JHEP01(2018)016 Springer January 3, 2018 December 8, 2017 : September 8, 2017 : : Published Accepted Received = 4 SYM as a differential [email protected] c N , Published for SISSA by https://doi.org/10.1007/JHEP01(2018)016 and Jaroslav Trnka b [email protected] , Hugh Thomas . a 3 1704.05069 The Authors. c Scattering Amplitudes, Supersymmetric Gauge Theory

We present new, fundamentally combinatorial and topological characteriza- , [email protected] CP 8888, Succ. Centre-ville, Montr´eal,QC,Center Canada for Quantum MathematicsDepartment and of Physics Physics, (QMAP), University1 of Shields California, Ave, Davis, CA 95616,E-mail: U.S.A. School of Natural Sciences, Institute1 for Einstein Advanced Dr, Study, Princeton, NJLaCIM, de `aMontr´eal, 08540, D´epartement Math´ematiques,Universit´edu Qu´ebec U.S.A. b c a Open Access Article funded by SCOAP ArXiv ePrint: understanding of the scatteringform super-amplitude on the in space planar of physical kinematical data. Keywords: “sign flips” of theeasily projected derived data. as The elementary locality consequencesnatural and “dual” of of unitarity this the of binary amplituhedron. scattering code.purely This amplitudes in Minimal picture the are gives winding configuration us space defines anphysics), of a avatar points a of in new the vector amplituhedron interpretation space of (momentum-twistor space the in canonical the amplituhedron form, and a direct bosonic Abstract: tions of the amplituhedron.the resulting Upon configuration projecting ofnumber”. external points Equivalently, the data has amplituhedron a through canjections be specified the of fully (and amplituhedron, the described maximal) geometry in down binary: generalized to “winding canonical one pro- dimension have a specified (and maximal) number of Nima Arkani-Hamed, Unwinding the amplituhedron in binary JHEP01(2018)016 3 7 23 31 15 33 29 36 15 26 and correlation functions 17 12 – 1 – M 34 = 4 super-Yang-Mills theory in the planar limit, the 21 37 N 17 M × Y 1 correct flips 18 ] provides an autonomous definition of scattering amplitudes in purely 5 · Z → ]. In the context of 10 4 C – 1 = Y 5.1 General positive5.2 projections and The relations positive5.3 between amplituhedra Grassmannian from flips The amplituhedron5.4 maximizes flips Recent years have revealedphysics a of fascinating particle andgeometry” scattering [ unexpected amplitudes connection and betweenAmplituhedron new the [ mathematical basic structuresgeometric in terms, “positive with no reference to quantum-mechanical evolution in space-time. The 1 The amplituhedron 13 A “dual” of the amplituhedron 14 Cutting out the amplituhedron with inequalities 15 Open problems and outlook 10 (Super)-amplitudes as differential forms on twistor11 space Parity 12 Different winding sectors, 7 Triangulations from sign flips 8 Loops 9 The amplituhedron in twistor space 6 Factorization 3 Winding 4 Crossings 5 The amplituhedron as binary code Contents 1 The amplituhedron 2 Projecting through JHEP01(2018)016 ) I a × 1) → 1 Z m a 2 − − (1.1) (1.2) c k α + ]. n C k 1) = as being , 17 I − 2 k α in ( Y Y )) is always C ) for the ex- Y m < a , n → + 1 ··· k is even, ( α ( ··· , < of the form C m ≥ 2 -planes 1 I , α k a n of the form Y I is (projectively) ( Y ). We think of a -polygon can be represented c 0 for n m > + for something else we will intro- ] I k a k . Note that if ( a , i.e. all a Z , I a ]), and a systematic mathematical C Z Z 16 ··· ··· – , I a 1 6 a Z 3. The convexity is reflected by positivity lives in the space of C , = 1 [ αa I C , ··· m , – 2 – = . Also, strong emphasis on positivity associated m,k,n + I ) dimensional which (since k α , we can’t easily check whether or not it is in the Y . We then consider all the A , for k = 1 Y k I a Y I − Z < a . Then the interior of the polygon can be thought of as ··· n , ( ’s in the polygon is obviously only 2-dimensional. More ··· since we are reserving is “positive” in the sense of the “positive Grassmannian”, k and Y I a a through < = 1 is matrix, which played a starring role in the story of on-shell Z Z , n a 1 α . The same is true for the a αa Z 1 αa a < b < c α ··· C C for , C 1 I α − that are also positive in the same sense: 0 for 0 for Y = 1 k which is the dimensionality of the tree amplituhedron. Concretely, a > > αa 1) odd the ordering is reflected in the amplituhedron geometry as well. ] ] C − which are in the convex hull of the c m ( m m for I Z + are positive, so are the minors of a new matrix where b k × I a vectors Y → a Z Z k a αa k = 1 and so the amplituhedron itself is invariant under an untwisted cyclic sym- αn Z 0. The (tree) amplituhedron C data in the first description of the amplituhedron, is essentially entirely absent 1 ···Z − a ,C > 1 m Z a a + c Z ··· k [ Note that this description of the amplituhedron is highly redundant. This is clear (We break slightly with earlier notation in the literature where the external data is The amplituhedron is a simple generalization of the notion of plane polygons into the , 3 1) α − already for thedimensional, polygon, while since the space thegenerally of space the space of of the the larger coefficients than this means that if we are given some diagrams, and was alreadyternal “demoted” to playing anin equal our role new with picture.Grassmannian the is Therefore, positivity assumed no of in familiarity ex- be what with introduced follows. the in non-trivial The a few aspects self-contained way “positive of as properties” the needed). we positive will use will referred to as non-caligraphic duce shortly, and whichwe will get make after a projecting morewith the ubiquitous the appearance positivity in of this the paper: the data ternal data, but theminors of notion of positivityC allows for a( “twisted” cyclic symmetry.metry, while If for the and we vary over and the simple ideathe “all-loop” of amplituhedron. “hiding particles” This gives definition a needs natural an extension ordering (1 of this geometry to the span of where the fixed external data Grassmannian. Thinking projectively, the vertices ofas a convex 3-vectors of minors [ all the points with dimensions. We have external data formulation of quantum fieldthe theory, positive to geometry. derivativeof This notions perspectives physics emerging and in hand-in-hand mathematicsexploration the from of has past the been notion few explored of years “positive from geometries” (see a has e.g. variety recently been [ initiated in [ principles of locality and unitarity are moved from their primary position in the usual JHEP01(2018)016 I,i W = 4 SYM, N 0? ≥ , there is an obvious = 2, this corresponds V + 1] m .) So it is natural to ask, i Y ii 14] (2.1) 1) dimensional vector space. Z Y 0, so we see that, unlike for − 0. for N < i 23] [ ≥ Y ] i 24] W is in the amplituhedron. Y [ Y , Y 0 etc. (Note that here, and sometimes in to get an ( 34] + [ ∗ 13] Y → V Y – 3 – -dimensional vector space 12] [ + 1] Y N jj 0. But then the Plucker relations tell us that > + 1 24] = [ Y ii Y , how can we check whether it is inside the amplituhedron? Y 14] all Y 13] [ Y 24] , which can be either both positive or both negative. The [ = 4, [ Y Y , [ [ , the polytope is defined as the convex hull of these points. This is , m I a 4). The inequalities associated with the codimension one boundaries 34] = 4 where the amplituhedron corresponds to the simplest positive , Z Y 13] = 2: is the amplituhedron characterized by [ (2 [ , n Y , + 0; for m description, which we directly generalize with the conventional definition of G 23] 00 a = 2 → Y Z [ a , c , m as certain differential forms on the (momentum-twistor) space of external kine- + 1] = 12] = 2 Y Y Let us start by defining the elementary notion of “projection”, which we will use repeat- The answer is easily seen to be “no”. The obstruction is a familiar one from the Can we extend this simple picture to the amplituhedron? We certainly know all the co- This is what we will do in this paper. We will give a radically different, more invari- Y ii k amplituhedron demands the choicepolygons, the where boundary [ inequalities are insufficient to define theedly space. in the rest ofnotion this of projection paper. through some Given fixed an vector are [ The right hand side is positivefix when the the boundary inequalities signs are of satisfied, but [ this doesn’t what follows, when itfor will instance not for cause confusion, we write usual story ofof the positive Grassmannian,Grassmannian and can be seen in the first non-trivial case Here we cut out theof polytope the by polytope, a i.e. collection of by inequalities imposing associated the with inequalities the [ dimension facets one boundaries of theto amplituhedron. [ For instance for Now for general convexpolytopes polytopes, can there be iscollection defined a of in standard points two answer differentthe to “ ways. this question. Thethe first Indeed, amplituhedron. is But “vertex-centered”: there is also given a a second, “face-centered” description of the polytope. 2 Projecting through We have posed a concreteamplituhedron: question given which some motivates the search for a new definition of the large number of examples, andnew picture will opens also up provide new proofsand avenues of in also investigation a into suggests the number a structure ofdirectly of striking special the new cases. amplituhedron, picture This matical of data. scattering super-amplitudes We will in investigations briefly to touch future on work. a number of these points, deferring more detailed be used to directly check whether or not aant given and intrinsic definitionrial/topological of the in amplituhedron, nature. which Whileof is this we essentially new do entirely definition not combinato- with yet the have usual a one, we complete have proof checked the of equivalence the numerically equivalence in a amplituhedron. We would like a different description of the amplituhedron, one which can JHEP01(2018)016 , . . a a V 1). Z Z , 0 ) and , ). The . Alge- m a } ξ | ··· . We can + V ) transfor- a , V k Z N , we translate = (0 = ( V | V ∈ ∗ a ∗ ] (2.2) V Z V m -plane to get to an ) transformation to a α -dimensional vectors K 1) dimensional plane m + m + − 2 2 ···Z {V , k ) dimensional vectors V V 1 N ,k a ) transformation on the m ] = ··· Z , in the form m V + Y ∗ ). Then the =1 [ k k V α × Y ≡ k 1) transformations on ). Note that those GL( ] 1) transformations on the projected 1 1 | − m − − m ), and we can associate the projected a ⇤ N ⇤ × N ; then given any vector 2 , ξ N k V V ∗ 1 ,V vectors V 0 V − ···Z k N 1 1 1 ··· = ( a 3 V , V V to get a configuration of Z till it intersects that plane, giving the point 1 V Y – 4 – ) V ··· ∗ k Y invariant act as GL( , V Y 1 we give in subsequent sections. 1 = ( Y V V 3 ··· 3 m V 1 V V ) transformation to put = ( -plane Y k N V as the span of in the form 1) plane act as GL( = [ ( ] with Y i Y V − dimensional space and project through a m itself is projected to the origin in the new space. The projection a N Z ∗ N V = 1, and see how the amplituhedron is specified by the elementary invariant simply act as GL( ··· ∗ m 1 V ) matrix a Z m h + k ( is then of the form = 2 and × ) transformations that leave V . The vector m k ) dimensional space. m V K + 1) dimensional vector [ We will spend the rest of this section examining what these projections look like for the We will be interested in taking the configuration of ( k − − dimensions. Representing . To repeat the general construction, we can always do a GL( a N N cases of notions of “winding” anddefinitions “crossings” for in general these even two and cases; odd this will motivate the analogous put the GL( There is also anm obvious relationship between the antisymmetric brackets in ( and projecting them through the Z also has an obviouspassing geometric through the description. originit We and in choose not the some containing direction parallelDifferent ( to choices the of vector thesimilarly ( start from an ( braically, we can always do a GL( A vector ( mations that leave vectors The vectors in the new space are just the equivalence classes [ JHEP01(2018)016 1 0 0, 24 − , k > + 2, > data ’s in 1) . For k a 1] Y k − winding Z 14] = Y n Y ] are fixed, n 0 =2 4 below: is odd, and , 4 = [ < k w i Y ab Y 3 1 1i ··· , n h (41) has a , Yn =4, = 1 h k 2 when k / (34) vectors in 2 dimensions. , Y + 2)-dimensional a k 0, so that all the segments Z + 1) (23) , k > = ( + 1] 0 w =2 2 . For the minimal value of > k w is mapped to the origin in this picture. Y ii Y 1i , 1 Y = [ Yn i =3 + 2 case, where the signs of all [ h 0, while in the second configuration they wind – 5 – k k 0. We can see that to characterize the + 1 < 2 , all of = i ii > Y n h i 24 h 3 , 24 i h 0 = 4, where the configurations come in two shapes: 0.) In the first picture, though, the line segments 13 =1 , i 13 < h , n Y ” with the same orientation. (Note that we have [ > = 2, we will project the external ( w Y 13 h Y 1i m 1] ] are fixed, and we show the pictures for = 2 k Yn Y n =2, 0), but also that the closed path (12) h Y ab 1). The necessary winding is . and draw the resulting configuration of n k Y > ( , Y is even. Note that these two cases are simply distinguished by ( 0. This is a reflection of the twisted cyclic symmetry for even k ··· + 1] > being in the amplituhedron is characterized by the winding number of the , 0 =1 Y 4 -plane Y ii 41] 1 k (23) > 14 “wind around , Y w , of 1 around Y 2) when = [ 1i we would have [ 34 i k/ , k Using the fact that we are in the We can easily repeat this exercise for general Since we are projecting through To start with the case Yn =1, 23 + 1 h = ( , k