About this Journal Submit a Manuscript Table of Contents
Journal of Applied Mathematics
Volume 2012 (2012), Article ID 925092, 18 pages
http://dx.doi.org/10.1155/2012/925092
Research Article

Zero Triple Product Determined Matrix Algebras

1Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2College of Science, Harbin Engineering University, Harbin 150001, China

Received 9 August 2011; Accepted 20 December 2011

Academic Editor: Xianhua Tang

Copyright © 2012 Hongmei Yao and Baodong Zheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let 𝐴 be an algebra over a commutative unital ring 𝒞. We say that 𝐴 is zero triple product determined if for every 𝒞-module 𝑋 and every trilinear map {,,}, the following holds: if {𝑥,𝑦,𝑧}=0 whenever 𝑥𝑦𝑧=0, then there exists a 𝒞-linear operator 𝑇𝐴3𝑋 such that {𝑥,𝑦,𝑧}=𝑇(𝑥𝑦𝑧) for all 𝑥,𝑦,𝑧𝐴. If the ordinary triple product in the aforementioned definition is replaced by Jordan triple product, then 𝐴 is called zero Jordan triple product determined. This paper mainly shows that matrix algebra 𝑀𝑛(𝐵), 𝑛3, where B is any commutative unital algebra even different from the above mentioned commutative unital algebra 𝒞, is always zero triple product determined, and 𝑀𝑛(𝐹), 𝑛3, where F is any field with ch𝐹2, is also zero Jordan triple product determined.

1. Introduction

Over the last couple of years, several papers characterizing bilinear maps on algebras through their action on elements whose certain product is zero have been written; see [16]. The philosophy in these papers is that certain classical problems concerning linear maps that preserve zero product, Jordan product, commutativity, and so forth can be sometimes effectively solved by considering bilinear maps that preserve certain zero product properties. For example, in [1], in order to determine whether a linear map preserving zero product (resp., zero Jordan product, zero Lie product) is “closed” to a homomorphism, the authors introduced the definitions of zero product (resp., zero Jordan product, zero Lie product) determined algebras. The core idea of these definitions is to answer the aforementioned questions by determining the bilinear maps preserving zero product (resp., zero Jordan product, zero Lie product). Furthermore, as the main task, they gave the positive answer that the matrix algebra 𝑀𝑛(𝐵) of 𝑛×𝑛 matrices over a unital algebra 𝐵 is zero product determined, and under some further restrictions on 𝐵, 𝑀𝑛(𝐵) is still zero Jordan (resp., Lie) product determined.

Meanwhile, in preserver problems, there have appeared many kinds of preserving product forms, such as the forms in papers [710]. Particularly, in [10], there appears a definition of Jordan triple product. This definition has applications not only in Banach algebra (see [11]) but also in generalized inverse of matrices (see [12]).

Inspired by the aforementioned, this gives rise to a question: whether can we generalize the products in paper [1] to more generalized forms, such as triple product and Jordan triple product? And immediately, it follows another interesting preserver problem: whether a linear map preserving zero triple product (resp., zero Jordan triple product) is still “closed” to a homomorphism? In order to answer the aforementioned questions, we introduce the following definitions. Let 𝒞 be a (fixed) commutative unital ring and 𝐴 an algebra over 𝒞. By 𝐴3, we denote the 𝒞-linear span of all elements of the form 𝑥𝑦𝑧, where 𝑥,𝑦,𝑧𝐴. X is a 𝒞-module and we denote by {,,}𝐴×𝐴×𝐴𝑋 a 𝒞-trilinear map. Consider the following conditions:(a)for all 𝑥,𝑦,𝑧𝐴 such that 𝑥𝑦𝑧=0, we have {𝑥,𝑦,𝑧}=0;(b)there exists a 𝒞-linear map 𝑇𝐴3𝑋 such that {𝑥,𝑦,𝑧}=𝑇(𝑥𝑦𝑧) for all 𝑥,𝑦,𝑧𝐴.

Trivially, (b) implies (a). We call that 𝐴 is a zero triple product determined algebra if for every 𝒞-module 𝑋 and every 𝒞-trilinear map {,,}, (a) implies (b). And we say that 𝐴 is a zero Jordan triple product determined algebra if we replace the above triple product by Jordan triple product (𝑥𝑦𝑧=𝑥𝑦𝑧+𝑧𝑦𝑥).

From the previous definitions, it is interesting to examine whether the matrix algebra 𝑀𝑛(𝐵) of 𝑛×𝑛 matrices over a unital algebra 𝐵 is still zero triple product (resp., zero Jordan triple product) determined. If the unital algebra 𝐵 has no further restrictions, this problem will be difficult. Therefore, the purpose of this paper is to characterize the zero triple product (resp., zero Jordan triple product) determined algebra under some additional restrictions on the unital algebra 𝐵. In Section 2, we show that the answer is “yes” for the triple product if 𝐵 is any commutative unital algebra even different from 𝒞. The Jordan triple product case, treated in Section 3, is more difficult; we only show that the matrix algebra 𝑀𝑛(𝐹), where 𝐹 is any field with ch𝐹2 (ch𝐹 stands for the characteristic of a field) is also zero Jordan triple product determined.

Finally, we end this section by giving an equivalent condition of (b) in the previous definition which is more convenient to use:

(b) if 𝑥𝑡,𝑦𝑡,𝑧𝑡A,𝑡=1,,𝑚, are such that 𝑚𝑡=1𝑥𝑡𝑦𝑡𝑧𝑡=0, then 𝑚𝑡=1{𝑥𝑡,𝑦𝑡,𝑧𝑡}=0.

2. Zero Triple Product Determined Matrix Algebras

In this part, we will consider the matrix algebra 𝑀𝑛(𝐵), where 𝐵 is any commutative unital algebra even different from the (fixed) commutative unital ring 𝒞 mentioned in Section 1. By 𝑏𝑒𝑖𝑗, where 𝑏𝐵, we denote the matrix whose (𝑖,𝑗) entry is 𝑏 and all other entries are 0.

Theorem 2.1. Let 𝐵 be a commutative unital algebra, then 𝑀𝑛(𝐵) is a zero triple product determined algebra for every 𝑛3.

Proof. In order to prove that (a) implies (b), we only prove that (a) implies the equivalent condition (b'). Set 𝐴=𝑀𝑛(𝐵), let 𝑋 be a 𝒞-module, and let {,,} be a 𝒞-trilinear map such that for all 𝑥,𝑦,𝑧𝐴, 𝑥𝑦𝑧=0 implies {𝑥,𝑦,𝑧}=0. Throughout the proof, 𝑎,𝑏,and𝑐 denote arbitrary elements in 𝐵 and 𝑖,𝑗,𝑘,𝑙,𝑟,and denote arbitrary indices.
First, since 𝑎𝑒𝑖𝑗𝑏𝑒𝑘𝑙𝑐𝑒𝑟=0 if 𝑗𝑘 or 𝑟𝑙, we get 𝑎𝑒𝑖𝑗,𝑏𝑒𝑘𝑙,𝑐𝑒𝑟=0if𝑗𝑘or𝑟𝑙.(2.1) Second, (𝑎𝑒𝑖𝑗+𝑎𝑏𝑒𝑖𝑘)(𝑏𝑒𝑗𝑙𝑒𝑘𝑙)𝑐𝑒𝑙=0, where 𝑗𝑘, and consequently 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑙,𝑐𝑒𝑙=𝑎𝑏𝑒𝑖𝑘,𝑒𝑘𝑙,𝑐𝑒𝑙,(2.2) replacing 𝑎 by 𝑎𝑏 and 𝑏 by 1 in (2.2), we have 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑙,𝑐𝑒𝑙=𝑎𝑏𝑒𝑖𝑗,𝑒𝑗𝑙,𝑐𝑒𝑙,(2.3) and using (2.3) and (2.2), this yields that 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑙,𝑐𝑒𝑙=𝑎𝑏𝑒𝑖1,𝑒1𝑙,𝑐𝑒𝑙.(2.4) Similarly, from 𝑎𝑒𝑖𝑗(𝑏𝑒𝑗𝑙+𝑏𝑐𝑒𝑗𝑘)(𝑐𝑒𝑙𝑒k)=0, where 𝑙𝑘, we obtain 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑙,𝑐𝑒𝑙=𝑎𝑒𝑖𝑗,𝑏𝑐𝑒𝑗1,𝑒1.(2.5) Finally, combining (2.5) and (2.4), it follows that 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑙,𝑐𝑒𝑙=𝑎𝑒𝑖𝑗,𝑏𝑐𝑒𝑗1,𝑒1=𝑎𝑏𝑐𝑒𝑖1,𝑒11,𝑒1.(2.6) Let 𝑥𝑡,𝑦𝑡,𝑧𝑡𝐴 be such that 𝑚𝑡=1𝑥𝑡𝑦𝑡𝑧𝑡=0; then we only need to show 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=0.(2.7) Writing 𝑥𝑡=𝑛𝑛𝑖=1𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑦𝑡=𝑛𝑛𝑘=1𝑙=1𝑏𝑡𝑘𝑙𝑒𝑘𝑙,𝑧𝑡=𝑛𝑛𝑟=1=1𝑐𝑡𝑟𝑒𝑟,(2.8) it follows, by examining the (𝑖,) entry of 𝑚𝑡=1𝑥𝑡𝑦𝑡𝑧𝑡, that for all 𝑖 and , we have 𝑚𝑛𝑡=1𝑛𝑗=1𝑙=1𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙=0.(2.9) Note that 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=𝑚𝑡=1𝑛𝑛𝑖=1𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑛𝑛𝑘=1𝑙=1𝑏𝑡𝑘𝑙𝑒𝑘𝑙,𝑛𝑛𝑟=1=1𝑐𝑡𝑟𝑒𝑟,(2.10) by (2.1), this summation reduces to 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙=1=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑙𝑒𝑗𝑙,𝑐𝑡𝑙𝑒𝑙,(2.11) and using (2.6) and (2.9), we obtain 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙=1=1𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑒𝑖1,𝑒11,𝑒1=𝑛𝑛𝑖=1=1𝑚𝑛𝑡=1𝑛𝑗=1𝑙=1𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑒𝑖1,𝑒11,𝑒1=0.(2.12) Therefore, the result of this theorem holds.

3. Zero Jordan Triple Product Determined Matrix Algebras

In this part, we only consider the matrix algebra 𝑀𝑛(𝐹), where 𝐹 is a field with ch𝐹2, and 𝒞 is still the (fixed) commutative unital ring mentioned in Section 1. Let 𝑏𝑒𝑖𝑗, where 𝑏𝐹, be the matrix whose (𝑖,𝑗) entry is 𝑏 and all other entries are 0.

Theorem 3.1. Let 𝐹 be a field with ch𝐹2; then 𝑀𝑛(𝐹) is a zero Jordan triple product determined algebra for every 𝑛3.

Proof. Set 𝐴=𝑀𝑛(𝐹), let 𝑋 be a 𝒞-module, and let {,,} be a 𝒞-trilinear map such that for all 𝑥,𝑦,𝑧𝐴, 𝑥𝑦𝑧=0 implies {𝑥,𝑦,𝑧}=0. Let 𝑎,𝑏,and𝑐 be arbitrary elements from 𝐹 and 𝑖,𝑗,𝑘,𝑙,𝑟,and denote arbitrary indices.
First, for 𝑗𝑘,𝑘or𝑗𝑘,𝑙𝑖or𝑙𝑟,𝑘or𝑙𝑟,and𝑙𝑖, we have 𝑎𝑒𝑖𝑗𝑏𝑒𝑘𝑙𝑐𝑒𝑟=0, and so 𝑎𝑒𝑖𝑗,𝑏𝑒𝑘𝑙,𝑐𝑒𝑟=0if𝑗𝑘,𝑘or𝑗𝑘,𝑙𝑖or𝑙𝑟,𝑘or𝑙𝑟,𝑙𝑖.(3.1) For any 𝑧A and 𝑖𝑘,𝑗𝑘, we have (𝑎𝑒𝑖𝑗+𝑎𝑏𝑒𝑖𝑘)(𝑏𝑒𝑗𝑘𝑒𝑘𝑘)𝑧=0 and 𝑧(𝑎𝑒𝑖𝑗+𝑎𝑏𝑒𝑖𝑘)(𝑏𝑒𝑗𝑘𝑒𝑘𝑘)=0, which implies 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑘=,𝑧𝑎𝑏𝑒𝑖𝑘,𝑒𝑘𝑘,𝑧if𝑖𝑘,𝑗𝑘,(3.2)𝑧,𝑎𝑒i𝑗,𝑏𝑒𝑗𝑘=𝑧,𝑎𝑏𝑒𝑖𝑘,𝑒𝑘𝑘if𝑖𝑘,𝑗𝑘.(3.3) Similarly, 𝑏𝑒𝑗𝑘,𝑎𝑒𝑖𝑗=𝑒,𝑧𝑘𝑘,𝑎𝑏𝑒𝑖𝑘,𝑧if𝑖𝑘,𝑗𝑘,(3.4)𝑧,𝑏𝑒𝑗𝑘,𝑎𝑒𝑖𝑗=𝑧,𝑒𝑘𝑘,𝑎𝑏𝑒𝑖𝑘if𝑖𝑘,𝑗𝑘.(3.5) For any 𝑧𝐴 and 𝑖𝑘, we have (𝑎𝑒𝑖𝑘𝑒𝑖𝑖)(𝑎𝑏𝑒𝑖𝑘+𝑏𝑒𝑘𝑘)𝑧=0, (𝑎𝑒𝑖𝑘𝑒𝑘𝑘)(𝑎𝑏𝑒𝑖𝑘+𝑏𝑒𝑖𝑖)𝑧=0, 𝑧(𝑎𝑒𝑖𝑘𝑒𝑖𝑖)(𝑎𝑏𝑒𝑖𝑘+𝑏𝑒𝑘𝑘)=0, and 𝑧(𝑎𝑒𝑖𝑘𝑒𝑘𝑘)(𝑎𝑏𝑒𝑖𝑘+𝑏𝑒𝑖𝑖)=0, and consequently, 𝑎𝑒𝑖𝑘,𝑏𝑒𝑘k=𝑒,𝑧𝑖𝑖,𝑎𝑏𝑒𝑖𝑘,𝑧if𝑖𝑘,(3.6)𝑎𝑒𝑖𝑘,𝑏𝑒𝑖𝑖=𝑒,𝑧𝑘𝑘,𝑎𝑏𝑒𝑖𝑘,𝑧if𝑖𝑘,(3.7)𝑧,𝑎𝑒𝑖𝑘,𝑏𝑒𝑘𝑘=𝑧,𝑒𝑖𝑖,𝑎𝑏𝑒𝑖𝑘if𝑖𝑘,(3.8)𝑧,𝑎𝑒𝑖𝑘,𝑏𝑒𝑖𝑖=𝑧,𝑒𝑘𝑘,𝑎𝑏𝑒𝑖𝑘if𝑖𝑘.(3.9) Similarly, 𝑏𝑒𝑘𝑘,𝑎𝑒𝑖𝑘=,𝑧𝑎𝑏𝑒𝑖𝑘,𝑒𝑖𝑖,𝑧if𝑖𝑘,(3.10)𝑏𝑒𝑖𝑖,𝑎𝑒𝑖𝑘=,𝑧𝑎𝑏𝑒𝑖𝑘,𝑒𝑘𝑘,𝑧if𝑖𝑘,(3.11)𝑧,𝑏𝑒𝑘𝑘,𝑎𝑒𝑖𝑘=𝑧,𝑎𝑏𝑒𝑖𝑘,𝑒𝑖𝑖if𝑖𝑘,(3.12)𝑧,𝑏𝑒𝑖𝑖,𝑎𝑒𝑖𝑘=𝑧,𝑎𝑏𝑒𝑖𝑘,𝑒𝑘𝑘if𝑖𝑘.(3.13) Since 𝑎𝑒𝑖(𝑒𝑖𝑖𝑒)(𝑒𝑖𝑖+𝑒)=0 and (𝑒𝑖𝑖𝑒)(𝑒𝑖𝑖+𝑒)𝑎𝑒𝑖=0, if 𝑖, it follows that 𝑎𝑒𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖=𝑎𝑒𝑖,𝑒,𝑒𝑒if𝑖,(3.14)𝑖𝑖,𝑒𝑖𝑖,𝑎𝑒𝑖=𝑒,𝑒,𝑎𝑒𝑖if𝑖.(3.15) Using (3.8), (3.11), (3.15), and (3.14), we arrive at 𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑒𝑖=𝑎𝑒𝑖,𝑒,𝑒=𝑒,𝑒,𝑎𝑒𝑖=𝑎𝑒𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖if𝑖.(3.16) Using (3.7) and (3.12), we have 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑖,𝑒𝑗𝑗=𝑎𝑏𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗if𝑖𝑗.(3.17) Since (𝑎𝑒𝑖𝑗+𝑏𝑒𝑗𝑖)(𝑎𝑒𝑖𝑗𝑏𝑒𝑗𝑖)𝑒𝑖𝑖=0 if 𝑖𝑗, then 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑖,𝑒𝑖𝑖=𝑏𝑒𝑗𝑖,𝑎𝑒i𝑗,𝑒𝑖𝑖if𝑖𝑗.(3.18) Using (3.18) and (3.17), it follows that 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑖,𝑒𝑖𝑖=𝑎𝑏𝑒𝑗𝑖,𝑒𝑖𝑗,𝑒𝑖𝑖if𝑖𝑗.(3.19) Using (3.6), (3.8), (3.18), and (3.17), we obtain 𝑒𝑖𝑖,𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑖=𝑏𝑎𝑒𝑗𝑖,𝑒𝑖𝑗,𝑒𝑖𝑖=𝑒𝑖𝑗,𝑏𝑎𝑒𝑗𝑖,𝑒𝑖𝑖if𝑖𝑗.(3.20) Similarly, using (3.10), (3.9), and (3.17), this yields 𝑒𝑖𝑖,𝑎𝑒𝑗𝑖,𝑏𝑒𝑖𝑗=𝑎𝑏𝑒𝑗𝑖,𝑒𝑖𝑗𝑒𝑖𝑖if𝑖𝑗.(3.21)
Further, we claim that 𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖=12𝑎𝑏𝑐𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖.(3.22) If 𝑖𝑗, then ((1/2)𝑎𝑏𝑒𝑖𝑖+𝑎𝑒𝑖𝑗(1/2)𝑎𝑏𝑒𝑗𝑗)(𝑏𝑒𝑗𝑖𝑒𝑖𝑖+𝑒𝑗𝑗)(𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗)=0; consequently, 12𝑎𝑏𝑒𝑖𝑖+𝑎𝑒𝑖𝑗12𝑎𝑏𝑒𝑗𝑗,𝑏𝑒𝑗𝑖𝑒𝑖𝑖+𝑒𝑗𝑗,𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗=0,if𝑖𝑗.(3.23) Using (3.8)–(3.13) and (3.16), we get 𝑎𝑒𝑖𝑗,𝑏𝑒𝑗𝑖,𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗=12𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+12𝑎𝑏𝑒𝑗𝑗,𝑒𝑗𝑗,𝑐𝑒𝑗𝑗,if𝑖𝑗.(3.24) From (𝑎𝑒𝑖𝑖+𝑏𝑒𝑗𝑖𝑏𝑒𝑖𝑗𝑎𝑒𝑗𝑗)(𝑏𝑒𝑖𝑖𝑎𝑒𝑖𝑗+𝑎𝑒𝑗𝑖𝑏𝑒𝑗𝑗)(𝑐𝑒𝑖𝑖+c𝑒𝑗𝑗)=0, where 𝑖𝑗, it follows that 𝑎𝑒𝑖𝑖+𝑏𝑒𝑗𝑖𝑏𝑒𝑖𝑗𝑎𝑒𝑗𝑗,𝑏𝑒𝑖𝑖𝑎𝑒𝑖𝑗+𝑎𝑒𝑗𝑖𝑏𝑒𝑗𝑗,𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗=0.(3.25) Using (3.3), (3.4), (3.6)–(3.13), and (3.16), we arrive at 𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑒𝑗𝑗,𝑏𝑒𝑗𝑗,𝑐𝑒𝑗𝑗=𝑏𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗+𝑏𝑒𝑗𝑖,𝑎𝑒𝑖𝑗,𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗.(3.26) Then, by (3.24), we have 𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑏𝑒𝑗𝑗,𝑒𝑗𝑗,𝑐𝑒𝑗𝑗=𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑒𝑗𝑗,𝑏𝑒𝑗𝑗,𝑐𝑒𝑗𝑗,where𝑖𝑗.(3.27) Since 𝑛3, we choose 𝑘 such that 𝑘𝑖,𝑗; applying (3.27), we get 𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑒𝑗𝑗,𝑏𝑒𝑗𝑗,𝑐𝑒𝑗𝑗+𝑎𝑒𝑘𝑘,𝑏𝑒𝑘𝑘,𝑐𝑒𝑘𝑘+𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖=𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑏𝑒𝑗𝑗,𝑒𝑗𝑗,𝑐𝑒𝑗𝑗+𝑎𝑏𝑒𝑘𝑘,𝑒𝑘𝑘,𝑐𝑒𝑘𝑘+𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖,𝑐𝑒𝑖𝑖=2𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖,𝑐𝑒𝑖𝑖+𝑎𝑒𝑘𝑘,𝑏𝑒𝑘𝑘,𝑐𝑒𝑘𝑘+𝑎𝑒𝑗𝑗,𝑏𝑒𝑗𝑗,𝑐𝑒𝑗𝑗,(3.28) then 𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑐𝑒𝑖𝑖=𝑎𝑏𝑒𝑖𝑖,𝑒𝑖𝑖𝑐𝑒𝑖𝑖.(3.29) Similarly, by (𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗)(𝑏𝑒𝑗𝑖𝑒𝑖𝑖+𝑒𝑗𝑗)((1/2)𝑎𝑏𝑒𝑖𝑖+𝑎𝑒𝑖𝑗(1/2)𝑎𝑏𝑒𝑗𝑗)=0 and (𝑐𝑒𝑖𝑖+𝑐𝑒𝑗𝑗)(𝑏𝑒𝑖𝑖𝑎𝑒𝑖𝑗+𝑎𝑒𝑗𝑖𝑏𝑒𝑗𝑗)(𝑎𝑒𝑖𝑖+𝑏𝑒𝑗𝑖𝑏𝑒𝑖𝑗𝑎𝑒𝑗𝑗)=0, where 𝑖𝑗, we get 𝑐𝑒𝑖𝑖,𝑏𝑒𝑖𝑖,𝑎𝑒𝑖𝑖=𝑐𝑒𝑖𝑖,𝑏𝑎𝑒𝑖𝑖,𝑒𝑖𝑖.(3.30) Then, combining (3.29) and (3.30), we have {𝑐𝑒𝑖𝑖,𝑎𝑒𝑖𝑖,𝑏𝑒𝑖𝑖}={𝑎𝑏𝑐𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖}. Hence, the claim (3.22) holds.
Finally, we claim that 𝑎𝑒𝑖𝑗,𝑏e𝑗𝑖,𝑐𝑒𝑖𝑗=𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑏𝑐𝑒𝑖𝑗,where𝑖𝑗.(3.31) For 𝑖𝑗, we have ((1/2)𝑏𝑐𝑒𝑖𝑗+𝑏1𝑐𝑒𝑗𝑖+𝑐𝑒𝑗𝑗)((1/2)𝑎𝑏𝑒𝑖𝑖+(1/2)𝑎𝑏𝑒𝑗𝑗+𝑎𝑒𝑗𝑖)(𝑒𝑖𝑖𝑒𝑗𝑗+𝑏𝑒𝑖𝑗)=0, and consequently {(1/2)𝑏𝑐𝑒𝑖𝑗+𝑏1𝑐𝑒𝑗𝑖+𝑐𝑒𝑗𝑗,(1/2)𝑎𝑏𝑒𝑖𝑖+(1/2)𝑎𝑏𝑒𝑗𝑗+𝑎𝑒𝑗𝑖,𝑒𝑖𝑖𝑒𝑗𝑗+𝑏𝑒𝑖𝑗}=0. Using (3.8)–(3.13), (3.16), (3.17), and (3.19), this can be reduced to 𝑎𝑏2𝑐𝑒𝑖𝑗,𝑒𝑖𝑖,𝑒𝑖𝑖+𝑎𝑏𝑐𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗=𝑐𝑒𝑗𝑗,12𝑎𝑏𝑒𝑗𝑗,𝑒𝑗𝑗+12𝑏𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑏𝑒𝑖𝑗.(3.32) Replacing 𝑏 by 1 and 2 in (3.32), respectively, we have 𝑎𝑐𝑒𝑖𝑗,𝑒𝑖𝑖,𝑒𝑖𝑖+𝑎𝑐𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗=𝑐𝑒𝑗𝑗,12𝑎𝑒𝑗𝑗,𝑒𝑗𝑗+12𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑒𝑖𝑗,(3.33)4𝑎𝑐𝑒𝑖𝑗,𝑒𝑖𝑖,𝑒𝑖𝑖+2a𝑐𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗=𝑐𝑒𝑗𝑗,𝑎𝑒𝑗𝑗,𝑒𝑗𝑗+𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,2𝑒𝑖𝑗.(3.34) Computing (3.34)–2(3.33), this yields 2𝑎𝑐𝑒𝑖𝑗,𝑒𝑖𝑖,𝑒𝑖𝑖=𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑒𝑖𝑗,(3.35) then taking (3.35) into (3.33), we derive 𝑎𝑐𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗=𝑐𝑒𝑗𝑗,12𝑎𝑒𝑗𝑗,𝑒𝑗𝑗,(3.36) replacing 𝑎 by 𝑎𝑏 in (3.36), we get 𝑎𝑏𝑐𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗=𝑐𝑒𝑗𝑗,12𝑎𝑏𝑒𝑗𝑗,𝑒𝑗𝑗,(3.37) then taking (3.37) into (3.32), it follows that 𝑎𝑏2𝑐𝑒𝑖𝑗,𝑒𝑖𝑖,𝑒𝑖𝑖=12𝑏𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑏𝑒𝑖𝑗,(3.38) and replacing 𝑏 by 𝑏+1 in (3.38), and using (3.38), (3.35), and (3.16), we get 𝑐𝑒𝑖𝑗,𝑎𝑒𝑗𝑖,𝑏𝑒𝑖𝑗=𝑒𝑖𝑖,𝑒𝑖𝑖,2𝑎𝑏𝑐𝑒𝑖𝑗=𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑏𝑐𝑒𝑖𝑗,where𝑖𝑗.(3.39) Therefore, (3.31) holds.
Let 𝑥𝑡,𝑦𝑡,and𝑧𝑡 be such that 𝑚𝑡=1𝑥𝑡𝑦𝑡𝑧𝑡=0; we have to prove that 𝑚𝑡=1{𝑥𝑡,𝑦𝑡,𝑧𝑡}=0. Writing 𝑥𝑡=𝑛𝑛𝑖=1𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑦𝑡=𝑛𝑛𝑘=1𝑙=1𝑏𝑡𝑘𝑙𝑒𝑘𝑙,𝑧𝑡=𝑛𝑛𝑟=1=1𝑐𝑡𝑟𝑒𝑟,(3.40) it follows, by examining the (𝑖,) entry of 𝑚𝑡=1𝑥𝑡𝑦𝑡𝑧𝑡=0, that for all 𝑖,, we have 𝑚𝑛𝑡=1𝑛𝑗=1𝑙=1𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙+𝑐𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑎𝑡𝑙=0.(3.41) First, by (3.1), we get 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑘𝑗𝑟=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘+(3.42)𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙=1=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑙𝑒𝑗𝑙,𝑐𝑡𝑙𝑒𝑙+(3.43)𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑟𝑖𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑟𝑗𝑒𝑟𝑗.(3.44) The summation (3.42) can be written as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟=1𝑎𝑡𝑖𝑖𝑒𝑖𝑖,𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘,(3.45) using (3.4), (3.5), and (3.21), the first summation can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟=1𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑒𝑘𝑗,𝑐𝑡𝑟𝑘𝑒𝑟𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟𝑗𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑒𝑘𝑗,𝑐𝑡𝑟𝑘𝑒𝑟𝑘+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑘𝑗𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑒𝑘𝑗,𝑐𝑡𝑗𝑘𝑒𝑗𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑘𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑐𝑡𝑗𝑘𝑒𝑘𝑗,𝑒𝑗𝑘,𝑒𝑗𝑗,(3.46) and using (3.8), (3.12), (3.9), (3.10), (3.4), (3.17), and (3.18), we split the second summation in three parts: 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟=1𝑎𝑡𝑖𝑖𝑒𝑖𝑖,𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟=1𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑒𝑘𝑘,𝑐𝑡𝑟𝑘𝑒𝑟𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑒𝑘𝑘,𝑐𝑡𝑟𝑘𝑒𝑟𝑘+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑒𝑘𝑘,𝑐𝑡𝑘𝑘𝑒𝑘𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘,𝑒𝑟𝑟+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑘𝑘𝑒𝑘𝑘=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘,𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑟𝑘𝑒𝑟𝑘,𝑒𝑟r+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑒𝑘𝑖,𝑐𝑡𝑖𝑘𝑒𝑖𝑘,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑘𝑘𝑒𝑘𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘,𝑖𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑖,𝑒𝑟𝑟+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑖𝑘𝑒𝑘𝑖,𝑒𝑖𝑘,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑘𝑘𝑒𝑘𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘,𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑘,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑖𝑘𝑒𝑘𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡i𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑘𝑘𝑒𝑘𝑖.(3.47) Therefore, the summation (3.42) can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑘𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑐𝑡𝑗𝑘𝑒𝑘𝑗,𝑒𝑗𝑘,𝑒𝑗𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑘𝑖𝑟𝑘,𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑘,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑖𝑘𝑒𝑘𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑘𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑘𝑖𝑐𝑡𝑘𝑘𝑒𝑘𝑖.(3.48) The summation (3.43) can be written as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑙𝑒𝑗𝑙,𝑐𝑡𝑙𝑒𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑗𝑒𝑗𝑗,𝑐𝑡𝑗𝑒𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑒𝑖,(3.49) using (3.2), (3.10), (3.8), (3.16), (3.17), (3.18), and (3.6), the first summation can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑙𝑒𝑗𝑙,𝑐𝑡𝑙𝑒𝑙=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗=1𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙e𝑖𝑙,𝑒𝑙𝑙,𝑐𝑡𝑙𝑒𝑙=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑒𝑙𝑙,𝑐𝑡𝑙𝑒𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑒𝑙𝑙,𝑐𝑡𝑙𝑙𝑒𝑙𝑙=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑐𝑡𝑙𝑒𝑙,𝑒+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑐𝑡𝑙𝑙𝑒𝑙𝑙=𝑚𝑛𝑡=1i𝑛=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙,𝑖𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑐𝑡𝑙𝑒𝑙,𝑒+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑒𝑖𝑙,𝑐𝑡𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙,𝑖𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑒𝑖,𝑒,𝑒+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑙,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙,𝑖,𝑗𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑙,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑙𝑐𝑡𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑙,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙c𝑡𝑙𝑙𝑒𝑖𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙,(3.50) and using (3.6), (3.3), (3.8), (3.16), and (3.20), we rewrite the second summation as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑗𝑒𝑗𝑗,𝑐𝑡𝑗𝑒𝑗=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖=1𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑒𝑖𝑗,𝑐𝑡𝑗𝑒𝑗=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑖𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑒𝑖𝑗,𝑐𝑡𝑗𝑒𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑒𝑖𝑗,𝑐𝑡𝑗𝑖𝑒𝑗𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑖,𝑗𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑒𝑖𝑗,𝑐𝑡𝑗𝑒𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑎𝑡𝑖j𝑏𝑡𝑗𝑗𝑒𝑖𝑗,𝑐𝑡𝑗𝑗𝑒𝑗𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑖𝑒𝑗𝑖,𝑒𝑖𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑖,𝑗𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑒𝑖,𝑒+𝑚𝑛𝑡=1i𝑛=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑖𝑒𝑗𝑖,𝑒𝑖𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑖,𝑗𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑖𝑒𝑗𝑖,𝑒𝑖𝑖,(3.51) and then using (3.3), (3.2), (3.8), (3.6), (3.20), (3.22), and (3.31), the third summation is split in four parts: 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑒𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑖𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗=1𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑖𝑒𝑖𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑖,𝑗𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑖𝑖𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑖=1𝑎𝑡𝑖𝑖𝑒𝑖𝑖,𝑏𝑡𝑖𝑖𝑒𝑖𝑖,𝑐𝑡𝑖𝑖𝑒𝑖𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑖,𝑗𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑒𝑗,𝑒+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑖𝑒𝑗𝑖+𝑚𝑛𝑡=1𝑖=112𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡𝑖𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑒𝑖,𝑒,𝑒+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡j𝑖𝑐𝑡𝑖𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑖𝑒𝑗𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑖=112𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡𝑖𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖.(3.52) Hence, the summation (3.43) can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙𝑖,𝑗𝑙,𝑖,𝑗𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑙,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑙𝑐t𝑙𝑖𝑒𝑙𝑖,𝑒𝑖𝑙,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑙𝑐𝑡𝑙𝑙𝑒𝑖𝑙+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑖,𝑗𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑖𝑒𝑗𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑖,𝑗𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑒𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑗𝑒𝑖𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑖𝑒𝑗𝑖,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑖=112𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡𝑖𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖.(3.53) Finally, using (3.5), (3.7), (3.8), (3.10), (3.12), (3.16), and (3.21), the summation (3.44) can be written as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑟𝑖𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑒𝑗𝑖,𝑐𝑡𝑟𝑗𝑒𝑟𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑟𝑖𝑎𝑡𝑖𝑖𝑒𝑖𝑖,𝑏𝑡𝑖𝑖𝑒𝑖𝑖,𝑐𝑡𝑟𝑖𝑒𝑟𝑖=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑟𝑖𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑐𝑡𝑟𝑗𝑒𝑟𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑟𝑖𝑎𝑡𝑖𝑖𝑒𝑖𝑖,𝑏𝑡𝑖𝑖𝑐𝑡𝑟𝑖𝑒𝑟𝑖,𝑒𝑟𝑟=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑟𝑖,𝑗𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑐𝑡𝑟𝑗𝑒𝑟𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑒𝑖𝑗,𝑏𝑡𝑗𝑖𝑐𝑡𝑗𝑗𝑒𝑗𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑟𝑖𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡r𝑖𝑒𝑟𝑖,𝑒𝑟𝑟,𝑒𝑟𝑟=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑟𝑖,𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑟𝑗𝑒𝑟𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑗𝑗𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑟𝑖𝑒𝑟𝑟,𝑒𝑟𝑟,𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡𝑟𝑖𝑒𝑟𝑖.(3.54) Consequently, rewriting the summation (3.44) as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑟𝑖,𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑟𝑗𝑒𝑟𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖c𝑡𝑗𝑗𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗+𝑚𝑛𝑡=1𝑛𝑖=1𝑟𝑖𝑒𝑟𝑟,𝑒𝑟𝑟,𝑎𝑡𝑖𝑖𝑏𝑡𝑖𝑖𝑐𝑡𝑟𝑖𝑒𝑟𝑖,(3.55) therefore, we get 𝑚𝑡=1𝑥𝑡,𝑦𝑡,𝑧𝑡=(3.47)+(3.52)+(3.54).(3.56)
Next, we only need to prove that (3.47)+(3.52)+(3.54)=0.
Replacing the indices 𝑟,𝑘,𝑖,and𝑗 by 𝑖,𝑗,𝑙,and, correspondingly, in the first summation of (3.48), and rewriting the summation as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑛𝑘𝑗𝑟𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑘𝑖𝑐𝑡𝑟𝑘𝑒𝑟𝑗=𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙=1𝑖,𝑗,𝑙𝑒,𝑒,𝑐𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑎𝑡𝑙𝑒𝑖,(3.57) then adding (3.57) and the first, the seventh, the tenth summations of (3.53), together, this yields 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑛𝑙=1𝑖,𝑗,𝑙𝑒,𝑒,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙+𝑐𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑎𝑡𝑙𝑒𝑖.(3.58) Replacing 𝑖and𝑘 by 𝑙and𝑖, respectively, in the fifth summation of (3.48), then by (3.16), this summation is further equal to 𝑚𝑛𝑡=1𝑛𝑖=1𝑙𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑙𝑙𝑏𝑡𝑖𝑙𝑐𝑡𝑖𝑖𝑒𝑖𝑙,(3.59) and replacing 𝑟,𝑘,and𝑖 by 𝑖,𝑗,and𝑙, correspondingly, in the third summation of (3.48), then using (3.16), the summation is further equal to 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑙𝑙𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑙.(3.60) Then adding (3.59), (3.60), and the fifth and the sixth summations of (3.53), together, this yields 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗=1𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑙+𝑎𝑡𝑙𝑙𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑙.(3.61) Replacing 𝑘 by 𝑗 in the fourth summation of (3.48), then adding the 12th summation of (3.53), we get 𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑖+𝑎𝑡𝑖𝑖𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑖.(3.62) Replacing 𝑖and𝑟 by 𝑗and𝑖, correspondingly, in the third summation of (3.55), then adding the 8th summation of (3.53), we have 𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑗+𝑎𝑡𝑗𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑖𝑗𝑒𝑖𝑗.(3.63) Replacing 𝑟,𝑗,and𝑖 by 𝑖,𝑗,and𝑙, respectively, in the first summation of (3.55), and rewriting this summation as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑗𝑗,𝑒𝑗𝑗,𝑎𝑡𝑙𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑗,(3.64) then adding (3.64) and the second summation of (3.53), together, and by (3.16), we derive 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑗+𝑎𝑡𝑙𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑗.(3.65) Replacing 𝑖and𝑙 by 𝑗and𝑖, respectively, in the fourth summation of (3.53), then adding the second summation of (3.55), we arrive at 𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑖j𝑏𝑡𝑗𝑖𝑐𝑡𝑗𝑗+𝑎𝑡𝑗𝑗𝑏𝑡𝑗𝑖𝑐𝑡𝑖𝑗𝑒𝑖𝑗,𝑒𝑗𝑖,𝑒𝑗𝑗.(3.66) Replacing 𝑖,𝑗,and𝑘 by 𝑙,𝑖,and𝑗, respectively, in the second summation of (3.48), and rewriting this summation as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑎𝑡𝑙𝑖𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑗,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑎𝑡𝑗𝑖𝑏𝑡𝑗𝑗𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑗,𝑒𝑖𝑖,(3.67) then using (3.19) and (3.18), the aforementioned summation can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗𝑎𝑡𝑙𝑖𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑗,𝑒𝑖𝑖+𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑗𝑖𝑏𝑡𝑗𝑗𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑖.(3.68) Adding the second summation of (3.68) and the ninth summation of (3.53), together, we get 𝑚𝑛𝑡=1𝑛𝑖=1𝑗𝑖𝑒𝑖𝑗,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑗𝑐𝑡𝑗𝑖+𝑎𝑡𝑗𝑖𝑏𝑡𝑗𝑗𝑐𝑡𝑖𝑗𝑒𝑗𝑖,𝑒𝑖𝑖.(3.69) By (3.36), we know that the first summation of (3.68) can be written as 𝑚𝑛𝑡=1𝑖=1n𝑛𝑗𝑖𝑙𝑖,𝑗12𝑎𝑡𝑙𝑖𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖,(3.70) similarly, using (3.36), the third summation of (3.53) can be written as 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗12𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑖𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖,(3.71) and then adding (3.70) and (3.71), together, we have 𝑚𝑛𝑡=1𝑛𝑖=1𝑛𝑗𝑖𝑙𝑖,𝑗12𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑖+𝑎𝑡𝑙𝑖𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑖,𝑒𝑖𝑖,𝑒𝑖𝑖.(3.72) Therefore, we derive +(3.47)+(3.52)+(3.54)=(3.52)+(3.60)+(3.62)+(3.64)+the11thsummationof(47)(3.61)+(3.65)+(3.68)+(3.71)+the13thsummationof(3.52).(3.73)
Our first goal is to show (3.52)+(3.60)+(3.62)+(3.64)+the11thsummationof(3.52)=0.(3.74) First, notice that by (3.16), the summation (3.61) can be rewritten as 𝑚𝑛𝑡=1𝑛𝑖𝑙𝑛𝑗𝑙𝑙=1𝑒𝑙𝑙,𝑒𝑙𝑙,𝑎𝑡𝑖𝑗𝑏𝑡𝑗𝑙𝑐𝑡𝑙𝑙+𝑎𝑡𝑙𝑙𝑏𝑡𝑗𝑙𝑐𝑡𝑖𝑗𝑒𝑖𝑙.(3.75) Then replacing 𝑙 by , we have 𝑚𝑛𝑡=1𝑛𝑖