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Let $(P,<)$ be a finite poset. Let $V$ be the free $\mathbb{R}$-vector space on $P \times \{0,1\}$; I'll write elements as sums of pairs of the form $(p,0)$ and $(0,q)$, so a general element is $$v = \sum_{p} \beta_{0,p}(p,0) + \beta_{1,p}(0,p).$$ Consider the subspace $B = \{v : \sum_p \beta_{0,p} - \beta_{1,p} = 0\}$. There are two cones in $B$ I'm interested in:

  1. The cone $C$ spanned by 'pure pairs' $(p_0,0) + (0,p_1)$, for all strict containments $p_0 \lneq p_1$ in $P$.

  2. The cone $C'$ cut out by the inequalities $\beta_{i,p} \geq 0$, plus the following inequalities, coming from antichains / order ideals in $P$: given an antichain $A$, let $A_\leq$ (resp. $A_\lneq$) be the set of elements dominated (resp. strictly dominated) by $A$. Let $\phi_A : B \to \mathbb{R}$ be defined by $$\phi_A(v) = \sum_{p \in A_\lneq} \beta_{0,p} - \sum_{p \in A_\leq} \beta_{1,p}.$$ Then we impose $\phi_A(v) \geq 0$ for all antichains $A$.

So $C'$ roughly corresponds to vectors $v$ with "smaller elements of $P$ on the left and larger elements on the right". It's automatic that $C \subseteq C'$. In fact, they are equal (this can be proven using a perfect-matching argument on an associated bipartite graph, giving a decomposition of $v$ into a sum of pure pairs).

Here is my question: which antichains correspond to facets of this cone? Are there conditions on $P$ that make this easier to answer?

Context: I am studying free resolutions of $GL(V)$-equivariant modules over certain rings. The free modules are indexed by partitions, so for me $P$ is (a finite truncation of) Young's lattice. The commutative algebra allows for a map of free modules $F_\lambda \to F_\mu$ if and only if $\mu \subseteq \lambda$, and the maps are only interesting when there is strict containment. The cone in question arises as a base case in the study of (the cones of) "equivariant Betti tables" in this setting, and I would like to identify its facets in the hope that they will correspond to identifiable objects in tableau combinatorics / Schubert calculus.

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