# Positive definiteness of fourth order three dimensional symmetric tensors

Yishenhg Song  School of Mathematical Sciences, Chongqing Normal University,
Chongqing 401331 P.R. China.
Email<EMAIL_ADDRESS>
###### Abstract

For a 4th order 3-dimensional symmetric tensor with its entries $1$ or $-1$,
we show the analytic sufficient and necessary conditions of its positive
definiteness. By applying these conclusions, several strict inequalities is
bulit for ternary quartic homogeneous polynomials.

###### keywords:

Positive definiteness, Fourth order tensors, Homogeneous polynomial.

††journal:  00footnotetext: The author’s work was supported by the National
Natural Science Foundation of P.R. China (Grant No.12171064), by The team
project of innovation leading talent in chongqing (No.CQYC20210309536) and by
the Foundation of Chongqing Normal univer- sity (20XLB009).

##  1 Introduction

One of the most direct applications of positive definite tensors is to verify
the vacuum stability of the Higgs scalar potential model [1, 2]. Qi [3] first
used the consept of positive definiteness for a symmetric tensor when the
order is even integer.

###### Definition 1.1

Let $\mathcal{T}=(t_{i_{1}i_{2}\cdots i_{m}})$ be an $m$th order $n$
dimensional symmetric tensor. $\mathcal{T}$ is called

  * (i)

positive semi-definite ([3]) if $m$ is an even number and in the Euclidean
space $\mathbb{R}^{n}$, its associated Homogeneous polynomial

$\mathcal{T}x^{m}=\sum\limits_{i_{1},i_{2},\cdots,i_{m}=1}^{n}t_{i_{1}i_{2}\cdots
i_{m}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\geq 0;$

  * (ii)

positive definite ([3]) if $m$ is an even number and $\mathcal{T}x^{m}>0$ for
all $x\in\mathbb{R}^{n}\setminus\\{0\\}$.

Clearly, a positive semi-definite tensor coincides with a positive semi-
definite matrix if $m=2$. It is well-known that Sylvester’s Criterion can
efficiently identify the positive (semi-)definiteness of a matrix. The
positive definiteness of a 4th order 2 dimensional symmetric tensor, (or
positivity condition of a quartic univariate polynomial) may trace back to
ones of Refs. Rees [4] , Lazard [5] Gadem-Li [6], Ku [7] and Jury-Mansour [8].
Untill to 2005, Wang-Qi [9] improved their proof, and perfectly gave analytic
necessary and sufficient conditions. However, the above result depends on the
discriminant of such a quartic polynomial. Hasan-Hasa [10] claimed that a
necessary and sufficient condition of positive definiteness was proved without
the discriminant. However, there is a problem in their argumentations. In
1998, Fu [11] pointed out that Hasan-Hasan’s results are sufficient only.
Recently, Guo[12] showed a new necessary and sufficient condition without the
discriminant. Very recently, Qi-Song-Zhang[13] gave a new necessary and
sufficient condition other than the above results. For more detail about
applications of these results, see Song-Qi [14] also.

In 2005, Qi [3] gave that the sign of all H-(Z-)eigenvalue of a even order
symmetric tensor can verify the positive definitiveness of such a higher order
tensor. Subsequently, Ni-Qi-Wang [15] provided a method of computing the
smallest eigenvalue for checking positive definiteness of a 4th order 3
dimensional tensor. Ng-Qi-Zhou [16] presented an algorithm of the largest
eigenvalue of a nonnegative tensor. For a 4th order 3 dimensional symmetric
tensor, Song [17] proved several sufficient conditions of its positive
definiteness. Until now, an analytic necessary and sufficient condition has
not been found for positive (semi-)definiteness for a 4th order 3 dimensional
symmetric tensor.

In this paper, we mainly dicuss analytic necessary and sufficient conditions
of positive definiteness of a class of 4th order 3-dimensional symmetric
tensors (Theorem 3.1). Furthermore, several strict inequalities of ternary
quartic homogeneous polynomial (Corollary 3.2) are built.

