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Consider the complex semiplane $\mathcal H=\{z\in\mathbb C\,:\,\Im(z)>0\}$, and lets indicate with $\tau$ the elements of $\mathcal H$. Serre's book "A course in arithmetic" says that the serie of function $$G_k(\tau)=\!\!\!\sum_{\substack{(m,n)\in\mathbb Z\times\mathbb Z,\\(m,n)\neq(0,0)}}\!\!\!{(m\tau+n)}^{-k}$$ converges normally on compact sets of $\mathcal H$ if $k\ge 3$. I have proved that:

$$\sum_{\substack{(m,n)\in\mathbb Z\times\mathbb Z,\\(m,n)\neq(0,0)}}\!\!\left|{(m\tau+n)}^{-k}\right|\le \frac{8}{r^k}\sum_{n=1}^{\infty}\frac{1}{n^{k-1}}\,\,<\infty$$ for all $\tau\in\mathcal H$ with $r\in\mathbb R$. So $G_k(\tau)$ is (pointwise) absolutely convergent in $\mathcal H$, but i don't understand why this serie is normally convergent on compact sets.

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Not sure you really want "Eisenstein" but "Eseinstein" seemed too unlikely. – Rudy the Reindeer Sep 10 '12 at 16:20
It is definitely “Eisenstein”. – Dylan Moreland Sep 10 '12 at 16:23
The idea is that on any compact subset of the halfplane the imaginary part of $\tau$ is bounded from below by some $\epsilon $ while the real part is bounded in magnitude by $M$ . This yields $|m\tau+n|^2\ge \epsilon^2( m^2+n^2/M^2)$ or something similar... the end result being uniform convergence on said subset. – user31373 Sep 12 '12 at 4:51
up vote 1 down vote accepted

Let $\rho = \frac{1 + \sqrt{-3}}{2}$. $\rho\bar\rho = 1$. $\rho + \bar\rho = 1$. Hence $|m\rho - n|^2 = (m\rho - n)(m\bar\rho - n) = m^2 -mn + n^2$ for $(m, n) \in \mathbb{Z}^2$.

Let $D = \{z \in \mathcal{H}; |z| \ge 1, |Re(z)| \le 1/2\}$.

Let $z \in D$. Let $(m, n) \in \mathbb{Z}^2$.

$|mz + n|^2 = (mz + n)(m\bar z + n) = m^2z\bar z + 2mnRe(z) + n^2 \ge m^2 -mn + n^2 = |m\rho - n|^2$ Hence $|mz + n|^{2k} \ge |m\rho - n|^{2k}$ for $k \ge 1$.

On the other hand, $\sum_{(m,n)\in \mathbb{Z}^2 - (0,0)} 1/|m\rho - n|^s$ converges for $s > 2$

Hence $G_{2k}(z) = \sum_{(m,n)\in \mathbb{Z}^2 - (0,0)} 1/|mz + n|^{2k}$ converges normally on $D$ for $k > 1$. We fix $k > 1$

Let $\sigma \in SL_2(\mathbb{Z})$. Let $K$ be a compact subset of $\sigma(D)$. Let $z \in K$. Suppose $\sigma^{-1} = \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$.

$m\sigma^{-1}(z) + n = m(az + b)/(cz + d) + n = ((ma + nc)z + mb + nd)/(cz + d)$ for $(m,n) \in \mathbb{Z}^2$.

Hence $G_{2k}(\sigma^{-1}(z)) = (cz + d)^{2k} G_{2k}(z)$.

Hence $G_{2k}(z) = (cz + d)^{-2k} G_{2k}(\sigma^{-1}(z))$.

$Im(\sigma^{-1}(z)) = Im(z)/|cz + d|^2$

Since $K$ is compact, there exists $M > 0$ such that $|Im(\sigma^{-1}(z))| \le M$ for every $z \in K$.

Since $K$ is compact, there exists $\delta > 0$ such that $|Im(z)| \ge \delta$ for every $z \in K$.

Hence $1/|cz + d|^2 \le \frac{M}{\delta}$ for every $z \in K$.

Hence $|G_{2k}(z)| \le (\frac{M}{\delta})^k |G_{2k}(\sigma^{-1}(z))|$ for every $z \in K$.

Hence $G_{2k}(z)$ converges normally on $K$.

Let $L$ be a compact subset of $\mathcal{H}$. By theorem 1 of chapter VII of Serre's book "A course in arithmetic", $L$ can be covered by a finite number of sets of the form $\sigma(D)$, where $\sigma \in SL_2(\mathbb{Z})$. Since $L \cap \sigma(D)$ is compact, $G_{2k}(z)$ converges normally on $L \cap \sigma(D)$. Hence $G_{2k}(z)$ converges normally on $L$.

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