# Basic Commutators in $n$-Lie Algebras

Seyedeh Nafiseh Akbarossadat

Abstract. In this paper, we give the structure of free $n$-Lie algebras. Next,
we introduce basic commutators in $n$-Lie algebras and generalize the Witt
formula to calculate the number of the basic commutators. Also, we prove that
the set of all of the basic commutators of weight $w$ and length
$n+(w-2)(n-1)$ is a basis for $F^{w}$, where $F^{w}$ is the $w$th term of the
lower central series in the free $n$-Lie algebra $F$.

Keywords: $n$-Lie algebra, Basic commutators, Free $n$-Lie algebras, Basic
product, String.

MSC: 17B05, 17B10, 17B99.

##  1\. Introduction and History

Lie polynomials appeared at the end of the $19$th century and the beginning of
the $20$th century in the work of Campbell [3], Baker [2], and Hausdorff [4]
on exponential mappings in a Lie group, which has led to the so-called
Campbell–Baker–Hausdorff formula. Around $1930$, Witt introduced the Lie
algebra of Lie polynomials. He showed that the Lie algebra of Lie polynomials
is actually the free Lie algebra and that its enveloping algebra is the
associative algebra of noncommutative polynomials. He proved what is now
called the Poincaré–Birkhoff–Witt theorem and showed how a free Lie algebra is
related lower central series of the free group. About at the same time, Hall
and Magnus, with their commutator calculus, opened the way to the bases of the
free Lie algebra. For more details about a historical review of free Lie
algebras, we refer the reader to the reference [8] and its references.

The concept of basic commutators is defined in groups and Lie algebras, and
there is also a way to construct and identify them. Moreover, a formula for
calculating their number is obtained.

In 1962, Shirshov [10] gave a method that generalizes Hall’s method for
choosing a basis in a free Lie algebra.

Basic commutators are of particular importance in calculating the dimensions
of different spaces and are, therefore, highly regarded. Niroomand and Parvizi
[7] investigated some more results about the $2$-nilpotent multiplier
$\mathcal{M}^{(2)}(L)$ of a finite-dimensional nilpotent Lie algebra $L$, and
by using the Witt formula, they calculated its dimension. Moreover, Salemkar,
Edalatzadeh, and Araskhan [9] introduced the concept of $c$-nilpotent
multiplier $\mathcal{M}^{(c)}(L)$ of a finite-dimensional Lie algebra $L$ and
obtained some bounds for $\mathcal{M}^{(c)}(L)$ by using the Witt formula and
basic commutators.

In 1985, Filippov [5] introduced the concept of $n$-Lie algebras as an $n$-ary
multilinear and skew-symmetric operation $[x_{1},\ldots,x_{n}]$, which
satisfies the following generalized Jacobi identity:

$[[x_{1},\ldots,x_{n}],y_{2},\ldots,y_{n}]=\sum_{i=1}^{n}[x_{1},\ldots,[x_{i},y_{2},\ldots,y_{n}],\ldots,x_{n}].$

Clearly, such an algebra becomes an ordinary Lie algebra when $n=2$.

In this paper, we are going to define basic commutators in $n$-Lie algebras
and give a formula to count the number of them.

##  2\. Free $n$-Lie algebras

For the first time, Akbarossadat and Saeedi [1] introduced the concept of free
$n$-Lie algebras. In the following subsection, we will review the definition
of this concept and some of its properties.

#### Free $n$-Lie Algebras

Given a set $X$, a free $n$-Lie algebra on $X$ over a field $\mathbb{F}$ is an
$n$-Lie algebra $L$ over $\mathbb{F}$, together with a mapping
$i:X\longrightarrow L$, with the following universal property:
For each $n$-Lie algebra, such as $M$, and each mapping $f:X\longrightarrow
M$, there exists a unique $n$-Lie algebra homomorphism $F:L\longrightarrow M$
such that $f:F\circ i$.

A standard argument shows that a free $n$-Lie algebra on $X$ is necessarily
unique, up to $n$-Lie algebra isomorphism. Also, its existence is shown as
follows:

Let $X$ be a set. We construct a Lie algebra generated by $X$, satisfying no
relations other than these:

(2.1) $\displaystyle[x,x,\ldots,x]=0,\qquad\qquad\text{for all }x\in X,$ (2.2)
$\displaystyle\sum_{i=1}^{n}[x_{1},\ldots,x_{i-1},[x_{i},x^{\prime}_{1},\ldots,x^{\prime}_{n-1}],x_{i+1},\ldots,x_{n}]=0,$
$\displaystyle\qquad\qquad\text{for all }x_{i},x_{j}^{\prime}\in X,\quad 1\leq
i\leq n,~{}1\leq j\leq n-1.$

Let $M(X)$ be the set inductively defined as follows:

  1. (1)

$X\subseteq M(X)$.

  2. (2)

If $x_{1},\ldots,x_{n}\in M(X)$, then also $(x_{1},\ldots,x_{n})\in M(X)$.

The set $M(X)$ is called the free magma on $X$. It consists of all bracketed
expressions on the elements of $X$. On $M(X)$ we define an $n$-ary operation

$\begin{array}[]{ccc}M(X)\times\cdots\times M(X)&\longrightarrow&M(X)\\\
x_{1}\cdot~{}\ldots~{}\cdot
x_{n}&\longmapsto&(x_{1},\ldots,x_{n}).\end{array}$

For $m\subseteq M(X)$, we define its degree recursively: $deg(m)=1$ if $m\in
X$ and $deg(m)=\sum_{i=1}^{n}m_{i}$ if $m=(m_{1},\ldots,m_{n})$. So the degree
of an element $m\in M(X)$ is just the number of elements of $X$ that occur in
$m$ (counted with multiplicities). For an integer $d\geq 1$, we set $M_{d}(X)$
as the subset of $M(X)$ consisting of all $m\in M(X)$ of degree $d$. Then

$M(X)=\bigcup_{d\geq 1}M_{d}(X).$

Let $\mathbb{F}$ be a field and let $A(X)$ be the vector space over
$\mathbb{F}$ spanned by $M(X)$. If we extend the $n$-ary operation on $M(X)$
$n$-linearly to $A(X)$, then $A(X)$ becomes an (nonassociative) algebra; it is
called the free $n$-algebra over $\mathbb{F}$ on $X$. Let $f\in A(X)$; if also
$f\in M(X)$, then $f$ is said to be a monomial. For $f\in A(X)$, we define
$deg(f)$ to be the maximum of the degrees of $m$, where $m$ runs over all
$m\in M(X)$ that occur in $f$ with nonzero coefficient.

Let $I_{0}$ be the ideal of $A(X)$ generated by all elements

$\displaystyle(m,m,\ldots,m),\qquad\text{for all }m\in M(X),$
$\displaystyle(m_{1},\ldots,m_{i},\ldots,m_{j},\ldots,m_{n})+(m_{1},\ldots,m_{j},\ldots,m_{i},\ldots,m_{n})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\text{for all }m_{i}\in
M(X),~{}\text{for all }1\leq i,j\leq n$
$\displaystyle\sum_{i=1}^{n}(m_{1},\ldots,m_{i-1},(m_{i},m^{\prime}_{1},\ldots,m^{\prime}_{n-1}),m_{i+1},\ldots,m_{n}),\qquad\text{for
all }m_{i},m^{\prime}_{i}\in M(X).$

Set $L(X)=A(X)/I_{o}$. Let $\mathcal{B}$ be a basis of $L(X)$ consisting of
(images of) elements of $M(X)$. Then it is immediate that we have

$(m,m,\ldots,m)=0,$
$(m_{1},\ldots,m_{i},\ldots,m_{j},\ldots,m_{n})+(m_{1},\ldots,m_{j},\ldots,m_{i},\ldots,m_{n})=0,$

and

$\sum_{i=1}^{n}(m_{1},\ldots,m_{i-1},(m_{i},m^{\prime}_{1},\ldots,m^{\prime}_{n-1}),m_{i+1},\ldots,m_{n})=0,$

for all $m,m_{i},m_{i}^{\prime}\in\mathcal{B}$. It is easy to check that the
relations (2.1) and (2.2) hold for all elements of $L(X)$, so $L(X)$ is an
$n$-Lie algebra. Therefore we will use a bracket to denote the product in
$L(X)$ and say $L(X)$ is the free Lie algebra on $X$.

In other words, an $n$-Lie algebra $L$ is free on $X$ if and only if it is
generated by all possible $n$-Lie bracketings of all elements of $X$. The only
existing and possible relations among these bracketing generators are the
consequence of $n$-linearity bracketings, generalized Jacobi identity, and

$[x_{1},x_{2},\dots,x_{i-1},x_{i},\dots,x_{j-1},x_{j},\dots,x_{n}]=-[x_{1},x_{2},\dots,x_{i-1},x_{j},\dots,x_{j-1},x_{i},\dots,x_{n}],$

for all $1\leq i,j\leq n$.

The following lemma is a main tool for the proof of main theorem of this
section.

######  Lemma 2.1.

Let $F$ be a free $n$-Lie algebra and let $i,j\in\mathbb{N}$. Then
$[F^{i},F^{j},\underbrace{F,\dots,F}_{n-times}]\subseteq F^{i+j}$.

###### Proof.

We prove this by induction on $i$ and $j$. For each $i\geq 1$ and $j=1$, by
the definition of lower central series terms, we have

$[F^{i},F^{j},F,\dots,F]=[F^{i},F^{1},F,\dots,F]=[F^{i},F,F,\dots,F]=F^{i+1}.$

Now, let $i$ be a fixed and arbitrary natural number and let the statement be
true for $j-1$; that is,

$[F^{i},F^{j-1},F,\dots,F]\subseteq F^{i+1}\qquad({\text{induction
hypothesis}}).$

By the definition of lower central series terms $F^{i}$’s, we know that
$F^{i}=[F^{i-1},F,F,\dots,F]$. Hence every element of $F^{i}$ such an element
$a$ is $a=[[a_{0},f_{2}^{\prime},\dots,f_{n}^{\prime}],f_{2},\dots,f_{n}]$,
where $a_{0}\in F^{i-1}$ and $f_{s},f_{r}^{\prime}\in F$, for each $2\leq
r,s\leq n$. Now, let $x=[a,b,f_{2},\dots,f_{n}]$ be an arbitrary element of
$[F^{i},F^{j},\underbrace{F,\dots,F}_{n-times}]$, where
$a=[a_{0},f_{2}^{\prime},\dots,f_{n}^{\prime}]\in F^{i}$, $b\in F^{j}$, and
$f_{s}^{\prime}\in F$, for all $2\leq s\leq n$. By the Jacobi identity, we
have

$\displaystyle x=[a,b,f_{2},\dots,f_{n}]$
$\displaystyle=[[a_{0},f_{2}^{\prime},\dots,f_{n}^{\prime}],b,f_{2},\dots,f_{n}]$
$\displaystyle=[[a_{0},b,f_{2},\dots,f_{n}],f_{2}^{\prime},\dots,f_{n}^{\prime}]+[a_{0},[f_{2}^{\prime},b,f_{2},\dots,f_{n}],f_{3}^{\prime},\dots,f_{n}^{\prime}]$
$\displaystyle\quad+\dots+[a_{0},f_{2}^{\prime},\dots,f_{n-1}^{\prime},[f_{n}^{\prime},b,f_{2},\dots,f_{n}]]$
$\displaystyle=-\underbrace{[\underbrace{[b,a_{0},f_{2},\dots,f_{n}]}_{\in
F^{j+i-1}},f_{2}^{\prime},\dots,f_{n}^{\prime}]}_{\in
F^{j+i}}-\underbrace{[\underbrace{[f_{2}^{\prime},b,f_{2},\dots,f_{n}]}_{\in
F^{j+i-1}},a_{0},f_{3}^{\prime},\dots,f_{n}^{\prime}]}_{\in F^{j+i}}$
$\displaystyle\quad+\dots+\underbrace{[a_{0},f_{2}^{\prime},\dots,f_{n-1}^{\prime},\underbrace{[f_{n}^{\prime},b,f_{2},\dots,f_{n}]}_{\in
F^{j+i-1}}]}_{\in F^{i+j}}$
$\displaystyle=\sum_{s=2}^{n}[a_{0},f_{2}^{\prime},\dots,[f_{s}^{\prime},b,f_{2},\dots,f_{n}],f_{s+1}^{\prime},\dots,f_{n}^{\prime}]\in
F^{i+j}.$

Thus $[F^{i},F^{j},\underbrace{F,\dots,F}_{n-times}]\subseteq F^{i+j}$. ∎

An ideal $I\unlhd L$ is called abelian, if $[I,I,L,\dots,L]=0$. Using this
fact, we prove the following theorem, which is an important result.

