# Gröbner-Shirshov bases for dialgebras111Supported by the NNSF of China
(Nos.10771077, 10911120389) and the NSF of Guangdong Province (No.06025062).

L. A. Bokut222Supported by RFBR 01-09-00157, LSS–344.2008.1 and SB RAS
Integration grant No. 2009.97 (Russia).
School of Mathematical Sciences, South China Normal University
Guangzhou 510631, P. R. China
Sobolev Institute of Mathematics, Russian Academy of Sciences
Siberian Branch, Novosibirsk 630090, Russia
Email<EMAIL_ADDRESS>

Yuqun Chen333Corresponding author. and Cihua Liu
School of Mathematical Sciences, South China Normal University
Guangzhou 510631, P. R. China
Email<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
Abstract: In this paper, we define the Gröbner-Shirshov basis for a dialgebra.
The Composition-Diamond lemma for dialgebras is given then. As results, we
give Gröbner-Shirshov bases for the universal enveloping algebra of a Leibniz
algebra, the bar extension of a dialgebra, the free product of two dialgebras,
and Clifford dialgebra. We obtain some normal forms for algebras mentioned the
above.

Key words: dialgebra; Gröbner-Shirshov basis; Leibniz algebra; Clifford
dialgebra.

AMS 2000 Subject Classification: 16S15, 13P10, 17A32, 17A99

##  1 Introduction

J.-L. Loday (1995, [11]) gave the definition of a new class of algebras,
dialgebras, which is closely connected to his notion of Leibniz algebras
(1993, [10]) in the same way as associative algebras connected to Lie
algebras. In the manuscript [12], J.-L. Loday found a normal form of elements
of a free dialgebra. Here we continue to study free dialgebras and prove the
Composition-Diamond lemma for dialgebras. As it is well known, this kind of
lemma is the cornerstone of the theory of Gröbner and Gröbner-Shirshov bases
(see, for example, [6] and cited literature). In commutative-associative case,
this lemma is equivalent to the Main Buchberger’s Theorem ([7, 8]). For Lie
and associative algebras, this is the Shirshov’s lemma [14] (see also L.A.
Bokut [3, 4], G. Bergman [2], L.A. Bokut and Y. Chen [5]). As results, we
obtain Gröbner-Shirshov bases for the universal enveloping algebra of a
Leibniz algebra, the bar extension of a dialgebra, the free product of two
dialgebras, and Clifford dialgebra. By using our Composition-Diamond lemma for
dialgebras (Theorem 3.9), we obtain some normal forms for algebras mentioned
the above. Moreover, we get another proof of the M. Aymon, P.-P. Grivel’s
result ([1]) on the Poincare-Birkhoff-Witt theorem for Leibniz algebras (see
P. Kolesnikov [9] for other proof).

##  2 Preliminaries

###### Definition 2.1

Let $k$ be a field. A $k$-linear space $D$ equipped with two bilinear
multiplications $\vdash$ and $\dashv$ is called a dialgebra, if both $\vdash$
and $\dashv$ are associative and

$\displaystyle a\dashv(b\vdash c)$ $\displaystyle=$ $\displaystyle a\dashv
b\dashv c$ $\displaystyle(a\dashv b)\vdash c$ $\displaystyle=$ $\displaystyle
a\vdash b\vdash c$ $\displaystyle a\vdash(b\dashv c)$ $\displaystyle=$
$\displaystyle(a\vdash b)\dashv c$

for any $a,\ b,\ c\in D$.

###### Definition 2.2

Let $D$ be a dialgebra, $B\subset D$. Let us define diwords of $D$ in the set
$B$ by induction:

  1. (i)

$b=(b),\ b\in B$ is a diword in $B$ of length $|b|=1$.

  2. (ii)

$(u)$ is called a diword in $B$ of length $|(u)|=n$, if $(u)=((v)\dashv(w))$
or $(u)=((v)\vdash(w))$, where $(v),\ (w)$ are diwords in $B$ of length $k,\
l$ respectively and $k+l=n$.

###### Proposition 2.3

([12]) Let $D$ be a dialgebra and $B\subset D$. Any diword of $D$ in the set
$B$ is equal to a diword in $B$ of the form

$(u)=b_{-m}\vdash\cdots\vdash b_{-1}\vdash b_{0}\dashv b_{1}\dashv\cdots\dashv
b_{n}$ (1)

where $b_{i}\in B,\ -m\leq i\leq n,\ m\geq 0,\ n\geq 0$. Any bracketing of the
right side of (1) gives the same result. $\square$

###### Definition 2.4

Let $X$ be a set. A free dialgebra $D(X)$ generated by $X$ over $k$ is defined
in a usual way by the following commutative diagram:

$D$$D(X)$$X$$i$$\forall\varphi$$\exists!\varphi^{*}$ (homomorphism)

where $D$ is any dialgebra.

In [12], a construction of a free dialgebra is given.

###### Proposition 2.5

([12]) Let $D(X)$ be a free dialgebra over $k$ generated by $X$. Any diword in
$D(X)$ is equal to the unique diword of the form

$[u]=x_{-m}\vdash\cdots\vdash x_{-1}\vdash x_{0}\dashv x_{1}\dashv\cdots\dashv
x_{n}\triangleq x_{-m}\cdots x_{-1}\dot{x_{0}}x_{1}\cdots x_{n}$ (2)

where $x_{i}\in X,\ m\geq 0,\ n\geq 0$, and $x_{0}$ is called the center of
the normal diword $[u]$. We call $[u]$ a normal diword (in $X$) with the
associative word $u,u\in X^{*}$. Clearly, if $[u]=[v]$, then $u=v$. In (2).
Let $[u],\ [v]$ be two normal diwords. Then $[u]\vdash[v]$ is the normal
diword $[uv]$ with the center at the center of $[v]$. Accordingly,
$[u]\dashv[v]$ is the normal diword $[uv]$ with the center at the center of
$[u]$. $\square$

###### Example 2.6

$(x_{-1}\vdash x_{0}\dashv x_{1})\vdash(y_{-1}\vdash y_{0}\dashv
y_{1})=x_{-1}\vdash x_{0}\vdash x_{1}\vdash y_{-1}\vdash y_{0}\dashv y_{1},$
$(x_{-1}\vdash x_{0}\dashv x_{1})\dashv(y_{-1}\vdash y_{0}\dashv
y_{1})=x_{-1}\vdash x_{0}\dashv x_{1}\dashv y_{-1}\dashv y_{0}\dashv y_{1}.\ \
\square$

##  3 Composition-Diamond lemma for dialgebras

Let $X$ be a well ordered set, $D(X)$ the free dialgebra over $k$, $X^{*}$ the
free monoid generated by $X$ and $[X^{*}]$ the set of normal diwords in $X$.
Let us define the deg-lex ordering on $[X^{*}]$ in the following way: for any
$[u],[v]\in[X^{*}]$,

$[u]<[v]\Longleftrightarrow wt([u])<wt([v])\ \ \mbox{lexicographicaly},$

where

$wt([u])=(n+m+1,m,x_{-m},\cdots,x_{0},\cdots,x_{n})$

if $[u]=x_{-m}\cdots x_{-1}\dot{x_{0}}x_{1}\cdots x_{n}$.

Throughout the paper, we will use this ordering.

It is easy to see that the ordering $<$ is satisfied the following properties:

$[u]<[v]\Longrightarrow x\vdash[u]<x\vdash[v],\ [u]\dashv x<[v]\dashv x,\
\mbox{for any }\ x\in X.$

Any polynomial $f\in D(X)$ has the form

$f=\sum_{[u]\in[X^{*}]}f([u])[u]=\alpha[\overline{f}]+\sum{\alpha}_{i}[u_{i}],$

where $[\overline{f}],\ [u_{i}]$ are normal diwords in $X$,
$[\overline{f}]>[u_{i}],\ \alpha,\ {\alpha}_{i},\ f([u])\in k,\ \alpha\neq 0$.
We call $[\overline{f}]$ the leading term of $f$. Denote $suppf$ by the set
$\\{[u]|f([u])\neq 0\\}$ and $deg(f)$ by $|[\overline{f}]|$. $f$ is called
monic if $\alpha=1$. $f$ is called left (right) normed if
$f=\sum{\alpha}_{i}u_{i}\dot{x_{i}}\ \ (f=\sum{\alpha}_{i}\dot{x_{i}}u_{i})$,
where each ${\alpha}_{i}\in k,\ x_{i}\in X$ and $u_{i}\in X^{*}$.

If $[u],\ [v]$ are both left normed or both right normed, then it is clear
that for any $[w]\in[X^{*}]$,

$\displaystyle[u]<[v]\Longrightarrow$
$\displaystyle[u]\vdash[w]<[v]\vdash[w],\ [w]\vdash[u]<[w]\vdash[v],$
$\displaystyle[u]\dashv[w]<[v]\dashv[w],\ [w]\dashv[u]<[w]\dashv[v].$

Let $S\subset D(X)$. By an $S$-diword $g$ we will mean a diword in $\\{X\cup
S\\}$ with only one occurrence of $s\in S$. If this is the case and $g=(asb)$
for some $a,b\in X^{*},\ s\in S$, we also call $g$ an $s$-diword.

From Proposition 2.3 it follows that any $s$-diword is equal to

$[asb]=x_{-m}\vdash\cdots\vdash x_{-1}\vdash x_{0}\dashv
x_{1}\dashv\cdots\dashv x_{n}|_{x_{k}\mapsto s}$ (3)

where $-m\leq k\leq n,\ s\in S,\ x_{i}\in X,\ -m\leq i\leq n$. To be more
precise, $[asb]=[a\dot{s}b]$ if $k=0$; $[asb]=[asb_{1}\dot{x_{0}}b_{2}]$ if
$k<0$ and $[asb]=[a_{1}\dot{x_{0}}a_{2}sb]$ if $k>0$. If the center of the
$s$-diword $[asb]$ is in $a$, then we denote it by
$[\dot{a}sb]=[a_{1}\dot{x_{0}}a_{2}sb]$. Similarly,
$[as\dot{b}]=[asb_{1}\dot{x_{0}}b_{2}]$ (of course, either $a_{i}$ or $b_{i}$
may be empty).

###### Definition 3.1

The $s$-diword (3) is called a normal $s$-diword if one of the following
conditions holds:

  1. (i)

$k=0$,

  2. (ii)

$k<0$ and $s$ is left normed,

  3. (iii)

$k>0$ and $s$ is right normed.

We call a normal $s$-diword $[asb]$ a left (right) normed $s$-diword if both
$s$ and $[asb]$ are left (right) normed. In particulary, $s$ is a left (right)
normed $s$-diword if $s$ is left (right) normed polynomial.

The following lemma follows from the above properties of the ordering $<$.

###### Lemma 3.2

For a normal $s$-diword $[asb]$, the leading term of $[asb]$ is equal to
$[a[\overline{s}]b]$, that is, $\overline{[asb]}=[a[\overline{s}]b]$. More
specifically, if

$[asb]=x_{-m}\vdash\cdots\vdash x_{-1}\vdash x_{0}\dashv
x_{1}\dashv\cdots\dashv x_{n}|_{x_{k}\mapsto s},$

then corresponding to $k=0,\ k<0,\ k>0$, respectively, we have

$\overline{x_{-m}\vdash\cdots\vdash x_{-1}\vdash s\dashv
x_{1}\dashv\cdots\dashv x_{n}}=x_{-m}\vdash\cdots\vdash
x_{-1}\vdash[\overline{s}]\dashv x_{1}\dashv\cdots\dashv x_{n},$
$\overline{x_{-m}\vdash\cdots\vdash s\vdash\cdots\vdash
x_{0}\dashv\cdots\dashv
x_{n}}=x_{-m}\vdash\cdots\vdash[\overline{s}]\vdash\cdots\vdash
x_{0}\dashv\cdots\dashv x_{n},$ $\overline{x_{-m}\vdash\cdots\vdash
x_{0}\dashv\cdots\dashv s\dashv\cdots\dashv x_{n}}={x_{-m}\vdash\cdots\vdash
x_{0}\dashv\cdots\dashv[\overline{s}]\dashv\cdots\dashv x_{n}}.\ \ \square$

Now, we define compositions of polynomials in $D(X)$.

###### Definition 3.3

Let the ordering $<$ be as before and $f,g\in D(X)$ with $f,g$ monic.

  1. 1)

Composition of left (right) multiplication.

Let $f$ be not a right normed polynomial and $x\in X$. Then $x\dashv f$ is
called the composition of left multiplication. Clearly, $x\dashv f$ is a right
normed polynomial (or 0).

Let $f$ be not a left normed polynomial and $x\in X$. Then $f\vdash x$ is
called the composition of right multiplication. Clearly, $f\vdash x$ is a left
normed polynomial (or 0).

  2. 2)

Composition of inclusion.

Let

$[w]=[\overline{f}]=[a[\overline{g}]b],$

where $[agb]$ is a normal $g$-diword. Then

$(f,g)_{[w]}=f-[agb]$

is called the composition of inclusion. The transformation $f\mapsto f-[agb]$
is called the elimination of leading diword (ELW) of $g$ in $f$, and $[w]$ is
called the ambiguity of $f$ and $g$.

  3. 3)

Composition of intersection.

Let

$[w]=[[\overline{f}]b]=[a[\overline{g}]],\ |\overline{f}|+|\overline{g}|>|w|,$

where $[fb]$ is a normal $f$-diword and $[ag]$ a normal $g$-diword. Then

$(f,g)_{[w]}=[fb]-[ag]$

is called the composition of intersection, and $[w]$ is called the ambiguity
of $f$ and $g$.

