ALAMT  Vol.7 No.4 , December 2017
Tight Monomials with t-Value ≤ 9 for Quantum Group of Type D4
Author(s) Yuwang Hu1*, Jifang Hu2, Qiongru Wu3
ABSTRACT
All monomials with t-value ≤9 in Canonical basis of quantum group for type D4 are determined in this paper.

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 , 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 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 = ( c i j ) i , j Γ 0 be a Cartan matrix of finite type such that c i i = 2 , c i j 0 for any i j , D = diag ( d i ) i Γ 0 be a diagonal matrix with integer entries making the matrix DC symmetric. Let g = g ( C ) be the complex semisimple Lie algebra associated with C, and U = U v ( g ) (here v is an indeterminate) be the corresponding quantized enveloping algebra, whose positive part U + = E i | i Γ 0 is the ( v ) -subalgebra of U , subject to the relations

r + s = 1 c i j ( 1 ) s E i ( s ) E j E i ( r ) = 0 , i , j Γ 0 , where E i ( s ) = E i s / [ s ] i ! , [ s ] i ! = [ 1 ] i [ s ] i ,

[ a ] i = ( v a d i v a d i ) / ( v d i v d i ) . Let A = [ v , v 1 ] , U + = E i ( s ) | i Γ 0 , s be the A -subalgebra of U + . Corresponding to every reduced expression i = ( i 1 , , i ν ) of the longest element w 0 = s i 1 s i ν of the Weyl group W = s i of g , one constructs a PBW basis B i of U + . Lusztig [4] proved that the [ v 1 ] -submodule L i = B i of U + is independent of the choice of i , write it L ; the image of B i under the canonical projection π : L L / v 1 L is independent of the choice of i , write it B; for any element b B there is a unique element b L which is fixed by the bar map of U + defined by v v 1 and satisfies π ( b ) = b . The set B = { b | b B } forms a [ v 1 ] -basis of L , an A -basis of U + and a ( v ) -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 ( a 1 ) E i t ( a t ) (1)

where i 1 , , i t Γ 0 , a 1 , , a t . When t = ν and s i 1 s i ν = 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 = ( Q 0 , Q 1 ) be a finite quiver with vertex set Q 0 and arrow set Q 1 . Write ρ Q 1 as t ρ ρ h ρ , where h ρ and t ρ denote the head and the tail of ρ respectively. An automorphism σ of Q is a permutation on the vertices of Q and on the arrows of Q such that σ ( h ρ ) = h σ ( ρ ) and σ ( t ρ ) = t σ ( ρ ) for any ρ Q 1 . Denote the quiver with automorphism σ as ( Q , σ ) . Attach to the pair ( Q , σ ) a valued quiver Γ = Γ ( Q , σ ) = ( Γ 0 , Γ 1 ) as follows. Its vertex set Γ 0 and arrow set Γ 1 are simply the sets of s-orbits in Q 0 and Q 1 , respectively. The valuation of Γ is given by d i = #{vertices in the s-orbit of i}, i Γ 0 ; m ρ = #{arrows in the σ-orbit of ρ}, ρ Γ 1 . The Euler form of Γ is defined to be the bilinear form , : [ Γ 0 ] × [ Γ 0 ] given by

X , Y = i Γ 0 d i x i y i ρ Γ 1 m ρ x t ρ y h ρ , where X = i Γ 0 x i i , Y = i Γ 0 y i i [ Γ 0 ] , so

X Y = X , Y + Y , X is the symmetric Euler form. The valued quiver Γ

defines a Cartan matrix C Γ = C Q , σ = ( c i j ) i , j Γ 0 , where

c i j = { 2 2 ρ Γ 1 h ρ = t ρ = i m ρ d i , i = j ; ρ Γ 1 { h ρ , t ρ } = { i , j } m ρ d i , i j .

For t , let i = ( i 1 , , i t ) Γ 0 t , a = ( a 1 , , a t ) t , and write E i ( a ) = E i 1 ( a 1 ) E i t ( a t ) U + . Define M i , a = { A = ( a r m ) t t | a r m , ro ( A ) = co ( A ) = a , a r m = 0 , i r i m } , where

ro ( A ) = ( m = 1 t a 1 m , , m = 1 t a t m ) , co ( A ) = ( r = 1 t a r 1 , , r = 1 t a r t ) . Obviously, D a = diag ( a 1 , , a t ) 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 , D n , E n , i Γ 0 t , a t as above. If the following quadratic form takes only values < 0 on M i , a \ { D a } (respectively, ≤ 0 on M i , a ), then monomial E i ( a ) is tight (respectively, semi-tight).

q i , a ( A ) = 1 m t 1 p < r t a p m a r m 1 p < r t 1 l < m t a p m a r l

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 , D n , E n , i Γ 0 t , a t as above, the monomial E i ( a ) is tight if and only if the following quadratic form takes only values <0 on M i , a \ { D a }

q i , a ( A ) = 1 m t 1 p < r t a p m a r m + 1 p < r t 1 l < m t ( i l i m ) a p m a r l + 1 r t 1 l < m t a r m a r l

In fact, q i , a ( A ) = q i , a ( A ) + q i , a ( A T ) (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 Γ 0 t , a t as above, the monomial E i ( a ) is tight if and only if the following quadratic form takes only values < 0 on M i , a \ { D a }

q i , a ( A ) = 1 m t 1 p < r t i m , i m a p m a r m + 1 p < r t 1 l < m t ( i l i m ) a p m a r l + 1 r t 1 l < m t i r , i r a r m a r l

By Theorem 2.3, we have the following Corollaries.

Corollary 2.4. When i 1 , , i t are mutually different, monomial E i 1 ( a 1 ) E i t ( a t ) is tight.

Proof: In fact, i 1 , , i t are mutually different, so M i , a = { D a } .

Corollary 2.5. If E i p + 1 ( a p + 1 ) E i p + q ( a p + q ) is tight, then for any mutually different i 1 , , i p { i p + 1 , , i p + q } and any mutually different i p + q + 1 , , i t { i 1 , , i p ; i p + 1 , , i p + q } , t l ( w 0 ) , E i ( a ) = E i 1 ( a 1 ) E i t ( a t ) is also tight.

Proof: Write j = ( i p + 1 , , i p + q ) , b = ( a p + 1 , , a p + q ) , i = ( i 1 , , i t ) , a = ( a 1 , , a t ) , then

E i ( a ) = E i 1 ( a 1 ) E i t ( a t ) , E j ( b ) = E i p + 1 ( a p + 1 ) E i p + q ( a p + q ) .

For any A ˜ M i , a , we have

A ˜ = ( D a 0 0 0 A 0 0 0 D a ) ,

where a = ( a 1 , , a p ) , a = ( a p + q + 1 , , a t ) . It is easy to see that A M j , b and q ( A ˜ ) = q ( A ) . Moreover, A ˜ = D a if and only if A = D b . Since E j ( b ) is tight, we get by Theorem 2.3 that q ( A ˜ ) = q ( A ) < 0 for all A ˜ M i , a \ { D a } , applying Theorem 2.3 again, we conclude that E i ( a ) 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 = ( i 1 , , i t ) Γ 0 t and a = ( a 1 , , a t ) t . If E i ( a ) is tight, then

(a) For 1 r s t , monomial E i r ( a r ) E i s ( a s ) is also tight;

(b) For 1 r < t , i r i r + 1 .

