Steps Toward a Two Loop Graphical Coproduct James Matthew in - PowerPoint PPT Presentation

Steps Toward a Two Loop Graphical Coproduct James Matthew in collaboration with Samuel Abreu, Ruth Britto, Claude Duhr and Einan Gardi Amplitudes in the LHC Era November 2018 James Matthew Two Loop Coproduct November 2018 1 / 19

Overview Background 1 Polylogarithms One Loop Diagrammatic Coaction Two Loop Coproducts 2 Coproduct of Hypergeometric Functions Two Loop Examples Conclusions 3 James Matthew Two Loop Coproduct November 2018 2 / 19

Polylogarithms The space of Goncharov polylogarithms A given by � z dt G ( a 1 , . . . , a n ; z ) = G ( a 2 , . . . , a n ; t ) , G (; z ) = 1 t − a 1 0 possesses a mapping ∆ : A → A ⊗ H called a coaction which encodes their analytic structure via the relations ∆ ◦ Disc = (Disc ⊗ 1) ◦ ∆ and ∆ ◦ ∂ = (1 ⊗ ∂ ) ◦ ∆, and allows easy derivation of functional relations. The coaction takes the form: � ∆ G ( a ; z ) = G ( b ; z ) ⊗ G b ( a ; z ) ∅⊆ b ⊆ a G ( b ; z ) has a modified integrand. G b ( a ; z ) denotes G ( a ; z ) with residues taken at poles b , so the integration contour is modified. James Matthew Two Loop Coproduct November 2018 3 / 19

Integrands and Contours in the Coproduct This points to a structure of the form � � � � ω i ⊗ ∆ ω = ω γ γ γ i i What is the relation between the { ω i } , { γ i } ? Let Γ b be the contour from 0 to z encircling poles in b and ω b be the integrand of G ( b ; z ), then:  b = a = ∅ z   (2 π i ) | a | � b = a � = ∅  ω a = (2 π i ) | b | G b ( a , z ) b � a Γ b   0 b �⊆ a  Normalise the contours ( γ ∅ = Γ ∅ / z , γ b = Γ b / (2 π i ) | b | ), then � P ss ω j = δ i , j γ i where P ss projects onto semisimple objects that obey ∆ x = x ⊗ 1. With this � γ ω = � � γ ω i ⊗ � normalisation, the coaction is given by ∆ γ i ω . i James Matthew Two Loop Coproduct November 2018 4 / 19

One Loop Graphs One loop Feynman integrals evaluate to polylogs, so what happens when we take the coaction of such an integral? Choose a basis of one loop integrals consisting of n d D k � 1 � n � ˆ � J E = e γǫ − 2 ǫ D = 2 ( k + q i ) 2 − m 2 i π D / 2 2 i i =1 where E is the set of edges of the graph. Then if we define a new set of graphs J normalised by leading singularity, we can write the coproduct in the form: One Loop Coproduct � 0 if | X | odd � � ∆ J E = ( J X + a X J X \ e ) ⊗ C X J E a X = 1 if | X | even 2 ∅ � x ⊆ E e ∈ X The cuts are computed as residues in complex kinematics [1702.03163]. James Matthew Two Loop Coproduct November 2018 5 / 19

Example 1 One Loop Coproduct � 0 if | X | odd � � ∆ J E = ( J X + a X J X \ e ) ⊗ C X J E a X = 1 if | X | even 2 ∅ � x ⊆ E e ∈ X First example: triangle with one external mass and one internal mass James Matthew Two Loop Coproduct November 2018 6 / 19

Example 2 One Loop Coproduct � 0 if | X | odd � � ∆ J E = ( J X + a X J X \ e ) ⊗ C X J E a X = 1 if | X | even 2 e ∈ X ∅ � x ⊆ E Second example: box with two adjacent external masses James Matthew Two Loop Coproduct November 2018 7 / 19

