If I have zeros at the vertices of an icosahedron, where should the poles go?

Question

I've been tinkering with Newton's method applied to polynomials. E.g., Newton's method for $z^5 - 1 = 0$ gives:

There aren't a lot of symmetric patterns of finite sets of points in the plane, so I decided to rerun the code on the Riemann sphere using a version of Newton's method with spherical symmetry. I thought this would allow me to use point patterns corresponding to all the Platonic solids. However, only the tetrahedral case works:

This image has tetrahedral symmetry using the identification of $\mathbb{C}_\infty$ with $S^2$. However, in order to get a rational function with tetrahedral symmetry, we need to decide where to put the poles. A polynomial no longer works: all of its poles are at $\infty$, which is not symmetric. Happily, for a tetrahedron there is a convenient place to put the poles: in the middle of each face. The face centers are at the antipodes $-1/\bar{z}$ of the vertices, so our rational function is

$$f(z) = \prod_{k=1}^4 \frac{z-z_k}{z+1/\bar{z_k}}$$

where $z_{1\ldots4}$ are the stereographically projected vertices of the tetrahedron. This rational function has tetrahedral symmetry, so a spherically symmetric version of Newton's method applied to it produces a tetrahedrally symmetric image.

Unfortunately, all of the other Platonic solids have vertices at the antipodes of their vertices (they are symmetric under $p \mapsto -p$). Let's consider the icosahedron specifically:

Question: If we put a simple zero at each of the 12 vertices of the icosahedron, what's the most symmetric place to put 12 poles (not necessarily simple)?

One answer is to inscribe a tetrahedron inside the icosahedron with tetrahedron vertices at face centers, and put order 3 poles at each tetrahedron vertex. This produces an image with tetrahedral symmetry:

Is that the best one can do?

Edit: Here's a version of the icosahedral symmetry one using @pregunton's linked rational function $F_{3,5}(z)$, but taking the fifth root to turn the zeros back into simple zeros and the order 3 poles into order $3/5$ singularities since that produces a slightly nicer picture:

Higher resolution versions of these images: quintic, tetrahedron, icosahedron, modified icosahedron.

Answer 1

You can color the icosahedron red and blue, such that three red faces and two non-adjacent blue faces meet at each vertex. Then you can put poles in the middle of the twelve red faces, which treats all of the vertices symmetrically.

Geoffrey's edit: The resulting render is

(Higher resolution)

Answer 2

MR1032073 Doyle, Peter; McMullen, Curt, Solving the quintic by iteration. Acta Math. 163 (1989), no. 3-4, 151–180.

Answer 3

In this answer from Math SE I give the inscription of a regular octahedron into a regular icosahedron by specifying the midpoints of certain icosahedral edges as the octahedral vertices.

We can then use this inscription in either of two ways to elegantly specify the poles. Both alternatives below will give the point group symmetry $T_h$, which has the same symmetry order as the usual tetrahedral symmetry $T_d$ but the symmetry elements are different.

1. Where an icosahedral edge has an octahedral vertex place a simple pole at the centers of the faces sharing that edge. The result is equivalent to Matt F's answer.

2. Place a second order pole directly at each octahedral vertex, coalescing the twelve proposed simple poles into six second-order ones. This arrangement gives a polar arrangement with octahedral symmetry, $O_h$; but only the $T_h$ component also includes the icosahedral vertices where the zeroes are located. It also has the property that the same set of poles would apply to a second icosahedron, obtained from the first by rotating 90° about the $D_{2h}$ axis passing through any pair of opposing edge midpoints. The two icosahedra taken together then duplicate the octahedral symmetry.

---

I also explore the remaining three centrosymmetric platonic solids. Somewhat unexpectedly, none of these allows preservation of any tetrahedral symmetry if we insist on simple zeroes at all the vertices. The regular icosahedron has twelve vertices, which is divisible by both three and four; this allows the poles to be placed in ways that preserve the tetrahedral symmetry without using the vertices or their antipodes (the latter also being vertices for centrosymmetric solids). With the different vertex counts of the cube, octahedron and dodecahedron we are less fortunate.

Start with the cube. With eight simple zeroes at the vertices and thus needing eight ($\equiv2\bmod3$) simple poles, we would have to place two simple poles (or a single second-order pole) directly on a threefold axis to preserve it. But such points project radially back onto the vertices where we need to have the zeroes, creating a contradiction if we want the poles actually on the cube.

So we give up the threefold symmetry component and settle for an arrangement that preserves what we can, $D_{4h}$ symmetry. There are several choices for this. Among them:

1. Place second-order poles at the centers of two pairs of opposing faces

2. Place simple poles on two of the three sets of parallel edges

3. Place a pole at each face center and render one opposing pair as second-order

4. Taking the vertices as $(\pm1,\pm1,\pm1)$ place a simple pole at $(1,0,\sqrt2/2)$ and its $D_{4h}$ equivalents with the $z$ axis as the $D_{4h}$ axis. This places the poles at the vertices of a smaller cube (but the symmetry is still only $D_{4h}$).

For the octahedron, it is the threefold symmetry we can preserve, having to give up symmetry around the fourfold octahedral axes. There are several choices giving $D_{3d}$ symmetry, which is what remains if we surrender the fourfold axes. One such arrangement is to place a simple pole at the midpoints of six edges, the edges being chosen so that traversing them gives a closed skew-hexagonal path. The poles themselves then lie at the vertices of a regular planar hexagon.

Finally for the dodecahedron, as for the cube, preserving threefold rotational symmetry will eclipse the desired zeroes at the vertices; so we reduce the symmetry to $D_{5d}$. A suitable arrangement with simple poles is to place ten of them at the edge midpoints of two opposing faces and the rest at the midpoints of the "equatorial" edges halfway between those opposing faces.

A summary of the obtained symmetries shows the value of having the right number of vertices when the vertices come in antipodal pairs:

Terahedron (4 vertices, no antipodal pairs) -- $\color{blue}{T_d}$, order $\color{blue}{24}$

Cube (8 vertices, yes antipodal pairs) -- $D_{4h}$, order $16$

Octahedron (6 vertices, yes antipodal pairs) -- $D_{3d}$, order $12$

Dodecahedron (20 vertices, yes antipodal pairs) -- $D_{5d}$, order $20$

Icosahedron (12 vertices, yes antipodal pairs) -- $\color{blue}{T_h}$, order $\color{blue}{24}$