Feynman Graphs and Periods Day Speakers:

Isabella Bierenbaum Christian Bogner Nehir Ikizlerli Kai Keller Organizers:

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Sector Decomposition and Hironaka's Polyhedra Game Christian Bogner , Durham 2007

Feynman Graphs and Periods Day

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Overview:

I: Basic facts about Feynman Integrals II: Sector Decomposition (Algorithm) III: Hironaka's Polyhedra Game IV: Comment on Periods

3

Part I: Basic facts about Feynman Integrals •

They are tools in particle physics.

•

They correspond to graphs:

•

Many

IG

are divergent

! 4

Regularization for

IG:

Introduce Parameter  such that IG() −→ IG

and

IG()

for

 −→ 0

well dened for some .

(We refer to so called Result is series in  :

Dimensional Regularization

IG() =

.)

PB i i=A Ci , A, B ∈

5

Z

We consider Feynman integrals of the following form: 

IG() =

Z x1 ≥0

dx1...

Z xn≥0

dxnδ 1 −

n X



r  Y

xi 

Pj (x1, ..., xn)

c +d j j

j=1

i=1

• n: # internal lines of graph G; cj , dj ∈ Z • : regularisation parameter; • Pj are homogeneous polynomials in x1, ..., xn; • Pj can depend on some momentum-squares

squares

m2i .

P

i pi


2

and mass-

• Pj can only vanish at borders of the integration domain

Qn

i=1 xi

6

=0




.

Example: Double Box

Z IG () =

dx1 ... x1 ≥0

n = 7;

Z

parameters:

dx7 δ x7 ≥0

1−

7 X

! xi

P11+3P2−3−2

i=1

x1 , ..., x7

P1 = (x1 + x2 + x3 ) (x5 + x6 + x7 ) + x4 (x1 + x2 + x3 + x5 + x6 + x7 ) , P2 = (x2 x3 (x4 + x5 + x6 + x7 ) + x5 x6 (x1 + x2 + x3 + x4 )

+x2 x4 x6 + x3 x4 x5 ) − (p1 + p2 )2 + x1 x4 x7 − (p2 + p3 )2

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Part II: Sector Decomposition

IG () =

step 1:

R

dx1 ... x1 ≥0

R

dxnδ 1 − xn ≥0

Pn

i=1 xi


Qr

j=1 (Pj

(x1 , ..., xn))cj +dj

step 2: iteration

step 3: expansion; perform divergent integrations analytically ; result

:

IG () =

PB

i i=A Ci 

; (with Ci nite, evaluable approximately). 8

Step 1:

Cutting the integration domain: Z

Z IG () =

=

dx1 ...

dxnδ

x1 ≥0

xn ≥0

n Z X

Z

l=1

x1 ≥0

dx1 ...

1−

! xi

i=1

dxnδ xn ≥0

n X

1−

r Y

(Pj (x1 , ..., xn))cj +dj

j=1 n X i=1

! xi

n Y

θ (xl ≥ xi)

r Y j=1

i=1 i 6= l

9

(Pj (x1 , ..., xn))

Transformation to cubes: In the l-th sector we xj =  substitute:  P xl x0j for j 6= l and perform xl integration, using δ 1 − xl 1 + i6=l x0i x and xl = 1+P . Result: x 0 l

0 i

i6=l

n Z X

1

Z\ Z 1 1 IG () = dx01 ... dx0n dx0l ... 0 | 0 {z } l=1 0 omitted   c n r   cj +dj Y Y 0 0 0 0 xj  · 1 + Pj x1 , ..., xb0l , ..., xn j=1

j6=l

| =

{z

=:Pr+1

n Z X l=1

0

1

}

Z 1 Z\ r+1 cj +dj 1 Y  0 0 0 0 0 dxn Pj x1 , ..., xb0l , ..., xn dx1 ... dx0l ... 0 j=1 | 0 {z } omitted

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Step 2: iterate

From now on we have the form

R1 0

dx1 ...

R1 0

dxn

Qn

ai +bi i=1 xi

Qr

j=1 (Pj )

Choose set

S = {α1 , ..., αk } ⊆ {1, ..., n} such that = 0, ..., xαk = 0) = 0 and cut along diagonals xαi = xαj

P (xα1

Z

1

Z dx1 ...

