# Problem 2.6c

A First Course in String Theory

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2.5 Constructing $\displaystyle{T^2/\mathbb{Z}_3}$ orbifold

(c) Determine the three fixed points of the $\displaystyle{\mathbb{Z}_3}$ action on the torus. Show that the orbifold $\displaystyle{T^2/\mathbb{Z}_3}$ is topologically a two-dimensional sphere, naturally presented as a triangular pillowcase with seamed edges and corners at the fixed points.

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[guess]

To find the fixed points, we consider the cases when

$\displaystyle{z + m + n e^{i \pi/3} = e^{2 \pi i/3} z}$,

where $\displaystyle{m,n \in \mathbb{Z}}$.

$\displaystyle{(e^{2 \pi i/3} - 1) z = m + n e^{i \pi/3}}$

$\displaystyle{z = \frac{m + n e^{i \pi/3}}{e^{2 \pi i/3} - 1}}$

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When $\displaystyle{m, n = 0}$,

$\displaystyle{z = 0}$

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When $\displaystyle{m = 0; n = 1}$,

$\displaystyle{z = \frac{e^{i \pi/3}}{e^{2 \pi i/3} - 1} = \frac{-i}{\sqrt{3}}}$

When $\displaystyle{m = 1; n = 0}$,

$\displaystyle{z = \frac{1}{e^{2 \pi i/3} - 1} = \frac{1}{\sqrt{3}} e^{- 5 i \pi/6}}$

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When $\displaystyle{m = -2; n = 1}$,

$\displaystyle{z = \frac{-2 + 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = 1}$

When $\displaystyle{m = -1; n = -1}$,

$\displaystyle{z = \frac{-1 - 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = e^{i \pi/3}}$

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In the fundamental domain, the 3 fixed points are:

$\displaystyle{z = 0}$

when $\displaystyle{z = R(z)}$;

$\displaystyle{z = 1}$

when $\displaystyle{T_2 \circ T_1^{-1} \circ T_1^{-1} (z) = R(z)}$;

$\displaystyle{z = e^{i \pi/3}}$

when $\displaystyle{T_2^{-1} \circ T_1^{-1} (z) = R(z)}$.

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Duplicate the fundamental triangle to create a fundamental parallelogram.

If we label the some edges as $B$ instead of $A$, the fundamental parallelogram will have a sphere topology $\displaystyle{ ABB^{-1}A^{-1} }$.

However, it is not exactly the same as a sphere topology, because a sphere topology would not have the $A=B$ identification.

[guess]

— Me@2021-02-23 03:44:57 PM

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