Theorem 2.7 ([Lusztig, 1990, Proposition 3.3 and Lusztig, 1993, §13]) [3] . Let Φ be the non-trivial automorphism of U + induced by Dynkin diagram automorphism φ of g , and Ψ : U + ( U + ) o p p be the unique ( v ) -algebra isomorphism such that E j E j . If E i ( a ) is tight, then Φ ( E i ( a ) ) and Ψ ( E i ( a ) ) are all tight.

A quadratic form f : n denoted by

f ( x 1 , , x n ) = i = 1 n x i 2 + i < j a i j x i x j

with a i j is called a unit form.

The symmetric matrix A f = ( a i j ) (when i > j , set a i j = a j i ) with a i i = 2 , defines a bilinear form f ( X , Y ) = X A f Y T , where

X = ( x 1 , , x n ) and Y = ( y 1 , , y n ) .

In particular, we have

f ( X ) = 1 2 f ( X , X ) and f ( X , Y ) = f ( X + Y ) f ( X ) f ( Y ) .

For a vector w n with non-negative coordinates, we write w 0 . The vector in n which has a 1 in the ith coordinate ( 1 i 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 { w n : 0 w and f ( w ) = 1 } . The linear transformation σ i : n n defined by σ i ( w ) = w f ( w , r i ) r i is called the reflection with respect to r i . The transformation σ i has the property that σ i 2 = id and f ( σ i ( w ) ) = f ( w ) for every w n .

Let f be a quadratic form, if 0 < f ( X ) for every 0 X n , then we call f weakly positive.

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

Let f : n ( n 3 ) be a unit form. First of all, we define

R 1 = { r i : 1 i n } .

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

R s = { z 1 , , z m } with s 1,

and that the process has not failed (to be defined subsequently). Then we construct R s + 1 as follows. Let z j R s . If either:

1) there is some 1 i n such that f ( z j , r i ) 2 , or

2) there is some 1 i n such that z j ( i ) 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 R s ). Let D s R s be the set of those roots z j with the property that there is some 1 i n such that f ( z j , r i ) = 1 . If D s = ϕ , then R s + 1 : = ϕ and the process is said to be successful. If D s ϕ , then

R s + 1 : = { σ i ( z j ) | z j D s and r i is such that f ( z j , r i ) = 1 } .

Remark: If R s + 1 ϕ , then we apply the algorithm again to obtain R s + 2 , etc. The roots in R s + 1 (if R s + 1 ϕ ) 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 = ϕ , then R 1 R s are all the positive roots of f .

3. Main Results

Let g be of type D 4 as follows.

Let i = ( i 1 , , i t ) Γ 0 t , a = ( a 1 , , a t ) t . For convenience, we abbreviate a monomial E i 1 ( a 1 ) E i t ( a t ) with any ( a 1 , , a t ) t as a word i 1 i t (1 as 0), an inequality a j 1 + + a j p a l 1 + + a l q as j 1 j p l 1 l q . For example, a monomial E 1 ( a 1 ) E 2 ( a 2 ) E 3 ( a 3 ) E 4 ( a 4 ) is abbreviated to a word 1234, and a monomial E 1 ( a 1 ) E 2 ( a 2 ) E 1 ( a 3 ) ( a 1 + a 3 a 2 ) to a word 121 (13 − 2), etc.

If s i 1 s i t is a reduced expression, we call i 1 i t a reduced word. By Theorem 2.6 (b), we only consider those reduced tight words i 1 i t with i r i r + 1 , 1 r < t , in this case, i 1 i t is called the word with t-value. If i r i r + 1 = 0 for some 1 r < t , we identify the word i 1 i r 1 i r i r + 1 i r + 2 i t with the word i 1 i r 1 i r + 1 i r i r + 2 i t . Denote the set of all words with t-value by M t . The non-trivial Dynkin diagram automorphism φ of g is φ : 1 3,3 4,4 1,2 2 . Let us present the so called M − S word-procedure from t-value to ( t + 1 ) -value.

Step 1. Take any i 1 i t M t , adding a number i t + 1 { 1,2,3,4 } 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 ( t + 1 ) -value from i 1 i t .

Step 2. Repeat step 1 until all words in M t are considered, deleting the non-tight words with ( t + 1 ) -value, get M t + 1 .

Step 3. Use Φ and Ψ , we have S t + 1 satisfying M t + 1 = Φ ( S t + 1 ) Ψ Φ ( S t + 1 ) .

For example, applying the M S word-procedure to M 1 = { 1,2,3,4 } , get M 2 = { 12 , 13 , 14 , 21 , 23 , 24 , 32 , 34 , 42 } . Considering Φ , Ψ , 24 φ 23 φ 21 Ψ 12 φ 32 φ 42 and 13 φ 34 φ 14 , so S 2 = { 12 , 13 } .

From now on, write M t = Φ ( S t ) Ψ Φ ( S t ) , t 2 .

Theorem 3.1. For the quantum group for type D 4 , we have the following results.

1) t = 0 , M 0 = { 0 } , tight monomial has only one.

2) t = 1 , M 1 = Φ ( S 1 ) , tight monomials have 4 families, where S 1 = { 1 , 2 } .

3) t = 2 , M 2 includes 9 families of tight monomials, where S 2 = { 12 , 13 } .

4) t = 3 , M 3 includes 19 families of tight monomials, where S 3 = S 3 1 S 3 2 ,

S 3 1 = { 123,132,134 } ; S 3 2 = { 121,212 ( 13 2 ) } .

5) t = 4 , M 4 includes 35 families of tight monomials, where S 4 = S 4 1 S 4 2 S 4 3 ,

S 4 1 = { 1234 , 1342 } ; S 4 2 = { 1213 , 1214 , 2123 , 2124 ( 13 2 ) } ; S 4 3 = { 2132 ( 14 23 ) } .

6) t = 5 , M5 includes 58 families of tight monomials, where S 5 = S 5 1 S 5 2 S 5 3 ,

S 5 1 = { 12134,24213 ( 13 2 ) ; 31214,12324 ( 24 3 ) ; 12342 ( 25 34 ) ; 21342 ( 15 234 ) } ;

S 5 2 = { 12312,12412 ( 14 2,25 34 ) ; 13213 ( 14 3,25 3 ) ; 12321,12421 ( 15 24 3 ) } ;

S 5 3 = { 21232 ( 13 2,35 4 ) } .

7) t = 6 , M 6 includes 93 families of tight monomials, where S 6 = S 6 1 S 6 2 S 6 3 S 6 4 ,

S 6 1 = { 123124,124123 ( 14 2,25 34 ) ; 123214,124213 ( 15 24 3 ) ; 132142,312342 ( 14 3,36 45 ) ; 132134 ( 14 3,25 3 ) ; 123421 ( 16 25 34 ) ; 123142 ( 14 2,26 345 ) } ;

S 6 2 = { 132132 ( 14 3,25 3,36 45 ) } ;

S 6 3 = { 212342 ( 13 2,36 45 ) ; 123242,124232 ( 24 3,46 5 ) } ;

S 6 4 = { 123121,124121 ( 14 2,46 5,25 34 ) ; 123212,124212 ( 46 5,15 24 3 ) } .

