Extensions of the conformal representations for orthogonal Lie algebras

Journal of Algebra 377 (2013) 97–124

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Journal of Algebra www.elsevier.com/locate/jalgebra

Extensions of the conformal representations for orthogonal Lie algebras ✩ Xiaoping Xu a,∗ , Yufeng Zhao b a

Hua Loo-Keng Key Mathematical Laboratory, Institute of Mathematics, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, Beijing 100190, PR China b LMAM, School of Mathematical Sciences, Peking University, Beijing 100871, PR China

a r t i c l e

i n f o

Article history: Received 9 July 2011 Available online 20 December 2012 Communicated by Leonard L. Scott, Jr. Keywords: Conformal transformation Polynomial algebra Tensor Pieri’s formula Generalizations of the conformal representation Orthogonal Lie algebra Irreducibility

a b s t r a c t The conformal transformations with respect to the metric defining o(n, C) give rise to an inhomogeneous polynomial representation of o(n + 2, C). Using Shen’s technique of mixed product, we generalize the above representation to an inhomogeneous representation of o(n + 2, C) on the tensor space of any finitedimensional irreducible o(n, C)-module with the polynomial space, where a hidden central transformation is involved. Moreover, we find a condition on the constant value taken by the central transformation such that the generalized representation is irreducible. In our approach, Pieri’s formulas, invariant operators and the idea of Kostant’s characteristic identities play key roles. The result could be useful in understanding higher-dimensional conformal field theory with the constant value taken by the central transformation as the central charge. Our representations virtually provide natural extensions of the conformal transformations on a Riemannian manifold to its vector bundles. © 2012 Elsevier Inc. All rights reserved.

1. Introduction A quantum field is an operator-valued function on a certain Hilbert space, which is often a direct sum of infinite-dimensional irreducible modules of a certain Lie algebra (group). The Lie algebra of two-dimensional conformal group is exactly the Virasoro algebra, which is infinite-dimensional. The minimal models of two-dimensional conformal field theory were constructed from direct sums

✩

*

Zhao’s research is supported by NSFC Grant 10701002; Xu’s research is supported by NSFC Grant 11171324. Corresponding author. E-mail address: [email protected] (X. Xu).

0021-8693/$ – see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2012.11.040

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of certain infinite-dimensional irreducible modules of the Virasoro algebra, where a distinguished module called, the vacuum module, gives rise to a vertex operator algebra. When n  3, n-dimension conformal groups (depending on the metric) are finite-dimensional. Higher-dimensional conformal field theory is not so well understood partly because we are lack of enough knowledge on the infinite-dimensional irreducible modules of orthogonal Lie algebras that are compatible to the natural conformal representations. This motivates us to study explicit infinite-dimensional irreducible representations of the orthogonal Lie algebra o (n + 2, C) by using the inhomogeneous polynomial representations arising from conformal transformations with respect to the metric defining o(n, C) and Shen’s technique of mixed product (cf. [S1,S2,S3]) (also known as Larsson functor (cf. [La])). It is well known that n-dimensional projective group gives rise to an inhomogeneous representation of the Lie algebra sl(n + 1, C) on the polynomial functions of the projective space. Using Shen’s mixed product for Witt algebras, the authors [ZX] generalized the above representation of sl(n + 1, C) to an inhomogeneous representation on the tensor space of any finite-dimensional irreducible gl(n, C)-module with the polynomial space. Moreover, the structure of such a representation was completely determined by employing projection operator techniques (cf. [Gm1]) and the wellknown Kostant’s characteristic identities (cf. [K]). In this paper, we generalize the conformal representation of o(n + 2, C) to an inhomogeneous representation of o(n + 2, C) on the tensor space of any finite-dimensional irreducible o(n, C)-module with the polynomial space by Shen’s idea of mixed product for Witt algebras. It turns out that a hidden central transformation is involved. More importantly, we find a condition on the constant value taken by the central transformation such that the extended representation is irreducible. In our approach, Pieri’s formulas, invariant operators and the idea of Kostant’s characteristic identities play key roles. The result could be useful in understanding higher-dimensional conformal field theory with the constant value taken by the central transformation as the central charge. Our representations virtually provide natural extensions of the conformal transformations on a Riemannian manifold to its vector bundles. The modules are also “modules of tensor densities” in the terminology of Rudakov [Ra]. A module of a finite-dimensional simple Lie algebra is called a weight module if it is a direct sum of its weight subspaces. A module of a finite-dimensional simple Lie algebra is called cuspidal if it is not induced from its proper parabolic subalgebras. Britten and Lemire [BL1] classified the weight modules of one-dimensional weight subspaces for finite-dimensional simple Lie algebras. Fernando [Fs] proved that any irreducible weight module of finite-dimensional weight subspaces for a finite-dimensional simple Lie algebra must be cuspidal or parabolically induced. Moreover, infinite-dimensional cuspidal irreducible weight modules of finite-dimensional weight subspaces exist only for special linear Lie algebras and symplectic Lie algebras. A similar result was independently obtained by Futorny [Fv]. Britten, Hooper and Lemire [BHL] showed that there exist exactly two nonequivalent irreducible infinite-dimensional highest-weight modules of one-dimensional weight subspaces for symplectic Lie algebras. Benkart, Britten and Lemire [BBL] proved that infinite-dimensional weight modules, whose weight-subspace dimensions are bounded, exist only for special linear Lie algebras and symplectic Lie algebras. Britten and Lemire [BL2] classified the weight modules of finite-dimensional weight subspaces for symplectic Lie algebras. Mathieu [M] gave a complete classification of infinite-dimensional cuspidal irreducible weight modules of finite-dimensional weight subspaces for special linear Lie algebras and symplectic Lie algebras. The irreducible modules explicitly constructed in this paper are weight modules of finite-dimensional weight subspaces. Characteristic identities have a long history. The first person to exploit them was Dirac [D], who wrote down what amounts to the characteristic identity for the Lie algebra o(1, 3). This particular example is intimately connected with the problem of describing the structure of relativistically invariant wave equations. It had been shown by Kostant [K] (also cf. [Gm2]) that the characteristic identities for semi-simple Lie algebras also hold for infinite-dimensional representations. The n-dimensional conformal group with respect to Euclidean metric (·, ·) is generated by the translations, rotations, dilations and special conformal transformations

x →

x − (x, x)b .   (b, b)(x, x) − 2(b, x) + 1

(1.1)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

99

n

Let A = C[x1 , . . . , xn ]. The Witt algebra W (n) = i =1 A∂xi with the commutator of differential operators as its Lie bracket. The conformal transformations with respect to the metric defining o(n, C) give rise to an inhomogeneous polynomial representation of ϑ : o(n + 2, C) → W (n), acting on A. Let E r ,s be the square matrix with 1 as its (r , s)-entry and 0 as the others. Acting on the entries of the elements of the Lie algebra gl(n, A), W (n) becomes a Lie subalgebra of the derivation Lie algebra of gl(n, A). In particular,

 (n) = W (n) ⊕ gl(n, A) W

(1.2)

becomes a Lie algebra with the Lie bracket

[d1 + A 1 , d2 + A 2 ] = [d1 , d2 ] + [ A 1 , A 2 ] + d1 ( A 2 ) − d2 ( A 1 )

(1.3)

(n) defined for d1 , d2 ∈ W (n) and A 1 , A 2 ∈ gl(n, A). Shen [S1] found a monomorphism  : W (n) → W by

 

n 

 =

f i ∂ xi

i =1

n  i =1

f i ∂ xi +

n 

∂ xi ( f j ) ⊗ E i , j ,

(1.4)

i , j =1

which can essentially be derived from the Lie derivative formula for the action of vector fields on the spaces of tensor fields. Moreover, he [S1,S2,S3] used the monomorphism  to develop a theory of mixed products for the modules of Lie algebras of Cartan type, which is also known as the Larsson functor (cf. [La]) in the case of Witt algebras. Rao [Rs] constructed some irreducible weight modules over the derivation Lie algebra of the algebra of Laurent polynomials based on Shen’s mixed product. Lin and Tan [LT] did the similar thing over the derivation Lie algebra of the algebra of quantum torus. The second author [Z] determined the module structure of Shen’s mixed product over Xu’s nongraded Lie algebras of Witt type in [X]. , C)-module and let b ∈ C. It can be verified that (ϑ(o(n + 2, C))) ⊂ ϑ(o(n + Let M be any o(n n = 2, C)) + o(n, A) + A i =1 E i ,i , which enable us to  define an o(n + 2, C)-module structure on M A ⊗C M, where the hidden central transformation ( ni=1 E i ,i )| M  takes the constant value b. Geometrically, the tensor modules yield natural extensions of the conformal transformations on a Riemannian manifold to its vector bundles. For any two integers m and n, we use the notion

 m, n =

{m, m + 1, . . . , n} if m  n, ∅

otherwise.

