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The recipes

From a polyhedron to a divisor

A rational function on the Riemann sphere is determined, up to a constant factor, by its divisor: a finite set of points, each carrying a nonzero integer order. Positive orders are zeros and negative orders are poles. The orders must sum to zero over the whole sphere, counting infinity.

A recipe places such points according to a convex polyhedron with its centre at the origin:

  • vertices at their directions from the centre;
  • faces at their outward normals, which are the vertex directions of the polar dual;
  • edges at the foot of the perpendicular from the centre, a direction shared with the dual edge.

The default recipe, R2, counts incidences:

\[ D_{R2} \;=\; \sum_{v} \operatorname{val}(v)\,[v] \;-\; \sum_{f} \operatorname{sides}(f)\,[f]. \]

Both sums equal \(2E\), so the divisor balances automatically. The alternative, R4, pairs cells with edges instead:

\[ D_{R4ve} = \sum_v \operatorname{val}(v)\,[v] - 2\sum_e [e], \qquad D_{R4fe} = \sum_f \operatorname{sides}(f)\,[f] - 2\sum_e [e]. \]

For these unreduced divisors, \(D_{R2} = D_{R4ve} - D_{R4fe}\). Polarity exchanges vertices and faces, so R2 on the polar dual gives exactly the reciprocal function, and R4ve and R4fe swap.

To see these divisors side by side in 3D, for any solid in the corpus, use Compare recipes.

Fixing the constant

The relief needs an absolute modulus, not one defined up to a constant. With \(\chi\) the chordal (straight-line) distance between points of the unit sphere, the modulus is

\[ \log|f(x)| \;=\; \sum_i m_i \log \chi(x, a_i). \]

It has three useful properties:

  • self-dual: the negated divisor gives \(1/|f|\);
  • rotation-covariant: rotating the polyhedron rotates the relief;
  • geometric mean 1 over the sphere: sea level, \(|f| = 1\), sits in the middle of the relief.

Worked example: the octahedron

step result
polyhedron regular octahedron: 6 vertices of valence 4, 8 triangular faces
R2 divisor zeros of order 4 at the 6 vertices; poles of order 3 at the 8 face normals, which point to the cube's vertices
degree \(6 \cdot 4 = 8 \cdot 3 = 24\)
function \(f(z) = C\, V(z)^4 / F(z)^3\), with Klein's octahedral forms \(V = z(z^4-1)\) and \(F = z^8 + 14z^4 + 1\)
constant \(C = 729/4\) exactly; the earlier sampled value was 0.55% off
object the Cube–Octahedron Dual 
from polyhedral_functions import fixtures
from polyhedral_functions.recipes import R2, apply

d = apply(R2, fixtures.load("octahedron")).divisor
print(d.summary())  # 6 zeros of order 4, 8 poles of order 3, degree 24

On the Platonic solids R2 reproduces Klein's classical invariants. The interesting cases are the solids beyond them, where recipes that agree on Platonic inputs come apart. That story, and the evidence for choosing R2 and keeping R4, is in the exploration.