On the description of Leibniz algebras with nilindex n−3

ON THE DESCRIPTION OF THE LEIBNIZ ALGEBRAS WITH NILINDEX n−3 J.M. CABEZAS, L.M. CAMACHO, J.R. G´ OMEZ, B.A. OMIROV Abs ac . In his pape we p esen he classi ica ion o a subclass o na u- ally g aded Leibniz algeb as. These n-dimensional Leibniz algeb as ha e he cha ac e is ic sequence equal o (n−3,3).Fo his pu pose we use he so wa e Ma hema ica. AMS Subjec Classi ica ions (2000): 17A32, 17A36, 17A60, 17B70. Key wo ds: Lie algeb a, Leibniz algeb a, nilpo ence, na u al g ada ion, cha ac e- is ic sequence, p- ili o mlici y. 1. In oduc ion Leibniz algeb as a e one o he new algeb as in oduced by Loday [11], [12] in connec ion wi h he s udy o pe iodici y phenomena in algeb aic K- heo y. Leibniz algeb as ha e been in oduced as a ”non-an isymme ic” analogue o Lie algeb as. A Leibniz algeb a Lis a ec o space equipped wi h a b acke [-,-] sa is ying he iden i y [x, [y, z]] = [[x, y], z]−[[x, z], y]. I he an isymme ic ela ion is assumed, his iden i y is equi alen o he Jacobi iden i y. Hence, a Lie algeb a is a Leibniz algeb a. I is well known ha he na u al g ada ion o nilpo en Lie and Leibniz algeb as is e y help ul in in es iga ing hei s uc u al p ope ies. A ema kable ac o he na u ally g aded algeb as is he ela i e simplici y o he s udy o he cohomological p ope ies, (see o example [6]- [10] and [13]). Recen ly, some pape s a e ocused o he s udy o some in e es ing amilies o Leibniz algeb as, such as p- ili o m and quasi- ili o m Leibniz algeb as. These al- geb as ha e hei cha ac e is ic sequences equal o (n−p, 1,1, ..., 1) and (n−2,2) wi h dim(L) = n, [4]–[5]. Na u ally g aded p- ili o m Leibniz algeb as a e al eady classi ied in [2] and [4]. The classi ica ion o na u ally g aded nul- ili o m and ili o m Leibniz alge- b as eade can ind in [1]. The quasi- ili o m n-dimensional Leibniz algeb as ha e cha ac e is ic sequence (n−2,1,1) ( he case o 2- ili o m) o (n−2,2) [3] and [5]. Fo a gi en Leibniz algeb a Lwe de ine he descending cen al se ies as ollows: L1=L, Lk+1 = [Lk, L], k ≥1. I he e exis s a na u al numbe ssuch ha Ls= 0, hen he Leibniz algeb a Lis said o be nilpo en and minimal such numbe is called he nilindex o he algeb a L. Bellow we p esen a g ada ion closely ela ed o he descending cen al se