Table Of ContentSTRUCTURE OF THE EXTENDED SCHRO¨DINGER-VIRASORO
LIE ALGEBRA sv∗
e
8
0 SHOULAN GAO, CUIPO JIANG†, AND YUFENG PEI
0
2
Abstract. In this paper, we study the derivations, the central extensions and the
n
automorphismgroupoftheextendedSchr¨odinger-VirasoroLiealgebrasv,introduced
a
J by J. Unterberger [25] in the context of two-dimensional conformal fieeld theory and
4 statisticalphysics. Moreover,weshowthatsvisaninfinite-dimensionalcompleteLie
1 algebra and the universal central extensioneof sv in the category of Leibniz algebras
is the same as that in the category of Lie algeberas.
]
A
R
.
h
t
a
1. Introduction
m
[ The Schr¨odinger-Virasoro Lie algebra sv, originally introduced by M. Henkel in [8]
1 during his study on the invariance of the free Schr¨odinger equation, is a vector space
v over the complex field C with a basis {L ,M ,Y | n ∈ Z} and the Lie brackets:
8 n n n+1
2
5
0 1−m
2 [L ,L ] = (n−m)L , [L ,M ] = nM , [L ,Y ] = (n+ )Y ,
. m n m+n m n m+n m n+21 2 m+n+12
1
0 [Mm,Mn] = 0, [Ym+1,Yn+1] = (m−n)Mm+n+1, [Mm,Yn+1] = 0,
8 2 2 2
0 for all m,n ∈ Z. It is easy to see that sv is a semi-direct product of the centerless
:
v Virasoroalgebra(Wittalgebra)Vir = CL ,whichcanberegardedastheLieal-
i 0 Ln∈Z n
X
gebra that consists of derivations on the Laurent polynomial ring [14], and the two-step
ar nilpotent infinite-dimensional Lie algebra h = Lm∈ZCYm+1 Lm∈ZCMm, which con-
2
tains the Schr¨odinger Lie algebra s spanned by {L ,L ,L ,Y ,Y ,M }. Clearly s is
−1 0 1 −1 1 0
2 2
isomorphictothesemi-direct productoftheLiealgebrasl(2)andthethree-dimensional
nilpotent Heisenberg Lie algebra hY ,Y ,M i. The structure and representation the-
−1 1 0
2 2
ory of sv have been extensively studied by C. Roger and J. Unterberger. We refer the
reader to [22] for more details. Recently, in order to investigate vertex representations
of sv, J. Unterberger [25] introduced a class of new infinite-dimensional Lie algebras sv
called the extended Schr¨odinger-Virasoro algebra (see section 2), which can be vieweed
as an extension of sv by a conformal current with conformal weight 1.
Keywords: Schro¨dinger-Virasoro algebra, central extension, derivation, automor-
phism.
∗ Supported in part by China NSF grant 10571119.
† Corresponding author: [email protected] .
1
2 GAO,JIANG, ANDPEI
Inthispaper,wegiveacompletedescriptionofthederivations, thecentralextensions
and the automorphisms for the extended Schr¨odinger-Virasoro Lie algebra sv. The
paper is organized as follows: e
In section 2, we show that the center of sv is zero and all derivations of sv are inner
derivations, i.e., H1(sv,sv) = 0, which impliees that sv is a complete Lie algeebra. Recall
that a Lie algebra isecaleled complete if its center isezero and all derivations are inner,
which was originally introduced by N. Jacobson in [10]. Over the past decades, much
progress has been obtained on the theory of complete Lie algebras ( see for example
[13, 26, 20]). Note that since the center of sv is non-zero and dim H1(sv,sv) = 3, sv is
not a complete Lie algebra.
In section 3, we determine the universal central extension of sv. The universal
covering algebra sv contains the twisted Heisenberg-Virasoro Lie algeebra, which plays
an important rolebin the representation theory of toroidal Lie algebras [2, 3, 11, 24], as
a subalgebra. Furthermore, in section 4, we show that there is no non-zero symmetric
invariant bilinear form on sv, which implies the universal central extension of sv in the
category of Leibniz algebraes is the same as that in the category of Lie algebraes [9].
