A matroid polytope approach to sharp affine isoperimetric inequalities
for volume decomposition functionals

Yude Liu Qiang Sun Ge Xiong

School of Mathematical Sciences, Key Laboratory of Intelligent Computing and
Applications

(Ministry of Education), Tongji University, Shanghai, 200092, P.R. China

††footnotetext: E-mail addresses: 1<EMAIL_ADDRESS>2.
<EMAIL_ADDRESS>3<EMAIL_ADDRESS>Research of the
authors was supported by NSFC No. 12271407.

Abstract New sharp affine isoperimetric inequalities for volume decomposition
functionals $X_{2}$ and $X_{3}$ in $\mathbb{R}^{n}$ are established. To fulfil
this task, we prove the recursion formulas for volume decomposition
functionals and find out the connection between the domains of these
functionals and matroid polytopes. Applications of matroid theory to convex
geometry are presented.

2020 Mathematics Subject Classification: 52B40, 52A40, 52B60.

Keywords: Volume decomposition functional; matroid polytope; affine
isoperimetric inequality

##  1\. Introduction

The setting for this paper is the $n$-dimensional Euclidean space,
$\mathbb{R}^{n}$. A _convex body_ is a compact convex set that has a nonempty
interior. Denote by $\mathcal{K}_{o}^{n}$ the set of convex bodies in
$\mathbb{R}^{n}$ with the origin $o$ in their interiors. A _polytope_ in
$\mathbb{R}^{n}$ is the convex hull of a finite set of points in
$\mathbb{R}^{n}$. A _face_ of a polytope $P$ is a convex subset $F\subseteq P$
such that $x,y\in P$ and $\frac{x+y}{2}\in F$ implies $x,y\in F.$ A face of
dimension $\dim P-1$ is called a _facet._ Write $\mathcal{P}_{o}^{n}$ for the
set of polytopes in $\mathbb{R}^{n}$ with the origin in their interiors. For
$P\in\mathcal{P}_{o}^{n}$, write $V_{n}(P)$ for its $n$-dimensional volume.

Suppose $P\in\mathcal{P}_{o}^{n}$ and $u$ is a unit outer normal vector to a
facet $F$ of $P$. The _cone-volume_ $V_{P}(\\{u\\})$ of $P$ associated with
$u$ is the volume of the convex hull of origin $o$ and facet $F$. The simplest
form of cone-volume is reduced to the area formula of triangles in ancient
geometry.

By using cone-volume and the idea of classification, Liu-Sun-Xiong [25]
introduced the so-called _volume decomposition functional_ of polytopes.

######  Definition 1.1.

Suppose $P\in\mathcal{P}_{o}^{n}$ and the unit outer normal vectors of $P$ are
$u_{1},u_{2},\dots,u_{N}$. The _$k$ th volume decomposition functional_
$X_{k}(P),$ $k=1,2,\ldots,n-1,n,$ is defined by

$X_{k}(P)^{n}=\sum_{\dim(\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\})=k}V_{P}(\\{u_{i_{1}}\\})V_{P}(\\{u_{i_{2}}\\})\cdots
V_{P}(\\{u_{i_{n}}\\}).$

Here $\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\}$ denotes the linear
subspace spanned by normal vectors $u_{i_{1}},\dots,u_{i_{n}}$. Obviously,
$X_{k}(P)^{n}$ is a homogeneous polynomial with degree $n$, $k=1,2,\ldots,n$;
$X_{k}(P)$ is centro-affine invariant, i.e., $X_{k}(TP)=X_{k}(P)$ for
$T\in\rm{SL}(n)$; and $X_{k}(\lambda P)=\lambda^{n}X_{k}(P)$ for $\lambda>0$.

It is interesting that volume decomposition functionals satisfy the regular
identity

(1.1) $V_{n}(P)^{n}=X_{1}(P)^{n}+X_{2}(P)^{n}+\cdots+X_{n}(P)^{n};$

and the $n$th volume decomposition functional

$X_{n}(P)=\big{(}\sum_{u_{i_{1}}\wedge\dots\wedge u_{i_{n}}\neq
0}V_{P}(\\{u_{i_{1}}\\})V_{P}(\\{u_{i_{2}}\\})\cdots
V_{P}(\\{u_{i_{n}}\\})\big{)}^{\frac{1}{n}}$

is precisely the functional $U$ introduced by E. Lutwak, D. Yang and G. Zhang
(LYZ) [28] to attack the longstanding Schneider projection problem. See [25,
Theorem 3.1] for the proof of identity (1.1).

In 2001, LYZ [28] conjectured that if $P$ is a polytope in $\mathbb{R}^{n}$
with its centroid at the origin, then

(1.2)
$\displaystyle\frac{X_{n}(P)}{V_{n}(P)}\geqslant\frac{(n!)^{\frac{1}{n}}}{n}$

with equality if and only if $P$ is a parallelotope. It took more than one
dozen years to completely settle this conjecture. Refer to [16], [35] and [17]
for its developments. In 2016, B$\rm{\ddot{o}}$r$\rm{\ddot{o}}$czky and Henk
[3] proved that LYZ’s conjecture is also affirmative for convex bodies.

In solving the LYZ conjecture, a “ _concentration phenomenon_ ” of cone-
volumes was discovered: If $P$ is a polytope in $\mathbb{R}^{n}$ with its
centroid at the origin and the unit outer normals of $P$ are
$u_{1},u_{2},\dots,u_{N}$, then

(1.3)
$\sum_{u_{i}\in\xi}V_{P}(\\{u_{i}\\})\leq\frac{\rm{dim}\xi}{n}V_{n}(P),\quad\quad\text{for
each subspace}\ \xi\subseteq\mathbb{R}^{n},$

with equality for a subspace $\xi$ if and only if there exists a subspace
$\xi^{{}^{\prime}}$ complementary to $\xi$ in $\mathbb{R}^{n}$, so that
$\\{u_{j}:u_{j}\not\in\xi\\}\subseteq\xi^{{}^{\prime}}$. In 2013, Böröczky and
LYZ [7] originally posed the subspace concentration condition (See Section 2
for details), and proved that it is this condition that guarantees the
existence of solutions to the even logarithmic Minkowski problem. For more
applications, see, e.g., [4, 5, 6, 19, 27, 32].

In light of the identity (1.1) and LYZ’s conjecture, Liu-Sun-Xiong [25] raised
the following problem.

Problem X. Let $P\in\mathcal{P}_{o}^{n}$ with its centroid at the origin. Does
there exist a constant $c(n,k)$ depending on $n$ and
$k,k\in\\{1,2,\dots,n-1\\},$ such that

$\frac{X_{k}(P)}{V_{n}(P)}\leq c(n,k)?$

The authors [25] solved this problem in $\mathbb{R}^{3}$ and established the
sharp affine isoperimetric inequalities

$\frac{X_{1}(P)}{V_{3}(P)}\leq\Big{(}\frac{1}{3}\Big{)}^{\frac{2}{3}},\ \
\frac{X_{2}(P)}{V_{3}(P)}\leq\Big{(}\frac{2}{3}\Big{)}^{\frac{1}{3}},\ \
\text{and}\ \ \
\frac{X_{3}(P)}{V_{3}(P)}\geq\frac{2^{\frac{1}{3}}}{3^{\frac{2}{3}}},$

and equality holds in each inequality if and only if $P$ is a parallelepiped.

In this article, we further attack Problem X and obtain the following results.

######  Theorem 1.1.

Let $P\in\mathcal{P}_{o}^{n}$ with its centroid at the origin and $n\geq 3$.
Then

$\frac{X_{2}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{2}\big{(}(\frac{2}{n})^{n}-\frac{2}{n^{n}}\big{)}}$

with equality if and only if $P$ is a parallelotope.

######  Theorem 1.2.

Let $P\in\mathcal{P}_{o}^{n}$ with its centroid at the origin and $n\geq 5$.
Then

$\frac{X_{3}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3(\frac{2}{n})^{n}+\frac{3}{n^{n}}\big{)}}$

with equality if and only if $P$ is a parallelotope.

Restricted to $\mathcal{P}_{4}^{4}$, i.e., the set of polytopes in
$\mathbb{R}^{4}$ whose any _four_ outer normal vectors (up to their antipodal
normal vectors) are linear independent, we prove the following.

######  Theorem 1.3.

Let $P\in\mathcal{P}_{4}^{4}$ with its centroid at the origin. Then

$\frac{X_{3}(P)}{V_{4}(P)}\leq\sqrt[4]{\frac{72}{125}}$

with equality if and only if $\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm
u_{1},\dots,\pm u_{5}\\}$, and $V_{P}(\\{\pm u_{i}\\})=\frac{V_{4}(P)}{5},$
$i=1,\dots,5.$

It is known that if $P$ is a polygon in $\mathbb{R}^{2}$ with its centroid at
the origin, then $\frac{X_{2}(P)}{V_{2}(P)}\geq\frac{\sqrt{2}}{2}$ with
equality if and only if $P$ is a parallelogram. See [35] or [27] for its
proof. Combining Theorem 1.2 in [25] and Theorem 1.1 together with the above
fact, Problem X for $X_{1}$ and $X_{2}$ are fully solved. Now, with Theorem
1.2 in hand, Problem X for $X_{3}$ is solved, except the _only_ case in
$\mathbb{R}^{4}$.

It is worth mentioning that Problem X for $X_{3}$ in $\mathbb{R}^{4}$ changes
_drastically_. On one hand, _parallelotopes_ are in the set
$\mathcal{P}_{4}^{4}$, but parallelotopes don’t satisfy the equality
conditions in Theorem 1.3; Meanwhile, _simplices_ with their centroids at the
origin attain the equality in Theorem 1.3. However, we illustrate that the
functional $X_{3}$ does _not_ attain its maximum at simplices in the set
$\mathcal{P}_{o}^{4}$. See Example 3.5 for details. So, it is a challenge to
find the extremal body for $X_{3}$ in $\mathbb{R}^{4}$.

This article is organized as follows.

After making some necessary preliminaries in Section 2, we establish the
recursion formulas for volume decomposition functionals in Section 3. These
recursion formulas are of the “dimension reduction” function such that we can
represent volume decomposition functionals as explicit polynomials.

In subsequent, we have to figure out the “effective domains” of these
polynomials. Capturing the essential attribute of cone-volumes, we prove that
these domains are indeed _relative interiors_ of a class of matroid polytopes,
which are important in matroid theory. Applications of matroid theory to
convex geometry are presented in Section 4.

In Section 5, we prove Theorems 1.1 and 1.2. The proof of Theorem 1.3 is
provided in Section 6.

##  2\. Preliminaries

Write $x\cdot y$ for the standard inner product of $x,y$ in $\mathbb{R}^{n}$.
For $u\in\mathbb{R}^{n}\backslash\\{o\\}$ and $\alpha\in\mathbb{R}$,

$H_{u,\alpha}=\\{x\in\mathbb{R}^{n}:x\cdot u=\alpha\\}$

denotes a _hyperplane_ in $\mathbb{R}^{n}$, which bounds a _closed halfspace_
$H_{u,\alpha}^{-}=\\{x\in\mathbb{R}^{n}:x\cdot u\leq\alpha\\}$.

For $u\in\mathbb{S}^{n-1}=\\{x\in\mathbb{R}^{n}:|x|=1\\}$, let $l_{u}$ be the
1-dimensional subspace spanned by $u$. Write $\mathrm{G}(n)$ for the set of
subspaces of $\mathbb{R}^{n}$, and $\mathrm{G}_{n,k}$ for the set of
$k$-dimensional subspaces of $\mathbb{R}^{n}$.

For $\mu\in\mathcal{B}(\mathbb{S}^{n-1})$, the set of finite Borel measures on
$\mathbb{S}^{n-1}$, let $\mathrm{supp}\mu$ denote its _support set_.

Let $\mathcal{P}_{c}^{n}$ and $\mathcal{P}_{os}^{n}$ be the class of polytopes
in $\mathbb{R}^{n}$ with centroid at the origin and the class of origin-
symmetric polytopes in $\mathbb{R}^{n}$, respectively.

For a polytope $P$ in $\mathbb{R}^{n}$, write $\mathcal{F}_{0}(P)$ for the set
of its vertices (i.e., $0$-dimensional faces), and $\mathcal{F}_{1}(P)$ for
the set of its edges (i.e., $1$-dimensional faces); write $\mathrm{relint}P$
for its relative interior and $\mathrm{relbd}P$ for its relative boundary,
respectively.

Let $Z$ be a finite set of unit vectors in $\mathbb{R}^{n}$, and
$Z\cup(-Z)=\\{\pm u_{1},\pm u_{2},\dots,\pm u_{N}\\}$. $Z$ is said to be in
$k$-general position, $k\in\\{1,2,\dots,n\\},$ if $Z$ is not contained in a
closed hemisphere of $\mathbb{S}^{n-1}$ and any $k$ elements of
$\\{u_{1},u_{2},\dots,u_{N}\\}$ are linearly independent.

A polytope $P$ in $\mathbb{R}^{n}$ is said to be in _$k$ -general position_,
if the set of unit outer normals of $P$ is in $k$-general position. Write
$\mathcal{P}_{k}^{n}$ for the set of polytopes in $\mathbb{R}^{n}$ which are
in $k$-general position and contain the origin in their interiors. See [27, p.
12] for more details about the definition. In [25, pp. 26-28], we also proved
that $\mathcal{P}_{k}^{n}$ is dense in $\mathcal{K}_{o}^{n}$ in the sense of
Hausdorff metric $\delta$, for $k=1,2,\dots,n,$.

Károlyi and Lovász [23] first posed the notion of general position. In fact,
the $n$-general position, up to antipodal unit outer normals, is indeed the
general position in the sense of Károlyi and Lovász.

Let $K$ be a convex body in $\mathbb{R}^{n}$. If $K=K_{1}+\cdots+K_{l}$ for
suitable $K_{i}$ lying in subspaces $\xi_{i}$ of $\mathbb{R}^{n}$ so that
$\mathbb{R}^{n}=\xi_{1}\oplus\cdots\oplus\xi_{l}$, write
$K=K_{1}\oplus\cdots\oplus K_{l}$. If $K=L\oplus M$ is only possible with
$\dim L=0$ or $\dim M=0$, then $K$ is called _directly indecomposable_. If
$K=L\oplus M$ with $\dim L>0$ and $\dim M>0$, then $K$ is called an
$(L,M)$-_cylinder_. Please refer to [31, p. 156] for details.

The following lemmas are needed in this article. See [25, p. 11] for details.

######  Lemma 2.1.

Let $P$ and $Q$ be polytopes in $\mathbb{R}^{n}$ with the origin in their
interiors. If $V_{P}(\\{\pm u\\})=V_{Q}(\\{\pm u\\})$ for any
$u\in\mathbb{S}^{n-1}$, then $X_{k}(P)=X_{k}(Q),$ $k=1,2,\dots,n.$

######  Lemma 2.2.

If $P$ is a polytope in $\mathbb{R}^{n}$ with its centroid at the origin, then
there exists an origin-symmetric polytope $Q$ in $\mathbb{R}^{n}$ so that
$X_{k}(Q)=X_{k}(P),$ $k=1,2,\dots,n.$

Cone-volume measure is a natural extension of cone-volume of polytopes to
convex bodies. For $K\in\mathcal{K}_{o}^{n}$, its _cone-volume measure_
$V_{K}$ is a finite Borel measure on $\mathbb{S}^{n-1}$, defined for each
Borel $\omega\subseteq\mathbb{S}^{n-1}$ by

(2.1)
$V_{K}(\omega)=\frac{1}{n}\int_{x\in\nu_{K}^{-1}(\omega)}x\cdot\nu_{K}(x)d\mathcal{H}^{n-1}(x),$

where $\nu_{K}:\partial^{\prime}K\rightarrow\mathbb{S}^{n-1}$ is the Gauss map
of $K$, defined on $\partial^{\prime}K$, the set of points of $\partial K$
that have a unique outer unit normal, and $\mathcal{H}^{n-1}$ is
$(n-1)$-dimensional Hausdorff measure.

Cone-volume measure has appeared in [2, 11, 15, 30, 31, 36], and been
intensively investigated in recent years. See, e.g., [3, 4, 5, 6, 17, 18, 20,
24, 35].

Following Böröczky and LYZ [7], we present the definition of subspace
concentration condition and the celebrated Böröczky-LYZ existence theorem on
solutions to the even logarithmic Minkowski problem, which will be used
repeatedly.

######  Definition 2.1.

A finite Borel measure $\mu$ on $\mathbb{S}^{n-1}$ is said to satisfy the
_subspace concentration inequality_ if, for every subspace $\xi$ of
$\mathbb{R}^{n}$, so that $0<\mathrm{dim}\xi<n$,

(2.2)
$\mu(\xi\cap\mathbb{S}^{n-1})\leq\frac{1}{n}\mu(\mathbb{S}^{n-1})\mathrm{dim}\xi.$

The measure is said to satisfy the _subspace concentration condition_ if in
addition to satisfying the subspace concentration inequality (2.2), whenever

$\mu(\xi\cap\mathbb{S}^{n-1})=\frac{1}{n}\mu(\mathbb{S}^{n-1})\mathrm{dim}\xi,$

for some subspace $\xi$, then there exists a subspace $\xi^{\prime}$, which is
complementary to $\xi$ in $\mathbb{R}^{n}$, so that also

$\mu(\xi^{\prime}\cap\mathbb{S}^{n-1})=\frac{1}{n}\mu(\mathbb{S}^{n-1})\mathrm{dim}\xi^{\prime},$

or equivalently so that $\mu$ is concentrated on
$\mathbb{S}^{n-1}\cap(\xi\cup\xi^{\prime})$.

######  Lemma 2.3.

_(Böröczky-Lutwak-Yang-Zhang,[7])_ A non-zero finite even Borel measure on the
unit sphere $\mathbb{S}^{n-1}$ is the cone-volume measure of an origin-
symmetric convex body in $\mathbb{R}^{n}$ if and only if it satisfies the
subspace concentration condition.

######  Lemma 2.4.

_(Henk-Linke,[17])_ Let $P$ be a polytope in $\mathbb{R}^{n}$ with its
centroid at the origin. Then its cone-volume measure $V_{P}$ satisfies the
subspace concentration condition.

##  3\. Recursion formulas

In this part, we find the recursion formulas of the volume decomposition
functionals $X_{k}$.

######  Definition 3.1.

Suppose $P\in\mathcal{P}_{o}^{n}$ and the unit outer normal vectors of $P$ are
$u_{1},u_{2},\dots,u_{N}$. Let $\xi$ be a subspace of $\mathbb{R}^{n}$ and
$k\in\\{1,2,\dots,n-1,n\\}.$ We define the quantity

$X_{k}(P;\xi)^{n}=\sum\limits_{\begin{subarray}{c}u_{i_{1}},\ldots,u_{i_{n}}\in\xi\\\
\mathrm{dim}(\mathrm{span}\\{u_{i_{1}},\ldots,u_{i_{n}}\\})=k\end{subarray}}V_{P}\left(\left\\{u_{i_{1}}\right\\}\right)\cdots
V_{P}\left(\left\\{u_{i_{n}}\right\\}\right).$

Note that if $\dim\xi<k$, then $X_{k}(P;\xi)=0$; If $\xi=\mathbb{R}^{n}$, then
$X_{k}(P;\mathbb{R}^{n})=X_{k}(P).$

For brevity, write $\mu(\xi)=\mu(\xi\cap\mathbb{S}^{n-1})$, for
$\mu\in\mathcal{B}(\mathbb{S}^{n-1})$ and $\xi\in\mathrm{G}(n)$. For instance,
if $\mu=V_{P}$, then $V_{P}(\xi)=V_{P}(\xi\cap\mathbb{S}^{n-1})$. It is
interesting that we establish the _local_ version of identity (1.1).

######  Lemma 3.1.

Let $\xi$ be a subspace of $\mathbb{R}^{n}$. Then

$V_{P}(\xi)^{n}=X_{1}(P;\xi)^{n}+X_{2}(P;\xi)^{n}+\cdots+X_{\mathrm{dim}\xi}(P;\xi)^{n}.$

###### Proof.

From the definitions of cone-volume measure and $X_{k}(P;\xi)$, it follows
that

$\displaystyle V_{P}(\xi)^{n}=$
$\displaystyle\big{(}\sum\limits_{\\{i:u_{i}\in\xi\\}}V_{P}\left(\left\\{u_{i}\right\\}\right)\big{)}^{n}=\sum\limits_{u_{i_{1}},\ldots,u_{i_{n}}\in\xi}V_{P}\left(\left\\{u_{i_{1}}\right\\}\right)\cdots
V_{P}\left(\left\\{u_{i_{n}}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{k=1}^{\mathrm{dim}\xi}\sum\limits_{\begin{subarray}{c}u_{i_{1}},\ldots,u_{i_{n}}\in\xi\\\
\mathrm{dim}(\mathrm{span}\\{u_{i_{1}},\ldots,u_{i_{n}}\\})=k\end{subarray}}V_{P}\left(\left\\{u_{i_{1}}\right\\}\right)\cdots
V_{P}\left(\left\\{u_{i_{n}}\right\\}\right)=\sum\limits_{k=1}^{\mathrm{dim}\xi}X_{k}(P;\xi)^{n}.$

That is,
$V_{P}(\xi)^{n}=X_{1}(P;\xi)^{n}+X_{2}(P;\xi)^{n}+\cdots+X_{\mathrm{dim}\xi}(P;\xi)^{n}.$
∎

In particular, if $\xi$ is a 1-dimensional subspace, say $\xi=l_{u_{i}}$, then

$X_{1}(P;l_{u_{i}})^{n}=V_{P}(\\{\pm u_{i}\\})^{n},\quad i=1,\dots,N;$

If $\xi=\mathbb{R}^{n}$, then

$V_{n}(P)^{n}=V_{n}(\mathbb{R}^{n})^{n}=X_{1}(P)^{n}+X_{2}(P)^{n}+\cdots+X_{n}(P)^{n}.$

Naturally, we pose the following

Problem Y. Let $P$ be a polytope in $\mathbb{R}^{n}$ with its centroid at the
origin. Do there exist constants $c_{1}$ and $c_{2}$ depending on $n$ and
$k,k\in\\{1,2,\dots,n-1\\},$ so that for $\xi\in\mathrm{G}_{n,k}$,

$\displaystyle X_{k}(P;\xi)\leq
c_{1}V_{P}(\xi)\quad\text{and}\quad\frac{X_{k}(P;\xi)}{X_{k}(P)}\leq c_{2}?$

For $k\in\\{1,\dots,n\\}$, let

$\\{\xi_{1}^{k},\xi_{2}^{k},\dots,\xi_{m_{k}}^{k}\\}=\\{\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\}:i_{1},\dots,i_{n}\in\\{1,2,\dots,N\\},\dim(\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\})=k\\}.$

That is, $\\{\xi_{1}^{k},\xi_{2}^{k},\dots,\xi_{m_{k}}^{k}\\}$ is the set of
$k$-dimensional subspaces spanned by unit outer normals of the polytope $P$,
and it has a total of $m_{k}$ elements.

Putting $\xi=\xi_{i}^{k}$ in Lemma 3.1, we obtain

(3.1)
$X_{k}(P;\xi_{i}^{k})^{n}=V_{P}(\xi_{i}^{k})^{n}-\sum\limits_{l=1}^{k-1}X_{l}(P;\xi_{i}^{k})^{n},\quad
i=1,2,\dots,m_{k},$

which suggests that $X_{k}(P;\xi_{i}^{k})$ can be represented by “lower order”
functionals $X_{l}(P;\xi_{i}^{k})$, $l=1,\dots,k-1$.

######  Lemma 3.2.

Let $\xi_{i}^{k}\in\\{\xi_{1}^{k},\xi_{2}^{k},\dots,\xi_{m_{k}}^{k}\\}$, and
$1\leq l<k\leq n$. Then

$X_{l}(P;\xi_{i}^{k})^{n}=\sum\limits_{\\{j\in\\{1,\dots,m_{l}\\}:\xi_{j}^{l}\subseteq\xi_{i}^{k}\\}}X_{l}(P;\xi^{l}_{j})^{n}.$

###### Proof.

By the Definition 3.1 and that $l<k$, it follows that

$\displaystyle X_{l}(P;\xi_{i}^{k})^{n}=$
$\displaystyle\sum\limits_{\begin{subarray}{c}u_{i_{1}},\ldots,u_{i_{n}}\in\xi_{i}^{k}\\\
\mathrm{dim}(\mathrm{span}\\{u_{i_{1}},\ldots,u_{i_{n}}\\})=l\end{subarray}}V_{P}\left(\left\\{u_{i_{1}}\right\\}\right)\cdots
V_{P}\left(\left\\{u_{i_{n}}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{\begin{subarray}{c}v_{1},\ldots,v_{n}\in\xi_{i}^{k}\cap\mathrm{supp}S_{P}\\\
\mathrm{dim}(\mathrm{span}\left\\{v_{1},\ldots,v_{n}\right\\})=l\end{subarray}}V_{P}\left(\left\\{v_{1}\right\\}\right)\cdots
V_{P}\left(\left\\{v_{n}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{\begin{subarray}{c}v_{1},\ldots,v_{n}\in\xi_{i}^{k}\cap\mathrm{supp}S_{P}\\\
\operatorname{span}\left\\{v_{1},\ldots,v_{n}\right\\}\in\\{\xi_{1}^{l},\dots,\xi_{m_{l}}^{l}\\}\end{subarray}}V_{P}\left(\left\\{v_{1}\right\\}\right)\cdots
V_{P}\left(\left\\{v_{n}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{\\{j\in\\{1,\dots,m_{l}\\}:\xi_{j}^{l}\subseteq\xi_{i}^{k}\\}}\sum\limits_{\begin{subarray}{c}v_{1},\ldots,v_{n}\in\xi_{i}^{k}\cap\mathrm{supp}S_{P}\\\
\operatorname{span}\left\\{v_{1},\ldots,v_{n}\right\\}=\xi_{j}^{l}\end{subarray}}V_{P}\left(\left\\{v_{1}\right\\}\right)\cdots
V_{P}\left(\left\\{v_{n}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{\\{j\in\\{1,\dots,m_{l}\\}:\xi_{j}^{l}\subseteq\xi_{i}^{k}\\}}\sum\limits_{\begin{subarray}{c}v_{1},\ldots,v_{n}\in\xi_{j}^{l}\cap\mathrm{supp}S_{P}\\\
\mathrm{dim}(\mathrm{span}\left\\{v_{1},\ldots,v_{n}\right\\})=l\end{subarray}}V_{P}\left(\left\\{v_{1}\right\\}\right)\cdots
V_{P}\left(\left\\{v_{n}\right\\}\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{\\{j\in\\{1,\dots,m_{l}\\}:\xi_{j}^{l}\subseteq\xi_{i}^{k}\\}}X_{l}(P;\xi^{l}_{j})^{n},$

as desired. ∎

Putting $k=n$ in Lemma 3.2, for $1\leq l<n$, we obtain

(3.2)
$X_{l}(P)^{n}=X_{l}(P;\mathbb{R}^{n})^{n}=X_{l}(P;\xi^{l}_{1})^{n}+X_{l}(P;\xi^{l}_{2})^{n}+\cdots+X_{l}(P;\xi^{l}_{m_{l}})^{n}.$

Combining Lemma 3.1 and Lemma 3.2, we derive the following recursion formulas.

(3.3) $\displaystyle X_{1}(P;\xi_{i}^{1})^{n}=V_{P}(\xi_{i}^{1})^{n},\quad
i=1,2,\dots,m_{1};$ (3.4) $\displaystyle
X_{1}(P;\xi)^{n}=\sum\limits_{\\{i:\xi_{i}^{1}\subseteq\xi\\}}X_{1}(P;\xi_{i}^{1})^{n},\quad\xi\in\mathrm{G}(n);$
(3.5) $\displaystyle
X_{k}(P;\xi^{k}_{i})^{n}=V_{P}(\xi^{k}_{i})^{n}-\sum\limits_{l=1}^{k-1}X_{l}(P;\xi_{i}^{k})^{n},\quad
i=1,2,\dots,m_{k},\quad k=2,3,\dots,n;$ (3.6) $\displaystyle
X_{l}(P;\xi_{i}^{k})^{n}=\sum\limits_{\\{j\in\\{1,\dots,m_{l}\\}:\xi_{j}^{l}\subseteq\xi_{i}^{k}\\}}X_{l}(P;\xi^{l}_{j})^{n},\quad
i=1,2,\dots,m_{k},\quad 1\leq l<k\leq n.$

These recursion formulas have the function of “dimension reduction”. As
applications, we work out the polynomial expressions of $X_{2}^{n}$ and
$X_{3}^{n}$.

######  Example 3.3.

Suppose $P\in\mathcal{P}_{o}^{n}$ and
$\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\pm u_{2},\dots,\pm
u_{N}\\}$. Then

$X_{2}(P)^{n}=\sum\limits_{i=1}^{m_{2}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}V_{P}(\\{\pm
u_{j}\\}))^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{)}.$

Indeed, by (3.2), (3.5), (3.4), (3.3) and the definition of $V_{P}$, we have

$\displaystyle X_{2}(P)^{n}=\sum_{i=1}^{m_{2}}X_{2}(P;\xi_{i}^{2})^{n}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{2}}\big{(}V_{P}(\xi_{i}^{2})^{n}-X_{1}(P;\xi_{i}^{2})^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{2}}\big{(}V_{P}(\xi_{i}^{2})^{n}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{i}^{2}\\}}X_{1}(P;\xi_{j}^{1})^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{2}}\big{(}V_{P}(\xi_{i}^{2})^{n}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{i}^{2}\\}}V_{P}(\xi_{j}^{1})^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{2}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}V_{P}(\\{\pm
u_{j}\\}))^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{)},$

as desired.

######  Example 3.4.

Suppose $P\in\mathcal{P}_{o}^{n}$ and
$\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\pm u_{2},\dots,\pm
u_{N}\\}$. Then

$X_{3}(P)^{n}=\sum\limits_{i=1}^{m_{3}}\big{\\{}V_{P}(\xi_{i}^{3})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}V_{P}(\xi_{l}^{2})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{\\}}.$

Indeed, by (3.2), (3.5), (3.6), (3.4), (3.3) and the definition of $V_{P}$, we
have

$\displaystyle X_{3}(P)^{n}=\sum_{i=1}^{m_{3}}X_{3}(P;\xi_{i}^{3})^{n}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\big{(}V_{P}(\xi_{i}^{3})^{n}-X_{2}(P;\xi_{i}^{3})^{n}-X_{1}(P;\xi_{i}^{3})^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\big{(}V_{P}(\xi_{i}^{3})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}X_{2}(P;\xi_{l}^{2})^{n}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{i}^{3}\\}}X_{1}(P;\xi_{j}^{1})^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\big{\\{}V_{P}(\xi_{i}^{3})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}V_{P}(\xi_{l}^{2})^{n}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{l}^{2}\\}}X_{1}(P;\xi_{j}^{1})^{n}\big{)}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{i}^{3}\\}}X_{1}(P;\xi_{j}^{1})^{n}\big{\\}}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\big{\\{}V_{P}(\xi_{i}^{3})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}V_{P}(\xi_{l}^{2})^{n}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{l}^{2}\\}}V_{P}(\xi_{j}^{1})^{n}\big{)}-\sum\limits_{\\{j:\xi_{j}^{1}\subseteq\xi_{i}^{3}\\}}V_{P}(\xi_{j}^{1})^{n}\big{\\}}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\big{\\{}V_{P}(\xi_{i}^{3})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}V_{P}(\xi_{l}^{2})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}V_{P}(\\{\pm
u_{j}\\})^{n}\big{\\}},$

where
$V_{P}(\xi_{i}^{3})=\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}V_{P}(\\{\pm
u_{j}\\})$,
$V_{P}(\xi_{l}^{2})=\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}V_{P}(\\{\pm
u_{j}\\})$, as desired.

In the following, we evaluate the $X_{3}$ of a cylinder in $\mathbb{R}^{4}$.

######  Example 3.5.

Let $u_{i}=(\cos\frac{2\pi i}{6},\sin\frac{2\pi i}{6},0,0),\ i=1,2,3;$
$u_{i}=(0,0,\cos\frac{2\pi i}{8},\sin\frac{2\pi i}{8}),\ i=4,5,6,7.$

Define an even discrete measure $\mu$ on $\mathbb{S}^{3}$ as the following

$\displaystyle\mu(\\{u_{i}\\})$
$\displaystyle=\mu(\\{-u_{i}\\})=\frac{1}{12},\quad i=1,2,3;$
$\displaystyle\mu(\\{u_{i}\\})$
$\displaystyle=\mu(\\{-u_{i}\\})=\frac{1}{16},\quad i=4,5,6,7.$

Then $\mu(\mathbb{S}^{3})=1$, and $\mathrm{supp}\mu=\\{\pm u_{1},\dots,\pm
u_{7}\\}$.

