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DAMTP/00-125 hep-th/0011040

Conformal Four Point Functions and the Operator Product Expansion

arXiv:hep-th/0011040v3 5 Apr 2001

F.A. Dolan and H. Osborn? Department of Applied Mathematics and Theoretical Physics, Silver Street, Cambridge, CB3 9EW, England

Various aspects of the four point function for scalar ?elds in conformally invariant theories are analysed. This depends on an arbitrary function of two conformal invariants u, v . A recurrence relation for the function corresponding to the contribution of an arbitrary spin ?eld in the operator product expansion to the four point function is derived. This is solved explicitly in two and four dimensions in terms of ordinary hypergeometric functions of variables z, x which are simply related to u, v . The operator product expansion analysis is applied to the explicit expressions for the four point function found for free scalar, fermion and vector ?eld theories in four dimensions. The results for four point functions obtained by using the AdS/CFT correspondence are also analysed in terms of functions related to those appearing in the operator product discussion. PACS no: 11.25.Hf Keywords: Conformal ?eld theory, Operator product expansion, Four point function

?

address for correspondence: Trinity College, Cambridge, CB2 1TQ, England emails: fad20@damtp.cam.ac.uk and ho@damtp.cam.ac.uk

1. Introduction Much work has been undertaken in the last few years based on the AdS/CFT correspondence in terms of understanding non trivial conformal ?eld theories in four, and also three and six, dimensions. In particular this has been applied to N = 4 supersymmetric SU (N ) gauge theories when supergravity on AdS5 determines the large N limit of the associated conformal ?eld theory which is de?ned on the boundary. The correlation functions of operators on the boundary in the AdS/CFT correspondence are then determined to leading order in 1/N by tree graphs with vertices given by the supergravity theory and with appropriate boundary/bulk and bulk/bulk propagators determined by the Green functions on the AdS space. The form of the two and three point functions are determined by conformal invariance while the four point function depends on an arbitrary function of two conformal invariants. The four point function is of particular interest since it is constrained by the operator product expansion for any two ?elds. For φ1 (x1 )φ2 (x2 )φ3 (x3 )φ4 (x4 ) , which depends on a function of two conformal invariant cross ratios u, v , the operator product expansion for φ1 (x1 )φ2 (x2 ) should provide in principle an expansion for the four point function which is convergent in some region when x1 ≈ x2 . Nevertheless this requires knowing the form of the contribution, as a function of u, v , corresponding to operators of arbitrary spin, including all their derivatives, which is analogous to determining an explicit expression for a partial wave expansion. In d dimensions, when, with a Euclidean metric, the conformal group is O(d + 1, 1), then for scalar ?elds we are concerned just with the contribution of ?elds belonging to (?, 0, . . .) representations of O(d), corresponding to ?elds O(?) ?1 ...?? which are totally symmetric traceless rank ? tensors. Although conformal partial wave expansions were obtained long ago [1,2,3] they have not been in a form which is easy to apply to disentangling the contributions of di?erent operators to the four point functions found through the AdS/CFT correspondence. This has necessitated approximate calculations of just the leading terms in the power expansion for the contribution of operators with non zero spin in the operator product expansion [4]. More recently [5] results have been obtained for the contributions of a conserved vector current and the energy momentum tensor, which correspond to ? = 1, 2 operators with dimensions d ? 1, d, to the scalar four point function. We follow a similar approach to ?nd a recurrence relation for the contribution of operators for any ?. This recurrence relation may be solved when d = 2, when the result follows from the special simpli?cations arising from the use of complex coordinates, and also when d = 4. In both cases the result is expressible in terms of products of ordinary hypergeometric functions. In two dimensions these depend on η, η ? which are related to the factors of u, v found by using complex coordinates, u = η η ?, v = (1 ? η )(1 ? η ?) while in four dimensions we similarly de?ne u = zx, v = (1 ? z )(1 ? x). The 1

permutation symmetries of the four point function may be translated into transformations of z, x in addition to the requirement of invariance under z ? x. The structure of this paper is thus that in section 2 we review the operator product expansion and obtain the recurrence relation for the contribution of a quasi-primary operator of spin ? in d dimensions to the scalar four point function in terms of the results for ? ? 1, ? ? 2 by using the corresponding relation for Gegenbaur polynomials. In section 3 we obtain compact explicit solutions for arbitrary ? when d = 2, 4. In section 4 we brie?y use these results to obtain the complete contribution arising from the energy momentum tensor, for which ? = 2. The overall normalisation is determined by Ward identities. In section 5 we analyse the form of the integrals arising in the AdS/CFT correspondence to the scalar four point function in some simple cases. The results are related to a two variable function H introduced by us earlier [6] which is related to the functions arising in the operator product analysis. In section 6 we consider the simple expressions for the four point function arising from free conformal ?eld theories and analyse their expansion in terms of the conformal partial wave expressions obtained here for appropriate scale dimension ? and spin ?. The results satisfy the positivity conditions required by unitarity. Some mathematical details are deferred to various appendices. In appendix A we construct the derivative operators which appear in the operator product expansion for ? = 1, 2 while in appendix B we ?nd, by direct calculation, the action of these di?erential operators to give the contribution of an ? = 1 operator to the operator product expansion. The result is in accord with the less direct discussion of section 2. In appendix C we describe some results for a function H which is obtained by the AdS/CFT integrals of section 5. This is obtained explicitly in various cases of interest. In appendix D we sketch some details of the derivation of the four point function for free vector theories.

2. Operator Product Expansion Analysis For scalar operators φi of scale dimension ?i the contribution of a spin ? operator of dimension ? to the operator product expansion, including all derivatives or descendants, may be written as O(?) ?1 ...?? φ1 (x1 )φ2 (x2 ) ? Cφ1 φ2 O(?) where xij = xi ? xj , rij = (xi ? xj )2 . (2.2) The derivative operator C (?) (s, ? ) is determined by the form of the associated three and 2 1 r12

1 2 (?1 +?2 ??)

C (?) (x12 , ?x2 )?1 ...?? O(?) ?1 ...?? ,

(2.1)

two point functions. For the former, conformal invariance requires φ1 (x1 )φ2 (x2 ) O(?) (x3 )· C = Cφ1 φ2 O(?) where Z? = 1 r12

1 2 (?1 +?2 ??+?)

r13

1 2 (?+?12 ??)

r23

1 2 (???12 ??)

Z?1 . . . Z?? C?1 ...?? ,

(2.3)

x12? x13? ? , r13 r12

Z2 =

r12 , r13 r23

(2.4)

transforms as a conformal vector at x3 and C?1 ...?? is an arbitrary symmetric traceless tensor, O(?) ·C = O(?) ?1 ...?? C?1 ...?? , and ?ij = ?i ? ?j . The two point function for O(?) is given by O(?) (x1 )· C O(?) (x2 )· C ′ = where I?ν (x) = δ?ν ? 2 1

? r12

(2.5)

C?1 ...?? I?1 ν1 (x12 ) . . . I?? ν? (x12 ) C ′ν1 ...ν? ,

(2.6)

x? xν , x2

(2.7)

is the inversion tensor. For completeness we assume the normalisation of the scalar ?elds is determined by 1 φi (x1 )φj (x2 ) = δij ?i . (2.8) r12 As a consequence of (2.3) and (2.6) we must therefore require in (2.1) C (?) (s, ? ) = 1 1 C 2 (?+?12 ??), 2 (???12 ??) (s, ? ) where C a,b (x12 , ?x2 )?1 ...?? 1 I (x23 ) . . . I?? ν? (x23 ) Cν1 ...ν? S ?1 ν1 r23 1 = a b Z?1 . . . Z?? C?1 ...?? , r13 r23

(2.9)

and S = a+b+?. (2.10)

We construct C a,b (s, ? ) explicitly in appendix A for ? = 1, 2 where it is given in terms of the known results for ? = 0. The generalisation to arbitrary ? is evident but is not needed here. 3

If (2.1) is applied in the four point function then the corresponding contribution has the form φ1 (x1 )φ2 (x2 )φ3 (x3 )φ4 (x4 ) ? 1 r12

1 2 (?1 +?2 ) 1 1 2 (?3 +?4 ) 1 2 ?12 1 2 ?34

r34

r24 r14

r14 r13

Cφ1 φ2 O(?) Cφ3 φ4 O(?)

(2.11)

× u 2 (???) G(?)

1 2 (?

? ?12 ? ?), 1 2 (? + ?34 ? ?), ?; u, v , r14 r23 . r13 r24

which depends on the two conformal invariants u= r12 r34 , r13 r24 v= (2.12)

The functions G(?) are determined by C where Y? = and, in addition to (2.10), we have S = e+f +?. (2.15)

a,b

(x12 , ?x2 )?1 ...??

1

e r f r23 24

Y?1 . . . Y??

1 = a b r14 r24

r14 r13

e

G(?) (b, e, S ; u, v ) ,

(2.13)

x32? x42? ? , r23 r24

(2.14)

We have undertaken a direct evaluation of (2.13) in appendix B for ? = 1 but for a discussion of arbitrary ? we establish a recurrence relation which may be used iteratively to determine G(?) from G(0) . We start from the integral representation C?1 ...?? 1

e r f r23 24

1

Y?1 . . . Y?? C?1 ...?? N?,e,f dd x 1 ( x2 ? x) 2 S ( x

3 2( 1 2 d?f ??) 2( 1 2 d?e??)

2 = r34

d?S

(2.16)

? x)

( x4 ? x)

× I?1 ν1 (x2 ? x) . . . I?? ν? (x2 ? x)X ′ν1 . . . X ′ν? , with X′ = x3 ? x x4 ? x ? . 2 ( x3 ? x) ( x4 ? x) 2 (2.17)

The general structure of (2.16) follows from conformal invariance assuming (2.15) and is a generalisation of the well known result when ? = 0 [7]. To obtain the overall constant in (2.16) we may de?ne y= x ? x2 , ( x ? x2 ) 2 yi = xi ? x2 , i = 2, 3 , ( xi ? x2 ) 2 4 Y′ = y3 ? y y4 ? y ? , (2.18) 2 ( y3 ? y ) ( y4 ? y ) 2

so that, for α2 + α3 + α4 + ? = d, 1 I? ν (x2 ? x) . . . I?? ν? (x2 ? x)X ′ν1 . . . X ′ν? C?1 ...?? dd x 2 α 2 ( x2 ? x) ( x3 ? x) 2 α 3 ( x4 ? x) 2 α 4 1 1 1 Y ′?1 . . . Y ′?? = y32α3 y42α4 C?1 ...?? dd y 2 α ( y3 ? y ) 3 ( y4 ? y ) 2 α4

??1

= π2d

r=0

1

(d ? 1 ? α2 + r )

1 Γ( 2 d ? αi ) Γ(αi + ?) i=1

3

× C?1 ...?? (y3 ? y4 )?1 . . . (y3 ? y4 )??

y32α3 y42α4 (y3 ? y4 )2( 2 d?α2 )

1

.

(2.19)

It is easy to see that (2.19) is in accord with (2.16) if we take N?,e,f = for (γ )n = 1 π2d

1

1 Γ( 1 1 2 d ? e)Γ( 2 d ? f )Γ(S + ?) , 1 (d ? 1 ? S )? Γ(e + ?)Γ(f + ?)Γ( 2 d ? S)

(2.20)

Γ(γ + n) . Γ(γ )

(2.21)

We may now use (2.16) in (2.13) and apply the de?nition (2.9) to give C a,b (x12 ,?x2 )?1 ...?? 1 I? ν (x2 ? x) . . . I?? ν? (x2 ? x)X ′ν1 . . . X ′ν? ( x2 ? x) 2 S 1 1 1 ′ ′ X? . . . X?? E (?) = ?1 ...?? ,ν1 ...ν? X ν1 . . . X ν? , 2 a ( x1 ? x) ( x2 ? x) 2 b 1

(2.22)

where E (?) ?1 ...?? ,ν1 ...ν? is the projector onto symmetric traceless rank ? tensors, and x1 ? x x2 ? x X= ? . (2.23) 2 ( x1 ? x) ( x2 ? x) 2 The contraction in (2.22) may further be evaluated as

′ ′ X?1 . . . X?? E (?) ?1 ...?? ,ν1 ...ν? X ν1 . . . X ν? =

?! X 2X ′2 1 ? 2 ( 2 d ? 1)? X ·X ′ t= 1 . 2 ′ 2 X X 2

1 2

1 2?

C?2

1

d?1

( t) , (2.24)

λ Cn (t) are Gegenbaur polynomials of order ? (for λ = als). In consequence (2.13) gives

1 2 d?S

these are just Legendre polynomi1 2?

r34

?! N?,e,f 1 2? ( 2 d ? 1)? = 1 a b r14 r24 r14 r13

e

d x

d

X 2X ′2

C?2

1

d?1

( t) (x4 ? x)2( 2 d?e??)

1

( x1 ? x) 2 a ( x2 ? x) 2 b ( x3 ? x)

1 d?f ??) 2( 2

G(?) (b, e, S ; u, v )

1 d + b ? S, 1 + Kb,e,S u 2 d?S G(?) ( 2 2 d + e ? S, d ? S ; u, v ) . (2.25) (?)

1

5

The last term on the right hand side of (2.25) is a so called shadow term which is non analytic at u = 0.1 This term may be neglected in our analysis of G(?) . To evaluate the left hand side of (2.25) we may note that X2 = and r12 , ( x1 ? x) 2 ( x2 ? x) 2 X ′2 = r34 , ( x3 ? x) 2 ( x4 ? x) 2 (2.26)

2X · X ′ = ?

r13 r23 + 2 2 ( x1 ? x) ( x3 ? x) ( x2 ? x) 2 ( x3 ? x) 2 r14 r24 + , ? 2 2 ( x2 ? x) ( x4 ? x) ( x1 ? x) 2 ( x4 ? x) 2

(2.27)

so that the integral in (2.25) is reducible to linear combinations of the form dd x 1 , ( x1 ? x) 2 α 1 ( x2 ? x) 2 α 2 ( x3 ? x) 2 α 3 ( x4 ? x) 2 α 4 αi = d ,

i

(2.28)

which are expressible in terms of functions of the conformal invariants u, v de?ned in (2.12). A useful recurrence relation may be obtained by using the recurrence relation for Gegenbaur polynomials,

λ λ λ ?C? (t) = 2(λ + ? ? 1)tC? ?1 (t) ? (2λ + ? ? 2)C??2 (t) .

(2.29)

Substituting this into (2.25) and with (2.20) we may then obtain using (2.26) and (2.27) a corresponding relation for G(?) , G(?) (b, e, S ; u, v ) = 1 S+??1 2d?S+??2 + ?

1 2d

?e?1 vG(??1) (b + 1, e + 1, S ; u, v ) ? G(??1) (b, e + 1, S ; u, v ) f +??1

1 2d

?f ?1 G(??1) (b, e, S ; u, v ) ? G(??1) (b + 1, e, S ; u, v ) e+??1

1 1 d ? e ? 1)( 2 d ? f ? 1) (2 1 (S + ? ? 1)(S + ? ? 2) (? ? 1)(d + ? ? 4) 1 1 4 (d ? S + ? ? 2)(d ? S + ? ? 3) (f + ? ? 1)(e + ? ? 1) ( 2 d + ? ? 2)( 2 d + ? ? 3)

× uG(??2) (b + 1, e + 1, S ; u, v ) ,

1

(2.30)

Both terms in (2.25) satisfy the recurrence relation derived for G(?) below. Using this

result and the evaluation of (2.25) for ? = 0 determines Kb,e,S =

(?) 1 d + b ? S + ?)Γ( 2 d + e ? S + ?)Γ(S + ?)Γ(S + ? ? 1) Γ( 1 2 Γ(b + ?)Γ(e + ?)

×

1 d ? b)Γ( 1 d ? e)Γ(S ? 1 d)Γ(d ? S ? 1) Γ( 2 2 2 . 1 Γ(S ? b)Γ(S ? e)Γ( 2 d ? S )Γ(S ? 1)

6

where f is determined by (2.15). The starting point in the iteration G(0) , corresponding to a scalar ?eld in the operator product expansion, has been obtained by various authors. In the short distance limit x1 → x2 we have u → 0, v → 1 and it is given as a power series in u, 1 ? v ,

1 d, S ; u, 1 ? v ) , G(0) (b, e, S ; u, v ) = G(b, e, S + 1 ? 2

(2.31)

where G(α, β, γ, δ ; u, 1 ? v ) = G(β, α, γ, δ ; u, 1 ? v ) = = (δ ? α)m (δ ? β )m (α)m+n (β )m+n m u (1 ? v )n m !( γ ) n !( δ ) m 2m+n m,n=0

(2.32) Γ(δ )Γ(δ ? α ? β ) F4 (α, β, γ, α + β + 1 ? δ ; u, v ) Γ(δ ? α)Γ(δ ? β ) Γ(δ )Γ(α + β ? δ ) δ?α?β + v F4 (δ ? α, δ ? β, γ, δ ? α ? β + 1; u, v ) , Γ(α)Γ(β )

with F4 one of the well documented Appell functions, ref.[8], p. 1080-1084. Two important symmetry relations may be obtained from the associated result demonstrated for G in [6] and checking consistency with (2.30). These correspond to letting x1 ? x2 , a ? b and also x3 ? x4 , e ? f , and using (2.10) and (2.15) we have G(?) (b, e, S ; u, v ) = (?1)? v ?e G(?) (a, e, S ; u′ , v ′ ) , = (?1)? v ?b G(?) (b, f, S ; u′, v ′ ) , u′ = u/v , v ′ = 1/v . (2.33)

Although (2.30) is rather involved it may be solved directly when u = 0 when the last term is absent. From (2.31) and (2.32) it is evident that G(0) (b, e, S ; 0, v ) = F (b, e; S ; 1 ? v ), an ordinary hypergeometric function and by using standard hypergeometric identities

1 G(?) (b, e, S ; 0, v ) = ? 2 (1 ? v ) F (b + ?, e + ?; S + ?; 1 ? v ) . ?

(2.34)

Other general results for arbitrary d are hard to ?nd but for v = 1 we may determine the leading behaviour for small u, ? 1 1 1 ? ? ? ? even, ? ? (?1) 2 g? u 2 , 2 1 G(?) (b, e, S ; u, 1) ? (? + 2 d ? 1)(a ? b)(e ? f ) 1 (?+1) 1 1 ? ? , ? odd, (?1) 2 (??1) g?+1 u2 ? +1 2 (S ? d ? ? + 2)(S + ?) (2.35) 1 1 where from (2.30) g? = (? ? 1)(? + d ? 4)g??2 /(? + 2 d ? 2)(? + 2 d ? 3) and starting from g0 = 1, 1 1 d ? 1) 2 ?! ( 2 ? . (2.36) g? = 1 1 ( 2 ?)! ( 2 d ? 1)? 7

3. Solutions in Two and Four Dimensions We show here how the recursion relation (2.30) may be solved explicitly in two and four dimensions. For d = 2, G(?) may be found directly using the simpli?cations obtained through using 1 complex coordinates z, z ? in this case. Letting xz ≡ z and, from x2 = z z ?, xz = 2 z ? then the inversion tensor de?ned in (2.7) is, in this complex basis, Izz ? ( x) = I z ?z (x) = 0 and 1 1 ?/z, Iz ?. Since (2.4) reduces to Izz (x) = ? 2 z ?z ?(x) = ? 2 z/z Zz = ? the equation (2.22) becomes C a,b (x12 , ?x2 )z...z

? 1 z12 1 = , S +? S ?? a+? b+? a b ?13 z ?23 z23 z ?23 z13 z23 z

z ?12 , z ?13 z ?23

? Zz =?

z12 z13 z23

(3.1)

(3.2)

together with its conjugate. The di?erential operator therefore factorises in the form[9]

? C a,b (x12 , ?x2 )z...z = z12 ?12 ?z 1 F1 (a + ?, S + ?; z12 ?z2 )1 F1 (a, S ? ?; z ?2 ) ,

(3.3)

where 1 F1 (α, β ; x) =

1F1 (a; S ; z12 ?z2 )

n (α)n /n!(β )n x

n

. Since

e

1

ez f z23 24

=

1 z14 a b z14 z24 z13 η=

F (b, e; S ; η ) ,

S =a+b =e+f,

(3.4)

where

z12 z34 , z13 z24

(3.5)

the de?nition (2.13) gives G(0) (b, e, S ; u, v ) = F (b, e; S ; η )F (b, e; S ; η ?) ,

1 ? η ) F (b + ?, e + ?; S + ?; η )F (b, e; S ? ?; η ?) G(?) (b, e, S ; u, v ) = (? 2

(3.6)

+ conjugate , for u = ηη ?,

? > 0,

v = (1 ? η )(1 ? η ?) .

(3.7)

For G(0) the validity of the result (3.6) depends, from (2.31), on the factorisation formula shown in [6] for the function G de?ned in (2.32). This is obtained from a similar reduction formula for F4 and takes the form G(α, β, γ, γ ; u, 1 ? v ) = F (α, β ; γ ; x)F (α, β ; γ ; z ) . 8 (3.8)

where u = xz , v = (1 ? x)(1 ? z ) . (3.9)

We have veri?ed that G(?) de?ned then through (2.30) for ? = 1, 2, . . . agrees with (3.6). The calculation is very similar to the d = 4 case which we describe next. When d = 4 the result for G(0) given by (2.31) and (2.32) may also be simpli?ed by use of a reduction formula extending (3.8) which was also obtained in [6], 1 zF (α ? 1, β ? 1; γ ? 1; x)F (α, β ; γ + 1; z ) z?x ? xF (α, β ; γ + 1; x)F (α ? 1, β ? 1; γ ? 1; z ) .

G(α, β, γ, γ + 1; u, 1 ? v ) =

(3.10)

The solution for any ? which follows from this is then G(?) (b, e, S ; u, v ) = 1 z )? zF (b ? 1, e ? 1; S ? 2 ? ?; x)F (b + ?, e + ?; S + ?; z ) (? 1 2 z?x

1 ? (? 2 x)? xF (b ? 1, e ? 1; S ? 2 ? ?; z )F (b + ?, e + ?; S + ?; x) .

(3.11) The veri?cation of (3.11) from (2.30) depends on zF (b ? 1, e ? 1; S ? 2 ? ?; x)F (b + ?, e + ?; S + ?; z ) S+??1 e?1 =? (1 ? z )(1 ? x)F (b, e; S ? 1 ? ?; x)F (b + ?, e + ?; S + ? ? 1; z ) S???2 f +??1 ? F (b ? 1, e; S ? 1 ? ?; x)F (b + ? ? 1, e + ?; S + ? ? 1; z ) + f ?1 F (b ? 1, e ? 1; S ? 1 ? ?; x)F (b + ? ? 1, e + ? ? 1; S + ? ? 1; z ) e+??1 ? F (b, e ? 1; S ? 1 ? ?; x)F (b + ?, e + ? ? 1; S + ? ? 1; z ) ? (S + ? ? 1)(S + ? ? 2) (e ? 1)(f ? 1) (S ? ? ? 2)(S ? ? ? 1) (f + ? ? 1)(e + ? ? 1) × xF (b, e; S ? ?; x)F (b + ? ? 1, e + ? ? 1; S + ? ? 2; z ) .

(3.12)

This corresponds exactly to the form of (2.30) for d = 4 except if ? = 1 when the last term, which matches the last term in (3.12), is missing. However in this case this piece times z is symmetric under z ? x and it is then cancelled by the other term in (3.11) which is obtained from the analogous result to (3.12) for z ? x. The justi?cation of (3.12) depends 9

on standard hypergeometric identities, ref.[8], p. 1071, in particular we use, (S + ? ? 1) (1 ? z )F (b + ?, e + ?; S + ? ? 1; z ) ? F (b + ? ? 1, e + ?; S + ? ? 1; z ) = ?(f + ? ? 1)zF (b + ?, e + ?; S + ?; z ) , (S + ? ? 1) F (b + ?, e + ? ? 1; S + ? ? 1; z ) ? F (b + ? ? 1, e + ? ? 1; S + ? ? 1; z ) = (e + ? ? 1)zF (b + ?, e + ?; S + ?; z ) , (e ? 1)(1 ? x)F (b, e; S ? 1 ? ?; x) + (f ? 1)F (b + ?, e + ? ? 1; S + ? ? 1; z ) = (S ? ? ? 2)F (b ? 1, e ? 1; S ? 2 ? ?; x) . For consistency we may check that (3.11) satis?es the consistency relations (2.33). It is crucial to recognise that (3.9) is invariant under x ? z , under which of course (3.11) is invariant. For the transformations corresponding to x1 ? x2 or x3 ? x4 we choose x → x′ = x , x?1 z → z′ = z z?1 ? u → u′ = u , v v → v′ = 1 . v (3.14) (3.13)

Using standard hypergeometric results, ref.[8], p. 1069, with (2.10), (2.15) we have F (b ? 1, e ? 1; S ? 2 ? ?; x) = (1 ? x)?e+1 F (a ? 1, e ? 1; S ? 2 ? ?; x′ ) = (1 ? x)?b+1 F (b ? 1, f ? 1; S ? 2 ? ?; x′ ) , F (b + ?, e + ?; S + ?; z ) = (1 ? z )?e?? F (a + ?, e + ?; S + ?; z ′ ) = (1 ? z )?b?? F (b + ?, f + ?; S + ?; z ′ ) . which in (3.11), with z ? x = ?(z ′ ? x′ )v , are su?cient to verify (2.33). Both results (3.6) and (3.11) are of course compatible with G(0) (0, 0, 0; u, v ) = 1, representing the contribution of the identity operator in the operator product expansion. (3.15)

4. Energy Momentum Tensor It is of interest to specialise the general discussion to the particular case of the energy momentum tensor T?ν which is an ? = 2 operator with dimension d satisfying the conservation equation ?? T?ν = 0. In this case the normalisation is not ?xed by (2.6) but instead through Ward identities. For the canonically normalised energy momentum tensor, but with the scalar ?eld φ still normalised as in (2.8), the φφT three point function obtained from (2.3) becomes [10] φ(x1 )φ(x2 ) T?ν (x3 ) = ? 1 1 ?d 1 1 1 d ? ? Sd d ? 1 r 2 2d 2d r13 r23 12 10 Z? Zν 1 ? δ?ν 2 Z d , (4.1)

for Sd = 2π 2 d /Γ( 1 2 d). The energy momentum tensor two point function is also of the form given by (2.6) and can be written T?1 ?2 (x1 )Tν1 ν2 (x2 ) = CT 1 d Sd2 r12

1 2

1

I?1 ν1 (x12 )I?2 ν2 (x12 ) + I?1 ν2 (x12 )I?2 ν1 (x12 ) ?

1 δ? ? δν ν , d 1 2 1 2

(4.2)

with the coe?cient CT > 0 depending on the particular conformal ?eld theory.2 The contribution to the four point function is then from (2.11)

1 1 ??′ d2 2 (d?2) G(2) u φ(x1 )φ(x2 )φ (x3 )φ (x4 ) ? ? ?′ r12 r34 CT (d ? 1)2

′

′

1 2 (d

1 ? 2), 2 (d ? 2), d; u, v .

(4.3)

Using (3.6) and (3.11) we can give complete expressions in two and four dimensions. If d = 2 then in (4.3) we have 1 G(2) (0, 0, 2; u, v ) = ?3 1 + (1 ? 1 2 η ) ln(1 ? η ) + conjugate , η while for d = 4, uG

(2) 1 1 2 1 z+ z (1 ? z + 6 z ) ln(1 ? z ) ? z ? x x 1? 2 (1, 1, 4; u, v ) = ?45 . z?x

(4.4)

(4.5)

5. AdS/CFT Integrals The metric on AdSd+1 de?nes a Weyl equivalence class of metrics on the boundary S while the isometry group SO (d + 1, 1) becomes the conformal group on the boundary. This is at the root of the AdS/CFT correspondence. Much work [11,12,13,14,15,16,17,18] has been undertaken investigating the structure of the conformally covariant correlation functions for boundary points xi obtained in terms Feynman graphs for propagators on AdSd+1 linking xi [19]. For vertices de?ned by IIB supergravity this is relevant to the large N limit of N = 4 SYM. The two and three point functions are dictated by conformal invariance while the four point function involves functions of the two invariants u, v , as de?ned in (2.12), which may be matched to the operator product expansion.

d

2

For d = 2 with T (z ) = ?2πTzz (x) (4.1) and (4.2) reduce to the well known results

1 ?(z12 /z13 z23 )2 |z12 |?2? 2 4 and T (z1 )T (z2 ) = CT /(4z12 ) so that 1 C 2 T

φ(x1 )φ(x2 )T (z3 ) =

= c

the Virasoro central charge.

11

We show here how the simplest graphs lead to integrals which may be reduced to the functions G de?ned earlier in (2.32) and discussed here and in [6]. With the usual metric on AdSd+1 and coordinates z = (z0 , x), x ∈ Rd , z0 ∈ R+ , ds2 = 1 2 dz0 + dx? dx? , 2 z0 (5.1)

the boundary corresponds to z0 = 0 together with the point at in?nity z0 = ∞. The bulk/boundary propagator is then [19] K? (z, x ) =

′

Γ(?) 2π 2 d Γ(? + 1 ? 1 2 d)

1

? ? (z, x′ ) , K

? ? (z, x′ ) = K

z0 2 z0 + (x ? x′ )2

?

. (5.2)

We are initially interested in integrals de?ning conformal N -point functions of the form, as de?ned in [13], D?1 ...?N (x1 , . . . , xN ) = 1 π2d

1

d

d+1

z

1

d+1 z0

N

? ? (z, xi ) . K i

i=1

(5.3)

2 Using standard integral representations for (z0 + (x ? xi )2 )??i the z -integration may be undertaken giving 1 Γ(Σ ? 2 d) D?1 ...?N (x1 , . . . xN ) = 2 i Γ(?i ) ∞ 0 N i=1 dλi

λi?i ?1

1 Λ

Σ

e

1 ?Λ

i<j

λi λj rij

,

(5.4)

where Λ =

i

λi and

N

Σ=

1 2 i=1

?i .

(5.5)

A crucial observation of Symanzik [20], recounted in [6], is that for integrals of the form (5.4), subject to (5.5), then Λ may be modi?ed, without changing the integral, to the form i κi λi for any κi , not all zero, satisfying κi ≥ 0. In particular we may choose Λ = λN and the integral may then be directly written in terms of conformally invariant cross ratios like u, v . For N = 3 we have

1 d) Γ(Σ ? 2 Γ(Σ ? ?1 )Γ(Σ ? ?2 )Γ(Σ ? ?3 ) . D?1 ?2 ?3 (x1 , x2 , x3 ) = 2Γ(?1 )Γ(?2 )Γ(?3 ) r23Σ??1 r13Σ??2 r12Σ??3

(5.6)

The result for N = 4 may also be expressed in term of a function of the conformal invariants u, v in the form

1 4 3 Γ(Σ ? 1 r14 r34 2 d) D?1 ?2 ?3 ?4 (x1 , x2 , x3 , x4 ) = Σ??4 ?2 2Γ(?1 )Γ(?2 )Γ(?3 )Γ(?4 ) r13 r24 × D?1 ?2 ?3 ?4 (u, v ) .

Σ?? ??

Σ?? ??4

(5.7)

12

Using Symanzik’s procedure for evaluating the integrals we have D?1 ?2 ?3 ?4 (u, v ) = H ?2 , Σ ? ?4 , ?1 + ?2 ? Σ + 1, ?1 + ?2 ; u, v , (5.8)

where the function H , de?ned in [6], is directly related to the function G given by the power series in u, 1 ? v in (2.32) through H (α, β, γ, δ ; u, v ) = H (β, α, γ, δ ; u, v ) = Γ(1 ? γ ) Γ(α)Γ(β )Γ(δ ? α)Γ(δ ? β ) G(α, β, γ, δ ; u, 1 ? v ) Γ(δ ) Γ(γ ? 1) Γ(α ? γ + 1)Γ(β ? γ + 1)Γ(δ ? γ ? α + 1)Γ(δ ? γ ? β + 1) + Γ(δ ? 2γ + 2) × u1?γ G(α ? γ + 1, β ? γ + 1, 2 ? γ, δ ? 2γ + 2; u, 1 ? v ) . The symmetry properties of the integral (5.4), for N = 4, are re?ected in various identities obeyed by H which are listed in appendix C. Besides (5.7) we may also consider the scalar exchange contribution to the four point function which is given by

? S? (x1 , x2 , x3 , x4 ) = 1 ?2 ?3 ?4

(5.9)

1 π

1 2d

dd+1 z

1

d+1 z0

? ? (z, x1 )K ? ? (z, x2 ) K 1 2 d+1 w0 (5.10) ? ? (w, x3 )K ? ? (w, x4 ) , × G? (z, w)K 3 4 dd+1 w

1

for G? the scalar Green function on AdSd+1 ,

d+1 d+1 ? ?2 + ?(? ? d) G? (z, w) = z0 δ (z ? w ) .

(5.11)

The explicit form for G? is unnecessary here except for

? G? (z, w) ? z0

1 2π

1 2d

Γ(?) ? ? (w, x) as z0 → 0 . K 1 d + 1) Γ(? ? 2

(5.12)

In consequence R(z ; x3 , x4 ) = satis?es ? ? (z, x3 )K ? ? (z, x4 ) , ? ?2 + ?(? ? d) R(z ; x3 , x4 ) = K 3 4 with the boundary condition, obtained by using (5.12) inside the integral

? R(z ; x3 , x4 ) ? z0

dd+1 w

1

d+1 w0

? ? (w, x3 )K ? ? (w, x4 ) , G? (z, w)K 3 4

(5.13)

(5.14)

Γ(?) D??3 ?4 (x, x3 , x4 ) as 1 2Γ(? ? 2 d + 1) 13

z0 → 0 .

(5.15)

To solve (5.14) with (5.15) we make use of

? ? (z, x3 )K ? ? (z, x4 ) ??2 + (?3 + ?4 )(?3 + ?4 ? d) K 3 4 ? ? +1 (z, x3 )K ? ? +1 (z, x4 ) r34 . = 4?3 ?4 K

3 4

(5.16)

With the aid of (5.16) we may then write a series solution for R as

R(z ; x3 , x4 ) = ? 1 4

( 1 (?3 s=0 2

(?3 )s (?4 )s 1 + ?4 ? ?))s+1 ( 2 (?3 + ?4 + ? ? d))s+1

? ? +s (z, x3 )K ? ? +s (z, x4 ) r s ×K 3 4 34 1 1 1 Γ( 2 (?3 + ?4 ? ?))Γ( 2 (? + ?34 ))Γ( 2 (? ? ?34 )) 1 1 + 4 Γ( 2 (?3 + ?4 + ? ? d)) 1 Γ(? ? 2 d + 1)Γ(?3 )Γ(?4 ) ×

s=0 1 (? + ?34 ))s ( 1 (2 2 (? ? ?34 ))s 1 s!(? ? 2 d + 1)s

2 ?1 ?1 (z, x3 )K (z, x4 ) r34 ×K 2 (?+?34 )+s 2 (???34 )+s 1

(???3 ??4 )+s

.

(5.17)

The series in (5.17) are convergent for su?ciently small z0 or r34 . The ?rst term in (5.17) satis?es the inhomogeneous equation (5.14) while the second term obeys the corresponding homogeneous equation but reproduces the required boundary behaviour (5.15) after using (5.6). This term generates the dominant contribution as z0 → 0 if ? < ?3 + ?4 which is necessary for the derivation of (5.15) to be valid. If ?3 + ?4 ? ? = 2n, n = 1, 2, . . . the two series cancel except for a ?nite number of terms and we get

(?3 )s (?4 )s ? ? +s (z, x3 )K ? ? +s (z, x4 ) r s , K 34 3 4 1 (n + s)!(? ? 2 d + n)s+1 s=?n (5.18) which coincides with the solution obtained in [12]. Using (5.17) in (5.10) we may obtain from (5.7) R(z ; x3 , x4 ) =

1 4 (n

?1

? 1)!

