Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 26 (Modular Forms)

Write $\Gamma =S{L}_2(\mathbb{Z})$, $\mathbb{H}=\{\tau :\mathrm{Im}\tau >0\}$ and $q=e^{2\pi i\tau }$. The space $M_k$ consists of holomorphic functions on the complex upper half-plane satisfyingand holomorphic at the cusp: their Fourier expansion of a modular form is $f={\sum }_{n\ge 0}{a}_n{q}^n$. The subspace $S_k$ of cusp forms has $a_0=0$. Since all cusps are equivalent at level one, this condition treats every cusp. The element $-I$ forces odd-weight forms to vanish.

Use the permitted properties of the normalized Eisenstein series and modular discriminant:They have integer Fourier coefficients and weights $4,6,12$. For even $k\ge 0$, the assumed dimension formula gives $d=\lfloor k/12\rfloor +1$ except when $k\equiv 2(\mathrm{mod}12)$, when $d=\lfloor k/12\rfloor$. For $0\le j<d$, the remaining weight $k-12j$ is nonnegative and is not two. Choose $b_j\in \{0,1\}$ and $a_j\ge 0$ with $4a_j+6b_j=k-12j$. Every allowable remaining weight admits such a choice. The formsgive an integral triangular basis of level-one modular forms: their distinct first exponents prove independence, and their number $d$ proves that they span.

Now take $f$ with integral coefficients. Subtract its constant coefficient times $F_0$. The remainder still has integral coefficients and has no constant term. Subtract its coefficient at $q$ times $F_1$, then its current coefficient at $q^2$ times $F_2$, and continue. Every subtraction uses an integer. After $d$ steps the remainder's first $d$ coefficients vanish; triangularity and the basis property force that remainder to be zero. ConsequentlyThis proves integral polynomial generation without dividing by $1728$. The relation $E_4^3-E_6^2=1728\Delta$ explains why the additional integral generator is useful, although over $\mathbb{C}$ two generators suffice. For $k=0$ the basis is the constant one; when $M_k=0$, including weight two and odd weights, the claim is vacuous.

Start from the integral triangular basis of level-one modular forms. Proceed downwards in $j$. Set $g_{d-1}=F_{d-1}$; once $g_{j+1},\dots ,g_{d-1}$ are constructed, subtract from $F_j$ its coefficients at $q^{j+1},\dots ,q^{d-1}$ times these respective forms. The already constructed rows have a single nonzero coefficient among those initial positions, so this removes every unwanted initial coefficient without changing the coefficient one at $q^j$. All subtractions use integers. The resulting integral echelon basis of level-one modular forms therefore hasIts first $d$ coefficient vectors are the standard coordinate vectors, hence it is a basis. If another form had the same first $d$ coefficients as $g_j$, their difference would have those coefficients all zero; the triangular basis proves that difference vanishes. This proves uniqueness of every $g_j$, not merely uniqueness up to basis change.

For any $f\in M_k$, its expansion in this basis is $f={\sum }_{j=0}^{d-1}{a}_j(f)g_j$, since its first $d$ coefficients determine the basis coordinates. Comparing subsequent coefficients givesThe coefficients of $f$ need not be integral for this last identity. When needed at earlier indices, extend the notation by $c_n(j)={\delta }_{nj}$ for $0\le n<d$. This agrees with the actual coefficients of the basis forms and is the convention needed for the next part's all-index wording.

