Yuwang Hu^1*, Jifang Hu², Qiongru Wu^3
1School of Mathematics and Statistics, Xinyang Normal University, Xinyang, China.
²School of Sciences, Chang’an University, Xi’an, China.
³School of Mathematics and Information, Xinyang University, Xinyang, China.
DOI: 10.4236/alamt.2017.74009   PDF    HTML   XML   1,386 Downloads   2,781 Views   Citations

Abstract

All monomials with t-value ≤9 in Canonical basis of quantum group for type D₄ are determined in this paper.

Keywords

Quantum Group, Canonical Basis, Tight Monomial

Share and Cite:

1. Introduction

Quantum group, also called quantized enveloping algebra, was introduced independently by (Drinfel’d, V. G., 1985) [1] and (Jimbo, M., 1985) [2] . It plays an important role in the study of Lie groups, Lie algebras, algebraic groups, Hopf algebras, etc. The positive part of a quantum group has a kind of important basis, i.e., Canonical basis $B$ introduced by (Lusztig, G., 1990) [3] , which plays an important role in the theory of quantum groups and their representations. Some efforts on $B$ have been done. (Lusztig, G., 1990) [3] introduced algebraic definition of $B$ for the quantum groups in the simply laced cases, and gave explicitly the longest monomials in $B$ for type $A_1\mathrm{,}{A}_2$ . Afterwards, (Lusztig, G., 1992) [4] extended algebraic definition of $B$ to the non-simply laced cases and gave 2 longest monomials in $B$ for type $B_2$ . Then, (Lusztig, G., 1993) [5] associated a quadratic form to every monomial, and proved that, given certain linear conditions, the monomial is tight (respectively, semi-tight) provided that this quadratic form satisfies a certain positivity condition (respectively, nonnegativity condition). He showed that the positivity condition always holds in type $A_3$ and computed 8 longest tight monomials for type $A_3$ , and he asked when we have (semi-) tightness in type $A_n$ . Based on Lusztig’s work, (Xi, N. H., 1999 [6] ; Xi, N. H., 1999 [7] ) found explicitly all 14 elements in $B$ for type $A_3$ and all 6 elements $B$ for type $B_2$ . For type $A_4$ , (Hu, Y. W., Ye, J. C., Yue, X. Q., 2003 [8] ; Hu, Y. W., Ye, J. C., 2005 [9] ; Li, X. C., Hu, Y. W., 2012 [10] ) determined all 62 longest monomials, all 144 polynomials with one-dimensional support, 112 polynomials with two-dimensional support in $B$ . (Marsh, R., 1998) [11] carried out thorough investigation for type $A_n$ . He showed that the positivity condition is always satisfied in type $A_4$ for a certain orientation of the Dynkin diagram, presented a semi-tight longest monomial for type $A_5$ , and exhibited a special longest monomial for type $A_r$ (for any $r\ge 6$ ) with a quadratic form that does not even satisfy the conditions for semi-tightness, for any orientation of the Dynkin diagram (although it may turn out that the corresponding monomial is still tight). (Bedard, R., 2004) [12] proved that all longest monomials of type $D_4$ are semi-tight. (Reineke, M., 2001) [13] associated a new quadratic form to every monomial, and gave a sufficient and necessary condition for the monomial to be tight for the simply laced cases. (Deng, B. M., Du, J., 2010) [14] proved that the Reineke’s criterion works also for any quantized enveloping algebra associated with a symmetrizable Cartan matrix, and they gave all monomials in $B$ for type $B_2$ , in which 2 longest monomials are the same as Lusztig and Xi’s results. By use of this criterion, (Wang, X. M., 2010) [15] listed all tight monomials for type $A_3$ and $G_2$ , in which 8 longest monomials for type $A_3$ are same as Lusztig and Xi’s results. (Hu,Y. W., Li, G. W., Wang, J., 2015) [16] determined all monomials with t-value ≤ 6 in $B$ for type $A_5$ , and (Hu, Y. W., Geng, Y. J., 2015) [17] determined all monomials t- value ≤ 6 in $B$ for type $B_3$ .

This paper computed all monomials with t-value ≤ 9 in $B$ for type $D_4$ .

2. Preliminaries

Let $C={\left(c_{ij}\right)}_{i,j\in {\Gamma }_0}$ be a Cartan matrix of finite type such that $c_{ii}=2,\mathrm{<mi>\; </mi>}{c}_{ij}\le 0$ for any $i\ne j$ , $D=\text{diag}{\left(d_i\right)}_{i\in {\Gamma }_0}$ be a diagonal matrix with integer entries making the matrix DC symmetric. Let $\mathfrak{g}=\mathfrak{g}\left(C\right)$ be the complex semisimple Lie algebra associated with C, and $U=U_v\left(\mathfrak{g}\right)$ (here v is an indeterminate) be the corresponding quantized enveloping algebra, whose positive part $U^{+}=\langle E_i|i\in {\Gamma }_0\rangle$ is the $\mathbb{Q}\left(v\right)$ -subalgebra of $U$ , subject to the relations

${\displaystyle \underset{r+s=1-c_{ij}}{\sum }}{\left(-1\right)}^s{E}_i^{\left(s\right)}{E}_j{E}_i^{\left(r\right)}=0,\forall i,j\in {\Gamma }_0$ , where $E_i^{\left(s\right)}=E_i^s/{\left[s\right]}_i^{\mathrm{<mo>\; !\; </mo>}}$ , ${\left[s\right]}_i^{\mathrm{<mo>\; !\; </mo>}}={\left[1\right]}_i\cdots {\left[s\right]}_i$ ,

${\left[a\right]}_i=\left(v^{a{d}_i}-v^{-a{d}_i}\right)/\left(v^{d_i}-v^{-d_i}\right)$ . Let $\mathcal{A}=\mathbb{Z}\left[v\mathrm{,}{v}^{-1}\right]$ , $U^{+}=\langle E_i^{\left(s\right)}|i\in {\Gamma }_0,s\in \mathbb{N}\rangle$ be the $\mathcal{A}$ -subalgebra of $U^{+}$ . Corresponding to every reduced expression $i=\left(i_1,\cdots ,i_{\nu }\right)$ of the longest element $w_0=s_{i_1}\cdots s_{i_{\nu }}$ of the Weyl group $W=\langle s_i\rangle$ of $\mathfrak{g}$ , one constructs a PBW basis $B_i$ of $U^{+}$ . Lusztig [4] proved that the $\mathbb{Z}\left[v^{-1}\right]$ -submodule ${\mathcal{L}}_i=\langle B_i\rangle$ of $U^{+}$ is independent of the choice of $i$ , write it $\mathcal{L}$ ; the image of $B_i$ under the canonical projection $\pi \mathrm{<mo>\; :\; </mo>}\mathcal{L}\to \mathcal{L}\mathrm{<mo>\; /\; </mo>}{v}^{-1}\mathcal{L}$ is independent of the choice of $i$ , write it B; for any element $b\in B$ there is a unique element $b\in \mathcal{L}$ which is fixed by the bar map of $U^{+}$ defined by $v\to v^{-1}$ and satisfies $\pi \left(b\right)=b$ . The set $B=\left\{b|b\in B\right\}$ forms a $\mathbb{Z}\left[v^{-1}\right]$ -basis of $\mathcal{L}$ , an $\mathcal{A}$ -basis of $U^{+}$ and a $\mathbb{Q}\left(v\right)$ -basis of $U^{+}$ , Lusztig calls $B$ Canonical basis of quantum group.

According to Lusztig, a monomial in $U^{+}$ is an element of the form

$E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}$ (1)

where $i_1\mathrm{,}\cdots \mathrm{,}{i}_t\in {\Gamma }_0\mathrm{,}{a}_1\mathrm{,}\cdots \mathrm{,}{a}_t\in \mathbb{N}$ . When $t=\nu$ and $s_{i_1}\cdots s_{i_{\nu }}=w_0$ is the longest element of Weyl group, the monomial (1) is called the longest monomial. We say that (1) is tight (or semi-tight) if it belongs to $B$ (or is a linear combination of elements in $B$ with constant coefficients).

Let $Q=\left(Q_0\mathrm{,}{Q}_1\right)$ be a finite quiver with vertex set $Q_0$ and arrow set $Q_1$ . Write $\rho \in Q_1$ as $t_{\rho }\stackrel{\rho }{\to }{h}_{\rho }$ , where $h_{\rho }$ and $t_{\rho }$ denote the head and the tail of $\rho$ respectively. An automorphism $\sigma$ of Q is a permutation on the vertices of Q and on the arrows of Q such that $\sigma \left(h_{\rho }\right)=h_{\sigma \left(\rho \right)}$ and $\sigma \left(t_{\rho }\right)=t_{\sigma \left(\rho \right)}$ for any $\rho \in Q_1$ . Denote the quiver with automorphism $\sigma$ as $\left(Q\mathrm{,}\sigma \right)$ . Attach to the pair $\left(Q\mathrm{,}\sigma \right)$ a valued quiver $\Gamma =\Gamma \left(Q,\mathrm{<mi>\; </mi>}\sigma \right)=\left({\Gamma }_0,\mathrm{<mi>\; </mi>}{\Gamma }_1\right)$ as follows. Its vertex set ${\Gamma }_0$ and arrow set ${\Gamma }_1$ are simply the sets of s-orbits in $Q_0$ and $Q_1$ , respectively. The valuation of $\Gamma$ is given by $d_i$ = #{vertices in the s-orbit of i}, $\forall i\in {\Gamma }_0$ ; $m_{\rho }$ = #{arrows in the σ-orbit of ρ}, $\forall \rho \in {\Gamma }_1$ . The Euler form of $\Gamma$ is defined to be the bilinear form $\langle \mathrm{,}\rangle \mathrm{<mo>\; :\; </mo>\; <mi>\; </mi>}\mathbb{Z}\left[{\Gamma }_0\right]\times \mathrm{<mi>\; </mi>}\mathbb{Z}\left[{\Gamma }_0\right]\to \mathbb{Z}$ given by

$\langle X,Y\rangle ={\displaystyle \underset{i\in {\Gamma }_0}{\sum }}{d}_i{x}_i{y}_i-{\displaystyle \underset{\rho \in {\Gamma }_1}{\sum }}{m}_{\rho }{x}_{t_{\rho }}{y}_{h_{\rho }}$ , where $X={\displaystyle \underset{i\in {\Gamma }_0}{\sum }}{x}_{i}i$ , $Y={\displaystyle \underset{i\in {\Gamma }_0}{\sum }}{y}_{i}i\in \mathbb{Z}\left[{\Gamma }_0\right]$ , so

$X\cdot Y=\langle X\mathrm{,}Y\rangle +\langle Y\mathrm{,}X\rangle$ is the symmetric Euler form. The valued quiver $\Gamma$

defines a Cartan matrix $C_{\Gamma }=C_{Q,\sigma }={\left(c_{ij}\right)}_{i,j\in {\Gamma }_0}$ , where

$c_{ij}=\{\begin{array}{l}2-2{\displaystyle \underset{\begin{array}{l}\rho \in {\Gamma }_1\\ h_{\rho }=t_{\rho }=i\end{array}}{\sum }}\frac{m_{\rho }}{d_i},\mathrm{<mi>\; </mi>}i=j;\\ -{\displaystyle \underset{\begin{array}{l}\rho \in {\Gamma }_1\\ \left\{h_{\rho },\mathrm{<mi>\; </mi>}{t}_{\rho }\right\}=\left\{i,j\right\}\end{array}}{\sum }}\frac{m_{\rho }}{d_i},\mathrm{<mi>\; </mi>}i\ne j.\end{array}$

For $t\in \mathbb{N}$ , let $i=\left(i_1\mathrm{,}\cdots \mathrm{,}{i}_t\right)\in {\Gamma }_0^t$ , $a=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathbb{N}}^t$ , and write $E_i^{\left(a\right)}=E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}\in U^{+}$ . Define ${\mathcal{M}}_{i,a}=\left\{A={\left(a_{rm}\right)}_{tt}|a_{rm}\in \mathbb{N},\mathrm{<mi>\; </mi>}\text{ro}\left(A\right)=\text{co}\left(A\right)=a,\mathrm{<mi>\; </mi>}{a}_{rm}=0,\forall i_r\ne i_m\right\}$ , where

$\text{ro}\left(A\right)=\left({\displaystyle \underset{m=1}{\overset{t}{\sum }}}{a}_{1m},\cdots ,{\displaystyle \underset{m=1}{\overset{t}{\sum }}}{a}_{tm}\right)$ , $\text{co}\left(A\right)=\left({\displaystyle \underset{r=1}{\overset{t}{\sum }}}{a}_{r1},\cdots ,{\displaystyle \underset{r=1}{\overset{t}{\sum }}}{a}_{rt}\right)$ . Obviously, $D_a=\text{diag}\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathcal{M}}_{i,a}$ .

Lusztig gave the following criterion for a monomial to be tight or semi-tight.

Theorem 2.1 ([Lusztig, 1993, §6 Theorem) [5] . Let $U$ be the quantum group of type $A_n\mathrm{,}{D}_n\mathrm{,}{E}_n$ , $i\in {\Gamma }_0^t\mathrm{,\; <mi>\; </mi>}a\in {\mathbb{N}}^t$ as above. If the following quadratic form takes only values < 0 on ${\mathcal{M}}_{i\mathrm{,}a}\backslash \left\{D_a\right\}$ (respectively, ≤ 0 on ${\mathcal{M}}_{i\mathrm{,}a}$ ), then monomial $E_i^{\left(a\right)}$ is tight (respectively, semi-tight).

