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This is a variation on my other question. A hyperbolic isometry group may be discrete and not finitely generated. What about the Euclidean case?


The generating set $S$ may contain an infinite number of translations, rotations, reflections, rotoreflections, and so on. If it contains one orientation-reversing isometry $f$, then any other orientation-reversing isometry $g$ may be composed with it to produce an orientation-preserving isometry $f\circ g$, which can replace $g$ as a generator. So, without loss of generality, $S$ contains no more than $1$ orientation-reversing isometry $f$. The other generators will be translations, rotations, and compositions of such.


Here's a proof that a discrete group's translation subgroup must be finitely generated.

Assume that $S$ contains an infinite number of translation vectors, $S\supset\{v_k\}$, which span a finite-dimensional Euclidean space $\mathbb E^n$. Take a basis for $\mathbb E^n$ from $S$, and label them $v_1,v_2,v_3,\cdots,v_n$. These determine a parallelotope, a "unit cube" in a different coordinate system. Any other vector $v_{n+k}$ (with $k>0$) may be replaced with its translate into the unit cube; if

$$v_{n+k} = c_1v_1+c_2v_2+\cdots+c_nv_n$$

and any coefficient $c$ is not in $[0,1)$, then it may be replaced with

$${v_{n+k}}' = v_{n+k}-\lfloor c_1\rfloor v_1-\lfloor c_2\rfloor v_2-\cdots-\lfloor c_n\rfloor v_n$$ $$= (c_1-\lfloor c_1\rfloor)v_1+(c_2-\lfloor c_2\rfloor)v_2+\cdots+(c_n-\lfloor c_n\rfloor)v_n$$

which does have coefficients in $[0,1)$. (Also, any ${v_{n+k}}'=\vec0$ may be removed from $S$.)

Now, the definition of "discrete isometry group" requires that any point $p\in\mathbb E^n$ has a neighbourhood disjoint from the other points $g(p)$ that the group sends it to. This means there is a ball of radius $\epsilon=\epsilon(p)>0$ centred on $p$ which does not contain any $p+v_k$, where $v_k\in S$. Equivalently, every point $p+v_k$ has a ball of radius $\epsilon/2$, and these balls must not intersect. Every such point is within the unit cube (translated to $p$). Each ball has finite volume (proportional to $\epsilon^n$), and is contained in the unit cube's expansion by $\epsilon$, which also has finite volume. So if there are infinitely many vectors $v_k\in S$, then there will be infinitely many $\epsilon/2$-balls within the expanded unit cube, and infinite volume. Contradiction.

So the group cannot be both discrete and infinitely generated by translations. It may still be infinitely generated by other isometries.


I tried to apply a similar argument to rotations around the origin.

Take any point on the unit sphere $p\in\mathbb S^{n-1}\subset\mathbb E^n$, and consider its rotations $r_k(p)$ by generators $r_k\in S$. All of the $\epsilon$-balls centred at these points will be contained in the expanded unit sphere with radius $1+\epsilon$. As before, if the balls are required to be disjoint, and there are infinitely many rotations $r_k$ (and an infinite subset send $p$ to different points $r_k(p)\neq p$) then there would be infinite volume within the radius $1+\epsilon$, a contradiction. But there is a complication.

There may be infinitely many rotations for which $r_k(p)=p$ is on the axis, and only finitely many distinct points on $\mathbb S^{n-1}$. This should be easy to get around, but I don't see how.


A complete answer could involve fundamental domains or generalized screw theory or something.

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  • $\begingroup$ I am not sure what you mean by "Euclidean isometry group" here. Do you mean the isometry group of a polyhedron in $\mathbb{R]^n$? A convex polyhedron? A closed flat manifold? $\endgroup$ – Qiaochu Yuan Nov 2 '18 at 20:05
  • $\begingroup$ A subgroup of the group of distance-preserving transformations of Euclidean space. en.wikipedia.org/wiki/Isometry_group $\endgroup$ – mr_e_man Nov 2 '18 at 21:12
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    $\begingroup$ The group $SO(n)$ of rotations in $n$-dimensional space is compact, so any discrete subset is finite. $\endgroup$ – Lukas Geyer Nov 5 '18 at 6:37
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The comment of Lukas Geyer about compactness is right, and it is essentially what you wanted to prove. You were in fact almost done : if there are infinately many distinct rotations fixing your chosen point $p$, then these stabilize the orthogonal hyperplane $\{p\}^\perp$. So you're left with infinitely many distinct rotations centered at the origin in $\{p\}^\perp$ : this is the same problem, but the dimension is $1$ less. Then you can reason by induction on the dimension, begining at dimension $2$ (which is the easy case), to show that a discrete group of rotations must be finite (indeed, that $SO(n)$ is compact).

The answer to your general question is much harder to tackle with such elementary methods, I think. The answer is no, thanks to Bieberbach's first theorem, which says in particular that the discrete subgroups of the euclidian group must be finitely generated, and says much more about their structure. In fact, if $G$ is a subgroup of the euclidian group $O(n) \ltimes \mathbb R^n$, then the subgroup $G \cap \mathbb R^n$ of translations that are in $G$ is an abelian finitely generated subgroup (as you showed). The quotient $G/(G \cap \mathbb R^n)$ is a subgroup of the orthogonal group $(O(n) \ltimes \mathbb R^n)/\mathbb R^n \cong O(n)$, and the hard part is to show that is has to be discrete, hence (as the above reasoning shows) finite. Then, choosing generators of this finite group, that you lift to elements of $G$, and adding generators of $G \cap \mathbb R^n$ gives you a finite set of generators of $G$. But this shows in fact that $G$ has a free abelian group of finite index, which is a much stronger result. There is a lot of litterature on crystallographic groups, that you should explore. See for instance this related question and the references given there.

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  • $\begingroup$ Thanks. This rules out infinite translations, and infinite rotations around one point. What about helical / screw motions, or rotations around different points? $\endgroup$ – mr_e_man Nov 5 '18 at 20:53
  • $\begingroup$ Sorry, I forgot that the bounty is halved when I don't award it manually. $\endgroup$ – mr_e_man Nov 10 '18 at 20:27
  • $\begingroup$ No problem ! It was my pleasure answering to your question. :) $\endgroup$ – J. Darné Nov 11 '18 at 1:06

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