ii we see that path (12) w factors associated with the twisted cyclic symmetry, which tells us that h number the signs of all of the [ and not [ odd wind oppositely and wein have the samethe direction amplituhedron, and we must require not only the correct orientation of the segments (i.e. Note that in both12 pictures we have through a We begin with the case JHEP01(2018)016 , : 1 n Z has sign n k 1) has . Here 2 n ,Z Y ( 4 , 3 ··· 4 , ) jumps over ··· 3flips 2 Y , n Z 1 Y , + 1) reveals the − k (23) 3 n , ( =3 , = ( Y k n ··· 1 4 , 0flips (23) , 3 ’s!). But let us follow the same k 2 = 1 amplituhedron if the sequence 2 m 2 0, and the path (12) 1 sign flips. = 2, but we can do something even more > k Y i m 2flips = 1 amplituhedron, i.e. it has precisely 0, which can’t cut out the amplituhedron (for 3 + 1 , 2 – 6 – is in the m > ii h ] Y even. As we will argue, this picture works for all =2 : 3 k Y n 1 k n has exactly [ , } 0 3 ] Y 0 for > Y n < [ 1] . , . Looking again at the case of minimal c Y 1] = 1? Here the only obvious co-dimension one boundary inequal- Y i} [ = 2 configuration. Then, if we project through e.g. the point 2 k ··· +1 n m , k h Y n m 1) , b 1] − 2 = [ = =2 flips Y ··· 2 [ i , w { k 1 i 1 n = 2, and ask what the picture looks like after we project through = h 1flip 4 4 {h 1 , Y m i} n h , =1 odd and k 1 ··· It is interesting to note that a natural relationship between the “winding” and “flip” Again this extends for general What happens for , k i (again mapped to the origin), or, equivalently, we can count the number of sign flips in is in the amplituhedron if and only if 1 pictures. Consider an to go down tothe one sign dimension, flip the pattern resulting compatibleflips. configuration with of the the projected {h pattern we are looking for: logic as for the final space is eventhe simpler—it notion is of only “winding 1-dimensional! number”primitive: Clearly as we we we can’t can did be look for talking atY about the number of timesthe the sequence path (12) Y winding number ities correspond to ( one thing they can’t even distinguish between different for JHEP01(2018)016 4 3 2 2 2 2 is not in 1 Y 1 2 3 4 , where we will use 2 Y Y Y Y m Y 2 1 2 3 4 , (and as always appropriately 1 2 3 1 4 a Z 3 . In the lefthand figure, k 4 . a Z – 7 – of the vertices 4 1 1 2 2 each 3 ); since this is a general topological notion we will do 1 m for the twisted cyclic symmetry). In the righthand figure . 1 3 3 4 k 1 − k 1) 2 2 4 Y Y − Y 2 2 3 Y Y 1 1 2 1 4 3 is in the amplituhedron, which can be verified either because it is on the right Y 4 Indeed, projecting down to one dimension from two dimensions gives us a way to Having motivated our approach to characterizingwe the now amplituhedron give with a simple more examples, a systematic account generalized starting notion with of thewinding “winding case of number number”. even (again Let for usthis even first for precisely a define completely generic what configuration we of mean by 3 Winding side of the boundariesone-dimensional and pictures, the has number thethe of correct flips amplituhedron, winding, equals which or canof be because verified the in either (34) each by boundary, ofpictures observing or is the that not that projected it always is the equal on number to the of wrong sign-flips side in the projected one-dimensional characterize the points inthe the right side amplituhedron of without the boundaries! explicitlyflip We pattern assuming can upon instead that simply projecting we demand through including that are the we on get factors the of correct sign ( below, JHEP01(2018)016 0: < (3.4) (3.2) + 1). linear i ii is even. +1 k ii + 1) if and h is obviously −} ii , ∗ positive Z + , slightly negative 0, is slightly positive {− +1 i 1) when → +1 i x or − . x 0 (3.1) + 1). Let’s follow what i +1 } i 0 with the twisted cyclic i x ii + > i , + 1 > 1 to be very slightly positive, 1), and only when ) (3.3) − +1 i ii ∗ +1 n i , − 0, and ( h i Z +1 i + i ( , x h > i Z +1 + 1 { i x ∗ i i w x x . This is both intuitively obvious ii , x = ∗ h ∗ will not change till the line pointing − × i Z +1 x i i} ) i ), so it is natural to ask about whether ii Z = intersects the boundary ( w i h +1 i 1 i ∗ +1 + i i should be expressible as a Z Z i 1 − with + 1) that is hit — where we can expand i Z ∗ i i ∗ Z − h∗ ii i Z Z h 0, and move i 1 h +1 , i h are the same, then when i – 8 – i > Z , i i i i Z sgn( ∗ 1 +1 Z Z +1 i +1 i i Z x , i − x h smoothly, the + + i i will intersect the interior of a given boundary ( X , ), and when we pass through to , x h ∗ ∗ x + i i i ∗ x i 1 Z = − ⇤ Z +1 , x Z i 1 +1 ∗ i − and i 0. This leads us to define Z = =2 Z x x i i i m i i i > Z i = , i.e. that we should be able to express 1 w i 1 i +1 ∗ {h Y − +1 with +1 i +1 Z ii − i i i h ∗ Z i Z Z is hitting the boundary of some interval ( x h i ∗ h∗ +1 i Z Z ∗ h h Z Z , = 2. We can count the winding number by asking whether or not a and 0 +1 i 0 otherwise i +1 if sgn m + 1) but not ( Z < x ( i ii + i i i +1 i Z Z ∗ Z ) = i i ∗ Z h x Z h Z ( i = Thus if the signs of The total winding number does not depend on Then we define the total winding number to sum all the boundaries that are hit in this This tells us that a vector in the direction Start with w ∗ Z ∗ we intersect ( in the direction of happens as wex start with somethen boundary zero, ( then slightly negative.also on Right the on boundary the ofor boundary the this where different interval interval is ( also hit. For small Note that in oursymmetry, applications, we where only we pick demand up a that minus sign for theand boundaries easy ( to prove. As we change way, with a factor of +1 when they are oriented as only if vector pointed in some direction This means that somecombination of positive multiple of JHEP01(2018)016 (3.6) i} +1 j 0. Again Z + 1) if and j > intervals are 0 (3.5) Z i i jj > +1 +1 +1 i j j produces points + 1 Z Z Z both +1 j j = 1 for both signs ∗ j Y ii Z Z 1 Z −} h , x − +1 +1 , i i , i j i Z Z + w i ∗ ) (3.7) , , x Z ∗ +1 Z , + h h j − i Z +1 i , , i ( Z 6 + 1) are zero for both signs w=2 0 j w +1 + 3 i,j j , x , Z ii i i < w of the intervals is hit. Thus the Z j Z i , x {− i ∗ × Z ∗ +1 Z ) x j or +1 h i j +1 Z is slightly positive +1 i , } ). Thus j ) and ( Z i i i j +1 Z Z + 1 j i 1 neither Z , +1 +1 with + 1), which demands that i 4 +1 i +1 Z Z j − i Z − j − x ∗ Z i Z , jj {h +1 i i Z Z j h Z 2 + intersects the boundary ( +1 h Z , +1 i ∗ i + 1 , Z − 0 +1 Z Z i – 9 – , ii j i Z x > + Z ∗ h = 4. Now, projecting through { +1 i i Z + +1 intersect ( h j = +1 +1 , m j j i 7 Z sgn( Z Z j do +1 j j x Z +1 Z i,j i +1 X Z + +1 +1 i i i Z = Z Z i ∗ +1 i i Z Z +1 h h Z =4 Z + 1) but ∗ , m 5 0 Z are opposite, then when ii +1 w h i 1 i x < crosses to be slightly negative if sgn 1 i + , i 0 otherwise + 1 j i 8 +1 +1 i j Z Z ii +1 x Z i h j +1 x i Z          Z intersects a given boundary ( i = +1 i Z ∗ and ∗ ∗ Z ) = Z i i Z + 1). To understand this winding very concretely, we ask whether a vector in the Z ∗ h i Z ∗ and the total winding number doesn’t change. On the other hand when the signs and again the total winding number doesn’t change. h Z 1 x ( jj − +1 +1 i,j We can immediately extend to i i in a four-dimensional vector space. In four dimensions, it is not meaningful to talk i + 1 w h x x a ii and we define thesign-weighted total by the winding orientation number of to the sum boundary in over the all same the way boundaries as above: hit in this way, this leads us to define direction This tells us that aonly vector if in the direction Z about the windingabout of the a winding curve ofsphere around some naturally present the topological in 3-sphere origin. the story:( around the the The piecewise origin linear obvious sphere instead. generalization formed from There is the is simplices to a ask 3- of of hit, while when sum of the contributionsof to the winding from ( we no longer intersect ( JHEP01(2018)016 . ), +1 m j (3.9) (3.8) +1). . For Z + 2 j c k x m/ i 2 +1) + 2 k = ( ( b n m/ +1 , by counting i i = Z m + 1), we define = 4. Now these ? = 1 the interval ··· ,k +1 ii i m m i c x =2 +1 m 1 + i w 1 i i Z was crossed in traversals i = 2, the winding numbers x Y m = ; the argument is exactly the , we only have to worry about ∗ ∗ + 1) are closely related to each ∗ ), and depending on the relative Z Z Z j m ∗ 1 ! x − c = 0 otherwise; if ! 1 j = 3, we look at the exactly the same i c − c lies in some two-dimensional boundary 2 2 m m 2 m 2 +3 ∗ + 1 k + Z k ii b and b and even ( hits the boundaries (

∗

. For m = 1, we have seen that the correct notion was −} = Z – 10 – m + ,k , either we pass from hitting one boundary to the m ) = { i + 1); i.e. when j =4 winding possible, so the amplituhedron maximizes 1 or 6, in general m jj 3, and in general are given by , m, k ). In other words, for any interval ( } − w , ( 6 3 Y , j + w + 1 , 3 = 2. In both cases, we look at the collection of simplices 2 , + 1) we consider for defining winding for ii , {− 3 + 1 , the correct topological notion characterizing the amplituhe- 2 , = 2. As we smoothly change jj , m , 1 ii = . How can we generalize this to general odd m 1 , h maximum } , m ] + 1 i} = 1 ; this is determined by looking at the number of sign flips in the ii n and Y n m + 1 [ i , i is the , the winding number is given by h 6 are 1 , , + 1 m i 5 ··· i , , m,k 4 jj {h w , 1] → · · · → 3 is smoothly changed with the net contribution to the winding equalling one, Y , + 1 2 [ ∗ 2 { , ii Z h → = 1 We can extend this idea to any odd In fact the topological notions for odd We have a simple proof of this fact for the positive Grassmannian case of This definition of winding generalizes in the obvious wayWhat for winding any even numbers define the amplituhedron? For k + 1). Looking at the number of sign flips simply counts how many of these intervals + 1) contains (or “crosses”) the origin. = 4, we have windings 1 = +1 if sgn i ii ii contain the origin (thec image of ( collection of simplices ( from 1 sequence other. Let’s consider ( 4 Crossings We have seen that for even dron is that ofthat “winding”. of Already counting for “crossings”, the number of times the origin and it is empiricallyto correct note in all that otherwinding; we examples will we not prove have thesestatements checked. statements about here, It instead sign giving is flip a also patterns simple proof in interesting of section analogous 5. For general even m other as or we go from hitting both to missing boththe with number of the times net a contribution line being in zero. the direction for same as we saw abovethe for situations where a pointof in the the three-dimensional direction cell of This ( boundary is also sharedsigns by of one other cell ( Once again this total winding number is independent of JHEP01(2018)016 ∗ Z = 1 (4.1) (4.2) m 1); this we have n , we have k dimensions m facets when i} 1) boundary. j the crossings . m 00 n c Z 2 n ) (4.3) +1 k 2) ··· i , / = 3 by looking at 1) boundaries, and Z 6 = 7 for large values sign flips. Of course i , all the boundaries n , + 1 to ∗ Z k 5 m (odd h m Z , + 1) , m 4 the boundary ( i k , } ! ) (4.4) 3 (( +1 . + k , 2 j b 4, while for odd , 2 1 − Z / , + 1. The reason is that if we etc − − 2 2 = m , except , m +1 + k i k + c (even i,j = 1 + 2) Z , c i k, m

k n Z k ! ( {− h 1 6= k 1 , defined to compute winding numbers. 1) amplituhedron after quotienting by 1) boundary — we would simply add i or − − X n − +1 = 2 2 m i,j i

k w Z , m all i m 2 m − Z – 11 – , h = = 2 , . First, most naively the crossing number for some + , c i 1 i k { we have − +1 c +1 j n = k k,m k,m 6= Z i X j . Again each winding “hit” contributes 2 to the crossing w w Z n = — but it is more natural not to include the ( we don’t get any correction from the ( − , however, the story is a little different. Already for and odd +1 = 2 k Z i k m =1 +1 Z , we expect + m m k c k,m 1 {h k,m c Z are analogous to the w i,j if sgn = 2 for odd , c , . For odd i ∗ , which can be unified in the expression c 0 otherwise 2 k,m Z +1 k,m c c 2) + 1 dimensions and project through some direction / m 12. In general for even      . Thus for odd , n + 1) 9 = Z , . We can be more precise by thinking about passing from k 6 ∗ i,j , should naively be double the winding number for Z c 4 It is straightforward to compute the number of crossings for There is a simple picture relating “crossing” and “winding” number that gives us an The objects . And again, analogous to the statement of maximal winding for even and , − k = (( 2 1 , k We have numerically checked theof validity of these expression upobserved to that this crossing number is maximized by the amplituhedron. On the other hand,we for find even by quotienting through number, however, we havecounting to the crossing correct number. forfor But the the winding these fact number facets if thatZ are we we go exactly down ignore telling into the us the ( about “1 what we get expression for k, m start from containing the origin willor be the ones that were intersected either in the direction + the case of1 the positive Grassmannian;c we find for there would have beenone no to harm the in “crossing” for includingWe odd the will follow ( this pattern for general odd In the windingindependent case, of we hadwe to saw sum that over it wasalready gave natural us to the characterization sum of over the all amplituhedron the in terms boundaries of of them contain the origin. We are then led to define 3-dimensional simplices are space-filling in JHEP01(2018)016 , , i 0 ]. is 1 a Z all Z 1 Z 15 = 1 = 1 Y > ’s to − k i Z m m 1 project 1) − − m ( + amplituhe- k a ≡ m i further ˆ Z should be in the ···Z data through 1 Y = 1 amplituhedron! a = 2 amplituhedron. Z Z m 1 is simply the same as m = 1, the obvious bound- hZ Y m by the requirement that = 2 dimensional data we project through, we end to get to a one-dimensional sign flips (5.2) m a 1 k Z Z sign flips (5.1) is in the ) obtained by projecting through k ;1 Y n Z we land a configuration in the , + 2) dimensional data, this is putting 1 = 1 amplituhedron recently given in [ k Z ··· , m fully determined 2;1 have precisely Z has precisely = 2 dimensional configuration of the vectors to – 12 – no matter which            i m i} i ]. n 1) ˆ 1 b i h 2 , Z n − ih a 1 n \ n Z ( ··· 2 , . . . n Y = 2 amplituhedron! Said more explicitly, we claim that · · · h i 0. But this is automatically a consequence of the sequence followed by a projection through , 2 amplituhedron is = 1 amplituhedron. The claim is that we are in the i h m ’s are also positive; this is because = 1 keeps us in the amplituhedron. But remarkably the oppo- = [ 1 > , i · · · h m i Z 12 i i · · · h = 2. Note that if we project the external Z i m m h 1 ˆ 1 23 n ab h h n {h m h . Then it is natural to ask that the projected h , 1 0 −            m > + since = 2 dimensions and projecting through i k to be in the 1 h Y m Y k < a 1) data characterizing the amplituhedron. We will now see that this information − ··· = 1 amplituhedron. Now, we claim that these give us necessary and sufficient a . Thus, we should demand that 00 1 Z < = 2 amplituhedron if and only if all the following sequences (where m Z 1 This statement is primary, but we can quickly derive some consequences of it that Let’s now look at Let’s begin with the m a = 1 amplituhedron with external data ( . We can phrase this purely as a statement about the 1 will lead to aWe first much observe more that efficientobvious the co-dimension check one sign-flip of boundaries conditions of whether thearies trivially amplituhedron. are reproduce For ( the correct signs of the Note as usual thatconstraints on in terms of the underlying ( conditions for the accounts for the twisted cyclic symmetry): space; thus it ishave natural the to property ask that foramplituhedron. when the projected Now through by thethe twisted “ cyclic symmetry, weup can in cycle any one of the for m Z since projecting through starting from This is equivalent to the characterization of the the rest of the projected dron, projecting down to site is also true: thepossible higher “positive projections” down to one dimension land us in the amplituhedron if and only if the sequence The “winding/crossing” description we have given capturesof a the “global”, topological property can even more efficiently bethrough captured some in of a the different external1 way. data dimension. The points, key It in idea is the very is only easy to to natural see way that possible, if to we get start with down some to point in a higher 5 The amplituhedron as binary code JHEP01(2018)016 0 k > (5.7) (5.8) (5.9) (5.4) (5.5) (5.6) i has i 2 i} ih . n 1 + 1 = 16 possible i 1 · · · h 4 are all positive. h , i i sign flips n 12 1 k h {h and is hence positive. 0 and i 0, it suffices to check the 0 (5.10) < is even; this tells us that sign flips. Now, since we and i 12 > > h i k k i i n + 1 = ). 12 1) i 1 +1 i 1 2 ih Z “flip both” ii − − 1 h ih i} 21 has precisely k i i − = 1! In other words, we claim that n ) = 2; we will show that the sign flip + 1 1 ˆ 1 k } ( 21 1) ] −h h b h a 2 h m , ii n 1) m − ) − − i h 1) 1 ( i} i i h − i i − b n 0 (5.3) a Y 1 n n 23 k = 1 ( [ h 1 2 − h i i , ( = ( > − + 1 + 1 k i , i n , 0. Suppose that i i i 2 ··· 2 1) 1) 1 2 , +1 12 ih . h > − ··· n − ( ( + 1 , – 13 – i h i h i · · · h i · · · h i Z i} 12] , i i i + 1 ii 1 2 0. Thus while in principle we have 2 Y 13 23 12 h 21 i h h 12 = 2 amplituhedron iff [ h h · · · h h 1 ··· { > {h , ( i h 1) i · · · m i i 0. Let’s start by showing that if − + 12 13 k 23 ( h h i − h odd. Thus, we see that the sign flip constraint forces the > +1 {h has the same sign as 1) i ii + k “don’t change” h + 1 − i of projections down to ( , i , i + 1 2 23 ) is in the h ih 12 ii one i Y h h ··· sign flips. Now let’s look at b b 1 a a , h i 0 (where as always k ( 23 > and the sequence {h i , 0 + 1 having are both positive. But now look at the next sequence > i ii i} h n n 1 h + 1] h , even this says that Y ii ··· as long as we have and [ Let us draw these two sequences one on top of the other, shifted in the natural way: We now show that all the sign flip patterns follow from just the one beginning with Having established this, we now show that so long as k , i i i 1 12 12 where we have used that sign patterns in the block,cannot the occur. 4 The combinations allowed where patterns can then be classified as The pattern of these signs cannot be arbitrary. Indeed by the Plucker relation which is clearly compatibleknow with what the the bottom ends of sequence the having sequences look like, let’s examine a block of signs in the middle, and let’s put in what we already know about the signs: h sign flips, so does The same argument works for boundaries sign flip pattern for only h For Continuing in this way we find that all of pattern forces Let’s start with the sequence Without loss of generality we can set {h JHEP01(2018)016 , i} i} i = 1 ji = 4. 1 m +1 m − jj j sign flips 2 1 k {h {h = 2, so if we , ) we just land is on the right m i} +1 i j Y Z j = 4 amplituhedron +1 are the same, we see Z jj m 1 i} i {h has precisely + 1 } ] n jj ), the remaining data will still 2 is in the +1 = 2, we will now show that this {h b 123 sign flips (5.11) 0; since we assume the flip pattern Y Z Y m b k “flip bottom when opposite to top” [ > 0 (5.12) Z and time there is a flip in one row but not i ) > ··· and demanding we end up in the i} , . Let us illustrate with the case +1 i are the same. Continuing in this way we a a a next m jj Z a a i} − + 1] + 1 i 1234] +1 ( = 4 amplituhedron if and only if, for all such jj jj , Y ii 1 134 – 14 – [ h = 2 amplituhedron; and as we have seen this in = 4 amplituhedron iff (5.13) h m { {h + 1 have precisely m m } and ] 0 follows from the sign flip pattern for Y ii . Since these have (23) in common, projecting through i [ is the + 1), data through any ( > i} i} i i i + 1 ) we have Z Y is in the +1 123 234 is in the + 1 j a, b, b {h {h Z ii Y Y abb Y j h 6= [ Z i { and = 2 where again we know the number of flips are equal. But then and the sequence = 2, we can see that this immediately implies that , i} m the result follows. And just as for 0 i m j are the same. But we can easily do this in two steps. First, let’s look at > are the obviously the same, since projecting through ( 123 i} {h i ··· + 1] , + 1 “flip top when same as bottom ” . As for = 2 problem for which we’ve already established this result. Very slightly more jj i} i jj ) m a, b 1 + 1 a ) lands us on +1 3 {h We can extend this analysis to any higher − Z jj 2 Y ii a a a 3 [ Z ( and the sequences ( from the fact that thethat number the of number flips of of flips of The proof is easy. First,{h the number of sign flips foron the the sequences non-trivially we need to show that the number of sign flips for the sequences projected through some ( must work for all further implies that we only have to check the sign flip pattern for a single sequence, that is for all side of the boundaries, i.e. so that the physicssince of we locality already follows saw from that the pattern of sign flips! This follows trivially amplituhedron. Thus, more explicitlyiff the the claim sequences is (for that all flip only when it has the opposite parity toFirst, the if we first. project thebe external positive. So,projections, we claim the that projected turn can be checked by projecting through any The crucial point issomewhere related we to have the a second flipany set in number of of the allowed flips top possibilities. of rowanother, both These but it rows tell not must thereafter, us be the the that that bottomthe the if one, first flip then row occurs can while in only the we second flip can row when have and it not has the the first! same This parity is as because the second, while the second can It is now trivialsame. to Obviously see the that “don’t the change” and number “flip of both” sign change flips the in number the of two flips sequences equally. must be the and JHEP01(2018)016 0, 0, > > is the i i (5.15) odd positive ’s. This Y is in the +1 +1 Z m ii jj Y all h data through also preserves = 0 sign flips, +1 a 0 for 1 k ii Z h Z > i has n = 1. We also make a 0 i} m n , the projected + 1) m h sign flips (5.14) + to be positive, this subset , k P 1 0 (5.16) so long as k picture, that if − 2 Z even > m ··· < a i i , a, b 1 has ·Z i m 0 this just tells us that the rest = 0 sign flips, and thus must all − 2 1) 2 ); that is, forcing these minors to ··· . It is then interesting to see that h k m C > i} a , i data. But much more non-trivially, − , n i n < ( i Z a = 1 n 0 for (2 1 1) 12 Z a = {h matrix are positive. We see in general ··· 1( + h . Y . To begin with, let us define a “positive > − a G i ) amplituhedron in the projected space. i} m 0 Z i m · · · h 0 for ( + 1) n Z , i 1 + 1) 123 > i × k, m ··· = 1. 1 ] 13 2 data is simply in the positive Grassmannian of m i {h 0 h – 15 – is independent of ( i 12 m m h = 2. Here our criterion is simply that h Z 2 ) plane, such that projecting the + = 1; further projecting through m , 0 , i} k i i i and a m ( m is the ( m , so that the ··· 0 +1 − ··· , is in the amplituhedron if, and only if, for + 1) Y Y i = 2 to ···Z > are projected to the origin). Doing this successively lets = 2 our conditions say that we should have 1 m 1 Y abb m i − 2 a m force all the ordered minors of m m + 1) +1 {h 1) , the remaining data is clearly positive. (This is a slightly = 2 or Z i b 0 1 + 1 1 i Z − +1 m − 2 1 m , b ii i b m → h ( have zero sign flips; since m Z i ), then projecting everything through h ( Z , m ( , the flip definition of the amplituhedron is then simply the space b P i} [ Z ··· n ··· m, k statement: to be some ( 1 manifestly , the projected 0 12 0 m m m, k, n {h + 1) · · · h = 1; here we say that the sequence → → , 1 i m i m m 1 only if P i P = 0 is interesting. Here the 12 are positive; so for = 0 we are declaring that certain minors have 1( k {h i h i ), and we don’t have any k leaves the data positive, that is k 1 h 0 1) ) amplituhedron associated with the projected 0 m ’s for which The “binary code” characterization specializes this fact for For general m, n − ( → ( Y + m k, m of the While this doesn’t of minors is very well-known to the a “cluster” of is trivial for which just says thatthat all for the entries ofhave the the 1 same sign. Let’sand now that look at 5.2 The positiveThe Grassmannian case from flips G our sign flip constraints give a different characterization of the positivity of the somewhat degenerate choice for theproject positive through projections, any making usedegenerate of choice the since fact thatus if project we down topositivity either and lets us get from Now, it is ratheramplituhedron trivial for to ( see,( directly from the we have an projections that relates amplituhedra with different valuesprojection” of P 5.1 General positiveOur projections “binary and code” characterization relations of between the amplituhedron amplituhedra generalizes to a deeper statement thus it suffices to only check the sequence of see that the number of flips of JHEP01(2018)016 is is Y 0 has i Y } > 31 (5.18) (5.17) ] h n 1) ), we force data can be − m are all 0 m Z , 0 also implies ( i + we would have , since the last > i} ) dimensional n k n i} > = 4 we have that ··· i} m 0 1) 21 + = ( (23) 14 h , n 12 + 1) > h , k , − = 0 sign flips forces a n , ( i 1 Y i [ − 2 m = 2 × k , 24 ( m (12) 13 h k i h ] ; it is obviously given (up , , , k 1 (5.19) , i with ··· ··· m are all i − 0 2 ) are the conjugate indices to , b 23 m ] (14) ; of course the boundary con- 12 m 12 = 2 i > i} , ( {h m ··· n Y = 0 sign flips. Then we quickly k, m 4). Conversely, obviously if {h , b m 1 , 1) 1) k · · · h ··· (34) (24) , (2 are negative, but then finally from − ··· Y b i − , + is forced to be positive while sign flips it must switch signs in every i 1 m m → G i b ( m k + 1) 24 ( ( 1) 0 } → h 1 32 i , ··· h + (14) together with (24) i 1 − > ··· , i , = 4: from ) = . Our constraint of ( 12 k 23 − m h 12 h ( , = , m Y odd we have that , n 14] i etc. all have (34) [ m + – 16 – , { Y { ··· − m · · · h [ i} n = 2 , , + 1) a = i 12 21 (23) 1 } h m m , {h 34] − 2 , ··· 1) m 1 Y 14] i [ . Let’s denote the minors of the a ··· 1 − , I a Y , − δ 2 minors. For instamce, for [ 0 and i ) works in exactly the same way. The external ), ( m m , = 0. k ( 23] i = m > 23 Y ( a k Y I a i + 13] {h [ ··· Z , k , ··· ··· Y [ 1 12 i} ). Thus our sign flip criterion successfully (and non-trivially) works , + 1) a +1, for this sequence to have = ( 12] n {h 1 2 k m + 1) Y n 12] i is in the positive Grassmannian [ ). Consider any object of the type [ 1 m, n i 2 k ( and m Y we have a contradiction since 1 · · · h a i [ Y i G , ( 1) = { i ). Now, since the sequence i} 1( ··· − h m even these are ··· 12 32 1 sgn h m , b a {h ( , m i 4) it will have the correct sign flips. For general − , + 1) ··· 0 and so 31 n , 1 h (2 1 i = 0, and again there are enough minors to guarantee all the minors are positive. , > b 1 + i The case where The story works the same way for any (We note paranthetically that here we are taking the “twisted” cyclic symmetry for i k ( G 34 h and of course(13) the positivity ofin (12) positivity on the ordered minorsfor that are the “conjugates” to the ones we described above straints also fix signs of the matrix as ( to sign) by the minor ( the ( length slot, and thus we have sign constraints on the minors of Quite beautifully, the positivityother minors of of these minorsfor suffice the most to trivial force case the of positivity of allset to the the identity matrix While for certain set of minors to be positive. For run into a contradictionto already say for that allsign these is minors negative are we (say){h would positive, have then to from sayforced to that be negative. Soof the sign twisted flips cyclic through symmetry any is projections). necessary to get the same number the support of the Plucker relations). granted, but if we backof up view. a step we Supposesequences can we actually didn’t see its have necessity the from twisted the sign cyclic flip symmetry, point but we ask that all the be positive automatically forces all the rest of the ordered minors to also be positive (on JHEP01(2018)016 ) ) − − m m ] m , the k gives + ( + j αa k Z k number C without ··· C 1) = ( C − 123 n ) vectors that Y ) (5.21) m 1 m ( Z + 1 maximum ··· sgn([ − for the k − m 12 sign flips (5.20) 1) ·Z Y k is in the amplituhedron positive matrix − C ( Y + 1) = ( = any ··· , since we can always use the + 1] ) = k Y m + α j ! Thus, a completely equivalent at most +1] can’t have the opposite sign ] = +[ ) block and vanishing elsewhere), 1 − 1) +1 K +1 +1 1) + ( k k m k j has × − j j as a linear combination of them; the − Z K } 1) m in the positive Grassmannian, we have ] ; every term on the right-hand side has data is positive, the − Z ] 1) ( +1 m 1 − n 1 j C j +1 Z − k m ··· 1) 1) Z j ( ( are ( Z − m − k ( 123 ··· with , in other words, the sequence 1) +1 m m – 17 – = 1 dimensions we have the maximum possible k ( Y k ( j − ··· 12 ·Z Z m Y ··· ··· , 12 + ( C Y 12 12 ) which correspond to positive matrices in a ( ··· = ··· Y Y , [ 1 [ + j Y k , k, n ( k Z gives the right sign flip pattern. 