Chapter 6

Determine all matrices \(M=\left(\begin{array}{c}a b \\ c & d\end{array}\right) \in \mathrm{SL}(2, \mathbb{R})\) with the fixed point i. Result. $$ M \mathrm{i}=\mathrm{i} \Longleftrightarrow M \in \mathrm{SO}(2, \mathbb{R}):=\left\\{M \in \mathrm{SL}(2, \mathbb{R}) ; \quad M^{\prime} M=E\right\\} $$

The group \(\operatorname{SL}(2, R)\) can be defined for any associative ring \(R\) with unit \(1=1_{R}\) Show that for the finite commutative ring \(R=\mathbb{Z} / q \mathbb{Z}\) the two matrices $$ \left(\begin{array}{rr} 0_{R} & -1_{R} \\ 1_{R} & 0_{R} \end{array}\right) \text { and }\left(\begin{array}{cc} 1_{R} & 1_{R} \\ 0_{R} & 1_{R} \end{array}\right) $$ are generating \(\operatorname{SL}(2, R)\).

2\. Show: (a) The group \(\mathrm{SL}(2, \mathbb{R})\) acts transitively on the upper half-plane \(\mathbb{H}\), i.e. for any two points \(z, w \in \mathbb{H}\) there exists a group element \(M \in \mathrm{SL}(2, \mathbb{R})\) with \(w=M z\). Hint. It is enough to consider the case \(w=\mathrm{i}\). Then we can even take \(c=0\) (b) The map $$ \begin{aligned} \mathrm{SL}(2, \mathbb{R}) / \mathrm{SO}(2, \mathbb{R}) & \longrightarrow \mathbb{H} \\ M \cdot \mathrm{SO}(2, \mathbb{R}) & \longrightarrow M \mathrm{i} \end{aligned} $$ is bijective. (It is even a homeomorphism, if we canonically organize its domain of definition with the quotient topology.)

Let \(M \in \mathrm{SL}(2, \mathbb{R})\), and let \(l\) be an integer number with the property \(M^{l} \neq \pm E\) The matrix \(M\) is elliptic, iff \(M^{l}\) is elliptic.

Let $$ f(z)=\sum_{n=1}^{\infty} a_{n} e^{2 \pi i n z} $$ be a cusp form of weight \(k\). Prove an estimation of the shape $$ \left|a_{n}\right| \leq C n^{k / 2} $$ $$ \left|a_{n}\right| \leq C n^{k / 2} $$ (E. HECKE, 1927 ) with a suitable constant \(C .\) Hint. Use the integral representation for the FOURIER coefficients, and apply the estimation $$ |f(z)| \leq C^{\prime} y^{-k / 2} $$ for the special value \(y=1 / n\). P. DELIGNE proved in 1974 the RAMANUJAN-PETERSSON conjecture, which insures the much stronger estimation $$ \left|a_{n}\right| \leq C(\varepsilon) n^{(k-1) / 2+\varepsilon} \text { for any } \varepsilon>0 $$.

Let \(p\) be a prime number. The group \(\operatorname{GL}(2, \mathbb{Z} / p \mathbb{Z})\) has \(\left(p^{2}-1\right)\left(p^{2}-p\right)\) elements. Hint. How many possibilities are there to fill in the first column of an "empty matrix" such that it still has chances to lie in \(\mathrm{GL}(2, \mathbb{Z} / p \mathbb{Z}) ?\) For such a fixed first column, how many possibilities are there to fill in the second column in order to obtain an invertible matrix ? Infer from this, that the group \(\operatorname{SL}(2, \mathbb{Z} / p \mathbb{Z})\) has \(\left(p^{2}-1\right) p\) elements.

Let \(G \subset \mathrm{SL}(2, \mathbb{R})\) be a finite subgroup, such that its elements admit a common fixed point in \(\mathbb{H}\). (One can show that any finite, or more general any compact, subgroup \(G \subset \operatorname{SL}(2, \mathbb{R})\) has this property !) Show that \(G\) is cyclic.

For any point \(a \in \mathbb{H}\) there exists an entire modular form (even of weight 12), which vanishes in \(a\), but does not vanish identically. Hint. Use the knowledge of the zeros of \(\Delta\).

