Eighth lecture: April 25, 2011
9 The structure of the shuffle and harmonic al- gebras
We shall see that the shuffle and harmonic algebras are polynomial algebras in the Lyndon words.
Consider the lexicographic order on X ∗ with x 0 < x 1 and denote by L the set of Lyndon words (see §3). We have seen that x 0 is the only Lyndon word which is not in H 1 x , while x 0 and x 1 are the only Lyndon words which are not in H 0 x .
For instance, there are 1 + 2 + 2 2 + 2 3 = 15 words of weight ≤ 3, and 5 among them are Lyndon words:
x 0 < x 2 0 x 1 < x 0 x 1 < x 0 x 2 1 < x 1 . We introduce 5 variables
T 10 , T 21 , T 11 , T 12 , T 01 , where T ij corresponds to x i 0 x j 1 .
9.1 Shuffle
The structure of the commutative algebra H x is given by Radford Theorem.
Theorem 33. The three shuffle algebras are (commutative) polynomial algebras
H x = K[ L ] x , H 1 x = K �
L \ { x 0 } �
x and H 0 x = K �
L \ { x 0 , x 1 } �
x .
We write the 10 non-Lyndon words of weight ≤ 3 as polynomials in these
Lyndon words as follows:
e = e,
x 2 0 = 1 2 x 0 xx 0 = 1 2 T 10 2 , x 3 0 = 1 3 x 0 xx 0 xx 0 = 1 3 T 10 3 ,
x 0 x 1 x 0 = x 0 xx 0 x 1 − 2x 2 0 x 1 = T 10 T 11 − 2T 21 , x 1 x 0 = x 0 xx 1 − x 0 x 1 = T 10 T 01 − T 11 ,
x 1 x 2 0 = 1 2 x 0 xx 0 xx 1 − x 0 xx 0 x 1 + x 2 0 x 1 = 1 2 T 10 2 T 01 − T 10 T 11 + T 21 , x 1 x 0 x 1 = x 0 x 1 xx 1 − 2x 0 x 2 1 = T 11 T 01 − 2T 12 ,
x 2 1 = 1 2 x 1 xx 1 = 1 2 T 01 2 ,
x 2 1 x 0 = 1 2 x 0 xx 1 xx 1 − x 0 x 1 xx 1 + x 0 x 2 1 = 1 2 T 10 T 01 2 − T 11 T 01 + T 12 , x 3 1 = 1 3 x 1 xx 1 xx 1 = 1 3 T 01 3 .
Corollary 34. We have
H x = H 1 x [x 0 ] x = H 0 x [x 0 , x 1 ] x and H 1 x = H 0 x [x 1 ] x .
9.2 Harmonic algebra
Hoffman’s Theorem gives the structure of the harmonic algebra H � :
Theorem 35. The harmonic algebras are polynomial algebras on Lyndon words:
H � = K[L] � , H 0 � = K �
L \ {x 0 , x 1 } �
� and H 1 � = K �
L \ {x 0 , x 1 } �
� .
For instance, the 10 non-Lyndon words of weight ≤ 3 are polynomials in the
5 Lyndon words as follows:
e = e,
x 2 0 = x 0 � x 0 = T 10 2 , x 3 0 = x 0 � x 0 � x 0 = T 10 3 , x 0 x 1 x 0 = x 0 � x 0 x 1 = T 10 T 11 , x 1 x 0 = x 0 � x 1 = T 10 T 01 , x 1 x 2 0 = x 0 � x 0 � x 1 = T 10 2 T 01 ,
x 1 x 0 x 1 = x 0 x 1 � x 1 − x 2 0 x 1 − x 0 x 2 1 = T 11 T 01 − T 21 − T 12 , x 2 1 = 1 2 x 1 � x 1 − 1 2 x 0 x 1 = 1 2 T 01 2 − 1 2 T 11 ,
x 2 1 x 0 = 1 2 x 0 � x 1 � x 1 − 1 2 x 0 � x 0 x 1 = 1 2 T 10 T 01 2 − 1 2 T 10 T 11 ,
x 3 1 = 1 6 x 1 � x 1 � x 1 − 1 2 x 0 x 1 � x 1 + 1 3 x 2 0 x 1 = 1 6 T 01 3 − 1 2 T 11 T 01 + 1 3 T 21 . In the same way as Corollary 34 follows from Theorem 33, we deduce from Theorem 35:
Corollary 36. We have
H � = H 1 � [x 0 ] � = H 0 � [x 0 , x 1 ] � and H 1 � = H 0 � [x 1 ] � . Remark. Consider the diagram
H x −→ K[ L ] x
� f
� g H � −→ K[ L ] �
The horizontal maps are just the identitication of H x with K[ L ] x and of H �
with K[L] � . The vertical map f is the identity map on H, since the algebras H x and H � have the same underlying set H (only the law differs). But the map g which makes the diagram commute is not a morphism of algebras: it maps each Lyndon word on itself, but consider, for instance, the image of the word x 2 0 , as a polynomial in K[ L ] x ,
x 2 0 = (1/2)x 0 xx 0 = (1/2)x x 0 2 , but as a polynomial in K[ L ] � ,
x 2 0 = x 0 � x 0 = x �2 0 ,
hence g(T 10 2 ) = 2T 10 2 .
10 Regularized double shuffle relations
10.1 Hoffman standard relations
The relation ζ(2, 1) = ζ(3) is the easiest example of a whole class of linear relations among MZV.
For any word w ∈ X ∗ , each of x 1 � x 0 wx 1 and x 1 xx 0 wx 1 is the sum of x 1 x 0 wx 1 with other words in x 0 X ∗ x 1 (i.e. convergent words):
x 1 xw − x 1 w ∈ H 0 , x 1 � w − x 1 w ∈ H 0 , Hence
x 1 xw − x 1 � w ∈ H 0 .
It turns out that these elements x 1 xw − x 1 � w are in the kernel of ζ � : H 0 → R.
Proposition 37. For w ∈ H 0 ,
ζ(x � 1 � w − x 1 xw) = 0.
Since x 1 xe = x 1 � e = e, these relations are can be written in an equivalent way ζ � (x 1 � y s − x 1 xy s ) = 0 whenever s 1 ≥ 2.
They are called Hoffman standard relations between multiple zeta values.
Example. Writing
x 1 � (x s 0 − 1 x 1 ) = y 1 � y s = y 1 y s + y s y 1 + y s+1 and
x 1 x(x s 0 − 1 x 1 ) = x 1 x s 0 − 1 x 1 + x 0 (x 1 xx s 0 − 2 x 1 ) =
� s
ν=1
y ν y s+1 − ν + y s y 1 , we deduce
x 1 � (x s 0 − 1 x 1 ) − x 1 x(x s 0 − 1 x 1 ) = x s 0 x 1 −
� s
ν=1
x ν 0 − 1 x 1 x s 0 − ν x 1 , hence
ζ(p) = �
s+s�=p s≥2, s�≥1