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Say we have a finite, connected, pure simplicial $n$-complex such that every $(n-1)$-simplex is the face of exactly two $n$-simplices and the link of every vertex is topologically equivalent to $S^{n-1}$. Is this simplicial complex topologically equivalent to some closed Riemann $n$-manifold? What about in the opposite direction? If we have a closed Riemann $n$-manifold, does there exist a finite, connected, pure simplicial $n$-complex in which every $(n-1)$-simplex is the face of exactly two $n$-simplices and the link of every vertex is topologically equivalent to $S^{n-1}$, such that the simplicial complex is topologically equivalent to the Riemann manifold (a triangulation of the manifold)?

Null Simplex
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    Perhaps you should investigate pseudomanifolds. – Ted Shifrin Jan 04 '23 at 19:49
  • Yes, every smooth manifold admits a triangulation (this is a somewhat non-trivial theorem due to Whitney). There are already 2-dimensional pseudomanifolds which are not topological manifolds. – Moishe Kohan Jan 04 '23 at 20:09
  • @TedShifrin you are correct, thank you. What if I add the condition that the link of every vertex in the simplicial complex must be topologically equivalent to $S^{n-1}$? – Null Simplex Jan 05 '23 at 02:10
  • There are nonsmoothable PL manifolds in dimension 8 and higher. – Moishe Kohan Jan 05 '23 at 04:28
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    https://math.stackexchange.com/questions/408221/the-easiest-non-smoothable-manifold – Moishe Kohan Jan 05 '23 at 04:52
  • @MoisheKohan is the example you gave in that response a compact manifold? The paper you linked says closed. I just want to make sure closed in this context means unbounded and compact and not just unbounded. – Null Simplex Jan 06 '23 at 06:40
  • I do not know what you mean by "unbounded": The usual terminology is "closed" which means compact and without boundary. Yes, this example is closed. – Moishe Kohan Jan 06 '23 at 07:40
  • @MoisheKohan My final question on this subject. Can the manifold you gave as an example be represented as a triangulation with finitely many vertices where the link of each vertex is $S^7$? – Null Simplex Jan 06 '23 at 08:18
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    Yes, every complex (and real!) projective variety admits a finite triangulation (and the given example is a complex projective variety). In Kuiper's example such triangulation even has links homeomorphic to spheres. None of this is an elementary fact. – Moishe Kohan Jan 06 '23 at 08:22
  • @NullSimplex: Yes, this is a very different story (approximating Riemannian metrics by simplicial structures): It is a geometric story as opposed to the topological one in the current question. I remember answering this as well in the past, I will check. – Moishe Kohan Jan 06 '23 at 08:45
  • @MoisheKohan I have been thinking about your response for a few years now. Say we are approximating a smooth manifold with triangulations. As the maximum edge length of the triangulation approaches 0, the angles at each vertex get “flatter” due to the smoothness. If we try to approximate a non-smoothable PL-manifold with triangulations, what happens to the angles at the vertices of the triangulations as the max edge length approaches 0? What is the lower and upper bound for how differentiable a non-smoothable PL manifold can be (as in differentiability class)? – Null Simplex May 21 '25 at 21:45

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