aa r X i v : . [ m a t h . R A ] J a n LATTICE ISOMORPHISMS OF LEIBNIZ ALGEBRASDAVID A. TOWERSDepartment of Mathematics and StatisticsLancaster UniversityLancaster LA1 [email protected]
Abstract
Leibniz algebras are a non-anticommutative version of Lie algebras.They play an important role in different areas of mathematics andphysics and have attracted much attention over the last thirty years.In this paper we investigate whether conditions such as being a Liealgebra, cyclic, simple, semisimple, solvable, supersolvable or nilpotentin such an algebra are preserved by lattice isomorphisms.
Mathematics Subject Classification 2000 : 17B05, 17B20, 17B30, 17B50.
Key Words and Phrases : Lie algebras, Leibniz algebras, cyclic, simple,semisimple, solvable, supersolvable, nilpotent, lattice isomorphism.
An algebra L over a field F is called a Leibniz algebra if, for every x, y, z ∈ L ,we have [ x, [ y, z ]] = [[ x, y ] , z ] − [[ x, z ] , y ]In other words the right multiplication operator R x : L → L : y [ y, x ] is aderivation of L . As a result such algebras are sometimes called right Leibniz1lgebras, and there is a corresponding notion of left
Leibniz algebras, whichsatisfy [ x, [ y, z ]] = [[ x, y ] , z ] + [ y, [ x, z ]] . Clearly the opposite of a right (left) Leibniz algebra is a left (right) Leibnizalgebra, so, in most situations, it does not matter which definition we use.Leibniz algebras which satisfy both the right and left identities are sometimescalled symmetric
Leibniz algebras.Every Lie algebra is a Leibniz algebra and every Leibniz algebra satis-fying [ x, x ] = 0 for every element is a Lie algebra. They were introduced in1965 by Bloh ([7]) who called them D -algebras, though they attracted morewidespread interest, and acquired their current name, through work by Lo-day and Pirashvili ([18], [19]). They have natural connections to a variety ofareas, including algebraic K -theory, classical algebraic topology, differentialgeometry, homological algebra, loop spaces, noncommutative geometry andphysics. A number of structural results have been obtained as analogues ofcorresponding results in Lie algebras.The Leibniz kernel is the set I = span { x : x ∈ L } . Then I is thesmallest ideal of L such that L/I is a Lie algebra. Also [
L, I ] = 0.We define the following series: L = L, L k +1 = [ L k , L ]( k ≥
1) and L (0) = L, L ( k +1) = [ L ( k ) , L ( k ) ]( k ≥ . Then L is nilpotent of class n (resp. solvable of derived length n ) if L n +1 =0 but L n = 0 (resp. L ( n ) = 0 but L ( n − = 0) for some n ∈ N . It isstraightforward to check that L is nilpotent of class n precisely when everyproduct of n + 1 elements of L is zero, but some product of n elements isnon-zero.The nilradical , N ( L ), (resp. radical , R ( L )) is the largest nilpotent(resp. solvable) ideal of L .The set of subalgebras of a nonassociative algebra forms a lattice underthe operations of union, ∪ , where the union of two subalgebras is the subal-gebra generated by their set-theoretic union, and the usual intersection, ∩ .The relationship between the structure of a Lie algebra L and that of thelattice L ( L ) of all subalgebras of L has been studied by many authors. Muchis known about modular subalgebras (modular elements in L ( L )) througha number of investigations including [1, 12, 14, 27, 28, 29]. Other latticeconditions, together with their duals, have also been studied. These includesemimodular, upper semimodular, lower semimodular, upper modular, lowermodular and their respective duals (see [8] for definitions). For a selectionof results on these conditions see [9, 13, 15, 16, 17, 20, 24, 26, 30, 31].2he subalgebra lattice of a Leibniz algebra, however, is rather differ-ent; in a Lie algebra every element generates a one-dimensional subalgebra,whereas in a Leibniz algebra elements can generate subalgebras of any di-mension. So, one could expect different results to hold for Leibniz algebrasanf this has been shown to be the case in [21]..Of particular interest is the extent to which important classes of Leibnizalgebras are determined by their subalgebra lattices. In order to investigatethis question we introduce the notion of a lattice isomorphism. If we denotethe subalgebra lattice of L by L ( L ), then a lattice isomorphism from L to L ∗ is a bijective map θ : L ( L ) → L ( L ∗ ) such that θ ( A ∪ B ) = θ ( A ) ∪ θ ( B )and θ ( A ∩ B ) = θ ( A ) ∩ θ ( B ) for all A, B ∈ L ( L ). If L is a Lie algebra overa field of characteristic zero the following were proved in [23]. Theorem 1.1 (i) If L is simple then either(a) L ∗ is simple, or(b) L is three-dimensional non-split simple and L ∗ is two-dimensional.(ii) If L is semisimple then either(a) L ∗ is semisimple, or(b) L is three-dimensional non-split simple and L ∗ is two-dimensional.(iii) If dim L, L ∗ > and R is the radical of L , then R ∗ is the radical of L ∗ .(iv) If L is supersolvable of dimension > , then L ∗ is supersolvable. In [15] the following was proved.
