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arXiv:1908.05057v1 [math.DG] 14 Aug 2019
Analytic Linear Lie rack Structures on Leibniz Algebras
Hamid Abchira, Fatima-Ezzahrae Abidb, Mohamed Boucettac
aUniversit´e Hassan II
Ecole Sup´erieure de Technologie
Route d’El Jadida Km 7, B.P. 8012, 20100 Casablanca, Maroc
e-mail: h abchir@yahoo.com
bUniversit´e Cadi-Ayyad
Facult´e des sciences et techniques
BP 549 Marrakech Maroc
e-mail: abid.fatimaezzahrae@gmail.com
cUniversit´e Cadi-Ayyad
Facult´e des sciences et techniques
BP 549 Marrakech Maroc
e-mail: m.boucetta@uca.ac.ma
Abstract
A linear Lie rack structure on a finite dimensional vector space Vis a Lie rack operation (x,y)7→
x⊲ypointed at the origin and such that for any x, the left translation Lx:y7→ Lx(y)=x⊲yis
linear. A linear Lie rack operation ⊲is called analytic if for any x,y∈V,
x⊲y=y+
∞
X
n=1
An,1(x,...,x,y),
where An,1:V×...×V−→ Vis an n+1-multilinear map symmetric in the nfirst arguments.
In this case, A1,1is exactly the left Leibniz product associated to ⊲. Any left Leibniz algebra
(h,[,]) has a canonical analytic linear Lie rack structure given by xc
⊲y=exp(adx)(y), where
adx(y)=[x,y].
In this paper, we show that a sequence (An,1)n≥1of n+1-multilinear maps on a vector space
Vdefines an analytic linear Lie rack structure if and only if [ ,] :=A1,1is a left Leibniz bracket,
the An,1are invariant for (V,[,]) and satisfy a sequence of multilinear equations. Some of
these equations have a cohomological interpretation and can be solved when the zero and the 1-
cohomology of the left Leibniz algebra (V,[,]) are trivial. On the other hand, given a left Leibniz
algebra (h,[,]), we show that there is a large class of (analytic) linear Lie rack structures on
(h,[,]) which can be built from the canonical one and invariant multilinear symmetric maps on
h. A left Leibniz algebra on which all the analytic linear Lie rack structures are build in this way
will be called rigid. We use our characterizations of analytic linear Lie rack structures to show
that sl2(R) and so(3) are rigid. We conjecture that any simple Lie algebra is rigid as a left Leibniz
algebra.
Keywords: Lie rack, Left Leibniz algebra, multilinear algebra, simple Lie algebra
Preprint submitted to Elsevier August 15, 2019
1. Introduction
In the 1980’s, Joyce [13] and Matveev [17] introduced the notion of quandle. This notion
has been derived from the knot theory, in the way that the axioms of a quandle are the algebraic
expressions of Reidemeister moves (I,II,III) for an oriented knot diagram [11]. The quandles
provide many knot invariants. The fundamental quandle or knot quandle was introduced by Joyce
who showed that it is a complete invariant of a knot (up to a weak equivalence). Racks which
are a generalization of quandles were introduced by Brieskorn [7] and Fenn and Rourke [12].
Recently (see [8, 9]), there has been investigations on quandles and racks from an algebraic point
of view and their relationship with other algebraic structures as Lie algebras, Leibniz algebras,
Frobenius algebras, Yang Baxter equation, and Hopf algebras etc..
A rack is a non-empty set Xtogether with a map ⊲:X×X−→ X, (a,b)7→ a⊲bsuch that,
for any a,b,c∈X, the map La:X−→ X,b7→ a⊲bis a bijection and
a⊲(b⊲c)=(a⊲b)⊲(a⊲c).(1)
A rack Xis called pointed if there exists a distinguished element e∈Xsuch that, for any
a∈X,
a⊲e=eand Le=IdX.(2)
A rack Xis called a quandle if, for any a∈X,a⊲a=a.
A Lie rack is a rack (X,⊲) such that Xis a smooth manifold, ⊲is a smooth map and the
left translations Laare diffeomorphisms. Any Lie group Ghas a Lie rack structure given by
g⊲h=g−1hg.
Leibniz algebras were first introduced and investigated in the papers of Bloh [6, 5] under
the name of D-algebras. Then they were rediscovered by Loday [14] who called them Leibniz
algebras. A left Leibniz algebra is an algebra (h,[,]) over a field Ksuch that, for every element
u∈h, adu:h−→ h,v7→ [u,v] is a derivation of h, i.e.,
[u,[v,w]] =[[u,v],w]+[v,[u,w]],v,w∈h.(3)
Any Lie algebra is a left Leibniz algebra and a left Leibniz algebra is a Lie algebra if and only if
its bracket is skew-symmetric. Many results of the theory of Lie algebras can be extended to left
Leibniz algebras (see [1, 2, 3]).
In 2004, Kinyon [15] proved that if (X,e) is a pointed Lie rack, TeXcarries a structure of left
Leibniz algebra. Moreover, in the case when the Lie rack structure is associated to a Lie group
Gthen the associated left Leibniz algebra is the Lie algebra of G.
Given a pointed Lie rack (X,e), for any a∈X, we denote by Ada:TeX=h−→ hthe
differential of Laat e. We have
La⊲b=La◦Lb◦L−1
aand Ada⊲b=Ada◦Adb◦Ad−1
a.
Thus Ad : X−→ GL(h) is an homomorphism of Lie racks. If we put
[u,v]⊲=d
dt |t=0
Adc(t)v,u,v∈h,c:] −ǫ, ǫ [−→ X,c(0) =e,c′(0) =u,
(h,[,]⊲) becomes a left Leibniz algebra.
2
Alinear Lie rack structure on a finite dimensional vector space Vis a Lie rack operation
(x,y)7→ x⊲ypointed at 0 and such that for any x, the map Lx:y7→ x⊲yis linear. A linear Lie
rack operation ⊲is called analytic if for any x,y∈V,
x⊲y=y+
∞
X
n=1
An,1(x,...,x,y),(4)
where for each n,An,1:V×...×V−→ Vis an n+1-multilinear map which is symmetric in the
nfirst arguments. In this case, A1,1is the left Leibniz bracket associated to ⊲.
If (h,[,]) is a left Leibniz algebra then the operation c
⊲:h×h−→ hgiven by
uc
⊲v=exp(adu)(v)
defines an analytic linear Lie rack structure on hsuch that the associated left Leibniz bracket on
T0h=his the initial bracket [ ,]. We call c
⊲the canonical linear Lie rack structure associated to
(h,[,]).
In this paper, we will study linear Lie rack structures with an emphasis on analytic linear Lie
rack structures.
