Icosahedral Symmetry

The icosahedral group is defined as the group of rotations that map the icosahedron into itself. Placing two antipodal vertices along the $\zeta$ axis at $(0,0,\pm 1)$, and one next-nearest vertex at ${\bf\hat n}_n = (-\sin\,2\alpha,0,\cos\,2\alpha)$, we find that there is a two-fold symmetry about the axis ${\bf \hat n}_2:=(-\sin\alpha,0,\cos\alpha)$, in addition to a five-fold symmetry about ${\bf \hat n}_5:=(0,0,1)$:

It can be shown that the above symmetry generators can be combined to reproduce the 60-fold symmetries of the icosahedral group. In particular, the three-fold symmetry of one of the icosahedron faces nearest to the top vertex can be written as

given an appropriate choice of ${\bf\hat n}_2$ (or $\alpha$) and ${\bf\hat n_3}$.

To determine $\alpha$, note that the $\{{\bf a}, d\}$ parametrizations of the above transformations are

Recalling the composition property of the $d$ parameter, $d_C = -a_{1A} a_{1B} - a_{2A} a_{2B} - a_{3A} a_{3B} + d_A d_B$, we find that

for the composed transformation. Therfore,

Similarly,

which together imply that

(${\bf\hat n}_3$ is guaranteed to be unit-normalized whenever $\alpha$ satisfies the condition above.)

Moving to stereographically-projected coordinates, the vertex at ${\bf\hat n}_n$ is given by

However, we will find that it is more convenient to instead write

since the other four vertices related to $z_n$ under the five-fold symmetry transformation $V$ are given by

The equivalence of the two expressions above follows from the defining relation

which together imply that

(as can be checked by squaring the second equality above). Noting that $\sin (\pi/5) = (\epsilon^2-\epsilon^3)/2i$, the previous condition on $\alpha$ becomes

and so

Across from $z_n$ is an antipodal vertex $z_m:= -1/z_n$ given by

In addition to the vertices $z_0 = 0$, $z_\infty = \infty$ and the five $z_k$ (where $z_5 = z_n$), the remaining five are related to $z_m$ under the five-fold symmetry transformation $V$ as follows:

(where $z_{10} = z_m$, similarly to $z_n$).

Relating $z_n$, $z_m$ back to to their explicit values, we see that

(where the last equality follows from the cosine half-angle formula). Recalling that $z_n$ and $z_m$ are antipodal, we see that $\phi$ must satisfy

Therefore, $\phi$ is the golden ratio: $\phi = (1+\sqrt{5})/2$ (since $\phi$, by definition, is positive). Additionally, since the $\epsilon$’s satisfy $\sum_{k=1}^5\epsilon^k = 0$, we find

which is consistent with the above.

Lastly, the homogeneous transformation corresponding to the symmetry transformation $IV$ is given by

whereas transformation $V$ corresponds to $u' = \epsilon^{1/2}\, u$, $v' = \epsilon^{-1/2}\, v$.

Icosahedral Polynomial Invariants¶

Multiplying together the homogeneous polynomials $(u-z_k\,v)$ for $k\in \{0,1,\ldots,10,\infty\}$, we find

Given the complexity of this expression, it makes sense to introduce a function which will efficiently expand out any polynomial and $\epsilon$ and enforce the conditions

such that the maximum power of $\epsilon$ appearing is 3. These considerations motivate the definition of the function e_simp() below:

As a second check, it’s worth pointing out that, for any real antipodal pair $(z_n, z_m)$ with $z_m=-1/z_n$, the following result holds:

$$

\begin{aligned} P(u,v,z_n,z_m) :=\prod_{k=1}^5\left(u - \epsilon^k,z_n,v\right)\left(u - \epsilon^k,z_m,v\right) &= \prod_{k=1}^5\left[u^2-\epsilon^k\left(z_n + z_m \right),uv-\epsilon^{2k},v^2\right] \ &= u^{10} - \left(z_n^5 + z_m^5\right)u^5v^5 - v^{10} ,. \end{aligned}

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This, combined with the product $(u-z_0\,v)(u-z_\infty\,v) = uv$, demonstrates the expected formula for the polynomial invariant $f(u,v)$ associated with the icosahedral vertices, since

and so

(The above follows from repeated use of the defining relations $z_n^2 = 1-z_n$, $z_n\,z_m = -1$, and $z_n+z_m=-1$.)

It should be straightforward to check that $f(u,v)$ is indeed a polynomial invariant under transformations $IV$, $V$:

Computing the Hessian of $f$, we find ${\rm Hess}[f](u,v) = -121\,H(u,v)$ where

We shall see later that this Hessian is precisely the polynomial invariant for the face centres of the icosahedron.

The Rotated Octahedron¶

Earlier, we demonstrated the that function $t(u,v) = u^5\,v-u\,v^5$ was an absolute invariant under tetrahedral symmetry, and that its square was an absolute invariant under octahedral symmetry. We now wish to consider how this particular product of roots changes under a rotation of the octahedron about its centre.

