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In an $n$-dimensional space I have a set of simplices where each simplex consists of facets. Some of the simplices are 'connected' by sharing facets. Each facet is made up on edges, each consisting two points. In 3D each simplex is a tetrahedron consisting of triangular facets. I would like to know, if the set of simplices I have is a 'non-manifold'.

In 3D I can check if my mesh is a 'non-manifold' as follows: - select all facets having more or less than two neighboring simplices - select all edges of these facets - per edge, count the number of facets connected to the edge - in case an edge has more of less than 2 neighboring facets, the edge is 'non-manifold'

This algorithm does only work in 3 dimension (3D). What about higher-dimensional sets of simplices? How can I determine, if an $n$-dimensional mesh of simplices is non-manifold?

Thank you!

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    $\begingroup$ A related question is this one: it is about CW complexes but some answers gives conditions on simplicial complexes. It has no answer to your question about algorithms, but some general conditions are given that could be of your interest. $\endgroup$
    – Dario
    Commented Aug 12, 2015 at 20:47
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    $\begingroup$ For $n \geq 6$ an algorithm is a logical impossibility. But I suppose you could do things very inefficiently -- if you want to know if something is an $n$-manifold, build a list of all triangulations of the $(n-1)$ sphere and then check to see your vertex links occur in that list. It's an infinite list, so membership tests take an unbounded amount of time... $\endgroup$
    – Ryan Budney
    Commented Aug 13, 2015 at 5:26

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In all dimensions, something is a manifold if the link of every cell is a sphere. This, sadly, is undecidable if the dimension of your complex is at least five. It is decidable (but not quickly) for complexes of dimension 4 (google "thin position", "Jaco-Rubinstein-Thompson"). In low dimensions ($1, 2, 3$) it is easy.

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  • $\begingroup$ Thanks Igor for your answer. Unfortunately I cannot mark your answer with the green check mark, because I cannot access the account the question is under. As soon as I can get accounts merged, I'll see about accepting this as the answer. $\endgroup$
    – Jörg
    Commented Aug 13, 2015 at 6:46

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