? S? (x1 , x2 , x3 , x4 ) 1 ?2 ?3 ?4

1 4 3 r14 r34 1 = Σ??4 ?2 8Γ(?1 )Γ(?2 )Γ(?3 )Γ(?4 ) r13 r24 ? × S? (u, v ) , 1 ?2 ?3 ?4

Σ?? ??

Σ?? ??4

(5.19)

14

where

? S? (u, v ) = ? 1 ?2 ?3 ?4 s=0

Γ(Σ ? 1 2 d + s) 1 1 ( 2 (?3 + ?4 ? ?))s+1 ( 2 (?3 + ?4 + ? ? d))s+1

× H (?2 , Σ ? ?4 , ?1 + ?2 ? Σ + 1 ? s, ?1 + ?2 ; u, v ) + ×

s=0 1 Γ( 1 2 (?3 + ?4 + ? ? d))Γ( 2 (?3 + ?4 ? ?)) 1 d + 1) Γ(? ? 2

(5.20)

(?1 + ?2 + ? ? d) + s) Γ( 1 2 s! (? ? 1 2 d + 1)s

1 × H (?2 , Σ ? ?4 , 2 (?1 + ?2 ? ?) + 1 ? s, ?1 + ?2 ; u, v ) .

As in (5.18) it is easy to verify that this reduces to a ?nite sum if ?3 + ?4 ? ? = 2n. For general ? we may use (5.9) and (2.32) to rewrite (5.20) as the sum of three terms with di?erent leading powers in u [21],

? u 2 (?1 +?2 ) S? (u, v ) 1 ?2 ?3 ?4 1 1 1 (?1 + ?2 ? ?))Γ( 2 (?3 + ?4 ? ?))Γ( 1 = Γ( 2 2 (?1 + ?2 + ? ? d))Γ( 2 (?3 + ?4 + ? ? d)) 1 1 1 1 × Γ( 1 1 2 (? + ?12 ))Γ( 2 (? ? ?12 ))Γ( 2 (? + ?34 ))Γ( 2 (? ? ?34 )) Γ(?)Γ(? + 1 ? 2 d)

1

× u 2 ?G ? for

1

1 1 1 ?? 2 ?12 , 1 ?+ 2 ?34 , ? + 1 ? 1 d, ?; u, 1 ? v 2 2 2 1 1 ? ? (u, v ) ? u 2 (?1 +?2 ) G? (u, v ) , u 2 (?3 +?4 ) G? 1 ?2 ?3 ?4 4 ?3 ?2 ?1

(5.21)

? G? (u, v ) 1 ?2 ?3 ?4

Γ(Σ ? 1 d) 2 = Γ(Σ ? ?3 ? ?4 )Γ(?3 )Γ(?4 )Γ(Σ ? ?1 )Γ(Σ ? ?2 ) Γ(?3 + ?4 ) (Σ ? ?2 )m (?4 )m G? m (?3 )m+n (Σ ? ?1 )m+n m u (1 ? v )n , × m !(Σ ? ? ? ? + 1) (? + ? ) 1 2 m 3 4 2m+n m,n=0

s

G? m

1 (Σ ? 2 d)s m! . = (?1) 1 (m ? s)! ( 2 (?3 + ?4 ? ?))s+1 ( 1 (?3 + ?4 + ? ? d))s+1 2 s=0

m

(5.22)

? To achieve the desired form for G? requires non trivial relations for 3 F2 func4 ?3 ?2 ?1 tions with argument 1. The ?rst term in (5.21) matches exactly the contribution of a scalar operator with dimension ? in the operator product expansion. From (5.22) ? ? ? G? (u, v ) = v ??3 G? (u/v, 1/v ) = v ?Σ+?1 G? (u/v, 1/v ). Further1 ?2 ?3 ?4 2 ?1 ?3 ?4 1 ?2 ?4 ?3 1 1 (? +? ) ? (? +? ) ? more u 2 1 2 S?1 ?2 ?3 ?4 (u, v ) = u 2 3 4 S?4 ?4 ?2 ?1 (u, v ) as well as satisfying relations for ?1 ? ?2 or ?3 ? ?4 and u → u/v, v → 1/v which follow from the preceding ? relations for G? . The result given by (5.21) and (5.22) provides a representation 1 ?2 ?3 ?4

15

valid for u ? 0, v ? 1 and corresponds essentially with that given by Liu [15]. In principle (5.20) may be used to ?nd a form appropriate for u ? 1, v ? 0 but, for the general case, the results are signi?cantly more complicated [18]. The last line in (5.21) corresponds to the contribution of operators with dimensions ?3 + ?4 + 2n, ?1 + ?2 + 2n, n = 0, 1, 2, . . ..

6. Results for Free Fields With the explicit formula (3.11) for the contribution for arbitrary spin operators to the four point function in four dimensions then it is natural to consider the associated partial wave expansion for some simple conformally invariant expressions for the four point function. We consider here some examples arising in free ?eld theories and verify consistency with the expected form of the operator product expansion. For the general case considered here we relax the normalisation assumption of section 2 and consider a quasi-primary scalar ?eld φ, with scale dimension ?φ , for which the two point function has the form Nφ φ(x1 )φ(x2 ) = ?φ . (6.1) r12 The corresponding four point function is then taken to be φ(x1 )φ(x2 )φ(x3 )φ(x4 ) = Nφ2 where gφ (u, v ) satis?es the symmetry conditions gφ (u, v ) = gφ (u/v, 1/v ) , 1 + gφ (u, v ) = u v

?φ

1

? r12 φ

r34 φ

?

1 + gφ (u, v ) ,

(6.2)

1 + gφ (v, u) ,

(6.3)

and we may assume gφ (0, v ) = 0. The 1 in (6.2) is then the leading singularity as r12 → 0 and represents just the contribution of the identity operator in the operator product expansion of φ(x1 )φ(x2 ). By virtue of (2.11) gφ (u, v ) should be expanded as gφ (u, v ) =

?,?

c?,? u 2 (???) G(?)

1

1 2 (?

? ? ), 1 2 (? ? ?), ?; u, v ,

(6.4)

with ? > 0. The set of ?, ? which are necessary in (6.4) determines the spectrum of operators which contribute to the operator product expansion of φ(x1 )φ(x2 ). As a consequence of (2.33) and the ?rst relation in (6.3) ? must be even. For free ?eld theories gφ (u, v ) is analytic in u, 1 ? v which corresponds to requiring that any ? contributing to the sum in (6.4) is an even integer as well. To apply the explicit result (3.11) we ?rst de?ne, with the de?nitions (3.9), hφ (z, x) = z?x gφ (u, v ) , u 16 (6.5)

where hφ (z, x) = ?hφ (x, z ) = ?hφ (z ′ , x′ ) , x′ =

x , x?1

z′ =

z . z?1

(6.6)

Writing ? = 2m, ? ? ? = 2(t + 1), then, with (3.11), (6.4) becomes hφ (z, x) =

t=0

Ht (z ) xt F (t, t; 2t; x) ? Ht (x) z t F (t, t; 2t; z ) ,

(6.7)

where, with c?,? → cmt for m, t integers, Ht (z ) =

m=0

cmt

1 22 m

z 2m+t+1 F (2m + t + 1, 2m + t + 1; 4m + 2t + 2; z ) .

(6.8)

By virtue of standard hypergeometric identities this satis?es Ht (z ) = (?1)t+1 Ht (z ′ ) . (6.9)

Furthermore if we compare with (4.3), with ? = ?′ = ?φ , d = 4, the contribution of the energy momentum tensor with ? = 2 corresponds to c10 = 16?φ2 . 9CT (6.10)

To obtain an algorithm for determining cmt it is convenient to write ?rstly ? φ (z, x) , hφ (z, x) = h0 (z ) ? h0 (x) + h (6.11)

? φ (z, x) is O(zx) and h0 (z ) = ?h0 (z ′ ), from (6.6). We also expand h ? φ (z, x) in where h powers of x, ? n ( z ) xn . ? φ (z, x) = h (6.12) h

n=1

The essential equation (6.7) may then be decomposed into independent equations for each power in x. For the terms of O(x0 ) we get h0 (z ) = H0 (z ) , which determines cm0 . For the terms O(xn ), n > 0, we have

n?1

(6.13)

? n (z ) ? h

t=1

(t)n?t Ht (z ) (n ? t)!(2t)n?t cmt (2m + t + 1)n?2m?t?1 z t F (t, t; 2t; z ) = Hn (z ) , 22m (n ? 2m ? t ? 1)!(4m + 2t + 1)n?2m?t?1 (6.14) 1 17

2

2

1 n?1 [ 2 (n?t?1)]

+

t=1 m=0

which expresses cmn recursively in terms of cmn′ , n′ < n. For n = 1 (6.14) becomes just ? 1 (z ) = H1 (z ) determining cm1 . With further manipulation this can be written as h

[1 2 n]?1

? (n) (z ) + h

m=0

cmt

1 22 m

z t F (t, t; 2t; z )

t=n?1?2m

= Hn (z ) ,

(6.15)

1 where, as also in (6.14), [ 2 n] denotes the integer part and n?1

? (n) (z ) = h ? n (z ) ? h

t=1

(t)n?t ? (t) (z ) = h (n ? t)!(2t)n?t

2

n?1

(n ? r )r ? n?r (z ) . (6.16) (?1) h r !(2 n ? r ? 1) r r=0

r

2

It is crucial that the left hand side of (6.15) is compatible with (6.9). For the terms involving z t F (t, t; 2t; z ) this is automatic with the restriction t = n ? 1 ? 2m. The condition ? (n) (z ) = (?1)n+1 h ? (n) (z ′ ) follows from (6.16) together with (6.6) which, with (6.11) and h ? n (z ) = ? n (?1)r (r )n?r h ? r (z ′ )/(n ? r )!. It is also necessary, although (6.12), implies h r=1 less evident, that the left hand side of (6.15) is O(z n+1 ) to match the leading term on the right hand side. The simplest case is that for a free scalar ?eld, φ → ? satisfying ? 2 ? = 0 and ?? = 1. With canonical normalisation in (6.1) we have N? = 1/4π 2 . For this case it is easy to calculate that in (6.2) u g? (u, v ) = u + , (6.17) v so that in (6.11) z ? ? (z, x) = 0 . , h (6.18) h0 (z ) = z + 1?z The only equation to solve is then (6.13). By using algebraic manipulation programmes we have veri?ed that 2 2m+1 (2m)! cm0 = 2 , (6.19) (4m)! is compatible with direct calculations for the ?rst 20 terms. Although a formal proof of (6.19) is doubtless possible we have not invested the e?ort necessary to achieve it. 4 and then (6.10) gives Reassuringly cm0 > 0 as required by unitarity. Furthermore c10 = 3 4 CT = 3 which is the correct value for the free scalar theory in four dimensions [10]. Since cmt = 0, t ≥ 1 only operators with twist (? ? ?) two are present in the operator product expansion. This is also as expected since they are just ???1 . . . ??? ? in free ?eld theory.

1 2 1 2 = 2 A further example in free scalar theory arises for φ → 2 ? , where we have ? 2 ? 4 and in (6.1) N 1 2 = 1/32π . In the corresponding four point expression (6.2) we then 2? have u u u2 u2 2 1 2 (u, v ) = u + = gC (u, v ) + C u + , (6.20) + C u + + g2 ? 2 v v v v

18

for

u2 u2 + C . (6.21) v2 v For a single scalar ?eld C = 4, more generally for Ns free scalar ?elds C = 4/Ns . It is convenient then to allow for arbitrary values of the paramter C . The terms which are O(u) give in this case z . (6.22) h0 (z ) = C z + 1?z gC (u, v ) = u2 + Applying (6.5) and (6.11) to gC (u, v ) as given in (6.21) gives ? C (z, x) = (z ? x)zx 1 + h 1 (1 ? z )2 (1 ? x) 2 + C . (1 ? z )(1 ? x) (6.23)

The expansion in powers of x is straightforward and from (6.16) we ?nd ? (n ? 1)!2 2 ? ? z + z ′2 + n(n ? 1) ? C (z + z ′ ) , n odd, ? (2 n ? 2)! ( n ) ? hC (z ) = (n ? 1)!2 2 ? ? ?? z ? z ′2 + n(n ? 1) + C (z ? z ′ ) , n even, (2n ? 2)!

