# The existence of balanced neighborly polynomials

Nguyen Thi Thanh Tam  Hung Vuong University, Phu Tho, Viet Nam
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###### Abstract.

Inspired by the definition of balanced neighborly spheres, we define balanced
neighborly polynomials and study the existence of these polynomials. The goal
of this article is to construct balanced neighborly polynomials of type
$(k,k,k,k)$ over any field $K$ for all $k\neq 2$, and show that a balanced
neighborly polynomial of type $(2,2,2,2)$ exists if and only if
$\operatorname{char}(K)\neq 2$. Besides, we also discuss a relation between
balanced neighborly polynomials and balanced neighborly simplicial spheres.

###### Key words and phrases:

Balanced neighborly polynomials, balanced neighborly simplicial spheres, upper
bound theorem.

This work was partially funded by Vingroup Joint Stock Company and supported
by the PhD Scholarship Programme of Vingroup Innovation Foundation (VINIF),
Vingroup Big Data Institute (VINBIGDATA)

##  1\. Introduction

Finding sharp upper bounds for the Hilbert function of certain graded algebras
is an important problem relating both commutative algebra and combinatorics.
One of the most famous results on this topic is Stanley’s solution of the
Upper Bound Conjecture, which gives an upper bound on the number of faces of
simplicial spheres by translating the problem into a problem on upper bounds
of Hilbert function of graded algebras. In this paper, we study upper bounds
for the Hilbert function of Gorenstein algebras defined by certain
polynomials, which we call balanced polynomials.

The motivation of this paper comes from the study of face numbers of balanced
simplicial complexes. Balanced simplicial complexes are defined by Stanley [3]
and their face number has been studied by many researchers [1, 2]. One
interesting problem on balanced simplicial complexes is to find the sharp
upper bound of the number of $i$-faces of balanced simplicial spheres with $n$
vertices for a fixed integers $i$ and $n$. Recently, to study this problem,
Zheng [5] defined the notion of balanced neighborly spheres and study their
existence. Balanced neighborly spheres are important ssince if a balanced
neighborly sphere of type $(k,k,...,k)$ exists then it maximizes the numbers
of faces of balanced simplicial $(d-1)$-spheres with $kd$ vertices.
Especially, the paper [5] proved that balanced neighborly spheres of type
$(2,2,2,2)$ do not exist, but type $(3,3,3,3)$ exist. Also, Zheng asked if
balanced neighborly spheres of type $(k,k,k,k)$ exist for all $k$ [5, Question
15]. Motivated by this work of Zheng, we define balanced neighborly
polynomials (see Section 2 for the definition) and study their existence. As
we will prove in Proposition 4.1, if balanced neighborly simplicial spheres of
type $(n_{1},\ldots,n_{d})$ exist, then balanced neighborly polynomials of
type $(n_{1},\ldots,n_{d})$ exist over any field $K.$ Thus, the study of the
existence problem of balanced neighborly polynomial can be considered as an
algebraic abstraction of the existence problem on balanced neighborly spheres.
On the other hand, the goal of this paper is to construct a balanced
neighborly polynomial of type $(k,k,k,k)$ over any field $K$ for all $k\neq
2$, (see Theorems 3.2, 3.3, and 3.4) and show that a balanced neighborly
polynomial of type $(2,2,2,2)$ exists only when $\operatorname{char}(K)\neq 2$
(see Theorem 3.1). Moreover, the results give an alternative proof for Zheng’s
result proving that balanced neighborly simplicial spheres of type $(2,2,2,2)$
do not exist.

The structure of this article is organized as follows. In section 2, we
discuss basic properties of balanced neighborly polynomials. In section 3, we
construct a balanced neighborly polynomial of type $(k,k,k,k)$ for any $k\neq
2$ and prove that there exists a balanced neighborly polynomial of type
$(2,2,2,2)$ only when $\operatorname{char}K\neq 2$. In section 4, we discuss a
relation between balanced neighborly polynomials and balanced neighborly
simplicial spheres.

##  2\. Balanced neighborly polynomials

In this section we introduce notations that will be used throughout this
article. Let $d,n_{1},\ldots,n_{d}$ be positive integers, and let
$R:=K[x_{ij}\mid 1\leq i\leq d,1\leq j\leq n_{i}]$ be a polynomial ring over a
field $K$. The ring $R$ has a $\mathbb{Z}^{d}$-graded structure by setting
$\operatorname{deg}(x_{ij})=\mathfrak{e}_{i}$ for $1\leq i\leq d$, where
$\mathfrak{e}_{1},\ldots,\mathfrak{e}_{d}$ are the unit vectors of
$\mathbb{Z}^{d}$. We say that $f\in R$ is a balanced polynomial if $f$ is
$\mathbb{Z}^{d}$-graded and $\operatorname{deg}(f)=(1,1,\ldots,1).$ For a
balanced polynomial $f\in R$ and $S\subseteq[d]=\\{1,2,\ldots,d\\}$, we set

$H(f,S)=\dim_{K}\\{g(\partial_{ij})f\mid g(x_{ij})\in
R_{\mathfrak{e}_{S}}\\},$

where $\partial_{ij}={\partial}/{\partial x_{ij}}$,
$\mathfrak{e}_{S}=\underset{i\in S}{\sum}\mathfrak{e}_{i}\in\mathbb{Z}^{d}$,
and $R_{\mathfrak{e}_{S}}$ is the graded component of $R$ of degree
$\mathfrak{e}_{S}$. Moreover, a balanced polynomial $f$ is said to be
neighborly of type $(n_{1},\ldots,n_{d})\in\mathbb{Z}^{d}$ if for every subset
$S\subseteq[d]$ with $|S|\leq\dfrac{d}{2}$, we have $H(f,S)=\underset{k\in
S}{\prod}n_{k}$.

For a graded polynomial
$f=f(x_{1},\ldots,x_{n})\in\mathfrak{R}=K[x_{1},\cdots,x_{n}],$ let

$\operatorname{ann}(f)=\\{g(x_{1},\ldots,x_{n})\in\mathfrak{R}\mid
g(\partial_{1},\cdots,\partial_{n})f=0\\},$

where $\partial_{i}=\dfrac{\partial}{\partial x_{i}}.$ We say that a
polynomial $f$ is squarefree if $f$ is a linear combination of squarefree
monomials. For $F\subseteq[n]$, let $x^{F}=\underset{i\in F}{\prod}x_{i}$. If

$f=\sum_{F\subset[n],|F|=d}\alpha_{F}x^{F},\quad
g=\sum_{F\subset[n],|F|=d}\beta_{F}x^{F},\text{ where }\alpha_{F},\beta_{G}\in
K$

are squarefree polynomials having the same degree, then
$g(\partial_{1},\cdots,\partial_{n})f=\underset{F\subset[n]}{\sum}\alpha_{F}\beta_{F}$.
We say that an Artinian graded $K$-algebra
$\mathfrak{R}/I=\displaystyle\bigoplus_{k=0}^{d}(\mathfrak{R}/I)_{k}$, where
$(\mathfrak{R}/I)_{d}\neq 0$ is Gorenstein of socle degree $d$ if
$0:_{\mathfrak{R}/I}(x_{1},\ldots,x_{n})=(\mathfrak{R}/I)_{d}$ has
$K$-dimension $1$, where $I$ is an ideal of $\mathfrak{R}$. Then we have the
following property.