Remark In the Definition 3.3, for the case of 2) or 3), we have
$\overline{(f,g)_{[w]}}<[w].$ For the case of 1), $deg(x\dashv f)\leq
deg(f)+1$ and $deg(f\vdash x)\leq deg(f)+1$.

###### Definition 3.4

Let the ordering $<$ be as before, $S\subset D(X)$ a monic set and $f,g\in S$.

  1. 1)

Let $x\dashv f$ be a composition of left multiplication. Then $x\dashv f$ is
called trivial modulo $S$, denoted by $x\dashv f\equiv 0\ mod(S)$, if

$x\dashv f=\sum{\alpha}_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S,\
[a_{i}s_{i}b_{i}]$ right normed $s_{i}$-diword and
$|[a_{i}[\overline{s_{i}}]b_{i}]|\leq deg(x\dashv f)$.

Let $f\vdash x$ be a composition of right multiplication. Then $f\vdash x$ is
called trivial modulo $S$, denoted by $f\vdash x\equiv 0\ mod(S)$, if

$f\vdash x=\sum{\alpha}_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S,\
[a_{i}s_{i}b_{i}]$ left normed $s_{i}$-diword and
$|[a_{i}[\overline{s_{i}}]b_{i}]|\leq deg(f\vdash x)$.

  2. 2)

Composition $(f,g)_{[w]}$ of inclusion (intersection) is called trivial modulo
$(S,[w])$, denoted by $(f,g)_{[w]}\equiv 0\ mod(S,[w])$, if

$(f,g)_{[w]}=\sum{\alpha}_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S,\
[a_{i}s_{i}b_{i}]$ normal $s_{i}$-diword, $[a_{i}[\overline{s_{i}}]b_{i}]<[w]$
and each $[a_{i}s_{i}b_{i}]$ is right (left) normed $s_{i}$-diword whenever
either both $f$ and $[agb]$ or both $[fb]$ and $[ag]$ are right (left) normed
$S$-diwords.

We call the set $S$ a Gröbner-Shirshov basis in $D(X)$ if any composition of
polynomials in $S$ is trivial modulo $S$ (and $[w])$.

The following lemmas play key role in the proof of Theorem 3.9.

###### Lemma 3.5

Let $S\subset D(X)$ and $[asb]$ an $s$-diword, $s\in S$. Assume that each
composition of right and left multiplication is trivial modulo $S$. Then,
$[asb]$ has a presentation:

$[asb]=\sum\alpha_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ s_{i}\in S,\ a_{i},b_{i}\in X^{*}$ and each
$[a_{i}s_{i}b_{i}]$ is normal $s_{i}$-diword.

Proof. Following Proposition 2.3, we assume that

$[asb]=x_{-m}\vdash\cdots\vdash x_{-1}\vdash x_{0}\dashv
x_{1}\dashv\cdots\dashv x_{n}|_{x_{k}\mapsto s}.$

There are three cases to consider.

Case 1. $k=0$. Then $[asb]$ is a normal $s$-diword.

Case 2. $k<0$. Then $[asb]=a\vdash(s\vdash x_{k+1})\vdash b,k<-1$ or
$[asb]=a\vdash(s\vdash x_{0})\dashv b$. If $s$ is left normed then $[asb]$ is
a normal $s$-diword. If $s$ is not left normed then for the composition
$s\vdash x_{k+1}\ \ (k<0)$ of right multiplication, we have

$s\vdash x_{k+1}=\sum{\alpha}_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S$ and
$[a_{i}s_{i}b_{i}]$ is left normed $s_{i}$-diword. Then

$[asb]=\sum{\alpha}_{i}(a\vdash[a_{i}s_{i}b_{i}]\vdash b)$

or

$[asb]=\sum{\alpha}_{i}(a\vdash[a_{i}s_{i}b_{i}]\dashv b)$

is a linear combination of normal $s_{i}$-diwords.

Case 3. $k>0$ is similar to the Case 2. $\square$

###### Lemma 3.6

Let $S\subset D(X)$ and each composition $(f,g)_{[w]}$ in $S$ of inclusion
(intersection) trivial modulo $(S,[w])$. Let $[a_{1}s_{1}b_{1}]$ and
$[a_{2}s_{2}b_{2}]$ be normal $S$-diwords such that
$[w]=[a_{1}[\bar{s_{1}}]b_{1}]=[a_{2}[\bar{s_{2}}]b_{2}]$, where
$s_{1},s_{2}\in S,\ a_{1},a_{2},b_{1},b_{2}\in X^{*}$. Then,

$[a_{1}s_{1}b_{1}]\equiv[a_{2}s_{2}b_{2}]\ \ mod(S,[w]),$

i.e., $[a_{1}s_{1}b_{1}]-[a_{2}s_{2}b_{2}]=\sum{\alpha}_{i}[a_{i}s_{i}b_{i}],$
where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S,\
[a_{i}s_{i}b_{i}]$ normal $s_{i}$-diword and
$[a_{i}[\overline{s_{i}}]b_{i}]<[w]$.

Proof. In the following, all letters $a,b,c$ with indexis are words and
$s_{1},s_{2},s_{j}\in S$.

Because $a_{1}\bar{s_{1}}b_{1}=a_{2}\bar{s_{2}}b_{2}$ as ordinary words, there
are three cases to consider.

Case 1. Subwords $\overline{s_{1}},\overline{s_{2}}$ have empty intersection.
Assume, for example, that $b_{1}=b\overline{s_{2}}b_{2}$ and
$a_{2}=a_{1}\overline{s_{1}}b$. Because any normal $S$-diword may be
bracketing in any way, we have

$[a_{2}s_{2}b_{2}]-[a_{1}s_{1}b_{1}]=(a_{1}s_{1}(b(s_{2}-[\overline{s_{2}}])b_{2}))-((a_{1}(s_{1}-[\overline{s_{1}}])b)s_{2}b_{2}).$

For any $[t]\in supp(s_{2}-[\overline{s_{2}}])$, we prove that
$(a_{1}s_{1}b[t]b_{2})$ is a normal $s_{1}$-diword. There are five cases to
consider.

1.1 $[w]=[\dot{a_{1}}[\overline{s_{1}}]b[\overline{s_{2}}]b_{2}]$;

1.2 $[w]=[a_{1}\dot{[\overline{s_{1}}]}b[\overline{s_{2}}]b_{2}]$;

1.3 $[w]=[a_{1}[\overline{s_{1}}]\dot{b}[\overline{s_{2}}]b_{2}]$;

1.4 $[w]=[a_{1}[\overline{s_{1}}]b\dot{[\overline{s_{2}}]}b_{2}]$;

1.5 $[w]=[a_{1}[\overline{s_{1}}]b[\overline{s_{2}}]\dot{b_{2}}]$.

For 1.1, since $[a_{1}s_{1}b_{1}]$ and $[a_{2}s_{2}b_{2}]$ are normal
$S$-diwords, both $s_{1}$ and $s_{2}$ are right normed by the definition, in
particular, $[t]$ is right normed. It follows that
$(a_{1}s_{1}b[t]b_{2})=[\dot{a_{1}}s_{1}b[t]b_{2}]$ is a normal
$s_{1}$-diword.

For 1.2, it is clear that $(a_{1}s_{1}b[t]b_{2})$ is a normal $s_{1}$-diword
and $[t]$ is right normed.

For 1.3, 1.4 and 1.5, since $[a_{1}s_{1}b_{1}]$ is normal $s_{1}$-diword,
$s_{1}$ is left normed by the definition, which implies that
$(a_{1}s_{1}b[t]b_{2})$ is a normal $s_{1}$-diword. Moreover, $[t]$ is right
normed, if 1.3, and left normed, if 1.5.

Clearly, for all cases, we have
$\overline{[a_{1}s_{1}b[t]b_{2}]}=[a_{1}[\overline{s_{1}}]b[t]b_{2}]<[a_{1}[\overline{s_{1}}]b[\overline{s_{2}}]b_{2}]=[w]$.

Similarly, for any $[t]\in supp(s_{1}-[\overline{s_{1}}])$,
$(a_{1}[t]bs_{2}b_{2})$ is a normal $s_{2}$-diword and
$[a_{1}[t]b[\overline{s_{2}}]b_{2}]<[w]$.

Case 2. Subwords $\overline{s_{1}}$ and $\overline{s_{2}}$ have non-empty
intersection $c$. Assume, for example, that $b_{1}=bb_{2},\ a_{2}=a_{1}a,\
w_{1}=\overline{s_{1}}b=a\overline{s_{2}}=acb$.

There are following five cases to consider:

2.1 $[w]=[\dot{a_{1}}[\overline{s_{1}}]bb_{2}]$;

2.2 $[w]=[a_{1}[\overline{s_{1}}]b\dot{b_{2}}]$;

2.3 $[w]=[a_{1}\dot{a}cbb_{2}]$;

2.4 $[w]=[a_{1}a\dot{c}bb_{2}]$;

2.5 $[w]=[a_{1}ac\dot{b}b_{2}]$.

Then

$[a_{2}s_{2}b_{2}]-[a_{1}s_{1}b_{1}]=(a_{1}([as_{2}]-[s_{1}b])b_{2})=(a_{1}(s_{1},s_{2})_{[w_{1}]}b_{2}),$

where $[w_{1}]=[acb]=[[\overline{s_{1}}]b]=[a[\overline{s_{2}}]]$ is as
follows:

2.1 $[w_{1}]$ is right normed;

2.2 $[w_{1}]$ is left normed;

2.3 $[w_{1}]=[\dot{a}cb]$;

2.4 $[w_{1}]=[a\dot{c}b]$;

2.5 $[w_{1}]=[ac\dot{b}]$.

Since each composition $(f,g)_{[w]}$ in $S$ is trivial modulo $(S,[w])$, there
exist $\beta_{j}\in k,\ u_{j},v_{j}\in X^{*},\ s_{j}\in S$ such that
$[s_{1}b]-[as_{2}]=\sum_{j}\beta_{j}[u_{j}s_{j}v_{j}]$, where each
$[u_{j}s_{j}v_{j}]$ is normal $S$-diword and
$[u_{j}[\overline{s_{j}}]v_{j}]<[w_{1}]=[acb]$. Therefore,

$[a_{2}s_{2}b_{2}]-[a_{1}s_{1}b_{1}]=\sum_{j}\beta_{j}(a_{1}[u_{j}s_{j}v_{j}]b_{2}).$

Now, we prove that each $(a_{1}[u_{j}s_{j}v_{j}]b_{2})$ is normal
$s_{j}$-diword and
$\overline{(a_{1}[u_{j}s_{j}v_{j}]b_{2})}<[w]=[a_{1}[[\overline{s_{1}}]b]b_{2}]$.

For 2.1, since $[\dot{a_{1}}s_{1}bb_{2}]$ and $[\dot{a_{1}}as_{2}b_{2}]$ are
normal $S$-diwords, both $[s_{1}b]$ and $[as_{2}]$ are right normed
$S$-diwords. Then, by definition, each $[u_{j}s_{j}v_{j}]$ is right normed
$S$-diword, and so each
$(a_{1}[u_{j}s_{j}v_{j}]b_{2})=[\dot{a_{1}}u_{j}s_{j}v_{j}b_{2}]$ is normal
$S$-diword.

For 2.2, both $[s_{1}b]$ and $[as_{2}]$ must be left normed $S$-diwords. Then,
by definition, each $[u_{j}s_{j}v_{j}]$ is left normed $S$-diword, and so each
$(a_{1}[u_{j}s_{j}v_{j}]b_{2})=[a_{1}u_{j}s_{j}v_{j}\dot{b_{2}}]$ is normal
$S$-diword.

For 2.3, 2.4 or 2.5, by noting that
$(a_{1}[u_{j}s_{j}v_{j}]b_{2})=((a_{1})\vdash[u_{j}s_{j}v_{j}]\dashv(b_{2}))$
and $[u_{j}s_{j}v_{j}]$ is normal $S$-diword, $(a_{1}[u_{j}s_{j}v_{j}]b_{2})$
is also normal $S$-diword.

Now, for all cases, we have
$\overline{[a_{1}u_{j}s_{j}v_{j}b_{2}]}=[a_{1}u_{j}[\overline{s_{j}}]v_{j}b_{2}]<[w]=[a_{1}[acb]b_{2}]$.

Case 3. One of the subwords $\overline{s_{1}}$ and $\overline{s_{2}}$ contains
another as a subword. Assume, for example, that $b_{2}=bb_{1},\ a_{2}=a_{1}a,\
w_{1}=\overline{s_{1}}=a\overline{s_{2}}b$.

Again there are following five cases to consider:

2.1 $[w]=[\dot{a_{1}}a[\overline{s_{2}}]bb_{1}]$;

2.2 $[w]=[a_{1}a[\overline{s_{2}}]b\dot{b_{1}}]$;

2.3 $[w]=[a_{1}\dot{a}[\overline{s_{2}}]bb_{1}]$;

2.4 $[w]=[a_{1}a\dot{[\overline{s_{2}}]}bb_{1}]$;

2.5 $[w]=[a_{1}a[\overline{s_{2}}]\dot{b}b_{1}]$.

