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 __________________________________________________