${q^{\prime }}_{i\mathrm{,}a}\left(A\right)={\displaystyle \underset{\begin{array}{c}1\le m\le t\\ 1\le p<r\le t\end{array}}{\sum }}{a}_{pm}{a}_{rm}-{\displaystyle \underset{\begin{array}{c}1\le p<r\le t\\ 1\le l<m\le t\end{array}}{\sum }}{a}_{pm}{a}_{rl}$

It should be noticed that the above theorem is sufficient but not necessary, M. Reineke gave a sufficient and necessary condition by symmetrizing Lusztig’s quadratic form.

Theorem 2.2 ([Reineke, 2001, Theorem 3.2]) [13] . Let $U$ be the quantum group of type $A_n\mathrm{,}{D}_n\mathrm{,}{E}_n$ , $i\in {\Gamma }_0^t\mathrm{,\; <mi>\; </mi>}a\in {\mathbb{N}}^t$ as above, the monomial $E_i^{\left(a\right)}$ is tight if and only if the following quadratic form takes only values <0 on ${\mathcal{M}}_{i\mathrm{,}a}\backslash \left\{D_a\right\}$

$q_{i\mathrm{,}a}\left(A\right)={\displaystyle \underset{\begin{array}{c}1\le m\le t\\ 1\le p<r\le t\end{array}}{\sum }}{a}_{pm}{a}_{rm}+{\displaystyle \underset{\begin{array}{c}1\le p<r\le t\\ 1\le l<m\le t\end{array}}{\sum }}\left(i_l\cdot i_m\right)a_{pm}{a}_{rl}+{\displaystyle \underset{\begin{array}{c}1\le r\le t\\ 1\le l<m\le t\end{array}}{\sum }}{a}_{rm}{a}_{rl}$

In fact, $q_{i\mathrm{,}a}\left(A\right)={q^{\prime }}_{i\mathrm{,}a}\left(A\right)+{q^{\prime }}_{i\mathrm{,}a}\left(A^T\right)$ (see [Reineke, 2001 [13] , Lemma 3.3]).

Deng and Du generalized the tight monomial criterion given by Reineke to any quantum group associated with symmetrizable matrices.

Theorem 2.3 ([Deng, Du, 2010, Theorem 2.5]) [14] . Let $U$ be the quantum group associated with any symmetrizable matrices, $i\in {\Gamma }_0^t\mathrm{,\; <mi>\; </mi>}a\in {\mathbb{N}}^t$ as above, the monomial $E_i^{\left(a\right)}$ is tight if and only if the following quadratic form takes only values < 0 on ${\mathcal{M}}_{i\mathrm{,}a}\backslash \left\{D_a\right\}$

$q_{i\mathrm{,}a}\left(A\right)={\displaystyle \underset{\begin{array}{c}1\le m\le t\\ 1\le p<r\le t\end{array}}{\sum }}\langle i_m,i_m\rangle a_{pm}{a}_{rm}+{\displaystyle \underset{\begin{array}{c}1\le p<r\le t\\ 1\le l<m\le t\end{array}}{\sum }}\left(i_l\cdot i_m\right)a_{pm}{a}_{rl}+{\displaystyle \underset{\begin{array}{c}1\le r\le t\\ 1\le l<m\le t\end{array}}{\sum }}\langle i_r,i_r\rangle a_{rm}{a}_{rl}$

By Theorem 2.3, we have the following Corollaries.

Corollary 2.4. When $i_1\mathrm{,}\cdots \mathrm{,}{i}_t$ are mutually different, monomial $E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}$ is tight.

Proof: In fact, $i_1\mathrm{,}\cdots \mathrm{,}{i}_t$ are mutually different, so ${\mathcal{M}}_{i\mathrm{,}a}=\left\{D_a\right\}$ .

Corollary 2.5. If $E_{i_{p+1}}^{\left(a_{p+1}\right)}\cdots E_{i_{p+q}}^{\left(a_{p+q}\right)}$ is tight, then for any mutually different $i_1\mathrm{,}\cdots \mathrm{,}{i}_p\notin \left\{i_{p+1}\mathrm{,}\cdots \mathrm{,}{i}_{p+q}\right\}$ and any mutually different $i_{p+q+1}\mathrm{,}\cdots \mathrm{,}{i}_t\notin \left\{i_1\mathrm{,}\cdots \mathrm{,}{i}_p\mathrm{<mo>\; ;\; </mo>}{i}_{p+1}\mathrm{,}\cdots \mathrm{,}{i}_{p+q}\right\}$ , $t\le l\left(w_0\right)$ , $E_i^{\left(a\right)}=E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}$ is also tight.

Proof: Write $j=\left(i_{p+1}\mathrm{,}\cdots \mathrm{,}{i}_{p+q}\right)$ , $b=\left(a_{p+1}\mathrm{,}\cdots \mathrm{,}{a}_{p+q}\right)$ , $i=\left(i_1\mathrm{,}\cdots \mathrm{,}{i}_t\right)$ , $a=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)$ , then

$E_i^{\left(a\right)}=E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}\mathrm{,}{E}_j^{\left(b\right)}=E_{i_{p+1}}^{\left(a_{p+1}\right)}\cdots E_{i_{p+q}}^{\left(a_{p+q}\right)}\mathrm{.}$

For any $\tilde{A}\in {\mathcal{M}}_{i\mathrm{,}a}$ , we have

$\tilde{A}=\left(\begin{array}{ccc}{D}_{a^{\prime }}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}A& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{D}_{a^{″}}\end{array}\right)\mathrm{,}$

where $a^{\prime }=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_p\right)$ , $a^{″}=\left(a_{p+q+1}\mathrm{,}\cdots \mathrm{,}{a}_t\right)$ . It is easy to see that $A\in {\mathcal{M}}_{j\mathrm{,}b}$ and $q\left(\tilde{A}\right)=q\left(A\right)$ . Moreover, $\tilde{A}=D_a$ if and only if $A=D_b$ . Since $E_j^{\left(b\right)}$ is tight, we get by Theorem 2.3 that $q\left(\tilde{A}\right)=q\left(A\right)<0$ for all $\tilde{A}\in {\mathcal{M}}_{i\mathrm{,}a}\backslash \left\{D_a\right\}$ , applying Theorem 2.3 again, we conclude that $E_i^{\left(a\right)}$ is tight.

The following two theorems are very useful in determining tight monomials.

Theorem 2.6 ([Deng & Du, 2010, Corollary 2.6, Theorem 6.2) [14] . Let $i=\left(i_1\mathrm{,}\cdots \mathrm{,}{i}_t\right)\in {\Gamma }_0^t$ and $a=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathbb{N}}^t$ . If $E_i^{\left(a\right)}$ is tight, then

(a) For $\forall \mathrm{<mi>\; </mi>1}\le r\le s\le t$ , monomial $E_{i_r}^{\left(a_r\right)}\cdots E_{i_s}^{\left(a_s\right)}$ is also tight;

(b) For $\forall \mathrm{<mi>\; </mi>1}\le r<t$ , $i_r\ne i_{r+1}$ .

Theorem 2.7 ([Lusztig, 1990, Proposition 3.3 and Lusztig, 1993, §13]) [3] . Let $\Phi$ be the non-trivial automorphism of $U^{+}$ induced by Dynkin diagram automorphism $\phi$ of $\mathfrak{g}$ , and $\Psi \mathrm{<mo>\; :\; </mo>}{U}^{+}\to {\left(U^{+}\right)}^{\text{o}pp}$ be the unique $\mathbb{Q}\left(v\right)$ -algebra isomorphism such that $E_j\to E_j$ . If $E_i^{\left(a\right)}$ is tight, then $\Phi \left(E_i^{\left(a\right)}\right)$ and $\Psi \left(E_i^{\left(a\right)}\right)$ are all tight.

A quadratic form $f\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}^n\to \mathbb{Z}$ denoted by

$f\left(x_1,\cdots ,x_n\right)={\displaystyle \underset{i=1}{\overset{n}{\sum }}}{x}_i^2+{\displaystyle \underset{i<j}{\sum }}{a}_{ij}{x}_i{x}_j$

with $a_{ij}\in \mathbb{Z}$ is called a unit form.

The symmetric matrix $A_f=\left(a_{ij}\right)$ (when $i>j$ , set $a_{ij}=a_{ji}$ ) with $a_{ii}=2$ , defines a bilinear form $f\left(X\mathrm{,}Y\right)=X{A}_f{Y}^{\text{T}}$ , where

$X=\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_n\right)\text{and}Y=\left(y_1\mathrm{,}\cdots \mathrm{,}{y}_n\right)\mathrm{.}$

In particular, we have

$f\left(X\right)=\frac{1}{2}f\left(X\mathrm{,}X\right)\text{and}f\left(X\mathrm{,}Y\right)=f\left(X+Y\right)-f\left(X\right)-f\left(Y\right)\mathrm{.}$

For a vector $w\in {\mathbb{Q}}^n$ with non-negative coordinates, we write $w\ge 0$ . The vector in ${\mathbb{Z}}^n$ which has a 1 in the i^th coordinate ( $1\le i\le n$ ) and 0’s elsewhere is denoted by $r_i$ .

Let $f$ be a unit form. We define the set of positive roots of $f$ as $\left\{w\in {\mathbb{Z}}^n\mathrm{<mo>\; :\; </mo>0}\le w\text{and}f\left(w\right)=1\right\}$ . The linear transformation ${\sigma }_i\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}^n\to {\mathbb{Z}}^n$ defined by ${\sigma }_i\left(w\right)=w-f\left(w\mathrm{,}{r}_i\right)r_i$ is called the reflection with respect to $r_i$ . The transformation ${\sigma }_i$ has the property that ${\sigma }_i^2=\text{id}$ and $f\left({\sigma }_i\left(w\right)\right)=f\left(w\right)$ for every $w\in {\mathbb{Z}}^n$ .

Let $f$ be a quadratic form, if $0<f\left(X\right)$ for every $0\ne X\in {\mathbb{N}}^n$ , then we call $f$ weakly positive.

The following algorithm and theorem are taken from (Blouin, Dean, Denver, Pershall, 1995) [18] :

Let $f\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}^n\to \mathbb{Z}\left(n\ge 3\right)$ be a unit form. First of all, we define

$R_1=\left\{r_i\mathrm{<mo>\; :\; </mo>1}\le i\le n\right\}\mathrm{.}$

Next we want to construct $R_j$ recursively. Assume that we have defined a set of positive roots of $f$ as

$R_s=\left\{z_1\mathrm{,}\cdots \mathrm{,}{z}_m\right\}\text{with}s\ge \mathrm{1,}$

and that the process has not failed (to be defined subsequently). Then we construct $R_{s+1}$ as follows. Let $z_j\in R_s$ . If either:

1) there is some $1\le i\le n$ such that $f\left(z_j\mathrm{,}{r}_i\right)\le -2$ , or

2) there is some $1\le i\le n$ such that $z_j\left(i\right)\ge 7$ ,

then the process is said to fail. Assume the process does not fail (so 1) and 2) do not occur for any $z_j\in R_s$ ). Let $D_s\subseteq R_s$ be the set of those roots $z_j$ with the property that there is some $1\le i\le n$ such that $f\left(z_j,r_i\right)=-1$ . If $D_s=\varphi$ , then $R_{s+1}:=\varphi$ and the process is said to be successful. If $D_s\ne \varphi$ , then

$R_{s+1}\mathrm{<mo>\; :\; </mo>}=\left\{{\sigma }_i\left(z_j\right)|z_j\in D_s\text{and}{r}_i\text{is}\text{such}\text{that}f\left(z_j\mathrm{,}{r}_i\right)=-1\right\}\mathrm{.}$

Remark: If $R_{s+1}\ne \varphi$ , then we apply the algorithm again to obtain $R_{s+2}$ , etc. The roots in $R_{s+1}$ (if $R_{s+1}\ne \varphi$ ) are all greater than the roots in $R_s$ . Thus condition 2) guarantees that this procedure is finite and if it is not successful then it will eventually fail.

Theorem 2.8 ([Blouin et al., 1995, Theorem 4]) [18] . The unit form $f$ is weakly positive if and only if the above process is successful. If the process is successful with $R_{s+1}=\varphi$ , then $R_1\cup \cdots \cup R_s$ are all the positive roots of $f$ .

3. Main Results

Let $\mathfrak{g}$ be of type $D_4$ as follows.

Let $i=\left(i_1\mathrm{,}\cdots \mathrm{,}{i}_t\right)\in {\Gamma }_0^t$ , $a=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathbb{N}}^t$ . For convenience, we abbreviate a monomial $E_{i_1}^{\left(a_1\right)}\cdots E_{i_t}^{\left(a_t\right)}$ with any $\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathbb{N}}^t$ as a word $i_1\cdots i_t$ (1 as 0), an inequality $a_{j_1}+\cdots +a_{j_p}\le a_{l_1}+\cdots +a_{l_q}$ as $j_1\cdots j_p-l_1\cdots l_q$ . For example, a monomial $E_1^{\left(a_1\right)}{E}_2^{\left(a_2\right)}{E}_3^{\left(a_3\right)}{E}_4^{\left(a_4\right)}$ is abbreviated to a word 1234, and a monomial $E_1^{\left(a_1\right)}{E}_2^{\left(a_2\right)}{E}_1^{\left(a_3\right)}\left(a_1+a_3\le a_2\right)$ to a word 121 (13 − 2), etc.