##  2 Copositivity of 4th order 2-dimensional symmetric tensors

Let $\mathcal{T}=(t_{ijkl})$ be a 4th-order 2-dimensional symmetric tensor.
Then for $x=(x_{1},x_{2})^{\top},$

$Tx^{4}=t_{1111}x_{1}^{4}+4t_{1112}x_{1}^{3}x_{2}+6t_{1122}x_{1}^{2}x_{2}^{2}+4t_{1222}x_{1}x_{2}^{3}+t_{2222}x_{2}^{4}.$
(2.1)

Let

$\displaystyle\Delta=$ $\displaystyle 4\times
12^{3}(t_{1111}t_{2222}-4t_{1112}t_{1222}+3t_{1122}^{2})^{3}$
$\displaystyle-72^{2}\times
6^{2}(t_{1111}t_{1122}t_{2222}+2t_{1112}t_{1122}t_{1222}-t_{1122}^{3}-t_{1111}t_{1222}^{2}-t_{1112}^{2}t_{2222})^{2}$
$\displaystyle=$ $\displaystyle 4\times 12^{3}(I^{3}-27J^{2}),$

where

$\displaystyle I=$ $\displaystyle
t_{1111}t_{2222}-4t_{1112}t_{1222}+3t_{1122}^{2},$ $\displaystyle J=$
$\displaystyle
t_{1111}t_{1122}t_{2222}+2t_{1112}t_{1122}t_{1222}-t_{1122}^{3}-t_{1111}t_{1222}^{2}-t_{1112}^{2}t_{2222}.$

and hence, the sign of $\Delta$ is the same as one of $(I^{3}-27J^{2})$.
Ulrich-Watson [18] presented the analytic conditions of the nonnegativity of a
quartic and univariate polynomial in $\mathbb{R}_{+}$. Qi-Song-Zhang [13] also
gave the nonnegativity and positivity of a quartic and univariate polynomial
in $\mathbb{R}$, which means the positive (semi-)definitiveness of 4th order
2-dimensional tensor [2].

######  Lemma 2.1 ([2, 13])

A 4th-order 2-dimensional symmetric tensor $\mathcal{T}=(t_{ijkl})$ is
positive definite if and only if

$\begin{cases}I^{3}-27J^{2}=0,\ \
t_{1112}\sqrt{t_{2222}}=t_{1222}\sqrt{t_{1111}},\\\
2t_{1112}^{2}+t_{1111}\sqrt{t_{1111}t_{2222}}=3t_{1111}t_{1122}<3t_{1111}\sqrt{t_{1111}t_{2222}};\\\
I^{3}-27J^{2}>0,\ \
|t_{1112}\sqrt{t_{2222}}-t_{1222}\sqrt{t_{1111}}|\leq\sqrt{6t_{1111}t_{1122}t_{2222}+2\sqrt{(t_{1111}t_{2222})^{3}}},\\\
(i)\ -\sqrt{t_{1111}t_{2222}}\leq 3t_{1221}\leq 3\sqrt{t_{1111}v_{2222}};\\\
(ii)\ t_{1221}>\sqrt{t_{1111}t_{2222}}\ \mbox{ and }\\\
|t_{1112}\sqrt{t_{2222}}+t_{1222}\sqrt{t_{1111}}|\leq\sqrt{6t_{1111}t_{1122}t_{2222}-2\sqrt{(t_{1111}t_{2222})^{3}}}.\end{cases}$
$None$

A 4th-order 2-dimensional symmetric tensor $\mathcal{T}=(t_{ijkl})$ is
positive semidefinite if and only if

$\begin{cases}I^{3}-27J^{2}\geq 0,\ \
|t_{1112}\sqrt{t_{2222}}-t_{1222}\sqrt{t_{1111}}|\leq\sqrt{6t_{1111}t_{1122}t_{2222}+2\sqrt{(t_{1111}t_{2222})^{3}}},\\\
(i)\ -\sqrt{t_{1111}t_{2222}}\leq 3t_{1122}\leq 3\sqrt{t_{1111}t_{2222}};\\\
(ii)\ t_{1122}>\sqrt{t_{1111}t_{2222}}\mbox{ and }\\\
|t_{1112}\sqrt{t_{2222}}+t_{1222}\sqrt{t_{1111}}|\leq\sqrt{6t_{1111}t_{1122}t_{2222}-2\sqrt{(t_{1111}t_{2222})^{3}}}.\end{cases}$
$None$

