Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 (Modular Forms)

Use the determinant-normalized slash operator and the usual Hecke operator normalization. A normalized eigenform is a nonzero cusp form $f={\sum }_{r\ge 1}{a}_r{q}^r$ with $a_1=1$ and $T_{n}f=\lambda (n)f$ for every $n$. We first establish the operator facts needed here, rather than assume the Hecke multiplication relations.

Integer matrices of determinant $n$ have the determinant-n matrix representatives for Hecke operators ${\gamma }_{a,b,d}=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $ad=n$ and $0\le b<d$. Integer row operations give these representatives: the Euclidean algorithm reduces the first column to $(a,0)$, and a shear reduces $b$ modulo $d$. Right multiplication by the modular group permutes the left cosets, so $T_{n}f=n^{k/2-1}{\sum }_{ad=n}{\sum }_{b\mathrm{mod}d}f{|}_k{\gamma }_{a,b,d}$ is modular. Rational changes of coordinate preserve cusp vanishing, so it remains a cusp form. Substituting its Fourier expansion of a modular form and summing $e^{2\pi irb/d}$ givesIn particular $a_1(T_{n}f)=a_n(f)$, and the normalization gives $\lambda (n)=a_n(f)$ for every $n$.

For coprime $u,v$, applying this coefficient formula twice amounts to choosing independently a divisor of $\gcd (m,u)$ and one of $\gcd (m,v)$. Their product runs exactly once through the divisors of $\gcd (m,uv)$, with the same factor $d^{k-1}$ and coefficient index. Thus $T_u{T}_v=T_{uv}$. For a prime $p$, put $P=p^{k-1}$ and $\nu =v_p(m)$. The coefficient of $q^m$ in $T_{p^e}f$ isThe coefficient in $T_p{T}_{p^e}f$ is $A_e(pm)+P{A}_e(m/p)$, the second summand being zero if $p\nmid m$. Expanding the two finite sums and matching their endpoints gives $A_{e+1}(m)+P{A}_{e-1}(m)$ for $e\ge 1$. Equality of all coefficients proves $T_p{T}_{p^e}=T_{p^{e+1}}+p^{k-1}{T}_{p^{e-1}}$. Applying these identities to the nonzero eigenvector $f$ yields

For a prime number $p$, the defining finite sum for the Hecke operator simplifies toThe first summand has no $q^1$ term. In the second, summing the $p$th roots of unity selects precisely the terms whose index is divisible by $p$, and its $q^1$ coefficient is $a_p$. Comparing with $T_{p}f=\lambda (p)f$, using $a_1=1$, gives $\boxed{\lambda (p)=a_p}$. This also follows from the general coefficient computation in part (a).

First the eigenvalues of Hecke operators are real. Here is the needed Petersson inner product argument: unfold the finite coset sum defining $T_n$ into its correspondence domain, then change variables by each matrix. The determinant-normalized slash action preserves $y^{k}dxdy/y^2$ in the paired integrand, and transfers a representative to its inverse. Multiplying that inverse by the positive scalar $n$ gives its adjugate without changing the slash action. Adjugation reverses the integer determinant-$n$ correspondence, so the resulting sum is again $T_n$. Consequently $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. Taking $g=f$ proves reality of $\lambda (n)$, since $⟨f,f⟩>0$.

Fix one prime $p$ and set $x=\lambda (p)/p^{(k-1)/2}$ and $b_e=\lambda (p^e)/p^{e(k-1)/2}$. The Hecke multiplication relations proved in (a) give $b_0=1$, $b_1=x$ and $b_{e+2}=x{b}_{e+1}-b_e$. The distinct-root assumption excludes $x=\pm 2$.