The set $\\{\xi_{1}^{3},\xi_{2}^{3},\dots,\xi_{m_{3}}^{3}\\}$ consists of 7
subspaces with dimension 3 as follows

$\mathrm{span}\\{u_{4},u_{5},u_{6},u_{7},u_{i}\\},\
i=1,2,3;\quad\quad\mathrm{span}\\{u_{1},u_{2},u_{3},u_{i}\\},\ i=4,5,6,7,$

and their mass are

$\displaystyle\mu(\mathrm{span}\\{u_{4},u_{5},u_{6},u_{7},u_{i}\\})$
$\displaystyle=8\times\frac{1}{16}+2\times\frac{1}{12}=\frac{2}{3}<\frac{3}{4},\quad
i=1,2,3;$ $\displaystyle\mu(\mathrm{span}\\{u_{1},u_{2},u_{3},u_{i}\\})$
$\displaystyle=6\times\frac{1}{12}+2\times\frac{1}{16}=\frac{5}{8}<\frac{3}{4},\quad
i=4,5,6,7.$

The set $\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}$ consists of 14
subspaces with dimension $2$ as follows

$\mathrm{span}\\{u_{1},u_{2},u_{3}\\};\
\mathrm{span}\\{u_{4},u_{5},u_{6},u_{7}\\};\ \mathrm{span}\\{u_{i},u_{j}\\},\
i=1,2,3,\ j=4,5,6,7,$

and their mass are

$\displaystyle\mu(\mathrm{span}\\{u_{1},u_{2},u_{3}\\})=6\times\frac{1}{12}=\frac{1}{2};\
\mu(\mathrm{span}\\{u_{4},u_{5},u_{6},u_{7}\\})=8\times\frac{1}{16}=\frac{1}{2};$
$\displaystyle\mu(\mathrm{span}\\{u_{i},u_{j}\\})=2\times\frac{1}{12}+2\times\frac{1}{16}=\frac{7}{24}<\frac{2}{4},\quad
i=1,2,3,\ j=4,5,6,7.$

Thus, $\mu$ satisfies subspace concentration inequalities. Since subspace
$\mathrm{span}\\{u_{1},u_{2},u_{3}\\}$ is complementary to subspace
$\mathrm{span}\\{u_{4},u_{5},u_{6},u_{7}\\}$ in $\mathbb{R}^{4}$, it follows
that $\mu$ satisfies subspace concentration condition. By the Böröczky-LYZ
existence theorem 2.3, there exists a cylinder $P\in\mathcal{P}_{os}^{4}$, so
that $V_{P}=\mu$.

Putting the above digits in Example 3.4, we obtain

$\displaystyle X_{3}(P)^{4}=$ $\displaystyle
3\big{\\{}(\frac{2}{3})^{4}-\big{[}\big{(}(\frac{1}{2})^{4}-4(\frac{1}{8})^{4}\big{)}+4\big{(}(\frac{7}{24})^{4}-(\frac{1}{6})^{4}-(\frac{1}{8})^{4}\big{)}\big{]}-\big{[}4(\frac{1}{8})^{4}+(\frac{1}{6})^{4}\big{]}\big{\\}}$
$\displaystyle+$ $\displaystyle
4\big{\\{}(\frac{5}{8})^{4}-\big{[}\big{(}(\frac{1}{2})^{4}-3(\frac{1}{6})^{4}\big{)}+3\big{(}(\frac{7}{24})^{4}-(\frac{1}{6})^{4}-(\frac{1}{8})^{4}\big{)}\big{]}-\big{[}3(\frac{1}{6})^{4}+(\frac{1}{8})^{4}\big{]}\big{\\}}$
$\displaystyle\approx$ $\displaystyle 0.613.$

Remark. In light of Theorem 1.3 and the fact that
$0.613>0.576=\frac{72}{125}$, it yields that

$\sup_{P\in\mathcal{P}_{4}^{4}}\frac{X_{3}(P)}{V_{4}(P)}=0.576<0.613\leq\sup_{P\in\mathcal{P}_{c}^{4}}\frac{X_{3}(P)}{V_{4}(P)},$

which suggests us that $\frac{X_{3}}{V_{4}}$ does _not_ attain its extremum at
_simplex_ in $\mathcal{P}_{c}^{4}$, i.e., the set of polytopes in
$\mathbb{R}^{4}$ with centroid at the origin.

Since $X_{k}(P)^{n}$ is a homogenous polynomial in $V_{P}(\\{\pm
u_{1}\\}),V_{P}(\\{\pm u_{2}\\}),\dots,V_{P}(\\{\pm u_{N}\\})$, to attack the
Problem X, we must characterize the domain of functional $X_{k}$

$\\{(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathbb{R}^{N}:P\in\mathcal{P}_{c}^{n},\
\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\dots,\pm u_{N}\\}\\}.$

Moreover, since $\frac{X_{k}}{V_{n}}$ is affine invariant, we focus on the
normalized domain

(3.7) $\displaystyle D(u_{1},\dots,u_{N})$ $\displaystyle=\\{(V_{P}$
$\displaystyle(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathbb{R}^{N}:P\in\mathcal{P}_{c}^{n},\ V_{n}(P)=n,\
\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\dots,\pm u_{N}\\}\\}.$

In Section 4, we prove that $D(u_{1},\dots,u_{N})$ is precisely the _relative
interior_ of the so-called _matroid polytope_. Please refer to Theorem 4.8 for
details.

In the following, we list some basic facts from matroid theory. For standard
reference, see [13, 29].

##  4\. Matroid polytope

###  4.1. Introduction of matroid and matroid polytope

Matroid theory dates from the 1930’s when van der Waerden in his “Moderne
Algebra” first approached linear and algebraic dependence axiomatically and
Whitney in his basic paper [34] first used the term matroid. In the past
several decades, matroid theory has been developing rapidly and witnessed
strong connections with other mathematical disciplines, such as algebraic
geometry [1, 10, 12, 21], lattice theory [8, 14] and graph theory [9, 33].

For a finite set $E$, write $|E|$ for the number of elements of $E$.

######  Definition 4.1.

A matroid is an ordered pair $(E,\mathcal{I})$ consisting of a finite set $E$
and a collection $\mathcal{I}$ of subsets of $E$ with the following three
properties:

$\mathrm{(i)}$ $\emptyset\in\mathcal{I}$.

$\mathrm{(ii)}$ If $I\in\mathcal{I}$ and $I^{\prime}\subseteq I$, then
$I^{\prime}\in\mathcal{I}$.

$\mathrm{(iii)}$ If $I_{1}$ and $I_{2}$ are in $\mathcal{I}$ and
$|I_{1}|<|I_{2}|$, then there exists an element $e\in I_{2}-I_{1}$ such that
$I_{1}\cup e\in\mathcal{I}$.

If $M$ is the matroid $(E,\mathcal{I})$, then $M$ is called a matroid on $E$.
The members of $\mathcal{I}$ are called the _independent sets_ of $M$; a
subset of $E$ that is not in $\mathcal{I}$ is called a _dependent set_ of $M$.
In the sense of inclusion relation of sets, we call a _maximal_ independent
set in $M$ a _basis_ of $M$, and a _minimal_ dependent set in $M$ a _circuit_
of $M$. Denote all bases and all circuits of $M$ by $\mathcal{B}(M)$ and
$\mathcal{C}(M)$, respectively.

For $x,y\in E$, $x$ and $y$ are _equivalent_ if there exists a circuit $C$
with $\\{x,y\\}\subseteq C.$ The equivalence classes are the _connected
components_ of $M$. Let $c(M)$ denote the number of connected components of
$M$. We say that $M$ is _connected_ if $c(M)=1.$

The _rank_ of $X\in 2^{E}$ is defined as

$r_{M}(X)=\max\\{|I|:I\subseteq X,I\in\mathcal{I}\\}.$

Let $\mathrm{cl}:2^{E}\to 2^{E}$, defined for all $X\in 2^{E}$ by

$\mathrm{cl}(X)=\\{x\in E:r_{M}(X\cup\\{x\\})=r_{M}(X)\\}.$

A subset $X$ of $E$ for which $\mathrm{cl}(X)=X$ is called a _flat_ or a
_closed set_ of $M$.

Clearly, once $\mathcal{I}$ has been specified, from the definition of bases,
$\mathcal{B}(M)$ is determined. Conversely, $\mathcal{I}$ can be determined
from the set $\mathcal{B}(M)$. In fact, from property $\mathrm{(ii)}$ in
Definition 4.1, we conclude that the members of $\mathcal{I}$ are precisely
all subsets of members of $\mathcal{B}(M)$. That is,

$\mathcal{I}=\bigcup\limits_{B\in\mathcal{B}(M)}2^{B}.$

Thus a matroid $M=(E,\mathcal{I})$ is uniquely determined by the set
$\mathcal{B}(M)$.

It is interesting that from $\mathcal{B}(M)$, the so-called _matroid polytope_
can be defined. In fact, let $E$ be a finite set, say, $E=\\{1,\dots,N\\}$.
Given a basis $B\in\mathcal{B}(M)$, the _indicator vector_ of $B$ is defined
as $e_{B}=\sum\limits_{\begin{subarray}{c}i=1\\\ i\in
B\end{subarray}}^{N}e_{i},$ where
$e_{i}=\\{\underbrace{0,\dots,0}_{i-1},1,\underbrace{0,\dots,0}_{N-i}\\}$.
Then, the associated matroid polytope of $M$ is given by

$P_{M}=\mathrm{conv}\\{e_{B}:B\in\mathcal{B}(M)\\}.$

Note that by property $\mathrm{(iii)}$, it follows that $|B_{1}|=|B_{2}|$ for
$B_{1},B_{2}\in\mathcal{B}(M)$. Assume $|B|=n$ for $B\in\mathcal{B}(M)$. Then,
each $e_{B}$ has $n$ coordinates $1$ and $(N-n)$ coordinates $0$. Refer to
[10, p. 440] and [12, p. 311] for more details on matroid polytopes.

######  Example 4.1.

Suppose $E=\\{1,2,3,4\\}$ and
$\mathcal{I}=\\{\emptyset,\\{1\\},\\{2\\},\\{3\\},\\{4\\},\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\}\\}$.
Then $M=(E,\mathcal{I})$ is a matroid. From the definition of the bases, we
have

$\mathcal{B}(M)=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\}\\}.$

Then the indicator vector set associated to $\mathcal{B}(M)$ is the following
set

$\displaystyle\\{e_{1}+e_{2},e_{1}+e_{3},e_{1}+e_{4},e_{2}+e_{3},e_{2}+e_{4}\\}.$

Therefore, the associated matroid polytope of $M$ is

$P_{M}=\mathrm{conv}\\{(1,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1)\\},$

which is a square pyramid in the $3$-dimension space
$x_{1}+x_{2}+x_{3}+x_{4}=2$.

So, $P_{M}\subseteq[0,1]^{4}\cap H_{(1,1,1,1),2}$. See figure (a) in the blow.

######  Example 4.2.

If $\\{3,4\\}$ is also an independent set of $M$, then

$\mathcal{B}(M)=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}\\},$

and the indicator vector set associated to $\mathcal{B}(M)$ is the following
set

$\displaystyle\\{e_{1}+e_{2},e_{1}+e_{3},e_{1}+e_{4},e_{2}+e_{3},e_{2}+e_{4},e_{3}+e_{4}\\}.$

Therefore, the associated matroid polytope of $M$ is

$P_{M}=\mathrm{conv}\\{(1,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1),(0,0,1,1)\\},$

which is an octahedron in the $3$-dimension space $x_{1}+x_{2}+x_{3}+x_{4}=2$.

So, $P_{M}=[0,1]^{4}\cap H_{(1,1,1,1),2}$. See figure (b) in the above.

######  Example 4.3.

In general, if $E=\\{1,\dots,N\\}$ and bases $\mathcal{B}(M)$ is the
collection of all $n$-element subsets of $\\{1,\dots,N\\}$, then
$\mathcal{B}(M)$ has $\tbinom{N}{n}$ elements, and therefore $P_{M}$ is the
convex hull of $\tbinom{N}{n}$ vectors, which have $n$ coordinates $1$ and
$(N-n)$ coordinates $0$. Such a matroid polytope $P_{M}$ is called the
_hypersimplex_ and denoted by $\bigtriangleup_{N}^{n}$, which is precisely the
intersection of the $N$-cube $[0,1]^{N}\subseteq\mathbb{R}^{N}$ with the
hyperplane $x_{1}+x_{2}+\cdots+x_{N}=n$. That is,
$\bigtriangleup_{N}^{n}=[0,1]^{N}\cap H_{\mathrm{I}_{N},n},$ where
$\mathrm{I}_{N}=(1,\dots,1)\in\mathbb{R}^{N}.$

Significantly, there are more exact geometric information on the hypersimplex
$\bigtriangleup_{N}^{n}$. For example, every edge of $\bigtriangleup_{N}^{n}$
is a translation of one of the vectors $e_{i}-e_{j}$, for $i\neq j$. Please
consult [22, p. 351] for details. In 1987, Gelfand, Goresky, MacPherson and
Serganova [12, p. 311] provided an interesting geometric characterization for
matroid polytope via $\bigtriangleup_{N}^{n}$.

######  Theorem 4.4.

_(Gelfand-Goresky-MacPherson-Serganova,[12])_ Suppose $P$ is a polytope
contained in the hypersimplex $\bigtriangleup_{N}^{n}$. Then there exists a
matroid $M$ such that $P=P_{M}$ if and only if the vertices of $P$ are a
subset of the vertices of $\bigtriangleup_{N}^{n}$ and each edge of $P$ is a
translation of one of the vectors $e_{i}-e_{j}$, for $i\neq j$.

In 2005, Feichtner and Sturmfels [10, pp. 441-442] obtained an explicit
representation for $P_{M}$ and its dimension by making use of Theorem 4.4.

######  Theorem 4.5.

_(Feichtner-Sturmfels,[10])_ Let $M=(E,\mathcal{I})$ be a matroid. Then

(4.1) $P_{M}=\\{(x_{1},\dots,x_{N})\in\bigtriangleup_{N}^{n}:\sum_{i\in
F}x_{i}\leq{r_{M}}(F)\>\text{for all flats}\>F\ \text{of}\ M\\},$

and its dimension $\dim P_{M}=n-c(M)$.

Remark. In our previous version of this paper submitted to a renowned jounal,
we defined the so-called “ _concentration polytope_ ” in terms of subspace
concentration condition, and expended much effort to characterize its vertices
and edges. See [26] for details.

We appreciate the anonymous reviewer very much for pointing out that our
concentration polytope is precisely the matroid polytope
$P_{M(u_{1},\dots,u_{N})}$ associated with a set of vectors
$u_{1},\dots,u_{N}$. See the following Section 4.2 for details. More
importantly, the reviewer provided two seminal papers [10] and [12], and told
us the vertices and edges of $P_{M(u_{1},\dots,u_{N})}$ have been fully
characterized as Theorem 4.4.

In the following, we return to the volume decomposition functional $X_{k}$ and
Problem X. To solve Problem X, we have to figure out the domain
$D(u_{1},\dots,u_{N})$ of $X_{k}$. Fortunately, we discover that
$D(u_{1},\dots,u_{N})$ is precisely the _relative interior_ of
$P_{M(u_{1},\dots,u_{N})}$.

###  4.2. The relative interior of matroid polytope

Suppose $u_{1},\dots,u_{N}\in\mathbb{S}^{n-1}$ are pairwise unparallel and
$\\{u_{1},\dots,u_{N}\\}$ is not contained in any subsphere of
$\mathbb{S}^{n-1}$. Via these $u_{i}$, $i=1,\dots,N$, a matroid
$M(u_{1},\dots,u_{N})$ is defined as the following.

Specifically, let $E=\\{1,2,\dots,N\\}$ and

$\mathcal{I}=\\{I\in 2^{E}:u_{i},\ i\in I,\ \text{are linear
independent}\\}\cup\\{\emptyset\\}.$

Then $M(u_{1},\dots,u_{N})=(E,\mathcal{I})$ is a matroid, and

$\displaystyle\mathcal{B}(M)=\\{B\in 2^{E}:u_{i},\ i\in B,\ \text{consist a
basis of }\mathbb{R}^{n}\\},$ $\displaystyle
r_{M(u_{1},\dots,u_{N})}(X)=\dim(\mathrm{span}\\{u_{i}:i\in X\\}),\ \text{for
}X\in 2^{E},$ $\displaystyle\mathrm{cl}(X)=\\{i\in
E:u_{i}\in\mathrm{span}\\{u_{i}:i\in X\\}\\},\ \ \,\text{for }X\in 2^{E}.$

By the definition of flat, $F$ is a flat of $M(u_{1},\dots,u_{N})$ if and only
if there exists a subspace $\xi$ of $\mathbb{R}^{n}$ such that $F=\\{i\in
E:u_{i}\in\xi\\}$. Write $F_{1},\dots,F_{m}$ for all the flats of
$M(u_{1},\dots,u_{N})$. Let

(4.2) $\xi_{j}=\mathrm{span}\\{u_{i}:i\in F_{j}\\}\quad\text{and}\quad
v_{j}=\sum_{i\in F_{j}}e_{i},\quad j=1,\dots,m.$

Then $r_{M(u_{1},\dots,u_{N})}(F_{j})=\dim\xi_{j}$, and

(4.3) $\sum_{i\in F_{j}}x_{i}=x\cdot v_{j},\quad\forall
x=(x_{1},\dots,x_{N})\in\mathbb{R}^{N}.$

Thus, by Theorem 4.5, (4.3) and Example 4.3, it follows that

$\displaystyle P_{M(u_{1},\dots,u_{N})}=$
$\displaystyle\\{(x_{1},\dots,x_{N})\in\bigtriangleup_{N}^{n}:\sum_{i\in
F}x_{i}\leq{r_{M(u_{1},\dots,u_{N})}}(F_{j}),\ j=1,\dots,m\\}$
$\displaystyle=$
$\displaystyle\\{(x_{1},\dots,x_{N})\in\bigtriangleup_{N}^{n}:x\cdot
v_{j}\leq\dim\xi_{j},\ j=1,\dots,m\\}$ $\displaystyle=$
$\displaystyle[0,1]^{N}\cap
H_{\mathrm{I}_{N},n}\cap(\cap_{j=1}^{m}H_{v_{j},\mathrm{dim}\xi_{j}}^{-})$
(4.4) $\displaystyle=$ $\displaystyle(\cap_{i=1}^{N}H_{-e_{i},0}^{-})\cap
H_{\mathrm{I}_{N},n}\cap(\cap_{j=1}^{m}H_{v_{j},\mathrm{dim}\xi_{j}}^{-}),$

where the last equality comes from that for $i\in\\{1,\dots,N\\}$, $\\{i\\}$
is indeed a flat of $M(u_{1},\dots,u_{N})$, say $F_{i}$, and therefore
$v_{i}=e_{i}$ and $\dim\xi_{i}=\dim(\mathrm{span}\\{u_{i}\\})=1$.
Consequently,

$\\{(x_{1},\dots,x_{N})\in\mathbb{R}^{N}:x_{i}\leq
1\\}=H_{e_{i},1}^{-}=H_{v_{i},\dim\xi_{i}}^{-}.$

It is the representation (4.4) of $P_{M(u_{1},\dots,u_{N})}$ that we fully
characterize the matroid polytope $P_{M(u_{1},\dots,u_{N})}$ in our previous
version [26].

In the following, we first characterize the relative interior of matroid
polytope $P_{M(u_{1},\dots,u_{N})}$.

Let $w_{i}\in\mathbb{S}^{N-1}$ and $a_{i}\in\mathbb{R}$, $i=1,2,\dots,M$.
Suppose the set $Q=\cap_{i=1}^{M}H_{w_{i},a_{i}}^{-}\subseteq\mathbb{R}^{N}$
is nonempty.

######  Lemma 4.6.

Let $x\in Q$. Then $x\in\mathrm{relint}Q$ if and only if there does not exist
an $i\in\\{1,\dots,M\\}$ such that $x\in H_{w_{i},a_{i}}$ and $Q\nsubseteq
H_{w_{i},a_{i}}.$

###### Proof.

Assume that

$Q\nsubseteq H_{w_{i},a_{i}},\ 1\leq i\leq j;\quad\text{and}\quad Q\subseteq
H_{w_{i},a_{i}},\ j+1\leq i\leq M,\quad j\in\\{0,1,\dots,M\\}.$

Then $Q$ is completely lying in an affine subspace
$\bar{S}=\cap_{i=j+1}^{M}H_{w_{i},a_{i}}$. If $j=0$, then $Q=\bar{S}$ is
itself an affine subspace; If $j=M$, then $Q$ is an $N$-dimensional polytope.
For these two cases, the lemma naturally holds. So assume that $1\leq j\leq
M-1$ in the following.

Suppose there does not exist an $i\in\\{1,\dots,M\\}$ so that $x\in
H_{w_{i},a_{i}},$ but $Q\nsubseteq H_{w_{i},a_{i}}$. Then
$x\in\mathrm{int}H_{w_{i},a_{i}}^{-}$, $1\leq i\leq j$. Let
$\varepsilon=\min\limits_{1\leq i\leq j}\\{a_{i}-x\cdot w_{i}\\}$. Then
$\varepsilon>0$, and $x+\varepsilon B\subseteq H_{w_{i},a_{i}}^{-}$, $1\leq
i\leq j$. So,

$(x+\varepsilon
B)\cap\bar{S}\subseteq(\cap_{i=1}^{j}H_{w_{i},a_{i}}^{-})\cap\bar{S}=(\cap_{i=1}^{j}H_{w_{i},a_{i}}^{-})\cap(\cap_{i=j+1}^{M}H_{w_{i},a_{i}})=Q,$

which implies that $x\in\mathrm{relint}Q$.

On the contrary, assume $x\in\mathrm{relint}Q$. We aim to show that $x\notin
H_{w_{i},a_{i}},\ 1\leq i\leq j$. If there exists an $i\in\\{1,\dots,j\\}$ so
that $x\in H_{w_{i},a_{i}}$, then there exists $y\in Q$ with
$y\in\mathrm{int}H_{w_{i},a_{i}}^{-}$ by that $Q\nsubseteq H_{w_{i},a_{i}}$.
From the facts that $x\in\mathrm{relint}Q$ and the segment $[x,y]\subseteq Q$,
there is a sufficiently small $\rho>0$ so that $x+\rho(x-y)\in Q$. Then
$(x+\rho(x-y))\cdot w_{i}=x\cdot w_{i}+\rho(x-y)\cdot w_{i}>a_{i},$ which
contradicts that $Q\subseteq H_{w_{i},a_{i}}^{-}$. ∎

######  Theorem 4.7.

Let $x\in\mathbb{R}^{N}$. Then $x\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$
if and only if there exists an origin-symmetric polytope $P$ in
$\mathbb{R}^{n}$ with $V_{n}(P)=n$ and $\mathrm{supp}S_{P}=\\{\pm
u_{1},\dots,\pm u_{N}\\}$, such that

$x=\big{(}V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\})\big{)}.$

###### Proof.

We first prove the sufficiency. Let $P$ be an origin-symmetric polytope with
$V_{n}(P)=n$ and $\mathrm{supp}S_{P}=\\{\pm u_{1},\dots,\pm u_{N}\\}$. Suppose
$x=\big{(}V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\})\big{)}$. Then

$x\in\mathrm{int}(H_{-e_{i},0}^{-}),\quad i=1,\dots,N.$

In light of formula (4.4) and Lemma 4.6, it suffices to show that there is the
following implication

$x\in H_{v_{j},\mathrm{dim}\xi_{j}}\ \Longrightarrow\
P_{M(u_{1},\dots,u_{N})}\subseteq H_{v_{j},\mathrm{dim}\xi_{j}},\quad
j=1,\dots,m,$

where $\xi_{j}$ and $v_{j}$ are defined by (4.2).

Assume that $x\in H_{v_{j},\mathrm{dim}\xi_{j}},$ for some
$j\in\\{1,\dots,m\\}$. By equation (4.3), it follows that

$\frac{V_{P}(\xi_{j}\cap\mathbb{S}^{n-1})}{V_{n}(P)}=\frac{1}{V_{n}(P)}\sum\limits_{\\{i:u_{i}\in\xi_{j}\\}}V_{P}(\\{\pm
u_{i}\\})=\frac{1}{n}\sum\limits_{\\{i:u_{i}\in\xi_{j}\\}}x_{i}=\frac{x\cdot
v_{j}}{n}=\frac{\mathrm{dim}\xi_{j}}{n}.$

From Lemma 2.4, there exists a subspace $\xi_{j}^{\prime}$ complementary to
$\xi_{j}$ such that

$\frac{1}{n}\sum\limits_{\\{i:u_{i}\in\xi_{j}^{\prime}\\}}x_{i}=\frac{V_{P}(\xi_{j}^{\prime}\cap\mathbb{S}^{n-1})}{V_{n}(P)}=\frac{\mathrm{dim}\xi_{j}^{\prime}}{n}=1-\frac{\mathrm{dim}\xi_{j}}{n},$

which suggests us that $\\{i:u_{i}\in\xi_{j}\\}$ and
$\\{i:u_{i}\in\xi_{j}^{\prime}\\}$ constitute a disjoint partition of
$\\{1,2,\dots,N\\}$. So, for each $y=(y_{1},\dots,y_{N})\in
P_{M(u_{1},\dots,u_{N})}$, it satisfies the equation

$\sum_{\\{i:u_{i}\in\xi_{j}\\}}y_{i}+\sum_{\\{i:u_{i}\in\xi_{j}^{\prime}\\}}y_{i}=\sum_{i=1}^{N}y_{i}=n.$

Meanwhile, by formula (4.4), it follows that

$\sum_{\\{i:u_{i}\in\xi_{j}\\}}y_{i}\leq\mathrm{dim}\xi_{j}\ \ \ \text{and}\ \
\sum_{\\{i:u_{i}\in\xi_{j}^{\prime}\\}}y_{i}\leq\mathrm{dim}\xi_{j}^{\prime}=n-\mathrm{dim}\xi_{j}.$

So, it is necessary that

$\mathrm{dim}\xi_{j}=\sum_{\\{i:u_{i}\in\xi_{j}\\}}y_{i}=y\cdot v_{j}.$

That is, $y\in H_{v_{j},\mathrm{dim}\xi_{j}}.$ Therefore,
$P_{M(u_{1},\dots,u_{N})}\subseteq H_{v_{j},\mathrm{dim}\xi_{j}}$. Second, we
prove the necessity. Assume
$x=(x_{1},\dots,x_{N})\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$. Let

$\mu=\sum_{i=1}^{N}\frac{x_{i}}{2}(\delta_{u_{i}}+\delta_{-u_{i}}).$

Then $\mu$ is a even discrete measure on $\mathbb{S}^{n-1}$, and
$\mu(\mathbb{S}^{n-1})=n$.

In the following, we verify that $\mu$ satisfies the subspace concentration
condition.

For each proper subspace $\xi$, the subspace
$\mathrm{span}(\xi\cap\\{u_{1},\dots,u_{N}\\})\cup\\{o\\}$ is exactly one of
elements of $\\{\xi_{1},\dots,\xi_{m}\\}$ by the definition (4.2), say
$\xi_{j}$. By formula (4.4), it follows that

(4.5)
$\frac{\mu(\xi\cap\mathbb{S}^{n-1})}{\mu(\mathbb{S}^{n-1})}=\frac{\sum\limits_{\\{i:u_{i}\in\xi_{j}\\}}x_{i}}{\sum_{i=1}^{N}x_{i}}\leq\frac{\dim\xi_{j}}{n}\leq\frac{\dim\xi}{n}.$

In addition, assume all the equalities hold in (4.5). Then $\xi=\xi_{j}$. We
aim to find a subspace $\xi_{j}^{\prime}$ complementary to $\xi_{j}$, such
that $\mu(\xi^{\prime}_{j}\cap\mathbb{S}^{n-1})=\mathrm{dim}\xi^{\prime}_{j}$.

First, since

$x\cdot v_{j}=\sum\limits_{\\{i:u_{i}\in\xi_{j}\\}}x_{i}=\mathrm{dim}\xi_{j},$

it follows that $x\in H_{v_{j},\mathrm{dim}\xi_{j}}.$ From the assumption that
$x\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$, and Lemma 4.6, it follows that
$P_{M(u_{1},\dots,u_{N})}\subseteq H_{v_{j},\mathrm{dim}\xi_{j}}$.

Second, since $\\{u_{1},\dots,u_{N}\\}$ is not to be contained in any
subsphere of $\mathbb{S}^{n-1}$, there exists an origin-symmetric polytope $Q$
in $\mathbb{R}^{n}$ with $V_{n}(Q)=n$ and $\mathrm{supp}S_{Q}=\\{\pm
u_{1},\dots,\pm u_{N}\\}$. By the proof of the sufficient condition, we
conclude that

(4.6) $\big{(}V_{Q}(\\{\pm u_{1}\\}),\dots,V_{Q}(\\{\pm
u_{N}\\})\big{)}\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}\subseteq
H_{v_{j},\mathrm{dim}\xi_{j}}.$

Thus,
$\frac{V_{Q}(\xi_{j}\cap\mathbb{S}^{n-1})}{V_{n}(Q)}=\frac{\dim\xi_{j}}{n}$.
By the Definition 2.1, there exists a subspace $\xi_{j}^{\prime}$
complementary to $\xi_{j}$, such that

$\frac{\sum_{\\{i:u_{i}\in\xi_{j}^{\prime}\\}}V_{Q}(\\{\pm
u_{i}\\})}{V_{n}(Q)}=\frac{V_{Q}(\xi_{j}^{\prime}\cap\mathbb{S}^{n-1})}{V_{n}(Q)}=\frac{\mathrm{dim}\xi_{j}^{\prime}}{n},$

which suggests that $\\{i:u_{i}\in\xi_{j}\\}$ and
$\\{i:u_{i}\in\xi_{j}^{\prime}\\}$ constitute a disjoint partition of
$\\{1,2,\dots,N\\}$. So,

$\frac{\mu(\xi^{\prime}_{j}\cap\mathbb{S}^{n-1})}{\mu(\mathbb{S}^{n-1})}=\frac{1}{n}\sum_{\\{i:u_{i}\in\xi_{j}^{\prime}\\}}x_{i}=\frac{1}{n}\sum_{i=1}^{N}x_{i}-\frac{1}{n}\sum_{\\{i:u_{i}\in\xi_{j}\\}}x_{i}=1-\frac{\mathrm{dim}\xi_{j}}{n}=\frac{\mathrm{dim}\xi_{j}^{\prime}}{n}.$

Therefore, $\mu$ satisfies the subspace concentration condition. By the
Böröczky-LYZ existence theorem 2.3, there exists an origin-symmetric polytope
$P$ such that $V_{P}=\mu$. So, $V_{n}(P)=\mu(\mathbb{S}^{n-1})=n$ and

$x=(x_{1},\dots,x_{N})=(\mu(\\{\pm u_{1}\\}),\dots,\mu(\\{\pm
u_{N}\\}))=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\})).$

Furthermore, in light of (4.6), it follows that
$P_{M(u_{1},\dots,u_{N})}\nsubseteq H_{-e_{i},0}$, $i\in\\{1,2,\dots,N\\}$.
Combining this fact with Lemma 4.6, it follows that
$x\in\mathrm{int}H_{-e_{i},0}^{-}$. So, $V_{P}(\\{\pm u_{i}\\})=x_{i}>0$,
$i=1,2,\dots,N$, which implies that $\mathrm{supp}S_{P}=\\{\pm u_{1},\dots,\pm
u_{N}\\}$. Consequently, the necessity part is derived. ∎

With Theorem 4.7 in hand, we show that the normalized domain
$D(u_{1},\dots,u_{N})$ of the functionals $X_{k}$ (see the formula (3.7)), is
precisely the relative interior of the matroid polytope.

######  Theorem 4.8.

$D(u_{1},\dots,u_{N})=\mathrm{relint}P_{M(u_{1},\dots,u_{N})}.$

###### Proof.

By Theorem 4.7, we obtain that $\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$ is
the following set

$\\{(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathbb{R}^{N}:P\in\mathcal{P}_{os}^{n},\ V_{n}(P)=n,\
\mathrm{supp}S_{P}=\\{\pm u_{1},\dots,\pm u_{N}\\}\\}.$

Comparing it with the definition of $D(u_{1},\dots,u_{N})$, it is obvious that
$\mathrm{relint}P_{M(u_{1},\dots,u_{N})}\subseteq D(u_{1},\dots,u_{N})$.

On the other hand, let $P\in\mathcal{P}_{c}^{n}$ with $V_{n}(P)=n$ and
$\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\dots,\pm u_{N}\\}$.
Let

$\mu=\sum_{i=1}^{N}\frac{V_{P}(\\{\pm
u_{i}\\})}{2}(\delta_{u_{i}}+\delta_{-u_{i}}).$

Then $\mu$ is an even discrete measure on $\mathbb{S}^{n-1}$. By Lemma 2.4,
$V_{P}$ satisfies the subspace concentration condition, and therefore $\mu$
satisfies the subspace concentration condition. By the Böröczky-LYZ existence
theorem 2.3, there is a polytope $P^{\prime}\in\mathcal{P}_{os}^{n}$ such that
$V_{P^{\prime}}=\mu$. Thus, $\mathrm{supp}S_{P^{\prime}}=\\{\pm
u_{1},\dots,\pm u_{N}\\}$ and $V_{n}(P^{\prime})=n$, and therefore
$(V_{P^{\prime}}(\\{\pm u_{1}\\}),\dots,V_{P^{\prime}}(\\{\pm
u_{N}\\}))\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$. So,

$(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\}))=(V_{P^{\prime}}(\\{\pm
u_{1}\\}),\dots,V_{P^{\prime}}(\\{\pm
u_{N}\\}))\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})},$

which yields that
$D(u_{1},\dots,u_{N})\subseteq\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$. ∎

Combining Theorem 4.8 with Theorem 4.4, the characterization for the _closure_
$\overline{D}(u_{1},\dots,u_{N})$ of $D(u_{1},\dots,u_{N})$ is immediately
obtained.