Then

$[a_{1}s_{1}b_{1}]-[a_{2}s_{2}b_{2}]=(a_{1}(s_{1}-as_{2}b)b_{1})=(a_{1}(s_{1},s_{2})_{[w_{1}]}b_{1}).$

It is similar to the proof of the Case 2 that we have
$[a_{1}s_{1}b_{1}]\equiv[a_{2}s_{2}b_{2}]\ \ mod(S,[w])$. $\square$

###### Definition 3.7

Let $S\subset D(X)$. Then

$Irr(S)\triangleq\\{u\in[X^{*}]|u\neq[a[\overline{s}]b],s\in S,a,b\in X^{*},\
[asb]\mbox{ is normal s-diword}\\}.$

###### Lemma 3.8

Let $S\subset D(X)$ and $h\in D(X)$. Then $h$ has a representation

$h=\sum_{I_{1}}\alpha_{i}[u_{i}]+\sum_{I_{2}}\beta_{j}[a_{j}s_{j}b_{j}]$

where $[u_{i}]\in Irr(S),\ i\in I_{1},[a_{j}s_{j}b_{j}]$ normal
$s_{j}$-diwords, $s_{j}\in S,\ j\in I_{2}$ with
$[a_{1}[\overline{s_{1}}]b_{1}]>[a_{2}[\overline{s_{2}}]b_{2}]>\cdots>[a_{n}[\overline{s_{n}}]b_{n}]$.

Proof. Let $h=\alpha_{1}[\overline{h}]+\cdots$. We prove the result by
induction on $[\overline{h}]$.

If $[\overline{h}]\in Irr(S)$, then take $[u_{1}]=[\overline{h}]$ and
$h_{1}=h-\alpha_{1}[u_{1}]$. Clearly, $[\overline{h_{1}}]<[\overline{h}]$ or
$h_{1}=0$.

If $[\overline{h}]\not\in Irr(S)$, then
$[\overline{h}]=[a_{1}[\overline{s_{1}}]b_{1}]$ with $[a_{1}s_{1}b_{1}]$ a
normal $s_{1}$-diword. Let $h_{1}=h-\beta_{1}[a_{1}s_{1}b_{1}]$. Then
$[\overline{h_{1}}]<[\overline{h}]$ or $h_{1}=0$. $\square$

The following theorem is the main result.

###### Theorem 3.9

(Composition-Diamond lemma) Let $S\subset D(X)$ be a monic set and the
ordering $<$ as before, $Id(S)$ is the ideal generated by $S$. Then
$(i)\Rightarrow(ii)\Leftrightarrow(ii)^{\prime}\Leftrightarrow(iii)$, where

  1. (i)

$S$ is a Gröbner-Shirshov basis in $D(X)$.

  2. (ii)

$f\in Id(S)\Rightarrow[\overline{f}]=[a[\overline{s}]b]$ for some $s\in S,\
a,b\in X^{*}$ and $[asb]$ a normal $S$-diword.

  3. $(ii)^{\prime}$

$f\in Id(S)\Rightarrow
f=\alpha_{1}[a_{1}s_{1}b_{1}]+\alpha_{2}[a_{2}s_{2}b_{2}]+\cdots+\alpha_{n}[a_{n}s_{n}b_{n}]$
with
$[a_{1}[\overline{s_{1}}]b_{1}]>[a_{2}[\overline{s_{2}}]b_{2}]>\cdots>[a_{n}[\overline{s_{n}}]b_{n}],$
where $[a_{i}s_{i}b_{i}]$ is normal $s_{i}$-diword, $i=1,2,\cdots,n$.

  4. (iii)

The set $Irr(S)$ is a linear basis of the dialgebra $D(X|S)=D(X)/Id(S)$
generated by $X$ with defining relations $S$.

Proof. $(i)\Rightarrow(ii)$. Let $S$ be a Gröbner-Shirshov basis and $0\neq
f\in Id(S)$. We may assume, by Lemma 3.5, that

$f=\sum_{i=1}^{n}\alpha_{i}[a_{i}s_{i}b_{i}],$

where each $\alpha_{i}\in k,\ a_{i},b_{i}\in X^{*},\ s_{i}\in S$ and
$[a_{i}s_{i}b_{i}]$ normal $S$-diword. Let

$[w_{i}]=[a_{i}[\overline{s_{i}}]b_{i}],\
[w_{1}]=[w_{2}]=\cdots=[w_{l}]>[w_{l+1}]\geq\cdots,\ l\geq 1.$

We will use induction on $l$ and $[w_{1}]$ to prove that
$[\overline{f}]=[a[\overline{s}]b]$ for some $s\in S\ \mbox{and}\ a,b\in
X^{*}$. If $l=1$, then
$[\overline{f}]=\overline{[a_{1}s_{1}b_{1}]}=[a_{1}[\overline{s_{1}}]b_{1}]$
and hence the result holds. Assume that $l\geq 2$. Then, by Lemma 3.6, we have
$[a_{1}s_{1}b_{1}]\equiv[a_{2}s_{2}b_{2}]\ \ mod(S,[w_{1}])$.

Thus, if $\alpha_{1}+\alpha_{2}\neq 0$ or $l>2$, then the result follows from
induction on $l$. For the case $\alpha_{1}+\alpha_{2}=0$ and $l=2$, we use
induction on $[w_{1}]$. Now, the result follows.

$(ii)\Rightarrow(ii)^{\prime}$. Assume (ii) and $0\neq f\in Id(S)$. Let
$f=\alpha_{1}[\overline{f}]+\sum_{[u_{i}]<[\overline{f}]}\alpha_{i}[u_{i}]$.
Then, by (ii), $[\overline{f}]=[a_{1}[\overline{s_{1}}]b_{1}]$, where
$[a_{1}s_{1}b_{1}]$ is a normal $S$-diword. Therefore,

$f_{1}=f-\alpha_{1}[a_{1}s_{1}b_{1}],\ [\overline{f_{1}}]<[\overline{f}]\mbox{
or }\ f_{1}=0,\ f_{1}\in Id(S).$

Now, by using induction on $[\overline{f}]$, we have $(ii)^{\prime}$.

$(ii)^{\prime}\Rightarrow(ii)$. This part is clear.

$(ii)\Rightarrow(iii)$. Assume $(ii)$. Then by Lemma 3.8, $Irr(S)$ spans
$D(X|S)$ as $k$-space.

Suppose that $0\neq\sum\alpha_{i}[u_{i}]\in Id(S)$ where
$[u_{1}]>[u_{2}]>\cdots,\ [u_{i}]\in Irr(S)$. Then by $(ii)$,
$[u_{1}]=[a_{1}[\overline{s_{1}}]b_{1}]$ where $[a_{1}s_{1}b_{1}]$ is a normal
$S$-diword, a contradiction.

This shows (iii).

$(iii)\Rightarrow(ii)$. Assume (iii). Let $0\neq f\in Id(S)$. Since the
elements in $Irr(S)$ are linearly independent in $D(X|S)$, by Lemma 3.8,
$[\bar{f}]=[a[\bar{s}]b]$, where $[asb]$ is a normal $S$-diword. Thus, (ii)
follows. $\square$

Remark: In general, $(iii)\not\Rightarrow(i).$ For example, it is noted that

$Irr(S)=\\{x_{j}\dashv x_{i_{1}}\dashv\dots\dashv x_{i_{k}}\ |\ j\in
I,i_{p}\in I-I_{0},\ 1\leq p\leq k,\ i_{1}\leq\dots\leq i_{k},\ k\geq 0\\}$

is a linear basis of $D(X|S)$ in Theorem 4.3. Let

$S_{1}=\\{x_{j}\vdash x_{i}-x_{i}\dashv x_{j}+\\{x_{i},x_{j}\\},\ x_{t}\dashv
x_{i_{0}},i,j,t\in I,i_{0}\in I_{0}\\}.$

Then $Irr(S_{1})=Irr(S)$ is a linear basis of $D(X|S)$. But in the proof of
Theorem 4.3, we know that $S_{1}$ is not a Gröbner-Shirshov basis of $D(X|S)$.

##  4 Applications

In this section, we give Gröbner-Shirshov bases for the universal enveloping
dialgebra of a Leibniz algebra, the bar extension of a dialgebra, the free
product of two dialgebras, and the Clifford dialgebra. By using our Theorem
3.9, we obtain some normal forms for dialgebras mentioned the above.

###### Definition 4.1

([10]) A k-linear space $L$ equipped with bilinear multiplication $[,]$ is
called a Leibniz algebra if for any $a,b,c\in L$,

$[[a,b],c]=[[a,c],b]+[a,[b,c]]$

i.e., the Leibniz identity is valid in $L$.

It is clear that if $(D,\dashv,\vdash)$ is a dialgebra then $D^{(-)}=(D,[,])$
is a Leibniz algebra, where $[a,b]=a\dashv b-b\vdash a$ for any $a,b\in D$.

If $f$ is a Leibniz polynomial in variables $X$, then by $f^{(-)}$ we mean a
dialgebra polynomial in $X$ obtained from $f$ by transformation $[a,b]\mapsto
a\dashv b-b\vdash a$.

###### Definition 4.2

Let $L$ be a Leibniz algebra. A dialgebra $U(L)$ together with a Leibniz
homomorphism $\varepsilon:L\rightarrow U(L)$ is called the universal
enveloping dialgebra for $L$, if the following diagram commute:

$\exists!f$$D$$U(L)$$L$$\varepsilon$$\forall\delta$

where $D$ is a dialgebra, $\delta$ is a Leibniz homomorphism and
$f:U(L)\rightarrow D$ is a dialgebra homomorphism such that
$f\varepsilon=\delta$ (i.e., $\varepsilon:L\rightarrow U(L)$ is a universal
arrow in the sense of S. MacLane [13], p55).

An equivalent definition is as follows: Let $L=Lei(X|S)$ is a Leibniz algebra
presented by generators $X$ and definition relations $S$. Then
$U(L)=D(X|S^{(-)})$ is the dialgebra with generators $X$ and definition
relations $S^{(-)}=\\{s^{(-)}|s\in S\\}.$

###### Theorem 4.3

Let $\mathcal{L}$ be a Leibniz algebra over a field $k$ with the product
$\\{,\\}$. Let $\mathcal{L}_{0}$ be the subspace of $\mathcal{L}$ generated by
the set $\\{\\{a,a\\},\\{a,b\\}+\\{b,a\\}\ |\ a,b\in\mathcal{L}\\}$. Let
$\\{x_{i}|i\in I_{0}\\}$ be a basis of $\mathcal{L}_{0}$ and $X=\\{x_{i}|i\in
I\\}$ a well ordered basis of $\mathcal{L}$ such that $I_{0}\subseteq I$. Let
$U(L)=D(X|x_{i}\dashv x_{j}-x_{j}\vdash x_{i}-\\{x_{i},x_{j}\\})$ be the
universal enveloping dialgebra for $L$ and the ordering $<$ on $[X^{*}]$ as
before. Then

  1. (i)

$D(X|x_{i}\dashv x_{j}-x_{j}\vdash x_{i}-\\{x_{i},x_{j}\\})=D(X|S)$, where $S$
consists of the following polynomials:

$\displaystyle(a)$ $\displaystyle f_{ji}=x_{j}\vdash x_{i}-x_{i}\dashv
x_{j}+\\{x_{i},x_{j}\\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ (i,j\in I)$ $\displaystyle(b)$ $\displaystyle f_{ji\vdash
t}=x_{j}\vdash x_{i}\vdash x_{t}-x_{i}\vdash x_{j}\vdash
x_{t}+\\{x_{i},x_{j}\\}\vdash x_{t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (i,j,t\in I,\
j>i)$ $\displaystyle(c)$ $\displaystyle h_{i_{0}\vdash t}=x_{i_{0}}\vdash
x_{t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i_{0}\in I_{0},\ t\in I)$
$\displaystyle(d)$ $\displaystyle f_{t\dashv ji}=x_{t}\dashv x_{j}\dashv
x_{i}-x_{t}\dashv x_{i}\dashv x_{j}+x_{t}\dashv\\{x_{i},x_{j}\\}\ \ \ \ \ \ \
\ \ \ \ \ \ \ (i,j,t\in I,\ j>i)$ $\displaystyle(e)$ $\displaystyle h_{t\dashv
i_{0}}=x_{t}\dashv x_{i_{0}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i_{0}\in I_{0},\
t\in I)$

  2. (ii)

$S$ is a Gröbner-Shirshov basis in $D(X)$.

  3. (iii)

The set

$\\{x_{j}\dashv x_{i_{1}}\dashv\dots\dashv x_{i_{k}}\ |\ j\in I,i_{p}\in
I-I_{0},\ 1\leq p\leq k,\ i_{1}\leq\dots\leq i_{k},\ k\geq 0\\}$

is a linear basis of the universal enveloping algebra $U(\mathcal{L})$. In
particular, $\mathcal{L}$ is a Leibniz subalgebra of $U(\mathcal{L})$.

Proof. (i) By using the following

$f_{ji\vdash t}=f_{ji}\vdash x_{t}\ \mbox{ and }\ f_{ji}\vdash
x_{t}+f_{ij}\vdash x_{t}=(\\{x_{i},x_{j}\\}+\\{x_{j},x_{i}\\})\vdash x_{t},$

we have (b) and (c) are in $Id(f_{ji})$. By symmetry, (d) and (e) are in
$Id(f_{ji})$. This shows (i).