Theorem 1.2 If L is a solvable Lie algebra over a perfect field of charac-teristic different from , , then either(i) L ∗ is solvable, or(ii) L ∗ is three-dimensional non-split simple. We say that a Lie algebra L is almost abelian if it is a split extension L = L ˙+ F a with ad a acting as the identity map on the abelian ideal L ; L is quasi-abelian if it is abelian or almost abelian. The quasi-abelian Liealgebras are precisely the ones in which every subspace is a subalgebra. Thefollowing is well-known and easy to show.3 roposition 1.3 If L is a quasi-abelian Lie algebra over a field of char-acteristic zero then L ∗ is quasi-abelian unless dim L = 2 and L ∗ is three-dimensional non-split simple. In this paper we consider corresponding results for Leibniz algebras.First, in section two, we show that cyclic Leibniz algebras are characterisedby their subalgebra lattice, and that a non-Lie Leibniz algebra cannot belattice isomorphic to a Lie algebra. In section three we see that if L is anon-Lie simple or semisimple Leibniz algebra then so is L ∗ . In section four,it is shown that if L is a non-Lie solvable or supersolvable Leibniz algebrathen so is L ∗ . It is also proved that the radical of a non-Lie Leibniz algebrais preserved by lattice isomorphisms. The final section is devoted to showingthat if L is a non-Lie nilpotent Leibniz algebra then so is L ∗ . Most of theabove results are over fields of characteristic zero.Throughout, L will denote a finite-dimensional Leibniz algebra over afield F . Algebra direct sums will be denoted by ⊕ , whereas vector spacedirect sums will be denoted by ˙+. The notation ‘ A ⊆ B ’ will indicate that A is a subset of B , whereas ‘ A ⊂ B ’ will mean that A is a proper subset of B . If A and B are subalgebras of L we will write h A, B i for A ∪ B .The centre of L is Z ( L ) = { z ∈ L | [ z, x ] = [ x, z ] = 0 for all x ∈ L } . TheFrattini ideal of L , φ ( L ), is the largest ideal of L contained in all maximalsubalgebras of L . The only previous paper that we are aware of on this topic is by Barnes ([5]).The following example shows that the Leibniz kernel of a non-Lie Leibnizalgebra is not necessarily preserved by a lattice isomorphism.
Example
Let L = F b + F a where the only non-zero products are [ b, b ] = a , [ a, b ] = a . Then the only subalgebras of L are , F a , F ( b − a ) and L , and I = F a . Then we can define a lattice automorphism of L which interchanges F a and F ( b − a ) , and the latter is not an ideal of L as [ b, b − a ] = a . Barnes called the above example the diamond algebra because of thestructure of its lattice of subalgebras as a Hasse diagram, but that name hassince been used for a different Leibniz algebra. He further showed that thisexample is exceptional in the following result.4 heorem 2.1 ([5, Theorem 3.1]) Let
L, L ∗ be Leibniz algebras with Leibnizkernels I, I ∗ respectively, and let θ : L → L ∗ be a lattice isomorphism.Suppose that dim L ≥ . Then θ ( I ) = I ∗ . However, this paper does not appear to have been followed by furtherinvestigations into the subalgebra structure of a Leibniz algebra. Theorem2.1, of course, has an immediate corollary.
Corollary 2.2
Let L be a non-Lie Leibniz algebra. Then L cannot be latticeisomorphic to a Lie algebra L ∗ . Proof.
If dim L ≥ I = 0 if and only if I ∗ = 0. If dim L = 2there are only two possibilities for L , both of them cyclic with basis x, x .In the first, [ x , x ] = 0 and the only proper subalgebra is F x , and in thesecond, [ x , x ] = x and the only proper subalgebras are F x and F ( x − x ).However, every Lie algebra of dimension greater than one has more thantwo proper subalgebras.There is no non-Lie Leibniz algebra of dimension one. (cid:3) A Leibniz algebra L is called cyclic if it is generated by a single element.In this case, L has a basis x, x , . . . , x n ( n >
1) and products [ x i , x ] = x i +1 for 1 ≤ i ≤ n −
1, [ x n , x ] = α x + . . . + α n x n , all other products being zero.Then we have the following. Theorem 2.3 If L is a cyclic Leibniz algebra over an infinite field F , then L ∗ is also a cyclic Leibniz algebra of the same dimension. Proof.
Over an infinite field a Leibniz algebra is cyclic if and only if it hasfinitely many maximal subalgebras, by [21, Corollary 2.3]. Moreover, thelength of a maximal chain of subalgebras of a cyclic algebra is equal to itsdimension. (cid:3)
Corollary 2.4 If L is a nilpotent cyclic Leibniz algebra, then L ∗ ∼ = L . Proof.
A nilpotent cyclic Leibniz algebra has only one maximal subalgebra,namely I , its Leibniz kernel. It follows that L ∗ is nilpotent of the samedimension. Note that the restriction on the field is unnecessary here, since,if M is the only maximal subalgebra of L and x ∈ L \ M , we must have L = h x i . (cid:3) Note that, in both of the above results, if L = h x i then L ∗ = h x ∗ i ,since x does not belong to any of the maximal subalgebras of L , and this isinherited by x ∗ in L ∗ . 5 roposition 2.5 Let L = A ˙+ F x be a non-Lie Leibniz algebra in which A is a minimal abelian ideal of L and x = 0 . Then L is cyclic and A = I . Proof.
Since L is not a Lie algebra, A = I , [ L, A ] = 0 and [
A, L ] = 0, so[ A, x ] = A . Let 0 = a ∈ A . Then ( x + a ) n = R n − x ( a ) for n ≥
2, whichimplies that [( x + a ) n , x ] = R nx ( a ) ∈ h x + a i for n ≥
1. Hence h x + a i ∩ A is anideal of L and so equals A or 0. However, the latter implies that [ a, x ] = 0,whence A = F a and [
A, x ] = 0, a contradiction. It follows that L = h x + a i . (cid:3) The following useful result was proved by Barnes in [4]. Note that we havemodified the statement to take account of the fact that Barnes’ result isstated for left Leibniz algebras and we are dealing with right Leibniz alge-bras.
Lemma 3.1
Let A be a minimal ideal of the Leibniz algebra L . Then [ L, A ] = 0 or [ x, a ] = − [ a, x ] for all a ∈ A , x ∈ L . A Leibniz algebra L is called simple if its only ideals of L are 0, I and L , and L = I . If L/I is a simple Lie algebra then L is not necessarilya simple Leibniz algebra. It is said to be semisimple if R ( L ) = I . Thisdefinition agrees with that of a semisimple Lie algebra, since, in this case, I = 0. Semisimple Leibniz algebras are not necessarily direct sums of simpleLeibniz algebras (see [10] or [2]).We have the following version of Levi’s Theorem. Theorem 3.2 (Barnes [3]) Let L be a finite-dimensional Leibniz algebraover a field of characteristic . Then there is a semisimple Lie subalgebra S of L such that L = S ˙+ R ( L ) . We shall need the following result which was proved by Gein in [11, p.23].
Lemma 3.3
Let S be a three-dimensional non-split simple Lie algebra, andlet R be an irreducible S -module. Then, for any s ∈ S , R has an ad s -invariant subspace of dimension less than or equal to two. If U is a subalgebra of L and 0 = U < U < . . . < U n = U is a maximalchain of subalgebras of U we will say that U has length n .6 heorem 3.4 Let L = S ˙+ A be a Leibniz algebra over a field of character-istic zero, where S is a three-dimensional non-split simple Lie algebra and A is a minimal abelian ideal of L . Then L ∗ has a simple Lie subalgebra. Proof.
