On generalized derivations of polynomial vector fields Lie algebras

In this paper, we study the generalized derivation of a Lie sub-algebra of the Lie algebra of polynomial vector fields on $\mathbb{R}^n$ where $n\geq1$, containing all constant vector fields and the Euler vector field, under some conditions on this Lie sub-algebra.


Introduction
In [2], we have studied the derivations of a Lie subalgebra P of the Lie algebra of vector fields χ (R n ) on R n for n ≥ 1.The subalgebra P contains all constant vector fields and the Euler vector field.In [1], we explore that each m-derivation D of P with m ≥ 2 is an endomorphism of P such that ∀X 1 , X 2 , . . ., X m ∈ P, we have It's clear that every derivation (i.e 2-derivation) of P is a m-derivation where m ≥ 2 but the converse is not true in general.We say that m-derivations for (m ≥ 2) are a generalization of derivations.Now, we study another generalization of derivations of P, that are called as generalized derivations in [3].However, in [3], only study for the finite dimensional Lie algebra has made, but study to the generalized derivations of finite and infinite dimensional Lie algebras can be made on Lie algebra P. We prove this fact in the form of examples for finite and infinite dimensional P, in the later sections.Let us recall that a monomial vector field is a vector field on R n having one and only one monomial component in one and only one element of the canonical basis of χ (R n ).A Lie algebra of polynomial vector fields P is called separated if P is spanned by monomial vector fields.Moreover, a diagonal linear vector fields is an element of the Cartan subalgebra x 1 ∂ ∂x 1 , . . ., x n ∂ ∂x n R where (x i ) i=1,...,n is the usual coordinate system of R n .In this research we show that if P is separated and contains all diagonal linear vector fields with some additional condition, then there are some examples of further Lie algebras structures that admit generalized derivations that were not shown in [3] even for the finite dimensional Lie algebras.The results present in this paper are explicit and advance in the sense that we completely describe the centroid, the quasicentroid and the quasiderivations that arise in the study of generalized derivations of Lie algebras.We adapt our results in [1] and [2], to study generalized derivations of infinite dimensional Lie algebras.We do not only present abstract theorems but we also give examples to illustrate all theorems of this paper.
This paper consists of four sections after the present section for introduction.The second section is a general result from [2] on P about its graded algebra structure and its first space of cohomology of Chevalley-Eilenberg.The third section focuses on the study of centroid and quasicentroid of P and proves that these two sets are equal.The next section describes the quasiderivation of P using the fact that this quasiderivation is a sum of homogeneous degree i ∈ Z quasiderivations of P. The final section uses the results from the previous sections and presents the main theorem of this paper.According to which, generalized derivations of P are sum of derivations, quasicentroid elements, and two special endomorphisms of P. As stated earlier, it would be of great interest to explore generalized derivations of more general subalgebra P.
Throughout this paper all Lie algebras are equipped as algebras by the usual bracket notation [, ].