(1.5)

Moreover, we denote by N the set of nonnegative integers. vector space over R with a basis {εi | i ∈ 1, m} and the inner product that E is a  Suppose m n ( m a ε , c ε ) = r r i i i =1 r =1 i =1 ai c i . As usual, we take the set of simple positive roots Π2m = {εi − εi +1 , εm−1 + εm | i ∈ 1, m − 1} of o(2m, C) and the set of simple positive roots Π2m+1 = {εi − εi+1 , εm | i ∈ 1, m − 1} of o(2m + 1, C). Set



Λn+ = μ ∈ E 2(μ, ν )/(ν , ν ) ∈ N for ν ∈ Πn ,

n = 2m, 2m + 1.

(1.6)

It is well known that any finite-dimensional irreducible o(n, C)-module is a highest-weight irreducible m module V (μ) with highest weight μ ∈ Λn+ (e.g., cf. [Hu]). For μ = i =1 μi εi ∈ Λn+ , we define ιμ ∈ 1, m by

μ1 = μ2 = · · · = μιμ = μιμ +1 .

(1.7)

Using Pieri’s formulas, invariant operators and the idea of Kostant’s characteristic identities, we prove:

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m

μi εi ∈ Λn+ with n = 2m, 2m + 1  3. The module V (μ) of o(2m + 2, C)  is irreducible if b ∈ C \ {m − 1 − N/2, μ1 + 2m − ιμ − 1 − N}. Moreover, the module V (μ) of o(2m + 3, C) is irreducible if b ∈ C \ {m − N/2, μ1 + 2m − ιμ − N}. Main Theorem. Let 0 = μ =

i =1

  (0) of o(n + 2, C) is irreducible if and only if b ∈ / −N. When b = 0, V (0) is isomorphic to The module V the natural conformal o(n + 2, C)-module A, on which A ( f ) = ϑ( A )( f ) for A ∈ o(n + 2, C) and f ∈ A. The subspace C forms a trivial o (n + 2, C)-submodule of the conformal module A and the quotient space A/C forms an irreducible o(n + 2, C)-module. In some singular cases of

μ = 0, the condition can be relaxed slightly. We speculate that the

 (0) may play the same role in higher-dimensional conformal field theory as that of the module V vacuum module of the Virasoro algebra plays in two-dimensional conformal field theory. The central charge in two-dimensional conformal field theory may be replaced by the constant b in the above theorem for higher-dimensional conformal field theory. The paper is organized as follows. In Section 2, we prove the main theorem for o(2m + 2, C) (m will be replaced by n customarily). Section 3 is devoted to the proof of the main theorem for o(2m + 3, C) (m will also be replaced by n customarily). 2. Extensions of the conformal representations for D n+1 Let n > 1 be an integer. The orthogonal Lie algebra



o(2n, C) =



C( E p ,n+q − E q,n+ p ) + C( E n+ p ,q − E n+q, p )

1 p
+

n 

C( E i , j − E n+ j ,n+i ).

(2.1)

i , j =1

We take the subspace

H=

n 

C( E i ,i − E n+i ,n+i )

(2.2)

i =1

as a Cartan subalgebra and define {εi | i ∈ 1, n} ⊂ H∗ by

εi ( E j, j − E n+ j,n+ j ) = δi, j .

(2.3)

The inner product (·, ·) on the Q-subspace

LQ =

n 

Qε i

(2.4)

i =1

is given by

(εi , ε j ) = δi , j for i , j ∈ 1, n.

(2.5)

Φ D n = {±εi ± ε j | 1  i < j  n}.

(2.6)

Then the root system of o(2n, C) is

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

101

We take the set of positive roots

Φ D+n = {εi ± ε j | 1  i < j  n}.

(2.7)

Π D n = {ε1 − ε2 , . . . , εn−1 − εn , εn−1 + εn } is the set of positive simple roots.

(2.8)

In particular,

Recall the set of dominant integral weights



Λ+ = μ ∈ L Q (εn−1 + εn , μ), (εi − εi +1 , μ) ∈ N for i ∈ 1, n − 1 .

(2.9)

According to (2.5),

 +

Λ = μ=

n 





μi εi μi ∈ Z/2; μi − μi+1 , μn−1 + μn ∈ N .

(2.10)

i =1

Note that if μ ∈ Λ+ , then {n0 , n1 , . . . , ns } such that

μn−1  |μn |. Given μ ∈ Λ+ , there exists a unique sequence S (μ) = 0 = n 0 < n 1 < n 2 < · · · < n s −1 < n s = n

(2.11)

and

μi = μ j

⇔

nr < i , j  nr +1

for some r ∈ 0, s − 1.

(2.12)

For λ ∈ Λ+ , we denote by V (λ) the finite-dimensional irreducible o(2n, C)-module with highest weight λ. The 2n-dimensional natural module of o(2n, C) is V (ε1 ) with the weights {±εi | i ∈ 1, n}. The following result is well known (e.g., cf. [FH]): Lemma 2.1 (Pieri’s formula). Given μ ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns },

V (ε1 ) ⊗C V (μ) ∼ =

s  



V (μ − εni ) ⊕ V (μ + ε1+ni−1 )

(2.13)

i =1

if μn−1 + μn > 0 and

V (ε1 ) ⊗C V (μ) ∼ =

s−2+δμn ,0



V (μ − εni ) ⊕

i =1

s 

V (μ + ε1+ni−1 )

(2.14)

i =1

when μn−1 + μn = 0. Denote by U (G ) the universal enveloping algebra of a Lie algebra G . The algebra U (G ) can be imbedded into the tensor algebra U (G ) ⊗ U (G ) by the associative algebra homomorphism d : U (G ) → U (G ) ⊗C U (G ) determined by

d (u ) = u ⊗ 1 + 1 ⊗ u

for u ∈ G .

(2.15)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Note that the Casimir element of o(2n, C) is

ω=

 

( E i ,n+ j − E j ,n+i )( E n+ j ,i − E n+i , j ) + ( E n+ j ,i − E n+i , j )( E i ,n+ j − E j ,n+i )

1i < j n

+

n 

  ( E i , j − E n+ j ,n+i )( E j ,i − E n+i ,n+ j ) ∈ U o(2n, C) .

(2.16)

i , j =1

Set

ω˜ =

     1 d(ω) − ω ⊗ 1 − 1 ⊗ ω ∈ U o(2n, C) ⊗C U o(2n, C) . 2

(2.17)

By (2.16),

 

ω˜ =

( E i ,n+ j − E j ,n+i ) ⊗ ( E n+ j ,i − E n+i , j ) + ( E n+ j ,i − E n+i , j ) ⊗ ( E i ,n+ j − E j ,n+i )

1i < j n

+

n 

( E i , j − E n+ j ,n+i ) ⊗ ( E j ,i − E n+i ,n+ j ).

(2.18)

i , j =1

Denote

ρ=

1  2

ν.

(2.19)

+

ν ∈Φ Dn

Then

(ρ , ν ) = 1 for ν ∈ Π D n

(2.20)

(e.g., cf. [Hu]). By (2.8),

ρ=

n −1 

(n − i )εi .

(2.21)

i =1

For any

μ ∈ Λ+ , we have

ω| V (μ) = (μ + 2ρ , μ)Id V (μ) .

(2.22)

+ (μ) = dim V (μ + εi ) if μ + εi ∈ Λ+ i

(2.23)

− (μ) = dim V (μ − εi ) if μ − εi ∈ Λ+ . i

(2.24)

Denote

and

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

103

Observe that

(μ + εi + 2ρ , μ + εi ) − (μ + 2ρ , μ) − (ε1 + 2ρ , ε1 ) = 2(μi + 1 − i )

(2.25)

(μ − εi + 2ρ , μ − εi ) − (μ + 2ρ , μ) − (ε1 + 2ρ , ε1 ) = 2(1 + i − 2n − μi )

(2.26)

and

for by

μ=

n

r =1

μr εr by (2.21). Moreover, the algebra U (o(2n, C))⊗C U (o(2n, C)) acts on V (ε1 )⊗C V (μ)

  (ξ1 ⊗ ξ2 )( v ⊗ u ) = ξ1 ( v ) ⊗ ξ2 (u ) for ξ1 , ξ2 ∈ U o(2n, C) , v ∈ V (ε1 ), u ∈ V (μ).

(2.27)

By Lemma 2.1, (2.17) and (2.23)–(2.26), we obtain:

n

Lemma 2.2. Let μ = i =1 μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }. If ˜ | V (ε1 )⊗C V (μ) is polynomial of ω

μn−1 + μn > 0, the characteristic

s  − + ( μ) (t − μ1+ni−1 + ni −1 ) 1+ni−1 (t + μni + 2n − ni − 1)ni (μ) .

(2.28)

i =1

˜ | V (ε1 )⊗C V (μ) is When μn−1 + μn = 0, the characteristic polynomial of ω

 s −1 

+ 1+n

(t − μ1+ni + ni )

i −1

( μ)

 s−2+δμ ,0 n

i =0

n−j (μ)

(t + μn j + 2n − n j − 1)

 .

(2.29)

j =1

We remark that the above lemma is equivalent to special detailed version of Kostant’s characteristic identity. Set

A = C[x1 , x2 , . . . , x2n ].

