Research topics in algebraic structures and graph theory have led to several exciting questions and results. Some papers have been written about assigning a graph to a group or ring and determining its algebraic properties. One can associate a graph to a group G by various ways. For example, zero-divisor graph of a commutative ring, non-commuting graph of a group and relative non-commuting graph of a group are defined (see [1, 2, 3]). One of the associated graphs to a group is the non-commuting graph, whose vertices are and two distinct vertices x and y are adjacent if and only if . Some results in group theory may be extended to Lie algebra. These conclusions are not always similar and even they are diferent in many situations. Recently, Lie algebras have attracted some authors due to wide applications in other sciences, especially physics. Understanding the structure of fundamental particles and the formulation of quantum theories in highenergy physics makes it inevitable for physicists to be familiar with Lie algebras.

In this paper, a graph is associated to a Lie algebra and the properties of this kind of graph are investigated. Let be a Lie algebra and be its center. We defined as follows: Take the vertices of and join two distinct vertices x and y whenever . We call the non-commuting graph of a Lie algebra L. All graphs are considered undirected and simple (with no loops or multiple edges). For every graph, we denote the set of vertices and the edges of by and , respectively. If is abelian, then is null graph. Our main goal is to detect the properties of a Lie algebra with its associated graph and vice verse.

In section 2, we study some properties of for a non-abelian Lie algebra. In addition, some results are obtained about connectivity, Hamiltonian, Eulerian properties, the girth and diameter of . In section 3, we determine under what conditions of c the graph is planar and outerplanar. In the last section, we consider isomorphism between two non-commuting graphs and show that under some conditions two Lie algebras have equal elements. In the rest of this section, we recall some concepts of graphs and Lie algebras that are used in the next sections.

Let be an arbitrary graph. The degree of a vertex denoted by deg(v), is the number of edges, which are incident to v. The minimum and maximum degrees are be denoted by and , respectively. Also, is called the size of . If the degree of all vertices is equal to , then is a complete graph. This graph with vertices is denoted by . A graph is regular if the degrees of all vertices of are the same. A path of length n consists of vertices from to and a sequence of distinct such that and A cycle is a path in a graph that starts and ends at the same vertex. A graph is connected when there is at least a path for every pair of vertices. If v and w are two vertices of , then is the length of the shortest path between v and w. The longest distance between all pairs of vertices is called the diameter of and denoted by diam(). The girth of , girth(), is the length of the shortest cycle in graph . In graph theory, a planar graph is a graph that can be drawn such that none of the edges cross each other. The following theorem is an important tool in proving some theorems related to the planar graphs in section 3.

Theorem 1.1. [4, Corollary 9.5.3] If is a simple and planar, then

If a graph can be drawn in the plane without any edge crossings, such that all vertices lie on the boundary of the outer face is called an outerplanar graph. Every outerplanar graph is planar, but the converse of fact is not true. For example, the complete graph with four vertices is the smallest planar graph that is not outerplanar.

Theorem 1.2. [5, Chaper 10] Every subgraph of an outerplanar contains a vertex with degree at most 2.

An Eulerian cycle is cycle that visits every edge of a graph exactly once. If graph has an Eulerian cycle, then it is called Eulerian. The next theorem states a condition equivalent to an Eulerian graph.

Theorem 1.3. [4, Theorem 4.1] A connected non-empty graph is Eulerian and only if it has no vertices of odd degree.

A Hamiltonian cycle is a cycle that visits each vertex exactly once. Every graph that has a Hamiltonian cycle is called Hamiltonian graph. The following theorem is an important equipment for detecting Hamiltonian graphs.

Theorem 1.4. (Dirac’s Theorem)[4], Theorem 4.3] Let be a simple graph of order n such that and the degree of every vertex is at least . Then is Hamiltonian.

Recall that a bipartite graph is a graph whose vertex set can be divided into two non-empty and disjoint subsets and V, every vertex in U is adjacent to every vertex in and two vertices within the same set are not adjacent. If and , then a bipartite graph is denoted . The next theorem is used to detect a planar graph.

Theorem 1.5. (Kuratowski’s Theorem)[4], Theorem 9.10] A graph is planar and only if it contains no subdivision of or

The neighbourhood of a vertex is the set of all vertices which are adjacent to x. A dominating set D for a graph is a subset of ) such that every vertices of is either or neighbourhood of . The domination number is the size of smallest dominating set of .

