Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 126 2 Solution Created 2026-10-03 Updated 2026-10-14
A weight- modular form for is a holomorphic function on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion , , near infinity. A cusp form has . The element makes a nonzero weight odd form impossible.
For weight zero, descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the expansion extends it over . The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve . The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.
The standard modular fundamental domain with paired vertical boundaries, paired circular arcs, and the added cusp at infinity
. Details. For positive even , define the unnormalized Eisenstein seriesThe real-linear map is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to . Thus the sum converges absolutely and locally uniformly for , proving holomorphy. For ,The map is a bijection on the nonzero integer pairs, giving .
For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of and its geometric-series expansion , valid for . It gives, for even ,The part of is , where is the Riemann zeta function; positive and negative contribute equally. Apply the kernel at for , and regroup the absolutely convergent sum according to . With the divisor sum , this provesThe expansion has no negative powers and converges for , so it also proves holomorphy at a cusp and completes the proof of modularity.
The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth and are modular forms of weight twelve. Their constant terms cancel, so their difference divided by is a cusp form. Integrality requires an argument: dividing an integral series by alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer , is divisible by twelve. Modulo three, ; modulo four, an even makes both powers divisible by four, while an odd has . Thus each coefficient is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 137 3 a Solution Created 2026-10-03 Updated 2026-10-05
The displayed definition uses iterated Eisenstein summation in weight two: the sum in is evaluated before the sum in , with the single term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.
For noninteger , the cosecant partial-fraction identity isFor completeness, apply the residue theorem to on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is . Its residues at the integers are and its residue at is , proving the formula. If , the geometric-series expression , differentiated termwise, gives the cotangent partial-fraction Fourier kernel
Put with in the complex upper half-plane. For positive , this gives . Negative gives the same value, by replacing with in its inner sum. The row is , by the Basel problem. The resulting series in does have absolute convergence, locally uniformly in , so collecting the coefficient at is legitimate:The coefficient is the sum-of-divisors function, since runs over the positive divisors of . ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 137 2 iii Solution Created 2026-10-03 Updated 2026-10-05
The original PDF has the summation condition ; the TeX's is a transcription error. There is also an actual missing hypothesis in the PDF's coefficient formula: that simplified formula requires a primitive Dirichlet character. We first derive a formula valid for every character, and then show both the primitive specialization and a counterexample to the unrestricted version.
For , the character-twisted Eisenstein series converges absolutely and locally uniformly on the complex upper half-plane. On a compact subset, is bounded below by a positive constant times , and the corresponding two-dimensional lattice sum converges. Changing to provesThe condition makes the terms for and equal. The terms with contribute , and all other terms are twice the sum over .
For , the cotangent identity andgive, after differentiations,Differentiation is justified by locally uniform convergence. This is the cotangent partial-fraction Fourier kernel.
Write , with running through the unit classes modulo , and define the finite Fourier transformHere is the Gauss sum of a Dirichlet character.
Applying the kernel with yieldsThe double series converges absolutely: and the exponential decay controls . Grouping the terms with proves the general Fourier expansion of a character-twisted Eisenstein series:
Applying the kernel with yieldsThe double series converges absolutely: and the exponential decay controls . Grouping the terms with proves the general Fourier expansion of a character-twisted Eisenstein series:
If is a unit modulo , substitution gives . Suppose now that is primitive and is a nonunit. Choose a prime . Reduction of units modulo onto units modulo is surjective. Primitivity means that is nontrivial on its kernel, so there is a unit with . Since , substitution by forces and thus . This proves the finite Fourier transform of a primitive Dirichlet character identity, and consequentlyThe inverse-character notation in the question is understood to mean , extended by zero on nonunits; literal inversion of would be undefined.
For a counterexample without primitivity, take , the principal character, and . Then , so but . At , the general formula gives , whereas the printed simplified formula, interpreted as zero on nonunits, gives . Hence the printed coefficient formula is false for arbitrary imprimitive characters; the general boxed formula above supplies the correction.
