2.1 General notation

Let $[n]=\{1,\ldots ,n\}$ , $\mathbb N=\{0,1,2,\ldots \}$ and ${\mathbb Z}_{\ge 1}=\{1,2,\ldots \}$ . For a subset $S\subseteq X$ and element $x\in X$ , we write $S+x:=S\cup \{x\}$ and $S-x:=S\smallsetminus \{x\}$ . For a sequence $\operatorname {\mathrm {\mathbf {a}}} =(a_1,\ldots ,a_m)$ , denote $|\operatorname {\mathrm {\mathbf {a}}}| := a_1 + \ldots + a_m$ . This sequence is log-concave if $a_i^2\ge a_{i-1} a_{i+1}$ for all $1< i < m$ .

2.2 Mixed volumes

Fix $n \geq 1$ . For two sets $A, B \subset \mathbb {R}^n$ and constants $\alpha ,\beta>0$ , denote by

$$\begin{align*}\alpha A + \beta B \, := \, \bigl\{ {\hskip.03cm} \alpha\operatorname{\mathrm{\mathbf{x}}} + \beta \operatorname{\mathrm{\mathbf{y}}} \, : \, \operatorname{\mathrm{\mathbf{x}}} \in A, \operatorname{\mathrm{\mathbf{y}}} \in B {\hskip.03cm} \bigr\} \end{align*}$$

the Minkowski sum of these sets. For a convex body $\mathrm {K} \subset \mathbb {R}^n$ with affine dimension d, denote by $\mathrm {Vol}_d(\mathrm {K})$ the volume of $\mathrm {K}$ . We drop the subscript when $d=n$ .

One of the basic result in convex geometry is Minkowski’s theorem (see e.g. [Reference Burago and ZalgallerBZ88, Section 19.1], 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).

For all convex bodies $\mathrm {K}_1, \ldots , \mathrm {K}_r \subset \mathbb {R}^n$ and $\lambda _1,\ldots , \lambda _r> 0$ , we have:

(2.1) $$ \begin{align} \mathrm{Vol}_d(\lambda_1 \mathrm{K}_1+ \ldots + \lambda_r \mathrm{K}_r) \ = \ \sum_{1 {\hskip.03cm} \le {\hskip.03cm} i_1{\hskip.03cm} ,{\hskip.03cm} \ldots {\hskip.03cm} , {\hskip.03cm} i_d{\hskip.03cm} \le {\hskip.03cm} r} \mathrm{V}\bigl(\mathrm{K}_{i_1},\ldots, \mathrm{K}_{i_d}\bigr) \, \lambda_{i_1} {\hskip.03cm}\cdots{\hskip.03cm} \lambda_{i_d}, \end{align} $$

where the functions $\mathrm {V}(\cdot )$ are nonnegative and symmetric, and where d is the affine dimension of $\lambda _1 \mathrm {K}_1+ \ldots + \lambda _r \mathrm {K}_r$ (which does not depend on the choice of $\lambda _1,\ldots , \lambda _r$ ).

The coefficients $\mathrm {V}(\mathrm {A}_{i_1},\ldots , \mathrm {A}_{i_d})$ are called mixed volumes of $\mathrm {A}_{i_1}, \ldots , \mathrm {A}_{i_d}$ . We refer to [Reference Hug and WeilHW20, Reference LeichtweissLei80, Reference SchneiderSchn14] for an accessible introduction to the subject.

2.3 Posets

For a poset $P=(X,\prec )$ and a subset $Y \subset X$ , denote by $P_Y=(Y,\prec )$ a subposet of P. We use $(P-z)$ to denote a subposet $P_{X-z}$ , where $z\in X$ . Element $x\in X$ is minimal in $ P$ if there exists no element $y \in X-x$ , such that $y \prec x$ . Define maximal elements similarly. Denote by $\min (P)$ and $\max (P)$ the set of minimal and maximal elements in P, respectively.

In a poset $P=(X,\prec )$ , elements $x,y\in X$ are called parallel or incomparable if $x\not \prec y$ and $y \not \prec x$ . We write $x\parallel y$ in this case. A comparability graph is a graph on X, with edges $(x,y)$ , where $x\prec y$ . Element $x\in X$ is said to cover $y\in X$ if $y\prec x$ and there are no elements $z\in X$ , such that $y\prec z \prec x$ .

A chain is a subset $C\subset X$ of pairwise comparable elements. The height of poset $P=(X,\prec )$ is the maximum size of a chain. An antichain is a subset $A\subset X$ of pairwise incomparable elements. The width of poset $P=(X,\prec )$ is the size of the maximal antichain.

A dual poset is a poset $P^\ast =(X,\prec ^\ast )$ , where $x\prec ^\ast y$ if and only if $y \prec x$ . A disjoint sum $P+Q$ of posets $P=(X,\prec )$ and $Q=(Y,\prec ')$ is a poset $(X\cup Y,\prec ^{\small {\diamond }})$ , where the relation $\prec ^{\small {\diamond }}$ coincides with $\prec $ and $\prec '$ on X and Y, and $x\| y$ for all $x\in X$ , $y\in Y$ . A linear sum $P\oplus Q$ of posets $P=(X,\prec )$ and $Q=(Y,\prec ')$ is a poset $(X\cup Y,\prec ^{\small {\diamond }})$ , where the relation $\prec ^{\small {\diamond }}$ coincides with $\prec $ and $\prec '$ on X and Y, and $x\prec ^{\small {\diamond }} y$ for all $x\in X$ , $y\in Y$ .

Posets constructed from one-element posets by recursively taking disjoint and linear sums are called series-parallel. Both n-chain $C_n$ and n-antichain $A_n$ are examples of series-parallel posets. A Forest is a series-parallel poset formed by recursively taking disjoint sums (as before), and linear sums with one element: $C_1 \oplus P$ . We refer to [Reference StanleySta12, Chapter 3] for an accessible introduction and to surveys [Reference Brightwell and WestBW00, Reference TrotterTro95] for further definitions and standard results.

2.4 Poset polytopes

Let $P=(X,\prec )$ be a poset with $|X|=n$ elements. The order polytope $\mathcal O_P\subset \mathbb R^n$ is defined as

(2.2) $$ \begin{align} 0\le \alpha_x \le 1 \quad \text{for all} \quad x\in X, \qquad \alpha_x \le \alpha_y \quad \text{for all} \quad x\prec y, \ \ x,y \in X. \end{align} $$

Similarly, the chain polytope (also known as the stable set polytope) $\mathcal S_P\subset \mathbb R^n$ is defined as

(2.3) $$ \begin{align} \beta_x \ge 0 \quad \text{for all} \quad x\in X, \qquad \beta_x + \beta_y + \ldots \le 1 \quad \text{for all} \quad x\prec y \prec \cdots\,, \ x,y,\ldots \in X. \end{align} $$

In [Reference StanleySta86], Stanley computed the volume of both polytopes:

(2.4) $$ \begin{align} \mathrm{Vol} (\mathcal O_P) \, = \, \mathrm{Vol} (\mathcal S_P) \, = \, \frac{e(P)}{n!}. \end{align} $$

This connection is the key to many applications of geometry to poset theory and vice versa.

2.5 Terminology

For functions $f,g: X\to \mathbb R$ , we write $f\geqslant g$ if $f(x) \ge g(x)$ for all $x\in X$ . For an inequality $f \geqslant g$ , the defect is a function $h:=f-g$ .

We use equality cases to describe the set of $x\in X$ , such that $f(x)=g(x)$ . Denote by $X_h :=\{x\in X : h(x)=0\} \subseteq X$ the subset of equality cases.

We use E $_h$ to denote the equality verification of $f(x)=g(x)$ , that is, the decision problem

where $x\in X$ is an input. Since E $_h=\big \{x\in ^? X_h\}$ , this is a special case of the inclusion problem. We use V $_h$ to denote the verification of $h(x)=a$ , that is, the decision problem

where $a\in \mathbb R$ and $x\in X$ are the input. Clearly, V $_h$ is a more general problem than E $_h$ .

For a subset $Y\subseteq X$ , we use description for an equivalent condition for the inclusion problem $\big \{x\in ^? Y\big \}$ , where $x\in X$ . We use equality conditions for a description of E $_h$ . We say that equality cases of $f\geqslant g$ have a description in the polynomial hierarchy if E $_h \in \mathrm{ {PH}}$ . In other words, there is a CNF Boolean formula $\Phi (y_1,y_2,y_3,\ldots ,x)$ , such that

2.6 Complexity

We assume that the reader is familiar with basic notions and results in computational complexity and only recall a few definitions. We use standard complexity classes: $\mathrm{ {P}}$ , $\mathrm{ {FP}}$ , $\mathrm{ {NP}}$ , $\mathrm{ {coNP}}$ , $\mathrm{ {\#P}}$ , $\Sigma ^{\text {p}}_m$ and $\mathrm{ {PH}}$ . The notation $\{a =^? b\}$ is used to denote the decision problem whether $a=b$ . We use the oracle notation R $^{\textsf {S}}$ for two complexity classes R, S $\subseteq \mathrm{ {PH}}$ and the polynomial closure ${\langle }$ A ${\rangle }$ for a problem A $\in \mathrm{ {PSPACE}}$ . We will also use less common classes

$$ \begin{align*}\mathrm{{GapP}}:= \{f-g \mid f,g\in \mathrm{{\#P}}\} \quad \text{and} \quad {\mathrm{{C}}_=\mathrm{{P}}} :=\{f(x)=^?g(y) \mid f,g\in \mathrm{{\#P}}\}. \end{align*} $$

Note that $\mathrm{ {coNP}} \subseteq \mathrm{ {C}}_=\mathrm{ {P}}$ .

We also assume that the reader is familiar with standard decision and counting problems: 3SAT, #3SAT and PERMANENT. Denote by #LE the problem of computing the number $e(P)$ of linear extensions. For a counting function $f\in \mathrm{ {\#P}}$ , the coincidence problem is defined as:

Note the difference with the equality verification problem E $_{f-g}$ defined above. Clearly, we have both and . Note also that is both $\mathrm{ {C}}_=\mathrm{ {P}}$ -complete and $\mathrm{ {coNP}}$ -hard.

The distinction between binary and unary presentation will also be important. We refer to [Reference Garey and JohnsonGJ78] and [Reference Garey and JohnsonGJ79, Section 4.2] for the corresponding notions of $\mathrm{ {NP}}$ -completeness and strong $\mathrm{ {NP}}$ -completeness. Unless stated otherwise, we use the word “reduction” to mean “polynomial Turing reduction”. We refer to [Reference Arora, Barak and ComplexityAB09, Reference GoldreichGol08, Reference PapadimitriouPap94] for definitions and standard results in computational complexity.

3.1 Geometric inequalities

The history of equality conditions of geometric inequalities goes back to antiquity, see, for example, [Reference BlåsjöBlå05, Reference PorterPor33], when it was discovered that the isoperimetric inequality

(Isop) $$ \begin{align} \ell(X)^2 \, \ge \, 4 \pi a(X) \end{align} $$

is an equality if and only if X is a circle. Here, $\ell (X)$ is the perimeter and $a(X)$ is the area of a convex $X\subset \mathbb R^2$ . This classical result led to numerous extensions and generalizations, leading to the Alexandrov–Fenchel inequality (AF). We refer to [Reference Burago and ZalgallerBZ88, Reference SchneiderSchn14] for a review of the literature.

Below, we highlight only the most important developments to emphasize how the equality conditions become more involved as one moves in the direction of the AF inequality (see also Sections 10.4 and 10.5). The celebrated Brunn–Minkowski inequality states that for all convex $\mathrm {K}, \mathrm {L} \subset \mathbb R^d$ , we have:

(BM) $$ \begin{align} \mathrm{Vol}(\mathrm{K}+\mathrm{L})^{1/d} \, \ge \, \mathrm{Vol}(\mathrm{K})^{1/d} + \mathrm{Vol}(\mathrm{L})^{1/d}, \end{align} $$

see, for example [Reference GardnerGar02] for a detailed survey. This inequality “plays an important role in almost all branches of mathematics” [Reference BarvinokBar07]. Notably, both Brunn and Minkowski showed the equality in (BM) holds if and only if $\mathrm {K}$ is an expansion of $ \mathrm {L}$ .

For the mean width inequality

(MWI) $$ \begin{align} s(\mathrm{K})^2 \, \ge \, 6 \pi w(\mathrm{K}) \mathrm{Vol}(\mathrm{K}), \end{align} $$

for all convex $ \mathrm {K} \subset \mathbb R^3$ , Minkowski conjectured (1903) the equality cases are the cap bodies (balls with attached tangent cones). Here, $s(\mathrm {K})$ is the surface area and $w(\mathrm {K})$ is the mean width of $ \mathrm {K}$ . Minkowski’s conjecture that was proved by Bol (1943), see, for example [Reference Bonnesen and FenchelBF34, Reference Burago and ZalgallerBZ88].

The Minkowski’s quadratic inequality for three convex bodies $\mathrm {K},\mathrm {L},\mathrm {M} \subset \mathbb R^3$ , states:

(MQI) $$ \begin{align} \mathrm{V}(\mathrm{K},\mathrm{L},\mathrm{M})^2 \, \ge \, \mathrm{V}(\mathrm{K},\mathrm{K},\mathrm{M}) \cdot \mathrm{V}(\mathrm{L},\mathrm{L},\mathrm{M}). \end{align} $$

This is a special case of (AF) for $n=d=3$ . When $\mathrm {L}=\mathrm {B}_1$ is a unit ball and $\mathrm {K}=\mathrm {M}$ , this gives (MWI). Favard [Reference FavardFav33, p. 248] wrote that the equality conditions for (MQI) “parait difficile à énonce” (“seem difficult to state”). There are even interesting families of convex polytopes that give equality cases (see, e.g. [Reference Shenfeld and van HandelSvH23, Figure 2.1]).

Shenfeld and van Handel [Reference Shenfeld and van HandelSvH22] gave a complete characterization of the equality cases of (MQI) as triples of convex bodies that are similarly truncated in a certain formal sense. Notably, for the full-dimensional H-polytopes in $\mathbb R^3$ , each with at most n facets, the equality conditions amount to checking $O(n)$ linear relations for distances between facet inequalities. This can be easily done in polynomial time.