######  Lemma 2.1.

We have

  * (i)

If $f$ is a squarefree polynomial of degree $d$, then
$x_{1}^{2},\ldots,x_{n}^{2}\in\operatorname{ann}(f)$ and
$\mathfrak{R}/(\operatorname{ann}(f))$ is an Artinian Gorenstein graded
$k$-algebra of socle degree $d$.

  * (ii)

If $I$ is a homogenerous ideal such that $\mathfrak{R}/I$ is an Artinian
Gorenstein graded $k$-algebra of socle degree $d$ and
$x_{1}^{2},\ldots,x_{n}^{2}\in I$, then there is a squarefree polynomial such
that $\operatorname{ann}(f)=I.$

###### Proof.

(i) Assume $f$ is squarefree. Then $\partial_{i}^{2}f=0,\forall
i=1,2,\ldots,n$. So $x_{1}^{2},\ldots,x_{n}^{2}\in\operatorname{ann}(f)$.
Next, we prove that
$0:_{\mathfrak{R}/\operatorname{ann}(f)}(x_{1},\ldots,x_{n})=(\mathfrak{R}/\operatorname{ann}(f))_{d}.$
Let $g\in\mathfrak{R}$ be a homogeneous element of degree $k<d$ with
$g\notin\operatorname{ann}(f)$. We must prove that $g$ is not an element of
$0:_{\mathfrak{R}/\operatorname{ann}(f)}(x_{1},\ldots,x_{n})$, that is,
$x_{j}g\notin\operatorname{ann}(f)$ for some $j$. Since
$g\notin\operatorname{ann}(f),$ we have
$g(\partial_{1},\cdots,\partial_{n})f\neq 0$. Since
$h=g(\partial_{1},\cdots,\partial_{n})f$ is a squarefree polynomial of degree
$\geq 1$, $\partial_{j}(h)\neq 0$ for some $j$. From this we obtain that
$x_{j}g\notin\operatorname{ann}(f).$ Now we prove
$\dim_{K}(\mathfrak{R}/\operatorname{ann}(f))_{d}=1.$ Indeed, let

$f=\underset{F\subset[n],|F|=d}{\sum}\alpha_{F}x^{F}\text{ for }\alpha_{F}\in
K,$

$E_{k}=\operatorname{span}_{K}\\{x^{F}:|F|=k\\}$ and $E=E_{0}\oplus
E_{1}\oplus\cdots\oplus E_{k}$ be the set of all squarefree polynomials at
most $k$. We consider the linear map

$\displaystyle\psi:E_{d}$ $\displaystyle\to K$ $\displaystyle
g=\sum_{F\subset[n],|F|=d}\beta_{F}x^{F}$
$\displaystyle\mapsto\sum_{F\subset[n],|F|=d}\beta_{F}\alpha_{F}=g(\partial_{1},\cdots,\partial_{n})f$

Then $\operatorname{Ker}\psi=\\{g\in E_{d}\mid
g(\partial_{1},\cdots,\partial_{n})f=0\\}=(\operatorname{ann}(f)\cap E)_{d}$.
Since $\psi$ is a non-zero map,
$\dim_{K}\operatorname{Ker}\psi=\dim_{K}E_{d}-1.$ Moreover,
$\operatorname{ann}(f)=(x_{1}^{2},\ldots,x_{n}^{2})\oplus(\operatorname{ann}(f)\cap
E)$ and $\mathfrak{R}=(x_{1}^{2},\ldots,x_{n}^{2})\oplus E$. Then

$\displaystyle\dim_{K}(\mathfrak{R}/\operatorname{ann}(f))_{d}$
$\displaystyle=\dim_{K}\mathfrak{R}_{d}-\dim_{K}\operatorname{ann}(f)_{d}$
$\displaystyle=\dim_{K}((x_{1}^{2},\ldots,x_{n}^{2})_{d}\oplus
E_{d})-\dim_{K}(x_{1}^{2},\ldots,x_{n}^{2})_{d}\oplus(\operatorname{ann}(f)\cap
E)_{d}$ $\displaystyle=\dim_{K}E_{d}-\dim_{K}\operatorname{Ker}\psi=1.$

(ii) Suppose $\mathfrak{R}/I$ is Artinian Gorenstein of socle degree $d$ and
$x_{1}^{2},\ldots,x_{n}^{2}\in I$. Then
$I_{d}=(x_{1}^{2},\ldots,x_{n}^{2})_{d}\bigoplus(E\cap I)_{d}$ as $K$-vector
space and

$\displaystyle\dim_{K}(E\cap I)_{d}$
$\displaystyle=\dim_{K}I_{d}-\dim_{K}(x_{1}^{2},\ldots,x_{n}^{2})_{d}$
$\displaystyle=(\dim_{K}R_{d}-1)-\dim_{K}(x_{1}^{2},\ldots,x_{n}^{2})_{d}$
$\displaystyle=\dim_{K}E_{d}-1.$

Since $(E\cap I)_{d}$ is a codimension $1$ subspace of $E_{d}$, it can be
written as a kernel of a $K$-linear function, that is,

$(E\cap I)_{d}=\left(\Big{\\{}\sum_{F\subset[n],|F|=d}\beta_{F}x^{F}\in
E_{d}\mid\sum_{F\subset[n],|F|=d}\alpha_{F}\beta_{F}=0\Big{\\}}\right)$

for some $\alpha_{F}\in K$ such that $|F|=d.$ Let
$f=\underset{F\subset[n],|F|=d}{\sum}\alpha_{F}x^{F}$. We claim that
$\operatorname{ann}(f)=I$. Indeed,

$(E\cap I)_{d}=\left(\Big{\\{}\sum_{F\subset[n],|F|=d}\beta_{F}x^{F}\in
E_{d}\mid\sum_{F\subset[n],|F|=d}\alpha_{F}\beta_{F}=g(\partial_{ij})f=0\Big{\\}}\right)=(\operatorname{ann}(f)\cap
E)_{d}.$

