On Drinfel'd twist deformation of the super-Virasoro algebra
aa r X i v : . [ m a t h . R A ] M a r On Drinfel’d twist deformation of the super-Virasoro algebra Hengyun Yang
Department of Mathematics, Shanghai Maritime University, Shanghai 201306, China
Abstract.
In this paper, we describe nonstandard quantum deformation of the super-Virasoroalgebra. Using the Drinfel’d twist quantization technique, we obtain the deformed coproduct andantipode. Hence we get a family of noncommutative and noncocommutative Hopf superalgebras.
Key words:
Quantization, Lie superbialgebra, Drinfel’d twist, the super-Virasoro algebra.
Mathematics Subject Classification (2000):
1. Introduction
Quantum groups, mathematically carry the structures of noncommutative and noncocom-mutative Hopf algebras, were first introduced by Drinfel’d and Jimbo. One of the mostimportant examples of quantum groups is deformation of the universal enveloping algebra ofa Lie algebra. This deformation induces a Lie bialgebra structure on the underlying Lie alge-bra. In [6], Drinfel’d posed the quantization problem of the Lie bialgebra. Lately, Etingof andKazhdan [7] settled this question. But a general formula of a quantization was not obtained.Many authors have made great efforts to quantize explicitly some Lie bialgebras.Inspired by the discovery of quantum groups, quantum supergroups, i.e., Hopf superal-gebras, have also been defined (c.f. [3, 4]), which provide a powerful tool for constructingtrigonometric solutions of the Z -graded Yang-Baxter equation. By extending Etingof andKazhdan’s work [7], Geer [8] recently proved that there exist a general quantizations of Liesuperbialgebras. Similar to Lie algebras cases, the deformations of Lie superalgebras are notunique. In [16], Zhang proved that there is a new Hopf superalgebra structure by the Drin-fel’d twist. Using this method, some good Hopf superalgebras have been found in recent years,e.g., Aizawa [2] and Celeghini et al [5] studied the drinfel’d twist deformations of sl (1 |
2) and osp (1 | F denotes a field of characteristic zero, all vector space and tensorproducts are over F . Let Z , Z + , N denote the sets of all integers, nonnegative integers,positive integers, respectively. We use the convention that if an undefined term appears in anexpression, we always treat it as zero; for instance, L = 0 if / ∈ Z . Supported by Science & Technology Program of Shanghai Maritime University.Corresponding E-mail: [email protected] . Main results Now let us start by recalling some definitions and preliminary results. A supervector space H is a Z -graded vector space, i.e., H = H ¯0 ⊕ H ¯1 . If an element x is in either H or H , we saythat it is Z -homogeneous. We assume that all elements below are Z -homogeneous, where Z = { ¯0 , ¯1 } . For x ∈ H , we always denote [ x ] ∈ Z to be its parity, i.e., x ∈ H [ x ] . We say that x is even (odd) if x ∈ H ¯0 (resp. x ∈ H ¯1 ). A superalgebra ( H, µ, τ ) over a commutative ring R is asupervector space equipped with an associative product µ : H ⊗ H → H respecting the gradingand a unit element 1 ∈ H ¯0 . A Hopf superalgebra ( H, µ, τ, ∆ , ǫ, S ) is a superalgebra equippedwith a coproduct ∆ : H → H ⊗ H , a counit ǫ : H → F , and an antipode S : H → H , satisfyingcompatibility conditions. Note that S satisfies S ( xy ) = ( − [ x ][ y ] S ( y ) S ( x ) for x, y ∈ H . Definition 2.1
A Drinfel’d twist F is an invertible element of H ⊗ H and satisfies ( F ⊗ ⊗ Id )( F ) = (1 ⊗ F )(1 ⊗ ∆)( F ) , (2.1)( ǫ ⊗ Id )( F ) = 1 ⊗ Id ⊗ ǫ )( F ) . (2.2)Write F = X f (1) ⊗ f (2) , F − = X f ′ (1) ⊗ f ′ (2) , and set u = µ · ( S ⊗ Id )( F − ) = X S ( f ′ (1) ) f ′ (2) . The property of S shows that u is invertible with inverse u − = µ · ( Id ⊗ S )( F ) = X f (1) S ( f (2) ) . The following result gives a method to construct new Hopf superalgebra from old ones(cf. [16]), the non-super case can be found in [6].
Lemma 2.2
The superalgebra ( H, µ, τ, ˜∆ , ˜ ǫ, ˜ S ) is a new Hopf superalgebra with ˜∆ = F ∆ F − , ˜ ǫ = ǫ, ˜ S = u − Su.
Now let us recall that the classical super-Virasoro algebra L without central extensionover F is defined as an infinite-dimensional Lie superalgebra generated by the generators { L i , G k | i ∈ Z , k ∈ Z } satisfying the defining relations[ L i , L j ] = ( j − i ) L i + j , [ L i , G k ] = ( k − i G i + k , [ G k , G l ] = 2 L k + l , for all i, j ∈ Z , k, l ∈ Z. Obviously, L contains the Witt algabra W as subalgebra. In thefollowing, we fix m ∈ Z / { } , α ∈ F . Denote X = 1 m ( L + αmL − m ) , Y = exp( α ad L − m )( L m ) .
2n [11], Su and Zhao proved that X and Y span a two-dimensional subalgebra of the Virasoroalgebra, i.e., [ X, Y ] = Y .Denote by U ( L ) the universal enveloping algebra of L . For any x ∈ U ( L ) , a ∈ F , r, k ∈ Z + , i ∈ Z , we set x
The superalgebra U ( L )[[ t ]] under the the coproduct ∆ , the counit ǫ , and theantipode S defined by (2.3-2.7) is a noncommutative and noncocommutative Hopf superalgebra. As a by-product of our proof, we obtain the following two combinatorial identities, whichdo not seem to be obvious to us: s X p =0 ( − p " i + mp m " i + ( r − p − mr m " i + ( r − p + 1) ms − p m = 0 , (2.10) s X p =0 ( − p " k + m p m " k + ( r − p − ) mr m " k + ( r − p + ) ms − p m = 0 , (2.11)where i ∈ Z , k ∈ Z , r ∈ Z + , s > r.
3. Proof of Theorem 2.3
We shall divide the proof of Theorem 2.3 into several lemmas. Set X ′ = 1 m L , Y ′ = L m . Let W = { X i ∈ Z a i L i | a i ∈ F , and a i = 0 for i ≪ } be the completed Witt Lie algebra . Then exp(ad L − m ) ∈ Aut( W ). Evidently, we haveexp( α ad L − m )( X ′ ) = X, exp( α ad L − m )( Y ′ ) = Y. emma 3.1 For any r ∈ Z + , i ∈ Z , we have (ad Y ) r ( L i ) = r X q =0 α q r ! a q ( r, i ) L i +( r − q ) m , (3.1)(ad Y ) r ( G k ) = r X q =0 α q r ! b q ( r, i ) G k +( r − q ) m . (3.2) Proof.
Note that exp( α ad L − m ) ∈ Aut( W ) with the inverse exp( − α ad L − m ). We have(ad Y ) r ( L i ) = exp( α ad L − m )exp( − α ad L − m )((ad Y ) r ( L i ))= exp( α ad L − m )(ad Y ′ ) r exp( − α ad L − m )( L i )= exp( α ad L − m )(ad Y ′ ) r ∞ X p =0 ( − α ) p " i + mp m L i − pm = exp( α ad L − m ) ∞ X p =0 ( − α ) p r ! " i + mp m " i + ( r − p − mr m L i +( r − q ) m = ∞ X p =0 ∞ X q =0 ( − p α q + p r ! " i + mp m " i + ( r − p − mr m ×× " i + ( r − p + 1) mq m L i +( r − p − q ) m = ∞ X q =0 q X p =0 ( − p α q r ! " i + mp m " i + ( r − p − mr m ×× " i + ( r − p + 1) mq − p m L i +( r − q ) m , (3.3)where the last equality follows by first setting q = q ′ − p and exchanging summands over q ′ and p and then replacing q ′ by q . From the fact that (ad Y ) r ( L i ) ∈ ⊕ i + rmq = i − rm F L q , we prove(3.1). Note that exp( α ad L − m )( G k ) = ∞ X p =0 ( − α ) p " k + mp m G k − pm , (ad Y ′ ) r ( G k ) = r ! " k + ( r − ) mr m G k + rm .
5e have (ad Y ) r ( G k ) = ∞ X q =0 q X p =0 ( − p α q r ! " k + m p m " k + ( r − p − ) mr m ×× " k + ( r − p + ) mq − p m G k +( r − q ) m , (3.4)Then (3.2) follows. (cid:3) As a by product of (3.3) and (3.4), we immediately obtain the combinatorial identities(2.10) and (2.11).
