# [r-t] Star polyhedra

Thu Apr 10 19:46:32 BST 2008

```Mark Davies said  on 09/04/2008 23:46:
>> Aren't the graphs of these isomorphic to the icosohedron or dodecahedron?
>>
>
> Ah, I think you were thinking just of the regular star polyhedra, aka
> Kepler-Poinsot polyhedra - there are only four of these. But the faces don't
> all have to be the same, since all we need is the vertices to be transitive.
> There are quite a few uniform nonconvex non-regular polyhedra - 54 or so I
> think - only some of which are edge & vertex equivalent to convex forms.
>
>
So, the general question is: which polyhedra can the vertices and edges
form the Cayley graph?

Equivalently (I think): is there a subgroup of the automorphism group
that is sharply vertex transitive?

For most of the Archimedean solids, the answer is trivially yes, since
they are not rotationally symmetric about a line through a vertex, and
hence either the rotation group (or in the cases where there are two
types of vertices, the full symmetry group) has the same order as the
number of vertices. The exceptions are the cuboctahedron and
icosidodecahedron.

For the Platonic solids we need to find a colouring of the edges that
gets rid of the excess symmetry.

The tetrahedron and cube are easy: three different coloured edges at
each vertex.

The octahedron: colour thedges of two opposite triangles with one
colour, and the remaining edges with another. Since there are two edges
of each colour at each vertex these must be directed. This gives a
six-cycle and a three-cycle for the two types of lead, with the
three-cycle equivalent to two or four times the six-cycle. A
representation on five bells has Plain lead 12453 and Single 21453 (note
that Decision (E)A.2 meands we can't have the six-cycle as the plain
lead, as the call doesn't move to a new course). Alternatively, use two
colours to join the triangles.

The icosahedron: since there is an odd number of edges at each vertex,
one must be non-directional. Give the two faces meeting at this edge the
same colour. Since no two adjcent edges are of this type, a face has one
or zero such edges, and there are 12 faces of this colour, three
adjacent to each vertex. Give the same colour to the faces that share a
vertex, but not an edge with these two. Extend to give twelve faces.
Colour the remaining eight faces as two sets of four, such that no faces
in the same set share a vertex. The rotation group is A4, and has a
representation as permutations of one of the sets of four. Double Bob
Doubles, with p=5432, b=3542, s=5324.

--
Regards
Philip