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While working on a project, I came across a paper ("On $C^2$-smooth Surfaces of Constant Width" by Brendan Guilfoyle and Wilhelm Klingenberg) that includes the definition $g ∈ 𝒢$ on page 15, where $𝒢$ is the "tetrahedral group" ("a discrete subgroup of isometries $𝒢 ⊂ O(3)$"). The paper then uses $g(z)$ as a function with a complex input, where $z$ is "the local complex coordinate on the unit 2-sphere in $𝔼^3$ obtained by stereographic projection from the south pole".

Is it possible to define an algebraic expression for $g(z)$ in a way that is understandable to a person with no background in group theory?

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2 Answers 2

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The group $O(3)$ is the group of isometric symmetries of the $2$-dimensional sphere $S^2$ in Euclidean 3-dimensional space. It has an index 2 subgroup of rotational symmetries denoted $SO(3)$.

When a regular tetrahedron $T$ is inscribed in $S^2$, the rotational symmetries of $T$ itself form a subgroup of $SO(3)$ with $12$ elements, called the tetrahedral group.

But, in order to write the elements of the tetrahedral group as functions of a complex variable, I have to guess a little bit at the intentions of the unknown author. (If this guess is wrong, then my whole answer is wrong).

My best guess is that the $2$-sphere $S^2$ is being transformed to the extended complex plane $\mathbb C \cup \{\infty\}$ using stereographic projection. When this is done, the individual elements of $SO(3)$ can all be written as fractional linear transformations of $\mathbb C \cup \{\infty\}$, meaning functions of the format $$g(z) = \frac{az+b}{cz+d} $$ where $a,b,c,d$ are complex numbers required to satisfy the normalization condition $ad-bc \ne 0$ (additional conditions are needed to pick out the case of a rotational symmetry of $S^2$ but that's beside the point).

It's common to write the coefficients $a,b,c,d$ in matrix format as $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, because composition of the functions $g(z)$ corresponds exactly to multiplication of the corresponding matrices, as long as you keep in mind that the matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix}$ represent the same function (for any nonzero complex number $k$). It is also common to require the more strict normalization $ad-bc=1$, as long as you keep in mind that the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and its negative $\begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix}$ represent the same function.

According to the paper of S. Ushiki entitled Julia Sets with Polyhedral Symmetry, the four vertices of the inscribed regular tetrahedron can be carefully chosen so that the stereographic transformation from $S^2$ to $\mathbb C \cup \{\infty\}$ takes those four vertices to the following four points in the complex plane $\mathbb C$, namely: $$0, \sqrt{2}, \sqrt{2} \omega, \sqrt{2} \omega^2 \qquad\text{where }\qquad \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \quad\text{hence}\quad \omega^3=1 $$ (and, of course, a real number $s$ -- such as $s=0$ or $s=\sqrt{2}$ --- is identified with the complex number $s+0i$).

And when the transformation is done in that manner, the $12$ elements of the tetrahedral subgroup of $O(3)$ correspond to the $12$ fractional linear transformations that permute the four points listed above. Writing those fractional linear transformations in matrix format, one gets the two matrices $$M_1 = \begin{pmatrix} \omega & 0 \\ 0 & \omega^2 \end{pmatrix} \qquad M_2 \begin{pmatrix} - 1/\sqrt{3} i & \sqrt{2}/\sqrt{3} i \\ \sqrt{2}/\sqrt{3} i & 1 / \sqrt{3} i \end{pmatrix} $$ as well as all the other matrices that can be obtained by multiplying these two in sequences (which is $24$ matrices altogether, occuring in $12$ negative pairs). In brief, those two matrices generate the tetrahedral subgroup, represented as a group of fractional linear transformations.

So for instance the first matrix above gives you the function $g_1(z)$ which simplifies to $$g_1(z) = \omega^2 z = (-\frac{1}{2} - \frac{\sqrt{3}}{2} i) z $$ When applying this fractional linear transformation to the four points above, you can see that it cyclically permutes the three vertices $\sqrt{2}$, $\sqrt{2}\omega$, $\sqrt{2}\omega^2$.