If $s_{i_1}\cdots s_{i_t}$ is a reduced expression, we call $i_1\cdots i_t$ a reduced word. By Theorem 2.6 (b), we only consider those reduced tight words $i_1\cdots i_t$ with $i_r\ne i_{r+1},\forall \mathrm{<mi>\; </mi>1}\le r<t$ , in this case, $i_1\cdots i_t$ is called the word with t-value. If $i_r\cdot i_{r+1}=0$ for some $1\le r<t$ , we identify the word $i_1\cdots i_{r-1}{i}_r{i}_{r+1}{i}_{r+2}\cdots i_t$ with the word $i_1\cdots i_{r-1}{i}_{r+1}{i}_r{i}_{r+2}\cdots i_t$ . Denote the set of all words with t-value by $M_t$ . The non-trivial Dynkin diagram automorphism $\phi$ of $\mathfrak{g}$ is $\phi \mathrm{<mo>\; :\; </mo>1}\mapsto \mathrm{3,3}\mapsto \mathrm{4,4}\mapsto \mathrm{1,2}\mapsto 2$ . Let us present the so called M − S word-procedure from t-value to $\left(t+1\right)$ -value.

Step 1. Take any $i_1\cdots i_t\in M_t$ , adding a number $i_{t+1}\in \left\{\mathrm{1,2,3,4}\right\}$ different from $i_1$ (or $i_t$ ) in the front (or behind) of $i_1$ (or $i_t$ ), deleting the those words with t-value, getting all words with $\left(t+1\right)$ -value from $i_1\cdots i_t$ .

Step 2. Repeat step 1 until all words in $M_t$ are considered, deleting the non-tight words with $\left(t+1\right)$ -value, get $M_{t+1}$ .

Step 3. Use $\Phi$ and $\Psi$ , we have $S_{t+1}$ satisfying $M_{t+1}=\Phi \left(S_{t+1}\right)\cup \Psi \Phi \left(S_{t+1}\right)$ .

For example, applying the $M-S$ word-procedure to $M_1=\left\{\mathrm{1,2,3,4}\right\}$ , get $M_2=\left\{12,13,14,21,23,24,32,34,42\right\}$ . Considering $\Phi \mathrm{,}\Psi$ , $24\stackrel{\phi }{\leftarrow }23\stackrel{\phi }{\leftarrow }21\stackrel{\Psi }{\leftarrow }12\stackrel{\phi }{\to }32\stackrel{\phi }{\to }42$ and $13\stackrel{\phi }{\to }34\stackrel{\phi }{\to }14$ , so $S_2=\left\{12,13\right\}$ .

From now on, write $M_t=\Phi \left(S_t\right)\cup \Psi \Phi \left(S_t\right)\mathrm{,}t\ge 2$ .

Theorem 3.1. For the quantum group for type $D_4$ , we have the following results.

1) $t=0$ , $M_0=\left\{0\right\}$ , tight monomial has only one.

2) $t=1$ , $M_1=\Phi \left(S_1\right)$ , tight monomials have 4 families, where $S_1=\left\{1,2\right\}$ .

3) $t=2$ , $M_2$ includes 9 families of tight monomials, where $S_2=\left\{12,13\right\}$ .

4) $t=3$ , $M_3$ includes 19 families of tight monomials, where $S_3=S_3^1\cup S_3^2$ ,

$S_3^1=\left\{\mathrm{123,132,134}\right\}$ ; $S_3^2=\left\{\mathrm{121,212}\left(13-2\right)\right\}$ .

5) $t=4$ , $M_4$ includes 35 families of tight monomials, where $S_4=S_4^1\cup S_4^2\cup S_4^3$ ,

$S_4^1=\left\{1234,1342\right\}$ ; $S_4^2=\left\{1213,1214,2123,2124\left(13-2\right)\right\}$ ; $S_4^3=\left\{2132\left(14-23\right)\right\}$ .

6) $t=5$ , M₅ includes 58 families of tight monomials, where $S_5=S_5^1\cup S_5^2\cup S_5^3$ ,

$\begin{array}{l}{S}_5^1=\{\mathrm{12134,24213}\left(13-2\right)\mathrm{<mo>\; ;\; </mo>31214,12324}\left(24-3\right)\mathrm{<mo>\; ;\; </mo>}\\ 12342\left(25-34\right)\mathrm{<mo>\; ;\; </mo>21342}\left(15-234\right)\};\end{array}$

$\begin{array}{l}{S}_5^2=\{\mathrm{12312,12412}\left(14-\mathrm{2,25}-34\right)\mathrm{<mo>\; ;\; </mo>13213}\left(14-\mathrm{3,25}-3\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12321,12421}\left(15-24-3\right)\};\end{array}$

$S_5^3=\left\{21232\left(13-\mathrm{2,35}-4\right)\right\}.$

7) $t=6$ , $M_6$ includes 93 families of tight monomials, where $S_6=S_6^1\cup S_6^2\cup S_6^3\cup S_6^4$ ,

$\begin{array}{l}{S}_6^1=\{\mathrm{123124,124123}\left(14-\mathrm{2,25}-34\right)\mathrm{<mo>\; ;\; </mo>123214,124213}\left(15-24-3\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{132142,312342}\left(14-\mathrm{3,36}-45\right)\mathrm{<mo>\; ;\; </mo>132134}\left(14-\mathrm{3,25}-3\right)\mathrm{<mo>\; ;\; </mo>}\\ 123421\left(16-25-34\right)\mathrm{<mo>\; ;\; </mo>123142}\left(14-\mathrm{2,26}-345\right)\};\end{array}$

$S_6^2=\left\{132132\left(14-\mathrm{3,25}-\mathrm{3,36}-45\right)\right\};$

$S_6^3=\left\{212342\left(13-\mathrm{2,36}-45\right)\mathrm{<mo>\; ;\; </mo>123242,124232}\left(24-\mathrm{3,46}-5\right)\right\};$

$\begin{array}{l}{S}_6^4=\{\mathrm{123121,124121}\left(14-\mathrm{2,46}-\mathrm{5,25}-34\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{123212,124212}\left(46-\mathrm{5,15}-24-3\right)\}.\end{array}$

8) $t=7$ , $M_7$ includes 133 families of tight monomials, where $S_7=S_7^1\cup S_7^2\cup S_7^3\cup S_7^4$ ,

$\begin{array}{l}{S}_7^1=\{\mathrm{1234213,1243214}\left(16-25-\mathrm{34,37}-5\right)\mathrm{<mo>\; ;\; </mo>}\\ 1321342\left(14-\mathrm{3,25}-\mathrm{3,37}-456\right)\mathrm{<mo>\; ;\; </mo>1342134}\left(15-\mathrm{4,26}-\mathrm{4,37}-4\right)\mathrm{<mo>\; ;\; </mo>}\\ 1234123\left(15-\mathrm{2,26}-\mathrm{345,37}-6\right)\mathrm{<mo>\; ;\; </mo>1324213}\left(16-35-\mathrm{4,27}-35\right)\mathrm{<mo>\; ;\; </mo>}\\ 1234234\left(36-\mathrm{5,47}-\mathrm{5,25}-34\right)\mathrm{<mo>\; ;\; </mo>1342132}\left(15-\mathrm{4,26}-\mathrm{4,47}-56\right)\};\end{array}$

$\begin{array}{l}{S}_7^2=\{\mathrm{1231242,1241232}\left(14-\mathrm{2,57}-\mathrm{6,25}-34\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{1231241,1241231}\left(25-\mathrm{34,14}-\mathrm{2,47}-5\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{1324212,3124232}\left(57-\mathrm{6,16}-35-4\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{1232142,1242132}\left(15-24-\mathrm{3,47}-56\right)\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}\mathrm{1232124,1242123}\left(46-\mathrm{5,15}-24-3\right)\mathrm{<mo>\; ;\; </mo>}\\ 1232421\left(17-\mathrm{246,24}-\mathrm{3,46}-5\right)\mathrm{<mo>\; ;\; </mo>1234212}\left(57-\mathrm{6,16}-25-34\right)\mathrm{<mo>\; ;\; </mo>}\\ 1231421\left(14-\mathrm{2,47}-\mathrm{6,26}-345\right)\mathrm{<mo>\; ;\; </mo>}2132142\left(25-\mathrm{4,14}-\mathrm{23,47}-56\right)\};\end{array}$

$S_7^3=\left\{2123242\left(13-\mathrm{2,35}-\mathrm{4,57}-6\right)\right\}\mathrm{<mo>\; ;\; </mo>}$

$S_7^4=\left\{\mathrm{2123212,2124212}\left(13-\mathrm{2,26}-35-\mathrm{4,57}-6\right)\right\}\mathrm{.}$

9) $t=8$ , $M_8$ includes 185 families of tight monomials, where $S_8=S_8^1\cup S_8^2\cup S_8^3\cup S_8^4$ ,

$\begin{array}{l}{S}_8^1=\{12134234\left(13-2,26-345,47-6,58-6\right);\\ 12342134\left(16-25-34,48-5,37-5\right);\\ 13421342\left(15-\mathrm{4,48}-\mathrm{567,26}-\mathrm{4,37}-4\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_8^2=\{\mathrm{12132142,12142132}\left(13-\mathrm{2,36}-\mathrm{5,25}-\mathrm{34,58}-67\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12142321,12132421}\left(13-\mathrm{2,38}-57-\mathrm{6,25}-34\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_8^3=\{\mathrm{12132423,12142324}\left(13-\mathrm{2,25}-\mathrm{34,48}-57-6\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12134213,12143214}\left(13-\mathrm{2,37}-\mathrm{6,26}-\mathrm{345,48}-6\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12321423,32123421}\left(15-24-\mathrm{3,38}-47-56\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12324213,12423214}\left(17-\mathrm{246,24}-\mathrm{3,38}-46-5\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12342132,12432142}\left(16-25-\mathrm{34,58}-\mathrm{67,37}-5\right)\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}\mathrm{13214232,31234212}\left(14-\mathrm{3,27}-36-\mathrm{45,68}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{13214213,31234231}\left(14-\mathrm{3,47}-\mathrm{6,28}-36-45\right)\mathrm{<mo>\; ;\; </mo>}\\ 13213242\left(14-\mathrm{3,36}-\mathrm{45,68}-\mathrm{7,25}-3\right)\mathrm{<mo>\; ;\; </mo>}\\ 13242132\left(16-35-\mathrm{4,58}-\mathrm{67,27}-35\right)\mathrm{<mo>\; ;\; </mo>}\\ 21321342\left(25-\mathrm{4,14}-\mathrm{23,36}-\mathrm{4,48}-567\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_8^4=\{\mathrm{12321242,12421232}\left(15-24-\mathrm{3,46}-\mathrm{5,68}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12324212,12423212}\left(17-\mathrm{246,24}-\mathrm{3,46}-\mathrm{5,68}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{12324232,12423242}\left(24-\mathrm{3,68}-\mathrm{7,37}-46-5\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{21232142,21242132}\left(13-\mathrm{2,58}-\mathrm{67,26}-35-4\right)\mathrm{<mo>\; ;\; </mo>}\\ 21234212\left(13-\mathrm{2,68}-\mathrm{7,27}-36-45\right)\}\mathrm{.}\end{array}$

10) $t=9$ , $M_9$ includes 265 families of tight monomials, where $S_9=S_9^1\cup S_9^2\cup S_9^3\cup S_9^4\cup S_9^5$ ,

