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Let $X = X_1 \cup \cdots X_n$ be a shellable complex, where the $X_i$ are the maximal faces, in the shelling order.

Question 1: Let $0 \leq k \leq n-1$. Then is $(X_1 \cup \cdots \cup X_k) \cap X_{k+1}$ shellable?

I want to say the answer is yes, but I can't find this in print anywhere, and I'm not sure.

Question 2: Same question, but for chain-lexicographic shellability.

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    $\begingroup$ The given intersection is a pure complex of codimension 1 of simplex $X_{k+1}$ hence it must be shellable. $\endgroup$
    – Wlod AA
    Commented Oct 29, 2022 at 8:26
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    $\begingroup$ @WlodAA Ah, thank you! I see now that the question is much simpler than I had appreciated. By definition of shellability, $(X_1 \cup \cdots \cup X_k) \cap X_{k+1} \subseteq X_{k+1} = \Delta^d$ is the union of some nonempty subset of the codimension-1 faces $\Delta^{d-1}$. To see that this is shellable (in any ordering), it suffices to observe that if $Y,Z \subseteq \Delta^d$ are any two codimension-1 faces, then $Y \cap Z$ is a simplex of dimension $d-2$. If you were to write this as answer, I would gladly accept it! $\endgroup$
    – Tim Campion
    Commented Oct 29, 2022 at 14:58
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    $\begingroup$ Tim, thank you. Sometimes one gets some kind of a block, looks at something and doesn't see it (e.g. that a simplex is a simplex), it happens (to me too, of course). I feel that I shouldn't call my comment an Answer. I hope that now you can go forward in your research smoothly for a while. Best regards. $\endgroup$
    – Wlod AA
    Commented Oct 29, 2022 at 17:09
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    $\begingroup$ For simplicial complexes the question is uninteresting, but for polyhedral complexes it makes more sense. The answer boils down to how one defines shellability. Usually the condition in Question1 is part of the definition. E.g., see Definition 2.3 at mi.fu-berlin.de/math/groups/discgeom/ziegler/Preprintfiles/…. $\endgroup$
    – Richard Stanley
    Commented Nov 1, 2022 at 14:29

1 Answer 1

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The answer is yes, as indicated in the comments. Thanks to WlodAA!

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