# On the dimension of $c$-nilpotent multiplier of $n$-Lie algebras

Seyedeh Nafiseh Akbarossadat

Abstract. Let $L$ be a finite-dimensional $n$-Lie algebra with free
presentation $F/R$. Then the concept of $c$-nilpotent multiplier of $L$,
denoted by $\mathcal{M}^{(c)}(L)$, is defined as follows:

$\mathcal{M}^{(c)}(L)=\dfrac{\gamma_{c+1}(F)\cap
R}{\gamma_{c+1}(R,F,\dots,F)}.$

In this paper, we obtain some inequalities and certain bounds for the
dimension of $\mathcal{M}^{(c)}(L)$ by using the basic commutators. Also, we
discuss the relationship between the dimension of the $c$-nilpotent multiplier
of $L$ and the $c$-nilpotent multiplier of some factor of $L$. We further
obtain an inequality between dimensions of $c$-nilpotent multiplier of $n$-Lie
algebra and non-abelian tensor (exterior) product of a central ideal by its
abelianized factor $n$-Lie algebra. Finally, we also determine the dimension
and structure of $c$-nilpotent multipliers Heisenberg $n$-Lie algebras, which
can be a useful tool for determining the dimension of the multiplier of
nilpotent $n$-Lie algebras of class $2$.

Key words: $n$-Lie algebra, Non-abelian tensor product, $c$-Nilpotent
multiplier, $c$-Capable.

MSC: 17B05, 17B30, 17B60.

##  1\. Introduction and Preparatory

Ellis [8] developed an analogous theory of non-abelian tensor products for Lie
algebras (see also [10]). Using tensor (exterior) products of Lie algebras,
Ellis described the universal central extension of Lie algebras.

In 1986, Filippov [11] introduced the notion of $n$-Lie algebras. An $n$-Lie
algebra over a field $\mathbb{F}$ is a vector space $L$ over $\mathbb{F}$
along with an anti-symmetric $n$-linear form $[x_{1},\ldots,x_{n}]$ satisfying
the Jacobi identity:

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

Clearly, $n$-Lie algebras are nothing but ordinary Lie algebras when $n=2$.
Studying $n$-Lie algebras is important because of their applications in
physics and geometry. So far, several papers have been published about their
classification.The author and Saeedi defined [3, 5, 4] the concepts of non-
abelian tensor (exterior) product of $n$-Lie algebras and proved some results
about them. Also, we introduced some bounds for the dimension of the non-
abelian tensor square and Schur multiplier of $n$-Lie algebras. So far,
various researches have been done in the field of multipliers (see [2, 6, 7,
14, 15, 16, 17]).

First, we will review some of the concepts required in this paper.

An $n$-Lie algebra $L$ is called nilpotent of class $s$ (for some positive and
integer number $s$), if $L^{s+1}=\gamma_{s+1}(L)=0$ and $L^{s}\neq 0$. This is
equivalent to $Z_{s-1}(L)\subsetneq Z_{s}(L)=L$. Then $s$ is said the
nilpotency class of $L$, and we write $cl(L)=s$. Note that the nilpotency
property of $n$-Lie algebras is closed under subalgebra, ideal, and
homomorphic image, but it is not closed under the extended property.

######  Definition 1.1.

Let $L$ and $P$ be two $n$-algebras over an arbitrary field $\mathbb{F}$ such
that $L$ and $P$ act on each other by the families of $n$-linear maps
$\\{f_{i}\\}_{1\leq i\leq n-1}$ and $\\{g_{i}\\}_{1\leq i\leq n-1}$,
respectively, and act on themselves by their brackets. Then $L\wedge P$ is
generated by all symbols

$l_{1}\wedge\cdots\wedge l_{i}\wedge p_{i+1}\wedge\cdots\wedge p_{n},$

for all $l_{i}\in L$, $p_{i}\in P$, and $\alpha\in\mathbb{F}$, in which the
following properties are hold:

$\displaystyle l_{1}$ $\displaystyle\wedge\cdots\wedge\alpha
l_{i}+l^{\prime}_{i}\wedge\cdots\wedge l_{j}\wedge p_{j+1}\wedge\cdots\wedge
p_{n}$ $\displaystyle=$ $\displaystyle\alpha(l_{1}\wedge\cdots\wedge
l_{i}\wedge\cdots\wedge l_{j}\wedge p_{j+1}\wedge\cdots\wedge p_{n})$
$\displaystyle+(l_{1}\wedge\cdots\wedge l^{\prime}_{i}\wedge\cdots\wedge
l_{j}\wedge p_{j+1}\wedge\cdots\wedge p_{n}),$ $\displaystyle l_{1}$
$\displaystyle\wedge\cdots\wedge l_{j}\wedge p_{j+1}\wedge\cdots\wedge\alpha
p_{j+i}+p^{\prime}_{j+i}\wedge\cdots\wedge p_{n}$ $\displaystyle=$
$\displaystyle\alpha(l_{1}\wedge\cdots\wedge l_{j}\wedge
p_{j+1}\wedge\cdots\wedge p_{j+i}\wedge\cdots\wedge p_{n})$
$\displaystyle+(l_{1}\wedge\cdots\wedge l_{j}\wedge p_{j+1}\wedge\cdots\wedge
p^{\prime}_{j+i}\wedge\cdots\wedge p_{n}),$ $\displaystyle[l_{1}$
$\displaystyle,\ldots,l_{n}]\wedge l^{\prime}_{2}\wedge\cdots\wedge
l^{\prime}_{i}\wedge p_{1}\wedge\cdots\wedge p_{n-i}$ (1.1) $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}(-1)^{n-j}l_{1}\wedge\cdots\wedge l_{j-1}\wedge
l_{j+1}\wedge\cdots\wedge l_{n}\wedge
f_{i}(l_{j},l^{\prime}_{2},\ldots,l^{\prime}_{i},p_{1},\ldots,p_{n-i}),$
$\displaystyle l_{1}$ $\displaystyle\wedge\cdots\wedge l_{i}\wedge
p_{1}\wedge\cdots\wedge p_{n-i-1}\wedge[p^{\prime}_{1},\ldots,p^{\prime}_{n}]$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{n}(-1)^{n-j-2}g_{i}(p^{\prime}_{j},l_{1},\ldots,l_{i},p_{1},\ldots,p_{n-i-1})\wedge
p^{\prime}_{1}\wedge\cdots\wedge p^{\prime}_{j-1}\wedge
p^{\prime}_{j+1}\wedge\cdots\wedge p^{\prime}_{n},$ $\displaystyle l_{1}$
$\displaystyle\wedge\cdots\wedge l_{j-1}\wedge l^{*}\wedge\dots\wedge
l_{k-1}\wedge l^{*}\wedge\cdots\wedge l_{i}\wedge p_{1}\wedge\cdots\wedge
p_{n-i}=0,$ $\displaystyle l_{1}$ $\displaystyle\wedge\cdots\wedge l_{i}\wedge
p_{1}\wedge\cdots\wedge p_{j-1}\wedge p^{*}\wedge\cdots\wedge p_{k-1}\wedge
p^{*}\wedge\cdots\wedge p_{n-i}=0.$ $\displaystyle{\small\begin{aligned}
\Big{[}&(l_{1}^{1}\wedge\cdots\wedge l_{i_{1}}^{1}\wedge
p_{1}^{1}\wedge\cdots\wedge p_{n-i_{1}}^{1}),(l_{1}^{2}\wedge\cdots\wedge
l_{i_{2}}^{2}\wedge p_{1}^{2}\wedge\cdots\wedge p_{n-i_{2}}^{2}),\ldots,\\\
&(l_{1}^{n-1}\wedge\cdots\wedge l_{i_{n-1}}^{n-1}\wedge
p_{1}^{n-1}\wedge\cdots\wedge
p_{n-i_{n-1}}^{n-1}),(l_{1}^{n}\wedge\cdots\wedge l_{i_{n}}^{n}\wedge
p_{1}^{n}\wedge\cdots\wedge p_{n-i_{n}}^{n})\Big{]}\\\
&=\frac{1}{2^{n-1}}\Big{(}(-1)^{\sum_{\begin{subarray}{c}s=1\end{subarray}}^{n-1}i_{s}(n-i_{s})}\big{(}g_{n-i_{1}}(p_{1}^{1},\ldots,p_{n-i_{1}}^{1},l_{1}^{1},\ldots,l_{i_{1}}^{1})\\\
&\hskip 48.36958pt\wedge
f_{i_{2}}(l_{1}^{2},\ldots,l_{i_{2}}^{2},p_{1}^{2},\ldots,p_{n-i_{2}}^{2})\\\
&\hskip 48.36958pt\wedge\cdots\wedge
f_{i_{n-1}}(l_{1}^{n-1},\ldots,l_{i_{n-1}}^{n-1},p_{1}^{n-1},\ldots,p_{n-i_{n-1}}^{n-1})\\\
&\hskip 48.36958pt\wedge
f_{i_{n}}(l_{1}^{n},\ldots,l_{i_{n}}^{n},p_{1}^{n},\ldots,p_{n-i_{n}}^{n})\big{)}\\\
&\quad+\sum_{r=2}^{n-1}(-1)^{\sum_{\begin{subarray}{c}s=1\\\ s\neq
r\end{subarray}}^{n-1}i_{s}(n-i_{s})}\big{(}g_{n-i_{1}}(p_{1}^{1},\ldots,p_{n-i_{1}}^{1},l_{1}^{1},\ldots,l_{i_{1}}^{1})\\\
&\hskip 28.45274pt\wedge
g_{n-i_{r}}(p_{1}^{r},\ldots,p_{n-i_{1}}^{r},l_{1}^{r},\ldots,l_{i_{1}}^{r})\wedge
f_{i_{2}}(l_{1}^{2},\ldots,l_{i_{2}}^{2},p_{1}^{2},\ldots,p_{n-i_{2}}^{2})\\\
&\hskip 28.45274pt\wedge\cdots\wedge
f_{i_{r-1}}(l_{1}^{r-1},\ldots,l_{i_{2}}^{r-1},p_{1}^{r-1},\ldots,p_{n-i_{2}}^{r-1})\\\
&\hskip 28.45274pt\wedge
f_{i_{r+1}}(l_{1}^{r+1},\ldots,l_{i_{2}}^{r+1},p_{1}^{r+1},\ldots,p_{n-i_{2}}^{r+1})\\\
&\hskip 28.45274pt\wedge\cdots\wedge
f_{i_{n-1}}(l_{1}^{n-1},\ldots,l_{i_{2}}^{n-1},p_{1}^{n-1},\ldots,p_{n-i_{2}}^{n-1})\\\
&\hskip 28.45274pt\wedge
f_{i_{n}}(l_{1}^{n},\ldots,l_{i_{2}}^{n},p_{1}^{n},\ldots,p_{n-i_{2}}^{n})\big{)}\\\
&\quad+\cdots+(-1)^{i_{1}(n-i_{1})}\big{(}g_{n-i_{1}}(p_{1}^{1},\ldots,p_{n-i_{1}}^{1},l_{1}^{1},\ldots,l_{i_{1}}^{1})\\\
&\hskip 28.45274pt\wedge
g_{n-i_{2}}(p_{1}^{2},\ldots,p_{n-i_{2}}^{2},l_{1}^{2},\ldots,l_{i_{2}}^{2})\\\
&\hskip 28.45274pt\wedge\cdots\wedge
g_{n-i_{n-1}}(p_{1}^{n-1},\ldots,p_{n-i_{n-1}}^{n-1},l_{1}^{n-1},\ldots,l_{i_{n-1}}^{n-1})\\\
&\hskip 28.45274pt\wedge
g_{n-i_{n}}(p_{1}^{n},\ldots,p_{n-i_{n}}^{n},l_{1}^{n},\ldots,l_{i_{n}}^{n})\big{)}\Big{)},\end{aligned}}$