$\begin{array}{l}{S}_9^1=\{\mathrm{121321423,121421324}\left(13-\mathrm{2,36}-\mathrm{5,25}-\mathrm{34,49}-58-67\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{121342132,121432142}\left(13-\mathrm{2,37}-\mathrm{6,26}-\mathrm{345,48}-\mathrm{6,69}-78\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{121324213,121423214}\left(13-\mathrm{2,38}-57-\mathrm{6,49}-\mathrm{57,25}-34\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{132142132,312342312}\left(14-\mathrm{3,47}-\mathrm{6,69}-\mathrm{78,28}-36-45\right)\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}\mathrm{124213214,123214213}\left(15-24-\mathrm{3,39}-47-\mathrm{56,58}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{124213213,123214214}\left(15-24-\mathrm{3,47}-\mathrm{56,58}-\mathrm{7,69}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ 132134213\left(14-\mathrm{3,25}-\mathrm{3,48}-\mathrm{7,59}-\mathrm{7,37}-456\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_9^2=\{\mathrm{123214234,124213243}\left(15-24-\mathrm{3,69}-\mathrm{7,38}-47-56\right)\mathrm{<mo>\; ;\; </mo>}\\ 121342134\left(13-\mathrm{2,37}-\mathrm{6,48}-\mathrm{6,59}-\mathrm{6,26}-345\right)\mathrm{<mo>\; ;\; </mo>}\\ 123421234\left(16-25-\mathrm{34,38}-57-\mathrm{6,49}-57\right)\mathrm{<mo>\; ;\; </mo>}\\ 123421324\left(16-25-\mathrm{34,37}-\mathrm{5,49}-58-67\right)\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}123421342\left(16-25-\mathrm{34,59}-\mathrm{678,37}-\mathrm{5,48}-5\right)\mathrm{<mo>\; ;\; </mo>}\\ 132134214\left(14-\mathrm{3,48}-\mathrm{7,37}-\mathrm{456,25}-\mathrm{3,69}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ 134213242\left(15-\mathrm{4,79}-\mathrm{8,26}-\mathrm{4,38}-47-56\right)\mathrm{<mo>\; ;\; </mo>}\\ 213421342\left(26-\mathrm{5,15}-\mathrm{234,59}-\mathrm{678,37}-\mathrm{5,48}-5\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_9^3=\{\mathrm{121324232,121423242}\left(13-\mathrm{2,25}-\mathrm{34,48}-57-\mathrm{6,79}-8\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{123212423,124212324}\left(15-24-\mathrm{3,46}-\mathrm{5,68}-\mathrm{7,39}-468\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{123214232,124213242}\left(15-24-\mathrm{3,38}-47-\mathrm{56,79}-8\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{123242132,124232142}\left(17-\mathrm{246,24}-\mathrm{3,69}-\mathrm{78,38}-46-5\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{123242321,124232421}\left(19-\mathrm{2468,24}-\mathrm{3,68}-\mathrm{7,37}-46-5\right)\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}\mathrm{123421232,123421242}\left(16-25-\mathrm{34,38}-57-\mathrm{6,79}-8\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{132421232,312423212}\left(16-35-\mathrm{4,57}-\mathrm{6,79}-\mathrm{8,28}-357\right)\mathrm{<mo>\; ;\; </mo>}\\ 123242123\left(17-\mathrm{246,24}-\mathrm{3,46}-\mathrm{5,39}-\mathrm{468,68}-7\right)\mathrm{<mo>\; ;\; </mo>}\\ \mathrm{212342132,212432142}\left(13-\mathrm{2,27}-36-\mathrm{45,48}-\mathrm{6,69}-78\right)\mathrm{<mo>\; ;\; </mo>}\\ 213242132\left(27-46-\mathrm{5,14}-\mathrm{23,69}-\mathrm{78,38}-46\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_9^4=\{\mathrm{121324212,121423212}\left(13-\mathrm{2,25}-\mathrm{34,38}-57-\mathrm{6,79}-8\right)\mathrm{<mo>\; ;\; </mo>}\\ 121321421\left(13-\mathrm{2,36}-\mathrm{5,69}-\mathrm{8,25}-\mathrm{34,58}-67\right)\mathrm{<mo>\; ;\; </mo>}\\ 123212421\left(15-24-\mathrm{3,59}-68-\mathrm{7,46}-5\right)\}\mathrm{<mo>\; ;\; </mo>}\end{array}$

$\begin{array}{l}{S}_9^5=\{\mathrm{212321242,212421232}\left(13-\mathrm{2,26}-35-\mathrm{4,57}-\mathrm{6,79}-8\right)\mathrm{<mo>\; ;\; </mo>}\\ 212324212\left(13-\mathrm{2,28}-\mathrm{357,35}-\mathrm{4,57}-\mathrm{6,79}-8\right)\}\mathrm{.}\end{array}$

4. Proof of Theorem 3.1

Consider the following quiver $Q=\left(Q_0,Q_1\right)$ of type $D_4$ where $Q_1=\left\{1\stackrel{{\rho }_1}{\to }\mathrm{2,\; <mi>\; </mi>3}\stackrel{{\rho }_2}{\to }\mathrm{2,\; <mi>\; </mi>4}\stackrel{{\rho }_3}{\to }2\right\}$ , $Q_0=\left\{1,2,3,4\right\}$ . Let $\sigma$ be the identity automorphism of Q such that $\sigma \left(i\right)=i,\sigma \left({\rho }_i\right)={\rho }_i,i=1,2,3,4$ , then the valued quiver of $\left(Q\mathrm{,}\sigma \right)$ is $\Gamma =\Gamma \left(Q,\sigma \right)=\left({\Gamma }_0,{\Gamma }_1\right)=\left(Q_0,Q_1\right)$ , the valuation is given by $d_i=1,m_{{\rho }_i}=1,i=1,2,3,4$ . For $X=x_{1}1+x_{2}2+x_{3}3+x_{4}4$ , $Y=y_{1}1+y_{2}2+y_{3}3+y_{4}4\in \mathbb{Z}\left[{\Gamma }_0\right]$ , Euler form $\langle ,\rangle$ on $\Gamma$ is

${\langle X,\mathrm{<mi>\; </mi>}Y\rangle }_{\Gamma }={\displaystyle \underset{i=1}{\overset{4}{\sum }}}{d}_i{x}_i{y}_i-{\displaystyle \underset{i=1}{\overset{3}{\sum }}}{m}_{{\rho }_i}{x}_{t_{{\rho }_i}}{y}_{h_{{\rho }_i}}=x_1{y}_1+x_2{y}_2+x_3{y}_3+x_4{y}_4-x_1{y}_2-x_3{y}_2-x_4{y}_2$ ,

symmetric Euler form $\cdot$ on $\Gamma$ is

$\begin{array}{l}X\cdot Y={\langle X,Y\rangle }_{\Gamma }+{\langle Y,\mathrm{<mi>\; </mi>}X\rangle }_{\Gamma }=2x_1{y}_1+2x_2{y}_2+2x_3{y}_3\\ +2x_4{y}_4-x_1{y}_2-x_2{y}_1-x_3{y}_2-x_2{y}_3-x_4{y}_2-x_2{y}_4\end{array}$ .

By simple computation, we have $\langle i,i\rangle =1,i=1,2,3,4$ ; $\langle 1,2\rangle =\langle 3,2\rangle =$ $\langle 4,2\rangle =-1$ , the other $\langle i,j\rangle =0,i\ne j$ . So $i\cdot i=2,i=1,2,3,4$ ; $1\cdot 2=2\cdot 3=2\cdot 4=$ $-1$ , $1\cdot 3=1\cdot 4=3\cdot 4=0$ .

Let us prove Theorem 3.1. For any $t\in \mathbb{N}$ , let $i=\left(i_1\mathrm{,}\cdots \mathrm{,}{i}_t\right)\in {\Gamma }_0^t$ , $a=\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_t\right)\in {\mathbb{N}}^t$ .

Case 1. $t\le 2$ . By Corollary 2.4, words with $t\le 2$ are all tight, so (1)~(3) hold.

Case 2. $t=3$ . Applying the $M-S$ word-procedure to $M_2$ , we get $S_3$ . Corollary 2.4 $\Rightarrow S_3^1$ . Consider $S_3^2$ , for $i\in \left\{\left(\mathrm{1,2,1}\right)\mathrm{,}\left(\mathrm{2,1,2}\right)\right\}$ , we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_x|x\in \mathbb{N}\right\}$ , where

$M=\left(\begin{array}{ccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x\\ 0& \mathrm{<mi>\; </mi>}{a}_2& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3-x\end{array}\right),$

and

$\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x-x^2\right)+2\langle i_3,i_3\rangle \left(a_{3}x-x^2\right)+\left(i_1\cdot i_3\right)x^2+\left(\left(i_1\cdot i_2\right)+\left(i_2\cdot i_3\right)\right)a_{2}x\\ =-2x^2+2\left(a_1+a_3-a_2\right)x.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $x=0$ , so two words in $S_3^2$ are all tight by Theorem 2.3, (4) holds.

Case 3. $t=4$ . Applying the $M-S$ word-procedure to $M_3$ , we get $S_4\cup \left\{1212\right\}$ . Corollary 2.4 $\Rightarrow S_4^1$ . Corollary 2.5 and $S_3^2\Rightarrow S_4^2$ . Consider words 2132, 1212. For word 2132, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_x|x\in \mathbb{N}\right\}$ , where

$M=\left(\begin{array}{cccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x\\ 0& \mathrm{<mi>\; </mi>}{a}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4-x\end{array}\right),$

and $\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{4}x-2x^2\right)+\left(i_1\cdot i_1\right)x^2+2\left(i_1\cdot i_2\right)a_{2}x+2\left(i_1\cdot i_3\right)a_{3}x\\ =-2x^2+2\left(a_1+a_4-a_2-a_3\right)x.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_4\le a_2+a_3$ . And $q\left(M\right)=0\iff a_1+a_4\le a_2+a_3$ , $x=0$ , so word 2132(14-23) is tight by Theorem 2.3.

For word 1212, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,y}|\left(x,y\right)\in {\mathbb{N}}^2\right\}$ , where

$M=\left(\begin{array}{cccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3-x& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4-y\end{array}\right),$

and $\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{3}x-2x^2\right)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{4}y-2y^2\right)\\ +\left(i_1\cdot i_1\right)x^2+\left(i_2\cdot i_2\right)y^2+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_{3}y-xy\right)\\ =-x^2-y^2-{\left(x-y\right)}^2+2\left(a_1+a_3-a_2\right)x+2\left(a_2+a_4-a_3\right)y.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_3\le a_2\mathrm{,}{a}_2+a_4\le a_3$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_2+a_4\le a_3$ , $x=y=0$ , but $a_1+a_3\le a_2$ , $a_2+a_4\le a_3\Rightarrow a_1+a_4\le 0$ , this is a contradiction, so word 1212 is not tight for any $a=\left(a_1\mathrm{,}{a}_2\mathrm{,}{a}_3\mathrm{,}{a}_4\right)\in {\mathbb{N}}^4$ , (5) holds.

Case 4. $t=5$ . Applying the $M-S$ word-procedure to $M_4$ , deleting words including subwords in $\Phi \left(1212\right)\cup \Psi \Phi \left(1212\right)$ , we get $S_5$ . By Corollary 2.5 and $S_3^2\mathrm{,}{S}_4^3$ , words in $S_5^1$ besides 21342 are all tight. For word 21342, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_x|x\in \mathbb{N}\right\}$ , where

$M=\left(\begin{array}{ccccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x\\ 0& \mathrm{<mi>\; </mi>}{a}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-x\end{array}\right),$

and $\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{5}x-2x^2\right)+\left(i_1\cdot i_1\right)x^2+2\left(i_1\cdot i_2\right)a_{2}x\\ +2\left(i_1\cdot i_3\right)a_{3}x+2\left(i_1\cdot i_4\right)a_{4}x\\ =-2x^2+2(a_1+a_5-a_2-a_3-a_4)x.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_5\le a_2+a_3+a_4$ . And $q\left(M\right)=0\iff a_1+a_5\le a_2+a_3+a_4$ , $x=0$ , so word 21342(15-234) is tight by Theorem 2.3.

Consider $S_5^2$ . For words 12312, 12412, 13213, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,y}|\left(x,y\right)\in {\mathbb{N}}^2\right\}$ , where

$M=\left(\begin{array}{ccccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& a_4-x& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-y\end{array}\right),$

and

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{4}x-2x^2\right)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{5}y-2y^2\right)+\left(i_1\cdot i_1\right)x^2\\ +\left(i_2\cdot i_2\right)y^2+2\left(i_1\cdot i_3\right)a_{3}x+2\left(i_2\cdot i_3\right)a_{3}y+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_{4}y-xy\right)\\ =\{\begin{array}{l}-{\left(x-y\right)}^2-x^2-y^2+2\left(a_1+a_4-a_2\right)x+2\left(a_2+a_5-a_3-a_4\right)y,12312,12412,\\ -2x^2-2y^2+2\left(a_1+a_4-a_3\right)x+2\left(a_2+a_5-a_3\right)y,13213.\end{array}\end{array}$

Obviously, $q\left(M\right)\le 0\iff \{\begin{array}{l}{a}_1+a_4\le a_2,a_2+a_5\le a_3+a_4,12312,12412,\\ a_1+a_4\le a_3,a_2+a_5\le a_3,13213\end{array}$ . And $q\left(M\right)=0\iff \{\begin{array}{l}{a}_1+a_4\le a_2,a_2+a_5\le a_3+a_4,12312,12412,\\ a_1+a_4\le a_3,a_2+a_5\le a_3,13213\end{array}$ and $x=y=0$ , so

words 12312, 12412(14-2, 25-34), 13213(14-3, 25-3) are tight by Theorem 2.3.

For words 12321, 12421, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,y}|\left(x,y\right)\in {\mathbb{N}}^2\right\}$ , where

$M=\left(\begin{array}{ccccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& a_4-y& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-x\end{array}\right),$

and $\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{5}x-2x^2\right)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{4}y-2y^2\right)+\left(i_1\cdot i_1\right)x^2\\ +\left(i_2\cdot i_2\right)y^2+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_{4}x\right)+2\left(i_1\cdot i_3\right)a_{3}x+2\left(i_2\cdot i_3\right)a_{3}y\\ =-2x^2-2y^2+2\left(a_1+a_5-a_2-a_4\right)x+2\left(a_2+a_4-a_3\right)y\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_5\le a_2+a_4\le a_3$ . And $q\left(M\right)=0\iff a_1+a_5\le a_2+a_4\le a_3$ and $x=y=0$ , so words 12321, 12421(14-25-3) are tight by Theorem 2.3.

Now let us consider $S_5^3$ , we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,x_1,x_2,x_3}|\left(x,x_1,x_2,x_3\right)\in {\mathbb{N}}^4\right\}$ , where

$M=\left(\begin{array}{ccccc}{a}_1-x-x_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1\\ 0& \mathrm{<mi>\; </mi>}{a}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3-x_2-x_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_3\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}\\ x+x_1-x_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2+x_3-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-x_1-x_3\end{array}\right),$

and

$\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle (a_{1}x+a_1{x}_1+a_3{x}_2+a_3{x}_3+a_5{x}_1+a_5{x}_3-2x^2-2x_1^2-2x_2^2\\ -2x_3^2-2x_{1}x+2x_{2}x+x_{3}x+x_1{x}_2-x_1{x}_3-2x_2{x}_3)+\left(i_1\cdot i_1\right)(a_{3}x\\ +2a_3{x}_1-a_3{x}_2+x^2+x_1^2+x_2^2+x_3^2+x_{1}x-x_{2}x-x_{3}x-x_1{x}_2+x_2{x}_3)\\ +2\left(i_1\cdot i_2\right)\left(a_{2}x+a_2{x}_1\right)+2\left(i_1\cdot i_4\right)\left(a_4{x}_1+a_4{x}_3\right)\end{array}$

$\begin{array}{c}=-{\left(x-x_2\right)}^2-x^2-2x_1^2-x_2^2-2x_3^2-2x_{1}x-2x_1{x}_3-2x_2{x}_3\\ +2\left(a_1+a_3-a_2\right)x+2\left(a_1+2a_3+a_5-a_2-a_4\right)x_1+2\left(a_3+a_5-a_4\right)x_3.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_3+a_5\le a_4$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_3+a_5\le a_4$ , $x=x_1=x_2=x_3=0$ , so word 21232(13-2, 35-4) is tight. So (6) holds.

Case 5. $t=6$ . Applying the $M-S$ word-procedure to $M_5$ , deleting words including subwords in $\Phi \left(1212\right)\cup \Psi \Phi \left(1212\right)$ , we get $S_6\cup \left\{123123\right\}$ .

Firstly, as $S_5^2\mathrm{,}{S}_5^3$ , we can prove that words in $S_6^1\mathrm{,}{S}_6^3$ are all tight.

Secondly, consider words 123123, 132132, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,y,z}|\left(x,y,z\right)\in {\mathbb{N}}^3\right\}$ , where

$M=\left(\begin{array}{cccccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3-z& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}z\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-y& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}z& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6-z\end{array}\right)$

and

$\begin{array}{c}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{4}x-2x^2\right)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{5}y-2y^2\right)\\ +2\langle i_3,i_3\rangle \left(a_{3}z+a_{6}z-2z^2\right)+\left(i_1\cdot i_1\right)x^2+\left(i_2\cdot i_2\right)y^2+\left(i_3\cdot i_3\right)z^2\\ +2\left(i_1\cdot i_2\right)\left(a_{2}x+a_{4}y-xy\right)+2\left(i_1\cdot i_3\right)\left(a_{3}x+a_{4}z-xz\right)\\ +2\left(i_2\cdot i_3\right)\left(a_{3}y+a_{5}z-yz\right)\end{array}$

$=\{\begin{array}{l}-{\left(x-y\right)}^2-{\left(z-y\right)}^2-x^2-z^2+2\left(a_1+a_4-a_2\right)x\\ +2\left(a_2+a_5-a_3-a_4\right)y+2\left(a_3+a_6-a_5\right)z,123123,\\ -{\left(x-z\right)}^2-{\left(z-y\right)}^2-x^2-y^2+2\left(a_1+a_4-a_3\right)x\\ +2\left(a_2+a_5-a_3\right)y+2\left(a_3+a_6-a_4-a_5\right)z,132132.\end{array}$

Obviously, $q\left(M\right)\le 0\iff \{\begin{array}{l}{a}_1+a_4\le a_2,a_2+a_5\le a_3+a_4,a_3+a_6\le a_5,123123,\\ a_1+a_4\le a_3,a_2+a_5\le a_3,a_3+a_6\le a_4+a_5,132132.\end{array}$ And $q\left(M\right)=0\iff \{\begin{array}{l}{a}_1+a_4\le a_2,a_2+a_5\le a_3+a_4,a_3+a_6\le a_5,123123,\\ a_1+a_4\le a_3,a_2+a_5\le a_3,a_3+a_6\le a_4+a_5,132132,\end{array}$ and

$x=y=z=0$ , but $a_1+a_4\le a_2$ , $a_2+a_5\le a_3+a_4$ , $a_3+a_6\le a_5\Rightarrow a_1+a_6\le 0$ , this is a contradiction, so word 123123 is not tight for any $\left(a_1\mathrm{,}\cdots \mathrm{,}{a}_6\right)\in {\mathbb{N}}^6$ , and word 132132(14-3, 25-3, 36-45) is tight by Theorem 2.3.

Lastly, let us consider $S_6^4$ . Take word 123121 as an example, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,x_1,x_2,x_3,y}|\left(x,x_1,x_2,x_3,y\right)\in {\mathbb{N}}^5\right\}$ , where

$M=\left(\begin{array}{cccccc}{a}_1-x-x_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4-x_2-x_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_3\\ 0& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-y& \mathrm{<mi>\; </mi>0}\\ x+x_1-x_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2+x_3-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6-x_1-x_3\end{array}\right)$

and

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle (a_{1}x+a_1{x}_1+a_4{x}_2+a_4{x}_3+a_6{x}_1+a_6{x}_3-2x^2-2x_1^2-2x_2^2-2x_3^2\\ -2x_{1}x+2x_{2}x+x_{3}x+x_1{x}_2-x_1{x}_3-2x_2{x}_3)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{5}y-2y^2\right)\\ +\left(i_1\cdot i_1\right)\left(a_{4}x+2a_4{x}_1-a_4{x}_2+x^2+x_1^2+x_2^2+x_3^2+x_{1}x-x_{2}x-x_{3}x-x_1{x}_2+x_2{x}_3\right)\\ +\left(i_2\cdot i_2\right)y^2+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_2{x}_1+a_5{x}_1+a_5{x}_3+a_{4}y-x_{2}y-x_{3}y\right)\\ +2\left(i_1\cdot i_3\right)\left(a_{3}x+a_3{x}_1\right)+2\left(i_2\cdot i_3\right)a_{3}y\end{array}$

$\begin{array}{l}=-{\left(x-x_2\right)}^2-\left(x_2-y\right){)}^2-{\left(x_3-y\right)}^2-x^2-2x_1^2-x_3^2-2x_{1}x-2x_1{x}_3-2x_2{x}_3\\ +2\left(a_1+a_4-a_2\right)x+2\left(a_1+2a_4+a_6-a_2-a_5\right)x_1+2\left(a_4+a_6-a_5\right)x_3\\ +2\left(a_2+a_5-a_3-a_4\right)y.\end{array}$

Obviously, $q\left(M\right)\le 0\iff a_1+a_4\le a_2$ , $a_4+a_6\le a_5$ , $a_2+a_5\le a_3+a_4$ . And $q\left(M\right)=0\iff a_1+a_4\le a_2$ , $a_4+a_6\le a_5$ , $a_2+a_5\le a_3+a_4$ , $x=x_1=x_2=x_3=y=0$ , so word 123121(14-2, 46-5, 25-34) is tight by Theorem 2.3. So (7) holds.

Case 6. $t=7$ . Applying the $M-S$ word-procedure to $M_6$ , deleting words including subwords in $\Phi \left(\left\{\mathrm{1212,123123}\right\}\right)\cup \Psi \Phi \left(\left\{\mathrm{1212,123123}\right\}\right)$ , we get $S_7\cup \left\{\mathrm{2132132,1232123}\right\}$ .

As $S_6^2\mathrm{,}{S}_6^4$ , we can prove $S_7^1\mathrm{,}{S}_7^2$ . Consider word 2123242 in $S_7^3$ , we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,x_1,\cdots ,x_8}|\left(x,x_1,\cdots ,x_8\right)\in {\mathbb{N}}^9\right\}$ , where

$M=\left(\begin{array}{ccccccc}{b}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2\\ 0& \mathrm{<mi>\; </mi>}{a}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{33}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_5\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_6& \mathrm{<mi>\; </mi>0}& x_7& \mathrm{<mi>\; </mi>0}& b_{55}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_8\\ \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6& \mathrm{<mi>\; </mi>0}\\ b_{71}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{73}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{75}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{77}\end{array}\right),$

$\begin{array}{l}{b}_{11}=a_1-x-x_1-x_2,b_{33}=a_3-x_3-x_4-x_5,b_{55}=a_5-x_6-x_7-x_8,\\ b_{77}=a_7-x_2-x_5-x_8,b_{71}=x+x_1+x_2-x_3-x_6,\\ b_{73}=x_3+x_4+x_5-x-x_7,b_{75}=x_6+x_7+x_8-x_1-x_4,\end{array}$

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle (a_{1}x+a_1{x}_1+a_1{x}_2+a_3{x}_3+a_3{x}_4+a_3{x}_5+a_5{x}_6+a_5{x}_7+a_5{x}_8\\ +a_7{x}_2+a_7{x}_5+a_7{x}_8-2x^2-2x_1^2-2x_2^2-2x_3^2-2x_4^2-2x_5^2-2x_6^2-2x_7^2-2x_8^2\\ -2x_{1}x-2x_{2}x+2x_{3}x+x_{4}x+x_{5}x+x_{6}x-x_{7}x-2x_1{x}_2+x_1{x}_3-x_1{x}_4+2x_1{x}_6\\ +x_1{x}_7+x_1{x}_8+x_2{x}_3-x_2{x}_5+x_2{x}_6-x_2{x}_8-2x_3{x}_4-2x_3{x}_5-x_3{x}_6+x_3{x}_7\\ -2x_4{x}_5+x_4{x}_6+2x_4{x}_7+x_4{x}_8+x_5{x}_7-x_5{x}_8-2x_6{x}_7-2x_6{x}_8-2x_7{x}_8)\end{array}$

$\begin{array}{l}+\left(i_1\cdot i_1\right)(a_{3}x+2a_3{x}_1+2a_3{x}_2-a_3{x}_3+a_5{x}_1+2a_5{x}_2-a_5{x}_6+a_5{x}_4+2a_5{x}_5-a_5{x}_7\\ +x^2+x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2+x_8^2+x_{1}x+x_{2}x-x_{3}x-x_{4}x-x_{5}x+x_{7}x\\ +x_1{x}_2-x_1{x}_3-x_1{x}_5-x_1{x}_6-x_1{x}_8-x_2{x}_3-x_2{x}_6+x_3{x}_4+x_3{x}_5-x_3{x}_7+x_4{x}_5-x_4{x}_6\\ -x_4{x}_7-x_4{x}_8-x_5{x}_6-x_5{x}_7+x_6{x}_7+x_6{x}_8+x_7{x}_8)+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_2{x}_1+a_2{x}_2\right)\\ +2\left(i_1\cdot i_4\right)\left(a_4{x}_1+a_4{x}_2+a_4{x}_4+a_4{x}_5\right)+2\left(i_1\cdot i_6\right)\left(a_6{x}_2+a_6{x}_5+a_6{x}_8\right)\end{array}$

$\begin{array}{l}=-2f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)+2\left(a_1+a_3-a_2\right)x+2\left(a_1+2a_3+a_5-a_2-a_4\right)x_1\\ +2\left(a_1+2a_3+2a_5+a_7-a_2-a_4-a_6\right)x_2+2\left(a_3+a_5-a_4\right)x_4\\ +2\left(a_3+2a_5+a_7-a_4-a_6\right)x_5+2\left(a_5+a_7-a_6\right)x_8\mathrm{,}\end{array}$

where $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)=\frac{1}{2}X{A}_f{X}^{\text{T}}$ , $X=\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)$ symmetric matrix $A_f$ is

as follows

$A_f=\left(\begin{array}{ccccccccc}2& 1& 1& -1& 0& 0& -1& 0& 0\\ 1& 2& 1& 0& 1& 1& -1& -1& 0\\ 1& 1& 2& 0& 0& 1& 0& 0& 1\\ -1& 0& 0& 2& 1& 1& 1& 0& 0\\ 0& 1& 0& 1& 2& 1& 0& -1& 0\\ 0& 1& 1& 1& 1& 2& 1& 0& 1\\ -1& -1& 0& 1& 0& 1& 2& 1& 1\\ 0& -1& 0& 0& -1& 0& 1& 2& 1\\ 0& 0& 1& 0& 0& 1& 1& 1& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,\cdots ,r_9\right\};\\ R_2=\left\{r_1+r_4,r_1+r_7,r_2+r_7,r_2+r_8,r_5+r_8\right\};\\ R_3=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)$ is weakly positive, i.e., $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)\ge 0$ for any $\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\right)\in {\mathbb{N}}^9$ , and $f\left(x,x_1,\cdots ,x_8\right)=0\iff$ $x=x_1=\cdots =x_8=0$ . So, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_5+a_7\le a_6$ , $a_3+a_5\le a_4$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_5+a_7\le a_6$ , $a_3+a_5\le a_4$ , $x=x_1=\cdots =x_8=0$ , so word 2123242(13-2, 57-6, 35-4) is tight by Theorem 2.3.

