Timeline for Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball
Current License: CC BY-SA 3.0
14 events
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| S Feb 18, 2015 at 12:47 | history | suggested | CommunityBot | CC BY-SA 3.0 |
fixed obvious typo
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| Feb 18, 2015 at 12:18 | review | Suggested edits | |||
| S Feb 18, 2015 at 12:47 | |||||
| Apr 27, 2012 at 12:32 | comment | added | Sam Nead | IL - regarding your third comment - I don't think such an example exists. See above. Regarding your fourth comment - I don't think such assumptions are needed. The problem is reduced to deciding if a given two-complex $Q$ is a sphere. First check that $Q$ is connected. Next check if $|Q|$ is a surface without boundary by checking that all vertex links are connected and are each circles. Finally check that the Euler characteristic $\chi(|Q|)$ is two, using the usual formula. | |
| Apr 27, 2012 at 12:20 | comment | added | Sam Nead | IL - regarding your first comment: I'd rather not think about $T'$ as I'm positive that checking my conditions is simpler to code and faster to boot -- not just linear, but linear with little overhead. Regarding your second comment: I think that you are correct here. That is, if $|\partial T|$ is homeomorphic to a two-sphere then $|T|$ is a manifold. (When this holds, $\partial |T|$ is homeomorphic to $|\partial T|$, which is nice.) | |
| Apr 27, 2012 at 11:40 | history | edited | Sam Nead | CC BY-SA 3.0 |
Many details added - perhaps too many!
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| Apr 26, 2012 at 18:54 | vote | accept | IL. | ||
| Apr 26, 2012 at 1:32 | comment | added | IL. | What if one additionally assumes that $T$ is a "sub-complex" of another tetrahedral complex $S$, and $|S|$ is a manifold? Will then the condition that $|\partial T|$ is 2-sphere guarantee that $T$ is 3d ball? By "sub-complex" I mean that one selects several tetrahedrons from $S$ and these tetrahedrons together with all their faces form $T$. | |
| Apr 26, 2012 at 1:24 | comment | added | IL. | Would you please give an example of finite tetrahedral complex $T$ embedded in $R\mathbb{R}^3$ such that $|\partial T|$ is homeomorphic to a 2-sphere, but $|T|$ is not homeomorphic to a 3d ball? If this example is written in some textbook, reference will be just fine. | |
| Apr 26, 2012 at 1:14 | comment | added | Igor Rivin | No, @Sam told you exactly the conditions you need (IF your complex is embedded in $\mathbb{R}^3,$ see my answer for the more general case). | |
| Apr 25, 2012 at 23:55 | comment | added | IL. | Let $\partial T$ be the boundary of simplicial complex $T$ ($\partial T$ consists of those 2-simplices which are on the boundary of only one tetrahedron). Let $|\partial T|$ be the underlying space. Suppose $|\partial T|$ is homeomorphic to a 2-sphere. Does it follow that $|T|$ is homeomorphic to a 3D ball? | |
| Apr 25, 2012 at 21:38 | comment | added | IL. | Can anything be said whether the conditions on Betti numbers of $T$ and $T'$ (see original post) guarantee that $T$ is homeomorphic to a ball? | |
| Apr 25, 2012 at 16:14 | history | edited | Sam Nead | CC BY-SA 3.0 |
Talking about the extra work.
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| Apr 25, 2012 at 15:18 | comment | added | Igor Rivin | I was aware of that, given that the OP asked about complexes in euclidean space, but since it was not clear that he knew what he wanted I gave a more general answer.... | |
| Apr 25, 2012 at 11:29 | history | answered | Sam Nead | CC BY-SA 3.0 |