Amplituhedron meets Jeffrey-Kirwan Residue

Tomasz ŁukowskiMathematical Institute, University of Oxford

Amplitudes 2018SLAC

19.06.2018

with Livia Ferro and Matteo Parisi – arXiv:1805.01301

Tomasz Łukowski (University of Oxford) 19.06.2018 1 / 18

Introduction

Scattering amplitudes in Quantum Field Theory ∫Dφ eiS[φ]

→ modern on-shell methods: generalized unitarity, recursion relations, . . .

Geometrization of scattering amplitudes: [Hodges]

“Amplitude = Volume of a geometric space”

Main example: N = 4 super Yang-Mills (SYM)

→ supersymmetric cousin of QCD

→ numerous dualities and correspondences

→ simplicity: “Hydrogen atom of 21st century”

Geometrization of amplitudes inN = 4 SYM: Amplituhedron A(m)n,k [Arkani-Hamed, Trnka]

Tomasz Łukowski (University of Oxford) 19.06.2018 2 / 18

Introduction

The Jeffrey-Kirwan residue is a powerful concept in

→ Mathematics: localization of non-abelian group actions in equivariant cohomology[Jeffrey, Kirwan]

→ Physics: (supersymmetric) localization in gauge theories [Witten]

Supersymmetric localization:

→ Gauge theories with fermionic symmetries: Q2 = 0

→ Supersymmetric path integrals can be evaluated explicitly

Example:

→ N = (2, 2) SUSY gauge theory in two dimensions with the gauge group: U(p)

partition function onCoulomb branch localization−−−−−−−−−−−−−→ 〈1〉 =

∫γ

dpσWX1 (σ)·...·WXs (σ)

The contour γ given by the Jeffrey-Kirwan residue — fixed by gauge charges of the fields Xi.Tomasz Łukowski (University of Oxford) 19.06.2018 3 / 18

Introduction

More generally: Jeffrey-Kirwan residue is an operation on rational forms [Brion, Vergne]

For tree-level Amplituhedron A(m)n,k : define a volume function Ω

(m)n,k . It can be obtained as

Ω(m)n,k =

∫γ

ω(m)n,k

Our work: [Ferro, TL, Parisi]

Apply the Jeffrey-Kirwan residue to find the Amplituhedron volume function

We showed that:

JK-residue gives the correct contour for cyclic polytopes, k = 1:

Ω(m)n,1 = JKResω(m)

n,1 for any n and any m

JK-residue gives the correct contour for conjugates to cyclic polytopes, k = n− m− 1:

Ω(m)n,n−m−1 = JKResω(m)

n,n−m−1 for any n and even m

Positivity plays a crucial role in getting the correct result!Tomasz Łukowski (University of Oxford) 19.06.2018 4 / 18

