Tetrahedral Symmetry - Introduction

A tetrahedron is formed by joining alternating vertices as one traverses the edges of a cube. Taking these four vertices to be

\begin{align*} {\bf V}_1 &= \frac1{\sqrt{3}}(1, 1, 1) \\ {\bf V}_2 &= \frac1{\sqrt{3}}(-1, -1, 1) \\ {\bf V}_3 &= \frac1{\sqrt{3}}(1, -1, -1) \\ {\bf V}_4 &= \frac1{\sqrt{3}}(-1, 1, -1) \,, \end{align*}

(1)

(such that they each lie on the unit sphere) we find that the corresponding spherically-projected coordinates are

k_1 = -k_2 = \frac{1+i}{\sqrt{3} -1} \,,\quad k_3 = -k_4 = \frac{1-i}{\sqrt{3}+1} \,.

(2)

(Note that an equivalent tetrahedron could be constructed using the points antipodal to the ones above; this is referred to as the “diametral tetrahedron”.)

The discrete rotational symmetries of the tetrahedron can be constructed by composing the two-fold rotation about the $z$ axis (which maps ${\bf V}_1 \leftrightarrow{\bf V}_2$ , ${\bf V}_3 \leftrightarrow{\bf V}_4$ ) with the three-fold rotation about the ${\bf V}_1$ axis (which maps ${\bf V}_2 \rightarrow {\bf V}_3$ , ${\bf V}_3 \rightarrow {\bf V}_4$ , ${\bf V}_4 \rightarrow {\bf V}_2$ ). On the complex plane, these correspond to

\begin{align*} I:\quad z \to z' &= -z \\ II:\quad z \to z' &= \frac{z+i}{z-i} \,, \end{align*}

(3)

respectively.

from sympy import I, Matrix, Rational, conjugate, cos, exp, pi, simplify, sqrt, symbols

# Check of the above:
sp1 = (1 + I) / (sqrt(3) - 1)
sp2 = -(1 + I) / (sqrt(3) - 1)
sp3 = (1 - I) / (sqrt(3) + 1)
sp4 = (-1 + I) / (sqrt(3) + 1)

# fmt: off
((sp1 + I) / (sp1 - I)).equals(sp1) and \
((sp2 + I) / (sp2 - I)).equals(sp3) and \
((sp3 + I) / (sp3 - I)).equals(sp4) and \
((sp4 + I) / (sp4 - I)).equals(sp2)
# fmt: on

True

In terms of the homogeneous coordinates $u$ and $v$ , these two transformations map directly to

\begin{aligned} I:&\quad \begin{array}{ccc} u' &=& i\,u \\ v' &=& -i\,v \end{array} \\ II:&\quad \begin{array}{ccc} u' &=& \tfrac12(1+i)\,u - \tfrac12(1-i)\,v \\ v' &=& \tfrac12 (1+i) u + \tfrac12(1-i)\,v \,. \end{array} \end{aligned}

(4)

Polynomial Invariants¶

Keeping in mind that all symmetry transformations of the tetrahedron merely exchange a number of vertices with one another, it’s expected that a polynomial with roots at the locations of the vertices in the complex plane will be invariant under such transformations. To see this, consider the product

\begin{align*} \Phi(u, v) &:= \prod_{i=1}^4 \left(u - k_i\, v\right) = (u^2-k_1^2)(v^2-k_3^2) \\ &= u^4 - 2i\sqrt{3}\,u^2v^2 + v^4 \end{align*}

(5)

and apply the transformation $(u, v) \to (u', v')$ as before. In the general case, this gives

\begin{align*} \Phi(u', v') &= \prod_{i=1}^4 \left(u' - k_i\, v'\right) \\ &= \prod_{i=1}^4 \left(A - k_i\, C\right) \prod_{i=1}^4 \left(u - k_i'\, v\right) \\ &= \prod_{i=1}^4 \left(A - k_i\, C\right) \, \Phi(u, v) \end{align*}

(6)

where we define

k'_i := \frac{k_i\, D - B}{- k_i\, C + A} \quad\leftrightarrow\quad k_i = \frac{k_i'\,A+B}{k_i'\,C+D} \,.

(7)

In the case of both transformations $I$ and $II$ , the product $\prod_{i=1}^4 \left(u - k_i'\, v\right)$ is clearly invariant, because the vertices all transform into one another under any of the transformations. For transformation $I$ , the product $\prod_{i=1}^4 \left(A - k_i\, C\right) = 1$ because $A=i$ and $C=0$ . For transformation $II$ , this product instead gives

\begin{align*} \prod_{i=1}^4 \left(A - k_i\, C\right) &= \left[\tfrac12 (1+i)\right]^4\prod_{i=1}^4 \left(1 - k_i\right) \\ &= -\tfrac14(1-k_1^2)(1-k_3^2) \\ &=-\tfrac14\left[1-i(2+\sqrt{3})\right]\left[1+i(2-\sqrt{3})\right] \\ &=-\frac12+i\frac{\sqrt{3}}2 = e^{2\pi i/3} \,. \end{align*}

(8)

The above guarantees that the function $\Phi^3(u,v)$ is an absolute invariant under the symmetries of the tetrahedral group.

