# Diameter of Commuting Graphs of Lie Algebras

Hieu V. Ha<EMAIL_ADDRESS>Hoa D. Quang<EMAIL_ADDRESS>Vu A. Le
<EMAIL_ADDRESS>Tuyen T. M. Nguyen<EMAIL_ADDRESS>University of
Economics and Law, Ho Chi Minh City, Vietnam  Vietnam National University, Ho
Chi Minh City, Vietnam  University of Finance - Marketing, Ho Chi Minh City,
Vietnam  Dong Thap University, Dong Thap Province, Vietnam

###### Abstract

In this paper, we study the connectedness of the commuting graph of a general
Lie algebra and provide a process to determine whether the commuting graph is
connected or not, as well as to compute an upper bound for its diameter. In
addition, we will examine the connectedness and diameter of the commuting
graphs of some remarkable classes of Lie algebras, including: (1) a class of
Lie algebras with one- or two-dimensional derived algebras; and (2) a class of
solvable Lie algebras over the real field of dimension up to $4$.

###### keywords:

Connected, Lie algebras, Commuting graphs, Diameter.

##  1 Introduction

The connections between graph theory and other mathematical structures, such
as group theory and ring theory, have become an important problem in recent
years. There are many papers on investing in these connections (for example,
see [1, 2, 3, 4, 5]). Let $\cal R$ be a noncommutative ring, and let $Z(\cal
R)$ denote the center $\cal R$, that is, $Z(\cal R)$ is the set of elements
$x\in\cal R$ such that $xy=yx$ for every $y\in\cal R$. Then the commuting
graph of $\cal R$ is defined as a simple undirected graph, denoted by
$\Gamma(\cal R)$, which has the vertex set $V=\mathcal{R}\setminus
Z(\mathcal{R})$ in which two vertices $x,\,y\in V$ are adjacent if and only if
$xy=yx$ [3]. For this graph, its diameter, denoted by
$\mathrm{diam}(\Gamma(\mathcal{R}))$, is defined as the supremum of the
distances of any two distinct vertices $u$ and $v$ in $V$. In [3, 4, 5, 7, 10,
11], the connectedness and diameter of the commuting graphs of some important
subsets of the ring $M_{n}(\mathbb{F})$ of all $n\times n$ matrices were
studied. Note that the matrix algebra $M_{n}(\mathbb{F})$ is also a Lie
algebra in which the Lie bracket is defined by $[A,B]=AB-BA$. In [14], Wang
and Xia extended these terms to Lie algebras and obtained the diameters of the
commuting graphs of simple, finite-dimensional Lie algebras over an
algebraically closed field of characteristic 0. More precisely, let
$\mathcal{L}$ be such a Lie algebra of rank $l$. If $l=1$ then the commuting
graph of $\mathcal{L}$ is disconnected. Otherwise, $\Gamma(\mathcal{L})$ is
connected and its diameter is bounded below and above by 3 and 6,
respectively.

In this paper, we extend these terms to more general classes of Lie algebras.
In particular, we prove that the diameter of the commuting graph of any direct
sum of two non-commutative Lie algebras is upper bounded by $3$ and give an
example to prove that this bound is sharp (Section 3). In this section, we
also give some remarkable conditions a Lie algebra to have a connected
commuting graph. Using these conditions, we obtain the connectedness of the
commuting graphs of all Lie algebras having one- or two-dimensional derived
algebras (Section 5). We also propose a new method, which is written as a
process, to obtain the connectedness and an upper bound for the diameter of
the commuting graph of a general Lie algebra (Section 4). Applying this
method, we determine the connectedness and diameter of all real solvable Lie
algebras of dimension up to $4$ (Section 6).

##  2 Preliminaries

In this paper, a graph $\Gamma=(V,E)$ is always a simple undirected graph with
vertex set $V$ and edge set $E$. For two distinct vertices $u,v\in V$, we will
denote the edge joined $u$ and $v$ by $\\{u,v\\}$. If such an edge exists, we
say that $u$ and $v$ are adjacent. Otherwise, two vertices $u$ and $v$ are
said to be non-adjacent. A path (joining $v_{1}$ and $v_{k}$) is a sequence of
pairwise distinct vertices $v_{1},v_{2},\ldots,v_{k}$ such that
$\\{v_{i},v_{i+1}\\}\in E$ for every $i=1,2,\ldots,k-1$. This path will be
written as $\\{v_{1},v_{2},\ldots,v_{k}\\}$. The length of a path $P$ is the
number of edges in $P$. The distance between two vertices $u$ and $v$, denoted
$\mathrm{dist}(u,v)$, is the length of the shortest path joining them. The
graph $\Gamma$ is said to be connected if every pair of distinct vertices is
joined by a path. Otherwise, it is disconnected.

######  Definition 1.

[6] The diameter of a connected graph $\Gamma$, denoted
$\mathrm{diam}(\Gamma)$, is defined by

$\mathrm{diam}(\Gamma)=\sup\\{\mathrm{dist}(u,v)\,|\,u\neq v\\}.$ (1)

It can be seen directly from the above definition that
$\mathrm{diam}(\Gamma)=1$ if and only if $\Gamma$ is a complete graph, i.e.,
every pair of distinct vertices is adjacent. Moreover, it should be noted that
$\mathrm{diam}(\Gamma)<\infty$ if $\Gamma$ is a finite graph. Otherwise, if
$\Gamma$ is an infinite graph, its diameter is not necessary to be less than
infinity.

Let $\mathcal{L}$ be a Lie algebra over a field $\mathbb{F}$. Then the center
of $\cal L$, denoted by $Z(\cal L)$, is defined as the set of all elements
$x\in\cal L$ such that $[x,\mathcal{L}]=0$, i.e.,

$Z(\mathcal{L}):=\\{x\in\mathcal{L}\,|\,[x,y]=0\text{ for every }y\in\cal
L\\}.$

It is clear that $Z(\cal L)$ is an ideal of $\cal L$. By definition, the
direct sum of two Lie algebras ${\cal L}_{1}$, ${\cal L}_{2}$, denoted by
${\cal L}_{1}\oplus{\cal L}_{2}$, is the direct sum of two vector spaces
${\cal L}_{1}$ and ${\cal L}_{2}$ equipped the following Lie bracket:

$[x_{1}+x_{2},y_{1}+y_{2}]:=[x_{1},y_{1}]+[x_{2},y_{2}],\,\forall
x_{1},y_{1}\in{\cal L}_{1};\,\forall x_{2},y_{2}\in{\cal L}_{2}.$ (2)

A Lie algebra $\cal L$ is said to be decomposable if $\mathcal{L}$ can be
decomposed into a direct sum of two non-trivial Lie subalgebras. Otherwise,
$\mathcal{L}$ is said to be indecomposable.

Besides, it is well-known that a Lie algebra of dimension $n$ can be defined
via its structure constants. More precisely, a Lie algebra of dimension $n$
over a field $\mathbb{F}$, spanned by $\\{e_{1},e_{2},\ldots,e_{n}\\}$, can be
defined directly by structure constants $\\{(c_{ij})^{k}\mid 1\leq i<j\leq
n,1\leq k\leq n\\}$ as follows:

$[e_{i},e_{j}]=c_{ij}^{1}e_{1}+c_{ij}^{2}e_{2}+\cdots+c_{ij}^{n}e_{n}.$ (3)

Of course, the structure constants are not completely arbitrary. They must
satisfy a relationship that is directly deduced from the Jacobi identity for
all the triples $(i,j,k),\,1\leq i<j<k\leq n$ [8]. Therefore, in this paper,
we will use the following notation

$\mathcal{L}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{n}}\right>\colon[e_{i},e_{j}]=c_{ij}^{1}e_{1}+c_{ij}^{2}e_{2}+\cdots+c_{ij}^{n}e_{n};\,(i,j)\in
M$ (4)

to indicate a Lie algebra of dimension $n$ spanned by
$\\{e_{1},\ldots,e_{n}\\}$ and having their non-trivial Lie brackets:

$[e_{i},e_{j}]=c_{ij}^{1}e_{1}+c_{ij}^{2}e_{2}+\cdots+c_{ij}^{n}e_{n};\,\forall(i,j)\in
M.$ (5)

Here, $M$ is some nonempty set of $\\{1,\cdots,n\\}\times\\{1,\cdots,n\\}$
such that if $(i,j)\in M$ then $i<j$. We refer the reader to [6] for more
details on graph theory and to [13] for more details about Lie algebras.

We next introduce a basic concept of this paper, that is, the commuting graph
of a noncommutative Lie algebra.

######  Definition 2.

[14] Let $\cal L$ be a noncommutative Lie algebra over a field $\mathbb{F}$
with center $Z(\cal L)$. Then the commuting graph $\Gamma(\cal L)$ $=(V_{\cal
L},E_{\cal L})$ of $\cal L$ is the simple undirected graph which have

  1. i)

the vertex set $V_{\cal L}:=\mathcal{L}\setminus Z(\cal L)$, and

  2. ii)

the edge set $E_{\cal
L}:=\mathopen{}\mathclose{{}\left\\{\\{x,y\\}\,|\,[x,y]=0}\right\\}$.

If $\cal S$ is a nonempty subset of $V_{\cal L}$ then we will denote by
$\Gamma(\mathcal{S})=(V_{\mathcal{S}},E_{\cal S})$ the subgraph of
$\Gamma(\cal L)$ induced by $\cal S$. That is,

$\mathopen{}\mathclose{{}\left\\{\begin{aligned} &V_{\cal S}:=\mathcal{S}\\\
&E_{\cal S}:=\mathopen{}\mathclose{{}\left\\{\\{x,y\\}\in E_{\cal
L}\,|\,x,y\in V_{\cal S}}\right\\}.\end{aligned}}\right.$ (6)

Note that the subgraph of $\Gamma(\mathcal{L})$ induced by any nonempty subset
of $\\{\lambda s\mid\lambda\in\mathbb{F},s\in V_{\cal L}\\}$ is connected.

######  Remark 3.

If the commuting graph of a noncommutative Lie algebra $\mathcal{L}$ is
connected, then its diameter is not less than $2$, i.e.,

$\mathrm{diam}(\Gamma(\mathcal{L}))\geq 2.$ (7)

In the rest of the paper, all considering Lie algebras are always assumed to
be noncommutative and therefore, the diameters of their commuting graphs, if
connected, are not less than $2$.

It is worth pointing out that the commuting graphs of two isomorphic Lie
algebras are obviously isomorphic. However, the converse is not true in
general. This is illustrated by the following example.

######  Example 4.

Let $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ be the two 3-dimensional Lie
algebras defined over the real field $\mathbb{R}$ as follows:

  * 1.

$\mathcal{L}_{1}=\mathopen{}\mathclose{{}\left<x_{1},x_{2},x_{3}}\right>:[x_{3},x_{1}]=x_{1},\,[x_{3},x_{2}]=x_{2};$

  * 2.

$\mathcal{L}_{2}=\mathopen{}\mathclose{{}\left<y_{1},y_{2},y_{3}}\right>:[y_{3},y_{1}]=y_{1},\,[y_{3},y_{2}]=y_{1}+y_{2}.$

According to the classification in [13, Chapter 16], $\mathcal{L}_{1}$ is not
isomorphic to $\mathcal{L}_{2}$. Now, it is easy to see that
$Z(\mathcal{L}_{1})=Z(\mathcal{L}_{2})=\\{0\\}.$ The vertex sets of
$\Gamma(\mathcal{L}_{1})$ and $\Gamma(\mathcal{L}_{2})$ are therefore
determined by:

$\displaystyle
V_{\mathcal{L}_{1}}=\mathopen{}\mathclose{{}\left\\{a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}\,|\,(a_{1},a_{2},a_{3})\neq(0,0,0)}\right\\},$
$\displaystyle
V_{\mathcal{L}_{2}}=\mathopen{}\mathclose{{}\left\\{a_{1}y_{1}+a_{2}y_{2}+a_{3}y_{3}\,|\,(a_{1},a_{2},a_{3})\neq(0,0,0)}\right\\}.$

Let $f:V_{\mathcal{L}_{1}}\rightarrow V_{\mathcal{L}_{2}}$be a function
defined by:

$f(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3})=a_{1}y_{1}+a_{2}y_{2}+a_{3}y_{3}.$

Then $f$ is a bijection. Moreover, it is elementary to check that

$\begin{array}[]{ll}&[f(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}),f(b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3})]=0\\\
\iff&[a_{1}y_{1}+a_{2}y_{2}+a_{3}y_{3},b_{1}y_{1}+b_{2}y_{2}+b_{3}y_{3}]=0\\\
\iff&a_{1}b_{3}-a_{3}b_{1}=a_{2}b_{3}-a_{3}b_{2}=0\\\
\iff&[a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3},b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3}]=0.\end{array}$

Therefore, $f$ is an isomorphism of graphs $\Gamma(\mathcal{L}_{1})$ and
$\Gamma(\mathcal{L}_{2})$.

