Polynomial Invariants
The goal of these notes is to work through several examples of polynomial
invariants, constructed from the symmetries of platonic solids.
After building up some preliminary results for the tetrahedron and octahedron,
corresponding expressions for the icosahedron and dodecahedron are constructed
using the former results as building blocks.
Preliminaries ¶ Rotation (in three dimensions) of the vector r {\bf r} r n ^ \hat{\bf n} n ^ θ \theta θ
r ′ = r cos θ + ( r ⋅ n ^ ) n ^ ( 1 − cos θ ) − r × n ^ sin θ {\bf r}' = {\bf r}\,\cos\theta + ({\bf r}\cdot\hat{\bf n}) \hat{\bf n}\, (1 - \cos\theta) - {\bf r} \times \hat{\bf n} \, \sin\theta r ′ = r cos θ + ( r ⋅ n ^ ) n ^ ( 1 − cos θ ) − r × n ^ sin θ or, in components,
r ′ i = [ δ i j cos θ + n i n j ( 1 − cos θ ) − ϵ i j k n k sin θ ] r j : = O i j r j . \begin{align*}
r'^i &= \left[{\delta^i}_j\,\cos\theta + n^in_j\,(1-\cos\theta)-{\epsilon^i}_{jk}\,n^k\,\sin\theta\right] r^j \\
:&= {\mathcal O^i}_j\,r^j \,.
\end{align*} r ′ i : = [ δ i j cos θ + n i n j ( 1 − cos θ ) − ϵ i jk n k sin θ ] r j = O i j r j . Introducing the variables
a i = n i sin ( θ / 2 ) , d : = cos ( θ / 2 ) , a_i = n_i\,\sin\left(\theta/2\right) \,,\quad d:= \cos\left(\theta /2\right) \,, a i = n i sin ( θ /2 ) , d := cos ( θ /2 ) , we see the rotation matrix O i j {{\mathcal O}^i}_j O i j
O i j = ( 2 d 2 − 1 ) δ i j + 2 ( a i a j − ϵ i j k a k d ) . {{\mathcal O}^i}_j = \left(2\,d^2 -1\right){\delta^i}_j+ 2\left( a^i\,a_j - {\epsilon^i}_{jk}\,a^k\,d\right) \,. O i j = ( 2 d 2 − 1 ) δ i j + 2 ( a i a j − ϵ i jk a k d ) . (Recall that cos θ = 2 cos 2 ( θ / 2 ) − 1 = 1 − 2 sin 2 ( θ / 2 ) \cos\theta = 2\,\cos^2\left(\theta/2\right) -1 = 1-2\,\sin^2\left(\theta/2\right) cos θ = 2 cos 2 ( θ /2 ) − 1 = 1 − 2 sin 2 ( θ /2 ) sin θ = 2 sin ( θ / 2 ) cos ( θ / 2 ) \sin\theta = 2\,\sin\left(\theta/2\right)\,\cos\left(\theta/2\right) sin θ = 2 sin ( θ /2 ) cos ( θ /2 )
The orthogonality of O {\mathcal O} O O k i O k j {\mathcal O_k}^i\,{\mathcal O^k}_j O k i O k j
( O T O ) i j = δ i j + 4 ( d 2 + a 2 − 1 ) ( d 2 δ i j + a i a j ) = δ i j {({\mathcal O}^T {\mathcal O})^i}_j = {\delta^i}_j + 4\left(d^2 + {\bf a}^2 -1\right)\left(d^2 \,{\delta^i}_j +
a^ia_j\right) = {\delta^i}_j ( O T O ) i j = δ i j + 4 ( d 2 + a 2 − 1 ) ( d 2 δ i j + a i a j ) = δ i j where the second equality follows from the definition of a {\bf a} a d d d sin 2 ( θ / 2 ) + cos 2 ( θ / 2 ) = 1 \sin^2\left(\theta/2\right) + \cos^2\left(\theta/2\right) = 1 sin 2 ( θ /2 ) + cos 2 ( θ /2 ) = 1