######  Theorem 4.9.

Each vertex of $\overline{D}(u_{1},\dots,u_{N})$ has $n$ coordinates $1$ and
$(N-n)$ coordinates $0$ and each edge of $\overline{D}(u_{1},\dots,u_{N})$ is
a translation of one of the vectors $e_{i}-e_{j}$, for $i\neq j$.

###  4.3. Applications of matroid theory to convex geometry

Define a mapping from the matroid polytope $P_{M(u_{1},\dots,u_{N})}$ to the
set of finite even discrete measures on $\mathbb{S}^{n-1}$ by

$x\longmapsto\mu_{x}=\sum_{i=1}^{N}\frac{x_{i}}{2}(\delta_{u_{i}}+\delta_{-u_{i}}),\quad\forall
x=(x_{1},\dots,x_{N})\in P_{M(u_{1},\dots,u_{N})}.$

Theorem 4.7 shows that for each $x\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$,
there exist an origin-symmetric polytope $P$ such that $\mu_{x}=V_{P}$. In the
following, we focus on the points on the relative boundary of
$P_{M(u_{1},\dots,u_{N})}$.

######  Theorem 4.10.

The relative boundary of $P_{M(u_{1},\dots,u_{N})}$ satisfies the following
properties.

$\mathrm{(i)}$ For each vertex $x$ of $P_{M(u_{1},\dots,u_{N})}$, there is an
origin-symmetric parallelotope $P$ such that $\mu_{x}=V_{P}$.

$\mathrm{(ii)}$ For any point $x$ in the relative interior of an edge of
$P_{M(u_{1},\dots,u_{N})}$, there does not exist an origin-symmetric polytope
$P$ such that $\mu_{x}=V_{P}$.

$\mathrm{(iii)}$ Let $\\{u_{1},\dots,u_{N}\\}$ be in general position. For
each $x\in\mathrm{relbd}P_{M(u_{1},\dots,u_{N})}$, there exists an origin-
symmetric polytope $P$ such that $\mu_{x}=V_{P}$ if and only if $x$ is a
vertex or $x\cdot e_{i}\in[0,1)$, $i=1,\dots,N$.

###### Proof.

$\mathrm{(i)}$ Suppose $x$ is a vertex of $P_{M(u_{1},\dots,u_{N})}$. By
Theorem 4.9, there exist $i_{1},\dots,i_{n}\in\\{1,\dots,N\\}$ such that
$x=\sum_{j=1}^{n}e_{i_{j}}$. From the definition of matroid polytope, it
follows that $\\{i_{1},\dots,i_{n}\\}$ is a basis of $M(u_{1},\dots,u_{N})$,
and therefore $\\{u_{i_{1}},\dots,u_{i_{n}}\\}$ is a basis of
$\mathbb{R}^{n}$. So, there is an origin-symmetric parallelotope $P$ with
volume $n$ and unit normal vector set $\\{\pm u_{i_{1}},\dots,\pm
u_{i_{n}}\\}.$ Hence,
$\mu_{x}=\frac{1}{2}\sum_{j=1}^{n}(\delta_{u_{i_{j}}}+\delta_{-u_{i_{j}}})=V_{P}$.

$\mathrm{(ii)}$ Suppose $x$ is a point in the relative interior of an edge of
$P_{M(u_{1},\dots,u_{N})}$. By Theorem 4.9, $x$ has $(n-1)$ coordinates $1$,
one coordinate $\lambda$ and one coordinate $(1-\lambda)$, where
$\lambda\in(0,1)$. W.l.o.g. assume
$x=(\underbrace{1,\dots,1}_{n-1},\lambda,1-\lambda,0,\dots,0)$. Then

$\mu_{x}=\frac{\lambda}{2}(\delta_{u_{n}}+\delta_{-u_{n}})+\frac{1-\lambda}{2}(\delta_{u_{n+1}}+\delta_{-u_{n+1}})+\frac{1}{2}\sum_{j=1}^{n-1}(\delta_{u_{j}}+\delta_{-u_{j}}).$

So,
$\frac{\mu_{x}(\mathrm{span}\\{u_{1},\dots,u_{n-1}\\}\cap\mathbb{S}^{n-1})}{\mu_{x}(\mathbb{S}^{n-1})}=\frac{n-1}{n}.$
Since $u_{n}$ is not parallel to $u_{n+1}$, it follows that $\\{\pm u_{n},\pm
u_{n+1}\\}$ does not lie in a $1$-dimensional subspace, and therefore
$\mu_{x}$ does not satisfy the subspace concentration condition. By Theorem
2.3, there does not exist an origin-symmetric polytope $P$ such that
$\mu_{x}=V_{P}$.

$\mathrm{(iii)}$ We first prove the sufficiency. According to $\mathrm{(i)}$,
it suffices to prove the case that $x\cdot e_{i}\in[0,1)$, $i=1,\dots,N$.
Since $u_{1},\dots,u_{N}$ are in general position, it follows that for each
proper subspace $\xi$ of $\mathbb{R}^{n}$, $\xi$ contains at most $\dim\xi$
vectors among $u_{1},\dots,u_{N}$. So,

$\frac{\mu_{x}(\xi\cap\mathbb{S}^{n-1})}{\mu_{x}(\mathbb{S}^{n-1})}=\frac{1}{n}\sum_{\\{i:u_{i}\in\xi\\}}x_{i}<\frac{\dim\xi}{n}.$

Hence, $\mu_{x}$ satisfies the subspace concentration condition. By Theorem
2.3, there exists an origin-symmetric polytope $P$ such that $\mu_{x}=V_{P}$

Second, we prove the necessity. Suppose that there exists an origin-symmetric
polytope $P$ such that $\mu_{x}=V_{P}$. It suffices to prove that if there
exists an $i\in\\{1,\dots,N\\}$ such that $x_{i}=1$, then $x$ is a vertex of
$P_{M(u_{1},\dots,u_{N})}$. Assume $x_{i}=1$. Then

$\frac{\mu_{x}(\pm u_{i})}{\mu_{x}(\mathbb{S}^{n-1})}=\frac{1}{n}.$

Since $\mu_{x}$ satisfies the subspace concentration condition, it follows
that there exists a $(n-1)$-dimensional subspace $\xi^{\prime}$ complementary
to $\mathrm{span}\\{u_{i}\\}$ so that $\mathrm{supp}\mu_{x}\setminus\\{\pm
u_{i}\\}\subseteq\xi^{\prime}$. Since $u_{1},\dots,u_{N}$ are in general
position, it follows that $\xi^{\prime}$ contains at most $(n-1)$ vectors
among $u_{1},\dots,u_{N}$. Thus, $\mathrm{supp}\mu_{x}$ contains precisely $n$
vectors among $u_{1},\dots,u_{N}$. Therefore, $x$ is a vertex of
$P_{M(u_{1},\dots,u_{N})}$ by the definition of $\mu_{x}$. ∎

In the following Theorem 4.11, we determine the dimension of
$P_{M(u_{1},\dots,u_{N})}$. By the Feichtner-Sturmfels theorem 4.5, it
suffices to determine $c(M(u_{1},\dots,u_{N}))$, i.e., the number of connected
components of $M(u_{1},\dots,u_{N})$. Let $P$ be any origin-symmetric polytope
such that $\mathrm{supp}S_{P}=\\{\pm u_{1},\dots,\pm u_{N}\\}$ and
$V_{n}(P)=n$. Suppose $P=P_{1}\oplus\cdots\oplus P_{l}$, $\dim P_{i}>0$ and
$P_{i}$ is directly indecomposable, $i=1,\dots,l$.

######  Theorem 4.11.

$\dim P_{M(u_{1},\dots,u_{N})}=N-l$.

###### Proof.

We complete the proof by dividing two cases: $l=1$ and $l>1$.

Case 1. Assume $l=1$. We aim to show that $P_{M(u_{1},\dots,u_{N})}$ contains
an $(N-1)$-dimensional ball. If so, then $\dim P_{M(u_{1},\dots,u_{N})}=N-1$
by the fact that $P_{M(u_{1},\dots,u_{N})}\subseteq H_{\mathrm{I}_{N},n}$.

Indeed, by Theorem 4.7 we have that $x=\big{(}(V_{P}(\\{\pm
u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}$. In the following, we
prove that there exists $\varepsilon>0$ such that

$(x+\varepsilon B)\cap H_{\mathrm{I}_{N},n}\subseteq
P_{M(u_{1},\dots,u_{N})},$

where $B$ is the unit ball in $\mathbb{R}^{N}$.

By Lemma 2.4 and the assumption that $l=1$ (i.e., $P$ is not a cylinder), it
follows that

$\frac{V_{P}(\xi\cap\mathbb{S}^{n-1})}{V_{n}(P)}<\frac{\dim\xi}{n},\quad\text{for
each proper subspace}\ \xi.$

Let

$\varepsilon=\frac{1}{N}\min\\{\dim\xi-
V_{P}(\xi\cap\mathbb{S}^{n-1}):\xi\in\\{\xi_{1},\dots,\xi_{m}\\}\text{ is a
proper subspace}\\},$

where $\\{\xi_{1},\dots,\xi_{m}\\}$ is defined by (4.2). Moreover, assume
$\varepsilon<V_{P}(\\{\pm u_{i}\\})$, $i=1,\dots,N$.

For each $y\in(x+\varepsilon B)\cap H_{\mathrm{I}_{N},n}$, on one hand we have

$y\cdot e_{i}\geq x\cdot e_{i}-\varepsilon=V_{P}(\\{\pm
u_{i}\\})-\varepsilon>0,\quad i=1,\dots,N.$

Meanwhile, since $|v_{j}|\leq N$, it follows that

$y\cdot v_{j}\leq(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\}))\cdot
v_{j}+N\varepsilon=V_{P}(\xi_{j}\cap\mathbb{S}^{n-1})+N\varepsilon\leq\dim\xi_{j},\quad
j=1,\dots,m.$

From formula (4.4), it follows that $y\in P_{M(u_{1},\dots,u_{N})}$, and
therefore $(x+\varepsilon B)\cap H_{\mathrm{I}_{N},n}\subseteq
P_{M(u_{1},\dots,u_{N})}.$

Case 2. Assume $l>1$. We aim to show that $c(M(u_{1},\dots,u_{N}))=l$.

By the definition of $M(u_{1},\dots,u_{N})$, it follows that $M(\phi
u_{1},\dots,\phi u_{N})=M(u_{1},\dots,u_{N})$ for $\phi\in\mathrm{GL}(n)$.
W.l.o.g., assume $P_{j}\subseteq\xi_{j}$, $j=1,\dots,l$, where
$\xi_{1},\dots,\xi_{l}$ are pairwise orthogonal subspaces of $\mathbb{R}^{n}$.

Let $F_{j}=\\{i:u_{i}\in\xi_{j}\\}$, $j=1,\dots,l.$ In the following, we prove
that all the connected components of $M(u_{1},\dots,u_{N})$ are precisely
$F_{1},\dots,F_{l}$, and therefore $c(M(u_{1},\dots,u_{N}))=l$.

First, we prove that $F_{j}$ is connected, for $j\in\\{1,\dots,l\\}$. In fact,
the unit outer normal vector set of $P_{j}$ in the subspace $\xi_{j}$ is
$\\{u_{i},-u_{i}:i\in F_{j}\\}$. By the fact that $P_{j}$ is directly
indecomposable and Case 1, it follows that $F_{j}$ is connected.

Second, we prove that for distinct $j_{1},j_{2}\in\\{1,\dots,l\\}$, $i_{1}$ is
not equivalent to $i_{2}$, for $i_{1}\in F_{j_{1}}$ and $i_{2}\in F_{j_{2}}.$
Otherwise, assume there exist $i_{1}\in F_{j_{1}}$ and $i_{2}\in F_{j_{2}}$
such that $i_{1}$ is equivalent to $i_{2}$. By the definition of equivalence,
there exists a circuit $C$ so that $\\{i_{1},i_{2}\\}\subseteq C$. Since a
circuit is a minimal dependent set, it follows that $C\setminus F_{j_{1}}$ and
$C\cap F_{j_{1}}$ are both independent sets. Since $\mathrm{span}\\{u_{i}:i\in
C\setminus F_{j_{1}}\\}$ is orthogonal to $\xi_{j_{1}}$, it follows that
$C=(C\setminus F_{j_{1}})\cup(C\cap F_{j_{1}})$ is an independent set, which
contradicts to that $C$ is a dependent set.

Consequently, $c(M(u_{1},\dots,u_{N}))=l$. ∎

##  5\. Proofs of Theorems 1.1 and 1.2

Now, with the aid of Theorem 4.9, we prove our main results. We first prove
Theorem 1.2.

######  Theorem 5.1.

Let $P$ be a polytope in $\mathbb{R}^{n}$ with its centroid at the origin and
$n\geq 5$. Then

$\frac{X_{3}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3(\frac{2}{n})^{n}+\frac{3}{n^{n}}\big{)}}$

with equality if and only if $P$ is a parallelotope.

###### Proof.

Assume that $\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\pm
u_{2},\dots,\pm u_{N}\\}$ and $V_{n}(P)=n.$

If $P$ is a parallelotope, by Example 3.4, it follows that

$\frac{X_{3}(P)^{n}}{V_{n}(P)^{n}}=\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3\big{(}(\frac{2}{n})^{n}-2\frac{1}{n^{n}}\big{)}-3\frac{1}{n^{n}}\big{)}=\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3(\frac{2}{n})^{n}+\frac{3}{n^{n}}\big{)}.$

In the following, we aim to show that

$\frac{X_{3}(P)^{n}}{V_{n}(P)^{n}}<\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3(\frac{2}{n})^{n}+\frac{3}{n^{n}}\big{)},$

as long as $P$ is _not_ a parallelotope.

Let

$\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}=\\{\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\}:i_{1},\dots,i_{n}\in\\{1,2,\dots,N\\},\dim(\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\})=2\\};$
$\\{\xi_{1}^{3},\xi_{2}^{3},\dots,\xi_{m_{3}}^{3}\\}=\\{\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\}:i_{1},\dots,i_{n}\in\\{1,2,\dots,N\\},\dim(\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\})=3\\}.$

That is, $\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}$ consists of
_all_ the $2$-dimensional subspaces spanned by $n$ outer normals of $P$;
$\\{\xi_{1}^{3},\xi_{2}^{3},\dots,\xi_{m_{3}}^{3}\\}$ consists of _all_ the
$3$-dimensional subspaces spanned by $n$ outer normals of $P$.

Let $a=(a_{1},\dots,a_{N})=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))$. From Example 3.4, it follows that

$X_{3}(P)^{n}=\sum\limits_{i=1}^{m_{3}}\Big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}a_{j})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}a_{j})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}a_{j}^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}a_{j}^{n}\Big{)}.$

In the rest, we investigate the maximum of the functional

$g(x_{1},\dots,x_{N})=\sum\limits_{i=1}^{m_{3}}\Big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j}^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j}^{n}\Big{)}$

in the matroid polytope $P_{M(u_{1},\dots,u_{N})}$. Obviously that
$g(a)=X_{3}(P)^{n}$.

Since $g$ is a polynomial and $P_{M(u_{1},\dots,u_{N})}$ is compact, it
follows that $g$ attains its maximum on $P_{M(u_{1},\dots,u_{N})}$. What
follows aims to show that $g$ attains its maximum precisely at _vertices_ of
the matroid polytope $P_{M(u_{1},\dots,u_{N})}$. We show this by
contradiction.

Step 1. Assume that $g$ attains its maximum at $x\in
P_{M(u_{1},\dots,u_{N})}\setminus\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.
By Theorem 4.9, there exist distinct $j_{1},j_{2}\in\\{1,\dots,N\\}$ and
sufficiently small $\varepsilon>0$ so that

$x+\varepsilon[e_{j_{2}}-e_{j_{1}},e_{j_{1}}-e_{j_{2}}]\subseteq
P_{M(u_{1},\dots,u_{N})}.$

W.l.o.g., assume $j_{1}=1,\ j_{2}=2$.

Let $G(t)=g(x+t(e_{1}-e_{2}))$, $t\in[-\varepsilon,\varepsilon]$. By the
assumption that $g$ attains its maximum at the point $x$, it implies that
$G(t)$ attains its maximum at $t=0$. Hence, it is necessary that

$G^{\prime}(0)=0,\ \ \ \ \text{and}\ \ \ \
G^{\prime\prime}(0)=(\frac{\partial^{2}g}{\partial
x_{1}^{2}}+\frac{\partial^{2}g}{\partial
x_{2}^{2}}-2\frac{\partial^{2}g}{\partial x_{1}\partial x_{2}})(x)\leq 0.$

However, in the following we show that $\frac{\partial^{2}g}{\partial
x_{1}^{2}}-\frac{\partial^{2}g}{\partial x_{1}\partial x_{2}}>0$ (Similarly,
that $\frac{\partial^{2}g}{\partial x_{2}^{2}}-\frac{\partial^{2}g}{\partial
x_{1}\partial x_{2}}>0$.), which is a contradiction. To prove it, we divide it
into four sub-steps.