A Lie algebra is a vector space over the field F together with a bilinear map called the Lie bracket with , satisfying the following axioms:

The Lie bracket is often referred to as the commutator of x and y. The second condition is known as the Jacobi identity. Since the Lie bracket is bilinear, we have . Hence the first condition implies for all . For example, every vector space with for all is a Lie algebra. This is called an abelian Lie algebra. A subalgebra of is a subspace such that is close under the Lie bracket. In other words, for all . For instance, is the centralizer of element x in and a subalgebra of . An ideal of is a subalgebra I with this condition for all and . Let and are two ideals of a Lie algebra . It is obvious that is an ideal of . In particular, the derived subalgebra is the ideal generates by all such that . Another important ideal is the centre of a Lie algebra and defined as for all . If is a Lie algebra over the field such that dim and is a basis of then every element x of are of the form , where and the number of elements is equal to . All notations and terminologies are standard and reader can refer to [6, 4]. Moreover, from now on all Lie algebras are considered finite-dimensional over the field unless otherwise stated.

In this section, we repeat the definition of non-commuting graph of Lie algebra over the field and then investigate about connectivity, diameter, girth, Hamiltonian, Eulerian, and domination number of this graph. First, we state the definition of the new graph as the following.

Definition 2.1.Let be a Lie algebra and be the centre of . The noncommuting graph of denoted by , is an undirected simple graph whose vertices are all elements in and two distinct vertices x and y are adjacent if and only

Lemma 2.2. Let be a Lie algebra and be the non-commuting graph associated to L. Then for every ).

PROOF. Since and , we have

Hence for all

Lemma 2.3. Let be a Lie algebra. Then does not have any isolated vertex.

PROOF. On the contrary, let x be an isolated vertex of . Then . We know that , by Lemma 2.2. Therefore and so . It is a contradiction with

In the following proposition, we show that the non-commuting graph is connected.

Proposition 2.4.Let be a Lie algebra. Then is connected.

PROOF. Let x and y be two arbitrary distinct vertices of . If they are adjacent, then is connected. Suppose that x and y are not adjacent. Then . Since , there exist such that and , by Lemma 2.3. If or , then or . Therefore is connected. Now, let . Then and and is a path. Hence is connected.

The next proposition states the girth of non-commuting graph is

Proposition 2.5. Let be a Lie algebra. Then

PROOF. Assume that x is an arbitrary vertex of . Then there exists such that x and y are adjacent, by Lemma 2.3. Since , we have and and , x and y are adjacent. Hence there is a cycle of length 3 among the vertices and the girth of is equal to 3.

In the next proposition, we obtain an upper bound for the diameter of noncommuting graph

Proposition 2.6.Let be a Lie algebra. Then diam

PROOF. Assume that . We show that . If x and y are adjacent, then . Suppose that x and y are not adjacent. Thus there are such that and , by Lemma 2.3. Now, we have the following cases:

Case 1.x is adjacent to . In this case,

Case 2.y is adjacent to . By a similar way, one can see

Case 3.x and y are not adjacent to y1 and x1, respectively. So, Since and , we have and

Hence diam , by the previous cases.

In the following, we investigate under what condition diam . First of all, we need to prove the next theorem.

Theorem 2.7. Let be a Lie algebra over the field and be complete. Then and

PROOF. Since is complete, then and so for all . On the other hand, we know that , by Lemma 2.2. Therefore . As x is a vertex of and , we have . Also, is a subalgebra of thus dim dim . Hence for some and so attains the maximum size if . It implies that and then

Hence . On the contrary, let , then is a non-empty subset of . Also, for all and so none of the vertices in A is adjacent to x. Therefore , which is a contradiction. Hence .

Corollary 2.8.Let be a Lie algebra over the field such that or . Then diam

In the next lemma and propositions, we obtain some properties of the noncommuting graph

Lemma 2.9. Let be a Lie algebra. Then deg for all

PROOF. First, we show that has at least three vertices. Since is non-abelian, then dim . Hence and so . Also, we know that thus . On the other hand, | for all by Lemma 2.2. Moreover, x is a vertex of and so x . We know that when . This leads to the inequality dim dim . Hence for some The centralizer attains its maximal possible size when , yielding the sharpest bound . In addition, the fact that and ensures that .

Corollary 2.10.Let be a Lie algebra and be the non-commuting graph of . Then is neither tree nor star graph.

Proposition 2.11.Let be a Lie algebra. Then is Hamiltonian.

PROOF. We know that de for all , by Lemma 2.2. and , by the proof of Lemma 2.9. Since for all and , then

Therefore | V ( G _ { L } ) | / 2 \end{document} ]]> for all and so is Hamiltonian, by Theorem 1.4.

Proposition 2.12.Let be a Lie algebra. Then is an Eulerian graph.

