On the cross-product conjecture for the number of linear extensions

We prove a weak version of the cross--product conjecture: ${F}(k+1,\ell) {F}(k,\ell+1) \geq (\frac12+\varepsilon) {F}(k,\ell) {F}(k+1,\ell+1)$, where ${F}(k,\ell)$ is the number of linear extensions for which the values at fixed elements $x,y,z$ are $k$ and $\ell$ apart, respectively, and where $\varepsilon>0$ depends on the poset. We also prove the converse inequality and disprove the {generalized cross--product conjecture}. The proofs use geometric inequalities for mixed volumes and combinatorics of words.


Introduction
This paper is centered around the cross-product conjecture (CPP) by Brightwell, Felsner and Trotter that gives the best known bound for the celebrated 1 3 -2 3 Conjecture [BFT95, Thm 1.3].Here we prove several weak versions of the conjecture, and disprove a stronger version we conjectured earlier in [CPP22a].
When F(k, ℓ) F(k + 1, ℓ + 1) = 0, the inequality (CPC) holds trivially.Curiously, the equality (1.4) does not hold in that case since the LHS can be strictly positive (Example 4.5).Except for the natural symmetry between (1.3) and (1.2), the proof of remaining three cases are quite different and occupies much of the paper.
Note that computing the number e(P ) of linear extensions of P is #P-complete [BW91], even for posets of height two or dimension two [DP18].Still, the vanishing assumptions which distinguish the cases in the Main Theorem 1.2, can be decided in polynomial time (see Theorem 4.2).
The proof of the Main Theorem 1.2 is a combination of geometric and combinatorial arguments.The former are fairly standard in the area, and used largely as a black box.The combinatorial part is where the paper becomes technical, as the translation of geometric ratios into the language of posets (following Stanley's pioneering approach in [Sta81]) leads to bounds on ratios of linear extensions that have not been investigated until now.Here we employ the combinatorics of words technology following our previous work [CPP22a,CPP22b,CPP23] (cf.§8.7) Let us emphasize that getting an explicit constant above 1/2 in the RHS is the main difficulty in the proof, as the 1/2 constant is relatively straightforward to obtain from Favard's inequality.This was noticed independently by Yair Shenfeld who derived it from Theorem 2.4 in the same way we did in the proof of Theorem 3.1. 1In another independent development, Julius Ross, Hendrik Süss and Thomas Wannerer gave a proof of the same 1/2 lower bound using the technology of Lorentzian polynomials [BH20] combined with a technical result from [BLP23]. 2  Our combinatorial tools also allow us to inch closer to the CPC for two classes of posets.Fix a subset A ⊆ X.We say that a poset P = (X, ≺) is t-thin with respect to A, if for every u ∈ X A there are at most t elements incomparable to u.For A = ∅, such posets are a subclass of posets of width t.Similarly, we say that a poset P = (X, ≺) is t-flat with respect to A, if for every u ∈ A there are at most t elements comparable to u.For A = X, such posets are a subclass of posets of height t.
Theorem 1.3.Let P = (X, ≺) be a finite poset.Fix distinct elements x, y, z ∈ X, and let A := {x, y, z}.Suppose that P is either t-thin with respect to A, or t-flat with respect to A. Then: (1.5) F(k + 1, ℓ) F(k, ℓ + 1) ≥ 1 2 + Our final result further confirms that CPC is somehow special among similar families of inequalities.While these other inequalities are not always true, they are not simultaneously too far off in the following sense.
Theorem 1.7.For every P = (X, ≺), every distinct x, y, z ∈ X, and every k, ℓ ≥ 1, at least two of the inequalities (CPC), (CPC1) and (CPC2) are true, where We prove that inequalities (CPC1) and (CPC2) hold for posets of width two (Corollary 7.3).However, they are false on infinite families of counterexamples (Proposition 7.1).By Theorem 1.7, this means that the CPC holds in all these cases.
Paper structure.We start with a short background Section 2 on mixed volumes and variations on the Alexandrov-Fenchel inequalities.This section is self-contained in presentation, and uses several well-known results as a black box.In a lengthy Section 3 we show how cross product inequalities arise as mixed volume, and make some useful calculations.We also prove Theorem 1.7.
