Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.