Classification of simple weight modules for the Neveu-Schwarz algebra with a finite-dimensional weight space

0.11 MB·English

by  Xiufu Zhang

Classification of simple weight modules for the
Neveu-Schwarz algebra with a finite-dimensional weight
space
2
1 Xiufu Zhang 1, Zhangsheng Xia 2
0
2 1. School ofMathematical Sciences,XuzhouNormalUniversity,Xuzhou221116,China

n
2. School ofSciences,HubeiUniversityforNationalities,Enshi445000, China
a
J

6

]
A Abstract
R
We show that the support of a simple weight module over the Neveu-
.
h Schwarz algebra, which has an infinite-dimensional weight space, coincides
t
a with the weight lattice and that all non-trivial weight spaces of such mod-
m
ule are infinite-dimensional. As a corollary we obtain that every simple
[
weight module over the Neveu-Schwarz algebra, having a non-trivial finite-

1 dimensional weight space, is a Harish-Chandra module (and hence is either a
v
highest or lowest weight module, or else a module of the intermediate series).
7
3 This result generalizes a theorem which was originally given on the Virasoro
3 algebra.
1
. 2000 Mathematics Subject Classification: 17B10, 17B65, 17B68
1
0
Keywords: Neveu-Schwarz algebra, weight module, Harish-Chandra mod-
2
1 ule.
:
v
i
X 1. Introduction
r
a It is well known that the Virasoro algebra Vir plays a fundamental role in two-
dimensional conformal quantum field theory. As the super-generalization, there
are two super-Virasoro algebras called the Neveu-Schwarz algebra and the Ramond
algebra correspondingtoN = 1 andN = 1super-conformalfieldtheoryrespectively.
2
Let θ = 1 or 0 which corresponds to the Neveu-Schwarz case or the Ramond case
2
respectively. The super-Virasoro algebra SVir(θ) is the Lie superalgebra SVir(θ) =
SVir ⊕ SVir where SVir has a basis {L ,c | n ∈ Z} and SVir has a basis
¯0 1¯, ¯0 n ¯1
* Supported by the National Natural Science Foundation of China (No. 10931006).
** Email: [email protected]; [email protected]
1
{G | r ∈ θ +Z}, with the commutation relations
r
m3 −m
[L ,L ] = (n−m)L +δ c,
m n n+m m+n,0
12
m
[L ,G ] = (r − )G ,
m r m+r
2
4r2 −1
[G ,G ] = 2L + δ c,
r s r+s r+s,0
12
[SVir ,c] = 0 = [SVir ,c],
¯0 ¯1
for m,n ∈ Z,r,s ∈ θ+Z.
The subalgebra h = CL ⊕ Cc is called the Cartan subalgebra of SVir(θ). An
0
h-diagonalizable SVir(θ)-module is usually called a weight module. If M is a weight
module, then M can be written as a direct sum of its weight spaces, M = M ,
λ
where M = {v ∈ M|L v = λ(L )v,cv = λ(c)v}. We call {λ|M 6= 0} the support of
λ 0 0 λ
L
M and is denoted by supp(M). A weight module is called Harish-Chandra module
if each weight space is finite-dimensional.
In [1], [2], [3], [7], the unitary modules, the Verma modules, the Fock modules
and the Harish-Chadra modules over SVir(θ) are studied. In this paper, we try to
give the classification of simple weight modules for the Neveu-Schwarz algebra with
a finite-dimensional weight space. Our main result generalizes a theorem which was
originally given on the Virasoro algebra in [6].
Forconvenience, wedenotetheNeveu-Schwarz algebrabyNS insteadofSVir(1)
2
and we define
L , if k ∈ Z,
E = k
k G , if k ∈ 1 +Z.
(cid:26) k 2
Suppose M is a simple weight NS−module, then c acts on M by a scalar and
M can be written as a direct sum of its weight spaces, M = M , where M =
λ λ
Z
{v ∈ M|L v = λv}. Obviously, if λ ∈ supp(M), then supp(M) ⊆ {λ+k|k ∈ }, the
weight lat0tice. Two elements i,j ∈ {λ+k|k ∈ Z} are called adjLacent if |i−2j| = 1,
2 2
otherwise, unadjacent.
In this paper, our main result is the following theorem:
Theorem 1. Let M be a simple weight NS−module. Assume that there exists
λ ∈ C such that dimM = ∞. Then supp(M) = λ+ Z and for each k ∈ Z we have
λ 2 2
dim(M ) = ∞.
λ+k
In [7], Y. Su proved the following result which generalizes a theorem originally
given as a conjecture by Kac in [4] on the Virasoro algebra and proved by Mathieu
in [5].
Theorem 2. A Harish-Chandra module over SVir(θ) is either a highest or lowest
weight module, or else a module of the intermediate series.
By Theorem 1 and 2, we get the following corollaries immediately:
2
Corrollary 3.LetM beasimpleweightNS−module. Assumethatthereexistsλ ∈
C such that 0 < dimM < ∞. Then M is a Harish-Chandra module. Consequently,
λ
M is either a highest or lowest weight module, or else a module of the intermediate
series.
A weight NS-module, M, is called mixed module if there exist λ ∈ C and k ∈ Z
2
such that dimM = ∞ and dimM < ∞.
λ λ+k
Corrollary 4. There are no simple mixed NS-modules.
2. Proof of the Theorem 1
Noting that {G−3,G−1,G1,G3} is a set of generators of NS, we have the fol-
2 2 2 2
lowing fact immediately:
Principal Fact: Assume that there exists µ ∈ C and 0 6= v ∈ M , such that
µ
G1v = G3v = 0 or G−1v = G−3v = 0. Then M is a Harish-Chandra NS-module.
2 2 2 2
Lemma 1. If there are µ,µ′ ∈ {λ+k|k ∈ Z2} such that dimMµ < ∞ and dimMµ′ <
′
∞, then µ,µ are adjacent.
Proof. Suppose that there exist two unadjacent elements in {λ + k|k ∈ Z} cor-
2
respond to finite-dimensional weight spaces or trivial vector spaces in M. Without
loss of generality, we may assume that dimMλ+1 < ∞ and dimMλ+k < ∞, where
2
k ∈ {3,2, 5,3,···}. Let V denote the intersection of the kernels of the linear maps:
2 2
G1 : Mλ → Mλ+1,
2 2
and
E : M → M .
k λ λ+k
Since dimMλ = ∞,dimMλ+1 < ∞, we know that the kernel kerG1 of G1 : Mλ →
2 2 2
Mλ+1 is infinite dimensional. Since dimMλ+k < ∞, we also have that the kernel of
2
Ek|kerG1 : kerG1 → Mλ+k is infinite dimensional. That is dimV = ∞.
2 2
Since
[G21,El] = ( 2(2Ll −12+l12,)G12+l, iiff ll ∈∈ Z21++\N{,1}
is nonzero, we get that
1 1
E V = 0, ∀l = ,k,k + ,k +1,··· (1)
l
2 2
If k = 3 then there exists 0 6= v ∈ V such that G3v = 0, and M would be
2 2
a Harish-Chandra module by the Principal Fact, a contradiction. Thus k > 3 and
2
dimG3V = ∞.SincedimMλ+1 < ∞,thereexists0 6= w ∈ G3V suchthatL−1w = 0.
2 2 2
3
Suppose w = G3u for some u ∈ V. For all l ≥ k > 3, using (1) we have
2 2
Elw = ElG32u = ( −GG233EElluu++[[EEll,,GG233]]uu == 00,, iiff kk ∈∈ Z12++.Z+,
2 2
Hence
1
E w = 0, ∀l = −1,k,k + ,k +1.···
l
2
Since [L−1,El] 6= 0 for all l ≥ 23, inductively, we get
1 3
E w = 0, ∀l = ,1, ,2,··· .
l
2 2
Hence M is a Harish-Chandra module by the Principal Fact. A contradiction. Then
(cid:3)
the lemma follows.
Z
By Lemma 1, we see that there exist at most two elements in {λ + k|k ∈ }
2
which correspond to finite-dimensional weight spaces or trivial vector spaces in M.
Moreover, if there are two, then they are adjacent.
Lemma 2. (i) Let 0 6= v ∈ M be such that G1v = 0. Then
2
1
( L1G1 −G3)G3v = 0.
2 2 2 2
(ii) Let 0 6= w ∈ M be such that G−1w = 0. Then
2
1
( L−1G−1 +G−3)G−3w = 0.
2 2 2 2
Proof. Note that L1v = G1G1v = 0,L−1w = G−1G−1w = 0, we can easily check
2 2 2 2 (cid:3)
by a direct calculation that Lemma 2 holds.
Lemma 3. Let M be a simple weight NS-module satisfying dimM < ∞ and
µ
dimMµ+21 < ∞. Then µ ∈ {−1, 12}.
Proof. Let V be the kernel of G1 : Mµ−1 → Mµ. Since dimMµ−1 = ∞ and
2 2 2
dimMµ < ∞, we see that dimV = ∞. For any 0 6= v ∈ V, consider the element G3v.
2
By the Principal Fact, G3v = 0 would imply that M is a Harish-Chandra module, a
2
contradiction. Hence G3v 6= 0, in particular, dimG3V = ∞. This implies that there
2 2
exists w ∈ G3V such that w 6= 0 and G−1w = 0 since dimMµ+1 < ∞. From Lemma
2 2 2
2 we have (12L1G21 −G32)w = 0. In particular, we have L−1G−12(21L1G12 −G32)w = 0.
By a direct calculation we obtain
1
L−1G−21(2L1G12 −G23) ≡ 2L20 −3L0 mod U(NS)G−12.
4
But w ∈ M which means L w = (µ+1)w and hence
µ+1, 0
2(µ+1)2 −3(µ+1) = 0.
So µ ∈ {−1, 1}. (cid:3)
2
From Lemma 3 we know that if M has two finite dimension weight spaces, then
they must be M−1, M−1 or M1,M1.
2 2
Lemma 4. µ 6= −1 and µ 6= 1.
2
Proof. By the symmetry, we need only to prove that µ 6= 1. Let V be the kernel
2
of G1 : M0 → M1. Then dimV = ∞. For v ∈ V, using G1v = L0v = 0, we have
2 2 2
G1G−1v = 2L0v −G−1G1v = 0. (2)
2 2 2 2
If G−1V were infinite-dimensional, there would exists 0 6= w ∈ G−1V such that
2 2
G1w = 0 (by (2)) and G3w = 0 (since dimM1 < ∞). Then the Principal Fact
2 2
would then imply that M is a Harish-Chandra module, a contradiction. Hence
dimG−1V < ∞.
2
This means that the kernel W of the linear map
G−1 : V → M−1
2 2
is infinite-dimensional. For every x ∈ W we have
G1G−3x = 2L−1x−G−3G1x = 0. (3)
2 2 2 2
If there exists 0 6= x ∈ W such that G−3x = 0, then we would have G−3x =
2 2
G−1x = 0 and the Principal Fact would imply that M is a Harish-Chandra module,
2
a contradiction. Thus
dimG−3W = ∞.
2
Let H denote the kernel of the linear map L2 : G−3W → M1. Since dimG−3W = ∞
2 2 2
and dimM1 < ∞, we have dimH = ∞. For every y ∈ H, we also have G1y = 0 by
2 2
(3), implying by induction that EkH = 0 for all k = 12,1,2, 25,3, 72,4,··· . If G32h = 0
for some 0 6= h ∈ H, then the Principal Fact implies that M is a Harish-Chandra
module, a contradiction. Hence
dimG3H = ∞.
2
For every h ∈ H and k ≥ 2, we have
EkG32h = ( (223L−23+kk2h)G−23+Gk23hG+khG=32L0k,h = 0, iiff kk ∈∈ Z12++.Z+,
5
Hence
5 7 9
EkG3h = 0, ∀k = 2, ,3, ,4, ,··· . (4)
2 2 2 2
Let, finally, K denote the infinite-dimensional kernel of the linear map
G1 : G3H → M1.
2 2 2
If G3z = 0 for some 0 6= z ∈ K, then the Principal Fact implies that M is a Harish-
2
Chandra module, a contradiction. Hence G3z 6= 0,∀0 6= z ∈ K. For every z ∈ K
2
and k ≥ 2, by (4), we have dimG3K = ∞ and
2
EkG32z = ( (232L−23+kk2z)G−32G+k23zG+kzG=32L0k,z = 0. iiff kk ∈∈ Z12++.Z+,
Hence EkG3K = 0 for all k ≥ 2. At the same time, since dimG3K = ∞ and
2 2
dimM1 < ∞, we can choose some 0 6= t ∈ G3K such that G−1t = 0, by induction,
2 2
we get E t = 0 forall i > 0 andthus M is a Harish-Chandra moduleby thePrincipal
i
(cid:3)
Fact. This last contradiction completes the proof of lemma 4.
By lemma 1, Lemma 3 and Lemma 4, we get the following lemma immediately:
Lemma 5. There is at most one element µ ∈ {λ+k|k ∈ Z} such that dimM < ∞.
2 µ
Now the proof of Theorem 1 follows from the following Lemma:
Lemma 6. There is no µ ∈ {λ+k|k ∈ Z} such that dimM < ∞.
2 µ
Proof. Suppose that dimM < ∞ and dimM = ∞ for all µ 6= ν ∈ {λ+k|k ∈ Z}.
µ ν 2
Define
V = ker(G1 : Mµ−1 → Mµ)∩ker(G1G−3G3 : Mµ−1 → Mµ)∩ker(L−1G3 : Mµ−1 → Mµ),
2 2 2 2 2 2 2 2
W = ker(G−1 : Mµ+1 → Mµ)∩ker(G−1G3G−3 : Mµ−1 → Mµ)∩ker(L1G−3 : Mµ−1 → Mµ).
2 2 2 2 2 2 2 2
Since dimMµ−1 = ∞ and dimMµ < ∞, V is a vector subspace of finite codimension
2
in Mµ−1. Since dimMµ+1 = ∞ and dimMµ < ∞, W is a vector subspace of finite
2 2
codimension in Mµ+1. In order not to get a direct contradiction using the Principal
2
Fact, we assume that G3v 6= 0 for all 0 6= v ∈ V and G−3w 6= 0 for all 0 6= w ∈ W.
2 2
Then dimG3V = ∞ and dimG−3W = ∞.
2 2
Claim µ 6= ±1. Moreover, for any v ∈ V,w ∈ W, we have
G1G−1G3v = τG3v, (5)
2 2 2 2
′
G−1G1G−3w = τ G−3w, (6)
2 2 2 2
where τ = (µ+1)(2µ−1) and τ′ = (µ−1)(2µ+1).
µ−1 µ+1
6
Proof of the Claim. It can be checked directly that
1
L−1G−21(2L1G12 −G32) ≡ 2L20 −3L0 −L0G21G−12 +2G21G−21 mod U(NS)L−1, (7)
and
1
L1G12(2L−1G−21 +G−23) ≡ −2L20−3L0+L0G−12G12 +2G−21G12 mod U(NS)L1. (8)
For any 0 6= v ∈ V, by Lemma 2(i) and (7), we have
2L20G3v −3L0G3v −L0G1G−1G3v +2G1G−1G3v = 0.
2 2 2 2 2 2 2 2
Then
(2(µ+1)2 −3(µ+1))G3v −(µ−1)G1G−1G3v = 0.
2 2 2 2
If µ = 1, then G3v = 0, a contradiction. Thus µ 6= 1, and (4) holds.
2
For any 0 6= w ∈ W, by Lemma 2(ii) and (8), we have
−2L20G−3w−3L0G−3w+L0G−1G1G−3w +2G−1G1G−3w = 0.
2 2 2 2 2 2 2 2
Then
(2(µ−1)2 +3(µ−1))G−3w−(µ+1)G−1G1G−3w = 0.
2 2 2 2
If µ = −1, then G−3w = 0, a contradiction. Thus µ 6= −1, and (6) holds. This
2
completes the proof of the Claim.
By (4) and (6), we see that G1G−1G3v = 0 if and only if µ = 1, G−1G1G−3w =
2 2 2 2 2 2 2
0 if and only if µ = −1. There only two cases may occur: µ ∈/ {1,−1, 1} or
2 2
µ ∈/ {1,−1,−1}. Because of the symmetry of our situation, to complete the proof
2
of the Theorem, it is enough to show that a contradiction can be derived when
µ ∈/ {1,−1, 1}.
2
Suppose µ ∈/ {1,−1, 1}, then τ 6= 0 and G−1G3v 6= 0 for any 0 6= v ∈ V. Set
2 2 2
S = G−1G3V ∩W.
2 2
Since both G−1G3V and W have finite codimension in Mµ+1, we have dimS = ∞.
2 2 2
Note that
1 1 2
G−3( L1G1 −G3) ≡ −G1G−1 − L1G1G−3 +G3G−3 − c mod U(NS)L−1.
2 2 2 2 2 2 2 2 2 2 2 3
For any v ∈ V, by Lemma 2(i), (4) and the definition of V, we have
2
G3G−3G3v = G1G−1G3v + cG3v = pG3v, for some p ∈ C. (9)
2 2 2 2 2 2 3 2 2
Similarly, For any w ∈ W, by
1 1 2
G3( L−1G−1 +G−3) ≡ G−1G1 − L−1G−1G3 −G−3G3 + c mod U(NS)L1,
2 2 2 2 2 2 2 2 2 2 2 3
7
Lemma 2(ii) , (6) and the definition of W, we have
2
G−3G3G−3w = G−1G1G−3w + cG−3w = qG−3w, for some q ∈ C. (10)
2 2 2 2 2 2 3 2 2
Choose v ∈ V such that 0 6= G−1G3v ∈ S. Set
2 2
x = G3v,y = G−1x,h = G−3x,z = G−3y. (11)
2 2 2 2
By the definitions of W,S and (10), we have
G−1(G3G−3y −qy) = 0−0 = 0
2 2 2
and
G−3(G3G−3y −qy) = G−3G3G−3y −qG−3y = 0.
2 2 2 2 2 2 2
So G3G−3y −qy = 0 by the Principal Fact, and we get the formula:
2 2
G3z = qy. (12)
2
Moreover, we have
G1z = G1G−3y = (2L−1−G−3G1)y = 2G2−1y−G−3G1G−1G3v = −τG−3x = −τh,
2 2 2 2 2 2 2 2 2 2 2
i.e.,
G1z = −τh. (13)
2
Let U+ and U− denote the subalgebras of U(NS), generated by G1,G3 and
2 2
G−1,G−3, respectively. We want to prove that the following vector space
2 2
N = U+x⊕U+y ⊕U−z ⊕U−h
is a proper weight submodule of M, which will derive a contradiction, as desired.
Since x,y,z and h are all eigenvectors for L , we see that N decomposes into a
0
direct sum of weight spaces which are obviously finite-dimensional. It remains to
show that N is stable under the action of the following four operators: G1,G3,G−1
2 2 2
and G−3. That GiU+x ⊂ U+x, GiU+y ⊂ U+y, G−iU−h ⊂ U−h and G−iU−z ⊂
2 2 2 2 2
U−z is clear for i = 1,3.
′ ′ ′ ′
For any a ∈ U+,a ∈ U−, there exists ai,j,bi,j,ci,j ∈ U+ and ai,j,bi,j,ci,j ∈ U−
such that
G−1a = aG−1 + ai,jLi0cj,
2 2
i,j
X
G−3a = aG−3 + ai,jLi0cj + bi,jLi0cjL−1 + ci,jLi0cjG−1,
2 2 2
i,j i,j i,j
X X X
G1a′ = a′G1 + a′i,jLi0cj,
2 2
i,j
X
8
G3a′ = a′G3 + a′i,jLi0cj + b′i,jLi0cjL1 + c′i,jLi0cjG1.
2 2 2
i,j i,j i,j
X X X
Thus, to show G−iU+x,G−iU+y,GiU−h,GiU−z ⊂ N, we need only to show that
2 2 2 2
G x,G y,G h,G z,G G x,G G y,G G h,G G z ∈ N for i,j = 1,3.
−i −i i i −i j −i j i −j i −j
2 2 2 2 2 2 2 2 2 2 2 2
Now we can check the following formulas one by one via the definitions of V,W,S
and (5)-(13):
G−1x = y,
2
G−1y = 0,
2
G−3x = h,
2
G−3y = z,
2
G1h = G1G−3G3v = 0,
2 2 2 2
G1z = −τh,
2
G3h = G3G−3G3v = pG3v = px,
2 2 2 2 2
G3z = qy,
2
G1G−3x = 0,
2 2
G3G−3x = px,
2 2
G1G−3y = −τh,
2 2
G3G−3y = qy,
2 2
G1G1G−3y = 0,
2 2 2
G3G1G−3y = −τpx,
2 2 2
L−1x = G2−1x = 0,
2
L−1y = G2−1y = 0,
2
G−1G3h = py,
2 2
G−3G3h = ph,
2 2
G−1G3z = 0,
2 2
G−3G3z = qz,
2 2
G−1G−1G3h = 0,
2 2 2
G−3G−1G3h = pz,
2 2 2
L h = G2h = 0,
1 1
2
L z = G2z = 0.
1 1
2
(cid:3)
This completes the proof of Lemma 6 and then of Theorem 1.
9
References
[1] Goddard, P., Kent, A. and Olive, D. Unitary representations of the Virasoro
and super-Virasoro algebras. Commun. Math. Phys. 103(1986): 105-119.
[2] Iohara, K., Koga, Y. Representation theory of Neveu-Schwarz and Remond
algebras I: Verma modules. Adv. Math. 177(2003): 61-69.
[3] Iohara, K., Koga, Y. Representation theory of Neveu-Schwarz and Remond
algebras II: Fock modules. Ann. Inst. Fourier 53(2003): 1755C1818.
[4] Kac, V. G.(1982).Some problems of infinite-dimensional Lie algebras and their
representations. lecture notes in math. 933, Springer, New York, pp. 117-126.
[5] Mathieu, O. Classification of Harish-Chandra modules over the Virasoro Lie
algebra. Invent. Math. 107(1992): 225-234.
[6] Mazorchuk, V. Zhao, K., Classification of simple weight Virasoro modules with
a finite-dimensional weight space. J. Algebra, 307(2007): 209-214.
[7] Su, Y. Classification of Harish-Chandra modules over the super-Virasoro alge-
bras. Comm. Algegra 23(1995): 3653-3675.
Note This artical has been accepted by Communications in algebra in
24/03/2011.
10