Octahedral Symmetry

Octahedral Symmetry#

The six vertices of the octahedron can be taken as identical to the six edge midpoints of the tetrahedron. As such they have $t (u, v)$ as their polynomial invariant.

The octahedral symmetry group is generated by transformations $I$ and $I I$ above, along with

I I I : u^{'} = e^{i π / 4} u, v^{'} = e^{- i π / 4} v .

Under transformation $I I I$ with $A = e^{i π / 4} = D^{*}$ , $B = C = 0$ , we find that

\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, only $t^{2} (u, v)$ is an absolute invariant under the octahedral symmetry group.

The 8 face centers of the octahedron are formed by combining the 8 tetrahedral roots ${k_{i}, {\bar{k}}_{i}}$ found before. As such, the octahedral polynomial invariant for the face centers is given by

W (u, v) := Φ (u, v) Ψ (u, v) = u^{8} + 14 u^{4} v^{4} + v^{8} .

# Check of the above:
from sympy import I, diff, simplify, sqrt, symbols
from tetra import Phi, Psi, t
from functools import reduce

u, v = symbols("u v")
W = simplify(Phi * Psi)
W

u^{8} + 14 u^{4} v^{4} + v^{8}

In order to work out the transformation properties of $W (u, v)$ , it is useful to note that $t$ and $W$ are inter-related as follows:

\begin{array}{r} H e s s [t] := det [\begin{array}{c} \frac{\partial^{2} t}{\partial u^{2}} & \frac{\partial^{2} t}{\partial u \partial v} \\ \frac{\partial^{2} t}{\partial v \partial u} & \frac{\partial^{2} t}{\partial v^{2}} \end{array}] = - 25 W (u, v) . \end{array}

In other words, $W (u, v)$ is the Hessian of $t (u, v)$ . As such, since $t \to - t$ under transformation $I I I$ (and since $W \sim t^{2}$ ), $W (u, v)$ is clearly invariant under all three octahedral symmetry generators.

# Check of the above:
def hessian(f, u, v):
    du2 = diff(diff(f, u), u)
    dudv = diff(diff(f, u), v)
    dv2 = diff(diff(f, v), v)
    return du2 * dv2 - dudv**2


hessian(t, u, v).equals(-25 * Phi * Psi)

True

The edge midpoints of the octahedron form a polynomial invariant, $χ (u, v)$ , which we will compute next.

The edge midpoints can be labeled as follows:

\begin{array}{r} \begin{array}{c} X_{1} = \frac{1}{\sqrt{2}} (1, 1, 0) & X_{2} = \frac{1}{\sqrt{2}} (- 1, 1, 0) & X_{3} = \frac{1}{\sqrt{2}} (- 1, - 1, 0) & X_{4} = \frac{1}{\sqrt{2}} (1, - 1, 0) \\ X_{5} = \frac{1}{\sqrt{2}} (1, 0, 1) & X_{6} = \frac{1}{\sqrt{2}} (0, 1, 1) & X_{7} = \frac{1}{\sqrt{2}} (- 1, 0, 1) & X_{8} = \frac{1}{\sqrt{2}} (0, - 1, 1) \\ X_{9} = \frac{1}{\sqrt{2}} (1, 0, - 1) & X_{10} = \frac{1}{\sqrt{2}} (0, 1, - 1) & X_{11} = \frac{1}{\sqrt{2}} (- 1, 0, - 1) & X_{12} = \frac{1}{\sqrt{2}} (0, - 1, - 1) \end{array} \end{array}

The corresponding stereographically-projected points in the complex plane are given by

\begin{array}{r} \begin{array}{c} l_{1} = \frac{1}{\sqrt{2}} (1 + i) & l_{2} = \frac{1}{\sqrt{2}} (- 1 + i) & l_{3} = \frac{1}{\sqrt{2}} (- 1 - i) & l_{4} = \frac{1}{\sqrt{2}} (1 - i) \\ l_{5} = 1 + \sqrt{2} & l_{6} = i (1 + \sqrt{2}) & l_{7} = - 1 - \sqrt{2} & l_{8} = - i (1 + \sqrt{2}) \\ l_{9} = - 1 + \sqrt{2} & l_{10} = i (- 1 + \sqrt{2}) & l_{11} = 1 - \sqrt{2} & l_{12} = i (1 - \sqrt{2}) \end{array} \end{array}

One way to compute $χ (u, v)$ is to blindly multiply together. Doing so gives

χ (u, v) := \prod_{i = 1}^{12} (u - l_{i} v) = u^{12} - 33 (u^{8} v^{4} + u^{4} v^{8}) + v^{12} .

