“NumberTheory & Discrete Mathematics”
International Conference in honour of Srinivasa Ramanujan Centre for Advanced Study in Mathematics
Panjab University, Chandigarh (October 2/6, 2000.)
Multiple Zeta Values
and
Euler-Zagier Numbers
by
Michel WALDSCHMIDT
http : //www.math.jussieu.fr/ ∼ miw/articles/ps/MZV.ps
2
1. Introduction
For s 1 , . . . , s k in Z with s 1 ≥ 2, ζ (s 1 , . . . , s k ) = X
n
1> ··· >n
k≥ 1
n − 1 s1 · · · n − k sk.
.
k = 1 integer values of Riemann zeta function ζ (s).
Euler: ζ (s)π − s ∈ Q for s even ≥ 2.
Fact: No known other algebraic relations between values of Riemann zeta function at positive integers.
Expected: there is no further relation : Are the numbers
π, ζ (3), ζ (5), . . . , ζ (2n + 1), . . . algebraically independent?
Means:
For n ≥ 0 and P ∈ Q [X 0 , X 1 , . . . , X n ] \ { 0 } , P ¡
π, ζ (3), ζ (5), . . . , ζ (2n + 1) ¢
6
= 0 ?
3
F. Lindemann (1882): π is transcendental.
R. Ap´ery (1978): ζ (3) is irrational.
T. Rivoal (2000): infinitely many irrational numbers among ζ (3), ζ (5), . . . , ζ (2n + 1), . . .
Theorem (T. Rivoal). Let ² > 0. For any suffi- ciently large n, the Q -vector space spanned by the n numbers
ζ (3), ζ (5), . . . , ζ (2n + 1) has dimension
≥ 1 − ²
1 + log 2 · log n.
The proof also yields:
There exists on odd integer j with 5 ≤ j ≤ 169 such that the three numbers
1, ζ (3), ζ (j )
are linearly independent over Q .
4
2. Sketch of Proof of Rivoal’s Theorem
Goal: Given a sufficiently large odd integer a, con- struct a sequence of linear forms in (a + 1)/2 vari- ables, with integer coefficients, such that the numbers
` n = p 0n +
(a − X 1)/2
i=1
p in ζ (2i + 1) satisfy, for n → ∞ ,
| ` n | = α − n+o(n) and
| p in | ≤ β n+o(n) with
α ' a 2a and β ' (2e) 2a . It will follow that the (a + 1)/2 numbers
1, ζ (3), ζ (5), . . . , ζ (a)
span a Q -vector space of dimension at least 1 + log α
log β ' log a
1 + log 2
(Nesterenko’s Criterion).
5
Explicit construction of the linear forms Previous works of R. Ap´ery, F. Beukers,
E. Nikishin, K. Ball, D. Vasilyev, . . .
Pochammer symbol: (m) 0 = 1 and, for k ≥ 1, (m) k = m(m + 1) . . . (m + k − 1).
Set r = £
a(log a) − 2 ¤
. Define
d m = l.c.m. of { 1, 2, . . . , m } ,
R n (t) = n! a − 2r (t − rn + 1) rn (t + n + 2) rn (t + 1) a n+1 , S n =
X ∞ k=0
R n (k), ` n = d a 2n S 2n . Write the partial fraction expansion
R n (t) =
X a
i=1
X n
j =0
c ijn (t + j + 1) i with
c ijn = 1 (a − i)!
µ d dt
¶ a − i ¡
R n (t)(t + j + 1) a ¢
| t= − j − 1 .
6
Set p in = d 2n q i,2n where q 0,n = −
X a
i=1
X n
j =1
c ijn
j − 1
X
k=0
1 (k + 1) i and
q in =
X n
j=0
c ijn (1 ≤ i ≤ a).
Estimate for | p in | : c ijn = 1
2πi Z
| t+j +1 | =1/2
R n (t)(t + j + 1) i − 1 dt.
Estimate for | ` n | : S n =
¡ (2r + 1)n + 1 ¢
!
n! 2r+1 · I n , I n =
Z
[0,1]
a+1F (x) · dx 1 dx 2 . . . dx a+1 (1 − x 1 x 2 · · · x a+1 ) 2 , F (x 1 , x 2 , . . . , x a+1 ) =
à Q a+1
i=1 x r i (1 − x i ) (1 − x 1 x 2 · · · x a+1 ) 2r+1
! n
.
7
3. Shuffle Product for Series Reflexion Formula:
ζ (s)ζ (s 0 ) = X
n ≥ 1
n − s · X
n
0≥ 1
(n 0 ) − s0
= X
n>n
0≥ 1
n − s (n 0 ) − s0 + X
n
0>n ≥ 1
n − s (n 0 ) − s0
+ X
n ≥ 1
n − s − s0
= ζ (s, s 0 ) + ζ (s 0 , s) + ζ (s + s 0 ).
Example:
ζ (s) 2 = 2ζ (s, s) + ζ(2s).
For s = 2: ζ (2) = π 2 /6, ζ (4) = π 4 /90, ζ (2, 2) = X
m>n ≥ 1
(mn) − 2 = π 4 120 · Other example:
ζ (2)ζ (3) = ζ (2, 3) + ζ (3, 2) + ζ (5).
8
Shuffle relations arising from the series representation.
ζ (s)ζ (s 0 ) = X
σ
ζ (σ),
where σ = (σ 1 , . . . , σ h ) ranges over the tuples ob- tained as follows:
s → ( s 1 0 s 2 · · · s k )
s 0 → ( 0 s 0 1 s 0 2 · · · 0 ) σ = ( s 1 s 0 1 s 2 + s 0 2 · · · s k ) Hence max { k, k 0 } ≤ h ≤ k + k 0 .
Example: k = k 0 = 1, s = s, s 0 = s 0 , then
s → (s 0) (0 s) s
s 0 → (0 s) (s 0 0) s 0
σ = (s s 0 ) (s 0 s) s + s 0
so that
{ σ 1 , σ 2 , σ 3 } = { (s, s 0 ), (s 0 , s), s + s 0 } .
9
Other Description:
Alphabet with two letters X = { x 0 , x 1 } . Words: X ∗ = { x a 01x b 11 · · · x a 0hx b 1h} . Non-commutative polynomials: Qh X i .
· · · x a 0hx b 1h} . Non-commutative polynomials: Qh X i .
} . Non-commutative polynomials: Qh X i .
For s ≥ 1 set y s = x s 0 − 1 x 1 .
For s = (s 1 , . . . , s k ) with s i ≥ 1, set x s = y s1 · · · y sk
= x s 01− 1 x 1 x s 0
2− 1 x 1 · · · x s 0
k− 1 x 1 .
The number k of factors x 1 is the depth of the word x s and of the tuple s.
The number p = s 1 + · · · + s k of letters is the weight . The set of such x s ’s is X ∗ x 1 togeether with the null word e (corresponds to ∅ with k = 0).
Convergent words: x 0 X ∗ x 1 S
{ e } .
Set ζ (w) = ζ (s) for w = x s with ζ ( ∅ ) = ζ (e) = 1.
Convergent polynomials: Qh X i conv ⊂ Qh X i .
Extend ζ by linearity to Qh X i conv .
10
Law ∗ on Qh X i conv :
e ∗ w = w for w ∈ X ∗ x 1
and, for s ≥ 1 and t ≥ 1, w and w 0 in X ∗ x 1 , (y s w ) ∗ (y t w 0 ) =
y s (w ∗ y t w 0 ) + y t (y s w ∗ w 0 ) + y s+t (w ∗ w 0 ).
Then
x s ∗ x s0 = X
σ
x σ .
Proposition. For w and w 0 in x 0 X ∗ x 1 , ζ (w)ζ (w 0 ) = ζ (w ∗ w 0 ).
Connection with quasi-symmetric functions Commutative infinite alphabet: t = { t 1 , t 2 , . . . } Formal power series: Q [[t]].
To w = x s ∈ X ∗ x 1 associate F w (t) = X
n
1> ··· >n
k≥ 1
t s n11 · · · t s nkk. Then for w and w 0 in X ∗ x 1 we have
. Then for w and w 0 in X ∗ x 1 we have
F w (t)F w0(t) = F w ∗ w0(t).
(t).
For w ∈ x 0 X ∗ x 1 , ζ (w) is the value of F w (t) with
t n = 1/n, (n ≥ 1).
11
4. Shuffle Product for Integrals
Let s = (s 1 , . . . , s k ) with s i ≥ 1; set p = s 1 + · · · + s k . Define ² i ∈ { 0, 1 } for 1 ≤ i ≤ p by
x s = x ²1 · · · x ²p.
.
For instance for s = (2, 3) with p = 5:
x (2,3) = x 0 x 1 x 2 0 x 1 , (² 1 , . . . , ² 5 ) = (0, 1, 0, 0, 1).
Define differential forms:
ω 0 (t) = dt
t et ω 1 (t) = dt 1 − t · Let ∆ p be the simplex in R p :
∆ p = { t ∈ R p ; 1 > t 1 > · · · > t p > 0 } . Proposition. For s 1 ≥ 2,
ζ (s) = Z
∆
pω ²1(t 1 ) · · · ω ²p(t p ).
(t p ).
Example:
ζ (2, 3) = Z
∆
5dt 1
t 1 · dt 2
1 − t 2 · dt 3
t 3 · dt 4
t 4 · dt 5 1 − t 5 · Proof. Expand 1/(1 − t) = P
s ≥ 0 t s .
12
Shuffle on X ∗ :
e tt w = w tt e = w,
and, for i and j in { 0, 1 } , u and v in X ∗ ,
(x i u) tt (x j v) = x i (u tt x j v) + x j (x i u tt v )
Example. Computation of y 2 tt y 3 = x 0 x 1 tt x 2 0 x 1 : get x 0 x 1 x 2 0 x 1 once, x 2 0 x 1 x 0 x 1 three times and x 3 0 x 2 1 six times. Hence
y 2 tt y 3 = y 2 y 3 + 3y 3 y 2 + 6y 4 y 1 . Corollary.
ζ (w)ζ (w 0 ) = ζ (w tt w 0 ) for w and w 0 in x 0 X ∗ x 1 .
Proof. The Cartesian product ∆ p × ∆ p0 is the union
of (p + p 0 )!/p!p 0 ! simplices.
13
Example. From
y 2 tt y 3 = y 2 y 3 + 3y 3 y 2 + 6y 4 y 1 we deduce
ζ (2)ζ(3) = ζ (2, 3) + 3ζ (3, 2) + 6ζ(4, 1).
On the other hand the shuffle relation for series gives ζ (2)ζ (3) = ζ (2, 3) + ζ (3, 2) + ζ (5),
hence
ζ (5) = 2ζ (3, 2) + 6ζ(4, 1).
There are further relations.
Example:
x 1 tt x 0 x 1 = x 1 x 0 x 1 + 2x 0 x 2 1 and
x 1 ∗ x 0 x 1 = x 1 x 0 x 1 + x 0 x 2 1 + x 2 0 x 1 , hence
x 1tt x 0 x 1 − x 1 ∗ x 0 x 1 = x 0 x 2 1 − x 2 0 x 1 . Fact: ζ (x 0 x 2 1 ) = ζ (x 2 0 x 1 ).
Euler : ζ (2, 1) = ζ (3).
14
Proposition. For w and w 0 in x 0 X ∗ x 1 , ζ (w)ζ (w 0 ) = ζ (w ∗ w 0 ),
ζ (w)ζ (w 0 ) = ζ (w tt w 0 ) and
ζ(x 1 tt w − x 1 ∗ w) = 0.
5. Symbolic Multizeta
Define Ze(s) for each s = (s 1 , . . . , s k ), with k ≥ 0 and s i ≥ 1. Next define Ze(w) for w in X ∗ x 1 by Ze(x s ) = Ze(s). Convergent symbols: Ze(s) with s 1 ≥ 2 or k = 0; these are the Ze(w) with w in x 0 X ∗ x 1 together with Ze(e) = Ze( ∅ ).
Algebra of Convergent MZV:
MZV conv is the commutative algebra over Q gener- ated by the convergent symbols Ze(s) with the rela- tions
Ze(w)Ze(w 0 ) = Ze(w ∗ w 0 ), Ze(w)Ze(w 0 ) = Ze(w tt w 0 ) and
Ze(x 1 tt w − x 1 ∗ w) = 0
for w and w 0 in x 0 X ∗ x 1 .
15
Main Diophantine Conjecture. The specializa- tion morphism from MZV conv into C which maps Ze(s) onto ζ (s) is injective.
Algebras MZV ∗ and MZV
tt: generators Ze(s) with s = (s 1 , . . . , s k ), k ≥ 0, s j ≥ 1, and ∗ (resp. tt ) defined by
Ze(w) ∗ Ze(w 0 ) = Ze(w ∗ w 0 ) resp.
Ze(w) tt Ze(w 0 ) = Ze(w tt w 0 ) for w ∈ X ∗ x 1 .
Remark.
x 1 ∗ x 1 = 2x 2 1 + x 0 x 1 and x 1 tt x 1 = 2x 2 1 , hence
Ze(x 1 ) ∗ Ze(x 1 ) = 2Ze(x 2 1 ) + Ze(x 0 x 1 ) while
Ze(x 1 ) tt Ze(x 1 ) = 2Ze(x 2 1 )
and ζ (x 0 x 1 ) = ζ (2) 6 = 0.
16
Conjecture of
Zagier, Drinfeld, Kontsevich and Goncharov.
For p ≥ 2 let d p denote the dimension of the Z - module in MZV conv spanned by the 2 p − 2 elements Ze(s) for s = (s 1 , . . . , s k ) of length p and s 1 ≥ 2.
Conjecture. We have
d 1 = 0, d 2 = d 3 = d 4 = 1 and
d p = d p − 2 + d p − 3 for p ≥ 4.
For each p ≥ 1, define Z p as the Q -vector space spanned by the Ze(s) with s convergent of weight p;
set Z 0 = Q . Then the sum of Z p (p ≥ 0) is direct, and the conjecture means
X
p ≥ 0
q p dim Q Z p = 1
1 − q 2 − q 3 ·
Remark: d p → ∞ by Rivoal’s result.
17
6. Further Results
Ecalle: for weight ´ ≤ 10, independent generators are Ze(2), Ze(3), Ze(5), Ze(7), Ze(9),
Ze(6, 2), Ze(8, 2).
Polylogarithms
Classical: for s ≥ 1 and | z | < 1, Li s (z) = X
n ≥ 1
z n n s .
Higher dimension: for s = (s 1 , . . . , s k ) with s i ≥ 1, Li s (z ) = X
n
1> ··· >n
k≥ 1
z n1
n s 11 · · · n s kk
Then
· · · n s kk
Then
Li uttv (z ) = Li u (z )Li v (z ) If s 1 ≥ 2, then Li s (1) = ζ (s).
Work of Petitot (Lille): the functions Li w , with w
in X ∗ , are linearly independent over C .