Introduction The MPLs Hopf algebra Application to Feynman integrals. Federico Granata. Ph.D. XXIX cicle 16/09/2014 Tutor: Carlo Oleari

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1 Federico Granata Ph.D. XXIX cicle 16/09/2014 Tutor: Carlo Oleari Seminar based on NNLO ante portas summer school June 2014, Debrecen Hungary)

2 Polylogarithms The analytic computation of Feynman scalar integrals requires the introduction of special functions One-loop integrals: Polylogarithms Li nz) = k=0 z k z k = n 0 dt t Li n 1t) = Li 1z) = log1 z) Li 2z) = z 0 log1 t) dt t Li n1) = ζ n : ζ 2 = π 2 /6 Multi-loop integrals: Multiple polylogarithms MPLs) Ga 1,..., a n; z) = z with G ; z) = 1; n weight 0 dt t a 1 Ga 2,..., a n; t) = G 0 n; z) = 1 n! logn z G a n; z) = 1 n! logn 1 z ) a ) z G 0 n 1, a; z) = Li n a

3 Polylogarithms Expressions involving multiple) polylogarithms can be simplified by means of functional equations: Li 2 z)+ Li 2 z) = 1 2 Li 2z 2 ) Li 2 z) Li 2 1 z) Li 21 z 2 ) = π2 log z log1 + z) 12 ) 1 Li 2 = Li 2 z) 12 z log2 z + iπ log z + π2 3 These equations are in general unknown, but they can be circumvented = HOPF ALGEBRA

4 Algebras and tensor products Algebra over field K K-vector space A with map m : A A A a, b) ma, b) a b With unit element: ɛ a = a ɛ = a Associative: a b c) = a b) c Distributive: a b + c) = a b + a c = m is bilinear Associative with respect to scalars: a kb) = ka b), k K Tensor product U, V, W vector spaces, define T U V :! T, τ : U V T bilinear β : U V W bilinear! µ linear such that β = µτ, that is, βa, b) = µa b)

5 Algebras and tensor products m bilinear A A A: same role as β! µ: A A A a b = ma, b) = µa b) Associativity: µid µ) = µµ id) µid µ)a b c) = µa µb c)) = µa b c)) = a b c) µµ id)a b c) = µµa b) c) = µa b) c) = a b) c Graded algebra Direct sum as a vector space: A = A n, A 0 = K n=0 such that A m A n A m+n

6 Coalgebras and Hopf algebras Coalgebra: dual of an algebra C A µ : C C C breaks into pieces Associativity: id ) = id) a a Hopf algebra H algebra + coalgebra: vector space with multiplication µ and comultiplication in general µ ) Multiplication and comultiplication are compatible with each other: a b) = a) b) a 1 a 2) b 1 b 2) = a 1 b 1) a 2 b 2) If H = H n, the coproduct acts as H n n=0 p+q=n H p H q

7 The multiple polylogarithms Hopf algebra The MPLs can be rewritten in another way: introducing an+1 dt Ia 0 ; a 1,..., a n ; a n+1 ) = Ia 0 ; a 1,..., a n 1 ; t) t a n a 0 we obtain Ga n,..., a 1 ; a n+1 ) = I0; a 1,..., a n ; a n+1 ) MPLs form a Hopf algebra Scalars: rational numbers Multiplication: shuffle product Ga; z)gb; z) = Ga, b; z) + Gb, a; z) Ga, b; z)gc; z) = Ga, b, c; z) + Ga, c, b; z) + Gc, a, b; z) The weight is preserved: algebra graded by the weight Coproduct: Ia 0 ; a 1,..., a n ; a n+1 )) = Ia 0 ; a i1,..., a ik ; a n+1 ) k Ia ip ; a ip+1,..., a ip+1 1; a ip+1 ) p=0 0=i 1<...<i k+1 =n

8 The multiple polylogarithms Hopf algebra Example Ga, b; z)) = 1 Ga, b; z) + Ga, b; z) 1 +Gb; z) [Ga; z) Ga; b)] + Ga; z) Gb; a) The coproduct preserves the weight = H is a graded Hopf algebra Particular cases log z) = 1 log z + log z 1 weight 1, no further decomposition) log x log y) = 1 log x log y +log x log y 1+log x log y +log y log x n ) n log n z) = log k z log n k z k k=0 n 1 Li n z)) = 1 Li n z)+ Li n z) 1 + Li n k z) logk z k! k=1

9 The π and ζ n problem Let s define a): a) 1 a + a 1 + a) ζ n ) = Li n 1)) = 1 ζ n + ζ n 1 ζ n ) = 0 ζ 4 = 2 5 ζ2 2 ζ 4 ) = 1 ζ 4 + ζ 4 1 ζ 4 ) = 2 5 ζ2 2) = 2 [ 1 ζ ζ2 2 ] 1 + 2ζ 2 ζ 2 Solution proposal) Work modulo π : the Hopf algebra is H = H/iπ = H = Q[iπ] H In this way : H H H : iπ) = iπ 1 ζ n ) = ζ n 1 ζ 4 ) = 2 5 ζ2 2) = 2 5 ζ 2 1) 2 = 2 5 ζ2 2 1) = ζ 4 1 We lose terms proportional to π 2n in...

