amplituhedron meets jeffrey-kirwan residue · 2018-11-22 · mathematics: localization of...
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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