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.

  • 1
    $\begingroup$ Perhaps you can use shellability and/or constructibility, both of which are sufficient conditions for the pseudomanifold to be homeomorphic to a ball. $\endgroup$ – Joseph O'Rourke Apr 25 '12 at 0:40
  • 2
    $\begingroup$ To expand on Joseph O'Rourke's comment: my favorite way to investigate whether a simplicial complex is a ball:<br><br> 1) Compute the $h$-vector, and see if any entries are negative. (If they are, then it is not Cohen-Macaulay, and in particular not a ball). 2) Search for a shelling. If the complex is shellable, it is quite easy to check if it is a ball. Not every triangulation of a 3-ball is shellable, but many small/nice/naturally occurring such triangulations are. (And if a shelling exists, it is often quick to find.) $\endgroup$ – Russ Woodroofe Apr 26 '12 at 18:18

$\newcommand{\RR}{\mathbb{R}}$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 following.

  • 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.

  • $\begingroup$ I was aware of that, given that the OP asked about complexes in euclidean space, but since it was not clear that he knew what he wanted I gave a more general answer.... $\endgroup$ – Igor Rivin Apr 25 '12 at 15:18
  • $\begingroup$ Can anything be said whether the conditions on Betti numbers of $T$ and $T'$ (see original post) guarantee that $T$ is homeomorphic to a ball? $\endgroup$ – IL. Apr 25 '12 at 21:38
  • 1
    $\begingroup$ Would you please give an example of finite tetrahedral complex $T$ embedded in $R\mathbb{R}^3$ such that $|\partial T|$ is homeomorphic to a 2-sphere, but $|T|$ is not homeomorphic to a 3d ball? If this example is written in some textbook, reference will be just fine. $\endgroup$ – IL. Apr 26 '12 at 1:24
  • 1
    $\begingroup$ IL - regarding your first comment: I'd rather not think about $T'$ as I'm positive that checking my conditions is simpler to code and faster to boot -- not just linear, but linear with little overhead. Regarding your second comment: I think that you are correct here. That is, if $|\partial T|$ is homeomorphic to a two-sphere then $|T|$ is a manifold. (When this holds, $\partial |T|$ is homeomorphic to $|\partial T|$, which is nice.) $\endgroup$ – Sam Nead Apr 27 '12 at 12:20
  • 1
    $\begingroup$ IL - regarding your third comment - I don't think such an example exists. See above. Regarding your fourth comment - I don't think such assumptions are needed. The problem is reduced to deciding if a given two-complex $Q$ is a sphere. First check that $Q$ is connected. Next check if $|Q|$ is a surface without boundary by checking that all vertex links are connected and are each circles. Finally check that the Euler characteristic $\chi(|Q|)$ is two, using the usual formula. $\endgroup$ – Sam Nead Apr 27 '12 at 12:32

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.

  • $\begingroup$ I second the suggestion about texts, especially regarding Hatcher's notes. $\endgroup$ – Sam Nead Apr 25 '12 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.