# Properties a triangulation must have in order to describe a manifold

I am mainly interested in the $$3$$-dimensional case. It is a well-known fact, following from the work of E. E. Moise and R. H. Bing in the 1950s, that every $$3$$-dimensional topological manifold (with or without boundary) admits a triangulation, i.e. its homeomorphic to (the geometric realization of) an abstract simplicial complex. Furthermore, it is a well known fact that a manifold is piecewise-linear if and only if it admits a combinatorial triangulation, i.e. a triangulation in which the link of each simplex is Pl-homeomorphic to a sphere, and that in $$d\leq 4$$, every triangulation of a manifold is combinatorial. In other words, every $$3$$-manifold admits a PL-structure.

I am interested in the other way round: Is there a bunch of properties an abstract simplicial complex has to have in order to define a topological manifold? Clearly, not all $$3$$-dimensional simplicial complexes which one can draw give rise to a manifold. The complex should be at least pure and non-branching, I guess. Is it maybe enough to assume that a complex is combinatorial?

In the literature, I also have found the notions of ''pseudo-manifolds'', which are abstract simplicial complexes, which are pure, non-branching and strongly-connected. How is this related to my question?

Any help is appreciated. If someone could provide some reference, I would be happy too.

• I think a 3-dim simplicial complex is a (purely 3-dimensional) topological manifold iff every every link is a 2-sphere (or 2-ball if boundary is allowed). And for a 2-dim simplicial complex, be a sphere/2-ball can be characterized (obvious local conditions + Euler characteristic).
– YCor
Sep 7, 2021 at 17:36
• For a simplicial complex, if all links are spheres or balls then you get a topological manifold (closed if you have only spheres). This is easy. What is not easy is to recognize a sphere, but this is not an issue in dimension 2. Also I'm not sure the converse holds in higher dimension (i.e., the links don't have to be topological spheres/balls). It might be in large dimension that checking whether a simplicial complex is a manifold is non-computable.
– YCor
Sep 7, 2021 at 18:18
• @YCor In dimension $n \leq 4$, a simplicial complex is a topological manifold if and only if all links of vertices are spheres if and only if all stars of vertices are balls. In dimension $n \geq 5$ this is false, because of Cannon's double suspension theorem: for every homology sphere $M$, the space $\Sigma^2 M$ is topologically a sphere. Because there is a smooth homology sphere of any dimension $n-2 \geq 3$ which has nontrivial fundamental group, triangulating it and taking the double suspension gives a triangulation of the sphere with a link homeomorphic to the non-manifold $\Sigma M$.
– mme
Sep 7, 2021 at 18:29
• In five dimensions we can have a triangulated topological manifold whose links are not even manifolds. Sep 7, 2021 at 18:29
• Your question has been answered in the comments. I would like to add, the combinatorial check that YCor describes is implemented in the software package "Regina". One of its functions is the generation of censuses of triangulated 3-manifolds with various properties. Basically it generates all simplicial complexes (with various constraints of your choosing) then checks to see if they are manifolds. It also implements a similar (but much slower) census algorithm to generate all triangulated 4-manifolds. Sep 7, 2021 at 20:15

1. Suppose that $$T$$ is a triangulation. If all vertex links are PL $$(n-1)$$-dimensional spheres then the realisation space of $$T$$ is a PL manifold and thus a topological manifold. (In the compact case this is equivalent to the usual definition.)
2. There are triangulations of topological manifolds (in fact, of $$S^5$$) that do not have this property. Examples come from the double suspension theorem of Cannon and also Edwards.