At last, let us see $S_7^4$ , for words 2123212, 2124212 in $S_7^4$ , we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,x_1,\cdots ,x_8,y}|\left(x,x_1,\cdots ,x_8,y\right)\in {\mathbb{N}}^{10}\right\}$ , where

$M=\left(\begin{array}{ccccccc}{b}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}\\ x_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{33}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_5\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_6& \mathrm{<mi>\; </mi>0}& x_7& \mathrm{<mi>\; </mi>0}& b_{55}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_8\\ \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6-y& \mathrm{<mi>\; </mi>0}\\ b_{71}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{73}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{75}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{77}\end{array}\right)$

$\begin{array}{l}{b}_{11}=a_1-x-x_1-x_2,b_{33}=a_3-x_3-x_4-x_5,b_{55}=a_5-x_6-x_7-x_8,\\ b_{77}=a_7-x_2-x_5-x_8,b_{71}=x+x_1+x_2-x_3-x_6,\\ b_{73}=x_3+x_4+x_5-x-x_7,b_{75}=x_6+x_7+x_8-x_1-x_4,\end{array}$

$\begin{array}{l}q\left(M\right)=2\langle i_1\mathrm{,}{i}_1\rangle (a_{1}x+a_1{x}_1+a_1{x}_2+a_3{x}_3+a_3{x}_4+a_3{x}_5+a_5{x}_6+a_5{x}_7+a_5{x}_8\\ +a_7{x}_2+a_7{x}_5+a_7{x}_8-2x^2-2x_1^2-2x_2^2-2x_3^2-2x_4^2-2x_5^2-2x_6^2-2x_7^2\\ -2x_8^2-2x_{2}x+2x_{3}x+x_{4}x+x_{5}x+x_{6}x-x_{7}x-2x_1{x}_2+x_1{x}_3-x_1{x}_4+2x_1{x}_6\\ +x_1{x}_7-2x_{1}x+x_1{x}_8+x_2{x}_3-x_2{x}_5+x_2{x}_6-x_2{x}_8-2x_3{x}_4-2x_3{x}_5-x_3{x}_6\\ +x_3{x}_7-2x_4{x}_5+x_4{x}_6+2x_4{x}_7+x_4{x}_8+x_5{x}_7-x_5{x}_8-2x_6{x}_7-2x_6{x}_8-2x_7{x}_8)\end{array}$

$\begin{array}{l}+2\langle i_2\mathrm{,}{i}_2\rangle \left(a_{2}y+a_{6}y-2y^2\right)+\left(i_1\cdot i_1\right)(a_{3}x+2a_3{x}_1+2a_3{x}_2-a_3{x}_3+a_5{x}_1\\ +2a_5{x}_2-a_5{x}_6+a_5{x}_4+2a_5{x}_5-a_5{x}_7+x^2+x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2\\ +x_7^2+x_8^2+x_{1}x+x_{2}x-x_{3}x-x_{4}x-x_{5}x+x_{7}x+x_1{x}_2-x_1{x}_3-x_1{x}_5-x_1{x}_6\\ -x_1{x}_8-x_2{x}_3-x_2{x}_6+x_3{x}_4+x_3{x}_5-x_3{x}_7+x_4{x}_5-x_4{x}_6-x_4{x}_7-x_4{x}_8\\ -x_5{x}_6-x_5{x}_7+x_6{x}_7+x_6{x}_8+x_7{x}_8)+\left(i_2\cdot i_2\right)y^2+2\left(i_1\cdot i_2\right)(a_{2}x+a_2{x}_1\end{array}$

$\begin{array}{l}+a_2{x}_2+a_6{x}_2+a_{3}y+a_{5}y+a_6{x}_5+a_6{x}_8-x_{3}y-x_{5}y-x_{6}y-x_{8}y)\\ +2\left(i_1\cdot i_4\right)\left(a_4{x}_1+a_4{x}_2+a_4{x}_4+a_4{x}_5\right)\\ =-2f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\mathrm{,}y\right)+2\left(a_1+a_3-a_2\right)x+2\left(a_1+2a_3+a_5-a_2-a_4\right)x_1\\ +2\left(a_1+2a_3+2a_5+a_7-a_2-a_4-a_6\right)x_2+2\left(a_3+a_5-a_4\right)x_4\\ +2\left(a_3+2a_5+a_7-a_4-a_6\right)x_5+2\left(a_5+a_7-a_6\right)x_8+2\left(a_2+a_6-a_3-a_5\right)y\mathrm{,}\end{array}$

where $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\mathrm{,}y\right)=\frac{1}{2}X{A}_f{X}^{\text{T}}$ , $X=\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_8\mathrm{,}y\right)$ , symmetric matrix

$A_f$ is as follows

$A_f=\left(\begin{array}{cccccccccc}2& 1& 1& -1& 0& 0& -1& 0& 0& 0\\ 1& 2& 1& 0& 1& 1& -1& -1& 0& 0\\ 1& 1& 2& 0& 0& 1& 0& 0& 1& 0\\ -1& 0& 0& 2& 1& 1& 1& 0& 0& -1\\ 0& 1& 0& 1& 2& 1& 0& -1& 0& 0\\ 0& 1& 1& 1& 1& 2& 1& 0& 1& -1\\ -1& -1& 0& 1& 0& 1& 2& 1& 1& -1\\ 0& -1& 0& 0& -1& 0& 1& 2& 1& 0\\ 0& 0& 1& 0& 0& 1& 1& 1& 2& -1\\ 0& 0& 0& -1& 0& -1& -1& 0& -1& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,\cdots ,r_{10}\right\};\\ R_2=\left\{r_1+r_4,r_1+r_7,r_2+r_7,r_2+r_8,r_4+r_{10},r_5+r_8,r_6+r_{10},r_7+r_{10},r_9+r_{10}\right\};\end{array}$

$\begin{array}{l}{R}_3=\left\{r_1+r_4+r_{10},r_1+r_7+r_{10},r_2+r_7+r_{10},r_4+r_9+r_{10}\right\};\\ R_4=\left\{r_1+r_4+r_7+r_{10},r_1+r_4+r_9+r_{10}\right\};\\ R_5=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_9\mathrm{,}y\right)$ is weakly positive, i.e., $f\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_9\mathrm{,}y\right)\ge 0$ for any $\left(x\mathrm{,}{x}_1\mathrm{,}\cdots \mathrm{,}{x}_9\mathrm{,}y\right)\in {\mathbb{N}}^{10}$ , and $f\left(x,x_1,\cdots ,x_9,y\right)=$ $$0\iff x=x_1=\cdots =x_8=y=0$$ . So, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_5+a_7\le a_6$ , $a_2+a_6\le a_3+a_5\le a_4$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_5+a_7\le a_6$ , $a_2+a_6\le a_3+a_5\le a_4$ and $x=x_1=\cdots =x_8=y=0$ , so words 2123212, 2124212(13-2, 57-6, 26-35-4) are all tight by Theorem 2.3. So (8) holds.

Case 7. $t=8$ . Applying the $M-S$ word-procedure to $M_7$ , deleting words including subwords in $\Phi \left(\left\{\mathrm{1212,123123,1232123,2132132}\right\}\right)\cup \Psi \Phi \left(\left\{\mathrm{1212,123123,1232123,2132132}\right\}\right)$ , we get $S_8$ .

As $S_7^4$ , we can prove that words in $S_8^4$ are all tight.

For $S_8^1$ , only consider word 12134234, we have ${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,y,z,u}|\left(x,y,z,u\right)\in {\mathbb{N}}^4\right\}$ , where

$M=\left(\begin{array}{cccccccc}{a}_1-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_3-x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4-z& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}z& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_5-u& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}u\\ \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}z& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_7-z& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}u& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_8-u\end{array}\right),$

and $\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{3}x-2x^2\right)+2\langle i_2,i_2\rangle \left(a_{2}y+a_{6}y-2y^2\right)\\ +2\langle i_4,i_4\rangle \left(a_{4}z+a_{7}z-2z^2\right)+2\langle i_5,i_5\rangle \left(a_{5}u+a_{8}u-2u^2\right)+\left(i_1\cdot i_1\right)x^2\\ +\left(i_2\cdot i_2\right)y^2+\left(i_4\cdot i_4\right)z^2+\left(i_5\cdot i_5\right)u^2+2\left(i_1\cdot i_2\right)\left(a_{2}x+a_{3}y-xy\right)\\ +2\left(i_2\cdot i_4\right)\left(a_{4}y+a_{6}z-yz\right)+2\left(i_2\cdot i_5\right)\left(a_{5}y+a_{6}u-yu\right)\\ =-2f\left(x,y,z,u\right)+2\left(a_1+a_3-a_2\right)x+2\left(a_2+a_6-a_3-a_4-a_5\right)y\\ +2\left(a_4+a_7-a_6\right)z+2\left(a_5+a_8-a_6\right)u,\end{array}$

where $f\left(x,y,z,u\right)=\frac{1}{2}X{A}_f{X}^{\text{T}},X=\left(x,y,z,u\right)$ , symmetric matrix $A_f$ is as

follows

$A_f=\left(\begin{array}{cccc}2& -1& 0& 0\\ -1& 2& -1& -1\\ 0& -1& 2& 0\\ 0& -1& 0& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,r_2,r_3,r_4\right\};\\ R_2=\left\{r_1+r_2,r_2+r_3,r_2+r_4\right\};\end{array}$

$\begin{array}{l}{R}_3=\left\{r_1+r_2+r_3,r_1+r_2+r_4,r_2+r_3+r_4\right\};\\ R_4=\left\{r_1+r_2+r_3+r_4\right\};\\ R_5=\left\{r_1+2r_2+r_3+r_4\right\};\\ R_6=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x\mathrm{,}y\mathrm{,}z\mathrm{,}u\right)$ is weakly positive, i.e., $f\left(x\mathrm{,}y\mathrm{,}z\mathrm{,}u\right)\ge 0$ for any $\left(x\mathrm{,}y\mathrm{,}z\mathrm{,}u\right)\in {\mathbb{N}}^4$ , and $f\left(x,y,z,u\right)=0\iff x=y=z=$ $u=0$ . So, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_4+a_7\le a_6$ , $a_5+a_8\le a_6$ , $a_2+a_6\le a_3+a_4+a_5$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_4+a_7\le a_6$ , $a_5+a_8\le a_6$ , $a_2+a_6\le a_3+a_4+a_5$ and $x=y=z=u=0$ , word 12134234(13-2, 47-6, 58-6, 26-345) is tight by Theorem 2.3.

Consider words $\mathrm{12132142,12142132}$ in $S_8^2$ , we have

${\mathcal{M}}_{i\mathrm{,}a}=\left\{M=M_{x,x_1,x_2,x_3,y,y_1,y_2,y_3}|\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)\in {\mathbb{N}}^8\right\}$ , where

$M=\left(\begin{array}{cccccccc}{b}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}x& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{b}_{22}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_1\\ x_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{33}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{y}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{55}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_3\\ \mathrm{<mi>\; </mi>}{b}_{61}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{63}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{66}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_7& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{b}_{82}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{85}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{88}\end{array}\right)$

$\begin{array}{l}{b}_{11}=a_1-x-x_1,b_{22}=a_2-y-y_1,b_{33}=a_3-x_2-x_3,b_{55}=a_5-y_2-y_3,\\ b_{61}=x+x_1-x_2,b_{63}=x_2+x_3-x,b_{66}=a_6-x_1-x_3,\\ b_{82}=y+y_1-y_2,b_{85}=y_2+y_3-y,b_{88}=a_8-y_1-y_3.\end{array}$

So, we have

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle (a_{1}x+a_1{x}_1+a_3{x}_2+a_3{x}_3+a_6{x}_1+a_6{x}_3-2x^2-2x_1^2-2x_2^2-2x_3^2\\ -2x_{1}x+2x_{2}x+x_{3}x+x_1{x}_2-x_1{x}_3-2x_2{x}_3)+2\langle i_2\mathrm{,}{i}_2\rangle (a_{2}y+a_2{y}_1+a_5{y}_2+a_5{y}_3\\ +a_8{y}_1+a_8{y}_3-2y^2-2y_1^2-2y_2^2-2y_3^2-2y_{1}y+2y_{2}y+y_{3}y+y_1{y}_2-y_1{y}_3-2y_2{y}_3)\\ +\left(i_1\cdot i_1\right)\left(a_{3}x+2a_3{x}_1-a_3{x}_2+x^2+x_1^2+x_2^2+x_3^2+x_{1}x-x_{2}x-x_{3}x-x_1{x}_2+x_2{x}_3\right)\\ +\left(i_2\cdot i_2\right)\left(a_{5}y+2a_5{y}_1-a_5{y}_2+y^2+y_1^2+y_2^2+y_3^2+y_{1}y-y_{2}y-y_{3}y-y_1{y}_2+y_2{y}_3\right)\end{array}$

$\begin{array}{l}+2\left(i_1\cdot i_2\right)\mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}_{2}x+a_2{x}_1+a_5{x}_1+a_5{x}_3+a_{3}y+a_3{y}_1+a_6{y}_1+a_6{y}_3-xy-x{y}_1-x_1{y}_1\\ -x_2{y}_2+x{y}_2-x_3{y}_2-x_1{y}_3-x_3{y}_3)+2\left(i_2\cdot i_4\right)\left(a_{4}y+a_4{y}_1\right)+2\left(i_2\cdot i_7\right)\left(a_7{y}_1+a_7{y}_3\right)\\ =-2f\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)+2\left(a_1+a_3-a_2\right)x+2\left(a_1+2a_3+a_6-a_2-a_5\right)x_1\\ +2\left(a_3+a_6-a_5\right)x_3+2\left(a_2+a_5-a_3-a_4\right)y+2(a_2+2a_5+a_8-a_3-a_4-a_6\\ -a_7)y_1+2\left(a_5+a_8-a_6-a_7\right)y_3\mathrm{,}\end{array}$

where $f\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)=\frac{1}{2}X{A}_f{X}^{\text{T}}$ , $X=\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)$ ,

symmetric matrix $A_f$ is as follows

$A_f=\left(\begin{array}{cccccccc}2& 1& -1& 0& -1& -1& 1& 0\\ 1& 2& 0& 1& 0& -1& 0& -1\\ -1& 0& 2& 1& 0& 0& -1& 0\\ 0& 1& 1& 2& 0& 0& -1& -1\\ -1& 0& 0& 0& 2& 1& -1& 0\\ -1& -1& 0& 0& 1& 2& 0& 1\\ 1& 0& -1& -1& -1& 0& 2& 1\\ 0& -1& 0& -1& 0& 1& 1& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,\cdots ,r_8\right\};\\ R_2=\left\{r_1+r_3,r_1+r_5,r_1+r_6,r_2+r_6,r_2+r_8,r_3+r_7,r_4+r_7,r_4+r_8,r_5+r_7\right\};\\ R_3=\left\{r_1+r_3+r_5,r_1+r_3+r_6,r_3+r_5+r_7,r_4+r_5+r_7\right\};\\ R_4=\left\{r_1+r_3+r_5+r_7\right\};\\ R_5=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)$ is weakly positive, i.e., $f\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)\ge 0$ for any $\left(x\mathrm{,}{x}_1\mathrm{,}{x}_2\mathrm{,}{x}_3\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\right)\in {\mathbb{N}}^8$ , and $f\left(x,x_1,x_2,x_3,y,y_1,y_2,y_3\right)=0\iff x=x_1=x_2=x_3=y=y_1=y_2=y_3=0$ . So, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_3+a_6\le a_5$ , $a_5+a_8\le a_6+a_7$ , $a_2+a_5\le a_3+a_4$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_3+a_6\le a_5$ , $a_5+a_8\le a_6+a_7$ , $a_2+a_5\le a_3+a_4$ and $x=x_1=x_2=x_3=y=y_1=y_2=y_3=0$ , words 12132142, 12142132(13-2, 36-5, 58-67, 25-34) are all tight by Theorem 2.3.