Structure of the One Loop Coproduct One Loop Coproduct � 0 if | X | odd � � J X \ e ) ⊗ C X J E ∆ J E = ( J X + a X a X = 1 if | X | even 2 ∅ � x ⊆ E e ∈ X � Where does the deformation term a X e ∈ X J X \ e come from? Can write γ ω = � n � � � coaction as ∆ γ ω i ⊗ γ i ω with contours that encircle poles of i =1 propagators as well as pole at ∞ These contours can then be replaced with ordinary cuts, and it can be verified that P ss � γ i ω j = δ i , j by using linear relations among the cuts [1703.05064]. James Matthew Two Loop Coproduct November 2018 8 / 19

Two Loop Coproducts The generalisation of the coaction beyond one loop is non-obvious due to: Topologies with multiple master integrals and so multiple cuts for a given collection of propagators. Non-polylogarithmic integrals. Take an expression for a Feynman integral to all orders in ǫ , e.g. the graph which evaluates to ( − p 2 1 ) − 2 ǫ 1 − ǫ, 1 − 2 ǫ ; 2 − 2 ǫ ; 1 − p 2 Γ 2 (1+ ǫ )Γ 4 (1 − ǫ ) � � e 2 γ E ǫ 1 2 F 1 1 + . . . ǫ 3 (1 − 2 ǫ ) Γ 2 (1 − 2 ǫ ) p 2 p 2 2 2 We can try to break this into pieces and find the coproduct using linearity of ∆ and ∆( ab ) = ∆( a )∆( b ). We can show ∆ z ǫ = z ǫ ⊗ z ǫ ( − ǫ ) k � ∞ ζ k = e γǫ Γ(1 + ǫ ) = e ⇒ ∆[ e γǫ Γ(1 + ǫ )] = e γǫ Γ(1 + ǫ ) ⊗ e γǫ Γ(1 + ǫ ) k =2 k What is the coproduct of the hypergeometric function part? James Matthew Two Loop Coproduct November 2018 9 / 19

2 F 1 Coproduct 1 ( c ) n n ! z n where a , b and c take Consider a function 2 F 1 ( a , b ; c ; z ) = � ∞ ( a ) n ( b ) n n =0 the form s + t ǫ for s , t ∈ Z : � ω γ � 1 duu m + a ǫ (1 − u ) n + b ǫ (1 − uz ) p + c ǫ = 0 =Γ(1 + m + a ǫ )Γ(1 + n + b ǫ ) 2 F 1 (1 + m + a ǫ, − p − c ǫ ; 2 + m + n + ( a + b ) ǫ ; z ) Γ(2 + m + n + ( a + b ) ǫ ) γ ω = � n � � � We will deduce the coproduct in the form ∆ γ ω i ⊗ γ i ω by i =1 � arranging for P ss γ i ω j = δ i , j There are two master integrals for the 2 F 1 function (due to contiguous relations), and two independent contours with endpoints at { 0 , 1 , 1 z , ∞} , so the system is two dimensional. Make the selections ω 1 = u a ǫ (1 − u ) − 1+ b ǫ (1 − uz ) c ǫ du Γ 1 = [0 , 1] ω 2 = u a ǫ (1 − u ) b ǫ (1 − uz ) − 1+ c ǫ du Γ 2 = [0 , 1 / z ] James Matthew Two Loop Coproduct November 2018 10 / 19

2 F 1 Coproduct 2 With this choice of { ω i } and { γ i } , we normalise the system ( γ 1 = b ǫ Γ 1 , � � γ 2 = c ǫ z Γ 2 ), then evaluating the integrals γ ω i and γ i ω produces the expression ∆ 2 F 1 ( α, β ; γ ; z ) = 2 F 1 (1 + a ǫ, − c ǫ ; 1 + ( a + b ) ǫ ; x ) ⊗ 2 F 1 ( α, β ; γ ; z ) c ǫ + z 1 − β 1 + ( a + b ) ǫ 2 F 1 (1 + a ǫ, 1 − c ǫ ; 2 + ( a + b ) ǫ ; z ) Γ(1 − α )Γ( γ ) ⊗ Γ(1 − α + β )Γ( γ − β ) 2 F 1 (1 + β − γ, β ; 1 − α + β ; 1 / z ) Given a 2 F 1 from a Feynman integral we apply this expression, then use identities on the space of 2 F 1 s to re-express the result using Feynman integrals and their cuts. Contiguous relations are encoded in the � n ∆ n , 0 part of the coproduct. The argument of ∆ is projected onto the basis of master integrands in the first entry, with coefficients that are determined by � γ i ω . James Matthew Two Loop Coproduct November 2018 11 / 19