The Lie algebra P
Let us denote the Lie algebra of vector fields on R n with the coordinates system (x i ) i , by χ (R n ).Throughout this section, we consider a Lie algebra P of polynomial vector fields on R n .This Lie algebra P contains the constant fields ∂ ∂x i for all i and the Euler vector field and let H i be the vector space of homogeneous polynomial vector fields of degree Denote H d 0 the Lie algebra generated by all diagonal linear fields.A diagonal linear field is identified by the diagonal matrix with coefficients in R of order n.We adopt the Einstein convention on the summation index unless expressly mentioned.
is the Lie algebra of vector fields on R n , given by: The following lemma is clear: Lemma 2.1.A Lie algebra of polynomial vector fields P = i≥−1 P i is graded with the following [P i , P j ] ⊂ P i+j , where We recall two theorems of [2] on the centralizer and the cohomology space of P: Proposition 2.2.Let C (P) be the centralizer of P, Here C (P) is zero.
Definition 3.2.The quasicentroid QC (L) of a Lie algebra L is the Lie algebra of endomorphism for all X, Y ∈ L.
Note that in general C (L) ⊂ QC (L).In this section, we prove that C (L) = QC (L) for L = P. Proposition 3.3.Recall that E is the Euler vector field and let Y ∈ P m with m ≥ −1.Each element f of the centroid C (P) is homogeneous of degree 0.
Proof.If f ∈ C (P), then we have (6) for X = E and Y an homogeneous element in P m with m ≥ −1: f Then, we have mf (Y ) = i≥−1 mY i = i≥−1 iY i .In term of homogeneous components mY i = iY i and for m = i, it yields Y i = 0. Hence, f (Y ) = Y m and f is homogeneous of degree 0. So, we get f (E) = E 0 (E of degree 0, then by f is homogeneous of degree 0, f (E) = E 0 ).Moreover, in term of homogeneous components of (8), we have In addition, we have for a constant field If m = 0, by the previous result: Y homogeneous of degree 0, then So by the Jacobi identity and (12),we have It yields [E 0 , Y ] = 0 for m = 0.Moreover, and null otherwise on P 0 where Id We can refer on the previous Proposition with its notions saying that if f ∈ C (P), then [E 0 , Y ] = 0 for m = 0.If we run Y on the set of linear diagonal vector fields (that is to say Y takes all values of all elements in this set), we have 0 for all Y ∈ P 0 is not satisfied.If we have this last condition on P 0 , we can conclude following the decomposition of E 0 .
Proof.This follows directly from the previous corollary.
Remark 3.8.In general, C (P) is neither Id R nor Id, Id 1 , . . ., Id n R , for which we can take an example of Proposition 3.9.The element f of the quasicentroid QC (P) is such that Proof.Following the above notations and using the second part of the Eq. ( 8), We obtain