Finallyinsection5, wegivetheautomorphismgroupsofsvanditsuniversal covering
algebra sv, which are isomorphic. e
Throubghout the paper, we denote by Z and C∗ the set of integers and the set of
non-zero complex numbers respectively, and all the vector spaces are assumed over the
complex field C.
2. The Derivation Algebra of sv
e
Definition 2.1. The extended Schr¨odinger-Virasoro Lie algebra sv is a vector space
spanned by a basis {L ,M ,N ,Y | n ∈ Z} with the following berackets
n n n n+1
2
[L ,L ] = (n−m)L , [M ,M ] = 0, [N ,N ] = 0, [Y ,Y ] = (m−n)M ,
m n m+n m n m n m+1 n+1 m+n+1
2 2
1−m
[L ,M ] = nM , [L ,N ] = nN , [L ,Y ] = (n+ )Y ,
m n m+n m n m+n m n+21 2 m+n+21
[N ,M ] = 2M , [N ,Y ] = Y , [M ,Y ] = 0,
m n m+n m n+1 m+n+1 m n+1
2 2 2
for all m,n ∈ Z.
It is clear that sv is a perfect Lie algebra, i.e., [sv,sv] = sv , which is finitely
generated with a seet of generators {L ,L ,L ,L ,Ne,Ye}. e
−2 −1 1 2 1 1
2
1
Define a Z-grading on sv by
2
e
1
deg(L ) = n, deg(M ) = n, deg(N ) = n, deg(Y ) = n+ ,
n n n n+21 2
for all n ∈ Z. Then
sev = Mn∈Zsevn2 = (Mn∈Zsevn)M(Mn∈Zsevn+12),
STRUCTURE OF THE EXTENDED SCHRO¨DINGER-VIRASORO LIE ALGEBRA 3
where sv = span{L ,M ,N } and sv = span{Y } for all n ∈ Z.
n n n n n+1 n+1
e e 2 2
Definition 2.2. ([5]) Let G be a commutative group, g = g a G-graded Lie algebra.
L g
g∈G
A g-module V is called G-graded, if
V = MVg∈G, ggVh ⊆ Vg+h, ∀ g,h ∈ G.
g∈G
Definition 2.3. ([5]) Let g be a Lie algebra and V a g-module. A linear map
D : g −→ V is called a derivation, if for any x,y ∈ g, we have
D[x,y] = x.D(y)−y.D(x).
If there exists some v ∈ V such that D : x 7→ x.v, then D is called an inner derivation.
Let g be a Lie algebra, V a module of g. Denote by Der(g,V) the vector space of
all derivations, Inn(g,V) the vector space of all inner derivations. Set
H1(g,V) = Der(g,V)/Inn(g,V).
Denote by Der(g) the derivation algebra of g, Inn(g) the vector space of all inner
derivations of g. We will prove that all the derivations of sv are inner derivations.
By Proposition 1.1 in [5], we have the following lemma.e
Lemma 2.4.
Der(sv) = MDer(sv)n,
2
e n∈Z e
where Der(sv)n(svm) ⊆ svm+n for all m,n ∈ Z.
e 2 e 2 e 2
(cid:3)
Lemma 2.5. H1(sv0,svn) = 0, ∀ n ∈ Z\{0}.
2
e e
Proof. We have to prove
H1(sv ,sv ) = 0, ∀n ∈ Z\{0},
0 n
e e
H1(sv ,sv ) = 0, ∀n ∈ Z.
0 n+1
e e 2
(1) For m 6= 0, let ϕ : sv −→ sv be a derivation. Assume that
0 m
e e
ϕ(L ) = a L +b M +c N ,
0 1 m 1 m 1 m
ϕ(M ) = a L +b M +c N ,
0 2 m 2 m 2 m
ϕ(N ) = a L +b M +c N ,
0 3 m 3 m 3 m
where a ,b ,c ∈ C,i = 1,2,3. Since
i i i
ϕ[L ,M ] = [ϕ(L ),M ]+[L ,ϕ(M )],
0 0 0 0 0 0
we have
a mL +(b m+2c )M +c mN = 0.
