[r-t] Star polyhedra

[Philip Saddleton]

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
http://myweb.tiscali.co.uk/saddleton/