Introduction to Multiple Zeta Values, I

M. Waldschmidt

August 22, 2001

▶ What is the arithmetic nature of a real number x ∈ ℝ? We can ask whether x is rational, or algebraic, or transcendental. Very little is known in terms of decimal expansions.

There is a recent conjecture of Kontsevich and Zagier in their joint paper “periods” (Mathematics Unlimited - 2001 and Beyond. Engquist, B.; Schmid, W., (Eds.), Springer (2000), 771–808). But it is too vague to be disproved by counterexamples.

Examples. x = ∑_{n≥0} \frac{P (n)}{Q(n)}, where P, Q ∈ ℤ[x], with deg Q ≥ deg P + 2. We omit those finitely

many n ≥ 0 with Q(n) = 0 in the sum.

Example 1. ∑_{n≥1} \frac{1}{n(n + 1)}· This is a telescoping series, where the general term can be written as

n^{−1} − (n + 1)^{−1}. So the sum converges to 1, a rational number.

Similarly, we can consider ∑ \frac{1}{(n + a)(n + b)}, with a, b ∈ ℤ, a ≠ b.

Example 2. ∑_{n≥0} \frac{1}{(2n + 1)(2n + 2)}· The general term is (2n + 1)^{−1} − (2n + 2)^{−1}. The series is now

seen to be convergent to log 2, which is transcendental by Hermite-Lindeman.

Example 3. ∑_{n=0}^{∞} \frac{1}{(n + 1)(2n + 1)(4n + 1)} = \frac{π}{3} ·

Example 4. ∑_{n=0}^{∞} ∏_{i=1}^{6} \frac{1}{6n + i} = \frac{1}{4320}(192 log 2 − 81 log 3 − 7π\sqrt{3})·

Write π as i log(−1), we can use Baker’s theorem to see that this number is transcendental. Baker’s theorem asserts that a finite sum ∑ β_{i} log α_{i}, with algebraic α_{i}, β_{i}, is transcendental whenever the sum is non-zero.

▶ The work of Adhikari, Saradha, Shorey, Tijdeman.

Consider x = ∑_{n≥0} \frac{P (n)}{Q(n)}, where Q has only simple rational zeros and deg Q ≥ 2 + deg P. Then

we can rewrite the series in the following form

x = \frac{a}{b} + ∑_{n≥0} ∑_{j=1}^{m} \frac{c_{j}}{k_{j}n + r_{j}} ,

with 0 < r_{j} ≤ k_{j}.

Define

f(z) = ∑_{n≥0} \frac{z^{n}}{kn + r} ·

This can be analyzed by considering g(z) = z^{r}f(z^{k}), then zg′(z) = z^{r}/(1 − z^{k}). We then get

x = \frac{a}{b} + ∑_{j=1}^{m} ∑_{l=0}^{k_{j}−1} \frac{c_{j}}{k_{j}} (1 − ζ_{k_{j}}^{r_{j}l}) log(1 − ζ_{k_{j}}^{l}),

where ζ_{k_{j}} = e(1/k_{j}), e(z) = exp(2πiz). Thus we conclude that x is either rational or transcendental. It can never be an algebraic irrationality. The proof also gives a way to decide whether x is rational or transcendental. If it is rational, it is just a/b.

Example. ∑ \frac{1}{(5n + 1)(5n + 3)(5n + 5)} is transcendental.

Example. ∑ (\frac{1}{4n + 1} − \frac{3}{4n + 2} + \frac{1}{4n + 3} + \frac{1}{4n + 4}) · This turns out to be rational as the “tran-

scendental part” is log 2−log(1+i)−log(1−i) is 0, but this is not apparent without going through the computations.

▶ Now consider the case that Q has distinct roots, but not rational.

Example. ∑_{n≥0} \frac{1}{n^{2} + 1} = \frac{1}{2} + \frac{π}{2} · \frac{e^{π} + e^{−π}}{e^{π} − e^{−π}} · This is transcendental by a theorem of Nesterenko: π

and e^{π} are algebraic independent.

Schanuel’s conjecture. Let x_{1}, …, x_{n} be complex numbers linearly independent over ℚ. Then the transcendence degree of ℚ(x_{1}, …, x_{n}, e^{x_{1}}, …, e^{x_{n}})/ℚ is at least n.

Question. What is a function field version of this conjecture related to Drinfeld modules?

Bundschuh’s formula. For s ∈ ℤ, s ≥ 2,

∑_{|n|≥2,n∈ℤ} \frac{1}{n^{s} − 1} = 2 − \frac{δ_{s}}{2s} − \frac{πi}{s} ∑_{σ} ζ_{σ} \frac{e(ζ_{s}^{σ}) + 1}{e(ζ_{s}^{σ}) − 1}·

Here, δ_{s} = 1 or 2 depending on s is odd or even, and ζ_{s} = e(1/s), and 1 ≤ σ ≤ s, σ ≠ s/2.

Example. Let s = 2, then we get 2 ∑_{n≥2} \frac{1}{n^{2} − 1} = \frac{3}{2} (telescoping).

Exercise. Let s = 4. We get 2 ∑_{n≥2} \frac{1}{n^{4} − 1} = \frac{7}{4} − \frac{π}{2} · \frac{e^{π} + e^{−π}}{e^{π} − e^{−π}} ·

Proof of Bundschuh’s formula. We use residue calculus. Consider

\frac{1}{2πi} ∫_{Γ_{N}} \frac{π cot(πz)}{z^{s} − 1} dz = I(N),

where I_{N} is the square with side length 2N + 1 centered at the origin. The poles are ℤ ∪ μ_{s}. Be careful at z = ±1, where we may have second order poles.

The case of multiple zeroes for Q. For example,

∑_{n≥1} \frac{1}{n^{2}} = \frac{π^{2}}{6} ·

This can be done by the same method of residue calculus. Here we give a proof of Calabi, which appeared in a recent Bourbaki talk (Cartier, 2001).

Consider

∫_{0}^{1} ∫_{0}^{1} \frac{dx dy}{1 − x^{2}y^{2}} = ∫_{0}^{1} ∫_{0}^{1} ∑_{n=0}^{∞}(x^{2}y^{2})^{n} dx dy = ∑_{n=0}^{∞} \frac{1}{(2n + 1)^{2}} ·

Now apply change of variables: x = sin u/ cos v, y = sin v/ cos u, so that dx dy/(1 − x^{2}y^{2}) = du dv. The new region of integration is 0 ≤ u ≤ π/2, 0 ≤ v ≤ π/2, u + v ≤ π/2. This gives that the integral is π^{2}/8, from where the desired result follows.

Riemann’s zeta.

ζ(s) = ∑_{n≥1} \frac{1}{n^{s}} ·

Consider the case that s is an integer ≥ 2. It was known to Euler that

ζ(2r) = (−1)^{r+1} \frac{2^{2r−1}B_{2r}}{(2r)!} π^{2r}

for r ≥ 1, where \frac{t}{e^{t} − 1} = ∑_{n=0}^{∞} \frac{B_{n}t^{n}}{n!} · In particular, B_{n} ∈ ℚ.

From here, we know that ζ(2r) are rational multiples of π^{2r}, hence are transcendental. But only relatively recently, it is shown that ζ(3) is irrational (Apéry). Now we know that there are infinitely many ζ(2r + 1), r ≥ 1 which are irrational (Rivoal, 2000, C.R.A.S.). In fact, the dimension of the

ℚ-vector space spanned by ζ(3), …, ζ(a), (a: odd) is ≥ \frac{log a}{1 + log 2}(1 + o(1)) (Ball, Rivoal, 2001,

Invent. Math.). Moreover, Rivoal and Zudilin showed that one of the 9 numbers ζ(5), …, ζ(21) is irrational in June 2001. In July, Zudilin improved 9 to 8, then recently to 4: one of the 4 numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational.

Expectation. The numbers π, ζ(3), ζ(5), …, ζ(2r + 1), … are algebraically independent.

Multiple zeta values. We want to consider

∑_{n_{1}>···>n_{k}≥1} \frac{1}{n_{1}^{s_{1}} ⋯ n_{k}^{s_{k}}} = ζ(s_{1}, …, s_{k}) = ζ(\underline{s}).

Here, s_{1}, …, s_{k} ∈ ℤ, s_{1} ≥ 2, s_{j} ≥ 1 for 2 ≤ j ≤ k. For k = 1, this reduces to Riemann’s zeta function. We call k the length or depth of \underline{s}, and |\underline{s}| = ∑ s_{j} the weight of \underline{s}.

Remark. These multiple zeta values have been considered by J. Écalle in the context of “resurgent series”where he introduced “mould calculus”. Roughly, for each \underline{s}, he introduced an object M\underline{s}, in a category, say of modules over a ring A, with lots of symmetries.

Multiple zeta values have also been considered by N. Nielsen, A.B. Goncharov and D. Zagier.

Fact. There are many algebraic relations among ζ(\underline{s}).

Now we feel that we have found all the relations and there should be no more. It is not easy to check whether a “new” relation is a consequence of known ones.

Simple example. ζ(s)ζ(s′) = ∑_{n≥1} \frac{1}{n^{s}} ∑_{m≥1} \frac{1}{m^{s′}} = ∑_{n>m} + ∑_{n<m} + ∑_{n=m} = ζ(s, s′) + ζ(s′, s) + ζ(s + s′).

This is an example of a quadratic relation, generally of the form

ζ(\underline{s})ζ(\underline{s}′) = ∑_{\underline{σ}} ζ(\underline{σ}).

Some non-commutative polynomial algebra ℚ⟨X⟩, X = {x_{0}, x_{1}}, and shuffle products, ⋯ will be used to describe the relations.

Introduction to Multiple Zeta Values, II

M. Waldschmidt

August 22, 2001

Remark. Adhikari-Saradha-Shorey-Tijdeman also considered sums of the form

∑_{n≥0} \frac{f(n)}{Q(n)},

where Q is a polynomial with simple rational zeros, and f is periodic modulo q > 0, such that the convergence condition is satisfied.

Corollary. Let χ be a Dirichlet character modulo q, not principal, then

L(χ, 1) = ∑_{n≥1} \frac{χ(n)}{n}

is transcendental.

Remark. But this is not a new way to prove the non-vanishing of L(χ, 1).

Free algebras.

Let K be a field (to be specialized to a subfield of ℝ later; three main examples: ℚ, ℚ̄ ∩ ℝ, ℝ). Let X be a set. Consider the universal problem of constructing a K-algebra K⟨X⟩ and a map

X → K⟨X⟩ such that for any K-algebra A, any map f : X → A, there is a unique extension f̄ making the diagram

\xymatrix{X \ar[r]^{f} \ar[dr] & A \\  & K⟨X⟩ \ar[u]_{f̄}}

commutative.

If X = {x}, the solution is just that K⟨x⟩ = K[x], the polynomial algebra in a single variable. In general polynomial algebras solve the problem if we replace algebras by commutative algebras.

Let X^{∗} be the set of words on X, that is, monomials x_{1} ⋯ x_{n}, n ≥ 0, x_{i} ∈ X. Denote by e the empty product. Take X^{∗} as the basis of K⟨X⟩ as a K-vector space. We can think of K⟨X⟩ as K^{(X^{∗})}, the space of functions from X^{∗} to K with finite supports.

Thus elements of K⟨X⟩ are finite sums s = ∑_{w∈X^{∗}}(s|w)w. The addition is defined componentwise, so (s + s′|w) = (s|w) + (s′|w). Multiplication is defined by concatenation, so (ss′|w) = ∑_{u,v∈X^{∗},uv=w}(s|u)(s′|v).

Let H = K⟨X⟩. This is a non-commutative graded algebra. The grading is defined by assigning weight n to a word x_{1} ⋯ x_{n}. Let H_{n} be the subspace spanned by words of length n. Clearly, H_{n}.H_{m} ⊂ H_{n+m}.

Most important to us is the case X = {x_{0}, x_{1}}. We will write H = K⟨x_{0}, x_{1}⟩.

Fact. The algebra K⟨Y ⟩ with Y = {y_{1}, y_{2}, …} is isomorphic to a subalgebra of K⟨x_{0}, x_{1}⟩.

This is very different from the case of free commutative algebras. To see this, we write a map Y → K⟨X⟩ by y_{s} ↦ x_{0}^{s−1}x_{1} for s ≥ 1. By the universal property, we get a morphism f : K⟨Y ⟩ → K⟨X⟩. It suffices to show that this is injective, which is not difficult.

The image H^{1} of f is a subalgebra of H, and is precisely the subalgebra spanned by words which end with x_{1}. The set of such words is denoted by X^{∗}x_{1}. We have H^{1} = K.e + Hx_{1}.

Notation. For \underline{s} = (s_{1}, …, s_{k}), we write y\underline{s} = y_{s_{1}} … y_{s_{k}}.

The words y_{s_{1}} ⋯ y_{s_{k}} with s_{1} ≥ 2 are the words which start with x_{0} and end with x_{1}.

We define the set of convergent words to be {e} ∪ (x_{0}X^{∗}x_{1}), and a subalgebra H^{0} = K.e + x_{0}.H.x_{1}. We have H^{0} ⊂ H^{1} ⊂ H.

Now assume that K ⊂ ℝ. We define ζ̂: H^{0} → ℝ to be the K-linear map defined by ζ̂(y\underline{s}) = ζ(\underline{s}) for \underline{s} = (s_{1}, …, s_{k}) with s_{1} ≥ 2.

Goal. Define a law on H^{0}, denoted by ∗, such that

ζ̂(y\underline{s})ζ̂(y_{\underline{s}}′) = ζ̂(y\underline{s} ∗ y_{\underline{s}}′).

Notice that the left hand side is simply ζ(\underline{s}) · ζ(\underline{s}′).

Now

ζ(\underline{s}) · ζ(\underline{s}′) = ∑_{n_{1}>···>n_{k}≥1} \frac{1}{n_{1}^{s_{1}} ⋯ n_{k}^{s_{k}}} · ∑_{n′_{1}>···>n′_{k′}≥1} \frac{1}{n′_{1}^{s′_{1}} ⋯ n′_{k′}^{s′_{k′}}} ·

We rewrite this as ∑\underline{σ} ζ(\underline{σ}) and define y\underline{s} ∗ y_{\underline{s}}′ = ∑\underline{σ} y\underline{σ}.

Definition. We define ∗ by induction on k + k′: for s ≥ 2, s′ ≥ 2, and u, v ∈ X^{∗}, we define

(y_{s}u) ∗ (y_{s′}v) = y_{s}(u ∗ (y_{s′}v)) + y_{s′}((y_{s}u) ∗ v) + y_{s+s′}(u ∗ v).

