Fermionic Superstring Loop Amplitudes in the Pure Spinor Formalism

Fermionic Superstring Loop Amplitudes in the Pure Spinor Formalism Christian Stahn (UNC) UNC String Theory Seminar, 3 May 2007 based on arxiv:0704.0015

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

Pure spinor quantisation of the superstring [Berkovits (2000)] Problems with superstring quantisation: RNS: Spacetime susy only after summation over spin structures GS: Need to go to light cone gauge Flat-space action: S = d 2 z ( 12 x m xm p α θ α pᾱ θᾱ + w α λ wᾱ λᾱ) α + Bosonic ghost λ subject to pure spinor constraint: Critical in ten dimensions. λγ m λ = 0 BRST operator Q = λ α d α : cohomology reproduces superstring spectrum Tree amplitudes agree with RNS [Berkovits & Vallilo (2000)] Relate to GS formalism in semi-lightcone gauge

Pure spinor loop amplitudes Multiloop amplitudes constructed in pure spinor formulation, and used to prove vanishing theorems [Berkovits (2004)]. Massless multiloop amplitude: Need at least four external states. One-loop amplitude: A = K K d 2 τ (Im τ) 5 d 2 z 2 d 2 z 3 d 2 z 4 G(z i, z j ) k i k j Kinematic factors expressed as pure spinor superspace integrals : K 1-loop = (λa)(λγ m W )(λγ n W )F mn i<j K 2-loop = (λγ mnpqr λ)(λγ s W )F mn F pq F rs Covariant zero mode integration: λ 3 θ 5 = 1. Superfields: A α (γ a θ) α ζ a + (uγ a θ)(γ a θ) α +... D β D β and A α Am W α D β F mn, where D α = α + 1 2 k a(γ a θ) α. Bosonic amplitudes agree with RNS [Berkovits & Mafra, Mafra (2005)]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

Superfield recursion relations [Ooguri et al. (2000), Grassi & Tamassia (2004), Policastro & Tsimpis (2006)] A α (x, θ) = 0 + 1 2 (θγa ) α ζ a + 1 3 (θγa ) α (θγ a u) +? A m, W α, F mn =? Choose a gauge θ α A α = 0. On-shell identities 2D (α A β) = γ m αβa m, D α W β = 1 4 (γmn ) α β F mn lead to recursion relations A (n) α = 1 n+1 (γm θ) α A (n 1) m A (n) m = 1 n (θγ mw (n 1) ) W α(n) = 1 2n (γmn θ) α m A (n 1) n where f (n) = 1 n! θαn θ α 1 (Dα1 D αn f ). Can solve: obtain e.g. A (n) m from ζ, u via derivative operator O m q = 1 2 (θγ m qp θ) p : A (2k) m = 1 (2k)! [Ok ] m q ζ q, A (2k+1) m = 1 (2k+1)! [Ok ] m q (θγ q u).

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

Pure spinor integrals: Tensorial formulae Recall: Lorentz invariant integration T αβγ,δ 1...δ 5 λ α λ β λ γ θ δ 1... θ δ5 = T (αβγ),[δ 1...δ 5] with normalisation (λγ m θ)(λγ n θ)(λγ p θ)(θγ mnp θ) = 1. Uniquely determined by symmetry λ (α λ β λ γ) : Sym 3 S + = [00003] [10001] θ [δ 1... θ δ5] : Alt 5 S + = [00030] [11010]. Can therefore derive tensorial expressions such as (λγ mnpqr θ)(λγ a θ)(λγ b θ)(θγ cde θ) ( = 1 42 δ mnpqr abcde + ) 1 5! εmnpqr abcde Generally, by Fierzing, reduce to three basic correlators: (λγ [5] λ)(λ{γ [1] or γ [3] or γ [5] }θ)(θγ [3] θ)(θγ [3] θ).

