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There are some discussion of shellable simplicial complexes here Testing simplicial complexes for shellability. My question is the following:

Assume that $\Delta$ is a pure simplicial complex on a finite vertex set $V$. Assume that $\Delta$ is Cohen-Macaulay, i.e., it has at most top-dimensional homology. (There are examples of CM but non-shellable simplicial complexes). Moreover, assume that the symmetric group $S_V$ acts transitively on $\Delta$. Could this possibly imply that the simplicial complex $\Delta$ is shellable?

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  • $\begingroup$ What do you mean by "the symmetric group $S_V$ acts transitively on $\Delta$"? Do you mean that the automorphism group of $\Delta$ acts transitively on $V$? $\endgroup$
    – Richard Stanley
    Commented Dec 3, 2013 at 18:09
  • $\begingroup$ Yes, the automorphism group of $\Delta$ does act transitively on $\Delta$. But maybe something stronger than that, but I'm not sure how to phrase it. An example that I have in mind is the matching complex of a complete graph. $\endgroup$
    – Thanh Vu
    Commented Dec 3, 2013 at 18:54
  • $\begingroup$ Take care that you want also links of faces to have at most top homology for Cohen-Macaulayness. $\endgroup$
    – Christian Stump
    Commented Dec 3, 2013 at 19:33
  • $\begingroup$ If $S_V$ acts transitively on $\Delta$, then I suppose $\Delta$ is some skeleton of a simplex, hence shellable. But that's almost surely not what you have in mind. $\endgroup$
    – Russ Woodroofe
    Commented Dec 5, 2013 at 1:16

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The Dissertation of Frank Lutz (http://page.math.tu-berlin.de/~lutz/dissertation.ps‎) has examples of Cohen-Macaulay vertex-transitive simplicial complexes that are not shellable. In particular, there is such a triangulation of Poincaré dodecahedral space $X$ with 17 vertices. This triangulation is Cohen-Macaulay since $X$ is a topological 3-manifold with the same homology groups as a 3-sphere. It is not shellable because it is not simply-connected.

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  • $\begingroup$ Why does simply-connected imply not shellable? Isn't the boundary of the triangle shellable, but not simply-connected? $\endgroup$
    – user2154420
    Commented Sep 18, 2019 at 18:59
  • $\begingroup$ I said that "it is not shellable because it is not simply-connected." A shellable manifold is simply-connected (and in fact must be a sphere if it without boundary and a ball otherwise), $\endgroup$
    – Richard Stanley
    Commented Sep 20, 2019 at 2:06
  • $\begingroup$ Yeah, typo on my part. I meant "Why does not simply-connected imply not shellable" $\endgroup$
    – user2154420
    Commented Sep 21, 2019 at 20:00
  • $\begingroup$ Every shellable simplicial complex has the homotopy type of a wedge of spheres. I don't have access to a reference at the moment. $\endgroup$
    – Richard Stanley
    Commented Sep 24, 2019 at 12:30

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