NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3
J.Q. ADASHEV, L.M. CAMACHO, S. G ´
OMEZ-VIDAL, I.A. KARIMJANOV
Abs ac . We p esen he classi ica ion o a subclass o n-dimensional na u ally g aded Zinbiel
algeb as. This subclass has he nilindex n−3 and he cha ac e is ic sequence (n−3,2,1).In ac ,
his esul comple es he classi ica ion o na u ally g aded Zinbiel algeb as o nilindex n−3.
Ma hema ics Subjec Classi ica ion 2010: 17A32.
Key Wo ds and Ph ases: Zinbiel algeb a, Leibniz algeb a, nilpo ency, cha ac e is ic sequence
1. In oduc ion.
In ensi e in es iga ion on Lie algeb as leads o he appea ance o a new algeb aic objec – Leibniz
algeb as. The Leibniz algeb as in oduced by Loday in [7] a e a ”non commu a i e” algeb as analogue
o Lie algeb as. I should be men ioned ha Leibniz algeb as inhe i an impo an Lie algeb a p ope y:
he ope a o o igh mul iplica ion on an elemen o an algeb a is a de i a ion.
Leibniz algeb as o m a Koszul ope ad in he sense o V. Ginzbu g and M. Kap ano [6]. Unde he
Koszul duali y he ope ad o Lie algeb as is dual he ope ad o associa i e and commu a i e algeb as.
The no ion o dual Leibniz algeb a de ined by J.-L. Loday [8] is p ecisely he dual ope ad o Leibniz
algeb as in his sense.
In his pape , we s udy algeb as which a e he dual o Leibniz algeb as in Koszul sense. J.-L. Loday
s udied in [8] ca ego ical p ope ies o Leibniz algeb as and conside ed in his connec ion a new objec
– Zinbiel algeb as (Leibniz is w i en in e e se o de ). Since he ca ego y o Zinbiel algeb as is Koszul
dual o he ca ego y o Leibniz algeb as, some imes hey a e also called dual Leibniz algeb as.
In [2, 5, 9] some c ucial p ope ies o Zinbiel algeb as we e ob ained. Pa icula ly, in [5], he au ho s
p o e ha e e y ini e-dimensional Zinbiel algeb a o e complex numbe s is nilpo en . Howe e , he
s udy o nilpo en algeb as is oo complex and should be ca ied ou wi h addi ional condi ions, such
as condi ions on nilindex, a ious ypes o g ada ions, cha ac e is ic sequence and o he s.
The aim o his wo k is o con inue he s udy o complex ini e-dimensional na u ally g aded Zinbiel
algeb as. The n-dimensional Zinbiel algeb as o nilindex kwi h n−2≤k≤na e classi ied in [1, 2].
The classi ica ion o complex n-dimensional na u ally g aded Zinbiel algeb as o nilindex n−3 is a
di icul p oblem and i should be di ided in o h ee cases. Namely, i is necessa y o conside he
possibili ies o he cha ac e is ic sequence o such algeb as: (n−3,3), (n−3,1,1,1) and (n−3,2,1).
The classi ica ion o complex na u ally g aded Zinbiel algeb as o nilindex n−3 wi h cha ac e is ic
sequence equal o (n−3,3) and (n−3,1,1,1) has been done in [1].
The knowledge o na u ally g aded algeb as o a ce ain amily o e s signi ican in o ma ion abou
hei s uc u al p ope ies.
In his pape we ob ain he classi ica ion o na u ally g aded Zinbiel algeb as o nilindex n−3 wi h
cha ac e is ic sequence (n−3,2,1).Thus, we comple e he s udy o he n−3 case. All he spaces
and he algeb as a e conside ed o e he ield o complex numbe s. We omi he p oduc s which a e
equal o ze o o con enience.
Th oughou all he wo k we use he so wa e Ma hema ica (see [3]) o compu e he Zinbiel iden-
i y in low dimensions and o o mula e he gene aliza ions o he calcula ions, which a e p o ed o
a bi a y dimension. Mo eo e , he p og am allows us o cons uc new bases using some gene al
ans o ma ion o he gene a o s o he algeb a.
Since he di ec sum o nilpo en Zinbiel algeb as is nilpo en , we shall conside only non spli
algeb as.
2. P elimina ies
In his sec ion we in oduce some de ini ions, no a ions and esul s, which a e necessa y o he
unde s anding o g aded Zinbiel algeb as.
1
2 J.Q. ADASHEV, L.M. CAMACHO, S. G´
OMEZ-VIDAL, I.A. KARIMJANOV
De ini ion 2.1. A ec o space Zo e a ield Kwi h a bilinea ope a ion “◦” is called Zinbiel algeb a
i o any x, y, z ∈ Z he ollowing iden i y
(2.1) (x◦y)◦z=x◦(y◦z) + x◦(z◦y)
holds.
Examples o Zinbiel algeb as can be ound in [2, 5, 8].
Z(a, b, c) deno es he ollowing polynomial:
Z(a, b, c) = (a◦b)◦c−a◦(b◦c)−a◦(c◦b).
Zinbiel algeb as a e de ined by he iden i y Z(a, b, c) = 0.
Fo a gi en Zinbiel algeb a Z he sequence o wo-sided ideals de ined ecu si ely as ollow:
Z1=Z,Zk+1 =Z ◦ Zk, k ≥1.
is said o be he lowe cen al se ies.
De ini ion 2.2. A Zinbiel algeb a Zis called nilpo en i he e exis s s∈Nsuch ha Zs6= 0 and
Zs+1 = 0.The minimal numbe ssa is ying his p ope y is called he index o nilpo ency o nilindex
o he algeb a Z.
Fo a gi en Zinbiel algeb a Zwe in oduce deno a ions:
R(Z) = {x∈ Z | y◦x= 0 o any y∈ Z} − − he igh annihila o o Z,
L(Z) = {x∈ Z | x◦y= 0 o any y∈ Z} − − he le annihila o o Z,
Cen (Z) = {x, y ∈ Z | x◦y=y◦x= 0 o any y∈ Z} − − he cen e o Z.
I is easy o see ha he cen e and he igh annihila o o Za e wo-sided ideals.
Le us deno e by Lx he ope a o o le mul iplica ion on elemen x, i.e. Lx:Z −→ Z such ha
Lx(y) = x◦y o any y∈ Z.
Le Zbe a complex n-dimensional Zinbiel algeb a and xbe an elemen o he se Z Z2. Fo he
ope a o Lxwe de ine a descending sequence C(x) = (n1, n2,...,nk) wi h n1+···+nk=n, which
consis s o he dimensions o he Jo dan blocks o he ope a o Lx. In he se o such sequences we
conside he lexicog aphic o de , ha is, C(x) = (n1, n2,...,nk)< C(y) = (m1, m2,...,ms) i he e
exis s isuch ha ni< miand nj=mj o j < i. Taking in o accoun he equali y n1+···+nk=
m1+···+mssuch compa ison is always applicable.
De ini ion 2.3. The sequence C(Z) = max{C(x) : x∈ Z Z2}is called he cha ac e is ic sequence
o he algeb a Z.
In [5], he au ho s p o e ha Zinbiel algeb as o ini e dimension a e nilpo en . Since we ocused
ou a en ion on ini e dimension complex nilpo en Zinbiel algeb as.
Le Zbe a ini e-dimensional nilpo en Zinbiel algeb a wi h nilindex equal o s. Fo i(1 ≤i≤s)
we pu Zi=Zi/Zi+1 and we ob ain he g aded Zinbiel algeb a
g (Z) = Z1⊕ Z2⊕...⊕ Zs,whe e Zi◦ Zj⊆ Zi+j.
An algeb a Zi called na u ally g aded i Z∼
=g (Z).I is no di icul o see ha Zi+1 =Z1◦ Zi
in he na u ally g aded algeb a Z.
Le Zbe a na u ally g aded Zinbiel algeb a wi h cha ac e is ic sequence (n−3,2,1).By de ini ion
o cha ac e is ic sequence he e exis s a basis {e1, e2,...,en}in he algeb a Zsuch ha he ope a o
Le1has one block Jn−3o size (n−3),one block J2o size 2 and one block J1o size one.
No e ha he e will be six possibili ies o he ope a o s Le1.By a change o basis i is easy o p o e
ha he six cases can be educed o he ollowing h ee cases:
I. 

