5
$\begingroup$

Let $\omega_1$ and $\omega_2$ be two complex numbers such that they don't lie along the same line through origin. Then we define the lattice $\Lambda = \{a\omega_1 + b\omega_2 : a,b\in \mathbb{Z}\}$.

I wish to prove the following claim (part of a big-proof in the text-book):

For $n\in \mathbb{N}$, we have $$ \sum_{\substack{\lambda\in\Lambda\\ n-1<|\lambda|\leq n}} \frac{1}{|\lambda|^r} \leq \frac{1}{n^{r-1}} $$

I understand that we are summing over annulus $n-1<|\lambda|\leq n$ and that the number of such $\lambda$'s will be finite, but why the number of $\lambda$ in that annulus has the order of magnitude $n$?

What I can see is that: $$\sum_{\substack{\lambda\in\Lambda\\ n-1<|\lambda|\leq n}} \frac{1}{|\lambda|^r} \leq \frac{d(n)}{(n-1)^r}$$ where $d(n)$ is the number of lattice points in the annulus. But how exactly do we reach the desired inequalty?

$\endgroup$
4
  • $\begingroup$ $\Lambda = W \mathbb{R}^2$. Let $\mu = \sup_{\|x\| = 1} \|W x\|$. Then $\displaystyle\sum_{\lambda\in\Lambda,|\lambda| \in [n,n+1]} \frac{1}{|\lambda|^r} \le \int_{x \in \mathbb{R}^2, \|x\| \in [n-2\mu,n+1]} \frac{1}{\|x\|^r} d (W^{-1} x) = \frac{1}{|\det(W)|} \int_{n-2\mu}^{n+1} \frac{1}{t^r}2\pi t dt$ $\endgroup$
    – reuns
    Dec 22, 2017 at 14:41
  • $\begingroup$ Sorry, I don't understand what you want to say. Can you please elaborate? Or consider answering the question completely? $\endgroup$
    – rationalbeing
    Dec 22, 2017 at 14:45
  • $\begingroup$ To upper bound the number of integer points in a thin ellipse, you can use a threshold to make the ellipse thicker and look at the area. $\endgroup$
    – reuns
    Dec 22, 2017 at 14:59
  • $\begingroup$ @reuns why do you answer in a comment? $\endgroup$
    – nilo de roock
    Sep 5, 2020 at 12:59

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .