Multiple Zeta Values [Lecture notes]

51 Pages·2017·0.508 MB·English

by  Michel Waldschmidt

Updated Mars 2, 2017 MZV IMSc 2011
Special year in number theory at IMSc
Lectures on Multiple Zeta Values
IMSC 2011
by
Michel Waldschmidt
Here are the notes of the lectures given in Chennai IMSc in April 2011 for
theSpecialYearonNumberTheory,basedonnotestakenduringthelecturesby
TapasChatterjee. TheauthorisalsothankfultoSanoliGunfortheorganization
of this special year and for taking care of the publication of these notes.
1 Introduction to the results of F. Brown on
Zagier’s Conjecture
1.1 Zeta values: main Diophantine conjecture
Riemann zeta function
(cid:88) 1
ζ(s)=
ns
n≥1
has been considered before Riemann by Euler for integer values of the variable
s, both positive and negative ones. Among the many results he proved are
π2 ζ(2n)
ζ(2)= and ∈Q
6 ζ(2)n
for any integer n≥1.
A quite ambitious goal is to determine the algebraic relations among the
numbers
π, ζ(3), ζ(5),...,ζ(2n+1),...
The expected answer is disappointingly simple: it is widely believed that there
are no relations, which means that these numbers should be algebraically inde-
pendent:
Conjecture 1. For any n≥0 and any nonzero polynomial P ∈Z[T ,...,T ],
0 n
(cid:0) (cid:1)
P π,ζ(3),ζ(5),...,ζ(2n+1) (cid:54)=0.
Iftrue,thispropertywouldmeanthatthereisnointerestingalgebraicstructure.
There are very few results on the arithmetic nature of these numbers, even
less on their independence: it is known that π is a transcendental numbers,
hence, so are all ζ(2n), n ≥ 1. It is also known that ζ(3) is irrational (Ap´ery,
1978), and that infinitely many ζ(2n+1) are irrational (further sharper results
have been achieved by T. Rivoal and others – see [9, 14, 5]). But so far, it has
not been disproved that all these numbers lie in the ring Q[π2] (see the Open
Problem 2).
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1.2 Multizeta values: Zagier’s conjecture
The situation changes drastically if we enlarge our set so as to include the so-
called Multiple Zeta Values (MZV, also called Polyzeta values, Euler-Zagier
numbers or multiple harmonic series):
(cid:88) 1
ζ(s ,...,s )=
1 k ns1···nsk
n1>n2>···>nk≥1 1 k
which are defined for k, s ,...,s positive integers with s ≥ 2. There are
1 k 1
plenty of relations between them, providing a rich algebraic structure.
We will call k the length of the tuple s=(s ,...,s ) and |s|:=s +···+s
1 k 1 k
the weight of this tuple. There are 2p−2 tuples s of weight p with s ≥ 2 and
1
s ≥1 for 2≤j ≤k. The length k and the weight p are related by k+1≤p.
j
One easily gets quadratic relations between MZV when one multiplies two
such series: it is easy to express the product as a linear combination of MZV.
We will study this phenomenon in detail, but we just give one easy example.
Splittingthesetof(n,m)withn≥1andm≥1intothreedisjointsubsetswith
respectively n>m, m>n and n=m, we deduce, for s≥2 and s(cid:48) ≥2,
(cid:88)n−s (cid:88) m−s(cid:48) = (cid:88) n−sm−s(cid:48) + (cid:88) m−s(cid:48)n−s+(cid:88)n−s−s(cid:48),
n≥1 m≥1 n>m≥1 m>n≥1 n≥1
which is the so–called Nielsen Reflexion Formula:
ζ(s)ζ(s(cid:48))=ζ(s,s(cid:48))+ζ(s(cid:48),s)+ζ(s+s(cid:48))
for s≥2 and s(cid:48) ≥2. For instance,
ζ(2)2 =2ζ(2,2)+ζ(4).
Suchexpressionsoftheproductoftwozetavaluesasalinearcombinationofzeta
values, arising from the product of two series, will be called “stuffle relations”
(see §8.1).
They show that the Q–vector space spanned by the multiple zeta values is
in fact an algebra: a product of linear combinations of numbers of the form
ζ(s) is again a linear combination of such numbers. As mentioned above, it has
not been disproved that this algebra is Q[π2]. While looking at the result on
the formal symbols representing the MZV, one should keep in mind that the
following Open Problem 2 is not yet solved. Many results on these symbols
are known, but almost nothing is known on the kernel of the corresponding
specialization, which maps a symbol onto the corresponding real number.
We denote by Z the Q–vector space spanned by the numbers ζ(s). For
p≥2, we denote by Z the Q-subspace of Z spanned by the numbers ζ(s) with
p
|s| = p. For k ≥ 1, we denote by FkZ the Q-subspace of Z spanned by the
numbers ζ(s) with s of length ≤ k. Finally, for p ≥ k+1 ≥ 2, we denote by
FkZ theQ-subspaceofZspannedbythenumbersζ(s)with|s|ofweightpand
p
length ≤ k. The inclusion FkZ ⊂ FkZ∩Z is plain, that there is equality is
p p
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only a conjecture. It is also conjectured but not proved that the weight defines
a graduation on Z. It is a fact that the subspaces FkZ define an increasing
filtration of the algebra Z (see §5.4), butit is not proved that this filtration is
not the trivial one: for instance, it could happen that
FkZ =Z =Z=Q[π2]
p p
for all k ≥ 1 and p ≥ 2. For p ≥ 2, the space F1Z has dimension 1, it is
p
spannedbyζ(p). FromRivoal’sresultweknowthatZ (cid:54)=Qforinfinitelymany
p
odd p.
Open Problem 2. Is–it true that Z(cid:54)=Q[π2]?
Allknowlinearrelationsthatexpressamultizetaζ(s)asalinearcombination
of such numbers are homogeneous for the weight. The next conjecture (which
is also still open) is that any linear relations among these numbers splits into
homogeneous linear relations.
Conjecture 3. The Q–subspaces Z of R are in direct sum:
p
(cid:77)
Z ⊂R.
p
p≥2
This is equivalent to saying that the weight defines a graduation (see §5.1)
onthealgebraZ. AveryspecialcaseofConjecture3,whichisopen,isZ ∩Z =
2 3
{0}, which means that the number ζ(3)/π2 should be irrational.
For p ≥ 1, we denote by d the dimension over Q of Z (with d = 0); we
p p 1
also set d = 1. It is clear that d ≥ 1 for p ≥ 2, because ζ(p) is not zero.
0 p
For p ≥ 1 and k ≥ 1, we denote by d the dimension of FkZ /Fk−1Z with
p,k p p
d = 1 for p ≥ 1. We also set d = 1 and d = 0 for k ≥ 1, d = 0 for
p,1 0,0 0,k p,0
p≥1. We have, for all p≥0,
(cid:88)
d = d (4)
p p,k
k≥0
and d =0 for k ≥p≥1.
p,k
We have d = 1, since Z is spanned by ζ(2). The relation ζ(2,1) = ζ(3),
2 2
which is again due to Euler, shows that d = 1. Also the relations, essentially
3
going back to Euler,
1 3 2
ζ(3,1)= ζ(4), ζ(2,2)= ζ(4), ζ(2,1,1)=ζ(4)= ζ(2)2,
4 4 5
show that d = 1. These are the only values of d which are known. It is not
4 p
yet proved that there exists a p ≥ 5 with d ≥ 2. The upper bound ζ(5) ≤ 2
p
follows from the fact that there are 6 independent linear relations among the 8
numbers
ζ(5), ζ(4,1), ζ(3,2), ζ(3,1,1), ζ(2,3), ζ(2,2,1), ζ(2,1,2), ζ(2,1,1,1),
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and Z is the Q-vector subspace of R spanned by ζ(2,3) and ζ(3,2):
5
4 6
ζ(5)= ζ(3,2)+ ζ(2,3)=ζ(2,1,1,1),
5 5
1 1
ζ(4,1)=− ζ(3,2)+ ζ(2,3)=ζ(3,1,1),
5 5
ζ(2,2,1)=ζ(3,2)
ζ(2,1,2)=ζ(2,3).
ThedimensionofZ is2ifζ(2,3)/ζ(3,2)isirrational(whichisconjectured,but
5
not yet proved), and 1 otherwise.
Similarly, the Q–space Z has dimension ≤ 2, as it is spanned by ζ(2,2,2)
6
and ζ(3,3):
16
ζ(6)= ζ(2,2,2)=ζ(2,1,1,1,1),
3
4
ζ(5,1)= ζ(2,2,2)−ζ(3,3)=ζ(3,1,1,1),
3
16
ζ(4,2)=− ζ(2,2,2)+2ζ(3,3)=ζ(2,2,1,1),
9
7
ζ(4,1,1)= ζ(2,2,2)−2ζ(3,3),
3
59
ζ(3,2,1)=− ζ(2,2,2)+6ζ(3,3),
9
59
ζ(2,4)= ζ(2,2,2)−2ζ(3,3)=ζ(2,1,2,1),
9
34
ζ(2,3,1)= ζ(2,2,2)−3ζ(3,3)=ζ(3,1,2),
9
ζ(2,1,2,1)=ζ(3,3).
Here is Zagier’s conjecture on the dimension d of the Q-vector space Z .
p p
Conjecture 5 (Zagier). For p≥3, we have
d =d +d .
p p−2 p−3
Since d =1, d =0 and d =1, this conjecture can be written
0 1 2
(cid:88) 1
d Xp = ·
p 1−X2−X3
p≥0
It has been proved independently by Goncharov and Terasoma that the
numbers defined by the recurrence relation of Zagier’s Conjecture 5 with initial
values d = 1, d = 0 provide upper bounds for the actual dimension d . This
0 1 p
shows that there are plenty of linear relations among the numbers ζ(s). For
each p from 2 to 11, we display the number of tuples s of length p which is
2p−2, the number d given by Zagier’s Conjecture 5 and the difference which is
p
(a lower bound for) the number of linear relations among these numbers.
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p 2 3 4 5 6 7 8 9 10 11
2p−2 1 2 4 8 16 32 64 128 256 512
d 1 1 1 2 2 3 4 5 7 9
p
2p−2−d 0 1 3 6 14 29 60 123 249 503
p
Since d grows like a constant multiple of rp, where r =1.324 717 957 244 7...
p
is the real root of x3−x−1, the difference 2p−2−d is asymptotic to 2p−2.
p
AccordingtoZagier’sConjecture5,abasisforZ shouldbegivenasfollows:
p
p=2, d =1, ζ(2);
2
p=3, d =1, ζ(3);
3
p=4, d =1, ζ(2,2);
4
p=5, d =2, ζ(2,3), ζ(3,2);
5
p=6, d =2, ζ(2,2,2), ζ(3,3);
6
p=7, d =3, ζ(2,2,3), ζ(2,3,2), ζ(3,2,2);
7
p=8, d =4, ζ(2,2,2,3), ζ(2,3,3), ζ(3,2,3), ζ(3,3,2);
8
p=9, d =5, ζ(2,2,2,3), ζ(2,2,3,2), ζ(2,3,2,2), ζ(3,2,2,2)ζ(3,3,3).
8
For these small values of p, the dimension d of FkZ /Fk−1Z is conjec-
p,k p p
turally given by the number of elements in the box (p,k) of the next figure,
whereconjecturalgeneratorsofFkZ /Fk−1Z shouldbegivenbytheclassesof
p p
the following MZV:
p
2 3 4 5 6 7 8 9
k
1 ζ(2) ζ(3) ζ(4) ζ(5) ζ(6) ζ(7) ζ(8) ζ(9)
ζ(8,1)
ζ(6,1) ζ(7,1)
2 ζ(4,1) ζ(5,1) ζ(7,2)
ζ(5,2) ζ(6,2)
ζ(6,3)
3 ζ(6,1,1) ζ(6,2,1)
d 1 1 1 2 2 3 4 5
p
The displayed elements should all be linearly independent over Q. The
numerical computations have been performed online thanks to the computer
program EZface [1].
1.3 Known results
AconjecturebyM.HoffmanisthatabasisofZ overQisgivenbythenumbers
p
ζ(s ,...,s ), s +···+s = p, where each s is either 2 or 3: the dimension
1 k 1 k i
agrees with Conjecture 5.
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As a side product of his main recent results, F. Brown [7, 8] obtains:
Theorem 6. The numbers ζ(s ,...,s ), with k ≥ 1 and s ∈ {2,3}, span the
1 k j
Q–space Z of multizeta.
Hence, Conjecture 5 is equivalent to saying that the numbers ζ(s ,...,s ),
1 k
withk ≥1ands ∈{2,3},occurringinTheorem6areQ–linearlyindependent.
j
OneoftheauxiliaryresultwhichwasneededbyF.Brownisaformulawhich
he conjectured and which has been established by D. Zagier (see §2).
1.4 Occurrences of powers of π2 in the set of generators
ThespaceF1Z isspannedbyπ2n overQ. ThenextProposition7showsthat
2n
the numbers π2n occur in the set of generators ζ(s) with s ∈ {2,3} given by
j
Theorem 6, by taking all s equal to 2.
j
For s≥2 and n≥1, we use the notation {s} for a string with n elements
a
all equal to s, that is {s} =(s ,...,s ) with s =···=s =s.
n 1 n 1 n
Proposition 7. For s≥2,
 
(cid:88)ζ(cid:0){s}n(cid:1)xn = (cid:89)(cid:18)1+ jxs(cid:19)=exp(cid:88)(−1)k−1kxkζ(sk).
n≥0 j≥1 k≥1
The proof will involve the infinite product
(cid:18) (cid:19)
(cid:89) x
F (x)= 1+ .
s js
j≥1
Proof. Expanding F (x) as a series:
s
(cid:88) 1 (cid:88) 1
F (x)=1+x +x2 +···
s js (j j )s
1 2
j≥1 j1>j2≥1
=1+xζ(s)+x2ζ(s,s)+···
yields the first equality in Proposition 7. For the second one, consider the
logarithmic derivative of F (x):
s
F(cid:48)(x) (cid:88) 1 (cid:88) 1 (cid:88) xk−1
s = = (−1)k−1
F (x) js+x js js(k−1)
s
j≥1 j≥1 k≥1
(cid:88) (cid:88) 1 (cid:88)
= (−1)k−1xk−1 = (−1)k−1xk−1ζ(sk).
jsk
k≥1 j≥1 k≥1
(cid:0) (cid:1)
Since F (0)=ζ {s} =1, Proposition 7 follows by integration.
s 0
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Lemma 8. We have
sin(πz)
F (−z2)= ·
2 πz
Proof. Lemma 8 follows from the product expansion of the sine function (see
for instance [18], Chap. 7, § 4.1, (27))
(cid:89)∞ (cid:18) z2 (cid:19)
sinz =z 1− .
n2π2
n=1
From Proposition 7 and Lemma 8 we deduce:
Corollary 9. For any n≥0,
(cid:0) (cid:1) π2n
ζ {2} = ·
n (2n+1)!
Proof. From the Taylor expansion of the sine function
(cid:88) z2k+1
sinz = (−1)k
(2k+1)!
k≥0
and from Lemma 8, we infer
sin(πz) (cid:88) π2kz2k
F (−z2)= = (−1)k ·
2 πz (2k+1)!
k≥0
Corollary 9 now follows from Proposition 7.
2 Zagier’s contribution to Brown’s proof
WeexplainthestrategyofZagier[21]forprovingtherelationwhichwasneeded
and conjectured by F. Brown [7, 8] concerning the numbers
H(a,b):=ζ({2} 3{2} )
b a
fora≥0andb≥0. InparticularH(0,0)=ζ(3). Bewarethatthenormalization
used by F. Brown and by D. Zagier [7, 8, 21] is the opposite of ours; this is why
a and b are in this reverse order here.
Another reference to this topic is the course [2] on MZV by J. Borwein and
W. Zudilin at the University of Newcastle in 2011–2012.
Set also (cf. Corollary 9)
π2n
H(n)=ζ({2} )=
n (2n+1)!
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for n≥0, with H(0)=1.
Consider the alphabet {2,3}; give to the letter 2 the weight 2 and to the
letter3theweight3,sothattheword2b·3·2a hasweight2a+2b+3,whilethe
word2n hasweight2n. Givealsotheweight(cid:96)toζ((cid:96))for(cid:96)≥2(thisisanabuse
oflanguage,sincewedonotknowwhethertheweightdefinesagraduation–see
Conjecture3). Lookingathomogeneousrelations,oneconsidersontheoneside
the numbers H(a,b) and on the other side the numbers ζ((cid:96))H(m) with (cid:96) ≥ 2
and m ≥ 0, restricted to the relation 2a+2b+3 = (cid:96)+2m. Notice that this
implies that the number (cid:96)=2a+2b+3−2m is odd.
Theorem 10 (Zagier, 2011). Let a and b be non–negative integers. Set k =
2a+2b+3. Then there exist a+b+1 rational integers c with m ≥ 0,
m,r,a,b
r ≥1, m+r =a+b+1, such that
(cid:88)
H(a,b)= c H(m)ζ(2r+1).
m,r,a,b
m+r=a+b+1
Conversely, given two integers r and m with r ≥1 and m≥0, there exist m+r
rational numbers c(cid:48) with a≥0, b≥0, a+b=m+r−1, such that
m,r,a,b
(cid:88)
H(m)ζ(2r+1)= c(cid:48) H(a,b).
m,r,a,b
a+b=m+r−1
The integers c are explicitly given:
m,r,a,b
(cid:20)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)(cid:21)
2r 1 2r
c =2(−1)r − 1− .
m,r,a,b 2a+2 22r 2b+1
It follows that the square matrices (cid:0)c (cid:1) and (cid:0)c(cid:48) (cid:1), of size a+b+1 =
m,r,a,b m,r,a,b
m+r, are inverse matrices.
We only give the sketch of proof of the first part of Theorem 10. Consider
the generating series
(cid:88)(cid:88)
F(x,y)= (−1)a+b+1H(a,b)x2a+2y2b+1
a≥0b≥0
and
(cid:88)(cid:88)
F(cid:98)(x,y)= (−1)a+b+1H(cid:98)(a,b)x2a+2y2b+1,
a≥0b≥0
where
a+b
(cid:88)
H(cid:98)(a,b)= cm,a,bH(m)ζ(k−2m).
m=0
The first step relates F(x,y) to a hypergeometric series F , namely F(x,y) is
3 2
the product of (1/π)sin(πy) by the z–derivative at z =0 of the function
(cid:18) (cid:12) (cid:19)
x, −x, z (cid:12)
3F2 1+y, 1−y(cid:12)(cid:12)1 .
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The second step relates F(cid:98)(x,y) to the digamma function ψ(x) = Γ(cid:48)(x)/Γ(x)
(logarithmicderivativeofΓ),namelyF(cid:98)(x,y)isalinearcombinationoffourteen
functions of the form
(cid:16) u(cid:17)sin(πv)
ψ 1+ with u∈{±x±y, ±2x±2y, ±2y} and v ∈{x,y}.
2 2π
The third step is the proof that F and F(cid:98) are both entire function on C×C,
they are bounded by a constant multiple of eπXlogX, when X =max{|x|,|y|}
tends to infinity, and also by a constant multiple of eπ|(cid:61)(y)|, when |y| tends to
infinity while x∈C is fixed.
The fourth step is about the diagonal: for z ∈C, we have
F(z,z)=F(cid:98)(z,z).