(2.30)

Then A forms an o(2n, C)-module with the action determined via

E i , j |A = x i ∂ x j

for i , j ∈ 1, 2n.

(2.31)

The corresponding Laplace operator and dual invariant are

=

n 

∂xr ∂xn+r ,

η=

r =1

n 

xi xn+i .

(2.32)

i =1

Denote

D=

2n 

xr ∂xr ,

J i = xi D − η∂xn+i ,

J n+i = xn+i D − η∂xi ,

(2.33)

r =1

A i , j = xi ∂x j − xn+ j ∂xn+i , for i , j ∈ 1, n. Then

B i , j = xi ∂xn+ j − x j ∂xn+i ,

C i , j = xn+i ∂x j − xn+ j ∂xi

(2.34)

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C2n = C D +

2n n   (C∂xr + C J r ) + C A i, j + r =1

i , j =1



is the Lie algebra of 2n-dimensional conformal group over C with Set

L0 =

n 



C A i, j +

i , j =1

(C B i , j + CC i , j )

(2.35)

1i < j n

(C B i , j + CC i , j ),

J =

1i < j n

η as the metric.

2n 

C Ji,

D=

i =1

2n 

C∂xi .

(2.36)

i =1

Then L0 = o(2n, C)|A . We can easily verify the following Lie brackets:

[J , J ] = {0},

[D, D] = {0},

[∂xk , J n+i ] = C i ,k ,

[∂xk , J i ] = δk,i D + A i ,k , [∂xk , A i , j ] = δk,i ∂x j ,

[∂xn+k , J n+i ] = δk,i D − Ak,i ,

[∂xk , B i , j ] = δk,i ∂xn+ j − δk, j ∂xn+i ,

[∂xn+k , A i , j ] = −δk, j ∂xn+i ,

[∂xn+k , B i , j ] = 0,

[ J k , A i , j ] = −δk, j J i ,

[ J k , B i , j ] = 0,

[ J n+k , A i , j ] = δk,i J n+ j , [D, Jk] = Jk,

[∂xn+k , J i ] = B i ,k , [∂xk , C i , j ] = 0,

[∂xn+k , C i , j ] = δk,i ∂x j − δk, j ∂xi , [ J k , C i , j ] = δk,i J n+ j − δk, j J n+i ,

[ J n+k , B i , j ] = δk,i J j − δk, j J i ,

[ D , J n+k ] = J n+k ,

[∂xk , D ] = ∂xk ,

[ J n+k , C i , j ] = 0, [∂xn+k , D ] = ∂xn+k

(2.37) (2.38) (2.39) (2.40) (2.41) (2.42) (2.43)

for i , j , k ∈ 1, n. Recall that the split



o(2n + 2, C) =



C( E r ,n+1+s − E s,n+1+r ) + C( E n+1+r ,s − E n+1+s,r )

1r
+

C( E i , j − E n+1+ j ,n+1+i ).

(2.44)

i , j =1

By (2.37)–(2.43), we have the Lie algebra isomorphism ϑ : o(2n + 2, F) → C2n determined by

ϑ( E i , j − E n+1+ j ,n+1+i ) = A i , j ,

ϑ( E r ,n+1+s − E s,n+1+r ) = B r ,s ,

(2.45)

ϑ( E n+1+r ,s − E n+1+s,r ) = C r ,s ,

ϑ( E n+1,n+1 − E 2n+2,2n+2 ) = − D ,

(2.46)

ϑ( E n+1,i − E n+1+i ,2n+2 ) = ∂xi ,

ϑ( E i ,2n+2 − E n+1,n+1+i ) = −∂xn+i ,

(2.47)

ϑ( E i ,n+1 − E 2n+2,i +n+1 ) = − J i ,

ϑ( E 2n+2,i − E n+1+i ,n+1 ) = J n+i

(2.48)

for i , j ∈ 1, n and 1  r < s  n. The above representation ϑ was derived from (1.1) with the product (x, y ) = ni=1 (xi yn+i + xn+i y i ) by using the derivatives of one-parameter conformal transformations just as one derives the Lie algebra of a Lie group. Analogous formulas for conformal symmetries of the nonlinear Dirac equations were on pages 39–42 of [KK]. given 2n Recall the Witt algebra W2n = i =1 A∂xi , and Shen [S1,S2,S3] found a monomorphism  from the Lie algebra W2n to the Lie algebra of semi-product W2n + gl(2n, A) defined by

 

2n  i =1

 f i ∂ xi

=

2n  i =1

f i ∂ xi +

n  i , j =1

∂ xi ( f j ) E i , j .

(2.49)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

105

Note that C2n ⊂ W2n and

( A i , j ) = A i , j + E i , j − E n+ j ,n+i ,

( B r ,s ) = B r ,s + E r ,n+s − E s,n+r ,

(C r ,s ) = C r ,s + E n+r ,s − E n+s,r ,

(∂xn+i ) = ∂xn+i ,

( D ) = D +

(∂xi ) = ∂xi , 2n 

(2.50)

E p, p ,

(2.51)

E p, p ,

(2.52)

p =1

( J i ) = J i +

n 

xn+ p ( E i ,n+ p − E p ,n+i ) +

p =1

( J n+i ) = J n+i +

n 

n 

xq ( E i ,q − E n+q,n+i ) + xi

q =1

xn+ p ( E n+i ,n+ p − E p ,i ) +

p =1

n 

2n  p =1

xq ( E n+i ,q − E n+q,i ) + xn+i

q =1

2n 

E p, p

(2.53)

p =1

for i , j ∈ 1, n and 1  r < s  n. Moreover,

C2n = C2n + o(2n, A) + A

2n 

E p, p

(2.54)

p =1

2n . In particular, the element forms a Lie subalgebra of W2n + gl(2n, A) and (C2n ) ⊂ C a hidden central element. Let M be an o(2n, C)-module and let b ∈ C be a fixed constant. Then

 = A ⊗C M M

2n

p =1

E p , p is

(2.55)

2n -module with the action: becomes a C

 d + f1 A + f2

2n 







E p , p ( g ⊗ v ) = d( g ) + bf 2 g ⊗ v + f 1 g ⊗ A ( v )

(2.56)

p =1

 a C2n -module with the action: for f 1 , f 2 , g ∈ A, A ∈ o(2n, C) and v ∈ M. Moreover, we make M

. ξ( w ) = (ξ )( w ) for ξ ∈ C2n , w ∈ M

(2.57)

 becomes an o(2n + 2, C)-module with the action Furthermore, M





A ( w ) =  ϑ( A ) ( w )

 for A ∈ o(2n + 2, C), w ∈ M

(2.58)

(cf. (2.45)–(2.48)). Lemma 2.3. If M is an irreducible o (2n, C)-module, then the space U (J )(1 ⊗ M ) is an irreducible o(2n +  2, C)-submodule of M. Proof. Recall o(2n, C)|A = L0 . Moreover, (2.41) and (2.42) give

[L0 , J ] = J . Note that D is the degree operator (cf. (2.33)) and

(2.59)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

[D, J ] ⊂ L0 + C D

(2.60)

by (2.37) and (2.38). According to (2.35) and (2.36),

  ϑ o(2n + 2, C) = C2n = L0 + D + J + C D .

(2.61)

 By (2.43) and (2.59)–(2.61), U (J )(1 ⊗ M ) forms an o(2n + 2, C)-submodule of M. Let W be a nonzero o (2n + 2, C)-submodule of U (J )(1 ⊗ M ). Note

∂ xi | M  = ∂xi ⊗ 1 for i ∈ 1, 2n.

(2.62)

By repeatedly applying the above operators to W , we can prove

W



(1 ⊗ M ) = {0}.

(2.63)



However, W (1 ⊗ M ) is a nonzero L0 -submodule of 1 ⊗ M, which is an irreducible L0 -module. Therefore, 1 ⊗ M ⊂ W . As a C2n -module, W ⊃ U (J )(1 ⊗ M ). 2 Write

xα =

2n  αi

Jα =

xi ,

i =1

2n  αi

Ji

for α = (α1 , α2 , . . . , α2n ) ∈ N2n .

(2.64)

i =1

For k ∈ N, we set





2n



Ak = SpanC xα α ∈ N2n , αi = k ,

 k = Ak ⊗C M M

(2.65)

i =1

and



  2n

 

α 2n U (J )(1 ⊗ M ) k = SpanC J (1 ⊗ M ) α ∈ N , αi = k .

(2.66)

i =1

Moreover,





  0 = 1 ⊗ M . U (J )(1 ⊗ M ) 0 = M

(2.67)

Furthermore,

= M

∞ 

 k , M

U (J )(1 ⊗ M ) =

k =0

k =0

Next we define a linear transformation



∞  





U (J )(1 ⊗ M ) k .

(2.68)

 determined by ϕ on M

ϕ xα ⊗ v = J α (1 ⊗ v ) for α ∈ N2n , v ∈ M .