If $|x|<2$, write $x=2\cos \theta$, $0<\theta <\pi$. Solving the linear recurrence gives $b_e=\sin ((e+1)\theta )/\sin \theta$. If $\theta /(2\pi )$ is rational, infinitely many $e$ satisfy $(e+1)\theta \equiv \theta (\mathrm{mod}2\pi )$, hence $b_e=1$. If it is irrational, density of the corresponding irrational rotation gives infinitely many $e$ in the open arc where $\sin ((e+1)\theta )>\sin \theta /2$.

If $|x|>2$, let $\beta ,{\beta }^{-1}$ be the real roots of $X^2-xX+1$, choosing $|\beta |>1$. Then $b_e=({\beta }^{e+1}-{\beta }^{-e-1})/(\beta -{\beta }^{-1})$. Its even-index terms are positive and tend to infinity, whether $\beta$ is positive or negative. Thus the prime-power coefficients of a level-one eigenform are often large and positive, and in either case

A weight-$2n$ meromorphic modular form satisfies $f(\gamma z)=(cz+d{)}^{2n}f(z)$, while $d(\gamma z)=(cz+d{)}^{-2}dz$. The factors cancel in $f(z)(dz{)}^n$, giving a $\Gamma$-invariant meromorphic n-differential on the complex upper half-plane.

Away from elliptic points and cusps, descend this differential through local inverse branches of the quotient map. At an elliptic point of effective stabilizer order $e$, choose a coordinate $t$ in which the stabilizer acts by $t\mapsto \zeta t$, and use $w=t^e$ downstairs. If the lifted coefficient has order $r$, invariance forces $r+n$ to be divisible by $e$. The descended differential has integer orderIndeed $F(w)(dw{)}^n$ pulls back to $e^n{t}^{n(e-1)}F(t^e)(dt{)}^n$. At a modular cusp of width $h$, the coordinate $q_c=e^{2\pi iz/h}$ gives $dz=(h/(2\pi i))d{q}_c/q_c$. ThereforeBoth formulas provide meromorphic extensions over the omitted points, producing the unique differential with the prescribed pullback on the compactified modular curve.

For the character-valued question, write $X=X_0(N)$ and let $f$ be the chosen nonzero meromorphic form. The quotient of two forms with the same weight and Dirichlet character is a meromorphic function on $X$. At an interior point $P$ with effective elliptic order $e_P$ (one at ordinary points), let $r_P$ be the vanishing order of the lifted $f$. At a cusp let $v_c$ be its order in the width coordinate; this can be fractional because of the character multiplier. Multiplying by a meromorphic function $\phi$ changes these orders by $e_P{\mathrm{ord}}_P\phi$ and ${\mathrm{ord}}_c\phi$, respectively. Interior holomorphy requires a nonnegative order, while cusp vanishing requires a strictly positive order. The exact divisor is consequentlyOnly finitely many coefficients are nonzero. The order inequalities are equivalent to $\mathrm{div}(\phi )+D(f)\ge 0$. Division by $f$ proves the reverse inclusion, so the Riemann-Roch space givesThis extends the cusp-form divisor presentation to elliptic points and character multipliers.

Choose a nonzero meromorphic one-differential on $X$, and let $g$ be its weight-two pullback. For $f_t=f{g}^t$, the interior order is $r_P(f)+t{r}_P(g)$ and the cusp order is $v_c(f)+t{v}_c(g)$. The elliptic orders are one, two or three, so adding six to $t$ changes each interior floor by $6r_P(g)/e_P$, and each cusp ceiling by $6v_c(g)$. The valence formula for the modular group for $g$ says ${\sum }_P{r}_P(g)/e_P+{\sum }_c{v}_c(g)=d/6$, where $d=[S{L}_2(\mathbb{Z}):{\Gamma }_0(N)]$. ThusThe degrees tend to infinity. For sufficiently large $t$, the Riemann-Roch theorem therefore givesThe periodic formula obtained this way is an eventual formula. For small $t$, the exact answer includes the Riemann-Roch correction $\ell (K_X-D(f_t))$; it should not silently be discarded without the large-degree hypothesis.

