J$. Equivalently, we have that for any $\alpha,\beta\in J$ and any index $i$
satisfying $\alpha_{i}>\beta_{i}$, there is an index $j$ satisfying
$\alpha_{j}<\beta_{j}$ and $\alpha-e_{i}+e_{j}\in J$ and $\beta-e_{j}+e_{i}\in
J$.

We will also not concern ourselves too much with $M$-convex sets. In this
thesis, the only fact that we use related to $M$-convex sets is that the set
of bases of a matroid form a $M$-convex set. Indeed, any M-convex set
restricted to the hypercube is a matroid. For more information about
$M$-convex sets, we refer the reader to [40]. We will consider the
$M$-convexity of the support of a homogeneous polynomial. For a polynomial
$f=\sum_{\alpha\in\Delta_{n}^{d}}c_{\alpha}x^{\alpha}$ in $H_{n}^{d}$, we
define the support of $f$ to be the exponents for which the corresponding
coefficient is non-zero. Specifically, we have

$\operatorname{supp}(f):=\\{\alpha\in\Delta_{n}^{d}:c_{\alpha}\neq 0\\}.$

For example, the Newton polygon of any polynomial will depend only on its
support. We let $M_{n}^{d}$ be the set of degree $d$ homogeneous polynomials
in $\mathbb{R}_{\geq 0}[x_{1},\ldots,x_{n}]$ whose support are $M$-convex.
With these notions on hands, we can now define Lorentzian polynomials
equivalently as the homogeneous polynomials with non-negative coefficients and
$M$-convex support such that all codegree $2$ partial derivatives are stable
polynomials. We write the full description of this set in Definition 2.24.

######  Definition 2.24 (Definition 2.6 in [12]).

We set $L_{n}^{0}=S_{n}^{0}$, $L_{n}^{1}=S_{n}^{1}$, and
$L_{n}^{2}=S_{n}^{2}$. For $d\geq 3$, we define

$\displaystyle L_{n}^{d}$ $\displaystyle=\\{f\in M_{n}^{d}:\partial_{i}f\in
L_{n}^{d-1}\text{ for all }i\in[n]\\}$ $\displaystyle=\\{f\in
M_{n}^{d}:\partial^{\alpha}f\in S_{n}^{2}\text{ for all
}\alpha\in\Delta_{n}^{d-2}\\}.$

######  Definition 2.25.

We call a degree $d$ homogeneous polynomial $f\in\mathbb{R}_{\geq
0}[x_{1},\ldots,x_{n}]$ a Lorentzian polynomial if it satisfies any of the
following equivalent conditions:

  1. (a)

$f\in L_{n}^{d}$.

  2. (b)

There exists a sequence of polynomials $f_{k}\in\underline{L}_{n}^{d}$ such
that $f_{k}\to f$ as $k\to\infty$.

  3. (c)

For any $\alpha\in\mathbb{N}^{n}$ with $|\alpha|\leq d-2$, the polynomial
$\partial^{\alpha}f$ is identically zero or log-concave at any
$a\in\mathbb{R}_{>0}^{n}$.

Polynomials which satisfy condition (c) in Definition 2.25 are also known as
strongly log-concave polynomials. These polynomials were studies in [3]. For
proofs of the equivalences of the conditions in Definition 2.25, we refer the
reader to [12]. As an application of this definition, we will now prove that
the basis generating polynomial of a matroid is Lorentzian.

######  Theorem 2.26.

Let $M$ be a matroid. Then the basis generating polynomial $f_{M}$ is
Lorentzian.

###### Proof.

We prove the result by induction on the size of our matroid. For matroids of
size $2$, the only possible basis generating polynomials are
$x_{1}x_{2},x_{1}+x_{2},x_{1},x_{2}$, and $0$. All of these polynomials are
Lorentzian because they are stable. Now, suppose that the claim holds for all
matroids of smaller size. Since the support of $f_{M}$ is the collection of
bases of a matroid, the support is $M$-convex. Hence, it suffices to prove
that all of its partial derivatives are Lorentzian. For all $i\in E$, we know
that $\partial_{i}f_{M}=f_{M/i}$ for all $i\in E$. From the inductive
hypothesis, the partial derivative is Lorentzian because it the basis
generating polynomial of a smaller matroid. This completes the induction and
suffices for the proof. ∎

In Proposition 2.27, we have compiled some properties of Lorentzian
polynomials. These properties will allow us to prove that certain polynomials
are Lorentzian by relating them to other Lorentzian polynomials.

######  Proposition 2.27.

Let $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ be a Lorentzian polynomial of degree
$d$. Then, $f$ satisfies the following properties:

  1. (a)

Let $A$ be any $n\times m$ matrix with nonnegative entries. Then
$f:\mathbb{R}^{m}\to\mathbb{R}$ defined by $f(y):=f(Ay)$ is Lorentzian.

  2. (b)

For any $a_{1},\ldots,a_{n}\geq 0$, the polynomial
$\sum_{i=1}^{n}a_{i}\partial_{i}f$ is Lorentzian.

  3. (c)

If $f\neq 0$, then $\operatorname{Hess}_{f}(a)$ has exactly one positive
eigenvalue for all $a\in\mathbb{R}_{>0}^{n}$.

###### Proof.

For a proof of these properties, see Theorem 2.10, Corollary 2.11, and
Proposition 2.14 in [12]. ∎

Lorentzian polynomials satisfy log-concavity inequalities similar to that of
Alexandrov’s inequality on mixed discriminants and the Alexandrov-Fenchel
inequality. We present the proofs of two analogs of these log-concavity
inequalities given in [12].

######  Proposition 2.28 (Proposition 4.4 in [12]).

If $f=\sum_{\alpha\in\Delta_{n}^{d}}\frac{c_{\alpha}}{\alpha!}x^{\alpha}$ is a
Lorentzian polynomial, then $c_{\alpha}^{2}\geq
c_{\alpha+e_{i}-e_{j}}c_{\alpha-e_{i}+e_{j}}$ for any $i,j\in[n]$ and any
$\alpha\in\Delta_{n}^{d}$.

###### Proof.

Consider the differential operator $\partial^{\alpha-e_{i}-e_{j}}$. Applying
$D$ to the polynomial $f$, we have

$\displaystyle Df$
$\displaystyle=\sum_{\beta\in\Delta_{n}^{d}}\frac{c_{\beta}}{\beta!}\partial^{\alpha-
e_{i}-e_{j}}x^{\beta}$
$\displaystyle=\frac{c_{\alpha+e_{i}-e_{j}}}{(\alpha+e_{i}-e_{j})!}\cdot\frac{(\alpha+e_{i}-e_{j})!}{2}x_{i}^{2}+\frac{c_{\alpha-
e_{i}+e_{j}}}{(\alpha-e_{i}+e_{j})!}\cdot\frac{(\alpha-
e_{i}+e_{j})!}{2}x_{j}^{2}+\frac{c_{\alpha}}{\alpha!}\cdot\alpha!\cdot
x_{i}x_{j}$
$\displaystyle=\frac{1}{2}\left(c_{\alpha+e_{i}-e_{j}}x_{i}^{2}+2c_{\alpha}x_{i}x_{j}+c_{\alpha-
e_{i}+e_{j}}x_{j}^{2}\right).$

From Proposition 2.27, the polynomial $Df$ is Lorentzian. This suffices for
the proof. ∎

Recall that the mixed discriminant $\mathsf{D}_{n}(A_{1},\ldots,A_{n})$ can be
defined as the polarization form of the polynomial $\det(X)$ where
$X\in\mathbb{R}^{n\times n}$. It happens to be the case that the polarization
of this polynomial satisfies a log-concavity inequality given by the
Alexandrov inequality for mixed discriminants. An analog of this phenomenon
occurs for general Lorentzian polynomials.

######  Proposition 2.29.

Let $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ be a Lorentzian polynomial of degree
$d$. Then, for any $v_{1},v_{2},\ldots,v_{d}\in\mathbb{R}_{\geq 0}^{n}$, we
have that

$F_{f}(v_{1},v_{2},v_{3},\ldots,v_{d})^{2}\geq
F_{f}(v_{1},v_{1},v_{3},\ldots,v_{d})\cdot
F_{f}(v_{2},v_{2},v_{3},\ldots,v_{d}).$

###### Proof.

The polarization identity in Theorem 1.28 gives us the equation

$f(x_{1}v_{1}+\ldots+x_{m}v_{m})=d!\sum_{\alpha\in\Delta_{m}^{d}}\frac{F_{f}(v_{1}[\alpha_{1}],\ldots,v_{m}[\alpha_{m}])}{\alpha!}x^{\alpha}$
(2.3)

The result follows from Proposition 2.28. ∎

######  Remark 3.

In Proposition 4.5 of [12], Bräden-Huh prove a stronger version of Proposition
2.29 where we allow the first vector $v_{1}$ to be any vector in
$\mathbb{R}^{n}$. The proof of this stronger fact uses the Cauchy interlacing
theorem from Section 2.2. This stronger fact also follows from the version we
proved in Proposition 2.29. Indeed, suppose that $v_{1}\in\mathbb{R}^{n}$ and
$v_{2}\in\mathbb{R}_{>0}^{n}$. Then, we can pick $\alpha>0$ sufficiently large
so that $v_{1}+\alpha v_{2}\in\mathbb{R}_{>0}^{n}$. To simplify notation, we
define

$\operatorname{F}(x,y):=F_{f}(x,y,v_{3},\ldots,v_{d}).$

From Proposition 2.29, we have that

$\operatorname{F}(v_{1}+\alpha
v_{2},v_{2})^{2}\geq\operatorname{F}(v_{1}+\alpha v_{2},v_{1}+\alpha
v_{2})\operatorname{F}(v_{2},v_{2}).$

This is equivalent to
$\operatorname{F}(v_{1},v_{2})\geq\operatorname{F}(v_{1},v_{1})\operatorname{F}(v_{2},v_{2})$.
From continuity, the inequality holds for all $v_{2}\in\mathbb{R}_{\geq
0}^{n}$.

######  Remark 4.

Through private discussions with Ramon van Handel, it may be possible to use
the ideas in [53] to give a characterization of the equality cases of
Proposition 2.29. In the expansion of Equation 2.3 for the case $m=n$, by
substituting $v_{i}=e_{i}$ for all $1\leq i\leq n$ we get the equation

$f(x_{1},\ldots,x_{n})=d!\sum_{\alpha\in\Delta_{n}^{d}}\frac{F_{f}(e_{1}[\alpha_{1}],\ldots,e_{n}[\alpha_{n}])}{\alpha!}x^{\alpha}.$

Thus a characterization for the Lorentzian analog of a characterization for
the extremals of the Alexandrov-Fenchel inequality (as given in [53]) would be
useful in characterizing equality in the inequalities given in Proposition
2.28. The characterization would involve checking when the vectors
$e_{1},\ldots,e_{n}$ satisfy the equaliy conditions. We plan to pursue this
direction in future work.

##  Chapter 3 Log-concavity Results for Posets and Matroids

In this chapter, we discuss proofs of several log-concavity results using
mixed discriminants, mixed volumes, and Lorentzian polynomials. In the first
section, we write about Stanley’s poset inequality where he proved that a
sequence which enumerates linear extensions based on where it sends a fixed
element is log-concave. We also sketch the proof of the combinatorial
characterization of Stanley’s inequality given in [53]. We then use the same
proof strategy to completely characterize the extremals of the Kahn-Saks
inequality. This was done in joint work with Ramon van Handel, Xinmeng Zeng,
and the author. In the final section, we study another inequality of Stanley
about the log-concavity of a basis counting sequence for regular matroids. By
combining perspectives from both mixed discriminants and mixed volumes, we
characterize the equality cases of a simplified version of Stanley’s matroid
inequality. We then generalize Stanley’s matroid inequality to all matroids
using the technology of Lorentzian polynomials.

###  3.1 Stanley’s Poset Inequality

Let $(P,\leq)$ be a finite poset on $n$ elements and let $x\in P$ be a
distinguished element in the poset. For every $k\in[n]$, let $N_{k}$ be the
number of linear extensions of $P$ which send $x$ to the index $k$. Then we
can consider properties of the sequence $N_{1},\ldots,N_{n}$. Intuitively, one
would expect that as a function in $[n]$, the sequence should be roughly
unimodal. The fact that the sequence is unimodal is originally a conjecture by
Ronald Rivest in his paper [16]. Rivest proves his conjecture in the case
where $P$ can be covered by two linear orders. In [56], Richard Stanley proves
the full conjecture by proving the stronger claim that the sequence is log-
concave.

######  Theorem 3.1 (Stanley’s Poset Inequality, Theorem 3.1 in [56]).

Let $x_{1}<\ldots<x_{k}$ be a fixed chain in $P$. If $1\leq
i_{1}<\ldots<i_{k}\leq n$, then define $N(i_{1},\ldots,i_{k})$ to be the
number of linear extensions $\sigma:P\to[n]$ satisfying $\sigma(x_{j})=i_{j}$
for $1\leq j\leq k$. Suppose that $1\leq j\leq k$ and
$i_{j-1}+1<i_{j}<i_{j+1}-1$, where $i_{0}=0$ and $i_{k+1}=n+1$. Then

$N(i_{1},\ldots,i_{k})^{2}\geq
N(i_{1},\ldots,i_{j-1},i_{j}-1,i_{j+1},\ldots,i_{k})N(i_{1},\ldots,i_{j-1},i_{j}+1,i_{j+1},\ldots,i_{k}).$

Rivest’s conjecture follows from Theorem 3.1 in the case $k=1$. We write the
statement of this simplified inequality in Corollary 3.2 and call it Stanley’s
Simple Poset Inequality.

######  Corollary 3.2 (Stanley’s Simple Poset Inequality).

The sequence $N_{1},\ldots,N_{n}$ is log-concave. That is, we have
$N_{k}^{2}\geq N_{k-1}N_{k+1}$ for all $k\in\\{2,\ldots,n-1\\}.$

We will present the proof to Corollary 3.2 rather than Theorem 3.1 since they
are essentially the same up to technical computations.

###### Proof of Corollary 3.2.

Define the polytopes $K,L\subseteq\mathbb{R}^{n}$ by

$\displaystyle K$ $\displaystyle:=\\{t\in\mathcal{O}_{P}:t_{\omega}=1\text{ if
}\omega\geq x\\}$ $\displaystyle L$
$\displaystyle:=\\{t\in\mathcal{O}_{P}:t_{\omega}=0\text{ if }\omega\leq
x\\}.$

For any $\lambda\in[0,1]$, we have that

$\operatorname{Vol}_{n-1}\left((1-\lambda)K+\lambda L\right)=\sum_{l\in
e(P)}\operatorname{Vol}_{n-1}\left(\Delta_{l}\cap\left((1-\lambda)K+\lambda
L\right)\right).$

Even though $K,L$ are in $\mathbb{R}^{n}$, they lie in parallel hyperplanes.
This allows us to take $(n-1)$ dimensional volumes and mixed volumes. Now
suppose that $l$ satisfies $l(x)=k$. Then, we have that
$\Delta_{l}\cap\left((1-\lambda)K+\lambda L\right)$ is equal to

$\\{0\leq t_{\pi^{-1}(1)}\leq\ldots\leq t_{\pi^{-1}(k-1)}\leq
1-\lambda\\}\cap\\{t_{\pi^{-1}(k)}=1-\lambda\\}\cap\\{1-\lambda\leq
t_{\pi^{-1}(k+1)}\leq\ldots\leq t_{\pi^{-1}(n)}\leq 1\\}.$

This is essentially the product of two simplices and has volume given by

$\operatorname{Vol}_{n-1}(\Delta_{l}\cap\left((1-\lambda)K+\lambda
L\right))=\frac{(1-\lambda)^{k-1}\lambda^{n-k}}{(k-1)!(n-k)!}.$

Thus, we have the equation

$\displaystyle\operatorname{Vol}_{n-1}\left((1-\lambda)K+\lambda L\right)$
$\displaystyle=\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{N_{k}}{(n-1)!}(1-\lambda)^{k-1}\lambda^{n-k}.$

From Theorem 1.53, we have for $1\leq i\leq n-1$ the equality

$\mathsf{V}_{n-1}(K[i-1],L[n-i])=\frac{N_{i}}{(n-1)!}\implies
N_{i}=(n-1)!\cdot\mathsf{V}_{n-1}(K[i-1],L[n-i]).$

The log-concavity of the sequence $N_{1},N_{2},\ldots,N_{n}$ follows from
Theorem 2.16. ∎

The proof of Corollary 3.2 illustrates one strategy for proving log-concavity
results. To prove that a sequence is log-concave, it is enough to associate
each element in the sequence to some mixed volume. Log-concavity then follows
immediately from the Alexandrov-Fenchel inequality. In the sequel, we will
discuss how to use the results about the extremals of the Alexandrov Fenchel
inequality from [53] to give a combinatorial characterization of the equality
cases of $N_{k}^{2}=N_{k-1}N_{k+1}$ in Stanley’s simple poset inequality.

####  3.1.1 Equality Case for the Simple Stanley Poset Inequality

The equality cases for the Stanley poset inequality was computed in [53] using
their results on the equality cases of the Alexandrov-Fenchel inequality for
polytopes. From the proof of Corollary 3.2, we have the equalities
$N_{k}=(n-1)!\mathsf{V}_{n-1}(K[k],L[n-k-1])$ where the polytopes $K$ and $L$
were defined in the proof. Thus, to find the equality cases of
$N_{k}^{2}=N_{k-1}N_{k+1}$, it is enough to find the equality cases of

$\mathsf{V}_{n-1}(K[k-1],L[n-k])^{2}\geq\mathsf{V}_{n-1}(K[k],L[n-k-1])\cdot\mathsf{V}_{n-1}(K[k-2],L[n-k+1])$
(3.1)

Our goal is to give combinatorial conditions to the poset $P$ so that the
corresponding polytopes $K$ and $L$ satisfy equality in Equation 3.1. We
sketch the proof given in Section 15 in [53] to give an idea on how the
argument works. The same type of argument will be used to charaterize the
extremals of the Kahn-Saks inequality. The combinatorial characterization
given in [53] is compiled in Theorem 3.3.

######  Theorem 3.3 (Theorem 15.3 in [53]).

Let $i\in\\{2,\ldots,n-1\\}$ be such that $N_{i}>0$. Then the following are
equivalent:

  1. (a)

$N_{i}^{2}=N_{i-1}N_{i+1}$.

  2. (b)

$N_{i}=N_{i-1}=N_{i+1}$.

  3. (c)

Every linear extension $\sigma:P\to[n]$ with $\sigma(x)=i$ assigns ranks $i-1$
and $i+1$ to elements of $P$ that are incomparable to $x$.

  4. (d)

$|P_{<y}|>i$ for all $y\in P_{>x}$, and $|P_{>y}|>n-i+1$ for all $y\in
P_{<x}$.

In Theorem 3.3, it is not difficult to prove that
$(d)\implies(c)\implies(b)\implies(a)$. The main difficulty lies in proving
that equality and positivity in $N_{i}^{2}=N_{i-1}N_{i+1}$ implies the
combinatorial conditions in (d). From here, let us assume that
$N_{i}^{2}=N_{i-1}N_{i+1}$ and $N_{i}>0$. From Lemma 1.57, we get necessary
and sufficient conditions to guarantee $N_{i}>0$.

######  Lemma 3.4 (Lemma 15.2 in [53]).

For any $i\in[n]$, we have that $N_{i}=0$ if and only if $|P_{<x}|>i-1$ or
$|P_{>x}|>n-i$.

###### Proof.

Recall that $N_{i}=\mathsf{V}_{n-1}(\underbrace{K,\ldots,K}_{i-1\text{
times}},\underbrace{L,\ldots,L}_{n-i\text{ times}})$. From Lemma 1.57, we know
that $N_{i}>0$ if and only if $\dim K\geq i-1$, $\dim L\geq n-i$, and
$\dim(K+L)\geq n-1$. From the definition of $K$ and $L$, we can compute

$\displaystyle\dim K$ $\displaystyle=n-1-|P_{>x}|$ $\displaystyle\dim L$
$\displaystyle=n-1-|P_{<x}|$ $\displaystyle\dim(K+L)$ $\displaystyle=n-1.$

To see the details behind this computation, see Lemma 15.7 in [53]. Thus,
$N_{i}=0$ if and only if $|P_{>x}|>n-i$ and $|P_{<x}|>i-1$. ∎

We get inequalities on the dimensions of $\dim K$ and $\dim L$ from the
assumption $N_{i}>0$. However, since $N_{i}^{2}=N_{i-1}N_{i+1}$, we also have
$N_{i-1},N_{i+1}>0$. This gives slightly stronger inequalities from Lemma 3.4
applied to $N_{i-1}$ and $N_{i+1}$. Specifically, we get the inequalities

$\displaystyle|P_{<x}|\leq i-2,\quad\text{and }\quad|P_{>x}|\leq n-i-1.$

With these inequalities and Theorem 2.20 we get a specialized version of
Corollary 2.16 in [53].

######  Lemma 3.5 (Lemma 15.8 in [53]).

Let $i\in\\{2,\ldots,n-1\\}$ be such that $N_{i}>0$ and
$N_{i}^{2}=N_{i+1}N_{i-1}$. Then $|P_{<x}|\leq i-2$, $|P_{>x}|\leq n-i-1$, and
there exist $a>0$ and $v\in\mathbb{R}^{n-1}$ so that

$h_{K}(x)=h_{aL+v}(x)\text{ for all
}x\in\operatorname{supp}S_{B,K[i-2],L[n-i-1]}.$

In the beginning of our analysis, the convex bodies $K$ and $L$ were contained
in parallel hyperplanes of $\mathbb{R}^{n-1}$. When studying equality via
Lemma 3.5, we project these bodies to the same copy of $\mathbb{R}^{n-1}$. The
vector $v\in\mathbb{R}^{n-1}$ will also lie in the same hyperplane as our
convex bodies $K$ and $L$. Lemma 3.5 will be the main tool to extract
combinatorial information about $P$ from the equality in the Alexandrov-
Fenchel inequality. Recall in Definition 1.58, we are giving a
characterization for vectors in the support of the mixed area measure. By
specializing this result in the Stanley case, we get the result in Lemma 3.6.

######  Lemma 3.6.

Let $u\in\mathbb{R}^{n-1}$ be a unit vector. Then the following are
equivalent.

  1. (a)

$u\in\operatorname{supp}S_{B,K[i-2],L[n-i-1]}$.

  2. (b)

$\dim F_{K}(u)\geq i-2$, $\dim F_{L}(u)\geq n-i-1$, and $\dim F_{K+L}(u)\geq
n-3$.

From here, the argument will involve finding suitable vectors
$u\in\mathbb{S}^{n-2}\subseteq\mathbb{R}^{n-1}$ and applying Lemma 3.5. The
suitability of the vector $u\in\mathbb{S}^{n-2}$ will depend on the
corresponding inequalities in Lemma 3.6. When we find a “suitable” vector
$u\in\mathbb{S}^{n-2}$, one of two things generally occur. In the first case,
the vector $u$ will unequivocally satisfy the dimension inequalities in Lemma
3.6 and from Lemma 3.5 we get the relation $h_{K}(u)=ah_{L}(u)+\langle
v,u\rangle$. In this case, the identity we get usually tells us something
about $v$ or $a$. In the second case, it is unclear whether or not the vector
$u$ will satisfy the dimension inequalities in Lemma 3.6. The dimension
inequalities will usually involve some statistics about the combinatorial
structure of our object. For example, it might involve the number of elements
greater than another, or the number of elements between two elements in a
poset. In this case, we know that if the dimension inequalities in Lemma 3.6
are satisfied then we get the identity in Lemma 3.5. If we are lucky, the
resulting identity we get will contradict some information that we know is
true from the vectors in the first case. This will imply that at least one of
the dimension inequalities are incorrect, giving a combinatorial condition
that must be satisfied for equality.

In the Stanley case, vectors in the first case are vectors of the form
$-e_{\omega}$ where $\omega$ is a minimal element, $e_{\omega}$ where $\omega$
is a maximal element, and some vectors of the form
$e_{\omega_{1},\omega_{2}}:=\frac{e_{\omega_{1}}-e_{\omega_{2}}}{\sqrt{2}}$
where $\omega_{1},\omega_{2}\in P$ are elements in the poset such that
$\omega_{1}\lessdot\omega_{2}$. From the first two vectors, we get that
$v_{\omega}=1-a$ for any maximal element $\omega$, the second vector gives
$v_{\omega}=0$ for any minimal element $\omega$. The third type of vector
tells us that in some situations where $\omega_{1}\lessdot\omega_{2}$ we have
$v_{\omega_{1}}=v_{\omega_{2}}$. From the proof of Corollary 15.11 of [53], it
is possible to create a chain from a minimal element to a maximal element such
as the adjacent relations are covering relations where the corresponding
vector $e_{\omega_{1},\omega_{2}}$ is in the support of the mixed area
measure. This implies that all of the coordinates of $v$ with respect to this
chain are equal to each other. In particular, the coordinate at the minimal
element will be equal to the coordinate at the maximal element (that is,
$a=1$). This result is interesting in its own right. We have just proved that
if $N_{i}^{2}=N_{i-1}N_{i+1}$ and $N_{i}>0$, then $N_{i+1}=N_{i}=N_{i-1}$.

Vectors in the second case consist of vectors of the form $-e_{\omega}$ where
$\omega$ is a minimal element in $P_{>x}$ and also vectors of the form
$e_{\omega}$ where $\omega$ is a maximal element in $P_{<x}$. From the
information that $a=1$, it is not difficult to show that the equality that we
would get as a result of these vectors being in the support of the mixed area
measure give contradictions. Thus, some of the dimension inequalities
corresponding to these vectors must be incorrect. This is enough to give the
combinatorial characterization in Theorem 3.3. During the present discussion,
we did not show any of the computations for the sake of time and space. To see
the computations in full detail and rigour, we refer the reader to the
original source [53].

###  3.2 Extremals of the Kahn-Saks Inequality

This section covers joint work by Ramon van Handel, Xinmeng Zeng, and the
author. We consider a slight generalization of the simple version of the
Stanley poset inequality called the Kahn-Saks Inequality. We also cover
aspects of the proof of our characterization of the equality cases. In the
paper [34], Jeff Kahn and Michael Saks prove that any finite poset contains a
pair of elements $x$ and $y$ such that the proportion of linear extensions of
$P$ in which $x$ lies below $y$ is between $\frac{3}{11}$ and $\frac{8}{11}$.
This fact has consequences in theoretic computer science and sorting
algorithms. The interested reader should consult the original source [34] for
these applications as they are tangential to our thesis. One ingredient in
their proof is a log-concavity inequality for a linear extension counting
sequence. Let $P$ be a finite poset and fix elements $x,y\in P$ with $x\leq
y$. Let $N_{k}$ be the number of linear extensions $f$ satisfying
$f(y)-f(x)=k$ for all $1\leq k\leq n-1$. The Kahn-Saks inequality written in
Theorem 3.7 states that the sequence $N_{1},\ldots,N_{n-1}$ is log-concave.

######  Theorem 3.7 (Kahn-Saks Inequality, Theorem 2.5 in [34]).

For all $k\in\\{2,\ldots,n-2\\}$, we have $N_{k}^{2}\geq N_{k-1}N_{k+1}$.

In the case where $x$ is an 0 element in $P$, Theorem 3.7 reduces to Corollary
3.2 since $x$ will be forced to be the lowest element in any linear extension.
The proof of Theorem 3.7 is similar to the proof of Stanley’s inquality. To
prove log-concavity, we will associate the sequence $N_{k}$ to a sequence of
mixed volumes of suitable polyopes. The polytopes that we use will be cross-
sections of the order polytope $\mathcal{O}_{P}$. For all
$\lambda\in\mathbb{R}$, we define the polytope $K_{\lambda}$ as

$K_{\lambda}:=\\{t\in\mathcal{O}_{P}:t_{y}-t_{x}=\lambda\\}.$

Each of these cross-sections can be written as a convex combination of $K_{0}$
and $K_{1}$. We defer the proof of $(1-\lambda)K_{0}+\lambda
K_{1}=K_{\lambda}$ to Lemma 5.2. From [34], we know that
$N_{k}=(n-1)!\mathsf{V}_{n-1}(K_{0}[n-k],K_{1}[k-1])$. The mixed volume is
well-defined since $K_{0}$ and $K_{1}$ lie in parallel hyperplanes. We give
the full computation of this fact in Lemma 5.3 in the appendix. The Kahn-Saks
inequality then follows immediately from the Alexandrov-Fenchel inequality. In
the next section, we begin our combinatorial characterization of the extremals
of the Kahn-Saks inequality. Our main results are Theorem 3.9, Theorem 3.10,
and Theorem 3.11.

####  3.2.1 Simplifications and Special Regions

Let $(P,\leq)$ be our poset with distinguished elements $x,y\in P$ satisfying
$x\not\geq y$. We can always add a $0$ element and $1$ element to our poset.
In any linear extension, these two elements will be forced to be placed in the
beginning and end of the linear extension. Hence, adding these two elements to
the poset will not change the values of the sequence $\\{N_{k}\\}$. If $x||y$,
Lemma 5.9 and Lemma 5.10 in the appendix prove that for the poset generated by
our original relations and the new relation $x\leq y$, our sequence $N_{k}$
will remain the same. Thus, we will assume that our poset $P$ has a $0$
element, has a $1$ elements, and satisfies $x\leq y$. In our poset, we will
define the several special regions which each need to be handled separately in
our analysis.

$\displaystyle\mathsf{END}_{x}$ $\displaystyle:=\\{\omega\in
P:\omega<x\\}=P_{<x}$ $\displaystyle\mathsf{END}_{y}$
$\displaystyle:=\\{\omega\in P:\omega>y\\}=P_{>y}$ $\displaystyle\mathsf{MID}$
$\displaystyle:=\\{\omega\in P:x<\omega<y\\}=P_{x<\cdot<y}$
$\displaystyle\mathsf{MID}_{x}$ $\displaystyle:=\\{\omega\in P:x<\omega\text{
and }\omega||y\\}=P_{\cdot>x,\cdot||y}$ $\displaystyle\mathsf{MID}_{y}$
$\displaystyle:=\\{\omega\in P:\omega<y\text{ and
}\omega||x\\}=P_{\cdot<y,\cdot||x}$

These regions are disjoint, but they do not account for all elements in $P$.
For a full partition of our poset $P$ we have

$P=\mathsf{END}_{x}\sqcup\mathsf{END}_{y}\sqcup\mathsf{MID}_{x}\sqcup\mathsf{MID}_{y}\sqcup\mathsf{MID}\sqcup\\{x,y\\}\sqcup
P_{\not\sim x,y}$

where the subset $P_{\sim x,y}$ consists of the elements in $P$ which are not
comparable to either $x$ or $y$.

####  3.2.2 Combinatorial characterization of the extremals

Some of our main results about the extremals of the Kahn-Saks inequality were
conjectured by Swee Hong Chan, Igor Pak, and Greta Panova in their paper [15].
In their paper, they define the notions of the midway and dual-midway
properties recorded in Definition 3.8.

######  Definition 3.8.

We say that $(x,y)$ satisfies the $k$-midway property if

  * •

$|P_{<z}|+|P_{>y}|>n-k$ for every $z\in P$ such that $x<z$ and $z\not>y$,

  * •

$|P_{z<\cdot<y}|>k$ for every $z<x$.

Similarly, we say that $(x,y)$ satisfies the dual $k$-midway property if:

  * •

$|P_{>z}|+|P_{<x}|>n-k$ for every $z\in P$ such that $z<y$ and $z\not<x$, and

  * •

$|P_{x<\cdot<z}|>k$ for every $z>y$.

For any $z\in P$, we say that $z$ satisfies the $k$-midway property or
satisfies the dual $k$-midway property if the relevant inequality for $z$ in
the corresponding property is satisfied. For example, if $z<x$, we would say
that $z$ satisfies the $k$-midway property if $|P_{z<\cdot<y}|>k$.

In Conjecture 8.7 in [15], Chan-Pak-Panova conjecture that when $N_{k}>0$, the
equality case $N_{k+1}=N_{k}=N_{k-1}$ occurs if and only if the midway
property holds or the dual midway property holds. Using purely combinatorial
methods, they prove this conjecture for width two posets where $x$ and $y$ are
elements in the same chain in the two chain decomposition. By appealing to the
extremals of the Alexandrov-Fenchel inequality, we are able to prove this
conjecture in its full generality. We are also able to find all possible
values of $a$ for which $N_{i+1}=aN_{i}=a^{2}N_{i-1}$, and give combinatorial
characterizations for the equality cases in all possible values of $a$. We now
state our main results.

######  Theorem 3.9.

If $N_{k}>0$ and $N_{k}^{2}=N_{k-1}N_{k+1}$, then we either have
$N_{k+1}=N_{k}=N_{k-1}$ or $N_{k+1}=2N_{k}=4N_{k-1}$.

######  Theorem 3.10.

If $N_{k}>0$, then the following are equivalent:

  1. (a)

$N_{k-1}=N_{k}=N_{k+1}$.

  2. (b)

$(x,y)$ satisfies the $k$-midway property or the dual $k$-midway property.

######  Theorem 3.11.

If $N_{k}>0$, then $N_{k+1}=2N_{k}=4N_{k-1}$ if and only if the following
properties hold:

  1. (i)

$k$-midway holds for $\mathsf{END}_{x}$ and dual $k$-midway holds for
$\mathsf{END}_{y}$.

  2. (ii)

Every $z\in P$ is comparable to $x$ and $y$.

  3. (iii)

$\mathsf{MID}$ is empty.

  4. (iv)

For every $z\in\mathsf{MID}_{y}$ and $z^{\prime}\in\mathsf{MID}_{x}$ with
$z<z^{\prime}$, we have that $|P_{z<\cdot<y}|+|P_{x<\cdot<z^{\prime}}|\geq
k-1$.

######  Remark 5.

Theorem 3.10 resolves Conjecture 8.7 in [15]. In their paper, Chan-Pak-Panova
prove the important implication $(b)\implies(a)$. Since this result is central
for the full characterization, we record it in Proposition 3.12.

######  Proposition 3.12 (Theorem 8.9, Proposition 8.8 in [15]).

If $(x,y)$ satisfies the $k$-midway property or the dual $k$-midway property,
then $N_{k-1}=N_{k}=N_{k+1}$.

The proof strategy of these results will be similar to characterization of the
equality cases of the Stanley inequality in [53]. In this discussion, let us
assume that $N_{k}^{2}=N_{k-1}N_{k+1}$ and $N_{k}>0$. Recall in the proof of
Theorem 3.7, we showed the identity

$N_{k}=(n-1)!\mathsf{V}_{n-1}\left(\underbrace{K_{0},\ldots,K_{0}}_{n-k\text{
times}},\underbrace{K_{1},\ldots,K_{1}}_{k-1\text{ times}}\right).$

Then, we have equality in $N_{k}^{2}=N_{k-1}N_{k+1}$ if and only if we have
equality in the corresponding Alexadrov-Fenchel inequality:

$\mathsf{V}_{n-1}(K_{0}[n-k],K_{1}[k-1])^{2}\geq\mathsf{V}_{n-1}(K_{0}[n-k-1],K_{1}[k])\cdot\mathsf{V}_{n-1}(K_{0}[n-k+1],K_{1}[k-2]).$

Using Lemma 1.57, we have the characterization of positivity given in
Proposition 3.13. We omit the proof since we have prove a similar positivity
result in the Stanley case.

######  Proposition 3.13.

The following are equivalent.

  1. (a)

$N_{k}=0$,

  2. (b)

$\dim K_{0}<n-k$ or $\dim K_{1}<k-1$,

  3. (c)

$|P_{<x}|+|P_{>y}|>n-k-1$ or $|P_{x<\cdot<y}|>k-1$.

Since $N_{k}^{2}=N_{k-1}N_{k+1}$ and $N_{k}>0$, we also have that $N_{k-1}>0$
and $N_{k+1}>0$. From Proposition 3.13 this gives us stronger inequalities on
$\dim K_{0}$ and $\dim K_{1}$. From Proposition 5.4, we know that

$\displaystyle\dim K_{0}$ $\displaystyle=n-|P_{x<\cdot<y}|-1$
$\displaystyle\dim K_{1}$ $\displaystyle=n-|P_{<x}|-|P_{>y}|-2$
$\displaystyle\dim(K_{0}+K_{1})$ $\displaystyle=n-1.$

Let $V=(e_{y}-e_{x})^{\perp}\cong\mathbb{R}^{n-1}$ be the orthogonal
complement of the vector $e_{y}-e_{x}$. The polytopes $K_{0}$ and $K_{1}$ are
contained in parallel copies of $V$. We can specialize Theorem 2.20 to the
Kahn-Saks case and prove Theorem 3.14. This will be the main workhorse theorem
to prove the combinatorial characterization for the equality case of the Kahn-
Saks inequality.

######  Theorem 3.14.

Let $k\in\\{2,\ldots,n-2\\}$ such that $N_{k}>0$ and
$N_{k}^{2}=N_{k-1}N_{k+1}$. Then

$1+|P_{x<\cdot<y}|<k<n-|P_{<x}|-|P_{>y}|-1$

and there exists a positive scalar $a>0$ and $v\in V$ such that
$h_{K_{0}}(u)=h_{aK_{1}+v}(u)$ for all
$u\in\operatorname{supp}S_{B,K_{0}[n-k-1],K_{1}[k-2]}$.

###### Proof.

Since $N_{k}>0$, we know that $N_{k-1},N_{k+1}>0$. Proposition 3.13 applied to
$N_{k-1}$ and $N_{k+1}$ implies that $\dim K_{0}\geq n-k+1$ and $\dim
K_{1}\geq k$. Thus, $\mathcal{P}:=(K_{0}[n-k-1],K_{1}[k-2])$ is supercritical.
The theorem then follows from Theorem 2.20. ∎

To slightly simplify our notation, we let $\mu:=S_{B,K_{0}[n-k-1],K_{1}[k-2]}$
be the mixed area measure on the copy of $\mathbb{S}^{n-2}$ in $V$. The next
lemma is a simple criterion for when a vector lies in the support of the mixed
area measure.

######  Lemma 3.15.

For any vector $u\in V$, we have that $u\in\operatorname{supp}\mu$ if and only
if

$\displaystyle\dim F(K_{0},u)$ $\displaystyle\geq n-k-1$ $\displaystyle\dim
F(K_{1},u)$ $\displaystyle\geq k-2$ $\displaystyle\dim F(K_{0}+F_{1},u)$
$\displaystyle\geq n-3.$

###### Proof.

This is an application of Lemma 2.3 in [53]. ∎

####  3.2.3 Transition Vectors

In this section, we assume that the hypothesis of Theorem 3.14 is true. That
is, we have $N_{k}>0$ and $N_{k}^{2}=N_{k-1}N_{k+1}$. We can label our poset
$(P,\leq)$ as $P=\\{z_{1},\ldots,z_{n}\\}$ where $x=z_{n-1}$ and $y=z_{n}$.
Let $v=(v_{1},\ldots,v_{n})$ where $v_{i}$ is the coordinate of $v$ with
respect to the coordinate indexed by $z_{i}$. When it is more convenient to
label the coordinate by the actual poset element itself, we use the notation
$v_{z_{i}}$. In Definition 3.16, we define the notion of transition vectors.
These are vectors in $\operatorname{supp}\mu$ which, from Corollary 3.17,
allows us to equate the two coordinates of $v$ of the corresponding pair of
poset elements.

######  Definition 3.16.

For $z_{i},z_{j}\in P\backslash\\{x,y\\}$, we define the vector

$e_{ij}:=\frac{1}{\sqrt{2}}(e_{i}-e_{j})\in V.$

We call $e_{ij}$ a transition vector if $z_{i}\lessdot z_{j}$ and
$e_{ij}\in\operatorname{supp}\mu$. We also use the notation $e_{z_{i},z_{j}}$
in some cases.

Transition vectors will imply that the coordinates of the corresponding pair
of poset elements are equal because the support function values of transition
vectors will always be $0$. Indeed, for $z_{i}\lessdot z_{j}$, we can compute

$\displaystyle h_{K_{0}}(e_{ij})$ $\displaystyle=\frac{1}{\sqrt{2}}\sup_{t\in
K_{0}}\langle t,e_{i}-e_{j}\rangle=\frac{1}{\sqrt{2}}\sup_{t\in
K_{0}}t_{i}-t_{j}\leq 0$ $\displaystyle h_{K_{1}}(e_{ij})$
$\displaystyle=\frac{1}{\sqrt{2}}\sup_{t\in K_{1}}\langle
t,e_{i}-e_{j}\rangle=\frac{1}{\sqrt{2}}\sup_{t\in K_{1}}t_{i}-t_{j}\leq 0$

where the last inequalities follow from the fact that $z_{i}\lessdot z_{j}$
and the parameter $t$ lies in the order polytope $\mathcal{O}_{P}$. Since we
have $(0,\ldots,0)\in K_{0}$ we have $h_{K_{0}}(e_{ij})=0$. To prove that
$h_{K_{1}}(e_{ij})=0$, note that $z_{i}\lessdot z_{j}$ implies that it cannot
be the case that $z_{i}<x$ and $z_{j}>y$. Since $\sum_{z_{i}\geq
y}e_{i},\sum_{z_{i}\not\leq x}e_{i}\in K_{1}$, we have $h_{K_{1}}(e_{ij})=0$.
The equality $v_{i}=v_{j}$ then follows from Theorem 3.14.