To this end, consider the rotated points $z'_k$, related to $z_k \in \{0, \infty, \pm 1,\pm i\}$ by

(The rotation above corresponds to a rotation about the $\eta$ axis by an angle $\alpha$, followed by a rotation about the $\zeta$ axis by an angle $\theta$.) Computing the product $t'(u,v) := \prod_{k}(u-z_k'\,v)$ we find

Note that the original expression for $t(u,v)$ arises only in the singular limit $\alpha\to 0$.

Icosahedral Edge Midpoints¶

If we now consider the product of 5 such functions, representing the 30 edge midpoints of the icosahedron, with each inscribed octahedron rotated by $e^{2\pi i/5}$ relative to one another, we find (introducing $w$ as a stand-in for $\tan(2\alpha)$)

Lastly, recognizing that the icosahedral edge midpoints are located at

we find that $w:=\tan(2\alpha) = 2\phi/(\phi^2-1) = 2$ and so

We also note in passing that $T(u,v)$ can also be expressed as

where

The factor $\epsilon^{15-3k}$ is engineered such that only non-negative integer powers of $\epsilon$ appear in the resulting expression, which allows more straightforward simplification. (Recall that $\epsilon^5=\epsilon^{15} = 1$.)

Icosahedral Face Centres¶

The icosahedral face centres nearest to the vertex ${\bf n}_5 = (0,0,+1)$ are given by

where we choose $w_1$ to be the ($6\pi/5$)-rotated stereographic projection of ${\bf n}_3$:

(The choice of rotating by $6\pi/5$ is made such that $w_1$ is positive and real.)

Defining $k:=\sqrt{3} / \cos\alpha$, and noting that

we find that we can simplify the expression for $w_1$ as follows:

where, in deriving the last equality, we make liberal use of the identity $z_n^2 = 1-z_n$ when rewriting the denominator.

Since, for arbitrary $A$ and $B$

we can also write $1/(1+z_n) = z_n$ (choosing $A=B=1$ in the above), which gives

This expression, with only positive powers of $k$ and $z_n$, guarantee that successive powers of $w_1$ can be easily simplified using the relations $k^2=3(3+z_n)$, $z_n^2 = 1-z_n$.

Using similar methods, one can easily show that the antipodal point ${\hat w}_1:= -1/w_1$ can be written

Therefore, the product of polynomial factors with antipodal pairs, as in (23), are found to be

However, the above expression only accounts for half of the face centres of the icosahedron.

The other 10 face centres can be found by using the twofold symmetry $(IV)$ of the icosahedron rotating $\hat w_1$ about ${\bf n}_2$ by $\pi$:

From here, the other four face centres (related by symmetry $V$) can be easily determined by applying more powers of $\epsilon$, and their antipodal pairs are given by

Computing $P(u,v,w_6, {\hat w}_6)$ as before, we find

Putting this all together, we find that the polynomial invariant resulting from the product of all 20 icosahedral face centres is given by

(The above results from the general product

where we take $X=64-100\,z_n$, $Y=64+100(1+z_n)$.

Lastly, we can verify that the polynomial invariants discussed above are interrelated as follows:

Edge Midpoints Redux¶

An entirely orthogonal approach to the computation above is to consider directly the complex coordinates of each midpoint. We introduce here a calculus which can be used to simplify expressions involving these quantities.

We begin by recalling that the icosahedral edge midpoint on the negative real axis is given by

where, in the last line, we define $j:=-\csc\alpha$ (recall that $z_n = -\cot\alpha$, as found earlier).

By directly computing the square of $j$, we can verify that $j^2=2-z_n$:

From here on out, it becomes quite straightforward to compute the invariant polynomial: for example, the antipodal point to $v_1$ is

The next rung of midpoints can be expressed in terms of $v_1$ by applying a five-fold rotation, a reflection in ${\bf \hat n}_2$, and then three more five-fold rotations (yielding the edge midpoint lying on the positive real axis, which we refer to as $v_9$):

We can also compute the antipodal point $\hat v_9$ by inverting the expression above:

Since the three pairs of antipodal points $(i,-i)$, $(v_1, \hat v_1)$ and $(v_9, \hat v_9)$ compromise one of the five octohedra, we now have a second way to compute the polynomial invariant $t_I(u,v)$: since pairs of real antipodal points $(z, \hat z)$ contribute a factor $(u^2 - (z+\hat z)uv - v^2)$, it’s sufficient to compute the sums

Therefore, the factor for this octohedron is given as before:

The contributions to $T(u,v)$ from the other 24 edge midpoints can be determined by applying successive five-fold rotations about the $\zeta$ axis:

For completeness, we also record for later use the corresponding Hessian of the octahedral vertex function $t_r(u,v)$, representing the polynomial invariant for the faces of each of the five inscribed octahedra: