifies the integral Hecke algebra of level-one cusp forms with the integral dual of the cusp-form lattice. An integral basis beginning $q,q^2,\dots ,q^m$ gives a unitriangular first-$m$ coefficient matrix, so $a_1,\dots ,a_m$ form a dual basis. Commutativity of the Hecke operators proves nondegeneracy on the algebra, and $T_1,\dots ,T_m$ then form its integral basis.