######  Corollary 3.17.

Suppose that the hypothesis of Theorem 3.14 is true and let $v\in V$ be the
vector in the theorem. If $e_{ij}$ is a support vector, then $v_{i}=v_{j}$.

###### Proof.

From Theorem 3.14, we have that
$h_{K_{0}}(e_{ij})=h_{aK_{1}+v}(e_{ij})=ah_{K_{1}}(e_{ij})+\langle
v,e_{ij}\rangle$. Since $h_{K_{0}}(e_{ij})=h_{K_{1}}(e_{ij})=0$ and $\langle
v,e_{ij}\rangle=v_{i}-v_{j}$, we have $v_{i}=v_{j}$. This proves the lemma. ∎

In Lemma 3.18, we compile a list of transition vectors.

######  Lemma 3.18.

Suppose that the hypothesis in Theorem 3.14 is true. Then, we have the
following transition vectors:

  1. (a)

Let $R=\mathsf{END}_{x}$ or $R=\mathsf{END}_{y}$. If $z_{i},z_{j}\in R$
satisfy $z_{i}\lessdot z_{j}$, then $e_{ij}$ is a transition vector.

  2. (b)

Let $R=\mathsf{MID}_{x}$ or $R=\mathsf{MID}_{y}$. If $z_{i},z_{j}\in R$
satisfy $z_{i}\lessdot z_{j}$, then $e_{ij}$ is a transition vector.

  3. (c)

If $z_{i}\in\mathsf{MID}_{x}$ and $z_{j}\in\mathsf{END}_{y}$ such that
$z_{i}\lessdot z_{j}$ and there does not exist $z_{l}\in\mathsf{MID}_{x}$ with
$z_{i}\lessdot z_{l}$, then $e_{ij}$ is a transition vector.

  4. (d)

If $z_{j}\in\mathsf{MID}_{y}$ and $z_{i}\in\mathsf{END}_{x}$ such that
$z_{i}\lessdot z_{j}$ and there does not exist $z_{l}\in\mathsf{MID}_{y}$ with
$z_{l}\lessdot z_{j}$, then $e_{ij}$ is a transition vector.

  5. (e)

If $z_{i},z_{j}\in\mathsf{MID}$ such that $z_{i}\lessdot z_{j}$, then $e_{ij}$
is a transition vector.

###### Proof.

We only prove the result when $R=\mathsf{END}_{x}$ since the proofs of the
other regions are relatively similar. A full proof of the result will be given
in an upcoming paper with Ramon van Handel and Xinmeng Zeng. Before computing
the dimensions of these faces, we first prove that there exists a linear
extension $f:P\to[n]$ with $f(z_{j})-f(z_{i})=1$ and
$f(y)-f(x)=1+|P_{x<\cdot<y}|$. From Proposition 5.6, there is a linear
extension $\tilde{f}:P\to[n]$ with
$\tilde{f}(y)-\tilde{f}(x)=|P_{x<\cdot<y}|+1$. Both $z_{i}$ and $z_{j}$ will
be located to the left of $x$ in the linear extension $\tilde{f}$. From
Proposition 5.8, we can modify $\tilde{f}$ by only changing the elements
between $z_{i}$ and $z_{j}$ in the linear extension $\tilde{f}$ to a linear
extension $f$ satisfying $f(z_{j})-f(z_{i})=1$. Clearly, we have the following
set inclusion

$F_{K_{0}}(e_{ij})\supseteq\Delta_{f}\cap\\{t_{i}=t_{j},t_{x}=t_{y}\\}.$

Taking the affine span of both sets, we have

$\displaystyle\operatorname{aff}F_{K_{0}}(e_{ij})$
$\displaystyle\supseteq\operatorname{aff}\Delta_{f}\cap\\{t_{i}=t_{j},t_{x}=t_{y}\\}$
$\displaystyle=\mathbb{R}\left[\sum_{\omega\in P_{x\leq\cdot\leq
y}}e_{\omega}\right]\oplus\mathbb{R}[e_{i}+e_{j}]\oplus\bigoplus_{\omega\in
P\backslash(P_{x\leq\cdot\leq
y}\cup\\{z_{i},z_{j}\\})}\mathbb{R}[e_{\omega}].$

This gives us the following bounds for the dimensions of $F_{K_{0}}(e_{ij})$
and $F_{K_{1}}(e_{ij})$:

$\displaystyle\dim F_{K_{0}}(e_{ij})$ $\displaystyle\geq
n-|P_{x<\cdot<y}|-2=\dim K_{0}-1$ $\displaystyle\dim F_{K_{1}}(e_{ij})$
$\displaystyle=\dim K_{1}.$

From Lemma 5.11, we have $\dim
F_{K_{0}+K_{1}}(e_{ij})\geq\dim(K_{1}+K_{0})-1=n-2$. From the bounds in
Theorem 3.14, we have

$\displaystyle\dim F_{K_{0}}(e_{ij})$ $\displaystyle=n-|P_{x<\cdot<y}|-2\geq
n-k$ $\displaystyle\dim F_{K_{1}}(e_{ij})$
$\displaystyle=n-|P_{<x}|-|P_{>y}|-2\geq k.$

From Lemma 3.15, we know that $e_{ij}\in\operatorname{supp}\mu$. This proves
that $e_{ij}$ is a transition vector. ∎

For any result about support vectors, the calculations needed to determine
whether or not certain vectors are support vectors look very similar to the
calculation performed in our proof of Lemma 3.18. The strategy is to find
“extremal linear extensions” where the intersection between the corresponding
simplex and the faces of our polytopes has large dimension. To avoid
reptitiveness and bogging the thesis with technical details, we omit a large
portion of the support vector calculations.

######  Lemma 3.19.

Suppose that the hypothesis of Theorem 3.14 holds. Let $z_{i}\in
P\backslash\\{x,y\\}$ be an element of the poset. If $z_{i}$ is minimal, then
$v_{i}=0$. If $z_{i}$ is maximal, then $v_{i}=1-a$.

###### Proof.

If $z_{i}\in P$ is a minimal element, then we can show that $-e_{i}$ is in the
support of the mixed area measure. If $z_{i}\in P$ is a maximal element, then
we can show that $e_{i}$ is in the support of the mixed area measure. These
facts will imply via Theorem 3.14 that $v_{i}=0$ when $z_{i}$ is minimal and
$v_{i}=1-a$ when $z_{i}$ is maximal. ∎

Using Lemma 3.18 and Lemma 3.19, we can compute some of the coordinates of the
vector $v$ depending on the region containing the element poset corresponding
to the coordinate.

######  Corollary 3.20.

Let $\omega\in P\backslash\\{x,y\\}$ be some element in our poset.

  1. (a)

If $\omega\in\mathsf{END}_{x}\cup\mathsf{MID}_{y}$, then $v_{\omega}=0$;

  2. (b)

If $\omega\in\mathsf{END}_{y}\cup\mathsf{MID}_{x}$, then $v_{\omega}=1-a$;

  3. (c)

If $\omega_{1},\omega_{2}\in\mathsf{MID}$ are comparable, then
$v_{\omega_{1}}=v_{\omega_{2}}$.

###### Proof.

Suppose that $\omega\in\mathsf{END}_{x}$. If $\omega$ is a minimal element,
then we already know that $v_{\omega}=0$ from Lemma 3.19. Suppose that
$\omega$ is not a minimal element. Then there is a sequence of poset elements
$\omega_{1},\omega_{2},\ldots,\omega_{l}$ for $l\geq 1$ such that
$\omega_{1}\lessdot\omega$, $\omega_{i+1}\lessdot\omega_{i}$ for all $i$, and
$\omega_{l}$ is a minimal element. From Lemma 3.18(a), we know that
$e_{\omega_{1},\omega}$ and $e_{\omega_{i+1},\omega_{i}}$ are transition
vectors for all $i$. From Corollary 3.17 and Corollary 3.19 applied to
$\omega_{l}$, we have $v_{\omega}=v_{\omega_{1}}=\ldots=v_{\omega_{l}}=0$. A
similar proof will work when $\omega\in\mathsf{END}_{y}$, but instead we built
a maximal chain to a maximal element.

Now, suppose that $\omega\in\mathsf{MID}_{x}$. If $\omega$ is maximal, then we
automatically know that $v_{\omega}=1-a$. If $\omega$ is not maximal, we can
build a chain
$\omega\lessdot\omega_{1}\lessdot\omega_{2}\lessdot\ldots\lessdot\omega_{l}$
where $\omega_{l}$ is a maximal element by using a special procedure to
guarentee that all adjacent elements correspond to transition vectors. To
illustrate this procedure, suppose that we have picked
$\omega\lessdot\omega_{1}\lessdot\ldots\lessdot\omega_{i}$ up to some $i$.
Then $\omega_{i}\in\mathsf{END}_{y}$ or $\omega_{i}\in\mathsf{MID}_{x}$. If
$\omega_{i}\in\mathsf{END}_{y}$, then we can continue to a maximal element
arbitrarily. Otherwise, if $\omega_{i}\in\mathsf{MID}_{x}$, then we pick
$\omega_{i+1}$ to be an element of $\mathsf{MID}_{x}$ which covers
$\omega_{i}$. If none exists, we then pick $\omega_{i+1}$ to be an element of
$\mathsf{END}_{y}$ which covers $\omega_{i}$. If this doesn’t exist again,
then we know that $\omega_{i}$ is a maximal element. By our construction and
Lemma 3.18, we know that the adjacent poset elements in the chain form
transition vectors. This implies that
$v_{\omega}=v_{\omega_{1}}=\ldots=v_{\omega_{l}}=1-a$. A similar proof works
for $\mathsf{MID}_{y}$ but instead we create the same type of chain to a
minimal element.

For the result on $\mathsf{MID}$, suppose that
$\omega_{1},\omega_{2}\in\mathsf{MID}$ are comparable. Since any two
comparable elements in $\mathsf{MID}$ can be connected by a chain completely
contained in $\mathsf{MID}$, it suffices to prove the result when
$\omega_{1}\lessdot\omega_{2}$. In this case, from Lemma 3.18(e), we know that
$v_{\omega_{1},\omega_{2}}$ is a transition vector. Hence
$v_{\omega_{1}}=v_{\omega_{2}}$. This suffices for the proof. ∎

To conclude the section on transition vectors, we give conditions for when a
vector of the form $e_{ij}$ is a transition vector from $\mathsf{MID}$ to
$\mathsf{MID}_{x}$ or $\mathsf{MID}_{y}$.

######  Proposition 3.21.

Let $z_{i}\in P\backslash\\{x,y\\}$ be a poset element satisfying $x<z_{i}<y$.

  1. (a)

If $z_{j}\in P$ satisfies $z_{j}\in\mathsf{MID}_{y}$ and $z_{j}\lessdot
z_{i}$, then $e_{ji}$ is a transition vector if $|\mathsf{MID}|+|S|+1\leq k$
where we define the set $S$ to be

$S=P_{z_{i}<\ldots<y}\backslash\mathsf{MID}=\\{z:z_{i}<z<y,z\notin\mathsf{MID}\\}.$

  2. (b)

If $z_{j}\in P$ satisfies $z_{j}\in\mathsf{MID}_{x}$ and $z_{j}\gtrdot z_{i}$,
then $e_{ij}$ is a transition vector if $|\mathsf{MID}|+|S|+1\leq k$ where we
define the set $S$ to be

$S=P_{x<\cdot<z_{i}}\backslash\mathsf{MID}=\\{z:x<z<z_{i},z\notin\mathsf{MID}\\}.$

###### Proof.

We only prove (a) as the proof of (b) is similar. Note that

$\displaystyle\operatorname{aff}F_{K}(e_{ji})$
$\displaystyle=\bigoplus_{\omega\notin\mathsf{MID}\sqcup
S\sqcup\\{x,y,z_{i}\\}}\mathbb{R}[e_{\omega}]\oplus\mathbb{R}\left[\sum_{\omega\in\mathsf{MID}\sqcup
S\sqcup\\{x,y,z_{i}\\}}e_{\omega}\right]$
$\displaystyle\operatorname{aff}F_{L}(e_{ji})$
$\displaystyle=\bigoplus_{\omega\notin\\{z_{i},z_{j}\\}\sqcup P_{\leq x}\sqcup
P_{\geq y}}\mathbb{R}[e_{\omega}]\oplus\mathbb{R}[e_{i}+e_{j}]+\sum_{\omega\in
P_{\geq y}}e_{\omega}$ $\displaystyle\operatorname{aff}F_{K+L}(e_{ji})$
$\displaystyle=\bigoplus_{\omega\notin\\{z_{i},z_{j},x,y\\}}\mathbb{R}[e_{\omega}]\oplus\mathbb{R}\left[\sum_{\omega\in\mathsf{MID}\sqcup
S\sqcup\\{x,y,z_{i}\\}}e_{\omega}\right]\oplus\mathbb{R}[e_{i}+e_{j}]\oplus\sum_{\omega\in
P_{\geq y}}e_{\omega}.$

The only obstruction to the vector $e_{ji}$ being in the support is the
dimension inequality corresponding to $\dim\operatorname{aff}F_{K}(e_{ji})$.
Specifically, we need the inequality

$n-|\mathsf{MID}|-|S|-2=\dim F_{K}(e_{ij})\geq
n-k-1\iff|\mathsf{MID}|+|S|+1\leq k.$

This suffices for the proof. ∎

####  3.2.4 More Vectors

In this section, we introduce a new type of vector not found in the Stanley
case that will allow us to extract the midway and dual-midway inequalities
from our poset.

######  Definition 3.22.

For $j\in\\{1,\ldots,n-2\\}$, define

$\displaystyle
u_{j}^{+}:=\sqrt{\frac{2}{3}}\left(e_{j}-\frac{1}{2}(e_{x}+e_{y})\right)$
$\displaystyle
u_{j}^{-}:=\sqrt{\frac{2}{3}}\left(\frac{1}{2}(e_{x}+e_{y})-e_{j}\right)$

where $e_{i}:=e_{z_{i}}$ is the basis vector corresponding to the poset
element $z_{i}\in P$.

We provide the following table which shows which vector we use in our analysis
based on the location of each element in the poset. We also display the
corresponding support values.

Region | Relation | Vector | $h_{K_{0}}(\cdot)$ | $h_{K_{1}}(\cdot)$
---|---|---|---|---
$z_{j}\in\mathsf{MID}$, | $z_{j}\lessdot y$ | $u_{j}^{+}$ | $0$ | $\sqrt{\frac{1}{6}}$
$z_{j}\in\mathsf{MID}$, | $z_{j}\gtrdot x$ | $u_{j}^{-}$ | $0$ | $\sqrt{\frac{1}{6}}$
$z_{j}\in\mathsf{MID}_{x}$, | $z_{j}\gtrdot x$ | $u_{j}^{-}$ | $0$ | $\sqrt{\frac{1}{6}}$
$z_{j}\in\mathsf{MID}_{y}$, | $z_{j}\lessdot y$ | $u_{j}^{+}$ | $0$ | $\sqrt{\frac{1}{6}}$
$z_{j}\in\mathsf{END}_{x}$, | $z_{j}\lessdot x$ | $u_{j}^{+}$ | $0$ | $-\sqrt{\frac{1}{6}}$
$z_{j}\in\mathsf{END}_{y}$, | $z_{j}\gtrdot y$ | $u_{j}^{-}$ | $0$ | $-\sqrt{\frac{1}{6}}$

After finding the conditions for the vectors in the table to be in the support
of the mixed area measure, we can prove Proposition 3.23. As discussed before,
we will omit the proofs.

######  Proposition 3.23.

Let $z_{i}\in P\backslash\\{x,y\\}$ be an element. Then

  1. (a)

If $z_{i}\in\mathsf{MID}$, $z_{i}\lessdot y$, then
$u_{i}^{+}\in\operatorname{supp}\mu$ if and only if $|P_{>z_{i}}|\leq
n-k-|P_{<x}|$.

  2. (b)

If $z_{i}\in\mathsf{MID}$, $z_{i}\gtrdot x$, then
$u^{-}_{i}\in\operatorname{supp}\mu$ if and only if $|P_{<z_{i}}|\leq
n-k-|P_{>y}|$.

  3. (c)

If $z_{i}\in\mathsf{MID}_{x}^{\text{min}}$, then
$u^{-}_{i}\in\operatorname{supp}\mu$ if and only if $|P_{<z_{i}}|\leq
n-k-|P_{>y}|$.

  4. (d)

If $z_{i}\in\mathsf{END}_{x}$, $z_{i}\lessdot x$, then
$u^{+}_{i}\in\operatorname{supp}\mu$ if and only if $|P_{z_{i}<\cdot<y}|\leq
k$.

  5. (e)

If $z_{i}\in\mathsf{MID}_{y}^{\text{max}}$, then
$u_{i}^{+}\in\operatorname{supp}\mu$ if and only if $|P_{>z_{i}}|\leq
n-k-|P_{<x}|$.

  6. (f)

If $z_{i}\in\mathsf{END}_{y}$, $z_{i}\gtrdot y$, then
$u^{-}_{i}\in\operatorname{supp}\mu$ if and only if $|P_{x<\cdot<z_{i}}|\leq
k$.

Using Theorem 3.14 and the table of support value computations, we can
translate Proposition 3.23 into information about $v$ and $a$. The result of
this application of Theorem 3.14 is given in Proposition 3.24.

######  Proposition 3.24.

Let $z_{i}\in P\backslash\\{x,y\\}$ be an element of our poset. Let
$v_{xy}:=\frac{1}{2}(v_{x}+v_{y})$.

  1. (a)

If $z_{i}\in\mathsf{MID}$ and $z_{i}\lessdot y$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{>z_{i}}|>n-k-|P_{<x}|$.

  2. (b)

If $z_{i}\in\mathsf{MID}$ and $z_{i}\gtrdot x$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{<z_{i}}|>n-k-|P_{>y}|$.

  3. (c)

If $z_{i}\in\mathsf{MID}_{x}^{\text{min}}$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{<z_{i}}|>n-k-|P_{>y}|$.

  4. (d)

If $z_{i}\in\mathsf{MID}_{y}^{\text{max}}$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{>z_{i}}|>n-k-|P_{<x}|$.

  5. (e)

If $z_{i}\in\mathsf{END}_{x}$ and $z_{i}\lessdot x$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{z_{i}<\ldots<y}|>k$.

  6. (f)

If $z_{i}\in\mathsf{END}_{y}$ and $z_{i}\gtrdot y$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{x<\cdot<z_{i}}|>k$.

In Proposition 3.24, the result was only stated for poset elements satisfying
certain minimal/maximal relations. But, the result is still true for all poset
elements in the corresponding region. This immediately gives Corollary 3.25.

######  Corollary 3.25.

Let $z_{i}\in P\backslash\\{x,y\\}$ be an element of our poset. Let
$v_{xy}:=\frac{1}{2}(v_{x}+v_{y})$. Then, the following statements are true.

  1. (a)

If $z_{i}\in\mathsf{MID}$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{>z_{i}}|>n-k-|P_{<x}|$.

  2. (b)

If $z_{i}\in\mathsf{MID}$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{<z_{i}}|>n-k-|P_{>y}|$.

  3. (c)

If $z_{i}\in\mathsf{MID}_{x}$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{<z_{i}}|>n-k-|P_{>y}|$.

  4. (d)

If $z_{i}\in\mathsf{MID}_{y}$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{>z_{i}}|>n-k-|P_{<x}|$.

  5. (e)

If $z_{i}\in\mathsf{END}_{x}$ and $z_{i}\lessdot x$, then $v_{i}\neq
v_{xy}+a/2\implies|P_{z_{i}<\ldots<y}|>k$.

