Modular Forms, PS 4

Bryden Cais

September 27, 2002

Exercise 1. Let e_{1} = ΔE_{4}^{3}, e_{2} = Δ^{2}. Then clearly e_{1}, e_{2} ∈ S_{24}(Γ). Moreover, we know that S_{24}(Γ) is two-dimensional. Moreover, e_{1}, e_{2} are linearly independent since E_{4}^{3}, Δ are not multiples of one another (recall the definition of Δ). Therefore, e_{1}, e_{2} is a basis for S_{24}(Γ). Recall that if

f = ∑_{n=1}^{∞} a_{n}q^{n} ∈ S_{24}(Γ)

then

T (m)f = ∑_{n=1}^{∞} b_{n}(m)q^{n} = ∑_{n=1}^{∞} ( ∑_{d|(m,n)} d^{23}a_{mn/d^{2}} ) q^{n}

is also a cusp form of weight 24. When m = 2, we have at once

b_{n}(2) = \begin{cases}a_{2n} & if n is odd \\ a_{2n} + 2^{23}a_{n/2} & otherwise \end{cases} .  (1)

Using PARI, we easily compute

e_{1} = q + 696q^{2} + 162252q^{3} + 12831808q^{4} + 34188270q^{5} − 882400608q^{6} − 833380744q^{7}

+ 25487347200q^{8} + ⋯

(2)

e_{2} = q^{2} − 48q^{3} + 1080q^{4} − 15040q^{5} + 143820q^{6} − 985824q^{7} + 4857920q^{8} + ⋯,

and using (1), we therefore have

T (2)e_{1} = 696q + 21220416q^{2} − 882400608q^{3} + 31325818368q^{4} + ⋯
T (2)e_{2} = q + 1080q^{2} + 143820q^{3} + 13246528q^{4} + · · · .

A quick computation now shows

T (2)e_{1} − 696e_{1} − 20736000e_{2} = O(q^{5})
T (2)e_{2} − e_{1} − 384e_{2} = O(q^{5}).

Observe that the left hand sides above are both cusp forms with a zero of order at least 5 at infinity. But we have

ord_{∞}(f) ≤ 24/6 = 4,

for any modular form of weight 24 that is not identically 0. It follows that

T (2)e_{1} = 696e_{1} + 20736000e_{2}
T (2)e_{2} = e_{1} + 384e_{2},

and hence that the matrix of T (2) in our basis is

T_{2} = (\begin{matrix}696 & 1 \\ 20736000 & 384 \end{matrix}) .

The characteristic polynomial of T_{2} is therefore

p(x) = x^{2} − 1080x − 20468736,

so that the eigenvalues of T_{2} are

λ_{1} = 540 + 12\sqrt{144169}
λ_{2} = 540 − 12\sqrt{144169}.

We then determine the corresponding eigenvectors:

v_{1} = (\begin{matrix}1 \\ −156 + 12\sqrt{144169} \end{matrix})
v_{2} = (\begin{matrix}1 \\ −156 − 12\sqrt{144169} \end{matrix}) .

We therefore set

f_{1} = ΔE_{4}^{3} − (156 − 12\sqrt{144169})Δ^{2}
f_{2} = ΔE_{4}^{3} − (156 + 12\sqrt{144169})Δ^{2}.

By our calculations above, we have

T (2)f_{1} = λ_{1}f_{1}
T (2)f_{2} = λ_{2}f_{2}.

Observe that λ_{1}, λ_{2} lie in a totally real quadratic field and are algebraic integers. Moreover, the galois group of this field has one nontrivial automorphism, σ, and

λ_{1} = σλ_{2},

from which it follows that

f_{1} = σf_{2},

since Δ, E_{4} have integer coefficients. Finally, we observe that since λ_{1} ≠ λ_{2}, we have found a Hecke eigenbasis for S_{24}(Γ).

Exercise 2. We approach this naively. Taking the usual region for Γ/ℍ, we have that the hyperbolic area of Γ/ℍ is

∫_{x=−1/2}^{1/2} ∫_{y=\sqrt{1−x^{2}}}^{∞} \frac{dy dx}{y^{2}} = ∫_{x=−1/2}^{1/2} \frac{dx}{\sqrt{1 − x^{2}}}

= 2 sin^{−1}(1/2)

= \frac{π}{3} .

Exercise 3. We have

∫_{ℝ} e^{−πx^{2}y}e^{−2πinx} dx = ∫_{ℝ} e^{−πy(x+in/y)^{2}−πn^{2}/y} dx

= e^{−πn^{2}/y} ∫_{ℝ} e^{−πyx^{2}} dx

= \frac{e^{−πn^{2}/y}}{\sqrt{πy}} ∫_{ℝ} e^{−x^{2}} dx

= \frac{e^{−πn^{2}/y}}{\sqrt{y}} .

The Poisson Summation formula then tells us (since e^{−πn^{2}y} is a rapidly decreasing function of n) that

∑_{n∈ℤ}e^{−πyn^{2}} = \frac{1}{\sqrt{y}} ∑_{n∈ℤ}e^{−πn^{2}/y}.

Analytic continuation then gives us the formula for all y with ℜ(y) > 0.

Exercise 4. We now give some intuition as to why the formula of Exercise 3 is likely. Let us compute the integral of e^{−πx^{2}} from x = −a to a as a Riemann sum. This is:

lim_{N→∞} \frac{a}{N} ∑_{j=−N}^{N} e^{−π(ja/N)^{2}}.

Now since e^{−πx^{2}} is rapidly decreasing, we may truncate the sum at some large value of N; the limit will then approximate the integral very closely. Therefore,

lim_{y→0} \sqrt{y} ∑_{j=−N}^{N} e^{−πj^{2}y} ≈ ∫_{−N}^{N} e^{−πx^{2}} dx

≈ ∫_{−∞}^{∞} e^{−πx^{2}} dx

= 1

= lim_{y→0} ∑_{n∈ℤ}e^{−πn^{2}/y}.

We therefore see that the transformation formula proved in Exercise 3 is likely, and may be interpreted as a statement about approximating integrals.