Amplitudes beyond four-dimensions

Transcription

1 Amplitudes beyond four-dimensions HKIAS-Nov-204

2 Why? why? QFT exists in D 4 Questions rephrased in terms of physical observables Allow for application of consistency conditions: Structures based on Unitarity + Locality, should be universal.

3 Why? why? QFT exists in D 4 Questions rephrased in terms of physical observables Allow for application of consistency conditions: Structures based on Unitarity + Locality, should be universal.

4 Why? why? QFT exists in D 4 Questions rephrased in terms of physical observables Allow for application of consistency conditions: Structures based on Unitarity + Locality, should be universal.

5 Where should we look? Spinor magic:

6 Where should we look? α SL(2, C) D = 4 : Little Group U()

7 Where should we look? α SL(2, C) D = 4 : Little Group U() D = 3 : α SL(2, R), D = 6 : A a A SU(4) a SU(2)

8 Complicated questions rephrased (D=3) Is the N = 8 interacting CFT unique? Four-point amplitude: (N = 8 SCS + factorization) A 4 = δ3 (P)δ 8 (Q) It is completely anti-symmetric in (, 2, 3, 4) must be dressed with f [234] (SU(2) SU(2)) The theory is unique if it has a perturbative S-matrix

9 Complicated questions rephrased (D=6) Can self-dual tensors interact? Difficult to non-abelianize δb µν = µλ ν+???? Difficult to construct self-dual cations H µνρ = µb ρ + = (H σρτ ) ) A ) Degenerate three-point kinematics: s 2 = 0 (λ a ( λ 2ȧ A Cheung, O Connell BBB = 0, BBO = 0 = u a ũ 2ȧ Three self-dual tensors cannot interact if the S-matrix for the theory exists

10 Test of structure beyond four-dimensions Where to search for new structures?

11 Test of structure beyond four-dimensions Consider scattering amplitudes in three-dimensions: Large class of pure SCFT with different supersymmetry (same physical singularities) L = L C S + L φ,kin + L ψ,kin + L 4φ 2 ψ 2 + L 6φ 6 Both gauge and gravity are perturbation of topological theories Chern Simons Matter Gravity+Matter AdA+AAA R Rich UV and IR-physics.

12 Test of structure beyond four-dimensions Simpler amplitudes: Odd-amplitudes vanishes to all orders. Vanishing soft-limits for gravity: M n+ (,, n, s) = SG 0 Mn(,, n) + S GMn(,, n) + S 0 = n a= ( ) µa 2 µs [sa] as [sa] 3 d kin n a= µa 2 µs 2 = All bosonic states of the theory satisfy a duality symmetry See Wei-Ming Chen s talk Leading UV and IR-divergences are absent from Odd-loops.

13 Test of structure beyond four-dimensions What structures should we search for? N = 4 Super Yang-Mills The planar theory enjoys SU(2,2 4) DSCI The string sigma model enjoys fermionic self T-duality The (super)amplitude is dual to a (super)wilson-loop The IR-divergence structure captured by BDS The leading singularities is given by residues of Gr(k, n) [dc] M δ(c Z ) The amplitude has uniform trancsendentality Geometrization of locality and unitarity

14 Test of structure beyond four-dimensions What structures should we search for? N = 4 Super Yang-Mills The planar theory enjoys SU(2,2 4) DSCI The string sigma model enjoys fermionic self T-duality The (super)amplitude is dual to a (super)wilson-loop The IR-divergence structure captured by BDS The leading singularities is given by residues of Gr(k, n) [dc] M δ(c Z ) The amplitude has uniform trancsendentality Geometrization of locality and unitarity

15 Test of structure beyond four-dimensions N = 6 Chern-Simons matter theory ABJM: The planar theory enjoys SU(2,2 4) DSCI OSp(6 4) The leading singularities is given by residues of Gr(k, n) OG(k,2k) S. Lee [dc] M δ(c T C)δ(C Z )

