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Mould calculus, resurgence and combinatorial Hopf algebras.

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Mould calculus, resurgence and combinatorial Hopf algebras.

Olivier Bouillot The 15 november 2011.

Table des mati` eres

1 Aim. 3

2 First expression of the horn map. 4

3 Evaluation of Γ · z. 5

4 Multitangent functions / Multizetas values. 6

5 Permutation of the two sums. 8

6 Calculation of Fourier coefficients of h + − id. 9

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During this talk, we set :

• f (z) = z + 1 + a(z) where a(z) = X

n≥3

a n

z n−1 ∈ 1 z 2 C

1 z

:

∃(C 0 ; C 1 ) ∈ ( R + ∗ ) 2 , |a n | ≤ C 0 C 1 n .

• t(z) = z + 1 .

• g(z) = f (z) − 1, so f = l ◦ g .

• N p = {n ∈ N ; n ≥ p}, where p ∈ N .

• seq(Ω) = {ω = (ω 1 ; · · · ; ω r ) ; r ∈ N , (ω 1 , · · · , ω r ) ∈ Ω r } , where Ω is an alphabet.

• ∀s = (s 1 , · · · , s r ) ∈ seq( N ),

||s|| = s 1 + · · · + s r

l(s) = r

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1 Aim.

We want to proof the following result :

Theorem : Let A defined by : ∀s ∈ seq( N 3 ) , A s = a s 1 · · · a s r . A defined by : ∀S ∈ seq seq( N 3 ) − {∅}

, A S = A s 1 · · · A s r .

then :

1. there exists an explicit mould τ defined on seq seq( N 3 ) − {∅})

such that :

π + = X

S∈seq(seq( N 3 )−{∅})

sg S τ S A S .

2. for all n ∈ Z , there exists an explicit mould τ b n , defined on seq seq( N 3 ) −{∅}

such that :

∀n ∈ Z , A + 2inπ = sg(n) X

S∈seq(seq( N 3 )−{∅})−{∅}

sg S τ b n S A S .

Remark : For all S ∈ seq seq( N 3 ) − {∅}

:

τ S ∈ MTGF ⊂ H( C − Z ) .

τ b n S ∈ MZV .

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2 First expression of the horn map.

We have : h + = v att ◦ (v rep ) −1 . In David’s talk, we have seen that :

v att (z) = z + X

k∈seq( N 1 )

ψ k (z) .

We can do the same with V att (ϕ) = ϕ ◦ v att , the substitution operator associate to v att . We obtain :

V att = X

n∈seq( Z )

U n + Γ n ,

where :

 

 

 

 

Γ n = T n ◦ (G − Id) ◦ T −n Γ n = Γ n r ◦ · · · ◦ Γ n 1

U n + =

1 if 0 ≤ n r < · · · < n 1 0 else

By the same (V rep ) −1 = X

n∈seq( Z )

U n Γ n where : U n =

1 if n r < · · · < n 1 < 0 0 else

So : H + =

X

n∈seq( Z )

U n + Γ n

 ◦

X

n∈seq( Z )

U n Γ n

= X

U + × U n

Γ n

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3 Evaluation of Γ · z.

Cf. transparents

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4 Multitangent functions / Multizetas values.

Let us make a little break with special functions...

Definition : Let S + ? =

s ∈ seq( N ); s 1 ≥ 2 S ? =

s ∈ seq( N ); s 1 ≥ 2 et s l(s) ≥ 2

So ∀s ∈ S + ? , Ze s = X

1≤n r <···<n 1

1 n s 1 1 · · · n s r r

.

∀s ∈ S ? , T e s = X

−∞<n r <···<n 1 <+∞

1

(n 1 + z) s 1 · · · (n r + z) s r . Why call we “multitangents functions” ?

Because T e 1 = π

tan(πz) ...

So the “multicotangents functions” · · · Notation : MZV = Vect Q (Z e s ) s∈S + ? .

MTGF = Vect Q (T e s ) s∈S ? .

Remark : 1. Ze and T e look simillar.

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Proposition 1 : ∀s ∈ S ? , ∃(z k ) 2≤k≤m ∈ MZV m−1 , T e =

m

X

k=2

z k T e k

Proof : 1. Partial fraction decomposition of 1

(n 1 + X) s

1

· · · (n r + X) s

r

. give :

T e =

m

X

k=1

z k T e k

2. z 1 = 0 because of iterate integral representation of multizetas values.

Proposition 2 : For all s ∈ S ? , we are able to calculate Fourier coefficients of T e s .

Proof : From Proposition 1, it suffice to evaluate Fourier coefficient of monotangent function.

T e s = (−1) s−1 (s − 1)!

s−1

∂z s−1

π tan(πz)

We know how to do for z 7−→ π tan(πz) It’s OK

Everything here is explicit !

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5 Permutation of the two sums.

Come back to the horn map h + : We have :

h + (z) − z = X

r≥1

X

n∈seq( Z )

X

S ∈ seq(seq( N 3 )−{∅}) l(S)=l(n)

l(s 1 )=1

sg S

X

i∈triangle(S)

B S,i

r

Y

p=1

1

(z + n p ) ||s p ||−l(s p )+diag p i

A S

So, we would like :

π + (z) − z = X

r≥1

X

S ∈ seq(seq( N 3 )−{∅}) l(S)=r

l(s 1 )=1

sg S

X

i∈triangle(S)

B S,i T e σ(S;i)

A S ,

where σ(S; i) = (||s p || − l(s p ) + diag p i ) 1≤k≤r

= X

r≥1

X

S ∈ seq(seq( N 3 )−{∅}) l(S)=r

sg S τ S A S ,

Here :

 id , if S = ∅

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6 Calculation of Fourier coefficients of h + − id.

Since h + (z) − z = X

seq( Z )

U Γ · z is normally convergent on every

compact

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