Let $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the modular group, left multiplication by powers of $T$ and by $S$ performs subtraction and exchange on its first-column entries. The Euclidean algorithm reduces $(a,c)$ to $(\pm 1,0)$, since the determinant condition makes them coprime. The resulting matrix is $\pm T^m$, and $-I=S^2$ is already generated. Hence $S,T$ generate $S{L}_2(\mathbb{Z})$.

The standard fundamental domain of the modular group is $D=\{z\in \mathbb{H}:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, with the vertical sides identified by $T$ and the circular sides by $S$. To prove every orbit meets it, choose a coprime integer pair $(c,d)$ minimizing $|cz+d|$. A minimum exists because only finitely many pairs have this quantity below any fixed bound: $|c|\mathrm{Im}z\le |cz+d|$ bounds $c$, then $d$. Complete the pair to a determinant-one matrix. Its image has maximal imaginary part in the orbit, since $\mathrm{Im}(\gamma z)=\mathrm{Im}z/|cz+d{|}^2$. Translate its real part into $[-1/2,1/2]$. If its modulus were below one, inversion would increase its imaginary part, a contradiction. This is the reduction to the standard modular region.

The normalized Eisenstein series satisfy $E_4=1+240{\sum }_{m\ge 1}{\sigma }_3(m)q^m$ and $E_6=1-504{\sum }_{m\ge 1}{\sigma }_5(m)q^m$. Their weights are four and six. The valence formula for the modular group makes the normalized weight-twelve cusp form $\Delta$ have one simple cusp zero and no zero in $\mathbb{H}$. Comparing its first coefficient gives $\Delta =(E_4^3-E_6^2)/1728$.

Odd weights vanish because $-I$ acts by $(-1{)}^k$. Weight two vanishes by the valence formula for the modular group: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge 0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^a{E}_6^b$. The remainder is a cusp form; division by $\Delta$ produces a holomorphic modular form of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight provesThe integral triangular basis of level-one modular forms is $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, with $4a_j+6b_j=k-12j\ge 0$ and $k-12j\ne 2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f={\sum }_j{c}_j{F}_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor (k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore all Fourier coefficients of $f$ are integers.

Set $C=AB=\left(\begin{array}{cc}0& -1\\ 1& 2\lambda \end{array}\right)$. It fixes $z=-\lambda +i\sqrt{1-{\lambda }^2}$ because $z^2+2\lambda z+1=0$. Write $\lambda =\cos \theta$, $0<\theta <\pi /2$; the eigenvalues of $C$ are $e^{\pm i\theta }$.

The stabilizer of an interior point of $\mathbb{H}$ in $PS{L}_2(\mathbb{R})$ is conjugate to the compact rotation group. Its intersection with a discrete subgroup is finite: otherwise compactness gives accumulating elements, and their quotients accumulate at the identity. Thus $C$ has finite order in the projective group. Its eigenvalues then satisfy $e^{2im\theta }=1$ for some positive integer $m$, forcing $\theta /\pi \in \mathbb{Q}$. Hence $\boxed{\lambda =\cos (\pi y)\text{ for some }y\in \mathbb{Q}}$. This is a necessary condition only.

