Linear reducibility of Feynman integrals

Transcription

1 Linear reducibility of Feynman integrals Erik Panzer Institute des Hautes Études Scientifiques RADCOR/LoopFest 2015 June 19th, 2015 UCLA, Los Angeles

2 Feynman integrals: special functions and numbers Some FI are expressible via multiple polylogarithms (MPL) k z 1 1 Li n1,...,n d (z 1,..., z d ) = zk d d k n 1 0<k 1 < <k d 1 kn d d Example (p1 2 = 1, p2 2 = z z, p3 2 = (1 z)(1 z)) ( ) Φ = 4i Im [Li 2(z) + log(1 z) log z ] z z 1 / 24

3 Feynman integrals: special functions and numbers Some FI are expressible via multiple polylogarithms (MPL) k z 1 1 Li n1,...,n d (z 1,..., z d ) = zk d d k n 1 0<k 1 < <k d 1 kn d d Example (massless propagators) ( ) ( ) Φ = 6ζ 3, Φ = 252ζ 3 ζ ζ 3, ζ4 2 1 / 24

4 Feynman integrals: special functions and numbers Some FI are expressible via multiple polylogarithms (MPL) k z 1 1 Li n1,...,n d (z 1,..., z d ) = zk d d k n 1 0<k 1 < <k d 1 kn d d Example (massless propagators) ( ) ( ) Φ = 6ζ 3, Φ = 252ζ 3 ζ ζ 3, ζ4 2 Questions 1 What comes beyond MPL? [Weinzierl s talk] 2 How to tell if a FI evaluates to MPL? What is the alphabet? 3 How to compute it explicitly? 1 / 24

5 Feynman integrals: special functions and numbers Some FI are expressible via multiple polylogarithms (MPL) k z 1 1 Li n1,...,n d (z 1,..., z d ) = zk d d k n 1 0<k 1 < <k d 1 kn d d Example (massless propagators) ( ) ( ) Φ = 6ζ 3, Φ = 252ζ 3 ζ ζ 3, ζ4 2 Questions 1 What comes beyond MPL? [Weinzierl s talk] 2 How to tell if a FI evaluates to MPL? What is the alphabet? 3 How to compute it explicitly in an automated way? 1 / 24

6 Tools to study Feynman integrals integral representations momentum space [IBP, unitarity] position space [Gardi s talk, Drummond & Schnetz graphical functions] twistor space [Trnka s talk] parametric space [this talk] Mellin-Barnes sum representations [Schneider s talk] differential equations [Smirnov s talk] 2 / 24

7 Tools to study Feynman integrals integral representations momentum space [IBP, unitarity] position space [Gardi s talk, Drummond & Schnetz graphical functions] twistor space [Trnka s talk] parametric space [this talk] Mellin-Barnes sum representations [Schneider s talk] differential equations [Smirnov s talk] Plan of the talk 1 predicting bounds on the singularities of an integral 2 nice (linearly reducible) integral representations can be integrated in terms of MPL 3 good parametrizations for FI via Laplace transform 2 / 24

8 Definite integral representations hypergeometric functions p F q ( ; z) 2F 1 ( a, b c ) z = Appell s functions F 1, F 2, F 3, F 4 ( α, α ) γ x F 3 y β, β = 1 1 v 0 0 Γ(c) 1 t b 1 (1 t) c b 1 (1 zt) a dt Γ(b)Γ(c b) 0 Γ(γ) Γ(β)Γ(β )Γ(γ β β ) u β 1 v β 1 (1 u v) γ β β 1 (1 ux) α (1 vy) α dudv Feynman integrals in Schwinger parameters Phase-space integrals... 3 / 24

9 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e 4 / 24

10 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = G = 2 1 p 3 ϕ = Φ(G) = dα2 dα 3 ψϕ α1 =1 4 / 24

11 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α1 + α 2 + α 3 G = 2 1 p 3 ϕ = Φ(G) = dα2 dα 3 ψϕ α1 =1 4 / 24