Step 1.1. To calculate and simplify $\frac{\partial^{2}g}{\partial
x_{1}^{2}}-\frac{\partial^{2}g}{\partial x_{1}\partial x_{2}}$. By calculating
directly, we obtain

$\displaystyle\frac{\partial g}{\partial
x_{1}}=n\sum\limits_{\\{i:u_{1}\in\xi_{i}^{3}\\}}\Big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n-1}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-1}-x_{1}^{n-1}\big{)}-x_{1}^{n-1}\Big{)},$
$\displaystyle\frac{\partial^{2}g}{\partial
x_{1}^{2}}=n(n-1)\sum\limits_{\\{i:u_{1}\in\xi_{i}^{3}\\}}\Big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n-2}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}-x_{1}^{n-2}\Big{)},$
$\displaystyle\frac{\partial^{2}g}{\partial x_{1}\partial
x_{2}}=n(n-1)\sum\limits_{\\{i:u_{1},u_{2}\in\xi_{i}^{3}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n-2}-(\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\\}}x_{j})^{n-2}\big{)},$

where
$\xi_{u_{1},u_{2}}=\mathrm{span}\\{u_{1},u_{2}\\}\in\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}$.
Hence,

$\displaystyle\frac{1}{n(n-1)}(\frac{\partial^{2}g}{\partial
x_{1}^{2}}-\frac{\partial^{2}g}{\partial x_{1}\partial x_{2}})$
$\displaystyle=$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\Big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n-2}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}-x_{1}^{n-2}\Big{)}$
$\displaystyle-\sum\limits_{\\{i:u_{1},u_{2}\in\xi^{3}_{i}\\}}\sum\limits_{\\{l:u_{1}\in\xi^{2}_{l}\subseteq\xi^{3}_{i},u_{2}\notin\xi_{l}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}$
$\displaystyle\triangleq$ $\displaystyle A-B.$

To further calculate A and B, w.l.o.g., assume
$\\{\xi_{i}^{2}:u_{1}\in\xi^{2}_{i}\\}=\\{\xi^{2}_{1},\xi^{2}_{2},\dots,\xi^{2}_{k}\\}$,
$k\leq m_{2}$, and $\xi_{1}^{2}=\xi_{u_{1},u_{2}}$. Two observations are in
order.

First,
$u_{1},u_{2}\in\mathrm{span}\\{u_{2},\xi_{l}^{2}\\}\in\\{\xi_{1}^{3},\xi_{2}^{3},\dots,\xi_{m_{3}}^{3}\\}$,
$l=2,3,\dots,k$.

Second, for distinct $i_{1},i_{2}\in\\{i:u_{1},u_{2}\in\xi^{3}_{i}\\}$,
$\xi_{i_{1}}^{3}\cap\xi_{i_{2}}^{3}=\xi_{u_{1},u_{2}}$.

Combining the above two observations, it follows that

(5.1)
$\bigcup_{i\in\\{i:u_{1},u_{2}\in\xi^{3}_{i}\\}}\\{l:u_{1}\in\xi^{2}_{l}\subseteq\xi^{3}_{i},~{}\text{and}~{}u_{2}\notin\xi_{l}^{2}\\}$

is indeed a _disjoint_ union of $\\{2,3,\dots,k\\}$. Hence,

$B=\sum\limits_{\\{i:u_{1},u_{2}\in\xi^{3}_{i}\\}}\sum\limits_{\\{l:u_{1}\in\xi^{2}_{l}\subseteq\xi^{3}_{i},u_{2}\notin\xi_{l}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}=\sum\limits_{l=2}^{k}\big{(}(\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}.$

Let

$y_{l}=\sum_{\\{j:u_{j}\in\xi_{l}^{2}\setminus\\{u_{1}\\}\\}}x_{j}=-x_{1}+\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j},\quad
l=1,2,\dots,k.$

Then $y_{l}\leq\sum_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j}\leq 2$,
$B=\sum\limits_{l=2}^{k}\big{(}(x_{1}+y_{l})^{n-2}-x_{1}^{n-2}\big{)}$ and

$A=\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\Big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n-2}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(x_{1}+y_{l})^{n-2}-x_{1}^{n-2}\big{)}-x_{1}^{n-2}\Big{)}.$

Observe that for each
$i\in\\{i:u_{1}\in\xi_{i}^{3},u_{2}\notin\xi_{i}^{3}\\},$

$\bigcup_{l\in\\{l:\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\\{j:u_{j}\in\xi_{l}^{2}\setminus\\{u_{1}\\}\\}$

is indeed a _disjoint_ union of
$\\{j\in\\{2,3,\dots,N\\}:u_{j}\in\xi_{i}^{3}\setminus\\{u_{1}\\}\\}$. We
obtain that

(5.2)
$\sum_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j}=x_{1}+\sum_{\\{j:u_{j}\in\xi_{i}^{3}\setminus\\{u_{1}\\}\\}}x_{j}=x_{1}+\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}y_{l},\quad\forall
i\in\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi_{i}^{3}\\}.$

Therefore,

$A=\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\Big{(}(x_{1}+\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}y_{l})^{n-2}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\big{(}(x_{1}+y_{l})^{n-2}-x_{1}^{n-2}\big{)}-x_{1}^{n-2}\Big{)}.$

Step 1.2. Varying $x_{1}$ in $A-B$, let

$\displaystyle h_{\alpha}(t)=$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\Big{(}(t+\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}y_{l})^{\alpha}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\big{(}(t+y_{l})^{\alpha}-t^{\alpha}\big{)}-t^{\alpha}\Big{)}$
$\displaystyle-\sum\limits_{l=2}^{k}\big{(}(t+y_{l})^{\alpha}-t^{\alpha}\big{)},\quad\quad\quad
t\in[0,\infty),\quad\alpha=3,4,\dots,n-2.$

Thus, $A-B=h_{n-2}(x_{1})$, and

$\frac{d^{n-2-\alpha}}{(dt)^{n-2-\alpha}}h_{n-2}=\frac{(n-2)!}{\alpha!}h_{\alpha},\quad\alpha=3,4,\dots,n-3.$

By the mean value theorem, there exists an $\eta\in(0,1)$ satisfying

$\displaystyle h_{n-2}(x_{1})=$ $\displaystyle
h_{n-2}(0)+(n-2)h_{n-3}(0)x_{1}+(n-2)(n-3)h_{n-4}(0)\frac{x_{1}^{2}}{2!}+\cdots$
$\displaystyle+\binom{n-2}{n-6}h_{4}(0)x_{1}^{n-6}+\binom{n-2}{n-5}h_{3}(\eta
x_{1})x_{1}^{n-5}.$

If we can show that $h_{n-2}(x_{1})>0,$ then $A-B>0$ is derived.

Step 1.3. To show that $h_{\alpha}(0)>0$, $\alpha=4,5,\dots,n-2$.

By $\alpha\geq 4$ and collecting the terms with the factor $\alpha
y_{l}^{\alpha-1}$, we obtain

$\displaystyle h_{\alpha}(0)=$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\big{\\{}(\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}y_{l})^{\alpha}-\sum_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}y_{l}^{\alpha}\big{\\}}-\sum_{l=2}^{k}y_{l}^{\alpha}$
$\displaystyle\geq$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\big{\\{}\sum_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\alpha
y_{l}^{\alpha-1}\sum\limits_{\\{s:u_{1}\in\xi_{s}^{2}\subseteq\xi^{3}_{i},s\neq
l\\}}y_{s}\big{\\}}-\sum\limits_{l=2}^{k}y_{l}^{\alpha}$ $\displaystyle=$
$\displaystyle\sum\limits_{l=2}^{k}\alpha
y_{l}^{\alpha-1}\sum\limits_{\big{\\{}s:u_{1}\in\xi_{s}^{2},u_{2}\notin\mathrm{span}\\{\xi^{2}_{s},\xi^{2}_{l}\\},s\neq
l\big{\\}}}y_{s}-\sum_{l=2}^{k}y_{l}^{\alpha}$ $\displaystyle=$
$\displaystyle\sum_{l=2}^{k}\alpha
y_{l}^{\alpha-1}\sum_{\big{\\{}s:u_{1}\in\xi_{s}^{2},\xi^{2}_{s}\nsubseteq\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\big{\\}}}y_{s}-\sum_{l=2}^{k}y_{l}^{\alpha}.$

Note that
$\bigcup_{l\in\\{l:u_{1}\in\xi_{l}^{2}\\}}\\{j:u_{j}\in\xi_{l}^{2}\setminus\\{u_{1}\\}\\}$
is indeed a _disjoint_ union of $\\{2,3,\dots,N\\}$. We obtain that

(5.3)
$\sum\limits_{l=1}^{k}y_{l}=\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\\}}\sum_{\\{j:u_{j}\in\xi_{l}^{2}\setminus\\{u_{1}\\}\\}}x_{j}=\sum_{j=2}^{N}x_{j}=n-x_{1}.$

From (5.3), (5.2), (4.4), and that $y_{l}\leq 2$, $\alpha\geq 4,n\geq 5$, it
follows that

$\displaystyle h_{\alpha}(0)\geq$ $\displaystyle\sum\limits_{l=2}^{k}\alpha
y_{l}^{\alpha-1}\big{(}n-x_{1}-\sum\limits_{\big{\\{}s:u_{1}\in\xi^{2}_{s}\subseteq\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\big{\\}}}\\!\\!y_{s}\big{)}-\sum\limits_{l=2}^{k}y_{l}^{\alpha}$
$\displaystyle=$ $\displaystyle\sum\limits_{l=2}^{k}\alpha
y_{l}^{\alpha-1}(n-x_{1}-(-x_{1}+\sum_{\\{j:u_{j}\in\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\\}}\\!\\!x_{j}))-\sum\limits_{l=2}^{k}y_{l}^{\alpha}$
$\displaystyle=$ $\displaystyle\sum\limits_{l=2}^{k}\alpha
y_{l}^{\alpha-1}(n-\sum_{\\{j:u_{j}\in\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\\}\\!\\!}x_{j})-\sum\limits_{l=2}^{k}y_{l}^{\alpha}$
$\displaystyle=$
$\displaystyle\sum\limits_{l=2}^{k}y_{l}^{\alpha-1}(n\alpha-\alpha\sum_{\\{j:u_{j}\in\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\\}}\\!\\!x_{j}-y_{l})$
$\displaystyle\geq$
$\displaystyle\sum\limits_{l=2}^{k}y_{l}^{\alpha-1}(n\alpha-3\alpha-2)>0.$

Step 1.4. To show that $h_{3}(t)>0$, for $t\in(0,x_{1})$.

By collecting the terms with the factor $3y_{l}^{2}$ and $y_{l}$,
respectively, (5.3) and (5.2), we obtain

$\displaystyle h_{3}(t)$ $\displaystyle=$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\\!\\!\\!\Big{(}(t+\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\\!\\!\\!y_{l})^{3}-\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\\!\\!\\!\big{(}(y_{l}+t)^{3}-t^{3}\big{)}-t^{3}\Big{)}-\sum\limits_{l=2}^{k}\big{(}(t+y_{l})^{3}-t^{3}\big{)}$
$\displaystyle\geq$
$\displaystyle\sum\limits_{\\{i:u_{1}\in\xi^{3}_{i},u_{2}\notin\xi^{3}_{i}\\}}\\!\\!\\!\big{(}\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\\!\\!\\!3y_{l}^{2}\sum\limits_{\\{s:u_{1}\in\xi_{s}^{2}\subseteq\xi^{3}_{i},s\neq
l\\}}\\!\\!\\!y_{s}+3t\\!\\!\\!\sum\limits_{\\{l:u_{1}\in\xi_{l}^{2}\subseteq\xi^{3}_{i}\\}}\\!\\!\\!y_{l}\sum\limits_{\\{s:u_{1}\in\xi_{s}^{2}\subseteq\xi^{3}_{i},s\neq
l\\}}\\!\\!\\!y_{s}\big{)}-\sum\limits_{l=2}^{k}\big{(}(t+y_{l})^{3}-t^{3}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{l=2}^{k}3y_{l}^{2}\sum\limits_{\big{\\{}s:u_{1}\in\xi_{s}^{2},\xi^{2}_{s}\nsubseteq\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\big{\\}}}\\!\\!\\!y_{s}+3t\sum\limits_{l=2}^{k}y_{l}\sum\limits_{\big{\\{}s:u_{1}\in\xi_{s}^{2},\xi^{2}_{s}\nsubseteq\mathrm{span}\\{u_{2},\xi^{2}_{l}\\}\big{\\}}}\\!\\!\\!y_{s}-\sum\limits_{l=2}^{k}(y_{l}^{3}+3y_{l}^{2}t+3y_{l}t^{2})$
$\displaystyle=$
$\displaystyle\sum\limits_{l=2}^{k}3y_{l}^{2}(n-\sum_{\\{j:u_{j}\in\mathrm{span}\\{u_{2},\xi_{l}^{2}\\}\\}}\\!\\!\\!x_{j})+3t\sum\limits_{l=2}^{k}y_{l}(n-\sum_{\\{j:u_{j}\in\mathrm{span}\\{u_{2},\xi_{l}^{2}\\}\\}}\\!\\!\\!x_{j})-\sum\limits_{l=2}^{k}(y_{l}^{3}+3y_{l}^{2}t+3y_{l}t^{2})$
$\displaystyle\geq$
$\displaystyle\sum\limits_{l=2}^{k}3y_{l}^{2}(n-3)+3t\sum\limits_{l=2}^{k}y_{l}(n-3)-\sum\limits_{l=2}^{k}(y_{l}^{3}+3y_{l}^{2}t+3y_{l}t^{2})$
$\displaystyle=$
$\displaystyle\sum\limits_{l=2}^{k}y_{l}^{2}(3n-9-y_{l}-3t)+3t\sum\limits_{l=2}^{k}y_{l}(n-3-t).$

Since $0<t<x_{1}\leq 1$, $y_{l}\leq 2$, and $n\geq 5$, we have

$h_{3}(t)>\sum\limits_{l=2}^{k}y_{l}^{2}(3n-9-2-3)+3t\sum\limits_{l=2}^{k}y_{l}(n-3-1)=\sum\limits_{l=2}^{k}y_{l}^{2}(3n-14)+3t\sum\limits_{l=2}^{k}y_{l}(n-4)>0.$

Combining the above four sub-steps, we have shown that

$\frac{\partial^{2}g}{\partial x_{j_{1}}^{2}}-\frac{\partial^{2}g}{\partial
x_{j_{1}}\partial x_{j_{2}}}=n(n-1)(A-B)={n(n-1)}h_{n-2}(x_{1})>0.$

Consequently, $g$ cannot attain its maximum in the region
$P_{M(u_{1},\dots,u_{N})}\setminus\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

Step 2. To evaluate the functional $g$ at points of
$\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

For each $x\in\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$, by Theorem 4.9,
w.l.o.g., assume

$x=(\underbrace{1,\dots,1}_{n},\underbrace{0,\dots,0}_{N-n}).$

By the definition of $P_{M(u_{1},\dots,u_{N})}$, it follows that
$u_{1},u_{2},\dots,u_{n}$ must be linearly independent. So, for each
$i\in\\{1,2,\dots,m_{3}\\}$, there are at most three elements of
$\\{u_{1},u_{2},\dots,u_{n}\\}$ lying in $\xi_{i}^{3}$. Moreover, if the
number of elements of $\\{u_{1},u_{2},\dots,u_{n}\\}$ lying in $\xi_{i}^{3}$
is strictly less than 3, i.e., $|\xi_{i}^{3}\cap\\{u_{1},\dots,u_{n}\\}|<3$,
then

$(\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j}^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j}^{n}=0$

by calculating directly when $|\xi_{i}^{3}\cap\\{u_{1},\dots,u_{n}\\}|=0,1,2$.

