[r-t] Rotations of a truncated icosahedron
Philip Saddleton
pabs at cantab.net
Sun Nov 14 21:29:36 UTC 2004
A truncated icosahedron has:
12 pentagonal faces
20 hexagonal faces
60 vertices
30 edges between two hexagons
60 edges between a hexagon and a pentagon
(Footballs often have the pentagons coloured black and the hexagons
white)
The rotation group can be represented as permutations of each of these
sets. In each case it is clearly transitive (a rotation can be found
that takes any element to any other), and in the case of the sets of
size 60 is sharply transitive (there is precisely one such rotation).
Opposite pairs must remain opposite in each case, so there are also
representations as permutations of 6, 10 and 15 objects.
Consider the pairs of pentagonal faces (or equivalently the six
diagonals of the icosahedron). We have a transitive group of order 60
that permutes these. There is only one such - the Hudson group. This is
isomorphic to A_5, so we should also be able to find a set of five
objects that are permuted.
The elements of order two are rotations of 180 degrees. The axes pass
through the centres of edges between hexagons: there are 15 pairs of
these. Any element of A_5 of order two fixes one object, so these must
split into five sets of three. These sets of three are orthogonal, and
their ends are the vertices of an octahedron (or the centres of the
faces of a cube). The rotation group of an octahedron is S_4 (sharply
transitive on the four diagonals of a cube), but the stabilizer within
our rotation group is only have half of this, since the edges must line
up (rotations through 90 degrees are not permitted): this is like an
old-fashioned football, where six square faces are each made up of three
rectangular strips.
Back to Cayley graphs: since the representation as permutations of
vertices is sharply transitive, we can label each vertex with a
permutation: the edges between two hexagons represent permutations of
order two (a), and those (directed) between hexagons and pentagons order
five (b). To traverse a hexagon correctly, ab has order 3. There are
several ways of achieving this, but the natural way seems to be to make
a correspond to pn 3 and b equivalent to two changes of plain hunt. The
rows of plain hunt are then alternate between the vertices of two
opposite pentagons.
--
Regards
Philip
http://www.saddleton.freeuk.com/
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