12 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α1 + α 2 + α 3 G = 2 1 p 3 ϕ = Φ(G) = dα2 dα 3 ψϕ α1 =1 4 / 24

13 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α1 + α 2 + α 3 G = 2 1 p 3 ϕ = Φ(G) = dα2 dα 3 ψϕ α1 =1 4 / 24

14 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α 1 + α 2 + α 3 G = 2 1 p 3 ϕ = p1α 2 2 α 3 + p2α 2 1 α 3 + p3α 2 1 α 2 dα2 dα 3 Φ(G) = ψϕ α1 =1 4 / 24

15 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α 1 + α 2 + α 3 G = 2 1 p 3 ϕ = p1α 2 2 α 3 + p2α 2 1 α 3 + p3α 2 1 α 2 dα2 dα 3 Φ(G) = ψϕ α1 =1 4 / 24

16 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α 1 + α 2 + α 3 G = 2 1 p 3 ϕ = p1α 2 2 α 3 + p2α 2 1 α 3 + p3α 2 1 α 2 dα2 dα 3 Φ(G) = ψϕ α1 =1 4 / 24

17 Schwinger parameters With the superficial degree of divergence sdd = E(G) D/2 loops(g), Φ(G) = Γ(sdd) ( ) 1 ψ sdd e Γ(a e) (0, ) E ψ D/2 δ(1 α N ) αe ae 1 dα e ϕ e Graph polynomials: ψ = α e T e / T ϕ = q 2 (T 1 ) α e + ψ e / F e F =T 1 T 2 m 2 eα e Example (D = 4) p 3 1 p 2 ψ = α 1 + α 2 + α 3 G = 2 1 p 3 ϕ = p1α 2 2 α 3 + p2α 2 1 α 3 + p3α 2 1 α 2 dα2 dα 3 Φ(G) = ψϕ α1 =1 4 / 24

18 Example: massless triangle ( ) Φ = dα 2 dα 3 (1 + α 2 + α 3 )(α 2 α 3 + z zα 3 + (1 z)(1 z)α 2 ) 5 / 24

19 Example: massless triangle ( ) Φ = dα 2 dα 3 (1 + α 2 + α 3 )(α 2 α 3 + z zα 3 + (1 z)(1 z)α 2 ) Singularities of the original integrand: S = {ψ, ϕ}, i.e. at α 3 = σ i for σ 1 = 1 α 2 and σ 2 = α 2(1 z)(1 z) α 2 + z z 5 / 24

20 Example: massless triangle ( ) Φ = = dα 2 dα 3 (1 + α 2 + α 3 )(α 2 α 3 + z zα 3 + (1 z)(1 z)α 2 ) dα 2 (z + α 2 )( z + α 2 ) log (1 + α 2)(z z + α 2 ) (1 z)(1 z)α 2 Singularities of the original integrand: S = {ψ, ϕ}, i.e. at α 3 = σ i for σ 1 = 1 α 2 and σ 2 = α 2(1 z)(1 z) α 2 + z z After integrating α 1 from 0 to, the integrand has singularities S 3 = { } 1 + α }{{ 2, α } 2, 1 z, 1 z, α }{{} 2 + z z, z + α }{{} 2, z + α }{{ 2 } σ 1 =0 σ 2 =0 σ 2 = σ 1 =σ 2 (pinch) 5 / 24

21 Example: massless triangle ( ) Φ = = dα 2 dα 3 (1 + α 2 + α 3 )(α 2 α 3 + z zα 3 + (1 z)(1 z)α 2 ) dα 2 (z + α 2 )( z + α 2 ) log (1 + α 2)(z z + α 2 ) (1 z)(1 z)α 2 Singularities of the original integrand: S = {ψ, ϕ}, i.e. at α 3 = σ i for σ 1 = 1 α 2 and σ 2 = α 2(1 z)(1 z) α 2 + z z After integrating α 1 from 0 to, the integrand has singularities S 3 = { } 1 + α }{{ 2, α } 2, 1 z, 1 z, α }{{} 2 + z z, z + α }{{} 2, z + α }{{ 2 } σ 1 =0 σ 2 =0 σ 2 = σ 1 =σ 2 (pinch) With the same logic, predict the possible singularities after 0 dα 2 : S 3,2 = {z, z, 1 z, 1 z, z z, z z 1} 5 / 24