Use the standard normalization of the Hecke operator, for whichThe Hecke multiplication relations make these operators commute; a coefficient-based proof is included in 3(i). For a cusp form, $a_0(f)=0$, so the preceding coefficient relation saysfor every $f\in S_k$. Denote the bracketed operator by $D_n$. A vanishing first coefficient alone would not prove $D_n=0$. Instead, for every $m\ge 1$, commute it past $T_m$ and apply the same relation to the cusp form $T_{m}f$:Its constant coefficient is zero too. Thus every Fourier coefficient of $D_{n}f$ vanishes, and holomorphy gives $D_{n}f=0$. We have provedFor $n\ge d$ these are the integers constructed in (ii); for $1\le n<d$, the extension $c_n(j)={\delta }_{nj}$ makes the formula tautological. Without that implicit extension the symbols in the printed all-index assertion are undefined. The $j=0$ term is excluded because its coefficient functional is zero on cusp forms. If $S_k=0$, the operator statement is vacuous. This also exhibits the perfect integral Hecke pairing: the first $d-1$ coefficient functionals on the cusp space are dual to $T_1,\dots ,T_{d-1}$.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe summands on a compact set away from the lattice are $O(|\omega {|}^{-3})$, and that lattice sum converges. Thus the series and its termwise derivative converge normally. The function is even, and ${\wp }^{\prime }=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is odd and periodic by reindexing its absolutely convergent series. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evaluation at $-{\omega }_j/2$ and evenness make this constant zero for each basis period. The only poles of wp are double lattice poles, with principal part $(z-\omega {)}^{-2}$.

Let $G_r={\sum }_{\omega \ne 0}{\omega }^{-r}$ for the convergent weights used here. Expanding around zero givesThereforePut $g_2=60G_4$ and $g_3=140G_6$. The elliptic function $({\wp }^{\prime }{)}^2-4{\wp }^3+g_2\wp +g_3$ has a removable singularity at zero and hence at every lattice point. It is entire and periodic, thus bounded on the plane and constant by the Liouville theorem. Its zero constant term makes that constant zero. This proves the Weierstrass elliptic differential equationThe three classes ${\omega }_i/2$, with ${\omega }_3=-{\omega }_1-{\omega }_2$, are the three distinct nonzero two-torsion classes. Oddness and periodicity give ${\wp }^{\prime }({\omega }_i/2)=-{\wp }^{\prime }({\omega }_i/2)$, so they are zeros of the derivative. Their half-period values of the Weierstrass elliptic function $e_i$ are therefore roots of $4X^3-g_{2}X-g_3$.

They are distinct. If $e_i=e_j=e$ with $i\ne j$, the elliptic function $\wp -e$ would have a zero of multiplicity at least two at each of those distinct point classes, since its derivative vanishes there. It has only one pole on the torus, of order two. The argument principle on a fundamental cell equates its total zero and pole orders, contradicting at least four zeros. Thus $e_1,e_2,e_3$ are distinct, and the cubic factors asFor the monic cubic $X^3-(g_2/4)X-g_3/4$, the polynomial discriminant is $(g_2^3-27g_3^2)/16={\prod }_{i<j}(e_i-e_j{)}^2$. Hence the nonvanishing discriminant of a complex lattice isHere $\Delta (\Lambda )$ is the lattice discriminant, not the normalized q-series of Question 1. For $\Lambda =\mathbb{Z}+\mathbb{Z}\tau$, the normalizations give $g_2=(4{\pi }^4/3)E_4$ and $g_3=(8{\pi }^6/27)E_6$, so $\Delta (\Lambda )=(2\pi {)}^{12}\Delta (\tau )$.

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.

State the characteristic convention explicitly: for real $\alpha ,\beta$, define the theta function with characteristics byBinary subscripts ${\vartheta }_{ab}$ often denote this function with $(\alpha ,\beta )=(a/2,b/2)$, $a,b\in \{0,1\}$. On compact subsets of the $(z,\tau )$ domain, a summand and any derivative are bounded by a polynomial in $|n|$ times $e^{-c{n}^2+C|n|}$. Normal convergence therefore gives an entire function of $z$, holomorphic dependence on $\tau$, and legitimate termwise differentiation. In particular the heat equation for theta functions with characteristics isCharacteristics are defined modulo integers with a phase: $\theta [\alpha +1,\beta ]=\theta [\alpha ,\beta ]$ and $\theta [\alpha ,\beta +1]=e^{2\pi i\alpha }\theta [\alpha ,\beta ]$. Shifting the summation index proves the spatial quasi-periodsThese multipliers describe a degree-one line bundle on the complex torus with periods $1,\tau$, rather than an ordinary elliptic function. The logarithmic derivative is unchanged under one and decreases by $2\pi i$ under tau. Pairing opposite sides of a fundamental cell therefore makes its contour integral $2\pi i$. The argument principle counts exactly one zero in a cell, with multiplicity.

