Take the 2-minute tour ×
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.

And what for a single finite order element $f$, i.e. $f:R^n\rightarrow R^n$ is a homeomorphism such that $f^d=id_{R^n}$, must $f$ have a fixed point?

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

Any continuous map, $f: \mathbb R^n \rightarrow \mathbb R^n$, such that $f^n = \text{ id}$ for any natural number $n$ must have a fixed point. The proof is not entirely trivial and there are two ways to do it: either using Smith theory or using algebraic topology (see Bredon, Geometry and Topology, for instance where a scheme for such a proof is laid out). In general, if a group acts freely and properly discontinuously on $\mathbb R^n$, it cannot have torsion. This is also the reason that classifying spaces of finite groups are infinite dimensional. For example, the classifying space of $Z/2$ is $R\mathbb P^\infty$.

share|improve this answer
    
Thanks for your answer. I fixed a small font issue in your answer: italicized text can be marked with single asterisks. –  user53153 Jan 3 '13 at 0:42
1  
That "also the reason that classifying spaces of finite groups are infinite dimensional" doesn't seem to hold water. There's no reason that the universal cover of such a group needs to look like $\mathbb{R}^n$. –  Qiaochu Yuan Jan 3 '13 at 1:05
    
I am actually more interested in the "Smith Theory"(actually the other proof was what I have in mind), but I can't find it anywhere. Could you supply some references? –  Ash GX Jan 4 '13 at 0:13
add comment

Your Answer

 
discard

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.