Geodesics on Regular Polyhedra II

In my previous post we explored the problem with drawing simple geodesics, beginning and ending at the same vertex, on the tetrahedron and octahedron. Today we tackle the beasts that are the cube and icosahedron. Relevant papers for this section are Fuchs & Fuchs 2007 [1] and Fuchs 2016 [2].

This post will be a little shorter as I’m writing it on my birthday.

The Cube

We go through what should now be a familiar method of labeling the vertices of the cube and constructing the corresponding planar development. Similar to before, we may represent simple geodesics on the cube starting from a vertex ${a}$ as line segments on the square lattice of the form ${p\pmb{u} + q\pmb{v}}$ where ${p}$ and ${q}$ are coprime integers. In the planar development of the cube, we may separate the 8 vertices of the cube into two subsets of equal size. Vertices on the lattice whose coordinates have the same parity are labelled from the set ${\lbrace a,b',c,d'\rbrace}$ while vertices whose coordinates have opposite parity are labelled from the set ${\lbrace a',b,c',d\rbrace}$ .

The straight line segment ${[(a,b),(a+p,b+q)]}$ where ${p}$ and ${q}$ are coprime integers corresponds to a simple geodesic segment on the cube joining two vertices. An example is given in the above figure of the straight segment ${[(0,0),(3,2)]}$ corresponding to a simple geodesic joining ${a}$ and ${a'}$ on the surface of the cube. For a simple geodesic segment starting from ${a}$ and represented by the line segment ${p\pmb{u}+q\pmb{v}}$ , denote by ${V(p,q)}$ the endpoint of this segment. Can we find coprime integers ${p}$ and ${q}$ such that ${V(p,q) = a}$ ?

To explore this further, we must define a closed geodesic. A closed geodesic is a geodesic on a manifold which returns to its start-point with the same tangent direction. If we shift a simple geodesic starting at ${a}$ and ending at ${V(p,q)}$ to a parallel geodesic segment with a start-point within some small neighbourhood of ${a}$ , such a segment is necessarily not closed. To obtain a closed geodesic from this segment, we must repeat the segment a number of times. Denote by ${n_C(p,q)}$ the minimum such number.

Now, Theorem 4.4 of [1] states:

For all coprime ${p}$ and ${q}$ , ${n_C(p,q) = 2}$ , ${3}$ , or ${4}$ .

Figure 2: The transformations of the square lattice, source [2]

The nail in the coffin for our magical simple geodesic beginning and ending at ${a}$ is Theorem 4.4 (2) of [2]:

Consider a simple geodesic segment on the cube beginning at ${a}$ and ending at ${V(p,q)}$ . The vertex ${V(p,q)}$ is one of ${a'}$ , ${b}$ , or ${d}$ if ${n_C(p,q) = 4}$ , is one of ${b'}$ , ${c}$ , or ${d'}$ if ${n_C(p,q) = 3}$ , and is ${c'}$ if ${n_C(p,q) = 2}$ .

While the result as a whole is interesting, we are interested only in the part which proves that a simple geodesic starting at ${a}$ cannot end at ${a}$ . This part of the proof uses the symmetries of the function ${V}$ . For example, one may replace the standard square of the lattice representation of the cube with one of the parallelograms in figure 2. For a simple line segment to end at ${a}$ , we would require ${p}$ and ${q}$ to be coprime and both odd. Through repeated applications of the invertible transformations in figure 4, and some sign changes where necessary, any line segment corresponding to coprime odd integers ${(p,q)}$ can be transformed to a line segment corresponding to ${(1,1)}$ . The line segment corresponding to ${(1,1)}$ begins at ${a}$ and ends at ${c}$ . We can express any line segment corresponding to odd coprime integers ${(p,q)}$ as some transformation of the line segment corresponding to ${(1,1)}$ . But via the above transformations in figure 2, we have the left transformation maps ${c}$ to ${c}$ and repeated use of the right transformation yields the loop ${c -> b' -> d' -> c}$ . Therefore we cannot have a simple geodesic segment beginning and ending at ${a}$ .

The Icosahedron

We begin, as ever, with a labelling of the twelve vertices of the icosahedron, labelling 6 vertices ${a}$ to ${f}$ , and their respective opposites ${a'}$ to ${f'}$ . As with the cases for the tetrahedron and octahedron, the surface of the icosahedron can be represented as a triangular tiling of the plane. Let a simple geodesic starting from ${a}$ be represented by ${p\pmb{u}+q\pmb{v}}$ , where ${p}$ and ${q}$ are coprime. Again any parallel segment within some small neighbourhood of ${a}$ is not closed. The number ${n_I(p,q)}$ for a geodesic segment on the icosahedron has an analogous definition to the number ${n_C(p,q)}$ for a geodesic segment on the cube. Well, with the necessary definitions, the case for the icosahedron follows the formula prescribed for the cube.

Theorem 6.1 of [1] states:

For all ${p}$ and ${q}$ coprime, the number ${n_I(p,q)}$ takes the value ${2}$ , ${3}$ , or ${5}$ .

Figure 4: The transformations of the triangular lattice, source [2]

With Theorem 5.4 (3) of [2] the case is closed for simple geodesics on the icosahedron.

Consider a simple geodesic on the icosahedron starting from ${a}$ and ending at some vertex ${V(p,q)}$ ; suppose it corresponds to some coprime ${p}$ and ${q}$ . The vertex ${V(p,q)}$ is one of ${b}$ , ${c}$ , ${d}$ , ${e}$ , or ${f}$ if ${n_I(p,q) = 5}$ , it is one of ${b'}$ , ${c'}$ , ${d'}$ , ${e'}$ , or ${f'}$ if ${n_I(p,q) = 3}$ , and it is ${a'}$ if ${n_I(p,q) = 2}$ .

The relevant details of this proof pertaining to the non-existence of simple geodesics beginning and ending at ${a}$ is, of course, left as an exercise to the reader 😉 .

In my next post we shall address the final boss of the platonic solids: the dodecahedron.

Bibliography

D. Fuchs, E. Fuchs, Closed geodesics on regular polyhedra, Mosc. Math. J., 7, 2007, 265-279
D. Fuchs, Geodesics on Regular Polyhedra with Endpoints at the Vertices, Arnold Math J. 2, 2016, 201–211

The Cube

The Icosahedron

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