###### Lemma 2.2

Let $\mathcal{T}=(t_{ijkl})$ be a 4th-order 2-dimensional symmetric tensor
with its entires $|t_{ijkl}|=1$ and $t_{1111}=t_{2222}=1.$ Then

  * (i)

$\mathcal{T}$ is positive semidefinite if and only if $t_{1122}=1;$

  * (ii)

$\mathcal{T}$ is positive definite if and only if $t_{1122}=1\mbox{ and
}t_{1112}t_{1222}=-1.$

Proof. (i) It follows from Lemma 2.1 (II) that $\mathcal{T}$ is positive
semidefinite if and only if

$I^{3}-27J^{2}\geq 0,\ \ |t_{1112}-t_{1222}|\leq\sqrt{6t_{1122}+2}\mbox{ and
}-1\leq 3t_{1122}\leq 3.$

Since $|t_{ijkl}|=1$, then which means $t_{1122}=1$ and either
$t_{1112}t_{1222}=1$,

$I^{3}-27J^{2}=(1-4+3)^{3}-27(1+2-1-1-1)^{2}=0,$
$|t_{1112}-t_{1222}|=0<\sqrt{6t_{1122}+2}=\sqrt{8};$

or $t_{1112}t_{1222}=-1,$

$I^{3}-27J^{2}=(1+4+3)^{3}-27(1-2-1-1-1)^{2}>0,$
$|t_{1112}-t_{1222}|=2<\sqrt{6t_{1122}+2}=\sqrt{8}.$

So $\mathcal{T}$ is positive semidefinite if and only if $t_{1122}=1.$

(ii) It follows from Lemma 2.1 (I) that $\mathcal{T}$ is positive definite if
and only if

$\displaystyle I^{3}-27J^{2}=$ $\displaystyle 0,\ \ t_{1112}=t_{1222},\ \
2t_{1112}^{2}+1=3t_{1122}<3;$ $\displaystyle I^{3}-27J^{2}>$ $\displaystyle
0,\ \ |t_{1112}-t_{1222}|\leq\sqrt{6t_{1122}+2}\mbox{ and }-1\leq
3t_{1122}\leq 3.$

Since $3=2t_{1112}^{2}+1=3t_{1122}<3$ can’t hold, then $\mathcal{T}$ is
positive definite if and only if $t_{1122}=1\mbox{ and }t_{1112}t_{1222}=-1.$
This completes the proof.

##  3 Positive definiteness of 4th order 3-dimensional symmetric tensors

###### Theorem 3.1

Let $\mathcal{T}=(t_{ijkl})$ be a 4th-order $3$-dimensional symmetric tensor
with

$|t_{iiij}|=|t_{iijk}|=t_{iiii}=1$ and $t_{ijjj}t_{iiij}=-1$ for all
$i,j,k\in\\{1,2,3\\}$, $i\neq j$, $i\neq k$, $j\neq k.$

  * (i)

If $t_{iijj}=2$ for all $i,j\in\\{1,2,3\\}$ and $i\neq j$, then $\mathcal{T}$
is positive definite if and only if

$\begin{cases}t_{1123}=t_{1223}=t_{1233}=1,\mbox{ or}\\\
t_{1123}=t_{1223}=t_{1233}=-1.\end{cases}$ $None$

  * (ii)

If $t_{iijj}=2.5$ for all $i,j\in\\{1,2,3\\}$ and $i\neq j$, then
$\mathcal{T}$ is positive definite if and only if

$\begin{cases}t_{1123}=t_{1223}=t_{1233}=1,\mbox{ or}\\\
t_{1123}=t_{1223}=t_{1233}=-1\mbox{ or}\\\ \mbox{two of
}\\{t_{1123},t_{1223},t_{1233}\\}\mbox{ are }-1.\end{cases}$ $None$

  * (ii)

If $t_{iijj}\geq 3$ for all $i,j\in\\{1,2,3\\}$ and $i\neq j$, then
$\mathcal{T}$ is positive definite if and only if

$\begin{cases}t_{1123}=t_{1223}=t_{1233}=1,\mbox{ or}\\\
t_{1123}=t_{1223}=t_{1233}=-1,\mbox{ or}\\\ \mbox{two of
}\\{t_{1123},t_{1223},t_{1233}\\}\mbox{ are }-1\mbox{ or}\\\ \mbox{one of
}\\{t_{1123},t_{1223},t_{1233}\\}\mbox{ are }-1.\end{cases}$ $None$