ies. Le Lbe a nilpo en Leibniz algeb a wi h nilindex s. We pu Li=Li/Li+1 o 1≤i≤s−1,and g L =L1⊕L2⊕ · · · ⊕ Ls−1.I is easy o check embedding 1 2 J.M. CABEZAS, L.M. CAMACHO, J.R. G´ OMEZ, B.A. OMIROV [Li, Lj]⊆Li+jand he e o e, he algeb a g L is g aded algeb a, which is called he na u ally g aded Leibniz algeb a. Le xbe a nilpo en elemen o he se L L2. Fo he nilpo en ope a o o igh mul iplica ion Rxwe de ine a dec easing sequence C(x) = (n1, n2,...,nk), which consis s o he dimensions o Jo dan blocks o he ope a o Rx. On he se o such sequences we conside he lexicog aphic o de , ha is, C(x) = (n1, n2,...,nk)≤ C(y) = (m1, m2, . . . , ms)⇐⇒ he e exis s i∈Nsuch ha nj=mj o any j < i and ni< mi. The sequence C(L) = max C(x)x∈L L2is called cha ac e is ic sequence o he algeb a L. I C(L) = (1,1,...,1) hen e iden ly, he algeb a Lis abelian. The se R(L) = {x∈L|[y, x] = 0 o any y∈L}is said o be a igh annihila o o he algeb a L. In his wo k we classi y a subclass o na u ally g aded Leibniz algeb as wi h nilindex n−3.In case o Leibniz algeb as wi h nilindex equal o n−3, o he cha ac e is ic sequence we ha e he ollowing ee possibili ies: (n−3,1,1,1),(n−3,2,1) and (n−3,3). The i s one is 3- ili o m case. We will ocus ou a en ion on he s udy o hose wi h cha ac e is ic sequence (n−3,3). Th oughou all he wo k, we use he so wa e Ma hema ica. Since in he case o non-Lie Leibniz algeb as he skew- symme ic iden i y is no alid, his classi ica ion is e y complex and we should o e come he di icul ies, which need a lo o compu a ions. Using compu e p o- g ams is e y help ul o compu ing he Leibniz iden i y in low dimension and o mula e he gene aliza ions o he calcula ions, which a e p o ed o a bi a y ini e dimension. The used p og am can be ind in [5]. Some examples o he p o- g ams o a ious ypes o Leibniz algeb as classes a e in he ollowing Web si e: h p://pe sonal.us.es/j gomez. 2. Na u ally g aded Leibniz algeb as wi h cha ac e is ic sequence (n−3,3). Le Lbe a na u ally g aded n-dimensional Leibniz algeb a which cha ac e is ic sequence equal o (n−3,3). F om he de ini ion o he cha ac e is ic sequence, i ollows he exis ence o a basis {e1, e2,...,en}such ha elemen e1∈L L2and he ope a o o igh mul iplica ion Re1has one o he ollowing o ms: Jn−30 0J3,J30 0Jn−3 De ini ion 2.1. A na u ally g