Timeline for Triangulations of exotic 4-spheres
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2023 at 7:44 | comment | added | S. Carnahan♦ | @DenisGorodkov Thank you for that interesting link! | |
| Sep 19, 2023 at 22:19 | comment | added | Denis Gorodkov | As a short update: explicit triangulations of $\mathbb{C}P^n$ were constructed in 2014 (arxiv.org/abs/1405.2568). | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Aug 12, 2010 at 16:36 | comment | added | S. Carnahan♦ | Upon further reflection, I think the vertex count isn't so bad, but the (dim/2)-simplices can crush lesser machines. | |
| May 30, 2010 at 7:55 | vote | accept | John Vrem | ||
| May 30, 2010 at 6:14 | comment | added | Victor Protsak | Thank you for clarifying! So if I understand correctly, they are trying to $\textit{glue}$ a manifold from small pieces, whereas the most natural way in the case of $\mathbb{CP}^n$ would seem to be to $\textit{subdivide}$ it until one gets convex polyhedra (and there may well be other ways). Of course, the point about the vertex count is a valid one. | |
| May 29, 2010 at 23:39 | comment | added | S. Carnahan♦ | The specific context in which I heard the claim was at a talk 3 weeks ago by John Palmieri on algebraic topology computations using SAGE. The speaker said that gluing triangulated manifolds in a way that guarantees you get the homeomorphism type you want is a process that tends to require subdivisions, and when dimension is big, this can make the vertex count very large. | |
| May 29, 2010 at 22:11 | comment | added | Victor Protsak | I somehow doubt that exhibiting vertices of an explicit triangulation of $\mathbb{CP}^3$ is hard. Of course, any statement of the form "no one has ever written blah" needs to be interpreted cautiously. | |
| May 29, 2010 at 19:34 | history | answered | S. Carnahan♦ | CC BY-SA 2.5 |