Let \(p\) be a prime, and let \(m\) be a natural number. The kernel of the natural homomorphism $$ \mathrm{GL}\left(2, \mathbb{Z} / p^{m} \mathbb{Z}\right) \longrightarrow \mathrm{GL}\left(2, \mathbb{Z} / p^{m-1} \mathbb{Z}\right) $$ is isomorphic to the additive group of all \(2 \times 2\) matrices with entries in \(\mathbb{Z} / p \mathbb{Z}\). Using this, show: $$ \begin{aligned} &\\# \mathrm{GL}\left(2, \mathbb{Z} / p^{m} \mathbb{Z}\right)=p^{4 m-3}\left(p^{2}-1\right)(p-1) \\ &\\# \mathrm{SL}\left(2, \mathbb{Z} / p^{m} \mathbb{Z}\right)=p^{3 m-2}\left(p^{2}-1\right) \end{aligned} $$

Let \(q_{1}\) and \(q_{2}\) be two relatively prime natural numbers. The Chinese Remainder Theorem claims that the natural homomorphism \(\mathbb{Z} / q_{1} q_{2} \mathbb{Z} \rightarrow \mathbb{Z} / q_{1} \mathbb{Z} \times \mathbb{Z} / q_{2} \mathbb{Z}\) is an isomorphism. Infer from this, that the natural homomorphism $$ \mathrm{GL}\left(2, \mathbb{Z} / q_{1} q_{2} \mathbb{Z}\right) \longrightarrow \mathrm{GL}\left(2, \mathbb{Z} / q_{1} \mathbb{Z}\right) \times \mathrm{GL}\left(2, \mathbb{Z} / q_{2} \mathbb{Z}\right) $$ is an isomorphism.

A subset \(\mathcal{F}_{0} \subset \mathbb{H}\) is a fundamental region of a congruence group \(\Gamma_{0}\), iff the following two conditions are satisfied: (a) There exists a subset \(S=S\left(\mathcal{F}_{0}\right) \subset \mathcal{F}_{0}\) of LEBESGUE measure 0, such that \(\mathcal{F}_{0} \backslash S\) is open, and any two points of \(\mathcal{F}_{0} \backslash S\) are inequivalent with respect to the action of \(\Gamma_{0}\) (b) The \(\Gamma_{0}\)-translates of \(\mathcal{F}_{0}\) are covering the upper half-plane, i.e. $$ \mathbb{H}=\bigcup_{M \in \Gamma_{0}} M \mathcal{F}_{0} $$ Let $$ \Gamma=\bigcup_{\nu=1}^{h} \Gamma_{0} M_{\nu} $$ be the decomposition of the full modular group in right congruence classes with respect to the subgroup \(\Gamma_{0}\), and let \(\mathcal{F}\) be the standard modular figure. Then $$ \mathcal{F}_{0}=\bigcup_{\nu=1}^{h} M_{\nu} \mathcal{F} $$ is a fundamental region of \(\Gamma_{0}\).

Let \(S=S^{(n)}\) be a positive, even, unimodular matrix. Then \(n \equiv 0 \bmod 8\) Hint. Use the relation $$ w:=1-\frac{1}{z}=\left(\frac{1}{1-z}-1\right)^{-1} $$ and transform \(\vartheta(S ; w)\) corresponding to these relations, by applying the formulas $$ \vartheta(S ; z+1)=\vartheta(S ; z), \quad \vartheta(S ;-1 / z)=\sqrt{\frac{z}{\mathrm{i}}}^{n} \vartheta(S ; z) $$ This gives the formula $$ \sqrt{z / \mathrm{i}}^{n}=\sqrt{z /(\mathrm{i}(1-z))}^{n} \sqrt{(z-1) / \mathrm{i}}^{n} $$ Now specialize \(z=\mathrm{i}\) in it to infer $$ 1=e^{2 \pi \mathrm{in} / 8}, \text { i.e. } n \equiv 0 \quad \bmod 8 $$.

Let \(p\) be a prime number. The group \(\Gamma_{0}[p]\) has exactly two cusp classes, which can be represented by 0 , and respectively i \(\infty\).