Moreover, it is worth noting that the connectedness of the commuting graph of
a Lie algebra substantially depends on the ground field. The example that
follows demonstrates this.

######  Example 5.

Let us consider a Lie algebra $\cal L$ defined over a field $\mathbb{F}$ by
the following structure constants:

$\mathcal{L}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{6}}\right>\colon[e_{1},e_{3}]=[e_{2},e_{4}]=e_{5},\,[e_{1},e_{4}]=-[e_{2},e_{3}]=-e_{6}.$
(8)

The center of is easily determined as follows
$Z(\mathcal{L})=\mathopen{}\mathclose{{}\left<e_{5},e_{6}}\right>.$ We shall
show that in the case of the real field $\mathbb{F}=\mathbb{R}$,
$\Gamma(\mathcal{L})$ is disconnected, whereas in another case,
$\mathbb{F}=\mathbb{F}_{2}$, is connected. Moreover, in the case where
$\mathbb{F}=\mathbb{F}_{2}$, we will show that

$\mathrm{diam}\mathopen{}\mathclose{{}\left(\Gamma(\mathcal{L})}\right)=3.$
(9)

Note that if the ground field is $\mathbb{R}$, the above Lie algebra
$\mathcal{L}$ is isomorphic to $\mathfrak{n}_{6,3}$ classified in [13, Chapter
19]. From the structure constants of $\mathcal{L}$, we easily observe that

$\displaystyle[a_{1}e_{1}+a_{2}e_{2},b_{1}e_{1}+b_{2}e_{2}+b_{3}e_{3}+b_{4}e_{4}]=0\iff
a_{1}b_{3}+a_{2}b_{4}=a_{2}b_{3}-a_{1}b_{4}=0$ (10)

We now consider the following two cases depending on the ground field as
follows.
Case 1. $\mathbb{F}=\mathbb{R}$. Let
$S=\\{a_{1}e_{1}+a_{2}e_{2}+a_{5}e_{5}+a_{6}e_{6}\mid(a_{1},a_{2})\neq(0,0)\\}$.
Then $S$ is a subset of the vertex set of $\Gamma(\cal L)$. It is easy to see
that for any $a_{1},a_{2}\in\mathbb{R}$ such that $a_{1}^{2}+a_{2}^{2}\neq 0$,
we have

$a_{1}b_{3}+a_{2}b_{4}=a_{2}b_{3}-a_{1}b_{4}=0\iff b_{3}=b_{4}=0.$ (11)

Therefore, the subgraph of $\Gamma(\mathcal{L})$ induced by $S$ is a connected
component of $\Gamma(\mathcal{L})$. It follows that $\Gamma(\mathcal{L})$ is
disconnected, as required.
Case 2. $\mathbb{F}=\mathbb{F}_{2}$. Then, by direct calculations, we may
verify that the graph in Figure 1 is a subgraph of $\Gamma(\mathcal{L})$.

$e_{1}+e_{2}$$e_{3}+e_{4}$$e_{1}+e_{2}+e_{3}+e_{4}$$e_{2}$$e_{1}$$e_{1}+e_{3}+e_{4}$$e_{2}+e_{3}+e_{4}$$e_{4}$$e_{3}$$e_{1}+e_{2}+e_{3}$$e_{1}+e_{2}+e_{4}$$e_{1}+e_{3}$$e_{1}+e_{4}$$e_{2}+e_{3}$$e_{2}+e_{4}$
Figure 1: A subgraph of the commuting graph of the Lie algebra defined in
Equation (8) over $\mathbb{F}_{2}$.

Since the vertex set of $\Gamma(\mathcal{L})$ is

$V_{\cal
L}=\mathopen{}\mathclose{{}\left\\{a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}+a_{5}e_{5}+a_{6}e_{6}\mid(a_{1},a_{2},a_{3},a_{4})\in\mathbb{F}_{2}^{4}\setminus(0,0,0,0)}\right\\},$
(12)

we can conclude from Figure 1 that $\Gamma(\mathcal{L})$ is connected and

$\mathrm{diam}(\Gamma(\mathcal{L}))\leq 3.$ (13)

Furthermore, it can be verified that

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
[e_{1}+e_{3},a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}]=0\\\
[e_{3},a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}]=0\end{aligned}}\right.\iff
a_{1}=a_{2}=a_{3}=a_{4}=0.$ (14)

Therefore, there is no vertex $v$ such that $\\{e_{1}+e_{3},v,e_{3}\\}$ is a
path. Equivalently,

$\mathrm{dist}(e_{1}+e_{3},e_{3})\geq 3.$ (15)

Besides, $\\{e_{1}+e_{3},e_{1}+e_{2}+e_{3}+e_{4},e_{3}+e_{4},e_{3}\\}$ is a
path in $\Gamma(\mathcal{L})$. It follows that
$\mathrm{dist}(e_{1}+e_{3},e_{3})=3.$ In combining with Equation (13) we get
$\mathrm{diam}(\Gamma(\mathcal{L}))=3$, as required.

##  3 Connectedness and diameters of commuting graphs of general Lie algebras

In this section, we will study the diameters of commuting graphs of
decomposable Lie algebras and give some equivalent conditions for a general
Lie algebra to have a connected commuting graph.

######  Proposition 6.

Let $\cal L$ be a decomposable Lie algebra over a field $\mathbb{F}$.

  * (i)

If $\mathcal{L}=\mathcal{L}_{1}\oplus\mathbb{F}^{k}$ for some noncommutative
Lie subalgebra $\mathcal{L}_{1}\subseteq\cal L$, then $\Gamma(\cal L)$ is
connected if and only if $\Gamma(\mathcal{L}_{1})$ is connected. Moreover,
$\mathrm{diam}(\Gamma(\mathcal{L}))=\mathrm{diam}(\Gamma(\mathcal{L}_{1})).$

  * (ii)

If $\mathcal{L}=\mathcal{L}_{1}\oplus\mathcal{L}_{2}$ for some noncommutative
Lie subalgebras $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ of $\cal L$, then
$\Gamma(\cal L)$ is connected. Moreover,
$\mathrm{diam}(\Gamma(\mathcal{L}))\leq 3.$

###### Proof.

Obviously, $\mathcal{L}$ can be decomposed into either
$\mathcal{L}_{1}\oplus\mathbb{F}^{k}$ or
$\mathcal{L}_{1}\oplus\mathcal{L}_{2}$.

  * $(i)$

Assume that $\mathcal{L}=\mathcal{L}_{1}\oplus\mathbb{F}^{k}$ for some
noncommutative subalgebra $\mathcal{L}_{1}\subseteq\cal L$. It is then trivial
to see that $Z(\mathcal{L})=Z(\mathcal{L}_{1})\oplus\mathbb{F}^{k}$. Then the
vertex set of the commuting graph of $\mathcal{L}$, $V_{\cal L}$, can be
expressed via the vertex set of the commuting graph of $\mathcal{L}_{1}$,
$V_{\mathcal{L}_{1}}$, as follows:

$V_{\mathcal{L}}=\\{x+y\,|\,x\in V_{\mathcal{L}_{1}},\,y\in\mathbb{F}^{k}\\}.$
(16)

Moreover, for every $x_{1},\ldots,x_{k}\in V_{\mathcal{L}_{1}}$ and for every
$y_{1},\ldots,y_{k}\in\mathbb{F}^{k}$, we easily see that
$\mathopen{}\mathclose{{}\left\\{x_{1}+y_{1},x_{2}+y_{2},\ldots,x_{k}+y_{k}}\right\\}$
is a path of $\Gamma(\mathcal{L})$ if and only if $\\{x_{1},\ldots,x_{k}\\}$
is a path of $\Gamma(\mathcal{L}_{1})$. This proves the first part of the
theorem.

  * $(ii)$

Assume that $\cal L$ is a direct sum of two noncommutative subalgebras
$\mathcal{L}_{1},\mathcal{L}_{2}$. First, we show that

$Z(\mathcal{L})=Z(\mathcal{L}_{1})\oplus Z(\mathcal{L}_{2}).$ (17)

Indeed, if $x\in Z(\cal L)$ then $x=x_{1}+x_{2}$ for some
$x_{1}\in\mathcal{L}_{1},x_{2}\in\mathcal{L}_{2}$ such that
$[x_{1}+x_{2},L]=0$. Since $[\mathcal{L}_{1},\mathcal{L}_{2}]=0$, we have
$[x_{1},\mathcal{L}_{1}]=[x_{2},\mathcal{L}_{2}]=0$. Therefore, $x\in
Z(\mathcal{L}_{1})+Z(\mathcal{L}_{2})$. Besides, if $x\in
Z(\mathcal{L}_{1})+Z(\mathcal{L}_{2})$ then there exist $x_{1}\in
Z(\mathcal{L}_{1})$ and $x_{2}\in Z(\mathcal{L}_{2})$ such that
$x=x_{1}+x_{2}$. Since $[\mathcal{L}_{1},\mathcal{L}_{2}]=0$, we get
$[x,\mathcal{L}_{1}]=[x,\mathcal{L}_{2}]=0$ which turns out that
$[x,\mathcal{L}]=0$ or $x\in Z(\mathcal{L})$. This proves Equation (17).

Now, let $x_{1},x_{2}$ be two arbitrary vertices of $\Gamma(\mathcal{L})$.
Then there exist $x_{11},x_{21}\in\mathcal{L}_{1}$, and
$x_{12},x_{22}\in\mathcal{L}_{2}$ such that $x_{1}=x_{11}+x_{12}$ and
$x_{2}=x_{21}+x_{22}$. Note that the vertex sets of both the commuting graphs
of $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are non-empty due to the fact that
$\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are noncommutative.

    * (a)

If $x_{11}\in Z(\mathcal{L}_{1})$ and $x_{21}\in Z(\mathcal{L}_{1})$ then for
every $t\in V_{\mathcal{L}_{1}}$, we have the path $\\{x_{1},t,x_{2}\\}$.

    * (b)

If $x_{11}\in Z(\mathcal{L}_{1})$ and $x_{21}\in V_{\mathcal{L}_{1}}$ then
$\\{x_{1},x_{21},x_{2}\\}$ is a path.

    * (c)

Similarly, if $x_{22}\in Z(\mathcal{L}_{2})$, then $x_{1}$ and $x_{2}$ are
joined by a path of length at most 2.

    * (d)

If $x_{11}\in V_{\mathcal{L}_{1}}$ and $x_{22}\in V_{\mathcal{L}_{2}}$ then we
can derive from Equation (17) that both $x_{11}$ and $x_{22}$ are the vertices
of $\Gamma(\mathcal{L})$. If so, it is elementary to see that
$\\{x_{1},x_{11},x_{22},x_{2}\\}$ is a path in $\Gamma(\mathcal{L})$.

In summary, we have proven that if $\mathcal{L}$ is a direct sum of two
noncommutative subalgebras then every pair of distinct vertices $x_{1},x_{2}$
are joined by a path of length at most $3$. This proves the second part of the
theorem.

∎

We shall give here an example to demonstrate Proposition 6.

######  Example 7.

Let
$\mathcal{L}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>$ be
the direct sum of two copies of the Lie algebra of dimension 2 over the binary
field $\mathbb{F}_{2}$. That is, $\mathcal{L}$ is the $4$-dimensional Lie
algebra defined over $\mathbb{F}_{2}$ as follows:

$\mathcal{L}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>\colon[e_{1},e_{2}]=e_{2},\,[e_{3},e_{4}]=e_{4}.$
(18)

Then we have $Z(\mathcal{L})=\\{0\\}$. Therefore, $\Gamma(\mathcal{L})$ has 15
vertices:

$V_{\mathcal{L}}=\\{e_{1},\ldots,e_{4},e_{1}+e_{2},\ldots,e_{3}+e_{4},e_{1}+e_{2}+e_{3},\ldots,e_{2}+e_{3}+e_{4},e_{1}+e_{2}+e_{3}+e_{4}\\}.$
(19)

According to the structure constants of $\mathcal{L}$, we easily observe that

$[a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4},b_{1}e_{1}+b_{2}e_{2}+b_{3}e_{3}+b_{4}e_{4}]=0\iff
a_{1}b_{2}-a_{2}b_{1}=a_{3}b_{4}-a_{4}b_{3}=0.$ (20)

Thus, we obtain the commuting graph of $\mathcal{L}$ as in Figure 2.

