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!