Modular Forms, PS 8

Bryden Cais

November 5, 2002

Exercise 1. Let a/b ∈ ℙ^{1}(ℚ) with (a, b) = 1 and 1/0 representing ∞. Since (a, b) = 1, we can find x, y ∈ ℤ so that xa + yb = 1. Without loss of generality, we may assume that x is odd: if it were even then b would have to be odd, in which case (x + b)a + (y − a)b = 1 also and x + b is odd. It is also clear that we must have (x, y) = 1, and since x is odd, (x, 4y) = 1. There therefore exist integers k, z so that xz − 4yk = 1. Then

γ = (\begin{matrix}x & y \\ 4k & z \end{matrix}) ∈ Γ_{0}(4)

and

γ \frac{a}{b} = \frac{1}{s}

for some integer s. Now let j ∈ ℤ be such that −2 ≤ s + 4j < 2. then

α_{j} = (\begin{matrix}1 & 0 \\ 4j & 1 \end{matrix}) ∈ Γ_{0}(4)

and

α_{j} \frac{1}{s} = \frac{1}{4j + s}.

Since ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) ∈ Γ_{0}(4), It follows that every a/b ∈ ℙ^{1}(ℚ) is equivalent under Γ_{0}(4) to one of

∞ = \frac{1}{0}, 1, \frac{1}{2}.

It is not difficult to see that these three rationals are inequivalent modulo Γ_{0}(4) by observing that the denominator of g(p/q) is congruent to q modulo 4 for any g ∈ Γ_{0}(4) and p/q ∈ ℙ^{1}(ℚ). Therefore, a set of representatives for the cusps of Γ_{0}(4)/ℍ is {∞, 1, 1/2}. It is easy to see that

γ_{1} = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix})
γ_{1/2} = (\begin{matrix}1 & 0 \\ 2 & 1 \end{matrix})

satisfy γ_{a}∞ = a.

Now recall that

Θ(z) := ∑_{n∈ℤ}e(n^{2}z)

obeys

Θ(z + 2) = Θ(z)
Θ(−1/z) = \sqrt{\frac{z}{i}} Θ(z).

Moreover, we have

Θ(z + 1) = ∑_{n∈ℤ}(−1)^{n}e(n^{2}z)

= 2 ∑_{n∈ℤ}e((2n)^{2}z) − ∑_{n∈ℤ}e(n^{2}z)

= 2Θ(4z) − Θ(z).

We therefore have

Θ(z)^{8}|_{γ_{1}} = (z + 1)^{−4}Θ (\frac{z}{z + 1})^{8}

= (z + 1)^{−4}Θ (\frac{−1}{−1 − \frac{1}{z}} )^{8}

= (z + 1)^{−4} (\sqrt{\frac{−1 − \frac{1}{z}}{i}} )^{8} Θ (−1 − \frac{1}{z})^{8}

= (z + 1)^{−4} (\sqrt{\frac{−1 − \frac{1}{z}}{i}} )^{8} (2Θ (−\frac{4}{z}) − Θ (−\frac{1}{z}))^{8}

= (z + 1)^{−4} (\sqrt{\frac{−1 − \frac{1}{z}}{i}} )^{8} (2 (\sqrt{\frac{z}{4i}})^{8} Θ (\frac{z}{4}) − (\frac{z}{i} )^{8} Θ (z))^{8}

= (z + 1)^{−4} (\sqrt{z + 1})^{8} (2Θ (\frac{z}{4}) − Θ(z))^{8}

= (2Θ (\frac{z}{4}) − Θ(z))^{8}

Now since

Θ(z) = 1 + 2 ∑_{n=1}^{∞} e(n^{2}z),

we see that the constant term in the fourier expansion of (2Θ (\frac{z}{4}) − Θ(z))^{8} is 1. It follows that the value of Θ(z)^{8} at the cusp 1 is 1. Similarly, we find

Θ(z)^{8}|_{γ_{1/2}} = (2z + 1)^{−4}Θ (\frac{z}{2z + 1})^{8}

= (2z + 1)^{−4}Θ (\frac{−1}{−2 − \frac{1}{z}} )^{8}

= (2z + 1)^{−4} (\sqrt{\frac{−2 − \frac{1}{z}}{i}} )^{8} Θ (−2 − \frac{1}{z})^{8}

= (2z + 1)^{−4} (\sqrt{\frac{−2 − \frac{1}{z}}{i}} )^{8} Θ (−\frac{1}{z})^{8}

= (2z + 1)^{−4} (\sqrt{\frac{−2 − \frac{1}{z}}{i}} )^{8} (\sqrt{\frac{z}{i}} i)^{8} Θ (z)^{8}

= (2z + 1)^{−4} (\sqrt{2z + 1})^{8} Θ (z)^{8}

= Θ (z)^{8} ,

so that the value of Θ(z) at the cusp 1/2 is also 1.

