Alternative Method of Complexifying the Tetrahedron

Aug 17—2026

IMPORTANT: this method of complexification will not result in the $S_4 \times \mathbb{Z}_2 \cong B_3$ group. Let’s see what group this complexification will lead to.

Again, we begin by considering the tetrahedron not in $\mathbb{R}^3$ , but in $\mathbb{RP}^2$ . In this space, we replace the tetrahedron with the 4 points, i.e. in $\mathbb{R}^3$ , these will be the lines passing through the 4 vertices of the tetrahedron.

If we identify $\mathbb{CP}^1$ with $S^2$ , then it is easy to see the double cover $\mathbb{RP}^2 \leftarrow \mathbb{CP}^1$ obtained by identifying the antipodal points on $S^1$ as the same point. Denote $\Psi: SU(2) \rightarrow SO(3)$ to be the double cover.

Thus, the 4 points in $\mathbb{RP}^2$ become 8 points on the sphere ( $\mathbb{CP}^1$ ). It is easy to see that these 8 points happen to be the vertices of the cube in which the initial tetrahedron is inscribed.

Key: $SU(2) / < \pm I > \cong SO(3)$ . The action of an element of $SU(2) / < \pm I >$ on $\mathbb{CP}^1$ can be interpreted as a rotation of $S^2$ . (Will post this separately)

The group of 3-dimensional rotations, say $G$ , that maps the set of 8 vertices of the cube to itself is isomorphic to $S_4$ . Now, let’s look at the pre-image of this group under $\Psi$ , $\Psi^{-1}(G)$ , in $SU(2)$ .

Since $\Psi$ is a double cover, $\Psi^{-1}(G)$ has order 48. The kernel of $\Psi$ is isomorphic to $\mathbb{Z}_2$ .

How do we know that $\Psi^{-1}(G)$ is not isomorphic to $S_4 \times \mathbb{Z}_2$ ?

The easiest way to see this is to look at the elements of order 2.

Claim: $\Psi^{-1}(G)$ has only one element of order 2.

Let $A \in \Psi^{-1}(G)$ such that $A^2 = I$ and $A \neq I$ . Since A is unitary, it is diagonalizable and its eigenvalues must satisfy $\lambda^2 = 1$ . Thus, its eigenvalues are among {1, -1}.

By definition, we also know that $\text{det}(A) = 1$ . Since $A$ is a 2×2 matrix, it follows that its 2 eigenvalues can either be: (1, 1) or (-1, -1).

Thus, $A = I$ or $A = -I$ . Since $A \neq I$ by assumption, $A = -I$ . This proves the claim.

$S_4 \times \mathbb{Z}_2$ has more than one element of order 2. In $S_4$ , there are 9 elements of order 2 (easy to see when we identify $S_4$ as the group of rotations of the cube). Every element $(g, e) \in S_4 \times \mathbb{Z}_2$ if and only if $g \in S_4$ has order 2, regardless of $e \in \mathbb{Z}_2$ . Thus, there are a total of 18 elements of order 2 in $S_4 \times \mathbb{Z}_2$ .