Tetrahedral Symmetry

Tetrahedral Symmetry#

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

\begin{array}{r} \begin{aligned} V_{1} & = \frac{1}{\sqrt{3}} (1, 1, 1) \\ V_{2} & = \frac{1}{\sqrt{3}} (- 1, - 1, 1) \\ V_{3} & = \frac{1}{\sqrt{3}} (1, - 1, - 1) \\ V_{4} & = \frac{1}{\sqrt{3}} (- 1, 1, - 1), \end{aligned} \end{array}

(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}, k_{3} = - k_{4} = \frac{1 - i}{\sqrt{3} + 1} .

(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 $V_{1} \leftrightarrow V_{2}$ , $V_{3} \leftrightarrow V_{4}$ ) with the three-fold rotation about the $V_{1}$ axis (which maps $V_{2} \to V_{3}$ , $V_{3} \to V_{4}$ , $V_{4} \to V_{2}$ ). On the complex plane, these correspond to

\begin{array}{r} \begin{aligned} I : z \to z^{'} & = - z \\ I I : z \to z^{'} & = \frac{z + i}{z - i}, \end{aligned} \end{array}

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{array}{r} \begin{matrix} \begin{matrix} \begin{aligned} I : & \begin{array}{ccc} u^{'} & = & i u \\ v^{'} & = & - i v \end{array} \\ I I : & \begin{array}{ccc} u^{'} & = & \frac{1}{2} (1 + i) u - \frac{1}{2} (1 - i) v \\ v^{'} & = & \frac{1}{2} (1 + i) u + \frac{1}{2} (1 - i) v . \end{array} \end{aligned} \end{matrix} \\ \begin{matrix}  \end{matrix} \end{matrix} \end{array}

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{array}{r} \begin{aligned} Φ (u, v) & := \prod_{i = 1}^{4} (u - k_{i} v) = (u^{2} - k_{1}^{2}) (v^{2} - k_{3}^{2}) \\ = u^{4} - 2 i \sqrt{3} u^{2} v^{2} + v^{4} \end{aligned} \end{array}

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

\begin{array}{r} \begin{aligned} Φ (u^{'}, v^{'}) & = \prod_{i = 1}^{4} (u^{'} - k_{i} v^{'}) \\ = \prod_{i = 1}^{4} (A - k_{i} C) \prod_{i = 1}^{4} (u - k_{i}^{'} v) \\ = \prod_{i = 1}^{4} (A - k_{i} C) Φ (u, v) \end{aligned} \end{array}

where we define

k_{i}^{'} := \frac{k_{i} D - B}{- k_{i} C + A} \leftrightarrow k_{i} = \frac{k_{i}^{'} A + B}{k_{i}^{'} C + D} .

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

\begin{array}{r} \begin{aligned} \prod_{i = 1}^{4} (A - k_{i} C) & = {[\frac{1}{2} (1 + i)]}^{4} \prod_{i = 1}^{4} (1 - k_{i}) \\ = - \frac{1}{4} (1 - k_{1}^{2}) (1 - k_{3}^{2}) \\ = - \frac{1}{4} [1 - i (2 + \sqrt{3})] [1 + i (2 - \sqrt{3})] \\ = - \frac{1}{2} + i \frac{\sqrt{3}}{2} = e^{2 π i / 3} . \end{aligned} \end{array}

The above guarantees that the function $Φ^{3} (u, v)$ 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)
)

- \frac{1}{2} + \frac{\sqrt{3} i}{2}

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 $Φ$ (since the face centers are in line with the antipodes of the vertices). This yields

\begin{array}{r} \begin{aligned} Ψ (u, v) & := \prod_{i = 1}^{4} (u + {\bar{k}}_{i}^{- 1} v) = \prod_{i = 1}^{4} (u - {\bar{k}}_{i} v) \\ = u^{4} + 2 i \sqrt{3} u^{2} v^{2} + v^{4} . \end{aligned} \end{array}

(The second-last equality follows since $k_{1} k_{4} = k_{2} k_{3} = - 1$ .) As one might expect, under transformation $I I$ the function $Ψ (u, v)$ becomes

Ψ (u^{'}, v^{'}) = e^{- 2 π i / 3} Ψ (u, v) .

Therefore, the combinations $Φ Ψ$ and $Ψ^{3}$ 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)]])

u^{4} - 2 \sqrt{3} i u^{2} v^{2} + v^{4}

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) = u v (u - v) (u + v) (u - i v) (u + i v) = u v (u^{4} - v^{4}) .

Under the transformation $I I$ (with $A = \frac{1}{2} (1 + i) = - B^{*} = C = D^{*}$ ), we see that

\begin{array}{r} \begin{aligned} t (u^{'}, v^{'}) & = (A u - A^{*} v) (A u + A^{*} v) \\ \times [{(A u - A^{*} v)}^{2} - {(A u + A^{*} v)}^{2}] [{(A u - A^{*} v)}^{2} + {(A u + A^{*} v)}^{2}] \\ = (A^{2} u^{2} - A^{* 2} v^{2}) [4 A A^{*} u v] [2 (A^{2} u^{2} + A^{* 2} v^{2})] \\ = \frac{i}{2} (u^{2} + v^{2}) [- 2 u v] [i (u^{2} - v^{2})] = t (u, v) \end{aligned} \end{array}

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{array}{r} \begin{aligned} t (u^{'}, v^{'}) & = A A^{*} u v (A^{2} u^{2} - A^{* 2} v^{2}) (A^{2} u^{2} + A^{* 2} v^{2}) \\ = t (u, v) . \end{aligned} \end{array}

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

By direct computation, one can evaluate the combination $Ψ^{3} - Φ^{3}$ . As it turns out, this expression has a particularly simple form:

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

# 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()

12 \sqrt{3} i u^{2} v^{2} {(u - v)}^{2} {(u + v)}^{2} {(u^{2} + v^{2})}^{2}

# 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

e_3, z_1 = symbols("e_3 z_1")

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

Tetrahedral Symmetry

Contents

Tetrahedral Symmetry#

Polynomial Invariants#