Lie-central derivations, Lie-centroids and Lie-stem Leibniz algebras
aa r X i v : . [ m a t h . R A ] J u l Lie -central derivations,
Lie -centroids and
Lie -stem Leibniz algebrasG. R. Biyogmam , J. M. Casas and N. Pacheco Rego Department of Mathematics, Georgia College & State UniversityCampus Box 17 Milledgeville, GA 31061-0490E-mail address: [email protected] Dpto. Matem´atica Aplicada I, Universidade de Vigo, E. E. ForestalCampus Universitario A Xunqueira, 36005 Pontevedra, SpainE-mail address: [email protected] IPCA, Dpto. de Ciˆencias, Campus do IPCA, Lugar do Ald˜ao4750-810 Vila Frescainha, S. Martinho, Barcelos, PortugalE-mail address: [email protected]
Abstract:
In this paper, we introduce the notion
Lie -derivation. This con-cept generalizes derivations for non-
Lie
Leibniz algebras. We study these
Lie -derivations in the case where their image is contained in the
Lie -center, call them
Lie -central derivations. We provide a characterization of
Lie -stem Leibniz al-gebras by their
Lie -central derivations, and prove several properties of the Liealgebra of
Lie -central derivations for
Lie -nilpotent Leibniz algebras of class 2. Wealso introduce ID ∗ - Lie -derivations. A ID ∗ - Lie -derivation of a Leibniz algebra g is a Lie -derivation of g in which the image is contained in the second term of the lower Lie -central series of g , and that vanishes on Lie -central elements. We provide anupperbound for the dimension of the Lie algebra ID Lie ∗ ( g ) of ID ∗ - Lie -derivationof g , and prove that the sets ID Lie ∗ ( g ) and ID Lie ∗ ( q ) are isomorphic for any two Lie -isoclinic Leibniz algebras g and q . Key words:
Lie -derivation;
Lie -center;
Lie -stem Leibniz algebra;
Lie -centralderivation;
Lie -centroid; almost inner
Lie -derivation.
Studies such as the work of Dixmier [13], Leger [16] and Tˆogˆo [20, 21, 22, 23] aboutthe structure of a Lie algebra L and its relationship with the properties of the Liealgebra of derivations of L have been conducted by several authors. A classicalproblem concerning the algebra of derivations consists to determine necessaryand sufficient conditions under which subalgebras of the algebra of derivations1oincide. For example, the coincidence of the subalgebra of central derivationswith the algebra of derivations of a Lie algebra was studied in [21]. Also cen-troids play important roles in the study of extended affine Lie algebras [2], in theinvestigations of the Brauer groups and division algebras, in the classification ofalgebras or in the structure theory of algebras. Almost inner derivations arise inmany contexts of algebra, number theory or geometry, for instance they play animportant role in the study of isospectral deformations of compact solvmanifolds[15]; the paper [6] is dedicated to study almost inner derivations of Lie algebras.Our aim in this paper is to conduct an analogue study by investigating variousconcepts of derivations on Leibniz algebras. Our study relies on the relativenotions of these derivations; derivations relative to the Liezation functor ( − ) Lie : Leib → Lie , which assigns to a Leibniz algebra g the Lie algebra g Lie , where
Leib denotes the category of Leibniz algebras and
Lie denotes the category of Liealgebras.The approached properties are closely related to the relative notions of centralextension in a semi-abelian category with respect to a Birkhoff subcategory [11,14]. A recent research line deals with the development of absolute propertiesof Leibniz algebras (absolute are the usual properties and it means relative tothe abelianization functor) in the relative setting (with respect to the Liezationfunctor); in general, absolute properties have the corresponding relative ones,but not all absolute properties immediately hold in the relative case, so newrequirements are needed as it can be seen in the following papers [3, 4, 5, 8, 10, 19].In order to develop a systematic study of derivation in the relative setting,we organize the paper as follows: in Section 2, we provide some backgroundon relative notions with respect to the Liezation functor. We define the sets of
Lie -derivations
Der
Lie ( g ) and central Lie -derivations
Der
Lie z ( g ) for a non-Lie Leib-niz algebra g . It is worth mentioning that the absolute derivations are also
Lie -derivations. In Section 3, we characterize
Lie -stem Leibniz algebras using their
Lie -central derivations. Using
Lie -isoclinism, we prove several results on the Liealgebra of
Lie -central derivations of
Lie -nilpotent Leibniz algebras of class two.In concrete, we prove that
Der
Lie z ( g ) is abelian if and only if Z Lie ( g ) = γ Lie ( g ),under the assumption that g is a finite-dimensional Lie -nilpotent Leibniz algebraof class 2. In Section 4, we define the
Lie -centroid Γ
Lie ( g ) of g and prove several ofits basic properties. In particular we study its relationship with the Lie -algebra
Der
Lie z ( g ) . In Section 5, we study the set ID ∗ ( g ) of ID ∗ - Lie -derivations of a Leib-niz algebra g and its subalgebra Der
Lie c ( g ) of almost inner Lie -derivations of g . Similarly to the result of Tˆogˆo [22] on derivations of Lie algebras, we providenecessary and sufficient conditions on a finite dimensional Leibniz algebra g forthe subalgebras Der
Lie z ( g ) and ID ∗ ( g ) to be equal. We also prove that if two Leib-niz algebras are Lie -isoclinic, then their sets of ID ∗ - Lie -derivations are isomorphic.This isomorphism also holds for their sets of almost inner
Lie -derivations. We es-tablish several results on almost inner
Lie -derivations, similar to the Lie algebracase [6]. Finally, we provide an upperbound of the dimension of ID ∗ ( g ) by means2f the dimension of [ g , g ] Lie . Let K be a fix ground field such that ∈ K . Throughout the paper, all vectorspaces and tensor products are considered over K .A Leibniz algebra [17, 18] is a vector space g equipped with a bilinear map[ − , − ] : g ⊗ g → g , usually called the Leibniz bracket of g , satisfying the Leibnizidentity : [ x, [ y, z ]] = [[ x, y ] , z ] − [[ x, z ] , y ] , x, y, z ∈ g . A subalgebra h of a Leibniz algebra g is said to be left (resp. right) ideal of g if [ h, g ] ∈ h (resp. [ g, h ] ∈ h ), for all h ∈ h , g ∈ g . If h is both left and right ideal,then h is called two-sided ideal of g . In this case g / h naturally inherits a Leibnizalgebra structure.Given a Leibniz algebra g , we denote by g ann the subspace of g spanned byall elements of the form [ x, x ], x ∈ g . It is clear that the quotient g Lie = g / g ann is a Lie algebra. This defines the so-called Liezation functor ( − ) Lie : Leib → Lie ,which assigns to a Leibniz algebra g the Lie algebra g Lie . Moreover, the canonicalepimorphism g ։ g Lie is universal among all homomorphisms from g to a Liealgebra, implying that the Liezation functor is left adjoint to the inclusion functor Lie ֒ → Leib .Given a Leibniz algebra g , we define the bracket[ − , − ] lie : g → g , by [ x, y ] lie = [ x, y ] + [ y, x ] , for x, y ∈ g . Let m , n be two-sided ideals of a Leibniz algebra g . The following notionscome from [10], which were derived from [11].The Lie -commutator of m and n is the two-sided ideal of g [ m , n ] Lie = h{ [ m, n ] lie , m ∈ m , n ∈ n }i . The
Lie -center of the Leibniz algebra g is the two-sided ideal Z Lie ( g ) = { z ∈ g | [ g, z ] lie = 0 for all g ∈ g } . The
Lie - centralizer of m and n over g is C Lie g ( m , n ) = { g ∈ g | [ g, m ] lie ∈ n , for all m ∈ m } . Obviously, C Lie g ( g ,