8) t = 7 , M 7 includes 133 families of tight monomials, where S 7 = S 7 1 S 7 2 S 7 3 S 7 4 ,

S 7 1 = { 1234213,1243214 ( 16 25 34,37 5 ) ; 1321342 ( 14 3,25 3,37 456 ) ; 1342134 ( 15 4,26 4,37 4 ) ; 1234123 ( 15 2,26 345,37 6 ) ; 1324213 ( 16 35 4,27 35 ) ; 1234234 ( 36 5,47 5,25 34 ) ; 1342132 ( 15 4,26 4,47 56 ) } ;

S 7 2 = { 1231242,1241232 ( 14 2,57 6,25 34 ) ; 1231241,1241231 ( 25 34,14 2,47 5 ) ; 1324212,3124232 ( 57 6,16 35 4 ) ; 1232142,1242132 ( 15 24 3,47 56 ) ;

1232124,1242123 ( 46 5,15 24 3 ) ; 1232421 ( 17 246,24 3,46 5 ) ; 1234212 ( 57 6,16 25 34 ) ; 1231421 ( 14 2,47 6,26 345 ) ; 2132142 ( 25 4,14 23,47 56 ) } ;

S 7 3 = { 2123242 ( 13 2,35 4,57 6 ) } ;

S 7 4 = { 2123212,2124212 ( 13 2,26 35 4,57 6 ) } .

9) t = 8 , M 8 includes 185 families of tight monomials, where S 8 = S 8 1 S 8 2 S 8 3 S 8 4 ,

S 8 1 = { 12134234 ( 13 2 , 26 345 , 47 6 , 58 6 ) ; 12342134 ( 16 25 34 , 48 5 , 37 5 ) ; 13421342 ( 15 4,48 567,26 4,37 4 ) } ;

S 8 2 = { 12132142,12142132 ( 13 2,36 5,25 34,58 67 ) ; 12142321,12132421 ( 13 2,38 57 6,25 34 ) } ;

S 8 3 = { 12132423,12142324 ( 13 2,25 34,48 57 6 ) ; 12134213,12143214 ( 13 2,37 6,26 345,48 6 ) ; 12321423,32123421 ( 15 24 3,38 47 56 ) ; 12324213,12423214 ( 17 246,24 3,38 46 5 ) ; 12342132,12432142 ( 16 25 34,58 67,37 5 ) ;

13214232,31234212 ( 14 3,27 36 45,68 7 ) ; 13214213,31234231 ( 14 3,47 6,28 36 45 ) ; 13213242 ( 14 3,36 45,68 7,25 3 ) ; 13242132 ( 16 35 4,58 67,27 35 ) ; 21321342 ( 25 4,14 23,36 4,48 567 ) } ;

S 8 4 = { 12321242,12421232 ( 15 24 3,46 5,68 7 ) ; 12324212,12423212 ( 17 246,24 3,46 5,68 7 ) ; 12324232,12423242 ( 24 3,68 7,37 46 5 ) ; 21232142,21242132 ( 13 2,58 67,26 35 4 ) ; 21234212 ( 13 2,68 7,27 36 45 ) } .

10) t = 9 , M 9 includes 265 families of tight monomials, where S 9 = S 9 1 S 9 2 S 9 3 S 9 4 S 9 5 ,

S 9 1 = { 121321423,121421324 ( 13 2,36 5,25 34,49 58 67 ) ; 121342132,121432142 ( 13 2,37 6,26 345,48 6,69 78 ) ; 121324213,121423214 ( 13 2,38 57 6,49 57,25 34 ) ; 132142132,312342312 ( 14 3,47 6,69 78,28 36 45 ) ;

124213214,123214213 ( 15 24 3,39 47 56,58 7 ) ; 124213213,123214214 ( 15 24 3,47 56,58 7,69 7 ) ; 132134213 ( 14 3,25 3,48 7,59 7,37 456 ) } ;

S 9 2 = { 123214234,124213243 ( 15 24 3,69 7,38 47 56 ) ; 121342134 ( 13 2,37 6,48 6,59 6,26 345 ) ; 123421234 ( 16 25 34,38 57 6,49 57 ) ; 123421324 ( 16 25 34,37 5,49 58 67 ) ;

123421342 ( 16 25 34,59 678,37 5,48 5 ) ; 132134214 ( 14 3,48 7,37 456,25 3,69 7 ) ; 134213242 ( 15 4,79 8,26 4,38 47 56 ) ; 213421342 ( 26 5,15 234,59 678,37 5,48 5 ) } ;

S 9 3 = { 121324232,121423242 ( 13 2,25 34,48 57 6,79 8 ) ; 123212423,124212324 ( 15 24 3,46 5,68 7,39 468 ) ; 123214232,124213242 ( 15 24 3,38 47 56,79 8 ) ; 123242132,124232142 ( 17 246,24 3,69 78,38 46 5 ) ; 123242321,124232421 ( 19 2468,24 3,68 7,37 46 5 ) ;

123421232,123421242 ( 16 25 34,38 57 6,79 8 ) ; 132421232,312423212 ( 16 35 4,57 6,79 8,28 357 ) ; 123242123 ( 17 246,24 3,46 5,39 468,68 7 ) ; 212342132,212432142 ( 13 2,27 36 45,48 6,69 78 ) ; 213242132 ( 27 46 5,14 23,69 78,38 46 ) } ;

S 9 4 = { 121324212,121423212 ( 13 2,25 34,38 57 6,79 8 ) ; 121321421 ( 13 2,36 5,69 8,25 34,58 67 ) ; 123212421 ( 15 24 3,59 68 7,46 5 ) } ;

S 9 5 = { 212321242,212421232 ( 13 2,26 35 4,57 6,79 8 ) ; 212324212 ( 13 2,28 357,35 4,57 6,79 8 ) } .

4. Proof of Theorem 3.1

Consider the following quiver Q = ( Q 0 , Q 1 ) of type D 4 where Q 1 = { 1 ρ 1 2, 3 ρ 2 2, 4 ρ 3 2 } , Q 0 = { 1 , 2 , 3 , 4 } . Let σ be the identity automorphism of Q such that σ ( i ) = i , σ ( ρ i ) = ρ i , i = 1 , 2 , 3 , 4 , then the valued quiver of ( Q , σ ) is Γ = Γ ( Q , σ ) = ( Γ 0 , Γ 1 ) = ( Q 0 , Q 1 ) , the valuation is given by d i = 1 , m ρ 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 [ Γ 0 ] , Euler form , on Γ is

X , Y Γ = i = 1 4 d i x i y i i = 1 3 m ρ i x t ρ i y h ρ 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 on Γ is

X Y = X , Y Γ + Y , X Γ = 2 x 1 y 1 + 2 x 2 y 2 + 2 x 3 y 3 + 2 x 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 .

By simple computation, we have i , i = 1 , i = 1 , 2 , 3 , 4 ; 1 , 2 = 3 , 2 = 4 , 2 = 1 , the other i , j = 0 , i j . So i i = 2 , i = 1 , 2 , 3 , 4 ; 1 2 = 2 3 = 2 4 = 1 , 1 3 = 1 4 = 3 4 = 0 .

Let us prove Theorem 3.1. For any t , let i = ( i 1 , , i t ) Γ 0 t , a = ( a 1 , , a t ) t .

Case 1. t 2 . By Corollary 2.4, words with t 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 S 3 1 . Consider S 3 2 , for i { ( 1,2,1 ) , ( 2,1,2 ) } , we have M i , a = { M = M x | x } , where

M = ( a 1 x 0 x 0 a 2 0 x 0 a 3 x ) ,

and

q ( M ) = 2 i 1 , i 1 ( a 1 x x 2 ) + 2 i 3 , i 3 ( a 3 x x 2 ) + ( i 1 i 3 ) x 2 + ( ( i 1 i 2 ) + ( i 2 i 3 ) ) a 2 x = 2 x 2 + 2 ( a 1 + a 3 a 2 ) x .