######  Theorem 2.2.

Let $F$ be a free $n$-Lie algebra and let $F^{i}$ be the $i$th term of the
lower central series of $F$, for each $i\in\mathbb{N}$. Then
$\dfrac{F^{i}}{F^{i+c}}$ is abelian, where $c=0,1,\ldots,i$.

###### Proof.

It is obvious for $c=0$, since
$\dfrac{F^{i}}{F^{i}}=0_{\frac{F^{i}}{F^{i}}}=F^{i}$. Let $c=1$. Since
$[F^{i},F^{i},F,\dots,F]\subseteq[F^{i},F,F,\dots,F]=F^{i+1}$, so
$Z\left(\dfrac{F^{i}}{F^{i+1}}\right)=\dfrac{F^{i}}{F^{i+1}}$, and hence
$\dfrac{F^{i}}{F^{i+1}}$ is abelian.

Now suppose that $0\leq c\leq i$. We show that $\dfrac{F^{i}}{F^{i+c}}$ is
abelian. In this regard, it is enough to show that $\dfrac{F^{i}}{F^{i+c}}$ is
an abelian ideal of $\dfrac{F}{F^{i+c}}$, which is equivalent to

$[F^{i},F^{i},F,\dots,F]+F^{i+c}=0_{\frac{F^{i}}{F^{i+c}}}=F^{i+c},\qquad{\text{or}}\qquad[F^{i},F^{i},F,\dots,F]\subseteq
F^{i+c}.$

By Lemma 2.1, we know that $[F^{i},F^{i},F,\dots,F]\subseteq F^{2i}$. Since
$0\leq c\leq i$, so $c+i\leq 2i$, and thus $F^{2i}\subseteq F^{i+c}$.
Therefore, $[F^{i},F^{i},F,\dots,F]\subseteq F^{i+c}$. ∎

##  3\. An introduction to basic commutators

In this section, our aim is to define basic commutators of $n$-Lie algebras.
For this, we need to introduce some concepts.

### Strings and collected parts

Let $X$ be a given set of noncommuting variables (also called letters). We
consider formal words or strings $b_{1}b_{2}\ldots b_{n}$, where each $b_{i}$
is one of the letters $x_{1},x_{2},\ldots,x_{r},\ldots$. In other words, the
word is a finite sequence of elements of $X$ or, equivalently, a
noncommutative monomial. We include the empty word denoted $1$. We denote by
$X^{*}$ the set of all words on $X$. With the product defined by the
concatenation of words, $X^{*}$ becomes a monoid, which is the free monoid on
$X$. Indeed, for any mapping $f:X\longrightarrow M$, where $M$ is any monoid,
there is a unique homomorphism of monoids $F:X^{*}\longrightarrow M$ such that
$f=F\circ i$, where $i$ is the natural injection $X\longrightarrow X^{*}$.

A noncommutative polynomial is a linear combination of words with coefficients
in the field $\mathbb{F}$. The set of all these noncommutative polynomials
becomes an algebra over $\mathbb{F}$ with the product inherited from the free
monoid $X^{*}$, which is denoted by $\mathbb{F}\langle X\rangle$ and is the
associative algebra $X$. Indeed, for any mapping $f:X\longrightarrow A$, where
$A$ is any $\mathbb{F}$-algebra, there is a unique homomorphism of algebras as
$F:\mathbb{F}\langle X\rangle\longrightarrow A$, such that $f=F\circ i$, where
$i$ is the natural injection from $X$ into $\mathbb{F}\langle X\rangle$.

Since $\mathbb{F}\langle X\rangle$ is an algebra, so from the product, we can
define a bracket on it and turn it into a Lie algebra. For this, let $P$ and
$Q$ be two polynomials in $\mathbb{F}\langle X\rangle$. Define
$[-,-]:\mathbb{F}\langle X\rangle\times\mathbb{F}\langle
X\rangle\longrightarrow\mathbb{F}\langle X\rangle$ by $[P,Q]:=PQ-QP$. It is
easy to check the bilinear and skew-symmetric properties and Jacobi identity.
Hence this defines a structure of Lie algebra on $\mathbb{F}\langle X\rangle$.

Now, let $\mathcal{B}_{\mathbb{F}}(X)$ denote the smallest Lie subalgebra of
$\mathbb{F}\langle X\rangle$ that contains each letter in $X$. It is easy to
prove that $\mathcal{B}_{\mathbb{F}}(X)$ is a free Lie algebra. In what
follows, we show that every element of the Lie subalgebra
$\mathcal{B}_{\mathbb{F}}(X)$ is a Lie polynomial and that the set of all
basic commutators on $X$ is a basis for $\mathcal{B}_{\mathbb{F}}(X)$. Hence
the dimension of it is the number of all basic commutators on $X$. Indeed, for
this purpose, we first need to introduce the concept of basic commutators and
some of their properties. In what follows, we discuss this topic.

Now, we introduce formal commutators $c_{i}$ and weights $\omega(c_{i})$ by
the rules:

1) $c_{i}=x_{i}$, $i=1,\ldots,r$, are the commutators of weight 1; that is,
$\omega(x_{i})=1$.

2) If $c_{i_{1}},\ldots,c_{i_{n}}$ are commutators, then
$c_{k}=(c_{i_{1}},\ldots,c_{i_{n}})$ is a commutator and
$\omega(c_{k})=\sum_{j=1}^{n}\omega(c_{i_{j}})$.

Note that these definitions yield only a finite number of commutators of any
given weight. We shall order the commutators by their subscripts, counting
$c_{i}=x_{i}$, $i=1,\ldots,r$, and listing in order of weight, but giving an
arbitrary ordering to commutators of the same weight.

A string $c_{i_{1}}\ldots c_{i_{r}}$ of commutators is said to be in collected
form if $i_{1}\leq\cdots\leq i_{r}$, that is, if the commutators are in order
read from left to right. An arbitrary string of commutators $c_{i_{1}}\ldots
c_{i_{r}}c_{i_{r+1}}\ldots c_{i_{s}}$ will, in general, has a collected part
$c_{i_{1}}\ldots c_{i_{r}}$ if $i_{1}\leq\cdots\leq i_{r}$ and if $i_{r}\leq
i_{j}$, $j=r+1,\ldots,s$. Moreover, this sting will have an uncollected part
$c_{i_{r+l}}\ldots c_{i_{s}}$, where $i_{r+1}$ is not the least of $i_{j}$,
$j=r+1,\ldots,s$. The collected part of a string $c_{i_{1}}\ldots c_{i_{s}}$
will be void unless $i_{1}$ is the least of the subscripts.

### Collecting process in strings

In this part, we define a collecting process for strings of commutators. If
$c_{u}$ is the earliest commutator in the uncollected part and if
$c_{i_{j}}=c_{u}$ is the leftmost uncollected $c_{u}$, then we replace
$c_{i_{1}}\ldots c_{i_{m}}\ldots c_{i_{j}-1}c_{i_{j}}\ldots c_{i_{n}}$ with
$-c_{i_{1}}\ldots c_{i_{m}}\ldots c_{i_{j}}c_{i_{j}-1}\ldots c_{i_{n}}$. This
has the effect of moving $c_{i_{j}}$ to the left as the formal coefficient of
$(-1)$. Thus $c_{i_{j}}$ is still the earliest commutator in the uncollected
part. Alter enough steps, $c_{i_{j}}$ will be moved to the $(m+1)$st position
and will become a part of the collection part. Note that in every step of
moving $c_{i_{j}}$ to the left, the coefficient of $(-1)$ is created that can
be abridged.

### Basic commutators and basic product

Suppose that sequences of basic commutators $c_{1},c_{2},\dots$ formed from
the generators $x_{1},x_{2},\dots,x_{d}$ are given. We call a product of basic
commutators $c_{i_{1}}c_{i_{2}}\dots c_{i_{s}}$, a basic product if it is in
collected order; that is, $i_{1}\leq i_{2}\leq\dots\leq i_{s}$. The number of
all basic products of weight $2$ and length $n$ formed from generators
$x_{1},x_{2},\dots,x_{d}$, is $d^{n}$.

In applying the collecting process to a positive word, not all commutators
will arise. Thus $(x_{n},\dots,x_{3},x_{2},x_{1})$ may arise but not
$(x_{n},\dots,x_{1},x_{2},x_{3})$, since $x_{3}$ is collected before $x_{1}$.
The commutators that may actually arise are called basic commutators. We give
a formal definition of the basic commutators for $n$-Lie algebras generated by
$x_{1},\dots,x_{d}$.

### Brief introduction of basic commutators in $n$-Lie algebras

In the previous sections, we completely and precisely define the basic
commutators by using the words. Indeed in this section, we are going to
briefly introduce the basic commutators for $n$-Lie algebra $L$ (especially)
with the free representation $\frac{F}{R}$ over the field $\mathbb{F}$, in the
bracket form. In other words, we introduce a method to determine the basic
commutators in $n$-Lie algebras. Note that two properties of $n$-Lie algebras,
as “skew-symmetric” and “Jacobian identity”, are important and necessary to
determine the basic commutators.

In general, it can be said that the set of all basic commutators $A$ is the
smallest subset of the set of all commutators of $L$, named $W$, such that $A$
can be produced $W$.

The basic commutators are characterized by two concepts as “weight” and
“length”. In the case when $n=2$, these two concepts are the same. The basic
commutator $0\neq c\in F$ has weight $w_{c}$, if $c\in F^{w_{c}}$ and
$c\not\in F^{w_{c}+1}$, where $F^{w_{c}}$ is $w_{c}$th terms of lower central
series.Also, the length of $c$ is the number of its components that are in
$X$, and we denote it by $m_{c}$ and calculate it as follows:

$m_{c}=n+(w_{c}-2)(n-1)=n+w_{c}n-2n-w_{c}+2=(w_{c}-1)n-(w_{c}-2).$

In the general case, weight and length are denoted by $w$ and $m$,
respectively. Thus, the general form of basic commutators with different
weights and lengths is shown in Table 1.

Table 1. Generally form of basic commutators Domain | weight($w$) | length ($m$) | generally form
---|---|---|---
$X$ | 1 | 1 | $x_{i}$
$F^{2}$ | 2 | $n$ | $[x_{i_{1}},\dots,x_{i_{n}}]$
$F^{3}$ | 3 | $2n-1$ | $[[x_{i_{1}},\dots,x_{i_{n}}],x_{j_{2}},\dots,x_{j_{n}}]$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$

With the above discussion, we have Table 2.

Table 2. The length of the basic commutators of weight $w$ in $n$-Lie algebras $w$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\cdots$
---|---|---|---|---|---|---|---|---|---
$n=2$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\cdots$
$n=3$ | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | $\cdots$
$n=4$ | 1 | 4 | 7 | 10 | 13 | 16 | 19 | 22 | $\cdots$
$n=5$ | 1 | 5 | 9 | 13 | 17 | 21 | 25 | 29 | $\cdots$
$n=6$ | 1 | 6 | 11 | 16 | 21 | 26 | 31 | 36 | $\cdots$
$n=7$ | 1 | 7 | 13 | 19 | 25 | 31 | 37 | 43 | $\cdots$
$n=8$ | 1 | 8 | 15 | 22 | 29 | 36 | 43 | 50 | $\cdots$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$

If in any $n$-Lie algebras (for each $n\in\mathbb{N}$), we denote the number
of length of the basic commutator of weight $w_{0}$ by $m_{w_{0}}^{n}$, then
we can obtain the following results:

  * •

$m_{w_{0}}^{n}=m_{n}^{w_{0}}$, $n\geq 2$, and $w_{0}\geq 2$.

  * •

$m_{w_{0}}^{3}=m_{w_{0}}^{2}+m_{w_{0}-1}^{2}$.

Let $X=\\{x_{1},x_{2},\ldots,x_{n},\ldots\\}$ be a basis set of $n$-Lie
algebra $L$. We define the relation ”$<$” on $X$ as
$x_{1}<x_{2}<\cdots<x_{n}<\cdots$. In what follows, we define the basic
commutators in $L$ as follows:

  1. (1)

Every basic element $c_{i}^{1}=x_{i}$ in $X$ is the basic commutator of weight
1.

  2. (2)

The element $c_{s}^{2}=[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}]$ (where
$x_{i_{j}}\in X$, for all $j=1,\ldots,n$) is a basic commutator of length $n$
and weight $2$. It belongs to $F^{2}$, if
$x_{i_{1}}>x_{i_{2}}>\cdots>x_{i_{n}}$; in other words,
$i_{1}>i_{2}>\cdots>i_{n}$. We can compare every two basic commutators
$c_{s}^{2}$ and $c_{k}^{2}$ with weight $2$ and length $n$. Suppose that
$c_{s}^{2}=[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}]$ and that
$c_{k}^{2}=[x^{\prime}_{j_{1}},x^{\prime}_{j_{2}},\ldots,x^{\prime}_{j_{n}}]$,
and let $r_{0}$ be the first index (in moving from right to left) of
$x_{i_{r}}$’s and $x^{\prime}_{j_{r}}$’s which $x_{i_{r_{0}}}\neq
x^{\prime}_{j_{r_{0}}}$. If $x_{i_{r_{0}}}>x^{\prime}_{j_{r_{0}}}$ or,
equivalently, $i_{r_{0}}>j_{r_{0}}$, then we say $c_{s}^{2}>c_{k}^{2}$, and
otherwise, we say $c_{s}^{2}<c_{k}^{2}$.

  3. (3)

Let $c_{t}^{w}=[c_{j_{1}}^{w_{1}},\ldots,c_{j_{n}}^{w_{n}}]$, where all of
$c_{j_{1}}^{w_{1}},\ldots,c_{j_{n}}^{w_{n}}$ are basic commutators of weights
$w_{1},\dots,w_{n}$ and lengths $n+(w_{1}-2)(n-1),\ldots,n+(w_{n}-2)(n-1)$,
respectively. Then $c_{t}^{w}$ is called a basic commutator if the following
conditions hold:

     * •

$w>w_{1}\geq w_{2}\geq\cdots\geq w_{n}$.

     * •

Whenever $w_{s}=w_{s+1}$, then $c_{j_{s}}^{w_{s}}>c_{j_{s+1}}^{w_{s+1}}$,
equivalently, $j_{s}>j_{s+1}$.

     * •

Whenever $w_{s}>w_{s+1}$ and
$c_{j_{s}}^{w_{s}}=[c_{j_{s_{1}}}^{w^{\prime}_{1}},\ldots,c_{j_{s_{n}}}^{w^{\prime}_{n}}]$,
then $c_{j_{s_{n}}}^{w^{\prime}_{n}}\leq
c_{j_{n}}^{w_{n}}<\cdots<c_{j_{s+2}}^{w_{s+2}}<c_{j_{s+1}}^{w_{s+1}}$.

In this case, $c_{t}^{w}$ is an basic element of $F^{w}$ of weight $w$ and
length $n+(w-2)(n-1)$.

Note that every basic commutator of weight $w$ is smaller than every basic
commutator of weight larger than $w$. Hence
$c_{i}^{1}<c_{j}^{2}<c_{k}^{3}<\cdots$, for all indices $i,j,k,\ldots$ .

### Collecting process in commutators of $n$-Lie algebras

It is obvious that there exist many commutators in generated free $n$-Lie
algebra $F$ over $X$ such that they does not have a basic form. Indeed it is
possible to choose an equivalent basic form for them and characterize them by
basic commutators. For this, we have to make the necessary changes to them by
the properties of $n$-Lie algebras.

Since $F$ is an $n$-Lie algebra, so every bracket with at least two same
components is equal to zero. In fact, there is no nonzero bracket such that it
has at least two same components in $F$, and all of the brackets in $F$ have
$n$ different components.

Suppose that
$c=[c_{j_{1}}^{w_{1}},c_{j_{2}}^{w_{2}},\dots,c_{j_{r}}^{w_{r}},c_{j_{r+1}}^{w_{r+1}},\dots,c_{j_{n}}^{w_{n}}]$
is a nonbasic commutator in $F$. Then there are the following two cases:

  1. (A)

Let $c_{j_{r}}^{w_{r}}<c_{j_{r+1}}^{w_{r+1}}$. Then it is enough to write it
as follows:

$c=[c_{j_{1}}^{w_{1}},c_{j_{2}}^{w_{2}},\dots,c_{j_{r}}^{w_{r}},c_{j_{r+1}}^{w_{r+1}},\dots,c_{j_{n}}^{w_{n}}]=-[c_{j_{1}}^{w_{1}},c_{j_{2}}^{w_{2}},\dots,c_{j_{r+1}}^{w_{r+1}},c_{j_{r}}^{w_{r}},\dots,c_{j_{n}}^{w_{n}}].$

  2. (B)

Let
$c_{j_{1}}^{w_{1}}=[{c^{\prime}}_{r_{1}}^{w^{\prime}_{1}},{c^{\prime}}_{r_{2}}^{w^{\prime}_{2}},\dots,{c^{\prime}}_{r_{n}}^{w^{\prime}_{n}}]$
be in the basic form. Then $c$ reads as follows:

$c=[c_{j_{1}}^{w_{1}},c_{j_{2}}^{w_{2}},\dots,c_{j_{n}}^{w_{n}}]=[[{c^{\prime}}_{r_{1}}^{w^{\prime}_{1}},{c^{\prime}}_{r_{2}}^{w^{\prime}_{2}},\dots,{c^{\prime}}_{r_{n}}^{w^{\prime}_{n}}],c_{j_{2}}^{w_{2}},\dots,c_{j_{n}}^{w_{n}}].$

We know that $c$ is a basic commutator if

${c^{\prime}}_{r_{n}}^{w^{\prime}_{n}}\leq
c_{j_{n}}^{w_{n}}<c_{j_{n-1}}^{w_{n-1}}<\dots<c_{j_{3}}^{w_{3}}<c_{j_{2}}^{w_{2}}.$

Otherwise, $c$ is not in a basic form. Thus we replace $c$ by using the Jacobi
identity and case (A).

Now, we state the theorems that are among the main results and our goals in
this paper.

######  Theorem 3.1.

Let $F$ be a free $n$-Lie algebra with the ordered basis set
$X=\\{x_{1},x_{2},x_{3},\dots\\}$, where $x_{1}<x_{2}<x_{3}<\dots$. The set of
all defined basic commutators of weight $w$ (denoted by $B(w)$) on $X$ is
linearly independent.

###### Proof.

It is obvious by the collecting process in commutators of $n$-Lie algebras.
Since $F$ is the generated free $n$-Lie algebra on $X$, then it has no other
relations except $n$-linearity, skew-symmetric properties, and Jacobi
identity. Therefore, all of the brackets in $F$ apply to these three
properties. On the other hand, in the process of determining and identifying
the basic commutators, we remove the commutators that are linearly dependent
under these three properties ($n$-linearity, skew-symmetric properties, and
Jacobi identity), and we separate only its linearly independent elements.
Hence, all of the basic commutators of arbitrary and fixed weight $w$ are
linearly independent. ∎

The following theorem is the main result of this section, which is similarly
discussed in group theory and Lie algebras.

######  Theorem 3.2.

Let $F$ be a free $n$-Lie algebra. Then $\dfrac{F^{i}}{F^{i+c}}$ is abelian
$n$-Lie algebras whose basis is the set of all basic commutators of weights
$i,i+1,i+2,i+3,\dots,i+c-1$ and lengths
$n+(i-2)(n-1),n+(i-1)(n-1),\ldots,n+(i+c-3)(n-1)$.

###### Proof.

This theorem is proved similar to the Lie algebra case. ∎

##  4\. Counting of basic commutators

It is necessary to note that counting of basic commutators on every arbitrary
basis set $X$ is not possible, in general. In fact, although it is possible to
define the concept of basic commutators for any countable set, counting their
number is only possible if the set $X$ is finite. The following formula, known
as the Witt formula, is provided in group theory and Lie algebras for counting
of all of the basic commutators:

(4.3) $l_{d}(w)=\dfrac{1}{w}\sum_{r|w}\mu(r)d^{\frac{w}{r}},$

where $d$ is the number of generators of given group/Lie algebra (the number
of members of $X$), $\mu$ is the Möbius function, which is defined for
positive integers by the rule $\mu(1)=1$, and for
$w=p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{s}^{a_{s}}$ that
$p_{1},p_{2},\dots,p_{s}$ are distinct primes numbers, by $\mu(p_{1}p_{2}\dots
p_{s})=(-1)^{s}$ and $\mu(p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{s}^{a_{s}})=0$,
if any $a_{i}>1$.

