Local derivations of conformal Galilei algebra

Abstract

The present paper is devoted to study local derivations on the conformal Galilei algebra. We prove that every local derivation on the conformal Galilei algebra is a derivation.

KEYWORDS:

• Conformal Galilei algebra
• derivations
• Li algebras
• local derivations

2020 Mathematics Subject Classification:

• 16E40
• 17B56
• 17B68

1 Introduction

Let A be an algebra (not necessary associative). Recall that a linear mapping $D:A\to A$ is said to be a derivation, if $D(\mathit{xy})=D(x)y+\mathit{xD}(y)$ for all $x,y\in A.$A linear mapping Δ is said to be a local derivation, if for every $x\in A$ there exists a derivation D_x on A (depending on x) such that $\Delta (x)=D_x(x).$

This notion was introduced and investigated independently by R.V. Kadison [17] and D.R. Larson and A.R. Sourour [18]. The above papers gave rise to a series of works devoted to the description of mappings which are close to automorphisms and derivations of $C^{\ast }$-algebras and operator algebras. R.V. Kadison set out a program of study for local maps in [17], suggesting that local derivations could prove useful in building derivations with particular properties. R.V.Kadison proved in [17, Theorem A] that each continuous local derivation of a von Neumann algebra M into a dual Banach M-bimodule is a derivation. This theorem gave way to studies on derivations on $C^{\ast }$-algebras, culminating with a result due to B.E. Johnson, which asserts that every local derivation of a $C^{\ast }$-algebra A into a Banach A-bimodule is automatically continuous, and hence is a derivation [15, Theorem 5.3].

Let us present a list of finite or infinite dimensional algebras for which all local derivations are derivations:

• $C^{\ast }$-algebras, in particular, the algebra $M_n(\mathbb{C})$ of all square matrices of order n over the field of complex numbers [15, 17];

• the complex polynomial algebra $\mathbb{C}[x]$ [17];

• finite dimensional simple Lie algebras over an algebraically closed field of characteristic zero [7];

• Borel subalgebras of finite-dimensional simple Lie algebras [22];

• infinite dimensional Witt algebras over an algebraically closed field of characteristic zero [12];

• Witt algebras over a field of prime characteristic [23];

• solvable Lie algebras of maximal rank [20];

• Cayley algebras [5];

• finite dimensional semi-simple Leibniz algebras over an algebraically closed field of characteristic zero [21];

• locally finite split simple Lie algebras over a field of characteristic zero [10];

• finite-dimensional semisimple Filippov n-ary (n > 1) algebras [14];

• the Schrödinger algebras [2].

On the other hand, some algebras (in most cases close to nilpotent algebras) admit pure local derivations, that is, local derivations which are not derivations. Below a short list of some classes of algebras which admit pure local derivations:

• the algebra $\mathbb{C}(x)$ of rational functions [17];

• finite dimensional filiform Lie algebras [7];

• p-filiform Leibniz algebras [9];

• direct sum null-filiform Leibniz algebras [1];

• solvable Leibniz algebras with abelian nilradicals, which have a one dimensional complementary space [6];

• the algebra of lower triangular n × n-matrices [13];

• real octonion algebra [8];

• the ternary Malcev algebra M₈ [14].

In this paper, we will study local derivations on the conformal Galilei algebra. We prove that every local derivation on the conformal Galilei algebra is a derivation.

2 Preliminaries

In this paper, we denoted by $\mathbb{Z},\mathbb{N}$, and $\mathbb{C}$ the sets of all integers, positive integers, and complex numbers, respectively. In this section we give some necessary notations, definitions and preliminary results.

A derivation on a Lie algebra $\mathfrak{g}$ is a linear map $D: \mathfrak{g}\to \mathfrak{g}$ satisfying $$D[x,y]=[D(x),y]+[x,D(y)]$$for all $x,y\in \mathfrak{g}$. Denote by $\mathit{Der}(\mathfrak{g})$ the set of all derivations of g. For all $a\in \mathfrak{g}$, the map ad(a) on $\mathfrak{g}$ defined as $\mathit{ad}(a)(x)=[a,x],x\in \mathfrak{g}$ is a derivation and derivations of this form are called inner derivation and denote by $\mathit{ad}(\mathfrak{g})$. Derivations which are not inner are called outer derivations.

A linear operator Δ is called a local derivation if for any $x\in \mathfrak{g},$ there exists a derivation $D_x:\mathfrak{g}\to \mathfrak{g}$ (depending on x) such that $\Delta (x)=D_x(x).$ we denote by $\text{LocDer}(g)$ the set of all local derivations on $\mathfrak{g}.$

In recent years Galilei groups and their Lie algebras have been intensively studied. The interest is due to an appearance of this kind of symmetries in very different areas of physics and mathematics [3, 16]. The conformal extension of the Galilei algebra is parameterized by a positive half-integer number l and is called the l-conformal Galilei algebra. For $l\in \mathbb{N}-\frac{1}{2}$, we denote the conformal Galilei algebra by $\mathfrak{g}$, which has a basis given by $$\{e,h,f,p_k,z|k=0,1,2,\dots ,2l\}$$and the Lie bracket given by: $$\begin{array}{c}[h,e]\mathrm{}=\mathrm{}\mathrm{}2e, [h,f]\mathrm{}=\mathrm{}\mathrm{}-2f, [e,f]\mathrm{}=\mathrm{}\mathrm{}h,\\ [h,p_k]\mathrm{}=\mathrm{}\mathrm{}2(l-k)p_k, [e,p_k]\mathrm{}=\mathrm{}\mathrm{}k{p}_{k-1}, [f,p_k]\mathrm{}=\mathrm{}\mathrm{}(2l-k)p_{k+1},\\ [z,\mathfrak{g}]\mathrm{}=\mathrm{}\mathrm{}0,\\ [p_k,p_{k\prime }]\mathrm{}=\mathrm{}\mathrm{}{\delta }_{k+k\prime ,2l}{(-1)}^{k+l+\frac{1}{2}}k!(2l-k)!z, \text{for} k,k\prime =0,1,2,\dots ,2l,\end{array}$$ where δ is the Kronecker Delta.

The conformal Galilei algebra can be viewed as a semidirect product $\mathfrak{g}=s{l}_2⋉H_l$ of two subalgebras. $s{l}_2=\text{span}\{e,h,f\}$ and Heisenberg subalgebra $H_l=\text{span}\{p_k,z|k=0,1,2,\dots ,2l\}$. The irreducible representations of the conformal Galilei algebra are studied in [4, 11, 19, 25].

The following lemma can be used to determine the derivation of $\mathfrak{g}$.

Let τ be an outer derivation of $\mathfrak{g}$ determined by $$\tau (h)=\tau (e)=\tau (f)=0, \tau (z)=z, \tau (p_k)=\frac{1}{2}{p}_k, k=0,1,\dots ,2l.$$

Lemma 2.1.

[24] $\text{Der}(\mathfrak{g})=\mathbb{C}\tau \oplus \text{ad}(\mathfrak{g})$.

3 Local derivations of conformal Galilei algebra

In this section we investigate local derivations on conformal Galilei algebra.

The following theorem is the main result of this section.

Theorem 3.1.

Every local derivation on l-conformal Galilei algebra is a derivation, where $l\ne \frac{1}{2},\frac{3}{2}$.

For any local derivation Δ on $\mathfrak{g}$ by Lemma 2.1 we have $$\Delta (x)=[b_x,x]+{\lambda }_x\tau (x).$$

Put $$b_x=b_{e,x}e+b_{h,x}h+b_{f,x}f+\sum _{k=0}^{2l}{b}_{p_i,x}{p}_i+b_{z,x}z\in \mathfrak{g}.$$

Then we obtain (3.1) $$\begin{array}{cc}\Delta (e)\hfill & =2b_{h,e}e-b_{f,e}h-\sum _{k=0}^{2l-1}(k+1)b_{p_{k+1},e}{p}_k,\hfill \\ \Delta (h)\hfill & =-2b_{e,h}e+2b_{f,h}f+\sum _{k=0}^{2l}2(k-l)b_{p_k,h}{p}_k,\hfill \\ \Delta (f)\hfill & =b_{e,f}h-2b_{h,f}f-\sum _{k=0}^{2l-1}(2l-k)b_{p_k,f}{p}_{k+1},\hfill \\ \Delta (p_0)\hfill & =2l{c}_{h,p_0}{p}_0+2l{b}_{f,p_0}{p}_1+{(-1)}^{l-\frac{1}{2}}(2l)!b_{p_{2l},p_0}z,\hfill \\ \Delta (p_k)\hfill & =k{b}_{e,p_k}{p}_{k-1}+2(l-k)c_{h,p_k}{p}_k+(2l-k)b_{f,p_k}{p}_{k+1}+\hfill \\ \hfill & +{(-1)}^{k+l-\frac{1}{2}}k!(2l-k)!b_{p_{2l-k},p_k}z,\hfill \\ \Delta (p_{2l})\hfill & =2l{b}_{e,p_{2l}}{p}_{2l-1}-2l{c}_{h,p_{2l}}{p}_{2l}+{(-1)}^{3l-\frac{1}{2}}{b}_{p_0,p_{2l}}z,\hfill \\ \Delta (z)\hfill & =b_{z,z}z,\hfill \end{array}$$(3.1) where $k=1,2,\dots ,2l-1$ and $$2(l-2i)c_{h,p_i}=\left(2(l-i)b_{h,p_i}+\frac{{\lambda }_{p_i}}{2}\right),\hspace{1em}i=0,1,\dots ,2l.$$

For the proof of this Theorem we need several Lemmata.

There exists an element $a=a_{e,x}e+a_{h,x}h+a_{f,x}f+{\sum }_{k=0}^{2l}{a}_{p_k,x}{p}_k+a_{z,x}z\in \mathfrak{g}$ such that $$\Delta (x)=[a,x]+{\lambda }_x\tau (x).$$

In the following lemmas, we assume that Δ be a local derivation on $\mathfrak{g}$ such that $\Delta (h+z)=0.$

Lemma 3.2.

$\Delta (h)=\Delta (z)=0.$

Proof.

We consider, $$0=\Delta (h+z)=\Delta (h)+\Delta (z)=-2b_{e,h}e+2b_{f,h}f+\sum _{k=0}^{2l}2(k-l)b_{p_k,h}{p}_k+b_{z,z}z,$$which implies $b_{e,h}=b_{f,h}=b_{p_k,h}=b_{z,z}=0,$ where $k\in \{0,1,2,\dots ,2l\}.$ Then $\Delta (h)=\Delta (z)=0.$ □

Lemma 3.3.

(3.2) $$\begin{array}{cc}\Delta (e)\hfill & =2b_{h,e}e-\sum _{k=0}^{2l-1}(k+1)b_{p_{k+1},e}{p}_k,\hfill \\ \Delta (f)\hfill & =-2b_{h,f}f-\sum _{k=0}^{2l-1}(2l-k)b_{p_k,f}{p}_{k+1}.\hfill \end{array}$$(3.2)

Proof.

We consider $$\begin{array}{cc}\Delta (h+e)\hfill & =[a,h+e]+{\lambda }_{h+e}\tau (h+e)=\hfill \\ \hfill & =\left[a_{e,h+e}e+a_{h,h+e}h+a_{f,h+e}f+\sum _{k=0}^{2l}{a}_{p_k,h+e}{p}_k+a_{z,x}z,h+e\right]+{\lambda }_{h+e}\tau (h+e)=\hfill \\ \hfill & =-2a_{e,h+e}e+2a_{f,h+e}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,h+e}{p}_k+\hfill \\ \hfill & +2a_{h,h+e}e-a_{f,h+e}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},h+e}{p}_k.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (h+e)\hfill & =\Delta (h)+\Delta (e)=2b_{h,e}e-b_{f,e}h-\sum _{k=0}^{2l-1}(k+1)b_{p_{k+1},e}{p}_k.\hfill \end{array}$$

Comparing the coefficients at the basis elements f and h, we get $b_{f,e}=0.$

Similarly, considering $$\begin{array}{cc}\Delta (h+f)\hfill & =[a,h+f]+{\lambda }_{h+f}\tau (h+f)=\hfill \\ \hfill & =\left[a_{e,h+f}e+a_{h,h+f}h+a_{f,h+f}f+\sum _{k=0}^{2l}{a}_{p_k,h+f}{p}_k+a_{z,x}z,h+f\right]\hfill \\ \hfill & +{\lambda }_{h+f}\tau (h+f)=\hfill \\ \hfill & =-2a_{e,h+f}e+2a_{f,h+f}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,h+f}{p}_k+\hfill \\ \hfill & +a_{e,h+f}h-2a_{h,h+f}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,h+f}{p}_{k+1}.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (h+f)\hfill & =\Delta (h)+\Delta (f)=b_{e,f}h-2b_{h,f}f-\sum _{k=0}^{2l-1}(2l-k)b_{p_k,f}{p}_{k+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements e and h, we get $b_{e,f}=0.$ □

Setting $$m=\frac{2l-1}{2}.$$

Lemma 3.4.

(3.3) $$\begin{array}{cc}\Delta (p_0)\hfill & =2l{c}_{h,p_0}{p}_0+2l{b}_{f,p_0}{p}_1,\hfill \\ \Delta (p_i)\hfill & =i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1},\hfill \\ \Delta (p_{2l})\hfill & =2l{b}_{e,p_{2l}}{p}_{2l-1}-2l{c}_{h,p_{2l}}{p}_{2l},\hfill \end{array}$$(3.3) where $i\in \{1,2,3,\dots ,2l-1\}\setminus \{m,m+1\}.$

Proof.

We consider $$\begin{array}{cc}\Delta (h+p_0)\hfill & =[a,h+p_0]+{\lambda }_{h+p_0}\tau (h+p_0)=\hfill \\ \hfill & =\left[a_{e,h+p_0}e+a_{h,h+p_0}h+a_{f,h+p_0}f+\sum _{k=0}^{2l}{a}_{p_k,h+p_0}{p}_k+a_{z,h+p_0}z,h+p_0\right]\hfill \\ \hfill & +{\lambda }_{h+p_0}\tau (h+p_0)=\hfill \\ \hfill & =-2a_{e,h+p_0}e+2a_{f,h+p_0}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,h+p_0}{p}_k+\hfill \\ \hfill & +\left(2l{a}_{h,h+p_0}+\frac{{\lambda }_{h+p_0}}{2}\right)p_0+2l{a}_{f,h+p_0}{p}_1+{(-1)}^{l-\frac{1}{2}}(2l)!a_{p_{2l},h+p_0}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (h+p_0)\hfill & =\Delta (h)+\Delta (p_0)=\hfill \\ \hfill & =2l{c}_{h,p_0}{p}_0+2l{b}_{f,p_0}{p}_1+{(-1)}^{l-\frac{1}{2}}(2l)!b_{p_{2l},p_0}z.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l}$ and z, we have $b_{p_{2l},p_0}=0.$

Now consider $i\in \{1,2,3,\dots ,2l-1\}\setminus \{m,m+1\}$ $$\begin{array}{cc}\Delta (h+p_i)\hfill & =[a,h+p_i]+{\lambda }_{h+p_i}\tau (h+p_i)=\hfill \\ \hfill & =\left[a_{e,h+p_i}e+a_{h,h+p_i}h+a_{f,h+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,h+p_i}{p}_k+a_{z,h+p_i}z,h+p_i\right]\hfill \\ \hfill & +{\lambda }_{h+p_i}\tau (h+p_i)=\hfill \\ \hfill & =-2a_{e,h+p_i}e+2a_{f,h+p_i}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,h+p_i}{p}_k+\hfill \\ \hfill & +i{a}_{e,h+p_i}{p}_{i-1}+\left(2(l-i)a_{h,h+p_i}+\frac{{\lambda }_{h+p_i}}{2}\right)p_i+(2l-i)a_{f,h+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},h+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (h+p_i)\hfill & =\Delta (h)+\Delta (p_i)=\hfill \\ \hfill & =k{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!b_{p_{2l-i},p_i}z.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l-i}$ and z, we get $b_{p_{2l-i},p_i}=0.$

Similarly, considering $$\begin{array}{cc}\Delta (h+p_{2l})\hfill & =[a,h+p_{2l}]+{\lambda }_{h+p_{2l}}\tau (h+p_{2l})=\hfill \\ \hfill & =\left[a_{e,h+p_{2l}}e+a_{h,h+p_{2l}}h+a_{f,h+p_{2l}}f+\sum _{k=0}^{2l}{a}_{p_k,h+p_{2l}}{p}_k+a_{z,h+p_{2l}}z,h+p_{2l}\right]\hfill \\ \hfill & +{\lambda }_{h+p_{2l}}\tau (h+p_{2l})=\hfill \\ \hfill & =-2a_{e,h+p_{2l}}e+2a_{f,h+p_{2l}}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,h+p_{2l}}{p}_k+\hfill \\ \hfill & +2l{a}_{e,h+p_{2l}}{p}_{2l-1}+\left(-2l{a}_{h,h+p_{2l}}+\frac{{\lambda }_{h+p_{2l}}}{2}\right)p_{2l}+{(-1)}^{3l-\frac{1}{2}}{a}_{p_0,h+p_{2l}}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (h+p_{2l})\hfill & =\Delta (h)+\Delta (p_i)=\hfill \\ \hfill & =2l{b}_{e,p_{2l}}{p}_{2l-1}-2l{c}_{h,p_{2l}}{p}_{2l}+{(-1)}^{3l-\frac{1}{2}}{b}_{p_0,p_{2l}}z.\hfill \end{array}$$

Comparing the coefficients at the basis elements p₀ and z, we obtain $b_{p_0,p_{2l}}=0.$ □

Lemma 3.5.

$$\begin{array}{cc}\Delta (e)\hfill & =2b_{h,e}e-m{b}_{p_m,e}{p}_{m-1}-(m+1)b_{p_{m+1},e}{p}_m,\hfill \\ \Delta (f)\hfill & =-2b_{h,f}f-(m+1)b_{p_m,f}{p}_{m+1}-m{b}_{p_{m+1},f}{p}_{m+2}.\hfill \end{array}$$

Proof.

We consider $$\begin{array}{cc}\Delta (e+p_0)\hfill & =[a,e+p_0]+{\lambda }_{e+p_0}\tau (e+p_0)=\hfill \\ \hfill & =\left[a_{e,e+p_0}e+a_{h,e+p_0}h+a_{f,e+p_0}f+\sum _{k=0}^{2l}{a}_{p_k,e+p_0}{p}_k+a_{z,e+p_0}z,e+p_0\right]+{\lambda }_{e+p_0}\tau (e+p_0)=\hfill \\ \hfill & =2a_{h,e+p_0}e-a_{f,e+p_0}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},e+p_0}{p}_k+\hfill \\ \hfill & +\left(2l{a}_{h,e+p_0}+\frac{{\lambda }_{e+p_0}}{2}\right)p_0+2l{a}_{f,e+p_0}{p}_1+{(-1)}^{l+\frac{1}{2}}(2l)!a_{p_{2l},e+p_0}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (e+p_0)\hfill & =\Delta (e)+\Delta (p_0)=2b_{h,e}e-\sum _{k=0}^{2l-1}(k+1)b_{p_{k+1},e}{p}_k+\hfill \\ \hfill & +2l{c}_{h,p_0}{p}_0+2l{b}_{f,p_0}{p}_1.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l-1}$ and z, we get $b_{p_{2l},e}=0.$

Now consider, $i\in \{1,2,3,\dots ,2l-1\}\setminus \{m,m+1\}$ $$\begin{array}{cc}\Delta (e+p_i)\hfill & =[a,e+p_i]+{\lambda }_{e+p_i}\tau (e+p_i)=\hfill \\ \hfill & =\left[a_{e,e+p_i}e+a_{h,e+p_i}h+a_{f,e+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,e+p_i}{p}_k+a_{z,a+p_i}z,e+p_i\right]\hfill \\ \hfill & +{\lambda }_{e+p_i}\tau (e+p_i)=\hfill \\ \hfill & =2a_{h,e+p_i}e-a_{f,e+p_i}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},e+p_i}{p}_k+\hfill \\ \hfill & +i{a}_{e,e+p_i}{p}_{i-1}+\left(2(l-i)a_{h,e+p_i}+\frac{{\lambda }_{e+p_i}}{2}\right)p_i+(2l-i)a_{f,e+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},e+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (e+p_i)\hfill & =\Delta (e)+\Delta (p_i)=2b_{h,e}e-\sum _{k=0}^{2l-1}(k+1)b_{p_{k+1},e}{p}_k+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l-i-1}$ and z, we have $b_{p_{2l-i},e}=0.$

Similarly, considering $i\in \{1,2,3,\dots ,2l-1\}\setminus \{m,m+1\}$ $$\begin{array}{cc}\Delta (f+p_i)\hfill & =[a,f+p_i]+{\lambda }_{f+p_i}\tau (f+p_i)=\hfill \\ \hfill & =\left[a_{e,f+p_i}e+a_{h,f+p_i}h+a_{f,f+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,f+p_i}{p}_k+a_{z,f+i}z,f+p_i\right]+{\lambda }_{f+p_i}\tau (f+p_i)=\hfill \\ \hfill & =a_{e,f+p_i}h-2a_{h,f+p_i}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,f+p_i}{p}_{k+1}+\hfill \\ \hfill & +i{a}_{e,f+p_i}{p}_{i-1}+\left(2(l-i)a_{h,f+p_i}+\frac{{\lambda }_{e+f_i}}{2}\right)p_i+(2l-i)a_{f,e+f_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},f+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (f+p_i)\hfill & =\Delta (f)+\Delta (p_i)=-2b_{h,f}f-\sum _{k=0}^{2l-1}(2l-k)b_{p_k,f}{p}_{k+1}+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{k-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l-i+1}$ and z, we get $b_{p_{2l-i},f}=0.$

We considering, $$\begin{array}{cc}\Delta (f+p_{2l})\hfill & =[a,f+p_{2l}]+{\lambda }_{f+p_{2l}}\tau (f+p_{2l})=\hfill \\ \hfill & =\left[a_{e,f+p_{2l}}e+a_{h,f+p_{2l}}h+a_{f,f+p_{2l}}f+\sum _{k=0}^{2l}{a}_{p_k,f+p_{2l}}{p}_k+a_{z,f+p_{2l}}z,f+p_{2l}\right]\hfill \\ \hfill & +{\lambda }_{f+p_{2l}}\tau (f+p_{2l})=\hfill \\ \hfill & =a_{e,f+p_{2l}}h-2a_{h,f+p_{2l}}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,f+p_{2l}}{p}_{k+1}+\hfill \\ \hfill & +2l{a}_{e,f+p_{2l}}{p}_{2l-1}+\left(-2l{a}_{h,f+p_{2l}}+\frac{{\lambda }_{f+p_{2l}}}{2}\right)p_{2l}+{(-1)}^{3l-\frac{1}{2}}{a}_{p_0,f+p_{2l}}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (f+p_{2l})\hfill & =\Delta (f)+\Delta (p_{2l})=-2b_{h,f}f-\sum _{k=0}^{2l-1}(2l-k)b_{p_k,f}{p}_{k+1}+\hfill \\ \hfill & +2l{b}_{e,p_{2l}}{p}_{2l-1}-2l{c}_{h,p_{2l}}{p}_{2l}.\hfill \end{array}$$

Comparing the coefficients at the basis elements p₁ and z, we have $b_{p_0,f}=0.$ □

Lemma 3.6.

Let $l\in \{\mathbb{N}+\frac{3}{2}\}.$ Then $$\begin{array}{cc}\Delta (p_i)\hfill & =(2l-i)c_{h,p_i}{p}_i,\hfill \\ \Delta (p_m)\hfill & =2(l-m)c_{h,p_m}{p}_m+m!(2l-m)!b_{p_{2l-m},p_m}z,\hfill \\ \Delta (p_{m+1})\hfill & =2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}-(m+1)!(2l-m-1)!b_{p_{2l-m-1},p_{m+1}}z,\hfill \end{array}$$where $i\in \{1,2,3,\dots ,2l\}\setminus \{m,m+1\}.$

Proof.

$l\in \{\mathbb{N}+\frac{3}{2}\}.$ We consider $$\begin{array}{cc}\Delta (e+p_{2l-1})\hfill & =[a,e+p_{2l-1}]+{\lambda }_{e+p_{2l-1}}\tau (e+p_{2l-1})=\hfill \\ \hfill & =\left[a_{e,e+p_{2l-1}}e+a_{h,e+p_{2l-1}}h+a_{f,e+p_{2l-1}}f+\sum _{k=0}^{2l}{a}_{p_k,e+p_{2l-1}}{p}_k+a_{z,e+p_{2l-1}}z,e+p_{2l-1}\right]+\hfill \\ \hfill & +{\lambda }_{e+p_{2l-1}}\tau (e+p_{2l-1})=\hfill \\ \hfill & =2a_{h,e+p_{2l-1}}e-a_{f,e+p_{2l-1}}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},e+p_{2l-1}}{p}_k+\hfill \\ \hfill & +(2l-1)a_{e,e+p_{2l-1}}{p}_{2l-2}+\left(2(1-l)a_{h,e+p_{2l-1}}+\frac{{\lambda }_{e+p_{2l-1}}}{2}\right)p_{2l-1}+a_{f,e+p_{2l-1}}{p}_{2l}+\hfill \\ \hfill & +{(-1)}^{3l-\frac{3}{2}}(2l-1)!a_{p_1,e+p_{2l-1}}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (e+p_{2l-1})\hfill & =\Delta (e)+\Delta (p_{2l-1})=2b_{h,e}e-m{b}_{p_m,e}{p}_{m-1}-(m+1)b_{p_{m+1},e}{p}_m+\hfill \\ \hfill & +(2l-1)b_{e,p_{2l-1}}{p}_{2l-2}+2(1-l)c_{h,p_{2l-1}}{p}_{2l-1}+b_{f,p_{2l-1}}{p}_{2l}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l}$ and h, we get $b_{f,p_{2l-1}}=0.$

We consider $i\in \{0,1,2,\dots ,2l-4\}\setminus \{m,m+1\},$ $$\begin{array}{cc}\Delta (p_{2l-1}+p_i)\hfill & =[a,p_{2l-1}+p_i]+{\lambda }_{p_{2l-1}+p_i}\tau (p_{2l-1}+p_i)=\hfill \\ \hfill & =[a_{e,p_{2l-1}+p_i}e+a_{h,p_{2l-1}+p_i}h+a_{f,p_{2l-1}+p_i}f\hfill \\ \hfill & +\sum _{k=0}^{2l}{a}_{p_k,p_{2l-1}+p_i}{p}_k+a_{z,p_{2l-1}+p_i}z,p_{2l-1}+p_i]+\hfill \\ \hfill & +{\lambda }_{p_{2l-1}+p_i}\tau (p_{2l-1}+p_i)=\hfill \\ \hfill & =(2l-1)a_{e,p_{2l-1}+p_i}{p}_{2l-2}+\left(2(1-l)a_{h,p_{2l-1}+p_i}+\frac{{\lambda }_{p_{2l-1}+p_i}}{2}\right)p_{2l-1}+a_{f,p_{2l-1}+p_i}{p}_{2l}+\hfill \\ \hfill & +{(-1)}^{3l-\frac{3}{2}}(2l-1)!a_{p_1,p_{2l-1}+p_i}z+\hfill \\ \hfill & +i{a}_{e,p_{2l-1}+p_i}{p}_{i-1}+\left(2(l-i)a_{h,p_{2l-1}+p_i}+\frac{{\lambda }_{p_{2l-1}+p_i}}{2}\right)p_i+(2l-i)a_{f,p_{2l-1}+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_{2l-1}+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_{2l-1}+p_i)\hfill & =\Delta (p_{2l-1})+\Delta (p_i)=(2l-1)b_{e,p_i}{p}_{2l-2}+2(1-l)c_{h,p_i}{p}_{2l-1}+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l}$ and $p_{i+1},$ we have $b_{f,p_i}=0.$

We consider $i\in \{m,m+1\},$ $$\begin{array}{cc}\Delta (p_{2l-1}+p_i)\hfill & =[a,p_{2l-1}+p_i]+{\lambda }_{p_{2l-1}+p_i}\tau (p_{2l-1}+p_i)=\hfill \\ \hfill & =[a_{e,p_{2l-1}+p_i}e+a_{h,p_{2l-1}+p_i}h+a_{f,p_{2l-1}+p_i}f\hfill \\ \hfill & +\sum _{k=0}^{2l}{a}_{p_k,p_{2l-1}+p_i}{p}_k+a_{z,p_{2l-1}+p_i}z,p_{2l-1}+p_i]+\hfill \\ \hfill & +{\lambda }_{p_{2l-1}+p_i}\tau (p_{2l-1}+p_i)=\hfill \\ \hfill & =(2l-1)a_{e,p_{2l-1}+p_i}{p}_{2l-2}+\left(2(1-l)a_{h,p_{2l-1}+p_i}+\frac{{\lambda }_{p_{2l-1}+p_i}}{2}\right)p_{2l-1}+a_{f,p_{2l-1}+p_i}{p}_{2l}+\hfill \\ \hfill & +{(-1)}^{3l-\frac{3}{2}}(2l-1)!a_{p_1,p_{2l-1}+p_i}z+\hfill \\ \hfill & +i{a}_{e,p_{2l-1}+p_i}{p}_{i-1}+\left(2(l-i)a_{h,p_{2l-1}+p_i}+\frac{{\lambda }_{p_{2l-1}+p_i}}{2}\right)p_i+(2l-i)a_{f,p_{2l-1}+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_{2l-1}+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_{2l-1}+p_i)\hfill & =\Delta (p_{2l-1})+\Delta (p_i)=(2l-1)b_{e,p_i}{p}_{2l-2}+2(1-l)c_{h,p_i}{p}_{2l-1}+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!b_{p_{2l-i},p_i}z.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l}$ and $p_{i+1},$ for $l\in \{\mathbb{N}+\frac{5}{2}\}$ we have $b_{f,p_m}=b_{f,p_{m+1}}=0$ and for $l=\frac{5}{2}$ we get only $b_{f,p_{m+1}}=0.$

We consider $i\in \{2l-3,2l-2\},$ $$\begin{array}{cc}\Delta (p_0+p_i)\hfill & =[a,p_0+p_i]+{\lambda }_{p_0+p_i}\tau (p_0+p_i)=\hfill \\ \hfill & =\left[a_{e,p_0+p_i}e+a_{h,p_0+p_i}h+a_{f,p_0+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,p_0+p_i}{p}_k+a_{z,p_0+p_i}z,p_0+p_i\right]\hfill \\ \hfill & +{\lambda }_{p_0+p_i}\tau (p_0+p_i)=\hfill \\ \hfill & =\left(2l{a}_{h,p_0+p_i}+\frac{{\lambda }_{p_0+p_i}}{2}\right)p_0+2l{a}_{f,p_0+p_i}{p}_1+{(-1)}^{l-\frac{1}{2}}(2l)!a_{p_{2l},p_0+p_i}z+\hfill \\ \hfill & +i{a}_{e,p_0+p_i}{p}_{i-1}+\left(2(l-i)a_{h,p_0+p_i}+\frac{{\lambda }_{p_0+p_i}}{2}\right)p_i+(2l-i)a_{f,p_0+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_0+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_0+p_i)\hfill & =\Delta (p_0)+\Delta (p_i)=2l{c}_{h,p_0}{p}_0+2l{b}_{f,p_0}{p}_1+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+(2l-i)b_{f,p_i}{p}_{i+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements f and $p_{i+1},$ we have $b_{f,p_i}=0.$

We consider $$\begin{array}{cc}\Delta (f+p_1)\hfill & =[a,f+p_1]+{\lambda }_{f+p_1}\tau (f+p_1)=\hfill \\ \hfill & =\left[a_{e,f+p_1}e+a_{h,f+p_1}h+a_{f,f+p_1}f+\sum _{k=0}^{2l}{a}_{p_k,f+p_1}{p}_k+a_{z,f+p_1}z,f+p_1\right]\hfill \\ \hfill & +{\lambda }_{f+p_1}\tau (f+p_1)=\hfill \\ \hfill & =a_{e,f+p_1}h-2a_{h,f+p_1}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,f+p_1}{p}_{k+1}+\hfill \\ \hfill & +a_{e,f+p_1}{p}_0+\left(2(l-1)a_{h,f+p_1}+\frac{{\lambda }_{f+p_1}}{2}\right)p_1+(2l-1)a_{f,f+p_1}{p}_2+\hfill \\ \hfill & +{(-1)}^{l+\frac{1}{2}}(2l-1)!a_{p_{2l-1},f+p_1}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (f+p_1)\hfill & =\Delta (f)+\Delta (p_1)=b_{h,e}f-(m+1)b_{p_m,f}{p}_{m+1}-m{b}_{p_{m+1},f}{p}_{m+2}+\hfill \\ \hfill & +b_{e,p_1}{p}_0+2(l-1)c_{h,p_1}{p}_1.\hfill \end{array}$$

Comparing the coefficients at the basis elements h and $p_0,$ we get $b_{e,p_1}=0.$

We consider $i\in \{4,5,\dots ,2l\}\setminus \{m,m+1\},$ $$\begin{array}{cc}\Delta (p_1+p_i)\hfill & =[a,p_1+p_i]+{\lambda }_{p_1+p_i}\tau (p_1+p_i)=\hfill \\ \hfill & =\left[a_{e,p_1+p_i}e+a_{h,p_1+p_i}h+a_{f,p_1+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,p_1+p_i}{p}_k+a_{z,p_1+p_i}z,p_1+p_i\right]\hfill \\ \hfill & +{\lambda }_{p_1+p_i}\tau (p_1+p_i)=\hfill \\ \hfill & =a_{e,p_1+p_i}{p}_0+\left(2(l-1)a_{h,p_1+p_i}+\frac{{\lambda }_{p_1+p_i}}{2}\right)p_1+(2l-1)a_{f,p_1+p_i}{p}_2+\hfill \\ \hfill & +{(-1)}^{l+\frac{1}{2}}(2l-1)!a_{p_{2l-1},p_1+p_i}z+\hfill \\ \hfill & +i{a}_{e,p_1+p_i}{p}_{i-1}+\left(2(l-i)a_{h,p_1+p_i}+\frac{{\lambda }_{p_1+p_i}}{2}\right)p_i+(2l-i)a_{f,p_1+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_1+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_1+p_i)\hfill & =\Delta (p_1)+\Delta (p_i)=b_{e,p_1}{p}_0+2(l-1)c_{h,p_1}{p}_1+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i.\hfill \end{array}$$

Comparing the coefficients at the basis elements p₀ and $p_{i-1},$ we have $b_{e,p_i}=0.$

We consider $i\in \{m,m+1\},$ $$\begin{array}{cc}\Delta (p_1+p_i)\hfill & =[a,p_1+p_i]+{\lambda }_{p_1+p_i}\tau (p_1+p_i)=\hfill \\ \hfill & =\left[a_{e,p_1+p_i}e+a_{h,p_1+p_i}h+a_{f,p_1+p_i}f+\sum _{k=0}^{2l}{a}_{p_k,p_1+p_i}{p}_k+a_{z,p_1+p_i}z,p_1+p_i\right]\hfill \\ \hfill & +{\lambda }_{p_1+p_i}\tau (p_1+p_i)=\hfill \\ \hfill & =a_{e,p_1+p_i}{p}_0+\left(2(l-1)a_{h,p_1+p_i}+\frac{{\lambda }_{p_1+p_i}}{2}\right)p_1+(2l-1)a_{f,p_1+p_i}{p}_2+\hfill \\ \hfill & +{(-1)}^{l+\frac{1}{2}}(2l-1)!a_{p_{2l-1},p_1+p_i}z+\hfill \\ \hfill & +i{a}_{e,p_1+p_i}{p}_{i-1}+\left(2(l-i)a_{h,p_1+p_i}+\frac{{\lambda }_{p_1+p_i}}{2}\right)p_i+(2l-i)a_{f,p_1+p_i}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_1+p_i}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_1+p_i)\hfill & =\Delta (p_1)+\Delta (p_i)=b_{e,p_1}{p}_0+2(l-1)c_{h,p_1}{p}_1+\hfill \\ \hfill & +i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!b_{p_{2l-i},p_i}z.\hfill \end{array}$$

Comparing the coefficients at the basis elements p₀ and $p_{i-1},$ for $l\in \{\mathbb{N}+\frac{5}{2}\}$ which implies $b_{e,p_i}=0.$

We consider $$\begin{array}{c}\Delta (p_1+p_{2l})=[a,p_1+p_{2l}]+{\lambda }_{p_1+p_{2l}}\tau (p_1+p_{2l})=\\ =\left[a_{e,p_1+p_{2l}}e+a_{h,p_1+p_{2l}}h+a_{f,p_1+p_{2l}}f+\sum _{k=0}^{2l}{a}_{p_k,p_1+p_{2l}}{p}_k+a_{z,p_1+p_{2l}},p_1+p_{2l}\right]\\ +{\lambda }_{p_1+p_{2l}}\tau (p_1+p_{2l})=\\ =a_{e,p_1+p_{2l}}{p}_0+\left(2(l-1)a_{h,p_1+p_{2l}}+\frac{{\lambda }_{p_1+p_{2l}}}{2}\right)p_1+(2l-1)a_{f,p_1+p_{2l}}{p}_2+\\ +{(-1)}^{l+\frac{1}{2}}(2l-1)!a_{p_{2l-1},p_1+p_{2l}}z+\\ +2l{a}_{e,p_1+p_{2l}}{p}_{2l-1}+\left(-2l{a}_{h,p_1+p_{2l}}+\frac{{\lambda }_{p_1+p_{2l}}}{2}\right)p_{2l}+{(-1)}^{3l-\frac{1}{2}}{a}_{p_1,p_1+p_{2l}}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_1+p_{2l})\hfill & =\Delta (p_1)+\Delta (p_i)=b_{e,p_1}{p}_0+2(l-1)c_{h,p_1}{p}_1+(2l-1)b_{f,p_1}{p}_2+\hfill \\ \hfill & +2l{b}_{e,p_{2l}}{p}_{2l-1}-2l{c}_{h,p_{2l}}{p}_{2l}.\hfill \end{array}$$

Comparing the coefficients at the basis elements p₀ and $p_{2l-1},$ we have $b_{e,p_{2l}}=0.$

We consider $i\in \{2,3\},$ $$\begin{array}{cc}\Delta (p_i+p_{2l})\hfill & =[a,p_i+p_{2l}]+{\lambda }_{p_i+p_{2l}}\tau (p_i+p_{2l})=\hfill \\ \hfill & =\left[a_{e,p_i+p_{2l}}e+a_{h,p_i+p_{2l}}h+a_{f,p_i+p_{2l}}f+\sum _{k=0}^{2l}{a}_{p_k,p_i+p_{2l}}{p}_k+a_{z,p_i+p_{2l}}z,p_i+p_{2l}\right]\hfill \\ \hfill & +{\lambda }_{p_i+p_{2l}}\tau (p_i+p_{2l})=\hfill \\ \hfill & =i{a}_{e,p_i+p_{2l}}{p}_{i-1}+\left(2(l-i)a_{h,p_i+p_{2l}}+\frac{{\lambda }_{p_i+p_{2l}}}{2}\right)p_i+(2l-i)a_{f,p_i+p_{2l}}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},p_i+p_{2l}}z+\hfill \\ \hfill & +2l{a}_{e,p_i+p_{2l}}{p}_{2l-1}+\left(-2l{a}_{h,p_i+p_{2l}}+\frac{{\lambda }_{p_i+p_{2l}}}{2}\right)p_{2l}+{(-1)}^{3l-\frac{1}{2}}{a}_{p_0,p_i+p_{2l}}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (p_i+p_{2l})\hfill & =\Delta (p_i)+\Delta (p_i)=i{b}_{e,p_i}{p}_{i-1}+2(l-i)c_{h,p_i}{p}_i-2l{c}_{h,p_{2l}}{p}_{2l}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{2l-1}$ and $p_{i-1},$ for $l\in \{\mathbb{N}+\frac{5}{2}\}$ we have $b_{e,p_2}=b_{f,p_3}=0$ and for $l=\frac{5}{2}$ we get only $b_{e,p_2}=0.$

Therefore, for case $l=\frac{5}{2}$ take an element $y=p_0+p_2+p_4$ $$\begin{array}{c}\Delta (y)=[a,p_0+p_2+p_4]+{\lambda }_y\tau (p_0+p_2+p_4)=\\ =\left[a_{e,y}e+a_{h,y}h+a_{f,y}f+\sum _{k=0}^{2l}{a}_{p_k,y}{p}_k+a_{z,y}z,p_0+p_2+p_4\right]+{\lambda }_y\tau (p_0+p_2+p_4)=\\ =\left(2l{a}_{h,y}+\frac{{\lambda }_y}{2}\right)p_0+2l{a}_{f,y}{p}_1+{(-1)}^{l+\frac{1}{2}}(2l)!a_{p_{2l},y}z+\\ +2a_{e,y}{p}_1+\left(2(l-2)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_2+(2l-2)a_{f,y}{p}_3+\\ +{(-1)}^{2+l-\frac{1}{2}}2!(2l-2)!a_{p_{2l-2},y}z+\\ +4a_{e,y}{p}_3+\left(2(l-4)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_4+(2l-4)a_{f,y}{p}_5+\\ +{(-1)}^{4+l-\frac{1}{2}}4!(2l-4)!a_{p_{2l-4},y}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (p_0)+\Delta (p_2)+\Delta (p_4)=2l{c}_{h,p_0}{p}_0+\hfill \\ \hfill & +2b_{e,p_2}{p}_1+2(l-2)c_{h,p_2}{p}_2+(2l-2)b_{f,p_2}{p}_3+\hfill \\ \hfill & +2(l-4)c_{h,p_4}{p}_4.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_1,p_3$ and $p_5,$ we have $b_{f,p_2}=0.$

Therefore, for case $l=\frac{5}{2}$ take an element $u=p_1+p_3+p_5$ $$\begin{array}{c}\Delta (u)=[a,p_1+p_3+p_5]+{\lambda }_y\tau (p_1+p_3+p_5)=\\ =\left[a_{e,u}e+a_{h,u}h+a_{f,u}f+\sum _{k=0}^{2l}{a}_{p_k,u}{p}_k+a_{z,u}z,p_1+p_3+p_5\right]+{\lambda }_u\tau (p_1+p_3+p_5)=\\ =a_{e,u}{p}_0+\left(2(l-1)a_{h,u}+\frac{{\lambda }_u}{2}\right)+(2l-1)a_{f,u}{p}_2+\\ +{(-1)}^{1+l-\frac{1}{2}}1!(2l-1)!a_{p_{2l-1},u}z+3a_{e,u}{p}_2+\left(2(l-3)a_{h,u}+\frac{{\lambda }_u}{2}\right)p_3+(2l-3)a_{f,u}{p}_4+\\ +{(-1)}^{3+l-\frac{1}{2}}3!(2l-3)!a_{p_{2l-3},u}z+5a_{e,u}{p}_4+\left(-5a_{h,p_5}+\frac{{\lambda }_{p_u}}{2}\right)p_5+{(-1)}^{3l-\frac{1}{2}}{b}_{p_0,u}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (u)\hfill & =\Delta (p_1)+\Delta (p_3)+\Delta (p_5)=3c_{h,p_1}{p}_1+\hfill \\ \hfill & +3b_{e,p_3}{p}_2-c_{h,p_3}{p}_3-5c_{h,p_5}{p}_5.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_0,p_2$, and $p_4,$ we have $b_{e,p_3}=0.$ □

Lemma 3.7.

$$\begin{array}{cc}\Delta (p_m)\hfill & =2(l-m)c_{h,p_m}{p}_m,\hfill \\ \Delta (p_{m+1})\hfill & =2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}.\hfill \end{array}$$

Proof.

We considering, take an element $y=h+p_m+p_{m+1}-m!(m+1)!z,$ (3.4) $$\begin{array}{cc}\Delta (y)\hfill & =[a,y]+{\lambda }_y\tau (y)=\hfill \\ \hfill & =\left[a_{e,y}e+a_{h,y}h+a_{f,y}f+\sum _{k=0}^{2l}{a}_{p_k,y}{p}_k,h+p_m+p_{m+1}-m!(m+1)!z\right]+\hfill \\ \hfill & +{\lambda }_y\tau (h+p_m+p_{m+1}-m!(m+1)!z)=\hfill \\ \hfill & =-2a_{e,y}e+2a_{f,y}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,y}{p}_k+\hfill \\ \hfill & +m{a}_{e,y}{p}_{m-1}+\left(2(l-m)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_m+(2l-m)a_{f,y}{p}_{m+1}+\hfill \\ \hfill & +{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!a_{p_{2l-m},y}z+\hfill \\ \hfill & +(m+1)a_{e,y}{p}_m+\left(2(l-m-1)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_{m+1}+\hfill \\ \hfill & +(2l-m-1)a_{f,y}{p}_{m+2}+{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!a_{p_{2l-m-1},y}z+\hfill \\ \hfill & -m!(m+1)!{\lambda }_{y}z.\hfill \end{array}$$(3.4)

On the other hand, (3.5) $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (h+p_m+p_{m+1}-m!(m+1)!z)=\Delta (p_m)+\Delta (p_{m+1})=\hfill \\ \hfill & =2(l-m)c_{h,p_m}{p}_m+{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!b_{p_{2l-m},p_m}z+\hfill \\ \hfill & +2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}+{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!b_{p_{2l-m-1},p_{m+1}}z.\hfill \end{array}$$(3.5)

Comparing the coefficients at the basis elements $e,f,p_m,p_{m+1}$ and z, (3.4) and (3.5) we obtain that (3.6) $$\begin{array}{c}-b_{p_{m+1},p_m}+b_{p_m,p_{m+1}}=-c_{h,p_m}+c_{h,p_{m+1}}.\hfill \end{array}$$(3.6)

Similarly, considering take an element $u=h+p_m-p_{m+1}+m!(m+1)!z,$ (3.7) $$\begin{array}{cc}\Delta (u)\hfill & =[a,u]+{\lambda }_u\tau (u)=\hfill \\ \hfill & =\left[a_{e,u}e+a_{h,u}h+a_{f,u}f+\sum _{k=0}^{2l}{a}_{p_k,u}{p}_k,h+p_m-p_{m+1}+m!(m+1)!z\right]+\hfill \\ \hfill & +{\lambda }_u\tau (h+p_m-p_{m+1}+m!(m+1)!z)=\hfill \\ \hfill & =-2a_{e,u}e+2a_{f,u}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,u}{p}_k+\hfill \\ \hfill & +m{a}_{e,u}{p}_{m-1}+\left(2(l-m)a_{h,u}+\frac{{\lambda }_u}{2}\right)p_m+(2l-m)a_{f,u}{p}_{m+1}+\hfill \\ \hfill & +{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!a_{p_{2l-m},u}z+\hfill \\ \hfill & -(m+1)a_{e,u}{p}_m-\left(2(l-m-1)a_{h,u}+\frac{{\lambda }_u}{2}\right)p_{m+1}-\hfill \\ \hfill & -(2l-m-1)a_{f,u}{p}_{m+2}-{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!a_{p_{2l-m-1},u}z-\hfill \\ \hfill & +m!(m+1)!{\lambda }_{u}z.\hfill \end{array}$$(3.7)

On the other hand, (3.8) $$\begin{array}{cc}\Delta (u)\hfill & =\Delta (h+p_m-p_{m+1}+m!(m+1)!z)=\Delta (p_m)-\Delta (p_{m+1})=\hfill \\ \hfill & =2(l-m)c_{h,p_m}{p}_m+{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!b_{p_{2l-m},p_m}z+\hfill \\ \hfill & -2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}-{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!b_{p_{2l-m-1},p_{m+1}}z.\hfill \end{array}$$(3.8)

Comparing the coefficients at the basis elements $e,f,p_m,p_{m+1}$ and z, by Equations (3.7)(3.7) $$\begin{array}{cc}\Delta (u)\hfill & =[a,u]+{\lambda }_u\tau (u)=\hfill \\ \hfill & =\left[a_{e,u}e+a_{h,u}h+a_{f,u}f+\sum _{k=0}^{2l}{a}_{p_k,u}{p}_k,h+p_m-p_{m+1}+m!(m+1)!z\right]+\hfill \\ \hfill & +{\lambda }_u\tau (h+p_m-p_{m+1}+m!(m+1)!z)=\hfill \\ \hfill & =-2a_{e,u}e+2a_{f,u}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,u}{p}_k+\hfill \\ \hfill & +m{a}_{e,u}{p}_{m-1}+\left(2(l-m)a_{h,u}+\frac{{\lambda }_u}{2}\right)p_m+(2l-m)a_{f,u}{p}_{m+1}+\hfill \\ \hfill & +{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!a_{p_{2l-m},u}z+\hfill \\ \hfill & -(m+1)a_{e,u}{p}_m-\left(2(l-m-1)a_{h,u}+\frac{{\lambda }_u}{2}\right)p_{m+1}-\hfill \\ \hfill & -(2l-m-1)a_{f,u}{p}_{m+2}-{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!a_{p_{2l-m-1},u}z-\hfill \\ \hfill & +m!(m+1)!{\lambda }_{u}z.\hfill \end{array}$$(3.7) and (3.8)(3.8) $$\begin{array}{cc}\Delta (u)\hfill & =\Delta (h+p_m-p_{m+1}+m!(m+1)!z)=\Delta (p_m)-\Delta (p_{m+1})=\hfill \\ \hfill & =2(l-m)c_{h,p_m}{p}_m+{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!b_{p_{2l-m},p_m}z+\hfill \\ \hfill & -2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}-{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!b_{p_{2l-m-1},p_{m+1}}z.\hfill \end{array}$$(3.8) we obtain that (3.9) $$\begin{array}{c}-b_{p_{m+1},p_m}-b_{p_m,p_{m+1}}=c_{h,p_m}-c_{h,p_{m+1}}.\hfill \end{array}$$(3.9)

Similarly, considering take an element $v=h-p_m+p_{m+1}+m!(m+1)!z,$ (3.10) $$\begin{array}{cc}\Delta (v)\hfill & =[a,v]+{\lambda }_v\tau (v)=\hfill \\ \hfill & =\left[a_{e,v}e+a_{h,v}h+a_{f,v}f+\sum _{k=0}^{2l}{a}_{p_k,v}{p}_k,h-p_m+p_{m+1}+m!(m+1)!z\right]+\hfill \\ \hfill & +{\lambda }_v\tau (h-p_m+p_{m+1}+m!(m+1)!z)=\hfill \\ \hfill & =-2a_{e,v}e+2a_{f,v}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,v}{p}_k+\hfill \\ \hfill & -m{a}_{e,v}{p}_{m-1}-\left(2(l-m)a_{h,v}+\frac{{\lambda }_v}{2}\right)p_m-(2l-m)a_{f,v}{p}_{m+1}+\hfill \\ \hfill & -{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!a_{p_{2l-m},v}z+\hfill \\ \hfill & +(m+1)a_{e,v}{p}_m+\left(2(l-m-1)a_{h,v}+\frac{{\lambda }_v}{2}\right)p_{m+1}+\hfill \\ \hfill & +(2l-m-1)a_{f,v}{p}_{m+2}+{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!a_{p_{2l-m-1},v}z+\hfill \\ \hfill & +m!(m+1)!{\lambda }_{v}z.\hfill \end{array}$$(3.10)

On the other hand, (3.11) $$\begin{array}{cc}\Delta (v)\hfill & =\Delta (h-p_m+p_{m+1}+m!(m+1)!z)=-\Delta (p_m)+\Delta (p_{m+1})=\hfill \\ \hfill & =-2(l-m)c_{h,p_m}{p}_m-{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!b_{p_{2l-m},p_m}z+\hfill \\ \hfill & +2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}+{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!b_{p_{2l-m-1},p_{m+1}}z.\hfill \end{array}$$(3.11)

Comparing the coefficients at the basis elements $e,f,p_m,p_{m+1}$ and z, (3.10) and (3.11) we obtain that (3.12) $$b_{p_{m+1},p_m}+b_{p_m,p_{m+1}}=c_{h,p_m}-c_{h,p_{m+1}}.$$(3.12)

Comparing (3.6), (3.9), and (3.12) we obtain that $$b_{p_{m+1},p_m}=b_{p_m,p_{m+1}}=0.$$

□

Lemma 3.8.

$$\begin{array}{cc}\Delta (e)\hfill & =2b_{h,e}e-(m+1)b_{p_{m+1},e}{p}_m,\hfill \\ \Delta (f)\hfill & =-2b_{h,f}f-(m+1)b_{p_m,f}{p}_{m+1}.\hfill \end{array}$$

Proof.

We consider $$\begin{array}{c}\Delta (e+p_{m+1})=[a,e+p_{m+1}]+{\lambda }_{e+p_{m+1}}\tau (e+p_{m+1})=\\ =\left[a_{e,e+p_{m+1}}e+a_{h,e+p_{m+1}}h+a_{f,e+p_{m+1}}f+\sum _{k=0}^{2l}{a}_{p_k,e+p_{m+1}}{p}_k+a_{z,e+p_{m+1}}z,e+p_{m+1}\right]+\\ +{\lambda }_{e+p_{m+1}}\tau (e+p_{m+1})=\\ =2a_{h,e+p_{m+1}}e-a_{f,e+p_{m+1}}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},e+p_{m+1}}{p}_k+\\ +(m+1)a_{e,e+p_{m+1}}{p}_m+\left(2(l-m-1)a_{h,e+p_{m+1}}+\frac{{\lambda }_{e+p_{m+1}}}{2}\right)p_{m+1}+\\ +(2l-m-1)a_{f,e+p_{m+1}}{p}_{m+2}+{(-1)}^{m+l+\frac{1}{2}}(m+1)!m!a_{p_{2l-m-1},e+p_{m+1}}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (e+p_{m+1})\hfill & =\Delta (e)+\Delta (p_{m+1})=2b_{h,e}e-m{b}_{p_m,e}{p}_{m-1}-(m+1)b_{p_{m+1},e}{p}_m+\hfill \\ \hfill & +2(l-m-1)c_{h,p_{m+1}}{p}_{m+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{m-1}$ and z, we get $b_{p_m,e}=0.$

Similarly, considering $$\begin{array}{c}\Delta (f+p_m)=[a,f+p_m]+{\lambda }_{f+p_m}\tau (f+p_m)=\\ =\left[a_{e,f+p_m}e+a_{h,f+p_m}h+a_{f,f+p_m}f+\sum _{k=0}^{2l}{a}_{p_k,f+p_m}{p}_k+a_{z+f+p_m}z,f+p_m\right]+\\ +{\lambda }_{f+p_m}\tau (f+p_m)=\\ =a_{e,f+p_m}h-2a_{h,f+p_m}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,f+p_m}{p}_{k+1}+\\ +m{a}_{e,f+p_m}{p}_{m-1}+\left(2(l-m)a_{h,f+p_m}+\frac{{\lambda }_{e+f_m}}{2}\right)p_m+(2l-m)a_{f,e+f_m}{p}_{m+1}+\\ +{(-1)}^{m+l-\frac{1}{2}}m!(2l-m)!a_{p_{2l-m},f+p_m}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (f+p_m)\hfill & =\Delta (f)+\Delta (p_m)=-2b_{h,e}f-(m+1)b_{p_m,f}{p}_{m+1}-m{b}_{p_{m+1},f}{p}_{m+2}+\hfill \\ \hfill & +m{b}_{e,p_m}{p}_{m-1}+2(l-m)c_{h,p_m}{p}_m+(2l-m)b_{f,p_m}{p}_{m+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $p_{m+2}$ and z, we get $b_{p_{m+1},f}=0.$ □

Lemma 3.9.

Let $l\in \{\mathbb{N}+\frac{3}{2}\}.$ Then $$\begin{array}{cc}\Delta (p_i)\hfill & =2(l-i)b_{h,f}{p}_i.\hfill \end{array}$$

Proof.

We consider take an element $y=h+p_i+p_{2l-i}+\frac{{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!}{2(i-l)}z,i=0,1,\dots ,m$ (3.13) $$\begin{array}{cc}\Delta (y)\hfill & =[a,y]+{\lambda }_y\tau (y)=\hfill \\ \hfill & =\left[a_{e,y}e+a_{h,y}h+a_{f,y}f+\sum _{k=0}^{2l}{a}_{p_k,y}{p}_k+a_{z,y}z,y\right]+\hfill \\ \hfill & +{\lambda }_y\tau (h+p_i+p_{2l-i}+\frac{{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!}{2(i-l)}z)=\hfill \\ \hfill & =-2a_{e,y}e+2a_{f,y}f+\sum _{k=0}^{2l}2(k-l)a_{p_k,y}{p}_k+\hfill \\ \hfill & +i{a}_{e,y}{p}_{i-1}+\left(2(l-i)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_i+(2l-i)a_{f,y}{p}_{i+1}+\hfill \\ \hfill & +{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},y}z+\hfill \\ \hfill & +(2l-i)a_{e,y}{p}_{2l-i-1}+\left(2(i-l)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_{2l-i}+i{a}_{f,y}{p}_{2l-i+1}+\hfill \\ \hfill & +{(-1)}^{3l-i-\frac{1}{2}}(2l-i)!i!a_{p_i,y}z+\frac{{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!}{2(i-l)}{\lambda }_{y}z.\hfill \end{array}$$(3.13)

On the other hand, (3.14) $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (h+p_i+p_{2l-i}+\frac{{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!}{2(i-l)}z)=\Delta (p_i)+\Delta (p_{2l-i})=\hfill \\ \hfill & =(2l-i)c_{h,p_i}{p}_i+i{c}_{h,p_{2l-i}}{p}_{2l-i}.\hfill \end{array}$$(3.14)

Comparing the coefficients at the basis elements $p_i,p_{2l-i}$, and z, we get $$\begin{array}{cc}2(i-l)a_{p_i,y}-2(i-l)a_{h,y}+\frac{{\lambda }_y}{2}\hfill & =-2(i-l)c_{h,p_i},\hfill \\ -2(i-l)a_{p_{2l-i},y}+2(i-l)a_{h,y}+\frac{{\lambda }_y}{2}\hfill & =-2(i-l)c_{h,p_{2l-i}},\hfill \\ {(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_{2l-i},y}-\hfill & \hfill \\ -{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_k,y}+\frac{i+l-\frac{1}{2}i!(2l-i)!}{2(i-l)}{\lambda }_y\hfill & =0,\hfill \end{array}$$which implies (3.15) $$c_{h,p_i}=c_{h,p_{2l-i}},\mathit{\hspace{1em}i}=0,1,\dots ,m.$$(3.15)

We consider take an element $t=f+p_{i+1}+p_{2l-i}+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i-1)!z,i=0,1,\dots ,m,$ $$\begin{array}{cc}\Delta (t)\hfill & =[a,t]+{\lambda }_t\tau (t)=\hfill \\ \hfill & =\left[a_{e,t}e+a_{h,t}h+a_{f,t}f+\sum _{k=0}^{2l}{a}_{p_k,t}{p}_k+a_{z,t}z,t\right]+\hfill \\ \hfill & +{\lambda }_t\tau (f+p_{i+1}+p_{2l-i}+{(-1)}^{i+l+\frac{1}{2}}i!(2l-i-1)!z)=\hfill \\ \hfill & =a_{e,t}h-2a_{h,t}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,t}{p}_{k+1}+\hfill \\ \hfill & +(i+1)a_{e,t}{p}_i+\left(2(l-i-1)a_{h,t}+\frac{{\lambda }_t}{2}\right)p_{i+1}+(2l-i-1)a_{f,t}{p}_{i+2}+\hfill \\ \hfill & +{(-1)}^{i+l+\frac{1}{2}}(i+1)!(2l-i-1)!a_{p_{2l-i-1},t}z+\hfill \\ \hfill & +(2l-i)a_{e,t}{p}_{2l-i-1}+\left(2(i-l)a_{h,t}+\frac{{\lambda }_t}{2}\right)p_{2l-i}+i{a}_{f,t}{p}_{2l-i+1}+\hfill \\ \hfill & +{(-1)}^{3l-i-\frac{1}{2}}(2l-i)!i!a_{p_i,t}z+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i-1)!{\lambda }_{t}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (f+p_{i+1}+p_{2l-i}+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i-1)!z)=\Delta (p_{i+1})+\Delta (p_{2l-i})=\hfill \\ \hfill & =(2l-i-1)c_{h,p_{i+1}}{p}_{i+1}+i{c}_{h,p_{2l-i}}{p}_{2l-i}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $f,p_{i+1},p_{2l-i}$, and z, we get (3.16) $$\begin{array}{cc}-2a_{h,t}\hfill & =-2b_{h,f},\hfill \\ (i-2l)a_{p_i,t}+2(l-i-1)a_{h,t}+\frac{{\lambda }_t}{2}\hfill & =2(l-i-1)c_{h,p_{i+1}},\hfill \\ -(i+l)a_{p_{2l-i-1},t}+2(i-l)a_{h,t}+\frac{{\lambda }_t}{2}\hfill & =2(i-l)c_{h,p_{2l-i}},\hfill \\ {(-1)}^{i+l+\frac{1}{2}}(i+1)!(2l-i-1)!a_{p_{2l-i-1},t}-\hfill & \hfill \\ -{(-1)}^{i+l-\frac{1}{2}}i!(2l-i)!a_{p_k,t}+{(-1)}^{i+l-\frac{1}{2}}i!(2l-i-1)!{\lambda }_y\hfill & =0,\hfill \end{array}$$(3.16) which implies (3.17) $$\begin{array}{cc}-2b_{h,f}\hfill & =2(l-i-1)c_{h,p_{i+1}}+2(i-l)c_{h,p_{2l-i}},\hspace{1em}i=0,1,\dots ,m-1,\hfill \end{array}$$(3.17) and i = m we obtain that $$b_{h,f}=c_{h,p_{m+1}}.$$

Comparing (3.15) and (3.17) we obtain that (3.18) $$-2(i-l)c_{h,p_i}=2(l-i-1)c_{h,p_{i+1}}+2b_{h,f},\hspace{1em}i=0,1,\dots ,m,$$(3.18) on recurrent equation instead of i, gradually substituting $m,m-1,\dots ,0$ we obtain the following (3.19) $$c_{h,p_i}=b_{h,f},\hspace{1em}i=0,1,\dots ,m.$$(3.19)

Comparing (3.15) and (3.19) we obtain that $$c_{h,p_i}=b_{h,f},\hspace{1em}i=0,1,\dots ,2l.$$

□

Lemma 3.10.

Let $l\in \{\mathbb{N}+\frac{3}{2}\}.$ Then $$\begin{array}{cc}\Delta (e)\hfill & =2b_{h,f}e,\hfill \\ \Delta (f)\hfill & =-2b_{h,f}f.\hfill \end{array}$$

Proof.

We consider $$\begin{array}{cc}\Delta (e+f)\hfill & =[a,e+f]+{\lambda }_{e+f}\tau (e+f)=\hfill \\ \hfill & =\left[a_{e,e+f}e+a_{h,e+f}h+a_{f,e+f}f+\sum _{k=0}^{2l}{a}_{p_k,e+f}{p}_k+a_{z,e+f}z,e+f\right]+{\lambda }_{e+f}\tau (e+f)=\hfill \\ \hfill & =2a_{h,e+f}e-a_{f,e+f}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},e+f}{p}_k+\hfill \\ \hfill & +a_{e,e+f}h-2a_{h,e+f}f-\sum _{k=0}^{2l-1}(2l-k)a{+}_{p_k,e+f}{p}_{k+1}.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (e+f)\hfill & =\Delta (e)+\Delta (f)=2b_{h,e}e-(m+1)b_{p_{m+1},e}{p}_m-2b_{h,f}f-(m+1)b_{p_m,f}{p}_{m+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements e and f, we get $b_{h,e}=b_{h,f}.$

We consider take an element $y=e+p_m-\frac{{(m!)}^2}{2}z,$ $$\begin{array}{cc}\Delta (y)\hfill & =\left[a,e+p_m-\frac{{(m!)}^2}{2}z\right]+{\lambda }_y\tau (e+p_m-\frac{{(m!)}^2}{2}z)=\hfill \\ \hfill & =\left[a_{e,y}e+a_{h,y}h+a_{f,y}f+\sum _{k=0}^{2l}{a}_{p_k,y}{p}_k+a_{z,y}z,y\right]+{\lambda }_y\tau (y)=\hfill \\ \hfill & =2a_{h,y}e-a_{f,y}h-\sum _{k=0}^{2l-1}(k+1)a_{p_{k+1},y}{p}_k+\hfill \\ \hfill & +m{a}_{e,y}{p}_{m-1}+\left(2(l-m)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_m+(2l-m)a_{f,y}{p}_{m+1}+\hfill \\ \hfill & +m!(2l-m)!a_{p_{2l-m},y}z-\frac{{(m!)}^2}{2}{\lambda }_{y}z.\hfill \end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (e)+\Delta (p_m)=2b_{h,e}e-(m+1)b_{p_{m+1},e}{p}_m+\hfill \\ \hfill & +2(l-m)b_{h,e}{p}_m.\hfill \end{array}$$

Comparing the coefficients at the basis elements $e,p_m$, and z, we get (3.20) $$\begin{array}{cc}{a}_{h,y}\hfill & =b_{h,e}\hfill \\ -(m+1)a_{p_{m+1},y}+a_{h,y}+\frac{{\lambda }_y}{2}\hfill & =-(m+1)b_{p_{m+1},e}+b_{h,e}\hfill \\ m!(m+1)!a_{p_{m+1},y}-\frac{{(m!)}^2}{2}{\lambda }_y\hfill & =0\hfill \end{array}$$(3.20) which implies $b_{p_{m+1},e}=0.$

We consider take an element $y=f+p_{m+1}+\frac{{(m!)}^2}{2}z,$ $$\begin{array}{c}\Delta (y)=\left[a,f+p_{m+1}+\frac{{(m!)}^2}{2}z\right]+{\lambda }_y\tau (f+p_{m+1}+\frac{{(m!)}^2}{2}z)=\\ =\left[a_{e,y}e+a_{h,y}h+a_{f,y}f+\sum _{k=0}^{2l}{a}_{p_k,y}{p}_k+a_{z,y}z,y\right]+{\lambda }_y\tau (y)=\\ =a_{e,y}h-2a_{h,y}f-\sum _{k=0}^{2l-1}(2l-k)a_{p_k,y}{p}_{k+1}+\\ +(m+1)a_{e,y}{p}_m+\left(2(l-m-1)a_{h,y}+\frac{{\lambda }_y}{2}\right)p_{m+1}+(2l-m-1)a_{f,y}{p}_{m+2}+\\ +{(-1)}^{m+l+\frac{1}{2}}(m+1)!(2l-m-1)!a_{p_{2l-m-1},y}z+\frac{{(m!)}^2}{2}{\lambda }_{y}z.\end{array}$$

On the other hand, $$\begin{array}{cc}\Delta (y)\hfill & =\Delta (f)+\Delta (p_{m+1})=-2b_{h,f}f-(m+1)b_{p_m,f}{p}_{m+1}+\hfill \\ \hfill & +2(l-m-1)b_{h,e}{p}_{m+1}.\hfill \end{array}$$

Comparing the coefficients at the basis elements $f,p_{m+1}$ and z, we get (3.21) $$\begin{array}{cc}{a}_{h,y}\hfill & =b_{h,f}\hfill \\ -(m+1)a_{p_m,y}-a_{h,y}+\frac{{\lambda }_y}{2}\hfill & =-(m+1)b_{p_m,e}-b_{h,e}\hfill \\ -(m+1)!(m)!a_{p_m,y}z+\frac{{(m!)}^2}{2}{\lambda }_y\hfill & =0\hfill \end{array}$$(3.21) which implies $b_{p_m,e}=0.$ □

Lemma 3.11.

Δ is an inner derivation on $\mathfrak{g}$ with $l\ne 1,5.$

Proof.

For any $x=x_{e}e+x_{h}h+x_{f}f+{\sum }_{k=0}^{2l}{x}_{p_k}{p}_k+x_{z}z\in \mathfrak{g}.$ We consider $$\begin{array}{cc}\Delta (x)-[b_{h,f}h,x]\hfill & =\Delta (x_{e}e+x_{h}h+x_{f}f+\sum _{k=0}^{2l}{x}_{p_k}{p}_k+x_{z}z)-\hfill \\ \hfill & -\left[b_{h,f}h,x_{e}e+x_{h}h+x_{f}f+\sum _{k=0}^{2l}{x}_{p_k}{p}_k+x_{z}z\right]=\hfill \\ \hfill & =2x_e{b}_{h,f}e-2x_f{b}_{h,f}f+\sum _{k=0}^{2l}{x}_{p_k}(2l-2k)b_{h,f}{p}_k-\hfill \\ \hfill & -2x_e{b}_{h,f}e+2x_f{b}_{h,f}f-\sum _{k=0}^{2l}{x}_{p_k}(2l-2k)b_{h,f}{p}_k=0.\hfill \end{array}$$

Then Δ is an inner derivation. □

Now we are in position to prove Theorem 3.1.

Proof of Theorem 3.1.

Let Δ be a local derivation of $\mathfrak{g}.$ Take a derivation D_x such that $$\Delta (x)=D_x(x).$$

Set ${\Delta }_1=\Delta -D_x.$ Then ${\Delta }_1$ is a local derivation such that ${\Delta }_1(h+z)=0.$ By Lemma 3.11, ${\Delta }_1(x)=a{d}_x.$ Thus $\Delta =D_x+a{d}_x.$ Then Δ is a derivation. The proof is complete. □

Acknowledgments

We thank professor K. K. Kudaybergenov for the helpful comments and suggestions that contributed to improving this paper.

Data availability statement

No data was used for the research described in the article.