Obviously, q ( M ) 0 a 1 + a 3 a 2 . And q ( M ) = 0 a 1 + a 3 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 { 1212 } . Corollary 2.4 S 4 1 . Corollary 2.5 and S 3 2 S 4 2 . Consider words 2132, 1212. For word 2132, we have M i , a = { M = M x | x } , where

M = ( a 1 x 0 0 x 0 a 2 0 0 0 0 a 3 0 x 0 0 a 4 x ) ,

and q ( M ) = 2 i 1 , i 1 ( a 1 x + a 4 x 2 x 2 ) + ( i 1 i 1 ) x 2 + 2 ( i 1 i 2 ) a 2 x + 2 ( i 1 i 3 ) a 3 x = 2 x 2 + 2 ( a 1 + a 4 a 2 a 3 ) x .

Obviously, q ( M ) 0 a 1 + a 4 a 2 + a 3 . And q ( M ) = 0 a 1 + a 4 a 2 + a 3 , x = 0 , so word 2132(14-23) is tight by Theorem 2.3.

For word 1212, we have M i , a = { M = M x , y | ( x , y ) 2 } , where

M = ( a 1 x 0 x 0 0 a 2 y 0 y x 0 a 3 x 0 0 y 0 a 4 y ) ,

and q ( M ) = 2 i 1 , i 1 ( a 1 x + a 3 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 4 y 2 y 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) y 2 + 2 ( i 1 i 2 ) ( a 2 x + a 3 y x y ) = x 2 y 2 ( x y ) 2 + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 2 + a 4 a 3 ) y .

Obviously, q ( M ) 0 a 1 + a 3 a 2 , a 2 + a 4 a 3 . And q ( M ) = 0 a 1 + a 3 a 2 , a 2 + a 4 a 3 , x = y = 0 , but a 1 + a 3 a 2 , a 2 + a 4 a 3 a 1 + a 4 0 , this is a contradiction, so word 1212 is not tight for any a = ( a 1 , a 2 , a 3 , a 4 ) 4 , (5) holds.

Case 4. t = 5 . Applying the M S word-procedure to M 4 , deleting words including subwords in Φ ( 1212 ) Ψ Φ ( 1212 ) , we get S 5 . By Corollary 2.5 and S 3 2 , S 4 3 , words in S 5 1 besides 21342 are all tight. For word 21342, we have M i , a = { M = M x | x } , where

M = ( a 1 x 0 0 0 x 0 a 2 0 0 0 0 0 a 3 0 0 0 0 0 a 4 0 x 0 0 0 a 5 x ) ,

and q ( M ) = 2 i 1 , i 1 ( a 1 x + a 5 x 2 x 2 ) + ( i 1 i 1 ) x 2 + 2 ( i 1 i 2 ) a 2 x + 2 ( i 1 i 3 ) a 3 x + 2 ( i 1 i 4 ) a 4 x = 2 x 2 + 2 ( a 1 + a 5 a 2 a 3 a 4 ) x .

Obviously, q ( M ) 0 a 1 + a 5 a 2 + a 3 + a 4 . And q ( M ) = 0 a 1 + a 5 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 M i , a = { M = M x , y | ( x , y ) 2 } , where

M = ( a 1 x 0 0 x 0 0 a 2 y 0 0 y 0 0 a 3 0 0 x 0 0 a 4 x 0 0 y 0 0 a 5 y ) ,

and

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 4 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 5 y 2 y 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) y 2 + 2 ( i 1 i 3 ) a 3 x + 2 ( i 2 i 3 ) a 3 y + 2 ( i 1 i 2 ) ( a 2 x + a 4 y x y ) = { ( x y ) 2 x 2 y 2 + 2 ( a 1 + a 4 a 2 ) x + 2 ( a 2 + a 5 a 3 a 4 ) y , 12312 , 12412 , 2 x 2 2 y 2 + 2 ( a 1 + a 4 a 3 ) x + 2 ( a 2 + a 5 a 3 ) y , 13213.

Obviously, q ( M ) 0 { a 1 + a 4 a 2 , a 2 + a 5 a 3 + a 4 , 12312 , 12412 , a 1 + a 4 a 3 , a 2 + a 5 a 3 , 13213 . And q ( M ) = 0 { a 1 + a 4 a 2 , a 2 + a 5 a 3 + a 4 , 12312 , 12412 , a 1 + a 4 a 3 , a 2 + a 5 a 3 , 13213 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 M i , a = { M = M x , y | ( x , y ) 2 } , where

M = ( a 1 x 0 0 0 x 0 a 2 y 0 y 0 0 0 a 3 0 0 0 y 0 a 4 y 0 x 0 0 0 a 5 x ) ,

and q ( M ) = 2 i 1 , i 1 ( a 1 x + a 5 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 4 y 2 y 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) y 2 + 2 ( i 1 i 2 ) ( a 2 x + a 4 x ) + 2 ( i 1 i 3 ) a 3 x + 2 ( i 2 i 3 ) a 3 y = 2 x 2 2 y 2 + 2 ( a 1 + a 5 a 2 a 4 ) x + 2 ( a 2 + a 4 a 3 ) y

Obviously, q ( M ) 0 a 1 + a 5 a 2 + a 4 a 3 . And q ( M ) = 0 a 1 + a 5 a 2 + a 4 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 M i , a = { M = M x , x 1 , x 2 , x 3 | ( x , x 1 , x 2 , x 3 ) 4 } , where

M = ( a 1 x x 1 0 x 0 x 1 0 a 2 0 0 0 x 2 0 a 3 x 2 x 3 0 x 3 0 0 0 a 4 0 x + x 1 x 2 0 x 2 + x 3 x 0 a 5 x 1 x 3 ) ,

and

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 1 x 1 + a 3 x 2 + a 3 x 3 + a 5 x 1 + a 5 x 3 2 x 2 2 x 1 2 2 x 2 2 2 x 3 2 2 x 1 x + 2 x 2 x + x 3 x + x 1 x 2 x 1 x 3 2 x 2 x 3 ) + ( i 1 i 1 ) ( a 3 x + 2 a 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 ( i 1 i 2 ) ( a 2 x + a 2 x 1 ) + 2 ( i 1 i 4 ) ( a 4 x 1 + a 4 x 3 )

= ( x x 2 ) 2 x 2 2 x 1 2 x 2 2 2 x 3 2 2 x 1 x 2 x 1 x 3 2 x 2 x 3 + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 1 + 2 a 3 + a 5 a 2 a 4 ) x 1 + 2 ( a 3 + a 5 a 4 ) x 3 .

Obviously, q ( M ) 0 a 1 + a 3 a 2 , a 3 + a 5 a 4 . And q ( M ) = 0 a 1 + a 3 a 2 , a 3 + a 5 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 Φ ( 1212 ) Ψ Φ ( 1212 ) , we get S 6 { 123123 } .

Firstly, as S 5 2 , S 5 3 , we can prove that words in S 6 1 , S 6 3 are all tight.

Secondly, consider words 123123, 132132, we have M i , a = { M = M x , y , z | ( x , y , z ) 3 } , where

M = ( a 1 x 0 0 x 0 0 0 a 2 y 0 0 y 0 0 0 a 3 z 0 0 z x 0 0 a 4 x 0 0 0 y 0 0 a 5 y 0 0 0 z 0 0 a 6 z )

and

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 4 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 5 y 2 y 2 ) + 2 i 3 , i 3 ( a 3 z + a 6 z 2 z 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) y 2 + ( i 3 i 3 ) z 2 + 2 ( i 1 i 2 ) ( a 2 x + a 4 y x y ) + 2 ( i 1 i 3 ) ( a 3 x + a 4 z x z ) + 2 ( i 2 i 3 ) ( a 3 y + a 5 z y z )

= { ( x y ) 2 ( z y ) 2 x 2 z 2 + 2 ( a 1 + a 4 a 2 ) x + 2 ( a 2 + a 5 a 3 a 4 ) y + 2 ( a 3 + a 6 a 5 ) z , 123123 , ( x z ) 2 ( z y ) 2 x 2 y 2 + 2 ( a 1 + a 4 a 3 ) x + 2 ( a 2 + a 5 a 3 ) y + 2 ( a 3 + a 6 a 4 a 5 ) z , 132132.

