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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?

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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 $R\mathbb P^\infty$.

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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
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
@QiaochuYuan : Do you know of a reference of the stated facts? Or can you briefly describe how it is proved? (Since the answerer is not active I ask you instead) – John Ma yesterday

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