In particular,

$I_{d}=(x_{1}^{2},\ldots,x_{n}^{2})_{d}+(E\cap
I)_{d}=(x_{1}^{2},\ldots,x_{n}^{2})_{d}+(E\cap\operatorname{ann}(f))_{d}=\operatorname{ann}(f)_{d}.$

Note that for any ideal $J$ of $\mathfrak{R}$, if $\mathfrak{R}/J$ is an
Artinian Gorenstein algebra of socle degree $d$, then $J_{d}$ determines the
whole $J$. In fact, for any polynomial $g\in\mathfrak{R}$ of degree $t<d$, one
has $g\in J$ if and only if $(x_{1},\ldots,x_{n})^{d-t}g\subset J_{d}$. Since
both $\mathfrak{R}/I$ and $\mathfrak{R}/\operatorname{ann}(f)$ are Artinian
Gorenstein algebras of socle degree $d$, $I_{d}=\operatorname{ann}(f)_{d}$
implies $I=\operatorname{ann}(f)$. ∎

From the definition of $H(f,S)$ and $\operatorname{ann}(f)$ as above, it is
easy to prove that $H(f,S)$ has the following symmetry.

######  Lemma 2.2.

If $f\in\mathfrak{R}$ is a balanced polynomial of degree $d$, then for any
$S\subseteq[d]$

  * (i)

$H(f,S)=\dim_{K}(\mathfrak{R}/\operatorname{ann}(f))_{\mathfrak{e}_{S}}$,

  * (ii)

$H(f,S)=H(f,[d]\backslash S)$.

###### Proof.

(i) Set $V_{\mathfrak{e}_{S}}(f)=\\{g(\partial_{1},\cdots,\partial_{n})f\mid
g(x_{1},\ldots,x_{n})\in\mathfrak{R}_{\mathfrak{e}_{S}}\\}$. Then
$H(f,S)=\dim_{K}V_{\mathfrak{e}_{S}}(f).$ Consider the short exact sequence

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where $\varphi(g)=g(\partial_{ij})f$ for all
$g\in\mathfrak{R}_{\mathfrak{e}_{S}}$. Note that
$\operatorname{Ker}\varphi=\operatorname{ann}(f)_{\mathfrak{e}_{S}}$.
Therefore

$H(f,S)=\dim_{K}\mathfrak{R}_{\mathfrak{e}_{S}}-\dim_{K}\operatorname{ann}(f)_{\mathfrak{e}_{S}}=\dim_{K}(\mathfrak{R}/\operatorname{ann}(f))_{\mathfrak{e}_{S}}.$

(ii) Set
$A=\mathfrak{R}/\operatorname{ann}(f)=\displaystyle\bigoplus_{k=1}^{d}A_{k}.$
Since $\mathfrak{R}/\operatorname{ann}(f)$ is a Gorenstein ring. The bilinear
map

$A_{\mathfrak{e}_{A}}\times A_{\mathfrak{e}_{[d]\backslash A}}\to
A_{\mathfrak{e}_{[d]}}\cong K$

is defined by the multiplication non-degenerate. Therefore,
$A_{\mathfrak{e}_{S}}\cong\operatorname{Hom}_{K}(A_{\mathfrak{e}_{[d]\backslash
S}},K)\cong A_{\mathfrak{e}_{[d]\backslash S}},$ and the statement follows
from (i). ∎

In the next section, we study balanced neighborly polynomials of degree $4$.
For this, the next Lemma says that we only need to consider balanced
neighborly polynomials of type $(k,k,k,k)$.

######  Lemma 2.3.

Let $f$ be a balanced neighborly polynomial of type
$(n_{1},\cdots,n_{d})\in\mathbb{Z}^{d}$. If $d$ is even then
$n_{1}=n_{2}=\cdots=n_{d}.$

###### Proof.

By the definition of a balanced polynomial $f\in\mathfrak{R}$, we see that the
$\mathbb{Z}^{d}$-degree of $f$ is $(1,1,\ldots,1)$. Since $d$ is even, we set
$d=2m$. Assume that $n_{1}\geq n_{2}\geq\cdots\geq n_{d}$. Then by Lemma 2.2
(ii), we have

$n_{1}n_{2}\cdots
n_{m}=H(f,\\{1,2,\cdots,m\\})=H(f,\\{m+1,\ldots,2m\\})=n_{m+1}n_{m+2}\ldots
n_{2m}.$

Since $n_{1}\geq n_{2}\geq\cdots\geq n_{d}$, we conclude that
$n_{1}=n_{2}=\cdots=n_{m}=n_{m+1}=\cdots=n_{2m}.$ ∎

In case $d$ is even, Lemma 2.3 for balanced simplicial spheres was proved by
Zheng in [5].

##  3\. The existence of balanced neighborly polynomials of type $(k,k,k,k)$

Throughout this section, we consider ring

$R=K[x_{1},\ldots,x_{k},y_{1},\ldots,y_{k},z_{1},\ldots,z_{k},w_{1},\ldots,w_{k}],$

where $\operatorname{deg}(x_{i})=\mathfrak{e}_{1}=(1,0,0,0)$,
$\operatorname{deg}(y_{i})=\mathfrak{e}_{2}=(0,1,0,0)$,
$\operatorname{deg}(z_{i})=\mathfrak{e}_{3}=(0,0,1,0)$,
$\operatorname{deg}(w_{i})=\mathfrak{e}_{4}=(0,0,0,1)$. So, if $f\in R$ is a
balanced polynomial, then we may assume that

$\displaystyle
f=\sum_{(i_{1},i_{2},i_{3},i_{4})\in{\\{1,2,\ldots,k\\}}^{4}}a_{i_{1},i_{2},i_{3},i_{4}}x_{i_{1}}y_{i_{2}}z_{i_{3}}w_{i_{4}},\quad\text{where
}a_{i_{1},i_{2},i_{3},i_{4}}\in K.$

Therefore,

$\displaystyle H(f,\\{1,2\\})$
$\displaystyle=H(f,\\{3,4\\})=\dim\operatorname{span}_{K}\left\\{\frac{\partial}{\partial
x_{i}}\frac{\partial}{\partial y_{j}}f\mid 1\leq i,j\leq k\right\\}$
$\displaystyle H(f,\\{1,3\\})$
$\displaystyle=H(f,\\{2,4\\})=\dim\operatorname{span}_{K}\left\\{\frac{\partial}{\partial
x_{i}}\frac{\partial}{\partial z_{j}}f\mid 1\leq i,j\leq k\right\\}$
$\displaystyle H(f,\\{1,4\\})$
$\displaystyle=H(f,\\{2,3\\})=\dim\operatorname{span}_{K}\left\\{\frac{\partial}{\partial
x_{i}}\frac{\partial}{\partial w_{j}}f\mid 1\leq i,j\leq k\right\\}.$

######  Theorem 3.1.

There exists a balanced neighborly polynomial of type $(2,2,2,2)$ if and only
if $\operatorname{char}K\neq 2$.

###### Proof.

We first prove that if $\operatorname{char}K\neq 2$ then there exists a
balanced neighborly polynomial of type $(2,2,2,2)$. Let

$f=-x_{1}y_{1}z_{2}w_{2}-x_{1}y_{2}z_{1}w_{1}+x_{1}y_{2}z_{2}w_{2}-x_{2}y_{1}z_{1}w_{2}+x_{2}y_{1}z_{2}w_{1}+x_{2}y_{2}z_{2}w_{1}.$

We have

$\displaystyle H(f,\\{1,2\\})=$
$\displaystyle\dim_{K}\operatorname{span}_{K}\\{-z_{2}w_{2},-z_{1}w_{1}+z_{2}w_{2},-z_{1}w_{2}+z_{2}w_{1},z_{2}w_{1}\\},$
$\displaystyle H(f,\\{1,3\\})=$
$\displaystyle\dim_{K}\operatorname{span}_{K}\\{-y_{2}w_{1},y_{2}w_{2}-y_{1}w_{2},-y_{1}w_{2},y_{2}w_{1}+y_{1}w_{1}\\},$
$\displaystyle H(f,\\{1,4\\})=$
$\displaystyle\dim_{K}\operatorname{span}_{K}\\{-y_{2}z_{1},y_{2}z_{2}-y_{1}z_{2},y_{2}z_{2}+y_{1}z_{2},-y_{1}z_{1}\\}.$

Then

$H(f,\\{1,2\\})=H(f,\\{1,3\\})=H(f,\\{1,4\\})=4.$

So, $f$ is a balanced neighborly polynomial of type $(2,2,2,2)$. Conversely,
suppose that $\operatorname{char}K=2$ and let

$\displaystyle f$
$\displaystyle=a_{1}x_{1}y_{1}z_{1}w_{1}+a_{2}x_{1}y_{1}z_{1}w_{2}+a_{3}x_{1}y_{1}z_{2}w_{1}+a_{4}x_{1}y_{1}z_{2}w_{2}+a_{5}x_{1}y_{2}z_{1}w_{1}+a_{6}x_{1}y_{2}z_{1}w_{2}$
$\displaystyle+a_{7}x_{1}y_{2}z_{2}w_{1}+a_{8}x_{1}y_{2}z_{2}w_{2}+a_{9}x_{2}y_{1}z_{1}w_{1}+a_{10}x_{2}y_{1}z_{1}w_{2}+a_{11}x_{2}y_{1}z_{2}w_{1}+a_{12}x_{2}y_{1}z_{2}w_{2}$
$\displaystyle+a_{13}x_{2}y_{2}z_{1}w_{1}+a_{14}x_{2}y_{2}z_{1}w_{2}+a_{15}x_{2}y_{2}z_{2}w_{1}+a_{16}x_{2}y_{2}z_{2}w_{2},$

where $a_{i}\in K,i=1,2,\ldots,16.$ We prove that $f$ is not a balanced
neighborly polynomial of type $(2,2,2,2)$. Let

$M=\begin{pmatrix}a_{1}&a_{2}&a_{3}&a_{4}\\\ a_{5}&a_{6}&a_{7}&a_{8}\\\
a_{9}&a_{10}&a_{11}&a_{12}\\\ a_{13}&a_{14}&a_{15}&a_{16}\\\ \end{pmatrix}.$

Since $H(f,\\{1,2\\})$ is the $K$-dimension of

$\displaystyle\operatorname{span}_{K}$
$\displaystyle\\{a_{1}z_{1}w_{1}+a_{2}z_{1}w_{2}+a_{3}z_{2}w_{1}+a_{4}z_{2}w_{2},a_{5}z_{1}w_{1}+a_{6}z_{1}w_{2}+a_{7}z_{2}w_{1}+a_{8}z_{2}w_{2},$
$\displaystyle
a_{9}z_{1}w_{1}+a_{10}z_{1}w_{2}+a_{11}z_{2}w_{1}+a_{12}z_{2}w_{2},a_{13}z_{1}w_{1}+a_{14}z_{1}w_{2}+a_{15}z_{2}w_{1}+a_{16}z_{2}w_{2}\\},$

$H(f,\\{1,2\\})=4$ if and only if $M$ is a nonsingular matrix. Let

$N=\begin{pmatrix}a_{1}&a_{2}&a_{5}&a_{6}\\\ a_{3}&a_{4}&a_{7}&a_{8}\\\
a_{9}&a_{10}&a_{13}&a_{14}\\\ a_{11}&a_{12}&a_{15}&a_{16}\\\
\end{pmatrix}\text{ and }P=\begin{pmatrix}a_{1}&a_{3}&a_{5}&a_{7}\\\
a_{2}&a_{4}&a_{6}&a_{8}\\\ a_{9}&a_{11}&a_{13}&a_{15}\\\
a_{10}&a_{12}&a_{14}&a_{16}\\\ \end{pmatrix}.$

A similar computation shows that $N$ (resp. $P$) is a non-singular matrix if
and only if $H(f,\\{1,3\\})=4$ (resp. $H(f,\\{1,4\\})=4$). Moreover, we have
$\det(M)-\det(N)=\det(P).$ It follows that one of $\det(M),\det(N),$ and
$\det(P)$ must be zero. So, $f$ is not a balanced neighborly polynomial of
type $(2,2,2,2)$. ∎

For an integer $i$, let $[i]_{k}\in\\{1,2,.\ldots,k\\}$ be the number such
that $[i]_{k}\equiv i(\text{mod}$ $k).$

######  Theorem 3.2.

Let $k\in\mathbb{N}$ be odd and $f=\sum\limits_{1\leq i,j\leq
k}x_{i}y_{j}z_{{[j-i]}_{k}}w_{{[i+j]}_{k}}.$ Then $f$ is a balanced neighborly
polynomial of type $(k,k,k,k).$

###### Proof.

Set $A=\\{z_{[j-i]_{k}}w_{[j+i]_{k}}\mid i,j=1,2,\ldots,k\\}$. We consider the
map

$\displaystyle\varphi:\\{1,2,\ldots,k\\}^{2}$
$\displaystyle\rightarrow\\{1,2,\ldots,k\\}^{2}$ $\displaystyle(i,j)$
$\displaystyle\mapsto([j-i]_{k},[j+i]_{k}).$

We prove that $\varphi$ is bijective. It suffices to prove the injectivity.
Suppose
$([j-i]_{k},[j+i]_{k})=([j^{\prime}-i^{\prime}]_{k},[j^{\prime}+i^{\prime}]_{k})$,
where $(i,j),(i^{\prime},j^{\prime})\in\\{1,2,\ldots,k\\}^{2}$. Then we have
$j-i\equiv j^{\prime}-i^{\prime}(\text{mod}$ $k)$, $j+i\equiv
j^{\prime}+i^{\prime}(\text{mod}$ $k)$. It follows that $2j\equiv
2j^{\prime}(\text{mod}$ $k)$ and $2i\equiv 2i^{\prime}(\text{mod}$ $k)$. Since
$k$ is odd, $j\equiv j^{\prime}(\text{mod}$ $k)$ and $i\equiv
i^{\prime}(\text{mod}$ $k)$. Then $(i,j)=(i^{\prime},j^{\prime})$ so,
$\varphi$ is an injective. The bijectivity of $\varphi$ implies that
$A=\\{z_{i}w_{j}\mid i,j=1,2,\ldots,k\\}$. Hence

$H(f,\\{1,2\\})=\dim_{K}\operatorname{span}_{K}A=k^{2}.$

In order to compute $\dim_{K}R_{\mathfrak{e}_{1}+\mathfrak{e}_{3}}$ and
$\dim_{K}R_{\mathfrak{e}_{1}+\mathfrak{e}_{4}}$, we rewrite $f$ as follows

$f=\sum\limits_{1\leq i,j\leq
k}x_{i}y_{[i+j]_{k}}z_{j}w_{{[2i+j]}_{k}}=\sum\limits_{1\leq i,j\leq
k}x_{i}y_{[j-i]_{k}}z_{[j-2i]_{k}}w_{j}.$

By a similar argument as computing $H(f,\\{1,2\\})$, we have

$H(f,\\{1,3\\})=\dim_{K}\operatorname{span}_{K}\\{y_{[i+j]_{k}}w_{{[2i+j]}_{k}}\mid
i,j=1,\ldots,k\\}=k^{2},$
$H(f,\\{1,4\\})=\dim_{K}\operatorname{span}_{K}\\{y_{[j-i]_{k}}z_{{[j-2i]}_{k}}\mid
i,j=1,\ldots,k\\}=k^{2}.$

So, $f$ is a balanced neighborly polynomial of type $(k,k,k,k)$. ∎

From now on we assume that $k$ is even. Let
$\overline{x}_{i}=x_{i+\frac{k}{2}},$ $\overline{y}_{i}=y_{i+\frac{k}{2}}$,
$\overline{z}_{i=z_{i}+\frac{k}{2}}$, $\overline{w}_{i}=w_{i+\frac{k}{2}}$ for
$i=1,2,\ldots,\frac{k}{2}.$

######  Theorem 3.3.

Suppose $k=4m,$ with $m\in\mathbb{N}$. Let

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}y_{j}z_{{[i+j-1]}_{2m}}w_{{[\frac{i-1}{2}+j]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}y_{j}\overline{z}_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-1}{2}+j+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}\overline{y}_{j}\overline{z}_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-1}{2}+j]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}\overline{y}_{j}z_{{[i+j-1]}_{2m}}w_{{[\frac{i-1}{2}+j+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}y_{j}z_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-2}{2}+j]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}y_{j}\overline{z}_{{[i+j-1]}_{2m}}w_{{[\frac{i-2}{2}+j+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}\overline{y}_{j}\overline{z}_{{[i+j-1]}_{2m}}w_{{[\frac{i-2}{2}+j]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}\overline{y}_{j}z_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-2}{2}+j+m]}_{2m}}.$

Then $f$ is a balanced neighborly polynomial of type $(k,k,k,k).$

###### Proof.

Set

$\displaystyle B_{1}:=$
$\displaystyle\\{z_{{[i+j-1]}_{2m}}w_{{[\frac{i-1}{2}+j]}_{2m}},\overline{z}_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-1}{2}+j]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:odd\\},$ $\displaystyle B^{\prime}_{1}:=$
$\displaystyle\\{\overline{z}_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-1}{2}+j+m]}_{2m}},z_{{[i+j-1]}_{2m}}w_{{[\frac{i-1}{2}+j+m]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:odd\\},$ $\displaystyle B_{2}:=$
$\displaystyle\\{z_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-2}{2}+j]}_{2m}},\overline{z}_{{[i+j-1]}_{2m}}w_{{[\frac{i-2}{2}+j]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:even\\},$ $\displaystyle B^{\prime}_{2}:=$
$\displaystyle\\{\overline{z}_{{[i+j-1]}_{2m}}w_{{[\frac{i-2}{2}+j+m]}_{2m}},z_{{[i+j-1]}_{2m}}\overline{w}_{{[\frac{i-2}{2}+j+m]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:even\\}.$

Then $B_{1}$, $B^{\prime}_{1}$, $B_{2}$ and $B^{\prime}_{2}$ are pairwise
disjoint. Hence

$|B_{1}\cup B^{\prime}_{1}\cup B_{2}\cup
B^{\prime}_{2}|=|B_{1}|+|B^{\prime}_{1}|+|B_{2}|+|B^{\prime}_{2}|.$

In the same way as the proof of Theorem 3.2, one can prove the injectivity of
the following maps:

$\displaystyle\\{1,3,\ldots,2m-1\\}\times\\{1,2,\ldots,2m\\}$
$\displaystyle\to\\{1,2,\ldots,2m\\}^{2}$ $\displaystyle(i,j)$
$\displaystyle\mapsto([i+j-1]_{2m},[\frac{i-1}{2}+j]_{2m}),$
$\displaystyle(i,j)$
$\displaystyle\mapsto([i+j-1]_{2m},[\frac{i-1}{2}+j+m]_{2m})$

and

$\displaystyle\\{2,4,\ldots,2m\\}\times\\{1,2,\ldots,2m\\}$
$\displaystyle\to\\{1,2,\ldots,2m\\}^{2}$ $\displaystyle(i,j)$
$\displaystyle\mapsto([i+j-1]_{2m},[\frac{i-2}{2}+j]_{2m}),$
$\displaystyle(i,j)$
$\displaystyle\mapsto([i+j-1]_{2m},[\frac{i-2}{2}+j+m]_{2m}).$

It follows that $|B_{1}|=|B^{\prime}_{1}|=|B_{2}|=|B^{\prime}_{2}|=4m^{2}$.
Therefore

