Timeline for Properties a triangulation must have in order to describe a manifold
Current License: CC BY-SA 4.0
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| Sep 2 at 12:55 | comment | added | AlpinistKitten | @G.Blaickner : thank you so much. The equivalence to the Poincare conjecture is honestly quite shocking since many people will intuitively take "combinatorial" for granted. | |
| Sep 2 at 7:40 | history | edited | G. Blaickner | CC BY-SA 4.0 |
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| Sep 2 at 7:35 | comment | added | G. Blaickner | @AlpinistKitten see the appendum of my question above. | |
| Sep 2 at 7:35 | history | edited | G. Blaickner | CC BY-SA 4.0 |
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| Sep 1 at 11:41 | comment | added | AlpinistKitten | Could you please provide a few references for the claims in your first paragraph? | |
| Sep 7, 2021 at 22:20 | vote | accept | G. Blaickner | ||
| Sep 7, 2021 at 20:33 | answer | added | Sam Nead | timeline score: 8 | |
| Sep 7, 2021 at 20:23 | comment | added | G. Blaickner | Thank you very much for all your helpful comments and answers! :-) | |
| Sep 7, 2021 at 20:15 | comment | added | Ryan Budney | Your question has been answered in the comments. I would like to add, the combinatorial check that YCor describes is implemented in the software package "Regina". One of its functions is the generation of censuses of triangulated 3-manifolds with various properties. Basically it generates all simplicial complexes (with various constraints of your choosing) then checks to see if they are manifolds. It also implements a similar (but much slower) census algorithm to generate all triangulated 4-manifolds. | |
| Sep 7, 2021 at 18:29 | comment | added | Richard Stanley | In five dimensions we can have a triangulated topological manifold whose links are not even manifolds. | |
| Sep 7, 2021 at 18:29 | comment | added | mme | @YCor In dimension $n \leq 4$, a simplicial complex is a topological manifold if and only if all links of vertices are spheres if and only if all stars of vertices are balls. In dimension $n \geq 5$ this is false, because of Cannon's double suspension theorem: for every homology sphere $M$, the space $\Sigma^2 M$ is topologically a sphere. Because there is a smooth homology sphere of any dimension $n-2 \geq 3$ which has nontrivial fundamental group, triangulating it and taking the double suspension gives a triangulation of the sphere with a link homeomorphic to the non-manifold $\Sigma M$. | |
| Sep 7, 2021 at 18:18 | comment | added | YCor | For a simplicial complex, if all links are spheres or balls then you get a topological manifold (closed if you have only spheres). This is easy. What is not easy is to recognize a sphere, but this is not an issue in dimension 2. Also I'm not sure the converse holds in higher dimension (i.e., the links don't have to be topological spheres/balls). It might be in large dimension that checking whether a simplicial complex is a manifold is non-computable. | |
| Sep 7, 2021 at 18:02 | comment | added | G. Blaickner | Thank you for your comment! So basically a combinatorial triangulation in my terminology above should always give rise to a topological manifold? | |
| Sep 7, 2021 at 17:37 | history | edited | YCor |
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| Sep 7, 2021 at 17:36 | comment | added | YCor | I think a 3-dim simplicial complex is a (purely 3-dimensional) topological manifold iff every every link is a 2-sphere (or 2-ball if boundary is allowed). And for a 2-dim simplicial complex, be a sphere/2-ball can be characterized (obvious local conditions + Euler characteristic). | |
| Sep 7, 2021 at 16:28 | history | asked | G. Blaickner | CC BY-SA 4.0 |