aa r X i v : . [ m a t h . R A ] J a n LR-ALGEBRAS
DIETRICH BURDE, KAREL DEKIMPE, AND SANDRA DESCHAMPS
Abstract.
In the study of NIL-affine actions on nilpotent Lie groups we introduced so calledLR-structures on Lie algebras. The aim of this paper is to consider the existence question ofLR-structures, and to start a structure theory of LR-algebras. We show that any Lie algebraadmitting an LR-structure is 2-step solvable. Conversely we find several classes of 2-stepsolvable Lie algebras admitting an LR-structure, but also classes not admitting such a structure.We study also ideals in LR-algebras, and classify low-dimensional LR-algebras over R . Introduction
LR-algebras and LR-structures on Lie algebras arise in the study of affine actions on nilpotentLie groups as follows. Let N be a real, connected and simply connected nilpotent Lie group.Denote by Aff( N ) = N ⋊ Aut( N ) the group of affine transformations of N , acting on N via ∀ m, n ∈ N, ∀ α ∈ Aut( N ) : ( m,α ) n = m · α ( n ) . Note that for the special case where N = R n , we obtain the usual group of affine transforma-tions Aff( R n ) of n -dimensional space. When N is not abelian, we sometimes talk about theNIL-affine group Aff( N ), or NIL-affine motions. Recently, there has been a growing interest inthose subgroups G ⊆ Aff( N ) which act either properly discontinuously (in case G is discrete)or simply transitively (in case G is a Lie group) on N (see for example [1], [6]). It is known thatall groups which appear as such a simply transitive NIL-affine group have to be solvable. Con-versely for any connected and simply connected solvable Lie group G , there exists a nilpotentLie group N for which one can find an embedding ρ : G → Aff( N ) realizing G as a subgroupof Aff( N ) acting simply transitively on N (see [6]).Nevertheless, it is still an open problem to determine for a given G all connected, simply con-nected nilpotent Lie groups N , on which G acts simply transitively via NIL-affine motions.Even for the case G = R n the problem is non-trivial and interesting. For this case we were ableto translate this question in [4] to the existence problem of an LR-structure on the Lie algebra n of N . Indeed, we showed the following result (for the definition of a complete LR-structuresee 1.2). Theorem 1.1. [4, Theorem 5.1]
Let N be a connected and simply connected nilpotent Lie groupof dimension n . Then there exists a simply transitive NIL-affine action of R n on N if and onlyif the Lie algebra n of N admits a complete LR-structure. Date : October 31, 2018.1991
Mathematics Subject Classification.
Primary 17B30, 17D25.The first author thanks the KULeuven Campus Kortrijk for its hospitality and support.The second author expresses his gratitude towards the Erwin Schr¨odinger International Institute for Mathe-matical Physics.Research supported by the Research Programme of the Research Foundation-Flanders (FWO): G.0570.06.Research supported by the Research Fund of the Katholieke Universiteit Leuven.
The aim of this paper is to begin a study of LR-algebras and LR-structures on Lie algebras.Although LR-algebras arose, as we just explained, in the context of Lie algebras over the field R , we will work over an arbitrary field k of characteristic zero. Definition 1.2.
An algebra ( A, · ) over k with product ( x, y ) x · y is called an LR-algebra ,if the product satisfies the identities x · ( y · z ) = y · ( x · z )(1) ( x · y ) · z = ( x · z ) · y (2)for all x, y, z ∈ A .Denote by L ( x ) , R ( x ) the left respectively right multiplication operator in the algebra ( A, · ).The letters LR stand for “left and right”, indicating that in an LR-algebra the left and rightmultiplication operators commute: [ L ( x ) , L ( y )] = 0 , (3) [ R ( x ) , R ( y )] = 0 . (4)LR-algebras are Lie-admissible algebras: Lemma 1.3.
The commutator [ x, y ] = x · y − y · x in an LR-algebra ( A, · ) defines a Lie bracket.Proof. We have, using the above identities for all x, y, z ∈ A ,[[ x, y ] , z ] + [[ y, z ] , x ] + [[ z, x ] , y ] = [ x, y ] · z − z · [ x, y ] + [ y, z ] · x − x · [ y, z ]+ [ z, x ] · y − y · [ z, x ]= 0 . This shows that the Jacobi identity is indeed satisfied. (cid:3)
The associated Lie algebra g then is said to admit an LR-structure: Definition 1.4. An LR-structure on a Lie algebra g over k is an LR-algebra product g × g → g satisfying [ x, y ] = x · y − y · x (5)for all x, y, z ∈ g . The LR-structure, resp. the LR-algebra is said to be complete , if all leftmultiplications L ( x ) are nilpotent. Remark . If g is abelian, then an LR-structure on g is commutative and associative. Indeed,then we have R ( x ) = L ( x ). Conversely, commutative, associative algebras form a subclass ofLR-algebras with abelian associated Lie algebra.To conclude this introduction, let us present some easy examples of LR-algebras. Denote by r ( k ) the 2-dimensional non-abelian Lie algebra over k with basis ( e , e ), and [ e , e ] = e . Example 1.6.
