New Operated Polynomial Identities and Gröbner-Shirshov Bases

Abstract

Twenty years ago, Rota posed the problem of finding all possible algebraic identities that can be satisfied by a linear operator on an algebra, named Rota’s Classification Problem later. Rota’s Classification Problem has proceeded two steps to understand it and has been studied actively recently. In particular, the method of Gröbner-Shirshov bases has been used successfully in the study of Rota’s Classification Problem. Quite recently, a new approach introduced to Rota’s Classification Problem and classified some (new) operated polynomial identities. In this paper, we prove that all operated polynomial identities classified via this new approach are Gröbner-Shirshov. This gives a partial answer of Rota’s Classification Problem.

1. Introduction

1.1. Rota’s Classification Problem

In the study of mathematics and mathematical physics, various linear operators—characterized by various operator identities—played crucial roles. Inspired by this, Rota [1] posed the problem of

finding all possible algebraic identities that can be satisfied by a linear operator on an algebra,

henceforth called Rota’s Classification Problem. Here, an algebra means an associative algebra. Such operator identities of interest to Rota included

$$\begin{array}{cc}\hfill d\left(x_1{x}_2\right)& =d\left(x_1\right)d\left(x_2\right),\hfill \\ \hfill d\left(x_1{x}_2\right)& =d\left(x_1\right)x_2+x_{1}d\left(x_2\right),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P\left(x_{1}P\left(x_2\right)\right),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P\left(P\left(x_1\right)x_2\right),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P(x_{1}P\left(x_2\right)+P\left(x_1\right)x_2+\lambda x_1{x}_2),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P(x_{1}P\left(x_2\right)+P\left(x_1\right)x_2-P\left(x_1\right)P\left(x_2\right)).\hfill \end{array}$$

After Rota posed Rota’s Classification Problem, more and more linear operators have appeared, such as

$$\begin{array}{cc}\hfill d\left(x_1{x}_2\right)& =d\left(x_1\right)x_2+x_{1}d\left(x_2\right)+\lambda d\left(x_1\right)d\left(x_2\right),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P(x_{1}P\left(x_2\right)+P\left(x_1\right)x_2-P\left(x_1{x}_2\right)),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P(x_{1}P\left(x_2\right)+P\left(x_1\right)x_2-x_{1}P\left(1\right)x_2),\hfill \\ \hfill P\left(x_1\right)P\left(x_2\right)& =P(x_{1}P\left(x_2\right)+P\left(x_1\right)x_2)+\lambda x_1{x}_2,\hfill \\ \hfill d\left(x_1{x}_2\right)& =d\left(x_1\right)x_2+x_{1}d\left(x_2\right)+\lambda x_1{x}_2.\hfill \end{array}$$

In particular, the endomorphism operator plays pivotal roles in Galois theory. The differential operator, an algebraic abstraction of derivation in analysis, has been used to lead to differential algebra as an algebraic study of differential equations and been widely and successfully applied in many important areas [2,3,4]. The understanding of Spitzer’s identity in fluctuation theory can be achieved by the Rota–Baxter operator, originated from [5] in 1960 based on the probability study. The broad connections of Rota–Baxter operators with many areas of mathematics and mathematical physics are remarkable, e.g., the classical Yang–Baxter equation, operads, combinatorics, Hopf algebra, and renormalization of quantum field theory [6,7,8,9,10,11]. Other linear operators have also been studied extensively [1,8,12,13,14,15,16,17].

1.2. History in Solving Rota’s Classification Problem

To understand Rota’s Classification Problem, we can take the following two steps. Firstly, we need to have the construction of the algebraic framework to consider algebraic identities in Rota’s Classification Problem. An easier case of algebraic identities satisfied by algebras is the noncommutative polynomial, which is free, and since then polynomial identity (PI) rings has been studied extensively in 1960s. Since algebraic identities in Rota’s Classification Problem involve linear operators, they become more complicated and are realized as free objects in the category of algebras together with linear operators, which are originated from Kurosh [18].

Secondly, in order to understand Rota’s Classification Problem, we need to consider a crucial problem, that is, what distinguishes the operated polynomial identities (OPIs) satisfied by these above listed linear operators from the arbitrary OPIs? That is to say, Rota believed that “good” OPIs needed to be identified for the purpose of further study. Roughly speaking, since algebras in Rota’s Classification Problem are associative algebras, such OPIs looked for by Rota are compatible with associativity. In the process of characterization of these compatibility, two special classes of OPIs are studied—differential type OPIs and Rota–Baxter type OPIs.

The study of differential type OPIs was carried out in [19], which includes the classical differential OPI. Therein theories of Gröbner-Shirshov bases and rewriting systems were applied successfully. Another important class of OPIs, namely Rota–Baxter type, was systematically studied in [20]. As to be expected from comparing integral calculus with differential calculus in analysis, the Rota–Baxter type OPIs are more challenging than the differential counterpart. In [20], Rota–Baxter type OPIs were also characterized by Gröbner-Shirshov bases and rewriting systems. An outstanding achievement of applications of Gröbner-Shirshov bases and rewriting systems in the characterization of differential type OPIs and Rota–Baxter type OPIs sheds a light to apply them to study general OPIs, which was carried out in [21].

Recently, a new approach to study Rota’s Classification Problem was brought forward by Bremner et al., based on the rank of matrices from OPIs [22]. They obtained six OPIs with degree 2 and multiplicity 1, and eighteen OPIs and two parametrized families with degree 2 and multiplicity 2. These operators include the derivation, average operator, inverse average operator, Rota–Baxter operator of weight zero, Nijenhuis operator and some new operators.

In the present paper, we prove that all OPIs classified in [22] are Gröbner-Shirshov in the framework of [21], via the method of Gröbner-Shirshov bases. In other words, we show that OPIs classified in [22] are “good” OPIs searched in Rota’s Classification Problem. Our study is a partial answer of Rota’s Classification Problem.

1.3. Outline of the Paper

In Section 2, the construction of free operated algebras is recalled firstly to understand Rota’s Classification Problem in the first step. We then review the notation of Gröbner-Shirshov OPIs. Section 3 is devoted to prove that all OPIs in [22] are Gröbner-Shirshov. Three monomial orders ${\le }_{\mathrm{dt}}$, ${\le }_{\mathrm{db}}$ and ${\le }_{\mathrm{Dl}}$ are recalled. Based on these three monomial orders, we first prove that all OPIs classified in [22] of degree 2 and multiplicity 1 are Gröbner-Shirshov (Theorem 1). Then, we show that all OPIs in [22] of degree 2 and multiplicity 2 are Gröbner-Shirshov (Theorem 2). In Section 4, we conclude the main results obtained in this paper and propose some further problems to study.

Notation. Throughout this paper, let $\mathbf{k}$ be a unitary commutative ring, which will be the base ring of all algebras and linear maps. Since the leading monomials of the OPIs considered in this paper are not fixed if we involve the unity $\mathbf{1}$, we consider the case not involving the unity $\mathbf{1}$ throughout the paper. See Remarks 1 and 2 for more details. An algebra means a non-unitary associative algebra. For a set X, $\mathbf{k}X$ is used to denote the free module on X. Denote $S\left(X\right)$ and $M\left(X\right)$ the free semigroup and free monoid on X, respectively. Write $\lfloor X\rfloor :=\{\lfloor x\rfloor \mid x\in X\},$ which is a disjoint copy of X.

2. Operated Algebras and Gröbner-Shirshov OPIs

A. G. Kurosh [18] introduced firstly the concept of algebras with linear operators. In [23], it was called operated algebras and the construction of free operated algebras was obtained. See also [24].

Definition 1

([23]).

(a)

A semigroup (resp. algebra) A together with a map (resp. linear map) $P_A:A\to A$ is called anoperated semigroup(resp.operated algebra);

(b)

Let $(A,P_A)$ and $(B,P_B)$ be two operated semigroups (resp. algebras). A map $f:A\to B$ is called amorphism of operated semigroups (resp. algebras)if it is a semigroup (resp. algebra) homomorphism such that $f\circ P_A=P_B\circ f$.