Suppose that L ∗ does not have a simple Lie subalgebra. Then R ( L ∗ ) = 0, by Theorem 3.2, and so L ∗ has a minimal abelian ideal B ∗ .As S ∗ is a maximal subalgebra of L ∗ we must have that L ∗ = S ∗ ˙+ B ∗ . Ifdim A = 1 we have that L = S ⊕ A is a Lie algebra and hence, so is L ∗ ,giving that S ∗ ∼ = S , by [23, Lemma 3.3] and contradicting our supposition.Hence dim A ≥ L are of two types: they are isomorphic to S , and so have length 2, or they are of the form F s ˙+ A , where s ∈ S , and soare solvable of length at least 3. Moreover, A is the intersection of those ofthe second type. The same must be true of the maximal subalgebras of L ∗ and so B ∗ = A ∗ and L ∗ = S ∗ ˙+ A ∗ . Also, dim S ∗ = 2, by Theorem 1.1. Now φ ( L ∗ ) = ( φ ( L )) ∗ = 0, so L ∗ = A ∗ ˙+ C ∗ , where C ∗ is abelian, by [6, Corollary2.9]. Since S ∗ ∼ = C ∗ , we have that S ∗ is abelian.Let 0 = s ∗ ∈ S ∗ , 0 = a ∗ ∈ A ∗ and let f ( θ ) be the polynomial of smallestdegree for which f ( R s ∗ )( a ∗ ) = 0. It follows from the fact that S ∗ is abelianthat { x ∗ ∈ A ∗ : f ( R s ∗ )( x ∗ ) = 0 } is an ideal of L ∗ , and hence that it coincideswith A ∗ . Clearly then f ( θ ) is the minimum polynomial of R s ∗ | A ∗ .Suppose that there is an s ∗ ∈ S ∗ for which the minimum polynomial for R s ∗ has degree two. and let this polynomial be f ( θ ) = θ − λ θ − λ . Pick s ∗ ∈ S ∗ linearly independent of s ∗ . Then[[ a ∗ , s ∗ ] , s ∗ ] = λ a ∗ + λ [ a ∗ , s ∗ ] and[[ a ∗ , s ∗ ] , s ∗ ] = α a ∗ + α [ a ∗ , s ∗ ] so[[ a ∗ , s ∗ ] , s ∗ ] = [ a ∗ , [ s ∗ , s ∗ ]] + [[ a ∗ , s ∗ ] , s ∗ ] = [[ a ∗ , s ∗ ] , s ∗ ]= β a ∗ + β [ a ∗ , s ∗ ] + β [ a ∗ , s ∗ ] , since [[ a ∗ , s ∗ + s ∗ ] , s ∗ + s ∗ ] ∈ F a ∗ + F [ a ∗ , s ∗ + s ∗ ]. Now[[[ a ∗ , s ∗ ] , s ∗ ] , s ∗ ] = λ [ a ∗ , s ∗ ] + λ [[ a ∗ , s ∗ ] , s ∗ ] , so ( β λ + β β ) a ∗ + ( β + β λ )[ a ∗ , s ∗ ] + β [ a ∗ , s ∗ ]= λ β a ∗ + λ β [ a ∗ , s ∗ ] + ( λ + λ β )[ a ∗ , s ∗ ] . Since f ( θ ) is irreducible, β = λ + λ β and so [ a ∗ , s ∗ ] = γ a ∗ + γ [ a ∗ , s ∗ ].Hence A ∗ is two dimensional. 7ut A = F a + F [ a, s ]. Choose s , s to be elements of S such that s, s , s are linearly independent. Then [ a, s ] = αa + β [ a, s ] and [ a, s ] = γa + δ [ a, s ]for some α, β, γ, δ ∈ F . Thus [ a, s − βs ] = αa and [ a, s − δs ] = γa . But s − βs and s − δs are linearly independent, so[ a, S ] = [ a, < s − βs, s − δs > ] ⊆ F a and A is one dimensional, a contradiction. (cid:3) Corollary 3.5
Let L be a non-Lie semisimple Leibniz algebra over a fieldof characteristic zero. Then L ∗ is a non-Lie semisimple Leibniz algebra. Proof.
We have that I = 0, so L = I ˙+ S where S is a semisimple Lie algebra,by Theorem 3.2. Then L ∗ /I ∗ is a semisimple Lie algebra or dim L ∗ /I ∗ = 2and S is 3-dimensional non-split simple, by Theorem 1.1(ii).Suppose that the latter holds, so L ∗ is solvable. Let A be a minimal idealof L inside I and put B = A ˙+ S . Then B ∗ has a simple Lie subalgebra, byTheorem 3.4 and L ∗ cannot be solvable. Hence the former holds and L ∗ isa non-Lie semsimple Leibniz algebra. (cid:3) A subalgebra U of L is called upper semi-modular if U is a maximalsubalgebra of h U, B i for every subalgebra B of L such that U ∩ B is maximalin B . Using this concept we have a further corollary. Corollary 3.6
Let L be a non-Lie simple Leibniz algebra over a field ofcharacteristic zero. Then L ∗ is a non-Lie simple Leibniz algebra. Proof.