We begin our combinatorial study of linear extensions in Section 4, where we give explicit conditions for vanishing of F(k, ℓ), and explore the consequences which include the equality (1.4).In Sections 5 and 6, we prove different cross product inequalities in the nonvanishing and vanishing case, respectively.We conclude with explicit examples (Section 7) and final remarks (Section 8).

Mixed volume inequalities
2.1.Alexandrov-Fenchel inequalities.Fix n ≥ 1.For two sets A, B ⊂ R n and constants a, b > 0, denote by aA + bB := a x +b y : x ∈ A, y ∈ B the Minkowski sum of these sets.For a convex body A ⊂ R n with affine dimension d, denote by Vol d (A) the volume of A. One of the basic result in convex geometry is Minkowski's theorem that the volume of convex bodies with affine dimension d behaves as a homogeneous polynomial of degree d with nonnegative coefficients: Theorem 2.1 (Minkowski, see e.g.[BuZ88, §19.1]).For all convex bodies A 1 , . . ., A r ⊂ R n and λ 1 , . . ., λ r > 0, we have: where the functions V(•) are nonnegative and symmetric, and where d is the affine dimension of λ 1 A 1 + . . .+ λ r A r (which does not depend on the choice of λ 1 , . . ., λ r ).
The coefficients V(A i 1 , . . ., A i d ) are called mixed volumes of A i 1 , . . ., A i d .We use d := d(A 1 , . . ., A r ) to denote the affine dimension of the Minkowski sum A 1 + . . .+ A r .
There are many classical inequalities concerning mixed volumes, and here we list those that will be used in this paper.
Theorem 2.3 (see e.g.[Sch14,Lemma 7.4.1]).We have 2.2.Favard's inequality for the cross-ratio.Towards proving the Main Theorem 1.2, we are most interested in bounds on the cross-ratio We start with the following well-known result which goes back to Favard (see §8.2).
Remark 2.5.From geometric point of view, the constant 1/2 in the inequality (2.3) is sharp.For example, take A and B non-collinear line segments, and C = A + B, see e.g.[AFO14, Prop.5.1] and [SZ16, Thm 6.1].However, for various families of convex bodies, it is possible to improve the constant perhaps, although not to 1 as one would wish.For example, when C is a unit ball in R 2 the constant can be improved to 2/π [AFO14, Prop.5.3].

Better cross-ratio inequalities.
The following two results follow from (2.2) by elementary arguments.They are variations on inequalities that are already known in the literature.We include simple proofs for completeness.
In this notation, we can rewrite (2.2) as Rearranging the terms, this gives: (2.5) By applying the AM-GM inequality to the terms α 2 1 β 2 + α 2 2 β 1 , we get Rearranging the terms, this gives: Since β 1 β 2 < 1, we can divide both side of the inequality above by 1 − √ β 1 β 2 and get This gives the desired (2.4).
We now present a variant of Proposition 2.6 in a degenerate case.
Proposition 2.7.Suppose that Proof.First note that (2.2) gives: We assume without loss of generality that In fact, otherwise, since the right side of (2.6) is at most 1 we immediately have (2.6).Now note that V Q (A, C) V Q (B, C) > 0 by (2.3) and by the assumption of the theorem.Taking the square root of (2.7) using (2.8), and then dividing by V Q (A, C) V Q (B, C), we get: This is equivalent to (2.6).

Poset inequalities via mixed volumes
3.1.Definitions and notation.We refer to [Tro95] for some standard posets notation.Let P = (X, ≺) be a poset with |X| = n elements.A dual poset is a poset P * = (X, ≺ * ), where x ≺ * y if and only if y ≺ x.
We somewhat change the notation and fix distinct elements z 1 , z 2 , z 3 ∈ X which we use throughout the paper.As in the introduction, for k, ℓ ≥ 1 let and let F(k, ℓ) := F(k, ℓ) .We will write F P,z 1 ,z 2 ,z 3 (k, ℓ) in place of F(k, ℓ) when there is a potential ambiguity in regards to the underlying poset P and the elements z 1 , z 2 , z 3 ∈ X .
