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I am trying to understand the article "A solution to a conjecture of Zeeman" by Akbulut, but I am not an expert in PL-geometry.

As far as I understand, two statements should be true, but I cannot find a reference:

Notation:

$B^n$ is the $n$-dimensional disk (with boundary).

Given PL manifolds $M^m,N^n $ with $m<n$, and a point $p \in M$, a PL map $f \colon M \to N$ is locally flat in p if there exist neighborhoods $p \in U \subset M$ and $f(p) \in V \subset N$ such that $(f(U),V)$ is homeomorphic to $(\mathbb{R}^m,\mathbb{R}^n)$.

Statements:

  1. Lat $M^4$ be a contractible 4-manifold. Let $f \colon B^2 \to M$ be a PL embedding which is not locally flat, such that $f(\partial B^2) \subset \partial M$. The set of points $p$ in $B^2$ such that $f$ is not locally flat in $p$ is finite.

  2. Given such an $f$, it is possible to build another PL embedding $g \colon B^2 \to M$ such that $g_{|\partial B^2} = f_{|\partial B^2}$ and $g$ has only one point where it is not locally flat (I refer to the sentence "By isotopy we can push these points together, that is we can assume that there is only one singular point $x_0$" in the article).

  3. The only possible singularity of such a map is a cone over a knot.

Attempts:

As far as 1) is concerned, I think that it is possible to get some results considering triangulations of $B^2$ and $M$ such that $f$ is simplicial with respect to these triangulations. If I can do this (and I am not sure about it), I hope that there is some way to prove that the singular points are contained in the set of vertices of the triangulation of $B^2$, but I do not know a simple way to prove it. Maybe this could help also for 2).

Any reference is appreciated. Thank you in advance.

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    $\begingroup$ Did you look at Chapter 9 in Akbulut's book "4-Manifolds"? Therein is a section called "A solution of a conjecture of Zeeman". $\endgroup$ Commented Nov 10, 2020 at 21:13
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    $\begingroup$ Yes, I actually read this section before the article. It is strictly contained in the article. Thank you anyway :). $\endgroup$
    – P. Tolo
    Commented Nov 10, 2020 at 22:21

1 Answer 1

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One reference for these matters is Rourke-Sanderson "Introduction to piecewise-linear topology". In particular, if $S\subset M$ is a simplical submanifold, then Corollary 4.2 in this book implies that $S$ is locally knotted in $M$ if and only if for each interior vertex of $S$ the pair $(\text{link of $x$ in $S$}, \text{link of $x$ in $M$})$ is a knotted sphere pair, and for each boundary vertex the same is true with "sphere pair" replaced by "ball pair".

This implies that if $\dim(S)=2$ and $\dim(M)=4$, then $S$ can only be non-locally-flat at vertices. Of course, if $S$ is compact, it has only finitely many vertices, which gives question 1.

We also get question 3 because a neighborhood of a vertex is a cone on its link.

Question 2 can be approached as follows: join two singular vertices by an arc in the $1$-skeleton, and take a regular neighborhoods of the arc in $S$ and $M$. It will be a ball pair because the arc is contractible. Proceeding inductively one can include all non-locally flat points in such a ball pair. Thus $M$ is PL homeomorphic to the union of two manifolds $M_1$ and $M_2$ along $S^3=M_1\cap M_2$, where $M_1\cap S$ is locally flat in $M_1$, and $(M_2, M_2\cap S)$ is a ball pair, and $S$ intersects $M_1\cap M_2$ along a knot.

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  • $\begingroup$ Thank you very much! It took a while to find and study the reference, but I think it works! Thank you again! $\endgroup$
    – P. Tolo
    Commented Nov 13, 2020 at 14:41
  • $\begingroup$ @P.Tolo: I am sure any book on PL topology covers the material, e.g. see Stallings' Tata notes "On polyhedral topology", which can be easily found online. $\endgroup$ Commented Nov 13, 2020 at 18:18
  • $\begingroup$ Is it clear that the regular neighborhood of the arc a ball? It seems not too far fetched, but does it already follows from your comment "because the arc is contractible"? It follows that the regular neighborhood is contractible, but why must it be a ball? $\endgroup$
    – M. Winter
    Commented Sep 1, 2022 at 22:27
  • $\begingroup$ @M.Winter: the short answer is "because it is a regular neighborhood". Take a look at the texts I mentioned. I think any two PL arcs must be ambiently isotopic, and hence so are their regular neighborhoods. $\endgroup$ Commented Sep 1, 2022 at 23:03

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