# The $2$-nilpotent multiplier of $n$-Lie algebras
and its applications

Seyedeh Nafiseh Akbarossadat

Abstract. In this paper, we first recall the concept of $c$-nilpotent
multiplier and $c$-capability of $n$-Lie algebras and also, recall the formula
for calculating the number of basic commutators in $n$-Lie algebras. Then we
give the structure of $2$-nilpotent multiplier of the direct sum of two
$n$-Lie algebras. Next, we calculate the dimension of $2$-nilpotent multiplier
of every abelian $n$-Lie algebras and Heisenberg $n$-Lie algebras $H(n,m)$.
Then we give a dimension of $2$-nilpotent multiplier of any nilpotent $n$-Lie
algebras of class $2$ by using the number of basic commutators.

Key words: $n$-Lie algebra, Basic commutators, Free $n$-Lie algebras,
$2$-nilpotent multiplier, Heisenberg $n$-Lie algebras.

MSC: 17B30; Secondary 17B05, 17B60.

##  1\. Introduction and Preliminary concepts

Lie polynomials appeared at the end of $19$th century and the beginning of the
$20$th century in the work of Campbell, Baker, and Hausdorff on exponential
mapping in a Lie group, which has lead to the so called the
Campbell–Baker–Hausdorff formula. Around $1930$, Witt introduced the Lie
algebra of Lie polynomials and showed that the Lie algebra of Lie polynomials
is actually a 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
[10, 11] and Magnus [12], with their commutator calculus, opened the way to
bases of the free Lie algebra. For more details about a historical review of
free Lie algebras, we refer the reader to the reference [18] and the
references therein.

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 [23] gave a method that generalizes Hall’s method [10] 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
[15] investigated some more results about $2$-nilpotent multiplier
$\mathcal{M}^{(2)}(L)$ of a finite-dimensional nilpotent Lie algebra $L$ and
by using the Witt formula, calculated its dimension. Moreover, Salemkar,
Edalatzadeh, and Araskhan [21] 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 [9] introduced the concept of $n$-Lie algebras, as an
$n$-ary multilinear and skew-symmetric operation $[x_{1},\ldots,x_{n}]$ that
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 $n=2$.

Let $L_{1},L_{2},\ldots,L_{n}$ be subalgebras of an $n$-Lie algebra $L$.
Denote by $[L_{1},L_{2},\ldots,L_{n}]$ the subalgebra of $L$ generated by all
vectors $[x_{1},x_{2},\ldots,x_{n}]$, where $x_{i}\in L_{i}$,
$i=1,2,\ldots,n.$ The subalgebra $[L,L,\ldots,L]$ is called the derived
algebra of $L$, and it is denoted by $L^{2}$. If $L^{2}=0$, then $L$ is called
an abelian algebra. An ideal $I$ of an $n$-Lie algebra $L$ is a subspace of
$L$ such that $[I,L,\ldots,L]\subseteq I.$ If $[I,I,L,\ldots,L]=0$, then $I$
is called an abelian ideal.

Let $L$ be an $n$-Lie algebra and let $z\in L$ such that $[z,L,\ldots,L]=0$.
We call the collection of all such elements in $L$ the center of $L$ and
denote it by $Z(L)$. One can check that $Z(L)$ is an ideal of $L$. Put
$Z_{0}(L)=Z(L)$, and define $Z(L/Z_{i-1}(L))=Z_{i}(L)/Z_{i-1}(L)$. Hence we
have $Z_{i-1}(L)\unlhd Z_{i}(L)$, for all $i\in\mathbb{N}$. So we can make the
following chain:

$0\unlhd Z_{0}(L)\unlhd Z_{1}(L)\unlhd Z_{2}(L)\unlhd\cdots\unlhd
Z_{i}(L)\unlhd\cdots.$

The above series is known as the upper central series.

Another well-known chain is the lower central series; Let $L^{1}=L$, let
$L^{2}=[L,L,\ldots,L,L]$, and let $L^{i+1}=[L^{i},L,\ldots,L,L]$. Then we have
$L^{i+1}\unlhd L^{i}$, for $i\geqslant 2$. Thus the following series, known as
the lower central series, can be formed as

$\cdots\subseteq L^{i}\subseteq L^{i-1}\subseteq\cdots\subseteq L^{3}\subseteq
L^{2}\subseteq L^{1}=L.$

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

For more information about nilpotency and solvability, we refer the interested
reader to [7].

Let $L$ be an $n$-Lie algebra over a field $\mathbb{F}$ with a free
presentation

$0\longrightarrow R\longrightarrow F\longrightarrow L\longrightarrow 0,$

where $F$ is a free $n$-Lie algebra. Then the $c$-nilpotent multiplier
$\mathcal{M}^{(c)}(L)$ of $L$ is defined as

$\mathcal{M}^{(c)}(L):=\frac{R\cap F^{c+1}}{\gamma_{c+1}[R,F,\ldots,F]}.$

So far, several studies have been done in the case $n=2$, that is, for Lie
algebras. For more information, we refer to [5, 8, 15, 17, 20, 21, 22].

The author and Saeedi [4] introduced the wedge (exterior) product of two
$n$-Lie algebra. Also, The author and Saeedi [2] defined the non-abelian
tensor products of two $n$-Lie algebras and the modular $n$-tensor products as
follows.

######  Definition 1.1 (Modular $n$-tensor product/$n$-tensor spaces).

Let $n\geqslant 1$ be a natural number. Let $V_{1}$ and $V_{2}$ be two vector
spaces over a field $\mathbb{F}$ of finite dimensions $d_{1}$ and $d_{2}$,
respectively. Also, let $V_{1}^{\times i}\times V_{2}^{\times(n-i)}$ denote
the Cartesian product

$\underbrace{V_{1}\times\cdots\times V_{1}}_{i\
\text{times}}\times\underbrace{V_{2}\times\cdots\times V_{2}}_{n-i\
\text{times}}$

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

A function $f$ from $V_{1}^{\times i}\times V_{2}^{\times(n-i)}$ to a vector
space $W$ is multilinear (or $n$-linear) if the restriction of $f$ on every
component of $V_{1}^{\times i}\times V_{2}^{\times(n-i)}$ is linear.
Let $\\{e_{ij}:1\leqslant j\leqslant d_{i}\\}$ be a basis of $V_{i}$ for
$i=1,2$. Then there exists a unique multilinear function $f:V_{1}^{\times
i}\times V_{2}^{\times(n-i)}\longrightarrow W$ admitting the legal values on
the elements of

(1.1) $\\{(e_{1j_{1}},\ldots,e_{1j_{i}},e_{2j_{i+1}},\ldots,e_{2j_{n}}):\\\
1\leqslant j_{k}\leqslant d_{1},\ 1\leqslant k\leqslant i,\ 1\leqslant
j_{s}\leqslant d_{2},\ i+1\leqslant s\leqslant n\\}.$

Note that the above set contains $d_{1}^{i}\times d_{2}^{n-i}$ elements, which
exceeds the dimension of $V_{1}^{\times i}\times V_{2}^{\times(n-i)}$ most of
the time.

A pair $(\mathbb{T}_{i},\Phi_{i})$ (where $\mathbb{T}_{i}$ is a vector space
and $\Phi_{i}$ is a multilinear function from $V_{1}^{\times i}\times
V_{2}^{\times(n-i)}$ to $\mathbb{T}_{i}$) satisfies the universal
factorization property if for each vector space $W$ and an $n$-linear function
$f:V_{1}^{\times i}\times V_{2}^{\times(n-i)}\longrightarrow W$, there is a
linear function $h_{i}:\mathbb{T}_{i}\longrightarrow W$ such that
$f=h_{i}\Phi_{i}$. The existence of the universal pair can be proved easily.
Also, up to isomorphism, there exists a unique universal pair
$(\mathbb{T}_{i},\Phi_{i})$ satisfying the universal factorization property.
Hence, $\mathbb{T}_{i}$ is a modular $i$-tensor product, which we may denote
it by

$\underbrace{V_{1}\otimes\cdots\otimes V_{1}}_{i\
\text{times}}\otimes\underbrace{V_{2}\otimes\cdots\otimes V_{2}}_{n-i\
\text{times}},$

or simply, by $V_{1}^{\otimes i}\otimes V_{2}^{\otimes(n-i)}$ for all
$1\leqslant i\leqslant n-1$. Now, let

$V\otimes_{\mathrm{mod}}^{n}W=\mathrm{span}\\{\mathbb{T}_{i}:1\leqslant
i\leqslant n-1\\}.$

The vector space $V\otimes_{\mathrm{mod}}^{n}W$ is called the modular tensor
product (or the abelian tensor product) of $V$ and $W$.

It is evident that
$\mathbb{T}_{i}=\mathbb{T}_{j}=V\otimes_{\mathrm{mod}}^{n}V$ for all
$1\leqslant i,j\leqslant n-1$ whenever $V=W$, and that
$V\otimes_{\mathrm{mod}}^{n}W$ coincides with the ordinary tensor product of
two vector spaces when $n=2$.

######  Remark 1.2.

For every finite-dimensional vector spaces $V$ and $W$ with $\dim V=d_{V}$ and
$\dim W=d_{W}$, we have

$\dim(V\otimes_{\mathrm{mod}}^{n}W)=\sum_{i=1}^{n-1}d_{V}^{i}d_{W}^{n-i},$

and also,

$\dim((V\otimes_{\mathrm{mod}}^{n}V)\otimes_{\mathrm{mod}}^{n}W)=\sum_{i=1}^{n-1}d_{V}^{ni}d_{W}^{n-i}.$

######  Definition 1.3 ($c$-Capable $n$-Lie algebras).

An $n$-Lie algebra $L$ is called $c$-capable, if there exists an $n$-Lie
algebra $H$ such that $L\cong H/Z_{c}(H)$.

######  Definition 1.4.

Let $L$ be an $n$-Lie algebra. We define $Z_{c}^{*}(L)$ to be the smallest
ideal as $M$ of $L$ such that the $n$-Lie algebra $L/M$ is $c$-capable.

It is easy to check that $Z_{c}^{*}(L)$ is a characteristic ideal of $L$ and
also, that $Z_{c}^{*}(L/Z_{c}^{*}(L))=0$. The notation $Z^{*}(L)$ was defined
for Lie algebras in [20], and it was shown that a Lie algebra $L$ is capable
if and only if $Z^{*}(L)=0$. Similarly, it can be proved for $n$-Lie algebras.

######  Lemma 1.5.

Let $L$ be an $n$-Lie algebra. Then $Z^{*}(L)=0$ if and only if $L$ is
capable.

###### Proof.

Let $Z^{*}(L)=0$. By the definition of $Z^{*}(L)$ we know that $L/Z^{*}(L)$ is
capable, that is, there is an $n$-Lie algebra $M$ such that $L/Z^{*}(L)\cong
M/Z(M)$. Since $Z^{*}(L)=0$, so $L/Z^{*}(L)\cong L$ and hence $L$ is capable.

On the other hand, if $L$ is capable, then there exists an $n$-Lie algebra $M$
such that $L\cong M/Z(M)$. Since $Z^{*}(L)$ is the smallest ideal of $L$ such
that $L/Z^{*}(L)$ is capable, we obtain that $Z^{*}(L)=0$. ∎

The following theorem was proved for Lie algebras in [22].

######  Theorem 1.6.

Let $L$ be an $n$-Lie algebra with the free representation $F/R$. If the map
$\pi:F/\gamma_{c+1}(R,F,\dots,F)\longrightarrow F/R$ is an epimorphism, then
$Z_{c}^{*}(L)=\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$.

###### Proof.

To prove this, we have to prove that
$L/\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$ is $c$-capable. Also,
we show that it is the smallest ideal of $L$ such that
$L/\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$ is $c$-capable. We know
that
$Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))=\dfrac{Z_{c}(F)+\gamma_{c+1}(R,F,\dots,F)}{\gamma_{c+1}(R,F,\dots,F)}$,
and hence $\pi(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F)))=Z_{c}(F)/R$. Therefore, by
choosing $H=F$, we have

$\dfrac{H}{Z_{c}(H)}=\dfrac{F}{Z_{c}(F)}\cong\dfrac{F/R}{Z_{c}(F)/R}=\dfrac{L}{\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)}.$

Let $M$ be an ideal of $L$ such that $L/M$ is $c$-capable and
$M\subseteq\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$. Since $L/M$ is
$c$-capable, there is an $n$-Lie algebra $N$ such that $L/M\cong N/Z_{c}(N)$.
We have $\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)/M\unlhd L/M$,
since $M\subseteq\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$.
Therefore,

$\dim\dfrac{L}{\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)}=\dim\dfrac{L/M}{\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)/M}\leqslant\dim
N/Z_{c}(N),$

and hence $M=\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$. Thus
$Z_{c}^{*}(L)=\pi\left(Z_{c}(F/\gamma_{c+1}(R,F,\dots,F))\right)$. ∎

######  Lemma 1.7.

Let $L$ be an $n$-Lie algebra and let $N$ be an ideal of $L$. Then $N\subseteq
Z^{*}(L)$ if and only if the natural map
$\mathcal{M}(L)\longrightarrow\mathcal{M}(L/N)$ is a monomorphism.

###### Proof.

According to the definition of multiplier of $n$-Lie algebras, we know that

$\mathcal{M}(L)\cong\dfrac{R\cap
F^{2}}{\gamma_{2}(R,F,\dots,F)},\qquad\mathcal{M}(L/N)\cong\dfrac{S\cap
F^{2}}{\gamma_{2}(S,F,\dots,F)},$

where $F/R$ and $S/R$ are free representations of $L$ and $N$, respectively.
Thus $R\subseteq S\unlhd F$, and hence $R\cap F^{2}\subseteq S\cap F^{2}$. If
$N\subseteq Z^{*}(L)$, then by Theorem 1.6,

$S/R\subseteq\pi(Z(F/\gamma_{2}(R,F,\dots,F))=\pi((Z(F)+\gamma_{2}(R,F,\dots,F))/\gamma_{2}(R,F,\dots,F)).$

Therefore, the natural map
$\alpha:\mathcal{M}(L)\longrightarrow\mathcal{M}(L/N)$ is a monomorphism.

Now conversely, let $\alpha$ be a monomorphism and let $N\not\subseteq
Z^{*}(L)$. Then there exists $n_{0}\in N$ such that $n_{0}\notin
Z^{*}(L)=\pi(Z(F/\gamma_{2}(R,F\dots,F)))$. Assume that $n_{0}=s_{0}+R\in
S/R=N$, for some $s_{0}\in S$. Then
$0\neq\pi^{-1}(s_{0}+R)=s_{0}+\gamma_{2}(R,F,\dots,F)\not\in
Z(F/\gamma_{2}(R,F,\dots,F))$. Therefore, there are
$f_{i}+\gamma_{2}(R,F,\dots,F)\in F/\gamma_{2}(R,F,\dots,F)$, for
$i=2,\ldots,n$, such that
$\bar{x}=[s_{0},f_{2},\dots,f_{n}]+\gamma_{2}(R,F,\dots,F)\neq 0$. Now, there
are two cases: First, if $[s_{0},f_{2},\dots,f_{n}]\not\in R$, then
$\alpha(\bar{x})$ is not defined, and hence the statement is hold. Second, if
$[s_{0},f_{2},\dots,f_{n}]\in R$, then

$\alpha(\bar{x})=\alpha([s_{0},f_{2},\dots,f_{n}]+\gamma_{2}(R,F,\dots,F))=[s_{0},f_{2},\dots,f_{n}]+\gamma_{2}(S,F,\dots,F)=0,$

and this contradicts with $\ker\alpha=0$. ∎

######  Theorem 1.8 ([6]).

The $d$-dimensional abelian $n$-Lie algebra $F(d)$ is capable if and only if
$d\geqslant n$.

The following proposition is similar to Lie algebras.

######  Proposition 1.9.

Let $L$ be an $n$-Lie algebra and let $M$ be an ideal of it, such that $L/M$
is capable. Then $Z^{*}(L)\subseteq M$.

###### Proof.

To prove this, it is enough to show that
$Z^{*}(L)=\bigcap\limits_{i=1}^{r}M_{i}$, where $M_{i}$’s are ideals of $L$
such that $L/M_{i}$ is capable, for each $i=1,2,\dots,r$. It is clear that
$Z^{*}(L)\subseteq\bigcap\limits_{i=1}^{r}M_{i}$, because $Z^{*}(L)$ is the
smallest ideal of $L$ such that $L/Z^{*}(L)$ is capable. Hence
$Z^{*}(L)\subseteq M_{i}$, for all $i$. On the other hand, since $L/Z^{*}(L)$
is capable, $Z^{*}(L)=M_{j}$, for some $1\leqslant j\leqslant r$. Thus
$\bigcap\limits_{i=1}^{r}M_{i}\subseteq M_{j}=Z^{*}(L)$. Therefore,
$Z^{*}(L)=\bigcap\limits_{i=1}^{r}M_{i}$. ∎

######  Proposition 1.10.

Let $L$ be a nonabelian nilpotent $n$-Lie algebra of finite dimension. Then
$Z^{*}(L)\subseteq L^{2}$.

###### Proof.

We know that $L/L^{2}$ is abelian with $\dim L/L^{2}\geqslant n$. Hence by
Theorem 1.8, $L/L^{2}$ is capable and so by Proposition 1.9, we have
$Z^{*}(L)\subseteq L^{2}$. ∎

The proof of the following lemmas is omitted because of their similarity with
Lie algebraic states. The exterior center of $n$-Lie algebra of $L$, denoted
by $Z^{\wedge}(L)$, is defined by

$Z^{\wedge}(L)=\\{l\in L;~{}l\wedge l_{2}\wedge\dots\wedge l_{n}=0,~{}\text{
for all }l_{2},\dots,l_{n}\in L\\}.$

######  Lemma 1.11.

Let $L$ be an $n$-Lie algebra and let $N$ be a central ideal of $L$. Then

$\mathcal{M}(L)\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}\mathcal{M}(L/N)\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}N\cap
L^{2}\longrightarrow 0.$

###### Proof.

The homomorphism $\alpha$ is defined in Lemma 1.7. Assume that $F/R$ and $S/R$
are free representations of $L$ and $N$, respectively, where $R\subseteq
S\unlhd F$. Then according to the definition of $c$-nilpotent multiplier, we
have $\mathcal{M}(L/N)=(S\cap F^{2})/\gamma_{2}(S,F,\dots,F)$ and $N\cap
L^{2}=S/R\cap(F^{2}+R)/R$. Consider

$\beta:\mathcal{M}(L/N)=(S\cap F^{2})/\gamma_{2}(S,F,\dots,F)\longrightarrow
N\cap L^{2}=S/R\cap(F^{2}+R)/R$

by $s+\gamma_{2}(S,F,\dots,F)\longmapsto s+R$. It is easy to check that
$\beta$ is an epimorphism and also that $\ker\beta=\mathrm{Im}\alpha$. ∎

######  Corollary 1.12.

Let $L$ be an $n$-Lie algebra. Then

$\mathcal{M}(L)\longrightarrow\mathcal{M}(L^{ab}){\longrightarrow}L^{2}\longrightarrow
0.$

###### Proof.

It is enough to put $N=L^{2}$ in Lemma 1.11. ∎

######  Proposition 1.13.

Let $L$ be a finite-dimensional $n$-Lie algebra and let $N$ be a central ideal
of $L$. Then

  1. (1)

$\dim(\mathcal{M}(L/N))\leqslant\dim(\mathcal{M}(L))+\dim(L^{2}\cap N)$ and

  2. (2)

$\dim(\mathcal{M}(L/N))=\dim(\mathcal{M}(L))+\dim(L^{2}\cap N)$ if and only if
$N\subseteq Z^{*}(L)$.

###### Proof.

According to Lemma 1.11, we have

$\mathcal{M}(L)\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}\mathcal{M}(L/N){\longrightarrow}N\cap
L^{2}\longrightarrow 0.$

Since $\alpha$ is not a monomorphism, s
$\mathcal{M}(L/N)\subseteq\mathcal{M}(L)\oplus N\cap L^{2}$, and hence

$\dim(\mathcal{M}(L/N))\leqslant\dim(\mathcal{M}(L))+\dim(L^{2}\cap N).$

If $N\subseteq Z^{*}(L)$, then by Lemma 1.7, $\alpha$ is a monomorphism. Hence

$0\longrightarrow\mathcal{M}(L)\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}\mathcal{M}(L/N){\longrightarrow}N\cap
L^{2}\longrightarrow 0.$

Also, $\mathcal{M}(L/N)=\mathcal{M}(L)\oplus N\cap L^{2}$. Therefore,

$\dim(\mathcal{M}(L/N))=\dim(\mathcal{M}(L))+\dim(L^{2}\cap N).$

∎

######  Corollary 1.14.

Let $L$ be an $n$-Lie algebra and let $N$ a central ideal of $L$. It holds
that $N\subseteq Z^{\wedge}(L)$ if and only if the natural map $L\wedge
L\longrightarrow L/N\wedge L/N$ is a monomorphism.

###### Proof.

Employing $L\wedge L\cong\dfrac{L\otimes L}{L\square L}$ and [3, Corollary
3.2] implies that there exists an exact series $N\otimes L\longrightarrow
L\otimes L\longrightarrow L/N\otimes L/N\longrightarrow 0$ such that
$N\subseteq L^{2}$. The definition of the nonabelian exterior product
completes the proof. ∎

######  Lemma 1.15.

Let $L$ be a finite-dimensional $n$-Lie algebra. Then
$Z^{\wedge}(L)=Z^{*}(L)$.

###### Proof.

It can be proved similar to [16], which was proved for Lie algebras. ∎

######  Corollary 1.16.

Let $L$ be an $n$-Lie algebra and let $N$ be an one-dimensional central ideal
of $L$. Then $L$ is capable if and only if
$\mathcal{M}(L)\longrightarrow\mathcal{M}(L/N)$ has a nontrivial kernel.

###### Proof.

It is easy to prove from Lemmas 1.5, 1.7, and 1.15. ∎

The Heisenberg Lie algebras are the ubiquities in the theory of Lie algebras,
in particular nilpotent Lie algebras and their classifications. An $n$-Lie
algebra $L$ is called Heisenberg if $L^{2}=Z(L)$ and $\dim(L^{2})=1$.

######  Theorem 1.17 ([8]).

  * (a).

Every Heisenberg Lie algebra is isomorphic to the following Lie algebra of odd
dimension:

$H(2,m)=\langle
x,x_{1},\ldots,x_{2m}:[x_{2i-1},x_{2i}]=x,i=1,\ldots,m\rangle.$

  * (b).

Let $H(n,m)$ be a Heisenberg $n$-Lie algebra of dimension $mn+1$. Then

$\dim(\mathcal{M}(H(n,m)))=\begin{cases}n,&m=1,\\\ \\\
\binom{mn}{n}-1,&m>1.\end{cases}$

The following lemma establishes the structure of every finite-dimensional
nilpotent $n$-Lie algebra $L$ satisfying $\dim(L^{2})=1$.

######  Lemma 1.18 ([8]).

Let $L$ be a nilpotent $n$-Lie algebra of dimension $d$ with $\dim(L^{2})=1$.
Then there exists $m\geqslant 1$ such that

$L\cong H(n,m)\oplus F(d-mn-1),$

where $F(d-mn-1)$ is an abelian $n$-Lie algebra of dimension $d-mn-1$.

######  Theorem 1.19 ([6]).

The $n$-Lie algebra $H(n,m)\oplus F(k)$ with dimension $d=mn+k+1$ is capable
if and only if $m=1$.

######  Lemma 1.20.

Let $H(n,m)$ be a Heisenberg $n$-Lie algebra with $m>1$. Then

$Z^{\wedge}(H(n,m))=H^{2}(n,m)=Z(H(n,m)).$

###### Proof.

Since $m>1$, by Theorem 1.19, $H(n,m)$ is not capable. Hence by Lemma 1.5,
$Z^{*}(H(n,m))\neq 0$. On the other hand, by Lemmas 1.10 and 1.15, we have
$Z^{\wedge}(H(n,m))=Z^{*}(H(,n,m))\subseteq H^{2}(n,m)=Z(H(n,m))$, because
$H(n,m)$ is nilpotent with dimension $mn+1<\infty$. Also, by [3, Proposition
4.1], we know that

$H(n,m)\otimes
H(n,m)\cong\dfrac{H(n,m)}{H^{2}(n,m)}\otimes\dfrac{H(n,m)}{H^{2}(n,m)},$

and so $H(n,m)\wedge
H(n,m)\cong\dfrac{H(n,m)}{H^{2}(n,m)}\wedge\dfrac{H(n,m)}{H^{2}(n,m)}$. Thus
by Corollary 1.14, we obtain

$Z^{\wedge}(H(n,m))=Z^{*}(H(n,m))=H^{2}(n,m)=Z(H(n,m)).$

∎

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

######  Theorem 1.21 ([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 $\alpha_{0}={{d-1}\choose{n-1}}$, $\alpha_{i}$, ($2\leqslant i\leqslant
w-1$) 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 ${{k-1}\choose{n-1}}+1\leqslant j\leqslant{{k}\choose{n-1}}$, (for
$k=n-1,n,n+1,n+2,\dots,d-1$), then $j^{*}={{k-1}\choose{n-1}}+1$ and
$\beta_{j^{*}}=(d-n-j^{*}+2)$.

######  Corollary 1.22 ([1]).

If $n=d$, then

  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)

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

######  Theorem 1.23 ([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 the abelian of dimension
$\sum\limits_{j=0}^{c-1}l_{d}^{n}(i+j)$, where $c=1,2,\ldots$.

##  2\. $2$-multipliers of $n$-Lie algebras

Salemkar and Aslizadeh [19] presented an explicit formula for the
$c$-nilpotent multipliers of the direct sum of Lie algebras whose
abelianisations are of finite dimension, and under some conditions, extended
it for arbitrary Lie algebras. They considered two Lie algebras $A$ and $B$
with dimension $d$ and $d^{\prime}$, respectively. Then they made the abelian
Lie algebra $\Gamma_{c+1}(A,B)$ with dimension
$l_{d+d^{\prime}}(c+1)-l_{d}(c+1)-l_{d^{\prime}}(c+1)$ (the symbol $\Gamma$
does not indicate the Whitehead quadratic functor or any generalization of
it). So they identified the structure of $\Gamma_{c+1}(A,B)$ and determined
its dimension. Thus they were able to calculate the $c$-nilpotent multiplier
of direct sum of two Lie algebras.

Niroomand and Parvizi [14] chose another method for identifying $2$-nilpotent
multipliers of a direct sum of Lie algebras. First, they considered a free
representation in the form $0\longrightarrow R_{1}\longrightarrow
F_{1}\longrightarrow L_{1}\longrightarrow 0$ and $0\longrightarrow
R_{2}\longrightarrow F_{2}\longrightarrow L_{2}\longrightarrow 0$ for each of
the Lie algebras $L_{1}$ and $L_{2}$, respectively. Then, with the help of the
free product of $F_{1}$ and $F_{2}$, they made a free representation
$F=F_{1}*F_{2}$ for $L_{1}\oplus L_{2}$, such that
$R=R_{1}+R_{2}+[F_{2},F_{1}]$. So, they computed the $2$-nilpotent multiplier
of $L_{1}\oplus L_{2}$ in terms of $F_{i}$’s and $R_{i}$’s as follows:

$\mathcal{M}^{(2)}(L_{1}\oplus L_{2})=\dfrac{R\cap
F^{3}}{[[R,F],F]}=\dfrac{(R_{1}+R_{2}+[F_{2},F_{1}])\cap(F_{1}*F_{2})^{3}}{[R_{1}+R_{2}+[F_{2},F_{1}],F_{1}*F_{2},F_{1}*F_{2}]}.$

They showed that $\mathcal{M}^{(2)}(L_{1})\oplus\mathcal{M}^{(2)}(L_{2})$ is a
direct summand of $\mathcal{M}^{(2)}(L_{1}\oplus L_{2})$, and so
$\mathcal{M}^{(2)}(L_{1}\oplus
L_{2})\cong\mathcal{M}^{(2)}(L_{1})\oplus\mathcal{M}^{(2)}(L_{2})\oplus{\bf
K}$, for some subalgebra ${\bf K}$ of $\mathcal{M}^{(2)}(L_{1}\oplus L_{2})$.
Furthermost, they proved that
$F^{3}=[F_{2},F_{1},F_{1}]+[F_{2},F_{1},F_{2}]+F_{1}^{3}+F_{2}^{3}$ and
$\ker\alpha\equiv[F_{2},F_{1},F_{1}]+[F_{2},F_{1},F_{2}](\mathrm{mod}[R,F,F])$,
where $\alpha:\mathcal{M}^{(2)}(L_{1}\oplus
L_{2})\longrightarrow\mathcal{M}^{(2)}(L_{1})\oplus\mathcal{M}^{(2)}(L_{2})$
is an epimorphism. Also, they proved $[F_{2},F_{1},F_{1}]+[F_{2},F_{1},F_{2}]$
(${\mathrm{mod}}F^{4}$) is an abelian Lie algebra generated by all basic
commutators of the form $[y_{i},x_{j},x_{k}]$ and $[y_{r},x_{s},y_{t}]$, where
$y_{i},y_{r},y_{t}$ and $x_{j},x_{k},x_{s}$ are taken from basic sets $Y$ and
$X$ of $F_{2}$ and $F_{1}$, respectively. Finally, they showed that the
following equation holds:

${\bf K}=[F_{2},F_{1},F_{1}]+[F_{2},F_{1},F_{2}]\equiv(L_{2}^{ab}\otimes
L_{1}^{ab}\otimes L_{1}^{ab})\oplus(L_{2}^{ab}\otimes L_{1}^{ab}\otimes
L_{2}^{ab})\quad({\mathrm{mod}}[R,F,F]).$

Indeed, here we use a new method to prove this decomposition for $n$-Lie
algebras. First, we have to prove the following lemma.

######  Lemma 2.1.

Let $L$ be an $n$-Lie algebra. Then there exists the following exact sequence:

$0\longrightarrow\mathcal{M}^{(2)}(L)\longrightarrow(L\wedge L)\wedge
L\longrightarrow L^{3}\longrightarrow 0.$

###### Proof.

Let $0\longrightarrow R\longrightarrow F\longrightarrow L\longrightarrow 0$ be
a free representation of $L$ ,and so $L\cong F/R$. We know that $(L\wedge
L)\wedge L\cong\left(\dfrac{F}{R}\wedge\dfrac{F}{R}\right)\dfrac{F}{R}$ and
that $\mathcal{M}^{(2)}(L)=\dfrac{R\cap F^{3}}{\gamma_{3}[R,F,\dots,F]}$.
Consider the following maps:

$\displaystyle\begin{array}[]{rcl}\Psi:\mathcal{M}^{(2)}(L)&\longrightarrow&\left(\dfrac{F}{R}\wedge\dfrac{F}{R}\right)\dfrac{F}{R}\\\
\\\
\left[\left[f_{1},\dots,f_{n}\right],f^{\prime}_{2},\dots,f^{\prime}_{n}\right]+\gamma_{3}\left[R,F,\dots,F\right]&\longmapsto&(\bar{f_{1}}\wedge\dots\wedge\bar{f_{n}})\wedge\bar{f^{\prime}_{2}}\wedge\dots\wedge\bar{f^{\prime}_{n}}\end{array}$
$\displaystyle\begin{array}[]{rcl}\Phi:\left(\dfrac{F}{R}\wedge\dfrac{F}{R}\right)\dfrac{F}{R}&\longrightarrow&\mathcal{M}^{(2)}(L)\\\
\\\
(\bar{f_{1}}\wedge\dots\wedge\bar{f_{n}})\wedge\bar{f^{\prime}_{2}}\wedge\dots\wedge\bar{f^{\prime}_{n}}&\longmapsto&[[f_{1},\dots,f_{n}],f^{\prime}_{2},\dots,f^{\prime}_{n}]+\gamma_{3}[R,F,\dots,F].\end{array}$

It is easy to check that $\Psi$ and $\Phi$ are monomorphism and epimorphism,
respectively. Also, $\Phi\Psi=0$ and $\ker\Phi\subseteq\mathrm{Im}\Psi$, and
hence $\ker\Phi=\mathrm{Im}\Psi$. ∎

The next theorem plays an essential role in proving the main results of this
paper. In this theorem, we get the $2$-nilpotent multiplier of the direct sum
of two $n$-Lie algebras.

