No dialgebra has Gelfand-Kirillov dimension strictly between 1 and 2

TL;DR

This paper investigates the Gelfand-Kirillov dimension of associative dialgebras, establishing an upper bound related to associated algebras and proving that no dialgebra has a dimension strictly between 1 and 2.

Contribution

It introduces bounds on the Gelfand-Kirillov dimension of dialgebras and proves the non-existence of dialgebras with dimensions between 1 and 2.

Findings

Gelfand-Kirillov dimension of dialgebras is at most twice that of associated algebras.

No associative dialgebra has Gelfand-Kirillov dimension strictly between 1 and 2.

Abstract

The Gelfand-Kirillov dimension measures the asymptotic growth rate of algebras. For every associative dialgebra $\mathcal{D}$, the quotient $\mathcal{A}_\mathcal{D}:=\mathcal{D}/\mathsf{Id}(S)$, where $\mathsf{Id}(S)$ is the ideal of $\mathcal{D}$ generated by the set $S:=\{x \vdash y-x\dashv y \mid x,y\in \mathcal{D}\}$, is called the associative algebra associated to $\mathcal{D}$. Here we show that the Gelfand--Kirillov dimension of $\mathcal{D}$ is bounded above by twice the Gelfand--Kirillov dimension of $\mathcal{A}_\mathcal{D}$. Moreover, we prove that no associative dialgebra has Gelfand-Kirillov dimension strictly between 1 and 2.

Equations

$$\{\begin{array}{l}x⊣(y⊢z)=x⊣(y⊣z),\\ (x⊣y)⊢z=(x⊢y)⊢z,\\ x⊢(y⊣z)=(x⊢y)⊣z,\end{array}$$

$$\mathsf{GKdim}({\mathcal{A}}_{\mathcal{D}})\le \mathsf{GKdim}(\mathcal{D})\le 2\mathsf{GKdim}({\mathcal{A}}_{\mathcal{D}}).$$

$$x_1⊢\cdots ⊢x_m⊣\cdots ⊣x_t$$

$$V_1⊢V_2={\mathsf{span}}_k\{x⊢y\mid x\in V_1,y\in V_2\}\text{mbox}and{V}_1⊣V_2={\mathsf{span}}_k\{x⊣y\mid x\in V_1,y\in V_2\}.$$

$$V^{\le n}:=V^1+V^2+\text{leavevmode}\cdot \cdot \cdot \text{leavevmode}+V^n.$$

$$V^n={\mathsf{span}}_k\{[x_1...x_n{]}_p\mid 1\le p\le n,\text{mbox}wheren,p\in \mathbb{N}\text{mbox}and{x}_1,...\text{leavevmode},x_n\in V\}$$

$$V^{\le n}={\mathsf{span}}_k\{[x_1...x_m{]}_p\mid 1\le p\le m\le n,\text{mbox}wherep,m,n\in \mathbb{N}\text{mbox}and{x}_1,...\text{leavevmode},x_m\in V\}.$$

$$\mathsf{GKdim}(\mathcal{D})=\underset{V}{\sup }\overline{\underset{n\to \infty }{\lim }}{\log }_n\mathsf{dim}(V^{\le n}),$$

$$\mathsf{GKdim}(\mathcal{D})=\underset{X^{\prime }\subseteq X}{\sup }\overline{\underset{n\to \infty }{\lim }}{\log }_n\mathsf{dim}((k{X}^{\prime }{)}^{\le n}),$$

$$\overline{\underset{n\to \infty }{\lim }}{\log }_n\mathsf{dim}(V^{\le n})$$

$$=\overline{\underset{t\to \infty }{\lim }}{\log }_t\mathsf{dim}((k{X}^{\prime }{)}^{\le t})\le \underset{X^{\prime }\subseteq X}{\sup }\overline{\underset{t\to \infty }{\lim }}{\log }_t\mathsf{dim}((k{X}^{\prime }{)}^{\le t})$$

$$\mathsf{GKdim}(\mathcal{D})=\underset{V}{\sup }\overline{\underset{n\to \infty }{\lim }}{\log }_n\mathsf{dim}(V^{\le n})\le \underset{X^{\prime }\subseteq X}{\sup }\overline{\underset{t\to \infty }{\lim }}{\log }_t\mathsf{dim}((k{X}^{\prime }{)}^{\le t}).$$

$$\mathsf{GKdim}(\mathcal{D})=\overline{\underset{n\to \infty }{\lim }}{\log }_n\mathsf{dim}((kX{)}^{\le n}).$$

$$[X^{+}{]}_{\omega }:=\{[a_1...a_n{]}_m\mid a_1,...\text{leavevmode},a_n\in X,m,n\in \mathbb{N},1\le m\le n\}$$

$$[u{]}_p<[v{]}_q\text{mbox}if(\ell (u),p,a_1,...\text{leavevmode},a_n)<(\ell (v),q,b_1,...\text{leavevmode},b_m)\text{mbox}lexicographically.$$

$$[u_3{]}_{m_3}⊣[u_1{]}_{m_1}<[u_3{]}_{m_3}⊣[u_2{]}_{m_2}\text{mbox}and[u_1{]}_{m_1}⊢[u_3{]}_{m_3}<[u_2{]}_{m_2}⊢[u_3{]}_{m_3}.$$

$$B_{\mathcal{D}}:=\{[u{]}_p\in [X^{+}{]}_{\omega }\mid$$

$$((z_j⊢(x_j⊢y_j-x_j⊣y_j))⊣z_j^{\prime })⊢y$$

$$=$$

$$[a_1{a}_2\cdots a_t{]}_p=[a_1...a_{p-1}{]}_1⊢[a_p...a_t{]}_1=0\text{mbox}in\mathcal{D}\text{mbox}andhencein{\mathcal{D}}^{\prime },$$

$$[a_1...a_t{]}_p-\sum _{1\le i\le n}{\alpha }_i[u_i{]}_{p_i}⊢[a_p{a}_{p+1}...a_t{]}_1=([a_1...a_{p-1}{]}_1-\sum _{1\le i\le n}{\alpha }_i[u_i{]}_{p_i})⊢[a_p{a}_{p+1}...a_t{]}_1=0$$

$$B_{\mathcal{D}}^n=\{[u{]}_p\in B_{\mathcal{D}}\mid \ell (u)=n\},B_{\mathcal{D}}^{\le n}=\{[u{]}_p\in B_{\mathcal{D}}\mid \ell (u)\le n\}.$$