The classification of non-isomorphic LR-algebras A with associated Lie algebra r ( k ) is given as follows: A Products A e · e = e , e · e = − e .A e · e = e .A e · e = − e . R-ALGEBRAS 3
The proof consists of an easy computation. The left multiplications defining an LR-algebrawith associated Lie algebra r ( k ) are of the following form: L ( e ) = (cid:18) α β (cid:19) , L ( e ) = (cid:18) β − γ (cid:19) , where αγ = β ( β − A , A , A . Notethat the algebra A is complete, whereas the algebras A and A are incomplete.2. Structural properties of LR-algebras
We just saw examples of LR-structures on a 2-step solvable Lie algebra, i.e., on r ( k ). Itturns out that all Lie algebras admitting an LR-structure are two-step solvable.
Proposition 2.1.
Any Lie algebra over k admitting an LR-structure is two-step solvable.Proof. For any x, y, u, v ∈ A we have the following symmetry relation:( x · y )( u · v ) = ( x · ( u · v )) · y = ( u · ( x · v )) · y = ( u · y ) · ( x · v )= x · (( u · y ) · v )= x · (( u · v ) · y )= ( u · v ) · ( x · y ) . Using this we obtain[[ x, y ] , [ u, v ]] = [ x · y − y · x, u · v − v · u ]= ( x · y − y · x ) · ( u · v − v · u ) − ( u · v − v · u ) · ( x · y − y · x )= ( x · y ) · ( u · v ) − ( x · y ) · ( v · u ) − ( y · x ) · ( u · v ) + ( y · x ) · ( v · u ) − ( u · v ) · ( x · y ) + ( v · u ) · ( x · y ) + ( u · v ) · ( y · x ) − ( v · u ) · ( y · x )= 0 . This shows that the associated Lie algebra is two-step solvable. (cid:3)
When we translate this result, using Theorem 1.1, in terms of NIL-affine actions, we find thefollowing:
Theorem 2.2.
Let N be a connected and simply connected nilpotent Lie group for which Aff( N ) contains an abelian Lie subgroup acting simply transitively on N , then N is two-step solvable.Remark . This result also explains Proposition 4.2 of [4] in a much more conceptual way.We now present some identities, which are useful when constructing LR-structures on a givenLie algebra. The first pair of identities remind one of the Jacobi identity for Lie algebras:
Lemma 2.4.
Let ( A, · ) be an LR-algebra. For all x, y, z ∈ A we have: [ x, y ] · z + [ y, z ] · x + [ z, x ] · y = 0 , (6) x · [ y, z ] + y · [ z, x ] + z · [ x, y ] = 0 . (7) D. BURDE, K. DEKIMPE, AND S. DESCHAMPS
Proof.
The first identity holds because we have[ x, y ] · z + [ y, z ] · x + [ z, x ] · y = ( x · y − y · x ) · z + ( y · z − z · y ) · x + ( z · x − x · z ) · y = (( x · y ) · z − ( x · z ) · y ) + (( y · z ) · x − ( y · x ) · z )+ (( z · x ) · y − ( z · y ) · x )= 0 . The second identity follows similarly. (cid:3)
We also have the following operator identities:
Lemma 2.5.
In an LR-algebra we have the following operator identities: ad([ x, y ]) − [ad( x ) , L ( y )] − [ L ( x ) , ad( y )] = 0 . (8) ad([ x, y ]) + [ad( x ) , R ( y )] + [ R ( x ) , ad( y )] = 0 . (9) Proof.
Using ad( x ) = L ( x ) − R ( x ) and (3) and (4) we obtainad([ x, y ]) = [ad( x ) , ad( y )]= [ L ( x ) − R ( x ) , L ( y ) − R ( y )]= [ L ( x ) , L ( y )] − [ R ( x ) , L ( y )] − [ L ( x ) , R ( y )] + [ R ( x ) , R ( y )]= [ − R ( x ) , L ( y )] + [ L ( x ) , − R ( y )]= [ad( x ) , L ( y )] + [ L ( x ) , ad( y )]This shows the first identity. The second identity follows similarly. (cid:3) We now study ideals of LR–algebras.
Lemma 2.6.
Let ( A, · ) be an LR-algebra and I, J be two-sided ideals of A . Then I · J is alsoa two-sided ideal of A .Proof. It is enough to show that for all a ∈ A , x ∈ I and y ∈ J , both a · ( x · y ) and ( x · y ) · a belong to I · J . But this is easy to see: a · ( x · y ) = x · ( a · y ) ∈ I · J, ( x · y ) · a = ( x · a ) · y ∈ I · J. (cid:3) Before continuing the study of ideals let us note the following:
Lemma 2.7.
