Recent developments of the Lauricella string scattering amplitudes and their exact SL(K+3,C) Symmetry
aa r X i v : . [ h e p - t h ] J a n Recent developments of the Lauricella string scatteringamplitudes and their exact SL ( K + 3 , C ) Symmetry
Sheng-Hong Lai, ∗ Jen-Chi Lee, † and Yi Yang ‡ Department of Electrophysics, National Chiao-Tung University, Hsinchu, Taiwan, R.O.C. (Dated: January 22, 2021)
Abstract
In this review we propose a new perspective to demonstrate Gross conjecture on high energysymmetry of string theory [1–5]. We review the construction of the exact string scattering am-plitudes (SSA) of three tachyons and one arbitrary string state, or the Lauricella SSA (LSSA), inthe 26 D open bosonic string theory. These LSSA form an infinite dimensional representation ofthe SL ( K + 3 , C ) group. Moreover, we show that the SL ( K + 3 , C ) group can be used to solve allthe LSSA and express them in terms of one amplitude. As an application in the hard scatteringlimit, the LSSA can be used to directly prove Gross conjecture which was previously corrected andproved by the method of decoupling of zero norm states (ZNS) [6–16]. Finally, the exact LSSA canbe used to rederive the recurrence relations of SSA in the Regge scattering limit with associated SL (5 , C ) symmetry and the extended recurrence relations (including the mass and spin dependentstring BCJ relations) in the nonrelativistic scattering limit with associated SL (4 , C ) symmetrydiscovered recently. ∗ Electronic address: [email protected] † Electronic address: [email protected] ‡ Electronic address: [email protected] ontents I. Introduction II. The exact LSSA and their SL ( K + 3 , C ) Symmetry SL (4 , C ) Symmetry 9E. The General SL( K + 3,C) Symmetry 12F. Discussion 15 III. Solving LSSA through Recurrence relations K + 3,C) Symmetry and Recurrence Relations 221. SL (4 , C ) Symmetry 222. SL ( K + 3 , C ) Symmetry 25E. Lauricella Zero Norm States and Ward Identities 271. The Lauricella zero norm states 282. The Lauricella Ward identities 30F. Summary 32 IV. Relations among LSSA in various scattering limits
Acknowledgments References I. INTRODUCTION
In contrast to low energy string theory, many issues of high energy behavior of stringtheory have not been well understood yet so far. Historically, it was first conjectured byGross [1–5] that there exist infinite linear relations among hard string scattering amplitudes(HSSA) of different string states. Moreover, these linear relations are so powerful that theycan be used to solve all HSSA and express them in terms of one amplitude. This conjecturewas later (slightly) corrected and proved by using the decoupling of zero norm states [6–9]in [10–16]. For more details, see the recent review articles [17, 18].In this paper, we review another perspective to understand high energy behavior of stringand demonstrate Gross conjecture on high energy symmetry of string theory. Since thetheory of string as a quantum theory consists of infinite number of particles with arbitraryhigh spins and masses, one first crucial step to uncover its high energy behavior is to exactlycalculate a class of SSA which contain the whole spectrum and are valid for all energies.Recently the present authors constructed a class of such exact SSA which contain threetachyons and one arbitrary string state in the spectrum, or the Lauricella SSA (LSSA), inthe 26 D open bosonic string theory.In chaper II of this review, we calculate the LSSA and express them in terms of the D -type Lauricella functions. As an application, we easily reproduce the string BCJ relation[19–22]. For illustration of LSSA, we give two simple examples to demonstrate the com-plicated notations. We then proceed to show that the LSSA form an infinite dimensionalrepresentation of the SL ( K + 3 , C ) group. For simplicity and as an warm up exercise, wewill begin with the case of K = 1 or the SL (4 , C ) group.In chapter III, we first show that there exist K + 2 recurrence relations among the D -typeLauricella functions. We then show that the corresponding K +2 recurrence relations amongthe LSSA can be used to reproduce the Cartan subalgebra and simple root system of the SL ( K + 3 , C ) group with rank K + 2. As a result, the SL ( K + 3 , C ) group can be used to3olve all the LSSA and express them in terms of one amplitude. We stress that these exact nonlinear relations among the exact LSSA are generalization of the linear relations amongHSSA in the hard scattering limit conjectured by Gross. Finally we show that, for the firstfew mass levels, the Lauricella recurrence relations imply the validity of Ward identitiesderived from the decoupling of Lauricella ZNS. However these Lauricella Ward identities are not good enough to solve all the LSSA and express them in terms of one amplitude.In chapter IV of this review, we calculate symmetries or relations among the LSSA ofdifferent string states at various scattering limits. These include the linear relations firstconjectured by Gross [1–5] and later corrected and proved in [10, 12–16] in the hard scat-tering limit, the recurrence relations in the Regge scattering limit with associated SL (5 , C )symmetry [23–25] and the extended recurrence relations (including the mass and spin depen-dent string BCJ relations) in the nonrelativistic scattering limit with associated SL (4 , C )symmetry [26] discovered recently. II. THE EXACT LSSA AND THEIR SL ( K + 3 , C ) SYMMETRYA. The exact LSSA
One important observation of calculating the LSSA is to first note that SSA of threetachyons and one arbitrary string state with polarizations orthogonal to the scattering planevanish. This observation will greatly simplify the calculation of the LSSA. In the CM frame,we define the kinematics as k = (cid:18)q M + | ~k | , −| ~k | , (cid:19) , (2.1) k = (cid:18)q M + | ~k | , + | ~k | , (cid:19) , (2.2) k = (cid:18) − q M + | ~k | , −| ~k | cos φ, −| ~k | sin φ (cid:19) , (2.3) k = (cid:18) − q M + | ~k | , + | ~k | cos φ, + | ~k | sin φ (cid:19) (2.4)with M = M = M = − φ is the scattering angle. The Mandelstam variables are s = − ( k + k ) , t = − ( k + k ) and u = − ( k + k ) . There are three polarizations on4he scattering plane and they are defined to be [10, 12] e T = (0 , , , (2.5) e L = 1 M (cid:18) | ~k | , q M + | ~k | , (cid:19) , (2.6) e P = 1 M (cid:18)q M + | ~k | , | ~k | , (cid:19) (2.7)where e P = M ( E , k ,
0) = k M the momentum polarization, e L = M (k , E ,
0) the longi-tudinal polarization and e T = (0 , ,
1) the transverse polarization. For later use, we alsodefine k Xi ≡ e X · k i for X = ( T, P, L ) . (2.8)We can now proceed to calculate the LSSA of three tachyons and one arbitrary stringstates in the 26 D open bosonic string theory. The general states at mass level M =2( N − N = P n,m,l> (cid:0) nr Tn + mr Pm + lr Ll (cid:1) with polarizations on the scattering plane areof the following form (cid:12)(cid:12) r Tn , r Pm , r Ll (cid:11) = Y n> (cid:0) α T − n (cid:1) r Tn Y m> (cid:0) α P − m (cid:1) r Pm Y l> (cid:0) α L − l (cid:1) r Ll | , k i . (2.9)The ( s, t ) channel of the LSSA can be calculated to be [27] A ( r Tn ,r Pm ,r Ll ) st = Y n =1 (cid:2) − ( n − k T (cid:3) r Tn · Y m =1 (cid:2) − ( m − k P (cid:3) r Pm Y l =1 (cid:2) − ( l − k L (cid:3) r Ll · B (cid:18) − t − , − s − (cid:19) F ( K ) D (cid:18) − t − R Tn , R Pm , R Ll ; u − N ; ˜ Z Tn , ˜ Z Pm , ˜ Z Ll (cid:19) (2.10)where we have defined R Xk ≡ (cid:8) − r X (cid:9) , · · · , (cid:8) − r Xk (cid:9) k with { a } n = a, a, · · · , a | {z } n . (2.11)and Z Xk ≡ (cid:2) z X (cid:3) , · · · , (cid:2) z Xk (cid:3) with (cid:2) z Xk (cid:3) = z Xk , · · · , z Xk ( k − . (2.12)In Eq.(2.12), we have defined z Xk = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) − k X k X (cid:19) k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , z Xkk ′ = z Xk e πik ′ k , ˜ z Xkk ′ ≡ − z Xkk ′ for k ′ = 0 , · · · , k − (cid:2) z Xk (cid:3) = z Xk , z Xk ω k , ..., z Xk ω k − k , ω k = e πik . (2.14)5he integer K in Eq.(2.10) is defined to be K = X j { for all r Tj =0 } + X j { for all r Pj =0 } + X j { for all r Lj =0 } . (2.15)The D -type Lauricella function F ( K ) D in Eq.(2.10) is one of the four extensions of theGauss hypergeometric function to K variables and is defined to be F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K )= ∞ X n , ··· ,n K =0 ( α ) n + ··· + n K ( γ ) n + ··· + n K ( β ) n · · · ( β K ) n K n ! · · · n K ! x n · · · x n K K (2.16)where ( α ) n = α · ( α + 1) · · · ( α + n −
1) is the Pochhammer symbol. There was an integralrepresentation of the Lauricella function F ( K ) D discovered by Appell and Kampe de Feriet(1926) [28] F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K )= Γ( γ )Γ( α )Γ( γ − α ) Z dt t α − (1 − t ) γ − α − · (1 − x t ) − β (1 − x t ) − β ... (1 − x K t ) − β K , (2.17)which was used to calculate Eq.(2.10). B. String BCJ relation as a by-product
Alternatively, by using the identity of the Lauricella function for b i ∈ Z − F ( K ) D ( a ; b , ..., b K ; c ; x , ..., x K ) = Γ ( c ) Γ ( c − a − P b i )Γ ( c − a ) Γ ( c − P b i ) · F ( K ) D (cid:16) a ; b , ..., b K ; 1 + a + X b i − c ; 1 − x , ..., − x K (cid:17) , (2.18)one can rederive the string BCJ relations [19–22] A ( r Tn ,r Pm ,r Ll ) st A ( r Tn ,r Pm ,r Ll ) tu = ( − ) N Γ (cid:0) − s − (cid:1) Γ (cid:0) s + 2 (cid:1) Γ (cid:0) u + 2 − N (cid:1) Γ (cid:0) − u − N (cid:1) = sin (cid:0) πu (cid:1) sin (cid:0) πs (cid:1) = sin ( πk · k )sin ( πk · k ) . (2.19)6his gives another form of the ( s, t ) channel amplitude A ( r Tn ,r Pm ,r Ll ) st = B (cid:18) − t − , − s − N (cid:19) Y n =1 (cid:2) − ( n − k T (cid:3) r Tn · Y m =1 (cid:2) − ( m − k P (cid:3) r Pm Y l =1 (cid:2) − ( l − k L (cid:3) r Ll · F ( K ) D (cid:18) − t − R Tn , R Pm , R Ll ; s − N ; Z Tn , Z Pm , Z Ll (cid:19) . (2.20)Similarly, the ( t, u ) channel amplitude can be calculated to be A ( r Tn ,r Pm ,r Ll ) tu = B (cid:18) − t − , − u − (cid:19) Y n =1 (cid:2) − ( n − k T (cid:3) r Tn · Y m =1 (cid:2) − ( m − k P (cid:3) r Pm Y l =1 (cid:2) − ( l − k L (cid:3) r Ll · F ( K ) D (cid:18) − t − R Tn , R Pm , R Ll ; s − N ; Z Tn , Z Pm , Z Ll (cid:19) . (2.21)To illustrate the complicated notations used in Eq.(2.10), we give two explicit examplesof the LSSA in the following subsection. C. Two simple examples of the LSSA
1. Example one
We take the tensor state of the second vertex to be | state i = (cid:0) α T − (cid:1) r T (cid:0) α P − (cid:1) r P (cid:0) α L − (cid:1) r L | , k i . (2.22)The LSSA in Eq.(2.10) can then be calculated to be A ( r T ,r P ,r Ll ) st = (cid:0) − k T (cid:1) r T (cid:0) − k P (cid:1) r P (cid:0) − k L (cid:1) r L B (cid:18) − t − , − s − (cid:19) · F (3) D (cid:18) − t − − r T , − r P , − r L ; u − N ; ˜ z T , ˜ z P , ˜ z L (cid:19) (2.23)7here the arguments in F (3) D are calculated to be R Tn = (cid:8) − r T (cid:9) , · · · , (cid:8) − r Tn (cid:9) k = (cid:8) − r T (cid:9) = − r T ,R Pm = (cid:8) − r P (cid:9) , · · · , (cid:8) − r Pm (cid:9) k = (cid:8) − r P (cid:9) = − r P ,R Ll = (cid:8) − r L (cid:9) , · · · , (cid:8) − r Ll (cid:9) k = (cid:8) − r L (cid:9) = − r L , (2.24)˜ Z Tn = (cid:2) ˜ z T (cid:3) , · · · , (cid:2) ˜ z Tn (cid:3) = (cid:2) ˜ z T (cid:3) = ˜ z T = 1 − z T = 1 − z Tk e πi = 1 − (cid:12)(cid:12)(cid:12)(cid:12) − k T k T (cid:12)(cid:12)(cid:12)(cid:12) , ˜ Z Pn = (cid:2) ˜ z P (cid:3) , · · · , (cid:2) ˜ z Pn (cid:3) = (cid:2) ˜ z P (cid:3) = ˜ z P = 1 − (cid:12)(cid:12)(cid:12)(cid:12) − k P k P (cid:12)(cid:12)(cid:12)(cid:12) , ˜ Z Ln = (cid:2) ˜ z L (cid:3) , · · · , (cid:2) ˜ z Ln (cid:3) = (cid:2) ˜ z L (cid:3) = ˜ z L = 1 − (cid:12)(cid:12)(cid:12)(cid:12) − k L k L (cid:12)(cid:12)(cid:12)(cid:12) (2.25)and the order K in Eq.(2.15) is K = X j { for all r Tj =0 } + X j { for all r Pj =0 } + X j { for all r Lj =0 } = 1 + 1 + 1 = 3 . (2.26)
2. Example two
We take the tensor state to be | state i = (cid:0) α T − (cid:1) r T (cid:0) α T − (cid:1) r T (cid:0) α T − (cid:1) r T (cid:0) α T − (cid:1) r T | , k i . (2.27)The LSSA in Eq.(2.10) can be calculated to be A ( r T ,r P ,r Ll ) st = (cid:0) − k T (cid:1) r T (cid:0) − k T (cid:1) r T (cid:0) − k T (cid:1) r T (cid:0) − k T (cid:1) r T B (cid:18) − t − , − s − (cid:19) · F (14) D − t − − r T , − r T , − r T | {z } , − r T , − r T , − r T , − r T , − r T | {z } , − r T , − r T , − r T , − r T , − r T , − r T | {z } ; u + 2 − N ; ˜ z T , ˜ z T , ˜ z T | {z } , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T | {z } , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T | {z } (2.28)where the arguments in F (14) D are calculated to be8 Tn = (cid:8) − r T (cid:9) , · · · , (cid:8) − r Tn (cid:9) k = (cid:8) − r T (cid:9) , (cid:8) − r T (cid:9) , (cid:8) − r T (cid:9) , (cid:8) − r T (cid:9) = − r T , − r T , − r T | {z } , − r T , − r T , − r T , − r T , − r T | {z } , − r T , − r T , − r T , − r T , − r T , − r T | {z } (2.29)˜ Z Tn = (cid:2) ˜ z T (cid:3) , · · · , (cid:2) ˜ z Tn (cid:3) = (cid:2) ˜ z T (cid:3) , (cid:2) ˜ z T (cid:3) , (cid:2) ˜ z T (cid:3) , (cid:2) ˜ z T (cid:3) = ˜ z T , ˜ z T , ˜ z T | {z } , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T | {z } , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T , ˜ z T | {z } (2.30)and K = X j { for all r Tj =0 } + X j { for all r Pj =0 } + X j { for all r Lj =0 } = (1 + 2 + 5 + 6) + 0 + 0 = 14 . (2.31)In the following subsections, we discuss the exact SL ( K + 3 , C ) symmetry of the LSSA. Forsimplicity, we will begin with the simple SL (4 , C ) symmetry with K = 1 . D. The SL (4 , C ) Symmetry
In this section, for illustration we first consider the simplest K = 1 case with SL (4 , C )symmetry. For a given K , there can be LSSA with different mass levels N . As an example,for the case of K = 1 there are three types of LSSA( α T − ) p , F (1) D (cid:18) − t − , − p , , u − p , (cid:19) , N = p , ( α P − ) q , F (1) D (cid:18) − t − , − q , u − q , (cid:2) ˜ z P (cid:3)(cid:19) , N = q , ( α L − ) r , F (1) D (cid:18) − t − , − r , u − r , (cid:2) ˜ z L (cid:3)(cid:19) , N = r . (2.32)To calculate the group representation of the LSSA for K = 1, we define [29] f bac ( α ; β ; γ ; x ) = B ( γ − α, α ) F (1) D ( α ; β ; γ ; x ) a α b β c γ . (2.33)We see that the LSSA in Eq.(2.10) for the case of K = 1 corresponds to the case a = 1 = c ,and can be written as A R X st = f − k X (cid:18) − t − R X ; u − N ; ˜ Z X (cid:19) . (2.34)9e can now introduce the ( K + 3) − − SL (4 , C ) group[29, 30] E α = a ( x∂ x + a∂ a ) ,E − α = 1 a [ x (1 − x ) ∂ x + c∂ c − a∂ a − xb∂ b ] ,E β = b ( x∂ x + b∂ b ) ,E − β = 1 b [ x (1 − x ) ∂ x + c∂ c − b∂ b − xa∂ a ] ,E γ = c [(1 − x ) ∂ x + c∂ c − a∂ a − b∂ b ] ,E − γ = − c ( x∂ x + c∂ c − ,E βγ = bc [( x − ∂ x + b∂ b ] ,E − β, − γ = 1 bc [ x ( x − ∂ x + xa∂ a − c∂ c + 1] ,E αγ = ac [(1 − x ) ∂ x − a∂ a ] ,E − α, − γ = 1 ac [ x (1 − x ) ∂ x − xb∂ b + c∂ c − ,E αβγ = abc∂ x ,E − α, − β, − γ = 1 abc [ x ( x − ∂ x − c∂ c + xb∂ b + xa∂ a − x + 1] ,J α = a∂ a ,J β = b∂ b ,J γ = c∂ c , (2.35)and calculate their operations on the basis functions [29, 30]10 α f bac ( α ; β ; γ ; x ) = ( γ − α − f bac ( α + 1; β ; γ ; x ) ,E β f bac ( α ; β ; γ ; x ) = βf bac ( α ; β + 1; γ ; x ) ,E γ f bac ( α ; β ; γ ; x ) = ( γ − β ) f bac ( α ; β ; γ + 1; x ) ,E βγ f bac ( α ; β ; γ ; x ) = βf bac ( α ; β + 1; γ + 1; x ) ,E αγ f bac ( α ; β ; γ ; x ) = ( β − γ ) f bac ( α + 1; β ; γ + 1; x ) ,E αβγ f bac ( α ; β ; γ ; x ) = βf bac ( α + 1; β + 1; γ + 1; x ) ,E − α f bac ( α ; β ; γ ; x ) = ( α − f bac ( α − β ; γ ; x ) ,E − β f bac ( α ; β ; γ ; x ) = ( γ − β ) f bac ( α ; β − γ ; x ) ,E − γ f bac ( α ; β ; γ ; x ) = ( α + 1 − γ ) f bac ( α ; β ; γ − x ) ,E − β, − γ f bac ( α ; β ; γ ; x ) = ( α − γ + 1) f bac ( α ; β − γ − x ) ,E − α, − γ f bac ( α ; β ; γ ; x ) = ( α − f bac ( α − β ; γ − x ) ,E − α, − β, − γ f bac ( α ; β ; γ ; x ) = ( − α + 1) f bac ( α − β − γ − x ) ,J α f bac ( α ; β ; γ ; x ) = αf bac ( α ; β ; γ ; x ) ,J β f bac ( α ; β ; γ ; x ) = βf bac ( α ; β ; γ ; x ) ,J γ f bac ( α ; β ; γ ; x ) = γf bac ( α ; β ; γ ; x ) . (2.36)It is important to note, for example, that since β is a nonpositive integer, the operationby E − β will not be terminated as in the case of the finite dimensional representation of acompact Lie group. Here the representation is infinite dimensional. On the other hand, asimple calculation gives [ E α , E − α ] = 2 J α − J γ , [ E β , E − β ] = 2 J β − J γ , [ E γ , E − γ ] = 2 J γ − ( J α + J β + 1) , which suggest the Cartan subalgebra[ J α , J β ] = 0 , [ J β , J γ ] = 0 , [ J α , J γ ] = 0 . (2.37)11ndeed, if we redefine J ′ α = J α − J γ ,J ′ β = J β − J γ ,J ′ γ = J γ −
