Octahedral Symmetry - Introduction

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 $II$ above, along with

III: \quad u'=e^{i\pi/4}\, u \,,\quad v' = e^{-i\pi/4} \,v \,.

(1)

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

\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*}

(2)

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) := \Phi(u,v)\, \Psi(u,v) = u^8 + 14\,u^4v^4+v^8\,.

(3)

# 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

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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:

{\rm Hess}[t] := \det\left[\begin{matrix}\frac{\partial^2t}{\partial u^2} & \frac{\partial^2t}{\partial u\partial v} \\[6pt] \frac{\partial^2t}{\partial v \partial u} & \frac{\partial^2t}{\partial v^2}\end{matrix}\right] = -25\,W(u,v)\,.

(4)

In other words, $W(u,v)$ is the Hessian of $t(u,v)$ . As such, since $t \to -t$ under transformation $III$ (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, $\chi(u,v)$ , which we will compute next.

The edge midpoints can be labeled as follows:

\begin{matrix} {\bf X}_1 = \tfrac{1}{\sqrt{2}}(1,1,0) \quad & {\bf X}_2 = \tfrac{1}{\sqrt{2}}(-1,1,0) \quad & {\bf X}_3 = \tfrac{1}{\sqrt{2}}(-1,-1,0) \quad & {\bf X}_4 = \tfrac{1}{\sqrt{2}}(1,-1,0) \\ {\bf X}_5 = \tfrac{1}{\sqrt{2}}(1,0,1) \quad & {\bf X}_6 = \tfrac{1}{\sqrt{2}}(0,1,1) \quad & {\bf X}_7 = \tfrac{1}{\sqrt{2}}(-1,0,1) \quad & {\bf X}_8 = \tfrac{1}{\sqrt{2}}(0,-1,1) \\ {\bf X}_9 = \tfrac{1}{\sqrt{2}}(1,0,-1) \quad & {\bf X}_{10} = \tfrac{1}{\sqrt{2}}(0,1,-1) \quad & {\bf X}_{11} = \tfrac{1}{\sqrt{2}}(-1,0,-1) \quad & {\bf X}_{12} = \tfrac{1}{\sqrt{2}}(0,-1,-1) \end{matrix}

(5)

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

\begin{matrix} l_1 = \tfrac{1}{\sqrt{2}}(1 + i) \quad & l_2 = \tfrac{1}{\sqrt{2}}(-1 + i) \quad & l_3 = \tfrac{1}{\sqrt{2}}(-1 - i) \quad & l_4 = \tfrac{1}{\sqrt{2}}(1 - i) \\ l_5 = 1 + \sqrt{2} \quad & l_6 = i(1 + \sqrt{2}) \quad & l_7 = -1 - \sqrt{2} \quad & l_8 = -i(1 + \sqrt{2}) \\ l_9 = -1 + \sqrt{2} \quad & l_{10} = i(-1 + \sqrt{2}) \quad & l_{11} = 1 - \sqrt{2} \quad & l_{12} = i(1 - \sqrt{2}) \end{matrix}

(6)

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

\chi(u,v) := \prod_{i=1}^{12}\left(u-l_i\,v\right) = u^{12} - 33\left(u^8\,v^4+u^4\,v^8\right) + v^{12} \,.

(7)

# 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)

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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)

(8)

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

\chi(u,v) = \chi_{\rm part}(u,v, l_1) \, \chi_{\rm part}(u,v, l_5) \, \chi_{\rm part}(u,v, l_9)

(9)

where each $\chi_{\rm part}(u,v,x)$ is given by

\chi_{\rm part}(u,v,x) = \prod_{k=1}^4\left(u-i^k\,x\,v\right) = u^4-x^4\,v^4 \,.

(10)

# 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)

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

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

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

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

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

{\textrm Jac}[{\bf V}](u,v) := \det \left[\begin{matrix} \frac{\partial t}{\partial u} & \frac{\partial t}{\partial v} \\[6pt] \frac{\partial W}{\partial u} & \frac{\partial W}{\partial v} \end{matrix}\right] = -8\,\chi(u,v) \,.

(11)

# 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 $\chi \sim t$ , and since $t\to-t$ under transformation $III$ , we deduce that $\chi^2$ is an absolute invariant of the octahedral symmetry group.

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

\chi^2 - W^3 + 108\, t^4 = 0 \,.

(12)

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

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

True