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!



What is the convex hull of the quaternionic symmetries of the 3 dimensional cube?
This is the [disphenoidal 288-cell](https://en.wikipedia.org/wiki/Truncated_24-cells#Disphenoidal_288-cell), which is the dual of the [bitruncated 24-cell](https://en.wikipedia.org/wiki/Truncated_24-cells#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).