Proof. Let $t_{1222}=t_{2333}=t_{1113}=1$ and $t_{1112}=t_{1333}=t_{2223}=-1$
without loss the generality. $\mathcal{T}x^{4}$ may be rewritten as follows,

$\displaystyle\mathcal{T}x^{4}=$
$\displaystyle(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})$
$\displaystyle+6(t_{1122}-1)x_{1}^{2}x_{2}^{2}+6(t_{1133}-1)x_{1}^{2}x_{3}^{2}+6(t_{2233}-1)x_{2}^{2}x_{3}^{2}$
$\displaystyle+12(t_{1123}-1)x_{1}^{2}x_{2}x_{3}+12(t_{1223}-1)x_{1}x_{2}^{2}x_{3}+12(t_{1233}-1)x_{1}x_{2}x_{3}^{2}$
$\displaystyle=$
$\displaystyle(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2})$
$\displaystyle+6(t_{1122}-1)x_{1}^{2}x_{2}^{2}+6(t_{1133}-1)x_{1}^{2}x_{3}^{2}+6(t_{2233}-1)x_{2}^{2}x_{3}^{2}$
$\displaystyle+12(t_{1123}+1)x_{1}^{2}x_{2}x_{3}+12(t_{1223}+1)x_{1}x_{2}^{2}x_{3}+12(t_{1233}-1)x_{1}x_{2}x_{3}^{2}$
$\displaystyle=$
$\displaystyle(x_{1}-x_{2}+x_{3})^{4}+8(x_{1}x_{2}^{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{3})$
$\displaystyle+6(t_{1122}-1)x_{1}^{2}x_{2}^{2}+6(t_{1133}-1)x_{1}^{2}x_{3}^{2}+6(t_{2233}-1)x_{2}^{2}x_{3}^{2}$
$\displaystyle+12(t_{1123}+1)x_{1}^{2}x_{2}x_{3}+12(t_{1223}-1)x_{1}x_{2}^{2}x_{3}+12(t_{1233}+1)x_{1}x_{2}x_{3}^{2}$
$\displaystyle=$
$\displaystyle(x_{2}+x_{3}-x_{1})^{4}+8(x_{1}^{3}x_{3}+x_{1}x_{2}^{3}-x_{2}^{3}x_{3})$
$\displaystyle+6(t_{1122}-1)x_{1}^{2}x_{2}^{2}+6(t_{1133}-1)x_{1}^{2}x_{3}^{2}+6(t_{2233}-1)x_{2}^{2}x_{3}^{2}$
$\displaystyle+12(t_{1123}-1)x_{1}^{2}x_{2}x_{3}+12(t_{1223}+1)x_{1}x_{2}^{2}x_{3}+12(t_{1233}+1)x_{1}x_{2}x_{3}^{2}.$

We claim that $\mathcal{T}x^{4}$ is bounded from below. In fact, we might take
$t_{1123}=t_{1223}=-1$ and $t_{1233}=1$. Since $t_{iijj}\geq 1$, then we have

$\mathcal{T}x^{4}\geq
f(x_{1},x_{2},x_{3})=(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2}).$

We only need show the boundedness from below of a polynomial
$f(x_{1},x_{2},x_{3})$. In fact, there are three cases.

Case 1. It follows from the conditions and Lemma 2.2 (ii) that
$f(x_{1},x_{2},x_{3})\geq 0$ if $x_{i}\equiv 0$ for some $i\in\\{1,2,3\\}.$

Case 2. $f(x_{1},x_{2},x_{3})\geq 0$ when $x_{1}+x_{2}-x_{3}=0$. Since
$(x_{1}+x_{2}-x_{3})^{4}=0$ implies $x_{1}+x_{2}-x_{3}=0$, then we claim that
$x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2}>0$ for all $x_{1},x_{2},x_{3}$
with $x_{3}=x_{1}+x_{2}-x_{3}$. In fact,

$\displaystyle x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2}=$ $\displaystyle
x_{1}^{3}(x_{1}+x_{2})+x_{2}(x_{1}+x_{2})^{3}-x_{1}^{3}x_{2}$ $\displaystyle=$
$\displaystyle
x_{1}^{4}+x_{1}^{3}x_{2}+3x_{1}^{2}x_{2}^{2}+3x_{1}x_{2}^{3}+x_{2}^{4},$

which is 4th order 2-dimensional symmetric tensor $\mathcal{A}=(a_{ijkl})$
with

$a_{1111}=a_{2222}=1,a_{1112}=\dfrac{1}{4},a_{1122}=\dfrac{1}{2},a_{1222}=\dfrac{3}{4}.$

Therefore, it follows from Lemma 2.1(I) that $\mathcal{A}$ is positive
definite, and so, $\mathcal{T}x^{4}\geq 0$ when $x_{1}+x_{2}-x_{3}=0$.

