An example of generalization of the Bernoulli numbers and polynomials to any dimension

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Olivier Bouillot,

Villebon-Georges Charpak institute, France

C.A.L.I.N. team seminary.

Tuesday, 29^thMarch 2016 .

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Introduction

Definition:

The numbersZe^s¹^,···^,s^r defined by Ze^s¹^,···^,s^r = X

0<n_r<···<n₁

1 n1s₁· · ·nrs_r ,

wheres1,· · ·,sr ∈Csuch that<(s1+· · ·+sk)>k,k∈[[ 1 ;r]], are called multiple zeta values.

Fact:There exists at least three different ways to renormalize multiple zeta values at negative integers.

Ze_MP^0,−2(0) = 7

720 , Ze_GZ^0,−2(0) = 1

120 , Ze^0,−2_FKMT(0) = 1 18 . Question:Is there a group acting on the set of all possible multiple zeta values renormalisations?

Main goal:Define multiple Bernoulli numbers in relation with this.

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Outline

1 Reminders

Reminders on Bernoulli Polynomials and Numbers

Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on Quasi-Symmetric Functions

2 Algebraic reformulation of the problem

3 The Structure of a Multiple Bernoulli Polynomial

4 The General Reflexion Formula of Multiple Bernoulli Polynomial

5 An Example of Multiple Bernoulli Polynomial

6 An algorithm to compute the double Bernoulli Numbers

7 Properties satisfied by Bernoulli polynomials and numbers

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Outline

1 Reminders

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Two Equivalent Definitions of Bernoulli Polynomials / Numbers

Bernoulli numbers: Bernoulli polynomials:

By a generating function: By a generating function:

t

e^t−1 =X

n≥0

bn

tⁿ

n! . te^xt

e^t−1 =X

n≥0

Bn(x)tⁿ n! . By a recursive formula: By a recursive formula:







b0= 1,

∀n∈N,

n

X

k=0

n+ 1 k

bk= 0.











B0(x) = 1,

∀n∈N, Bn+1⁰ (x) = (n+ 1)Bn(x),

∀n∈N^∗, Z 1

0

Bn(x)dx= 0.

First examples: First examples:

bn= 1,−1 2,1

6,0,−1 30,0, 1

42,· · · B0(x) = 1, B1(x) = x−1

2 , B2(x) = x²−x+1

6 , ..

.

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Elementary properties satisfied by the Bernoulli polynomials and numbers

P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.

P3

m

X

k=0

m+ 1 k

bk= 0, m>0.

P4







B_n⁰(z) =nBn−1(z) ifn>0.

Bn(x+y) =

n

X

k=0

n k

Bk(x)y^n−k for alln.

P5 Bn(x+ 1)−Bn(x) =nxⁿ⁻¹, for alln.

P6 (−1)ⁿBn(1−x) =Bn(x), for alln.

P7

N−1

X

k=0

kⁿ=Bn+1(N)−Bn+1(0)

n+ 1 .

P8 Z x

a

Bn(t)dt= Bn+1(x)−Bn+1(a)

n+ 1 .

P9 Bn(mx) =mⁿ⁻¹

m−1

X

k=0

Bn

x+ k

m

for allm>0 andn≥0.

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







Have to be extended, but is not restritive enough.

P3

m

X

k=0

m+ 1 k

bk= 0, m>0.

Has to be extended,but too particular.

P4







Bn(x+y) =

n

X

k=0

n k

Important property,but turns out to have a generalization with a corrective term...

Has to be extended, but how???

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 .

Does not depend of the Bernoulli numbers...

P8 Z x

a

n+ 1 .

Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials.

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

bk= 0, m>0. Has to be extended,but too particular.

P4







Bn(x+y) =

n

X

k=0

n k

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 .

P8 Z x

a

n+ 1 .

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 .

P8 Z x

a

n+ 1 .

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P5 Bn(x+ 1)−Bn(x) =nxⁿ⁻¹, for alln. Has to be extended, but how???

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 .

P8 Z x

a

n+ 1 .

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P6 (−1)ⁿBn(1−x) =Bn(x), for alln. Has to be extended, but how???

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 .

P8 Z x

a

n+ 1 .

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

n+ 1 . Does not depend of the Bernoulli numbers...

P8 Z x

a

n+ 1 .

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

P8 Z x

a

n+ 1 . Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials.

m−1

X

k=0

Bn

x+ k

m

???

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P1 b2n+1= 0if n>0.

P2 Bn(0) =Bn(1)if n>1.