a 2 + d 2 = 1 . {\bf a}^2 + d^2 = 1 \,. a 2 + d 2 = 1 . # Imports
from functools import reduce
from sympy import I, Matrix, conjugate, expand, simplify, sqrt, symbols
# Check of the above:
a1, b1, c1, d1 = symbols("a1 b1 c1 d1")
def o_mat(a, b, c, d):
return Matrix(
[
[2 * d**2 - 1 + 2 * a**2, 2 * (a * b - c * d), 2 * (a * c + b * d)],
[2 * (a * b + c * d), 2 * d**2 - 1 + 2 * b**2, 2 * (b * c - a * d)],
[2 * (a * c - b * d), 2 * (b * c + a * d), 2 * d**2 - 1 + 2 * c**2],
]
)
O1 = o_mat(a1, b1, c1, d1)
expand((O1 * O1.transpose()).subs(d1, sqrt(1 - a1**2 - b1**2 - c1**2)))Two such rotations can be composed as follows:
O C : = O B ⋅ O A \mathcal O_C := \mathcal O_B \cdot \mathcal O_A O C := O B ⋅ O A where, if O A = O A ( a A , d A ) \mathcal O_A = \mathcal O_A(\mathbf a_A, d_A) O A = O A ( a A , d A ) O B = O B ( a B , d B ) \mathcal O_B = \mathcal O_B(\mathbf a_B, d_B) O B = O B ( a B , d B ) O C = O C ( a C , d C ) \mathcal O_C = \mathcal O_C(\mathbf a_C, d_C) O C = O C ( a C , d C )
a 1 C = a 1 A d B + a 1 B d A − a 2 A a 3 B + a 2 B a 3 A a 2 C = a 2 A d B + a 2 B d A − a 3 A a 1 B + a 3 B a 1 A a 3 C = a 3 A d B + a 3 B d A − a 1 A a 2 B + a 1 B a 2 A d C = − a 1 A a 1 B − a 2 A a 2 B − a 3 A a 3 B + d A d B . \begin{align*}
a_{1C} &= a_{1A} d_B + a_{1B} d_A - a_{2A} a_{3B} + a_{2B} a_{3A} \\
a_{2C} &= a_{2A} d_B + a_{2B} d_A - a_{3A} a_{1B} + a_{3B} a_{1A} \\
a_{3C} &= a_{3A} d_B + a_{3B} d_A - a_{1A} a_{2B} + a_{1B} a_{2A} \\
d_C &= -a_{1A} a_{1B} - a_{2A} a_{2B} - a_{3A} a_{3B} + d_A d_B \,.
\end{align*} a 1 C a 2 C a 3 C d C = a 1 A d B + a 1 B d A − a 2 A a 3 B + a 2 B a 3 A = a 2 A d B + a 2 B d A − a 3 A a 1 B + a 3 B a 1 A = a 3 A d B + a 3 B d A − a 1 A a 2 B + a 1 B a 2 A = − a 1 A a 1 B − a 2 A a 2 B − a 3 A a 3 B + d A d B . # Check of the above:
a2, b2, c2, d2 = symbols("a2 b2 c2 d2")
O2 = o_mat(a2, b2, c2, d2)
expand(
(
O2 * O1
- o_mat(
a1 * d2 + a2 * d1 - b1 * c2 + b2 * c1,
b1 * d2 + b2 * d1 - c1 * a2 + c2 * a1,
c1 * d2 + c2 * d1 - a1 * b2 + a2 * b1,
-a1 * a2 - b1 * b2 - c1 * c2 + d1 * d2,
)
)
.subs(d1, sqrt(1 - a1**2 - b1**2 - c1**2))
.subs(d2, sqrt(1 - a2**2 - b2**2 - c2**2))
)Stereographic Projection ¶ In what follows, it’s helpful to map the rotations on the sphere to the complex