In what follows, we recall the definition of crossed modules of $n$-Lie
algebras.

######  Definition 1.2 (Crossed module).

A crossed module is a homomorphism of Lie $n$-algebras $\mu:L\longrightarrow
P$ together with an action $\\{g_{i}\\}_{1\leq i\leq n-1}$ of $P$ on $L$
satisfying the following conditions:

  1. (1)

$\mu$ is compatible with the action of $P$ on $L$, that is,

$\mu\left(g_{i}(p_{1},\ldots,p_{i},l_{1},\ldots,l_{n-i})\right)=[p_{1},\ldots,p_{n-i},\mu(l_{1}),\ldots,\mu(l_{i})],$

for all $1\leq i\leq n-1$.

  2. (2)

For all $1\leq i\leq n-1$,

$g_{i}(\mu(l_{1}),\ldots,\mu(l_{i}),l_{i+1},\ldots,l_{n})=[l_{1},\ldots,l_{i},l_{i+1},\ldots,l_{n}].$

  3. (3)

For all $1\leq i\leq n-1$ and $i+1\leq j\leq n$,

$g_{i}(p_{1},\ldots,p_{i},l_{i+1},\ldots,l_{n})=(-1)^{j-i-1}g_{i+1}(p_{1},\ldots,p_{i},\mu(l_{j}),l_{i+1},\ldots,l_{j-1},l_{j+1},\ldots,l_{n}).$

It is easy to check that the kernel of $\mu$ is in the center of $L$ and that
its image is an ideal in $P$. Furthermore, the $n$-Lie algebra
$\mathrm{Im}(\mu)$ acts trivially on the center $Z(L)$, and so trivially on
$\ker\mu$. Hence $\ker\mu$ inherits an action of $P/\mathrm{Im}(\mu)$ making
$\ker\mu$ a representation of the $n$-Lie algebra $P/\mathrm{Im}(\mu)$.

For the first time, the author and Saeedi [4] introduced the concept of free
$n$-Lie algebras and, then in [1], defined the concept of basic commutators of
weight $w$ in $d$-dimensional $n$-Lie algebras. Also, we proved some of its
properties and the formula to calculate the number of them. In what follows,
we review it.

######  Theorem 1.3 ([1]).

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

(1.2)
$l_{d}^{n}(w)=\sum_{j=1}^{\alpha_{0}}\beta_{j^{*}}\left(\sum_{i=2}^{w-1}\alpha_{i}{{{{d}\choose{n-1}}}\choose{w-i}}\right),$

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
$(a+b)^{w-3}$, $({\text{i.e.}}~{}\alpha_{i}={{w-3}\choose{i-2}})$ 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
$\beta_{j^{*}}=(d-n-j^{*}+2)$.

The following theorem expresses one of the most important and main
applications of basic commutators of $n$-Lie algebras.

######  Theorem 1.4 ([1]).

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
$\sum\limits_{j=0}^{c-1}l_{d}^{n}(i+j)$, where $c=1,2,\ldots$.

The next proposition determines the dimension of $c$-nilpotent multiplier of
all abelian $n$-Lie algebras.

######  Proposition 1.5 ([2]).

Let $L$ be an abelian $n$-Lie algebra with dimension finite $d$. Then
$\dim\mathcal{M}^{(c)}(L)=l^{n}_{d}(c+1)$. In particular,
$\dim\mathcal{M}(L)=l_{d}^{n}(2)=\dfrac{1}{2}d(d-1)$.

##  2\. Preliminary results

In this section, we try to present some important and practical tools in order
to prove the main results of this paper.

######  Lemma 2.1.

Let $0\longrightarrow R\longrightarrow F\longrightarrow L\longrightarrow 0$ be
a free presentation of the $n$-Lie algebra $L$ and let $M$ be an ideal of $L$
with free presentation $S/R$, for some ideal $S$ of $F$. Then

$\gamma_{c+1}(L)\cap M\cong\dfrac{(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)}{\mathcal{M}^{(c)}(L)}.$

###### Proof.

By the definition of the lower central series, we know that

$\displaystyle\gamma_{c+1}(L)\cap M=\gamma_{c+1}(F/R)\cap(S/R)$
$\displaystyle=\dfrac{\gamma_{c+1}(F)+R}{R}\cap\dfrac{S}{R}$
$\displaystyle=\dfrac{(\gamma_{c+1}(F)+R)\cap S}{R}$
$\displaystyle=\dfrac{(\gamma_{c+1}(F)\cap S)+(R\cap S)}{R}$
$\displaystyle=\dfrac{(\gamma_{c+1}(F)\cap S)+R}{R}$
$\displaystyle\cong\dfrac{(\gamma_{c+1}(F)\cap S)}{(\gamma_{c+1}(F)\cap S)\cap
R}$ $\displaystyle=\dfrac{(\gamma_{c+1}(F)\cap S)}{\gamma_{c+1}(F)\cap R}$
$\displaystyle\cong\dfrac{(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)}{(\gamma_{c+1}(F)\cap
R)/\gamma_{c+1}(R,F,\dots,F)}$ $\displaystyle=\dfrac{(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)}{\mathcal{M}^{(c)}(L)}.$

∎

######  Lemma 2.2.

Let $L$ be an $n$-Lie algebra with a free presentation $L\cong F/R$, and let
$M$ be its central ideal and $M\cong S/R$, for some ideals $S$ of the free
$n$-Lie algebra $F$. Then
$\gamma_{c+1}(S,F,\dots,F)/(\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S))$ is a
homomorphic image of

$(L/M)^{ab}\ {}^{c}\otimes
M=\underbrace{(L/M)^{ab}\otimes\dots\otimes(L/M)^{ab}}_{c-times}\otimes M,$

where $(L/M)^{ab}=\dfrac{L/M}{(L/M)^{2}}$ and the tensor product is in sense
of tensor product of $n$-Lie algebras.

###### Proof.

Since $F/R$ is a free presentation of $L$, so there exists an epimorphism
$\pi:F\longrightarrow L$ such that $\ker\pi=R$. Consider the following map:

$\theta:\begin{cases}\underbrace{(L/M)^{ab}\times\dots\times(L/M)^{ab}}_{(c-1)(n-1)-times}\times
M\longrightarrow\gamma_{c+1}(S,F,\dots,F)/(\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S))\\\
(\bar{l}_{1},\dots,\bar{l}_{n},\bar{l}_{n+1},\dots,\bar{l}_{2n-1},\dots,\bar{l}_{(c-1)(n-1)})\longmapsto\\\
\\\
\overline{[\dots[[\bar{f}_{1},\dots,\bar{f}_{n-1},m],\bar{f}_{n},\bar{f}_{n+1},\dots,\bar{f}_{2n-2}],\dots],\bar{f}_{(c-2)(n-1)+1},\dots,\bar{f}_{(c-1)(n-1)}],\dots]},\end{cases}$

where $\pi^{-1}(l_{i})=\bar{f}_{i}$, for all $i$. It is easy to check that
$\theta$ is linear, and hence it induces the following epimorphism:

$(L/M)^{ab}\ {}^{c}\otimes
M=\underbrace{(L/M)^{ab}\otimes...\otimes(L/M)^{ab}}_{c-times}\otimes
M\rightarrow\gamma_{c+1}(S,F,...,F)/(\gamma_{c+1}(R,F,...,F)+\gamma_{c+1}(S)).$

∎

The author and Saeedi [3] proved a result similar to the following lemma for
non-abelian tensor product of $n$-Lie algebras. By a similar method and
reasoning, the following lemma can be proved for the non-abelian exterior
(wedge) product of $n$-Lie algebras as well.