0

1

dxn = 0

In sector l: substitute x0α factors out.

k Z X l=1

1

1

Z dx1 ...

0

xαl = x0αl

dxn 0

and

Y

:

θ (xαl ≥ xαi ) .

i=1 i 6= l

xαi = x0αl x0αi

for

i 6= l

and

l

Iterate until constant term appears:

cj +dj

P =

n x1m1 ...xm n

1 + P˜


11

=⇒

.

Example: Z

1

Z dx1

0

P = x1 + x2x3; S = {1, 2}

1

Z dx2

0

1

Z dx3 P

=

0

1

Z

1

Z

1

dx1 dx2 dx3 θ (x1 ≥ x2 ) P 0 Z 1 0Z 1 0Z 1 + dx1 dx2 dx3 θ (x2 ≥ x1 ) P 0

0

0

sector

l = 1: x1 = x01 , x2 = x01 x02
0

 R R1 0 R1 0  0 R1 0 R1 0 R1 0 0 0 0 0 x0 x0 = 1 dx0 0 x0 x + x dx dx x 1 + x dx dx ≥ x x dx θ x 1 2 3 1 0 2 0 3 1 0 2 0 3 1 2 3 0 0 | 1 {z 1 2 } 1 =1

sector

l = 2: x1 = x01 x02 , x2 = x02



 R1 0

dx01

R1 0

dx02

R1 0




R1 R1 R1 dx03 θ x02 ≥ x01 x02 x01 x02 + x02 x03 = 0 dx01 0 dx02 0 dx03 x02 x01 + x03  {z } | {z } | P0

=1

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step 3: • After step 2 in every
sector the polynomials are monomialized : mn 1 + P 1 ˜ P = xm 1 ...xn

.

• The singular behaviour is completely controlled by the pre-

factors

n x1m1 ...xm n .

• Taylor expansions around xj = 0. • Integrations leading to singularities can be done analytically. • -expansion =⇒ result: IG() =

PB

i i=A Ci  ,

A, B ∈ Z.

• Finite integrals in Ci are calculated numerically. 13

Note: •

•

•

In order to get monomialized polynomials we choose
S = {α1, ..., αk } such that P xα1 = 0, ..., xαk = 0 = 0. This still leaves us a certain freedom! Given P = x1 + x2x3 we can choose for example S = {1, 2} or S = {1, 3}. Wether the algorithm terminates depends on the choice of S ! =⇒ Additional restrictions are needed. 14

Example:

2 P (x1 , x 2 , x 3 ) = x1 x2 3 + x2 + x2 x3

We might choose S = {1, 2} as P (x1 = 0, x2 = 0, x3) = 0. First sector

(l = 1): x1 = x01, x2 = x01x02, x3 = x03

02 x02 + x0 x0 x0 + x P (x1, x2, x3) = x01x02 3 1 2 1 2 3 0 02 0 02 = x1 x3 + x1x2 + x02x03

    0 0 0 0 0 0 0 0 0 = x1 P x1 , x 2 , x 3 = x1 P x1 , x 3 , x 2

Problem: We have not changed the situation. Innitely many repetitions possible

=⇒no

termination. 15

How can we guarantee the termination of the decomposition? How shall we choose S ? Problem:

Interpret Sector Decomposition as special case of Hironaka's Polyhedra Game. Choose S according to a winning strategy for this game.

Solution:

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Part III: Hironaka's Polyhedra Game Starting point: • two players A and B; • a given set of k points in

Nn+

:

m(i)

=



(i) , ..., m m(i) n 1



; i = 1, ..., k;

• ∆ ⊂ Rn+ the positive convex hull of this set;

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One move: A chooses non-empty set

S ⊆ {1, ..., n}.

B chooses one element l of this set S . For each of the k given points m(i), is replaced: m(i)0 l

=

X

i = 1, ..., k,

the l-th coordinate

m(i) j −1

j∈S

m(i)0 j

= m(i) j

A wins, if after nitely many moves

if

j 6= l

∆

is generated by one point:

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Example: for n = 2: given (black) points (1, 2), (1, 5), (3, 3), (4, 1), (6, 2) rst move: A chooses S = {1, 2}, B chooses l = 1 (i) (i) (i)0 (i) m(i)0 2 = m2 ; m1 = m1 + m2 − 1; i = 1, ..., 5

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new (red) points:

(2, 2), (4, 1), (5, 3), (5, 5), (7, 2)

second move: A chooses S = {1, 2}, B chooses l = 2 (i) (i) (i)0 (i) m(i)0 1 = m1 ; m2 = m1 + m2 − 1; i = 1, ..., 5

∆ = (2, 3) + R2+ =⇒ Player

A wins the game.

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Sector Decomposition

Polyhedra Game

polynomial with q terms with n parameters P =

m(i) m(i) m(i) 1 2 n c x x ...x i n 1 2 i=1

Pq

S = {α1 , ..., αk }

look at sector

l

xαl = x0αl , xαj = x0αl x0αj , j 6= l

factor out

xα l

q m(i)

=



points in

(i) , ..., m m(i) n 1

Nn+



; i = 1, ..., q

S = {α1 , ..., αk }

player B chooses

l

(i) m(i)0 = m j j , j 6= l

m(i)0 = l

(i) m j −1 j∈S

P

The transformation is the same. 21

Example: P = x1 + x2 x23

m(1) = (1, 0, 0), m(2) = (0, 1, 2)

S = {1, 3}

S = {1, 3}

look at sector 1

B chooses

l=1

x1 = x01

(i) (i) m(i)0 1 = m1 + m3 − 1

x2 = x02

(i) m(i)0 2 = m2

x3 = x03 x01

(i) m(i)0 3 = m3 , i = 1, 2

m(1) = (1 − 1, 0, 0) = (0, 0, 0) 0 02 0 0 0 P 0 = x01 + x02 1 x2 x3 = x1 1 + x1 x2 x3


02 m(2) = (1, 1, 2)

22

Sector Decomposition

polynomial with constant term after factorizations: mn 1 + P 1 ˜ P = xm 1 ...xn




Polyhedra Game

monomialized

∆:

∆ = m + Rn+

termination of the algorithm:

player A wins

constant term in every sector

for every possible choice of B

A

winning strategy

for player A would

guarantee the termination of the algorithm!

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winning strategy:

•

•

tells player A how to choose situation;

S

for every possible

guarantees A to win the game.

Luckily M. Spivakovsky, H. Hauser have found some winning strategies!

and D. Zeilinger

We just had to add them to the algorithm in order to guarantee termination. 24

Part IV: Comment on Periods

Denition (Kontsevich and Zagier):

A period is a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coecients, over domains in Rn given by polynomial inequalities with rational coecients.

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Remember: We have considered 

IG() =

Z x1 ≥0

dx1...

Z xn≥0

dxnδ 1 −

n X i=1

with the Pj depending on (

P 2 i pi)



xi 

r  Y

Pj (x1, ..., xn)

j=1

and mass-squares m2i .

Assume: • ( i pi)2 < 0 ⇒Pj > 0 P

•

all

P ( i pi)2 , m2 i ∈

inside integration domain;

Q; 26

c +d j j

Theorem:

The coecients of the Laurent expansion IG() = are periods.

PB i i=A Ci

A constructive proof uses Sector Decomposition. (C.B., S. Weinzierl, (2007), 0711.4863v1 [hep-th])

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Checklist for the proof:

After sector decomposition we have coecients Ci with •

•

•

all integration domains being unit-cubes;

√

all integrands consisting of rational functions and √ logarithms of rational functions; all integrals being absolutely convergent.

√

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Summary:

•

Sector Decomposition is used for an algorithm to calculate Feynman integrals numerically.

•

It gives a

•

The combinatorics is the same as in Hironaka's Polyhedra Game.

•

Winning strategies for this game guarantee the termination of the algorithm.

•

Guaranteed termination is needed for the periods-proof.

resulution

of the appearing singularities.

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Literature: Sector Decomposition: • T. Binoth and G. Heinrich, Nucl. Phys. B585, 741 (2000),

hep-ph/0004013.

Winning strategy for the game: • M. Spivakovsky, Progr. Math. 36, 419 (1983).

All put together / periods-proof: • C. B. and S. Weinzierl, (2007), hep-ph/0709.4092v1. • C. B. and S. Weinzierl, (2007), hep-th/0711.4863v1. 30