plane, by performing an equatorial spherical projection from the point where
n ^ = ( 0 , 0 , 1 ) \hat{\bf n} = (0, 0, 1) n ^ = ( 0 , 0 , 1 ) r ^ = ( ξ , η , ζ ) \hat{\bf r} = (\xi,\eta,\zeta) r ^ = ( ξ , η , ζ ) z = x + i y z = x + iy z = x + i y ( ξ η ) (\xi \eta) ( ξ η )
z = x + i y = ξ + i η 1 − ζ . z = x + i\, y = \frac{\xi +i\,\eta}{1-\zeta} \,. z = x + i y = 1 − ζ ξ + i η . Inverting, we find
ξ = z + z ˉ r , η = i ( z ˉ − z ) r , ζ = z z ˉ − 1 r \xi = \frac{z + \bar{z}}r \,,\quad \eta = \frac{i(\bar{z} - z)}r \,,\quad
\zeta = \frac{z\bar{z} -1}r ξ = r z + z ˉ , η = r i ( z ˉ − z ) , ζ = r z z ˉ − 1 where we define r : = 1 + z z ˉ r := 1 + z\bar{z} r := 1 + z z ˉ
Rotations about the ζ \zeta ζ z z z
z → z ′ = e i θ z ↔ ξ ′ = ξ cos θ − η sin θ η ′ = ξ sin θ + η cos θ . z \to z' = e^{i\theta} \, z \quad\leftrightarrow\quad \begin{array}{cc}\xi' &=
\xi\,\cos\theta
- \eta\,\sin\theta\\
\eta' &= \xi\,\sin\theta + \eta\,\cos\theta
\end{array}\,. z → z ′ = e i θ z ↔ ξ ′ η ′ = ξ cos θ − η sin θ = ξ sin θ + η cos θ . If we wish to write an arbitrary rotation of a point on the sphere about some
direction specified by n ^ \hat{\bf n} n ^ θ \theta θ a {\bf a} a d d d z 1 = 0 z_1=0 z 1 = 0 z 2 = ∞ z_2=\infty z 2 = ∞ n ^ \hat{\bf n} n ^
z 1 = n 1 + i n 2 1 − n 3 , z 2 = − n 1 − i n 2 1 + n 3 , z_1 = \frac{n_1+i\,n_2}{1-n_3} \,,\quad z_2 = \frac{-n_1-i\,n_2}{1+n_3} \,, z 1 = 1 − n 3 n 1 + i n 2 , z 2 = 1 + n 3 − n 1 − i n 2 , which satisfy the antipodal relation z 1 z ˉ 2 = − 1 z_1\,\bar z_2 = -1 z 1 z ˉ 2 = − 1
z 1 + z 2 z 1 − z 2 = n 3 , 2 z 1 z 2 z 2 − z 1 = n 1 + i n 2 , 2 z 1 − z 2 = n 1 − i n 2 . \tfrac{z_1 + z_2}{z_1 - z_2} = n_3 \,,\quad \tfrac{2 z_1 z_2}{z_2 - z_1} = n_1 +
i\,n_2 \,,\quad \frac2{z_1 - z_2} = n_1 -i\,n_2\,. z 1 − z 2 z 1 + z 2 = n 3 , z 2 − z 1 2 z 1 z 2 = n 1 + i n 2 , z 1 − z 2 2 = n 1 − i n 2 . # Check of the above
n1, n2, n3 = symbols("n1 n2 n3", real=True)
z1 = (n1 + I * n2) / (1 - n3)
z1A = (-n1 - I * n2) / (1 + n3)
simplify((z1 * conjugate(z1A)).subs(n3, sqrt(1 - n1**2 - n2**2)))simplify((z1 + z1A) / (z1 - z1A))simplify((2 * z1 * z1A) / (z1A - z1))simplify((2 / (z1 - z1A)).subs(n3, sqrt(1 - n1**2 - n2**2)))As such, the rotation leaving such points fixed is
z ′ − z 2 z ′ − z 1 = e i θ z − z 2 z − z 1 . \frac{z'-z_2}{z'-z_1} = e^{i\theta} \,\frac{z-z_2}{z-z_1} \,. z ′ − z 1 z ′ − z 2 = e i θ z − z 1 z − z 2 . Solving for z ′ z' z ′
z → z ′ = z [ ( z 1 − z 2 ) cos ( θ / 2 ) + i ( z 1 + z 2 ) sin ( θ / 2 ) ] − 2 i sin ( θ / 2 ) z 1 z 2 2 z i sin ( θ / 2 ) + [ ( z 1 − z 2 ) cos ( θ / 2 ) − i ( z 1 + z 2 ) sin ( θ / 2 ) ] = ( d + i a 3 ) z − ( a 2 − i a 1 ) ( a 2 + i a 1 ) z + ( d − i a 3 ) \begin{aligned}
z \to z' &= \frac{z\left[(z_1 - z_2)\cos(\theta/2) + i
(z_1+z_2)\,\sin(\theta/2)\right]-2i\sin(\theta/2)z_1 z_2}{2z\,i\sin(\theta/2) +
\left[(z_1 - z_2)\cos(\theta/2)-i(z_1+z_2)\sin(\theta/2)\right]} \\
&= \frac{(d+i\,a_3)z - (a_2-i\,a_1)}{(a_2+i\,a_1)z + (d-i\,a_3)}
\end{aligned} z → z ′ = 2 z i sin ( θ /2 ) + [ ( z 1 − z 2 ) cos ( θ /2 ) − i ( z 1 + z 2 ) sin ( θ /2 ) ] z [ ( z 1 − z 2 ) cos ( θ /2 ) + i ( z 1 + z 2 ) sin ( θ /2 ) ] − 2 i sin ( θ /2 ) z 1 z 2 = ( a 2 + i a 1 ) z + ( d − i a 3 ) ( d + i a 3 ) z − ( a 2 − i a 1 ) where we make use of the notation introduced above which interrelates
{ n ^ , θ } \{\hat{\bf n}, \theta\} { n ^ , θ } { a , d } \{{\bf a}, d\} { a , d } r ^ ′ = ( ξ ′ , η ′ , ζ ′ ) = O ⋅ r ^ \hat {\bf r}' =(\xi',\eta',\zeta') = {\mathcal O} \cdot
\hat{\bf r} r ^ ′ = ( ξ ′ , η ′ , ζ ′ ) = O ⋅ r ^ z ′ = ( ξ ′ + i η ′ ) / ( 1 − ζ ′ ) z' = (\xi' + i\,\eta')/(1-\zeta') z ′ = ( ξ ′ + i η ′ ) / ( 1 − ζ ′ )
Fractionalized Coordinates ¶ Lastly, to re-interpret the above transformation in the complex plane as a
homogeneous one, it is conventional to write z z z
z = u v . z = \frac uv \,. z = v u . In terms of u u u v v v
u ′ = ( d + i a 3 ) u − ( a 2 − i a 1 ) v v ′ = ( a 2 + i a 1 ) u + ( d − i a 3 ) v . \begin{align*}
u' &= (d+i\,a_3)u - (a_2-i\,a_1)v \\
v' &= (a_2+i\,a_1)u + (d-i\,a_3) v \,.
\end{align*} u ′ v ′ = ( d + i a 3 ) u − ( a 2 − i a 1 ) v = ( a 2 + i a 1 ) u + ( d − i a 3 ) v . In this space, the rotations described above are a subgroup of G L ( 2 , C ) \mathbb{GL}(2,
\mathbb{C}) GL ( 2 , C )
O = ( A B C D ) such that ( u ′ v ′ ) = O ( u v ) {\mathcal O} = \begin{pmatrix} A & B \\ C & D\end{pmatrix} \quad\textrm{such that}\quad
\begin{pmatrix}u' \\ v'\end{pmatrix} = \mathcal O \begin{pmatrix}u \\ v\end{pmatrix} O = ( A C B D ) such that ( u ′ v ′ ) = O ( u v ) satisfies d e t O = 1 {\rm det}\, {\mathcal O} = 1 det O = 1 D = A ∗ D = A^* D = A ∗ B = − C ∗ B = -C^* B = − C ∗
O − 1 = ( D − B − C A ) . {\mathcal O}^{-1} = \begin{pmatrix} D & -B \\ -C & A\end{pmatrix} \,. O − 1 = ( D − C − B A ) .