$$\mathsf{GKdim}(\mathcal{D})=\overline{\underset{n\to \infty }{\lim }}{\log }_n|B_{\mathcal{D}}^{\le n}|.$$

$$\mathsf{GKdim}({\mathcal{A}}_{\mathcal{D}})\le \mathsf{GKdim}(\mathcal{D})\le 2\mathsf{GKdim}({\mathcal{A}}_{\mathcal{D}}).$$

$$|B_{{\mathcal{D}}^{\prime }}^{\le n}|=\sum _{1\le t\le n}|B_{{\mathcal{D}}^{\prime }}^t|\le \sum _{1\le t\le n}\sum _{1\le p\le t}|B_{{\mathcal{A}}^{\prime }}^{p-1}||X^{\prime }||B_{{\mathcal{A}}^{\prime }}^{t-p}|=|X^{\prime }|\sum _{1\le t\le n}\sum _{\begin{array}{c}i\ge 0;j\ge 0;\\ i+j=t-1\end{array}}|B_{{\mathcal{A}}^{\prime }}^i||B_{{\mathcal{A}}^{\prime }}^j|$$

$$=$$

$$\le$$

$$\mathsf{GKdim}({\mathcal{D}}^{\prime })=\overline{\underset{n\to \infty }{\lim }}{\log }_n|B_{{\mathcal{D}}^{\prime }}^{\le n}|\le \overline{\underset{n\to \infty }{\lim }}{\log }_n(|X^{\prime }|(|B_{{\mathcal{A}}^{\prime }}^{\le n}|+1{)}^2)=2\mathsf{GKdim}({\mathcal{A}}^{\prime })\le 2\mathsf{GKdim}({\mathcal{A}}_{\mathcal{D}}).$$

$$B_1=\{[a_1...a_n{]}_p\in B\mid [a_1...a_{p-1}{]}_1\in B_{\mathcal{A}}\text{mbox}ifp>1,\text{mbox}and[a_{p+1}...a_n{]}_1\in B_{\mathcal{A}}\text{mbox}ifp<n\}.$$

$$[a_1...a_n{]}_p=[a_1...a_{p-1}{]}_1⊢[a_p{a}_{p+1}..a_n{]}_1=\sum _{1\le i\le n}{\alpha }_i[u_i{]}_{p_i}⊢[a_p{a}_{p+1}\cdots a_n{]}_1.$$

$$|B_1^t|\le \sum _{\begin{array}{c}1\le p\le m;\\ p\le t\end{array}}|B_{\mathcal{A}}^{p-1}||X||B_{\mathcal{A}}^{t-p}|+\sum _{\begin{array}{c}t-m+1\le p\le t;\\ p\ge 1\end{array}}|B_{\mathcal{A}}^{p-1}||X||B_{\mathcal{A}}^{t-p}|=2\sum _{\begin{array}{c}1\le p\le m;\\ p\le t\end{array}}|B_{\mathcal{A}}^{p-1}||X||B_{\mathcal{A}}^{t-p}|,$$

$$|B_1^{\le n}|=\sum _{1\le t\le n}|B_1^t|\le 2\sum _{1\le t\le n}\sum _{\begin{array}{c}1\le p\le m;\\ p\le t\end{array}}|B_{\mathcal{A}}^{p-1}||X||B_{\mathcal{A}}^{t-p}|=2|X|\sum _{1\le t\le n}\sum _{\begin{array}{c}0\le i\le m-1;j\ge 0;\\ i+j=t-1\end{array}}|B_{\mathcal{A}}^i||B_{\mathcal{A}}^j|$$

Full text

‡ Supported by the NNSF of China (11571121), the NSF of Guangdong Province (2017A030313002) and the Science and Technology Program of Guangzhou (201707010137)

∗ Supported by the Innovation Project of Graduate School of South China Normal University

♯ Corresponding author

1. Introduction

The Gelfand-Kirillov dimension has become one of the important tools in the study of algebras. There are many well known results on Gelfand-Kirillov dimensions of associative algebras and modules [3]. For instance, the Gelfand-Kirillov dimension of a finitely generated commutative algebra $\mathcal{A}$ over a field is the classical Krull dimension of $\mathcal{A}$ [3, Theorem 4.5]. In particular, the Gelfand-Kirillov dimension of the free commutative algebra generated by $n$ elements is exactly $n$, where $n$ is a positive integer. Here we shall investigate the Gelfand-Kirillov dimensions of associative dialgebras.

Recall that an associative dialgebra (dialgebra for short) $(\mathcal{D},\vdash,\dashv)$ over a field $k$ is a $k$-vector space equipped with two bilinear operations $\vdash:\mathcal{D}\otimes\mathcal{D}\rightarrow\mathcal{D}$ and $\dashv\ :\mathcal{D}\otimes\mathcal{D}\rightarrow\mathcal{D}$ such that $(\mathcal{D},\vdash)$ and $(\mathcal{D},\dashv)$ are associative algebras and the following identities hold:

[TABLE]

for all $x,y$ and $z$ in $\mathcal{D}$. For instance, let $(\mathcal{A},\partial)$ be a differential associative algebra satisfying $\partial^{2}=0$, and define $x\vdash y=\partial(x)y$ and $x\dashv y=x\partial(y)$. Then $(\mathcal{A},\vdash,\dashv)$ becomes a dialgebra [4].

Let $\mathcal{D}$ be a dialgebra and let $\mathcal{A}_{\mathcal{D}}=\mathcal{D}/\mathsf{Id}(\SS)$, where $\mathsf{Id}(\SS)$ is the ideal of $\mathcal{D}$ generated by the set $\SS:=\{x\vdash y-x\dashv y\mid x,y\in\mathcal{D}\}$. Then $\mathcal{A}_{\mathcal{D}}$ is an associative algebra, which is called the associative algebra associated to $\mathcal{D}$. We find that the Gelfand-Kirillov dimension $\mathsf{GKdim}(\mathcal{D})$ of $\mathcal{D}$ and the Gelfand-Kirillov dimension $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ of $\mathcal{A}_{\mathcal{D}}$ have the following relation:

Theorem A**.**

Let $\mathcal{D}$ be a dialgebra and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Then we have

[TABLE]

In particular, the inequality $\mathsf{GKdim}(\mathcal{D})<\infty$ holds if and only if $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})<\infty$ holds.

For every associative algebra $(\mathcal{A},\cdot)$, we define $x\vdash y=x\dashv y=x\cdot y$ for all $x,y$ in $\mathcal{A}$. Then it is clear that $(\mathcal{A},\vdash,\dashv)$ becomes a dialgebra. It is well-known that for every real number $r\geq 2$, there exists an associative algebra $\mathcal{A}$ satisfying $\mathsf{GKdim}(\mathcal{A})=r$, so for every real number $r\geq 2$, there exists a dialgebra $\mathcal{D}$ satisfying $\mathsf{GKdim}(\mathcal{D})=r$. Bergman [1] proved that there is no associative algebra having Gelfand-Kirillov dimension in the open interval $(1,2)$. C. Martinez and E. Zelmanov [5] proved an analogous theorem for Jordan algebras. However, for Lie algebras and for Jordan superalgebras, the situations are quite different: the Gelfand–Kirillov dimension of a finitely generated Lie algebra [6] or of a Jordan superalgebra [7] can be an arbitrary number in $\{0\}\cup[1,\infty]$. This leaves the following problem open: Does there exist a dialgebra of Gelfand-Kirillov dimension in the open interval $(1,2)$? Based on Bergman’s result, we prove the following result:

Theorem B**.**

No dialgebra has Gelfand-Kirillov dimension strictly between 1 and 2.

Although Theorem B is based on the corresponding result of Bergman, the extension is not obvious because really different techniques are needed here, among which constructing connected linear bases for $\mathcal{D}$ and $\mathcal{A}_{\mathcal{D}}$ simultaneously plays an important role. In general, assuming $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=1$ does not imply $\mathsf{GKdim}(\mathcal{D})=1$. We shall provide a sufficient condition in Lemma 3.11 ensuring $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. Moreover, we shall show that, for $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=1$, the sufficient condition is satisfied.

The paper is organized as follows. In Section 2, we recall some basic properties of a dialgebra and the definition of the Gelfand-Kirillov dimension. We then provide a formula, involving the generators of a dialgebra, for calculating the Gelfand-Kirillov dimension (Lemma 2.4). In Section 3, we introduce how to construct a shortest-middle-lexicographic (linear) basis of a dialgebra (Definition 3.3). Such (linear) bases turn out to be very useful in connecting the Gelfand-Kirillov dimension of a dialgebra and that of its associated associative algebra, and they play important roles in the proofs of Theorem A and Theorem B. Finally, to prove Theorem B, we also need to investigate the middle entries (Definition 2.1) of monomials in the constructed shortest-middle-lexicographic (linear) bases.

2. The Gelfand-Kirillov dimension of a dialgebra

Our aim in this section is to recall the notion of the Gelfand-Kirillov dimension of a dialgebra. We shall first recall some basic notations and properties of a dialgebra. Then we shall introduce several formulas for the Gelfand-Kirillov dimension of a dialgebra.

Recall that for every dialgebra $\mathcal{D}$, for all $x_{1},...\leavevmode\hbox{}\hskip 0.56905pt,x_{t}$ in $\mathcal{D}$, every parenthesizing of

[TABLE]

gives the same element in $\mathcal{D}$ [4], which we denote by $[x_{1}...x_{t}]_{m}$ or ${x_{1}\vdash\cdots\vdash x_{m}\dashv\cdots\dashv x_{t}}$. The following definition of middle submonomials of a monomial resembles that of subwords of a word.

Definition 2.1**.**

[4] For all $x_{1},...\leavevmode\hbox{}\hskip 0.56905pt,x_{t}$ in $\mathcal{D}$, the element $x_{m}$ in the monomoial $[x_{1}...x_{t}]_{m}$ is called the middle entry of $[x_{1}...x_{t}]_{m}$, and every monomial $[x_{p}...x_{q}]_{m-p+1}$ satisfying $1\leq p\leq m\leq q\leq t$ is called a middle submonomial of $[x_{1}...x_{t}]_{m}$.

The following formulas for calculating the product of two monomials in a dialgebra will be very useful in the sequel. Roughly speaking, the middle entry of the product of two monomials is the middle entry that the operation $\vdash$ or $\dashv$ points to. More precisely, we have the following lemma:

Lemma 2.2**.**

[4]* For all integers $m,n,p,q$ such that $1\leq p\leq n<m$ and $1\leq q\leq m-n$, for all $x_{1},...\leavevmode\hbox{}\hskip 0.56905pt,x_{m}$ in a dialgebra $\mathcal{D}$, we have the following formulas:*

(i)* $[x_{1}...x_{n}]_{p}\vdash[x_{n+1}...x_{m}]_{q}=[x_{1}...x_{m}]_{n+q}$.*

(ii)* $[x_{1}...x_{n}]_{p}\dashv[x_{n+1}...x_{m}]_{q}=[x_{1}...x_{m}]_{p}$.*

Before introducing a general definition of the Gelfand-Kirillov dimension of a dialgebra $\mathcal{D}$ over a field $k$, we have to introduce some more notations. Let $V,V_{1}$ and $V_{2}$ be vector subspaces of $\mathcal{D}$. We first define

[TABLE]

Then we define $V^{1}=V$ and $V^{n}=\sum_{1\leq i\leq n-1}(V^{i}\vdash V^{n-i}+V^{i}\dashv V^{n-i})$ for every integer number $n\geq 2$. Finally, we define

[TABLE]

By Lemma 2.2, we easily obtain

[TABLE]

and

[TABLE]

Now we are ready to introduce the Gelfand-Kirillov dimension of a dialgebra.

Definition 2.3**.**

Let $\mathcal{D}$ be a dialgebra over a field $k$, the Gelfand-Kirillov dimension of a dialgebra $\mathcal{D}$ is defined to be

[TABLE]

where the supremum is taken over all finite dimensional subspaces $V$ of $\mathcal{D}$.

For instance, let $\mathcal{D}$ be the free dialgebra generated by a letter $a$. Then it is easy to see that $\{[a_{1}...a_{n}]_{p}\mid a_{1}=\mathrel{\leavevmode\hbox{}\hskip 0.56905pt{\cdot}{\cdot}{\cdot}\leavevmode\hbox{}\hskip 0.56905pt}=a_{n}=a,1\leq p\leq n$, and $p,n\in~{}\mathbb{N}\}$ is a linear basis of $\mathcal{D}$. So by direct calculation, we obtain $\mathsf{GKdim}(\mathcal{D})=2$.

By Definition 2.3, if $\mathcal{D}^{\prime}$ is a subalgebra of $\mathcal{D}$ or a homomorphic image of $\mathcal{D}$, then we have $\mathsf{GKdim}(\mathcal{D}^{\prime})\leq\mathsf{GKdim}(\mathcal{D})$. And since we often consider generating sets of a dialgebra, we shall also use the following formula for calculating the Gelfand-Kirillov dimension in the sequel.

Lemma 2.4**.**

Let $\mathcal{D}$ be a dialgebra generated by a set $X$. Then we have

[TABLE]

where the supremum is taken over all finite nonempty subsets $X^{\prime}$ of $X$, and $kX^{\prime}$ is the vector subspace of $\mathcal{D}$ spanned by $X^{\prime}$.

Proof.

Let $V$ be a finite dimensional subspace of $\mathcal{D}$. Then for some finite subset $X^{\prime}$ of $X$ and for some integer number $m$, we have $V\subseteq(kX^{\prime})^{\leq m}$ and $V^{\leq n}\subseteq((kX^{\prime})^{\leq m})^{\leq n}\subseteq(kX^{\prime})^{\leq mn}$ for every integer number $n\geq 1$. Consequently, we deduce

[TABLE]

and

[TABLE]

On the other hand, for every finite subset $X^{\prime}$ of $X$, the subspace $kX^{\prime}$ is a finite dimensional subspace of $\mathcal{D}$. By Definition 2.3, the lemma follows. ∎

As a consequence, when one applies Equation (2.1) to calculate the Gelfand-Kirillov dimension of a dialgebra $\mathcal{D}$, it is independent of the choices of the sets of generators of $\mathcal{D}$. In particular, when $\mathcal{D}$ is finitely generated, Equation (2.1) becomes easier:

Corollary 2.5**.**

Let $\mathcal{D}$ be a dialgebra generated by a finite set $X$. Then we have

[TABLE]

In particular, for every dialgebra $\mathcal{D}$, we have $\mathsf{GKdim}(\mathcal{D})=\sup_{\mathcal{D}^{\prime}}\mathsf{GKdim}(\mathcal{D}^{\prime})$, where the supremum is taken over all finitely generated subalgebras $\mathcal{D}^{\prime}$ of $\mathcal{D}$.

3. No dialgebra has Gelfand-Kirillov dimension in the open interval $(1,2)$

Our aim in this section is to prove our main results on Gelfand-Kirillov dimensions of dialgebras. We shall prove the inequality $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})\leq\mathsf{GKdim}(\mathcal{D})\leq 2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ and construct dialgebras such that $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=\mathsf{GKdim}(\mathcal{D})$ or $\mathsf{GKdim}(\mathcal{D})=2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. Finally, we shall conclude the paper with Theorem B.

We first recall the linear basis of a free dialgebra generated by an arbitrary well-ordered set $X$ constructed by Loday [4]. Let $(X,<)$ be a fixed well-ordered set. Then we use $X^{+}$ for the free semigroup generated by $X$. For every $u=a_{1}...a_{n}$ in $X^{+}$, where $a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{n}$ lie in $X$, we define the length $\ell(u)$ of $u$ to be $n$. Finally, let $\mathcal{D}(X)$ be the free dialgebra generated by $X$. Then the following set

[TABLE]

forms a linear basis of $\mathcal{D}(X)$ [4]. For a nonempty sequence $u=a_{1}...a_{n}$ over $X$, we shall also use the notation $[u]_{m}$ for $[a_{1}...a_{n}]_{m}$, and we call $[a_{1}...a_{n}]_{m}$ a disequence over $X$. Obviously, two disequences $[a_{1}...a_{n}]_{m}$ and $[b_{1}...b_{p}]_{q}$ are the same only if $n=p$, $m=q$ and $a_{i}=b_{i}$ for every $i\leq n$. Finally, for a subset $Y$ of $[X^{+}]_{\omega}$, we define $|Y|$ to be cardinality of the set $Y$.

Now we recall a well-ordering on $[X^{+}]_{\omega}$ introduced in [2]. Here we use a slightly different name for the ordering to make it easy to remember how we compare two monomials.

Definition 3.1**.**

[2] We recall that the length-middle-lexicographic ordering $<$ on $[X^{+}]_{\omega}$ is defined as follows: For all $[u]_{p}=[a_{1}...a_{n}]_{p}$ and $[v]_{q}=[b_{1}...b_{m}]_{q}$ in $[X^{+}]_{\omega}$ such that $a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{n}$ and $b_{1},...\leavevmode\hbox{}\hskip 0.56905pt,b_{m}$ lie in $X$, we define

[TABLE]

For instance, for all $a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{4}$ in $X$ such that $a_{1}<\mathrel{\leavevmode\hbox{}\hskip 0.56905pt{\cdot}{\cdot}{\cdot}\leavevmode\hbox{}\hskip 0.56905pt}<a_{4}$, we have $[a_{4}a_{3}a_{2}]_{1}<[a_{1}a_{3}a_{2}]_{2}$ and $[a_{1}a_{2}a_{3}a_{4}]_{1}>[a_{1}a_{2}a_{4}]_{2}$. Note that the above ordering is not compatible with the products in general. For instance, we have $[a_{1}a_{2}]_{2}>[a_{2}a_{1}]_{1}$ and $[a_{1}a_{2}]_{2}\vdash a_{3}<[a_{2}a_{1}]_{1}\vdash a_{3}$.

However, the length-middle-lexicographic ordering still has some good properties as follows:

Lemma 3.2**.**

Let $[u_{1}]_{m_{1}}$, $[u_{2}]_{m_{2}}$ and $[u_{3}]_{m_{3}}$ be monomials in $[X^{+}]_{\omega}$ with $[u_{1}]_{m_{1}}<[u_{2}]_{m_{2}}$. Then we have $[u_{3}]_{m_{3}}\vdash[u_{1}]_{m_{1}}<[u_{3}]_{m_{3}}\vdash[u_{2}]_{m_{2}}$ and $[u_{1}]_{m_{1}}\dashv[u_{3}]_{m_{3}}<[u_{2}]_{m_{2}}\dashv[u_{3}]_{m_{3}}$; if, in addition, $m_{2}=1$, then we also have

[TABLE]

Proof.

Note that $[u_{1}]_{m_{1}}<[u_{2}]_{1}$ implies $(\ell(u_{1}),m_{1})\leq(\ell(u_{2}),1)$. Moreover, if $\ell(u_{1})=\ell(u_{2})$, then we obtain $m_{1}=1$. The remain of the proof is straightforward. ∎

Let $\mathcal{D}$ be a dialgebra generated by a well-ordered set $X$. Then $[X^{+}]_{\omega}$ is a linear generating set of $\mathcal{D}$. Moreover, there is no harm to assume that $\mathcal{A}_{\mathcal{D}}$ is also generated by $X$ and  $[X^{+}]_{\omega}$ is also a linear generating set of $\mathcal{A}_{\mathcal{D}}$. To avoid possible confusions, whenever we write down a linear combination $f$ of elements in $[X^{+}]_{\omega}$, we shall declare $f$ in $\mathcal{D}$ or $\mathcal{A}_{\mathcal{D}}$. In this way, we are able to use the same notation for an element in $\mathcal{D}$ or in $\mathcal{A}_{\mathcal{D}}$.

We are now ready to introduce a general way of constructing a subset $B_{\mathcal{D}}$ of $[X^{+}]_{\omega}$ for a dialgebra $\mathcal{D}$ with respect to a generating set $X$, and we shall see in Lemma 3.4 that $B_{\mathcal{D}}$ turns out to be a linear basis of $\mathcal{D}$. By convention, we shall assume that the empty set is a linear basis of the trivial vector space $\{0\}$.

Definition 3.3**.**

Let $\mathcal{D}$ be a dialgebra generated by a well-ordered set $X$. We call

[TABLE]

the shortest-middle-lexicographic basis of $\mathcal{D}$ with respect to $X$.

By Definition 3.3, we know that $B_{\mathcal{D}}$ depends on the generating set $X$. However, to make the notation and formulas simple, we shall still use the notation $B_{\mathcal{D}}$ but not $B_{\mathcal{D},X}$. (No confusions arise in the sequel.)

Lemma 3.4**.**

Let $\mathcal{D}$ be a dialgebra generated by a well-ordered set $X$. Then the shortest-middle-lexicographic basis $B_{\mathcal{D}}$ of $\mathcal{D}$ with respect to $X$ is a linear basis of $\mathcal{D}$. In particular, let $\mathcal{A}=\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$, and let $B_{\mathcal{A}}$ be the shortest-middle-lexicographic basis of $\mathcal{A}$ with respect to $X$. Then $B_{\mathcal{A}}$ is a linear basis of $\mathcal{A}$.

Proof.

If $\mathcal{D}=\{0\}$, then $B_{\mathcal{D}}$ is an empty set. Now we assume $\mathcal{D}\neq\{0\}$. To show that $B_{\mathcal{D}}$ is a linear generating set of $\mathcal{D}$, it suffices to show that every $[u]_{p}$ in $[X^{+}]_{\omega}$ lies in the subspace $\mathsf{span}_{k}(B_{\mathcal{D}})$ of $\mathcal{D}$ spanned by $B_{\mathcal{D}}$. We use induction on $[u]_{p}$ with respect to the length-middle-lexicographic ordering in Definition 3.1. The minimal element of $X$ that is not [math] in $\mathcal{D}$, say $a$, lies in $B_{\mathcal{D}}$ by the definition of $B_{\mathcal{D}}$. Suppose $[u]_{p}>a$. If $[u]_{p}$ lies in $B_{\mathcal{D}}$ or $[u]_{p}=0$ in $\mathcal{D}$, then there is nothing to prove. Otherwise, say $[u]_{p}=\sum_{1\leq i\leq n}\alpha_{i}[u_{i}]_{p_{i}}$ for some elements $\alpha_{1},...\leavevmode\hbox{}\hskip 0.56905pt,\alpha_{n}$ in $k$ and for some monomials $[u_{1}]_{p_{1}},...\leavevmode\hbox{}\hskip 0.56905pt,[u_{n}]_{p_{n}}$ in $[X^{+}]_{\omega}$ such that $[u_{i}]_{p_{i}}<[u]_{p}$ for every $i\leq n$, then by induction hypothesis, we have $[u]_{p}\in\mathsf{span}_{k}(B_{\mathcal{D}})$. Finally, by the construction of $B_{\mathcal{D}}$, it is clear that the set $B_{\mathcal{D}}$ is linear independent in $\mathcal{D}$. ∎

Now we show that, every $x\neq 0$ in $\mathcal{D}$ such that $x=0$ in $\mathcal{A}_{\mathcal{D}}$ is a left zero divisor with respect $\vdash$ and a right zero divisor with respect to $\dashv$ in $\mathcal{D}$. To simplify the formulas, we define $\varepsilon$ to be an empty disequence, and for every $y$ in $\mathcal{D}$, we define $\varepsilon\vdash y=y\dashv\varepsilon=y$.

Lemma 3.5**.**

Let $\mathcal{D}$ be a dialgebra and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Then for every $x$ in $\mathcal{D}$ such that $x=0$ in $\mathcal{A}_{\mathcal{D}}$, for every $y$ in $\mathcal{D}$, we have $x\vdash y=y\dashv x=0$ in $\mathcal{D}$.

Proof.

If $x=0$ in $\mathcal{A}_{\mathcal{D}}$, then we have $x=\sum\alpha_{j}(z_{j}\vdash(x_{j}\vdash y_{j}-x_{j}\dashv y_{j}))\dashv z_{j}^{\prime}\mbox{ in }\mathcal{D}$ by Lemma 2.2, where each $x_{j}$ and each $y_{j}$ lie in $\mathcal{D}$, each $z_{j}$ and each $z_{j}^{\prime}$ may be elements of $\mathcal{D}$ or be the empty disequence $\varepsilon$. So for every $y$ in $\mathcal{D}$, by Lemma 2.2 again, we obtain

[TABLE]

Similar to the above reasoning, we obtain $y\dashv x=0$. ∎

Before going further, we shall observe some more properties on the shortest-middle-lexicographic basis of a dialgebra. In Lemma 3.6, we shall use the shortest-middle-lexicographic basis of an associative subalgebra $\mathcal{A}^{\prime}$ of $\mathcal{A}_{\mathcal{D}}$ generated by $X^{\prime}$, where $X^{\prime}$ is a nonempty subset of $X$. By changing every occurrence of $\mathcal{D}$ into $\mathcal{A}^{\prime}$ and every occurrence of $X$ into $X^{\prime}$ in (3.1), we shall obtain the corresponding set $B_{\mathcal{A}^{\prime}}$.

Lemma 3.6**.**

Let $\mathcal{D}$ be a dialgebra generated by a well-ordered set $X$, and let $\mathcal{A}=\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Let $\mathcal{D}^{\prime}$ be a subalgebra of $\mathcal{D}$ generated by a subset $X^{\prime}$ of $X$, and let $\mathcal{A}^{\prime}$ be a subalgebra of $\mathcal{A}$ generated by $X^{\prime}$. Then for every monomial $[a_{1}...a_{t}]_{p}$ in the shortest-middle-lexicographic basis $B_{\mathcal{A}^{\prime}}$ of $\mathcal{A}^{\prime}$ with respect to $X^{\prime}$, where the letters $a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{t}$ lie in $X^{\prime}$, we have $p=1$. Moreover, if $[a_{1}...a_{t}]_{p}$ lies in the shortest-middle-lexicographic basis $B_{\mathcal{D}^{\prime}}$ of $\mathcal{D}^{\prime}$ with respect to $X^{\prime}$ and $p>1$, then $[a_{1}...a_{p-1}]_{1}$ lies in $B_{\mathcal{A}^{\prime}}$; if $[a_{1}...a_{t}]_{p}$ lies in $B_{\mathcal{D}^{\prime}}$ and $p<t$, then $[a_{p+1}...a_{t}]_{1}$ lies in $B_{\mathcal{A}^{\prime}}$.

Proof.

By the definition of $\mathcal{A}$, for every integer number $p$ such that $1\leq p\leq t$, we have $[a_{1}...a_{t}]_{p}=[a_{1}...a_{t}]_{1}$ in $\mathcal{A}$ and thus in $\mathcal{A}^{\prime}$. So $[a_{1}...a_{t}]_{p}\in B_{\mathcal{A}^{\prime}}$ forces $p=1$. For the second claim, we may assume that $p>1$ without loss of generality. Suppose that $[a_{1}...a_{p-1}]_{1}\notin B_{\mathcal{A}^{\prime}}$.

If $[a_{1}...a_{p-1}]_{1}=0$ in $\mathcal{A}^{\prime}$ and thus in $\mathcal{A}$, then by Lemma 3.5, we have

[TABLE]

which contradicts with the fact that $[a_{1}...a_{t}]_{p}$ lies in $B_{\mathcal{D}^{\prime}}$.

If $[a_{1}...a_{p-1}]_{1}\neq 0$ in $\mathcal{A}^{\prime}$, then we have $[a_{1}...a_{p-1}]_{1}=\sum_{1\leq i\leq n}\alpha_{i}[u_{i}]_{p_{i}}$ in $\mathcal{A}^{\prime}$ and thus in $\mathcal{A}$ for some monomials $[u_{1}]_{p_{1}},...\leavevmode\hbox{}\hskip 0.56905pt,[u_{n}]_{p_{n}}$ in $[{X^{\prime}}^{+}]_{\omega}$ satisfying $[u_{i}]_{p_{i}}<[a_{1}...a_{p-1}]_{1}$ for every integer $i\leq n$ and for some elements $\alpha_{1},...\leavevmode\hbox{}\hskip 0.56905pt,\alpha_{n}$ in the field $k$. In particular, by Lemma 3.5 again, we obtain

[TABLE]

in $\mathcal{D}$, and thus in $\mathcal{D}^{\prime}$. Finally, by Lemma 3.2, we obtain $[u_{i}]_{p_{i}}\vdash[a_{p}a_{p+1}...a_{t}]_{1}<[a_{1}...a_{t}]_{p}$, which contradicts with the construction of $B_{\mathcal{D}^{\prime}}$. ∎

Note that the length-middle-lexicographic ordering on $[X^{+}]_{\omega}$ plays an important role in Lemma 3.6. Using another way of constructing linear bases for $\mathcal{D}$ and $\mathcal{A}_{\mathcal{D}}$ simultaneously may not have the claimed properties in the lemma.

Remark 3.7**.**

With the notations of Lemma 3.6, if $X\neq X^{\prime}$, then $\mathcal{A}^{\prime}$ is not necessary the associative algebra associated to $\mathcal{D}^{\prime}$ in general. For instance, let $X=\{a,b\}$, and let $\mathcal{D}(a,b)$ be the free dialgebra generated by $X$. Then we define $\mathcal{D}$ to be the quotient of $\mathcal{D}(a,b)$ by the ideal generated by the set $\{b=a\vdash a-a\dashv a\}$. It is clear that $\mathcal{D}$ is isomorphic to the free dialgebra generated by $\{a\}$ and $\mathcal{A}$ is isomorphic to the free associative algebra generated by $\{a\}$. Finally, for $X^{\prime}=\{b\}$, we deduce that,  $\mathcal{D}^{\prime}$ and the associative algebra associated to $\mathcal{D}^{\prime}$ are the linear space spanned by $b$ while $\mathcal{A}^{\prime}$ is the 0 space.

Let $\mathcal{D}$ be a dialgebra generated by a finite well-ordered set $X$ and let $B_{\mathcal{D}}$ be the shortest-middle-lexicographic basis of $\mathcal{D}$ with respect to $X$. For every positive integer $n$, we define

[TABLE]

Then we have $\mathsf{dim}((kX)^{\leq n})=|B_{\mathcal{D}}^{\leq n}|$ for every $n\geq 1$, where $|B_{\mathcal{D}}^{\leq n}|$ is the cardinality of the set $B_{\mathcal{D}}^{\leq n}$. In particular, by Corollary 2.5, when $\mathcal{D}$ is generated by a finite well-ordered set $X$, we deduce the following formula for calculating the Gelfand-Kirillov dimension of $\mathcal{D}$:

[TABLE]

Now we are ready to prove Theorem A of the introduction.

Theorem 3.8**.**

Let $\mathcal{D}$ be a dialgebra and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Then we have

[TABLE]

In particular, the inequality $\mathsf{GKdim}(\mathcal{D})<\infty$ holds if and only if $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})<\infty$ holds.

Proof.

We use the notations of Lemma 3.6. By Definition 2.3, it is easy to see that, for every homomorphic image $\mathcal{D}_{1}$ of $\mathcal{D}$, we have $\mathsf{GKdim}(\mathcal{D}_{1})\leq\mathsf{GKdim}(\mathcal{D})$. In particular, we obtain $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})\leq\mathsf{GKdim}(\mathcal{D})$.

Now we turn to the inequality $\mathsf{GKdim}(\mathcal{D})\leq 2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. For finitely generated algebras $\mathcal{D}^{\prime}$ and $\mathcal{A}^{\prime}$ as in Lemma 3.6, we define $|B_{\mathcal{A}^{\prime}}^{0}|=|B_{\mathcal{D}^{\prime}}^{0}|=1$ for convenience. Then by Lemma 3.6, for every integer number $t\geq 1$, we obtain  $|B_{\mathcal{D}^{\prime}}^{t}|\leq\sum\limits_{1\leq p\leq t}|B_{\mathcal{A}^{\prime}}^{p-1}||X^{\prime}||B_{\mathcal{A}^{\prime}}^{t-p}|$ and

[TABLE]

Consequently, we obtain

[TABLE]

By Corollary 2.5, we deduce $\mathsf{GKdim}(\mathcal{D})=\sup_{\mathcal{D}^{\prime}}\mathsf{GKdim}(\mathcal{D}^{\prime})\leq 2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$, where the supremum is taken over all finitely generated subalgebras $\mathcal{D}^{\prime}$ of $\mathcal{D}$. ∎

Corollary 3.9**.**

Let $\mathcal{D}$ be a dialgebra. Then $\mathsf{GKdim}(\mathcal{D})<1$ implies that $\mathsf{GKdim}(\mathcal{D})=0$.

Proof.

Let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Then $\mathsf{GKdim}(\mathcal{D})<1$ implies that $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=0$. So we obtain $0=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})\leq\mathsf{GKdim}(\mathcal{D})\leq 2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=0$. ∎

Now we construct examples to show that the boundaries in Theorem 3.8 are the best that one can expect. We shall describe a sufficient condition ensuring $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ in Lemma 3.11.

Example 3.10**.**

Let $\mathcal{D}$ be the free dialgebra generated by a letter $a$, and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. Then we have $\mathsf{GKdim}(\mathcal{D})=2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=2$.

Proof.

Note that $B_{\mathcal{D}}=\{[a_{1}...a_{n}]_{p}\mid a_{1}=\mathrel{\leavevmode\hbox{}\hskip 0.56905pt{\cdot}{\cdot}{\cdot}\leavevmode\hbox{}\hskip 0.56905pt}=a_{n}=a,1\leq p\leq n$, and $p,n\in~{}\mathbb{N}\}$, and $\mathcal{A}_{\mathcal{D}}$ is the free commutative algebra generated by $a$. We have $\mathsf{GKdim}(\mathcal{D})=2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=2$ immediately by Equation (3.2). ∎

Lemma 3.11**.**

Let $\mathcal{D}$ be a dialgebra generated by a finite well-ordered set $X$, and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. If there exists a positive integer $m$ such that $\mathcal{D}$ has a linear generating set $B=\{[a_{1}...a_{n}]_{p}\mid a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{n}\in X,1\leq p\leq m$ or $0\leq n-p\leq m-1$, where $p,n$ lies in $\mathbb{N}$ and $p\leq n\}$. Then we obtain $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$.

Proof.

For simplicity, we denote $\mathcal{A}_{\mathcal{D}}$ by $\mathcal{A}$. Since $\mathsf{GKdim}(\mathcal{A})\leq\mathsf{GKdim}(\mathcal{D})$, it suffices to show $\mathsf{GKdim}(\mathcal{D})\leq\mathsf{GKdim}(\mathcal{A})$. Let $B_{\mathcal{A}}$ be the shortest-middle-lexicographic basis of $\mathcal{A}$ with respect to $X$ and let $B_{1}$ be the subset of $B$ defined as follows

[TABLE]

We claim that $B_{1}$ is a linear generating set of $\mathcal{D}$. We use induction with respect to the length-middle-lexicographic ordering on $[X^{+}]_{\omega}$ to show that every element $[a_{1}...a_{n}]_{p}$ of $B$ can be written as a linear combination of elements of $B_{1}$. It is clear that $X$ is a subset of $B_{1}$. Without loss of generality, we assume that $p>1$ and $[a_{1}...a_{p-1}]_{1}\notin B_{\mathcal{A}}$. The following proof is similar to Lemma 3.6. If $[a_{1}...a_{p-1}]_{1}=0$ in $\mathcal{A}$, then we have $[a_{1}...a_{n}]_{p}=0$ in $\mathcal{D}$. If $[a_{1}...a_{p-1}]_{1}\neq 0$ in $\mathcal{A}$, then we may assume that $[a_{1}...a_{p-1}]_{1}=\sum_{1\leq i\leq n}\alpha_{i}[u_{i}]_{p_{i}}$ in $\mathcal{A}$ for some monomials $[u_{1}]_{p_{1}},...\leavevmode\hbox{}\hskip 0.56905pt,[u_{n}]_{p_{n}}$ in $[{X}^{+}]_{\omega}$ satisfying $[u_{i}]_{p_{i}}<[a_{1}...a_{p-1}]_{1}$ for every integer $i$ such that $1\leq i\leq n$ and for some elements $\alpha_{1},...\leavevmode\hbox{}\hskip 0.56905pt,\alpha_{n}$ in the field $k$. By Lemma 3.2, we obtain $[u_{i}]_{p_{i}}\vdash[a_{p}a_{p+1}\cdots a_{n}]_{1}<[a_{1}...a_{n}]_{p}$. Finally, we have

[TABLE]

Since $\ell(u_{i})\leq p-1<m$, each monomial $[u_{i}]_{p_{i}}\vdash[a_{p}a_{p+1}\cdots a_{n}]_{1}$ lies in $B$. By induction hypothesis, each monomial $[u_{i}]_{p_{i}}\vdash[a_{p}a_{p+1}\cdots a_{n}]_{1}$, and thus $[a_{1}...a_{n}]_{p}$, can be written as a linear combination of elements of $B_{1}$.

For every $t\geq 1$, define $B_{1}^{t}=\{[a_{1}...a_{t}]_{p}\in B_{1}\mid a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{t}\in X,p\in~{}\mathbb{N}\}$, $B_{1}^{\leq t}=\cup_{1\leq i\leq t}B_{1}^{i}$ and define $|B_{1}^{0}|=1$. Then for every $t\geq 1$, we have

[TABLE]

and

[TABLE]

Finally, we obtain

[TABLE]

By Corollary 2.5, we obtain $\mathsf{GKdim}(\mathcal{D})\leq\mathsf{GKdim}(\mathcal{A})$. ∎

There are many dialgebras satisfying the assumptions of Lemma 3.11. In particular, certain identities ensure the validity of the assumptions of Lemma 3.11.

Corollary 3.12**.**

Let $\mathcal{D}$ be a dialgebra generated by a finite set $X$, and let $\mathcal{A}_{\mathcal{D}}$ be the associative algebra associated to $\mathcal{D}$. If $\mathcal{D}$ satisfies one of the identities: (i) $x\vdash y=y\vdash x$ for all $x,y\in\mathcal{D}$; (ii) $x\dashv y=y\dashv x$ for all $x,y\in\mathcal{D}$; (iii) $x\vdash y=y\dashv x$ for all $x,y\in\mathcal{D}$, then we have $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})\leq|X|$, in particular, $\mathsf{GKdim}(\mathcal{D})$ is a nonnegative integer.

Proof.

We shall show that, if one of the identities (i)-(iii) holds, then the assumptions of Lemma 3.11 are valid. Without loss of generality, assume that identity (i) holds in $\mathcal{D}$. Then for every $[a_{1}...a_{t}]_{p}$ in $\mathcal{D}$ such that $p>1$, we have

[TABLE]

In particular, the set $B_{1}=\{[a_{1}...a_{n}]_{p}\mid a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{n}\in X$, $p=1$ or $p=n$, $n\in\mathbb{N}\}$ is a linear generating set of $\mathcal{D}$. By Lemma 3.11, we obtain $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. Moreover, since $\mathcal{A}_{\mathcal{D}}$ is a commutative algebra generated by $X$, the Gelfand-Kirillov dimension $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ is a nonnegative integer satisfying $\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})\leq|X|$. ∎

Recall that a dialgebra $(\mathcal{D},\vdash,\dashv)$ is commutative [8] if both $\vdash$ and $\dashv$ are commutative, that is, for all $x$ and $y$ in $\mathcal{D}$, we have $x\vdash y=y\vdash x$ and $x\dashv y=y\dashv x$.

Corollary 3.13**.**

Let $\mathcal{D}$ be a finitely generated commutative dialgebra. Then $\mathsf{GKdim}(\mathcal{D})$ is a nonnegative integer number. In particular, if $\mathcal{D}$ is a free commutative dialgebra generated by $n$ letters, then we have $\mathsf{GKdim}(\mathcal{D})=n$.

Now we shall conclude the paper with our main result (Theorem B of the introduction), which is based on Bergman’s corresponding result for associative algebras.

Theorem 3.14**.**

No dialgebra has Gelfand-Kirillov dimension strictly between 1 and 2.

Proof.

We first assume that $\mathcal{D}$ is a dialgebra generated by a finite well-ordered set $X$ satisfying the inequality $\mathsf{GKdim}(\mathcal{D})<2$, and we shall show that $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. Let $B_{\mathcal{D}}$ be the shortest-middle-lexicographic basis of $\mathcal{D}$ with respect to $X$. Then for some positive integer $m$, we have $|B_{\mathcal{D}}^{m}|<m$, for otherwise we would have $|B_{\mathcal{D}}^{\leq n}|\geq 1+2+\cdots+n$ for all $n$ and thus $\mathsf{GKdim}(\mathcal{D})\geq 2$. Define

[TABLE]

We shall show that $B_{\mathcal{D}}$ is a subset of $B$ and thus by Lemma 3.11 we deduce $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$.

For every $[a_{1}...a_{t}]_{p}$ in $B_{\mathcal{D}}$, if $t\leq 2m$, then it is clear that $[a_{1}...a_{t}]_{p}$ lies in $B$. Assume now that $t>2m$ and $m<p<t-m+1$, then we deduce $p+m-1<t$ and $p-(m-1)>1$. Therefore, for every integer $j$ such that $0\leq j\leq m-1$, the monomial

[TABLE]

is a middle submonomial of $[a_{1}...a_{t}]_{p}$, and we have $[u_{i}]_{m-i}<[u_{j}]_{m-j}$ if $i>j$. Since ${|B_{\mathcal{D}}^{m}|<m}$, the set $\{[u_{j}]_{m-j}\mid 0\leq j\leq m-1\}\cup B_{\mathcal{D}}^{\leq m-1}$ is linear dependent in $\mathcal{D}$. In other words, for some integer $n$ such that $0\leq n\leq m-1$, we have

[TABLE]

for some elements $\alpha_{n+1},...\leavevmode\hbox{}\hskip 0.56905pt,\alpha_{m-1},\beta_{1},...\leavevmode\hbox{}\hskip 0.56905pt,\beta_{l}$ in $k$ and for some monomials $[w_{1}]_{q_{1}},...\leavevmode\hbox{}\hskip 0.56905pt,[w_{l}]_{q_{l}}$ in $[X^{+}]_{\omega}$ such that $[w_{i}]_{q_{i}}<[u_{n}]_{m-n}$ for every $i\leq l$. Finally, we obtain

[TABLE]

Since each $[v]_{q}$ on the right hand side of Equation (3.4) satisfies ${[v]_{q}<[a_{1}...a_{t}]_{p}}$, we obtain a contradiction with the definition of $B_{\mathcal{D}}$. So we have $B_{\mathcal{D}}\subseteq B$ and $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$. Finally, the inequality $1\leq\mathsf{GKdim}(\mathcal{D})<2$ forces $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})=1$.

Assume now that $\mathcal{D}$ is not finitely generated. It is clear that for every finitely generated subalgebra $\mathcal{D}^{\prime}$ of $\mathcal{D}$, we have $\mathsf{GKdim}(\mathcal{D}^{\prime})\leq\mathsf{GKdim}(\mathcal{D})<2$. By the above reasoning, we deduce $\mathsf{GKdim}(\mathcal{D}^{\prime})=0$ or $1$. Moreover, by Corollary 2.5, we obtain $\mathsf{GKdim}(\mathcal{D})\leq 1$. Finally, the inequality $1\leq\mathsf{GKdim}(\mathcal{D})<2$ forces $\mathsf{GKdim}(\mathcal{D})=1$. ∎

For summary, the set of all the possible Gelfand–Kirillov dimensions of dialgebras is $\{0,1\}\cup[2,\infty)$.

What remains open in view of the previous results is the question of whether the boundaries of inequation (3.3) are the only possible values of the Gelfand-Kirillov dimension  of a dialgebra, that is, can one show that either $\mathsf{GKdim}(\mathcal{D})=\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ or $\mathsf{GKdim}(\mathcal{D})=2\mathsf{GKdim}(\mathcal{A}_{\mathcal{D}})$ holds? We have a feeling that this is true, but so far, we have no proof in this direction.