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Mar 6, 2020 at 16:58 history edited Moishe Kohan CC BY-SA 4.0
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Mar 6, 2020 at 4:48 comment added Moishe Kohan @RichardStanley: Interesting. This means that $k(d-1)\ge [d/2]$, at least if $d$ is high enough. I will have to read more about this counting result.
Mar 6, 2020 at 0:28 comment added Richard Stanley There are superexponentially many (in both $n$ and $d$) triangulations of a $(d-1)$-dimensional sphere (or even a simplicial $d$-polytope) with $n$ vertices such that every $\lfloor d/2 \rfloor$-element set of vertices form a face. On the other hand, if every ($\lfloor d/2\rfloor +1$)-element set of vertices form a face, then $n=d+1$ and the triangulation has just one maximal face. See en.wikipedia.org/wiki/Neighborly_polytope. Some results are also known for other manifolds.
Mar 5, 2020 at 23:59 history edited Moishe Kohan CC BY-SA 4.0
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Mar 5, 2020 at 23:41 history asked Moishe Kohan CC BY-SA 4.0