In term of homogeneous components
Because only E 0 is the only term of f (E), so Y m is the only term of f (Y ) and for m = 0, Using Jacobi identity and the last equality, we get (12) for f in the quasicentroid and it yields the last assertion of our theorem.Now, we can conclude for the centroid and the quasicentroid of P: Corollary 3.10.For all P, C (P) = QC (P).
4 Quasiderivation of P Definition 4.1.The set of quasiderivation QDer (L) of a Lie algebra L is the Lie algebra of endomorphism f in L such that there exists another endomorphism g of L such that It is more convenient to denote f ∈ QDer (L) by (f, f, g) and f ∈ QC (L) by (f, −f, 0).A quasiderivation of P is a generalization of a derivation of P. We compute QDer (P), it is clear that: where QDer i (P) is the set of quasiderivation of degree i. Lemma 4.1 of [3] can be adapted to our P taking account that it doesn't need to specify the dimension of P and P contains a torus E R .So Proposition 4.2.We have the following description of QDer (P): In the rest of this paper, we assume n ≥ 2.
Proof.Let us reason by induction knowing that f is homogeneous of degree 0. If X ∈ P −1 then f (X) = 0 by hypothesis, by ( 16) where Y = E, we have: For all X ∈ P 0 by ( 16) where From the previous results on g and f , the second term of the equality is null and then [f (X) , Y ] = 0 for all Y ∈ P −1 .Thus f (X) ∈ P −1 ∩ P 0 = {0} for all X ∈ P 0 .For X ∈ P 0 ∩ [P, P], we have two cases.
1.The first is X = [Y, Z] with (Y, Z) ∈ P 0 × P 0 where we deduce by definition 4.1 and the previous result that 0 It yields g (X ∈ P 0 [P, P]) is arbitrary.For all X ∈ P 1 by ( 16) where Y runs in P −1 : From g ([P −1 , P 1 ]) = 0, the second term of the above equality is null.Because of , then g(X) = 0 also.We set k ≥ 2, we suppose that f |Pt ≡ g |Pt ≡ 0 for all t ≥ k − 1.For X ∈ P k≥2 , we write (16) with Y ∈ P −1 .Then we obtain f (X) 16) with Y = E, we have 0 = −kf (X) = −kg(X) = 0 which yields g(X) = 0 by f (X) = 0. Proposition 4.4.Suppose that all elements of [P 1 , P −1 ] are also elements of [P 0 , P 0 ].The (f, f, g) ∈ QDer 0 (P) vanishing on E is of the following form: (L X , L X , L X + k), where X ∈ H 0 and k ∈ End (P, P) such that k is arbitrary on P 0 [P, P] and null elsewhere.
Proof.We prove this Proposition by induction.Let (f, f, g) ∈ QDer i≥1 (P).By (16) where X = E and Y runs in P −1 , we can say that g Then, in the turn of g, if X ∈ [P 0 , P 0 ], then Eq. (16) yields g(X) = 0.In addition, we have g But the degree of f is not -2, then f (X) = 0.It follows that the equation ( 16) with Y = E and this X permit us to say that g(X) = 0. Now, we suppose that f = g = 0 on P t≤l with l ≥ 1 and t ≥ 1.Consider X ∈ P l+1 , if we run Y ∈ P −1 on (16), we have [f (X) , If we write Y = E with this X on (16), we have −(l + 1)g(X) = [f (X), E] = −(i + l + 1)f (X) which is equal to 0 by the above result on f .Then, we get g(X) = 0 because l + 1 = 0.For (f, f, g) ∈ QDer −1 (P), it is obvious that f |P −1 = g |P −1 ≡ 0. The rest of the proof is the same as for (f, f, g) ∈ QDer i≥1 (P).
Proposition 4.6.Consider all hypothesis of Proposition 4.4.Every (f, f, g) in the set QDer i≥−1, =0,1 (P) is of the form , where f (E) i the homogeneous term f (E) of degree i and k is an endomorphism of P, arbitrary on P 0 [P, P] and null elsewhere.
Proof.Let (f, f, g) ∈ QDer i≥−1, =0,1 (P) be such quasiderivation, we have . Moreover, by using X = E and Y ∈ P 0 in Eq. ( 16), we have This yields 0 = [E, f (Y )] and f (Y ) = 0 because the degree of f (Y ) is nonzero.Then g |[P 0 ,P 0 ] = {0} from (16).Knowing that all elements of [P 1 , P −1 ] are elements of [P 0 , P 0 ], we have g [P 1 ,P −1 ] ≡ 0. Next, we will check whether f ∂x t where all P s is a polynomial of degree i, in which all P s doesn't depend on x s because of the hypothesis that each linear diagonal field doesn't belong to [P, P].When Y = x t ∂ ∂x t in Eq. ( 16), 0 = g ∂ ∂x t , x l ∂ ∂x l = f (X), x l ∂ ∂x l with a fixed l = t.
On generalized derivations of polynomial vector fields Lie algebras 59 Proposition 4.7.We suppose all hypothesis of Proposition 4.4.Every (f, f, g) in the set QDer 1 (P) is of the form where f (E) 1 is the homogeneous term of f (E) of degree 1, k is arbitrary on P 0 [P, P] and null elsewhere, f |P l =−1 is null and f (P −1 ) is a subset of x 1 ∂ ∂x 1 , . . ., x n ∂ ∂x n R .Proof.Let (f, f, g) ∈ QDer 1 (P) be such quasiderivation, we write for i = 1: We can assume all results of the first part of the previous proof ending on g |[P 1 ,P −1 ] ≡ 0 with i = 1.In this result, we have f |P 0 ≡ 0.
Let us take ), E] = 0 because f (X) of degree 0 and f (E) = 0. Let us precise X = ∂ ∂x j , ∂ ∂x j , x j ∂ ∂x j = X and by Eq. ( 16) with Thus, let f ∈ End (P) be such that f is null except on P −1 where the value is on Definition 4.8.For a fixed i 0 , j 1 < j 2 • • • < j p with 1 ≤ p ≤ n, α j 1 , . . ., α jp ∈ N p and j l ∈ {1, . . ., n}, the vector field (x j 1 ) α j 1 . . .(x jp ) α jp ∂ ∂x i 0 is called monomial.We recall that P is separated, if all homogeneous elements of degree k ≥ −1 are spanned by monomial vector field in P.
Remark 4.9.In the proof of the previous theorem, the following fact is true, x j ∂ ∂x j / ∈ [P, P] if we adopt the hypothesis on P: [P 1 , P −1 ] ⊂ [P 0 , P 0 ].In the same proof, we take X = ∂ ∂x j .It is supposed that P is separated.We can reason as the following, if there exists for all X ∈ P −(i+1) and g |(P 0 [P,P]) is arbitrary.
Next, by using result on odd-derivations in [1], we can prove that (f, f, g) ∈ QDer −2 (P) is an odd-derivations of P in the case Proposition 4.13.Let (f, f, g) ∈ QDer −2 (P).Then g |P 0 ∩[P,P] ≡ 0 and g |(P 0 [P,P]) is arbitrary.In addition, Proof.If f have degree −2, then f |P i≤0 ≡ 0. Thus, applying (16) with (X, Y ) ∈ P 1 × P −1 and (X, Y ) ∈ P 0 × P 0 resp., knowing that f is of degree −2, we obtain that g [X, Y ] = 0 but g |(P 0 [P,P]) is arbitrary.Now, let's take X ∈ P 1 and Y = E in (16), we deduce that g (X) = −f (X) for f of degree −2.Next, we consider X ∈ P 2 and Y = E in Eq. ( 16), we find g(X) = 0 with f (X) is determined by the same relation, if X ∈ P 2 and Y ∈ P −1 , we find that f [Y, X] = [f (X), Y ] taking account the value of f and g on P 1 .If X ∈ P k≥3 and Y runs in P −1 , by induction we have f (X) = 0 and g (X) = 0 by Eq. ( 16) when Y = E.
Let us recall the Theorem 3.10 of [1]: Theorem 4.14.We suppose that m ≥ 3 is odd.A linear map D of degree −2 on P, null on P P 1 is an m-derivation, if and only if: [D (P 1 ) , [P, P] where i is the first index with X i ∈ P 1 with the existence of 1 ≤ j < i such that X j ∈ P −1 ∪ P t≥2 , then Proposition 4.15.If (f, f, g) ∈ QDer −2 (P) and if for all X ∈ P 2 there exists X ∈ P −1 such that [X, X ] = 0.Moreover, suppose that P is separated and Then the endomorphism f is an odd-derivation of P, the endomorphism g = −f on P 1 and vanishes otherwise except on E R where g is arbitrary.
Proof.The f is an R-linear map of degree −2 on P with (f, f, g) ∈ QDer −2 (P).For all 16) and Proposition 4.13 saying f (E) = 0 and g(X ) = 0.It yields that there exists c ∈ R such that f (X ) = cE.In addition, for all (X , Y ) ∈ P 2 × P −1 , we have f [Y, X ] = [f X , Y ] using Proposition 4.13.By hypothesis, there is a Y ∈ P −1 such that [Y, X ] = 0. We can choose Y = ∂ ∂x j ∈ P −1 for a fixed j.It conducts to 0 = [f X , Y ] = −c ∂ ∂x j and c = 0. We can conclude that f |P 2 ≡ 0. Now, if we look at Theorem 4.14, we must check the 4 relations in that theorem for f .Let us take X ∈ P 1 , Y ∈ P i =0 , Z ∈ P 0 : 1 ] by Proposition 4.13 saying g |P 0 ∩[P,P] ≡ 0 and )] = 0 by Proposition 4.13 and the above result on 3.