######  Theorem 2.2.

Let $L$ and $M$ be two finite-dimensional $n$-Lie algebras. Then

$\mathcal{M}^{(2)}(L\oplus
M)\cong\mathcal{M}^{(2)}(L)\oplus\mathcal{M}^{(2)}(M)\oplus((L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab})\oplus((M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab}).$

###### Proof.

By Lemma 2.1, we have

(2.3) $\displaystyle 0\longrightarrow\mathcal{M}^{(2)}(L\oplus
M)\longrightarrow((L\oplus M)\wedge(L\oplus M))\wedge(L\oplus
M)\longrightarrow(L\oplus M)^{3}\longrightarrow 0,$ (2.4) $\displaystyle
0\longrightarrow\mathcal{M}^{(2)}(L)\longrightarrow(L\wedge L)\wedge
L\longrightarrow L^{3}\longrightarrow 0,$ (2.5) $\displaystyle
0\longrightarrow\mathcal{M}^{(2)}(M)\longrightarrow(M\wedge M)\wedge
M\longrightarrow M^{3}\longrightarrow 0.$

On the other hand,

$\displaystyle((L\oplus M)\wedge(L\oplus M))\wedge(L\oplus M)$
$\displaystyle\cong\left((L\wedge
L)\oplus(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\oplus(M\wedge
M)\right)\wedge(L\oplus M)$ $\displaystyle\cong((L\wedge L)\wedge(L\oplus
M))\oplus\left(\left(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab}\right)\wedge(L\oplus
M)\right)\oplus\left((M\wedge M)\wedge(L\oplus M)\right)$
$\displaystyle\cong(((L\wedge L)\wedge L)\oplus((L\wedge L)\wedge
M)\oplus\left(((L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\wedge
L)\oplus((L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\wedge M)\right)$ (2.6)
$\displaystyle\quad\oplus(((M\wedge M)\wedge L)\oplus((M\wedge M)\wedge
M))={\mathbf{K}}.$

Since $L$ and $M$ act trivially on $L^{ab}$ and $M^{ab}$, respectively (and
conversely), there are the following isomorphims:

$\displaystyle(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\wedge
L^{ab}\cong(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab}\cong(L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab},$
(2.7) $\displaystyle(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\wedge
M^{ab}\cong(L^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab}\cong(M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab}.$

We know that

(2.8) $(L\oplus M)^{3}\cong L^{3}\oplus M^{3}.$

Moreover, we have the following exact sequences:

(2.9) $\displaystyle
0\rightarrow(L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab}\rightarrow((L\wedge
L)\wedge
M)\oplus(L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab}\rightarrow
0\rightarrow 0,$ (2.10) $\displaystyle
0\rightarrow(M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab}\rightarrow((M\wedge
M)\wedge
L)\oplus(M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab}\rightarrow
0\rightarrow 0.$

Now, by calculating the direct sum of the corresponding sentences in sequences
(2.4), (2.5), (2.9), and (2.10) and applying the isomorphisms (2.7) and
equation (2.8), we obtain the following exact sequence:

(2.11) $0\rightarrow{\mathbf{P}}\rightarrow\mathbf{K}\rightarrow L^{3}\oplus
M^{3}\rightarrow 0,$

where
${\mathbf{P}}=\mathcal{M}^{(2)}(L)\oplus\mathcal{M}^{(2)}(M)\oplus((L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab})\oplus((M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab})$.
Since the sequences (2.3) and (2.11) are exact, by comparing them with each
other, we conclude that

$\mathcal{M}^{(2)}(L\oplus
M)\cong\mathcal{M}^{(2)}(L)\oplus\mathcal{M}^{(2)}(M)\oplus((L^{ab}\otimes_{\mathrm{mod}}^{n}L^{ab})\otimes_{\mathrm{mod}}^{n}M^{ab})\oplus((M^{ab}\otimes_{\mathrm{mod}}^{n}M^{ab})\otimes_{\mathrm{mod}}^{n}L^{ab}).$

∎

Note that since every one-dimensional $n$-Lie algebra $L$ is abelian and hence
isomorphic to $A(1)$, by the definition of $c$-nilpotent multiplier of $n$-Lie
algebras, $\mathcal{M}^{(2)}(L)=0$.

In what follows, we state a result that is also proved in Lie algebras.

######  Theorem 2.3.

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

###### Proof.

Let $F$ be a free $n$-Lie algebra on $d$ elements. By Theorem 1.23, $F/F^{2}$
is an abelian $n$-Lie algebra of dimension $d$, and so it is isomorphic to $L$
and hence $R=F^{2}$. Thus

$\mathcal{M}^{(c)}(L)=\dfrac{R\cap\gamma_{c+1}(F)}{\gamma_{c+1}(R,F)}=\dfrac{F^{2}\cap\gamma_{c+1}(F)}{\gamma_{c+1}(F^{2},F)}=\dfrac{\gamma_{c+1}(F)}{\gamma_{c+2}(F)}.$

Hence
$\dim\mathcal{M}^{(c)}(L)=\dim\gamma_{c+1}(F)/\gamma_{c+2}(F)=l_{d}^{n}(c+1)$,
which gives the result. ∎

Eshrati, Saeedi, and Darabi [8] proved that every finite-dimensional nilpotent
$n$-Lie algebra can be decomposed into the direct sum of one Heisenberg
$n$-Lie algebra and one abelian $n$-Lie algebra. So in order to get the
$2$-nilpotent multiplier of each $n$-Lie algebra, we must first calculate the
$2$-multiplier of each Heisenberg $n$-Lie algebra. In the following two
theorems, we identify its $2$-nilpotent multiplier.

######  Theorem 2.4.

Let $H(n,1)$ be a Heisenberg $n$-Lie algebra of dimension $n+1$. Then

$\mathcal{M}^{(2)}(H(n,1))\cong A\left(\dfrac{n^{2}+3n}{2}\right).$

###### Proof.

Since $H(n,1)$ is a nilpotent of class $2$, its free representation is
$F/F^{3}$ and so $R=F^{3}$, where $F$ is a free $n$-Lie algebra on $n$
letters. By the definition of $2$-nilpotent multiplier of $n$-Lie algebras, we
have

$\displaystyle\mathcal{M}^{(2)}(H(n,1))=\dfrac{R\cap
F^{3}}{\gamma_{3}(R,F,\dots,F)}=\dfrac{F^{3}}{[[F^{3},F,\dots,F],F,\dots,F]}=\dfrac{F^{3}}{F^{5}}.$

On the other hand, by Theorem 1.23, we know that
$\dim(F^{3}/F^{5})=l_{n}^{n}(3)+l_{n}^{n}(4)$, where $l_{n}^{n}(i)$ ($i=3,4$)
is the number of basic commutators of weight $i$ on $n$ letters in free
$n$-Lie algebras. Since $\mathcal{M}^{(2)}(H(n,1))$ is abelian,

$\displaystyle\mathcal{M}^{(2)}(H(n,1))\cong A(l_{n}^{n}(3)+l_{n}^{n}(4))$
$\displaystyle=A\left(n+{{n}\choose{2}}+n\right)$
$\displaystyle=A\left(2n+\dfrac{n(n-1)}{2}\right)$
$\displaystyle=A\left(\dfrac{n^{2}+3n}{2}\right).$

∎

We know that a Heisenberg $n$-Lie algebra $H(n,m)$ is capable if and only if
$m=1$ and that if $m\geqslant 2$, then $H(n,m)$ is not capable. Thus the
following theorem can be proved for noncapable Heisenberg $n$-Lie algebras.

######  Theorem 2.5.

Let $H(n,m)$ be a Heisenberg $n$-Lie algebra of dimension $mn+1$ with
$m\geqslant 2$. Then

$\mathcal{M}^{(2)}(H(n,m))\cong A(l_{mn}^{n}(3)).$

###### Proof.

Since $m\geqslant 2$, so $H(n,m)$ is not capable, and hence by Lemma 1.20, we
have $Z^{\wedge}(H(n,m))=H^{2}(n,m)=Z(H(n,m))$. On the other hand, if put
$I=Z^{\wedge}(H(n,m))$, then by Lemma 2.8, we have
$\mathcal{M}^{(2)}(H(n,m))\cong\mathcal{M}^{(2)}(H(n,m)/H^{2}(n,m))$. Since
$H(n,m)/H^{2}(n,m)$ is abelian of dimension $mn$, by Theorem 2.3 for $c=2$, we
have

$\mathcal{M}^{(2)}(H(n,m)/H^{2}(n,m))\cong A(l_{mn}^{n}(3)).$

Therefore,

$\mathcal{M}^{(2)}(H(n,m))\cong A(l_{mn}^{n}(3)).$

∎

In the following theorem, we determine the $2$-nilpotent multiplier of $n$-Lie
algebras with one-dimensional derived subalgebra.