Use the normalized Fricke involution $W_N=f\mapsto f{|}_k\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$, and let $D_{\ell }=\left(\begin{array}{cc}\ell & 0\\ 0& 1\end{array}\right)$. Choose integers $m,n$ with $m\ell -Nn=1$. The matrix $\gamma =\left(\begin{array}{cc}1& n\\ N& m\ell \end{array}\right)$ lies in ${\Gamma }_1(N)$, and ${\gamma }^{\prime }=\left(\begin{array}{cc}\ell & n\\ N& m\end{array}\right)$ has determinant one and lies in ${\Gamma }_0(N)$, with $m\equiv {\ell }^{-1}(\mathrm{mod}N)$. The displayed identity in the PDF reads $D_{\ell }{\gamma }^{-1}=\mathrm{diag}(1,\ell ){\gamma }^{\prime }$. Taking double cosets therefore identifies the reversed correspondence with the diamond operator $⟨{\ell }^{-1}⟩$ times the original one. Conjugation by $W_N$ reverses the two diagonal entries, soThe same finite-correspondence change of variables in the Petersson inner product transfers a matrix to its inverse and yields $T_{\ell }^{\ast }=⟨{\ell }^{-1}⟩T_{\ell }$. The diamond operators commute with $T_{\ell }$, so $T_{\ell }{T}_{\ell }^{\ast }=T_{\ell }^{\ast }{T}_{\ell }$. Thus $T_{\ell }$ is a normal operator. Multiplying $W_N$ by a nonzero scalar does not change the conjugation formula.

The commuting normal operators $T_{\ell }$, together with the finite-order diamond operators, admit simultaneous diagonalization for the Petersson inner product. Their restrictions to the new subspace of cusp forms have one-dimensional common eigenspaces by the newform multiplicity-one theorem. This additional theorem is the essential newform input: normality alone would allow higher multiplicities. Since every $U_p$ commutes with the good-prime operators and preserves the new space, it preserves each such line and is scalar on it. Normalize its nonzero first Fourier coefficient to one to obtain a newform. This gives the requested simultaneous eigenbasis of the new space.

By definition, every old form is a linear combination of lower-level oldforms by argument dilation. Repeatedly split each lower-level cusp space into its old space and its Petersson inner product orthogonal new space; induction on the level shows that the specified dilations of newforms span the full space. To prove independence, separate distinct primitive eigenforms by their good-prime eigenvalues, using the newform multiplicity-one theorem. For a fixed normalized form, its dilations $f(tz)$ are independent: order the distinct positive integers $t$ increasingly, and the coefficient of the least occurring $q^t$ kills its coefficient in any vanishing linear combination. Repeat for the next least $t$. Thus these dilations are indeed a basis.

For trivial character, part (a) shows that $W_N$ commutes with all the good-prime operators, so it preserves the one-dimensional primitive line. Also $W_N^2=(-1{)}^k$ for the determinant-normalized slash operator; a nonzero trivial-character form has even $k$. ThereforeThe power of $N$ in an unnormalized Fricke eigenvalue depends on the convention. The plain slash action $(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{-k/2}$, while the Hecke-style action $(\det \gamma {)}^{k-1}(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{k/2-1}$. The PDF prints $\pm N^{k-1}$; that value requires the deliberate rescaling $w_N=N^{k-1}{W}_N$ and does not follow under either usual convention. Under that rescaling its displayed eigenvalue is $\epsilon N^{k-1}$; all the preceding eigenspace and conjugation statements are unchanged.

Let $f_j(z)=f(p^{j}z)$ for $0\le j\le r$, and write $a=a_p(f)$ and $W_{M}f=\epsilon f$. The Hecke operator $U_p$ acts on the Fourier expansion by $\sum a_n{q}^n\mapsto \sum a_{pn}{q}^n$. Since $p\mid M$, the primitive relation is $a_{pn}=a_p{a}_n$. Hence $U_p{f}_0=a{f}_0$ and $U_p{f}_j=f_{j-1}$ for $j\ge 1$. Substituting the argument directly in the Fricke involution givesThus, in the basis $e_j=p^{jk/2}{f}_j$, put $b=p^{k/2}$ and let $R$ be the reversal matrix, $R{e}_j=e_{r-j}$. The matrices areIf the exam's rescaling $w_N=N^{k-1}{W}_N$ is used, multiply the second matrix by $N^{k-1}$.

