MODULAR FORMS

Problem sheet 2 KMB, Michaelmas 1997

I keep forgetting to say: you can get copies of these problem sheets on the web, they’re

at http://www.dpmms.cam.ac.uk/˜ buzzard/maths/modforms.html, all being well.

1. This question is a thorough examination of the modular curve asociated to Γ_{2}.

(a) A fundamental domain F for a congruence subgroup Γ is connected subset of ℌ equal to the closure of its interior, such that any element of ℌ is Γ-equivalent to a point in F, and no two points in the interior of F are Γ-equivalent. Show that each of the numbered regions below is a fundamental domain for SL_{2}(ℤ), and that their union is a fundamental domain for Γ_{2}.

(b) What are the cusps and the elliptic points for Γ_{2}?

(c) What is the genus of X(Γ_{2})?

2. This is a similar sort of question for the congruence subgroup

Γ^{0}(11) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ)|b ≡ 0 mod 11 \} .

(a) Show that the region below is a fundamental domain for Γ^{0}(11). (Hint: show that the

matrices ( \begin{matrix}1 & r \\ 0 & 1 \end{matrix} ) for 0 ≤ r ≤ 10 and ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) form a set of representatives for the

coset space Γ^{0}(11)\SL_{2}(ℤ).)

(b) What is the genus of X(Γ^{0}(11))? To answer this question you have to work out how the boundary of the fundamental domain gets glued together. Use the matrix

( \begin{matrix}1 & 11 \\ 0 & 1 \end{matrix} ) to do the vertical lines, use ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) to do the region near the cusp 0,

and use appropriate matrices of the form ( \begin{matrix}s & b \\ 1 & −r \end{matrix} ) to glue the arcs |z − r| = 1 and

|z − s| = 1 for appropriate 1 ≤ r, s ≤ 10.

3. Various results about cusps and elliptic points.

(a) Show that if p is a prime then X_{0}(p) := X(Γ_{0}(p)) has two cusps.

(b) Show that if N ≥ 2 then Γ_{N} has no elliptic points. Similarly show that if N ≥ 4 then Γ_{1}(N) has no elliptic points. This means in practice that for these Γ, the curve X(Γ) is easier to study (although one can always deal with elliptic points, one just has to work harder.)

(c) Show that if N ≥ 5 then Γ_{1}(N) has \frac{1}{2} ∑_{0<d|N} ϕ(d)ϕ(N/d) cusps.

The next three questions analyse carefully the case Γ = SL_{2}(ℤ), following Serre’s lowlevel approach in his book “A Course in Arithmetic”. By doing some cunning integration, Serre manages to explicitly evaluate the dimension of M_{k} = M_{k}(SL_{2}(ℤ)) and hence works out the ring ⊕_{k}M_{k}. As I mentioned in lectures, the “correct” way to work out the dimension of M_{k}(Γ), which works for all Γ, is via the Riemann-Roch theorem, but I don’t want to pursue that line in this course because it needs more geometry than I am prepared to assume. These questions are long, and the first one is so analytic that I decided not to do this stuff in lectures. The reward for doing them is that at the end of it all one has a very explicit description of the space of modular forms of level 1. If you get stuck then all the answers are in Serre’s book “A Course in Arithmetic.”

4. This question gets the analysis out of the way.

(a) Let Γ = SL_{2}(ℤ.) Say 0 ≠ f ∈ M_{k}(Γ). Although f cannot be regarded as a function X(Γ) → ℂ, one can still talk about the order of the zero of f at a point of X(Γ), because f vanishes to the same order at z ∈ ℌ and γz for γ ∈ Γ. Verify that there are only finitely many points of X(Γ) where f vanishes. Write v_{x}(f) for the order of vanishing of f at a point x ∈ Γ\ℌ, and write v_{∞}(f) for the order of vanishing of f at infinity (with the obvious definition: look at the q-expanaion.)