3.2.Half CPC.We first prove that (CPC) holds up to a factor of 2. Formally, start with the following weak version of the Main Theorem 1.2: Theorem 3.1.For every k, ℓ ≥ 1, we have: To prove Theorem 3.1, we will first interpret the quantity F(k, ℓ) as in the language of mixed volumes.Here we follow Stanley's approach in [Sta81] (see also [KS84]).
Fix a poset P = (X, ≺), and let R X be the space of real vectors v that are indexed by elements x ∈ X.Throughout this section, the entries of the vector v that corresponds to x ∈ X will be denoted by v(x), to maintain legibility when x are substituted with elements z i .The order polytope K := K(P ) ⊂ R X is defined as follows: for all x ≺ y, x, y ∈ X , and 0 ≤ v(x) ≤ 1 for all x ∈ X .
Let K 1 , K 2 , K 3 ⊆ K be the slices of the order polytope defined as follows: Note that all Minkowski sums of these three polytopes have affine dimension ).
This lemma follows by a variation on the argument in the proof of [Sta81, Thm 3.2] and [KS84, Thm 2.5].
Proof.For 0 < s, t < 1, 0 < s + t < 1, define Note that K (s,t) = sK 1 + tK 2 + (1 − s − t)K 3 .Let us now compute the volume of K (s,t) .For every L ∈ E(P ) we denote by ∆ L ⊂ K (s,t) the polytope Note that K (s,t) is the union of ∆ L 's over all linear extensions L such that L(z 1 ) < L(z 2 ) < L(z 3 ), and furthermore all ∆ L 's have pairwise disjoint interiors.Hence it remains to compute the volume of ∆ L 's.Let L ∈ F(k, ℓ) for some k, ℓ ≥ 1, let h := L(z 1 ), and let x i (i ∈ {1, . . ., n}) be the i-th smallest element under the total order of L. Note that z 1 = x h , z 2 = x h+k , and z 3 = x h+k+ℓ .Then ∆ L consists of v ∈ R X that satisfies these three inequalities: 0 Denote by Φ : R X → R X the (volume preserving) transformation defined as follows: Φ(v) = w, where Then the image Φ(∆ L ) is the set of w ∈ R X that satisfies This set is the direct product of three simplices and has volume It follows from here that Since the choice of s, t is arbitrary, equation (3.2) follows from the Minkowski Theorem 2.1.
The theorem now follows by applying Lemma 3.2 into Theorem 2.4.

3.3.
Applications to cross products.We now quickly derive the key applications of mixed volume cross-ratio inequalities for the cross product inequalities.Proposition 3.3.Suppose that F(k, ℓ) F(k + 1, ℓ + 1) > 0. Then: The conclusion of the proposition now follows from Lemma 3.2 and Proposition 2.6.
3.4.More half-CPC inequalities.We start with the following half-versions of (CPC1) and (CPC2).The proofs follow the proof of Theorem 3.1 given above.
Note that (CPC1) is a dual inequality to (CPC2) in the following sense.Let P * := (X, ≺ * ) be the dual poset of P , i.e. x ≺ * y if and only if ) by the maps that send linear extensions of P to linear extensions of P * by reversing the total order.
On the other hand, one can think of (CPC1) and (CPC2) as negative variants of (CPC), in the following sense.Let Let k ′ := −k − 1 and ℓ ′ := ℓ + k.Under this change of variable, (CPC) then becomes which coincides with (CPC2) in this case.
Note, however, that (CPC) does not imply (CPC1) and vice versa, since k ′ are necessarily negative under this transformation.In fact, as mentioned in the introduction, we will present counterexamples to (CPC1) in §7.2.

Variations on the theme. The following three inequalities are variations on (CPC).
Lemma 3.6.For every k, ℓ ≥ 1 we have: By applying Lemma 3.2, we get the desired inequalities.