aded Leibniz algeb a Lwhich cha ac e is ic sequence is equal o (n−3,3), is called algeb a o he second ype i he e exis s a basic elemen e1∈L L2such ha he ope a o Re1has he o m: Jn−30 0J3; i Re1has he o he o m, hen i is called algeb a o he second ype. Since he classi ica ion o Leibniz algeb as o he second ype is mo e complica ed and i needs o use mo e o iginal echnics, i s we p esen he desc ip ion o he second ype. ON THE DESCRIPTION OF THE LEIBNIZ ALGEBRAS WITH NILINDEX n−3 3 Theo em 2.1. Le Lbe an n-dimensional na u ally g aded Leibniz algeb a o he second ype (n≥9). Then i is isomo phic o one o he ollowing pai wise non- isomo phic algeb as: λ µ dim(L) L0,1 (0,0,0,0,0) odd o e en L0,2 (0,0,0,λ,−1) λ∈ {0,1}odd o e en L0,3 (1,0,0,λ,−1) λ∈Codd o e en L0,4 (1,0,1/4,λ,−1) λ∈Codd o e en L0,5 (0,0,1,λ,−1) λ∈Codd o e en L0,6 (0,1,0,λ,−1) λ∈ {0,1}odd o e en L0,6 (µ,1,0,λ,−1) λ∈Cµ∈ {1,2}odd o e en L0,7 (0,1,µ,λ,−1) λ∈Cµ∈C {0}odd o e en L0,8 (−2λ,1,−λ,2,−1) λ∈ {−2,−4/3}odd o e en L0,9 (2λ,1,λ,0,−1) λ∈C {0,1}odd o e en L0,10 (1,1,1/4,1/4,−1) odd o e en L0,10 (1,1,1/4,1/2,−1) odd o e en L0,10 (2,1,1,1,−1) odd o e en L0,10 (2,1,1,0,−1) odd o e en L0,11 (1,λ,1/4,0,−1) λ∈C {0,1/2}odd o e en L1,2 (0,0,0,λ,−1) λ∈ {0,1}e en L1,3 (1,0,0,λ,−1) λ∈Ce en L1,4 (1,0,1/4,λ,−1) λ∈Ce en L1,6 (µ,1,0,λ,−1) λ∈Cµ∈Ce en L1,7 (0,γ,µ,λ,−1) λ∈Cγ, µ ∈C {0}e en L1,9 (−2λ,1,λ,µ,−1) λ∈C {0,1}µ∈Ce en L1,11 (λ,1,λ2/4,µ,−1) λ∈C {−2,0}µ∈Ce en L1,12 (−1,0,0,λ,−1) λ∈ {0,1}e en L1,13 (−2,0,1,λ,−1) λ∈Ce en L1,14 (−4,0,2,λ,−1) λ∈Ce en L1,15 (0,0,−1,λ,−1) λ∈Ce en L1,16 (−2,0,−1,λ,−1) λ∈Ce en L1,17 (0,−1,0,λ,−1) λ∈ {0,1}e en L1,18 (−1,−1,0,λ,−1) λ∈Ce en L1,19 (−2,−1,0,1,−1) e en L1,20 (1,−1,0,λ,−1) λ∈C {−1/2}e en L1,21 (1,1/3,0,λ,−1) λ∈Ce en L1,22 (−2,−1,1,λ,−1) λ∈ {0,1}e en L1,23 (1,1/2,1/4,λ,−1) λ∈Ce en 4 J.M. CABEZAS, L.M. CAMACHO, J.R. G´ OMEZ, B.A. OMIROV λ γ, µ dim(L) L1,24 (−4,−1,2,λ,−1) λ∈Ce en L1,25 (−3,−4/3,2,λ,−1) λ∈Ce en L1,26 (2/5,2,2/5,λ,−1) λ∈Ce en L1,27 (2/λ,λ,1,µ,−1) λ∈C {−1,0,1}µ∈Ce en L1,28 (8/5,1/2,−4/5,λ,−1) λ∈Ce en L1,29 (λ,−1,λ2/4,0,−1) λ∈C {−2,0}e en L1,30 (1,−1,1/4,λ,−1) λ∈ {−1/2,1/4}e en L1,31 (−8,2,16,λ,−1) λ∈Ce en L1,32 (−2,λ,1,0,−1) λ∈C {−1,0}e en L1,33 (−2,1,1,λ,−1) λ∈ {−1,1}e en whe e he algeb a Lǫ,j (α1,α2,α3,α4,β):ǫ∈ {0,1},1≤j≤33, β ∈ {−1,0} has he ollowing mul iplica