######  Lemma 2.3.

Let $L$, $P$, $K$, and $Q$ be $n$-Lie algebras such that $L$ and $K$ act
compatibly on each other by $\\{f_{i}\\}_{1\leq i\leq n-1}$ and
$\\{g_{i}\\}_{1\leq i\leq n-1}$, and so they act $P$ and $Q$ by
$\\{h_{i}\\}_{1\leq i\leq n-1}$ and $\\{s_{i}\\}_{1\leq i\leq n-1}$,
respectively. Moreover, suppose that $\sigma_{1}:L\longrightarrow P$ and
$\sigma_{2}:K\longrightarrow Q$ are $n$-Lie homomorphisms preserving the
actions in the sense that

$\displaystyle\sigma_{1}(g_{i}(a_{1},\ldots,a_{n}))=s_{i}(\sigma_{r}(a_{1}),\ldots,\sigma_{r}(a_{n})),$
$\displaystyle\sigma_{2}(f_{i}(a_{1},\ldots,a_{n}))=h_{i}(\sigma_{r}(a_{1}),\ldots,\sigma_{r}(a_{n})),$

for all $a_{1},\ldots,a_{n}\in L\cup K$, where $r=1$ if $a_{i}\in L$, and
$r=2$ if $a_{i}\in K$. Then there is a unique homomorphism (up to sign)

$\begin{array}[]{rcl}\sigma_{1}\wedge\sigma_{2}:L\wedge
K&\longrightarrow&P\wedge Q\\\
\sum\limits_{i=1}^{n-1}l_{1}^{i}\wedge\cdots\wedge l_{i}^{i}\wedge
k_{i+1}^{i}\wedge\cdots\wedge
k_{n}^{i}&\longmapsto&\sum\limits_{i=1}^{n-1}\sigma_{1}(l_{1}^{i})\wedge\cdots\wedge\sigma_{1}(l_{i}^{i})\wedge\sigma_{2}(k_{i+1}^{i})\wedge\cdots\wedge\sigma_{2}(k_{n}^{i}).\end{array}$

###### Proof.

The proof is similar to [3, Proposition 4.5]. ∎

######  Lemma 2.4.

Let $F$ be a free $n$-Lie algebra and let $S$ and $I$ be two ideals of $F$.
Then

$[I,\gamma_{j}(S,F,\dots,F),F,\dots,F]\subseteq\gamma_{j}(I,F,\dots,F),$

for all $j\geq 1$.

###### Proof.

We prove this by induction on $j$. The result is trivially true for $j=1$.
Assume that the result holds for $j\geq 1$. Then for any ideal $I$ in $F$, by
the definition of $\gamma_{j+1}$ and the Jacobi identity, we have

$\displaystyle[I,\gamma_{j+1}(S,F,\dots,F),F,\dots,F]$
$\displaystyle=[\gamma_{j+1}(S,F,\dots,F),I,F,\dots,F]$
$\displaystyle=[[\gamma_{j}(S,F,\dots,F),\underbrace{F,\dots,F}_{n-1}],I,\underbrace{F,\dots,F}_{n-2}]$
$\displaystyle=[[\gamma_{j}(S,F,\dots,F),I,\underbrace{F,\dots,F}_{n-2}],\underbrace{F,\dots,F}_{n-1}]$
$\displaystyle\quad+[[\gamma_{j}(S,F,\dots,F),[F,I,\underbrace{F,\dots,F}_{n-2}],\underbrace{F,\dots,F}_{n-2}]$
$\displaystyle\quad+\dots$
$\displaystyle\quad+[\gamma_{j}(S,F,\dots,F),\underbrace{F,\dots,F}_{n-2},[F,I,\underbrace{F,\dots,F}_{n-2}]]$
$\displaystyle\subseteq[\gamma_{j}(I,F,\dots,F),F,\dots,F]+[\gamma_{j}(S,F,\dots,F),I,F,\dots,F]$
$\displaystyle\subseteq\gamma_{j}([I,F,\dots,F],F,\dots,F)+\gamma_{j+1}(I,F,\dots,F)$
$\displaystyle=\gamma_{j+1}(I,F,\dots,F).$

∎

Let $L$ be an $n$-Lie algebra with a free presentation $F/R$, for some free
$n$-Lie algebra $F$ and an ideal $R$ of $F$ with the $n$-Lie algebra
epimorphism $\pi$, and let $M$ be an ideal of $L$ with a free presentation
$S/R$, for some ideal $S$ of $F$. Since $R\subseteq S\subseteq F$, so
$\gamma_{j}(R,F,\dots,F)\subseteq\gamma_{j}(S,F,\dots,F)$, and hence we can
put $T_{j}=\dfrac{\gamma_{j}(S,F,\dots,F)}{\gamma_{j}(R,F,\dots,F)}$, for all
$j\geq 1$. The following lemma shows that $T_{j}$ and $L$ act on each other
compatibly.

######  Lemma 2.5.

By the above assumptions and notations, $T_{j}$ and $L$ act compatibly on each
other.

###### Proof.

Define the two family of maps $\\{g_{i}\\}$ and $\\{h_{i}\\}$ as follows:

$\begin{cases}g_{i}:L^{\times i}\times T_{j}^{\times n-i}\longrightarrow
T_{j}\\\
g_{i}\left(l_{1},\dots,l_{i},\bar{t_{i+1}},\dots,\bar{t_{n}}\right)=[f_{1},\dots,f_{i},t_{i+1},\dots,t_{n}]+\gamma_{j}(R,F,\dots,F)\end{cases}$

and

$\begin{cases}h_{i}:T_{j}^{\times i}\times L^{\times n-i}\longrightarrow L\\\
h_{i}\left(\bar{t_{1}},\dots,\bar{t_{i}},l_{i+1},\dots,l_{n}\right)=\pi([t_{1},\dots,t_{i},l_{i+1},\dots,l_{n}])\end{cases}$

where $l_{p}\in L$, $x_{q}\in\gamma_{j}(S,F,\dots,F)$, and
$f_{p}=\pi^{-1}(l_{p})$, for all $p$ and $q$. By Lemma 2.4, it is easy to
check that $T_{j}$ and $L$ act compatibly on each other. ∎

The following proposition is useful in some of our results and future
investigation.

######  Proposition 2.6.

Let $L$ be an $n$-Lie algebra and let $M$ be its ideal. Then

  1. (1)

$T_{j+1}$ is an isomorphic image of $M\wedge^{j}L$.

  2. (2)

if $M$ is an $r$-central ideal of $L$, then $T_{j+1}$ is an isomorphic image
of $M\wedge^{j}K$, where $K=\dfrac{L}{\gamma_{r+1}(L)}$.

###### Proof.

  1. (1)

Consider the map $\lambda:T_{j}\longrightarrow L$ by
$\lambda(\bar{t})=\pi(t)$. Since $\pi$ is a well-defined $n$-Lie homomorphism,
so $\lambda$ is too. The $n$-Lie homomorphism $\lambda$ together with the
defined action of $L$ on $T_{j}$ in Lemma 2.5 and also, the identity map
$id_{L}:L\longrightarrow L$ are crossed modules of $n$-Lie algebras. The
family of bilinear maps $\phi_{j}^{i}:T_{j}^{\times i}\times L^{\times
n-i}\stackrel{{\scriptstyle epi.}}{{\longrightarrow}}T_{j+1}$ given by
$\phi_{(}\bar{x}_{1},\dots,\bar{x}_{i},l_{i+1},\dots,l_{n})=[x_{1},\dots,x_{i},f_{i+1},\dots,f_{n}]+\gamma_{j+1}(R,F,\dots,F)$,
is an $n$-multiplying. By Theorem 3.5 of [3] consequently, this family induces
an epimorphism $\Phi_{j}:T_{j}\wedge L\longrightarrow T_{j+1}$. Thus, by using
Lemma 2.3, we obtain an epimorphism $\Phi_{j}\wedge id_{L}:(T_{j}\wedge
L)\wedge L\longrightarrow T_{j+1}\wedge L$. Now, we continue by induction on
$j$.

  2. (2)

By the assumption, $\gamma_{n+1}(S,F,\dots,F)\subseteq R$ and then

$[\gamma_{j}(S,F,\dots,F),\gamma_{n+1}(F),F,\dots,F]\subseteq\gamma_{j+n+1}(S,F,\dots,F)\subseteq\gamma_{j+1}(R,F,\dots,F).$

So, the action of $\gamma_{n+1}(F)$ on $T_{j}$ is trivial and the $n$-Lie
algebras $L/\gamma_{n+1}(L)$ and $T_{j}$ act compatibly on each other by the
induced actions $\bar{g}_{i}$’s and $\bar{h}_{i}$’s of defined $g_{i}$’s and
$h_{i}$’s in Lemma 2.5; that is,

$\begin{cases}\bar{g}_{i}:L/\gamma_{n+1}(L)^{\times i}\times T_{j}^{\times
n-i}\longrightarrow T_{j}\\\
\bar{g}_{i}\left(\bar{l}_{1},\dots,\bar{l}_{i},\bar{t}_{i+1},\dots,\bar{t}_{n}\right)=g_{i}\left(l_{1},\dots,l_{i},\bar{t}_{i+1},\dots,\bar{t}_{n}\right)\\\
\qquad\qquad\qquad\qquad\qquad\quad=[f_{1},\dots,f_{i},t_{i+1},\dots,t_{n}]+\gamma_{j}(R,F,\dots,F)\end{cases}$

and

$\begin{cases}\bar{h}_{i}:T_{j}^{\times i}\times L/\gamma_{n+1}(L)^{\times
n-i}\longrightarrow L/\gamma_{n+1}(L)\\\
\bar{h}_{i}\left(\bar{t_{1}},\dots,\bar{t_{i}},\bar{l}_{i+1},\dots,\bar{l}_{n}\right)=\pi([t_{1},\dots,t_{i},l_{i+1},\dots,l_{n}])\end{cases}$

where $l_{p}\in L$, $x_{q}\in\gamma_{j}(S,F,\dots,F)$, and
$f_{p}=\pi^{-1}(l_{p})$, for all $p$ and $q$.

