Positivity of Scattering Amplitudes in the Amplituhedron

Jaroslav Trnka California Institute of Technology

Montreal, July 30, 2015

In collaboration with Nima Arkani-Hamed and Andrew Hodges, 1412.8478

You already heard about Amplituhedron todayfrom Nima and Hugh

Scattering Amplitudes

✤ Basic objects in Quantum Field Theory (QFT)

✤ Predictions for colliders: cross-sections

✤ My motivation: new ideas in QFT

Perturbative QFT

✤ Loop expansion

✤ Integrand: rational function before integration

sum of Feynman diagrams

⌦ = d4`1 . . . d4`L I(`j , ki, si)

I(`j , ki, si)

A =

Z

`j2R⌦

Integrand form

Special case: Planar N=4 SYM

✤ “Simplest Quantum Field Theory”

✤ Yangian symmetry of the integrand

✤ Toy model for QCD: tree-level amplitudes identical

✤ Formulation using on-shell diagrams, positive Grassmannian and the Amplituhedron

The Amplituhedron

(Arkani-Hamed, JT 2013)

Definition of the Amplituhedron

✤ The full definition that Nima and Hugh described fits on one slide:

Definition of the Amplituhedron

✤ The full definition that Nima and Hugh described fits on one slide:

✤ The full definition that Nima and Hugh described fits on one slide:

Definition of the Amplituhedron

⌦n,k,`

Form with logarithmic singularities on the boundaries of Amplituhedron

Integrand of amplitudes in planar N=4 SYM can be extracted from it

`nk

- number of loops- number of external legs- SU(4) R-charge

Amplituhedron form

✤ Geometry set-up for

Projective space

points in this space

-dimensional projective plane

lines

⌦n,k,`,m

Zj

Y

n

k

` Lk = (AB)k

Pm+k�1

⌦n,k,`,m(Y, Zj , ABk)

Amplituhedron form

✤ Geometry set-up for

Projective space

points in this space

-dimensional projective plane

lines

⌦n,k,`,m

Zj

Y

n

k

` Lk = (AB)k

Simple case:m = 2 k = 1 ` = 0

points

Point

Pm+k�1 n Z1, . . . , Zn 2 P2

Y 2 P2

2

1

54

3

Y

⌦n,k,`,m(Y, Zj , ABk)

Simplest example: triangle

✤ Polygon for n=3:

✤ We can expand:

✤ First boundaries:

✤ Differential form:

Y, Z1, Z2, Z3

Y = Z1 + c2Z2 + c3Z3

Y 2 (23) : c2, c3 ! 1,c2c3

fixed

Y 2 (12) : c3 = 0Y 2 (13) : c2 = 0

⌦ =dc2c2

dc3c3

Y

1

2

3

Simplest example: triangle

✤ Polygon for n=3:

✤ We can expand:

✤ We can solve in terms of SL(3) invariants

and we get

Y, Z1, Z2, Z3

Y = Z1 + c2Z2 + c3Z3

c2 =hY 13ihY 23i , c3 =

hY 12ihY 23i

⌦ =hY d2Y ih123i2

hY 12ihY 23ihY 13i

hX1X2X3i = ✏abcXa1X

b2X

c3

1

2

3

Y

Polygon

✤ General polygon

✤ We can again expand

✤ The result is not

✤ It is some non-trivial two-form

Y = Z1 + c2Z2 + c3Z3 + · · ·+ cnZn

⌦ =dc2c2

dc3c3

. . .dcncn

1

2 3

45

Y

Triangulation of polygon

✤ We can triangulate

✤ In each triangle

and we have

✤ The form for polygon:

Y = Z1 + ciZi + ci+1Zi+1

(Z1, Zi, Zi+1)

⌦i =dcici

dci+1

ci+1=

hY d2Y ih1 i i+ 1i2

hY 1 iihY 1 i+ 1ihY i i+ 1i

⌦ =n�1X

i=2

⌦i

1

i

i+ 1

Y

Triangulation of polygon

✤ For n=4 we have

⌦ =hY d2Y ih123i2

hY 12ihY 23ihY 13i +hY d2Y ih134i2

hY 13ihY 14ihY 34i

hY 13i is spurious

1

23

4

Triangulation of polygon

✤ For n=4 we have

⌦ =hY d2Y i (hY 23ih134ih124i � hY 41ih123ih234i)

hY 12ihY 23ihY 34ihY 41i

1

23

4

Triangulation of polygon

✤ For n=4 we have

⌦ =hY d2Y i N (Y, Z)

hY 12ihY 23ihY 34ihY 41i

We want to fix the numerator

from singularities1

23

4

Positivity in the Amplituhedron

(Arkani-Hamed, Hodges, JT 2014)

Numerator of the form

✤ First singularities OK

✤ Second singularities: some are bad

⌦ =hY d2Y i N (Y, Z)

hY 12ihY 23ihY 34ihY 41i

hY 12i = hY 34i = 0

hY 23i = hY 14i = 0Y = X24 = (23) \ (41) = Z2h341i � Z3h241i

Y = X13 = (12) \ (34) = Z1h234i � Z2h134ihY 12i = hY 23i = 0

Y = Z2

GOOD BAD

etc.