22 Example: massless triangle ( ) Φ = = 1 z z dα 2 dα 3 (1 + α 2 + α 3 )(α 2 α 3 + z zα 3 + (1 z)(1 z)α 2 ) ( dα2 α 2 + z dα ) 2 log (α 2 + 1)(α 2 + z z) α 2 + z (1 z)(1 z)α 2 Singularities of the original integrand: S = {ψ, ϕ}, i.e. at α 3 = σ i for σ 1 = 1 α 2 and σ 2 = α 2(1 z)(1 z) α 2 + z z After integrating α 1 from 0 to, the integrand has singularities S 3 = { } 1 + α }{{ 2, α } 2, 1 z, 1 z, α }{{} 2 + z z, z + α }{{} 2, z + α }{{ 2 } σ 1 =0 σ 2 =0 σ 2 = σ 1 =σ 2 (pinch) With the same logic, predict the possible singularities after 0 dα 2 : S 3,2 = {z, z, 1 z, 1 z, z z, z z 1} 5 / 24

23 Polynomial reduction [F. Brown] Definition Let S denote a set of polynomials, then S e are the irreducible factors of { } lead e (f ), f, D αe=0 e(f ): f S and {[f, g] e : f, g S}. Lemma If the singularities of F are cointained in S, then the singularities of 0 F dα e are contained in S e. 6 / 24

24 Polynomial reduction [F. Brown] Definition Let S denote a set of polynomials, then S e are the irreducible factors of { } lead e (f ), f, D αe=0 e(f ): f S and {[f, g] e : f, g S}. Lemma If the singularities of F are cointained in S, then the singularities of 0 F dα e are contained in S e. This gives only very coarse upper bounds. For example, z z 1 is spurious: It drops out in S 2,3 S 3,2 = {z, z, 1 z, 1 z, z z} because S 2,3 = {z, z, 1 z, 1 z, z z, z z z z}. Improvements Fubini: intersect over different orders Compatibility graphs 6 / 24

25 Compatibility graphs Keep track of compatibilities C ( S 2) between polynomials for S e, we only need the resultants [f, g] e for compatible {f, g} C only pairs [f, g] e, [g, h] e (with one polynomial in common) become compatible in C e Example (triangle) 1+α 2 α 2 (1 z)(1 z) z +α 2 z +α 2 z z +α 2 Since z zα 1 + α 2 and α 1 + α 2 are not compatible, their resultant 1 z z does not occur. 7 / 24

26 HyperInt: massive triangle Graph polynomials: > E:=[[2,3],[1,3],[1,2]]: > M:=[[1,s1],[2,s2],[3,s3]],[m1ˆ2,m2ˆ2,m3ˆ3]: > psi:=graphpolynomial(e): > phi:=secondpolynomial(e,m): Polynomial reduction: > L[{}]:=[{psi,phi}, {{psi,phi}}]: > cgreduction(l, {s1,s2,s3,m1,m2,m3}, 2): > L[{x[1],x[2]}][1]; s i + (m j ± m k ) 2, si 2 s i s j i i j s 1s 2 s 3 si 2 mi 2 + s i (mi 2 mj 2 )(mi 2 mk) 2 + i i i<j s i s j (mi 2 + mj 2 ) 8 / 24

27 Linear reducibility Definition If for some order of variables (edges), all S 1,...,k are linear in α k+1, then S (the Feynman graph G with S = {ψ, ϕ}) is called linearly reducible. Lemma If S is linearly reducible, the integral e dα e f of any rational function f with singularities in S is a MPL with symbol letters in S 1,...,N. 0 9 / 24