$H(f,\\{1,2\\})=\dim_{K}\operatorname{span}_{K}(B_{1}\cup B^{\prime}_{1}\cup
B_{2}\cup B^{\prime}_{2})=(4m)^{2}=k^{2}.$

In order to compute $\dim_{K}R_{\mathfrak{e}_{1}+\mathfrak{e}_{3}}$, we can
rewrite $f$ as follows

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}y_{{[j+1-i]}_{2m}}z_{j}w_{{[j-\frac{i-1}{2}]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}y_{[j+1-i]_{2m}}\overline{z}_{j}\overline{w}_{{[j-\frac{i-1}{2}+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}\overline{y}_{[j+1-i]_{2m}}\overline{z}_{j}\overline{w}_{{[j-\frac{i-1}{2}]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}\overline{y}_{[j+1-i]_{2m}}z_{j}w_{{[j-\frac{i-1}{2}+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}y_{[j+1-i]_{2m}}z_{j}\overline{w}_{{[j-\frac{i}{2}]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}y_{[j+1-i]_{2m}}\overline{z}_{j}w_{{[j-\frac{i}{2}+m]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}\overline{y}_{[j+1-i]_{2m}}\overline{z}_{j}w_{{[j-\frac{i}{2}]}_{2m}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}\overline{y}_{[j+1-i]_{2m}}z_{j}\overline{w}_{{[j-\frac{i}{2}+m]}_{2m}}.$

Set

$\displaystyle B_{3}:=$
$\displaystyle\\{y_{{[j+1-i]}_{2m}}w_{{[j-\frac{i-1}{2}]}_{2m}},\overline{y}_{{[j+1-i]}_{2m}}\overline{w}_{{[j-\frac{i-1}{2}]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:odd\\},$ $\displaystyle B^{\prime}_{3}:=$
$\displaystyle\\{y_{{[j+1-i]}_{2m}}\overline{w}_{{[j-\frac{i-1}{2}+m]}_{2m}},\overline{y}_{{[j+1-i]}_{2m}}w_{{[j-\frac{i-1}{2}+m]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:odd\\},$ $\displaystyle B_{4}:=$
$\displaystyle\\{y_{{[j+1-i]}_{2m}}\overline{w}_{{[j-\frac{i}{2}]}_{2m}},\overline{y}_{{[j+1-i]}_{2m}}w_{{[j-\frac{i}{2}]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:even\\},$ $\displaystyle B^{\prime}_{4}:=$
$\displaystyle\\{y_{{[j+1-i]}_{2m}}w_{{[j-\frac{i}{2}+m]}_{2m}},\overline{y}_{{[j+1-i]}_{2m}}\overline{w}_{{[j-\frac{i}{2}+m]}_{2m}}\mid
i,j=1,2,\ldots,2m,i:even\\}.$

Then $H(f,\\{1,3\\})=\dim\operatorname{span}_{K}(B_{3}\cup B^{\prime}_{3}\cup
B_{4}\cup B^{\prime}_{4})$. A similar argument as computing $H(f,\\{1,2\\})$,
we have $H(f,\\{1,3\\})=(4m)^{2}=k^{2}.$ Similarly, we can rewrite $f$ as
follows

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}y_{[j-\frac{i-1}{2}]_{2m}}z_{[j+\frac{i-1}{2}]_{2m}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}y_{[j-\frac{i-1}{2}-m]_{2m}}\overline{z}_{[j+\frac{i-1}{2}-m]_{2m}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}x_{i}\overline{y}_{[j-\frac{i-1}{2}]_{2m}}\overline{z}_{[j+\frac{i-1}{2}]_{2m}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:odd}\overline{x}_{i}\overline{y}_{[j-\frac{i-1}{2}-m]_{2m}}z_{[j+\frac{i-1}{2}-m]_{2m}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}y_{[j-\frac{i-2}{2}]_{2m}}z_{[j+\frac{i}{2}]_{2m}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}y_{[j-\frac{i-2}{2}-m]_{2m}}\overline{z}_{[j+\frac{i}{2}-m]_{2m}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}x_{i}\overline{y}_{[j-\frac{i-2}{2}]_{2m}}\overline{z}_{[j+\frac{i}{2}]_{2m}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m,i:even}\overline{x}_{i}\overline{y}_{[j-\frac{i-2}{2}-m]_{2m}}z_{[j+\frac{i}{2}-m]_{2m}}\overline{w}_{j}.$

By similar computing as above, we get that $H(f,\\{1,4\\})=(4m)^{2}=k^{2}.$
So, $f$ is a balanced neighborly polynomial of type $(k,k,k,k)$. ∎

######  Theorem 3.4.

Suppose $k=4m+2,$ with $m\in\mathbb{N}$. Let

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}y_{j}z_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-1}{2}+j]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}y_{j}\overline{z}_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-1}{2}+j+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}\overline{y}_{j}\overline{z}_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-1}{2}+j]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}\overline{y}_{j}z_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-1}{2}+j+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}y_{j}z_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-2}{2}+j]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}y_{j}\overline{z}_{{[i+j-1]}_{2m+1}}w_{{[\frac{i}{2}+j+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}\overline{y}_{j}\overline{z}_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-2}{2}+j]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}\overline{y}_{j}z_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i}{2}+j+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}y_{j}\overline{z}_{j}w_{j}+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}\overline{y}_{j}z_{j}\overline{w}_{j}.$

Then $f$ is a balanced neighborly polynomial of type $(k,k,k,k).$

###### Proof.

Set

$\displaystyle C_{1}:=$
$\displaystyle\\{z_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-1}{2}+j]}_{2m+1}},\overline{z}_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-1}{2}+j]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:odd\\},$ $\displaystyle C^{\prime}_{1}:=$
$\displaystyle\\{\overline{z}_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-1}{2}+j+m]}_{2m+1}},z_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-1}{2}+j+m]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:odd,i\neq 2m+1\\},$ $\displaystyle C_{2}:=$
$\displaystyle\\{z_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i-2}{2}+j]}_{2m+1}},\overline{z}_{{[i+j-1]}_{2m+1}}w_{{[\frac{i-2}{2}+j]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:even\\},$ $\displaystyle C^{\prime}_{2}:=$
$\displaystyle\\{\overline{z}_{{[i+j-1]}_{2m+1}}w_{{[\frac{i}{2}+j+m]}_{2m+1}},z_{{[i+j-1]}_{2m+1}}\overline{w}_{{[\frac{i}{2}+j+m]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:even\\},$ $\displaystyle C_{3}:=$
$\displaystyle\\{\overline{z}_{j}w_{j}+\overline{z}_{[2m+j]_{2m+1}}\overline{w}_{[2m+j]_{2m+1}},z_{j}\overline{w}_{j}+z_{[2m+j]_{2m+1}}w_{[2m+j]_{2m+1}}\mid
j=1,2,\ldots,2m+1\\},$ $\displaystyle C^{\prime}_{3}:=$
$\displaystyle\\{\overline{z}_{j}w_{j},z_{j}\overline{w}_{j}\mid
j=1,2,\ldots,2m+1\\}.$

