From:           John Conway <conway@math.princeton.edu>
Date:           Wed, 16 Mar 94 18:36:15 EST
To:             burgiel@math.washington.edu, conway@math.princeton.edu
Subject:        Re:  Archimedean Solids

Of course it's only  n-gon x n-gon whose squares realise a regular
map - but  n-gon x m-gon is still "Archimedean".

   There are certainly some quite interesting maps among the other
Archimedean polytopes - for instance if you take the polytope

                  3       4       3
               .------O-------O-------.

(the mid-truncation, or  1 1/2 'th ambo version of the 24-cell),
then the cells are truncated cubes, which touch along their
octagonal faces in the twisted way, and you get a nice map of
octagons, 4 to a vertex.

   However, there aren't too many such cases, and I expect they've
already been "mined".  It was only a month or two ago that I re-listed
all the vertex-figures of the archimedean 4-polytopes (not that I
have the list here now, of course), and it's easy to see that not
too many of them include regular skew polygons among the edges of
their vertex figures.  

   I see the English of that got a bit mixed up, since "vertex figures"
was already the subject of the subordinate clause, but you see what I
mean anyway.

   I should imagine that a better way to find examples would be to
first find the maps in abstracto, then do the eigenvalue-type graph
theory to find their embeddings, which might be in quite high-dimensional
spaces.

   For example, letting  1,A,B,C  be the maps that take the typical
vertex of an icosahedron to itself, the sum of its neighbors, the
sum of the neighbors of its antipode, and its antipode, we find:

     A^2 = 5 + 2A + 2B,  BA = 2A + 2B + 5C,  CA = B

so  (A^3 - 2A^2 - 5A) = 2BA = 2A + (A^2-5-2A) + C,   so

No!  error!                 = 4A + 2(A^2-5-2A) + 10C, so

    (A^3 - 4A^2 - 5A + 10) = 10C  if I'm not mistaken, so

    A^4 - 4A^3 - 5A^2 + 10A = 10B = 5A^2 - 10A - 25,  or

      (1,-4,-10,20,25)(A) = 0,  which I factorize:
       1 -5
          1  -5
             -5 25
                -5 25    as  (A-5)(A^3 + A^2 - 5A - 5)

    ie.,   (A-5)(A+1)(A^2-5) = 0

So the eigenvalues of  A  are   5, -1, r5, -r5      (r5 = sqrt(5))

with multiplicities found as    1   5   3    3

   So the icosahedron has 4 irreducible "regular representations",
namely the trivial one, the one in which opposite vertices coincide
to give the vertices of a regular 5-simplex, the standard 3-dimensional
one, and the conjugate 3-dimensional one in which it becomes the
great icosahedron.

     You could now mix these a bit, for instance getting an 8-dimensional
representation by adding a 5 and a 3.  (Since there's a scale parameter
involved, this actually gives a 1-parameter family of such representations.)

    In general, if the permutation character involves k distinct
irreducibles, you get k basic representations this way (which, as in the
above case, may not all be faithful), and then as far as I can see
you can just take any linear combination of them to get the most
general regular embedding.  However, I'm not quite sure what happens
when the character isn't multiplicity-free.  I must work out the
dodecahedron as prototype.

            JHC