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).