From:John Conway <conway@math.princeton.edu>Date:Wed, 16 Mar 94 18:36:15 ESTTo:burgiel@math.washington.edu, conway@math.princeton.eduSubject: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