(2.69)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

107

2n

Note that A1 = i =1 Cxi forms the 2n-dimensional natural L0 -module (equivalently o (2n, C)module). According to (2.41) and (2.42), J forms an L0 -module with respect to the adjoint representation, and the linear map from A1 to J determined by xi → J i for i ∈ 1, 2n gives an L0 -module  to U (J )(1 ⊗ M ). isomorphism. Thus ϕ can also be viewed as an L0 -module homomorphism from M Moreover,





 k ) = U (J )(1 ⊗ M ) ϕ (M

k

for k ∈ N.

(2.70)

˜ )| M Lemma 2.4. We have ϕ | M  1 = (b + ω  1 (cf. (2.16)–(2.18)). Proof. Let i ∈ 1, n and v ∈ M. Expressions (2.52), (2.53) and (2.56)–(2.58) give

ϕ (xi ⊗ v ) = J i (1 ⊗ v ) =

n 

xn+ p ⊗ ( E i ,n+ p − E p ,n+i )( v )

p =1

+

n 

xq ⊗ ( E i ,q − E n+q,n+i )( v ) + bxi ⊗ v ,

(2.71)

q =1

ϕ (xn+i ⊗ v ) = J n+i (1 ⊗ v ) =

n 

xn+ p ⊗ ( E n+i ,n+ p − E p ,i )( v )

p =1

+

n 

xq ⊗ ( E n+i ,q − E n+q,i )( v ) + bxn+i ⊗ v .

(2.72)

q =1

On the other hand, (2.18) and (2.31) yield



ω˜ (xi ⊗ v ) =



( E p ,n+q − E q,n+ p ) ⊗ ( E n+q, p − E n+ p ,q )(xi ⊗ v )

1 p
+ ( E n+q, p − E n+ p ,q ) ⊗ ( E p ,n+q − E q,n+ p )(xi ⊗ v )

+

n 

( E r ,s − E n+s,n+r ) ⊗ ( E s,r − E n+r ,n+s )(xi ⊗ v )

r , s =1

=

n 

xn+ p ⊗ ( E i ,n+ p − E p ,n+i )( v ) +

p =1



ω˜ (xn+i ⊗ v ) =

n 

xr ⊗ ( E i ,r − E n+r ,n+i )( v ),

(2.73)

r =1



( E p ,n+q − E q,n+ p ) ⊗ ( E n+q, p − E n+ p ,q )(xn+i ⊗ v )

1 p
+ ( E n+q, p − E n+ p ,q ) ⊗ ( E p ,n+q − E q,n+ p )(xn+i ⊗ v )

+

n 

( E r ,s − E n+s,n+r ) ⊗ ( E s,r − E n+r ,n+s )(xn+i ⊗ v )

r , s =1

=

n  p =1

x p ⊗ ( E n+i , p − E n+ p ,i )( v ) +

n 

xn+s ⊗ ( E n+i ,n+s − E s,i )( v ).

s =1

Comparing the above four expressions, we get the conclusion in the lemma.

2

(2.74)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

For f ∈ A, we define the action

f (g ⊗ v) = f g ⊗ v

for g ∈ A, v ∈ M .

(2.75)

Then we have the o(2n, C)-invariant operator

T=

n  ( J i xn+i + J n+i xi )| M .

(2.76)

i =1

Lemma 2.5. We have T | M  k = (2b + 2 − 2n + k)η| M  k . Proof. Let f be a homogeneous polynomial with degree k and let v ∈ M. By (2.52), (2.53) and (2.57), we have

T ( f ⊗ v) =

n n    ( J i xn+i + J n+i xi )( f ⊗ v ) = J i (xn+i f ⊗ v ) + J n+i (xi f ⊗ v ) i =1

=

i =1



n 

J i (xn+i f ) ⊗ v +

n 

(xn+ p xn+i f ) ⊗ ( E i ,n+ p − E p ,n+i )( v )

p =1

i =1

n  (xq xn+i f ) ⊗ ( E i ,q − E n+q,n+i )( v ) + (xi xn+i f ) ⊗ + q =1 n 

 E p, p ( v )

(xn+ p xi ) f ⊗ ( E n+i ,n+ p − E p ,i )( v )

p =1 n  (xq xi f ) ⊗ ( E n+i ,q − E n+q,i )( v ) + (xn+i xi f ) ⊗ + q =1 n  

2n  p =1

+ J n +i ( xi f ) ⊗ v +

=





2n 





E p, p ( v )

p =1

n  (xn+ p xn+i f ) ⊗ E i ,n+ p ( v ) − (xn+ p xn+i f ) ⊗ E p ,n+i ( v ) + (xq xn+i f )

i , p =1

i ,q=1

⊗ ( E i ,q − E n+q,n+i )( v ) +

n 

(xn+ p xi ) f ⊗ ( E n+i ,n+ p − E p ,i )( v ) +

i , p =1

n  

(xq xi f )

i ,q=1

  n   ⊗ E n+i ,q ( v ) − (xq xi f ) ⊗ E n+q,i ( v ) + 2bη f + J i (xn+i f ) + J n+i (xi f ) ⊗ v  =

n 





J i (xn+i f ) + J n+i (xi f ) + 2bη f

i =1

 ⊗ v.

(2.77)

i =1

According to (2.33), we find n  



J i (xn+i f ) + J n+i (xi f ) =

i =1

n  

(xi D − η∂xn+i )(xn+i f ) + (xn+i D − η∂xi )(xi f )

i =1

= 2(k + 1)η f − 2nη f − η D ( f ) = (k + 2 − 2n)η f . So the lemma holds.

2

(2.78)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

For 0 = μ =

n

i =1

109

μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }, we define 

Θ(μ) =

μ1 + n − 1 − N if μn−1 = −μn > 0 and s = 2, μ1 + 2n − n1 − 1 − N otherwise.

(2.79)

Theorem 2.6. For 0 = μ ∈ Λ+ , the o(2n + 2, C)-module V (μ) defined by (2.45)–(2.58) is irreducible if b ∈ C \ {n − 1 − N/2, Θ(μ)}. Proof. By Lemma 2.3, it is enough to prove that the homomorphism

ϕ ( V (μ)) = V (μ). According to (2.70), we only need to prove 

ϕ defined in (2.69) satisfies



ϕ V (μ)k = V (μ)k

(2.80)

for any k ∈ N. We will prove it by induction on k. n When k = 0, (2.80) holds by the definition (2.69). Consider k = 1. Write μ = i =1 μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }. According to Lemma 2.2 and Lemma 2.4 with M = V (μ), the eigenvalues of ϕ | V are (μ) 1

b + μ1+ni−1 − ni −1 , b − μni − 2n + ni + 1 for i ∈ 1, s if μn−1 + μn > 0

(2.81)

b + μ1+ni−1 − ni −1 , b − μnr − 2n + nr + 1 for i ∈ 1, s, r ∈ 1, s − 2 + δmn ,0

(2.82)

and

when

μn−1 + μn = 0. Recall that μr ∈ Z/2 for r ∈ 1, n,

μι+1 − μι ∈ N for ι ∈ 1, n − 1,

μn−1  |μn |

(2.83)

and (2.12) holds. So

−μ1+ni−1 + ni −1 , μni + 2n − ni − 1 ∈ μ1 + 2n − n1 − 1 − N for i ∈ 1, s. If b ∈ /

μ1 + 2n − n1 − 1 − N, then all the eigenvalues of ϕ | V (μ)

1

−μn−1 > 0 and s = 2, μ1 = μn−1 = −μn and the eigenvalues ϕ | V (μ)

1

are nonzero. In the case

(2.84)

μn =

are b + μ1 and b + μn − n + 1 =

b − μ1 − n + 1, which are not equal to 0 because of b ∈ / Θ(μ) = μ1 + n − 1 − N. Thus (2.80) holds for k = 1. Suppose that (2.80) holds for k   with   1. Consider k =  + 1. Note that





ϕ V (μ)+1 =

2n  



ϕ xi V (μ) =

i =1

2n   

Ji



ϕ V (μ) =

i =1

2n  

J i V (μ)



(2.85)

i =1

by the inductional assumption. To prove (2.81) with k =  + 1 is equivalent to prove 2n  



J i V (μ) = V (μ)+1 .

i =1

(2.86)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

For any u ∈ V (μ)−1 , Lemma 2.5 says that n  



J i (xn+i u ) + J n+i (xi u ) = (2b + 1 − 2n + )ηu .

(2.87)

i =1

Since b ∈ / n − 1 − N/2, 2b + 1 − 2n +  = 0 and (2.87) gives

ηu ∈

2n  

J i V (μ)



for u ∈ V (μ)−1 .

