Kempner Colloquium of the Department of Mathematics
University of Colorado at Boulder. February 28, 2005
Multiple Zeta Values
Michel Waldschmidt
Periods
M. Kontevich and D. Zagier (2000) – Periods.
A period is a complex number whose real and imaginary parts
are values of absolutely convergent integrals of rational functions
with rational coefficients over domains of R
ngiven by polynomials
(in)equalities with rational coefficients.
Examples:
√
2 = Z
2x2≤1
dx,
π = Z
x2+y2≤1
dxdy,
log 2 = Z
1<x<2
dx x ,
ζ (2) = Z
1>t1>t2>0
dt
1t
1· dt
21 − t
2= π
26 ·
Relations between periods 1 Additivity
Z
ba
f (x) + g(x)
dx =
Z
ba
f (x)dx +
Z
ba
g(x)dx and
Z
ba
f (x)dx =
Z
ca
f (x)dx +
Z
bc
f (x)dx.
2 Change of variables
Z
ϕ(b)f (t)dt =
Z
bf ϕ(u)
ϕ
0(u)du.
3 Newton–Leibniz–Stokes
Z
ba
f
0(t)dt = f (b) − f (a).
3 Newton–Leibniz–Stokes
Z
ba
f
0(t)dt = f (b) − f (a).
Conjecture (Kontsevich–Zagier). If a period has two
representations, then one can pass from one formula to another
using only rules 1 , 2 and 3 in which all functions and domains
of integrations are algebraic with algebraic coefficients.
Example:
π =
Z
x2+y2≤1
dxdy
= 2
Z
1−1
p 1 − x
2dx
=
Z
1−1
√ dx
1 − x
2=
Z
∞−∞
dx
1 + x
2·
Far reaching consequences:
No “new” algebraic dependence relation among classical
constants from analysis.
Zeta Values – Euler Numbers
ζ (s) = X
n≥1
1
n
sfor s ≥ 2.
These are special values of the Riemann Zeta Function: s ∈ C.
Zeta Values – Euler Numbers
ζ (s) = X
n≥1
1
n
sfor s ≥ 2.
These are special values of the Riemann Zeta Function: s ∈ C.
For s ∈ Z with s ≥ 2, ζ (s) is a period:
ζ (s) = Z
1>t1>···>ts>0
dt
1t
1· · · dt
s−1t
s−1· dt
s1 − t
s·
Zeta Values – Euler Numbers
ζ (s) = X
n≥1
1
n
sfor s ≥ 2.
These are special values of the Riemann Zeta Function: s ∈ C.
Euler: π
−2kζ (2k) ∈ Q for k ≥ 1 (Bernoulli numbers).
Zeta Values – Euler Numbers
ζ (s) = X
n≥1
1
n
sfor s ≥ 2.
These are special values of the Riemann Zeta Function: s ∈ C.
Euler: π
−2kζ (2k) ∈ Q for k ≥ 1 (Bernoulli numbers).
Diophantine Question: Describe all the algebraic relations among the numbers
ζ (2), ζ (3), ζ (5), ζ (7), . . .
Conjecture. There is no algebraic relation at all: these numbers
ζ (2), ζ (3), ζ (5), ζ (7), . . .
are algebraically independent.
Conjecture. There is no algebraic relation at all: these numbers
ζ (2), ζ (3), ζ (5), ζ (7), . . . are algebraically independent.
Known:
• Hermite-Lindemann: π is transcendental, hence ζ (2k ) also for
k ≥ 1.
Conjecture. There is no algebraic relation at all: these numbers
ζ (2), ζ (3), ζ (5), ζ (7), . . . are algebraically independent.
Known:
• Hermite-Lindemann: π is transcendental, hence ζ (2k ) also for k ≥ 1.
• Ap´ ery (1978): ζ (3) is irrational.
Conjecture. There is no algebraic relation at all: these numbers
ζ (2), ζ (3), ζ (5), ζ (7), . . . are algebraically independent.
Known:
• Hermite-Lindemann: π is transcendental, hence ζ (2k ) also for k ≥ 1.
• Ap´ ery (1978): ζ (3) is irrational.