28 Linear reducibility Definition If for some order of variables (edges), all S 1,...,k are linear in α k+1, then S (the Feynman graph G with S = {ψ, ϕ}) is called linearly reducible. Lemma If S is linearly reducible, the integral e dα e f of any rational function f with singularities in S is a MPL with symbol letters in S 1,...,N. Integration of linearly reducible integrands in terms of MPL can be automatized via hyperlogarithms. Definition (Hyperlogarithms, Lappo-Danilevsky 1927) G(σ 1,..., σ w ; z) := z 0 0 dz 1 z 1 σ 1 G(σ 2,..., σ w ; z 1 ) 9 / 24

29 Linear reducibility Definition If for some order of variables (edges), all S 1,...,k are linear in α k+1, then S (the Feynman graph G with S = {ψ, ϕ}) is called linearly reducible. Lemma If S is linearly reducible, the integral e dα e f of any rational function f with singularities in S is a MPL with symbol letters in S 1,...,N. Integration of linearly reducible integrands in terms of MPL can be automatized via hyperlogarithms. 0 Definition (Hyperlogarithms, Lappo-Danilevsky 1927) G(σ 1,..., σ w ; z) := z 0 dz 1 z 1 σ 1 z1 0 dz 2 z 2 σ 2 zw 1 0 dz w z w σ w 9 / 24

30 Symbolic integration: Rewriting hyperlogarithms Problem: Given G( σ(α); z), write it as hyperlogarithm with constant letters and final argument α. Recursive solution via dg( σ; z) = n i=1 G(, σ i, ; z) d log σ i σ i 1 σ i σ i+1 σ 0 := z, σ n+1 := 0 Example α G (0, α; 1) = 1 G( α; 1) α 10 / 24

31 Symbolic integration: Rewriting hyperlogarithms Problem: Given G( σ(α); z), write it as hyperlogarithm with constant letters and final argument α. Recursive solution via dg( σ; z) = n i=1 G(, σ i, ; z) d log σ i σ i 1 σ i σ i+1 σ 0 := z, σ n+1 := 0 Example α G (0, α; 1) = 1 [G(0; α) G( 1; α)] α 10 / 24

32 Symbolic integration: Rewriting hyperlogarithms Problem: Given G( σ(α); z), write it as hyperlogarithm with constant letters and final argument α. Recursive solution via dg( σ; z) = n i=1 G(, σ i, ; z) d log σ i σ i 1 σ i σ i+1 σ 0 := z, σ n+1 := 0 Example α G (0, α; 1) = 1 [G(0; α) G( 1; α)] α G (0, α; 1) = G(0, 0; α) + G(0, 1; α) 10 / 24

33 Symbolic integration: Rewriting hyperlogarithms Problem: Given G( σ(α); z), write it as hyperlogarithm with constant letters and final argument α. Recursive solution via dg( σ; z) = n i=1 G(, σ i, ; z) d log σ i σ i 1 σ i σ i+1 σ 0 := z, σ n+1 := 0 Example α G (0, α; 1) = 1 [G(0; α) G( 1; α)] α G (0, α; 1) = G(0, 0; α) + G(0, 1; α) + ζ 2 10 / 24

34 Symbolic integration: Rewriting hyperlogarithms Problem: Given G( σ(α); z), write it as hyperlogarithm with constant letters and final argument α. Recursive solution via dg( σ; z) = n i=1 G(, σ i, ; z) d log σ i σ i 1 σ i σ i+1 σ 0 := z, σ n+1 := 0 Example α G (0, α; 1) = 1 [G(0; α) G( 1; α)] α G (0, α; 1) = G(0, 0; α) + G(0, 1; α) + ζ 2 Computation of integration constants relies on shuffle algebra G( σ; z) G( τ; z) = G( σ τ; z). All this is implemented in the Maple program HyperInt. Similar things can be done with other programs. 10 / 24