Proof.Taking the product of (LogC-1), (LogC-2) and (LogC-3), we have: By the assumptions, this implies the result.3 Proof of Theorem 1.7.First, assume that both (CPC1) and (CPC2) are false: Taking the product of both inequalities, we then get which contradicts (LogC-1).The proofs for the other cases are analogous.By analogy, let B(x, y) = {z ∈ X : x z y} be the interval between x and y, and let b(x, y) = |B(x, y)|.Without loss of generality we can always assume that z 1 ≺ z 2 ≺ z 3 , since otherwise these relations can be added to the poset.We then have b Let x, y ∈ X be two incomparable elements in P , write y x.Define Similarly, define Finally, let t(x) := max u(x, y) : y ∈ X, y x and t * (x) := max u * (x, y) : y ∈ X, y x , and we define t(x) := 1, t * (x) := 1 if every element y ∈ X is comparable to x.Clearly, t(x) ≤ b(x) and t * (x) ≤ b * (x), by definition.In this notation, recall that a poset P = (X, ≺) is t-thin with respect to A, if for every u ∈ X A we have n − b(u) − b * (u) ≤ t − 1.Similarly, recall that a poset P = (X, ≺) is t-flat with respect to A, if for every u ∈ A we have b(u) + b * (u) ≤ t + 1.Note that t(u), t * (u) ≤ t in either case.4.2.Vanishing conditions.Recall the following conditions for existence of restricted linear extensions.
Theorem 4.1 ([CPP22b, Thm 1.12]).Let P = (X, ≺) be a poset with |X| = n elements, and let z 1 , . . ., z r ∈ X be distinct elements such that Then there exists a linear extension L ∈ E(P ) with L(z i ) = a i for all 1 ≤ i ≤ r if and only if We apply this result to determine the vanishing conditions for F(k, ℓ).
Theorem 4.2.Let P = (X, ≺) be a poset with |X| = n elements, and let Note that conditions in the theorem can be viewed as 6 linear inequalities for (k, ℓ) ∈ N 2 .These inequalities determine a convex polygon in R 2 (see below).
Proof.We have that F(k, ℓ) > 0 if and only if there exists an integer a, such that the conditions of Theorem 4.1 are satisfied for the elements Rewriting the inequalities we obtain the following conditions The integer a exists if and only if the last inequalities are consistent, which leads to Noting that b(z i ) + b * (z i ) ≤ n + 1 for all i, the second inequality translates to 6 unconditional linear inequalities for k and ℓ, which can be written as Finally, since |X| = n, we also have: Combining with the previous inequalities, we obtain the desired conditions.
4.3.Cross product equality in the vanishing case.We are now ready to prove (1.4) in the main theorem.
5. Cross product inequalities in the nonvanishing case 5.1.Algebraic setup.We employ the algebraic framework from [CPP23,§6].With every linear extension L ∈ E(P ) we associate a word x L = x 1 . . .x n ∈ X * , such that L(x i ) = i for all 1 ≤ i ≤ n.In the notation of the previous section, this says that X = {x 1 , . . ., x n } is a natural labeling corresponding to L.
We can now define the following action of the group G n on E(P ) as the right action on the words x L , L ∈ E(P ).For x L = x 1 . . .x n as above, let Lemma 5.1.We have: The idea and basic setup of the proof will be used throughout.
Proof.Consider the first inequality.The main idea is to construct an explicit injection φ : N k → N k−1 ×I, where I := {1, . . ., t(a)}.This will show that We identify a linear extension L where L(a) = k with a word x ∈ N k where x k = a.Let x i be the last element in x appearing before a which is incomparable to a, that is set i := max{i : i < k, x i ≺ x k }.Such element exists because N k−1 > 0 implies that b(a) ≤ k − 1 and so among x 1 , . . ., x k−1 there is at least one x i ≺ a.Moreover, since i is maximal, we must have x j ≺ x k for j ∈ [i + 1, k].Also, for j ∈ [i + 1, k] we must have x j x i , as otherwise we would have x i ≺ x j ≺ x k = a.Thus, we have x j ∈ U (a, x i ) for i < j < k and so 1 ≤ k − i ≤ t(a).
