This picture shows the Binary Tetrahedral Group
embedded as a subgroup of the matrix group SU(2).
SU(2) is geometrically a 3-dimensional sphere of
radius one; here we see a conformal projection of that
sphere into 3-space. The point 1 is at the intersection of the three black circles of longitude. The six points $\pm {\bf i}, {\bf j}, {\bf k}$ at the intersections of these circles with the yellow equatorial sphere. Eight other group points lie inside the sphere, at a distance $\frac{\pi}{3}$ from 1. They are $a,b,c,d$ (colored violet) and $-a^2,-b^2,-c^2,-d^2$. Six of the nine points outside the equatorial sphere are visible here. They appear larger because they are closer to the projection point (near $-1$). The image of the point $-1$ itself has been suppressed. It would occupy the whole background. This image was produced, using GeomView modules developed at the Geometry Center, by Summer Institute student Rebecca Frankel. |

The *Tetrahedral Group* is the group of orientation-
preserving symmetries
of an equilateral tetrahedron.
If the vertices of the tetrahedron are labeled A,B,C,D, each
of the symmetries may be represented as a permutation of
these four symbols. Each even permutation corresponds to
an orientation-preserving symmetry. The twelve are

- the identity
- eight rotations about axes going through a vertex and
the opposite face. There are four vertices; each rotation can be
by 120$^{\circ}$ or 240$^{\circ}$

(ABC), (ACB), (ABD), (ADB), (ACD), (ADC), (BCD), (BDC), - three 180$^{\circ}$ rotations about axes joining
the midpoints of two opposite edges:

(AB)(CD), (AC)(BD), (AD)(BC).

The Special Unitary group SU(2) consists of
2 by 2 complex matrices of the form
$$
\left ( \begin{array}{cc} (x+iy) & (z+iw)\\(-z+iw)&(x-iy)\end{array}\right )
$$
with determinant $x^2 + y^2 + z^2 + w^2 = 1$.
There is a group homomorphism SU(2) $\rightarrow$ SO(3)
which can be described in terms of $x,y,z,w$ but which
can be more geometrically defined using *quaternions*.
A quaternion can be identified with a matrix like the
one above, but without the determinant condition
in general. The quaternionic notation for that matrix
is ${\bf 1}x + {\bf i}y + {\bf j}z + {\bf k}w$,
where ${\bf 1},{\bf i}, {\bf j}, {\bf k}$ correspond to the matrices
$$\left ( \begin{array}{cc}1&0\\0&1\end{array}\right ),
\left ( \begin{array}{cc}i&0\\0&-i\end{array}\right ),
\left ( \begin{array}{cc}0&1\\-1&0\end{array}\right ),
\left ( \begin{array}{cc}0&i\\i&0\end{array}\right )$$
respectively. The rules for multiplying quaternions
follow from the rules for matrix multiplication;
equivalently: ${\bf i}^2 = {\bf j}^2 = {\bf k}^2 = -{\bf 1}$, ${\bf i}{\bf j} = -{\bf j}{\bf i} = {\bf k},$
${\bf j}{\bf k} = -{\bf k}{\bf j} = {\bf i}$, ${\bf k}{\bf i} = -{\bf i}{\bf k} = {\bf j}$, extended linearly.
The condition $x^2 + y^2 + z^2 + w^2 = 1$ defines
a * unit quaternion*. Since the matrix corresponding
to the unit quaternion
${\bf 1}x + {\bf i}y + {\bf j}z + {\bf k}w$
has unit determinant, its inverse
is exactly
$$
\left ( \begin{array}{cc} (x-iy) & (-z-iw)\\(z-iw)&(x+iy)\end{array}\right )
$$
so the multiplicative
inverse of the unit quaternion ${\bf 1}x + {\bf i}y + {\bf j}z + {\bf k}w$
is ${\bf 1}x - {\bf i}y - {\bf j}z - {\bf k}w$.

The vector $(a,b,c)$ in ${\bf R}^3$ can be identified with
the *pure imaginary* quaternion $a{\bf i} + b{\bf j} + c{\bf k}$;
then each quaternion ${\bf q}$ defines a linear map
$L({\bf q}): {\bf R}^3 \rightarrow {\bf R}^3$ by $L({\bf q})(a{\bf i} + b{\bf j} + c{\bf k}) =
{\bf q} \times (a{\bf i} + b{\bf j} + c{\bf k}) \times {\bf q}^{-1}$, using quaternionic
multiplication. It is straightforward to check
that $L({\bf q})$ is also pure imaginary; the fact that
the determinant of a product is the product of the
determinants then guarantees that $L({\bf q})$ is a
length-preserving, and therefore orthogonal,
transformation; the question of orientation
can be answered by noting that $L(1)$ is the
identity and that SU(2) is connected. Also,
the definition yields immediately that $L$ is
a homomorphism from SU(2) to SO(3):
$L({\bf pq}) = L({\bf p})L({\bf q})$ for any
two unit quaternions ${\bf p}$ and ${\bf q}$. Finally,
two unit quaternions ${\bf p}$ and ${\bf q}$ give the same SO(3)
element if and only if ${\bf p} = \pm {\bf q}$; the condition
is equivalent to the three equations $L({\bf q}^{-1}{\bf p}){\bf i} = {\bf i}$,
$L({\bf q}^{-1}{\bf p}){\bf j} = {\bf j}$, $L({\bf q}^{-1}{\bf p}){\bf k} = {\bf k}$;
writing ${\bf q}^{-1}{\bf p} = {\bf 1}x + {\bf i}y + {\bf j}z + {\bf k}w$
in these three equations
yields $x^2 = 1$, $y = z = w = 0$, so ${\bf q}^{-1}{\bf p} = \pm 1$,
i.e. ${\bf p} = \pm {\bf q}$.

For obvious reasons, then, the map $L$ is called the
double-covering homomorphism from SU(2) to SO(3).
Since the tetrahedral group is a 12-element
subgroup of SO(3), the SU(2) matrices which map
to elements of the tetrahedral group will form a
24-element subgroup of SU(2). This is the
*Binary Tetrahedral Group*.

Click here for an explicit description of the elements of the Binary Tetrahedral Group in terms of quaternions.

Reset using LaTeX, 3/6/23.

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