Jn−30 0
0J20
0 0 J1

, II. 

J20 0
0Jn−30
0 0 J1

, III.

J10 0
0Jn−30
0 0 J2

.
De ini ion 2.4. A Zinbiel algeb a Zis called ei he o i s ype ( ype I), second ype ( ype II) o hi d
ype ( ype III) i he ope a o Le1has he o m:
I. 

Jn−30 0
0J20
0 0 J1

, II. 

J20 0
0Jn−30
0 0 J1

, III.

J10 0
0Jn−30
0 0 J2


NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 3
espec i ely.
F om now on we deno e by Cj
i he combina o ial numbe s Cj
i=i
j.
The ollowing esul holds:
Lemma 2.5. [4] Le Zbe a Zinbiel algeb a such ha e1◦ei=ei+1 o 1≤i≤k−1,wi h espec o
he adap ed basis {e1,...,ek, ek+1,...,en}.Then
ei◦ej=Cj
i+j−1ei+j, o 2≤i+j≤k
3. Main Resul
3.1. Type I. Algeb as o ype I wi h n≥8.So, we ha e he ollowing b acke s:











e1◦ei=ei+1,1≤i≤n−4,
e1◦en−3= 0,
e1◦en−2=en−1,
e1◦en−1= 0,
e1◦en= 0.
I is easy o see ha Zi⊇ heiiwhe e 1 ≤i≤n−3. I is e iden ha dim(Z1)>1.In ac , i
dim(Z1) = 1, hen he algeb a Zis one-degene a ed and he e o e i is a ze o- ili o m algeb a, bu i
is no an algeb a o nilindex n−3.Le us assume ha en−2∈ Z 1and en∈ Z 2, hen en−1∈ Z 1+1.
We can dis inguish he ollowing cases:
Case I. I 1= 2= 1.
Then we ha e ha
Z1=< e1, en−2, en>, Z2=< e2, en−1>, Z3=< e3>, . . . , Zn−3=< en−3>
and he ollowing p oduc s:
e1◦e1=e2, e1◦en−2=en−1, en−2◦e1=α1e2+α2en−1,
en−2◦en−2=α3e2+α4en−1, en−2◦en=α5e2+α6en−1, en◦e1=β1e2+β2en−1,
en◦en−2=β3e2+β4en−1, en◦en=β5e2+β6en−1, e1◦e2=e3,
en−2◦e2=γ1e3, en−2◦en−1=γ2e3, en◦e2=γ3e3,
en◦en−1=γ4e3.
F om he equali y Z(e1, en, e1) = Z(e1, en, en) = 0 we ha e β1=β5= 0.
Le us conside he equali ies Z(e1, en−2, e1) = Z(e1, en−1, e1) = 0 hen i ollows α1= 0.
F om he equali ies
Z(e1, e1, en−2) = Z(en−2, e1, e1) = Z(en−2, en−1, e1) = Z(e1, e1, en) = 0
Z(e1, en, e2) = Z(en, en−1, e1) = Z(e1, en−2, en−2) = Z(e1, en−1, en−2) = 0
Z(e1, en−2, en) = Z(e1, en, en−1) = Z(e1, e1, en−1) = Z(e1, en−2, e2) = 0
Z(en−2, en, e1) = Z(en−2, en−2, e1) = 0
we ob ain
γ1=γ2=γ3=γ4=α3=α5=β3= 0,
and
e2◦en−2=en−2◦e2=e2◦en−1=en−1◦e2=e2◦en=en◦e2= 0.
Now, by ma hema ical induc ion me hod, we p o e ha en−1◦ek= 0 and ek◦en−1= 0 wi h
2≤k≤n−3.
•I k= 2, hen we ha e en−1◦e2=e2◦en−1= 0.
•Le us suppose ha o some k he equali ies en−1◦ek= 0 and ek◦en−1= 0 a e ue. We
p o e i o k+ 1.
en−1◦ek+1 =en−1◦(e1◦ek) = (en−1◦e1)◦ek−en−1◦(ek◦e1) =
=−C1
ken−1◦ek+1 =−ken−1◦ek+1, en−1◦ek+1 = 0.
ek+1 ◦en−1= (e1◦ek)◦en−1=e1◦(ek◦en−1) + e1◦(en−1◦ek) =
= 0
4 J.Q. ADASHEV, L.M. CAMACHO, S. G´
OMEZ-VIDAL, I.A. KARIMJANOV
As in p e ious cases, i easy o see ha ek◦en−2=en−2◦ek= 0 and ek◦en=en◦ek= 0 o
2≤k≤n−3.
Thus, we ha e ob ained he ollowing amily o algeb as:
Z(a1, a2, a3, a4, a5, a6) :