(2.88)

i =1

Let g ⊗ v ∈ V (μ) . According to (2.52)–(2.57) and Lemma 2.4,

J i ( g ⊗ v ) = xi D ( g ) ⊗ v − η∂xn+i ( g ) ⊗ v +

n 

xn+ p g ⊗ ( E i ,n+ p − E p ,n+i )( v )

p =1

+

n 



xq g ⊗ ( E i ,q − E n+q,n+i )( v ) + xi g ⊗

q =1

 = −η∂xn+i ( g ) ⊗ v + g ( + b + ω˜ )(xi ⊗ v )

2n 

 E p, p ( v )

p =1

(2.89)

and

J n+i ( g ⊗ v ) = xn+i D ( g ) ⊗ v − η∂xi ( g ) ⊗ v +

n 

xn+ p g ⊗ ( E n+i ,n+ p − E p ,i )( v )

p =1

+

n 



xq g ⊗ ( E n+i ,q − E n+q,i )( v ) + xn+i g ⊗

q =1

2n 

 E p, p ( v )

p =1

 = −η∂xi ( g ) ⊗ v + g ( + b + ω˜ )(xn+i ⊗ v )

(2.90)

for i ∈ 1, n. Since

η∂xi ( g ) ⊗ v , η∂xn+i ( g ) ⊗ v ∈

2n 



J r V (μ)



for i ∈ 1, n

(2.91)

r =1

by (2.88), expressions (2.89) and (2.90) show





˜ )(xi ⊗ v ) ∈ g ( + b + ω

2n 



J r V (μ)



for i ∈ 1, 2n, g ∈ A .

(2.92)

r =1

˜ )| V According to Lemmas 2.2 and 2.4, the eigenvalue of ( + b + ω (μ)

1

are among

{b +  + μ1+ni−1 − ni −1 , b +  − μni − 2n + ni + 1 | i ∈ 1, s}. Again

(2.93)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

111

− − μ1+ni−1 + ni −1 , − − μni + 2n − ni − 1 ∈ μ1 + 2n − n1 − 1 − N for i ∈ 1, s. If b ∈ / μ1 + 2n − n1 − 1 − N, then all the eigenvalues of ( + b + ω˜ )| V (μ)

1

(2.94)

are nonzero. In the case

−μn−1 > 0 and s = 2, μ1 = μn−1 = −μn and the eigenvalues ( + b + ω˜ )| V (μ)

1

μn =

are b + μ1 +  and

b + μn − n + 1 +  = b − μ1 − n + 1 + , which are not equal to 0 because of b ∈ / Θ(μ) = μ1 + n − 1 − N. Hence

  ( + b + ω˜ ) V (μ)1 = V (μ)1 .

(2.95)

By (2.92) and (2.95),





g V (μ)1 ⊂

2n 



J r V (μ)



for g ∈ A ,

(2.96)

r =1

equivalently, (2.80) holds for k =  + 1. By induction, (2.80) holds for any k ∈ N.

2

We remark that the o(2n + 2, C)-module V (μ) is o(2n, C)-finite, that is, V (μ) is of (G , K)-type with G = o(2n + 2, C) and K = o(2n, C). Up to this stage, we do not know if the condition in Theorem 2.6 is necessary for the o(2n + 2, C)-module V (μ) to be irreducible if μ = 0. In the case μ = 0, the situation becomes clear.

  (0) is irreducible if and only if b ∈ / −N. When b = 0, V (0) is isomorTheorem 2.7. The o(2n + 2, C)-module V phic to the natural conformal o(2n + 2, C)-module A, on which A ( f ) = ϑ( A )( f ) for A ∈ o(2n + 2, C) and f ∈ A (cf. (2.45)–(2.48)). The subspace C forms a trivial o (2n + 2, C)-submodule of the conformal module A and the quotient space A/C forms an irreducible o(2n + 2, C)-module.  Proof. Pick 0 = v 0 ∈ V (0). Then V (0) = A ⊗ v 0 . Since V (0) is the trivial o(2n, C)-module, (2.50)– (2.57) yield

ξ( f ⊗ v 0 ) = ξ( f ) ⊗ v 0 for f ∈ A, ξ ∈ L0 + D

(2.97)

D ( f ⊗ v 0 ) = (b + D )( f ) ⊗ v 0 ,

(2.98)

J i ( f ⊗ v 0 ) = xi ( D + b) − η∂xn+i ( f ) ⊗ v 0 ,

(2.99)

J n+i ( f ⊗ v 0 ) = xn+i ( D + b) − η∂xi ( f ) ⊗ v 0

(2.100)

(cf. (2.36)) and









for f ∈ A and i ∈ 1, 2n (cf. (2.33)).

 (0) by (2.58). In particular, the map f ⊗ v 0 → f for f ∈ A Recall the action of o(2n + 2, C) on V

 gives an o(2n, C)-module isomorphism from V (0 ) to A. Remember the o(2n, C)-invariant differenn n tial operator  = i =1 ∂xi ∂xn+i and its dual η = i =1 xi xn+i . Moreover, Ak denotes the subspace of homogeneous polynomials in A with degree k. Set



Hk = f ∈ Ak ( f ) = 0 for k ∈ N.

(2.101)

Then Hk ⊗ v 0 is an irreducible o(2n, C)-submodule with the highest-weight vector xk1 ⊗ v 0 . Indeed,

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

∞ 

 V (0) =

ηm Hk ⊗ v 0

(2.102)

m,k=0

is a direct sum of irreducible o(2n, C)-submodules. On the other hand, U (J )(1 ⊗ v 0 ) forms an irre (0) (cf. Lemma 2.3). By (2.99) and (2.100), a necessary condition ducible o(2n + 2, C)-submodule of V  (0) = U (J )(1 ⊗ v 0 ) is b ∈ / −N. for V  / −N − 1. Let W be a nonzero o(2n + 2, C)-submodule of V (0) such that Next we assume b ∈

W ⊂ C ⊗ v 0

if b = 0.

(2.103)

By repeatedly acting D on W if necessary, we have 1 ⊗ v 0 ∈ W . Note

J i (1 ⊗ v 0 ) = bxi ⊗ v 0

for i ∈ 1, 2n.

(2.104)

Thus

 V (0)1 ⊂ W

(2.105)

if b = 0. When b = 0, (2.105) also holds because of (2.103), D ( W ) ⊂ W and the irreducibility of

 V (0)1 as an o(2n, C)-submodule. Suppose that

 V (0)k ⊂ W

for k < ,

(2.106)

ηm H−2m ⊗ v 0 .

(2.107)

where 2   ∈ N. According to (2.102),

 V (0) =

J/2K



m =0

Moreover,

[, η] = n + D .

(2.108)

Set

 V (0),r  =

r 

ηm H−2m ⊗ v 0

(2.109)

m =0

for r ∈ 0, J/2K. Then





  V (0),r  = w ∈ V (0) r +1 ( w ) = 0

(2.110)

and

   r V (0),r  = H−2r ⊗ v 0 by (2.108).

(2.111)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

113

Since





J 1 x1−1 ⊗ v 0 = (b +  − 1)x1 ⊗ v 0 ∈ W

(2.112)

x1 ⊗ v 0 ∈ W .

(2.113)

 V (0),0 = H ⊗ v 0 ⊂ W

(2.114)

and b ∈ / −N − 1, we have

Hence

because H ⊗ v 0 is an irreducible o(2n, C)-submodule generated by x1 ⊗ v 0 . Recall n  2 by our assumption. For r ∈ 1, J/2K,





J 2 x1−r −1 xnr +1 ⊗ v 0 = (b +  − 1)x1−r −1 x2 xnr +1 ⊗ v 0 ∈ W .

(2.115)

So x2 xnr +1 ⊗ v 0 ∈ W . Moreover, r



−r −1

 x1

x2 xnr +1





⊗ v 0 = r!

r 

 ( − r − s) x1−2r −1 x2 ⊗ v 0 ∈ H−2r .

(2.116)

s =1

Observe that (2.107) is a direct sum of irreducible o(2n, C)-submodules with distinct highest weights. So

 V (0)



J/2K

W =



ηm H−2m ⊗ v 0



W.

(2.117)

m =0



By (2.109)–(2.111) and (2.116), (ηr H−2r ⊗ v 0 ) W is a nonzero o(2n, C)-submodule. Since ηr H−2r ⊗ r r v 0 is an irreducible o(2n, C)-module, we have η H−2r ⊗ v 0 = (η H−2r ⊗ v 0 ) W . Therefore,

   V (0) ⊂ W . By induction, V (0)k ⊂ W for any k ∈ N, equivalently, W = V (0). This proves the theorem. 2 3. Extensions of the conformal representations for B n+1 Let n  1 be an integer. The orthogonal Lie algebra

o(2n + 1, C) = o(2n, C) +

n  

C( E 0,i − E n+i ,0 ) + C( E 0,n+i − E i ,0 )

(3.1)

i =1

(cf. (2.1)). We take (2.2) as a Cartan subalgebra and use the settings in (2.3)–(2.5). Then the root system of o(2n + 1, C) is

Φ B n = {±εi ± ε j , ±εr | 1  i < j  n; r ∈ 1, n}.

(3.2)

We take the set of positive roots

Φ B+n = {εi ± ε j , εr | 1  i < j  n, r ∈ 1, n}.

(3.3)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

In particular,

Π B n = {ε1 − ε2 , . . . , εn−1 − εn , εn } is the set of positive simple roots.

(3.4)

Recall the set of dominant integral weights



Λ+ = μ ∈ L Q 2(εn , μ), (εi − εi +1 , μ) ∈ N for i ∈ 1, n − 1 .

(3.5)

According to (2.5),

 +

Λ = μ=

n 





μi εi μi ∈ N/2; μi − μi+1 ∈ N for i ∈ 1, n − 1 .

(3.6)

i =1

n

Given μ = i =1 μi εi ∈ Λ+ , there exists a unique sequence S (μ) = {n0 , n1 , n2 , . . . , ns } such that (2.11) and (2.12) hold. Denote

ρ=

1  2

ν.

(3.7)

ν ∈Φ B+n

Then

2(ρ , ν )

(ν , ν )

= 1 for ν ∈ Π B n

(3.8)

(e.g., cf. [Hu]). By (3.4),

ρ=

n −1  (n − i + 1/2)εi .

(3.9)

i =1

For λ ∈ Λ+ , we denote by V (λ) the finite-dimensional irreducible o(2n + 1, C)-module with highest weight λ. The (2n + 1)-dimensional natural module of o(2n + 1, C) is V (ε1 ) with weights {0, ±εi | i ∈ 1, n}. The following result is well known (e.g., cf. [FH]): Lemma 3.1 (Pieri’s formula). Given μ ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns },

V (ε1 ) ⊗C V (μ) ∼ = (1 − δμn ,0 ) V (μ) ⊕

s−δμn ,0 −δμn ,1/2

 i =1

V (μ − εni ) ⊕

s 

V (μ + ε1+nr −1 ).

(3.10)

r =1

Note that the Casimir element of o(2n + 1, C) is

ω=

 

( E i ,n+ j − E j ,n+i )( E n+ j ,i − E n+i , j ) + ( E n+ j ,i − E n+i , j )( E i ,n+ j − E j ,n+i )

1i < j n

+

n  

( E 0,i − E n+i ,0 )( E i ,0 − E 0,n+i ) + ( E i ,0 − E 0,n+i )( E 0,i − E n+i ,0 )

i =1

+

n 

  ( E i , j − E n+ j ,n+i )( E j ,i − E n+i ,n+ j ) ∈ U o(2n + 1, C) .

i , j =1

(3.11)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

115

Set

ω˜ =

     1 d(ω) − ω ⊗ 1 − 1 ⊗ ω ∈ U o(2n + 1, C) ⊗C U o(2n + 1, C) . 2

(3.12)

By (3.11),

ω˜ =

 

( E i ,n+ j − E j ,n+i ) ⊗ ( E n+ j ,i − E n+i , j ) + ( E n+ j ,i − E n+i , j ) ⊗ ( E i ,n+ j − E j ,n+i )

1i < j n

+

n  

( E 0,i − E n+i ,0 ) ⊗ ( E i ,0 − E 0,n+i ) + ( E i ,0 − E 0,n+i ) ⊗ ( E 0,i − E n+i ,0 )

i =1 n 

+

( E i , j − E n+ j ,n+i ) ⊗ ( E j ,i − E n+i ,n+ j ).

(3.13)

i , j =1

Moreover, (2.22) also holds. Take the settings in (2.23) and (2.24). Denote

(μ) = dim V (μ).

(3.14)

(μ + 2ρ , μ) − (μ + 2ρ , μ) − (ε1 + 2ρ , ε1 ) = −2n,

(3.15)

(μ + εi + 2ρ , μ + εi ) − (μ + 2ρ , μ) − (ε1 + 2ρ , ε1 ) = 2(μi + 1 − i )

(3.16)

(μ − εi + 2ρ , μ − εi ) − (μ + 2ρ , μ) − (ε1 + 2ρ , ε1 ) = 2(i − 2n − μi )

(3.17)

Observe that

and

n

for μ = r =1 μr εr by (3.9). Moreover, the algebra U (o (2n + 1, C)) ⊗C U (o (2n + 1, C)) acts on V (ε1 ) ⊗C V (μ) by

  (ξ1 ⊗ ξ2 )( v ⊗ u ) = ξ1 ( v ) ⊗ ξ2 (u ) for ξ1 , ξ2 ∈ U o(2n + 1, C) , v ∈ V (ε1 ), u ∈ V (μ).

(3.18)

By Lemma 3.1, (3.12), (2.22) and (3.15)–(3.17), we get: Lemma 3.2. Let ω˜ | V (ε1 )⊗C V (μ) is

μ=

n

i =1

 s−δμ (μ)(1−δμn ,0 )

(t + n)

μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }. The characteristic polynomial of

n ,0 −δμn ,1/2



+ 1+n

(t − μ1+ni + ni )

i −1

( μ)

 s−1 

i =0

n−j (μ)

(t + μn j + 2n − n j − 1)

 .

j =1

(3.19) We remark that the above lemma is also equivalent to special detailed version of Kostant’s characteristic identity. Set

A = C[x0 , x1 , x2 , . . . , x2n ].

(3.20)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Then A forms an o(2n + 1, C)-module with the action determined via

E i , j |A = x i ∂ x j

for i , j ∈ 0, 2n.

(3.21)

The corresponding Laplace operator and dual invariant are

 = ∂x20 + 2

n 

1

η = x20 +

∂xr ∂xn+r ,

2

r =1

n 

xi xn+i .

(3.22)

i =1

Denote

D=

2n 

xr ∂xr ,

J i = xi D − η∂xn+i ,

J n+i = xn+i D − η∂xi ,

(3.23)

r =0

J 0 = x0 D − η∂x0 ,

K i = x0 ∂xi − xn+i ∂x0 ,

A i , j = xi ∂x j − xn+ j ∂xn+i ,

K n+i = x0 ∂xn+i − xi ∂x0 ,

B i , j = xi ∂xn+ j − x j ∂xn+i ,

(3.24)

C i , j = xn+i ∂x j − xn+ j ∂xi

(3.25)

for i , j ∈ 1, n. Then

C2n+1 = C D +

2n 

CKs +

s =1

2n n   (C∂xr + C J r ) + C A i, j + r =0

i , j =1



is the Lie algebra of (2n + 1)-dimensional conformal group over C with Moreover, the subspace  C2n = CD +

2n n   (C∂xr + C J r ) + C A i, j + r =1

i , j =1

(C B i , j + CC i , j )

(3.26)

1i < j n



η in (3.22) as the metric.

(C B i , j + CC i , j )

(3.27)

1i < j n

forms a Lie subalgebra of C2n+1 which is isomorphic to C2n in (2.35). Set

L0 =

2n  s =1

CKs +

n 

C A i, j +

i , j =1



(C B i , j + CC i , j ),

1i < j n

J =

2n 

C Ji,

D=

i =0

2n 

C∂xi .

i =0

(3.28) Then L0 = o(2n + 1, C)|A and (2.37)–(2.43) hold. Moreover,

[∂x0 , J 0 ] = D ,

[∂x0 , J n+i ] = − K i ,

[∂x0 , K n+i ] = ∂xn+i ,

[∂xi , J 0 ] = K i ,

[∂xi , K j ] = [∂xn+i , K n+ j ] = 0, [ J 0 , K i ] = J n +i ,

[∂x0 , J i ] = − K n+i ,

[∂x0 , K i ] = ∂xi ,

[∂xn+i , J 0 ] = K n+i ,

[∂xi , K n+ j ] = [∂xn+i , K j ] = −δi , j ∂x0 ,

[ J 0 , K n +i ] = J i , [ J i , K j ] = [ J n+i , K n+ j ] = −δi , j J 0 ,   [ J i , K n + j ] = [ J n + i , K j ] = 0, ∂x0 , o(2n, C)|A = J 0 , (2n, C)|A = {0}

for i , j ∈ 1, n (cf. (3.21)).

(3.29) (3.30) (3.31) (3.32) (3.33)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

117

Recall that the split

o(2n + 3, C) = o(2n + 2, C) +

n +1 



C( E 0,i − E n+i +1,0 ) + ( E 0,n+1+i − E i ,0 )

(3.34)

i =1

(cf. (2.44)). By (3.29)–(3.33), we have the Lie algebra isomorphism ϑ : o(2n + 3, F) → C2n+1 determined by (2.45)–(2.48) and

ϑ( E 0,i − E n+1+i ,0 ) = K i ,

ϑ( E 0,n+1+i − E i ,0 ) = K n+i for i ∈ 1, n,

ϑ( E 0,2n+2 − E n+1,0 ) = −∂x0 ,

ϑ( E 0,n+1 − E 2n+2,0 ) = J 0 .

(3.35) (3.36)

2n

Recall the Witt algebra W2n+1 = i =0 A∂xi , and Shen [S1,S2,S3] found a monomorphism  from the Lie algebra W2n to the Lie algebra of semi-product W2n+1 + gl(2n + 1, A) defined by

 

2n 

 f i ∂ xi

i =0

=

2n 

n 

f i ∂ xi +

i =0

∂ xi ( f j ) E i , j .

(3.37)

i , j =0

Note that C2n+1 ⊂ W2n+1 , and (2.50) and (2.51) except the last equation hold. Moreover,

( K i ) = K i + E 0,i − E n+i ,0 ,

( D ) = D +

2n 

E p, p ,

(3.38)

p =0

( K n+i ) = K n+i + E 0,n+i − E i ,0 , ( J 0 ) = J 0 +

n  

(∂x0 ) = ∂x0 ,

xs ( E 0,s − E n+s,0 ) + xn+s ( E 0,n+s − E s,0 ) + x0

s =1

( J i ) = J i +

(3.39) 2n 

E p, p ,

(3.40)

p =0 n 

xn+ p ( E i ,n+ p − E p ,n+i ) +

p =1

n 

xq ( E i ,q − E n+q,n+i )

q =1

+ x0 ( E i ,0 − E 0,n+i ) + xi

2n 

E p, p ,

(3.41)

p =0

( J n+i ) = J n+i +

n 

xn+ p ( E n+i ,n+ p − E p ,i ) +

p =1

+ x0 ( E n+i ,0 − E 0,i ) + xn+i

n 

xq ( E n+i ,q − E n+q,i )

q =1 2n 

E p, p .

(3.42)

p =0

Furthermore,

C2n+1 = C2n+1 + o(2n + 1, A) + A

2n 

E p, p

(3.43)

p =0

2n+1 . In particular, the element forms a Lie subalgebra of W2n+1 + gl(2n + 1, A) and (C2n+1 ) ⊂ C 2n E is a hidden central element. p , p p =0

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Let M be an o(2n + 1, C)-module and let b ∈ C be a fixed constant. Then

 = A ⊗C M M

(3.44)

2n+1 -module with the action: becomes a C

 d + f1 A + f2



2n 





E p , p ( g ⊗ v ) = d( g ) + bf 2 g ⊗ v + f 1 g ⊗ A ( v )

(3.45)

p =0

 a C2n+1 -module with the action: for f 1 , f 2 , g ∈ A, A ∈ o(2n + 1, C) and v ∈ M. Moreover, we make M

. ξ( w ) = (ξ )( w ) for ξ ∈ C2n+1 , w ∈ M

(3.46)

 becomes an o(2n + 3, C)-module with the action Furthermore, M





A ( w ) =  ϑ( A ) ( w )

 for A ∈ o(2n + 2, C), w ∈ M

(3.47)

(cf. (2.45)–(2.48), (3.35) and (3.36)). By a proof similar to that of Lemma 2.3, we have: Lemma 3.3. If M is an irreducible o (2n + 1, C)-module, then the space U (J )(1 ⊗ M ) is an irreducible o(2n +  3, C)-submodule of M. Write

xα =

2n  αi

Jα =

xi ,

i =0

2n  αi

for α = (α0 , α1 , . . . , α2n ) ∈ N2n+1 .

Ji

(3.48)

i =0

For k ∈ N, we set





2n



Ak = SpanC xα α ∈ N2n+1 , αi = k ,

 k = Ak ⊗C M M

(3.49)

i =0

and



  2n

 

α 2n+1 U (J )(1 ⊗ M ) k = SpanC J (1 ⊗ M ) α ∈ N , αi = k .

(3.50)

i =0

Moreover,





  0 = 1 ⊗ M . U (J )(1 ⊗ M ) 0 = M

(3.51)

Furthermore,

= M

∞  k =0

 k , M

U (J )(1 ⊗ M ) =

∞   k =0



U (J )(1 ⊗ M ) k .

(3.52)

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Next we define a linear transformation



119

 determined by ϕ on M



ϕ xα ⊗ v = J α (1 ⊗ v ) for α ∈ N2n+1 , v ∈ M .

(3.53)

2n

Note A1 = i =0 Cxi forms the (2n + 1)-dimensional natural L0 -module (equivalently o(2n + 1, C)module). According to (2.41), (2.42), (3.32) and (3.33), J forms an L0 -module with respect to the adjoint representation, and the linear map from A1 to J determined by xi → J i for i ∈ 0, 2n gives  to an L0 -module isomorphism. Thus ϕ can also be viewed as an L0 -module homomorphism from M U (J )(1 ⊗ M ). Moreover,





 k ) = U (J )(1 ⊗ M ) ϕ (M

k

for k ∈ N.

(3.54)

˜ )| M Lemma 3.4. We have ϕ | M  1 = (b + ω  1 (cf. (3.11)–(3.13)).  1 = A1 ⊗C M. Let i ∈ 1, n and v ∈ M. Expressions (3.40), (3.45) and (3.46) give Proof. Recall M

ϕ (x0 ⊗ v ) =

n  



xs ⊗ ( E 0,s − E n+s,0 )( v ) + xn+s ⊗ ( E 0,n+s − E s,0 )( v ) + bx0 ⊗ v .

(3.55)

s =1

Moreover, (3.41), (3.45) and (3.46) imply

ϕ ( xi ⊗ v ) =

n 

xn+ p ⊗ ( E i ,n+ p − E p ,n+i )( v ) + x0 ⊗ ( E i ,0 − E 0,n+i )( v )

p =1 n 

+

xq ⊗ ( E i ,q − E n+q,n+i )( v ) + bxi ⊗ v

(3.56)

q =1

for i ∈ 1, n. Furthermore, (3.42), (3.45) and (3.46) yield

ϕ (xn+i ⊗ v ) =

n 

xn+ p ⊗ ( E n+i ,n+ p − E p ,i )( v ) + x0 ⊗ ( E n+i ,0 − E 0,i )( v )

p =1 n 

+

xq ⊗ ( E n+i ,q − E n+q,i )( v ) + bxn+i ⊗ v

(3.57)

q =1

for i ∈ 1, n. On the other hand, (3.13) and (3.21) yield

ω˜ (x0 ⊗ v ) =

n  

−xn+i ⊗ ( E i ,0 − E 0,n+i )( v ) + xi ⊗ ( E 0,i − E n+i ,0 )( v ) ,

(3.58)

i =1

ω˜ (xi ⊗ v ) =

n 

xn+ p ⊗ ( E i ,n+ p − E p ,n+i )( v ) + x0 ⊗ ( E i ,0 − E 0,n+i )( v )

p =1

+

n  r =1

xr ⊗ ( E i ,r − E n+r ,n+i )( v ),

(3.59)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

ω˜ (xn+i ⊗ v ) =

n 

x p ⊗ ( E n+i , p − E n+ p ,i )( v ) − x0 ⊗ ( E 0,i − E n+i ,0 )( v )

p =1

+

n 

xn+s ⊗ ( E n+i ,n+s − E s,i )( v ).

(3.60)

s =1

Comparing the above six expressions, we get the conclusion in the lemma.

2

We use the definition (2.76) and we have the o(2n + 1, C)-invariant operator

 T=



n



J 0 x0 + ( J i xn+i + J n+i xi ) .

 i =1

(3.61)

M

Lemma 3.5. We have T | M  k = (2b − 2n + k + 1)η . Proof. Let f ∈ Ak and v ∈ M. According to (3.24) and (3.40),

J 0 x0 ( f ⊗ v ) = x0

n  



xs f ⊗ ( E 0,s − E n+s,0 )( v ) + xn+s f ⊗ ( E 0,n+s − E s,0 )( v )

s =1

 + (k + 1 + b)x20 − η f ⊗ v − η x0 ∂x0 ( f ) ⊗ v .

(3.62)

Moreover, (2.78), (3.23), (3.41) and (3.42) give n 

( J i xn+i + J n+i xi )( f ⊗ v )

i =1

= x0

n  



xi f ⊗ ( E n+i ,0 − E 0,i )( v ) + xn+i f ⊗ ( E i ,0 − E 0,n+i )( v )

i =1





+ 2 (b + k + 1)

n 

 xi xn+i

 − nη ( f ) ⊗ v −

i =1

2n 

η x i ∂ xi ( f ) ⊗ v .

(3.63)

i =1

Thus



T ( f ⊗ v ) = 2(b + k + 1) − 2n − 1



η( f ) ⊗ v − η D ( f ) ⊗ v

= (2b + k + 1 − 2n)η f ⊗ v . So the lemma holds. For 0 = μ =

n

i =1

(3.64)

2

μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }, we define  Θ(μ) =

∅

n

if μ = (

i =1

μ1 + 2n − n1 − N otherwise.

εi )/2,

(3.65)

Theorem 3.6. For 0 = μ ∈ Λ+ , the o (2n + 3, C)-module V (μ) defined by (2.50), (2.51) except the last equation, and (3.37)–(3.47) is irreducible if b ∈ C \ {n − N/2, Θ(μ)}.