(Cf. the quadratic relation computation for ζ(s)ζ(s′)).

Now we have defined ∗ on H^{0}. We will extend it to H^{1} and H. We do so by defining x_{0}^{m} ∗ u = u ∗ x_{0}^{m} = ux_{0}^{m}. Then H is a commutative algebra for (+, ∗). The commutativity is obvious. This commutative algebra will be denoted by H_{∗}, and called the harmonic algebra. The subspace H^{0} and H^{1} are subalgebras, denoted by H_{∗}^{0} and H_{∗}^{1}. By definition, ζ̂ : H_{∗}^{0} → ℝ is a morphism of K-algebras.

The relation

ζ̂(u ∗ v) = ζ̂(u) · ζ̂(v) (stuffle product)

is a quadratic relation among zeta values.

Examples.

ζ(2)^{2} = 2ζ(2, 2) + ζ(4),
ζ(2)^{3} = 6ζ(2, 2, 2) + 3ζ(2, 4) + 3ζ(4, 2) + ζ(6).

Quasi-symmetric functions. Consider infinitely many commuting variables t_{1}, …, t_{n}, … and power series in these variables (with coefficients in K). The algebra of such power series is denoted by K[[\underline{t}]].

The symmetric functions form a subalgebra Sym of K[[\underline{t}]], which is spanned by

M\underline{s}(\underline{t}) = ∑_{n_{1}≥1,...,n_{k}≥1} t_{n_{1}}^{s_{1}} ⋯ t_{n_{k}}^{s_{k}}.

In fact, the M\underline{s}’s with (s_{1} ≥ ⋯ ≥ s_{k} ≥ 1) form a basis Sym.

The algebra Q Sym of quasi-symmetric functions is spanned by

QM\underline{s}(\underline{t}) = ∑_{n_{1}>···>n_{k}≥1} t_{n_{1}}^{s_{1}} ⋯ t_{n_{k}}^{s_{k}},

where s_{1}, …, s_{k} ≥ 1. This contains the symmetric functions as M\underline{s} ∼ ∑\underline{t} QM\underline{t}, where \underline{t} range over permutations of \underline{s}.

There is an isomorphism of algebras from H_{∗}^{1} and Q Sym, such that y\underline{s} ↦ QM\underline{s}(\underline{t}) = F (y\underline{s}), where F (y\underline{s}) = ∑_{n_{1}>···>n_{k}≥1} t_{n_{1}}^{s_{1}} ⋯ t_{n_{k}}^{s_{k}} extended by linearity. We have

F (y\underline{s}) · F (y_{\underline{s}}′) = F (y\underline{s} ∗ y_{\underline{s}}′).

Hoffman shows that H^{∗} is a polynomial algebra on the so-called Lyndon words.

We put an order on X^{∗} by using the lexicographic order with x_{0} < x_{1}. For example, for words of length ≤ 2:

x_{0} < x_{0}^{2} < x_{0}x_{1} < x_{1} < x_{1}x_{0} < x_{1}^{2}.

Then a Lyndon word is a word w which is smaller than each proper right factor: if w = uv then w < v. Lyndon words of length ≤ 2 are x_{0}, x_{0}x_{1}, x_{1}.

Each Lyndon word different from x_{0}, x_{1} starts with x_{0} and ends with x_{1}.

Consequence. Using this product, we can not get algebraic relations between ζ̂(x_{0}^{2}x_{1}) and ζ̂(x_{0}x_{1}^{2}) (x_{0}^{2}x_{1} and x_{0}x_{1}^{2} are Lyndon words). But the first one is ζ(3) and the second one ζ(2, 1), and they are actually the same. Thus we have not captured all relations.

Exercises. Check ζ(2, …, 2) (k copies of 2’s) is equal to \frac{π^{2k}}{(2k + 1)!}·

Hint. ∑_{k≥0} \frac{π^{2k}}{(2k + 1)!}z^{2k}(−1)^{k} = \frac{sin(πz)}{πz} = ∏_{n≥1} (1 − \frac{z^{2}}{n^{2}} ) ·

∏_{n≥1}(1 − zt_{n}) = ∑_{k≥0}(−1)^{k}λ_{k}(\underline{t})z^{n},

where λ_{k}(t) = ∑_{n_{1}>···>n_{k}≥1} t_{n_{1}} ⋯ t_{n_{k}} = QM_{(1,...,1)} (k copies of 1’s).

Introduction to Multiple Zeta Values, III

M. Waldschmidt August 24, 2001

▶ Chen’s iterated integrals.

Let φ_{1}, …, φ_{p} be holomorphic differential forms on some open set of ℂ. We define

∫_{a}^{b} φ_{1} ⋯ φ_{p}

by induction: (when p = 1, it is the usual integral)

∫_{a}^{b} φ_{1} ⋯ φ_{p} = ∫_{a}^{b} φ_{1}(t) ∫_{a}^{t} φ_{2} ⋯ φ_{p}.

By means of a change of variable we may assume that a = 0 and b = 1.

We have the following explicit formula:

∫_{0}^{1} φ_{1} ⋯ φ_{p} = ∫_{1>t_{1}>···>t_{p}>0} φ_{1}(t_{1}) ⋯ φ_{p}(t_{p}).

These iterated integrals occur naturally when one studies product of integrals:

∫_{0}^{1} φ_{1} ∫_{0}^{1} φ_{2} = ∫_{1>t>0,1>u>0} φ_{1}(t)φ_{2}(u) = ∫_{1>u>t>0} + ∫_{1>t>u>0} = ∫_{0}^{1} φ_{1}φ_{2} + ∫_{0}^{1} φ_{2}φ_{1}.

Lemma. Define

S_{p,q} = {σ ∈ S_{p+q} : σ(1) < ⋯ < σ(p), σ(p + 1) < ⋯ < σ(p + q)}.

Then

∫_{0}^{1} φ_{1} ⋯ φ_{p} ∫_{0}^{1} φ_{p+1} ⋯ φ_{p+q} = ∑_{σ∈S_{p,q}} ∫_{0}^{1} φ_{σ(1)} ⋯ φ_{σ(p+q)}.

The subset S_{p,q} ⊂ S_{p+q} is called the set of (p, q)-shuffles. Proof. The idea is in the above simple example.

LHS = ∫_{1>t_{1}>···>t_{p}>0,1>t_{p+1}>···>t_{p+q}} φ_{1}(t_{1}) ⋯ φ_{p+q}(t_{p+q}).

Now use the fact that the domain of integration is a product of two simplices, and decomposes into the disjoint union of {1 > t_{σ(1)} > ⋯ > t_{σ(p+q)}} for σ ∈ S_{p,q}. Lemma.

∫_{x_{0}}^{x_{1}} φ_{1} ⋯ φ_{p} = ∑_{j=0}^{p} ∫_{x}^{x_{1}} φ_{1} ⋯ φ_{j} ∫_{x_{0}}^{x} φ_{j+1} ⋯ φ_{p}.

Proof. This is an exercise. Also an exercise: this is not true with ∫_{x_{0}}^{x} φ_{1} ⋯ φ_{j} ∫_{x}^{x_{1}} φ_{j+1} ⋯ φ_{p}.

▶ Polylogarithms.

