Let \(G = \Gamma_\theta \subset \operatorname{SL}_2(\mathbb{Z})\) denote the theta subgroup generated by \(S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) and \(T^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
Suppose that \(f : \mathbb{H} \to \mathbb{C}\) is holomorphic, bounded, and that there exists a sequence of points \(\tau_k = x_k + i y_k \in \mathbb{H}\) such that \[ f(\tau_k) = 0, \quad \sum_{k=1}^\infty y_k = \infty, \quad 0 < y_k \le 1, \quad \text{and} \quad |x_k| \le 1. \] Prove that \(f \equiv 0\).
Given two relatively prime integers \(c\) and \(d\) of different parity, show that there exist integers \(a\) and \(b\) such that \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G\).
Prove that \(\sum \frac{1}{c^2 + d^2} = \infty\), where the sum is taken over all pairs \((c, d)\) of relatively prime integers with opposite parity.
Prove that if \(F : \mathbb{H} \to \mathbb{C}\) is holomorphic, bounded, and invariant under \(G\), then \(F\) is constant.
Solution
1. (a) If \(f\) is bounded and holomorphic on \(\mathbb H\) and vanishes at points \(\tau_k=x_k+iy_k\) with \(\abs{x_k}\le1\), \(0<y_k\le1\), and \(\sum y_k=\infty\), then \(f\equiv0\).
Let \(\Phi(z) = \frac{z - i}{z + i}\) be the Cayley transform \(\mathbb H\to\mathbb D\) and \(g=f\circ\Phi^{-1}\), a bounded holomorphic function on \(\mathbb D\) vanishing at \(\alpha_k = \Phi(\tau_k)\). Then \[1 - |\alpha_k|^2 = \frac{(x_k^2 + (y_k+1)^2) - (x_k^2 + (y_k-1)^2)}{x_k^2 + (y_k+1)^2} = \frac{4 y_k}{x_k^2 + (y_k + 1)^2}\ge\frac{4y_k}{5},\] since \(x_k^2 + (y_k + 1)^2 \le 1 + 4\). Using \(1 - r \ge \frac{1}{2}(1 - r^2)\) for \(0\le r<1\), \(\sum_k (1 - |\alpha_k|) \ge \frac{2}{5} \sum_k y_k = \infty\). By E-SS5.PR-1, the zeros of a bounded holomorphic function on \(\mathbb D\) that is not identically zero satisfy \(\sum(1-\abs{\alpha_k})<\infty\). Hence \(g \equiv 0\) and \(f \equiv 0\).
2. (b) If \(\gcd(c,d)=1\) and \(c\not\equiv d\pmod2\), there are \(a,b\in\ZZ\) with \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G\).
By Bézout’s identity there are \(a_0, b_0\) with \(a_0 d - b_0 c = 1\), and \(a = a_0 + c t\), \(b = b_0 + d t\) also satisfy \(ad-bc=1\) for every \(t \in \ZZ\). A matrix of \(\operatorname{SL}_2(\ZZ)\) lies in \(G\) if and only if \(a \equiv d\) and \(b \equiv c \pmod 2\). If \(c\) is even and \(d\) odd, then \(ad\) is odd, so \(a\) is odd for every \(t\), and \(t \equiv b_0 \pmod 2\) makes \(b\) even. If \(c\) is odd and \(d\) even, then \(bc\) is odd, so \(b\) is odd for every \(t\), and \(t \equiv a_0 \pmod 2\) makes \(a\) even.
3. (c) \(\sum \frac{1}{c^2 + d^2} = \infty\) over coprime pairs \((c,d)\) of opposite parity.
A coprime pair \((a,b)\) either has opposite parity or has both entries odd. For \(a,b\) coprime and odd, \(c = \frac{a+b}{2}\) and \(d = \frac{a-b}{2}\) are coprime of opposite parity with \(c^2 + d^2 = \frac{a^2 + b^2}{2}\), and \((a,b)\mapsto(c,d)\) is injective. So if the sum in (c) were finite, \(\sum_{\gcd(a,b)=1} \frac{1}{a^2 + b^2}\) would be finite. Writing each nonzero \((k,\ell)\) as \(n(a,b)\) with \(n=\gcd(k,\ell)\), \[\sum_{(k, \ell) \neq (0, 0)} \frac{1}{k^2 + \ell^2} = \Bigl(\sum_{n=1}^\infty \frac{1}{n^2}\Bigr) \Bigl(\sum_{\gcd(a, b)=1} \frac{1}{a^2 + b^2}\Bigr)\] would be finite. But for each \(m\ge1\) there are \(8m\) pairs with \(\max(\abs k,\abs\ell)=m\), each with \(k^2+\ell^2\le2m^2\), so the left side is at least \(\sum_{m\ge1}\frac{8m}{2m^2}=\infty\).
4. (d) A bounded holomorphic \(G\)-invariant \(F\) on \(\mathbb H\) is constant.
Put \(F_0 = F - F(i)\), which is bounded, holomorphic, \(G\)-invariant, and vanishes at \(i\). For each coprime pair \((c,d)\) of opposite parity, step 2 gives \(g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G\), and since \(ad - bc = 1\), \[g(i) = \frac{(ai+b)(-ci+d)}{c^2 + d^2} = \frac{ac+bd}{c^2+d^2} + \frac{i}{c^2+d^2}.\] Applying a power of \(T^2\in G\) gives a point \(\tau_{c,d}\) of the \(G\)-orbit of \(i\) with \(\abs{\Re\tau_{c,d}} \le 1\) and \(\Im\tau_{c,d} = \frac{1}{c^2 + d^2} \in (0, 1]\), and \(F_0(\tau_{c,d}) = F_0(i) = 0\) by invariance. If two pairs give the same point, the corresponding elements \(h,h'\) of \(G\) (the matrices \(g\) composed with the translations) satisfy \(h^{-1}h'(i)=i\), so \(h^{-1}h'\in\{\pm I,\pm S\}\) and the bottom row of \(h'\) is \(\pm(c,d)\) or \(\pm(d,-c)\); thus each point arises from at most four pairs. By step 3, \(\sum \Im\tau_{c,d}=\infty\), so the sum over distinct points \(\tau_{c,d}\) is also infinite, and step 1 gives \(F_0 \equiv 0\), that is, \(F\equiv F(i)\).