Isometries and fixed points I am new to geometric group theory and I am trying to read a bit to expand my horizons. I have encountered the following theorem: Suppose that $G$ is a group that has a free action by isometries on $\mathbb{R}^n$ with the Euclidean metric. Then $G$ is torsion free. I am wondering about generalisations of it:


*

*I think it follows from the proof that if an isometry (that is it the cyclic group generated by it) acts on $\mathbb{R}^n$ and it has a finite orbit, then it has a fixed point. Am I right?

*I saw somewhere that the theorem holds if one can replace isometry by a continuous function. I think I can prove it for $n=1$ because then a convex set is the same as a connected set, namely, an interval. Does anyone know a reference for the proof?

*I assume the theorem fails if $\mathbb{R}$ is replaced by $\mathbb{Q}$. Can anyone give a counter example?

*What happens if $\mathbb{R}$ is replaced by the $p$-adics?

*Is there a version in characteristic $p$, but maybe instead of torsion free you cannot have orders that are coprime to $p$?
 A: An isometry must be an affine-linear map.  If it has a finite orbit
then the average of the orbit is a fixed point.  This works equally over
$\bf R$ or over $\bf Q$, and should answer questions 1 and 3.
Over ${\bf Q}_p$, if we limit to affine-linear maps, the same argument
works, as it does in characteristic $p$ for elements order coprime to $p$
(a necessary hypothesis because translations have order $p$).  This would
also answer questions 4 and 5.  But vector spaces over ${\bf Q}_p$ 
and over finite fields can have isometries that are not affine-linear, 
and such an isometry might fail to have a fixed point whether or not
it has order coprime to $p$.
Question 2 is of a rather different flavor.  I see that Lee Mosher
gave a cohomological proof in the comments.
A: Let isometry $\ F:\mathbb R^n\rightarrow \mathbb R^n\ $ and
$\ a\in \mathbb R^n\ $ be such that the $t$-fold composition $\ F^t(a)=a\ $ brings $\ a\ $ back to itself for a positive integer $\ t.\ $ Remember that $\ F\ $ is affine. Thus
$$ b\ :=\ \frac{\sum_{s=0}^{t-1}F^s(a)}t $$
is a fixed point, $\ F(b)=b.$
A: There is a non-trivial generalization of the result you mention to a whole class of Riemannian manifolds called the Cartan-Hadamard theorem https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem. A very nice proof is given in the book of P. Petersen, Riemannian geometry.