(ii) We will prove that all compositions in $S$ are trivial modulo $S$ (and
$[w]$). For convenience, we extend linearly the functions $f_{ji},\
f_{ji\vdash t},\ f_{t\dashv ji},\ h_{i_{0}\vdash t}$ and $h_{t\dashv i_{0}}$
to $f_{j\\{p,q\\}}\ (f_{\\{p,q\\}i}),\ f_{ji\vdash\\{p,q\\}}$ and
$h_{\\{p,q\\}\dashv i_{0}}$, etc respectively. For example, if
$\\{x_{p},x_{q}\\}=\sum\alpha_{pq}^{s}x_{s}$, then

$\displaystyle f_{j\\{p,q\\}}$ $\displaystyle=$ $\displaystyle
x_{j}\vdash\\{x_{p},x_{q}\\}-\\{x_{p},x_{q}\\}\dashv
x_{j}+\\{\\{x_{p},x_{q}\\},x_{j}\\}=\sum\alpha_{pq}^{s}f_{js},$ $\displaystyle
f_{ji\vdash\\{p,q\\}}$ $\displaystyle=$
$\displaystyle\sum\alpha_{pq}^{s}(x_{j}\vdash x_{i}\vdash x_{s}-x_{i}\vdash
x_{j}\vdash x_{s}+\\{x_{i},x_{j}\\}\vdash
x_{s})=f_{ji}\vdash\\{x_{p},x_{q}\\},$ $\displaystyle h_{\\{p,q\\}\dashv
i_{0}}$ $\displaystyle=$ $\displaystyle\sum\alpha_{pq}^{s}h_{s\dashv i_{0}}.$

By using the Leibniz identity,

$\\{\\{a,b\\},c\\}=\\{a,\\{b,c\\}\\}+\\{\\{a,c\\},b\\},$ (4)

we have

$\displaystyle\\{a,\\{b,b\\}\\}=0\ \ and\ \ \\{a,\\{b,c\\}+\\{c,b\\}\\}=0$

for any $a,b,c\in\mathcal{L}$. It means that for any $i_{0}\in I_{0},\ j\in
I$,

$\\{x_{j},x_{i_{0}}\\}=0$ (5)

and by noting that
$\\{x_{i_{0}},x_{j}\\}=\\{x_{j},x_{i_{0}}\\}+\\{x_{i_{0}},x_{j}\\}$, we have

$\\{x_{i_{0}},x_{j}\\}\in\mathcal{L}_{0}.$ (6)

This implies that $\mathcal{L}_{0}$ is an ideal of $\mathcal{L}$. Clearly,
$\mathcal{L}/\mathcal{L}_{0}$ is a Lie algebra.

The formulas (4), (5) and (6) are useful in the sequel.

In $S$, all the compositions are as follows.

1) Compositions of left or right multiplication.

All possible compositions in $S$ of left multiplication are ones related to
(a), (b) and (c).

By noting that for any $s,i,j,t\in I$, we have

$\displaystyle x_{s}\dashv f_{ji}$ $\displaystyle=$ $\displaystyle f_{s\dashv
ji}\ \ \ \ \ \ \ (j>i),$ $\displaystyle x_{s}\dashv f_{ji}$ $\displaystyle=$
$\displaystyle-f_{s\dashv
ij}+x_{s}\dashv(\\{x_{i},x_{j}\\}+\\{x_{j},x_{i}\\})\ \ \ \ \ \ \ (j<i),$
$\displaystyle x_{s}\dashv f_{ii}$ $\displaystyle=$ $\displaystyle
x_{s}\dashv\\{x_{i},x_{i}\\},$ $\displaystyle x_{s}\dashv f_{ji\vdash t}$
$\displaystyle=$ $\displaystyle f_{s\dashv ji}\dashv x_{t}\ \ (j>i)\ \ \ \ \
\mbox{and}$ $\displaystyle x_{s}\dashv h_{i_{0}\vdash t}$ $\displaystyle=$
$\displaystyle h_{s\dashv i_{0}}\dashv x_{t},$

it is clear that all cases are trivial modulo $S$.

By symmetry, all compositions in $S$ of right multiplication are trivial
modulo $S$.

2) Compositions of inclusion and intersection.

We denote, for example, $(a\wedge b)$ the composition of the polynomials of
type $(a)$ and type $(b)$. It is noted that since (b) and (c) are both left
normed, we have to prove that the corresponding compositions of the cases of
$(b\wedge b),\ (b\wedge c),\ (c\wedge c)$ and $(c\wedge b)$ must be a linear
combination of left normed $S$-diwords in which the leading term of each
$S$-diword is less than $w$. Symmetrically, we consider the cases for the
right normed (d) and (e).

All possible compositions of inclusion and intersection are as follows.

  1. ($a\wedge c$)

$[w]=x_{i_{0}}\vdash x_{i}\ (i_{0}\in I_{0})$. We have, by (5),

$\displaystyle(f_{i_{0}i},h_{i_{0}\vdash i})_{[w]}=-x_{i}\dashv
x_{i_{0}}+\\{x_{i},x_{i_{0}}\\}=-h_{i\dashv{i_{0}}}\equiv 0\ \ mod(S,[w]).$

  2. ($a\wedge d$)

$[w]=x_{j}\vdash x_{i}\dashv x_{q}\dashv x_{p}\ \ (q>p)$. We have

$\displaystyle(f_{ji},f_{i\dashv qp})_{[w]}$ $\displaystyle=$ $\displaystyle-
x_{i}\dashv x_{j}\dashv x_{q}\dashv x_{p}+\\{x_{i},x_{j}\\}\dashv x_{q}\dashv
x_{p}+x_{j}\vdash x_{i}\dashv x_{p}\dashv x_{p}-x_{j}\vdash
x_{i}\dashv\\{x_{p},x_{q}\\}$ $\displaystyle=$ $\displaystyle-x_{i}\dashv
f_{j\dashv qp}+f_{\\{i,j\\}\dashv qp}+f_{ji}\dashv x_{p}\dashv
x_{q}-f_{ji}\dashv\\{x_{p},x_{q}\\}$ $\displaystyle\equiv$ $\displaystyle 0\ \
mod(S,[w]).$

  3. $(a\wedge e)$

$[w]=x_{j}\vdash x_{i}\dashv x_{i_{0}}\ (i_{0}\in I_{0})$. We have

$\displaystyle(f_{j{i}},h_{{i}\dashv{i_{0}}})_{[w]}=-x_{i}\dashv x_{j}\dashv
x_{i_{0}}+\\{x_{i},x_{j}\\}\dashv x_{i_{0}}=-x_{i}\dashv h_{j\dashv
i_{0}}+h_{\\{i,j\\}\dashv i_{0}}\equiv 0\ \ mod(S,[w]).$

  4. ($b\wedge a$)

There are two cases to consider: $[w]=x_{j}\vdash x_{i}\vdash x_{t}$ and
$[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash x_{p}$.

For $[w]=x_{j}\vdash x_{i}\vdash x_{t}\ \ (j>i)$, by (4), we have

$\displaystyle(f_{ji\vdash t},f_{it})_{[w]}$ $\displaystyle=$ $\displaystyle-
x_{i}\vdash x_{j}\vdash x_{t}+\\{x_{i},x_{j}\\}\vdash x_{t}+x_{j}\vdash
x_{t}\dashv x_{i}-x_{j}\vdash\\{x_{t},x_{i}\\}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash f_{jt}+f_{\\{i,j\\}t}+f_{jt}\dashv
x_{i}-f_{j\\{t,i\\}}+f_{i\\{t,j\\}}-f_{it}\dashv x_{j}+f_{t\dashv ji}$
$\displaystyle\equiv$ $\displaystyle 0\ \ mod(S,[w]).$

For $[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash x_{p}\ \ (j>i)$, we have

$\displaystyle(f_{ji\vdash t},f_{tp})_{[w]}$ $\displaystyle=$ $\displaystyle-
x_{i}\vdash x_{j}\vdash x_{t}\vdash x_{p}+\\{x_{i},x_{j}\\}\vdash x_{t}\vdash
x_{p}+x_{j}\vdash x_{i}\vdash x_{p}\dashv x_{t}-x_{j}\vdash
x_{i}\vdash\\{x_{p},x_{t}\\}$ $\displaystyle=$ $\displaystyle-x_{i}\vdash
x_{j}\vdash f_{tp}+\\{x_{i},x_{j}\\}\vdash f_{tp}+f_{ji\vdash p}\dashv
x_{t}-f_{ji\vdash\\{p,t\\}}$ $\displaystyle\equiv$ $\displaystyle 0\ \
mod(S,[w]).$

  5. $(b\wedge b$)

There are two cases to consider: $[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash
x_{s}\vdash x_{p}$ and $[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash x_{p}$.

For $[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash x_{s}\vdash x_{p}\ \ (j>i,t>s)$,
we have

$\displaystyle(f_{ji\vdash t},f_{ts\vdash p})_{[w]}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash x_{j}\vdash x_{t}\vdash x_{s}\vdash
x_{p}+\\{x_{i},x_{j}\\}\vdash x_{t}\vdash x_{s}\vdash x_{p}+x_{j}\vdash
x_{i}\vdash x_{s}\vdash x_{t}\vdash x_{p}$ $\displaystyle-x_{j}\vdash
x_{i}\vdash\\{x_{s},x_{t}\\}\vdash x_{p}$ $\displaystyle=$ $\displaystyle-
x_{i}\vdash x_{j}\vdash f_{ts\vdash p}+\\{x_{i},x_{j}\\}\vdash f_{ts\vdash
p}+f_{ji\vdash s}\vdash x_{t}\vdash x_{p}-f_{ji\vdash\\{s,t\\}}\vdash x_{p}$
$\displaystyle\equiv$ $\displaystyle 0\ \ mod(S,[w])$

since it is a combination of left normed $S$-diwords in which the leading term
of each $S$-diword is less than $w$.

For $[w]=x_{j}\vdash x_{i}\vdash x_{t}\vdash x_{p}\ \ (j>i>t)$, suppose that

$\\{x_{i},x_{j}\\}=\sum_{m\in
I_{1}}\alpha_{ij}^{m}x_{m}+\alpha_{ij}^{t}x_{t}+\sum_{n\in
I_{2}}\alpha_{ij}^{n}x_{n}\ (m<t<n).$

Denote

$B_{t\vdash\\{i,j\\}\vdash p}=x_{t}\vdash\\{x_{i},x_{j}\\}\vdash
x_{p}-\\{x_{i},x_{j}\\}\vdash x_{t}\vdash
x_{p}-\\{x_{t},\\{x_{i},x_{j}\\}\\}\vdash x_{p}.$

Then

$B_{t\vdash\\{i,j\\}\vdash p}=\sum_{m\in I_{1}}\alpha_{ij}^{m}f_{tm\vdash
p}-\sum_{n\in I_{2}}\alpha_{ij}^{n}f_{nt\vdash p}-\sum_{q\in
I_{0}}\beta_{q}h_{q\vdash p}$

is a linear combination of left normed $S$-diwords of length 2 or 3, where

$\sum_{q\in I_{0}}\beta_{q}x_{q}=\sum_{m\in
I_{1}}\alpha_{ij}^{m}(\\{x_{t},x_{m}\\}+\\{x_{m},x_{t}\\})+\alpha_{ij}^{t}\\{x_{t},x_{t}\\}.$

Denote

$\sum_{l\in
I_{0}}\gamma_{l}x_{l}=-(\\{x_{j},\\{x_{t},x_{i}\\}\\}+\\{\\{x_{t},x_{i}\\},x_{j}\\})+(\\{x_{i},\\{x_{t},x_{j}\\}\\}+\\{\\{x_{t},x_{j}\\},x_{i}\\}).$

Now, by (4), we have

$\displaystyle(f_{ji\vdash t},f_{it\vdash p})_{[w]}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash x_{j}\vdash x_{t}\vdash
x_{p}+\\{x_{i},x_{j}\\}\vdash x_{t}\vdash x_{p}+x_{j}\vdash x_{t}\vdash
x_{i}\vdash x_{p}-x_{j}\vdash\\{x_{t},x_{i}\\}\vdash x_{p}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash f_{jt\vdash p}-B_{t\vdash\\{i,j\\}\vdash
p}+f_{jt\vdash i}\vdash x_{p}-B_{j\vdash\\{t,i\\}\vdash p}+\sum_{l\in
I_{0}}\gamma_{l}h_{{l}\vdash p}$ $\displaystyle+B_{i\vdash\\{t,j\\}\vdash
p}-f_{it\vdash j}\vdash x_{p}+x_{t}\vdash f_{ji\vdash p}$
$\displaystyle\equiv$ $\displaystyle 0\ \ mod(S,[w])$

since it is a combination of left normed $S$-diwords in which the leading term
of each $S$-diword is less than $w$.

  6. $(b\wedge c)$

There are three cases to consider: $[w]=x_{j}\vdash x_{i_{0}}\vdash x_{t}\
(i_{0}\in I_{0}),\ [w]=x_{j_{0}}\vdash x_{i}\vdash x_{t}\ (j_{0}\in I_{0})$
and $[w]=x_{j}\vdash x_{i}\vdash x_{t_{0}}\vdash x_{n}\ (t_{0}\in I_{0})$.

