Derivations of First Type of Algebra of Second Class Filiform Leibniz Algebras of Dimension Derivation (n+2)
ALNashri ALHossain Ahmad
Department of Mathematics, AL Qunfudha University College, Umm AL Qura University, Makkah, Saudia Arabia
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To cite this article:
ALNashri ALhossain 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. 2331. 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(L_{n}(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 Klinear 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:
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=n1 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