P3

m

X

k=0

m+ 1 k

P4







Bn(x+y) =

n

X

k=0

n k

P7

N−1

X

k=0

k^p=Bn+1(N)−Bn+1(0)

P8 Z x

a

n+ 1 . Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials.

m−1

X

k=0

Bn

x+ k

m

for allm>0 andn≥0. ???

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Outline

1 Reminders

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On the Hurwitz Zeta Function

Definition:

The Hurwitz Zeta Function is defined, for<e s>1, andz∈C−N^<0, by:

ζ(s,z) =X

n>0

1 (n+z)^s .

Property:

H1











∂ζ

∂z(s,z) =−sζ(s+ 1,z).

ζ(s,x+y) =X

n≥0

−s n

ζ(s+n,x)yⁿ. H2 ζ(s,z−1)−ζ(s,z) =z^−s.

H3 ζ(s,mz) =m^−s

m−1

X

k=0

ζ

s,z+ k m

ifm∈N^∗.

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Link between the Hurwitz Zeta Function and the Bernoulli polynomials

Property:

s7−→ζ(s,z) can be analytically extend to a meromorphic function onC, with a simple pole located at 1.

Remark: ζ(−n,z) = −Bn+1(z)

n+ 1 for alln∈N.

ζ(−n,0) = −bn+1

n+ 1 for alln∈N. Related properties:

Hurwitz zeta function Bernoulli polynomials

Derivative property H1 P4

Difference equation H2 P5

Multiplication theorem H3 P9

Consequence:

The extension from Bernoulli to multiple Bernoulli polynomials will be done using a generalization of the Hurwitz zeta function: theHurwitz multiple zeta functions.

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On Hurwitz Multiple Zeta Functions

Definition of Hurwitz Multiple Zeta Functions He^s¹^,···^,s^r(z) = X

0<n_r<···<n₁

1

(n1+z)^s¹· · ·(nr+z)^s^r ,ifz∈C−N^<0and (s1,· · ·,sr)∈(N^∗)^r, such thats1≥2 .

Lemma 1: (B., J. Ecalle, 2012)

For all sequences (s1,· · ·,sr)∈(N^∗)^r,s1≥2, we have:

He^s¹^,···^,s^r(z−1)− He^s¹^,···^,s^r(z) =He^s¹^,···^,s^r−1(z)·z^−s^r .

Lemma 2:

The Hurwitz Multiple Zeta Functions multiply by the stuffle product (ofN^∗).

Reminder:If (Ω,+) is a semi-group, the stuffle is defined over Ω^?by:

ε u = u ε = u .

ua vb = (u vb)a+ (ua v)b+ (u v)(a+b).

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Outline

1 Reminders

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Reminders on Quasi-symmetric functions

Definition:

Let x ={x1,x2,x3,· · · }be an infinite commutative alphabet.

A series is said to be quasi-symmetric when its coefficient ofx₁^s¹· · ·x_r^s^r is equals to this ofx_i^s¹

1 · · ·x_i^s_r^r for alli1<· · ·<ir.

Example : M2,1(x1,x2,x3, . . .) =x₁²x2+x₁²x3+· · ·+x₁²xn+· · ·+x₂²x3+· · · x1x₂²is not inM2,1but inM1,2.

Fact 1: •Quasi-symmetric functions span a vector space: QSym.

•A basis ofQSymis given by the monomialsMI, for composition I= (i1,· · ·,ir):

Mi₁,···,i_r(X) = X

0<n₁<···<n_r

x_nⁱ¹₁· · ·x_nⁱ^r_r

Fact 2: •QSymis an algebra whose product is the stuffle product.

•QSymis also a Hopf algebra whose coproduct ∆ is given by:

∆Mi1,···,ir(x)

=

r

X

k=0

Mi1,···,i_k(x)⊗Mi_k+1,···,ir(x).

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Outline

1 Reminders

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Main Goal

Heuristic:

B^s¹^,···^,s^r(z) = Multiple (Divided) Bernoulli Polynomials =He^−s¹^,···^,−s^r(z). b^s¹^,···^,s^r= Multiple (Divided) Bernoulli Numbers =He^−s¹^,···^,−s^r(0).

We want to defineB^s¹^,···^,s^r(z) such that:

their properties are similar to Hurwitz Multiple Zeta Functions’ properties.

their properties generalize these of Bernoulli polynomials.

Main Goal:

Find some polynomialsB^s¹^,···^,s^r such that:











Bⁿ(z) = Bn+1(z)

n+ 1 , wheren≥0,

Bⁿ¹^,···^,n^r(z+ 1)−Bⁿ¹^,···^,n^r(z) =Bⁿ¹^,···^,n^r−1(z)zⁿ^r , for n1,· · ·,nr≥0, theBⁿ¹^,···^,n^r multiply by the stuffle product.

