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Let $\mathcal{C}$ be an abstract simplicial complex on some finite set $\Omega$. I say that a subset $\Lambda\subset\Omega$ is minimally non-simplicial if it is not a simplex, but all of its subsets are. Write $h(\mathcal{C})$ for the size of the largest minimally non-simplicial subset of $\Omega$.

I need to calculate $h(\mathcal{C})$ for a particular family of abstract simplicial complexes. I'm wondering if this parameter can be seen somehow using algebraic topology -- via some (co)homology calculation or suchlike. Also, if this parameter has been studied somewhere in the literature, then I would be very interested to know about that.

I apologise if my question is vague/ ill-posed/ inappropriate -- I am not at all an expert on abstract simplicial complexes. For those who are interested, this family of complexes arose when studying some properties of permutation groups.

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    $\begingroup$ A trivial observation: Your parameter is not a topological invariant of the geometric realization of your simplicial complex. So no (co)homology calculation on the space as a whole can give you this parameter. (That doesn't exclude the possibility of getting the parameter by (co)homology calculations on suitable parts of the space, like stars or links.) $\endgroup$
    – Andreas Blass
    Commented Nov 9, 2017 at 15:24
  • $\begingroup$ @AndreasBlass, thanks for your comment which is very interesting. However, forgive my total ignorance -- what is trivial for you is not so much for me! Is it easy for you to give an example that illustrates why the parameter is not a topological invariant? $\endgroup$
    – Nick Gill
    Commented Nov 9, 2017 at 16:12
  • $\begingroup$ Consider a 3-point path the complex with three vertices, called 1,2,3, and with edges $\{1,2\}$ and $\{2,3\}$. So $\{1,2\}$ is a hole of cardinality 2. Compare with the 2-point path (vertices 1 and 2; edge $\{1,2\}$). It has no hole (of cardinality 2 or otherwise). $\endgroup$
    – Andreas Blass
    Commented Nov 9, 2017 at 16:16
  • $\begingroup$ Ah, I think I understand. Did you mean to write that the first example -- a 3-point path -- has a hole equal to $\{1,3\}$? If so, then I get what you are saying, thank you. $\endgroup$
    – Nick Gill
    Commented Nov 9, 2017 at 16:35
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    $\begingroup$ In fact, for any finite simplicial complex $\mathcal{C}$ there is a finite simplicial complex $\mathcal{D}$ with the same geometric realization and with $h(\mathcal{D})=2$. For instance, take $\mathcal{D}$ to be the first barycentric subdivision of $\mathcal{C}$. $\endgroup$
    – Richard Stanley
    Commented Nov 9, 2017 at 19:02

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