0) = Z Lie ( g ).The right-center of a Leibniz algebra g is the two-sided ideal Z r ( g ) = { x ∈ g | [ y, x ] = 0 for all y ∈ g } . The left-center of a Leibniz algebra g is the set Z l ( g ) = { x ∈ g | [ x, y ] = 0 for all y ∈ g } , which might not even be a subalgebra. Z ( g ) = Z l ( g ) ∩ Z r ( g ) is called the center of g , which is a two-sided ideal of g .3 efinition 2.1 [10] Let n be a two-sided ideal of a Leibniz algebra g . The lower Lie -central series of g relative to n is the sequence · · · E γ Lie i ( g , n ) E · · · E γ Lie ( g , n ) E γ Lie ( g , n ) of two-sided ideals of g defined inductively by γ Lie ( g , n ) = n and γ Lie i ( g , n ) = [ γ Lie i − ( g , n ) , g ] Lie , i ≥ . We use the notation γ Lie i ( g ) instead of γ Lie i ( g , g ) , ≤ i ≤ n .If ϕ : g → q is a homomorphism of Leibniz such that ϕ ( m ) ⊆ n , where m is a two-sided ideal of g and n a two-sided ideal of q , then ϕ ( γ Lie i ( g , m )) ⊆ γ Lie i ( q , n ) , i ≥ Definition 2.2
The Leibniz algebra g is said to be Lie -nilpotent relative to n ofclass c if γ Lie c +1 ( g , n ) = 0 and γ Lie c ( g , n ) = 0 . Definition 2.3 [10] The upper
Lie -central series of a Leibniz algebra g is thesequence of two-sided ideals, called i - Lie centers, i = 0 , , , . . . , Z Lie ( g ) E Z Lie ( g ) E · · · E Z Lie i ( g ) E · · · defined inductively by Z Lie ( g ) = 0 and Z Lie i ( g ) = C Lie g ( g , Z Lie i − ( g )) , i ≥ . Theorem 2.4 [10, Theorem 4] A Leibniz algebra g is Lie -nilpotent of class c ifand only if Z Lie c ( g ) = g and Z Lie c − ( g ) = g . Definition 2.5 [8, Definition 2.8] Let m be a subset of a Leibniz algebra g . The Lie -normalizer of m is the subset of g : N g ( m ) = { g ∈ g | [ g, m ] , [ m, g ] ∈ m , for all m ∈ m } Definition 2.6 [10, Proposition 1] An exact sequence of Leibniz algebras → n → g π → q → is said to be Lie -central extension if [ g , n ] Lie = 0 , equivalently n ⊆ Z Lie ( g ) . Definition 2.7
A linear map d : g → g of a Leibniz algebra ( g , [ − , − ]) is said tobe a Lie -derivation if for all x, y ∈ g , the following condition holds: d ([ x, y ] lie ) = [ d ( x ) , y ] lie + [ x, d ( y )] lie We denote by
Der
Lie ( g ) the set of all Lie -derivations of a Leibniz algebra g ,which can be equipped with a structure of Lie algebra by means of the usualbracket [ d , d ] = d ◦ d − d ◦ d , for all d , d ∈ Der ( g ).4 xample 2.8 The absolute derivations, that is linear maps d : g → g such that d ([ x, y ]) = [ d ( x ) , y ] + [ x, d ( y )] , are also Lie -derivations, since: d ([ x, y ] lie ) = d ([ x, y ] + [ y, x ]) = [ d ( x ) , y ] lie + [ x, d ( y )] lie , for all x, y ∈ g . (1) In particular, for a fixed x ∈ g , the inner derivation R x : g → g , R x ( y ) = [ y, x ] ,for all y ∈ g , is a Lie -derivation, so it gives rise to the following identity: [[ y, z ] lie , x ] = [[ y, x ] , z ] lie + [ y, [ z, x ]] lie , for all x, y ∈ g . However there are
Lie -derivations which are not derivations. For instance,every linear map d : g → g is a Lie -derivation for any Lie algebra g , but it is nota derivation in general. Definition 2.9 A Lie -derivation d : g → g of a Leibniz algebra g is said to be Lie -central derivation if its image is contained in the
Lie -center of g . Remark 2.10
The absolute notion corresponding to Definition 2.9 is the socalled central derivations, that is derivations d : g → g such that its image iscontained in the center of g . Obviously, every central derivation is a Lie -centralderivation. However the converse is not true as the following example shows:let g be the two-dimensional Leibniz algebra with basis { e, f } and bracket opera-tion given by [ e, f ] = − [ f, e ] = e [12]. The inner derivation R e is a Lie -centralderivation, but it is not central in general.
We denote the set of all
Lie -central derivations of a Leibniz algebra g by Der
Lie z ( g ). Obviously Der
Lie z ( g ) is a subalgebra of Der
Lie ( g ) and every element of Der
Lie z ( g ) annihilates γ Lie ( g ) = [ g , g ] Lie . Der
Lie z ( g ) = C Der
Lie ( g ) (( R + L )( g )), where L ( g ) = { L x | x ∈ g } , L x denotes the left multiplication operator L x ( y ) = [ x, y ], R ( g ) = { R x | x ∈ g } and C g ( m ) = { x ∈ g | [ x, y ] = 0 = [ y, x ] , for all y ∈ m } , theabsolute centralizer of an ideal m over the Leibniz algebra g .Let A and B be two Leibniz algebras and T ( A, B ) denotes the set of alllinear transformations from A to B . Clearly, T ( A, B ) endowed with the bracket[ f, g ]( x ) = [ f ( x ) , g ( x )] is an abelian Leibniz algebra if B is an abelian Leibnizalgebra too.Consider the Lie -central extensions ( g ) : 0 → n χ → g π → q → g i ) : 0 → n i χ i → g i π i → q i → , i = 1 , . Let be C : q × q → [ g , g ] Lie given by C ( q , q ) = [ g , g ] lie , where π ( g j ) = q j , j = 1 ,
2, the
Lie -commutator map associated to the extension ( g ). In a similarway are defined the Lie -commutator maps C i corresponding to the extensions( g i ) , i = 1 , q is a Lie algebra, then π ([ g , g ] Lie ) = 0, hence [ g , g ] Lie ⊆ n ≡ χ ( n ).5 efinition 2.11 [3, Definition 3.1] The Lie -central extensions ( g ) and ( g ) aresaid to be Lie -isoclinic when there exist isomorphisms η : q → q and ξ :[ g , g ] Lie → [ g , g ] Lie such that the following diagram is commutative: q × q C , η × η (cid:12) (cid:18) [ g , g ] Lie ξ (cid:12) (cid:18) q × q C , [ g , g ] Lie (2)
The pair ( η, ξ ) is called a Lie -isoclinism from ( g ) to ( g ) and will be denotedby ( η, ξ ) : ( g ) → ( g ) . Let g be a Leibniz algebra, then we can construct the following Lie -centralextension ( e g ) : 0 → Z Lie ( g ) → g pr g → g /Z Lie ( g ) → . (3) Definition 2.12 [3, Definition 3.3] Let g and q be Leibniz algebras. Then g and q are said to be Lie -isoclinic when ( e g ) and ( e q ) are Lie -isoclinic
Lie -centralextensions.A
Lie -isoclinism ( η, ξ ) from ( e g ) to ( e q ) is also called a Lie -isoclinism from g to q , denoted by ( η, ξ ) : g ∼ q . Proposition 2.13 [3, Proposition 3.4] For a
Lie -isoclinism ( η, ξ ) : ( g ) ∼ ( g ) ,the following statements hold:a) η induces an isomorphism η ′ : g /Z Lie ( g ) → g /Z Lie ( g ) , and ( η ′ , ξ ) is a Lie -isoclinism from g to g .b) χ ( n ) = Z Lie ( g ) if and only if χ ( n ) = Z Lie ( g ) . Definition 2.14 [19, Definition 4] A
Lie -stem Leibniz algebra is a Leibniz algebra g such that Z Lie ( g ) ⊆ [ g , g ] Lie . Theorems 1 and 2 in [19] prove that every
Lie -isoclinic family of Leibniz algebrascontains at least one
Lie -stem Leibniz algebra, which is of minimal dimension ifit has finite dimension. Lie -stem Leibniz algebras and
Lie -central deriva-tions
Proposition 3.1 If g is a Lie -stem Leibniz algebra, then
Der
Lie z ( g ) is an abelianLie algebra. roof. Since
Der
Lie z ( g ) is a subalgebra of Der
Lie ( g ) , it is enough to show that[ d , d ] = 0 for all d , d ∈ Der
Lie z ( g ) . First, we notice that if d ∈ Der
Lie z ( g ) , then d ([ x, y ] lie ) = 0 for all x, y ∈ g since d ( x ) , d ( y ) ∈ Z Lie ( g ) . So in particular, d ( Z Lie ( g )) = 0 since Z Lie ( g ) ⊆ [ g , g ] Lie as g is a Lie -stem Leibniz algebra. Nowlet d , d ∈ Der
Lie z ( g ) and x ∈ g . Then d ( x ) , d ( x ) ∈ Z Lie ( g ) , which implies that[ d , d ]( x ) = d ( d ( x )) − d ( d ( x )) = 0 . Hence [ d , d ] = 0 . The converse of the above result is not true in general. Indeed, let g be anyLie algebra. Then Z Lie ( g ) = g and so Der
Lie z ( g ) is an abelian Lie algebra. However g is not a Lie -stem Leibniz algebra since Z Lie ( g ) = g g , g ] Lie . Proposition 3.2
Let g be a Lie -nilpotent finite dimensional Leibniz algebra suchthat γ Lie ( g ) = 0 . Then Der
Lie z ( g ) is abelian if and only if g is a Lie -stem Leibnizalgebra.
Proof.