ei◦ej=Cj
i+j−1ei+j,2≤i+j≤n−3,
e1◦en−2=en−1,
en−2◦e1=a1en−1,
en−2◦en−2=a2en−1,
en−2◦en=a3en−1,
en◦e1=a4en−1,
en◦en−2=a5en−1,
en◦en=a6en−1,
whe e we omi he p oduc s ha a e equal o ze o.
Theo em 3.1. An a bi a y Zinbiel algeb a o he amily Z(a1, a2, a3, a4, a5, a6)is isomo phic o one
o he ollowing pai wise non-isomo phic algeb as:
Z1(1,0,0,0,1,0), Z2(0,0,0,0,1,0), Z3(0,1,0,1,0,0),
Z4(0,0,0,1,0,0), Z5(0,1,0,0,0,0), Z6(1,1,0,0,0,0),
Z7(λ, 0,0,0,0,0), λ ∈C, Z8(0, λ, 1,0,0,1), λ ∈C {0}, Z9(α, −α
(α−1)2,1,0,0,1), α ∈C {0,1},
Z10(0,0,1,0,1,1), Z11(1,0,1,0,1,1), Z12(0,0,1,1,0,0),
Z13(0,0,1,0,0,0), Z14(λ, 1,1,0,1,1), λ ∈C, Z15(0,1,1,−1,1,1),
Z16(1,1,1,0,1,1).
P oo . Le Zbe sa is ying o he hypo hesis o he heo em. Due o he p ope y o na u al g ada ion
o he algeb a i is enough o conside he ollowing change o gene a o s:
e′
1=P1e1+Pn−2en−2+Pnen,
e′
n−2=Q1e1+Qn−2en−2+Qnen,
e′
n=R1e1+Rn−2en−2+Rnen.
Making he gene al change o basis in he amily Z(a1, a2, a3, a4, a5, a6),we de i e he exp essions
o he new pa ame e s in he new basis (1):
a′
1=a1P1Qn−2+a2Pn−2Qn−2+a3PnQn−2+a4P1Qn+a5Pn−2Qn+a6PnQn
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
a′
2=a2Q2
n−2+a3Qn−2Qn+a5Qn−2Qn+a6Q2
n
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
a′
3=a2Qn−2Rn−2+a3Qn−2Rn+a5QnRn−2+a6QnRn
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
a′
4=a1P1Rn−2+a2Pn−2Rn−2+a3PnRn−2+a4P1Rn+a5Pn−2Rn+a6PnRn
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
a′
5=a2Qn−2Rn−2+a3QnRn−2+a5Qn−2Rn+a6QnRn
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
a′
6=a2R2
n−2+a3Rn−2Rn+a5Rn−2Rn+a6R2
n
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn
,
and he ollowing es ic ions:
(2) 








Q1=R1= 0,
P1Rn−2+a2Pn−2Rn−2+a3Pn−2Rn+a5PnRn−2+a6PnRn= 0,
P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn6= 0,
P1(Qn−2Rn−QnRn−2)6= 0.
We can dis inguish wo cases:
Case 1. Le en∈R(Z) be, hen a3=a6= 0.
NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 5
F om he es ic ions,
(P1+a2Pn−2+a5Pn)Rn−2= 0,
(P1+a2Pn−2+a5Pn)Qn−26= 0,
P1(Qn−2Rn−QnRn−2)6= 0.



⇒Rn−2= 0.
i ollows ha P1Qn−2Rn6= 0.Thus, he new pa ame e s a e:
a′
1=a1P1Qn−2+a2Pn−2Qn−2+a4P1Qn+a5Pn−2Qn
Qn−2(P1+a2Pn−2+a5Pn),
a′
2=a2Qn−2+a5Qn
P1+a2Pn−2+a5Pn
,
a′
4=Rn(a4P1+a5Pn−2)
Qn−2(P1+a2Pn−2+a5Pn),
a′
5=a5Rn
P1+a2Pn−2+a5Pn
,
We obse e ha he nulli y o a5is in a ian . Mo eo e , i is easy o check ha he nulli y o he
ollowing exp ession
a′
2a′
4−a′
1a′
5=(a2a4−a1a5)P1Rn
(P1+a2Pn−2+a5Pn)2
is in a ian . Thus, we can dis inguish he ollowing non-isomo phic cases:
Case 1.1. Le a56= 0 be. Then choosing
Rn=P1+a2Pn−2+a5Pn
a5
, Pn−2=−a4P1
a5
, Qn=−a2Qn−2
a5
we ha e
a′
5= 1, a′
4= 0, a′
2= 0, a′
1=(a2a4−a1a5)P1
(a2a4−a5)P1−a2
5Pn
and he de e minan is o med by he po encies o he ollowing non-ze o ac o s: P1Qn−2a5((a2a4−
a5)P1−a2
5Pn).
a) I a2a4−a1a56= 0, choosing Pn=P1(a1−1)
a5we ecei e a′
1= 1.I ollows he algeb a
Z1(1,0,0,0,1,0).
b) I a2a4−a1a5= 0, hen we ob ain a′
1= 0 and we ha e he algeb a Z2(0,0,0,0,1,0).
Case 1.2. Le a5= 0 be. Then, a′
5= 0 and we ha e
a′
1=a1P1Qn−2+a2Pn−2Qn−2+a4P1Qn
Qn−2(P1+a2Pn−2),
a′
2=a2Qn−2
P1+a2Pn−2
,
a′
4=a4P1Rn
Qn−2(P1+a2Pn−2).
wi h P1Qn−2Rn(P1+a2Pn−2)6= 0.
We obse e ha he nulli ies o a2and a4a e in a ian , so we can dis inguish he ollowing cases:
a) Le a46= 0 be. Then, choosing
Rn=Qn−2(P1+a2Pn−2)
a4P1
, Qn=−Qn−2(a1P1+a2Pn−2)
a4P1
,
we ge a′
4= 1 and a′
1= 0.
a.1) I a26= 0, hen choosing Qn−2=P1+a2Pn−2
a2
,we ob ain a′
2= 1 and he algeb a
Z3(0,1,0,1,0,0).
a.2) I a2= 0, hen we ha e a′
2= 0 and he algeb a Z4(0,0,0,1,0,0).

6 J.Q. ADASHEV, L.M. CAMACHO, S. G´
OMEZ-VIDAL, I.A. KARIMJANOV
b) Le a4= 0 be. Then a′
4= 0 and we ha e
a′
1=a1P1+a2Pn−2
P1+a2Pn−2
, a′
2=a2Qn−2
P1+a2Pn−2
.
We ha e ha he nulli y o he ollowing exp ession:
a′
1−1 = P1(a1−1)
P1+a2Pn−2
.
is in a ian .
b.1) Le a26= 0 be. Then, choosing Qn−2=P1+a2Pn−2
a2
,we ob ain a′
2= 1.
•I a1−16= 0, hen pu ing Pn−2=−a1P1
a2,we ha e a′
1= 0 and he algeb a Z5(0,1,0,0,0,0).
The de e minan o change o basis consis s o he po encies o he ollowing non-ze o ac o s
a2(a1−1)P1Rn.
•I a1−1 = 0, hen a′
1= 1 and we ob ain Z6(1,1,0,0,0,0).
b.2) Le a2= 0 be. Then, we ha e a′
2= 0, a′
1=a1=λ∈Cand he amily Z7(λ, 0,0,0,0,0),wi h
λ∈C.
Case 2. Le en/∈R(Z) be, hen (a3, a6)6= (0,0).We can suppose ha a36= 0,in ano he case, a3= 0
and a66= 0 we make he ollowing change o basis ′
1= 1+ 3.Thus, a36= 0.Taking in o accoun he
exp essions gi en in (1), he es ic ions (2) and he ollowing exp ession:
∆ = a3
3a4+a2
3a4a5−a1a2
3a4a5+a2a3a2
4a5−a1a3a4a2
5−
−a1a2
3a6−3a2a3a4a6+a1a2a3a4a6−a2
2a2
4a6+
+a3a5a6+a2
1a3a5a6+a2a4a5a6+a1a2a4a5a6−
−a1a2
5a6−a2a2
6+ 2a1a2a2
6−a2
1a2a2
6,
he nulli y o he ollowing exp essions a e in a ian
a′2
3−a′
3a′
5+a′2
5−a′
2a′
6=(a2
3−a3a5+a2
5−a2a6)(Qn−2Rn−QnRn−2)2
P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn
,
a′
3a′
5−a′
2a′
6=(a3a5−a2a6)(Qn−2Rn−QnRn−2)2
P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn
,
∆′=∆P2
1(Qn−2Rn−QnRn−2)4
(a2Qn−2Rn−2+a5QnRn−2+a3Qn−2Rn+a6QnRn)2,
a′
3−a′
5=(a3−a5)(Qn−2Rn−QnRn−2)
P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn
.
We can dis inguish he ollowing non isomo phic cases:
Case 2.1. Le a3a5−a2a66= 0 be. Then, choosing
Pn−2=−(a6P1QnRn−2+a2a5Qn−2R2
n−2+a2
5QnR2
n−2−a6P1Qn−2Rn+
+a3a5Qn−2Rn−2Rn+a2a6Qn−2Rn−2Rn+ 2a5a6QnRn−2Rn+a3a6Qn−2R2
n+
+a6QnR2
n)1
(a3a5−a2a6)(Qn−2Rn−QnRn−2)
Pn=−(−a3P1QnRn−2−a2
2Qn−2Rn−2−a2a5QnR2
n−2+a3P1Qn−2Rn−
−2a2a3Qn−2Rn−2Rn−a3a5QnRn−2Rn−a2a6QnRn−2Rn−a2
3Qn−2R2
n−
−a3a6QnR2
n)1
(a3a5−a2a6)(Qn−2Rn−QnRn−2)
and using he es ic ion (2), we ob ain a′
3= 1.
NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 7
a) Le a3−a56= 0 be. Then, choosing
Qn=−a2Qn−2Rn−2+a5Qn−2Rn
a3Rn−2+a6Rn
, Rn−2=a3Qn−2−a5Qn−2−a6Rn
a3
we ge a′
5= 0, a′
6= 1 and a′
2=λ∈C {0}.The de e minan o he change o basis is o med by he
po encies o he ollowing non-ze o ac o s:
(a3Rn−2+a6Rn)P1Qn−2(a2R2