3.2 Alexandrov–Fenchel inequality

For the AF inequality (AF), the equality conditions have long been believed to be out of reach, as these would be generalized for (MWI) and (MQI). Alexandrov made a point of this in his original 1937 paper:

Half a century later, Burago and Zalgaller reviewed the literature and summarized:

Schneider made a case for perseverance:

The AF inequality has a number of proofs using ideas from convex geometry, analysis and algebraic geometry, going back to two proofs by Alexandrov (Fenchel’s full proof never appeared). We refer to [Reference Burago and ZalgallerBZ88, Reference SchneiderSchn14] for an overview of the older literature, especially [Reference SchneiderSchn14, p. 398] for historical remarks, and to [Reference Brändén and LeakeBL23, Reference Chan and PakCP22, Reference Cordero-Erausquin, Klartag, Merigot and SantambrogioCKMS19, Reference Kaveh and KhovanskiiKK12, Reference Shenfeld and van HandelSvH19, Reference WangWang18] for some notable recent proofs. All these proofs use a limit argument at the end, which can create new equality cases that do not hold for generic convex bodies. This partially explains the difficulty of the problem (cf. Section 10.2 and [Reference Shenfeld and van HandelSvH22, Remark 3.7]).

In [Reference AlexandrovAle37], Alexandrov gave a description of equality cases for combinatorially isomorphic polytopes. This is a large family of full-dimensional polytopes, for which every convex body is a limit. In particular, he showed that for the full-dimensional axis-parallel boxes $[\ell _1\times \ldots \times \ell _n]$ , the equality in (AF) is equivalent to $\mathrm {K}$ and $\mathrm {L}$ being homothetic (cf. Section 10.6).

In the pioneering work [Reference SchneiderSchn85], Schneider published a conjectural description of the equality cases, corrected later by Ewald [Reference EwaldEwa88], see also [Reference SchneiderSchn14]. After many developments, this conjecture was confirmed for all smooth (full-dimensional) convex bodies $\mathrm {Q}_i$ [Reference SchneiderSchn90a], and for all (not necessarily full-dimensional) convex bodies $\textrm {Q}_1=\ldots =\mathrm {Q}_{n-2}$ , by Shenfeld and van Handel [Reference Shenfeld and van HandelSvH23]. Closer to the subject of this paper, in a remarkable development, the authors gave a geometric description of the equality cases for all convex polytopes. They explain:

The geometric description of the equality cases in [Reference Shenfeld and van HandelSvH23] is indirect, technically difficult to prove and computationally hard in the degenerate cases.Footnote ³ While we will not quote the full statement (Theorem 2.13 in [Reference Shenfeld and van HandelSvH23]), let us mention the need to find witnesses polytopes $\mathrm {M}_i , \mathrm {N}_i \subset \mathbb R^n$ which must satisfy certain conditions (Definition 2.10, ibid.). The second of these conditions is an equality of certain mixed volumes (Equation (2.4), ibid.).

In [Reference Shenfeld and van HandelSvH23, Section 2.2.3], the authors write: “Condition (2.4) should be viewed merely as a normalization”.Footnote ⁴ From the computational complexity point of view, asking for the equality of mixed volumes (known to be hard to compute, see Section 3.8), lifts the problem outside of the polynomial hierarchy, to a hard coincidence problem (see Section 2.6). This coincidence problem eventually percolated into [Reference Ma and ShenfeldMS24], see (3.3) below, which, in turn, led directly to this work.

3.3 Stability

Bonnesen’s inequality is an extension of the isoperimetric inequality (Isop), which states that for every convex $X\subset \mathbb R^2$ , we have:

(Bon) $$ \begin{align} \ell(X)^2 - 4 \pi a(X) \, \geq \, 4 \pi (R-r)^2, \end{align} $$

where R is the smallest radius of the circumscribed circle, and r is the maximal radius of the inscribed circle.Footnote ⁵ Moreover, Bonnesen proved [Reference BonnesenBon29] that there is an annulus (thin shell) U between concentric circles of radii $R\ge r$ , such that $\partial X\subseteq U$ and (Bon) holds. Note that the optimal such annulus can be computed in polynomial time (see [Reference Agarwal, Aronov, Har-Peled and SharirAAHS99]).

Bonnesen’s inequality (Bon) was an inspiration for many Bonnesen type inequalities [Reference OssermanOss78, Reference OssermanOss79, Reference GroemerGro90] (see also discrete versions in Section 10.4, and applications in computational geometry in [Reference Kumar and SivakumarKS99]). There is now an extensive literature on stability inequalities in geometric and more general context (see, e.g. [Reference FigalliFig13, Reference GroemerGro93]).

There is an especially large literature on the stability of the Brunn–Minkowski inequality (BM). For major early advances by Diskant (1973), Groemer (1988) and others, see, e.g. [Reference GroemerGro93] and references therein. We refer to [Reference FigalliFig14] for an overview of more recent results, including [Reference Figalli, Maggi and PratelliFMP09, Reference Figalli, Maggi and PratelliFMP10] (see also [Reference Eldan and KlartagEK14] for the thin shell type bounds, and [Reference Figalli and JerisonFJ17] for the stability of (BM) for nonconvex sets).

For the Alexandrov–Fenchel inequality (AF), there are very few stability results, all for the full dimensional convex bodies with various regularity conditions, see e.g. [Reference Martinez-MaureMar17, Reference SchneiderSchn90b].

3.4 Linear extensions

Linear extensions play a central role in enumerative combinatorics and order theory. They appear in connection with saturated chains in distributive lattices, standard Young tableaux and P-partitions (see, e.g. [Reference StanleySta12]).

The world of inequalities for linear extensions has a number of remarkable results, some with highly nontrivial equality conditions. Notably, the Björner–Wachs inequality for $e(P)$ is an equality if and only if P is a forest [Reference Björner and WachsBW89, Theorem 6.3] (see also [Reference Chan, Pak and PanovaCPP23b]). On the other hand, the celebrated XYZ inequality established by Shepp in [Reference SheppShe82], see also [Reference Alon and SpencerAS16, Section 6.4], has no nontrivial equality cases [Reference FishburnFis84]. An especially interesting example is the Sidorenko inequality

(3.1) $$ \begin{align} e(P) \cdot e(P^\circ) \ge n! \end{align} $$

for posets $P, P^\circ $ on the same ground set with n elements, which have complementary comparability graphs [Reference SidorenkoSid91] (other proofs are given in [Reference Chan, Pak and PanovaCPP23b, Reference Gaetz and GaoGG22]). Sidorenko [Reference SidorenkoSid91] also proved that the series-parallel posets are the only equality cases. This solves the equality verification problem of (3.1), since the recognition problem of series-parallel posets is in $\mathrm{ {P}}$ (see [Reference Valdes, Tarjan and LawlerVTL82]).

It was noticed in [Reference Bollobás, Brightwell and SidorenkoBBS99], that the Sidorenko inequality follows from Mahler’s conjecture, which states that for every convex centrally symmetric body $\mathrm {K}\subset \mathbb R^n$ , we have:

(3.2) $$ \begin{align} \mathrm{Vol}(\mathrm{K}) \cdot \mathrm{Vol}(\mathrm{K}^{o}) \, \ge \, \frac{4^n}{n!}\,. \end{align} $$

To derive (3.1) from (3.2), take $\mathrm {K}$ to be the union of all axis reflections of the chain polytope $\mathcal S_P$ defined in (2.3). Mahler’s conjecture (3.2) is known for all axis symmetric convex bodies [Reference Saint-RaymondStR81], but remains open in full generality [Reference Artstein-Avidan, Sadovsky and SanyalAASS20], in part due to the many equality cases [Reference TaoTao08, Section 1.3].

3.5 Stanley inequality

Stanley’s inequality (Sta) is of independent interest in order theory, having inspired a large literature, especially in the last few years. The case $k=0$ is especially interesting. The unimodality, in this case, was independently conjectured by Kislitsyn [Reference KislitsynKis68] and Rivest, while the log-concavity was conjectured by Chung et al. [Reference Chung, Fishburn and GrahamCFG80], who established both conjectures for posets of width two. Stanley proved them in [Reference StanleySta81] in full generality.Footnote ⁶ The authors of [Reference Chung, Fishburn and GrahamCFG80] called Rivest’s conjecture “tantalizing” and Stanley’s proof “very ingenious”.

The Kahn–Saks inequality is a generalization of the $k=0$ case of (Sta), and is also proved from the AF inequality. This inequality was used to obtain the first positive result in the direction of the $\frac 13-\frac 23$ conjecture [Reference Kahn and SaksKS84]. For posets of width two, both the $k=0$ case of the Stanley inequality and the Kahn–Saks inequality have natural q-analogues [Reference Chan, Pak and PanovaCPP23a]. A generalization of Stanley’s inequality to marked posets was given in [Reference Liu, Mészáros and StLMS19].

For all $k\ge 0$ , the vanishing conditions $\{\textrm {N}_{\operatorname {\mathrm {\mathbf {z}}}\operatorname {\mathrm {\mathbf {c}}}}(P, x,a)=^?0\}$ are in $\mathrm{ {P}}$ . This was shown by Daykin and Daykin in [Reference Daykin and DaykinDD85, Theorem 8.2], via explicit necessary and sufficient conditions. Recently, this result was rediscovered in [Reference Chan, Pak and PanovaCPP23b, Theorem 1.11] and [Reference Ma and ShenfeldMS24, Theorem 5.3]. Similarly, the uniqueness conditions $\{\textrm {N}_{\operatorname {\mathrm {\mathbf {z}}}\operatorname {\mathrm {\mathbf {c}}}}(P, x,a)=^?1\}$ are in ${\mathrm{ {P}}}$ by the result of Panova et al. [Reference Chan, Pak and PanovaCPP23b, Theorem 7.5], where we gave explicit necessary and sufficient conditions. Both the vanishing and the uniqueness conditions give examples of equality cases of the Stanley inequality, which remained a “major challenge” in full generality [Reference Chan, Pak and PanovaCPP23b, Section 9.10].

As we mentioned in the Introduction, Shenfeld and van Handel resolved the $k=0$ case of Stanley equality conditions by giving explicit necessary and sufficient conditions, which can be verified in polynomial time (see [Reference Shenfeld and van HandelSvH23]). Similar explicit necessary and sufficient conditions for the Kahn–Saks inequality were conjectured in [Reference Chan, Pak and PanovaCPP23a, Conjecture 8.7] and proved for posets of width two. Building on the technology in [Reference Shenfeld and van HandelSvH23], van Handel et al. gave the proof of this conjecture in [Reference van Handel, Yan and ZengvHYZ23].

In [Reference Chan and PakCP24a], we gave a new proof of the $k=0$ case of (Sta), using a combinatorial atlas technology. This is an inductive self-contained linear algebraic approach (see [Reference Chan and PakCP22] for the introduction). We also gave a new proof of the Shenfeld and van Handel equality conditions and generalized both results to weighted linear extensions (see Sections 1.16–18 in [Reference Chan and PakCP24a]).

In an important development, Ma and Shenfeld [Reference Ma and ShenfeldMS24] advanced the technology of [Reference Shenfeld and van HandelSvH23] to give a clean, albeit ineffective combinatorial description of the equality cases in full generality. In particular, they showed that (AF) is an equality if and only if

(3.3) $$ \begin{align} \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a-1)\, = \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a)\, = \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a+1). \end{align} $$

They proceeded to give explicit necessary and sufficient conditions for these equalities in some cases (see Section 10.11). About the remaining cases that they called critical, see Section 9.2, they write: “It is an interesting problem to find $[$ an explicit description $]$ for critical posets” [Reference Ma and ShenfeldMS24, Remark 1.6]. Our Theorem 1.3 implies that such a description is unlikely, as it would imply a disproof of a major conjecture in computational complexity (see also Section 10.12).

3.6 Complexity aspects

There are two standard presentations of polytopes: H-polytopes described by the inequalities and V-polytopes described by the vertices. These two presentation types have very different natures in higher dimensions (see, e.g. [Reference Dyer, Gritzmann and HufnagelDGH98]). We refer to [Reference Gritzmann and KleeGK94, Reference Gritzmann and KleeGK97] for an overview of standard complexity problems in geometry and to [Reference SchrijverSchr86, Section 19], [Reference SchrijverSchr03, Section 5.16] for the background on totally unimodular matrices and $\textrm {TU}$ -polytopes. Note also that testing whether matrix A is totally unimodular can be done in polynomial time (see [Reference SeymourSey80]).

When the dimension n is bounded, H-polytopes and V-polytopes have the same complexity, so the volume and the mixed volumes are in $\mathrm{ {FP}}$ . Thus, the dimension n is unbounded throughout the paper. The volume of TU-polytopes is $\mathrm{ {\#P}}$ -hard via reduction to KNAPSACK [Reference Dyer and FriezeDF88]. Note that for rational H-polytopes in $\mathbb R^n$ , the volume denominators can be doubly exponential [Reference LawrenceLaw91], thus not in $\mathrm{ {PSPACE}}$ . This is why we constrain ourselves to TU-polytopes which is a subclass of H-polytopes that includes all order polytopes (see Section 5.1).

The mixed volume $\mathrm {V}(\mathrm {Q}_1,\ldots ,\mathrm {Q}_n)$ coincides with the permanent when all $\mathrm {Q}_i$ are axis parallel boxes (see [Reference van LintvL82] and Section 10.6). Thus, computing the mixed volume is $\mathrm{ {\#P}}$ -hard even for the boxes (see [Reference Dyer, Gritzmann and HufnagelDGH98]). For rational H-polytopes, the vanishing problem $\{\mathrm {V}(\cdot )=^?0\}$ can be described combinatorially and is thus in $\mathrm{ {NP}}$ (see [Reference Dyer, Gritzmann and HufnagelDGH98, Reference EsterovEst10]). It is equivalent to computing the rank of intersection of two geometric matroids (with a given realization), which is in $\mathrm{ {P}}$ (see [Reference SchrijverSchr03, Section 41]). For TU-polytopes in $\mathbb R^n$ , the uniqueness problem $\big \{\mathrm {V}(\cdot )=^?\frac {1}{n!}\big \}$ is in $\mathrm{ {NP}}$ by a result in [Reference Esterov and GusevEG15].