Case 1. $[w]=x_{j}\vdash x_{i_{0}}\vdash x_{t}\ \ (j>i_{0},\ i_{0}\in I_{0})$.
By (6), we can assume that $\\{x_{i_{0}},x_{j}\\}=\sum_{l\in
I_{0}}\gamma_{l}x_{l}$. Then, we have

$(f_{ji_{0}\vdash t},h_{i_{0}\vdash t})_{[w]}=-x_{i_{0}}\vdash x_{j}\vdash
x_{t}+\\{x_{i_{0}},x_{j}\\}\vdash x_{t}=-h_{i_{0}\vdash j}\vdash
x_{t}+\sum_{l\in I_{0}}\gamma_{l}h_{l\vdash t}\equiv 0\ \ mod(S,[w]).$

Case 2. $[w]=x_{j_{0}}\vdash x_{i}\vdash x_{t}\ \ (j_{0}>i,j_{0}\in I_{0})$.
By (5), we have

$(f_{j_{0}i\vdash t},h_{j_{0}\vdash i})_{[w]}=-x_{i}\vdash x_{j_{0}}\vdash
x_{t}+\\{x_{i},x_{j_{0}}\\}\vdash x_{t}=-x_{i}\vdash h_{{j_{0}}\vdash t}\equiv
0\ \ mod(S,[w]).$

Case 3. $[w]=x_{j}\vdash x_{i}\vdash x_{t_{0}}\vdash x_{n}\ \ (j>i,t_{0}\in
I_{0})$. We have

$\displaystyle(f_{ji\vdash{t_{0}}},h_{{t_{0}}\vdash n})_{[w]}$
$\displaystyle=$ $\displaystyle-x_{i}\vdash x_{j}\vdash x_{t_{0}}\vdash
x_{n}+\\{x_{i},x_{j}\\}\vdash x_{t_{0}}\vdash x_{n}$ $\displaystyle=$
$\displaystyle(-x_{i}\vdash x_{j}+\\{x_{i},x_{j}\\})\vdash h_{{t_{0}}\vdash
n}$ $\displaystyle\equiv$ $\displaystyle 0\ \ mod(S,[w]).$

  7. $(b\wedge d)$

$[w]=x_{j}\vdash x_{i}\vdash x_{t}\dashv x_{q}\dashv x_{p}\ \ (j>i,q>p)$. We
have

$\displaystyle(f_{ji\vdash t},f_{t\dashv qp})_{[w]}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash x_{j}\vdash x_{t}\dashv x_{q}\dashv
x_{p}+\\{x_{i},x_{j}\\}\vdash x_{t}\dashv x_{q}\dashv x_{p}$
$\displaystyle+x_{j}\vdash x_{i}\vdash x_{t}\dashv x_{p}\dashv
x_{q}-x_{j}\vdash x_{i}\vdash x_{t}\dashv\\{x_{p},x_{q}\\}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash x_{j}\vdash f_{t\dashv qp}+\\{x_{i},x_{j}\\}\vdash
f_{t\dashv qp}+f_{ji\vdash t}\dashv x_{p}\dashv x_{q}-f_{ji\vdash
t}\dashv\\{x_{p},x_{q}\\}$ $\displaystyle\equiv$ $\displaystyle 0\ \
mod(S,[w]).$

  8. ($b\wedge e$)

$[w]=x_{j}\vdash x_{i}\vdash x_{t}\dashv x_{n_{0}}\ \ (j>i,n_{0}\in I_{0})$.
We have

$\displaystyle(f_{ji\vdash t},h_{t\dashv{n_{0}}})_{[w]}$ $\displaystyle=$
$\displaystyle-x_{i}\vdash x_{j}\vdash x_{t}\dashv
x_{n_{0}}+\\{x_{i},x_{j}\\}\vdash x_{t}\dashv x_{n_{0}}$ $\displaystyle=$
$\displaystyle(-x_{i}\vdash x_{j}+\\{x_{i},x_{j}\\})\vdash h_{t\dashv{n_{0}}}$
$\displaystyle\equiv$ $\displaystyle 0\ \ mod(S,[w]).$

  9. $(c\wedge a)$

There are two cases to consider: $[w]=x_{n_{0}}\vdash x_{t}\ (n_{0}\in I_{0})$
and $[w]=x_{n_{0}}\vdash x_{t}\vdash x_{s}\ (n_{0}\in I_{0})$.

For $[w]=x_{n_{0}}\vdash x_{t}\ (n_{0}\in I_{0})$, we have

$(h_{{n_{0}}\vdash t},f_{{n_{0}}t})_{[w]}=x_{t}\dashv
x_{n_{0}}-\\{x_{t},x_{n_{0}}\\}=h_{t\dashv{n_{0}}}\equiv 0\ \ mod(S,[w]).$

For $[w]=x_{n_{0}}\vdash x_{t}\vdash x_{s}\ (n_{0}\in I_{0})$, we have

$\displaystyle(h_{{n_{0}}\vdash t},f_{ts})_{[w]}=x_{n_{0}}\vdash x_{s}\dashv
x_{t}-x_{n_{0}}\vdash\\{x_{s},x_{t}\\}=h_{{n_{0}}\vdash s}\dashv
x_{t}-h_{{n_{0}}\vdash\\{s,t\\}}\equiv 0\ \ mod(S,[w]).$

  10. $(c\wedge b)$

$[w]=x_{n_{0}}\vdash x_{t}\vdash x_{s}\vdash x_{p}\ \ (t>s,n_{0}\in I_{0})$.
We have

$\displaystyle(h_{{n_{0}}\vdash t},f_{ts\vdash p})_{[w]}$ $\displaystyle=$
$\displaystyle x_{n_{0}}\vdash x_{s}\vdash x_{t}\vdash
x_{p}-x_{n_{0}}\vdash\\{x_{s},x_{t}\\}\vdash x_{p}$ $\displaystyle=$
$\displaystyle h_{{n_{0}}\vdash s}\vdash x_{t}\vdash
x_{p}-h_{{n_{0}}\vdash\\{s,t\\}}\vdash x_{p}$ $\displaystyle\equiv$
$\displaystyle 0\ \ mod(S,[w]).$

  11. $(c\wedge c)$

$[w]=x_{n_{0}}\vdash x_{t_{0}}\vdash x_{r}\ (n_{0},t_{0}\in I_{0})$. We have

$\displaystyle(h_{{n_{0}}\vdash{t_{0}}},h_{{t_{0}}\vdash r})_{[w]}=0.$

  12. $(c\wedge d)$

$[w]=x_{n_{0}}\vdash x_{t}\dashv x_{q}\dashv x_{p}\ \ (q>p,n_{0}\in I_{0})$.
We have

$\displaystyle(h_{{n_{0}}\vdash t},f_{t\dashv qp})_{[w]}$ $\displaystyle=$
$\displaystyle x_{n_{0}}\vdash x_{t}\dashv x_{p}\dashv x_{q}-x_{n_{0}}\vdash
x_{t}\dashv\\{x_{p},x_{q}\\}$ $\displaystyle=$ $\displaystyle h_{{n_{0}}\vdash
t}\dashv(x_{p}\dashv x_{q}-\\{x_{p},x_{q}\\})$ $\displaystyle\equiv$
$\displaystyle 0\ \ mod(S,[w]).$

  13. $(c\wedge e)$

$[w]=x_{n_{0}}\vdash x_{t}\dashv x_{s_{0}}\ ({n_{0}},{s_{0}}\in{I_{0}})$. We
have

$\displaystyle(h_{{n_{0}}\vdash t},h_{t\dashv{s_{0}}})_{[w]}=0.$

Since $(d\wedge d)$, $(d\wedge e)$, $(e\wedge d)$, $(e\wedge e)$ are symmetric
with $(b\wedge b)$, $(b\wedge c)$, $(c\wedge b)$, $(c\wedge c)$ respectively,
they have the similar representations. We omit the details.

So, we show that $S$ is a Gröbner-Shirshov basis.

(iii) Clearly, the mentioned set is just the set $Irr(S)$. Now, the results
follow from Theorem 3.9. $\square$

A Gröbner-Shirshov basis $S$ is called reduced if $S$ is a monic set and no
monomial in any element of the basis contains the leading words of the other
elements of the basis as subwords.

Remark: Let the notation be in Theorem 4.3. Let $S^{red}$ consist of the
following polynomials:

$\displaystyle(a)$ $\displaystyle f_{ji}=x_{j}\vdash x_{i}-x_{i}\dashv
x_{j}+\\{x_{i},x_{j}\\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ (i\in I,j\in I-I_{0})$ $\displaystyle(b)$ $\displaystyle f_{ji\vdash
t}=x_{j}\vdash x_{i}\vdash x_{t}-x_{i}\vdash x_{j}\vdash
x_{t}+\\{x_{i},x_{j}\\}\vdash x_{t}\ \ \ \ \ \ \ \ \ \ \ \ (i,j\in I-I_{0},\
j>i,t\in I)$ $\displaystyle(c)$ $\displaystyle h_{i_{0}\vdash
t}=x_{i_{0}}\vdash x_{t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i_{0}\in I_{0},\
t\in I)$ $\displaystyle(d)$ $\displaystyle f_{t\dashv ji}=x_{t}\dashv
x_{j}\dashv x_{i}-x_{t}\dashv x_{i}\dashv x_{j}+x_{t}\dashv\\{x_{i},x_{j}\\}\
\ \ \ \ \ \ \ \ \ \ \ (i,j\in I-I_{0},\ j>i,t\in I)$ $\displaystyle(e)$
$\displaystyle h_{t\dashv i_{0}}=x_{t}\dashv x_{i_{0}}\ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ (i_{0}\in I_{0},\ t\in I)$

Then $S^{red}$ is a reduced Gröbner-Shirshov basis for $D(X|S)$.

We have the following corollary.

###### Corollary 4.4

([1]) Let the notation be as in Theorem 4.3. Then as linear spaces,
$U(\mathcal{L})$ is isomorphic to $\mathcal{L}\otimes U(\mathcal{L/L}_{0})$,
where $U(\mathcal{L/L}_{0})$ is the universal enveloping of the Lie algebra
$\mathcal{L/L}_{0}$.

Proof. Clearly, $\\{x_{j}\ |\ j\in I-I_{0}\\}$ is a $k$-basis of the Lie
algebra $\mathcal{L/L}_{0}$. It is well known that the universal enveloping
$U(\mathcal{L/L}_{0})$ of the Lie algebra $\mathcal{L/L}_{0}$ has a $k$-basis

$\\{x_{i_{1}}x_{i_{2}}\dots x_{i_{k}}\ |\ i_{1}\leq\dots\leq i_{k},\ i_{p}\in
I-I_{0},\ 1\leq p\leq k,\ k\geq 0\\}.$

By using (iii) in Theorem 4.3, the result follows. $\square$

###### Definition 4.5

Let $D$ be a dialgebra. An element $e\in D$ is called a bar unit of $D$ if
$e\vdash x=x\dashv e=x$ for any $x\in D$.

###### Theorem 4.6

Each dialgebra has a bar unit extension.

Proof. Let $(D,\vdash,\dashv)$ be an arbitrary dialgebra over a field $k$ and
$A$ the ideal of $D$ generated by the set $\\{a\dashv b-a\vdash b|\ a,b\in
D\\}$. Let $X_{0}=\\{x_{i_{0}}|{i_{0}}\in I_{0}\\}$ be a $k$-basis of $A$ and
$X=\\{x_{i}|i\in I\\}$ a well ordered $k$-basis of $D$ such that
$I_{0}\subseteq I$. Then $D$ has a presentation by the multiplication table
$D=D(X|S)$, where $S=\\{x_{i}\vdash x_{j}-\\{x_{i}\vdash x_{j}\\},\
x_{i}\dashv x_{j}-\\{x_{i}\dashv x_{j}\\},\ i,j\in I\\}$, where
$\\{x_{i}\vdash x_{j}\\}$ and $\\{x_{i}\dashv x_{j}\\}$ are linear
combinations of $x_{t},t\in I$.

Let $D_{1}=D(X\cup\\{e\\}|S_{1})$, where $S_{1}=S\cup\\{e\vdash y-y,\ y\dashv
e-y,\ e\dashv x_{0},\ x_{0}\vdash e\ |\ y\in X\cup\\{e\\},x_{0}\in X_{0}\\}$.
Then $D_{1}$ is a dialgebra with a bar unit $e$.

Denote

$\displaystyle 1.$ $\displaystyle f_{i\vdash j}=x_{i}\vdash
x_{j}-\\{x_{i}\vdash x_{j}\\},$ $\displaystyle 2.$ $\displaystyle f_{i\dashv
j}=x_{i}\dashv x_{j}-\\{x_{i}\dashv x_{j}\\},$ $\displaystyle 3.$
$\displaystyle g_{e\vdash y}=e\vdash y-y,$ $\displaystyle 4.$ $\displaystyle
g_{y\dashv e}=y\dashv e-y,$ $\displaystyle 5.$ $\displaystyle
h_{x_{i_{0}}\vdash e}=x_{i_{0}}\vdash e,$ $\displaystyle 6.$ $\displaystyle
h_{e\dashv x_{i_{0}}}=e\dashv x_{i_{0}},$

where $i,j\in I,\ i_{0}\in I_{0},\ y\in X\cup\\{e\\}$.

We show that $\\{x_{t}\dashv x_{i_{0}}\\}=0$ and $\\{x_{i_{0}}\vdash
x_{t}\\}=0$ for any $t\in I,\ i_{0}\in I_{0}$.

Since $x_{i_{0}}\in A$, we have $x_{i_{0}}=\sum\alpha_{i}(c_{i}f_{i}d_{i})$,
where $f_{i}=a_{i}\dashv b_{i}-a_{i}\vdash b_{i}$, $\alpha_{i}\in k$,
$a_{i},b_{i}\in D$ and $c_{i},d_{i}\in X^{*}$.

Since $x_{t}\dashv(c_{i}(a_{i}\dashv b_{i}-a_{i}\vdash b_{i})d_{i})=0$, we
have $\\{x_{t}\dashv\\{c_{i}\\{a_{i}\dashv b_{i}-a_{i}\vdash
b_{i}\\}d_{i}\\}\\}=0$ for each $i$. Then $\\{x_{t}\dashv x_{i_{0}}\\}=0$.

By symmetry, we have $\\{x_{i_{0}}\vdash x_{t}\\}=0$.