n−2+a3Rn−2Rn+a5Rn−2Rn+a6R2
n).
a.1) Le ∆ 6= 0 be. Then, we choose
Rn=−(a2a3a4−a3a5−a2a4a5+a1a2
5+a2a6−a1a2a6)P1
(a3−a5)(a3a5−a2a6),
Qn−2= (a2
3a4−a1a2
3a6−2a2a3a4a6+a3a5a6+a1a3a5a6+
+a2a4a5a6−a1a2
5a6−a2a2
6+a1a2a2
6)P1
(a3−a5)2(a3a5−a2a6),
and we ge a′
1=a′
4= 0 and he amily Z8(0, λ, 1,0,0,1), λ ∈C {0}.The de e minan o change
o basis is o med by he non-ze o po encies o he ollowing ac o s (a3−a5)(a3a5−a2a6)∆P1.
a.2) Le ∆ = 0 be. Then, we ha e
∆′=a′
5= 0, a′
3=a′
6= 1,
a′
4−a′
1−3a′
2a′
4+a′
2a′
1a′
4−a′2
2a′2
4−a′
2(a′
1−1)2= 0.
Thus, we ob ain he ollowing amily
ei◦ej=Cj
i+j−1ei+j,2≤i+j≤n−3,
e1◦en−2=en−1,
en−2◦e1=αen−1,
en−2◦en=en−1,
en−2◦en−2=βen−1,wi h β6= 0
en◦en=en−1,
en◦e1=γen−1,wi h γ−α−3βγ +αβγ −β2γ2−β(1 −α)2= 0
Now, we make he gene ic change o basis
e′
1=P1e1+Pn−2en−2+Pnen,
e′
n−2=Q1e1+Qn−2en−2+Qnen,
e′
n=R1e1+Rn−2en−2+Rnen.
and we ha e he exp essions o he new pa ame e s and he new es ic ions:
α′=αP1Qn−2+βPn−2Qn−2+γP1Qn+PnQn−2+Pn−2Qn+PnQn
P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn
,
β′=βQ2
n−2+ 2Qn−2Qn+Q2
n
P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn
,
γ′=αP1Rn−2+βPn−2Rn−2+γP1Rn+PnRn−2+Pn−2Rn+PnRn
P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn
,
1 = βQn−2Rn−2+Qn−2Rn+QnRn−2+QnRn
P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn
,
1 = βR2
n−2+ 2Rn−2Rn+R2
n
P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn
,
0 = P1Rn−2+βPn−2Rn−2+Pn−2Rn+PnRn−2+PnRn= 0.(∗∗)
8 J.Q. ADASHEV, L.M. CAMACHO, S. G´
OMEZ-VIDAL, I.A. KARIMJANOV
Pu ing
Pn−2=−P1QnRn−2−P1Qn−2Rn+βQn−2Rn−2Rn+Qn−2R2
n+QnR2
n
β(QnRn−2−Qn−2Rn),
Pn= (P1QnRn−2+β2Qn−2R2
n−2−P1Qn−2Rn+ 2βQn−2Rn−2Rn+
+βQnRn−2Rn+Qn−2R2
n+QnR2
n)1
β(QnRn−2−Qn−2Rn),
Qn−2=Rn−2+Rn,
Qn=−βQn−2Rn−2
Rn−2+Rn
=−βRn−2,
we ge
α′=−(−P1Rn−2+ 2βP1Rn−2−αβP1Rn−2+β2γP1Rn−2+
+βR2
n−2−β2R2
n−2−P1Rn+βP1Rn−αβP1Rn+
+Rn−2Rn−βRn−2Rn+R2
n−βR2
n)1
β(βR2
n−2+Rn−2Rn+R2
n),
γ′= (P1Rn−2−βP1Rn−2+αβP1Rn−2−βR2
n−2+P1Rn+
+βγP1Rn−Rn−2Rn−R2
n)1
β(βR2
n−2+Rn−2Rn+R2
n),
β′=β6= 0
wi h P1(βR2
n−2+Rn−2Rn+R2
n)6= 0.I is easy o p o e ha :
γ′−α′−3β′γ′+α′β′γ′−β′2γ′2−β′(α′−1)2=γ−α−3βγ +αβγ −β2γ2−β(α−1)2
βR2
n−2+Rn−2Rn+R2
n
P2
1= 0.
Now, i we choose
Rn=1
2(P1+βγP1−Rn−2)−
−q(P1+βγP1−Rn−2)2+ 4(P1Rn−2−βP1Rn−2+αβP1Rn−2−βR2
n−2)
2
we ge γ′= 0, β′=β6= 0, α′+β′(α′−1)2= 0 wi h α′∈C {0,1}, hus β′=−α′
(α′−1)2
and we ha e he amily Z9(α, −α′
