# What is the convex hull of the quaternionic symmetries of the 3 dimensional cube?

It is well known that there are exactly five 3-dimensional regular convex polyhedra, known as the Platonic solids. In 1852 the Swiss mathematician Ludwig Schlafli found that there are exactly six regular convex 4-polytopes (the generalization of polyhedra to 4 dimensions) and that, for dimensions 5 and above, there are only three!

The six regular 4-polytopes are:

NAME                    VERTEXES   EDGES   FACES  CELLS
Hypertetrahedron              5      10      10      5
Hypercube                    16      32      24      8
Hyperoctahedron               8      24      32     16
24-cell                      24      96      96     24
Hyperdodecahedron           600    1200     720    120
Hypericosahedron            120     720    1200    600


The easiest ones to be described are the first two: a model for the hypertetrahedron may be obtained as the convex hull of the canonical basis in $$\mathbb R^5$$ (hence a 4-dimensional object), while a model for the hypercube is the Cartesian product $$[0, 1]\times[0, 1]\times[0, 1]\times[0, 1]$$.

As in the case of 3 dimensions, the dual of a regular 4-polytope is also a regular 4-polytope and it turns out that the six regular 4-polytopes found by Schlafli are related to each other via duality as follows.

Hypetetrahedron    <->   Itself
24-cell            <->   Itself
Hypercube          <->   Hyperoctahedron
Hyperdodecahedron  <->   Hypericosahedron


This means that one needs only describe the 24-cell and the hypericosahedron for all of them to be known. In other words:

Hypertetrahedron    =  convex hull of the canonical basis in 5 dimensions
Hypercube           =  [0,1]x[0,1]x[0,1]x[0,1]
Hyperoctahedron     =  dual of the hypercube
Hyperdodecahedron   =  dual of the hypericosahedron
24-cell                ???
Hypericosahedron       ???


The description of the last two 4-polytopes above may be obtained by considering the quaternions $$\mathbb H$$. Viewing $$\mathbb R^3$$ within $$\mathbb H$$ via the map $$(x,y,z)\mapsto xi+yj+zk,$$ it is well known that every quaternion $$q$$, with $$\Vert q\Vert=1$$, gives a rotation $$R_q$$ on $$\mathbb R^3$$ via the formula $$R_q(v) = qvq^{-1}, \quad \forall v \in \mathbb R^3.$$

In fact the correspondence $$q\mapsto R_q$$ is a two-fold covering of $$SO(3)$$ by the unit sphere in $$\mathbb H$$.

Letting $$P_{20}$$ be the icosahedron in $$\mathbb R^3$$, consider the quaternionic symmetries of $$P_{20}$$, which I will write as $$\mathbb {HS}(P_{20})$$, defined to be the set of all unit quaternions $$q$$ such that $$R_q$$ leaves $$P_{20}$$ invariant. In symbols $$\mathbb {HS}(P_{20}) =\{q\in \mathbb H: \Vert q\Vert=1,\ R_q(P_{20})=P_{20}\}.$$ Well, the convex hull of $$\mathbb {HS}(P_{20})$$ in $$\mathbb R^4$$ turns out to be a model for the hypericosahedron!

Since the symmetries of a regular polyhedron are the same as the symmetries of its dual, it is clear that the symmetries of $$P_{12}$$, the dodecahedron, gives nothing new: the convex hull of $$\mathbb {HS}(P_{12})$$ is just another model for the hypericosahedron.

Passing to the (self dual) tetrahedron, call it $$P_4$$, the convex hull of $$\mathbb {HS}(P_{4})$$ gives a model for the remaining 4-polytope, namely the 24-cell, completing the description of the six Schlafli's 4-polytopes.

Question: What is the convex hull of the quaternionic symmetries of the 3 dimensional cube?

If I am not mistaken, this 4-polytope has 48 vertexes and 144 edges, so it is not in Schlafli's list and hence cannot be regular.

EDIT: Yes I was mistaken about the number of edges which is in fact 336 according to M. Winter's answer below!

• How you came to the number of 144 edges? Jan 27, 2020 at 13:33
• I found the vertices numerically, computed the minimum distance among them, and checked how many pairs share that distance. I guess this is consistent with your answer (below).
– Ruy
Jan 27, 2020 at 14:47

This is the disphenoidal 288-cell, which is the dual of the bitruncated 24-cell. This is also mentioned in the "Geometry" section of the Wikipedia article on the 288-cell.

It has 48 vertices, and 336 edges. However, 144 of these are of the shortest length, and I suppose you have counted these.

The symmetry group of the tetrahedron is "the half" of the symmetry group of the cube (as the tetrahedron is the 3-dimensional demicube). You already know that the tetrahedral symmetries give you the 24-cell. In the same way, the 24-cell is "the half" of the disphenoidal 288-cell: the latter is the convex hull of the union of a 24-cell and its dual (which is a 24-cell as well, but differently oriented).

• Thanks for your quick and thorough answer! What puzzled me in the first place is that, since the cube is so regular, why wouldn't its quaternionic symmetries also give a regular 4-polytope?
– Ruy
Jan 27, 2020 at 14:49
• @Ruy To be honest, I am more amazed by the fact that the other two actually give regular polytopes. At the moment, I see no a priori reason for why the resulting polytope should have such an exceptionally high symmetry. On the other hand, in low-dimensional geometry many coincidences happen. Jan 27, 2020 at 14:51
• Explicitly, the vertices are: eight vertices with one coordinate $\pm1$ and all others zero; 24 vertices with two coordinates $\pm1/\sqrt2$ and two others zero; and sixteen vertices $(\pm1/2,\pm1/2,\pm1/2,\pm1/2)$ Jan 27, 2020 at 15:38
• If $G$ is a non-cyclic subgroup of the quaternions of order $n$ containing $-1$, the action of $G\times G$ on the set $G$ has kernel of order 2. The map $g\mapsto g^{-1}$ also permutes the set $G$. These together give $n^2$ symmetries for the polytope defined as the convex hull of $G$. A polytope is regular if and only if it has the same number of full flags as symmetries. In the case when $G$ is the group of symmetries of the cube, there are too many flags for it to be regular despite having $2304=48^2$ symmetries.
– IJL
Jan 27, 2020 at 17:11