35 HyperInt open source Maple program integration of hyperlogarithms transformations of MPL to G( ; z) Note: No further simplification (e.g. rewrite as Li 2,2 ) provided! polynomial reduction graph polynomials symbolic computation of constants (no numerics) Example > read "HyperInt.mpl": > hyperint(polylog(2,-x)*polylog(3,-1/x)/x,x=0..infinity): > fibrationbasis(%); 8 7 ζ3 2 computes 0 Li 2 ( x) Li 3 ( 1/x)dx = 8 7 ζ / 24

36 HyperInt: propagator > E := [[2,1],[2,3],[2,5],[5,1],[5,3],[5,4],[4,1],[4,3]]: > psi := graphpolynomial(e): > phi := secondpolynomial(e, [[1,1], [3,1]]): > add((epsilon*log(psiˆ5/phiˆ4))ˆn/n!,n=0..2)/psiˆ2: > hyperint(eval(%,x[8]=1), [seq(x[n],n=1..7)]): > collect(fibrationbasis(%), epsilon); ( 254ζ ζ 5 200ζ 2 ζ 5 196ζ ζ ) ε 2 ( + 28ζ ζ ) 7 ζ3 2 ε + 20ζ ζ2 2ζ 3 12 / 24

37 HyperInt: triangle Graph polynomials: > E:=[[1,2],[2,3],[3,1]]: > M:=[[3,1],[1,z*zz],[2,(1-z)*(1-zz)]]: > psi:=graphpolynomial(e): > phi:=secondpolynomial(e,m): Integration: > hyperint(eval(1/psi/phi,x[3]=1),[x[1],x[2]]): > factor(fibrationbasis(%,[z,zz])); (Hlog (1; z) Hlog (0; zz) Hlog (0; z) Hlog (1; zz) + Hlog (0, 1; zz) Hlog (1, 0; zz) + Hlog (1, 0; z) Hlog (0, 1; z))/(z zz) Polynomial reduction: > L[{}]:=[{psi,phi}, {{psi,phi}}]: > cgreduction(l): > L[{x[1],x[2]}][1]; { 1 + z, 1 + zz, zz + z} 13 / 24

38 Linearly reducible families (fixed loop order) 1 all 4 loop massless propagators (Panzer) 14 / 24

39 Linearly reducible families (fixed loop order) 1 all 4 loop massless propagators (Panzer) 2 all 3 loop massless off-shell 3-point (Chavez & Duhr, Panzer) 14 / 24

40 Linearly reducible families (fixed loop order) 1 all 4 loop massless propagators (Panzer) 2 all 3 loop massless off-shell 3-point (Chavez & Duhr, Panzer) 14 / 24

41 Linearly reducible families (fixed loop order) 1 all 4 loop massless propagators (Panzer) 2 all 3 loop massless off-shell 3-point (Chavez & Duhr, Panzer) 3 all 2 loop massless on-shell 4-point (Lüders) 14 / 24

42 Linearly reducible massive graphs (some examples) 15 / 24

43 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

44 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

45 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

46 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

47 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

48 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

49 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Example 16 / 24

50 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Theorem (Panzer) All ɛ-coefficients of these graphs (off-shell) are MPL over the alphabet {z, z, 1 z, 1 z, z z, 1 z z, 1 z z, z z z z}. 16 / 24

51 Linearly reducible families (infinite) 3-constructible graphs (3-point functions) [Brown, Schnetz, Panzer] Theorem (Panzer) All ɛ-coefficients of these graphs (off-shell) are MPL over the alphabet {z, z, 1 z, 1 z, z z, 1 z z, 1 z z, z z z z}. minors of ladder-boxes (up to 2 legs off-shell) Theorem (Panzer) All ɛ-coefficients of these graphs are MPL. For the massless case, the alphabet is just {x, 1 + x} for x = s/t. 16 / 24

