Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 / 3

Let $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the modular group, left multiplication by powers of $T$ and by $S$ performs subtraction and exchange on its first-column entries. The Euclidean algorithm reduces $(a,c)$ to $(\pm 1,0)$, since the determinant condition makes them coprime. The resulting matrix is $\pm T^m$, and $-I=S^2$ is already generated. Hence $S,T$ generate $S{L}_2(\mathbb{Z})$.

The standard fundamental domain of the modular group is $D=\{z\in \mathbb{H}:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, with the vertical sides identified by $T$ and the circular sides by $S$. To prove every orbit meets it, choose a coprime integer pair $(c,d)$ minimizing $|cz+d|$. A minimum exists because only finitely many pairs have this quantity below any fixed bound: $|c|\mathrm{Im}z\le |cz+d|$ bounds $c$, then $d$. Complete the pair to a determinant-one matrix. Its image has maximal imaginary part in the orbit, since $\mathrm{Im}(\gamma z)=\mathrm{Im}z/|cz+d{|}^2$. Translate its real part into $[-1/2,1/2]$. If its modulus were below one, inversion would increase its imaginary part, a contradiction. This is the reduction to the standard modular region.

The normalized Eisenstein series satisfy $E_4=1+240{\sum }_{m\ge 1}{\sigma }_3(m)q^m$ and $E_6=1-504{\sum }_{m\ge 1}{\sigma }_5(m)q^m$. Their weights are four and six. The valence formula for the modular group makes the normalized weight-twelve cusp form $\Delta$ have one simple cusp zero and no zero in $\mathbb{H}$. Comparing its first coefficient gives $\Delta =(E_4^3-E_6^2)/1728$.

Odd weights vanish because $-I$ acts by $(-1{)}^k$. Weight two vanishes by the valence formula for the modular group: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge 0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^a{E}_6^b$. The remainder is a cusp form; division by $\Delta$ produces a holomorphic modular form of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight provesThe integral triangular basis of level-one modular forms is $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, with $4a_j+6b_j=k-12j\ge 0$ and $k-12j\ne 2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f={\sum }_j{c}_j{F}_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor (k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore all Fourier coefficients of $f$ are integers.

Set $C=AB=\left(\begin{array}{cc}0& -1\\ 1& 2\lambda \end{array}\right)$. It fixes $z=-\lambda +i\sqrt{1-{\lambda }^2}$ because $z^2+2\lambda z+1=0$. Write $\lambda =\cos \theta$, $0<\theta <\pi /2$; the eigenvalues of $C$ are $e^{\pm i\theta }$.

The stabilizer of an interior point of $\mathbb{H}$ in $PS{L}_2(\mathbb{R})$ is conjugate to the compact rotation group. Its intersection with a discrete subgroup is finite: otherwise compactness gives accumulating elements, and their quotients accumulate at the identity. Thus $C$ has finite order in the projective group. Its eigenvalues then satisfy $e^{2im\theta }=1$ for some positive integer $m$, forcing $\theta /\pi \in \mathbb{Q}$. Hence $\boxed{\lambda =\cos (\pi y)\text{ for some }y\in \mathbb{Q}}$. This is a necessary condition only.