Then this expression becomes
and then and by the Jacobi identity, we have This right term of the above equality becomes [[f X , X] , X ] because the degree of f is −2.Thus [[f X , X] , X ] = 0 by (1).So we conclude that [f (P 1 ) , [P, P] ∩ P 0 ] = {0}.
We conclude that f is a (2k + 1)-derivation on P like in Theorem 4.14.We have f = −g except on E R where g is arbitrary.
Proposition 4.17.If we have the hypothesis of Proposition 4.4 and if all linear diagonal fields are in P, then every elements of (f, f, g) ∈ QDer 0 (P) is a sum of an element of Id, Id 1 , . . ., Id n R and a quasiderivation of the type Proposition 4.4.
Proof.If we take such f , we can apply Eq.( 16) with X = E and Y a linear diagonal fields and find that [f (E) , Y ] = 0 for all Y .It conducts to the following facts, there exists Y a non-null linear diagonal field such that f (E) = Y .We take h ∈ Id, Id 1 , . . ., Id n R with Id, Id 1 , . . ., Id n R ⊃ QC (P) by corollary 3.4 such that h (E) = Y .This implies that (f − h, f + h, g) ∈ QDer 0 (P) and by Proposition 4.4, we have the result.By [3], It is clear that QDer (L) and GenDer (L) are two sets of Lie algebras and satisfy QDer (L) ⊂ GenDer (L).Moreover by assuming that a quasiderivation is a generalization of derivation and a generalized derivation is a generalization of quasiderivation, we have following result for a centerless Lie algebra L: ad L ⊆ Der (L) ⊆ QDer (L) ⊆ GenDer (L) ⊆ gl (L) .
As we know that the center of P is null by Proposition 2.2, it is interesting to find whether above inclusions verifies for P or not?.
For our convenience, we denote f ∈ GenDer (L) by (f, h, g), f ∈ QDer (L) by (f, f, g) and f ∈ QC (L) by (f, −f, 0).Study in [3], holds only for finite dimensional Lie algebras.But the following equation is always holds for our L = P.It further yields the following results.
Proposition 5.3.GenDer (P) = QDer (P) + QC (P).Proof.By corollary 3.4, Propositions 5.3, 4.2, 4.4, 4.6, 4.7, 4.12 and 4.17, we have the final result taking into account that X ∈ H 0 must be in the normalizer N of P, because L X is stable in P. By Theorem 2.4, N = P because all linear diagonal vector fields are in P which is also separated.Where G is the space of f in Proposition 4.7 and K is the space of quasiderivations of the form (0, 0, g), where g is arbitrary and g vanishes except on x ∂ ∂x , y ∂ ∂y R .
Example 5.6.In R 2 , we consider the habitual coordinate system (x, y).Where K is same as in the previous example.