We only need to prove the converse of Proposition 3.1. Assume that g is not a Lie -stem Leibniz algebra. Then, there is some z ∈ Z Lie ( g ) such that z / ∈ [ g , g ] Lie . Since g is a Lie -nilpotent Leibniz algebra and γ Lie ( g ) = 0, it followsthat Z Lie ( g ) ∩ [ g , g ] Lie = 0 . Let z ∈ Z Lie ( g ) ∩ [ g , g ] Lie , z = 0 , and consider thefollowing maps: d : g → g , d ( z ) = ( z if z = z z = z and d : g → g , d ( z ) = ( z if z = z z = z . Clearly, d and d are Lie -central derivations, and d and d do not commute,since [ d , d ]( z ) = d ( d ( z )) − d ( d ( z )) = − z = 0 . Therefore
Der
Lie z ( g ) is notabelian. This completes the proof. Lemma 3.3
Let ( η, ξ ) be a Lie -isoclinism between the Leibniz algebras g and q .If g is a Lie -stem Leibniz algebra, then ξ maps Z Lie ( g ) onto Z Lie ( q ) ∩ [ q , q ] Lie . Proof.
Since Z Lie ( g ) ⊆ [ g , g ] Lie , then an element z of Z Lie ( g ) can be written as z = n P i =1 λ i [ x i , y i ] lie , with λ i ∈ K and x i , y i ∈ g , i = 1 , . . . , n .Let η ′ : g /Z Lie ( g ) −→ q /Z Lie ( q ), η ′ ( x i + Z Lie ( g )) = η ( x i ) + Z Lie ( q ) and η ′ ( y i + Z Lie ( g )) = η ( y i ) + Z Lie ( q ) , i = 1 , . . . , n, the isomorphism provided by [3, Proposi-7ion 3.4]. Then ξ ( z ) + Z Lie ( q ) = ξ (cid:18) n P i =1 λ i [ x i , y i ] lie (cid:19) + Z Lie ( q )= n P i =1 λ i ξ [ x i , y i ] lie + Z Lie ( q )= n P i =1 λ i [ η ( x i ) , η ( y i )] lie + Z Lie ( q )= η ′ (cid:18) n P i =1 λ i [ x i , y i ] lie + Z Lie ( g ) (cid:19) = Z Lie ( q ) . The surjective character can be easily established.
Proposition 3.4
Let g and q be two Lie -isoclinic Leibniz algebras and g be a Lie -stem Leibniz algebra. Then every d ∈ Der
Lie z ( g ) induces a Lie -central derivation d ∗ of q . Moreover, the map d d ∗ is a monomorphism from Der
Lie z ( g ) into Der
Lie z ( q ) . Proof.
Let ( η, ξ ) be a
Lie -isoclinism between g and q , and let d ∈ Der
Lie z ( g ) . Thenfor any y ∈ q , we have y + Z Lie ( q ) = η ( x + Z Lie ( g )) for some x ∈ g , since η isbijective. Now consider the map d ∗ : q → q defined by d ∗ ( y ) = ξ ( d ( x )) , whichis well-defined since d ( Z Lie ( g )) = 0 as Z Lie ( g ) ⊆ [ g , g ] Lie . Moreover, d ∗ ∈ Der
Lie z ( q )since d ( x ) ∈ Z Lie ( g ) and ξ ( d ( x )) ∈ Z Lie ( q ) ∩ [ q , q ] Lie by Lemma 3.3. d ∗ is a Lie -derivation since d ∗ ([ y , y ] lie ) = ξ ( d ([ x , x ] lie )) = ξ ([ d ( x ) , x ] lie + [ x , d ( x )] lie ) = ξ (0 + 0) = 0 and [ y , d ∗ ( y )] lie + [ d ∗ ( y ) , y ] lie = 0 since d ∗ ( y ) , d ∗ ( y ) ∈ Z Lie ( q ) . Clearly, the map φ : d → d ∗ is linear and one-to-one since ξ an isomorphism.To show that φ is compatible with the Lie-bracket, let d , d ∈ Der
Lie z ( g ) . Thenfor i, j = 1 , , d i ( g ) ⊆ Z Lie ( g ) ⊆ [ g , g ] Lie and d j ([ g , g ] Lie ) = 0 . So on one hand[ d , d ] = d d − d d = 0 and thus [ d , d ] ∗ = 0 as ξ is an isomorphism. Onthe other hand, d ∗ i ( q ) ⊆ Z Lie ( q ) ∩ [ q , q ] Lie . So d ∗ j ( d ∗ i ( q )) = 0 , by definition of d ∗ j , and thus [ d ∗ , d ∗ ] = d ∗ d ∗ − d ∗ d ∗ = 0 . Therefore φ ([ d , d ]) = [ φ ( d ) , φ ( d )] . Thiscompletes the proof.
Lemma 3.5
For any
Lie -stem Leibniz algebra g , there is a Lie algebra isomor-phism Der
Lie z ( g ) ∼ = T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) . Proof.
Let d ∈ Der
Lie z ( g ) be, then d ( g ) ⊆ Z Lie ( g ) , and thus d ([ g , g ] Lie ) = 0 . So d induces the map g [ g , g ] Lie α d −→ Z Lie ( g ) defined by α d ( x + [ g , g ] Lie ) = d ( x ). Now definethe map β : Der
Lie z ( g ) −→ T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) by β ( d ) = α d . Clearly, β is a linearmap, which is one-to-one by definition of α d .β is onto since for a given d ∗ ∈ T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) , there exists a linearmap d : g → Z Lie ( g ), d = d ∗ ◦ π , where π : g → g [ g , g ] Lie is the canonical8rojection, such that β ( d ) = d ∗ . Finally, d ∈ Der
Lie z ( g ) since d ([ x, y ] lie ) = d ∗ ([ π ( x ) , π ( y )] lie ) = d ∗ (0) = 0; on the other hand, [ d ( x ) , y ] lie + [ x, d ( y )] lie =[ d ∗ ( π ( x )) , y ] lie + [ x, d ∗ ( π ( y ))] lie = 0, since d ∗ ( π ( x )) , d ∗ ( π ( y )) ∈ Z Lie ( g ). To fin-ish, we show that β ([ d , d ]) = [ β ( d ) , β ( d )] for all d , d ∈ Der
Lie z ( g ) . Indeed,let x ∈ g . It is clear that β ([ d , d ])( π ( x )) = α [ d ,d ] ( π ( x )) = [ d , d ]( x ) = 0since d ( g ) , d ( g ) ⊆ Z Lie ( g ) ⊆ [ g , g ] Lie and d ([ g , g ] Lie ) = d ([ g , g ] Lie ) = 0 . On theother hand, [ β ( d ) , β ( d )]( π ( x )) = [ α d , α d ]( π ( x )) = α d ( d ( x )) − α d ( d ( x )) = 0since α d ([ g , g ] Lie ) = 0 = α d ([ g , g ] Lie ) . Hence β ([ d , d ]) = [ β ( d ) , β ( d )] . Thiscompletes the proof.
Corollary 3.6
For any arbitrary Leibniz algebra q , the Lie algebra Der
Lie z ( q ) hasa central subalgebra n isomorphic to T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) for some Lie -stem Leibnizalgebra g Lie -isoclinic to q . Moreover, each element of n sends Z Lie ( q ) to the zerosubalgebra. Proof.
By [5, Corollary 4.1], there is a
Lie -stem Leibniz algebra g Lie -isoclinicto q . Denote this
Lie -isoclinism by ( η, ξ ) . Now, by the proof of Proposition 3.4, n := { d ∗ | d ∈ Der
Lie z ( g ) } is a subalgebra of Der
Lie z ( q ) isomorphic to Der
Lie z ( g ) . n is a central subalgebra of Der
Lie z ( q ) . Indeed, let d ∈ n and d ∈ Der
Lie z ( q ) . Thenfor any y ∈ q , we have by definition, d ∗ ( y ) = ξ ( d ( x )) with π ( y ) = η ( π ( x )) . So d ( d ∗ ( y )) = 0 since d ∗ ( q ) ⊆ Z Lie ( q ) ∩ [ q , q ] Lie by Lemma 3.3, and d ([ q , q ] Lie ) =0 . Also, d ∗ ( Z Lie ( q )) = 0 since η is one-to-one and ξ is a homomorphism. Inparticular, d ∗ ( d ( y )) = 0 since d ( q ) ⊆ Z Lie ( q ) . Therefore [ d ∗ , d ] = 0 . Moreover,for any d ∗ ∈ n , d ∗ ( Z Lie ( q )) = 0 as mentioned above. To complete the proof, noticethat Der
Lie z ( g ) ∼ = T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) thanks to Lemma 3.5. Lemma 3.7
Let g and q be two Lie -isoclinic Leibniz algebras. If g is Lie -nilpotentof class c , then so is q . Proof.
