Finite-dimensional Leibniz algebra representations of sl 2
aa r X i v : . [ m a t h . R T ] A p r FINITE-DIMENSIONAL LEIBNIZ ALGEBRA REPRESENTATIONS OF sl T. KURBANBAEV AND R. TURDIBAEV
Abstract.
All finite-dimensional Leibniz algebra bimodules of a Lie algebra sl over a field of charac-teristic zero are described. Introduction
Finite-dimensional representations of a finite-dimensional semisimple Lie algebra is a well-studiedbeautiful classical theory. There is a Weyl’s theorem on complete reducibility that claims that any finite-dimensional module over a semisimple Lie algebra is a direct sum of simple modules. A textbook approachstarts with the finite-dimensional representations of the simple Lie algebra sl . In this work we find allfinite-dimensional representations of sl in a larger category – Leibniz algebra representations of sl .The notion of a Leibniz algebra first appeared under the name of D-algebra, introduced by A. Blohin [1] as one of the generalizations of Lie algebras, in which multiplication by an element is a derivation.Later, they were discovered independently by J.-L. Loday [2] and gain popularity under the name ofLeibniz algebras. Given a Leibniz algebra L there is a two-sided ideal Leib( L ) = Span { [ x, x ] | x ∈ L } ,associated to it, also known as the Leibniz kernel by some authors. The canonical Lie algebra L/ Leib( L ) is called the liezation of L . Due to Leibniz kernel, there are no simple non-Lie Leibniz algebras. However,by abuse of standard terminology a simple Leibniz algebra is introduced in [5] as an algebra with simpleliezation and simple Leibniz kernel. All such algebras are described via irreducible representations ofsimple Lie algebras.While originally defined differently (cf. [3], [2]), the representation of a Leibniz algebra is given in [4]as a K -module M with two actions - left and right, satisfying compatibility conditions coming from aso called square-zero construction. It is known that the category of Leibniz representations of a givenLeibniz algebra is not semisimple and any non-Lie Leibniz algebra admits a representation, which isneither simple, nor completely reducible [9, Proposition 1.2]. In [7] the indecomposable objects of thecategory of Leibniz representations of a Lie algebra are studied and for sl the indecomposable objectsin that category are described as extensions (see Theorem 2.5 below). Our goal in the current work is tobuild these extensions explicitly. Remarkably, the authors of [7] prove that for sl n ( n ≥ ) the categoryof Leibniz representations is of wild type.This work is a direct continuation of an investigation started in [9]. If M is an irreducible Leibnizrepresentation of sl , by Weyl’s result the left action on M as a Lie algebra representation decomposesinto a direct sum of irreducible Lie representations of sl . Hence, the problem of description reducesto the study of the right action. In case the number of such irreducible Lie representations is two,up to a Leibniz algebra representation isomorphism there are exactly two types of irreducible Leibnizrepresentations, whose actions are described in [9, Theorem 3.1]. In the current work, we establish thedescription in full generality in Theorem 3.5.All representations and algebras in this work are finite-dimensional over a field of characteristic zero.2. Preliminaries
Definition 2.1.
An algebra ( L, [ − , − ]) over a field K is called a (right) Leibniz algebra if for all x, y, z ∈ L the following identity holds: [ x, [ y, z ]] = [[ x, y ] , z ] − [[ x, z ] , y ] . Mathematics Subject Classification.
Key words and phrases.
Leibniz algebra, Leibniz algebra bimodule.The authors were supported by the project Ë Φ A- Φ TEX-2018-79.
In case the bracket is skew-symmetric, the identity above, called Leibniz identity transforms intoJacobi identity. One can establish that category of Lie algebras constitute a full subcategory of categoryof Leibniz algebras.Next we use definition from [4] to define representation of a Leibniz algebra.