# Check of the above:
def stereo(v1, v2, v3):
    return (v1 + I * v2) / (1 - v3)


ls = []

ls.append(stereo(1 / sqrt(2), 1 / sqrt(2), 0))
ls.append(stereo(1 / sqrt(2), -1 / sqrt(2), 0))
ls.append(stereo(-1 / sqrt(2), 1 / sqrt(2), 0))
ls.append(stereo(-1 / sqrt(2), -1 / sqrt(2), 0))
ls.append(stereo(0, 1 / sqrt(2), 1 / sqrt(2)))
ls.append(stereo(0, 1 / sqrt(2), -1 / sqrt(2)))
ls.append(stereo(0, -1 / sqrt(2), 1 / sqrt(2)))
ls.append(stereo(0, -1 / sqrt(2), -1 / sqrt(2)))
ls.append(stereo(1 / sqrt(2), 0, 1 / sqrt(2)))
ls.append(stereo(1 / sqrt(2), 0, -1 / sqrt(2)))
ls.append(stereo(-1 / sqrt(2), 0, 1 / sqrt(2)))
ls.append(stereo(-1 / sqrt(2), 0, -1 / sqrt(2)))

chi = reduce(lambda x, y: x * y, [u - x * v for x in ls])
simplify(chi)

u^{12} - 33 u^{8} v^{4} - 33 u^{4} v^{8} + v^{12}

Alternatively, we can make use of the four-fold symmetry relating the midpoints

(l_{1}, l_{2} = i l_{1}, l_{3} = i^{2} l_{1}, l_{4} = i^{3} l_{1})

and similarly for $(l_{5}, l_{6}, l_{7}, l_{8})$ , $(l_{9}, l_{10}, l_{11}, l_{12})$ . Grouping in this way, we can write

χ (u, v) = χ_{p a r t} (u, v, l_{1}) χ_{p a r t} (u, v, l_{5}) χ_{p a r t} (u, v, l_{9})

where each $χ_{p a r t} (u, v, x)$ is given by

χ_{p a r t} (u, v, x) = \prod_{k = 1}^{4} (u - i^{k} x v) = u^{4} - x^{4} v^{4} .

# Check of the above
x = symbols("x")


def reduce_multiply(any_list):
    return reduce(lambda x, y: x * y, any_list)


expr = reduce_multiply([u - x * I**k * v for k in range(4)])
simplify(expr)

u^{4} - v^{4} x^{4}

chi_part1 = expr.subs(x, 1 / sqrt(2) * (1 + I)).expand()
chi_part1

u^{4} + v^{4}

chi_part5 = expr.subs(x, 1 + sqrt(2)).expand()
chi_part5

u^{4} - 17 v^{4} - 12 \sqrt{2} v^{4}

chi_part9 = expr.subs(x, -1 + sqrt(2)).expand()
chi_part9

u^{4} - 17 v^{4} + 12 \sqrt{2} v^{4}

chi = (chi_part1 * chi_part5 * chi_part9).expand()
chi

u^{12} - 33 u^{8} v^{4} - 33 u^{4} v^{8} + v^{12}

As it turns out, $χ (u, v)$ is also related to $t (u, v)$ and $W (u, v)$ : $χ$ is proportional to the Jacobian of $V (u, v) := [t (u, v), W (u, v)]$ :

\begin{array}{r} J a c [V] (u, v) := det [\begin{array}{c} \frac{\partial t}{\partial u} & \frac{\partial t}{\partial v} \\ \frac{\partial W}{\partial u} & \frac{\partial W}{\partial v} \end{array}] = - 8 χ (u, v) . \end{array}

# Check of the above:
def jacobian(f, g, u, v):
    return diff(f, u) * diff(g, v) - diff(f, v) * diff(g, u)


jacobian(t, W, u, v).equals(-8 * chi)

True

Since $χ \sim t$ , and since $t \to - t$ under transformation $I I I$ , we deduce that $χ^{2}$ is an absolute invariant of the octahedral symmetry group.

Lastly, by direct computation we can verify that the absolute invariants $χ^{2}$ , $W^{3}$ , and $t^{4}$ (all sharing the same order of homogeneity) are inter-related:

χ^{2} - W^{3} + 108 t^{4} = 0 .

# Check of the above:
simplify(chi**2 - W**3).factor()

- 108 u^{4} v^{4} {(u - v)}^{4} {(u + v)}^{4} {(u^{2} + v^{2})}^{4}

# Second check:
(chi**2 - W**3).equals(-108 * t**4)

True