Therefore,

$\displaystyle g(x)=$
$\displaystyle\sum\limits_{i=1}^{m_{3}}\Big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j}^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j}^{n}\Big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{\\{i:|\xi_{i}^{3}\cap\\{u_{1},\dots,u_{n}\\}|=3\\}}\Big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j})^{n}-\sum\limits_{\\{l:\xi_{l}^{2}\subseteq\xi_{i}^{3}\\}}\big{(}(\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j})^{n}-\sum\limits_{\\{j:u_{j}\in\xi_{l}^{2}\\}}x_{j}^{n}\big{)}-\sum\limits_{\\{j:u_{j}\in\xi_{i}^{3}\\}}x_{j}^{n}\Big{)}$
$\displaystyle=$ $\displaystyle\binom{n}{3}\big{(}3^{n}-3(2^{n}-2)-3\big{)}$
$\displaystyle=$ $\displaystyle\binom{n}{3}\big{(}3^{n}-3\times
2^{n}+3\big{)}.$

Step 3. Combining Step 1 and Step 2, we obtain that for $x\in
P_{M(u_{1},\dots,u_{N})}$,

$g(x)\leq\binom{n}{3}\big{(}3^{n}-3\times 2^{n}+3\big{)},$

with equality if and only if $x\in\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

If $P$ is _not_ a parallelotope, then $N>n$. By Theorem 4.8, we obtain that

$a=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}.$

Hence,

$X_{3}(P)^{n}=g(a)<\max_{x\in
P_{M(u_{1},\dots,u_{N})}}g(x)=\binom{n}{3}\big{(}3^{n}-3\times
2^{n}+3\big{)}.$

To sum up, for a polytope $P$ with its centroid at the origin, it follows that

$\frac{X_{3}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{3}\big{(}(\frac{3}{n})^{n}-3(\frac{2}{n})^{n}+\frac{3}{n^{n}}\big{)}},$

with equality if and only if $P$ is a parallelotope. ∎

Now, we prove Theorem 1.1.

######  Theorem 5.2.

Let $P$ be a polytope in $\mathbb{R}^{n}$ with its centroid at the origin and
$n\geq 3$. Then

$\frac{X_{2}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{2}\big{(}(\frac{2}{n})^{n}-\frac{2}{n^{n}}\big{)}},$

with equality if and only if $P$ is a parallelotope.

###### Proof.

Assume that $\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\pm
u_{2},\dots,\pm u_{N}\\}$ and $V_{n}(P)=n.$

If $P$ is a parallelotope, then by Example 3.3, it follows that

$X_{2}(P)^{n}=\binom{n}{2}\big{(}2^{n}-2\big{)}.$

In the following, we aim to show that

$X_{2}(P)^{n}<\binom{n}{2}\big{(}2^{n}-2\big{)},$

as long as $P$ is _not_ a parallelotope.

Let

$\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}=\\{\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\}:i_{1},\dots,i_{n}\in\\{1,2,\dots,N\\},\dim(\mathrm{span}\\{u_{i_{1}},\dots,u_{i_{n}}\\})=2\\}.$

That is, $\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}$ consists of
_all_ the $2$-dimensional subspaces spanned by $n$ unit outer normals of the
polytope $P$.

Let $a=(a_{1},\dots,a_{N})=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))$. From the assumption that $V_{n}(P)=n$ and Example 3.3, it follows
that

$\frac{X_{2}(P)^{n}}{V_{n}(P)^{n}}=\sum\limits_{i=1}^{m_{2}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}a_{j})^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}a_{j}^{n}\big{)}.$

In the rest, we investigate the maximum of the functional

$f(x_{1},\dots,x_{N})=\sum\limits_{i=1}^{m_{2}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j})^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j}^{n}\big{)}$

in the set $P_{M(u_{1},\dots,u_{N})}$. Obviously that
$f(a)=X_{2}(P)^{n}/V_{n}(P)^{n}$.

Since $P_{M(u_{1},\dots,u_{N})}$ is compact and $f$ is a polynomial, it
follows that $f$ attains its maximum on $P_{M(u_{1},\dots,u_{N})}$. What
follows aims to show that $f$ attains its maximum precisely at the _vertices_
of the polytope $P_{M(u_{1},\dots,u_{N})}$. We show this by contradiction.

Step 1. Assume that $f$ attains its maximum at $x\in
P_{M(u_{1},\dots,u_{N})}\setminus\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.
By Theorem 4.9, there exist distinct $j_{1},j_{2}\in\\{1,\dots,N\\}$ and
sufficiently small $\varepsilon>0$ so that

$x+\varepsilon[e_{j_{2}}-e_{j_{1}},e_{j_{1}}-e_{j_{2}}]\subseteq
P_{M(u_{1},\dots,u_{N})}.$

W.l.o.g., assume $j_{1}=1,\ j_{2}=2$.

Let

$F(t)=f(x+t(e_{1}-e_{2})),\quad t\in[-\varepsilon,\varepsilon].$

By the assumption that $f$ attains its maximum at the point $x$, it implies
that $F(t)$ attains its maximum at $t=0$. Hence, it is necessary that

$F^{\prime}(0)=0,\ \ \ \ \text{and}\ \ \ \
F^{\prime\prime}(0)=(\frac{\partial^{2}f}{\partial
x_{1}^{2}}+\frac{\partial^{2}f}{\partial
x_{2}^{2}}-2\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}})(x)\leq 0.$

However, in the following we show that $\frac{\partial^{2}f}{\partial
x_{1}^{2}}-\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}>0$ (Similarly,
that $\frac{\partial^{2}f}{\partial x_{2}^{2}}-\frac{\partial^{2}f}{\partial
x_{1}\partial x_{2}}>0$.), which is a contradiction. To prove it, we divide it
into two sub-steps.

Step 1.1. By calculating directly, we obtain

$\displaystyle\frac{\partial f}{\partial
x_{1}}=n\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{2}\\}}x_{j})^{n-1}-x_{1}^{n-1}\big{)},$
$\displaystyle\frac{\partial^{2}f}{\partial
x_{1}^{2}}=n(n-1)\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}\big{)},$
$\displaystyle\frac{\partial^{2}f}{\partial x_{1}\partial
x_{2}}=n(n-1)(\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\\}}x_{j})^{n-2},$

where
$\xi_{u_{1},u_{2}}=\mathrm{span}\\{u_{1},u_{2}\\}\in\\{\xi_{1}^{2},\xi_{2}^{2},\dots,\xi_{m_{2}}^{2}\\}$.

Since $n\geq 3$, it follows that

$\displaystyle~{}~{}~{}~{}\frac{1}{n(n-1)}\big{(}\frac{\partial^{2}f}{\partial
x_{1}^{2}}-\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\big{)}$
$\displaystyle=\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}-\big{(}\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\\}}x_{j}\big{)}^{n-2}$
$\displaystyle=-x_{1}^{n-2}+\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2},u_{2}\notin\xi_{i}^{2}\\}}\big{(}(\sum_{\\{j:u_{j}\in\xi_{i}^{2}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}$
$\displaystyle=-x_{1}^{n-2}+\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2},u_{2}\notin\xi_{i}^{2}\\}}\big{(}(x_{1}+\sum_{\\{j:u_{j}\in\xi_{i}^{2}\setminus\\{u_{1}\\}\\}}x_{j})^{n-2}-x_{1}^{n-2}\big{)}$
$\displaystyle\geq-
x_{1}^{n-2}+\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2},u_{2}\notin\xi_{i}^{2}\\}}\big{(}x_{1}^{n-2}+(n-2)x_{1}^{n-3}(\sum_{\\{j:u_{j}\in\xi_{i}^{2}\setminus\\{u_{1}\\}\\}}x_{j})-x_{1}^{n-2}\big{)}$
(5.4)
$\displaystyle=-x_{1}^{n-2}+(n-2)x_{1}^{n-3}\sum\limits_{\\{i:u_{1}\in\xi_{i}^{2},u_{2}\notin\xi_{i}^{2}\\}}\sum_{\\{j:u_{j}\in\xi_{i}^{2}\setminus\\{u_{1}\\}\\}}x_{j}.$

Step 1.2. Since
$x\pm\varepsilon(e_{1}-e_{2})=(x_{1}\pm\varepsilon,x_{2}\mp\varepsilon,x_{3},\dots,x_{N})\in
P_{M(u_{1},\dots,u_{N})}$, it follows that

(5.5) $0<x_{1}<1.$

Combining (5.4), (5.3), (5.5) and the definition of
$P_{M(u_{1},\dots,u_{N})}$, we obtain that

$\displaystyle~{}~{}~{}~{}\frac{1}{n(n-1)}\big{(}\frac{\partial^{2}f}{\partial
x_{1}^{2}}-\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\big{)}$
$\displaystyle\geq-
x_{1}^{n-2}+(n-2)x_{1}^{n-3}(n-x_{1}-\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\setminus\\{u_{1}\\}\\}}x_{j})$
$\displaystyle=-x_{1}^{n-2}+(n-2)x_{1}^{n-3}(n-x_{1}-\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\\}}x_{j}+x_{1})$
$\displaystyle=x_{1}^{n-3}\big{(}-x_{1}+(n-2)(n-\sum_{\\{j:u_{j}\in\xi_{u_{1},u_{2}}\\}}x_{j})\big{)}$
$\displaystyle>x_{1}^{n-3}\big{(}-1+(n-2)(n-2)\big{)}$
$\displaystyle=(n-1)(n-3)x_{1}^{n-3}\geq 0.$

Consequently, $f$ cannot attain its maximum in the region
$P_{M(u_{1},\dots,u_{N})}\setminus\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

Step 2. To evaluate the functional $f$ at points of
$\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

For each $x\in\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$, by Theorem 4.9,
w.l.o.g., assume

$x=(\underbrace{1,\dots,1}_{n},\underbrace{0,\dots,0}_{N-n}).$

By the definition of $P_{M(u_{1},\dots,u_{N})}$, it follows that
$u_{1},u_{2},\dots,u_{n}$ must be linearly independent. Thus, for each
$i\in\\{1,2,\dots,m_{2}\\}$, there are at most two elements of
$\\{u_{1},u_{2},\dots,u_{n}\\}$ lying in $\xi_{i}^{2}$. Moreover, if the
number of elements of $\\{u_{1},u_{2},\dots,u_{n}\\}$ lying in $\xi_{i}^{2}$
is strictly less than 2, i.e., $|\xi_{i}^{2}\cap\\{u_{1},\dots,u_{n}\\}|<2$,
then

$\big{(}\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j}\big{)}^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j}^{n}=0$

by calculating directly when $|\xi_{i}^{2}\cap\\{u_{1},\dots,u_{n}\\}|=0,1$.

Therefore,

$\displaystyle f(x)=$
$\displaystyle\sum\limits_{i=1}^{m_{2}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j})^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j}^{n}\big{)}$
$\displaystyle=$
$\displaystyle\sum\limits_{\\{i:|\xi_{i}^{2}\cap\\{u_{1},\dots,u_{n}\\}|=2\\}}\big{(}(\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j})^{n}-\sum\limits_{\\{j:{u_{j}}\in\xi^{2}_{i}\\}}x_{j}^{n}\big{)}$
$\displaystyle=$ $\displaystyle\binom{n}{2}\big{(}2^{n}-2\big{)}.$

Step 3. Combining Step 1 and Step 2, we obtain that for $x\in
P_{M(u_{1},\dots,u_{N})}$,

$f(x)\leq\binom{n}{2}\big{(}2^{n}-2\big{)},$

with equality if and only if $x\in\mathcal{F}_{0}(P_{M(u_{1},\dots,u_{N})})$.

If $P$ is _not_ a parallelotope, then $N>n$. By Theorem 4.8, we obtain that

$a=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\}))\in\mathrm{relint}P_{M(u_{1},\dots,u_{N})}.$

Hence,

$X_{2}(P)^{n}=f(a)<\max_{x\in
P_{M(u_{1},\dots,u_{N})}}f(x)=\binom{n}{2}\big{(}2^{n}-2\big{)}.$

To sum up, for a polytope $P$ with its centroid at the origin, it follows that

$\frac{X_{2}(P)}{V_{n}(P)}\leq\sqrt[n]{\binom{n}{2}\big{(}(\frac{2}{n})^{n}-\frac{2}{n^{n}}\big{)}},$

with equality if and only if $P$ is a parallelotope. ∎

##  6\. Proof of Theorem 1.3

Let $P\in\mathcal{P}_{4}^{4}$ with its centroid at the origin. Assume that

$V_{4}(P)=4,\quad\text{and}\quad\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm
u_{1},\dots,\pm u_{N}\\}.$

Recall that $\mathcal{P}_{4}^{4}$ is the set of polytopes in $\mathbb{R}^{4}$
which are in $4$-general position and contain the origin in their interiors.
In this setting, any $4$ elements of $\\{u_{1},\dots,u_{N}\\}$ are linearly
independent. So, the bases $\mathcal{B}(M(u_{1},\dots,u_{N}))$ is the
collection of all $4$-element subsets of $\\{1,\dots,N\\}$, and therefore
$P_{M(u_{1},\dots,u_{N})}$ is the hypersimplex
$\bigtriangleup_{N}^{4}=[0,1]^{4}\cap H_{\mathrm{I}_{N},4}$.

Write $a=(a_{1},\dots,a_{N})=(V_{P}(\\{\pm u_{1}\\}),\dots,V_{P}(\\{\pm
u_{N}\\})).$ By Example 3.4, it follows that

$\frac{X_{3}(P)^{4}}{V_{4}(P)^{4}}=\frac{1}{256}\sum\limits_{1\leq i<j<k\leq
N}\big{(}(a_{i}+a_{j}+a_{k})^{4}-(a_{i}+a_{j})^{4}-(a_{i}+a_{k})^{4}-(a_{j}+a_{k})^{4}+a_{i}^{4}+a_{j}^{4}+a_{k}^{4}\big{)}.$

For $x\in[0,1]^{4}\cap H_{\mathrm{I}_{N},4}$, let

$f(x)=\frac{1}{256}\sum\limits_{1\leq i<j<k\leq
N}\big{(}(x_{i}+x_{j}+x_{k})^{4}-(x_{i}+x_{j})^{4}-(x_{i}+x_{k})^{4}-(x_{j}+x_{k})^{4}+x_{i}^{4}+x_{j}^{4}+x_{k}^{4}\big{)}.$

Obviously that $f$ is _symmetric_ in $x_{i}$, $i\in{1,2,\dots,N}$, and
$f(a)=X_{3}(P)^{4}/V_{4}(P)^{4}$.

In the following, we study the extremal problem for $X_{3}$ by two cases of
quantity $N$, i.e., the number of outer normal vector pairs of $P$. To prove
Theorem 1.3, we first show two lemmas.

######  Lemma 6.1.

If $N=5$, then

$f(a)\leq\frac{72}{125},$

with equality if and only if $a=\frac{4}{5}(1,1,1,1,1).$

###### Proof.

If $a=\frac{4}{5}(1,1,1,1,1),$ then
$f(a)=\binom{5}{3}\big{(}(\frac{3}{5})^{4}-3(\frac{2}{5})^{4}+3(\frac{4}{5})^{4}\big{)}=\frac{72}{125}.$

Since $f$ is a polynomial, it follows that $f$ attains its maximum on
$[0,1]^{5}\cap H_{(1,1,1,1,1),4}$. It suffices to show that if
$a\neq\frac{4}{5}(1,1,1,1,1)$, then $f$ _cannot_ attain its maximum at $a$.

Otherwise, assume $f$ attains its maximum at

$a\in[0,1]^{5}\cap H_{(1,1,1,1,1),4}\setminus\\{\frac{4}{5}(1,1,1,1,1)\\},$

w.l.o.g., say $a=(a_{1},a_{2},a_{3},a_{4},a_{5})$ satisfying that

$0\leq a_{1}\leq a_{2}\leq a_{3}\leq a_{4}\leq a_{5}\leq 1.$

Since $a\neq\frac{4}{5}(1,1,1,1,1)$, it yields that $0\leq a_{1}<a_{5}\leq 1$.
Let $\varepsilon=\min\\{a_{5},1-a_{1}\\}$. Then $\varepsilon>0$ and

$a+te_{1}-te_{5}\in[0,1]^{5}\cap H_{(1,1,1,1,1),4},\quad t\in[0,\varepsilon].$

Let $F(t)=f(a+te_{1}-te_{5}),\ t\in[0,\varepsilon].$ Since $f$ attains its
maximum at $a$, it follows that $F(t)$ attains its maximum at $t=0$. Hence, it
is necessary that $F_{+}^{\prime}(0)\leq 0.$ However, we will show that
$F_{+}^{\prime}(0)>0$, which is a contradiction.

By directly calculating, we obtain

$\displaystyle\frac{\partial f}{\partial x_{1}}$
$\displaystyle=\frac{1}{64}\sum\limits_{2\leq j<k\leq
5}\big{(}(x_{1}+x_{j}+x_{k})^{3}-(x_{1}+x_{j})^{3}-(x_{1}+x_{k})^{3}+x_{1}^{3}\big{)}$
$\displaystyle=\frac{3}{64}\sum\limits_{2\leq j<k\leq
5}(2x_{1}x_{j}x_{k}+x_{j}^{2}x_{k}+x_{j}x_{k}^{2})$
$\displaystyle=\frac{3}{64}\sum\limits_{2\leq j\leq
4}(2x_{1}x_{j}x_{5}+x_{j}^{2}x_{5}+x_{j}x_{5}^{2})+\frac{3}{64}\sum\limits_{2\leq
i<j\leq 4}(2x_{1}x_{i}x_{j}+x_{i}^{2}x_{j}+x_{i}x_{j}^{2});$
$\displaystyle\frac{\partial f}{\partial x_{5}}$
$\displaystyle=\frac{3}{64}\sum\limits_{2\leq j\leq
4}(2x_{1}x_{j}x_{5}+x_{j}^{2}x_{1}+x_{j}x_{1}^{2})+\frac{3}{64}\sum\limits_{2\leq
i<j\leq 4}(2x_{5}x_{i}x_{j}+x_{i}^{2}x_{j}+x_{i}x_{j}^{2}).$

Since $0\leq a_{1}\leq a_{2}\leq a_{3}\leq a_{4}\leq a_{5}$, and
$\sum_{j=1}^{5}a_{j}=4$, it follows that

$\displaystyle F_{+}^{\prime}(0)=$ $\displaystyle(\frac{\partial f}{\partial
x_{1}}-\frac{\partial f}{\partial x_{5}})(a)$ $\displaystyle=$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})\sum\limits_{2\leq j\leq
4}a_{j}(a_{1}+a_{5}+a_{j})+\frac{3}{32}(a_{1}-a_{5})\sum\limits_{2\leq i<j\leq
4}a_{i}a_{j}$ $\displaystyle=$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})\sum\limits_{2\leq j\leq
4}a_{j}(a_{1}+a_{5}+a_{j})-\frac{3}{32}(a_{5}-a_{1})\frac{1}{2}\sum\limits_{2\leq
i,j\leq 4,i\neq j}a_{i}a_{j}$ $\displaystyle=$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})\sum_{j=2}^{4}a_{j}(a_{1}+a_{5}+a_{j})-\frac{3}{64}(a_{5}-a_{1})\sum_{j=2}^{4}a_{j}(a_{2}+a_{3}+a_{4}-a_{j})$
$\displaystyle=$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})\sum_{j=2}^{4}a_{j}(a_{1}+a_{5}+2a_{j}-a_{2}-a_{3}-a_{4})$
$\displaystyle\geq$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})\sum_{j=2}^{4}a_{2}(a_{1}+a_{5}+2a_{j}-a_{2}-a_{3}-a_{4})$
$\displaystyle=$
$\displaystyle\frac{3}{64}(a_{5}-a_{1})a_{2}(3a_{1}+3a_{5}-a_{2}-a_{3}-a_{4})\geq
0,$

with equality if and only if $a_{1}=0$ and $a_{2}=a_{3}=a_{4}=a_{5}=1$.

However,

$f(0,1,1,1,1)=\binom{4}{3}\big{(}(\frac{3}{4}\big{)}^{4}-3(\frac{2}{4})^{4}+3(\frac{1}{4})^{4}\big{)}=\frac{9}{16}=\frac{1125}{2000}<\frac{1152}{2000}=\frac{72}{125}=f(\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5}),$

which implies that $f$ _cannot_ attain its maximum at $(0,1,1,1,1)$. So,
$a\neq(0,1,1,1,1)$, and therefore $F_{+}^{\prime}(0)>0$, which contradicts to
that $F_{+}^{\prime}(0)\leq 0$. ∎

######  Lemma 6.2.

If $N\geq 6$ and $x\in(0,1]^{N}\cap H_{\mathrm{I}_{N},4},$ then $x$ is not the
maximal point of $f$ in $[0,1]^{N}\cap H_{\mathrm{I}_{N},4}$.

###### Proof.

Assume $f$ attains its maximum at $x\in(0,1]^{N}\cap H_{\mathrm{I}_{N},4}.$
W.o.l.g., say

$0<x_{1}\leq x_{2}\leq\cdots\leq x_{N}\leq 1.$

Step 1. We prove that there exists a sufficiently small $\varepsilon>0$, so
that

$x+t(e_{1}-e_{2})\in[0,1]^{N}\cap H_{\mathrm{I}_{N},4},\quad
t\in[-\varepsilon,\varepsilon].$

If $x_{2}=1$, then $x_{3}=\cdots=x_{N}=1$. However, by $N\geq 6$, it yields
that

$4=\sum_{i=1}^{N}x_{i}>\sum_{i=3}^{N}x_{i}=N-2\geq 4,$

which is a contradiction. So, $x_{2}<1$. Let

$\varepsilon=\min\\{x_{1},x_{2},1-x_{1},1-x_{2}\\}.$

Then $\varepsilon>0$ and satisfies the above inclusion.

Step 2. Let $F(t)=f(x+te_{1}-te_{2}),\ t\in[-\varepsilon,\varepsilon].$ By the
assumption that $f$ attains its maximum at $x$, it yields that $F(t)$ attains
its maximum at $t=0$. Hence, it is necessary that

$F^{\prime}(0)=0,\quad\text{ and }\quad
F^{\prime\prime}(0)=\big{(}\frac{\partial^{2}f}{\partial
x_{1}^{2}}+\frac{\partial^{2}f}{\partial
x_{2}^{2}}-2\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\big{)}(x)\leq
0.$

In the following, it suffices to show that $F^{\prime\prime}(0)>0$ to get a
contradiction.

Step 2.1. we calculate $\frac{\partial^{2}f}{\partial
x_{1}^{2}},\frac{\partial^{2}f}{\partial x_{2}^{2}}$ and
$\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}$.

$\displaystyle\frac{\partial f}{\partial x_{1}}(x)=$
$\displaystyle\frac{1}{64}\sum_{2\leq j<k\leq
N}\big{(}(x_{1}+x_{j}+x_{k})^{3}-(x_{1}+x_{j})^{3}-(x_{1}+x_{k})^{3}+x_{1}^{3}\big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{64}\sum_{k=3}^{N}\big{(}(x_{1}+x_{2}+x_{k})^{3}-(x_{1}+x_{2})^{3}-(x_{1}+x_{k})^{3}+x_{1}^{3}\big{)}+\frac{3}{64}\sum\limits_{3\leq
j<k\leq N}x_{j}x_{k}(x_{j}+x_{k}+2x_{1});$ $\displaystyle\frac{\partial
f}{\partial x_{2}}(x)=$
$\displaystyle\frac{1}{64}\sum_{k=3}^{N}\big{(}(x_{1}+x_{2}+x_{k})^{3}-(x_{1}+x_{2})^{3}-(x_{2}+x_{k})^{3}+x_{2}^{3}\big{)}+\frac{3}{64}\sum_{3\leq
j<k\leq N}x_{j}x_{k}(x_{j}+x_{k}+2x_{2}).$

Thus,

$\displaystyle\frac{\partial^{2}f}{\partial x_{1}^{2}}(x)=$
$\displaystyle\frac{3}{64}\sum_{k=3}^{N}\big{(}(x_{1}+x_{2}+x_{k})^{2}-(x_{1}+x_{2})^{2}-x_{k}(2x_{1}+x_{k})\big{)}+\frac{3}{32}\sum\limits_{3\leq
j<k\leq N}x_{j}x_{k};~{}$ $\displaystyle\frac{\partial^{2}f}{\partial
x_{2}^{2}}(x)=$
$\displaystyle\frac{3}{64}\sum_{k=3}^{N}\big{(}(x_{1}+x_{2}+x_{k})^{2}-(x_{1}+x_{2})^{2}-x_{k}(2x_{2}+x_{k})\big{)}+\frac{3}{32}\sum\limits_{3\leq
j<k\leq N}x_{j}x_{k};$ $\displaystyle\frac{\partial^{2}f}{\partial
x_{1}x_{2}}(x)=$
$\displaystyle\frac{3}{64}\sum_{k=3}^{N}\big{(}(x_{1}+x_{2}+x_{k})^{2}-(x_{1}+x_{2})^{2}\big{)}.$

Step 2.2. Since $\sum_{j=1}^{N}x_{j}=4$ and that $x_{1}\leq
x_{2}\leq\cdots\leq x_{N-1}\leq x_{N}$, it follows that

$\displaystyle F^{\prime\prime}(0)$
$\displaystyle=\big{(}\frac{\partial^{2}f}{\partial
x_{1}^{2}}+\frac{\partial^{2}f}{\partial
x_{2}^{2}}-2\frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\big{)}(x)$
$\displaystyle=\frac{3}{16}\sum_{3\leq j<k\leq
N}x_{j}x_{k}-\frac{3}{64}\sum_{k=3}^{N}x_{k}(2x_{1}+2x_{2}+2x_{k})$
$\displaystyle=\frac{3}{16}\times\frac{1}{2}\sum_{3\leq j,k\leq N,j\neq
k}x_{j}x_{k}-\frac{3}{32}\sum_{k=3}^{N}x_{k}(x_{1}+x_{2}+x_{k})$
$\displaystyle=\frac{3}{32}\sum_{k=3}^{N}x_{k}\sum_{3\leq j\leq N,j\neq
k}x_{j}-\frac{3}{32}\sum_{k=3}^{N}x_{k}(x_{1}+x_{2}+x_{k})$
$\displaystyle=\frac{3}{32}\sum_{k=3}^{N}x_{k}(4-x_{1}-x_{2}-x_{k})-\frac{3}{32}\sum_{k=3}^{N}x_{k}(x_{1}+x_{2}+x_{k})$
$\displaystyle=\frac{3}{32}\sum_{k=3}^{N}x_{k}(4-2x_{1}-2x_{2}-2x_{k})$ (6.1)
$\displaystyle\geq\frac{3}{32}\big{(}x_{N}(4-2x_{1}-2x_{2}-2x_{N})+\sum_{k=3}^{N-1}x_{k}(4-2x_{1}-2x_{2}-2x_{N-1})\big{)},$

with equality if and only if $x_{3}=x_{4}=\cdots=x_{N-1}$.

Since $N\geq 6$, it follows that

$2x_{1}+2x_{2}+x_{N-1}+x_{N}\leq x_{1}+x_{2}+x_{3}+x_{4}+x_{N-1}+x_{N}\leq 4.$

Hence,

(6.2) $4-2x_{1}-2x_{2}\geq x_{N}+x_{N-1},$

with equality if and only if $N=6,$ and $x_{1}=x_{2}=x_{3}=x_{4}$.

Combining (6.1), (6.2), that $\sum_{k=1}^{N}x_{k}=4$, and that $x_{1},x_{2}<1$
and $x_{N}\leq 1$, we obtain

$\displaystyle F^{\prime\prime}(0)$
$\displaystyle\geq\frac{3}{32}\big{(}x_{N}(x_{N}+x_{N-1}-2x_{N})+\sum_{k=3}^{N-1}x_{k}(x_{N}+x_{N-1}-2x_{N-1})\big{)}$
$\displaystyle=\frac{3}{32}\big{(}-x_{N}(x_{N}-x_{N-1})+(x_{N}-x_{N-1})\sum_{k=3}^{N-1}x_{k}\big{)}$
(6.3)
$\displaystyle=\frac{3}{32}(x_{N}-x_{N-1})(4-x_{1}-x_{2}-x_{N}-x_{N})\geq 0,$

where the last equality holds if and only if $x_{N}=x_{N-1}$.

Combining (6.1) (6.2) and (6.3), it concludes that

$F^{\prime\prime}(0)\geq 0,$

with equality if and only if $N=6$, and
$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=x_{6}=\frac{4}{6}$.

However,

$f(\frac{4}{6},\frac{4}{6},\frac{4}{6},\frac{4}{6},\frac{4}{6},\frac{4}{6})=\binom{6}{3}\big{(}(\frac{3}{6})^{4}-3(\frac{2}{6})^{4}+3(\frac{1}{6})^{4}\big{)}=\frac{5}{9}=\frac{625}{1125}<\frac{648}{1125}=\frac{72}{125}=f(\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5},0),$

which implies $f$ _cannot_ attain its maximum at $\frac{4}{6}(1,1,1,1,1,1)$.
So, $x\neq\frac{4}{6}(1,1,1,1,1,1)$, and therefore $F^{\prime\prime}(0)>0$,
which contradicts to that $F^{\prime\prime}(0)\leq 0$.

Consequently, if $x\in(0,1]^{N}\cap H_{\mathrm{I}_{N},4}$, then $x$ is not the
maximal point of $f$ in $[0,1]^{N}\cap H_{\mathrm{I}_{N},4}$. ∎

Combining Lemma 6.1 and Lemma 6.2, we immediately obtain the following result.

######  Theorem 6.3.

Let $N\geq 5$ and $x\in[0,1]^{N}\cap H_{\mathrm{I}_{N},4}.$ Then

$f(x)\leq\frac{72}{125},$

with equality if and only if there exists
$\\{i_{1},i_{2},i_{3},i_{4},i_{5}\\}\subseteq\\{1,2,\dots,N\\}$, so that
$x_{i_{j}}=\frac{4}{5}$, $j=1,2,3,4,5.$

###### Proof.

Since $f$ is a polynomial, it follows that $f$ attains its maximum on
$[0,1]^{N}\cap H_{\mathrm{I}_{N},4}$. Assume $f$ attains its maximum at
$\bar{x}$. Since $f$ is symmetric in $x_{i}$, $i\in{1,2,\dots,N}$, w.l.o.g.,
say

$\bar{x}=(\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{M},0,\dots,0),\quad\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{M}>0,\
4\leq M\leq N.$

Since

$f(1,1,1,1,0,\dots,0)<f(\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5},\frac{4}{5},0,\dots,0),$

it follows that $M\geq 5$.

For $x\in[0,1]^{M}\cap H_{\mathrm{I}_{M},4}$, let

$f_{M}(x)=\frac{1}{256}\sum\limits_{1\leq i<j<k\leq
M}\big{(}(x_{i}+x_{j}+x_{k})^{4}-(x_{i}+x_{j})^{4}-(x_{i}+x_{k})^{4}-(x_{j}+x_{k})^{4}+x_{i}^{4}+x_{j}^{4}+x_{k}^{4}\big{)}.$

Then $f_{M}(\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{M})=f(\bar{x})$. Note that

$f_{M}(\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{M})\leq\max_{x\in[0,1]^{M}\cap
H_{\mathrm{I}_{M},4}}f_{M}(x)\leq\max_{x\in[0,1]^{N}\cap
H_{\mathrm{I}_{N},4}}f(x)=f(\bar{x}).$

Hence, $f_{M}$ attains its maximum at
$(\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{M})$. Combining Lemma 6.2 and Lemma
6.1, it follows that $M=5$ and
$\bar{x}_{1}=\bar{x}_{2}=\bar{x}_{3}=\bar{x}_{4}=\bar{x}_{5}=\frac{4}{5}.$ ∎

Now, we finish the proof of Theorem 1.3.

######  Theorem 6.4.

Let $P\in\mathcal{P}_{4}^{4}$ with its centriod at the origin. Then

$\frac{X_{3}(P)}{V_{4}(P)}\leq\sqrt[4]{\frac{72}{125}},$

with equality if and only if $\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm
u_{1},\dots,\pm u_{5}\\}$, and $V_{P}(\\{\pm u_{i}\\})=\frac{V_{4}(P)}{5},$
$i=1,\dots,5.$

###### Proof.

Assume that $\mathrm{supp}S_{P}\cup\mathrm{supp}S_{-P}=\\{\pm u_{1},\pm
u_{2},\dots,\pm u_{N}\\}$ and $V_{4}(P)=4.$

If $P$ is a parallelotope, then by Example 3.4, it follows that

$\frac{X_{3}(P)^{4}}{V_{4}(P)^{4}}=\binom{4}{3}\big{(}(\frac{3}{4})^{4}-3(\frac{2}{4})^{4}+3(\frac{1}{4})^{4}\big{)}=\frac{9}{16}<\frac{72}{125}.$

If $P$ is _not_ a parallelotope, then $N\geq 5$. Let $a=(V_{P}(\\{\pm
u_{1}\\}),\dots,V_{P}(\\{\pm u_{N}\\}))$. From Theorem 6.3, we obtain

$\frac{X_{3}(P)^{4}}{V_{4}(P)^{4}}=f(a)\leq\max_{x\in[0,1]^{N}\cap
H_{\mathrm{I}_{N},4}}f(x)=\frac{72}{125},$

with equality if and only if $N=5$ and $V_{P}(\\{\pm u_{i}\\})=\frac{4}{5},\
i=1,2,3,4,5.$ ∎

In particular, if $P$ is a _simplex_ in $\mathbb{R}^{4}$ with its centroid at
the origin, then the equality holds in the above theorem. Moreover, by Lemma
2.1 and Lemma 2.2, there exists an origin-symmetric _decahedron_
$Q\in\mathcal{P}_{4}^{4}$, so that the equality also holds. However, according
to Example 3.5, it is striking that $\frac{X_{3}}{V_{4}}$ does _not_ attain
its extremum at the simplex or decahedron $Q$ in $\mathcal{P}_{c}^{4}$.

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