######  Theorem 2.6.

Let $L$ be an $n$-Lie algebra of dimension $d$ and with $\dim L^{2}=m$. Then

$\mathcal{M}^{(2)}(L)=\begin{cases}{{A\left(\dfrac{n^{2}+3n}{2}+l_{d-n-1}^{n}(3)+\displaystyle\sum_{i=1}^{n-1}n^{ni}(d-n-1)^{n-i}+(d-n-1)^{ni}n^{n-i}\right)}},\\\
\qquad\hskip 327.20668ptm=1,\\\
{{A\left(l_{mn}^{n}(3)+l_{d-mn-1}^{n}(3)+\displaystyle\sum_{i=1}^{n-1}(mn)^{ni}(d-mn-1)^{n-i}+(d-mn-1)^{ni}(mn)^{n-i}\right)}},\\\
\qquad\hskip 327.20668ptm\geqslant 2.\end{cases}$

###### Proof.

Since $\dim L^{2}=1$, by Lemma 1.18, $L\cong H(n,m)\oplus A(d-mn-1)$. Thus by
Lemma 2.2, we have

$\displaystyle\mathcal{M}^{(2)}(L)$
$\displaystyle\cong\mathcal{M}^{(2)}(H(n,m)\oplus A(d-mn-1))$
$\displaystyle\cong\mathcal{M}^{(2)}(H(n,m))\oplus\mathcal{M}^{(2)}(A(d-mn-1))$
$\displaystyle\quad\oplus\left((H^{ab}(n,m)\otimes_{\mathrm{mod}^{n}}A(d-mn-1))\otimes_{\mathrm{mod}^{n}}H^{ab}(n,m)\right)$
(2.12)
$\displaystyle\quad\oplus\left((H^{ab}(n,m)\otimes_{\mathrm{mod}^{n}}A(d-mn-1))\otimes_{\mathrm{mod}^{n}}A(d-mn-1)\right).$

Now, we consider two following cases:

  1. (a).

Assume that $m=1$. By Theorems 2.3 and 2.4, we have

$\displaystyle\mathcal{M}^{(2)}(L)$ $\displaystyle\cong
A(l_{n}^{n}(3)+l_{n}^{n}(4))\oplus A(l_{d-n-1}^{n}(3))$
$\displaystyle\quad\oplus A\left(\sum_{i=1}^{n-1}n^{ni}(d-n-1)^{n-i}\right)$
$\displaystyle\quad\oplus A\left(\sum_{i=1}^{n-1}(d-n-1)^{ni}n^{n-i}\right)$
$\displaystyle\cong
A\left(l_{n}^{n}(3)+l_{n}^{n}(4)+l_{d-n-1}^{n}(3)+\sum_{i=1}^{n-1}n^{ni}(d-n-1)^{n-i}+(d-n-1)^{ni}n^{n-i}\right)$
$\displaystyle\cong
A\left(\dfrac{n^{2}+3n}{2}+l_{d-n-1}^{n}(3)+\sum_{i=1}^{n-1}n^{ni}(d-n-1)^{n-i}+(d-n-1)^{ni}n^{n-i}\right).$

  2. (b).

Now, suppose that $m\geqslant 2$. By Theorems 2.3 and 2.5, we have

$\displaystyle\mathcal{M}^{(2)}(L)$ $\displaystyle\cong
A\left(l_{mn}^{n}(3)\right)\oplus A\left(l_{d-mn-1}^{n}(3)\right)$
$\displaystyle\quad\oplus
A\left(\sum_{i=1}^{n-1}(mn)^{ni}(d-mn-1)^{n-i}\right)$
$\displaystyle\quad\oplus
A\left(\sum_{i=1}^{n-1}(d-mn-1)^{ni}(mn)^{n-i}\right)$ $\displaystyle\cong
A\left(l_{mn}^{n}(3)+l_{d-mn-1}^{n}(3)+\sum_{i=1}^{n-1}(mn)^{ni}(d-mn-1)^{n-i}+(d-mn-1)^{ni}(mn)^{n-i}\right).$

∎

######  Proposition 2.7.

Let $L$ be an $n$-Lie algebra with free representation $F/R$ and let $M$ be an
ideal of $L$, such that $M\cong S/R$, for some ideal $S$ of $F$. Then
$Q_{3}=\dfrac{\gamma_{3}(S,F,\dots,F)}{\gamma_{3}(R,F,\dots,F)}$ is an
isomorphic image of $(M\wedge L)\wedge L$. Moreover, if $M$ is an $r$-central
ideal, that is, $M\subseteq Z_{r}(L)$, then $Q_{3}$ is the isomorphic image of
$(M\wedge K)\wedge K$, where $K=L/\gamma_{r+1}(L)$.

###### Proof.

We know that

$Q_{3}=\dfrac{\gamma_{3}(S,F,\dots,F)}{\gamma_{3}(R,F,\dots,F)}=\dfrac{[[S,F,\dots,F],F,\dots,F]}{[[R,F,\dots,F],F,\dots,F]}.$

Since $L\cong F/R$, there exists an epimorphism $g:F\longrightarrow L$, with
$\ker g=R$, and hence $\bar{g}:F/R\longrightarrow L$ is an isomorphism.
Consider then map $\alpha:(M\wedge L)\wedge L\rightarrow Q_{3}$, such that
$\alpha$ maps every element $\left(\sum_{j=1}^{n-1}(m_{1}\wedge\dots\wedge
m_{j}\wedge l_{j+1}\wedge\dots\wedge l_{n})\wedge
l_{i+1}^{\prime}\wedge\dots\wedge l_{n}^{\prime}\right)$ from $(M\wedge
L)\wedge L$ to
$\sum_{j=1}^{n-1}[[s_{1},\dots,s_{j},f_{j+1},\dots,f_{n}],f_{i+1}^{\prime},\dots,f_{n}^{\prime}]+\gamma_{3}(R,F,\dots,F)$,
where $\bar{g}^{-1}(m_{i})=s_{i}+R$, $\bar{g}^{-1}(l_{k})=f_{k}+R$, for
$1\leqslant i\leqslant j$, $i+1\leqslant k\leqslant n$, $1\leqslant j\leqslant
n-1$. Since $\bar{g}$ is an isomorphism and the bracket of $n$-Lie algebras is
an $n$-linear, $\alpha$ is a well-defined homomorphism of $n$-Lie algebras.
Also, it is easy to check that $\alpha$ is onto.
The second part can be proved with a similar argument. ∎

The following proposition is an important tool to prove the main theorem of
this section.

######  Proposition 2.8.

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

(2.13)
$\dim\mathcal{M}^{(2)}(L/M)\leqslant\dim\mathcal{M}^{(2)}(L)+\dim\dfrac{M\cap
L^{2}}{\gamma_{3}(M,L,\dots,L)}.$

Moreover, if $M$ is a $2$-central subalgebra (i.e., $M\subseteq Z_{2}(L)$),
then

(2.14) $\displaystyle(M\wedge L)\wedge
L\longrightarrow\mathcal{M}^{(2)}(L)\longrightarrow\mathcal{M}^{(2)}(L/M)\longrightarrow
M\cap L^{3}\longrightarrow 0,$ (2.15)
$\displaystyle\dim\mathcal{M}^{(2)}(L)+\dim M\cap
L^{3}\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes L/L^{3})\otimes
L/L^{3}.$

###### Proof.

By attention to the notations of Proposition 2.7 and
$Q_{3}\subseteq\mathcal{M}^{(2)}(L)$, we have the following exact sequence:

$(M\wedge L)\wedge
L\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}\mathcal{M}^{(2)}(L)\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}\mathcal{M}^{(2)}(L/M)\stackrel{{\scriptstyle\gamma}}{{\longrightarrow}}\dfrac{M\cap
L^{3}}{\gamma_{3}(M,L,\dots,L)}\longrightarrow 0,$

where $\alpha$ is defined in Proposition 2.7. Since $R\subseteq S$, so

$R\cap F^{3}\subseteq S\cap
F^{3},\qquad\gamma_{3}(R,F,\dots,F)\subseteq\gamma_{3}(S,F,\dots,F).$