1) j , + ··· − 1) G , ( ·Z − ] − Z + 1) sign flips in the sequence, and that they occur at the slots m + 1)] to our sequence, and since we already have the maximum C m ··· ( k ] , contradicting a sign flip at +1 n Z k 1) = correct flips j , +1 ] + k Z 1 − ), we have also established the right flip pattern for the image of those 1)( Y j − ’s fixes the signs in the expansion as described in section 6: ··· m k , i.e. that we have sgn([ m − j 1) , Z ( 1 = + + columns. But we would like to show that for Z +1 − m Z k · Z → k ( 1) n ··· ( sign flips. +1 with the correct flips, then we can always add zero columns to , j m − C ( k k ··· 12 +1 ) matrix being set to the identity in some ( of flips we can’t have any more. In this way, by adding a zero column at the end k m 1 , j j Y n > = ( N ··· ·Z k [ Z Y matrix in the positive Grassmannian, including generic points in the interior (or the { × dimensional cells of Y ··· C 12 ··· , K N n Here we make use of a simple but non-trivial fact about positive matrices, which tells ] ). Then, 2 Y α = 12 × j , j × Y 1 K “top cell”), can be constructedthe ( starting from some zero-dimensionaland cell recursively (corresponding shifting to the columnsneighboring of (non-vanishing) the columns. matrix by positive multiples of its immediately case, for k subset of the projection through us how to systematically build more complicated positive matrices from simpler ones. Any cyclic symmetry to puta the last zero column [ atnumber the very end. Butand then then cyclically we shifting, are we cantotal merely add number adding zeroes of in sign any flips. columns we Since like we’ve without already changing the proven than 5.4 We’d like to now showthe that correct for sign-flipY pattern. First wechanging the show conclusions. that The if argument is we’ve trivial already for shown even some as [ way of characterizing theif amplituhedron and is only simply to ifnumber say under of that any projection to But then we[ can computethe same that sign as [ the first term, and so [ that there are atj least ( 1) give a basis for thepositivity space, of so the we can expand We pause to noteof that, flips so for long our as sequences the is external also given by The proof uses the same simple observations exploited in the previous subsection. Suppose 5.3 The amplituhedron maximizes flips JHEP01(2018)016 ) is k → i f and y αa k ; since C ···Z ). Since +1 1 k i is a zero- a , i Z where = 4, on the Z C y +1 1. We can see m ··· f a = ( , y − x 1 i k = Y + a = Y ), k R k , i → Z by adjacent columns that + a L . In the particular case of ··· C Z 1 C C A , k m 1 i 1) plane. This implies that the “factorize” into two lower-point to “factorize” in the form − R k C ’s corresponding to zero-dimesional k, n 0, we can write + 2 and C C n block, and then repeatedly shifting a → k ⇤ = is a ( +1 R j × y , by beginning with zero dimensional cells + 1] ⇤ n k C j + jj L L – 18 – ) and n ). + 1 C +1 j , again the (maximized) number of flips can not be ) and the argument follows as before; the number of ⇤ k, n +1 Y ii Z ( i Z ” description of the amplituhedron, in the form of the , + is the same as if we shifted R ) with j k, n ⇤ i ( R G Z ·Z Y , C + , n C G +1 R i k Z , i 0 B B @ Z other columns, with tiny values for positive co-ordinates chosen so to be in the positive Grassmannian associated only with columns ) and ( m by positive multiples of its neighbors. But note that under L C , n , the effect on C ] vanish and so there is ambiguity in how to assign the signs and decide i L ) to any point in +1 k 1) matrix is the identity in some αa − k, n C C ( ) and any m + + 1). But then, remarkably, positivity forces k +1 ( to a generic point in a G is very close to the zero-dimensional cell which is the identity in ( , i x jj C ··· C + The only subtlety in this argument is that at the starting point, where Thus, we can make any positive matrix ··· + 1 12 matrix should have a representation where the top row is non-zero only in the entries , = 4 we expect to see the amplituhedron with some -matrices on co-dimension one boundaries of the space. For instance when 1 αa . This is expected to be a feature of amplituhedra for all Y ii i ( an avatar of factorization inC “ co-dimension-one boundaries where [ a point in theC span of ( One of the central features ofone amplituhedron boundaries geometry is of the the way amplituhedron inn which are the closely co-dimension related tom amplituhedra with lower amplituhedra ( Starting from this point, wetakes do exactly the shifts offlips columns of is preserved in every step and we etablish6 the claimed result. Factorization We simply choose ( that we have already established thatdone we what get was the needed correct — sign-flip pattern find for a this slight case, deformation we that have has the correct sign flip pattern. cells of dimensional cell fixedis to also the on identity a[ matrix zero-dimensional in boundary columns ofwhether the ( the amplituhedron, starting and flip many pattern of in the correct. brackets But there is a very easy fix to this problem. where the given column of C this preserves the positivityaltered. of the In this way we can work our way up from JHEP01(2018)016 - } 0. = 1) ] C n } − < 2 (6.1) k is the ]) Y i 0. The } [ 1 13] ] , y ), we can < n Y is a ( 2 [ 2 0. Further- y ··· +1 x j Y , > 13] [ . But this can 0, but we also 0. Now we are Z , and here the , } sign flips; even , Y k ) is still positive. and the “right” 24] j sgn( b sign flips, as well < ]) 12] { k i i Z Y Z ··· x > 0, and k n,f 2 , [ Y , , Z , y = Z , 13] +1 , i y has places to the right of 3 } 23] has ··· x > 23] Z , ) ··· [ 2 k ]) } Y } 2 , ··· , i Y ] i ] b [ ] x k i 1 [ k sgn([ 1 { n ,f . Z Z 1 +1 { b { 2 1 Y , 1) } j . Then if we project through 1 3 Y space. As we will now see, the j Z ) with n 0, then necessarily [ Y Z − = [ I 2 Y 1 [ Z ( , { [ i Y , , } Z + Y I → 2 sgn([ x 1 amplituhedron. Said in terms of i [ ]) 0, we must have [ data. Z { I i ··· + ··· ··· , − 2 , 1 Z 1 12] ··· of) ≡ → k ] Y , ; in other words we must have the signs Z Z 2 Y ) f k b 13] . (It is easy to see that this projected y 0; thus we must have 1 + j +1 1) flips, as desired. 13] , b 12] Y 0 we have that = ( j Y [ > ,I Y sgn([ Y 0, the sequence − f Z , j [ { < ··· y { k ][ picture; for instance it is not obvious that the Z , is in the +1 1 → , 1 23] j 0, it was approached from [ 12] 0. Then we must have b y b y – 19 – α 13] 1 − ·Z Y > [ → ··· Y 12] Y + C { , where j Y y which is in the span of ( Z +1 13] 12] f i j y Y 13] [ Y α Z Y + must have ( Y = are individually positive. This is strongly suggestive of in 0; without loss of generality we will consider the boundary } , the resulting projected data ( ) + ( ] Y , then if [ f f R n y → y y +1 ) + 12] [ 1 C i 2 0 in terms of the basis of Y Y Z ) splits are all non-overlapping in Z [ [ 1 = 2. The factorization picture we expect is the following. The x R b + 1] +1 and ( i Z ··· + , n m 0, we started this sequence with a single flip. Therefore, the rest of α , L 0 which means that [ 1 R 1 data will be important, since it implies a fixed pattern of signs in this Y ii + k C < Z i > 13] Z Z 0. We can set ); ( Y i = ( L [ 13] . Thus, the number of sign flips of the sequence α 23] { → } Y , n Y ] Y i L = ( 2 k 12] 1) flips. f Y y Y − k Suppose to the contrary that [ Now we know that the sequence So we now simply have to prove that as [ The heart of the matter will be to show that if [ Let’s start with sgn([ , the geometry should consist of “left” and “right” amplituhedron, where the external f We will now expand positivity of the expansion. proof will importantly use bothas the sign fact patterns that associated the with sequence the positivity of the where the sign flips occurred, let’s call then though on the boundary wemore have since [ [ the sequence know that [ interested in the sign patternclearly of be the related sequence to−{ the sign patternsame of as the counting sequence the number of sign flips of sign flips, this means thathas as ( we take [ Let us assume this forwrite the again moment and show how our desired result follows from it. Let’s boundaries are at [ where [ plane. If we project through The “factorization” statement is then that amplituhedron has external data data is positive).matrix, While it is this not fact easydifferent is to ( prove strongly from the suggested byfactorization the structure of “factorization” the ofour amplituhedron the point boundary of view, follows as simply an and elementary provably consequence from of the “binary code” of sign-flip patterns. factorization for the amplituhedronprecisely. geometry itself. Given the Letexpand point us examine thisy geometry more data of the “left” are (the projections through where the blocks JHEP01(2018)016 5 a 1) 0. Z +2] − , < (6.5) (6.6) (6.2) (6.3) (6.4) j k can be } +1 ··· l 13] 1 (6.7) , a 1 jj Y + 1) to get 5 Z a +1] is a ( − a 12 Z k y jj ] Z y ] 0, we conclude k [ njj 4 b = 1 2 a 1 > − 3 R Y k Y a [ k [ 2 , where 1) k a + have the same sign we +1] 1 y − 1) ) L a ··· , any given [ jj k +1 − , j +1 +1 j − 12 K , x 4 + ( in a basis of the rest: Z Y j +1] a k [ with b 1 = 4 and any five 0, and must have [ x +1] = ( , +1 , Z a ··· jj ] j Z 0 0 < a dimensions, with all ordered mi- 5 > Z k jj x K = 4. Suppose we are sitting on a > > 3 1) K ··· ··· + ] + +2) 1) sign flips and since we have seen ··· m a ] 1 3 − j +2) 13] j 2 j b + in + Z j − ] a 0; since 1 Z 1 < Y 2( 2 a 5 1 j k + ( 2( + +2 − sign flips sign flips − a Y a 1 x l l Y Z > + (+) 4 [ j 2 a a Y a L R a + ··· + [ 3 k k − 12 ··· 2 a 3 ] +2] + − Z − Z − Z Y +2] have the same sign. But this means that 2 a Z 2 j +1] ; [ 3 a j 3 1 b x Z [ has b has ] +1 vectors − 1 Z l l – 20 – jj 5 +1 a + a +1 n Y a } + } 5 1) 1 4 Z Z +1] and [ jj − Z a jj a Z 2 − + + 2 jj + 1) we can conclude that the sequence 12 b + 1] have the same sign, there is no sign flip at those j 12 a 0. By projecting through either (12) or ( + 1] + 1] y Z 1) − Z 1 y = [ = ( 2( jj jj j a 4 + 12] + [ l − [ jj → a Y x a +1) of these vectors. Then we can expand any one of them j Y njj Y [ having the same sign. Also, from what we’ve just learned and any ordered 1) Z − Z 2 = [ , K + 2) 2 1 12( Y − + Z a − b +1 j 0. Thus we can’t have [ + 1] [ y j 3 j ··· , Z [ 0 and Z a 2( ] , 1 < jj 5 2( , x +1] and [ 13] − Y > j ; the positivity of the ordered minors implies certain sign patterns ] a k ··· Y 12 1 4 x jj Y , − Z [ b K +1] a . Then, we can for instance expand 1 Y 1) , 1 2 3 5 a jj +1] − − a = (+) + (0) + (+) + (+) + Y 1 j Z ] = [ + 1] ; [ + 1] ··· jj 24 a < a 1 , [ 1 12 b Y jj jj 0, with y 1 [ − = = + ··· , j 1) Y + 1] 1 [ +1] = ( < 12 a − + 2) x > y jj 2 j jj +1] Z j [ = 2, projecting through ( 1) sign flips. But from the facts that [ 2( 1) 23 2( jj +1 m < a − j − Y Y Y = 2, we can conclude that 23 j [ k [ 1 x Let us now move on to the more interesting case Applying this general fact to our case of interest we have simply Indeed let us consider more generally { { a Y 2( m [ { Y have the same sign. Giventhat the [ above sequence has ( slots, so we conclude that has ( that conclude that [ [ the boundary where [ to plane and about which contradicts [ But using this expansion we can compute expanded in terms ofleft the and others, right starting and with alternating + signs signs both for to its the immediate left neighbors and to to the the right: where in the secondMore expression generally, we for are positive only keeping track of the signs of the coefficients. nors positive. Let us considerin ( a basis of theon other the coefficients of this expansion.for For instance consider JHEP01(2018)016 ; + 0; < + 0. B 2 k j Z j > jj (7.2) Z B + 1] . [ x > x x )( ··· +1)] , 1 + +1) − ijj 1 j +1 j A j ( 