For $S_8^3$ , only consider word 12132423, we have ${\mathcal{M}}_{i,a}=\left\{M=M_{x,y,y_1,y_2,y_3,z}|\left(x,y,y_1,y_2,y_3\mathrm{,}z\right)\in {\mathbb{N}}^6\right\}$ , where

$M=\left(\begin{array}{cccccccc}{a}_1-x& 0& x& 0& 0& 0& 0& 0\\ 0& a_2-y-y_1& 0& 0& y& 0& y_1& 0\\ x& 0& a_3-x& 0& 0& 0& 0& 0\\ 0& 0& 0& a_4-z& 0& 0& 0& z\\ 0& y_2& 0& 0& a_5-y_2-y_3& 0& y_3& 0\\ 0& 0& 0& 0& 0& a_6& 0& 0\\ 0& y+y_1-y_2& 0& 0& y_2+y_3-y& 0& a_7-y_1-y_3& 0\\ 0& 0& 0& z& 0& 0& 0& a_8-z\end{array}\right),$

and

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle \left(a_{1}x+a_{3}x-2x^2\right)+2\langle i_2,i_2\rangle (a_{2}y+a_2{y}_1+a_5{y}_2+a_5{y}_3+a_7{y}_1+a_7{y}_3\\ -2y^2-2y_1^2-2y_2^2-2y_3^2-2y_{1}y+2y_{2}y+y_{3}y+y_1{y}_2-y_1{y}_3-2y_2{y}_3)+2\langle i_4,i_4\rangle \\ \times \left(a_{4}z+a_{8}z-2z^2\right)+\left(i_1\cdot i_1\right)x^2+\left(i_2\cdot i_2\right)(a_{5}y+2a_5{y}_1-a_5{y}_2+y^2+y_1^2+y_2^2+y_3^2\\ +y_{1}y-y_{2}y-y_{3}y-y_1{y}_2+y_2{y}_3)+\left(i_4\cdot i_4\right)z^2+2\left(i_1\cdot i_2\right)(a_{2}x+a_{3}y+a_3{y}_1-xy\\ -x{y}_1)+2\left(i_2\cdot i_4\right)\left(a_{4}y+a_4{y}_1+a_{5}z+a_{7}z-yz-y_{1}z\right)+2\left(i_2\cdot i_6\right)\left(a_6{y}_1+a_6{y}_3\right)\end{array}$

$\begin{array}{l}=-2f\left(x,y,y_1,y_2,y_3,z\right)+2\left(a_1+a_3-a_2\right)x+2\left(a_2+a_5-a_3-a_4\right)y+2(a_2+2a_5\\ +a_7-a_3-a_4-a_6)y_1+2\left(a_5+a_7-a_6\right)y_3+2\left(a_4+a_8-a_5-a_7\right)z,\end{array}$

where $f\left(x\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\mathrm{,}z\right)=\frac{1}{2}X{A}_f{X}^{\text{T}}\mathrm{,}X=\left(x\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\mathrm{,}z\right)$ , symmetric matrix

$A_f$ is as follows

$A_f=\left(\begin{array}{cccccc}2& -1& -1& 0& 0& 0\\ -1& 2& 1& -1& 0& -1\\ -1& 1& 2& 0& 1& -1\\ 0& -1& 0& 2& 1& 0\\ 0& 0& 1& 1& 2& 0\\ 0& -1& -1& 0& 0& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,\cdots ,r_6\right\};\\ R_2=\left\{r_1+r_2,r_1+r_3,r_2+r_4,r_2+r_6,r_3+r_6\right\};\\ R_3=\left\{r_1+r_2+r_4,r_1+r_2+r_6,r_1+r_3+r_6,r_2+r_4+r_6\right\};\\ R_4=\left\{r_1+r_2+r_4+r_6,r_1+r_2+r_3+r_6\right\};\\ R_5=\left\{r_1+2r_2+r_4+r_6,r_1+r_2+2r_3+r_6,r_1+r_2+r_3+r_4+r_6\right\};\end{array}$

$\begin{array}{l}{R}_6=\left\{2r_1+r_2+2r_3+r_6,r_1+r_2+2r_3+r_4+r_6,r_1+r_2+2r_3+2r_6\right\};\\ R_7=\left\{2r_1+r_2+2r_3+r_4+r_6,2r_1+r_2+2r_3+2r_6,r_1+r_2+2r_3+r_4+2r_6\right\};\\ R_8=\left\{2r_1+r_2+2r_3+r_4+2r_6\right\};\\ R_9=\left\{2r_1+2r_2+2r_3+r_4+2r_6\right\};\\ R_{10}=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\mathrm{,}z\right)$ is weakly positive, i.e., $f\left(x\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\mathrm{,}z\right)\ge 0$ for any $\left(x\mathrm{,}y\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}{y}_3\mathrm{,}z\right)\in {\mathbb{N}}^6$ , and $f\left(x,y,y_1,y_2,y_3,z\right)=0\iff x=y=y_1=y_2=y_3=z=0$ . So, $q\left(M\right)\le 0\iff a_1+a_3\le a_2$ , $a_2+a_5\le a_3+a_4$ , $a_4+a_8\le a_5+a_7\le a_6$ . And $q\left(M\right)=0\iff a_1+a_3\le a_2$ , $a_2+a_5\le a_3+a_4$ , $a_4+a_8\le a_5+a_7\le a_6$ and $x=y=y_1=y_2=y_3=z=0$ , word 12132423(13-2, 25-34, 48-57-6) is tight by Theorem 2.3. So (9) holds.

Case 8. $t=9$ . Applying the $M-S$ word-procedure to $M_8$ , deleting words including subwords in $\Phi \left(\left\{\mathrm{1212,123123,1232123,2132132}\right\}\right)\cup \Psi \Phi \left(\left\{\mathrm{1212,123123,1232123,2132132}\right\}\right)$ , we get $S_9\cup \Phi \left(\left\{121342342\right\}\right)\cup \Psi \Phi \left(\left\{121342342\right\}\right)$ .

As $S_8^2\mathrm{,}{S}_8^3\mathrm{,}{S}_8^4$ , words in $S_9^1\mathrm{,}{S}_9^2\mathrm{,}{S}_9^3$ are all tight.

Consider $S_9^4$ , it is found that there is four 2, three 1 (or 2), one 3, and one 4 in every word, only consider word 121324212, we have ${\mathcal{M}}_{i,a}=\left\{M=M_{x_1,\cdots x_4,y_1,\cdots ,y_9}|\left(x_1,\cdots x_4,y_1,\cdots \mathrm{,}{y}_9\right)\in {\mathbb{N}}^{13}\right\}$ , where

$M=\left(\begin{array}{ccccccccc}{b}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{b}_{22}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_3\\ x_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{33}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_4& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{y}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& b_{55}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_5& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_6\\ \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{y}_7& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_8& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{77}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{y}_9\\ b_{81}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{83}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{88}& \mathrm{<mi>\; </mi>0}\\ 0& \mathrm{<mi>\; </mi>}{b}_{92}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{95}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{97}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{99}\end{array}\right)$

$\begin{array}{l}{b}_{11}=a_1-x_1-x_2,b_{22}=a_2-y_1-y_2-y_3,b_{33}=a_3-x_3-x_4,\\ b_{55}=a_5-y_4-y_5-y_6,b_{77}=a_7-y_7-y_8-y_9,b_{81}=x_1+x_2-x_3,\\ b_{83}=x_3+x_4-x_1,b_{88}=a_8-x_2-x_4,b_{92}=y_1+y_2+y_3-y_4-y_7,\\ b_{95}=y_4+y_5+y_6-y_1-y_8,b_{97}=y_7+y_8+y_9-y_2-y_5,\\ b_{99}=a_9-y_3-y_6-y_9,\end{array}$

and

$\begin{array}{l}q\left(M\right)=2\langle i_1,i_1\rangle (a_1{x}_1+a_1{x}_2+a_3{x}_3+a_3{x}_4+a_8{x}_2+a_8{x}_4-2x_1^2-2x_2^2-2x_3^2-2x_4^2\\ -2x_1{x}_2+2x_1{x}_3+x_1{x}_4+x_2{x}_3-x_2{x}_4-2x_3{x}_4)+2\langle i_2,i_2\rangle (a_2{y}_1+a_2{y}_2+a_2{y}_3+a_5{y}_4\\ +a_5{y}_5+a_5{y}_6+a_7{y}_7+a_7{y}_8+a_7{y}_9+a_9{y}_3+a_9{y}_6+a_9{y}_9-2y_1^2-2y_2^2-2y_3^2-2y_4^2\\ -2y_5^2-2y_6^2-2y_7^2-2y_8^2-2y_9^2-2y_1{y}_2-2y_1{y}_3+2y_1{y}_4+y_1{y}_5+y_1{y}_6+y_1{y}_7\\ -y_1{y}_8-2y_2{y}_3+y_2{y}_4-y_2{y}_5+2y_2{y}_7+y_2{y}_8+y_2{y}_9+y_3{y}_4-y_3{y}_6+y_3{y}_7-y_3{y}_9\end{array}$

$\begin{array}{l}-2y_4{y}_5-2y_4{y}_6-y_4{y}_7+y_4{y}_8-2y_5{y}_6+y_5{y}_7+2y_5{y}_8+y_5{y}_9+y_6{y}_8-y_6{y}_9\\ -2y_7{y}_8-2y_7{y}_9-2y_8{y}_9)+\left(i_1\cdot i_1\right)(a_3{x}_1+2a_3{x}_2-a_3{x}_3+x_1^2+x_2^2+x_3^2+x_4^2+x_1{x}_2\\ +x_3{x}_4-x_1{x}_3-x_1{x}_4-x_2{x}_3)+\left(i_2\cdot i_2\right)(a_5{y}_1+2a_5{y}_2+2a_5{y}_3-a_5{y}_4+a_7{y}_2+2a_7{y}_3\\ -a_7{y}_7+a_7{y}_5+2a_7{y}_6-a_7{y}_8+y_1^2+y_2^2+y_3^2+y_4^2+y_5^2+y_6^2+y_7^2+y_8^2+y_9^2+y_1{y}_2\\ +y_1{y}_3-y_1{y}_4-y_1{y}_5-y_1{y}_6+y_1{y}_8+y_2{y}_3-y_2{y}_4-y_2{y}_6-y_2{y}_7-y_2{y}_9-y_3{y}_4\end{array}$

$\begin{array}{l}-y_3{y}_7+y_4{y}_5+y_4{y}_6-y_4{y}_8+y_5{y}_6-y_5{y}_7-y_5{y}_8-y_5{y}_9-y_6{y}_7-y_6{y}_8+y_7{y}_8\\ +y_7{y}_9+y_8{y}_9)+2\left(i_1\cdot i_2\right)(a_2{x}_1+a_2{x}_2+a_5{x}_2+a_5{x}_4+a_7{x}_2+a_7{x}_4+a_3{y}_1+a_3{y}_2\\ +a_3{y}_3+a_8{y}_3+a_8{y}_6+a_8{y}_9-x_1{y}_1-x_1{y}_2-x_1{y}_3+x_1{y}_4+x_1{y}_7-x_2{y}_3-x_2{y}_6\\ -x_2{y}_9-x_3{y}_4-x_3{y}_7-x_4{y}_4-x_4{y}_6-x_4{y}_7-x_4{y}_9)+2\left(i_2\cdot i_4\right)\left(a_4{y}_1+a_4{y}_2+a_4{y}_3\right)\\ +2\left(i_2\cdot i_6\right)\left(a_6{y}_2+a_6{y}_3+a_6{y}_5+a_6{y}_6\right)\end{array}$