12 ( J α + J β + 1) , we discover that each of the following six triplets [29, 30] (cid:8) J + , J − , J (cid:9) ≡ { E α , E − α , J ′ α } , (cid:8) E β , E − β , J ′ β (cid:9) , (cid:8) E γ , E − γ , J ′ γ (cid:9) , (cid:8) E α,β,γ , E − α, − β, − γ , J ′ α + J ′ β + J ′ γ (cid:9) , (cid:8) E αγ , E − α, − γ , J ′ α + J ′ γ (cid:9) , (cid:8) E αβ , E − α, − β , J ′ α + J ′ β (cid:9) constitutes the well known commutation relations (cid:2) J , J ± (cid:3) = ± J ± , (cid:2) J + , J − (cid:3) = 2 J . (2.38) E. The General SL( K + 3 ,C) Symmetry We are now ready to generalize the calculation of the previous section and calculate thegroup representation of the LSSA for general K . We first define [29] f b ··· b K ac ( α ; β , · · · , β K ; γ ; x , · · · , x K )= B ( γ − α, α ) F ( K ) D ( α ; β , · · · , β K ; γ ; x , · · · , x K ) a α b β · · · b β K K c γ . (2.39)Note that the LSSA in Eq.(2.10) corresponds to the case a = 1 = c , and can be written as A ( r Tn ,r Pm ,r Ll ) st = f − ( n − k T , − ( m − k P , − ( l − k L (cid:18) − t − R Tn , R Pm , R Ll ; u − N ; ˜ Z Tn , ˜ Z Pm , ˜ Z Ll (cid:19) . (2.40)It is possible to extend the calculation of the SL (4 , C ) symmetry group for the K = 1 casediscussed in the previous section to the general SL ( K + 3 , C ) group. We first introduce the12 K + 3) − SL ( K + 3 , C ) group ( k = 1 , , ...K ) [29, 30] E α = a X j x j ∂ j + a∂ a ! ,E β k = b k ( x k ∂ k + b k ∂ b k ) ,E γ = c X j (1 − x j ) ∂ x j + c∂ c − a∂ a − X j b j ∂ b j ! ,E αγ = ac X j (1 − x j ) ∂ x j − a∂ a ! ,E β k γ = b k c [( x k − ∂ x k + b k ∂ b k ] ,E αβ k γ = ab k c∂ x k ,E α = 1 a "X j x j (1 − x j ) ∂ x j + c∂ c − a∂ a − X j x j b j ∂ b j ,E β k = 1 b k " x k (1 − x k ) ∂ x k + x k X j = k (1 − x j ) x j ∂ x j + c∂ c − x k a∂ a − X j b j ∂ u j ,E γ = − c X j x j ∂ x j + c∂ c − ! ,E αγ = 1 ac "X j x j (1 − x j ) ∂ x j − X j x j b j ∂ b j + c∂ c − ,E β k γ = 1 b k c " x k ( x k − ∂ x k + X j = k ( x j − x j ∂ x j + x k a∂ a − c∂ c + 1 ,E αβ k γ = 1 ab k c "X j x j ( x j − ∂ x j − c∂ c + x k a∂ a + X j x j b j ∂ b j − x k + 1 ,E β k β p = b k b p [( x k − x p ) ∂ z k + b k ∂ b k ] , ( k = p ) ,J α = a∂ a ,J β k = b k ∂ b k ,J γ = c∂ c . (2.41)Note that we have used the upper indices to denote the ”raising operators” and the lowerindices to denote the ”lowering operators”. The number of generators can be counted bythe following way. There are 1 E α , K E β k , 1 E γ ,1 E αγ , K E β k γ and K E αβ k γ which sumup to 3 K + 3 raising generators. There are also 3 K + 3 lowering operators. In addition,13here are K ( K − E β k β p and K + 2 J , the Cartan subalgebra. In sum, the total numberof generators are 2(3 K + 3) + K ( K −
1) + K + 2 = ( K + 3) −
1. It is straightforward tocalculate the operation of these generators on the basis functions ( k = 1 , , ...K ) [29] E α f b ··· b K ac ( α ) = ( γ − α − f b ··· b K ac ( α + 1) ,E β k f b ··· b K ac ( β k ) = β k f b ··· b K ac ( β k + 1) ,E γ f b ··· b K ac ( γ ) = γ − X j β j ! f b ··· b K ac ( γ + 1) ,E αγ f b ··· b K ac ( α ; γ ) = X j β j − γ ! f b ··· b K ac ( α + 1; γ + 1) ,E β k γ f b ··· b K ac ( β k ; γ ) = β k f b ··· b K ac ( β k + 1; γ + 1) ,E αβ k γ f b ··· b K ac ( α ; β k ; γ ) = β k f b ··· b K ac ( α + 1; β k + 1; γ + 1) ,E α f b ··· b K ac ( α ) = ( α − f b ··· b K ac ( α − ,E β k f b ··· b K ac ( β k ) = γ − X j β j ! f b ··· b K ac ( β k − ,E γ f b ··· b K ac ( γ ) = ( α − γ + 1) f b ··· b K ac ( γ − ,E αγ f b ··· b K ac ( α ; γ ) = ( α − f b ··· b K ac ( α − γ − ,E β k γ f b ··· b K ac ( β k ; γ ) = ( α − γ + 1) f b ··· b K ac ( β k − γ − ,E αβ k γ f b ··· b K ac ( α ; β k ; γ ) = (1 − α ) f b ··· b K ac ( α − β k − γ − ,E β k β p f b ··· b K ac ( β k ; β p ) = β k f b ··· b K ac ( β k + 1; β p − ,J α f b ··· b K ac ( α ; β k ; γ ) = αf b ··· b K ac ( α ; β k ; γ ) ,J β k f b ··· b K ac ( α ; β k ; γ ) = β k f b ··· b K ac ( α ; β k ; γ ) ,J γ f b ··· b K ac ( α ; β k ; γ ) = γf b ··· b K ac ( α ; β k ; γ ) (2.42)where, for simplicity, we have omitted those arguments in f b ··· b K ac which remain the same afterthe operation. The commutation relations of the SL ( K + 3) Lie algebra can be calculated inthe following way. In addition to the Cartan subalgebra for the K +2 generators { J α , J β k , J γ } ,14et’s redefine J ′ α = J α − J γ ,J ′ β k = J β k − J γ + X j = k J β j ,J ′ γ = J γ − J α + X j J β j + 1 ! . (2.43)We discover that each of the following seven triplets [29] (cid:8) J + , J − , J (cid:9) ≡ { E α , E α , J ′ α } , (cid:8) E β k , E β k , J ′ β k (cid:9) , (cid:8) E γ , E γ , J ′ γ (cid:9) , (cid:8) E αβ k γ , E αβ k γ , J ′ α + J ′ β k + J ′ γ (cid:9) , (cid:8) E αγ , E αγ , J ′ α + J ′ γ (cid:9) , (cid:8) E αβ k , E αβ k , J ′ α + J ′ β k (cid:9) , n E β l β p , E β p β l , J ′ β l − J ′ β p o (2.44)satisfies the commutation relations in Eq.(2.38).Finally, in addition to Eq.(2.44), there is another compact way to write down the Liealgebra commutation relations of SL ( K + 3 , C ) . Indeed, one can check that the Lie algebracommutation relations of SL ( K + 3 , C ) can be written as [29][ E ij , E kl ] = δ jk E il − δ li E kj (2.45)with the following identifications E α = E , E α = E , E β k = E k +3 , , E β = E ,k +3 ,E γ = E , E γ = E , E αγ = E , E αγ = E ,E β k γ = −E k +3 , , E β k γ = −E ,k +3 , E αβ k γ = −E k +3 , ,E αβ k γ = −E ,k +3 , J ′ α = 12 ( E − E ) , J ′ β k = 12 ( E k +3 ,k +3 − E ) , J ′ γ = 12 ( E − E ) . (2.46) F. Discussion
There are some special properties in the SL ( K + 3 , C ) group representation of the LSSA,which make it different from the usual symmetry group representation of a physical system.First, the set of LSSA does not fill up the whole representation space V . For example, states f b ··· b K ac ( α ; β , · · · , β K ; γ ; x , · · · , x K ) in V with a = 1 or c = 1 are not LSSA.15ndeed, there are more states in V with K ≥ K = 2 there are six types of LSSA ( ω = − α T − ) p ( α P − ) q , F (2) D ( a, − p , − q , c − p − q , , (cid:2) ˜ z P (cid:3) ), N = p + q , (2.47)( α T − ) p ( α L − ) r , F (2) D ( a, − p , − r , c − p − r , , (cid:2) ˜ z L (cid:3) ), N = p + r , (2.48)( α P − ) q ( α L − ) r , F (2) D ( a, − q , − r , c − q − r , (cid:2) ˜ z P (cid:3) , (cid:2) ˜ z L (cid:3) ), N = q + r , (2.49)( α T − ) p , F (2) D ( a, − p , − p , c − p , ,
1) , N = 2 p , (2.50)( α P − ) q , F (2) D ( a, − q , − q , c − q , − z P , − ωz P ) , N = 2 q , (2.51)( α L − ) r , F (2) D ( a, − r , − r , c − r , − z L , − ωz L ) , N = 2 r . (2.52)One can show that those states obtained from the operation by E β on either states inEq.(2.50) to Eq.(2.52) are not LSSA. However, it can be shown in chap III that all statesin V including those ”auxiliary states” which are not LSSA stated above can be exactlysolved by recurrence relations or the SL ( K + 3 , C ) group and express them in terms of oneamplitude. These ”auxiliary states” and states with a = 1 or c = 1 in V may represent otherSSA, e.g. SSA of two tachyon and two arbitrary string states etc. which will be consideredin the near future. III. SOLVING LSSA THROUGH RECURRENCE RELATIONS
In the previous section, the string scattering amplitudes of three tachyons and one ar-bitrary string states in the 26D open bosonic string theory.has been obtained in term ofthe D -type Lauricella functions, i.e. LSSA in Eq.(2.10). The symmetry of the LSSA wasalso discussed by constructing the SL ( K + 3 , C ) group for the D -type Lauricella functions F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ). It is natural to suspect that the LSSA are dependent eachother due to the symmetry among them. In fact, we are able to show that all the LSSA arerelated to a single LSSA by the recurrence relations of the D -type Lauricella functions.To solve all the LSSA, a key observation is that all arguments β m in the Lauricellafunctions F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ) in the LSSA (2.10) are nonpositive integers. Wewill see that this plays a key role to prove the solvability of all the LSSA.The generalization of the 2 + 2 recurrence relations of the Appell functions to the K + 2recurrence relations of the Lauricella functions was given in [31]. One can use these K + 2recurrence relations to reduce all the Lauricella functions F ( K ) D in the LSSA (2.10) to the16auss hypergeometry functions F ( α, β, γ ). Then all the LSSA can be solved by derivinga multiplication theorem for the Gauss hypergeometry functions.In this section, we will review these steps constructed in [31]. A. Recurrence Relations of the LSSA