Obviously, q ( M ) 0 { a 1 + a 4 a 2 , a 2 + a 5 a 3 + a 4 , a 3 + a 6 a 5 , 123123 , a 1 + a 4 a 3 , a 2 + a 5 a 3 , a 3 + a 6 a 4 + a 5 , 132132. And q ( M ) = 0 { a 1 + a 4 a 2 , a 2 + a 5 a 3 + a 4 , a 3 + a 6 a 5 , 123123 , a 1 + a 4 a 3 , a 2 + a 5 a 3 , a 3 + a 6 a 4 + a 5 , 132132 , and

x = y = z = 0 , but a 1 + a 4 a 2 , a 2 + a 5 a 3 + a 4 , a 3 + a 6 a 5 a 1 + a 6 0 , this is a contradiction, so word 123123 is not tight for any ( a 1 , , a 6 ) 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 M i , a = { M = M x , x 1 , x 2 , x 3 , y | ( x , x 1 , x 2 , x 3 , y ) 5 } , where

M = ( a 1 x x 1 0 0 x 0 x 1 0 a 2 y 0 0 y 0 0 0 a 3 0 0 0 x 2 0 0 a 4 x 2 x 3 0 x 3 0 y 0 0 a 5 y 0 x + x 1 x 2 0 0 x 2 + x 3 x 0 a 6 x 1 x 3 )

and

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 1 x 1 + a 4 x 2 + a 4 x 3 + a 6 x 1 + a 6 x 3 2 x 2 2 x 1 2 2 x 2 2 2 x 3 2 2 x 1 x + 2 x 2 x + x 3 x + x 1 x 2 x 1 x 3 2 x 2 x 3 ) + 2 i 2 , i 2 ( a 2 y + a 5 y 2 y 2 ) + ( i 1 i 1 ) ( a 4 x + 2 a 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 ) + ( i 2 i 2 ) y 2 + 2 ( i 1 i 2 ) ( a 2 x + a 2 x 1 + a 5 x 1 + a 5 x 3 + a 4 y x 2 y x 3 y ) + 2 ( i 1 i 3 ) ( a 3 x + a 3 x 1 ) + 2 ( i 2 i 3 ) a 3 y

= ( x x 2 ) 2 ( x 2 y ) ) 2 ( x 3 y ) 2 x 2 2 x 1 2 x 3 2 2 x 1 x 2 x 1 x 3 2 x 2 x 3 + 2 ( a 1 + a 4 a 2 ) x + 2 ( a 1 + 2 a 4 + a 6 a 2 a 5 ) x 1 + 2 ( a 4 + a 6 a 5 ) x 3 + 2 ( a 2 + a 5 a 3 a 4 ) y .

Obviously, q ( M ) 0 a 1 + a 4 a 2 , a 4 + a 6 a 5 , a 2 + a 5 a 3 + a 4 . And q ( M ) = 0 a 1 + a 4 a 2 , a 4 + a 6 a 5 , a 2 + a 5 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 Φ ( { 1212,123123 } ) Ψ Φ ( { 1212,123123 } ) , we get S 7 { 2132132,1232123 } .

As S 6 2 , S 6 4 , we can prove S 7 1 , S 7 2 . Consider word 2123242 in S 7 3 , we have M i , a = { M = M x , x 1 , , x 8 | ( x , x 1 , , x 8 ) 9 } , where

M = ( b 11 0 x 0 x 1 0 x 2 0 a 2 0 0 0 0 0 x 3 0 b 33 0 x 4 0 x 5 0 0 0 a 4 0 0 0 x 6 0 x 7 0 b 55 0 x 8 0 0 0 0 0 a 6 0 b 71 0 b 73 0 b 75 0 b 77 ) ,

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 ,

q ( M ) = 2 i 1 , i 1 ( 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 2 x 2 2 x 1 2 2 x 2 2 2 x 3 2 2 x 4 2 2 x 5 2 2 x 6 2 2 x 7 2 2 x 8 2 2 x 1 x 2 x 2 x + 2 x 3 x + x 4 x + x 5 x + x 6 x x 7 x 2 x 1 x 2 + x 1 x 3 x 1 x 4 + 2 x 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 2 x 3 x 4 2 x 3 x 5 x 3 x 6 + x 3 x 7 2 x 4 x 5 + x 4 x 6 + 2 x 4 x 7 + x 4 x 8 + x 5 x 7 x 5 x 8 2 x 6 x 7 2 x 6 x 8 2 x 7 x 8 )

+ ( i 1 i 1 ) ( a 3 x + 2 a 3 x 1 + 2 a 3 x 2 a 3 x 3 + a 5 x 1 + 2 a 5 x 2 a 5 x 6 + a 5 x 4 + 2 a 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 ( i 1 i 2 ) ( a 2 x + a 2 x 1 + a 2 x 2 ) + 2 ( i 1 i 4 ) ( a 4 x 1 + a 4 x 2 + a 4 x 4 + a 4 x 5 ) + 2 ( i 1 i 6 ) ( a 6 x 2 + a 6 x 5 + a 6 x 8 )

= 2 f ( x , x 1 , , x 8 ) + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 1 + 2 a 3 + a 5 a 2 a 4 ) x 1 + 2 ( a 1 + 2 a 3 + 2 a 5 + a 7 a 2 a 4 a 6 ) x 2 + 2 ( a 3 + a 5 a 4 ) x 4 + 2 ( a 3 + 2 a 5 + a 7 a 4 a 6 ) x 5 + 2 ( a 5 + a 7 a 6 ) x 8 ,

where f ( x , x 1 , , x 8 ) = 1 2 X A f X T , X = ( x , x 1 , , x 8 ) symmetric matrix A f is

as follows

A f = ( 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 )

Using the above algorithm in §2, we have

R 1 = { r 1 , , r 9 } ; R 2 = { r 1 + r 4 , r 1 + r 7 , r 2 + r 7 , r 2 + r 8 , r 5 + r 8 } ; R 3 = ϕ .