The second matrix above gives you the function $g_2(z)$ which simplifies to $$g_2(z) = \frac{-z + \sqrt{2}}{\sqrt{2} z + 1} $$ When applying this transformation to the four vertices listed above you can see that it transposes the two vertices $0$ and $\sqrt{2}$ and that it transposes the two vertices $\sqrt{2}\omega$ and $\sqrt{2}\omega^2$.

You might find it instructive to pick up a regular tetrahedron (for example a 4-sided gaming die) in order to visualize these symmetries.

If you want to see the formulas for all 12 of the different functions in the tetrahedral group, start multiplying $g_1$ and $g_2$ in all possible ways (or, for less work, multiply their matrices) in order to generate all the other elements of the tetrahedral group.

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  • $\begingroup$ I added to my question a line from the paper explaining that the complex input of $g(z)$ is "the local complex coordinate on the unit 2-sphere in $𝔼^3$ obtained by stereographic projection from the south pole", if that affects your answer at all. $\endgroup$
    – Lawton
    Feb 13 at 17:42
  • $\begingroup$ I'm a bit confused about how to go from the two matrices you give to the full set of 12. Can you elaborate on that? I know matrix multiplication is non-commutative so the two orderings give two different results, but that only gives a total of four, not 12. I'm also a little confused about going from the matrix representation to the function representation. You say that $g_1(z) = ω z$, but by my interpretation it should be $g_1(z) = \frac{ω z + 0}{0 z + ω^2} = \frac{z}{ω}$. Am I missing something, or is that a typo? $\endgroup$
    – Lawton
    Feb 13 at 18:06
  • $\begingroup$ Thanks for that first comment. In fact that confirms my guess. $\endgroup$
    – Lee Mosher
    Feb 13 at 18:08
  • $\begingroup$ For part of your second comment, you are right: using that $\omega^3=1$ it follows that $\frac{1}{\omega} = \omega^2$ and $g_1(z)=\omega^2 z$ $\endgroup$
    – Lee Mosher
    Feb 13 at 18:11
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    $\begingroup$ For the other part of your second comment: you can multiply $M_1 \cdot M_1$, and $M_1 \cdot M_2$, and $M_2 \cdot M_1$, and $M_1 \cdot M_1 \cdot M_2$, and $M_1 \cdot M_2 \cdot M_1$, and $M_2 \cdot M_1 \cdot M_1$, and so on. After a while you will start seeing repetitions, and in the end you will get only 24 matrices in all, occurring in 12 inverse pairs. $\endgroup$
    – Lee Mosher
    Feb 13 at 18:13
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To add to the previous answer, you can give a fairly straight-forward description of the tetrahedron on the Riemann sphere if you note that

$$ T=\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\}=\{(\epsilon_1,\epsilon_2,\epsilon_3): \epsilon_i \in \{\pm 1\}, \epsilon_1\epsilon_2\epsilon_3 =1\} $$

form the vertices of a regular tetrahedron - although they lie on the sphere of radius $\sqrt{3}$ rather than the unit sphere. Now stereographic projection from $C=(-1,-1,-1)$ takes a vector $v$ on the sphere to the vector $S(v)$ on the plane perpendicular to $C$ which is also on the line through $C$ and $v$. Thus $S(v) = t.v +(1-t)C$ where $t$ is determined by the condition that the dot product $S(v)\cdot C =0$. Thus $tv\cdot C +3(1-t)=0$, that is, $t = 3/(3-v\cdot C)$. It follows that for the projection of $(-1,-1,1)$ we must take $t= 3/(3-1)=3/2$, and hence the projection is $(-1,-1,2)$. Now rescaling by $1/\sqrt{3}$ we see that $(1,1,1)$ is sent to $0$, while the other 3 vertices are sent to the vectors $$ \frac{1}{\sqrt{3}}(-1,-1,2), \frac{1}{\sqrt{3}}(-1,2,-1),\frac{1}{\sqrt{3}}(2,-1,-1). $$

These are vectors of length $\sqrt{2}$ forming the vertices of an equilateral triangle, and hence can be rotated in the complex plane (identified with the plane $\{(x,y,z): x+y+z=0\}\subseteq \mathbb R^3$) so as to coincide with $\sqrt{2}, \sqrt{2}\omega, \sqrt{2}\omega^2$ where $\omega$ is the cube root of unity $\frac{-1}{2}+\frac{\sqrt{3}i}{2}$

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