52 4-point recursions Start with the box and repeat, in any order: Appending a vertex: v 1 v 4 v 1 v 4 G = G = e v 2 v 3 v 2 v 3 Adding an edge: v 1 v 4 G = v 1 v 3 or v 1 v 4 v 2 v 3 v 4 e v 4 G = v 2 v 3 v 2 v 3 e Example v v 4 v 1 v 4 v 1 v 4 v 1 v 4 v 2 v 3 v 2 v 3 v 2 v 3 v 2 v 3 17 / 24

53 Forest polynomials Let f 3, f 4, f 12 and f 14 denote the spanning forest polynomials such that ϕ = F = (p 1 + p 2 ) 2 f 12 + (p 1 + p 4 ) 2 f 14 + p 2 3f 3 + p 2 4f 4 Example v v 4 v 2 v 3 ψ = α 1 + α 2 + α 3 + α 4 f 12 = α 2 α 4 f 14 = α 1 α 3 f 3 = α 2 α 3 f 4 = α 3 α 4 18 / 24

54 Forest polynomials Let f 3, f 4, f 12 and f 14 denote the spanning forest polynomials such that ϕ = F = (p 1 + p 2 ) 2 f 12 + (p 1 + p 4 ) 2 f 14 + p 2 3f 3 + p 2 4f 4 Definition F (G; z) := R E + ψ D/2 G δ (4) ( f ψ z ) e E dα ae 1 e α e ( R 4 + R + ) Example v v 4 v 2 v 3 ψ = α 1 + α 2 + α 3 + α 4 f 12 = α 2 α 4 f 14 = α 1 α 3 f 3 = α 2 α 3 f 4 = α 3 α 4 18 / 24

55 Forest polynomials Let f 3, f 4, f 12 and f 14 denote the spanning forest polynomials such that ϕ = F = (p 1 + p 2 ) 2 f 12 + (p 1 + p 4 ) 2 f 14 + p 2 3f 3 + p 2 4f 4 Definition F (G; z) := R E + ψ D/2 G δ (4) ( f ψ z ) e E dα ae 1 e α e ( R 4 + R + ) Example 1 v 1 v 4 (D = 4) z 4 3 z 4 F 1 3 ; z 2 = z 12 [z v 2 v 12 (z 14 + z 3 + z 4 ) + z 3 z 4 ] 3 2 (D = 6) }{{} Q 18 / 24

56 Forest polynomials Let f 3, f 4, f 12 and f 14 denote the spanning forest polynomials such that ϕ = F = (p 1 + p 2 ) 2 f 12 + (p 1 + p 4 ) 2 f 14 + p 2 3f 3 + p 2 4f 4 Definition Φ(G) = Γ(sdd) e Γ(a e) 0 F (G; z) Ω [ (p1 + p 2 ) 2 z 12 + (p 1 + p 4 ) 2 z 14 + p 2 3 z 3 + p2 4 z 4] sdd Example 1 v 1 v 4 (D = 4) z 4 3 z 4 F 1 3 ; z 2 = z 12 [z v 2 v 12 (z 14 + z 3 + z 4 ) + z 3 z 4 ] 3 2 (D = 6) }{{} Q 18 / 24

57 Appending a vertex G = v 1 v 2 v 3 v 4 G = v 1 v 4 e v 2 v 3 Using (f 12, f 14, f 3, f 4, ψ ) = (f 12, f 14, f 3, f 4 + α e ψ, ψ), F (G ; z) = z4 0 v 3 or v 1 v 4 v 2 v 3 F (G; z 12, z 14, z 3, z 4 α e ) αe ae 1 dα e e v 4 19 / 24