In group theory, the strategy of the proof of the above formula is to remove
nonbasic commutators from all the commutators. That is, it first counts all
possible commutators on $X$ and then removes nonbasic members and their
periods. Indeed in $n$-Lie algebras, with this method, only a bound can be
found for the number of basic commutators.

Now, let $X=\\{x_{1},x_{2},\dots,x_{d}\\}$ with the ordered relation
$x_{d}>x_{d-1}>\dots>x_{2}>x_{1}$. We are going to determine the number of all
of basic commutators of weight $w$ and length $m=n+(w-2)(n-1)$. Now, consider
the following string of circles:

$\underbrace{\bigcirc\bigcirc\bigcirc\dots\bigcirc\bigcirc\bigcirc}_{m-times}$

We want to put $x_{i}$’s in each circle such that the final result is a basic
commutator. First, we consider all of the cases for putting them. It is easy
to see that the number of these cases is $d^{m}$ (every circle has $d$
choices). Obviously, all of them are not basic, and hence we remove nonbasic
cases. For this, we remove duplicate and linear dependent commutators by using
two properties, skew-symmetric and Jacobi identity of $n$-Lie algebras. This
method is similar to the method used in the group theory (see [6]), and also,
Shirshov [10] provided it for Lie algebras. We know that

$m=n+(w-2)(n-1)=n+wn-2n-w+2=(w-1)n-(w-2),$

and so $n$ is a divisor of $m$ if $w\stackrel{{\scriptstyle n}}{{\equiv}}2$;
in other words, $w=qn+2$, for some integer positive $q$. Thus

$m=(w-1)n-(qn+2-2)=(w-1)n+qn=(w+q-1)n,$

and hence $\dfrac{m}{n}=w+q-1$ is integer. Therefore, in general, $n\not|m$,
but $n-1|m-1$, because

$m=n+(w-2)(n-1)=n+wn-2n-w+2,$

and so

$\displaystyle m-1=n+wn-2n-w+2-1$
$\displaystyle=\underline{n-2n+1}+\underline{wn-w}$
$\displaystyle=-(n-1)+w(n-1)$ $\displaystyle=(w-1)(n-1).$

We call a word $a_{1}a_{2}\dots a_{s}$ circular if $a_{1}$ is regarded as
$a_{s}$, and also, $a_{1}a_{2}\dots a_{s}$, $a_{2}a_{3}\dots a_{s}a_{1}$,
$a_{3}\dots a_{n-1}a_{n}a_{2}$, $\dots$, $a_{n}a_{1}a_{2}\dots a_{n-1}$ are
regarded as the same word. A circular word $C$ of weight $w$ and length
$m=n+(w-2)(n-1)$ may conceivably be given by repeating a segment of $q$
letters $m/q$ times, where $q$ is some divisor of $m$. We say that $C$ is of
period $q$ if we have

$\underbrace{(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})\dots(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})}_{m-{\mbox{circles
and}}~{}m/q-{\mbox{parts}}}.$

So by the above, circular commutators must be removed from all of the
generated commutators on $X$. Also, in weights that smaller than $w$, we have

$\underbrace{\underbrace{\bigotimes\bigotimes\bigotimes\dots\bigotimes}_{i-times}(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})\dots(\underbrace{\bigcirc\bigcirc\dots\bigcirc}_{q-times})}_{(m-i)-{\mbox{circles
and}}~{}(m-i)/q-{\mbox{parts}}}$ (4.4) $\displaystyle l_{d}^{n}(w)$
$\displaystyle\leq\dfrac{1}{m}\left(\sum_{r|m}\mu(r)d^{\frac{m}{r}}\right).$

Note that the equality holds if and only if $n=2$.

In the next section, we introduce the recursive formula.

##  5\. New method of counting of basic commutators

In the previous section, we provide the generalized Witt formula for counting
basic commutators in $n$-Lie algebras with finite basis set $X$. Indeed, it
could not discuss the relationship between the numbers of basic commutators of
different weights. Now, in this section, we introduce some formulas in special
cases with the new method and show their relationship.

First, note the following theorem in which we give the number of all of the
basic commutators of weight 2, in general $n$-Lie algebras with $d$
generators.

######  Theorem 5.1.

Let $L$ be an $n$-Lie algebra of dimension $d$. Then

$l_{d}^{n}(2)={{d}\choose{n}}.$

###### Proof.

Let $L$ be an $n$-Lie algebra of dimension $d$ with the basis
$X=\\{x_{1},x_{2},\dots,x_{d}\\}$ and put $w=2$. We are going to count the
number of all basic commutators of weight $2$ on $d$ letters. It is clear that
the number of all commutators of weight $2$ is equal to $d^{n}=n^{n}$. By the
skew-symmetric property in $n$-Lie algebras, we have

$[x_{i_{1}},x_{i_{2}},\dots,x_{i_{j}},\dots,x_{i_{k}},\dots,x_{i_{n}}]=-[x_{i_{1}},x_{i_{2}},\dots,x_{i_{k}},\dots,x_{i_{j}},\dots,x_{i_{n}}],$

and hence
$[x_{i_{1}},x_{i_{2}},\dots,x_{i_{j-1}},x,x_{i_{j+1}},\dots,x_{i_{k-1}},x,x_{i_{k+1}},\dots,x_{i_{n}}]=0$.
Thus just independent and nonzero commutators of weight $2$ are as
$[x_{i_{1}},x_{i_{2}},\dots,x_{i_{n}}]$, where all of $x_{i_{j}}$’s are
distinct. So $l_{d}^{n}(2)={{d}\choose{n}}$. ∎

Remark. Since every $n$-Lie algebra has the skew-symmetric property, it is not
possible to exist at least two same components. Hence the number of basic
commutators of weights greater than or equal to $2$ is equal to zero, if
$d<n$.

######  Corollary 5.2.

If $n=d$, then the following properties hold:

  1. (1)

$l_{n}^{n}(1)=n$.

  2. (2)

$l_{n}^{n}(2)={{n}\choose{n}}=1$.

  3. (3)

$l_{n}^{n}(3)={{n}\choose{n-1}}=n$.

  4. (4)

$l_{n}^{n}(4)={{n}\choose{2}}+n$.

  5. (5)

$l_{n}^{n}(5)={{n}\choose{3}}+2{{n}\choose{2}}+n$.

  6. (6)

$l_{n}^{n}(6)={{n}\choose{4}}+3{{n}\choose{3}}+3{{n}\choose{2}}+n$.

  7. (7)

$l_{n}^{n}(7)={{n}\choose{5}}+4{{n}\choose{4}}+6{{n}\choose{3}}+4{{n}\choose{2}}+n$.

  8. (8)

$l_{n}^{n}(8)={{n}\choose{6}}+5{{n}\choose{5}}+10{{n}\choose{4}}+10{{n}\choose{3}}+5{{n}\choose{2}}+n$.

  9. (9)

$l_{n}^{n}(9)={{n}\choose{7}}+6{{n}\choose{6}}+15{{n}\choose{5}}+20{{n}\choose{4}}+15{{n}\choose{3}}+6{{n}\choose{2}}+n$.

  10. (10)

$l_{n}^{n}(10)={{n}\choose{8}}+7{{n}\choose{7}}+21{{n}\choose{6}}+35{{n}\choose{3}}+35{{n}\choose{4}}+21{{n}\choose{3}}+7{{n}\choose{2}}+n$.

###### Proof.

Let $L$ be an $n$-Lie algebra of dimension $d=n$ with the basis
$X=\\{x_{1},x_{2},\dots,x_{n}\\}$ such that $x_{n}>x_{n-1}>\dots>x_{2}>x_{1}$.

(1). It is clear from the definition of basic commutators of weight $1$.

(2). By Theorem 5.1, we have

$l_{d}^{n}(2)=l_{n}^{n}(2)={{n}\choose{n}}=\dfrac{n!}{(n-n)!\times n_{!}}=1.$

(3). We know that the general form of the basic commutators of weight $3$ is
as follows:

$[[x_{11},\dots,x_{1n}],x_{22},\dots,x_{2n}]\in L^{3},$

where $x_{ij}\in X$, for all $1\leq j\leq n$ and $i=1,2$. By part $(2)$, we
know that at $n=d$, there is only one basic commutator of weight $2$ as
$[x_{n},x_{n-1},\dots,x_{2},x_{1}]$. Therefore, it is enough to determine only
the status of the outer components of the second bracket. Since $n=d$, so
$(n-1)$-components $x_{22},\dots,x_{2n}$ are equal to $(n-1)$-components of
inner bracket. Without reducing the totality, assume that $x_{1j}=x_{2j}$, for
all $2\leq j\leq n$. Hence in this situation, according to the Jacobi identity
and skew-symmetric property of $n$-Lie algebras, we have

$\displaystyle[[x_{11},\dots,x_{1n}],x_{22},\dots,x_{2n}]$
$\displaystyle=[[x_{11},\dots,x_{1n}],x_{12},\dots,x_{1n}]$
$\displaystyle=[[x_{11},\dots,x_{1n}],x_{12},\dots,x_{1n}]+[x_{11},\cancelto{0}{[x_{12},x_{12}\dots,x_{1n}]},x_{13},\dots,x_{1n}]$
$\displaystyle\quad+[x_{11},x_{12},\cancelto{0}{[x_{13},x_{12}\dots,x_{1n}]},x_{14},\dots,x_{1n}]+\dots+[x_{11},x_{12},\dots,\cancelto{0}{[x_{1n},x_{12}\dots,x_{1n}]}]$
$\displaystyle=[[x_{11},\dots,x_{1n}],x_{12},\dots,x_{1n}].$

Thus, to count the number of all basic commutators of weight $3$ on $d$
letters (which $d=n$), it is enough to calculate the number of all possible
modes to select $(n-1)$ distinct letters from $n$ letters (in other words, it
is equal to the number of all basic commutators of weight $2$ on $d$ letters
in $(n-1)$-Lie algebras), which according to Theorem 5.1, is equal to

$l_{n}^{d}(3)=l_{n}^{n}(3)={{n}\choose{n-1}}=n.$

So the number of all basic commutators of weight $w=3$, where $n=d$, is equal
to $n$.