$e_{1}$$e_{2}$$e_{1}+e_{2}$$e_{3}$$e_{4}$$e_{3}+e_{4}$$e_{1}+e_{3}$$e_{1}+e_{3}+e_{4}$$e_{1}+e_{4}$$e_{1}+e_{2}+e_{3}$$e_{1}+e_{2}+e_{3}+e_{4}$$e_{1}+e_{2}+e_{4}$$e_{2}+e_{3}$$e_{2}+e_{3}+e_{4}$$e_{2}+e_{4}$

Figure 2: The commuting graph of the Lie algebra $\mathcal{L}$ defined in
Equation (18) with the ground field $\mathbb{F}_{2}$.

.

It is clear that the diameter of the graph in Figure 2 is 3. Therefore,

$\mathrm{diam}\mathopen{}\mathclose{{}\left(\Gamma(\mathcal{L})}\right)=3.$
(21)

######  Remark 8.

The equality (21) still holds for any field $\mathbb{F}$. Indeed,

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
&[e_{1}+e_{3},a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}]=0,\\\
&[e_{2}+e_{4},a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}]=0\end{aligned}}\right.\iff\mathopen{}\mathclose{{}\left\\{\begin{aligned}
&a_{2}=a_{4}=0,\\\ &a_{1}=a_{3}=0.\end{aligned}}\right.$ (22)

Therefore, there is no vertex in $\Gamma(\mathcal{L})$ which is adjacent to
both $e_{1}+e_{3}$ and $e_{2}+e_{4}$. It follows that
$\mathrm{diam}(\Gamma(\mathcal{L}))\geq 3.$ Applying Proposition 6, we observe
that $\mathrm{diam}(\Gamma(\mathcal{L}))=3$, as required.

######  Proposition 9.

Let $\cal L$ be an indecomposable Lie algebra. Then $\Gamma(\cal L)$ is
disconnected if and only if there is a nonempty set $\mathcal{S}\subsetneq
V_{\cal L}$ such that $\ker(\mathrm{ad}_{x})\subseteq\mathcal{S}\cup Z(\cal
L)$ for every $x\in\cal S$.

###### Proof.

Suppose that there is a nonempty set $\mathcal{S}\subsetneq V_{\cal L}$ such
that $\ker(\mathrm{ad}_{x})\subseteq\mathcal{S}\cup Z(\cal L)$ for every
$x\in\mathcal{S}$. Let $y$ be any vertex in $\Gamma(\cal L)$ which does not
belong to $\mathcal{S}$. Such $y$ exists because $\mathcal{S}\subsetneq
V_{\cal L}$. Since $\ker(\mathrm{ad}_{x})\subseteq\mathcal{S}\cup Z(\cal L)$
for every $x\in\mathcal{S}$, we have $y\notin\ker(\mathrm{ad}_{x})$ for every
$x\in\mathcal{S}$. Therefore, $[y,x]\neq 0$ for every $x\in\mathcal{S}$. It
follows that $y$ is not adjacent to any vertex in $\mathcal{S}$. Thus, there
is no path between any pair of vertices $x\in\mathcal{S}$ and $y\in
V_{\mathcal{L}}\setminus\mathcal{S}$. Therefore, $\Gamma(\cal L)$ is
disconnected.

Conversely, suppose that $\Gamma(\cal L)$ is disconnected. Let $G$ be a
connected component of $\Gamma(\cal L)$. Let $\mathcal{S}=V_{G}$ be the vertex
set of $G$. Then $\mathcal{S}\subsetneq V_{\cal L}$. Let $x$ be any element in
$\cal S$ and let $y$ be any element in $\ker(\mathrm{ad}_{x})$. Then
$[x,y]=0$. Thus, $x$ is adjacent to any element in
$\ker(\mathrm{ad}_{x})\setminus Z(\cal L)$. Since $G$ is a connected component
of $\Gamma(\cal L)$, we should have $\ker(\mathrm{ad}_{x})\setminus
Z(\mathcal{L})\subseteq\cal S$. In other words,
$\ker(\mathrm{ad}_{x})\subseteq\mathcal{S}\cup Z(\cal L)$ for every $x\in\cal
S$. This completes the proof. ∎

In particular, we may obtain the connectedness of the commuting graph of a Lie
algebra $\mathcal{L}$ by examining a basis of $\mathcal{L}$ as in Corollary 10
and the dimension of $\ker(\mathrm{ad}_{x})$ for some $x\in\mathcal{L}$ as in
Corollary 11.

######  Corollary 10.

Let $\cal L$ be an indecomposable, finite-dimensional Lie algebra. Suppose
that $Z^{c}(\cal L)$ is a complementary subspace of $Z(\cal L)$ in $\cal L$
and $\boldsymbol{b}$ is a basis of $Z^{c}(\cal L)$. Moreover, assume that
there is a nonempty proper subset $\boldsymbol{b}_{1}$ of $\boldsymbol{b}$
such that $[x_{1},x_{2}]\neq 0$ for every
$x_{1}\in\mathopen{}\mathclose{{}\left<\boldsymbol{b}_{1}}\right>\setminus\\{0\\}$
and
$x_{2}\in\mathopen{}\mathclose{{}\left<\boldsymbol{b}\setminus\boldsymbol{b}_{1}}\right>\setminus\\{0\\}$.
Then $\Gamma(\cal L)$ is disconnected.

###### Proof.

Let
$\mathcal{S}=\mathopen{}\mathclose{{}\left\\{\sum_{x\in\boldsymbol{b}_{1}}\lambda_{x}x+y\,|\,\lambda_{x}\in\mathbb{F},y\in
Z(\mathcal{L})}\right\\}\setminus Z(\cal L)$. Then
$\emptyset\subsetneq\mathcal{S}\subsetneq V_{\mathcal{L}}$ and
$\ker(\mathrm{ad}_{x})\subseteq\mathcal{S}\cup Z(\cal L)$ for every $x\in\cal
S$. According to Proposition 9, we conclude that $\Gamma(\cal L)$ is
disconnected, as required. ∎

######  Corollary 11.

Let $\cal L$ be an indecomposable, finite dimensional Lie algebra with center
$Z(\cal L)$. if there exists $x$ such that
$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{x})}\right)=\dim
Z(\mathcal{L})+1$ then $\Gamma(\mathcal{L})$ is disconnected.

###### Proof.

Since $\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{x})}\right)=\dim
Z(\mathcal{L})+1$, we have $x\notin Z(\mathcal{L})$. Indeed, if $x\in
Z(\mathcal{L})$ then $\dim
Z(\mathcal{L})=\dim(\ker(\mathrm{ad}_{x}))-1=\dim\mathcal{L}-1$ which turns
out that $\mathcal{L}$ is commutative, a contradiction. Therefore, the
dimension of the direct sum of two vector spaces $Z(\mathcal{L})$ and
$\mathopen{}\mathclose{{}\left<x}\right>$ can be determined by the following
formula:

$\dim\mathopen{}\mathclose{{}\left(Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<x}\right>}\right)=\dim
Z(\mathcal{L})+1.$ (23)

Furthermore, it is clear that
$Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<x}\right>$ is contained in the
vector space $\ker(\mathrm{ad}_{x})$. We then conclude from the equality
$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{x})}\right)=\dim
Z(\mathcal{L})+1$ that

$\ker(\mathrm{ad}_{x})=Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<x}\right>.$
(24)

Besides, it is easy to see that
$\ker(\mathrm{ad}_{ax+y})=\ker(\mathrm{ad}_{x})$ for every $0\neq
a\in\mathbb{F}$ and $y\in Z(\mathcal{L})$. Therefore, by letting
$\mathcal{S}=\mathopen{}\mathclose{{}\left\\{ax+y\,|\,0\neq
a\in\mathbb{F},y\in Z(\mathcal{L})}\right\\}$, we observe from Equality (24)
that $\ker(\mathrm{ad}_{s})=Z(\mathcal{L})\cup\mathcal{S}$ for every
$s\in\mathcal{S}$. Applying Proposition 9, we get the proof. ∎

######  Example 12.

Let $\mathfrak{n}_{6,3}$ be the real Lie algebra defined in [13, Chapter 19]
as follows:

$\mathfrak{n}_{6,3}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{6}}\right>\colon[e_{1},e_{3}]=[e_{2},e_{4}]=e_{5},\,[e_{1},e_{4}]=-[e_{2},e_{3}]=-e_{6}.$
(25)

Since
$Z(\mathfrak{n}_{6,3})=\mathopen{}\mathclose{{}\left<e_{5},e_{6}}\right>$, a
complementary of $Z(\mathfrak{n}_{6,3})$ in the vector space
$\mathfrak{n}_{6,3}$ can be chosen to be
$Z^{c}(\mathfrak{n}_{6,3})=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>$.
Let $\boldsymbol{b}_{1}=\\{e_{1},e_{2}\\}$. Then

$[\lambda_{1}e_{1}+\lambda_{2}e_{2},\mu_{1}e_{3}+\mu_{2}e_{4}]=0\iff\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2}=\lambda_{1}\mu_{2}-\lambda_{2}\mu_{1}=0\iff\mathopen{}\mathclose{{}\left[\begin{aligned}
\lambda_{1}=\lambda_{2}=0,\\\ \mu_{1}=\mu_{2}=0.\end{aligned}}\right.$ (26)

Therefore, by applying Corollary 10, $\Gamma(\mathfrak{n}_{6,3})$ is
disconnected.

By definition, the vertex set of the commuting graph of a Lie algebra $\cal L$
is $V_{\cal L}=\mathcal{L}\setminus Z(\mathcal{L})$. Therefore, when the
center of $\mathcal{L}$ is “big enough”, the commuting graph
$\Gamma(\mathcal{L})$ should be disconnected. This remarkable point is given
in the following proposition.

######  Proposition 13.

Let $\cal L$ be an indecomposable, finite-dimensional Lie algebra over a field
$\mathbb{F}$ such that $\dim\mathcal{L}-\dim Z(\mathcal{L})\leq 3$. Then
$\Gamma(\cal L)$ is disconnected.

###### Proof.

Since $\mathcal{L}$ is noncommutative, we must have $\dim\mathcal{L}-\dim
Z(\mathcal{L})\geq 2$. Therefore, $\dim\mathcal{L}-\dim
Z(\mathcal{L})\in\\{2,\,3\\}$.
Case 1. $\dim\mathcal{L}-\dim Z(\mathcal{L})=2$. If so, there are
$e_{1},e_{2}\in\mathcal{L}$ such that the vector space $\mathcal{L}$ is
decomposed into the following direct sum:
$\mathcal{L}=Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>$.
It follows that the vertex set of $\Gamma(\mathcal{L})$ is

$V_{\cal
L}=\mathopen{}\mathclose{{}\left\\{a_{1}e_{1}+a_{2}e_{2}+x\,|\,(0,0)\neq(a_{1},a_{2})\in\mathbb{F}^{2},\,x\in
Z(\mathcal{L})}\right\\}.$ (27)

Since $\mathcal{L}$ is noncommutative, we have $[e_{1},e_{2}]\neq 0$.
Therefore, for every $x,y\in Z(\mathcal{L})$ and for every
$a_{1},a_{2},b_{1},b_{2}\in\mathbb{F}$, we have

$[a_{1}e_{1}+a_{2}e_{2}+x,b_{1}e_{1}+b_{2}e_{2}+y]=0\iff
a_{1}b_{2}-a_{2}b_{1}=0.$ (28)

Thus, $\mathopen{}\mathclose{{}\left\\{a_{1}e_{1}+x\,|\,0\neq
a_{1}\in\mathbb{F},\,x\in Z(\mathcal{L})}\right\\}$ and
$\mathopen{}\mathclose{{}\left\\{a_{2}e_{2}+y\,|\,0\neq
a_{2}\in\mathbb{F},\,y\in Z(\mathcal{L})}\right\\}$ are two different
connected components of $\Gamma(\mathcal{L})$. It follows that
$\Gamma(\mathcal{L})$ is disconnected, as required.
Case 2. $\dim\mathcal{L}-\dim Z(\mathcal{L})=3$. Without loss of generality,
we assume that the vector space $\mathcal{L}$ has a decomposition
$\mathcal{L}=Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3}}\right>$
for some independent vectors $\\{e_{1},e_{2},e_{3}\\}$ in $\mathcal{L}$. If
so, the derived algebra of $\mathcal{L}$, denoted by $\mathcal{L}^{1}$, is
determined by:

$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{1},e_{2}],[e_{1},e_{3}],[e_{2},e_{3}]}\right>.$
(29)

First, we prove that $\dim\mathcal{L}^{1}\geq 2$. By contradiction, suppose
that $\dim\mathcal{L}^{1}=1$. Then we may assume (after a re-indexing if
necessary) that
$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{1},e_{2}]}\right>$. Thus,
there are $a$ and $b$ in $\mathbb{F}$ such that $[e_{1},e_{3}]=a[e_{1},e_{2}]$
and $[e_{2},e_{3}]=b[e_{1},e_{2}]$. It follows that

$[e_{1},e_{3}-ae_{2}+be_{1}]=[e_{2},e_{3}-ae_{2}+be_{1}]=[e_{3},e_{3}-ae_{2}+be_{1}]=0.$
(30)

This turns out that $e_{3}-ae_{2}+be_{1}$ belongs to $Z(\mathcal{L})$, a
contradiction. Therefore, $\dim\mathcal{L}^{1}\geq 2$. Without loss of
generality, we may assume that $[e_{1},e_{2}]$ and $[e_{1},e_{3}]$ are
linearly independent. Thus, for every $x\in Z(\mathcal{L})$, we have

$[e_{1},a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+x]=0\iff a_{2}=a_{3}=0.$ (31)

In other words,
$\ker(\mathrm{ad}_{e_{1}})=Z(\mathcal{L})+\mathopen{}\mathclose{{}\left<e_{1}}\right>$.
According to Corollary 11, we observe that $\Gamma(\mathcal{L})$ is
disconnected, as required. ∎

The following corollary follows immediately from Propositions 6 and 13.