The classical polylogarithms are

Li_{s}(z) = ∑_{n≥1} \frac{z^{n}}{n^{s}} ·

Here, s ≥ 1 is an integer, z ∈ ℂ is such that |z| < 1 (or s ≥ 2 and |z| ≤ 1). We have ζ(s) = Li_{s}(1). Now

Li_{1}(z) = ∑_{n≥1} \frac{z^{n}}{n} = − log(1 − z) = ∫_{0}^{z} \frac{dt}{1 − t}

z \frac{d}{dz} Li_{s}(z) = Li_{s−1}(z), s ≥ 2.

So

Li_{s}(z) = ∫_{0}^{z} Li_{s−1}(t) \frac{dt}{t}, s ≥ 2.

By induction,

Li_{s}(z) = ∫_{0}^{z} \frac{dt_{1}}{t_{1}} ⋯ \frac{dt_{s−1}}{t_{s−1}} \frac{dt_{s}}{1 − t_{s}} ·

Put ω_{0} = dt/t, ω_{1} = dt/(1 − t), so that

Li_{s}(z) = ∫_{0}^{z} ω_{0}^{s−1}ω_{1} = ∫_{0}^{z} ω_{s},

where ω_{s} = ω_{0}^{s−1}ω_{1} for s ≥ 1. Multiple polylogarithms in a single variable. Let \underline{s} = (s_{1}, …, s_{k}). Define

Li\underline{s}(z) = ∑_{n_{1}>···>n_{k}≥1} \frac{z^{n_{1}}}{n_{1}^{s_{1}} ⋯ n_{k}^{s_{k}}} ,

where s_{i} ≥ 1 and |z| < 1. Also we can allow |z| ≤ 1 if s_{1} ≥ 2. We have Li\underline{s}(1) = ζ(\underline{s}). Observe

\frac{d}{dz} Li\underline{s}(z) = \frac{1}{z} Li_{(s_{1}−1,s_{2},...,s_{k})}(z)

if s_{1} ≥ 2;

\frac{d}{dz} Li\underline{s}(z) = \frac{1}{1 − z} Li_{(s_{2},...,s_{k})}(z)

if s_{1} = 1. So

Li\underline{s}(z) = ∫_{0}^{z} ω_{0}^{s_{1}−1}ω_{1} Li_{(s_{2},...,s_{k})}(t) = ∫_{0}^{z} ω_{0}^{s_{1}−1}ω_{1} ⋯ ω_{0}^{s_{k}−1}ω_{1}.

For \underline{s} = (s_{1}, …, s_{k}), define ω\underline{s} = ω_{s_{1}} ⋯ ω_{s_{k}}, so

Li\underline{s}(z) = ∫_{0}^{z} ω\underline{s}.

Recall that ℚ ⊂ K ⊂ ℝ and we have defined H = K⟨x_{0}, x_{1}⟩, y_{s} = x_{0}^{s−1}x_{1} for s ≥ 1, y\underline{s} = y_{s_{1}} ⋯ y_{s_{k}} for \underline{s} = (s_{1}, …, s_{k}). We have

H^{1} = Ke + Hx_{1} = \{c_{0} + ∑_{\underline{s}} c\underline{s}y\underline{s} : c_{0}, c\underline{s} ∈ K\}.

Now define

L̂i_{y\underline{s}} = Li\underline{s}

and extend the definition by linearity:

L̂i_{c_{0}+∑ c\underline{s}y\underline{s}} = c_{0} + ∑_{\underline{s}} c\underline{s} Li\underline{s} .

Hence L̂i_{u}(z) is well-defined for u ∈ H^{1}, |z| < 1.

We want to investigate Li\underline{s}(z) · Li_{\underline{s}}′(z). For \underline{s} = (s_{1}, …, s_{k}), recall ω\underline{s} = ω_{0}^{s_{1}−1}ω_{1} ⋯ ω_{0}^{s_{k}−1}ω_{1}. We can rewrite this as ω_{ϵ_{1}} ⋯ ω_{ϵ_{p}} where ϵ_{i} ∈ {0, 1}, where p = ∑ s_{i} and #{i : ϵ_{i} = 1} = k.

Recall the lemma:

∫_{0}^{z} ω_{ϵ_{1}} ⋯ ω_{ϵ_{p}} ∫_{0}^{z} ω_{ϵ_{p+1}} ⋯ ω_{ϵ_{p+q}} = ∑_{σ∈S_{p,q}} ∫_{0}^{z} ω_{ϵ_{σ(1)}} ⋯ ω_{ϵ_{σ(p+q)}}.

Definition of Shuffle � on H by induction

e � w = w � e = w
x_{ϵ}u � x_{ϵ′}v = x_{ϵ}(u � x_{ϵ′}v) + x_{ϵ′}(x_{ϵ}u � v)

and distributivity. It follows by induction that the shuffle product is commutative. Denote the vector spaces H ⊃ H^{1} ⊃ H^{0} with the shuffle product by

H_{�} ⊃ H1_{�} ⊃ H0_{�} .

Proposition. For u, v ∈ H^{1}, L̂i_{u}(z)L̂i_{v}(z) = L̂i_{u� v}(z).

Corollary. For u, v ∈ H^{0}, ζ̂(u)ζ̂(v) = ζ̂(u � v).

Examples. Li_{1}^{n}(z) = n! Li_{{1}_{n}}(z), where {1}_{n} = (1, …, 1) (n copies).

Li_{1}(z) Li_{2}(z) = 2 Li_{(2,1)}(z) + Li_{(1,2)}(z).

Let u = x_{1}, v = x_{0}x_{1}. Then u � v = 2x_{0}x_{1}^{2} + x_{1}x_{0}x_{1}. This gives the above formula.

Li_{2}(z)^{2} = 4 Li_{(3,1)}(z) + 2 Li_{(2,2)}(z).

Collorary. ζ(2)^{2} = 4ζ(3, 1) + 2ζ(2, 2). (Shuffle).

Compare with stuffle: ζ(2)^{2} = 2ζ(2, 2) + ζ(4). We get ζ(4) = 4ζ(3, 1).

The algebra H_{�} was studied by Radford. It is a polynomial algebra with a basis the set of Lyndon words ℒ.

\xymatrix{ & H_{∗} \\ K[ℒ] \ar@{-->}[ur]^{∼} \ar[dr]^{∼} \\  & H_{�}}

Take a simple non-Lyndon word x_{0}^{2}:

\xymatrix{ & x_{0}^{∗2} = x_{0} ∗ x_{0} = x_{0}^{2} \\ x_{0}^{2} \ar[ur] \ar[dr] \\  & x_{0}^{� 2} = x_{0} � x_{0} = 2x_{0}^{2}.}

Generating series for L̂i_{w}(z), w ∈ X^{∗}. We need to define L̂i_{w}(z) for w = ux_{0}^{m}, m ≥ 1. The trick is

“ ∫_{0}^{z} \frac{dz}{z} ” ↦ ∫_{1}^{z} \frac{dz}{z} ·

So we set

L̂i_{x_{0}^{m}}(z) = \frac{1}{m!}(log z)^{m}

and define L̂i_{w}(z) by induction: for i = 0, 1,

L̂i_{x_{i}u}(z) = ∫_{0}^{z} ω_{i}(t) Li_{u}(t).

The generating series for L̂i_{w}:

L̂i(z) = ∑_{w∈X^{∗}} L̂i_{w}(z) · w

is a (non-commutative) power series in x_{0}, x_{1}.

Differential equation. On a simply connected open set in ℙ^{1} not containing 0, 1, ∞:

\frac{d}{dz} L̂i(z) = (\frac{x_{0}}{z} + \frac{x_{1}}{1 − z} ) L̂i(z).

