I know some manifolds can be given the structure of simplicial complex by triangulation, but what about the other way around? Can every simplicial complex be given the structure of manifolds? If so, how do we do this? (how do we resolve the "sigularities" of the corners?)

Specifically, if we have a simplicial complex $A$ whose manifold structure is given,denoted by $M$, do we have the following relation?

  1. $A$ has no boundary $\rightarrow$ $M$ has no boundary

  2. $A$ is oriented (which is always true) $\rightarrow$ $M$ is orientable

Also, I wonder if the coefficient here is important, but I basicially want to understand the simplest case, i.e. the $\mathbb{Z}$-coefficients.

Edits: I'm asking the above question becuase I read the following arguments but did not really understand what it means (that's the reason wht my above question is confusing, sorry for that):

Every 2 dimensional integral homology class $A$ of a smooth manifold $X$ of dimension dim $X \geq 3$ by definition can be represented by a continuous map defined on a compact 2-dimensional simplicial complex without boundary. Every such complex can be given the structure of a smooth compact manifold without boundary (which in the case of integer coefficients is orientable).

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    $\begingroup$ What do you mean by "give the structure of a manifold"? You have already pointed out that not all simplicial complexes are literally manifolds, so what is it you are after? $\endgroup$ Apr 23, 2021 at 15:39
  • $\begingroup$ It's really hard (if even possible) to understand what you are asking. Could you please elaborate more your question? $\endgroup$
    – user126154
    Apr 23, 2021 at 15:44
  • $\begingroup$ One sufficient condition involves a shelling. More generally, showing complexes to be manifolds often involves some recursive requirement on the relationships between the maximal faces. $\endgroup$
    – William
    Apr 23, 2021 at 15:50
  • $\begingroup$ This is really meaningless as written: There is no notion of boundary or orientation on general simplicial complexes. Just take a simplicial graph and try to define its orientation or/and boundary. I am voting to close for now. $\endgroup$ Apr 23, 2021 at 15:54
  • $\begingroup$ @user126154 I did make the edits. Thanks for the comments. $\endgroup$
    – Lamda8
    Apr 25, 2021 at 23:33

1 Answer 1


Not every simplicial complex can be given the structure of a manifold. For example, a 1-dimensional complexes might have more than two 1-cells incident to a 0-cell. A more elaborate example is the cone of a torus, which fails to be a manifold at the cone point.

Suppose your simplicial complex is a topological space with a manifold structure. Ideally, the simplicial complex would in some way generate the manifold structure in that the link of every vertex (i.e., the union of the simplices incident to it) would be homeomorphic to a ball, but it turns out that there are some issues in dimensions greater than 4, and there are simplicial complexes that are homeomorphic to manifolds but which fail to be "PL manifolds" themselves https://en.wikipedia.org/wiki/Triangulation_(topology)

Whether or not the simplicial complex that is homeomorphic to a manifold is itself a PL manifold, to answer your questions:

  1. The boundary of $A$ is the boundary of $M$.
  2. $A$ is orientable if and only if $M$ is orientable. (It is not always true that a simplicial complex is orientable! For example, it's not so hard to make a simplicial complex that's homeomorphic to $\mathbb{R}P^2$. Indeed, every manifold is homeomorphic to some simplicial complex.)

You are probably thinking about homology when you mention coefficients.


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