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Define the transpose product of a partial order $P$ over a set $S$ in the following way. The direct product of a partial order $P \subseteq S \times S$ and its converse, $P^{op}$, gives a partial ordering of $S \times S$, defined by $(v,w) \leq (x,y)$ when $v \leq_P x$ and $w \leq_{P^{op}} y$. Note that the converse to this partial order is the partial order generated by the direct product of $P^{op}$ and $P$. It can be easily proved that a restriction of this partial ordering to the pairs present in $P$ provides an inclusion order on the intervals of $P$. Moreover, in the incidence algebra of $P$, over any field, the two-sided ideals correspond precisely to the distributive lattice generated by this inclusion order (as join-irreducibles). And there is nothing more central to an incidence algebra than the structure of its intervals. The question is : what is behind this transpose product? Richard Stanley's Enumerative Combinatorics, vol 1, Chapter 3 exercise 7c (1st ed) refers to a 'twisted directed product' of posets. Is there a connection?

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  • $\begingroup$ I am not sure if this answers your question. Let $A$ be the incidence algebra of $P$. Then incidence algebra of $P\times P^{op}$ is the tensor product $A\otimes A^{op}$ and hence $A$-$A$-bimodules are the same as modules for the incidence algebra of $P\times P^{op}$. As ideals of $A$ are just sub-bimodules of $A$, this may have something to do with the connection with ideals. $\endgroup$ Oct 17, 2015 at 19:49
  • $\begingroup$ @Benjamin Steinberg: This is useful -- thank you. However there seems to be more at work than the tensor product. The transpose product appears to also be playing a role in the ordering of the meet irreducibles of the supersolvable lattice of subposets of $P$. And the elements of the inverted order of intervals are themselves fragments of the transpose product. I suspect a specific but unknown (to me) effect of the ternary relation of transitivity imposed by the binary relation of partial order. $\endgroup$ Oct 18, 2015 at 8:40

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