(4). We know that the general form of basic commutators of weight $4$ can be
considered as
$[[[x_{11},x_{12},\dots,x_{1n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}]\in
L^{4}$. Note that $x_{22},\dots,x_{2n}$ and $x_{32},\dots,x_{3n}$ can be
considered as strings such as basic commutators of weight $2$ on $n$ letters
and with length $n-1$. In this case, according to the definition of the basic
commutators,
$(x_{12},\dots,x_{1n})\leq(x_{22},\dots,x_{2n})\leq(x_{32},\dots,x_{3n})$
should be hold. Given that at $n=d$, there is only one basic commutator of
weight $2$ as $[x_{1},x_{2},\dots,x_{n}]$, it suffices to consider the
possible choices for $(x_{22},\dots,x_{2n})$ and $(x_{32},\dots,x_{3n})$.
Obviously, for each $2\leq j\leq n$, $x_{2j}$ and $x_{3j}$ are equal to one of
the $[x_{1},x_{2},\dots,x_{n}]$ components, and $(x_{2},\dots,x_{n})$ is
smaller than or equal to $(x_{22},\dots,x_{2n})$ and $(x_{32},\dots,x_{3n})$.
At $n=d$, the general form of the base shifters of weight $4$ can be rewritten
as follows:

$[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}]\in
L^{4},\qquad x_{2j},x_{3j}\neq x_{1},~{}~{}\text{for all }2\leq j\leq n.$

According to the above description, it can be divided into the following two
categories:

(5.5)
$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{22},\dots,x_{2n}],\quad(x_{22},\dots,x_{2n})=(x_{32},\dots,x_{3n})$
$\displaystyle\Longrightarrow{\text{the number of all basic commutators in
this form is equal to~{}}}n.$ (5.6)
$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}],\quad(x_{22},\dots,x_{2n})\neq(x_{32},\dots,x_{3n}),$
$\displaystyle\Longrightarrow{\text{the number of all basic commutators in
this form is equal to~{}}}{{n}\choose{2}}.$

Note that the number of all basic commutator of weight $2$ on $d=n$ letters in
$(n-1)$-Lie algebras is equal to $n$. Therefore, the number of all basic
commutators of weight $4$ on $d=n$ letters is equal to the sum of the above
two cases (5.5) and (5.6); that is,

$l_{n}^{n}(4)={{l_{n}^{n-1}(2)}\choose{2}}+n={{{{n}\choose{n-1}}}\choose{2}}+n={{n}\choose{2}}+n.$

(5). Similar to part (4), we have the following categories:

$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{22},\dots,x_{2n}],x_{22},\dots,x_{2n}],$
$\displaystyle{\text{where}}~{}(x_{22},\dots,x_{2n})=(x_{i2},\dots,x_{in}),\quad
i=3,4,$ (5.7) $\displaystyle\Longrightarrow{\text{the number of all basic
commutators in this form is equal to~{}}}n.$
$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}]\quad{\text{and}}$
$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}],x_{32},\dots,x_{3n}],$
$\displaystyle{\text{where}}~{}(x_{22},\dots,x_{2n})\neq(x_{32},\dots,x_{3n}),$
(5.8) $\displaystyle\Longrightarrow{\text{the number of all basic commutators
in this form is equal to~{}}}2\times{{n}\choose{2}}.\qquad\qquad\qquad$
$\displaystyle[[[x_{1},x_{2},\dots,x_{n}],x_{22},\dots,x_{2n}],x_{32},\dots,x_{3n}],x_{42},\dots,x_{4n}],$
$\displaystyle{\text{where}}~{}(x_{i2},\dots,x_{in})\neq(x_{j2},\dots,x_{jn}),\quad
i,j=2,3,4,$ (5.9) $\displaystyle\Longrightarrow{\text{the number of all basic
commutators in this form is equal
to~{}}}{{n}\choose{3}}.\qquad\qquad\qquad\qquad\ $

Hence, $l_{n}^{n}(5)$ is equal to the sum of the three cases (5.7), (5.8), and
(5.9) as follows:

$\displaystyle l_{n}^{n}(5)$
$\displaystyle={{l_{n}^{n-1}(2)}\choose{3}}+2{{l_{n}^{n-1}(2)}\choose{2}}+n$
$\displaystyle={{{{n}\choose{n-1}}}\choose{2}}+2{{{n}\choose{n-1}}\choose{2}}+n$
$\displaystyle={{n}\choose{3}}+2{{n}\choose{2}}+n.$

By a similar process to the previous one, it can be proved that

$\displaystyle l_{n}^{n}(6)$
$\displaystyle={{{{n}\choose{n-1}}}\choose{4}}+3{{{n}\choose{n-1}}\choose{3}}+3{{{n}\choose{n-1}}\choose{2}}+n={{n}\choose{4}}+3{{n}\choose{3}}+3{{n}\choose{2}}+n,\qquad\qquad\
$ $\displaystyle l_{n}^{n}(7)$
$\displaystyle={{{{n}\choose{n-1}}}\choose{5}}+4{{{n}\choose{n-1}}\choose{4}}+6{{{{n}\choose{n-1}}}\choose{3}}+4{{{{n}\choose{n-1}}}\choose{2}}+n$
$\displaystyle={{n}\choose{5}}+4{{n}\choose{4}}+6{{n}\choose{3}}+4{{n}\choose{2}}+n,$
$\displaystyle l_{n}^{n}(8)$
$\displaystyle={{{{n}\choose{n-1}}}\choose{6}}+5{{{{n}\choose{n-1}}}\choose{5}}+10{{{{n}\choose{n-1}}}\choose{4}}+10{{{{n}\choose{n-1}}}\choose{3}}+5{{{n}\choose{n-1}}\choose{2}}+n$
$\displaystyle={{n}\choose{6}}+5{{n}\choose{5}}+10{{n}\choose{4}}+10{{n}\choose{3}}+5{{n}\choose{2}}+n,$
$\displaystyle l_{n}^{n}(9)$
$\displaystyle={{{{n}\choose{n-1}}}\choose{7}}+6{{{{n}\choose{n-1}}}\choose{6}}+15{{{{n}\choose{n-1}}}\choose{5}}+20{{{{n}\choose{n-1}}}\choose{4}}+15{{{{n}\choose{n-1}}}\choose{3}}+6{{{n}\choose{n-1}}\choose{2}}+n$
$\displaystyle={{n}\choose{7}}+6{{n}\choose{6}}+15{{n}\choose{5}}+20{{n}\choose{4}}+15{{n}\choose{3}}+6{{n}\choose{2}}+n,$
$\displaystyle l_{n}^{n}(10)$
$\displaystyle={{{{n}\choose{n-1}}}\choose{8}}+7{{{{n}\choose{n-1}}}\choose{7}}+21{{{{n}\choose{n-1}}}\choose{6}}+35{{{{n}\choose{n-1}}}\choose{5}}+35{{{{n}\choose{n-1}}}\choose{4}}+21{{{{n}\choose{n-1}}}\choose{3}}$
$\displaystyle\quad+7{{{n}\choose{n-1}}\choose{2}}+n$
$\displaystyle={{n}\choose{8}}+7{{n}\choose{7}}+21{{n}\choose{6}}+35{{n}\choose{3}}+35{{n}\choose{4}}+21{{n}\choose{3}}+7{{n}\choose{2}}+n.$

∎

According to Corollary 5.2, we obtain
$l_{n}^{n}(w)=\displaystyle\sum_{i=1}^{w-2}a_{i}{{n}\choose{i}}$, and in Table
3 (for $i\geq 1$), see the order between the coefficients.

Table 3. coefficients ${{n}\choose{i}}$’s | $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$
---|---|---|---|---|---|---|---|---
$l_{n}^{n}(4)$ | 1 | 1 | – | – | – | – | – | –
$l_{n}^{n}(5)$ | 1 | 2 | 1 | – | – | – | – | –
$l_{n}^{n}(6)$ | 1 | 3 | 3 | 1 | – | – | – | –
$l_{n}^{n}(7)$ | 1 | 4 | 6 | 4 | 1 | – | – | –
$l_{n}^{n}(8)$ | 1 | 5 | 10 | 10 | 5 | 1 | – | –
$l_{n}^{n}(9)$ | 1 | 6 | 15 | 20 | 15 | 6 | 1 | –
$l_{n}^{n}(10)$ | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1

According to Table 3, we assume that the following relations are established.

  * •

$a_{1}=a_{w-2}=1$, $a_{2}=a_{w-3}=w-3$, and $a_{i}=a_{w-i-1}$, for each $i$.

  * •

Comparing the coefficients in Table 3 with the values of the Khayyam triangle,
it is observed that to calculate the coefficients $a_{i}$ in $l_{n}^{n}(w)$,
it is sufficient to obtain the $(w-3)$th power coefficients of the binomial,
that is, $(x+y)^{w-3}$. In other words, $a_{i}$ is the coefficient
$x^{w-3-(i-1)}y^{i-1}$. Each $a_{i}$, for $1\leq i\leq w-2$, can also be
considered the $i$th digit in the answer of $(11)^{w-3}$.

Table 4 denotes the number of basic commutators in different weights with
$n=d$, which can be obtained by the formulas in Corollary 5.2 and manual
calculations.

Table 4. Number of basic commutators in different weights $n=d$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---
$w=1$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
$w=2$ | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1
$w=3$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
$w=4$ | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55
$w=5$ | 6 | 10 | 20 | 35 | 56 | 84 | 120 | 165 | 220
$w=6$ | 9 | 15 | 35 | 70 | 126 | 210 | 330 | 495 | 715
$w=7$ | 18 | 21 | 56 | 126 | 252 | 462 | 792 | 1287 | 2002
$w=8$ | 30 | 28 | 84 | 210 | 462 | 924 | 1716 | 3003 | 5005
$w=9$ | 56 | 36 | 120 | 330 | 792 | 1716 | 3432 | 6435 | 11440
$w=10$ | 99 | 45 | 165 | 495 | 1287 | 3003 | 6435 | 12870 | 24310

We obtain the following simple formulas that are easily proved by induction:

  * •

$l_{n}^{n}(4)=l_{n}^{n}(3)+(l_{n}^{n}(3)-1)+(l_{n}^{n}(3)-2)+\dots+2+1=\sum\limits_{i=0}^{l_{n}^{n}(3)-1}(l_{n}^{n}(3)-i)=\Big{(}l_{n}^{n}(3)\times
l_{n}^{n}(3)\Big{)}+\sum\limits_{i=1}^{l_{n}^{n}(3)}i=\Big{(}l_{n}^{n}(3)\Big{)}^{2}+\displaystyle\dfrac{l_{n}^{n}(3)(l_{n}^{n}(3)-1)}{2}=\Big{(}l_{n}^{n}(3)\Big{)}^{2}+{{l_{n}^{n}(3)}\choose{2}},$
where $n\geq 2$.

  * •

$l_{n}^{n}(w)=\sum\limits_{w^{\prime}=2}^{w}l_{n-1}^{n-1}(w^{\prime})=l_{n}^{n}(w-1)+l_{n-1}^{n-1}(w),$
where $n>3$.

  * •

$l_{3}^{3}(w)=l_{3}^{3}(w-1)+(w-1)$, where $w\geq 3$, $l_{3}^{3}(1)=3$, and
$l_{3}^{3}(2)=1$.

  * •

$l_{4}^{4}(w)=\begin{cases}l_{4}^{4}(w-1)+(w-1)r,\quad w=2r,\\\ \\\
l_{4}^{4}(w-1)+wr,\quad w=2r+1.\end{cases}$.