Hence $\beta:\mathcal{M}^{(2)}(L)=\dfrac{R\cap
F^{3}}{\gamma_{3}(R,F,\dots,F)}\longrightarrow\mathcal{M}^{(2)}(L/M)=\dfrac{S\cap
F^{3}}{\gamma_{3}(S,F,\dots,F)}$. It is easy to check that $\beta$ is a Lie
homomorphism with $\ker\beta=Q_{3}$. Also, since $\mathcal{M}^{(2)}(L/M)=S\cap
F^{3}/\gamma_{3}(S,F,\ldots,F)$, we define
$\gamma:\mathcal{M}^{(2)}(L/M)\longrightarrow M\cap
L^{3}/\gamma_{3}(M,L,\ldots,L)$ such that
$s+\gamma_{3}(S,F,\dots,F)\longmapsto g(s)+\gamma_{3}(M,L,\dots,L)$. It is
obvious that since $g$ is an epimorphism so is $\gamma$ as well. Therefore,
the inequality (2.13) is obtained. Furthermore, if $M$ is $2$-central, then
$\gamma_{3}(M,L,\dots,L)=0$ and $(M\wedge L)\wedge L\cong(M\otimes
L/L^{3})\otimes L/L^{3}$. Hence we have the exact sequence (2.14) and the
inequality (2.15). ∎

The next theorem is the main result of this section.

######  Theorem 2.9.

Let $L$ be a nilpotent $n$-Lie algebra of dimension $d$, with $\dim
L^{2}=k\geqslant 1$. Then

$\dim\mathcal{M}^{(2)}(L)\leqslant\dfrac{n^{2}+3n}{2}+l_{d-n-k}^{n}(3)+\displaystyle\sum_{i=1}^{n-1}n^{ni}(d-n-k)^{n-i}+(d-n-k)^{ni}n^{n-i}+(d-k)^{2n-2}-k+1.$

###### Proof.

In Theorem 2.6, it can be seen that in the first case, that is, $m=1$, the
dimension of $\mathcal{M}^{(2)}(L)$ is larger than the second case, that is,
$m>1$. Also, in the case $m>1$, the dimension of $\mathcal{M}^{(2)}(L)$ is
decreasing with respect to $m$. Now, we do induction on $k$. For $k=1$, the
result follows from Theorem 2.6. Now, let $k=2$ and let $I$ be a one-
dimensional central ideal of $L$. Then $\dim L/M=d-1$ and $\dim(L/M)^{2}=\dim
L^{2}/M=1$. Also, since $M\subseteq Z(L)\subseteq Z_{2}(L)$, thus $M$ and
$L/L^{3}$ act on each other trivially. Hence $(M\otimes L/L^{3})\otimes
L/L^{3}\cong(M\otimes_{\mathrm{mod}}^{n}(L/L^{3})^{ab})\otimes_{\mathrm{mod}}^{n}(L/L^{3})^{ab}$.
So by equation (2.15) of Proposition 2.8 and the first step of induction, we
have

$\displaystyle\dim\mathcal{M}^{(2)}(L)$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes
L/L^{3})\otimes L/L^{3}-\dim M\cap L^{3}$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes_{\mathrm{mod}}^{n}(L/L^{3})^{ab})\otimes_{\mathrm{mod}}^{n}(L/L^{3})^{ab}-\dim
M\cap L^{3}$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes_{\mathrm{mod}}^{n}\dfrac{L/L^{3}}{(L/L^{3})^{2}})\otimes_{\mathrm{mod}}^{n}\dfrac{L/L^{3}}{(L/L^{3})^{2}}-\dim
M\cap L^{3}$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes_{\mathrm{mod}}^{n}\dfrac{L/L^{3}}{L^{2}/L^{3}})\otimes_{\mathrm{mod}}^{n}\dfrac{L/L^{3}}{L^{2}/L^{3}}-\dim
M\cap L^{3}$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)+\dim(M\otimes_{\mathrm{mod}}^{n}L/L^{2})\otimes_{\mathrm{mod}}^{n}L/L^{2}-\dim
M\cap L^{3}$ $\displaystyle\leqslant\dfrac{n^{2}+3n}{2}+l_{(d-1)-n-1}^{n}(3)$
$\displaystyle\quad+\displaystyle\sum_{i=1}^{n-1}n^{ni}((d-1)-n-1)^{n-i}+((d-1)-n-1)^{ni}n^{n-i}+(d-2)^{2n-2}-1.$

Now suppose that the inequality holds for $k-1$, and let $M$ be a one-
dimensional central ideal of $L$, with $\dim L/M=d-1$ and $\dim L^{2}=k$. Then

$\displaystyle\dim\mathcal{M}^{(2)}(L)$
$\displaystyle\leqslant\dim\mathcal{M}^{(2)}(L/M)-\dim(M\cap
L^{3})+\dim(M\otimes_{\mathrm{mod}}^{n}L/L^{2})\otimes_{\mathrm{mod}}^{n}L/L^{2}$
$\displaystyle\leqslant\dfrac{n^{2}+3n}{2}+l_{(d-1)-n-(k-1)}^{n}(3)$
$\displaystyle\quad+\displaystyle\sum_{i=1}^{n-1}n^{ni}((d-1)-n-(k-1))^{n-i}+((d-1)-n-(k-1))^{ni}n^{n-i}$
$\displaystyle\quad-\dim(M\cap
L^{3})+\dim(M\otimes_{\mathrm{mod}}^{n}L/L^{2})\otimes_{\mathrm{mod}}^{n}L/L^{2}$
$\displaystyle\leqslant\dfrac{n^{2}+3n}{2}+l_{d-n-k}^{n}(3)$
$\displaystyle\quad+\displaystyle\sum_{i=1}^{n-1}n^{ni}(d-n-k)^{n-i}+(d-n-k)^{ni}n^{n-i}-k+1+(d-k)^{2n-2}.$

∎

##  3\. $c$-Capability of $n$-Lie algebras

Detection of $2$-capable $n$-Lie algebras is one of the applications of
$2$-nilpotent multipliers. Hence in this section, we are going to state the
conditions for detecting $2$-capable $n$-Lie algebras.

######  Proposition 3.1.

Let $L$ be an $n$-Lie algebra. Then $L$ is $2$-capable if and only if
$Z_{2}^{*}(L)=0$.

###### Proof.

If $L$ is $2$-capable, then there exists an $n$-Lie algebra $H$ such that
$L\cong H/Z_{2}(H)$. Assume that $F^{\prime}/R^{\prime}$ is a free
representation of $H$. Then $Z_{2}(H)\cong S/R^{\prime}$, where
$R^{\prime}\subseteq S\unlhd F^{\prime}$. Thus

$L\cong\dfrac{H}{Z_{2}(H)}\cong\dfrac{F^{\prime}/R^{\prime}}{S^{\prime}/R^{\prime}}\cong\dfrac{F^{\prime}}{S^{\prime}}.$

On the other hand, since $S^{\prime}\unlhd F^{\prime}$, so
$\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})\subseteq S^{\prime}$,
$\dfrac{F^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})}{S^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})}\cong
F^{\prime}/S^{\prime}$, and also the map
$\pi:F^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})\rightarrow
F^{\prime}/S^{\prime}$ is the natural epimorphism. Hence by Theorem 1.6,
$Z_{2}^{*}(L)=\pi\left(Z_{2}(F^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime}))\right)$.
Therefore, it is enough to show that
$Z_{2}(F^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})\subseteq\ker\pi$.
Since
$S/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})=Z_{2}(F^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime}))$,
so
$S^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime})\subseteq\ker\pi$,
and hence
$\pi(S^{\prime}/\gamma_{3}(S^{\prime},F^{\prime},\dots,F^{\prime}))=0$. Thus
$Z_{2}^{*}(L)=0$.