######  Corollary 14.

Let $\mathcal{L}$ be a Lie algebra of dimension not exceeding $3$. Then the
commuting graph of $\mathcal{L}$ is disconnected.

##  4 A Process to determine the Connectedness of the Commuting Graph of a
Noncommutative Lie Algebra

In this section, we will give a process to determine the connectedness, as
well as an upper bound for the diameter of the commuting graph of any Lie
algebra $\cal L$. Let ${S}$ be a subset of the vertex set $V_{\cal L}$ of
$\Gamma(\mathcal{L})$, the closed neighborhood of ${S}$, denoted by $N({S})$,
is defined as the union of ${S}$ with the set of all vertices of
$\Gamma(\mathcal{L})$ adjacent to a vertex in ${S}$. That is,

$N({S}):=\mathopen{}\mathclose{{}\left\\{x\in\mathcal{L}\setminus
Z(\mathcal{L})\,|\,[x,s]=0\text{ for some }s\in{S}}\right\\}.$

######  Definition 15.

Let $\cal L$ be a noncommutative Lie algebra and let ${S}$ be a non-empty
subset of the vertex set $V_{\cal L}$. For each natural number $k\geq 1$, we
define by induction, the $k$-th closed neighborhood of $S$ as:
$N^{k}(S):=N\mathopen{}\mathclose{{}\left(N^{k-1}(S)}\right)$ where
$N^{0}(S):=S$. The series of neighborhoods of ${S}$ is then defined as
follows:

$N^{0}({S})\subseteq N^{1}({S})\subseteq N^{2}({S})\subseteq
N^{3}({S})\subseteq N^{4}({S})\subseteq\cdots$ (32)

If there is a natural number $k$ such that

$N^{k-1}(S)\subsetneq N^{k}(S)=N^{k+1}(S),$ (33)

we say that the series of neighborhoods of $S$ terminates at $k$.

For convenience, in the case where $S$ contains only one vertex, say
${S}:=\\{s\\}$, the $k$-th neighborhood of $S$ will be also denoted as
$N^{k}(s)$. And the series of neighborhoods of $\\{s\\}$ is defined as a
consequence.

######  Theorem 16.

Let $\cal L$ be a Lie algebra and let ${S}$ be a non-empty subset of the
vertex set $V_{\cal L}$ such that the subgraph of $\Gamma(\mathcal{L})$
induced by $S$, denoted by $\Gamma(S)$, is connected. Suppose that the series
of closed neighborhoods of ${S}$ terminates at a positive integer $d$. Then
$\Gamma(\mathcal{L})$ is connected if and only if $N^{d}({S})=V_{\cal L}$.
Moreover, if $N^{d}({S})=V_{\cal L}$ then
$\mathrm{diam}(\Gamma(\mathcal{L}))\leq 2d+\mathrm{diam}(\Gamma(S))$.

###### Proof.

  * $(\Rightarrow)$

Assume that $N^{d}({S})=V_{\cal L}$. Let $u,v$ be two distinct vertices in
$V_{\cal L}$. We need to prove that there is path joining $u$ and $v$ having
distance at most $2d+\mathrm{diam}(\Gamma(S))$. Indeed, let $k$ and $l$ be the
smallest indices such that $u\in N^{k}(S)$ and $v\in N^{l}(S)$, respectively.
Obviously, $k$ and $l$ are both upper bounded by $d$. Because $k$ is the
smallest index such that $u\in N^{k}(S)$, we have $u\in N^{k}(S)\setminus
N^{k-1}(\mathcal{S})$. If so, there is a vertex $u_{1}\in N^{k-1}(S)$ such
that $u$ and $u_{1}$ are adjacent. Moreover, we observe that $u_{1}$ must be
an element in $N^{k-1}(S)\setminus N^{k-2}(S)$ (otherwise, $u\in N^{k-1}(S)$,
a contradiction to the definition of $k$). Therefore, $\Gamma(\mathcal{L})$
contains a path $\\{u,\,u_{1},\,\ldots,\,u_{k}\\}$ where $u_{k}\in S$.

Similarly, $\Gamma(\mathcal{L})$ contains a path $\\{v,v_{1},\ldots,v_{l}\\}$
with $v_{l}\in S$. Since $\Gamma(S)$ is connected, there is a path joined
$u_{k}$ and $v_{l}$. Therefore, $u$ and $v$ are joined by a path of length
bounded above by $k+l+\mathrm{diam}(\Gamma(S))\leq
2d+\mathrm{diam}(\Gamma(S))$, as required.

  * $(\Leftarrow)$

Assume that $N^{d}({S})\subsetneq V_{\cal L}$. We need to prove that
$\Gamma(\mathcal{L})$ is disconnected. By contradiction, suppose that
$\Gamma(\mathcal{L})$ is connected. Then, for any $u\in N^{d}({S})$ and $v\in
V_{\cal L}\setminus N^{d}({S})$, there is a path
$P:=\\{u,u_{1},\ldots,u_{k},v\\}$ in $\Gamma(\mathcal{L})$. It follows that
$v\in N^{d+k+1}(S)$. Since the series of closed neighborhoods of ${S}$
terminates at $d$, we obtain $v\in N^{d}(S)$, a contradiction. So,
$\Gamma(\mathcal{L})$ is disconnected, as required.

∎

######  Corollary 17.

Let $\cal L$ be a Lie algebra and let $s$ be a vertex of its commuting graph.
Suppose that the series of closed neighborhoods of $s$ terminates at a
positive integer $d$. Then $\Gamma(\mathcal{L})$ is connected if and only if
$N^{d}({s})=V_{\cal L}$. Moreover, if $N^{d}({s})=V_{\cal L}$ then
$\mathrm{diam}(\Gamma(\mathcal{L}))\leq 2d$.

######  Example 18.

Let $\mathcal{L}=\mathrm{M}_{2}(\mathbb{F})$ be the usual Lie algebra of all
$2\times 2$ matrices over the field $\mathbb{F}$. Then

$Z(\mathcal{L})=\\{\lambda I:\lambda\in\mathbb{F}\\}.$ (34)

Moreover, by letting $A=\begin{bmatrix}1&0\\\ 0&0\end{bmatrix}$, we can check
easily that

$\mathopen{}\mathclose{{}\left[\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix},A}\right]=0\iff\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix}A-A\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix}=0\iff a_{12}=a_{21}=0.$ (35)

Therefore,

$\displaystyle
N^{1}(A)=\mathopen{}\mathclose{{}\left\\{\begin{bmatrix}a_{11}&0\\\
0&a_{22}\end{bmatrix}\,|\,a_{11},\,a_{22}\in\mathbb{R};\,a_{11}\neq
a_{22}}\right\\}.$ (36)

Furthermore, it is elementary to check that

$\begin{bmatrix}b_{11}&b_{12}\\\
b_{21}&b_{22}\end{bmatrix}\begin{bmatrix}a_{11}&0\\\
0&a_{22}\end{bmatrix}-\begin{bmatrix}a_{11}&0\\\
0&a_{22}\end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\\\
b_{21}&b_{22}\end{bmatrix}=0\iff\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}(a_{11}-a_{22})b_{12}=0,\\\
(a_{11}-a_{22})b_{21}=0.\end{array}}\right.$ (37)

It thus follows that

$N^{2}(A)=N^{1}(A).$ (38)

In the other words, the series of neighborhoods of $A$ terminates at $k=1$.
Moreover, $N^{1}(A)\subsetneq V_{\cal L}$. By Theorem 16, we conclude that the
commuting graph of $\mathrm{M}_{2}(\mathbb{F})$ is disconnected.

Remark this result was proven in 2004 by Akbari [3]. Akbari also showed that
the commuting graph of $\mathrm{M}_{2}(\mathbb{F})$ is disconnected for every
field $\mathbb{F}$ and each connected component of
$\Gamma(\mathrm{M}_{2}(\mathbb{F}))$ is a complete graph. Moreover, in the
case whether $\mathbb{F}$ is a finite field, the number of connected component
if $|\mathbb{F}|^{2}+|\mathbb{F}|+1$, and its connected component is of the
size $|\mathbb{F}|^{2}-|\mathbb{F}|$.

Based on Corollary 17, we propose a process to determine the connectedness of
$\Gamma(\mathcal{L})$ and an upper bound for the diameter of its commuting
graph.

1:The structure constants of the Lie brackets

2:The connectedness and an upper bound for the diameter of the commuting graph
Step 1: Set ${S}:=\\{\text{ }\\}$ and $d:=0$. Choose any vertex in the
commuting graph, that is, an element $x\in\mathcal{L}\setminus Z(\mathcal{L})$
and append this element to $S$. Step 2: Use the structure constants to
determine the closed neighborhood of ${S}$:
$N({S})=\\{y\in\mathcal{L}\setminus Z(\mathcal{L}):[y,s]=0;\,\text{ for some
}s\in{S}\\}.$ (39) Step 3:

  * 1.

If $N({S})={S}\subsetneq\mathcal{L}\setminus Z(\mathcal{L})$ then return
“disconnected”.

  * 2.

If $N({S})=\mathcal{L}\setminus Z(\mathcal{L})$ then return “connected” and
upper bound $2(d+1)$.

  * 3.

If ${S}\subsetneq N({S})\subsetneq\mathcal{L}\setminus Z(\mathcal{L})$ then
append $N({S})$ to ${S}$, set $d:=d+1$, and repeat from Step 2.

Process 1 ConnectednessDiameter (Determining the connectedness and an upper
bound for the diameter of the commuting graph of a Lie algebra $\mathcal{L}$)

######  Remark 19.

The process may not stop if we are not sure that the series of neighborhoods
of ${\cal S}$ terminates. However, if $\cal L$ is finite-dimensional over a
finite field $\mathbb{F}$, then the commuting graph is finite, and therefore
Process 1 stops at a finite iteration.

