Barrett-Johnson inequalities for totally nonnegative matrices

Given a matrix $A$, let $A_{I,J}$ denote the submatrix of $A$ determined by rows $I$ and columns $J$. Fischer's Inequalities state that for each $n \times n$ Hermitian positive semidefinite matrix $A$, and each subset $I$ of $\{1,\dotsc,n\}$ and its complement $I^c$, we have $\det(A) \leq \det(A_{I,I})\det(A_{I^c,I^c})$. Barrett and Johnson (Linear Multilinear Algebra 34, 1993) extended these to state inequalities for sums of products of principal minors whose orders are given by nonincreasing integer sequences $(\lambda_1,\dotsc,\lambda_r)$, $(\mu_1,\dotsc,\mu_s)$ summing to $n$. Specifically, if $\lambda_1+\cdots+\lambda_i\leq \mu_1+\cdots+\mu_i$ for all $i$, then $$\lambda_1!\cdots\lambda_r! \sum_{(I_1,\dotsc,I_r)} \det(A_{I_1,I_1}) \cdots \det(A_{I_r,I_r}) ~\geq~ \mu_1!\cdots\mu_s! \sum_{(J_1,\dotsc,J_s)} \det(A_{J_1,J_1}) \cdots \det(A_{J_s,J_s}),$$ where sums are over sequences of disjoint subsets of $\{1,\dotsc,n\}$ satisfying $|I_k| = \lambda_k$, $|J_k| = \mu_k$. We show that these inequalities hold for totally nonnegative matrices as well

Monomial Nonnegativity and the Bruhat Order

We show that five nonnegativity properties of polynomials coincide when restricted to polynomials of the form x1, pi(1) ... xn,pi(n) - x1, sigma(1) ... xn, sigma(n), where $\pi and sigma are permutations in Sn. In particular, we show that each of these properties may be used to characterize the Bruhat order on Sn

Enumerating (2+2)-free posets by the number of minimal elements and other statistics

An unlabeled poset is said to be (2+2)-free if it does not contain an induced subposet that is isomorphic to 2+2, the union of two disjoint 2-element chains. Let $p_n$ denote the number of (2+2)-free posets of size $n$. In a recent paper, Bousquet-M\'elou et al.\cite{BCDK} found, using so called ascent sequences, the generating function for the number of (2+2)-free posets of size $n$: $P(t)=\sum_{n \geq 0} p_n t^n = \sum_{n\geq 0} \prod_{i=1}^{n} (1-(1-t)^i)$. We extend this result in two ways. First, we find the generating function for (2+2)-free posets when four statistics are taken into account, one of which is the number of minimal elements in a poset. Second, we show that if $p_{n,k}$ equals the number of (2+2)-free posets of size $n$ with $k$ minimal elements, then $P(t,z)=\sum_{n,k \geq 0} p_{n,k} t^n z^k = 1+ \sum_{n \geq 0} \frac{zt}{(1-zt)^{n+1}} \prod_{i=1}^n (1-(1-t)^i)$. The second result cannot be derived from the first one by a substitution. On the other hand, $P(t)$ can easily be obtained from $P(t,z)$ thus providing an alternative proof for the enumeration result in \cite{BCDK}. Moreover, we conjecture a simpler form of writing $P(t,z)$. Our enumeration results are extended to certain restricted permutations and to regular linearized chord diagrams through bijections in \cite{BCDK,cdk}. Finally, we define a subset of ascent sequences counted by the Catalan numbers and we discuss its relations with (2+2)- and (3+1)-free posets