1 2 ( j 0 (7.1) Z 0. Similarly Y Y Z Y [ [ 1 ] = ≥ j , 0, so then the = > matrices of the x 0 x k 1 +1) dimensions. Y > + Y j C x x − k > 1 j as . Then we can also ] 0 but [ )] ’th slot, it is clearly Z 1 j +1 Y has one sign flip, let’s +2 < ). j ] = α flips in slots n Y j j i ] = ( } as Z ) with ] Z Z } 2 , 0 will belong to this “cell” Y ] B α Y j x +1 + 1 -plane in ( x k Y n Z > j k [ , + j ··· Y n , 2 + Z , A [ 2 k j 12      α Z , x j x is a Z 1 ··· +1) ··· jy for which [ ( Z j , this precisely checks membership , x [ + ( Y α 1 2 + 1 amplituhedron with external data j i 1] +1 j k I Z 1] = x j 1 for which [ α , j Y i L Y = α 1 Z j [ I k [ Z ( Y . Then we see that [ { a = 1 and { Z j )] tells us we must have , description of the amplituhedron; we can 1 a +1 ] = ··· Z 1 j j i ) 0 will belong to this cell. We can proceed in +1 00 α Z Z 2 1 0 otherwise with in some basis j x Z +1 = 1, where α 2 Z = + 1) Y j      – 21 – C. 2 x > Y m x Z j ; since 12 = 1 B = 2 = } ), the first sign sequence above is precisely what of the form with is covered by + x ] Z } j 12( 2 k n Y j Y + ,i jI Y Z 2 ··· ( j , and 1 + 1) points of i Z +1 j { 0 tells us that [ αa j 1 )( k j Z C , where = 1. Since we know < a Z +1 12( 1 k 1 0. Again we can conveniently expand Z = = 1 “cell” where the sign flip occurs in the j j + 1) = ( Y [ Z Z > αa of this form with A , m [ 1 = 1 amplituhedron where jj C 1 + 1)] x Z Y x m 1 ··· = + j , + 1] . Thus to match the sign pattern in this cell we must have ( +1)] 1 α = 2. Here we can characterize the sign flips completely by specifying the = 2, keeping track of the sign flip pattern give us a natural triangulation 2 j ] = = 1 amplituhedron is then covered for the collection of these regions for all Y Y j k 1 jj Z = 1 we can always expand ( = 2 amplituhedron. In general then, we find that + 2) m . Note also that [ , k k Y j j Y j I [ = ( , k , Z as the span of the 0 = 1 , Y amplituhedron with external data ( do have the right flip count, we know that all the image of all the j = 1 Y + 1)( < m Z ). Now [ L + 1] = [ j , ] ·Z = 1 and k m 2 +1 0. And again conversely, every C 2 Strictly speaking, this argument tells us that every point on the boundary of the am- Y j m 12( j ··· Y j [ > The region in the , = Z [ , Y . Now, with 2 2 2 , [ think of recognize this as We can trivially relate this to the “ x x of the 0 positivity of [ convenient to choose 1] and conversely, every the same way to two slots in the flips took place; so there is some Start with the easiest case focus on the place this flipThe takes full place; there is some m in order to describe the 7 Triangulations from signFor flips of the amplituhedron. Let’s consider first plituhedron belongs to the factorized productis of left the open lower amplituhedra, that butof the the lower possibility amplituhedron amplituhedra boundary and is doesY only not a fully subset coverfactorized it. of form the However, will since have sum the we of correct have the flip shown counts product on that both all the left and right, and we are done. we would lookZ at toThus check the membership number of in{ sign the flips of thein second the sequence is exactly the same as the sequence Now note that since ( JHEP01(2018)016 + to 2 ∗ j (7.5) (7.3) (7.4) +1). + 1). Z , and Z 1 1 + 1) is k of this i 2 i 2 1 Z x k i i i Y 2 ( i + , language as 1 we hit one of , we find that +1) boundary ··· -matrix are all 1 1 , ·Z 1 −Z i Z C 1 Z )( + 2) dimensions, = 1, since we have i C + 1) k = 1, the pattern of k +1 = 1 1 j i  m 1 ); here the alternating Y Z ) where the sequence i ] 1 k ] y +1 , j α k +1 (12) or miss both of them. i j α + we must hit two boundaries , i          Z Z 1 1 Z ··· k 1) j Y , y n [ -plane in ( Y } 1 Z α 0, and conversely any + 1 [ k j k 1 some other boundary ( −Z j + ,i x α  1 ≥ = 1 k j = ··· j , boundaries ( Z matrix of the form + a 1 α ; equivalently we are looking at which a Z = i log is a k 1 1 { k and the origin. For d C , y a x Z Z Ω 1 α 1 Y k =1 x Z 1) + must intersect a single boundary ( − Y α − 1) are (12) are hit; a small variation means α 1 1 = + α α n + 1) is hit. This pattern obviously continues n 0 otherwise ( = = (+ = 2. The winding number here is again 1, and y Z x 1) ∗ 3 ≤ k i α k Z k − −Z 3 i – 22 – Y ( x 1) i X ··· + 1). But now in the direction + −          < 1 log 1 , giving us a i (( i d , and 1 = ≤ i 1 +1 1 } ··· k =1 α forces all the k j Y ) α 0. Note that the ordered minors of this ,j −Z k Z -plane as ··· +1 = ≥ α , Ω = , j 3 k = 1 y j } j α to the origin intersects ∗ k { αa Z + Z ,i ··· 3 1 Z C , y α ··· Z 1 , j 1 j . Next let’s look at i + Z { 1 α 3 has its sign flips. Now j Ω are chosen for convenience. Then just as for x is degenerate since both ( Z } −Z 1 again hits some ( + 3 ] 1 Z x 1 1 n Z = 3 with winding number 2, in direction + 1), while in (a small deformation of) the direction + 1 Z + 2 k Y −Z 1) i 1 = 2 amplituhedron can be triangulated in precisely the same way. The [ 2 − , i Z α form associated with this cell is ( ( , m k ··· 1) : the line joining )(+ , k − +1 (12), and thus one more boundary ( + 1) 2 , We can also see this triangulation very naturally from the winding picture. Let us char- The The 12] j 1 = ( i 1) Y Z 1 α [ 2 i n hit. Then for ( ( for all This picture corresponds precisely to what we would see by projecting through The direction + only one of the two ishit hit. in the So direction it is usefulthe to direction characterize a cell justa by the small ( variation willThus either winding cause number the 1 line means that to in hit the both direction of + ( acterize the winding pattern by lookingpoint at in the the boundaries two that directions areboundaries hit when are intersected we by choose the full linewinding joining number 1 the line in the direction with manifestly positive minors. signs forced by flipsform at has flips inY these slots. We can think of these “cells” in the only difference is{ that weand we have can to parametrizey any mark thesigns slots in front ( of and the full form is positive. with the positive variables JHEP01(2018)016 k  for k i (7.6) (7.7) α ) ··· 1 4, there α ,  ] Y AB = 3 +1 k i m Z 3 k i i ), and any one sign Z 1 +1 Z 3 k, n i k +1 ( + 2)-planes ( ] α 1 +
k i 1 1 +1 i G α − i k . We can describe any of the Z Y α Y ( } i k Z [ ,i Y ··· , ··· 1 +1 ][ i ] 2 { -plane 2 i i +1 k Ω +1 α 2 i i 1)  loops, we have ( ] Z − ] Z 1 n 2 L ( i Y +1 2 amplituhedra. Starting with Z +1 α ≤ , Z i α Y k i 1 5 i – 23 – +1 Z i X Z Z ][ 5 ··· α = 1 1 i i α 2 < i 1 α Z Z i m = 4; at Z
Y ≤ 1 1 Y 1 [ [ − Z m k 4  Y i Y [ +1 Ω = α 4 log α i ][ d y Q  +1 1 ] log i d ] Z +1 α 1 i α α i i x Z Z Z 1 1 α i Z Z 1 log 1 d Z Y α [ Y
α )) [ 1 Y , all of which intersect on a common form associated with this cell is k Y −  k = k ,L +2) Y } log k k ( d [ ,i ··· k  , d ··· , This most direct connection between sign patterns and triangulations of the ampli- The 2 α Vol(GL( 1 Y × i { = 1 = = Ω 8 Loops We now move onloop-level to amplituhedron. loops, We beginning fix i with a quick review of the usual definition of the This form also has spurious poles that cancel betweentuhedron the is restricted terms. to theisn’t simplest a simple relation betweenpattern. the image of a particular cell of As usual the full form arise from summing over the form for each piece of the triangulation the cells correspond to exactly the same one we arrived at from the flip picture. JHEP01(2018)016 , . , ) b 0, Y to l , i αa ( a γ > ). It C , the (8.3) (8.1) (8.2) ) i -plane -plane ) ,D 2 k k AB AB + 1] A a, b ··· 1 , jj A ) 1 i 0. ( + 1 by ( D > ) is in the 1-loop i ] , with the ) ρ Y ii ’s, γ ) [ ) ) the 2-dimensional , (8.5) (8.4) (8.6) D should land us in the 0 0 AB AB ρ ( > ) > γ Y AB Y AB sign flips for all Y AB, Y ) of the k AB + 1] + 1] AB sign flips L ); together with a redundancy ( ii ) k has jj of the form ≤ Y I a i l ; AB } Z ] + 2) sign flips I σ + 1 ) i has ≤ i k A ( σa Y AB = 2 amplituhedron, with the . If we denote ( +2)-planes ( } +1 D ] Y ii Y k m [ n , and = has ( 0       I α 123 Y abb } ) σ,i ) > [ ] l Y 1 i i Y { . . . n ( ( [ C ,A , D )1 D I a, a for which projects down to a 2-plane we can call ( – 24 – which are defined up to translations by the + 1] + 2) amplituhedron, while projecting through ) is in the Z       ··· ) Y ii k i , ( αa ) σ,a ( = 1 amplituhedron, which tells us that for fixed = 0” amplituhedron. None of this is obvious from , ), Y AB C D [( Y AB m Y AB m , amplituhedron as usual. As before, we can phrase this = , 1234] = 2 Y AB “tree” amplituhedra! That is, ( I Y α together with a 2-plane ( Y Y AB , k [( [ ··· m ) in any direction of Y , { , k = 4 amplituhedron. And for any number of loops we have Y = 4 AB m )12] = 4 ) matrices m n +2) plane ( m k × +2) sign flips for all Y AB k [( , these satisfy an extended positivity constraint of the “loop-positive { αa ) is in the as the span of C has ( i ) } ] Y AB ai + 2) and the ) k Y AB ( , Note that this definition gives us an extremely simple picture for the loop ampli- Let us turn to extending our topological characterization of the amplituhedron to loop Y AB = 0 amplituhedron”, which is just the condition that [ inside ( = 2 common to all of them. The conditions are exactly the same as the above for each loop [( -loop amplituhedron is defined to be all the m { tuhedron. At one loop,m we simply have thatamplituhedron, if the the amplituhedron ( isY the intersection of the the further intersection with the “ When there is more than oneY loop, we have several ( separately. But it isget also to natural a to 2-dimensional demand space, after“ that projecting further through projecting any through of ( the ( Again as before, this hasand the effect if of we requiringthrough that assume [( one these set conditions, of1-loop then amplituhedron projections. it as those suffices This ( to leads check to the the most sign efficient flip characterization pattern of only the is then natural todata conjecture should the correspond following: to projectingwe the should through end ( upin in terms the of projections down to the are all positive. level. When we project through Together with Grassmannian”, which say that, forthe any collection ordered of minors 0 of the matrix that allows us to translateL ( where we have new (2 planes ( JHEP01(2018)016 + × Y 4) 3 0 is = 0. × Z (8.9) > i = 0; so 0 (8.7) = 0. For 2 45 i ) C > 5) positive 12 , ) is a plane CD i AB 4-form”: one h ( (3 , 1 AB + × ) h G Y AB = 2. In the old = 0 AB and also normalize ABCD i 0 (8.8) = 2 amplituhedron, h ,L 1 34 m Z ) u,v,x,y,z,α,β > ; z > y ) − = 5 CD ) x + h 2 5 x 4 Z , n βZ + (1 + ) ) in the zZ + v x are in the usual 1-loop (same as 3 + (1 + = 0 4 0, and 2 Z . We first have to demand that 2 − 2 k uv , ) = 5. In the new picture, we simply αZ Y AB 1 − yZ y (1 + = 1). Projectively ( AB + + 4 v D i → + , n 1 + y k Z 3 ) Z 12 x Z v and − = = 1 ), are positive. This certainly implies that 1 2 AB 5 ( h (1 + ) inside the aforementioned pentagon. ,L Z (1 + = ,D AB – 25 – 1 uv ,B = 1 D ” matrices 2 + Y AB k D y < z < ) xZ v = 3 amplituhedron where it is just the 0 on ( + , 5) “ 1 0 , k Y Z ) are in the 1-loop amplituhedra tells us we can parametrize 0. There are two one-loop cells which cover × = = 5 = (1 + AB ( ,A i , 5 , n ) u, v > only belongs to a single cell, and can be put in the form ). In the new picture, it is more naturally expressed as “6 form vZ = 5 amplitude where = 2 x, y, u, v > CD + AB 4 n CD are in the 1-loop amplituhedron, then demanding ( 5) matrix stacking ( ABCD with m Z h dependence corresponding to the “R-invariant”), multiplied by another 0, but seems to demand even more. However, our new claim is that, 2 5 × ) = Y > vZ ,D i AB picture of the loop amplituhedron, and this even suggests new approaches to + 4 = 1. Given the 3-term triangulation of the 1-loop amplituhedron, it is easy to ( 2 ) , 00 4 ) in the i 1 ) are positive (which means that 4 Z uZ ) ) is forced to lie inside this pentagon! Note this also suggests a different way of 2 Z , AB + 3 = 1 ( 3 1 AB Y AB Z C,D D ) Z 2 It is straightforward to check this picture by computing the full 2-loop amplitude, As another example, let us look at the case of Consider for example the case of Y AB ,D Z = = 2 tree) amplituhedron), together with the requirement that all the ordered (4 4 1 AB C ( Z and thus we have the inequalities Now in eachinstance of in these the cells, first cell, we we have have the additional condition that see that on this cut uZ just demanding that ( but in order to“cut” illustrate of the the methods 2-loop inFor a simplicity simpler we non-trivial use example,h positive let data us where compute a m minors of the (4 h once ( enough to enforce being in the 2-loop amplituhedron. 