  6. (f)

If $z_{i}\in\mathsf{END}_{y}$ and $z_{i}\gtrdot y$, then $v_{i}\neq
v_{xy}-a/2\implies|P_{x<\cdot<z_{i}}|>k$.

###### Proof.

We only prove (a) since the rest are similar. Let $z_{i}\in\mathsf{MID}$. Then
there is some $z_{0}\in\mathsf{MID}$ so that $z_{i}\leq z_{0}\lessdot y$.
Then, since $v_{z_{i}}=v_{z_{0}}$, we have that

$|P_{>z_{i}}|\geq|P_{>z_{0}}|>n-k-|P_{<x}|.$

This suffices for the proof. ∎

From Corollary 3.20, we already know some of the values of $v_{i}$ in
Proposition 3.24. This immediately implies Corollary 3.26.

######  Corollary 3.26.

Let $z_{i}\in P\backslash\\{x,y\\}$ be an element of our poset. Let
$v_{xy}:=\frac{1}{2}(v_{x}+v_{y})$. Then, the following statements are true.

  1. (a)

Suppose that $z_{i}\in\mathsf{MID}_{x}$. Then
$a\neq\frac{2}{3}(1-v_{xy})\implies|P_{<z_{i}}|>n-k-|P_{>y}|$.

  2. (b)

Suppose that $z_{i}\in\mathsf{MID}_{y}$. Then $a\neq
2v_{xy}\implies|P_{>z_{i}}|>n-k-|P_{<x}|$.

  3. (c)

Suppose that $z_{i}\in\mathsf{END}_{x}$. Then
$a\neq-2v_{xy}\implies|P_{z_{i}<\cdot<y}|>k$.

  4. (d)

Suppose that $z_{i}\in\mathsf{END}_{y}$. Then $a\neq
2(1-v_{xy})\implies|P_{x<\cdot<z_{i}}|>k$.

Based on Corollary 3.26, we define four statements which are stand-ins for
midway and dual-midway in the regions
$\mathsf{END}_{x},\mathsf{END}_{y},\mathsf{MID}_{x}$, and $\mathsf{MID}_{y}$.
Consider the four statements

$\displaystyle M_{x}$
$\displaystyle:=\left\\{a\neq\frac{2}{3}(1-v_{xy})\right\\},$ $\displaystyle
M_{y}$ $\displaystyle:=\left\\{a\neq 2v_{xy}\right\\},$ $\displaystyle E_{x}$
$\displaystyle:=\\{a\neq-2v_{xy}\\},$ $\displaystyle E_{y}$
$\displaystyle:=\\{a\neq 2(1-v_{xy})\\}.$

From Corollary 3.26, we know that $M_{x}$ implies that midway holds for
$\mathsf{MID}_{x}$, $M_{y}$ implies that dual-midway holds for
$\mathsf{MID}_{y}$, $E_{x}$ implies that midway holds for $\mathsf{END}_{x}$,
and $E_{y}$ implies that dual-midway holds for $\mathsf{END}_{y}$.

######  Lemma 3.27.

If $a\neq 1/2$, then either $M_{x}$ and $E_{x}$ hold simultaneously or $M_{y}$
and $E_{y}$ hold simultaneously. As a consequence, dual-midway holds for
$\mathsf{END}_{y}$ and $\mathsf{MID}_{y}$ or midway holds for
$\mathsf{END}_{x}$ and $\mathsf{MID}_{x}$.

###### Proof.

Suppose that $M_{x}$ and $M_{y}$ do not hold simultaneously. Then, we have

$2v_{xy}=\frac{2}{3}(1-v_{xy})\implies v_{xy}=\frac{1}{4}.$

This would imply that $a=1/2$, which contradicts the hypothesis. Thus $M_{x}$
and $M_{y}$ cannot be false simultaneously. Suppose that $M_{x}$ and $E_{y}$
do not hold simultaneously. Then,

$2(1-v_{xy})=\frac{2}{3}(1-v_{xy})\implies v_{xy}=1.$

But then $a=0$ which contradicts $N_{i}>0$. Thus $M_{x}$ and $E_{y}$ cannot be
false simultaneously. Suppose that $E_{x}$ and $M_{y}$ do not hold
simultaneously. Then we run into the same contradiction $a=0$. Now, suppose
that $E_{x}$ and $E_{y}$ do not hold simultaneously. Then we get an obvious
contradiction. This suffices for the proof. ∎

####  3.2.5 Understanding MID

We now introduce some results that will give us a better understand of the
region $\mathsf{MID}$.

######  Lemma 3.28.

Suppose that the hypothesis in Theorem 3.14 holds. Let $z,w\in P$ be two
comparable elements in $\mathsf{MID}$. Then, the following are true.

  1. (a)

If $z$ violates midway, then $w$ cannot violate dual-midway. In other words,
the inequality $|P_{>y}|+|P_{<z}|\leq n-k$ implies $|P_{<x}|+|P_{>w}|>n-k$.

  2. (b)

If $z$ violates dual-midway, then $w$ cannot violate midway. In other words,
the inequality $|P_{<x}|+|P_{>z}|\leq n-k$ implies $|P_{>y}|+|P_{<w}|>n-k$.

###### Proof.

We only prove (a) as (b) is symmetric. Since $|P_{>y}|+|P_{<z}|\leq n-k$, we
have that $v_{z}=v_{xy}+a/2$ from Proposition 3.24. Suppose for the sake of
contradiction that $|P_{<x}|+|P_{>w}|\leq n-k$ as well. Then Proposition 3.24
would imply that $v_{w}=v_{xy}-a/2$. But Corollary 3.20 gives $v_{z}=v_{w}$.
This would give $a=0$ which contradicts $N_{k}>0$. This suffices for the
proof. ∎

######  Proposition 3.29.

Let $z\in P$ be an element satisfying $x<z<y$. Then, the following are true.

  1. (a)

If $|P_{<x}|+|P_{>z}|\leq n-k$, then there is a chain from $z$ to a minimal
element of $P_{x<\cdot<y}$ where the minimal element covers an element outside
of $P_{x<\cdot<y}$.

  2. (b)

If $|P_{>y}|+|P_{<z}|\leq n-k$, then there is a chain from $z$ to a maximal
element of $P_{x<\cdot<y}$ which is covered by an element outside of
$P_{x<\cdot<y}$.

  3. (c)

We cannot have both $|P_{<x}|+|P_{>z}|\leq n-k$ and $|P_{>y}|+|P_{<z}|\leq
n-k$.

###### Proof.

Part (c) follows from Lemma 3.28. We only prove (a) since (b) is symmetric.
There is some $z^{\prime}\in\mathsf{MID}$ so that $x\lessdot z^{\prime}\leq
z<y$. Suppose that we cannot extend $z^{\prime}$ to a minimal element outside
of $P_{x<\cdot<y}$. Then, we would have $|P_{<z^{\prime}}|=1+|P_{<x}|$. From
(b) of Lemma 3.28, we have the inequality

$n-k<|P_{>y}|+|P_{<z^{\prime}}|=|P_{>y}|+|P_{<x}|+1<n-k$

where the last inequality follows from Theorem 3.14. This is a contradiction
and suffices for the proof of the proposition. ∎

######  Lemma 3.30.

Let $z_{i}\in\mathsf{MID}$ be an element in the poset. If $z_{i}$ violates
midway then $M_{x}$ is false. If $z_{i}$ violates dual midway, then $M_{y}$ is
false. In other words, we have the following implications:

$\displaystyle|P_{>y}|+|P_{<z_{i}}|\leq n-k$ $\displaystyle\implies
a=\frac{2}{3}(1-v_{xy})$ $\displaystyle|P_{<x}|+|P_{>z_{i}}|\leq n-k$
$\displaystyle\implies a=2v_{xy}.$

###### Proof.

Suppose that $z_{j}\in\mathsf{MID}$ violates midway. That is, we have that
$|P_{>y}|+|P_{<z_{j}}|\leq n-k$. Then, from Proposition 3.29, we know that
there is an element $z^{\prime}$ satisfying $x<z_{j}\leq z^{\prime}\lessdot y$
so that there exists $z^{\prime\prime}\in\mathsf{MID}_{x}$ where
$z^{\prime}\lessdot z^{\prime\prime}$. From Proposition 3.28, we have the
inequality $|P_{<x}|+|P_{>z^{\prime}}|\geq n-k+1$. Let

$S=P_{x<\cdot<z^{\prime\prime}}\backslash\mathsf{MID}=\\{z:x<z<z^{\prime\prime},z\notin\mathsf{MID}\\}.$

Then, we have that

$|S|+|\mathsf{MID}|+|P_{<x}|+|P_{>z^{\prime}}|\leq n$

since these sets do not over lap. We can conclude that

$|S|+|\mathsf{MID}|+1\leq n-(|P_{<x}|-|P_{>z^{\prime}}|)+1\leq k.$

From Proposition 3.21, this implies that $e_{z^{\prime},z^{\prime\prime}}$ is
a transition vector. Hence $v_{z_{j}}=1-a$. Now consider $z_{0}\in P$
satisfying $x\lessdot z_{0}\leq z_{j}<y$. We have that

$|P_{<z_{0}}|+|P_{>y}|\leq|P_{<z_{j}}|+|P_{>y}|\leq n-k.$

This implies that $v_{z_{0}}=v_{xy}+a/2$ from Proposition 3.24. But then this
implies that $1-a=v_{xy}+a/2$ or $a=\frac{2}{3}(1-v_{xy})$. This means that
$M_{x}$ is false. Similarly, if $z_{j}$ violates dual-midway, then we would
know that $a=2v_{xy}$ or $M_{y}$ is false. ∎

######  Lemma 3.31.

If $a\neq 1/2$, then the hypothesis in Theorem 3.14 implies that $(x,y)$
satisfies $k$-midway or dual $k$-midway.

###### Proof.

From Lemma 3.28, we know that every element either satisfies dual-midway or
satisfies midway (or possibly both). Suppose that there is an element which
violates midway and another element which violates dual-midway. Then we must
have $M_{x}$ and $M_{y}$ are both false, which cannot happen since $a\neq
1/2$. Thus every element in $\mathsf{MID}$ either all satisfies midway or all
satisfies dual-midway or all satisfies both. If it all satisfies both, then we
are done. If there is at least one element in $\mathsf{MID}$ which violates
midway, then all elements in MID satisfies dual-midway. Moreover, it also
means that $M_{x}$ is false from Lemma 3.30 and $M_{y},E_{y}$ are true. This
means that $\mathsf{MID}$, and $\mathsf{MID}_{y}$ and $\mathsf{END}_{y}$
satisfies dual-midway, i.e. $P$ satisfies dual $k$-midway. By similar
reasoning, if there is at least one element which violates dual-midway, then
$P$ satisfies $k$-midway. This suffices for the proof. ∎

We can now prove Theorem 3.9 and Theorem 3.10.

###### Proof of Theorem 3.10.

If $N_{k-1}=N_{k}=N_{k+1}$, that means that $a=1$. From the previous theorem,
this implies that $P$ satisfies $k$-midway or dual $k$-midway. The reverse
direction was proved in [15]. ∎

###### Proof of Theorem 3.9.

If $a\neq 1/2$, then we have that $P$ satisfies $k$-midway or dual $k$-midway.
But this implies that $N_{k}=N_{k-1}=N_{k+1}$, or $a=1$. Hence the only two
choices of $a$ are $a=1,1/2$. This corresponds to $N_{k+1}=N_{k}=N_{k-1}$ and
$N_{k+1}=2N_{k}=4N_{k-1}$. ∎

The only case that is remaining is the case $N_{k+1}=2N_{k}=4N_{k-1}$. To show
that this case actually occurs, we give an example in Figure 3.1. In this
poset, we can compute explicitly compute $N_{1}=1,N_{2}=2$ and $N_{3}=4$. This
satisfies $N_{3}=2N_{2}=4N_{1}$.

Figure 3.1: Example with $N_{i+1}=2N_{i}=4N_{i-1}$

###### Proof of Theorem 3.11..

We first prove that if $N_{k+1}=2N_{k}=4N_{k-1}$ and $N_{k}>0$, then
conditions (i)-(iv) are true in Theorem 3.11. Since $a=1/2$, we either have
that both $M_{x}$ and $M_{y}$ are true or both $E_{x}$ and $E_{y}$ are true.
Suppose that both $M_{x}$ and $M_{y}$ are true. Then every element in
$\mathsf{MID}$ does not violate midway or dual midway. Moreover, at least one
of $E_{x}$ and $E_{y}$ must be true as well. This proves that $P$ satisfies
midway and dual midway. But this would imply that $a=1$, which contradicts
$a=1/2$. Hence, it must be the case that $E_{x}$ and $E_{y}$ are true. In
other words, $\mathsf{END}_{x}$ and $\mathsf{END}_{y}$ both satisfy the midway
and dual-midway property, respectively. This is exactly the property that (i)
describes. The fact that $\mathsf{MID}$ is empty follows from the fact that
$\mathsf{END}_{x}$ satisfies midway and $\mathsf{END}_{y}$ satisfies dual
midway. Consider an arbitrary $z\in\mathsf{MID}$. Then, we can pick
$z_{1},z_{2}\in P$ so that $x\lessdot z_{1}\leq z\leq z_{2}\lessdot y$. Since
$v_{z_{1}}=v_{z_{2}}=v_{z}$, if both of the inequalities

$\displaystyle|P_{>y}|+|P_{<z_{1}}|$ $\displaystyle\leq n-k$
$\displaystyle|P_{<x}|+|P_{>z_{2}}|$ $\displaystyle\leq n-k$

are true, then Corollary 3.25 implies that $a=0$, which cannot happen. This
means that one of the inequalities must be wrong. Without loss of generality,
suppose that $|P_{>y}|+|P_{<z_{1}}|>n-k$. Note that
$\mathsf{MID}_{x}\sqcup\mathsf{MID}\sqcup\\{y\\}\sqcup P_{<z_{1}}\sqcup
P_{>y}\subseteq P$ is a disjoint union of sets. Thus, we have that

$|\mathsf{MID}_{x}|+|\mathsf{MID}|+|P_{<z_{1}}|+|P_{>y}|+1\leq
n\implies|\mathsf{MID}_{x}|+|\mathsf{MID}|\leq k-2.$

Let $z^{\prime\prime}\gtrdot y$. This element exists because $P$ has a $1$
element. Then, we have that

$|P_{x<\cdot<z^{\prime\prime}}|\leq|\mathsf{MID}_{x}|+|\mathsf{MID}|\leq k-2.$

But this is a contradiction since $|P_{x<\cdot<z^{\prime\prime}}|>k$ from the
fact that $\mathsf{END}_{y}$ satisfies dual-midway. This proves (iii). To
prove (ii), suppose that $z\in P$ is incomparable to $x$ and $y$. Then, we can
build a chain from $z$ to a maximal element by picking only elements
incomparable to $x$ and $y$ until we are forced to pick elements in
$\mathsf{MID}_{x}$ or $\mathsf{END}_{y}$. We can do the same to reach a
minimal element. It is not hard to prove that all of these vectors are
transition vectors. Hence, we get that $0=1-a$. But this cannot be true since
$a=1/2$. This proves (ii). To prove (iv), suppose that $z\in\mathsf{MID}_{y}$
and $z^{\prime}\in\mathsf{MID}_{x}$ so that $z<z^{\prime}$. Then, we can find
$z_{1}\in\mathsf{MID}_{y}$ and $z_{2}\in\mathsf{MID}_{x}$ such that $z\leq
z_{1}\lessdot z_{2}\leq z^{\prime}$ since $\mathsf{MID}=\emptyset$. We analyze
the conditions needed for $e_{z_{1},z_{2}}$ to be a transition vector. Note
that

$\displaystyle\operatorname{aff}F_{K_{0}}(e_{z_{1},z_{2}})$
$\displaystyle=\bigoplus_{\omega\notin P_{z_{1}\leq\cdot\leq y}\cup
P_{x\leq\cdot\leq
z_{2}}}\mathbb{R}[e_{\omega}]\oplus\mathbb{R}\left[\sum_{\omega\in
P_{z_{1}\leq\cdot\leq y}\cup P_{x\leq\cdot\leq z_{2}}}e_{\omega}\right]$
$\displaystyle\operatorname{aff}F_{K_{1}}(e_{z_{1},z_{2}})$
$\displaystyle=\sum_{\omega\geq y}e_{\omega}+\bigoplus_{\omega\notin P_{\leq
x}\cup P_{\geq y}\cup\\{z_{1},z_{2}\\}}\mathbb{R}[e_{\omega}]$
$\displaystyle\operatorname{aff}F_{K_{0}+K_{1}}(e_{z_{1},z_{2}})$
$\displaystyle=\sum_{\omega\geq
y}e_{\omega}+\bigoplus_{\omega\notin\\{z_{1},z_{2},x,y\\}}\mathbb{R}[\omega]\oplus\mathbb{R}\left[\sum_{\omega\in
P_{z_{1}\leq\cdot\leq y}\cup P_{x\leq\cdot\leq z_{2}}}e_{\omega}\right].$

The only obstruction to $e_{z_{1},z_{2}}\in\operatorname{supp}\mu$ is the
dimension of $F(K_{0},e_{z_{1},z_{2}})$. We have that
$e_{z_{1},z_{2}}\in\operatorname{supp}\mu$ if and only if

$n-|P_{z_{1}<\cdot<y}|-|P_{x<\cdot<z_{2}}|-3\geq n-k-1$

or $|P_{z_{1}<\cdot<y}|+|P_{x<\cdot<z_{2}}|\leq k-2$. But we know that
$e_{z_{1},z_{2}}$ cannot be in the support because that would imply that
$0=1-a$. Hence, we have that

$|P_{z<\cdot<y}|+|P_{x<\cdot<z^{\prime}}|\geq|P_{z_{1}<\cdot<y}|+|P_{x<\cdot<z_{2}}|\geq
k-1.$

To finish the proof, we must prove the reverse implication. Specifically,
given the conditions that (0) $N_{k}>0$, (1) $\mathsf{END}_{x}$ and
$\mathsf{END}_{y}$ satisfy midway and dual-midway, (2) $\mathsf{MID}$ is
empty, (3) every element is comparable to either $x$ or $y$, and (4) for every
$z\in\mathsf{MID}_{y}$ and $w\in\mathsf{MID}_{x}$ satifsying $z<w$, we have
$|P_{z<\cdot<y}|+|P_{x<\cdot<w}|\geq k-1$, we want to prove that
$N_{k+1}=2N_{k}=4N_{k-1}$. To prove this, we prove Claim 3.32.

######  Claim 3.32.

Let $L_{x}$ be the number of linear extensions of the poset
$\mathsf{MID}_{y}\sqcup\mathsf{END}_{x}$ and let $L_{y}$ be the number of
linear extensions of the poset $\mathsf{MID}_{x}\sqcup\mathsf{END}_{y}$. Then,
we have that $N_{m}=2^{m-1}L_{x}L_{y}$ for $m\in\\{k-1,k,k+1\\}$.

###### Proof.

Any linear extension $f$ satisfying $f(y)-f(x)=k$ induces a total ordering on
$\mathsf{MID}_{x}\sqcup\mathsf{END}_{y}$ and
$\mathsf{MID}_{y}\sqcup\mathsf{END}_{x}$ as well as a binary string in
$\\{0,1\\}^{m-1}$ where a $0$ in the $i$th coordinate means that the $i$th
element between $x$ and $y$ is in $\mathsf{MID}_{x}$ and a $1$ in the $i$th
coordinate menas that the $i$th element between $x$ and $y$ is in
$\mathsf{MID}_{y}$. Given this binary string and the two total orderings, we
can reconstruct the linear extension $f$. This is because after specifying the
positions between $x$ and $y$ that will be in $\mathsf{MID}_{x}$ and
$\mathsf{MID}_{y}$, the exact placement of the elements will be completely
determined by the total orderings. In order to prove the claim, it suffices to
prove that if we try to do this reconstruction process with any two linear
extensions on $\mathsf{MID}_{x}\sqcup\mathsf{END}_{y}$ and
$\mathsf{MID}_{y}\sqcup\mathsf{END}_{x}$ and any binary string in
$\\{0,1\\}^{m-1}$, we will get a linear extension of $P$ with $f(y)-f(x)=k$.
There are three possibilities that may stop $f$ from being a linear extension:

  1. (1)

$\mathsf{MID}_{y}$ may be too small. For example, we may not be able to
accommodate the string of all zeros $0\ldots 0$. To overcome this problem we
must prove that $|\mathsf{MID}_{y}|\geq m-1$.