Tensorial formulae: example application Example from two-loop four-fermion amplitudes: F = (λγ mnpq[r λ)(λγ s] u 1 )(θγ n ab θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ). After Fierz transformations: ( F = (λγ mnpq[r λ)(λγ c θ)(θγ ab n θ)(θγ jkl θ) (u 1 γ s] γ c γ b u 2 ) 1 16 + 1 3! 16 + 1 2 5! 16 (λγ mnpq[r λ)(λγ cde θ)(θγ ab n θ)(θγ jkl θ) (u 1 γ s] γ cde γ b u 2 ) (λγ mnpq[r λ)(λγ cdefg θ)(θγ ab n θ)(θγ jkl θ) (u 1 γ s] γ cdefg γ b u 2 )) 1 3! 16 (u 3γ q γ jkl γ s u 4 ), Use correlator formulae such as (λγ mnpqr λ)(λγ abcde θ)(θγ fgh θ)(θγ jkl θ) ( = 16 7 δ mnpqr [ ] δ m n p abc δf j δg d δ q k ( δe h δ r l +2δl e δ r h)+δ m n ab δfg cd δ p q jk (δe h δ r l 3δl e δ r h) Fairly complicated m n p q r + 1 5! εmnpqr m n p q r ) [abcde][fgh][jkl](fgh jkl)

Reduction to traces and bilinears Covariant expression for correlator, using gamma matrices: λ α λ β λ γ θ δ 1... θ δ5 = N 1 [ (γ m ) αδ 1 (γ n ) βδ 2 (γ p ) γδ 3 (γ mnp ) δ 4δ 5 ] (αβγ)[δ 1...δ 5] Normalisation constant N fixed by N (λγ x θ)(λγ y θ)(λγ z θ)(θγ xyz θ) = [ (γ m ) αδ 1 (γ n ) βδ 2 (γ p ) γδ 3 (γ mnp ) δ 4δ 5 ](αβγ)[δ 1...δ 5] (γ x ) αδ1 (γ y ) βδ2 (γ z ) γδ3 (γ xyz ) δ4 δ 5 = 1 60 Tr(γ xγ m ) Tr(γ y γ n ) Tr(γ z γ p ) Tr(γ xyz γ pnm ) +... 1 60 Tr(γ zγ pnm γ zyx γ n γ x γ m γ y γ p ) (60 terms) = 5160960. Use computer algebra to carry out spinor index symmetrisations simplify γ products and traces (e.g. Mathematica with GAMMA)

Reduction to traces and bilinears (2) Allows for easier evaluation of fermionic correlators: N (λγ mnpq[r λ)(λγ s] u 1 )(θγ n ab θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) = 1 60 Tr(γ xγ ab nγ y γ mnpq[r )(u 3 γ q γ xyz γ s u 4 )(u 1 γ s] γ z γ b u 2 ) +... 1 30 (u 2γ b γ xyz γ q u 3 )(u 1 γ s γ y γ ab nγ x γ mnpq[r γ z γ s] u 4 ), (24 terms) Simpler because: no Fierzing necessary fewer open indices fewer ε terms easier to put on computer

Component approach Can also use gamma matrix expression λ α λ β λ γ θ δ 1... θ δ5 = N 1 [ (γ m ) αδ 1 (γ n ) βδ 2 (γ p ) γδ 3 (γ mnp ) δ 4δ 5 ] to compute correlators of particular spinor components. Example: Check coefficients in t mnm 1n 1...m 4 n 4 (αβγ)[δ 1...δ 5] 10 (λγ p γ m 1n 1 θ)(λγ q γ m 2n 2 θ)(λγ r γ m 3n 3 θ)(θγ m γ n γ pqr γ m 4n 4 θ) ( = 2 45 η mn t m 1n 1...m 4 n 4 ) 8 1 2 εmnm 1n 1...m 4 n 4 Choose particular index values. Expand integrand and evaluate correlators on all monomials (here, about 10 5 ) : (λγ p γ 12 θ)(λγ q γ 21 θ)(λγ r γ 34 θ)(θγ 0 γ 0 γ pqr γ 43 θ) = 12 λ 1 λ 1 λ 1 θ 1 θ 9 θ 10 θ 11 θ 12 + + 12 λ 16 λ 16 λ 16 θ 5 θ 6 θ 7 θ 8 θ 16 = 1 45. Probably unsuitable for more complex applications.

