Table Of ContentThe derivation algebra and automorphism group
of the twisted Schr¨odinger-Virasoro algebra1
Junbo Li∗,†), Yucai Su‡)
∗)Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
†)Department of Mathematics, Changshu Institute of Technology, Changshu 215500, China
‡)Department of Mathematics, University of Science and Technology of China, Hefei 230026, China
E-mail: sd [email protected], [email protected]
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0
0 Abstract. In this article, we determine the derivation algebra DerL and the automor-
2
phism group AutL of the twisted Schr¨odinger-Virasoro algebra L.
n
a Keywords: twistedSchr¨odinger-Virasoroalgebra;derivationalgebra; automorphismgroup.
J
5 MR(2000) Subject Classification: 17B05, 17B40, 17B65, 17B68.
1
]
A
R 1. Introduction
.
h
It iswell known that theinfinite-dimensional Schr¨odinger LiealgebrasandVirasoroalge-
t
a
braplayimportant rolesinmanyareasofmathematics andphysics (e.g., statisticalphysics).
m
[ The original Schr¨odinger-Virasoro Lie algebra was introduced in [5], in the context of non-
1 equilibrium statistical physics, containing as subalgebras both the Lie algebra of invariance
v
7 of the free Schr¨odinger equation and the Virasoro algebra. The infinite-dimensional Lie al-
0
gebra discussed in this article called the twisted Schr¨odinger-Virasoro algebra is the twisted
2
2
deformation of the original Schr¨odinger-Virasoro Lie algebra.
.
1
0 Both original and twisted Schr¨odinger-Virasoro Lie algebras are closely related to the
8
0 Schr¨odinger Lie algebras and the Virasoro Lie algebra (see [6], [7], [14] and [12]). They
:
v should consequently play a role akin to that of the Virasoro Lie algebra in two-dimensional
i
X
equilibrium statistical physics.
r
a
Motivated by the research for deformations and central extensions of both original and
twistedSchr¨odinger-VirasoroLiealgebras,thesetsofgeneratorsprovidedbythecohomology
classes of the cocycles were presented in [10]. In [13], the author constructed vertex algebra
representations of the Schr¨odinger-Virasoro Lie algebras out of a charged symplectic boson
and a free boson with its associated vertex operators. The purpose of this article is to
determine the derivation algebra and the automorphism group of the twisted Schr¨odinger-
Virasoro Lie algebras L. We show that the derivation algebra of L is the direct sum of
three linear independent outer derivations and the inner derivation algebra. Finally, we
characterize the automorphism group AutL of L.
Now we give the definition of the Lie algebra L. A Lie algebra L is called a twisted
1
SupportedbyNSFgrants10471091,10671027ofChina,OneHundredTalentsProgramfromUniversity
of Science and Technology of China.
1
Schr¨odinger-Virasoro Lie algebra (see [10]), if L has the C-basis
{L ,Y ,M ,C|n ∈ Z}
n n n
with the Lie brackets
n3 −n
[Ln,Ln′] = (n′ −n)Ln+n′ +δn,−n′ C, (1.1)
12
n
[L ,Y ] = (m− )Y , (1.2)
n m n+m
2
[L ,M ] = pM , (1.3)
n p n+p
[Ym,Ym′] = (m′ −m)Mm+m′, (1.4)
[Y ,M ] = [M ,M ] = [L,C] = 0. (1.5)
m p n p
ThetwistedSchr¨odinger-VirasoroLiealgebrahasaninfinite-dimensionaltwisted Schr¨odinger
subalgebra denoted by SS with the C-basis {Y ,M |n ∈ Z} and a Virasoro subalgebra de-
n n
noted by V with the C-basis {L ,C|n ∈ Z}. The action of the Virasoro subalgebra on the
n
Schr¨odinger subalgebra is natural. The center of L, denoted by C(L), is two-dimensional,
spanned by {M ,C}. Introduce a Z-gradation on L by degL = degY = degM = n,
0 n n n
degC = 0 and decompose L with respect to the following gradation:
L = L , L = Span {L ,Y ,M ,C}, L = Span {L ,Y ,M }, n ∈ Z\{0}.
n 0 C 0 0 0 n C n n n
n∈Z
M
Throughout the article, we denote by Z∗ the set of all nonzero integers and C∗ the set
of all nonzero complex numbers.