There are three singularities 0, 1, ∞, and are all regular.

Since the residues at z = 0 and z = 1 are the variables x_{0} and x_{1} up to sign, one can think of this differential equation as a “universal” first order differential equation with regular singularities at 0, 1, ∞.

Near 0, we have

∑_{m≥0} \frac{1}{m!}(log z)^{m}x_{0}^{m} = e^{x_{0} log z}.

Initial condition:

lim_{z→0} L̂i(z)e^{−x_{0} log z} = 1.

Near 1, we have

lim_{z→1} e^{−x_{1} log(1−z)}L̂i(z) = ϕ_{KZ}(x_{0}, x_{1}).

This is the associator of Knizhnik-Zamolodchikov, introduced by Drinfeld.

We can study the monodromy of the differential equation. Minh and Petitot proved:

Theorem. {L̂i_{w}(z)}_{w∈X^{∗}} are linearly independent over ℝ(z).

Introduction to Multiple Zeta Values, IV

M. Waldschmidt

August 25, 2001

Conjectural description of ker ζ̂. We first review the known linear relations among zeta values. Take u, v ∈ H^{0} (with k = ℚ), then

ζ̂(u ∗ v − u � v) = 0.

Proof. ζ̂(u) · ζ̂(v) = ζ̂(u ∗ v) = ζ̂(u � v).

We also have the following relation, due to Euler:

ζ(3) = ζ(2, 1).

Therefore, y_{3} −y_{2}y_{1} ∈ ker ζ̂. But the preceeding system of relations all have weights ≥ 4. So Euler’s relation is not spanned by the previous ones.

We have

x_{1} ∗ (x_{0}x_{1}) = y_{1} ∗ y_{2} = y_{1}y_{2} + y_{2}y_{1} + y_{3},
x_{1} � (x_{0}x_{1}) = 2x_{0}x_{1}^{2} + x_{1}x_{0}x_{1}.

Notice that x_{1} ∉ H^{0}. Still, the above computation suggests that

ζ(1, 2) + ζ(2, 1) + ζ(3) = 2ζ(2, 1) + ζ(1, 2).

However, ζ(1, 2) is not well-defined. If it were, we would have got Euler’s relation legitimately. We have (from Chen integrals)

Li_{1}(z) · Li_{2}(z) = Li_{x_{1}� x_{0}x_{1}}(z) = 2 Li_{2,1}(z) + Li_{1,2}(z),

valid for |z| < 1. If we look at the series,

Li_{1}(z) · Li_{2}(z) = ∑_{n≥1} \frac{z^{n}}{n} ∑_{m≥1} \frac{z^{m}}{m^{2}}

= ∑_{n>m} \frac{z^{n}z^{m}}{nm^{2}} + ∑_{m>n} + ∑_{n=m} \frac{z^{2n}}{n^{3}}

= Li_{1,2}(z, z) + Li_{2,1}(z, z) + Li_{3}(z^{2}).

Here, we have used

Definition.

Li\underline{s}(z_{1}, …, z_{k}) = ∑_{n_{1}>···>n_{k}≥1} \frac{z_{1}^{n_{1}} ⋯ z_{k}^{n_{k}}}{n_{1}^{s_{1}} ⋯ n_{k}^{s_{k}}} ·

We now have

Li_{3}(z^{2}) − 2 Li_{2,1}(z) + Li_{2,1}(z, z) = Li_{1,2}(z, 1) − Li_{1,2}(z, z).

The LHS is ζ(3) − ζ(2, 1) when z = 1. It is an exercise that lim_{z→1} RHS = 0.

This proves the Euler relation, which is an example of the third standard relations:

ζ(x_{1} ∗ w − x_{1} � w) = 0 for all w ∈ H^{0}.

Algebraic relations among zeta values.

Linear relations ↔ ker ζ̂.

Conjecture 1. (Hoffman). ker ζ̂ is spanned by elements in H^{0} homogeneous for weight.

Denote by Z_{p} the ℚ-space spanned by ζ̂(w), w ∈ H_{p}^{0} (homogeneous elements in H^{0} of weight p).

Conjecture 1 implies that the ℚ-algebra spanned by Z_{p}, p ≥ 0 is ⨁_{p≥0} Z_{p}. That is, all algebraic relations are given by homogeneous relation. This statement was conjectured by Goncharov.

We have:

Z_{0} = ℚ,
Z_{1} = 0,
Z_{2} = ⟨ζ(2)⟩_{ℚ} = ⟨π^{2}⟩,
Z_{3} = ⟨ζ(3)⟩_{ℚ},
Z_{4} = ⟨ζ(4)⟩_{ℚ} = ⟨π^{4}⟩_{ℚ},
Z_{5} = ⟨ζ(2)ζ(3), ζ(5)⟩_{ℚ}.

So if we put d_{p} = dim Z_{p}, we have d_{1} = 0, d_{0} = d_{2} = d_{3} = d_{4} = 1, 1 ≤ d_{5} ≤ 2. We do not know whether ζ(2)ζ(3)/ζ(5) ∈ ℚ.

There is a program for finding relations, called “EZ-face”, available on the internet

                    http://www.cecm.sfu.ca/projects/EZFace/index.html

Exercise. ζ(3, 1) = \frac{1}{4}ζ(4), ζ(2, 2) = \frac{3}{4}ζ(4), ζ(2, 1, 1) = ζ(4).

Conjecture. (Zagier). For all p ≥ 0, d_{p} = d_{p−2} + d_{p−3}. Hence

∑_{p≥0} d_{p}t^{p} = \frac{1}{1 − t^{2} − t^{3}} ·

The “exercise” of checking the compatibility of conjectures of Zagier and Goncharov is not an easy one, and has not yet been done completely.

The number of ζ(\underline{s}) of weight s_{1} + ⋯ + s_{k} = p (with s_{1} ≥ 2) is 2^{p−2}.

The number of linear relations is 2^{p−5}(p + 1). Here we are not counting independent relations just the ways of getting relations. It is not easy to see what are the independent linear relations among these relations.

Each ζ(\underline{s}) with |\underline{s}| ≤ 11 are homogeneous polynomials in

ζ(2), ζ(3), ζ(5), ζ(7), ζ(9), ζ(11), ζ(6, 2), ζ(8, 2), ζ(8, 2, 1).

Broadhurst has given a conjectural description for such a list for |\underline{s}| ≤ p. (This is a problem in algebra).

Suggestion of Hoffman. A basis for Z_{p} may be

ζ(s_{1}, …, s_{k}), s_{j} ∈ {2, 3}.

Conjecture. The three standard relations (shuffle, stuffle, and the third, which are respectively quadratic, quadratic, and linear) generate the ideal of all algebraic relations among zeta values.

More precisely, take independent variables Z\underline{s}, \underline{s} = (s_{1}, …, s_{k}) with s_{1} ≥ 2 and consider the map ℚ[Z\underline{s} : s_{1} ≥ 2] → ℝ, Z\underline{s} ↦ ζ(\underline{s}). Then the conjecture is that the kernel of this map is the ideal spanned by

Z_{u}Z_{v} − Z_{u� v}, Z_{u}Z_{v} − Z_{u∗v}, Z_{x_{1}∗w−x_{1}� w},

where u, v, w ∈ H^{0}, and Z_{u} means Z\underline{s} if u = y\underline{s}.

Consequence(?). π, ζ(3), ζ(5), … are algebraically independent.