Let ( A, · ) be an LR-algebra with associated Lie algebra g , and a ∈ A . Then alloperators L ( a ) and R ( a ) are Lie derivations of g , i.e., for any x, y ∈ A , the following identitieshold: a · [ x, y ] = [ a · x, y ] + [ x, a · y ] , [ x, y ] · a = [ x · a, y ] + [ x, y · a ] . Proof.
We have a · [ x, y ] = a · ( x · y ) − a · ( y · x )= x · ( a · y ) − y · ( a · x ) − ( a · y ) · x + ( a · x ) · y = [ a · x, y ] + [ x, a · y ] . The second identity follows similarly. (cid:3)
R-ALGEBRAS 5
The above lemma implies the following result:
Corollary 2.8.
Let ( A, · ) be an LR-algebra and assume that I, J are two-sided ideals of A .Then [ I, J ] is also a two-sided ideal of A . In particular, [
A, A ] is a two-sided ideal in A . Let γ ( A ) = A and γ i +1 ( A ) = [ A, γ i ( A )] for all i ≥ Corollary 2.9.
Let A be an LR-algebra. Then all γ i ( A ) are two-sided ideals of A . Lemma 2.10.
Let A be an LR-algebra. Then we have γ i +1 ( A ) · γ j +1 ( A ) ⊆ γ i + j +1 ( A ) for all i, j ≥ .Proof. We will use induction on i ≥
0. The case i = 0 follows from the fact that γ j +1 ( A ) is anideal of A . Now assume i ≥ γ k ( A ) · γ j +1 ( A ) ⊆ γ k + j ( A ) for all k = 1 , . . . , i .Let x ∈ γ ( A ), y ∈ γ i ( A ) and z ∈ γ j +1 ( A ). We have to show that [ x, y ] · z ∈ γ i + j +1 ( A ). UsingLemma 2.7 and the induction hypothesis, we see that[ x, y ] · z = [ x · z, y ] + [ x, y · z ] ∈ γ i + j +1 ( A ) . (cid:3) It is natural to introduce the center of an LR-algebra ( A, · ) by Z ( A ) = { x ∈ A | x · y = y · x for all y ∈ A } . Clearly Z ( A ) coincides with Z ( g ), the center of the associated Lie algebra g . Lemma 2.11.
Let ( A, · ) be an LR-algebra. Then Z ( A ) · [ A, A ] = [
A, A ] · Z ( A ) = 0 .Proof. Let a, b ∈ A and z ∈ Z ( A ). By (7) we have z · [ a, b ] + a · [ b, z ] + b · [ z, a ] = 0 . Since z ∈ Z ( g ), where g is the associated Lie algebra of A , we obtain z · [ a, b ] = 0. Analogouslyone shows that [ a, b ] · z = 0. (cid:3) Lemma 2.12.
Let A be an LR-algebra. Then Z ( A ) is a two-sided ideal of A .Proof. Let z ∈ Z ( A ). We have to show that [ a · z, b ] = [ z · a, b ] = 0 for all a, b ∈ A . UsingLemma 2.7 we see that a · [ z, b ] = [ a · z, b ] + [ z, a · b ][ z, b ] · a = [ z · a, b ] + [ z, b · a ] . Since z ∈ Z ( A ) the claim follows. (cid:3) Let Z ( A ) = Z ( A ) and define Z i +1 ( A ) by the identity Z i +1 ( A ) /Z i ( A ) = Z ( A/Z i ( A )). Notethat the Z i ( A ) are the terms of the upper central series of the associated Lie algebra g . As animmediate consequence of the previous lemma, we obtain Corollary 2.13.
Let A be an LR-algebra. Then all Z i ( A ) are two-sided ideals of A . D. BURDE, K. DEKIMPE, AND S. DESCHAMPS Classification of LR-structures
A classification of LR-structures in general is as hopeless as a classification of Lie algebras.However, one can study such structures in low dimensions. We will give here a classificationof complete LR-structures on real nilpotent Lie algebras of dimension n ≤
4. The restrictionto complete structures reduces the computations a lot, in particular for abelian Lie algebras.Nevertheless, we have classified also incomplete LR-structures in some cases.If the Lie algebra is abelian then LR-structures, and also LSA-structures, are just given bycommutative and associative algebras. Here a classification in terms of polynomial rings andtheir quotients is well known for n ≤
6, see [9] and the references cited therein. We wouldlike, however, to have explicit lists in terms of algebra products. This seems only availablein dimension n ≤ R and C , see [7]. For n = 4, there is an explicit list of nilpotent commutative, associative algebras (see the references in [9]), but not for all algebras. Suchnilpotent, commutative, associative algebras correspond exactly to complete left-symmetricalgebras with abelian associated Lie algebra. As [8] gives a list of all complete LSAs with anilpotent associated Lie algebra in dimension 4, one can easily extract those with an abelianassociated Lie algebra from that list, and so one obtains the complete abelian LR-structures indimension 4.It remains to classify all complete LR-structures on a non-abelian nilpotent Lie algebra ofdimension n ≤ R , which is one of the following: g Lie brackets n ( R ) [ e , e ] = e n ( R ) ⊕ R [ e , e ] = e n ( R ) [ e , e ] = e , [ e , e ] = e Proposition 3.1.