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

Purely bosonic expressions Easy to algorithmically deal with bosonic on-shell identitites: momentum conservation i k i = 0 and masslessness ki 2 = 0: eliminate one momentum (e.g. k 4 ) and set all ki 2 to zero eliminate one quadratic invariant (e.g. s 23 s 12 s 13 ) equation of motion for polarisation vector k i ζ i = 0: set all k i ζ i to zero replace one extra k ζ, e.g. k 3 ζ 4 ( k 1 k 2 ) ζ 4 For gauge invariant expressions: start with Fi ab = 2k [a i ζ b] i, e.g. gauge invariant k 4 ζ 1 ζ 2 ζ 3 ζ 4 scalars: Tr(F 1 F 2 F 3 F 4 ) Tr(F 1 F 2 F 4 F 3 ) Tr(F 1 F 3 F 2 F 4 ) Tr(F 1 F 2 ) Tr(F 3 F 4 ) Tr(F 1 F 3 ) Tr(F 2 F 4 ) Tr(F 1 F 4 ) Tr(F 2 F 3 ) gauge invariant k 6 ζ 1 ζ 2 ζ 3 ζ 4 scalars: additionally terms like k 3 F 1 F 2 k 3 Tr(F 3 F 4 ), k 4 F 1 F 3 k 2 Tr(F 2 F 4 )

Fermions: Fierz identities Spinors more complicated. Example question: which scalars from u 1 u 2 u 3 u 4? start from T 1 (1234) = (u 1 γ a u 2 )(u 3 γ a u 4 ), T 3 (1234) = (u 1 γ abc u 2 )(u 3 γ abc u 4 ). Fierz transformations T 3 (1234) = 12T 1 (1234) 24T 1 (1324) (γ a ) (αβ (γ a ) γ)δ = 0 T 1 (1234) + T 1 (1324) + T 1 (1423) = 0 leaves two scalars T 1 (1234) and T 1 (1324) agrees with rep theory, (S + ) 4 = 2 1 +... Easy to computerise in matrix representation for γ a : Reduce to independent monomials u α 1 uβ 2 uγ 3 uδ 4, e.g. T 1 (1234) = 2u 9 1 u9 2 u1 3u 1 4 + 2u 10 1 u10 2 u1 3u 1 4 +...

Fermions: Dirac equation In one- and two-loop amplitudes: need (k 2 or k 4 ) u 1 u 2 u 3 u 4. Spinors subject to /k i u i = 0 Useful observation: only need one k into (u i γ [n] u j ) bilinears. Thus consider (k 2 or k 4 ) (u 1 u 2 u 3 u 4 scalar or 2-tensor) Computer treatment: solve Dirac equation with SO(8) spinors ( ) ( ) ( ) u s u = u c with γ 1...8 u s +u s u c = u c. Obtain coupled equations (k 0 + k 9 )u s (σ k)u c = 0, (k 0 k 9 )u c (σ T k)u s = 0 with solution u s = (σ k)u c / 2k +. Everything reduced to independent monomials in (u c i )1...8

Using particular momenta This algorithm reduces all scalars containing k i and u i to fαβγδ (k i )(u c 1 )α (u c 2 )β (u c 3 )γ (u c 4 )δ Problem: manifest Lorentz invariance broken by γ a matrix rep. difficult to deal with the f αβγδ (k i ) (e.g. terms proportional to ki 2 = 0) One solution: substitute sets of particular vectors for k i, e.g. k µ 1 = (5, 3, 0,..., 0, 4) k µ 2 k µ 3 = (5, 3, 0,..., 4, 0) k µ 4 = ( 5, 0, 3,..., 4, 0) = ( 5, 0, 3,..., 0, 4) Results: Algorithms to find independent scalars, decomposition algorithms, e.g. for k 2 u 1... u 4 scalars, (u 1 /k 3 u 2 )(u 3 /k 1 u 4 ) s 12 T 1 (1234) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) s 13 T 1 (1234) (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) s 13 T 1 (1324)

The problems addressed in this talk Evaluation of pure spinor superspace integrals in kinematic factors: K 2-loop = (λγ mnpq[r λ)(λγ s] W (θ))f mn (θ)f pq (θ)f rs (θ) Expand superfields, saturate θ 5 +ka 2 kmk 2 p 3 kr 4 (λγ mnpq[r λ)(λγ s] u 1 )(θγ ab n θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) + Evaluate correlator = (u 1 γ mpr u 2 )(u 3 γ a u 4 ) + (u 1 γ ai 1i 2 u 2 )(u 3 γ mpr i 1 i 2 u 4 ) + +ε ampr i 1...i 6 (u 1 γ i 3...i 9 u 2 )(u 3 γ i 1i 2 i7i 8 i 9 u 4 ) + Reduce to kinematic basis (Use on-shell identities) K = s 12 [ (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ]

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

One-loop massless four-point amplitude Derived in pure spinor formalism [Berkovits (2004)]: A = K K Kinematic factor: Saturate θ 5 : d 2 τ (Im τ) 5 d 2 z 2 d 2 z 3 d 2 z 4 G(z i, z j ) k i k j K = (λa 1 )(λγ m W 2 )(λγ n W 3 )F 4,mn + ( cycl(234) ) X ABCD = i<j (λa (A) 1 )(λγm W (B) 2 )(λγ n W (C) 3 )F (D) 4,mn with A + B + C + D = 5. Parities of A, B, C, D bosonic or fermionic external states

Four bosons (Review of [Mafra (2005)]) Can evaluate correlator for one labelling, and then symmetrise: K (4B) 1-loop = (λa 1 )(λγ m W 2 )(λγ n W 3 )F 4B 4,mn + ( cycl (234) ) = X 3110 + X 1310 + X 1130 + X 1112 + ( cycl (234) ) Result must be a linear combination of single trace Tr(F (1 F 2 F 3 F 4) ) and double trace Tr(F (1 F 2 ) Tr(F 3 F 4) ). Pure spinor integrals: X 3110 = 15 64 F mnf 1 pqf 2 rsf 3 tu 4 (λγ [t γ pq θ)(λγ u] γ rs θ)(λγ a θ)(θγ amn θ), X 1112 = 15 16 k mζ 4 nf 1 pqf 2 rsf 3 tu 4 (λγ [m γ pq θ)(λγ a] γ rs θ)(λγ n θ)(θγ tu a θ), X 1310 = 5 16 k mζ 3 nf 1 pqf 2 rsf 3 tu 4 (λγ [t γ ma θ)(λγ u] γ rs θ)(λγ n θ)(θγ pq a θ). Reduce to traces: X 3110 = N 1[ 1 60 Tr(γ aγ z ) Tr(γ xyz γ anm ) Tr(γ x γ qp γ [t ) Tr(γ y γ sr γ u] ) + + 1 60 Tr(γ[u γ rs γ zyx γ qp γ t] γ x γ a γ y γ mna γ z ) = ( 1 35 δmp rs δ nq tu 1 630 δmn tu δrs pq 1 90 δmn rs δ pq tu + 26 315 δmn pr δ qs tu ] (60 terms) ) [mn][pq][rs][tu](pq rs)

Four bosons (2) Contract with field strengths, momenta and polarisations, and symmetrise: X 3110 + ( cycl(234) ) = 11 112 Tr(F (1F 2 F 3 F 4) ) 1 56 Tr(F (1F 2 ) Tr(F 3 F 4) ) X 1112 + ( cycl(234) ) = 19 448 Tr(F (1F 2 F 3 F 4) ) 31 1792 Tr(F (1F 2 ) Tr(F 3 F 4) ) X 1310 + ( cycl(234) ) ( = 3 512 4 Tr(F(1 F 2 F 3 F 4) ) Tr(F (1 F 2 ) Tr(F 3 F 4) ) ) Total: K 4B 1-loop = 3 64 ( 4 Tr(F(1 F 2 F 3 F 4) ) Tr(F (1 F 2 ) Tr(F 3 F 4) ) ) = 1 128 t 8F 4 Agrees with Green-Schwarz and RNS results

Four fermions Could similarly evaluate: K (4F) 1-loop = (λa 1 )(λγ m W 2 )(λγ n W 3 )F 4,mn 4F + ( cycl (234) ) Note the symmetry: cycl(234) implies complete antisymmetry in u 2, u 3, u 4 on k 2 u 1 u 2 u 3 u 4 scalars, [234] is equivalent to [1234] there is only one antisymmetric k 2 u 1 u 2 u 3 u 4 scalar Thus the result is fixed by symmetry: K 4F 1-loop (u 1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) Again, agrees with previously known expressions

Symmetrisations more tricky now: Two bosons, two fermions K1-loop(f 2B2F 1 f 2 b 3 b 4 ) = (1 π 34 ) (λa (even) 1 )(λγ m W (even) 2 )(λγ n W (odd) 3 )F (even) 4,mn + (λa (even) 1 )(λγ m W (odd) 3 )(λγ n W (odd) 4 )F (odd) 2,mn. First line has wrong [34] symmetry and must give zero. Fermionic expansion: K 2B2F 1-loop = (1 π 34 ) (X 4010 + X 2210 + X 2030 + X 2012 ) + X 2111, Typical correlator: X 4010 = 1 60 k 1 q k 3 b ζ3 c k 4 mζ 4 n Reduce to u 1 u 2 bilinears: (λγ a θ)(θγ a pq θ)(θγ p u 1 )(λγ [m u 2 )(λγ n] γ bc θ) X 4010 = 1 360 δq[b mn (u 1 γ c] u 2 ) + 1 90 δq[m bc (u 1γ n] u 2 ) + 1 720 δbc mn(u 1 γ q u 2 ) 1 2520 δ[m q (u 1 γ n]bc u 2 ) 1 720 δ[b q (u 1 γ c]mn u 2 ) 1 1260 δ[b [m (u 1γ c]q n] u 2 )+ 1 3360 (u 1γ bcmnq u 2 )