Now, as part (1), there exist epimorphisms $T_{j}\wedge K\longrightarrow
T_{j+1}$ and $M\wedge^{j}K\longrightarrow T_{j+1}$ and hence the proof is
completed.

∎

######  Corollary 2.7.

By the assumptions and notations in Proposition 2.6, there exists an
epimorphism $\kappa$ from $\ker\mu^{j}_{M}$ onto
$\dfrac{R\cap\gamma_{j+1}(S,F,\dots,F)}{\gamma_{j+1}(R,F,\dots,F)}$.

###### Proof.

The proof is similar to [16, Corollary 1.5]. ∎

##  3\. On the $c$-nilpotent multiplier of $n$-Lie algebras

In this section, first we introduce some inequalities for the dimension of the
$c$-nilpotent multiplier of the $n$-Lie algebras and their factors. Then, we
show that every $c$-perfect $n$-Lie algebra has at least one cover. Also, we
give some upper bounds for the dimension of $n$-nilpotent multiplier of
$n$-Lie algebras using the basic commutators.

The following lemma is one of the important tools for obtaining the required
inequalities, which is similar to the work of Araskhan and Rismanchian [7] for
the Lie algebras and the work of Jones [12] for the group case.

######  Lemma 3.1.

Let $L$ be an $n$-Lie algebra with dimension finite, and let $M$ be its ideal.
Then there exists an $n$-Lie algebra $K$ with its ideal $S$ such that

  1. (1)

$\gamma_{c+1}(L)\cap M=K/S$;

  2. (2)

$S\cong\mathcal{M}^{(c)}(L)$;

  3. (3)

$\mathcal{M}^{(c)}(L/M)$ is a homomorphic image of $K$.

  4. (4)

If, in addition, $M$ is a $c$-central ideal of $L$, then $\gamma_{c+1}(L)\cap
M$ is a homomorphic image of $\mathcal{M}^{(c)}(L/M)$.

###### Proof.

  1. (1)

Since $\gamma_{c+1}(F)\cap R\unlhd\gamma_{c+1}(F)\cap S$, so by Lemma 2.1, it
is enough to put $K/M=\dfrac{(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)}{(\gamma_{c+1}(F)\cap
R)/\gamma_{c+1}(R,F,\dots,F)}$.

  2. (2)

By the definition of the $c$-nilpotent multiplier of $L$ and Lemma 2.1, we
obtain $S=\mathcal{M}^{(c)}(L)$.

  3. (3)

Consider the free presentation

$L/M=\dfrac{F/R}{S/R}\cong F/S,$

and hence

$\mathcal{M}^{(c)}(L/M)=\dfrac{\gamma_{c+1}(F)\cap
S}{\gamma_{c+1}(S,F,\dots,F)}.$

Now, consider the map $\alpha:(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)\longrightarrow\dfrac{\gamma_{c+1}(F)\cap
S}{\gamma_{c+1}(S,F,\dots,F)}$. It is easy to check that $\alpha$ is an
epimorphism of $n$-Lie algebras with the kernel ideal
$\gamma_{c+1}(S,F,\dots,F)/\gamma_{c+1}(R,F,\dots,F)$.

  4. (4)

Since $M$ is a $c$-central ideal, so $\gamma_{c+1}(M,L,\dots,L)=1$. Hence
$\dfrac{\gamma_{c+1}(S,F,\dots,F)+R}{R}=R$. Thus
$\gamma_{c+1}(S,F,\dots,F)\subseteq R$. Therefore,
$\gamma_{c+1}(S,F,\dots,F)\subseteq\gamma_{c+1}(F)\cap R$ and
$\gamma_{c+1}(S,F,\dots,F)\subseteq\gamma_{c+1}(F)\cap S$. Also, by the proof
of Lemma 2.1, we know that

$\displaystyle\gamma_{c+1}(L)\cap M\cong\dfrac{\gamma_{c+1}(F)\cap
S}{\gamma_{c+1}(F)\cap R}\cong\dfrac{\dfrac{\gamma_{c+1}(F)\cap
S}{\gamma_{c+1}(S,F,\dots,F)}}{\dfrac{\gamma_{c+1}(F)\cap
R}{\gamma_{c+1}(S,F,\dots,F)}}=\dfrac{\mathcal{M}^{(c)}(L/M)}{\dfrac{\gamma_{c+1}(F)\cap
R}{\gamma_{c+1}(S,F,\dots,F)}}.$

Thus the proof is completed by considering the epimorphism
$\alpha:\mathcal{M}^{(c)}(L/M)\longrightarrow\gamma_{c+1}(L)\cap
M=\dfrac{\mathcal{M}^{(c)}(L/M)}{\dfrac{\gamma_{c+1}(F)\cap
R}{\gamma_{c+1}(S,F,\dots,F)}}$.

∎

The following corollary follows immediately from the previous lemma.

######  Corollary 3.2.

Let $L$ be an $n$-Lie algebra with dimension finite, and let $M$ be its ideal.
Then we have the following inequality:

$\dim\mathcal{M}^{(c)}(L/M)\leq\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap
M).$

###### Proof.

From Lemma 3.1, we have $\gamma_{c+1}(L)\cap M=K/S$ and
$S\cong\mathcal{M}^{(c)}(L)$. Thus

$\dim(\gamma_{c+1}(L)\cap M)=\dim K-\dim S,\qquad\dim
S=\dim\mathcal{M}^{(c)}(L).$

Hence,

$\dim(\gamma_{c+1}(L)\cap M)=\dim K-\dim\mathcal{M}^{(c)}(L).$

On the other hand, since $\mathcal{M}^{(c)}(L/M)$ is a homomorphic image of
$K$, so $\dim\mathcal{M}^{(c)}(L/M)\leq\dim K$. Therefore,

$\dim\mathcal{M}^{(c)}(L/M)\leq\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap
M).$

∎

######  Theorem 3.3.

Let $L$ be an $n$-Lie algebra with dimension finite and free presentation
$F/R$, and let $M$ be a central ideal of $L$ with free presentation $S/R$, for
some ideal $S$ of $F$. Then

$\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap
M)\leq\dim\mathcal{M}^{(c)}(L/M)+\dim\mathcal{M}^{(c)}(M)+\dim((L/M)^{ab}\
{}^{c}\otimes M).$

###### Proof.

Since $M$ is central, so $\gamma_{2}(S,F,\dots,F)\subseteq R$. By Corollary
3.2 and the proof of part (3) of Lemma 3.1, we know that
$\mathcal{M}^{(c)}(L/M)$ is an isomorphic image of $(\gamma_{c+1}(F)\cap
S)/\gamma_{c+1}(R,F,\dots,F)$ with the kernel ideal
$\gamma_{c+1}(S,F,\dots,F)/\gamma_{c+1}(R,F,\dots,F)$. Also, we have

$\dim\mathcal{M}^{(c)}(L/M)+\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}=\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap
M).$

On the other hand, we know that

$\dfrac{\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}}{\dfrac{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}{\gamma_{c+1}(R,F,\dots,F)}}=\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}.$

Hence

$\dim\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}-\dim\dfrac{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}{\gamma_{c+1}(R,F,\dots,F)}=\dim\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}.$

Also, by this fact that $M$ is a central ideal, we have

(3.3) $\displaystyle\gamma_{c+1}(S)\subseteq\gamma_{2}(S,F,\dots,F)\subseteq
R$ (3.4) $\displaystyle\gamma_{2}(R,F,\dots,F)\subseteq R.$

Therefore, it follows from equations (3.3) and (3.4) that

(3.5) $\gamma_{2}(R,F,\dots,F)+\gamma_{c+1}(S)\subseteq R.$

On the other hand, since $R\subseteq S$, so

(3.6) $\gamma_{2}(R,F,\dots,F)\subseteq\gamma_{c+1}(S),$

and hence equations (3.5) and (3.6) imply that

(3.7) $\gamma_{2}(R,F,\dots,F)+\gamma_{c+1}(S)\subseteq\gamma_{c+1}(S)\cap R.$

Thus

$\displaystyle\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap M)$
$\displaystyle=\dim\mathcal{M}^{(c)}(L/M)+\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dim\mathcal{M}^{(c)}(L/M)+\dim\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}$
$\displaystyle\quad+\dim\dfrac{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle(\text{by Lemma \ref{lem2.3(2)}})\qquad\qquad$
$\displaystyle\leq\dim\mathcal{M}^{(c)}(L/M)+\dim\left((L/M)^{ab}\
{}^{c}\otimes M\right)$
$\displaystyle\quad+\dim\dfrac{\gamma_{c+1}(R,F,\dots,F)+\gamma_{c+1}(S)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle(\text{by \eqref{equation(5)}})\qquad\qquad\qquad\ $
$\displaystyle\leq\dfrac{\gamma_{c+1}(S)\cap R}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dim\mathcal{M}^{(c)}(L/M)+\dim\mathcal{M}^{(c)}(M)$
$\displaystyle\quad+\dim\left((L/M)^{ab}\ {}^{c}\otimes M\right).$

∎

The following proposition plays a fundamental and important role in obtaining
some main results of this section.

