# Most symmetric triangulation of $\mathbb{R}^n$

Consider the $n$-dimensional Euclidean space with its standart metric. It is well known that it admits a triangulation by regular simplices only if $n\le 2$. So let us consider less regular triangulations. Lets call such a triangulation $T$ symmetric if the group of isomoetries of $\mathbb{R}^n$ that preserve $T$ acts transitively on the set of $n$-dimensional simplices.

For any $n$-simplex $X$ we can then consider the subgroup of those isometries that map the simplex to itself. It permutes the vertices of the simplex and hence is a subgroup of the symmetric group $\Sigma_{n+1}$. If it was the entire group (or even if it acts transtively on the set of edges) the simplex would be regular. So this cannot be the case.

What is the biggest group that can occur this way? (for fixed $n\ge 3$) and how does the correspoonding triangulation look like ? I would consider such a triangulation an analogue of the tesselation of the Euclidean 2-plane with regular triangles. So it is interesting to wonder how they look like.

• Hi Henrik. Could you give me a reference for the fact that $\mathbb{R}^n$ cannot be triangulated with regular simplices if $n > 2$? Thanks. – yasmar Oct 30 '15 at 1:04
• Never mind, I've got it. The dihedral angles of a regular $n$-simplex are given by $\cos^{-1}(1/n)$. This can be seen by observing that the foot of an altitude is at the barycentre of the facet. If $n>2$, then $2\pi$ is not an integer multiple of this angle. – yasmar Oct 30 '15 at 3:50

## 1 Answer

There are five known space-filling tetrahedral families. It may be a bit hard to decide which has the greatest isometry group. A regular tetrahedron has a symmetry group of order 24 including reflections.

Conway and Torquato wrote a 2006 article Packing, tiling, and covering with tetrahedra which addresses among other things the question "What are the “closest-to-regular” tetrahedra that will tile 3-space?". Several meanings of "closest to regular" are considered, including a minimum ratio of longest to shortest edge, and a minimum difference between greatest and least angles between edges.

In any case there are families (see "Scottish irregulars" in the above linked article) where the individual tetrahedra are fixed by eight symmetries of the whole tessellation.

According to the Wikipedia article Tetrahedron there are seven possibilities for the isometry group of an irregular tetrahedron, with the maximum order being eight. This occurs when the four sides are congruent isoceles triangles.

Added: I learned about the Coxeter simplex $A_3$ which illustrates the above symmetry group of order 8 from a 2003 paper by Warren D. Smith, Pythagorean triples, rational angles, and space-filling simplices. Here's a write-up adapted from a sci.math post I made in 2009.

The construction gives a 1-parameter tiling of 3D by tiling the plane with unit equilateral triangles and then erecting prisms (of infinite length) perpendicular to the plane based on those triangles. Going cyclically around the vertices of one triangle we mark points at heights $na, (n+1)a, (n+2)a, (n+3)a$ along the lines rising perpendicular to those three vertices.

Note that the last marked point is at distance 3a over the first. These four points are vertices of a tetrahedron with edge lengths 3a,b,b,b,c,c where:

$$b^2 = a^2 + 1 \;\; \mathscr{and} \;\; c^2 = 4a^2 + 1$$

The construction continues to fill the prism by adding the next simplex among corresponding points at heights $(n+1)a, (n+2)a, (n+3)a,$ and $(n+4)a,$ with each successive simplex sharing a triangular face with the preceding one. It is clear the simplexes are congruent and fill the prism, so that space can be filled by a like filling of each prism.

If $a$ is chosen so that $b = 3a$, i.e. $a = 1/\sqrt{8}$, then the resulting prism is Coxeter's $A_3$, a "maximally symmetric" member of the 1-parameter family with all four faces congruent.

At the time I speculated that the solution in higher dimensions would be Coxeter simplex $A_n$, but I haven't yet taken the time to revisit that.