Definition 2.2. A K -vector space M with two bilinear maps [ − , − ] : L × M → M and [ − , − ] : M × L → M is called representation of Leibniz algebra L if the following holds: [ m, [ x, y ]] = [[ m, x ] , y ] − [[ m, y ] , x ] , (1) [ x, [ m, y ]] = [[ x, m ] , y ] − [[ x, y ] , m ] , (2) [ x, [ y, m ]] = [[ x, y ] , m ] − [[ x, m ] , y ] . (3)Note that, these are exactly the conditions for a direct sum of K -vector spaces L ⊕ M to be Leibnizalgebra, where L and M are contained as subalgebra and abelian ideal, correspondingly. Such constructionis called square-zero construction. Adding identities (2) and (3) we obtain [ x, [ m, y ] + [ y, m ]] = 0 (4)which is often used instead of identity (3).Given a representation M of a Leibniz algebra L , one defines linear maps λ x , ρ x : M → M by λ x ( m ) = [ x, m ] and ρ x ( m ) = [ m, x ] for every x ∈ L, m ∈ M . Defining relations of Leibniz representationyield for all x, y ∈ L the following: ρ [ x,y ] = ρ y ◦ ρ x − ρ x ◦ ρ y , (5) λ [ x,y ] = ρ y ◦ λ x − λ x ◦ ρ y , (6) λ x ◦ ( ρ y + λ y ) = 0 . (7)A representation of a Leibniz algebra L is called symmetric ( anti-symmetric ) if ρ x = − λ x ( respectively, λ x = 0 ) for all x ∈ L . Considering a Lie algebra g as a Leibniz algebra, equation (5) (equation (1) formodule argument) shows that the map ρ : g → End( M ) defined by ρ ( x ) = ρ x coincides with Lie algebrarepresentation of the Lie algebra g . Moreover, it is known from [4] that the category of symmetric, aswell as, the category of anti-symmetric representations of a given Leibniz algebra L is equivalent to thecategory of Lie algebra representations of the liezation of L .For the sake of convenience, throughout this work, for a Leibniz algebra L we call representation M a bimodule M , and a Lie algebra module N an L -module N or simply a module N . Given a module M over a Lie algebra g , one can introduce symmetric and antisymmetric Leibniz bimodules M s and M a , bytaking the left action to be negative of the right action for the first, and identically zero for the secondbimodule, correspondingly.A bimodule is called simple or irreducible , if it does not admit non-trivial subbimodules. It is well-known that the simple objects in the category of Leibniz representations of a given Leibniz algebra areexactly symmetric and anti-symmetric representations [6, Lemma 1.9].A bimodule is called indecomposable , if it is not a direct sum of its subbimodules. Obviously, a simplebimodule is indecomposable, while the converse is not necessarily true (see [9, Proposition 1.2] and aparagraph that follows). To study bimodules it suffices to study indecomposable ones. In case Leibnizalgebra is a Lie algebra, we utilize the following Weyl’s result on complete reducibility of the right actionof the bimodule. Theorem 2.3. ( [8]) If g is a finite-dimensional semi-simple Lie algebra over a field of characteristiczero, then every finite-dimensional module over g is completely reducible. In order to describe all finite-dimensional indecomposable Leibniz bimodules of a Lie algebra sl overa field of characteristic zero with basis { e, f, h } and the products [ e, f ] = h, [ e, h ] = 2 e, [ f, h ] = − f, we use the following well-known description of simple sl -modules. Theorem 2.4. ( [8]) For every non-negative integer m there exists up to an sl -module isomorphismone and only one irreducible sl -module V ( m ) of dimension m + 1 . The module V ( m ) admits a basis { v , v , . . . , v m } in which the following holds for all k = 0 , . . . , m : INITE-DIMENSIONAL LEIBNIZ ALGEBRA REPRESENTATIONS OF sl [ h, v k ] = ( m − k ) v k , [ f, v k ] = v k +1 , [ e, v k ] = − k ( m + 1 − k ) v k − . J.