######  Proposition 3.4.

Let $L$ be an $n$-Lie algebra and let $M$ be its ideal. Then we have the
following exact sequences:

(3.8)
$\displaystyle\ker(\mu_{M}^{c})\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)\longrightarrow\dfrac{M\cap\gamma_{c+1}(L)}{\gamma_{c+1}(M,L,\dots,L)}\longrightarrow
0;$ (3.9) $\displaystyle
M\wedge^{c}\dfrac{L}{\gamma_{r+1}(L)}\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)\longrightarrow
M\cap\gamma_{c+1}(L)\longrightarrow 0,$

with the condition that $M$ is $r$-central and $r\leq c$.

###### Proof.

Let $0\longrightarrow R\longrightarrow F\longrightarrow L\longrightarrow 0$ be
a free presentation of $L$ and let $M\cong S/R$, for some ideal $S$ of $F$.
Then we have the following exact sequence:

$\displaystyle
0\longrightarrow\dfrac{R\cap\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\longrightarrow\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle\longrightarrow\dfrac{S\cap\gamma_{c+1}(F)}{\gamma_{c+1}(S,F,\dots,F)}$
(3.10)
$\displaystyle\longrightarrow\dfrac{(S+\gamma_{c+1}(F))+R}{\gamma_{c+1}(S,F,\dots,F)+R}\longrightarrow
0.$

Since $r\leq c$ and $M$ is $c$-central, so

(3.11) $\gamma_{c+1}(S,F,\dots,F)\subseteq\gamma_{r+1}(S,F,\dots,F)\subseteq
R=0_{S/R}.$

We know that $L\cong F/R$ and $M\cong S/R$, and hence $F/S$ is a free
presentation of $L/M$. Thus according to the definition of $c$-nilpotent
multiplier of $L/M$, we have

(3.12)
$\displaystyle\mathcal{M}^{(c)}(L)=\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)},\qquad\mathcal{M}^{(c)}(L/M)=\dfrac{S\cap\gamma_{c+1}(F)}{\gamma_{c+1}(S,F,\dots,F)}.$

Also,

$\displaystyle\dfrac{M\cap\gamma_{c+1}(L)}{\gamma_{c+1}(M,L,\dots,L)}$
$\displaystyle=\dfrac{(S/R)\cap\gamma_{c+1}(F/R)}{\gamma_{c+1}(S/R,F/R,\dots,F/R)}$
$\displaystyle=\dfrac{\dfrac{S+R}{R}\cap\dfrac{\gamma_{c+1}(F)+R}{R}}{\dfrac{\gamma_{c+1}(S,F,\dots,F)+R}{R}}$
$\displaystyle=\dfrac{\dfrac{(S\cap\gamma_{c+1}(F))+R}{R}}{\dfrac{\gamma_{c+1}(S,F,\dots,F)+R}{R}}$
(3.13)
$\displaystyle=\dfrac{(S\cap\gamma_{c+1}(F))+R}{\gamma_{c+1}(S,F,\dots,F)+R}.$

By Corollary 2.7, there exists an epimorphism
$\psi:\ker(\mu_{M}^{c})\longrightarrow\dfrac{R\cap\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$,
and so

(3.14)
$\dfrac{R\cap\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\cong\dfrac{\ker(\mu_{M}^{c})}{\ker\psi}.$

Therefore, by replacing the terms of sequence (3.10) by (3.11), (3.12),
(3.13), and (3.14), the sequence (3.8) is obtained.

On the other hand, from (3.11), we have

(3.15)
$\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle\theta}}{{\longrightarrow}}\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}.$

Moreover, since $M$ is $c$-central, so by part $(2)$ of Proposition 2.6, there
is an epimorphism
$\lambda:M\wedge^{c}\dfrac{L}{\gamma_{c+1}(L)}\longrightarrow
T_{c+1}=\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$. Hence

(3.16)
$\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\cong\dfrac{M\wedge^{c}\dfrac{L}{\gamma_{c+1}(L)}}{\ker\lambda}.$

Now, by putting (3.12), (3.13), (3.14), (3.15), and (3.16) in the sequence
(3.10), we obtain the sequence (3.9). ∎

The following corollary is an immediate consequence of Proposition 3.4 and
helps us to prove the useful Proposition 3.7.

######  Corollary 3.5.

Let $L$ be a finite-dimensional $n$-Lie algebra and let $M$ be its ideal. Then
the following properties hold:

  1. (1).

The $n$-Lie algebra $\mathcal{M}^{(c)}(L)$ has dimension finite.

  2. (2).

$\dim\mathcal{M}^{(c)}(L/M)\leq\mathcal{M}^{(c)}(L)+\dim\left(\dfrac{M\cap\gamma_{c+1}(L)}{\gamma_{c+1}(M,L,\dots,L)}\right)$.

  3. (3).

$\dim\mathcal{M}^{(c)}(L)+\dim(M\cap\gamma_{c+1}(L))=\dim\mathcal{M}^{(c)}(L/M)+\dim\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$,
where $F/R$ and $S/R$ are free presentations of $L$ and $M$, respectively. In
particular,

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)=\dim\left(\gamma_{c+1}\left(\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\right)\right).$

  4. (4).

If $\mathcal{M}^{(c)}(L)=0$, then
$\mathcal{M}^{(c)}(L/M)\cong\dfrac{M\cap\gamma_{c+1}(L)}{\gamma_{c+1}(M,L,\dots,L)}$.

  5. (5).

If $M$ is an $r$-central ideal, that is, $M\subseteq Z_{r}(L)$ and $r\leq c$,
then

$\dim\mathcal{M}^{(c)}(L)+\dim(M\cap\gamma_{c+1}(L))\leq\dim\mathcal{M}^{(c)}(L/M)+\dim\left(M\wedge^{c}\dfrac{L}{\gamma_{r+1}(L)}\right).$

###### Proof.

  1. (1).

It is clear according to the short exact sequence (3.10) in the proof of
Proposition 3.4.

  2. (2).

According to part $(1)$ of Proposition 3.4, this is clear.

  3. (3).

Since $\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$ is an
isomorphic image of $M\wedge^{c}\dfrac{L}{\gamma_{r+1}(L)}$, so by replacing
$M\wedge^{c}\dfrac{L}{\gamma_{r+1}(L)}$ with
$\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$ in equation
(3.9), we have the following sequence:

$0\longrightarrow\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)\longrightarrow
M\cap\gamma_{c+1}(L)\longrightarrow 0.$

Hence

$\dim\mathcal{M}^{(c)}(L)+\dim(M\cap\gamma_{c+1}(L))=\dim\mathcal{M}^{(c)}(L/M)+\dim\dfrac{\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}.$

If $M=L$, then $\dim\mathcal{M}^{(c)}(L/M)=0$ and

$\gamma_{c+1}\left(\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\right)=\dfrac{\gamma_{c+1}(F,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}.$

Therefore

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)=\dim\left(\gamma_{c+1}\left(\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\right)\right).$

  4. (4).

If $\mathcal{M}^{(c)}(L)=1$, then by part $(1)$ of Proposition 3.4, the proof
is completed.

  5. (5).

According to part $(2)$ of Proposition 3.4, the result is obtained
immediately.

∎

The short exact sequence $0\longrightarrow M\longrightarrow
L^{*}\longrightarrow L\longrightarrow 0$ of $n$-Lie algebras is called
$c$-stem cover, if $M\subseteq Z_{c}(L^{*})\cap\gamma_{c+1}(L^{*})$ and
$M\cong\mathcal{M}^{(c)}(L)$. Then $L^{*}$ is called $c$-cover of $L$.
Specially, if $c=1$, then it is the definition of cover.

Moreover, an $n$-Lie algebra $L$ is called $c$-perfect, if
$\gamma_{c+1}(L)=L$.

######  Corollary 3.6.

Let $L$ be a $c$-perfect finite-dimensional $n$-Lie algebra with $c$-cover
$L^{*}$. Then $\mathcal{M}^{(c)}(L)$ is trivial.

###### Proof.

By the definition of $c$-cover, there exists an ideal $M$ of $L$ such that

$L^{*}/M\cong L,\qquad M\subseteq
Z_{c}(L^{*})\cap\gamma_{c+1}(L^{*}),\qquad\mathcal{M}^{(c)}(L)\cong M.$

Since $M$ is $c$-central and $L$ is $c$-perfect, that is, $L=\gamma_{c+1}(L)$,
so by part (5) of 3.5, we have

$\dim\mathcal{M}^{(c)}(L^{*})+\dim(M\cap\gamma_{c+1}(L^{*}))\leq\dim\mathcal{M}^{(c)}(L^{*}/M)+\underbrace{\dim(M\wedge^{c}\dfrac{L^{*}}{\gamma_{c+1}(L^{*})})}_{0}.$

Thus

$\dim\mathcal{M}^{(c)}(L^{*})\leq\dim\mathcal{M}^{(c)}(L^{*}/M)=\dim\mathcal{M}^{(c)}(L).$

Hence $\dim\mathcal{M}^{(c)}(L^{*})\leq 0$, which implies that
$\dim\mathcal{M}^{(c)}(L^{*})=0$. ∎

The author in [2, Paroposition 1.6] proved that if $L$ is an $n$-Lie algebra
with the free representation $F/R$ and the map
$\pi:F/\gamma_{c+1}(R,F,\dots,F)\longrightarrow F/R$ is an epimorphism, then
$Z_{c}^{*}(L)=\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$. Using this
fact, we prove the following practical theorem.