(6.24)

for z ′ = z/(z ? 1). The required symmetry under z ? z ′ is evident. The result for cm0 obtained from (6.13) is clearly C times that given by (6.19), which if C = 4 gives the correct value of c10 for the same value of CT . A similar approach for solving (6.15) to that leading to (6.19) suggests, for any C , cmt = 2

2m (2m

+ t ? 1)! (2m + t)! (t ? 1)! (4m + 2t ? 1)! (2t ? 2)!

2

2(2m + 1)(m + t) + (?1)t+1 C ,

t ≥ 1.

(6.25) This is positive for ?2 < C < 4. If C = 4 then c02 = 0 which re?ects the fact that the only potential operator with ? = 0, ? = 6, ?2 ??·?? is a descendant of ?4 . ? , which is a scalar For the case of a free fermion ?eld ψ we may consider φ → ψψ operator with ?ψψ ? = 3. the basic two point function is ?(x2 ) = ψ ( x1 ) ψ

4 from which in (6.1) Nψψ ? = 1/π . In (6.2) 3 gψψ ? (u, v ) = u +

γ ·x12 2 , 2π 2 r12

(6.26)

u3 u 1 u3 u ( v ? 1 ? u ) + + (1 ? u ? v ) + (u ? 1 ? v ) . v3 4 v2 v2 z3 1 (2 ? z ) , 4 (1 ? z )2 19

(6.27)

From (6.5) and (6.11), h0 (z ) = (6.28)

and ? ? (z, x) = (z ? x)z 2 x2 1 + h ψψ 1 (1 ? z )3 (1 ? x) 3 ? 1 1 ? . 2 4(1 ? z ) (1 ? x) 4(1 ? z )(1 ? x)2 (6.29)

From (6.12) and (6.16) in this case ? (n ? 1)!n! ? ? 4(z 3 + z ′3 ) + n(n ? 1) ? 3 (z 2 + z ′2 ) , n odd, ?? 8(2 n ? 3)! ( n ) ? (z ) = h (n ? 1)!n! ? ? 4(z 3 ? z ′3 ) + n(n ? 1) ? 1 (z 2 ? z ′2 ) , n even. ? 8(2n ? 3)! cm0 = 22m?2 (2m)!(2m + 1)! , (4m ? 1)!

(6.30)

In a similar fashion as previously we have found for t = 0 c00 = 0 , and for t = 1, 2, . . ., cmt = 22m?3

m = 1, 2, . . . ,

(6.31)

(2m + t + 1)! (2m + t)! (t ? 1)! t! 2(2m + 1)(m + t) + (?1)t+1 . (4m + 2t ? 1)! (2t ? 3)!

(6.32)

? (? ?? . . . ?? ) ψ and the leading operator which The operators for t = 0 are of the form ψγ 2 1 ? contributes is the ? = 2 energy momentum tensor. In this theory c10 = 2 so that (6.10) gives CT = 8, as expected for free Dirac fermions. From (6.32) we may note that cm1 = 0 which follows from (6.29) since the leading term is O(z 2 x2 ). This may be explained since ? r ψ ψ? ? s ψ with ? = 6 + r + s and r + s ≥ ? so that the minimal the relevant operators are ψ? t for operators of this form is t = 2.

1 2 F = We may also consider the case of free vector ?elds when we may take φ → 4 1 1 2 = 4. The essential two point F F , the lowest dimension scalar operator with ? 4 F 4 ?ν ?ν function is

F?ν (x1 )Fσρ (x2 ) =

1 2 I?σ (x12 )Iνρ (x12 ) ? I?ρ (x12 )Iνσ (x12 ) , π 2 r12

(6.33)

4 1 2 = 3/π . In where the inversion tensor is given by (2.7). It is easy to ?nd in (6.1) N 4 F this case we may calculate in (6.2) (the derivation is sketched in appendix D), 4 2 u u4 u4 2 2 2 u 2 2 u (u, v ) = u + 4 + u(v ?1?u) + 3 (1?u?v ) + 3 (u?1?v ) ?u ? 2 ? 2 . (6.34) v 9 v v v v 4

g

1 2 4F

From (6.5) and (6.11), h0 (z ) = z3 2 3 z + 9 (1 ? z )3 20 . (6.35)

and ? 1 2 (z, x) = (z ? x)z 3 x3 1 + h 4F 1 (1 ? z )4 (1 (6.36) .

? x) 4 2 1 1 1 + + + 3 2 2 3 9 (1 ? z )(1 ? x) (1 ? z ) (1 ? x) (1 ? z ) (1 ? x)

As previously we may calculate ? (n ? 1)!(n + 1)! ? ? 9(z 4 + z ′4 ) + n(n ? 1) ? 8 (z 3 + z ′3 ) , n odd, ? (n ? 2) 72(2 n ? 3)! (n) ? h (z ) = (n ? 1)!(n + 1)! ? ? 9(z 4 ? z ′4 ) + n(n ? 1) ? 4 (z 3 ? z ′3 ) , n even. ? ?(n ? 2) 72(2n ? 3)! (6.37) Just as before we then ?nd c00 = 0 , and for t = 1, 2, . . ., cmt = (t ? 2)22m?4 (2m + t + 2)! (2m + t)! (t ? 1)! (t + 1)! (2m + t ? 1) 9(4m + 2t ? 1)! (2t ? 3)!

t+1

cm0 = 22m?1 (2m ? 1)

(2m)!(2m + 2)! , 9(4m ? 1)!

m = 1, 2, . . . ,

(6.38)

(6.39)

× (2m + 1)(m + t) + (?1)

.

For t = 0 the relevant operators are F(?1 |ν | ??2 . . . ????1 F?? )ν , with traces subtracted. For ? = 2, giving the energy momentum tensor, we have c10 = 16/9 so that (6.10) gives CT = 16, in accord with the required result for free vector theories. It should also be noted that 1 2 1 2 the operator 4 F does not appear in the operator product expansion of 4 F ( x1 ) 1 F 2 (x2 ). 4 We also have in (6.39) cm1 = cm2 = 0 which has a similar explanation as for the fermion case, the F 4 operators which are present have ? ≥ ? + 8. Finally we consider an example based of a four point function formed by scalar opera?ij which, with i, j = 1, 2 SU (2)R indices, have dimensions tors T ij = T ji and its conjugate T ?T = ?T ? = 2 and are the lowest dimension scalar operators in a N = 2 supersymmetric theory [22]. The relevant four point function has the general form ?i j (x2 )T i3 j3 (x3 )T ?i j (x4 ) T i1 j2 (x1 )T 2 2 4 4 1 1 = δ (i1 i2 δ j1 ) j2 δ (i3 i4 δ j3 ) j4 2 2 a(u, v ) + δ (i1 i4 δ j1 ) j4 δ (i3 i2 δ j3 ) j2 2 2 b(u, v ) r12 r34 r14 r23 1 + δ (i1 (i2 δ j1 ) (i4 δ (i3 j4 ) δ j3 ) j2 ) c(u, v ) , (6.40) r12 r23 r34 r14 21

with a(u, v ) = b(v, u), c(u, v ) = c(v, u). In the free case a, b, c are constants and setting, ? with suitable normalisation a = b = 1, the conformal invariant four point functions for T T projected on R = 0, 1, 2 representations are A0 (u, v ) = 1 +

1 3

u2 1 u +2 c , 2 v v

A1 (u, v ) =

u2 u + , 2 v v

A2 (u, v ) =

u2 . v2

(6.41)

For this case there is no symmetry under u ? u/v, v ? 1/v but it is straightforward to decompose each term arising in (6.41) into even and odd pieces. For the even pieces, u2 /v 2 + u2 and u/v + u, the relevant expansion coe?cients are given by (6.25), for C = 0, and (6.19).3 For the odd pieces only odd values of ? contribute and, with the same de?nition of t as previously we can then write u2 ? u2 = ? dmt ut+1 G(2m+1) (t + 1, t + 1, 2t + 2m + 3; u, v ) , v2 m,t (6.42)

where the negative sign is a consequence of the fact that the operators occurring in the operator product expansion are anti-hermitian for odd ? in this example. As before we determine dmt = 2

2m+2 (2m

+ t + 1)! (2m + t)! (t ? 1)! (4m + 2t + 1)! (2t ? 2)!

2

(2m + 2t + 1)(m + 1) .

(6.43)

A similar equation to (6.42) may be written for u/v ? v but in this case only t = 0 contributues. This equation may be reduced to 1 z ?z = dm 2m+1 z 2m+2 F (2m + 2, 2m + 2; 4m + 4; z ) , 1?z 2 m=0 which determines dm = 2

3

(6.44)

2m+2

(2m + 1)! . (4m + 2)!

(n)

2

(6.45)

Perhaps we may note that for (u/v )n + un the expansion coe?cients cmt are given by 2m (t ? 1)! (n ? 1)!(n ? 2)!

2

cmt = 2(2m + 1)(m + t)

(n)

(2m + t ? 1)! (2m + t)! (2m + t + n ? 2)! (t + n ? 3)! . (4m + 2t ? 1)! (2m + t ? n + 2)! (2t ? 2)! (t ? n + 1)!

This is zero for t = 0, 1, . . . n ? 2, n ≥ 2. This formula is also correct for n = 1 when only t = 0 contributes.

22

7. Conclusion A crucial result of this paper is that it is possible to ?ne a simple closed form expression for the contribution of an arbitrary spin operator to the four point function. For simplicity, taking ?1 = ?2 , ?3 = ?4 , the result from (2.11) and (3.11) is u 2 (???) G(?)

1 1

1 (? 2

1 ? ? ), 2 (? ? ?), ?; u, v 1 (? 2

(zx) 2 (???) 1 ? = (? 2 z ) zF z?x ×F

1 (? 2

+ ? ), 1 (? + ?); ? + ?; z 2

(7.1)

1 ? ? ? 2), 2 (? ? ? ? 2); ? ? ? ? 2; x ? z ? x ,

for u = zx, v = (1 ? z )(1 ? x), as in (3.9). In the previous section we have shown how this may be applied to identify the relevant operators in some simple cases based on free ?eld theories. The variables z, x play an essential role in the expression (7.1) and it would be desirable to ?nd a more direct justi?cation of this result in which the signi?cance of such a parameterisation of u, v was perhaps more transparent. Of course individual contributions of the form (7.1) do not have the required form for v ? 0, u ? 1 and this constrains the contributions of di?erent operators although, as yet, there is no organising principle as in two dimensions. The critical unitarity constraint that the invariant function of u, v which describes the four point function in conformal ?eld theories should be expandible in terms of contributions of the form (7.1), for suitable ? and ? and with positive coe?cients in appropriate cases, should become easier to analyse with the explicit expressions for G(?) obtained here and exhibited in (7.1). These results should allow further extension of the analysis of four point functions obtained through the AdS/CFT correspondence in terms of the operator product expansion in the large N limit of N = 4 supersymmetric gauge theories [23,5]. To see the simpli?cations obtained by using the variables z, x in another context we mention also some recent results [22] found through the use of superconformal Ward identities, using the harmonic superspace formalism, for the four point function exhibited ?ij are scalar ?elds which are the lowest components of N = 2 in (6.40). Although T ij , T hypermultiplets the superconformal Ward identity lead to constraints on the dependance of a, b, c on the invariants u, v , ? c= ?u ? c= ?v v ? ? 1 u ? a? b? 1? + b, u ?v ?v v v ?u u ? ? 1 v ? b? a? 1? + a. v ?u ?u u u ?v ? 1?x ? x ? c= a+ b. ?z x ?z 1 ? x ?z 23

(7.2)

Rewriting these equations in terms of z, x gives 1?z ? z ? ? c= a+ b, ?x z ?x 1 ? z ?x (7.3)

The solutions are then clearly c? z 1?z a? b = f (z ) , z 1?z c? 1?x x a? b = f ( x) , x 1?x (7.4)

where we have imposed the essential symmetry under z ? x. Eliminating c or a gives 1 1 f ( z ) ? f ( x) a? b= , u v z?x 1 1?v?u c? b= v v2

z 1?z f (z )

?

x 1?x f (x)

z?x

,

(7.5)

which are equivalent to the solutions found in [22]. To satisfy the symmetry under u ? v we must have f (z ) = f (1 ? z ). For the free case discussed in section 6 f (z ) = 2+ c ? 1/z (1 ? z ). Eden et al [22] have argued that there are no higher order corrections in the interacting theory. It would be interesting to see the implications in the context of the operator product expansion. Acknowledgements One of us (FAD) would like to thank the EPSRC, the National University of Ireland and Trinity College, Cambridge for support. He is also very grateful to David Grellscheid for assistance with Mathematica. HO would like to thank Anastasios Petkou for email correspondence and very useful comments.

24

Appendix A. Di?erential Operators for the Operator Product Expansion We describe here how to construct di?erential operators satisfying (2.9) for ? = 1, 2. These cases have been considered previously for a = b and S = d ? 1, d respectively in [24,5] and results for any ? were given in [25]. For ? = 1 and S = a + b + 1 we use the de?nition of Z in (2.4) to write 1 1 1 Zν = a b S B (a + 1, b + 1) r13 r23 1 1 = S B (a + 1, b + 1) × where y = x23 + αx12 , A = ?α(1 ? α)r12 , (A.2) so that αr13 + (1 ? α)r23 = y 2 + A, and we have used, for ? = 0, 1,

1 2 n 1 yν 4? y 2S 1 1

dα

0 1

αa?1 (1 ? α)b?1 αr13 + (1 ? α)r23

S

bα x13 ? a(1 ? α) x23

n

ν

dα αa?1 (1 ? α)b?1

0

1 1 2 A4 ?x2 n ! n=0

1 y 2S

1 S?1 α(1 ? α)x12ν + bα ? a(1 ? α) yν 1 1 d)n (S + 1 ? 2 d)n (S ? 2

, (A.1)

. . . yν? Cν1 ...ν? = (S )n (S + 1 ? 1 2 d ? ? )n

yν y 2(S +n) 1

1

. . . yν? Cν1 ...ν? .

(A.3)

We now employ

1 1 d 1 S?1 1 x12ν = 2S x12? I?ν (y ) ? yν , 2 S S y y S dα y 2S

(A.4)

and, after integrating by parts, the relation 1 1 (a + n)(1 ? α) ? (b + n)α + bα ? a(1 ? α) 1 (S + 1 ? 2 d)n (S ? 1 2 d)n 1 1 = ?n (a + 1 ? 1 1 2 d)α ? (b + 1 ? 2 d)(1 ? α) , (S ? 2 d)n+1 to obtain 1 1 Z = ν a b B (a + 1, b + 1) r13 r23 × + r12 B (a + 1, b + 1) ×

1

(A.5)

dα αa (1 ? α)b

0

1 1 1 2 A4 ?x2 1 n! (S + 1 ? 2 d)n n=0

1 0

n

1 y 2S

x12? I?ν (y )

1 dα αa (1 ? α)b (a + 1 ? 1 2 d)α ? (b + 1 ? 2 d)(1 ? α) n

1 1 1 2 A4 ?x2 1 n ! ( S + 1 ? d ) n +1 2 n=0 25

yν y 2(S +1)

1

.

(A.6)

Since ??

I?ν (y ) y 2S

1

1

= 2(S ? d + 1)

yν y 2(S +1)

1

,

(A.7)

we may therefore write for the ? = 1 case C a,b (s, ? )? = 1 B (a + 1, b + 1) × dα αa (1 ? α)b eαs·?

0

1 1 2 2 ?α(1 ? α) 4 s ? n ! n=0

n

1 s? 1 d)n (S + 1 ? 2 d)α ? (b + 1 ? 1 d)(1 ? α) (a + 1 ? 1 s2 2 2 + ?? . 1 2(S ? d + 1) d)n+1 (S + 1 ? 2

(A.8)

For ? = 2, S = a + b + 2 the calculation is similar although more tedious. Following the same route as led to (A.1) we have, using (A.3) for ? = 0, 1, 2 and for Cν1 ν2 an arbitrary symmetric traceless tensor, 1 Zν1 Zν2 Cν1 ν2 ar b r13 23 = × 1 1 S (S + 1) B (a + 2, b + 2)

1

dα αa?1 (1 ? α)b?1

0

1 1 2 A4 ?x2 n ! n=0

n

1 y 2S

Cν1 ν2

S (S ? 1) α2 (1 ? α)2 x12ν1 x12ν2 1 (S + 1 ? 2 d)n 2(S ? 1) α(1 ? α) (b + 1)α ? (a + 1)(1 ? α) x12ν1 yν2 + (S ? 1 2 d)n +

1 b(b + 1)α2 + a(a + 1)(1 ? α)2 ? 2(b + 1)(a + 1)α(1 ? α) yν1 yν2 , d ) (S ? 1 ? 1 2 n (A.9) We may now write S?1 1 1 x x = x12?1 x12?1 I?1 ν1 (y )I?2 ν2 (y ) 12 ν 12 ν 1 1 S + 1 y 2S y 2S S?1 d 1 (x12ν1 yν2 + x12ν2 yν1 ) ? S (S + 1) dα y 2S 1 2 d2 1 r12 yν yν . ? y ν1 y ν2 ? 2 2 S 2( S (S + 1) dα y S + 1 y S +1) 1 2

(A.10)

The resulting expression has three pieces, the ?rst of which comes from the ?rst line of (A.10) in (A.9) and is readily seen to be 1 B (a + 2, b + 2)

1

dα αa+1 (1 ? α)b+1

0 n

1 1 2 A1 × 1 4 ?x2 n ! ( S + 1 ? d ) 2 n n=0

1 y 2S

(A.11) x12?1 x12?1 I?1 ν1 (y )I?2 ν2 (y )Cν1 ν2 .

26

After integrating by parts the remaining terms in (A.10) then in addition to (A.11) we have 1 S?1 2r12 1 dα αa+1 (1 ? α)b+1 (a + 1 ? 2 d)α ? (b + 1 ? 1 2 d)(1 ? α) S + 1 B (a + 2, b + 2) 0 1 1 1 1 2 n x y C A4 × ?x2 1 2( S +1) 12ν1 ν2 ν1 ν2 n ! y d ) ( S + 1 ? n +1 2 n=0

1 1 r12 + dα αa (1 ? α)b S + 1 B (a + 2, b + 2) 0 1 2 1 × J? α(1 ? α) 1 1 n ! ( S ? d ) ( S + 1 ? d ) n +2 n 2 2 n=0 1 d) (S + b ? 1 d)n + 2(S ? 2 d)(b + 1) α2 J = (a + 1 ? 1 2 2 2 1 1 d) (S + a ? 1 + (b + 1 ? 2 2 d)n + 2(S ? 2 d)(a + 1) (1 ? α)

1 2 A4 ?x2

n

yν yν Cν ν y 2(S +1) 1 2 1 2

1

,

? 2 ((S ? 1 d)(S ? 1 d ? 1) ? (a + 1)(b + 1))n 2 2

1 + (S ? 2 d)((S ? 1 2 d ? 1)(S + 1) ? 2(a + 1)(b + 1)) α(1 ? α) .

(A.12)

We may now use, similarly to (A.4), 1 S?1 1 x12ν1 yν2 Cν1 ν2 = 2(S +1) x12?1 I?1 ν1 (y )yν2 Cν1 ν2 2( S +1) S+1y y 1 d 1 ? yν yν Cν ν , 2( S S + 1 dα y +1) 1 2 1 2 so that the ?rst term in (A.12) gives a contribution, r12 B (a + 2, b + 2)

1 0 1 dα αa+1 (1 ? α)b+1 (a + 1 ? 2 d)α ? (b + 1 ? 1 2 d)(1 ? α) n

(A.13)

1 1 2 × A1 4 ?x2 1 n ! d ) ( S + 1 ? n +1 2 n=0 r122 B (a + 2, b + 2)

1

1 y 2(S +1)

(A.14) x12?1 I?1 ν1 (y )yν2 Cν1 ν2 .

After another integration by parts the ?nal contribution becomes dα αa+1 (1 ? α)b+1

0 1 × (a + 1 ? 1 d)(a + 2 ? 2 d)α2 + (b + 1 ? 1 d)(b + 2 ? 1 d)(1 ? α)2 2 2 2 1 ? 2(a + 1 ? 1 2 d)(b + 1 ? 2 d)α(1 ? α) 1 1 1 n × A1 ?x2 y y C . 1 4 2 2( S +2) ν1 ν2 ν1 ν2 n ! y ( S + 1 ? d ) n +2 2 n=0

(A.15)

Just as with (A.7) we may now use 1 1 ??2 2S I?1 ν1 (y )I?2 ν2 (y )Cν1 ν2 = 2(S ? d) 2(S +1) I?1 ν1 (y )yν2 Cν1 ν2 , y y 1 1 ??1 ??2 2S I?1 ν1 (y )I?2 ν2 (y )Cν1 ν2 = 2(S ? d)(S ? d + 1) 2(S +1) yν1 yν2 Cν1 ν2 , y y 27

(A.16)

in (A.14) and (A.15) to write the sum of (A.11), (A.14) and (A.15) in the form C a,b (x12 , ?x2 )?1 ?2 de?ning C a,b (s, ? ) for ? = 2. Both this result and (A.8) may be expressed in terms of the di?erential operators

a,b Cκ (s, ? )

1 I (x23 )I?2 ν2 (x23 )Cν1 ν2 , S ?1 ν1 r23

(A.17)

1 = B (a, b)

1

dα αa?1 (1 ? α)b?1 eαs·?

0

1 2 2 ?α(1 ? α) 1 4s ? n !( κ ) n n=0

n

.

(A.18)

a+?,b+? For any ?, neglecting terms O(s2 ), C a,b (s, ? )·C has the form CS (s, ? )s?1 . . . s?? C?1 ...?? . +1? 1 d

2

Appendix B. Calculation for ? = 1 Here we consider (2.13) for ? = 1, C a,b (x12 , ?x2 )? 1 Y = f ? 1 a b r14 r24 r14 r13

e

e r23 r24

G(1) (b, e, S ; u, v ) ,

(B.1)

with C a,b (x12 , ?x2 )? given by (A.8), and evaluate directly G(1) (b, e, S ; u, v ) following a similar route to that described for ? = 0 in [26] and sketched in [6]. Using an integral representation similar to (A.1) the ?rst term (A.8) leads to a contribution,

1 1 1 dα αa (1 ? α)b S B (a + 1, b + 1)B (e + 1, f + 1) 0 1 (S )n 2 m × Am B n 1 z 2 ?z 1 4 m ! n ! (S + 1 ? 2 d)m m,n=0 1

dβ β e?1 (1 ? β )f ?1

0

(B.2)

× for

1 z 2(S +n)

x12 · (S ? 1)β (1 ? β )x34 + e(1 ? β ) ? f β z ,

z = x24 + αx12 ? βx34 ,

A = ?α(1 ? α)r12 ,

B = ?β (1 ? β )r34 .