Other relations. Is it true that ζ̂(u ∗ v − u � v) = 0 with u, v ∈ H^{1} but u ∗ v − u � v ∈ H^{0}? The answer is no. There exist u, v ∈ H^{1} such that u ∗ v − u � v ∈ H^{0} but ζ(u ∗ v − u � v) ≠ 0.

Example. u = x_{1}, v = x_{1}, u � v = 2x_{1}^{2} and u∗v = x_{1} ∗x_{1} = y_{1} ∗y_{1} = 2y_{1}^{2} +y_{2} = 2x_{1}^{2} +y_{2}. Therefore, u ∗ v − u � v = y_{2} ∈ H^{0} but ζ(y_{2}) = ζ(2) ≠ 0.

The map ζ̂: H^{0} → ℝ is a morphism of algebras for ∗ and � . We want to extend it to H^{1}. If ζ̂: H^{1} → ℝ were a morphism for ∗ and � , we would have

ζ̂(x_{1})^{2} = ζ̂(x_{1} ∗ x_{1}) = ζ̂(x_{1} � x_{1}).

But the 2nd equality is not true.

Introduction to Multiple Zeta Values, V

M. Waldschmidt

August 25, 2001

Extension of ζ̂ to H^{1} ⊃ H^{0}. Recall that

H = ℚ⟨x_{0}, x_{1}⟩  ⊃  H^{1} = ℚ.e + H.x_{1}  ⊃  H^{0} = ℚ.e + x_{0}.H.x_{1}.

Also, recall that L̂i_{w}(z) is defined for w ∈ H^{1}, and ζ̂(w) is defined for w ∈ H^{0}.

We also have

H_{�} = ℚ[ℒ]_{�}  ⊃  H1_{�} = ℚ[ℒ ∖ {x_{0}}]_{�}  ⊃  H0_{�} = ℚ[ℒ ∖ {x_{0}, x_{1}}]_{�} .

Also ζ̂: H0_{�} → ℝ and H1_{�} = H0_{�} [x_{1}]. So we only have to decide what ζ(1) is. The difficulty is that ζ(1) is not defined (possible candidates: 0 or the Euler constant γ). An easy way out, since there are more than one reasonable choices, is to make no choice:

Denote by Ẑ_{�} the unique morphism of algebras H1_{�} → ℝ[T] such that Ẑ_{�} (w) = ζ̂(w) for w ∈ H0_{�} , and Ẑ_{�} (x_{1}) = T, where T is an indeterminate. We do the same with H_{∗}. We have

Ẑ_{∗} : H_{∗}^{1} → ℝ[T], x_{1} ↦ T.

Theorem. (Boutet de Mouvel, Zagier). There is an isomorphism of ℝ-vector spaces ρ : ℝ[T] → ℝ[T] such that

\xymatrix{ & ℝ[T] \\ H^{1} \ar@{-->}[ur]^{Ẑ_{�}} \ar[dr]_{Ẑ_{∗}} \\  & ℝ[T] \ar[uu]_{ρ}}

is commutative.

In fact, ρ is determined by the formula:

∑_{ℓ≥0} \frac{1}{ℓ!}ρ(T^{ℓ})t^{ℓ} = exp(T t + ∑_{n=2}^{∞}(−1)^{n} \frac{ζ(n)}{n}t^{n})

explicit formula is more interesting.

Example. ρ(T^{0}) = 1, ρ(T) = T, ρ(T^{2}) = T^{2} + ζ(2), ρ(T^{3}) = T^{3} + 3ζ(2)T − 2ζ(3), ρ(T^{4}) = T^{4} − 6ζ(2)T^{2} − 8ζ(3)T + 6ζ(4) + 3ζ(2)^{2}.

We can check: x_{1} � x_{1} = 2x_{1}^{2}, so Ẑ_{�} (2x_{1}^{2}) = T^{2}. On the other hand, x_{1} ∗ x_{1} = 2x_{1}^{2} + x_{2}. So Ẑ_{∗}(2x_{1}^{2}) = T^{2} − ζ(2).

We have for s ≥ 2,

exp(∑_{n=1}^{∞}(−1)^{n} \frac{t^{n}}{n} ζ(sn)) = ∏_{j≥1}(1 + \frac{t}{j^{s}} )^{−1}.

The RHS reminds us of the product formual for Γ-function. In fact, we should compare the above with

exp(−γt + ∑_{n=2}^{∞}(−1)^{n} \frac{t^{n}}{n} ζ(sn)) = Γ(1 + t).

We also have

ρ(e^{T t}) = e^{T t} · exp(∑_{n≥2}(−1)^{n} \frac{ζ(n)}{n}t^{n}).

So we can consider ρ as a differential operator

ρ = exp (∑_{n≥2}(−1)^{n} \frac{ζ(n)}{n}(\frac{∂}{∂T} )^{n}) .

Hint for the proof of Zagier-Boutet de Mouvel. Recall

H_{∗}^{1} ⟶^{∼} Q Sym, y\underline{s} ↦ QM\underline{s}(\underline{t}) = ∑_{n_{1}>···>n_{k}≥1} t_{n_{1}}^{s_{1}} ⋯ t_{n_{k}}^{s_{k}}.

The specialization t_{n} = 1/n, n ≥ 1 gives ζ̂ on H^{0}.

Take N large and consider the specialization \underline{t} ↦ (1, 1/2, 1/3, …, 1/N, 0, …), we get asymptotic expansions. Moreover, consider

exp_{�} (ty_{1}) = ∑_{n≥0} \frac{t^{n}}{n!}y_{1}^{� n} = ∑_{n≥0} t^{n}y_{1}^{n},

exp_{∗}(ty_{1}) = ∑_{n≥0} \frac{t^{n}}{n!}y_{1}^{∗n}.

We then use Newton’s formulae.

Next step. From H^{1} to H. We will extend the map

ζ̂_{�} : H1_{�} → ℝ[T] ⟶^{T ↦0} ℝ

to H_{�} . Recall that H_{�} = H0_{�} [x_{0}, x_{1}] (commutative polynomial algebra). So given w ∈ H_{�} , we can associate the constant term (as a polynomial in x_{0}, x_{1}) in H0_{�} . Call this map Reg_{�} : H_{�} → H0_{�} (the notation means: regularize a possibly non-convergent word to a convergent word). We now define

ζ̂_{�} = ζ̂ ∘ Reg_{�} : H_{�} → ℝ.

Proposition. (K. Ihara and M. Kaneko). Let w ∈ X^{∗}, w = x_{1}^{m}w_{0}x_{0}^{n} with w_{0} ∈ H^{0}. Then

reg_{�} (w) = ∑_{i=0}^{m} ∑_{j=0}^{n}(−1)^{i+j}x_{1}^{i} � (x_{1}^{m−i}w_{0}x_{0}^{n−j}) � x_{0}^{j}.

Moreover,

w = ∑_{i=0}^{m} ∑_{j=0}^{n} Reg_{�} (x_{1}^{m−i}w_{0}x_{0}^{n−j}) � x_{1}^{i} � x_{0}^{j}.

Examples. Reg_{�} (x_{0}^{n}) = Reg_{�} (x_{1}^{m}) = 0 for n ≥ 1, m ≥ 1. Also Reg_{�} (x_{1}^{m}x_{0}^{n}) = (−1)^{n+m−1}x_{0}^{n}x_{1}^{m}.

