JHEP01(2021)205 Springer , August 20, 2020 October 24, 2020 January 29, 2021 : : December 14, 2020 : : a Revised Received Published Accepted Published for SISSA by [email protected] https://doi.org/10.1007/JHEP01(2021)205 Andrew J. McLeod, , c a [email protected] , and Cristian Vergu [email protected] c . Holmfridur Hannesdottir, 3 , a,b 2007.13747 The Authors. c Scattering Amplitudes, Supersymmetric Gauge Theory

We generalize the relation between discontinuities of scattering amplitudes , [email protected] Cambridge, MA 02138, U.S.A. E-mail: [email protected] Niels Bohr International AcademyUniversity and of Discovery Center, Copenhagen, Blegdamsvej Niels 17, BohrInstitute DK-2100, Institute, for Copenhagen Gravitation Ø, and the Denmark University Cosmos, Park, Department PA of 16892, Physics, U.S.A. PennsylvaniaDepartment of State University, Physics, Harvard University, b c a Open Access Article funded by SCOAP izing the analytic properties ofof Feynman our integrals. new We formulas carry in out polylogarithmic a examples, number in of some cross-checks Keywords: cases to all loop orders. ArXiv ePrint: and cut diagrams toarbitrary cover sequential momentum discontinuities channels. (discontinuitiesturbation of The theory, discontinuities) in and new holdsimultaneously relations at accessible. are phase-space As derived points partdiscontinuities using of where as this time-ordered all monodromies analysis, and cut per- we explore momentum explain the channels use how of to are the compute monodromy sequential group in character- Abstract: Jacob L. Bourjaily, Matthew D. Schwartz Sequential discontinuities of Feynmanthe integrals monodromy and group JHEP01(2021)205 73 36 34 68 66 74 8 – i – 18 5 17 78 31 81 72 45 15 33 21 5 39 45 11 57 54 51 59 11 loops 1 62 41 L 40 6.2.5 6.2.1 Triangle6.2.2 kinematics One6.2.3 loop Two6.2.4 loops Three loops 6.2 Triangles and boxes 5.1 Sequential5.2 discontinuities in the Sequential same5.3 discontinuities channel in different channels Steinmann relations 6.1 Bubbles 4.1 Warm-up:4.2 the natural logarithm The4.3 monodromy group Monodromies of propagators 2.2 Time-ordered perturbation theory 3.1 Covariant3.2 approach Discontinuities in TOPT 2.1 Cutkosky, ’t Hooft and Veltman F Cuts of the three-loopG triangle Massless three-point vertices C Single-valued polylogarithms D Permutation symmetry of the triangleE integral Variation matrix of the two-loop box 7 Conclusions A The coproduct from variation matrices B The monodromy and fundamental groups 6 Examples 4 Discontinuities as monodromies 5 Sequential discontinuities 3 Discontinuities Contents 1 Introduction 2 Cutting rules: a review JHEP01(2021)205 ]). 16 – ], which provide a 13 9 – 5 ], these branch cuts were 17 ]. However, these works are 22 , 21 ]. However, these integrals have also 1 ]. ’t Hooft and Veltman later gave a simple , which is appropriate in the physical region ∗ 20 ) ]. These functions are under good theoretical 4 – 1 – i – discontinuity of Feynman integrals. 2 + ]). Cutkosky subsequently gave a general formula s ( 19 F , single − 18 ) i + s ( F 1 ) = ]; however, no coaction has been worked out for the types of worse-than-elliptic i 12 − – s 10 ( F − ) i In this paper, we are interested in studying the cutting rules for discontinuities of That Feynman integrals have branch cut singularities has been known since the early Knowing the analytic structure of polylogarithms has proven especially useful in the One of the aspects of Feynman integrals that has become better understood in recent + This constraint on the symbol can also be extended to Feynman integrals that evaluate to elliptic s 1 ( F where the cut runs along the positive kinematicpolylogarithms invariants and [ theintegrals amplitude that satisfies appear in a Feynman integrals in integer dimensions (see for instance [ discontinuities: is there awith way cut to diagrams, compute as sequential theretouched discontinuities is on of for this Feynman single integrals topic, discontinuities?plicated but Cutkosky computing than and sequential computing his discontinuities a contemporaries is single significantly discontinuity. more Usually com- cuts are computed as differences propagators coalesce around thetween integration the singularities contour, (see sorelating also that the [ discontinuity the across contour these branchpropagators is cuts are pinched to replaced “cut be- by graphs” deltadiagrammatic in functions which derivation [ some of Feynman Cutkosky’smostly cutting confined to rules the [ study of a momenta. This places strong constraintsthese on integrals the produce. symbol and coaction of the polylogarithms days of quantum field theory.shown In to a be seminal associated paper with by Landau regions [ of external momenta where the poles in Feynman computation of Feynman integrals andof scattering these amplitudes, quantities as is the constrainedexample, branch by in cut physical the structure principles Euclidean suchintegrals region as can where locality only all and have Mandelstam causality. logarithmic invariants For branch are points negative, at Feynman the vanishing loci of sums of external and numerical control,systematic due way to in understand their part analyticIn structure to and particular, to the exploit identities arbitrarily symbol amongbuilding them. complicated and blocks coaction polylogarithms such [ can asintegration be boundary logarithms data. broken and Riemann down zeta into values, simpler at the cost of losing only exhibiting a variety of geometric, analytic, and number-theoretic properties. years is the classparticular, of at transcendental low functions loopin they terms order evaluate of to and generalized in low polylogarithms integer particle [ dimensions. multiplicity, they In can often be expressed Feynman integrals — integrals over Feynmantum propagators appearing field in theory perturbative quan- about calculations particle — physics are experiments.most primarily Famously, precise useful they predictions have for in beenincreasingly making the used become observable history to predictions recognized make of some science as of [ interesting the mathematical objects in their own right, 1 Introduction JHEP01(2021)205 . ]; iε iπ ± 28 , 27 ]. While ], where a 26 23 ]. Our approach was also influenced more 24 – 2 – . However, as can be observed in explicit examples, ∗ ], these relations follow from causality and express ) ]). Steinmann originally studied these relations for the i 29 30 + s . In this paper, our main goal was to come up with a pre- ( F ]. This has been worked out explicitly at one loop [ While this monodromy-based approach does not appear to be 4.2 25 2 ) = i − s ]. Later, it was shown that the Steinmann relations imply scattering ( F 34 – 31 -products (as defined in [ ]. The Steinmann relations have also been studied directly from the point of view ]. R 35 36 Steinmann-type constraints have proven extremely useful for the modern amplitude One set of constraints on sequential discontinuities are the Steinmann relations. As Sequential cuts of Feynman integrals can also be computed using the multivariate Some progress on the study of sequential discontinuities was made in [ -matrix theory, without reference to local fields and their commutators; for a review, Note that in this approach the choice for where to place the branch cut is not important. S 2 of see [ bootstrap program, which attempts to determine the functional forms of Feynman inte- called case of four localmultiplicity gauge-invariant operators; [ they wereamplitudes subsequently generalized cannot to have higher doublenels discontinuities [ in partially overlapping momentum chan- scription for computing sequential discontinuities thatthan was existing more approaches. computationally tractable originally studied bylinear Steinmann relations [ between vacuum expectation values of certain types of operator products residue calculus of Leray [ this approach is bothtionally general onerous. and mathematically More rigorous, Hodge-theoreticwe it comment approaches quickly on have becomes also thepaper computa- been difference at between developed the these in end approaches [ of and section the methods of the present in several ways.by One numerical of matrices the whose simplificationsUsing entries these is methods, are that we integer theincluding are (or monodromy discontinuities able in rational) group to the multiples compute is same discontinuities of represented channel. in powers any of sequence of channels, branch point and returns toevaluated the physical on region. a Now different thevalue function sheet and can and the be final the thought as value. widely discontinuity being used, is it the goes differenceby back between recent at mathematical the least literature initial as on far polylogarithms, as where [ the general theory simplifies conjectured. Drawing inspiration from this work,theory we (TOPT) make use to of derive time-orderedFeynman perturbation more integrals general and relations cut integrals. betweenprescription the In in particular, sequential our cut discontinuities method integrals, of clarifiesaround and the emphasizes branch role the points of importance the ratherwe of than can considering analytically discontinuities monodromies across continue from branch the cuts. physical region, In along this a approach, path that goes around a when taking cuts of cutsencounters the complex situations branch is points, morethis and complicated. reality the In condition same particular, clashes reality one with condition generically holomorphy). does not holdformula (in relating fact, sequential discontinuities in different channels to a sum over cuts was reality condition JHEP01(2021)205 , ], m 44 48 – (1.2) (1.1) 37 ]. 63 , channel by taking s + s R  -cuts k M  M k − m ) m s 1) − ( M in a momentum channel corresponding to supersymmetric Yang-Mills theory [ ) − ]) or remainder functions, defined as ratios of discontinuities in the k m M 1 60 ( – 3 – m – = 4 . We write this as m = ( s ∞ = 56 N X , are the Stirling numbers of the second kind. On k ! M k 48 ` m m s ) ]. However, similar analytic constraints and bootstrap = m ` ( 52 s ` Disc – R −  satisfies the relation m 49 s M 1) m s − ( -matrix (for example, through the construction of an infrared-finite =1 S Disc m `  , instead of studying the amplitude itself, one typically studies finite P ! ]), there remains some uncertainty over how and when constraints like 1 m = 4 65 ]). , While there has been some progress in systematically extracting the infrared- = 55 3 64 N – } ]. k m 53 { 62 , discontinuity of the Feynman integral The main practical results of this paper take the form of relations between disconti- More broadly, in this paper we set out to provide some clarity on how to think about Scattering amplitudes in Yang-Mills theories necessarily involve massless particles, The Steinmann relations were first used to analyze these amplitudes in the multi-Regge limit, where it 61 3 , th -matrix [ the left side ofmonodromies the around equation, a we branch compute point in was also pointed out that normalizing by the BDS ansatz did not preserve these relations [ the Mandelstam invariant where integrals with the usedescribe of in variation some matrices detail. and the monodromy group, bothnuities of of which Feynman we integralsm and cuts of those integrals. For example, we show that the constraints these sequential discontinuities satisfy.of We cut treat integrals this and problem at bothtime-ordered the at perturbation level the of theory level polylogarithmic (TOPT) functions.discontinuities to In of prove particular, Feynman new we integrals make relationstinuities use and between of can their the be cuts. sequential computed We systematically also describe from how polylogarithmic these representations discon- of these Steinmann relations should hold.assumptions One used goal in of the this axiomatic paperamplitudes field is in theory to mass-gapped approach. pry theories, Thus, away we some rather study of than Feynman the full integrals strong scattering directly. and compute sequential discontinuities of Feynman integrals, and to study the types of amplitudes or ratios of amplitudesthese to the types exponentiation of of remainder lower-order amplitudes.45 functions that It is Steinmann-type to constraintsfinite are content often of applied the [ S stance [ so the Steinmann relations,necessarily apply. originally Indeed, massless derived particles in engenderIn infrared field divergences planar in theories these theories. withFeynman integrals a (see mass for instance gap, [ do not properties, and factorization in certainmostly kinematic applied limits). to So processes far,where in these there methods is planar have rich been theoreticalcrucial data available consistency and checks integrability-based computations [ techniques provide are expected to extend to non-supersymmetric quantities as well (see for in- grals or scattering amplitudes from their general properties (such as symmetries, analytic JHEP01(2021)205 , ? 2 R ], but is 67 , . 66 7 . A careful treatment iε denotes the sum over all + -cuts k discusses the relation between M . We also derive similar relations A 5 discusses the relationship between B , where the sets of summed-over external i E , where the + subscript indicates that all the J s + ∈ i R -loop order. A summary and discussion of some – 4 – P L to introduce the main mathematical tools we use times, with positive energy flowing across all cuts. 4 k which is real and positive, and the Euclidean region we work through some explicit examples that illustrate ) is derived in section s 6 1.1 we use these tools to prove our main results for sequential disconti- , which we define to be the region in which all Mandelstam invariants s 5 R that appear in a given diagram are strict subsets or supersets of each J in the cut diagrams is essential to have a sensible (and correct) formula relating iε ± by reviewing first the cutting rules and then the discontinuities of integrals in both We also include in this paper a number of appendices with some technical details This is a long paper, partly because we wanted to give a pedagogical introduction to One thing that our analysis makes clear is that sequential discontinuities can only 3 not needed for thethe main variation results matrix of and thethe paper. the monodromy coproduct. group Appendix associated Appendix manifold with on a which polylogarithm it and is the defined, fundamental and group explicitly works of out the the relation between these groups literature. In section nuities and cuts of Feynman integrals.Steinmann A relations. corollary is In a section these new integral-by-integral new proof relations of the betweenbubble, the triangle, cuts and box and diagrams discontinuitiespossible up of to implications Feynman of integrals, our work including and future directions are given in section for computing sequential discontinuities.ation Here, of we polylogarithmic discuss functionsthen the and show maximal introduce how the analytic the formalism continu- discontinuitiesthe of of monodromy variation polylogarithms group. matrices. can We Our beintended to treatment computed be of using introductory these the since topics these action topics draws of have heavily not from featured prominently [ in the physics various subjects relevant for the mainand results in a uniform language.covariant perturbation We theory begin and in TOPT. sections Theseto sections understand essentially review and what prove isoptical needed theorem. the We relation proceed between in section single discontinuities and cuts, as in the tum channels. Thistheory. amounts We to emphasize a that newcuts the proof hold relations for we of individual derive the Feynmanbe between integrals, Steinmann sequential obeyed and relations discontinuities by as and individual in such the Feynman perturbation integrals. Steinmann relations must also corresponding to each cut momentumthe channel. cut in When this one channel oftake vanishes. the these Since form energies of the is sums energies not ofparticle that present, external appear indices energies in TOPT diagramsother, always TOPT graphs never have sequential discontinuities in partially-overlapping momen- Feynman propagators in these cutof the diagrams should bediscontinuities assigned and cuts. Eq.between ( cuts and discontinuities in different channels. be nonzero when there exists at least one TOPT diagram that depends on the energies between the region are real and negative, exceptwhere for all invariants are negative.ways On of the cutting right-hand the side, FeynmanThese integral cuts must be computed in the region These monodromies are taken by analytically continuing along a closed contour that goes JHEP01(2021)205 µ i /i p 1 I , we (2.1) (2.2) ∈ i D P -matrix. S ≡ . µ ) I P B ← ]. In this section , but since they are X 20 µ ( , p internal line. We do ? , where 2 . I X A th ) P shows how single-valued j iε indexes the internal lines. X integrals. Throughout this = j C + I 0 ` 2 j → s k m A at any order. However, it does ( , and − ` 1 A presents the variation matrix for A 2 k ) )] X E p k, p − ( j A q [ p ( 4 j δ Y 4 – 5 – ) d ` ) π ’s generated by the loop momenta k give some details of the calculation of cuts of the i π d (2 appearing in the two-loop ladder triangle and box L d (2 G i X ) ¯ Π z ` d Y z, and ( Z Z 2 F indexes Φ X ` X ) = i p ( in the numerators of the propagators, but include a factor of is given by a sum over intermediate states denote the momentum and mass of the ) = i denote the collective set of loop and external momenta, respectively, M A j B p m → ], and later Cutkosky described how to compute discontinuities across -loop triangle diagrams. and A L 17 ( and k A ) Im -matrix theory. Landau described how to compute the location of these branch k, p ( S j q ]. Feynman integrals are defined in terms of external four-momenta One can derive stronger results than the optical theorem by directly studying indi- 68 not include factors of per loop integral inpaper, anticipation we of take the incoming particles to have positive energy. Lorentz invariant they depend only on invariants of the form In our notation, the integer The variables while theorem, including examples where disconnectedin diagrams play [ a crucial role, can be found vidual Feynman integrals.propagators, These and integrals take are the Lorentz-invariant form integrals over Feynman This optical theorem isBy non-perturbative expanding and each follows side fromon order-by-order the the in unitarity any sum of coupling, ofnot the the provide all theorem any Feynman constraints implies diagrams on a contributing individual constraint diagrams. to Some nontrivial checks on the optical 2.1 Cutkosky, ’t HooftWe begin and with Veltman the generalizedscattering optical amplitude theorem, which states that the imaginary part of a hypersurfaces [ these hypersurfaces, using Feynmanwe integrals review with the cut cuttingimaginary propagators part rules [ of and a scattering the amplitude. relationship between cuts, discontinuities, and the three-loop and 2 Cutting rules: aThe review branch points and branchdays cuts of of Feynman integrals have been studied since the early functions can be easilyprovide constructed in details the on variation how matrixon the formalism. its permutation In rational symmetry appendix andthe of transcendental transcendental the parts. function one-loopdiagrams. triangle Appendix integral Finally, appendices acts in the case of the triangle and box ladder integrals. Appendix JHEP01(2021)205 . 2 j (2.3) (2.4) m . The to be a I = s M 2 )] (that are in . 2 k directly. They K k, p m ( j 1 , and consider all − M q external momenta [ 2 k invariants M q n n J 2 ∈ Y k /   , but the Landau equations ) 2 0 j p q . are highly interdependent. The M )Θ( ]. If the singularity is associated depends on I 2 j in the propagators. For physical K can be thought of as the endpoints s . He argued that the discontinuity ) 20 ∪ m 0 p J 0 ( L M − go on-shell, then the discontinuity is → J 2 j → M L J q iε ( Disc δ iε ∈ ]. Their approach sidestepped the Landau ) = j πi 22 2 , M – 6 – − J 21 ( L J are real, but we can analytically continue ∈ Y I j s   Disc d K ` ) L k π d d . In 1959, Landau derived a set of equations whose solutions (2 i Disc M A simplified treatment of cuts and discontinuities was provided ` Y independent quantities, while there are ]. The Landau surface may be disconnected, but each connected Z n 17 [ 4 . Then the singularities as may become singular as are easiest to derive using their expression in terms of four-momenta. I = ) going on-shell is given by where the propagators s I J s J M M L J L Shortly after Landau’s paper, Cutkosky gave a prescription for computing the Disc Landau surface Unfortunately, Cutkosky’s results are phrased entirely in terms of discontinuities across The integral ’t Hooft and Veltman. in the 1970’s byequations ’t and Hooft analytic and continuation entirely,start Veltman to [ with provide the a Feynman constraint graphpossible on associated colorings of with the the vertices Feynman of integral this graph as either black or white. The following rules given set of propagatorsbubbles in depends the only on space a of singleidentify external independent a kinematic different invariants. invariant branch For hypersurface example,Thus, when a the Cutkosky’s string propagators formula in of gives differentchannel, bubbles a no are central constraint cut. focus for of this sequential paper. discontinuities in the same This is the type of sequential discontinuity we focusregions on of in this the paper. Landaugenerally not surface possible where to particular isolate a propagators region go corresponding on-shell. to the singularity However, locus it of is (just) a Cutkosky also considered theacross singularities a of region Disc ofthe the complement Landau of surface associated with a set of propagators with the region given by as the component corresponds to some set of propagatorsCutkosky. becoming singular: discontinuity across one region of the Landau surface [ momenta the Mandelstam invariants function of complex of branch cuts oncontinuation) a associated Riemann to surface (moreindicate generally the a regions hypersurface of of momenta maximal where analytic these singularities may reside, collectively known instance, in four dimensionsand a hence Feynman (at integral number most) of independent invariants ison-shell condition further for reduced each by externalconstraints momentum on particle. conservation the and Thus, the the denotes a sum of external momenta. These invariants cannot all be independent. For JHEP01(2021)205 ) M 0 p (2.5) (2.6) (2.7) ) can is the , with , ◦ 2.7 )Θ( . M . It does 2 L iε , where M m M − − M 2 j 2 1 − m p M − cutting rules ( − 2 j ) is agnostic to the πiδ q 2 . 2.3 − Y iε ◦−◦ ≡ ) − 0 covariant as an analytic function, so q ) is derived algebraically, as 2 j ) we note that the covariant ) results. m 2.7 M )Θ( 2.3 2.1 2 j − 1 2 m )] (where the numerators and vertices − ? 2 j ) is the imaginary part of k, p q ( ( iε M ), and the numerators of cut propagators j δ 2.7 q iε ) propagators. Since the Feynman integrals [ iε + around a branch point, this has to be done 2 j − πi j iε 1 2 2 Y m M − 1 − propagators, while eq. ( → − d – 7 – m ( ) − ` π iε 2 − j iε k q d Y 2 + •−◦ (2 − d i p × − Y •−• ≡ d and ` ` ) Y k π iε d Z ) is the discontinuity across a Landau surface defined by the d (2 + i 2.3 ` ) = Y p ( . Although we would like to view Z ◦ M L iε M 1) + − 2 ( 1 ◦ m , • X − 2 is related to the discontinuity of = p ≡ propagators to an integral with all M M propagators, while the graph with all white vertices corresponds to iε iε Finally, because the ’t Hooft-Veltman derivation of the cutting rules builds on a single Third, if we compare to Cutkosky’s formula in eq. ( Second, even in a non-unitary theory the covariant cutting rules relate an integral with There are a few important aspects of this equation to note. First, the covariant cutting is related to the complex-conjugated integral + + M − are then assigned to the edges between these colored vertices: cut propagators while thehowever left-hand make it side difficult of to eq. explicitlybe verify ( verified Cutkosky’s in equation. a In straightforward contrast, manner eq. in ( any numberconstraint of among examples. all the diagrams (the largest time equation), it is hard to break it down that Im with some care. The covariant cutting rules directly letcutting us rules compute involve only mixed pole positions ofsince the left-hand-side propagators. of eq. ( This does not make the two equations inconsistent, cross section, and the generalized optical theorem in eq.all ( we consider have alldirectly the compute other Im sources of imaginary parts stripped out, the cutting rules rules (like Cutkosky’s rules) doa not constraint require among unitarity. integrals Eq.M over ( propagators andare delta complex functions. conjugated in In additioncorrespond to a to unitary a sum theory, over physical spins. Then the sum over cuts gives the total scattering where the sum isnumber of over loops all connecting diagrams exclusively white with vertices. mixed black and white vertices and of these rules, ’t Hooftwhite and and black Veltman showed vertices that is the zero. sum This over implies all what possible we assignments call of the is defined by Propagators connecting black and whiteon-shell vertices are and said positive to be energy cut, flows meaning from these lines black are to white. Using the position-space version The graph with allall black vertices is the original time-ordered Feynman integral JHEP01(2021)205 ) ) ). 2.8 2.9 2.8 (2.8) (2.9) (2.10) . n C 1 − , then eq. ( n ) are related by 2 j A j p ( C ]). For example, the ··· ) reduces to eq. ( 1 propagators and delta πiδ 65 2.9 A 2 and iε − propagators and one with + j from each propagator with − , with its energy subtracted q = B iε , 1 ω ··· depending on the number of j . M j 2 + ) , the left hand side of eq. ( C + A 2 j j p M n C ( ] (see also [ B + and 69 πiδ , ··· 2 and iε M j 3 ) and three-momentum is conserved at 1 − − j 2 2 j C B p or 2 B m = , = j C ) can be applied, but even then it produces = ), we use TOPT. Recall that covariant Feyn- + j 1 also acquires a iε propagators, some A M j 2 A B conserved at each vertex. The positive sign is 2.9 – 8 – ). 1 − 2.9 ~q B M iε + 2 j − , p . + p 2.7 j n iε not M − iε A 1 = + B − 2 j + p q iε ··· ω = 2 1 on the right hand side, and so eq. ( + j = B j 2 j 1 A p 0 B q C the number of vertices. In a time-ordered diagram, the internal or = v j n A B ··· 1 B − , and loop three-momenta are integrated over. A detailed discussion, and a q n . In general, with this choice of A , there are no iε propagators, which is either 1 + ··· 2 j p = 1 1 iε To derive the cutting rules using eq. ( For example, if we take n A − = TOPT diagrams, with ! contributes an energy denominator tofrom the the TOPT initial amplitude energy, along withmomentum a derivation of the TOPTscalar Feynman loop rules, can is be given written in as [ kept separate and no energyv integral is introduced.lines Each are Feynman diagram on-shell is (meaning each the sum vertex, of but energyalways taken is for in the energy general (inFock-state front elements of of the physical square root), on-shell because positive-energy all particles. intermediate states Each are intermediate state some combination of propagatorsfunctions with with no clear relation to eq. ( man rules aregenerate derived expressions for in time-ordered terms of products Feynman by propagators. inserting In extra TOPT, the energy time-orderings integrals are to To be clear, thisM is an identity incorresponds the to sense the of difference distributions;all between it an is the integral with cuttingloops. equation all For for an even number of loops, eq. ( corresponds to the familiar relation then For To prove the cuttingfollowing rules simple mathematical in identity. time-ordered If perturbation some functions theory (TOPT) we exploit the might be possible, we findwhere it more the transparent cutting to work rules incuts can time-ordered and be perturbation discontinuities theory derived more in straightforward. a way that makes2.2 generalizations to sequential Time-ordered perturbation theory further to derive constraints on individual Feynman diagrams. Although such a dissection JHEP01(2021)205 2 2 0 k ), m k − + , 2.9 p . 