Exercise 2. Let Γ be a congruence subgroup of SL_{2}(ℤ) and write

SL_{2}(ℤ) = ⋃_{j=1}^{n} α_{j}Γ.

Suppose that f ∈ M_{0}(Γ) and let

g(z) = ∏_{j=1}^{n} (f(α_{j}^{−1}z) − f(z_{0})) .

We show that g ∈ M_{0}(SL_{2}(ℤ)). Observe that for γ ∈ SL_{2}(ℤ) we have

g(z)|_{γ} = ∏_{j=1}^{n} (f(α_{j}^{−1}γz) − f(z_{0}))

= ∏_{i=1}^{n} (f(α_{i}^{−1}z) − f(z_{0}))

= g(z)

since as α_{j} runs over a complete set of representatives of Γ/SL_{2}(ℤ), so does γ^{−1}α_{j} for any γ ∈ SL_{2}(ℤ). That g is holomorphic at the cusps follows directly from the fact that f ∈ M_{0}Γ. We therefore see that g ∈ M_{0}(SL_{2}(ℤ)).

Exercise 3. Using the same techniques as in exercise 1, it may be shown that the cusps of Γ_{0}(2) are represented by ∞, 0. Moreover, the cusp 0 has width 2 while ∞ has width 1. Recall that Δ(z) ∈ S_{12}(SL_{2}(ℤ)) has no zeroes in ℍ and a simple 0 at ∞. Moreover, suppose that

γ = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ Γ_{0}(2).

Then

Δ(2z)|_{γ} = (cz + d)^{−12}Δ (\frac{a(2z) + 2b}{(c/2)(2z) + d}) = Δ(2z).

It follows that

Δ(z)Δ(2z) ∈ S_{24}(Γ_{0}(2)).

Moreover, since

Δ(z) = q ∏_{n=1}^{∞}(1 − q^{n})^{24},

we see that Δ(z)Δ(2z) = q^{3} + ⋯ and therefore has a zero of order 3 at infinity. Let

γ = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}) ∈ SL_{2}(ℤ).

It is clear that γ interchanges 0 and ∞. Moreover, we have

Δ(z)Δ(2z)|_{γ} = (z)^{−24}Δ(−2/z)Δ(−1/z)
= z^{−24}(z/2)^{12}Δ(z/2)z^{12}Δ(z)
= 2^{−12}Δ(z/2)Δ(z)
= 2^{−12}q^{3/2} + ⋯,

so that, since 0 has width 2, Δ(z)Δ(2z) has a zero of order 3 at 0. Since Δ(z) is never 0 on ℍ, we can define a function f(z) := (Δ(z)Δ(2z))^{1/3} that is holomorphic on ℍ. We claim that f ∈ S_{8}(Γ_{0}(2)). Indeed, recall that

Δ(z) = η(z)^{24}

and that

η(z + 1) = e^{π/12}η(z)
η(−1/z) = e^{−iπ/4}\sqrt{z}η(z).

Moreover, the group Γ_{0}(2) is generated by the transformations z ↦ z + 1 and

z ↦ \frac{z − 1}{2z − 1} = \frac{−1}{−2 − \frac{1}{z−1}} .

We have

f(z + 1) = (η(z + 1)η(2z + 2))^{8}
= (e^{πi/12}η(z)e^{2πi/12}η(2z))^{8}
= (η(z)η(2z))^{8}
= f(z),

and

f (\frac{−1}{−2 − \frac{1}{z−1}} ) = (η (\frac{−1}{−2 − \frac{1}{z−1}} ) η (\frac{−1}{−1 − \frac{1}{2z−2}} ))^{8}

= (e^{−iπ/4}\sqrt{−2 − \frac{1}{z − 1}}η (−2 − \frac{1}{z − 1}) e^{−iπ/4}\sqrt{−1 − \frac{1}{2z − 2}}η (−1 − \frac{1}{2z − 2}))^{8}

= (e^{−iπ/4−2πi/12}\sqrt{−2 − \frac{1}{z − 1}}η (\frac{−1}{z − 1}) e^{−iπ/4−iπ/12}\sqrt{−1 − \frac{1}{2z − 2}}η (\frac{−1}{2z − 2}))^{8}

= (e^{−9iπ/12}\sqrt{−2 − \frac{1}{z − 1}}e^{−iπ/4}\sqrt{z − 1}η (z − 1) \sqrt{−1 − \frac{1}{2z − 2}}e^{−iπ/4}\sqrt{2z − 2}η (2z − 2))^{8}

= (e^{−15iπ/12}\sqrt{−2z + 1}e^{−iπ/12}η (z) \sqrt{−2z + 1}e^{−2iπ/12}η (2z))^{8}

= (e^{−18iπ/12}(−2z + 1)η (z) η (2z))^{8}

= (2z − 1)^{8}f(z).

It follows that f ∈ M_{8}(Γ_{0}(2)). Since we have determined that f^{3} has a zero or order 3 at each of the cusps 0, ∞, we must have that f has a simple zero at each cusp. Therefore f ∈ S_{8}(Γ_{0}(2)).

Exercise 4. In this exercise we determine the fourier expansion for the unrestricted Eisenstein series

G_{k}(z; u, v; N) := ∑_{\begin{matrix}c≡u (N) \\ d≡v (N) \\ (c,d)≠(0,0) \end{matrix}} \frac{1}{(cz + d)^{k}} ,

for k ≥ 3. Recall that we have the formula

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

We have

G_{k}(z; u, v; N) = ∑_{\begin{matrix}c≡u (N) \\ d≡v (N) \\ (c,d)≠(0,0) \end{matrix}} \frac{1}{(cz + d)^{k}}

= ∑_{j∈ℤ} ∑_{l∈ℤ} \frac{1}{((lN + u)z + (jN + v))^{k}}

= N^{−k} ∑_{j∈ℤ} ∑_{l∈ℤ} \frac{1}{(j + ((lN + u)z + v)/N)^{k}}

= N^{−k} ∑_{j∈ℤ} \frac{1}{(j + (uz + v)/N)^{k}}

+ N^{−k} ∑_{j∈ℤ} ∑_{l=1}^{∞} (\frac{1}{(j + ((lN + u)z + v)/N)^{k}} + (−1)^{k} \frac{1}{(j + ((lN − u)z − v)/N)^{k}} )

= \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{n=1}^{∞} n^{k−1}e (\frac{uz + v}{N} n)

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{n=1}^{∞} ∑_{l=1}^{∞} n^{k−1} (e (\frac{(lN + u)z + v}{N} n) + (−1)^{k}e (\frac{(lN − u)z − v}{N} n))

= \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} ∑_{n=1}^{∞}(nN + r)^{k−1}e (\frac{uz + v}{N} (nN + r))

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} ∑_{n=1}^{∞} ∑_{l=1}^{∞}(nN + r)^{k−1}e (\frac{(lN + u)z + v}{N} (nN + r))

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} ∑_{n=1}^{∞} ∑_{l=1}^{∞}(nN + r)^{k−1}e (\frac{(lN − u)z − v}{N} (nN + r))

= \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(vr/N) ∑_{n=1}^{∞}(nN + r)^{k−1}e (\frac{uz}{N} (nN + r))

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(vr/N) ∑_{n=1}^{∞} ∑_{l=1}^{∞}(nN + r)^{k−1}e (\frac{(lN + u)(nN + r)z}{N} )

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(−vr/N) ∑_{n=1}^{∞} ∑_{l=1}^{∞}(nN + r)^{k−1}e (\frac{(lN − u)(nN + r)z}{N} )

= \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(vr/N) ∑_{\begin{matrix}n≡r (N) \\ n>0 \end{matrix}} ∑_{\begin{matrix}l≡u (N) \\ l>0 \end{matrix}}^{∞} n^{k−1}e (\frac{lnz}{N} )

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(−vr/N) ∑_{\begin{matrix}n≡r (N) \\ n>0 \end{matrix}} ∑_{\begin{matrix}l≡−u (N) \\ l>0 \end{matrix}}^{∞} n^{k−1}e (\frac{lnz}{N} )

= \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(vr/N) ∑_{\begin{matrix}n≡ru (N) \\ n>0 \end{matrix}} ( ∑_{\begin{matrix}d|n \\ d≡r (N) \end{matrix}} d^{k−1}) e (\frac{nz}{N} )

+ \frac{(−2πi/N)^{k}}{(k − 1)!} ∑_{r=0}^{N−1} e(−vr/N) ∑_{\begin{matrix}n≡−ru (N) \\ n>0 \end{matrix}} ( ∑_{\begin{matrix}d|n \\ d≡r (N) \end{matrix}} d^{k−1}) e (\frac{nz}{N} )

It seems that this should have a nicer form, but I can’t seem to find it.