Two bosons, two fermions (2) Contract with k and ζ, and expand in kinematic basis C 1... C 5 : (1 π 34 )X 4010 = 1 60480 ( 1, 12, 12, 1, 2) C 1...C 5 (1 π 34 )X 2210 = 1 60480 (31, 30, 30, 1, 2) C 1...C 5 (1 π 34 )X 2030 = 1 60480 (0, 21, 21, 0, 0) C 1...C 5 (1 π 34 )X 2012 = 1 60480 ( 30, 39, 39, 0, 0) C 1...C 5 X 2111 = 1 480 (1, 0, 4, 1, 0) C 1...C 5 Terms with the wrong [34] symmetry cancel: (1 π 34 ) (X 4010 + X 2210 + X 2030 + X 2012 ) = 0 Remaining part agrees with GS and RNS: ( ) K1-loop 2B2F = X 2111 = 1 480 s 13 (u 2 /ζ 3 (/k 2 +/k 3 )/ζ 4 u 1 )+s 23 (u 2 /ζ 4 (/k 1 +/k 3 )/ζ 3 u 1 )

Outline Introduction Methods for kinematic factor evaluation Superfield components Integration in pure spinor superspace Reduction to kinematic bases Application to one- and two-loop amplitudes One loop Two loops Discussion

Two-loop massless four-point amplitude A = 4 d 2 Ω 11 d 2 Ω 12 d 2 Ω 22 i=1 ( exp ) d 2 i,j k i k j G(z i, z j ) z i (det Im Ω) 5 K (k i, z i ) Fermionic zero mode integrals: K = 12 34 (λγ mnpqr λ)(λγ s ( ) W 1 )F 2,mn F 3,pq F 4,rs + perm(1234) 12 34 K 12 + 13 24 K 13 + 14 23 K 14 Kinematic factors K ij : K 12 = 4 W [1 F 2] F [3 F 4] + 4 W[3 F 4] F [1 F 2] Biholomorphic one-form, ij = (z i, z j ) = ω 1 (z i )ω 2 (z j ) ω 2 (z i )ω 1 (z j )

Four bosons (Review of [Berkovits & Mafra (2005)]) All three kinematic factors equivalent; consider K 4B 12 = 4 W [1 F 2] F [3 F 4] 4B + 4 W [3 F 4] F [1 F 2] 4B = (1 π 12 )(1 π 34 )(1 + π 13 π 24 ) W 1 F 2 F 3 F 4 4B Fermionic expansion: W1 F 2 F 3 F 4B 4 = Y 5000 + Y 1400 + Y 1040 + Y 1004 + + Y 1022 = (1 + π 23 )(1 π 24 ) ( 1 3 Y ) 5000 + Y 1400 + Y 3200 + Y 1022 Introduce symmetrisation operator: K12 4B = ( S 1 4B 3 Y ) 5000 + Y 1400 + Y 3200 + Y 1022, S 4B = (1 π 12 )(1 π 34 )(1 + π 13 π 24 )(1 + π 23 )(1 π 24 ). Non-zero correlators: Y 3200 = 1 192 k a 1 Fcdk 1 mf 2 ef 2 F pqf 3 rs 4 (λγ mnpqr λ)(λγ s γ ab θ)(θγ cd b θ)(θγ ef n θ), Y 1022 = 1 64 F abf 1 mnk 2 p 3 Fcd 3 k r 4 Fef 4 (λγ mnpq[r λ)(λγ s] γ ab θ)(θγ cd q θ)(θγ ef s θ).

