I was hoping you would write a post on the ‘uninteresting case’ of p=5 in this context. Note that the truncated tetrahedron has (V,E,F)=(12,18,8) which is a triple that appears in the ternary (cyclic) geometry for the cube. This triple can be 4 hexagons and 4 triangles (the truncated tetrahedron) OR 4 pentagons and 4 squares!
Kea commented and I didnt know the answer to the ‘obvious’ question :
how can one get the truncated tetrahedron from either of the two conjugacy classes of order 5 elements in 
, each consisting of 12 elements.
Fortunately the groups involved are small enough to enable hand-calculations. Probably there is a more elegant way to do this, but I was already happy to find this construction…
This time, there is just one conjugacy class of subgroups isomorphic to 
(the symmetry group of the (truncated) tetrahedron) in 
. Take one of the two conjugacy classes C of 5-cycles in 
and use the following notation for its 12 elements :
A=(1,2,3,4,5), B=(1,2,4,5,3), C=(1,2,5,3,4), D=(1,3,5,4,2), E=(1,3,2,5,4), F=(1,3,4,2,5), G=(1,5,4,3,2), H=(1,5,3,2,4), I=(1,5,2,4,3), J=(1,4,2,3,5), K=(1,4,5,2,3), L=(1,4,3,5,2)
We’d like to view these elements as the vertices of a truncated tetrahedron, so we need to find the 4 triangles and the 6 connecting edges between them. The first task calls for order 3 elements, the second one for order two elements.
Take a conjugacy class of order 3 elements in say 
and observe that when one computes the products of T with a fixed 5-cycle in the conjugacy class C there is a unique element among the four obtained that belongs to the conjugacy class C. This gives a cyclic action on C with orbits of length 3 (the triangles). Here they are :
A–> J –> F –> A, B–>C–>H–>B, D–>G –> E–>D, I–>L–>K–>I
For the edges, take the conjugacy class 
of order two elements in and compute for any 5-cycle c in C the products cSc and observe that among the elements obtained there is again one element belonging to C. This gives the following pairing
A<-->C, B<-->I, D<-->F, E<-->H, G<-->L and J<-->K and a bit of puzzling shows that all this can indeed be realized within a truncated tetrahedron (on the right). As to her other request
… and how about a post on how 1 + 4 + 9 + … + 24^2 = 70^2 is REALLY a statement about unifying cusps and holes (genus) as degrees of freedom in quantum geometry.
The scarecrow will need to take some time to think before giving his answer…
geometry, groups, symmetry
1 comment
Posted in groups
Written on Fri, 04 July 2008 at 8:32 pm
Tags: geometry, groups, symmetry
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July 5th, 2008 at 6:52 am
Heh, cool! Thanks a lot!