(B.3)

Carrying out the di?erentiation according to (A.3) and using 2(S + m + n ? 1) 1 z 2(S +m+n) x12 ·z = ? d 1 , 2( S + m +n?1) dα z (B.4)

28

allows (B.2) to be rewritten as

1 1 x12 ·x34 S?1 dα αa (1 ? α)b dβ β e (1 ? β )f S B (a + 1, b + 1)B (e + 1, f + 1) 0 0 1 m n ( S ) ( S + 1 ? d ) 1 A B m+n 2 m+n × 1 1 2( m!n! (S + 1 ? 2 d)m (S + 1 ? 2 d)n z S +m+n) m,n=0 1 1 1 1 a?1 b?1 dβ β e?1 (1 ? β )f ?1 dα α (1 ? α) + 2S (S ? 1) B (a + 1, b + 1)B (e + 1, f + 1) 0 0 × a(1 ? α) ? bα e(1 ? β ) ? f β

× +

1 d)m+n Am B n 1 (S ? 1)m+n (S ? 2 1 m!n! (S ? 2 z 2(S ?1+m+n) d)m (S ? 1 d) 2 n m,n=0 1 1

r12 1 2S B (a + 1, b + 1)B (e + 1, f + 1) 1 m!n! m,n=0

dα αa (1 ? α)b

dβ β e?1 (1 ? β )f ?1 ? α) e(1 ? β ) ? f β (B.5)

×

0 0 1 1 × (a ? 2 d + 1)α ? (b ? 2 d + 1)(1 (S )m+n (S ? 1 d) Am B n 2 m+n+1 . 1 z 2(S +m+n) d)m+2 (S ? 1 (S ? 2 2 d)n

The term in (A.8) involving ?? also gives a contribution r12 1 ? 2S B (a + 1, b + 1)B (e + 1, f + 1) × (a ?

1 2d 1 1 a b 0 1 2d

dα α (1 ? α)

0

dβ β e?1 (1 ? β )f ?1

+ 1)α ? (b ?

+ 1)(1 ? α) e(1 ? β ) ? f β

1 d)m+n (S )m+n (S + 1 ? 2 1 Am B n × m!n! (S + 1 ? 1 d) (S + 1 ? 1 d) z 2(S +m+n) 2 m+1 2 n m,n=0

r12 r34 1 e?f + 2 S ? d + 1 B (a + 1, b + 1)B (e + 1, f + 1)

1

1

(B.6)

dα αa (1 ? α)b

0 0

dβ β e (1 ? β )f

1 1 d + 1)α ? (b ? 2 d + 1)(1 ? α) × (a ? 2

×

1 d)m+n 1 (S + 1)m+n (S + 2 ? 2 Am B n . 1 m!n! (S + 1 ? 1 z 2(S +1+m+n) 2 d)m+1 (S + 2 ? 2 d)n m,n=0

To obtain the form shown in (B.1) requires three critical steps. First, writing z 2 = A + B + C where C = α(1 ? β )r14 + β (1 ? α)r23 + αβr13 + (1 ? α)(1 ? β )r24 , we use 1 1 (λ)m+n (κ)m+n Am B n = λ 2( λ + m + n ) m!n! (κ)m (κ)n C z m,n=0 29 (λ)2m m!(κ)m m=0 AB C2

m

(B.7)

.

(B.8)

Secondly we require the α, β integrals to be of the form

1

dα αa?1 (1 ? α)b?1

0

1 C a+b 1

= B (a, b)

1 s a tb

, (B.9)

s = βr13 + (1 ? β )r14 , t = βr23 + (1 ? β )r24 ,

1

dβ β

0

e?1

(1 ? β )

f ?1

1 = B (e, f ) e f , e + f ? C s ? t ? = αr14 + (1 ? α)r24 , s ? = αr13 + (1 ? α)r23 , t

and then ?nally, for λ = a + b = e + f ,

1

dβ β

0 1

e?1

(1 ? β )

f ?1

1 1 = B (e, f ) a b a b s t r14 r24 1 1 = B (a, b) a b e f ? r14 r24 s ? t

r14 r13 r14 r13

e

F (b, e; λ; 1 ? v ) ,

e

(B.10) F (b, e; λ; 1 ? v ) .

dα αa?1 (1 ? α)b?1

0

To apply these results to (B.5) and (B.6) requires some manipulation. We may immediately apply (B.8) to the ?rst two terms in (B.5). In the ?rst term we write 2αβ x12 ·x34 = ?C + r24 + α(r14 ? r24 ) + β (r23 ? r24 ) , (B.11)

and the ?C piece may be combined with the second term in (B.5), using a(1 ? α) ? bα = (S ? 1)(1 ? α) ? b, e(1 ? β ) ? f β = (S ? 1)(1 ? β ) ? f , so that either the α or β integration may be carried out using (B.9). After further algebra these two terms become 1 a b r14 r24 r14 r13

e

S (S ? 1) 1 G(b, e, S ? 2 d, S ? 1; u, 1 ? v ) 2ae d, S ; u, 1 ? v ) ? G(b + 1, e, S + 1 ? 1 2

1 r12 r34 dα αa (1 ? α)b+1 B (a + 1, b + 1)B (e + 1, f + 1) 0 1 1 (S + 1)2m AB m × (S ? d + 1) S +1 1 C m! (S ? 2 d)m+2 C 2 m=0 1

?1 2 (S ? 1)

dβ β e (1 ? β )f +1

0

+ r12 r34

1 1 d+1 S?1 a? 2 1 dβ β e (1 ? β )f +1 a+1 b+1 2a B (e + 1, f + 1) 0 s t 1 (a)m+1 (b + 1)m r12 r34 β (1 ? β ) m × m! (S ? 1 st 2 d)m+2 m=0 1 1 S?1 e? 1 2d + 1 dα αa (1 ? α)b+1 a+1 b+1 ? 2e B (a + 1, b + 1) 0 s ? t 1 (e)m+1 (f + 1)m r12 r34 α(1 ? α) × ? m! (S ? 1 s ?t 2 d)m+2 m=0

+ r12 r34

m

,

(B.12)

30

with G de?ned by the series in (2.32). Combining the last term in (B.5) with (B.6) leads to

1 1 1 S?1 r12 r34 dα αa (1 ? α)b dβ β e (1 ? β )f 2 S ? d + 1 B (a + 1, b + 1)B (e + 1, f + 1) 0 0 1 1 1 1 d + 1)(1 ? β ) × (a ? 2 d + 1)α ? (b ? 2 d + 1)(1 ? α) (e ? 2 d + 1)β ? (f ? 2

d) 1 (S + 1)m+n (S ? 1 Am B n 2 m+n+2 . × 1 m!n! (S ? 2 z 2(S +1+m+n) d)m+2 (S ? 1 2 d)n+2 m,n=0

(B.13)

This is symmetric and the summation formula (B.8) may now be applied and the result combined with the corresponding term in (B.12) using

1 1 1 d + 1)α ? (b ? 2 d + 1)(1 ? α) (e ? 2 d + 1)β ? (f ? 2 d + 1)(1 ? β ) (a ? 1 2

? (S ? d + 1)2 (1 ? α)(1 ? β )

1 = (a ? 1 2 d + 1)(e ? 2 d + 1) 1 d + 1)(1 ? β ) + (e ? 1 d + 1)(1 ? α) . ? (S ? d + 1) (a ? 2 2

(B.14)

The integrals arising from the terms in the last line involving 1 ? α, 1 ? β then cancel the remaining α, β integrals in (B.12) leaving just the ?rst term in (B.14) for which the associated integral may be evaluated giving 1 a b r14 r24 r14 r13

e 1 (S 2 1 (a ? 1 2 d + 1)(e ? 2 d + 1) 1 (S ? d + 1)(S ? 1 d)(S ? 2 d + 1) 2 1 d 2

? 1)

(B.15)

× uG(b + 1, e + 1, S ?

+ 2, S + 1; u, 1 ? v ) .

In consequence from (B.12) and (B.15) we have ?nally altogether G(1) (b, e, S ; u, v ) = S (S ? 1) 1 d, S ? 1; u, 1 ? v ) ? G(b + 1, e, S + 1 ? 2 d, S ; u, 1 ? v ) G(b, e, S ? 1 2 (B.16) 2ae 1 1 (S ? 1)(a ? 2 d + 1)(e ? 2 d + 1) + uG(b + 1, e + 1, S ? 1 2 d + 2, S + 1; u, 1 ? v ) . 1 d )( S ? d + 1) 2(S ? d + 1)(S ? 1 2 2

The result (B.16) is di?erent from that which is given by (2.30) for ? = 1 together with (2.31). To show the equivalence of the two expressions it is su?cient to use the following 31

relations for the function G de?ned by (2.32), α vG(α + 1, β + 1, γ, δ + 1; u, 1 ? v ) + (δ ? α ? β )G(α, β + 1, γ, δ + 1; u, 1 ? v ) ? (δ ? β )G(α, β, γ, δ + 1; u, 1 ? v ) = α(δ ? β )(δ ? α ? β ) uG(α + 1, β + 1, γ + 1, δ + 2; u, 1 ? v ) , γ (δ + 1) ? δG(α, β, γ ? 1, δ ; u, 1 ? v ) αβ (δ ? β ) uG(α + 1, β + 1, γ + 1, δ + 2; u, 1 ? v ) , = ?(δ ? α ? γ + 1) (γ ? 1)γ (δ + 1) αG(α + 1, β, γ, δ + 1; u, 1 ? v ) ? βG(α, β + 1, γ, δ + 1; u, 1 ? v ) ? (α ? β )G(α, β, γ, δ + 1; u, 1 ? v ) = ?(α ? β ) αβ uG(α + 1, β + 1, γ + 1, δ + 2; u, 1 ? v ) . γ (δ + 1)

βG(α, β + 1, γ, δ + 1; u, 1 ? v ) + (δ ? β )G(α, β, γ, δ + 1; u, 1 ? v ) (B.17)

These may be obtained from relations given in [6] which express G(α, β, γ, δ ; u, 1 ? v ) in terms of G(α′ , β ′ , γ ′ , δ ′ ; u, 1 ? v ) with δ ′ ? γ ′ = δ ? γ + 1.

Appendix C. Identities for H From the de?nition (5.9) and properties of the function G the following relations were obtained in [6], H (α, β, γ, δ ; u, v ) = v ?α H (α, δ ? β, γ, δ ; u/v, 1/v ) = v δ?α?β H (δ ? α, δ ? β, γ, δ ; u, v ) = H (α, β, α + β ? δ + 1, α + β ? γ + 1; v, u) = u1?γ H (α ? γ + 1, β ? γ + 1, 2 ? γ, δ ? 2γ + 2; u, v ) . In terms of the result (5.8) for the four point function these give D?1 ?2 ?3 ?4 (u, v ) = D Σ??3 Σ??4 Σ??1 Σ??2 (u, v ) = v ??2 D?1 ?2 ?4 ?3 (u/v, 1/v ) = v ?4 ?Σ D ?2 ?1 ?3 ?4 (u/v, 1/v ) = v ?1 +?4 ?Σ D?2 ?1 ?4 ?3 (u, v ) = D ?3 ?2 ?1 ?4 (v, u) = u?3 +?4 ?Σ D?4 ?3 ?2 ?1 (u, v ) , 32 (C.2) (C.1a) (C.1b) (C.1c) (C.1d)

which re?ect the symmetry under the interchanges xi , ?i ? xj , ?j for various i, j . Results obtained in [6] for G also imply (α ? β )H (α, β, γ, δ ; u, v ) = H (α + 1, β, γ, δ + 1; u, v ) ? H (α, β + 1, γ, δ + 1; u, v ) , (δ ? α ? β )H (α, β, γ, δ ; u, v ) = H (α, β, γ, δ + 1; u, v ) ? vH (α + 1, β + 1, γ, δ + 1; u, v ) , (1 ? γ )H (α, β, γ, δ ; u, v ) = H (α, β, γ ? 1, δ ; u, v ) ? uH (α + 1, β + 1, γ + 1, δ + 2; u, v ) , (δ ? γ ? β + 1)H (α, β, γ, δ ; u, v ) = H (α, β, γ ? 1, δ ; u, v ) + H (α + 1, β, γ, δ + 1; u, v ) + H (α, β, γ, δ + 1; u, v ) . The ?rst three relations imply (?2 + ?4 ? Σ)D?1 ?2 ?3 ?4 (u, v ) = D?1 ?2 +1 ?3 ?4 +1 (u, v ) ? D ?1 +1 ?2 ?3 +1 ?4 (u, v ) , (?1 + ?4 ? Σ)D?1 ?2 ?3 ?4 (u, v ) = D?1 +1 ?2 ?3 ?4 +1 (u, v ) ? vD?1 ?2 +1 ?3 +1 ?4 (u, v ) , (?3 + ?4 ? Σ)D?1 ?2 ?3 ?4 (u, v ) = D?1 ?2 ?3 +1 ?4 +1 (u, v ) ? uD ?1 +1 ?2 +1 ?3 ?4 (u, v ) , (C.4) which are equivalent to results obtained in [13]. The last relation in (C.3) also gives ?4 D?1 ?2 ?3 ?4 (u, v ) = D?1 ?2 ?3 +1 ?4 +1 (u, v ) + D?1 ?2 +1 ?3 ?4 +1 (u, v ) + D ?1 +1 ?2 ?3 ?4 +1 (u, v ) . We also have ?v H (α, β, γ, δ ; u, v ) = ? H (α + 1, β + 1, γ, δ + 1; u, v ) , ?u H (α, β, γ, δ ; u, v ) = ? H (α + 1, β + 1, γ + 1, δ + 2; u, v ) , or ?v D?1 ?2 ?3 ?4 (u, v ) = ? D?1 ?2 +1 ?3 +1 ?4 (u, v ) ?u D?1 ?2 ?3 ?4 (u, v ) = ? D?1 +1 ?2 +1 ?3 ?4 (u, v ) . (C.7) (C.6) (C.5) (C.3)