Here is an explicit irreducibility argument for the prime-power old block at a bad prime. Any nonzero common invariant complex subspace $V$ contains an eigenvector of the restricted $U_p$. Its eigenvalue is either $a$ or zero. If $a\ne 0$, the $a$-eigenspace is $\mathbb{C}{e}_0$; containing it gives $e_r$ by reversal, then every $e_j$ by successive powers of $U_p$. The kernel of $U_p$ is $\mathbb{C}(e_1-(b/a)e_0)$. If $V$ instead contains this vector, reversal and $U_p^r$ giveThis is nonzero: the bad-prime eigenvalue of a trivial-character newform satisfies $a^2=p^{k-2}$ when $p\parallel M$, and $a=0$ when $p^2\mid M$, whereas $b^2=p^k$. If $a=0$, the kernel is already $\mathbb{C}{e}_0$, and the same reversal-and-shift argument works. Therefore the block has no nonzero proper subspace invariant under $U_p$ and the Fricke operator, and hence none invariant under the full ring in the question.

For cusps $\alpha ,\beta \in {\mathbb{P}}^1(\mathbb{Q})$, the modular symbol $\{\alpha ,\beta \}$ is the oriented path class in $H_1(X(\Gamma ),C;\mathbb{Z})$, where $C$ is the finite cusp set. Reversing paths changes the sign, concatenating paths gives $\{\alpha ,\beta \}+\{\beta ,\gamma \}=\{\alpha ,\gamma \}$, and the group identifies translated paths. A continued fraction expansion subdivides any rational-endpoint path into determinant-one edges. Representatives of the finitely many cosets therefore supply Manin symbols $[g]=\{g0,g\infty \}$. With inversion $S$ and translation $T$, reversing an edge and taking the boundary of an ideal triangle giveThese relations give a finite rational linear algebra presentation. Integral computation additionally removes the elliptic torsion in this naive presentation; the actual relative homology group of the compact surface is torsion-free.

The boundary of $\{\alpha ,\beta \}$ is $[\beta ]-[\alpha ]$ in $H_0(C)$. Its kernel is $H_1(X;\mathbb{Q})$ by the long exact sequence of a pair. A Hecke operator acts by summing the images of each path under its finite determinant-$n$ correspondence. Rational endpoints are again rational, so continued fractions reduce each image back to Manin symbols. Finite row reduction produces rational matrices for the relative symbols and for the boundary kernel. Integration of a weight-two cusp form against these absolute cycles is compatible with the Hecke action. The resulting decompositionexplains the relation between these matrices and cusp forms. Strictly, a rational matrix on absolute homology is not automatically a rational matrix on the holomorphic summand in an arbitrary complex basis. For subgroups preserved by real conjugation, the conjugation eigenspaces provide rational models of the cusp-form representation. More generally one should specify a compatible rational model or use the doubled homological representation. This distinction matters in interpreting the essay's rationality claim for an arbitrary subgroup; the algorithm on homology is always rational.

For the supplied subgroup, use $P=(1736)(254)$ for translation and $Q=(17)(26)(34)$ for inversion, following the PDF's naming. Translation has two cycles, so there are two cusps of widths four and three. Inversion fixes just coset five, giving one elliptic orbit of order two. The order-three product acts with cycles $(1)(253)(467)$ in one of the two equivalent composition conventions, giving one elliptic orbit of order three. Since $Q^2=1$, the central $-I$ acts trivially and the effective index is seven. The genus formula for a modular curve givesThe long exact sequence of a pair now givesOne can see this directly from the symbols. Over $\mathbb{Q}$, $Q$ gives $m_7=-m_1$, $m_6=-m_2$, $m_4=-m_3$ and $m_5=0$. The fixed order-three coset gives $m_1=0$, and the other triangle relations give $m_3=-m_2$. Thus $(m_1,\dots ,m_7)=(0,t,-t,t,0,-t,0)$. After removing elliptic torsion the integral group is free of rank one, agreeing with the surface calculation. Since $g=0$, the absolute boundary kernel and $S_2(\Gamma )$ both vanish: the cusp-form Hecke matrices for this particular example are the empty $0\times 0$ matrices.