The problem #LE is proved $\mathrm{ {\#P}}$ -complete by Brightwell and Winkler [Reference Brightwell and WinklerBW91, Theorem 1], and this holds even for posets of height two [Reference Dittmer and PakDP20]. Linial noticed [Reference LinialLin86], that this result and (2.4) together imply that the volume of H-polytopes is $\mathrm{ {\#P}}$ -hard even when the input is in unary. Linial also observed that the number of vertices of order polytopes is $\mathrm{ {\#P}}$ -complete (ibid.).

Now, fix $k\ge 0$ , $x\in X$ and $\operatorname {\mathrm {\mathbf {z}}}\in X^k$ . Clearly, we have:

$$ \begin{align*} e(P) \ = \ \sum_{a\in [n]} \, \sum_{\operatorname{\mathrm{\mathbf{c}}} \in [n]^k} \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a), \end{align*} $$

where the summation has size $O(n^{k+1})$ . Thus, computing $\textrm {N}_{\operatorname {\mathrm {\mathbf {z}}}\operatorname {\mathrm {\mathbf {c}}}}(P, x,a)$ is also $\mathrm{ {\#P}}$ -complete.

Finally, it was proved in [Reference Chan and PakCP23a] that , and are not in $\mathrm{ {PH}}$ , unless $\mathrm{ {PH}}$ collapses to a finite level. The proof idea of Theorem 1.3 is inspired by these results.

3.7 Combinatorial interpretations

Finding a combinatorial interpretation is a standard problem throughout combinatorics whenever a positivity phenomenon or an inequality emerges. Having a combinatorial interpretation allows one to deeper understand the underlying structures, give asymptotic and numerical estimates, as well as analyze certain algorithms. We refer to [Reference HuhHuh18, Reference StanleySta89, Reference StanleySta00] for an overview of inequalities in algebraic combinatorics and matroid theory, and to [Reference PakPak22] for a recent survey from the complexity point of view.

Recall that $\mathrm{ {GapP}}:=\mathrm{ {\#P}}-\mathrm{ {\#P}}$ is the class of difference of two $\mathrm{ {\#P}}$ functions, and let $\mathrm{ {GapP}}_{\ge 0}$ be a subclass of $\mathrm{ {GapP}}$ of nonnegative functions. Thus, for every inequality $f\geqslant g$ of counting functions $f,g\in \mathrm{ {\#P}}$ , we have $(f-g) \in \mathrm{ {GapP}}_{\ge 0}$ . It was shown in [Reference Ikenmeyer and PakIP22, Proposition 2.3.1] that $\mathrm{ {GapP}}_{\ge 0}\ne \mathrm{ {\#P}}$ , unless $\mathrm{ {PH}} = \Sigma ^{\text {p}}_2$ . The key example is

$$ \begin{align*}\big(\text{\#3SAT}(F) - \text{\#3SAT}(F')\big)^2, \end{align*} $$

(see also the first function in (⊛)). The other two functions in (⊛) were given in [Reference Chan and PakCP23a]. A natural $\mathrm{ {GapP}}_{\ge 0}$ problem of computing $S_n$ character squared: $[\chi ^\lambda (\mu )]^2$ , was proved not in $ \mathrm{ {\#P}}$ (in unary), under the same assumptions [Reference Ikenmeyer, Pak and PanovaIPP22].

The idea that some natural combinatorial inequalities can have no combinatorial interpretations appeared in [Reference PakPak19]. A number of interesting examples were given in [Reference Ikenmeyer and PakIP22, Section 7], including the Cauchy, Minkowski, Hadamard, Karamata and Ahlswede–Daykin inequalities, all proved not in $\mathrm{ {\#P}}$ under varying complexity assumptions.

Closer to the subject of this paper, Ikenmeyer and the second author showed that the AF defect $\delta (\cdot )$ is not in $\mathrm{ {\#P}}$ (unless $\mathrm{ {PH}}=\Sigma ^{\text {p}}_2$ ), even for axis parallel rectangles in $\mathbb R^2$ whose edge lengths are given by #3SAT formulas [Reference Ikenmeyer and PakIP22, Theorem 7.1.5]. This is a nonstandard model of computation. One can think of our Main Theorem 1.1 as a tradeoff: in exchange for needing a higher dimension, we now have unary input and the standard model of computation.

3.8 Complexity assumptions

The results in the paper use different complexity assumptions, and navigating between them can be confusing. Here is a short list of standard implications:

$$ \begin{align*}\mathrm{{PH}} \ne \Sigma^{\text{p}}_m \ \ \text{for all} \ \ m\ge 2 \quad \Longrightarrow \quad \mathrm{{PH}} \ne \Sigma^{\text{p}}_2 \quad \Longrightarrow \quad \mathrm{{PH}} \ne \mathrm{{NP}} \quad \Longrightarrow \quad \mathrm{{P}} \ne \mathrm{{NP}}. \end{align*} $$

In other words, the assumption in Theorems 1.1 and 1.3 is the strongest, while $\mathrm{ {P}}\ne \mathrm{ {NP}}$ is the weakest. Proving either of these would be a major breakthrough in theoretical computer science. Disproving either of these would bring revolutionary changes to the way the computational complexity understands the nature of computation. We refer to [Reference AaronsonAar16, Reference WigdersonWig19] for an extensive discussion, philosophy and implications in mathematics and beyond.

4.1 Around Stanley equality

First, we show that Theorem 1.1 follows from Theorem 1.3. Recall the notation from the Introduction. Let $P=(X,\prec )$ be a poset on $|X|=n$ elements. As before, let $x\in X$ , $a\in [n]$ , $\operatorname {\mathrm {\mathbf {z}}} \in X^k$ and $\operatorname {\mathrm {\mathbf {c}}} \in [n]^k$ . Recall also

where $\textrm {N}_{\operatorname {\mathrm {\mathbf {z}}} \operatorname {\mathrm {\mathbf {c}}}}(P,x,a)$ are defined in Section 1.4.

The proof of the proposition given in Section 5 is very close to Stanley’s original proof of the inequality (Sta). The key difference is the observation that slices of order polytopes are TU-polytopes. Next, we need a simple technical result.

Proof. Let $P=(X,\prec )$ be a poset on n elements, and let $\operatorname {\mathrm {\mathbf {z}}}\in X^k$ , $\operatorname {\mathrm {\mathbf {c}}}\in [n]^k$ , $x\in X$ , $a\in [n]$ be as in Section 1.4. Denote by $P':=P+A_{k-\ell }$ a poset obtained by adding $(k-\ell )$ independent elements $z^{\prime }_1,\ldots ,z_{k-\ell }^{\prime }$ . Let $c_i^{\prime }:=n+i$ for all $1\le i\le k-\ell $ . For $\operatorname {\mathrm {\mathbf {z}}}':=\big (z_1,\ldots ,z_\ell ,z_1^{\prime },\ldots ,z_{k-\ell }^{\prime }\big )$ and $\operatorname {\mathrm {\mathbf {c}}}':=\big (c_1,\ldots ,c_\ell ,c_1^{\prime },\ldots ,c_{k-\ell }^{\prime }\big )$ , we have:

$$ \begin{align*}\textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}'\operatorname{\mathrm{\mathbf{c}}}'}(P', x,a) \, = \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a). \end{align*} $$

Varying a, we conclude that EqualityStanley $_k$ is equivalent to EqualityStanley $_\ell $ in this special case. This gives the desired reduction.

Next, we simplify the Stanley equality problem to the following flatness problem:

The idea is to ask whether a is in the flat part of the distribution of $f(x)$ (cf. Figure 15.1 in [Reference Shenfeld and van HandelSvH23]).

We prove Lemma 4.3 in Section 6.

4.2 Relative numbers of linear extensions

Let $P=(X,\prec )$ be a poset on $|X|=n$ elements, and let $\min (P)\subseteq X$ be the set of minimal elements of P. For every $x \in \min (P)$ , define the relative number of linear extensions:

(4.1) $$ \begin{align} \rho(P,x) \, := \, \frac{e(P)}{e(P-x)}. \end{align} $$

In other words, $\rho (P,x) = \mathbb P[f(x)=1]^{-1}$ , where $f\in \mathcal {E}(P)$ is a uniform random linear extension of P. Denote by #RLE the problem of computing $\rho (P,x)$ .

Proof. By definition, reduces to . In the opposite direction, let $P=(X,\prec )$ be a poset on $|X|=n$ elements. Fix a linear extension $g\in \mathcal {E}(P)$ , and let $x_i:=g^{-1}(i)$ , $1\le i \le n$ . Denote by $P_i$ a subposet of P restricted to $x_i,\ldots ,x_n$ , and observe that $x_i \in \min (P_i)$ . We have:

$$ \begin{align*}e(P) \ = \ \frac{e(P_1)}{e(P_2)} \cdot \frac{e(P_2)}{e(P_3)} \, \cdots \ = \ \rho(P_1,x_1) \cdot \rho(P_2,x_2) \, \cdots \,, \end{align*} $$

which gives the desired reduction from to .

We relate RLE to flatness equality through the following series of reductions. Consider the following coincidence problem:

where $P=(X,\prec )$ , $Q=(Y,\prec ')$ are posets and $x\in \min (P)$ , $y \in \min (Q)$ .

Next, consider the following decision problem:

where $P_1,P_2,P_3,P_4$ are finite posets and $x_i\in \min (P_i)$ for all $1\le i \le 4$ .

4.3 Verification lemma

Let $P=(X,\prec )$ be a poset on $|X|=n$ elements, and let $x\in \min (P)$ . Consider

(4.2)

where $ A,B $ are coprime integers with $1\le B \le A\leq n!.$ We need the following:

Note that the opposite direction “ $\supseteq $ ” is also true and easy to prove. Indeed, suppose you have an oracle VerRLE. Guess the values $a_i:=\rho (P_i,x_i)\in \mathbb Q$ , verify that they are correct and check that $a_1\cdot a_2 = a_3\cdot a_4.$ This gives QuadRLE. We will only need the direction in the lemma, which is highly nontrivial.

4.4 Putting everything together

We can now obtain all the results stated in the Introduction, except for Theorem 1.4, which uses different tools and is postponed until Section 9.

Proof of Theorem 1.3.

Recall that #LE is $\mathrm{ {\#P}}$ -complete [Reference Brightwell and WinklerBW91] (see also Section 3.6). By Lemma 4.4, we conclude that #RLE is $\mathrm{ {\#P}}$ -hard. We then have:

(4.3)

where the first inclusion is Toda’s theorem [Reference TodaToda91], the second inclusion is because #RLE is $\mathrm{ {\#P}}$ -hard and the third inclusion is because one can simulate #RLE by first guessing and then verifying the answer.

Fix $k\ge 2$ . Combining Lemmas 4.2, 4.3, 4.5 and 4.6, we conclude that QuadRLE reduces to EqualityStanley $_k$ . We have:

(4.4)

where the first inclusion is the Verification Lemma 4.7. Now, suppose . Then EqualityStanley $_k \in \Sigma ^{\text {p}}_m$ for some m. Combining (4.3) and (4.4), this implies:

(4.5)

as desired.

As a byproduct of the proof, we get the same conclusion for the intermediate problems. This result is potentially of independent interest (cf. [Reference Chan and PakCP23a]).

Proof of Corollary 1.2.

By the “Bonnesen type” assumption, we have

Since computing $\xi $ is in $\mathrm{ {FP}}$ , we have EqualityAF $\in \mathrm{ {P}}$ . Then (4.5) for $k=2$ , and Proposition 4.1 gives:

(4.6)

as desired.

Proof of Corollary 1.5.

Suppose $\phi _k \in \mathrm{ {\#P}}$ . By definition, we have:

$$ \begin{align*} \big\{ \Phi_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P, x,a) \, \ne^? \, 0\big\} \in \mathrm{{NP}}. \end{align*} $$

In other words, we have EqualityStanley $_k \in \mathrm{ {coNP}}$ . Then (4.5) gives:

as desired.

5.1 Slices of order polytopes

Let $P=(X,\prec )$ be a poset on $|X|=n$ elements. Recall the construction of order polytopes $\mathcal O_P\subseteq [0,1]^n$ given in (2.2). Fix $z_1\prec \ldots \prec z_k$ and $1 \le c_1 < \ldots < c_k\le n$ . Denote $Z:=\{z_1,\ldots ,z_k\}$ , and let $Y:=X\smallsetminus Z$ . For all $0\le i \le k$ , consider the following slices of the order polytopes:

$$ \begin{align*}\mathrm{S}_i \, := \, \mathcal O_P \cap \{\alpha_x=0 \, : \, x \preccurlyeq z_i, x \in X \} \cap \{\alpha_x=1 \, : \, x \succcurlyeq z_{i+1}, x \in X \}. \end{align*} $$

Here, the conditions $x \preccurlyeq z_i$ and $x \succcurlyeq z_{i+1}$ are vacuous when $i=0$ and $i=k$ , respectively. Note that $\dim \mathrm {S}_i \le n-k$ for all $0\le i \le k$ , since $\alpha _x$ is a constant on $\mathrm {S}_i$ for all $x\in Z$ .Footnote ⁷ The same argument implies that these slices are themselves order polytopes of subposets of P, a fact we do not need. Instead, we need the following simple result:

5.2 Proof of Proposition 4.1

Denote by $\mathcal {E}_{\operatorname {\mathrm {\mathbf {z}}} \operatorname {\mathrm {\mathbf {c}}}}(P)$ the set of all linear extensions $f\in \mathcal {E}(P)$ , such that $f(z_i)=c_i$ for all i, and let $\textrm {N}_{\operatorname {\mathrm {\mathbf {z}}}\operatorname {\mathrm {\mathbf {c}}}}(P):=|\mathcal {E}_{\operatorname {\mathrm {\mathbf {z}}} \operatorname {\mathrm {\mathbf {c}}}}(P)|$ .