Amplituhedron A(m)n,k : the space

A(2)3,1 : Triangle

YA =3∑

i=1ci ZA

i

A = 1, 2, 3

1

23

Y

more points

−−−−−−−→1

2

3 4

5

6

Y

A(2)n,1 : Polygon

YA =n∑

i=1ci ZA

i

A = 1, 2, 3

ci > 0

higherdimension

−−−−−−−→

k = 1

higherdimension

−−−−−−−→

A(m)m+1,1: Simplex

YA =m+1∑i=1

ci ZAi

A = 1, ...,m + 1

1

2

3

4Y

more points

−−−−−−−→

1

2

3

4

56

Y

A(m)n,1 : Cyclic polytope

YA =n∑

i=1ci ZA

i

A = 1, ...,m + 1

Tomasz Łukowski (University of Oxford) 19.06.2018 5 / 18

Amplituhedron A(m)n,k : the space

A(2)3,1 : Triangle

YA =3∑

i=1ci ZA

i

A = 1, 2, 3

1

23

Ymore points

−−−−−−−→1

2

3 4

5

6

Y

A(2)n,1 : Polygon

YA =n∑

i=1ci ZA

i

A = 1, 2, 3

ci > 0

higherdimension

−−−−−−−→

k = 1

higherdimension

−−−−−−−→

A(m)m+1,1: Simplex

YA =m+1∑i=1

ci ZAi

A = 1, ...,m + 1

1

2

3

4Y

more points

−−−−−−−→

1

2

3

4

56

Y

A(m)n,1 : Cyclic polytope

YA =n∑

i=1ci ZA

i

A = 1, ...,m + 1

Tomasz Łukowski (University of Oxford) 19.06.2018 5 / 18

Amplituhedron A(m)n,k : the space

A(2)3,1 : Triangle

YA =3∑

i=1ci ZA

i

A = 1, 2, 3

1

23

Ymore points

−−−−−−−→1

2

3 4

5

6

Y

A(2)n,1 : Polygon

YA =n∑

i=1ci ZA

i

A = 1, 2, 3

ci > 0

higherdimension

−−−−−−−→

k = 1

higherdimension

−−−−−−−→

A(m)m+1,1: Simplex

YA =m+1∑i=1

ci ZAi

A = 1, ...,m + 1

1

2

3

4Y

more points

−−−−−−−→

1

2

3

4

56

Y

A(m)n,1 : Cyclic polytope

YA =n∑

i=1ci ZA

i

A = 1, ...,m + 1

Tomasz Łukowski (University of Oxford) 19.06.2018 5 / 18

Amplituhedron A(m)n,k : the space

A(2)3,1 : Triangle

YA =3∑

i=1ci ZA

i

A = 1, 2, 3

1

23

Ymore points

−−−−−−−→1

2

3 4

5

6

Y

A(2)n,1 : Polygon

YA =n∑

i=1ci ZA

i

A = 1, 2, 3

ci > 0

higherdimension

−−−−−−−→

k = 1

higherdimension

−−−−−−−→

A(m)m+1,1: Simplex

YA =m+1∑i=1

ci ZAi

A = 1, ...,m + 1

1

2

3

4Y

more points

−−−−−−−→

1

2

3

4

56

Y

A(m)n,1 : Cyclic polytope

YA =n∑

i=1ci ZA

i

A = 1, ...,m + 1

Tomasz Łukowski (University of Oxford) 19.06.2018 5 / 18

Amplituhedron A(m)n,k : the space

1

2

3

4

56

Yhigher helicities

−−−−−−−−−−−→

A(m)n,1 : Cyclic polytope

YA =n∑

i=1ci ZA

i

A = 1, ...,m + 1

A(m)n,k : Amplituhedron

YAα =

n∑i=1

cαi ZAi

A = 1, ...,m + k

α = 1, ..., k

[Illustration by Andy Gilmore]

The Amplituhedron

Y = c · Z

Z ∈ M+(k + m, n) fixed positive external data

c ∈ G+(k, n) vary over all positive matrices

Y ∈ G(k, k + m)Tomasz Łukowski (University of Oxford) 19.06.2018 6 / 18

Amplituhedron: the volume form

For each amplituhedron A(m)n,k one defines a volume form Ω

(m)n,k =

∏α

〈Y1...Yk dmYα〉Ω(m)n,k

Ω(m)n,k has logarithmic singularities at all boundaries of A(m)

n,k

→ Example:1

23

Y → Y = c1Z1 + c2Z2 + c3Z3 → Ω

(2)3,1 =

dc2

c2

dc3

c3=〈Y dY dY〉〈123〉2

〈Y12〉〈Y23〉〈Y31〉

→ Compatible with triangulations

Ω

[ ]= Ω

[ ]+ Ω

[ ]

Why do we care?

Tree amplitudes inN = 4 SYM can be extracted from Ω(4)n,k

space-time dimensionhelicity sectornumber of particles

Tomasz Łukowski (University of Oxford) 19.06.2018 7 / 18

Amplituhedron: the volume form

How to find the volume function?