######  Proposition 3.7.

Let $L$ be a finite-dimensional $n$-Lie algebra and let $M$ be an ideal of $L$
with $M\subseteq Z_{c}(L)$. Then the following statements are equivalent:

  1. (1)

$M\subseteq Z^{*}_{c}(L)$;

  2. (2)

The natural map
$\alpha:\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)$ is a
monomorphism;

  3. (3)

$\dim\mathcal{M}^{(c)}(L/M)=\dim\mathcal{M}^{(c)}(L)+\dim(M\cap\gamma_{c+1}(L))$.

###### Proof.

Similar to [2, Lemma 1.7], it can be proved that $M\subseteq Z^{*}_{c}(L)$ if
and only if $\alpha:\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)$
is a monomorphism.

If $M\subseteq Z^{*}_{c}(L)$, then by part (4) of Lemma 3.1 and part (2), we
have the following sequence:

$0\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/M)\longrightarrow
M\cap\gamma_{c+1}(L)\longrightarrow 0.$

Hence

$\dim\mathcal{M}^{(c)}(L/M)=\dim\mathcal{M}^{(c)}(L)+\dim(M\cap\gamma_{c+1}(L)).$

Conversely, if part (3) is hold, then by comparing part (3) of Corollary 3.5
and part (2), we get
$\gamma_{c+1}(S,F,\dots,F)\subseteq\gamma_{c+1}(R,F,\dots,F)$. On the other
hand, since $R\subseteq S$, so
$\gamma_{c+1}(R,F,\dots,F)\subseteq\gamma_{c+1}(S,F,\dots,F)$. Therefore
$\gamma_{c+1}(S,F,\dots,F)=\gamma_{c+1}(R,F,\dots,F)$, and this fact is
equivalent to $S/\gamma_{c+1}(R,F,\dots,F)\subseteq
Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))$. From [2, Proposition 1.6] we know that
$Z_{c}^{*}(L)=\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$, and hence
$\pi(S/\gamma_{c+1}(R,F,\dots,F))\subseteq Z_{c}^{*}(L)$. Therefore,
$M\subseteq Z_{c}^{*}(L)$, which completes the proof. ∎

The following proposition states that every $c$-perfect $n$-Lie algebra has at
least one cover.

######  Proposition 3.8.

Let $L$ be a $c$-perfect $n$-Lie algebra. Then $L$ has at least one cover.

###### Proof.

Let $\pi:F\longrightarrow L$ be an epimorphism with $\ker\pi=R$, and hence
$F/R$ be a free presentation of $L$. Then

$0\longrightarrow\dfrac{R}{\gamma_{c+1}(R,F,\dots,F)}\longrightarrow\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle\bar{\pi}}}{{\longrightarrow}}L\longrightarrow
0,$

be the free $c$-central extension of $L$. Since
$\bar{\pi}(\gamma_{c+1}(F)/\gamma_{c+1}(R,F,\dots,F))=L$, so by restricting
$\pi$ to $\gamma_{c+1}(F)/\gamma_{c+1}(R,F,\dots,F)$, we have

$\displaystyle\ker\bar{\pi}=\ker\pi\bigcap\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dfrac{R}{\gamma_{c+1}(R,F,\dots,F)}\bigcap\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}=\mathcal{M}^{(c)}(L).$

Thus we get the following $c$-central extension:

(3.17)
$0\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\longrightarrow
L\longrightarrow 0.$

On the other hand, $L$ is $c$-perfect, so

$L\cong\dfrac{F}{R}\longrightarrow F\cong L\oplus R\longrightarrow
F=\gamma_{c+1}(F)+R.$

Thus

$\displaystyle\gamma_{c+1}\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\right)$
$\displaystyle=\dfrac{\gamma_{c+1}(\gamma_{c+1}(F),F,\dots,F)+\gamma_{c+1}(R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dfrac{\gamma_{c+1}(\gamma_{c+1}(F)+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}.$

Therefore, $\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$ is $c$-perfect
and so (3.17) is a stem $c$-central extension of $L$. ∎

In the following theorem, we give upper and lower bounds for the dimension of
$\mathcal{M}^{(c)}(L)$.

######  Theorem 3.9.

Let $L$ be a finite-dimensional nilpotent $n$-Lie algebra of class $m$, let
$d=d(L)$, and let $\dim\gamma_{i+1}(L)=a_{i}$, for all $i\geq 1$. Then the
following inequalities are hold:

$l_{d}^{n}(c+1)\leq\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)\leq
l_{d}^{n}(c+1)+\sum_{i=1}^{c}a_{i}d^{c(n-1)-i+1}.$

If $L$ is abelian, then the lower and upper bounds are attained.

###### Proof.

Since $L$ is nilpotent, so $\Phi(L)=L^{2}$, and hence $L/L^{2}$ is an abelian
$n$-Lie algebra of dimension $d$. If we put $M=L^{2}$, then by Proposition 1.5
we have

(3.18)
$\dim\mathcal{M}^{(c)}(L/M)=\dim\mathcal{M}^{(c)}(L/L^{2})=l_{d}^{n}(c+1).$

Thus by putting $M=L^{2}$ in part $(2)$ of Corollary 3.5, we obtain

$\displaystyle\dim\mathcal{M}^{(c)}(L/L^{2})\leq\dim\mathcal{M}^{(c)}(L)+\dim\dfrac{L^{2}\cap\gamma_{c+1}(L)}{\gamma_{c+1}(L^{2},L,\dots,L)}$
$\displaystyle=\dim\mathcal{M}^{(c)}(L)+\dim\dfrac{\gamma_{c+1}(L)}{\gamma_{c+2}(L)}$
(3.19) $\displaystyle\leq\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L).$

Therefore, by (3.18) and (3.19), we have

(3.20) $l_{d}^{n}(c+1)\leq\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L).$

Now let $0\longrightarrow R\longrightarrow F\longrightarrow L\longrightarrow
0$ be a free presentation of $L$, and put $M=L^{2}$, in part $(1)$ of
Proposition 2.6. Then

$L=\dfrac{F}{R}\longrightarrow
M=L^{2}=\left(\dfrac{F}{R}\right)^{\prime}=\dfrac{F^{2}+R}{R},$

and hence
$T_{c+1}=\dfrac{\gamma_{c+1}(F^{2}+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$,
and it is an isomorphic image of $L^{2}\wedge^{c}L$, by part $(1)$ of
Proposition 2.6. That is, there is an epimorphism
$\psi:L^{2}\wedge^{c}L\longrightarrow T_{c+1}$ such that

(3.21) $\dfrac{L^{2}\wedge^{c}L}{\ker\psi}\cong
T_{c+1}=\dfrac{\gamma_{c+1}(F^{2}+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}.$

Now, if $M=L=F/R$, then
$T^{\prime}_{c+1}=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$. Thus

(3.22)
$\displaystyle\dfrac{T^{\prime}_{c+1}}{T_{c+1}}=\dfrac{\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}}{\dfrac{\gamma_{c+1}(F^{2}+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}}$
$\displaystyle\cong\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2}+R,F,\dots,F)}.$

On the other hand, by (3.22) we know that there exists the following
epimorphism:

$\displaystyle\theta:T^{\prime}_{c+1}=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle
epi.}}{{\longrightarrow}}\dfrac{T^{\prime}_{c+1}}{T_{c+1}}=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2}+R,F,\dots,F)}.$

Then
$\ker\theta=\dfrac{\gamma_{c+1}(F^{2}+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$,
and so we obtain the following short exact sequence:

$0\longrightarrow\ker\theta\stackrel{{\scriptstyle
inc.}}{{\longrightarrow}}T^{\prime}_{c+1}\stackrel{{\scriptstyle\theta}}{{\longrightarrow}}\dfrac{T^{\prime}_{c+1}}{T_{c+1}}\longrightarrow
0.$

That is,

(3.23)
$0\longrightarrow\dfrac{\gamma_{c+1}(F^{2}+R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle
inc.}}{{\longrightarrow}}\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle\theta}}{{\longrightarrow}}\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2}+R,F,\dots,F)}\longrightarrow
0.$

Thus by (3.21) and (3.23), we have

(3.24)
$L^{2}\wedge^{c}L\longrightarrow\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\stackrel{{\scriptstyle\theta}}{{\longrightarrow}}\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2}+R,F,\dots,F)}\longrightarrow
0.$

Now, by part (3) of Corollary 3.5, and then using (3.24) and Theorem 1.4, we
have

$\displaystyle\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)$
$\displaystyle=\dim\left(\gamma_{c+1}\left(\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\right)\right)$
$\displaystyle=\dim\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}\right)$
$\displaystyle\leq\dim(L^{2}\wedge^{c}L)+\dim\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(\underbrace{F^{2}+R}_{\supseteq
F^{2}},F,\dots,F)}\right)$
$\displaystyle\leq\dim(L^{2}\wedge^{c}L)+\dim\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2},F,\dots,F)}\right)$
(3.25)
$\displaystyle=\dim(L^{2}\wedge^{c}L)+\dim\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+2}(F)}\right).$

We know that $L^{2}\wedge^{c}L=(\dots((L^{2}\wedge L)\wedge L)\dots)\wedge L$
and also that the generators of $L^{2}\wedge L$ are as $a_{1}\wedge
l_{2}\wedge\dots\wedge l_{n}$, $a_{1}\wedge a_{2}\wedge l_{3}\wedge\dots\wedge
l_{n},\ldots,$ and $a_{1}\wedge a_{2}\wedge\dots\wedge a_{n-1}\wedge l_{n}$,
where $a_{i}\in L^{2}$, and $l_{j}\in L$, for all $1\leq i\leq n-1$ and $2\leq
j\leq n$. Indeed by using equation (1.1), one can easily check that
$a_{1}\wedge a_{2}\wedge l_{3}\wedge\dots\wedge l_{n}\in L^{3}\wedge L$, and
similarly, $a_{1}\wedge\dots\wedge a_{i}\wedge l_{i+1}\wedge\dots\wedge
l_{n}\in L^{i+1}\wedge L$. Thus

(3.26) $\dim(L^{2}\wedge^{c}L)\leq\sum_{i=1}^{c}a_{i}d^{c(n-1)-i+1}.$

Therefore, by (3.25) and (3.26), we obtain

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)\leq
l_{d}^{n}(c+1)+\sum_{i=1}^{c}a_{i}d^{c(n-1)-i+1}.$

∎

The next theorem states an upper bound for the dimension of $c$-nilpotent
multiplier of finite-dimensional nilpotent $n$-Lie algebras of class $m$.

######  Theorem 3.10.

Let $L$ be a $d$-dimensional nilpotent $n$-Lie algebra of class $m$. Then

(3.27)
$\dim\mathcal{M}^{(c)}(L)\leq\displaystyle\sum_{k=0}^{c}l_{d}^{n}(m+k),$

when $m\leq c$, and we have

(3.28)
$\dim\mathcal{M}^{(c)}(L)\leq\displaystyle\sum_{k=1}^{m}l_{d}^{n}(c+k),$

for all $c$ in which $m\geq c+1$, and if $L$ is abelian, then the equality is
attained.

###### Proof.

Let $L\cong\dfrac{F}{R}$. First assume that $m\leq c$. Since
$\gamma_{m+1}(L)=0$, so

$\displaystyle 0=\gamma_{m+1}(L)=\gamma_{m+1}\left(\dfrac{F}{R}\right)$
$\displaystyle=\dfrac{\gamma_{m+1}(F)+R}{R}\cong\dfrac{R}{\gamma_{m+1}(F)\cap
R},$

which implies that

$\gamma_{m+1}(F)+R=\gamma_{m+1}(F)\cap R=R,$

that is ,

$\gamma_{m+1}(F)\subseteq\gamma_{m+1}(F)\cap R\subseteq R.$

Also, we know that $\gamma_{m+1}(F)\subseteq\gamma_{m}(F)$. Thus
$\gamma_{m+1}(F)\subseteq\gamma_{m}(F)\cap R$. Therefore,

$\gamma_{c+m+1}(F)\subseteq\gamma_{c+m}(F)\subseteq\dots\subseteq\gamma_{m+2}(F)\subseteq\gamma_{m+1}(F)\subseteq
R\cap\gamma_{m}(F).$

Hence

(3.29)
$\dim\dfrac{R\cap\gamma_{m}(F)}{\gamma_{c+m+1}(F)}\leq\dim\dfrac{\gamma_{m}(F)}{\gamma_{c+m+1}(F)}=\sum_{k=0}^{c}l_{d}^{n}(m+k).$

On the other hand, we know that

$\displaystyle\dim\mathcal{M}^{(c)}(L)=\dim\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle\leq\dim\dfrac{R\cap\gamma_{m}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle\leq\dim\dfrac{R\cap\gamma_{m}(F)}{\gamma_{c+1}(F)}$ (3.30)
$\displaystyle\leq\dim\dfrac{R\cap\gamma_{m}(F)}{\gamma_{c+m+1}(F)}.$

Thus by (3.29) and (3.30), we obtain that

$\dim\mathcal{M}^{(c)}(L)\leq\sum_{k=0}^{c}l_{d}^{n}(m+k).$

Now, suppose that $c+1\leq m$. Then

$\displaystyle\dim\mathcal{M}^{(c)}(L)=\dim\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F,\dots,F)}$
$\displaystyle=\dim\dfrac{\Big{(}R\cap\gamma_{c+1}(F)\Big{)}/\gamma_{m+1}(F)}{\gamma_{c+1}(R,F,\dots,F)/\gamma_{m+1}(F)}$
$\displaystyle\leq\dim\dfrac{\gamma_{c+1}(F)/\gamma_{m+1}(F)}{\gamma_{c+1}(R,F,\dots,F)/\gamma_{m+1}(F)}$
$\displaystyle\leq\dim\dfrac{\gamma_{c+1}(F)}{\gamma_{m+1}(F)}-\dim\dfrac{\gamma_{c+1}(R,F,\dots,F)}{\gamma_{m+1}(F)}$
$\displaystyle\leq\dim\dfrac{\gamma_{c+1}(F)}{\gamma_{m+1}(F)}$
$\displaystyle\leq\dim\dfrac{\gamma_{c+1}(F)}{\gamma_{c+m+1}(F)}=\sum_{k=1}^{m}l_{d}^{n}(c+k).$

∎

######  Corollary 3.11.

Let $K$ be an $n$-Lie algebra over a field $\mathbb{F}$ of characteristic
zero, and also suppose that the algebra $L=\dfrac{K}{Z_{c}(K)}$ satisfies in
the stated conditions of Theorem 3.9. Then

$\dim\gamma_{c+1}(K)\leq l_{d}^{n}(c+1)+\sum_{i=1}^{c}a_{i}d^{c(n-1)-i+1}.$

###### Proof.

Let $F/R$ and $S/R$ be a free presentations of $K$ and $Z_{c}(K)$,
respectively, where $S$ is an ideal of $F$. Therefore, according to Theorem
3.9, we have

(3.31)
$\displaystyle\dim\gamma_{c+1}(L)=\dim\gamma_{c+1}(K/Z_{c}(K))=\dim\gamma_{c+1}(F/S)=\dfrac{\gamma_{c+1}(F)+S}{S}=\dim\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F)\cap
S}.$

Hence

$\displaystyle\dim\gamma_{c+1}(L)=\dim\gamma_{c+1}(K/Z_{c}(K))$
$\displaystyle=\dim\dfrac{\gamma_{c+1}(K)+Z_{c}(K)}{Z_{c}(K)}$ (3.32)
$\displaystyle=\dim\gamma_{c+1}(K)+\dim Z_{c}(K)-\dim
Z_{c}(K)=\dim\gamma_{c+1}(K).$

Therefore, by equations (3.31) and (3.32), we have

$\displaystyle\dim\gamma_{c+1}(K)=\dim\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(F)\cap
S}$
$\displaystyle=\dim\left(\dfrac{\gamma_{c+1}(F)/\gamma_{c+1}(S,F,\dots,F)}{\gamma_{c+1}(F)\cap
S/\gamma_{c+1}(S,F,\dots,F)}\right)$
$\displaystyle\leq\dim\left(\dfrac{\gamma_{c+1}(F)}{\gamma_{c+1}(S,F,\dots,F)}\right)$
$\displaystyle=\dim\mathcal{M}^{(c)}(K/Z_{c}(K))+\dim\gamma_{c+1}(K/Z_{c}(K))$
$\displaystyle=\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)$
$\displaystyle\leq l_{d}^{n}(c+1)+\sum_{i=1}^{c}a_{i}d^{c(n-1)-i+1}.$

∎

In the next proposition, we give another upper bound for the dimension of
terms of lower central series, which is similar to [13, Lemma 14 ].

######  Proposition 3.12.

Let $L$ be an $n$-Lie algebra over a field $\mathbb{F}$ such that
$\dim(L/Z_{c}(L))=d$. Then $\dim\gamma_{c+1}(L)\leq l_{d}^{n}(c+1)$.

###### Proof.

Let $X=\\{\bar{x}_{1},\bar{x}_{2},\dots,\bar{x}_{d}\\}$ be a basis set for
$L/Z_{c}(L)$. Then each member of $L$ can be shown as
$\left(\sum\limits_{i=1}^{d}c_{i}\bar{x}_{i}\right)+z$, where
$c_{i}\in\mathbb{F}$ and $z\in Z_{c}(L)$, for all $i=1,2,\dots,d$. If $F/R$ is
a free presentation of $L$, then

$\displaystyle\dim\gamma_{c+1}(L)=\dim\gamma_{c+1}(F/R)=\dim\dfrac{F^{c+1}+R}{R}=\dim\dfrac{F^{c+1}}{F^{c+1}\cap
R}$ $\displaystyle=\dim\dfrac{F^{c+1}/F^{c+2}}{(F^{c+1}\cap R)/F^{c+2}}$
$\displaystyle\leq\dim\dfrac{F^{c+1}}{F^{c+2}}=l_{d}^{n}(c+1).$

∎

The following corollary shows that, under some conditions, the converse of
Proposition 1.5 is hold.

######  Corollary 3.13.

Let $L$ be an $n$-Lie algebra with dimension $d$. Then the following
properties hold:

  1. (1).

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)\leq l_{d}^{n}(c+1)$. In
particular,

$\dim\mathcal{M}(L)\leq\dim\mathcal{M}(L)+\dim L^{2}\leq l_{d}^{n}(2).$

  2. (2).

If $\dim\mathcal{M}^{(c)}(L)=l_{d}^{n}(c+1)$, then

    1. (a)

There exists an epimorphism
$\theta:\mathcal{M}^{(c)}(L)\longrightarrow\dim\mathcal{M}^{(c)}(L/L^{2})$,
which is obtained by choosing $M=L^{2}$ in the second part of Proposition 3.4.

    2. (b)

If $\ker\theta=0$, then $L$ is abelian.

###### Proof.

  1. (1).

