Subject: Dunwoody's approach to the PC
Date: Wed, 15 May 2002 15:55:15 +0200 (MSZ)
From: "Simon A. King"
John Stallings wrote:
> I haven't heard anything recently about Dunwoody's proof of the PC.
> ... I wonder where really the simple connectivity is used
> seriously.
Probably most of you know that Dunwoody has put a question mark on his
announcement, version#8. I will roughly describe his approach and show
a gap in the proof.
The approach is based on the theory of normal and almost normal surfaces.
A normal surface in a triangulated 3-manifold is an embedded surface that
intersects each tetrahedron in a disjoint union of normal triangles (= a
disc whose boundary intersects exactly three different edges of the
tetrahedron) and normal squares (= a disc whose boundary intersects
exactly four different edges of the tetrahedron). The most famous
application of normal surfaces is a classification algorithm for knots and
links.
Let M be a homotopy sphere with triangulation T. By old results of
Kneser, there is a normal surface S in M whose connected components are
2-spheres, such that any normal 2-sphere in M-S is parallel to a connected
component of S.
By the Rubinstein-Thompson algorithm, M is homeomorphic to S^3 if and only
if any connected component of M-S contains a vertex of T or has
disconnected boundary or contains an "almost normal" 2-sphere. This is an
embedded sphere formed by normal triangles and squares, and exactly one
normal octagon (= a disc in a tetrahedron whose boundary meets two
opposite edges twice and each of the four remaining edges once. Sorry for
not drawing a picture...).
This yields an obvious strategy for a proof of the Poincare conjecture:
Take any connected component N of M-S that doesn't contain a vertex and is
bounded by a single 2-sphere, and show that N contains an almost normal
2-sphere.
Since M is a homotopy sphere, N is a homotopy ball. Thus there is a
homotopy H: S^2 x I -> N that relates the boundary of N with a point (this
is where one uses simple connectivity). Of course, in general
H_t=H(S^2x{t}) is not embedded.
One can assume that H is in "thin position" (a notion introduced by
Gabai). Under this hypothesis, Dunwoody shows that there is some u in
I=[0,1] so that H_u is an "almost normal immersed" 2-sphere. This means
that H_u is formed by intersecting and self intersecting normal triangles,
squares and exactly one normal octagon.
It is not difficult to prove that there is a unique almost normal embedded
surface F that intersects the edges of T exactly as H_u does. Moreover the
Euler characteristic of F is two. It remains to show that F is connected
(=> it is a 2-sphere, hence M is homeomorphic to S^3 by
Rubinstein-Thompson, hence the Poincare conjecture is true). This is
certainly non-trivial, since a priori F might be a disjoint union of
copies of the boundary of N and an almost normal surface of higher genus.
To prove that F is a 2-sphere, Dunwoody wants to relate H_u with F by a
normal homotopy, i.e. by a homotopy that respects the simplices of T. He
uses a technique called "unravelling". It seems to me that the
description of this technique is not complete.
The idea of unravelling is roughly like this. The points of intersection
of H_u with the edges of T are removed in pairs. We number the pairs
according to the order in which they disappear. Self intersections shall
be removed by exchanging the points of pairs in ascending order (see
Figure 1 in the attachment). Dunwoody explains why one can remove all self
intersections of H_u by repeated use of this exchanging process.
There are situations in which it is impossible to remove intersections in
that way (see Figure 2). It is not clear (at least to me) how Dunwoody
proceeds in this situation; note that it is not allowed to "unravel" in
backwards direction, i.e., to exchange the points of pair 1. In Figure 3,
I have drawn a configuration that might occur as part of the intersection
of H_u with the 2-skeleton of T. If H_u contains a configuration as in the
figure, then it is not normally homotopic to an embedded almost normal
surface; in particular unravelling doesn't work. As yet, Dunwoody can not
exclude such a configuration. So far as I know, this problem was pointed
out to Dunwoody by Colin Rourke.
Dunwoody still tries to close the gaps. I expect that this will be
hard work, perhaps impossible.
Best regards
Simon King
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