Constructing Cones as Orbifolds: Fundamental Domain for an Identification

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Homework Statement

Consider the ##(x,y)## plane and the complex coordinate ##z=x+iy##. The identification ##z \sim z\ exp^{(\frac{2 \pi i}{N})}##, with ##N## an integer greater than 2, can be used to construct a cone.

Examine now the identification ##z\sim z\ e^{2 \pi i \frac{M}{N}}, N>M \geq 2,## where ##M## and ##N## are relatively prime integers (that is, the greatest common divisor of M and N is 1). Determine a fundamental domain for the identification.

Homework Equations



The Attempt at a Solution



A fundamental domain for the identification is ##0 \leq arg(z) < 2 \pi \frac{Ma+Nb}{N},## because ##Ma+Nb=1## for two integers ##a## and ##b##.

Would you say that my answer is correct?
 
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bumpp!
 
You haven't given any reasons for your answer so I doubt that many teachers would accept it.
 
Here's my explanation:

One may guess that the fundamental domain is provided by the points ##z## that satisfy ##0 \leq arg(z) < 2 \pi \frac{M}{N}##. However this is not true. Take, for example. ##M = 2## and ##N = 3##. In this case, the identification becomes

##z \sim \exp^{(2 \pi i \frac{2}{3})}z ##

and the fundamental domain becomes

##0 \leq arg(z) < 2 \pi \frac{2}{3}##.

Therefore, the fundamental domain covers ##\frac{2}{3}## of the complex plane.
Therefore, not all the points of the fundamental domain can be identified with points on the remaining ##\frac{1}{3}## of the complex plane.

However, this is a possible fundamental domain for the identification:

##0 \leq arg(z) < 2 \pi \frac{Ma+Nb}{N}##

By Bezout's identity, if ##M## and ##N## are non-zero integers, and if ##d## is their greatest common divisor, then there exist integers ##a## and ##b## such that ##aM+bN=d##.

In our case, ##M## and ##N## are relatively prime integers, so that their greatest common divisor is 1. Therefore, there exist integers ##a## and ##b## such that ##aM+bN=1##.

This is very helpful because, now, the entire complex plane can be broken down into the fundamental domain and other regions whose points can be identified with points in the fundamental domain.

What do you think? :frown:
 
bumpp!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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