-L. Loday and T. Pirashvili described the Gabriel quiver of Leibniz representations of sl usingClebsch-Gordon formula in [7], and citing results of [10] and [11] they found all indecomposable objects inthe category of Leibniz representations of sl as extensions of simple objects. For the sake of convenience,we express their result as the following Theorem 2.5. [7] For every non-negative integers n and k ≤ ⌊ n/ ⌋ + 1 there are exactly two indecom-posable sl -bimodules M and M determined uniquely by the following extensions: −→ M ≤ i< k V ( n − i − a −→ M −→ M ≤ i ≤ k − V ( n − i ) s −→ , −→ M ≤ i ≤ k − V ( n − i ) a −→ M −→ M ≤ i< k V ( n − i − s −→ , where V ( d ) s and V ( d ) a are irreducible symmetric and antisymmetric Leibniz representations of sl , cor-respondingly and V (0) a = V (0) s is a trivial one-dimensional representation. Our goal is to build these extensions explicitly. Let M be a finite-dimensional Leibniz bimodule of sl . As a right module, by Theorem 2.3 it is completely reduces into a direct sum of simple sl -modules V ⊕ · · · ⊕ V k , the right action on each simple submodule being described by Theorem 2.4. Hence, thestudy is reduced to the left action only. In the case k = 1 it is V ( d ) s and V ( d ) a , i.e. simple objects inthe category of Leibniz representation of sl . In [9, Theorem 3.1] the case k = 2 is exploited: Theorem 2.6. An sl -module M = V ( n ) ⊕ V ( m ) is indecomposable as a Leibniz sl -bimodule if and onlyif m = n − . For any integer n ≥ , up to sl -bimodule isomorphism there are exactly two indecomposablebimodules. The non-zero brackets of the left action are either [ h, v i ] = − ( n − i ) v i − iw i − [ h, w j ] = 2( m − j + 1) v j +1 − ( m − j ) w j [ f, v i ] = − v i +1 + w i or [ f, w j ] = v j +2 − w j +1 [ e, v i ] = i ( n − i + 1) v i − + i ( i − w i − [ e, w j ] = ( m − j + 1)(( m − j + 2) v j + iw j − ) corresponding to two bimodules, where { v , . . . , v n } and { w , . . . , w n − } are bases of V ( n ) and V ( n − of the Theorem 2.4. Note that in the first case
M/V ( n − is symmetric and V ( n − is anti-symmetric, while in the secondone M/V ( n ) is symmetric and V ( n ) is anti-symmetric bimodules, that is in accordance with Theorem2.5. In the current work, we use results on the left action established in [9] using only equality (2). Tillthe rest of the section let M be an sl -bimodule that decomposes as a right module into the direct sum M = V ( n ) ⊕ V ( m ) . The next statements shed light on the general form of the left action depending on n and m , that satisfies only identity (2), where { v , . . . , v n } and { w , . . . , w m } are bases of V ( n ) and V ( m ) of the Theorem 2.4. Proposition 2.7. [9, Proposition 2.6] Let n = m . Then identity (2) implies the following: [ h, v i ] = ( n − i )( ψ v i + ψ w i ) , ≤ i ≤ n, [ f, v i ] = ψ v i +1 + ψ w i +1 , ≤ i ≤ n − , [ e, v i ] = − i ( n − i + 1)( ψ v i − + ψ w i − ) , ≤ i ≤ n, [ h, w i ] = ( n − i )( ψ v i + ψ w i ) , ≤ i ≤ n, [ f, w i ] = ψ v i +1 + ψ w i +1 , ≤ i ≤ n − , [ e, w i ] = − i ( n − i + 1)( ψ v i − + ψ w i − ) , ≤ i ≤ n. T. KURBANBAEV AND R. TURDIBAEV
Proposition 2.8. [9, Proposition 2.5] Let n = m − . Then identity (2) implies the following: [ h, v i ] = ( n − i ) φ v i − iφ w i − , ≤ i ≤ n, [ f, v i ] = φ v i +1 + φ w i , ≤ i ≤ n − , [ e, v i ] = − i ( n − i + 1) φ v i − + i ( i − φ w i − , ≤ i ≤ n, [ h, w i ] = 2( m − i + 1) φ v i +1 + ( m − i ) φ w i , ≤ i ≤ m, [ f, w i ] = φ v i +2 + φ w i +1 , ≤ i ≤ m, [ e, w i ] = ( m − i + 1)(( m − i + 2) φ v i − iφ w i − ) , ≤ i ≤ m. Proposition 2.9.