Actually, there is a large class of linear Lie rack structures on (h,[,]) containing the canon-
ical one. This class was suggested to us by an example sent to us by Martin Bordemann. The
proof of the following proposition will be given in Section 2.
Proposition 1.1. Let (h,[,]) be a left Leibniz algebra, F :R−→ Ra smooth function and
P:h×...×h−→ Ra symmetric multilinear p-form such that, for any y,x1. . . , xp∈h,
p
X
i=1
P(x1,...,[y,xi],...,xp)=0.
Then the operation ⊲given by
x⊲y=exp(F(P(x,...,x))adx)(y)
is a linear Lie rack structure on hand its associated left Leibniz bracket is [,]⊲=F(0)[ ,].
Moreover, if F is analytic then ⊲is analytic .
This proposition shows that a left Leibniz algebra might be associated to many non equivalent
pointed Lie rack structures. For instance if one takes F(0) =0 in Proposition 1.1, the two pointed
Lie rack structures
x⊲0y=yand x⊲1y=exp(F(P(x,...,x)))adx)(y)
are two pointed Lie rack structures on hwhich are not equivalent (even locally near 0) and have
the same left Leibniz algebra, namely, the abelian one. This contrasts with the theory of Lie
groups where two Lie groups are locally equivalent near their unit elements if and only if they
have the same Lie algebra. Moreover, this proposition motivates the study of linear Lie rack
structures and gives a sense to the following definition.
3
Definition 1.1. A left Leibniz algebra (h,[,]) is called rigid if any analytic linear Lie rack
structure ⊲on hsuch that [,]⊲=[,]is given by
x⊲y=exp(F(P(x,...,x)))adx)(y),
where F :R−→ Ris analytic with F (0) =1and P :h×...×h−→ Ris a symmetric multilinear
p-form such that, for any y,x1. . . , xp∈h,
p
X
i=1
P(x1,...,[y,xi],...,xp)=0.
Remark 1. We have seen that the abelian left Leibniz algebra is not rigid.
This paper is an introduction to the study of the rigidity of left Leibniz algebras. Our approach
was suggested to us by the one used in the study of linearization of Poisson structures (see [10]).
One of our main results is the following theorem.
Theorem 1.1. Let V be a real finite dimensional vector space and (An,1)n≥1a sequence of n +1-
multilinear maps symmetric in the n first arguments. We suppose that the operation ⊲given
by
x⊲y=y+
∞
X
n=1
An,1(x,...,x,y)
converges. Then ⊲is a Lie rack structure on V if and only if for any p,q∈N∗and x,y,z∈V,
Ap,1(x,Aq,1(y,z)) =X
s1+...+sq+k=p
Aq,1(As1,1(x,y),...,Asq,1(x,y),Ak,1(x,z)),(5)
where for sake of simplicity Ap,1(x,y) :=Ap,1(x,...,x,y).
In particular, if p =q=1we get that [,] :=A1,1is a left Leibniz bracket which is actually
the left Leibniz bracket associated to (V,⊲).
Remark 2. When p =1and q ∈N∗, the relation (5) becomes
LxAq,1(y1,...,yq+1) :=[x,Aq,1(y1,...,yq+1)] −
q+1
X
i=1
Aq,1(y1,...,[x,yi], . . . , yq+1)=0.(6)
A multilinear map on a left Leibniz algebra satisfying (6) will be called invariant. Thus
Theorem 1.1 reduces the study of analytic linear Lie rack structures to the study of the datum
of a left Leibniz algebra with a sequence of invariant multilinear maps satisfying a sequence of
multilinear equations. Even though equations (5) are complicated, we will see in this paper that
they are far more easy to handle than the distributivity condition (1). In Section 2 we will give
the proofs of Proposition 1.1 and Theorem 1.1 and we will show that there is a large class of non
rigid left Leibniz algebras (see Corollary 2.1). On the other hand, when q=1 the equation (5)
has a cohomological interpretation with respect to the cohomology of the left Leibniz algebra
(V,[,]). When H0=H1=0 we can deduce a refined expression of the (An,1) (see Theorem
3.1 in Section 3). By using Theorems 1.1 and 3.1 we will prove that sl2(R) and so(3) are rigid
(see Sections 4). As the reader will see, the proof of the rigidity of sl2(R) and so(3) based on
Theorems 1.1 and 3.1 is quite difficult and has needed a deep understanding of the structure of
these Lie algebras as a simple Lie algebras. We think that the study of the following conjecture
can be a challenging mathematical problem.
Conjecture 1. Every simple Lie algebra is rigid in the sense of Definition 1.1.
4
2. Some classes of non rigid left Leibniz algebras, proofs of Proposition 1.1 and Theorem
1.1
The proof of Proposition 1.1 is a consequence of the following well-known result.
Proposition 2.1. Let (X,⊲)be a rack and J :X−→ X a map such that, for any x,y∈X,
J(x⊲y)=x⊲J(y), i.e., J ◦Lx=Lx◦J for any x. Then the operation
x⊲Jy=J(x)⊲y
defines a rack structure on X.
Proof. We have, for any x,y,z∈X,
x⊲J(y⊲Jz)=J(x)⊲(J(y)⊲z)
=(J(x)⊲J(y)) ⊲(J(x)⊲z)
=(J(J(x)⊲y)) ⊲(x⊲Jz)
=(x⊲Jy)⊲J(x⊲Jz).
Proof of Proposition 1.1.
Proof. We consider the map J:h−→ hgiven by J(x)=F(P(x,...,x))x. Since Pis invariant,
we have P(exp(adx)(y), . . . , exp(adx)(y)) =P(y,...,y) and hence J(xc
⊲y)=xc
⊲J(y) and one
can apply Proposition 2.1 to conclude.
The following proposition shows that the class of non rigid left Leibniz algebras is large.
Recall that if his a left Leibniz algebra then its center Z(h)={a∈h,[a,h]=[h,a]=0}.
Proposition 2.2. Let (h,[,]) be a left Leibniz algebra. Choose a scalar product h,ion h,
(a1,b1,...,ak,bk)a family of vectors in [h,h]⊥∩Z(h)⊥,(z1,...,zk)a family of vector in Z(h)
and f1,..., fk:R−→ Rwith f j(0) =0for j =1,...,k. If c
⊲is the canonical linear Lie rack
operation on hthen
x⊲y=xc
⊲y+
k
X
j=1
hy,bjifj(hx,aji)zj
is a linear Lie rack operation pointed at 0. Moreover, if f ′
j(0) =0for j =1,...,k then [,]⊲=
[,].
