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I want to know the topological relation between a manifold and a simplicial complex. I know that a simplicial complex cannot be a manifold since its a union of simplices which are manifolds of distinct dimensions. Now can we say that every manifold can be given a structure of a simplicial complex? and more informally, which of the two spaces is "nicer" and more regular topologically?

rschwieb
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palio
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    Simplicial complexes are "nicer" in that you can work with them combinatorially, but it might be hard to determine if, for example, they represent the same manifold. – MartianInvader Aug 28 '12 at 20:38

1 Answers1

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A simplicial complex can be (homeomorphic to) a manifold. For example, the tetrahedron (minus its interior) is homeomorphic to $S^2$. A choice of such a homeomorphism is called a triangulation of a manifold. The fact that there are lower-dimensional simplices lying around is not an issue; the natural topology does not have open sets which are somehow restricted to lower-dimensional simplices.

Manifolds in dimensions $d \le 3$ can be triangulated, but not in general.

My impression is that simplicial complexes are generally easier to work with than arbitrary manifolds. For example, it is straightforward to define simplicial homology, although not straightforward to prove that it is homotopy invariant. However, some natural statements one might hope to be true about triangulations are false.

Qiaochu Yuan
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  • thank you for your answer! just i don't understand the meaning of open sets being "restricted" to lower-dimensional simplices. – palio Aug 28 '12 at 20:42
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    The point is that they're not ("restricted to" means "contained in" in this case), so your argument that simplicial complexes aren't manifolds isn't valid. – Qiaochu Yuan Aug 28 '12 at 20:46
  • For an example of a simplicial complex that is not a manifold, see here https://math.stackexchange.com/a/4113713/81250 – apg Jan 18 '23 at 20:14