Unitary representations for the Schrödinger-Virasoro Lie algebra

0.15 MB·English

by  Xiufu Zhang

Unitary representations for the Schr¨odinger-Virasoro
Lie algebra
Xiufu Zhang 1,2, Shaobin Tan 1
2
1
1. SchoolofMathematical Sciences, XiamenUniversity,Xiamen361005, China
0
2 2. School ofMathematical Sciences,XuzhouNormalUniversity,Xuzhou221116,China

n
a
J

6
Abstract

]
A
Inthispaper,conjugate-linearanti-involutionsandunitaryHarish-Chandra
R modules over the Schro¨dinger-Virasoro algebra are studied. It is proved that
.
h thereareonlytwoclassesconjugate-linearanti-involutionsovertheSchro¨dinger-
t Virasoro algebra. The main result of this paper is that a unitary Harish-
a
m Chandra module over the Schro¨dinger-Virasoro algebra is simply a unitary
[ Harish-Chandra module over the Virasoro algebra.

1 2000 Mathematics Subject Classification: 17B10, 17B40, 17B68
v
2 Keywords: Schro¨dinger-Virasoro algebra, Harish-Chandra module, unitary
4
module.
3
1
.
1
0
2
1 Introduction
1
:
v
i The Schro¨dinger-Virasoro algebra sv is defined to be a Lie algebra with C-basis
X
r {Ln,Mn,Yn+21,c | n ∈ Z} subject to the following Lie brackets:
a
m3 m
[L ,L ] = (n m)L +δ − c,
m n n+m m+n,0
− 12
[L ,M ] = nM ,
m n n+m
1 m
[Lm,Yn+21] = (n+ −2 )Ym+n+12,
[Ym+12,Yn+21] = (n−m)Mm+n+1,
[Mm,Mn] = 0 = [Mm,Yn+1] = [sv,c].
2
* Supported by the National Natural Science Foundation of China (No. 10931006).
** Email: [email protected]; [email protected]
1
It was introduced by M. Henkel in Ref. [7] by looking at the invariance of the free
Schro¨dinger equation. Due to its important roles in mathematics and statistical
physics, it has been studied extensively by many authors. In Refs. [17, 19, 20], the
twisted Schro¨dinger-Virasoro algebra, ε-deformation Schro¨dinger-Virasoro algebra,
thegeneralizedSchro¨dinger-Virasoroalgebraandtheextended Schro¨dinger-Virasoro
algebras are introduced. These Lie algebras are all natural deformations of the
Schro¨dinger-Virasoro algebra sv. In Refs [6, 17, 19, 21], the derivations, the 2-
cocycles, the central extensions and the automorphisms for these algebras have been
studied.
It is well known that the Virasoro Lie algebra is an important Lie algebra, whose
representation theory plays a crucial role in many areas of mathematics and physics.
Manyrepresentations, suchastheHarish-Chandra modules, theVermamodulesand
the Whittaker modules, of it have been well studied (cf. Refs. [1, 3, 5, 8, 11, 13-16,
18]). The Virasoro Lie algebra is a subalgebra of the Schro¨dinger-Virasoro algebra,
it is natural to consider those representations for the Schro¨dinger-Virasoro algebra.
In Refs[12, 19, 20, 22], the Harish-Chandra modules, theVerma modules, the vertex
algebra representations and the Whittaker modules over the Schro¨dinger-Virasoro
algebra are studied.
In Ref. [1, 2-5], the nontrivial unitary irreducible unitary modules are classified
and the unitary highest weight modules of the Virasoro algebra are well studied.
Motivated by these work, we consider the unitary modules over the Schro¨dinger-
Virasoro algebra in this paper. The paper is organized as follows. In section 2,
we prove that there are only two classes conjugate-linear anti-involutions over the
Schro¨dinger-Virasoroalgebrasv. Insection3,weprovethataunitaryweightmodule
over the Schro¨dinger-Virasoro algebra is simply a unitary weight module over the
Virasoro algebra. Then the unitary weight modules over sv are classified since that
the ones over the Virasoro algebra are classified.
Throughout this paper we make a convention that the weight modules over the
Schro¨dinger-Virasoro algebra and Virasoro algebra are all with finite dimensional
weight spaces, i.e., the Harish-Chandra modules. The symbols C,N, Z, Z and Z
+ −
represent for the complex field, the set of nonnegative integers, the set of integers,
the set of positive integers and the set of negative integers respectively.
2 Conjugate-linear anti-involution of sv
It is easy to see the following facts about sv :
(i) C := CM Cc is the center of sv.
0
⊕
(ii) If x sv acts semisimply on sv by the adjoint action, then x h, where
∈ ∈
h := span L ,M ,c is the unique Cartan subalgebra of sv.
C 0 0
{ }
(iii) sv has a weight space decomposition according to the Cartan subalgebra h :
sv = Msvn ⊕Msv12+n,
n∈Z n∈Z
2
wheIrfewsevnde=nostpeanVCi{rL=n,⊕Mnn∈}Z,CsLvn21+⊕nC=c,spMan=C{⊕Y21n+∈nZ}C,Mnn∈,YZ.= ⊕n∈ZCY12+n. Then we
have the following lemma:
Lemma 2.1. M Y Cc is the unique maximal ideal of sv.
⊕ ⊕
Proof. The proof is similar as that for Lemma 2.2 in Ref. [19]. (cid:3)
Definition 2.2. Let g be a Lie algebra and θ be a conjugate-linear anti-involution
of g, i.e. θ is a map g g such that
→
θ(x+y) = θ(x)+θ(y), θ(αx) = α¯θ(x),
θ([x,y]) = [θ(y),θ(x)], θ2 = id
for all x,y g, α C, where id is the identity map of g. A module V of g is called
∈ ∈
unitary if there is a positive definite Hermitian form , on V such that
h i
xu,v = u,θ(x)v
h i h i
for all u,v V,x g.
∈ ∈
Lemma 2.3. (Proposition 3.2 in Ref. [1]) Any conjugate-linear anti-involution of
Vir is one of the following types:
(i) θ+(L ) = αnL , θ+(c) = c, forsome α R×,the set of nonzero real number.
α n −n α ∈
(ii) θ−(L ) = αnL , θ−(c) = c, for some α S1, the set of complex number
α n − n α − ∈
of modulus one.
Lemma 2.4. Let θ be an arbitrary conjugate-linear anti-involution of sv. Then
(i) θ(M Y) = M Y.
⊕ ⊕
(ii) θ(h) = h.
(iii) θ(c) = λc+λ′M , θ(M ) = µM , where λ,µ S1,λ′ C.
0 0 0
∈ ∈
Proof. (i), x sv,y M Y, the identity [x,θ(y)] = θ([y,θ(x)]) means that
∀ ∈ ∈ ⊕
θ(M Y) is an ideal of sv. Thus θ(M Y) M Y Cc by Lemma 2.1. Assume
⊕ ⊕ ⊆ ⊕ ⊕
that
θ(Y1+n) = anx+βnc,
2
where an,βn ∈ C,x ∈ M⊕Y. Then by [L0,Y12+n] = (21+n)Y12+n, we see that βn = 0.
Moreover, θ(M ) M Y since M = [Y,Y].
n
⊆ ⊕
For (ii), x sv, [x,θ(M )] = θ([M ,θ(x)]) = 0. So θ(M ) C. Similarly,
0 0 0
∀ ∈ ∈
θ(c) C. The identities
∈
[θ(L ),θ(L )] = nθ(L ), [θ(L ),θ(M )] = nθ(M ),
0 n n 0 n n
− −
and
1
[θ(L0),θ(Y21+n)] = −(2 +n)θ(Y12+n)
3
imply that θ(L ) acts semisimply on sv. Thus θ(L ) h.
0 0
∈
For (iii), note that C is the center of sv, we have θ(C) = C, so we can assume
′
θ(c) = λc+λ M .
