Here are necessary and sufficient conditions for an abstract, finite simplicial complex $mathcal{S}$ to be the order complex of some partially ordered set.
(i) $mathcal{S}$ has no missing faces of cardinality $geq 3$; and
(ii) The graph given by the edges (=$1$-dimensional simplices) of $mathcal{S}$ is a comparability.
[Definitions.
(a) A missing face of $mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that
$M not in mathcal{S}$, but all proper subsets $Psubseteq M$ satisfy $Pin mathcal{S}$.
(b) A graph (=undirected graph with no loops nor multiple edges) is a comparability if its
edges can be transitively oriented, meaning that whenever edges ${p, r_1}, {r_1, r_2},ldots, {r_{u−1}, r_u}, {r_u, q}$ are oriented as $(p, r_1), (r_1, r_2),ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge ${p, q}$ oriented as $(p, q)$.]
This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in
M. M. Bayer, Barycentric subdivisions. Pacific J. Math. 135 (1988), no. 1, pp. 1-16.
As Bayer points out, the result was first observed in
R. Stanley, Balanced Cohen-Macaulay complexes, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.
@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities must be used. For this, first check that the barycentric subdivision of any simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.
I hope this helps.
No comments:
Post a Comment