We know that the number of all commutators (all of the basic and nonbasic
commutators) with weight $w$ is $d^{m_{w}}$. Hence,

$\underbrace{\stackrel{{\scriptstyle d}}{{\bigcirc}}\stackrel{{\scriptstyle
d}}{{\bigcirc}}\dots\stackrel{{\scriptstyle
d}}{{\bigcirc}}\stackrel{{\scriptstyle d}}{{\bigcirc}}}_{n-times}.$

Also, it is obvious that the number of all nonbasic commutators of weight $w$,
denoted by $L_{d}^{n}(w)$, is equal to

(5.10) $L_{d}^{n}(w)=d^{m_{w}}-l_{d}^{n}(w).$

On the other hand, by Theorem 5.1, we have $l_{d}^{n}(2)={{d}\choose{n}}$, and
so the number of all nonbasic commutators of weight 2 is equal to

(5.11) $L_{d}^{n}(2)=d^{m_{2}}-l_{d}^{n}(2)=d^{n}-{{d}\choose{n}}.$

Now, we calculate the number of all basic commutators of weight 3. Every
commutator of weight 3 has $n+(n-1)=2n-1$ components, in which $n$ of them
form the commutators of weight 2, and others are weight 1 and form the
exterior part in the bracket. It is obvious that every nonbasic commutator of
weight 2 forms a nonbasic commutator of weight 3, independent of the selection
of the $n-1$ components in external parts. So these must be removed, too. We
denote the number of all of such cases by ${L^{\prime}}_{d}^{n}(3)$, which is
equal to

(5.12) ${L^{\prime}}_{d}^{n}(3)=L_{d}^{n}(2)\times d^{n-1}.$

On the other hand, note that if a basic commutator of weight 2 is arranged in
a nonbasic form way with the external part, then the result will be nonbasic
again. So we should remove such cases. The number of these, denoted by
${L^{\prime\prime}}_{d}^{n}(3)$, is calculated as follows:

(5.13) $\displaystyle{L^{\prime\prime}}_{d}^{n}(3)$
$\displaystyle=l_{d}^{n}(2)\times L_{d}^{n-1}(2)$ (5.14)
$\displaystyle={{d}\choose{n}}\times\left(d^{n-1}-{{d}\choose{n-1}}\right).$

Thus,

(5.15) $L_{d}^{n}(3)={L^{\prime}}_{d}^{n}(3)+{L^{\prime\prime}}_{d}^{n}(3).$

Moreover, it may be that two parts are basic, but the third condition of the
definition of basic commutators is not true. Hence commutators in this
situation must be replaced by terms with the basic form, and the number of
them must be decreased. In this case, we can replace it by using the Jacobi
identity, such that every term of the Jacobi identity has the basic form, or
otherwise, repeat this replacing for nonbasic terms in the Jacobi identity.
Note that if any term of Jacobi identity does not have a basic form, then we
use the Jacobi identity next-handed to replace it (i.e., if the first time, we
write the right Jacobi identity, then next time, must be used the left Jacobi
identity to replace the other terms). Note that by attention to the collecting
process, all of the above terms can be assumed. We denote the number of them
with $L_{w}^{*}(n,d,m_{w-1},n-1)$, where $m_{w}=m_{w-1}+n-1$. It is important
that if $n=d$, then each product of two basic commutators as $ab$, in which
$a$ is a basic commutator of weight $w-1$ and $b$ is a basic string of length
$n-1$, has a basic form with weight $w$. Thus

(5.16) $L_{w}^{*}(n,n,m_{w-1},n-1)=0.$

In general,

(5.17) $L_{w}^{*}(n,d,m_{w-1},n-1)=l_{d}^{n}(w-1)\times
l_{d}^{n-1}(2)-\kappa=l_{d}^{n}(w-1)\times{{d}\choose{n-1}}-\kappa.$

We are going to find $\kappa$ by induction on $w$.
Let us first show that the number of all basic commutators of weight $w$ is
equal to $\kappa$. According to the above explanation, it is clear that the
number of all of the nonbasic commutators is equal to the sum of both
$L^{n}_{d}(w)$ and $L_{w}^{*}(n,d,m_{w-1},n-1)$, and so other commutators are
basic. Thus the number of all basic commutators of weight $w$ is

$\displaystyle l_{d}^{n}(w)$
$\displaystyle=d^{m_{w}}-L_{d}^{n}(w)-L_{w}^{*}(n,d,m_{w-1},n-1)$
$\displaystyle=d^{m_{w}}-({L^{\prime}}_{d}^{n}(w)+{L^{\prime\prime}}_{d}^{n}(w))-L_{w}^{*}(n,d,m_{w-1},n-1)$
$\displaystyle=d^{m_{w}}-\left(L_{d}^{n}(w-1)\times
d^{n-1}\right)-\left(l_{d}^{n}(w-1)\times
L_{d}^{n-1}(2)\right)-\left(l_{d}^{n}(w-1)\times l_{d}^{n-1}(2)-\kappa\right)$
$\displaystyle=d^{m_{w}}-\left((d^{m_{w-1}}-l_{d}^{n}(w-1))\times
d^{n-1}\right)-\left(l_{d}^{n}(w-1)\times(d^{n-1}-l_{d}^{n-1}(2))\right)$
$\displaystyle\quad-\left(l_{d}^{n}(w-1)\times l_{d}^{n-1}(2)-\kappa\right)$
$\displaystyle=d^{m_{w}}-\underbrace{(d^{m_{w-1}}\times
d^{n-1})}_{d^{m_{w}}}+(l_{d}^{n}(w-1)\times
d^{n-1})-\left(l_{d}^{n}(w-1)\times d^{n-1}\right)$
$\displaystyle\quad+(l_{d}^{n}(w-1)\times
l_{d}^{n-1}(2))-\left(l_{d}^{n}(w-1)\times
l_{d}^{n-1}(2)\right)+\kappa=\kappa.$

Now, we calculate $\kappa$ by induction on $w$. It is obvious that the
smallest value of $w$ must be $3$, and so, for $w=3$, we have
$l_{d}^{n}(2)={{d}\choose{n}}$, $l_{d}^{n-1}(2)={{d}\choose{n-1}}$, and

(5.18)
$L_{3}^{*}(n,d,m_{2},n-1)={{d}\choose{n}}{{d}\choose{n-1}}-\sum_{i=1}^{d-n+1}\
\sum_{j=i+1}^{d-1}(d-j)\left[{{d-i+1}\choose{n-1}}-j+i+1\right].$

Therefore, by (5.11), (5.12), (5.14), and (5.18), we have

$\displaystyle l_{d}^{n}(3)=\kappa$ $\displaystyle=\sum_{i=1}^{d-n+1}\
\sum_{j=i+1}^{d-1}(d-j)\left[{{d-i+1}\choose{n-1}}-j+i+1\right].$

A comparison of two equations (5.19) and (5.18) shows that there are no other
basic commutators of weight $3$ except equation (5.19). Therefore, we obtain a
formula to count basic commutators of weight $3$ in any given $n$-Lie algebras
with arbitrary dimension $d\geq n$, and hence the following theorem is proved.

######  Theorem 5.3.

Let $F$ be a free $n$-Lie algebra over the ordered set
$X=\\{x_{i}|x_{i+1}>x_{i},~{}i=1,2,\dots,d\\}$. Then the number of basic
commutators of weight $3$ is

(5.19) $l_{d}^{n}(3)=\sum_{i=1}^{d-n+1}\
\sum_{j=i+1}^{d-1}(d-j)\left[{{d-i+1}\choose{n-1}}-j+i+1\right].$

By a similar way and induction, we can rewrite (5.12), (5.13), and (5.15) and
obtain the following equations:

$\displaystyle{L^{\prime}}_{d}^{n}(w)=L_{d}^{n}(w-1)\times d^{n-1},\ \ \ $
$\displaystyle{L^{\prime\prime}}_{d}^{n}(w)=l_{d}^{n}(w-1)\times
L_{d}^{n-1}(2),\ $ $\displaystyle
L_{d}^{n}(w)={L^{\prime}}_{d}^{n}(w)+{L^{\prime\prime}}_{d}^{n}(w).$

Now, in another way, we calculate the number of basic commutators of weights
$4$ and $5$ on $d\geq 4$ letters. Then we generalize it to the state of
arbitrary $w$ and obtain the final and general formula for the various
arbitrary values of $n$, $d$, and $w$.

######  Theorem 5.4.

Let $F$ be a free $n$-Lie algebra over the ordered set
$X=\\{x_{i}|x_{i+1}>x_{i},~{}i=1,2,\dots,d\\}$. Then the number of basic
commutators of weight $4$ is

(5.20)
$l_{d}^{n}(4)=\sum_{j=1}^{\alpha_{0}}\beta_{j^{*}}\left({{{{d}\choose{n-1}}}\choose{2}}+{{d}\choose{n-1}}\right),$

where $\alpha_{0}={{d-1}\choose{n-1}}$, and if ${{k-1}\choose{n-1}}+1\leq
j\leq{{k}\choose{n-1}}$ (for $k=n-1,n,n+1,n+2,\dots,d-1$), then
$j^{*}={{k-1}\choose{n-1}}+1$ and $\beta_{j^{*}}=(d-n-j^{*}+2)$.

###### Proof.

Let $d\geq n$, let $X=\\{x_{i}~{}|~{}x_{i}>x_{i-1},~{}i=1,2,3,4,\dots,d\\}$,
and let $w=4$. We know that $l_{d-1}^{n-1}(2)={{d-1}\choose{n-1}}$ is the
number of collected strings of length $n-1$, too. We calculate the collected
strings of length $n-1$ and arrange them as follows. Then we number them from
right to left (We use the $j$ counter to count them. Obviously,
$j=1,2,\dots,l_{d-1}^{n-1}(2)$). Note that here we display $n=3$ just to make
it easier to display the components, but the same procedure applies to the
general case $n$. Hence

(5.21) $x_{d-1}x_{d-2}$ | $x_{d-1}x_{d-3}$ | $\dots$ | $x_{4}x_{3}$ | $x_{4}x_{2}$ | $x_{4}x_{1}$ | $x_{3}x_{2}$ | $x_{3}x_{1}$ | $x_{2}x_{1}$
---|---|---|---|---|---|---|---|---
$\downarrow$ | $\downarrow$ |  | $\downarrow$ | $\downarrow$ | $\downarrow$ | $\downarrow$ | $\downarrow$ | $\downarrow$
$j={{d-1}\choose{n-1}}$ | $j={{d-1}\choose{n-1}}-1$ | $\dots$ | $j=6$ | $j=5$ | $j=4$ | $j=3$ | $j=2$ | $j=1$

The general form of the basic commutators of weight $4$ is as follows:

$(((a_{1},a_{2},a_{3}),b_{2},b_{3}),c_{2},c_{3}),\qquad a_{r},b_{s},c_{q}\in
X.$

First, we select the first collected string from the chain (5.21) (for $j=1$)
and replace $a_{2},a_{3}$ with it; see Figure 1.

Figure 1.

We have to calculate the number of all basic commutators by counting the
number of allowed modes for cells such as $x$, $y$, and $z$. Since the
corresponding term $j=1$ in the chain (5.21) is the first and smallest
sentence, the allowed states for selecting $x$ are
$x_{3},x_{4},\ldots,x_{d-1}$, and $x_{d}$, so the number of these states is
equal to $d-(n-1)=d-2$.

