Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 24 / 2 / i

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 $\tau (p^2)=\tau (p{)}^2-p^{11}$.