Solution (source code)

= Solution

A weight-$k$ <modular form> for $\Gamma=SL_2(\mathbb Z)$ is a <holomorphic function> $f$ on the <complex upper half-plane> satisfying
$$
f\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^kf(\tau)
\qquad\left(\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma\right),
$$
and <holomorphic at a cusp>. Since the full <modular group> has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau)=\sum_{n\geq0}a_nq^n$, $q=e^{2\pi i\tau}$, near infinity. A <cusp form> has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a <holomorphic function> on the quotient of the <complex upper half-plane> by the <modular group>. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The <standard fundamental domain of the modular group>, with its boundary identified and its cusp added, is a compact <Riemann surface>, the <compactified modular curve> $X(1)$. The <maximum modulus principle> now proves
$$
\boxed{M_0(SL_2(\mathbb Z))=\mathbb C.}
$$
Holomorphy at the cusp is essential to this conclusion.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-126-modular-domain.png]
{title=The standard modular fundamental domain with paired vertical boundaries, paired circular arcs, and the added cusp at infinity}

For positive even $k\geq4$, define the unnormalized <Eisenstein series>
$$
G_k(\tau)=\sum_{(m,n)\in\mathbb Z^2\setminus\{(0,0)\}}(m\tau+n)^{-k}.
$$
The real-linear map $(m,n)\mapsto m\tau+n$ is invertible, and on a compact subset of the <complex upper half-plane> its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving <holomorphy>. For $\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$,
$$
m\gamma\tau+n=\frac{(ma+nc)\tau+(mb+nd)}{c\tau+d}.
$$
The map $(m,n)\mapsto(ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma\tau)=(c\tau+d)^kG_k(\tau)$.

For the expansion, the <cotangent partial-fraction Fourier kernel> follows by differentiating the partial-fraction expansion of $\pi\cot\pi z$ and its geometric-series expansion $-\pi i(1+2\sum_{r\geq1}e^{2\pi i rz})$, valid for $\operatorname{Im}z>0$. It gives, for even $k\geq2$,
$$
\sum_{n\in\mathbb Z}(z+n)^{-k}
=\frac{(2\pi i)^k}{(k-1)!}\sum_{r\geq1}r^{k-1}e^{2\pi i rz}.
$$
The $m=0$ part of $G_k$ is $2\zeta(k)$, where $\zeta$ is the <Riemann zeta function>; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the <divisor sum> $\sigma_{k-1}(N)=\sum_{r\mid N}r^{k-1}$, this proves
$$
\boxed{G_k(\tau)=2\zeta(k)+\frac{2(2\pi i)^k}{(k-1)!}\sum_{N\geq1}\sigma_{k-1}(N)q^N.}
$$
The expansion has no negative powers and converges for $|q|<1$, so it also proves <holomorphy at a cusp> and completes the proof of modularity.

The supplied zeta values yield the <Fourier expansion of a normalized Eisenstein series>
$$
E_4=1+240S_3,\qquad E_6=1-504S_5,\qquad
S_j=\sum_{n\geq1}\sigma_j(n)q^n.
$$
Both $E_4^3$ and $E_6^2$ are <modular forms> of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a <cusp form>. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the <integrality identity for the modular discriminant>
$$
\Delta=\frac{5S_3+7S_5}{12}+100S_3^2+8000S_3^3-147S_5^2.
$$
For every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d\pmod4$. Thus each coefficient $(5\sigma_3(n)+7\sigma_5(n))/12$ is integral. The other three series visibly have integral coefficients, proving
$$
\boxed{\Delta\in S_{12}(SL_2(\mathbb Z)),\qquad\Delta\in q\mathbb Z[[q]],\qquad a_1(\Delta)=1.}
$$
This proves integrality without assuming a product expansion for the <modular discriminant>.