By Theorem 2.8, the unit form f ( x , x 1 , , x 8 ) is weakly positive, i.e., f ( x , x 1 , , x 8 ) 0 for any ( x , x 1 , , x 8 ) 9 , and f ( x , x 1 , , x 8 ) = 0 x = x 1 = = x 8 = 0 . So, q ( M ) 0 a 1 + a 3 a 2 , a 5 + a 7 a 6 , a 3 + a 5 a 4 . And q ( M ) = 0 a 1 + a 3 a 2 , a 5 + a 7 a 6 , a 3 + a 5 a 4 , x = x 1 = = 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 M i , a = { M = M x , x 1 , , x 8 , y | ( x , x 1 , , x 8 , y ) 10 } , where

M = ( b 11 0 x 0 x 1 0 x 2 0 a 2 y 0 0 0 y 0 x 3 0 b 33 0 x 4 0 x 5 0 0 0 a 4 0 0 0 x 6 0 x 7 0 b 55 0 x 8 0 y 0 0 0 a 6 y 0 b 71 0 b 73 0 b 75 0 b 77 )

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 ,

q ( M ) = 2 i 1 , i 1 ( 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 2 x 2 2 x 1 2 2 x 2 2 2 x 3 2 2 x 4 2 2 x 5 2 2 x 6 2 2 x 7 2 2 x 8 2 2 x 2 x + 2 x 3 x + x 4 x + x 5 x + x 6 x x 7 x 2 x 1 x 2 + x 1 x 3 x 1 x 4 + 2 x 1 x 6 + x 1 x 7 2 x 1 x + x 1 x 8 + x 2 x 3 x 2 x 5 + x 2 x 6 x 2 x 8 2 x 3 x 4 2 x 3 x 5 x 3 x 6 + x 3 x 7 2 x 4 x 5 + x 4 x 6 + 2 x 4 x 7 + x 4 x 8 + x 5 x 7 x 5 x 8 2 x 6 x 7 2 x 6 x 8 2 x 7 x 8 )

+ 2 i 2 , i 2 ( a 2 y + a 6 y 2 y 2 ) + ( i 1 i 1 ) ( a 3 x + 2 a 3 x 1 + 2 a 3 x 2 a 3 x 3 + a 5 x 1 + 2 a 5 x 2 a 5 x 6 + a 5 x 4 + 2 a 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 ) + ( i 2 i 2 ) y 2 + 2 ( i 1 i 2 ) ( a 2 x + a 2 x 1

+ 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 ( i 1 i 4 ) ( a 4 x 1 + a 4 x 2 + a 4 x 4 + a 4 x 5 ) = 2 f ( x , x 1 , , x 8 , y ) + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 1 + 2 a 3 + a 5 a 2 a 4 ) x 1 + 2 ( a 1 + 2 a 3 + 2 a 5 + a 7 a 2 a 4 a 6 ) x 2 + 2 ( a 3 + a 5 a 4 ) x 4 + 2 ( a 3 + 2 a 5 + a 7 a 4 a 6 ) x 5 + 2 ( a 5 + a 7 a 6 ) x 8 + 2 ( a 2 + a 6 a 3 a 5 ) y ,

where f ( x , x 1 , , x 8 , y ) = 1 2 X A f X T , X = ( x , x 1 , , x 8 , y ) , symmetric matrix

A f is as follows

A f = ( 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 )

Using the above algorithm in §2, we have

R 1 = { r 1 , , r 10 } ; R 2 = { 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 } ;

R 3 = { r 1 + r 4 + r 10 , r 1 + r 7 + r 10 , r 2 + r 7 + r 10 , r 4 + r 9 + r 10 } ; R 4 = { r 1 + r 4 + r 7 + r 10 , r 1 + r 4 + r 9 + r 10 } ; R 5 = ϕ .

By Theorem 2.8, the unit form f ( x , x 1 , , x 9 , y ) is weakly positive, i.e., f ( x , x 1 , , x 9 , y ) 0 for any ( x , x 1 , , x 9 , y ) 10 , and f ( x , x 1 , , x 9 , y ) = 0 x = x 1 = = x 8 = y = 0 . So, q ( M ) 0 a 1 + a 3 a 2 , a 5 + a 7 a 6 , a 2 + a 6 a 3 + a 5 a 4 . And q ( M ) = 0 a 1 + a 3 a 2 , a 5 + a 7 a 6 , a 2 + a 6 a 3 + a 5 a 4 and x = x 1 = = 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 Φ ( { 1212,123123,1232123,2132132 } ) Ψ Φ ( { 1212,123123,1232123,2132132 } ) , 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 M i , a = { M = M x , y , z , u | ( x , y , z , u ) 4 } , where

M = ( a 1 x 0 x 0 0 0 0 0 0 a 2 y 0 0 0 y 0 0 x 0 a 3 x 0 0 0 0 0 0 0 0 a 4 z 0 0 z 0 0 0 0 0 a 5 u 0 0 u 0 y 0 0 0 a 6 y 0 0 0 0 0 z 0 0 a 7 z 0 0 0 0 0 u 0 0 a 8 u ) ,

and q ( M ) = 2 i 1 , i 1 ( a 1 x + a 3 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 6 y 2 y 2 ) + 2 i 4 , i 4 ( a 4 z + a 7 z 2 z 2 ) + 2 i 5 , i 5 ( a 5 u + a 8 u 2 u 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) y 2 + ( i 4 i 4 ) z 2 + ( i 5 i 5 ) u 2 + 2 ( i 1 i 2 ) ( a 2 x + a 3 y x y ) + 2 ( i 2 i 4 ) ( a 4 y + a 6 z y z ) + 2 ( i 2 i 5 ) ( a 5 y + a 6 u y u ) = 2 f ( x , y , z , u ) + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 2 + a 6 a 3 a 4 a 5 ) y + 2 ( a 4 + a 7 a 6 ) z + 2 ( a 5 + a 8 a 6 ) u ,

where f ( x , y , z , u ) = 1 2 X A f X T , X = ( x , y , z , u ) , symmetric matrix A f is as

follows

A f = ( 2 1 0 0 1 2 1 1 0 1 2 0 0 1 0 2 )

Using the above algorithm in §2, we have

R 1 = { r 1 , r 2 , r 3 , r 4 } ; R 2 = { r 1 + r 2 , r 2 + r 3 , r 2 + r 4 } ;

R 3 = { r 1 + r 2 + r 3 , r 1 + r 2 + r 4 , r 2 + r 3 + r 4 } ; R 4 = { r 1 + r 2 + r 3 + r 4 } ; R 5 = { r 1 + 2 r 2 + r 3 + r 4 } ; R 6 = ϕ .

By Theorem 2.8, the unit form f ( x , y , z , u ) is weakly positive, i.e., f ( x , y , z , u ) 0 for any ( x , y , z , u ) 4 , and f ( x , y , z , u ) = 0 x = y = z = u = 0 . So, q ( M ) 0 a 1 + a 3 a 2 , a 4 + a 7 a 6 , a 5 + a 8 a 6 , a 2 + a 6 a 3 + a 4 + a 5 . And q ( M ) = 0 a 1 + a 3 a 2 , a 4 + a 7 a 6 , a 5 + a 8 a 6 , a 2 + a 6 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 12132142,12142132 in S 8 2 , we have

M i , a = { M = M x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 | ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) 8 } , where

M = ( b 11 0 x 0 0 x 1 0 0 0 b 22 0 0 y 0 0 y 1 x 2 0 b 33 0 0 x 3 0 0 0 0 0 a 4 0 0 0 0 0 y 2 0 0 b 55 0 0 y 3 b 61 0 b 63 0 0 b 66 0 0 0 0 0 0 0 0 a 7 0 0 b 82 0 0 b 85 0 0 b 88 )

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 .