We now define φ(x To see this is an injection we construct φ −1 , if it exists.Namely, φ −1 (x ′ , r) moves the element . This completes the proof of the first inequality.The second inequality follows by applying the same argument to the dual poset P * .
Corollary 5.2.We have: Note that the inequalities in the corollary are tight, see Proposition 7.4.
Proof.Observe that t(a) ≤ k − 1 since there are at most (k − 1) elements less than or equal to a by the assumption that N k−1 > 0. Similarly, observe that t * (a) ≤ n − k since there are at most (n − k) elements greater than or equal to a by the assumption that N k+1 > 0. These imply the result.

Double element ratio bounds.
We now give bounds for nonzero ratios of F(k, ℓ).For the degenerate case, see Section 4.
Lemma 5.3.Suppose that F(k, ℓ + 2) > 0. Then we have: Similarly, suppose that F(k + 2, ℓ) > 0. Then we have: Proof.For the first inequality, we construct an injection ψ : where I = I 1 ⊔ I 2 ⊔ I 3 and I i are intervals of lengths given by the RHS (see below).We use notation [p, q] = {i ∈ N : p ≤ i ≤ q} to denote the integer interval.Let x ∈ F(k + 1, ℓ + 1) be a word, such that x i = z 1 , x i+k+1 = z 2 and x i+k+ℓ+2 = z 3 .We consider several cases.Case 1: Suppose there exists an element x j ≺ z 2 for some j ∈ [i + 1, i + k].Let j be the maximal such index.Then for every r ∈ [j + 1, i + k] we have that x r ∈ U (z 2 , x j ).Set ψ(x ) = (x τ j • • • τ i+k , i+k+1−j), i.e. ψ moves x j to the position after z 2 , so that z 2 is now in position i+k.Observe that the inverse of ψ exists for all y ∈ F(k, ℓ + 2), since y i+k = z 2 y i+k+1 .Note that i + k + 1 − j ≤ min{u(z 2 , x j ), k}.Thus, we can record the value (i + k + 1 − j) in the first interval Case 2: Suppose that we have x j ≺ z 2 for all j ∈ [i, i+k].Then there exists an element x j ≻ z 1 .Indeed, otherwise x j ∈ B(z 1 , z 2 ) for all j ∈ [i, i+k+1], which gives k+2 ≤ |B(z 1 , z 2 )| and implies F(k, ℓ + 2) = 0 contradicting the assumption.As above, let j be the smallest possible index such that x j z 1 , so we can move x j in front of z 1 .Note that j − i ≤ min{b(z 1 , z 2 ) − 2, u * (z 1 , x j )}.We now have a word x ′ ∈ F(k, ℓ + 1).We split this case into two subcases.Subcase 2.1: Suppose there exists x r ≻ z 3 for r > i + k + ℓ + 2. Let r be the minimal such index, and move x r in front of z 3 , creating a word x ′′ ∈ F(k, ℓ + 2).Note that r − (k + ℓ + 2 + i) ≤ u * (z 3 , x r ).Thus, we can record the value (j − i, r − (k + ℓ + 2 + i)) in the second interval Subcase 2.2: Suppose x s ≻ z 3 for all s > i + k + ℓ + 2.Then, since F(k, ℓ + 2) = 0, there must be some x s ≺ z 2 , for s < i + k + 1.Since we are in Case 2, we have s < i.Let s be the largest such index.Thus x s+1 , . . ., x i+k ≺ x i+k+1 = z 2 .We can then move x s past all these entries to right past z 2 and obtain a word in F(k, ℓ + 2).Note that i − s ≤ u(z 2 , x s ).Thus, we can record the value (j − i, i − s) in the third interval Gathering these cases, and noting that t(x) ≥ u(x, y) and t * (x) ≥ u * (x, y) for all x, y ∈ X, we obtain the desired first inequality.For the second inequality, we apply the analogous argument to the dual poset P * .5.4.Bounds on cross product ratios.We can now bound the cross product ratios in the nonvanishing case.