######  Example 20.

Let $\mathfrak{s}_{4,11}$ be a Lie algebra defined over a field $\mathbb{F}$
as follows:

$\mathfrak{s}_{4,11}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>\colon[e_{2},e_{3}]=e_{1},\,[e_{4},e_{1}]=e_{1},\,[e_{4},e_{2}]=e_{2}.$
(40)

It is easy to see that the the center of $\mathfrak{s}_{4,11}$ is just the
trivial vector space. Therefore, each non-zero vector of $\mathfrak{s}_{4,11}$
is a vertex of the corresponding commuting graph. Choose the vertex $e_{4}$ as
the starting point to apply Process 1. Since

$[a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4},e_{4}]=0\iff a_{1}=a_{2}=0,$
(41)

we have

$N^{1}(e_{4})=\mathopen{}\mathclose{{}\left\\{a_{3}e_{3}+a_{4}e_{4}\mid(a_{3},a_{4})\neq(0,0)}\right\\}.$
(42)

Besides, it is elementary to check that

$[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},a_{3}e_{3}+a_{4}e_{4}]=0\iff-
x_{1}a_{4}+x_{2}a_{3}=x_{2}a_{4}=0.$ (43)

Remark that

$-x_{1}a_{4}+x_{2}a_{3}=x_{2}a_{4}=0\iff\mathopen{}\mathclose{{}\left[\begin{array}[]{ll}(x_{1}=0,x_{2}=0),&\text{if
}a_{4}\neq 0,\\\ (x_{1}=x_{1},x_{2}=0),&\text{if }a_{3}\neq 0=a_{4},\\\
(x_{1}=x_{1},x_{2}=x_{2}),&\text{if }a_{3}=0=a_{4}.\\\ \end{array}}\right.$
(44)

Therefore, we observe that

$N^{2}(e_{4})=\mathopen{}\mathclose{{}\left\\{a_{1}e_{1}+a_{3}e_{3}+a_{4}e_{4}\mid(a_{1},a_{3},a_{4})\neq(0,0,0)}\right\\}$
(45)

Similarly, we also have

$[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},a_{1}e_{1}+a_{3}e_{3}+a_{4}e_{4}]=0\iff-
x_{1}a_{4}+x_{2}a_{3}+x_{4}a_{1}=x_{2}a_{4}=0.$ (46)

And for any $(a_{1},a_{3},a_{4})\neq(0,0,0)$, we may see that

$\mathopen{}\mathclose{{}\left\\{\begin{array}[]{r}-x_{1}a_{4}+x_{2}a_{3}+x_{4}a_{1}=0\\\
x_{2}a_{4}=0\end{array}}\right.\iff\mathopen{}\mathclose{{}\left[\begin{array}[]{ll}(x_{1}=\dfrac{x_{4}a_{1}}{a_{4}},x_{2}=0,x_{4}=x_{4}),&\text{if
}a_{4}\neq 0,\\\
(x_{1}=x_{1},x_{2}=x_{2},x_{4}=\dfrac{-x_{2}a_{3}}{a_{1}}),&\text{if
}a_{1}\neq 0=a_{4},\\\
(x_{1}=x_{1},x_{2}=\dfrac{-x_{4}a_{1}}{a_{3}},x_{4}=x_{4}),&\text{if
}a_{3}\neq 0=a_{4},\\\ \end{array}}\right.$ (47)

Therefore, $x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4}$ is adjacent to
$e_{1}-\dfrac{x_{4}}{x_{2}}e_{3}$ for any $(x_{1},x_{2},x_{3},x_{4})$ with
$x_{2}\neq 0$. Combining with Equation (45), we conclude that

$N^{3}(e_{4})=\Gamma\mathopen{}\mathclose{{}\left(\mathfrak{s}_{4,11}}\right).$
(48)

According to Theorem 16, we observe that $\Gamma(\mathfrak{s}_{4,11})$ is
connected and

$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))\leq 6.$ (49)

We now prove that the diameter of $\Gamma(\mathfrak{s}_{4,11})$ is equal to
$3$. Indeed, it is easy to verify that

$\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},e_{2}]=0\\\
{}[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},e_{4}]=0\end{array}}\right.\iff\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}x_{3}e_{1}+x_{4}e_{2}=0\\\
x_{1}e_{1}+x_{2}e_{2}=0\end{array}}\right.\iff\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}x_{3}=x_{4}=0\\\
x_{1}=x_{2}=0\end{array}}\right.$ (50)

In other words, there is no vertex adjacent to both $e_{2}$ and $e_{4}$.
Therefore, the length of any path joining $e_{2}$ and $e_{4}$ must be at least
3. It follows that

$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))\geq 3.$ (51)

Besides, we observe from Equations (43) and (47) that

$\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}[x_{1}e_{1}+x_{3}e_{3}+x_{4}e_{4},e_{3}]=0,\\\
{}[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},x_{2}e_{1}-x_{4}e_{3}]=0\end{array}}\right.$
(52)

for every $x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}$. Therefore, for two arbitrary
vertices $u$ and $v$ in $\Gamma(\mathfrak{s}_{4,11})$, there are two vertices
of the forms $ae_{1}+be_{3}$ and $ce_{1}+de_{3}$ such that
$\\{u,ae_{1}+be_{3},ce_{1}+de_{3},v\\}$ is a path of
$\Gamma(\mathfrak{s}_{4,11})$. That is,

$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))\leq 3.$ (53)

Combining this with Equation (51), we get

$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))=3.$ (54)

To close this section, we remark that the speed of Process 1 depends on the
starting set of vertices. For instance, in Example 20, we may verify by the
same method that $N^{2}(e_{1})=\Gamma(\mathfrak{s}_{4,11})$. Therefore, if we
choose $S=\\{e_{1}\\}$, we just need two iterations to obtain the desired
equality $N^{2}(e_{1})=\Gamma(\mathfrak{s}_{4,11})$. Thus, it gives a better
upper bound: $\mathrm{diam}(\mathfrak{s}_{4,11})\leq 4$ compared to Equation
(49). This results comes from a fact that
$\dim(\ker(\mathrm{ad}_{e_{1}}))=3>\dim(\ker(\mathrm{ad}_{e_{4}}))=2$.

##  5 Connectedness and Diameters of commuting graphs of Lie algebras having
small-dimensional derived algebras

In this section, we will determine the diameters of commuting graphs of
finite-dimensional Lie algebras having either one- or two-dimensional derived
algebras. For convenience, we will denote by $\mathrm{Lie}(n,k)$ the family of
all Lie algebras of dimension $n$, which have derived algebras of dimension
$k$. As usual, the derived Lie algebra of a Lie algebra $\mathcal{L}$ will be
denoted by $\mathcal{L}^{1}$. We first show in the following proposition that
most of the commuting graphs of Lie algebras having small-dimensional derived
algebras have diameter 2.

######  Proposition 21.

Let $\cal L$ be a Lie algebra of dimension $n$ such that $\dim
Z(\mathcal{L})=m$ and $\dim\mathcal{L}^{1}=k$.

  * $(i)$

If $k<m$ then $\cal L$ is decomposable.

  * $(ii)$

If $m<n-2k$ then $\Gamma(\mathcal{L})$ is connected, and
$\mathrm{diam}(\Gamma(\mathcal{L}))=2$.

###### Proof.

If $k<m$ then $Z(\mathcal{L})\setminus\mathcal{L}^{1}\neq\emptyset$. This
yields the existence of an element $x\in\mathcal{L}\setminus\mathcal{L}^{1}$
such that $[x,L]=0$. In the other words, $\mathcal{L}$ is decomposed into
$\mathcal{L}^{\prime}\oplus\mathopen{}\mathclose{{}\left<x}\right>$, as
required.

Now, suppose that $m<n-2k$. Let $u$ and $v$ be two arbitrary vertices in
$\Gamma(\mathcal{L})$. Then

$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{u})}\right)\geq
n-\dim\mathopen{}\mathclose{{}\left(\mathrm{Im}(\mathrm{ad}_{u})}\right)\geq
n-\dim\mathcal{L}^{1}=n-k.$ (55)

Similarly,
$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{v})}\right)\geq n-k$.
Therefore,

$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{u})\cap\ker(\mathrm{ad}_{v})}\right)\geq
n-2k>m.$ (56)

It follows that there exists a vertex $w\in\Gamma(\mathcal{L})$ such that
$w\in\ker(\mathrm{ad}_{u})\cap\ker(\mathrm{ad}_{v})$. In other words,
$\\{u,w,v\\}$ is a path, which has distance $2$. Therefore,

$\mathrm{dist}(u,v)\leq 2\text{ for every }u,v\in\Gamma(\mathcal{L}).$ (57)

This proves the second part of the proposition. ∎

According to the above proposition, the majority of commuting graphs of Lie
algebras with small-dimensional derived algebras are connected and have a
diameter of two. For example, $\Gamma(\mathcal{L})$ is connected, and
$\mathrm{diam}(\Gamma(\mathcal{L}))=2$ if either $\dim\mathcal{L}^{1}=1$ and
$\dim Z(\mathcal{L})<n-2$, or $\dim\mathcal{L}^{1}=2$ and $\dim
Z(\mathcal{L})<n-4$.

######  Lemma 22.

Let $\mathcal{L}$ be a noncommutative Lie algebra of finite dimension. Then
$\dim Z(\mathcal{L})\leq n-2.$

###### Proof.

By contradiction, suppose that $\dim Z(\mathcal{L})=\dim\mathcal{L}-1$. If so,
there is a basis $e_{1},\ldots,e_{n}$ of $\mathcal{L}$ such that
$Z(\mathcal{L})=\mathopen{}\mathclose{{}\left<e_{1},\ldots,e_{n-1}}\right>$.
It follows that $[e_{i},e_{j}]=0$ for every $1\leq i\leq j\leq n$. That is,
$\mathcal{L}$ is commutative, a contradiction. ∎

######  Lemma 23.

Let $\mathcal{L}$ be a Lie algebra in the family $\mathrm{Lie}(n,1)$ such that
$\dim Z(\mathcal{L})=n-2$. Then $\Gamma(\mathcal{L})$ is disconnected.

###### Proof.

It is direct from Proposition 13. ∎

Besides, it is proven in [12, Theorem 1] that an indecomposable Lie algebra
over the real field in the family $\mathrm{Lie}(n,1)$ is isomorphic to one of
the followings:

$\displaystyle\mathfrak{n}_{2,1}:=\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>\colon$
$\displaystyle[e_{1},e_{2}]=e_{1},$ (58)
$\displaystyle\mathfrak{h}_{2m+1}:=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},\ldots,e_{2m+1}}\right>\colon$
$\displaystyle[e_{1},e_{1+m}]=[e_{2},e_{2+m}]=\cdots=[e_{m},e_{2m}]=e_{2m+1}.$
(59)

The proof given in [12, Theorem 1] is also available on every field.
Therefore, any Lie algebra in the family $\mathrm{Lie}(n,1)$ must be
isomorphic to one of the Lie algebras given in Equations (58) and (59). Note
that $Z(\mathfrak{h}_{2m+1})=\mathopen{}\mathclose{{}\left<e_{2m+1}}\right>$.
Therefore, by directly applying Proposition 21 and Lemma 23, we get the
following theorem.

######  Theorem 24.

Let $\mathcal{L}$ be an indecomposable Lie algebra in the family
$\mathrm{Lie}(n,1)$. Then the commuting graph of $\mathcal{L}$ is connected if
and only if $\mathcal{L}$ is isomorphic to $\mathfrak{h}_{2m+1}$ for some
$m\geq 2$. Moreover, the following equality holds for every $m\geq 2$:

$\mathrm{diam}\mathopen{}\mathclose{{}\left(\Gamma(\mathfrak{h}_{2m+1})}\right)=2.$
(60)

######  Remark 25.

Note that the centers of $\mathfrak{n}_{2,1}$ and $\mathfrak{h}_{3}$ defined
in Equations (58) and (59) are:

$Z(\mathfrak{n}_{2,1})=\\{0\\},\quad\text{ and }\quad
Z(\mathfrak{h}_{3})=\mathopen{}\mathclose{{}\left<e_{3}}\right>.$ (61)

  * 1.

Let $a_{1}e_{1}+a_{2}e_{2}$ and $b_{1}e_{1}+b_{2}e_{2}$ be two vertices of
$\Gamma(\mathfrak{n}_{2,1})$. Then

$[a_{1}e_{1}+a_{2}e_{2},b_{1}e_{1}+b_{2}e_{2}]=0\iff a_{1}b_{2}-a_{2}b_{1}=0.$
(62)

Let $S_{1,a}:=\\{\lambda(e_{1}+ae_{2})\mid 0\neq\lambda\in\mathbb{F}\\}$ and
$S_{2}:=\\{\lambda e_{2}\mid 0\neq\lambda\in\mathbb{F}\\}$. Then we may
conclude from Equation (62) that each subgraph of $\Gamma(\mathfrak{n}_{2,1})$
induced by $S_{1,a}$ (for some $a\in\mathbb{F}$) or by $S_{2}$ is a connected
component of $\Gamma(\mathfrak{n}_{2,1})$. In particular, if $\mathbb{F}$ is
finite, then the number of connected components of
$\Gamma(\mathfrak{n}_{2,1})$ is $(|\mathbb{F}|+1)$ and each connected
component is a complete graph of $|\mathbb{F}|-1$ vertices.

  * 2.

