Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Not sure you really want "Eisenstein" but "Eseinstein" seemed too unlikely. –  Matt N. 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
add comment

1 Answer 1

up vote 0 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 $L$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.