Proof.We proceed as in the proof of Lemma 5.3, constructing an injection ψ : , where I = I 1 ⊔ I 2 ⊔ I 3 ⊔ I 4 are intervals of lengths specified by the RHS, each of them given in the corresponding case below.Let x ∈ F(k + 1, ℓ) be a word (corresponding to a linear extension) with x i = z 1 , x i+k+1 = z 2 and x i+k+ℓ+1 = z 3 .We consider several independent cases, which correspond to different parts of the interval I: Case 1: Suppose that there exists x j z 1 with j ∈ [i + 1, i + k] and let j be the minimal such index.Then {x i , . . ., x j−1 } ⊆ U * (z 1 , x j ) and j − i ≤ min{k, t * (z 1 )}.Take x τ j−1 • • • τ i , which moves x j to position i and z 1 to position i + 1.Then the resulting word is in F(k, ℓ), and we record the value (j − i) in the first interval Case 2: Suppose that x j ≻ z 1 for all j ∈ [i + 1, i + k].Furthermore, suppose that x r ≻ z 1 and x r ≺ z 3 for all r ∈ [i + k + 2, i + k + ℓ].These assumptions imply that there exists j ∈ [i+1, i+k] such that x j z 3 , as otherwise we have {x i , . . ., x i+k+ℓ+1 } ∈ B(z 1 , z 3 ), contradicting the assumption that F(k, ℓ) > 0. Assume that j is the maximal such index j.It then follows that {x j+1 , . . ., x i+k+ℓ+1 } ⊆ U (z 3 , x j ).This implies that i Case 3: Suppose again that x j ≻ z 1 for all j ∈ [i + 1, i + k], but now that there exists r ∈ [i + k + 2, i + k + ℓ] such that either x r z 1 or x r z 3 .The first condition implies that there exists x j z 2 with j ∈ [i + 1, i + k], as otherwise we would have F(k, ℓ) = 0. Let j be the maximal such index.Then {x j+1 , . . ., x i+k+1 } ⊆ B(z 1 , z 2 ) − z 1 , and thus i + k + 1 − j ≤ b(z 1 , z 2 ) − 1.Also note that {x j+1 , . . ., x i+k+1 } ⊆ U (z 2 , x j ), and thus i + k + 1 − j ≤ t(z 2 ).Move x j right past z 2 via x τ j • • • τ i+k+1 and record that move with s := i + k + 1 − j ≤ min{b(z 1 , z 2 ) − 1, t(z 2 )}.We now consider the new word x ′ ∈ F(k, ℓ + 1).We split this case into two subcases.Subcase 3.1: Suppose that there exists an element x ′ r = x r z 3 for some r r is moved past z 3 .We record the pair (s, i Subcase 3.2: Suppose that there exists r to the position before z 1 and record the pair (s, r − i − k − 1) in the product of intervals Gathering these cases we obtain the desired inequality in the lemma.Lemma 6.2.Suppose that F(k + 2, ℓ) > 0. Then: Proof.We proceed as in the proof of Lemma 5.3, constructing an injection ψ : , where I = I 1 ⊔ I 2 ⊔ I 3 are intervals of lengths specified by the RHS corresponding to each case below.Let x ∈ F(k + 1, ℓ) be a word (corresponding to a linear extension) with x i = z 1 , x i+k+1 = z 2 and x i+k+ℓ+1 = z 3 .We consider three independent cases, which correspond to different intervals I i (see below).
Case 1: Suppose that there exists x j z 1 with j ∈ [1, i − 1], and let j be the maximal such index.Then {x j+1 , . . ., x i } ⊂ U (z 1 , x j ) and so i − j ≤ t(z 1 ).We take x τ j • • • τ i−1 , which moves x j to position i and z 1 to position i − 1.Then the resulting word is in F(k + 2, ℓ), and we record the value (i − j) in the first interval Case 2: Suppose that x j ≺ z 1 for all j ∈ [1, i − 1].Since F(k + 2, ℓ) > 0, there exists x j z 2 with j ∈ [i + k + 2, n].Let j be the minimal such index.Then {x i+k+1 , . . ., x j−1 } ⊂ U * (z 2 , x j ), and thus j − i − k − 1 ≤ t * (z 2 ).Move x j to the front of z 2 via x τ j−1 • • • τ i+k+1 to get a new word x ′ .We split this case into two subcases: . By the assumption of Case 2 and the fact that F(k + 2, ℓ) > 0, there exists r ∈ r to the front of z 3 to obtain a new word x ′′ ∈ F(k + 2, ℓ), and we record the value (j Gathering these cases we obtain the desired inequality in the lemma.
