It is well known that there are exactly five regular three-dimensional convex polyhedra, called *Platonic solids*.

In 1852 the Swiss mathematician Ludwig Schlafli discovered that there were exactly six regular 4 convex polytopes (the generalization of

4-dimensional polyhedra) and that, for dimensions 5 and above, there are only three!

The six 4 regular 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 simplest to describe are the first two:

a model for the hypertetrahedron can be obtained as the convex shell of the canonical base 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 regular six

The 4-polytopes found by Schlafli are linked to each other via duality as follows.

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

This means that it is enough to describe the 24 cells and the hypericosahedron to be all 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 can be obtained by considering the quaternions $ mathbb Q $.

visualization $ mathbb R ^ 3 $ in $ mathbb Q $ via the card

$$ (x, y, z) mapsto xi + yj + zk, $$

it is well known that each quaternion

$ q $, with $ Green q Green = 1 $, give a rotation $ R_q $ sure $ 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 double coating of $ N / A (3) $ by the unitary sphere $ mathbb Q $.

leasing $ P_ {20} $ be the icosahedron $ mathbb R ^ 3 $, Take into account *quaternionic symmetries* of $ P_ {20} $, which I will write as

$ mathbb {QS} (P_ {20}) $, defined as the

set of all unitary quaternions $ q $ such as $ R_q $ leaves $ P_ {20} $

invariant. In symbols

$$

mathbb {QS} (P_ {20}) = {q in mathbb Q: Vert q Vert = 1, R_q (P_ {20}) = P_ {20} }.

$$

Well, the convex shell of $ mathbb {QS} (P_ {20}) $ in $ mathbb R ^ 4 $ proves 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, is nothing new: the convex shell of $ mathbb {QS} (P_ {12}) $ is just another model for the

hypericosaèdre.

Go to the tetrahedron (double auto), call it $ P_4 $, the convex shell of $ mathbb {QS} (P_ {4}) $ give a model for the

4 remaining polytopes, namely the 24 cells, completing the description of the six 4 Schlafli polytopes.

**Question**: What is the convex shell of the quaternion symmetries of the 3-dimensional cube?

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