2 m 2 1 m 2 m
4 GAO,JIANG, ANDPEI
1
So a = 0,c = − b m,c = 0 and ϕ(M ) = b M . Since
2 1 2 2 0 2 m
2
ϕ[L ,N ] = [ϕ(L ),N ]+[L ,ϕ(N )],
0 0 0 0 0 0
we have
a mL +(b m−2b )M +c mN = 0.
3 m 3 1 m 3 m
1
Then a = 0,b = b m,c = 0 and ϕ(N ) = b M . Therefore, we get
3 1 3 3 0 3 m
2
1 1
ϕ(L ) = a L + b mM − b mN , ϕ(M ) = b M , ϕ(N ) = b M .
0 1 m 3 m 2 m 0 2 m 0 3 m
2 2
a 1 1
1
Let X = L + b M − b N , we have
m m 3 m 2 m
m 2 2
ϕ(L ) = [L ,X ], ϕ(M ) = [M ,X ], ϕ(N ) = [N ,X ].
0 0 m 0 0 m 0 0 m
Then ϕ ∈ Inn(sv ,sv ). Therefore,
0 m
e e
H1(sv ,sv ) = 0, ∀ m ∈ Z\{0}.
0 m
e e
(2) For all m ∈ Z, let ϕ : sv −→ sv be a derivation. Assume that
0 m+1
e e 2
ϕ(L ) = aY , ϕ(M ) = bY , ϕ(N ) = cY ,
0 m+1 0 m+1 0 m+1
2 2 2
for some a,b,c ∈ C. Because ϕ[L ,M ] = [ϕ(L ),M ]+[L ,ϕ(M )], we have
0 0 0 0 0 0
1
0 = [L ,ϕ(M )] = [L ,bY ] = b(m+ )Y .
0 0 0 m+21 2 m+12
So b = 0 and ϕ(M ) = 0. Since ϕ[L ,N ] = [ϕ(L ),N ]+[L ,ϕ(N )], we have
0 0 0 0 0 0 0
1
0 = [aY ,N ]+[L ,cY ] = (c(m+ )−a)Y .
m+21 0 0 m+12 2 m+12
Then a = c(m+ 1). Hence, we get
2
1
ϕ(L ) = c(m+ )Y , ϕ(M ) = 0, ϕ(N ) = cY .
0 2 m+12 0 0 m+21
Letting X = cY , we obtain
m+1 m+1
2 2
ϕ(L ) = [L ,X ], ϕ(M ) = [M ,X ], ϕ(N ) = [N ,X ].
0 0 m+1 0 0 m+1 0 0 m+1
2 2 2
Then ϕ ∈ Inn(sv ,sv ). Therefore,
0 m+1
e e 2
H1(sv ,sv ) = 0, ∀ m ∈ Z.
0 m+1
e e 2
(cid:3)
Lemma 2.6. Homsev0(svm2 ,svn2) = 0 for all m,n ∈ Z, m 6= n.
e e
STRUCTURE OF THE EXTENDED SCHRO¨DINGER-VIRASORO LIE ALGEBRA 5
Proof. Let f ∈ Homsev0(svm2 ,svn2), where m 6= n. Then for any E0 ∈ sv0,Em2 ∈ svm2 ,
we have e e e e
f([E0,Em]) = [E0,f(Em)].
2 2
Then f([L0,Em]) = [L0,f(Em)], i.e.,
2 2
m n
f(Em) = [L0,f(Em)] = f(Em).
2 2 2 2 2
So we have f(Em) = 0 for all m 6= n. Therefore, we have f = 0. (cid:3)
2
By Lemma 2.5-2.6 and Proposition 1.2 in [5], we have the following Lemma.
Lemma 2.7. Der(sv) = Der(sv) +Inn(sv). (cid:3)
0
e e e
Lemma 2.8. For any D ∈ Der(sv) , there exist some a,b,c ∈ C such that
0
e c a
D = ad(aL − M +(b− )N ).