6.2.Bounds on cross product ratios.We are now ready to obtain bounds on the cross product ratios in the vanishing case.
We then have: Lemmas 6.1 and 6.2 now give: as desired.
We also need the following variation on this proposition.
Proposition 6.4.Let P = (X, ≺) be either a t-thin or t-flat poset with respect to {z 1 , z 2 , z 3 }.Suppose also that F(k, ℓ) F(k + 2, ℓ) > 0. Then we have: Proof.We follow the proof of the proposition above with the following adjustments.For the first inequality in the maximum, we replace the bound (6.2) with the following: Now the first inequality follows from Lemmas 6.1 and 6.2, with the parameters bounded by (6.1) and (6.3).
For the second inequality in the maximum, we replace the bound (6.1) and (6.2) with the following: Now the second inequality follows from Lemmas 6.1 and 6.2, with the parameters bounded by (6.4) and (6.5).
Proof of Theorem 1.4.Lemma 6.1 combined with (6.1), gives Similarly, Lemma 6.2 for k ′ = k − 1 and ℓ ′ = ℓ + 1, combined with (6.2), gives: Multiplying these inequalities, we obtain the first term in the minimum of the desired upper bound.Via poset duality, see the proof of Theorem 1.2 above, we can exchange the k and ℓ terms and obtain the other inequality.

Examples and counterexamples
7.1.Inequalities (CPC1) and (CPC2).Recall that by Theorem 1.7 at least one of these two inequalities must hold.We now show that for some posets (CPC2) does not hold.By the poset duality, the inequality (CPC1) also does not hold.
Proposition 7.1.The inequality (CPC2) fails for an infinite family of posets of width three.
7.2.Counterexamples to the generalized CPC.We now show that the examples in proof of Proposition 7.2 are also counterexamples to Conjecture 1.5, thus proving Theorem 1.6.
This follows from Proposition 7.2 and Theorem 3.3 in [CPP22a] which proves (GCPC) for posets of width two.7.3.Stanley ratio.It follows from Corollary 5.2, the following bound on the Stanley ratio: whenever the LHS is well defined.The following example shows that both the inequality (7.1) and Corollary 5.2 are tight.
In the notation of §5.2, fix 1 ≤ k ≤ n.Let P k := (X, ≺) be the width two poset given by Thus, for posets P k the inequality (7.1) is an equality.
In summary, for the poset P k,ℓ , we have: as desired.

Final remarks
8.1.The cross-product conjecture (Conjecture 1.1) has been a major open problem in the area for the past three decades, albeit with relatively little progress to show for it (see [CPP22a] for the background).The following quote about a closely related problem seems applicable: "As sometimes happens, we cannot point to written evidence that the problem has received much attention; we can only say that a number of conversations over the last 10 years suggest that the absence of progress on the problem was not due to absence of effort."[KY98, p. 87] 8.2.Theorem 2.4 is well-known in the area and can be traced back to the works of Jean Favard in the early 1930s. 4Of course, this is not the only Favard's inequality known in the literature.In fact, Theorem 2.3 which goes back to Matsumura (1932) and Fenchel (1936), seem to also have been inspired by Favard's work. 5In a closely related context of Lorentzian polynomials, Favard's inequality appears in [BH20, Prop.2.17].For more on Theorem 2.3, see [BF87, §51] and references therein.