We have that L = I ˙+ S where S is a simple Lie subalgebra of L and I = 0. If L ∗ /I ∗ is not simple then S must be three-dimensional non splitsimple, by Theorem 1.1(i), and we get a contradiction as in the previouscorollary.Let 0 = A ∗ be an ideal of L ∗ . Suppose first that A ∗ ⊆ I ∗ . Then A is an upper semi-modular subalgebra of L with A ⊆ I . Let s ∈ S .Then A ∩ F s = 0 is a maximal subalgebra of
F s . Hence A is a maximalsubalgebra of C = h A, s i . Now A ⊆ C ∩ I ⊂ C , so A = C ∩ I . Thus[ s, A ] , [ A, s ] ⊆ C ∩ I = A , so A is an ideal of L , whence A = I . It followsthat A ∗ = I ∗ .Next, suppose that A ∗ I ∗ . Then I ∗ + A ∗ = L ∗ and I ∗ ∩ A ∗ = I ∗ or0, by the previous paragraph. The former implies that A ∗ = L ∗ ; the lattergives that L ∗ = I ∗ ⊕ A ∗ giving I ∗ = 0 and L ∗ = A ∗ again.Clearly ( L ∗ ) = I ∗ , so L ∗ is a non-Lie simple Leibniz algebra. (cid:3) Solvable and supersolvable Leibniz algebras
Proposition 4.1
Let L be a non-Lie solvable Leibniz algebra over a field ofcharacteristic zero. Then L ∗ is a non-Lie solvable Leibniz algebra. Proof.
Let L be a minimal counter-example. Then L ∗ has a semisimpleLie subalgebra S ∗ , and so S ( = L ) must be two dimensional and S ∗ must bethree-dimensional non-split simple. Moreover, L ∗ = S ∗ ˙+ A ∗ , where A ∗ is aminimal ideal of L ∗ , since, otherwise, this is a smaller counter-example. Butthen L has a simple subalgebra, by Theorem 3.4, a contradiction. (cid:3) Lemma 4.2
Let L be a Leibniz algebra over a field of characteristic zero.Then the radical, R , of L is the intersection of the maximal solvable subal-gebras of L . Proof.
Let Γ be the intersection of the maximal solvable subalgebras of L .Then R ⊆ Γ. Furthermore, Γ is invariant under all automorphisms of L ,and hence is invariant under all derivations of L , by [22, Corollary 3.2]. Itfollows that Γ is a right ideal of L . But [ x, y ] + [ y, x ] ∈ I ⊆ Γ for all x ∈ L , y ∈ Γ, so Γ is an ideal of L , whence Γ ⊆ R . (cid:3) Then we have the following corollaries to Proposition 4.1.
Corollary 4.3
Let L be a non-Lie Leibniz algebra over a field of character-istic zero, and let R be the radical of L . Then R ∗ is the radical of L ∗ . Proof.
Let U be a maximal solvable subalgebra of L . If U is non-Lie then U is solvable, by Proposition 4.1. If U is Lie, then U ∗ is solvable, unlessdim U = 2 and U ∗ is three-dimensional non-split simple. If R ∗ = 0 then L ∗ ,and hence L , is a Lie algebra, a contradiction. Hence dim R ∗ = 0.Moreover, R ⊆ U . If R = U then R is a maximal solvable subalgebraof L , which is impossible unless R = L . But then the result follows fromProposition 4.1. So suppose dim R = 0 ,
1. The former implies that L is asemisimple Lie algebra, which is impossible. The latter implies that L = S ⊕ F a , where S is a semisimple Lie algebra. But this is also a Lie algebraand so is impossible.It follows that U ∗ must be a maximal solvable subalgebra of L ∗ . Theresult now follows from Lemma 4.2. (cid:3) A subalgebra U of L is called lower semi-modular in L if U ∩ B is maximalin B for every subalgebra B of L such that U is maximal in h U, B i . We saythat L is lower semi-modular if every subalgebra of L is lower semi-modularin L . 9 orollary 4.4 Let L be a non-Lie supersolvable Leibniz algebra over a fieldof characteristic zero. Then L ∗ is supersolvable. Proof.
We have that L is solvable and lower semi-modular, by [21, Proposi-tion 5.1]. It follows from Proposition 4.1 that the same is true of L ∗ . Hence L ∗ is supersolvable, by [21, Proposition 5.1] again. (cid:3) A Lie algebra L is callled almost nilpotent of index n if it has a basis { x ; e , . . . , e r ; . . . ; e n , . . . , e nr n } such that − [ e ij , x ] = [ x, e ij ] = e ij + e i +1 ,j for 1 ≤ i ≤ n − , ≤ j ≤ r i , − [ e nj , x ] = [ x, e nj ] = e nj and r j ≤ r j +1 for 1 ≤ j ≤ n − Theorem 5.1
Let L be a nilpotent Lie algebra of index n and of dimensiongreater than two for which L ∗ is not nilpotent, over a field of characteris-tic zero. Then L ∗ is almost nilpotent of index n . Moreover, every almostnilpotent Lie algebra is lattice isomorphic to a nilpotent Lie algebra. For non-Lie Leibniz algebras we have the following result.
Theorem 5.2
Let L be a nilpotent non-Lie Leibniz algebra over a field ofcharacteristic zero. Then L ∗ is a non-Lie nilpotent Leibniz algebra. First we need a lemma.
Lemma 5.3
Let L be a nilpotent Leibniz algebra and let W = F w be aminimal ideal of L contained in the Leibniz kernel, I , of L . Then W ∗ is aminimal ideal of L ∗ and W ∗ ⊆ Z ( L ∗ ) . Proof.
Suppose that x / ∈ I , where x n = 0 but x n − = 0. Then S = h x, W i = h x i + W and h x i ∩ W = 0 or 1. The former implies that h x i is amaximal subalgebra of S , whence h x ∗ i is a maximal subalgebra, and hence10n ideal, of S ∗ = h x ∗ , W ∗ i . The latter implies that W ⊆ h x i , whence W ∗ ⊆h x ∗ i . In either case, [ w ∗ , x ∗ ] ∈ h x ∗ i ∩ I ∗ . Hence [ w ∗ , x ∗ ] = P ni =2 λ i ( x ∗ ) i .Suppose that λ = 0 and consider h λ x − w i . If W ⊆ h x i then W = F x n and h λ x − w i = h x i . If W
6⊆ h x i then ( λ x − w ) k = λ k x k − λ k − µ k − w , where[ w, x ] = µw . In either case, h λ x − w i is a cyclic subalgebra of dimension n .However,( λ x ∗ − w ∗ ) = λ ( x ∗ ) − λ n X i =2 λ i ( x ∗ ) i = λ n X i =3 λ i ( x ∗ ) i , so h λ x ∗ − w ∗ i is a cyclic subalgebra of dimension n −
1, contradictingCorollary 2.4. It follows that λ = 0. A similar argument shows that λ i = 0for all 2 ≤ i ≤ n , so [ w ∗ , x ∗ ] = 0. Also, [ x ∗ , w ∗ ] = 0, since w ∗ ∈ I ∗ , fromwhich the result follows. (cid:3) Now we can prove Theorem 5.2.
Proof.