PROOF. Suppose that x is an arbitrary vertex of . Since is a subalgebra of it follows that where m \end{document} ]]> , and the order of divides the order of . On the other hand d , by Lemma 2.2. Thus d . Hence if is either even or odd, then in both cases deg(x) is even. Therefore the degree of all vertices of is even. It implies that is Eulerian, by Theorem 1.3.

Proposition 2.13.Let be an n-dimensional Lie algebra over the field with the derived subalgebra of dimension 1. Then is -regular.

PROOF. We claim that dim for all . By Rank-Nullity Theorem, we have dim ker ad + dim Im , where is a linear transformation given by for every . Since ker and dim , then dim dim Im . On the other hand, Im and dim thus dim Im for all . Therefore dim for all . Hence for all and so is -regular.

Proposition 2.14.Let be a Lie algebra. Then is not complete bipartite.

PROOF. On the contrary, let be complete bipartite and the vertex set of be divided to two distinct parts A and B such that . Then

and so . Assume that such that . Then and we have . Therefore

Since , we may assume that for some Hence by (1). It implies that . Hence , which is a contradiction.

In the following proposition, we prove that if the domination number of noncommuting graph associated to a Lie algebra over the field is 1, then the center of Lie algebra L is zero and

Proposition 2.15. Let be a Lie algebra over the field and be a noncommuting graph. , then and

PROOF. Since , there exists such that Also, there is a vertex y such that , by Lemma 2.3. On the contrary, let . Then we can consider a non-zero element . On the other hand, thus , but . It is a contradiction when de . Therefore . Now, we prove . On the contrary, let . Since x is a vertex and , we have Also, none of the vertices in A is adjacent to x. So, and it is a contradiction. Hence

Note that one can prove Theorem 2.7 in another way by the above proposition. In the following example, we show that the converse of previous proposition is not true.

Example 2.16.Let is a Lie algebra over the field F . It is obvious and and . According to the representation of L and the vertices of we can see that every vertex is not adjacent to at least a vertex besides itself. Hence

Theorem 2.17. Let be a Lie algebra and be the non-commuting graph associated to . Then if and only if there is such that

PROOF. Let . Then there is such that On the other hand, , by Lemma 2.2. and , we have , by Proposition 2.15, hence . Conversely, let for some . Then . Hence x is adjacent to all vertex of besides itself and .

In this section, we classify all planar and outerplanar non-commuting graphs of a Lie algebra . First, we prove the following propositions that play an important role to determine planar graphs.

Lemma 3.1. There is no 3-dimensional Lie algebra over the field with the derived subalgebra of dimension 1 and

PROOF. On the contrary, let be a 3-dimensional Lie algebra over the field such that dim and . Assume that is a basis of . Then the set generates . Without loss of generality, let be a basis of . Then and for some . Since , then at least one of the coeficients α and is non-zero. If and , then we can check . It is a contradiction. Also, when and and so we have a contradiction. Finally, if , then . Again, it is a contradiction. Hence there is no such Lie algebra.

Proposition 3.2. Let be a 3-dimensional Lie algebra over the field with and dim . Then is isomorphic to the following graph:

PROOF. First, let the set be a basis of and expand it to of a basis of Then generates . Since is abelian by [6, Lemma we have . On the other hand, dim thus is another basis of . Hence the set is linearly independent. Also, we know that . So, one see that the brackets and are non-zero. Therefore x is adjacent to and . By a similar method, the adjacency of other vertices can be found. Hence is isomorphic to Figure 1.

Proposition 3.3. Let be a 3-dimensional Lie algebra over the with and dim . Then is isomorphic to the following graph:

PROOF. Let be a basis of . Since , so . Also, dim and generates , then is a basis of . Hence all elements of are adjacent and is isomorphic to Figure 2.

Proposition 3.4. Let be a 3-dimensional Lie algebra over the field with dim . Then dim and is isomorphic to the following graph:

PROOF. Suppose that thus we expand to a basis of . Thus generates . On the other hand, and so . It implies and so . Additionally, . Since and are non-zero, then x is adjacent to the vertices and . By a similar way, the adjacency of other vertices are obtained . Hence is isomorphic to Figure 3.

Theorem 3.5. Let be a 3-dimensional Lie algebra over the field . Then is isomorphic to one of the Figure 1, Figure 2 or Figure 3.

PROOF. Since is non-abelian, we have dim and dim . Also, dim thus 1 ≤ dim and dim . Now, the proof is completed by Lemma 3.1, Propositions 3.2, 3.3 and 3.4.

Theorem 3.6. Let be the non-commuting associated graph to a Lie algebra . Then is planar if and only if it is isomorphic to one of the following graphs:

PROOF. Suppose that is a planar graph thus there is such that , by Theorem 1.1. Since , by Lemma 2.2. and , then . On the other hand, ) for all is a descending function and thus . It is obvious that |L| is or . Consider the following cases:

Case 1. . Since is a non-abelian Lie algebra of dimension 2, we have dim and so | can be 1. Let be a basis of . Then and . Then and . Hence is isomorphic to Figure 4.