Several equivalent explicit formula for this function are given. Let
π
A(z):=− F(z,z).
sin(πz)
Then
(cid:88)∞ (cid:88)∞ z2
A(z)= ζ(2r+1)z2r =
n(n2−z2)
r=1 n=1
and
∞ (cid:18) (cid:19)
1 (cid:88) 1 1 2 1(cid:0) (cid:1)
A(z)= + − =ψ(1)− ψ(1+z)+ψ(1−z) .
2 n−z n+z n 2
n=1
ThefifthstepshowsthatF(n,y)andF(cid:98)(n,y)areequalwhenn∈Nandy ∈C,
an explicit formula is given.
ThesixthstepshowsthatF(x,k)andF(cid:98)(x,k)areequalwhenk ∈Nandx∈C,
an explicit formula is given.
Now comes the conclusion, which rests on the next lemma:
Lemma 11. Anentirefunctionf :C→Cthatvanishesatallrationalintegers
and satisfies
(cid:16) (cid:17)
f(z)=O eπ|(cid:61)(z)|
is a constant multiple of sin(πz).
A proof of this lemma, using a Theorem of Phragm´en–Lindel¨of, is given by
Zagier (Lemma 2 in [21]), but he also notices that other references have been
given subsequently to him, in particular by F. Gramain who pointed out that
this lemma is known since the work of P´olya and Valiron; see [6]
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3 Lyndon words: conjectural transcendence ba-
sis
D. Broadhurst considered the question of finding a transcendence basis for the
algebra of MZV, he suggested that one should consider Lyndon words. We first
give the definitions.
Here, we use the alphabet alphabet A = {a,b}. The set of words on A is
denoted by A∗ (see §6.2). The elements of A∗ can be written
an1bm1···ankbmk
with k ≥ 0, n ≥ 0, m ≥ 0 and m ≥ 1 for 1 ≤ i < k, and with n ≥ 1
1 k i i
for 2 ≤ i ≤ k. We endowed A∗ with the lexicographic order with a < b. A
Lyndon word is a non–empty word w ∈ A∗ such that, for each decomposition
w = uv with u (cid:54)= e and v (cid:54)= e, the inequality w < v holds. Denote by L the
set of Lyndon words. Examples of Lyndon words are a, b, abk (k ≥ 0), a(cid:96)b
((cid:96) ≥ 0), a2b2, a2bab. Let us check, for instance, that a2bab is a Lyndon word:
this follows from the observation that a2bab is smaller than any of
abab, bab, ab, b.
But a2ba2b is not a Lyndon word, since a2ba2b>a2b.
Any Lyndon word other than b starts with a and any Lyndon word other
than a ends with b.
Herearethe21Lyndonwordsonthealphabet{a,b}withweight≤15,when
a has weight 2 and b weight 3:
a<a6b<a5b<a4b<a4b2 <a3b<a3bab<a3b2 <a3b3 <a2b<a2bab
<a2bab2 <a2b2 <a2b2ab<a2b3 <ab<abab2 <ab2 <ab3 <ab4 <b.
We list them according to their weight p = 2,...,15, we display their number
N(p)andthecorrespondingmultiplezetavalues,wherethewordan isreplaced
by the tuple {2} and bn by {3} :
n n
p=2; N(2)=1; a; ζ(2)
p=3; N(3)=1; b; ζ(3)
p=4; N(4)=0
p=5; N(5)=1; ab; ζ(2,3)
p=6; N(6)=0
p=7; N(7)=1; a2b; ζ(2,2,3)
p=8; N(8)=1; ab2; ζ(2,3,3)
p=9; N(9)=1; a3b; ζ(2,2,2,3)
p=10; N(10)=1; a2b2; ζ(2,2,3,3)
p=11; N(11)=2; a4b, ab3; ζ(2,2,2,2,3), ζ(2,3,3,3)
p=12; N(12)=2; a3b2, a2bab; ζ(2,2,2,3,3), ζ(2,2,3,2,3)
p=13; N(13)=3; a5b, a2b3, abab2;
ζ(2,2,2,2,2,3), ζ(2,2,3,3,3), ζ(2,3,2,3,3)
10