Scattering-Equations-based Formulasfor Massless Bosons

Ellis Ye Yuan

Perimeter Institute for Theoretical PhysicsDepartment of Physics & Astronomy, University of Waterloo

yyuan@pitp.ca

IAS Program on ScatteringAmplitudes, HKUSTNovember 18, 2014

collaboration with Freddy Cachazo, Song He

Goalintroducing

an integral representation oftree-level amplitudes for massless particles

preliminaries forCachazo, He, EYY, arXiv:1411.xxxx

Table of Contents

What Is It

How It Works for Massless Bosons

How It Makes Life Amusing

Where It Goes

I

What Is It

Motivation

An({k, ε}) = An({k , ε}) δD(n∑

a=1

ka). (1)

I Determined by Feynman rules.I A rational function at tree level.

How is this complicated?

I Gauge redundancies.I Complicated kinematic space.

Compact, closed formulas?

I Introducing auxiliary objects,taking care of universal properties of amplitudes.

Locality & Unitarity: Four Points

No “codim 2” factorizations:Only need one variable to dictate the singularity structure.

Any way to realize this? (answer in the picture)

Moduli space of n-punctured Riemann spheresM0,n

&A map ϕ : KD,n −→M0,n

Introducing a Meromorphic FormInhomogeneous coordinates: {σ1, σ2, . . . , σn}, up to SL(2,C).A Lorentz-vector-valued meromorphic form:

ωµ(z) :=n∑

a=1

kµaz − σa

dz . (2)

Connection to kinematic data:

kµa =

∮|z−σa|=ε

ωµ(z), ∀a ∈ {1, 2, . . . , n}. (3)

No pole at∞ (momentum conservation⇔ residue theorem).Consider a degenerate Riemann sphere:σa → σ∗ ∀a ∈ S ⊂ {1, . . . , n}

kµS =

∮|z−σ∗|=ε

ωµ︸ ︷︷ ︸internalparticle

= −∑

a∈{1,...,n}\S

∮|z−σa|=ε

ωµ︸ ︷︷ ︸kµa

. (4)

Construct a Quadratic Differential

How to associate degenerate CP1 with singular kinematics?

Define:Q(z) := ω(z) · ω(z). (5)

ka · ka = 0 kills double poles =⇒ only simple poles at z = σa.Consider simplest examples:

I n = 3: Q ≡ 0.I n = 4: fix SL(2,C) by {σ2, σ3, σ4} = {0, 1,∞}.

Require: σ1 → 0 (k1 + k2)2 → 0,

σ1 → 1 (k1 + k3)2 → 0,

σ1 →∞ (k1 + k4)2 → 0.

(6)

The only solution: σ1 = − (k1+k2)2

(k1+k4)2. =⇒ Q ≡ 0.

Scattering Equations [Cachazo, He , EYY, ‘13]

Proposal: Q ≡ 0, i.e., ∮|z−σa|=ε

Q(z)

dz= 0. (7)

⇓n∑

b=1b 6=a

sa,bσa − σb

= 0, ∀a ∈ {1, 2, . . . , n}. (8)

I (n − 3) independent equations:∑na=1

∑b 6=a

σia sa,b

σa−σb = 0 for i = 0, 1, 2.I (n − 3)! isolated solutions.I Generic {ka} =⇒ non-degenerate solutions

(all σ’s are distinct).

Generic Factorization ChannelsApproach the channel (k1 + k2 + · · ·+ knL)2 −→ 0.Two types of solutions:

(n−3)! −→

(nL + 1− 3)!︸ ︷︷ ︸

left

× (n − nL + 1− 3)!︸ ︷︷ ︸right

degenerate

(n − 3)!− (nL − 2)!× (n − nL − 2)! non-degenerate(9)

Formulate amplitudes for massless particles:I As a rational function evaluated on the solutions:

M =∑

σ solutions

F ({ka, εa}, {σa}). (10)

I Leading terms come from only the degenerate solutions.I Non-degenerate solutions only have

sub-leading contributions.

General FormulationContour integral inMn,0 / summation over solutions∫

dnσavol.

∏a

′δ(∑b 6=a

sa,bσa − σb

) I({k , ε, σ}) =∑

(n−3)!solns.

IJ({k , σ})

. (11)

I dnσavol. := σa1,a2 σa2,a3 σa3,a1

∏a/∈{a1,a2,a3} dσa.

I∏′

aδ(∑b 6=a

sa,bσa,b

) := σa′1,a′2 σa′2,a′3 σa′3,a′1

∏a/∈{a′1,a′2,a′3}

δ(∑b 6=a

sa,bσa,b

).