For K = 2, the Lauricella functions D -type F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ) reduce tothe type-1Appell functions F ( α ; b , β ; γ, x, y ). The four fundamental recurrence relationswhich link the contiguous functions are( α − β − β ) F ( α ; β , β ; γ, x, y ) − αF ( α + 1; β , β ; γ, x, y )+ β F ( α ; β + 1 , β ; γ, x, y ) + β F ( α ; β , β + 1; γ, x, y ) = 0 , (3.1) γF ( α ; β , β ; γ, x, y ) − ( γ − α ) F ( α ; β , β ; γ + 1 , x, y ) − αF ( α + 1; β , β ; γ + 1 , x, y ) = 0 , (3.2) γF ( α ; β , β ; γ, x, y ) + γ ( x − F ( α ; β + 1 , β ; γ, x, y ) − ( γ − α ) xF ( α ; β + 1 , β ; γ + 1 , x, y ) = 0 , (3.3) γF ( α ; β , β ; γ, x, y ) + γ ( y − F ( α ; β , β + 1; γ, x, y ) − ( γ − α ) yF ( α ; β , β + 1; γ + 1 , x, y ) = 0 . (3.4)17t is straightforward to generalize the above relations and prove the following K +2 recurrencerelations for the D -type Lauricella functions [31] α − X i β i ! F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ) − αF ( K ) D ( α + 1; β , ..., β K ; γ ; x , ..., x K )+ β F ( K ) D ( α ; β + 1 , ..., β K ; γ ; x , ..., x K ) + ... + β K F ( K ) D ( α ; β , ..., β K + 1; γ ; x , ..., x K ) = 0 , (3.5) γF ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ) − ( γ − α ) F ( K ) D ( α ; β , ..., β K ; γ + 1; x , ..., x K ) − αF ( K ) D ( α + 1; β , ..., β K ; γ + 1; x , ..., x K ) = 0 , (3.6) γF ( K ) D ( α ; β , ..., β m , ..., β K ; γ ; x , ..., x m , ..., x K )+ γ ( x m − F ( K ) D ( α ; β , ..., β m + 1 , ..., β K ; γ ; x , ..., x m , ..., x K )+( α − γ ) x m F ( K ) D ( α ; β , ..., β m + 1 , ..., β K ; γ + 1; x , ..., x m , ..., x K ) = 0 , (3.7)where m = 1 , , ..., K . In the case of K = 2, Eq.(3.7) reduces to the Appell recurrencerelations in Eq.(3.3) and Eq.(3.4).To simplify the notation, we will omit those arguments of F ( K ) D which remain the samein the rest of the paper. Then the above K + 2 recurrence relations can be expressed as α − X i β i ! F ( K ) D − αF ( K ) D ( α + 1) + β F ( K ) D ( β + 1) + ... + β K F ( K ) D ( β K + 1) = 0 , (3.8) γF ( K ) D − ( γ − α ) F ( K ) D ( γ + 1) − αF ( K ) D ( α + 1; γ + 1) = 0 , (3.9) γF ( K ) D + γ ( x m − F ( K ) D ( β m + 1) + ( α − γ ) x m F ( K ) D ( β m + 1 , ; γ + 1) = 0 . (3.10)To proceed, we first consider the two recurrence relations from Eq.(3.10) for m = i , j with i = j , cF ( K ) D + γ ( x i − F ( K ) D ( β i + 1) + ( α − γ ) x i F ( K ) D ( β i + 1; γ + 1) = 0 , (3.11) γF ( K ) D + γ ( x j − F ( K ) D ( β j + 1) + ( α − γ ) x j F ( K ) D ( β j + 1; γ + 1) = 0 , (3.12)By shifting β i,j to β i,j − F ( K ) D ( c + 1)term, we obtain the following key recurrence relation [31] x j F ( K ) D ( β i − − x i F ( K ) D ( β j −
1) + ( x i − x j ) F ( K ) D = 0 . (3.13)18 a) (b) FIG. 1: The neighborhood points in the figures are related by the recurrence relations.One can repeatly apply Eq.(3.13) to the Lauricella functions in the LSSAin Eq.(2.10) and end up with an expression which expresses F ( K ) D ( β , β , ..β K )in terms of F ( K − D ( β , ..β i − , β i +1 ..β ′ j , ..β K ), β ′ j = β j , β j − , .., β j − | β i | or F ( K − D ( β , ..β ′ i , ..β j − , β j +1 , ..β K ), β ′ i = β i , β i − , .., β i − | β j | (assume i < j ). We can re-peat the above process to decrease the value of K and reduce all the Lauricella functions F ( K ) D in the LSSA to the Gauss hypergeometry functions F (1) D = F ( α, β, γ, x ) as shown inFig.1. B. Solving all the LSSA
In the last subsection, we have expressed all the LSSA in terms of the Gauss hyperge-ometry functions F (1) D = F ( α, β, γ, x ). In this subsection, we further reduce the Gausshypergeometry functions by deriving a multiplication theorem for them, and solve all theLSSA in terms of one single amplitude.We begin with the Taylor’s theorem f ( x + y ) = ∞ X n =0 y n n ! d n dx n f ( x ) . (3.14)By replacing y by ( y − x , we get the identity f ( xy ) = ∞ X n =0 ( y − n x n n ! d n dx n f ( x ) . (3.15)19ne can now use the derivative relation of the Gauss hypergeometry function d n dx n F ( α, β, γ, x ) = ( α ) n ( β ) n ( γ ) n F ( α + n, β + n, γ + n, x ) , (3.16)where ( α ) n = α · ( α + 1) · · · ( α + n −
1) is the Pochhammer symbol, to obtain the followingmultiplication theorem F ( α, β, γ, xy ) = | β | X n =0 ( y − n x n n ! ( α ) n ( β ) n ( γ ) n F ( α + n, β + n, γ + n, x ) . (3.17)It is important to note that the summation in the above equation is up to a finite integer | β | given β is a nonpositive integer for the cases of LSSA.In particular if we take x = 1 in Eq.(3.17), we get the following relation F ( α, β, γ, y ) = | β | X n =0 ( y − n n ! ( α ) n ( β ) n ( γ ) n F ( α + n, β + n, γ + n, | β | X n =0 ( y − n n ! ( α ) n ( β ) n ( γ ) n ( − ) n ( γ ) n ( γ − α − β ) n F ( α, β, γ, . (3.18)By using the following one of the 15 Gauss contiguous relations { γ − β + ( β − α ) x } F + β (1 − x ) F ( β + 1) + ( β − γ ) F ( β −
1) = 0 , (3.19)and set x = 1 which kills the second term of Eq.(3.19), we can reduce the argument β in F ( α, β, c,
1) to β = − C. Examples of solving LSSA
For illustration, in this subsection, we calculate the Lauricella functions which correspondto the LSSA for levels K = 1 , , K = 1 there are three type of LSSA ( α = − t − , γ = u + 2)( α T − ) p , F (1) D ( α, − p , γ − p , N = p , (3.20)( α P − ) q , F (1) D ( α, − q , γ − q , (cid:2) ˜ z P (cid:3) ), N = q , (3.21)( α L − ) r , F (1) D ( α, − r , γ − r , (cid:2) ˜ z L (cid:3) ), N = r . (3.22)20or K = 2 there are six type of LSSA ( ω = − α T − ) p ( α P − ) q , F (2) D ( α, − p , − q , γ − p − q , , (cid:2) ˜ z P (cid:3) ), N = p + q , (3.23)( α T − ) p ( α L − ) r , F (2) D ( α, − p , − r , γ − p − r , , (cid:2) ˜ z L (cid:3) ), N = p + r , (3.24)( α P − ) q ( α L − ) r , F (2) D ( α, − q , − r , γ − q − r , (cid:2) ˜ z P (cid:3) , (cid:2) ˜ z L (cid:3) ), N = q + r , (3.25)( α T − ) p , F (2) D ( α, − p , − p , γ − p , , N = 2 p , (3.26)( α P − ) q , F (2) D ( α, − q , − q , γ − q , − Z P , − ωZ P ), N = 2 q , (3.27)( α L − ) r , F (2) D ( α, − r , − r , γ − r , − Z L , − ωZ L ), N = 2 r . (3.28)For K = 3, there are ten type of LSSA ( ω = − , ω = ( − i √ ) / )( α T − ) p ( α P − ) q ( α L − ) r , F (3) D ( α, − p , − q , − r , γ − p − q − r , , (cid:2) ˜ z P (cid:3) , (cid:2) ˜ z L (cid:3) ), N = p + q + r , (3.29)( α T − ) p ( α P − ) q , F (3) D ( α, − p , − p , − q , γ − p − q , , , (cid:2) ˜ z P (cid:3) ), N = 2 p + q , (3.30)( α T − ) p ( α L − ) r , F (3) D ( α, − p , − p , − r , γ − p − r , , , (cid:2) ˜ z L (cid:3) ), N = 2 p + r , (3.31)( α T − ) p ( α P − ) q , F (3) D ( α, − p , − q , − q , γ − q − p , , − Z P , − ω Z P ), N = 2 q + p , (3.32)( α P − ) q ( α L − ) r , F (3) D ( α, − q , − q , − r , γ − q − r , − Z P , − ω Z P , (cid:2) ˜ z L (cid:3) ), N = 2 q + r , (3.33)( α T − ) p ( α L − ) r , F (3) D ( α, , − p , − r , − r , γ − r − p , , − Z L , − ω Z L ), N = 2 r + p . (3.34)( α P − ) q ( α L − ) r , F (3) D ( α, , − q , − r , − r , γ − r − q , (cid:2) ˜ z P (cid:3) , − Z L , − ω Z L ), N = 2 r + q . (3.35)( α T − ) p , F (3) D ( α, − p , − p , − p , γ − p , , , N = 3 p , (3.36)( α P − ) q , F (3) D ( α, − q , − q , − q , γ − q , − Z P , − ω Z P , − ω Z P ), N = 3 q , (3.37)( α L − ) r , F (3) D ( α, − r , − r , − r , γ − r , − Z L , − ω Z L , − ω Z L ), N = 3 r . (3.38)21ll the LSSA for K = 2 , K = 1 . Furthermore, all resulting LSSA for K = 1 canbe further reduced by applying Eq.(3.18) to Eq.(3.19) and finally expressed in terms of onesingle LSSA. D. SL( K + 3 ,C) Symmetry and Recurrence Relations In this subsection, we are going to use the recurrence relations of the D -type F ( K ) D ( α ; β , ..., β K ; γ ; x , ..., x K ) to reproduce the Cartan subalgebra and simple root sys-tem of SL ( K + 3 , C ) with rank K + 2. We will first review the case of SL (4 , C ) symmetrygroup, and then extend it to the general case of SL ( K + 3 , C ) Symmetry. SL (4 , C ) Symmetry
We first relate the SL (4 , C ) group to the recurrence relations of F (1) D ( α ; β ; γ ; x ) or of theLSSA in Eq.(2.32). For our purpose, there are K + 2 = 1 + 2 = 3 recurrence relations among F (1) D ( α ; β ; γ ; x ) or Gauss hypergeometry functions( α − β ) F (1) D − αF (1) D ( α + 1) + βF (1) D ( β + 1) = 0 , (3.39) γF (1) D − ( γ − α ) F (1) D ( γ + 1) − αF (1) D ( α + 1; γ + 1) = 0 , (3.40) γF (1) D + γ ( x − F (1) D ( β + 1) − ( γ − α ) xF (1) D ( β + 1; γ + 1) = 0 , (3.41)which can be used to reproduce the Cartan subalgebra and simple root system of the SL (4 , C ) group with rank 3.With the identification in Eq.(2.33), the first recurrence relation in Eq.(3.39) can berewritten as( α − β ) f bac ( α ; β ; γ ; x ) B ( γ − α, α ) a α b β c γ − αf bac ( α + 1; β ; γ ; x ) B ( γ − α − , α + 1) a α +1 b β c γ + βf bac ( α ; β + 1; γ ; x ) B ( γ − α, α ) a α b β +1 c γ = 0 . (3.42)By using the identity B ( γ − α − , α + 1) = Γ ( γ − α −
1) Γ ( α + 1)Γ ( γ ) = αγ − α − γ − α ) Γ ( α )Γ ( γ ) , (3.43)22he recurrence relation then becomes( α − β ) f bac ( α ; β ; γ ; x ) − γ − α − a f bac ( α + 1; β ; γ ; x ) + βb f bac ( α ; β + 1; γ ; x ) = 0 , (3.44)or (cid:18) α − β − E α a + E β b (cid:19) f bac ( α ; β ; γ ; x ) = 0 , (3.45)which means [ α − β − ( x∂ x + a∂ a ) + ( x∂ x + b∂ b )] f bac ( α ; β ; γ ; x ) = 0 , (3.46)or [( α − J α ) − ( β − J β )] f bac ( α ; β ; γ ; x ) = 0 . (3.47)Similarly for the second recurrence relation in Eq.(3.40), we obtain (cid:20) c ( γ − β ) − E γ + E αγ a (cid:21) f bac ( α ; β ; γ ; x ) = 0 . (3.48)which means [( γ − c∂ c ) − ( β − b∂ b )] f bac ( α ; β ; γ ; x ) = 0 , (3.49)or [( γ − J γ ) − ( β − J β )] f bac ( α ; β ; γ ; x ) = 0 . (3.50)Finally the third recurrence relation in Eq.(3.41) can be rewritten as (cid:20) bβ + ( x − E β − xE βγ c (cid:21) f bac ( α ; β ; γ ; x ) = 0 , (3.51)which gives after some computation( β − J β ) f bac ( α ; β ; γ ; x ) = 0 . (3.52)It is easy to see that Eq.(3.47), Eq.(3.50) and Eq.(3.52) imply the last three equations ofEq.(2.36) or the Cartan subalgebra in Eq.(2.37) as expected.In addition to the Cartan subalgebra, we need to derive the operations of the { E α , E β , E γ } from the recurrence relations. With the operations of Cartan subalgebra and { E α , E β , E γ } ,one can reproduce the whole SL (4 , C ) algebra.We first use the operation of E α,β in Eq.(2.36) to express Eq.(3.39) in the following twoways, (cid:18) α − β − E a a (cid:19) f bac ( α ; β ; γ ; x ) + βb f bac ( α ; β + 1; γ ; x ) = 0 , (3.53) (cid:18) α − β + E β b (cid:19) f bac ( α ; β ; γ ; x ) − ( γ − α − a f bac ( α + 1; β ; γ ; x ) = 0 , (3.54)23hich, by using the definition of E α,β in Eq.(2.35), become (cid:18) α − β − a ( x∂ x + a∂ a ) a (cid:19) f bac ( α ; β ; γ ; x ) = − βf bac ( α ; β + 1; γ ; x ) b , (3.55) (cid:18) α − β + b ( x∂ x + b∂ b ) b (cid:19) f bac ( α ; β ; γ ; x ) = ( γ − α − f bac ( α + 1; β ; γ ; x ) a , (3.56)which in turn imply[ b ( b∂ b + x∂ x )] f bac ( α ; β ; γ ; x ) = E β f bac ( α ; β ; γ ; x ) = βf bac ( α ; β + 1; γ ; x ) , (3.57)[ a ( a∂ a + x∂ x )] f bac ( α ; β ; γ ; x ) = E α f bac ( α ; β ; γ ; x ) = ( γ − α − f bac ( α + 1; β ; γ ; x ) , (3.58)The above Eq.(3.57) and Eq.(3.58) are consistent with the operation of E α,β in Eq.(2.36).Finally we check the operation of E γ . Note that Eq.(3.40) can be written as γf bac ( α ; β ; γ ; x ) B ( γ − α, α ) a α b β c γ − ( γ − α ) f bac ( α ; β ; γ + 1; x ) ( γ − α ) γ B ( γ − α, α ) a α b β c γ +1 − αf bac ( α + 1; β ; γ + 1; x ) αγ B ( γ − α, α ) a α +1 b β c γ +1 = 0 , (3.59)which gives f bac ( α ; β ; γ ; x ) − c f bac ( α ; β ; γ + 1; x ) − ac f bac ( α + 1; β ; γ + 1; x ) = 0 . (3.60)Using the definition and operation of E αγ in Eq.(2.35), we obtain f bac ( α ; β ; γ ; x ) − c f bac ( α ; β ; γ + 1; x ) − E αγ ac ( β − γ ) f bac ( α ; β ; γ ; x ) = 0 , which gives f bac ( α ; β ; γ ; x ) − ac [(1 − x ) ∂ x − a∂ a ] f bac ( α ; β ; γ ; x ) ac ( β − γ ) = f bac ( α ; β ; γ + 1; x ) c . (3.61)After some simple computation, we get − c [ b∂ b − c∂ c − (1 − x ) ∂ x + a∂ a ] f bac ( α ; β ; γ ; x ) = E γ f bac ( α ; β ; γ ; x ) = ( γ − β ) f bac ( α ; β ; γ + 1; x ) , which is consistent with the operation of E γ in Eq.(2.36).We thus have shown that the extended LSSA f bac ( α ; β ; γ ; x ) in Eq.(2.33) with arbitrary a and c form an infinite dimensional representation of the SL (4 , C ) group. Moreover, the 3recurrence relations among the LSSA can be used to reproduce the Cartan subalgebra andsimple root system of the SL (4 , C ) group with rank 3. The recurrence relations are thusequivalent to the representation of the SL (4 , C ) symmetry group.24 . SL ( K + 3 , C ) Symmetry
The K + 2 fundamental recurrence relations among F ( K ) D ( α ; β ; γ ; x ) or the Lauricellafunctions.have been listed in Eqs.(3.8-3.10). In the following we will show that the threetypes of recurrence relations above imply the Cartan subalgebra of the SL ( K + 3 , C ) groupwith rank K + 2.With the identification in Eq.(2.39), the first type of recurrence relation in Eq.(3.8) canbe rewritten as α − X j β j ! f b ··· b K ac − E α f b ··· b K ac ( α ) a + X j E β j f b ··· b K ac ( β j ) b j = 0 , (3.62)which gives α − X j β j ! f b ··· b K ac − X j x j ∂ j + a∂ a ! f b ··· b K ac + X j (cid:0) x j ∂ j + b j ∂ b j (cid:1) f b ··· b K ac = 0 (3.63)or " ( α − a∂ a ) + X j (cid:0) β j − b j ∂ b j (cid:1) f b ··· b K ac = 0 , (3.64)which means " ( α − J α ) + X j (cid:0) β j − J β j (cid:1) f b ··· b K ac = 0 . (3.65)The second type of recurrence relation in Eq.(3.9) can be rewritten as f b ··· b K ac − E γ f b ··· b K ac ( γ ) c γ − X j β j ! − E αγ f b ··· b K ac ( α ; γ ) ac X j β j − γ ! = 0 , (3.66)which gives " γ − X j β j − X j (1 − x j ) ∂ x j + c∂ c − a∂ a − X j b j ∂ b j ! + X j (1 − x j ) ∂ x j − a∂ a ! f b ··· b K ac = 0(3.67)or " ( γ − c∂ c ) − X j (cid:0) β j − b j ∂ b j (cid:1) f b ··· b K ac = 0 . (3.68)Eq.(3.68) can be written as " ( γ − J γ ) − X j (cid:0) β j − J β j (cid:1) f b ··· b K ac = 0 . (3.69)25he third type of recurrence relation in Eq.(3.10) can be rewritten as ( m = 1 , , ...K ) f b ··· b K ac + ( x m − E β m f b ··· b K ac b m β m − x m E β m γ f b ··· b K ac b m cβ m = 0 , (3.70)which gives β m f b ··· b K ac + ( x m −
1) ( x m ∂ m + b m ∂ b m ) f b ··· b K ac − x m [( x m − ∂ x m + b m ∂ b m ] f b ··· b K ac = 0 (3.71)or ( β m − b m ∂ b m ) f b ··· b K ac = 0 . (3.72)In the above calculation, we have used the definition and operation of E β m γ in Eq.(2.41)and Eq.(2.42), respectively.Eq.(3.72) can be written as( β m − J β m ) f b ··· b K ac = 0 , m = 1 , , ...K. (3.73)It is important to see that Eq.(3.65), Eq.(3.69) and Eq.(3.73) imply the last three equa-tions of Eq.(2.42) or the Cartan subalgebra of SL ( K + 3 , C ) as expected.In addition to the Cartan subalgebra, we need to derive the operations of the { E α , E β k , E γ } from the recurrence relations. With the operations of Cartan subalgebraand { E α , E β k , E γ } , one can reproduce the whole SL ( K + 3 , C ) algebra. The calculationsof E α and E γ are straightforward and are similar to the case of SL (4 , C ) in the previoussection. Here we present only the calculation of E β k . The recurrence relation in Eq.(3.8)can be rewritten as α − X j β j ! f b ··· b K ac − E α f b ··· b K ac ( α ) a + X j = k E β j f b ··· b K ac ( β j ) b j + β k f b ··· b K ac ( β k + 1) b k = 0 . (3.74)After operation of E β j , we obtain α − X j β j ! f b ··· b K ac − X j x j ∂ j + a∂ a ! f b ··· b K ac + X j = k (cid:0) x j ∂ j + b j ∂ b j (cid:1) f b ··· b K ac = − β k f b ··· b K ac ( β k + 1) b k , which gives the consistent result b k ( b k ∂ b k + x k ∂ k ) f b ··· b K ac ( β k ) = E β k f b ··· b K ac = β k f b ··· b K ac ( β k + 1) , k = 1 , , ...K. (3.75)In the above calculation, we have used the definitions and operations of E β k and E α inEq.(2.41) and Eq.(2.42), respectively. 26he K + 2 equations in Eq.(3.65), Eq.(3.69) and Eq.(3.73) together with K + 2 equationsfor the operations { E α , E β k , E γ } are equivalent to the Cartan subalgebra and the simpleroot system of SL ( K + 3 , C ) with rank K + 2 . With the Cartan subalgebra and the simpleroots, one can easily write down the whole Lie algebra of the SL ( K + 3 , C ) group. So onecan construct the Lie algebra from the recurrence relations and vice versa.In the previous subsections, it was shown that [32] the K + 2 recurrence relations among F ( K ) D can be used to derive recurrence relations among LSSA and reduce the number ofindependent LSSA from ∞ down to 1. We conclude that the SL ( K + 3 , C ) group can beused to derive infinite number of recurrence relations among LSSA, and one can solve allthe LSSA and express them in terms of one amplitude. E. Lauricella Zero Norm States and Ward Identities
In addition to the recurrence relations among LSSA, there are on-shell stringy Wardidentities among LSSA. These Ward identities can be derived from the decoupling of twotype of zero norm states (ZNS) in the old covariant first quantized string spectrum. However,as we will see soon that these Lauricella zero norm states (LZNS) or the correspondingLauricella Ward identities are not good enough to solve all the LSSA and express them interms of one amplitude.On the other hand, in the last section, we have shown that by using (A) Recurrencerelations of the LSSA, (B) Multiplication theorem of Gauss hypergeometry function and(C) the explicit calculation of four tachyon amplitude, one can explicitly solve and calculateall LSSA. This means that the solvability of LSSA through the calculations of (A), (B) and(C) imply the validity of Ward identities. Ward identities can not be identities independentof recurrence relations we used in the last section. Otherwise there will be a contradictionwith the solvabilibity of LSSA.In this section, we will study some examples of Ward identities of LSSA from this point ofview. Incidentally, high energy zero norm states (HZNS) [10, 12–16] and the correspondingstringy Ward identities at the fixed angle regime, and Regge zero norm states (RZNS)[24, 25] and the corresponding Regge Ward identities at the Regge regime have been studiedpreviously. In particular, HZNS at the fixed angle regime can be used to solve all the highenergy SSA [10, 12–16]. 27 . The Lauricella zero norm states
We will consider the set of Ward identities of the LSSA with three tachyons and onearbitrary string states. Thus we only need to consider polarizations of the tensor states onthe scattering plane since the amplitudes with polarizations orthogonal to the scatteringplane vanish.There are two types of zero norm states (ZNS) in the old covariant first quantum stringspectrum,Type I : L − | x i , where L | x i = L | x i = 0 , L | x i = 0; (3.76)Type II : (cid:18) L − + 32 L − (cid:19) | ˜ x i , where L | ˜ x i = L | ˜ x i = 0 , ( L + 1) | ˜ x i = 0 . (3.77)While type I ZNS exists at any spacetime dimension, type II ZNS only exists at D = 26.We begin with the case of mass level M = 2. There is a type II ZNS (cid:20) α − · α − + 52 k · α − + 32 ( k · α − ) (cid:21) | , k i , (3.78)and a type I ZNS [ θ · α − + ( k · α − )( θ · α − )] | , k i , θ · k = 0 . (3.79)The three polarizations defined in Eq.(2.5) to Eq.(2.7) of the 2nd tensor state with momen-tum k on the scattering plane satisfy the completeness relation η µν = X α,β e αµ e βν η αβ = diag ( − , ,
1) (3.80)where µ, ν = 0 , , α, β = P, L, T . and α T − = P µ e Tµ α µ − , α T − α L − = P µ,ν e Tµ e Lν α µ − α ν − etc.The type II ZNS in Eq.(3.78) gives the LZNS (cid:18) √ α P − + α P − α P − + 15 α L − α L − + 15 α T − α T − (cid:19) | , k i . (3.81)Type I ZNS in Eq.(3.79) gives two LZNS( α T − + √ α P − α T − ) | , k i , (3.82)28 α L − + √ α P − α L − ) | , k i . (3.83)where α T − = P µ e Tµ α µ − , α T − α L − = P µ,ν e Tµ e Lν α µ − α ν − etc. LZNS in Eq.(3.82) and Eq.(3.83)correspond to choose θ µ = e T and θ µ = e L respectively. In conclusion, there are 3 LZNS atthe mass level M = 2.At the second massive level M = 4 , there is a type I scalar ZNS (cid:20)
174 ( k · α − ) + 92 ( k · α − )( α − · α − ) + 9( α − · α − ) + 21( k · α − )( k · α − ) + 25( k · α − ) (cid:21) | , k i , (3.84)a symmetric type I spin two ZNS[2 θ µν α ( µ − α ν ) − + k λ θ µν α λµν − ] | , k i , k · θ = η µν θ µν = 0 , θ µν = θ νµ , (3.85)where α λµν − ≡ α λ − α µ − α ν − and two vector ZNS (cid:20)(cid:18) k µ k ν θ ′ λ + η µν θ ′ λ (cid:19) α ( µνλ ) − + 9 k µ θ ′ ν α ( µν ) − + 6 θ ′ µ α µ − (cid:21) | , k i , θ · k = 0 , (3.86) (cid:20)(cid:18) k µ k ν θ λ + 2 η µν θ λ (cid:19) α ( µνλ ) − + 9 k µ θ ν α [ µν ] − − θ µ α µ − (cid:21) | , k i , θ · k = 0 . (3.87)Note that Eq.(3.86) and Eq.(3.87) are linear combinations of a type I and a type II ZNS.This completes the four ZNS at the second massive level M = 4.The scalar ZNS in Eq.(3.84) gives the LZNS (cid:2) α P − ) + 9 α P − ( α L − ) + 9 α P − ( α T − ) + 9 α L − α L − + 9 α T − α T − + 75 α P − α P − + 50 α P − (cid:3) | , k i . (3.88)For the type I spin two ZNS in Eq.(3.85), we define θ µν = X α,β e αµ e βν u αβ . (3.89)The transverse and traceless conditions on θ µν then implies u P P = u P L = u P T = 0 and u P P − u LL − u T T = 0 , (3.90)which gives two LZNS( α L − α L − + α P − α L − α L − − α T − α T − − α P − α T − α T − ) | , k i , (3.91)( α ( L − α T ) − + α P − α L − α T − ) | , k i . (3.92)29he vector ZNS in Eq.(3.86) gives two LZNS[6 α T − + 18 α ( P − α T ) − + 9 α P − α P − α T − + α L − α L − α T − + α T − α T − α T − ] | , k i , (3.93)[6 α L − + 18 α ( P − α L ) − + 9 α P − α P − α L − + α L − α L − α L − + α L − α T − α T − ] | , k i . (3.94)The vector ZNS in Eq.(3.87) gives two LZNS[3 α T − − α [ P − α T ] − − α L − α L − α T − − α T − α T − α T − ] | , k i , (3.95)[3 α L − − α [ P − α L ] − − α L − α L − α L − − α L − α T − α T − ] | , k i . (3.96)In conclusion, there are totally 7 LZNS at the mass level M = 4.It is important to note that there are 9 LSSA at mass level M = 2 with only 3 LZNS,and 22 LSSA at mass level M = 4 with only 7 LZNS. So in constrast to the recurrencerelations calculated in Eq.(3.13) and Eq.(3.17), these Ward identities are not enough to solveall the LSSA and express them in terms of one amplitude.
2. The Lauricella Ward identities
In this subsection, we will explicitly verify some examples of Ward identities throughprocesses (A),(B) and (C). Process (C) will be implicitly used through the kinematics.Ward identities can not be identities independent of recurrence relations we used in processes(A),(B) and (C) in the last section.For M = 2,we define the following kinematics variables, α = − t − M k P − N + 1 = √ k P − , (3.97) γ = s − N = − M k P = −√ k P , (3.98) d = (cid:18) − k L k L (cid:19) , − (cid:18) − k P k P (cid:19) = α − γ + 1 α + 1 , (3.99)then u − N = α − γ + 1 − N = α − γ − . (3.100)30s the examples, we calculate the Ward identities associated with the LZNS in Eq.(3.82)and Eq.(3.83). The calculation is based on processes (A) and (B). By using Eq.(2.10), theWard identities we want to prove are (cid:0) − k T (cid:1) F (2) D α ; − , − α − γ −
1; 1 − (cid:18) − k T k T (cid:19) , (cid:18) − k T k T (cid:19) ! + √ (cid:0) − k P (cid:1) (cid:0) − k T (cid:1) F (2) D (cid:18) α ; − , − α − γ −
1; 1 − (cid:18) − k P k P (cid:19) , − (cid:18) − k T k T (cid:19)(cid:19) = 0 , (3.101) (cid:0) − k L (cid:1) F (2) D α ; − , − α − γ −
1; 1 − (cid:18) − k L k L (cid:19) , (cid:18) − k L k L (cid:19) ! + √ (cid:0) − k P (cid:1) (cid:0) − k L (cid:1) F (2) D (cid:18) α ; − , − α − γ −
1; 1 − (cid:18) − k P k P (cid:19) , − (cid:18) − k L k L (cid:19)(cid:19) = 0 (3.102)or, using the kinematics variables we just defined, F (2) D ( a ; − , − α − γ −
1; 1 , − ( α + 1) F (2) D (cid:18) α ; − , − α − γ − α − γ + 1 α + 1 , (cid:19) = 0 , (3.103) F (2) D ( α ; − , − α − γ −
1; 1 − d, d ) − ( α + 1) F (2) D (cid:18) α ; − , − α − γ − α − γ + 1 α + 1 , − d (cid:19) = 0 . (3.104)The Eq.(3.103) and Eq.(3.104) can be explicitly proved as F (2) D ( α ; − , − α − γ −
1; 1 , − ( α + 1) F (2) D (cid:18) α ; − , − α − γ − α − γ + 1 α + 1 , (cid:19) = F (1) D ( α ; − α − γ −
1; 1) − ( α + 1) α − γ +1 α +1 F (1) D ( α ; − α − γ −
1; 1)+ γα +1 F (1) D ( α ; − α − γ −
1; 1) (3.105)= ( γ − α ) F (1) D ( α ; − α − γ −
1; 1) − γF (1) D ( α ; − α − γ −
1; 1)= 0 , (3.106)31nd F (2) D ( α ; − , − α − γ −
1; 1 − d, d ) − ( α + 1) F (2) D (cid:18) α ; − , − α − γ − α − γ + 1 α + 1 , − d (cid:19) = 1 − d d F (1) D ( α ; − α − γ −
1; 1 + d ) − d d F (1) D ( α ; − α − γ − , d ) − ( α + 1) α − γ +1( α +1)(1 − d ) F (1) D ( α ; − α − γ −
1; 1 − d )+ (cid:16) α − γ +1( α +1)(1 − d ) − (1 − d ) (cid:17) F (1) D ( α ; − α − γ −
1; 1 − d ) (3.107)= 1 − d d (cid:18) − αdγ − α ( α + 1) ( γ − γ − (cid:19) F (1) D ( α ; − α − γ −
1; 1) − d d (cid:18) − αdγ (cid:19) F (1) D ( α ; − α − γ −
1; 1) − ( α + 1) α − γ +1( α +1)(1 − d ) (cid:16) αd γ − + α ( α +1) d ( γ − γ − (cid:17) F (1) D ( α ; − α − γ −
1; 1)+ (cid:16) α − γ +1( α +1)(1 − d ) − (1 − d ) (cid:17) (cid:16) αd γ (cid:17) F (1) D ( α ; − α − γ −
1; 1) (3.108)= 0 , (3.109)where we used of Eq.(3.13) in the process (A) to get Eq.(3.105) and Eq.(3.107), and Eq.(3.18)in the process (B) to get Eq.(3.108). The last last lines of the above equations are obtainedby using Eq.(3.19). F. Summary
In this section we have shown that there exist infinite number of recurrence relations validfor all energies among the LSSA of three tachyons and one arbitrary string state. Moreover,these infinite number of recurrence relations can be used to solve all the LSSA and expressthem in terms of one single four tachyon amplitude. In addition, we find that the K + 2recurrence relations among the LSSA can be used to reproduce the Cartan subalgebra andsimple root system of the SL ( K +3 , C ) group with rank K +2. Thus the recurrence relationsare equivalent to the representation of SL ( K + 3 , C ) group of the LSSA. As a result, the SL ( K + 3 , C ) group can be used to solve all the LSSA and express them in terms of oneamplitude [32].We have also shown that for the first few mass levels the solvability of LSSA through thecalculations of recurrence relations implies the validity of Ward identities derived from thedecoupling of LZNS. However the Lauricella Ward identities are not good enough to solveall the LSSA and express them in terms of one amplitude.32 V. RELATIONS AMONG LSSA IN VARIOUS SCATTERING LIMITS
In this section, we will show that there exist relations or symmetries among SSA ofdifferent string states at various scattering limits. In the first subsection, we will show thatthe linear relations [1–5] conjectured by Gross among the hard SSA (HSSA) at each fixedmass level in the hard scattering limit can be rederived from the LSSA. These relationsreduce the number of independent HSSA from ∞ down to 1.In the second subsection, we will show that the Regge SSA (RSSA) in the Regge scatteringlimit can be rederived from the LSSA. All the RSSA can be expressed in terms of the Appellfunctions with associated SL (5 , C ) symmetry [23–25]. Moreover, the recurrence relations ofthe Appell functions can be used to reduce the number of independent RSSA from ∞ downto 1.Finally, in the nonrelativistic scattering limit, we show that the nonrelativistic SSA(NSSA) and various extended recurrence relations among them an be rederived from theLSSA. In addition, we will also derive the nonrelativistic level M dependent string BCJrelations which are the stringy generalization of the massless field theory BCJ relation [33]to the higher spin stringy particles. These NSSA can be expressed in terms of the Gausshypergeometry functions with associated SL (4 , C ) symmetry [23–25]. A. Hard scattering limit–Proving Gross conjecture from LSSA
In this subsection, we will show that the linear relations conjectured by Gross [1–5] in thehard scattering limit can be rederived from the LSSA. First, we briefly review the resultsdiscussed in [17, 18] for the linear relations among HSSA. It was first observed that for eachfixed mass level N with M = 2( N −
1) the following states are of leading order in energyat the hard scattering limit [14, 15] | N, m, q i ≡ ( α T − ) N − m − q ( α L − ) m ( α L − ) q | , k i . (4.1)Note that in Eq.(4.1) only even powers 2 m in α L − [10–12] survive and the naive energy orderof the amplitudes will drop by an even number of energy powers in general. The HSSA withvertices corresponding to states with an odd power in ( α L − ) m +1 turn out to be of subleadingorder in energy and can be ignored. By using the stringy Ward identities or decoupling of33wo types of zero norm states (ZNS) in the hard scattering limit, the linear relations amongHSSA of different string states at each fixed mass level N were calculated to be [14, 15] A ( N, m,q ) st A ( N, , st = (cid:18) − M (cid:19) m + q (cid:18) (cid:19) m + q (2 m − . (4.2)Exactly the same result can be obtained by using two other techniques, the Virasoro con-straint calculation and the corrected saddle-point calculation [14, 15]. The calculation ofof Eq.(4.2) was first done for one high energy vertex in Eq.(4.1) and can then be easilygeneralized to four high energy vertices. In the decoupling of ZNS calculations at the masslevel M = 4, for example, there are four leading order HSSA [10, 12] A T T T : A LLT : A ( LT ) : A [ LT ] = 8 : 1 : − − A T T T ∝ A [ LT ] , and A LLT = 0 which are inconsistent with the decoupling of ZNS or unitarity ofthe theory. Indeed, a sample calculation was done [10, 12] to explicitly verify the ratios inEq.(4.3).One interesting application of Eq.(4.2) was the derivation of the ratio between A ( N, m,q ) st and A ( N, m,q ) tu in the hard scattering limit [19] A ( N, m,q ) st ≃ ( − ) N sin( πk · k )sin( πk · k ) A ( N, m,q ) tu (4.4)where A ( N, m,q ) tu is the corresponding ( t, u ) channel HSSA.Eq.(4.4) was shown to be valid for scatterings of four arbitrary string states in the hardscattering limit and was obtained in 2006. This result was obtained earlier than the discoveryof four-point field theory BCJ relations in [33] and ”string BCJ relations” in Eq.(2.19) [20–22]. In contrast to the the calculation of string BCJ relations in [21, 22] which was motivatedby the field theory BCJ relations in [33], the result of Eq.(4.4) was inspired by the calculationof hard closed SSA [19] by using KLT relation [34]. More detailed discussion can be foundin [18, 19].Now we are ready to rederive Eq.(4.1) and Eq.(4.2) from the LSSA in Eq.(2.10). The34elevant kinematics are k T = 0, k T ≃ − E sin φ, (4.5) k L ≃ − p M ≃ − E M , (4.6) k L ≃ E M sin φ . (4.7)where E and φ are CM frame energy and scattering angle respectively. One can calculate˜ z Tkk ′ = 1 , ˜ z Lkk ′ = 1 − (cid:16) − st (cid:17) /k e i πk ′ k ∼ O (1) . (4.8)The LSSA in Eq.(2.10) reduces to A ( r Tn ,r Ll ) st = B (cid:18) − t − , − s − (cid:19) · Y n =1 [( n − E sin φ ] r Tn Y l =1 (cid:20) − ( l − E M sin φ (cid:21) r Ll · F ( K ) D (cid:18) − t − R Tn , R Ll ; u − N ; (1) n , ˜ Z Ll (cid:19) . (4.9)As was mentioned earlier that, in the hard scattering limit, there was a difference betweenthe naive energy order and the real energy order corresponding to the (cid:0) α L − (cid:1) r L operator inEq.(2.9). So let’s pay attention to the corresponding summation and write A ( r Tn ,r Ll ) st = B (cid:18) − t − , − s − (cid:19) · Y n =1 [( n − E sin φ ] r Tn Y l =1 (cid:20) − ( l − E M sin φ (cid:21) r Ll · X k r (cid:0) − t − (cid:1) k r (cid:0) u + 2 − N (cid:1) k r (cid:0) − r L (cid:1) k r k r ! (cid:16) st (cid:17) k r · ( · · · ) (4.10)where we have used ( a ) n + m = ( a ) n ( a + n ) m and ( · · · ) are terms which are not relevant tothe following discussion. We then propose the following formula r L X k r =0 (cid:0) − t − (cid:1) k r (cid:0) u + 2 − N (cid:1) k r (cid:0) − r L (cid:1) k r k r ! (cid:16) st (cid:17) k r =0 · (cid:18) tus (cid:19) + 0 · (cid:18) tus (cid:19) − + · · · + 0 · (cid:18) tus (cid:19) − (cid:20) rL (cid:21) − + C r L (cid:18) tus (cid:19) − (cid:20) rL (cid:21) + O (cid:18) tus (cid:19) − (cid:20) rL (cid:21) +1 . (4.11)35here [ ] stands for Gauss symbol, C r L is independent of energy E and depends on r L andpossibly the scattering angle φ . For r L = 2 m being an even number, we further proposethat C r L = (2 m )! m ! and is φ independent. We have verified Eq.(4.11) for r L = 0 , , , · · · , s → ∞ with t fixed) and setting r L = 2 m , m X k r =0 (cid:0) − t − (cid:1) k r (cid:0) − s (cid:1) k r ( − m ) k r k r ! (cid:16) st (cid:17) k r ≃ m X k r =0 ( − m ) k r (cid:18) − t − (cid:19) k r ( − /t ) k r k r != 0 · ( − t ) + 0 · ( − t ) − + · · · + 0 · ( − t ) − m +1 + (2 m )! m ! ( − t ) − m + O ((cid:18) t (cid:19) m +1 ) , (4.12)which was proposed in [23] and proved in [35].It was demonstrated in [23] that the ratios in the hard scattering limit in Eq.(4.2) can bereproduced from a class of Regge string scattering amplitudes presented in Eq.(4.20). Thekey of the proof of this relationship between HSSA and RSSA was the new Stirling numberidentity proposed in Eq.(4.12) and mathematical proved in [35]. On the other hand, themathematical proof of Eq.(4.11), which is a generalization of the identity in Eq.(4.12), is anopen question and may be an interesting one to study.The 0 terms in Eq.(4.11) correspond to the naive leading energy orders in the HSSAcalculation. In the hard scattering limit, the true leading order SSA can then be identified A ( r Tn ,r Ll ) st ≃ B (cid:18) − t − , − s − (cid:19) · Y n =1 [( n − E sin φ ] r Tn Y l =1 (cid:20) − ( l − E M sin φ (cid:21) r Ll · C r L ( E sin φ ) − (cid:20) rL (cid:21) · ( · · · ) ∼ E N − P n ≥ nr Tn − (cid:18) (cid:20) rL (cid:21) − r L (cid:19) − P l ≥ lr Ll , (4.13)which means that SSA reaches its highest energy when r Tn ≥ = r Ll ≥ = 0 and r L = 2 m , aneven number. This result is consistent with the previous result presented in Eq.(4.1) [10–16].Finally, the leading order SSA in the hard scattering limit, i.e. r T = N − m − q ,36 L = 2 m and r L = q , can be calculated to be A ( N − m − q, m,q ) st ≃ B (cid:18) − t − , − s − (cid:19) ( E sin φ ) N (2 m )! m ! (cid:18) − M (cid:19) m + q = (2 m − (cid:18) − M (cid:19) m + q (cid:18) (cid:19) m + q A ( N, , st (4.14)which reproduces the ratios in Eq.(4.2), and is consistent with the previous result [10–16]. B. Regge scattering limit
There is another important high energy limit of SSA, the RSSA in the Regge scatteringlimit. The relevant kinematics in the Regge limit are k T = 0, k T ≃ −√− t, (4.15) k P ≃ − s M , k P ≃ − ˜ t M = − t − M − M M , (4.16) k L ≃ − s M , k L ≃ − ˜ t ′ M = − t + M − M M . (4.17)One can easily calculate ˜ z Tkk ′ = 1 , ˜ z Pkk ′ = 1 − (cid:16) − s ˜ t (cid:17) /k e i πk ′ k ∼ s /k (4.18)and ˜ z Lkk ′ = 1 − (cid:16) − s ˜ t ′ (cid:17) /k e i πk ′ k ∼ s /k . (4.19)In the Regge limit, the SSA in Eq.(2.20) reduces to A ( r Tn ,r Pm ,r Ll ) st ≃ B (cid:18) − t − , − s − (cid:19) Y n =1 (cid:2) ( n − √− t (cid:3) r Tn · Y m =1 (cid:20) ( m − t M (cid:21) r Pm Y l =1 (cid:20) ( l − t ′ M (cid:21) r Ll · F (cid:18) − t − − q , − r ; − s s ˜ t , s ˜ t ′ (cid:19) . (4.20)where F is the Appell function. Eq.(4.20) agrees with the result obtained in [25] previously.The recurrence relations of the Appell functions can be used to reduce the number ofindependent RSSA from ∞ down to 1. One can also calculate the string BCJ relation in theRegge scattering limit, and study the extended recurrence relation in the Regge limit [26].37 . Nonrelativistic Scattering Limit and Extended Recurrence Relations In this section, we discuss nonrelativistic string scattering amplitudes (NSSA) and theextended recurrence relations among them. In addition, we will also derive the nonrelativisticlevel M dependent string BCJ relations which are the stringy generalization of the masslessfield theory BCJ relation [33] to the higher spin stringy particles.We will take the nonrelativistic string scattering limit or | ~k | << M limit to calculate themass level and spin dependent low energy SSA. In constrast to the zero slope α ′ limit used inthe literature to calculate the massless Yang-Mills couplings [37, 38] for superstring and thethree point ϕ scalar field coupling [39–41] for the bosonic string, we found it appropriateto take the nonrelativistic limit in calculating low energy SSA for string states with bothhigher spins and finite mass gaps.
