Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I don't know how to show this.

Do I assume $G$ acts on $S^{2n}$ by homeomorphisms? Then, since $S^{2n}$ is Hausdorff I'd know $G$ acts freely and properly discontinuously, and since $\pi_1(S^{2n})={1}$ I'd have $\pi_1(X/G)\cong G$. But I'm not sure whether this is useful.

share|cite|improve this question
Why does $S^{2n}$ being Hausdorff imply that $G$ acts freely? Why does $\pi_1(X/G) \cong G$? I don't see how either of your conclusions follow. Anyway, this can be solved by considering the degree of the maps in $G$. – Carl Apr 24 '12 at 6:11
Euler characteristic is multiplicative for covers. – user641 Apr 24 '12 at 6:20
Use Lefschetz fixed point theorem, you'll get all fixed-point-free homomorphism from $S^{2n}$ to itself must be orientation reversing. – Yuchen Liu Apr 24 '12 at 7:58
Any map $S^{2n}\to S^{2n}$ without fixed points is homotopic to the antipodal map $x\mapsto -x$, and this map has degree $-1$. Since degrees are multiplicative, the only way how all non-$1$ elements of $G$ have degree $-1$ is that there is at most one such element. – user8268 Apr 24 '12 at 10:02

(Of course the trivial group acts freely on all spheres! Let's assume $G$ is finite and nontrivial.)

Since $G$ is finite, the action is certainly properly discontinuous, and since the action is given to be free, it follows that the quotient $q: X \rightarrow X/G$ is a finite covering map. Recalling that in any $n$-sheeted covering map $q: X \rightarrow Y$ we have $\chi(X) = n \chi(Y)$ and that the Euler characteristic must be an integer, we are almost done.

share|cite|improve this answer
Note: this expands out Steve D's answer from almost a year ago. The homework assignment was probably due before now! – Pete L. Clark Mar 20 '13 at 16:11

Do you know the definition of degree? You can arguing as follows. Suppose a group G act on 2n-sphere. Then there is a homomorphism from G to {1,-1} which sends an element g to the degree of g-action. But if g is not identity, then g(x) is not x, thus the g-action is homotopic to the antipodal map. And in sphere of even dimension, the antipodal map is of degree -1(because it reverse the orientation). Let's see what do we have:(1) a homomorphism p:G->$C_2$; (2)p(g)=-1 if g is not identity. So now you can conclude that G actually has only two element.(So I don't understand why you assume that G is a finite group.)

share|cite|improve this answer
This is already covered in the comments above. – user641 Jan 15 '13 at 15:31
I didn't see the comments at first. You can take it as a expand. – lee Jan 16 '13 at 1:12

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.