# Is the face poset a Heyting algebra?

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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Can you tell me what happens in the case of the face lattice of some well-known convex polytope? – Tomaž Pisanski Mar 18 '10 at 14:45
@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. – Mikael Vejdemo-Johansson Mar 18 '10 at 16:46
@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. – Tomaž Pisanski Mar 18 '10 at 22:33
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? – François G. Dorais Mar 19 '10 at 5:15
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. – Michał Kukieła Sep 13 '10 at 10:49