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An algebraic construction

Notation 1:

Let X ={X1,· · ·,Xn,· · · }be a (commutative) alphabet of indeterminates.

We denotes:

B^Y¹^,···^,Y^r(z) = X

n₁,···,n_r≥0

Bⁿ¹^,···^,n^r(z)Y₁ⁿ¹ n1! · · ·Yrⁿ^r

nr! , for allr∈N^∗,Y1,· · ·,Yr ∈X.

Remark:B^Y¹^,···^,Y^r(z+ 1)− B^Y¹^,···^,Y^r(z) =B^Y¹^,···^,Y^r−1(z)e^zY^r . Notation 2:

Let A ={a1,· · ·,an,· · · }be a non-commutative alphabet.

We denotes:

B(z) = 1 +X

r>0

X

k1,···,kr>0

B^X^k¹^,···^,X^kr(z)ak₁· · ·ak_r ∈C[z][[X]]hhAii.

Remark:B(z+ 1) =B(z)· 1 +X

k>0

e^zX^kak

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The abstract construction in the case of quasi-symmetric functions

Let see an analogue ofB(z) where the multiple Bernoulli polynomials are replaced with the monomial functionsMI(x) ofQSym:

M^Y¹^,···^,Y^r(x) := X

n₁,···,n_r≥0

Mn₁+1,···,nr+1(x)Y₁ⁿ¹ n1! · · ·Yrⁿ^r

nr! , for allY1,· · ·,Yr ∈X. M := 1 +X

r>0

X

k₁,···,k_r>0

M^X^k¹^,···^,X^kr(x)ak₁· · ·ak_r

= 1 +X

r>0

X

0<p₁<···<p_r r

Y

i=1

1 +X

k>0

xne^xⁿ^X^kak

M^X^k¹^,···^,X^kr(x)

=

−→

Y

n>0

1 +X

k>0

xne^xⁿ^X^kak

∈C[[x]][[X]]hhAii.

Computation of the coproduct ofM:(which does not act on theX’s)

∆M^Y¹^,···^,Y^r(x) =

r

X

k=0

M^Y¹^,...,Y^k(x)⊗M^Y^k+1^,...,Y^r(x).

∆M=M⊗M.

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Transcription of the multiplication by the stuffle

Property: (J. Y. Thibon, F. Chapoton, J. Ecalle, F. Menous, D. Sauzin, ...) A family of objects (Bⁿ¹^,···^,n^r)n₁,n₂,n₃,···≥0multiply by the stuffle product if, and only if, there exists a characterχz ofQSymsuch that

χz Mn₁+1,···,n_r+1(x)

=Bⁿ¹^,···^,n^r(z) (1)

Consequences:

1.χz can be extended toQSym[[X]], applying it terms by terms.

χz

M^Y¹^,···^,Y^r(x)

=B^Y¹^,···^,Y^r(z) , for allY1,· · ·,Yr ∈X.

2. IfBⁿ¹^,···^,n^r multiply the stuffle,B=χz(M) is “group-like” inC[z][[X]]hhAii.

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Reformulation of the main goal

Find some polynomialsBⁿ¹^,···^,n^r such that:











hB(z)|aki= e^zX^k e^X^k−1− 1

Xk

,

B(z+ 1) =B(z)·E(z) , whereE(z) = 1 +X

k>0

e^zX^kak , Bis a “group-like” element ofC[z][[X]]hhAii.

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Outline

1 Reminders

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A singular solution

Remainder:E(z) = 1 +X

k>0

e^zX^kak.

From a false solution to a singular solution...

S(z) =

←−

Y

n>0

E(z−n) = 1 +X

r>0

X

k1,···,kr>0

e^z(X^k¹^+···+X^kr⁾

r

Y

i=1

(e^X^k¹^+···+X^ki −1)

ak₁· · ·ak_r is a

false solution to system











hB(z)|aki= e^zX^k e^X^k−1− 1

Xk

, B(z+ 1) =B(z)·E(z),

Bis a “group-like” element ofC[z][[X]]hhAii. Explanations: 1. B(z) =· · ·= B(z−n)·E(z−n)· · ·E(z−1)

=· · ·=

n−→+∞lim B(z−n)

·

←−

Y

n>0

E(z−n).

2. S(z)∈C[z]((X))hhAii,S(z)6∈C[z][[X]]hhAii.

Heuristic:Find a correction ofS, to send it intoC[z][[X]]hhAii.