Math 678, Lecture 3

Notes by Bryden Cais Sept. 9, 2002

1 Fourier Expansions

The main goal today will be to compute the fourier expansions of the modular forms G_{k}(z). Let f(w) be holomorphic on ℍ with period 1. Recall that we can then write

f(z) = ∑_{n∈ℤ}a(n)e(nz),

where

a(n) = ∫_{0+iy}^{1+iy} f(z)e(−nz) dz,

for any y. Observe also that

G_{k}(z) = ∑_{n∈ℤ\0}n^{−k} + ∑_{m∈ℤ\0} ∑_{n∈ℤ} \frac{1}{(mz + n)^{k}}

= 2ζ(k) + 2 ∑_{m=1}^{∞} ∑_{n∈ℤ} \frac{1}{(mz + n)^{k}} .

So in order to determine the fourier expansion of G_{k}(z), it suffices to determine that of

f(w) = ∑_{n∈ℤ} \frac{1}{(w + n)^{k}} .

Write

f(w) = ∑_{n∈ℤ}a(n)e(nw),

where as usual,

a(n) = ∫_{0+iv}^{1+iv} f(w)e(−nw) dw

= ∑_{k=−∞}^{∞} ∫_{k+iv}^{k+1+iv} \frac{e(−nw)}{w^{k}} dw

= ∫_{−∞+iv}^{∞+iv} \frac{e(−nw)}{w^{k}} dw.

We now split into three cases:

1. n = 0: Letting v → ∞ we see that

∫_{−∞+iv}^{∞+iv} \frac{dw}{w^{k}} → 0

since the minimum value of the norm of w tends to ∞ and the contour does not include 0.

2. n < 0: In this case we have |e(−nw)| = |e(−2πin(u + iv))| = e^{2πnv}. Since n < 0, we can again take the limit as v → ∞ and we see that the integrand again tends to 0 and that the contour does not contain 0, so that the integral is 0 also.

3. n > 0: This is the only tricky case. We find that

∫_{−∞+iv}^{∞+iv} \frac{e(−nw)}{w^{k}} dw = ∫_{C} \frac{e(−nw)}{w^{k}} dw,

where C is the contour with vertices at −∞ + iv, ∞ + iv, −∞ − it, ∞ − it for v, t > 0 oriented clockwise. Observe that the integral along the vertical sides is 0. We now consider the limit of this integral as t → ∞. As before, the integral along the bottom horizontal segment of C tends to 0 as t → ∞. Cauchy’s Theorem then tells us that

∫_{−∞+iv}^{∞+iv} \frac{e(−nw)}{w^{k}} dw = −2πi Res_{0} (\frac{e(−nw)}{w^{k}}) = \frac{(−2πi)^{k}n^{k−1}}{(k − 1)!}

We have therefore shown that

f(w) = ∑_{n∈ℤ} \frac{1}{(n + w)^{k}} = ∑_{n=1}^{∞} \frac{(−2πi)^{k}n^{k−1}}{(k − 1)!} e(nw).

Exercise 1. Use the above formula to show that

ζ(k) = −\frac{(2πi)^{k}B_{k}}{2k!} ,

where B_{k} is defined by

\frac{x}{e^{x} − 1} = ∑_{k=0}^{∞} B_{k} \frac{x^{k}}{k!} .

We then have

G_{k}(z) = 2ζ(k) + \frac{2(−2πi)^{k}}{(k − 1)!} ∑_{m=1}^{∞} ∑_{n=1}^{∞} n^{k−1}e(mnz)

= 2ζ(k) + \frac{2(−2πi)^{k}}{(k − 1)!} ∑_{r=1}^{∞} σ_{k−1}(r)e(rz).

Exercise 2. Show that σ_{k−1} is multiplicative; more generally, show that

σ_{k−1}(m)σ_{k−1}(n) = ∑_{d|(m,n)} d^{k−1}σ_{k−1}(\frac{mn}{d^{2}}).

Exercise 3. Show that

n^{k−1} < σ_{k−1}(n) < ζ(k − 1)n^{k−1}

and that these are the sharpest asymptotics one can hope for in the sense that

lim inf \frac{σ_{k−1}(n)}{n^{k−1}} = 1

and

lim sup \frac{σ_{k−1}(n)}{n^{k−1}} = ζ(k − 1).

Now let l be any integer. Then we claim that for k even, almost all σ_{k−1}(n) are divisible by l in the sense that

lim_{x→∞} \frac{# \{n ≤ x : l|σ_{k−1}(n)\}}{x} = 1.

To see this, observe that if p ≡ −1 mod l and p|n but p^{2} |̸n then

σ_{k−1}(n) = σ_{k−1}(p)σ_{k−1}(n/p),

and l|σ_{k−1}(p) since σ_{(}k − 1)(p) = p^{k−1} + 1 ≡ 0 mod l. Now notice that for large x one has

# \{p ≤ x : p|x, p^{2} |̸x\} ≃ x (\frac{1}{p} − \frac{1}{p^{2}} ) ,

and therefore the probability that an arbitrary integer n has no prime factors congruent to −1 mod l that divide n exactly once is

∏_{p≡−1 mod l} (1 − \frac{1}{p} + \frac{1}{p^{2}} ) = 0

by Dirichlet’s Theorem (which tells us that # {p ≤ x : p ≡ a mod q} ≃ x/(φ(q) log(x)) if (p, a) = 1).