Case 2. . By a similar method of case 1, we can see that is 1 or 3. Let be a basis of L such that . Since is non-abelian, then and so . Then . So, the graph contains the subgraph as following:

Hence is not planar, by Theorem 1.5.

Case 3. . In this case, is isomorphic to Figure 1, Figure 2 or Figure 3, by theorem 3.5 if GL is isomorphic to figure 3 then we can draw the graph by another way and see that it is planar. So, is isomorphic to Figure 5. Suppose that is isomorphic to Figure 1. Then the graph has the following subgraph that is isomorphic to .

Hence is not planar, by Theorem 1.5. Again, is not planar, by Theorem 1.5 when it is isomorphic to Figure 2. Therefore is isomorphic to Figure 4 or Figure 5.

In the next theorem, we study the existence of non-commuting outerplanar graph associated to a Lie algebra .

Theorem 3.7.Let be a Lie algebra and be the non-commuting associated graph to . Then is outerplanar if and only is isomorphic to Figure 4.

PROOF. Let be an outerplanar graph. Then there is a vertex x such that , by Theorem 1.2. On the other hand, we know that and . It implies that for all Since for all is descending function, we have is or , or 3, then is abelian and it is a contradiction. Therefore . Now, is isomorphic to Figure 4 ,by a similar way in the case 1 of the proof of Theorem 3.6. The converse is obvious.

In this section, we are going to see that if for two Lie algebras and then what properties can be deduced from this isomorphism. In the following theorem, we show that if there exists a vertex of degree prime power number in and non-commuting graphs of two Lie algebras and are isomorphic, then their fields have the same characteristic.

Theorem 4.1. Let and be two finite-dimensional Lie algebras over fields and , respectively. Also, suppose that and there exists a vertex such that deg(x) is a prime power number. Then

PROOF. Since , we have and

by Lemma 2.2. Let by Lemma 2.2. Let , where m _ { 1 } \end{document} ]]> and m _ { 2 } \end{document} ]]>. Then . On the other hand, for some prime and thus and , which imply that and the proof follows.

Theorem 4.2.Let and be Lie algebras and . If there exists a vertex such that , where q is a prime number, then

PROOF. We know that and are the Lie algebras over the by the previous theorem. Since , we have and so . One can see that and . On the other hand, and imply that . In addition, , by the existence of isomorphism between 1 and and |. Therefore , as required.

Theorem 4.3.Let and be Lie algebras and . If there exists a vertex such that , where p and q are distinct prime numbers and q \end{document} ]]> , then

PROOF. By a similar way of the previous theorem, we have

and so there are three possibilities:

and . In this case, Thus the only possibility is when and . Therefore and . Then and so and . Since q . \end{document} ]]> , then it is impossible. pq and . Thus , which is not possible.

Hence the only possibility is case (i) when . Similar to the proof of previous theorem, , where is an isomorphism from to . Thus , which implies that . Since up to isomorphism there is a unique 2-dimensional non-abelian Lie algebra, then

Theorem 4.4.Let and be Lie algebras and . If there exists a vertex such that , where p and q are prime numbers and q \end{document} ]]> . Then

PROOF. We know that 1). Thus there are the following possibilities:

(i) . We have and so Since must be a prime power number, this is impossible.

(ii) . Then and again it is impossible.

(iii) . Since is not a prime power, this case does not occur. . We have and the only possibility is when and is an isomorphism between and , then and , by considering the above cases. Hence . Since, there is only a 2-dimensional non-abelian Lie algebra, by [6, Theorem 3.1], then

Corollary 4.5.Let and be Lie algebras and . If there is a vertex such that , where p and q are prime numbers, q \end{document} ]]> and , then

PROOF. The proof is similar to the above theorems.

Proposition 4.6.Let and be Lie algebras over the field such that . Then

PROOF. We know that when . Then . Suppose that and . Thus . We can assume that . So,

On the other hand, is odd and so is odd only if Hence and so . It implies that

It is a common question to ask what properties of Lie algebras are preserved under isomorphism between the non-commuting graphs of them. The next example shows that it is not true for the property of nilpotency.

Example 4.7. Assume that and We can see that the non-commuting graph associated to and is isomorphic to Figure 3. Also, is nilpotent, but does not have such property. So, the nilpotency property is not kept under isomorphism of the non-commuting graphs.

It is interesting to see that under what conditions we will have if and only if . We left it as a conjecture here.

Conjecture. Find conditions such that if and only if