Weak Faddeev-Takhtajan-Volkov algebras; Lattice $W_n$ algebras

TL;DR

This paper explores the construction of lattice $W_n$ algebras using Poisson brackets, universal variables, and computational tools, advancing the understanding of algebraic structures in mathematical physics.

Contribution

It introduces a new Poisson bracket on $sl_2$, constructs lattice $W_n$ algebras systematically, and provides computational methods for their analysis.

Findings

Construction of a new Poisson bracket on $sl_2$

Development of lattice $W_n$ algebra structures

Implementation of Mathematica code for algebra analysis

Abstract

In this paper, we will start by looking through our project's historical general view and then we will try to construct a new Poisson bracket on our simplest example $sl_2$ and then we will try to give a universal construction based on our universal variables and then will try to construct lattice $W_2$ algebras which will play a key role in our other constructions on lattice $W_3$ algebras and finally we will try to find the only nontrivial dependent generator of our lattice $W_4$ algebras and so on for lattice $W_n$ algebras. And at the end of this paper, we will have appendix A, which will contain some parts of the Mathematica coding which we have used and have made for to find our algebra structures.

Equations

$$U_q(n)\hookrightarrow U_q(n)\bar{\otimes }{\mathbb{C}}_q[X_l^{jl}]\hookrightarrow U_q(n)\bar{\otimes }{\mathbb{C}}_q[X_l^{jl}]\bar{\otimes }{\mathbb{C}}_q[X_m^{jm}]$$

$$\{\begin{array}{ll}\{X_i,X_j\}:=2X_i{X}_j& \text{if }i<j\\ \{X_i,X_i\}:=0& \end{array}$$

$$\{\begin{array}{l}{\mathfrak{D}}_x^{(n)}◃{\tau }_1=0\\ H_x^{(n)}◃{\tau }_1=0\end{array}$$

$$⋮⋮⋮⋮⋮⋮$$

$$\cdots X_1^{(11)}{X}_1^{(21)}{X}_1^{(31)}{X}_1^{(41)}\cdots$$

$$\cdots X_2^{(12)}{X}_2^{(22)}{X}_2^{(32)}{X}_2^{(42)}\cdots$$

$$\cdots X_3^{(13)}{X}_3^{(23)}{X}_3^{(33)}{X}_3^{(43)}\cdots$$

$$\cdots X_4^{(14)}{X}_4^{(24)}{X}_4^{(34)}{X}_4^{(44)}\cdots$$

$$⋮⋮⋮⋮⋮⋮$$

$$\{S_{X_i^{ji}},{\tau }_1{\}}_p=S_{X_i^{ji}}{\tau }_1-p^{\text{deg}{\tau }_1<{\alpha }_i,{\alpha }_j>}{\tau }_1{S}_{X_i^{ji}}$$

$${\tau }_2={\tau }_1[X_1^{(11)}\to X_1^{(21)},X_1^{(21)}\to X_1^{(31)},\cdots ],$$

$${\tau }_3={\tau }_2[X_1^{(21)}\to X_1^{(31)},X_1^{(31)}\to X_1^{(41)},\cdots ]$$

$$⋮$$

$${\tau }_i:={\text{Inv}}_{U_q(n_{+})}({\mathbb{C}}_q[X_i^{ji}|i\in \mathbb{Z}])$$

$$H_{X_i^{ji}}({\tau }_i)=0\text{and}{D}_{X_i^{ji}}({\tau }_i)=0$$

$$=(D_{X_1}+D_{X_2}+D_{X_3}){\tau }_1$$

$$(\text{ref}Equ7)=\sum _j(\{X_1,X_j^n\}+\{X_2,X_j^n\}+\{X_3,X_j^n\})$$

$$D_X^{(2)}=X_1(X_1+2X_2+2X_3)\frac{\partial }{\partial X_1}+X_2(X_2+2X_3)\frac{\partial }{\partial X_2}+X_3^2\frac{\partial }{\partial X_3}$$

$$H_X^{(2)}:=\sum _i{X}_i\frac{\partial }{\partial X_i}$$

$$\{\begin{array}{ll}(X_1(X_1+2X_2+2X_3)\frac{\partial }{\partial X_1}+X_2(X_2+2X_3)\frac{\partial }{\partial X_2}+X_3^2\frac{\partial }{\partial X_3}){\tau }_1[\cdots ,X_1,X_2,X_3,& \\ \cdots ]=0,& \text{mbox}\\ (X_1\frac{\partial }{{\partial }_{X_1}}+X_2\frac{\partial }{{\partial }_{X_2}}+X_3\frac{\partial }{{\partial }_{X_3}}){\tau }_1[\cdots ,X_1,X_2,X_3,\cdots ]=0.& \text{mbox}\end{array}$$

$${\tau }_1^{(2)}[X_1,X_2,X_3]=\frac{(X1+X2)(X2+X3)}{X2(X1+X2+X3)}=\frac{({\sum }_{\begin{array}{c}1\le i_1\le 2\end{array}}{X}_{i_1}^{(1)})({\sum }_{\begin{array}{c}1\le i_1\le 2\end{array}}{X}_{i_1+1}^{(1)})}{X_2^{(1)}({\sum }_{\begin{array}{c}1\le i_1\le 3\end{array}}{X}_{i_1}^{(1)})}.$$

$$\{\begin{array}{l}{\tau }_1^{(2)}[X_1,X_2,X_3]=\frac{({\sum }_{\begin{array}{c}1\le i_1\le 2\end{array}}{X}_{i_1}^{(1)})({\sum }_{\begin{array}{c}1\le i_1\le 2\end{array}}{X}_{i_1+1}^{(1)})}{X_2^{(1)}({\sum }_{\begin{array}{c}1\le i_1\le 3\end{array}}{X}_{i_1}^{(1)})};\\ {\tau }_2^{(2)}[X_2,X_3,X_4]=\frac{({\sum }_{\begin{array}{c}2\le i_1\le 3\end{array}}{X}_{i_1}^{(1)})({\sum }_{\begin{array}{c}2\le i_1\le 3\end{array}}{X}_{i_1+1}^{(1)})}{X_2^{(1)}({\sum }_{\begin{array}{c}2\le i_1\le 4\end{array}}{X}_{i_1}^{(1)})};\\ {\tau }_3^{(2)}[X_3,X_4,X_5]=\frac{({\sum }_{\begin{array}{c}3\le i_1\le 4\end{array}}{X}_{i_1}^{(1)})({\sum }_{\begin{array}{c}3\le i_1\le 4\end{array}}{X}_{i_1+1}^{(1)})}{X_2^{(1)}({\sum }_{\begin{array}{c}3\le i_1\le 5\end{array}}{X}_{i_1}^{(1)})}.\end{array}$$

$$F_j^{(n)}:=\{{\tau }_i^{(n)},{\tau }_j^{(n)}\}=\sum _i\frac{\partial {\tau }_i^{(n)}}{\partial X_i}\sum _j\frac{\partial {\tau }_j^{(n)}}{\partial X_j}\{X_i,X_j\},$$

$$\{\tau_{1}^{(2)},\tau_{2}^{(2)}\}=\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{1}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{1},X_{2}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}\{X_{1},X_{3}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{1},X_{4}\}\Big{)}$$

$$\hskip 48.36958pt+\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{2}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{2},X_{2}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}\{X_{2},X_{3}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{2},X_{4}\}\Big{)}$$

$$\hskip 48.36958pt+\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{3}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{3},X_{2}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}\{X_{3},X_{3}\}+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}\{X_{3},X_{4}\}\Big{)}$$

$$\hskip 48.36958pt=\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{1}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(2X_{1}X_{2})+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}(2X_{1}X_{3})+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(2X_{1}X_{4})\Big{)}$$

$$\hskip 17.07182pt+\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{2}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(0)+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}(2X_{2}X_{3})+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(2X_{2}X_{4})\Big{)}$$

$$\hskip 25.6073pt+\Big{(}\frac{\partial\tau_{1}^{(2)}}{\partial X_{3}}\Big{)}\Big{(}\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(-2X_{3}X_{2})+\frac{\partial\tau_{2}^{(2)}}{\partial X_{3}}(0)+\frac{\partial\tau_{2}^{(2)}}{\partial X_{2}}(2X_{3}X_{4})\Big{)}$$

$$=2\frac{X_1{X}_2^2{X}_3^2{X}_4(X_1+X_2+X_3+X_4)}{(X_1+X_2{)}^2(X_2+X_3{)}^3(X_3+X_4{)}^2}$$

Full text

And at the end of this paper we will have appendix A, which will contain some parts of the Mathematica coding which we have used and have made for to find our algebra structures.

1. Introduction

There is an old problem which has been considered and introduced by Boris Feigin in 1992. It has been born in its new formulation; on quantum Gelfand-Kirillov conjecture; in a public talk at RIMS in 1992 based on the nilpotent part of $U_{q}(g)$ i.e. $U_{q}(\mathfrak{n})$ for $g$ a simple Lie algebra.

Now, this problem is known as “Feigin’s Conjecture”.

In the mentioned talk, Feigin proposed the existence of a certain family of homomorphisms on the quantized enveloping algebra $U_{q}(g)$ which will led us to a deffinition of lattice $W-$algebras.

These “homomorphisms” has been turned to a very useful tool for to study the fraction field of quantized enveloping algebras. [6]

There have been many attempts to construct lattice $W$-algebras in Feigin’s sence, which ensures the simplicity of the construction process of lattice $W$-algebra; for example the best known articles in the subject has been written by Kazuhiro Hikami and Rei Inoue who tried to obtain the algebra structure by using lax operators and generalized R matrices. [7] [8]

Or Alexander Belov and Alexander Antonov and Karen Chaltikian, who first tried to follow Feigin’s construction but finaly they also solved part of the conjecture by getting help of lax operators, and it made it very difficult to follow their publication.[9] [10]

But here, in this paper, we will proceed and will introduce the simplest way of constructing such kind of algebras by just employing Feigin’s homomorphisms and screening operators by defining a Poisson bracket on our variables just based on our Cartan matrix. [1] [2]

We have to note that in [2], Yaroslav Pugai has constructed lattice $W_{3}$ algebras already, but here we will introduce its weaker version based on our newly defined Poisson bracket, constructed just based on the Cartan matrix $A_{n}$, which will make our job easier and more elegant.

For to do this, let us set $C$ an arbitrary symmetrizable Cartan matrix of rank $r$ and let $n=n_{+}$ be the standard maximal nilpotent sub-algebra of the Kac-Moody algebra associated with $C$.

So $n$ is generated by elements $E_{1},\ldots,E_{r}$ which are satisfying in Serre relations, [11] Where $r$ stands for $\text{rank}(C)$.

In [1], we proved that screening operators $S_{X_{i}^{ji}}=\sum\limits_{\begin{subarray}{c}j\in\mathbb{Z}\\ \text{for}~{}i~{}\text{fixed}\end{subarray}}^{n}X_{i}^{ji}$; for $X_{i}^{ji}$ generators of the $q-$commutative ring $\mathbb{C}_{q}[X_{i}^{ji}]:=\frac{\mathbb{C}[X_{i}^{ji}]}{\left\langle X_{i}^{ji}X_{k}^{jk}-q^{<\alpha_{i},\alpha_{j}>}X_{k}^{jk}X_{i}^{ji}\right\rangle}$ and for $<\alpha_{i},\alpha_{j}>=a_{ij}$ the $ij$’s components of our Cartan matrix $C$; are satisfying in quantum Serre relations $\text{ad}_{q}(X_{i})^{1-a_{ij}}(X_{j})$ for adjoint action $\text{ad}_{q}(X_{i})(X_{j})=X_{i}X_{j}-q^{a_{ij}}X_{j}X_{i}$ and $X_{i}\in(U_{q})_{\alpha}$, $X_{j}\in(U_{q})_{\beta}$ [5], for $(U_{q})_{\alpha}=\{u\in U_{q}(g)|q^{\mathfrak{h}}uq^{-\mathfrak{h}}=q^{\alpha(\mathfrak{h})}u~{}~{}~{}~{}\text{for all}~{}\mathfrak{h}\in\overset{\vee}{P}\}$ and $U_{q}(g)=\underset{\alpha~{}in~{}Q}{\oplus}(U_{q})_{\alpha}$, for $Q=\underset{i\in I}{\oplus}\mathbb{Z}_{\alpha_{i}}$ the root lattice and for $\overset{\vee}{P}$ a free Abelian group of rank $2|I|-\text{rank}C$ with $\mathbb{Z}-$basis $\{h_{i}|i\in I\}\cup\{d_{s}|s=1,\cdots,|I|-\text{rank}C\}$ and $\mathfrak{h}={\rm I\!F}\otimes_{\mathbb{Z}}\overset{\vee}{P}$ be the ${\rm I\!F}-$linear space spanned by $\overset{\vee}{P}$. [5] $\overset{\vee}{P}$ will be called dual weight lattice and $\mathfrak{h}$ the Cartan subalgebra. And ${\rm I\!F}$ will stand for our ground field.[5]

Here for our Cartan matrix $C$, the quantum Serre relation will be

                          $\text{ad}_{q}(X_{i})^{1-(-1)}(X_{j})=\text{ad}_{q}^{2}(X_{i})(X_{j})$

                                                    $=X_{i}^{2}X_{j}-[2]_{q}X_{i}X_{j}X_{i}+X_{j}X_{i}^{2}$

                                                    $=X_{i}^{2}X_{j}-(q+q^{-1})X_{i}X_{j}X_{i}+X_{j}X_{i}^{2}$

Where $[2]_{q}$ stands for quantum number $[2]_{q}=\frac{q^{2}-q^{-2}}{q-q^{-1}}$.

And again as what we had in [1], we can define

                        U_{q}(n):=\big{\langle}S_{X_{i}^{ji}},S_{X_{k}^{jk}}\mid(\text{ad}_{q}(S_{X_{i}^{ji}}))^{2}(S_{X_{k}^{jk}})=0\big{\rangle},

and for $\mathbb{C}_{q}[X]$ the quantum polynomial ring in one variable and twisted tensor product $\bar{\otimes}$, we can define

               U_{q}(n)\bar{\otimes}\mathbb{C}_{q}[X_{l}^{jl}]:=\big{\langle}S_{X_{i}^{ji}},S_{X_{k}^{jk}},X_{l}^{jl}\mid(\text{ad}_{q}(S_{X_{i}^{ji}}))^{2}(S_{X_{k}^{jk}})=0

                                        ,S_{X_{i}^{ji}}X_{l}^{jl}=q^{2}X_{l}^{jl}S_{X_{i}^{ji}},S_{X_{k}^{jk}}X_{l}^{jl}=q^{-1}X_{l}^{jl}S_{X_{k}^{jk}}\big{\rangle}

such that we have the following embeding

[TABLE]

where \mathbb{C}_{q}[X_{l}^{jl}]\bar{\otimes}\mathbb{C}_{q}[X_{m}^{jm}]=\mathbb{C}\big{\langle}X_{l}^{jl},X_{m}^{jm}\mid X_{l}^{jl}X_{m}^{jm}=q^{a_{lm}}X_{m}^{jm}X_{l}^{jl}.[1]

Which will ensure the well-definedness of our definition of lattice $W-$algebras.

2. Weak Faddeev-Takhtajan-Volkov algebras

As it has been mentioned already in [1], the main tools which we will use, are difference equations, screening operators, Feigin’s homomorphisms, adjoint actions, partial differential equations, and Cartan matrices.

We know that from an abstract view $g=sl_{m+1}$ is an algebra related to the Cartan matrix $(a_{ij})_{i,j}$, for $a_{ij}=\begin{cases}2&\text{if }i=j\\ -1&\text{if }|i-j|=1\\ 0&\text{if }|i-j|>1\par\end{cases}$ and so for $sl_{2}$ it will consist of just one row and one column, i.e. we have $A_{1}=(2)$ and let us denote by $C\langle X\rangle$ the skew polynomial ring on generators $X=(X_{i})_{i}$ labeled by $i\in\{-\infty,\cdots-1,0,1,\cdots,+\infty\}$ and the defining $q-$commutation relations $X_{i}X_{j}=q^{2}X_{j}X_{i}~{}~{}\text{ for if}~{}~{}i\leq j$ with all having the same color.

Definition 2.1**.**

Let’s define our Poisson bracket as follows in the case of $sl_{2}$:

[TABLE]

The main problem is to find solutions of the system of difference equations from infinite number of non-commutative variables in quantum case and commutative variables in classical case. It is significant that commutation relations (2.1) depend just on the sign of the difference $(i-j)$ and is based on our Cartan matrix. We should try to find all solutions of the system:

[TABLE]

Let us define our system of variables as follows

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

And let us equip this system of variables with lexicographic ordering, i.e. $j_{k_{m}}i<j_{k_{n}}i$ if $j_{k_{m}}<j_{k_{n}}$ and $ji_{k_{m}}<ji_{k_{n}}$ if $i_{k_{m}}<i_{k_{n}}$. And we need this kind of ordering because we have different kind of set of variables with a proper coloring such that each set has its own color different from its neighbors.

We have $\tau_{1}:=\tau_{1}[\cdots,X_{1}^{(11)},X_{1}^{(21)},X_{1}^{(31)},\cdots,X_{2}^{(12)},X_{2}^{(22)},X_{2}^{(32)},\cdots]$, a multi-variable function depend on $\{X_{i}^{(ji)}\}$’s for $i,j\in\{-\infty,\cdots,1,\cdots,n,\cdots,+\infty\}$ and $\mathfrak{D}_{x}^{(n)}$ comes from

[TABLE]

where $<\alpha_{i},\alpha_{j}>=a_{ij}$ is related to our Cartan matrix and $S_{X_{i}^{ji}}$ is a screening operator on one of our variable sets, i.e. $S_{X_{i}^{ji}}=\sum\limits_{j\in\mathbb{Z}}X_{i}^{ji}$. Then we will obtain the whole set of solutions by using the following shift operator:

[TABLE]

[TABLE]

[TABLE]

Definition 2.2**.**

Let us define our lattice W-algebra based on its generators according to [2] [1].

Generators of lattice W-algebra associated with simple Lie algebra $g$ constitute of the functional basis of the space of invariants

[TABLE]

with additional requirements

[TABLE]

where $H_{X_{i}^{ji}}$ and $D_{X_{i}^{ji}}$ will be specified later.

Equation (2.4) means that the generators have to satisfy in quantum Serre relations and the first equation in (2.6) means that they should have zero degree.

Here in this paper we just will work on the case where $g=\text{sl}_{n}$ and we will use $\tau_{i}^{(n)}$ instead of $\tau_{i}$. Where $(n)$ stands for $n$ in $\text{sl}_{n}$.

2.1. Lattice $W_{2}$ algebra

Let us first consider the $sl_{2}$ case and to simplifying the notations, let us consider our set of variables as $X_{i}:=X_{i}^{ji}$.

And as it has shown in [1], it is enough just to work with $S_{X_{i}^{ji}}=:S_{X_{i}}=\sum\limits_{i=1}^{3}X_{i}$, because the other parts for $i>3$ and $i<1$ will tend to zero.

By setting $q=e^{-\mathfrak{h}}$, for the Planck constant $\mathfrak{h}$, we will try to find generators of our lattice $W_{2}$-algebra, in the case of $\text{sl}_{2}$.

**First step:

**First let us try to find $D_{X}^{(2)}$.

For to do this and for simplicity, we will set $\tau_{1}:=\tau_{1}[\cdots,X_{1},X_{2},X_{3},,\cdots]$. And as it has been defined already, we have

                                    $D_{X}^{(2)}:=\{S_{X_{i}},\tau_{1}\}$

                                           $=\{X_{1}+X_{2}+X_{3},\tau_{1}\}$

                                           $=\{X_{1},\tau_{1}\}+\{X_{2},\tau_{1}\}+\{X_{3},\tau_{1}\}$

[TABLE]

Now for to understand what is (2.7), we note that $D_{X_{i}}=\{X_{i},\tau_{1}\}$ and also we note that our function $\tau_{1}[\cdots,X_{1},X_{2},X_{3},\cdots]$ is a polynomial function consist of powers of $X_{i}$. What I mean is that, it is enough to find $D_{X_{i}}$ on just powers of $X_{j}$ for different values of $j\in\mathbb{Z}$.

So

[TABLE]

Where according to rules which has been pointed out in [1], we have

                                    $\{X_{1},X_{j}^{n}\}=X_{1}X_{j}^{n}-q^{2n}X_{j}^{n}X_{1}$

                                                 $=\begin{cases}0,&\mbox{if }j>1\\ (1-q^{4n})X_{1}X_{j}^{n},&\mbox{if }j<1\\ (1-q^{2n})X_{1}X_{j}^{n},&\mbox{if }j=1\end{cases}$

Where by setting $q=e^{-\mathfrak{h}}$ and letting $\mathfrak{h}=1$ at the end, we will have:

          First case: $j>1$;

                                    $\{X_{1},X_{j}^{n}\}=0$;

          Second case: $j<1$;

                                    $\{X_{1},X_{j}^{n}\}=(1-e^{-4n\mathfrak{h}})X_{1},X_{j}^{n}$

                                                 $\sim(1-(1-4n\mathfrak{h}))X_{1},X_{j}^{n}$

                                                 $=4n\mathfrak{h}X_{1},X_{j}^{n}\sim 4nX_{1},X_{j}^{n}$

                                                 $=4X_{1}X_{j}\frac{\partial X_{j}^{n}}{\partial X_{j}}$.

          Third case: $j=1$;

                                    $\{X_{1},X_{1}^{n}\}=(1-q^{2n})X_{1}X_{1}^{n}$

                                                 $=(1-e^{-2n\mathfrak{h}})X_{1}X_{1}^{n}$

                                                 $\sim(1-(1-2n\mathfrak{h}))X_{1}X_{1}^{n}$

                                                 $=2n\mathfrak{h}X_{1}X_{1}^{n}\sim 2nX_{1}X_{1}^{n}$

                                                 $=2X_{1}^{2}\frac{\partial X_{1}^{n}}{\partial X_{1}}$.

And so we have

                                    $(\ref{Equ8})=\{X_{1},X_{1}^{n}\}+\sum\limits_{j<1}\{X_{1},X_{j}^{n}\}+\sum\limits_{j>1}\{X_{1},X_{j}^{n}\}$

                                           $+\{X_{2},X_{2}^{n}\}+\sum\limits_{j<2}\{X_{2},X_{j}^{n}\}+\sum\limits_{j>2}\{X_{2},X_{j}^{n}\}$

                                           $+\{X_{3},X_{3}^{n}\}+\sum\limits_{j<3}\{X_{3},X_{j}^{n}\}+\sum\limits_{j>3}\{X_{3},X_{j}^{n}\}$

                                           $=2X_{1}^{2}\frac{\partial}{\partial X_{1}}+0+0$

                                           $+2X_{2}^{2}\frac{\partial}{\partial X_{2}}+4X_{2}X_{1}\frac{\partial}{\partial X_{1}}+0$

                                           $+2X_{3}^{2}\frac{\partial}{\partial X_{3}}+4X_{3}X_{2}\frac{\partial}{\partial X_{2}}+4X_{3}X_{1}\frac{\partial}{\partial X_{1}}$

                                           $=2X_{1}(X_{1}+2X_{2}+2X_{3})\frac{\partial}{\partial X_{1}}+2X_{2}(X_{2}+2X_{3})\frac{\partial}{\partial X_{2}}$

                                           $+2X_{3}^{2}\frac{\partial}{\partial X_{3}}.$

So we found $D_{X}^{(2)}$ which is as follows and we can omit 2, because finally we will make the action equal to zero and we can cancel 2 from both sides. So we have

[TABLE]

**Second step:

**Now we will try to find $H_{X}^{(2)}$.

For to find $H_{X}^{(2)}$, we note that it resembles the degree of our polynomial function. So if for example $H_{X}^{(2)}$ acts on $X_{1}^{n}X_{2}^{m}X_{3}^{l}$, then we should get $(n+m+l)$.

So let us define:

[TABLE]

and then we have;

                  $H_{X}^{(2)}(X_{1}^{n}X_{2}^{m}X_{3}^{l})=(\sum_{i}X_{i}\frac{\partial}{\partial_{X_{i}}})(X_{1}^{n}X_{2}^{m}X_{3}^{l})$

                                        $=\sum_{i}X_{i}\frac{\partial X_{1}^{n}X_{2}^{m}X_{3}^{l}}{\partial_{X_{i}}}$

                                         $=X_{1}\frac{\partial X_{1}^{n}X_{2}^{m}X_{3}^{l}}{\partial_{X_{1}}}+X_{2}\frac{\partial X_{1}^{n}X_{2}^{m}X_{3}^{l}}{\partial_{X_{2}}}+X_{3}\frac{\partial X_{1}^{n}X_{2}^{m}X_{3}^{l}}{\partial_{X_{3}}}$

                                        $=nX_{1}^{n}X_{2}^{m}X_{3}^{l}+mX_{1}^{n}X_{2}^{m}X_{3}^{l}+lX_{1}^{n}X_{2}^{m}X_{3}^{l}$

                                        $=(n+m+l)X_{1}^{n}X_{2}^{m}X_{3}^{l}$.

Which gives us

                                    $H_{X}^{(2)}(X_{1}^{n}X_{2}^{m}X_{3}^{l})=(n+m+l)X_{1}^{n}X_{2}^{m}X_{3}^{l}$

and on the other side we have

   $(n+m+l)X_{1}^{n}X_{2}^{m}X_{3}^{l}=nX_{1}X_{1}^{n-1}X_{2}^{m}X_{3}^{l}+mX_{1}^{n}X_{2}X_{2}^{m-1}X_{3}^{l}+lX_{1}^{n}X_{2}^{m}x_{3}X_{3}^{l-1}$

                               $=X_{1}\frac{X_{2}^{m}X_{3}^{l}\partial X_{1}^{n}}{\partial_{X_{1}}}+X_{2}\frac{X_{1}^{n}X_{3}^{l}\partial X_{2}^{m}}{\partial_{X_{2}}}+X_{3}\frac{X_{1}^{n}X_{2}^{m}\partial X_{3}^{l}}{\partial_{X_{3}}}$

                              $=X_{1}\frac{\partial}{\partial_{X_{1}}}+X_{2}\frac{\partial}{\partial_{X_{2}}}+X_{3}\frac{\partial}{\partial_{X_{3}}}$

Which gives us

                                    $(n+m+l)X_{1}^{n}X_{2}^{m}X_{3}^{l}=\sum\limits_{i}X_{i}\frac{\partial}{\partial X_{i}}$.

And it shows that $(\ref{Equ10})$ is well defined.

Now the only thing that remains is just to find the solutions of the following system of 2-linear homogeneous equations in one unknown $\tau_{1}$:

[TABLE]

Now the goal is to find such $\tau_{1}[\cdots,X_{1},X_{2},X_{3},\cdots]$ which satisfies in our system of equations (2.11).

The second equation ensures that the solution has degree 0 and also the partial differentials will give us a multi-variable function dependent on just $X_{1},X_{2},X_{3}$.

The system of PDEs (2.11) can be solved using the procedure described in Chapter V, Section IV of [3].

And after doing some calculation in mathematica it become clear that the system (2.11) has only one functional dependent nontrivial solution:

[TABLE]

And again as before, $(2)$ goes back to 2 in $Sl_{2}$ and $1$ is a default index which will be used later for to employ shifting operator.

According to the number of variables, we will have two shifts and then everything will be in a loop.

So here in $sl_{2}$ case we have three solutions for our system of linear equations $(\ref{Equ11})$ which belong to the fraction ring of polynomial functions:

[TABLE]

We go to define our non-commutative Poisson algebra according to definition of Poisson brackets given by Poisson himself [4] with the difference that here we work on $q-$commutative ring $\frac{\mathbb{C}[X_{i}^{ji}]}{X_{i}^{ji}X_{k}^{jk}-q^{<\alpha_{i},\alpha_{k}>}X_{k}^{jk}X_{i}^{ji}}$, based on the generators which are the solutions of PDEs system $(\ref{Equ2})$.

For to do this we will use the following bracket:

[TABLE]

where $\{X_{i},X_{j}\}$ is our previously defined Poisson bracket on our set of variables.

For instance in the case of $sl_{2}$ we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

So we have

[TABLE]

And it is enough to find our brackets just based on the first generator, because after that we are able to find other brackets based on the other generators, so for $\tau_{3}^{(2)}$ in an almost same process we have:

          $F_{3}^{(2)}=\{\tau_{1}^{(2)},\tau_{3}^{(2)}\}$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We have to note that we almost are done with our Poisson algebra in $\text{sl}_{2}$ case, but for the further plan i.e. to find our Volterra system, the differential-difference chain of non-linear equations

[TABLE]

where $\Gamma_{i}$ stands for $\frac{{\tau_{1},\tau_{i}}}{\tau_{1}\tau_{i}}$ [2], we have to write down the brackets $\{\tau_{1},\tau_{i}\}$ in terms of their decompositions to $\tau_{j}$’s for $1\leq j\leq i$.

So we need to write it as the decomposition of our generators and it will be done by using the Mathematica coding which we have produced in Appendix A.

And the result is as follows:

[TABLE]

This result is weaker than the Faddeev-Takhtajan-Volkov algebra which has been mentioned in [2] and if we continue this for $sl_{3}$, then we will have again a weaker version of what which has been mentioned in [2].

2.2. Lattice $W_{3}$ algebra

In this case we will use the following defined Poisson bracket based on Cartan matrix A_{2}=\left[\begin{array}[]{cc}2&-1$\\$-1&2\\ \end{array}\right], but for to do this according to our previous ordering and list of variables, let us for simplicity set our variables as follows

Set $X_{i}^{(1i)}:=X_{i}$ and $X_{i}^{(2i)}:=Y_{i}$.

Definition 2.3**.**

Let’s define our Poisson bracket as follows in the case of $sl_{3}$:

[TABLE]

And instead of $(\ref{Equ1})$ we will have the following $q-$commutation relations

[TABLE]

And we will get the following equations in a same manner as in $\text{sl}_{2}$:

          First case: $i<j$;

                                    $\{X_{i},Y_{j}^{n}\}=X_{i}Y_{j}^{n}-q^{-n}Y_{j}^{n}X_{i}$

                                                   $=X_{i}Y_{j}^{n}-q^{0}X_{i}X_{j}^{n}$

                                                   $=0$

          Second case: $i\geq j$;

$\{X_{i},Y_{j}^{n}\}=X_{i}Y_{j}^{n}-q^{-n}Y_{j}^{n}X_{i}$

                                                   $=(1-q^{-2n})X_{i}Y_{j}^{n}$

                                                   $=(1-e^{2n\mathfrak{h}})X_{i}Y_{j}^{n}$

                                                   $\sim(1-(1+2n\mathfrak{h}))X_{i}Y_{j}^{n}$

                                                   $=-2n\mathfrak{h}X_{i}Y_{j}^{n}$

                                                   $\sim-2nX_{i}Y_{j}^{n}$

                                                   $=-2X_{i}Y_{j}\frac{\partial Y_{j}^{n}}{\partial Y_{j}}$

[TABLE]

As in $sl_{2}$ case we will try to find $H_{X}^{(3)}$ as follows:

$\{X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}},X_{0}\}$

$=X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}X_{0}-X_{0}X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}$

$=(1-q^{2\alpha_{1}+2\alpha_{2}+2\alpha_{3}-\beta_{1}-\beta_{2}-\beta_{3}})X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}X_{0}$

$\sim(1-(1-n\mathfrak{h}(2\alpha_{1}+2\alpha_{2}+2\alpha_{3}-\beta_{1}-\beta_{2}-\beta_{3})))X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}X_{0}$

$=(2\alpha_{1}+2\alpha_{2}+2\alpha_{3}-\beta_{1}-\beta_{2}-\beta_{3})n\mathfrak{h}X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}X_{0}$

$\sim(2\alpha_{1}+2\alpha_{2}+2\alpha_{3}-\beta_{1}-\beta_{2}-\beta_{3})nX_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}X_{3}^{\alpha_{3}}Y_{1}^{\beta_{1}}Y_{2}^{\beta_{2}}Y_{3}^{\beta_{3}}X_{0}$

$=(2X_{1}\frac{\partial}{\partial X_{1}}+2X_{2}\frac{\partial}{\partial X_{2}}+2X_{3}\frac{\partial}{\partial X_{3}}-Y_{1}\frac{\partial}{\partial Y_{1}}-Y_{2}\frac{\partial}{\partial Y_{2}}-Y_{3}\frac{\partial}{\partial Y_{3}})\tau_{1}^{(3)}$.

Now let us as usual suppose $i>j$ and then we will define the following quantities.

Here for $X_{i}$s we have:

                               ${}_{X_{j}}D_{X_{i}}:=\{X_{i},X_{j}^{n}\}$

                                         $=X_{i}X_{j}^{n}-q^{2n}X_{j}^{n}X_{i}$

                                          $=(1-q^{4n})X_{i}X_{j}^{n}$

                                          $=(1-e^{-4n\mathfrak{h}})X_{i}X_{j}^{n}$

                                         $\sim(1-(1-4n\mathfrak{h}))X_{i}X_{j}^{n}$

                                         $=4n\mathfrak{h}X_{i}X_{j}^{n}$

                                         $\sim 4nX_{i}X_{j}^{n}$

                                          $=4nX_{i}X_{j}\frac{\partial X_{j}^{n}}{\partial X_{j}}$.

And the same will be for $Y_{i}$s.

And for the different quantities $X_{i}$ and $Y_{j}$s we have:

          First case: for $i>j$ we have

                                    ${}_{Y_{j}}D_{X_{i}}:=\{X_{i},Y_{j}^{n}\}$

                                             $=X_{i}Y_{j}^{n}-q^{-n}Y_{j}^{n}X_{i}$

                                             $=(1-q^{-2n})X_{i}Y_{j}^{n}$

                                             $=(1-e^{-2n\mathfrak{h}})X_{i}Y_{j}^{n}$

                                             $\sim(1-(1-2n\mathfrak{h}))X_{i}Y_{j}^{n}$

                                             $=2n\mathfrak{h}X_{i}Y_{j}^{n}$

                                             $\sim 2nX_{i}Y_{j}^{n}$

                                              $=2X_{i}Y_{j}\frac{\partial Y_{j}^{n}}{\partial Y_{j}}$.

          Second case: for $i\leq j$ we have

According to what has just mentioned we have

${}_{Y_{j}}D_{1}^{Y}:=_{Y_{j}}D_{Y_{1}}$

                                               $=4Y_{1}Y_{j}\frac{\partial Y_{j}^{n}}{\partial Y_{j}}$.

And

                                      ${}_{Y_{1}}D_{1}^{Y}:=_{Y_{1}}D_{Y_{1}}$

                                               $=2Y_{1}^{2}\frac{\partial Y_{1}^{n}}{\partial Y_{1}}.$

And in a same way we can find the desired results for ${}_{Y_{j}}D_{2}^{Y}$ and ${}_{Y_{j}}D_{3}^{Y}$.

So let us define

[TABLE]

And then we will have

[TABLE]

And

[TABLE]

And

[TABLE]

And finally we get

                              ${}_{Y}D_{Y}^{(3)}:=_{Y}D_{1}+_{Y}D_{2}+_{Y}D_{3}$

                                       $=Y_{1}(Y_{1}+2Y_{2}+2Y_{3})\frac{\partial}{\partial Y_{1}}+Y_{2}(Y_{2}+2Y_{3})\frac{\partial}{\partial Y_{2}}+Y_{3}^{2}\frac{\partial}{\partial Y_{3}}$.

For $j\geq 1$ we have

                             ${}_{X_{1}}D_{Y_{j}}:=\{Y_{j},X_{1}^{n}\}$

                                        $=Y_{j}X_{1}^{n}-q^{-n}X_{1}^{n}Y_{j}$

                                        $=(1-q^{-2n})Y_{j}X_{1}^{n}$

                                        $=(1-e^{2n\mathfrak{h}})Y_{j}X_{1}^{n}$

                                        $\sim(1-(1+2n\mathfrak{h}))Y_{j}X_{1}^{n}$

                                        $=-2n\mathfrak{h}Y_{j}X_{1}^{n}$

                                        $\sim-2nY_{j}X_{1}^{n}$

                                        $=-2Y_{j}X_{1}\frac{\partial}{\partial X_{1}}$

                                        $=-2X_{1}(Y_{1}+Y_{2}+Y_{3})\frac{\partial}{\partial X_{1}}.$

For $j\geq 2$ we have

                             ${}_{X_{2}}D_{Y_{j}}:=\{Y_{j},X_{2}^{n}\}$

                                        $=Y_{j}X_{2}^{n}-q^{-n}X_{2}^{n}Y_{j}$

                                        $=(1-q^{-2n})Y_{j}X_{2}^{n}$

                                         $=(1-e^{2n\mathfrak{h}})Y_{j}X_{2}^{n}$

                                        $\sim(1-(1+2n\mathfrak{h}))Y_{j}X_{2}^{n}$

                                        $=-2n\mathfrak{h}Y_{j}X_{2}^{n}$

                                        $\sim-2nY_{j}X_{2}^{n}$

                                        $=-2Y_{j}X_{2}\frac{\partial}{\partial X_{2}}$

                                        $=-2X_{2}(Y_{2}+Y_{3})\frac{\partial}{\partial X_{2}}.$

For $j\geq 3$ we have

                              ${}_{X_{3}}D_{Y_{j}}:=\{Y_{j},X_{3}^{n}\}$

                                         $=Y_{j}X_{3}^{n}-q^{-n}X_{3}^{n}Y_{j}$

                                         $=(1-q^{-2n})Y_{j}X_{3}^{n}$

                                         $=(1-e^{2n\mathfrak{h}})Y_{j}X_{3}^{n}$

                                         $\sim(1-(1+2n\mathfrak{h}))Y_{j}X_{3}^{n}$

                                         $=-2n\mathfrak{h}Y_{j}X_{3}^{n}$

                                         $\sim-2nY_{j}X_{3}^{n}$

                                         $=-2Y_{j}X_{3}\frac{\partial}{\partial X_{3}}$

                                         $=-2X_{3}Y_{3}\frac{\partial}{\partial X_{3}}.$

And after all these, let us define

                              ${}_{X}D_{Y}^{(3)}:=_{X_{1}}D_{Y_{j}}+_{X_{2}}D_{Y_{j}}+_{X_{3}}D_{Y_{j}}$

                                         $=-2X_{1}(Y_{1}+Y_{2}+Y_{3})\frac{\partial}{\partial X_{1}}-2X_{2}(Y_{2}+Y_{3})\frac{\partial}{\partial X_{2}}$

                                            $-2X_{3}Y_{3}\frac{\partial}{\partial X_{3}}$.

And finally let us define

                                  $D_{Y}^{(3)}:=_{Y}D_{Y}^{(3)}+_{X}D_{Y}^{(3)}$

                                          $=Y_{1}(Y_{1}+2Y_{2}+2Y_{3})\frac{\partial}{\partial Y_{1}}+Y_{2}(Y_{2}+2Y_{3})\frac{\partial}{\partial Y_{2}}$

                                            $+Y_{3}^{2}\frac{\partial}{\partial Y_{3}}-2X_{1}(Y_{1}+Y_{2}+Y_{3})\frac{\partial}{\partial X_{1}}-2X_{2}(Y_{2}$

                                            $+Y_{3})\frac{\partial}{\partial X_{2}}-2X_{3}Y_{3}\frac{\partial}{\partial X_{3}}$.

**Next step:

**Now let us try to find $D_{X}^{(3)}$:

For $i>1$, let us define ${}_{Y_{1}}D_{X_{i}}$ as follows:

                                 ${}_{Y_{1}}D_{X_{i}}:=\{X_{i},Y_{1}^{n}\}$

                                          $=X_{i}Y_{1}^{n}-q^{-n}Y_{1}^{n}X_{i}$

                                          $=(1-q^{-2n})X_{i}Y_{1}^{n}$

                                          $=(1-e^{2n\mathfrak{h}})X_{i}Y_{1}^{n}$

                                          $\sim(1-(1+2n\mathfrak{h}))X_{i}Y_{1}^{n}$

                                          $=-2n\mathfrak{h}X_{i}Y_{1}^{n}$

                                          $\sim-2nX_{i}Y_{1}^{n}=-2X_{i}Y_{1}\frac{\partial}{\partial Y_{1}}$

                                          $=-2Y_{1}(X_{2}+X_{3})\frac{\partial}{\partial Y_{1}}.$

For $i>2$ we have

                                ${}_{Y_{2}}D_{X_{i}}:=\{X_{i},Y_{2}^{n}\}$

                                          $=X_{i}Y_{2}^{n}-q^{-n}Y_{2}^{n}X_{i}$

                                          $=(1-q^{-2n})X_{i}Y_{2}^{n}$

                                         $=(1-e^{2n\mathfrak{h}})X_{i}Y_{2}^{n}$

                                          $\sim(1-(1+2n\mathfrak{h}))X_{i}Y_{2}^{n}$

                                          $=-2n\mathfrak{h}X_{i}Y_{2}^{n}$

                                          $\sim-2nX_{i}Y_{2}^{n}$

                                          $=-2X_{i}Y_{2}\frac{\partial}{\partial Y_{2}}$

                                          $=-2Y_{2}X_{3}\frac{\partial}{\partial Y_{2}}.$

For $i>3$ we have 0.

Let us again have the following definitions

[TABLE]

[TABLE]

[TABLE]

Now let us define

[TABLE]

And now as before we have

                                ${}_{X_{j}}D_{1}^{X}:=_{X_{j}}D_{X_{1}}$

                                          $=4X_{1}X_{j}\frac{\partial X_{j}^{n}}{\partial X_{j}}$.

                                ${}_{X_{1}}D_{1}^{X}:=_{X_{1}}D_{X_{1}}$

                                          $=2X_{1}^{2}\frac{\partial X_{1}^{n}}{\partial X_{1}}.$

And in a same way we are able to define ${}_{X_{j}}D_{2}^{X}$ and ${}_{X_{j}}D_{3}^{X}$. So let us define

[TABLE]

Then we will have

[TABLE]

And

[TABLE]

And

[TABLE]

So we will have

                                ${}_{X}D_{X}^{(3)}:=_{X}D_{1}+_{X}D_{2}+_{X}D_{3}$

                                          $=X_{1}(X_{1}+2X_{2}+2X_{3})\frac{\partial}{\partial X_{1}}+X_{2}(X_{2}+2X_{3})\frac{\partial}{\partial X_{2}}$

                                             $+X_{3}^{2}\frac{\partial}{\partial X_{3}}$.

And therefore as in (2.11) we will have the following system of $PDE$s

[TABLE]

And according to appendix A, we have the following functional dependent nontrivial solution for the whole system of $PDE$s (2.24)

[TABLE]

And again as before, $(3)$ goes back to 3 in the $Sl_{3}$ and $1$ is a default index which later we will use it for to employ our shifting operators.

According to the number of variables, we will have 6 shifts and then after that it will be in a loop.

So here in $sl_{3}$ case we have six solutions which belong to the fraction ring of polynomial functions.

[TABLE]

Where $\tau_{1}^{(3)}:=\tau_{1}^{(3)}[\cdots,X_{1},Y_{1},X_{2},Y_{2},X_{3},Y_{3}\cdots]$.

Again by setting $X_{i}^{(1i)}:=X_{i}$ and $X_{i}^{(2i)}:=Y_{i}$ and $X_{i}^{(3i)}:=Z_{i}$ and according to $(\ref{Equ17})$ we have to write down the following brackets as a composition of $\tau_{i}^{(3)}$s, because of the algebra structure and it will be done by using Mathematica coding in appendix A.

[TABLE]

2.3. Lattice $W_{4}$ algebra; main generator

In this case we will use the following defined Poisson bracket (2.28) based on Cartan matrix

                                   A_{3}=\left[\begin{array}[]{ccc}2&-1&0$\\$-1&2&-1$\\$0&-1&2\\ \end{array}\right].

But for to do this according to our previous ordering and list of variables, and the same as what we have done in $sl_{3}$ case, let us for simplicity of the calculations, order the set of our variables as follows:

Set $X_{i}^{(1i)}:=X_{i}$ and $X_{i}^{(2i)}:=Y_{i}$ and $X_{i}^{(3i)}:=Z_{i}$ and so on.

Definition 2.4**.**

Let’s define our Poisson bracket as follows in the case of $sl_{4}$:

[TABLE]

And instead of (2.1) we will have the following $q-$commutation relations for $j\in\{1,2,3\}$ and as always $i\in\{1,2,3\}$:

[TABLE]

And by using the same approach as in the $sl_{2}$ and $sl_{3}$ case, it become clear that the equations $D_{X}^{(4)}$, $D_{Y}^{(4)}$ and $D_{Z}^{(4)}$ and also $H_{X}^{(4)}$, $H_{Y}^{(4)}$ and $H_{Z}^{(4)}$ will have the following forms:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

And the functional dependent nontrivial solutions for the whole system of first order partial differential equation is as follows:

[TABLE]

And again as before, $(4)$ goes back to 4 in the $Sl_{4}$ and $1$ is a default index which later we will use it for to employ our shifting operators.

According to the number of variables, we will have 9 shifts and then after that it will be in a loop.

So here in $sl_{4}$ case we have nine solutions:

$\tau_{1}^{(4)}:=\tau_{1}^{(4)}[X_{1},Y_{1},Z_{1},X_{2},Y_{2},Z_{2},X_{3},Y_{3},Z_{3}]$;

$\tau_{2}^{(4)}:=\tau_{1}^{(4)}[X_{1}\rightarrow Y_{1},Y_{1}\rightarrow Z_{1},Z_{1}\rightarrow X_{2},X_{2}\rightarrow Y_{2},Y_{2}\rightarrow Z_{2},Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3}];$

$\tau_{3}^{(4)}:=\tau_{2}^{(4)}[Y_{1}\rightarrow Z_{1},Z_{1}\rightarrow X_{2},X_{2}\rightarrow Y_{2},Y_{2}\rightarrow Z_{2},Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4}];$

$\tau_{4}^{(4)}:=\tau_{3}^{(4)}[Z_{1}\rightarrow X_{2},X_{2}\rightarrow Y_{2},Y_{2}\rightarrow Z_{2},Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4}];$

$\tau_{5}^{(4)}:=\tau_{4}^{(4)}[X_{2}\rightarrow Y_{2},Y_{2}\rightarrow Z_{2},Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4},Y_{4}\rightarrow Z_{4}];$

$\tau_{6}^{(4)}:=\tau_{5}^{(4)}[Y_{2}\rightarrow Z_{2},Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4},Y_{4}\rightarrow Z_{4},Z_{4}\rightarrow X_{5}];$

$\tau_{7}^{(4)}:=\tau_{6}^{(4)}[Z_{2}\rightarrow X_{3},X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4},Y_{4}\rightarrow Z_{4},Z_{4}\rightarrow X_{5},X_{5}\rightarrow Y_{5}];$

$\tau_{8}^{(4)}:=\tau_{7}^{(4)}[X_{3}\rightarrow Y_{3},Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4},Y_{4}\rightarrow Z_{4},Z_{4}\rightarrow X_{5},X_{5}\rightarrow Y_{5},Y_{5}\rightarrow Z_{5}];$

$\tau_{9}^{(4)}:=\tau_{8}^{(4)}[Y_{3}\rightarrow Z_{3},Z_{3}\rightarrow X_{4},X_{4}\rightarrow Y_{4},Y_{4}\rightarrow Z_{4},Z_{4}\rightarrow X_{5},X_{5}\rightarrow Y_{5},Y_{5}\rightarrow Z_{5},Z_{5}\rightarrow X_{6}];$

which belong to the fraction ring of polynomial functions.

2.4. Lattice $W_{5}$ algebra; main generator

In this case we will use the following defined Poisson bracket based on Cartan matrix

                                   A_{4}=\left[\begin{array}[]{cccc}2&-1&0&0$\\$-1&2&-1&0$\\$0&0&-1&2\\ \end{array}\right].

But for to do this according to our previous ordering and list of variables, and the same as what we have done in $sl_{4}$ case, let us for simplicity in the calculations, order the set of our variables as follows:

Set $X_{i}^{(1i)}:=X_{i}$ and $X_{i}^{(2i)}:=Y_{i}$ and $X_{i}^{(3i)}:=Z_{i}$ and $X_{i}^{(4i)}:=K_{i}$.

Definition 2.5**.**

Let’s define our Poisson bracket as follows in the case of $sl_{5}$:

[TABLE]

And instead of $(\ref{Equ1})$ we will have the following $q-$commutation relations for $j\in\{1,2,3\}$ and as always $i\in\{1,2,3\}$:

[TABLE]

And by using the same approach as $sl_{2}$, $sl_{3}$ and $sl_{4}$ case , it become clear that the equations $D_{X}^{(5)}$, $D_{Y}^{(5)}$, $D_{Z}^{(5)}$ and $D_{K}^{(5)}$ and also $H_{X}^{(5)}$, $H_{Y}^{(5)}$, $H_{Z}^{(5)}$ and $H_{K}^{(5)}$ will have the following forms:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

And the functional dependent nontrivial solution for the whole system of first order partial differential equation is as follows:

[TABLE]

And again as before, $(5)$ goes back to 5 in the $Sl_{5}$ and $1$ is a default index which later we will use it for to employ our shifting operators.

According to the number of variables, we will have 12 shifts and then after that it will be in a loop.

So here in $sl_{5}$ case we will have twelve solutions just as what we did in the $sl_{4}$ case, and here we skip to write them down.

2.5. Lattice $W_{n}$ algebra; main generator

Here for $sl_{n}$, we skip writing down all steps which we have been done in the previous sections and we just will write down the main generator of the lattice $W_{n}$ algebra.

The functional dependent nontrivial solution for the whole system of the first order partial differential equations will be as what comes in follow:

[TABLE]

[TABLE]

We should notice that $x_{i_{j}}^{(j)}$s are different of each other for any $j\in\{1,\cdots n-1\}$

Acknowledgement

The research in this paper would have taken far longer to complete without the encouragement from many others. It is a delight to acknowledge those who have supported me over the last three years during the preparation of this paper.

I would like to thank Prof. Yaroslav Pugai, for his guidance and relaxed, thoughtful insight.

I thank all of the Institute for information transmission problems (Kharkevich Institute)’s staff for their hospitality over the last year.

I am particularly thankful for the help and advice of Prof. Brendan Godfrey, without whom the learning Mathematica would have been very much steeper and unimaginable.

And I would like to thank Prof. Boris Feigin for suggesting me this interesting problem and enlightening discussions during the preparation for my first paper which was my first step in this subject!

And finally, I would like to thank Prof. Alexei Kanel-Belov for his support and for whom I will always be indebted for being a constant source of inspiration and for the great talent and patience which he has always guided me with.

3. Appendix A

*Brendan B. Godfrey *       and       *Farrokh Razavinia *

                                    brendan.godfrey@ieee.org

                                    f.razavinia@phystech.edu

This section has been completed by getting help from professor Brendan B. Godfrey from the Institute for Research in Electronics and Applied Physics (The University of Maryland), indirect communications and discussions through email, and also through a series of questions and discussions in Mathematica StackExchange.

And I have to say that without his great Mathematica skills, it nearly was impossible to get such interesting results!

In this appendix, you will find some parts of Mathematica codings which we have used to obtain our algebra structures.

And we believe that what is written in this appendix can open a new approach in solving the following system of $q-$linear homogeneous equations in one unknown $f$.

[TABLE]

Where the coefficients $a_{ik}$ are functions of $n$ independent variables $x_{1},\cdots,x_{n}$ and do not contain the unknown function $f$. [3]

And we have to mention that, to reach to this point was impossible without using Mathematica!

3.1. Lattice $W_{3}$ algebra

.

As you see DSolve returns un-evaluated i.e. it means that it is not able to solve our system of first order partial differential equations.

Consequently, the dimensionality of this problem can be reduced from six to four.

Although DSolve cannot solve these equations as a pair either. But it can solve each separately.

The first results indicates that $g$ is a function of

and also

The second list of functions can be simplified by

Now the next step is to combine the previous two expressions for $g$ for to obtain a single expression, presumably as a function of two variables.

The system of PDEs above can be solved using the procedure described in Chapter V, Sec IV of Goursat’s Differential Equations [3].

The first step is to find the complete, non-commutative group of differential operators that includes $equ5$ and $equ6$.

which by inspection is independent of $equ5$ and $equ6$. On the other hand, $comm[equ5,equ7]$ and $comm[equ6,equ7]$ do not yield independent equations, again by inspection. Thus $\{equ5,equ6,equ7\}$ is a complete group of three operators in four independent variables. From this information alone, we know that $g$ is an arbitrary function of precisely one first integral. This first integral can be obtained by systematically eliminating variables and equations, one pair at a time, until a single equation of two variable remains. We start by solving any one of the equations.

and use the solution as the basis for a change of variables:

indicating that $h$ is independent of $w2$. This leaves us with two equations in three variables

Proceeding as before, we next solve one of $equ10$ and $equ11$. (We choose the simpler one.)

and use it as the basis for a further change of variables.

indicating that $k$ is independent of $b2$. This leaves us with one equation in two variables.

Finally, $DSolve$ yields

Transforming back to the original independent variables gives

Finally, designating the solution for $g$ as $ansg$,

3.2. Lattice $W_{4}$ algebra

.

Again $DSolve$ returns un-evaluated, meaning that it can not solve the system of equations.

As before, this computation can be simplified by the substitution,

in which case the six equations become

As before, this system of first-order $PDEs$ can be solved by using the procedure described in Chapter V, Sec IV of Goursat’s Differential Equations.

The first step is to find the complete, non-commutative group of differential operators that includes $equ4$, $equ5$, and $equ6$. To do so, we use the function $comm$, generalized from $W_{3}$

which are independent of the first three operators, increasing the size of the group to five. $comm[equ4,equ6]$ vanishes identically and so does not add an operator. On the other hand, the seven additional commutators involving $equ7$ and $equ8$ yield expressions that are linear combinations of $\{equ4,equ5,equ6,equ7,equ8\}$. Thus, these five operators comprise the entire group.

From this information alone, we know that $g$ is an arbitrary function of precisely one first integral. This first integral can be obtained by systematically eliminating variables and equations, one pair at a time, until a single equation of two variable remains. Start by solving any one of the equations.

Final solution

3.3. Expressing a fractional multivariate polynomials to its low-order polynomial decomposition

Suppose we have given the following question.

Question:

Let $f2$ be fractional multivariate polynomial as follows

and also let $k1$ and $k2$ be given as follows

then express $f2$ as a low - order polynomial in $k1$ and $k2$.

This can be done as follows.

First, generate a generic low order polynomial.

and then use SolveAlways. After about twenty seconds we will gwt result

And we have the final solution

which is the desired result.

And for completeness we have

Also here we have much faster alternative:

Because SolveAlways determines the coefficients $c$ for any $\{x1,x2,x3,x4,y1,\\ y2,y3\}$, Solve must be able to obtain the same values for the coefficients $c$ for specific values of $\{x1,x2,x3,x4,y1,y2,y3\}$, and much faster. As before we do have $f2$ and $k1$ and $k2$.

Question:

Let $f6$ be fractional multivariate polynomial as follows

and also let $k1$, $k2$, $k3$, $k4$, $k5$ and $k6$ be given as follows

then express $f6$ as a low - order polynomial in $k1$, $k2$, $k3$, $k4$, $k5$ and $k6$.

Which is the desired result.

And as before for completeness we have

By using Groebner Basis:

Also there is another way for to reach to the solution by using Groebner Basis. But this approach is very slow!

Now let us compute Groebner Basis

The remainder $r$ gives a representation of poly in terms of $K1$ , $K2$, $K3$, $K4$, $K5$ and $K6$.

Where $r$ is our solution in $K1$ , $K2$, $K3$, $K4$, $K5$ and $K6$. And the following code validates correctness:

And please note that, this may take a while. (May be more than a while! It depends on how powerful is your computer. )

3.4. Checking symmetries in our shift operators

. First, before starting, we need to know which variables are employed in our functions. For to do this we employ the following code:

Set

and

Then by using the following code we will obtain the set of our variables which have been employed

Now in what comes below, we specifically mean that for example in $sl_{4}$ in a process for finding $F_{9}^{(4)}$, the substitution

[TABLE]

instead of

[TABLE]

transforms $\tau_{9}^{(4)}$ to $\tau_{1}^{(4)}$, $\tau_{8}^{(4)}$ to $\tau_{2}^{(4)}$, $\tau_{7}^{(4)}$ to $\tau_{3}^{(4)}$, and $\tau_{6}^{(4)}$ to $\tau_{4}^{(4)}$ while leaving $F_{9}^{(4)}$ unchanged.

Therefore, those four pairs must enter the expression for $F_{9}^{(4)}$ symmetrically.

As a result, the generic polynomials we have been using above, can be reduced greatly in numbers of terms, a factor of $(\frac{2}{3})^{4}$. Corresponding running time then should be reduced by a factor of $(\frac{2}{3})^{8}$, other things being equal. It is possible that additional symmetries exist! It needs to be checked!

Here for simplification and for to be able in codding them, we write instead $F_{9}^{(4)}:=F9$ and $\tau_{7}^{(4)}:=Ki$;

**Checking symmetries in F6: **

We can find the set of variables in a same way as what we did for $F_{9}^{(4)}$ and so here we omit most of the calculations.

Again as in $F_{9}^{(4)}$ , here we specifically mean that the substitution

$\{x1,x2,x3,x4,x5,x6,y1,y2,y3,y4,y5\}$ instead of $\{x6,x5,x4,x3,x2,x1,$

$y5,y4,y3,y2,y1\}$ transforms $\tau_{6}^{(4)}$ to $\tau_{1}^{(4)}$, $\tau_{5}^{(4)}$ to $\tau_{2}^{(4)}$, $\tau_{4}^{(4)}$ to $\tau_{3}^{(4)}$ while leaving $F_{6}^{(4)}$ unchanged.

Therefore, those three pairs must enter the expression for $F_{6}^{(4)}$ symmetrically.

555Farrokh Razavinia, Department of discrete mathematics, Moscow Institute of Physics and Technology, Institutskiy per., 9, Dolgoprudny, Moscow Oblast, Russia

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666Brendan B. Godfrey, Institute for Research in Electronics and Applied Physics (The University of Maryland), College Park, United States of America

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