# Check of the above
simplify(
    (Rational(1, 2) * (1 + I)) ** 4 * (1 - sp1) * (1 - sp2) * (1 - sp3) * (1 - sp4)
)

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The above can also be done for the edge midpoints, and face centers, of the tetrahedron. In the latter case, it’s as simple as trading $k_i \to -1/\bar k_i$ in the above expression for $\Phi$ (since the face centers are in line with the antipodes of the vertices). This yields

\begin{align*} \Psi(u, v) &:= \prod_{i=1}^4\left(u + \bar k_i^{-1}\, v\right) = \prod_{i=1}^4\left(u - \bar k_i\, v\right) \\ &= u^4 + 2i\sqrt{3}\,u^2v^2 + v^4 \,. \end{align*}

(9)

(The second-last equality follows since $k_1k_4 = k_2k_3 =-1$ .) As one might expect, under transformation $II$ the function $\Psi(u,v)$ becomes

\Psi(u',v') = e^{-2\pi i/3} \, \Psi(u,v) \,.

(10)

Therefore, the combinations $\Phi \Psi$ and $\Psi^3$ are also absolute invariants.

orth = Rational(1, 2) * Matrix([[1 + I, -1 + I], [1 + I, 1 - I]])
u, v = symbols("u v")
U, V = symbols("U V")
uv = Matrix([u, v])

uv_prime = orth * uv
print(f"{uv_prime=}")

Phi = (u - sp1 * v) * (u - sp2 * v) * (u - sp3 * v) * (u - sp4 * v)

# print(Phi.subs(u, uv_prime[0]))
Phi_prime = Phi.subs([(u, U), (v, V)]).subs([(U, uv_prime[0]), (V, uv_prime[1])])
(Phi_prime / exp(2 * pi * I / 3)).rewrite(cos).radsimp().expand()

uv_prime=Matrix([
[u*(1/2 + I/2) + v*(-1/2 + I/2)],
[ u*(1/2 + I/2) + v*(1/2 - I/2)]])

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In contrast, the six midpoints are at $z=0$ , $z=\infty$ , $z=\pm 1$ , and $z=\pm i$ . In this case, the invariant polynomial is given by

t(u, v) = uv(u-v)(u+v)(u-iv)(u+iv) = uv(u^4-v^4) \,.

(11)

Under the transformation $II$ (with $A=\tfrac12(1+i)=-B^*=C=D^*$ ), we see that

\begin{align*} t(u',v') &= \left(A\,u-A^*\,v\right)\left(A\,u+A^*\,v\right) \\ &\times\left[\left(A\,u-A^*\,v\right)^2 - \left(A\,u+A^*\,v\right)^2\right]\left[\left(A\,u-A^*\,v\right)^2+\left(A\,u+A^*\,v\right)^2\right] \\ &=\left(A^2\,u^2 - A^{*2}\,v^2\right)\left[4\,AA^*\,uv\right]\left[2\left(A^2\,u^2+A^{*2}\,v^2\right)\right] \\ &=\frac i2\left(u^2 + v^2\right)\left[-2\,uv\right]\left[i\left(u^2-v^2\right)\right] = t(u,v) \end{align*}

(12)

where, in the last line, we make use of $A^2 = i/2 = -A^{*2}$ . Similarly, $t(u',v') = t(u,v)$ under transformation $I$ where $A=i=D^*$ , $B=C=0$ , we find

\begin{align*} t(u',v') &= AA^*\,uv(A^2\,u^2-A^{*2}\,v^2)(A^2\,u^2+A^{*2}\,v^2) \\ &= t(u,v) \,. \end{align*}

(13)

Therefore, $t(u,v)$ is an absolute invariant as well.

By direct computation, one can evaluate the combination $\Psi^3- \Phi^3$ . As it turns out, this expression has a particularly simple form:

\Psi^3(u, v) -\Phi^3(u, v) = 12\sqrt{3}\,i\,t^2(u, v) \,.

(14)

# Check of the above:
Phi = (u - sp1 * v) * (u - sp2 * v) * (u - sp3 * v) * (u - sp4 * v)
Psi = (
    (u - conjugate(sp1) * v)
    * (u - conjugate(sp2) * v)
    * (u - conjugate(sp3) * v)
    * (u - conjugate(sp4) * v)
)
simplify(Psi**3 - Phi**3).factor()

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# Second check:
t = u * v * (u - v) * (u + v) * (u**2 + v**2)
(Psi**3 - Phi**3).equals(12 * sqrt(3) * I * t**2)

True

from functools import reduce
from sympy import Poly


e_3, z_1 = symbols("e_3 z_1")


def e3_simp(expr, e, full=True, cleanup=True):
    result = Poly(expr, e)
    e_coeffs = reversed(result.all_coeffs())
    simp_coeffs = [0] * 5
    # enforce e**3 = 1
    for i, coeff in enumerate(e_coeffs):
        simp_coeffs[i % 3] += coeff
    # enforce e**2 = - e - 1
    if full:
        simp_coeffs = [simp_coeffs[i] - simp_coeffs[2] for i in range(2)]
    if cleanup:
        return simplify(Poly.from_list(reversed(simp_coeffs), e).as_expr())
    return Poly.from_list(reversed(simp_coeffs), e).as_expr()


zs = [0, z_1, z_1 * e_3, z_1 * e_3**2]

e3_simp(reduce(lambda x, y: x * y, [u - x * v for x in zs]), e_3)

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