Case 3. Let $x_{i}\neq 0$ for all $i\in\\{1,2,3\\}$ with
$x_{1}+x_{2}-x_{3}\neq 0$. Then

$\displaystyle f(x_{1},x_{2},x_{3})=$
$\displaystyle(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2})$
$\displaystyle=$ $\displaystyle
x_{3}^{4}\left(\left(\dfrac{x_{1}}{x_{3}}+\dfrac{x_{2}}{x_{3}}-1\right)^{4}+8\left(\left(\dfrac{x_{1}}{x_{3}}\right)^{3}+\dfrac{x_{2}}{x_{3}}-\left(\dfrac{x_{1}}{x_{3}}\right)^{3}\left(\dfrac{x_{2}}{x_{3}}\right)\right)\right)$
$\displaystyle=$ $\displaystyle
x_{3}^{4}((y_{1}+y_{2}-1)^{4}+8(y_{1}^{3}+y_{2}-y_{1}^{3}y_{2}))$
$\displaystyle=$ $\displaystyle x_{3}^{4}f(y_{1},y_{2},1).$

Therefore, the boundedness from below of $\mathcal{T}x^{4}$ if and only if the
boundedness from below of $f(y_{1},y_{2},1)$. The stationary points of the
function $f(y_{1},y_{2},1)$ is the solution of the systems

$\begin{cases}4(y_{1}+y_{2}-1)^{3}+8(3y_{1}^{2}-3y_{1}^{2}y_{2})=0\\\
4(y_{1}+y_{2}-1)^{3}+8(1-y_{1}^{3})=0,\end{cases}$

which implies

$1-y_{2}=\dfrac{1-y_{1}^{3}}{3y_{1}^{2}}\ \Rightarrow\
\left(y_{1}-\dfrac{1-y_{1}^{3}}{3y_{1}^{2}}\right)^{3}+2(1-y_{1}^{3})=0.$

So we have

$10y_{1}^{9}+6y_{1}^{6}+12y_{1}^{3}-1=0.$

Then it may be regarded as a univariate cubic equation

$f(s)=10s^{3}+6s^{2}+12s-1=0,\ \ s=y_{1}^{3},$

which has one real root $y_{1}^{*}$ and a dual complex conjugate roots since
its discriminant $\Delta>0,$

$\Delta=B^{2}-4AC\ \mbox{ and }\begin{cases}A=6^{2}-3\times 10\times
12=-324\\\ B=6\times 12-9\times 10\times(-1)=162\\\ C=12^{2}-3\times
6\times(-1)=180\end{cases}$

Since $f(\frac{1}{8})>0$ and $f(\frac{1}{27})<0$, then the unique real root

$s^{*}=y_{1}^{*3}\mbox{ with
}\frac{1}{27}<s^{*}<\dfrac{1}{8}\Rightarrow\frac{1}{3}<y_{1}^{*}<\dfrac{1}{2}\mbox{
and }-\dfrac{17}{9}<y_{2}^{*}<-\dfrac{1}{6}<1.$

Clearly, $y_{1}^{*}+y_{2}^{*}<\dfrac{1}{3},$ and then the Hessian matrix of
$\mathcal{T}y^{4}$ is positive definite at a point $(y_{1}^{*},y_{2}^{*})$,

$H=12\begin{pmatrix}(y_{1}+y_{2}-1)^{2}+4y_{1}(1-y_{2})&(y_{1}+y_{2}-1)^{2}-2y_{1}^{2}\\\
(y_{1}+y_{2}-1)^{2}-2y_{1}^{2}&(y_{1}+y_{2}-1)^{2}\end{pmatrix}.$

Therefore, $(y_{1}^{*},y_{2}^{*})$ is unique local minimum point of
$f(y_{1},y_{2},1)$, and so, it is global also. Thus $f(x_{1},x_{2},x_{3})$ is
bounded from below

As a result, $\mathcal{T}x^{4}$ is bounded from below, and the proof is
similar under other conditions. So, the claim is proved.