Proof. Note first that for any z∈Z(h) and for any x∈h,x⊲z=zand z⊲x=x. Moreover,
hbj,x⊲yi=hbj,xc
⊲yi=hbj,yiand (x⊲y)c
⊲z=(xc
⊲y)c
⊲zfor any x,y,z∈h. So, for any
5
x,y,z∈h,
x⊲(y⊲z)=x⊲(yc
⊲z)+
k
X
j=1
hz,bjifj(hy,aji)x⊲zj
=xc
⊲(yc
⊲z)+
k
X
j=1
hyc
⊲z,bjifj(hx,aji)zj+
k
X
j=1
hz,bjifj(hy,aji)zj,
=xc
⊲(yc
⊲z)+
k
X
j=1
hz,bjifj(hx,aji)zj+
k
X
j=1
hz,bjifj(hy,aji)zj,
(x⊲y)⊲(x⊲z)=(x⊲y)⊲(xc
⊲z)+
k
X
j=1
hz,bjifj(hx,aji)zj
=(x⊲y)c
⊲(xc
⊲z)+
k
X
j=1
hxc
⊲z,bjifj(hx⊲y,aji)zj+
k
X
j=1
hz,bjifj(hx,aji)zj
=(xc
⊲y)c
⊲(xc
⊲z)+
k
X
j=1
hz,bjifj(hy,aji)zj+
k
X
j=1
hz,bjifj(hx,aji)zj.
This proves the proposition.
Corollary 2.1. 1. Let hbe a left Leibniz algebra which is a Lie algebra such that [h,h]+
Z(h),h, Z(h),{0}. Then his not rigid.
2. Let hbe a left Leibniz algebra such that [h,h]+Z(h),hand Z(h)is not contained in [h,h].
Then his not rigid.
Proof. 1. By virtue of Definition 1.1, if his rigid then any linear analytic rack structure ⊲on
hsatisfies x⊲x=xfor any x∈h. Choose z∈Z(h)\ {0}, a scalar product h,ion hand
a∈[h,h]⊥∩Z(h)⊥with a,0. According to Proposition 2.2, the operation
x⊲y=xc
⊲y+hx,ai2hy,aiz
is an analytic linear Lie rack structure on hsatisfying [ ,]⊲=[,]. However, this
operation satisfies a⊲a=a+|a|6z,aand hence his not rigid.
2. We have also that if his rigid then any linear analytic rack structure ⊲on hsatisfies
x⊲y−xc
⊲y∈[h,h]. We proceed as the first case and we consider the same Lie rack
operation on hwith a∈Z(h) and a<[h,h] and we get a contradiction.
Remark 3. There is a large class of left Leibniz algebras satisfying the hypothesis of Corollary
2.1, for instance, any 2-step nilpotent Lie algebra belongs to this class.
Proof of Theorem 1.1.
Proof. Put A0,1(x,y)=y. We have
x⊲(y⊲z)=X
n∈N
An,1(x,...,x,y⊲z)
=X
n,p∈N
An,1(x,...,x,Ap,1(y,...,y,z)),
6
(x⊲y)⊲(x⊲z)=
∞
X
n=0
An,1(x⊲y,...,x⊲y,x⊲z)
=X
n,s1,...,sn,k
An,1(As1,1(x,y),...,Asn,1(x,y),Ak,1(x,z)).
By identifyin g the homogeneous com ponent of degree nin xand of degree pin yin both x⊲(y⊲z)
and (x⊲y)⊲(x⊲z) we get the desired relation.
The following result is an immediate and important consequence of Theorem 1.1.
Corollary 2.2. Let (h,[,]) be a left Leibniz algebra and c
⊲its canonical linear Lie rack opera-
tion. Then
xc
⊲y=
∞
X
n=0
A0
n,1(x,...,x,y)
where
A0
0,1(x,y)=y and A0
n,1(x1,...,xn,y)=1
(n!)2X
σ∈Sn
adxσ(1) ◦...◦adxσ(n)(y),
and S nis the group of permutations of {1,...,n}. Furthermore, the A0
n,1n∈Nsatisfy the sequence
of equations (5).
3. Analytic linear Lie racks structures over left Leibniz algebras with trivial 0-cohomology
and 1-cohomology
In this section, we recall the definition of the cohomology of a left Leibniz algebra. We will
give an important expression of the An,1defining an analytic linear Lie rack structure on a left
Leibniz algebra hwhen H0(h)=H1(h)=0.
Let (h,[,]) be a left Leibniz algebra. For any n≥0, the operator δ:Hom(⊗nh,h)−→
Hom(⊗n+1h,h) given by
δ(ω)(x0,...,xn)=
n−1
X
i=0
[xi, ω(x0,..., ˆxi,...,xn)] +(−1)n−1[ω(x0, . . . , xn−1),xn]
+X
i<j
(−1)i+1ω(x0,..., ˆxi,...,xj−1,[xi,xj],xj+1,...,xn),
satisfies δ2=0 and then defines a cohomology Hp(h) for p∈N. For any x∈hand F,G:h−→ h,
we have
δ(x)(m)=−[x,m] and δ(F)(y,z)=[y,F(z)] +[F(y),z]−F([y,z])
and one can see easily that
δ(F◦G)(y,z)=δ(F)(y,G(z)) +δ(F)(G(y),z)+F◦δ(G)(y,z)−[F(y),G(z)] −[G(y),F(z)].(7)
Remark 4. Let (h,[,]) be a left Leibniz algebra which is a Lie algebra. The cohomology of h
as left Leibniz algebra is different from its cohomology as a Lie algebra, however H0and H1are
the same for both cohomologies.
7
Now let’s take a closer look to equations (5) when q=1. Let (h,[,]) be a left Leibniz
algebra and (An,1)p∈Na sequence of (n+1)-multilinear maps on hwith values in hsymmetric in
the nfirst arguments and such that A0,1(x,y)=yand A1,1(x,y)=[x,y] . For sake of simplicity
we write An,1(x,y)=An,1(x,...,x,y).
Equation (5) for q=1 can be written for any x,y,z∈h,
Ap,1(x,[y,z]) =[y,Ap,1(x,z)] +[Ap,1(x,y),z]+
p−1
X
r=1
[Ar,1(x,y),Ap−r,1(x,z)].
Thus
δ(ix...ixAp,1)(y,z)=−
p−1
X
r=1
[Ar,1(x,y),Ap−r,1(x,z)],(8)
where ix...ixAp,1:h−→ h,y7→ Ap,1(x,...,x,y).
On the other hand, the sequence (A0
n,1)n∈Ndefining the canonical linear Lie rack structure of
h(see Corollary 2.2) satisfies (5) and hence
δ(ix...ixA0
p,1)(y,z)=−
p−1
X
r=1
[A0
r,1(x,y),A0
p−r,1(x,z)].(9)
If p=2, since A0,1=A0
0,1and A1,1=A0
1,1, Equations (8) and (9) implies that, for any x∈h,
δ(ixixA2,1−ixixA0
2,1)=0.