We have
L/L is abelian and L = φ ( L ), so L ∗ /φ ( L ∗ ) is almostabelian or three-dimensional non-split simple. The latter is impossible, as itwould imply that L ∗ = φ ( L ∗ ) ˙+ S ∗ = S ∗ , where S ∗ is three-dimensional non-split simple, by Theorem 3.2. But then L is a two-dimensional Lie algebra,by Theorem 2.2, a contradiction. It follows that L ∗ /φ ( L ∗ ), and hence L ∗ , issupersolvable (see [5, Theorems 3.9 and 5.2]) and has nilradical N ∗ = φ ( L ∗ ) + F e ∗ + · · · + F e ∗ r . Let L be a minimal counter-example, so L is non-Lie and nilpotent, but L ∗ is not nilpotent.Now I is non-zero, so choose a minimal ideal W = F w of L inside I .We have that W ∗ is a minimal ideal of L ∗ inside Z ( L ∗ ), by Lemma 5.3.Then L ∗ /W ∗ is not nilpotent, so L/W is a Lie algebra and L ∗ /W ∗ is almostnilpotent. Hence there is a basis { x ∗ ; e ∗ , . . . , e ∗ r ; . . . ; e ∗ n , . . . , e ∗ nr n , w ∗ } for L ∗ such that[ x ∗ , e ∗ ij ] = e ∗ ij + e ∗ i +1 ,j + λ ij w ∗ for 1 ≤ i ≤ n − , ≤ j ≤ r i , [ x ∗ , e ∗ nj ] = e ∗ nj + λ nj w ∗ and r j ≤ r j +1 for 1 ≤ j ≤ n − , where λ ij ∈ F , I ∗ = F w ∗ and ( N ∗ ) ⊆ W ∗ . Let M ∗ be spanned by all ofthe basis vectors for L ∗ apart from e ∗ . Then M ∗ is not nilpotent and has11ilradical F ∗ spanned by all of the basis vectors apart from e ∗ and x ∗ . Bythe minimality, we must have that M is Lie and M ∗ is almost nilpotent, so( F ∗ ) = 0, ( x ∗ ) = 0 and [ e ∗ ij , x ∗ ] = − [ x ∗ , e ∗ ij ] for all of the e ∗ ij ’s apart from e ∗ . But also [ e ∗ , x ∗ ] = − e ∗ − e ∗ + µw ∗ for some µ ∈ F , so[ e ∗ , x ∗ ] = [[ x ∗ , e ∗ ] , x ∗ ] − [ e ∗ , x ∗ ]= [ x ∗ , [ e ∗ , x ∗ ]] + [( x ∗ ) , e ] + [ x ∗ , e ∗ ]= − [ x ∗ , e ∗ ] − [ x ∗ , e ∗ ] + [ x ∗ , e ∗ ] = − [ x ∗ , e ∗ ]We now claim that ( N ∗ ) = 0. It suffices to show that [ N ∗ , e ∗ ] = 0,which we do by a backwards induction argument. We have, for any f ∗ ∈ F ∗ ,[ f ∗ , e ∗ ] = [ f ∗ , [ x ∗ , e ∗ ] − e ∗ − λ w ∗ ] = [ f ∗ , [ x ∗ , e ∗ ]]= [[ f ∗ , x ∗ ] , e ∗ ] − [[ f ∗ , e ∗ ] , x ∗ ] = [[ f ∗ , x ∗ ] , e ∗ ] , (1)since [ f ∗ , e ∗ ] ∈ W ⊆ Z ( L ∗ ). Now putting f ∗ = e ∗ nj gives[ e ∗ nj , e ∗ ] = [[ e ∗ nj , x ∗ ] , e ∗ ] = − [ e ∗ nj , e ∗ ] , whence [ e ∗ nj , e ∗ ] = 0. So now suppose that [ e ∗ ij , e ∗ ] = 0 for some 2 ≤ i ≤ n .Putting f ∗ = e ∗ i − ,j (( i − , j ) = (2 , e ∗ i − ,j , e ∗ ] = [[ e ∗ i − ,j , x ∗ ] .e ∗ ] = − [ e ∗ i − ,j , e ∗ ]which, again, yields that [ e ∗ i − ,j , e ∗ ] = 0. Finally, note that, if we now put f ∗ = e ∗ , then (1) remains valid, so ( e ∗ ) = 0 and ( N ∗ ) = 0.Now replace e ∗ nj by e ∗ nj + λ nj w ∗ , e ∗ ij by e ∗ ij + ( − n − i λ i +1 ,j w ∗ to see that L ∗ is almost nilpotent and L is a Lie algebra, a contradiction. Hence theresult holds. (cid:3) References [1]
R.K. Amayo and J. Schwarz , ‘Modularity in Lie Algebras’,
Hi-roshima Math. J. (1980), 311-322.[2] Sh. Ayupov, K. Kudaybergenov, B. Omirov andK. Zhao , ‘Semisimple Leibniz algebras, their derivationsand automorphisms’,
Linear and Multilinear Alg. (2019),https://doi.org/10.1080/03081087.2019.1567674.[3]
D.W. Barnes , ‘On Levi’s Theorem for Leibniz algebras’,
Bull. Aust.Math. Soc. (2012), 184-185. 124] D.W. Barnes , ‘Some theorems on Leibniz algebras’,
Comm. Alg. (2011), 2463-2472.[5]
D.W. Barnes , ‘Lattices of subalgebras of Leibniz algebras’,
Comm.Alg.