58 Appending a vertex G = v 1 v 2 v 3 v 4 G = v 1 v 4 e v 2 v 3 Using (f 12, f 14, f 3, f 4, ψ ) = (f 12, f 14, f 3, f 4 + α e ψ, ψ), F (G ; z) = z4 0 v 3 or v 1 v 4 v 2 v 3 F (G; z 12, z 14, z 3, z 4 α e ) αe ae 1 dα e e v 4 Example (D = 6 and a e = 1) v 4 ; z F v 1 v 2 v 3 = z3 0 F v v 4 v 2 v 3 ; z 12, z 14, z 3, z 4 dz 3 19 / 24

59 Appending a vertex G = v 1 v 2 v 3 v 4 G = v 1 v 4 e v 2 v 3 Using (f 12, f 14, f 3, f 4, ψ ) = (f 12, f 14, f 3, f 4 + α e ψ, ψ), F (G ; z) = z4 0 v 3 or v 1 v 4 v 2 v 3 F (G; z 12, z 14, z 3, z 4 α e ) αe ae 1 dα e e v 4 Example (D = 6 and a e = 1) v 1 v 4 z3 z F ; z 12 dz 3 = [z 12 (z 14 + z 3 + z 4) + z 3 z 4] 2 = z 3 (z 14 + z 4 ) Q v 2 v / 24

60 Appending a vertex G = v 1 v 2 v 3 v 4 G = v 1 v 4 e v 2 v 3 Using (f 12, f 14, f 3, f 4, ψ ) = (f 12, f 14, f 3, f 4 + α e ψ, ψ), F (G ; z) = z4 0 v 3 or v 1 v 4 v 2 v 3 F (G; z 12, z 14, z 3, z 4 α e ) αe ae 1 dα e e v 4 Example (D = 6 and a e = 1) v 1 v 4 z3 z F ; z 12 dz 3 = [z 12 (z 14 + z 3 + z 4) + z 3 z 4] 2 = z 3 (z 14 + z 4 ) Q v 2 v 3 v 4 ; z F v 1 v 2 v 3 = 0 z4 0 F v 1 v 4 ; z 12, z 14, z 3, z 4 dz 4 v 2 v 3 19 / 24

61 Appending a vertex G = v 1 v 2 v 3 v 4 G = v 1 v 4 e v 2 v 3 Using (f 12, f 14, f 3, f 4, ψ ) = (f 12, f 14, f 3, f 4 + α e ψ, ψ), F (G ; z) = z4 0 v 3 or v 1 v 4 v 2 v 3 F (G; z 12, z 14, z 3, z 4 α e ) αe ae 1 dα e e v 4 Example (D = 6 and a e = 1) v 1 v 4 z3 z F ; z 12 dz 3 = [z 12 (z 14 + z 3 + z 4) + z 3 z 4] 2 = z 3 (z 14 + z 4 ) Q v 2 v 3 v 4 ; z F v 1 v 2 v 3 = 0 1 ln z 12(z 3 + z 14 )(z 4 + z 14 ) z 12 z 14 z 14 Q 19 / 24

62 Adding an edge v 1 v 4 G = v 1 v 4 G = v 2 v 3 v 2 v 3 e z12 F G (z) [ = Q ae+sdd D x D/2 2 Q D/2 sdd F G ]z dx 12 =z 12 x 0 20 / 24

63 Adding an edge v 1 v 4 G = v 1 v 4 G = v 2 v 3 v 2 v 3 e z12 F G (z) [ = Q ae+sdd D x D/2 2 Q D/2 sdd F G ]z dx 12 =z 12 x Example (D = 6 and a e = 1) v 1 v 4 F ; z = 1 z12 Q 2 F v 2 v v 1 v 4 v 2 v 3 ; z 12 x, z 14, z 3, z 4 xdx 20 / 24