When γ is an integer the de?nition (5.9) gives ln u terms as well as an expansion in powers in u, 1 ? v . For γ = n = 1, 2, . . . we easily ?nd the terms involving ln u as (?1)n Γ(α)Γ(β )Γ(δ ? α)Γ(δ ? β ) G(α, β, n, δ ; u, 1 ? v ) . (n ? 1)! Γ(δ ) (C.8) 1?n If γ = n = 0, ?1, ?2, . . . then the leading log term is u ln u and the corresponding formula may be obtained from (C.8) using (C.1d). By virtue of (C.1c) there are similarly terms involving ln v in an expansion for v ? 0 when α + β ? δ is an integer. For γ = 1, which is of relevance in (5.7) when ?1 + ?2 = ?3 + ?4 , a complete formula for the additional H (α, β, n, δ ; u, v )log.terms = ln u 33

power terms as well as the log. terms displayed in (C.8) is given in [6]. For γ = 0 the corresponding formula is H (α, β, 0, δ ; u, v ) = 1 Γ(α)Γ(β )Γ(δ ? α)Γ(δ ? β ) F (α, β ; δ ; 1 ? v ) Γ(δ ) (δ ? α)m (δ ? β )m (α)m+n (β )m+n ? ln u + fmn um (1 ? v )n , ? ( m ? 1)! m ! n ! ( δ ) 2m+n m=1,n=0 ? ψ (α + m + n) ? ψ (β + m + n) . (C.9)

fmn = ψ (1 + m) + ψ (m) + 2ψ (δ + 2m + n) ? ψ (δ ? α + m) ? ψ (δ ? β + m)

With the aid of the above relations we may determine Dn1 n2 n3 n4 (u, v ) for ni = 1, 2, . . ., i ni even, in terms of D 1111 (u, v ). For example if ni = n = 1, 2, . . . then, from (5.8), Dnnnn (u, v ) = H (n, n, 1, 2n; u, v ), where H (n, n, 1, 2n; u, v ) = v ?n H (n, n, 1, 2n; u/v, 1/v ) = H (n, n, 1, 2n; v, u) , and using (C.6) and (C.1b, d) we may obtain the recurrence relation H (n + 1, n + 1, 1, 2n + 2; u, v ) = ?u u?u H (n, n, 1, 2n; u, v ) = ?v v?v H (n, n, 1, 2n; u, v ) . (C.11) For the starting point D1111 , with the de?nitions in (3.9) for z, x (note that (z ? x)2 = 1 + u2 + v 2 ? 2u ? 2v ? 2uv ), we have D1111 (u, v ) = H (1, 1, 1, 2; u, v ) = with Φ(z, x) = ln zx ln 1 Φ(z, x) , z?x (C.12) (C.10)

1?z ? 2Li2 (x) + 2Li2 (z ) , 1?x

(C.13)

where Li2 is the dilogarithm function. The result (C.12) with (C.13) was derived in [6] and is equivalent to results [27] known for some time for the integral in (2.28) with αi = 1, d = 4. The function Φ in (C.13) satis?es the following critical identities z , z?1 (C.14) which depend on standard results for the dilogarithm function. The results (C.14) are required in order to satisfy (C.10) for n = 1. Φ(z, x) = ?Φ(x, z ) = ?Φ(1 ? z, 1 ? x) = ?Φ(z ′ , x′ ) , x′ = z′ = For n = 2 we may use ?u Φ(z, x) = 1?u?v 1 ln v + 2 ln u , z?x u 34 ?v Φ(z, x) = 1?u?v 1 ln u + 2 ln v , z?x v (C.15) x , x?1

to obtain H (2, 2, 1, 4; u, v ) = 12uv 1+u+v Φ(z, x) + 5 ( z ? x) ( z ? x) 3 6 + (1 + u ? v )v ln v + (1 ? u + v )u ln u ( z ? x) 4 2 + ln uv + 1 . ( z ? x) 2

(C.16)

Also for n = 3 we have 1680u2 v 2 240uv 24 4 H (3, 3, 1, 6; u, v ) = + + (1 + u + v ) + Φ(z, x) 9 7 5 ( z ? x) ( z ? x) ( z ? x) ( z ? x) 3 100 480uv 840u + v 2 (1 + u ? v ) + + 8 6 ( z ? x) ( z ? x) ( z ? x) 6 1 12(1 + u) + 76v ln v + u ? v + ( z ? x) 4 260uv 26 + + (1 + u + v ) . (C.17) 6 ( z ? x) ( z ? x) 4 For recent applications [28] it is necessary to know D?1 ?2 ?3 ?4 for other small integer values of ?i . To this end we ?rst determine H (1, 1, 1, 3; u, v ) = v ?1 H (1, 2, 1, 3; u/v, 1/v ) = vH (2, 2, 1, 3; u, v ) = vH (2, 2, 2, 4; v, u) = H (1, 1, 0, 2; v, u) , (C.18)

which follow from (C.1,) and correspond to D2112 (u, v ) = v ?1 D2121 (u/v, 1/v ) = vD1221 (u, v ) = vD 2211 (v, u) = D1122 (v, u). From (C.6) we may ?nd 1+u?v 1 H (1, 1, 1, 3; u, v ) = ?v Φ(z, x) ? (1 ? u ? v ) ln u + 2v ln v . (C.19) 3 ( z ? x) ( z ? x) 2 Similarly for D2233 (u, v ) it is necessary to determine H (2, 2, 1, 5; u, v ) = v ?2 H (2, 3, 1, 5; u/v, 1/v ) = vH (3, 3, 1, 5; u, v ) = vH (3, 3, 2, 6; v, u) = H (2, 2, 0, 4; v, u) , and explicitly we have H (2, 2, 1, 5; u, v ) =? 6v 4v 60uv 2 Φ(z, x) (1 + u ? v ) + v (1 + u ? v ) + 4u + 7 5 ( z ? x) ( z ? x) ( z ? x) 3 2v 120uv 2 + 6(1 + u) + 5v ln v ? (C.21) 6 ( z ? x) ( z ? x) 4 2 1 60uv (1 ? u ? v ) + 1 ? u + v ? 9uv ? 2v 2 ? ln u ? 6 4 ( z ? x) ( z ? x) ( z ? x) 2 10v 2 ? (1 + u ? v ) ? . 4 ( z ? x) ( z ? x) 2 35 (C.20)

Appendix D. Vector Four Point Function We describe here a few details concerning the derivation of the result (6.34) where maintaining manifest conformal invariance simpli?es the calculation (a related calculation is described by Herzog [4]). The disconnected graphs contributing to 1 2 1 2 F ( x1 ) 1 F 2 ( x2 ) 4 F ( x3 ) 1 F 2 (x4 ) are of course straightforward while the connected 4 4 4 graphs, using (6.33), give 1 1 1 2 tr(I (x12 )I (x24 )I (x43 )I (x31 )) 8 2 π (r12 r24 r34 r13 ) 2 (D.1) ? tr I (x12 )I (x24 )I (x43 )I (x31 )I (x12 )I (x24 )I (x43 )I (x31 ) , together with two other permutations. To evaluate the traces of the inversion tensors in (D.1) we use [10] xj 1 rij xi1 2 ? , X1( , (D.2) I (x1i )I (xij )I (xj 1 ) = I (X1(ij ) ) , X1(ij ) = ij ) = r1 i r1 j r1 i r2 j where X1(ij ) transforms as a conformal vector at x1 . With (D.2) X1(24) ·X1(43) tr I (x12 )I (x24 )I (x43 )I (x31 ) = tr I (X1(24) )I (X1(43) ) = 4 2 2 X1(24) X1(43) Similarly tr I (x12 )I (x24 )I (x43 )I (x31 )I (x12 )I (x24 )I (x43 )I (x31 ) = tr I (X1(24) )I (X1(43) )I (X1(24) )I (X1(43) ) 2 X1(24) ·I (X1(43) )I (X1(24) )I (X1(43) ) ·X1(24) = tr I (X1(43) )I (X1(24) )I (X1(43) ) ? 2 X1(24) (X1(24) ·X1(43) )2 =4 1?2 2 2 X1(24) X1(43) Since 2X1(24) ·X1(43) = we have tr(I (x12 )I (x24 )I (x43 )I (x31 )) X1(24) ·X1(43) = 16 2 2 X1(24) X1(43) and (D.1) becomes

2 2 2 2

.

(D.3)

. r23 1 u 1? ? r12 r14 v v

(D.4)

(D.5)

? tr I (x12 )I (x24 )I (x43 )I (x31 )I (x12 )I (x24 )I (x43 )I (x31 ) (D.6)

1 ? 4 = 4 ( v ? 1 ? u) 2 ? 4 , u

2 1 u( v ? 1 ? u) 2 ? u2 . (D.7) 8 4 π (r12 r34 ) The other two contributions may be obtained similarly or by applying permutations to the results (D.7). 36

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[18] L.C. Ho?mann, A.C. Petkou and W. R¨ uhl, Phys. Lett. B478 (2000) 320, hepth/0002025; L.C. Ho?mann, A.C. Petkou and W. R¨ uhl, Adv. Theor. Math. Phys. 4 (2000), hep-th/0002154. [19] E. Witten, Adv. Theor. Math. Phys. 2 (1998) 253, hep-th/9802150. [20] K. Symanzik, Lett. al Nuovo Cimento 3 (1972) 734. [21] L. Ho?mann, L. Mesref and W. R¨ uhl, Nucl. Phys. B589 (2000) 337. [22] B. Eden, P.S. Howe, A. Pickering, E. Sokatchev and P.C. West, Nucl. Phys. B581 (2000) 71, hep-th/0001138; B. Eden, A.C. Petkou, C. Schubert and E. Sokatchev, Nucl. Phys., to be published, hep-th/0009106. [23] E. D’Hoker, S.D. Mathur, A. Matsusis and L. Rastelli, Nucl.Phys. B589 (2000) 38, hep-th/9911222. [24] K. Lang and W. R¨ uhl, Nucl. Phys. B402 (1993) 573. [25] S. Ferrara, A.F. Grillo and R. Gatto, Ann. Phys. 76 (1973) 161. [26] A.C. Petkou, Ann. Phys. (N.Y.) 249 (1996) 180, hep-th/9410093. [27] N.I. Ussyukina and A.I. Davydychev, Phys. Lett. B298 (1993) 363; B305 (1993) 136; A.I. Davydychev and J.B. Tausk, Nucl. Phys. B397 (1993) 133. [28] G. Arutyunov and S. Frolov, Phys. Rev. D62 (2000) 064016, hep-th/0002170.

38

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