positivity of scattering amplitudes in the …positivity of scattering amplitudes in the...

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!