16 Known unknowns:. The amplitudes appears to be uniform transcendental (proof?) 2. Why is the IR-divergence (Dual conformal anomaly equation) the same? Y-t, W. Chen, S. Caron-Huot A 2 loop 6 = ( N k ) 2 { A tree 6 2 A 2 loop 4 = ( ) N 2 A tree 4 k [BDS 6 + R 6 ] + Atree 2 BDS 4 6,shifted 4i [ ln u ]} 2 ln χ + cyclic 2 u 3 At four-point to all orders in ɛ M. Bianchi, M. Leoni, S Penati, exponentiation verified at three-loops M. Bianchi, M. Leoni

17 Known unknowns: 3. Why is the amplitude non-analytic? a b 5 2 a b 6 2 a b 5 tree i cyclic

18 Let us see how far can we get by understanding the amplitude through Grassmannian

19 Orthogonal Grasmmannian Consider k-planes in n-dimensional space equipped with a symmetric bi-linear Q ij The orthogonal grassmannian Q ij C αi C βj = 0 Consider n = 2k and Q ij = η ij signature (+, +, +,, +) L n = res k =, C αi = (, ±i) ( ) ±i cos z 0 i sin z k = 2, C αi = 0 ±i sin z i cos z S. Lee, D. Gang, E. Koh, E. Koh, A. Lipstein, Y-t dc ( k) (k n ) δ(qij C αi C βj )δ 2k (C λ)δ 3k (C η)

20 Orthogonal Grasmmannian Consider k-planes in n-dimensional space equipped with a symmetric bi-linear Q ij The orthogonal grassmannian Q ij C αi C βj = 0 Consider n = 2k and Q ij = η ij signature (+, +, +,, +) L n = res k =, C αi = (, ±i) ( ) ±i cos z 0 i sin z k = 2, C αi = 0 ±i sin z i cos z S. Lee, D. Gang, E. Koh, E. Koh, A. Lipstein, Y-t dc ( k) (k n ) δ(qij C αi C βj )δ 2k (C λ)δ 3k (C η)

21 Positive Orthogonal Grasmmannian Positivity: (i, i +,, i + k) > 0 Q ij = η ij signature (+, +, +,, +) k =, C αi = (, ±i) ( ±i cos z 0 i sin z k = 2, C αi = 0 ±i sin z i cos z )

22 Positive Orthogonal Grasmmannian Positivity: (i, i +,, i + k) > 0 Q ij = η ij signature (+, +, +,, +) k =, C αi = (, ±i) ( ±i cos z 0 i sin z k = 2, C αi = 0 ±i sin z i cos z )

23 Positive Orthogonal Grasmmannian Positivity: ordered (i,, j) > 0 Q ij = η ij signature (+,, +,, ) k =, C αi = (, ) ( cos z 0 sin z k = 2, C αi = 0 sin z cos z ) Positive for 0 z π/2 Volume form w. logarithmic singularity at the boundary: z = π/2, z = 0 dz = d log tan z cos z sin z d log tan δ 4 (C λ)δ 6 (C η) This is not the amplitude A 4!

24 Positive Orthogonal Grasmmannian Positivity: ordered (i,, j) > 0 Q ij = η ij signature (+,, +,, ) k =, C αi = (, ) ( cos z 0 sin z k = 2, C αi = 0 sin z cos z ) Positive for 0 z π/2 Volume form w. logarithmic singularity at the boundary: z = π/2, z = 0 dz = d log tan z cos z sin z d log tan δ 4 (C λ)δ 6 (C η) This is not the amplitude A 4!

25 Positive Orthogonal Grasmmannian Positivity: ordered (i,, j) > 0 Q ij = η ij signature (+,, +,, ) k =, C αi = (, ) ( cos z 0 sin z k = 2, C αi = 0 sin z cos z ) Positive for 0 z π/2 Volume form w. logarithmic singularity at the boundary: z = π/2, z = 0 dz = d log tan z cos z sin z d log tan δ 4 (C λ)δ 6 (C η) This is not the amplitude A 4!

26 Branches of Positive Orthogonal Grasmmannian ( cos z 0 sin z k = 2, C αi = 0 sin z cos z ( cos z 0 sin z k = 2, C αi = 0 sin z cos z For 0 z π/2 Positivity: (i,, j) > 0 and ±(i,, 2k) > 0 A 4 = d log tan δ 4 (C λ)δ 6 (C η) + (OG 2+ ) ) ) The four-point amplitude is given by the sum of two branches in OG 2+

27 Branches of Positive Orthogonal Grasmmannian ( cos z 0 sin z k = 2, C αi = 0 sin z cos z ( cos z 0 sin z k = 2, C αi = 0 sin z cos z For 0 z π/2 Positivity: (i,, j) > 0 and ±(i,, 2k) > 0 A 4 = d log tan δ 4 (C λ)δ 6 (C η) + (OG 2+ ) ) ) The four-point amplitude is given by the sum of two branches in OG 2+

28 Why Two Branches of Positive Orthogonal Grasmmannian k = 2, C αi = ( cos z 0 sin z 0 sin z cos z ) δ 4 (C λ) λ + cos zλ 2 sin zλ 4 = 0 λ 3 + sin zλ 2 + cos zλ 4 = 0 k = 2, C αi = ( cos z 0 sin z 0 sin z cos z δ 4 (C λ) λ + cos zλ 2 + sin zλ 4 = 0 λ 3 + sin zλ 2 cos zλ 4 = 0 There are two branches in the kinematics as well: 34 2 = s 34 = s 2 = = 2 ) 34 = 2

29 Why Two Branches of Positive Orthogonal Grasmmannian 3D- kinematics is topologically a circle p i = (, cos θ i, sin θ i ) p p p 2 p 2 p n p p n p 3 p4 p 5 p 4 i i+ i+ i i i+2 i i+2 n n

30 On-shell diagrams in Orthogonal Grasmmannian z z 2 z 3 z 3 Cαi(z) z 5 z 7 z 6 z4 z z 2 Are these diagrams related to A n? 2 2 δ 4 (C λ) λˆ + sec zλ + tan zλ 2 λˆ2 tan zλ sec zλ 2 n n Arkani-Hamed, J. Bourjaily, F. Cachazo, A. Goncharov, A. Postnikov, J. Trnka

31 On-shell diagrams in Orthogonal Grasmmannian Are these diagrams related to A n? A 6 = d log tan d log tan 2 d log tan 3 δ 2k (C λ)δ 3k (C η) branch No z 2 z 3 y 2 y 3 z y 6 6

32 On-shell diagrams in Orthogonal Grasmmannian Are these diagrams related to A n? A 6 = branch d log tan d log tan 2 d log tan 3 ( + sin sin 2 sin 3 )δ 2k (C λ)δ 3k (C η) Yes z 2 z 3 y 2 y 3 z y 6 6 A 6 = d log tan d log tan 2 d log tan 3 ( + cos cos 2 cos 3 )δ 2k (C λ)δ 3k (C η) branch No new singularities 0 z π/2.

33 On-shell diagrams in Orthogonal Grasmmannian In general i i 2k A n = branch dia k d log tan i J δ 2k (C λ)δ 3k (C η) i= How to get J?

34 On-shell diagrams in Orthogonal Grasmmannian A 6 = branch d log tan d log tan 2 d log tan 3 ( + sin sin 2 sin 3 )δ 2k (C λ)δ 3k (C η) z 2 z 3 y 2 y 3 f f 2 z y 6 6 A 6 = d log tan d log tan 2 d log tan 3 ( + cos cos 2 cos 3 )δ 2k (C λ)δ 3k (C η) branch J is naturally associated with faces!

35 On-shell diagrams in Orthogonal Grasmmannian J = + J + J 2 + J 3 + J 3 + J 23 J : J = J i + J i J j + J i J j J k +... single disjoint pairs disjoint triples J 2 : Two closed loops sharing a single vertex J 3 : Two closed loops sharing two vertices without sharing an edge. J 3 and J 23 : The effect of the bigger loop from J (a) (b)

36 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian The loop-level recurssion Arkani-Hamed, J. Bourjaily, F. Cachazo, A. Goncharov, A. Postnikov, J. Trnka i A l n = n 2 l +l 2 =l i=4 2 + n n

37 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian The loop-level recurssion = A l= 4

38 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian The loop-level recurssion = A l= S = 0 4 S = S = S =

39 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian Using reduction: We can separate out the d log measure

40 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian Using reduction: We can separate out the d log measure X 0 dx 0 dx 0 dx 0 dx 0 I 4 = X 2 0 =0 X 2 0 0,, 2, 3, 4 (0.)(0.2)(0.3)(0.4), (i.j) X i X j X 0 = a X + a 2 X 2 + a 3 X 3 + a 4 X 4 + a ɛ,, 2, 3, 4. I 4 = d log a 2 d log a 3 d log a 4.

41 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian Using reduction: We can separate out the d log measure X 0 dx 0 dx 0 dx 0 dx 0 I 4 = X 2 0 =0 X 2 0 0,, 2, 3, 4 (0.)(0.2)(0.3)(0.4), (i.j) X i X j X 0 = a X + a 2 X 2 + a 3 X 3 + a 4 X 4 + a ɛ,, 2, 3, 4. I 4 = d log a 2 d log a 3 d log a 4.

42 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian 6 6 A loop 6 =

43 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian A 2 loop 4 = + (i i+2) A 2 loop 6 = + (i i+2)

44 Loop-amplitude and on-shell diagrams in Orthogonal Grasmmannian The solution to BCFW is manifestly cylic i i + 2 For each cell, a single chart covers all singularities All loop: 4 and 6-point amplitudes is a product of independent d log Proved all physical sing present, spurious cancels

45 Any hint on the close tie to N = 4 SYM?

46 Embedding OG(k, 2k) into G(k, 2k) f 2 f f 4 0 f 2 4 f 3 3 ( ) /f 0 f C = 4. 0 f 2 /f 3

47 Embedding OG(k, 2k) into G(k, 2k) f 2 f f 4 0 f 2 4 f 3 3 ( ) /f 0 f C = 4. 0 f 2 /f 3 f = c, f 4 = s, f 2 = s, f 3 = c OG 2+ has an image in Gr(2, 4) + 2 C S C 2 2 S S 4 3 C

48 Embedding OG(k, 2k) into G(k, 2k) Cluster transformation: ( ) /f 0 f C = 4. 0 f 2 /f 3 2 C C 2 S 2 C 2 S S S 2 S 2 C S 4 C 3 4 C 3 c, s c, s

49 Embedding OG(k, 2k) into G(k, 2k) 2 f 2 6 f 6 f a f f 5 5 fc f 0 f 4 f b 4 f 3 (f a, f b, f c) = (c 2 /s2, c2 2 /s2 2, c2 3 /s2 3 ), f 0 = c c 2 c 3 f = c, f 2 = s s 2, f 3 = c 3, f 4 = s 2 s 3, f 5 = c 3, f 6 = s s 3 The variable for the k new faces is simply f = c 2 /s 2. Take a clockwise orientation on each face. The contribution from each vertex is /c if one first encounters the black vertex, otherwise the contribution is s.

50 Embedding OG(k, 2k) into G(k, 2k) 2 f 2 6 f 6 f a f f 5 5 fc f 0 f 4 f b 4 f 3 (f a, f b, f c) = (c 2 /s2, c2 2 /s2 2, c2 3 /s2 3 ), f 0 = c c 2 c 3 f = c, f 2 = s s 2, f 3 = c 3, f 4 = s 2 s 3, f 5 = c 3, f 6 = s s 3 The variable for the k new faces is simply f = c 2 /s 2. Take a clockwise orientation on each face. The contribution from each vertex is /c if one first encounters the black vertex, otherwise the contribution is s.

51 In terms of the Grassmannian coordinates, ABJM is at the boundary of N = 4 SYM: C T C = 0 To see the IR-divergence, we need the integrated answer

52 In terms of the Grassmannian coordinates, ABJM is at the boundary of N = 4 SYM: C T C = 0 To see the IR-divergence, we need the integrated answer

53 All multiplicity integrands for ABJM Song He, Y-t Huang: -loop: A loop n = i<j<k ( ) C + i,j,k I+ (i, j, k) + C i,j,k I (i, j, k) A tree n n ( ) i I(i, i, i+) i=

54 All multiplicity integrands for ABJM Song He, Y-t Huang: -loop: A loop n = i<j<k ( ) C + i,j,k I+ (i, j, k) + C i,j,k I (i, j, k) A tree n n ( ) i I(i, i, i+) I ± ɛ(a, i, j, k, X) (i j)(j k)(k i) (i, j, k) = ± a 2(a i)(a j)(a k)(a X) (a i)(a j)(a k) i= C + i,j,k k j Res + i,j,k I+ (i, j, k) = Res i,j,k I (i, j, k) =, Res + i,j,k I (i, j, k) = Res i,j,k I+ (i, j, k) = 0. i