The classification of LR-algebra structures on the Heisenberg Lie algebra n ( R ) is given as follows: A Products A ( α ) , α ∈ R e · e = e , e · e = e , e · e = αe .A ( β ) , β ∈ R e · e = βe , e · e = ( β − e , e · e = e .A e · e = e , e · e = − e .A e · e = − e , e · e = e ,e · e = e , e · e = e All LR-algebras are complete, except for A .Proof. Any LR-algebra structure on n ( R ) is isomorphic to one of the following, written downwith left multiplication operators for the basis ( e , e , e ): L ( e ) = α γ β δ γ , L ( e ) = λ γ µ δ − ν µ , L ( e ) = γ µ , R-ALGEBRAS 7 satisfying the following polynomial equations: αλ = 0 γλ = 0 γ − αµ = 0 γ (2 δ − − αν − βµ = 0 βλ = 0A straightforward case by case analysis yields the result. We have A ( α ) ≃ A ( α ′ ) if and only α ′ = α , and the same result for A ( β ). (cid:3) Similarly we obtain the following result.
Proposition 3.2.
The classification of LR-algebra structures on g = n ( R ) is given as follows: A Products A ( α ) e · e = α ( α − e , e · e = αe , e · e = αe ,α ∈ R e · e = ( α − e , e · e = e , e · e = ( α − e .A e · e = e , e · e = − e ,e · e = e , e · e = − e .A e · e = e , e · e = e ,e · e = e , e · e = e .A ( α, β, γ ) e · e = αe , e · e = βe + γe , e · e = βe ,α, β, γ ∈ { , } e · e = ( β − e + γe , e · e = ( β − e .A ( α ) e · e = αe , e · e = − e , e · e = e ,α ∈ { , } e · e = e , e · e = − e , e · e = e .A e · e = − e , e · e = e , e · e = e , e · e = e ,e · e = − e , e · e = e , e · e = e , e · e = e . The algebra A is not complete. All the other ones are complete. The family A ( α, β, γ ) consists of 8 different algebras. Proposition 3.3.
The classification of complete LR-algebra structures on g = n ( R ) ⊕ R isgiven as follows: A Products A ( α ) e · e = αe , e · e = ( α − e ,α ∈ R e · e = e , e · e = e .A ( α ) e · e = αe , e · e = e , e · e = − e + e ,α ∈ { , } e · e = e , e · e = αe , e · e = αe .A ( α, β ) e · e = αe , e · e = ( α − e , e · e = e ,α ∈ R , β ∈ { , } e · e = βe , e · e = βe .A ( α ) e · e = e , e · e = e , e · e = − e ,α ∈ { , } e · e = αe , e · e = e .A ( α ) e · e = e , e · e = − e ,α ∈ { , } e · e = αe , e · e = e . D. BURDE, K. DEKIMPE, AND S. DESCHAMPS A Products A ( α ) e · e = αe , e · e = − e ,α ∈ R e · e = e , e · e = e .A ( α ) e · e = αe , e · e = − e ,α ≤ e · e = − e , e · e = e .A e · e = e , e · e = − e ,e · e = e .A ( α ) e · e = e , e · e = αe ,α ≥ e · e = ( α − e , e · e = e .A ( α ) e · e = e , e · e = αe ,α ≥ e · e = ( α − e , e · e = − e .A ( α ) e · e = e , e · e = αe ,α ∈ R e · e = ( α − e .A e · e = e , e · e = − e ,e · e = e .A ( α ) e · e = e , e · e = − e ,α ∈ R e · e = αe .A e · e = e , e · e = − e .A ( α ) e · e = e , e · e = − e ,α ≥ e · e = αe − e . Remark . The computations for the above result are quite complicated, but do not givemuch insight. Therefore we have omitted them here. However, we did the computationsindependently to be sure that they are correct.4.
LR-structures on nilpotent Lie algebras
We know that any Lie algebra admitting an LR-structure must be 2-step solvable. Converselywe can ask which 2-step solvable Lie algebras admit an LR-structure. We start with 2-stepsolvable, filiform nilpotent Lie algebras f n of dimension n . There exists a so called adaptedbasis ( e , . . . , e n ) of f n such that the Lie brackets are given as follows:[ e , e i ] = e i +1 , ≤ i ≤ n − , [ e , e i ] = n X k = i +2 c i,k e k , ≤ i ≤ n − , [ e i , e j ] = 0 , ≤ i ≤ j. The Jacobi identity is satisfied if and only if c i +1 ,k = c i,k − for all 6 ≤ i + 3 ≤ k ≤ n . Fordetails, see for example [2]. Lemma 4.1.