Now, we want to form a chain of all basic commutators of weight $2$ (with
length $n-1$) on $d$ letters $x_{1},x_{1},\dots,x_{d}\in X$. We know that
$l_{d-1}^{n-1}(2)={{d-1}\choose{n-1}}$ is the number of all basic commutators
of weight $2$ and length $n-1$. We calculate the basic commutators of weight
$2$ and then arrange them by the defined relation “$<$” on basic commutators
as follows. Then we label them from right to left (We use the counter $i$, to
count them. Obviously, $i=1,2,\dots,l_{d}^{n-1}(2)$). Hence

(5.22)
${\footnotesize{\underbrace{x_{d}x_{d-1}}_{\begin{subarray}{c}\downarrow\\\
i={{d}\choose{n-1}}\end{subarray}}>\underbrace{x_{d}x_{d-2}}_{\begin{subarray}{c}\downarrow\\\
i={{d}\choose{n-1}}-1\end{subarray}}>\underbrace{x_{d-1}x_{d-2}}_{\begin{subarray}{c}\downarrow\\\
i={{d}\choose{n-1}}-2\end{subarray}}>\dots>\underbrace{x_{d}x_{2}}_{\begin{subarray}{c}\downarrow\\\
i=2d-3\end{subarray}}>\dots>\underbrace{x_{4}x_{2}}_{\begin{subarray}{c}\downarrow\\\
i=d+1\end{subarray}}>\underbrace{x_{3}x_{2}}_{\begin{subarray}{c}\downarrow\\\
i=d\end{subarray}}>\underbrace{x_{d}x_{1}}_{\begin{subarray}{c}\downarrow\\\
i=d-1\end{subarray}}>\dots>\underbrace{x_{3}x_{1}}_{\begin{subarray}{c}\downarrow\\\
i=2\end{subarray}}>\underbrace{x_{2}x_{1}}_{\begin{subarray}{c}\downarrow\\\
i=1\end{subarray}}.}}$

Also, since the corresponding sentence $j=i=1$ in the chain (5.22) is the
smallest sentence, all the sentences in the chain (5.22) can be placed after
it. On the other hand, both $z\geq y$ and $y\geq x_{2}x_{1}$ conditions must
be met. Obviously, both sentences we choose from the chain (5.22) for $y$ and
$z$, one of which is smaller than the other, which can be considered as $y$
and another as $z$. Therefore, the number of choice states $y$ and $z$,
provided that $y$ and $z$ are distinct, is equal to
${{{{d}\choose{n-1}}}\choose{w-2}}={{{{d}\choose{n-1}}}\choose{2}}$. If $y=z$,
then we have ${{d}\choose{n-1}}$ choices for them. Thus, in this case $j=1$,
the number of basic commutators is equal to
$(d-2)\times\left[{{{{d}\choose{n-1}}}\choose{2}}+{{d}\choose{n-1}}\right]$.

By the same process, we also calculate the number of basic commutators in the
cases $j=2,3,4,\dots,{{d-1}\choose{n-1}}$, and the sum of the number of basic
commutators in the various cases $j$ is equal to the number of all the basic
commutators of weight $4$ on $d$ letters. Moreover,

$l_{d}^{n}(4)=\sum_{j=1}^{\alpha_{0}}\beta_{j^{*}}\left({{{{d}\choose{n-1}}}\choose{2}}+{{d}\choose{n-1}}\right),$

where $\alpha_{0}={{d-1}\choose{n-1}}$, and if ${{k-1}\choose{n-1}}+1\leq
j\leq{{k}\choose{n-1}}$, (for $k=n-1,n,n+1,n+2,\dots,d-1$), then put
$j^{*}={{k-1}\choose{n-1}}+1$ and $\beta_{j^{*}}=(d-n-j^{*}+2)$. Therefore it
is clear that $1\leq j^{*}\leq d-n+1$. ∎

In the following, we want to get a formula for calculating the number of basic
commutators of any weight $w$ in every $d$-dimensional $n$-Lie algebras. For
this purpose, it is necessary to note that in the case of duplicate selection,
the placement of strings can also have several different modes, the number of
which must also be calculated. For example, suppose $w=5$. We have to count
the number of allowed choices for $x$, $y$, and $z$. If all three strings $x$,
$y$, and $z$ are distinct, then they are arranged as follows:

$\displaystyle((((a_{1},\underbrace{a_{2},\dots,a_{n}}_{a}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{x}),\underbrace{y_{2},y_{3},\dots,y_{n}}_{y}),\underbrace{z_{2},z_{3},\dots,z_{n}}_{z}),\qquad
z>y>x\geq a.$

Indeed if two of the strings $x$, $y$, and $z$ are the same, then the
following two states must be considered and counted:

$\displaystyle((((a_{1},\underbrace{a_{2},\dots,a_{n}}_{a}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{x}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{y}),\underbrace{z_{2},z_{3},\dots,z_{n}}_{z}),\qquad
z>y=x\geq a,$
$\displaystyle((((a_{1},\underbrace{a_{2},\dots,a_{n}}_{a}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{x}),\underbrace{y_{2},y_{3},\dots,y_{n}}_{y}),\underbrace{y_{2},y_{3},\dots,y_{n}}_{z}),\qquad
z=y>x\geq a.$

Also, if $x=y=z$, then

$((((a_{1},\underbrace{a_{2},\dots,a_{n}}_{a}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{x}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{y}),\underbrace{x_{2},x_{3},\dots,x_{n}}_{z}),\qquad
z=y=x\geq x.$

In the weight $w=4$, there exist two outer strings $x$ and $y$, and so there
were two choice modes for $x$ and $y$, which are as follows: $(1)$-A state
where both $x$ and $y$ are distinct (Equivalently, $x$ and $y$ are replaced by
exactly two distinct strings). $(2)$-The state that $x$ and $y$ are equal
(Equivalently, $x$ and $y$ are replaced by only one string). Also, in the case
$w=5$, there exist three outer strings $x$, $y$, and $z$. Hence, there were
three choice modes for $x$, $y$, and $z$, which are as follows: $(1)$-A state
that all three strings $x$, $y$, and $z$ are distinct (Equivalently, $x$, $y$,
and $z$ are replaced by exactly three distinct strings, in which case there is
only one arrangement for these three strings as $z>y>x$). $(2)$-The state that
two strings of $x$, $y$, and $z$ are equal (Equivalently, $x$, $y$, and $z$
are replaced by exactly two distinct strings, in which case there is only two
arrangement for these strings as $z=y>x$ and $z>y=x$). $(3)$-The state where
all of the three strings are equal together (Equivalently, $x$, $y$, and $z$
are replaced by only one string, in which case there is only one arrangement
for strings as $z=y=x$).

In general, for an arbitrary $w$, the number of outer strings is equal to
$w-2$, called $y_{1},y_{2},y_{3},\dots,y_{w-2}$ (where it can replaced by
basic strings of length $n-1$, or in other words, by basic commutators of
weight $2$ in $(n-1)$-Lie algebras) on $d$ letters, in which the number of all
them is equal to $l_{d}^{n-1}(2)={{d}\choose{n-1}}$). If we want to replace
them with $w-2$ distinct strings (a state in which they are all distinct),
then we have only one arrangement as $y_{1}<y_{2}<y_{3}<\dots<y_{w-2}$. If we
replace all of them with just one string (a state in which they are all
repetitive), then we have one arrangement $y_{1}=y_{2}=y_{3}=\dots=y_{w-2}$.
If we replace them with $w-3$ distinct strings (a state in which exactly two
strings are same) or replace them with $2$ distinct strings (a state in which
some of the $y_{i}$ are equal to string $1$ and others are equal to string
$2$), then the number of arrangements of them is equal to
$\alpha_{2}=\alpha_{w-1}={{w-3}\choose{0}}={{w-3}\choose{w-3}}$.

Continuing this process, we can say that if we want to replace them with $i$
distinct strings or replace them with $w-i-1$ (a state in which exactly $i$
strings are the same), then we have $\alpha_{i}$ different arrangements, where
$\alpha_{i}=\alpha_{w-i+1}={{w-3}\choose{i-2}}={{w-3}\choose{w-i+1}}$ and
$2\leq i\leq w-1$.

Now, considering the above explanations and the proof of the previous theorem,
we are ready to introduce the formula for calculating the number of basic
commutators.

######  Theorem 5.5.

Let $F$ be a free $n$-Lie algebra over the ordered set
$X=\\{x_{i}|x_{i+1}>x_{i}:~{}i=1,2,\dots,d\\}$ and let $w$ be a positive
integer number. Then the number of basic commutators of weight $w$ is

(5.23)
$l_{d}^{n}(w)=\sum_{j=1}^{\alpha_{0}}\beta_{j^{*}}\left(\sum_{i=2}^{w-1}\alpha_{i}{{{{d}\choose{n-1}}}\choose{w-i}}\right),$

where $\alpha_{0}={{d-1}\choose{n-1}}$, $\alpha_{i}$, ($2\leq i\leq w-1$) is
the coefficient of the $(i-2)$th sentence in Newton’s binomial expansion
$(a+b)^{w-3}$, and if ${{k-1}\choose{n-1}}+1\leq j\leq{{k}\choose{n-1}}$, (for
$k=n-1,n,n+1,n+2,\dots,d-1$), then $j^{*}={{k-1}\choose{n-1}}+1$ and
$\beta_{j^{*}}=(d-n-j^{*}+2)$.

###### Proof.

According to the recent explanation, it is proved by a similar process of the
proof of Theorem 5.20. ∎

##  6\. Relationship between the number of basic commutators of $n$-Lie
algebras and Lie algebras

We know that the $n$-Lie algebras are independent of the selection of the
bracket, and the concept of basic commutators is also related to the free
$n$-Lie algebras. In addition, if $L$ is a Lie algebra over the field
$\mathbb{F}$ with the bracket $[-,-]:L\times L\longrightarrow L$, then for
each natural number $n\in\mathbb{N}$, $L$ can be equipped with an $n$-linear
function $[-,\dots,-]:\underbrace{L\times L\times\dots\times
L}_{n-times}\longrightarrow L$, such that $L$ forms an $n$-Lie algebra with
the bracket $[-,\dots,-]$ on the field $\mathbb{F}$ in which

$\displaystyle[-,\dots,-]:\underbrace{L\times L\times\dots\times L}_{n-times}$
$\displaystyle\longrightarrow L$ $\displaystyle(x_{1},x_{2},\dots,x_{n})$
$\displaystyle\longmapsto[x_{1},x_{2},\dots,x_{n}]:=[[\dots[[x_{1},x_{2}],x_{3}],\dots]].$

The above explanation may raise the question of whether there is a connection
between $l_{d}(w)$ and $l^{n}_{d}(w)$, which are the number of basic
commutators of weight $w$ and on $d$ generators in Lie algebras and $n$-Lie
algebras?

In this section, we intend to answer this question and examine the
relationship between them. By using (4.3), we have

$l^{2}_{d}(w)=l_{d}(w)=\dfrac{1}{w}\sum_{s|w}\mu(s)d^{\frac{w}{s}},$

for all $w,d\in\mathbb{N}$.