Now, let $Z_{2}^{*}(L)=0$, and suppose that $F/R$ is a free representation of
$L$. We have $\gamma_{3}(R,F,\dots,F)\subseteq R\unlhd F$, and hence
$\pi:F/\gamma_{3}(R,F,\dots,F)\rightarrow F/R$ is an epimorphism with
$\ker\pi=S/\gamma_{3}(R,F,\dots,F)$. Also,

(3.16)
$\dfrac{F/\gamma_{3}(R,F,\dots,F)}{\ker\pi}\cong\dfrac{F/\gamma_{3}(R,F,\dots,F)}{R/\gamma_{3}(R,F,\dots,F)}\cong\dfrac{F}{R}=\mathrm{Im}\pi=L.$

Put $H=F/\gamma_{3}(R,F,\dots,F)$. Using the definition of $Z_{2}(H)$, it is
easy to check that

(3.17)
$Z_{2}(H)=Z_{2}(F/\gamma_{3}(R,F,\dots,F))=\dfrac{R}{\gamma_{3}(R,F,\dots,F)}.$

Therefore, $L\cong H/Z_{2}(H)$ and hence is $2$-capable. ∎

######  Theorem 3.2.

Let $L$ be an $n$-Lie algebra and let $M$ be its ideal such that $M\subseteq
Z_{2}^{*}(L)$. Then the natural homomorphism
$\psi:\mathcal{M}^{(2)}(L)\longrightarrow\mathcal{M}^{(2)}(L/M)$ is one-to-
one.

###### Proof.

Suppose that $F/R$ and $S/R$ are free representations of $L$ and $M$,
respectively, for some $S\unlhd F$ with $R\subseteq S$. Then $L/M\cong F/S$,
and we have

$\mathcal{M}^{(2)}(L)=\dfrac{R\cap
F^{3}}{\gamma_{3}(R,F,\dots,F)},\qquad\mathcal{M}^{(2)}(L/M)=\dfrac{S\cap
F^{3}}{\gamma_{3}(S,F,\dots,F)}.$

Since $R\subseteq S$, so $R\cap F^{3}\subseteq S\cap F^{3}$ and

(3.18) $\gamma_{3}(R,F,\dots,F)\subseteq\gamma_{3}(S,F,\dots,F)\subseteq S.$

Hence we define the homomorphism $\psi:(R\cap
F^{3})/\gamma_{3}(R,F,\dots,F)\longrightarrow(S\cap
F^{3})/\gamma_{3}(S,F,\dots,F)$ by
$\psi(x+\gamma_{3}(R,F,\dots,F))=x+\gamma_{3}(S,F,\dots,F)$.

On the other hand, since $I\subseteq Z_{2}^{*}(L)$, so $S/R\subseteq
Z_{2}(F/R)$, or, equivalently,

(3.19) $[[S,F,\dots,F],F,\dots,F]=\gamma_{3}(S,F,\dots,F)\subseteq R.$

By equations (3.18) and (3.19), we obtain

$\gamma_{3}(R,F,\dots,F)\subseteq\gamma_{3}(S,F,\dots,F)\subseteq R\cap S=R.$

Therefore one of the two situations $\gamma_{3}(S,F,\dots,F)=R$ or
$\gamma_{3}(R,F,\dots,F)=\gamma_{3}(S,F,\dots,F)$ can occur. If
$\gamma_{3}(S,F,\dots,F)=R$, then $S=R$ must be, which means $I=S/R=0$, a
contradiction. Thus $\gamma_{3}(R,F,\dots,F)=\gamma_{3}(S,F,\dots,F)$. Hence
for arbitrary $x+\gamma_{3}(R,F,\dots,F)\in\ker\psi$, we have

$\psi(\gamma_{3}(R,F,\dots,F))=x+\gamma_{3}(S,F,\dots,F)=\gamma_{3}(S,F,\dots,F).$

So $x\in\gamma_{3}(S,F,\dots,F)=\gamma_{3}(R,F,\dots,F)$. Thus $\ker\psi=0$. ∎

In the following theorem, we prove that the necessary condition for $H(n,m)$
to be a $2$-capable $n$-Lie algebra is that $m=1$.

######  Theorem 3.3.

If the Heisenberg $n$-Lie algebra $H(n,m)$ is $2$-capable, then $m=1$.

###### Proof.

Suppose that $H(n,m)$ is a $2$-capable Heisenberg $n$-Lie algebra. Then there
is an $n$-Lie algebra $K$ such that $H(n,m)\cong K/Z_{2}(K)$, and we have

$H(n,m)\cong
K/Z_{2}(K)\cong\dfrac{K/Z(K)}{Z_{2}(K)/Z(K)}=\dfrac{K/Z(K)}{Z(K/Z(K))}.$

Thus putting $P=K/Z(K)$, we conclude that $H(n,m)=P/Z(P)$. Hence $H(n,m)$ is
capable, and so by Theorem 1.19, $m=1$. ∎

In the following result, we show that there is no $2$-capable Heisenberg
$n$-Lie algebras, for $n>2$.

######  Theorem 3.4.

The Heisenberg $n$-Lie algebra $H(n,1)$ is not $2$-capable, for $n\geqslant
3$.

###### Proof.

Assume that the Heisenberg $n$-Lie algebra $H(n,1)$ is $2$-capable, and $n\neq
2$. So there is an $n$-Lie algebra $H$, such that $H(n,1)\cong H/Z_{2}(H)$ and
$Z_{2}^{*}(H(n,1))=0$. By Theorem 1.6,
$\pi(Z_{2}(F/\gamma_{3}(R,F,\dots,F))=0$. That is,

$Z_{2}(F/\gamma_{3}(R,F,\dots,F)\subseteq\ker\pi=K/\gamma_{3}(R,F,\dots,F),$

for some ideal $K$ of $F$. Hence
$Z_{2}(F)+\gamma_{3}(R,F,\dots,F)/\gamma_{3}(R,F,\dots,F)\subseteq
K/\gamma_{3}(R,F,\dots,F)$, and so $Z_{2}(F)\subseteq K$.

On the other hand, $Z_{2}^{*}(H(n,1))=\pi(Z_{2}(F/\gamma_{3}(R,F,\dots,F))=0$
implies that

$\pi(Z_{2}(F/\gamma_{3}(R,F,\dots,F))=\pi(Z_{2}(F)+\gamma_{3}(R,F,\dots,F))+\gamma_{3}(R,F,\dots,F)=R.$

Since $c$-central ideals are characteristic, thus
$\pi(Z_{2}(F))=Z_{2}(F)\subseteq R$.

Therefore, $Z_{2}(F)\subseteq K\cap R$ and that is $Z_{2}(F/R)=0$, which is a
contradiction. ∎

According to the above discussion, we have the following corollary.

######  Corollary 3.5.

The Heisenberg $n$-Lie algebra $H(1)$ is the only $2$-capable Heisenberg
$n$-Lie algebra.

###### Proof.

It is proved that $H(2,1)=H(1)$ is $2$-capable; see [15, Theorem 3.3]. Also,
by attention to Theorems 3.4 and 3.3, the proof is completed. ∎

The following theorem is similar to [13, Lemma 4.2]. Moneyhun proved that if
$\dim L/Z(L)=d$, then $\dim
L^{2}=\dim\gamma_{2}(L)\leqslant\dfrac{1}{2}d(d-1)=l_{d}(2)$.

######  Theorem 3.6.

Let $L$ be an $n$-Lie algebra such that $\dim L/Z_{2}(L)=d$. Then the
dimension of $\gamma_{3}(L)$ is at most $l_{d}^{n}(3)$.

###### Proof.

Suppose that $F/R$ is the free representation of $L$ and that
$\\{x_{1},x_{2},\dots,x_{d}\\}$ is a basis for $L/Z_{2}(L)$. Thus every
element of $L/Z_{2}(L)$ can be represented as $z+\sum_{i=1}^{d}c_{i}x_{i}$,
where $c_{i}$s are scalers. Hence

$\displaystyle\dim\gamma_{3}(L)=\dim\gamma_{3}(F/R)$
$\displaystyle=\dim\dfrac{F^{3}+R}{R}=\dim\dfrac{F^{3}}{F^{3}\cap
R}=\dim\dfrac{F^{3}/F^{4}}{(F^{3}\cap
R)/F^{4}}\leqslant\dim\dfrac{F^{3}}{F^{4}}.$

By Theorem 1.23, we know that the basis of $F^{3}/F^{4}$ is the set of all
basic commutators of weight $3$, and so $\dim\gamma_{3}(L)\leqslant
l_{d}^{n}(3)$. ∎

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