  2. (2)

$\mathsf{MID}_{x}$ may be too small for the same reason. Similarly, we must
prove that $|\mathsf{MID}_{x}|\geq m-1$.

  3. (3)

Assuming that (1) and (2) are non-issues, the final issue is that the final
ordering may not be a linear extension. This will happen when there is a
$z\in\mathsf{MID}_{x}$ and $w\in\mathsf{MID}_{y}$ satisfying $w<z$ where $z$
lies between $x$ and $w$ in the final ordering, and $w$ lies between $z$ and
$y$ in the final ordering.

Note that (1) and (2) are not issues after applying the midway property and
dual midway property of $\mathsf{END}_{x}$ and $\mathsf{END}_{y}$ to
$z_{x}\lessdot x$ and $z_{y}\gtrdot y$. These elements exists because of the
assumption of a $0$ element and $1$ element. To prove that (3) is not an
issue, suppose for the sake of contradiction that we had elements
$z\in\mathsf{MID}_{x}$ and $w\in\mathsf{MID}_{y}$ that satisfy the problem in
(3). Then, we would have

$\displaystyle m-1$
$\displaystyle\geq\left|P_{x<\cdot<z}\sqcup\\{z\\}\sqcup|P_{w<\cdot<y}|\sqcup\\{w\\}\right|=2+|P_{x<\cdot<z}|+|P_{w<\cdot<y}|\geq
k+1.$

But we know that $m\leq k+1$, which is a contradiction. This suffices for the
proof of the claim. ∎

From Claim 3.32, we have that

$\frac{N_{k+1}}{4}=\frac{N_{k}}{2}=N_{k-1}=2^{k-2}L_{x}L_{y}.$

This completes the proof. ∎

###  3.3 Stanley’s Matroid Inequality

Stanley’s poset inequality was proven by Richard Stanley in [56]. We have
already written about this inequality extensively in Section 3.1. In the same
paper, Stanley proves an inequality associated to the bases of a matroid. Let
$M=(E,\mathcal{B})$ be a matroid of rank $r$ with ground set $E$ and bases
$\mathcal{B}$. For any subsets $T_{1},\ldots,T_{r}\subseteq E$, we can let
$B(T_{1},\ldots,T_{r})$ denote the number of sequences $(y_{1},\ldots,y_{r})$
where $y_{i}\in T_{i}$ for $i\in[r]$ such that
$\\{y_{1},\ldots,y_{r}\\}\in\mathcal{B}$. Explicitly, we have that

$B(T_{1},\ldots,T_{r}):=\\#\\{(y_{1},\ldots,y_{r})\in T_{1}\times\ldots\times
T_{r}:\\{y_{1},\ldots,y_{r}\\}\in\mathcal{B}(M)\\}.$

Stanley’s inequality associated to this sequence is written in Theorem 3.33.
The inequality describes the log-concavity of the basis counting sequence
$B(T_{1},\ldots,T_{r})$.

######  Theorem 3.33 (Theorem 2.1 in [56]).

Let $M=(E,\mathcal{B})$ be a regular matroid of rank $r$. Let
$\mathcal{T}=(T_{1},\ldots,T_{r-m})$ be a collection $r-m$ subsets of $E$ and
let $X,Y\subseteq E$. Then,

$B(\mathcal{T},\underbrace{X,\ldots,X}_{k\text{
times}},\underbrace{Y,\ldots,Y}_{m-k\text{ times}})^{2}\geq
B(\mathcal{T},\underbrace{X,\ldots,X}_{k-1\text{
times}},\underbrace{Y,\ldots,Y}_{m-k+1\text{
times}})B(\mathcal{T},\underbrace{X,\ldots,X}_{k+1\text{
times}},\underbrace{Y,\ldots,Y}_{m-k-1\text{ times}})$

for all $1\leq k\leq m-1$.

Before we present the proof of Theorem 3.33, we will first go over two
different ways to view the basis counting numbers $B(T_{1},\ldots,T_{r})$. The
first way to view these numbers is to associate them with a mixed volume of
suitable polytopes. The second way to view these numbers is to associate them
with a mixed discriminant of suitable matrices. Both the mixed volume and
mixed discriminant perspective will lead to immediate proofs of the theorem.
When we study the equality case of the inequality, it is also useful to have
both perspectives. If we wish to prove Theorem 3.33 with the regularity
assumption removed, the mixed volume and mixed discriminant perspectives will
prove to be insufficient. At the core of the mixed discriminant and mixed
volume perspective is that the matroid has a unimodular coordinatization. In
Section 3.3.5 we will remove the regularity assumption and prove Stanley’s
matroid inequality in its full generality using the technology of Lorentzian
polynomials.

####  3.3.1 Mixed Volume Perspective of Basis Counting Number

Since our matroid $M=(E,\mathcal{B})$ is unimodular, there is a unimodular
coordinatization $v:E\to\mathbb{R}^{r}$. This implies that to every subset
$T\subseteq E$, we can associate the zonotope

$Z(T):=Z(v(e):e\in T)=\sum_{e\in T}[0,v(e)].$

Since $v:E\to\mathbb{R}^{r}$ is a unimodular coordination, any determinant of
$r$ of these vectors will lie in the set $\\{0,\pm 1\\}$. From Example 1.54,
this gives us the equation

$\operatorname{Vol}_{r}(Z(T))=\sum_{I\subseteq T:|I|=r}|\text{Det}(v(e):e\in
I)|=\sum_{I\subseteq T:|I|=r}\mathds{1}_{I\text{ is a
basis}}=\\#\mathcal{B}\left(M|_{T}\right).$ (3.2)

Given subsets $T_{1},\ldots,T_{r}\subseteq E$, consider the Minkowski sum
$\sum_{i=1}^{r}\lambda_{i}Z(T_{i})$. From Equation 3.2, we have that

$\displaystyle\operatorname{Vol}_{r}\left(\sum_{i=1}^{r}\lambda_{i}Z(T_{i})\right)$
$\displaystyle=\operatorname{Vol}_{n}\left(\sum_{i=1}^{r}\sum_{e\in
T_{i}}[0,\lambda_{i}v(e)]\right)=\sum_{a_{1}+\ldots+a_{r}=r}C(a_{1},\ldots,a_{r})\lambda_{1}^{a_{1}}\ldots\lambda_{r}^{a_{r}}$

where $C(a_{1},\ldots,a_{r})$ is the number of ways to pick subsets
$Q_{i}\subseteq T_{i}$ and $|Q_{i}|=a_{i}$ for $i\in[r]$ where
$Q_{1}\cup\ldots\cup Q_{r}$ is a basis of $M$. According to Theorem 1.53, we
have that

$C(a_{1},\ldots,a_{r})=\binom{r}{a_{1},\ldots,a_{r}}\mathsf{V}_{r}(Z(T_{1})[a_{1}],\ldots,Z(T_{r})[a_{r}])$

and combinatorially we have $B(T_{1},\ldots,T_{r})=C(1,\ldots,1)$. Thus, we
have the equation

$B(T_{1},\ldots,T_{r})=\binom{r}{1,\ldots,1}\mathsf{V}_{r}(Z(T_{1}),\ldots,Z(T_{r}))=r!\mathsf{V}_{r}(Z(T_{1}),\ldots,Z(T_{r})).$
(3.3)

Equation 3.3 immediately implies Theorem 3.33 via the Alexandrov-Fenchel
inequality (Theorem 2.12).

####  3.3.2 Mixed Discriminant Perspective of the Basis Counting Number

From Section 3.3.1, we saw that the regularity condition on our matroid $M$
implies that there exists a unimodular coordinatization
$v:E\to\mathbb{R}^{r}$. From [9], this implies that we can view the basis
counting sequence in terms of mixed discriminants. To see this, consider
$T_{1},\ldots,T_{s}\subseteq E$. Recall that for $r_{1},\ldots,r_{s}\geq 0$
satisfying $r_{1}+\ldots+r_{s}=r$, we defined
$B(T_{1}[r_{1}],\ldots,T_{s}[r_{s}])$ as the number of sequences in
$T_{1}^{r_{1}}\times\ldots\times T_{s}^{r_{s}}$ which forms a basis of $M$. We
define a different sequence
$N(r_{1},\ldots,r_{s}):=N^{T_{1},\ldots,T_{s}}(r_{1},\ldots,r_{s})$ which is
the number of ways to pick $Q_{i}\subseteq T_{i}$ with $|Q_{i}|=r_{i}$ such
that $Q_{1}\cup\ldots\cup Q_{s}$ is a basis of $M$. Then, we have that

$B(T_{1}[r_{1}],\ldots,T_{s}[r_{s}])=r_{1}!\ldots
r_{s}!N^{T_{1},\ldots,T_{s}}(r_{1},\ldots,r_{s}).$ (3.4)

For $1\leq i\leq s$, let $X_{i}$ be the matrix with colums $v(e)$ for $e\in
T_{i}$. Let $A_{i}=X_{i}X_{i}^{T}$. Then, we know that

$\mathsf{D}_{r}(A_{1}[r_{1}],\ldots,A_{s}[r_{s}])=\frac{1}{r!}B(T_{1}[r_{1}],\ldots,T_{s}[r_{s}])=\frac{N^{T_{1},\ldots,T_{s}}(r_{1},\ldots,r_{s})}{\binom{r}{r_{1},\ldots,r_{s}}}$
(3.5)

from Lemma 1.32 and Equation 3.4. Equation 3.5 immediately implies Theorem
3.33 via Alexandrov’s Inequality on mixed discriminants (Theorem 2.6).

####  3.3.3 Equality cases of Stanley’s Matroid Inequality for Graphic
Matroids

At the moment, we have represented Stanley’s basis counting sequence in terms
of mixed volumes and in terms of mixed discriminants. Based on our knowledge
of the equality cases of the Alexandrov inequality for mixed discrminants and
the Alexandrov-Fenchel inequality, we can hope to understand the combinatorial
conditions on regular matroids $M$ which give equality in the log-concavity
inequality. To simplify the problem, we only consider the case where we have
two subsets $T_{1},T_{2}\subseteq E$ which partition our ground set. In other
words, we have some set $R\subseteq E$. We define $N_{i}$ to be the number of
bases of $M$ which intersect $R$ at $k$ points. From our analysis in previous
sections, we know that

$N_{i}=C^{R,E\backslash
R}(i,r-i)=\binom{r}{i}\mathsf{V}_{r}(Z(R)[k],Z(E\backslash R)[r-k]).$

Thus, the sequence $N_{i}$ is ultra-log-concave. Let
$\widetilde{N}_{i}=N_{i}/\binom{r}{i}$ be the log-concave normalization of
$N_{i}$. We study the equality cases of
$\widetilde{N}_{i}^{2}=\widetilde{N}_{i-1}\widetilde{N}_{i+1}$. From the mixed
volume perspective, Stanley in [56] is able to give the equality cases for the
Minkowski inequality analog for his basis counting sequence. However, we will
later see that in the strictly positive case, the extremals of the Minkowski
inequality analog are exactly the extremals of
$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for some $k$.
This fact comes from viewing the sequence as mixed discriminants. It is
unknown to the author whether or not Stanley knew about this fact at the time,
but it is an interesting fact nonetheless.

######  Theorem 3.34 (Theorem 2.8 in [56]).

Let $M$ be a loopless regular matroid of rank $n$ on the finite set $E$, and
let $R\subseteq S$. Then, the following two conditions are equivalent:

  1. (a)

$\widetilde{N}_{1}^{n}=\widetilde{N}_{0}^{n-1}\widetilde{N}_{n}$.

  2. (b)

One of the following two conditions hold:

    1. (i)

$\widetilde{N}_{1}=0$.

    2. (ii)

There is some rational number $q\in(0,1)$ such that for every point $x\in E$,
we have that $|\overline{x}\cap R|=q|\overline{x}|$ for all $x\in E$. Here
$\overline{x}$ is the closure of the point $x$.

Before we present Stanley’s proof of Theorem 3.34 for regular matroids, we
first present the proof for a combinatorial characterization of the equality
cases of Stanley’s matroid inequality for graphic matroids. This proof will
use the mixed discriminant perspective to find the extremals of
$\widetilde{N}_{k}^{2}\geq\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ in the
strictly positive case. In general, the combinatorial meaning of the matrices
which show up in the mixed discriminant perspective are difficult to decipher.
However, in the case of graphic matroids, the matrices in the mixed
discriminant are related to the Laplacian of the underlying graph. Suppose
that $M$ is a graphic matroid. Proposition 1.12 implies that there is a
connected graph $G=(V,E)$ such that $M\cong M(G)$. Then, in the case of
graphic matroids, we have Theorem 3.35.

######  Theorem 3.35.

Suppose that $M$ is a graphic matroid corresponded to a connected graph
$G=(V,E)$. Let $n=|V|$ and let $E=R\sqcup Q$ be a partition of the edge set so
that both $R$ and $Q$ contain at least one spanning tree of $G$. Then, the
following conditions are equivalent:

  1. (a)

$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for some
$k\in\\{1,\ldots,n-1\\}$.

  2. (b)

$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for all
$k\in\\{1,\ldots,n-1\\}$.

  3. (c)

For any two distinct vertices $v,w\in V(G)$, the ratio between the number of
edges between $v$ and $w$ that are in $R$ to the number of edges between $v$
and $w$ that are in $Q$ is some positive number which does not depend on our
choice of $v,w\in V(G)$.

###### Proof.

The proof of $(c)\implies(b)$ follows from Theorem 3.41. It suffices to prove
$(a)\implies(c)$. Suppose that $n\geq 4$. Consider the edges of a $R$ spanning
tree. Without loss of generality, we get let the vertices $n-1$ and $n$ be two
leaves of this spanning tree. Since $n\geq 4$, there exists a $R$-edge
connecting two vertices each of which neither $n-1$ nor $n$. .

Let $B\in\mathbb{R}^{n\times|E|}$ be the incidence matrix of $G$. We can
partition the columns of $B=[B_{R}|B_{Q}]$ where $B_{R}$ are the columns
corresponding to the edges in $R$ and $B_{Q}$ are the columns corresponding to
the edges in $Q$. We can further partition $B_{R}$ and $B_{Q}$ into

$\displaystyle B_{R}$ $\displaystyle=\begin{bmatrix}X_{R}\\\
v_{R}^{T}\end{bmatrix},\quad\text{ and }\quad B_{Q}=\begin{bmatrix}X_{Q}\\\
v_{Q}^{T}\end{bmatrix}$

where $X_{R}$ and $X_{Q}$ are $B_{R}$ and $B_{Q}$ with the bottom row removed.
We know that the reduced incidence matrix $C=[X_{R}|X_{Q}]$ is a unimodular
coordinatization of our graphic matroid. From the equality cases of
Alexandrov’s inequality for mixed discriminants, the equality in (a) implies
that $X_{R}X_{R}^{T}=\alpha X_{Q}X_{Q}^{T}$ for some $\alpha\in\mathbb{R}$. We
can apply the mixed discriminant equality characterization because
$X_{R},X_{Q}$ are both full rank and $X_{R}X_{R}^{T}$, $X_{Q}X_{Q}^{T}$ are
positive definite. Let $L_{R}$ be the Laplacian of the subgraph on $V$ with
$R$-edges and $L_{Q}$ be the Laplacian of the subgraph on $V$ with $Q$-edges.
Proposition 1.8 gives us

$\displaystyle L_{G|_{R}}$
$\displaystyle=B_{R}B_{R}^{T}=\begin{bmatrix}X_{R}X_{R}^{T}&\bullet\\\
\bullet&\bullet\end{bmatrix},$ $\displaystyle L_{G|_{Q}}$
$\displaystyle=B_{Q}B_{Q}^{T}=\begin{bmatrix}X_{Q}X_{Q}^{T}&\bullet\\\
\bullet&\bullet\end{bmatrix}.$

where we only care about the top left $(n-1)\times(n-1)$ principal submatrix.
This tells us the off-diagonal entries of $X_{R}X_{R}^{T}$ and
$X_{Q}X_{Q}^{T}$ encode exactly the number of $R$-edges and $Q$-edges between
vertices in $V\backslash n$. This implies that

$e_{R}(v,w)=\alpha\cdot e_{Q}(v,w)\text{ for all distinct
}v,w\in[n]\backslash\\{n\\}.$

From the assumption that $R$ and $Q$ contain spanning trees, it must be the
case that $\alpha>0$. If we repeat the same argument except we delete the row
corresponding to $n-1$, we get a similar result:

$e_{R}(v,w)=\beta\cdot e_{Q}(v,w)\text{ for all distinct
}v,w\in[n]\backslash\\{n-1\\}.$

From the same reasoning, we must have $\beta>0$. From our construction, there
is a $R$-edge connecting two vertices in $[n]\backslash\\{n-1,n\\}$. This
implies that $\alpha=\beta$. The only thing we must check is that
$e_{R}(n-1,n)=\alpha\cdot e_{Q}(n-1,n)$. But this follows
$\deg_{R}(v)=\alpha\deg_{Q}(v)$ for all $v\in[n]$. This proves (c) for $n\geq
4$. We defer the case $n=3$ to Lemma 5.12 in the appendix. The case $n<3$ is
vacously true. This suffices for the proof. ∎

####  3.3.4 Equality cases of Stanley’s Matroid Inequality for Regular
Matroids

Before we prove Theorem 3.34 for regular matroids, we give necessary and
sufficient conditions for when two zonotopes are homothetic to each other. For
any $v_{1},\ldots,v_{m}\in\mathbb{R}^{n}$, we define the centrally symmetric
zonotope generated by $v_{1},\ldots,v_{m}$ as the convex body given by

$Z_{0}(v_{1},\ldots,v_{m})=\sum_{i=1}^{m}[-v_{i},v_{i}]=\left\\{\sum_{i=1}^{m}\lambda_{i}v_{i}:\lambda_{i}\in[-1,1]\text{
for all }1\leq i\leq m\right\\}.$

This is related to our definition of a zonotope by the relation

$Z_{0}(v_{1},\ldots,v_{m})=2Z(v_{1},\ldots,v_{m})-\sum_{i=1}^{m}v_{i}.$

In the literature, a zonotope is generally defined to be simply the Minkowski
sum of a collection of segments. However, it is common to see the study of
zonotopes restricted to either centrally symmetric zonotopes or zonotopes in
the sense we have defined them. For examples of such conventions, we refer the
reader to [49, 54, 56]. We call a convex body a (centrally symmetric) zonoid
if it is the limit, in the sense of Hausdorff distance, of a sequence of
(centrally symmetric) zonotopes. From Theorem 3.36, to every centrally
symmetric zonoid we can associate an even measure on $\mathbb{S}^{n-1}$. From
Theorem 3.37, this even measure is unique. Both of these facts are found in
[49].

######  Theorem 3.36.

A convex body $K\subseteq\mathbb{R}^{n}$ is a centrally symmetric zonoid if
and only if its support function can be represented in the form

$h_{K}(x)=\int_{\mathbb{S}^{n-1}}|\langle x,v\rangle|\,\rho(dv)\text{ for all
$x\in\mathbb{R}^{n}$}$

where $\rho$ is some even measure on $\mathbb{S}^{n-1}$.

###### Proof.

See Theorem 3.5.3 in [49]. ∎

######  Theorem 3.37.

If $\rho$ is an even signed measure on $\mathbb{S}^{n-1}$ with

$\int_{\mathbb{S}^{n-1}}|\langle u,v\rangle|\,\rho(dv)=0\text{ for
}u\in\mathbb{S}^{n-1},$

then $\rho=0$.

###### Proof.

See Theorem 3.5.4 in [49]. ∎

For a centrally symmetric zonotope $Z_{0}=Z_{0}(v_{1},\ldots,v_{m})$, the
support function is given by the equation
$h_{Z_{0}}(x)=\sum_{i=1}^{k}\left\lVert v_{i}\right\rVert\cdot|\langle
x,u_{i}\rangle|$ where $u_{i}:=v_{i}/\left\lVert v_{i}\right\rVert$. Thus, the
corresponding even signed measure will be given by

$\rho_{Z_{0}}=\sum_{i=1}^{k}\frac{\left\lVert
v_{i}\right\rVert}{2}(\delta_{u_{i}}+\delta_{-u_{i}}).$ (3.6)

With the description and the uniqueness of the even measure corresponding to
centrally symmetric zonoids, we immediately get the result in Lemma 3.38.