55 All multiplicity integrands for ABJM Song He, Y-t Huang: -loop: A loop n = i<j<k ( ) C + i,j,k I+ (i, j, k) + C i,j,k I (i, j, k) A tree n n ( ) i I(i, i, i+) i= ɛ(a, i, i, i+, X) I(i, i, i+) = a 2(a i)(a i )(a i+)(a X) x i+ x i+ x i x a x i x a x i x i

56 All multiplicity integrands for ABJM Song He, Y-t Huang: 2-loop: -loop -loop I A (i, j) a,b I ± B (r; i, j, k) ɛ(a, i, i, i+, ) ɛ(b, j, j, j+, ) (a i)(b j)(i i+)(j j+), 2(a i )(a i)(a i+)(a b)(b j )(b j)(b j+) a,b I B (i; j, k) ɛ(a, r, r, r+, ) ɛ(b, i, j, k, ) ± 2(i j)(j k)(k i)ɛ(a, r, r, r+, b) 2(a r )(a r)(a r+)(a b)(b i)(b j)(b k) a,b ɛ(a, i, i, i+, j) (i j)(a i )(a i)(a i+)(a b)(b j)(b k). +..

57 All multiplicity integrands for ABJM Song He, Y-t Huang: 2-loop: -loop -loop A 2 loop n,soft = (C + i,j,k ( I + B (r; i, j, k) + I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) i<j<k i<r<j +C i,j,k ( I B (r; i, j, k) I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) ) i<r<j A tree n ( ) i+j I A (i, j). i,j,j i>

58 All multiplicity integrands for ABJM Song He, Y-t Huang: 2-loop: -loop -loop A 2 loop n,soft = (C + i,j,k ( I + B (r; i, j, k) + I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) i<j<k i<r<j +C i,j,k ( I B (r; i, j, k) I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) ) i<r<j A tree n ( ) i+j I A (i, j). i,j,j i> Also need to include pure two-loop leading singularity 3 5 a + b+ 7 A 2 loop 8 = A 2 loop 8,soft + Integrating... 4 i= C + i,i+2,i+4;i+4,i 2,i I+ (i, i+2, i+4; i+4, i 2, i) + (+ ) C

59 All multiplicity integrands for ABJM Song He, Y-t Huang: 2-loop: -loop -loop A 2 loop n,soft = (C + i,j,k ( I + B (r; i, j, k) + I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) i<j<k i<r<j +C i,j,k ( I B (r; i, j, k) I B (i; j, k) + {(i, j, k) (j, k, i), (k, i, j)}) ) i<r<j A tree n ( ) i+j I A (i, j). i,j,j i> Also need to include pure two-loop leading singularity 3 5 a + b+ 7 A 2 loop 8 = A 2 loop 8,soft + Integrating... 4 i= C + i,i+2,i+4;i+4,i 2,i I+ (i, i+2, i+4; i+4, i 2, i) + (+ ) C

60 Conclusion: There are rich structures for scattering amplitudes in D 4: D=3 Chern-Simons matter theory For N = 6 on-shell diagrams are applicable: Grassmannian representation reflects the topology of kinematics. The Grassmannian representation exposes its close tie to N = 4 SYM One- two-loop all multiplicity integrand under control.

61 Outlook Extension to N < 6, no new physical singularities. Unlike N < 4 SYM A (N ) n,2, = a a 2 4 N (N =4) A n,2, + (4 N ) a a 2 2 N A chiral n,2,, A chiral n,2, = D(a, a 2 )D(a 2, a ) D n(a, b) = b a{i i+}b AB ABii+ i=a What about N = 8? What is the dual object? + b i=a+ UV : Res AB I A (N ) n,k, lim AB I AB 2 A (N ) n,k,, dcδ 2 3 (CΛ) (2)(23) dcδ 2 3 (CΛ) (23)(234)(345) dcδ 2 4 (CΛ) (2)(23)(3) dcδ 2 4 (CΛ) (23)(234)(345)(35) ai bi ABi i+ ABi i ABii+