Notice that for all g ∈ g and x , x , . . . , x i ∈ g , and setting ¯ t := t + Z Lie ( g ) ,t = g, x , x , . . . , x i , we have[[[¯ g, ¯ x ] lie , ¯ x ] lie , . . . , ¯ x i ] lie = [[[ g, x ] lie , x ] lie , . . . , x i ] lie + Z Lie ( g ) . So g ∈ Z Lie i +1 ( g ) ⇐⇒ g + Z Lie ( g ) ∈ Z Lie i ( g /Z Lie ( g )) . Therefore Z Lie i +1 ( g ) /Z Lie ( g ) = Z Lie i ( g /Z Lie ( g )) . If ( η, ξ ) is the
Lie -isoclinism between g and q , we have as η is anisomorphism, η ( Z Lie i +1 ( g ) /Z Lie ( g )) = η ( Z Lie i ( g /Z Lie ( g ))) = Z Lie i ( q /Z Lie ( q )) . It follows that g / Z Lie i +1 ( g ) ∼ = g /Z Lie ( g ) Z Lie i +1 ( g ) /Z Lie ( g ) ∼ = q /Z Lie ( q ) Z Lie i +1 ( q ) /Z Lie ( q ) ∼ = q / Z Lie i +1 ( q ) . g is Lie -nilpotent of class c. Then Z Lie c ( g ) = g . So q / Z Lie c ( q ) ∼ = g / Z Lie c ( g ) = 0 , implying that Z Lie c ( q ) = q . Also g / Z Lie c − ( g ) = 0 ⇐⇒ q / Z Lie c − ( q ) =0 . Hence q is also Lie-nilpotent of class c. Corollary 3.8
Let q be a Lie -nilpotent Leibniz algebra of class 2. Then
Der
Lie z ( q ) has a central subalgebra isomorphic to T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) containing ( R + L )( q ) . Proof.
By [5, Corollary 4.1], there is a
Lie -stem Leibniz algebra g Lie -isoclinicto q . Denote this
Lie -isoclinism by ( η, ξ ) . Since q is Lie -nilpotent Leibniz algebraof class 2, so is g , thanks to Lemma 3.7. Then Z Lie ( g ) = [ g , g ] Lie ξ ∼ = [ q , q ] Lie , and g [ g , g ] Lie ∼ = g Z Lie ( g ) η ∼ = q Z Lie ( q ) . So T (cid:16) g [ g , g ] Lie , Z
Lie ( g ) (cid:17) ∼ = T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) . Therefore
Der
Lie z ( q ) has a central subalgebra n isomorphic to T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) , thanks toCorollary 3.6. Moreover, the map ζ : q Z Lie ( q ) → T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) defined by x + Z Lie ( q ) ζ ( x + Z Lie ( q )) : q Z Lie ( q ) → [ q , q ] Lie with ζ ( x + Z Lie ( q ))( y + Z Lie ( q )) =[ x, y ] lie , is a well-defined one-to-one linear map, since for all x, x ′ ∈ q ,x − x ′ ∈ Z Lie ( q ) ⇐⇒ [ x − x ′ , y ] lie = 0 for all y ∈ q ⇐⇒ [ x, y ] lie = [ x ′ , y ] lie for all y ∈ q ⇐⇒ ζ ( x )( y + Z Lie ( q )) = ζ ( x ′ )( y + Z Lie ( q )) for all y ∈ q ⇐⇒ ζ ( x ) = ζ ( x ′ ) . Here we are using the notation x = x + Z Lie ( q ).( R + L )( q ) = Im ( ζ ) ⊆ T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) since ζ ( x )( y ) = [ x, y ] lie = [ x, y ] +[ y, x ] = L x ( y ) + R x ( y ).For any Leibniz algebra g with γ Lie ( g ) abelian, we denote K ( g ) := \ Ker (cid:0) f : g → γ Lie ( g ) (cid:1) Lemma 3.9
Let q be a Lie -nilpotent Leibniz algebra of class 2. Then γ Lie ( q ) = K ( q ) . Proof.
Let f : q → γ Lie ( q ) be a homomorphism of Leibniz algebras. Then for all q , q ∈ q , f ([ q , q ] lie ) = [ f ( q ) , f ( q )] lie ∈ [ γ Lie ( q ) , γ Lie ( q )] Lie ⊆ γ Lie ( q ) = 0 as q is Lie -nilpotent of class 2. So γ Lie ( q ) ⊆ Ker ( f ) . Therefore γ Lie ( q ) ⊆ K ( q ) since f isarbitrary.For the reverse inclusion, we proceed by contradiction. Let x ∈ K ( q ) suchthat x / ∈ γ Lie ( q ) , and let h be the complement of h x i in q . Then h is a maximalsubalgebra of q . So either h + γ Lie ( q ) = h or h + γ Lie ( q ) = q . The latter is notpossible. Indeed, if h + γ Lie ( q ) = q then γ Lie ( q ) = γ Lie ( h + γ Lie ( q )) ⊆ γ Lie ( h ) +10 Lie ( q ) . But since q is a Lie -nilpotent Leibniz algebra of class 2, then γ Lie ( q ) = 0 , which implies that γ Lie ( q ) = γ Lie ( h ) , and thus q = h + γ Lie ( q ) = h + γ Lie ( h ) = h . A contradiction. So we have h + γ Lie ( q ) = h , and thus γ Lie ( q ) ⊆ h , whichimplies that h is a two-sided ideal of q . Now, choose q ∈ γ Lie ( q ) and consider themapping f : q → γ Lie ( q ) defined by h + αx αq . Clearly, f is a well-definedhomomorphism of Leibniz algebras, and Ker ( f ) = h . This is a contradiction since x ∈ K ( q ) and x / ∈ h . Thus K ( q ) ⊆ γ Lie ( q ) . Theorem 3.10
Let q be a Lie -nilpotent Leibniz algebra of class 2. Then Z (cid:0) Der
Lie z ( q ) (cid:1) ∼ = T (cid:18) q Z Lie ( q ) , [ q , q ] Lie (cid:19) . Proof.
By the proof of Corollary 3.8,
Der
Lie z ( q ) has a central subalgebra n iso-morphic to T (cid:16) q Z Lie ( q ) , [ q , q ] Lie (cid:17) , where n := { d ∗ | d ∈ Der
Lie z ( g ) } for some Lie -stemLeibniz algebra g Lie -isoclinic to q . Denote this
Lie -isoclinism by ( η, ξ ) . It remains to show that Z (cid:0) Der
Lie z ( q ) (cid:1) ⊆ n , that is, given T ∈ Z (cid:0) Der
Lie z ( q ) (cid:1) , we must find d ∈ Der
Lie z ( g ) such that T = d ∗ . First, we claim that T ( q ) ⊆ K ( q ) . Indeed, let f : q → [ q , q ] Lie be a ho-momorphism of Leibniz algebras. Then consider the mapping t f : q → q de-fined by t f ( z ) = f ( z ) . Clearly, t f ∈ Der
Lie z ( q ) since t f ( q ) ⊆ [ q , q ] Lie = Z Lie ( q )as q is a Lie -nilpotent Leibniz algebra of class 2. So [
T, t f ] = 0 and thus f ( T ( z )) = t f ( T ( z )) = T ( t f ( z )) = 0 for all z ∈ q since t f ( z ) ∈ [ q , q ] Lie and T ([ q , q ] Lie ) = 0 as T ∈ Der
Lie z ( q ) . Therefore T ( q ) ⊆ Ker ( f ) . Hence T ( q ) ⊆ K ( q )since f is arbitrary, which proves the claim.It follows by Lemma 3.9 that T ( q ) ⊆ [ q , q ] Lie . Now, for any x ∈ g , we have x + Z Lie ( g ) = η − ( y + Z Lie ( q )) for some y ∈ q , since η is bijective. Consider themap d : g → g defined by x ξ − ( T ( y )) . Clearly d is well-defined, and it is easyto show that d ∈ Der
Lie z ( g ) since T ( q ) ⊆ [ q , q ] Lie = Z Lie ( q ) . Hence T = d ∗ . Thiscompletes the proof.
Corollary 3.11
Let q be a finite-dimensional Lie -nilpotent Leibniz algebra ofclass 2. Then
Der
Lie z ( q ) is abelian if and only if γ Lie ( q ) = Z Lie ( q ) . Proof.