1 0
we write the following (f, f, g) − (L X , L X , L X ) ∈ QDer 0 (P) with f (E) = L X (E) = 0.If we denote f = f − L X and g = g − L X , we have f (E) = 0, f = P −and g |[P 1 ,P −1 ] ≡ 0 by the first hypothesis in the statement of the present proposition.That is to say, if X 2 ∈ [P 1 , P −1 ], then there are (Y, Z) 0 and g |[P 1 ,P −1 ] ≡ 0. Then we can use Proposition 4.5 and have the following results: f = 0, g |P t =0 +(P 0 ∩[P,P]) ≡ 0 and g |(P 0 [P,P]) is arbitrary.It follows the final results.

Example 4. 10 .
We are in R 2 with the habitual coordinates (x, y),P = ∂ ∂x , ∂ ∂y , x ∂ ∂x , y ∂ ∂y R.Define an endomorphism f on P, such that f∂ ∂x = x ∂ ∂x , f ∂ ∂y= y ∂ ∂y and null otherwise.Then (f , f , 0) is an example of quasiderivation of type Proposition 4.7, which is neither a derivation nor an element of QC (P).

5
The main result on generalized Lie derivations of P and examples Definition 5.1.The set of all generalized derivations of a Lie algebra L is denoted by GenDer (L).It is the Lie algebra of endomorphisms f in L such that for every f there exists another two endomorphisms g, h in L such that[f (X) , Y ] + [X, h (Y )] = g [X, Y ](21)for all X, Y ∈ L. Definition 5.2.The center of a Lie algebra L is the set Z(L) = {X ∈ L | [X, L] = {0}}.

Example 5. 5 .
If we take the example of Example 4.10, we have GenDer (P) = Id, Id 1 , Id 2 R + ad P + G + K.
P, it is called inner derivation.ad P is the Lie algebra of inner derivations of P. The first Chevalley-Eilenberg cohomology space is denoted H 1 (P) = Der (P) /ad P .
Definition 2.3.We denote Der (P) the Lie algebra of derivations of P, L X = [X, .] is the Lie derivative with respect to X ∈ Theorem 5.4.If P verifies all hypotheses in the Proposition 4.12, then all generalized derivations of P are of the formh + (L X+Y , L X+Y , L X+Y + k) + (f , f , 0) with h ∈ QC (P) ⊂ Id, Id 1 , . .., Id n R if x 1 ∂ ∂x 1 , . .., x n ∂ ∂x n R ⊂ P 0 and h ∈ Id R if H 0 = P 0 .Where X, Y ∈ P, k ∈ End (P) is null except on P 0 [P, P] where k is arbitrary, f is a homogeneous map of degree 1, given in Proposition 4.7.f is null if P 0 contains a non diagonal linear vector field.
The Lie algebra P is the infinite dimensional vector space R, thenGenDer (P) = Id R + ad P + K.