# Does there exist a highly counterintuitive space group?

Has anyone ever proved or disproved the existence of a subgroup of the group of all distance preserving operations on R3 under the operation of composition that satisfies the following properties?

• The only translation in that subgroup is the identity transformation.
• There exists a point in R3 and a positive real number r such that every open disc of radius r contains at least 1 and only finitely many points from that set of all points that are an image of that point under one of the transformations in that subgroup.

If it's already been proven that there are 219 space groups, does that mean what I'm describing is not a space group because the existence of a unit cell is a requirement of a space group?

Your question is not well written and has several possible interpretations. Here is one:

Does there exist a subgroup $G$ of $R^3\rtimes O(3)$ (containing no translations), a point $p\in R^3$ and a number $r>0$ such that for every $q\in R^3$ the intersection $B(q,r)\cap G\cdot p$ is finite and nonempty?

Lemma. Such subgroups do not exist. Proof. Suppose that $G$ is such a group. Then $G$ has to be discrete (the finiteness assumption on $B(q,r)\cap G\cdot p$). If $G$ is crystallographic, i.e., $R^3/G$ is compact, then $G$ contains nontrivial translations. The alternative is that $G$ either has an invariant line $L$ or fixed point in $R^3$. Consider the first case. Then $G\cdot p$ is contained in $R$-neighborhood of $L$. Therefore, for arbitrary $r>0$, take $q$ such that $d(q, L)>r+R$. Then $B(q,r)$ contains no points from $G\cdot p$.

If $G$ has a fixed point $a$, then $G\cdot p$ is contained in the $R$-ball $B(a,R)$ for some $R$. Now, take $q$ such that $d(q,a)>R+r$. Again a contradiction. qed

A side remark: Every group of isometries of $R^3$ which contains no translations either fixes a point in $R^3$ or has an invariant line in $R^3$. One can ask about groups which contain no skew translations (isometries without fixed points). Then such subgroup of isometric motions of $R^3$ has to fix a point in $R^3$. Hence, it is conjugate to a subgroup of $O(3)$. Interestingly enough, there are isometry groups of $R^4$ which contain no (skew) translations but are not conjugate to subgroups of $O(4)$. (A proof that I know uses the fact that $U(2)$ contains free nonabelian subgroups.)

Edit: Here's the answer I should have written (after initially screwing it up, and taking a subsequent trip to the bunny planet; thanks to Studiosus for pointing out the error).

It is possible to use the classification of space groups if you add one more hypothesis to your two bullet points:

• Cocompactness: There exists a compact set that has nonempty intersection with each orbit of the group action. (This is equivalent to the existence of a compact "fundamental domain", which I believe is the same as what you call a "unit cell" although one should be careful about whether compactness is a requirement of a unit cell.)

If you add the cocompactness hypothesis then no such groups exist: cocompactness, together with your second bullet point, and together with an application of the Bieberbach theorems in dimension 3, implies that the group is one of the space groups, and that it has a rank 3 translation subgroup, contradicting your first bullet point.

If you do not require cocompactness then there are still things to say. And as Studiosus points out in his comment there are indeed noncocompact examples satisfying your two bullet points.

A group of operations that you describe in your second bullet point is called a "discrete subgroup of isometries of $\mathbb{R}^3$". Let me use a notation $\Gamma$ for such a group. The main theorem about this situation is as follows:

• There exists an affine subspace (a "hyperplane") $P \subset \mathbb{R}^n$ of some dimension $d \in \{0,1,2,3\}$ such that $P$ is invariant under $\Gamma$, and such that the restriction of $\Gamma$ to $P$ is a discrete and cocompact group (i.e. a "crystallographic group" of dimension $d$). However, elements of $\Gamma$ may also have components acting in the directions orthogonal to $P$.
• If $d = 0$ then $P$ is a point fixed by $\Gamma$ and $\Gamma$ is finite.
• If $d = 1$ then $P$ is a line preserved by $\Gamma$ and the restriction of $\Gamma$ to $P$ is one of the two 1-dimensional "crystallographic groups", an infinite cyclic translation group or an infinite dihedral group. However, elements of $\Gamma$ may also a rotational component around the linking circle of the line $P$, hence the possibility that $\Gamma$ is an infinite cyclic group with a "skew motion" as mentioned in Studiosus' comment.
• If $d=2$ then $P$ is a plane preserved by $\Gamma$, and the restriction of $\Gamma$ to $P$ is one of the seventeen 2-dimensional crystallographic groups, also known as wallpaper groups. An element of $\Gamma$ may have a reflectional component across the plane $P$, but $\Gamma$ has a subgroup of index at most 2 with no reflectional component, and that subgroup contains a rank 2 abelian group of translations.
• If $d=3$ then $P=\mathbb{R}^3$ and as said earlier $\Gamma$ is one of the two hundred and nineteen 3-dimensional crystallographic groups, also known as space groups, each of which has a rank 3 abelian group of translations.

And, by the way, the obvious generalization of this discussion to all dimensions is true. The general theorems along these lines go under the names of the "Bieberbach Theorems".

• Lee: what is your definition of a translation? With the standard def, theorem you quoted is false (cyclic group of skew motions). – Moishe Kohan May 3 '14 at 16:48
• Oops, good point. I will have to correct this, although I cannot do it right now. – Lee Mosher May 3 '14 at 17:27