2 (2.13) (2.11) (2.12) # p ) ω k ~ iε and − k + 2 ~p ) + ( k p − ω q p ω = propagators gives + 2 k ~ + k k − − k iε ]). Keep in mind that − p 1 . ~p propagators gives the p ~p ω − and those after the cut ) ω ω j 70 , iε , iε 2 1 ω iε ~p + + 69 m + + k − k ~ 2 2 ω j + ( E m 2 ( + − k ~ 1 − p q 2 πiδ E ) 2 ~p k = − + k − ) can be verified by performing the ω = iε p k ~ integration contour, it cannot be removed ( k ~ 0 − iε 2.11 iε k and ) + originates from the fact that particles move ~p k − µ + − j iε p p factors which impose energy conservation at an 2 1 – 9 – 1 ω ~p + integrals to reduce Feynman diagrams to TOPT ω 1 ) m j 1 − 0 + ω k j − k 2 E − ω = ): k ( j − 4 E ) − ; in the second diagram, the intermediate state includes ( k π 2.10 k iε 4 δ p p d − (2 E p + i " is the energy of ω j k 1 2 k ω Z + k − − m p − 1 k = p ω j ω + ) all the propagators before the cut have , while the sum of these diagrams with all 2 E 2 k ~p 2.9 M p p 1 ω p 2 is necessary to determine the 3 = ) iε k k 3 π p + k − d ω (2 p = . Thus the cut TOPT diagram is one particular time-ordering of a white/black Z p iε p − E − = Now for each term in the TOPT decomposition we can apply the identity in eq. ( It is often difficult to perform the . The remaining terms have in TOPT). By eq. ( have partition, which is onecut time-ordering TOPT of diagrams a gives cutand all Feynman therefore the diagram. reproduces possible The the time-orderings sum full of cut of the Feynman all black diagram possible and and confirms white the vertices, cutting rules. The sum of allFeynman TOPT diagram diagrams with aM given topology andintermediate all time. These diagramsautomatically neatly flowing split across in the two cutintermediate along (because time) the TOPT and cut, diagrams have where with positive positive all energy energy cut at particles any are on-shell (since all particles are on-shell using the TOPT analog of eq. ( theory, since both compute the same time-orderedalthough products the (cf. [ after the integration isforward done. in Indeed time the withpropagator positive in energy TOPT. and is an essential part of the Lippmann-Schwinger (where time flows to thelines, right). In so the its first energy diagramalso this is the state energy includes of only the the initial and final states,diagrams. and Their thus equivalence is its easiest energy to show is from more general principles of quantum field The intermediate state in the TOPT diagrams changes as each vertex is passed in time where are the energies ofintegral, the which virtual picks up particles. two of Eq. the ( four poles. In terms of diagrams, we have JHEP01(2021)205 ) ) and 2.9 2.9 (2.16) (2.17) (2.14) (2.15) . ) 0 0 . k . iε iε . )Θ( − 2 2 − n ) gives what we 1 n m ω 1 ω )) − − k 2.9 )) = 0 ) − k 2 2 1 − n 0 p − n k m E p ω ( ) E ω δ j − + ) ω ··· 2 + k ··· πi k k − ω iε ( 2 ( ω iε j δ − ) − E − + − )( ( πi p 0 p δ 1 +1 2 ) k j 1 E ω ω ( 1 − ω πi δ (2 Θ( − )( ), there is only one intermediate state 2 ) 0 − 1 − k − × πi ( E p 2 +1 − 2.12 j E iε − ≡ ( k E ( δ + k ) − × M 1 − p p πi 1 − ( j 2 – 10 – ω 4 1 2 ω ) to eq. ( − δ ( 3 k , − ) k 1 . ω 2.9 1 j π iε − 2 p 1 − ω 3 j + ω ) k : ) (2 2 E 3 π n d k 1 πi ω and (2 1 2 ω ··· j 2 − − Z E ( 3 n iε ) k 4 = E 0 3 π ) + d k π 1 (2 4 ··· 1 ω d (2 i Z iε − 1 Z = + E 4 1 ) k 1 ω π j 4 X d − (2 i 1 = E Z time-ordered cutting rules M ≡ = − In fact, we have derived something stronger than the covariant cutting rules: the More broadly, the key reason why the cutting rules can be derived diagrammatically in For example, when we apply eq. ( M M When the loop momentathis are equation integrated holds over, for this any equation implies the cutting rules, but Then, by putting incall the the explicit form of the TOPT propagators, eq. ( it holds point-by-point in phase space.holds Indeed, at the equation the that we integrand usemomenta level. to as prove it, Let eq. ( us define an individual TOPT integrand for fixed loop particles. So adiagram cut, in two, which ordered replaces byjust time, a opens in up TOPT contrast a propagator to loop. by Feynman diagrams, a where delta usingconstraint function, eq. on ( the splits amplitude the holds for each time-ordered Feynman diagram separately This is typical ofwith TOPT vertices graphs: in reversed when time order one cannot time-ordering be can cut. TOPT be is that cut, cuts the in TOPT same are associated diagram with internal multiparticle states, not individual So this diagram alone givesis the cut zero, of since the energy Feynman conservation diagram. at The the cut cut of is the impossible other diagram to satisfy: in each diagram to cutpropagators (in to cut). contrast to Cutting the the Feynman first diagram, diagram which gives has two intermediate JHEP01(2021)205 are can along M M , which M and depends. As into a regime region M M M around this branch . More specifically, we M s . For example, suppose are separated by a branch is a Feynman integral with M propagators, multiplied by a M from the positive or negative M are complex conjugates of each iε 0 ), − , viewed as an analytic function . Then we can continue and across some region of its Landau M 0 → 2.6 M M M iε and s > ) and ( M , which can be computed using the covariant ). The second concept is the discontinuity of 2.2 before and after analytic continuation along a M 2.7 – 11 – (but no other branch points). Since Mandelstam M starts at a branch point (more generally, a branch , and take s ) M − iε M − s and − can be thought of as the discontinuity of a single function M is the same integral with all M . From this point of view, = ln( M iε , as this will allow us to connect the total discontinuity computed ). As defined in eqs. ( and M M 2.7 M and ) as the difference between . At any real phase-space point, iε L , and may have a finite difference as + M 1) s s − = 0 ( − propagators and iε iε The branch cut between There are two related concepts that we will discuss, and which we want to connect. = ln( + . We would like to understand the analytic continuation contour along which hypersurface) somewhere in the spacesuch, of the Mandelstam discontinuity can invariants on bepoint which computed to by analytically the continuing otherwhere side it of is the analytic, branchM and cut. then To to do the this, region we where can it continue matches difference between M be transformed into by the covariant cuttingMandelstam rules invariants. to the notion of discontinuities with respect to particular factor of other for finite values of cut at direction. In contrast,evaluations of viewed the as same function an at analytic different points function on of a complex the manifold. momenta, Thus the finite 3.1 Covariant approach We begin with the totalcutting discontinuity rules in eq. ( all compute the total discontinuity usinga eq. Feynman ( integral with respect todefine a Disc particular kinematic invariant path that encircles the branchinvariants are point in not all independent, this has to be done with some care. The first isis the computed total by discontinuity the ofof covariant the a cutting signs Feynman rules. of integralare the A in incoming Mandelstam region and a invariants, in which particular and this are the context outgoing), signs is if of necessary. the the specification Once energies the (which signs particles are specified, we can surface by summing overHowever, to integrals provide in practical which constraints onexplicit different amplitudes than sets we Cutkosky’s. need of a For propagatorsare prescription example, much probing have how more been from do knowledge cut. wewe of identify actually which what perform Feynman region the propagators of analytic have the continuation been around surface cut? the we relevant And branch how points? do Having understood the cutting rulesperturbation in theory, covariant perturbation we theory canplitudes. and now in As time-ordered proceed discussed toof above, Mandelstam connect the invariants, cuts Feynman is toshowed integral a that the multi-valued one discontinuities function can of on compute am- a the complex discontinuity manifold. of Cutkosky 3 Discontinuities JHEP01(2021)205 4 is = (3.3) (3.1) (3.2) , the M M . The j on the iε tot + M , is analytic. 2 Disc i for all . , which them- j ? = 2 p 0 M T R 2 , these relations will ≥ M 6∈T 5 X j s is now manifestly a 2 j h m M = . ) 2 2 T . From this region we can L − and ( 2 0 L P 2 2 − n T − is n F s < U F
]. This is done by using Feynman j , iε, s x for all 71 i , or keep going to arrive at j − 0 s x X 2 j . Since the integration region corresponds 1 iα to indicate a region in which all positive < − m − 6∈T j Y ) e . 1 j = 0 + 2 x 0 h T R δ . We can also continue increasing the phase of ( F → s 1 =1 – 12 – P j X j T s X s iα s > e dx is U = j U to the region where → Y , which correspond to tree diagrams that connect all U ) + 1 ) s 0 π using T 2 ≥ T Z j ( ≤ x P M s α − ) = ( ≤ p i ( j 0 x increases, we end up on higher and higher sheets of the Riemann ) can be written as M can only arise when 2 . In equations, for the logarithm we have Disc α 2.2 6∈T Y in the region where j M with = 0 h s s . A single discontinuity corresponds to the single monodromy around the M 2 ) iα T X s e Im − = i → and in the Euclidean region are the masses of the internal lines and the sum is over 2-trees ln( F s 0 = 2 j m ≥ M . We use the more precise notation i We denote generic regions, in which kinematic invariants can be positive or negative, Singularities in A useful concept for studying the analytic properties of Feynman integrals is the Eu- Due to momentum conservation, the Euclidean region will not exist for all Feynman integrals. However, R x 4 in this manner: as invariants are slightly above the associated branch cut, i.e. allwhile the propagators Euclidean have regionnot is useful depend for on motivating theallowed the to existence relations be we positive of in derive this theseare in relations region. being section in addition taken. to In the particular, Mandelstams with any respect number to which of discontinuities Mandelstam invariants will be Note that the Euclideananalytic; regime it is requires identified that with allwith a Mandelstam 2-trees stronger invariants from are requirement a negative, than particular not that graph. just those We denote associated theby Euclidean region by original graph. The nicefunction thing of about Mandelstam invariants. this parametrization is that to denominator will never vanish and the result will be analytic in the external momenta. where selves correspond to pairs of disconnected tree diagrams that involve all vertices of the Here, the first Symanzik polynomial where the sum isvertices over in all the 1-trees graph. The second Symanzik polynomial amplitude as in eq. ( M − clidean region. In thisTo region, see all that Mandelstam integrals invariants areFeynman are negative integral analytic and in in the theparameters Euclidean Symanzik and region, representation then it [ integrating is over helpful the to loop write a momenta. general The result is that a Feynman either go back andother reproduce side of the branchs cut using surface of branch point at the path JHEP01(2021)205 to R (3.4) ). Then the iε , this rotation u − (where all Man- − due to the additional − t − as there are no energies R − R iε = . in s − L → − R 1) iε . Eventually, at some point − ( 1 − M all take non-exceptional values, + R R α < has been changed to M iε = with + R j i αE Thus, we write M 5 . – 13 – → on the opposite side of the real energy axis. This to the corresponding region − j + . There are many ways to do this. The precise path R M − E h − R = 1 R = α R ] . We construct a number of similar paths for the examples , and M ? . Unfortunately, since the invariants are not all independent, 1 while leaving the negative invariants stationary. This puts us tot R − , R + and the amplitude is nonsingular. We can then keep going, and . Disc R [ 0 6 , will be denoted and return to < j iε < α < π I E − from being positive to negative while holding the other invariants fixed, s 0 s min is Lorentz invariant, it may seem most natural to continue the invariants α with , to end up in M π I 2 s iα e , where all all the invariants become negative. One can then rotate the energies in the complex ? Let us now assume that an appropriate analytic continuation in the energies has been In general, there are many ways to rotate external energies to get from a region In this paper, we will restrict ourselves to analytic continuations in external energies Since With Feynman propagators, the amplitude in this region also has a → R 5 < α < min I chosen, which takes us fromdelstam a invariants have region the same sign, but each rotation of the energies inin the the loop loop integrals. integrals. With TOPT, we simply flip procedure respects energy-momentum conservationhas everywhere to along be the careful path.example that involving One the three only momenta invariants that do follows not aparagraph path encircle is homotopic their to shown branch the in one pointswe figure described twice. in consider this A in concrete section the Euclidean region. Forone example, can if uniformly the lowerα momenta the in energies plane around that respect overall energyaddition conservation to and avoiding leaverelations the all between issue external sequential described three-momenta discontinuitiespredictions. fixed. above, and In cut this In addition, integrals, choice rotating andtion facilitates the ensures leads energies our that to while we derivation unambiguous respecting always of satisfy four-momentum any conserva- Gram determinant constraints. we could try theintegral to above depend analytic just continuation onwould path. the seem other to But invariants have using ifthat no the respects we relation effect. the rewrite Thus, reparameterization our one invariance amplitude of must the or be integrals. careful to do the rotation in a manner in analytically continue all the0 invariants that were originallythis positive procedure further can by be extending want ambiguous. to For rotate example, in massless four-particle kinematics, if we should not affect thethat answer the for path exists, the andare discontinuity. encircled. having an It explicit is path nevertheless can important help to determine which know themselves. branch For points example, wes can rotate all the positive invariants to negative values via propagators have To compute the right handthe side, amplitude we between would like to understand how to analytically continue region in which all positive invariants are instead below the associated branch cut, and all JHEP01(2021)205 , . j E M )] = 0 from (3.5) (3.6) (3.7) 2 1 πs p M cos( 40 iπs e . 9 20 . . The small gaps , where we have |− iε + 0) 0 , R 1 + 0 0 , and the cuts of . , M 8 [0 s M − s -20 ) |− , → 0 7 + p j R X − ( -40 θ E cuts in ) M 2 = ( circles its branch point at = m -60 2 1 X cuts , we must show that the analytic 3 |− p p − for all propagators after the cut, as + , while the other invariants remain = 2 M R -80 iε s iε and all other kinematic invariants are p using the covariant cutting rules: − ( . − M R 0 − R iε 0 s , and πiδ ± -100 60 40 20 X 2 cuts 0) s > -60 -20 -40 , − M 0 = , + 6 s − corresponds to analytically continuing → − R – 14 – , R is positive in this region, all the nonzero cuts in s iε M s in this region encircles a branch point in only in + = = (3 − M 2 tot + 1 s 2 + m R . We rotate the energies by . p ] R , but changing the sign of the corresponding -channel. As a result, we have 15 − 1 |− , s 5 M M 2 − + R 0) p ). When we cut a set of propagators, we replace each one by , R tot = 0 = : → , to being evaluated at s 2.7 ? 2 3 2 R with respect to , p ] R Disc cut iε [ before and after this analytic continuation should match the total M 0 → + M 1 = (4 s M tot ) are in the , and R 1 27 p 3.5 . As only the invariant − Disc [ . During this rotation the positive invariant s 1 for all propagators before the cut and R = -5 2 2 iε . Example analytic continuation involving three external energies. We start at the s < + , p ≤ 0 We would now like to derive a concrete relation between Disc = 12 5 0 -10 , and in no other invariants. This turns out to be easiest to see in TOPT, which we 2 1 -5 To further connect thiscontinuation sum corresponding of cut to integrals Disc s to Disc turn to now. being evaluated at unchanged. Let us denote thenegative region by in which the sum in eq. ( and use implied by the subscript on The discontinuity of We emphasize the right sideas of explicitly this given equation in involves eq. a ( sum over all cuts (in all channels), thus taking us from at the beginning and end of the paths represent the difference between discontinuity of a Feynman integral in the region Figure 1 kinematic point p with JHEP01(2021)205 and does (3.9) (3.8) if the 0 (3.11) (3.10) P 0 > E for some 0 . . < 3 . q 2 j ω 2 iε iε . p m + . We use the 2 3 2 + ) + ) + E E ω 3 3 2 + j ω ω ~q 3 + : + + E 1 √ 0 → − as incoming lines with 2 2 ω − = ω ω 3 5 3 > 1 1 ( ( + p j E i E 5 ω − − E − E ) ) 1 ). 3 3 and while in the second propagator + E E 2 2 3.8 2 p ω ,E + − E 3 4 5 3 + 1 1 p p p E 1 + E E ω ( ( − 1 5 E = iε iε E q appearing in the amplitude have a natural − ω 1 I ) + ) + corresponds to a sum over external energies, 2 2 E P ω ω and E ,E – 15 – + + 1 1 3 5 1 1 E E ω ω . If we had drawn + ( ( − 2 2 1 ω − − E 1 2 , where 1 1 p p + ) values as we move forward in time is = 2 E E ,E iε 5 1 ω P and the energy past the first vertex (i.e. the energies of the + = = E 2 3 E E = p p 2 q − = q ω E 1 3 ω q P − + E 1 P ,E 5 E and E 2 ( E ; this first propagator depends on the difference between these energies 1 2 q / j q 2 − 1 3 E p ω 1 p is a sum over particles in internal lines, where = j 3 1 ω 1 p p ,E j E 5 . The sequence of P 5 E − E 1 = − ). External lines, however, have no such restriction; they can have E − q 0 ω For a general TOPT graph, the energies Consider for example the one-loop TOPT graph, with all Because we are ultimately interested in the analytic properties of Feynman integrals = = ≥ P P 2 The initial energy is states that cross theinternal first energies vertical dashed line)E is sequence. We begin withconnecting the to total an initial-state external energyis momentum on incoming, is the or far passed, subtracted, left. the ifnot external change. it Each energy is time For example, outgoing. a is consider vertex either If this added, the graph: vertex if is it purely internal, then The value ofconvention the that diagram all lines is have the positive energies, same as since in we eq. ( have flipped In the first propagator, E negative energy, the diagram wouldfound have an been equivalent more expression: awkward to draw, but we would have as functions of externalpropagators energies, from it internal is and external helpfulgator lines. to in In the separate form particular, out we theand can contributions put to each TOPT TOPT propa- In TOPT all internalq lines are on-shell withdiagram positive is energy meant and to realin be deep-inelastic masses embedded scattering ( is in spacelike), aenergy and if larger incoming they external diagram particles correspond (for can to example, have outgoing negative the particles. off-shell photon 3.2 Discontinuities in TOPT JHEP01(2021)205 , . . ? I ) q 0 so 1 E ω 5 E > E = (3.13) (3.14) (3.12) I . The − s 4 . So we but can I 2 j | E ~q E at the end I ← before and j ~ , and P ? I − 2 4 , P E 3 . M ) ≥ | q E ,E = ← iε 3 ω ? I I − 3 E E ~ − − P 4 n I E ← ω 1 E − around the pole of this ( 1 − I . Thus there is a one-to- = n µ i E πiδ ← I 5 p 2 E , namely E 5 ± 6 − q I ω ∈ ··· is the same as the sum of the = i , which only happens if ) q . To take the discontinuity in the j I must be timelike when P ω iε ) , with the same three-momentum. ~ ω P . ) µ I iε = I − = 4 − ~ P P q µ around the branch point I I + j , 1 using a path that encircles the branch ω I → q I P q in any TOPT propagator corresponding E E and these energies. A TOPT propagator ω ω 2 E − ( R iε 2 δ I I − , since it corresponds to the sum of four- ) → + = ( P 0 E I → 3 πi µ E ? 2 = − q ( > – 16 – / R → − I ( 1 iε 2 s 1 q and → + ··· → ) q I R 1 for a given external momentum. This branch point ω 5 iε ~ . Thus the TOPT propagators can go on-shell only in P q q , − . + ω I ω I 1 5 E , we want to analytically continue E ω 1 6= I E − I = ( = 1 E µ I I ) would be positive. The corresponding sequence is iε P E + should have all the energies pass around their branch points, holding is the energy of a four-vector j q 3.11 then must be timelike, X − 1 I ω 0 can only become singular when µ R E = − q ) is the sum of the (positive) energies of the on-shell internal lines. Thus, < I iε M q E 2 I and ω . As a corollary, we can drop the + tot P 0 associated with tot + q I R ω > = s Disc I I − Disc s s I Taking the difference between a single TOPT propagator before and after this analytic Now let us discuss how to take the discontinuity of a TOPT graph. A TOPT graph Each energy Note that we do not require the external masses to remain fixed, so in general external particles will E 6 ( / points in all of the energies, we get not remain on-shell during this analytic continuation. continuation gives as expected. Similarly, taking theafter difference analytically between a continuing generic from TOPT graph is at least asbe large strictly as larger, the for magnitude example,between of if the the momentum internal in linesthe the are three-momenta channel, massive. fixed and The respecting analytic energy continuation conservation. is a product of propagatorschannel of the form propagator. More precisely, we wantof to continue the line of possible values of momenta of physical on-shellSo particles. if Therefore, the kinematical regions wherewhen there are singularities into the a full negative Feynman integral, invariant. namely To check this claim,three-momenta of note all that the thehave internal two particles three-momentum four-vectors, contributing to Recall that the four-vector fact that the energies appearingthat in appeared each the successive preceding propagatorSteinmann propagators are relations (or a in vice subset section versa) of will the be energies important toone proving correspondence the between invariants 1 all the signs in eq. ( If we use energy conservationthe to sequence rewrite would the be energy the sum (i.e. same, in the opposite direction: If we took all momenta to be incoming, then we would flip the sign of JHEP01(2021)205 . , s iε R M − and ? (3.15) . R s there can R to s s + R R . |− + is only defined for R ) . This is what we set s M s in the region s − Θ( M | X notation is not sufficient to s | ), in the region tot cuts in 0 iε ln = ± > πi |− 2 + P R The = 7 M ) = 4 s s . Let us now focus on the region ) using TOPT. ( R 2 3.5 X . What is then the right way to cut a ln and all cuts iε 0 = − – 17 – > is positive, and all other invariants are negative. s R 0 to s ] . In other words, as we continue from P s ) holds with the left-hand side explicitly written as a iε M + 2.7 tot prescription. We also discuss how these types of analytic Disc iε ± = [ ) hold in any region s R ] 3.14 M s Disc , where the branch cut is. In addition, when we take a cut, we turn all the [ is homotopic to the path used to compute Disc s ) and ( , we can only pass around branch points associated with 3.5 M s − s notation in Feynman propagators is sufficient to compute single discontinuities R iε ± We would next like to generalize this formula to the case of sequential discontinuities, in Eqs. ( While more general types of functions are known to appear in Feynman integrals, we leave these 7 be analytically continued beyondcan the be related principal back branch,continuations to can and the be how carried the out resulting on TOPT functions propagators. generalizations to future work. of Feynman integrals, because thisvalue first of discontinuity computes the the integralwe difference between must on the explore different a sidesfunctions larger that of appear swath a in ofdescribe branch a the this given cut. analytic structure. Feynman integral. structure Thus, For in of this sequential the section discontinuities, various we review polylogarithmic how polylogarithmic functions can 4 Discontinuities as monodromies The negative real propagators beyond the cutpropagator? from To proceed, wetools will that now will describe allowThis us a will to more allow analytically sophisticated us continue to set Feynman take of integrals sequential mathematical beyond discontinuities the of Feynman cut integrals. plane. the same or different channels.allowed Unfortunately, us we to cannot simply compute repeattakes the the the first procedure difference discontinuity. that of The twoon problem functions is the on that the branch this branch cut cut, first itself. and discontinuity thus seems For to example, be Disc only defined Stated more formally, what wepute have Disc shown is that the analytic continuation used to com- Since we have shown thatcorresponding singularities invariant in are TOPT positive diagrams ( only only be arise when singularities the associated energy with back and to out to show at the end of the last subsection. As a result, we can now write ant cutting rules forThat is, the we total have shown discontinuitydiscontinuity, and that of have eq. the thereby ( rederived corresponding eq. Feynman ( integral where only the Mandelstam invariant If we sum over all TOPT diagrams with the same topology, this reproduces the covari- JHEP01(2021)205 s ln with (4.2) (4.3) (4.1) s ln ) around by the sum 4.1 1 demonstrates s < 2 | . , since otherwise 1 ) 0 ln s − , − s | ) s Θ( s < | . − s | 1 Θ( along the negative real axis, ln < | πi s 1 πi 2 4 ln − pd s 7−→ | pd 7−→ ) ) iε for only makes sense when one restricts to iε illustrates the ambiguity in defining − s − s , one can series expand eq. ( s of the logarithm. 1 s ln ( n s ln( 2 ) s < ln − – 18 – | is only nonzero for real − ) 1 − s iε ) (1 − 2 iε 1 n is usually defined to mean the function produced by + s ln | s s notation and the associated non-analytic theta function + =1 ∞ s X n ( iε . The standard branch cut choice for the logarithm, going axis, is consistent with this requirement. We call 2 ) ± s = ln( s principal branch ( ≡ − s f = ln s that are within the original region of convergence to find sum ln s agree. But if this discontinuity is only nonzero on the negative real s ln ) = 2 ) s (¯ ln iε = 1 f s Disc − s s to the entire complex plane, excluding a curve going from the origin to ( for negative real 2 Disc s outside the region s along the real ln notation is sufficient for indicating which side of a branch cut we are on ln s ln , the discontinuity of s indicates that this value for Disc iε ln and −∞ ± ln ) pd iε 7−→ + The With the standard placement of the branch cut for s ( 2 As with ln axis, further analytic continuationsdiscontinuities. are ambiguous, and correspondingly so are sequential when we restrict ourselvesadditional to discontinuities, the the principalare branch problematic. of a The single function.the logarithm problem. However, is Its when a discontinuity bit taking in too the cut simple, complex but plane already picture is discontinuity is nonzero fordirectly negative on values the of branch cut.in This the is previous consistent section, withsame the point as way in the discontinuities the were only cut calculated way complex to plane is analytically if continue we a start function and end back on to the the cut. where infinitesimally-separated points that are both in the principle domain. The fact that this the value of analytic continuation going counterclockwisediscontinuity from of the the positive logarithm,function real which just axis. above computes and the below Moreover, difference the the negative between real the axis, value is of given by this branch cut has beensome chosen. of While this the arbitrariness shapethe can reality of be property this removed branch iffrom cut 0 we is to ask in that principlethis the arbitrary, choice continued of logarithm branch satisfy cut the To define points other than representations that are valid beyondthe this region. function Iterating thisinfinity procedure, (the one branch can cut). extend Thisis is simply called connected, the cut this complex analytic plane. continuation is Since uniquely the defined, cut once complex plane the location of the Consider first the natural logarithm. It can be defined in the region 4.1 Warm-up: the natural logarithm JHEP01(2021)205 n γ and (4.6) (4.4) (4.5) ) s iε for the Re is defined + 0 s , where the γ = 0 s x dx s γ = ln( R 1 s = 0 γ ), which apparently s ln γ s 4.5 πi , ln axis = 2 s ) beyond the cut complex s in eq. ( 1 x Im 4.4 0 dx − 0 γ γ 0 Z depends on the number of times the , the discontinuity will not be con- s ) s s = s γ ( , . 2 ln x dx x x dx dx iε − s γ s 1 Z 1 Z Z = s = – 19 – − s s γ Re x dx ln ln . The value of iε s + s = 1 Z 1 x or the dilogarithm Li = s s 0 3 2 γ − ln ln γ s s and end at Disc Im . ) = 1 1 . This definition agrees with the series definition and analytic continuation. − iε s x γ − to s s 1 is the infinitesimal contour that wraps around the origin once counterclockwise. . The logarithm can be defined as an integral along a path 1 denotes the number of times the path circles the origin. The family labelled 0 γ )) is through the contour integral = ln( n s 4.1 We can extend the definition of the logarithm in eq. ( A clue to how to proceed is given by the closed contour To proceed, we note that an alternative way to define the logarithm (other than 1 − γ is immaterial. Thelogarithm. only This invariant is is the the unmovable singularity location of of the integrand. theplane by branch simply point, writing at to the cut complex plane. passes right through thediscontinuity cut. across Indeed, the although cut,value the what of integral it the computation is function agrees actually on with computing two the sheets is of the a difference Riemann between surface; the the location of the branch cut where For other functions, like stant. In such casesthe we maximal can analytic continuation consider of further our functions, discontinuities. in To which we do do so not restrict we their need domain to consider The integration is togoes be from performed along anyThe contour discontinuity across within the the branch cut cut complex can plane then that be computed as eq. ( paths begin at integration contour wraps around thewhere branch point at theto origin. give We define the families principalln of branch paths of by the logarithm. On the negative real Figure 2 JHEP01(2021)205 ) s − s 4.7 axis (4.8) (4.9) (4.7) s (4.11) (4.10) ln( . This axis. πi 2 s and ] ) πi iε 2 . + i − s 2 ) π iε 4 ln( ] + − πi ) s 12 iε s . 1 − that does not pass through + − ) γ s s axis (and everywhere else) as . The domain of the maximal ) [2 ln( iε s s . πi ln( to πi , + ) ln s 1 ) s ) = ln πi − 6 . With this identification, eq. ( = 2 iε that differ by multiples of iε s can be thought of as being generated s ) = (2 − Θ( − − x s [6 ln( dx ) iε s ln s 2 ( 0 iε ln equivalence classes of paths that end at ) 3 − ln( n Z + s ln πi γ ( s 2 = − ( 2 ) ln s – 20 – 1 iε ) = (2 − − s , 3 ln γ ) as analytic generalizations of 0 ) h + iε around the origin, and constitutes the only element γ ln ) iε s 4.8 + s ( times in the counterclockwise direction. The principal − πi 3 + . We then find s ln n ( s s can be any path from iε 3 0 ( ) = ln γ 2 − γ ln = (2 iε ln s ) = ln becomes + is the monodromy. To connect the monodromy picture to the iε s s is analytic on the negative real 0 ) up to the theta function. Indeed, the discontinuity defined in ) = ln Disc + ln s . These additional copies can be accessed by integration contours s ln( , where we denote by iε s ( 4.2 γ on the left side of these equations means we approach the real 2 3 + ) ln = 0 Disc is deformed smoothly to change s ln iε switched to ( s s 2 γ ± iε ln s s + Disc ln( Disc ). For instance, 4.7 If we adopt the relations in eq. ( The infinite tower of values associated with , we can easily compute discontinuities of powers of logarithms by simply substituting ) We can now proceed tovalue take with additional discontinuities all by subtracting from the function its and terms of the monodromybranch is of an the analytic logarithm comes function, with while a the non-analytic differenceiε using the principal in eq. ( To be clear, from above or belowThe on logarithms the on principal the branchcut right of — hand the the side logarithm function are onlong defined the as through cut the contours complex path then and plane. agrees have no with branch eq. ( where the integral over cut-plane picture, we now identify of the natural logarithm’s monodromycomputed group. in terms The of discontinuities their of monodromies; polylogarithmsacross for can instance, the be in branch our cut new notation of the discontinuity is illustrated in figure after wrapping around the origin branch of the logarithm corresponds to paths that neverby cross the the closed negative integration real referred contour to around as the the branch monodromy point of at the origin. This integral is the origin. This isanalytic the continuation maximal in analytic this continuation casea of is branch an point infinite number at ofthat copies wrap of the around complexpaths, this plane we with branch obtain point an infinite a number given of number values for of times. By considering all such where the integration contour JHEP01(2021)205 , s for n , we s ln (4.16) (4.14) (4.15) (4.12) (4.13) ln = 3 n into a vector s 3 ln . 3 ) πi A. , . involve lower powers of  s s ds s n Disc 3 ! ln we get zero. It is worth emphasizing ln B ) = 6(2 s for convenience. Let’s take ) 1 1)! 3! 1 + iε ! iε − ω, n − s n B + 2 + n s ln ( V· ( s ln 3 (

1 2 Disc 3 = prescription with integration contours that end ln – 21 – s A ln = s V iε d  6= ± s ) 1 ln ! ) can then be put in the matrix form Disc n  s n ln AB by a factor of ( 4.14  Disc s V ≡ d s n Disc simultaneously. The total differential of these functions is ln Disc n . As the discontinuities of in general satisfy the product rule n Let us first return to the example of powers of logarithms not In summary, we have seen that for powers of logarithms, we can compute sequen- The differential relations in eq. ( where we have normalized concreteness and collect the functions that appear in the derivatives of by working through some examples. One branch point. for any positive integer consider all powers up to cally continuing around one ofencircles its this branch branch points point. by integrating Themonodromies along integrals of a along the closed these function, contour closed that of contours and are form this referred a group, to group. we aslytically the can By continued computing compute domain. an the We explicit value illustrate representation of how this this group function can anywhere be in systematically its computed, maximally ana- procedure for computing the generatorspolylogarithmic of the function. monodromy group associated with a general 4.2 The monodromy group Given a function defined by a contour integral, we can determine the effect of an analyti- on different Riemann sheets,around the and branch the point. discontinuity Ining across general, amplitudes the the are transcendental more cut functionsinvariants that complicated with with show than many the up logarithms, branch in monodromy and scatter- points.complicated depend functions Understanding on the will many monodromy help Mandelstam compute group us their of untangle these sequential their more discontinuities. analytic structure, Correspondingly, and we thereby now help turn us to a systematic The discontinuity operator computesnot a an finite infinitesimal differential difference in around any a sense. branch point,tial which discontinuities by is identifying the If we take anyhere further that discontinuities Disc of does and JHEP01(2021)205 . A , the s (4.21) (4.19) (4.20) (4.17) (4.18) ). Since and ending 4.16 1 whose entries implies matrix } 0 ω \{ will mix with other ◦ ) is given by C V ω ≡ 4.16 starting at ∗ ). These other functions C γ ··· . . 4.16 , which we normalize to have ) + 2 t       ω ( s ), we can group these solutions s ◦ 3 kj 2 s , ω is a matrix, ω ln ln ) b 4.16 .  1 a 1 1 2 3! ω t ω Z (       γ s s ik 0 Z ds + 2 s ω  ) beyond the cut complex plane, we need ω ln b s 0 0 ds b ≤ 1 2 matrix defined on a variation matrix , the vector of functions 2 t Z exp s 4.18 s ds ≤ 1 P – 22 – ) will contain all the possible information about t = 0 0 0 0 0 00 0 0 0 0 + 1) ≤ s       1 ln 0 10 0 ln 0 0 1 0 ln 1 a n ( ) = Z = 1 +       4.16 s = , the path-ordered exponential is defined by ( × b = ! γ called the ω ω ) = ω to b M s around a ◦ ( M + 1) a Z 0 s n ω γ

n ( b a M ln Z exp independent solutions to eq. ( is an P ω , satisfy the differential equation in eq. ( V + 1 is a path-ordered exponential along the path n ) denotes an iterated integral. Since ω γ ω R ◦ connection ω exp( b a R P can be written as . Thus, general solutions to eq. ( The most general solution to the differential equation in eq. ( To extend the variation matrix in eq. ( As there are . For a given contour from s iπ = 3 ’s along the diagonal. The variation matrix on the principal branch of the logarithm for where multiplication: where at variation matrix, we canvariation interpret matrix the by action another matrix, of the the monodromy monodromy matrix. as multiplication of the to determine the effecttheir branch of points. deforming thedifferentials Although of integration this the contour extension function will changes defining stillthe the be its general value related solution entries by of to the around this the differential differential equation function eq. equation at ( are linear combinations of the rows of the Variation matrices have aFeynman integral close calculations. Further connection discussion of to this connection the is given coproduct in appendix structure often utilized in into an upper-triangular matrix 1 n As we analytically continue functions that, like have lower transcendental weight, and the mixingof coefficients will be proportional tothe powers monodromies of the function. are one-forms: where the JHEP01(2021)205 , , . ) 0 is 0 t x ω γ ( γ i γ and ξ ∗ (4.23) (4.24) (4.22) + γ γ , where between 0 γ along this γ ◦ followed by . ω ) 2 1 0 E γ γ − 0 and defined by R γ start and end at M , followed by . ), and a series of ◦ 1] 2 ) − in the variables , ω that goes from the k γ exp n γ 0 i can be computed by t ) [0 ( 0 γ P x ) k γ or ( ◦ x is to be summed over. ξ ∈ k and ( ∗ 0 lies on the negative real 0 + γ n t k γ 0 γ 1 i γ . A very useful feature of γ s ( γ b γ M γ M , these iterated integrals are ··· M · ) , . . . , ξ ) n ) 1 n i t x ··· γ ( ◦ · · · ◦ ( ) is exactly 1 1 1 0 ) is nilpotent. i ξ ξ 1 γ M ∗ denotes the composition of paths in ω − 0 . γ , . . . , x M 4.18 0 γ b into a contour 1 ··· 1 ). We can split up any path · γ γ x ◦ 1 ≤ x 0 i ◦ k ◦ 1 , the value of γ γ , except when 0 t i but starts and ends at the same basepoint s a x 4.19 γ − = ( parametrized by k γ where the paths M M γ i ◦ and encircle one of the branch points of x in eq. ( γ to γ 1 . In summary, to compute the monodromy ≤···≤ 0 = = γ 1 0 x x γ t 1 γ 0 . Hence, we can also write n to go just above the branch point at zero. Then 0 – 23 – ≤ i γ M . This convention ensures that the monodromy we first multiply on the left by x 0 γ ◦ 0 M x = ( is finite since ◦ · · · ◦ Z n 0 2 γ i , where γ γ γ − , whereupon they become differential forms ω ◦···◦ n and then along the path 0 i γ γ 1 0 i ) = γ a that ends at γ ◦···◦ x = ◦ γ 1 ( γ i 0 k γ γ γ ξ . closed contour integrals from a common basepoint onto the M M x or ◦ · · · ◦ into some number of iterations of 0 1 = = 0 i γ γ followed by γ γ γ ◦ ◦ · · · ◦ 0 ◦ 1 there is only a single branch point at the origin. The contour ) M M γ + 0 x s γ γ ( n 1 ξ = ln prepending ) below). Since there is only one branch point, we define paths encircles the same poles as γ γ parameterizing the path. The iterated integral of these forms along Z . Then, given some differential forms 1 without encircling any of branch points (the poles in ◦ · · · ◦ − t 0 )) that each begin and end at 4.29 t γ + x ( ◦ } γ n 0 j k 0 i γ = γ { ◦ γ along the path 0 γ , . . . , x rather than starting and ending at x ) In the case of Given an integration contour For differential forms in several variables t 0 that encircle the origin counterclockwise or clockwise with unit radius. We can thus = ( γ 1 k − i x one can check that thepath variation (see matrix eq. ( γ decompose a general path namely from multiplication on the left ofthe the basepoint result independent by of can be taken toaxis, be in the which straight case path we deform from the path where we are now integration contour before wematrices arrive are at independent of the endpoint Now, this contour can alsoγ be written as which we first run alongdefining the matrices path as path-orderedto exponentials matrix is multiplication. that So composing two paths corresponds integrating the path-ordered exponential in eq.the ( basepoint (where thebasepoint integration to starts) and contours That is, we have in the variable defined as We discuss how to evaluate integrals of this type in more detail in appendix Note that the expansion in powers of defined as follows.( First, we choosewe a can path pull them back to the path Here we have made the matrix indices explicit for clarity, and JHEP01(2021)205 . . ] π       2 3 , times ) (4.30) (4.28) (4.29) (4.25) (4.26) (4.27) [0 πi k 3! 2 ∈       − ( 3 2 θ ) ) 2 , ) + πi πi πi πi       s 2 2 for 2 2 2 s s − ) − − ln − 3 2 s ( can be computed ( ( 2 iθ 1 1 2 ln 3! ) ln − πi + − γ 1 1 2 2 πi 3! 2 2 s 2 ) − M − s ln ( πi πi 2 s 2 2 πi + ln − = exp( − and 2 s 1 2 ( ) s 1 2 − 2 s s ( ln . . 0 by πi . γ j ) πi 2 ) ) 2 s 0 10 0 ln 0 0 1 0 ln 1 1 ln 2 − ( s M − ! − γ ! πi ( 0       one more time. Such representations j j j γ 2 0 00 0 1 0 1 1 0 1 2 = ) − ln       M ( = 0 πi 2 · ω 2 = = s k ◦ = −
} ( ω } s ω , we find − ds s ◦ s γ ds πi + ◦ 2 3 ω s M ω – 24 – − ln ◦ j 0 {z ln j {z γ ω ◦ · · · ◦ Z just return the logarithms we started with, namely πi ◦ · · · ◦ − ) = 2 0 γ + s ) for s times the top-right entry of this matrix, in agreement ds γ Z s | ( − ds | ω k 0 γ 3! + γ − ◦ 4.17 πi Z γ ω M 2 ω Z ◦ 0 − γ ω ). Z , we parametrize the path 0 0 0 0 00 0 0 0 0 − is then −       γ + γ of the equivalence class of contours that encircle the origin Z , and thus s 4.18 − ω 3 , we have k M + 0 s γ ln idθ γ ω Z ), we can then compute the effect of going around the branch point by ) = − s − + can thus be seen to be a generator of the monodromy group, since it maps ( γ = 0 Z 1 γ − 4.25 s γ ds ). + M 1 M · ) = s ) 4.10 ( − 0 γ ) = γ s ( M M to its value after encircling the branch point − Using eq. ( Since we have specified their integration contours, Given a member γ s − n 1 M ( The discontinuity of with eq. ( in agreement with eq. ( multiplying these matrices. For example, we can calculate the first discontinuity by and Expanding the path-ordered exponentials and evaluatingabove the on iterated the integrals as connection described in eq. ( The analogous set of integrals over of the monodromy group generators are sometimes called monodromydirectly. matrices. To calculate This gives us clockwise and end at The matrix ln JHEP01(2021)205 . − γ M (4.39) (4.34) (4.35) (4.36) (4.37) (4.38) (4.31) (4.32) (4.33) − 1 k n , as we did non-empty ] ) πi πi πi m k 2 4 2 ) − − − πi . ) ) )( 2 k iε ) iε iε − ! + )( πi + + k s 2 iε s s . ( − n ( k ] + k ) ) − elements into s . )! πi s n ( ( πi k 2 k k + [ln( k ) = ln( ln k . ` k 2 act on a function rather than ) − n ` iε − − 1 ] n ! − ! πi − − n k ! n k γ n 2 − )( ln ) πi < ( ln ` s 2

m − | ! iε iε ! M ( s n

| k n − ) n =1 ln( + + ` )! X s ) k n =1 − s ( s k X k ( for iε k − ) m k − − − k − + m [ln( 1) n n n the entry in the first row and last column of s ( ( − ) = = ln ! ( ln 2 − ` n s iε m n =1 γ s n n n ! =0 =1 m X – 25 – k

` X X k − 2[ln( k n ! M ` ! s =1

∞ ) ln 1 m = − ( X n 1) m − n ) m ) 2) γ − ≡ s ln iε ( = ! 1) ( − ) M n n s − − + =0 m ) k ( ` X − ln ) s 2 k m (2 ( − 1 = ( iε n ( Li γ Let us now consider an example involving two branch points, discontinuities is n =1 ) = X + k M = ( iε s m = ( s + n n ) = ln s ), which gives us ln ( iε n s 4.9 + ln ) = ln s ( m s Disc iε n is the number of ways of partitioning a set of . + ln } n s s Disc k m ( { n ln Disc transforms as s s ) s ( Further discontinuities can be computed by acting with the same operator More generally, under the action of 0 Disc Disc γ sets. Multiple branch points. the dilogarithm where are the Stirling numbers ofpretation: second kind. These numbers have a useful combinatorial inter- Similarly, the formula for for arbitrary For later reference, we listsubstitution here method some general or formulas with that the can use be of derived monodromy either matrices: using the for instance in eq. ( the variation matrix in which it is the upper-right entry. Thus, the discontinuity is This agrees with what we get using the substitution M Here we are generalizing notation slightly by having JHEP01(2021)205 has a (4.40) (4.41) (4.42) (4.43) ) s ( , diverting 2 s , ) s s ds − , with a branch cut ◦ . in the analytic integral s s = 1 0 ln(1 − − γ s ds ds to make the singularities 1 − ω 1 · ) s 0 − = s  γ ) R x − axis. The standard placement s Re ( ) = s − 2 dx s 1 Li ln(1 s ) = − s ) s 0 . The principal branch of Li − ), this power series definition is only s ( Z (    2 1 ln(1 s ≡ s = 1 4.1 0 ds and proceeds in a straight line to Li ) Li s s 0 s is consistent with the standard branch cut 1 ( 1 − ds  1 1 ∞ and ε > , d = . The standard contour 0 0 0 0 0 0 ) – 26 – s =     ∞ rather than ) s = 0 s + ) − = both have branch points at ( s 0 s 2 ( ) < s < to ω 1 s s Li ( 1 ln(1 along the positive real with Li 2 s ) , but can be uniquely continued to the rest of the cut s Im = 1 s 1 ds ( ∞ 1 s − < , Li to , from ) | and Li ) 1 x 1 s ) s | ( ) =  ( s ), we have 1 s ( n d ( . 1 2 Li 0 . 4.40 Li x } dx d 1 s < s , begins at a basepoint at 0 has branch points at 0 Z ) ) s \{ s ( ≡ ( 2 C 2 ) . The dilogarithm can also be given by an integral definition, s ( ∞ 2 . Li Li to 1 Using equation ( is defined on We can again put these relations in a matrix form where conventionally going from of the branch cutfor for the Li logarithm, from We write the integral inmore transparent, terms as of Li Li definition of Li in a counterclockwise path around the branch points whenSimilar necessary. to the definitionconvergent of in the the logarithm region incomplex eq. plane, ( where the branch cut is usually placed on the positive real axis running Figure 3 branch cut on the real line from JHEP01(2021)205 x , ) with s (4.47) (4.48) (4.44) (4.45) (4.46) ( 1 2 and , we find 0 0 that result (see ε s s , ε on the positive real ln ) ln ) ln  defining its principal s s , . 1 0 − − γ    . Note that this contour ⊗ ) ) s 3 s . Then it is straightforward s ( ln(1 ln(1 0 ( 2 2 − − Li → . Li ) ) : ) s s 0 ε s ( ( +    γ ( 2 2 1 s πi , but for all the other functions we Li Li Li 1 ln and the path − − . When the function under study only 1 00 1 0 ln 1 x p ⊗ along = ∼ = 1 0 0 0 1 2 0 0 1    . In contrast, we denote the path around ) ω x p s 00 = 0    ( 00 = 1 s s ds =  0 Li 0 s ω ε – 27 – 0 , which avoids the branch points at Z γ s ) + 0 Z M s s 0 (  to 2 = − , which diverges at the lower integration bound. This ds are dropped and then 0 0 Li . 1 s ds exp ε s ⊗ on the contour. For the monodromy around A ε P ) Z M , we will often drop the index indicating which variable the ω x = γ ) = ) = 1 R s s s ( counterclockwise by ( 0 − 2 ds γ p we took the basepoint to be 1 exp( Li M s ◦ P ∆ For example, n s ds 8 ln s ]). 0 , we take the basepoint to be Z ) 72 s ( 2 on the lower integration limit and dropping the powers of means terms divergent in ε ∼ = that starts and ends at the basepoint by . As there are multiple branch points, we should in general be careful to distinguish ) p The contribution from moving along any contour is computed by evaluating the path- We would now like to extend this construction to the maximal analytic continuation of For Li s In more detail, this regularization implies a choice of parametrization for the path in which the tangent ( 8 2 = vector to the path atthe the paths basepoint that is satisfy ofcontinuously this length deformed constraint. one. into In The one other monodromy another words, group with we is the consider then tangent homotopy defined at classes with the of respect basepoint paths to being which can kept constant. be contour is taken in. ordered exponential study in this paper we willaxis, take when the regularization basepoint to is be required).that 0 encircles We (or the denote a small point the value infinitesimalx contour in a variable depends on a single variable Li between infinitesimal contoursthat that not encircle only these wrap aroundintegration. branch these points, For points but and also the start and full end contours at our chosen basepoint of Note that the this variation matrix encodes precisely the coproduct structure of Li as discussed further in appendix where to compute the variation matrix by integrating for instance [ branch to be the straighta line counterclockwise from detour ifis necessary. problematic This for is the showncan differential in be dealt figure with usinga tangential cutoff basepoint regularization, which amounts to introducing JHEP01(2021)205 · ) 1 while (4.54) (4.51) (4.52) (4.53) (4.49) (4.50) . 0 M and then − = 1 1 s ( , we first use = 1 1 s , and compute the . 1 i , . . ) z    (¯       , 2 1 s and 0 )    Li 0 ln πi , 0 πi − 2 (2 πi    ) does not have a discontinuity πi 2 − s − z 2 πi ) ( ) 1 0 10 0 0 1 − s 1 s − ( ) (    2 0 0 0 00 0 0 0 + 2 Li s 2 , and then moving back to 1 0 10 0 0 1 ( h 1 s =    ) 2    Li along the real line, rotating counter- ¯ z 1 Li 1 = z = − )
0 1 γ 1 s to πi ln( ( 2 → 1 0 M 0 − . Similarly, we can compute that M only have a monodromy around · − Li 2 ) ) ) ) M ) z , as expected. We can also now compute the 0 1 00 1 0 ln 1 (¯ s s , – 28 – ( πi 2 ( ·    2 1 1    (2 Li allows us to compute any sequence of monodromies = 0 2 Li = 2 6 − M 0 π 0 s γ 1 00 1 0 ln 1 M − γ − . For instance, prepending a monodromy around 0 to ·    ) M 0 and Li by first taking the monodromy around ) = 1 1 z M 1 0 0 1 0 0 0 1 γ s ( ( ) · = ) ( → 2 ·    s s 0 2 as before. We find 0 M 0 ( ( ) γ Li 1 2 1 = 1 Li M 1 1 M · = M → 0 1 ) = 2 1 M Let us finally turn to an example involving multiple variables. We Disc ¯ z 0 , consistent with the fact that Li M − z, ( 1 M M 1 ( = 0 Φ 0 γ M . . As we prepend these contours, this corresponds to · gives ) 0 0 = 0 = 0 γ ) computes the effect of moving from ) to determine the contribution from the path between s s M 4.50 4.46 only has a monodromy around Acting with these matrices on − s 1 around Multiple variables. consider the two-variable function which tells us that Disc ( These matrices imply thatln Li sequential discontinuity of Li around while prepending a contour around 1 gives on the functions appearingthe in path We highlight again that theeq. action ( of the monodromy matricesclockwise proceeds around from an left infinitesimal to contour right; centered at where we have droppedbasepoint all regularization. logarithmically-divergent terms The complete in path accordance thus with gives tangential eq. ( infinitesimal contour around To compute the monodromy matrix associated with the branch point at JHEP01(2021)205 0 ). γ and 6.2 M = 0 0 (4.59) (4.57) (4.58) (4.55) (4.56) ω ∧ ω . −       . ) . z dω )       (¯ ¯ z . 1 z πi ) ) ) 2 Li are on the real line ¯ z ¯ z in the same way we ln( − ¯ z z, z z, ) z/ (¯  ( 0 0 ) + , so there are now four ¯ z 0 ) on cycles that encircle 1 ¯ z z } or γ . . ( ¯ z, z ) and flat ( πi 1 ( z − d M E       = 1 4.20 Li 1 ¯ z ¯ z 0 00 1 0 0 1 1 2 0 1 0 0 = 0 ¯ − z d − ¯ z 1 − , ¯ and       z d )Φ 1) = z z dω + = 0 (¯ 0) − − = 0 1 z , , − ¯ z 0 dz , namely 1 z ¯ − only depend on the homotopy class z dz dz 1 Li (0 1 , , , where 1) 1  ω M − , ω − γ ) + · − , = 1 ¯ z R = ¯ z z 1 0
− is closed ( ( d ¯ z 0 , γ ) 1 1 , z 1 z ω exp (¯ , M diag(1 Li + 1 · P ) = 0 z M = ) Li ¯ – 29 – z − ¯ dz z z in its arguments is encoded in the matrices 0 z 1 γ , ) diag(1 z, \{ ) + ¯ z ¯ ( ¯ z z 2 M d z 0       ( d C γ 1 ln( 00 10 0 0 0 1 0 ln( 1 z, 1 πi ( +       1 M Li z as independent variables, so this function is analytic for · dz Φ 0 0 = ¯ z  0 00 00 0 0 0 0 0 0 1) between the basepoint  ¯ z πi ¯ z −       0 ω d , γ . Following the same steps as in our previous examples, we first 0 1 and = γ 1 2 1 2 0 10 0 0 0 0 1 2 0 0 1 − , Z 1 z       ω  < , z dz | = 1 2  z 0 exp is well-defined in − = by evaluating the path-ordered exponential ( P ¯ diag(1 z 1 ω ¯ z | M , when we go counterclockwise around the branch points. Integrating along Φ , namely we use straight line paths, except when d = 1) ) = ) , ¯ z 0 z and s and ( (0 z 1 2 2 z, ( 0 < M γ by the action of conjugation by | 1 2 ω M We now compute the monodromy matrices associated with the branch points at We can define a path . Further discussion of this point can be found in appendix γ − in both z guarantee that the functionsof appearing in 1 each of these four poles. First, we compute Further, it can be checked that theThese connection requirements were trivially satisfied in the preceding one-variable examples, but Note that the antisymmetryand of did for Li outside of this path gives the variation matrix on the principal branch. The result is The connection codimension-one branching varieties. This can be put in the matrix form below). Here we| treat compute This function arises in the one-loop triangle and one-loop box integrals (see section JHEP01(2021)205 (4.64) (4.65) (4.60) (4.61) (4.62) (4.63) and then 1 . ,       . we need       πi , around 1       0 2 commute with the . πi 0 . z       − πi z       0 2 2       πi 0 ) 2 and the other functions − πi πi ) 2 2 ) πi ¯ z πi − − πi πi (2 2 2 z, − ( − − 1 1 0 0 10 0 00 2 1 0 01 0 0 0 1 10 0 00 1 0 0 0 1 Φ             0 0 0 1 1 0 0 10 0 0 1 0 1 0 0 10 0 00 2 1 0 0 0 1 0 0 0 (2 0 0 00 0 0 0 0 0 0 0 0 = = 0 0 0 0 0 00 0 0 0 0 0 0 0 0                   ¯ z 1 z 1       = . = = ¯ z = 1 1 M M 0 − − 0 γ )
γ 1 1 M ¯ z z M → → · 0 0 · ) ) z 1 M M z 0 – 30 – . · · B ¯ z 1 z 1 . We construct this monodromy matrix and discuss M M , ∞ − , − M M       1 · · 2 1 (       ζ 1 1 ( · 2 2 ¯ · z z ζ → → ) ) 0 0 − z 0 z 1 M M (and correspondingly the kinematic invariants of the triangle or = = M M 0 0 10 0 0 0 1 1 0 0 0 1 00 0 0 10 0 0 0 1 1 0 0 2 0 1 0 0 ¯ ¯ z z 1 z 1 ) anywhere in their domain, and therefore to compute sequential −             − or 1 1 ( = = 4.57 ( z M M 1 1 ¯ z z → → 0 0 M M These matrices allow us to compute monodromies of , we would evaluate It is also possiblemonodromy matrix to associated take with athe discontinuity full around monodromy both group in branch appendix points by considering the Taking these discontinuities in a different order, we get a different result 0 matrices that encode monodromies in the variable appearing in eq.discontinuities ( in box diagrams). For example, to compute a sequential discontinuity in Note that the matrices that encode monodromies in the variable Putting these paths together, we find To compute the monodromy matrices associated with contours around JHEP01(2021)205 ] can (4.66) (4.67) (4.68) ]. The 28 , finding , ) 28 , 27 iε 27 . Using this − . . With these s s s 2 2 displacement of − ln( γ iε ln ± − s lower-diagonal matrices is given by a path that 1 , πi . − 1 s γ 1 − , but also − γ ) is applicable. Recall however γ ln once versus twice. ] = 2 iε for all values of ln several iε = s πi ] + 2 ± ≡ s πi πi , and 2 ) − s 2 , as shown in figure ln( iε ) s − − iε , we are not computing the difference ) − ) ) iε s + iε iε s + − ln( − . Rather, we are computing the difference s s ) ln( s [ iε s [ln( ln( – 31 – + , s − s Disc 0 ) = ln( ) we describe a general construction that builds a around the origin of γ iε iε ln( s = ln C 0 ) γ − − ≡ iε s s ln ) − using different integration contours, we should be able to iε ln( s ln( ) notation is good for describing where we are on the principal + iε = ln( s ) s iε ± iε ± ln( s − Disc ln( s ln( displacements in Feynman integrals originate in the s iε ± Disc is homotopic to the straight path from 1 to 0 γ For sequential discontinuities, the contour definitions are particularly helpful as they In the case of the logarithm, we recall that this amounts to identifying Finally, let us highlight that the variation matrices we associate with Feynman integrals As long as we analytically continue along paths which are fully contained in the Eu- allow us to migrate away fromthat the principal all branch the where the poles in TOPTnumber propagators. versions of Thus, just as we were able to identify higher winding Thus, when we takebetween the its discontinuity value of andbetween the analytically continuing value of identifications, we have that identity, we can compute the discontinuity of not only This can be rewritten in a more suggestive manner: where first crosses the real negative axis before ending at 4.3 Monodromies ofWe propagators have seen that the branch of multivalued functions, wherecut. they describe We being havemonodromies on also around opposite the seen sides branch that of point a where discontinuities branch the across cut the begins. branch cut can be recast using we consider are useful fortransform determining how under the monodromies. transcendental functions Itbe appearing is in used unclear them in to us thethey if same furnish. the way, and matrices described if in they [ can, what representations of the monodromy group using the above construction arematrices considered not in the those same works as lookin the similar, which a insofar matrices as Feynman considered integral they inintegral sits are appear [ in lower-diagonal matrices in the other bottom-left entries. entryare However, and in associated various that cuts with approach of each thetriangle diagram, Feynman graph. labeled by spanning For trees example, whose edges they are ordered. associate The six variation matrices such matrices with the clidean region, we neversingle-valued. encounter branch The singularities variation and matrixvalued approach the lends functions functions, itself we and well consider tosingle-valued in are version the appendix of description any of generalized single- polylogarithm from its variation matrix. JHEP01(2021)205 is ), iε ) (4.73) (4.74) (4.75) (4.76) (4.69) (4.70) (4.71) (4.72) ± n 4.35 D 2 − n and the P ( . . ··· . ±∞  , ) iε 1  1) − . The propagators we = j . − D 1) iEt n 1 ( ω t k ) 2 − − P k E l j − ( ( to dt e k E i 1 +1 P j n P 0 Y Z = πiδ −∞ . = +1 2 k = 0 i , j n 1) 1) t  Y − j = − perms − − j ( k ( j j D = ) ) + ( D + =  P P j n n j  n ω ) to express propagators with nonzero n iε =1 D iEt D + P Y (0) j k  − 1 and ) − − P iε j k = ··· l dt e 4.72 1 n ( iε E E k + =1 − P ( + Y j +1 k P ∞ M ( j j Z k 1 1 1 ω  −∞ ω i P =1 πiδ − − Y ··· k , j k – 32 – j 2 − n − =1 , ) X j E  D j 1 − (0) j = E iEt D = P n = =1 = ··· X j ) become iε j − 1 = n =1 M (0) M Y j = j dt e ) 1 1 D − D n ) P h P − 2.17 ∞ ( tot j j 0 = Z l E ( 2) 2( i P j M P − − M − Disc − = k n = =1 iε Y − j ) (2 1 n M + k iε tot P M 1 E 1) tot + tot − ··· E ( Disc 1 Disc k P Disc X = = ( = is a linear operator, this can also be generalized to products of propagators M tot 2 tot In this notation, a TOPT amplitude and its conjugate are Disc To take further discontinuities, wewinding just number use in eq. terms ( of propagators with winding number 0. Then, as in eq. ( Since Disc with arbitrary winding numbers: The TOPT cutting rules in eq. ( where we call this distributionare a used propagator to with seeing winding correspond number to of products of propagatorsnotation in the same way as for logarithms. To do so, weand introduce the so that Thus, for the propagatorshorthand the for integration this path integration path. goes We from can correspondingly take sequential discontinuities comes originally from a semi-infinite integral over time identify higher winding number versions of propagators. To do so, recall that propagator JHEP01(2021)205 . µ i I 4 P P I (5.1) = ∈ (4.77) i s ), these P P . choices for , and each = i 4.77 n ) µ I and I ) is n k P E I ( P perms E 4.37 . with =  2 ) + I s n ) and eq. ( P iε E i D ) → · · · → ` + = n ( 1 4.76 I n ). We then saw in section I D s 1 0 ω ` ··· E ) = − ≥ − s s 2 I +1 n k E P P channel by ( D s ` = ··· ( I ··· k s ) iε P 1 takes the form + D s ··· ` 1 and R 1 1 ω 0 − P h − 1 ≥ – 33 – s P n I ( =1 E X h k E  in all TOPT propagators not involving the energy ! ! i ` ` ω m m iε

− 1 ` ` i P 1) 1) E − − , where only a single invariant ( ( s s P propagator to a family of propagators with additional winding =0 =0 m m R 6= ` ` X X Y i P iε = = + = n that an advantage of TOPT over the covariant formalism is that one notation is sufficient to identify the two sides of a branch cut for taking P M 3 language was insufficient. For a single discontinuity, one can compare a iε ··· ± 1 iε P ± m ) propagators to replace with delta functions. The analog of eq. ( tot k If we work in a region Let us try to briefly summarize this section. We found that to take sequential discon- Disc ( associated with this invariant. Todenote make the the energy equations and in momentum this associatedIn with section this the more notation, transparent, a we generic TOPT diagram in a single discontinuity, forbranch sequential points discontinuities and it monodromies. provesthe sequential useful We discontinuities now to of make think Feynman use integrals in of in terms terms these of of tools cuts. is to positive, derive formulas then for we can drop the can directly identify thea given origin TOPT of diagrampropagator singularities depend will in on only a a leadenergy particular sequence and to channel. invariant of are a non-negative energies, singularity ( Propagatorsthat, in while in the the integration region if the corresponding 5 Sequential discontinuities We saw in section of projecting ontofunction. a complex Moreover, monodromies plane) canmatrix and be and the computed a in discontinuity connection. angeneralization is algebraic of automatically Finally, way the using we an a sawnumbers. analytic variation that These propagators the will monodromy becuts picture used and in led sequential the to discontinuities, derivation to of a the which natural relation we between now multiple return. tinuities the function on two sides ofdiscontinuities, a one branch needs cut an on analytic theway function principal to defined branch. do away However, from to thatIn the take is the additional cut itself. to monodromy language, A treat natural there the is discontinuity no as branch a cut monodromy at around all the (the branch branch point. cut is an artifact Although the winding numbersequations have are been valid left for implicit any assignment in of eq. winding ( numbers. which where the sum over permutations in the last bracket corresponds to the JHEP01(2021)205 , . On (5.3) (5.2) iε can be M ) to all − ¯ z n 1 ω 2.17 ) . 1 and ) − − ) n ) only holds in z k s 1 ω ω E − 5.2 j − − that appears in this ω s ··· s s propagators, denoted ) − E E j ( ( s , as all the delta func- iε ω s 2) 1) E − − ( E − − ( ( ones. s (0) P P , to E 2) P ( − ··· δ (0) ( ··· ) ) ) ··· P P ) πi +1 +1 1 2 k j ). ω − ω ω ( − 5.2 − − s s s ··· E E E ( and cut integrals in eq. ( ( iε ( 2) (0) 1) + − M − P ( 1 ( s – 34 – 1 ω P channel is the same as the total discontinuity: P k ) ) X propagators into s j k − ω ω s j 1) X E − propagators, denoted − − ( i s s ω j P iε E E X ( ( + − i 1 δ δ i ω ) ) s P R πi πi − E ] channel we can simply rotate all the energies around the same 1 , we will see there are multiple ways of encircling branch points ). 2 2 in arbitrary regions, these discontinuities must be evaluated in i s s P M P − − ). In principle, the discontinuity operator Disc are used to take this discontinuity. This requirement, that the ( ( , where it must satisfy eq. ( 6= E 6.2 Y i 3.15 s tot M s P ? × × 5.2 R P R 6= = Y i Disc s P to ) in the variables used there that correspond to encircling the branch point in a R are zero. However, it is perfectly valid for us to analytically continue the i = [ = s ¯ z in eq. ( 5.2 s M R from eq. ( s R M 2 s ] s R R and can be computed (and will in general be nonzero) in the Euclidean region, where M ] s Disc z h M . The second cut turns the M s s Before taking further discontinuities, let us pause to clarify the role being played by 1) Disc [ − ( Disc In words, the first cut turnsP the We are now ready tonuity consider of discontinuities eq. of ( discontinuities.path To take as a for second the disconti- first discontinuity. This gives the cuts of discontinuity that has been computed usingregion the right to monodromy the matrices in region the Euclidean 5.1 Sequential discontinuities in the same channel given Mandelstam invariant; however, onlyaccessed some via of paths these that pass monodromiesthe through in the discontinuities appropriate of regions.the Thus, while appropriate we region can and compute usingDisc appropriate contours to be related to cuts. For instance, analytic continuation path startsonly passes in through an the adjacent region, regionsequential will where discontinuities become even below. the more important For cuts when instance,will we are consider in compute in being the section triangle computedin and and box the ladder integrals we the region equation can be applied anywherethe in other the hand, maximal the analyticregions domain relation where of between the Disc these function paths cuts are from allowed, and only when appropriate analytic continuation The second equalitypropagators, comes or from equivalently applying justtions the to involving TOPT the other sums cutting propagatorswith of involving rules a energies given in evaluate topology eq. to then[ zero. gives ( the Summing discontinuity of over the all corresponding TOPT Feynman graphs integral, In this region, the discontinuity in the JHEP01(2021)205 . , i = 0 (5.6) (5.7) (5.4) (5.5) (5.8) s . indicates ), are taken perms 6 s + + , R . The result 4.72 ) -cuts iε s + k E R ( + i , which should be , M δ n 0 ) γ 1 s + ω πi -cuts R M 2 k − # , s − times, and s + ( M E k h R -cuts − # k k . ··· − ) iε M m 1 + -cuts iε E 3) k ( E 1) δ + exactly ) − − M ( k = πi +1 2) M ) ) 2 1 k 2 · ω k E m − − ( ( ): k ( ) and directly compute the cut Feynman − 2 1) . Explicitly, we get + 3 . Examples are given in section s m (2 − s , as the cuts in all other channels vanish in iε − ( 5.5 k k ∞ = E 2) s 4.37 X 3 R k ) + P -channel propagators, in which each non-cut P 1) ! iε k − s − ( − ω . We have also included the definition of the k m 1 – 35 – + ( k , which returns the monodromy around iε s 0 − (2 = 1) E =2 k + s X k are the Stirling numbers of the second kind. We − " E 1) ( . Each cut should split the diagram in two, and the M ( = k propagators, we rewrite them using eq. ( M ) − ` δ iε = m =3 ( ) ) ) ) X k E s + 0 s m ` ( " E n =2 πi R ( ( X k 2) ` i 2 = 2) j − − − ( cuts is as in eq. ( − s ( M ω m M ( 2 s R P i P 1) − m − 1 ··· i − ) 1 P M ( 1 Disc 3 s h E propagators. Doing so, we get ω =1 = ( m ` P − iε ). s 6= Disc P s i Y R + h ] ! P E 1 m ( 4.76 is the sum over all possible ways to cut = M δ ) s m s = R πi i -cuts 2 } k Disc − m k M [ ( { 2 s M h The formula for the triple discontinuity can be computed the same way, giving Summing over the double discontinuities of all TOPT diagrams with the same topology, To make sense of the × . Disc s h More precisely, this monodromy matrixcomputed acts on along the paths variation from matrix the basepoint to and the generalization to where that all uncut propagatorssum have of momenta flowing acrossR it should be can extract the combinatorial factordiagram from with eq. all ( similar to eq. ( we get the double discontinuityfunction of in the a associated TOPT Feynman diagram integral. directly Recall corresponds that to each a delta Feynman diagram cut. As such, we propagator is in the region corresponding to To avoid any ambiguity,is we a also sum substitute over cutting different numbers of emphasize again that this relationto holds when be all non-cut in propagatorsdiscontinuity the in operator region in corresponding terms to of where JHEP01(2021)205 t s 2 ) E R I P (5.9) . We → (5.10) before J , where } channel = ( P s M I s,t s R { = s . t R = . P iε s iε , since these will − t t + n ` 1 , and or ω appearing in any 1 ω J s P k − , where − E ω t t M E E = t branch cut. Thus we must t t ··· =1 n propagators in the E Y ` ) Disc = 0 , ` s s I t iε ω n P has the form . The difference between + − iε s } . t = k } E + 1 ω s,t , and all other Mandelstam invariants E s { ( k s,t 0 P { δ − 1 ω R ) , s is the number of external particles. One R I − t > independent parameters that we can vary πi P E n s 2 , → E ), we have not found these constraints to be 1 s 0 E − =1 s n Y ( 6 k = − R s i – 36 – =1 n s Y n s > k ω ··· E → i channel will remain unaffected since our analytic − iε ). Let us assume such a path exists. This path will } ω 1 channel, we want to pass around the branch point at i t t 6 s,t P are different momentum invariants. We abbreviate the + − 1 { on the other side of the i 1 E J R P 1 } ω } , located at the smallest value of t E t s,t − ,P } { E s t and t P R channel in the region Y ,P E I s / t ∈{ P i t Y =1 P n from all propagators in channels other than ` X / ∈{ i = P iε × the region in which } for sets } = s,t 2 { } s,t ) { R s,t J ] { R P R M ] [ . = ( and then back to M t J B 0 s propagators in the t = Disc and no other branch points. We can do this by passing through the region [ s > n t To take the discontinuity in the To fix our notation, suppose we want to compute Disc = 0 Again, the propagatorscontinuation not path has in gone the from (some examples are givenencircle in the section branch pointpropagator, for but will not encircleand the branch after point analytic for continuation along this path is thus never go on-shell. t only find a path rotating the energies, respecting energy conservation, to go from and We have dropped the and associated energies and momentaalso with denote by are real and negative. A general TOPT amplitude with class of the integration contour,of iterated integrals the in integration general depend path.operators on the that This homotopy give class means theThis that highlights same one the importance can first of in discontinuity,continuing our through prescription but general specific for different find taking kinematic discontinuities sequential multiple regions. byappendix discontinuity analytically discontinuities. We discuss this ambiguity in more detail in along each analytic continuationalso path, must where make sure thatthe the examples relevant invariants we only have encircle exploredoverly their restrictive. (see branch section points Nevertheless, once. choosingresidue In paths theorem has guarantees to that be normal done contour carefully. integrals While only Cauchy’s depend on the homology Next, let us consider how tocase take of sequential discontinuities sequential in discontinuities differentalong in channels. the Unlike at the same least channel, twoexternal we different energies must now paths. while analytically leavingmomentum continue As the conservation. external before, This three-momenta we gives fixed insist us and on respecting using energy- paths that rotate the 5.2 Sequential discontinuities in different channels JHEP01(2021)205 i → i } (5.12) (5.11) (5.13) s,t { perms R . perms + iε cuts in the iε + propagators 1 − iε − s iε iε + ≥ t n 1 } n + s ω ` + 1 and which should n ω s,t t 1 { + − ω n 1 R − iε s ) to get a Feynman ω # t − + and others at a phase E } t s − E s t E 5.11 ··· while going from R E ··· . ) s ) k ··· ` E cuts in ··· ω ω ` = 0 iε -channel and -channel propagators are unaf- , s t − t iε s − + s t . This gives + E E +1 ( . You can use a region other than ( 1 k δ } +1 δ 5.1 ) ` 1 cuts in ω ) s,t ω k { πi πi − { 2 2 R − s t − . cuts in the M − E ( ` ( E ) iε 1 + ) k k ` + ··· ω ··· ≥ ω channel in an analogous way, using an analytic – 37 – 1) iε k − iε i i s − − s ( ω ω t + + E 1 E =1 ∞ 1 ( − − 1 ( 1 ` X 1 ω 1 δ i ω i δ ) as applying at an implicit phase-space point in the P P =1 − ∞ − X ··· E k E ··· s t -channel discontinuity, the ) " t t 5.13 ) s 1 E E 1 ,P ,P ω = s s ω s t P P =1 } =1 n n Y Y − X ` k X 6= 6= − i i s,t s t { P P × × E R E ] ( ( = = ), as long as the paths in analytic continuation between these regions δ δ } k } } ` M ) ) t s,t s,t . Thus, our formula is derived assuming we want to relate cuts and { { s,t,u t πi πi { R R 2 2 ] ] R Disc R − − s ( ( M M t h t h ` . This allow us to compute k } Disc 1) 1) [ s,t Disc Disc − { − s . To remedy the problem, we rewrite each diagram in terms of all s ( ( (such as R t iε s =1 } n =1 We cannot immediately sum over TOPT diagrams in eq. ( We can take a discontinuity in the One should think of eq. ( n ` X − X k → Disc Disc s,t [ [ t { × channel, and all propagators are assigned × After summing over all TOPT diagrams with the same topology, we get fected since we have not gone around the branchintegral, point since at it is notget clear which Feynman propagators shouldas get we did for the sequential discontinuities in section Like before, when we take the continuation path in energy that encircles theR branch point for one cannot evaluate somespace of the point cuts in atdiscontinuities a at phase a space single pointR in phase space pointexist. in physical region where theresulting cut cuts graphs are to to anywill region not be one be computed. wants, the suchregion. same as One as the This evaluating can Euclidean is the region, analytically cut becausewhether but graphs continue the the the cut at the theta result vanishes, a functions rather phase-space than associated point by with in the the kinematics the of Euclidean original the region new determine region. In other words, where the sum is overt all diagrams with JHEP01(2021)205 ) } not n 6.43 ,s (5.17) (5.18) (5.14) (5.16) (5.15) ··· , without 1 i s i { + where the ∪S + R are positive i i . Then the R i S i }  S j R 1 n . Alternatively, S s s except for those } s, t i { i S + and then continue s,t . { R . . . ? ∪S i = n , R i R cuts in cuts in R k i 1 n S ∈ S M + k k ) j s 0 s  cuts from set being individual invariants ··· j M k i + h s ending in M 1 M . ) cuts in should be evaluated along paths k γ h ? − n n R M k ) = 1 m h M n n discontinuities in channel )( ( k t 0 m j i n cuts X ( m m N n ∞ = = X n m M k i ∞ = where all invariants in any set X M − n ∪S i ) k ··· and i R 1 ] S ∪S ) – 38 – ··· 1 R 1 M n = ( k ) } n m . The monodromy matrices are computed from the } m 1 n } M 1 m ( ,s s,t ) k m 1 + { s,t − n ··· { , R ( m S ] 1 1 1 ∞ R = s ··· X { 1 m k M R + ∞ t ! = ] = ( Disc X 1 ( n 1 i k S M m m R n Disc ··· ] s cuts m 1 = ··· ) N m M ! n − i ) 1 s 1 S Disc S [ discs m is computed by taking the monodromy from a region where all the invariants have the same sign as in discs N i N Disc j cuts Disc ( S S [ are positive through the Euclidean region and back. Then Disc N 1) / i i [( − − ··· is taken between the region S ! ∪S 1 n j 1 discs R S m = ( m ) N 1 ; since we are prepending the monodromies, whether we continue before or after s 1) } ··· − s,t ! , which are negative. An example of this type of set discontinuity is given in eq. ( { 1 One can generalize this formula to apply to In terms of monodromy matrices, this sequential discontinuity can be computed as One can even go one step farther and generalize from j Disc 1 R S = ( m [( where Disc to a region in below. The generalization to multiple sets and multiple discontinuities is additional complication: basepoint and the monodromies are prependedone to can the apply path the monodromyto matrices in some other region,we such prepend as them gives thetrue same of answer. cuts However, — we highlight for again instance, that all the cuts same evaluate is to zero in where we recall that(unlike the discontinuity action operators). of The these variationfrom monodromy matrix the matrices basepoint should to be the read region left to right discontinuity in invariants in This is the masterchannels. formula for Note computing that any thediscontinuities number right of are side sequential of taken discontinuities this on inthat equation the any the does left discontinuities not of side, depend these which on functions the points must order to satisfy. in ato which non-obvious being set sets of of identities invariants. For example, we might have a set where JHEP01(2021)205 ] t 3 or E 23 are t iε + prop- and (5.19) 2 + s iε E and − + s 1 E to begin with. = } s before computing s,t { E } R s,t I. { R should be computed by 6⊂ J M t will vanish whenever } and and then extending the path into Disc s s,t in the region t { t R J R ] 6⊂ . Then, M 2 t I ] for how to compute sequential cuts in- ) i P 23 , where all invariants other than if } J Disc s ∈ i s,t { – 39 – P appear. However, it is a general feature of TOPT R Disc = 0 t [ appear in the propagators of a single diagram, one depend only on the external three-momenta, which } = ( can only be nonzero when there exists at least one E s t s,t k } { E ω s,t R ] { and to a path going into R and s ] and M 2 E , and then analytically continue to t t ) t = 0 M i t E R ] for how to choose paths for analytic continuation. t P I Disc 23 in s ∈ i Disc s P M Disc t [ ). It follows that 2 = ( Disc [ E s + 1 E ], a different prescription for calculating sequential discontinuities in different ) is that = 23 t is empty. In such a case, we cannot immediately apply our formulas. . Since the monodromy matrix is independent of the endpoint of the integration, this . They defined these discontinuities as the difference between a function on different E 5.13 } } should be chosen. For the examples they considered, this algorithm worked. However, s Second, we go around the poles in the TOPT propagators by continuing the external It is worth emphasizing two conditions that are necessary for our proof of the Stein- As for the cuts, the prescription given in [ In [ s,t s,t iε { { when there are masslessR external particles, the on-shell constraint may meanenergies, holding the the region externallarities, three-momenta since fixed. the This internal energies allowed us to isolate the singu- tinuity in such overlapping channelsof must Feynman integrals. vanish, which we have thus proven atmann the relations level to hold. First,negative the and region all momenta arewith real, the must assumptions exist. of The axiomatic existence field of theory, such where regions all is consistent particles are massive; however, this section that This is a version of the Steinmann relations, which state that the double sequential discon- TOPT diagram in which both that whenever two energies must depend on a subsetand of the energies thatinvolve appear partially in overlapping the sets other of (e.g. energies. More precisely, recall from the beginning of restrictions. 5.3 Steinmann relations Finally, let us connecteq. to ( the Steinmann relations. One of the important implications of − in more complicated cases, itagators may that not appear correctly after accountexcluded for a from the first consideration discontinuity discontinuity. cases of where Thechannel. sequential main discontinuities Our difference, were formulas however, taken is allow in that the for same [ any number of discontinuities in any channels, with no monodromy matrix around R is the same asNo simply details computing were the given discontinuity in in [ volves an algorithm with tuples of black and white dots that determines whether channels was proposed.first Their calculating Disc proposal wasDisc that Disc sides of a branchthan cut. discontinuities across Using branch the cuts, language this of can be monodromies interpreted around to a mean branch first point prepending a rather JHEP01(2021)205 ) ) ]. t − 74 5.8 (5.20) ) only ) ln( s 5.15 −  ln( ) ), should still ε can only simul- ( t O 5.19 ) do not restrict all + and 2 5.19 π s term has a . With massless external ) does not know anything − ) t 0 t 2 ε ]. Thus, in many cases the ( µ − 5.15 and O 74 ln s s 2 µ − particles is that the massless condition ] + 2 ln 73  t 2 -matrix theory; since µ − S – 40 – external + ln partially overlap. The s 2 2 µ − ) 3 ln p  + 2 ε 2 p − -matrix theory, external states are stable and massless three- S = ( 2 4 ε t does not exist, so there is no contradiction with our formula. This  } . In 1 st and s,t { 2 G = ) R 2 particles are massless, our sequential discontinuity formulas in eq. ( m p 0 dimensions it has the expansion [ ), and correspondingly the Steinmann relations in eq. ( + 1 ε kinematic points, in the physical region. This subtlety appears, for instance, in M p 2 5.13 − = ( real internal s ]). When two overlapping momentum channels only depend on a single common = 4 Finally, let us highlight the fact that the right side of eq. ( If Because of these preconditions, the Steinmann relations in eq. ( 36 d not immediately apply. 6 Examples In this section,relations we between consider cuts and a discontinuities number developed in of the examples previous sections. in which we can check the general channels to vanish, evenlong if sequence these of partially-overlapping unrelatedgoverns discontinuities discontinuities. discontinuities are This that separated are is by computed relatedcuts at a to are a the phase-space accessible, fact point that andrelevant in region eq. holding which may ( all all not the correspond other relevant to variables real fixed kinematics, [ in which case this restriction does given in appendix point vertices do not appear. about the order ofSteinmann the relations discontinuities force begin any taken on sequence the of left discontinuities side. involving partially-overlapping This implies that the no such constraint.to Nevertheless, be with treated massless with internal carein particles, when [ applying certain the cuts Steinmannmomentum, also relations (as cutting have explained, both for channels instance, state can decays lead into to a a pair three-point of vertex internal in which physical an states. external Some discussion of these vertices is observation is consistent with resultstaneously from vanish outside of the physical region, the Steinmannand relations do eq. not apply ( [ apply. The keyconstrains problem the with surface of massless maximal analytic continuation; massless internal particles impose where component that has a nonzerolines, sequential the discontinuity region in possible double discontinuities in partially-overlapping channels.apply to In particular, discontinuities they on do sheetshold not at that are far removedthe from one-loop the box physical with sheet;In massless they internal only and external legs. This box is infrared divergent. of the external momenta,one must such also as rotate fixingcases, the their finding external the masses momenta singular to to varietyderivation maintain for zero also the the does or mass-shell TOPT not condition. some propagators immediately other is In apply. more such value, complicated then and our are held fixed during the analytic continuation. If one tries to impose a constraint on some JHEP01(2021)205 (6.9) (6.5) (6.6) (6.7) (6.8) (6.1) (6.2) (6.3) (6.4) . ) s , contribution), so that we can iε Θ(  2 2 π i + m 8 − 2 − ) 1 has a branch cut on k  1 ) = iε 0 − k M p − , ( − s 2 iε 0 e µ . cut 1 − p 1 +   M . 2 ]Θ( iε k 2 ,  Now we consider an example that ) , d − + ln ) = ) iε k k s π s π i  2 1 d = 0 8 − − µ d − (2 s − p i s 2  ) Θ(  = e [( µ − 2 3 1 δ Z πi π  ) 2 1 ln d M 2 − − 16 ln ) πi 4 (  s 0 2 2 µ − 2 2 – 41 – − π . 1 m m π 1 m = = 0 , this cut vanishes but the cut with energy flowing )( 1 16 . The chain of three bubbles is given by 16 M 0 0 − s − k − s dimensions evaluates to − s > < √ . The counterterm graph is analytic, so we add it to =   ABC 0 1 = 2 )Θ( 2 ( p 1 3 µ 2 1 ) k k E − 2 m > ( M γ k − M π plane. The discontinuity across this branch cut is . If δ − 1 s p → 0 , where ) s 16 s = 4 p πe πi > − R d ( p 2 Disc 0 p − = 4 = = ( 2 3 4 ) e µ = k π 4 M d (2 → i bare 1 and p 2 Z M p = = = cut 1 s M Sequential discontinuities in thehas same a channel. nonzero sequential discontinuityloops in massless, a but single give channel. theignore We internal keep their lines the discontinuities connecting propagators for the in bubbles the a mass monodromy computed around the branch point at gives the same answer in in the opposite direction compensatesthe positive and real gives the line same in result. the in agreement with the covariant cutting rules and the optical theorem. Similarly, the Here we have assumed The cut through the bubble is finite in four dimensions: where remove the UV divergencegiving and a the simpler answer algebraic for part the of renormalized amplitude: the integral (the The first examples wemassless consider internal are lines sequences in of bubbles. The single bubble integral with 6.1 Bubbles JHEP01(2021)205 . AB , cut 3 (6.16) (6.13) (6.14) (6.15) (6.10) (6.11) (6.12)  2 M π 4 = − BC  cut 3 iε M − s = 2 . µ − , are simple to calculate   AC .  cut 3 s iε 2 ln ). πi  M − πi 2 . s 5.8 2 m µ  − + 12 1 + 6 iε  . There are also three diagrams −   2 s iε − cut AC iε  s 2 − ) ln 3 3 − µ M − ) ) s propagators, the cut through loop A is 2 πi 2 s 2  µ 2 − πi = π . µ − iε −  ln 2 (  + C 2 6(2 (16  cut 3 2 2  2  6 ln 2 = m m M 2  m – 42 – 3 ln m m 3 1 m m m 2  m 1 = − 1  2 − M s 2 − s  B s  s cut 3 2 ABC m  1 ABC 3  ABC m 3 ) − M 3 ) 1 Disc 3 2 2 ) s s ) ) − 2 π , and using all 2 πi s  π → 0 2 πi π 16 → 1 3 2  p → − 2 16 Disc ) − 16 p ( ) s 3 − 2 p ( − ) ( s > − π ( 2 πi ( = = πi π (2 Disc = = 2 (16 = = and (16 − A 0 AB ABC cut 3 cut 3 cut 3 = = > 3 3 M . Thus, we have M 0 M p ). We find iε M M s s + 4.37 Disc Disc s Assuming Disc The other diagrams involvingThe two triple cuts cut give is identical results: The cuts of thepropagators second and third loopinvolving give two cuts. identical results Cutting since loops we A always and assign B gives uncut We expect these discontinuities to be related to cutsgiven by by eq. ( and using eq. ( Since this is just a product of logarithms, the discontinuities in JHEP01(2021)205 ) . In 6.11 . (6.24) (6.22) (6.23) (6.17) (6.18) (6.19) (6.20) (6.21)  = 0 2 π t 4 , discontinuity ABC , − cut 3 .        th  ). ) t M we get iε iε and at m . − 5.7 ABC ) ) − cut 3 − t s       ) + t = 3 s 2 2 − − = 0 ) ln( M πi 0 µ µ − − s BC 6 m s cut 3   . − − πi M ln ln ) and  + ABC ) ln( πi BC cut 3 t , we get iε cut 3 0 1 00 0 2 10 0 0 0 1 1 0 2 − We now turn to an example in- AC = 2 (3-cuts) 3 ) holds for the + 6 M − cut 3       M = 1 s 2  m M 5.8 = + M µ = 6 − ) ln( m iε t 0 s + +  − AC − cut 3 ln s 2 . -cuts) AB M 4 µ cut 3 k − M ( 3 ). For 4 π 1 ln( 00 10 0 0 0 1 0 ln( ln( 1 (2-cuts) 3  1 + M 2       M ( 5.8 256 – 43 – , ) AB = − 3 ln cut 3 3 − M k = )       0  γ ( t C 2 M πi P cut 3 k  M 2 1) M (1-cut) 3 0 0 − = 2( m ( + 1 M πi − , =3 B ∞ X s = k cut 3 , these branch points correspond to one-dimensional complex       1 2 0 10 0 0 0 0 1 2 0 0 1 -cuts) t  3! . This function has branch points at k t s 0 dt ds       M ( 3 , as expected. Similarly, for 2 3 t 3 ) ) − P + -cuts) t = 2 M dt and k πi π M ( 3 s 0 A s s = ) 2 s cut 3 s ds 2 16 k P t − M ( ( − M 0 00 0 0 0 0 0 0 0 0 ( ) = M k       1 k are st and 1) = ( = ( 0 = − 2 k ). s M ( ω P 1) t < =2 ∞ − X k 6.12 = ( -loop bubble chain. This is not particularly surprising, since the algebra involved 2 s n =1 and ∞ X k The connection and variation matrix for this function in the Euclidean region where One can similarly check that the relation in eq. ( 0 − and the monodromy matrices are s < where the space of complex hypersurfaces. We have depicted this in figure volving discontinuities in different channels. We consider the diagram of the is essentially the same as the algebraSequential used discontinuities to in derive equations different like channels. eq. ( and It can be checked that theseand quantities eq. agree with ( the discontinuities computed in eq. ( This agrees with Disc We can now compute the right side of eq. ( JHEP01(2021)205 → and )  s s (6.28) (6.25) (6.26) (6.27) iε − + ln( t 2 µ −  = 0 , shown in black. st ln = 0 , . s R } t 2  = 0 ) . 4 s,t t iε { π πi  R 2 + iε and 256 − s t 2 ( − µ − R t 2 = 0 =  s is the same with µ − s R st ln 0  .  − M 2 ln ) ) iε 4  t 0 4 t > where neither cut vanishes. There, π πi π πi + iε } 2 2 ? t 2 . We can compute discontinuities in − 256 − s,t − R µ 256 M − 0 ( { t 2  R gives − µ − = = and/or t < s 1  } st 0 )( s,t + ln ln s 0 { M and )   R – 44 – ] 0 s 0 . iε iε s > st ) M − − iε s < M by computing monodromies around the branch points − M s s 2 2 st − 4 has branch hypersurfaces at µ µ 1 st − − t − of this quantity gives ) t   − st t 1 M Cut − = ( [ t ln ln ln( M   st = ( 2 ) ln( 4 4 − s Disc → st M π π πi s s st ) − 1 2 t M corresponds to − s 256 256 − ln( M ? plane. Disc 0 t R = = = ln( Disc s, t . First, the discontinuity in } s,t Disc { = 0 R ] t -2 st and/or ) M and . The function ). iε tot by rotating around the branch points as indicated by the curves on the right. These curves − -4 s 5.13 = 0 0 4 2 We can also compute the total discontinuity of this function in We can compute Disc M -4 -2 − s Disc [ of We see that theeq. cut ( and the sequential discontinuity agree, as they should according to To compute the cuts,we we find must be in the region at Computing the discontinuity in pass out of the real The variation matrix inln( a region with Figure 4 The Euclidean region t JHEP01(2021)205 . , ) 0 } t 5 − = 0 s,t < { t . We 2 3 (6.32) (6.33) (6.34) (6.29) (6.30) (6.31) ln( . These , p 2 3 2 2 < z and p p 1 . . Finally, we = 0 1 , . The result is and  s ¯ z < , and  4 0 2 2 πi p πi < > , + 2 0 2 1 ], and can be treated < z < 2 1 + 2 p . p  , 0 75 ,  v v iε ]. Since all internal lines ) is flipped on the , defined as 2 2 iε ¯ z ¯ z − ¯ . Similarly, we denote the 76 z < − − iε − − t 2 u u t 2 µ − 2 2 < z µ − )(1  , and 1 z − −  z ] and [ , − 23 v uv uv + ln , which implies 2 2 , 0 + ln  u while st − −  = (1 < iε . 0 2 2 M iε 2 3 2 1 v v for real kinematics. The triangle ladders 2 2 ) − p p t 0 = 0 − z , p + + s 2 2 2 ¯ z < 2 1 µ i 2 2 ≡ s 2 µ − p , and here we have ≤ . p u u µ 2 − v ¯ z  M – 45 – ¯ z cut. We can also write this in our standardized R  P s 0 ln ↔ and 1 + 1 + s  ln by z  . We denote the region where 0 √ √ 4 and 2 3 M 0 4 π πi p + − > 2 π πi , as depicted by the green curve in figure − < ¯ v v z 2 2 3 } − 256 z p 2 3 1 are real, and − − − 256 s,t { = = , p ¯ z , and u u 2 1 = ( = 2 2 R } 2 2 2 1 p ). p symmetry } p p s,t , { 1 + 1 + → 2 s,t and 2 1 R { ≡ ? Z , this should match the function returned by the operator Disc p ] 6.29 and z R = = ] u st R 0 ¯ z z st 5.2 M → are homogeneous: > M } 2 2 tot iε p s,t { the region where Disc [ R Disc 3 [ R . In this region, 1 For these integrals, it is possible to find real phase-space points with any pattern of R signs for the invariants by region in which denote by This corresponds to theare convention invariant that under the where we choose 6.2.1 Triangle kinematics For the triangle integrals, we followare the massless, conventions the of amplitude [ dependskinematics only can on be ratios parametrized of using the the invariants variables massive external lines. These integralssimultaneously are known because to they all loop giveFor orders rise [ simplicity, to we the concentrate sameladders mostly transcendental at on the function end the at of triangleAll each the momenta order. ladders, section. are and Our incoming, momentum comment and labeling we on convention have is the shown box in figure in agreement with eq. ( 6.2 Triangles and boxes Next we consider the triangle and box ladder integrals, with massless internal lines and which corresponds to analytically continuing aroundalong both a the branch path points propagator because it comesform, after where the the According to section in agreement with the standard cut prescription, where the JHEP01(2021)205 3 0 p < . 2 2 1 ) 3 p p and also + 2 p 123 are real. The invariants (or R invariant (and , where corresponds to 2 i = ( are all analytic 2 j p . The Euclidean t v 23 p ? ? B ¯ z R n R R , which corresponds and work in a frame and and ? A 2 2 and holding all three- ) u ~p R we need to analytically 2 1 2 1 2 p p p p p − 1 + R 1 , and Li = 0  ~p that are related by complex 1 z 2 1 p n leaves p 3 − ¯ z E j , and region = = ( 0 , Li s + 3 Disc z and  ~p 2 z ln ¯ E for all z, , and the remaining unfixed momentum and + 2 j . A summary of the regions is shown in ¯ z has the same form in a given region and 2 ln p 1 ? C , ¯ z < z < ··· v E z E ··· R , in which all the invariants ¯ z − → − . It is nevertheless important to distinguish a 1 1 and 2 j – 46 – and E R p and u direction. Then, rescaling these momenta so that − ? B z x R = and , 3 involves complex in terms of ? A E ? 2 I 23 R E R R . The functions 1 . All of these regions correspond to two-dimensional slices of and corresponds to real ∗ 1 z ? C and back. Our formula relating cuts and discontinuities assumes E ). Of particular importance is the region 3 4 3 R = ? ¯ p p p z ), any function of ¯ z R ¯ z ¯ z < z < to and < 1 -loop triangle and box ladder integrals. We take all momenta incoming with and z L 0 R . Finally, region z , and so on. Since taking 0 > ¯ z < z . The 2 3 take the same form in . v , p < , we can solve for 2 2 6 1 p To take sequential discontinuities of Feynman integrals, we analytically continue around The Euclidean region, where all invariants are negative, is denoted = 1 and x 1 continue from that we rotate themomenta fixed. energies Thus, while we canwhere preserving set all momenta are alignedp in the has subregions corresponding to figure branch points where Mandelstaminto invariants different vanish. kinematic This regions; analytic for continuation example, takes to us take in this region.real Region conjugation, namely the four-dimensional space of complex dual of the Euclidean region, where all invariants are negative, is denoted region from its dual because cuts can only be nonzero forregion positive has invariants. aequivalently number of of subregions,to based real on values the relative sizes of the also consider dual regionsand in which two invariants aretherefore positive, also such as the dual region inu which all invariants have the opposite sign. For example, functions of Figure 5 along the long direction of the triangle. For the box ladders, JHEP01(2021)205 , 2 1 /p 2 3 p = v the invariants and 1 , has four further R 2 1 j /p 2 2 p = for all u 0 < space correspond to regions in 2 j p ¯ z z, and u, v – 47 – . The Euclidean region, where 0 < 2 3 p . The different regions in ¯ z , and and 0 z < 2 2 p , 0 > . The triangle ladder integrals we consider depend only on 2 1 p which the Mandelstam invariantssatisfy have different relative signs.subregions, described For in instance, the in text. Figure 6 or equivalently on JHEP01(2021)205 . 2 1 2 j ? A p p R , and . The . The (6.35) (6.36) . The 0 to (6.37c) (6.37a) (6.37b) ¯ z 23 ∞ 1 to get to R . is crossed. R 2 3 = ↔ z < 0 p ¯ z z → z + ¯ < ’s rest frame, 3 j x 2 . We also show x 2 . . For example, R p and ? A and 2 1 7 p 2 2 R z ) → < p . Thus, we cannot p ¯ z 1 and 1 . 23 R − − 0 and ) R , z 1 to an acceptable path in z )(1 R + ¯ ¯ z z z > ∞ z z ( − + M x 2 − x 2 p and . When this path intersects the , by the same type of argument − p ¯ z , this constraint is impossible to ? C ∗ B , or as going around z + 0 1 = (1 R R ¯ z 2 3 > = z = 1 2 to x 2 z 1 z p 2 = , p R . In fact, we need 2 1 2 must be positive in 0 and M ¯ zp 2 1 · , . z > p ¯ ¯ z 0 z 0 z 1 2 1 = 0 = p , we must have – 48 – 2 2 . It is also possible to construct a path from z ,E 2 ) M M M z R < z ). This choice amounts to permuting − − − , p 2 1 + ¯ ) 1 1 1 → R z z z ( = = = , after passing through − ¯ ? A z < + ¯ → − 1 3 1 2 2 ¯ z x 2 R ) R R R x 2 < R . It turns out, however, that an analytic continuation path p ] ] ] ∗ C ¯ z p x 2 2 3 2 1 2 2 0 )( 2 R p p p p → corresponds to going around infinity counterclockwise, where − , which implies z − ? A z 2 z → + ( R = Disc Disc Disc R ∞ x 2 [ [ [ 2 1 p where in R M E 1 4( 0 R or . are negative. For a fixed value of = ? A 2 . Conversely, no path exists from 2 j 2 1 ¯ z > R 2 p p R : would not be an acceptable path, as it would encircle the branch point in plane, the branch cut in the square root that distinguishes R x 2 to p j ¯ z → z > 1 → E . But this is a contradiction, since ∗ A R 0 πi ? C Re 2 R R e , all the < = ? A Having constructed these paths, we can enumerate the monodromies corresponding to Some paths that satisfy all of these constraints are shown in figure In addition to making sure the path exists, one must check that the path only encircles 2 1 → z → → . But then, in p R 2 2 ? A j R monodromy matrix infinity is approached alongthe some contour angle around infinity that crosseson goes the the below branch cut positive the on real real the axis. line. negative real This This axis choice before implies to the that go one below the real axis corresponds to taking In each case, there are two choices of Euclidean region that we can pass through (e.g. existence of such paths is required to take sequentialeach discontinuities of in the discontinuities we’rea interested single in channel, computing. we find For sequential discontinuities in that showed the impossibility of analyticallya continuing path between that starts andRe ends in This path can be viewedright as side going of around this figure shows paths between other regions, such as the desired branch pointsE in the invariants once.twice. For example, in particle we show aR path from satisfy, as R so go from To see this, note that in these coordinates, the invariants are given by In One can use these equationsenergy to for translate a a given givencannot value path be of in found between any pair of regions. For example, we cannot go from component JHEP01(2021)205 are equal ¯ z . The figure 3 or R . z → B

? B

) z ¯ + R z ( Im → 3

R

23

)

z ¯ + z ( Im ¯ /R z 1 and R 2 Re with a small negative imaginary R z 12 → z¯

. These paths each encircle some combi-

1 = 0 z ? A /R 3 ¯ R z 23 = 1 R

Re R

z R = 1 Re → ¯ z 3 → 13 2 R 3

R z /R 2 R , 2 = 0 1 R R z

– 49 – Re R → B . Sample paths of analytically continuing the energies R 23 → ¯ z A R ? C R R and C and → z R 1 13 R R → 3 R ). This monodromy matrix is computed in appendix → 13 R 6.33 , 23 R → . Analytic continuation of three-point and four-point ladder diagrams takes place in the 2 R To compute sequential discontinuities in different channels, we consider analytic con- → 23 to have a small positivepart, imaginary as per part, eq. which endows ( tinuation paths from regionsthese with invariants has multiple the positive opposite invariants sign. to To regions construct the in discontinuity which operator correspond- one of on the right depictsR paths relevant fornation computing of sequential the discontinuities branch into hypersurfaces different either shown channels: 0 as or black 1. lines, corresponding to where Figure 7 four-dimensional space of complex are shown. The figurecontinuities on in a the single left channel: depicts contours that are relevant for computing sequential dis- JHEP01(2021)205 ) 1 123 C R 6.33 (6.43) (6.38) (6.39) (6.40) (6.41) (6.42c) (6.42a) (6.42b) → 1 ¯ z < z < R < , so → . We first take . This suggests 0 2 , , . 12 πi . ¯ z 1 z 1 z 0 = 1 123 C R 2 R z R . − in 7 1 M M M . , since . and ? A , . 1 z − − − )) = 2 1 2 1 R ¯ z = 0 1 1 1 p p 2 3 = 0 − . = = = < z < p ln ln M z 0 z · 0 d d 23 13 12 ln must satisfy ¯ z 0 in the region R R R ln(1 − − d ] ] ] γ 2 1 2 3 2 3 2 2 d 2 2 2 3 γ ¯ M z < p p p p p p I · and the other squared momenta are M z 1 ln ln ) + = 0 z − Disc Disc Disc d d 2 2 [ [ [ > p 1 − = , while M 2 i ln ) = = p z ) if both of these branch points are encircled 12 , our path ¯ z − d 2 1 R ln(1 12 ln ¯ γ p 1 − R I d d – 50 – ] 6.41 ( in 2 1 = γ , , p + z 1 z 1 I . This path encircles ln(1 z 123 C } < z d are analytic there. To evaluate the matrices for real 2 3 R ln Disc ] 1 [ ) πi, d M M , p 23 z ) + 2 2 · · , where only S (¯ , i should encircle 0 along this path. We see that this can be z p ¯ ¯ ¯ n z 0 z 0 z 0 = 2 ¯ πi , z < { R ¯ z 2 − 2 1 < Disc = p − [ 0 M M M ): ln 23 ln(1 and Li S d − − − ) = d ) z γ a small positive imaginary part. It can be checked using eqs. ( 6.32 1 1 1 z I ( ) then imply that 2 i ln ¯ = = = n p d Li , which computes a discontinuity in 13 12 23 below 0 or above 1, we need to be careful about which side of the branch 6.39 + R R R 12 z, ] ] ] ¯ z z 2 1 2 1 2 2 R p p p ln ¯ ln the path encircles. Let us illustrate how this can be done for the path from z, d and → ( ) and ( ¯ z Disc Disc Disc z γ 2 ln [ [ [ I R should encircle 1 while 6.38 z and → It is easiest to compute the monodromy matrices in one region and then continue the One can also consider other analytic continuation paths, such as Using similar reasoning, we compute the discontinuity operators in each of the regions z 12 result to the otherso regions. all of The mostvalues of natural region tocuts use we is are on.negative, In we assign the region to the pair of invariants Other paths that encircle the branch points of more than one invariant are also possible. In contrast to theso first discontinuity, there the is region thatcorresponding only we to a pass these single through discontinuity operators is correct are completely monodromy depicted fixed, matrix in figure in(not shown each in of the these figure). cases. Such a The path paths exists and gives us the discontinuity with respect that achieved in a manner consistentclockwise. with eq. Thus, ( we conclude that involving two positive invariants to be Eqs. ( We furthermore have that Since we are considering a discontinuity in in R the differential of eq. ( ing to each of these analytic continuations, we need to determine which branch points JHEP01(2021)205 ). , the (6.50) (6.46) (6.47) (6.48) (6.49) (6.45) ? A on the (6.44a) (6.44b) R 6.44 ¯ z iε , i + z and 2 ) ln ¯ , 1 z k . .  − i + ) z       ¯ z 3 ) πi z p z ln − ) . ( (¯ ¯ z h ). We find 1 + 2  . In the region z iε 1)( ) − ) πi ¯ iε , iε . z z z ¯ z ¯ z ln ¯ + − 6.37 (1 + − − − ln ¯ z, 2 z, z = 2 − ) + Li ( ) ¯ ¯ ( 1 1 z z ( 1 1 − z 1 k ), we can calculate the differ- z   ( z , to 1 → → − )Φ ln ) z 1 ¯ ¯ z z ¯ 2 Li z 4.62 ) ln 0) − p ¯ , the Feynman integral is z , ): ( − z, ∗ z ( = ) + ln z (0 )Φ 1 iε M z iε, iε, the following small imaginary parts in z (¯ Φ (¯ 4.57 = πi 1 + 1 ¯ − means replacing 1 z ¯ z − + ) and ( ¯ z 2 1 1 Li z z 1 ) + ln( k − ( R z ] 4 ) + 2 − z (¯ and 4.59 1 ) → → 2 ) k ) + Li 2 1 iε T π z 4 z z z p z 1 2 Li ( – 51 – d ( p 2 + , (2 1 2 1 1 i π : : i z ) Li Li − 16 h z Z Disc 23 ) (¯ ln(¯ [ z 123 z 1 πi = = ( 2 ,R − 2 Li ,R 3 was given in eq. ( z − 2 3 Li + ln ¯ 13 p p − 1 ,R = z ) 2 Φ 3 ,R z 1 q ( R 12 1 ) = 2 ) + ln )Φ 00 10 0 0 0 1 0 ln 1 1 ln ¯ z z , this function is analytic. z Li (¯ ,R       ? A h 1 2 1 z, ∞ 1 q q ( R πi = 1 Li R M Φ ? A − 0 − R = 2 γ ) and 1 1 1 p ? I iε ( M R )Φ − 0 z z ( 1 = , where all invariants are negative and ) that this corresponds to giving 1 ? I is the straight-line path from the basepoint Li M T R 0 − γ 6.34 Using the monodromy matrices in eqs. ( The variation matrix for 1 ( Rewriting these results inrelevant terms region, of which in logarithms the with case manifestly of positive arguments in the ences of paths relevant to evaluating the discontinuities in eq. ( and Here variation matrix is analytic.appropriate sides In of other their regions, branch it cuts has as the determined by same the form displacements with in eq. ( where In the regions the different regions. 6.2.2 One loop The one-loop triangle withregion all massless internal lines is finite in four dimensions. In the these regions: These assignments allow us to evaluate the variation matrix and monodromy matrices in and ( JHEP01(2021)205 ]. 23 . In (6.53) (6.54) (6.55) (6.56) (6.52) (6.51c) (6.51a) (6.51b) = 0 )] 123 C iε . R ) ] + 2 1 1 . p T z  in the logarithms, ). This gives Θ( 123 ¯ z z ln(¯ ¯ z S o , all the cuts vanish ] − 6.43 − , , iε ) 123 A  1 + ln  Disc . A path between these iε − ) [ R ) ¯ z z ¯ z ) ¯ z → z − ¯ z z 123 A − ) − − − ) ¯ z z , ¯ which can be deduced from z R ¯ z − 1 1  1)( 2 3 1)( − ¯ z z − (1 p ln → . and − z  − (1 − − , using eq. (  (1 z z ? A 3 ¯ z − ) + ln( z ¯ ( 1 1 z z ( p z R    and − πi (¯  − ln[ 1 2 ]. . 1 2 z ln ln ln ln → − p Li  and 2 1 23 ¯ ¯ ¯ z z z ] ¯ z ) ) → p 2 ¯ − z z 2 πi πi πi iε − − − πi p z 123 A 1 − ) 2 2 2 π − − 2 z z z z − R z ( )( )( 2 1 2 1 2 1 16 ) 2 1 1 ¯ z z p p p z – 52 – p 2 2 2 2 Li = ) as well as the corresponding expression in [ 1 1 1 − − 1 π π π − π + (1 (1 16 16 16 123 C (1 16 , we have  R ¯ z πi 3 6.51a i = = = = − 1 p [2 ln 2 3 1 T 1 ¯ ¯ z z R R R 2 3 ln[ R ↔ p i i i i πi πi 1 1 1 − − n 2 1 2 2 ¯ T T T z p , and evaluates to z z T 2 2 2 3 2 1 2 1 Cut 0 2 1 2 1 p p p πi p − p p 2 + > 2 2 z in the rational prefactor. The action of this symmetry is discussed 1 1 can similarly be computed, and agree with the discontinuities in 1 . π π 2 1 2 1 ) ), and with the results of [ ): Cut Disc Disc Disc T p ¯ p z 2 3 h h h h 2 2 16 16 2 D p p 1 − π z = = 6.51c 6.51c ( 16 Cut and h 123 C = 2 2 R requires ] 1 → − p , this can be written 1 2 1 T 1 ) T ) and ( ) and ( 2 1 p ¯ z p R 23 − S z Cut ( the sum of the cuts also vanishes, although each individual cut does not. In other 6.51b 6.51b A similar example involves going from We can also compute the discontinuity in The corresponding cuts must be computed in the appropriate region. For example, Disc [ 123 A regions exists that doesR not go aroundwords, total any discontinuity branch in the points.separate dual channels Euclidean So do region vanishes, not.individually but (and the In the discontinuities contrast, in total in discontinuity is the still Euclidean zero, region using the same path). eqs. ( Again, we see the discontinuities and cuts agree. We should compare to the sum of the cuts in matching the discontinuity inThe eq. cuts ( in eqs. ( In region while in detail in appendix the cut in As an initial crossdihedral check, symmetry we that note permutes that theand these legs the discontinuities of transcendental map the part one-loop tothe of triangle. each legs; these other Both for functions the under instance, pick rational the under up part a sign under odd permutations of we have JHEP01(2021)205 ). As (6.62) (6.60) (6.61) (6.57) (6.58) (6.59) ); that 6.58 . by passing 5.13 ] . . πi 123 A 2 ¯ 2 z we would have , ) . ¯ z ) for computing ) R 2 2 1 2 ¯ ¯ − z z πi ¯ ) ) z − ) πi T ) + 2 → z 2 2 (2 z − − z πi πi 6.42 − (2 p πi 2 1 z 23 z 2 1 z (2 (2 p (2 ln(¯ p 2 R 2 1 2 1 can be computed in the 2 1 2 1 Disc p π p 1 p − π 2 3 2 3 2 2 2 → p p ) 1 1 1 16 π π π 16 z 16 16 − 123 = 16 A and 1 − R − j . = = 2 2 T ∀ ) p 1 1 = = ) + ln( ¯ z 0 T T z ) ) 12 (¯ 13 ¯ ¯ z 0 z 1 1 R . Similarly, we find R ] ] 1 Li 23 M 1 T R T = 0 ] 2 1 − − 2 1 M M p 1 p j ) 1 T z R − − 2 3 ( i )( p 1 . Using the discontinuity operators defined in 1 1 1 z 1 Disc Disc 0 T 2 2 )( )( 2 3 Li ). In the case of Disc 2 j – 53 – p [ ¯ ¯ z 1 z 0 p p < Disc ¯ z 2 2 2 1 M p 6.37 p πi − Disc Disc 2 Disc − M M one can easily go from z 2 j p 1 = [ 2 1 − − Disc ¯ = [ z p 1 1 2 , and 12 13 = ( 1 = [ 0 π R Disc R ] ] and h 1 23 = ( = ( 23 1 > 16 z T R R 3 2 T ] 2 3 ] 2 2 2 3 R R 1 p = 1 p ] ] , p T T 2 3 1 2 2 0 . The discontinuity along this path is , these relative signs are expected from the invariance of the triangle 2 2 R R T p p ] ] . The corresponding double cut in ) Disc 1 1 1 D Disc > z 1 2 1 2 1 T T = 1 T p 2 2 p 2 2 2 3 2 3 Disc p Disc p p z p 2 3 2 3 p p M Disc Disc · [ [ Disc Disc Disc and ¯ z 0 [ [ 2 2 Disc 2 3 2 2 Disc p [ [ p p , where = 0 23 ), we find ). We find that these double discontinuities vanish in all channels, M ¯ z Disc Disc R [ [ − 6.42 6.37 1 Let us also reiterate that all the discontinuities we consider are computed along paths To illustrate the importance of using the specific operators in eq. ( To take sequential discontinuities in a single channel, we iterate the monodromies in ( seem more natural, by continuingFor the example, Lorentz by invariants directly, continuing one canaround run into trouble. The results differ by a sign.by This analytically highlights the continuing importance ofadjacent from computing regions. the the discontinuities region in which thein cuts external are energies being such that computed energy into is conserved. If one tries instead to do what may found discussed in appendix integral under permutations of its external legs. sequential discontinuities in different channels,the we discontinuity can operators see from what eq. happens ( if we instead use and Notice the additional minus sign in both of these expressions compared to eq. ( Notice that we couldboth have sequences equivalently of taken discontinuities theseis, are discontinuities we related in have to the the other same order, cut as integrals by eq. ( such as Disc region eq. ( eq. ( This is consistent withchannel. our We can expectations, also consider since sequential discontinuities the of the triangle triangle in has different channels, at most one cut in each JHEP01(2021)205 . πi 2 , (6.63) (6.64) o (6.66c) (6.65c) (6.65a) (6.66a) (6.66b) (6.65b)
, z , )] ]. o ):
z ) (¯ 23 )] 2 z z + ln ¯ (¯ (¯ 2 6.37 2 Li ) 2 z 2 ). The variation )). The resulting Li − Li k ln ) , − ) by the extra z − −
( 6.34 o ) ) 6.44 1 2
z πi z k ( ( ( Li 6.52 2 πi 2 2 2 )[ ) + 2 Li ¯ Li z 2 [ z − z k

), and ( ( z 2 ln ¯ iπ iπ − in eq. ( ln ln ¯ 3 − − + 6.33 p 2 1 z
( z z 123 A 2 2 ), ( πi ln k R 1 )] + 2 . z ) + ln ¯ + ln ¯ 1 2 + 2 o in (¯ 1 6.32 ,
z z 3 p z 1 z , where the relevant monodromy matrices o Li T ) ) + ln ln + ln ¯ 2 1 E ln ¯ z z p 1 − (¯ (¯ takes the form − 2 2 k ) − − ( ) z z ) + ) ¯ z Li Li 1 2 z ( 2 p p – 54 – z z ) 3 ln ( (¯ 1 ln z, − − 3 3 ( Li k ) ) 6 2 z )[ z z Li Li z − ( ( ¯ ) Φ z 3 3 2 2 ¯ z ln z 3 ln ¯ p − − 2 1 Li Li z z, ( 4 5 ) ) ( n n 2 1 2 z z 2 2 ln + 3 ln( k (¯ ( ) ) 3 Φ 3 3 1 2 4 − ) 2 ) πi πi in these expressions can be absorbed into polylogarithms that . ¯ z Li Li k )] π − 3 3 4 z 1 iπ − (¯ n n n d (2 1 2 i 4 = (2 = (2 = 0 the function , but differs from Cut z × × × ( 1 2 3 satisfy the relations in eqs. ( Li ? A Z ) ) ) 2 3 123 A R R R ¯ z p − R i i i 4 R πi πi πi 2 1 1 ) 2 2 2 ) p k π z Φ Φ Φ 4 1 3 ( 4 2 1 2 2 2 3 p and ) 4 d (2 p p p = (2 = (2 = (2 i π z Li 1 2 3 (4 Z R R R Disc Disc Disc i i i 2 1 2 2 2 3 2 2 2 = = = p p p Φ Φ Φ 2 ) = 6[ 2 2 2 3 2 1 ¯ z T p p p Disc Disc Disc z, The sequential discontinuities in these channels can be computed using the same mon- We first compute the single discontinuities, using the operators in eq. ( h h h ( 2 Disc Disc Disc h h h Φ are manifestly real in theexpressions appropriate region agree (taking with into the account eq. cuts ( computed in eqs.odromy (5.26), matrices. (5.37) and We find (5.41) of [ All the explicit factors of are also presented. where in the region and as before matrix for this integral is described in appendix Thus, specifying the regions ofto interest connect is them. not in general enough:6.2.3 one must also Two know loops how Next, we considermassless. the The two-loop Feynman integral triangle. evaluates to As before, all internal lines are taken to be This is analytic in JHEP01(2021)205 . In , and (6.73) (6.70) (6.71) (6.72) (6.67) (6.68) (6.69) 1 reg R m in ) /z (1 2 Li . , cut , , 2 − T cut cut ) 2 2 . This is consistent with 2 ¯ z T T ¯ z / = 0 = 0 − / 1 2 . = = 1 2 p p 1 (1 p p 2 1 = 2 ¯ 1 2 z z p p p p , 1 2 6 → 6 p p ln 6 6 ¯ z 6 = 0 3 2 2 reg 5 4 5 4 p j /z, 4 5 m R 1 5 4 3 3 i 4 5 3 2 3 ln , and work to leading power in → get mapped to minus each other under 3 Φ ) 2 j z ¯ reg 1 z 2 1 2 1 p R 2 1 2 1 m  − 1 2 2 z Disc ( Φ 2 j 2 2 2 – 55 – 3 3 1 p p p p π 3 3 3 2 p p 3 p p 1 2 ) can be rewritten as Li Disc Disc p 2 j 2 2 = = p p 64 = = 3 3 . The triple discontinuities all vanish, = R R 3 3 6.66a = i i 3 D R R Disc Disc  i i h R i cut 2 (234) (135) , , T (135) (234) (45) and , , , which corresponds to , 2 cut cut (12) (12) 1 p cut cut T T (45) (45) cut R (12) h h  T T T 2 ↔ h h h Φ 1 2 1 p p Disc 2 1 p , we find 3 in the figure below with a small mass R 5 Disc  These sequential cuts in the same channel have not been computed before to our and and where The other cuts give multiples of this expression. In particular, we find in accordance with the fact that there aren’t threeknowledge. cuts To in do any so, of we4 the regulate channels. the IR divergenceregion of the cuts by giving the lines labeled thus the permutation what we expect from appendix Note that the right side of eq. ( JHEP01(2021)205 ). ). 4 2 − d , 6.77a (6.76) (6.74) (6.75) x ). (6.77c) (6.77a) o ) (6.77b) z x . ( ) both in- ln ) 6.77b z (¯ dxδ iπ 2 6.73 , ) by the same R − ) would vanish. convention, the 2 Li ¯ 2 z z R i iε ). , − − 6.66a ln ¯ 6.73 such cut graphs may + o ) z z . z z 5.6 4 (136) ( , ]. However, summing channel are related by o 2 ln ) and Eq ( z 23 2 1 + ln ] uses a different cut pre- cut Li (46) d > p iπ ln 2 z T 23 h 6.72 2 3 2 − iπ p ) 2 1 ln ), and eq. ( − p , which agrees with eq. ( iε = 2 1 2 2 1 ) o 2 π 2 + , 6 − iε 6.72 R π z i channel, and thus the sum of double ) ) − 256 − 2 2 ¯ iε z = 0 z ln(¯ z p − 3 z, − z Recall that [ ln ¯ ( R ln(¯ = ), eq. ( z is exactly zero, z 2 9 # ( z T 2 3 2 ]. 2 2 propagators, and that they use dimensional + ln T p ln R 1 2 Li + ln 6.71 23 p p z z iε − 2 – 56 – − 2 6 + z (they contain integrals of the form ) Disc ln ln 2 2 2 z X 1 2 p (¯ 1 2 2 ln ), eq. ( − = 4 5 4 and − double cuts 1 2 Li z " d 3 Disc n n n iε h 6.70 2 2 2 ) ) ) − ) would give a non-vanishing result in dimensional regular- 1) ln = πi πi πi 2 1 2 − ). Note that the diagrams in eq. ( R z ). The sum of double cuts in the 6.69 # 2 = (2 = (2 = (2 5.6 ), we then find T 3 6.66c 13 12 23 p ) + ln( R R R z i i i 2 2 2 6.65b − to the sum of double cuts in the Φ Φ Φ X , we find that summing their expressions with some typos corrected gives = (1 2 1 2 1 2 3 ¯ z 2 2 p p p for more details. 2 / Φ double cuts Li 1 2 1 R " p G i Disc Disc Disc ↔ is related to the sequential discontinuity computed in eq. ( ) + 2 3 2 2 2 2 z p p p , there is only one diagram. We find Disc ¯ z (¯ , we believe it is safer to use a mass regulator. With our 1 (136) 2 2 2 2 , , p R R Li G /z Disc Disc Disc cut n (46) 1 h h h 2 Finally, we compute the sequential discontinuities in different channels. We find In T ) These equations differ slightly from eqs. (6.4) and (6.5) in [ h 9 ↔ πi regularization and so masslesspendix three-point vertices vanish. For reasons discussedthe in ap- resultsFor from Disc their(2 appendix D, we believe their (6.4) should agree with our eq. ( We believe these agreescription, with the which results involves in both [ in agreement with eq.z ( cuts in combinatorial factor. These provide highly nontrivial checks of eq. ( Comparing to eq. ( be zero, while they areIf nonzero we in were to setthe them example to above, zero, as we eq.ization, would ( while get the the wrong graphsSee answer. in appendix This eq. can ( easily be seen in which agrees with eq. ( volve an isolated three-point vertex with only massless lines. For It follows that the sum of all double cuts in JHEP01(2021)205 . G . o (6.81) (6.79) (6.80) (6.78) z o 2 1 } ln ¯ z p p z πi ln 6 ln ¯ 2 iπ − − z − z 5 4 2 1 1 2 p p z p p ln 3 ln { 6 ln ¯ 6 3 iπ z ) 2 1 + πi 5 4 2 z 5 4 + ln R 3 3 z i ln ¯ = (2 2 z 1 2 3 p p p ln 2 1 2 1 1 2 6 (3-cuts) + ln 3 − T z + ) 6 2 5 4 3 2 1 ) tells us we should find iε 3 p p p ln − p 3 ). In particular, the three-cut diagrams − 1 2 5.6 6 z ( − 2 + ), we must in our analysis add the three-cut 2 1 ) + 6.77b (2-cuts) Li 5 4 3 1 2 iε 1 2 p p T p p 3 – 57 – − 2 − ) 6 h 6.77b 6 z z 3 (¯ p ( = 2 2 2 1 2 5 4 Li Li R 5 4 n i 3 + − 2 3 3 ) ) T 1 2 3 2 2 z p p p πi 2 p (¯ ). We find 2 1 channel, where eq. ( 2 2 1 6 2 2 Li p give = (2 Disc n 5.15 + h 2 . ) 12 12 5 4 3 ) the sum of all cuts gives 3 p R p R G πi 3 i 5.15 = (2 = 2 1 = (3-cuts) 2 12 12 Φ R R i i − 3 p containing massless three-point vertices must be added to get the correct result. We (2-cuts) 2 (3-cuts) 2 Φ (2-cuts) 2 Φ h Φ h (3-cuts) 2 h 6.2.4 Three loops It is instructive totwo continue cuts to are three taken loops. in the The most interesting case is the one in which have verified this result using aNote mass that regulator, while and the these techniquenaïvely diagrams discussed be add in set up appendix to toto zero a a in finite wrong dimensional result result. regularization in Furtherbe as this discussion discussed found case, on in earlier, each how appendix which to diagram would calculate would lead massless three-point vertices can in agreement with theΦ discontinuity in eq. ( Inserting into eq. ( To match onto the discontinuitygraphs in according eq. to ( eq. ( double-cut graphs in JHEP01(2021)205 , ! ) 2 2 , 1 p m , F i o 6.82 o ) (6.87) (6.85) (6.86) (6.82) (6.83) (6.84) z )] (¯ ) z ’s in the . 2 2 . We can (¯ ¯ z 2 2 4 iε o z Li p ( − Li in eq. ( 2 )] − z 2 (¯ − ) C 6 )] z . ) )] ln 1 ( z ’s to z 3 z C (¯ Li ( 2 2 C (¯ , 4 iε 4 4 2 − Li C + 3 Li Li h . ) Li is the number of cuts prescription involving T z 1 2 4 [ − ( )] − iε 6 m ) z ) − − (¯ z + 2 2 1 z Li 3 1 and summing over the one- 3 ] ( ( k k 3 )] p p 1 4 4 C z Li , πi C + + (¯ 1 Li Li C ) gives 120 [ 3 3 3 − 3 p p T ) 3 [ Li − z 12 [ ) + 2 ( ) − ¯ 6.37 z to correct the − 3 ¯ z − z ) in a way that allows us to recycle + 2 n ) z ) 3 Li ) z [ ¯ z C ( ¯ z 5.6 z ) 3 ( ¯ z − )] [ln( 3 )] ln( (3-cuts) Li 3 z z [ z on r.h.s. − 1 (¯ (¯ T ( 3 5 z 6 ! iε 4 3 )] ln , with the traditional ] to agree with the discontinuity in ( 2 p
z − Li 2 3 − Li 2 4 3

2 1 2 3 (¯ p i 23 m 2 C p – 58 – 3 1 p − − C 2 1 2 , p k , 4 − ) p Li ) 2 2 1 k π 3 z C z 3 , we show that one gets an additional factor of

− ( ( , p − π 1 T 3 2 5 2 1 channel using eq. ( ) G p . p z Li 2 2 Li 2048 [ ( 512 + ln iε 1 p 2 3 = 2 C n − + T = 2 ) 1 Li [ ¯ z k 3 R = C + 60 [  , n 3 3 − 1 ) C T T 3 ¯ z z 2 2 ( 2 T 3 p 2 − 4 3  p 2 2 + p 1 z p 2 1 ( p Disc 4 3 4 p 1 π 2 1 Disc p   6 ) 1024 π . The result we want to verify is therefore ’s to the right of the cut, adding a triple cut term to compensate for it. When 2 = on r.h.s. of cut C iε 6 (4 3 1 iε C − T 3 − − 2 2

T 2 p 2 2 = C , The three-loop triangle ’s to the right of the cut, was shown in [ 3 1 C 3 T Disc iε T The first two terms in this expressioncuts correspond of to the cutting two-loop in triangle.and The the details result of is the calculation are worked out in appendix doing so, we must bethree-point careful with vertices, the as combinatorial factors theseintegration that endpoints. cut come integrals along In with involvecompared appendix massless delta to naïvely functions evaluating with thesebeing delta support taken. functions only Thus, to we at 1,term must where absorb a term results for the singlechannel of cuts the of two-loop triangle the− two-loop triangle. Theuse sum these results of if all weto single rewrite have cuts the term in corresponding the to the double cut while taking three discontinuities results in To facilitate the cut computation, we rewrite eq. ( Taking two discontinuities in the is given by when we assign all propagators JHEP01(2021)205 , . )] ) . z (6.93) (6.90) (6.91) (6.92) (6.88) (6.89) z (¯ ( 1)! iε j 1 . − − , which is Li + o j o , we get j iε 2 )] ( )] z ln )] − ) z z 1 + z 1)! = 1 ) (¯ ) L 1 (¯ (¯ − ¯ z z 2 k − 3 4 )! n z 4 ( ) z n j ( 1 ( ln j 1 Li Li − Li k j − 3 Li − πi − − [ p − − L 2 ( L ) ) ) ) 2 2 z z ¯ z z )! ( ( p ( z ( j 4 ··· 3 )! (2 ( ! ln 4 j 1 j Li L − iε gives Li 1 2 iε k [ j Li − p p [ 4 [ L − L + 1 # − + 1) 2 − 2 j 2 L 2 − = 1 1 πi − ( p ) k k  ] ( )! (2 1 ! ln z ) 1 j k − L z L πi 1 L (¯ j 3 ··· = 2 k X ) + 2 p − j 2 ] − ¯ 1) z 3 iε 1 2 j z Li 2 75 p k − ( ) + 2 − k ( ). Using the fact that the discontinuity 1 2 ( 1 + ¯ z L −
used to regulate the IR divergence of the z 2 2 L 3 iε L − 2 3 k p = 2 k p X ) 6.92 + ln ( j + ··· z iε 1 channel) gives zero. To see this, we note that ( 2 1 2 ! )] [ln( ) − p 2 1 2 1 2 3 + 2 3 z L L p – 59 – L (¯ p 2 1
Li m k 1 2 3 2 3 k ) 1 2 3 ). p − − 4 C π Li  , ) L

1 2 2 1 π C k − + 3 − p 1 ! (4 6.84 4 ln ) L T (2 d L L z k i 2 − factor in eq. ( ( ( ¯ ) z 3 " πi ) + 2 π 2 z n − ··· Li 3 ··· ( [ ) corresponds to taking a monodromy around C − ) and ( z 4 L j , ! (4 ¯ z ) n k 1 1 is given by iε 2 3 L C -loop triangle integral, ) k π 3 p = − 3 ¯ 4 z ¯ z L + T C 6.86 3 (2 d z 2 2 ( i − R 1 C − 3 ) 3 4 3 p 2 z 1 T + 2 z i p ( Z k 1 2 3 2 1 4 3 − p − = = p , p 1 4 2 1 2 2 1 =  π p k 2 3 4  ( , p on r.h.s. defined as before. 2 3 2 1 π , p p iε ¯ 2 2 × z 2048 , p  loops − 2 2

corresponding to encircling the branch point at , p is a small mass of the line labelled as L 2 = 1024 2 1 , p L ) C T p , and 2 1 3 2 3 z m 1  C p = p ( , C z 3 2 n  L One thing we can immediately observe about this expression is that taking two or more T C 3 T L 2 T T Disc h only nonvanishing for the Li of Li with discontinuities along the long axistaking (in a the discontinuity in The result after performing the loop integration is [ in agreement with eqs. ( 6.2.5 Let us now consider the The sum of all cuts is therefore cut graphs. The cut where JHEP01(2021)205  1  0 L 0 L k −  k cuts  1 L (6.97) (6.98) (6.99) (6.94) (6.95) (6.96) k L − · Θ on-shell 2 1  disconti- . . p − 2 L L 2 2 2 0 L k L ) ) k  − L . L  . 2 2 k δ k . ···C 1 p )] 1 Θ 2 )] + 1  z + , and get i 3 z ··· C R (¯ δ
2 3 (¯ p  ) L 2 p ( L L i L 0 1 ( ) L k 2 3 k Li k Li ···  πi , p δ ··· − 2 . Thus, further discon- − 2 2 − (2 ) Θ ¯ z ) 2 1 1 ) , p ) ··· z  L − k z ( 2 1 2 k 2 1
( or 1 L p k 2 1 k ω L + k L L  k 1 z  2 3 ( k + Li p h 3 L δ 3 [ Li ( ) 3 δ [ T δ d L 1 p ) π L 1 ) is indeed satisfied. 2 ( channel amounts to taking − L C ) − , (2 ··· 2 L . k 2 2 L ) ·
πi ··· p 
1 , 2 3 1 6.95 2 0 2 2 3 ··· 1 k p − C k p = 0 2 − 1 L . The result of the integral, worked out k ( + 2 1 ! 3 -loop triangle along the long axis must 1 k − 2 1 1 1 ω p L 4 3 R 2 L 2 p Cut 0 ) 2 1 i p L k L ! ( k 3 L − π k ) 3 L ) ) ( L  4 d T π h (2 π δ π d × 2 i Θ ) – 60 – (2 (8 = 2 3 ! (8 i ··· p 2 1 ¯ z 2 ) L ) ··· k 2 R 2 L ¯ z ω 4 k − i 1 k · ) 2 1 , evaluating the integrals over these remaining delta L i Disc k are all collinear. We therefore get a product of z − 1 i k π 3 ( 2 3 3 − ) k 4 h z L d − G ), we see that eq. ( (2 1 d π 2 , p k in the expression above. Only the first term in the sum i discontinuities in the k 2 2 − = − ( (2 ) ( − h L , p 2 6.96 ¯ = z Z δ 2 1 1 δ Z R z p  − ( ×  = 2 3  = L i 0 j 1 2 3  k k L −  , p 2 3 2 3 L 2 2 T , is , p 2 − L 2 2 ··· , p ,p , p
0 2 , 2 2 ln 2 2 G 2 1 2 2 2 p , p , contributes to this discontinuity. The result is p p 2 1 , p k  ,p  2 1 L p 2 1 gives the following expression: p Θ −  p L  =  i Disc 1 2 T L 2 vanish, k L j L L T ) R h 2 2 T 2 3 T 1 L p p ···C L k L 1 C C , , C and − ··· ··· , , Disc 2 L 1 1 h p k , where C C Cut ( j h We now show that taking ··· δ , Cut Cut 1 × in detail in appendix Comparing this result to eq. ( The remaining delta functionsk show that thismassless cut vertices. only has ThisTOPT. support As configuration when explained is the in singularfunctions appendix momenta gives and rise must to a be combinatorial treated factor of with care, using We perform the energy integrals using the delta functions Next, we calculate thein cuts. the Putting region the lines corresponding to the cuts nuities of the factor over in the same channel, i.e. We start by computing the sequential discontinuity, which amounts to taking tinuities in The sum ofcorrespondingly taking also two vanish. and more cuts of the In this expression, there are no longer branch points at 1 in JHEP01(2021)205 1 2 p p 2 . (6.102) (6.100) (6.101) , which x 2 x ) 4 1 for each loop ` x i x ` 1 − x to infinity. In L 4 ` x x ( 2 ) +1 ··· i variables for the box are for external points, with ` , 1 3 x ¯ j z Translating to dual space, 1 x x 2 3 x We finally comment on the − p 10 st 2 1 i ` ]. p and x 1 z ( 77 − 1 L ) = ` − =1 ¯ z Y i , and by x L i ` − 2 ) x L ` 1 ` x ) (1 x z 1 − 4 − 2 x x (1 ( – 61 – . All of the analysis for the triangle integrals 2 2 4 3 ) ) -loop box ladders. p p 3 3 L p x =  , + − 2 2 ) i -loop box ladder, ) 2 2 4 2 3 ` 3 4 i p p p . This is precisely the triangle ladder in dual space. The L ` x p x 2 2 2 1 1 ( x p p 2 + 4 − ) 2 = ( d → 1 2 = p x ( 2 t x 2 ¯ ) z , ( ) 4 2 z 4 − L x ) x ) i 2 − ` -loop box ladders. 2 − and p ) L 2 x ` L 3 x ( + x 2 ( x ( ) 1 2 p L =1 Y − i ( p , 1 for each region between two external lines. Then, once we pick an orientation on each of 2 4 + Z x channels, which is expected to vanish due to the Steinmann relations. To . The ladder integral is then given by i 1 ,p } x (( t × 2 3 p 4 , ∝ ,p 3 . By a combination of translation and inversion we can send 2 2 , 2 L = ( µ 2 x x ,p ]). , B 2 1 s 1 p 78 We can also compute the sequential discontinuity of the box ladder integrals in the Dual space can be defined as follows: for any planar diagram, we associate a variable →  and then 10 L ∈ { µ and a variable the edges, we take theon momentum the flowing through right that andsome edge the to cases, be dual the the variable dual difference(see on between [ variables the the make dual left. manifest variable hidden This symmetries, ensures momentum such conservation as at the each dual vertex. conformal symmetry In with therefore extends to s box and triangle integrals thereforeexact give transformation the between same the analytic two, expression,given one and in can working terms show out that the of the Mandelstams as This integral is invariantcan under conformal be transformations shown of usingx the Lorentz dual invariance and variables thethis (less limit obvious) we have invariance under inversion j These ladder integrals yieldThis the is easiest same to transcendental seewe functions in label dual the as space, dual the as points triangle first corresponding considered integrals. to in loops [ by B sequential discontinuities of the Sequential discontinuities of the JHEP01(2021)205 , , 3 0 E ) was (6.103) (6.104) and . In the displace- > α > 6.93 2 . We then 1 2 E iε . A similar = 0 E − by s 2 = 1 and and z . αE 1 , iε E 0) → , + 0) 7 2 , , . Since this sequence of E 5 while all other invariants prescription is in general 4 , 0 6 − iε , ]. = 1 − 3 and ± , 81 z 1 − – ) confirms that the Steinmann along with s, t > 79 3 αE = (1 = ( 6.94 E 2 4 → − 1 2 E , where E } − s,t { 1 R E – 62 – − = can be constructed by instead rescaling 4 by rescaling ], where the expression that appears in eq. ( E t 55 = 0 , p , p R t 0) 0) , , 6 6 propagators that appear after the first discontinuity by analyt- − − , , iε 7 5 channel of the triangle, eq. ( , , − 3 2 p = (1 = (1 and 11 3 1 p p fixed and varying iε by the reverse path, after encircling the branch point at 3 + } E s,t ], which also appear in Feynman integral calculations. It would be interesting to see if it { 12 R – variables, this corresponds to analytically continuing around 10 ¯ z . For concreteness, we consider the phase-space point 0 It should be possible to extend the variation and monodromy matrix construction to elliptic polylog- < and 11 i 2 have also described how discontinuitiesmonodromy can group. be computed using variation matrices and the arithms [ could also be used in conjunction with the diagrammatic coaction [ be carried out by considering monodromiesIn around particular, the branch by points understanding ofhomogenize Feynman the discontinuities integrals. in termsically of continuing them monodromies, into we same cut areto complex be able plane. taken. to This allows For subsequent integrals discontinuities that are expressible in terms of generalized polylogarithms, we These traditional approaches arecuts based in on physical the regions,ments idea and are that that on Feynman integrals opposite integralsto over sides have propagators extend of branch with these these methods branchinsufficient to cuts. for sequential computing The discontinuities. more main than The focus one of discontinuity, but this the paper relevant has computations can been In this paperseveral we points have of analyzed view. theintegrals We first in discontinuities described terms and how of cuts to cuts,also compute reviewing of described the the Feynman the imaginary work integrals analogous part of relations of from Cutkosky Feynman in and ’t non-covariant time-ordered Hooft perturbation and Veltman, theory. and also shown to reducehas to slightly a different simpler rational functional normalization). form (given as eq. (19) of that paper, which 7 Conclusions and also corresponds todiscontinuity operators computing is a identical to monodromy thediscontinuities around sequence in of operators the used torelations compute are sequential satisfiedSteinmann by analysis the carried out box in ladder [ integral at all loop orders. This matches the We can analytically continue into while keeping return to z path around the branch point at p compute this quantity, we go to the region JHEP01(2021)205 (7.1) . -matrix } ) hold in S n ,s to cuts: ··· 5.13 j , ) are those in 1 s s { + R 5.13 i This amounts to a } i ) and ( s 12 5.8 cuts in channel i k { } M n h i ,s k ··· , 1 P s − evaluated at the same physical value of real { i R ] m – 63 – M P M n 1) m ). In addition, the derivation of our formulas is made ) − ( n B s ) n n , insofar as these integrals appear in the space of Steinmann-satisfying k m Disc ( = 4 ( n N ··· m ]. 1 ∞ = X m n 82 ) k , 1 s 48 ··· , ) 44 Disc 1 1 [( k -matrix program, this was shown to hold for full non-perturbative m ! S ( n 1 1 m m ∞ = X 1 ··· k ! 1 Of course, the constraint that all external lines be massive is a strong one, and excludes An important class of sequential discontinuities described by eq. ( An important consideration that we have spent considerable time exploring is that the The main result of this paper is a formula relating the sequential discontinuities in the 1 = It had previously been observed that the Steinmann relations were obeyed by many of the Feynman m 12 many theories of physical interest. As such, it wouldintegrals be that appear good in to planar hexagon understand functions the [ massless elements in a mass-gappedSteinmann relations scalar in quantum factproof field hold of theory. for the individual Steinmann Ourrequires Feynman relations analysis only integrals. in that implies perturbation the theory, thatand region diagram that the by where the diagram. both external momenta channels Our are can proof not be constrained simultaneously (for cut instance must by exist, being massless). explicit path in energy between regions (so that ourwhich formulas the do discontinuity not channels apply). encodes are the partially Steinmann overlapping. relations, originallywhich In state derived that these using sequential discontinuities axiomatic cases, in quantumthe partially this field overlapping original channels equation theory, must vanish. In computed, but there issometimes no fails). guarantee of While agreementconstraints with there on cuts the is (and paths, undoubtedlywe in have in a fact, found every more the agreement case between agreement covariantconversely, discontinuities where way in and we to cuts cases have according understand where found to the our our an formulas, formulas explicit and seem path to in fail, energy we have not been able to find an fixed and respecting energy conservation.cut We and have discontinuity presented computations, manythese nontrivial and examples. examples have of For checked each thatconnecting example, eqs. the we ( relevant have regions, beenencircled. and sure used If to one the find just path an picks to an explicit arbitrary determine path path which in between branch energies regions, points the are discontinuity can still be analytic continuations by whichcare. these discontinuities Paths are that computed areanswers must (as homologous be discussed but in chosen not appendixassuming with in a the path same exists homotopy which class continues may the give external different energies, holding the three-momenta It is crucial that theseof relations interest are are understood to nonvanishing.always apply taken only In as in particular, regions theexternal where we difference momenta all emphasize on between the different that cuts Riemann these sheets. discontinuities are same or different channels around branch points associated with invariants JHEP01(2021)205 , . 4 ) ) ) l l j p p p · · · 2 j k i p µ p p d > )( )( − j k p p · · i i ln( p p ( ( i , we get the (2 . R . 0 1 πi, 1 2 and we get , − and R 1 M M 0 R · · = 0 1 0 ln 1 diag − R πi R M , we have M 1 . + 2 ln = = 2 >    ) π |       →∞ R . R 2 R πi | R ( πi 0 π 2 = 2 2    4 − 2 Li lim πi π πi . πiu , ) 2 2 | 2 πi 8 R πiu − R 2 | πi Re 1 2 0 1 0 0 1 − 1 0 10 2 0 1 2 Re    −    − 1 = 1 0 10 2 0 1 ln(1 = – 69 – . Up to conjugation by    R du R − (so that the contour circles the branch point at 1 , we get 1 t 1 → 0 0 → − 1 00 1 0 ln 1 = 0 − 0 0 Z = 1    . There, we see that the choice to take Im ∞ M z M dt 8 · = · R > 1 R < 0 M ∞ R ∞ Z and in the monodromy matrix associated with infinity is also 2 → ) 0 ) M M ): 2 · πi · M π ), we can compute the monodromy matrix from an infinitesimal = 0 ( R with Im R with Im z 4.46 O → → = (2 0 0 4.43 M s M ds → ∞ → ∞ = ◦ = R R s ∞ ∞ − ds is a continuous function for large M M 1 1 R − R γ R Z . If we compute the matrix along a straight line path between 1 ln This ambiguity at Note that this connection to the fundamental group remains valid if we compactify and The result is thearound product infinity is of illustrated the on 0 the and right 1 in figure monodromiespresent in for the the opposite other order. monodromy matrices. This path For example, we could have computed the at 1. If wefirst), take the monodromy matrix differs in the top-right entry This monodromy around infinity canand be 1, written since as the theillustrated path product in around of the a infinity left monodromy is partwhat around homotopic of determined 0 to figure that a we encircled path the around branch 0 point then at around 0 1, first, as and then the branch point Then, if we take Note that going around infinity clockwise0 corresponds to a counterclockwise contour around variation matrix in eq. ( The full matrix can be computed to be a circular path that starts and ends at a complex point Since monodromy matrices give us an explicit representation of thisthe group. complex plane bythe considering connection the in monodromy eq. ( matrixcontour associated encircling infinity. with For infinity. instance, Using if we integrate the dilogarithm integrand around classes of paths around JHEP01(2021)205 ) ¯ z γ } s { ), as ( (B.9) (B.8) f ) to be B.2 is called ln to obtain 1 d 1 4.62 − γ − b H b 1 1 − while holding − in eq. ( z ) and ( which we can take 1 . aba aba . } j          s 4.59 2 πi { 0 M π 4 , we can find another path πi . πi 2 ) πi 2 that appears in the one-loop − π − ) ( ¯ z 1 0 10 0 0 1 O terms are different. 0 10 0 00 2 1 0 0 0 1 1 0 ) = 2 z, )    ( }       2 1 j π = s Φ ( = . However, as there are many paths { 0 ( z 1 O πi f . ∞ . The result would have been ln M M d · 8 were computed in eqs. ( ) = 2 γ 1 M } H of the fundamental group, we can quotient the , j , for some set of variables – 70 – , the elements of the monodromy group depend s b = 1 γ       { ( M satisfy the same relation; however, the paths in this z πi · f , and ) = 0 1 γ 1 } and ln j − 0 0 and 0 s d a M { γ ( πi , H 0 f = 0 M ambiguities more formally, consider a codimension-one branch 1 2 0 10 0 0 0 0 1 2 0 0 1 z . ) M ), which satisfy a single multiplicative identity. Note, however,       = 2 γ 1 π , = ( 0 B.6 z 0 using a contour that first crosses the negative real axis before going O such that 1 γ M M ), and ( which satisfies the condition B.2 γ terms in this monodromy matrix are the same as for that also satisfies this relation. Moreover, as this new path belongs to a different 1 ) , as illustrated on the right in figure ), ( − 1 π b ( 1 B.1 O − For a multivariable function, like the function Despite these ambiguities, any choice of closed contours around 0, 1, and infinity The fundamental group and first homology group are related by Hurewicz theorem, To describe these that the contours used mustmatrices all corresponding start to at the same basepoint, so we cannot usetriangle the and rational box, we canfixed. carry The out contours the around same analysis for the contours in homotopy class, it yields a different monodromy beyond will furnish us withwith three a marked representation points. ofeqs. For instance, ( the we fundamental can group choose on the rational the matrices Riemann appearing sphere in fundamental group by the commutator subgroupthe generated by homology elements group.a The Pochhammer contour contour, correspondingevery and to path corresponds the commutator toγaba a trivial element in homology. Thus, for depends only on theon homology the class homotopy class of of which states that the firstThat homology is, group is given the any abelianization two of elements the fundamental group. to be Mandelstam invariants. Tofind compute a the closed monodromy path aroundthat this satisfy branch this variety, we requirement, therepaths is in the some same ambiguity homology inhomology class this class of choice. may still In be particular, in all different the homotopy classes. While the integral The expected from Cauchy’s residue theorem, but the variety defined by an equation around monodromy around JHEP01(2021)205 n s gener- (B.10) (B.11) Re 1 − . n ∞ 1 , ∞ γ 1 0 = 1 s γ , z 1 . z 1 0 s M · Im M z 0 · z 0 M M = =       2       π 2 πi πi can be computed in a similar fashion, and 4 . Like for the case of the dilogarithm, each πi πi π z − 4 ¯ z s – 71 – πi πi Re 2 2 − − . We depict two possible contours that go around infinity, 1 πi πi ∞ γ 1 ∞ γ 1 2 0 10 0 00 0 2 1 2 0 1 1 2 0 10 0 00 0 1 2 0 2 1             with the points 0, 1, and infinity in each variable removed. ∞ and ¯ z = = 1 z z , ∞ 0 ∞ and 0 M M z s 0 Im γ ∞ γ . Paths around More generally, the monodromy group describing the discontinuity of a set of polylog- manifold on which thesethat only polylogarithms depend are on defined. amarked single points variable, and When the the we relevant fundamentalators. manifold consider group is However, corresponds polylogarithms the the to Riemann fundamental the group spheremore free with of complicated. group higher-dimensional with manifolds will in general be monodromy matrix gives rise toator an of associated the rational fundamental matrix group, thatthe corresponds which space to in of a this complex gener- case describes the manifold corresponding to arithms also furnishes us with a representation of the fundamental group describing the The monodromy matrices forcommute with contours the in monodromy matrices in For the contours around infinity, a calculation analogous to the dilog case gives Figure 8 starting at points inand the 1, upper but or inThe different lower right orders. half-plane. panel The shows These two that are contours the each around path homotopic infinity ambiguity are to is not paths present homotopically also around equivalent. for 0 paths around JHEP01(2021)205 (C.6) (C.2) (C.3) (C.4) (C.5) (C.1) ], string indexed , namely } 1 − ,k sv 113 , is an element M M 54 k M = M {  , we want to find a ] we show here that M · , and M . 67 , πi D    2 · 66 z k and M ) ln · z . 1) . ( , 1 1 ,k 1 − 1 z (which encodes the relative weight , M − − D · D, , ln when analytically continued around D  ) is equivalent to the map defined in that depends on any number of vari- · M · D M − · ) + Li k · · z have additionally been replaced by their M D C.4 ,k ( M 1 1 M · j 2 · − − z 1 ]. Motivated by [ 1 Li D − = diag(1 − M M · − D 115 D . 1 D ) – 72 – · 1 − · z . Having made this observation, we define the =  ( 1 k → → 1 6= − D ,k M · Li ,k D ≡ 1 ], the infrared structure of gauge theory [ 1 theory [ − − k − ]. In these new single-valued functions, all contributions sv M M 4 1 00 1 0 1 φ M M 112    M · M , · 110 1 = ,k 1 − 111 1 − − D in which all variables · M 1 . Under the action of the monodromy group, this pair of matrices M ), we see that j M − z M A.1 . We note that the definition (  ] using the coproduct formalism. ∗ j ], and massless ) does not depend on z 112 = M 114 → j in eq. ( z . In particular, we have is a diagonal matrix whose entries are integer powers of M sv B D M . In order to construct a single-valued version of the matrix Let us see how this works in the case of the dilogarithm. Referring to its variation This mismatch can be fixed by decomposing the monodromy matrices as discussed in We begin by considering a variation matrix k matrix This matrix invariant under the action of the monodromy group, since whenever eq. (3.82) of [ of the generalmatrices linear preserves transcendental group weight, the withthe matrix rational rows entries. of single-valued Since matrix the action of the monodromy continuations, because section where transform as Thus, the product of these two matrices is not quite invariant under arbitrary analytic by matrix that transformsbranch in points. the A opposite naturalthe object way inverse to as matrix consider of is thecomplex inverse conjugates conjugate matrix such as multi-Regge limits [ amplitudes [ the same map can be constructed in terms ofables, whose variation discontinuities matrices. are described by a set of monodromy matrices Using the Knizhnik-Zamolodchikovvalued equation, avatars polylogarithms of themselves can [ generated be by mapped analytically toout continuing single- by around new branch functional pointsfunction. dependence This are on type systematically variables of conjugate single-valued cancelled map to has the proven useful variables in of a the variety of original physics contexts, C Single-valued polylogarithms JHEP01(2021)205 . ) 3 1 z ) − ¯ 1) z p ¯ z , 1 ↔ − − )(1 ↔ (C.7) (D.4) (D.2) (D.3) (D.1) z E z z z ( 1 ( 2 1 − p 2 1 p = ( p (1 1 p ↔ ¯ z corresponds to . z 3 p    ) . ) z ↔ ), where in opposite directions ¯ (¯ z 1 2 ¯ z p − 6.35 , these momenta become z ) Li ( must correspond to x ¯ z and 2 1 e . z 2 p ) z | p | σ ) z , we need to find the action of | . | ¯ z ) and ↔ x ¯ z z 1) , given by e t symmetry under the permutation 1 − ) + ln( e z, p ) ln( ln( ∧ 3 ( z ) z (¯ 1 t ¯ permutation we have S z − 2 e Φ 3 − ) 6.2.2 z, ) changes sign. Similarly, under p Li ¯ z ( )) = ( 1 z − − (¯ ↔ Φ D.2 ) σ z ¯ 2 z z ( p ( 2 1 − 1 2 − p ) ) + arg(1 z 2 1 – 73 – z z ( 2 1 ( ) Li − σ p 2 ), respects an z ( 2 (¯ 1 = π Li 1 , and we have (1) 2 x is the sign representation. . We conclude that the representation of the symmetric 2 σ = 6.46 p 16 e 2 p ) 1 x 2 p ¯ z ∧ p − ∧ ) + Li 1 ) = − 1 + p z z ( p )) = ( t z . From the above, we know that 1 ), that under the ¯ z ( e ¯ z − D 2 1 2 p − E = . Similarly, one can show that 0 0 1 1 Li 0 6.32 z ¯ z 2 ( and the left-hand side of eq. ( =    p 2 1 2 − 2 z p ∧ p ( p = 1 3 σ p . In terms of a pair of basis vectors sv ↔ ) ↔ x 2 1 and was defined in eq. ( ↔ M ¯ z p , p x ) 2 2 ¯ e p z E . The action of the remaining permutations can be determined from these two ∧ and + z, ¯ when acting on 1 z 1 ( t z p 1 3 e = ( symmetry on 1 . We work in the coordinate system described above eq. ( Φ S ↔ − 2 2 3 E 1 p ¯ p z We are now ready to study the symmetry of the transcendental part of the triangle Before moving on to discuss the symmetries of We first discuss the rational prefactor. To determine how the quantity S = ∧ . These constraints can be solved with the unique solution that ↔ ) 1 1 ¯ and transformations. function. It turns out that this is related to the Bloch-Wigner function We also know, from eq. ( z z we have group the p We can correspondingly use this quantity toClearly, study under the transformation properties of collusion of this integral’s rational and transcendental parts. transforms under the permutationp of external momenta,and we consider the wedge product where of its external legs. This symmetry turns out to be realized in an interesting way, by the around any branch point cancel out in the entries ofD this matrix, as expected. Permutation symmetry of theThe one-loop triangle triangle integral integral considered in section It is not hard to check that all effects of analytically continuing The single-valued matrix is thus given by JHEP01(2021)205 (E.1) (D.5) (D.6) (D.7) (D.8) (D.9) . )] z (¯ 2 Li − ) z ( . , the transcendental 2 ) ∗ ∗ z Li )[ zz ¯ 6= z z ln( ¯ z ( 2 ,  . ∗ ln )) z z ∗  1 2 . By the argument above, under z invariant under the permutation − 1 z − ( 2 πi 1  1 2 )] + Li  , not D z ∗ (¯ − z − z 3 , ) ) − z Li ∗ − ( 1 ) = . 2 1 − )) + ln zz z ¯ z ∗ ) ln z z (Li − ln( ( ( i i 2 – 74 – 3 1 1 2 2 1 2 and (1 Li Li z D = )[ − | ) = ) = ¯ − z z ) z z z | ( z ( 2 − ln 2 ) = Li z 3 ln( ( , this gives precisely the transcendental part of the one-loop = ∗ − D z arg(1 )] z = (¯ ) = 2(Li 4 ¯ z z ( Li iD − 4 ) z ( . , where ) 4 ∗ I ¯ z Li R z, ( 1 Φ ) = 6[ ¯ z z, How is the symmetry realized on the cuts? It is instructive to consider the example The Bloch-Wigner function satisfies ( 2 Φ In this appendix wetriangle/box present function the connection and variation matrix for the two-loop ladder A similar statement holds for the rest of the cuts. E Variation matrix of the two-loop box the action of theHence, permutation the of residue external on legs,group. the a rational given However, prefactor each leading may leading singularityresidue, pick as singularity is required up locus by a is global sign. takes residue paired on theorems. all with It the another leading follows one that singularities is the with invariant set under opposite of the values action the of residue the permutation group. part should be thought ashold a function under of the two transformation independent of variables. both Still, the same relations of a leading singularity,prefactor, where while the the transcendental only part dependence is on a the power of kinematics is in the rational These signs preciselygroup compensate on the the signs rational arising prefactor. from In the the action other of regions, the where permutation In the region triangle, we have In particular, using JHEP01(2021)205 (E.9) (E.5) (E.6) (E.7) (E.8) (E.2) (E.3) (E.4) , , we also have                   0 0 ). 0 0 0 0 0 0 0 = 0 ω ω . n 1 0 σ 0 = 0 ω . We encounter inte- ω γ ω 0 ¯ ω z 0 Z ∧ γ 1 + ω 0 , . ∧ Z ω . 0 M ω 1 1 ··· − 0 ω ω ¯ z 2  ω − − dz d . σ 2 − satisfies ¯ − z ¯ z z σ γ ) ¯ z . 0 2 0 Z Z γ 2 ω σ 0 1 dω = = 1 Z ω 1 2 ln ω ω σ 0 0 000 0 0 0 00 0 0 1 1 − + ¯ ω γ 1 2 − − + 0   and 0 Z 0 1 ω ω + 1 σ ω ( ) along a concretely chosen contour, it = 0 σ z ¯ γ z ◦ ω n Z j ) ln ¯ Z 0  σ + 0 4.22 0 z 0 ω ω ω z ) = + 1 ◦ · · · 1 + ln 0 Z 0 – 75 – σ 2 ω ω j z ω ( ◦ + , and using the fact that − σ 2 ¯ z 2 0 ◦ 0 ln z, σ ω 1 1 = 0 ω 0 0 0 00 00 00 0 00 0 0 0 0 0 0 0 0 0 1 2 j z ω Z + − σ Z 2 ) as } = dω = 0 σ 0 n , , ω 6 ω 6 , ω , ··· ◦ ¯ E.5 1 , z } 1 ¯ z z z 1 1 + n d dz { σ ,j M 0 M Z ( = = ω ··· γ 1 2 , X 0 0 Z 1 1 j ) can be iterated, to give us ¯ ω ω { = ω 6 perms of , − E.6 0 0 000 0 0 0 0 0 0 0 0 0 0 0 1 ∈ 1 γ ω M Z − . Thus, the connection has zero curvature ( 0 0 0 0 0 0 0 0 0 ]. As an example, consider                   = 0 ω = 106 , ∧ ω 2 ω Using this connection, we can compute the variation matrix The relation in eq. ( These integrals are much simpler to evaluate, and we get is easier to use the fact that any pair of one-forms This allows us to rewrite eq. ( grals [ While this integral can be computed using eq. ( The connection trivially satisfies that grals of one-forms, whichcan are familiar, be but easily also evaluated iterated by integrals of leveraging higher the weight. shuffle These algebra associated with iterated inte- where The two-loop connection is JHEP01(2021)205 (E.20) (E.14) (E.15) (E.16) (E.17) (E.18) (E.19) (E.10) (E.11) (E.12) (E.13) ¯ z 2 ) ln , z .                   3 − 9 9 9 9 9 9 9 9 σ , , , , , , , , 1 2 3 4 5 6 7 8 ◦ 1 2 M M M M M M M M ln(1 σ 8 8 8 8 8 8 γ , , , , , , − 1 2 3 4 5 6 Z ) 1 . z M M M M M M σ ) , 7 7 7 7 7 7 γ z , , , , , , ( Z } 0 1 2 3 4 5 6 , 10 0 0 1 0 + ln ¯ 2 ω i z M M M M M M z . Li 2 ) = 6 6 z 1 , , 1 1 3 σ · · · ◦ 0 ln ) ln ) (ln − {z , , 0 ◦ ) = 0 n z z M M ) ) 0 (¯ − ω 3 z z 2 5 5 5 − ω ( ( ) , , , σ ◦ 1 2 3 2 2 z Li 0 ◦ ( + 2 Li Li ω | 2 M M M 3 0 ln(1 σ ◦ − ω 4 4 4 Li , , , ( 1 ) − 1 2 3 ) + ) + + 10 00 1 0 0 0 0 0 0 1 0 0 0 0 0 0 z ω z z – 76 – ) ∧ 3 = 2 M M M z z ) σ ( 0 0 0 3 3 Z ◦ , ω ω + ln ¯ ¯ 1 z, + ln ¯ + ln ¯ 1 Li − 2 ◦ z σ z z + , and is sometimes called integrability condition. In our M 1 ) ln 0 ◦ + 6 z 2 ω ω , ω 2 ) = 1 2 1 ( z ) (ln ◦ z σ 00 1 0 0 0 0 0 0 0 0 0 0 0 1 0 ) (ln ) (ln − ( z ¯ z z 0 + ( M ln ¯ + n ω 2 z z − − 3 0 0 0 0 0 0 0 1 0 z Li ln(1 Li , along with relations such as 0 σ ln ¯ 4 n                   Z ◦ z ) ln − ln(1 z − 2 , σ ) ¯ σ z ) z, ) = − − z ¯ z ◦ + ln ) + ln(1 ) (¯ ··· − 3 1 z z , z z, (¯ (¯ 1 σ ( + ln ¯ 2 2 2 Li ( σ 0 6 γ γ z ln(1 2 ln(1 ln Li Li h Z 1 2 1 2 − − − M − = ln = = = = = 5 6 7 2 3 4 , , , , , , 1 1 1 1 1 1 one-forms M M M M M M The result of performing the integrations is n where This condition is trivialnon-trivial when restrictions both when two forms or only more depend variables are on involved. a single variable, but imposes The iterated integrals we study have thedeformations special of property the that integration they contour are whichof independent preserve of its the small endpoints. flatness This of isexample, the a the connection consequence integrability condition reads along with integrals that can easily be performed, such as Using these kindsvariation of matrix formulas, to familiar we integrals, can such reduce as the expressions in the calculation of the for JHEP01(2021)205 z (E.34) (E.35) (E.24) (E.25) (E.26) (E.27) (E.28) (E.29) (E.30) (E.31) (E.32) (E.33) (E.21) (E.22) (E.23) 2 is ) ln ¯ z = 1 z, z, − z , ) ln ¯ ln z . (¯ z z 2 z,                   , ln ln ¯ ) 2 Li ) z ( − − ) + ln(1 − + ln ¯ 3 πi z πi 0 0 0 0 0 0 1 ¯ z z ) 2 z Li 2 2 . (2 z ( 3 1 2 ln ln 2                   + ln ¯ − 2 1 2 1 2 0 0 1 0 0 0 0 0 0 Li (but not two) sequential cuts = ln ) z, z ) , 0 6 z i , = = πi = πi πi ln ¯ 3 ¯ z 2 2 9 9 9 z (2 2 , , , πi ) (ln 4 5 6 − − 1 2 three M 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 z + ln ¯ ( , − ) ln − M M 2 M ) z ¯ + ln z , z ) Li ( z πi 00 0 0 0 0 01 0 0 0 0 1 0 z 2 − 0 0 2 2 ( ) (ln , 2 Li 2 ) ln z 2 ) πi (¯ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 ¯ ) + 2 z Li , z, 1 2 2 2 ) z − πi πi 0 ¯ − z − ln Li 0 ) ) + (2 z z, – 77 – ) + ln(1 − z 1 2 = − − z + ln ¯ πi z, (¯ ) ln z. 8 z ln(1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 2 , z 3 z 2 πi − − + ln ¯ − Li + ln ¯ z M 6 = ln 0 0 0 0 0 0 0 0 1 = z z, = ln ¯ = 1 00 10 0 00 0 00 1 00 0 0 0 0 0 0 0 2 0 0 2 = ln ¯ = ln(1 ) (ln + ln ¯ 5 8 , − 9 5 z                   8 8 , , , ln ¯ , , 2 (¯ z 4 8 3 πi 5 6 z 2 is ) (ln z = . This is consistent with ) (ln z M 2 M ¯ z M M ln ¯ M M Li z 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 1 z, ln ) (ln z − − = 0 = 0 0 0 0 0 0 0 1 0 0 z 1 2 ln ( 3 z 2 )                   ) + 4 z ) ln M , z 0 − ln(1 z ¯ z ) Li ( = ¯ ln ¯ z z , 3 z, − − ) 2 z 0 − ) z − Li ln ¯ ) + ln(1 ) + M ln z ¯ 6 z z z (¯ z, − z, z (¯ (¯ − 2 2 − 2 3 − M , z z, ln ¯ ln(1 ln ¯ ln ln 1 Li 2 h Li Li ( 2 1 2 1 2 1 − − − − + 2 ln(1 = ln = = = ln = = ln(1 = = 2 = 3 = = = = Φ channel of the 2-loop triangle vanishing. The monodromy around 7 9 8 9 7 7 7 9 4 7 9 4 8 , , , , , , , , , , , , , 2 2 6 7 3 3 4 5 2 3 3 1 2 1 2 p M M M M M M M M M M M M M in the We note that The monodromy around JHEP01(2021)205 (F.1) (E.38) (E.36) (E.37)                   ) 2 0 1 ) . We start by k πi πi 0 0 0 0 0 0 1 − , we can use the 2 (2 2 0 ¯ z sequential cuts in 1 2 p 6.2.4 ) via conjugation by 2 Θ( . ) 1 two i 2 πi − πi πi E.2 ) 0 0 0 2 2 1 C (2 z 1 k − − 1 2 . − −                   2 3 1 2 M ) p ) ( − C h πi πi , πi πi 1 0 0 100 0 1 0 δ 2 = (2 2 i ) (2 − − 1 2 2 3 ¯ z 1 − πi 1 0 − , p 0 2 2 1 − 2 and write − ) M , k 1 )( πi 2 πi 0 C 1 0 0 1 ) . Thus, we also have that k 1 (2 1 − 1 2 k − – 78 – , 1 0 0 Θ( 1 0 0 0 1 + . This is consistent with −  πi − 0 0 2 3 2 p C CωC ( m z 0 2 = 0 h 0 1 1 0 ) 2 − 2 πi 1 0 2 10 00 10 0 00 0 0 1 0 0 0 0 0 0 . πi ) ) = T 2 1 2 z 1 2 1 M                   k (2 k  3 = 0 C z, z z 1 C = (¯ δ , πi 3 1 ) = ω ) M 2 Cut C C 3 z ¯ z 0 πi T − M → 2 πi X · 10 0 0 1 0 0 0 0 0 0 0 0 0 0 0 + 2 ) M − (1 2 z 0 ¯ z ( ). − ) M − 1 4 z, 1 0 0 0 0 0 0 0 0 ) k 1 ( can be exchanged in the connection from eq. ( k π (1 1 6.42 ω M                   4 + ¯ z on r.h.s. (2 d 3 = = i iε p ( z Z −