The numerator should vanish on X13, X24

Numerator of the form

⌦ =hY d2Y ihY X13X24i

hY 12ihY 23ihY 34ihY 41i

1

2

3

4

X13

X24

Pentagon

✤ For n=5 the form is

✤ The numerator N (Y, Z) = CIJYIY J

1

2 3

45

Y

⌦ =hY d2Y iN (Y, Z)

hY 12ihY 23ihY 34ihY 45ihY 51i

Pentagon

1

2 3

4

5

X13

X24

X35X41

X52

✤ Five bad points:

✤ Numerator vanishes

Pentagon

X13 = (12) \ (34) X24 = (23) \ (45)

X35 = (34) \ (51) X41 = (45) \ (12)

X52 = (51) \ (23)

N (Y = X) = CIJXIXJ = 0

Pentagon

1

2 3

4

5

X13

X24

X35X41

X52

Pentagon

Numerator vanishes if these six points are on the conic

N = ✏I1J1,I2J2,I3J3,I4J4,I5J5,I6J6YI1Y J1XI2

13XJ213X

I324X

J324X

I435X

J435X

I541X

J541X

I652X

J652

General polygon

✤ General polygon:

✤ We have forbidden points

✤ Points lie on algebraic curve of degree n-3

⌦ =hY d2Y iN (Y, Z)

hY 12ihY 23ihY 34i . . . hY n1i

n(n� 3)

2Xij = (i i+ 1) \ (j j + 1)

Y,Xij

Lesson from polygon

✤ The numerator is fully specified by a set of its zeroes

✤ These zeroes: illegal singularities from denominator

✤ Illegal: points outside the polygon

✤ Outside = they can not be written as

Trivial: it is a polynomial in Y

Y = c1Z1 + c2Z2 + · · ·+ cnZn ci > 0

Positivity

✤ Interesting property: positive if Y inside polygon

✤ Denominator manifestly positive

✤ Numerator also positive (zeroes outside polygon)

✤ This simple case: area of polygon = positive

⌦ =hY d2Y iN (Y, Z)

hY 12ihY 23ihY 34i . . . hY n1i

Other example

✤ We consider but

✤ We have a line in and n points

✤ These n points are positive

✤ Line where

✤ Boundaries of this space

m = 2, ` = 0 k = 2

P3Y ↵� Zj

hijkli > 0

Y = C · Z C 2 G+(2, n) Z 2 M+(4, n)

hY i i+ 1i

Four point example

✤ For n=4 we have

✤ The logarithmic form on the boundaries is just

✤ Rewrite using SL(4) invariants

Y = (C · Z) 2 G+(2, 4)

C =

✓1 c1 0 �c20 c3 1 c4

◆

⌦ =dc1c1

dc2c2

dc3c3

dc4c4

Four point example

✤ We can choose two points on

✤ Then we can solve for

C =

✓1 c1 0 �c20 c3 1 c4

◆

Y = Y1Y2

Y1 = Z1 + c1Z2 � c2Z4

Y2 = c3Z2 + Z3 + c4Z4

ci

c2 =hY 12ihY 24i c3 =

hY 34ihY 24ic1 =

hY 41ihY 24i c4 =

hY 23ihY 24i

dc1dc2 =hY d2Y1ih1234i

hY 24i2 dc3dc4 =hY d2Y2ih1234i

hY 24i2

Four point example

✤ The form is then

✤ Manifestly positive

✤ No numerator in this case

⌦ =dc1c1

dc2c2

dc3c3

dc4c4

=hY d2Y1ihY d2Y2ih1234i2

hY 12ihY 23ihY 34ihY 41i

General case

✤ In general,

✤ The differential form is

✤ The numerator

Y = (C · Z) : G+(2, n) ! G(2, 4)

⌦ =hY d2Y1ihY d2Y2i N (Y, Z)

hY 12ihY 23ihY 34i . . . hY n1i

N (Y ) = Ca1b1...an�4bn�4Ya1b1 . . . Y an�4bn�4 ⌘ (C · Y Y . . . Y )

General case

✤ Four-dimensional object

✤ Legal boundaries (consistent with positivity):

✤ Illegal: all others

hY i i+ 1i = 0Level 1:

Level 2: hY i i+ 1i = hY j j + 1i = 0

Level 3: hY i� 1 ii = hY i i+ 1i = hY j j + 1i = 0

hY i� 1 ii = hY i i+ 1i = hY j � 1 ji = hY j j + 1i = 0Level 4:

hY 12i = hY 34i = hY 56i = 0

E.g.

but also Level 3:

hY 12i = hY 34i = hY 56i = hY 67i = 0

General case

✤ In general we have many illegal Level 3 boundaries

✤ The line Y is then given by

✤ Vanishing of the numerator

hY i i+ 1i = hY j j + 1i = hY k k + 1i = 0

X = Zi + ↵Zi+1whereY = (X j j + 1) \ (X k k + 1) = X1 + ↵X2 + ↵2X3

N (Y ) =2n�8X

k=0

C(k)↵k = 0

2n� 7conditions

Fixing the numerator

✤ Number of all independent conditions

✤ All illegal points/lines outside the Amplituhedron

✤ Numerator completely fixed by vanishing there

2

✓n4

◆�

✓n3

◆� 1

Number of degrees of freedomin the numerator

Overall constant

More cases

✤ Next example:

✤ The form is

✤ Double poles in the denominator

m = 3, k = 1, ` = 0 Point Y in P3

⌦ =hY d3Y iN (Y, Z)

hY 123ihY 125ihY 145ihY 235ihY 345ihY 134i

hY 123i = hY 125i = 0 Y = Z1 + ↵Z2We have:

hY 145i = ↵h2145iOn that configuration: hY 134i = ↵h2134i

⌦ ⇠ d↵N (↵)

↵2Legal boundary but double pole

N (↵) ⇠ ↵

More cases

✤ Numerator specified by:

✤ We also checked another case

✤ This construction makes manifest:

Vanishing on all illegal boundariesPreserve logarithmic singularities on all legal boundaries

m = 4, k = 1, ` = 0

“The form” is positive in the Amplituhedron

Numerical checks

✤ Other cases: too complicated to construct

✤ Check: positivity of the form (without measure)

✤ Evidence for such construction for all cases

✤ Form = Amplitude: No physics interpretation

N

Evidence for dual picture

✤ Definition: is a differential form with special properties

✤ Positivity: evidence of (dual) volume interpretation

…… Nima might say more on Monday……

⌦

Beyond the Amplituhedron

✤ Differential form on the Amplituhedron

✤ Final amplitude: integrate over loop momenta

✤ This translates into some complicated complex contour in the Amplituhedron

⌦

= Integrand of scattering amplitudes

A =

Z

`2R3,1

d4` I(`)

Beyond the Amplituhedron

✤ We do not know what it means geometrically

✤ Also IR divergencies, we take ratios of amplitudes

✤ Ratio function

✤ The result depends on Y, Z.

R6 =A(k=1)

6

A(k=0)6

n = 6, ` = 1,m = 4

Beyond the Amplituhedron

✤ The result is

✤ Rational prefactors

✤ Transcendental functions

R6 = H1 · [(2)� (3) + (4)] +H2 · [(3)� (4) + (5)] +H3 · [(4)� (5) + (6)]

(1) =hY d4Y ih12345i4

hY 1234ihY 2345ihY 3451ihY 4512ihY 5123i

H1 = Li2(1� u1) + Li2(1� u2) + Li2(1� u3) + log(u3) log(u1)� 2⇣2

u1 =hY 1234ihY 4561ihY 1245ihY 3461i , u2 =

hY 2345ihY 5612ihY 2356ihY 4512i , u3 =

hY 3456ihY 6123ihY 3461ihY 5623i

Beyond the Amplituhedron

✤ The result is positive

✤ Rational prefactors

✤ Transcendental functions

R6 = H1 · [(2)� (3) + (4)] +H2 · [(3)� (4) + (5)] +H3 · [(4)� (5) + (6)]

(1) =hY d4Y ih12345i4

hY 1234ihY 2345ihY 3451ihY 4512ihY 5123i

H1 = Li2(1� u1) + Li2(1� u2) + Li2(1� u3) + log(u3) log(u1)� 2⇣2

u1 =hY 1234ihY 4561ihY 1245ihY 3461i , u2 =

hY 2345ihY 5612ihY 2356ihY 4512i , u3 =

hY 3456ihY 6123ihY 3461ihY 5623i

Conclusion

✤ The Amplituhedron: generalization of

✤ Differential forms -> amplitudes in planar N=4 SYM

✤ Interesting properties of this form

✤ Extends beyond the integrand form

No triangulation, fixing the form from geometryPositivity of the form

G+(k, n)

Thank you!