The characteristic function is a translate of the unshifted theta function:At $z=(1+\tau )/2$, the unshifted series cancels in pairs $n,-n-1$. Hence the zeros of theta functions with characteristics occur exactly and simply atReindexing also gives $\theta [\alpha ,\beta ](-z)=\theta [-\alpha ,-\beta ](z)$. For half-integer characteristics the parity is $e^{4\pi i\alpha \beta }$, so only $(1/2,1/2)$ is odd and vanishes at zero; the other three theta constants are nonzero.

The modular transformations of theta characteristics are obtained from Gaussian Poisson summation and elementary phase rearrangement. With the holomorphic branch of $\sqrt{-i\tau }$ positive at $\tau =i$,For the first formula, apply Poisson summation to $F(t)=e^{\pi i\tau (t+\alpha {)}^2+2\pi i(t+\alpha )(z+\beta )}$. Completing the square in its Fourier transform givesRearranging ${\sum }_{n}F(n)={\sum }_m\hat{F}(m)$ proves the stated inversion law. For translation by one, use $e^{\pi i{n}^2}=e^{\pi in}$ in the defining series; the remaining characteristic phase is precisely the displayed factor.

Write $Q=e^{\pi i\tau }$, so the modular-form variable of Question 1 is $q=Q^2$, and put ${\Theta }_2=\theta [1/2,0](0,\tau )$, ${\Theta }_3=\theta [0,0](0,\tau )$ and ${\Theta }_4=\theta [0,1/2](0,\tau )$. Inversion interchanges ${\Theta }_2,{\Theta }_4$ and fixes ${\Theta }_3$, each with factor $\sqrt{-i\tau }$. Translation interchanges ${\Theta }_3,{\Theta }_4$ and multiplies ${\Theta }_2$ by $e^{\pi i/4}$. The Jacobi triple product givesIt also gives the conventional odd theta function ${\theta }_1=-\theta [1/2,1/2]$ its product $2Q^{1/4}\sin (\pi z){\prod }_{m\ge 1}(1-Q^{2m})(1-2Q^{2m}\cos (2\pi z)+Q^{4m})$. Differentiate at zero and multiply the three even products to obtain the Jacobi derivative formulaIn the constant-product simplification, ${\prod }_m(1+Q^{2m})(1-Q^{4m-2})=1$, by separating even and odd positive exponents. The products prove nonvanishing of all three even constants as well.

The Jacobi abstruse identity can be proved directly, without assuming it from the products. Splitting pairs of summation indices according to whether their sum and difference are even or odd givesFor example, $u=(n+m)/2,v=(n-m)/2$ are either both integers or both half-integers, and $n^2+m^2=2u^2+2v^2$. The alternating signs give the second identity; shifting both original indices by one half gives the third. Squaring the first two and subtracting then proves ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$.

These identities connect theta theory to modular forms and elliptic curves. The fourth powers are weight-two forms on the principal level-two congruence subgroup. The modular lambda function $\lambda ={\Theta }_2^4/{\Theta }_3^4$ obeys $\lambda (-1/\tau )=1-\lambda (\tau )$ and $\lambda (\tau +1)=\lambda (\tau )/(\lambda (\tau )-1)$; the characteristic permutation kernel is $\Gamma (2)$. It parametrizes an elliptic curve with its ordered two-torsion in Legendre form $y^2=x(x-1)(x-\lambda )$. The corresponding Klein j-invariant is $256(1-\lambda +{\lambda }^2{)}^3/[{\lambda }^2(1-\lambda {)}^2]$. Finally the same products giveIndeed ${\Theta }_2{\Theta }_3{\Theta }_4=2Q^{1/4}{\prod }_m(1-Q^{2m}{)}^3$, whose eighth power divided by $256$ is $q{\prod }_m(1-q^m{)}^{24}$. This recovers the normalized modular discriminant of Question 1 and displays the bridge between Gaussian theta sums and the lattice discriminant of Question 2.