8.3.As we mentioned in the introduction, the Υ ≥ 1/2 lower bound derived from Favard's inequality (Theorem 2.4) easily implies the 1/2 lower bound on the cross product.Given the straightforward nature of this implication, one can think of this paper as the first attempt to finding the best ε ≥ 0, such that In this notation, the CPC states that ε = 1 2 .Our Main Theorem 1.2 and especially the "t-thin or t-flat" Theorem 1.3 are the first effective bounds for ε > 0.More precisely, here we prove ε = Ω 1 n for all posets, and a constant lower bound on ε for posets with bounded parameter t.Improving these bounds seems an interesting challenging problem even if the CPC ultimately fails.
We should mention that from the Analysis point of view, the first explicit bound on ε > 0 is typically an important step, no matter how small.We refer to [KLT00] and [CGMS23] for especially remarkable breakthroughs of this type.8.4.The constant 1/2 in Favard's inequality has the same nature as the constant 2 in [RSW23, Cor.1.5] which also follows from Favard's inequality written in terms of the Lorentzian polynomials technology.The relationships to the constant 2 in [CP22b, Thm 1.1] and [HSW22, Thm 5] are more distant, but fundamentally of the same nature.While in the former case it is tight, in the latter is likely much smaller, see [Huh18, §2.3].
8.5.The reader might find surprising the discrepancy between the vanishing and the nonvanishing cases in the Main Theorem 1.2.Note that the vanishing case actually implies a worse bound (1.2) compared to the bound (1.1) in the nonvanishing case, instead of making things simpler.This is an artifact of the mixed volume inequalities and combinatorial ratios.Proposition 6.3 gives a better bound than Proposition 6.4 simply because the ratio of F(•, •)'s in the former is under a square root which decreases the order.However, these combinatorial bounds can only be applied when the corresponding terms are nonzero.
Clearly, there is no way to justify this discrepancy, as otherwise we would know how to disprove the CPC.Still, one can ask if there is another approach to the vanishing case which would improve the bound?We caution the reader that sometimes nonvanishing does indeed make a difference (see e.g.Example 4.5).8.6.Theorem 4.1 gives the vanishing conditions for the generalized Stanley inequalities.It was first stated without a proof in [DD85, Thm 8.2], and it seems the authors were aware of a combinatorial proof by analogy with their proof of the corresponding results for the order polynomial.The theorem was rediscovered in [CPP22b, Thm 1.12], where it was proved via combinatorics of words.Independently, it was also proved in [MS22, Thm 5.3] by a geometric argument.8.7.There is a large literature on the negative dependence in a combinatorial context, see e.g.[BBL09,Huh18,Pem00], and in the context of linear extensions [KY98,She82].When it comes to correlation inequalities for linear extensions of posets, this paper can be viewed as the third in a series after [CP22b] and [CP23] by the first two authors.These papers differ by the tools involved.In [CP22b], we use the combinatorial atlas technology (see [CP21,CP22a]), while in [CP23] we use the FKG-type inequalities.
The idea of this paper was to use geometric inequalities for mixed volumes, to obtain new cross product type inequalities.As we mentioned in the introduction, it transfers the difficulty to the combinatorics of words.This is the approach introduced in [Hai92, MR94] (see also [Sta09]), and advanced in [CPP22a,CPP22b,CPP23] in a closely related context.8.8.Despite the apparent symmetry between the t-thin and t-flat notions, there is a fundamental difference between them.For posets P = (X, ≺) which are t-thin with respect to a set A of bounded size, the number e(P ) of linear extension can be computed in polynomial time, since P ′ := (X A, ≺) has width at most t.On the other hand, for posets which have bounded height, computing e(P ) is #P-complete [BW91, DP18], and the same holds for posets which are t-flat with respect to a set of bounded size.

4.
Vanishing of poset inequalities 4.1.Poset parameters.For an element x ∈ X, let B(x) := y ∈ X : y x denote the lower order ideal generated by x, and let b(x) := |B(x)|.Similarly, let B * (x) := y ∈ X : y x denote the upper order ideal generated by x, and let b * (x) := |B * (x)|.

5. 2 .
Single element ratio bounds.Let P = (X, ≺) be a poset with |X| = n elements, and fix an element a ∈ X of the poset.Let N k be the set of linear extensions L ∈ E(P ) such that L(a) = k, and let N k := | N k |.