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Are there explicit examples of triangulations of exotic 4-spheres?

flag	asked May 29 2010 at 7:38 John Vrem 244●5

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I was unaware that exotic 4-spheres existed. That seems to be the smooth Poincaré conjecture. – S. Carnahan♦ May 29 2010 at 8:14

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en.wikipedia.org/wiki/Exotic_sphere#Gluck_twists – Spinorbundle May 29 2010 at 8:25

OK, thanks. And what about higher-dimensional exotic spheres? – John Vrem May 29 2010 at 15:03

The section on "Explicit examples of exotic spheres" in the wikipedia article describes Milnor's 7-sphere. – jc May 29 2010 at 17:21

I thought "exotic sphere" just meant a differentiable structure on the standard topological sphere which is inequivalent to the standard differentiable structure. So all you have to do is triangulate the standard topological sphere. Am I making a dumb mistake? – Todd Trimble May 29 2010 at 18:59

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S. Carnahan · Accepted Answer · 2010-05-29 19:34:06Z UTC

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Here is my comment expanded to answer form: The question of existence of exotic 4-spheres (i.e., the smooth Poincaré conjecture) is still open, and (according to Wikipedia) the existence of exotic PL structures is equivalent to it. Therefore, the answer is that no such explicit triangulations are known.

In general, explicit triangulations of higher dimensional manifolds seem to be difficult to write down. I've heard from computer algebra specialists that no one has even written an explicit triangulation of $\mathbb{CP}^3$. The chaos surrounding this earlier question might suggest that the problem is subtle.

answered May 29 2010 at 19:34

S. Carnahan♦
25.6k●41●90

	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. – Victor Protsak May 29 2010 at 22:11
	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. – S. Carnahan♦ May 29 2010 at 23:39
	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. – Victor Protsak May 30 2010 at 6:14
	Upon further reflection, I think the vertex count isn't so bad, but the (dim/2)-simplices can crush lesser machines. – S. Carnahan♦ Aug 12 2010 at 16:36

Torsten Asselmeyer-Maluga · Answer 2 · 2010-08-11 10:33:31Z UTC

The current status of the smooth Poincare conjecture in dimension 4 is presented in the paper: Michael Freedman, Robert Gompf, Scott Morrison and Kevin Walker "Man and machine thinking about the smooth 4-dimensional Poincare conjecture" in Quantum Topology, Volume 1, Issue 2 (2010), pp. 171–208 (arXiv) Thus the Cappel-Shaneson approach seem to fail by Akbuluts work. Now there is only possible construction via the Gluck twist with a real 2-knot (i.e. knotted 2-sphere), i.e. a knot not coming from a 3-dimensional (classical) knot.

Triangulations of exotic 4-spheres

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