Let f n be given as above. Then the identities ad( e ) ad( e ) = ad( e ) ad( e ) ad( e ) , (10) ad( e ) ad( e ) = ad( e ) ad( e ) ad( e ) , (11) ad( e i +2 ) = ad( e ) i ad( e ) − ad( e ) ad( e ) i , i ≥ . (12) hold. R-ALGEBRAS 9
Proof.
The identity (10) is equivalent to0 = [ad( e ) , ad( e )] ad( e ) = ad( e ) ad( e ) . But this follows from [ e , [ e , e k ]] = 0 for all k ≥
1. Similarly, (11) is equivalent to ad( e ) ad( e ) =0, which follows again by definition. The identity (12) is proved by induction on i ≥
1. For i = 1 we have ad( e ) = [ad( e ) , ad( e )] = ad( e ) ad( e ) − ad( e ) ad( e ) . By induction hypothesis, ad( e i +1 ) = ad( e ) i − ad( e ) − ad( e ) ad( e ) i − . Then, using (10) re-peatedly, we obtain for i ≥ e i +2 ) = ad( e ) ad( e i +1 ) − ad( e i +1 ) ad( e )= ad( e ) i ad( e ) − ad( e ) ad( e ) ad( e ) i − − ad( e ) i − ad( e ) ad( e ) + ad( e ) ad( e ) i = ad( e ) i ad( e ) − ad( e ) ad( e ) i − ad( e ) ad( e ) i + ad( e ) ad( e ) i = ad( e ) i ad( e ) − ad( e ) ad( e ) i . (cid:3) Proposition 4.2.
Any -step solvable filiform nilpotent Lie algebra f n admits a complete LR-structure.Proof. Define an LR-structure on f n as follows: L ( e ) = 0 ,L ( e i ) = ad( e ) i − ad( e ) , ≤ i ≤ n. In particular, this means e · e j = 0 , e · e j = [ e , e j ] , ≤ j ≤ n,e j · e = [ e j , e ] , e j · e = 0 , ≤ j ≤ n, so that R ( e ) = − ad( e ) and R ( e ) = 0. Furthermore, we have e i · e j = [ e , e i + j − ] , ≤ i, j ≤ n. To see this, note that e · e j = ad( e ) ad( e )( e j ) = [ e , e j +1 ] for j ≥
3. Then the result for i ≥ e i · e j − e j · e i = [ e i , e j ] , ≤ i ≤ j ≤ n. The cases i = 1 and i = 2 are obvious. For j ≥ i ≥ e i · e j − e j · e i = 0 = [ e i , e j ] . In particular it follows R ( e i ) = L ( e i ) − ad( e i ). The formula (12) then implies R ( e i ) = ad( e ) ad( e ) i − , i ≥ . (13)It remains to show that all operators L ( e i ) commute, and all R ( e i ) commute, i.e., L ( e i ) L ( e j ) = L ( e j ) L ( e i ) , ≤ i < j ≤ n,R ( e i ) R ( e j ) = R ( e j ) R ( e i ) , ≤ i < j ≤ n. The first identity is obvious for i = 1. For 2 ≤ i < j ≤ n use (10) and (11) repeatedly to obtain L ( e i ) L ( e j ) = ad( e ) i − ad( e ) ad( e ) j − ad( e )= ad( e ) j − ad( e ) ad( e ) i − ad( e )= L ( e j ) L ( e i ) . This argument also shows R ( e i ) R ( e j ) = R ( e j ) R ( e i ) for 3 ≤ i < j ≤ n , because of (13). For i = 2 this is trivially true since R ( e ) = 0. For i = 1 and j ≥ e ) R ( e j ) = R ( e j ) ad( e ). This follows again from (10). It is obvious that all L ( e i ) arenilpotent, hence the LR-structure is complete. (cid:3) It is natural to ask which other nilpotent Lie algebras admit LR-structures. We first observethe following fact.
Proposition 4.3.
Every -step nilpotent Lie algebra g admits a complete LR-structure.Proof. For x ∈ g define an LR-structure by L ( x ) = 12 ad( x ) . Indeed, for all x, y, z ∈ g we have x · y − y · x = 12 [ x, y ] −
12 [ y, x ] = [ x, y ] ,x · ( y · z ) = 0 = y · ( x · z ) , ( x · y ) · z = 0 = ( x · z ) · y. Finally L ( x ) is a nilpotent derivation for all x ∈ g , since g is nilpotent. (cid:3) Proposition 4.4.