Similarly, for two distinct vertices $a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}$ and
$b_{1}e_{1}+b_{2}e_{2}+b_{3}e_{3}$ of $\Gamma(\mathfrak{h}_{3})$, we can check
that

$[a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3},b_{1}e_{1}+b_{2}e_{2}+b_{3}e_{3}]=0\iff
a_{1}b_{2}-a_{2}b_{1}=0.$ (63)

Now, let $S_{1,a}:=\\{\lambda(e_{1}+ae_{2}+be_{3})\mid
0\neq\lambda\in\mathbb{F},b\in\mathbb{F}\\}$, and
$S_{2}:=\\{\lambda(e_{2}+be_{3})\mid
0\neq\lambda\in\mathbb{F},b\in\mathbb{F}\\}$ for some $a\in\mathbb{F}$. Then
the subgraph induced by one of the sets $S_{1,a}$ and $S_{2}$, is a connected
component of $\Gamma(\mathfrak{h}_{3})$. In particular, if $\mathbb{F}$ is
finite, then $\Gamma(\mathfrak{h}_{3})$ has $(|\mathbb{F}|+1)$ connected
components, and each connected component is a complete graph of
$|\mathbb{F}|(|\mathbb{F}|-1)$ vertices.

Next, we study the case where $\mathcal{L}$ belongs to the family
$\mathrm{Lie}(n,2)$. If $\dim Z(\mathcal{L})<n-4$ then $\Gamma(\mathcal{L})$
is connected and $\mathrm{diam}(\Gamma(\mathcal{L}))=2$. By Lemma 22, $\dim
Z(\mathcal{L})\leq n-2$ unless $\mathcal{L}$ is commutative. If $\dim
Z(\mathcal{L})=n-2$ then there is a basic $\\{e_{1},\ldots,e_{n}\\}$ of
$\mathcal{L}$ such that $[e_{i},e_{j}]=0$ for every $i<j$ with
$(i,j)\neq(n-1,n).$ It follows that $\dim\mathcal{L}^{1}\leq 1$, a
contradiction. This proves the following lemma.

######  Lemma 26.

Let $\mathcal{L}$ be a noncommutative Lie algebra in the family
$\mathrm{Lie}(n,2)$. Then $\dim Z(\mathcal{L})\leq n-3$.

Now, applying directly Proposition 13, we get the following result.

######  Lemma 27.

Let $n$ be an integer such that $n\geq 3$ and let $\cal L$ be a Lie algebra in
the family $\mathrm{Lie}(n,2)$ such that $\dim Z(\mathcal{L})=n-3$. Then
$\Gamma(\mathcal{L})$ is disconnected.

Next, we will classify all the Lie algebras in the family $\mathrm{Lie}(n,2)$
having centers of codimension 4 and determine the diameters of their commuting
graphs.

######  Theorem 28.

Let $n$ be an integer such that $n\geq 4$ and let $\cal L$ be a Lie algebra in
the family $\mathrm{Lie}(n,2)$ such that $\dim Z(\mathcal{L})=n-4$. Then
$\mathcal{L}=\mathcal{M}\oplus\mathbb{F}^{n-k}$ for some integer $4\leq k\leq
6$, and some Lie algebra, $\mathcal{M}$, belonging to the family
$\mathrm{Lie}(k,2)$. Moreover, $\mathcal{M}$ is isomorphic to one of the
following Lie algebras:

$\displaystyle\mathcal{M}_{1,k}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{k}}\right>\colon$
$\displaystyle[e_{1},e_{2}]=\sum_{i=1}^{k}a_{i}e_{i},\,[e_{3},e_{4}]=\sum_{i=1}^{k}b_{i}e_{i}$
(64)
$\displaystyle\mathcal{M}_{2,k}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{k}}\right>\colon$
$\displaystyle[e_{1},e_{2}]=\sum_{i=1}^{k}a_{i}e_{i},\,[e_{1},e_{3}]=\sum_{i=1}^{k}b_{i}e_{i},\,[e_{3},e_{4}]=\sum_{i=1}^{k}(a_{i}+\alpha
b_{i})e_{i}$ (65)
$\displaystyle\mathcal{M}_{3,k}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},\ldots,e_{k}}\right>\colon$
$\displaystyle[e_{1},e_{2}]=[e_{2},e_{4}]=\sum_{i=1}^{k}a_{i}e_{i},\,[e_{1},e_{3}]=\sum_{i=1}^{k}b_{i}e_{i},\,[e_{3},e_{4}]=\sum_{i=1}^{k}(\beta
a_{i}+\gamma b_{i})e_{i}$ (66)

for some $a_{i},b_{i},\alpha,\beta,\gamma\in\mathbb{F}$ such that
$(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ are linearly independent
over $\mathbb{F}$ and that $x^{2}+\gamma x-\beta$ has no root in $\mathbb{F}$.
Furthermore, the commuting graphs of $\mathcal{M}_{1,k}$ and
$\mathcal{M}_{2,k}$ are connected and both have diameter $3$, while the
commuting graph of $\mathcal{M}_{3,k}$ is disconnected.

###### Proof.

Since $\dim Z(\mathcal{L})=n-4$, there is a basic $\\{e_{1},\ldots,e_{n}\\}$
of $\mathcal{L}$ such that

$Z(\mathcal{L})=\mathopen{}\mathclose{{}\left<e_{5},\ldots,e_{n}}\right>,\text{
and }\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{i},e_{j}]:1\leq i<j\leq
4}\right>.$

Because $\dim\mathcal{L}^{1}=2$, we have only two possibilities for the
spanning set of $\mathcal{L}^{1}$ as follows:
Case 1. There is no $i,j,k$ such that
$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{i},e_{j}],[e_{i},e_{k}]}\right>$.
In this case, by changing the indices if necessary, we may assume that
$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{1},e_{2}],[e_{3},e_{4}]}\right>$
and

$[e_{1},e_{3}]=[e_{1},e_{4}]=[e_{2},e_{3}]=[e_{2},e_{4}]=0.$ (67)

It follows that

$\displaystyle[c_{1}e_{1}+c_{2}e_{2},d_{3}e_{3}+d_{4}e_{4}]=0\quad\text{for
every }c_{1},c_{2},d_{3},d_{4}\in\mathbb{F}.$ (68)

Therefore, for arbitrary two vertices $u=\sum_{i=1}^{n}c_{i}e_{i}$ and
$v=\sum_{i=1}^{n}d_{i}e_{i}$,

  * 1.

$\\{u,e_{1}+e_{2},v\\}$ is a path of $\Gamma(\mathcal{L})$ provided that
$(c_{1},c_{2})=(0,0)$ and $(d_{1},d_{2})=(0,0)$,

  * 2.

$\\{u,d_{1}e_{1}+d_{2}e_{2},v\\}$ is a path of $\Gamma(\mathcal{L})$ provided
that $(c_{1},c_{2})=(0,0)$ and $(d_{1},d_{2})\neq(0,0)$,

  * 3.

$\\{u,c_{1}e_{1}+c_{2}e_{2},e_{3}+e_{4},v\\}$ is a path of
$\Gamma(\mathcal{L})$ provided that $(c_{1},c_{2})\neq(0,0)$ and
$(d_{3},d_{4})=(0,0)$,

  * 4.

$\\{u,c_{1}e_{1}+c_{2}e_{2},d_{3}e_{3}+d_{4}e_{4},v\\}$ is a path of
$\Gamma(\mathcal{L})$ provided that $(c_{1},c_{2})\neq(0,0)$ and
$(d_{3},d_{4})\neq(0,0)$.

Therefore, any pair of vertices of $\Gamma(\mathcal{L})$ is joined by a path
of length at most $3$. Moreover, it can be seen that there is no vertex in
$\Gamma(\mathcal{L})$ adjacency to both $e_{1}+e_{3}$ and $e_{2}+e_{4}$. We
thus conclude that $\Gamma(\mathcal{L})$ is connected and
$\mathrm{diam}(\Gamma(\mathcal{L}))=3$ in this case.

Furthermore, if we denote by $\cal H$ the subalgebra of $\mathcal{L}$
generated by $\\{e_{1},e_{2},e_{3},e_{4},[e_{1},e_{2}],[e_{3},e_{4}]\\}$, then
$\mathcal{L}$ is isomorphic to
$\mathcal{H}\oplus\mathbb{F}^{n-\dim\mathcal{H}}$. This Lie algebra
$\mathcal{H}$ is clearly isomorphic to $\mathcal{M}_{1,k}$ where
$k=\dim\mathcal{H}$. Therefore, in this case, $\mathcal{L}$ is isomorphic to
$\mathcal{M}_{1,k}\oplus\mathbb{F}^{n-k}$, and
$\dim(\Gamma(\mathcal{L}))=\dim(\Gamma(\mathcal{M}_{1,k})=3.$
Case 2. There is $i,j,k$ such that
$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{i},e_{j}],[e_{i},e_{k}]}\right>$.
Without loss of generality, assume that
$\mathcal{L}^{1}=\mathopen{}\mathclose{{}\left<[e_{1},e_{2}],[e_{1},e_{3}]}\right>$.
If so, there are $c_{1},\ldots,c_{6}\in\mathbb{F}$ such that

$\displaystyle[e_{1},e_{4}]=c_{1}[e_{1},e_{2}]+c_{2}[e_{1},e_{3}],$ (69)
$\displaystyle[e_{2},e_{3}]=c_{3}[e_{1},e_{2}]+c_{4}[e_{1},e_{3}],$ (70)
$\displaystyle[e_{2},e_{4}]=c_{5}[e_{1},e_{2}]+c_{6}[e_{1},e_{3}].$ (71)

Let $e_{2}^{\prime}=e_{2}-c_{4}e_{1}$, $e_{3}^{\prime}=e_{3}+c_{3}e_{1}$ and
$e_{4}^{\prime}=e_{4}-(c_{1}c_{4}-c_{5})e_{1}-c_{1}e_{2}^{\prime}-c_{2}e_{3}^{\prime}$.
Then

$\displaystyle[e_{2}^{\prime},e_{3}^{\prime}]=[e_{1},e_{4}^{\prime}]=0,\text{
and
}[e_{2}^{\prime},e_{4}^{\prime}]=(c_{6}-c_{2}c_{4})[e_{1},e_{3}^{\prime}].$
(72)

Therefore, without loss of generality, we may assume that

$[e_{1},e_{4}]=[e_{2},e_{3}]=0,[e_{2},e_{4}]=\alpha[e_{1},e_{3}],\text{ and
}[e_{3},e_{4}]=\beta[e_{1},e_{2}]+\gamma[e_{1},e_{3}].$ (73)

for some $\alpha,\beta,\gamma$ in $\mathbb{F}$. We split this case into three
subcases as follows:

Case 2.1.

$\alpha=\beta=0$. If so, $\gamma\neq 0$ holds because $e_{4}\notin
Z(\mathcal{L})$. By letting $e_{1}^{\prime}=\gamma e_{1}+e_{4}$, we easily see
that

$[e_{1}^{\prime},e_{3}]=[e_{1}^{\prime},e_{4}]=0.$

Therefore, $\mathcal{L}$ is isomorphic to
$\mathcal{M}_{1,k}\oplus\mathbb{F}^{n-k}$. As a consequence, its commuting
graph, $\Gamma(\mathcal{L})$, is connected and has diameter $3$.

Case 2.2.

$\alpha=0$, and $\beta\neq 0$. Let $u=\sum_{i=1}^{n}c_{i}e_{i}$ and
$v=\sum_{i=1}^{n}d_{i}e_{i}$ be two arbitrary vertices in
$\Gamma(\mathcal{L})$.

  * 1.

Case 2.2.1. $c_{1}\neq 0$. If so, by letting

$u_{1}=(c_{1}\gamma)e_{1}+(c_{2}\gamma-c_{3}\beta)e_{2}+c_{1}e_{4},$ (74)

we observe that $u_{1}$ is an adjacent vertex of $u$ in $\Gamma(\mathcal{L})$.
Moreover, by letting
$u_{2}=x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4}\in\mathcal{L}$, we get

$\mathopen{}\mathclose{{}\left\\{\begin{aligned} =0\\\
{}[v,u_{2}]=0\end{aligned}}\right.\Longleftrightarrow\mathopen{}\mathclose{{}\left\\{\begin{aligned}
a_{1}\gamma x_{2}-(a_{2}\gamma-a_{3}\beta)x_{1}-a_{1}\beta x_{3}&=0\\\
a_{1}x_{2}-a_{2}x_{1}+\beta(a_{3}x_{4}-a_{4}x_{3})&=0\\\
a_{1}x_{3}-a_{3}x_{1}+\gamma(a_{3}x_{4}-a_{4}x_{3})&=0\end{aligned}}\right.$
(75)

The above homogeneous system must has a non-trivial solution. Therefore, there
is a vertex $u_{2}$ in $\Gamma(\mathcal{L})$ such that $\\{u,u_{1},u_{2},v\\}$
is a path.

  * 2.