ion:                            [ei, e1] = ei+1,1≤i≤n−1, i 6= 3 [e1, e4] = α1e2+βe5, [e2, e4] = α2e3, [e4, e4] = α3e2, [e5, e4] = α4e3, [e1, e5] = (α1−α2)e3−e6, [e4, e5] = (α3−α4)e3, [e1, ei] = βei+1,6≤i≤n−1, [ei, en+3−i] = ǫ(−1)ien,4≤i≤n−1. P oo . F om he condi ion o he heo em we ha e he ollowing mul iplica ion o he basic elemen e1on he igh side: [ei, e1] = ei+1,1≤i≤n−1, i 6= 3,[e3, e1] = [en, e1] = 0. F om hese p oduc s we conclude ha L1=< e1, e4>, L2=< e2, e5>, L3=< e3, e6>, Li=< ei+3 >, 4≤i≤n−3 and e2, e3∈R(L). Le us in oduce deno a ions [e1, e4] = α1e2+β1e5,[e2, e4] = α2e3+β2e6,[e3, e4] = β3e7, [e4, e4] = α3e2+β4e5,[e5, e4] = α4e3+β5e6, [ei, e4] = βiei+1,6≤i≤n−1,[en, e4] = 0. The equali ies [ei, e5] = [[ei, e4], e1]−[[ei, e1], e4],1≤i≤nde i e [e1, e5] = (α1−α2)e3+ (β1−β2)e6,[e2, e5] = (β2−β3)e7,[e3, e5] = β3e8, [e4, e5] = (α3−α4)e3+ (β4−β5)e6,[e5, e5] = (β5−β6)e7, [ei, e5] = (βi−βi+1)ei+2,6≤i≤n−2 [en−1, e5] = [en, e5] = 0. Using induc ion on j o any alue ii can be p o ed ha [ei, ej] = j−4 X k=0 (−1)kj−4 kβi+k!ei+j−3,5≤i≤n−3,6≤j≤n+ 3 −i. In he case o e4∈R(L) we ob ain he algeb a L0,1 (0,0,0,0,0). ON THE DESCRIPTION OF THE LEIBNIZ ALGEBRAS WITH NILINDEX n−3 5 Le now e4/∈R(L).Then we conside he ollowing cases: e5∈R(L) Then ei∈R(L) o 2 ≤i≤n, i 6= 4. F om he equali ies [[ei, e1], e4] = [[ei, e4], e1],1≤i≤n, we ha e α2=α1, α4=α3, β3=β2=β1, βi=β4,5≤i≤n−1. Fo n≥8 we ha e also β1= 0. The change o basis aken as e′ i=ei,1≤i≤n, i 6= 4,5,6, e′ j=ej−β4ej−3,4≤j≤6 deduces β4= 0. I we ake he change o basis in he ollowing way: e′ 1=Ae1+Be4, e′ n−2=e1, e′ j= [e′ j−1, e′ 1],2≤j≤n, j 6=n−2 wi h condi ion AB(A+α1B)6= 0, hen we ob ain he algeb a o he i s ype. The e o e, his case is impossible o he algeb a o he second ype. e5/∈R(L) The embedding [e4, e4]∈R(L) implies β4= 0 and om [ei,[e4, e1]] = −[ei,[e1, e4]], wi h 1 ≤i≤nwe ob ainβ1=−1. I e6∈R(L), hen o n≥9 i ollows β1= 0,which is a con adic ion wi h he condi ion β1=−1. The e o e, e6/∈R(L). I is easy o check ha [ei, ej] + [ej, ei]∈R(L) o any alues o i, j. Applying his o i= 1 and j= 5 we ob ain β2= 0. The ollowing equali ies: [e1, ei] = −ei+1,[e2, ei] = [e3, ei] = 0,6≤i≤n−1 a e p o ed by induc ion on i. F om [e1,[e4, e2j+1]] = −[e5, e2j+1] + [e2j+2, e4], j ≥2, we ha e ha 2β2j+2 =β5+β2j+1 + 2j−4 X k=1 (−1)k2j−3 k(β5+k−β4+k), j ≥2. Simila as in [5] we de i e βj=β5,6≤j≤n−1, o n odd, βj=β5,6≤j≤n−2, o n e en and [e4, en−1] = −β5en o nodd, [e4, en−1] = (βn−1−2β5)en o ne en, [ei, en+3−i] = (−1)i(βn−1−β5)en,5≤i≤n−2, o ne en. I βn−1=β5, hen by he change o basis de ined as e′ i=ei,1≤i≤n, i 6= 4,5,6, and e′ i=ei−β5ei−3,4≤i≤6 we can assume β5= 0. I βn−16=β5( he case o ne en), hen by using he change o basis:            e′ i= (βn−1−β5)iei,1≤i≤3, e′ 4=e4−β5e1, e′ 5= (βn−1−β5)(e5−β5e2), e′ 6= (βn−1−β5)2(e6−β5e3), e′ i= (βn−1−β5)i−4ei,7≤i≤n 6 J.M. CABEZAS, L.M. CAMACHO, J.R. G´ OMEZ, B.A. OMIROV we ob ain [ei, en+3−i] = (−1)ien o 4 ≤i≤n−1.Thus, mul iplica ion in Lis as ollows:                            [ei, e1] = ei+1,1≤i≤n−1, i 6= 3, [e1, e4] = α1e2−e5, [e2, e4] = α2e3, [e4, e4] = α3e2, [e5, e4] = α4e3, [e1, e5] = (α1−α2)e3−e6, [e4, e5] = (α3−α4)e3, [e1, ei] = −ei+1,6≤i≤n−1, [ei, en+3−i] = ǫ(−1)ien,4≤i≤n−1, ǫ ∈ {0,1}. Case 1. ǫ= 0 (nodd o e en) Applying he gene al change o gene a o s o he basis: e′ 1= n X i=1 Aiei, e′ n−2= n X i=1 Biei, we de e mine he o he elemen s o he new basis and he p oduc s in his basis. Then he new pa ame e s a e he ollowing: α′ 1=(α1A1+ 2α3A4)B4 A2 1+α1A1A4+α3A2 4 , α′ 2=α2B4 A1+α2A4 , α′ 3=α3B2 4 A2 1+α1A1A4+α3A2 4 , α′ 4=(α4A1+α2α3A4)B2 4 (A1+α2A4)(A2 1+α1A1A4+α3A2 4), sa is ying he es ic ion A1(A1+α2A4)(A2 1+α1A1A4+α3A2 4)B46= 0. No e ha o new pa ame e s we ha e α′2 1−4α′ 3=(α2 1−4α3)A2 1B2 4 (A2 1+α1A1A4+α3A2 4)2, α′ 1α′ 2−2α′ 3=(α1α2−2α3)A1B2 4 (A1+α2A4)(A2 1+α1A1A4+α3A2 4), α′ 1α′ 2−2α′ 4=(α1α2−2α4)A1B2 4 (A1+α2A4)(A2 1+α1A1A4+α3A2 4). Consequen ly, he nulli y o α2 1−4α3is in a ian in he ollowing sense: i α2 1−4α3= 0, hen α′2 1−4α′3= 0 and i α2 1−4α36= 0, hen α′2 1−4α′36= 0. Analogously, he exp essions α1α2−2α3and α1α2−2α4a e nulli y in a ian s. Conside he ollowing subcases: α2= 0,α3= 0 Then, α′ 1=α1B4 A1+α1A4 , α′ 2= 0, α′ 3= 0 and α′ 4=α4B2 4 A1(A1+α1A4). •α1= 0. I α4= 0, hen he algeb a L0,2 (0,0,0,λ,−1) wi h λ= 0 is ob ained. I α46= 0, hen we ob ain he algeb a L0,2 (0,0,0,λ,−1) wi h λ= 1. •α16= 0. I α4= 0, hen we easily ob ain α′ 1= 1. Thus, we ha e he algeb a L0,3 (1,0,0,λ,−1) wi h λ= 0. ON THE DESCRIPTION OF THE LEIBNIZ ALGEBRAS WITH NILINDEX n−3 7 I α46= 0, hen choosing app op ia e alues o A4and B4we de i e α′ 1=α′ 4= 1. Hence, he