[GAP Forum] Re: icosahedral group question

[Mathieu Dutour]

Dear All,

just for your information. I have written an unfinished package
named plangraph for dealing with planar graphs (you can get it
from http://www.liga.ens.fr/~dutour/PlanGraph/plangraph.tar.gz).

It could be useful for such routine computations:
----
RequirePackage("plangraph");
PL:=ArchimedeanPolyhedra("TruncatedIcosahedron");
GRPinfo:=GroupPlan(PL);
GRP:=GRPindo.SymmetryGroup;
LN:=NormalSubgroups(GRP);
A5:=First(LN, x->Order(x)=60);
----
Feel free to ask me if you have some questions.

  Mathieu


On Wed, May 17, 2006 at 03:59:37PM +0200, Rudolf Zlabinger wrote:
> Dear Walter Becker,
> 
> attached to this message a GAP file generating a rotation group "fullerene".
> 
> Its the rotation group A5 (icosahedral group) acting on the 60 points of
> your "buckyball".
> 
> This group supports a special labelling of the original icosahedron as
> already outlined in my first message "icosahedron exercises". I repeat this
> here:
> 
> Having a blueprint in mind of icosahedron, there is a top vertex, two
> pentagons one on top of another, the second pentagon rotated by 36 degrees
> clockwise against the first one, and a bottom vertex.
> 
> As we label the top vertex by 1, the two pentagons by 2,3,4,5,6 and
> 7,8,9,10,11 clockwise looking in bottom direction respectively and the
> bottom vertex by 12, we have following picture:
> 
> The respective opposite vertices are: (1,12), (2,9), (3,10), (4,11), (5,7),
> (6,8).
> 
> The edges are:
> 
> (1,2),(1,3),(1,4),(1,5),(1,6),
> from the top vertex to first pentagon
> (12,7),(12,8)(12,9)(12,10)(12,11)
> from the bottom vertex to second pentagon
> (2,3)(3,4)(4,5)(5,6)(6,2)
> the top pentagon
> (7,8)(8,9)(9,10)(10,11)(11,7)
> the bottom pentagon
> (2,11)(2,7)(3,7)(3,8)(4,8)(4,9)(5,9)(5,10)(6,10)(6,11)
> between the pentagons
> 
> Following this labels of the vertices of the icosahedron you have to derive
> the labels of the 60 vertex "buckyball" as follows:
> label a pentagon by 1,2,3,4,5 and then the other pentagons by (x-1)*5 + y,
> where x is the original vertex label of the icosahedron and y the numbers
> 1,2,3,4,5. The respective adjacent pentagons have to be labelled such as
> after rotating a pentagon into an adjacent (or also another) pentagon the
> respective target and image labels are congruent modulo 5.
> 
> Using the rotation group fullerene, you can derive the positions of the such
> labelled "buckyball" vertices directly as result of a rotation permuation,
> where the group contains all possible rotations of them.
> 
> best regards, Rudolf Zlabinger
> PS.: by the way, I am not graduated, so to be correct, I have to tell it.

-- 
Mathieu Dutour Sikiric		Researcher in Math
Tel. (+972)2 65 84 103		and Computer Science
Fax. (+972)2 56 30 702		Einstein Institute of Mathematics
E-mail: Mathieu.Dutour at ens.fr	Hebrew University of Jerusalem
http://www.liga.ens.fr/~dutour	Israel