Case 2.2.2. $c_{1}=d_{1}=0$. If so, we observe from the assumption
$[e_{2},e_{3}]=[e_{2},e_{4}]=0$ in Equation (73) that $\\{u,e_{2},v\\}$ is a
path in $\Gamma(\mathcal{L})$.

  * 3.

Case 2.2.3. $c_{1}=0$ and $d_{1}\neq 0$. Similar to Case 2.2.2 above, we
observe that any pair of vertices, $u$ and $v$, is joined by a path of length
$3$.

Therefore, in this Case 2.2, we observe that $\Gamma(\mathcal{L})$ is
connected and

$\mathrm{diam}(\Gamma(\mathcal{L}))\leq 3.$ (76)

Besides, according to the structure constants given in Equation (73), we
observe that

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
&\mathopen{}\mathclose{{}\left[e_{1},\sum_{i=1}^{n}c_{i}e_{i}}\right]=0\\\
&\mathopen{}\mathclose{{}\left[e_{3},\sum_{i=1}^{n}c_{i}e_{i}}\right]=0\end{aligned}}\right.\iff\mathopen{}\mathclose{{}\left\\{\begin{aligned}
c_{2}=c_{3}=0\\\ c_{1}=c_{4}=0\end{aligned}}\right.$ (77)

Therefore, there is no vertex $w$ in $\Gamma(\mathcal{L})$ such that $w$ is
adjacent to both the vertices $e_{1}$ and $e_{3}$. In combining with Equation
(76), we get $\mathrm{diam}(\Gamma(\mathcal{L}))=3$.

Now, by basis change $\frac{1}{\beta}e_{4}\rightarrow e_{4}$, we easily see
that $\mathcal{L}$ is isomorphic to $\mathcal{M}_{2,k}$ where $k$ equals to
the dimension of the vector space generated by
$\\{e_{1},e_{2},e_{3},e_{4},[e_{1},e_{2}],[e_{1},e_{3}]\\}$.

Case 2.3.

$\alpha\neq 0$. Since
$[e_{2},\frac{1}{\alpha}e_{4}]=\frac{1}{\alpha}[e_{2},e_{4}]=[e_{1},e_{3}]$,
we may assume (after a change of basis if necessary) that $\alpha=1$.

  * 1.

Case 2.3.1. The quadratic $x^{2}+\gamma x-\beta$ has no root in $\mathbb{F}$.
Let $b_{1},\,b_{2}$ be two elements in $\mathbb{F}$ such that
$(b_{1},b_{2})\neq(0,0)$ and let $V_{b_{1},b_{2}}$ be the set of all vertices
in $\Gamma(\mathcal{L})$ of the form $\lambda b_{1}e_{1}+\lambda
b_{2}e_{2}+\delta b_{2}e_{3}-\delta b_{1}e_{4}+\sum_{i=5}b_{i}e_{i}$ where
$(0,0)\neq(\lambda,\delta)\in\mathbb{F}^{2}$ and $b_{5},\ldots,b_{n}$ in
$\mathbb{F}$. That is,

$V_{b_{1},b_{2}}=\mathopen{}\mathclose{{}\left\\{\lambda b_{1}e_{1}+\lambda
b_{2}e_{2}+\delta b_{2}e_{3}-\delta
b_{1}e_{4}+\sum_{i=5}b_{i}e_{i}:(\lambda,\delta)\neq(0,0)}\right\\}\subseteq
V_{\cal L}.$ (78)

We will prove that the subgraph of $\Gamma(\cal L)$ induced by
$V_{b_{1},b_{2}}$, denoted by $\Gamma(V_{b_{1},b_{2}})$, is a connected
component of $\Gamma(\mathcal{L})$. Indeed, let $u=\sum_{i=1}^{n}a_{i}e_{i}$
and $v=\lambda b_{1}e_{1}+\lambda b_{2}e_{2}+\delta b_{2}e_{3}-\delta
b_{1}e_{4}+\sum_{i=5}^{n}b_{i}e_{i}$ be two arbitrary vertices in
$\Gamma(\mathcal{L})$ and $V_{b_{1},b_{2}}$, respectively. Then

$[u,v]=0\iff\mathopen{}\mathclose{{}\left\\{\begin{aligned}
&\lambda(a_{1}b_{2}-a_{2}b_{1})-\beta\delta(a_{3}b_{1}+a_{4}b_{2})=0\\\
&\delta a_{1}b_{2}-\lambda a_{3}b_{1}-\delta a_{2}b_{1}-\lambda
a_{4}b_{2}-\gamma\delta(a_{3}b_{1}+a_{4}b_{2})=0\end{aligned}}\right.$ (79)

It follows that

$[u,v]=0\iff\mathopen{}\mathclose{{}\left\\{\begin{aligned}
&\lambda(a_{1}b_{2}-a_{2}b_{1})-\beta\delta(a_{3}b_{1}+a_{4}b_{2})=0\\\
&\delta(a_{1}b_{2}-a_{2}b_{1})-(\gamma\delta+\lambda)(a_{3}b_{1}+a_{4}b_{2})=0\end{aligned}}\right.$
(80)

It is elementary to check that

$-\lambda(\gamma\delta+\lambda)+\beta\delta^{2}=-\lambda^{2}-\gamma(\lambda\delta)+\beta\delta^{2}.$
(81)

Since $x^{2}+\gamma x-\beta$ has no root in $\mathbb{F}$, Equation (81)
implies that $-\lambda(\gamma\delta+\lambda)+\beta\delta^{2}\neq 0$. Thus, we
observe from Equation (80) that

$[u,v]=0\iff a_{1}b_{2}-a_{2}b_{1}=a_{3}b_{1}+a_{4}b_{2}=0.$ (82)

Therefore, $u$ is adjacent to a vertex in $\Gamma(V_{b_{1},b_{2}})$ if and
only if $u\in V_{b_{1},b_{2}}$. In other words, the induced subgraph of
$\Gamma(\mathcal{L})$ generated by $V_{b_{1},b_{2}}$ is a connected component
of $\Gamma(\mathcal{L})$. It follows that $\Gamma(\mathcal{L})$ is
disconnected. Moreover, from the above proof, we may see that each connected
component is a complete graph. In the case where $\mathbb{F}$ is a finite
field, then the number of connected components is $(|\mathbb{F}|+1)$ and each
connected component is a complete graph of
$(|\mathbb{F}|^{2}-1)|\mathbb{F}|^{n-4}$ vertices.

Note that this Lie algebra is isomorphic to
$\mathcal{M}_{3,k}\oplus\mathbb{F}^{n-k}$, where $k$ equals to the dimension
of the vector space generated by
$\\{e_{1},e_{2},e_{3},e_{4},[e_{1},e_{2}],[e_{1},e_{3}]\\}$.

  * 2.

Case 2.3.2. The quadratic $x^{2}+\gamma x-\beta$ has a root $\zeta$ in
$\mathbb{F}$. In this case, $\beta=\zeta(\gamma+\zeta)$. Therefore,

$\displaystyle[e_{3}+\zeta e_{2},e_{4}+(\gamma+\zeta)e_{1}]$ $\displaystyle=$
$\displaystyle[e_{3},e_{4}]+\zeta[e_{2},e_{4}]-(\gamma+\zeta)[e_{1},e_{3}]-\zeta(\gamma+\zeta)[e_{1},e_{2}]$
$\displaystyle=$
$\displaystyle\beta[e_{1},e_{2}]+\gamma[e_{1},e_{3}]+\zeta[e_{1},e_{3}]-(\gamma+\zeta)[e_{1},e_{3}]-\zeta(\gamma+\zeta)[e_{1},e_{2}]=0.$
(83)

Moreover, we may check directly the following equality:

$[e_{2},e_{4}+(\gamma+\zeta)e_{1}]=[e_{1},e_{3}+\zeta
e_{2}]-(\gamma+2\zeta)[e_{1},e_{2}].$ (84)

Thus, this Lie algebra is isomorphic to
$\mathcal{M}_{2,k}\oplus\mathbb{F}^{n-k}$. Therefore, the commuting graph of
Lie algebras in this case is connected and has diameter $3$.

∎

######  Remark 29.

In the case where the characteristic of $\mathbb{F}$ is different from $2$,
the polynomial $x^{2}+\gamma x-\beta$ has no root in $\mathbb{F}$ if and only
if $\gamma^{2}+4\beta$ is not a square in $\mathbb{F}$. Otherwise,
$x^{2}+\gamma x-\beta$ does not have root in a field $\mathbb{F}$ of
characteristic $2$ if and only if $\gamma\neq 0$ and
$\dfrac{-\beta}{\gamma}+\mathopen{}\mathclose{{}\left(\dfrac{-\beta}{\gamma}}\right)^{2}+\cdots+\mathopen{}\mathclose{{}\left(\dfrac{-\beta}{\gamma}}\right)^{2^{n-1}}=1,$
where $n=[\mathbb{F}:\mathbb{F}_{2}]$ is the degree of extension of the field
of $\mathbb{F}$ over $\mathbb{F}_{2}$, the field of 2 elements [9, Corollary
3.79].

##  6 Diameters of Commuting graphs of Lie algebras of small dimension

In this section, we will use results in previous sections, especially Process
in Section 4 to study the diameters of the commuting graphs of all real
solvable Lie algebras of dimension up to 4. These Lie algebras are classified
in [13, Chapters 16, 17, 18]. Note that these results are applied similarly to
the complex case. Recall that the commuting graph of any Lie algebra of
dimension $\leq 3$ is disconnected (see Corollary 14).

######  Theorem 30.

Let $\cal L$ be a solvable Lie algebra of dimension 4 over $\mathbb{R}$. Then
$\Gamma(\mathcal{L})$ is connected if and only if $\mathcal{L}$ is isomorphic
to one of the following Lie algebras:
$\mathfrak{n}_{2,1}\oplus\mathfrak{n}_{2,1}$ and $\mathfrak{s}_{4,11}$, where
$\mathfrak{n}_{2,1}$ is defined in Equation (58), and $\mathfrak{s}_{4,11}$ is
defined as follows:

$\mathfrak{s}_{4,11}=\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>\colon[e_{2},e_{3}]=e_{1},\,[e_{4},e_{1}]=e_{1},\,[e_{4},e_{2}]=e_{2}.$
(85)

Moreover,
$\mathrm{diam}\mathopen{}\mathclose{{}\left(\Gamma(\mathfrak{n}_{2,1}\oplus\mathfrak{n}_{2,1})}\right)=\mathrm{diam}\mathopen{}\mathclose{{}\left(\Gamma(\mathfrak{s}_{4,11})}\right)=3.$

###### Proof.

According to Proposition 13, we observe that $\Gamma(\mathcal{L})$ is
disconnected if $\dim(Z(\mathcal{L}))\geq 1$. Therefore, in the case where
$\mathcal{L}$ is decomposable, the commuting graph of $\mathcal{L}$ is
connected only if $\mathcal{L}$ is isomorphic to
$\mathcal{L}_{1}\oplus\mathcal{L}_{2}$ where $\mathcal{L}_{1}$ and
$\mathcal{L}_{2}$ are two noncommutative subalgebras of $\mathcal{L}$
satisfying $\dim\mathcal{L}_{1}=\dim\mathcal{L}_{2}=2$. In this case,
$\Gamma(\mathcal{L})$ is connected, and $\mathrm{diam}(\Gamma(\mathcal{L}))=3$
(see Remark 8). Note that any noncommutative Lie algebra of dimension 2 is
isomorphic to the Lie algebra $\mathfrak{n}_{2,1}$ defined in Equation (58).
Therefore, in the case where $\mathcal{L}$ is decomposable, $\mathcal{L}$ is
connected if and only if $\mathcal{L}$ is isomorphic to
$\mathfrak{n}_{2,1}\oplus\mathfrak{n}_{2,1}$.

Also, if $\mathcal{L}$ is indecomposable, then $\mathcal{L}$ must be
isomorphic to one of the nonisomorphic $13$ families: $\mathfrak{n}_{4,1}$,
$\mathfrak{s}_{4,1},\ldots,\mathfrak{s}_{4,12}$ (see [13, Chapter 17]). Note
that the families $\mathfrak{n}_{4,1},\,\mathfrak{s}_{4,1}$,
$\mathfrak{s}_{4,6}$, and $\mathfrak{s}_{4,7}$ have a non-trivial center.
Therefore, the commuting graphs of $\mathfrak{n}_{4,1},\,\mathfrak{s}_{4,1}$,
$\mathfrak{s}_{4,6}$, and $\mathfrak{s}_{4,7}$ are disconnected. These
families of Lie algebras and their centers are given in Table 1.

