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Assume that the cyclic group $C_{n}$ of order $n$ acts on $T^2=S^1\times S^1$ by rotating each factor, i.e. a generator of $C_{n}$ acts as $$ (x,y)\mapsto (e^{\frac{2\pi i}{n}}x,e^{\frac{2\pi i}{n}}y). $$ What is the quotient $T^2/C_{n}$ topologically? I initially thought it would be again $T^2$, but things seem not that easy.

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2 Answers 2

up vote 3 down vote accepted

One way to identify it is the following:

The map $T^2\to T^2/C$ is a covering map, becase the action of the group is properly discontinuous, and the quotient is a compact orientable surface. The fundamental group of $T^2$ is isomorphic to a subgroup of finite index of the fundamental group of $T^2/C$, so the latter is virtually abelian. The only compact orientable surface with a virtually abelian fundamental group is the torus itself.

Alternatively, observe as aboce that the quotient is an orientable surface. Now its rational homology $H_\bullet(T^2/C,\mathbb Q)$ is isomorphic to the subspace of the rational homology $H_\bullet(T^2,\mathbb Q)$ which are fixed under the natural action of $C$. Since each element of $C$ acts by a map homotopic to the identity of $T^2$, the group $C$ acts trivially on $H_\bullet(T^2,\mathbb Q)$. Therefore $H_\bullet(T^2/C,\mathbb Q)\cong H_\bullet(T^2,\mathbb Q)$. Now compact orientable surfaces are classified under homeomorphism by their rational homology, so we must have $T^2/C\cong T^2$.

Finally, it is easy to see that the composition of the usual map $\mathbb R^2\to T^2$ with the quotient $T^2\to T^2/C$ is an identification map $q:\mathbb R^2\to T^2/C$, and it is not difficult to find explicitely a $q$ is in fact the quotient map for a specific action of a group isomorphic to $\mathbb Z^2$ on $\mathbb R^2$.&c.

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Thank you for the response. I am now convinced that $T^2?C_{n}$ is a torus, reading your answer with "modern technique" (except the last one). I need some time to really see what is going on. Many thanks again! –  Pooya Sep 29 '12 at 5:10

Hint: draw the torus as a quotient of $[0,1]\times [0,1]$ and then draw what the group action does. If you're still confused, can you see a way to simplify your picture, perhaps by cutting and pasting?

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This elementary observation works well! One can take $[0,1]\times [0,\frac{1}{n}]$ as a fundamental domain of this action. Thanks you for your help! –  Pooya Sep 29 '12 at 5:13

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