1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cones as orbifolds

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data


    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.

    2. Relevant equations

    3. 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?
     
  2. jcsd
  3. May 16, 2015 #2
  4. May 16, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You haven't given any reasons for your answer so I doubt that many teachers would accept it.
     
  5. May 16, 2015 #4
    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:
     
  6. May 18, 2015 #5
    bumpp!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Cones as orbifolds
  1. Cone edge (Replies: 3)

  2. Volume of a cone (Replies: 5)

  3. Approximation a cone (Replies: 4)

  4. Parametrizeing A Cone (Replies: 2)

  5. Geodesics on a cone (Replies: 0)

Loading...