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Is the face poset of a simplicial or nice enough cellular complex a Heyting algebra in some natural way?

Edited to add: For the benefit of illustration, here's a few face posets:

  • the boundary of a square: Elements $a,b,c,d$ of dimension 0 and $x,y,z,w$ of dimension 1; poset relations $a,b < x$; $b,c < y$; $c,d < z$; $a,d < w$.

  • Two 2-simplices joined along a common edge: Elements $a,b,c,d$ of dimension 0, $j,k,l,m,n$ of dimension 1 and $x,y$ of dimension 2. Poset relations are: $a,b < j$; $a,c < k$; $b,c < l$; $b, d < m$; $c,d < n$ and $j,k,l < x$ and $l,m,n < y$.

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  • $\begingroup$ Can you tell me what happens in the case of the face lattice of some well-known convex polytope? $\endgroup$ Mar 18, 2010 at 14:45
  • $\begingroup$ @Tomaž Is this you leading me to the answer, or you asking me what the face lattice looks like? I'd start out with considering the face lattice of the simplex, which is the boolean algebra of corresponding size, and therefore - I think - a Heyting algebra from its boolean structure. $\endgroup$ Mar 18, 2010 at 16:46
  • $\begingroup$ @Mikael: No, No, I just wanted to see if there is an example of a face lattice with non-boolean Heyting algebra structure. I am not really used to thinking in terms of Heyting algebras. $\endgroup$ Mar 18, 2010 at 22:33
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    $\begingroup$ I'm not used to face posets, so I'll only give some pointers. I presume you formally add a top and bottom element to the poset, otherwise you certainly won't have a Heyting algebra. A finite lattice is a Heyting algebra if and only if it is distributive. Are you also considering infinite face posets? $\endgroup$ Mar 19, 2010 at 5:15
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    $\begingroup$ Consider the face poset of the 3-element discrete space (with top and bottom elements added). It is the diamond M3 lattice, which is not distributive. $\endgroup$ Sep 13, 2010 at 10:49

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