Since A2,1and A0
2,1are symmetric in the two first arguments this is equivalent to
δ(ixiyA2,1−ixiyA0
2,1)=0,for any x,y∈h.
This is a cohomological equation and if H1(h)=0 then there exists B2:h×h−→ hsuch that,
for any x,y,z∈h,
A2,1(x,y,z)=A0
2,1(x,y,z)+[B2(x,y),z].(10)
Moreover, if H0(h)=0 then B2is unique and symmetric and one can check easily that A2,1is
invariant if and only if B2is invariant.
We have triggered an induction process and, under the same hypothesis, the (Ap,1)p≥2satisfy
a similar formula as (10). This is the purpose of the following theorem.
Theorem 3.1. Let (h,[,]) be a left Leibniz algebra such that H0(h)=H1(h)=0. Let (An,1)n≥0
be a sequence where A0,1(x,y)=y and A1,1(x,y)=[x,y]and, for any n ≥2, An,1:h×...×h−→ h
is multilinear invariant and symmetric in the n first arguments. We suppose that the An,1satisfy
(8). Then there exists a unique sequence (Bn)n≥2of invariant symmetric multilinear maps Bn:
h×...×h−→ hsuch that, for any x,y∈h,
An,1(x,y)=A0
n,1(x,y)+X
1≤k≤hn
2i
s=l1+...+lk≤n
A0
k,1(Bl1(x),...,Blk(x),A0
n−s,1(x,y)),(11)
where Ap,1(x,y)=Ap,1(x,...,x,y)and Bl(x)=Bl(x,...,x).
8
Remark 5. Formula (11) deserves some explications. As any formula depending inductively on
n, to find the general form one needs to check it for the first values of n and it is what we have
done. There are the formulas we found directly and which helped us to establish the expression
(11).
A3,1(x,y)=A0
3,1(x,y)+[B2(x),A0
1,1(x,y)] +[B3(x),A0
0,1(x,y)]
=A0
3,1(x,y)+A0
1,1(B2(x),A0
1,1(x,y)) +A0
1,1(B3(x),A0
0,1(x,y)),
A4,1(x,y)=A0
4,1(x,y)+[B4(x),A0
0,1(x,y)] +[B3(x),A0
1,1(x,y)] +[B2(x),A0
2,1(x,y)]
+1
2[B2(x),[B2(x),A0
0,1(x,y)]],
A5,1(x,y)=A0
5,1(x,y)+[B5(x),y]+[B4(x),A0
1,1(x,y)] +[B3(x),A0
2,1(x,y)] +[B2(x),A0
3,1(x,y)]
1
2([B2(x),[B3(x),y]] +[B3(x),[B2(x),y]])+1
2[B2(x),[B2(x),A0
1,1(x,y)]].
To prove Theorem 3.1, we will proceed by induction. The proof is rather technical and needs
some preliminary formulas.
Fix n≥2 and x∈h. For any 1 ≤k≤hn+1
2iand s=l1+...+lk≤n+1, in the proof of
Theorem 3.1, we will need to compute δ(Fk◦Gs) where Fk,Gs:h−→ hare given by
Fk(y)=A0
k,1(Bl1(x),...,Blk(x),y),Gs(y)=A0
n+1−s,1(x,y).
This is straightforward from (7) and the formula
δ(ix1...ixkA0
k,1)(y,z)=−1
k!
k−1
X
p=1X
σ∈Sk
[A0
p,1(xσ(1),...,xσ(p),y),A0
k−p,1(xσ(p+1),...,xσ(k),z)]
whose polar form is (9). We use here the well-know fact that two symmetric multilinear forms
are equal if and only if their polar forms are equal. For sake of simplicity put Q(k,s)=δ(Fk◦
Gs)(y,z).
Proposition 3.1. We have
Q(1,s)=−
n−s
X
r=0
[A0
1,1(Bs(x),A0
r,1(x,y)),A0
n+1−s−r,1(x,z)] −
n+1−s
X
r=1
[A0
r,1(x,y),A0
1,1(Bs(x),A0
n+1−s−r,1(x,z))],s≤n,
Q(k,n+1) =−1
k!
k−1
X
h=1X
σ∈Sk
[A0
h,1(Blσ(1)(x),...,Blσ(h)(x),y),A0
k−h,1(Blσ(h+1)(x),..., Blσ(k)(x),z)],k≥2,
Q(k,s)=−1
k!
n+1−s
X
r=0
k−1
X
p=1X
σ∈Sk
[A0
p,1(Blσ(1)(x),...,Blσ(p)(x),A0
r,1(x,y)),A0
k−p,1(Blσ(p+1)(x),..., Blσ(k)(x),A0
n+1−s−r,1(x,z))],
−
n+1−s
X
r=1
[A0
r,1(x,y),A0
k,1(Bl1(x),...,Blk(x),A0
n+1−s−r,1(x,z))] −
n−s
X
r=0
[A0
k,1(Bl1(x),...,Blk(x),A0
r,1(x,y)),A0
n+1−s−r,1(x,z)],
k≥2,s≤n.
Proof of Theorem 3.1.
Proof. We prove the formula by induction on n. For n=2, the formula has been established in
(10).
9
Suppose that there exists a family (B2,...,Bn) where Bkis an invariant symmetric k-from on
hwith values in hsuch that for any 2 ≤r≤n,
Ar,1(x,y)=A0
r,1(x,y)+X
1≤k≤hr
2i
l1+...+lk=s≤r
A0
k,1(Bl1(x),...,Blk(x),A0
r−s,1(x,y)).(12)
We look for Bn+1:h×...×h−→ hsymmetric and invariant such that
An+1,1(x,y)=A0
n+1,1(x,y)+X
1≤k≤hn+1
2i
l1+...+lk=s≤n+1
A0
k,1(Bl1(x),...,Blk(x),A0
n+1−s,1(x,y))
=[Bn+1(x),y]+A0
n+1,1(x,y)+X
1≤k≤hn+1
2i
l1+...+lk=s≤n+1
l1,...,lk≤n
A0
k,1(Bl1(x),...,Blk(x),A0
n+1−s,1(x,y))
=[Bn+1(x),y]+R(x)(y),
where R(x) depends only on (B2,...,Bn).
The idea of the proof is to show that, for any x∈h,δ(D(x)) =0 where D(x) : h−→ his given
by D(x)(y)=An+1,1(x,y)−R(x)(y). Then since H0(h)=H1(h)=0 there exists a unique Bn+1
satisfying D(x)(y)=[Bn+1(x),y]. By using the fact that D(x) is the polar form of a symmetric
form and H0(h)=0 one can see that Bn+1(x) is the polar form of a symmetric form which is also
invariant.