Every free -step nilpotent Lie algebra g on n generators x , . . . , x n admitsa complete LR-structure.Proof. A vector space basis of g is given by x , x , · · · , x n y i,j = [ x i , x j ] , ≤ i < j ≤ n,z i,j,k = [ x i , y j,k ] . An LR-structure on g is defined as follows: x j · x i = − y i,j , ≤ i < j ≤ nx i · y j,k = z i,j,k , ≤ j < k ≤ i = z k,j,i , j < i < ky j,k · x i = z k,j,i − z i,j,k , j < i < k = − z i,j,k , i ≤ j < k and all other products equal to zero. (cid:3) Example 4.5.
Let f be the free -step nilpotent Lie algebra with generators. Then there is abasis ( x , . . . , x ) of f with generators ( x , x , x ) and Lie brackets R-ALGEBRAS 11 x = [ x , x ] x = [ x , x ] x = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x − x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] An LR-structure is given by x .x = − x , x .x = x ,x .x = x , x .x = − x ,x .x = x , x .x = − x ,x .x = − x , x .x = x − x ,x .x = − x , x .x = x − x ,x .x = x , x .x = − x .x .x = x , Proposition 4.4 implies, in the same way as for Novikov structures (see [3]), the followingcorollary.
Corollary 4.6.
Any -generated -step nilpotent Lie algebra admits a complete LR-structure. One might ask whether or not all 3-step nilpotent Lie algebras admit an LR-structure. Thisturns out to be not the case. To find a counterexample we have to look at Lie algebras with atleast 4 generators.
Proposition 4.7.
Let g be the following -step nilpotent Lie algebra on generators of dimen-sion , with basis ( x , . . . , x ) and non-trivial Lie brackets [ x , x ] = x , [ x , x ] = − x , [ x , x ] = x , [ x , x ] = − x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = − x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x + x . [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , This -step solvable Lie algebra does not admit an LR-structure.Proof. We will assume that g admits an LR-structure and show that this leads to a contra-diction. We denote by ad( x i ) the adjoint operators, by L ( x i ) the left multiplications, and by R ( x i ) the right multiplication operators with respect to the basis ( x , x , . . . , x ). The op-erators ad( x i ) are given by the Lie brackets of g , while the left multiplication operators are unknown. We denote the ( j, k )-th entry of L ( x i ) by L ( x i ) j,k = x ij,k . The j -th column of L ( x i ) gives the coordinates of L ( x i )( x j ). We have to satisfy the identities(1), (2) and (5), where x , y and z run over all basis vectors. This leads to a huge system ofquadratic equations in the variables x ij,k for 1 ≤ i, j, k ≤
13, summing up to a total of 13 = 2197variables. It is quite impossible to solve these equations without further information. However,we can use our knowledge on ideals in LR-algebras to conclude that a lot of unknowns x ij,k already have to be zero. This, together with Lemmas 2.4 and 2.5, simplifies the system ofequations considerably. In this way we can show that the equations are contradictory. Thisworks exactly as in the proof of proposition 3 . (cid:3) In [3] we showed that the 2-generated, free 4-step nilpotent Lie algebra (which is 2-stepsolvable) does not admit any Novikov structure. It turns out however, that this example doesadmit an LR-structure:Let g be the free 4-step nilpotent Lie algebra on 2 generators x and x . Let ( x , . . . , x ) be abasis of g with the following Lie brackets: x = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , x ]] = [ x , x ] x = [ x , [ x , [ x , x ]]] = [ x , x ] x = [ x , [ x , [ x , x ]]] = [ x , x ]= [ x , [ x , [ x , x ]]] = [ x , x ] x = [ x , [ x , [ x , x ]]] = [ x , x ] Proposition 4.8.
