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Suppose you are given a 3-ball $B$ in $\mathbb{R}^3$ that is bounded by a PL sphere, a triangulation $T$ of $B$ by Euclidean tetrahedra. Is that triangulation necessarily shellable?

I know that if $T$ can be lifted to a convex hypersurface in $\mathbb{R}^4$, then it is shellable; this applies if $T$ comes from a Delaunay triangulation. But I believe most triangulations are not liftable.

I suspect the answer is "no", that $T$ is not necessarily shellable. The first obstruction I found was the existence of a knotted arc made out of two edges of the triangulation; this obviously cannot happen for triangulations of the kind that I described.

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up vote 8 down vote accepted

I haven't had time to check through the details of the construction, but the example B_3_9_18 found in the proof of Theorem 2 here by Frank Lutz appears to be embeddable in 3-space.

Frank specializes in creating wickedly tiny polytopes with unexpected properties, his website may also be useful for those studying similar questions: http://page.math.tu-berlin.de/~lutz/ (see in particular "the manifold page").

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Thanks! According to that page, there were earlier examples by Grünbaum and by Ziegler. – Dylan Thurston Feb 27 at 4:11
    
@DylanThurston you're welcome! There are older examples, but the fact that this one has only 9 vertices makes it easy to check that it can sit inside 3-space... – Vidit Nanda Feb 27 at 4:28

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