Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 4

Write $w=2k$. For a positive-determinant real matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, extend the slash operator for modular forms byLet ${\mathcal{M}}_n$ be the integral matrices of determinant $n>0$. Each left $S{L}_2(\mathbb{Z})$ orbit has a unique representative $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $ad=n$, $a,d>0$, $0\le b<d$. To obtain it, the Euclidean algorithm on the first column produces $(a,0)$ with $a>0$; the determinant fixes $d$, and a row shear reduces $b$ modulo $d$. This also proves uniqueness.

Define the normalized Hecke operatorRight multiplication by $S{L}_2(\mathbb{Z})$ permutes these left orbits, and the slash operation is associative. Hence the sum satisfies the same modular form transformation law. It is holomorphic on $\mathbb{H}$ since each argument stays there.

For $f={\sum }_{r\ge 0}c(r)q^r$, summing the phase $e^{2\pi irb/d}$ over $b$ gives zero unless $d\mid r$, and gives $d$ when $d\mid r$. For the resulting term to contribute to $q^m$, one needs $a\mid m$ and $r=dm/a=nm/a^2$. Its scalar factor is $n^{w-1}{d}^{1-w}=a^{w-1}$. Thuswhere $\gcd (n,0)=n$. All powers are nonnegative, so the sum is holomorphic at infinity and belongs to $M_w$; level one has only this modular cusp class. Its constant coefficient is ${\sigma }_{w-1}(n)c(0)$, so modular cusp forms are preserved. This is the full Fourier coefficients of a composite-index Hecke operator formula, including the constant term.

An eigenform is understood to be nonzero; the zero form would put no constraint on the alleged eigenvalues. The coefficient of $q$ in $T_w(n)f$ is $c(n)$, since $\gcd (n,1)=1$. Comparing with the coefficient in $\lambda (n)f$ givesThe constant coefficient comparison gives $\lambda (n)c(0)={\sigma }_{w-1}(n)c(0)$, so

For completeness derive the operator multiplication law underlying the last relation. On formal Fourier series defineThese auxiliary maps need not separately preserve modularity, but coefficient identities are valid and $U_p{V}_p=I$. Part (a) givesMultiplication on the left by $U_p+A{V}_p$ yields the $T_w(p^{e+1})$ sum and, from $U_p{V}_p^j=V_p^{j-1}$ for $j\ge 1$, the remaining sum $A{T}_w(p^{e-1})$. HenceInduction on the smaller exponent, applying this recurrence and cancelling its two overlapping sums, now givesTo make the induction explicit, write $T_r=T_w(p^r)$ and assume $2\le r\le s$. Substituting the formulas already established for $r-1$ and $r-2$ into $T_r{T}_s=(T_1{T}_{r-1}-A{T}_{r-2})T_s$ givesThe last term is exactly the missing $j=r$ term. The cases $r=0,1$ are the identity operator and the prime recurrence above.

For distinct primes, their $U$ and $V$ maps commute, and the divisor formula in (a) factors independently at each prime. It follows that $T_w(m)T_w(n)=T_w(mn)$ for coprime $m,n$. Multiplying the prime-power identities therefore gives the general Hecke multiplication relationsApply this identity to the same nonzero eigenform and cancel the vector $f$, obtainingThis reasoning does not silently divide by $c(1)$, which is important when considering noncuspidal weight-zero forms.

For a nonzero Hecke eigenform that is a cusp form, $c(1)\ne 0$: otherwise part (b) would make all its positive Fourier coefficients zero, and cuspidality already gives $c(0)=0$, forcing $f=0$. Since $T_w(1)=I$, $\lambda (1)=1$. Part (b) gives multiplicativity at coprime integers and the prime-power recurrenceMultiply its generating series by the corresponding quadratic to obtainUnique prime factorization and multiplicativity giveFor a normalized eigenform $c(1)=1$, the prefactor disappears. It must be retained for an arbitrary scalar multiple of that eigenform.

This product identity holds analytically in an absolute-convergence half-plane. The invariant norm of a positive-weight cusp form is bounded, as proved in Question 6. For a period-one Fourier expansion of a modular form,Choosing $y=1/n$ gives $c(n)=O(n^{w/2})$, so $\mathrm{Re}s>w/2+1$ suffices. Absolute convergence then justifies expansion and rearrangement of the Euler product of a Hecke eigenform; no sharper coefficient estimate is needed here.