(i) Necessity. Suppose the conditions (III) can’t hold when $\mathcal{T}$ is
positive definite, then there may be two cases.

Case 1. There is two $-1$ in $\\{t_{1123},t_{1223},t_{1233}\\}$. We might as
well take $t_{1123}=t_{1223}=-1$ and $t_{1233}=1$. Take
$x=(\frac{1}{5},-\frac{1}{5},1)^{\top}$. Then we have

$\displaystyle\mathcal{T}x^{4}=$
$\displaystyle(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2})+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})$
$\displaystyle=$ $\displaystyle
1+8\left(\dfrac{1}{5^{3}}-\dfrac{1}{5}+\dfrac{1}{5^{4}}\right)+6\left(\dfrac{1}{5^{4}}+\dfrac{1}{5^{2}}+\dfrac{1}{5^{2}}\right)=-\dfrac{21}{625}<0;$

Case 2. There is only one $-1$ in $\\{t_{1123},t_{1223},t_{1233}\\}$. We might
as well take $t_{1123}=t_{1233}=1$ and $t_{1223}=-1$. For
$x=(\frac{1}{2},-\frac{1}{2},1)^{\top}$, we have

$\displaystyle\mathcal{T}x^{4}=$
$\displaystyle(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})-24x_{1}x_{2}^{2}x_{3}$
$\displaystyle\ =$ $\displaystyle
1-8\left(-\dfrac{1}{2^{4}}+\dfrac{1}{2}-\dfrac{1}{2^{3}}\right)+6\left(\dfrac{1}{2^{4}}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{2}}\right)-3=-\dfrac{9}{8}<0.$

This is a contradiction to the positive definiteness of $\mathcal{T}$, and
hence, the conditions (III) are necessary.

Sufficiency. $t_{1123}=t_{1223}=t_{1233}=1$. Rewriting $\mathcal{T}x^{4}$ as
follow,

$\mathcal{T}x^{4}=(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2}).$

Then the stationary points of the function $\mathcal{T}x^{4}$ is the solutions
of the system,

$\mathcal{T}x^{3}=\dfrac{1}{4}\nabla\mathcal{T}x^{4}=0,\mbox{ i.e.,
}\begin{cases}(x_{1}+x_{2}+x_{3})^{3}-2(3x_{1}^{2}x_{2}+x_{3}^{3})+12(x_{1}x_{2}^{2}+x_{1}x_{3}^{2})=0,\\\
(x_{1}+x_{2}+x_{3})^{3}-2(3x_{2}^{2}x_{3}+x_{1}^{3})+12(x_{1}^{2}x_{2}+x_{2}x_{3}^{2})=0,\\\
(x_{1}+x_{2}+x_{3})^{3}-2(3x_{1}x_{3}^{2}+x_{2}^{3})+12(x_{1}^{2}x_{3}+x_{2}^{2}x_{3})=0.\end{cases}$

This yields $x_{1}=x_{2}=x_{3}=0$. Therefore, this unique stationary point
$O(0,0,0)$ is is unique minimum point of $\mathcal{T}x^{4}.$ That is,
$\mathcal{T}x^{4}=0$ implies $x=0.$ So, $\mathcal{T}$ is positive definite.

$t_{1123}=t_{1223}=t_{1233}=-1$. Rewriting $\mathcal{T}x^{4}$ as follow,

$\mathcal{T}x^{4}=(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2})+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})-24x_{1}x_{2}x_{3}^{2}.$

Then the solutions of the system
$\mathcal{T}x^{3}=\dfrac{1}{4}\nabla\mathcal{T}x^{4}=0$ is stationary points
of the function $\mathcal{T}x^{4}$. That is,

