Realizing not-quite-barycentric subdivision of a polytope - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:49:00Z http://mathoverflow.net/feeds/question/102696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102696/realizing-not-quite-barycentric-subdivision-of-a-polytope Realizing not-quite-barycentric subdivision of a polytope Alexander Woo 2012-07-19T19:48:12Z 2012-07-20T04:46:16Z <p>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 $r([a,b])=r(b)-r(a)$).</p> <p>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?</p> <p>Notes:</p> <p>1) The 2-faces will always be quadrilaterals.</p> <p>2) The underlying cellular complex is not the barycentric subdivision, whose faces are the chains of $S$, not the intervals.</p> <p>3) If you apply the construction to a simplex, you should get a cube (of one higher dimension).</p> <p>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 $I(P)$.</p> <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.</p> http://mathoverflow.net/questions/102696/realizing-not-quite-barycentric-subdivision-of-a-polytope/102711#102711 Answer by Patricia Hersh for Realizing not-quite-barycentric subdivision of a polytope Patricia Hersh 2012-07-20T02:41:01Z 2012-07-20T04:46:16Z <p>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.</p> <p>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. </p> <p>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. </p>