Assume that γ Lie ( q ) = Z Lie ( q ) , then by Proposition 3.2, Der
Lie z ( q ) isabelian since q is a Lie -stem Leibniz algebra. Conversely, suppose that
Der
Lie z ( q )is an abelian Lie algebra. Then, again by Proposition 3.2, q is a Lie -stem Leibnizalgebra. This implies by Lemma 3.5 that
Der
Lie z ( q ) ∼ = T (cid:16) q γ Lie ( q ) , Z Lie ( q ) (cid:17) . Also,by Theorem 3.10,
Der
Lie z ( q ) = Z (cid:0) Der
Lie z ( q ) (cid:1) ∼ = T (cid:16) q Z Lie ( q ) , γ Lie ( q ) (cid:17) . It follows that T (cid:16) q γ Lie ( q ) , Z Lie ( q ) (cid:17) ∼ = T (cid:16) q Z Lie ( q ) , γ Lie ( q ) (cid:17) . Now, let K be the K -vector subspace11omplement of Z Lie ( q ) in γ Lie ( q ). We claim that K = 0 . Indeed, since as vectorspaces Z Lie ( q ) ⊕ K = γ Lie ( q ), then T (cid:18) q Z Lie ( q ) , γ Lie ( q ) (cid:19) = T (cid:18) q Z Lie ( q ) , Z Lie ( q ) (cid:19) ⊕ T (cid:18) q Z Lie ( q ) , K (cid:19) . As q Z Lie ( q ) ։ q γ Lie ( q ) by the Snake Lemma, it follows that T (cid:16) q Z Lie ( q ) , γ Lie ( q ) (cid:17) ∼ = T (cid:16) q γ Lie ( q ) , Z Lie ( q ) (cid:17) is isomorphic to a subalgebra of T (cid:16) q Z Lie ( q ) , Z Lie ( q ) (cid:17) . Hence T (cid:16) q γ Lie ( q ) , K (cid:17) = 0 . This completes the proof.
Example 3.12
The following is an example of Leibniz algebra satisfying the re-quirements of Corollary 3.11.Let q be the three-dimensional Leibniz algebra with basis { a , a , a } and bracketoperation given by [ a , a ] = [ a , a ] = a and zero elsewhere (see algebra 2 (c) in[9]). It is easy to check that γ Lie ( q ) = Z Lie ( q ) = < { a } > . Lie -Central derivations and
Lie -centroids
Definition 4.1
The
Lie -centroid of a Leibniz algebra g is the set of all linearmaps d : g → g satisfying the identities d ([ x, y ]) lie = [ d ( x ) , y ] lie = [ x, d ( y )] lie for all x, y ∈ g . We denote this set by Γ Lie ( g ) . Proposition 4.2
For any Leibniz algebra g , Γ Lie ( g ) is a subalgebra of End ( g ) such that Der
Lie z ( g ) = Der
Lie ( g ) ∩ Γ Lie ( g ) . Proof.
Assume that d ∈ Der
Lie ( g ) ∩ Γ Lie ( g ). For all x, y ∈ g , we have that d ([ x, y ] lie ) = [ d ( x ) , y ] lie + [ x, d ( y )] lie ; on the other hand, d ([ x, y ] lie ) = [ x, d ( y )] lie ,hence [ d ( x ) , y ] lie = 0 for any y ∈ g , that is d ( x ) ∈ Z Lie ( g )Conversely, Der
Lie z ( g ) is a subalgebra of Der
Lie ( g ) and for any d ∈ Der
Lie z ( g ) , wehave d ([ x, y ] lie ) = [ d ( x ) , y ] lie + [ x, d ( y )] lie = 0, since [ d ( x ) , y ] lie = 0, [ x, d ( y )] lie = 0,for any x, y ∈ g , hence d ∈ Γ Lie ( g ). Proposition 4.3
Let g be a Leibniz algebra. For any d ∈ Der
Lie ( g ) and φ ∈ Γ Lie ( g ) , the following statements hold:(a) Der
Lie ( g ) ⊆ N Der
Lie ( g ) (Γ Lie ( g )) .(b) d ◦ φ ∈ Γ Lie ( g ) if and only if φ ◦ d ∈ Der
Lie z ( g ) .(c) d ◦ φ ∈ Der
Lie ( g ) if and only if [ d, φ ] ∈ Der
Lie z ( g ) . roof. (a) Straightforward verification. (b)
Assume d ◦ φ ∈ Γ Lie ( g ). Then[ φ, d ]([ x, y ] lie ) = ( φ ◦ d )([ x, y ] lie ) − ( d ◦ φ )([ x, y ] lie )= [( φ ◦ d ) ( x ) , y ] lie + [ x, ( φ ◦ d ) ( y )] lie − [( d ◦ φ ) ( x ) , y ] lie = [[ φ, d ]( x ) , y ] lie + [ x, ( φ ◦ d ) ( y )] lie = [ φ, d ]([ x, y ] lie ) + [ x, ( φ ◦ d ) ( y )] lie Therefore [ x, ( φ ◦ d ) ( y )] lie = 0. Similarly [( d ◦ φ ) ( x ) , y ] lie = 0.Conversely, assume φ ◦ d ∈ Der
Lie z ( g ), then [ d, φ ]([ x, y ] lie ) = ( d ◦ φ )([ x, y ] lie ) − ( φ ◦ d )([ x, y ] lie ), hence ( d ◦ φ )([ x, y ] lie ) = [ d, φ ]([ x, y ] lie ) , since ( φ ◦ d )([ x, y ] lie ) = 0.Now it is a routine task to check that [ d, φ ] ∈ Γ Lie ( g ), which completes the proof. (c) Assume d ◦ φ ∈ Der
Lie ( g ) . A direct computation shows that [ φ, d ] ∈ Γ Lie ( g ).On the other hand, it is easy to check that [ d, φ ] ∈ Der
Lie ( g ), therefore [ φ, d ] = − [ d, φ ] ∈ Γ Lie ( g ) ∩ Der
Lie ( g ). Proposition 4.2 completes the proof.Conversely, assume [ d, φ ] ∈ Der
Lie z ( g ), then ( d ◦ φ ) ([ x, y ] lie ) = [ d, φ ] ([ x, y ] lie ) +( φ ◦ d ) ([ x, y ] lie ) = ( φ ◦ d ) ([ x, y ] lie ). Now it is easy to check that φ ◦ d is a Lie -derivation of g . Definition 4.4
Let m be a two-sided ideal of a Leibniz algebra g . Then m is saidto be Γ Lie ( g ) -invariant if ϕ ( m ) ⊂ m for all ϕ ∈ Γ Lie ( g ) . Proposition 4.5
Let g be a Leibniz algebra and m be a two-sided ideal of g . Thefollowing statements hold:(a) C Lie g ( m , is invariant under Γ Lie ( g ) .(b) Every Lie -perfect two-sided ideal m ( m = γ Lie ( m ) ) of g is invariant under Γ Lie ( g ) . Proof. (a)
Let g ∈ C Lie g ( m ,
0) and ϕ ∈ Γ Lie ( g ) be, then ϕ ( g ) ∈ C Lie g ( m , ϕ ( g ) , m ] lie = ϕ [ g, m ] lie = 0 , for all m ∈ m (b) Let m be a Lie -perfect two-sided ideal of g and let ϕ ∈ Γ Lie ( g ) be, thenany x ∈ m can be written as x = n P i =1 λ i [ m i , m i ] lie , m i , m i ∈ m , hence ϕ ( x ) = n P i =1 λ i [ ϕ ( m i ) , m i ] lie ∈ m . Theorem 4.6
Let m be a nonzero Γ Lie ( g ) -invariant two-sided ideal of a Leibnizalgebra g , V ( m ) = { ϕ ∈ Γ Lie ( g ) | ϕ ( m ) = 0 } and W = Hom (cid:0) gm , C Lie g ( m , (cid:1) be.Define T ( m ) = { f ∈ W | f [ x, y ] lie = [ f ( x ) , y ] lie = [ x, f ( y )] lie } with x = x + m and y = y + m . Then the following statements hold:(a) T ( m ) is a vector subspace of W isomorphic to V ( m ) . b) If Γ Lie ( m ) = K . Id m , then Γ Lie ( g ) = K . Id g ⊕ V ( m ) as vector spaces. Proof. (a)
Define α : V ( m ) −→ T ( m ) by α ( ϕ ) ( x + m ) = ϕ ( x ).Obviously, α is an injective well-defined linear map and it is onto, since forevery f ∈ T ( m ), set ϕ f : g −→ g , ϕ f ( x ) = f ( x + m ), for all x ∈ g . It iseasy to check that ϕ f ∈ Γ Lie ( g ) and ϕ f ( m ) = 0 , so ϕ f ∈ V ( m ). Moreover, α ( ϕ f ) ( x + m ) = ϕ f ( x ) = f ( x + m ). (b) If Γ
Lie ( m ) = K . Id m , then for all ψ ∈ Γ Lie ( g ), ψ | m = λ. Id m , for some λ ∈ K .If ψ = λ. Id g , define ϕ : g → g by ϕ ( x ) = λx , then ϕ ∈ Γ Lie ( g ) and ψ − ϕ ∈ V ( m ). Clearly, ψ = ϕ + ( ψ − ϕ ) ∈ K . Id g + V ( m ). Furthermore it is evident that K . Id g ∩ V ( m ) = 0, which completes the proof. Corollary 4.7 If K is a field of zero characteristic, then the following equalitieshold: Der
Lie z ( g ) = V ( γ Lie ( g )) = T ( γ Lie ( g )) Proof. If d ∈ Der
Lie z ( g ), then d ∈ Der
Lie ( g ) ∩ Γ Lie ( g ) by Proposition 4.2, hence[ d ( x ) , y ] lie = [ x, d ( y )] lie = 0, so d ∈ V ( γ Lie ( g )).Conversely, if d ∈ V ( γ Lie ( g )), then d ∈ Γ Lie ( g ) and d ([ x, y ] lie ) = 0, so d ([ x, y ] lie )= [ d ( x ) , y ] lie = [ x, d ( y )] lie = 0 . Hence d ([ x, y ] lie ) = [ d ( x ) , y ] lie + [ x, d ( y )] lie = 0,which implies that d ∈ Der
Lie z ( g ) . The second equality is provided by Theorem 4.6 since γ Lie ( g ) is Γ Lie ( g )-invariant. Theorem 4.8
Let g be a Leibniz algebra such that g = g ⊕ g , where g , g aretwo-sided ideals of g . Then the following isomorphism of K -vector spaces holds: Γ Lie ( g ) ∼ = Γ Lie ( g ) ⊕ Γ Lie ( g ) ⊕ C ⊕ C where C i = { ϕ ∈ Hom ( g i , g j ) | ϕ ( g i ) ⊆ Z Lie ( g j ) and ϕ ( γ Lie ( g i )) = 0 for ≤ i = j ≤ } . Proof.