0
Since θ(M ) (M Y) C, we can assume θ(M ) = µM .So M = θ2(M ) = µµ¯M ,
0 0 0 0 0 0
thus µ S1.∈Simila⊕rly, ∩we have λ S1. (cid:3)
∈ ∈
Proposition 2.5. Any conjugate-linear anti-involution of sv is one of the following
types:
n+1 n 1
(i) : θα+,β,µ(Ln) = αnL−n +( 2 αn−1β + −2 αn−1µβ)M−n,
θ+ (c) = c,
α,β,µ
θ+ (M ) = µαnM ,
α,β,µ n −n
θα+,β,µ(Y1+n) = µ21α21+nY−1−n
2 2
for some α R×,µ S1,β C.
∈ ∈ ∈
n+1 n 1
(ii) : θ− (L ) = αnL +( αn+1µr − αn−1µr )M ,
α,r1,r2,µ n − n 2 1 − 2 2 n
−
θ (c) = c,
α,r1,r2,µ −
θ− (M ) = µαnM ,
α,r1,r2,µ n n
θα−,r1,r2,µ(Y21+n) = (−µ)21α12+nY21+n
for some α,µ S1,r ,r R.
1 2
∈ ∈
Proof. Let θ be any conjugate-linear anti-involution of sv. By Lemma 2.4 (i), we
have the induced conjugate-linear anti-involution of sv/(M Y) Vir :
⊕ ≃
θ¯: sv/(M Y) sv/(M Y).
⊕ → ⊕
¯
Thus by Lemma 2.3 we see that θ is one of the following types:
(a) θ¯+(L¯ ) = αnL¯ , θ¯+(c¯) = c¯, for some α R×.
(b) θ¯α−(L¯n) = αn−L¯n , θ¯α−(c¯) = c¯, for some∈α S1.
¯ α n − n α − ∈
If θ is of type (a), we can assume
θ(Ln) = αnL−n +Xβn,iMi +Xγn,jY1+j, (2.1)
2
i j
where β ,γ ,a C. By (2.1) and Lemma 2.4 (ii), we have
n,i n,j n
∈
θ(L ) = L +β M +a c.
0 0 0,0 0 0
Then by [θ(L ),θ(L )] = 2θ(L ), we deduce that a = 0. Thus
−1 1 0 0
−
θ(L ) = L +β M . (2.2)
0 0 0,0 0
4
By (2.1), (2.2) and the identity [θ(L ),θ(L )] = nθ(L ), it can be deduced easily
n 0 n
that β = 0 unless i = n, γ = 0 for all j Z, i.e.,
n,i n,j
− ∈
θ(L ) = αnL +β M . (2.3)
n −n n,−n −n
By (2.3) and the identity [θ(L ),θ(L )] = (n m)θ(L )+δ n−n3θ(c), we can
n m − m+n m+n,0 12
get
((n m)β nβ αm +mαnβ )M
m+n,−(m+n) n,−n m,−m −(m+n)
− −
n n3
′
= δ − ((1 λ)c λ M ) (2.4)
m+n,0 0
12 − −
Let m = n = 1,0,1 in (2.4), we see that
− 6 −
λ = 1. (2.5)
Let m = n = 1 in (2.4), we have
−
α−1 α
β = β + β . (2.6)
0,0 1,−1 −1,1
2 2
Let m = n = 2 in (2.4), we have
−
λ′ = 8β 4α2β 4α−2β . (2.7)
0,0 −2,2 2,−2
− −
Let m = 2,n = 1 and m = 2,n = 1 in (2.4) respectively , we have
− −
3α α3 3α−1 α−3
β = β β , β = β β . (2.8)
2,−2 1,−1 −1,1 −2,2 −1,1 1,−1
2 − 2 2 − 2
By (2.6)-(2.8), we have
′
λ = 0. (2.9)
By (2.4), (2.5), (2.9) and Lemma 2.4 (iii), we have
θ(c) = c, (2.10)
and
(n m)β = nβ αm mαnβ . (2.11)
m+n,−(m+n) n,−n m,−m
− −
Let n = 1 in (2.11), we have
(1 m)β +mαβ αmβ = 0.
m+1,−(m+1) m,−m 1,−1
− −
Then using induction, we can prove that, for m 1,
≥
β = (m 2)αm−1β +(m 1)αm−2β . (2.12)
m,−m 1,−1 2,−2
− − −
5
Then (2.8) and (2.12) give us that
m+1 m 1
βm,−m = αm−1β1,−1 − αm+1β−1,1, ( m Z+). (2.13)
2 − 2 ∀ ∈
Let n = 1 in (2.11) and by a similar argument as above, we can prove that
−
m+1 m 1
βm,−m = αm−1β1,−1 − αm+1β−1,1,( m Z−). (2.14)
2 − 2 ∀ ∈
Then by (2.6), (2.13) and (2.14), we see that
m+1 m 1
βm,−m = αm−1β1,−1 − αm+1β−1,1,( m Z). (2.15)
2 − 2 ∀ ∈
Now by (2.3) and (2.15), we have
n+1 n 1
θ(Ln) = αnL−n +( αn−1β1,−1 − αn+1β−1,1)M−n. (2.16)
2 − 2
By Lemma 2.4 (i), we can assume that
θ(Mm) = Xζm,iMi +Xξm,jY1+j, (2.17)
2
i j
θ(Y1+n) = Xλn,iMi +Xµn,jY1+j, (2.18)
2 2
i j
where ζ , ξ , λ , µ C. By (2.16), (2.17), Lemma 2.4 (iii) and the identity
m,i m,j n,i n,j
∈
[θ(M ),θ(L )] = nθ(M )
−n n 0
−
we get that ζ = 0 for all i = n, αnζ = µ, ξ = 0 for all j Z. Thus
−n,i −n,n −n,j
6 ∈
θ(M ) = αnµM . (2.19)
n −n
By (2.3) and (2.19) we have
L = θ2(L ) = L +(αβ +α−1µβ )M .
1 1 1 −1,1 1,−1 1
Thus
β = α−2µβ (2.20)
−1,1 1,−1
−
By (2.16) and (2.20), we have
n+1 n 1
θ(Ln) = αnL−n +( αn−1β + − αn−1µβ)M−n, (2.21)
2 2
where β = β .
1,−1
6
By (2.16), (2.18) and the identity [θ(Y21+m),θ(L0)] = (12 + m)θ(Y21+m), we get
that λ = 0 for all i Z, µ = 0 unless j = (m+1). Thus
m,i m,j
∈ −
θ(Y1+m) = a1+mY−1−m, (2.22)
2 2 2
where a12+m = µm,−m−1. If n 6= 1, by (2.16), (2.22) and the identity [θ(Y12),θ(Ln)] =
(1−2n)θ(Y21+n), we have
1 n 1 n
−2 a12αnY−21−n = −2 a12+nY−21−n.
Then a12+n = αna21,(n 6= 1). By [θ(Y23),θ(L−2)] = θ([L−2,Y23]), we can easily get
that a1 = αa1. So we have
+1
2 2
a12+n = αna21,∀n ∈ Z. (2.23)
By (2.19), (2.22) and the identity [θ(Y1+m),θ(Y−1−m)] = θ([Y−1−m,Y1+m]), we have
2 2 2 2
a1+ma−1−m(2m+1)M0 = µ(2m+1)M0.
2 2
Thus
a1+ma−1−m = µ. (2.24)
2 2
By (2.23) and (2.24), we see that
a1 = √µα. (2.25)
2
By (2.22), (2.23) and (2.25), we have
θ(Y1+m) = √µα12+mY−1−m. (2.26)
2 2
Now (i) follows from (2.10), (2.19), (2.21) and (2.26).
¯
If θ is of type (b), by a similar discussion in the way of (2.1)-(2.16), we can prove
that
n+1 n 1
θ(Ln) = αnLn +( αn−1β1 − αn+1β−1)Mn. (2.27)
− 2 − 2
θ(c) = c, (2.28)
−
where α S1,β ,β C. By a similar discussion in the way of (2.17)-(2.19) and
1 −1
∈ ∈
(2.22)-(2.26), we have
θ(M ) = µαnM , (2.29)
n n
θ(Y12+n) = (−µ)21α21+nY21+n, (2.30)
where µ S1. By (2.27), (2.29) and the identities θ2(L ) = L and θ2(L ) = L ,
1 1 −1 −1
∈
we see that
αβ = β αµ, αβ = β αµ.
1 1 −1 −1
7
If we set α = eiσ,µ = eiτ,β = β eix, then by αβ = β αµ, we see that
1 1 1 1
| |
β = β ei(2σ+τ) or β ei(2σ+τ).
1 1 1
| | −| |
Similarly,
β = β ei(−2σ+τ) or β ei(−2σ+τ).
−1 −1 −1
| | −| |
We set r ,r R such that r = β and r = β . Then
1 2 1 1 2 −1
∈ | | | | | | | |
β = r α2µ, β = r α2µ. (2.31)
1 1 −1 2
(cid:3)
Thus (ii) follows from (2.27)-(2.31).
The following Lemma is crucial for the proof of Proposition 3.4.
Lemma 2.6. Letθ beaconjugate-linearanti-involutionoftheSchro¨dinger-Virasoro
algebra sv.
(i) If θ = θ+ , we denote by Vir′ the subalgebra of sv generated by
α,β,µ
n 1
c,L′ := L − α−1βM n Z .
{ n n − 2 n | ∈ }
Then Vir′ Vir and θ+ (L′ ) = αnL′ ,θ+ (c) = c.
≃ α,β,µ n −n α,β,µ
(ii) If θ = θ− , we denote by Vir′ the subalgebra of sv generated by
α,r1,r2,µ
c,L′ := L +x M n Z ,
{ n n n n | ∈ }
θw−here x(nL′∈) =C saαtinsLfy′i,nθg−xnµ21(+c)x=nµ−c21. = n−21r2 − n+21r1. Then Vir′ ≃ Vir and
α,r1,r2,µ n − n α,r1,r2,µ −
Proof. It can be checked directly, we omit the details. (cid:3)
Lemma 2.7. (Proposition 3.4 in Ref. [1]) Let V be a nontrivial irreducible weight
Vir-module.
(i) If V is unitary for some conjugate-linear anti-involution θ of Vir, then θ = θ+
α
for some α > 0.
(ii) If V is unitary for θ+ for some α > 0, then V is unitary for θ+.
α 1
Proposition 2.8. Let V be a nontrivial irreducible weight sv-module.
(i)IfV isunitaryforsomeconjugate-linearanti-involutionθ ofsv, thenθ = θ+
α,β,µ
for some α > 0.
(ii) If V is unitary for θ+ for some α > 0, then V is unitary for θ+ .
α,β,µ 1,β,µ
Proof. (i) Suppose V is unitary for some conjugate-linear anti-involution θ of sv.
′
By Lemma 2.6, V can be viewed as a unitary Vir -module for the conjugate-linear
′
anti-involution θ ′. Then V is a direct sum of irreducible unitary Vir -modules
|Vir
since any unitary weight Vir-module is complete reducible. We claim that V is a
8
nontrivial Vir′-module. Otherwise, for any 0 = v V,n Z 0,m Z, we have
6 ∈ ∈ \ ∈
1
′ ′
M v = (L M M L )v = 0,
0 −n n −n − −n n
1
′ ′
M v = (L M M L )v = 0,
n n 0 n − n 0
2
′ ′
Y21+mv = 1+2m(L0Y21+m −Y21+mL0)v = 0.
′
So sv.V = 0, a contradiction. Thus there is a nontrivial irreducible unitary Vir -
submodule of V for conjugate-linear anti-involution θ ′. By Lemma 2.7, θ ′ =
|Vir |Vir
θ+ for some α > 0. Then by Proposition 2.5, we have θ = θ+ for some µ S1,β
α α,,β,µ ∈ ∈
C.
(ii) Suppose V is unitary for θ+ for some α R×,µ S1,β C and .
α,,β,µ ∈ ∈ ∈ h iα
′
is the Hermitian form on V. We can assume V is generated by a L -eigenvector v
0 0
with eigenvalue a C since V is irreducible weight sv-module. By
∈
′ ′
L v ,v = v ,L v ,
h 0 0 0iα h 0 0 0iα
we see that a R. Then L′-eigenvalues on V are of the form a+ n, n Z. Define
∈ 0 2 ∈
a new form , on V by
h i
n
−n ′
v,w = α 2 v,w , v,w v L v = (a+ )v .
h i h iα ∀ ∈ { | 0 2 }
It is easy to check that this form makes V unitary with the conjugate-linear anti-
involution θ+ . (cid:3)
1,β,µ
3 Unitary representations for sv
In this section, we study the unitary weight modules for sv. By Prop. 2.8, we
see that the conjugate-linear anti-involution is of the form θ+ for some α > 0. For
α,β,µ
the sake of simplicity, we write θ+ by θ.
α,β,µ
It is known that a unitary weight module over Virasoro algebra is completely
reducible. This result also holds for sv :
Lemma 3.1. If V is a unitary weight module for sv, then V is completely reducible.
Proof. Let N be a submodule. Then N+ := v V v,N = 0 is a submodule
{ ∈ |h i }
of V since for any v N+, x.v,N = v,θ(x)N = 0. It is well known that any
∈ h i h i
submodule of a weight module is a weight module. For any weight λ of V, denote by
V ,N ,N+ theweightspacewithweightλofV,N,N+ respectively. Itisobviousthat
λ λ λ
dim(V ) < since V is a Harish-Chandra module, so we can extend an orthogonal
λ
∞
basis of N as an orthogonal basis of V , thus we have
λ λ
V = N N+,
λ λ ⊕ λ
9
which means V = N N+. (cid:3)
⊕
Lemma 3.2. (Theorem 1.3 (i) in [12]) An irreducible weight module over sv is
either a highest/lowest weight module or a uniformly bounded one.
It is well known (See Refs. [10] and [11]) that there are three types modules of
the intermediate series over Vir, denoted respectively by A ,A ,B , they all have
a,b α β
basis v k Z such that c acts trivially and
k
{ | ∈ }
A : L v = (a+k +nb)v ;
a,b n k n+k
A : L v = (n+k)v if k = 0, L v = n(n+a)v ;
α n k n+k n 0 n
6
B : L v = kv if k = n, L v = n(n+a)v .
β n k n+k n −n 0
6 − −
for all n,k Z. For the irreducible modules of the intermediate series of type A ,
a,b
∈
we have fact that: A and A are isomorphic if and only if a c Z and b = d
a,b c,d
− ∈
or 1 d.
−
Lemma 3.3. (Theorem 0.5 in [1]) Let V be an irreducible unitary module of Vir
with finite-dimensional weight spaces. Then either V is highest or Lowest weight,
or V is isomorphic to A for some a R,b 1 +√ 1R.
a,b ∈ ∈ 2 −
Proposition 3.4. A unitary weight module over sv is simply a unitary weight
module over Vir. That is, if V is a unitary weight module over sv, then M.V =
Y.V = 0.
Proof. Let V be a unitary weight module over sv for a conjugate-linear anti-
′
involution θ. By Lemma 2.6 (i), V is also unitary for Vir , thus the well known
result for the unitary modules over Virasoro Lie algebra can be used freely. By
Lemma 3.1, we may assume that V is irreducible. By Lemma 3.2, it is sufficient to
consider the following two cases:
Case 1. V isaunitaryirreduciblehighest/lowest weightmodule. Letv beahighest
λ
weight vector. For n Z , we have M v ,M v = v ,µM M v = 0, thus
+ −n λ −n λ λ −n n λ
∈ h i h i
M v = 0. Furthermore,
−n λ
L′ v ,M v = v ,α−nL′ M v = nα−nλ(M ) v ,v = 0.
h −n λ −n λi h λ n −n λi − 0 h λ λi
So M v = 0. Thus
0 λ
M.V = 0.
For n N, note that M v = 0, we have
0 λ
∈
1
hY−21−nvλ,Y−21−nvλi = hvλ,µ2Y21+nY−21−nvλi = 0.
Thus Y−1−nvλ = 0, which means that
2
Y.V = 0.
10