Let us compute now δ(D(x)). According to (8), we have
δ(ix...ixAn+1,1)(y,z)=−
n
X
r=1
[Ar,1(x,...,x,y),An+1−r,1(x,...,x,z)].
By expanding this relation using our induction hypothesis given in (12), we get that
δ(ix. . . ixAn+1,1)(y,z)=δ(ix...ixA0
n+1,1)(y,z)+S+T+U,
where
S=−
n−1
X
r=1X
1≤k≤hn+1−r
2i
s=l1+...+lk≤n+1−r
[A0
r,1(x, . . . , x,y),A0
k,1(Bl1(x), . . . , Blk(x),A0
n+1−r−s,1(x,z))],
T=−
n
X
r=2X
1≤k≤hr
2i
s=l1+...+lk≤r
[A0
k,1(Bl1(x), . . . , Blk(x),A0
r−s,1(x,y)),A0
n+1−r,1(x,...,x,z)],
U=−
n−1
X
r=2X
1≤k≤hr
2i
s1=l1+...+lk≤r
X
1≤h≤hn+1−r
2i
s2=p1+. . . +ph≤n+1−r
[A0
k,1(Bl1(x), . . . , Blk(x),A0
r−s1,1(x,y)),A0
h,1(Bp1(x), . . . , Bph(x),A0
n+1−r−s2,1(x,z))].
10
On the other hand, if we denote
Dk,s(x)(y)=A0
k,1(Bl1(x),...,Blk(x),A0
n+1−s,1(x,y)),
we remark that the computation of δ(R(x)) is based on Proposition 3.1 where we have computed
the δ(Dk,s(x)).
To conclude, we need to show that
S+T+U=X
1≤k≤hn+1
2i
l1+...+lk=s≤n+1
l1,...,lk≤n
Q(k,s),
where Q(k,s) is given in Proposition 3.1. Let S1and T1be the terms in Sand Tcorresponding
to k=1. We have
S1=−
n−1
X
r=1X
2≤s≤n+1−r
[A0
r,1(x,...,x,y),A0
1,1(Bs(x),A0
n+1−r−s,1(x,z))],
T1=−
n
X
r=2X
2≤s≤r
[A0
1,1(Bs(x),A0
r−s,1(x,y)),A0
n+1−r,1(x,...,x,z)].
On the other hand,
X
2≤s≤n
Q(1,s)=−X
2≤s≤n
n−s
X
r=0
[A0
1,1(Bs(x),A0
r,1(x,y)),A0
n+1−s−r,1(x,z)] −X
2≤s≤n
n+1−s
X
r=1
[A0
r,1(x,y),A0
1,1(Bs(x),A0
n+1−s−r,1(x,z))].
Since
{(r,s),1≤r≤n−1,2≤s≤n+1−r}={(r,s),1≤r≤n+1−s,2≤s≤n},
{(r−s,s),2≤r≤n,2≤s≤r}={(r,s),0≤r≤n−s,2≤s≤n},
S1+T1=P2≤s≤nQ(1,s). In the same way, one can see easily that
S−S1+T−T1=−X
2≤k≤hn+1
2i
l1+...+lk=s≤n+1
n+1−s
X
r=1
[A0
r,1(x,y),A0
k,1(Bl1(x), . . . , Blk(x),A0
n+1−s−r,1(x,z))] −
n−s
X
r=0
[A0
k,1(Bl1(x), . . . , Blk(x),A0
r,1(x,y)),A0
n+1−s−r,1(x,z)].
To conclude, we must show that Q1=Q2where
Q1=−X
2≤k≤[n+1
2]
l1+...+lk=s≤n+1
1
k!
n+1−s
X
r=0
k−1
X
p=1X
σ∈Sk
[A0
p,1(Blσ(1) (x), . . . , Blσ(p)(x),A0
r,1(x,y)),A0
k−p,1(Blσ(p+1) (x),...,Blσ(k)(x),A0
n+1−s−r,1(x,z))].
11
Q2=−
n−1
X
r=2X
1≤p≤hr
2i
s1=l1+...+lp≤r
X
1≤h≤hn+1−r
2i
s2=m1+...+mh≤n+1−r
[A0
p,1(Bl1(x),...,Blp(x),A0
r−s1,1(x,y)),A0
h,1(Bm1(x),...,Bmh(x),A0
n+1−r−s2,1(x,z))].
Denote by N2={2,3, . . .}, for any l=(l1,...,lk)∈Nk,|l|=l1+... +lk, for any σ∈Sk,
lσ=(lσ(1),...,lσ(k)) and for k∈n1,...,hn+1
2ioand s≤n+1 put
S(k,s)=n(l, σ, r,p)∈Nk
2×Sk×N×N,|l|=s,0≤r≤n+1−s,1≤p≤k−1o
and for any (l, σ, r,p)∈ S(k,s) put
Φ(l, σ, r,p)=[A0
p,1(Blσ(1)(x),...,Blσ(p)(x),A0
r,1(x,y)),A0
k−p,1(Blσ(p+1) (x),...,Blσ(k)(x),A0
n+1−s−r,1(x,z))].
Thus
Q1=−X
2≤k≤hn+1
2i,s≤n+1
1
k!X
(l,σ,r,p)∈S(k,s)
Φ(l, σ, r,p).
The map Sk× S(k,s)−→ S(k,s), (µ, (l, σ, r,p)) 7→ (lµ, σ ◦µ−1,r,p) defines a free action of
Skand the map [l, σ, r,p]7→ (lσ,r,p) identifies the quotient ^
S(k,s) to
^
S(k,s)=n(l,r,p)∈Nk
2×N×N,|l|=s,0≤r≤n+1−s,1≤p≤k−1o.
Moreover, Φ(l, σ, r,p)= Φ(lµ, σ ◦µ−1,r,p) so
Q1=−X
2≤k≤hn+1
2i,s≤n+1
X
(l,r,p)∈]
S(k,s)
Φ(l,Id,r,p).
On the other hand, put
T=((l,m,p,q,r)∈Np
2×Nq
2×N×N×N,2≤r≤n−1,|l| ≤ r,|m| ≤ n+1−r,1≤p≤r
2,1≤q≤"n+1−r
2#).
We have
Q2=X
(l,m,p,q,r)∈T
Ψ(l,m,p,q,r),
where
Ψ(l,m,p,q,r)=[A0
p,1(Bl1(x),...,Blp(x),A0
r−|l|,1(x,y)),A0
h,1(Bm1(x),...,Bmh(x),A0
n+1−r−|m|,1(x,z))].
We consider now
J:T−→ [
2≤k≤hn+1
2i,s≤n+1
^
S(k,s)
12
given by
J(l,m,p,q,r)=((l,m),r− |l|,p)∈^
S(p+q,|l|+|m|).
Indeed, it is obvious that 2 ≤p+q≤hn+1
2i, 1 ≤p≤p+q−1. Moreover, since |l| ≤ rand
|m| ≤ n+1−rthen 0 ≤r− |l| ≤ n+1−(|l|+|m|) and hence ((l,m),r− |l|,p)∈^
S(p+q,|l|+|m|)..
Jis a bijection since we have
J−1(l,p,r)=((l1,...,lp),(lp+1,...,lk),p,k−p,s=r+l1+...+lp)∈T.
Indeed, we have
2≤k≤"n+1
2#,1≤p≤k−1 and 0 ≤r≤n+1− |l|.
This implies obviously that 1 ≤pand 1 ≤k−p. Now 2p≤l1+...+lpand hence p≤hs
2i. This
with k≤hn+1
2iimply that k−p≤hn+1−s
2i. It is obvious that s≥2 and from 0 ≤r≤n+1− |l|,
we get
s≤n+1−(lp+1+...lk)≤n−1 and lp+1+...+lk≤n+1−s.
This completes the proof.
4. Analytic linear Lie rack structures on sl2(R) and so(3)
We denote by sl2(R) the Lie algebra of traceless real 2 ×2-matrices and by so(3) the Lie
algebra of skew-symmetric real 3 ×3-matrices. We consider them as left Leibniz algebras and
the purpose of this section is to prove that they are rigid in the sense of Definition 1.1. Namely,
we will prove the following theorem.
Theorem 4.1. Let hbe either sl2(R)or so(3) and ⊲an analytic linear Lie rack structure on h
such that [,]⊲is the Lie algebra bracket of h. Then there exists an analytic function F :R−→ R
given by
F(u)=1+
∞
X
k=1
akuk
such that, for any x,y∈h,
x⊲y=exp(F(hx,xi)adx)(y),
where hx,xi=1
2tr(adx◦adx). So his rigid.
The proof of this theorem is based on Theorem 3.1. So the first step is the determination
of symmetric invariant multilinear maps on h=sl2(R) and so(3). To achieve that, we use the
Chevalley restriction theorem for vector-valued functions proved in [16]. We recall its statement
as explained in [4].
Let gbe a complex semi-simple Lie algebra, ha Cartan subalgebra, Gthe connected and
simply connected Lie group of gand Hthe maximal torus in Ggenerated by expG(h). We denote
by NG(H) the normalizer of Hin G. Note that for any a∈NG(H), Adaleaves hinvariant and
W={Ada|h,a∈NG(H)}is the Weyl group of h. Let B:g×... ×g−→ gbe a symmetric
n-multilinear map which is g-invariant, i.e., for any a∈Gand any x1,...,xn∈g,
B(Adax1,...,Adaxn)=AdaB(x1,...,xn).(13)
13
This is equivalent to
[y,B(x1, . . . , xn)] =
n
X
i=1
B(x1,...,[y,xi],...,xn),y,x1,...,xn∈g.(14)
We denote by Sg
n(g,g) the vector space of g-invariant n-multilinear symmetric forms on gwith
values in g.
Let B∈Sg
n(g,g) and we denote by e
Bits restriction to h. From (14), we get that for any
y,x1,...,xn∈h,
[y,e
B(x1,...,xn)] =0
and hence e
B(x1,...,xn)∈h(since his a maximal abelian subalgebra). So e
Bdefines a n-
multilinear symmetric map e
B:h×...×h−→ hwhich is W-invariant. If we denote by SW
n(h,h)
the vector space of G-invariant n-multilinear symmetric forms on hwith values in h, we get a
map Res : Sg
n(g,g)−→ SW
n(h,h).
Theorem 4.2 ([16]).Res is injective.
Let gbe a real semi-simple Lie algebra. The definition of Sg
n(g,g) is similar to the complex
case. The complexified Lie algebra gC=g⊕igof gis also semi-simple and we have an injective
map Sg
n(g,g)−→ SgC
n(gC,gC), which assigns to each g-invariant n-multilinear invariant form B
on gthe unique C-multilinear map BCfrom gC×...×gCto gCwhose restriction to gis B. By
using (14) one can see easily that since Bis g-invariant then BCis gC-invariant.
We will now apply Theorem 4.2 and the embedding above to compute Sg
n(g,g) for any n∈N∗
when g=sl2(C), g=sl2(R) or g=so(3).
Let g=sl2(C), g=sl2(R) or g=so(3). For any n∈N∗, we define P:g2n−→ K(K=R,C)
by
Pn(x1,...,x2n)=1
(2n)! X
σ∈S2n
hxσ(1),xσ(2) i...hxσ(2n−1),xσ(2n)iand P0=1,
where hx,xi=1
2tr(ad2
x). This defines a symmetric invariant form on gand the map Bg
n:g2n+1−→
ggiven by
Bg
n(x1,...,x2n+1)=
2n+1
X
k=1
Pn(x1,..., ˆxk,...,x2n+1)xk
is symmetric and invariant.
Theorem 4.3. Let g=sl2(C). Then, for any n ∈N∗, we have
Sg
2n(g,g)=0and S g
2n+1(g,g)=CBg
n.
Proof. A Cartan subalgebra of gis h=C 1 0
0−1!which is one dimensional and hence, for any
n∈N∗, the dimension of SW
n(h,h) is less than or equal to ≤1. By virtue of Theorem 4.2 we get
dim Sg
n(g,g)≤1. Moreover, the associated Lie group to his H=( z0
0z−1!,z∈C∗)and one
can see easily that a= 0 1
−1 0!∈NG(H) and hence Ada|h∈W. Now
Ada 1 0
0−1!=a 1 0
0−1!a−1=− 1 0
0−1!.
14
Thus for any B∈SW
n(h,h), the invariance by Adaimplies that, for any x1,...,xn∈h,
(−1)nB(x1,...,xn)=−B(x1,...,xn).
So if nis even then B=0 and hence Sg
n(g,g)=0. If n=2p+1 is odd, the restriction theorem
shows that dim Sg
2p+1(g,g)≤1 and since Bg
p∈Sg
2p+1(g,g) we get the result.
If g=sl2(R) or g=so(3) then gCis isomorphic to sl2(C) and since the invariants of gare
embedded in the invariants of gCwe get the following corollary.
Corollary 4.1. If g=sl2(R)or g=so(3) then, for any n ∈N∗, we have
Sg
2n(g,g)=0and S g
2n+1(g,g)=RBg
n.
Let us pursue our preparation of the proof of Theorem 4.1. Let h=sl2(R) or so(3,R) and
x∈h. Then
x= a b
c−a!or x=
0a b
−a0c
−b−c0
.
Put
hx,xi=(1
2tr(ad2
x)=2tr(x2)=4(a2+bc) if h=sl2(R),
1
2tr(ad2
x)=1
2tr(x2)=−a2−b2−c2if h=so(3).
The following formula which is true in both sl2(R) and so(3) is easy to check and will play a
crucial role in the proof of Theorem 4.1. Indeed, for any x,y∈h,
adx◦adx(z)=−hx,zix+hx,xiz.(15)
This implies easily, by virtue of Corollary 2.2, that
A0
2n,1(x,y)=hx,xin−1
(2n)! ad2
x(y)=hx,xin
(2n)! y−hx,xin−1hx,yi
(2n)! x,n≥1,
A0
2n+1,1(x,y)=hx,xin
(2n+1)! [x,y],n≥0.
(16)
Proposition 4.1. Let hbe either sl2(R)or so(3) and ⊲an analytic linear Lie rack product on
hsuch that [,]⊲is the Lie algebra bracket of h. Then there exists a sequence (Un)n∈N∗with
U1=1, U2=1
2, for any x,y∈h,
x⊲y=y+
∞
X
n=0
U2n+1hx,xin
[x,y]+
∞
X
n=1
U2nhx,xin−1
ad2
x(y)
and for any n ∈N∗,
U2n=1
2
n−1
X
r=0
U2r+1U2(n−r)−1−
n−1
X
r=1
U2rU2(n−r).
15
Proof. By virtue of Theorem 1.1, x⊲y=
∞
X
n=0
An,1(x,y) where the sequence (An,1) satisfies (5).
Moreover, since his simple H0(h)=H1(h)=0 and we can apply Theorem 3.1. Thus
An,1(x,y)=A0
n,1(x,y)+X
2k≤s=l1+...+lk≤n
1≤k≤hn
2i
A0
k,1(Bl1(x),...,Blk(x),A0
n−s,1(x,y)),
and the Blare symmetric invariant. By virtue of Corollary 4.1,
B2l=0 and B2l+1(x)=clhx,xilx.
Thus
An,1(x,y)=A0
n,1(x,y)+X
2k≤s=2l1+...+2lk+k≤n
1≤k≤hn
2i
cl1...clkhx,xil1+...+lkA0
k,1(x,A0
n−s,1(x,y)).
But by using Corollary 2.2, we have A0
k,1(x,A0
n−s,1(x,y)) =(n+k−s)!
k!(n−s)! A0
n−2(l1+...+lk)(x,y) and
hence we can write
An,1(x,y)=
[n−1
2]
X
l=0
Kn,lhx,xilA0
n−2l(x,y),
where Kn,lare constant such that Kn,0=1. Note that in particular A2,1(x,y)=A0
2,1(x,y). Now by
using (16), we get
A2n+1,1(x,y)=
n
X
l=0
K2n+1,lhx,xilA0
2(n−l)+1(x,y)
=
n
X
l=0
K2n+1,lhx,xil1
(2(n−l)+1)! hx,xin−l[x,y]
=U2n+1hx,xin[x,y]=C2n+1A0
2n+1,1(x,y),
where U2n+1=C2n+1
(2n+1)! are constant. In the same way, one can show that there exists constants
U2n=C2n
(2n)! such that
A2n,1(x,y)=U2nhx,xin−1ad2
x(y)=C2nA0
2n,1(x,y),
and get the desired expression of x⊲y.
On the other hand, The equation (5) for q=1 and p=2nholds for both the Anand the A0
nso
we get
A2n,1(x,[y,z]) =C2nA0
2n,1(x,[y,z])
=C2n[y,A0
2n,1(x,z)] +C2n[A0
2n,1(x,y),z]+C2n
2n−1
X
r=1
[A0
r,1(x,y),A0
2n−r,1(x,z)]
=C2n[y,A0
2n,1(x,z)] +C2n[A0
2n,1(x,y),z]+
2n−1
X
r=1
CrC2n−r[A0
r,1(x,y),A0
2n−r,1(x,z)].
16
Thus 2n−1
X
r=1
(C2n−CrC2n−r)[A0
r,1(x,y),A0
2n−r,1(x,z)] =0
and hence
0=
n−1
X
r=1
(C2n−C2rC2(n−r))[A0
2r,1(x,y),A0
2(n−r),1(x,z)]+
n−1
X
r=0
(C2n−C2r+1C2(n−r)−1)[A0
2r+1,1(x,y),A0
2(n−r−1)+1,1(x,z)].
By using (16) we get
0=
n−1
X
r=1
(C2n−C2rC2(n−r))hx,xin−2
(2r)!(2(n−r)!) [ad2
x(y),ad2
x(z)] +
n−1
X
r=0
(C2n−C2r+1C2(n−r)−1)hx,xin−1
(2r+1)!(2(n−r)−1)! [[x,y],[x,y]].
One can show easily by using (15) that
[ad2
x(y),ad2
x(z)] +hx,xi[[x,y],[x,z]] =0
and deduce that
n−1
X
r=1
1
(2r)!(2(n−r)!) (C2n−C2rC2(n−r))=
n−1
X
r=0
1
(2r+1)!(2(n−r)−1)! (C2n−C2r+1C2(n−r)−1).
On the other hand
0=(1 −1)2n=
n
X
r=0
(2n)!
(2r)!(2(n−r))! −
n−1
X
r=0
(2n)!
(2r+1)!(2(n−r)−1)! ,
and finally,
C2n
(2n)! =1
2
n−1
X
r=0
1
(2r+1)!(2(n−r)−1)! C2r+1C2(n−r)−1−
n−1
X
r=1
1
(2r)!(2(n−r)!) C2rC2(n−r).
To complete the proof, it suffices to replace Cr
r!by Ur.
Proof of Theorem 4.1.
Proof. According to Proposition 4.1, there exists a sequence (Un)n∈N∗with U1=1, U2=1
2, for
any x,y∈h,
x⊲y=y+
∞
X
n=0
U2n+1hx,xin
[x,y]+
∞
X
n=1
U2nhx,xin−1
ad2
x(y)
and for any n∈N∗,
U2n=1
2
n−1
X
r=0
U2r+1U2(n−r)−1−
n−1
X
r=1
U2rU2(n−r).(17)
17
We will show that there exists a unique sequence (an)n≥1such that the function F(t)=1+
P∞
t=1antnsatisfies
x⊲y=exp(F(hx,xi)adx)(y)=y+
∞
X
n=0
F(hx,xi)2n+1A0
2n+1,1(x,y)+
∞
X
n=1
F(hx,xi)2nA0
2n,1(x,y).