I J is the Jacobian from solving the equations:

1

J:=

σa1,a2 σa2,a3 σa3,a1 σa′1,a′2 σa′2,a′3 σa′3,a′1∣∣∣ ∂∂σ (∑

b 6=asa,bσa,b

)∣∣∣ . (12)

I L.h.s. for studying properties;R.h.s. for actual computation.

Constraints on IWays to determine I:

I Correct SL(2,C) weight.I Multi-linearity in polarization vectors/tensors.

(correct behavior under little group)I Correct mass dimensionality.I Various symmetries.

Consistency checks:I Soft limits.I Factorizations.I Other known properties (e.g., vanishing in certain sectors).I Comparison with known formulas

(mostly numerical & in 4d).Rigorous proof:

I BCFW [Britto, Cachazo, Feng, Witten, ‘04][Bo’s lecture].

II

How It Works for Massless Bosons

First Building BlockConsider I = I({σ}).Under σa −→ (ασa + β)/(γ σa + δ), the measure becomes∫

dnσavol.

∏a

′δ(∑b 6=a

sa,bσa − σb

)n∏

a=1

1

(γ σa + δ)2. (13)

Require:

I SL(2,C)−−−−→ In∏

a=1

(γ σa + δ)2. (14)

A natural ingredient:

1

σa − σbSL(2,C)−−−−→ (γ σa + δ) (γ σb + δ)

σa − σb. (15)

Need a ratio function of (σa − σb), each σa having weight −2.

Fundamental ExampleIn many situations, natural to consider certain planar ordering,e.g., partial amplitude in Yang–Mills.Define Parke–Taylor factor:

V [π] :=1

(σπ(1) − σπ(2)) (σπ(2) − σπ(3)) · · · (σπ(n) − σπ(1)). (16)

Construct the following integral [Cachazo, He, EYY, ‘13]:

m[π|ρ] :=

∫dnσavol.

∏a

′δ(∑b 6=a

sa,bσa − σb

)V [π]V [ρ]. (17)

I SL(2,C) neutral.I Mass dimension [M]2(3−n).

What does this compute?

Examples at Five Points

m[1, 2, 3, 4, 5|1, 2, 3, 4, 5] =1

s1,2 s3,4+ (cycl. perm.), (18)

m[1, 2, 3, 4, 5|1, 2, 5, 4, 3] =1

s1,2 s3,4+

1

s1,2 s4,5. (19)

1

2

3

4

5 ï

1

2

3

4

5 +

1

2

3

4

5

I m[π|ρ] computes the sum of all trivalent scalar diagramsthat are consistent with both π, ρ orderings.

I Backwards: any trivalent scalar diagram can be translatedinto m[π|ρ] for some π and ρ.(e.g., m[1, 2, 3, 4, 5|1, 2, 5, 3, 4] for the last graph)

φ3 & MoreIntegrand for φ3 [Dolan, Goddard, ‘13]

Iφ3 :=1

2n−2

∑π∈Sn−1/Z2

V [π]V [π]. (20)

More generally,

I({k, ε, σ}) =∑i

Ci ({k , ε})Ri ({σ}). (21)

Ri : ratio function of (σa − σb), with correct SL(2,C) weights.I If the answer for I = R is known in general, no need of

solving the scattering equations.Homework: Any interpretation for R beyond V [π]V [ρ]?

Second Building Block, for Yang–Mills [Cachazo, He, EYY, ‘13]

Still need V [π] for the planar ordering of partial amplitudes.The remaining factor should

I Carry the same SL(2,C) weight as V [π],and mass dimension [M]n−2.

I Be multi-linear in {ε}, and fulfil gauge invariance.Introduce a 2n × 2n matrix Ψ, consisting of four n × n blocks

Ψ :=

(A −CT

C B

), (22)

where

Aa,b :=

{2 ka·kbσa,b

a 6= b

0 a = b, Ba,b :=

{2 εa·εbσa,b

a 6= b

0 a = b,

Ca,b :=

{2 εa·kbσa,b

a 6= b

−∑

c 6=a Ca,c a = b.

(23)

Integrand for Pure Yang–MillsCorankΨ = 2. Null vectors (m = 0, 1):

(σm1 , σm2 , . . . , σ

mn , 0, 0, . . . , 0︸ ︷︷ ︸

n

)T (24)

Define reduced Pfaffian:

Pf′Ψ :=(−1)i+j

σi − σjPf Ψ(i ,j). (25)

(i , j): deleting rows/columns labeled by 1 ≤ i < j ≤ n.This is independent of the choice.

Integrand for pure Yang–Mills partial amplitude:

IYM := V [π] Pf′Ψ. (26)

Integrand for Pure Gravity

Identify εµν = εµεν .We need to two copies of εµ for each particle.

And so,

IGR := (Pf′Ψ)2 = det′Ψ. (27)

Integrand for Single-Trace Amplitudes inEinstein–Yang–Mills [Cachazo, He, EYY, ‘14]

I Cyclic symmetry among gluon labels.I Permutation symmetry among graviton labels.

Define (for r > 1):

V [i1, i2, . . . , ir ] :=1

(σi1 − σi2) (σi2 − σi3) · · · (σir − σi1). (28)

Consider the minor [Ψ]S , where entry index runs in S .Matrix [Ψ]S is not degenerate =⇒ Pf [Ψ]S .

Integrand for single-trace amplitudes in EYM:

IEYM := V [i1, i2, . . . , ir ] Pf [Ψ]{1,2,...,n}\{i1,i2,...,ir} Pf′Ψ. (29)

III

How It Makes Life Amusing

Gauge Invariance in Yang–Mills Amplitudes

0 · · ·∑n

b=22 k1·kbσ1,b

· · ·2 k2·k1σ2,1

· · · 2 k2·k1σ2,1

· · ·...

...2 kn·k1σ2,1

· · · 2 kn·k1σ2,1

· · ·−∑n

b=22 k1·kbσ1,b

· · · 0 · · ·2 ε2·k1σ2,1

· · · 2 ε2·k1σ2,1

· · ·...

...2 εn·k1σ2,1

· · · 2 εn·k1σ2,1

· · ·

, (30)

Substituting ε1 → k1 =⇒ one more independent null vector.

Simplification in Four DimensionsRecall

ωµ(z) =n∑

a=1

kµaz − σa

dz =Pµ(z)∏n

a=1(z − σa)dz . (31)

Pµ(z): degree (n − 2).Applying spin-helicity formalism: Pα,α̇(z) := σµα,α̇ Pµ(z).

Q ≡ 0 =⇒ P1,1̇ P2,2̇ = P1,2̇ P2,1̇. (32)

Define λα(z) and λ̃α̇(z) s.t.:

I Any root of λα is the root of Pα,1̇ and Pα,2̇ simultaneously.

I Any root of λ̃α̇ is the root of P1,α̇ and P2,α̇ simultaneously.

ThenPα,α̇(z) = λα(z) λ̃α̇(z). (33)

Simplification in Four Dimensions (continued)deg λα(z) = (k − 1), deg λ̃α̇(z) = (n − k − 1), withk = 2 (“MHV”), 3 (“NMHV”), · · · , n − 2 (“MHV”).

Solutions fall into sectors!

Fact: For kinematics of helicity sector Nk−2MHV,Pf′Ψ 6= 0 only in solution sector “Nk−2MHV”.

AYM =∑all

solutions

IYM

J=

∑“Nk−2MHV”

solutions

IYM

J. (34)

I Another taste of why Yang–Mills is simpler than scalar.

Kawai–Lewellen–Tye (KLT) Relation [Kawai, Lewellen, Tye, ‘86]

Why Ψ applies to both YM and GR?

KLT orthogonality [Cachazo, Geyer, ‘12][Cachazo, He, EYY ‘13]:∑π,ρ∈Sn−3

V [1, π, n − 1, n]i S[π|ρ]V [1, ρ, n, n − 1]j = δi ,j Ji , (35)

where subscript i denotes evaluation on the i th solution,and S the KLT momentum kernel [Bjerrum-Bohr et al, ‘10].

M =∑π,ρ

A(1, π, n − 1, n)S A(1, ρ, n, n − 1)

=∑π,ρ

∑i

V [1, π, n − 1, n]i (Pf Ψ)iJi

S∑j

V [1, ρ, n, n − 1]j (Pf Ψ)jJj

.

In fact, S[π|ρ] = m[π|ρ]−1 =⇒ “GR× scalar = YM2”.

IV

Where It Goes

More Homework ...

I Corank of the Ψ matrix:all + 1 − MHV NMHV N2MHV · · ·

“MHV” 6 4 2 4 6“NMHV” 8 6 4 2 4“N2MHV” 10 8 6 4 2“N3MHV” 12 10 8 6 4

...

Why is it so?

I Examples of formula for fermions[Mason, Skinner, ‘13][Bjerrum-Bohr et al, ‘14].No closed formula using scattering equations is known.

Compact integrand for theories involving fermions?

Outlook

I A deeper understanding of the building blocks,mainly the Ψ matrix.

I How these formulas tell about the Lagrangian?(particularly the contact terms)

I Situation with fermions, supersymmetries...I How close to/different from ordinary strings?I Extension to loop level.

Thank you very much!questions & comments are welcome