Let $\mathrm {S}_0,\ldots ,\mathrm {S}_k\subset \mathbb R^n$ be the slices defined above, and note that the dimension $\dim {\langle }\mathrm {S}_0,\ldots ,\mathrm {S}_k{\rangle } $ of the subspace spanned by vectors in $\mathrm {S}_0,\ldots , \mathrm {S}_k$ is equal to $n-k$ . Stanley’s original proof of (Sta) is based on the following key observation:

Lemma 5.2 [Reference StanleySta81, Theorem 3.2].

Let $z_1\prec \ldots \prec z_k$ and $1 \le c_1 < \ldots < c_k\le n$ . We have:

(5.1) $$ \begin{align} \mathrm{V}\big(\underbrace{\mathrm{S}_0 , \ldots , \mathrm{S}_0}_{c_1-1 \text{ times}}, \underbrace{\mathrm{S}_1 , \ldots , \mathrm{S}_1}_{c_2-c_1-1\text{ times}}, \ldots , \underbrace{\mathrm{S}_k , \ldots , \mathrm{S}_k}_{n-c_k \text{ times}}\big) \ = \ \frac{1}{(n-k)!} \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P). \end{align} $$

Now let $z_i \gets x$ and $c_i\gets a$ for some i, such that $1 \le c_1 < \ldots < c_k\le n$ . By Lemma 5.2, the AF inequality (AF) becomes (Sta). By Lemma 5.1, slices $\mathrm {S}_i\subset \mathbb R^n$ are TU-polytopes defined by $O(n^2)$ inequalities. This gives the desired reduction.

6.1 Ma–Shenfeld poset notation

Recall the following terminology from [Reference Ma and ShenfeldMS24]. Let $s\in \{-1,0,1\}$ . For any $f \in \mathcal {E}_{\operatorname {\mathrm {\mathbf {z}}}\operatorname {\mathrm {\mathbf {c}}}}(P,x,a+s)$ , the companions in $ f$ are the elements in

$$\begin{align*}\textrm{Com} (f) \, := \, \big\{f^{-1}(a-1), f^{-1}(a), f^{-1}(a+1) \big\} - x. \end{align*}$$

Note that $| \textrm {Com} (f)| =2$ for all s as above. Let the lower companion $\textrm {lc} (f)\in \textrm {Com} (f)$ be the companion with the smaller of the two values in f. Similarly, let the upper companion $\textrm {uc} (f)\in \textrm {Com} (f)$ be the companion with the larger of the two values in f. Denote by $\mathcal C(x)\subset X$ the set of elements $y\in X$ comparable to x, that is, $\mathcal C(x) := \{y \in X \, : \, x \prec y \ \text {or} \ x \succ y\}$ .

6.2 Proof of Lemma 4.3

Let $P=(X,\prec )$ , and let x, a, $\operatorname {\mathrm {\mathbf {z}}}=(z_1,\ldots ,z_k)$ and $\operatorname {\mathrm {\mathbf {c}}}=(c_1,\ldots ,c_k)$ be an instance of FlatLE $_k$ as in Lemma 4.3. To prove the reduction in the lemma, we construct a poset $Q=(Y,\prec )$ for which P is a subposet, and x, b, $\operatorname {\mathrm {\mathbf {y}}}$ and $\operatorname {\mathrm {\mathbf {x}}}$ , which give the desired instance EqualityStanley $_{k+2}$ .

Without loss of generality, we can assume that $\min (P)=\{z_0\}$ and $\max (P)=\{z_{k+1}\}$ . In other words, assume that there are elements $z_0,z_{k+1} \in X$ , such that $z_0 \preccurlyeq y \preccurlyeq z_{k+1}$ for all $y \in X$ .

Let $\textrm {M}_1,\textrm {M}_2, \textrm {M}_3$ be given by

$$ \begin{align*} \textrm{M}_1 \ &:= \ \big|\big\{ f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a) \, : \, f^{-1}(a+1) \succ x \big\} \big|, \\ \textrm{M}_2 \ &:= \ \big|\big\{ f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \, : \, f^{-1}(a) \prec x \big\} \big|,\\ \textrm{M}_3 \ &:= \ \big|\big\{ f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a) \, : \, f^{-1}(a+1) \| x \big\} \big| \\ &\ \,{=} \ \big|\big\{ f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \, : \, f^{-1}(a) \| x \big\} \big|. \end{align*} $$

Note that the two sets in the definition of $\textrm {M}_3$ are in bijection with each other via the map that swaps $f(a)$ with $f(a+1)$ . It then follows from here that

$$\begin{align*}\textrm{N}_{\operatorname{\mathrm{\mathbf{z}}} \operatorname{\mathrm{\mathbf{c}}}}(P,x,a) \, = \, \textrm{M}_1 + \textrm{M}_3 \qquad \text{and} \qquad \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}} \operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \, = \, \textrm{M}_2 + \textrm{M}_3. \end{align*}$$

This implies that

(6.1) $$ \begin{align} \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}} \operatorname{\mathrm{\mathbf{c}}}}(P,x,a) \, = \, \textrm{N}_{\operatorname{\mathrm{\mathbf{z}}} \operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \qquad \Longleftrightarrow \qquad \textrm{M}_1 \ = \ \textrm{M}_2. \end{align} $$

Now, let $Q=(Y,\prec )$ be the poset $P+C_3$ , that is, $Y:=X \cup \{u,v,w\}$ and with the additional relation $u \prec v \prec w$ and $\{u,v,w\}$ is incomparable to all elements in X. Let $\ell :=\max \{i : c_i< a\}$ be the maximal index, such that the corresponding element in $\operatorname {\mathrm {\mathbf {z}}}$ is less than a. Let $b := a+2$ , and let

$$ \begin{align*} &\operatorname{\mathrm{\mathbf{y}}} \ := \ (z_1,\ldots, z_{\ell}, u, w, z_{\ell+1}, \ldots, z_{k}) \in Y^{k+2}, \\ & \operatorname{\mathrm{\mathbf{b}}} \ := \ (c_1,\ldots, c_\ell, a, a+4, c_{\ell+1}+3, \ldots, c_{k}+3) \in \mathbb N^{k+2}. \end{align*} $$

In the notation above, for $s\in \{-1,0,1\}$ and $f \in \mathcal {E}_{\operatorname {\mathrm {\mathbf {y}}}\operatorname {\mathrm {\mathbf {b}}}}(Q,x,b+s)$ , the companions in $ f$ are the elements in

$$\begin{align*}\textrm{Com} (f) \, := \, \big\{f^{-1}(b-1), f^{-1}(b), f^{-1}(b+1) \big\} - x. \end{align*}$$

LetFootnote ⁸

$$ \begin{align*} \mathcal F(b,\operatorname{\mathrm{\text{com}}},\operatorname{\mathrm{\text{inc}}}) \ &:= \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \, : \, \textrm{lc} (f)\in \mathcal C(x), \textrm{uc} (f)\not\in \mathcal C(x) \big\},\\ \mathcal F(b,\operatorname{\mathrm{\text{inc}}},\operatorname{\mathrm{\text{com}}}) \ &:= \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \, : \, \textrm{lc} (f)\notin \mathcal C(x), \textrm{uc} (f)\in \mathcal C(x) \big\},\\ \mathcal F(b,\operatorname{\mathrm{\text{com}}},\operatorname{\mathrm{\text{com}}}) \ &:= \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \, : \, \textrm{lc} (f)\in \mathcal C(x), \textrm{uc} (f)\in \mathcal C(x) \big\},\\ \mathcal F(b,\operatorname{\mathrm{\text{inc}}},\operatorname{\mathrm{\text{inc}}}) \ &:= \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \, : \, \textrm{lc} (f)\not\in \mathcal C(x), \textrm{uc} (f)\not\in \mathcal C(x) \big\}, \end{align*} $$

and we write $\operatorname {\mathrm {\mathrm {F}}}(b,\cdot , \cdot ) := |\mathcal F(b,\cdot ,\cdot )|$ . Note that by construction, it follows that for all $f \in \mathcal F(b,\cdot ,\cdot )$ , we have

$$\begin{align*}b-2 \ = \ f(u) \ < \ f(v) \ < \ f(w) \ = \ b+2, \end{align*}$$

so $f(v) \in \{b-1,b,b+1\}$ , and thus v will always be a companion in f. Sets $\mathcal F(b+1,\ast ,\ast )$ and $\mathcal F(b-1,\ast ,\ast )$ are defined analogously.

Claim 6.1. We have:

$$ \begin{align*} & \operatorname{{\mathrm{F}}}(b,{{\mathrm{com}}},{{{\rm inc}}}) \ = \ \mathrm{M}_2\,, \qquad && {\mathrm{{F}}}(b,{\mathrm{{inc}}},{\mathrm{{com}}}) \ = \ \mathrm{M}_1,\\ & {\mathrm{{F}}}(b,{\mathrm{{com}}},{\mathrm{{com}}}) \ = \ 0\,, \qquad && {\mathrm{{F}}}(b,{\mathrm{{inc}}},{\mathrm{{inc}}}) \ = \ 2\mathrm{M}_3,\\ & {\mathrm{{F}}}(b+1,{\mathrm{{com}}},{\mathrm{{inc}}}) \ = \ \mathrm{M}_2\,, \qquad && {\mathrm{{F}}}(b+1,{\mathrm{{inc}}},{\mathrm{{com}}}) \ = \ \mathrm{M}_2,\\ & {\mathrm{{F}}}(b+1,{\mathrm{{com}}},{\mathrm{{com}}}) \ = \ 0\,, \qquad && {\mathrm{{F}}}(b+1,{\mathrm{{inc}}},{\mathrm{{inc}}}) \ = \ 2\mathrm{M}_3,\\ & {\mathrm{{F}}}(b-1,{\mathrm{{com}}},{\mathrm{{inc}}}) \ = \ \mathrm{M}_1\,, \qquad && {\mathrm{{F}}}(b-1,{\mathrm{{inc}}},{\mathrm{{com}}}) \ = \ \mathrm{M}_1,\\ & {\mathrm{{F}}}(b-1,{\mathrm{{com}}},{\mathrm{{com}}}) \ = \ 0\,, \qquad && {\mathrm{{F}}}(b-1,{\mathrm{{inc}}},{\mathrm{{inc}}}) \ = \ 2\mathrm{M}_3. \end{align*} $$

Proof. We only compute the values $\operatorname {\mathrm {\mathrm {F}}}(b,\ast ,\ast )$ , as proof of the other cases is analogous. Denote by $\mathcal {E}_{\operatorname {\mathrm {\mathbf {z}}} \operatorname {\mathrm {\mathbf {c}}}}(P)$ the set of all linear extensions $f\in \mathcal {E}(P)$ , such that $f(z_i)=c_i$ for all i.

Let $\psi :\mathcal {E}_{\operatorname {\mathrm {\mathbf {y}}} \operatorname {\mathrm {\mathbf {b}}}}(Q) \to \mathcal {E}_{\operatorname {\mathrm {\mathbf {z}}} \operatorname {\mathrm {\mathbf {c}}}}(P)$ be the map given by $\psi (f) = g$ , where

$$ \begin{align*} g(s) \ := \ \begin{cases} f(s) & \text{ if } \ \ f(s)<f(u),\\ f(s)-1 & \text{ if } \ \ f(u) < f(s) < f(v),\\ f(s)-2 & \text{ if } \ \ f(v) < f(s) < f(w),\\ f(s)-3 & \text{ if } \ \ f(s)>f(w) \end{cases} \end{align*} $$

for all $s\in X$ . It follows from the definition of $\textrm {lc} (f)$ and $\textrm {uc} (f)$ that

$$\begin{align*}\mathcal F(b,\operatorname{\mathrm{\text{com}}},\operatorname{\mathrm{\text{inc}}}) \ = \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \, : \, f^{-1}(b-1) \prec x, f^{-1}(b+1)=v \big\}. \end{align*}$$

It then follows that $\varphi $ restricted to $\mathcal F(b,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}})$ is a bijection onto

$$\begin{align*}\big\{g \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \, : \, g^{-1}(a+1) \prec x \big\}, \end{align*}$$

which gives us $\operatorname {\mathrm {\mathrm {F}}}(b,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}})=\textrm {M}_2$ . Similar arguments give $\operatorname {\mathrm {\mathrm {F}}}(b,\operatorname {\mathrm {\text {inc}}},\operatorname {\mathrm {\text {com}}})=\textrm {M}_1$ . Note that $\operatorname {\mathrm {\mathrm {F}}}(b,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {com}}})=0$ , because v is always a companion in f but $v \| x$ by definition. Note also that

$$ \begin{align*} \begin{aligned} \mathcal F(b,\operatorname{\mathrm{\text{inc}}},\operatorname{\mathrm{\text{inc}}}) \ & = \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \ : \ f^{-1}(b-1) \| x, f^{-1}(b+1)=v \big\} \\ & \hskip1.cm \cup \ \big\{f \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \ : \ f^{-1}(b+1) \| x, f^{-1}(b-1)=v \big\}. \end{aligned} \end{align*} $$

It then follows that $\psi $ restricted to $\mathcal F(b,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}})$ is a bijection onto

$$\begin{align*}\big\{g \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a+1) \, : \, g^{-1}(a) \| x \big\} \ \cup \ \big\{g \in \mathcal{E}_{\operatorname{\mathrm{\mathbf{z}}}\operatorname{\mathrm{\mathbf{c}}}}(P,x,a) \, : \, g^{-1}(a+1) \| x \big\}, \end{align*}$$

which gives $\operatorname {\mathrm {\mathrm {F}}}(b,\operatorname {\mathrm {\text {inc}}},\operatorname {\mathrm {\text {inc}}})=2\textrm {M}_3$ . This finishes proof of the claim.