$\mathcal{T}x^{3}=\dfrac{1}{4}\nabla\mathcal{T}x^{4}=\left(\begin{aligned}
(x_{1}+x_{2}-x_{3})^{3}+2(3x_{1}^{2}x_{3}-3x_{1}^{2}x_{2})+12(x_{1}x_{2}^{2}+x_{1}x_{3}^{2})-24x_{2}x_{3}^{2}\\\
(x_{1}+x_{2}-x_{3})^{3}+2(x_{3}^{3}-x_{1}^{3})+12(x_{1}^{2}x_{2}+x_{2}x_{3}^{2})-24x_{1}x_{3}^{2}\\\
-(x_{1}+x_{2}-x_{3})^{3}+2(x_{1}^{3}+3x_{2}x_{3}^{2})+12(x_{1}^{2}x_{3}+x_{2}^{2}x_{3})-48x_{1}x_{2}x_{3}\end{aligned}\right)=\left(\begin{aligned}
0\\\ 0\\\ 0\end{aligned}\right),$

and hence, $x_{1}=x_{2}=x_{3}=0,$ and so, the function $\mathcal{T}x^{4}$ has
a unique stationary point $O(0,0,0)$, which is unique minimum point. So,
$\mathcal{T}x^{4}>0$ for all $x\neq 0.$ That is, $\mathcal{T}$ is positive
definite.

(ii) Assume $t_{iijj}=2.5$ for all $i,j\in\\{1,2,3\\}$ and $i\neq j$. If the
conditions (IV) can’t hold when $\mathcal{T}$ is positive definite, then there
is only one case: one of $\\{t_{1123},t_{1223},t_{1233}\\}$ is $-1$. We might
be $t_{1123}=t_{1233}=1$ and $t_{1223}=-1$. Then for
$x=(\frac{1}{4},-\frac{1}{4},1)^{\top}$, we have

$\displaystyle\mathcal{T}x^{4}=$
$\displaystyle(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})+9(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})-24x_{1}x_{2}^{2}x_{3}$
$\displaystyle=$ $\displaystyle
1-8\left(-\dfrac{1}{4^{4}}+\dfrac{1}{4}-\dfrac{1}{4^{3}}\right)+9\left(\dfrac{1}{4^{4}}+\dfrac{1}{4^{2}}+\dfrac{1}{4^{2}}\right)-24\times\dfrac{1}{4^{3}}=-\dfrac{15}{256}<0.$

This obtains a contracdiction, and so, the conditions (IV) is necessary.

Now we show the sufficiency. The first and second conditions easily
established by the proof of (i), we only show the conclusion holds under the
condition: $\mbox{two of }\\{t_{1123},t_{1223},t_{1233}\\}\mbox{ are }-1$.
Without loss the generality, take $t_{1123}=t_{1223}=-1$ and $t_{1233}=1.$
Then we have

$\mathcal{T}x^{4}=(x_{1}+x_{2}-x_{3})^{4}+8(x_{1}^{3}x_{3}+x_{2}x_{3}^{3}-x_{1}^{3}x_{2})+9(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2}),$

and then the solutions of the system
$\mathcal{T}x^{3}=\dfrac{1}{4}\nabla\mathcal{T}x^{4}=0$ is stationary points
of the function $\mathcal{T}x^{4}$. That is,

$\mathcal{T}x^{3}=\dfrac{1}{4}\nabla\mathcal{T}x^{4}=\left(\begin{aligned}
(x_{1}+x_{2}-x_{3})^{3}+2(3x_{1}^{2}x_{3}-3x_{1}^{2}x_{2})+18(x_{1}x_{2}^{2}+x_{1}x_{3}^{2})\\\
(x_{1}+x_{2}-x_{3})^{3}+2(x_{3}^{3}-x_{1}^{3})+18(x_{1}^{2}x_{2}+x_{2}x_{3}^{2})\\\
-(x_{1}+x_{2}-x_{3})^{3}+2(x_{1}^{3}+3x_{2}x_{3}^{2})+18(x_{1}^{2}x_{3}+x_{2}^{2}x_{3})\end{aligned}\right)=\left(\begin{aligned}
0\\\ 0\\\ 0\end{aligned}\right),$

and hence, $x_{1}=x_{2}=x_{3}=0,$ and so, the function $\mathcal{T}x^{4}$ has
a unique stationary point $O(0,0,0)$, which is unique minimum point. So,
$\mathcal{T}x^{4}>0$ for all $x\neq 0.$ That is, $\mathcal{T}$ is positive
definite.