40 (11) (2012), 4330-4335.[6]
C. Batten Ray, L. Bosko-Dunbar, A. Hedges, J.T. Hird, K.Stagg and E. Stitzinger , ‘A Frattini theory for Leibniz algebras’,
Comm. Alg. (2013), 1547–1557.[7]
A. Bloh . ‘On a generalization of the concept of Lie algebra’.
Dokl. Akad.Nauk SSSR. (1965), 471–473.[8]
K. Bowman and D.A.Towers , ‘Modularity conditions in Lie alge-bras’,
Hiroshima Math. J. (1989), 333-346.[9] K. Bowman and V.R. Varea , ‘Modularity* in Lie algebras’,
Proc.Edin. Math. Soc. (1997), 99-110.[10]
I. Demir, K.C. Misra and E. Stitzinger , ‘On some structures ofLeibniz algebras’, in Recent Advances in Representation Theory, Quan-tum Groups, Algebraic Geometry, and Related Topics,
ContemporaryMathematics , . Amer. Math. Soc., Providence, RI, (2014), 41–54.[11] A.G. Gein , ‘Projections of a Lie algebra of characteristic zero’,
Izvestijavys˜s. ucebn. Zaved. Mat. (1978), 26-31.[12]
A.G. Gein , ‘Modular rule and relative complements in the lattice ofsubalgebras of a Lie algebra’,
Sov. Math. (1987), 22-32; translatedfrom
Izv. Vyssh. Uchebn. Zaved. Mat. (1987), 18-25.[13] A.G. Gein , ‘Semimodular Lie algebras’,
Siberian Math. J. (1976),243-248; translated from Sibirsk Mat. Z. (1976), 243-248.[14] A.G. Gein , ‘On modular subalgebras of Lie algebras’,
Ural Gos. Univ.Mat. Zap. (1987), 27-33.[15] A.G. Gein and V.R. Varea , ‘Solvable Lie algebras and their subalge-bra lattices’,
Comm. Alg. (1992), 2203-2217. Corrigenda:
Comm.Alg. (1995), 399-403.[16]
B. Kolman , ’Semi-modular Lie algebras’,
J. Sci. Hiroshima Univ. Ser.A-I (1965), 149-163. 1317] A.A. Lashi , ‘On Lie algebras with modular lattices of subalgebras’,
J.Algebra (1986), 80-88.[18] J.-L. Loday , ‘Une version non commutative des alg`ebres de Lie: lesalg`ebres de Leibniz’.
Enseign. Math. (2)
39 (3–4) (1993), 269–293.[19]
J.-L. Loday and T. Pirashvili , ‘Universal enveloping algebras ofLeibniz algebras and (co)homology’,
Math. Annalen
296 (1) (1993) 139–158.[20]
C. Scheiderer , ‘Intersections of maximal subalgebras in Lie algebras’,
J. Algebra (1987), 268-270.[21]
S. Siciliano and D.A. Towers , ‘On the subalgebra lattice of a Leib-niz algebra’, arXiv:2010.12254 (2020).[22]
D.A. Towers , ‘A Frattini theory for algebras’,
Proc. London Math.Soc. (3) (1973), 440–462.[23] D.A. Towers , ‘Lattice isomorphisms of Lie algebras’,
Math. Proc.Camb. Phil. Soc. (1981), 285-292.[24] D.A. Towers , ‘Semimodular subalgebras of a Lie algebra’,
J. Algebra (1986), 202-207.[25]
D.A. Towers , ‘Almost nilpotent Lie algebras’,
Glasgow Math. J. (1987), 7-11.[26] D.A. Towers , ‘On modular* subalgebras of a Lie algebra’,
J. Algebra (1997), 461-473.[27]
V.R. Varea , ‘Modular subalgebras, Quasi-ideals and inner ideals inLie Algebras of prime characteristic’,
Comm. Alg. (1993), 4195–4218.[28]
V.R. Varea , ‘The subalgebra lattice of a supersolvable Lie algebra’,
In Lecture Notes in Mathematics . Springer-Verlag: New York, 1989; Vol.1373, 81-92.[29]
V.R. Varea , ‘On modular subalgebras in Lie algebras of prime char-acteristic,,
Contemporary Math. (1990), 289-307.[30]
V.R. Varea , ‘Lower Semimodular Lie algebras’,
Proc. Edin. Math.Soc. (1999), 521-540. 1431] V.R. Varea , ‘Lie algebras whose maximal subalgebras are modular’,
Proc. Roy. Soc. Edinburgh Sect. A94