→ Geometrically: Triangulate A(m)n,k and sum over the known volumes of each triangles

e.g.:

1

2

3 4

5

=

1

2

3 4

5

Ω(2)5,1 = [123] + [134] + [145]

→ Analytically: Evaluate the contour integral

Ω(m)n,k =

∫γ

dk·nc(12 . . . k) . . . (n1 . . . k − 1)

∏α,A

δ(YAα −

∑i

cαi ZAi ) =

∫γ

ω(m)n,k

e.g.: γ = c4 = c5 = 0 ∪ c2 = c5 = 0 ∪ c2 = c3 = 0

different contours ↔ different triangulations

Finding γ is a highly non-trivial task!Tomasz Łukowski (University of Oxford) 19.06.2018 8 / 18

Jeffrey-Kirwan residue: definition

The Jeffrey-Kirwan residue is an operation on rational differential forms

ω =dx1 ∧ . . . ∧ dxr

β1(x) . . . βn(x), βi(x) = βi · x + αi

Ingredients

B = βi is the set of r-dimensional vectors: charges

A cone is the positive span of r charges

Fixed reference vector η ∈ Rr

Jeffrey-Kirwan residue

JKResB,η ω =∑

cone3η

Rescone ω

→ Rescone ω is the multivariate residue, up to a sign

→ the signs are determined by the orientation of cones

Example

β1

β2

β4

β5

β3

Tomasz Łukowski (University of Oxford) 19.06.2018 9 / 18

Jeffrey-Kirwan residue: definition

The Jeffrey-Kirwan residue is an operation on rational differential forms

ω =dx1 ∧ . . . ∧ dxr

β1(x) . . . βn(x), βi(x) = βi · x + αi

Ingredients

B = βi is the set of r-dimensional vectors: charges

A cone is the positive span of r charges

Fixed reference vector η ∈ Rr

Jeffrey-Kirwan residue

JKResB,η ω =∑

cone3η

Rescone ω

→ Rescone ω is the multivariate residue, up to a sign

→ the signs are determined by the orientation of cones

Example

β1

β2

β4

β5

β3

C14

Tomasz Łukowski (University of Oxford) 19.06.2018 9 / 18

Jeffrey-Kirwan residue: definition

The Jeffrey-Kirwan residue is an operation on rational differential forms

ω =dx1 ∧ . . . ∧ dxr

β1(x) . . . βn(x), βi(x) = βi · x + αi

Ingredients

B = βi is the set of r-dimensional vectors: charges

A cone is the positive span of r charges

Fixed reference vector η ∈ Rr

Jeffrey-Kirwan residue

JKResB,η ω =∑

cone3η

Rescone ω

→ Rescone ω is the multivariate residue, up to a sign

→ the signs are determined by the orientation of cones

Example

β1

β2

β4

β5

β3

C12

Tomasz Łukowski (University of Oxford) 19.06.2018 9 / 18

Jeffrey-Kirwan residue: definition

The Jeffrey-Kirwan residue is an operation on rational differential forms

ω =dx1 ∧ . . . ∧ dxr

β1(x) . . . βn(x), βi(x) = βi · x + αi

Ingredients

B = βi is the set of r-dimensional vectors: charges

A cone is the positive span of r charges

Fixed reference vector η ∈ Rr

Jeffrey-Kirwan residue

JKResB,η ω =∑

cone3η

Rescone ω

→ Rescone ω is the multivariate residue, up to a sign

→ the signs are determined by the orientation of cones

Example

β1

β2

β4

β5

β3

η

Tomasz Łukowski (University of Oxford) 19.06.2018 9 / 18

Jeffrey-Kirwan residue: definition

The Jeffrey-Kirwan residue is an operation on rational differential forms

ω =dx1 ∧ . . . ∧ dxr

β1(x) . . . βn(x), βi(x) = βi · x + αi

Ingredients

B = βi is the set of r-dimensional vectors: charges

A cone is the positive span of r charges

Fixed reference vector η ∈ Rr

Jeffrey-Kirwan residue

JKResB,η ω =∑

cone3η

Rescone ω

→ Rescone ω is the multivariate residue, up to a sign

