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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:

  1. 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?

  2. 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?

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

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

  5. Is there a version in characteristic $p$, but maybe instead of torsion free you cannot have orders that are coprime to $p$?

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    $\begingroup$ Regarding question 2: Every nontrivial, finite, cyclic group $G$ has infinite cohomological dimension (proof: compute the cohomology of the quotient of $S^\infty$ by a free action of $G$). It follows that $G$ has no free action on any contractible space $X$ of finite dimension, because if it did then the cohomology of $G$ would be isomorphic to the cohomology of $X/G$ which has only finitely many nontrivial dimensions. And from this it follows that there does not exist a free action of a non-(torsion-free) group on a contractible space of finite dimension. $\endgroup$ – Lee Mosher Jun 4 '18 at 17:10
  • $\begingroup$ @LeeMosher Thanks! I am afraid that cohomological arguments are beyond me. It feels to me that such claims should be elementary, but this is probably wishful thinking. $\endgroup$ – Yiftach Barnea Jun 4 '18 at 17:49
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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.

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  • $\begingroup$ Noam TODA. Once you have said isometries are affine-linear everything became easy. $\endgroup$ – Yiftach Barnea Jun 4 '18 at 16:55
  • $\begingroup$ You may like to know that all Banach spaces have all their self-isometries affine. $\endgroup$ – Wlod AA Jun 4 '18 at 17:04
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    $\begingroup$ Y.Barnea -- LABRIUT. I see, though, that ${\bf Q}_p$ can have non-affine isometries, because the $p$-adic digits can be permuted arbitrarily -- and this yields counterexamples if the permutation is not cyclic but has no fixed points. So then I ask whether you meant arbitrary isometries or only affine-linear ones in that case (and also mod $p$). @Wlod AA: For real Banach spaces, yes; complex ones, not necessarily: there can be conjugate-linear isometries, and even more for inner-product spaces. $\endgroup$ – Noam D. Elkies Jun 4 '18 at 17:33
  • $\begingroup$ @NoamD.Elkies Nice! I think the proof that every isometry in $\mathbb{R}^n$ is affine-linear follows from the fact that if you fix $n+1$ points not contained in a hyperplane, then any other point is determined by its distances from these points. So I guess this fails for $\mathbb{Q}_p$. Can you see an example? $\endgroup$ – Yiftach Barnea Jun 4 '18 at 17:43
  • $\begingroup$ @NoamD.Elkies I was more interested in arbitrary isometry. $\endgroup$ – Yiftach Barnea Jun 4 '18 at 17:44
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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, $\ F(b)=b.$

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  • $\begingroup$ why this is more general? $\endgroup$ – Yiftach Barnea Jun 4 '18 at 16:56
  • $\begingroup$ $F\ $ from my answer doesn't have to belong to the torsion group (its order doesn't have to be finite). $\endgroup$ – Wlod AA Jun 4 '18 at 17:00
  • $\begingroup$ but that exactly what I have said, that you only need a finite orbit. $\endgroup$ – Yiftach Barnea Jun 4 '18 at 17:03
  • $\begingroup$ Thank you. I'll remove "more generally". $\endgroup$ – Wlod AA Jun 4 '18 at 17:07
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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.

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  • $\begingroup$ Sorry, this far from my background. Could you please explian what is the connection beween the result and this theorem? $\endgroup$ – Yiftach Barnea Jun 5 '18 at 12:45
  • $\begingroup$ Let me take a simple example. The hyperbolic space $\mathbb{H}^n$ is defined as the hypersurface $\{ -(x_0)^2 + (x_1)^2 + \ldots + (x_n)^2 = -1, x_0 > 0\}$ in $\mathbb{R}^{n+1}$ endowed with the quadratic form $\eta = -(dx_0)^2 + (dx_1)^2 + \ldots + (dx_n)^2$. This is the "negatively curved" model of Riemannian geometry, the other ones being the flat one $\mathbb{R}^n$ and the positively curved one $\mathbb{S}^n$ (the unit sphere in $\mathbb{R}^{n+1}$ with the standard metric). The group of isometries of $\mathbb{H}^n$ is the connected component of the identity of $O(n, 1)$. $\endgroup$ – Romain Gicquaud Jun 7 '18 at 14:50
  • $\begingroup$ ($O(n, 1)$ is the isometry group of $\eta$ which has signature $(n, 1)$). And your theorem is true in this setting: if there exists a finite orbit for the action of a subgroup of isometries, then there exists a fixed point. You can read more about the geometry of the hyperbolic space and subgroups of isometries in the book of R. Benedetti and C. Petronio, Lectures on Hyperbolic Geometry. This might be more manageable if you have no a priori knowledge of Riemannian geometry. $\endgroup$ – Romain Gicquaud Jun 7 '18 at 14:59
  • $\begingroup$ Something you should also look at is the fundamental group which is the "simplest" construction you can do in algebraic topology. This sets up the framework to study discrete groups acting on a topological space. A standard reference to this is the book of A. Hatcher, Algebraic topology. $\endgroup$ – Romain Gicquaud Jun 7 '18 at 15:02

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