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" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
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 <[EMAIL PROTECTED]>
To: Beth Holmes <[EMAIL PROTECTED]>
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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