Pj are homogeneous polynomials in x1, ..., xn;. ⢠Pj can depend on some momentum-squares (â i pi. ) 2 and mass- squa
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
2
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
10
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 everysector 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)
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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