We can see that
$\overline{z}_{[2m+j]_{2m+1}}\overline{w}_{[2m+j]_{2m+1}},z_{[2m+j]_{2m+1}}w_{[2m+j]_{2m+1}}$
belong to $C_{1}$. Thus

$\operatorname{span}(C_{1}\cup C^{\prime}_{1}\cup C_{2}\cup C^{\prime}_{2}\cup
C_{3})=\operatorname{span}(C_{1}\cup C^{\prime}_{1}\cup C_{2}\cup
C^{\prime}_{2}\cup C^{\prime}_{3}).$

Moreover, $\overline{z}_{j}w_{j},z_{j}\overline{w}_{j}$ do not belong to
$C_{2}\cup C^{\prime}_{2}$, so
$C_{1},C^{\prime}_{1},C_{2},C^{\prime}_{2},C^{\prime}_{3}$ are pairwise
disjoint. Hence $|C_{1}\cup C^{\prime}_{1}\cup C_{2}\cup C^{\prime}_{2}\cup
C^{\prime}_{3}|=|C_{1}|+|C^{\prime}_{1}|+|C_{2}|+|C^{\prime}_{2}|+|C^{\prime}_{3}|$.
In the same way as in the proof of Theorem 3.3, one can show
$|C_{1}|=2(m+1)(2m+1)$, $|C^{\prime}_{1}|=2m(2m+1)$,
$|C_{2}|=|C^{\prime}_{2}|=2m(2m+1)$ and $|C^{\prime}_{3}|=2(2m+1)$. Then
$|C_{1}\cup C^{\prime}_{1}\cup C_{2}\cup C^{\prime}_{2}\cup
C^{\prime}_{3}|=(4m+2)^{2}=k^{2}$. Therefore,

$H(f,\\{1,2\\})=\dim_{K}\operatorname{span}_{K}C_{1}\cup C^{\prime}_{1}\cup
C_{2}\cup C^{\prime}_{2}\cup C^{\prime}_{3}=(4m+2)^{2}=k^{2}.$

On the other hand, we can rewrite $f$ as follows

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}y_{{[j+1-i]}_{2m+1}}z_{j}w_{{[j-\frac{i-1}{2}]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}y_{[j+1-i]_{2m+1}}\overline{z}_{j}\overline{w}_{{[j-\frac{i-1}{2}+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}\overline{y}_{[j+1-i]_{2m+1}}\overline{z}_{j}\overline{w}_{{[j-\frac{i-1}{2}]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}\overline{y}_{[j+1-i]_{2m+1}}z_{j}w_{{[j-\frac{i-1}{2}+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}y_{[j+1-i]_{2m+1}}z_{j}\overline{w}_{{[j-\frac{i}{2}]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}y_{[j+1-i]_{2m+1}}\overline{z}_{j}w_{{[j-\frac{i-2}{2}+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}\overline{y}_{[j+1-i]_{2m+1}}\overline{z}_{j}w_{{[j-\frac{i}{2}]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}\overline{y}_{[j+1-i]_{2m+1}}z_{j}\overline{w}_{{[j-\frac{i-2}{2}+m]}_{2m+1}}$
$\displaystyle+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}y_{j}\overline{z}_{j}w_{j}+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}\overline{y}_{j}z_{j}\overline{w}_{j}.$

Set

$\displaystyle C_{4}:=$
$\displaystyle\\{y_{[j+1-i]_{2m+1}}w_{{[j+\frac{1-i}{2}]}_{2m+1}},\overline{y}_{[j+1-i]_{2m+1}}\overline{w}_{{[j+\frac{1-i}{2}]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:odd\\},$ $\displaystyle C^{\prime}_{4}:=$
$\displaystyle\\{y_{[j+1-i]_{2m+1}}\overline{w}_{[\frac{1-i}{2}+j+m]_{2m+1}},\overline{y}_{[j+1-i]_{2m+1}}w_{{[\frac{1-i}{2}+j+m]}_{2m}}\mid
i,j=1,2,\ldots,2m+1,i:odd,i\neq 2m+1\\},$ $\displaystyle C_{5}:=$
$\displaystyle\\{y_{[j+1-i]_{2m+1}}\overline{w}_{{[j-\frac{i}{2}]}_{2m+1}},\overline{y}_{[j+1-i]_{2m+1}}w_{{[j-\frac{i}{2}]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:even\\},$ $\displaystyle C^{\prime}_{5}:=$
$\displaystyle\\{y_{[j+1-i]_{2m+1}}w_{{[j-\frac{i-2}{2}+m]}_{2m+1}},\overline{y}_{[j+1-i]_{2m+1}}\overline{w}_{{[j-\frac{i-2}{2}+m]}_{2m+1}}\mid
i,j=1,2,\ldots,2m+1,i:even\\},$ $\displaystyle C_{6}:=$
$\displaystyle\\{y_{j}w_{j}+y_{[j+1]_{2m+1}}\overline{w}_{j},\overline{y}_{j}\overline{w}_{j}+\overline{y}_{[j+1]_{2m+1}}w_{j}\mid
j=1,2,\ldots,2m+1\\}.$ $\displaystyle C^{\prime}_{6}:=$
$\displaystyle\\{y_{j}\overline{w}_{j},\overline{y}_{j}w_{j}\mid
j=1,2,\ldots,2m+1\\}.$

A similar argument as computing $H(f,\\{1,2\\})$, we have

$H(f,\\{1,3\\})=\dim\operatorname{span}_{K}(C_{4}\cup C^{\prime}_{4}\cup
C_{5}\cup C^{\prime}_{5}\cup C^{\prime}_{6})=(4m+2)^{2}=k^{2}.$