64 Adding an edge v 1 v 4 G = v 1 v 4 G = v 2 v 3 v 2 v 3 e z12 F G (z) [ = Q ae+sdd D x D/2 2 Q D/2 sdd F G ]z dx 12 =z 12 x Example (D = 6 and a e = 1) v 1 v 4 F ; z = 1 z12 Q 2 F v 2 v 3 = z 12 z 14 Q 2 + z 12 z 14 Q 2 Li 2 [ ln Q ( z3 z 4 Q 0 0 v 1 v 4 v 2 v 3 ; z 12 x, z 14, z 3, z 4 xdx ln (z ( )] 14 + z 3 )(z 14 + z 4 ) z 3 z 4 z 14 (z 14 + z 3 + z 4 ) Li z3 z 4 (z 14 z 12 ) 2 z 14 Q ) + z 12 Q 2 ln z 14 z 3 z 4 z 12 (z 14 + z 3 )(z 14 + z 4 ) ln(z 3z 4 /Q) Q(z 14 + z 3 + z 4 ) 20 / 24

65 Compatibility graph of box-ladders z 12 z 14 z 12 +z 3 Q z 12 +z 4 z 14 +z 3 +z 4 z 14 +z 3 z 14 +z 4 21 / 24

66 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) 22 / 24

67 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult 22 / 24

68 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results 22 / 24

69 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations 22 / 24

70 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) 22 / 24

71 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph 22 / 24

72 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! 22 / 24

73 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! Polynomial reduction can be a guideline to find parametrizations 22 / 24

74 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! Polynomial reduction can be a guideline to find parametrizations not restricted to FI 22 / 24

75 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! Polynomial reduction can be a guideline to find parametrizations not restricted to FI for divergent integrals, choose finite masters 22 / 24

76 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! Polynomial reduction can be a guideline to find parametrizations not restricted to FI for divergent integrals, choose finite masters 22 / 24

77 Summary Holy grail: Infer analytic structure of a FI from combinatorial structure, i.e. by looking at the graph (vs. case-by-case) extremely difficult so far, only very few results Problem: Find linearly reducible integral representations Schwinger parameters are not always optimal (e.g. for K 4 ) try to tailor the variables towards the graph Once a linearly reducible representation is found, programs can do all the work for you! Polynomial reduction can be a guideline to find parametrizations not restricted to FI for divergent integrals, choose finite masters Thank you! 22 / 24

78 Massless φ 4 theory: primitive sixth roots of unity P 7,11 = is not linearly reducible! Tenth denominator: d 10 = α 2 α 2 4α 1 + α 2 α 2 4α 3 α 1 α 2 α 3 α 4 + α 2 2α 4 α 1 + α 2 2α 4 α 3 2α 2 α 2 3α 4 α 2 2α 2 3 2α 2 2α 3 α 1 2α 2 α 2 3α 1 α 2 3α 2 4 2α 2 3α 4 α 1 α 2 2α 2 1 2α 2 α 3 α 2 1 α 2 3α 2 1. Change variables: α 3 = α 3 α 1 α 1 +α 2 +α 4, α 4 = α 4 (α 2 + α 3 ) and α 1 = α 1 α 4, d 10 = (α 2 + α 3)(α 2 + α 2 α 4 α 1)(α 1α 4 + α 2 + α 2 α 4 + α 3α 4) Final result: not a multiple zeta value, instead MPL at e iπ/3 23 / 24

79 3 P(P7,11 ) ( ) = Im ζ 9 Li ζ 7 Li ζ3( Li 2,3 7ζ 3 Li 2 ) 9675 π3 ζ 3, (36 Im Li 2,2,2,5 +27 Li 2,2,3,4 +9 Li 2,2,4,3 +9 Li 2,3,2,4 +3 Li 2,3,3,3 ) 43ζ 3 (Li 2,3,3 +3 Li 2,2,4 ) π Im ( Li 8 ζ Li 2, π2 Im Li 6 ζ Li 3,8 ) Li 4, Li 5,6 ( Li 2, Li 6 ζ Li 4, Li 3,6 ) ( + π ζ ζ ,7 215 Re(3 Li 4,6 +10 Li 3,7 ) ) Abbreviation: Li n1,...,n r := Li n1,...,n r (e iπ/3 ) 24 / 24