Lemma 3.2
For any a ∈ F , r, s ∈ Z + , i ∈ Z , and k ∈ Z , the following equations hold in U ( L ) . L i X
We prove (3.5) and (3.6) by induction on r . The case of r = 1 follows from the formula L i X = ( X − im ) L i − α ( m + i ) L i − m and G k X = ( X − km ) G k − α ( m + k ) G k − m . Let a rp = α p r !( r − p )! " ( p − m − ip m X
For any x ∈ U ( L ) , a, d ∈ F and r, s, m ∈ Z + , we have x a + r , (3.12) x [ r + s ] a = x [ r ] a x [ s ] a − r , (3.13) x [ r ] a = x d = (cid:18) a − dm (cid:19) , (3.15) X r + s = m ( − s r ! s ! x [ r ] a x [ s ] d − r = (cid:18) a − d + m − m (cid:19) . (3.16)For a ∈ F , we set F a = ∞ X r =0 ( − r r ! X [ r ] a ⊗ Y r t r , F a = ∞ X r =0 r ! X
For a, d ∈ F , we have F a F d = 1 ⊗ (1 − Y t ) a − d , v a u d = (1 − Y t ) − ( a + d ) . Therefore F a , F a , u a , v a are all invertible, and F − a = F a , u − a = v − a .Proof. Using (3.15), we have F a F d = ( ∞ X r =0 ( − r r ! X [ r ] a ⊗ Y r t r ) · ( ∞ X s =0 s ! X d ⊗ Y s t s )= ∞ X r,s =0 ( − r r ! s ! X [ r ] a X d ⊗ Y r + s t r + s = ∞ X p =0 ( − p (cid:18) a − dp (cid:19) ⊗ Y p t p = 1 ⊗ (1 − Y t ) a − d . v a u d = ( ∞ X r =0 r ! X [ r ] a Y r t r ) · ( ∞ X s =0 ( − s s ! X [ s ] − d Y s t s )= ∞ X r,s =0 r ! ( − s s ! X [ r ] a Y r X [ s ] − d Y s t r + s = ∞ X p =0 (cid:18) a + d + p − p (cid:19) Y p t p = (1 − Y t ) − ( a + d ) . Then this Lemma follows. (cid:3)
Lemma 3.5.
For a ∈ F , i ∈ Z , we have ( L i ⊗ F a = ∞ X s =0 α s " ( s − m − is m F a − im + s (cid:0) L i − sm ⊗ Y s t s (cid:1) , (3.17)( G k ⊗ F a = ∞ X s =0 α s " ( s − ) m − ks m F a − km + s (cid:0) G k − sm ⊗ Y s t s (cid:1) , (3.18)(1 ⊗ L i ) F a = ∞ X s =0 ( − s F a + s (cid:18) s X p =0 α p a p ( s, i ) X a ⊗ L i +( s − p ) m t s (cid:19) , (3.19)(1 ⊗ G k ) F a = ∞ X s =0 ( − s F a + s (cid:18) s X p =0 α p b p ( s, i ) X a ⊗ G k +( s − p ) m t s (cid:19) , (3.20) L i u a = u a + im ∞ X s =0 ∞ X p =0 2 p X q =0 ( − s α s + q " ( s − m − is m × (3.21) × a q ( p, i − sm ) X [ p ] − a − im L i +( p − s − q ) m Y s t s + p ,G k u a = u a + km ∞ X s =0 ∞ X p =0 2 p X q =0 ( − s α s + q " ( s − ) m − ks m × (3.22) × b q ( p, k − sm ) X [ p ] − a − km G k +( p − s − q ) m Y s t s + p . Proof.
From (3.5) and the definition of F a , we have( L i ⊗ F a = ∞ X r =0 r ! L i X a ⊗ L i +( s − p ) m t s (cid:19) , So (3.19) is right. (3.20) is obtained from (3.12) and (3.11). Now we prove (3.21). From(3.13), (3.7), (3.9) and (3.10), we get L i u a = ∞ X r =0 ( − r r ! L i X [ r ] − a Y r t r = ∞ X r =0 ( − r r ! r X s =0 α s r !( r − s )! " ( s − m − is m X [ r − s ] − a − im L i − sm Y r t r = ∞ X r =0 ∞ X s =0 ( − r + s α s r ! " ( s − m − is m X [ r ] − a − im L i − sm Y r + s t r + s = ∞ X r =0 ∞ X s =0 ( − r + s α s r ! " ( s − m − is m X [ r ] − a − im ×× r X p =0 2 p X q =0 ( − p α q r !( r − p )! a q ( p, i − sm ) Y r − p L i +( p − s − q ) m Y s t r + s = ∞ X s =0 ∞ X p =0 (cid:18) ∞ X r =0 ( − r r ! X [ r ] − a − im Y r t r (cid:19) ( − s " ( s − m − is m X [ p ] − a − im ×× p X q =0 α s + q a q ( p, i − sm ) L i +( p − s − q ) m Y s t s + p u a + im ∞ X s =0 ∞ X p =0 2 p X q =0 ( − s α s + q " ( s − m − is m a q ( p, i − sm ) X [ p ] − a − im L i +( p − s − q ) m Y s t s + p . Thus (3.21) holds. Similarly, (3.22) can be proved from (3.8), (3.9),(3.11), and (3.13). (cid:3)
Now we give the proof of Theorem 2.3.
Proof of Theorem 2.3.
Sice F is a Drinfel’d twist, according to Lemma 2.2, we only need todetermine the action of ∆ and S on L i , G k ∈ L , i ∈ Z , k ∈ Z . From (3.17), (3.19), andLemma 3.4, we have∆( L i ) = F ∆ ( L i ) F − = F ( L i ⊗ F − + F (1 ⊗ L i ) F − = F ( L i ⊗ F + F (1 ⊗ L i ) F = F ∞ X r =0 α r " ( r − m − ir m F − im + r (cid:0) L i − rm ⊗ Y r t r )+ F ∞ X r =0 ( − r F r (cid:18) r X s =0 α s a s ( r, i ) X
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