0 0 0
2 2
Therefore, Der(sv) ⊆ Inn(sv).
0
e e
Proof. For any D ∈ Der(sv) , assume that for all m ∈ Z,
0
e
D(L ) = amL +amM +amN ,
m 11 m 12 m 13 m
D(M ) = amL +amM +amN ,
m 21 m 22 m 23 m
D(N ) = amL +amM +amN ,
m 31 m 32 m 33 m
D(Ym+1) = bm+12Ym+1,
2 2
where amij, bm+12 ∈ C, i,j = 1,2,3. For any Em ∈ svm, En ∈ svn, we have
2 2 2 2
e e
D[Em,En] = [D(Em),En]+[Em,D(En)].
2 2 2 2 2 2
In particular, D[L0,Em] = [D(L0),Em]+[L0,D(Em)]. Then
2 2 2
m m
D(Em) = [D(L0),Em]+ D(Em).
2 2 2 2 2
So
[D(L0),Em] = 0.
2
Since [D(L ),L ] = 0,[D(L ),M ] = 0 and [D(L ),N ] = 0, we can deduce that
0 1 0 0 0 0
a0 = a0 = a0 = 0.
11 12 13
So D(L ) = 0. Because D[M ,L ] = [D(M ),L ]+[M ,D(L )], we have
0 0 m 0 m 0 m
a0 mL −2amM = 0.
21 m 13 m
Then
a0 = 0, am = 0, D(M ) = a0 M +a0 N .
21 13 0 22 0 23 0
By the fact that D[M ,N ] = [D(M ),N ]+[M ,D(N )], we have
0 m 0 m 0 m
amL +amM +amN = (a0 +am)M .
21 m 22 m 23 m 22 33 m
6 GAO,JIANG, ANDPEI
Therefore,
am = 0, am = a0 +am, am = 0.
21 22 22 33 23
Then
a0 = 0, D(M ) = amM = (a0 +am)M .
33 m 22 m 22 33 m
Since D[N ,L ] = [D(N ),L ]+[N ,D(L )], we have
0 m 0 m 0 m
a0 mL +2amM = 0.
31 m 12 m
Then a0 m = 0,2am = 0 for all m ∈ Z. Therefore,
31 12
a0 = 0, am = 0, D(N ) = a0 M .
31 12 0 32 0
According to D[N ,N ] = [D(N ),N ]+[N ,D(N )], we obtain
m n m n m n
(amn−an m)N +2(an −am)M = 0.
31 31 m+n 32 32 m+n
Then amn = an m,an = am for all m,n ∈ Z. Hence, we have
31 31 32 32
am = a1 m, am = a0 , ∀ m ∈ Z.
31 31 32 32
By the following two relations
D[Y ,Y ] = [D(Y ),Y ]+[Y ,D(Y )],
m+1 n+1 m+1 n+1 m+1 n+1
2 2 2 2 2 2
D[L ,L ] = [D(L ),L ]+[L ,D(L )],
m n m n m n
we have
am+n+1 = bm+12 +bn+21, m 6= n. (2.1)
22
am+n = am +an , m 6= n. (2.2)
11 11 11
It is easy to see that a−m = −am for all m ∈ Z. Let n = 1 in (2.2), then
11 11
am+1 = am +a1 , m 6= 1.
11 11 11
By induction on m ∈ Z+ and m ≥ 3, we have
am = a2 +(m−2)a1 , m ≥ 3.
11 11 11
Let m = 4,n = −2 in (2.2), then we have a2 = 2a1 . Therefore,
11 11
am = ma1 , ∀ m ∈ Z.
11 11
Since D[L ,M ] = [D(L ),M ]+[L ,D(M )] and
m n m n m n
D[L ,N ] = [D(L ),N ]+[L ,D(N )],
m n m n m n
we get
am+n = ma1 +an , n 6= 0, (2.3)
22 11 22
a1 (m+n) = a1 (n−m), am+n = ma1 +an , n 6= 0.
31 31 33 11 33
Then we can deduce that
a1 = 0, am+1 = ma1 +a1 .
31 33 11 33
STRUCTURE OF THE EXTENDED SCHRO¨DINGER-VIRASORO LIE ALGEBRA 7
Note a0 = 0, then a1 = a1 . Therefore,
33 11 33
am = ma1 , am = a0 +ma1 , ∀ m ∈ Z.
33 11 22 22 11
Because D[L ,Y ] = [D(L ),Y ]+[L ,D(Y )], we have
m n+1 m n+1 m n+1
2 2 2
1−m
bm+n+12 = ma1 +bn+21, n+ 6= 0.
11 2
Let m = 1, then
bn+1+12 = a1 +bn+12, n 6= 0.
11
By (2.1), we get
a0 +(m+n+1)a1 = bm+12 +bn+12, m 6= n. (2.4)
22 11
Let m = 0,n = 1 in (2.4), then we have
a0 +a1 = 2b12.
22 11
Let n = 0 in (2.4), then we obtain bm+12 = b12 + ma1 for all m ∈ Z. Set
11
a1 = a,b12 = b,a0 = c, then a0 = 2b−a and
11 32 22
D(L ) = maL , D(M ) = (2b−a+ma)M ,
m m m m
D(N ) = cM +maN , D(Y ) = (b+ma)Y ,
m m m m+1 m+1
2 2
for all m ∈ Z. Then we can deduce that
c a
D = ad(aL − M +(b− )N ).
0 0 0
2 2
(cid:3)
From the above lemmas, we obtain the following theorem.
Theorem 2.9. Der(sv) = Inn(sv), i.e., H1(sv,sv) = 0.
e e e e
Lemma 2.10. C(sv) = {0}, where C(sv) is the center of sv.
e e e
Proof. For any En ∈ svn, we have
2 2
e
n
[L0,En] = En.
2 2 2
It forces x ∈ sv , for any x ∈ C(sv), since [L ,x] = 0. Let x = aL +bN +cM , where
0 0 0 0 0
a,b,c ∈ C. Theen e
[x,L ] = [aL +bN +cM ,L ] = [aL ,L ] = aL = 0.
1 0 0 0 1 0 1 1
So a = 0. By the following relations,
[x,Y ] = [bN +cM ,Y ] = [bN ,Y ] = bY = 0,
1 0 0 1 0 1 1
2 2 2 2
[x,N ] = [cM ,N ] = −2cM = 0,
0 0 0 0
we have b = 0 and c = 0. Therefore, x = 0. (cid:3)
By Lemma 2.10 and Theorem 3.1, we have
8 GAO,JIANG, ANDPEI
Corollary 2.11. sv is an infinite-dimensional complete Lie algebra.
e
3. The Universal Central Extension of sv
e
In this section, we discuss the structure of the universal central extension of sv.
Let us first recall some basic concepts. Let g be a Lie algebra. A bilinear functioen
ψ : g × g −→ C is called a 2-cocycle on g if for all x,y,z ∈ g, the following two
conditions are satisfied:
ψ(x,y) = −ψ(y,x),
ψ([x,y],z)+ψ([y,z],x)+ψ([z,x],y) = 0. (3.1)
For any linear function f : g −→ C, one can define a 2-cocycle ψ as follows
f
ψ (x,y) = f([x,y]), ∀ x,y ∈ g.
f
Such a 2-cocycle is called a 2-coboudary on g. Denote by C2(g,C) the vector space of
2-cocycles on g, B2(g,C) the vector space of 2-coboundaries on g. Then the quotient
space H2(g,C) = C2(g,C)/B2(g,C) is called the second cohomology group of g.
Theorem 3.1. dimH2(sv,C) = 3.
e
Proof. Let ϕ : sv×sv −→ C be a 2-cocycle on sv. Let f : sv −→ C be a linear function
defined by e e e e
1 1
f(L ) = − ϕ(L ,L ), f(L ) = ϕ(L ,L ), m 6= 0,
0 1 −1 m 0 m
2 m
1
f(M ) = −ϕ(L ,M ), f(M ) = ϕ(L ,M ), m 6= 0,
0 1 −1 m 0 m
m
1
f(N ) = −ϕ(L ,N ), f(N ) = ϕ(L ,N ), m 6= 0,
0 1 −1 m 0 m
m
1
f(Y ) = ϕ(L ,Y ), ∀ m ∈ Z.
m+21 m+ 1 0 m+21
2
Let ϕ = ϕ−ϕ , where ϕ is the 2-coboundary induced by f, then
f f
ϕ(x,y) = ϕ(x,y)−f([x,y]), ∀ x,y ∈ sv.
e
By the known result on the central extension of the classical Witt algebra ( see [1] or
[15] ), we have
ϕ(L ,L ) = αδ (m3 −m), ∀ m,n ∈ Z, α ∈ C.
m n m+n,0
By the fact that ϕ([L ,L ],M )+ϕ([L ,M ],L )+ϕ([M ,L ],L ) = 0, we get
0 m n m n 0 n 0 m
(m+n)ϕ(L ,M ) = nϕ(L ,M ).
m n 0 m+n
So
n
ϕ(L ,M ) = ϕ(L ,M ), m+n 6= 0. (3.2)
m n 0 m+n
m+n
Then it is easy to deduce that
ϕ(L ,M ) = 0, m+n 6= 0.
m n
STRUCTURE OF THE EXTENDED SCHRO¨DINGER-VIRASORO LIE ALGEBRA 9
Furthermore,
ϕ(L ,M ) = ϕ(L ,M )+mf(M ) = ϕ(L ,M )−mϕ(L ,M ). (3.3)
m −m m −m 0 m −m 1 −1
Similarly, denote ϕ(L ,N ) = c(m) for all m ∈ Z, then we have
m −m
ϕ(L ,N ) = δ c(m), ϕ(L ,N ) = ϕ(L ,N )−mϕ(L ,N ). (3.4)
m n m+n,0 m −m m −m 1 −1
By (3.1), ϕ([L ,L ],Y )+ϕ([L ,Y ],L )+ϕ([Y ,L ],L ) = 0, so
0 m n+1 m n+1 0 n+1 0 m
2 2 2
1 1−m
(m+n+ )ϕ(L ,Y ) = (n+ )ϕ(L ,Y ).
2 m n+21 2 0 m+n+21
Then for all m,n ∈ Z, we get
n+ 1−m
ϕ(L ,Y ) = 2 ϕ(L ,Y ).
m n+21 m+n+ 1 0 m+n+21
2
Consequently, we have
ϕ(L ,Y ) = 0, m,n ∈ Z.
m n+1
2
From the relation that
ϕ([N ,Y ],Y )+ϕ([Y ,Y ],N )+ϕ([Y ,N ],Y ) = 0, m ∈ Z,
0 m+1 −m−1 m+1 −m−1 0 −m−1 0 m+1
2 2 2 2 2 2
we have
1
ϕ(Y ,Y ) = (m+ )ϕ(N ,M ), m ∈ Z. (3.5)
m+21 −m−21 2 0 0
By (3.1), ϕ([L ,Y ],Y )+ϕ([Y ,Y ],L )+ϕ([Y ,L ],Y ) = 0, then
m p+1 q+1 p+1 q+1 m q+1 m p+1
2 2 2 2 2 2
1−m 1−m
(p+ )ϕ(Y ,Y )+(p−q)ϕ(M ,L )−(q+ )ϕ(Y ,Y ) = 0,
2 m+p+21 q+21 p+q+1 m 2 m+q+12 p+21
for all m,p,q,∈ Z. Let m = −p−q −1, then for all p,q ∈ Z, we have
p+q p+q
(p+1+ )ϕ(Y ,Y )+(p−q)ϕ(M ,L )−(q+1+ )ϕ(Y ,Y ) = 0.
2 −q−21 q+21 p+q+1 −p−q−1 2 −p−21 p+21
(3.6)
Using (3.5), we get
p+q +1
ϕ(L ,M ) = ϕ(N ,M ), p 6= q. (3.7)
−p−q−1 p+q+1 0 0
2
Letting p = −2,q = 0 in (3.7), we have
1
ϕ(L ,M ) = − ϕ(N ,M ) = −ϕ(Y ,Y ). (3.8)
1 −1 2 0 0 21 −21
It follows from (3.5) and (3.7) that
ϕ(Y ,Y ) = −(2m+1)ϕ(L ,M ), m ∈ Z, (3.9)
m+1 −m−1 1 −1
2 2
ϕ(L ,M ) = −mϕ(Y ,Y ) = mϕ(L ,M ), m ∈ Z. (3.10)
m −m 1 −1 1 −1
2 2
By (3.3) and (3.10), we have ϕ(L ,M ) = 0 for all m ∈ Z. Therefore,
m −m
ϕ(L ,M ) = 0, ∀ m,n ∈ Z.
m n
10 GAO,JIANG, ANDPEI
Accordingto(3.1),ϕ([L ,L ],N )+ϕ([L ,N ],L )+ϕ([N ,L ],L ) = 0,
m n −m−n n −m−n m −m−n m n
then we have
(n−m)ϕ(L ,N )−(m+n)ϕ(N ,L )+(m+n)ϕ(N ,L ) = 0.
m+n −m−n −m m −n n
By (3.4), we get
(n−m)c(m+n)+(m+n)c(m) = (m+n)c(n). (3.11)
Let n = 1 in (3.11) and note c(0) = c(1) = 0, then we obtain
(m−1)c(m+1) = (m+1)c(m). (3.12)
Let n = 1−m in (3.11), then
c(m) = c(1−m), ∀ m ∈ Z.
By induction on m, we deduce that c(−1) = c(2) determines all c(m) for m ∈ Z. On
the other hand, c(m) = m2 −m is a solution of equation (3.12). So
c(m) = β(m2 −m), β ∈ C,
is the general solution of equation (3.12). Therefore,
ϕ(L ,N ) = δ β(m2 −m), ∀ m,n ∈ Z, β ∈ C. (3.13)
m n m+n,0
By (3.1), we have
ϕ([M ,Y ],Y )+ϕ([Y ,Y ],M )+ϕ([Y ,M ],Y ) = 0.
p m+1 n+1 m+1 n+1 p n+1 p m+1
2 2 2 2 2 2
Then(m−n)ϕ(M ,M ) = 0.Letm+n+1 = q,thenwehave(2m+1−q)ϕ(M ,M ) = 0.
m+n+1 p q p
This means that
ϕ(M ,M ) = 0, ∀ p,q ∈ Z.
p q
Therefore, we have
ϕ(M ,M ) = ϕ(M ,M ) = 0, ∀ m,n ∈ Z. (3.14)
m n m n
By (3.1), ϕ([L ,M ],N )+ϕ([M ,N ],L )+ϕ([N ,L ],M ) = 0, we have
0 m n m n 0 n 0 m
(m+n)ϕ(M ,N ) = −2ϕ(L ,M ). (3.15)
m n 0 m+n
Then for m+n 6= 0, we have
2
ϕ(M ,N ) = ϕ(M ,N )+ ϕ(L ,M ) = 0.
m n m n 0 m+n
m+n
On the other hand,
ϕ(M ,N ) = ϕ(M ,N )+2f(M ) = ϕ(M ,N )−2ϕ(L ,M ).
m −m m −m 0 m −m 1 −1
By (3.1), ϕ([N ,Y ],Y )+ϕ([Y ,Y ],N )+ϕ([Y ,N ],Y ) = 0, we have
p m+1 n+1 m+1 n+1 p n+1 p m+1
2 2 2 2 2 2
ϕ(Y ,Y )+(m−n)ϕ(M ,N )−ϕ(Y ,Y ) = 0.
p+m+1 n+1 m+n+1 p p+n+1 m+1
2 2 2 2
Let n = −m−p−1, then we have
ϕ(Y ,Y )+(2m+p+1)ϕ(M ,N )−ϕ(Y ,Y ) = 0.
p+m+1 −p−m−1 −p p −m−1 m+1
2 2 2 2