To prove the theorem, by using our Theorem 3.9, it suffices to prove that with
the ordering on $[(X\cup\\{e\\})^{*}]$ as before, where $x<e,\ x\in X$,
$S_{1}$ is a Gröbner-Shirshov basis in $D(X\cup\\{e\\})$. Now, we show that
all compositions in $S_{1}$ are trivial.

All possible compositions of left and right multiplication are: $z\dashv
f_{i\vdash j}$, $z\dashv g_{e\vdash y}$, $z\dashv h_{x_{i_{0}}\vdash e}$,
$f_{i\dashv j}\vdash z$, $g_{y\dashv e}\vdash z$, $h_{e\dashv x_{i_{0}}}\vdash
z$, $z\in X\cup\\{e\\}$.

For $z\dashv f_{i\vdash j},\ z=x_{t}\in X$, since $(x_{t}\dashv x_{i})\dashv
x_{j}=x_{t}\dashv(x_{i}\vdash x_{j})$, we have $\\{\\{x_{t}\dashv
x_{i}\\}\dashv x_{j}\\}=\\{x_{t}\dashv\\{x_{i}\vdash x_{j}\\}\\}$ and

$\displaystyle x_{t}\dashv f_{i\vdash j}$ $\displaystyle=$ $\displaystyle
x_{t}\dashv x_{i}\dashv x_{j}-x_{t}\dashv\\{x_{i}\vdash x_{j}\\}$
$\displaystyle=$ $\displaystyle f_{t\dashv i}\dashv x_{j}+f_{\\{t\dashv
i\\}\dashv j}-f_{t\dashv\\{i\vdash j\\}}+\\{\\{x_{t}\dashv x_{i}\\}\dashv
x_{j}\\}-\\{x_{t}\dashv\\{x_{i}\vdash x_{j}\\}\\}$ $\displaystyle=$
$\displaystyle f_{t\dashv i}\dashv x_{j}+f_{\\{t\dashv i\\}\dashv
j}-f_{t\dashv\\{i\vdash j\\}}$ $\displaystyle\equiv$ $\displaystyle 0\
mod(S_{1}).$

For $z\dashv f_{i\vdash j},z=e$, let $\\{x_{i}\dashv x_{j}\\}-\\{x_{i}\vdash
x_{j}\\}=\sum\alpha_{i_{0}}x_{i_{0}}$. Then

$\displaystyle e\dashv f_{i\vdash j}$ $\displaystyle=$ $\displaystyle e\dashv
x_{i}\dashv x_{j}-e\dashv\\{x_{i}\vdash x_{j}\\}$ $\displaystyle=$
$\displaystyle e\dashv(x_{i}\dashv x_{j}-\\{x_{i}\dashv
x_{j}\\})+e\dashv\\{x_{i}\dashv x_{j}\\}-e\dashv\\{x_{i}\vdash x_{j}\\}$
$\displaystyle=$ $\displaystyle e\dashv f_{i\dashv
j}+\sum\alpha_{i_{0}}h_{e\dashv x_{i_{0}}}$ $\displaystyle\equiv$
$\displaystyle 0\ mod(S_{1}).$

For $z\dashv g_{e\vdash y}$, we have

$\displaystyle z\dashv g_{e\vdash y}=z\dashv e\dashv y-z\dashv y=(z\dashv
e-z)\dashv y=g_{z\dashv e}\dashv y\equiv 0\ mod(S_{1}).$

For $z\dashv h_{x_{i_{0}}\vdash e}$, we have

$\displaystyle z\dashv h_{x_{i_{0}}\vdash e}=z\dashv x_{i_{0}}\dashv e=z\dashv
g_{x_{i_{0}}\dashv e}+z\dashv x_{i_{0}}.$

It is clear that $z\dashv x_{i_{0}}=h_{e\dashv x_{i_{0}}}$ if $z=e$ and
$z\dashv x_{i_{0}}=x_{t}\dashv x_{i_{0}}-\\{x_{t}\dashv
x_{i_{0}}\\}=f_{t\dashv i_{0}}$ if $z=x_{t}\in X$, since $\\{x_{t}\dashv
x_{i_{0}}\\}=0$. This implies that $z\dashv h_{x_{i_{0}}\vdash e}\equiv 0\
mod(S_{1})$.

Thus we show that all compositions of left multiplication in $S_{1}$ are
trivial modulo $S_{1}$. By symmetry, all compositions of right multiplication
in $S_{1}$ are trivial modulo $S_{1}$.

Now, all possible ambiguities $[w]$ of compositions of intersection in $S_{1}$
are:

$1\wedge 1$, $[x_{i}x_{j}\dot{x_{t}}]$; $1\wedge 2$,
$[x_{i}\dot{x_{j}}x_{t}]$; $1\wedge 4$, $[x_{i}\dot{x}_{j}e]$; $1\wedge 5$,
$[x_{i}x_{i_{0}}\dot{e}]$.

$2\wedge 2$, $[\dot{x_{i}}x_{j}x_{t}]$; $2\wedge 4$, $[\dot{x}_{i}x_{j}e]$.

$3\wedge 1$, $[ex_{i}\dot{x}_{j}]$; $3\wedge 2$, $[e\dot{x}_{i}x_{j}]$;
$3\wedge 3$, $[ee\dot{y}]$; $3\wedge 4$, $[e\dot{y}e]$; $3\wedge 5$,
$[ex_{i_{0}}\dot{e}]$; $3\wedge 6$, $[e\dot{e}x_{i_{0}}]$.

$4\wedge 4$, $[\dot{y}ee]$; $4\wedge 6$, $[\dot{y}ex_{i_{0}}]$.

$5\wedge 3$, $[x_{i_{0}}e\dot{y}]$; $5\wedge 4$, $[x_{i_{0}}\dot{e}e]$;
$5\wedge 6$, $[x_{i_{0}}\dot{e}x_{j_{0}}]$.

$6\wedge 2$, $[\dot{e}x_{i_{0}}x_{j}]$; $6\wedge 4$, $[\dot{e}x_{i_{0}}e]$.

In the above, all $i,j,t\in I$, $i_{0},j_{0}\in I_{0}$ and $y\in
X\cup\\{e\\}$.

There is no composition of inclusion in $S_{1}$.

We will show that all compositions of intersection in $S_{1}$ are trivial. We
check only the cases of $1\wedge 2$, $1\wedge 5$ and $4\wedge 6$. Others can
be similarly proved.

For $1\wedge 2$, $[w]=[x_{i}\dot{x_{j}}x_{t}]$, since $(x_{i}\vdash
x_{j})\dashv x_{t}=x_{i}\vdash(x_{j}\dashv x_{t})$, we have $\\{\\{x_{i}\vdash
x_{j}\\}\dashv x_{t}\\}=\\{x_{i}\vdash\\{x_{j}\dashv x_{t}\\}\\}$ and

$\displaystyle(1\wedge 2)_{[w]}$ $\displaystyle=$
$\displaystyle-\\{x_{i}\vdash x_{j}\\}\dashv x_{t}+x_{i}\vdash\\{x_{j}\dashv
x_{t}\\}$ $\displaystyle=$ $\displaystyle-f_{\\{i\vdash j\\}\dashv
t}+f_{i\vdash\\{j\dashv t\\}}-\\{\\{x_{i}\vdash x_{j}\\}\dashv
x_{t}\\}+\\{x_{i}\vdash\\{x_{j}\dashv x_{t}\\}\\}$ $\displaystyle=$
$\displaystyle-f_{\\{i\vdash j\\}\dashv t}+f_{i\vdash\\{j\dashv t\\}}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S_{1},{[w]}).$

For $1\wedge 5$, $[w]=[x_{i}x_{i_{0}}\dot{e}]$, since $x_{i}\vdash
x_{i_{0}}\in A$, we have $\\{x_{i}\vdash
x_{i_{0}}\\}=\sum\alpha_{j_{0}}x_{j_{0}}$ and

$\displaystyle(1\wedge 5)_{[w]}=\\{x_{i}\vdash x_{i_{0}}\\}\vdash
e=\sum\alpha_{j_{0}}h_{x_{j_{0}}\vdash e}\equiv 0\ mod(S_{1},{[w]}).$

For $4\wedge 6$, $[w]=[\dot{y}ex_{i_{0}}]$, we have $(4\wedge
6)_{[w]}=-h_{e\dashv x_{i_{0}}}$ if $y=e$ and $(4\wedge 6)_{[w]}=-f_{t\dashv
i_{0}}$ if $y=x_{t}\in X$ since $\\{x_{t}\dashv x_{i_{0}}\\}=0$. Then
$(4\wedge 6)_{[w]}\equiv 0\ mod(S_{1},[w])$.

Then all the compositions in $S_{1}$ are trivial.

The proof is complete. $\square$

Remark: Let the notation be as in the proof of Theorem 4.6. Let
$D^{\prime}=D(X\cup\\{e_{j}\\}_{J}|S^{\prime})$ be a dialgebra, where
$S^{\prime}=S\cup\\{e_{j}\vdash y-y,y\dashv e_{j}-y,\ e_{j}\dashv
x_{0},x_{0}\vdash e_{j}\ |\ y\in X\cup\\{e_{j}\\}_{J},x_{0}\in X_{0},\ j\in
J\\}$. Let $J$ be a well ordered set. Then with the ordering on
$[(X\cup\\{e_{j}\\}_{J})^{*}]$ as before, where $x_{i}<e_{j}$ for all $i\in
I,\ j\in J$, by a similar proof of Theorem 4.6, $S^{\prime}$ is a Gröbner-
Shirshov basis in $D(X\cup\\{e_{j}\\}_{J})$. It follows from Theorem 3.9 that
$D$ can be embedded into the dialgebra $D^{\prime}$ while $D^{\prime}$ has bar
units $\\{e_{j}\\}_{J}$.

###### Definition 4.7

Let $D_{1},D_{2}$ be dialgebras over a field $k$. The dialgebra $D_{1}*D_{2}$
with two dialgebra homomorphisms $\varepsilon_{1}:D_{1}\rightarrow
D_{1}*D_{2}$, $\varepsilon_{2}:D_{2}\rightarrow D_{1}*D_{2}$ is called the
free product of $D_{1},D_{2}$, if the following diagram commute:

$D$$\exists!f$$D_{1}*D_{2}$$D_{1}$$D_{2}$$\varepsilon_{1}$$\varepsilon_{2}$$\forall\delta_{1}$$\forall\delta_{2}$

where $D$ is a dialgebra, $\delta_{1},\delta_{2}$ are dialgebra homomorphisms
and $f:D_{1}*D_{2}\rightarrow D$ is a dialgebra homomorphism such that
$f\varepsilon_{1}=\delta_{1},f\varepsilon_{2}=\delta_{2}$ (i.e.,
$(\varepsilon_{1},\varepsilon_{2}):(D_{1},D_{2})\rightarrow(D_{1}*D_{2},D_{1}*D_{2})$
is a universal arrow in the sense of S. Maclane [13]).

An equivalent definition is as follows: Let $D_{i}=D(X_{i}|S_{i})$ be a
presentation by generators and defining relations with $X_{1}\cap
X_{2}=\varnothing$, $i=1,2$. Then $D_{1}*D_{2}=D(X_{1}\cup X_{2}|S_{1}\cup
S_{2})$.

Let $(D_{1},\vdash,\dashv)$, $(D_{2},\vdash,\dashv)$ be two dialgebras over a
field $k$, $A_{1}$ the ideal of $D_{1}$ generated by the set $\\{a\dashv
b-a\vdash b|\ a,b\in D_{1}\\}$ and $A_{2}$ the ideal of $D_{2}$ generated by
the set $\\{c\dashv d-c\vdash d|\ c,d\in D_{2}\\}$. Let
$X_{0}=\\{x_{i_{0}}|{i_{0}}\in I_{0}\\}$ be a $k$-basis of $A_{1}$ and
$X=\\{x_{i}|i\in I\\}$ a well ordered $k$-basis of $D_{1}$ such that
$I_{0}\subseteq I$. Let $Y_{0}=\\{y_{l_{0}}|{l_{0}}\in J_{0}\\}$ be a
$k$-basis of $A_{2}$ and $Y=\\{y_{l}|l\in J\\}$ a well ordered $k$-basis of
$D_{2}$ such that $J_{0}\subseteq J$. Then $D_{1}$ and $D_{2}$ have
multiplication tables:

$\displaystyle D_{1}=D(X|S_{1}),\ \ \ S_{1}=\\{x_{i}\vdash
x_{j}-\\{x_{i}\vdash x_{j}\\},\ x_{i}\dashv x_{j}-\\{x_{i}\dashv x_{j}\\},\
i,j\in I\\},$ $\displaystyle D_{2}=D(Y|S_{2}),\ \ \ S_{2}=\\{y_{l}\vdash
y_{m}-\\{y_{l}\vdash y_{m}\\},\ y_{l}\dashv y_{m}-\\{y_{l}\dashv y_{m}\\},\
l,m\in J\\}.$

The free product $D_{1}*D_{2}$ of $D_{1}$ and $D_{2}$ is

$D_{1}*D_{2}=D(X\cup Y|S_{1}\cup S_{2}).$

We order $X\cup Y$ by $x_{i}<y_{j}$ for any $i\in I,j\in J$. Then we have the
following theorem.