• Rivoal (2000) + Ball, Zudilin. . . Infinitely many ζ (2k + 1)
are irrational + lower bound for the dimension of the Q-space
T. Rivoal: Let > 0. For any sufficiently large odd integer a, the dimension of the Q-space spanned by 1, ζ (3), ζ (5), · · · , ζ (a) is at least
1 −
1 + log 2 log a.
T. Rivoal: Let > 0. For any sufficiently large odd integer a, the dimension of the Q-space spanned by 1, ζ (3), ζ (5), · · · , ζ (a) is at least
1 −
1 + log 2 log a.
W. Zudilin:
• One at least of the four numbers
ζ (5), ζ (7), ζ (9), ζ (11) is irrational.
• There is an odd integer j in the range [5, 69] such that the
three numbers 1, ζ (3), ζ (j ) are linearly independent over Q.
Linearization of the problem (Euler). The product of two zeta
values is a sum of multiple zeta values.
Linearization of the problem (Euler). The product of two zeta values is a sum of multiple zeta values.
From
X
n1≥1
n−s1 1 X
n2≥1
n2−s2 = X
n1>n2≥1
n−s1 1n2−s2 + X
n2>n1≥1
n−s2 2n1−s1 + X
n≥1
n−s1−s2
one deduces, for s
1≥ 2 and s
2≥ 2,
ζ (s
1)ζ (s
2) = ζ (s
1, s
2) + ζ (s
2, s
1) + ζ (s
1+ s
2) with
ζ (s , s ) = X
n
−s1n
−s2.
For instance
ζ (2)
2= X
n1≥1
n
−21X
n2≥1
n
2−2= X
n1>n2≥1
n
−21n
2−2+ X
n2>n1≥1
n
−22n
1−2+ X
n≥1
n
−4= 2ζ (2, 2) + ζ (4).
For k, s
1, . . . , s
kpositive integers with s
1≥ 2, define s = (s
1, . . . , s
k) and
ζ (s) = X
n1>n2>···>nk≥1
1
n
s11· · · n
skk·
For k, s
1, . . . , s
kpositive integers with s
1≥ 2, define s = (s
1, . . . , s
k) and
ζ (s) = X
n1>n2>···>nk≥1
1
n
s11· · · n
skk·
For k = 1 one recovers Euler’s numbers ζ (s).
For k, s
1, . . . , s
kpositive integers with s
1≥ 2, define s = (s
1, . . . , s
k) and
ζ (s) = X
n1>n2>···>nk≥1
1
n
s11· · · n
skk·
Fact: These Multiple Zeta Values are periods
For k, s
1, . . . , s
kpositive integers with s
1≥ 2, define s = (s
1, . . . , s
k) and
ζ (s) = X
n1>n2>···>nk≥1
1
n
s11· · · n
skk·
Fact: These Multiple Zeta Values are periods Example:
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3·
Notation: Define
ω
0= dt
t , ω
1= dt 1 − t · Then for s ≥ 2 write the relation
ζ (s) = Z
1>t1>···>ts>0
dt
1t
1· · · dt
s−1t
s−1· dt
s1 − t
sas
ζ (s) =
Z
10
ω
0s−1ω
1.
This defines a non-commutative product of differential forms.
Chen Iterated Integrals For a holomorphic 1-form ϕ,
Z
z0
ϕ
is the primitive of ϕ which vanishes at z = 0.
For 1-forms ϕ
1, . . . , ϕ
k, define inductively Z
z0
ϕ
1· · · ϕ
k:=
Z
z0
ϕ
1(t)
Z
t0
ϕ
2· · · ϕ
k.
Chen Iterated Integrals Z
z0
ϕ
1· · · ϕ
k:=
Z
z0
ϕ
1(t)
Z
t0
ϕ
2· · · ϕ
k.
If ϕ
1(t) = ψ
1(t)dt, then d
dz
Z
z0
ϕ
1· · · ϕ
k= ψ
1(z )
Z
z0
ϕ
2· · · ϕ
k.
For s = (s
1, . . . , s
k), define
ω
s= ω
s1· · · ω
sk= ω
0s1−1ω
1· · · ω
0sk−1ω
1. Then
ζ (s) =
Z
10
ω
s.
For s = (s
1, . . . , s
k), define
ω
s= ω
s1· · · ω
sk= ω
0s1−1ω
1· · · ω
0sk−1ω
1. Then
ζ (s) =
Z
10
ω
s.