→ the signs are determined by the orientation of cones

Example

β1

β2

β4

β5

β3

η

Tomasz Łukowski (University of Oxford) 19.06.2018 9 / 18

Jeffrey-Kirwan residue: properties

JKResB,η ω =

∑cone3η

Rescone ω

The JK-residue provides a particular contour on which we integrate ω

1 Any η

JKResB,η ω = ResC25 ω + ResC45 ω + ResC23 ω

2 For η′ close to η

JKResB,η′ ω = ResC25 ω + ResC45 ω + ResC23 ω = JKResB,η ω

→ A chamber Λ is a maximal non-empty intersection of cones

→ JK-residue gives the same form of the answer in each chamber

Example

ω =dx1 ∧ dx2

β1(x)...β5(x)

β1

β2

β4

β5

β3

η

Tomasz Łukowski (University of Oxford) 19.06.2018 10 / 18

Jeffrey-Kirwan residue: properties

JKResB,η ω =

∑cone3η

Rescone ω

The JK-residue provides a particular contour on which we integrate ω

1 Any η

JKResB,η ω = ResC25 ω + ResC45 ω + ResC23 ω

2 For η′ close to η

JKResB,η′ ω = ResC25 ω + ResC45 ω + ResC23 ω = JKResB,η ω

→ A chamber Λ is a maximal non-empty intersection of cones

→ JK-residue gives the same form of the answer in each chamber

Example

ω =dx1 ∧ dx2

β1(x)...β5(x)

β1

β2

β4

β5

β3

η

η′

Tomasz Łukowski (University of Oxford) 19.06.2018 10 / 18

Jeffrey-Kirwan residue: properties

JKResB,η ω =

∑cone3η

Rescone ω

The JK-residue provides a particular contour on which we integrate ω

1 Any η

JKResB,η ω = ResC25 ω + ResC45 ω + ResC23 ω

2 For η′ close to η

JKResB,η′ ω = ResC25 ω + ResC45 ω + ResC23 ω = JKResB,η ω

→ A chamber Λ is a maximal non-empty intersection of cones

→ JK-residue gives the same form of the answer in each chamber

Example

ω =dx1 ∧ dx2

β1(x)...β5(x)

β1

β2

β4

β5

β3

Λ

Tomasz Łukowski (University of Oxford) 19.06.2018 10 / 18

Jeffrey-Kirwan residue: properties

JKResB,η ω =

∑cone3η

Rescone ω

3 JKRes gives different form of the answer in various chambers

JKResB,η1 ω = ResC25 ω + ResC45 ω + ResC23 ω

=

JKResB,η2 ω = ResC45 ω + ResC12 ω + ResC42 ω

However, the function is independent of the chamber

→ This equality follows from the global residue theorem

∫Γ

dx1 ∧ dx2

β1(x)...β5(x)= 0 , with Γ :

|β2(x)| = ε

|β1(x)β3(x)β4(x)β5(x)| = ε

Example

ω =dx1 ∧ dx2

β1(x)...β5(x)

β1

β2

β4

β5

β3

Λ1

Λ2

η2η1

Tomasz Łukowski (University of Oxford) 19.06.2018 11 / 18

Amplituhedron meets Jeffrey-Kirwan residue

Apply Jeffrey-Kirwan residue to: Ω(m)n,1 =

∫γ

ω(m)n,1

E.g. n = 5: positivity of Z implies that vectors

β1,−β2, β3,−β4, β5,−β1, β2,−β3, β4,−β5, β1

are ordered clockwise

Each Rescone ω(2)5,1 computes the volume of a triangle

ResC25 ω(2)5,1 = [134]

Each chamber provides a triangulation

JKResB,η ω(2)5,1 = [123] + [134] + [145]

Adjacent chambers: related by bistellar flip

→ Geometrically: =

→ Analytically: Global residue theorem

β1

β2

β4

β5β3

η

1

2

3 4

5

1

2

3 4

5

1

2

3 4

51

2

3 4

5=

Tomasz Łukowski (University of Oxford) 19.06.2018 12 / 18

Amplituhedron meets Jeffrey-Kirwan residue

Apply Jeffrey-Kirwan residue to: Ω(m)n,1 =

∫γ

ω(m)n,1

E.g. n = 5: positivity of Z implies that vectors

β1,−β2, β3,−β4, β5,−β1, β2,−β3, β4,−β5, β1

are ordered clockwise

Each Rescone ω(2)5,1 computes the volume of a triangle

ResC25 ω(2)5,1 = [134]