Let n − m ≥ . Then identity (2) implies the following: [ f, v i ] = φ v i +1 ≤ i ≤ n − f, w j ] = φ w j +1 ≤ j ≤ m − e, v i ] = − i ( n − i + 1) φ v i − ≤ i ≤ n [ e, w j ] = − j ( m − j + 1) φ w j − ≤ j ≤ m [ h, v i ] = ( n − i ) φ v i ≤ i ≤ n [ h, w i ] = ( m − i ) φ w i ≤ i ≤ m . Proof.
From [9, Proposition 2.2, 2.3, 2.4 ] we have the following table of brackets: [ f, v i ] = φ v i +1 ≤ i ≤ n − f, w j ] = φ w j +1 ≤ j ≤ m − e, v i ] = i ( n − i + 1) n ǫ v i − ≤ i ≤ n [ e, w j ] = j ( m − j + 1) m ǫ w j − ≤ j ≤ m [ h, v i ] = ( η − iφ ) v i ≤ i ≤ n [ h, w i ] = ( η − iφ ) w i ≤ i ≤ m . Considering identity (2) for triples ( f, v , e ) and ( f, w , e ) one obtains η = nφ and η = mφ ,correspondingly. Analogously, identity (2) for ( h, v i , e ) and ( h, w i , e ) implies ǫ = − nφ and ǫ = − mφ , correspondingly. This completes the proof. (cid:3) Main Results
Throughout this section M is an sl -bimodule that decomposes into the direct sum of simple sl -modules M = V ( n ) ⊕ V ( n ) ⊕ · · · ⊕ V ( n k ) . Without loss of generality one can assume that n ≥ n ≥· · · ≥ n k . By Theorem 2.4 each simple module V p ( ≤ p ≤ k ) admits basis { v p , v p , . . . , v pn p } such thatfor ≤ i ≤ n p the following holds: [ v pi , h ] = ( n p − i ) v pi , [ v pi , f ] = v pi +1 , [ v pi , e ] = − i ( n p + 1 − i ) v pi − . In general [ sl , V p ] ⊆ M and let us set the following for all ≤ p ≤ k : [ h, v pi ] = k X q =1 n q X j =0 η pqij v qj , [ f, v pi ] = k X q =1 n q X j =0 φ pqij v qj , [ e, v pi ] = k X q =1 n q X j =0 ǫ pqij v qj . The description of Leibniz bimodules over sl is reduced to simplify the left action. As the followingproposition shows, most of the coefficients above are annihilated. Proposition 3.1.
Set l pq = ( n p − n q ) . Then [ h, v pi ] = k X q =1 η pqi v qi − l pq , [ f, v pi ] = k X q =1 φ pqi v qi +1 − l pq , [ e, v pi ] = k X q =1 ǫ pqi v qi − − l pq , where η pqi = φ pqi = ǫ pqi = 0 if l pq / ∈ Z . INITE-DIMENSIONAL LEIBNIZ ALGEBRA REPRESENTATIONS OF sl Proof.