(iii) Assume $t_{iijj}\geq 3$ for all $i,j\in\\{1,2,3\\}$ and $i\neq j$. We
need only show the conclusion holds under the condition: one of
$\\{t_{1123},t_{1223},t_{1233}\\}$ is $-1$. Other three cases directly follow
from (i) and (ii). We might take $t_{1123}=t_{1223}=1$ and $t_{1233}=-1$. Then
we have

$\mathcal{T}x^{4}\geq(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})+12(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})-24x_{1}x_{2}x_{3}^{2},$

and then the solutions of the system $\dfrac{1}{4}\nabla
g(x_{1},x_{2},x_{3})=0$ is stationary points of the function $g$,

$g(x_{1},x_{2},x_{3})=(x_{1}+x_{2}+x_{3})^{4}-8(x_{1}^{3}x_{2}+x_{1}x_{3}^{3}+x_{2}^{3}x_{3})+12(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})-24x_{1}x_{2}x_{3}^{2}.$

That is,

$\begin{cases}(x_{1}+x_{2}+x_{3})^{3}-2(3x_{1}^{2}x_{2}+x_{3}^{3})+24(x_{1}x_{2}^{2}+x_{1}x_{3}^{2})-24x_{2}x_{3}^{2}=0,\\\
(x_{1}+x_{2}+x_{3})^{3}-2(3x_{2}^{2}x_{3}+x_{1}^{3})+24(x_{1}^{2}x_{2}+x_{2}x_{3}^{2})-24x_{1}x_{3}^{2}=0,\\\
(x_{1}+x_{2}+x_{3})^{3}-2(3x_{1}x_{3}^{2}+x_{2}^{3})+24(x_{1}^{2}x_{3}+x_{2}^{2}x_{3})-48x_{1}x_{2}x_{3}=0.\end{cases}$

and hence, $x_{1}=x_{2}=x_{3}=0,$ and so, the function $g(x_{1},x_{2},x_{3})$
has a unique stationary point $O(0,0,0)$, which is unique minimum point. So,
$\mathcal{T}x^{4}\geq g(x_{1},x_{2},x_{3})>0$ for all $x\neq 0.$ That is,
$\mathcal{T}$ is positive definite. This completes the proof.

By applying Theorems 3.1, the following strict inequalities are established
easily for ternary quartic homogeneous polynomials.

###### Corollary 3.2

If $(x_{1},x_{2},x_{3})\neq(0,0,0)$, then

  * (i)

$(x_{1}+x_{2}+x_{3})^{4}+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})>8(x_{1}x_{3}^{3}+x_{1}^{3}x_{2}+x_{2}^{3}x_{3});$

  * (ii)

$(x_{1}+x_{2}+x_{3})^{4}+12(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})>8(x_{1}x_{3}^{3}+x_{1}^{3}x_{2}+x_{2}^{3}x_{3})+24x_{1}x_{2}x_{3}^{2};$

  * (iii)

$(x_{1}+x_{2}-x_{3})^{4}+6(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})>8(x_{1}^{3}x_{2}-x_{1}^{3}x_{3}-x_{2}x_{3}^{3})+24x_{1}x_{2}x_{3}^{2};$

  * (iv)

$(x_{1}+x_{2}-x_{3})^{4}+9(x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{3}^{2}+x_{2}^{2}x_{3}^{2})>8(x_{1}^{3}x_{2}-x_{1}^{3}x_{3}-x_{2}x_{3}^{3}).$

Furthermore, these strict inequalities still hold if $x_{1}^{3}x_{2}$ and
$x_{1}x_{2}^{3}$ are exchangeable, or $x_{1}^{3}x_{3}$ and $x_{1}x_{3}^{3}$
are exchangeable, or $x_{2}x_{3}^{3}$ and $x_{2}^{3}x_{3}$ are exchangeable,
or $x_{1}x_{2}x_{3}^{2}$ and $x_{1}x_{2}^{2}x_{3}$ are exchangeable, or
$x_{1}x_{2}x_{3}^{2}$ and $x_{1}^{2}x_{2}x_{3}$ are exchangeable.

##  4 Conclusions

For a 4th order 3-dimensional symmetric tensor with its entries $1$ or $-1$,
the analytic necessary and sufficient conditions are established for its
positive definiteness. Several strict inequalities of ternary quartic
homogeneous polynomial are built by means of these analytic conditions.

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