The -step solvable Lie algebra from the above example does admit an LR-structure.Proof. We define the left multilications by L ( x ) = 0, L ( x ) = ad( x ) and, if x i , i ≥ x and x , then L ( x i ) is the corresponding composition of ad( x ) and ad( x ): L ( x ) = L ([ x , x ]) = ad( x ) ad( x ) ,L ( x ) = L ([ x , [ x , x ]]) = ad( x ) ad( x ) L ( x ) = L ([ x , [ x , x ]]) = ad( x ) ad( x ) ad( x ) L ( x ) = L ([ x , [ x , [ x , x ]]]) = ad( x ) ad( x ) L ( x ) = L ([ x , [ x , [ x , x ]]]) = ad( x ) ad( x ) ad( x )= L ([ x , [ x , [ x , x ]]]) = ad( x ) ad( x ) ad( x ) ad( x ) L ( x ) = L ([ x , [ x , [ x , x ]]]) = ad( x ) ad( x ) ad( x ) . In fact this really defines an LR-structure. Note that L ( x ) = L ( x ) = L ( x ) = 0. (cid:3) Construction of LR-structures via extensions
In the following we will consider Lie algebras g which are an extension of a Lie algebra b byan abelian Lie algebra a . Hence we have a short exact sequence of Lie algebras0 → a ι −→ g π −→ b → . Since a is abelian, there exists a natural b -module structure on a . We denote the action of b on a by ( x, a ) ϕ ( x ) a , where ϕ : b → End( a ) is the corresponding Lie algebra representation. R-ALGEBRAS 13
We have(14) ϕ ([ x, y ]) = ϕ ( x ) ϕ ( y ) − ϕ ( y ) ϕ ( x )for all x, y ∈ b . The extension g is determined by a two-cohomology class. Let Ω ∈ Z ( b , a ) bea 2-cocycle describing the extension g . This implies that Ω : b × b → a is a skew-symmetricbilinear map satisfying(15) ϕ ( x )Ω( y, z ) − ϕ ( y )Ω( x, z ) + ϕ ( z )Ω( x, y ) = Ω([ x, y ] , z ) − Ω([ x, z ] , y ) + Ω([ y, z ] , x ) , such that the Lie algebra with underlying vector space a × b and Lie bracket given by(16) [( a, x ) , ( b, y )] := ( ϕ ( x ) b − ϕ ( y ) a + Ω( x, y ) , [ x, y ])for a, b ∈ a and x, y ∈ b , is isomorphic to g . As a shorthand, we will use g = ( a , b , ϕ, Ω) to saythat g is the extension determined by this specific data.Note that the Lie algebras g we are interested in, are all 2-step solvable Lie algebras andhence can be obtained as extensions of two abelian Lie algebras a = [ g , g ] and b = g / [ g , g ]. So,although we will treat extensions in the general case, we pay special attention to this specificsituation where both a and b are abelian. In this specific case, the Lie bracket of g = a × b isgiven by [( a, x ) , ( b, y )] := ( ϕ ( x ) b − ϕ ( y ) a + Ω( x, y ) , ϕ and Ω are now given as follows: since a and b are abelian, ϕ is just alinear map satisfying ϕ ( x ) ϕ ( y ) = ϕ ( y ) ϕ ( x )for all x, y ∈ b . On the other hand, Ω : b × b → a is a skew-symmetric bilinear map satisfying ϕ ( x )Ω( y, z ) − ϕ ( y )Ω( x, z ) + ϕ ( z )Ω( x, y ) = 0 . Now, let us return to the more general case (i.e., b does not have to be abelian), and tryto construct LR-structures on Lie algebras g which are given as extension g = ( a , b , ϕ, Ω) ofa Lie algebra b by an abelian Lie algebra a . Suppose that we have already an LR-product( a, b ) a · b on a and an LR-product ( x, y ) x · y on b . In other words, we have x · y − y · x = [ x, y ] x · ( y · z ) = y · ( x · z )( x · y ) · z = ( x · z ) · ya · b = b · aa · ( b · c ) = b · ( a · c )( a · b ) · c = ( a · c ) · b for all x, y, z ∈ b and for all a, b, c ∈ a . (In fact the product on a has to be commutative andassociative). We want to lift these LR-products to g . Consider ω : b × b → a ϕ , ϕ : b → End( a )where ω is a bilinear map and ϕ , ϕ are linear maps. We will define a bilinear product g × g → g by(17) ( a, x ) ◦ ( b, y ) := ( a · b + ϕ ( y ) a + ϕ ( x ) b + ω ( x, y ) , x · y ) Proposition 5.1.
The above product defines an LR-structure on g if and only if the followingconditions hold: ω ( x, y ) − ω ( y, x ) = Ω( x, y )(18) ϕ ( x ) − ϕ ( x ) = ϕ ( x )(19) ϕ ( x ) ω ( y, z ) − ϕ ( y ) ω ( x, z ) = ω ( y, x · z ) − ω ( x, y · z )(20) a · ω ( y, z ) + ϕ ( y · z ) a = ϕ ( y ) ϕ ( z ) a (21) [ ϕ ( x ) , ϕ ( y )] = 0(22) ϕ ( y )( a · c ) = a · ( ϕ ( y ) c )(23) a · ( ϕ ( z ) b ) = b · ( ϕ ( z ) a )(24) ϕ ( z ) ω ( x, y ) − ϕ ( y ) ω ( x, z ) = ω ( x · z, y ) − ω ( x · y, z )(25) ω ( x, y ) · c + ϕ ( x · y ) c = ϕ ( y ) ϕ ( x ) c (26) [ ϕ ( x ) , ϕ ( y )] = 0(27) ϕ ( z )( a · b ) = ( ϕ ( z ) a ) · b (28) ( ϕ ( x ) c ) · b = ( ϕ ( x ) b ) · c (29) for all a, b, c ∈ a and x, y, z ∈ b .Proof. Let u = ( a, x ) , v = ( b, y ) , w = ( c, z ) denote three arbitrary elements of g . Let us firstconsider the equation (5) for the product, i.e., [ u, v ] = u ◦ v − v ◦ u . Using (16), (17) and thecommutativity of the LR-product in a we obtain[ u, v ] = ( ϕ ( x ) b − ϕ ( y ) a + Ω( x, y ) , [ x, y ]) u ◦ v − v ◦ u = (( ϕ ( x ) − ϕ ( x )) b − ( ϕ ( y ) − ϕ ( y )) a + ω ( x, y ) − ω ( y, x ) , [ x, y ]) . Suppose that the two expressions are equal for all a, b ∈ a and x, y ∈ b . For a = b = 0 we obtain ω ( x, y ) − ω ( y, x ) = Ω( x, y ). Taking this into account, a = 0 implies ϕ ( x ) − ϕ ( x ) = ϕ ( x ).Conversely, equations (18) and (19) imply (5).A similar computations shows that (1) corresponds to the equations 20 , . . . ,
24, and (2) corre-sponds to 25 , . . . , (cid:3)
Corollary 5.2.
