5
$\begingroup$

There are two different notions of triangulations for a manifold $\mathcal{M}$:

  1. A (simplicial) triangulation is an abstract simplicial complex $\Delta$ such that $\vert\Delta\vert\cong \mathcal{M}$.
  2. A combinatorial triangulation is an abstract simplicial complex $\Delta$ such that $\vert\Delta\vert\cong \mathcal{M}$ with the extra property that the link of every vertex is PL-homeomorphic to the standard PL-sphere, i.e. the boundary of a single $d$-simplex. (or in the case of manifolds with boundary, the link of a vertex is also allowed to be PL-homeomorphic to the standard PL-ball, i.e. a single $d$-simplex; In this case, the corresponding vertex is on the boundary).

Now, in sufficient high dimensions there are well-known examples of triangulations of manifolds, which are not combinatorial, like the famous example of $S^{5}$ constructed using the suspension theorem.

However, I have often found the following claim (i.e. on wikipedia):

Every triangulation of a $d$-dimensional manifold with $d\leq 4$ is combinatorial.

For $d=1$ the claim is of course trivial. Now, I have in essence two (very closely related) questions:

  1. Often, it is stated that the case $d=2,3$ is a consequence of the famous triangulation theorems of Rado and Moise-Bing. However, I can't understand why. These theorems "just" say that every $2$ (resp. 3)-dimensional manifold has a piecewise-linear structure (equivalently a combinatorial triangulation), which is unique up to PL-homeomorphism. Why does this imply that every triangulation is a combinatorial one? Do I miss something, or are there other, stricter results of these Theorems of which I am not aware of?

  2. What about the case $d=4$? Does anyone know a reference for this? I have heard somewhere that this case is equivalent to Poincare's conjecture, however, I do no remember where. I know that not every $4$-manifolds admits a triangulation (like $E_{8}$). Furthermore, I know that in $d=4$, the smooth and PL-category are equivalent, but I do not know if this helps.

$\endgroup$
3
  • 2
    $\begingroup$ Your definition of combinatorial triangulation is not quite exactly the same as the one that I have seen, which is a definition by induction: assuming $\mathcal M$ has dimension $n$, the extra property is that the link of every vertex is simplicially isomorphic to combinatorial triangulation of $S^{n-1}$. $\endgroup$
    – Lee Mosher
    Jan 11, 2022 at 19:47
  • $\begingroup$ I don't quite see whats the difference. $\endgroup$
    – B.Hueber
    Jan 12, 2022 at 10:08
  • $\begingroup$ Ah, I think I know what you mean. In my definition, I also meant of course "the link of every vertex is PL-homeomorphic to the standard PL-sphere, i.e. the boundary of a single $d$-simplex (and for manifolds with boundary, also the standard PL-ball, i.e. a single $d$-simplex). Of course, this has to be added, since it is for example an open problem wether $S^{4}$ has a unique PL-structure or not. Is this what you meant? $\endgroup$
    – B.Hueber
    Jan 12, 2022 at 10:10

1 Answer 1

Reset to default
2
$\begingroup$

Here is a proof; it requires the 3-dimensional Poincare Conjecture (PC3). Suppose that $M$ is a triangulated 4-dimensional manifold. We need to show that links of vertices are 3-dimensional spheres, equivalently (in view of PC3), that they are simply connected. Suppose that $L_v$ is the link of a vertex $v$; let $B_v$ denote the component of $M\setminus L_v$ containing $v$. Then $B_v$ is homeomorphic to the open cone over $L_v$; it is a contractible 4-dimensional topological (in fact, smoothable) manifold. If $L_v$ is not simply-connected, then so is $B_v\setminus \{v\}$. Thus, removing a point from $B_v$ changes the fundamental group. It is a nice exercise to show that this cannot happen to any topological manifold of dimension $\ge 3$. Hence, all links are simply-connected. qed

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .