[r-t] Methods as polyhedra
Philip Saddleton
pabs at cantab.net
Wed Apr 9 19:39:29 BST 2008
Mark Davies said on 08/04/2008 22:18:
> Hugh Pumphrey:
>
>
>> After making the original post I discovered that double court minor also
>> makes a truncated isosahedron.
>>
>
> Truncated icosahedron, presumably!
>
> It's interesting that all of these you've named so far are Archimedean
> solids, presumably this is because these solid have the property that all
> vertices radiate the same number of edges: in ringing terms, every leadhead
> has the same number of connections via plains and calls.
We need more than that - the graph has to be vertex-transitive (in other
words there is an automorphism taking any vertex to any other, so that
we can call the same touch starting from any vertex), hence so does the
polyhedron.
> Can we construct
> methods to match other such solids? According to Wikipedia, ones with
> sensible numbers of vertices (factorials or factorials/2) are:
>
We don't need this restriction - there are methods whose lead heads
generate a group of different order (e.g. Scientific Triples).
But restricting ourselves to graphs formed by the edges of polyhedra,
means we miss out on the best property of the truncated Icosohedron -
its symmetry group can be represented as acting on six elements, the
Hudson course heads for Stedman Triples.
In fact a rotation of the solid gives us natural permutations of
5 objects (tetrahedra with vertices at the centre of hexagons)
6 objects (diagonals with vertices at the centre of pentagons)
10 objects (the pentagons)
15 objects (diagonals with vertices at the centre of hexagons)
20 objects (the hexagons)
30 objects (the edges between two hexagons)
60 objects (the vertices)
Take the five hexagons which adjoin one of the pentagons. Number these,
clockwise, 1-5. For the other pentagon adjoining those numbered 1 & 2,
number the five hexagons 2, 1, 3, 5, 4. Extend this numbering so that
each pentagon adjoins one hexagon with each number, and the permutation
when read in order is in-course (there is now a unique way to complete
the numbering).
Each of the 60 in-course permutations occurs precisely once with this
arrangement. Label each vertex with the clockwise row surrounding its
pentagon and starting with the two numbers in the adjoining hexagons.
Now the centres of the four hexagons with a given number are the
vertices of a regular tetrahedron.
Because the rotation group is sharply transitive (there is precisely one
rotation that transforms any given vertex to any other), there is a 1-1
correspondence between elements of the group and vertices of the solid.
A rotation of the solid permutes the five tetrahedra, the permutation
being given by the vertex occupying the same position as rounds did
originally.
The edges between two hexagons correspond to place notation 3:
alternatively consider this as a rotation of the solid about an axis
through the centre of the edge of this type from rounds. Lines
corresponding to place notations 1 and 5 are not edges of the solid, but
can still be defined to give a graph (not including the edges of the
pentagons) that can be used to generate touches of Doubles. The
corresponding rotations are again about an axis through the centre of
the relevant graph edge from rounds.
The construction for Hudson's course heads is analogous. Label two
opposite pentagons 6, and the five closest to one of these as 1-5
reading clockwise. Their opposites then read 5-1 around the other 6. The
sixty course heads each appear once as five pentagons read clockwise
followed by the one in their centre. In each case label the vertex of
the central pentagon that has an edge connecting it to e.g. the leading
bell with that course head, and define edges and rotations corresponding
to S, H, L, Q. The axes for S, L, Q are orthogonal, since the
corresponding permutations commute.
--
Regards
Philip
http://myweb.tiscali.co.uk/saddleton/
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