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It should be true that $n$ $2(n-1)$-simplices can be glued together, such that $k$ of them intersect in a common $(2n -k -1)$-cell and the resulting object is a "convex $2(n-1)$-disc" whose boundary $(2n-3)$-sphere contains $n^2$ $(2n-3)$-cells. I am not sure how to make this more precise. For $n=1$ it's just a point, for $n=2$ we have two triangles glued together on a common side, and so on. Does the resulting object have a name?

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  • $\begingroup$ You write "...can be glued together on common $2(n-1)$-cells", but in your $n=2$ example they are glued along a $1$-simplex, not a $2$-simplex. Maybe I'm misunderstanding, but I can't extrapolate from the two examples you're giving. $\endgroup$ Commented May 20 at 12:12
  • $\begingroup$ Ah, I thought that $n$-simplex = $(n+1)$-cell. Is that not true? $\endgroup$ Commented May 20 at 12:16
  • $\begingroup$ Not to my understanding (both an $n$-simplex and a $n$-cell are $n$-dimensional objects), but regardless of conventions, there seems to be an inconsistency either way (since you also write "they all intersect in a $(2n-3)$-cell", and in your $n=2$ example they do intersect in a $1$-simplex). But these problems aside, it's not clear how to extrapolate from the two examples you give. $\endgroup$ Commented May 20 at 12:27
  • $\begingroup$ Sorry, there was a mistake in the "they all intersect"-part. I changed it and also the convention. $\endgroup$ Commented May 20 at 13:12

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I wonder whether you're looking for the following. Interpret simplicial complexes as determined by subsets of vertices that correspond to simplices, in the usual fashion. Take a set $S$ of $2n$ vertices, and partition them into two subsets, $A$ and $B$ of size $n$. Take for your $n$ simplices of dimension $2n−2$ the $n$ subsets of size $2n−1$ that contain $A$ and are missing one element of $B$. Your boundary of dimension $2n−3$ consists of the $n^2$ subsets of size $2n−2$ missing one element from each of $A$ and $B$. This accounts for your numbers, but is it what you're looking for? The name of this object is the join $\Delta^{(n-1)} * \smash{\dot{\Delta}}^{(n-1)}$. It's contractible, and the boundary is $\smash{\dot\Delta}^{(n-1)} * \smash{\dot{\Delta}}^{(n-1)}$ which is a $(2n-3)$-sphere.

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  • $\begingroup$ Yes, it's what I was looking for! Is the delta dot just the boundary of a simplex? $\endgroup$ Commented May 20 at 15:31
  • $\begingroup$ @BipolarMinds Yes it is. $\endgroup$ Commented May 20 at 15:32
  • $\begingroup$ I have to say that it was an interesting challenge to guess what you were looking for. The numerology helped. $\endgroup$ Commented May 20 at 21:06
  • $\begingroup$ Thx for accepting the challenge :) my construction was basically the same as yours but I didn't see the join.. $\endgroup$ Commented May 21 at 12:48
  • $\begingroup$ Can you actually describe both simplices in $\Delta^{(n-1)}*\Delta^{(n-1)}$ separetely in terms of your sets $A$ and $B$? $\endgroup$ Commented May 31 at 16:57

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