Theorem. (Ihara and Kaneko). For w ∈ H^{1} and w_{0} ∈ H^{0},

reg_{�} (w ∗ w_{0} − w � w_{0}) ∈ ker ζ̂.

That is, ζ̂_{�} (w ∗ w_{0} − w � w_{0}) = 0.

Conjecture. (Ihara and Kaneko) As a ℚ-vector space, ker ζ̂ is spanned by these elements.

The third standard relations can be obtained from w∗w_{0} −w � w_{0} by taking w = x_{1}. In general, we must regularize. This conjecture is probably more reasonable (more generators for the kernel), but is perhaps equivalent to the previous conjecture.

Example. x_{1}^{2} ∗ x_{0}x_{1} − x_{1}^{2} � x_{0}x_{1} = x_{1}x_{0}x_{1}^{2} − x_{1}x_{0}^{2}x_{1} − x_{0}^{2}x_{1}^{2} is not in H^{0}.

Œsterlé shows: The conjecture of Ihara and Kaneko is equivalent to the following: ker ζ̂ is spanned by

w_{1} ∗ w_{2} − w_{1} � w_{2}, with w_{1}, w_{2} ∈ H^{0} and

Reg_{�} (x_{1}^{m} ∗ w_{0} − x_{1}^{m} � w_{0}) with w_{0} ∈ H^{0} and m ≥ 1.

For each m ≥ 1, consider the map

θ_{m} : H → H, m ↦ (−1)^{m}Reg_{�} (x_{1}^{m} ∗ w − x_{1}^{m} � w).

– For w_{0} ∈ H^{0}, it is known that θ_{m}(w_{0}) ∈ ker(ζ̂).

– For m = 1, w ↦ x_{1} � w − x_{1} ∗ w (no need to regularize).

– Reg_{�} (x_{1}^{m}) = 0 ⟹ θ_{m}(w) = (−1)^{m}Reg_{�} (x_{1}^{m} ∗ w).

Definition. We define ζ̂_{∗} as ζ̂ ∘ Reg_{∗}, where Reg_{∗} : H_{∗} = H_{∗}^{0}[x_{0}, x_{1}] → H_{∗}^{0} is again “taking the constant term”.

We have

Reg_{∗}(x_{1}^{m}w_{0}) = ∑_{i=0}^{m}(−1)^{i} \frac{x_{1}^{∗i}}{i!} ∗ (x_{1}^{m−i}w_{0})

for w_{0} ∈ H^{0}.

Other linear relations on the multiple zeta values. Duality. Start with

∫_{x_{0}}^{x_{1}} φ_{1} ⋯ φ_{p} = ∫_{x_{1}>t_{1}>···>t_{p}>x_{0}} φ_{1}(t_{1}) ⋯ φ_{p}(t_{p}) = ∫_{x_{1}}^{x_{0}} φ_{p} ⋯ φ_{1}(−1)^{p}.

So

∫_{0}^{1} ω_{ϵ_{1}} ⋯ ω_{ϵ_{p}} = ∫_{0}^{1} ω_{1−ϵ_{p}} ⋯ ω_{1−ϵ_{1}}.

Denote (1, …, 1) (ℓ times) by {1}_{ℓ}. Write

\underline{s} = (s_{1}, …, s_{k}) = (t_{1} + 2, \{1\}_{r_{1}}, t_{2} + 2, \{1\}_{r_{2}}, …, t_{m} + 2, \{1\}_{r_{m}})

with t_{1}, …, t_{m}, r_{1}, …, r_{m} ≥ 0. Let τ : H → H be the anti-isomorphism exchanging x_{0} and x_{1}. Then τy\underline{s} = y_{\underline{s}}′ where \underline{s}′ = (r_{m} + 2, \{1\}_{t_{m}}, …, r_{1} + 2, \{1\}_{t_{1}}). The duality relation is:

ζ(τy\underline{s}) = ζ(y\underline{s}).

Introduction to Multiple Zeta Values, VI

M. Waldschmidt August 28, 2001

Recall that we have

[figure: H = ℚ⟨x_{0}, x_{1}⟩; H^{1} = ℚ.e + H.x_{1}; H^{0} = ℚ.e + x_{0}.H.x_{1}; ζ̂; ℝ]

ker ζ̂ ↔ linear dependence relations among multiple zeta values.

We have the basic double shuffle relations:

ζ̂(u)ζ̂(v) = ζ̂(u � v) = ζ̂(u ∗ v) for all u, v ∈ H^{0},

⟹ u � v − u ∗ v ∈ ker ζ̂.

Ihara and Kaneko give the more general regularized double shuffle relations: (RDSR) Reg_{�} (w � w_{0} − w ∗ w_{0}) ∈ ker ζ̂, w_{0} ∈ H^{0}, w ∈ H^{1}, where

Reg_{�} : H = H^{0}[x_{0}, x_{1}]_{�} → H^{0}

sends a polynomial to its constant term. Remark.

– For w = x_{1}, x_{1} � w_{0} − x_{1} ∗ w_{0} = d_{1}(w_{0}) ∈ H^{0} for w_{0} ∈ H^{0}, there is no need to regularize.

– For w = x_{1}^{n}, n ≥ 1, RDSR can be written as

∑_{i=0}^{n}(−1)^{i}x_{1}^{i} � (x_{1}^{n−i} ∗ w_{0}) ∈ ker ζ̂.

– The function d_{1} : H^{0} → H^{0} above is the derivation (satisfying d(uv) = udv + (du)v) on H with d_{1}(x_{0}) = x_{0}x_{1}, d_{1}(x_{1}) = −x_{0}x_{1}. We have

d_{1}(x_{0}^{s−1}x_{1}) = ∑_{i=1}^{s−1} x_{0}^{i}x_{1}x_{0}^{s−1−i}x_{1} − x_{0}^{s}x_{1}.

– Consequence of d_{1}(y_{s}) ∈ ker ζ̂:

ζ(s + 1) = ∑_{i=1}^{s−1} ζ(i + 1, s − i). (Euler)

This is a special case of the more general sum formula: for p ≥ k + 1, k ≥ 1 fixed,

∑_{\underline{s}=(s_{1},...,s_{k}):s_{1}+···+s_{k}=p} ζ(\underline{s}) = ζ(p).

This reduces to Euler’s relation when k = 2. For k = 3, it was proved by Hoffman and Moen (1996). The general formula for k ≥ 2 is due to Granville-Zagier (1997). They were often first observed numerically on computers.

Examples.

– ζ(3) = ζ(2, 1), and more generally,

– ζ(2, {1}_{k−1}) = ζ(k + 1) (take p = k + 1 in the sum formula).

Proof of the sum formula from RDSR. Granville-Zagier proved the formula by analytic means. We would like to see how to get the formula from RDSR.

For m ≥ 0, w ∈ H, let

θ_{m}(w) = (−1)^{m}Reg_{�} (x_{1}^{m} ∗ w).

So θ_{m}(w) ∈ ker ζ̂ for m ≥ 1, w ∈ H^{0}.

Lemma. (Ihara-Kaneko) For p > k + 1, k ≥ 1,

θ_{k}(x_{0}^{p−k−1}x_{1}) = x_{0}(x_{1}^{k} � x_{0}^{p−k−2})x_{1} − x_{0}(x_{1}^{k−1} � x_{0}^{p−k−1})x_{1}.

The proof is just a simple induction.