algeb a L0,3 (1,0,0,λ,−1) wi h λ= 1 is ob ained. α2= 0,α36= 0 Then, α′ 1=(α1A1+ 2α3A4)B4 A2 1+α1A1A4+α3A2 4 , α′ 2= 0, α′ 3=α3B2 4 A2 1+α1A1A4+α3A2 4 , α′ 4=α4B2 4 A2 1+α1A1A4+α3A2 4 . •I α2 1−4α3= 0, hen aking adequa e alue o B4we ob ain α′ 1= 1, α′ 3= 1/4 and α′ 4=α4 α2 1 =λ. So, we ob ain he amily o algeb as L0,4 (1,0,1/4,λ,−1) wi h λ∈C. •I α2 1−4α36= 0, hen aking sui able alues o A4and B4we deduce α′ 1= 0, α′ 3= 1 and α′ 4=α4 α3 =λ. The amily L0,5 (0,0,1,λ,−1),λ∈Cis ob ained. α26= 0,α3= 0 Then, α′ 1=α1B4 A1+α1A4 , α′ 2=α2B4 A1+α2A4 , α′ 3= 0, α′ 4=α4B2 4 (A1+α1A4)(A1+α2A4). •α1= 0. I α4= 0, hen he choosing app op ia e B4leads α′ 2= 1. Thus, we ob ain L0,6 (0,1,0,λ,−1), λ = 0. I α46= 0, hen aking adequa e A4and B4we de i e α′ 2=α′ 4= 1. The algeb a L0,6 (0,1,0,λ,−1), λ = 1 is ob ained. •α16= 0. Xα4= 0. I α1−α2= 0, hen o sui able B4we ha e α′ 1=α′ 2= 1,i.e. we ob ain he algeb a L0,6 (µ,1,0,λ,−1) wi h µ= 1, λ = 0. I α1−α26= 0, hen o adequa e A4and B4i ollows ha α′ 1= 2, α′ 2= 1. The algeb a L0,6 (µ,1,0,λ,−1),wi h µ= 2, λ = 0 is ob ained. Xα46= 0. I α1−α2= 0, hen o app op ia e alue o B4we ha e α′ 1=α′ 2= 1 and α′ 4=α4 α2 1 =λ. The e o e, we ob ain he amily o algeb as L0,6 (µ,1,0,λ,−1), whe e µ= 1, λ∈C {0}. I α1−α26= 0, hen aking sui able alues o A4and B4we ob ain α′ 1= 2, α′ 2= 1, α′ 4=2α4 α1α2 =λ, i.e., he amily L0,6 (µ,1,0,λ,−1), µ= 2, λ ∈C {0}is ob ained. α26= 0,α36= 0 •α2 1−4α36= 0, α1α2−2α36= 0. Taking app op ia e A4and B4we de i e α′ 1= 0, α′ 2= 1, α′ 3=−(α1α2−2α3)2 α2 2(α2 1−4α3)=µ, 8 J.M. CABEZAS, L.M. CAMACHO, J.R. G´ OMEZ, B.A. OMIROV α′ 4=−(α1α2−2α3)(α1α2−2α4) α2 2(α2 1−4α3)=λ. Hence, we ob ain he amily o algeb as L0,7 (0,1,µ,λ,−1), whe e µ∈C {0}, λ ∈ C. •α2 1−4α36= 0, α1α2−2α3= 0. I yields α′ 3−α′ 4=(α3−α4)α2A1B2 4 (A1+α2A4)(α2A2 1+ 2α3A1A4+α2α3A2 4), 2α′ 3α′ 4−α′2 2α′ 3−α′2 4=(2α3α4−α2 2α3−α2 4)α2 2A2 1B4 4 (A1+α2A4)2(α2A2 1+ 2α3A1A4+α2α3A2 4)2. Xα3−α4= 0. The e o e, 2α3α4−α2 2α3−α2 46= 0 and aking he sui able alues o A4 and B4we ob ain α′ 1= 4, α′ 2= 1, α′ 3= 2, α′ 4= 2. Thus, he algeb a L0,8 (−2λ,1,−λ,2,−1) wi h λ=−2 is ob ained. Xα3−α46= 0. I 2α3α4−α2 2α3−α2 4= 0, hen α46= 0, α3=α2 4 2α4−α2 2 ,α46=α2 2 2. Choos- ing adequa e alues o A4and B4we ob ain α′ 1= 8/3, α′ 2= 1, α′ 3= 4/3, α′ 4= 2, i.e., we de i e he algeb a L0,8 (−2λ,1,−λ,2,−1) wi h λ=−4/3. I 2α3α4−α2 2α3−α2 46= 0, hen as be o e we deduce α′ 1= 2α′ 3, α′ 2= 1, α′ 3=−(α3−α4)2 2α3α4−α2 2α3−α2 4 =λ,α′ 4= 0 and he amily L0,9 (2λ,1,λ,0,−1) wi h λ∈C {0,1}is ob ained. •α2 1−4α3= 0, α1α2−2α36= 0. Then, α16= 2α2,α′2 1−4α′ 4=(α2 1−4α4)4A1B2 4 (2A1+α1A4)2(A1+α2A4). Xα2 1−4α4= 0. Then, α1α2−2α46= 0 and om he abo e we deduce α′ 1= 1, α′ 2= 1, α′ 3= 1/4, α′ 4= 1/4.So, we ob ain he algeb a L0,10 (λ,1,λ2/4,µ,−1) wi h λ= 1, µ = 1/4. Xα2 1−4α46= 0, α1α2−2α4= 0 ⇒α′ 1= 1, α′ 2= 1, α′ 3= 1/4, α′ 4= 1/2, i.e., we ob ain L0,10 (λ,1,λ2/4,µ,−1) wi h λ= 1, µ = 1/2. Xα2 1−4α46= 0, α1α2−2α46= 0 ⇒α′ 1= 1, α′ 2=α1α2−2α4 α2 1−4α4 , α′ 3= 1/4, α′ 4= 0. The amily L0,11 (1,λ,1/4,0,−1), whe e λ∈C {0,1/2}is ob ained. •α2 1−4α3= 0, α1α2−2α3= 0. ON THE DESCRIPTION OF THE LEIBNIZ ALGEBRAS WITH NILINDEX n−3 9 Then, α1= 2α2, α3=α2 2, α′2 2−α′ 4=(α2 2−α4)A1B2 4 (A1+α2A4)3. Xα2 2−α4= 0. Taking an app op ia e alue o B4i ollows ha α′ 1= 2, α′ 2= 1, α′ 3= 1, α′ 4= 1. Hence, we ob ain L0,10 (λ,1,λ2/4,µ,−1) wi h λ= 2, µ = 1. Xα2 2−α46= 0. Choosing adequa e A4and B4=(α2 2−α4)A1 α3 2 yields α′ 1= 2, α′ 2= 1, α′ 3= 1 and α′ 4= 0. Thus, he algeb a L0,10 (λ,1,λ2/4,µ,−1) wi h λ= 2, µ = 0 is ob ained. Now, we conside he o he case. Case 2. ǫ= 1 (ne en) Simila o he case 1, we apply he gene al change o gene a o s o basis. Then, we ob ain all p oduc s and he ollowing exp essions o α′ i,1≤i≤4: α′ 1=(A1−A4)(α1A1+ 2α3A4) A2 1+α1A1A4+α3A2 4 , α′ 2=α2(A1−A4) A1+α2A4 , α′ 3=α3(A1−A4)2 A2 1+α1A1A4+α3A2 4 , α′ 4=(A1−A4)2(α4A1+α2α3A4) (A1+α2A4)(A2 1+α1A1A4+α3A2 4), e i ying he es ic ion A1(A1−A4)(A1+α2A4)(A2 1+α1A1A4+α3A2 4)6= 0. No e ha o hese pa ame e s we ha e α′2 1−4α′ 3=(α2 1−4α3)A2 1(A1−A4)2 (A2 1+α1A1A4+α3A2 4)2, α′ 1α′ 2−2α′ 3=−(α1α2−2α3)A1(A1−A4)2 (A1+α2A4)(A2 1+α1A1A4+α3A2 4), α′ 1α′ 2−2α′ 4=(α1α2−2α4)A1(A1−A4)2 (A1+α2A4)(A2 1+α1A1A4+α3A2 4), α′ 1+ 2α′ 3=(α1+ 2α3)(A1−A4)A1 A2 1+α1A1A4+α3A2 4 . Consequen ly, he nulli y o he exp essions α2 1−4α3, α1α2−2α3, α1α2−2α4, α1+ 2α3a e in a ian s. Applying a gumen s as in he case 1 o he ollowing subcases: α2= 0 α3= 0 , α2= 0,α36= 0 , α26= 0 α3= 0 , α26= 0,α36= 0 we ob ain he es algeb as and amilies o he heo em.  The nex heo em comple es he classi ica ion o na u ally g aded Leibniz alge- b as wi h cha ac e is ic sequence (n−3,3).