######  Lemma 3.38.

Let $v_{1},\ldots,v_{r},w_{1},\ldots,w_{s}\in\mathbb{R}^{n}$ be non-zero
vectors such that none of the $v_{i}$’s are parallel and none of the $w_{i}$’s
are parallel. Then, there exists $\lambda\in\mathbb{R}$ and
$x\in\mathbb{R}^{n}$ such that

$Z(v_{1},\ldots,v_{r})=\lambda Z(w_{1},\ldots,w_{s})+x$

if and only if $r=s$, and up to a permutation of the vectors $v_{i}=\lambda
w_{i}$ for all $i$.

###### Proof.

It is clear that the conditions in the lemma imply that
$Z(v_{1},\ldots,v_{r})$ and $Z(w_{1},\ldots,w_{s})$ are homothetic. For the
other direction, note that $Z(v_{1},\ldots,v_{r})$ and $Z(w_{1},\ldots,w_{s})$
being homothetic implies that $Z_{0}(v_{1},\ldots,v_{r})$ and
$Z_{0}(w_{1},\ldots,w_{s})$ are also homothetic. Hence, we must have

$Z_{0}(v_{1},\ldots,v_{r})=\lambda Z_{0}(w_{1},\ldots,w_{s})$

where $\lambda$ is the same dilation factor in the homothety for
$Z(v_{1},\ldots,v_{r})$ and $Z(w_{1},\ldots,w_{s})$. In the homothety between
centrally symmetric zonotopes, there is only a dilation factor since they both
have center of mass at $0$. From the uniqueness and description in Equation
3.6 of the even measure corresponding to centrally symmetric zonoids, we can
conclude that $r=s$ and $v_{i}=\lambda s_{i}w_{i}$ for all $i$ up to a
reordering of the vectors and some choice of $s_{i}\in\\{\pm 1\\}$. This
suffices for the proof of the lemma. ∎

We can now begin the proof of Theorem 3.34.

###### Proof of Theorem 3.34.

The fact that (b) implies (a) is the subject of Theorem 3.41. For the other
direction, let $Q=E\backslash R$ and let $v:E\to\mathbb{R}^{n}$ be a
unimodular coordinatization of $M$. Since negating vectors in our
coordinatization will not break the unimodularity, we can assume that whenever
two vectors are parallel, they are positive scalar multiples of each other.
Since our coordinatization is unimodular, we know that two vectors are equal
to each other if and only if they are parallel. Moreover, since our matroid is
loopless, all of the vectors are non-zero. So, for every parallel class
$\overline{x}$ there is a vector $v_{\overline{x}}\in\mathbb{R}^{n}$ such that
$v_{y}=v_{\overline{x}}$ if and only if $y\in\overline{x}$. Recall that for
all $0\leq k\leq n$, we have that

$N_{k}=\binom{n}{i}\mathsf{V}_{n}\left(\underbrace{Z(R),\ldots,Z(R)}_{k\text{
times}},\underbrace{Z(Q),\ldots,Z(Q)}_{n-k\text{ times}}\right).$

Thus the equality
$\widetilde{N}_{1}^{n}=\widetilde{N}_{0}^{n-1}\widetilde{N}_{n}$ is equivalent
to the equality in Minkowski’s inequality (Theorem 2.18):

$\mathsf{V}_{n}\left(Z(R),\underbrace{Z(Q),\ldots,Z(Q)}_{n-1\text{
times}}\right)^{n}\geq\operatorname{Vol}_{n}(Z(R))^{n-1}\cdot\operatorname{Vol}_{n}(Z(Q)).$

From the equality case of Theorem 2.18, we know that one of the following
three things occur:

  1. (1)

$\dim Z(R)\leq n-2$.

  2. (2)

$\dim Z(R)$ and $\dim Z(Q)$ lie in parallel hyperplanes.

  3. (3)

$Z(R)$ and $Z(Q)$ are homothetic.

If $\dim Z(R)\leq n-2$ or $\dim Z(R)$ and $\dim Z(Q)$ lie in parallel
hyperplanes, then we know that
$\operatorname{Vol}_{n}(Z(R))=\operatorname{Vol}_{n}(Z(Q))=0$. This implies
that $\widetilde{N}_{1}=0$. Now suppose that $Z(R)$ and $Z(Q)$ are homothetic.
Let $R=\\{v_{1},\ldots,v_{r}\\}$ and $Q=\\{w_{1},\ldots,w_{s}\\}$ with
$r+s=|E|$ be the vectors in the coordinations. Let
$R^{\prime}=\\{v_{1}^{\prime},\ldots,v_{t}^{\prime}\\}$ and
$Q^{\prime}=\\{w_{1}^{\prime},\ldots,w_{u}^{\prime}\\}$ be the result of
adding parallel vectors together. The collection $R^{\prime}$ consists of
vectors of the form $|R\cap\overline{x}|v_{\overline{x}}$ for all $x$
satisfying $|R\cap\overline{x}|>0$. The same is true for the collection
$Q^{\prime}$. Since these zonotopes are homothetic, we know from Lemma 3.38
that $t=u$ and there exists a $\lambda>0$ so that after a relabeling of
indices we have $v_{i}^{\prime}=\lambda s_{i}w_{i}^{\prime}$ for all $1\leq
i\leq t$ where $s_{i}\in\\{\pm 1\\}$. From our construction, we are forced to
have $s_{i}=1$. Thus, it must be the case that equality holds if and only if
there exists $\lambda>0$ so that
$|R\cap\overline{x}|=\lambda|Q\cap\overline{x}|$ for all $x\in E$. This
suffices for the proof. ∎

######  Remark 6.

In his paper [56], Stanley gives a combinatorial proof of Theorem 3.34 by
proving Lemma 3.38 with the additional hypothesis that all vectors lie in the
same orthant. Because the vectors are forced to be in the same positive
orthant, he can conclude that $v_{i}=\lambda w_{i}$ without any sign
constraints from the $s_{i}$. However, it is not clear to us why he can apply
this fact to Theorem 3.34. In particular, Stanley did not prove that for any
regular matroid, it is possible to find a unimodular coordinatization such
that all of the vectors lie in the same orthant.

With the mixed discriminant perspective, we can extend Theorem 3.34 slightly
to the case where we are only guarenteed that one of the inequalities
$\widetilde{N}_{k}^{2}\geq\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ is equality.
Indeed, by viewing the sequence $\widetilde{N}_{k}$ as a sequence of mixed
discriminants, we can prove Lemma 3.39.

######  Lemma 3.39.

Let $M$ be a regular, loopless matroid of rank $n$. Suppose that
$\widetilde{N}_{0}>0$ and $\widetilde{N}_{n}>0$. Then
$\widetilde{N}_{1}^{n}=\widetilde{N}_{0}^{n-1}\widetilde{N}_{n}$ if and only
if $\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for some
$k\in\\{1,\ldots,n-1\\}$.

###### Proof.

Assuming that
$\widetilde{N}_{1}^{n}=\widetilde{N}_{0}^{n-1}\widetilde{N}_{n}$, we know that
$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for all $k$
from our proof of equality conditions in Theorem 2.18. For the converse,
suppose that $\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$
for some $k$. Recall that we can view

$\widetilde{N}_{k}=\mathsf{D}(A_{R}[k],A_{Q}[n-k])$

where $A_{R}=X_{R}X_{R}^{T}$, $A_{Q}=X_{Q}X_{Q}^{T}$, and $X_{R},X_{Q}$ are
the matrices which consist of the unimodular representations of the elements
in $R$ and $Q$ as columns. From Corollary 2.11, we get that
$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ holds for all
$k\in\\{1,\ldots,n-1\\}$. We can apply Corollary 2.11 because the condition
$\widetilde{N}_{0}>0$ and $\widetilde{N}_{n}>0$ implies that the matrices
$X_{R}X_{R}^{T}$ and $X_{Q}X_{Q}^{T}$ are positive definite. ∎

From Lemma 3.39, whenever $\widetilde{N}_{0},\widetilde{N}_{n}>0$ (which
translate to $R$ and $Q$ both having full rank), we get that the necessary and
sufficient conditions for equality in Theorem 3.34 are exactly the necessary
and sufficient conditions to guarentee equality at a single instance of the
log-concavity inequality. Thus, we make the following conjecture for all
matroids (not necessarily regular).

######  Conjecture 3.40.

Let $M$ be a loopless matroid of rank $n$ on a set $E$. Let $R\subseteq E$ be
a subset and let $Q=E\backslash R$. For $k$, $0\leq k\leq n$, we define
$N_{k}$ to be the number of bases of $M$ with $k$ elements in $R$. Define
$\widetilde{N}_{k}=\frac{N_{k}}{\binom{n}{k}}$. Suppose that $R$ and $Q$ both
have rank $n$. Then, the following are equivalent.

  1. (a)

$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for some
$k\in\\{1,\ldots,n-1\\}$.

  2. (b)

$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for all
$k\in\\{1,\ldots,n-1\\}$.

  3. (c)

There are positive integers $r,q\geq 1$ such that $|\overline{x}\cap
R|r=|\overline{x}\cap Q|q$ for all $x\in E$.

To show some evidence of Conjecture 3.40, we will prove one of the directions
of Conjecture 3.40. This result is written in the statement of Theorem 3.41.

######  Theorem 3.41.

Suppose that there are positive integers $r,s\geq 1$ such that
$|\overline{x}\cap R|r=|\overline{x}\cap Q|q$. Then,
$\widetilde{N}_{k}^{2}=\widetilde{N}_{k-1}\widetilde{N}_{k+1}$ for all
$k\in\\{1,\ldots,n-1\\}$. In particular, we have that
$\widetilde{N}_{k}=\frac{q}{r}\widetilde{N}_{k-1}$.

###### Proof.

For all $x$, we have that $|\overline{x}\cap R|r=|\overline{x}\cap Q|q$. This
implies that for $x\in E$, we can define an isomorphism
$\varphi_{x}:(\overline{x}\cap R)\times[r]\to(\overline{x}\cap Q)\times[q]$
where $\varphi_{x}=\left(\varphi_{x}^{1},\varphi_{x}^{2}\right)$. Let the
inverse map be $\psi_{x}=(\psi_{x}^{1},\psi_{x}^{2})$. For all
$k\in\\{1,\ldots,n\\}$, define the sets

$\displaystyle\Omega_{k}$ $\displaystyle:=\\{(a,i,U):U\in\mathcal{B}(M),|U\cap
R|=k,i\in[r],a\in U\cap R\\}$ $\displaystyle\Omega_{k-1}$
$\displaystyle:=\\{(b,j,V):V\in\mathcal{B}(M),|V\cap R|=k-1,j\in[q],b\in V\cap
Q\\}.$

To count the number of elements in $\Omega_{k}$ and $\Omega_{k-1}$, we count
the number of elements in each set with a fixed $U$ and $V$. Then, we sum over
all possible choices of $U$ and $V$. This gives the identity

$\displaystyle|\Omega_{k}|$ $\displaystyle=N_{k}\cdot k\cdot r$
$\displaystyle|\Omega_{k-1}|$ $\displaystyle=N_{k-1}\cdot(n-k+1)\cdot q.$

I claim that there is one-to-one correspondence between $\Omega_{k}$ and
$\Omega_{k-1}$. We define maps
$\varphi_{\downarrow}:\Omega_{k}\to\Omega_{k-1}$ and
$\varphi_{\uparrow}:\Omega_{k-1}\to\Omega_{k}$ given by

$\displaystyle\varphi_{\downarrow}(a,i,U)$
$\displaystyle=(\varphi_{1}^{x}(a,i),\varphi_{2}^{x}(a,i),(U\backslash
a)\cup\varphi_{1}^{x}(a,i))$ $\displaystyle\varphi_{\uparrow}(b,j,V)$
$\displaystyle=(\psi_{1}^{x}(b,j),\psi_{2}^{x}(b,j),(V\backslash
b)\cup\psi_{1}^{x}(b,j)).$

These two maps are well-defined because of interchangeability of parallel
elements in indepenent sets (and in particular bases). From construction, the
maps $\varphi_{\downarrow}$ and $\varphi_{\uparrow}$ are two-sided inverses of
each other. This gives a one-to-one correspondence between $\Omega_{k}$ and
$\Omega_{k-1}$. From our computation of the cardinalities of these sets, we
have that

$N_{k}\cdot k\cdot r=|\Omega_{k}|=|\Omega_{k-1}|=N_{k-1}(n-k+1)q.$

This implies that $\widetilde{N}_{k}=\frac{q}{r}\widetilde{N}_{k-1}$. This
proves the theorem. ∎

In the subsequent subsection, we prove Stanley’s matroid inequality without
the regularity condition using the technology of Lorentzian polynomials.

####  3.3.5 Lorentzian Perspective of the Basis Counting Number

One essential hypothesis in Stanley’s matroid inequality was that the matroid
$M=(E,\mathcal{B})$ had to be regular. This condition was needed in order to
have a unimodular coordinatization. The unimodular coordinatization was at the
heart of the mixed discriminant and mixed volume constructions. In this
section, we generalize Stanley’s matroid inequality by removing the hypothesis
of unimodularity using the theory of Lorentzian polynomials. Our proof will be
based on the fact that the basis generating polynomial of a matroid is
Lorentzian. For simplicity, suppose that the ground set of $M=(E,\mathcal{I})$
is $E=[n]$ and $\operatorname{rank}(M)=r$. Recall that the basis generating
polynomial of $M$ is defined as

$f_{M}(x_{1},\ldots,x_{n})=\sum_{B\in\mathcal{B}}x^{B}=\sum_{\begin{subarray}{c}1\leq
i_{1}<\ldots<i_{r}\leq n\\\
\\{i_{1},\ldots,i_{r}\\}\in\mathcal{B}(M)\end{subarray}}x_{i_{1}}\ldots
x_{i_{r}}.$

For any sequence of subsets $\mathcal{T}:=(T_{1},\ldots,T_{m})\subseteq E$, we
can define a modification of the basis generating polynomial which gives us a
better handle on Stanley’s basis counting numbers.

######  Definition 3.42.

Let $\mathcal{T}:=(T_{1},\ldots,T_{m})$ be a sequence of subsets of $E$. Then,
consider the polynomial

$g_{M}^{\mathcal{T}}(y_{1},\ldots,y_{m}):=f_{M}\left(x_{e}=\sum_{i:e\in
T_{i}}y_{i}\right)$

where in the right hand side, we consider the basis generating polynomial
where we replace each instance of the coordinate $x_{e}$ with the linear form
$\sum_{i:e\in T_{i}}y_{i}$.

For $a_{1},\ldots,a_{m}\geq 0$ satisfying $a_{1}+\ldots+a_{m}=r$ we define the
number $N(a_{1},\ldots,a_{m})$ as the number of ways to pick subsets
$Q_{i}\subset T_{i}$ with $|Q_{i}|=a_{i}$ such that $Q_{1}\cup\ldots\cup
Q_{m}$ is a basis of $M$.

######  Lemma 3.43.

For $T_{1},\ldots,T_{m}\subseteq E$ and $a_{1},\ldots,a_{m}\geq 0$ satisfying
$a_{1}+\ldots+a_{m}=r$, we have that

$N(a_{1},\ldots,a_{m})=\frac{B(T_{1}[a_{1}],\ldots,T_{m}[a_{m}])}{a_{1}!\ldots
a_{m}!}.$

###### Proof.

Any choice of subsets $Q_{i}\subseteq T_{i}$ with $|Q_{i}|=a_{i}$ and
$Q_{1}\cup\ldots\cup Q_{r}$ a basis of $M$ gives $a_{1}!\ldots a_{m}!$
sequences which are included in the count of
$B(T_{1}[a_{1}],\ldots,T_{m}[a_{m}])$. Conversely, every sequence determines
subsets $Q_{i}\subseteq T_{i}$ with $|Q_{i}|=a_{i}$ and $Q_{1}\cup\ldots\cup
Q_{r}$ a basis of $M$. It is clear that this is a $1:a_{1}!\ldots a_{m}!$
correspondence. This suffices for the proof of the lemma. ∎

######  Proposition 3.44.

For $T_{1},\ldots,T_{m}\subseteq E$, we have that

$g_{M}^{T_{1},\ldots,T_{m}}(y_{1},\ldots,y_{m})=\sum_{a_{1}+\ldots+a_{m}=r}\frac{B(T_{1}[a_{1}],\ldots,T_{m}[a_{m}])}{a_{1}!\ldots
a_{m}!}\cdot y_{1}^{a_{1}}\ldots y_{m}^{a_{m}}.$

###### Proof.

By substituting $x_{e}=\sum_{i:e\in\mathcal{T}_{i}}y_{i}$ in the formula for
the basis generating polynomial, we get that

$\displaystyle g_{M}^{\mathcal{T}}(y_{1},\ldots,y_{m})$
$\displaystyle=\sum_{\begin{subarray}{c}1\leq i_{1}<\ldots<i_{r}\leq m\\\
\\{i_{1},\ldots,i_{r}\\}\in\mathcal{B}(M)\end{subarray}}\prod_{k=1}^{r}\left(\sum_{i_{k}\in
T_{j}}y_{j}\right)$ $\displaystyle=\sum_{\begin{subarray}{c}1\leq
i_{1}<\ldots<i_{r}\leq m\\\
\\{i_{1},\ldots,i_{r}\\}\in\mathcal{B}(M)\end{subarray}}\left(\sum_{a_{1}+\ldots+a_{m}=r}N^{\\{i_{1},\ldots,i_{r}\\}}(a_{1},\ldots,a_{m})\right)y_{1}^{a_{1}}\ldots
y_{m}^{a_{m}}$

where $N^{B}(a_{1},\ldots,a_{m})$ is the number of ways to pick subsets
$Q_{i}\subseteq T_{i}$ with $|Q_{i}|=a_{i}$ and $Q_{1}\cup\ldots\cup Q_{m}=B$.
In particular, we have that

$\sum_{B\in\mathcal{B}(M)}N^{B}(a_{1},\ldots,a_{m})=N(a_{1},\ldots,a_{m}).$

This allows us to simplify the equation after changing the order of
summations:

$\displaystyle g_{M}^{\mathcal{T}}(y_{1},\ldots,y_{m})$
$\displaystyle=\sum_{a_{1}+\ldots+a_{m}=r}\left(\sum_{B\in\mathcal{B}(M)}N^{B}(a_{1},\ldots,a_{m})\right)y_{1}^{a_{1}}\ldots
y_{m}^{a_{m}}$
$\displaystyle=\sum_{a_{1}+\ldots+a_{m}=r}N(a_{1},\ldots,a_{m})\cdot
y_{1}^{a_{1}}\ldots y_{m}^{a_{m}}.$

From Lemma 3.43, we have proven the lemma. ∎

######  Lemma 3.45.

For $T_{1},\ldots,T_{m}\subseteq E$, the polynomial
$g_{M}^{T_{1},\ldots,T_{m}}$ is Lorentzian.

###### Proof.

This follows from Theorem 2.26 and Proposition 2.27. ∎

We can now prove the main theorem for this section. This theorem will be a
generalization of Stanley’s matroid inequality in Theorem 3.33.

######  Theorem 3.46.

Let $M=(E,\mathcal{B})$ be any matroid of rank $r$. Let
$\mathcal{T}=(T_{1},\ldots,T_{r-m})$ be a sequence of subsets in $E$ and let
$Q,R\subseteq E$ be subsets. Define the sequence

$B_{k}(\mathcal{T},Q,R):=B(T_{1},\ldots,T_{r-m},Q[k],R[m-k]).$

Then the sequence $B_{0}(\mathcal{T},Q,R),\ldots,B_{m}(\mathcal{T},Q,R)$ is
log-concave.

###### Proof.

Consider the Lorentzian polynomial $g_{M}^{T_{1},\ldots,T_{r-m},Q,R}$. The
result then follows from Proposition 2.28. ∎

Using Theorem 3.46, we can also remove the regularity hypothesis from
Corollary 2.4 in [56].

######  Corollary 3.47.

Let $M=(E,\mathcal{B})$ be any matroid of rank $n$ and let
$T_{1},\ldots,T_{r},Q,R$ be pairwise disjoint subsets of $E$ whose union is
$E$. Fix non-negative integers $a_{1},\ldots,a_{r}$ such that
$m=n-a_{1}-\ldots-a_{r}\geq 0$, and for $0\leq k\leq m$ define $f_{k}$ to be
the number of bases $B$ of $M$ such that $|B\cap T_{i}|=a_{i}$ for $1\leq
i\leq r$, and $|B\cap R|=k$ (so $|B\cap Q|=m-k$). Then the sequence
$f_{0},\ldots,f_{m}$ ultra-log-concave.

