A triangulation of a topological manifold $\mathcal{M}$ possibly with boundary is an abstract simplicial complex $\Delta$ together with a homeomorphism $\varphi:\vert\Delta\vert\to\mathcal{M}$, where $\vert\Delta\vert$ denotes the geometric realization and where an abstract simplicial complex is a collection of simplices with the property that $\sigma\in\Delta$ and $\tau\subset\sigma$ implies $\tau\in\Delta$.

Now, for simplicial complexes one can define the following extra properties:

  1. $\Delta$ is "pure", i.e. every simplex $\sigma\in\Delta$ of dimension $<d$ is the face of some $d$-simplex.
  2. $\Delta$ is "non-branching", i.e. every $(d-1)$-simplex is face of exactly one or two $d$-simplices.
  3. $\Delta$ is "strongly-connected", i.e. for every pair of $d$-simplices $\sigma,\tau\in\Delta_{d}$, there is a sequence of $d$-simplices $\sigma=\sigma_{1},\sigma_{2},\dots,\sigma_{k}=\tau$ such that the intersection $\sigma_{l}\cap\sigma_{l+1}$ is a $(d-1)$-simplex for every $l\in\{1,\dots,k-1\}$.

An abstract simplicial complex with these properties is usually called a "pseudomanifold (possibly with boundary)". Now, I know that every piecewise-linear manifold, i.e. a manifold with a triangulation satisfying the extra property that the links of every vertex is a sphere or ball (or equivalently a manifold with a piecewise-linear atlas), is a pseudomanifold. This is stated in many resources and is clear I think.

However, on the other side, there are examples of triangulation for dimension $d>4$, which are not piecewise-linear. In fact, there are even triangulations of manifolds (like for $S^{5}$) whose links are not even manifolds. However, do these complexes still have the properties above? I think they should at least be pure, since this property is needed in order to define the boundary of a simplicial complex and I think this should be possible for arbitrary triangulations of manifold with boundary. On the other hand, triangulations of manifolds should be non-branching, I guess, since this is needed in order to define orientability of simplicial complexes.

In some sense, I am asking if the concept of pseudomanifold generalizes the concept of PL-manifolds or of general triangulazible manifolds.

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    $\begingroup$ I think it should be pure. Otherwise there is a $d'$- simplex which is not in the boundary of an higher dimensional simplex. Thus its midpoint would have a open neighborhood homeomorphic to $\mathbb{R}^{d'}$. By the well definedness of the dimension of a topological manifold, we would then get $d=d'$. $\endgroup$ Jan 5, 2022 at 19:24
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    $\begingroup$ Non branching should also be clear. We now only have to exclude that a $d-1$ simplex is a face of three or more $d$-simplices. It should be possible to show that a point in the $d-1$-simplex cannot have a neighborhood homeomorphic to $R^d$. $\endgroup$ Jan 5, 2022 at 19:29
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    $\begingroup$ You need to assume that $M$ is connected (and thus path connected) to obtain "strongly connected" for $\Delta$. $\endgroup$
    – Sam Nead
    Jan 6, 2022 at 10:00

1 Answer 1


Suppose that $M$ is a connected $d$-dimensional topological manifold without boundary. (We make the last assumption to simplify matters.) Let $\Delta$ be the given pseudo-triangulation. So the realisation $|\Delta|$ has the same local homology groups as $M$. These are $H_k(M, M - x) \cong \mathbb{Z}$ if $k = d$ and are zero if $k \neq d$.

If $\Delta$ is not pure then there is a $\ell$-simplex, say $\sigma$, with $\ell < d$ which is not the face of a $\ell+1$-simplex. Thus, for a point $x$ in the interior of $\sigma$ we have $H_\ell(M, M - x) \cong \mathbb{Z}$, a contradiction.

If $\Delta$ is branching then there is a $(d-1)$-simplex, say $\sigma$, which is the face of $n > 2$ top-dimensional simplices. Thus, for a point $x$ in the interior of $\sigma$ we have $H_d(M, M - x) \cong \mathbb{Z}^{n-1}$, a contradiction.

If $\Delta$ is not strongly connected, then there is a simplex, say $\sigma$, whose link has $n > 1$ connected components. Thus, for a point $x$ in the interior of $\sigma$ we have $H_0(M, M - x) \cong \mathbb{Z}^{n-1}$, a contradiction.


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