From [ h, [ m, h ]] = [[ h, m ] , h ] we get ( n p − i ) k X q =1 n q X j =0 η pqij v qj = ( n p − i )[ h, v pi ] = [ h, [ v pi , h ]]= [[ h, v pi ] , h ] = [ k X q =1 n q X j =0 η pqij v qj , h ] = k X q =1 n q X j =0 ( n q − j ) η pqij v qj . Thus η pqij = 0 unless j = ( n q − n p ) + i . Denote by η pqi := η pqi,i − l pq .From [ f, [ m, h ]] = [[ f, m ] , h ] − f, m ] , as above we obtain ( n p − i ) k X q =1 n q X j =0 φ pqij v qj = ( n p − i )[ f, v pi ] = [ f, [ v pi , h ]] = k X q =1 n q X j =0 ( n q − j − φ pqij v qj . Therefore φ pqij = 0 unless j = ( n q − n p ) + i + 1 . Denote by φ pqi := φ pqi,i +1 − l pq From [ e, [ m, h ]] = [[ e, m ] , h ] − e, m ] we get ( n p − i ) k X q =1 n q X j =0 ǫ pqij v qj = k X q =1 n q X j =0 ( n q − j − ǫ pqij v qj . Hence, ǫ pqij = 0 unless j = ( n q − n p ) + i − and denote by ǫ pqi := ǫ pqi,i − − l pq . (cid:3) Next proposition is the main tool in partially reducing the general case to the case k = 2 . Proposition 3.2.
For any x ∈ sl and ≤ i ≤ j ≤ k , the restriction of the left action λ x on V ( n i ) ⊕ V ( n j ) coincides with the left action described in Propositions 2.7–2.9.Proof. Let x ∈ sl and for any i, j from { , . . . , k } let us denote by π i,j the linear projection from M to V ( n i ) ⊕ V ( n j ) . Consider m = v m + · · · + v km ∈ ⊕ ki =1 V ( n i ) . Using the fact that ρ x ( V ( n i )) ⊆ V ( n i ) for all ≤ i ≤ k we have π ij ( ρ x ( m )) = ρ x ( v im ) + ρ x ( v jm ) = ρ x ( v im + v jm ) = ρ x ( π ij ( m )) . This implies that π ij and ρ x commute. Moreover, using equality (6) we have π ij ( λ x ◦ ρ y ) = π ij ( ρ y ◦ λ x − λ [ x,y ] ) = ρ y ◦ ( π ij ◦ λ x ) − π ij ◦ λ [ x,y ] . Denote by λ ijx := π ij ◦ λ x . Then λ ijx ρ y = ρ y λ ijx − λ ij [ x,y ] that shows that λ ijx satisfies equation (2). However,for fixed i and j linear maps satisfying such condition are studied in Section (2) of [9] and are describedin Propositions 2.7-2.9. (cid:3) Although it is known from Theorem 2.5 that for a bimodule M to be indecomposable the sequence n ≥ n ≥ · · · ≥ n k must decrease by 2, there is a direct proof why M is decomposable if the sequencementioned is stable. Proposition 3.3.
Let M = ⊕ ki =1 V i , where dim V i = n + 1 . Then bimodule M is decomposable.Proof. By Proposition 3.1 for all ≤ i ≤ n + 1 , ≤ p ≤ k we have [ h, v pi ] = k X q =1 η pqi v qi , [ f, v pi ] = k X q =1 φ pqi v qi +1 , [ e, v pi ] = k X q =1 ǫ pqi v qi − . Furthermore, by Proposition 3.2 for (1 ≤ s, j ≤ k ) and Proposition 2.7 we get the following for all ≤ i ≤ n + 1 , ≤ p ≤ k : [ h, v pi ] = ( n − i ) k X q =1 φ pq v qi , [ f, v pi ] = k X q =1 φ pq v qi +1 , [ e, v pi ] = − i ( n − i + 1) k X q =1 φ pq v qi +1 . (8) T. KURBANBAEV AND R. TURDIBAEV
In the matrix form, we can write the first equality of (8) as follows: [ h, v i ][ h, v i ] ... [ h, v ki ] T = ( n − i ) φ φ . . . φ k φ φ . . . φ k . . . . . .φ k φ k . . . φ kk · v i v i ... v ki T = ( n − i ) · [ v i v i ... v ki ]Φ T , where Φ = ( φ ij ) ≤ i,j ≤ k is a matrix. Verifying identity (4) for h and v pi , (1 ≤ p ≤ k ) we φ φ . . . φ k φ φ . . . φ k . . . . . .φ k φ k . . . φ kk · Φ · v i v i ... v ki = ... . Hence, ( I + Φ)Φ = O and therefore, Φ is diagonalizable. Let ~x = k P q =1 x q v qi ∈ M (0 ≤ i ≤ n ) be aneigenvector of Φ T with an eigenvalue λ . Then [ h, ~x ] = [ h, x v i + x v i + ... + x k v ki ] = [[ h, v i ] [ h, v i ] ... [ h, v ki ]] · x x ... x k == n [ v i v i ... v ki ] · Φ T · x x ... x k = ( n − i )[ v i v i ... v ki ] λ x x ... x k = ( n − i ) λ~x. Consequently, this implies that [ sl , V i ] ⊆ V i , which means the module M is decomposable. (cid:3) The following statement describes all subbimodules of M when all n i ’s are different. Proposition 3.4.