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Proof. By Lemma 3.3, it is enough to prove that the homomorphism

ϕ ( V (μ)) = V (μ). According to (3.54), we only need to prove 

121

ϕ defined in (3.53) satisfies



ϕ V (μ)k = V (μ)k

(3.66)

for any k ∈ N. We will prove it by induction on k. n When k = 0, (3.66) holds by the definition (3.53). Consider k = 1. Write μ = i =1 μi εi ∈ Λ+ with S (μ) = {n0 , n1 , . . . , ns }. According to Lemma 3.2 and Lemma 3.4 with M = V (μ), the eigenvalues of ϕ | V are among (μ) 1

{b − n, b + μni−1 − ni −1 , b − μni − 2n + ni for i ∈ 1, s}. Recall that

(3.67)

μr ∈ N/2 for r ∈ 1, n,

μι+1 − μι ∈ N for ι ∈ 1, n − 1

(3.68)

−μ1+ni−1 + ni −1 , μni + 2n − ni ∈ μ1 + 2n − n1 − N for i ∈ 1, s.

(3.69)

and (2.12) holds. So

If b ∈ /

μ1 + 2n − n1 − N and b = n, then all the eigenvalues of ϕ | V are nonzero. In the case (μ)1 n μ = ( i=1 εi )/2, the eigenvalues ϕ | V are b − n and b + 1/2, which are not equal to 0 because of (μ) 1

/ n − N/2. Thus (3.66) holds for k = 1. b∈ Suppose that (3.66) holds for k   with   1. Consider k =  + 1. Note that





ϕ V (μ)+1 =

2n  



ϕ xi V (μ) =

i =0

2n   

Ji



ϕ V (μ) =

i =0

2n  

J i V (μ)



(3.70)

i =0

by the inductional assumption. To prove (3.66) with k =  + 1 is equivalent to prove 2n  



J i V (μ) = V (μ)+1 .

(3.71)

i =0

For any u ∈ V (μ)−1 , Lemma 3.5 says that

J 0 (x0 u ) +

n  



J i (xn+i u ) + J n+i (xi u ) = (2b − 2n + )ηu .

(3.72)

i =1

Since b ∈ / n − N/2, we have 2b − 2n +  = 0, and so (3.72) gives

ηu ∈

2n  

J i V (μ)

i =0



for u ∈ V (μ)−1 .

(3.73)

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X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

Let g ⊗ v ∈ V (μ) . According to (3.40)–(3.46) and Lemma 3.4,





˜ )(x0 ⊗ v ) , J 0 ( g ⊗ v ) = −η∂x0 ( g ) ⊗ v + g ( + b + ω

(3.74)

 ˜ )(xi ⊗ v ) , J i ( g ⊗ v ) = −η∂xn+i ( g ) ⊗ v + g ( + b + ω  ˜ )(xn+i ⊗ v ) J n+i ( g ⊗ v ) = −η∂xi ( g ) ⊗ v + g ( + b + ω

(3.75) (3.76)

for i ∈ 1, n (cf. (2.89) and (2.90)). Since

η ∂ xi ( g ) ⊗ v ∈

2n 



J r V (μ)



for i ∈ 0, 2n

(3.77)

r =0

by (3.73), expressions (3.74)–(3.76) show





˜ )(xi ⊗ v ) ∈ g ( + b + ω

2n 



J r V (μ)



for i ∈ 0, 2n, g ∈ A .

(3.78)

r =0

˜ )| V According to Lemmas 3.2 and 3.4, the eigenvalue of ( + b + ω (μ)

1

are among

{b +  − n, b +  + μ1+ni−1 − ni −1 , b +  − μni − 2n + ni | i ∈ 1, s}.

(3.79)

− − μ1+ni−1 + ni −1 , − − μni + 2n − ni ∈ μ1 + 2n − n1 − N for i ∈ 1, s.

(3.80)

Again

If b ∈ / {n − N/2, μ1 + 2n − n1 − N}, then all the eigenvalues of ( + b + ω˜ )| V (μ) case

n

μ=(

1

˜ )| V i =1 εi )/2, the eigenvalues of ( + b + ω (μ)

1

not equal to 0 because of b ∈ / n − N/2. Hence

are nonzero. In the

are b +  − n and b +  + 1/2, which are

  ( + b + ω˜ ) V (μ)1 = V (μ)1 .

(3.81)

By (3.78) and (3.81),





g V (μ)1 ⊂

2n 



J r V (μ)



for g ∈ A ,

(3.82)

r =0

equivalently, (3.66) holds for k =  + 1. By induction, (3.66) holds for any k ∈ N.

2

We remark that the o(2n + 3, C)-module V (μ) is o(2n + 1, C)-finite, that is, V (μ) is of (G , K)type with G = o(2n + 3, C) and K = o(2n + 1, C). Up to this stage, we do not know if the condition in Theorem 3.6 is necessary for the o(2n + 3, C)-module V (μ) to be irreducible if μ = 0. In the case μ = 0, the situation becomes clear.

  Theorem 3.7. The o(2n + 3, C)-module V (0) is irreducible if and only if b ∈ / −N. When b = 0, V (0) is isomorphic to the natural conformal o(2n + 3, C)-module A, on which A ( f ) = ϑ( A )( f ) for A ∈ o(2n + 3, C) and f ∈ A (cf. (2.45)–(2.48), (3.35) and (3.36)). The subspace C forms a trivial o (2n + 3, C)-submodule of the conformal module A and the quotient space A/C forms an irreducible o(2n + 3, C)-module.

X. Xu, Y. Zhao / Journal of Algebra 377 (2013) 97–124

123

 Proof. Pick 0 = v 0 ∈ V (0). Then V (0) = A ⊗ v 0 . We only list some facts different from the proof of Theorem 2.7. Since V (0) is the trivial o(2n + 1, C)-module,





J 0 ( f ⊗ v 0 ) = x0 ( D + b) − η∂x0 ( f ) ⊗ v 0 .

(3.83)



Hk = f ∈ Ak ( f ) = 0 for k ∈ N.

(3.84)

Recall (3.22) and (3.49). Set

Then Hk ⊗ v 0 is an irreducible o(2n + 1, C)-submodule also with the highest-weight vector xk1 ⊗ v 0 . Moreover,

[, η] = 1 + 2n + D .

(3.85)

Furthermore,





J 0 x1−r −1 xnr +1 ⊗ v 0 = (b +  − 1)x1−r −1 x0 xnr +1 ⊗ v 0

(3.86)

and x1−2r −1 x0 ∈ H−2r . By the similar arguments as those in the proof of Theorem 2.7, we can prove the conclusion in Theorem 3.7. 2 Acknowledgments Part of this work was done during the first author’s visit to The University of Sydney, under the financial support from Prof. Ruibin Zhang’s ARC research grant. We thank Prof. Zhang for his invitation, hospitality and helpful mathematical discussion. References [BBL] G. Benkart, D. Britten, F.W. Lemire, Modules with bounded multiplicities for simple Lie algebras, Math. Z. 225 (1997) 333–353. [BHL] D. Britten, J. Hooper, F.W. Lemire, Simple C n -modules with multiplicities 1 and applications, Canad. J. Phys. 72 (1994) 326–335. [BL1] D. Britten, F.W. Lemire, A classification of simple Lie modules having 1-dimensional weight space, Trans. Amer. Math. Soc. 299 (1987) 683–697. [BL2] D. Britten, F.W. Lemire, On modules of bounded multiplicities for symplectic algebras, Trans. Amer. Math. Soc. 351 (1999) 3413–3431. [D] P.A.M. Dirac, Relativistic wave equations, Proc R. Soc. Lond. Ser. A 155 (1936) 447–459. [Fs] S.L. Fernando, Lie algebra modules with finite-dimensional weight spaces, I, Trans. Amer. Math. Soc. 322 (1990) 757–781. [FH] W. Fulton, J. Harris, Representation Theory: A First Course, Grad. Texts Math. Read. Math., vol. 129, Springer-Verlag, New York, Berlin, Heidelberg, London, Paris, Tokyo, Hong Kong, Barcelona, Budapest, 1991. [Fv] V. Futorny, The weight representations of semisimple finite-dimensional Lie algebras, PhD thesis, Kiev University, 1987. [Gm1] M.D. Gould, Tensor operators and projection techniques in infinite dimensional representations of semi-simple Lie algebras, J. Phys. A: Math. Gen. 17 (1984) 1–17. [Gm2] M.D. Gould, Characteristic identities for semi-simple Lie algebras, J. Aust. Math. Soc. Ser. B Appl. Math. 26 (1985) 257– 283. [Hu] J.E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer-Verlag, New York, Heidelberg, Berlin, 1972. [K] B. Kostant, On the tensor product of a finite and an infinite dimensional representation, J. Funct. Anal. 20 (1975) 257– 285. [KK] I. Krasil’shchik, P. Kersten, Symmetries and Recursion Operators for Classical Supersymmetric Differential Equations, Math. Appl., Kluwer Academic Publishers, 2000. [La] T. Larsson, Conformal fields: A class of representations of Vect( N )[ J ], Internat. J. Modern Phys. A 7 (26) (1992) 6493– 6508. [LT] W. Lin, S. Tan, Representations of the Lie algebra for quantum torus, J. Algebra 275 (2004) 250–274. [M] O. Mathieu, Classification of irreducible weight modules, Ann. Inst. Fourier (Grenoble) 50 (2000) 537–592.

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