## Realizing not-quite-barycentric subdivision of a polytope

Given a poset $S$, one can form a new poset $I(S)$ whose elements are intervals in $S$ (i.e. either $\emptyset$ or $[a,b]$ for some $a\leq b\in S$) with ordering by (set) inclusion. If $S$ is ranked, then $I(S)$ will also be ranked (by $r([a,b])=r(b)-r(a)$).

If $S$ is the face lattice of a $d$-dimensional polytope $P$, is there a canonical way to construct a $d+1$-dimensional polytope $I(P)$ with face lattice $I(S)$? Is there a name for this construction?

Notes:

1) The 2-faces will always be quadrilaterals.

2) The underlying cellular complex is not the barycentric subdivision, whose faces are the chains of $S$, not the intervals.

3) If you apply the construction to a simplex, you should get a cube (of one higher dimension).

4) Of course the best construction should preserve symmetries and intertwine the inclusion of a face $F$ into $P$ with that of $I(F)$ into $I(P)$.

5) The only polytope I actually need an answer for right now is the regular 3-dimensional cube. If this construction only works for, say, simple polytopes, I'm fine with that.

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 Note that posets are enriched in themselves and that $I(S)$ is representable; it is just $\text{Hom}(0 \to 1, S)$. I don't think a similar construction will work for polytopes though... – Qiaochu Yuan Jul 19 at 20:12

Edit: I was including the empty set as a face in my earlier answer, which gives the wrong poset I(S). I have now corrected my answer below so as not to include the empty set, significantly changing the answer.

The poset I(S) cannot be the face poset of any polytope, because it will have multiple maximal elements of the form $[v,P]$ for the various vertices of P. I(S) should be the face poset of a subdivision of P. This subdivision is less refined than the barycentric subdivision of the polytope. Its vertices are the barycenters of faces, but its edges only connect barycenters of faces of consecutive dimensions, in contrast to the barycentric subdivision of the original polytope where they would come from all face inclusion pairs. One can continue upward in dimension, likewise describing the faces of the subdivison by progressively filling in its lower skeleta.

You also ask about the cube specifically. Your subdivision in that case is cubical, breaking each i-dimensional cubical face of the original cube into 2^i cubical faces of dimension i.

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 I meant for the empty set to be a face - precisely to avoid the problem you mentioned. – Alexander Woo Jul 20 at 17:14 In that case you have a different problem, which I addressed in my original answer -- if the empty set is a face, then the face poset for the boundary of your desired polytope will be the poset I discussed above. Hence the face poset of the boundary of your desired polytope will be the face poset of a regular CW complex which is homeomorphic to a ball. This is impossible, since the boundary of a polytope is homeomorphic to a sphere, and you cannot have two non-homeomorphic regular CW complexes with the same face poset. – Patricia Hersh Jul 20 at 22:05 Sorry for the confusion. Are you assuming that the original relations in the face poset of $P$ are supposed to also hold in the face poset of $I(P)$? My intention is that they do NOT. I know there is such a construction that turns simplices into cubes - each face of the simplex becomes a vertex; each incidence relation between a $d$-face and $d+1$-face becomes an edge, and so on. You can visualize that the Hasse diagram for a the boolean lattice (the face lattice of a simplex) becomes the edge graph of a cube. The problem is that I don't know how this works in general if it does at all. – Alexander Woo Jul 21 at 1:18 No, I'm not. I am using the order you describe by interval containment. This means (1) that each element of your original poset gives a minimal element in $I(S)$, since that gives a closed interval having a single element, (2) that each cover relation $a\prec b$ of the original poset gives an element of $I(S)$ which covers two of the minimal elements, namely covers $[a,a]$ and $[b,b]$ since $[a,b]$ is the set of size two with these two elements, (3) each $[a,c]$ where there exists $b$ with $a\prec b \prec c$ in the original poset will be at rank 2 with $[a,b]$ and $[b,c]$ just below it, etc. – Patricia Hersh Jul 22 at 18:07 Ah - reading more closely... there is a second set of maximal faces in the boundary, namely $[\emptyset, F]$ where $F$ is a facet of the original polytope. I'm not confident that makes the boundary into a sphere always, but it does if we start with a simplex or a 2-dimensional polygon. – Alexander Woo Jul 22 at 22:14