Table 1: 4-dimensional solvable Lie algebras having non-trivial centers. $\mathcal{L}$ | Non-trivial Lie brackets | $Z(\mathcal{L})$ | $\dim\mathcal{L}-\dim Z(\mathcal{L})$ | Connectedness
---|---|---|---|---
$\mathfrak{n}_{4,1}$ | $[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2}$ | $\mathopen{}\mathclose{{}\left<e_{1}}\right>$ | $3$ | Disconnected
$\mathfrak{s}_{4,1}$ | $[e_{4},e_{2}]=e_{1},[e_{4},e_{3}]=e_{3}$ | $\mathopen{}\mathclose{{}\left<e_{1}}\right>$ | $3$ | Disconnected
$\mathfrak{s}_{4,6}$ | $[e_{2},e_{3}]=e_{1},[e_{4},e_{2}]=e_{2},[e_{4},e_{3}]=-e_{3}$ | $\mathopen{}\mathclose{{}\left<e_{1}}\right>$ | $3$ | Disconnected
$\mathfrak{s}_{4,7}$ | $[e_{2},e_{3}]=e_{1},[e_{4},e_{2}]=-e_{3},[e_{4},e_{3}]=e_{2}$ | $\mathopen{}\mathclose{{}\left<e_{1}}\right>$ | $3$ | Disconnected

Now, we will show that each of the families
$\mathfrak{s}_{4,2},\,\mathfrak{s}_{4,3},\,\mathfrak{s}_{4,4},\,\mathfrak{s}_{4,5},\,\mathfrak{s}_{4,8},\,\mathfrak{s}_{4,9},\,\mathfrak{s}_{4,10}$
is disconnected by using Corollary 11. These families are defined in Table 2.
Let us first consider the algebra $\mathfrak{s}_{4,2}$. In this algebra, we
easily see that

$\begin{array}[]{ll}[x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4},e_{4}]=0&\iff-
x_{1}e_{1}-x_{2}(e_{1}+e_{2})-x_{3}(e_{2}+e_{3})=0\\\ &\iff
x_{1}=x_{2}=x_{3}=0.\end{array}$ (86)

Therefore,
$\ker(\mathrm{ad}_{e_{4}})=\mathopen{}\mathclose{{}\left<e_{4}}\right>$. It
follows that
$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{e_{4}})}\right)=1=\dim
Z(\mathfrak{s}_{4,2})+1$. Applying Corollary 11, the commuting graph of
$\mathfrak{s}_{4,2}$ is disconnected. Using the same method, we may show that,
for a Lie algebra $\mathcal{L}$ in
$\mathfrak{s}_{4,2},\,\mathfrak{s}_{4,3},\,\mathfrak{s}_{4,4},\,\mathfrak{s}_{4,5},\,\mathfrak{s}_{4,8},\,\mathfrak{s}_{4,9},\,\mathfrak{s}_{4,10}$,
there is a nonzero vector $x\in\mathcal{L}$ or a vertex $x\in V_{\mathcal{L}}$
such that
$\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{x})}\right)=\dim
Z(\mathcal{L})+1$ (see Table 2), and thus the commuting graph of each Lie
algebra in $\mathfrak{s}_{4,2}$, $\mathfrak{s}_{4,3}$, $\mathfrak{s}_{4,4}$,
$\mathfrak{s}_{4,5}$, $\mathfrak{s}_{4,8},$ $\mathfrak{s}_{4,9},$
$\mathfrak{s}_{4,10}$, is disconnected.

Table 2: Solvable Lie algebras, $\mathcal{L}$, of dimension $4$ satisfying $\dim\mathopen{}\mathclose{{}\left(\ker(\mathrm{ad}_{x})}\right)=\dim Z(\mathcal{L})+1$ for some $x\in V_{\mathcal{L}}$. Here, all Lie algebras are assumed to be spanned by $\\{e_{1},e_{2},e_{3},e_{4}\\}$. $\mathcal{L}$ | Non-trivial Lie brackets | $x$ | $\ker(\mathrm{ad}_{x})$ | Connectedness
---|---|---|---|---
$\mathfrak{s}_{4,2}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=e_{1}+e_{2},\\\ &[e_{4},e_{3}]=e_{2}+e_{3}\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,3}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=\alpha e_{2},[e_{4},e_{3}]=\beta e_{3}\\\ &\mathopen{}\mathclose{{}\left(0<|\beta|\leq|\alpha|\leq 1,(\alpha,\beta)\neq(-1,-1)}\right)\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,4}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=e_{1}+e_{2},\\\ &[e_{4},e_{3}]=\alpha e_{3}\quad(\alpha\neq 0)\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,5}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{4},e_{1}]=\alpha e_{1},[e_{4},e_{2}]=\beta e_{2}-e_{3},\\\ &[e_{4},e_{3}]=e_{2}+\beta e_{3}\quad(\alpha>0)\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,8}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=(1+\alpha)e_{1},[e_{4},e_{2}]=e_{2},\\\ &[e_{4},e_{3}]=\alpha e_{3}\quad(-1<\alpha\leq 1)\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,9}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=2\alpha e_{1},[e_{4},e_{2}]=\alpha e_{2}-e_{3},\\\ &[e_{4},e_{3}]=e_{2}+\alpha e_{3}\quad(\alpha>0)\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected
$\mathfrak{s}_{4,10}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=2e_{1},\\\ &[e_{4},e_{2}]=e_{2},[e_{4},e_{3}]=e_{2}+e_{3}\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{4}}\right>$ | Disconnected

For the family $\mathfrak{s}_{4,11}$, a calculation similar to Example 20
gives us the series of closed neighborhoods of $e_{4}$ in
$\Gamma(\mathfrak{s}_{4,11})$ as presented in Table 3. This series satisfies
the equality $N^{3}(e_{4})=\Gamma(\mathfrak{s}_{4,11})$. It follows that
$\Gamma(\mathfrak{s}_{4,11})$ is connected and
$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))\leq 6$ (see Theorem 16). Using the
same argument as in Example 20, the equations (51) and (53) both hold for the
case where $\mathbb{F}=\mathbb{R}$ also. Therefore,
$\mathrm{diam}(\Gamma(\mathfrak{s}_{4,11}))=3$.

Finally, for the family $\mathfrak{s}_{4,12}$, one can check that

$[e_{1},x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4}]=0\iff-
x_{3}e_{1}+x_{4}e_{2}=0\iff x_{3}=x_{4}=0,$ (87)

and

$[a_{1}e_{1}+a_{2}e_{2},x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4}]=0\iff\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}a_{1}x_{3}+a_{2}x_{4}=0\\\
a_{1}x_{4}-a_{2}x_{3}=0\end{array}}\right.\iff x_{3}=x_{4}=0,$ (88)

for every $(0,0)\neq(a_{1},a_{2})\in\mathbb{R}^{2}$. Therefore,

$\displaystyle
N^{1}(e_{1})=\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>\setminus\\{0\\},$
(89) $\displaystyle
N^{2}(e_{1})=\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>\setminus\\{0\\}=N^{1}(e_{1}).$
(90)

By Theorem 16, the commuting graph of $\mathfrak{s}_{4,12}$ is disconneted.
This completes the proof.

Table 3: A series of closed neighborhoods of $\mathfrak{s}_{4,11}$ and $\mathfrak{s}_{4,12}$. Here, we denote by $\\{e_{1},e_{2},e_{3},e_{4}\\}$ a basis for those Lie algebras, and by $U^{*}:=\\{x\in U\mid x\neq\mathbf{0}_{U}\\}$ the set of non-zero elements of a vector space $U$. $\cal L$ | Non-trivial Lie brackets | $S$ | $N^{1}(S)$ | $N^{2}(S)$ | $N^{3}(S)$
---|---|---|---|---|---
$\mathfrak{s}_{4,11}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=e_{1},\\\ &[e_{4},e_{2}]=e_{2}\end{aligned}}\right.$ | $e_{4}$ | $\mathopen{}\mathclose{{}\left<e_{3},e_{4}}\right>^{*}$ | $\mathopen{}\mathclose{{}\left<e_{1},e_{3},e_{4}}\right>^{*}$ | $\mathopen{}\mathclose{{}\left<e_{1},e_{2},e_{3},e_{4}}\right>^{*}=V_{\mathfrak{s}_{4,11}}$
$\mathfrak{s}_{4,12}$ | $\mathopen{}\mathclose{{}\left\\{\begin{aligned} &[e_{3},e_{1}]=e_{1},[e_{3},e_{2}]=e_{2}\\\ &[e_{4},e_{1}]=-e_{2},[e_{4},e_{2}]=e_{1}\end{aligned}}\right.$ | $e_{1}$ | $\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>^{*}$ | $\mathopen{}\mathclose{{}\left<e_{1},e_{2}}\right>^{*}\subsetneq V_{\mathfrak{s}_{4,12}}$ |

∎

##  7 Conclusion

This study has explored several interesting properties related to the
connectedness and diameter of the commuting graph of a general Lie algebra
over any field. Because the diameter of the commuting graph of a decomposable
Lie algebra can be easily computed via its subalgebras, it is sufficient to
investigate only indecomposable Lie algebras. For indecomposable Lie algebras,
we obtained an equivalent condition for these Lie algebras to have connected
commuting graphs and proposed a process to determine an upper bound for their
diameters. To demonstrate the theory as well as the process, we have
determined the connectedness of: (1) the families of all Lie algebras having
one- or two-dimensional derived algebras, and (2) real solvable Lie algebras
of dimension up to 4. The diameters of these Lie algebras were also computed.
Therefore, it is worth extending these results to all Lie algebras and using
their commuting graphs to obtain a new classification of Lie algebras.
Besides, if $\mathbb{F}$ is finite then the commuting graph of any finite-
dimensional Lie algebra over $\mathbb{F}$ is finite which thus implies that
the process 1 terminates for this case. To close, we propose a conjecture that
this result is still valid if $\mathbb{F}$ is either the complex or the real
fields.

## Acknowledgements

This research is supported by the project B2024-SPD-07.

## References

  * [1] D. F. Anderson and A. Badawi, The total graph of a commutative ring, J. of Algebra, 320 (2008): 2706–2719. https://doi.org/10.1016/j.jalgebra.2008.06.028
  * [2] D. F. Anderson and P. S. Livingston, The zero-divisor graph of a commutative ring, J. of Algebra, 217 (1999): 434–447. https://doi.org/10.1006/jabr.1998.7840
  * [3] S. Akbari, M. Ghandehari, M. Hadian, A. Mohammadian, On commuting graphs of semisimple rings, Linear Algebra Appl. 390 (2004): 345–355. https://doi.org/10.1016/j.laa.2004.05.001
  * [4] S. Akbari, H. Bidkhori, and A. Mohammadian, Commuting Graphs of Matrix Algebras, Comm. Algebra, 36 (2008): 11, 4020–4031. https://doi.org/10.1080/00927870802174538
  * [5] S. Akbari and P. Raja, Commuting graphs of some subsets in simple rings, Linear Algebra Appl., 416 (2006): 1038–1047. https://doi.org/10.1016/j.laa.2006.01.006
  * [6] G. Chartrand and L. Lesniak, Graphs and Digraphs, Chapman and Hall, fourth ed., 2004.
  * [7] H. R. Dorbidi, The diameter of commuting graph of matrix ring of division rings, J. Algebra Appl. (2024), available online at https://doi.org/10.1142/S021949882550121X
  * [8] A. A. Kirillov, Elements of the Theory of Representaions, Springer - Verlag, Berlin - Heidenberg - New York, 1976.
  * [9] R. Lidl and H. Niederreiter, Finite Fields, 2nd edition, Cambridge University Press, 1996. https://doi.org/10.1017/CBO9780511525926
  * [10] C. Miguel, On the diameter of the commuting graph of the matrix ring over a centrally finite division ring, Linear Appl. 509 (2016): 276–285. https://doi.org/10.1016/j.laa.2016.08.001
  * [11] C. M. Nam, M. H. Bien, and B. X. Hai, On diameters of commuting graphs of matrix algebras over division rings, J. Algebra Appl. (2024), available online at https://doi.org/10.1142/S0219498825501488
  * [12] C. Schöbel, A classification of real finite-dimensional Lie algebras with a low-dimensional derived algebra, Rep. Math. Phys. 33 (1993): 175–186. https://doi.org/10.1016/0034-4877(93)90053-H
  * [13] L. Šnobl, and P. Winternitz, Classification and Identification of Lie algebras, American Mathematical Society, 2014.
  * [14] D. Wang, C. Xia, Diameters of the Commuting Graphs of Simple Lie Algebras, J. Lie Theory 27 (2017): 139–154.