Four bosons (2) Symmetriser S 4B projects out k F terms: S 4B (Y 3200 +Y 1022 ) = 1 120 (s 13 s 23 ) ( 4 Tr(F (1 F 2 F 3 F 4) ) Tr(F (1 F 2 ) Tr(F 3 F 4) ) ) Very simple result: K 4B 12 = 1 720 (s 13 s 23 )t 8 F 4 Trivially obtain K 13 and K 14. Total: K2-loop 4B ( ) = 1 720 (s13 s 23 ) 12 34 +(s 12 s 23 ) 13 24 +(s 12 s 13 ) 14 23 t8 F 4 Agrees with (relatively recent!) RNS result [d Hoker, Phong (2001)]. Note product structure, with one half being the one-loop kinematic factor

Very similar calculation: Four fermions K 4F 12 = (1 π 12 )(1 π 34 )(1 + π 13 π 24 ) W 1 F 2 F 3 F 4 4F = 4(1 π 12 ) W 1 F 2 F 3 F 4 4F = S 4F ( 1 3 Y 2111 + Y 0311 ) with symmetrisation S 4F = 4(1 π 12 )(1 π 23 )(1 + π 24 ). Two correlators have to be computed: Y 2111 = ( 2)k 1 a k 2 mk 3 p k 4 r (λγ mnpq[r λ)(λγ s] γ ab θ)(θγ b u 1 )(θγ n u 2 )(θγ q u 3 )(θγ s u 4 ) = 1 5040 ( 19, 21, 21, 19, 17, 17, 0, 0, 0, 0) B 1...B 10, Y 0311 = ( 2 3 )k 2 a k 2 mk 3 p k 4 r (λγ mnpq[r λ)(λγ s] u 1 )(θγ n ab θ)(θγ b u 2 )(θγ q u 3 )(θγ s u 4 ) = 1 15120 ( 14, 16, 0, 2, 8, 8, 0, 5, 5, 0) B 1...B 10.

Four fermions (2) Act with symmetriser S 4F and obtain the one-loop u 4 scalar: K 4F 12 = S 4F ( 1 3 Y 2111 + Y 0311 ) = 1 45 ( 1, 0, 1, 0, 1, 0, 0, 0, 0, 0) B 1...B 10 = 1 45 s 12 ( (u1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ). Total result, including K 13 and K 14 : K 4F 2-loop = 1 45 (s 12 12 34 s 13 13 24 + s 14 14 23 ) ( (u 1 /k 3 u 2 )(u 3 /k 1 u 4 ) (u 1 /k 2 u 3 )(u 2 /k 1 u 4 ) + (u 1 /k 2 u 4 )(u 2 /k 1 u 3 ) ) Again simple product structure, with one-loop factor

Two bosons, two fermions Exchange symmetry prevents simple product structure with one-loop factor: K2-loop 2B2F 1j kl [ f j (s 12, s 13 )K1-loop(f 2B2F 1 f 2 b 3 b 4 ) ]. j=1,2,3 (Note that any f (s 12, s 13 ) has identical symmetry under 1 2 and 3 4, and is symmetric under (1, 3) (2, 4)). Instead, find with K12 2B2F = (1 π 12 )(1 π 34 ) K K13 2B2F = (2 1 + π 12 + π 34 + 2π 12 π 34 ) K K14 2B2F = (1 + 2π 12 + 2π 34 + π 12 π 34 ) K K = (λ 3 W (even) 1 )F (odd) 2 F (even) 3 F (even) 4 + (λ 3 W (odd) 3 )F (even) 4 F (odd) 1 F (odd) 2

Two bosons, two fermions (2) Correlators in K : combination of ten k 3 u 1 u 2 F 3 F 4 scalars. Images of the symmetrisers are spanned by s 12, s 13 times (u 1 γ a u 2 ) ( ka 3 Fbc 3 F bc 4 2Fab 3 F bck 4 c 1 + 2FabF 4 bc 3 k c 1 ) Tensor structure thus fixed up to three constants: K2-loop 2B2F = [ c 1 s 12 12 34 + (c 2 s 12 + c 3 s 13 ) 13 24 ] +((c 3 c 2 )s 12 + c 3 s 13 ) 14 23 (u 1 γ a u 2 ) ( ka 3 Fbc 3 F bc 4 2Fab 3 F bck 4 c 1 + 2FabF 4 bc 3 k c 1 )

Discussion Summary: The pure spinor formalism yields manifestly supersymmetric loop amplitudes Superspace integration not (much) more complicated for fermions than for bosons Some amplitudes fixed or highly constrained by exchange symmetries Outlook: Improve computer algebra: custom-made Mathematica algorithms vs. specialised program (e.g. Cadabra) Methods applicable to future higher-loop amplitudes