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Shellability of a simplicial sphere tells us that we can build up the complex one facet at a time such that at each step (except the last step) the complex is a PL-ball. At the last step it is of course a PL-sphere.

Lickorish showed that there exist non-shellable d-spheres for $d \geq 3$.

My question, perhaps a cynical one, is 'Are there any practical implications for why someone would care to have a shellable simplicial sphere, rather than just a PL-sphere?

user80533
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  • Well, it rules out a naive approach to the PL Poincare conjecture. You might have hoped to prove it by showing that a PL manifold that had the same homotopy type as the sphere was shellable, but you can't even do this for certain triangulations of the sphere. – Jim Conant Sep 21 '15 at 13:54
  • There may be nice uses for a non-shellable sphere, but I think the main way one would use the theorem in practice is as an affirmation that you can't assume that some random triangulation that someone hands you is shellable. – Jim Conant Sep 21 '15 at 14:03
  • Ah, my comments above were directed at why Lickorish's theorem is important, but I see that this is not the question. – Jim Conant Sep 21 '15 at 15:12

2 Answers2

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Read the first paragraph of Gunter Ziegler's nice paper.

Igor Rivin
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I think the question is slightly missing the point, in a way that it may be useful to explain. Frequently, the reason you would want to show something is a shellable simplicial sphere is in order to be able to deduce that it is a sphere at all. Given some crazy simplicial complex, it is not obvious whether it is a sphere or not, but for a shellable complex, it's much easier to tell.

Hugh Thomas
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