[GAP Forum] van Kampen diagrams

[chris chalk]

Beth,
 thanks for this lovely question. Its got a strong 4-color theorem flavour 
about it. So how about this as an approach? Consider the dual D of the 
spherical VK diagram. So faces become vertices and the 3 degree vertices 
become triangles and we have a triangulation of the sphere, where, from 
Euler, the average degree of  the vertices of D must be < 6. Your question 
is then, is it always possible somehow to enumerate or order the vertices of 
D so that in this 'ordering'
  I) each vertex is connected to its predecessor/successor by an edge of D, 
and
  II)  the average degree of  the vertices of every initial subset of this 
ordering is < 6.
 eg if the degrees of vertices of  an 'ordering'  begin  as 
(5,6,6,5,7,6,...) then this satisfies this 'ordering condition' (so far at 
least), but if they begin  (5,6,6,7,5,6,..) then they don't.

So here's a  sketch of a start of a possible proof....!??
 Following the 4 Colour Thm proof, assume there exist one or more spherical 
diagrams which don't allow this 'ordering' of vertices. Choose one, M, 
which has the smallest number of vertices.  Now pick any vertex V of degree 
5 or less in M. There then exists a maximal connected planar subdiagram S of 
M containing V in its interior, such that all other interior vertices of S 
are of degree 6 or less. For example it could be that V is connected to 
another vertex W of degree 5  but that all the other (6?) vertices connected 
to V and W have degree > 6. In this case S would a 6 sided subdiagram of M. 
We can then shrink this subdiagram to a point or vertex to  form a spherical 
diagram  M' with fewer vertices than M for which, by assumption, an 
'ordering' of the vertices exist. eg S, above would shrink to a vertex of 
degree 6 and M' would have 2 fewer vertices than M.
Could it be that from the 'ordering' of M' we could construct an 'ordering' 
of the larger diagram M and so derive a contradiction to the assumption that 
no such 'ordering' exists.?
Its of course possible that S wraps round the sphere and so isn't planar 
then cant just shrink S to a vertex without forming two spheres - maybe it 
isn't necessary to let vertices of degrees 6 belong to S....
 I hope this might be of some help.
 regards Chris

----- Original Message ----- 
From: "Petra Holmes" <holmespe at for.mat.bham.ac.uk>
To: <forum at gap-system.org>
Sent: Monday, January 08, 2007 10:33 AM
Subject: [GAP Forum] van Kampen diagrams


> Dear Group Pub Forum,
>
> I have been told to send this to everyone I can think of.  Any ideas on it 
> are welcome.
>
> Beth
>
> ---------- Forwarded message ----------
> Date: Mon, 8 Jan 2007 08:26:37 +0000
> From: Richard Parker <RParker at amadeuscapital.com>
> To: Beth Holmes <P.E.Holmes at dpmms.cam.ac.uk>
> Subject: The central problem
>
>
> Let P be a polyhedral ball which topologically is a sphere,
> and where precisely three faces meet at every vertex.
>
> For each face with n sides, define the "curvature" of that
> face to be 6-n, so a pentagon has curvature 1, an octagon
> has curvature -2 and so on.  The faces do not need to be
> regular.
>
> Euler's formula, F + V = E + 2, implies that the total curvature
> over the whole of P is 12.
>
> A "patch" of f faces is a subset of the faces of P that has
> no holes in it.  Technically it is simply connected.
>
> I need to prove or disprove the following result.
>
> For every such P there is a sequence of patches, each containing
> one more face than the previous, such that the sum of the curvature
> of all the faces in each patch is greater than zero.  In other words
> we can build P one face at a time such that the curvature of the
> patch we have made so far is always positive.
>
> This result has huge implications for an algorithm for finitely
> presented groups.  P is a van Kampen diagram, and I want to know
> whether a short relator with a long proof can be built up relator
> by relator looking only at sensible things along the way.
>
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