###### Proof.

Let $\mathcal{T}=(T_{1}[a_{1}],\ldots,T_{r}[a_{r}])$. Then we have
$B_{k}(\mathcal{T},Q,R)=a_{1}!\ldots a_{r}!k!(m-k)!f_{k}$ where we use the
same notation as in Theorem 3.46. Thus, we have that

$\frac{f_{k}}{\binom{m}{k}}=\frac{B_{k}(\mathcal{T},Q,R)}{a_{1}!\ldots
a_{r}!}.$

The ultra-log-concavity of $f_{k}$ then follows from Theorem 3.46. ∎

According to Stanley in [56], Theorem 3.46 would imply the first Mason
conjecture. Indeed, this is the content of Theorem 2.9 in [56]. We rewrite the
proof now for the sake of completeness.

######  Theorem 3.48 (Mason’s Conjecture).

Let $M=(E,\mathcal{I})$ be a matroid of rank $n$. For all $0\leq k\leq n$, let
$I_{k}$ be the number of independent sets of $M$ of rank $k$. Then
$I_{k}^{2}\geq I_{k-1}I_{k+1}$ for all $2\leq k\leq n-1$.

###### Proof.

Let $B_{n}$ be the boolean matroid on $n$ elements. Consider $T^{n}(B_{n}+M)$
the level $n$ truncation of the matroid sum $B_{n}+M$. Let $f_{k}$ be the
number of bases of $T^{n}(B_{n}+M)$ which shares $k$ elements with $E(M)$.
Then, we have that $f_{k}=I_{k}\binom{n}{n-k}$. From Corollary 3.47, we have
that $I_{k}$ is log-concave. This suffices for the proof. ∎

######  Remark 7.

In [12], Bräden-Huh prove the strongest Mason conjecture using Lorentzian
polynomials. Their proof involves proving that the homogeneous multivariate
Tutte polynomial of a matroid is Lorentzian.

######  Remark 8.

In hindsight, it seems that Stanley would have been unable to prove his
matroid inequality for arbitrary matroids using mixed volumes without the
hypothesis that the matroid is regular. Indeed, from Remark 4.3 in [12], the
basis generating polynomial of a matroid on $[n]$ is the volume polynomial of
$n$ convex bodies precisely when the matroid is regular.

##  Chapter 4 Hodge Theory for Matroids

In this chapter we will study cohomology rings associated to matroids. In the
general case, our cohomology ring will be a graded $\mathbb{R}$-algebra of the
form $A^{\bullet}=\bigoplus_{i=0}^{d}A^{i}$ with a top-degree isomorphism:
$\deg:A^{d}\to\mathbb{R}$. If in addition, the natural pairing $A^{i}\times
A^{d-i}\to A^{d}\to\mathbb{R}$ is non-degenerate for all $i$, we will call our
graded algebra $A^{\bullet}$ a Poincaré duality algebra. In the finite
dimensional case, this will automatically imply that $\dim A^{k}=\dim A^{d-k}$
for all $k$. Cohomology rings with these properties show up naturally in
topology and algebraic geometry. For an overview of such examples, we refer
the reader to [31]. Given a cohomology ring associated to a matroid, we will
study when this graded object satisfies Poincaré duality, Hard Lefschetz, and
Hodge-Riemann relations. If it satisfies all three properties in all degrees
($\leq\frac{d}{2}$) we say that it satisfies the Kähler package. For the
development of the Hodge theory of matroids, we refer the reader to the papers
[1, 11, 5, 20, 21] where the notions of the Chow ring, intersection
cohomology, and conormal Chow ring are studied. Through this development, many
difficult and long standing conjectures in combinatorics have been solved by
proving that a suitable matroid cohomology satisfies the Kähler package. In
the next three sections, we define the notions of the graded Möbius algebra,
the Chow ring of a matroid, and the augmented Chow ring of a matroid. We
discuss these notions at a surface level, and the sections primarily serve as
a place to write down definitions and semi-small decomposition results for
future sections (see Section 4.6). In subsequent sections, we define the
Gorenstein ring associated to the basis generating polynomial of a matroid. It
is known from [39] that this ring satisfies the Hard Lefschetz property and
Hodge-Riemann relations of degree $1$ on the positive orthant. We prove
necessary and sufficient conditions for these properties to hold on the
boundary of the positive orthant. We also make some progress in proving the
complete Kähler package. Our progress is summarized in Section 4.6.

###  4.1 Cohomology Rings for Matroids

In this section, we will review the definitions and properties of three graded
commutative algebras associated to matroids: the graded Möbius algebra, the
Chow ring, and the augmented Chow ring. The Chow ring and the augmented Chow
ring of a matroid are automatically Poincaré duality algebras. The graded
Möbius algebra is in general not a Poincaré duality algebra.

####  4.1.1 Graded Möbius Algebra

Let $M$ be a matroid on ground set $E$ and let $\mathcal{L}$ be its lattice of
flats. For $k\geq 0$, let $\mathcal{L}^{k}(M)$ be the set of flats of rank
$k$. For every $k$, we can define the real vector space
$\operatorname{H}^{k}(M)$ given by

$\operatorname{H}^{k}(M):=\bigoplus_{F\in\mathcal{L}^{k}(M)}\mathbb{R}y_{F}$

where we have a variable $y_{F}$ for every flat $F\in\mathcal{L}^{k}(M)$. We
can then define a graded multiplicative structure
$\operatorname{H}^{k}(M)\times\operatorname{H}^{l}(M)\to\operatorname{H}^{k+l}(M)$
where for $F_{1}\in\mathcal{L}^{k}(M)$ and $F_{2}\in\mathcal{L}^{l}(M)$ we
have

$y_{F_{1}}\cdot y_{F_{2}}=\begin{cases}y_{F_{1}\vee F_{2}}&\text{if
$\operatorname{rank}_{M}(F_{1})+\operatorname{rank}_{M}(F_{2})=\operatorname{rank}_{M}(F_{1}\vee
F_{2})$},\\\ 0&\text{if
$\operatorname{rank}_{M}(F_{1})+\operatorname{rank}_{M}(F_{2})>\operatorname{rank}_{M}(F_{1}\vee
F_{2})$}.\end{cases}$ (4.1)

######  Definition 4.1.

For every matroid $M$, we define the graded Möbius algebra to be the graded
ring $\operatorname{H}(M)=\bigoplus_{k\geq 0}\operatorname{H}^{k}(M)$ equipped
with the multiplicative structure defined in Equation 4.1.

For every $e\in E$ in the ground set, we can define a graded ring homomorphism
$\theta_{e}:\operatorname{H}(M\backslash e)\to\operatorname{H}(M)$ where
$\theta_{e}(y_{F})=y_{\text{clo}_{M}(F)}$ for all $F\in\mathcal{L}(M\backslash
e)$. Explicitly, we have that $\operatorname{clo}_{M}(F)$ is $F$ if
$F\in\mathcal{L}(M)$ and $F\cup\\{e\\}$ if $F\notin\mathcal{L}(M)$. Since the
flats of $M\backslash e$ are of the form $F-\\{e\\}$ where $F$ is a flat of
$M$, this map is well-defined. It is also a degree preserving map since
$\operatorname{rank}(S)=\operatorname{rank}(\operatorname{clo}(S))$. It is
also easy to see that $\theta_{i}$ is injective. We prove that $\theta_{e}$ is
a ring homomorphism in Lemma 4.2.

######  Lemma 4.2.

For every $e\in M$, the map $\theta_{e}:\operatorname{H}(M\backslash
i)\to\operatorname{H}(M)$ is a ring homomorphism.

###### Proof.

Let $F_{1}\in\mathcal{L}^{k}(M\backslash e)$ and
$F_{2}\in\mathcal{L}^{l}(M\backslash e)$ be flats $M\backslash e$. We want to
prove that $\theta_{e}(y_{F_{1}}\cdot
y_{F_{2}})=\theta_{e}(y_{F_{1}})\theta_{e}(y_{F_{2}})$. Since the
multiplicative structure of our ring depends on the structure of our flats, we
will prove the homomorphism property in separate cases.

  1. (a)

Suppose that $y_{F_{1}}\cdot y_{F_{2}}=0$, then we have
$\operatorname{rank}_{M}(F_{1}\vee
F_{2})<\operatorname{rank}_{M}(F_{1})+\operatorname{rank}_{M}(F_{2})$.

    1. (i)

Additionally, suppose that $F_{1},F_{2}\notin\mathcal{L}(M)$. Then
$\theta(y_{F_{1}})=y_{F_{1}\cup i}$ and $\theta(y_{F_{2}})=y_{F_{2}\cup i}$.
To prove that $\theta_{y_{F_{1}}}\theta_{y_{F_{2}}}=0$, it is enough to prove

$\operatorname{rank}_{M}(F_{1}\cup F_{2}\cup
e)<\operatorname{rank}_{M}(F_{1}\cup e)+\operatorname{rank}_{M}(F_{2}\cup e).$

If $e\in\operatorname{clo}_{M}(F_{1}\cup F_{2})$, then this result follows
from $\operatorname{rank}_{M}(F_{1}\vee
F_{2})<\operatorname{rank}_{M}(F_{1})+\operatorname{rank}_{M}(F_{2})$. If
$e\notin\operatorname{clo}_{M}(F_{1}\cup F_{2})$, then there is a basis of
$F_{1}\cup F_{2}$ such that $I\cup e$ is a basis of $F_{1}\cup F_{2}\cup e$.
We then have

$\displaystyle\operatorname{rank}_{M}(F_{1}\cup F_{2}\cup e)=1+|I|\leq
1+|I\cap F_{1}|+|I\cap
F_{2}|<\operatorname{rank}_{M}(F_{1})+\operatorname{rank}_{M}(F_{2}).$

This proves the homomorphism property for $F_{1},F_{2}\notin\mathcal{L}(M)$.

    2. (ii)

Now, suppose that $F_{1}\notin\mathcal{L}(M)$ and $F_{2}\in\mathcal{L}(M)$.
Then $\operatorname{rank}(F_{1})=\operatorname{rank}(F_{1}\cup i)$ and
$\operatorname{rank}(F_{1}\cup
F_{2})<\operatorname{rank}(F_{1})+\operatorname{rank}(F_{2})$. We have that

$\displaystyle\operatorname{rank}\left((F_{1}\cup e)\cup F_{2}\right)$
$\displaystyle=\operatorname{rank}(F_{1}\cup F_{2})$
$\displaystyle<\operatorname{rank}(F_{1})+\operatorname{rank}(F_{2})$
$\displaystyle=\operatorname{rank}(F_{1}\cup e)+\operatorname{rank}(F_{2}).$

This proves the homomorphism property for $F_{1}\notin\mathcal{L}(M)$ and
$F_{2}\in\mathcal{L}(M)$.

    3. (iii)

Suppose that $F_{1},F_{2}\in\mathcal{L}(M)$. Then we automatically get the
homomorphism property.

  2. (b)

Now, suppose that $y_{F_{1}}\cdot y_{F_{2}}=y_{F_{1}\vee F_{2}}$. Then, we
have $\operatorname{rank}(F_{1}\vee
F_{2})=\operatorname{rank}(F_{1})+\operatorname{rank}(F_{2})$.

    1. (i)

Suppose that $F_{1},F_{2}\notin\mathcal{L}(M)$. Then, we have that
$\operatorname{rank}(F_{1})=\operatorname{rank}(F_{1}\cup e)$ and
$\operatorname{rank}(F_{2})=\operatorname{rank}(F_{2}\cup e)$. This implies
that $\operatorname{rank}(F_{1}\cup F_{2}\cup e)=\operatorname{rank}(F_{1}\cup
F_{2})$. Thus, we have that

$\displaystyle\operatorname{rank}\left((F_{1}\vee e)\vee(F_{2}\vee e)\right)$
$\displaystyle=\operatorname{rank}(F_{1}\cup F_{2})$
$\displaystyle=\operatorname{rank}(F_{1})+\operatorname{rank}(F_{2})$
$\displaystyle=\operatorname{rank}(F_{1}\vee e)+\operatorname{rank}(F_{2}\vee
e).$

This proves the homomorphism property for $F_{1},F_{2}\notin\mathcal{L}(M)$.

    2. (ii)

Suppose that $F_{1}\notin\mathcal{L}(M)$ and $F_{2}\in\mathcal{L}(M)$. Then we
have that $\operatorname{rank}(F_{1}\cup e)=\operatorname{rank}(F_{1})$ and
$\operatorname{rank}(F_{2}\cup e)=1+\operatorname{rank}(F_{2})$. We also have
$\operatorname{rank}(F_{1}\cup F_{2}\cup i)=\operatorname{rank}(F_{1}\cup
F_{2})$. Thus, we have

$\displaystyle\operatorname{rank}((F_{1}\cup e)\cup F_{2})$
$\displaystyle=\operatorname{rank}(F_{1}\cup F_{2})$
$\displaystyle=\operatorname{rank}(F_{1})+\operatorname{rank}(F_{2})$
$\displaystyle=\operatorname{rank}(F_{1}\cup e)+\operatorname{rank}(F_{2}).$

This proves the homomorphism property ofr $F_{1}\notin\mathcal{L}(M)$ and
$F_{2}\in\mathcal{L}(M)$.

    3. (iii)

Finally, suppose that $F_{1},F_{2}\in\mathcal{L}(M)$. Then the homomorphism
property follows automatically.

This suffices for the proof. ∎

####  4.1.2 Chow Ring of a Matroid

In this section, we give a brief overview of the Chow ring associated to a
matroid. We will not prove many of the claims we make, and refer the reader to
[11] for the details. Let $M=(E,\mathcal{I})$ be a loopless matroid. According
to [11], when $M$ is realizable over a field $k$, then the Chow ring of $M$ is
isomorphic to the Chow ring of a smooth projective variety over $k$. The Chow
ring of a matroid was originally introduced by Feichtner and Yuzvinsky in the
paper [21]. It is defined as a quotient ring of the polynomial ring

$\underline{S}_{M}:=\mathbb{R}\left[x_{F}:F\text{ is a nonempty proper flat of
}M\right].$

We can define the two ideals of $\underline{S}_{M}$ given by

$\displaystyle\underline{I}_{M}$ $\displaystyle:=\left\langle\sum_{i_{1}\in
F}x_{F}-\sum_{i_{2}\in F}x_{F}:\text{ for all }i_{1},i_{2}\in
E\right\rangle=\left\langle\sum_{i_{1}\notin F}x_{F}-\sum_{i_{2}\notin
F}x_{F}:\text{ for all }i_{1},i_{2}\in E\right\rangle,$
$\displaystyle\underline{J}_{M}$ $\displaystyle:=\left\langle
x_{F_{1}}x_{F_{2}}:\text{ $F_{1}$ and $F_{2}$ are incomparable}\right\rangle.$

######  Definition 4.3.

For a matroid $M$, we define the Chow ring of $M$ to be the quotient algebra

$\underline{CH}(M):=\frac{\underline{S}_{M}}{\underline{I}_{M}+\underline{J}_{M}}.$

The Chow ring of a matroid is a graded algebra where the grading is inherited
from the grading on $\underline{S}_{M}$. When $M$ has rank $d$, the top
dimension of $\underline{CH}(M)$ is $d-1$. There exists an isomorphism

$\underline{deg}:\underline{CH}(M)^{d-1}\longrightarrow\mathbb{R},\quad\prod_{F\in\mathcal{F}}x_{F}\longmapsto
1$

whenever $\mathcal{F}$ is any complete flag of nonempty proper flats. Similar
to the graded Möbius algebra, there exists a natural embedding
$\underline{$\theta$}_{i}:\underline{CH}(M\backslash e)\to\underline{CH}(M)$
which gives $\underline{CH}(M)$ a $\underline{CH}(M\backslash i)$-module
structure. From [6], the category of graded $\underline{CH}(M\backslash i)$
modules is a Krull-Schmidt category. Therefore, there is a unique way to
decompose $\underline{CH}(M)$ into indecomposable $\underline{CH}(M\backslash
i)$ modules. Theorem 1.2 in [11] gives the decomposition in terms of smaller
matroids.

######  Theorem 4.4 (Theorem 1.2 in [11]).

If $i$ is not a coloop of $M$, there is a decomposition of
$\underline{$\operatorname{CH}$}(M)$ into indecomposable graded
$\underline{$\operatorname{CH}$}(M\backslash i)$-modules such that

$\underline{\operatorname{CH}}(M)=\theta_{i}\left(\underline{\operatorname{CH}}(M\backslash
i)\right)\oplus\bigoplus_{F\in\underline{\mathcal{S}}_{i}}x_{F\cup
i}\cdot\theta_{i}\left(\underline{\operatorname{CH}}(M\backslash i)\right).$

Here, $\underline{$\mathcal{S}$}_{i}$ consists of all non-empty proper flats
$F$ of $E\backslash i$ such that $i$ is a coloop of $F\cup i$.

The semi-small decomposition in Theorem 4.4 provides an avenue to inductively
prove that the Chow ring satisfies the Kähler package.

####  4.1.3 Augmented Chow Ring of a Matroid

The augmented Chow ring of a matroid is defined in [11]. This ring is
intimately related to the Chow ring of matroid. The augmented Chow ring is
defined to be a quotient ring of the polynomial ring

$\operatorname{S}_{M}:=\mathbb{R}\left[x_{F},y_{i}:i\in E,F\text{ a proper
flat of }M\right].$

Let $\operatorname{J}_{M}^{(1)}$ be the ideal generated by
$x_{F_{1}}x_{F_{2}}$ where $F_{1}$ and $F_{2}$ are incomparable proper flats.
Let $\operatorname{J}_{M}^{(2)}$ be the ideal generated by $y_{i}x_{F}$ where
$i\in E$ and $F$ is a proper flat not containing $i$. Then, we have the two
ideals which are used to define the augmented Chow ring.

$\displaystyle\operatorname{I}_{M}$ $\displaystyle:=\left\langle
y_{i}-\sum_{i\notin F}x_{F}:\text{ for all }i\in E\right\rangle,$
$\displaystyle\operatorname{J}_{M}$
$\displaystyle:=\operatorname{J}_{M}^{(1)}+\operatorname{J}_{M}^{(2)}.$

######  Definition 4.5.

For a matroid $M$, we define the augmented Chow ring of $M$ to be the quotient
algebra

$\operatorname{CH}(M):=\frac{\operatorname{S}_{M}}{\operatorname{I}_{M}+\operatorname{J}_{M}}.$

Like the Chow ring of a matroid, the augmented Chow ring of a matroid is a
graded algebra with the grading inherited from the polynomial ring
$\operatorname{S}_{M}$. The top dimension of $\operatorname{CH}(M)$ will be
the rank of the matroid $M$, and there is an isomorphism
$\deg:\operatorname{CH}^{d}(M)\longrightarrow\mathbb{R}$. Note that the
subring generated by $\\{y_{i}\\}_{i\in E}$ is isomorphic to the graded Möbius
algebra. Hence, there is a embedding
$\operatorname{H}(M)\to\operatorname{CH}(M)$ which also gives
$\operatorname{CH}(M)$ a $\operatorname{H}(M)$-module structure. There is also
a natural injection $\theta_{i}:\operatorname{CH}(M\backslash
i)\to\operatorname{CH}(M)$ for any $i\in E$ which makes $\operatorname{CH}(M)$
into a $\operatorname{CH}(M\backslash i)$ module. The Krull-Schmidt
decomposition of $\operatorname{CH}(M)$ into $\operatorname{CH}(M\backslash
i)$ modules when $i$ is not a coloop is given by Theorem 1.5 in [11].

######  Theorem 4.6 (Theorem 1.5 in [11]).

If $i$ is not a coloop of $M$, then we can decompose $\operatorname{CH}(M)$
into indecomposable graded $\operatorname{CH}(M\backslash i)$-modules such
that

$\operatorname{CH}(M)=\theta_{i}(\operatorname{CH}(M\backslash
i))\oplus\bigoplus_{F\in\mathcal{S}_{i}}x_{F\cup
i}\cdot\theta_{i}(\operatorname{CH}(M\backslash i)).$

This decomposition is similar to that of Theorem 4.4. This allows us to hope
that similar decompositions hold for other cohomology rings associated to
matroids. We discuss this point in Section 4.6 when we discuss possible
directions for future research.

###  4.2 Gorenstein Ring associated to a polynomial

In this section, we examine a Poincaré algebra associated to a homogeneous
polynomial. We call this ring the Gorenstein ring associated to a polynomial.
As the name suggests, there is such a ring for every homogeneous polynomial
$f\in\mathbb{R}[x_{1},\ldots,x_{n}]$. The case where the polynomial is the
basis generating polynomial of a matroid was studied by Toshiaki Maeno and
Yasuhide Numata in their paper [37]. In the case where the polynomial is the
volume polynomial of a polytope, the ring was studied in [35] and [59]. The
idea of associating a Poincaré duality algebra to a polynomial is an old one
due to Macaulay.