By the claim, we have:

$$ \begin{align*} \begin{aligned} \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}} \operatorname{\mathrm{\mathbf{b}}}}(Q,x,b) \ &= \ \operatorname{\mathrm{\mathrm{F}}}(b,\operatorname{\mathrm{\text{com}}},\operatorname{\mathrm{\text{inc}}}) \ + \ \operatorname{\mathrm{\mathrm{F}}}(b,\operatorname{\mathrm{\text{inc}}},\operatorname{\mathrm{\text{com}}}) \ + \ \operatorname{\mathrm{\mathrm{F}}}(b,\operatorname{\mathrm{\text{com}}},\operatorname{\mathrm{\text{com}}}) \ + \ \operatorname{\mathrm{\mathrm{F}}}(b,\operatorname{\mathrm{\text{inc}}},\operatorname{\mathrm{\text{inc}}})\\ \ &= \ \textrm{M}_2 \ + \ \textrm{M}_1 \ + \ 2\textrm{M}_3. \end{aligned} \end{align*} $$

Similarly, we have:

$$ \begin{align*} \begin{aligned} \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}} \operatorname{\mathrm{\mathbf{b}}}}(Q,x,b+1) \ &= \ 2 \textrm{M}_2 \ + \ 2\textrm{M}_3, \\ \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b-1) \ &= \ 2 \textrm{M}_1 \ + \ 2\textrm{M}_3. \end{aligned} \end{align*} $$

We conclude:

$$ \begin{align*} & \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b)^2 - \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}} \operatorname{\mathrm{\mathbf{b}}}}(Q,x,b+1) \cdot \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b-1) \\ & \hskip1.cm = (\textrm{M}_1+\textrm{M}_2+2\textrm{M}_3)^2 - 4(\textrm{M}_1+\textrm{M}_3)(\textrm{M}_2+\textrm{M}_3) \, = \, (\textrm{M}_1-\textrm{M}_2)^2. \end{align*} $$

This implies that

(6.2) $$ \begin{align} \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b)^2 \, = \, \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b+1) \cdot \textrm{N}_{\operatorname{\mathrm{\mathbf{y}}}\operatorname{\mathrm{\mathbf{b}}}}(Q,x,b-1) \quad \Longleftrightarrow \quad \textrm{M}_1=\textrm{M}_2. \end{align} $$

Lemma 4.3 now follows by combining (6.1) and (6.2).

7.1 One poset from two

The following results give a quantitative versionFootnote ⁹ of Lemma 4.5.

Proof. Let $P^\ast =(X,\prec ^\ast )$ be the dual poset of P. Define $R=(Z,\prec ^\circ )$ to be a poset on

$$\begin{align*}Z \ := \ (X-x) \cup (Y-y) \cup \{w,z\},\end{align*}$$

where $w,z$ are two new elements. Let the partial order $\prec ^\circ $ coincide with $\prec ^\ast $ on $(X-x)$ , and with $\prec '$ on $(Y-y)$ , with additional relations

(7.3) $$ \begin{align} &\!\! p \prec^\circ z \prec^\circ q, \quad \text{ for all } \ \ p \in X-x, \ q \in Y-y,\qquad \end{align} $$

(7.4) $$ \begin{align} & p \prec^\circ w \ \ \text{ if and only if } \ \ x \prec p , \quad \text{ for all } \ \ p \in X-x, \end{align} $$

(7.5) $$ \begin{align} & w \prec^\circ q \ \ \text{ if and only if } \ \ y \prec' q, \quad \text{ for all } \ \ q \in Y-y. \end{align} $$

That is, we are taking the series sum $(P^*-x) \oplus \{z\} \oplus (Q-y)$ , then adding an element w to emulate x in P for $f(w) <f(z)$ , where $f \in \mathcal {E}(R)$ , and emulate y in Q when $f(w)> f(z)$ . It then follows from a direct calculation that

$$\begin{align*}\textrm{N}(R,z,n+1) \, = \, e(P) \cdot e(Q-y). \end{align*}$$

Indeed, by (7.3), for every $f \in \mathcal {N}(R,z,n+1)$ , we have:

$$\begin{align*}\big\{f^{-1}(1),\ldots, f^{-1}(n) \big\} \, = \, X-x +w, \qquad \big\{f^{-1}(n+2),\ldots, f^{-1}(m+n) \big\} \, = \, Y-y. \end{align*}$$

These two labelings define a linear extension of $P^*$ after a substitution $w\gets x$ given by (7.4), and a linear extension of $Q-y$ , and it is clear that this construction defines a bijection. By an analogous argument, we have:

$$\begin{align*}\textrm{N}(R,z,n) \, = \, e(P^*-x) \cdot e(Q). \end{align*}$$

Set $c\gets n$ . Combining these two observations, we get

$$\begin{align*}\frac{\textrm{N}(R,z,c+1)}{\textrm{N}(R,z,c)} \ = \ \frac{\textrm{N}(R,z,n+1)}{\textrm{N}(R,z,n)} \ = \ \frac{\rho(P,x)}{\rho(Q,y)}\,, \end{align*}$$

which gives the desired reduction and proves the result.

7.2 Two posets from four

Now recall the decision problem

(7.6)

The following result give a quantitative version of Lemma 4.6.

We now build toward the proof of this theorem.

Lemma 7.3. Let $P=(X,\prec )$ and $Q=(Y,\prec ')$ be posets with $m=|X|$ and $n=|Y|$ elements, respectively. Let $x \in \min (P)$ and $y\in \min (Q)$ . Then there exists a poset $R=(Z,\prec ^\circ )$ and $z \in \min (P)$ , such that $|Z| = m+n+1$ and

$$\begin{align*}\rho(R,z) \ = \ m \ + \ \left(1 + \tfrac{\rho(Q,y)}{\rho(P,x)}\right)^{-1}. \end{align*}$$

Proof. Let $P^\ast = (X,\prec ^\ast )$ denote the dual poset to P. Let $R:=(Z,\prec ^\circ )$ be given by

$$\begin{align*}Z \ := \ (X-x) \cup (Y-y) \cup \{v,w,z\}, \end{align*}$$

where $\prec ^\circ $ inherits the partial order $\prec ^*$ on $X-x$ , the partial order $\prec '$ on $Y-y$ and with additional relations:

$$ \begin{align*} & p \, \prec^\circ \, v \prec^\circ q \qquad \forall \, p \in X-x, \, \ q \in Y-y, \\ & p \, \prec^\circ \, w \ \ \, \Leftrightarrow \ \ \, p \prec^* x \quad \forall \, p \in X-x,\\ & q \, \succ^\circ \, w \ \ \, \Leftrightarrow \ \ \, q \succ' y \quad \forall \, y \in Y-y,\\ & z \, \|_{\prec^\circ} \ p \ \ \, \forall \ p \in X-x, \quad z \, \prec^\circ \ q \quad \forall \, q \in Y-y, \\ & z \, \prec^\circ \, v, \quad z \, \|_{\prec^\circ} \ w. \end{align*} $$

That is, we are taking the series sum $(P^*-x) \oplus \{v\} \oplus (Q-y)$ , then adding an element w to emulate x in P for all $f(w) <f(v)$ , emulate y in Q for all $f(w)> f(v)$ and finally adding z to track the value of $f(v)$ . Here is the linear extension $f \in \mathcal {E}(R)$ in each case. By construction, we have either $f(v)=m+1$ or $f(v)=m+2$ .

Claim. We have:

(7.7) $$ \begin{align} e(R) \ = \ m e(P-x) e(Q) + (m+1) e(P) e(Q-y). \end{align} $$

Proof of claim.

Let us show that the first term $m e(P-x) e(Q) $ is the number of linear extensions $f \in \mathcal {E}(R)$ s.t. $f(v)=m+1$ . For such f, we have:

$$\begin{align*}\begin{aligned} \big\{f^{-1}(1),\ldots, f^{-1}(m) \big\} \ & = \ X- x + z, \\ \big\{f^{-1}(m+2),\ldots, f^{-1}(m+n+1) \big\} \ & = \ Y -y + w. \end{aligned} \end{align*}$$

Note that the restriction of f to $\big \{f^{-1}(1),\ldots , f^{-1}(m) \big \}$ defines a linear extension of $(P^*-x +z)$ . Additionally, note that the restriction of f to $\big \{f^{-1}(m+2),\ldots , f^{-1}(m+n+1) \big \}$ defines a linear extension of Q. It is also clear that this construction defines a bijection. In total, we have $e(P^*-x +z) e(Q) = m e(P-x) e(Q)$ linear extensions $ f$ as above.

Similarly, let us show that the second term $(m+1) e(P) e(Q-y)$ is the number of linear extensions $f \in \mathcal {E}(R)$ s.t. $f(v)=m+2$ . For such f, we have:

$$\begin{align*}\begin{aligned} \big\{f^{-1}(1),\ldots, f^{-1}(m+1) \} \ & = \ X-x + w +z, \\ \big\{f^{-1}(m+3),\ldots, f^{-1}(m+n+1) \big\} \ & = \ Y -y. \end{aligned}\end{align*}$$

Note that the restriction of f to $\big \{f^{-1}(1),\ldots , f^{-1}(m+1) \big \}$ defines a linear extension of $(P^* + z)$ . Additionally, note that the restriction of f to $\big \{f^{-1}(m+3),\ldots , f^{-1}(m+n+1) \big \}$ defines a linear extension of $(Q-y)$ . It is also clear that this construction defines a bijection. In total, we have $e(P^* + z) e(Q-y) = (m+1) e(P) e(Q-y)$ linear extensions $ f$ as above. This completes the proof.

Following the argument in the claim, we similarly have:

(7.8) $$ \begin{align} e(R-z) \ = \ e(P-x) e(Q) + e(P) e(Q-y). \end{align} $$

Indeed, the term $e(P-x) e(Q)$ is the number of linear extensions $f \in \mathcal {E}(R)$ for which $f(v)=m$ , and the term $e(P) e(Q-y)$ is the number of linear extensions $f \in \mathcal {E}(R)$ for which $f(v)=m+1$ . We omit the details.

Combing (7.7) and (7.8), we have:

$$ \begin{align*} \rho(R,z) \ = \ m + \frac{e(P) e(Q-y)}{e(P-x) e(Q) + e(P) e(Q-y)} \ = \ m \ + \ \left(1 + \frac{\rho(Q,y)}{\rho(P,x)}\right)^{-1}, \end{align*} $$

as desired.

Lemma 7.4. Let $P=(X,\prec )$ be a poset on $n=|X|$ elements, and let $x\in \min (P)$ . Then there exists a poset $Q=(Y,\prec ')$ and an element $y \in \min (Q)$ , such that $|Y|=n+1$ and

$$\begin{align*}\rho(Q,y) \ = \ 1 \, + \, \frac{1}{\rho(P,x)}\,.\end{align*}$$

Proof. Let $Y:=X+z$ , and let $\prec '$ coincide with $\prec $ on P, with added relations

$$ \begin{align*} z \prec' u \quad \text{for all} \ \ u \in X-x , \quad \text{and} \quad z \| x. \end{align*} $$

Note that $z\in \min (Q)$ . Note also that

$$\begin{align*}e(Q-z) \, = \, e(P) \quad \text{and} \quad e(Q) \, = \, e(P) + e(P-x), \end{align*}$$

since for every $f\in \mathcal {E}(Q)$ , we either have $f(z)=1$ , or $f(z)=2$ , and thus $f(x)=1$ . We now take $y \gets z$ , and observe that

$$\begin{align*}\rho(Q,y) \ = \ \frac{e(Q)}{e(Q-z)} \ = \ \frac{e(P) + e(P-x)}{e(P)} \ = \ 1 \, + \, \frac{1}{\rho(P,x)} \,, \end{align*}$$

as desired.

Lemma 7.5. Let $P=(X,\prec )$ be a poset on $n=|X|$ elements, and let $x\in \min (P)$ . Then there exists a poset $Q=(Y,\prec ')$ , and $y \in \min (Q)$ , such that $|Y|=n+1$ and

$$\begin{align*}\rho(Q,y) \ = \ 1 + \rho(P,x).\end{align*}$$

Proof. Let Q be as in the proof of Lemma 7.4. Note that $x\in \min (Q)$ and that

$$\begin{align*}e(Q-x) \, = \, e(P-x), \end{align*}$$

since z is the unique minimal element in $Q-x$ . We now take $y \gets x$ , and observe that

$$\begin{align*}\rho(Q,y) \ = \ \frac{e(Q)}{e(Q-x)} \ = \ \frac{e(P)+e(P-x)}{e(P-x)} \ = \ 1 + \rho(P,x), \end{align*}$$

as desired.

Proof of Theorem 7.2.

By symmetry, we will, without loss of generality, assume that $n_2 \geq n_3$ . By applying Lemma 7.3 followed by applying Lemma 7.5 for $n_2-n_3$ many times, we get a poset $P=(X,\prec )$ and $x \in \min (P)$ , such that

$$\begin{align*}\rho(P,x) \ = \ (n_2- n_3) \ + \ \Big( n_3 \ + \ \left(1 + \tfrac{\rho(P_1,x_1)}{\rho(P_3,x_3)}\right)^{-1} \Big) \ = \ n_2 + \left(1 + \tfrac{\rho(P_1,x_1)}{\rho(P_3,x_3)}\right)^{-1}. \end{align*}$$

Additionally, poset P has

$$\begin{align*}|X| \ = \ n \ = \ (n_1 + n_3+1) + n_2-n_3 \ = \ n_1+ \max\{n_2,n_3\}+1 \quad \text{elements}. \end{align*}$$

On the other hand, by Lemma 7.3, we get a poset Q and $y \in \min (Q)$ , s.t., such that

$$\begin{align*}\rho(Q,y) \ = \ n_2 \ + \ \left(1 + \tfrac{\rho(P_4,x_4)}{\rho(P_2,x_2)}\right)^{-1}, \end{align*}$$

and with

$$\begin{align*}m \ = \ n_2 + n_4+1 \ = \ n_4+ \max\{n_2,n_3\}+1. \end{align*}$$

It now follows that

$$\begin{align*}\rho(P,x) = \rho(Q,y) \qquad \Longleftrightarrow \qquad \frac{\rho(P_1,x_1)}{\rho(P_3,x_3)} = \frac{\rho(P_4,x_4)}{\rho(P_2,x_2)}\,, \end{align*}$$

as desired.

8.1 Continued fractions

Given $a_0\geq 0$ , $a_1, \ldots , a_s \in \mathbb Z_{\ge 1} $ , where $s \geq 0$ , the corresponding continued fraction is defined as follows:

$$\begin{align*}[a_0; a_1,\ldots, a_s] \ := \ a_0 + \cfrac{1}{a_1 + \cfrac{1}{\ddots \ + \frac{1}{a_s}}}. \end{align*}$$

Numbers $a_i$ are called quotients (see, e.g. [Reference Hardy and WrightHW08, Section 10.1]). We refer to [Reference KnuthKnu98, Section 4.5.3] for a detailed asymptotic analysis of the quotients in connection with the Euclidean algorithm and further references. The following technical result is key in the proof of the Verification Lemma.

Proposition 8.1 (cf. [Reference Kravitz and SahKS21, Section 3]).

Let $a_0, \ldots , a_s \in \mathbb Z_{\ge 1} $ . Then there exists a poset $P=(X,\prec )$ of width two on $|X|=a_0+\ldots +a_s$ elements, and element $x \in \min (P)$ , such that

$$\begin{align*}\rho(P,x) \ = \ [a_0; a_1,\ldots,a_s]. \end{align*}$$

Corollary 8.2. Let $a_1,\ldots a_s \in \mathbb Z_{\ge 1}$ . Then there exists a width two poset $P=(X,\prec )$ on $|X|=a_1+\ldots +a_s$ elements, and element $x \in \min (P)$ , such that

$$\begin{align*}\frac{1}{\rho(P,x)} \ = \ [0; a_1,\ldots,a_s].\end{align*}$$

We now present the proof of Proposition 8.1, which uses the following corollary of Lemmas 7.4 and 7.5.

Corollary 8.4. Let $P=(X,\prec )$ be a width two poset on $n=|X|$ elements, let $x\in \min (P)$ , and let $a \in \mathbb Z_{\ge 1}$ . Then there exists a width two poset $Q=(Y,\prec ')$ and $y \in \min (Q)$ , such that $|Y|=n+a$ and

$$\begin{align*}\rho(Q,y) \ = \ a + \frac{1}{\rho(P,x)}.\end{align*}$$

Proof of Proposition 8.1.

We use induction on s. For $s=0$ , let $P := C_{a_0-1} + \{x\}$ be a disjoint sum of two chains, and observe that $\rho (P,x) = a_0$ .

Suppose the claim holds for $s-1$ , that is, there exists a poset $P_1$ on $n=a_1+\ldots +a_s$ elements and $x_1 \in \min (P_1)$ , such that $\rho (P_1,x_1) = [a_1; a_2,\ldots , a_s]$ , and with $|P_1|=a_1+\ldots +a_s$ . By Corollary 8.4, there exists a poset Q on $a_0+n$ elements, and $x \in \min (P)$ , such that

$$ \begin{align*} \rho(P,x) \, = \, a_0 \, + \, \frac{1}{\rho(P_1,x_1)} \, = \, a_0 \, + \, \frac{1}{[a_1; a_2,\ldots, a_s]} \, = \, [a_0; a_1,\ldots,a_s]. \end{align*} $$

This completes the proof.

8.2 Number theoretic estimates

For $A\in \mathbb Z_{\geq 1}$ and $m \in [A]$ , consider the quotients in the continued fraction of $m/A$ and their sum:

$$\begin{align*}\frac{m}{A} \, = \, [0; a_1(m),\ldots, a_s(m)] \quad \text{and} \quad S_A(m) \, := \, \sum_{i=1}^s a_i(m).\end{align*}$$

Note that every rational number can be represented by continued fractions in two ways (depending if the last quotient is strictly greater than $1$ , or is equal to $1$ ), and $S_{A}(m)$ are equal for both representations. Also note that

(8.1) $$ \begin{align} S_{A}(m) \, = \, S_{A'}(m'), \qquad \text{where} \quad A' := \frac{A}{\gcd(A,m)} \quad \text{and} \quad m' := \frac{m}{\gcd(A,m)} \end{align} $$

are normalized to be coprime integers. The following technical result will also be used in the proof of the Verification Lemma 4.7.

Proposition 8.5. There exists a constant $C>0$ , such that for all coprime integers $A,B$ which satisfy $C < B < A < 2B$ , there exists an integer $m:=m(A,B)$ , such that $m<B$ ,

$$\begin{align*}S_A(m) \ \leq \ 2 (\log A)^2 \quad \text{ and } \quad S_B(m) \ \leq \ 2 (\log B)^2. \end{align*}$$

We now build toward the proof of this result. We need the following technical result.

Lemma 8.6 (Yao–Knuth [Reference Yao and KnuthYK75]).

We have:

$$\begin{align*}\frac{1}{n} \sum_{m \in [n]} S_n(m) \ = \ \frac{6}{\pi^2} (\log n)^2 + O\big((\log n) (\log \log n)^2\big) \quad \text{as } \ \ n\to \infty. \end{align*}$$

By the Markov inequality, it follows from Lemma 8.6 that

(8.2) $$ \begin{align} \begin{aligned} \big|\big\{ m \in [n] \, : \, S_n(m)> 2 (\log n)^2 \big\}\big| & \ \leq \ \frac{3}{\pi^2} n (1+o(1)). \end{aligned} \end{align} $$

Proof of Proposition 8.5.

Denote

$$ \begin{align*} \vartheta(A,B) \, := \, \big|\{ m \in [B] \, : \, S_A(m) \leq 2 (\log A)^2, \ S_B(m) \leq 2 (\log B)^2 \ \} \big|. \end{align*} $$

To prove the result, it sufficed to show that

$$ \begin{align*}\vartheta(A,B) \, = \, \Omega\big(B\big) \quad \text{as } \ \ C \to \infty. \end{align*} $$

Now, it follows from the inclusion-exclusion principle that

$$\begin{align*}\vartheta(A,B) \ \ge \ B - \big|\big\{ m \in [B] \, : \, S_A(m)> 2 (\log A)^2 \ \big\} \big| - \big|\big\{ m \in [B] \, : \, \ S_B(m) > 2 (\log B)^2 \big\} \big|. \end{align*}$$

On the other hand, we have:

$$ \begin{align*} \big|\{ m \in [B] \, : \, S_A(m)> 2 (\log A)^2 \ \} \big| \ \leq \ \big|\{ m \in [A] \, : \, S_A(m) > 2 (\log A)^2 \ \} \big| \ \leq_{(8.2)} \ \tfrac{3}{\pi^2} A (1+o(1)), \end{align*} $$

and

$$ \begin{align*} \big|\{ m \in [B] \, : \, S_B(m)> 2 (\log B)^2 \ \} \big| \ \leq_{(8.2)} \ \tfrac{3}{\pi^2} B (1+o(1)). \end{align*} $$

Combining these inequalities, we get

$$ \begin{align*} \vartheta(A,B) \ \geq \ B - \tfrac{3}{\pi^2} (A +B) \big(1+o(1)\big) \ \geq \ B \big(1-\tfrac{9}{\pi^2} \big) \big(1-o(1)\big), \end{align*} $$

and the result follows since $\big (1-\frac {9}{\pi ^2}\big )> 0$ .

8.3 Bounds on relative numbers of linear extensions

The following simple bound is the final ingredient we need for the proof of the Verification Lemma.

The lower bound holds for all $x\in X$ (see, e.g. [Reference Edelman, Hibi and StanleyEHS89]). The upper bound is a special case of [Reference Chan, Pak and PanovaCPP24, Lemma 5.1]. We include a short proof for completeness.

8.4 Proof of Verification Lemma 4.7

Recall the decision problem

where $P=(X,\prec )$ is a poset on $n=|X|$ elements, $x\in \min (P)$ and $A,B$ are coprime integers with $B< A\leq n!$ . We simulate VerRLE with an oracle for QuadRLE as follows.

By Proposition 8.8, we need only to consider the cases $1 < \frac {A}{B} \leq n$ . Indeed, when $\rho (P,x)<1$ or $\rho (P,x)>n!$ , VerRLE does not hold. Additionally, when $\rho (P,x)=1$ , VerRLE holds if and only if P is a chain. Let $k := \left \lfloor \tfrac {A}{B} \right \rfloor $ . As in the $s=0$ part of the proof of Proposition 8.1, there exists a poset $P_3=(X_3,\prec _3)$ with $|X_3| = k \le n$ , and an element $x_3 \in \min (P_3)$ , such that $\rho (P_3,x_3) = k$ .

Let $A',B'$ be coprime integers, such that

$$ \begin{align*}\frac{A}{B} \, = \, k \frac{A'}{B'}\,. \end{align*} $$

Then we have $B\le B'<A'<2B'$ , $A'\le A$ , and thus $\log A' = O(n \log n)$ . By Proposition 8.5, there is a positive integer $m \in [B']$ , such that

$$\begin{align*}S_{A'}(m) \ \leq \ 2 (\log A')^2 \quad \text{ and } \quad S_{B'}(m) \ \leq \ 2 (\log B')^2. \end{align*}$$

At this point, we guess such m. Since computing the quotients of $m/A'$ can be done in polynomial time, we can verify in polynomial time that m satisfies the inequalities above.

By Corollary 8.2, we can construct posets $P_2=(X_2,\prec _2)$ , $P_4=(X_4,\prec _4)$ with $x_2\in \min (P_2)$ , $x_4\in \min (P_2)$ , such that

$$\begin{align*}\rho(P_2,x_2) \, = \, \frac{B'}{m} \qquad \text{and} \qquad \rho(P_4,x_4) \, = \, \frac{A'}{m} \,. \end{align*}$$

The corollary also gives us

$$\begin{align*}|X_2| \, \leq \, S_{B'}(m) \, \leq \, 2 (\log B')^2 \, = \, O\big(n^2 (\log n)^2\big), \end{align*}$$

and we similarly have $|X_4| = O\big (n^2 (\log n)^2\big )$ . Since posets $P_2,P_3$ and $P_4$ have polynomial size, we can call QuadRLE to check

$$ \begin{align*} \big\{ \rho(P,x) \cdot \rho(P_2,x_2) \, =^? \, \rho(P_3,x_3) \cdot \rho(P_4,x_4) \big\}. \end{align*} $$

Observe that

$$ \begin{align*}\frac{\rho(P_3,x_3) \cdot \rho(P_4,x_4)}{\rho(P_2,x_2)} \ = \ \frac{m}{B'} \cdot k \cdot \frac{A'}{m} \ = \ \frac{A}{B}\,. \end{align*} $$

Thus, in this case, QuadRLE is equivalent to VerRLE, as desired.

9.1 Explicit equality conditions

For $k=1$ , the equality cases of Stanley’s inequality (Sta) are tuples $(P,x,z,a,c)$ , where $P=(X,\prec )$ is a poset on $n=|X|$ elements, $x,z\in X$ , $a,c \in [n]$ and the following holds:

(9.1) $$ \begin{align} \textrm{N}_{z c}(P, x,a)^2 \, = \, \textrm{N}_{z c}(P, x,a+1)\cdot \textrm{N}_{z c}(P, x,a-1). \end{align} $$

The subscripts here and throughout this section are no longer bold, to emphasize that $k=1$ . Recall also both the notation in Section 1.4 and the Ma–Shenfeld poset notation in Section 6.1.

We prove Lemma 9.1 later in this section.

Proof of Theorem 1.4.

As before, let $P=(X,\prec )$ be a poset on $n=|X|$ elements, let $x,y,z\in X$ and $a,b,c\in [n]$ . Denote by $\textrm {N}_{z c}(P,x,a,y,b)$ the number of linear extensions $f \in \mathcal {E}_{z c}(P,x,a)$ that additionally satisfy $f(y)=b$ .

Now, condition $(\divideontimes )$ in Lemma 9.1, can be rewritten as follows:

(9.2) $$ \begin{align} \textrm{N}_{z c}(P,x,a',y,b') = 0 \quad \text{for all} \quad y \in \mathcal C(x) \ \ \ \text{and} \ \ \ a',b' \in \{a-1,a,a+1\}. \end{align} $$

Indeed, each vanishing condition in (9.2) is checking whether there exists a companion of x in a linear extension that is comparable to x.

Recall that each vanishing condition in (9.2) is in $\mathrm{ {P}}$ (see references in Section 3.5). There are at most $6n$ instances to check, since for all $y\in X$ , there are at most six choices of distinct $a',b'$ in $\{a-1,a,a+1\}$ . Therefore, EqualityStanley $_{1} \in \mathrm{ {P}}$ .

9.2 Ma–Shenfeld theory

We now present several ingredients needed to prove Lemma 9.1. We follow closely the Ma–Shenfeld paper [Reference Ma and ShenfeldMS24], presenting several results from that paper.

In [Reference Ma and ShenfeldMS24], Ma–Shenfeld defined the notions of subcritical, critical and supercritical posets, which are directly analogous to the corresponding notions for polytopes given in [Reference Shenfeld and van HandelSvH23], cf. Section 3.2. As the precise definitions are rather technical, we will not state them here while still including key properties of those families that are needed to prove Lemma 9.1.

We start with the following hierarchical relationship between the three families:

$$\begin{align*}\{\text{subcritical posets}\} \quad \supseteq \quad \{\text{critical posets}\} \quad \supseteq \quad \{\text{supercritical posets}\}. \end{align*}$$

A poset that is subcritical but not critical is called sharp subcritical, and a poset that is critical but not super critical is called sharp critical.

The equality conditions for (9.1) are directly determined by the classes to which the poset P belongs, as we explain below. We note that these families depend on the choices of $P,x,a,z,c$ , which we omit from the notation to improve readability. Furthermore, without loss of generality, we can assume that $z \notin \{a-1,a,a+1\}$ , as otherwise one of the numbers in (9.1) is equal to $0$ , making the problem in $\mathrm{ {P}}$ (see above).

We now state two other properties of these families, which require the following definitions. Following [Reference Ma and ShenfeldMS24], we add two elements $z_0,z_{k+1}$ into the poset, such that $z_0 \preccurlyeq y \preccurlyeq z_{k+1}$ for all $y \in X$ , and we define $c_0:=0$ and $c_{k+1}:=n+1$ . A splitting pair is a pair of integers $(r,s)$ in $\{0,\ldots , k+1\}$ , such that $(r,s)\neq (0,k+1)$ .Footnote ¹⁰

Lemma 9.3 [Reference Ma and ShenfeldMS24, Lemma 5.10].

Let $P=(X,\prec )$ be a sharp subcritical poset. Then there exists a splitting pair $(r,s)$ , such that

(9.3) $$ \begin{align} \big|\big\{ u \in X \, : \, z_r \prec u \prec z_s \big\}\big| \, = \, c_s - c_r - 1. \end{align} $$

We say that poset P is split indecomposable if, for every splitting pair $(r,s)$ ,

$$\begin{align*}\big|\big\{ u \in X \, : \, z_r \prec u \prec z_s \big\}\big| \, \leq \, c_s - c_r - 2. \end{align*}$$

In particular, by Lemma 9.3, every sharp subcritical poset is not split indecomposable.

It was shown in [Reference Ma and ShenfeldMS24] that we can, without loss of generality, assume that poset P is split indecomposable. Indeed, otherwise checking (9.1) can be reduced to checking the same problem for a smaller poset: either restricting to the set in (9.3), or removing this set from the poset (see [Reference Ma and ShenfeldMS24, Section 6] for details. Thus, we can, without loss of generality, assume that P is a critical poset.

Lemma 9.4 [Reference Ma and ShenfeldMS24, Lemma 5.11].

Let P be a split indecomposable sharp critical poset. Then there exists a splitting pair $(r,s)$ , such that $c_r<a<c_s$ and

(9.4) $$ \begin{align} \big|\big\{ u \in X \, : \, z_r \prec u \prec z_s \big\}\big| \, = \, c_s - c_r - 2. \end{align} $$

Recall from Section 6.2 that $\mathcal F(a,\operatorname {\mathrm {\text {com}}}, \operatorname {\mathrm {\text {com}}})$ is the set of linear extensions in $\mathcal {E}_{z c}(P,x,a)$ , such that both the lower and upper companions of x are incomparable to x. Next, $\mathcal F(a,\operatorname {\mathrm {\text {com}}}, \operatorname {\mathrm {\text {inc}}})$ is the set of linear extensions in $\mathcal {E}_{z c}(P,x,a)$ , such that the lower companion is comparable to x, but the upper companion is incomparable to x. Similarly, $\mathcal F(a,\operatorname {\mathrm {\text {inc}}}, \operatorname {\mathrm {\text {com}}})$ is the set of linear extensions in $\mathcal {E}_{z c}(P,x,a)$ , such that the lower companion is incomparable to x, but the upper companion is comparable to x. Let $\mathcal F(a-1,\cdot , \cdot )$ and $\mathcal F(a+1,\cdot , \cdot )$ be defined analogously. Finally, let $\operatorname {\mathrm {\mathrm {F}}}(a+s,\cdot ,\cdot ) := |\mathcal F(a+s,\cdot ,\cdot )|$ , where $s\in \{0,\pm 1\}$ , be the numbers of these linear extensions.

Lemma 9.6 [Reference Ma and ShenfeldMS24, Theorem 1.5].

Let P be a critical poset. Then (9.1) holds if and only if

(9.5) $$ \begin{align} & \textrm{F}(a-1,\mathrm{com},\mathrm{com})\ = \ \textrm{F}(a,\mathrm{com},\mathrm{com}) \ = \ \textrm{ F}(a+1,\mathrm{com},\mathrm{com}) \, = \, 0 \quad \ \text{ and} \end{align} $$

(9.6) $$ \begin{align} &\!\!\!\!\!\!\!\! \textrm{F}(a-1,\mathrm{com},\mathrm{inc}) \ = \ \textrm{F}(a-1,\mathrm{inc},\mathrm{com}) \ = \ \textrm{F}(a,\mathrm{com},\mathrm{inc})\notag\\ & \hskip1.cm = \ \textrm{F}(a,\mathrm{inc},\mathrm{com}) \ = \ \textrm{F}(a+1,\mathrm{com},\mathrm{inc}) \ = \ \textrm{F}(a+1,\mathrm{inc},\mathrm{com}). \end{align} $$

Now note that $\operatorname {\mathrm {\mathrm {F}}}(a-1,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}}) \ \leq \ \operatorname {\mathrm {\mathrm {F}}}(a-1,\operatorname {\mathrm {\text {inc}}},\operatorname {\mathrm {\text {com}}})$ , with the equality if and only if every upper companion of x is always incomparable to the lower companion of x. By an analogous argument applied to $\operatorname {\mathrm {\mathrm {F}}}(a,\cdot , \cdot )$ and $\operatorname {\mathrm {\mathrm {F}}}(a+1,\cdot , \cdot )$ , we get the following corollary.

Corollary 9.7. Let P be a critical poset. Suppose

$$\begin{align*}\textrm{N}_{z c}(P, x,a)^2 \, = \, \textrm{N}_{z c}(P, x,a+1) \cdot \textrm{N}_{z c}(P, x,a-1) \ \neq \ 0. \end{align*}$$

Then, for every linear extension $f\in \mathcal {E}(P)$ counted by (9.6), the upper companion is incomparable to the lower companion: $\textrm {uc} (f) \| \, {} \textrm {lc} (f)$ .

Finally, we have equality conditions for supercritical posets.

Lemma 9.8 [Reference Ma and ShenfeldMS24, Theorem 1.3].

Let P be a supercritical poset. Then (9.1) holds if and only if equalities (9.5) and (9.6) hold, and additionally

(9.7) $$ \begin{align} \text{all numbers in}\ (9.6)\ \text{are equal to } 0. \end{align} $$

9.3 Proof of Lemma 9.1

Note that (9.5), (9.6) and (9.7) are equivalent to requiring that x is incomparable to both $\textrm {lc} (f)$ and $\textrm {uc} (f)$ . Thus, it suffices to show that if P is a critical poset, then (9.7) holds.

Suppose to the contrary that $P=(X,\prec )$ is a counterexample, and let $n:=|X|$ . Then P is a sharp critical poset. By taking the dual poset if necessary, we can assume, without loss of generality, that $c<a$ . It then follows that the splitting pair $(r,s)$ in Lemma 9.4 is $(1,2)$ . This means that $c_r=c$ and $c_s=n+1$ , so we have from (9.4) that

(9.8) $$ \begin{align} \left|\{ u \in P \, : \, z \prec u \}\right| \ = \ n -c-1. \end{align} $$

Since (9.7) does not hold, there exist $f \in \mathcal F(a,\operatorname {\mathrm {\text {com}}}, \operatorname {\mathrm {\text {inc}}})$ and $h \in \mathcal F(a-1,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}})$ . Let ${y_1:=f^{-1}(a-1)}$ (i.e. the lower companion in f) and $y_2:=h^{-1}(a)$ (i.e. the lower companion in h). Note that we have $y_1 \prec x \prec y_2$ . Let $m=f(y_2)$ , and note that $m\geq a+2$ by definition.

We claim: There exists a new linear extension $g\in \mathcal {E}(P)$ , such that $g(y_2)=m-1$ , and such that $g \in \mathcal F(a,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {inc}}})$ if $m>a+2$ , and $g \in \mathcal F(a,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {com}}})$ if $m=a+2$ . Note that this suffices to prove the lemma, as by replacing f with g and decreasing m repeatedly, we get that $\operatorname {\mathrm {\mathrm {F}}}(a,\operatorname {\mathrm {\text {com}}},\operatorname {\mathrm {\text {com}}})>0$ , which contradicts (9.5).

We now prove the claim. Since $h(y_2)=a <m=f(y_2)$ , there exists $w \in X$ , such that $f(w) < m$ and $w \| y_2$ . Suppose w is such an element that maximizes $f(w)$ . There are three cases:

First, suppose that $f(w)> a$ . By the maximality assumption, every element ordered between w and $y_2$ according to f, is incomparable to w. Then we can promote w to be larger than $ y_2$ . Note that the resulting linear extension $g\in \mathcal {E}(P)$ satisfies $g(y_2)=m-1$ , $g(y_1)=a-1$ and $g(x)=a$ , as desired.

Second, suppose that $c<f(w)<a$ . By the maximality assumption, every element ordered between w and $y_2$ according to f is incomparable to $ w$ . Then we can promote w to be larger than $ y_2$ . The resulting linear extension $g'\in \mathcal {E}(P)$ satisfies $g'(y_2)=m-1$ . Note, however, that we have $g'(y_1)=a-2$ and $g'(x)=a-1$ . In order to fix this, let $v:=f^{-1}(a+1)$ . It follows from Corollary 9.7 that v is incomparable to $y_1$ and x. Thus, we can demote v to be the smaller than $y_1$ . We obtain a new linear extension $g\in \mathcal {E}(P)$ that satisfies $g(y_1)=a-1$ and $g(x)=a$ , as desired.

Third, suppose that $f(w)<c$ . Then, every element ordered between z and $y_2$ according to $ f$ is less than $ y_2 .$ Note that there are $m-c-1$ many such elements. On the other hand, it follows from (9.8) that there is exactly one element in $\{f^{-1}(c+1), f^{-1}(c+2), \ldots , f^{-1}(n) \}$ that is incomparable to $ z$ . It then follows that there are at least $m-c-2$ elements that are greater than z and less than $ y_2$ , that is

(9.9) $$ \begin{align} \big|\big\{ u \in X \, : \, z \prec u \prec y_2 \big\}\big| \, \geq \, m-c-2. \end{align} $$

On the other hand, the existence of h implies that

(9.10) $$ \begin{align} \big|\big\{ u \in X \, : \, z \prec u \prec y_2 \big\}\big| \, \leq \, h(y_2)-c-1 \, = \, a-c-1 \, \leq \, m-c-3, \end{align} $$

a contradiction. This finishes the proof of the claim.

10.1 The basis of our work

Due to the multidisciplinary nature of this paper, we make a special effort to simplify the presentation. Namely, the proofs of our main results (Theorems 1.1 and 1.3) are largely self-contained in a sense that we only use standard results in combinatorics (Stanley’s theorem in Section 5.2 and the Brightwell–Winkler’s [Reference Brightwell and WinklerBW91] theorem in Section 3.6), computational complexity (Toda’s [Reference TodaToda91] theorem in Section 4.4) and number theory (Yao–Knuth’s [Reference Yao and KnuthYK75] theorem in Section 8.2). In reality, the paper freely uses tools and ideas from several recent results worth acknowledging.

First, we heavily build on the recent paper by Shenfeld and van Handel [Reference Shenfeld and van HandelSvH23], and the follow up by Ma and Shenfeld [Reference Ma and ShenfeldMS24]. Without these results, we would not know where to look for “bad posets” and “bad polytopes”. Additionally, the proof in Section 6.2 is a reworking and simplification of many technical results and ideas in [Reference Ma and ShenfeldMS24].

Second, in Section 8.1, we use and largely rework the continued fraction approach by Kravitz and Sah [Reference Kravitz and SahKS21]. There, the authors employ the Stern–Brocot and Calkin–Wilf tree notions, which we avoid in our presentation as we aim for different applications.

Third, in the heart of our proof of Theorem 1.3 in Section 4.4, we follow the complexity roadmap championed by Ikenmeyer et al. in [Reference Ikenmeyer and PakIP22, Reference Ikenmeyer, Pak and PanovaIPP22]. Same for the heart of the proof of the Verification Lemma 4.7 in Section 8.4, which follows the approach in our companion paper [Reference Chan and PakCP23a].

On the other hand, the proof of Theorem 1.4 given in Section 9 is the opposite of self-contained, as we rely heavily on both results and ideas in [Reference Ma and ShenfeldMS24]. We also use properties of the promotion and demotion operations on linear extensions that were introduced by Schützenberger in the context of algebraic combinatorics (see [Reference SchützenbergerSchü72]).Footnote ¹¹ Panova et al. employed this approach in a closely related setting in [Reference Chan, Pak and PanovaCPP23a, Reference Chan, Pak and PanovaCPP23b, Reference Chan, Pak and PanovaCPP24]. We emphasize once again that our proof of Theorem 1.4 is independent of the rest of the paper and is the only part that uses results in [Reference Ma and ShenfeldMS24].

10.2 Equality cases

The reader unfamiliar with the subject may wonder whether equality conditions of known inequalities are worth an extensive investigation. Here is how Gardner addresses this question:

Closer to the subject of this paper, Shenfeld and van Handel explain the difficulty of finding equality conditions for (MQI) and (AF):

10.3 Polytopes

The family of TU-polytopes that we chose is very special in that these H-polytopes have integral vertices (but not a description in $\mathrm{ {P}}$ , as V-polytopes are defined to have). In [Reference Chan and PakCP24+], we consider a family of axis-parallel boxes which have similar properties. Clearly, for general convex bodies, there is no natural way to set up a computational problem that would not be immediately intractable (unless one moves to a more powerful computational model; see, e.g. [Reference Blum, Cucker, Shub and SmaleBCSS98]).

10.4 Discrete isoperimetric inequality

For a discrete version of the isoperimetric inequality in the plane, one can consider convex polygons with given normals to edges. In this case, L’Huilier (1775) proved that the isoperimetric ratio is minimized for circumscribed polygons (see, e.g. [Reference TóthFej72, Section I.4]). In the 1860s, Steiner and Lindelöf [Reference TóthFej72] studied a natural generalization of this problem in $\mathbb R^3$ but were unable to solve it in full generality.

At the turn of the 20th century, Minkowski developed the theory of mixed volumes, motivated, in part, to resolve the Steiner–Lindelöf problem. He showed that among all polytopes with given normals, the isoperimetric ratio is minimized on circumscribed polytopes (see, e.g. [Reference TóthFej72, Section V.7].

There are several Bonnesen type and stability versions of the discrete isoperimetric inequality (see, e.g. [Reference Fisher, Ruoff and ShilletoFRS85, Reference Indrei and NurbekyanIN15, Reference ZhangZhang98]). Let us single out a hexagon version used by Hales in his famous proof of the honeycomb conjecture [Reference HalesHal01, Theorem 4].

10.5 Brunn–Minkowski inequality

There are several proofs of the Brunn–Minkowski inequality (BM), but some of them do not imply the equality conditions, such as, for example, the “brick-by-brick” inductive argument in [Reference MatoušekMat02, Section 12.2]. Note also that Alexandrov’s proof of the Minkowski uniqueness theorem (of polytopes with given facet volumes and normals) relies on the equality conditions for the Brunn–Minkowski inequality (see [Reference AlexandrovAle50]). This is essential for Alexandrov’s “topological method” and is the basis for the variational principle approach (see, e.g. [Reference PakPak09]).

10.6 van der Waerden conjecture

The Alexandrov–Fenchel inequality (AF) came to prominence in combinatorics after Egorychev [Reference EgorychevEgo81] used it to prove the van der Waerden conjecture that was proved earlier by Falikman [Reference FalikmanFal81]Footnote ¹² (see [Reference KnuthKnu81, Reference van LintvL82] for friendly expositions). This development set the stage for Stanley’s paper [Reference StanleySta81]. The conjecture states that for every bistochastic $n\times n$ matrix A, we have

(vdW) $$ \begin{align} \mathrm{per}(A) \, \ge \, \frac{n!}{n^n}, \end{align} $$

and the equality holds only if $A=(a_{ij})$ has uniform entries: $a_{ij}=\frac {1}{n}$ for all $1\le i,j\le n$ .

Note that Egorychev’s proof of the equality conditions for (vdW) used Alexandrov’s equality conditions (AF) for nondegenerate boxes (see Section 3.2 (cf. [Reference KnuthKnu81, p. 735] and [Reference van LintvL82, Section 7]). In a follow-up paper [Reference Chan and PakCP24+], we analyze the complexity of the Alexandrov–Fenchel equality condition for degenerate boxes. Note also that Knuth’s exposition in [Reference KnuthKnu81] is essentially self-contained, while Gurvits’s proof of (vdW) completely avoids (AF) (see [Reference GurvitsGur08, Reference Laurent and SchrijverLS10]).

10.7 Matroid inequalities

Of the several log-concavity applications of the AF inequality given by Stanley in [Reference StanleySta81], see also [Reference StanleySta86, Section 6], one stands out as a special case of a Mason’s conjecture (Theorem 2.9 in [Reference StanleySta81]). The strongest of the three Mason’s conjectures states that the numbers $ \textrm {I} (M,k)/\binom {n}{k}$ are log-concave, where $\textrm {I} (M,k)$ is the number of independent sets of size k in a matroid M on n elements. These Mason’s conjectures were recently proved in a long series of spectacular papers, culminating with [Reference Adiprasito, Huh and KatzAHK18, Reference Anari, Liu, Gharan and VinzantALOV24, Reference Brändén and HuhBH20] (see also an overview in [Reference HuhHuh18, Reference KalaiKal22]).

Curiously, the equality cases for these inequalities are rather trivial and can be verified in polynomial time [Reference Murai, Nagaoka and YazawaMNY21] (see also [Reference Chan and PakCP24a, Section 1.6]). Here, we assume that the matroid is given in a concise presentation (such presentations include graphical, bicircular and representable matroids). Curiously, for the weighted extension of Mason’s third conjecture given in [Reference Chan and PakCP24a, Theorem 1.6], the equality cases are more involved. It follows from [Reference Chan and PakCP24a, Theorem 1.9], however, that this problem is in $\mathrm{ {coNP}}$ . In other words, Theorem 1.3 shows that is likely much more powerful.

Note that the defect $\psi (M,k):=\textrm {I} (M,k)^2-\textrm {I} (M,k+1)\cdot \textrm {I} (M,k-1)$ is conjectured to be not in $\mathrm{ {\#P}}$ (see [Reference PakPak22, Conjecture 5.3]). Clearly, the argument in the proof of Corollary 1.5 does not apply in this case. Thus, another approach is needed to prove this conjecture, just as another approach is needed to prove that $\phi _0 \notin \mathrm{ {\#P}}$ (see Section 1.4).

10.8 Complexity of equality cases

Recall that Theorem 1.1 does not imply that is $\mathrm{ {NP}}$ -hard or $\mathrm{ {coNP}}$ -hard, more traditional measures of computational hardness. This remains out of reach. Note, however, that is naturally in the class $\mathrm{ {C}}_=\mathrm{ {P}}$ (see Section 2.6).

If this holds for all $k\ge 2$ , this would imply a remarkable dichotomy with $k \le 1$ (see Theorem 1.4). To motivate the conjecture, recall from Section 3.6 that $\mathrm{ {C}}_=\mathrm{ {P}}$ -complete problem is $\mathrm{ {coNP}}$ -hard (see [Reference Chan and PakCP23a] for more on the complexity of combinatorial coincidence problems).

Note that the proof of implies that , even when at most four polytopes, are allowed to be distinct. It would be interesting to decide if this number can be reduced down to three. It is known that two distinct TU-polytopes are not enough. This follows from a combination of our arguments that for supercritical cases (in the sense of [Reference Shenfeld and van HandelSvH23]), we have , and an argument that for two polytopes, the equality cases are supercritical.Footnote ¹³

10.9 Injective proofs

In enumerative combinatorics, whenever one has an equality between the numbers counting certain combinatorial objects, one is tempted to find a direct bijection between the sides (see, e.g. [Reference LoehrLoe11, Reference PakPak05, Reference StanleySta12]). Similarly, when presented an inequality $f \geqslant g$ , one is tempted to find a direct injection (see, e.g. [Reference PakPak19, Reference StanleySta89]). In the context of linear extensions, such injections appear throughout the literature (see, e.g. [Reference BrentiBre89, Reference Brightwell and TrotterBT02, Reference Chan, Pak and PanovaCPP23a, Reference Daykin and DaykinDD85, Reference Gaetz and GaoGG22, Reference Lam and PylyavskyyLP07]).

Typically, a direct injection and its inverse are given by simple polynomial time algorithms, thus giving a combinatorial interpretation for the defect $(f-g)$ . Therefore, if a combinatorial inequality is not in $\mathrm{ {\#P}}$ , it is very unlikely that there is a proof by a direct injection. In particular, Corollary 1.5 implies that the Stanley inequality (Sta) most likely cannot be proved by a direct injection. This confirms an old speculation:

Similarly, Corollary 1.5 suggests that the strategy in [Reference Chan, Pak and PanovaCPP23b, Section 9.12] is unlikely to succeed, at least for $k\ge 2$ .Footnote ¹⁴

To fully appreciate how delicate is Corollary 1.5, compare it with a closely related problem. It is known that for all $k\ge 0$ , the analogue of the Stanley inequality (Sta) holds for the number $\Omega (P,t)$ of order preserving maps $X\to [t]$ for all $t \in \mathbb N$ . This was conjectured by Graham in [Reference GrahamGra82, p. 129], see also [Reference GrahamGra83, p. 233], motivated by the proof of the XYZ inequality [Reference SheppShe82] (cf. Section 3.4). The result was proved in [Reference Daykin, Daykin and PatersonDDP84, Theorem 4] by a direct injection (see also [Reference DaykinDay84, Section 4.2] for additional details of the proof). In other words, in contrast with $ \phi _k$ , the defect of the analogue of (Sta) for order preserving maps has a combinatorial interpretation. Note that it is not known whether the defect of the XYZ inequality is in $\mathrm{ {\#P}}$ (see [Reference PakPak22, Conjecture 6.4]).

10.10 Stability proofs

By analogy with the injective proofs, Corollary 1.2 suggests that certain proofs of the Alexandrov–Fenchel inequality are likely not possible. Here, we are thinking of the mass transportation proof of characterization of the isoperimetric sets given in [Reference Figalli, Maggi and PratelliFMP10, Appendix], following Gromov’s approach in [Reference GromovGro86]. It would be interesting to make this idea precise.

10.11 Dichotomy of the equality cases

As we discuss in Section 9.2, it follows from the results in [Reference Ma and ShenfeldMS24] that the equality verification of the Stanley inequality (Sta) can be decided in polynomial time for supercritical posets. In contrast, by Theorem 1.3, the problem is not in $\mathrm{ {PH}}$ for critical posets.Footnote ¹⁵ We believe that this dichotomy also holds for the equality cases of the Alexandrov–Fenchel inequality (AF) for classes of H-polytopes for which the scaled mixed volume is in $\mathrm{ {\#P}}$ .

10.12 The meaning of it all

Finding the equality conditions of an inequality may seem like a straightforward, unambiguous problem, but the case of the Alexandrov–Fenchel inequality shows that it is nothing of the kind. Even the words “equality conditions” are much too vague for our taste. What the problem asks is a description of the equality cases. But since many geometric and combinatorial inequalities have large families of equalities cases, the word “description” becomes open-ended (cf. Section 2.5). How do you know when you are done? At what point are you satisfied with the solution and do not need further details?

These are difficult questions which took many decades to settle, and the answers depend heavily on the area. In the context of geometric inequalities discussed in Section 3.1, the meaning of “description” starts out simple enough. There is nothing ambiguous about discs as equality cases of the isoperimetric inequality in the plane (Isop), or pairs of homothetic convex bodies for the Brunn–Minkowski inequality (BM) or circumscribed polygons with given normals for the discrete isoperimetric inequality (see Section 10.4). Arguably, Bol’s [Reference BolBO43] equality cases of (MWI) are also unambiguous — in $\mathbb R^3$ , you literally know the cap bodies when you see them. However, when it comes to Minkowski’s quadratic inequality (MQI), the exact meaning of “description” is no longer obvious. Shenfeld and van Handel write, “The main results of this paper will provide a complete solution to this problem” [Reference Shenfeld and van HandelSvH22]. Indeed, their description of $3$ -dimensional triples of convex bodies cannot be easily improved upon, at least not in the case of convex polytopes (see Section 3.1). Some questions may still linger, but they are on the structure of the equality cases rather than on their recognition.Footnote ¹⁶

What Shenfeld and van Handel did is finished off the geometric approach going back to Brunn, Minkowski, Favard, Fenchel, Alexandrov and others, further formalized by Schneider. “Maybe a published conjecture will stimulate further study of this question”, Schneider wrote in [Reference SchneiderSchn85]. This was prophetic, but that conjecture was not the whole story, as it turned out.

In [Reference Shenfeld and van HandelSvH23], the authors write again: “We completely settle the extremals of the Alexandrov–Fenchel inequality for convex polytopes.” Unfortunately, their description is extraordinary complicated in higher dimensions, so the problem of recognizing the equality cases is no longer easy (see Section 3.2). And what good is a description if it cannot be used to recognize the equality cases?

In combinatorics, the issue of “description” has also been a major problem for decades, until it was fully resolved with the advent of computational complexity. For example, consider the following misleadingly simple description: “Let G be a planar cubic Hamiltonian graph.” Is that good enough? How can you tell if a given graph G is as you describe? We now know that the problem whether G is planar, cubic and Hamiltonian is $\mathrm{ {NP}}$ -complete [Reference Garey, Johnson and TarjanGJT76]. But if you only need the “planar” condition, the problem is computationally easy, while the “cubic” condition is trivial. Consequently, “planar cubic Hamiltonian” should not be viewed as a “good” description, but if one must consider the whole class of such graphs, this description is (most likely) the best one can do.

Going over equality cases for various inequalities on the numbers of linear extensions already gives an interesting picture. For the Björner–Wachs inequality, see Section 3.4, the recognition problem of forests is in $\mathrm{ {P}}$ , of course. On the other hand, as we explain in Section 3.4, for the Sidorenko inequality (3.1), the recognition problem of series-parallel posets is in $\mathrm{ {P}}$ for a more involved reason. On the opposite end of the spectrum, for the (rather artificial) inequality $(e(P)-e(Q))^2\ge 0$ , the equality verification is not in $\mathrm{ {PH}}$ , unless $\mathrm{ {PH}}$ collapses (see Section 3.7 and [Reference Chan and PakCP23a, Theorem 1.4]).

In this language, for the $k=0$ case of the Stanley inequality (Sta), the description of equality cases given in [Reference Shenfeld and van HandelSvH23] is trivially in $\mathrm{ {P}}$ . Similarly, for the $k=1$ case, the description of equality cases is also in $\mathrm{ {P}}$ by Theorem 1.4. On the other hand, Theorem 1.3 shows that for $k\ge 2$ , the description in [Reference Ma and ShenfeldMS24] is (very likely) not in $\mathrm{ {P}}$ . Under standard complexity assumptions, there is no description of the equality cases in $\mathrm{ {P}}$ at all, or even in $\mathrm{ {PH}}$ for that matter.

Now, the problem of counting the equality cases brings a host of new computational difficulties, making seemingly easy problems appear hard when formalized (see [Reference PakPak22]). Even for counting nonisomorphic forest posets on n elements, to show that this function in $\mathrm{ {\#P}}$ , one needs to define a canonical labeling to be able to distinguish the forests, to make sure each is counted exactly once (see, e.g. [Reference Schweitzer and WiebkingSW19]).

In this language, Corollary 1.5 states that there are no combinatorial objects that can be counted to give the number of nonequality cases of the Stanley inequality, neither the nonequality cases themselves, nor anything else. The same applies to the equality cases. Fundamentally, this is because you should not be able to efficiently tell if the instances you are observing are the ones you should be counting.

Back to the Alexandrov–Fenchel inequality (AF), the description of equality cases by Shenfeld and van Handel is a breakthrough in convex geometry and gives a complete solution for a large family of (n-tuples of) convex polytopes (see Section 10.11). However, our Theorem 1.1 says that from the computational point of view, the equality cases are intractable in full generality. Colloquially, this says that there is no good description of the equality cases of the Alexandrov–Fenchel inequality, unless the world of computational complexity is not what we think it is. As negative as this may seem, this is what we call a complete solution indeed.