$\begin{array}{l}=-2f\left(x_1,\cdots ,x_4,y_1,\cdots ,y_9\right)+2\left(a_1+a_3-a_2\right)x_1+2\left(a_1+2a_3+a_8-a_2-a_5-a_7\right)x_2\\ +2\left(a_3+a_8-a_5-a_7\right)x_4+2\left(a_2+a_5-a_3-a_4\right)y_1+2(a_2+2a_5+a_7-a_3-a_4\\ -a_6)y_2+2\left(a_2+2a_5+2a_7+a_9-a_3-a_4-a_6-a_8\right)y_3+2\left(a_5+a_7-a_6\right)y_5\\ +2\left(a_5+2a_7+a_9-a_6-a_8\right)y_6+2\left(a_7+a_9-a_8\right)y_9,\end{array}$

where $f\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_4\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_9\right)=\frac{1}{2}X{A}_f{X}^{\text{T}}$ , $X=\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_4\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_9\right)$ symmetric

matrix $A_f$ is as follows

$A_f=\left(\begin{array}{ccccccccccccc}2& 1& -1& 0& -1& -1& -1& 1& 0& 0& 1& 0& 0\\ 1& 2& 0& 1& 0& 0& -1& 0& 0& -1& 0& 0& -1\\ -1& 0& 2& 1& 0& 0& 0& -1& 0& 0& -1& 0& 0\\ 0& 1& 1& 2& 0& 0& 0& -1& 0& -1& -1& 0& -1\\ -1& 0& 0& 0& 2& 1& 1& -1& 0& 0& -1& 0& 0\\ -1& 0& 0& 0& 1& 2& 1& 0& 1& 1& -1& -1& 0\\ -1& -1& 0& 0& 1& 1& 2& 0& 0& 1& 0& 0& 1\\ 1& 0& -1& -1& -1& 0& 0& 2& 1& 1& 1& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 1& 2& 1& 0& -1& 0\\ 0& -1& 0& -1& 0& 1& 1& 1& 1& 2& 1& 0& 1\\ 1& 0& -1& -1& -1& -1& 0& 1& 0& 1& 2& 1& 1\\ 0& 0& 0& 0& 0& -1& 0& 0& -1& 0& 1& 2& 1\\ 0& -1& 0& -1& 0& 0& 1& 0& 0& 1& 1& 1& 2\end{array}\right)$

Using the above algorithm in §2, we have

$\begin{array}{l}{R}_1=\left\{r_1,\cdots ,r_{13}\right\};\\ R_2=\{r_1+r_3,r_1+r_5,r_1+r_6,r_1+r_7,r_2+r_7,r_2+r_{10},r_2+r_{13},r_3+r_8,r_3+r_{11},\\ r_4+r_8,r_4+r_{10},r_4+r_{11},r_4+r_{13},r_5+r_8,r_5+r_{11},r_6+r_{11},r_6+r_{12},r_9+r_{12}\};\\ R_3=\{r_1+r_3+r_5,r_1+r_3+r_6,r_1+r_3+r_7,r_1+r_6+r_{12},r_3+r_5+r_8,r_3+r_5+r_{11},\\ r_3+r_6+r_{11},r_4+r_5+r_8,r_4+r_8+r_{13},r_4+r_5+r_{11},r_4+r_6+r_{11}\};\end{array}$

$\begin{array}{l}{R}_4=\{r_1+r_3+r_5+r_8,r_1+r_3+r_5+r_{11},r_1+r_3+r_6+r_{11},r_1+r_3+r_6+r_{12},\\ r_3+r_5+r_8+r_{11},r_4+r_5+r_8+r_{11},r_4+r_5+r_8+r_{13}\};\\ R_5=\left\{r_1+r_3+r_5+r_6+r_{11},r_3+r_4+r_5+r_8+r_{11}\right\};\\ R_6=\varphi .\end{array}$

By Theorem 2.8, the unit form $f\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_4\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_9\right)$ is weakly positive, i.e., for any $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_4\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_9\right)\in {\mathbb{N}}^{13}$ , $f\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_4\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_9\right)\ge 0$ , and $f\left(x_1,\cdots ,x_4,y_1,\cdots ,y_9\right)=0\iff x_1=\cdots =x_4=y_1=\cdots =y_9=0$ . So, we have q(M) ≤ 0 if and only if

$a_1+a_3\le a_2,a_3+a_8\le a_5+a_7\le a_6,a_7+a_9\le a_8,a_2+a_5\le a_3+a_4.$ (2)

$q\left(M\right)=0$ if and only if (2) hold and $x_1=\cdots =x_4=y_1=\cdots =y_9=0$ , so word 121324212(13-2, 38-57-6, 79-8, 25-34) is tight by Theorem 2.3.

At last, let us see $S_9^5$ , we find that there is five 2, two 1, one 2, and one 4 in every word, so it suffices to consider word 212321242, we have ${\mathcal{M}}_{i,a}=\left\{M=M_{x_1,\cdots x_{16},y}|\left(x_1,\cdots ,x_{16},y\right)\in {\mathbb{N}}^{17}\right\}$ , where

$M=\left(\begin{array}{ccccccccc}{b}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_1& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_2& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_3& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_4\\ 0& \mathrm{<mi>\; </mi>}{a}_2-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_5& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{33}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_6& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_7& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_8\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_4& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_9& \mathrm{<mi>\; </mi>0}& x_{10}& \mathrm{<mi>\; </mi>0}& b_{55}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_{11}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_{12}\\ \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_6-y& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}\\ x_{13}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_{14}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_{15}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{77}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{x}_{16}\\ 0& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{a}_8& \mathrm{<mi>\; </mi>0}\\ b_{91}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{93}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{95}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{97}& \mathrm{<mi>\; </mi>0}& \mathrm{<mi>\; </mi>}{b}_{99}\end{array}\right)$

$\begin{array}{l}{b}_{11}=a_1-x_1-x_2-x_3-x_4,b_{33}=a_3-x_5-x_6-x_7-x_8,\\ b_{55}=a_5-x_9-x_{10}-x_{11}-x_{12},b_{77}=a_7-x_{13}-x_{14}-x_{15}-x_{16},\\ b_{91}=x_1+x_2+x_3+x_4-x_5-x_9-x_{13},b_{93}=x_5+x_6+x_7+x_8-x_1-x_{10}-x_{14},\\ b_{95}=x_9+x_{10}+x_{11}+x_{12}-x_2-x_6-x_{15},b_{97}=x_{13}+x_{14}+x_{15}+x_{16}-x_3-x_7-x_{11},\\ b_{99}=a_9-x_4-x_8-x_{12}-x_{16}.\end{array}$

So, we have

$\begin{array}{l}q\left(M\right)=2\langle i_1\mathrm{,}{i}_1\rangle (a_1{x}_1+a_1{x}_2+a_1{x}_3+a_1{x}_4+a_3{x}_5+a_3{x}_6+a_3{x}_7+a_3{x}_8+a_5{x}_9+a_5{x}_{10}\\ +a_5{x}_{11}+a_5{x}_{12}+a_7{x}_{13}+a_7{x}_{14}+a_7{x}_{15}+a_7{x}_{16}+a_9{x}_4+a_9{x}_8+a_9{x}_{12}+a_9{x}_{16}-2x_1^2\\ -2x_2^2-2x_3^2-2x_4^2-2x_5^2-2x_6^2-2x_7^2-2x_8^2-2x_9^2-2x_{10}^2-2x_{11}^2-2x_{12}^2-2x_{13}^2\\ -2x_{14}^2-2x_{15}^2-2x_{16}^2-2x_1{x}_2-2x_1{x}_3-2x_1{x}_4+2x_1{x}_5+x_1{x}_6+x_1{x}_7+x_1{x}_8+x_1{x}_9\\ -x_1{x}_{10}+x_1{x}_{13}-x_1{x}_{14}-2x_2{x}_3-2x_2{x}_4+x_2{x}_5-x_2{x}_6+2x_2{x}_9+x_2{x}_{10}+x_2{x}_{11}\end{array}$

$\begin{array}{l}+x_2{x}_{12}+x_2{x}_{13}-x_2{x}_{15}-2x_3{x}_4+x_3{x}_5-x_3{x}_7+x_3{x}_9-x_3{x}_{11}+2x_3{x}_{13}+x_3{x}_{14}\\ +x_3{x}_{15}+x_3{x}_{16}+x_4{x}_5-x_4{x}_8+x_4{x}_9-x_4{x}_{12}+x_4{x}_{13}-x_4{x}_{16}-2x_5{x}_6-2x_5{x}_7\\ -2x_5{x}_8-x_5{x}_9+x_5{x}_{10}-x_5{x}_{13}+x_5{x}_{14}-2x_6{x}_7-2x_6{x}_8+x_6{x}_9+2x_6{x}_{10}+x_6{x}_{11}\\ +x_6{x}_{12}+x_6{x}_{14}-x_6{x}_{15}-2x_7{x}_8+x_7{x}_{10}-x_7{x}_{11}+x_7{x}_{13}+2x_7{x}_{14}+x_7{x}_{15}+x_7{x}_{16}\\ +x_8{x}_{10}-x_8{x}_{12}+x_8{x}_{14}-x_8{x}_{16}-2x_9{x}_{10}-2x_9{x}_{11}-2x_9{x}_{12}-x_9{x}_{13}+x_9{x}_{15}-2x_{10}{x}_{11}\end{array}$

$\begin{array}{l}-2x_{10}{x}_{12}-x_{10}{x}_{14}+x_{10}{x}_{15}-2x_{11}{x}_{12}+x_{11}{x}_{13}+x_{11}{x}_{14}+2x_{11}{x}_{15}+x_{11}{x}_{16}\\ +x_{12}{x}_{15}-x_{12}{x}_{16}-2x_{13}{x}_{14}-2x_{13}{x}_{15}-2x_{13}{x}_{16}-2x_{14}{x}_{15}-2x_{14}{x}_{16}-2x_{15}{x}_{16})\\ +2\langle i_2\mathrm{,}{i}_2\rangle \left(a_{2}y+a_{6}y-2y^2\right)+2\left(i_1\cdot i_2\right)(a_2{x}_1+a_2{x}_2+a_2{x}_3+a_2{x}_4+a_6{x}_3\\ +a_6{x}_4+a_6{x}_7+a_6{x}_8+a_6{x}_{11}+a_6{x}_{12}+a_{3}y+a_{5}y-x_{5}y-x_{7}y-x_{8}y-x_{9}y\\ -x_{11}y-x_{12}y)+2\left(i_1\cdot i_4\right)\left(a_4{x}_2+a_4{x}_3+a_4{x}_4+a_4{x}_6+a_4{x}_7+a_4{x}_8\right)\\ +2\left(i_1\cdot i_8\right)\left(a_8{x}_4+a_8{x}_8+a_8{x}_{12}+a_8{x}_{16}\right)\end{array}$

$\begin{array}{l}=-2f\left(x_1,\cdots ,x_{16},y\right)+2\left(a_1+a_3-a_2\right)x_1+2\left(a_1+2a_3+a_5-a_2-a_4\right)x_2\\ +2\left(a_1+2a_3+2a_5+a_7-a_2-a_4-a_6\right)x_3+2(a_1+2a_3+2a_5+2a_7+a_9\\ -a_2-a_4-a_6-a_8))x_4+2\left(a_3+a_5-a_4\right)x_6+2\left(a_3+2a_5+a_7-a_4-a_6\right)x_7\\ +2\left(a_3+2a_5+2a_7+a_9-a_4-a_6-a_8\right)x_8+2\left(a_5+a_7-a_6\right)x_{11}+2(a_5\\ +2a_7+a_9-a_6-a_8)x_{12}+2\left(a_7+a_9-a_9\right)x_{16}+2\left(a_2+a_6-a_3-a_5\right)y,\end{array}$

where $f\left(x_1,\cdots ,x_{16},y\right)=\frac{1}{2}X{A}_f{X}^{\text{T}},X=\left(x_1,\cdots ,x_{16},y\right)$ , symmetric matrix $A_f$ is

as follows

$A_f=(\begin{array}{ccccccccccccccccc}2& 1& 1& 1& -1& 0& 0& 0& -1& 0& 0& 0& -1& 0& 0& 0& 0\\ 1& 2& 1& 1& 0& 1& 1& 1& -1& -1& 0& 0& -1& -1& 0& 0& 0\\ 1& 1& 2& 1& 0& 0& 1& 1& 0& 0& 1& 1& -1& -1& -1& 0& 0\\ 1& 1& 1& 2& 0& 0& 0& 1& 0& 0& 0& 1& 0& 0& 0& 1& 0\\ -1& 0& 0& 0& 2& 1& 1& 1& 1& 0& 0& 0& 1& 0& 0& 0& -1\\ 0& 1& 0& 0& 1& 2& 1& 1& 0& -1& 0& 0& 0& -1& 0& 0& 0\\ 0& 1& 1& 0& 1& 1& 2& 1& 1& 0& 1& 1& 0& -1& -1& 0& -1\\ 0& 1& 1& 1& 1& 1& 1& 2& 1& 0& 0& 1& 1& 0& 0& 1& -1\\ -1& -1& 0& 0& 1& 0& 1& 1& 2& 1& 1& 1& 1& 1& 0& 0& -1\\ 0& -1& 0& 0& 0& -1& 0& 0& 1& 2& 1& 1& 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 1& 0& 1& 1& 2& 1& 0& 0& -1& 0& -1\\ 0& 0& 1& 1& 0& 0& 1& 1& 1& 1& 1& 2& 1& 1& 0& 1& -1\\ -1& -1& -1& 0& 1& 0& 0& 1& 1& 0& 0& 1& 2& 1& 1& 1& 0\\ 0& -1& -1& 0& 0& -1& -1& 0& 1& 1& 0& 1& 1& 2& 1& 1& 0\\ 0& 0& -1& 0& 0& 0& -1& 0& 0& 0& -1& 0& 1& 1& 2& 1& 0\\ 0& 0& 0& 1& 0& 0& 0& \end{array}$