######  Definition 4.7.

Let $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ be a homomgeneous polynoimal and let
$S:=\mathbb{R}[\partial_{1},\ldots,\partial_{n}]$ be the polynomial ring of
differentials where $\partial_{i}:=\partial_{x_{i}}$. Let
$A_{f}^{\bullet}:=S/\operatorname{Ann}_{S}(f)$. Then, we call
$A_{f}^{\bullet}$ the Gorenstein ring associated to the polynomial $f$.
Alternatively, we can view $\mathbb{R}[x_{1},\ldots,x_{n}]$ as a
$\mathbb{R}[X_{1},\ldots,X_{n}]$ modules where the action is defined by the
relation

$p(X_{1},\ldots,X_{n})\cdot
q(x_{1},\ldots,x_{n}):=p(\partial_{1},\ldots,\partial_{n})q(x_{1},\ldots,x_{n}).$

Then the Gorenstein ring is exactly
$\mathbb{R}[X_{1},\ldots,X_{n}]/\operatorname{Ann}(f)$ where the annihilator
is with respect to the $\mathbb{R}[X_{1},\ldots,X_{n}]$ action.

The Gorenstein ring associated to a polynomial has a natural grading with
respect to the degree of the differential form. Before we prove that this
actually gives $A_{f}^{\bullet}$ a graded ring structure, we first prove Lemma
4.8.

######  Lemma 4.8.

Let $\xi\in\mathbb{R}[\partial_{1},\ldots,\partial_{n}]$ and
$f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ be a homogeneous polynomial. We can
decompose $\xi=\xi_{0}+\xi_{1}+\ldots$ into its homogeneous parts. If
$\xi(f)=0$, then $\xi_{d}(f)=0$ for all $d\geq 0$.

###### Proof.

Let $d=\deg(f)$. If $\xi_{i}(f)\neq 0$, then $i\leq d$ and $\xi_{i}(f)$ is a
homomgeneous polynomial of degree $d-i$. Thus,
$\xi(f)=\xi_{0}(f)+\xi_{1}(f)+\ldots$ will be the homomgeneous decomposition
of the polynomial $\xi(f)$. Since this is equal to $0$, all components of the
decomposition are equal to zero. This suffices for the proof. ∎

######  Proposition 4.9.

The ring $A_{f}^{\bullet}$ is a graded $\mathbb{R}$-algebra where $A_{f}^{k}$
consists of the forms of degree $k$.

###### Proof.

Let us define $A_{f}^{k}$ as in the statement of the lemma. Let $d=\deg(f)$ be
the degree of the homomgeneous polynomial. Whenver $k>d$, the ring $A_{f}^{k}$
is clearly trivial. From Lemma 4.8, we have the direct sum decomposition

$A_{f}^{\bullet}=\bigoplus_{k=0}^{d}A_{f}^{k}.$

It is also clear that multiplication induces maps $A_{f}^{r}\times
A_{f}^{s}\to A_{f}^{r+s}$ for all $r,s\geq 0$. ∎

Proposition 4.10 states that the natural pairing in $A_{f}^{\bullet}$ equips
the ring with a Poincaré duality algebra structure. The proposition follows
from Theorem 2.1 in [38].

######  Proposition 4.10.

Let $f$ be a homogeneous polynomial of degree $d$. Then, the ring
$A_{f}^{\bullet}$ is a Poincaré-Duality algebra. That is, the ring satisfies
the following two properties:

  1. (a)

$A_{f}^{d}\simeq A_{f}^{0}\simeq\mathbb{R}$;

  2. (b)

The pairing induced by multiplication $A_{f}^{d-k}\times A_{f}^{k}\to
A_{f}^{d}\simeq\mathbb{R}$ is non-degenerate for all $0\leq k\leq d$.

In Lemma 4.11, we give an characterization of non-degeneracy. From this
characterization, it follows that the Hilbert polynomial of any Poincaré
duality algebra is palindromic. In other words, $\dim A_{f}^{k}=\dim
A_{f}^{d-k}$ for all $k$.

######  Lemma 4.11.

Let $B:V\times W\to k$ be a bilinear pairing between two finite-dimensional
$k$-vector spaces $V$ and $W$. Then, any two of the following three conditions
imply the third.

  1. (i)

The map $B_{V}:V\to W^{*}$ defined by $v\mapsto B(v,\cdot)$ has trivial
kernel.

  2. (ii)

The map $B_{W}:W\to V^{*}$ defined by $w\mapsto B(\cdot,w)$ has trivial
kernel.

  3. (iii)

$\dim V=\dim W$.

###### Proof.

Condition (i) implies $\dim V\leq\dim W$ and Condition (ii) implies $\dim
W\leq\dim V$. Thus (i) and (ii) both imply (iii). Now, suppose that (i) and
(iii) are true. Then $B_{V}$ is an isomorphism between $V$ and $W^{*}$ (see
3.69 in [7]). Let $v_{1},\ldots,v_{n}$ be a basis for $V$. Then
$B_{V}(v_{1}),\ldots,B_{V}(v_{n})$ is a basis of $W^{*}$. Let
$w_{1},\ldots,w_{n}$ be the dual basis in $W$ with respect to this basis of
$W^{*}$. Suppose that $\sum\lambda_{i}w_{i}\in\ker B_{W}$. Then for all $v\in
V$, we have

$\sum_{i=1}^{n}\lambda_{i}B_{V}(v)(w)=B\left(v,\sum_{i=1}^{n}\lambda_{i}w_{i}\right)=0.$

By letting $v=v_{1},\ldots,v_{n}$, we get $\lambda_{i}=0$ for all $i$. ∎

We say a pairing between finite-dimensional vector spaces is non-degenerate
whenever all three conditions in Lemma 4.11 hold. As a consequence, we have
Corollary 4.12.

######  Corollary 4.12.

Let $f$ be a homogeneous polynomial of degree $d\geq 2$ and let $k$, $0\leq
k\leq d$, by a non-negative integer. Then
$\dim_{\mathbb{R}}A_{f}^{k}=\dim_{\mathbb{R}}A_{f}^{d-k}$.

Given a homogeneous polynomial $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ of degree
$d$, we can define an isomorphism $\deg_{f}:A_{f}^{d}\to\mathbb{R}$ given by
evaluation at $f$. This means that for any differential $d$-form $\xi\in
A_{f}^{d}$, we have that $\deg_{f}(\xi):=\xi(f)$. Since $\xi$ and $f$
homogeneous of the same degree, the value of $\xi(f)$ will be a real number.
Following the terminology in [1], we give the following definition.

######  Definition 4.13.

Let $f$ be a homogeneous polynomial of degree $d$ and let $k\leq d/2$ be a
non-negative integer. For an element $l\in A_{f}^{1}$, we define the following
notions:

  1. (a)

The Lefschetz operator on $A_{f}^{k}$ associated to $l$ is the map
$L_{l}^{k}:A_{f}^{k}\to A_{f}^{d-k}$ defined by $\xi\mapsto l^{d-2k}\cdot\xi$.

  2. (b)

The Hodge-Riemann form on $A_{f}^{k}$ associated to $l$ is the bilinear form
$Q_{l}^{k}:A_{f}^{k}\times A_{f}^{k}\to\mathbb{R}$ defined by
$Q_{l}^{k}(\xi_{1},\xi_{2})=(-1)^{k}\deg(\xi_{1}\xi_{2}l^{d-2k})$.

  3. (c)

The primitive subspace of $A_{f}^{k}$ associated to $l$ is the subspace

$P_{l}^{k}:=\\{\xi\in A_{f}^{k}:l^{d-2k+1}\cdot\xi=0\\}\subseteq A_{f}^{k}.$

######  Definition 4.14.

Let $f$ be a homogeneous polynomial of degree $d$, let $k\leq d/2$ be a non-
negative integer, and let $l\in A_{f}^{1}$ be a linear differential form. We
define the following notions:

  1. (a)

(Hard Lefschetz Property) We say $A_{f}$ satisfies $\mathsf{HL}_{k}$ with
respect to $l$ if the Lefschetz operator $L_{l}^{k}$ is an isomorphism.

  2. (b)

(Hodge-Riemann Relations) We say $A_{f}$ satisfies $\mathsf{HRR}_{k}$ with
respect to $l$ if the Hodge-Riemann form $Q_{l}^{k}$ is positive definite on
the primitive subspace $P_{l}^{k}$.

We say that a homogeneous polynomial $f$ satisifes $\mathsf{HL}$ or
$\mathsf{HRR}$ if the associated ring $A_{f}^{\bullet}$ satisfies
$\mathsf{HL}$ or $\mathsf{HRR}$. For any $a\in\mathbb{R}^{n}$, we can define
the linear differential form
$l_{a}:=a_{1}\partial_{1}+\ldots+a_{n}\partial_{n}$. We say that $f$ satisfies
$\mathsf{HL}$ or $\mathsf{HRR}$ with respect to $a$ if and only if it
satisfies $\mathsf{HL}$ or $\mathsf{HRR}$ with respect to $l_{a}$. Since we
have shown that $A_{f}^{\bullet}$ automatically satisfies Poincaré duality, we
say that $f$ satisfies the Kähler package for a linear form $l$ if it
satisfies $\mathsf{HL}_{k}$ and $\mathsf{HRR}_{k}$ with respect to $l$ for all
$k\leq\frac{d}{2}$.

######  Proposition 4.15 (Lemma 3.4 in [39]).

Let $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ be a homogeneous polynomial of degree
$d\geq 2$ and $a\in\mathbb{R}^{n}$. Assume that $f(a)>0$. Then,

  1. (a)

$A_{f}$ has $\mathsf{HL}_{1}$ with respect to $l_{a}$ if and only
$Q_{l_{a}}^{1}$ is non-degenerate.

  2. (b)

Suppose that $A_{f}$ satisfies $\mathsf{HL}_{1}$. Then $A_{f}$ has
$\mathsf{HRR}_{1}$ with respect to $l_{a}$ if and only if $-Q_{l_{a}}^{1}$ has
signature $(+,-,\ldots,-)$.

###### Proof.

We include a proof for completeness. We first prove the statement in (a).
Suppose that $A_{f}$ has $\mathsf{HL}_{1}$ with respect to $l_{a}$. We have
the following commutative diagram:

${{A_{f}^{1}\times A_{f}^{1}}}$${{A_{f}^{1}\times
A_{f}^{d-1}}}$${{\mathbb{R}}}$$\scriptstyle{\text{id}\times
L_{l_{a}}^{1}}$$\scriptstyle{-Q_{l_{a}}^{1}}$

where the missing mapping is multiplication. If $A_{f}$ has $\mathsf{HL}_{1}$
with respect to $l_{a}$, then the top map between $A_{f}^{1}\times
A_{f}^{1}\to A_{f}^{1}\times A_{f}^{d-1}$ is a bijection. Thus the non-
degeneracy of $Q_{l_{a}}^{1}$ follows from the non-degeneracy of the
multiplication pairing as stated in Proposition 4.10. Now, suppose that
$Q_{l_{a}}^{1}$ is non-degenerate. Then, the map
$B:A_{f}^{1}\to(A_{f}^{1})^{*}$ defined by $\xi\mapsto-
Q_{l_{a}}^{1}(\xi,\cdot)$ is given by $m(L_{l_{a}}^{1}\xi,\cdot)$ where
$m:A_{f}^{1}\to A_{f}^{d-1}\to\mathbb{R}$ is the multiplication map. This is
the composition of $A_{f}^{1}\to A_{f}^{d-1}\to(A_{f}^{1})_{*}$ where the
first map is $L_{l_{a}}^{1}$ and the second map is injective from the non-
degeneracy of the multiplication map. This proves that $L_{l_{a}}^{1}$ is
injective. From Corollary 4.12, the map $L_{l_{a}}^{1}$ is an isomorphism.
This suffices for the proof of (a).

To prove (b), consider the commutative diagram

${{\mathbb{R}}}$${{A^{0}_{f}}}$${{A_{f}^{1}}}$${{A_{f}^{d-1}}}$${{A_{f}^{d}}}$${{\mathbb{R}}}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{L_{l_{a}}^{1}}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{L_{l_{a}}^{0}}$

Note that $L_{l_{a}}^{0}$ is an isomorphism because

$\deg L_{l_{a}}^{0}(1)=l_{a}^{d}(f)=d!f(a)\neq 0.$

Thus, we have $A_{f}^{1}=\mathbb{R}l_{a}\oplus P_{l}^{1}$ where the direct sum
is orthogonal over the Hodge-Riemann form by definition of the primitive
subspace. Now, note that

$-Q_{l_{a}}^{1}(l_{a},l_{a})=l_{a}^{d}(f)=-d!f(a)>0.$

Thus, the signature of $-Q_{l_{a}}^{1}$ is $(+,-,\ldots,-)$ if and only if
$Q_{l_{a}}^{1}$ is positive definite over the primitive subspace if and only
if $A_{f}$ satisfies $\mathsf{HRR}_{1}$ with respect to $l_{a}$. ∎

Recall from our definition of Lorentzian polynomials, we know that non-zero
Lorentzian polynomials are log-concave at any point in
$a\in\mathbb{R}_{>0}^{n}$. When $f(a)>0$, the polynomial $f$ is log-concave at
$a$ if and only if its Hessian has exactly one positive eigenvalue. Hence, for
any element in the non-negative orthant, we know that the Hessians of
Lorentzian polynomials have at most one positive eigenvalue. This fact and
Sylvester’s Law of Intertia gives a proof of Lemma 4.16.

######  Lemma 4.16.

If $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ is Lorentzian, then for any
$a\in\mathbb{R}_{\geq 0}^{n}$ with $f(a)>0$, $A_{f}^{1}$ has $\mathsf{HL}_{1}$
with respect to $l_{a}$ if and only if $f$ has the $\mathsf{HRR}_{1}$ with
respect to $l_{a}$.

###### Proof.

See Lemma 3.5 in [39]. ∎

###  4.3 Local Hodge-Riemann Relations

This section illustrates an example of a general inductive technique to prove
Hodge-Riemann relations. We define a local version of the Hodge-Riemann
relations. If this local version is satisfied, then this will imply that the
Hard Lefschetz property will be satisfied. In some situations, this is enough
to imply that the original Hodge-Riemann relations are satisfied. We give an
example of this inductive process in Lemma 4.18. We take our definition of the
local Hodge-Riemann relations from [39].

######  Definition 4.17.

A homogeneous polynomial $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ of degree $d\geq
2k+1$ satisfies the local $\mathsf{HRR}_{k}$ with respect to a form $l\in
A_{f}^{1}$ if for all $i\in[n]$, either $\partial_{i}f=0$ or $\partial_{i}f$
satisfies $\mathsf{HRR}_{k}$ with respect to $l$.

######  Lemma 4.18 (Lemma 3.7 in [39]).

Let $f\in\mathbb{R}_{\geq 0}[x_{1},\ldots,x_{n}]$ be a homogeneous polynomial
of degree $d$ and $k$ a positive integer with $d\geq 2k+1$, and
$a=(a_{1},\ldots,a_{n})\in\mathbb{R}^{n}$. Suppose that $f$ has the local
$\mathsf{HRR}_{k}$ with respect to $l_{a}$.

  1. (i)

If $a\in\mathbb{R}_{>0}^{n}$, then $A_{f}$ has the $\mathsf{HL}_{k}$ with
respect to $l_{a}$.

  2. (ii)

If $a_{1}=0,a_{2},\ldots,a_{n}>0$ and $\\{\xi\in
A_{f}^{k}:\partial_{i}\xi=0\text{ for $i=2,\ldots,n$}\\}=\\{0\\}$, then
$A_{f}$ has the $\mathsf{HL}_{k}$ with respect to $l_{a}$.

With Lemma 4.18, we can inductively prove that all Lorentzian polynomials
satisfy $\mathsf{HRR}_{1}$ with respect to $l_{a}$ for all
$a\in\mathbb{R}_{>0}^{n}$. We can prove this directly for all small Lorentzian
polynomials. For an arbitrary Lorentzian polynomial, we know that all of its
partial derivatives are Lorentzian. Thus, by induction, all of the partial
derivatives satisfy $\mathsf{HRR}_{1}$. This means that $f$ satisfies the
local $\mathsf{HRR}_{1}$. From lemma 4.18, we know that $f$ satisfies
$\mathsf{HL}_{1}$. But we have shown that $\mathsf{HL}_{1}$ and
$\mathsf{HRR}_{1}$ are equivalent for Lorentzian polynomials. This gives a
proof of Theorem 4.19 using the inductive procedure that we alluded to.

######  Theorem 4.19 (Theorem 3.8 in [39]).

If $f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ is Lorentzian, then $f$ has
$\mathsf{HRR}_{1}$ with respect to $l_{a}$ for any $a\in\mathbb{R}_{>0}^{n}$.

The next result in Corollary 4.20 gives a condition for general homogeneous
polynomials to satisfy $\mathsf{HRR}_{1}$ in terms of the signature of its
Hessian. This result will hold for any $a\in\mathbb{R}^{n}$ satisfying the
conditions in the statement.

######  Corollary 4.20.

Let $f$ be a homogeneous polynomial of degree $d\geq 2$. If
$\partial_{1}f,\ldots,\partial_{n}f$ are linearly independent in
$\mathbb{R}[x_{1},\ldots,x_{n}]$ and $f(a)>0$ for some $a\in\mathbb{R}^{n}$,
then $A_{f}$ satisfies $\mathsf{HRR}_{1}$ with respect to $l_{a}$ if and only
if $\operatorname{Hess}_{f}|_{x=a}$ has signature $(+,-,\ldots,-)$.

###### Proof.

Since $\partial_{1}f_{1},\ldots,\partial_{n}f_{n}$ are linearly independent,
the partials $\partial_{i}$ form a basis for $A_{f}^{1}$. Thus, the signature
of $Q_{l_{a}}^{1}$ is actually the signature of matrix of $Q_{l_{a}}^{1}$ with
respect to the set $\\{\partial_{1},\ldots,\partial_{n}\\}.$ We have

$-Q_{l_{a}}^{1}(\partial_{i},\partial_{j})=\partial_{i}\partial_{j}l_{a}^{d-2}f=l_{a}^{d-2}\partial_{i}\partial_{j}f=(d-2)!\partial_{i}\partial_{j}f(a).$

This proves that the signature of $-Q^{1}_{l_{a}}$ is the same as the
signature of $\operatorname{Hess}_{f}|_{x=a}$. We are done from Lemma 4.15(b).
∎

###  4.4 The Gorenstein Ring associated to the Basis Generating Polynomial of
a Matroid

In this section, we specialize the Gorenstein ring associated to a polynomial
to the Gorenstein ring associated to the basis generating polynomial of a
matroid. This graded $\mathbb{R}$-algebra is another cohomology ring
associated to a matroid. Thus, it is interesting study the Hard Lefschetz
property and Hodge-Riemann relations on this algebra. The Gorenstein ring
associated to the basis generating polynomial of a matroid is intimately
related to graded Möbius algebra. In particular, we can realize the Gorenstein
ring associated to the basis generating polynomial as a quotient ring of
$\operatorname{H}(M)$ by quotienting out the kernel of the Poincaré pairing.

######  Definition 4.21.

Let $M=(E,\mathcal{I})$ be a matroid. We let $\operatorname{A}(M)$ be the
Gorenstein ring associated to the basis generating polynomial of $M$.
Explicitly, we define the ring $\operatorname{S}_{M}:=\mathbb{R}[X_{e}:e\in
E]$ and the ideal
$\operatorname{Ann}_{M}=\operatorname{Ann}_{\operatorname{S}_{M}}(f_{M})$
where $f_{M}$ is the basis generating polynomial of $M$. Then, the ring
$\operatorname{A}(M)$ is equal to
$\operatorname{A}(M)=\operatorname{S}_{M}/\operatorname{Ann}_{M}$.

For a matroid $M$, we will prove that the ring $\operatorname{A}(M)$ will
depend only on its simplification. This is a reasonable claim because whenever
we have two elements $e,f\in E$ which are parallel, the differential
$\partial_{e}-\partial_{f}$ will be in the annihilator of $f_{M}$. Indeed, the
elements $e$ and $f$ are interchangeable in any independent set. Moreover, if