Equivalence of triangulations and piecewise-linear triangulations in dimension $d\leq 4$

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.

• 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 $S^{n-1}$. Jan 11, 2022 at 19:47
• I don't quite see whats the difference. Jan 12, 2022 at 10:08
• 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? Jan 12, 2022 at 10:10

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