###### Theorem 4.8

(i) $S$ is a Gröbner-Shirshov basis of $D_{1}*D_{2}=D(X\cup Y|S_{1}\cup
S_{2})$, where $S$ consists of the following relations:

$\displaystyle 1.$ $\displaystyle f_{x_{i}\vdash x_{j}}=x_{i}\vdash
x_{j}-\\{x_{i}\vdash x_{j}\\},\ \ \ \ i,j\in I,$ $\displaystyle 2.$
$\displaystyle f_{x_{i}\dashv x_{j}}=x_{i}\dashv x_{j}-\\{x_{i}\dashv
x_{j}\\},\ \ \ \ i,j\in I,$ $\displaystyle 3.$ $\displaystyle f_{y_{l}\vdash
y_{m}}=y_{l}\vdash y_{m}-\\{y_{l}\vdash y_{m}\\},\ \ \ \ l,m\in J,$
$\displaystyle 4.$ $\displaystyle f_{y_{l}\dashv y_{m}}=y_{l}\dashv
y_{m}-\\{y_{l}\dashv y_{m}\\},\ \ \ \ l,m\in J,$ $\displaystyle 5.$
$\displaystyle h_{x_{i_{0}}\vdash y_{l}}=x_{i_{0}}\vdash y_{l},\ \ \ \ \ \ \ \
i_{0}\in I_{0},l\in J,$ $\displaystyle 6.$ $\displaystyle h_{y_{l}\dashv
x_{i_{0}}}=y_{l}\dashv x_{i_{0}},\ \ \ \ \ \ \ \ i_{0}\in I_{0},l\in J,$
$\displaystyle 7.$ $\displaystyle h_{y_{l_{0}}\vdash x_{i}}=y_{l_{0}}\vdash
x_{i},\ \ \ \ \ \ \ \ i\in I,l_{0}\in J_{0},$ $\displaystyle 8.$
$\displaystyle h_{x_{i}\dashv y_{l_{0}}}=x_{i}\dashv y_{l_{0}},\ \ \ \ \ \ \ \
i\in I,l_{0}\in J_{0}.$

(ii) $Irr(S)$, which is a $k$-linear basis of $D_{1}*D_{2}$, consists of all
elements $z_{-m}\cdots z_{-1}\dot{z}_{0}z_{1}\cdots z_{n}$, where $m,n\geq
0,z_{0}\in X\cup Y,z_{i}\in(X\setminus X_{0})\cup(Y\setminus Y_{0}),-m\leq
i\leq n,i\neq 0,\mbox{neither }\\{z_{j},z_{j+1}\\}\subseteq X\ \mbox{nor
}\\{z_{j},z_{j+1}\\}\subseteq Y,-m\leq j\leq n-1$.

Proof. By the proof of Theorem 4.6, we have $\\{x_{i}\dashv x_{i_{0}}\\}=0,\
\\{x_{i_{0}}\vdash x_{i}\\}=0,\ \\{y_{l}\dashv y_{l_{0}}\\}=0$ and
$\\{y_{l_{0}}\vdash y_{l}\\}=0$ for any $i\in I,\ i_{0}\in I_{0},\ l\in J,\
l_{0}\in J_{0}$.

Firstly, we prove that $h_{y_{l}\dashv x_{i_{0}}}\in Id(S_{1}\cup S_{2})$ for
any $i_{0}\in I_{0},\ l\in J$.

Since $y_{l}\dashv(c_{i}(\\{a_{i}\dashv b_{i}\\}-\\{a_{i}\vdash
b_{i}\\})d_{i})=y_{l}\dashv(c_{i}((a_{i}\dashv b_{i}-\\{a_{i}\dashv
b_{i}\\})-(a_{i}\vdash b_{i}-\\{a_{i}\vdash b_{i}\\})d_{i})\in Id(S_{1}\cup
S_{2})$, we have $y_{l}\dashv\\{c_{i}\\{a_{i}\dashv b_{i}-a_{i}\vdash
b_{i}\\}d_{i}\\}\in Id(S_{1}\cup S_{2})$ for all $i,l$. Then $h_{y_{l}\dashv
x_{i_{0}}}\in Id(S_{1}\cup S_{2})$.

Similarly, we have $h_{x_{i_{0}}\vdash y_{l}},\ h_{y_{l_{0}}\vdash x_{i}},\
h_{x_{i}\dashv y_{l_{0}}}\in Id(S_{1}\cup S_{2})$ for any $i\in I,\ i_{0}\in
I_{0},\ l\in J,\ l_{0}\in J_{0}$.

Secondly, we will show that all compositions in $S$ are trivial.

All possible compositions of left and right multiplication are: $z\dashv
f_{x_{i}\vdash x_{j}},\ z\dashv f_{y_{l}\vdash y_{m}},\ z\dashv
h_{x_{i_{0}}\vdash y_{l}},\ z\dashv h_{y_{l_{0}}\vdash x_{i}},\ f_{x_{i}\dashv
x_{j}}\vdash z,\ f_{y_{l}\dashv y_{m}}\vdash z,\ h_{y_{l}\dashv
x_{i_{0}}}\vdash z,\ h_{x_{i}\dashv y_{l_{0}}}\vdash z,$ where $z\in X\cup Y$.

By a similar proof in Theorem 4.6, all compositions of left and right
multiplication mentioned the above are trivial modulo $S$.

Now, all possible ambiguities $[w]$ of compositions of intersection in $S$
are:

$\displaystyle 1\wedge 1,[x_{i}x_{j}\dot{x}_{t}];1\wedge
2,[x_{i}\dot{x}_{j}x_{t}];1\wedge 5,[x_{i}x_{i_{0}}\dot{y}_{l}];1\wedge
8,[x_{i}\dot{x}_{j}y_{l_{0}}].$ $\displaystyle 2\wedge
2,[\dot{x}_{i}x_{j}x_{t}];2\wedge 8,[\dot{x}_{i}x_{j}y_{l_{0}}].$
$\displaystyle 3\wedge 3,[y_{l}y_{m}\dot{y}_{t}];3\wedge
4,[y_{l}\dot{y}_{m}y_{t}];3\wedge 6,[y_{l}\dot{y}_{m}x_{i_{0}}];3\wedge
7,[y_{m}y_{l_{0}}\dot{x}_{i}].$ $\displaystyle 4\wedge
4,[\dot{y}_{l}y_{m}y_{t}];4\wedge 6,[\dot{y}_{l}y_{m}x_{i_{0}}].$
$\displaystyle 5\wedge 3,[x_{i_{0}}y_{l}\dot{y}_{t}];5\wedge
4,[x_{i_{0}}\dot{y}_{l}y_{t}];5\wedge
6,[x_{i_{0}}\dot{y}_{l}x_{j_{0}}];5\wedge 7,[x_{i_{0}}y_{l_{0}}\dot{x}_{t}].$
$\displaystyle 6\wedge 2,[\dot{y}_{l}x_{i_{0}}x_{t}];6\wedge
8,[\dot{y}_{m}x_{i_{0}}y_{l_{0}}].$ $\displaystyle 7\wedge
1,[y_{l_{0}}x_{i}\dot{x}_{j}];7\wedge 2,[y_{l_{0}}\dot{x}_{i}x_{j}];7\wedge
5,[y_{l_{0}}x_{i_{0}}\dot{y}_{m}];7\wedge 8,[y_{l_{0}}\dot{x}_{i}y_{m_{0}}].$
$\displaystyle 8\wedge 4,[\dot{x}_{i}y_{l_{0}}y_{t}];8\wedge
6,[\dot{x}_{i}y_{l_{0}}x_{i_{0}}].$

There is no composition of inclusion in $S$.

We will show that all compositions of intersection in $S$ are trivial. We
check only the cases of $1\wedge 5$ and $2\wedge 8$. Others can be similarly
proved.

For $1\wedge 5$, $[w]=[x_{i}x_{i_{0}}\dot{y}_{l}]$, let $\\{x_{i}\vdash
x_{i_{0}}\\}=\sum\alpha_{t_{0}}x_{t_{0}}$. Then

$(1\wedge 5)_{[w]}=-\\{x_{i}\vdash x_{i_{0}}\\}\vdash
y_{l}=-\sum\alpha_{t_{0}}h_{x_{t_{0}}\vdash y_{l}}\equiv 0\ \ mod(S,[w]).$

For $2\wedge 8$, $[w]=[\dot{x}_{i}x_{j}y_{l_{0}}]$, let $\\{x_{i}\dashv
x_{j}\\}=\sum\alpha_{t}x_{t}$. Then

$(2\wedge 8)_{[w]}=-\\{x_{i}\dashv x_{j}\\}\dashv
y_{l_{0}}=-\sum\alpha_{t}h_{x_{t}\dashv y_{l_{0}}}\equiv 0\ \ mod(S,[w]).$

Then all the compositions in $S$ are trivial. This show (i).

(ii) follows from our Theorem 3.9. $\square$

###### Definition 4.9

Let $X=\\{x_{1},\dots,x_{n}\\}$ be a set, $k$ a field of characteristic $\neq
2$ and $(a_{ij})_{n\times n}$ a non-zero symmetric matrix over $k$. Denote

$D(X\cup\\{e\\}\ |\ x_{i}\vdash x_{j}+x_{j}\dashv x_{i}-2a_{ij}e,\ e\vdash
y-y,\ y\dashv e-y,\ x_{i},x_{j}\in X,\ y\in X\cup\\{e\\})$

by $C(n,f)$. Then $C(n,f)$ is called a Clifford dialgebra.

We order $X\cup\\{e\\}$ by $x_{1}<\dots<x_{n}<e$.

###### Theorem 4.10

Let the notation be as the above. Then

  1. (i)

$S$ is a Gröbner-Shirshov basis of Clifford dialgebra $C(n,f)$, where $S$
consists of the following relations:

$\displaystyle 1.\ f_{x_{i}x_{j}}=x_{i}\vdash x_{j}+x_{j}\dashv
x_{i}-2a_{ij}e,$ $\displaystyle 2.\ g_{e\vdash y}=e\vdash y-y,$ $\displaystyle
3.\ g_{y\dashv e}=y\dashv e-y,$ $\displaystyle 4.\ f_{y\dashv
x_{i}x_{j}}=y\dashv x_{i}\dashv x_{j}+y\dashv x_{j}\dashv x_{i}-2a_{ij}y,\ \
(i>j),$ $\displaystyle 5.\ f_{y\dashv x_{i}x_{i}}=y\dashv x_{i}\dashv
x_{i}-a_{ii}y,$ $\displaystyle 6.\ f_{x_{i}x_{j}\vdash y}=x_{i}\vdash
x_{j}\vdash y+x_{j}\vdash x_{i}\vdash y-2a_{ij}y,\ \ (i>j),$ $\displaystyle
7.\ f_{x_{i}x_{i}\vdash y}=x_{i}\vdash x_{i}\vdash y-a_{ii}y,$ $\displaystyle
8.\ h_{x_{i}e}=x_{i}\vdash e-e\dashv x_{i},$

where $x_{i},x_{j}\in X,y\in X\cup\\{e\\}$.

  2. (ii)

A $k$-linear basis of $C(n,f)$ is a set of all elements of the form
$\dot{y}x_{i_{1}}\cdots x_{ik}$, where $y\in X\cup\\{e\\}$, $x_{ij}\in X$ and
$i_{1}<i_{2}<\cdots<i_{k}\ \ (k\geq 0)$.

Proof. Let $S_{1}=\\{f_{x_{i}x_{j}},g_{e\vdash y},g_{y\dashv e}\ |\
x_{i},x_{j}\in X,y\in X\cup\\{e\\}\\}.$

Firstly, we will show that $f_{y\dashv x_{i}x_{j}},\ f_{y\dashv x_{i}x_{i}},\
f_{x_{i}x_{j}\vdash y},\ f_{x_{i}x_{i}\vdash y},\ h_{x_{i}e}\in Id(S_{1})$.

In fact, $f_{y\dashv x_{i}x_{j}}=y\dashv f_{x_{i}x_{j}}+2a_{ij}g_{y\dashv e}$
implies $f_{y\dashv x_{i}x_{j}},\ f_{y\dashv x_{i}x_{i}}\in Id(S_{1})$. By
symmetry, we have $f_{x_{i}x_{j}\vdash y},\ f_{x_{i}x_{i}\vdash y}\in
Id(S_{1})$.

If there exists $t$ such that $a_{it}\neq 0$, then

$2a_{it}h_{x_{i}e}=f_{x_{i}x_{i}\vdash x_{t}}-x_{i}\vdash f_{x_{i}\vdash
x_{t}}+f_{x_{i}\vdash x_{t}}\dashv x_{i}-f_{x_{t}\dashv x_{i}x_{i}}\in
Id(S_{1}).$

Otherwise, $a_{it}=0$ for any $t$. Since $(a_{ij})\neq 0$, there exists $j\neq
i$ such that $a_{jt}\neq 0$ for some $t$. Then

$\displaystyle 2a_{jt}h_{x_{i}e}$ $\displaystyle=$ $\displaystyle
f_{x_{i}x_{j}\vdash x_{t}}-x_{i}\vdash f_{x_{j}\vdash x_{t}}-x_{j}\vdash
f_{x_{i}\vdash x_{t}}+f_{x_{i}\vdash x_{t}}\dashv x_{j}+f_{x_{j}\vdash
x_{t}}\dashv x_{i}-f_{x_{t}\dashv x_{i}x_{j}}\in Id(S_{1}).$

This shows that $h_{x_{i}e}\in Id(S_{1}).$

Secondly, we will show that all compositions in $S$ is trivial.

