Pure and Applied Mathematics Journal
Volume 5, Issue 1, February 2016, Pages: 23-31

Derivations of First Type of Algebra of Second Class Filiform Leibniz Algebras of Dimension Derivation (n+2)

AL-Nashri AL-Hossain Ahmad

Department of Mathematics, AL Qunfudha University College, Umm AL Qura University, Makkah, Saudia Arabia

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AL-Nashri AL-hossain Ahmad. Derivations of First Type of Algebra of Second Class Filiform Leibniz Algebras of Dimension Derivation (n+2). Pure and Applied Mathematics Journal. Vol. 5, No. 1, 2016, pp. 23-31. doi: 10.11648/j.pamj.20160501.14


Abstract: This paper describes the derivations of first type of algebra from the second class filiform Leibniz algebras of dimension derivation (n+2). The set of all derivations of an algebra L is denoted by Der(L) From the description of the derivations, we found the basis of the space Der(Ln(a)) of the algebra.

Keywords: Filiform Leibniz Algebra, Leibniz Algebra, Gradation, Natural Gradation, Derivation


1. Introduction and Preliminaries

In mathematics and in particular in the theory of Lie algebra, a Leibniz algebra (which was first introduced by L. Loday in 1993, [5]) is an algebra L over a field K satisfying the following Leibniz identity

for any , where [.,.] denotes the multiplication in L. Leibniz algebra is a generalization of Lie algebra. Onwards, all algebras are assumed to be over the field of complex numbers C. Now, let L be a Leibniz algebra, we put:  for .

Following [7], a Leibniz algebra  is said to be nilpotent if there exists  such that . A Leibniz algebra  is said to be filiform if  where and , see [1]. Let d be a K-linear transformation of an algebra L, d is called a derivation of L ([6]) if

The set of all derivations of an algebra  is denoted by  We also, denote by Leibniz the set of all (n+1)-dimensional filiform Leibniz algebras. We now look at the following theorem from [3] which splits the set of fixed dimension filiform Liebniz algebras into three disjoint subsets. However we just take the result of this theorem regarding only .

Theorem 1.1 Any -dimensional complex filiform Leibniz algebra  admits a basis  called adapted, such that the table of multiplication of  has the following forms, where non defined products are zero:

for .

Lemma 1.1 [6] Let . In this case  where  and  for

The purpose of this paper is to study the low dimension of algebras in order to get the basis of the space . We attempt to find the basis of the derivation for this algebra and the relationship between the algebra and its derivations, by studying the table of this algebra from dimension 5 to 15.

2. Algebra  of the Second Class Filiform Leibniz Algebras

We observe the derivations of this type of algebra in low dimension in the following table:

Table 1. Derivations of  with low dimension.

dimension equation(dim Der) dim Der No. of equations
5

6 14
6

7 19

8 25
8

9 32
9

10 40
11

12 59
12

13 70
13

14 82
14

15 95
15

16 109

Notes:

From Table 1, we obtain the main result of this paper: We are able to find a basis of the space  and this is given in the proposition 2.1. Therefore we can find the dim Der of this algebra for any dimension by using this rule:  for . Finally, we observe that the number of equations that can be calculated for any dimension obtained from derivations the following rule: = .

Let  By using Lemma , we have

In Lemma , we give definition for

In Lemma , we give definition for

In Lemma , we give definition for

We are going to find the basis of algebra .

For this purpose, we present the following Lemma ( 2.1, 2.2 and 2.3 ).

Lemma 2.1 Let . Then, we have

Proof. Consider  which is defined by

(1)

where , , are scalars. Consider the family of derivations

We now look at the problem case by case. In each case, we repeatedly use algebra  and

Case 1: if  and , then

and so,

which implies

and thus

(2)

Case 2: if  then

so that,

which implies

and so,

(3)

If i=2 in (3), then . By (2) we obtain

(4)

If i=3 in (3) then . By (2) and (4), we obtain

(5)

Similarly, if i=n-1 in (3) then

But . Thus,

(6)

From (2), (4), (5) and (6) we obtain

(7)

Thus, using (1) and (7), we get

From this we obtain

(8)

Lemma 2.1 gives some vectors of the basis. In the next we will complete this basis.

Lemma 2.2 Consider  such that  is defined as

(9)

where  and  are constants.

Then  and  for  and .

Proof. Consider the family of derivations

Similarly, we calculate case by case. In which we repeatedly  and .

Case 1: if , then

Then

which implies

to obtain

(10)

Case 2: if  and , then

Thus,

(11)

From (11), if , then . Also, if  then . Similarly, if , then  From (10) and (11) this implies

(12)

and hence, by (9) and (7), we get

Thus,

(13)

This conclude the proof.

Lemma 2.2 gives a second part of the basis of Algebra. Up to now, we obtained two parts of the basis. This imcomplete basis will be completed by other vectors will be given Lemma 2.3

Lemma 2.3 Let . Then

Proof. Consider  where  is defined by

(14)

where  is a constant. Consider the family of derivations

Case 1: if  then

by  and (13), thus

If  then

Hence, using (14), we get

and we obtain

(15)

This conclude the proof of Lemma 2.3.

In Lemma 2.4 Fullfilment to conclude our study we will now turn to show that the set of vectors given in Lemma 2.1, Lemma 2.2 and Lemma2.3 form a basis of Algebra.

For that, we will show that the set vectors are linearly independent.

Lemma 2.4. The mappings  and  for  are linearly independent.

Proof. Consider that

(16)

where  We will show that  for

This implies

We thus have

Here we have these following results:

1.  which implies .

2.  which implies , but since  then .

3.  which implies , but since  then .

4.  which implies  but since , then .

5.  which implies , but since  then .

6. Similarly,  which implies  but since . Then, we get .

From of the above we will obtain

(17)

This conclude our proof and the mappings are linearly independent.

Lemma 2.5 The linear mappings  defined by (1), (9) and (14) are linearly composition.

Proof. Let

First we observe that by (1), (9) and (14),

(18)

Hence by using  and (12 ),

Thus

The linear composition of the mappings is proved.

Lemma 2.6 The mappings  for  are derivations.

Proof. Consider that

and

Then

)

and so,

(19)

Thus,

Hence

(20)

and thus,

Therefore,

(21)

By adding  to  we will obtain . This implies  is a derivation.

We now show that  is also a derivation.

and

From easy calculation we have ,  and  and thus  is a derivation.

Now, since

thus,

then,

(22)

In addition,

and we have

(23)

Also,

(24)

By adding  to  we will get , thus  is a derivation.

This complete the proof of the derivation.

We recall that the  and  for  are defined in Lemma ,…, Lemma .

We will conclude our discussion with the following Proposition.

Proposition 2.1 Let  be  and . Then  and  for  form a basis of the space Der().

Proof.: The proof follows from Lemma  to Lemma

3. Conclusion

Finally, the purpose of this manuscript is fourfold:

1. This algebra , is nilpotent, but it is not characteristically nilpotent.

2. This algebra work with basis derivations from five dimension and above.

3. We can find number derivations of this algebra on any dimension by this rule:

4. We can determine the number of equations from the result of derivations by this rule, i.e, the number of equations is equal to .


References

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