Put $K=F/\gamma_{c+1}(R,F,\dots,F)$, in Proposition 3.12, where $F/R$ is a
free presentation of $L$. Then

$\displaystyle\dim\dfrac{K}{Z_{c}(K)}$
$\displaystyle=\dim\left(\dfrac{F/\gamma_{c+1}(R,F,\dots,F)}{Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))}\right)$
$\displaystyle=\dim\left(\dfrac{F/\gamma_{c+1}(R,F,\dots,F)}{(Z_{c}(F)+\gamma_{c+1}(R,F,\dots,F))/\gamma_{c+1}(R,F,\dots,F)}\right)$
$\displaystyle=\dim\left(\dfrac{F}{Z_{c}(F)+\gamma_{c+1}(R,F,\dots,F)}\right)$
$\displaystyle\leq\dim\dfrac{F}{R}=d.$

Therefore,

(3.33) $\dim\gamma_{c+1}(F/\gamma_{c+1}(R,F,\dots,F))\leq l_{d}^{n}(c+1).$

On the other hand, according to part $(3)$ of Corollary 3.5, we have

(3.34)
$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)=\dim\gamma_{c+1}\left(\dfrac{F}{\gamma_{c+1}(R,F,\dots,F)}\right).$

Hence by (3.33) and (3.34), we have

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)\leq l_{d}^{n}(c+1).$

In particular, if $c=1$, then $\dim\mathcal{M}(L)+\dim L^{2}\leq
l_{d}^{n}(c+1)$, and so $\dim\mathcal{M}(L)\leq l_{d}^{n}(c+1)$.

  2. (2).

According to Proposition 3.4, $\theta$ exists. Thus it is enough to show that
$\theta$ is onto. By part $(1)$, we have

$\dim\mathcal{M}^{(c)}(L)+\dim\gamma_{c+1}(L)\leq l_{d}^{n}(c+1).$

On the other hand, we know that $\dim\mathcal{M}^{(c)}(L)=l_{d}^{n}(c+1)$. So
$\dim\gamma_{c+1}(L)=0$. Now, if we put $M=L^{2}$, in part $(2)$ of
Proposition 3.4, then we obtain

$\dim M\cap\gamma_{c+1}(L)=\dim L^{2}\cap\gamma_{c+1}(L)=0,$

and thus $\theta$ is an epimorphism.

Now, let $\ker\theta=0$. Since $\theta$ is onto, so by putting $M=L^{2}$ in
part $(2)$ of Proposition 3.4, we get the following exact sequence:

$0\longrightarrow\mathcal{M}^{(c)}(L)\longrightarrow\mathcal{M}^{(c)}(L/L^{2})\longrightarrow
0,$

that is, $\mathcal{M}^{(c)}(L)=\mathcal{M}^{(c)}(L/L^{2})$. Since $L/L^{2}$ is
abelian, so $L^{2}=0$ and hence $L$ is abelian.

∎

The following corollary determines the dimension of $c$-nilpotent multiplier
of nilpotent $n$-Lie algebras of maximal class. We recall that a
$d$-dimensional $n$-Lie algebra $L$ is nilpotent of maximal class $c+1$ when
$\dim Z_{c}(L)=\dim\gamma_{2}(L)=d-n$ and also $\dim(Z_{i}(L)/Z_{i-1}(L))=1$.

######  Corollary 3.14.

Let $L$ be an $n$-Lie algebra of maximal nilpotency class $c+1$ with dimension
finite $d$. Then

$\dim\mathcal{M}^{(c)}(L)\leq l_{d-1}^{n}(c+1)+n^{c(n-1)}.$

###### Proof.

Since $L$ is an $n$-Lie algebra of maximal class $c+1$, then
$Z(L)=\gamma_{c+1}(L)$, and hence $\dim(\gamma_{c+1}(L))=1$. By using Theorem
3.3, if we choose $M=Z(L)$, then we obtain

$\displaystyle\dim\mathcal{M}^{(c)}(L)+\dim(\gamma_{c+1}(L)\cap Z(L))\leq$
$\displaystyle\dim\mathcal{M}^{(c)}(L/Z(L))+\dim\mathcal{M}^{(c)}(Z(L))$
$\displaystyle+\dim((L/Z(L))^{ab}\,{}^{c}\otimes Z(L)).$

On the other hand, we know that $Z_{c+1}(L)=L$ and that $\dim
Z_{c}(L)=\dim\gamma_{2}(L)=d-n=c$. Moreover, for all $1\leq i\leq c$, we have
$\dim Z_{i}(L)-Z_{i-1}(L)=1$, and $\dim Z(L)=1$. Also, by Proposition 1.5,
$\dim\mathcal{M}^{(c)}(Z(L))=l_{1}^{n}(c+1)=0$, and also
$\dim(\gamma_{c+1}(L)\cap Z(L))=1$. Thus $Z(L)\subseteq\gamma_{2}(L)$. Hence,

$\displaystyle\dim((L/Z(L))^{ab}=\dim\dfrac{L/Z(L)}{(L/Z(L))^{2}}$
$\displaystyle=\dim\dfrac{L/Z(L)}{(\gamma_{2}(L)+Z(L))/Z(L)}$
$\displaystyle=\dim\dfrac{L/Z(L)}{\gamma_{2}(L)/Z(L)}$ $\displaystyle=\dim
L/Z(L)-\dim\gamma_{2}(L)/Z(L)$ $\displaystyle=(d-1)-((d-n)-1)=n.$

Therefore,

$\displaystyle\dim\mathcal{M}^{(c)}(L)+1$
$\displaystyle\leq\dim\mathcal{M}^{(c)}(L/Z(L))+0+n^{((c+1)-1)(n-1)}\times 1$
$\displaystyle\leq l_{d-1}^{n}(c+1)+n^{c(n-1)}.$

∎

Eshrati, Saeedi, and Darabi [9] proved that every $d$-dimensional nilpotent
$n$-Lie algebra of class $2$ with $\dim\gamma^{2}(L)=1$ has the structure
$H(n,m)\oplus A(d-mn-1)$, where $H(n,m)$ is a Heisenberg $n$-Lie algebra of
dimension $mn+1$ and $A(d-mn-1)$ is an abelian $n$-Lie algebra of dimension
$d-mn-1$. Also, they gave the dimension of the Schur multiplier of $H(n,m)$ as
follows:

$\mathcal{M}(H(n,m))=\begin{cases}n,&m=1,\\\ {{mm}\choose{n}}-1,&m\geq
2.\end{cases}$

Therefore, according to the above explanation and Proposition 1.5, if we get
the dimension of the $c$-nilpotent multiplier of $H(n,m)$, then we will be
able to calculate the dimension of the $c$-nilpotent multiplier of every
finite-dimensional nilpotent $n$-Lie algebra of class $2$ with
$\dim\gamma_{2}(L)=1$. In the following theorem, we get the structure and
dimension of the $c$-nilpotent multiplier of $H(n,m)$.

######  Theorem 3.15.

If $L\cong H(n,m)$, then for all $c\geq 2$, we have

$\mathcal{M}^{(c)}(H(n,m))\cong\begin{cases}A(l^{n}_{m}(c+1)+l^{n}_{m}(c+2)),&m=1,\\\
A(l^{n}_{mn}(c+1)),&m\geq 2.\end{cases}$

###### Proof.

Assume first $m=1$, and let $F/R$ be a free presentation of $H(n,1)$. Since
$cl(H(n,1)=2$, so $\gamma_{3}(H(n,1)=1$. Hence

$1=\gamma_{3}(H(n,1))=\gamma_{3}(F/R)=\dfrac{\gamma_{3}(F)+R}{R}\cong\dfrac{R}{\gamma_{3}(F)\cap
R},$

which implies that $R=\gamma_{3}(F)$. Thus

$\mathcal{M}^{(c)}(H(n,1))=\dfrac{\gamma_{c+1}(F)\cap
R}{\gamma_{c+1}(R,F,\dots,F)}=\dfrac{\gamma_{c+1}(F)\cap\gamma_{3}(F)}{\gamma_{c+1}(\gamma_{3}(F),F,\dots,F)}=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+3}(F)}.$

Therefore by Theorem 1.4, we obtain

$\dim\mathcal{M}^{(c)}(H(n,1))=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+3}(F)}=l^{n}_{m}(c+1)+l^{n}_{m}(c+2).$

Since the $c$-nilpotent multiplier of every $n$-Lie algebra is abelian, so
$\mathcal{M}^{(c)}(H(n,1))\cong A(l^{n}_{m}(c+1)+l^{n}_{m}(c+2))$.

Now, let $m\geq 2$. Since $m\neq 1$, so $H(n,m)$ is not capable, by [2,
Theorem 1.19] and hence by [2, Lemma 1.5], we have $Z^{*}(H(n,m))\neq 0$. Thus
$Z^{*}(H(n,m))=\gamma_{2}(H(n,m))$, by [2, Proposition 1.10]. Also, it is easy
to check that $Z^{*}(H(n,m))\subseteq Z_{c}^{*}(H(n,m))$, and hence
$\gamma_{2}(H(n,m))\subseteq Z_{c}^{*}(H(n,m))$. Thus by using Proposition 3.7
and bearing in mind that $\gamma_{c+1}(H(n,m))=0$, so we have
$\dim\mathcal{M}^{(c)}(H(n,m))=\dim\mathcal{M}^{(c)}(H(n,m))^{ab}$. Therefore,
by Theorem 1.5,
$\mathcal{M}^{(c)}(H(n,m))\cong\mathcal{M}^{(c)}(H(n,m))^{ab}\cong
A(l_{mn}^{n}(c+1))$. ∎

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