All possible compositions of left and right multiplication are: $z\dashv
f_{x_{i}x_{j}},\ z\dashv g_{e\vdash y},\ z\dashv f_{x_{i}x_{j}\vdash y},\
z\dashv f_{x_{i}x_{i}\vdash y},\ z\dashv h_{x_{i}e},\ f_{x_{i}x_{j}}\vdash z,\
g_{y\dashv e}\vdash z,\ f_{y\dashv x_{i}x_{j}}\vdash z,\ f_{y\dashv
x_{i}x_{i}}\vdash z,\ h_{x_{i}e}\vdash z,$ where $z\in X\cup\\{e\\}$. We just
check the cases of $f_{y\dashv x_{i}x_{j}}\vdash z$ and $h_{x_{i}e}\vdash z$.
Others can be similarly proved.

For $f_{y\dashv x_{i}x_{j}}\vdash z$, we have

$\displaystyle f_{y\dashv x_{i}x_{j}}\vdash z=y\vdash x_{i}\vdash x_{j}\vdash
z+y\vdash x_{j}\vdash x_{i}\vdash z-2a_{ij}y\vdash z=y\vdash
f_{x_{i}x_{j}\vdash z}\equiv 0\ mod(S).$

For $h_{x_{i}e}\vdash z$,

$\displaystyle h_{x_{i}e}\vdash z=x_{i}\vdash e\vdash z-e\vdash x_{i}\vdash
z=x_{i}\vdash g_{e\vdash z}-g_{e\vdash x_{i}}\vdash z\equiv 0\ mod(S).$

Now, all possible ambiguities $[w]$ of compositions of intersection in $S$
are:

$\displaystyle 1\wedge 3,[x_{i}\dot{x}_{j}e];1\wedge
4,[x_{i}\dot{x}_{j}x_{m}x_{n}]\ (m>n);1\wedge 5,[x_{i}\dot{x}_{j}x_{n}x_{n}].$
$\displaystyle 2\wedge 1,[ex_{i}\dot{x}_{j}];2\wedge 2,[ee\dot{y}];2\wedge
3,[e\dot{y}e];2\wedge 4,[e\dot{y}x_{i}x_{j}]\ (i>j);$ $\displaystyle\ \ \ \
2\wedge 5,[e\dot{y}x_{i}x_{i}];2\wedge 6,[ex_{i}x_{j}\dot{y}]\ (i>j);2\wedge
7,[ex_{i}x_{i}\dot{y}];2\wedge 8,[ex_{i}\dot{e}].$ $\displaystyle 3\wedge
3,[\dot{y}ee];3\wedge 4,[\dot{y}ex_{i}x_{j}]\ (i>j);3\wedge
5,[\dot{y}ex_{i}x_{i}].$ $\displaystyle 4\wedge 3,[\dot{y}x_{i}x_{j}e]\
(i>j);4\wedge 4,[\dot{y}x_{i}x_{j}x_{m}x_{n}]\
(i>j,m>n),[\dot{y}x_{i}x_{j}x_{t}]\ (i>j>t);$ $\displaystyle\ \ \ \ 4\wedge
5,[\dot{y}x_{i}x_{j}x_{t}x_{t}]\ (i>j),[\dot{y}x_{i}x_{j}x_{j}]\ (i>j).$
$\displaystyle 5\wedge 3,[\dot{y}x_{i}x_{i}e];5\wedge
4,[\dot{y}x_{i}x_{i}x_{m}x_{n}]\ (m>n),[\dot{y}x_{i}x_{i}x_{j}]\ (i>j);$
$\displaystyle\ \ \ \ 5\wedge
5,[\dot{y}x_{i}x_{i}x_{m}x_{m}],[\dot{y}x_{i}x_{i}x_{i}].$ $\displaystyle
6\wedge 1,[x_{i}x_{j}x_{m}\dot{x}_{n}]\ (i>j);6\wedge 2,[x_{i}x_{j}e\dot{y}]\
(i>j);6\wedge 3,[x_{i}x_{j}\dot{y}e]\ (i>j);$ $\displaystyle\ \ \ \ 6\wedge
4,[x_{i}x_{j}\dot{y}x_{m}x_{n}]\ (i>j,m>n);6\wedge
5,[x_{i}x_{j}\dot{y}x_{m}x_{m}]\ (i>j);$ $\displaystyle\ \ \ \ 6\wedge
6,[x_{i}x_{j}x_{m}x_{n}\dot{y}]\ (i>j,m>n),[x_{i}x_{j}x_{t}\dot{y}]\ (i>j>t);$
$\displaystyle\ \ \ \ 6\wedge 7,[x_{i}x_{j}x_{m}x_{m}\dot{y}]\
(i>j),[x_{i}x_{j}x_{j}\dot{y}]\ (i>j);6\wedge 8,[x_{i}x_{j}x_{t}\dot{e}]\
(i>j).$ $\displaystyle 7\wedge 1,[x_{i}x_{i}x_{m}\dot{x}_{n}];7\wedge
2,[x_{i}x_{i}e\dot{y}];7\wedge 3,[x_{i}x_{i}\dot{y}e];7\wedge
4,[x_{i}x_{i}\dot{y}x_{m}x_{n}]\ (m>n);$ $\displaystyle\ \ \ \ 7\wedge
5,[x_{i}x_{i}\dot{y}x_{m}x_{m}];7\wedge 6,[x_{i}x_{i}x_{m}x_{n}\dot{y}]\
(m>n),[x_{i}x_{i}x_{t}\dot{y}]\ (i>t);$ $\displaystyle\ \ \ \ 7\wedge
7,[x_{i}x_{i}x_{m}x_{m}\dot{y}],\ [x_{i}x_{i}x_{i}\dot{y}];7\wedge
8,[x_{i}x_{i}x_{j}\dot{e}].$ $\displaystyle 8\wedge 3,[x_{i}\dot{e}e];8\wedge
4,[x_{i}\dot{e}x_{m}x_{n}]\ (m>n);8\wedge 5,[x_{i}\dot{e}x_{m}x_{m}].$

All possible ambiguities $[w]$ of compositions of inclusion in $S$ are:

$\displaystyle 6\wedge 1,[x_{i}x_{j}\dot{x}_{t}]\ (i>j);\ 6\wedge
8,[x_{i}x_{j}\dot{e}]\ (i>j).$ $\displaystyle 7\wedge
1,[x_{i}x_{i}\dot{x}_{j}];\ 7\wedge 8,[x_{i}x_{i}\dot{e}].$

We just check the cases of intersection $1\wedge 4,4\wedge 4,6\wedge 4,6\wedge
8,8\wedge 4$ and of inclusion $6\wedge 1,6\wedge 8$. Others can be similarly
proved.

For $1\wedge 4$, $[w]=[x_{i}\dot{x}_{j}x_{m}x_{n}]\ (m>n)$, we have

$\displaystyle(1\wedge 4)_{[w]}$ $\displaystyle=$ $\displaystyle x_{j}\dashv
x_{i}\dashv x_{m}\dashv x_{n}-2a_{ij}e\dashv x_{m}\dashv x_{n}-x_{i}\vdash
x_{j}\dashv x_{n}\dashv x_{m}+2a_{mn}x_{i}\vdash x_{j}$ $\displaystyle=$
$\displaystyle x_{j}\dashv f_{x_{i}\dashv x_{m}x_{n}}-2a_{ij}f_{e\dashv
x_{m}x_{n}}-f_{x_{i}x_{j}}\dashv x_{n}\dashv x_{m}+2a_{mn}f_{x_{i}x_{j}}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w]).$

For $4\wedge 4$, there are two cases to consider:
$[w_{1}]=[\dot{y}x_{i}x_{j}x_{m}x_{n}]\ (i>j,m>n)$ and
$[w_{2}]=[\dot{y}x_{i}x_{j}x_{t}]\ (i>j>t)$. We have

$\displaystyle(4\wedge 4)_{[w_{1}]}$ $\displaystyle=$ $\displaystyle y\dashv
x_{j}\dashv x_{i}\dashv x_{m}\dashv x_{n}-2a_{ij}y\dashv x_{m}\dashv
x_{n}-y\dashv x_{i}\dashv x_{j}\dashv x_{n}\dashv x_{m}+2a_{mn}y\dashv
x_{i}\dashv x_{j}$ $\displaystyle=$ $\displaystyle y\dashv x_{j}\dashv
f_{x_{i}\dashv x_{m}x_{n}}-2a_{ij}f_{y\dashv x_{m}x_{n}}-f_{y\dashv
x_{i}x_{j}}\dashv x_{n}\dashv x_{m}+2a_{mn}f_{y\dashv x_{i}x_{j}}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w_{1}])\ \ \ \ \ \mbox{and }$
$\displaystyle(4\wedge 4)_{[w_{2}]}$ $\displaystyle=$ $\displaystyle y\dashv
x_{j}\dashv x_{i}\dashv x_{t}-2a_{ij}y\dashv x_{t}-y\dashv x_{i}\dashv
x_{t}\dashv x_{j}+2a_{jt}y\dashv x_{i}$ $\displaystyle=$ $\displaystyle
y\dashv f_{x_{j}\dashv x_{i}x_{t}}-f_{y\dashv x_{j}x_{t}}\dashv
x_{i}-f_{y\dashv x_{i}x_{t}}\dashv x_{j}+y\dashv f_{x_{t}\dashv x_{i}x_{j}}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w_{2}]).$

For $6\wedge 4$, $[w]=[x_{i}x_{j}\dot{y}x_{m}x_{n}]\ (i>j,m>n)$, we have

$\displaystyle(6\wedge 4)_{[w]}$ $\displaystyle=$ $\displaystyle x_{j}\vdash
x_{i}\vdash y\dashv x_{m}\dashv x_{n}-2a_{ij}y\dashv x_{m}\dashv
x_{n}-x_{i}\vdash x_{j}\vdash y\dashv x_{n}\dashv x_{m}+2a_{mn}x_{i}\vdash
x_{j}\vdash y$ $\displaystyle=$ $\displaystyle x_{j}\vdash x_{i}\vdash
f_{y\dashv x_{m}x_{n}}-2a_{ij}f_{y\dashv x_{m}x_{n}}-f_{x_{i}x_{j}\vdash
y}\dashv x_{n}\dashv x_{m}+2a_{mn}f_{x_{i}x_{j}\vdash y}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w]).$

For $6\wedge 8$, $[w]=[x_{i}x_{j}x_{t}\dot{e}]\ (i>j)$, we have

$\displaystyle(6\wedge 8)_{[w]}$ $\displaystyle=$ $\displaystyle x_{j}\vdash
x_{i}\vdash x_{t}\vdash e-2a_{ij}x_{t}\vdash e+x_{i}\vdash x_{j}\vdash e\dashv
x_{t}$ $\displaystyle=$ $\displaystyle x_{j}\vdash x_{i}\vdash
h_{x_{t}e}-2a_{ij}h_{x_{t}e}+f_{x_{i}x_{j}\vdash e}\dashv x_{t}$
$\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w]).$

For $8\wedge 4$, $[w]=[x_{i}\dot{e}x_{m}x_{n}]\ (m>n)$, we have

$\displaystyle(8\wedge 4)_{[w]}$ $\displaystyle=$ $\displaystyle-e\dashv
x_{i}\dashv x_{m}\dashv x_{n}-x_{i}\vdash e\dashv x_{n}\dashv
x_{m}+2a_{mn}x_{i}\vdash e$ $\displaystyle=$ $\displaystyle-e\dashv
f_{x_{i}\dashv x_{m}x_{n}}-h_{x_{i}e}\dashv x_{n}\dashv
x_{m}+2a_{mn}h_{x_{i}e}$ $\displaystyle\equiv$ $\displaystyle 0\ mod(S,[w]).$

Now, we check the compositions of inclusion $6\wedge 1$ and $6\wedge 8$.

For $6\wedge 1$, $[w]=[x_{i}x_{j}\dot{x}_{t}]\ (i>j)$, we have

$\displaystyle(6\wedge 1)_{[w]}$ $\displaystyle=$ $\displaystyle x_{j}\vdash
x_{i}\vdash x_{t}-2a_{ij}x_{t}-x_{i}\vdash x_{t}\dashv
x_{j}+2a_{jt}x_{i}\vdash e$ $\displaystyle=$ $\displaystyle x_{j}\vdash
f_{x_{i}x_{t}}-f_{x_{i}x_{t}}\dashv
x_{j}+2a_{jt}h_{x_{i}e}-f_{x_{j}x_{t}}\dashv x_{i}+f_{x_{t}\dashv
x_{i}x_{j}}+2a_{it}h_{x_{j}e}$ $\displaystyle\equiv$ $\displaystyle 0\
mod(S,[w]).$

For $6\wedge 8$, $[w]=[x_{i}x_{j}\dot{e}]\ (i>j)$, we have

$\displaystyle(6\wedge 8)_{[w]}$ $\displaystyle=$ $\displaystyle x_{j}\vdash
x_{i}\vdash e-2a_{ij}e+x_{i}\vdash e\dashv x_{j}$ $\displaystyle=$
$\displaystyle x_{j}\vdash h_{x_{i}e}+h_{x_{i}e}\dashv x_{j}+h_{x_{j}e}\dashv
x_{i}+f_{e\dashv x_{i}x_{j}}$ $\displaystyle\equiv$ $\displaystyle 0\
mod(S,[w]).$

Then all the compositions in $S$ are trivial. We have proved (i).

For (ii), since the mentioned set is just the set $Irr(S)$, by Theorem 3.9 the
result holds.

The proof is complete. $\square$

Remark: In the Theorem 4.10, if the matrix $(a_{ij})_{n\times n}=0$, then
Clifford dialgebra $C(n,f)$ has a Gröbner-Shirshov basis $S^{\prime}$ which
consists of the relations 1–7.

Acknowledgement: The authors would like to thank P.S. Kolesnikov who gives
some valuable remarks for this paper.

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