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!

  • 1
    $\begingroup$ How you came to the number of 144 edges? $\endgroup$
    – M. Winter
    Jan 27, 2020 at 13:33
  • $\begingroup$ 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). $\endgroup$
    – Ruy
    Jan 27, 2020 at 14:47

1 Answer 1


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).

  • $\begingroup$ 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? $\endgroup$
    – Ruy
    Jan 27, 2020 at 14:49
  • 3
    $\begingroup$ @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. $\endgroup$
    – M. Winter
    Jan 27, 2020 at 14:51
  • $\begingroup$ 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 $(\pm1/2,\pm1/2,\pm1/2,\pm1/2)$ $\endgroup$ Jan 27, 2020 at 15:38
  • $\begingroup$ 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. $\endgroup$
    – IJL
    Jan 27, 2020 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.