(b) Evaluate the limit of the integral ∫_{C_{ϵ}} \frac{df}{f} as ϵ tends to zero, by working it out explicitly along each line/arc. Here C is the contour

and all the small arcs (whose positions are explained below) have radius ϵ. The small arcs in C_{ϵ} are around i, ω = e^{2πi/3}, −ω^{2} = e^{πi/3} and the (finitely many) points on the lines AB and D′E where f has a zero. Of course, by Cauchy’s theorem, for ϵ sufficiently small the integral is independent of ϵ. Here’s a hint for doing the integral: the integrals along AB and D′E cancel, and the only hard bits left are B′C and C′D. Use the fact that f|_{k}S = f

for S = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) to do these bits. For more hints see Serre’s book.

(c) By using the standard result that the integral above is just 2πi times the sum of the orders of vanishing of the zeros of f inside C_{ϵ}, deduce the key formula

v_{∞}(f) + ∑_{x∈X(Γ)} \frac{1}{e_{x}} v_{x}(f) = k/12.

Here e_{x} is defined to be 1 unless x is an elliptic point, e_{i} = 2 and e_{ω} = 3. Note that e_{x} is the ramification index of the natural map from ℌ to X(Γ) at a point above x.

5. This question, a continuation of question 4, shows how one can work out dim_{ℂ}(M_{k}) for all k. The only thing that we need from question 4 is the formula proved in part (c). Note that all the v_{x}(f) occurring in the formula are non-negative integers.

(a) Set k = 4 in the formula and deduce that E_{4} has a simple zero at ω and no other zeros in X(Γ) (use question 3 of example sheet 1). Set k = 6 and deduce that E_{6} has a simple zero at i and no other zeros in X(Γ). Here E_{k} = 1 + … is the Eisenstein series G_{k} normalised so that its constant term is 1.

(b) Conversely show that if f ∈ M_{4} is a non-zero modular form then f must have a simple zero at ω and no other zeros, and hence f is a constant multiple of E_{4}. Do a similar thing for M_{6}. Hence dim_{ℂ}(M_{4}) = dim_{ℂ}(M_{6}) = 1.

(c) Show that M_{2} = 0. Show that M_{8} is 1-dimensional and spanned by (E_{4})^{2}, and that M_{10} is 1-dimensional and spanned by E_{4}E_{6}. Deduce that E_{8} = (E_{4})^{2} and E_{10} = E_{4}E_{6}.

(d) For k ≥ 12 these tricks don’t work, because the formula proved in question 4 has more than one solution. But we can still work out the dimension of M_{k} thus. Let Δ = \frac{(E_{4})^{3}−(E_{6})^{2}}{1728} = q + … and note that Δ is a cusp form of level 12. Use the formula to deduce that Δ does not vanish anywhere on ℌ.

(e) If k ≥ 4 is even, show that the space S_{k} has dimension one less than the dimension of M_{k}. Show moreover that multiplication by Δ induces an isomorphism M_{k} → S_{k+12}. Deduce that for k ≥ 12 an even integer, we have dim_{ℂ}(M_{k}) = 1 + dim_{ℂ}(M_{k−12}). Deduce that for k ≥ 0 an even integer, we have

dim_{ℂ}(M_{k}) = [k/12] + 1

unless k ≡ 2 mod 12, in which case dim_{ℂ}(M_{k}) = [k/12]. Here [x] denotes the integer part of x. Finally note that because −I ∈ SL_{2}(ℤ), the dimension of M_{k} is zero if k is odd, so this question has completely determined the dimension of M_{k} for all k.

6. This question uses question 5 to evaluate explicitly the ring ⊕_{k}M_{k}.

(a) Let k ≥ 0 be an even integer. Show that the number of solutions to 4a + 6b = k with a, b ∈ ℤ_{≥0} is equal to dim_{ℂ}(M_{k}).

(b) Our candidate for a basis of M_{k} is the set (E_{4})^{a}(E_{6})^{b} for a, b as in part (a). By considering the order of vanishing of (E_{4})^{a}(E_{6})^{b} at ω show that these modular forms are ℂ-linearly independent, and deduce that they are a basis for M_{k}.

(c) Deduce that ⊕_{k}M_{k} ≅ ℂ[X, Y] via the map sending X to E_{4} and Y to E_{6}.