1. Nonrelavistic LSSA
In this subsection, we first calculate the NSSA from the LSSA. In the nonrelativistic limit | ~k | ≪ M , we have k T = 0 , k T = − (cid:20) ǫ M + M ) M M ǫ | ~k | (cid:21) sin φ, (4.21) k L = − M + M M | ~k | + O (cid:16) | ~k | (cid:17) , (4.22) k L = − ǫ φ + M + M M | ~k | + O (cid:16) | ~k | (cid:17) , (4.23) k P = − M + O (cid:16) | ~k | (cid:17) , (4.24) k P = M + M − ǫ M cos φ | ~k | + O (cid:16) | ~k | (cid:17) (4.25)where ǫ = p ( M + M ) − M and M = M = M = M tachyon . One can easily calculate z Tk = z Lk = 0 , z Pk ≃ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) M M + M (cid:19) k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (4.26)38he SSA in Eq.(2.20) reduces to A ( r Tn ,r Pm ,r Ll ) st ≃ Y n =1 h ( n − ǫ φ i r Tn Y m =1 (cid:20) − ( m − M + M (cid:21) r Pm · Y l =1 h ( l − ǫ φ i r Ll B (cid:18) M M , − M M (cid:19) · F ( K ) D (cid:18) M M R Pm ; M M ; (cid:18) M M + M (cid:19) m (cid:19) (4.27)where K = X m { for all r Pm =0 } . (4.28)
2. Nonrelativistic string BCJ relations
Note that for string states with r Pk = 0 in Eq.(2.20) for all k ≥
2, one has K = 1and the Lauricella functions in the low energy nonrelativistic SSA reduce to the Gausshypergeometric functions F (1) D = F with the associated SL (4 , C ) symmetry. In particular,for the case of the leading trajectory string state in the second vertex with mass level N = N + N + N where r T = N , r P = N , r L = N , and r Xk = 0 for all k ≥
2, the SSAreduces to A ( N ,N ,N ) st = (cid:16) ǫ φ (cid:17) N (cid:16) ǫ φ (cid:17) N · (cid:18) − M + M (cid:19) N B (cid:18) M M , − M M (cid:19) · F (cid:18) M M − N ; M M ; 2 M M + M (cid:19) , (4.29)which agrees with the result obtained in [20] previously. Similarly, one can calculate thecorresponding nonrelativistic t − u channel amplitude as A ( N ,N ,N ) tu = ( − N (cid:16) ǫ φ (cid:17) N (cid:16) ǫ φ (cid:17) N · (cid:18) − M + M (cid:19) N B (cid:18) M M , M M (cid:19) · F (cid:18) M M − N ; M M ; 2 M M + M (cid:19) . (4.30)Finally the ratio of s − t and t − u channel amplitudes is [20]39 ( N ,N ,N ) st A ( N ,N ,N ) tu = ( − N B (cid:0) − M M + 1 , M M (cid:1) B (cid:0) M M , M M (cid:1) = ( − N Γ ( M M ) Γ ( − M M + 1)Γ (cid:0) M M (cid:1) Γ (cid:0) − M M + 1 (cid:1) ≃ sin π ( k · k )sin π ( k · k ) (4.31)where, in the nonrelativistic limit, we have k · k ≃ − M M , (4.32a) k · k ≃ ( M + M ) M . (4.32b)We thus have ended up with a consistent nonrelativistic level M dependent string BCJrelations. Similar relations for t − u and s − u channel amplitudes can be calculated. Westress that the above relation is the stringy generalization of the massless field theory BCJrelation [33] to the higher spin stringy particles. Moreover, as we will show in the nextsubsection, there exist much more relations among the NSSA.
3. Extended recurrence relations in the nonrelativistic scattering limit a. Leading trajectory string states
In this subsection, we derive two examples of ex-tended recurrence relations among NSSA. We first note that there existed a recurrencerelation of Gauss hypergeometry function, F ( a ; b ; c ; z ) = c − b + 2 + ( b − a − z ( b − z − F ( a ; b − c ; z ) + b − c − b − z − F ( a ; b − c ; z ) , (4.33)which can be used to derive the recurrence relation, (cid:18) − M + M (cid:19) A ( p,r,q ) st = M ( M M + 2 q + 2)( q + 1) ( M − M ) (cid:16) ǫ φ (cid:17) p − p ′ (cid:16) ǫ φ (cid:17) p ′ − p +1 A ( p ′ ,p + r − p ′ − ,q +1) st + 2 ( M M + q + 1)( q + 1) ( M − M ) (cid:16) ǫ φ (cid:17) p − p ′′ (cid:16) ǫ φ (cid:17) p ′′ − p +2 A ( p ′′ ,p + r − p ′′ − ,q +2) st (4.34)where p ′ and p ′′ are the polarization parameters of the second and third Amplitudes on therhs of Eq.(4.34). For example, for a fixed mass level N = 4 , one can derive many recurrencerelations for either s − t channel or t − u channel amplitudes with q = 0 , , . For say q = 2 , ( p, r ) = (2 , , (1 , , (0 , . We have p ′ = 0 , p ′′ = 0 . We can thus derive, for example40or ( p, r ) = (2 ,
0) and p ′ = 1, the recurrence relation among amplitudes A (2 , , st A (1 , , st A (0 , , st as following (cid:18) − M + M (cid:19) A (2 , , st = M ( M M + 6)3 ( M − M ) (cid:16) ǫ φ (cid:17) A (1 , , st + 2 ( M M + 4)3 ( M − M ) (cid:16) ǫ φ (cid:17) A (0 , , st . (4.35)Exactly the same relation can be obtained for t − u channel amplitudes since the F ( a ; b ; c ; z )dependence in the s − t and t − u channel amplitudes calculated above are the same. Moreover,we can for example replace A (2 , , st amplitude above by the corresponding t − u channelamplitude A (2 , , tu through Eq.(4.31) and obtain( − N πM M (cid:18) − M + M (cid:19) A (2 , , tu = M ( M M + 6)3 ( M − M ) (cid:16) ǫ φ (cid:17) A (1 , , st + 2 ( M M + 4)3 ( M − M ) (cid:16) ǫ φ (cid:17) A (0 , , st , (4.36)which relates higher spin nonrelativistic string amplitudes in both s − t and t − u channels.Eq.(4.36) is one example of the extended recurrence relations in the nonrelativistic stringscattering limit . b. General string states Eq.(4.36) relates NSSA of different polarizations of a fixedleading trajectory string state. In the next sample calculation, we will calculate one exampleof extended recurrence relation which relates NSS amplitudes of different higher spin particlesfor each fixed mass level M . In particular, the s − t channel of NSS amplitudes of threetachyons and one higher spin massive string state at mass level N = 3 p + q +3 correspondingto the following three higher spin string states A ˜ (cid:0) i∂ X T (cid:1) p (cid:0) i∂X P (cid:1) (cid:0) i∂X L (cid:1) q +2 , (4.37) A ˜ (cid:0) i∂ X T (cid:1) p (cid:0) i∂X P (cid:1) (cid:0) i∂X L (cid:1) p + q +1 , (4.38) A ˜ (cid:0) i∂X T (cid:1) p (cid:0) i∂X P (cid:1) (cid:0) i∂X L (cid:1) p + q (4.39)41an be calculated to be A = h ǫ φ i p (cid:20) − (1 − M + M (cid:21) h ǫ φ i q +2 × B (cid:18) M M , − M M (cid:19) F (cid:18) M M , − , M M , − M M + M (cid:19) , (4.40) A = h ǫ φ i p (cid:20) − (2 − M + M (cid:21) h ǫ φ i p + q +1 × B (cid:18) M M , − M M (cid:19) F (cid:18) M M , − , M M , − M M + M (cid:19) , (4.41) A = h ǫ φ i p (cid:20) − (3 − M + M (cid:21) h ǫ φ i p + q × B (cid:18) M M , − M M (cid:19) F (cid:18) M M , − , M M , − M M + M (cid:19) . (4.42)To apply the recurrence relation in Eq.(4.33) for Gauss hypergeometry functions, we choose a = M M , b = − , c = M M , z = − M M + M . (4.43)One can then calculate the extended recurrence relation16 (cid:18) M M + M + 1 (cid:19) (cid:18) − M + M (cid:19) (cid:16) ǫ φ (cid:17) p A = 8 · P (cid:18) M M (cid:19) (cid:18) M M + M + 2 (cid:19) (cid:18) − M + M (cid:19) (cid:16) ǫ φ (cid:17) p +1 A − P ( M M + 2) (cid:16) ǫ φ (cid:17) A (4.44)where p is an arbitrary integer. More extended recurrence relations can be similarly derived.The existence of these low energy stringy symmetries comes as a surprise from Gross’shigh energy symmetries [1, 3, 5] point of view. Finally, in contrast to the Regge stringspacetime symmetry which was shown to be related to SL (5 , C ) of the Appell function F ,here we found that the low energy stringy symmetry is related to SL (4 , C ) [30] of the Gausshypergeometry functions F . D. Summary
In this section, we rederive from the LSSA the relations or symmetries among SSA ofdifferent string states at three different scattering limits. We first reproduce the linearrelations [14, 15] of the HSSA from the LSSA in the hard scattering limit. We also obtain42ppell functions F and Gauss hypergeometric functions F with SL (5 , C ) and SL (4 , C )symmetry in the Regge and the nonrelativistic limits respectively. In contrast to the linearrelations in the hard scattering limit, we obtain extended recurrence relations for the casesof RSSA and NSSA. These two classes of recurrence relations are closely related to thoseof the LSSA with K = 2 and K = 1 respectively. In the end, we also show that with thenonrelativistic string BCJ relations [20], the extended recurrence relations we obtained canbe used to connect SSA with different spin states and different channels. Acknowledgments
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