F 4 Coproduct 1 Now consider the function F 4 ( a , b ; c , d ; X , Y ) = � ∞ ( a ) m + n ( b ) m + n ( c ) m ( d ) n m ! n ! X m Y n m , n =0 with a , b , c and d written as s + t ǫ for s , t ∈ Z . The relevant integrand and contour are: � ω γ � 1 � 1 u m + a ǫ v n + b ǫ (1 − u ) p + c ǫ (1 − v ) q + d ǫ (1 − ux − vy ) r + g ǫ � = du dv 0 0 (1 − ux ) s + h ǫ (1 − vy ) t + j ǫ � with X = x (1 − y ), Y = y (1 − x ). But we cannot replicate the 2 F 1 construction on this integrand = ⇒ expand the integrand by adding extra factor (1 − x − vy ) w + k ǫ generated from 1 1 � 1 � x (1 − u )(1 − ux − vy ) = 1 − u − 1 − x − vy 1 − ux − vy James Matthew Two Loop Coproduct November 2018 12 / 19

F 4 Coproduct 2 General ω is now: u m + a ǫ v n + b ǫ (1 − u ) p + c ǫ (1 − v ) q + d ǫ (1 − ux − vy ) r + g ǫ (1 − ux ) s + h ǫ (1 − vy ) t + j ǫ (1 − x − vy ) w + k ǫ Obtain integrands by fixing integer parts of the exponents: m n p q r s t w ω 1 0 0 − 1 − 1 0 0 0 0 ω 2 0 0 − 1 0 0 0 − 1 0 ω 3 0 0 0 − 1 0 0 0 − 1 ω 4 0 0 0 − 1 − 1 0 0 0 − 1 − 1 ω 5 0 0 0 0 0 0 ω 6 0 0 0 0 − 1 0 0 − 1 ω 7 0 0 0 − 1 0 − 1 0 0 − 1 − 1 ω 8 0 0 0 0 0 0 ω 9 0 0 0 0 0 − 1 0 − 1 Select corresponding contours: � 1 � 1 � 1 / y � 1 � 1 − x � 1 � � � γ 1 = γ 2 = γ 3 = y 0 dv 0 du dv 0 du dv 0 du 0 0 � 1 � 1 / y � 1 − x � 1 − yv � 1 − yv � 1 − yv � � � γ 4 = γ 5 = γ 6 = y 0 dv x du dv x du dv x du 0 0 0 0 0 � 1 − x � 1 � 1 / x � 1 / y � 1 / x � 1 / x � � � γ 7 = γ 8 = γ 9 = y 0 dv du dv du dv du 0 0 0 0 0 James Matthew Two Loop Coproduct November 2018 13 / 19

F 4 Coproduct 3 Diagonalise and normalise the system, then need to reduce from the full space of 9 terms to the F 4 case: Eliminate extra factor (1 − x − vy ) w + k ǫ by putting k → 0. System develops linear relations that lower number of degrees of freedom. Implement constraints among the parameters. The diagonalised system contains terms proportional to vanishing combinations of the parameters. Result is a system depending on 4 linear combinations of the integrands, and 4 dual combinations of the contours. Coproduct encodes contiguous relations on for F 4 functions in the same way as for the 2 F 1 . James Matthew Two Loop Coproduct November 2018 14 / 19

Double Triangle Graph Consider a double triangle graph with p 2 1 � = 0, p 2 2 � = 0 and p 2 3 = 0. Taking coproducts of each term and manipulating the hypergeometric part produces: where each graph is chosen in a suitable number of dimensions. There are no deformation terms for any of the graphs. James Matthew Two Loop Coproduct November 2018 15 / 19