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This is equivalent to my earlier question A question about something like "shelling" in a PL manifold, but maybe more comprehensible and to the point.

Given a triangulation of the PL sphere $S^n$, is there always a subdivision (a.k.a. refinement, a.k.a. finer triangulation) that makes it shellable?

Put this way, I'm guessing that the answer is well-known.

EDIT: I quickly got two different answers, each of which seems to give just what I need. I'm more or less arbitrarily accepting Allan's.

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According to the reviewer of Bruggesser, H.; Mani, P., Shellable decompositions of cells and spheres. Math. Scand. 29 (1971), 197–205 (1972), MR0328944, "The authors provide a rather ingenious proof of the following proposition: For every triangulation of an n-cell and every triangulation of an n-sphere there exists a subdivision of the triangulation that is sellable."

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  • $\begingroup$ Is the difference with the Adiprasito and Benedetti theorem in the nature of the allowed subdivisions? They only look at derived subdivisions. $\endgroup$
    – Russ Woodroofe
    Commented Feb 5, 2015 at 23:02
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In "Subdivisions, shellability, and collapsibility of products", Karim Adiprasito and Bruno Benedetti claim to show that "a triangulation of a sphere or ball is PL if and only if it becomes shellable after sufficently many derived subdivisions."

The paper doesn't seem to have appeared in print yet, but according to the webpage of one of the authors, it is to appear in Combinatorica.

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    $\begingroup$ In a later paper Adiprasito and Izmestiev prove the following stronger result: "a triangulation of a sphere or ball is PL if and only if it becomes polytopal after sufficently many derived subdivisions." arxiv.org/abs/1311.2965 $\endgroup$
    – Francisco Santos
    Commented Feb 17, 2015 at 10:01

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