Each chamber provides a triangulation

JKResB,η ω(2)5,1 = [123] + [134] + [145]

Adjacent chambers: related by bistellar flip

→ Geometrically: =

→ Analytically: Global residue theorem

β1

β2

β4

β5

β3

Tomasz Łukowski (University of Oxford) 19.06.2018 12 / 18

General Statement: Cyclic Polytopes

For cyclic polytopes (k = 1, any m, any n):

Ω(m)n,1 = JKResB,ηω

(m)n,1

For each chamber

→ Geometrically: different triangulations of A(m)n,1

→ Analytically: different representation of Ω(m)n,1

For adjacent chamber

→ Geometrically: triangulations related by bistellar flip

→ Analytically: representations related by global residue theorem

Tomasz Łukowski (University of Oxford) 19.06.2018 13 / 18

Secondary polytope [Gelfand, Kapranov, Zelevinsky]

Vertices of the secondary polytope Σ(P) are triangulations of a given polytope P

Examples of secondary polytopesFor m = 1: combinatorially equivalent to the hypercube in n− 2 dimensions

1 2 3 1 3

1 2 3 4

1 2 4 1 3 4

1 4

12345

12 45

1 3 5

1 5

123 5 1 345

1 4512 5

For an n-gon, m = 2: AssociahedronIn general very complicated

→ e.g. physical case, eight particles: 3-dimensional with 40 vertices

→ e.g. physical case, nine particles: 4-dimensional with 357 vertices

It is fully captured by the Jeffrey-Kirwan residue applied to ω(m)n,1 !

Tomasz Łukowski (University of Oxford) 19.06.2018 14 / 18

General statement: Conjugates to cyclic polytopes

Parity conjugation

→ Amplitudes: conjugate the helicities of the particles

→ Amplituhedron: k↔ n− m− k

Conjugates to cyclic polytopes are not polytopes!

Jeffrey-Kirwan residue prescription for conjugates to cyclic polytopes:

Ω(m)n,n−m−1 = JKResB,ηω

(m)n,n−m−1 , for even m

Cyclic polytopes and their conjugates

Y ∈ G(1,m + 1) Y ∈ G(n− m− 1, n− 1)

Triangles in A(m)n,1 ↔ Triangles in A(m)

n,n−m−1 all m

Triangulations of A(m)n,1 ↔ Triangulations of A(m)

n,n−m−1 even m

Σ(A(m)n,1 ) ↔ Σ(A(m)

n,n−m−1) even m

Tomasz Łukowski (University of Oxford) 19.06.2018 15 / 18

Conclusions

The Jeffrey-Kirwan residue

provides a triangulation-independent formula for the volume forms Ω(m)n,1 and Ω

(m)n,n−m−1

strengthen the connection between

→ the complicated analytic/algebraic nature of scattering amplitudes

→ the rich geometric/combinatorial structure of the amplituhedron

naturally leads to the study of the secondary amplituhedron→ encodes all triangulations of the amplituhedron

→ generalises the notion of secondary polytope

Tomasz Łukowski (University of Oxford) 19.06.2018 16 / 18

Outlook

Generalisations:

What is the generalisation of the Jeffrey-Kirwan residue for all other amplituhedra?

Can we find the secondary amplituhedron in all these cases?

→ many new representations of scattering amplitudes

→ new and rich structure of all triangulations

The origin:

Why Jeffrey-Kirwan residue provides the correct answer?

Can one understand the amplituhedron from some underlying localization perspective?

Can we learn some new lessons for supersymmetric localization?

Tomasz Łukowski (University of Oxford) 19.06.2018 17 / 18

Thank you for your attention!

Tomasz Łukowski (University of Oxford) 19.06.2018 18 / 18