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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2021.02.24 Wednesday (c) All rights reserved by ACHK