Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball - MathOverflow

Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball

2012-04-24T21:12:50Z

Dear experts,

let $T$ be finite tetrahedral complex in flat 3-dimensional euclidean space. Additionally, let $T$ be 'homogeneous' in a sense that each simplex in $T$ is a face of some tetrahedron from $T$. Are there any known sufficient conditions (say in terms of Betti numbers of $T$ and possibly Betti numbers of some other complex derived from $T$) for $T$ to be homeomorphic to a 3D ball?

My understanding is that the condition on Betti numbers of $T$ itself is necessary but not sufficient (consider two tetrahedrons touching at a vertex or an edge, but not sharing any 2-simplex).

Consider another complex $T'$, which is dual to $T$ in the following sense. Each 'vertex' (0-cell) of $T'$ corresponds to a tetrahedron in $T$. Each 'edge' (1-cell) of $T'$ corresponds to an interior (i.e., not lying on the global boundary of $T$) 2-simplex in $T$. Each 2-cell of $T'$ corresponds to an interior 1-simplex in $T$. Each 3-cell of $T'$ corresponds to an interior vertex of $T$. The boundary operation on $T'$ is defined by 'transposing' the boundary operation in $T$. For example, if a vertex in $T$ is part of the boundary of an edge in $T$, then the corresponding 3-cell in $T'$ has corresponding 2-cell as part of its boundary, with the same sign (orientation).

Suppose both $T$ and $T'$ have 1 connected component (that is, corresponding Betti number equals one), and all other Betti numbers of $T$ and $T'$ vanish. Does this imply that $T$ is homeomorphic to a ball? If yes, would you please provide me with a reference?

For example, if $T$ contains one tetrahedron (plus all its faces), $T'$ will consist of a single vertex. For two tetrahedrons touching each other at edge or vertex (but not sharing a 2-simplex), $T'$ will consist of two isolated points.

Another example: suppose one takes a ball, picks two diametrically opposite points on the surface and pushes them inside until they meet. Then the deformed ball (with the two opposite points glued) is "triangulated with tetrahedrons". My understanding is that the obtained tetrahedral complex will have the same Betti numbers as the 3D ball, however 'dual' complex will have 1-cycle which will not be a boundary of any 2-chain.

Update: I'm looking for (reasonably) fast algorithms to check whether a given set of tetrahedrons is or is not homeomorphic to a 3d ball. The tetrahedrons are picked from a tetrahedral mesh of some nice domain. I have been looking into Betti numbers because there exist fast (linear or almost linear complexity) algorithms for computing Betti numbers of simplicial complexes embeddable in $\mathbb{R}^3$, see here. Admittedly, I do not know whether these algorithms will (provably) work for the complex $T'$, which is not necessarily simplicial.

Answer by Igor Rivin for Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball

2012-04-25T03:26:18Z

First you check that your complex is a PL manifold with boundary (this is easy, the hardest part is checking that the links of interior vertices are 2-spheres (which is an euler characteristic argument. Then, as @Anton suggests, you can attach a ball along the boundary and check that the resulting thing is $S^3.$ Checking that the fundamental group is trivial is not the way to do this, unless you can wait for a couple of millenia.On the other hand, in this paper Saul Schleimer shows that sphere recognition is in NP, which is pretty darn good. On another hand, your examples show that you are a novice in the ways of topology, so you might want to read a book (e.g., Rourke and Sanderson's PL topology book, followed by Hempel's or Jaco's three-manifold texts, or Hatcher's 3-manifold notes), before trying any of the above.

Answer by Sam Nead for Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball

2012-04-25T11:29:08Z

$\newcommand{\RR}{\mathbb{R}}$The other answers are completely general, but there is simpler way if we use the (given) hypothesis that all the action is taking place in $\RR^3$. So, suppose that $T$ is a finite triangulation contained in $\RR^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

the space $|T|$ is a manifold and
the boundary $\partial\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

EDIT x2 - Here is a discussion of the "bit of work". Suppose that $C = |T|$ is a manifold and $S = \partial C$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B \subset \RR^3$. We need to show that $B$ is equal to $C$. By the Jordan–Brouwer separation theorem there are two possibilities. Either $S$ separates $B$ from $C$ or it does not.

In the separating case form $M = B \cup C$. Thus $M$ is a compact three-manifold without boundary, embedded in $\RR^3$. This contradicts invariance of domain. See Corollary 2B.4 of Hatcher's Algebraic topology.

Suppose instead that $B$ and $C$ are on the same side of $S$. It follows that $C \subset B$. We must prove the opposite inclusion. Suppose that $p$ is a point of $B$. Let $r$ be any point of $S$ that is as close as possible to $p$. Let $I = [p,r]$ be the line segment from $p$ to $r$. So $I \subset B$. Order the points of $I$, from $p$, to $r$. Note that $r \in S$ so $r \in C$. Let $J = I \cap C$. Let $q = \inf J$. Since $C$ is a closed subset of $\RR^3$ the set $J$ is closed and thus $q$ lies in $C$. Since $C$ is a manifold there is a neighborhood $V \subset C$ so that $q \in V$. Show that $V \cap I$ is a neighborhood of $q$ in $I$. Thus $q = p$ and we are done.

I don't see how to do the second half with "invariance of domain" directly. I'll also remark that the "bit of work" has now been greatly expanded, and perhaps unnecessarily so. One is supposed to do this sort of thing once and then not worry about it ever again.