2 form”, where theand 6-form the is “2-form” the is canonical the form one for for ( definition, we lookboth at two (2 and the intersection with thison polytope ( is just aexpressing the pentagon loop on integrand/amplituhedron this form plane.triangulation. than the And Traditionally for usual the this one case point coming we4-form from would write BCFW (for the the form as “4-form 4-form for the loop ( triangulating the amplituhedron. have ( Grassmannian. Now thisgiven plane by slices the through convex the hull tree of amplituhedron the (just external the data polytope for the “( JHEP01(2018)016 ). )- y m = 4, + (8.10) (8.11) (8.12) (8.13) + x , n k ) x (1+ = 0 components / (of course in k ) ) 5 z z k (1+ ) z x βZ (1+ )+ + x y 4 (1+     v + = ) ) αZ ) x − x y x ) gives an expression for + + , β y ) 1 x (1+ (1+ y ) + Z β ) x (1) v x (1+ x (1 + + − )+ + β v ) x uv ) in these two co-ordinates gives (1+ y = (1+ x + 1 v + 1 v v (1+ y + ) ) + B AB v x (1+ v uv (1+ (1 + x + x/ + (1+ v y (1+ − − v or as ) y − v 4 α = z ’s looks like after projecting the ( )+ and adding Ω zZ − α (1 + − Z x xz + 1 α 1 z )((1+ 2 (2) + ). Matching ( z – 26 –     z yZ + dα dz + (1+ y < α < 3 z v v + ) dv dv 0 Z y x , y, z, α, β u u 0 = du + du 0; this is present even for the simplest case of , is more mysterious, related to a “bosonization” of the . For the case of relevance to scattering amplitudes with x y B β dy Y > Y dβ )(1+ y i x x (1+ ρ dx dx + ) x = = AB -plane uvxy x, β, u, v > ( as either k = 4” part of course has to do with the physics of tree amplitudes. (1) (2) (1+ γ ) = 4-dimensional momentum-twistors have a manifest importance as Ω Ω B uv m m AB : data through = 2” part is the physics of the leading quantum corrections. ( in the two cells, we have to make the co-ordinate change between them. We h (2) m Z B + Ω ’s, and the = 0” part of the geometry is about understanding the geometry of mutual positivity Z (1) dxdydzdudv m Our new description of the full amplituhedron for both trees and loops now has a = 4, this means that everything can be described as a property of the configuration Ω= of bosonic momentum-twistorview, data! while the Thisspecifying is the pleasing, externalof since kinematical the data, from a thesupersymmetry. physical introduction This point of structure of the is extra needed since the canonical amplituhedron form lives 9 The amplituhedron inAs twistor we space have remarked,reference only it to is whatdimensional striking the that configuration our of m new picture of the amplituhedron makes between loops and is associated with thedimension. physics of The the “ universalFinally IR the divergences and “ the cusp-anomalous This expression precisely (and2-loop highly amplitude. non-trivially) matches the corresponding cut of the satisfyingly strong resonance withThe “ three central aspects of scattering amplitude physics. Inserting this intoΩ = the Ω expression for Ω Now we simply addparametrize the two forms.can always expand Of course sincegeneral with we no sign have restriction on usedus different the relationship variables between to the parameters as and the form Exactly the same exercise for the second cell gives us the inequalities and the corresponding form is JHEP01(2018)016 ) a k n i in ∗ or k Z ∗ + Y (9.1) (9.2) Z ) of m m, n ( -plane M m components as dimensions. m ⇤ n Z dimensional. On the -dimensional space, it matrix of data, n  m n ) where the configuration × × Y ]. But given that our new ne Subspace dimensional, and the subset m ) 5 m dimensional m, n n ( m . ⇥ -plane ∆ in × + M k Y k A Translate Subspace k m α a ) is ∆ -plane ! that can be obtained starting from dimensional space). of the form i α k a α a y i i − a i a ∗ m, n Z ∆ Z Z ( m + a

i ∗ M ); we can think of the first – 27 – Z ) that can be obtained from some fixed ( = dimensional and has lower dimension. We would m I a -dimensional space, without ever referring to = data. + m Z , i.e. all i a m m, n α a m-plane k-plane Z k ( Z × ( ). We can think of this as giving a fixed ⇤ , k M Z ··· m, n , ( M = 1 ). Now, the space I ) dimensional by projecting though some k m, n is the vector index on the ( a + Z m dimensions. Let us define the subspace of , m ⊂ M m runs from ) ··· , in I ) is clearly a top-dimensional subspace and is also a = 1 Z i k, m, n space, and the super-amplitude is extracted from it [ ( Note that such a subspace is specified by giving a ( In equations, we look at the space of all But this is both natural and trivial. Suppose we begin with some fixed set of vectors This is very easy to do. We are working with the configuration space that give us a point in k, m, n Y W a ( dimensions. Now, let us consider the affine subspaces which are linear translates of this -plane, by translating in directions lying in some fixed ∗ which is what we thinkthe of index as “fixed external data” in the usual amplituhedron story. Here and translating in the direction of ∆ (Here dimensional data Z n m vectors has the correct “winding” oras “flip” pattern we have discussedW earlier appropriate to some other hand, the amplituhedronthus is like to identify subspaces in definition of the amplitudewould seems be very to pleasing make ifcould reference the be geometry only of directly to the associated amplituhedron the the with as underlying the well ( as the superamplitude in JHEP01(2018)016 ) )- n m × (9.3) data. space + = 2 for m indices. Z k m ( α m × follows un- k i by a positive more rows so )-dimensional Z as that subset i k L m Z = 4 of relevance ) positive + which are 2-planes m k AB ) from a α ] (9.5) -dimensional charac- ( ) )-dimensional data is L m ; × + 2)-dimensional space ) as the affine subspace m ) k m, n AB ( + : rescaling , k, n i m, n k M , k, n ( Z ] (9.4) = 4 M = 4 planes ( ) dimensional picture — we must m ’s in the ( β [ α L m m δ ( Z m, k, n + = [ ∩ W k β
⊂ M = L I ) ∩ W α ) = 2 dimensional data we have with correct ] L ; ,Y ? Said more prosaically: given some ( Z ’s satisfying the right winding condition. Is it m components as corresponding to the [ AB i α ( Y we also have Z ) dimensions, where y Y k – 28 – a ]: . But these moves on the projected − × m , k, n Z ] Z +1 [ ] = = + i +2) dimensional data. Now, if we have two different Z [ i = 4 Z Z k k [ -plane = Y k I m +1 α m,k,n i ( in ( Y , and this operation preserves the positivity of the x A m,k,n I matrix is positive? W α ] = + +1 A i Y 1 n Z [ Z − i × Z +1 ) ,k,n i 1 x m − in the direction of either of its neighbors; i.e. a series of operations of =4 i -plane indices and the last -dimensional point of view, specifying ( i + x k m i + above are precisely what we get by projecting the ( through some 1 Z m A + k a − i i Z Z Z Z with the same orientation for the windings for both curves, then we should i 1 i x − i Z -dimensional configuration of ’s that satisfies the winding/flip criteria, can we always add x → a m i + Z matrix are positive. It is interesting to ask the extent to which we can remove i Z +1) are also the same, then it seems plausible that such a deformation can be gen- Z i Z ii x Finally this picture clearly extends to all loop orders. For the case While we do not have a general proof of this statement, we suspect that the answer is In this picture, there is one last vestige of the ( For any of these affine subspaces, we can look at the part of the subspace which is Thus from the ]. data through the → Z [ i a to scattering amplitudes, aside from the in 4-dimensions. We can define that has the correct winding properties at loop level. Then, the loop-level amplituhedron is erated by a combination ofconstant, elementary or moves moving on the vertices the form der projection from exactly theZ same operation on the simplicity. We’d like to showwinding, that we whatever can uplift itcollections to of positive ( be able to smoothly deformment one ( configuration into the other. If the orientations of each seg- dimensional data matrix of that the resulting ( likely “yes”. A sketch of an approach to a proof might be the following, setting demand that this affineof subspace the be “positive”even in this restriction. the To sense beginhave that with some we all can the ask the orderedguaranteed following that obvious minors we question. can Suppose think we of having obtained this data by projecting terization of the amplituhedron actually picking out a particular translationof of possible the translations. subspace of We canY refer to this translation of compatible with the correct winding/flip pattern, and gives us the Furthermore, the Z corresponding to the JHEP01(2018)016 ; - Y m , let Y (10.2) (10.3) (10.4) (10.1) space, Y ) of the = 1, which are m, n ( 4 , k ). We can describe M 5123] Y -invariant in standard = 4 [ R m, n ) (10.5) m ··· ] [ 12345] + a i i Y Z k 4 ( ( i x G Y d 1234] 4512 5123 -planes that are constrained by i [ h h = i Y k [ i ! We will shortly understand why I a x log = 5123 Z 5123 + cyclic) + cyclic) dZ d 5 5 η → i ··· dZ i · · · h i -invariant, if the (anti-commuting) super- i · · · h I a i i 4512] 5123] η R Y Y 1234 [ [ h 1234 1234 1234 ( h 1234 5123 – 29 – h h h h 4 ( has solutions log δ 4 Y , δ log d ; this will generically intersect the subspace in points; Z d = 0 Z ··· dimensional subspace of ) = form on the Grassmannian, and let us consider the pull- I a a . Starting from this form there is a simple prescription for k k ). Now, consider those Z -dimensional way — as we will see the super-amplitude is Z ) = ) Y k -invariants”. Let’s first describe the a i × × m × x (12345) = Ω( Z R 1234] 5123] ( m -plane, . It is then very natural to look at a 4-form, not on m m Y Y is to remind us that we are to take the external data as fixed, Ω( αa m Y , x [ [ differential form on the configuration space C Y k d log ··· , Y × 1 d x space, that is -dimensional image of the amplituhedron without any reference to m ( m Y αa ) = ’s. Before showing how this works in generality, let’s start with a simple a a C Z Z are replaced by the differentials I a Y, η a degree Ω( Suppose we have some back of this formthis to by some some being orthogonal to some concretely we are just saying that the equations Note that remarkably, thisvariables is precisely the this happens on general grounds, but let us first make some general comments. Now, one can directly verify that this form can be re-written as thus we can interpret themwhen all projecting as through 4-brackets onbut the on space momentum-twistor of space, momentum-twistors as obtained Here the subscript on with the differentials acting on extracting the superamplitude, butNote we that will of present course a all more the direct brackets and occurring striking as connection. arguments of the dlog’s above contain Now in the language of thesingularities amplituhedron, on we instead talk about a 4-form with logarithmic example familiar from thebuilt simplest out of scattering the amplitudes well-knownterms, “ with as a super-amplitude, which we can write as Having seen the us go further and discuss howamplitude to in think about an the intrinsically canonicalliterally form and the scattering (super)- dimensional 10 (Super)-amplitudes as differential forms on twistor space JHEP01(2018)016 - m (10.6) (10.7) (10.13) (10.10) (10.11) 2-planes L = 2 ampli- I a , we have ) ··· m = 1, from the (10.9) a k dZ ] space, in other 2 + k a ] ) a dZ i forms on the a → 1 ) 4 Z ) (10.8) dZ I a x a k dZ ( dZ η dZ αa i 4 . For i × dZ αa αa C = 0, to find a [ +1 C C i 14 αa Z I [ k a m ( , we also have Z k × i C Z i k a ] with ( × m a × form for any AB Z ABdZ Z k +1 αa δ h m i Z m ih ] k × δ C ) δ Z 3 a ) form. For instance, even the ] + in terms of the wedge products i y m × a space. Note that in this setting ( i Z L δ I × a Z i · · · h k ) dZ Z 1 } 3 y αa × ) ih + = dZ 12 ( dZ α − y i m C ) ) k k ( [ +1 αa i x =2 × k +1  dx ( i αa AB C Z I × m a h [ ! AB ,m 1 Z C ABdZ αa ( m [ k Z k 1 h ··· | Z , dx =1 ) C 2 k k × a Z 1 k k i ih x × × − m i Z ( i Ω ··· δ dx m m 12 { Z − h 1 = δ δ αa k 1 ∂x = i k k × – 30 – Z dx +1 AB h × × i ∂C m -space picture. Let us return to the h dx =2  m m I a  Y dy I dZ a + } i Z dy dy k,m Z i ) i ··· Ω Z αa ··· ) form on αI,i x 1 ··· ··· 1 (10.12) { ( i L ∂x 1 1 Z + ∂C dy = h αa ( 2 ∂x + dy dy det i i i  Z k ∂C i ( Z Z 14 13 i X = =  data exposes some remarkable relationships between forms that × 1234 = = } 14 k = AB AB a ih k h h × Z × αI,i B m =2 AB + 1) of the polygon we have { 2 m ’s. The result is simple; we will show that log ,m ii dx a d power of both sides we find dx det (1 =1 k i dZ k i · · · h ··· ABd ··· ··· 1 Ω × i i ih P 12 1 A dx 12 13 m 2 = 4 1-loop amplitude corresponds to the 4-form dx AB of the h n AB AB k h h , and we have a 4 ABd h α × ) This understanding of the scattering amplitude as a differential form obviously extends Thinking of the canonical amplituhedron forms instead as log d = m AB simplest to loop level as( well. In addition to“the the loop external integrand” twistor is data just one component of the 4( But there is now a beautifully simple expression for the 2 dimensional space of are not evident from thetuhedron conventional for which we gavere-interpret a these triangulation as and forms determinedtriangulation on the the form in space section of 7. 