What Is the Fundamental Domain in a ##\mathbb Z_{\frac{m}n}## Orbifold?

[rbwang1225]

$\mathbb Z_{\frac{m}n}$ orbifold

Homework Statement

Consider the identification $z\sim ze^{2\pi i \frac{m}n}$, where $m$ and $n$ are relatively prime integers. Determine a fundamental domain for the identification.

Homework Equations

Given two relatively prime integers $a$ and $b$, there exist integers $m$ and $n$ s.t. $ma+nb=1$.

The Attempt at a Solution

By considering cases of small $m,n$'s, I conclude that the fundamental domain is $0\le arg(z)<gcd(\frac{2m}n,2)$, but I can't give a more rigorous proof.
I guess that needs some knowledge of basic number theory.
Any advices would be very appreciated.

Regards.

 

[anathema]

Hello! Thank you for bringing this interesting problem to our attention. I am always fascinated by the intricate patterns and structures found in mathematics, and the concept of orbifolds is no exception.

First, let's define what an orbifold is. An orbifold is a generalization of a manifold, which is a geometric space that looks like Euclidean space at a small scale. In an orbifold, we allow for certain points or regions to have a "twist" or "fold" in them, which can be described by an identification rule. In this case, the identification rule is given by $z\sim ze^{2\pi i \frac{m}n}$, which means that if we rotate a point by $2\pi \frac{m}n$ radians, we will end up at the same point. This creates a twist or fold in the space, and the orbifold is the resulting space after all of these identifications are made.

Now, let's consider the fundamental domain for this orbifold. A fundamental domain is a region in the space that contains all of the information needed to reconstruct the entire space. In other words, if we take this region and make all of the necessary identifications, we will end up with the entire space. In this case, our fundamental domain will be a region in the complex plane.

To find this fundamental domain, we can use the fact that $m$ and $n$ are relatively prime. This means that there exists integers $p$ and $q$ such that $pm+qn=1$. Now, let's consider the point $z=re^{i\theta}$ in the complex plane. If we rotate this point by $2\pi \frac{m}n$ radians, we will end up at $ze^{2\pi i \frac{m}n}=re^{i(\theta+2\pi \frac{m}n)}$. However, since $pm+qn=1$, we can write $\theta+2\pi \frac{m}n=2\pi q+\theta$, which means that the point $ze^{2\pi i \frac{m}n}=re^{2\pi iq}$ is just a rotation of the original point by $2\pi q$ radians.

Now, let's consider the points that are rotated by ##2\pi q