Remark on ω
0s1−1ω
1· · · ω
0sk−1ω
1:
• Ends with ω
1• Starts with ω
0(s
1≥ 2).
For s = (s
1, . . . , s
k), define
ω
s= ω
s1· · · ω
sk= ω
0s1−1ω
1· · · ω
0sk−1ω
1. Then
ζ (s) =
Z
10
ω
s.
Example:
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3=
Z
10
ω
0ω
12·
For s = (s
1, . . . , s
k), define
ω
s= ω
s1· · · ω
sk= ω
0s1−1ω
1· · · ω
0sk−1ω
1. Then
ζ (s) =
Z
10
ω
s.
Example:
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3=
Z
10
ω
0ω
12·
Main Fact: The product of two Multiple Zeta Values is a
linear combination, with integer coefficients, of Multiple Zeta
Values.
Main Fact: The product of two Multiple Zeta Values is a linear combination, with integer coefficients, of Multiple Zeta Values.
Moreover there are two kinds of such quadratic equations: one arising from the definition as series
ζ (s) = X
n1>n2>···>nk≥1
1
n
s11· · · n
skk, the other from the integrals
ζ (s) =
Z
1ω
s.
These two collections of quadratic equations are essentially
distinct. Consequently the Multiple Zeta Values satisfy many
linear relations with rational coefficients.
These two collections of quadratic equations are essentially distinct. Consequently the Multiple Zeta Values satisfy many linear relations with rational coefficients.
Example:
Product of series: ζ (2)
2= 2ζ (2, 2) + ζ (4)
Product of integrals: ζ (2)
2= 2ζ (2, 2) + 4ζ (3, 1)
Hence ζ (4) = 4ζ (3, 1).
These two collections of quadratic equations are essentially distinct. Consequently the Multiple Zeta Values satisfy many linear relations with rational coefficients.
A complete description of these relations would in principle settle the problem of the algebraic independence of
π, ζ (3), ζ (5), . . . , ζ (2k + 1).
These two collections of quadratic equations are essentially distinct. Consequently the Multiple Zeta Values satisfy many linear relations with rational coefficients.
A complete description of these relations would in principle settle the problem of the algebraic independence of
π, ζ (3), ζ (5), . . . , ζ (2k + 1).
Goal: Describe all linear relations among Multiple Zeta Values.
Further example of linear relation.
Euler:
ζ (2, 1) = ζ (3).
Further example of linear relation.
Euler:
ζ (2, 1) = ζ (3).
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3·
Further example of linear relation.
Euler:
ζ (2, 1) = ζ (3).
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3·
ζ (3) = Z
1>t1>t2>t3>0
dt
1t
1· dt
2t
2· dt
31 − t
3·
Further example of linear relation.
Euler:
ζ (2, 1) = ζ (3).
ζ (2, 1) = Z
1>t1>t2>t3>0
dt
1t
1· dt
21 − t
2· dt
31 − t
3·
ζ (3) = Z
1>t1>t2>t3>0
dt
1t
1· dt
2t
2· dt
31 − t
3·
Denote by Z
pthe Q-vector subspace of R spanned by the real numbers ζ (s) with s of weight s
1+ · · · + s
k= p, with Z
0= Q and Z
1= {0}.
Here is Zagier’s conjecture on the dimension d
pof Z
p. Conjecture (Zagier). For p ≥ 3 we have
d
p= d
p−2+ d
p−3.
(d
0, d
1, d
2, . . .) = (1, 0, 1, 1, 1, 2, 2, . . .).
Exemples
d
0= 1 ζ (s
1, . . . , s
k) = 1 for k = 0.
Exemples d
0= 1 ζ (s
1, . . . , s
k) = 1 for k = 0.
d
1= 0 {(s
1, . . . , s
k) ; s
1+ · · · + s
k= 1, s
1≥ 2} = ∅.
Exemples d
0= 1 ζ (s
1, . . . , s
k) = 1 for k = 0.
d
1= 0 {(s
1, . . . , s
k) ; s
1+ · · · + s
k= 1, s
1≥ 2} = ∅.
d
2= 1 ζ (2) 6= 0
Exemples d
0= 1 ζ (s
1, . . . , s
k) = 1 for k = 0.
d
1= 0 {(s
1, . . . , s
k) ; s
1+ · · · + s
k= 1, s
1≥ 2} = ∅.
d
2= 1 ζ (2) 6= 0
d
3= 1 ζ (2, 1) = ζ (3) 6= 0
Exemples d
0= 1 ζ (s
1, . . . , s
k) = 1 for k = 0.
d
1= 0 {(s
1, . . . , s
k) ; s
1+ · · · + s
k= 1, s
1≥ 2} = ∅.
d
2= 1 ζ (2) 6= 0
d
3= 1 ζ (2, 1) = ζ (3) 6= 0 d
4= 1 ζ (3, 1) = (1/4)ζ (4),
ζ (2, 2) = (3/4)ζ (4),
ζ (2, 1, 1) = ζ (4) = (2/5)ζ (2)
2Question: d
5= 2 ? Since
ζ (2, 1, 1, 1)= ζ (5),
ζ (3, 1, 1) =ζ (4, 1)= 2ζ (5) − ζ (2)ζ (3), ζ (2, 1, 2) =ζ (2, 3)= 9
2 ζ (5) − 2ζ (2)ζ (3), ζ (2, 2, 1) =ζ (3, 2)=3ζ (2)ζ (3) − 11
2 ζ (5), we have d
5∈ {1, 2}.
Further, d
5= 2 if and only if the number
ζ (2)ζ (3)/ζ (5)
Zagier’s conjecture can be written
X
p≥0
d
pX
p= 1
1 − X
2− X
3·
M. Hoffman conjectures: a basis of Z
pover Q is given by the numbers ζ (s
1, . . . , s
k), s
1+ · · · + s
k= p, where each s
iis either 2 or 3.
True for p ≤ 16 (Hoang Ngoc Minh)
A.G. Goncharov (2000) – Multiple ζ -values, Galois groups and Geometry of Modular Varieties.
T. Terasoma (2002) – Mixed Tate motives and Multiple Zeta Values.
The numbers defined by the recurrence relation of Zagier’s Conjecture
d
p= d
p−2+ d
p−3.
with initial values d
0= 1, d
1= 0 are actual upper bounds for
the actual dimension of Z
p.
A.G. Goncharov (2000) – Multiple ζ -values, Galois groups and Geometry of Modular Varieties.
T. Terasoma (2002) – Mixed Tate motives and Multiple Zeta Values.
The numbers defined by the recurrence relation of Zagier’s Conjecture
d
p= d
p−2+ d
p−3.
with initial values d
0= 1, d
1= 0 are actual upper bounds for the actual dimension of Z
p.
To prove a lower bound is the main Diophantine conjecture!
Algebraic description of the quadratic relations among MZV
1 Integrals:
Shuffle product of differential forms
ϕ
1· · · ϕ
nx ψ
1· · · ψ
k= ϕ
1(ϕ
2· · · ϕ
nx ψ
1· · · ψ
k) + ψ
1(ϕ
1· · · ϕ
nx ψ
2· · · ψ
k).
ϕ
1x ψ
1= ϕ
1ψ
1+ ψ
1ϕ
1.
Product of iterated integrals:
Let ϕ
1, . . . , ϕ
n, ψ
1, . . . , ψ
kbe differential forms with n ≥ 0 and k ≥ 0. Then
Z
z0
ϕ
1· · · ϕ
nZ
z0
ψ
1· · · ψ
k=
Z
z0
ϕ
1· · · ϕ
nx ψ
1· · · ψ
k.
Product of iterated integrals:
Let ϕ
1, . . . , ϕ
n, ψ
1, . . . , ψ
kbe differential forms with n ≥ 0 and k ≥ 0. Then
Z
z0
ϕ
1· · · ϕ
nZ
z0
ψ
1· · · ψ
k=
Z
z0
ϕ
1· · · ϕ
nx ψ
1· · · ψ
k.
Proof. Assume z > 0. Decompose the Cartesian product
{t ∈ Rn ; z ≥ t1 ≥ · · · ≥ tn ≥ 0} × {u ∈ Rk ; z ≥ u1 ≥ · · · ≥ uk ≥ 0}
into a disjoint union of simplices (up to sets of zero measure) {v ∈ Rn+k ; z ≥ v1 ≥ · · · ≥ vn+k ≥ 0}.
Example.
ab x cd = abcd + acbd + acdb + cabd + cadb + cdab
Example.
ab x cd = abcd + acbd + acdb + cabd + cadb + cdab
ω
0ω
1x ω
0ω
1= 4ω
02ω
12+ 2ω
0ω
1ω
0ω
1Example.
ab x cd = abcd + acbd + acdb + cabd + cadb + cdab
ω
0ω
1x ω
0ω
1= 4ω
02ω
12+ 2ω
0ω
1ω
0ω
1Z
10
ω
0ω
1·
Z
10
ω
0ω
1= 4
Z
10
ω
02ω
12+ 2
Z
10
ω
0ω
1ω
0ω
1Example.
ab x cd = abcd + acbd + acdb + cabd + cadb + cdab
ω
0ω
1x ω
0ω
1= 4ω
02ω
12+ 2ω
0ω
1ω
0ω
1Z
10
ω
0ω
1·
Z
10
ω
0ω
1= 4
Z
10
ω
02ω
12+ 2
Z
10
ω
0ω
1ω
0ω
1ζ (2)
2= 4ζ (3, 1) + 2ζ (2, 2).
Next goal: Extend the definition of Multiple Zeta Values to linear combinations of ω
s, so that the product of two Multiple Zeta Values is a Multiple Zeta Value.
Write ζ b (ω
s) in place of ζ (s) and define more generally
ζ b X
c
sω
s= X
c
sζ (s) so that
ζ (s)ζ (s
0) = ζ ω b
sx ω
s0.
Tool: Free algebra on {ω , ω }.
The free monoid X
∗Let X = {x
0, x
1} denote the alphabet with two letters x
0, x
1and X
∗the free monoid on X . The elements of X
∗are words.
A word can be written
x
1· · · x
kwith k ≥ 0 and where each
jis 0 or 1.
This law is called concatenation. It is not commutative:
x
0x
16= x
1x
0.
The Algebra H = Qhx
0, x
1i
The free Q-vector space with basis X
∗is the free algebra on X , denoted by H = QhX i. Its elements are non commutative polynomials in the two variables x
0, x
1.
Its unit is the empty word e.
The words which end with x
1are the elements of X
∗x
1.
The words which end with x
1are the elements of X
∗x
1.
Let w ∈ X
∗x
1. Write w = x
1· · · x
pwhere each
iis 0 or 1 and
p= 1.
If k is the number of x
1, we define positive integers s
1, . . . , s
kby
w = x
s01−1x
1· · · x
s0k−1x
1.
The words which end with x
1are the elements of X
∗x
1.
Let w ∈ X
∗x
1. Write w = x
1· · · x
pwhere each
iis 0 or 1 and
p= 1.
If k is the number of x
1, we define positive integers s
1, . . . , s
kby w = x
s01−1x
1· · · x
s0k−1x
1.
For s ≥ 1 define y
s= x
s−10x
1. For s = (s
1, . . . , s
k) with s
i≥ 1, set
y
s= y
s1· · · y
sk= x
s01−1x
1· · · x
s0k−1x
1.
y
sis a word on the alphabet
Y = {y
1, y
2, . . . , y
s, . . .}.
y
sis a word on the alphabet
Y = {y
1, y
2, . . . , y
s, . . .}.
The free monoid Y
∗on Y
Y
∗= {y
s; s = (s
1, . . . , s
k), k ≥ 0, s
j≥ 1 (1 ≤ j ≤ k)}
is the set {e} ∪ X
∗x
1of words which do not end with x
0, hence Y
∗is a submonoid of X
∗.
Any message can be coded with only two letters.
The Subalgebra H
1= Qe + H x
1of H
The free Q-vector space with basis Y
∗is the free algebra H
1= QhY i
on Y . Its elements are non commutative polynomials in the
variables {y
1, . . . , y
s, . . .}. It is a subalgebra of H .
The Subalgebra H
0= Qe + x
0H x
1of H
The set of words in X
∗which start with x
0and end with x
1is x
0X
∗x
1.
The set of words in X
∗which do not start with x
1and do not
end with x
0is {e} ∪ x
0X
∗x
1.
The Subalgebra H
0= Qe + x
0H x
1of H
The set of words in X
∗which start with x
0and end with x
1is x
0X
∗x
1.
The set of words in X
∗which do not start with x
1and do not end with x
0is {e} ∪ x
0X
∗x
1.
This is NOT the same as the free monoid on the infinite alphabet {y
2, y
3, . . .}.
Example: y
2y
1∈ x
0X
∗x
1.
The Subalgebra H
0= Qe + x
0H x
1of H
The set of words in X
∗which start with x
0and end with x
1is x
0X
∗x
1.
The set of words in X
∗which do not start with x
1and do not end with x
0is {e} ∪ x
0X
∗x
1.
The Q-vector subspace of H
1spanned by {e} ∪ x
0X
∗x
1is the sub-algebra
H
0= Qe + x
0H x
1⊂ H
1⊂ H .
Multizeta values associated to words
For w ∈ x
0X
∗x
1, write w = y
swith s = (s
1, . . . , s
k) and s
1≥ 1, and define
ζ b (w) = ζ (s).
Define also ζ b (e) = 1 and extend by Q-linearity the definition of ζ b to H
0. Hence we get a mapping
ζ b : H
0−→ R.
Shuffle relations among MZV
For w and w
0in H
0, the shuffle product w x w
0belongs to H
0. Furthermore,
ζ b (w) ζ b (w
0) = ζ b (w x w
0) for any w and w
0in H
0.
Proposition. The map ζ b : H
0→ R is a morphism of algebras
of H
0xinto R.
2 Series:
The Harmonic Algebra
The product ζ (s) · ζ (s
0):
X
n1>n2>···>nk≥1
1
n
s11· · · n
skk· X
n01>n02>···>n0
k0≥1
1 n
0s0 1
1
· · · n
0sk00k0is a linear combination of MZV.
Shuffle like product (stuffle) on the alphabet Y .
The map ? : Y
∗× Y
∗→ H is defined by induction
y
su ? y
tv = y
s(u ? y
tv ) + y
t(y
su ? v ) + y
s+t(u ? v ) for u and v in Y
∗, s and t positive integers.
This defines Hoffman’s harmonic algebra denoted by H
?.
Examples.
y
2?2= y
2? y
2= 2y
22+ y
4.
y
2?3= y
2? y
2? y
2= 6y
23+ 3y
2y
4+ 3y
4y
2+ y
6.
Quadratic relations arising from the product of series The map ζ b : H
0→ R is a morphism of algebras of H
0?into R:
ζ b (u ? v ) = ζ b (u) ζ b (v).
for u and v in H
0.
Quadratic relations arising from the product of series The map ζ b : H
0→ R is a morphism of algebras of H
0?into R:
ζ b (u ? v ) = ζ b (u) ζ b (v).
for u and v in H
0.
Consequence of the two sets of quadratic relations:
ζ b (u x v − u ? v) = 0
for u and v in H
0.
Hoffman Third Standard Relations For any w ∈ H
0, we have x
1x w − x
1? w ∈ H
0and
ζ b (x
1x w − x
1? w) = 0.
Hoffman Third Standard Relations For any w ∈ H
0, we have x
1x w − x
1? w ∈ H
0and
ζ b (x
1x w − x
1? w) = 0.
Example. For w = x
0x
1,
x
1x x
0x
1= x
1x
0x
1+ 2x
0x
21= y
1y
2+ 2y
2y
1, x
1? x
0x
1= y
1? y
2= y
1y
2+ y
2y
1+ y
3, hence
y
2y
1− y
3∈ ker ζ b
Euler’s proof with divergent series:
Product of series: ζ (1)ζ (2) = ζ (1, 2) + ζ (2, 1) + ζ (3) Product of integrals: ζ (1)ζ (2) = ζ (1, 2) + 2ζ (2, 1)
Hence ζ (3) = ζ (2, 1).
Diophantine Conjecture (simple form)
Conjecture (Petitot, Hoang Ngoc Minh. . . ). The kernel of ζ b is spanned by the standard relations
ζ b (u x v − u ? v ) = 0 and ζ b (x
1x w − x
1? w ) = 0 for u, v and w in x
0X
∗x
1.
Minh, H.N, Jacob, G., Oussous, N. E., Petitot, M. –
Aspects combinatoires des polylogarithmes et des sommes d’Euler-Zagier.
J. ´Electr. S´em. Lothar. Combin. 43 (2000), Art. B43e, 29 pp.
Regularized Double Shuffle Relations
The map ζ b : H
0→ R is a morphism of algebras for x and for ?:
ζ b (u x v) = ζ b (u) ζ b (v) and ζ b (u ? v) = ζ b (u) ζ b (v).
Question: Is-it possible to extend ζ b to H
1into a morphism of
algebras both for x and ??
Regularized Double Shuffle Relations
The map ζ b : H
0→ R is a morphism of algebras for x and for ?:
ζ b (u x v) = ζ b (u) ζ b (v) and ζ b (u ? v) = ζ b (u) ζ b (v).
Question: Is-it possible to extend ζ b to H
1into a morphism of algebras both for x and ??
Answer: NO!
x
1x x
1= 2x
21, x
1? x
1= y
1? y
1= 2x
21+ y
2ζ (y ) = ζ (2) 6= 0.
Radford’s Theorem:
H
x= H
1x[x
0]
x= H
0x[x
0, x
1]
xand H
1x= H
0x[x
1]
x.
Radford’s Theorem:
H
x= H
1x[x
0]
x= H
0x[x
0, x
1]
xand H
1x= H
0x[x
1]
x.
Hoffman’s Theorem:
H
?= H
1?[x
0]
?= H
0?[x
0, x
1]
?and H
1?= H
0?[x
1]
?.
Radford’s Theorem:
H
x= H
1x[x
0]
x= H
0x[x
0, x
1]
xand H
1x= H
0x[x
1]
x. Hoffman’s Theorem:
H
?= H
1?[x
0]
?= H
0?[x
0, x
1]
?and H
1?= H
0?[x
1]
?.
From H
1x= H
0x[x
1]
xand H
1?= H
0?[x
1]
?we deduce that there are two uniquely determined algebra morphisms
Z b
x: H
1x −→ R[T ] and Z b
?: H
1?−→ R[T ]
Theorem (Boutet de Monvel, Zagier). There is a R-linear isomorphism % : R[T ] → R[X ] which makes commutative the following diagram:
R[X ] Z b
x%
H
1↑
%Z b
?&
R[T ]
An explicit formula for % is given by means of the generating series
X
`≥0
%(T
`) t
``! = exp Xt +
∞
X
n=2
(−1)
nζ (n) n t
n! .
Compare with the formula giving the expansion of the logarithm of Euler Gamma function:
Γ(1 + t) = exp −γt +
∞
X
n=2
(−1)
nζ (n) n t
n!
.
One may see % as the differential operator of infinite order
exp
∞
X
n=2
(−1)
nζ (n) n
∂
∂T
n!
(just consider the image of e
tT).
Denote by reg
xthe Q-linear map H → H
0which maps w ∈ H onto its constant term when w is written as a polynomial in x
0, x
1in the shuffle algebra H
0[x
0, x
1]
x. Then reg
x
is a morphism of algebras H
x→ H
0x
.
Theorem.(Regularized Double Shuffle Relations – Ihara and Kaneko). For w ∈ H
1and w
0∈ H
0,
reg
x(w x w
0− w ? w
0) ∈ ker ζ. b
0
Diophantine Conjectures
Conjecture (Zagier, Cartier, Ihara-Kaneko,. . . ). All existing algebraic relations between the real numbers ζ (s) are in the ideal generated by the ones described above.
Petitot and Hoang Ngoc Minh: up to weight s
1+ · · · s
k≤ 16, the three standard relations for u, v and w in x
0X
∗x
1ζ b (u) ζ b (v) = ζ b (u x v), ζ b (u) ζ b (v ) = ζ b (u ? v ), ζ b (x
1x w − x
1? w) = 0
suffice.
Goncharov’s Conjecture
Let Z denote the Q-vector space spanned in C by the numbers
(2iπ )
−|s|ζ (s)
s = (s
1, . . . , s
k) ∈ N
kwith k ≥ 1, s
1≥ 2, s
i≥ 1 (2 ≤ i ≤ k).
Hence Z is a Q-subalgebra of C bifiltered by the weight and by
the depth.
For a graded Lie algebra C
•denote by U C
•its universal envelopping algebra and by
U C
•∨= M
n≥0