2-dimensional vectors We can as desired. so that The proof is easy. Let’s start with taking the differential of Taking the words we would like toof re-write the measure We would like to push forward any form from the Grassmannian onto JHEP01(2018)016 ) ) ’s > k d m i − − in the . Note n + 1 m i} ); but in jj 4), which − n k . Now, for 1 also − n n − ( + 1 k which are the \ n i.e. we flip the ii · · · h = ( h − ↔ , a i aI a 2), and is ( n ’s are Z k ↔ ( W a 13 − h W k b , 1) n ↔ i − . There is a more non- k ( a 12 . As we will now see, our {h W = 1 and → m a a Z ). This has long cried out for a , n , and the last term, where the ] naturally associated with cluster a (4 1) for Z 18 − M sign flips, then the ) for general m k ), parity interchanges = 2) the sequence k b − ) and the n, k m has – 31 – ), is the familiar 1-loop integrand; then we have − AB ! i} k n n ( , we have the parity conjugates ’s are in the amplituhedron, i.e. that AB I a symmetries that in concert give us the physical parity. Z Z ↔ 123 2 h k Z , = 4 SYM amplitudes — the arguments of the polylogs are which knows nothing about the ’s hit ( ··· 2 d N , : if the Z i a ’s hitting the ( W d 1234 {h ); with an additional factor of ( . Now consider (e.g. for ) dimensional while obviously the cluster structure in integrated results +1 Z k a )! Z ] for the external kinematical data in . k a a . Now, obviously if two consecutive signs agree before this transformation, Z Z − 20 1 ’s in superamplitudes obscure this connection. The bosonization of the inte- , m this is the rather more peculiar looking interchange of 00 − m a η 19 k Z − n ( This is a rather trivial The are in fact two different Our new picture of scattering amplitudes as differential forms is very satisfying. The → k trivial fact featuring the 0 and the sequence amplituhedron with the same value of The first one issymmetry extremely should simple be expected. butsign already of Suppose ever shows other we strikingly simplythat why change the a number of possiblefor positions of general sign-flips of thisthey sequence will is disagree ( afterwards, and vice-versa. So this changes the number of sign flips from the full scattering amplitudesterms labeled of by ( presumably reflects a symmetry “winding” picture gives a beautifully simple understanding of these symmetries. Parity is a fundamentalpletely symmetry obscured of in scattering momentum-twistor amplitudes space.to which The see: is bosonic conventionally given action com- ofplanes momentum the ( twistors symmetry is easy external data in thecombinatorial/topological 4-dimensional “winding/flip” criteria. space It is willthese not be may merely fascinating be to “positivity”, reflected understand but in how the involves transcendental further functions appearing after11 loop integration. Parity grand afforded by thedata amplituhedron that improves is the (4 + situation,only but knows about leaves 4-dimensional us momentum withof twistor data. external amplitudes as But finally forms, withon integrand the the and new same amplitudes picture variables. are As on we a have fully seen, equal however, the footing, “positive depending geometry” associated with In particular, recent years hasthe seen polylogarithms a found fascinating in emergenceexpressed of as cluster cross-ratios algebra of structurealgebras momentum in [ twistor data [ link with the positive Grassmannian/amplituhedronbut structure at the the “ level of the integrand, terms with a mixture of hit only the “super”-part of superamplitudes has always presentederties an of obstruction between the linking integrand prop- on the one hand, and the final integrated amplitudes on the other. The first terms, where all 4 JHEP01(2018)016 , ··· ··· , , i ’s are i (11.1) (11.3) (11.5) 5 Z 0, since . 13 W i 3 0. > + 1] . The W i AB +1 > i 2 h jj , i +1 1 jj i W j +2 1 j 1 − , and so on. Now 12 W Z + 1 − i j W i h +1 bb , W nj j AB i h Y nj 4 ih 145 Z +1 h j i + 1 . The third row has the sign flips, as desired. W 12 Z i sign flips (11.4) i 3 sign flips (11.2) W i 1 ) k to both sets of brackets, 12] [ i aa k − n h W k j AB 2 W 1 Y h 134 h Z h has i i i W +1)’st rows must be the same. + has − ih has 1 i = 0, this is the statement that Y AB i n n n i ) n . . . k W +2 i} n sign flips. The second row has the i} i 2 n ) +1 123 134 145 1 {h Z , the sequence k n = 2 amplituhedron is independent of − W jj − 1 1 +1 n 1 i m AB n + 1] + [ ’th and ( − Z − 1( i i 2 0, then also j n jj Z 1 − 1 1 2 ABW > h ih − n i ··· i , − − 2 Z i · · ·i · · ·i · · · h h i · h · · h 1( j , which has +1 n h sign flips. For j – 32 – 1 ih , i} . . . 1( Z k i i · · · Y h ABn j +1 ’s, we have to add a 3 1236 1346 1456 1456 , j Z ··· Y W Z , 2] [ 123 i h i h i h i h i j 1 i−h +1 ··· i {h Z , . . . i Z i +1 +1 1345 1235 1345 1345 i Z h ABW jj Y ABn Z 1(345) h i [ i h i h i h i h 1 h , , Z i 1(345) − i − h 2 . . . h j , , though we have checked these statements numerically for a large 2 1234 1234 1234 1234 i = W k h h h h + 2 sign flips. Now, we would like to see what happens when we sign flips. A short computation of these four brackets turns this 1 ih i and thus have a fixed sign, this means that the number of sign flips + 1] 1 k k i 1(234) +1 ; our claim is that loop-level parity is the statement that the sequence j i n jj {h 1(234) 1 ABW W W has ABn j {h h − are positive, a statement we will say more about in section 14. We don’t {h . + 11 to W j i i} = i i 2 ii n h +1 i Z i Y j W W 1 3 W k, n, L should have W i 1] [ 1 W i W 2 n h 1 The statement of parity at loop level is more interesting. Let’s work at one-loop to Now, we only have to show that the sequence First note that as long as ABW W h 1 ABW h Y ABn AB W [ claim is that thisall sequence should have have a proof for general range of In general, we can expand If we want to write this back with that the sequence begin with. We know that whenh we project through dualize the the point we project through, namely,flips that by in Plucker, successive we slots, know that or eitherjust both there in flip, are one no or row, sign if it the mustof top occur the row in form flips, the bottom the next row.must slot be Since where we equal. a know that flip In all occurs this the last way entries we are work our way from the top to bottom rows, and conclude The first row is our usualsame sequence first entry as thesame first first row, two and entries thereafter as the is secondit of row, is the and easy form is to thereafter see of thatThe the the first form number parts of of sign the flips twowe rows of used coincide; the to thereafter show the that argument the is number exactly of the sign same flips as for what the into the pretty statement that the sequence This statement is easy to prove. Let us consider the following sequences of minors Note that for this conclusionpositive we (which don’t they have aren’t!), to only assume that that the all minors the ofh minors the of form these JHEP01(2018)016 M (12.3) (12.4) (12.1) (12.2) M × i i = 0. When 5123 k ih 5123 ··· ih ··· + 4512 ) + 1 2 ih 4512 0 but where we have ih M 34 M 1 > 4 i i MHV integrand with the 34 g AB M 51 51 0 (12.5) + AB MHV 1-loop integrand. We + 1 + 1] > AB AB 1 , written in a loop expansion as i − h i M ih ih j 2 ) M Y ii i − h M g 1 45 45 2345 AB M ih ( AB 2345 AB i ) + ih ih ih = 2 amplitudes as the 1-loop integrand and correlation functions = 1 + 2 34 34 1234 m AB M ( M ih disjoint, in the sense that they don’t touch 1234 M h , AB AB , + ih 51 0 2 ih ih – 33 – 51 ··· = 5 we have > 23 23 AB M i n + ( M × almost AB 4 2 ); doing this takes us from the integrand for MHV g AB AB + 1 i − h +1 ih ih M a 4 ii i − h ) + j 12 12 g Z MHV amplitudes to be 1 ) , we either get winding number 0 or 1. If we define the a 3451 i + ) Z M ih AB AB 1 1 4512 AB h h − ( + ih a h AB M 1 2 Z (234) g MHV amplitudes. So it is natural to conjecture that the canonical M 2345 = 2. The amplituhedron corresponds to winding number 1, but is ∩ ( = ( 2 ih k g a 13 = 1 + W (512) AB = 2, M h = 1 + AB m h with = M a = Z M MHV We can continue in this way to discuss any number of loops, still with It is interesting to note a feature of the geometry also reflected in the forms. The MHV M M is naturally decomposing the space defined simply by the boundary inequalities into the pieces with different windingis numbers. a It well-defined is interesting form to with note that logarithmic at singuarties 1-loop there on these boundaries. Interestingly, it then At each loop order, we are adding over all the possible winding numbers, and thus boundaries match. projecting through each ( all-loop integrand for MHV and on co-dimension one boundaries.correspond Only when to we collinear go regionswhile to do both higher-dimensional forms the boundaries have that two theon regions same the touch. “physical co-dimension poles”, This one theand is forms boundaries going are reflected are different, to in also and different. the high the residues forms: enough But upon co-dimension taking boundaries, enough the residues forms match when the shared while winding number 1 and 0 regions are canonical form with logarithmicare singularities of on course the different, minimally for winding instance space. for The forms for the MHV amplitudesreplace suggests an obvious candidate.amplitudes We to know that that for byform parity associated we can withhave winding verified 0 empirically sector that this corresponds is to correct, the by identifying the Given that the amplituhedronis has any maximal meaning winding, tothe it sectors case is of with natural differentthere to winding/flip some ask patterns. meaning whether towinding Let there the number us 0? sector start where The again we interpretation with still of have the [ 12 Different winding sectors, JHEP01(2018)016 i L ) for i L Y 0 (12.6) of these lines , we can do the k > n = 2, the canonical ] β ) m AB ( + 4) dimensions ( α MHV integrands,but to the ) k . Might any of these objects = 4 but is non-vanishing for ,L AB n ( =0 Y [ n,k . But it would be much more satis- , 2 0 ] for some interesting attempts along = 1; while this doesn’t hold true for M M / . It is therefore tempting to associate > 21 k 0 (12.7) i L > n,k,L ’th power form” is interesting, for instance ] + 1] j + 2)-planes in ( k of the MHV and ii L M k i α – 34 – ) L ) on the same footing; we can indeed think of a sum Y [ , in such a way the planes are “mutually positive” AB AB ( Y cyclically. The correlation function itself is however a Y [ = 4 SYM. For MHV amplitudes, it is natural to think , way of associating a form with this space. One obvious 0 +1 a N > L . Going to the light-like limit simply picks some L 0 with the correlation function. For general + 1] + ’th power of the form for ,n k jj > ··· , i ) and the loops ( 1 j + 1 L L +1 to intersect ), which overlap on i i is the L a Z Y ii i hL [ L k + Z n ( , . n ··· , There are a number of other interesting objects closely related to the amplitudes and It is interesting that this space defined by the “obvious physical boundaries” inequal- = 4, we have nonetheless observed this “ = 1 be associated with different winding sectors? 13 A “dual” ofContinuing the in amplituhedron the veinnatural of to exploring ask the about significance a natural of counterpart different to winding/flip the amplituhedron: sectors, it what space is do we define the amplituhedron canonical form.form for For any instancem we saw thatit for non-trivially has onlynatural simple “ratio poles. function” which Even is more the interestingly, at ratio loop level we have the forms associated with thefying square if of the this amplitude couldthese be lines. done Obviously more any such intrinsically; picturewith see must the contain [ all topology the intricate of information the associated amplituhedron. correlators themselves shows that they doresidues not we have logarithmic encounter singularities double-poles — uponnating that taking to ruin the nonetheless logarithmic find strategy property. is It simply would to beknow ask fasci- that the the form geometry to does become decompose logarithmic into in different the winding lightlike limit, sectors where with we well-defined same thing; we consideri some number of ( This is a perfectlya well-defined form space, with but this the geometry to crucial reproduce question the is, correlation how functions? can we An associate inspection of the of the lines ( collection of lines out and asks fully permutation invariant functionthe of geometry all the does not give“square” us of the the amplituhedron, amplitude,recent but and years it have simplified may seen even aof find further, beautiful stress-tensor a to connection correlators natural correlation between in functions. meaning amplitudes and related Indeed the to light-like the limit parity odd difference betweenhigher them, which vanishes for ities has a niceby physical noting interpretation. that the Indeed space we we began get our simply investigations from in imposing this the paper obvious physics boundaries does not correspond to the parity even JHEP01(2018)016 ) y m m + and ˜ (13.1) (13.2) k ; again ˜ e Y . So for Y in ( a Z ˜ Y ) dimensions m inside + 1 dimensions, y + m = 0. The space of k 0 m = 0 amplituhedron, the n k = 4. At one-loop, we have 0 (13.5) < x k > m dimensional data that simply -planes in plane given by the bottom -planes in ( i ··· k down to m < a − -dimensional planes m +2 k < m m a =0 , whose facets are the ··· 0 1 = m x m < 0 ’s are P ]. → k 1 ···Z , the same as the amplituhedron. Now, ˜ gives one natural working definition for Y m a 0 (13.4) 1 k a 17 with k P )-dimensional data and projecting through , > × will give us Z to be an m ˜         i y 10 in the amplituhedron (13.3) h m e a , the Y 1 + e 0 for Y , Z − k m 0 Y · – 35 – n 2 n > m . . . 1 x x inside it has the same dimensionality as the 1-loop + i e > Y k n k h do have such a property. Note that in the language y i a x k space is a e Y . . . e Y ··· ··· ··· ··· 0 for all ···Z . Then the inequalities -plane, the resulting data has “minimal” winding? It is in ) dimensions; but we also have a 2-plane ˜ 1 1 ···Z 0 > a k − m 1 k = 1 for any i 1 2 1 a Z m ’s defined in this way has the property that . . . P 1 x x Y e + k Z ˜ Y Y k 1 · h . But if we take e Y x n rows of the matrix which are still positive. h e Y with a 2-plane ˜ h k         = 4. Requiring minimal winding when projecting through e are “positive projections” Y < x e m Y ··· < 1 x This definition can naturally be extended to loops when For the special case of Note that the space of data that has the form of the “moment curve” which is a four-plane in (4 + in the amplituhedron would have left us with positive data. As we have seen this is wrong, = 1, this space can be identified with the dual of the (cyclic) polytope coming from the ’s of this form is not empty, easy examples are afforded by looking at a matrix of positive a e e Y this space of 4-planes amplituhedron for requires are the equations definingk a polytope inexternal data. the The dual obviousa extension dual to of general the amplituhedron, as described in [ which are points in the dual All the ordered minorspositive of for this matrixrows of are this given matrix, then bycorresponds projecting Vandermonde to through determinants the and top are but it is natural toof ask section what 5.1, planes the Y Z which satisfy A related motivation for defining thisnaively space thought is that simply starting the with following. positive ( Y We may have extremely particular natural to ask this fordimensions; projecting the through dimensionality of “minimal flips” has a meaning;positive as we Grassmannian saw itself in gives our us discussionone of configurations dimension. the of So, “zero it flips” is via natural projection to down define to the subspace of if, in projecting through some JHEP01(2018)016 Y = 1 = 2. )! k m . Then, spanned k, n a ( V + not G )-dimensional k in a point. But 0 (13.6) Y ρ > 0, and also that the + 2) dimensions. Just > data is positive, k = 1, we can hope for a 0 (14.1) with . Let us start with Z , k > 4 Y I + 1]