10 Practical examples Strategy Two functions with the same are equal apart from terms in π 2n : Compute up to logarithms Use functional equations among logarithms much easier) and simplify the expression Find a function with the same Fix the missing terms by comparing the results at some points Example 1: Li 2 z)+ Li 2 1 z) Li 2 z)+ Li 2 1 z)) = log1 z) log z log z log1 z) = log z log1 z)) = Li 2 z)+ Li 2 1 z) = log z log1 z) + c z = 0: Li 2 1) = c = π 2 /6

11 Practical examples Drop the log symbol Fix log x) = log x + iπ Example 2: Li 2 1 z)+ Li 2 1 1/z) Li 2 1 z)+ Li 2 1 1/z)) = z) 1 z) 1/z 1 1/z) 1 z = z 1 z) + z = z z z iπ = 1 z 2 log 2 z) = Li 2 1 z)+ Li ) = 1 z 2 log2 z + c = z = 1: c = 0 Example 3: Li 2 1/z), 0 < z < 1 ) 1 z Li 2 1/z)) = Li 1 1/z) 1/z = 1 1/z) z = z z = 1 z) z z z iπ z = Li 2 z) 1 ) 2 log2 z iπ log z ) 1 = Li 2 = Li 2 z) 1 z 2 log2 z iπ log z + c = z = 1: c = π 2 /3

12 The three-mass triangle integral p 2 1 C 0 = d D k 2π) D 1 k 2 k + p 1 ) 2 k p 2 ) 2 pi 2 > 0, u = p2 1 p3 2, v = p2 2 p3 2 p 2 2 p 2 3 The integral is finite = D = 4 α-parameter representation 1 D ν1 1 Dν2 2 Dν3 3 = 0 α ν1 1 1 α ν2 1 2 α ν3 1 3 dα 1 dα 2 dα 3 Γν 1 )Γν 2 )Γν 3 ) δ 1 ) Γν1 + ν 2 + ν 3 ) α i α i D i ) ν1+ν2+ν3 After integrating over d 4 k and using the δ function: C 0 = i 1 dα 2 dα 3 4π) 2 p α 2 + α 3 )α 2 α 3 + α 2 u + α 3 v) i 1 4π) 2 p3 2 I

13 The three-mass triangle integral I = I = 0 0 I = 1 z z dα 3 log1 + α 3 ) log v + logu + α 3 ) log α 3 α u v)α 3 + u G 1; α 3 ) + logu/v) + G u; α 3 ) G0; α 3 ) dα 3 α 3 + z)α 3 + z) [ G z, u; α 3 ) logu/v)g z; α 3 ) ] G z, 1; α 3 ) + G z, 0; α 3 ) z z) α 3 a u = z z v = 1 z)1 z) Iα 3 ): use inversion relations to bring G... ; α 3 ) in G ) 1... ; 1 α 3

14 Inversion relations for the MPLs Reminder Ga, b; z)) = Gb; z) [Ga; z) Ga; b)] + Ga; z) Gb; a) G 0 n ; z) = 1 n! logn z G a n ; z) = 1 n! logn 1 z ) a G 1 )) ) [ z, 0; 1 1 = G 0; G 1 ) α 3 α 3 z ; 1 G 1z )] α ; 0 3 G 1 )) z, 0; 1 = + G 1 ) α 3 z ; 1 G 0; 1 ) α 3 z G 1 )) z, 0; 1 = 1 ) ) 1 + zα zα3 1 ) α 3 α 3 z G 1 )) z, 0; 1 = α 3 z + α 3 ) + α 3 α 3 z + α 3 ) z + α 3 z α 3

15 Inversion relations for the MPLs G z, 0; α 3 )) = G 0; α 3 ) [G z; α 3 ) G z; 0)]+G z; α 3 ) G 0; z) G z, 0; α 3 )) = α α ) α ) 3 z) z z G z, 0; α 3 )) = α 3 z+α 3 ) α 3 z+z+α 3 ) z z z+α 3 α 3 α 3 α 3 G z, 0; α 3 )) = G 1 )) z, 0; 1 1 α 3 2 log 2 z ) log 2 ) α 3 G z, 0; α 3 ) = G 1 ) z, 0; 1 1 α 3 2 log2 z log2 α 3 + c The terms independent of z cancel when subtracting G z, 0; α 3 ) G 1 ) z, 0; 1 vanishes when taking the limit α 3 α 3

16 The three-mass triangle integral I = 1 [ z z log u Gu; z) + Gu, 0; z) + log u v log z 1 2 log2 z ] I = 1 z z [+ Li 2z) + log1 z) log z z z) Applying again the procedure of calculating the result can be simplified: Analytical result I = 2 [ Li 2 z) Li 2 z) logz z) log 1 z ] z z 1 z

17 Conclusions Results Method for computing and simplifying analytical results of Feynman integrals based on an algebraic approach The Hopf algebra of MPLs allows to simplify results involving polylogarithms without knowing all the functional relations among them Can be used to complete the analytic computation of HVj virtual corrections in my research activity This procedure is more powerful beyond one-loop order H + 2-loop: from more than 7 pages to 10 lines)