The expression x_{1}^{k−1} � x_{0}^{p−k−1} is of length k − 1 and weight p − 2 (recall that the length is the number of x_{1}’s, and the weight is the number of letters). Let

S(p, k) = x_{0}(x_{1}^{k−1} � x_{0}^{p−k−1})x_{1}.

This is the sum of all convergent words of length k and weight p. Thus

θ_{k}(x_{0}^{p−k−1}x_{1}) = S(p, k + 1) − S(p, k) ∈ ker ζ̂, 1 ≤ k < p − 1.

It follows that S(p, k)−S(p, 1) ∈ ker ζ̂, for 1 ≤ k ≤ p−1. But S(p, 1) is just x_{0}^{p−1}x_{1}, so ζ̂(S(p, 1)) = ζ(p). On the other hand,

ζ̂(S(p, k)) = ∑_{s_{1}≥2,s_{1}+···+s_{k}=p} ζ(s_{1}, …, s_{k}).

This proves the sum formula.

Ohno’s relations. These contain all the relations given earlier.

Let’s first recall the duality relation: let τ : H → H be the anti-automorphism exchanging x_{0} and x_{1} (anti-automorphism means that τ(uv) = τ(v)τ(u)). Then (1 − τ)w_{0} ∈ ker ζ̂ for w ∈ H^{0}; this follows from the identity

∫_{0}^{1} ω_{0}^{t_{1}+1}ω_{1}^{r_{1}+1} ⋯ ω_{0}^{t_{m}+1}ω_{1}^{r_{m}+1} = ∫_{0}^{1} ω_{0}^{r_{m}+1}ω_{1}^{t_{m}+1} ⋯ ω_{0}^{r_{1}+1}ω_{1}^{t_{1}+1}

which can be checked by a change of variables. Fix \underline{s} = (s_{1}, …, s_{k}) and ℓ ≥ 0. Write τy\underline{s} = y_{\underline{s}}′, \underline{s}′ = (s′_{1}, …, s′_{k′}). Then Ohno’s relation is

∑_{e_{1}+···+e_{k}=ℓ,e_{i}≥0} ζ(s_{1} + e_{1}, …, s_{k} + e_{k}) = ∑_{e′_{1}+···+e′_{k′}=ℓ,e′_{j}≥0} ζ(s′_{1} + e′_{1}, …, s′_{k′} + e′_{k′}).

This contains the duality relation: just set ℓ = 0. It also contains the sum relation, which is

ζ(p) = ∑_{|\underline{σ}|=p,length(\underline{σ})=h} ζ(\underline{σ}).

We can take k = 1, s_{1} = h + 1, e = p − h − 1.

Ihara-Kaneko: Ohno’s formula is equivalent to: for n ≥ 1, w ∈ H^{0}, d_{n}(w) ∈ ker ζ̂, where d_{n} is the derivation on H such that

d_{n}(x_{0}) = x_{0}(x_{0} + x_{1})^{n−1}x_{1},
d_{n}(x_{1}) = −x_{0}(x_{0} + x_{1})^{n−1}x_{1}.

Notice that (x_{0} + x_{1})^{n−1} is the sum of all monomials of weight n − 1, and d_{n}(x_{0}) is the sum of all monomials of weight n + 1 in H^{0}.

People are working on deriving Ohno’s relation from RDSR.

A conjecture of Zagier. The conjecture was given in 1994, and solved by Borwein in 1997. Bradley and Minh-Petitot (2000) proved it using syntaxic identities:

4^{k}ζ({3, 1}_{k}) = ζ({4}_{k}).

Definition. For S ∈ ℚ⟨⟨x_{0}, x_{1}⟩⟩ such that (S|e) = 0,

S^{∗} = e + S + S^{2} + ⋯

is the unique solution to either (1 − S)S^{∗} = e or S^{∗}(1 − S) = e. We have (y_{2}^{∗}) ∗ (−y_{2})^{∗} = (−y_{4})^{∗}.

Idea of proof: consider H_{∗}^{0} ⟶^{∼} Q Sym^{0}, sending y\underline{s} to

QM(\underline{s}) = ∑_{n_{1}>···>n_{k}≥1} t_{n_{1}}^{s_{1}} ⋯ t_{n_{k}}^{s_{k}}

and

∑_{n_{1}>···>n_{k}≥1} t_{n_{1}}^{2} ⋯ t_{n_{k}}^{2} = ∏_{n≥1}(1 + t_{n}^{2}) ⟷ y_{2}^{∗}.

So the formula y_{2}^{∗} ∗ (−y_{2})^{∗} = −y_{4}^{∗} amounts to

∏(1 + t_{n}^{2}) ∏(1 − t_{n}^{2}) = ∏(1 − t_{n}^{4}).

(x_{0}x_{1})^{∗} � (−x_{0}x_{1})^{∗} = (−4x_{0}^{2}x_{1}^{2})^{∗}. (1)

We replace x_{0} by tω_{0} = tdz/z, x_{1} by tω_{1} = tdz/(1 − z). We get

∫_{0}^{1}(−t^{4}ω_{0}^{3}ω_{1})^{∗} = ∑_{k≥0}(−t)^{4k}ζ({4}_{k}).

∫_{0}^{1}(−4t^{4}ω_{0}^{2}ω_{1}^{2})^{∗} = ∑_{k≥0}(−4t^{4})^{k}ζ({3, 1}_{k}).

Thus Zagier’s relation is that the difference of the LHS’s is in the kernel of ζ̂.

Example. Consider the following automatom:

[figure: (1); x_{0}; (2); x_{1}; gives S_{1} = 1 + x_{0}x_{1} + (x_{0}x_{1})^{2} + ⋯ = (x_{0}x_{1})^{∗}.]

[figure: (A); −x_{0}; (B); x_{1}; gives S_{1} = 1 − x_{0}x_{1} + (x_{0}x_{1})^{2} + ⋯ = (−x_{0}x_{1})^{∗}.]

Values of Li\underline{s}(\underline{z}) at roots of unity. Say we consider N-th roots of unity.

When N = 1, this is the classical theory of multiple zeta values. We have Zagier’s conjecture on the dimension spanned by zeta values:

d_{p} = d_{p−2} + d_{p−3}.

When N = 2,

d_{p} = d_{p−1} + d_{p−2}.

When N = 3,

d_{p} = 2^{p}.

N = 4, complicated...

Half integers. For example,

Li_{2}(1/2) = \frac{π^{2}}{12} − \frac{1}{2}(log 2)^{2},

ζ(3) = \frac{1}{12}π^{2} log 2 + Li_{(2,1)}(1/2) + Li_{3}(1/2).

This last formula is due to Ramanujan:

∑_{n=1}^{∞} \frac{1}{2^{n}n^{2}} ∑_{j=1}^{n} \frac{1}{j} = ζ(3) − \frac{1}{12}π^{2} log 2.

Generalized Hurwitz zeta function.

∑_{n_{1}>···>n_{k}≥1} \frac{z_{1}^{n_{1}}}{(n_{1} − t_{1})^{s_{1}}} ⋯ \frac{z_{k}^{n_{k}}}{(n_{k} − t_{k})^{s_{k}}} ·

What about the transcendence of ζ(1/2) or ζ(i)? The solution may not have immediate consequence but certainly requires new ideas.

                  http://www.institut.math.jussieu.fr/∼miw/articles/ps/ncts.ps
                  http://math.cts.nthu.edu.tw/Mathematics/english/lecture.html
see also:  http://www.institut.math.jussieu.fr/∼miw//articles/ps/mpl.ps