Every partially ordered set gives a triangulation of (the geometric realisation of) its order complex. (The n-simplices of the order complex are the chains $x\_0\leq x\_1\leq\cdots\leq x\_n$.) However, there are triangulations of topological spaces that do not arise this way.


Is there a name for triangulations having this special property of "coming from a poset?"


EDIT: Apparently, the following formulation of my question is cleaner: *what conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?*



Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?
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 $P\subseteq M$ satisfy $P\in \mathcal{S}$.
 (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



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> M. M. Bayer, *Barycentric subdivisions*. Pacific J. Math. 135 (1988), no. 1, pp. 1-16. 
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As Bayer points out, the result was first observed in 



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> R. Stanley, *Balanced Cohen-Macaulay complexes*, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.
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@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.