Let π i : g −→ g i be the canonical projection for i = 1 ,
2. Then π , π ∈ Γ Lie ( g ) and π + π = Id g .So we have for ϕ ∈ Γ Lie ( g ) that ϕ = π ◦ ϕ ◦ π + π ◦ ϕ ◦ π + π ◦ ϕ ◦ π + π ◦ ϕ ◦ π Note that π i ◦ ϕ ◦ π j ∈ Γ Lie ( g ) for i, j = 1 ,
2. So, by the above equality itfollows thatΓ
Lie ( g ) = π Γ Lie ( g ) π ⊕ π Γ Lie ( g ) π ⊕ π Γ Lie ( g ) π ⊕ π Γ Lie ( g ) π as vector spaces. Indeed, it is enough to show that π i Γ Lie ( g ) π k ∩ π l Γ Lie ( g ) π j = 0 for i, j, k, l = 1 ,
2, such that ( i, j ) = ( k, l ). For instance π Γ Lie ( g ) π ∩ π Γ Lie ( g ) π = 0,since for any β ∈ π Γ Lie ( g ) π ∩ π Γ Lie ( g ) π , there are some f , f ∈ Γ Lie ( g ) suchthat β = π ◦ f ◦ π = π ◦ f ◦ π , then β ( x ) = π ◦ f ◦ π ( x ) = π ◦ f ◦ π ( π ( x )) =14 ◦ f ◦ π ( π ( x )) = π ◦ f (0) = 0, for all x ∈ g . Hence β = 0. Other cases canbe checked in a similar way.Now let’s denote Γ Lie ( g ) ij = π i Γ Lie ( g ) π j , i, j = 1 ,
2. We claim that the follow-ing isomorphisms of vector spaces hold:Γ
Lie ( g ) ∼ = Γ Lie ( g ) , Γ Lie ( g ) ∼ = Γ Lie ( g ) , Γ Lie ( g ) ∼ = C , Γ Lie ( g ) ∼ = C For ϕ ∈ Γ Lie ( g ) , ϕ ( g ) = 0 so ϕ | g ∈ Γ Lie ( g ). Now, considering Γ Lie ( g ) as asubalgebra of Γ Lie ( g ) such that for any ϕ ∈ Γ Lie ( g ), ϕ vanishes on g , that is, ϕ ( x ) = ϕ ( x ), ϕ ( x ) = 0, for all x ∈ g and x ∈ g . Then ϕ ∈ Γ Lie ( g )and ϕ ∈ Γ Lie ( g ) . Therefore, Γ Lie ( g ) ∼ = Γ Lie ( g ) by means of the isomorphism σ : Γ Lie ( g ) −→ Γ Lie ( g ), σ ( ϕ ) = ϕ | g , for all ϕ ∈ Γ Lie ( g ) .The isomorphism Γ Lie ( g ) ∼ = Γ Lie ( g ) can be proved in an analogous way.Γ Lie ( g ) ∼ = C . Indeed, for any ϕ ∈ Γ Lie ( g ) there exists a ϕ ∈ Γ Lie ( g ) suchthat ϕ = π ◦ ϕ ◦ π . For x k = ( x k , x k ) ∈ g , where x ik ∈ g i , i = 1 , k = 1 ,
2, wehave ϕ ([ x , x ] lie ) = π ◦ ϕ ◦ π ([ x , x ] lie ) = π ◦ ϕ ◦ π ([( x , x ) , ( x , x )] lie )= π ϕ ([ x , x ] lie ) = π ([ ϕ ( x ) , x ] lie ) = 0hence ϕ ( γ Lie ( g )) = 0. On the other hand, [ ϕ ( x ) , x ] lie = ϕ ([ x , x ] lie ) = 0, so, ϕ ( g ) ⊆ Z Lie ( g ) and ϕ ( γ Lie ( g )) = 0.It follows that ϕ | g ( g ) ⊆ Z Lie ( g ) and ϕ | g ( γ Lie ( g )) = 0, hence ϕ | g ∈ C . Conversely, for ϕ ∈ C , expanding ϕ on g by ϕ ( g ) = 0, we have π ◦ ϕ ◦ π = ϕ and so ϕ ∈ Γ Lie ( g ) . Hence Γ Lie ( g ) ∼ = C , by means the isomorphism τ :Γ Lie ( g ) −→ C , τ ( ϕ ) = ϕ | g for all ϕ ∈ Γ Lie ( g ) .Similarly, can be proved that Γ Lie ( g ) ∼ = C , which completes the proof. ID Lie -derivations
Definition 5.1 A Lie -derivation d : g → g is said to be an ID - Lie -derivation if d ( g ) ⊆ γ Lie ( g ) . The set of all ID - Lie -derivations of g is denoted by ID Lie ( g ) .An ID - Lie -derivation d : g → g is said to be ID ∗ - Lie -derivation if d vanisheson the Lie -central elements of g . The set of all ID ∗ - Lie -derivations of g is denotedby ID Lie ∗ ( g ) . It is obvious that ID Lie ( g ) and ID Lie ∗ ( g ) are subalgebras of Der
Lie ( g ) and Der
Lie c ( g ) ⊆ ID Lie ∗ ( g ) ⊆ ID Lie ( g ) (4)where Der
Lie c ( g ) is the subspace of Der
Lie ( g ) given by { d ∈ Der
Lie ( g ) | d ( x ) ∈ [ x, g ] lie , ∀ x ∈ g } . These kinds of derivations are called almost inner Lie -derivationsof g . 15 xample 5.2 Let g be the three-dimensional Leibniz algebra with basis { a , a , a } and bracket operation given by [ a , a ] = [ a , a ] = a and zero elsewhere (algebra2 (c) in [9]). The right multiplications Lie -derivations R x , x ∈ g , are examples ofalmost inner Lie -derivations.
Definition 5.3
An almost inner
Lie -derivation d is said to be central almostinner Lie -derivation if there exists an x ∈ Z l ( g ) such that ( d − R x )( g ) ⊆ Z Lie ( g ) .We denote the K -vector space of all central almost inner Lie -derivation by
Der
Lie cz ( g ) . Theorem 5.4
Let g and q be two Lie -isoclinic Leibniz algebras. Then ID Lie ∗ ( g ) ∼ = ID Lie ∗ ( q ) . Proof.
Let ( η, ξ ) be the
Lie -isoclinism between g and q and let α ∈ ID Lie ∗ ( g ) . Consider the map ζ α : q → q defined by ζ α ( y ) := ξ ( α ( x )) , where y + Z Lie ( q ) = η ( x + Z Lie ( g )) . Clearly ζ α is a well-defined linear map since α and ξ are linearmaps, and if y ∈ Z Lie ( q ) , then x ∈ Z Lie ( g ) and thus ζ α ( y ) = ξ ( α ( x )) = ζ (0) = 0 . To show that ζ α is a Lie -derivation, let y , y ∈ q and x , x ∈ g such that y i + Z Lie ( q ) = η ( x i + Z Lie ( g )) , i = 1 , . Then ζ α ([ y , y ] lie ) = ξ ( α ([ x , x ] lie ))= ξ ([ α ( x ) , x ] lie ) + ξ ([ x , α ( x )] lie ) by [3 , P rop. . ξ ( α ( x )) , y ] lie + [ y , ξ ( α ( x ))] lie = [ ζ α ( y ) , y ] lie + [ y , ζ α ( y )] lie . Moreover, since α ( g ) ⊆ γ Lie ( g ) and ξ is an isomorphism, it follows that ζ α ( q ) ⊆ γ Lie ( q ) . Therefore ζ α ∈ ID Lie ∗ ( q ) . Now consider the map ζ : ID Lie ∗ ( g ) → ID Lie ∗ ( q )defined by ζ ( α ) = ζ α . We claim that ξ is a Lie -homomorphism. Indeed, for α , α ∈ ID Lie ∗ ( g ) , we have for all y ∈ q , and x ∈ g such that y + Z Lie ( q ) = η ( x + Z Lie ( g )) ,ζ ([ α , α ])( y ) = ζ [ α ,α ] ( y ) = ξ ([ α , α ]( x )) = ξ ( α ( α ( x )) − α ( α ( x )))= ξ ( α ( α ( x ))) − ξ ( α ( α ( x )))= ζ α ( ξ ( α ( x )) − ζ α ( ξ ( α ( x )) by [3 , P rop. . ζ α ( ζ α ( y )) − ζ α ( ζ α ( y )) = [ ζ α , ζ α ]( y ) = [ ζ ( α ) , ζ ( α )]( y ) . Hence ζ ([ α , α ]) = [ ζ ( α ) , ζ ( α )] . Conversely, let β ∈ ID Lie ∗ ( q ) . By using theinverse
Lie -isoclinism ( η − , ξ − ) , we similarly construct a homomorphism ζ ′ : ID Lie ∗ ( q ) → ID Lie ∗ ( g ) defined by ζ ′ ( β ) = ζ ′ β where ζ ′ β ( x ) = ξ − ( β ( y )) with y + Z Lie ( q ) = η ( x + Z Lie ( g )) . It is clear that ( ζ ′ ◦ ζ )( α )( x ) = ζ ′ ( ζ ( α ))( x ) = ζ ′ ζ ( α ) ( x ) = ξ − ( ζ ( α )( y )) = ξ − ( ζ α ( y )) = ξ − ( ξ ( α ( x ))) = α ( x ) . So ζ ′ ◦ ζ = Id ID Lie ∗ ( g ) . Similarly,one shows that ζ ◦ ζ ′ = Id ID Lie ∗ ( q ) . Therefore ID Lie ∗ ( g ) ∼ = ID Lie ∗ ( q ).16 orollary 5.5 Let g and q be two Lie -isoclinic Leibniz algebras. Then
Der
Lie c ( g ) ∼ = Der
Lie c ( q ) . Proof.
Let ( η, ξ ) be the
Lie -isoclinism between g and q and let α ∈ Der
Lie c ( g ) . Consider again the map ζ α : q → q defined by ζ α ( y ) := ξ ( α ( x )) , where y + Z Lie ( q ) = η ( x + Z Lie ( g )) , given in the proof of Theorem 5.4. Since α ( x ) ∈ [ x, g ] lie and ξ is an isomorphism, it is clear that ζ α ( y ) ∈ [ y, q ] lie for all y ∈ q . So ζ α ∈ Der
Lie c ( q ) . So the restriction ζ | Der
Lie c ( g ) : Der
Lie c ( g ) → Der
Lie c ( q ) of the map ζ inthe proof of Theorem 5.4 to Der
Lie c ( g ) is a homomorphism. Similarly, by usingthe inverse Lie -isoclinism ( η − , ξ − ) , one obtains a homomorphism by taking therestriction ζ ′| Der
Lie c ( q ) : Der
Lie c ( q ) → Der
Lie c ( g ) of the map ζ ′ in the proof of Theorem5.4 to Der
Lie c ( q ) . It is clear that ζ ◦ ζ ′| Der
Lie c ( q ) = Id Der
Lie c ( q ) and ζ ′ ◦ ζ | Der
Lie c ( g ) = Id Der
Lie c ( g ) . Therefore
Der
Lie c ( g ) ∼ = Der
Lie c ( q ).For any d ∈ Der
Lie z ( g ), the map ψ d : g γ Lie ( g ) → Z Lie ( g ) given by ψ d ( g + γ Lie ( g )) = d ( g ) is a linear map. It is easy to show that the linear map ψ : Der
Lie z ( g ) → T (cid:16) g γ Lie ( g ) , Z Lie ( g ) (cid:17) , ψ ( d ) = ψ d is bijective. Therefore, dim (cid:0) Der
Lie z ( g ) (cid:1) = dim (cid:16) T (cid:16) g γ Lie ( g ) , Z Lie ( g ) (cid:17)(cid:17) for any finite-dimensional Leibniz algebra g . Corollary 5.6
Let g be a finite-dimensional Leibniz algebra such that [ g , g ] = γ Lie ( g ) and Z Lie ( g ) ⊆ Z r ( g ) . Then ID Lie ∗ ( g ) = Der
Lie z ( g ) if and only if γ Lie ( g ) = Z Lie ( g ) . Proof.
Assume that γ Lie ( g ) = Z Lie ( g ). It is clear that for all d ∈ Der
Lie z ( g ) ,d ( g ) ⊆ Z Lie ( g ) ⇐⇒ d ( g ) ⊆ γ Lie ( g ) and d ( Z Lie ( g )) = d ( γ Lie ( g )) = 0 . Therefore ID Lie ∗ ( g ) = Der
Lie z ( g ) . Conversely, assume that ID Lie ∗ ( g ) = Der
Lie z ( g ) . Then since [ g , g ] = γ Lie ( g ) and Z Lie ( g ) ⊆ Z r ( g ), it follows that the map R x : g → g , R x ( y ) = [ y, x ] , is a Lie -derivation; moreover it is easy to check that R x ∈ ID Lie ∗ ( g ) = Der
Lie z ( g ), hence R x ( y ) ∈ Z Lie ( g ) , for all y ∈ g . Therefore Z Lie ( g ) = g , and thus g is Lie -nilpotentof class 2 by Theorem 2.4. Now, by [5, Corollary 4.1], there is a
Lie -stem Leib-niz algebra q Lie -isoclinic to g . Denote this
Lie -isoclinism by ( η, ξ ) . Since g is Lie -nilpotent Leibniz algebra of class 2, so is q , thanks to Lemma 3.7. This im-plies that [ g , g ] Lie ξ ∼ = [ q , q ] Lie = Z Lie ( q ) , and g Z Lie ( g ) η ∼ = q Z Lie ( q ) ∼ = q [ q , q ] Lie . It follows byTheorem 5.4, the first implication and Lemma 3.5 thatdim(
Der
Lie z ( g )) = dim( ID Lie ∗ ( g )) = dim( ID Lie ∗ ( q )) = dim( Der
Lie z ( q ))= dim (cid:18) T (cid:18) q [ q , q ] Lie , Z
Lie ( q ) (cid:19)(cid:19) = dim (cid:18) T (cid:18) q Z Lie ( q ) , [ q , q ] Lie (cid:19)(cid:19) = dim (cid:18) T (cid:18) g Z Lie ( g ) , [ g , g ] Lie (cid:19)(cid:19) = dim (cid:0) Z ( Der
Lie z ( g )) (cid:1) . g is Lie -nilpotent of class 2.Therefore
Der
Lie z ( g ) is abelian. We now conclude by Corollary 3.11 that γ Lie ( g ) = Z Lie ( g ). Remark 5.7
Let us observe that the requirements [ g , g ] = γ Lie ( g ) and Z Lie ( g ) ⊆ Z r ( g ) in Corollary 5.6 are not needed in the absolute case, but in our relative set-ting they are absolutely necessary as the following counterexample shows: let g bethe four-dimensional complex Leibniz algebra with basis { a , a , a , a } and bracketoperation given by [ a , a ] = − [ a , a ] = a ; [ a , a ] = a and zero elsewhere (class R in [1, Theorem 3.2]). It is easy to check that [ g , g ] = h{ a }i = γ Lie ( g ) , Z Lie ( g ) = h{ a , a , a }i and Z r ( g ) = h{ a }i .Consider the Lie -derivation R a , which belongs to Der
Lie z ( g ) . However R a / ∈ ID Lie ∗ ( g ) since R a doesn’t vanish on Z Lie ( g ) . Example 5.8
The three-dimensional complex Leibniz algebra with basis { a , a , a } and bracket operation given by [ a , a ] = γa , γ ∈ C ; [ a , a ] = [ a , a ] = a andzero elsewhere (class 2 (a) in [9]) satisfies the requirements of Corollary 5.6, since [ g , g ] = γ Lie ( g ) = Z Lie ( g ) = Z r ( g ) = h{ a }i . Theorem 5.9
Let g be a Leibniz algebra such that γ Lie ( g ) is finite dimensionaland g Z Lie ( g ) is generated by p elements. Then dim( ID Lie ∗ ( g )) ≤ p · dim( γ Lie ( g )) Proof.
Consider the map α : ID Lie ∗ ( g ) → T (cid:16) g Z Lie ( g ) , γ Lie ( g ) (cid:17) defined by d d ∗ such that d ∗ ( x + Z Lie ( g )) = d ( x ) . Then α is a well-defined injective linear map.It follows that dim( ID Lie ∗ ( g )) ≤ dim (cid:16) T (cid:16) g Z Lie ( g ) , γ Lie ( g ) (cid:17)(cid:17) = p · dim( γ Lie ( g )) Example 5.10
Now we present two examples illustrating the inequality in The-orem 5.9.(a) Let g be the three-dimensional Leibniz algebra with basis { a , a , a } andbracket operation given by [ a , a ] = − [ a , a ] = a , [ a , a ] = a and zeroelsewhere (class 2 (f ) in [9]).It is an easy task to check that g Z Lie ( g ) = h{ a }i , hence the number of gener-ators is p = 1 . Moreover γ Lie ( g ) = h{ a }i . Also it can be checked that anelement d ∈ ID Lie ∗ ( g ) is represented by a matrix of the form a .Hence dim( ID Lie ∗ ( g )) = 1 ≤ · . b) Let g be the four-dimensional Leibniz algebra with basis { a , a , a , a } andbracket operation given by [ a , a ] = a , [ a , a ] = a and zero elsewhere(class R in [7, Theorem 2.7]).It is an easy task to check that g Z Lie ( g ) = h{ a , a , a }i , hence the numberof generators is p = 3 . Moreover γ Lie ( g ) = h{ a , a }i . Also it can bechecked that an element d ∈ ID Lie ∗ ( g ) is represented by a matrix of the form a a a a . Hence dim( ID Lie ∗ ( g )) = 4 ≤ · . Corollary 5.11
Let g be a Leibniz algebra such that Z r ( g ) = Z Lie ( g ) , [ g , g ] = γ Lie ( g ) is finite dimensional and g Z Lie ( g ) is generated by p elements. Then dim (cid:18) g Z Lie ( g ) (cid:19) ≤ p · dim( γ Lie ( g )) Proof.
Under these hypothesis, we have from the proof of Corollary 5.6 that R x ∈ ID Lie ∗ ( g ) for all x ∈ g . Now, consider the K -linear map β : g Z Lie ( g ) → ID Lie ∗ ( g )defined by x + Z Lie ( g ) R x , which is an injective well-defined linear map, since Ker ( β ) = Z r ( g ) Z Lie ( g ) = 0. Hence dim (cid:16) g Z Lie ( g ) (cid:17) ≤ dim (cid:0) ID Lie ∗ ( g ) (cid:1) . Now Theorem 5.9completes the proof. Example 5.12
The three-dimensional non-Lie Leibniz algebra with basis { a , a , a } and bracket operation [ a , a ] = a and zero elsewhere, satisfies the requirementsof Corollary 5.11. Definition 5.13
A Leibniz algebra g of dimension n is said to be Lie -filiform (or - Lie -filiform) if dim( γ Lie i ( g )) = n − i, ≤ i ≤ n . Lie -filiform Leibniz algebras are
Lie -nilpotent of class n − Corollary 5.14
Let g be an n -dimensional Leibniz algebra such that Z r ( g ) = Z Lie ( g ) ⊆ Z l ( g ) and it attains the upper bound of Corollary 5.11. If g is Lie -filiform, then n = 3 . Proof. If g is Lie -filiform, then dim( γ Lie ( g )) = n − , n ≥ . By the assumptionon Corollary 5.11, p = dim (cid:16) g Z Lie ( g ) (cid:17) = p · dim( γ Lie ( g )) = p ( n − , which impliesthat n = 3. Remark 5.15
Example 5.12 provides a
Lie -filiform Leibniz algebra which illus-trates Corollary 5.14. roposition 5.16 Let g be a Leibniz algebra. Then the following statementshold:(a) Let d ∈ Der
Lie c ( g ) be. Then d ( g ) ⊆ γ Lie ( g ) , d ( Z Lie ( g )) = 0 and d ( n ) ⊆ n forevery two-sided ideal n of g .(b) For d ∈ Der
Lie cz ( g ) there exists an x ∈ Z l ( g ) such that d | γ Lie ( g ) = R x | γ Lie ( g ) .(c) If g is 2-step Lie -nilpotent, then
Der
Lie cz ( g ) = Der
Lie c ( g ) .(d) If Z Lie ( g ) = 0 , then Der
Lie cz ( g ) ⊆ R ( g ) and R ( Z l ( g )) ⊆ Der
Lie cz ( g ) .(e) If g is Lie -nilpotent, then
Der
Lie c ( g ) is Lie -nilpotent and all d ∈ Der
Lie c ( g ) arenilpotent.(f ) Der
Lie c ( g ⊕ g ′ ) = Der
Lie c ( g ) ⊕ Der
Lie c ( g ′ ) , for any Leibniz algebras g and g ′ . Proof. (a)
For any x ∈ g , d ( x ) ∈ [ x, g ] Lie ⊆ [ g , g ] Lie ; if x ∈ Z Lie ( g ), then d ( x ) = [ x, y ] lie = 0, for all y ∈ g ; d ( n ) ⊆ [ n , g ] Lie ⊆ n . (b) Let d ∈ Der
Lie cz ( g ), then there exists x ∈ Z l ( g ) such that ( d − R x )( g ) ⊆ Z Lie ( g ). Since d − R x is a Lie -derivation we have( d − R x )([ y, z ] lie ) = [( d − R x )( y ) , z ] lie + [ y, ( d − R x )( z )] lie = 0and thus d ([ y, z ] lie ) = R x ([ y, z ] lie ), for all y, z ∈ g . Hence d | γ Lie ( g ) = R x | γ Lie ( g ) . (c) Notice that if g is 2-step Lie -nilpotent, then γ Lie ( g ) ⊆ Z Lie ( g ) . So for all d ∈ Der
Lie c ( g ) , any x ∈ Z l ( g ) and y ∈ g , we have d ( y ) ∈ [ y, g ] Lie ⊆ γ Lie ( g ) ⊆ Z Lie ( g )and R x ( y ) = [ y, x ] = [ y, x ] lie ∈ γ Lie ( g ) ⊆ Z Lie ( g ). Therefore ( d − R x )( g ) ⊆ Z Lie ( g ) , and thus d ∈ Der
Lie cz ( g ) . (d) Assume that Z Lie ( g ) = 0 . Then for all d ∈ Der
Lie cz ( g ) , there exists an x ∈ Z l ( g ) such that ( d − R x )( g ) = 0 , i.e. d = R x ∈ R ( g ) . So Der
Lie cz ( g ) ⊆ R ( g ) . The second inclusion can be easily checked. (e) If g is Lie -nilpotent of class c , then γ Lie c +1 ( g ) = 0 . So for any d ∈ Der
Lie c ( g ) ,d ( x ) ∈ [ x, g ] Lie ⊆ γ Lie ( g ) . One inductively proves that d c ( x ) ⊆ γ Lie c +1 ( g ), d c ( x ) = d ( d c − ( x )) ∈ [ d c − ( x ) , g ] Lie ⊆ γ Lie c +1 ( g ) = 0 . So d is nilpotent.Also, a routine inductive argument shows that γ Lie c +1 ( Der
Lie c ( g ))( g ) ⊆ γ Lie c +1 ( g ) =0 . So γ Lie c +1 ( Der
Lie c ( g )) = 0 and thus Der
Lie c ( g ) is Lie -nilpotent. (f )
For any d ∈ Der
Lie c ( g ⊕ g ′ ) , it is clear that d | g ∈ Der
Lie c ( g ) and d | g ′ ∈ Der
Lie c ( g ′ ) . Conversely, for d ∈ Der
Lie c ( g ) and d ′ ∈ Der
Lie c ( g ′ ) , one easily showsthat the mapping d ′′ : g ⊕ g ′ → g ⊕ g ′ defined by d ′′ ( x, x ′ ) := ( d ( x ) , d ′ ( x ′ )) isa Lie -derivation such that for x, x ′ ∈ g , g ′ we have d ′′ ( x, x ′ ) = ( d ( x ) , d ′ ( x ′ )) ∈ ([ x, g ] Lie , [ x ′ , g ′ ] Lie ) = [( x, x ′ ) , g ⊕ g ′ ] Lie by definition of the bracket of g ⊕ g ′ . cknowledgements Second author was supported by Agencia Estatal de Investigaci´on (Spain), grantMTM2016-79661-P (AEI/FEDER, UE, support included).
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