∞ and 1 2 2 × C z , M 1 = C 3 channel (the long direction) of the 2-loop triangle vanishing. Finally, the clockwise T 2 3 To compute the monodromy matrices associated with contours in p F Cuts of theIn three-loop this triangle appendix, we workcomputing out the the sum details of of the the two cuts calculations involving in section These are the lastoperators in monodromy eq. matrices ( that are needed to construct the discontinuity fact that the matrix and again we get the monodromy around infinity (where we approach infinity above the real line) is In this case we have That is, we have JHEP01(2021)205 , o θ . (F.9) (F.5) (F.6) (F.7) (F.8) (F.2) (F.3) (F.4) . The )] x G cos (¯ to have a d 2 1 1 1 . Changing Li − k 2 Z / − 2 ) π x i , m ( 16 2 )] = is small, and hence θ Li | 1 ¯ ) = )[ x k to be small. The final ~ 0 1 ¯ | x k x ¯ x = − − 1 , or 0 2 k ], p ) through the two-loop triangle ln( ) (1 + cos ω θ x θ ¯ z 2 1 , . 23 − ¯ z Θ( k 2 2 )] − i cos ) ) cos − 2 x 1 1 ) (¯ p 1 − k k 1 dx . 2 3 2 3 k 1 4 − p + + ) m Li 1 2 2 3 3 1 ) + (1 − 3 3 becomes θ − k ω p p and momentum of the outgoing particles 2 ) pp ( + 1 ( ( p 2 + x 2 2 1 k ( θ ( p cos h 3 = m m 3 − z + p δ 2 ) = 2 3 − ¯ ( x 3 ) p Li ¯ m − x − x p 2 – 79 – − + 2 3 πi 3[ 1 2 2 − p ) (1 2 gives a Jacobian of n = m 1 + cos and hence used that z , we can take either arises because the mass regulator does not capture ) massless vertices, as worked out in appendix − 2 = 2 1 2 θ 2 ¯ ) x . We take the particle with momentum is the sum of cuts in ) (1 1 2 = − 1 1 2 2 3 )( m k m x i ) − 2 3 3 p 0 = 1 p 2 3 1 − cos + x k ω [(1 3 k = − x d p L , p 2 1 )( ( i is 2 1 2 1 Θ( p + 1 ¯ 2 3 (1 x and − k 1 2 3  3 − 1 , k 1 2 − 1 ,p 2 p p 2 2 k ( ∼ = ) m , 1 )(1 2 = 1 2 ,m k ) − ) x 2 1 x 1 2 ) 1 , the propagator in + k − k 1 2 k 3 . The factor of k p to  (1 + + p 2 , which gives δ ( ) + 4 θ 3 3 h ) ¯ m z ) 3 p p 1 ( 2 p ( k ( πi − cos T h 1 to regulate the IR divergences that arise in the cut calculations, and work to 2 2 1 + z 2 k ( 3 − . m T ( p ) 2 2 1 ( 2 1 x 4 k m p Cut ) 4 1 − x 2 is the magnitude of the three-momentum of the outgoing particles, and where we 2 3 k π π (1 p arising from a product of p P − 4 πi 2 Cut (2 ! d 2 m 1 Using the delta functions, and performing the integral over the azimuthal angle, the i L = 256 with masses Z X p = = 2 ¯ and hence we have In this frame, the energyis of where have dropped power corrections in variables from phase space can be written as In the rest frame of Working to leading poweranswer in is independent ofx which one is picked, so we assume that with T small mass leading power in the sum of the cuts through the two-loop triangle is given by [ where JHEP01(2021)205 . ) ¯ z . o − = 1  ) 2 z (F.15) (F.13) (F.14) (F.10) (F.11) (F.12) 2 1 θ x can be k ( ( , p 2 1 2 − 2 1 2 cos m k p . We write ) 3 2 Li )] 2 p 1 ) iε + z k − 2 p , for ¯ − x (¯ ) + 3 0 1 x 4 + p = z k 3 πi − Li 3 p → 1 p = p + ( 2 2 − , ln ( 3 + 2 k x + 2 ) m p ω  z ( 1 − (
2 , k ) k ) ) k ) 2 2 0 1 4 ) ω and θ x p θ k 2 ( Li  ) 1 3 ) − cos θ . cos 3 [ B − p 0 2 Li 2 p ) p 3 − − s 2 cos 2 2 3 = i o n C k k p p − ) ω n Θ( ( θ − − z 3 ) x 2 2 3 − − i (¯ can be taken to be a straight line dx ¯ k z ω p 2 ω 2 2 1 1 ( ( ω ( z ) , 2 cos 2 2 ¯ k k z 1 − 2 Li i R Z − k m z 2 3 3 2 1 m Cut ( 4 3 − p − for ) − 2 3 ( 4 3 , p ) 0 2 − p 1 1 2 2 2 2 p z p p ( z p 1 k k 2 3 2 1 p 2 ( 2 2 ( p p , k 2 2 m = ¯ m h Θ( m 2 2 4 2 δ ) Li 4  x π 2 − ) h ) – 80 – 2 m k 1  π 2 1 m πi 1 + 2 , with all other propagators having a 2048 in the region − log − 3 3 − that are subleading in the limit C 2 p 2 16 (4 z . π )] = 2 ( 2 2 ¯ k )( 1 x z ) h C p k 3 0 1 (¯  16 = x to C 2 . Integrating this expression and using k , 3 δ 3 1 1 T ¯ ) z = = C − C , 3 2 Li 2 Θ( 1 k  T 1 channel cut through a box with one massive internal line, πi C  (1 3 1 2 − 2 2 1 − T p + ) k s , p − 2 Cut z (  < z < + ( 2 2 . This shows that m ) ) 4 δ 3 0 2 2 ) 2 2 ) k k Li k π dx m is an [ 4 πi 1 z + (2 + d ¯ 2 z ! B i 3 Z 3 2 2 − 2 3 p ) p , with ( p ( π 2 ( m Z z on r.h.s. of cut i , k 4 2 ) 2 1 2 1 16 − − × iε on r.h.s. of cut 2 k π to

p − , k 4 − = (

iε 2 2 (2 0 d 2 − p 3 i = C

, C  , 2 1 + ln 2 C C Z , 3 B C ¯ z < 3 1 s T C T 3 T Cut where Cut the cut as Next, we calculate the double cut The integration contour from from gives Then, dropping polylogarithms in Putting everything together,written phase as space along with the propagator in to leading power in JHEP01(2021)205 ξ z to . As 0 θ (F.16) (F.20) (F.21) (F.22) (F.17) (F.18) (F.19) < cos 2 ) )] . 2 z 2 (¯ k  and . ) 3 , + θ ) 2 ) Li k , 3 ξ θ p ω 2 − ( cos − ) integration are at ξ ) results in 2 p cos z 2 3 k ) p x ( 2 3 − ¯ z p 3 + pp ln (1 3 −  2 3 − Li 2 3 ω 3 ¯ # ξ ξ [ p ( p m z 2 1 ω ( # ( 2 ( − − 2 2 xp m πi = 1 1 2 1 , ) m p m  2  x − ξ ! ¯ ξ . 2 3 2 3 θ θ − − ln − p ) + 2 p −  2 3 2 2  2 ¯ z ∂ξ = ∂x cos cos p ) p 1 (1 z )] 2 ∂ ∂ p  z 2 = k − (¯ ) 2 1 m θ ξ 4 k + 0 ∂ξ − 3 + ln ( Li − = 1 ∂ω " ∂x p cos ( !

− (1 ∂ ln 2 )  2 3 , while the limits of the . The cut of the three mass triangle is given by ¯ = ξ z p x 2 m – 81 – ( dξ ) k 4 − − x θ i 0 ξ

Li ) Z [  and cos 2 3 ln π , − , and using that 3 ξ, x 8 2 x − ( dx p  k ) " xpp ∂ , x ω = , z , ( z (¯  2 1 ¯ 2 and it can be shown that for these cuts, z ) ∂ ) ) and 2 2 ) Z 2 θ T ¯ z 2 0 ξ 2 2 k 2 3 Li k 2 3 k 2 2 p − 1 2 < is small. We make a change of variables from p 4 cos + 2 k + ) z 2 3 ¯ ξ 3 p ( 2 − 3 m p Cut p k 4 3 4 p ) 1 ( , − ( p z π 1 0 2 1 + 3 ( integrals are at 0 and − p = 2 2 3 ω 4 > ξ ( p ¯ ξ π ( Li 2 3 4096 2 2 ξ p 1 2 = 1 k 2  = ξ − 2048 3 + C , 2 = = C 3 is now the angle between 3 2 T C k θ , , defined by . Putting everything together, we get 2 ∂ξ ¯ C z x When evaluating diagrams with massless three-point vertices in dimensional regular- 3 ∂ω T involved in computing thesein cut dimensional subgraphs. regularization, and The theendpoints first second of relates comes integration. from to delta evaluating functions the evaluated diagrams at the ization using the covariant cutting rules, one gets a delta function in the angle between the G Massless three-point vertices When calculating cut graphs, we sometimeson encounter subgraphs either with side cuts of of massless lines a three-point vertex. This appendix discusses two important subtleties Performing the integrals in and where The limits of the with Jacobian where we take before, we assume that and with where JHEP01(2021)205 ] it , the (G.2) (G.1) 23 = 4 d to 1, where θ , ) θ cos d L ) k cos is not smooth. For θ − 4 cos ]. However, it may give -loop triangle: L (1 C L it is divergent. In [ → − 23 δ 4 . Although masses are not 4 d (1 0 2 − δ d is the angle between internal 1 . Naïvely, using the covariant  1 1 → − d < θ θ R − 2 L reg k cos m − − L 1  , and we cut the massless propagators with θ 2 , . . . , k m result and set all such graphs to zero. Indeed, cos 1 – 82 – k d = 4 1 2 1 − − p it is finite, and for Z 2 d > k 2 ∝ C 2 . The dashed lines correspond to cuts and the circled and take the limit = 4 1 2 k 2 p p d Φ , and reg 2 6 ) 1 2 2 m k p 1 C − 5 4 1 1 p k k 3 Disc ( , ). An alternative to using dimensional regularization is to give the − L p 2 1 6.80 p , . . . , k 1 k 3 compared to the naïve expectation of setting p ! ) and ( 1 L the integral is zero, for is the number of massless three point vertices in the cut diagram. 6.74 4 1 To see how the combinatorial factor arises, we calculate the The second problem is that, even if a graph or sum of graphs is IR finite in The first problem with this expression is that the limit − The incoming particle is massivemomentum with cutting rules, one would put all the cut particles on-shell and the diagram above would be delta function of the angle betweenendpoints the of two the particles may limits need ofmore to integration. be careful evaluated analysis Such at is expressions one needed. are offactor not the As of generally we well-defined, will and show,L this ultimately results in a combinatorial eqs. ( internal lines a smallgreat mass regulators in general,invariance, for particularly the in Feynman integrals gaugeresults we for theories consider the where in cuts this they consistent paper with can they the violate always discontinuities. seems gauge to give d > was argued that onesuch should an use approach the seems toresults work for in the the cut graphs examples that considered are in inconsistent [ with the discontinuities, as discussed below which contributes to subgraph is theparticles problematic 1 three-point and 3 vertex. in the Here, diagram. graph two particles multiplied with its argument raised to a power. For example, consider the JHEP01(2021)205  ) 0 1 2 k − ) as 1 − (G.8) (G.6) (G.7) (G.3) (G.4) (G.5) ω 0 G.7 p . )] −  2 , 2  1 Θ ω L θ i ,k 2 − 1 ) − 1 1 cos L k ω k − θ . −  − p  (1 ( 2 . m 0 L h 2 m cos ) ( k δ ω − L ε − ) 1 − δ 1 π ω ) too early. For massless 1 1 ) ω 2  . − > ω (2 π  0 ) L δ −  [ G.6 k δ , E 0 L δ  ··· L ) (2 ( , the above expression can be , and denote the angle between 2 ··· k ) 1 0 p i ε ) Θ  > π ω dω k 2 i , πiδ 2 + ε Θ 1 2 2 1 (2 2 ) ≡ − 2 θ  −  L − L i 2 L 1 E ω k ω > ω k i 0 m = cos  2 ( Z − 1 mω δ 1 ε ω  1 − − 2 1 ) δ − − − iε ) π L = (1 L − L k π 1 δ (2 ( − 2 ω , define )] Θ ( iε h dω L (2 m – 83 – δ ,L E i,j ··· ,k 2 2 1 − 1  1 ) θ −  − − − π L δ − 0 1 1 1 E L L ω ) k ω k by (2 0 θ  iε π θ 2 Z − j 1 k 1 ··· Θ (2 + cos 1 ω iε   ··· cos L 2 0 2 2 1 E − 1 d 2 ω + 2 after all the integrals have been evaluated. To shorten our k 2 k and  L 1 2 1  ω ω k i (1 − 0 0 E dω − 3 2 2 δ 3 0 k ) 1 Z -loop case is worked out analogously afterwards; it involves the k L 1 → ) 3 d k lim π ε 3 ω ) L → ω π  0 d 1 π ··· (2 ε Z (2 − Θ 2 . In the center of mass frame of (2 1 , L 4 i θ 1 L ) ··· 2 ω θ k ) dω Z π 2 1 4 2 1 d ω k − (2 ∞ [ cos 1 ω 2 0 − 1 k δ 2 Z d 3 1 ··· 3 k For extra clarity, we now show how the correct combinatoric factor results ) ) 3 1 1 k 3 d ) 4 π -axis as π m 1 − ( ) d π . z h k Z L (2 π (2 2 4 L δ ) θ ε (2 i d ) (2 π ε + Z π 2 Z (8 x L cos Z i 2 (2 × · · · 1 π d i L = i × 1 = 1 and the = ≡ − T = T Z 1 ε T in the two-loop case. The same ideas but with longer expressions. The two-loop TOPT diagram is given by and only take theequations, limit we define theδ expression that appears onTwo the loops. right-hand side of eq. ( Thus, when we encounterthis a implies delta we function havethree-point used that vertices, the is we distributional should evaluated identity instead at in use an the eq. integration expression ( endpoint, for the corresponding diagram,products of where delta we functions. have Namely, a we know handle that on they how arise to when using make the sense relation of these This integral is ambiguous,integration endpoints. since the To evaluate delta it functions properly, we of must the go angles back are to the evaluated TOPT at expression the Extracting the Jacobian factors results in We label the angle between k written T given by JHEP01(2021)205 , 2 1 2 2 1 . in − . 1 ) 1 2 2 ω (G.9) −  (G.12) (G.13) (G.14) (G.10) (G.11) 1 to 2 ω ω , 2 ε , 2 − 1 2 2 θ − 2 π function, has ω  ε = cos − δ . 1 ∞ 0 ω #  2 arctan 2 − ω and a  − 2 m  2 (  ε x ω − ε 1 ε 2 δ 1  ) ω ε )  . Although this factor of − 1 ω 2 1 2 1 2 2 arctan ε ω  ω ω − 2 ε . 2 2 − 2 − m − − arctan m   dω m 1 π  m ( arctan ( Z ε ε = δ − θ δ 1 2 1  dx arctan − arctan ω 1 1 cos arctan dω 2  ω ε ) − d 2 − 1 dω Z ε ε +  . ω 2 Z -loop case, where we will see that we encounter 2  − 1 2 2 2 ω – 84 – L 1 1 L ω 2 ), we learn that we must multiply the right hand x ω + m dω m ω − 1 − 2 -loop case. The combinatorial factor arises because 2 when writing the last equation. The factor of . We have already imposed three-momentum con- 2 2 1 ω ε  2 i Z π L Z , ε ω − m G.5 + 7 1 − Z ( θ 1 π 0 2 θ 2 ε m m δ = =  1 → cos dω cos  arctan d ε 2 T by a combinatorial factor of dω dx Z  ω ) , we get " Z 1 θ rather than x ) Z 4 ( ω 1 2 ! x ? 2 arctan ε 1 = 1 2 2 ( L arctan case of eq. (  δ factor in the cos 2 ) ε  = everywhere except at singular points, to get 2 ) dω ) ! − 2 d ) to get i , 1 1 Z L 1 π π 2 1 2 πδ 2 m ω Z θ Z = 2 ω (2 2 θ × 1 = = G.12 L + −
) only have support on the endpoint of the sequential delta function. We cos = 1 dω 2 ε 1 x cos ω m ω − T d ( Z G.9 q ε 4 (1 Z 2 δ πδ = 3 2 1 2 ) 1 π i 1 2 1 arctan − 6 ) is only integrated up to its endpoint, and hence should be evaluated to give π − R 2 dω d 1 dx (2 s in eq. ( ω ε G.5 ∞ = δ 0 = Z T T side of could potentially be justifiedeq. in ( the two-loopthat case argument by does claiming not that generalizea the to combinatorial the delta factor function of in plug this into eq. ( Comparing to the where we have taken thethis limit equation, arising fromthe the same integral origin over as the thethe product of We can plug in Since We now use that to write servation. We perform theto azimuthal integrals, get and change variables from with JHEP01(2021)205  ) 2 m . The 1 (G.18) (G.19) (G.20) (G.16) (G.15) (G.17) ω − ) 2 +1) 2 ) i 0 ( 2 x − , x − ) 1 i  1 L − x m ω ε ω − ( ) 1 δ 2 3 − ε 1) x = L 1 δ − 0 − − i 2 L − x L . x ( dω ω 3 m ) 1 ω dω − ( ω 1 − − 1 3 ε L − 2 − − δ ) ω ··· ω L ) with L − 3 , ) 0 1 1 x 1 ω 2 − Z − 1 π 0 1 ω ω ω 2 − , 1 − Z 1 2 1 − 1 (2 ω , m L − ) − . − 1 ω ( ! − 1 1 − L L  − 3 − ε 1 L , x − − m i,j 1 − L L δ +1 ω 2 ω ( ω +1 L i m θ . 2 dω i ,L − − · · · − 1) , k ( ε − L 1 k ω 1 − dω 1 2 ··· 1 − δ 2 1)  ε − x ω i , 2 − L δ ω 1 i x L cos 2 − dω k ( 1 dx L L − L ω j ω − − L ω x ω ω ( 1 dx − − L ,k ω 0 0 1) 1 2 x ω ω ω 1 i 3 − 3

1 Z x x 0 − 3 + 1 ω − ω x ω 3 + ω Z L − L − + 2 m 3 ( ··· ω − k L 2 − ( ··· ω − − ω θ 2 + ω ω 2 = ε − L ··· 2 2 − 0 ω 1 L 2 − δ 2 ω ω 2 j 1 Z Z 2 1 ω 1 ω ω ) ω 2 − · · · − cos dω ) − ω ω Z − π ω 1 – 85 – − 2 d dω 2 0 − ! ··· ) x + x − 1 1 variables to 1 2 1 1 1 m 0 1 . We now have a product of delta functions where m 2 i ) (2 x − · · · − ( − ω +1 1 ω ω +1 − Z ( 1 ω 2 i 0 ω Z ε 2 2 dω i,i ε k 2 +1 ω 0 Z − − δ 2 − 0 − θ δ − 1 L − q ω 1 i,i 1 1 0 i ω x ) ··· 1 = ω dx ω θ − k L x m ω 2 1 2 − L = ω 2 ( , 0 cos dω ω ω ,L − k − 1 ∞ 1 3 j 0 ε 1 − − 1 x ∂ω 3 θ ) 0 ∂ cos 2 δ Z ω − − 2 ∞ ω d i Z L π 1 ω 0 1 dx

1 ω ω ω θ − − Z − 2 (2 cos ω and L − 2 = − dω θ − 2 − d − m 1 i 1 i − ,L ω cos ( 1 1 dx 1 ··· x ω ,L 1 J ω m ε 1 x d − ω cos ( L 1 − m x δ Z + L 1 + ∞ ε ω − 1 1 1  1 ω 1 + 0 − ω L δ θ − + − 1 2 ε is given by L Z ω ω 1 1 − Z m Z k δ dx ω ··· 3 1 i ( m − + 3 -loop TOPT diagram is given by L ) ) 2 L − 1 − − L L dx cos  ε ) d 2 ) 1 ω π + L ) π L − ω δ 0 ε π d ω 0 − ω π L + π m 1 x δ ) (2 x (2 L L 1 1 − ω Z ( 1 L (2 x i + π 1 (2 ω ω + (2 i − ε L 2 ω 2 Z i L Z The δ Z ω L (2 = ω Z 2 +1 2 − L 2 × × × · · · 2 i L ,i × × · · · Z × × · · · 2 1 = 2 x = × · · · × = T T = T loops. T where each is evaluated at the endpoint of the previous one. To handle this more carefully, we Shifting the integrals gives so We change variables from the Jacobian for each where Preforming the azimuthal integrals gives L JHEP01(2021)205 ) 1 − 1 , ,L ) . − 2 1 1 , then ) (G.25) (G.23) (G.24) (G.21) (G.22) y ,L 1 − 1 y − ,L 1 . 1 dy ) ,L + − 2 1 1 (1 + a y ,L . − π 1 1 ,L = L 1 − dy  ω , . . . , x x L 2 . (1 + y 2 , ! )) + π 1 1 0 2 L + . We keep only the 2 2 y − , x L − − 1 = ,L ω ,L 1 ,L = 0 ∞ 1 y 1 , x y y  − 0 Z Z ∞ −∞ x 2  ( L ··· ) ··· F 2 arctan ( 1 )) ) ) ) x 0 1 2 2 , ) − , y 2 − 1 2 1 L . 2 − 2 π y integrations is , y , ,L ) ω ( 1 1 0 , . . . , m/ 1 2 1 2 x
, x y function raised to a power each time, − dy dy 2 0 1 − y − 2 (1 + (1 + − dy y π 2 L π 0 (1 + ω 2 arctan ( x + 3 1 + ∞ 1 π  ) ω 3 y vanishes, as the upper and lower integration arctan − ) − 2 − × L Z ω ) L L 1 + = ω π L 2 ) m 1 , π − ω ( 0 , . . . , m/ − 2 ( ω y 1 1 1 ,i 2 − ε y – 86 – y 1 L x 2 dx − dy 1)! δ ω 1 Z − ) 2 y π 1 2 ! a ) − − 1 − dy 2 1 m L 0 dx , m/ ( − y (1 + 2 all strictly positive, then the upper integration limits L L x 1 ε 1 0 ( 0 2 π i x δ x x ( 1 y 1  − ω dy +  + ∞ 0 − − 2 , m/ 3 L δ 2 , . . . , m/ (1 + y + L ω 2 , . . . , m/ 2 ω ω ,L Z π limit. Hence, we obtain 2 2 1) 2 − 1 ω ) Z ω 2 ω − − + m/ 2 limit of this integral. We use the fact that if 0  ω ( dx − ∞ ( 0 2  y − −∞ − m 2 0 + 2 F + Z ( × 2 2 − 2 2 0 m 0 ε → − 1 dy y 0 ,L δ = y 1  − ,L 2 − (1 + → , 1 x , m/ Z is smooth, we can series expand it around L I 1 π , m/ x 0 , m/ 2  + ) 0 2 + x + 1) L 2 ). The result after performing 0 F 0 dx L ω  y ∞ y − m/ ω 1 0 −∞ lim → ( + ε ( m/ 1 x  in the 1 Z − ( δ x 1 + dy F − 0 × G.13 3 2 + + F − L × (1 + ∞ 3 ω L ω = dx ω π ω + = + m/ distributions. In particular, we investigate the expression 2 − Z I ε ( ∞ 2 ω m I 2 vanishes, then the integral over ε is a test function, which we take to be a smooth function of compact support. We 0 + ∞ ω F δ i Z 0 + − Z ··· . 0 Z F ω lim → × + =  lim → 0  If = I I → The last integral evaluates to just as in eq. ( Performing the integrals one by one, we get an  limits are coincident. Ifall all become of the Since the function zeroth-order terms in the expansion; the higher-order terms do not contribute in the limit Using this formula repeatedly, we find where aim to compute the use the JHEP01(2021)205 , , ] . ] (2018) (G.26) 98 . Math. Res. , . (2018) 014 L ]. 08 dω Independence of arXiv:1101.4497 1 arXiv:1006.5703 Phys. Rev. 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