Similarly, we can rewrite $f$ as follows

$\displaystyle f$ $\displaystyle=\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}y_{[j-\frac{i-1}{2}]_{2m+1}}z_{[j+\frac{i-1}{2}]_{2m+1}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}\overline{y}_{[j-\frac{i-1}{2}-m]_{2m+1}}z_{[j+\frac{i-1}{2}-m]_{2m+1}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}x_{i}\overline{y}_{[j-\frac{i-1}{2}]_{2m+1}}\overline{z}_{[j+\frac{i-1}{2}]_{2m+1}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:odd}\overline{x}_{i}y_{[j-\frac{i-1}{2}-m]_{2m+1}}\overline{z}_{[j+\frac{i-1}{2}-m]_{2m+1}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}\overline{y}_{[j-\frac{i-1}{2}]_{2m+1}}\overline{z}_{[j+\frac{i-1}{2}]_{2m+1}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}y_{[j-\frac{i-1}{2}-m]_{2m+1}}\overline{z}_{[j+\frac{i-1}{2}-m]_{2m+1}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}x_{i}y_{[j-\frac{i-1}{2}]_{2m+1}}z_{[j+\frac{i-1}{2}]_{2m+1}}w_{j}$
$\displaystyle+\sum\limits_{1\leq i,j\leq
2m+1,i:even}\overline{x}_{i}\overline{y}_{[j-\frac{i-1}{2}-m]_{2m+1}}z_{[j+\frac{i-1}{2}-m]_{2m+1}}\overline{w}_{j}$
$\displaystyle+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}y_{j}\overline{z}_{j}w_{j}+\sum\limits_{1\leq j\leq
2m+1}\overline{x}_{2m+1}\overline{y}_{j}z_{j}\overline{w}_{j},$

and we can see $H(f,\\{1,4\\})=(4m+2)^{2}=k^{2}.$ So, $f$ is a balanced
neighborly polynomial of type $(k,k,k,k)$. ∎

##  4\. Balanced neighborly polynomials and balanced neighborly spheres

In this section, we present a relation between balanced neighborly polynomials
and balanced neighborly spheres. We first recall basic definitions on
simplicial spheres. Let $\Delta$ be a finite simplicial complex on the vertex
set $V=\\{x_{1},x_{2},\ldots,x_{n}\\}$. Thus $\Delta$ is a nonempty collection
of subsets of $V$ satisfying that $F\in\Delta$ and $G\subset F$ imply
$G\in\Delta.$ Elements of $\Delta$ are called faces of $\Delta$ and maximal
faces (under inclusion) are called facets. The dimension of a face
$F\in\Delta$ is $\dim F=|F|-1,$ and the dimension of $\Delta$ is the maximum
dimension of its faces. Faces of dimension $0$ are called vertices and faces
of dimension $1$ are called edges. Let $R=K[x_{1},\ldots,x_{n}]$ be a
polynomial ring over an infinite field $K.$ The ring $K[\Delta]=R/I_{\Delta}$,
where $I_{\Delta}=(\underset{x_{i}\in F}{\prod}x_{i}\mid F\subset
V,F\notin\Delta)$, is called the Stanley–Reisner ring of $\Delta$. It is well-
known that the Krull dimension of $K[\Delta]$ equals $\dim\Delta+1$ [4, II
Theorem 1.3]. If $\dim K[\Delta]=d,$ and a sequence
$\Theta=\theta_{1},\ldots,\theta_{d}$ of linear forms such that
$\dim_{K}K[\Delta]/(\Theta)<\infty$ then $\Theta$ is called a linear system of
parameters (l.s.o.p. for short) of $K[\Delta].$

A map $\kappa:V\to[d]$ is called a proper $d$-coloring map of $\Delta,$ if we
have $\kappa(i)\neq\kappa(j)$ for any edge $\\{i,j\\}\in\Delta.$ We say that a
$(d-1)$-dimensional simplicial complex $\Delta$ on $V$ is balanced (completely
balanced in some literature) if its graph is $d$-colorable. Let $\Delta$ be a
$d-1$ dimensional balanced simplicial complex and let $\kappa:V\to[d]$ be a
proper $d$-coloring map of $\Delta.$ Set $V_{k}=\\{v\in V\mid\kappa(v)=k\\}.$
Let $\theta_{k}=\underset{v\in V_{k}}{\sum}x_{v}$ for $k=1,2,\ldots,d$. Then
$\Theta$ is a l.s.o.p of $K[\Delta]$ [4, III Proposition 4.3]. Recall that a
simplicial sphere is a simplicial complex which is homeomorphic to a sphere.
That means if $\Delta$ is a simplicial sphere then $K[\Delta]$ is a Gorenstein
ring and $K[\Delta]/(\Theta)$ is an Artinian Gorenstein algebra of socle
degree $d$ [4, II Corollary 5.2]. A balanced simplicial sphere of dimension
$d-1$ is called neighborly of type
$(n_{1},n_{2},\ldots,n_{d})\in\mathbb{Z}^{d}$ if
$\dim_{K}(K[\Delta]/(\Theta))_{\mathfrak{e}_{S}}=\underset{k\in
S}{\prod}n_{k}$ for all $S\subset[d]$.

From now on, we assume that $R=K[x_{ij}\mid 1\leq i\leq d,1\leq j\leq n_{i}$]
and $\Delta$ has vertex set $\\{x_{ij}\mid 1\leq i\leq d,1\leq j\leq
n_{i}\\}$.

######  Proposition 4.1.

If a balanced neighborly simplicial sphere of type $(n_{1},\ldots,n_{d})$
exists, then a balanced neighborly polynomial of type $(n_{1},\ldots,n_{d})$
exists over any field $K$.

###### Proof.

Set $R^{\prime}=K[x_{ij}\mid 1\leq i\leq d,1\leq j\leq n_{i}-1$]. Then
$R/(\Theta)\cong R^{\prime}$, so

$R/{(I_{\Delta}+(\Theta))}\cong R^{\prime}/J\text{ for some ideal }J.$

By Proposition 4.3 in [3], we have $x_{ij}x_{ik}=0$ in
$R/(I_{\Delta}+(\Theta))$ for all $1\leq i\leq d$ and $1\leq j\leq k\leq
n_{i}$. It follows that $x_{ij}x_{ik}=0$ in $S^{\prime}/J$ for all $1\leq
i\leq d$ and $1\leq j\leq k\leq n_{i}-1$. Since $R/(I_{\Delta}+\Theta)$ is
Artinian Gorenstein, Lemma 2.1 shows that there is a balanced polynomial $f\in
R$ such that $\operatorname{ann}(f)=J$. This $f$ must be neighborly by Lemma
2.2, (i). ∎

By Proposition 4.1 and Theorem 3.1, we can recover the following a result of
Zheng [5], Proposition 6.

######  Proposition 4.2.

There are no balanced neighborly simplicial spheres of type $(2,2,2,2).$

Acknowledgement. The authors would like to thank Professor Satoshi Murai for
his interesting discussions on this subject. The authors would like to thank
the referee for the valuable comments to improve this article.

## References

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