[GAP Forum] Indexing of Ikosaedrons vertices

[Rudolf Zlabinger]

Dear MCKAY john,

Thank you for your hint to the finite rotation groups, you are right, the
icosahedron (dodecahedron) has 3 pole classes, with 2,3,5 cycle rotations,
resulting in 30,20,12 cosets.

My theoretical problem was, to translate this structure to mere permutation
representations. Distributing numbers 1 to 12 to the vertices has to some
degree influence to the selection of distinct permutation groups
representing the icosahedron indexed by these numbers.

As in our example indexing the icosahedrons vertices from top to down as 1,
(2,3,4,5,6),(7,8,9,10,11), 12 lead to the request, that a representing group
has to contain the permutation (2,3,4,5,6)(7,8,9,10,11) as the rotation
around the axis 1,12.

The problem is, that there is no formal way to describe a special indexing
of the vertices, as to do it intuitively as above. So i am looking for a
method to derive the indexing(s) from the selected permutation group(s) in
reverse order if this is possible.

One way to do it is to derive the vertex indices from one of the 5 cycles,
taking the 2 missing points (vertices) as axis. This results in 4 different
possible kinds of indexing the vertices, these are the 5 cycle permutation ^
1,2,3,4. In this way, the indexing is (partially) determined and described
by a selected special permutation group.

Thank you so far, best wishes, Rudolf Zlabinger

----- Original Message ----- 
From: "MCKAY john" <mckay at encs.concordia.ca>
To: "Rudolf Zlabinger" <Rudolf.Zlabinger at chello.at>
Cc: "Thomas Breuer" <thomas.breuer at math.rwth-aachen.de>;
<forum at gap-system.org>
Sent: Tuesday, May 09, 2006 3:38 PM
Subject: Re: [GAP Forum] Numberings of Ikosaeders vertices



May I put in a word? I have worked with the groups:

[a,b,c] = <x,y,z: x^a=y^b=z^c=xyz=1> 1/a+1/b+1/c > 1.
These are finite subgps of SO3.
[a,b,c] is the icosahedral group = dodecahedral group.

use the flags. One has a vertex incident with an edge incident
with a face. The operations x,y,z fix each of these.

For your icosahedron we have [2,3,5]. You can make a
dodecahedron by constructing a pentagon (fold a strip of paper)
and five others adjacent to it. Repeat for the
other half. Put them on top of each other with corners
alternating and an elastic band round them. You can easily
number vertices, edges, and faces. V-E+F=2 30-20+12=2.

Unlike the symmetric group, as you say, here we have two
choices for the element of order 5. Take either z or z^2
of order 5 and they are in distinct conjugacy classes in
[2,3,5].

Good luck!

By the way, the figures in stone were known to late neolithic
man in Scotland (Skara Brae) thousands of years before Plato.

John McKay



On Tue, 9 May 2006, Rudolf Zlabinger wrote:

> Dear Thomas Breuer,
>
> Thank you at least for the hint to ATLAS, but also for the rich material
> given.
>
> To introduce Ikosaeder itself as group only its suffficient to begin with
> A5, it can be mirrored, as desired, to a 12 point permutation group by
> Isomorphic Subgroups to s12.
>
> My question mainly was, as there is a direct method to find out a group
out
> of isomophic subgroups from A5 to S12, containing a special permutation.
In
> our case it was the permutation (2,3,4,5,6)(7,8,9,10,11). I expected
> intuitively, that such a group recognizes a special numbering of the
> vertices of Ikosaeder.
>
> More general: As the action of all possible numberings, that is S12 in our
> case, on a starting set of one representative group for each  conjugacy
> class of groups (isomorphic to A5 in our case) is injective, one can
expect,
> that there is one group for each conjugacy class fitting to a special
> numbering.
>
> But in using numberings of vertices only, there is the problem to uniquely
> describe this numbering formally up to a selection of a starter set of
> groups belonging to a initial numbering, that has to be induced
intuitively
> from outside the formalism, as to, for example, requesting the inclusion
of
> special permutations in the starter group(s) (see above).
>
> I didnt explore yet, whether the reverse does hold, in order to have a
> unique description of a special numbering itself by the groups selected,
> because this relation above seems not to be bijective. That is, one group
> out of a conjugacy class may not uniquely describe a special numbering,
but
> a distinct manyfold of numberings. In this case one had to look, what
> manyfold it is, maybe it is determined as a structure quite well, and
maybe
> the numbering is uniquely determined by the full set of the selected
groups,
> one group per conjugacy class. I will try to explore this myself as an
> exercise, perhaps this problem reveales to be trivial at all.
>
> So the method given by you (out of ATLAS) has the advantage of uniquely
> describing the vertices by vectors, independent from a selected group
> representation.
>
> Thank you, and best regards, Rudolf Zlabinger
>
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