Let N be a subbimodule of M and n i = n j for all ≤ i = j ≤ k . Then N is expressedas N = V n i ⊕ V n i ⊕ · · · ⊕ V n it for some ≤ i < i < · · · < i t ≤ k .Proof. Let N be a subbimodule of M and u = ( α v i p + . . . ) + ( α v i p + . . . ) + · · · + ( α t v i t p t + . . . ) ∈ N with α α . . . α t = 0 . Acting with f from the right ( n i − p ) -times on u we obtain α v i n i + ( α v i q + . . . ) + · · · + ( α t v i t q t + . . . ) ∈ N (9)If all the brackets vanish, then v i n i ∈ N and acting from the right with e consecutively, one has V n i ⊆ N .Therefore, u mod V n i ∈ N and recursively, the process continues.If some of the brackets are non-zero, apply h from the right to expression (9) and add it to expression(9) multiplied by n i , to deduce ( α ( n i + n i − q ) v i q + . . . ) + · · · + ( α t ( n + n i t − q t ) v i t q t + . . . ) ∈ N. Note that due to n > n > · · · > n k , none of the first coefficients is equal to zero in the brackets thatdid not vanish in expression (9). Hence, we reduce the number of components to one less and recursivelywe obtain v i t t ∈ N . Applying e from the right continuously one has V i t ⊆ N . Therefore, u mod V i p ∈ N and applying the arguments recursively from the start we are done. (cid:3) Note that if n i = n j for some i and j , the result of Proposition 3.4 is not true (cf. there are twosubbimodules constructed in Case 1 of the proof of [9, Proposition 3.1]). Theorem 3.5.
Let M be an sl -bimodule and as a right sl -module let it decompose as M = V ( n ) ⊕ V ( n ) ⊕· · ·⊕ V ( n k ) , where V ( n i ) are simple sl -modules of Theorem 2.4 with base { v i , . . . , v in i } , ≤ i ≤ k and n ≥ n ≥ · · · ≥ n k . Then M is an indecomposable Leibniz sl -bimodule only if n i − n i +1 = 2 for INITE-DIMENSIONAL LEIBNIZ ALGEBRA REPRESENTATIONS OF sl all ≤ i ≤ k − . Moreover, up to sl -bimodule isomorphism there are exactly two indecomposable sl -bimodules. The non-zero brackets of the left action is either [ h, v pi ] = 2( n − p − i + 3) v p − i +1 − ( n − p − i + 2) v pi +1 − iv p +1 i − , [ f, v pi ] = v p − i +2 − v pi +1 + v p +1 i , [ e, v pi ] = ( n − p − i + 3)(( n − p − i + 4) v p − i + iv pi − ) + i ( i − v p +1 i − , for all ≤ p ≤ k/ or [ h, v i ] = − ( n − i ) v i − iv i − , [ f, v i ] = − v i +1 + v i , [ e, v i ] = i ( n − i + 1) v i − + i ( i − v i − , [ h, v p +1 i ] = ( n − p − i + 1) v pi +1 − ( n − p − i ) v p +1 i +1 − iv p +2 i − , [ f, v p +1 i ] = v pi +2 − v p +1 i +1 + v p +2 i , [ e, v p +1 i ] = ( n − p − i + 1)(( n − p − i + 2) v pi + iv p +1 i − + i ( i − v p +2 i − , for all ≤ p ≤ ( k − / , where n = n .Proof. By Theorem 2.5 it is clear that the sequence { n i | ≤ i ≤ k } must decrease by two. Let us denoteby n = n and for the sake of convenience, denote by V i = V ( n − i +2) = { v i , v i , . . . , v in − i +2 } , ≤ i ≤ k .First we use Proposition 3.2 for pair ( j, j + 1) for all ≤ j ≤ k − and Proposition 2.8, then we useProposition 3.2 for pairs ( j, s ) , ≤ j ≤ k − , j + 2 ≤ s ≤ k and Proposition 2.9 to obtain the following: [ h, v i ] = ( n − i ) φ , v i − iφ , v i − , ≤ i ≤ n, [ f, v i ] = φ , v i +1 + φ , v i , ≤ i ≤ n − , [ e, v i ] = − i ( n − i + 1) φ , v i − + i ( i − φ , v i − , ≤ i ≤ n, ≤ j ≤ k − , ≤ i ≤ n − j + 2 :[ h, v ji ] = 2( n − j − i + 3) φ j,j − v j − i +1 + ( n − j − i + 2) φ j,j v ji − iφ j,j +1 v j +1 i − , [ f, v ji ] = φ j,j − v j − i +2 + φ j,j v ji +1 + φ j,j +1 v j +1 i , [ e, v ji ] = ( n − j + 3 − i )(( n − j + 4 − i ) φ j,j − v j − i − iφ j,j v ji − ) + i ( i − φ j,j +1 v j +1 i − , ≤ i ≤ n − k + 2 :[ h, v ki ] = 2( n − k + 3 − i ) φ k,k − v k − i +1 + ( n − k + 2 − i ) φ k,k v ki , [ f, v ki ] = φ k,k − v k − i +2 + φ k,k v ki +1 , [ e, v ki ] = ( n − k − i + 3)(( n − k + 4 − i ) φ k,k − v k − i − iφ k,k v ki − . Consider identity (4) for corresponding triples: • For ( f, v , f ) we have
1) (1 + φ , + φ , ) φ , = 0 ,
2) (1 + φ , ) φ , + φ , φ , = 0 , φ , φ , = 0 . (10) • For ( f, v , h ) we get (
1) (1 + φ , ) φ , = 0 ,
2) (1 + φ , ) φ , = 0 . (11) • For ( f, v j , f ) , ≤ j ≤ k − we obtain
1) (1 + φ j − ,j − + φ j,j ) φ j,j − = 0 ,
2) (1 + φ j,j ) φ j,j + φ j,j − φ j − ,j + φ j,j +1 φ j +1 ,j = 0 ,
3) (1 + φ j,j + φ j +1 ,j +1 ) φ j,j +1 = 0 . (12) T. KURBANBAEV AND R. TURDIBAEV • For ( f, v j , h ) , ≤ j ≤ k − we have (1 + φ j,j ) φ j,j +1 = 0 . (13) • For ( f, v j , e ) , ≤ j ≤ k − we get φ j,j − φ j − ,j − = φ j,j − φ j − ,j − = φ j,j − φ j − ,j = 0 . (14)Suppose k is odd and consider the following cases (the case when k is an even is carried out analogously). Case 1 . Let φ , = 0 . Then by (11) we have φ , = 0 , hence [ sl , V ] = 0 .If φ , = φ , = 0 , then bimodule M is decomposable. Thus φ , = and φ , = 0 . Hence, from 1) of(12) and (14) we get φ , = − and φ , = 0 , respectively. Since φ , = 0 , then from equality 3) of (12)we have φ , = 0 , hence from (13) we obtain φ , = 0 . Thus [ sl , V ] = 0 .Let φ , = 0 , φ , = 0 , otherwise M is decomposable. Then in (12) equalities 1) and 3) imply φ , = − and