3 ] ] I 2 I . Said in the (4+ +1 +1 Y ii 1 j natural geometric “dual” of the j I e Y inside e Z Y Z j a k j J Z in points that we will call Z 1 ··· 1 1 − − Y J j , which lives in ( j 4 ), multiplied by some appropriate factors I Z Z Y a ··· b +1 1 Z a I Z is in the positive Grassmannian Z represents contraction with the antisymmetric ρ ][ ][ a a – 36 – V ], there are further motivations to find a dual ii Z ; in addition to the above constraints, we must +1 -plane = +1 + 2) dimensions will intersect i i i 1 k 4 4), a direct extension of the analogy with y − Z Z I k 17 , i i a 3 , I abcd Z Z (2 Z ) 10 1 1 j hh + y − − i i (˜ G 2 Z Z I b a satisfies this condition is interesting. These inequalities are 1 2-planes ˜ I Z Z ) [ [ = 2 the claim is that as long as the i L

y ·Z k (˜ 0, where k C J det . These 3-planes intersect > ··· = 1 , n ii J -dimensional vectors 4 j k Y I ˜ y , i is in the amplituhedron if and only if [ ··· = 1 ˜ . Amusingly, the picture of inequalities is defined precisely in the opposite y 1 n I a hh Y  Y for 1 − -plane k k ] for the simplest case of 1) -loop order, we have Checking that As we have seen, the “winding” picture becomes natural when projecting through The space we have described certainly gives us 17 − L satisfied due to somewhatminors”. magical For positivity instance properties for of the following “determinants of what natural 3-planes can we consider?is natural Given the to cyclic consider structureof the inherent in ( 3-planes the ( set-up, it we claim that configuration of — we are focusingby on the the information thatway, by is looking contained at in an all interestingWe configuration the would of like directions points to identifyby points dimension in counting, the a 3-plane in ( ones to determine thedescribes amplituhedron. the amplituhedron Here purely by we cuttingwith describe it a out an brief with alternative description inequalities; description of wethis content these which ourselves subject inequalities to here, future leaving work. a more complete investigation of 14 Cutting out the amplituhedronWe began with by asking inequalities whether there wasinequalities a that way of cut defining out the a amplituhedron polytopeare analogous to but not the immediately enough. saw the We obvious have boundary seen inequalities that these conditions must be supplemented by topological in [ already involves novel featuresdefinition not seen of for the polytopes. dualimportant role amplituhedron It in will we determining be the have interesting canonical given to form will for see nonetheless both if tree end the and up loops. playing an amplituhedron. Asamplituhedron; described by in analogy [ direct with and the intrinsic well-understood definitionamplituhedron case of of expressed the canonical as form an with integral logarithmic over singularities the on the dual geometry. As already described also have that tensor on the 4-dimensional space definedterms, by this the says 4-plane At JHEP01(2018)016 , ), is − } − +1) +2 1 m a a < b , and i 0. We Z − γ is in the , k, k 3 in points > +1 1 a ≤ ] ) intersect i < j < k . These 5- α , β Y , β ) − i Z n ) a +2 β k a ( Z ··· , 0 (14.3) Z 1 , ≤ AB − ( Y AB i > a +1 α α a +1) + 1 )

Z ] ] ] Z 2 1 0, and the configura- a < b < c a − β +1 +1 AB +1 a , and choose any , j, j Z > ( k k k } 1 1 with Z 1 Z Z Z Y − ’s that correspond to 0- − [ . Now, consider the set of 3 k k k a i , j } − { − Z + 1] ( Z Z β Z Z 0 1 3 , 1 1 2 1 +2)-planes ( < γ − − jj − − ≤ − > , β k 0 (14.2) k k a k i 2 3 = 4 SYM scattering amplitudes, +1) Z Z Z Z α + 1 > b c c ··· , α

N , Z Z + 1] Z < γ ] ] b , i, i a a 3 3 Y ii 1 1 ii ··· Z Z Z with α β for indices that wrap past γ + 1 α , − 2 2 ; ) 3 1 i ][ ][ ][ 1 3 β α β − 1 1 ’s by shifting adjacent columns; it is easy + 1 { +1 +1 +1 k j j j < β Z 1 - bβ bα < β 1) Z Z Y AB Z α 2 } ). The extension to the all-loop amplituhedron j j j . Once again, we conjecture that ) intersect the ( ][ 2 − [( ][ 1 a 3 , Z Z 3 Z – 37 – 1 1 +1 U 1 0 α β < β − k, n a 2 2 − − − ). ( < β } − { 3 1 j j j Z β > α + β 1 a 1 1 Z Z Z , n , α b c c β ), and the 5-planes ( G k, n Z = 4, we consider the 5-planes ( ; ( aβ . The obvious generalization of these statements holds 1 Z aα Z Z [ [ 3 b , n a a + 1] + − ··· ···

m a , , Z Z Z , and G is in in points 1 3 jj Z + 2 { < α a ][ ][ ][ det + 1 Y k U 2 +1 +1 ( +1 1 + 1 < α i i i + α a < b < c 2 . This inequality follows non-trivially as a consequence of the Z Z Z { i i i G < α Y ii - Z Z Z 1 1 1 1 < α } -plane α − − − with k i i 1 i i < j = 3, the analog of this statement is that for any , n } data; indeed it is a consequence of a more general interesting statement. α Z Z Z b c 1 c k Z Z Z ··· Z which belong to − and b a a , 3 2 a Z Z Z , [ [ [ U . 1 , γ

{ k a < b ··· -dimensional vectors det , k are chosen from the set A great deal remains to be understood both about the mathematics and physics associ- The extension of the inequalities cutting out the tree amplituhedron to any Similarly for which are belong to + 1 in points 1 α a γ associated with it, all thecan non-trivial ultimately physics determined of by planar sign nothing flips. more than specifying a simple pattern ofated + with this and new picture. Most pressingly, we would like to fully establish the equivalence Y 15 Open problems andWe outlook have presented an essentiallytuhedron. combinatorial/topological characterization It of the is ampli- remarkable that the rich, intricate geometry of the amplituhedron, and amplituhedron if and only iftion the of obvious boundaries [ then follows. We havealso [ demand that the 3-planes ( V for higher straightforward. For instance for again multiplied by appropriate factorsplanes of intersect ( the This follows from a more generalare statement where replaced ( by any a, b, c { we have These statements candimensional be cells of proven the external recursively,and data positive building starting Grassmannian up where from to theyto are a easily show general verified, that configuration the of shifts push all such determinants to be positive. positivity of the Let’s consider any all indices in this set. Then the claim is that for any JHEP01(2018)016 = Y ”’s = 4 η N m ]. 26 , 25 can be written in ’s to get the right Y − and s = 4 SYM, where “winding” 0 N ], and ABJM theory [ ’s into + 00 24 – 38 – forms on this kinematical space, which have log- k that change “0 × = 2 amplituhedra, we saw that an exhaustive account Y m . , rather than exploiting polarization vectors, or using the “ = 1 and n, k, L m and correlation functions; but what is the invariant property of the dimensional image of the amplituhedron made possible by our new ]). This should be especially useful in the context of the powerful new ’s are in fact legal boundaries of the amplituhedron? ]. The winding/flip picture of the amplituhedron should reduce this on the physical (momentum-twistor) data determining the momenta of 22 Y m M 23 ” form. At an even more basic level, we would like to have a proof of the , ·Z 20 differential form , solely M × C , but is there any topological interpretation of the known triangulations of 19 = m Y But more ambitiously, the notion of combining all helicity information together in one It is also exciting to have a new picture of the integrand of scattering amplitudes, which Finally, the Do the sectors with different winding numbers have role to play in the physics? We For the simplest We have largely focused on describing points on the interior of the amplituhedron, but in ordinary momentum space (or ordinaryshould twistor plausibly space) make for contact with twistor-strings [ object as a of supersymmetric theories, and fixing this form by singularities determined by topological depending the particles. We havearithmic seen singularities that on 4 regionssupersymmetric with amplitudes. correct It winding would numbers, beamples determine fascinating of the amplitudes to which extend maximally are this known picture to to be the connected other to ex- positive geometry — for instance methods being developed,constrain using (and the perhaps amplituhedron determine) together4 the with SYM “symbol” Landau [ of equations,program multiloop to to MHV perfectly amplitudes well-definedfor in geometry amplitudes with problems, all not just for MHV amplitudes but correlators from the geometry? picture seems important fromcan a finally treat number the of geometryfooting points of (see “the of also integrand” view. and [ “the amplitude” In on one exactly obvious the same direction, we have seen that thegeneralized space defined to purely simply by mutualrelated the positivity obvious to between physical the inequalities, 2-planescanonical even defining form further loops, generalizing the seems notion to of be “logarithmic singularities” which can determine of the sign flip/windingtermination patterns of directly their led associatedhigher to canonical forms. triangulations of This the pictureamplituhedra, spaces does and and not if trivially the“winding/flip” not, extend de- point to are of there view? new triangulations that are more natural from the the same lines.vanish On and, boundaries for of instance,ask the the if amplituhedron, “sign there many flip” of arepattern criterion of the perturbations sign becomes brackets to flips, involving ill-defined. but isconfigurations there of We a can more efficient of combinatorial check course of whether degenerate shown is that satisfyingthe correct “ winding or flipequivalence patterns between implies “sign-flip” that and “winding/crossing number” pictures. it is desirable to find an characterization of all the boundaries of the amplituhedron along of our new formulation of the amplituhedron with the usual one; all that remains to be JHEP01(2018)016 ]. 03 JHEP 343 , (2015) . SPIRE JHEP IN 02 [ , (1998) 70 JHEP 2 , (2014) 182 12 Commun. Math. Phys. (2014) 030 , math.CO/0609764 , 10 JHEP , ]. , Cambridge University Press, Represent. Theory JHEP , , Lie theory and geometry. In honor of forms only, are there natural forms of all Anatomy of the amplituhedron k SPIRE , in IN × Positive amplitudes in the amplituhedron ]. ][ ]. – 39 – ]. SPIRE IN SPIRE SPIRE ][ IN ]. The amplituhedron Into the amplituhedron IN ][ ][ , Birkh¨auser,Boston U.S.A., (1994), pg. 531. ), which permits any use, distribution and reproduction in ]. ]. SPIRE 123 arXiv:1212.5605 IN ][ SPIRE SPIRE IN IN ][ ][ The amplituhedron from momentum twistor diagrams CC-BY 4.0 arXiv:1412.8478 This article is distributed under the terms of the Creative Commons Total positivity, Grassmannians and networks [ arXiv:1408.5531 Progr. Math. arXiv:1408.3410 Total positivity in partial flag manifolds Total positivity in reductive groups [ , [ Amplituhedron cells and Stanley symmetric functions arXiv:1408.2459 [ (2015) 030 arXiv:1312.7878 arXiv:1312.2007 065 (2016) 1025 08 [ (2015) 128 B. Kostant [ Scattering amplitudes and theCambridge positive U.K., Grassmannian (2016) [ N. Arkani-Hamed, A. Hodges and J. Trnka, T. Lam, N. Arkani-Hamed and J. Trnka, S. Franco, D. Galloni, A. Mariotti and J. Trnka, Y. Bai and S. He, G. Lusztig, A. Postnikov, N. Arkani-Hamed and J. Trnka, N. Arkani-Hamed, J.L. Bourjaily, F. Cachazo, A.B. Goncharov, A. Postnikov and J. Trnka, G. Lusztig, [9] [6] [7] [8] [3] [4] [5] [1] [2] [10] Open Access. Attribution License ( any medium, provided the original author(s) and source areReferences credited. We would like toWilliams for thank useful Yuntao discussions. Bai,DOE The DE-SC0009988. Song work The He, of work NAHChairs of Thomas program. is HT Lam, supported We is gratefully supported by Steve recognizeof by the the Karp hospitality NSERC this DOE of and work and under the the was Fields grant Lauren Canada carried Institute, Research out. where some conformal invariance; are there newAnd notions it of is “winding” peculiar associated to withdegrees, restrict these our which structures? attention would to certainly 4 be associated with less (orAcknowledgments non)supersymmetric theories? now depends only oncombinatorial imaginations the are momenta no longer of necessarilyinvariance external shackled and to particles, supersymmetry, toy and worlds our with weto geometric, conformal can the topological hope real and to worldled in describe to this scattering ask: language. amplitudes closer what As happens some when first we steps have in additional this data, direction, like we lines are at naturally infinity that break properties, is a simple and powerful idea that begs for generalization. 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