So, we have

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 1 x 1 + a 3 x 2 + a 3 x 3 + a 6 x 1 + a 6 x 3 2 x 2 2 x 1 2 2 x 2 2 2 x 3 2 2 x 1 x + 2 x 2 x + x 3 x + x 1 x 2 x 1 x 3 2 x 2 x 3 ) + 2 i 2 , i 2 ( a 2 y + a 2 y 1 + a 5 y 2 + a 5 y 3 + a 8 y 1 + a 8 y 3 2 y 2 2 y 1 2 2 y 2 2 2 y 3 2 2 y 1 y + 2 y 2 y + y 3 y + y 1 y 2 y 1 y 3 2 y 2 y 3 ) + ( i 1 i 1 ) ( a 3 x + 2 a 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 ) + ( i 2 i 2 ) ( a 5 y + 2 a 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 )

+ 2 ( i 1 i 2 ) ( 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 x y 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 ( i 2 i 4 ) ( a 4 y + a 4 y 1 ) + 2 ( i 2 i 7 ) ( a 7 y 1 + a 7 y 3 ) = 2 f ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 1 + 2 a 3 + a 6 a 2 a 5 ) x 1 + 2 ( a 3 + a 6 a 5 ) x 3 + 2 ( a 2 + a 5 a 3 a 4 ) y + 2 ( a 2 + 2 a 5 + a 8 a 3 a 4 a 6 a 7 ) y 1 + 2 ( a 5 + a 8 a 6 a 7 ) y 3 ,

where f ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) = 1 2 X A f X T , X = ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) ,

symmetric matrix A f is as follows

A f = ( 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 )

Using the above algorithm in §2, we have

R 1 = { r 1 , , r 8 } ; R 2 = { 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 } ; R 3 = { r 1 + r 3 + r 5 , r 1 + r 3 + r 6 , r 3 + r 5 + r 7 , r 4 + r 5 + r 7 } ; R 4 = { r 1 + r 3 + r 5 + r 7 } ; R 5 = ϕ .

By Theorem 2.8, the unit form f ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) is weakly positive, i.e., f ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) 0 for any ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) 8 , and f ( x , x 1 , x 2 , x 3 , y , y 1 , y 2 , y 3 ) = 0 x = x 1 = x 2 = x 3 = y = y 1 = y 2 = y 3 = 0 . So, q ( M ) 0 a 1 + a 3 a 2 , a 3 + a 6 a 5 , a 5 + a 8 a 6 + a 7 , a 2 + a 5 a 3 + a 4 . And q ( M ) = 0 a 1 + a 3 a 2 , a 3 + a 6 a 5 , a 5 + a 8 a 6 + a 7 , a 2 + a 5 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 M i , a = { M = M x , y , y 1 , y 2 , y 3 , z | ( x , y , y 1 , y 2 , y 3 , z ) 6 } , where

M = ( 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 ) ,

and

q ( M ) = 2 i 1 , i 1 ( a 1 x + a 3 x 2 x 2 ) + 2 i 2 , i 2 ( a 2 y + a 2 y 1 + a 5 y 2 + a 5 y 3 + a 7 y 1 + a 7 y 3 2 y 2 2 y 1 2 2 y 2 2 2 y 3 2 2 y 1 y + 2 y 2 y + y 3 y + y 1 y 2 y 1 y 3 2 y 2 y 3 ) + 2 i 4 , i 4 × ( a 4 z + a 8 z 2 z 2 ) + ( i 1 i 1 ) x 2 + ( i 2 i 2 ) ( a 5 y + 2 a 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 ) + ( i 4 i 4 ) z 2 + 2 ( i 1 i 2 ) ( a 2 x + a 3 y + a 3 y 1 x y x y 1 ) + 2 ( i 2 i 4 ) ( a 4 y + a 4 y 1 + a 5 z + a 7 z y z y 1 z ) + 2 ( i 2 i 6 ) ( a 6 y 1 + a 6 y 3 )

= 2 f ( x , y , y 1 , y 2 , y 3 , z ) + 2 ( a 1 + a 3 a 2 ) x + 2 ( a 2 + a 5 a 3 a 4 ) y + 2 ( a 2 + 2 a 5 + a 7 a 3 a 4 a 6 ) y 1 + 2 ( a 5 + a 7 a 6 ) y 3 + 2 ( a 4 + a 8 a 5 a 7 ) z ,

where f ( x , y , y 1 , y 2 , y 3 , z ) = 1 2 X A f X T , X = ( x , y , y 1 , y 2 , y 3 , z ) , symmetric matrix

A f is as follows

A f = ( 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 )

Using the above algorithm in §2, we have

R 1 = { r 1 , , r 6 } ; R 2 = { r 1 + r 2 , r 1 + r 3 , r 2 + r 4 , r 2 + r 6 , r 3 + r 6 } ; R 3 = { r 1 + r 2 + r 4 , r 1 + r 2 + r 6 , r 1 + r 3 + r 6 , r 2 + r 4 + r 6 } ; R 4 = { r 1 + r 2 + r 4 + r 6 , r 1 + r 2 + r 3 + r 6 } ; R 5 = { r 1 + 2 r 2 + r 4 + r 6 , r 1 + r 2 + 2 r 3 + r 6 , r 1 + r 2 + r 3 + r 4 + r 6 } ;

R 6 = { 2 r 1 + r 2 + 2 r 3 + r 6 , r 1 + r 2 + 2 r 3 + r 4 + r 6 , r 1 + r 2 + 2 r 3 + 2 r 6 } ; R 7 = { 2 r 1 + r 2 + 2 r 3 + r 4 + r 6 , 2 r 1 + r 2 + 2 r 3 + 2 r 6 , r 1 + r 2 + 2 r 3 + r 4 + 2 r 6 } ; R 8 = { 2 r 1 + r 2 + 2 r 3 + r 4 + 2 r 6 } ; R 9 = { 2 r 1 + 2 r 2 + 2 r 3 + r 4 + 2 r 6 } ; R 10 = ϕ .

By Theorem 2.8, the unit form f ( x , y , y 1 , y 2 , y 3 , z ) is weakly positive, i.e., f ( x , y , y 1 , y 2 , y 3 , z ) 0 for any ( x , y , y 1 , y 2 , y 3 , z ) 6 , and f ( x , y , y 1 , y 2 , y 3 , z ) = 0 x = y = y 1 = y 2 = y 3 = z = 0 . So, q ( M ) 0 a 1 + a 3 a 2 , a 2 + a 5 a 3 + a 4 , a 4 + a 8 a 5 + a 7 a 6 . And q ( M ) = 0 a 1 + a 3 a 2 , a 2 + a 5 a 3 + a 4 , a 4 + a 8 a 5 + a 7 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 Φ ( { 1212,123123,1232123,2132132 } ) Ψ Φ ( { 1212,123123,1232123,2132132 } ) , we get S 9 Φ ( { 121342342 } ) Ψ Φ ( { 121342342 } ) .

As S 8 2 , S 8 3 , S 8 4 , words in S 9 1 , S 9 2 , 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 M i , a = { M = M x 1 , x 4 , y 1 , , y 9 | ( x 1 , x 4 , y 1 , , y 9 ) 13 } , where

M = ( b 11 0 x 1 0 0 0 0 x 2 0 0 b 22 0 0 y 1 0 y 2 0 y 3 x 3 0 b 33 0 0 0 0 x 4 0 0 0 0 a 4 0 0 0 0 0 0 y 4 0 0 b 55 0 y 5 0 y 6 0 0 0 0 0 a 6 0 0 0 0 y 7 0 0 y 8 0 b 77 0 y 9 b 81 0 b 83 0 0 0 0 b 88 0 0 b 92 0 0 b 95 0 b 97 0 b 99 )

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 ,

and

q ( M ) = 2 i 1 , i 1 ( a 1 x 1 + a 1 x 2 + a 3 x 3 + a 3 x 4 + a 8 x 2 + a 8 x 4 2 x 1 2 2 x 2 2 2 x 3 2 2 x 4 2 2 x 1 x 2 + 2 x 1 x 3 + x 1 x 4 + x 2 x 3 x 2 x 4 2 x 3 x 4 ) + 2 i 2 , i 2 ( 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 2 y 1 2 2 y 2 2 2 y 3 2 2 y 4 2 2 y 5 2 2 y 6 2 2 y 7 2 2 y 8 2 2 y 9 2 2 y 1 y 2 2 y 1 y 3 + 2 y 1 y 4 + y 1 y 5 + y 1 y 6 + y 1 y 7 y 1 y 8 2 y 2 y 3 + y 2 y 4 y 2 y 5 + 2 y 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

2 y 4 y 5 2 y 4 y 6 y 4 y 7 + y 4 y 8 2 y 5 y 6 + y 5 y 7 + 2 y 5 y 8 + y 5 y 9 + y 6 y 8 y 6 y 9 2 y 7 y 8 2 y 7 y 9 2 y 8 y 9 ) + ( i 1 i 1 ) ( a 3 x 1 + 2 a 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 ) + ( i 2 i 2 ) ( a 5 y 1 + 2 a 5 y 2 + 2 a 5 y 3 a 5 y 4 + a 7 y 2 + 2 a 7 y 3 a 7 y 7 + a 7 y 5 + 2 a 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

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 ( i 1 i 2 ) ( 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 ( i 2 i 4 ) ( a 4 y 1 + a 4 y 2 + a 4 y 3 ) + 2 ( i 2 i 6 ) ( a 6 y 2 + a 6 y 3 + a 6 y 5 + a 6 y 6 )

= 2 f ( x 1 , , x 4 , y 1 , , y 9 ) + 2 ( a 1 + a 3 a 2 ) x 1 + 2 ( a 1 + 2 a 3 + a 8 a 2 a 5 a 7 ) x 2 + 2 ( a 3 + a 8 a 5 a 7 ) x 4 + 2 ( a 2 + a 5 a 3 a 4 ) y 1 + 2 ( a 2 + 2 a 5 + a 7 a 3 a 4 a 6 ) y 2 + 2 ( a 2 + 2 a 5 + 2 a 7 + a 9 a 3 a 4 a 6 a 8 ) y 3 + 2 ( a 5 + a 7 a 6 ) y 5 + 2 ( a 5 + 2 a 7 + a 9 a 6 a 8 ) y 6 + 2 ( a 7 + a 9 a 8 ) y 9 ,

where f ( x 1 , , x 4 , y 1 , , y 9 ) = 1 2 X A f X T , X = ( x 1 , , x 4 , y 1 , , y 9 ) symmetric

matrix A f is as follows

A f = ( 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 )

Using the above algorithm in §2, we have

R 1 = { r 1 , , r 13 } ; 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 } ;

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 = { r 1 + r 3 + r 5 + r 6 + r 11 , r 3 + r 4 + r 5 + r 8 + r 11 } ; R 6 = ϕ .

By Theorem 2.8, the unit form f ( x 1 , , x 4 , y 1 , , y 9 ) is weakly positive, i.e., for any ( x 1 , , x 4 , y 1 , , y 9 ) 13 , f ( x 1 , , x 4 , y 1 , , y 9 ) 0 , and f ( x 1 , , x 4 , y 1 , , y 9 ) = 0 x 1 = = x 4 = y 1 = = y 9 = 0 . So, we have q(M) ≤ 0 if and only if

a 1 + a 3 a 2 , a 3 + a 8 a 5 + a 7 a 6 , a 7 + a 9 a 8 , a 2 + a 5 a 3 + a 4 . (2)

q ( M ) = 0 if and only if (2) hold and x 1 = = x 4 = y 1 = = 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 M i , a = { M = M x 1 , x 16 , y | ( x 1 , , x 16 , y ) 17 } , where

M = ( b 11 0 x 1 0 x 2 0 x 3 0 x 4 0 a 2 y 0 0 0 y 0 0 0 x 5 0 b 33 0 x 6 0 x 7 0 x 8 0 0 0 a 4 0 0 0 0 0 x 9 0 x 10 0 b 55 0 x 11 0 x 12 0 y 0 0 0 a 6 y 0 0 0 x 13 0 x 14 0 x 15 0 b 77 0 x 16 0 0 0 0 0 0 0 a 8 0 b 91 0 b 93 0 b 95 0 b 97 0 b 99 )

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 .

So, we have

q ( M ) = 2 i 1 , i 1 ( 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 2 x 1 2 2 x 2 2 2 x 3 2 2 x 4 2 2 x 5 2 2 x 6 2 2 x 7 2 2 x 8 2 2 x 9 2 2 x 10 2 2 x 11 2 2 x 12 2 2 x 13 2 2 x 14 2 2 x 15 2 2 x 16 2 2 x 1 x 2 2 x 1 x 3 2 x 1 x 4 + 2 x 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 2 x 2 x 3 2 x 2 x 4 + x 2 x 5 x 2 x 6 + 2 x 2 x 9 + x 2 x 10 + x 2 x 11

+ x 2 x 12 + x 2 x 13 x 2 x 15 2 x 3 x 4 + x 3 x 5 x 3 x 7 + x 3 x 9 x 3 x 11 + 2 x 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 2 x 5 x 6 2 x 5 x 7 2 x 5 x 8 x 5 x 9 + x 5 x 10 x 5 x 13 + x 5 x 14 2 x 6 x 7 2 x 6 x 8 + x 6 x 9 + 2 x 6 x 10 + x 6 x 11 + x 6 x 12 + x 6 x 14 x 6 x 15 2 x 7 x 8 + x 7 x 10 x 7 x 11 + x 7 x 13 + 2 x 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 2 x 9 x 10 2 x 9 x 11 2 x 9 x 12 x 9 x 13 + x 9 x 15 2 x 10 x 11

2 x 10 x 12 x 10 x 14 + x 10 x 15 2 x 11 x 12 + x 11 x 13 + x 11 x 14 + 2 x 11 x 15 + x 11 x 16 + x 12 x 15 x 12 x 16 2 x 13 x 14 2 x 13 x 15 2 x 13 x 16 2 x 14 x 15 2 x 14 x 16 2 x 15 x 16 ) + 2 i 2 , i 2 ( a 2 y + a 6 y 2 y 2 ) + 2 ( i 1 i 2 ) ( 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 ( i 1 i 4 ) ( a 4 x 2 + a 4 x 3 + a 4 x 4 + a 4 x 6 + a 4 x 7 + a 4 x 8 ) + 2 ( i 1 i 8 ) ( a 8 x 4 + a 8 x 8 + a 8 x 12 + a 8 x 16 )

= 2 f ( x 1 , , x 16 , y ) + 2 ( a 1 + a 3 a 2 ) x 1 + 2 ( a 1 + 2 a 3 + a 5 a 2 a 4 ) x 2 + 2 ( a 1 + 2 a 3 + 2 a 5 + a 7 a 2 a 4 a 6 ) x 3 + 2 ( a 1 + 2 a 3 + 2 a 5 + 2 a 7 + a 9 a 2 a 4 a 6 a 8 ) ) x 4 + 2 ( a 3 + a 5 a 4 ) x 6 + 2 ( a 3 + 2 a 5 + a 7 a 4 a 6 ) x 7 + 2 ( a 3 + 2 a 5 + 2 a 7 + a 9 a 4 a 6 a 8 ) x 8 + 2 ( a 5 + a 7 a 6 ) x 11 + 2 ( a 5 + 2 a 7 + a 9 a 6 a 8 ) x 12 + 2 ( a 7 + a 9 a 9 ) x 16 + 2 ( a 2 + a 6 a 3 a 5 ) y ,

where f ( x 1 , , x 16 , y ) = 1 2 X A f X T , X = ( x 1 , , x 16 , y ) , symmetric matrix A f is

as follows

A f = ( 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