φ , = 0 . Hence by (13) and (14) we obtain φ , = 0 and φ , = 0 , correspondingly. This means that [ sl , V ] = 0 . Continuing this process we will get the following: [ f, v i ] = 0 , [ f, v i ] = φ , v i +2 − v i +1 + φ , v i , [ f, v i ] = 0 , [ f, v i ] = φ , v i +2 − v i +1 + φ , v i , [ f, v i ] = 0 ,. . . . . . . . [ f, v pi ] = φ p, p − v p − i +2 − v pi +1 + φ p, p +1 v p +1 i , [ f, v p +1 i ] = 0 , where ≤ p ≤ k − . Make a basis change ( v i ) ′ = v i , ( v i ) ′ = 1 φ , v i , ( v i ) ′ = φ , φ , v i , ( v i ) ′ = φ , φ , φ , v i , ( v i ) ′ = φ , φ , φ , φ , v i , ( v i ) ′ = φ , φ , φ , φ , φ , v i , . . . , ( v pi ) ′ = φ , φ , . . . φ p − , p − φ , φ , . . . φ p − , p − φ p, p − v pi , ( v p +1 i ) ′ = φ , φ , . . . φ p, p +1 φ , φ , . . . φ p, p − v p +1 i to obtain the following [ f, v i ] = 0 , [ f, v i ] = v i +2 − v i +1 + v i , [ f, v i ] = 0 ,. . . . . . . . [ f, v pi ] = v p − i +2 − v pi +1 + v p +1 i , [ f, v p +1 i ] = 0 . Thus for all ≤ p ≤ k − we obtain [ h, v pi ] = 2( n − p − i + 3) v p − i +1 − ( n − p − i + 2) v pi +1 − iv p +1 i − , ≤ i ≤ n − p + 2 , [ f, v pi ] = v p − i +2 − v pi +1 + v p +1 i , ≤ i ≤ n − p + 2 , [ e, v pi ] = ( n − p − i + 3)(( n − p − i + 4) v p − i + iv pi − ) + i ( i − v p +1 i − , ≤ i ≤ n − p + 2 , [ h, v p +1 i ] = 0 , ≤ i ≤ n − p, [ f, v p +1 i ] = 0 , ≤ i ≤ n − p, [ e, v p +1 i ] = 0 , ≤ i ≤ n − p. Using Proposition 3.4 it is easy to see that M is indecomposable. INITE-DIMENSIONAL LEIBNIZ ALGEBRA REPRESENTATIONS OF sl Case 2 . Let φ , = 0 . Then by (11) we have φ , = − . Hence in (14) we get φ , = 0 . If φ , = 0 ,then bimodule M is decomposable. So we may assume that φ , = 0 . Then by equations 1) and 3) of(10) one has φ , = 0 and φ , = 0 . Hence [ sl , V ] = 0 . Continuing a similar reasoning as in the Case 1,we obtain for all ≤ p ≤ k − the following: [ h, v i ] = − ( n − i ) v i − iv i − , ≤ i ≤ n, [ f, v i ] = − v i +1 + v i , ≤ i ≤ n, [ e, v i ] = i ( n − i + 1) v i − + i ( i − v i − , ≤ i ≤ n, [ h, v pi ] = 0 , ≤ i ≤ n − p + 2 , [ f, v pi ] = 0 , ≤ i ≤ n − p + 2 , [ e, v pi ] = 0 , ≤ i ≤ n − p + 2 , [ h, v p +1 i ] = ( n − p − i + 1) v pi +1 − ( n − p − i ) v p +1 i +1 − iv p +2 i − , ≤ i ≤ n − p, [ f, v p +1 i ] = v pi +2 − v p +1 i +1 + v p +2 i , ≤ i ≤ n − p, [ e, v p +1 i ] = ( n − p − i + 1)(( n − p − i + 2) v pi + iv p +1 i − + i ( i − v p +2 i − , ≤ i ≤ n − p, Once again, indecomposability is proved using Proposition 3.4. (cid:3)
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