2. The derivation algebra DerL of L
Let V be a L-module. A linear map ϕ from L to V is called a derivation, if for any
x,y ∈ L, we have
ϕ[x,y] = x.ϕ(y)−y.ϕ(x).
For v ∈ V, the map φ : x → x.v is called an inner derivation.
Denote by Der(L,V) the vector space of all derivations, Inn(L,V) the vector space of
all inner derivations. Then the first cohomology group of L with coefficients in V is
H1(L,V) = DerC(L,V)/InnC(L,V). (2.1)
The right-hand side is also called the space of outer derivations.
By definition, the algebra L is a semidirect product of Z-graded algebras L = V⋉SS,
with V := CL ⊕CC being the Virasoro algebra and SS := (CY ⊕CM ) being
n∈Z n n∈Z n n
L L
2
the twisted Schr¨odinger algebra. Clearly, V is Z-graded by V = CL ⊕CC and V = CL
0 0 n n
for n ∈ Z \{0} while SS is Z-graded by SS = CY ⊕ CM for n ∈ Z. By [8], the short
n n n
exact sequence
{0} −→ SS −→ L −→ L/SS −→ {0},
induces a long exact sequence
{0} −→ H0(L,SS) −→ H0(L,L) −f→ H0(L,L/SS) −→
H1(L,SS) −→ H1(L,L) −→ H1(L,L/SS) −→ ··· (2.2)
of Z-graded vector spaces. Note that H0(L,SS) = SSL = CM , while H0(L,L) =
0
LL ∼= CM +CC, and H0(L,L/SS) = (L/SS)L ∼= CC. From this we see that the map
0
f is surjective. Thus (2.2) gives the following exact sequence
{0} −→ H1(L,SS) −→ H1(L,L) −→ H1(L,L/SS). (2.3)
The right-hand side of (2.3) can be computed from the exact sequence
{0} −→ H1(L/SS,L/SS) −→ H1(L,L/SS) −→ H1(SS,L/SS)L. (2.4)
Note thatH1(L/SS,L/SS) = H1(V,V) is thealgebra of outer derivations of the Virasoro
algebra V, which is equal to zero by [14], while H1(SS,L/SS)L can be embeded into
HomU(V)(SS/[SS,SS],V)(whereU(V)istheuniversalenveloping algebraofV). Therefore,
by (2.3) and (2.4), H1(L,L) will follow from the result of the computation of H1(L,SS)
and HomU(V)(SS/[SS,SS],V).
Lemma 2.1. Der(L,SS) = Der(L,SS) +Inn(L,SS), where
0
Der(L,SS) = {D ∈ Der(L,SS) | D(L ) ⊆ SS , for any n ∈ Z}.
0 n n
(cid:3)
Proof. This follows immediately from (1.2) of [1].
Lemma 2.2. D ∈ Der(L,SS) if and only if D is a linear map from L to SS satisfying
0
the following conditions:
(i) D(L ) = dn+d M ,
n 1 n
(ii) D(M ) =(cid:0)2g M , (cid:1)
n 0 n
(iii) D(Y ) = g Y ,
n 0 n
(iv) D(C) = 0,
for any n ∈ Z and some d, d , g ∈ C.
1 0
3
Proof. LetD bealinearmapformL toSS satisfying(i)–(iv). ItisclearthatD(L ) ⊆ SS ,
n n
and it is easy to check that D is a derivation from L to SS.
Let D ∈ Der(L,SS) . For any n ∈ Z, assume
0
D(L ) = c Y +d M , c , d ∈ C. (2.5)
n n n n n n n
Applying D to [L ,L ] = (n−1)L and [L ,L ] = (n+2)L , we obtain
1 n n+1 −1 n+1 n
n 1
d −nd +(n−1)d M = ( −1)c +(n− )c −(n−1)c Y , (2.6)
1 n n+1 n+1 1 n n+1 n+1
2 2
(cid:0) (cid:1) (cid:0) n+3 3 (cid:1)
(n+2)d −d −(n+1)d M = c −(n+2)c +(n+ )c Y . (2.7)
n −1 n+1 n −1 n n+1 n
2 2
(cid:0) (cid:1) (cid:0) (cid:1)
Comparing the coefficients of M ,Y in (2.6) and M ,Y in (2.7), one has
n+1 n+1 n n
(n−1)d = nd −d , (2.8)
n+1 n 1
n 1
(n−1)c = ( −1)c +(n− )c , (2.9)
n+1 1 n
2 2
(n+1)d = (n+2)d −d , (2.10)
n+1 n −1
3 n+3
(n+ )c = (n+2)c − c . (2.11)
n+1 n −1
2 2
Subtracting (2.8) from (2.10), and (2.9) from (2.11), we respectively obtain
d −d
1 −1
d −d = , (2.12)
n+1 n
2
n 1
c −c = − (c +c )+ (2c −3c ). (2.13)
n+1 n −1 1 1 −1
5 5
Denoting d1−d−1 = d and using induction on n in (2.12), one can deduce
2
d = d +(n−1)d, ∀ n ∈ Z. (2.14)
n 1
Taking n = −1 in (2.11), one obtains c = 0. Then taking n = 0 in (2.13) gives c = −c .
0 1 −1
From this, (2.13) can be rewritten as c − c = c . Thus c = nc . This together with
n+1 n 1 n 1
(2.9) gives
c = 0, ∀ n ∈ Z. (2.15)
n
According to the equations (2.14) and (2.15), for any n ∈ Z, we can rewrite (2.5) as
D(L ) = d +(n−1)d M , where d, d ∈ C. (2.16)
n 1 n 1
(cid:0) (cid:1)
Re-denoting d −d by d , we obtain (i).
1 1
4
Applying D to [L ,L ] = 6L −2C, one has
−3 3 0
6(d −d)M −2D(C) = 6D(L )−2D(C) = D [L ,L ]
1 0 0 −3 3
= [(d −4d)M ,L ]+[(cid:0)L ,(d +(cid:1)2d)M ]
1 −3 3 −3 1 3
= 3(d −4d)M +3(d +2d)M
1 0 1 0
= 6(d −d)M .
1 0
This gives D(C) = 0. Thus (iv) follows.
For any n ∈ Z, write
D(Y ) = g Y +h M , g , h ∈ C. (2.17)
n n n n n n n
Applying D to [L ,Y ] = (n− 1)Y and comparing the coefficients of Y and M , we
1 n 2 n+1 n+1 n+1
obtain
1 1 1
(n− )g = (n− )g , nh = (n− )h . (2.18)
n n+1 n n+1
2 2 2
Applying D to [L ,Y ] = 3Y and comparing the coefficients of Y and M , we obtain
−1 1 2 0 0 0
g = g , h = 3h . Then using induction on n in (2.18), one obtains
1 0 1 2 0
g = g , h = 0, ∀ n ∈ Z. (2.19)
n 0 n
Therefore, for any n ∈ Z, one can rewrite (2.17) as D(Y ) = g Y , i.e., (iii) follows.
n 0 n
For any m,n ∈ Z, applying D to [Y ,Y ] = (n−m)M and using (iii), one can easily
m n n
(cid:3)
obtain (ii).
Set d = 1, d = g = 0; d = 1, d = g = 0; g = 1, d = d = 0 in D ∈ Der(L,SS)
1 0 1 0 0 1 0
respectively, we obtain three derivations from L to SS:
D : D (L ) = M , D (M ) = D (Y ) = D (C) = 0;
1 1 n n 1 n 1 n 1
D : D (L ) = nM , D (M ) = D (Y ) = D (C) = 0;
2 2 n n 2 n 2 n 2
D : D (Y ) = Y , D (M ) = 2M , D (L ) = D (C) = 0;
3 3 n n 3 n n 3 n 3
where n ∈ Z. It is straightforward to verify that D , i = 1,2,3, are linear independent outer
i
derivations. Thus (2.1) and Lemma 2.2 prove Lemma 2.3 below.
Lemma 2.3. H1(L,SS) = CD +˙CD +˙CD .
1 2 3
Lemma 2.4. Hom (SS/[SS,SS],V) = 0.
U(V)
5
Proof. By the definition of SS, we have [SS,SS] = Span {M |n ∈ Z}. For any f ∈
C n
HomU(V)(SS/[SS,SS],V) and any n ∈ Z, one can assume
f(Y ) = p(n)L +c(n)C, for some p(n), c(n) ∈ C.
n k k k
k
P
For any m,n ∈ Z, applying f to both sides of [L ,Y ] = (n−m)Y , using f([L ,Y ]) =
m n m+n m n
[L ,f(Y )], one has
m n
m3 −m
(k −m)p(n)L +p(n) C = (n−m) p(m+n)L +c(m+n)C . (2.20)
k k+m −m 12 k k
k k
P (cid:0)P (cid:1)
Comparing the coefficients of C and L in (2.20), one has
k
m3 −m
p(n) = (n−m)c(m+n), (2.21)
−m 12
(m+n) (n)
(n−m)p = (k −2m)p . (2.22)
k k−m
Taking m = 0, n 6= 0 and m = 1,n = −1 in (2.21) respectively, one immediately obtains
C(n) = 0 for n ∈ Z. Using this in (2.21), one has
(n)
p = 0 for all k ∈/ {−1,0,1}. (2.23)
k
If k ∈ {−1,0,1}, by choosing some m such that 2n−m 6= 0 and k − m ∈/ {−1,0,1} and
(n)
replacing n by n− m in (2.22), and using (2.23), one again obtains p = 0. This proves
k
f(Y ) = 0 for all n ∈ Z. (cid:3)
n
Theorem 2.5. DerL = CD +˙CD +˙CD +˙adL.
1 2 3
Proof. By Lemma 2.4, H1(SS,L/SS)L = 0. According to the exact sequence (2.4) and
the fact H1(L/SS,L/SS) = 0, we have H1(L,L/SS) = 0. So by Lemma 2.3 and the
exact sequence (2.3), one has
H1(L,L) = CD +˙CD +˙CD .
1 2 3
Hence the theorem follows from H1(L,L) = DerL/adL. (cid:3)
3. The automorphism group AutL of L
Throughout this section, we denote by Inn(L) the set of inner automorphisms. Then
Inn(L) is a normal subgroup of AutL and
Inn(L) is generated by exp(kadL ), exp(kadY ) and exp(kadM ), (3.1)
0 n n
where n ∈ Z,k ∈ C.
Denote J = Span {M ,C|n ∈ Z} and J = Span {Y ,M ,C|n ∈ Z}. They both are
1 C n 2 C n n
ideals of L. And the following lemma is clear.
6
Lemma 3.1. For any ß ∈ AutL, one has ß(J ) ⊆ J (i = 1,2) and ß C(L) ⊆ C(L).
i i
(cid:0) (cid:1)
For any a ∈ C∗, we can define the following maps on L.
ß : L −→ anL , Y −→ anY , M −→ anM , C −→ C; (3.2)
a n n n n n n
ε : L −→ −L , Y −→ −Y , M −→ −M , C −→ −C; (3.3)
n −n n −n n −n
ψ : L −→ L , Y −→ aY , M −→ a2M , C −→ C. (3.4)
a n n n n n n
It is easy to see that ß ∈ InnL, ε,ψ ∈ AutL \InnL.
a a
Lemma 3.2. For any ß ∈ Aut(L), there exists an automorphism ξ ∈ Inn(L) of the form
exp b′adY + c′adM such that ξ−1σ(L ) = a(L +b Y +c M +d C), where
j6=0 j j k6=0 k k 0 0 0 0 0 0 0
a ∈ C∗,b ,c ,d ∈ C.
(cid:0)P 0 0 0 P (cid:1)
Proof. For any ß ∈ Aut(L), we can assume that
σ(L ) = a L + b Y + c M +a(L +b Y +c M +d C), (3.5)
0 i i j j k k 0 0 0 0 0 0
i6=0 j6=0 k6=0
P P P
where a ,b ,c ∈ C, a ∈ C∗ (we must have a 6= 0, otherwise, σ(L ) would be ad-locally
i j k 0
nilpotent while L is ad-semi-simple).
0
We shall find some ξ = exp b′adY + c′adM ∈ Inn(L), such that
j j k k
j6=0 k6=0
(cid:0) P P (cid:1)
σ(L ) = exp b′adY + c′adM a′(L +b′Y +c′M +d′C) . (3.6)
0 j j k k 0 0 0 0 0 0
j6=0 k6=0
(cid:0)P P (cid:1)(cid:0) (cid:1)
for some a′ ∈ C∗,b′,c′,d′ ∈ C. We only need to solve a′,b′,c′,d′ by comparing the
j k 0 j k 0
coefficients of L ,Y ,M and C, which gives
i j k
a = a′, b = b′, c = c′, d = d′, b′ = −a−1j−1b , (3.7)
0 0 0 0 0 0 j j
l(l−i)k−1a−2i−1l−1b b
i l
c′ = −b b′ −a−1k−1c − i+l=k . (3.8)
k 0 k k P 2
Applying ξ−1 to both sides of (3.5) and using (3.7), we obtain
ξ−1σ(L ) = a(L +b Y +c M +d C). (3.9)
0 0 0 0 0 0 0
Lemma 3.3. b = 0.
0
7
Proof. Since L is an ad-semi-simple element in L, then ξ−1σ(L ) = a(L +b Y +c M +
0 0 0 0 0 0 0
d C) is also ad-semi-simple in L. By linear algebra, the following matrix (given by the
0
adjoint action of ada(L +b Y +c M +d C) on the basis {L ,Y ,M ,C} of L )
0 0 0 0 0 0 n n n n
an 0 0 0
ab0n an 0 0
2
0 ab n an 0
0
0 0 0 0
can be diagonalized, which forces b = 0. This lemma follows. (cid:3)
0
According to Lemma 3.2 and 3.3, for any ß ∈ Aut(L), by replacing ß with ξß for some
ξ ∈ Inn(L), we can write
ß(L ) = a(L +c M +d C), (3.10)
0 0 0 0 0
where a ∈ C∗,c ,d ∈ C.
0 0
Lemma 3.4. For any ß ∈ Aut(L), by replacing ß with εß if necessary where ε is defined
in (3.3), we can write
ß(L ) = L +c M +d C, ß(Y ) = e′ Y +e′′M +δ e C, ß(M ) = f′M +δ f C,
0 0 0 0 0 n n n n n n,0 0 n n n n,0 0
for some n ∈ Z, e′ , f′ ∈ C∗, e ,f ,e′′ ∈ C.
n n 0 0 n
Proof. By Lemma 3.1, we can write
ß(Y ) = e′Y + e′′M +e C, (3.11)
n i i j j n
i j
P P
ß(M ) = f′M +f C, (3.12)
n i i n
i
P
where e′ ∈ C∗,f′ ∈ C∗(i 6= 0), e′′,e ,f′,f ∈ C, n ∈ Z.
i i j n 0 n
Applying ß to [L ,Y ] = nY and [L ,M ] = nM , we obtain
0 n n 0 n n
a(L +c M +d C), e′Y + e′′M +e C = n( e′Y + e′′M +e C),
0 0 0 0 i i j j n i i j j n
i j i j
h i
P P P P
a(L +c M +d C), f′M +f C = n( f′M +f C).
0 0 0 0 i i n i i n
i i
h i
P P
That is,
(ai−n)e′Y + (aj −n)e′′M −ne C = 0, (3.13)
i i j j n
i j
P P
(ai−n)f′M −nf C = 0. (3.14)
i i n
i
P
8
Comparing the coefficients of Y , M and C in the above two equations, we obtain
i j
n
i = ∈ Z, (aj −n)e′′ = 0, e = δ e and f = δ f , (3.15)
a j n n,0 0 n n,0 0
for all n ∈ Z. It follows that
a = ±1, i = an and e′′ = 0 if j 6= an. (3.16)
j
According to (3.15) and (3.16), one can respectively rewrite (3.11) and (3.12) as
ß(Y ) = e′ Y +e′′ M +δ e C, (3.17)
n an an an an n,0 0
ß(M ) = f′ M +δ f C. (3.18)
n an an n,0 0
For one case a = 1, the lemma is right. For the other case a = −1, replacing ß with εß, we
(cid:3)
can rewrite (3.10), (3.17) and (3.18) as those in the lemma. Then this lemma follows.