Thus
exp(F(hx,xi)adx)(y)=y+
∞
X
n=0
[F(hx,xi)]2n+1hx,xin
(2n+1)!
[x,y]+
∞
X
n=1
[F(hx,xi)]2nhx,xin−1
(2n)!
ad2
x(y).
Put [F(hx,xi)]n=P∞
m=0Bn,mhx,ximand compute the coefficients Bn,m. Indeed,
[F(hx,xi)]n=1+a1hx,xi+a2hx,xi2+...+amhx,xim+Rn
=1+a1hx,xi+a2hx,xi2+...+amhx,ximn+P,
where Pcontains terms of degree ≥m+1. The multinomial theorem gives
1+a1hx,xi+a2hx,xi2+...+amhx,ximn=X
k0+...+km=n
n!
k0!k1!...km!ak1
1...akm
mhx,xik1+2k2+...+mkm.
Thus
Bn,0=1 and Bn,m=X
k1+2k2+...+mkm=m,k0+k1+...+km=n
n!
k0!k1!...km!ak1
1...akm
mfor m≥1.
So
∞
X
n=0
F(hx,xi)2n+1hx,xin
(2n+1)! =
∞
X
n=0
∞
X
m=0
B2n+1,mhx,xim+n
(2n+1)! =
∞
X
n=0
n
X
p=0
B2p+1,n−p
(2p+1)!
hx,xin,
∞
X
n=1
F(hx,xi)2nhx,xin−1
(2n)! =
∞
X
n=1
∞
X
m=0
B2n,mhx,xim+n−1
(2n)! =
∞
X
n=1
n
X
p=1
B2p,n−p
(2p)!
hx,xin−1.
For sake of simplicity and clarity, put
Vn,m(a1,...,am)=Bn,m
n!=X
k1+2k2+...+mkm=m,k0+k1+...+km=n
ak1
1...akm
m
k0!k1!...km!.
To prove the theorem we need to show that there exists a unique sequence (an)n≥1such that
U2n+1=
n
X
p=0
V2p+1,n−p(a1,...,an−p),n≥1,(18)
U2n=
n
X
p=1
V2p,n−p(a1,...,an−p),n≥1.(19)
18
Note first that the relation (17) and the fact that U2=1
2defines the sequence (U2n)n≥1entirely in
function of the sequence (U2n+1)n≥0. On the other hand, since V1,n(a1,...,an)=anand U1=1
then
U3=a1+1
3! and U2n+1=an+
n
X
p=1
V2p+1,n−p(a1,...,an−p),n≥2.
Since the quantity Pn
p=1V2p+1,n−p(a1,...,an−p) depends only on (a1,...,an−1), these relations
define inductively and uniquely the sequence (an)n≥1in function of (U2n+1)n≥0. To achieve the
proof we need to prove (19). We will proceed by induction and we will use the following relation
∂Vn,m
∂al
(a1,...,am)=Vn−1,m−l(a1,...,am−l),l=1,...,m.(20)
Indeed,
∂Vn,m
∂al
(a1,...,am)=X
k1+2k2+...+mkm=m,k0+k1+...+km=n,kl≥1
ak1
1...akl−1
l...akm
m
k0!k1!...(kl−1)! ...km!
k′
l=kl−1
=X
k1+2k2+...+lk′
l+...+mkm=m−l,k0+k1+...+k′
l+...+km=n−1
ak1
1...ak′
l
l...akm
m
k0!k1!...(k′
l)! ...km!.
To conclude, it suffices to remark that in the relation
k1+2k2+...+lk′
l+...+mkm=m−l
the left side is a sum of nonnegative numbers and the right side is nonnegative so (m−l+
1)km−l+1=...=mkm=0 and hence the relation is equivalent to
k1+2k2+...+(m−l)km−l=m−l.
Now, we are able to prove (19). We proceed by induction. For n=1, we have U2=1
2and
V2,0=1
2. Suppose that the relation holds from 1 to n−1. By virtue of (17), we have
U2n=1
2
n−1
X
r=0
U2r+1U2(n−r)−1−
n−1
X
r=1
U2rU2(n−r)
and all the Urappearing in this formula are given by (18) and (19) this implies that U2nis a
function of (a1,...,an−1) and we can put U2n=H(a1,...,an−1). We can also put
n
X
p=1
V2p,n−p(a1,...,an−p)=G(a1,...,an−1).
To show that U2nsatisfies (19) is equivalent to showing
H(0) =G(0) and ∂H
∂al
=∂G
∂al
,l=1,...n−1.
19
But Vn,m(0) =0 if m≥1 and Vn,0(0) =1
n!. Hence
H(0) =1
2
n−1
X
r=0
1
(2r+1)!(2(n−r)−1)! −
n−1
X
r=1
1
(2r)!(2(n−r))!
=1
2
n−1
X
r=0
1
(2r+1)!(2(n−r)−1)! −
n
X
r=0
1
(2r)!(2(n−r))!
+1
(2n)!
=−1
2(1 −1)2n+1
(2n)! =1
(2n)! ,
G(0) =V2n,0(0) =1
(2n)! =H(0).
For r=0,...,n−1, by induction hypothesis U2r+1is given by (18) and by using (20) one can see
easily that ∂U2r+1
∂al
=U2(r−l)if l=1,...,rand 0 if l≥r+1. Similarly, we have ∂U2r
∂al
=U2(r−l)−1
if l=1,...,r−1 and 0 if l≥r. For sake of simplicity, we put
∂U2r+1
∂al
=U2(r−l)and ∂U2r
∂al
=U2(r−l)−1
with the convention U0=1 and Us=0 if sis negative. Then, for l=1,...,n−1, we have
∂H
∂al
=1
2
n−1
X
r=0 ∂U2r+1
∂al
U2(n−r)−1+∂U2(n−r)−1
∂al
U2r+1!−
n−1
X
r=1 ∂U2r
∂al
U2(n−r)+∂U2(n−r)
∂al
U2r!
=1
2
n−1
X
r=0U2(r−l)U2(n−r)−1+U2(n−r−l−1)U2r+1−
n−1
X
r=1U2(r−l)−1U2(n−r)+U2(n−r−l)−1U2r
=1
2
n−1−l
X
r=0
U2rU2(n−r−l)−1+1
2
n−1
X
r=0
U2(n−r−l−1)U2r+1−1
2
n−l−2
X
r=0
U2r+1U2(n−r−l−1) −1
2
n−1
X
r=1
U2(n−r−l)−1U2r
=1
2U2(n−l)−1+1
2
n−1
X
r=n−l−1
U2(n−r−l−1)U2r+1−1
2
n−1
X
r=n−l
U2(n−r−l)−1U2r
=U2(n−l)−1.
This completes the proof.
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