Let X be a set. We recall the construction of the free operated semigroup on X, defined recursively as follows. For the initial step, let

$${\mathfrak{S}}_0\left(X\right):=S\left(X\right)\mathrm{and}{\mathfrak{S}}_1\left(X\right):=S(X\bigsqcup \lfloor {\mathfrak{S}}_0\left(X\right)\rfloor ).$$

Since $X\hookrightarrow X\bigsqcup \lfloor {\mathfrak{S}}_0\rfloor$, we have a semigroup monomorphism

$$i_0:{\mathfrak{S}}_0\left(X\right)=S\left(X\right)\hookrightarrow {\mathfrak{S}}_1\left(X\right)=S(X\bigsqcup \lfloor {\mathfrak{S}}_0\rfloor ),$$

which identifies ${\mathfrak{S}}_0\left(X\right)$ with its image in ${\mathfrak{S}}_1\left(X\right)$. Suppose we have defined ${\mathfrak{S}}_{n-1}\left(X\right)$ and obtained the embedding

$$i_{n-2,n-1}:{\mathfrak{S}}_{n-2}\left(X\right)\hookrightarrow {\mathfrak{S}}_{n-1}\left(X\right)\mathrm{for}n\ge 2.$$

Consider the inductive step of n. Let

$${\mathfrak{S}}_n\left(X\right):=S\left(X\bigsqcup \lfloor {\mathfrak{S}}_{n-1}\left(X\right)\rfloor \right).$$

Notice that ${\mathfrak{S}}_{n-1}\left(X\right)=S\left(X\bigsqcup \lfloor {\mathfrak{S}}_{n-2}\left(X\right)\rfloor \right)$ is the free semigroup on $X\bigsqcup \lfloor {\mathfrak{S}}_{n-2}\left(X\right)\rfloor$. So we obtain a semigroup embedding

$${\mathfrak{S}}_{n-1}\left(X\right)=S\left(X\bigsqcup \lfloor {\mathfrak{S}}_{n-2}\left(X\right)\rfloor \right)\hookrightarrow {\mathfrak{S}}_n\left(X\right)=S\left(X\bigsqcup \lfloor {\mathfrak{S}}_{n-1}\left(X\right)\rfloor \right)$$

in terms of the injection

$$X\bigsqcup \lfloor {\mathfrak{S}}_{n-2}\left(X\right)\rfloor \hookrightarrow X\bigsqcup \lfloor {\mathfrak{S}}_{n-1}\left(X\right)\rfloor .$$

Finally, we define

$${\displaystyle \mathfrak{S}\left(X\right):=\underset{⟶}{lim}{\mathfrak{S}}_n\left(X\right)=\bigcup _{n\ge 0}{\mathfrak{S}}_n\left(X\right).}$$

Denote by $\mathbf{k}\mathfrak{S}\left(X\right)$ the free module spanned by $\mathfrak{S}\left(X\right)$. By linearity, the concatenation product on $\mathfrak{S}\left(X\right)$ can be extended to a multiplication on $\mathbf{k}\mathfrak{S}\left(X\right)$, making $\mathbf{k}\mathfrak{S}\left(X\right)$ into an algebra. Next, we define an operator

$$\lfloor \rfloor :\mathfrak{S}\left(X\right)\to \mathfrak{S}\left(X\right),u\mapsto \lfloor u\rfloor .$$

Again via linearly, this operator $\lfloor \rfloor :\mathfrak{S}\left(X\right)\to \mathfrak{S}\left(X\right)$ can be extended to a linear operator $\lfloor \rfloor :\mathbf{k}\mathfrak{S}\left(X\right)\to \mathbf{k}\mathfrak{S}\left(X\right)$, which turns $\left(\mathbf{k}\mathfrak{S}\right(X),\lfloor \rfloor )$ into an operated algebra.

Lemma 1

([23,24]). Let X be a set. Let $i:X\to \mathfrak{S}\left(X\right)$ and $j:X\to \mathbf{k}\mathfrak{S}\left(X\right)$ be the natural embedding.

(a)

The triple $\left(\mathfrak{S}\right(X),\lfloor \rfloor ,i)$ is the free operated semigroup on X;

(b)

The triple $\left(\mathbf{k}\mathfrak{S}\right(X),\lfloor \rfloor ,j)$ is the free operated algebra on X.

For $\varphi (x_1,\dots ,x_k)\in \mathbf{k}\mathfrak{S}\left(X\right)$ with $k\ge 1$ and $x_1,\dots ,x_k\in X$, the $\varphi (x_1,\dots ,x_k)=0$ (or simply $\varphi (x_1,\dots ,x_k)$) is called an operated polynomial identity (OPI). Let $\mathsf{\Phi }\subseteq \mathbf{k}\mathfrak{S}\left(X\right)$ and Y be a set. Define the substitution set

$$S_{\mathsf{\Phi }}\left(Y\right):=\{\varphi (r_1,\dots ,r_k)\mid r_1,\dots ,r_k\in \mathfrak{S}\left(Y\right),\varphi (x_1,\dots ,x_k)\in \mathsf{\Phi }\},$$

where $\varphi (r_1,\dots ,r_k)$ is the substitution of $\varphi (x_1,\dots ,x_k)$ at the point $(r_1,\dots ,r_k)$ [20].

The following is the concept of Gröbner-Shirshov bases. For more details of the notations of Gröbner-Shirshov base, the readers are referred to [19,24,25]. Any bracketed word in $\mathfrak{S}\left(Y^{\star }\right)$ with exactly one occurrence of ⋆, counting multiplicities, is called a ⋆-bracketed word on Y, where Y is a set and ⋆ is a symbol out of Y and $Y^{\star }:=Y\bigsqcup \{\star \}$. Let us denote ${\mathfrak{S}}^{\star }\left(Y\right)$ the set of all ⋆-bracketed words on Y. Let $q\in {\mathfrak{S}}^{\star }\left(X\right)$ and $u\in \mathfrak{S}\left(Y\right)$. Define ${q|}_u{:=q|}_{\star \mapsto u}\in \mathfrak{S}\left(Y\right)$, replacing the symbol ⋆ in q by u.

Definition 2

([20,24]). Let Y be a set and ≤ be a monomial order on $\mathfrak{S}\left(Y\right)$. Let $S\subseteq \mathbf{k}\mathfrak{S}\left(Y\right)$ be monic.

(a)

We call an element $f\in \mathbf{k}\mathfrak{S}\left(Y\right)$trivial modulo $(S,w)$with $w\in \mathfrak{S}\left(Y\right)$ if

$$f=\sum _i{c}_i{q}_i{|}_{s_i}with\overline{q_i{|}_{s_i}}<w,\mathit{where}{c}_i\in \mathbf{k},q_i\in {\mathfrak{S}}^{\star }\left(Y\right),s_i\in S.$$

In this case, we denote by $f\equiv 0mod(S,w)$. We write $f\equiv gmod(S,w)$ if $f-g\equiv 0mod(S,w)$;

(b)

We call S aGröbner-Shirshov basisin $\mathbf{k}\mathfrak{S}\left(Y\right)$ with respect to ≤ if, for any $f,g\in S$, every intersection composition of the form ${(f,g)}_w^{u,v}$ is trivial modulo $(S,w)$, and every including composition of the form ${(f,g)}_w^q$ is trivial modulo $(S,w)$.

We end this section with a characterization of Rota’s Classification Problem via the method of Gröbner-Shirshov bases.

Definition 3

([21]). Let X be a set and $\mathsf{\Phi }\subseteq \mathbf{k}\mathfrak{S}\left(X\right)$ be a system of OPIs. Let Y be a set and ≤ be a monomial order on $\mathfrak{S}\left(Y\right)$. The Φ is called Gröbner-Shirshov on Y with respect to ≤ if $S_{\mathsf{\Phi }}\left(Y\right)$ is a Gröbner-Shirshov basis in $\mathbf{k}\mathfrak{S}\left(Y\right)$ with respect to ≤.

3. Gröbner-Shirshov Operated Polynomial Identities

In this section, we prove that all OPIs classified in [22] are Gröbner-Shirshov. In the rest of the paper, in order to be consistent with the notations in [22], we use L to denote the operator $\lfloor \rfloor$.

3.1. OPIs of Degree 2 and Multiplicity 1

In this subsection, we prove that all OPIs of degree 2 and multiplicity 1 classified in [22] are Gröbner-Shirshov. Here the degree means the number of variables in each term and the multiplicity is the the number of operators L in each term. For example, the OPI

$$L\left(x_1{x}_2\right)-x_{1}L\left(x_2\right)-L\left(x_1\right)x_2$$

has degree 2 and multiplicity 1. Let us now recall these OPIs.

Lemma 2

([22] [Theorem 5.1]). Let X be a set. Let

$$R:=aL\left(x_1{x}_2\right)+bL\left(x_1\right)x_2+c{x}_{1}L\left(x_2\right)\in \mathbf{k}\mathfrak{S}\left(X\right)\mathit{with}a,b,c\in \mathbf{k}\mathit{and}{x}_1,x_2\in X$$

be a nonzero OPI. The matrix of consequences $M\left(R\right)$ has rank 14 or 17. Rank 14 occurs if, and only if, the values of the coefficients (up to nonzero scalar multiples) correspond to one of the following six OPIs:

$$\begin{array}{cc}\hfill & L\left(x_1{x}_2\right),L\left(x_1{x}_2\right)-x_{1}L\left(x_2\right),L\left(x_1{x}_2\right)-L\left(x_1\right)x_2,\hfill \\ \hfill & L\left(x_1{x}_2\right)-x_{1}L\left(x_2\right)-L\left(x_1\right)x_2,L\left(x_1\right)x_2,x_{1}L\left(x_2\right).\hfill \end{array}$$

Next, we recall the monomial order ${\le }_{\mathrm{dt}}$ on $\mathfrak{S}\left(Y\right)$ [19]. Let $(Y,\le )$ be a well-ordered set. For $u\in \mathfrak{S}\left(Y\right)$, denote by ${deg}_Y\left(u\right)$ the number of $y\in Y$ in u with repetition. For any

$$u=u_1\cdots u_m\mathrm{and}v=v_1\cdots v_n,$$

where $u_i$ and $v_j$ are prime. Define $u{<}_{\mathrm{dt}}v$ inductively on $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)\ge 0$. For the initial step of $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)=0$, we have $u,v\in S\left(Y\right)$ and define $u{<}_{\mathrm{dt}}v$ if $u{<}_{\mathrm{deg}-\mathrm{lex}}v$, that is,

$$({deg}_Y\left(u\right),u_1,\dots ,u_m)<({deg}_Y\left(v\right),v_1,\dots ,v_n)\mathrm{lexicographically}.$$

Then, ${\le }_{\mathrm{dt}}$ is a monomial order on $S\left(Y\right)$ [26]. For the induction step, if $u=\lfloor u^{\prime }\rfloor$ and $v=\lfloor v^{\prime }\rfloor$, define $u{<}_{\mathrm{dt}}v\mathrm{if}{u}^{\prime }{<}_{\mathrm{dt}}{v}^{\prime }.$ If $u\in Y$ and $v=\lfloor v^{\prime }\rfloor$, define $u{<}_{\mathrm{dt}}v$. Otherwise, define

$$u{<}_{\mathrm{dt}}v\mathrm{if}({deg}_Y\left(u\right),u_1,\dots ,u_m)<({deg}_Y\left(v\right),v_1,\dots ,v_n)\mathrm{lexicographically}.$$

Then, ${\le }_{\mathrm{dt}}$ is a monomial order on $\mathfrak{S}\left(Y\right)$ [19].

Now, we arrive at our first main result of this paper.

Theorem 1.

Let X and Y be sets. The six OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ of degree 2 and multiplicity 1 classified in [22] listed in Lemma 2 are, respectively, Gröbner-Shirshov on Y with respect to the monomial order ${\le }_{\mathrm{dt}}$.

Proof. 

It is trivial that three monomial OPIs are, respectively, Gröbner-Shirshov on Y with respect to the monomial order ${\le }_{\mathrm{dt}}$. Other three OPIs are differential type OPIs, which are Gröbner-Shirshov on Y with respect to the monomial order ${\le }_{\mathrm{dt}}$ [25]. □

3.2. OPIs of Degree 2 and Multiplicity 2

In this subsection, we turn to prove that all OPIs of degree 2 and multiplicity 2 classified in [22] are Gröbner-Shirshov. Let us first review these OPIs.

Lemma 3

([22] [Theorem 6.12]). Let X be a set. The matrix $M{\left(R\right)}^t$ has rank 16 if, and only if, the parameters $a,b,c,d,e,f$ correspond to one of the following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$:

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)-L^2\left(x_1\right)x_2-x_1{L}^2\left(x_2\right),L^2\left(x_1{x}_2\right)-L^2\left(x_1\right)x_2,\hfill \\ \hfill & L^2\left(x_1{x}_2\right)-x_1{L}^2\left(x_2\right),L^2\left(x_1{x}_2\right),L^2\left(x_1\right)x_2,x_1{L}^2\left(x_2\right),\hfill \end{array}$$

where $x_1,x_2\in X$. The matrix $M{\left(R\right)}^t$ has rank 19 if, and only if, the parameters $a,b,c,d,e,f$ correspond to one of the following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$:

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)+L\left(x_{1}L\left(x_2\right)\right)+x_1{L}^2\left(x_2\right)(\mathit{New}\mathit{identity}\mathit{A}\left(\mathit{right}\right)),\hfill \\ \hfill & L^2\left(x_1{x}_2\right)+L\left(L\left(x_1\right)x_2\right)+L^2\left(x_1\right)x_2(\mathit{New}\mathit{identity}\mathit{A}\left(\mathit{left}\right)),\hfill \\ \hfill & L^2\left(x_1{x}_2\right)+dL\left(x_{1}L\left(x_2\right)\right)-(d+1)x_1{L}^2\left(x_2\right)\mathit{with}d\in \mathbf{k}\setminus \left\{0\right\}(\mathit{New}\mathit{identity}\mathit{B}\left(\mathit{right}\right)),\hfill \\ \hfill & L^2\left(x_1{x}_2\right)+bL\left(L\left(x_1\right)x_2\right)-(b+1)L^2\left(x_1\right)x_2\mathit{with}b\in \mathbf{k}\setminus \left\{0\right\}(\mathit{New}\mathit{identity}\mathit{B}\left(\mathit{left}\right)),\hfill \\ \hfill & L^2\left(x_1{x}_2\right)+L^2\left(x_1\right)x_2+x_1{L}^2\left(x_2\right)+2L\left(x_1\right)L\left(x_2\right)-2L\left(L\left(x_1\right)x_2\right)-2L\left(x_{1}L\left(x_2\right)\right)(\mathit{New}\mathit{identity}\mathit{C}),\hfill \\ \hfill & L\left(L\left(x_1\right)x_2\right)-L^2\left(x_1\right)x_2\left(P_1\right),\hfill \\ \hfill & L\left(L\left(x_1\right)x_2\right)\left(P_2\right),\hfill \\ \hfill & L\left(x_{1}L\left(x_2\right)\right)-x_1{L}^2\left(x_2\right)\left(P_3\right),\hfill \\ \hfill & L\left(x_{1}L\left(x_2\right)\right)\left(P_4\right),\hfill \\ \hfill & L\left(x_1\right)L\left(x_2\right)\left(P_5\right),\hfill \\ \hfill & L\left(x_1\right)L\left(x_2\right)-L\left(L\left(x_1\right)x_2\right)-L\left(x_{1}L\left(x_2\right)\right)(\mathit{Rota}-\mathit{Baxter}),\hfill \\ \hfill & L\left(x_1\right)L\left(x_2\right)-L\left(L\left(x_1\right)x_2\right)-L\left(x_{1}L\left(x_2\right)\right)+L^2\left(x_1{x}_2\right)\left(\mathit{Nijenhuis}\right),\hfill \\ \hfill & L\left(x_1\right)L\left(x_2\right)-L\left(L\left(x_1\right)x_2\right)(\mathit{inverse}\mathit{average}),\hfill \\ \hfill & L\left(x_1\right)L\left(x_2\right)-L\left(x_{1}L\left(x_2\right)\right)\left(\mathit{average}\right),\hfill \end{array}$$

where $x_1,x_2\in X$.

Notice that the average (resp. inverse average) OPI was named the right (resp. left) average OPI in [22].

Next, we recall two monomial orders ${\le }_{\mathrm{db}}$ [20] and ${\le }_{\mathrm{Dl}}$ [27], which will be used frequently in the remainder of the paper. Let $(Y,\le )$ be a well-ordered set. Extend the well order ≤ on Y to the degree lexicographical order ${\le }_{\mathrm{deg}-\mathrm{lex}}$ on $M\left(Y\right)$. Further, we extend ≤ on $\mathfrak{S}\left(Y\right)$. Every $u\in \mathfrak{S}\left(Y\right)$ may be uniquely written as a product in the form

$$u=u_{0}L\left(u_1^{*}\right)u_{1}L\left(u_2^{*}\right)u_2\cdots L\left(u_r^{*}\right)u_r$$

where $u_0,\dots ,u_r\in M\left(Y\right)$, $u_1^{*},\dots ,u_r^{*}\in {\mathfrak{S}}_{n-1}\left(Y\right)$. Denote by ${deg}_L\left(u\right)$ the number of occurrence of L, and define the L-breadth ${\mathrm{bre}}_L\left(u\right)$ of u to be r. For example, if $u=x_{0}L\left(x_1\right)x_{2}L\left(x_3{x}_4\right)x_5$ with $x_0,\dots ,x_5\in Z$, we have ${deg}_L\left(u\right)=2$ and ${\mathrm{bre}}_L\left(u\right)=2$. Let $u,v\in \mathfrak{S}\left(Y\right)$ and write them uniquely in the form:

$$u=u_{0}L\left(u_1^{*}\right)u_{1}L\left(u_2^{*}\right)u_2\cdots L\left(u_r^{*}\right)u_r\mathrm{and}v=v_{0}L\left(v_1^{*}\right)v_{1}L\left(v_2^{*}\right)v_2\cdots L\left(v_s^{*}\right)v_s,$$

where $u_0,\dots ,u_r$, $v_0,\dots ,v_s\in M\left(Y\right)$ and $u_1^{*},\dots ,u_r^{*},v_1^{*},\dots ,v_s^{*}\in \mathfrak{S}\left(Y\right)$. We define $u{\le }_{\mathrm{db}}v$ by induction on $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)\ge 0$. For the initial step of $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)=0$, we have $u,v\in S\left(Y\right)$ and use the degree lexicographical order. For the induction step of $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)\ge 1$, we define

$$u{\le }_{\mathrm{db}}v\iff \left\{\begin{array}{c}{deg}_L\left(u\right)<{deg}_L\left(v\right),\hfill \\ \mathrm{or}{deg}_L\left(u\right)={deg}_L\left(v\right)\mathrm{and}{\mathrm{bre}}_L\left(u\right)<{\mathrm{bre}}_L\left(v\right),\hfill \\ \mathrm{or}{deg}_L\left(u\right)={deg}_L\left(v\right),{\mathrm{bre}}_L\left(u\right)={\mathrm{bre}}_L\left(v\right)(=r)\mathrm{and}\hfill \\ (u_1^{*},\cdots ,u_r^{*},u_0,\cdots ,u_r)\le (v_1^{*},\cdots ,v_r^{*},v_0,\cdots ,v_r)\mathrm{lexicographically}.\hfill \end{array}\right.$$

Here, $u_i^{*}{\le }_{\mathrm{db}}{v}_i^{*}$ and $u_i{\le }_{\mathrm{db}}{v}_i$ are compared by the induction hypothesis. Then, ${\le }_{\mathrm{db}}$ is a monomial order on $\mathfrak{S}\left(Y\right)$ [20].

Let $(Y,\le )$ be a well-ordered set, and let $u=u_1\cdots u_m$ and $v=v_1\cdots v_n$ be in $\mathfrak{S}\left(Y\right)$, where $u_i$ and $v_j$ are prime. We define $u{<}_{\mathrm{Dl}}v$ by induction on $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)\ge 0$. If $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)=0$, we obtain $u,v\in S\left(Y\right)$ and define $u{<}_{\mathrm{Dl}}v$ by $u{<}_{\mathrm{deg}-\mathrm{lex}}v$, that is,

$$u{<}_{\mathrm{Dl}}v\mathrm{if}({\mathrm{deg}}_Y\left(u\right),u_1,\dots ,u_m)<({\mathrm{deg}}_Y\left(v\right),v_1,\dots ,v_n)\mathrm{lexicographically}.$$

Suppose $\mathrm{dep}\left(u\right)+\mathrm{dep}\left(v\right)\ge 1$. If $u=\lfloor u^{\prime }\rfloor$ and $v=\lfloor v^{\prime }\rfloor$, define $u{<}_{\mathrm{Dl}}v\mathrm{if}{u}^{\prime }{<}_{\mathrm{Dl}}{v}^{\prime }.$ Otherwise, define

$$u{<}_{\mathrm{Dl}}v\mathrm{if}({\mathrm{deg}}_Y\left(u\right),\left|u\right|,u_1,\dots ,u_m)<({\mathrm{deg}}_Y\left(v\right),\left|v\right|,v_1,\dots ,v_n)\mathrm{lexicographically}.$$

Then, ${\le }_{\mathrm{Dl}}$ is a monomial order on $\mathfrak{S}\left(Y\right)$ [27].

As a preparation, we obtain the following result.

Proposition 1.

Let X and Y be sets. Then, the following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ are, respectively, Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$:

$$\begin{array}{}\mathrm{(1)}& \hfill & L^2\left(x_1{x}_2\right)-L^2\left(x_1\right)x_2-x_1{L}^2\left(x_2\right),\hfill \mathrm{(2)}& \hfill & L^2\left(x_1{x}_2\right)-L^2\left(x_1\right)x_2,\hfill \mathrm{(3)}& \hfill & L^2\left(x_1{x}_2\right)-x_1{L}^2\left(x_2\right),\hfill \end{array}$$

where $x_1,x_2\in X$.

Proof. 

We first consider Equation (1). Denote by

$$S=\{L^2\left(xy\right)-L^2\left(x\right)y-x{L}^2\left(y\right)\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

With respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$, the leading monomial of $L^2\left(xy\right)-L^2\left(x\right)y-x{L}^2\left(y\right)$ is $L^2\left(xy\right)$. Since the breadth of the leading monomial $L^2\left(xy\right)$ is $\mathrm{bre}\left(L^2\left(xy\right)\right)=1$, there are no intersection ambiguities. Further there are two cases of including ambiguities to consider.

Case 1. We have

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(p{|}_{L^2\left(xy\right)}z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(pz\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(p{|}_{L^2\left(xy\right)}z\right)-L^2\left(p{|}_{L^2\left(xy\right)}\right){z-p|}_{L^2\left(xy\right)}{L}^2\left(z\right),\hfill \\ \hfill g=& L^2\left(xy\right)-L^2\left(x\right)y-x{L}^2\left(y\right).\hfill \end{array}$$

The corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2\left(p{|}_{L^2\left(xy\right)}\right){z-p|}_{L^2\left(xy\right)}{L}^2\left(z\right)+L^2\left(p{|}_{L^2\left(x\right)y}z\right)+L^2\left(p{|}_{x{L}^2\left(y\right)}z\right)\hfill \\ \hfill \equiv & -L^2\left(p{|}_{L^2\left(x\right)y}\right)z-L^2\left(p{|}_{x{L}^2\left(y\right)}\right)z-p{{|}_{L^2\left(x\right)y}{L}^2\left(z\right)-p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)\hfill \\ \hfill & +L^2\left(p{|}_{L^2\left(x\right)y}\right)z+p{{|}_{L^2\left(x\right)y}{L}^2\left(z\right)+L^2\left(p{|}_{x{L}^2\left(y\right)}\right)z+p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 2. We have

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(xp{|}_{L^2\left(yz\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(xp\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(xp{|}_{L^2\left(yz\right)}\right)-L^2\left(x\right){p|}_{L^2\left(yz\right)}-x{L}^2\left(p{|}_{L^2\left(yz\right)}\right),\hfill \\ \hfill g=& L^2\left(yz\right)-L^2\left(y\right)z-y{L}^2\left(z\right).\hfill \end{array}$$

Then, the including composition of this ambiguity is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2{\left(x\right)p|}_{L^2\left(yz\right)}-x{L}^2\left(p{|}_{L^2\left(yz\right)}\right)+L^2\left(xp{|}_{L^2\left(y\right)z}\right)+L^2\left(xp{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & -L^2\left(x\right)p{{|}_{L^2\left(y\right)z}-L^2\left(x\right)p|}_{y{L}^2\left(z\right)}-x{L}^2\left(p{|}_{L^2\left(y\right)z}\right)-x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill & +L^2\left(x\right)p{{|}_{L^2\left(y\right)z}+x{L}^2\left(p{|}_{L^2\left(y\right)z}\right)+L^2\left(x\right)p|}_{y{L}^2\left(z\right)}+x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

So the OPI in Equation (1) is Gröbner-Shirshov.

Next, we consider Equation (2). Let

$$S=\{L^2\left(xy\right)-L^2\left(x\right)y\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

With a similar argument to the case of Equation (1), there are two cases to consider.

Case 3. We have

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(p{|}_{L^2\left(xy\right)}z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(pz\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(p{|}_{L^2\left(xy\right)}z\right)-L^2\left(p{|}_{L^2\left(xy\right)}\right)z,\hfill \\ \hfill g=& L^2\left(xy\right)-L^2\left(x\right)y.\hfill \end{array}$$

The corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2\left(p{|}_{L^2\left(xy\right)}\right)z+L^2\left(p{|}_{L^2\left(x\right)y}z\right)\hfill \\ \hfill \equiv & -L^2\left(p{|}_{L^2\left(x\right)y}\right)z+L^2\left(p{|}_{L^2\left(x\right)y}\right)z\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 4. We have

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(xp{|}_{L^2\left(yz\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(xp\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(xp{|}_{L^2\left(yz\right)}\right)-L^2\left(x\right){p|}_{L^2\left(yz\right)},\hfill \\ \hfill g=& L^2\left(yz\right)-L^2\left(y\right)z.\hfill \end{array}$$

In this case, the corresponding including composition is trivial mod $(S,w)$ as follows:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2{\left(x\right)p|}_{L^2\left(yz\right)}+L^2\left(xp{|}_{L^2\left(y\right)z}\right)\hfill \\ \hfill \equiv & -L^2\left(x\right)p{{|}_{L^2\left(y\right)z}+L^2\left(x\right)p|}_{L^2\left(y\right)z}\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

So the OPI in Equation (2) is Gröbner-Shirshov. The proof of the OPI in Equation (3) is similar to the one in Equation (2). □

Remark 1.

If we consider the case involving the unity $\mathbf{1}$, then the leading monomial of the OPI

$$L^2\left(x_1{x}_2\right)-L^2\left(x_1\right)x_2-x_1{L}^2\left(x_2\right)$$

in Equation (1) is not necessary $L^2\left(x_1{x}_2\right)$. For example, if $x_2=\mathbf{1}$, then Equation (1) is $-x_1{L}^2\left(\mathbf{1}\right)$ whose leading monomial is not $L^2\left(x_1{x}_2\right)=L^2\left(x_1\right)$.

Next, we turn to consider the new identity A.

Proposition 2.

Let X and Y be sets. The following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ are, respectively, Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$:

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)+L\left(x_{1}L\left(x_2\right)\right)+x_1{L}^2\left(x_2\right)(\mathit{New}\mathit{identity}\mathit{A}\left(\mathit{right}\right)),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)+L\left(L\left(x_1\right)x_2\right)+L^2\left(x_1\right)x_2(\mathit{New}\mathit{identity}\mathit{A}\left(\mathit{left}\right)),\hfill \end{array}$$

(5)

where $x_1,x_2\in X$.

Proof. 

We only prove Equation (4), as the case of Equation (5) is a similar one. Let

$$S=\{L^2\left(xy\right)+L\left(xL\left(y\right)\right)+x{L}^2\left(y\right)\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

Notice that the leading monomial of $L^2\left(xy\right)+L\left(xL\left(y\right)\right)+x{L}^2\left(y\right)$ is $L^2\left(xy\right)$ with respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$. There are no intersection compositions, and there are two including compositions.

Case 1. The ambiguity is

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(p{|}_{L^2\left(xy\right)}z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(pz\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(p{|}_{L^2\left(xy\right)}z\right)+L\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right){+p|}_{L^2\left(xy\right)}{L}^2\left(z\right),\hfill \\ \hfill g=& L^2\left(xy\right)+L\left(xL\left(y\right)\right)+x{L}^2\left(y\right),\hfill \end{array}$$

whose including composition is trivial mod $(S,w)$ as follows:

$$\begin{array}{cc}\hfill {f-q|}_g=& L\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right){+p|}_{L^2\left(xy\right)}{L}^2\left(z\right)-L^2\left(p{|}_{L\left(xL\right(y\left)\right)}z\right)-L^2\left(p{|}_{x{L}^2\left(y\right)}z\right)\hfill \\ \hfill \equiv & -L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)-L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right)-p{{|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)-p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)\hfill \\ \hfill & +L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)+p{{|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)+L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right)+p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 2. The ambiguity is

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(xp{|}_{L^2\left(yz\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(xp\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(xp{|}_{L^2\left(yz\right)}\right)+L\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right)+x{L}^2\left(p{|}_{L^2\left(yz\right)}\right),\hfill \\ \hfill g=& L^2\left(yz\right)+L\left(yL\left(z\right)\right)+y{L}^2\left(z\right).\hfill \end{array}$$

Then, the including composition is trivial mod $(S,w)$ as follows:

$$\begin{array}{cc}\hfill {f-q|}_g=& L\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right)+x{L}^2\left(p{|}_{L^2\left(yz\right)}\right)-L^2\left(xp{|}_{L\left(yL\right(z\left)\right)}\right)-L^2\left(xp{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & -L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)-L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)-x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)-x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill & +L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)+x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)+L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)+x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & 0,\hfill \end{array}$$

as needed. □

The following focuses on the new identity B.

Proposition 3.

Let X and Y be sets. The following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ are, respectively, Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$:

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)+dL\left(x_{1}L\left(x_2\right)\right)-(d+1)x_1{L}^2\left(x_2\right)\mathit{with}d\in \mathbf{k}\setminus \left\{0\right\}(\mathit{New}\mathit{identity}\mathit{B}\left(\mathit{right}\right)),\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill & L^2\left(x_1{x}_2\right)+bL\left(L\left(x_1\right)x_2\right)-(b+1)L^2\left(x_1\right)x_2\mathit{with}b\in \mathbf{k}\setminus \left\{0\right\}(\mathit{New}\mathit{identity}\mathit{B}\left(\mathit{left}\right)),\hfill \end{array}$$

(7)

where $x_1,x_2\in X$.

Proof. 

By symmetry, it is enough to prove Equation (6). Let

$$S=\{L^2\left(xy\right)+dL\left(xL\left(y\right)\right)-(d+1)x{L}^2\left(y\right)\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

The leading monomial of $L^2\left(xy\right)+dL\left(xL\left(y\right)\right)-(d+1)x{L}^2\left(y\right)$ is $L^2\left(xy\right)$ with respect to ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$. There are no intersection compositions, and there are two including compositions.

Case 1. The ambiguity is

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(p{|}_{L^2\left(xy\right)}z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(pz\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(p{|}_{L^2\left(xy\right)}z\right)+dL\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right)-(d+1){p|}_{L^2\left(xy\right)}{L}^2\left(z\right),\hfill \\ \hfill g=& L^2\left(xy\right)+dL\left(xL\left(y\right)\right)-(d+1)x{L}^2\left(y\right),\hfill \end{array}$$

whose including composition is trivial mod $(S,w)$ as follows:

$$\begin{array}{cc}\hfill {f-q|}_g=& dL\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right){-(d+1)p|}_{L^2\left(xy\right)}{L}^2\left(z\right)-d{L}^2\left(p{|}_{L\left(xL\right(y\left)\right)}z\right)+(d+1)L^2\left(p{|}_{x{L}^2\left(y\right)}z\right)\hfill \\ \hfill \equiv & -d^{2}L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)+d(d+1)L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right){+d(d+1)p|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)\hfill \\ \hfill & -{(d+1)}^{2}p{{|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)+d^{2}L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)-d(d+1)p|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)\hfill \\ \hfill & -d(d+1)L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right)+{(d+1)}^2{p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 2. The ambiguity is

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(xp{|}_{L^2\left(yz\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(xp\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(xp{|}_{L^2\left(yz\right)}\right)+dL\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right)-(d+1)x{L}^2\left(p{|}_{L^2\left(yz\right)}\right),\hfill \\ \hfill g=& L^2\left(yz\right)+dL\left(yL\left(z\right)\right)-(d+1)y{L}^2\left(z\right).\hfill \end{array}$$

The including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& dL\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right)-(d+1)x{L}^2\left(p{|}_{L^2\left(yz\right)}\right)-d{L}^2\left(xp{|}_{L\left(yL\right(z\left)\right)}\right)+(d+1)L^2\left(xp{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & -d^{2}L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)+d(d+1)L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)+d(d+1)x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill & -{(d+1)}^{2}x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)+d^{2}L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)-d(d+1)x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill & -d(d+1)L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)+{(d+1)}^{2}x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)\hfill \\ \hfill \equiv & 0,\hfill \end{array}$$

as required. □

Now, we are in a position to consider the new identity C.

Proposition 4.

Let X and Y be sets. The following OPI in $\mathbf{k}\mathfrak{S}\left(X\right)$ is Gröbner-Shirshov on Y with respect to the monomial order ${\le }_{\mathrm{dt}}$:

$$\begin{array}{c}\hfill L^2\left(x_1{x}_2\right)+L^2\left(x_1\right)x_2+x_1{L}^2\left(x_2\right)+2L\left(x_1\right)L\left(x_2\right)-2L\left(L\left(x_1\right)x_2\right)-2L\left(x_{1}L\left(x_2\right)\right)(\mathit{New}\mathit{identity}\mathit{C}).\end{array}$$

(8)

where $x_1,x_2\in X$.

Proof. 

Let

$$S=\{L^2\left(xy\right)+L^2\left(x\right)y+x{L}^2\left(y\right)+2L\left(x\right)L\left(y\right)-2L\left(L\left(x\right)y\right)-2L\left(xL\left(y\right)\right)\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

With respect to ${\le }_{\mathrm{dt}}$, the leading monomial of

$$L^2\left(xy\right)+L^2\left(x\right)y+x{L}^2\left(y\right)+2L\left(x\right)L\left(y\right)-2L\left(L\left(x\right)y\right)-2L\left(xL\left(y\right)\right)$$

is $L^2\left(xy\right)$. Further there are no intersection compositions, and there are two including compositions.

Case 1. The ambiguity is of the form

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(p{|}_{L^2\left(xy\right)}z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(pz\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(p{|}_{L^2\left(xy\right)}z\right)+L^2\left(p{|}_{L^2\left(xy\right)}\right){z+p|}_{L^2\left(xy\right)}{L}^2\left(z\right)+2L\left(p{|}_{L^2\left(xy\right)}\right)L\left(z\right)-2L\left(L\left(p{|}_{L^2\left(xy\right)}\right)z\right)-2L\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right),\hfill \\ \hfill g=& L^2\left(xy\right)+L^2\left(x\right)y+x{L}^2\left(y\right)+2L\left(x\right)L\left(y\right)-2L\left(L\left(x\right)y\right)-2L\left(xL\left(y\right)\right).\hfill \end{array}$$

Then, the including intersection is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& L^2\left(p{|}_{L^2\left(xy\right)}\right){z+p|}_{L^2\left(xy\right)}{L}^2\left(z\right)+2L\left(p{|}_{L^2\left(xy\right)}\right)L\left(z\right)-2L\left(L\left(p{|}_{L^2\left(xy\right)}\right)z\right)-2L\left(p{|}_{L^2\left(xy\right)}L\left(z\right)\right)\hfill \\ \hfill & -L^2\left(p{|}_{L^2\left(x\right)y}z\right)-L^2\left(p{|}_{x{L}^2\left(y\right)}z\right)-2L^2\left(p{|}_{L\left(x\right)L\left(y\right)}z\right)+2L^2\left(p{|}_{L\left(L\right(x\left)y\right)}z\right)+2L^2\left(p{|}_{L\left(xL\right(y\left)\right)}z\right)\hfill \\ \hfill \equiv & -L^2\left(p{|}_{L^2\left(x\right)y}\right)z-L^2\left(p{|}_{x{L}^2\left(y\right)}\right)z-2L^2\left(p{|}_{L\left(x\right)L\left(y\right)}\right)z+2L^2\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z+2L^2\left(p{|}_{L\left(xL\right(y\left)\right)}\right)z\hfill \\ \hfill & {-p|}_{L^2\left(x\right)y}{L}^2{\left(z\right)-p|}_{x{L}^2\left(y\right)}{L}^2{\left(z\right)-2p|}_{L\left(x\right)L\left(y\right)}{L}^2\left(z\right)+2p{{|}_{L\left(L\right(x\left)y\right)}{L}^2\left(z\right)+2p|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)\hfill \\ \hfill & -2L\left(p{|}_{L^2\left(x\right)y}\right)L\left(z\right)-2L\left(p{|}_{x{L}^2\left(y\right)}\right)L\left(z\right)-4L\left(p{|}_{L\left(x\right)L\left(y\right)}\right)L\left(z\right)+4L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)L\left(z\right)+4L\left(p{|}_{L\left(xL\right(y\left)\right)}\right)L\left(z\right)\hfill \\ \hfill & +2L\left(L\left(p{|}_{L^2\left(x\right)y}\right)z\right)+2L\left(L\left(p{|}_{x{L}^2\left(y\right)}\right)z\right)+4L\left(L\left(p{|}_{L\left(x\right)L\left(y\right)}\right)z\right)-4L\left(L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z\right)-4L\left(L\left(p{|}_{L\left(xL\right(y\left)\right)}\right)z\right)\hfill \\ \hfill & +2L\left(p{|}_{L^2\left(x\right)y}L\left(z\right)\right)+2L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right)+4L\left(p{|}_{L\left(x\right)L\left(y\right)}L\left(z\right)\right)-4L\left(p{|}_{L\left(L\right(x\left)y\right)}L\left(z\right)\right)-4L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)\hfill \\ \hfill & +L^2\left(p{|}_{L^2\left(x\right)y}\right){z+p|}_{L^2\left(x\right)y}{L}^2\left(z\right)+2L\left(p{|}_{L^2\left(x\right)y}\right)L\left(z\right)-2L\left(L\left(p{|}_{L^2\left(x\right)y}\right)z\right)-2L\left(p{|}_{L^2\left(x\right)y}L\left(z\right)\right)\hfill \\ \hfill & +L^2\left(p{|}_{x{L}^2\left(y\right)}\right){z+p|}_{x{L}^2\left(y\right)}{L}^2\left(z\right)+2L\left(p{|}_{x{L}^2\left(y\right)}\right)L\left(z\right)-2L\left(L\left(p{|}_{x{L}^2\left(y\right)}\right)z\right)-2L\left(p{|}_{x{L}^2\left(y\right)}L\left(z\right)\right)\hfill \\ \hfill & +2L^2\left(p{|}_{L\left(x\right)L\left(y\right)}\right){z+2p|}_{L\left(x\right)L\left(y\right)}{L}^2\left(z\right)+4L\left(p{|}_{L\left(x\right)L\left(y\right)}\right)L\left(z\right)-4L\left(L\left(p{|}_{L\left(x\right)L\left(y\right)}\right)z\right)-4L\left(p{|}_{L\left(x\right)L\left(y\right)}L\left(z\right)\right)\hfill \\ \hfill & -2L^2\left(p{|}_{L\left(L\right(x\left)y\right)}\right){z-2p|}_{L\left(L\right(x\left)y\right)}{L}^2\left(z\right)-4L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)L\left(z\right)+4L\left(L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z\right)+4L\left(p{|}_{L\left(L\right(x\left)y\right)}L\left(z\right)\right)\hfill \\ \hfill & -2L^2\left(p{|}_{L\left(xL\right(y\left)\right)}\right){z-2p|}_{L\left(xL\right(y\left)\right)}{L}^2\left(z\right)-4L\left(p{|}_{L\left(xL\right(y\left)\right)}\right)L\left(z\right)+4L\left(L\left(p{|}_{L\left(xL\right(y\left)\right)}\right)z\right)+4L\left(p{|}_{L\left(xL\right(y\left)\right)}L\left(z\right)\right)\hfill \\ \hfill & \equiv 0.\hfill \end{array}$$

Case 2. The ambiguity is of the form:

$$\begin{array}{cc}\hfill w=& \overline{f}=L^2\left(xp{|}_{L^2\left(yz\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L^2\left(xp\right)\mathrm{and}\hfill \\ \hfill f=& L^2\left(xp{|}_{L^2\left(yz\right)}\right)+L^2\left(x\right){p|}_{L^2\left(yz\right)}+x{L}^2\left(p{|}_{L^2\left(yz\right)}\right)+2L\left(x\right)L\left(p{|}_{L^2\left(yz\right)}\right)-2L\left(L\left(x\right)p{|}_{L^2\left(yz\right)}\right)-2L\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right),\hfill \\ \hfill g=& L^2\left(yz\right)+L^2\left(y\right)z+y{L}^2\left(z\right)+2L\left(y\right)L\left(z\right)-2L\left(L\left(y\right)z\right)-2L\left(yL\left(z\right)\right).\hfill \end{array}$$

The corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& L^2{\left(x\right)p|}_{L^2\left(yz\right)}+x{L}^2\left(p{|}_{L^2\left(yz\right)}\right)+2L\left(x\right)L\left(p{|}_{L^2\left(yz\right)}\right)-2L\left(L\left(x\right)p{|}_{L^2\left(yz\right)}\right)-2L\left(xL\left(p{|}_{L^2\left(yz\right)}\right)\right)\hfill \\ \hfill & -L^2\left(xp{|}_{L^2\left(y\right)z}\right)-L^2\left(xp{|}_{y{L}^2\left(z\right)}\right)-2L^2\left(xp{|}_{L\left(y\right)L\left(z\right)}\right)+2L^2\left(xp{|}_{L\left(L\right(y\left)z\right)}\right)+2L^2\left(xp{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill \equiv & -L^2{\left(x\right)p|}_{L^2\left(y\right)z}-L^2{\left(x\right)p|}_{y{L}^2\left(z\right)}-2L^2{\left(x\right)p|}_{L\left(y\right)L\left(z\right)}+2L^2\left(x\right)p{{|}_{L\left(L\right(y\left)z\right)}+2L^2\left(x\right)p|}_{L\left(yL\right(z\left)\right)}\hfill \\ \hfill & -x{L}^2\left(p{|}_{L^2\left(y\right)z}\right)-x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)-2x{L}^2\left(p{|}_{L\left(y\right)L\left(z\right)}\right)+2x{L}^2\left(p{|}_{L\left(L\right(y\left)z\right)}\right)+2x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill & -2L\left(x\right)L\left(p{|}_{L^2\left(y\right)z}\right)-2L\left(x\right)L\left(p{|}_{y{L}^2\left(z\right)}\right)-4L\left(x\right)L\left(p{|}_{L\left(y\right)L\left(z\right)}\right)+4L\left(x\right)L\left(p{|}_{L\left(L\right(y\left)z\right)}\right)+4L\left(x\right)L\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill & +2L\left(L\left(x\right)p{|}_{L^2\left(y\right)z}\right)+2L\left(L\left(x\right)p{|}_{y{L}^2\left(z\right)}\right)+4L\left(L\left(x\right)p{|}_{L\left(y\right)L\left(z\right)}\right)-4L\left(L\left(x\right)p{|}_{L\left(L\right(y\left)z\right)}\right)-4L\left(L\left(x\right)p{|}_{L\left(yL\right(z\left)\right)}\right)\hfill \\ \hfill & +2L\left(xL\left(p{|}_{L^2\left(y\right)z}\right)\right)+2L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)+4L\left(xL\left(p{|}_{L\left(y\right)L\left(z\right)}\right)\right)-4L\left(xL\left(p{|}_{L\left(L\right(y\left)z\right)}\right)\right)-4L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)\hfill \\ \hfill & +L^2{\left(x\right)p|}_{L^2\left(y\right)z}+x{L}^2\left(p{|}_{L^2\left(y\right)z}\right)+2L\left(x\right)L\left(p{|}_{L^2\left(y\right)z}\right)-2L\left(L\left(x\right)p{|}_{L^2\left(y\right)z}\right)-2L\left(xL\left(p{|}_{L^2\left(y\right)z}\right)\right)\hfill \\ \hfill & +L^2{\left(x\right)p|}_{y{L}^2\left(z\right)}+x{L}^2\left(p{|}_{y{L}^2\left(z\right)}\right)+2L\left(x\right)L\left(p{|}_{y{L}^2\left(z\right)}\right)-2L\left(L\left(x\right)p{|}_{y{L}^2\left(z\right)}\right)-2L\left(xL\left(p{|}_{y{L}^2\left(z\right)}\right)\right)\hfill \\ \hfill & +2L^2{\left(x\right)p|}_{L\left(y\right)L\left(z\right)}+2x{L}^2\left(p{|}_{L\left(y\right)L\left(z\right)}\right)+4L\left(x\right)L\left(p{|}_{L\left(y\right)L\left(z\right)}\right)-4L\left(L\left(x\right)p{|}_{L\left(y\right)L\left(z\right)}\right)-4L\left(xL\left(p{|}_{L\left(y\right)L\left(z\right)}\right)\right)\hfill \\ \hfill & -2L^2{\left(x\right)p|}_{L\left(L\right(y\left)z\right)}-2x{L}^2\left(p{|}_{L\left(L\right(y\left)z\right)}\right)-4L\left(x\right)L\left(p{|}_{L\left(L\right(y\left)z\right)}\right)+4L\left(L\left(x\right)p{|}_{L\left(L\right(y\left)z\right)}\right)+4L\left(xL\left(p{|}_{L\left(L\right(y\left)z\right)}\right)\right)\hfill \\ \hfill & -2L^2{\left(x\right)p|}_{L\left(yL\right(z\left)\right)}-2x{L}^2\left(p{|}_{L\left(yL\right(z\left)\right)}\right)-4L\left(x\right)L\left(p{|}_{L\left(yL\right(z\left)\right)}\right)+4L\left(L\left(x\right)p{|}_{L\left(yL\right(z\left)\right)}\right)+4L\left(xL\left(p{|}_{L\left(yL\right(z\left)\right)}\right)\right)\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

This completes the proof. □

Remark 2.

(a)

If we involve the unity $\mathbf{1}$, then the leading monomial of the OPI

$$L^2\left(x_1{x}_2\right)+L^2\left(x_1\right)x_2+x_1{L}^2\left(x_2\right)+2L\left(x_1\right)L\left(x_2\right)-2L\left(L\left(x_1\right)x_2\right)-2L\left(x_{1}L\left(x_2\right)\right)$$

in Equation (8) is not necessary $L^2\left(x_1{x}_2\right)$ with respect to the order ${\le }_{\mathrm{dt}}$. For example, taking $x_1=\mathbf{1}$, then the above OPI in Equation (8) is

$$L^2\left(\mathbf{1}\right)x_2+2L\left(\mathbf{1}\right)L\left(x_2\right)-2L\left(L\left(\mathbf{1}\right)x_2\right),$$

whose leading monomial is not $L^2\left(x_1{x}_2\right)=L^2\left(x_2\right)$;

(b)

In Proposition 4, if we apply the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$, then the leading monomial of

$$L^2\left(xy\right)+L^2\left(x\right)y+x{L}^2\left(y\right)+2L\left(x\right)L\left(y\right)-2L\left(L\left(x\right)y\right)-2L\left(xL\left(y\right)\right)$$

is $L\left(x\right)L\left(y\right)$. It induces a rewriting rule

$$L\left(x\right)L\left(y\right)\to -\frac{1}{2}{L}^2\left(xy\right)-\frac{1}{2}{L}^2\left(x\right)y-\frac{1}{2}x{L}^2\left(y\right)+L\left(L\left(x\right)y\right)+L\left(xL\left(y\right)\right).$$

Taking y to be $L\left(y\right)$, we obtain an infinite rewriting process:

$$\begin{array}{ccc}\hfill L\left(x\right)L^2\left(y\right)& \to & -\frac{1}{2}{L}^2\left(xL\left(y\right)\right)-\frac{1}{2}{L}^2\left(x\right)L\left(y\right)-\frac{1}{2}x{L}^3\left(y\right)+L\left(L\left(x\right)L\left(y\right)\right)+L\left(x{L}^2\left(y\right)\right)\hfill \\ \hfill & \hfill \to & -\frac{1}{2}{L}^2\left(xL\left(y\right)\right)+\frac{1}{4}{L}^2\left(L\left(x\right)y\right)+\frac{1}{4}{L}^3\left(x\right)y+\frac{1}{4}L\left(x\right)L^2\left(y\right)-\frac{1}{2}L\left(L^2\left(x\right)y\right)\hfill \\ \hfill & \hfill & -\frac{1}{2}L\left(L\left(x\right)L\left(y\right)\right)-\frac{1}{2}x{L}^3\left(y\right)+L\left(L\left(x\right)L\left(y\right)\right)+L\left(x{L}^2\left(y\right)\right),\hfill \\ \hfill & \hfill \to & \cdots .\hfill \end{array}$$

Notice that the term $L\left(x\right)L^2\left(y\right)$ appears again in the right-hand side.

The following result is needed.

Proposition 5.

Let X and Y be sets. The following OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ are, respectively, Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$:

$$\begin{array}{cc}\hfill & L\left(L\left(x_1\right)x_2\right)-L^2\left(x_1\right)x_2\left(P_1\right),\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & L\left(x_{1}L\left(x_2\right)\right)-x_1{L}^2\left(x_2\right)\left(P_3\right),\hfill \end{array}$$

(10)

where $x_1,x_2\in X$.

Proof. 

By symmetry, it suffices to prove the case of Equation (9). Let

$$S=\{L\left(L\left(x\right)y\right)-L^2\left(x\right)y\mid x,y\in \mathfrak{S}\left(Y\right)\}.$$

The leading monomial of $L\left(L\left(x\right)y\right)-L^2\left(x\right)y$ is $L\left(L\right(x\left)y\right)$ with respect to ${\le }_{\mathrm{db}}$ or ${\le }_{\mathrm{Dl}}$. There are no intersection compositions, and there are three including compositions.

Case 1. The ambiguity is of the form

$$\begin{array}{cc}\hfill w=& \overline{f}=L\left(L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L\left(L\left(p\right)z\right)\mathrm{and}\hfill \\ \hfill f=& L\left(L\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z\right)-L^2\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z,\hfill \\ \hfill g=& L\left(L\left(x\right)y\right)-L^2\left(x\right)y.\hfill \end{array}$$

Then, the corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2\left(p{|}_{L\left(L\right(x\left)y\right)}\right)z+L\left(L\left(p{|}_{L^2\left(x\right)y}\right)z\right)\hfill \\ \hfill \equiv & -L^2\left(p{|}_{L^2\left(x\right)y}\right)z+L^2\left(p{|}_{L^2\left(x\right)y}\right)z\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 2. The ambiguity is of the form

$$\begin{array}{cc}\hfill w=& \overline{f}=L\left(L\left(x\right)p{|}_{L\left(L\right(y\left)z\right)}\right){=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),p,q\in {\mathfrak{S}}^{\star }\left(Y\right),q=L\left(L\left(x\right)p\right)\mathrm{and}\hfill \\ \hfill f=& L\left(L\left(x\right)p{|}_{L\left(L\right(y\left)z\right)}\right)-L^2{\left(x\right)p|}_{L\left(L\right(y\left)z\right)},\hfill \\ \hfill g=& L\left(L\left(y\right)z\right)-L^2\left(y\right)z.\hfill \end{array}$$

The corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2{\left(x\right)p|}_{L\left(L\right(y\left)z\right)}+L\left(L\left(x\right)p{|}_{L^2\left(y\right)z}\right)\hfill \\ \hfill \equiv & -L^2\left(x\right)p{{|}_{L^2\left(y\right)z}+L^2\left(x\right)p|}_{L^2\left(y\right)z}\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

Case 3. The ambiguity is of the form

$$\begin{array}{cc}\hfill w=& \overline{f}{=L\left(L\left(L\left(x\right)y\right)z\right)=q|}_{\overline{g}},\mathrm{where}x,y,z\in \mathfrak{S}\left(Y\right),q=L(\star z)\in {\mathfrak{S}}^{\star }\left(Y\right)\mathrm{and}\hfill \\ \hfill f=& L\left(L\left(L\left(x\right)y\right)z\right)-L^2\left(L\left(x\right)y\right)z,\hfill \\ \hfill g=& L\left(L\left(x\right)y\right)-L^2\left(x\right)y,\hfill \end{array}$$

whose corresponding including composition is trivial mod $(S,w)$:

$$\begin{array}{cc}\hfill {f-q|}_g=& -L^2\left(L\left(x\right)y\right)z+L\left(L^2\left(x\right)yz\right)\hfill \\ \hfill \equiv & -L\left(L^2\left(x\right)y\right)z+L^3\left(x\right)yz\hfill \\ \hfill \equiv & -L^3\left(x\right)yz+L^3\left(x\right)yz\hfill \\ \hfill \equiv & 0.\hfill \end{array}$$

This completes the proof. □

In summary, we conclude the second main result of this paper.

Theorem 2.

Let X and Y be sets. All OPIs in $\mathbf{k}\mathfrak{S}\left(X\right)$ of degree 2 and multiplicity 2 classified in [22] listed in Lemma 3 are, respectively, Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$, ${\le }_{\mathrm{Dl}}$ or ${\le }_{\mathrm{dt}}$.

Proof. 

The Rota–Baxter OPI, Nijenhuis OPI, average OPI and inverse average OPI are Rota–Baxter type OPIs, which are, respectively, Gröbner-Shirshov on Y with respect to the monomial order ${\le }_{\mathrm{db}}$ [20]. Further, the monomial OPIs are respectively Gröbner-Shirshov on Y with respect to the monomial orders ${\le }_{\mathrm{db}}$, ${\le }_{\mathrm{Dl}}$ and ${\le }_{\mathrm{dt}}$. Finally, the remainder follows from Propositions 1–5. □

4. Conclusions

As recalled earlier, Rota twenty years ago posed the Rota’s Classification Problem: finding all possible algebraic identities that can be satisfied by a linear operator on an algebra. In this paper, we verify that all operated polynomial identities classified by Bremner et al. [22] are Gröbner-Shirshov. These operated polynomial identities include some new ones. So our results answer in part Rota’s Classification Problem, as explained in the Introduction. Rota’s Classification Problem is for associative algebras originally. However, it can also be applied to other algebras, such as Lie algebras. This necessitates the following further tasks:

• Introduce the Lie version of the new operated polynomial identities in [22] and prove that they are Gröbner-Shirshov in the context of operated Lie algebras;
• Use the method in [22] to find more operated polynomial identities;
• In general, study the operadic version of Rota’s Classification Problem.

We plan to address these and related items in the future.