Lemma 3.5. Let ß ∈ Aut(L) be such that
ß(L ) = L +c M +d C, ß(Y ) = e′ Y +e′′M +δ e C, ß(M ) = f′M +δ f C,
0 0 0 0 0 n n n n n n,0 0 n n n n,0 0
where n ∈ Z, e′ , f′ ∈ C∗, e ,f ,e′′ ∈ C. Then by replacing ß by ψ ß ß for some x,y ∈ Z∗
n n 0 0 n x y
(where ß and ψ are defined in (3.2) and (3.4) respectively), one can suppose ß = δ for
y x α,β,γ
some α,β,γ ∈ C, where δ is defined by
α,β,γ
δ (C) = C, (3.19)
α,β,γ
δ (M ) = M , (3.20)
α,β,γ n n
δ (Y ) = Y +2αnM , (3.21)
α,β,γ n n n
δ (L ) = L +αnY + α2n2 +βn+γ M . (3.22)
α,β,γ n n n n
(cid:0) (cid:1)
Proof. For any n ∈ Z∗, we can write
σ(L ) = u L + v Y + w M +C , (3.23)
n i i j j k k n
i j k6=0
P P P
where u ∈ C∗,v ,w ∈ C, C ∈ C(L).
i j k n
For any n ∈ Z∗, applying σ to nL = [L ,L ], we obtain
n 0 n
nu L + nv Y + nw M +nC = nσ(L ) = σ [L ,L ]
i i j j k k n n 0 n
i j k6=0
P P P (cid:0) (cid:1)
= iu L + jv Y + kw M .
i i j j k k
i j k6=0
P P P
Comparing the coefficients of L , Y , M and C , one has
i i i n
u = v = w = 0 if i 6= n, and C = 0 for any n ∈ Z∗.
i i i n
9
Hence (3.23) can be rewritten as
σ(L ) = u L +v Y +w M where n ∈ Z∗. (3.24)
n n n n n n n
For n 6= 0,±1, applying σ to (n−1)L = [L ,L ], we obtain
n+1 1 n
(n−1)σ(L ) = (n−1)(u L +v Y +w M ) = [σ(L ),σ(L )]
n+1 n+1 n+1 n+1 n+1 n+1 n+1 1 n
1 n
= (n−1)u u L + (n− )u v +( −1)v u Y
1 n n+1 1 n 1 n n+1
2 2
(cid:0) (cid:1)
+ nu w +(n−1)v v −w u M .
1 n 1 n 1 n n+1
(cid:0) (cid:1)
Comparing the coefficients of L , Y and M , one has
n+1 n+1 n+1
u = u u , (3.25)
n+1 1 n
1 n
(n−1)v = (n− )u v +( −1)v u , (3.26)
n+1 1 n 1 n
2 2
(n−1)w = nu w +(n−1)v v −w u . (3.27)
n+1 1 n 1 n 1 n
Applying σ to [L ,L ] = −2L , we obtain
1 −1 0
−2σ(L ) = −2(L +c M +d C) = ß [L ,L ]
0 0 0 0 0 1 −1
3 (cid:0) (cid:1)
= −2u u L − (u v +v u )Y −(u w +2v v +w u )M .
1 −1 0 1 −1 1 −1 0 1 −1 1 −1 1 −1 0
2
Comparing the coefficients of L , Y , M and C, one has
0 0 0
u = u−1, (3.28)
−1 1
v = −u−2v , (3.29)
−1 1 1
1
c = (u w +2v v +w u−1), d = 0. (3.30)
0 2 1 −1 1 −1 1 1 0
By (3.10), we can rewrite ßL as
0
1
ß(L ) = L + (u w +2v v +w u−1)M .
0 0 2 1 −1 1 −1 1 1 0
Applying σ to [L ,L ] = −4L + 1C, we obtain
2 −2 0 2
1
−4(L +c M )+ ßC = ß [L ,L ]
0 0 0 2 −2
2
(cid:0) (cid:1) 1
= −4u u L −3(u v +v u )Y + u u C −2u w M
2 −2 0 2 −2 2 −2 0 2 −2 2 −2 0
2
−4v v M −2w u M .
2 −2 0 2 −2 0
10