Assume that g = a ⋊ ϕ b is a semidirect product of an abelian Lie algebra a and a Lie algebra b by a representation ϕ : b → End( a ) = Der( a ) , i.e., we have a split exactsequence → a ι −→ g π −→ b → . If b admits an LR-structure ( x, y ) x · y such that ϕ ( x · y ) = 0 for all x, y ∈ b , then also g admits an LR-structure.Proof. Because the short exact sequence is split, the 2-cocycle Ω in the Lie bracket of g istrivial, i.e., Ω( x, y ) = 0. Let a · b = 0 be the trivial product on a and take ϕ = 0, ϕ = ϕ and ω ( x, y ) = 0. Assume that ( x, y ) x · y is an LR-product. Then all conditions of Proposition5.1 are satisfied except for (26) and (22) requiring ϕ ( x · y ) = 0[ ϕ ( x ) , ϕ ( y )] = 0 . But we have (26) by assumption, and since ϕ is a representation it follows0 = ϕ ( x · y − y · x ) = ϕ ([ x, y ]) = [ ϕ ( x ) , ϕ ( y )] . R-ALGEBRAS 15
Hence (17) defines an LR-structure on g , given by( a, x ) ◦ ( b, y ) = ( ϕ ( x ) b, x · y ) . (cid:3) Corollary 5.3.
Suppose that the LR-products on a and b are trivial. Hence b is also abelian.Then (17) defines an LR-structure on g if and only if the following conditions hold: ω ( x, y ) − ω ( y, x ) = Ω( x, y )(30) ϕ ( x ) − ϕ ( x ) = ϕ ( x )(31) ϕ ( x ) ω ( y, z ) = ϕ ( y ) ω ( x, z )(32) ϕ ( x ) ϕ ( y ) = 0(33) [ ϕ ( x ) , ϕ ( y )] = 0(34) ϕ ( z ) ω ( x, y ) = ϕ ( y ) ω ( x, z )(35) ϕ ( x ) ϕ ( y ) = 0(36) [ ϕ ( x ) , ϕ ( y )] = 0(37)We can apply this corollary as follows: Proposition 5.4.
Let g = ( a , b , ϕ, Ω) be an extension in which both a and b are abelian. If thereexists an e ∈ b such that ϕ ( e ) ∈ End( a ) is an isomorphism, then g admits an LR-structure. Infact, in that case (17) defines an LR-product, where ϕ = 0 , ϕ = ϕ , the product on a and b istrivial, and ω ( x, y ) = ϕ ( e ) − ϕ ( x )Ω( e, y ) . Proof.
We have to show that the above conditions of corollary 5.3 are satisfied. Applying ϕ ( e ) − to (15) with z = e it follows Ω( x, y ) − ϕ ( e ) − ϕ ( x )Ω( e, y ) + ϕ ( e ) − ϕ ( y )Ω( e, x ) = 0. This justmeans that Ω( x, y ) = ω ( x, y ) − ω ( y, x ). Furthermore we have, since ϕ ( x ) ϕ ( y ) = ϕ ( y ) ϕ ( x ) forall x, y ∈ b , ϕ ( x ) ω ( y, z ) − ϕ ( y ) ω ( x, z ) = ϕ ( x ) ϕ ( e ) − ϕ ( y )Ω( e, z ) − ϕ ( y ) ϕ ( e ) − ϕ ( x )Ω( e, z )= ϕ ( e ) − ( ϕ ( x ) ϕ ( y ) − ϕ ( y ) ϕ ( x ))Ω( e, z )= 0 . All the other conditions follow trivially from ϕ = 0. Hence the product defines an LR-structure. (cid:3) References [1] O. Baues:
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Complete left-invariant affine structures on nilpotent Lie groups . J. Differential Geom. (1986),no. 3, 373–394.[9] Bjorn Poonen: Isomorphism types of commutative algebras of finite rank over an algebraically closed field .Preprint (2006).
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