Integration-by-parts reductions via algebraic geometry Kasper J. Larsen University of Southampton Amplitudes in the LHC era GGI Florence, 29th of October 2018 Based on PRD 93 (2016)041701, PRD 98 (2018)025023, JHEP 09 (2018)024 with J. B¨ ohm, A. Georgoudis, H. Sch¨ onemann, M. Schulze, Y. Zhang Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 1 / 20
Overview Motivation 1 IBP identities on unitarity cuts 2 Syzygy equations and their solution 3 Main example: 4 Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 2 / 20
Integration-by-parts reductions IBP identities arise from the vanishing integration of total derivatives, [Chertyrkin, Tchakov, Nucl. Phys. B 192 , 159 (1981)] � v µ L L � d D ℓ i � j P ∂ = 0 ∂ℓ µ D a 1 1 · · · D a k π D / 2 j k i =1 j =1 where P and v µ j are polynomials in ℓ i , p j , and a i ∈ N . Role in perturbative QFT calculations: Reduction. Reduce number of contributing loop integrals by factor of O (10 2 ) − O (10 6 ) to basis. Computing master integrals. Enable setting up differential equations for basis integrals I j : [Gehrmann and Remiddi, Nucl. Phys. B 580 , 485 (2000)] [Henn, PRL 110 (2013) 251601] ∂ I ( x , ǫ ) = A m ( x , ǫ ) I ( x , ǫ ) ∂ x m where x m denotes a kinematical invariant. Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 3 / 20
IBP reductions on unitarity cuts Standard approach: enumerate all linear relations and apply Gauss-Jordan elimination to large linear systems [Laporta, Int.J.Mod.Phys. A 15 (2000) 5087-5159] Idea here: use unitarity cuts to block-diagonalize system � k = L ( L +1) We use the Baikov representation � , + LE 2 � d z 1 · · · d z k L d D ℓ j � N D − L − E − 1 � I ( N ; a ) ≡ = Gram p , ℓ ) ( z ) N 2 D a 1 1 · · · D a k z a 1 1 · · · z a k i π D / 2 ( � k k j =1 [Baikov, Phys.Lett. B 385 (1996) 404-410] in which cuts are straightforward to apply, � d z i � d z i cut − → i ∈ S cut z a i z a i Γ ǫ (0) i i Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 4 / 20
Example: Zurich-flag cut Let us construct IBP identities on the Zurich-flag cut Define S cut = { 1 , 2 , 4 , 5 , 7 } and G = Gram ( � p , ℓ ) . On S cut , the double-box integral takes the form � � � D − 6 � d � z i G ( � z ) 2 I DB cut [ P ] = d � z j P ( � z ) z i � z 3 � � z 6 Γ ǫ (0) i ∈ S cut j / ∈ S cut Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 5 / 20
Example: Zurich-flag cut Let us construct IBP identities on the Zurich-flag cut Define S cut = { 1 , 2 , 4 , 5 , 7 } and G = Gram ( � p , ℓ ) . On S cut , the double-box integral takes the form � � � D − 6 � d � z i G ( � z ) 2 I DB cut [ P ] = d � z j P ( � z ) z i � z 3 � � z 6 Γ ǫ (0) i ∈ S cut j / ∈ S cut Relabeling z { 1 , 2 , 3 , 4 } = � z { 3 , 6 , 8 , 9 } , this becomes � d z 1 d z 2 d z 3 d z 4 D − 6 I DB 2 P ( z ) cut [ P ] = G ( z ) z 1 z 2 Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 5 / 20
Generic total derivative on cut Need to find IBP identities which involve � d z 1 d z 2 d z 3 d z 4 D − 6 I DB 2 P ( z ) cut [ P ] = G ( z ) z 1 z 2 Total derivatives − → IBP identities. Generic total derivative on cut: 4 D − 6 � � � a i ( z ) G ( z ) �� ∂ 2 � 0 = d z 1 · · · d z 4 ∂ z i z 1 z 2 i =1 4 D − 6 � � ∂ z i + D − 6 � G ( z ) � ∂ a i a i ∂ G a j � 2 � � = − d z 1 · · · d z 4 2 G ∂ z i z j z 1 z 2 i =1 j =1 , 2 The red term corresponds to an integral in ( D − 2) dimensions, and the purple term in general produces doubled propagators. Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 6 / 20
IBP identities from syzygies To avoid dimension shifts and doubled propagators in � � � G ( z ) � ∂ a i � 4 D − 6 � � + D − 6 ∂ G a j 2 0 = a i − d z 1 · · · d z 4 ∂ z i 2 G ∂ z i z j z 1 z 2 i =1 j =1 , 2 we demand that each term is polynomial, 4 � ∂ G a i + bG = 0 ∂ z i i =1 a j + b j z j = 0 with a i , b i , b polynomials in z . Such eqs. are known as syzygy equations . [Gluza, Kajda, Kosower, PRD 83 (2011)045012], [Schabinger, JHEP 01 (2012)077], [Ita, PRD 94 (2016)116015] Obtain IBPs by plugging ( a i , b ) into the top equation. Note: ( qa i , qb ) is also a solution, for polynomial q . Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 7 / 20
Strategy to solve syzygy equations Solve syzygy equations with c cuts a j + b j z j = 0 , j = 1 , . . . , k − c (1) m − c � ∂ G a j + bG = 0 (2) ∂ z k j =1 as follows. 1) The generators of (1) are trivial: M 1 = � z 1 e 1 , . . . , z k e k , e k +1 , . . . , e m � � � 2) Generators M 2 = ( a 1 , . . . , a m , b ) , . . . of (2) for the off-shell case c = 0 can be explicitly found: � E + L � � ∂ z α ( a α , b ) = (1+ δ ik ) x jk , 2 δ ij ∂ x ik k =1 where x ij = v i · v j with v i , j ∈ { p 1 , . . . , p E , ℓ 1 , . . . , ℓ L } . [B¨ ohm, Georgoudis, KJL, Schulze, Zhang, PRD 98 (2018)025023] � � � � 3) Take module intersection M 1 cut ∩ M 2 cut Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 8 / 20
Example 1: syzygies of planar double box Set P 12 = p 1 + p 2 and z 2 = ( ℓ 1 − p 1 ) 2 , z 1 = ℓ 2 z 3 = ( ℓ 1 − P 12 ) 2 1 , z 4 = ( ℓ 2 + P 12 ) 2 , z 5 = ( ℓ 2 − p 4 ) 2 , z 6 = ℓ 2 2 z 7 = ( ℓ 1 + ℓ 2 ) 2 , z 8 = ( ℓ 1 + p 4 ) 2 , z 9 = ( ℓ 2 + p 1 ) 2 0 0 0 0 0 0 1 0 0 − 2 0 0 0 0 0 1 0 0 − 2 0 − 2 0 0 0 1 0 0 Only need to find explicit relation z = Ax + B . Here 0 2 0 2 0 0 0 0 1 A = 0 2 0 2 0 2 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 2 1 − 2 0 − 2 0 − 2 0 1 0 0 0 2 0 0 0 0 0 0 1 Set t i , j = ( a α , b ). The syzygy generators are linear in the z k t 4 , 1 = ( z 1 − z 2 , z 1 − z 2 , − s + z 1 − z 2 , 0 , 0 , 0 , z 1 − z 2 − z 6 + z 9 , t + z 1 − z 2 , 0 , 0) t 4 , 2 = ( s + z 2 − z 3 , z 2 − z 3 , z 2 − z 3 , 0 , 0 , 0 , z 2 − z 3 + z 4 − z 9 , − t + z 2 − z 3 , 0 , 0) t 4 , 3 = ( − s + z 3 − z 8 , t + z 3 − z 8 , z 3 − z 8 , 0 , 0 , 0 , z 3 − z 4 + z 5 − z 8 , z 3 − z 8 , 0 , 0) t 4 , 4 = (2 z 1 , z 1 + z 2 , − s + z 1 + z 3 , 0 , 0 , 0 , z 1 − z 6 + z 7 , z 1 + z 8 , 0 , − 2) t 4 , 5 = ( − z 1 − z 6 + z 7 , − z 1 + z 7 − z 9 , s − z 1 − z 4 + z 7 , 0 , 0 , 0 , − z 1 + z 6 + z 7 , − z 1 − z 5 + z 7 , 0 , 0) t 5 , 1 = (0 , 0 , 0 , s − z 6 + z 9 , − t − z 6 + z 9 , z 9 − z 6 , z 1 − z 2 − z 6 + z 9 , 0 , z 9 − z 6 , 0) t 5 , 2 = (0 , 0 , 0 , z 4 − z 9 , t + z 4 − z 9 , − s + z 4 − z 9 , z 2 − z 3 + z 4 − z 9 , 0 , z 4 − z 9 , 0) t 5 , 3 = (0 , 0 , 0 , z 5 − z 4 , z 5 − z 4 , s − z 4 + z 5 , z 3 − z 4 + z 5 − z 8 , 0 , − t − z 4 + z 5 , 0) t 5 , 4 = (0 , 0 , 0 , s − z 3 − z 6 + z 7 , − z 6 + z 7 − z 8 , − z 1 − z 6 + z 7 , z 1 − z 6 + z 7 , 0 , − z 2 − z 6 + z 7 , 0) t 5 , 5 = (0 , 0 , 0 , − s + z 4 + z 6 , z 5 + z 6 , 2 z 6 , − z 1 + z 6 + z 7 , 0 , z 6 + z 9 , − 2) Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 9 / 20
Example 2: syzygies of non-planar double pentagon Set P i , j ≡ p i + p j and z 2 = ( ℓ 1 − p 1 ) 2 , z 3 = ( ℓ 1 − P 1 , 2 ) 2 , z 1 = ℓ 2 1 , z 4 = ( ℓ 2 − P 3 , 4 ) 2 , z 5 = ( ℓ 2 − p 4 ) 2 , z 6 = ℓ 2 2 , z 7 = ( ℓ 1 + ℓ 2 ) 2 , z 8 = ( ℓ 1 + ℓ 2 + p 5 ) 2 , z 9 = ( ℓ 1 + p 3 ) 2 , z 10 = ( ℓ 1 + p 4 ) 2 , z 11 = ( ℓ 2 + p 1 ) 2 Here z = Ax + B with 0 0 0 0 0 0 0 0 1 0 0 − 2 0 0 0 0 0 0 0 1 0 0 − 2 0 − 2 0 0 0 0 0 1 0 0 0 0 0 0 0 − 2 0 − 2 0 0 1 0 0 0 0 0 0 0 − 2 0 0 1 0 0 0 0 0 0 0 0 0 0 1 A = , 0 0 0 0 0 0 0 0 1 2 1 − 2 − 2 − 2 − 2 − 2 − 2 − 2 − 2 1 2 1 0 0 0 0 2 0 0 0 1 0 0 0 0 0 0 0 0 2 0 1 0 0 0 2 0 0 0 0 0 0 0 0 1 and the syzygy generators are again compact: Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 10 / 20
Computing module intersections Given M 1 = � v 1 , . . . , v p � and M 2 = � w 1 , . . . , w q � with v i , w j m -tuples of polynomials. Let Q denote the m × ( p + q ) matrix . . . . . . . . . . . . Q = v 1 v p w 1 w q · · · · · · . . . . . . . . . . . . Then compute wrt. POT and variable order [ z 1 , . . . , z m ] ≻ [ s ij ] Q 1 0 � h 1 , . . . , h t � ≡ Gr¨ obner basis of column space of ... 0 1 m � �� � Selecting h i = ( 0 , . . . , 0 , x 1 , . . . , x p , y 1 , . . . , y q ), we have p q � � 0 = x j v j + y k w k j =1 k =1 Kasper J. Larsen University of Southampton IBP reductions via algebraic geometry 11 / 20
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