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My question concerns PL-homeomorphism and shellability.

I see from a previous question here that bistellar flips do not preserve shellability. However, the following two results are corollaries of Pachner's Theorem:

  1. Every simplicial PL-sphere is the boundary of a shellable simplicial ball.
  2. Bistellar equivalence implies PL-homeomorphism.

These results seem to be incompatible with the existence of non-shellable simplicial spheres, so I'm trying to find the flaw in my understanding.

If every simplicial PL-sphere is the boundary of a shellable ball, then how can we ever obtain non-shellable objects via bistellar flips? Is every triangulated sphere the boundary of a shellable object, while not being shellable itself?

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up vote 3 down vote accepted

My understanding is that the proof of shellability uses Pachner moves in one dimension down: if $S$ is a $d$-dimensional simplicial PL-sphere, then its boundary can be partitioned into two $(d-1)$-dimensional balls (e.g. one of the simplices is a ball itself, and all of the other simplices also form a ball), there exists a sequence of Pachner moves from one triangulated ball to the other, each individual Pachner move can be interpreted geometrically as a $d$-dimensional simplex, and the sequence of Pachner moves forms a shelling of a triangulation of $S$.

So the fact that there also exists a sequence of Pachner moves from a shellable triangulation of $S$ to some other non-shellable triangulation of $S$ is irrelevant; that sequence isn't even in the same dimension as the one used to prove shellability.

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Many thanks! I understand now. –  manifold-destiny Oct 27 '11 at 6:05
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