On the other hand, we know that

$\displaystyle{{d}\choose{2}}$
$\displaystyle=\dfrac{d(d-1)}{2!}=\dfrac{d^{2}-d}{2!}=\dfrac{1}{2}\sum_{s|2}\mu(s)d^{\frac{2}{s}}=l_{d}^{2}(2).$

Now, let $n=3$. Then

$\displaystyle{{d}\choose{3}}$
$\displaystyle=\dfrac{d^{3}-d}{3!}-\dfrac{3d^{2}-3d}{3!}$
$\displaystyle=\dfrac{d^{3}-d}{3\times 2!}-\dfrac{d^{2}-d}{2!}$
$\displaystyle=\dfrac{1}{2!}\left(\dfrac{1}{3}\sum_{s|3}\mu(s)d^{\frac{3}{s}}\right)-\dfrac{1}{2}\sum_{s|2}\mu(s)d^{\frac{2}{s}}$
$\displaystyle=\dfrac{1}{2!}l_{d}(3)-l_{d}(2).$

By continuing this process, Table 5 can be obtained for different values of
$n\in\mathbb{N}$.

Now, we assume that the following equation is true for each natural number
smaller than or equal to $n_{0}$:

(6.24)
${{d}\choose{n}}=\sum_{s=2}^{n}(-1)^{n-s}\left(\dfrac{a_{s}}{(s-1)!\times(s+1)(s+2)\ldots(n-1)n}\right)l_{d}(s),$

where, $a_{s}>0$, for all $2\leq s\leq n$. Then for $n_{0}+1$, we have

$\displaystyle{{d}\choose{n_{0}+1}}$
$\displaystyle=\dfrac{\prod\limits_{i=0}^{d-1}(d-i)}{(d-(n_{0}+1))!\times(n_{0}+1)!}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
$\displaystyle=\dfrac{\prod\limits_{i=0}^{n_{0}}(d-i)}{(n_{0}+1)!}$
$\displaystyle=\dfrac{(d-n_{0})\prod\limits_{i=0}^{n_{0}-1}(d-i)}{(n_{0}+1)\times
n_{0}!}$
$\displaystyle=\dfrac{(d-n_{0})}{n_{0}+1}\times\dfrac{\prod\limits_{i=0}^{n_{0}-1}(d-i)}{n_{0}!}$
$\displaystyle=\dfrac{(d-n_{0})}{n_{0}+1}\times\left(\sum_{s=2}^{n_{0}}(-1)^{n_{0}-s}\left(\dfrac{a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}}\right)l_{d}(s)\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}(d-n_{0})a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}l_{d}(s)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}(d-n_{0})a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(r)d^{s/r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}(d-n_{0})a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)d^{r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{(d-n_{0})}{s}\sum_{r|s}\mu(s/r)d^{r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)(d-n_{0})d^{r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)d^{r+1}\right)$
$\displaystyle\quad+\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)(-n_{0})d^{r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)d^{r+1}\right)$
$\displaystyle\quad+\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}(-n_{0})a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)d^{r}\right)$
$\displaystyle=\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}\left(\dfrac{1}{s}\sum_{r|s}\mu(s/r)d^{r+1}\right)$
$\displaystyle\quad-\sum_{s=2}^{n_{0}}\dfrac{(-1)^{n_{0}-s}n_{0}a_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}l_{d}(s)$
$\displaystyle=\sum_{s=2}^{n_{0}+1}\dfrac{(-1)^{n_{0}-s+1}b_{s}}{(s-1)!\times(s+1)(s+2)\dots(n_{0}-1)n_{0}(n_{0}+1)}l_{d}(s).$

Thus equation (6.24) holds for each natural number $n$, and we have

(6.25)
${{d}\choose{n}}=\sum_{s=2}^{n}(-1)^{n-s}\left(\dfrac{a_{s}}{(s+1)(s+2)\ldots(n-1)n}\right)l_{d}(s).$

Moreover,

$\displaystyle{{d}\choose{n}}$
$\displaystyle=\sum_{s=2}^{n}(-1)^{n-s}\left(\dfrac{a_{s}}{3\times
4\times\dots\times(s-1)(s+1)(s+2)\ldots(n-1)n}\right)l_{d}(s)$ (6.26)
$\displaystyle=\sum_{s=2}^{n}(-1)^{n-s}\left(\dfrac{a_{s}}{\prod\limits_{\begin{subarray}{c}r=3\\\
r\neq s\end{subarray}}^{n}r}\right)l_{d}(s).$

| $l_{d}(2)$ | $l_{d}(3)$ | $l_{d}(4)$ | $l_{d}(5)$ | $l_{d}(6)$ | $l_{d}(7)$ | $l_{d}(8)$ | $l_{d}(9)$ | $l_{d}(10)$
---|---|---|---|---|---|---|---|---|---
${{d}\choose{2}}=\dfrac{1}{2!}\prod\limits_{i=0}^{1}(d-i)$ | $\frac{1}{1}$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$
${{d}\choose{3}}=\dfrac{1}{3!}\prod\limits_{i=0}^{2}(d-i)$ | $\frac{-3}{3}$ | $\frac{1}{2!}$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$
${{d}\choose{4}}=\dfrac{1}{4!}\prod\limits_{i=0}^{3}(d-i)$ | $\frac{12}{4\times 3}$ | $\frac{-6}{4\times 2!}$ | $\frac{1}{3!}$ | $-$ | $-$ | $-$ | $-$ | $-$ | $-$
${{d}\choose{5}}=\dfrac{1}{5!}\prod\limits_{i=0}^{4}(d-i)$ | $\frac{-60}{5\times 4\times 3}$ | $\frac{35}{5\times 4\times 2!}$ | $\frac{-10}{5\times 3!}$ | $\frac{1}{4!}$ | $-$ | $-$ | $-$ | $-$ | $-$
${{d}\choose{6}}=\dfrac{1}{6!}\prod\limits_{i=0}^{5}(d-i)$ | $\frac{360}{6\times 5\times 4\times 3}$ | $\frac{-224}{6\times 5\times 4\times 2!}$ | $\frac{85}{6\times 5\times 3!}$ | $\frac{-15}{6\times 4!}$ | $\frac{1}{5!}$ | $-$ | $-$ | $-$ | $-$
${{d}\choose{7}}=\dfrac{1}{7!}\prod\limits_{i=0}^{6}(d-i)$ | $\frac{-2520}{7\times 6\times\dots\times 3}$ | $\frac{1603}{7\times 6\times 5\times 4\times 2!}$ | $\frac{-735}{7\times 6\times 5\times 3!}$ | $\frac{175}{7\times 6\times 4!}$ | $\frac{-21}{7\times 5!}$ | $\frac{1}{6!}$ | $-$ | $-$ | $-$
${{d}\choose{8}}=\dfrac{1}{8!}\prod\limits_{i=0}^{7}(d-i)$ | $\frac{20160}{8\times 7\times\dots\times 3}$ | $\frac{-12810}{8\times 7\times\dots\times 4\times 2!}$ | $\frac{6770}{8\times 7\times 6\times 5\times 3!}$ | $\frac{-1960}{8\times 7\times 6\times 4!}$ | $\frac{322}{8\times 7\times 5!}$ | $\frac{-28}{8\times 6!}$ | $\frac{1}{7!}$ | $-$ | $-$
${{d}\choose{9}}=\dfrac{1}{9!}\prod\limits_{i=0}^{8}(d-i)$ | $\frac{-181440}{9\times 8\times\dots\times 3}$ | $\frac{113589}{9\times 8\times\dots\times 4\times 2!}$ | $\frac{-67320}{9\times 8\times\dots\times 5\times 3!}$ | $\frac{22449}{9\times 8\times 7\times 6\times 4!}$ | $\frac{-4536}{9\times 8\times 7\times 5!}$ | $\frac{546}{9\times 8\times 6!}$ | $\frac{-36}{9\times 7!}$ | $\frac{1}{8!}$ | $-$
${{d}\choose{10}}=\dfrac{1}{10!}\prod\limits_{i=0}^{9}(d-i)$ | $\frac{1814400}{10\times 9\times\dots\times 3}$ | $\frac{-1109472}{10\times 9\times\dots\times 4\times 2!}$ | $\frac{724550}{10\times 9\times\dots\times 5\times 3!}$ | $\frac{-269324}{10\times 9\times\dots\times 6\times 4!}$ | $\frac{63273}{10\times 9\times 8\times 7\times 5!}$ | $\frac{-9450}{10\times 9\times 8\times 6!}$ | $\frac{870}{10\times 9\times 7!}$ | $\frac{-45}{10\times 8!}$ | $\frac{1}{9!}$
Table 5. The connection between $l_{d}^{2}(w)$ and
$l^{n}_{d}(2)={{d}\choose{n}}$

According to the above description, the following theorem is proved easily.

######  Theorem 6.1.

The number of basic commutators of weight $2$ on $d$ generators in $n$-Lie
algebras can be concluded by the number of basic commutators of weights $w\leq
n$ in Lie algebras.

###### Proof.

It is easy to check from equation (6.25) and Theorem 5.1. ∎

The following corollary expresses the relationship between the number of basic
commutators in Lie algebras and $n$-Lie algebras. The following formula is
Witt’s generalized formula.

######  Corollary 6.2.

The number of basic commutators of arbitrary weight $w$ on $d$ generators in
$n$-Lie algebras, can be concluded by the number of basic commutators in Lie
algebra. We have

$l_{d}^{n}(w)=\sum_{j=1}^{\alpha_{0}}\beta_{j^{*}}\left(\sum_{i=2}^{w-1}\alpha_{i}\left(\sum_{s=2}^{w-i}(-1)^{w-i-s}\left(\dfrac{a_{s}}{\prod\limits_{\begin{subarray}{c}r=3\\\
r\neq s\end{subarray}}^{w-i}r}\right)l_{d^{*}}(s)\right)\right),$

where $d^{*}={{d}\choose{n-1}}$, $\alpha_{0}={{d-1}\choose{n-1}}$,
$\alpha_{i}$ ($2\leq i\leq w-1$) is the coefficient of the $(i-2)$th sentence
in Newton’s binomial expansion $(a+b)^{w-3}$. If ${{k-1}\choose{n-1}}+1\leq
j\leq{{k}\choose{n-1}}$, (for $k=n-1,n,n+1,n+2,\dots,d-1$), then
$j^{*}={{k-1}\choose{n-1}}+1$ and $\beta_{j^{*}}=(d-n-j^{*}+2)$.

###### Proof.

It is proved from equation (5.23) in Theorem 5.5 and Theorem 6.1. ∎

The following theorem is the most important and practical result of this
paper.

######  Theorem 6.3.

Let $F$ be a free $n$-Lie algebra and let $F^{i}$ be the $i$th term of the
lower central series of $F$, for each $i\in\mathbb{N}$. Then
$\dfrac{F^{i}}{F^{i+c}}$ is abelian of dimension
$l_{d}^{n}(i)+l_{d}^{n}(i+1)+\dots+l_{d}^{n}(i+c-1)$, where
$c=0,1,2,\ldots,i$.

###### Proof.

It is proved similar to its version in the theory of Lie algebras. ∎

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Seyedeh Nafiseh Akbarossadat
Department of Mathematics, Mashhad Branch, Islamic Azad University, Mashhad,
Iran.
E-mail address<EMAIL_ADDRESS>