(α′−1)2,1,0,0,1),wi h α∈C {0,1}.
b) Le a3−a5= 0 be. Then, we ha e a′
3=a′
5= 1, a2
3−a2a66= 0 and
a′
1=((a1−1)Qn−2+a4Qn)P1
a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn
+ 1,
a′
2=a2Q2
n−2+ 2a3Qn−2Qn+a6Q2
n
a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn
,
a′
4=((a1−1)Rn−2+a4Rn)P1
a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn
,
a′
6=a2R2
n−2+ 2a3Rn−2Rn+a6R2
n
a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn
,
wi h
P1(QnRn−2−Qn−2Rn)(a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn)6= 0,
P1Rn−2+a2Pn−2Rn−2+a3Pn−2Rn+a5PnRn−2+a6PnRn= 0.
We can suppose a′
2= 0 ,
•a26= 0,i we choose Qn−2=−a3±pa2
3−a2a6
a2
Qn,we ge a′
2= 0,
•a2= 0,i we choose Qn−2=−a6Qn
2a3
we ha e a′
2= 0.
NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 9
Analogously, we can suppose ha a′
4= 0 , using Rn.
Now, we ha e he new amily:
ei◦ej=Cj
i+j−1ei+j,2≤i+j≤n−3,
e1◦en−2=en−1,
en−2◦e1=a′
1en−1,
en−2◦en=en−1,
en◦en−2=en−1,
en◦en=a′
6en−1.
and we make a gene ic change o basis. Choosing Pnand Pn−2(as in p e ious cases) we ge a′′
3=a′′
5= 1.
Pu ing Rn−2= 0, Qn−2=−a6Qn
2we ha e a′′
4=a′′
2= 0 and
a′′
1=(1 −a′
1)P1+Rn
Rn
, a′′
6=2Rn
Qn
I is easy o check ha he nulli y o a′′
1−1 is in a ian because
a′′
1−1 = (a′
1−1)P1Qn−2
(Qn−2+a6Qn)Rn
Mo eo e , choosing Qn= 2Rnwe ob ain a′
6= 1.The de e minan o change o basis is o med by he
non-ze o po encies o he ollowing ac o s a6P1QnRn.
Now, we can dis inguish wo cases:
b.1) I a′
1−16= 0,choosing Rn= (a′
1−1)P1we ge a′′
1= 0 and he algeb a Z10(0,0,1,0,1,1).
b.2) I a′
1−1 = 0, hen a′′
1= 1 and we ha e Z11(1,0,1,0,1,1).
Case 2.2. Le a3a5−a2a6= 0 be.
As a36= 0 ⇒a5=a2a6
a3.We subs i u e in (1) and in (2) and we choose
Pn−2=−a3P1Rn−2+a2a6PnRn−2+a3a6PnRn
a3(a2Rn−2+a3Rn).
Now, aking in o accoun he new pa ame e s (1), we ge o:
a′
2=(a2Qn−2+a3Qn)(a3Qn−2+a6Qn)(a2Rn−2+a3Rn)
a2
3P1(Qn−2Rn−QnRn−2),
a′
3=(a3Qn−2+a6Qn)(a2Rn−2+a3Rn)2
a2
3P1(Qn−2Rn−QnRn−2)6= 0.
wi h a3P1(QnRn−2−Qn−2Rn)(a2R1+a3Rn)6= 0 and ha he nulli y o he ollowing exp essions:
a′
1a′
6−a′
3a′
4=a3(a1a6−a3a4)(a2Rn−2+a3Rn)(Qn−2Rn−QnRn−2)P1
(a3a6PnQn+a3P1Qn−2+a2a3Pn−2Qn−2+a2a6PnQn−2+a2
3Pn−2Qn)2,
a′2
3−a′
2a′
6=(a2
3−a2a6)(a3Qn−2+a6Qn)(a2Rn−2+a3Rn)(Qn−2Rn−QnRn−2)
(a3a6PnQn+a3P1Qn−2+a2a3Pn−2Qn−2+a2a6PnQn−2+a2
3Pn−2Qn)2,
a e in a ian . Thus, we can dis inguish he non isomo phic cases: