# Eisenstein Series, discriminant and cusp forms

I'm working on the following exercise:

Let $E_k=\frac{G_k(z)}{2\zeta(k)}$ be the normalized Eisenstein series of weight $k$.

Calculate a)$E_4=1+240\sum_{n=1}\sigma_3(n)q^n$ and b)$E_6=1-504\sum_{n=1}\sigma_5(n)q^n$, where $q=\exp(2\pi i z)$

c) Show that $\sigma_3(n)=\sigma_5(n)\,\text{mod}(12)$, where $\sigma_k(n)=\sum_{d|n}d^k$

d) Finally show that $\Delta=\frac{1}{1728}(E^3_4-E^2_6)$ is a non-zero cusp form of weight $k=12$ and integer fourier coefficients.

My solution attempt:

a&b)The Fourier transform for $G_k$ is

$G_k(z)=2\zeta(k)+\frac{2(2\pi i)^k}{(k-1)!}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^{n}$, where $q=\exp(2\pi i z)$. Further $\zeta(2k)=\sum_{n=1}n^{-2k}=\frac{(-1)^{k+1}(2\pi)^{2k}}{2(2k)!}B_{2k}$ from the Euler formula, which gives $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(6)=\frac{\pi^6}{945}$. Hence $$E_4=\frac{G_4}{2\zeta(4)}=\left(\frac{\pi^4}{45}+\frac{2(2\pi i)^4}{3!}\sum_{n=1}^{\infty}\sigma_3(n)q^n\right)\frac{\pi^4}{45}=1+240\sum_{n=1}\sigma_3(n)q^n$$

Similarly for $E_6$.

c) is where I'm starting to have problems, I have honestly no idea how to approach this part.

d) First we have that for $G_{k}(\frac{az+b}{cz+d})=(cz+d)^{k}G_k(z)$ and further that $G_k$ is analytic and possesses a Fourier transform, as such it is a modular form of weight $k$. By definition of $E_k$ this property is "inherited" obviously and we also immediately get that $E_k^j$ is a modular form of weight $kj$. As such $\Delta'=E^3_4-E^2_6$ is the difference of two modular forms of weight $kj=12$ and hence a modular form of weight $12$ itself. Thus $\Delta$ is one as well.

To verify that $\Delta$ is indeed a cusp form we have to examine the behavior at $\infty$. We have that $\lim_{z\to\infty}G_k(z)=2\zeta(k)$ and hence that $\Delta=\frac{1}{1728}(E^3_4-E^2_6)=\frac{1}{1728}\left(\left(\frac{G_4(z)}{2\zeta(4)}\right)^3-\left(\frac{G_6(z)}{2\zeta(6)}\right)^2\right)$ tends to $0$ as $z\to\infty$, as such $\Delta$ is a cusp form of weight $12$.

For the fourier coefficients I'm having problems again. Calculating some of the coefficients, it's somewhat obvious that they have to be whole numbers, but I'm unsure how to prove this.

As such I guess I'm hoping for some help with c) and the second part of d). Thanks in advance!

Edit: And if someone could confirm that my reasoning at all is sound would be amazing as well! I'm not entirely sure with my arguments in d).

Edit2: I think I've solved it:

\begin{align*} E_4^3-E_6^2&=(1+240\sum_{n=1}\sigma_3(n)q^n)^3-(1-504\sum_{n=1}\sigma_5(n) q^n)^2\\ &=12^2\left(5\sum_{n=1}\sigma_3(n)q^n+7\sum_{n=1}\sigma_5(n)q^n\right)+12^3\left(100(\sum_{n=1}\sigma_3(n)q^n)^2-147(\sum_{n=1}\sigma_5(n) q^n)^2+8000(\sum_{n=1}\sigma_3(n) q^n)^3\right) \\ &=12^2\left(\left(5\sum_{n=1}\sigma_3(n)q^n+7\sum_{n=1}\sigma_5(n)q^n\right)\right)+\left(12\left(100(\sum_{n=1}\sigma_3(n)q^n)^2-147(\sum_{n=1}\sigma_5(n) q^n)^2+8000(\sum_{n=1}\sigma_3(n) q^n)^3\right)\right) \end{align*}

With $\Delta=\frac{1}{1728}(E^3_4-E^2_6)=\frac{1}{2^63^3}(E^3_4-E^2_6)$ and the above it remains to show that $5\sum_{n=1}\sigma_3(n)q^n-7\sum_{n=1}\sigma_5(n) q^n$ is divisible by 12. We have $$5\sum_{n=1}\sigma_3(n)q^n-7\sum_{n=1}\sigma_5(n) q^n=\sum_{n=1}(5\sigma_3(n) +7\sigma_5(n))q^n=\sum_{n=1}\sum_{d|n}(5d^3+7d^5)q^n$$ for $d\in\mathbb{N}$.

This leaves to prove $5d^3+7d^5$ is divisible by 12. We have $$5d^3+7d^5=d^3(5+7d^2)= \begin{cases} d^3(1-d^2)&=0 (\text{ mod }4)\\ d^3(-1+d^2)&=0 (\text{ mod }3) \end{cases}$$

This proves c) and what was left of d)

This is another way to do part $c$. For a fixed integer $n$, $$\sigma_3(n)-\sigma_5(n)=\sum_{d|n}(n^3-n^5)=\sum_{d|n}n^3(n-1)(n+1)$$
Whatever the parity of $n$, $n^3(n-1)(n+1)$ is always divisible by $4$ and $3$ and hence by $12$. This is why $\sigma_3(n)$ and $\sigma_5(n)$ are always congruent modulo $12$.
Here is how I would do it. First we need to prove d) and then using d) and your calculations for part c) it follows that $5\sigma_3(n)+7\sigma_5(n)$ is divisible by $12$ which is same as saying that $5(\sigma_3(n)-\sigma_5(n))$ is divisible by $12$ and we are done for part c).
For d) we need to establish the famous identity $$E_{4}^3(q)-E_{6}^2(q)=1728q\prod_{n=1}^{\infty}(1-q^n)^{24}\tag{1}$$ To prove this note that each of the LHS and RHS is a modular form of weight $12$ and if one performs the calculation one can easily see that the ratio of LHS to RHS, say $\phi(q)$, has the Fourier series $$1+c_1q+c_2q^2+\dots$$ Thus $\phi(q)$ is a modular form of weight $0$ and therefore a constant and thus equal to $1$ identically so that the above identity $(1)$ holds. This shows that the coefficients of $\Delta$ are integers.