III.3 The elliptic Kashiwara-Vergne Lie algebras
III.3.5 Comparison of the two independently defined elliptic Kashiwara-Vergne Lie
This last part follows the exposition given by Leila Schneps and the author given in the article [RS2].
After defining the other elliptic construction by Alekseev-Kawazumi-Kuno-Naef from [AKKN2], we reformulate their defining properties to reformulate it in terms similar to the ones use in Chapter II Section 3.
The elliptic Kashiwara-Vergne Lie algebra from Aleksee-Kawazumi-Kuno-Naef
Let𝔩𝔦𝔢(1,1) be the free Lie algebra on two generators Lie[𝑥, 𝑦], to be thought of as the Lie algebra associated to the fundamental group of the once-punctured torus.
We set𝑐 = [𝑥, 𝑦]so that the relation[𝑥, 𝑦] = 𝑐holds in𝔩𝔦𝔢(1,1). Let𝔩𝔦𝔢(1,1)𝑛 denote the weight𝑛part of𝔩𝔦𝔢(1,1), where the weight is the total degree in𝑥and𝑦, and let𝔩𝔦𝔢(1,1)𝑛,𝑟 denote the weight𝑛, depth𝑟 part of𝔩𝔦𝔢(1,1), where the depth is the𝑦-degree.
For any element𝑓 ∈𝔩𝔦𝔢(1,1), we decompose it as
𝑓 =𝑓𝑥𝑥+𝑓𝑦𝑦=𝑥𝑓𝑥+𝑦𝑓𝑦=𝑥𝑓𝑥𝑥𝑥+𝑥𝑓𝑦𝑥𝑦+𝑦𝑓𝑥𝑦𝑥+𝑦𝑓𝑦𝑦𝑦.
Let𝔡𝔢𝔯(1,1)denote the Lie subalgebra ofDer𝔩𝔦𝔢(1,1)of derivations𝑢such that𝑢(𝑐) = 0. Let𝔡𝔢𝔯(1,1)𝑛 denote the subspace of𝔡𝔢𝔯(1,1)of derivations𝑢such that𝑢(𝑥), 𝑢(𝑦) ∈𝔩𝔦𝔢(1,1)𝑛 .
Let𝑓 ∈ 𝔩𝔦𝔢(1,1)𝑛 for𝑛 > 1. Then𝑓 is the value on𝑥of a derivation𝑢∈ 𝔡𝔢𝔯(1,1)𝑛 if and only if𝑓 is push-invariant, in which case𝑢(𝑦)is uniquely defined. If𝑢∈ 𝔡𝔢𝔯(1,1)and𝑢(𝑥) = 𝑓,𝑢(𝑦) = 𝑔, we sometimes write𝑢=𝐷𝑓 ,𝑔.
Let𝔱𝔯2be the quotient of𝔩𝔦𝔢(1,1)by the equivalence relation: two words𝑤and𝑤′are equivalent if one can be obtained from the other by cyclic permutation of the letters.
Theelliptic divergence𝑑𝑖𝑣∶𝔡𝔢𝔯(1,1) →𝔱𝔯2is defined by 𝑑𝑖𝑣(𝑢) =𝑡𝑟(𝑓𝑥+𝑔𝑦) where𝑢=𝐷𝑓 ,𝑔. Since𝑢([𝑥, 𝑦]) = [𝑥, 𝑔] + [𝑓 , 𝑦] = 0, we have
𝑥𝑔𝑥𝑥+𝑥𝑔𝑦𝑦−𝑥𝑔𝑥𝑥−𝑦𝑔𝑦𝑦=𝑦𝑓𝑥𝑥+𝑦𝑓𝑦𝑦−𝑥𝑓𝑥𝑦−𝑦𝑓𝑦𝑦.
Comparing the terms on both sides that start with𝑥and end with𝑦shows that𝑔𝑦 = −𝑓𝑥. Thus we can write the divergence condition as a function of just𝑓:
𝑑𝑖𝑣(𝑢) =𝑡𝑟(𝑓𝑥−𝑓𝑥).
Definition III.3.27. The elliptic Kashiwara-Vergne Lie algebra𝔨𝔯𝔳(1,1)defined in ([AKKN2]) is the ℚ-vector space spanned by the derivations𝑢∈𝔡𝔢𝔯(1,1)𝑛 ,𝑛≥3, such that
𝑑𝑖𝑣(𝑢) = {
𝐾 𝑡𝑟( 𝑐𝑛−12 )
for some𝐾 ∈ℚ if𝑛is odd
0 if𝑛is even. (1)
It is closed under the bracket of derivations.
A reformulation of the𝑑𝑖𝑣condition.
Observe that the Lie algebra𝔨𝔯𝔳(1,1) is bigraded by the weight and depth. We write𝔨𝔯𝔳(1,1)𝑛,𝑟 for the vector subspace of𝔨𝔯𝔳𝑒𝑙𝑙 spanned by derivations𝑢such that𝑢(𝑥) ∈ 𝔩𝔦𝔢(1,1)𝑛,𝑟 and𝑢(𝑦) ∈ 𝔩𝔦𝔢(1,1)𝑛,𝑟+1. Let 𝑢∈𝔨𝔯𝔳(1,1)𝑛,𝑟 , let𝑓 =𝑢(𝑥)and𝑔=𝑢(𝑦), and write
𝑓 = ∑
𝑖=(𝑖0,...,𝑖𝑟)
𝑐𝑖𝑥𝑖0𝑦𝑥𝑖1𝑦⋯𝑦𝑥𝑖𝑟. Then
𝑓𝑥= ∑
𝑖s.t. 𝑖𝑟≠0
𝑐𝑖𝑥𝑖0𝑦𝑥𝑖1𝑦⋯𝑦𝑥𝑖𝑟−1
and
𝑓𝑥= ∑
𝑖s.t. 𝑖0≠0
𝑐𝑖𝑥𝑖0−1𝑦𝑥𝑖1𝑦⋯𝑦𝑥𝑖𝑟.
For any word 𝑤in𝑥, 𝑦, let𝐶(𝑤) denote the trace class of𝑤, i.e. the set of words obtained by cyclically permuting the letters of𝑤. The trace of a polynomialℎis given by
∑
𝐶(𝑤)
(𝑡𝑟(ℎ)|𝐶(𝑤))
⋅𝐶(𝑤) where the sum runs over the trace classes of words and the coefficient(
𝑡𝑟(ℎ)|𝐶(𝑤))
and extend the operators𝑝𝑢𝑠ℎand𝑝𝑢𝑠ℎ𝑠𝑦𝑚to polynomials by linearity. If𝑢is a word of weight𝑛− 2 and depth𝑟− 1such that𝑦𝑢∈𝐶𝑦(𝑤), then𝑢𝑦∈𝐶𝑦(𝑤)and we have
{𝐶𝑦(𝑤) = {𝑦 𝑝𝑢𝑠ℎ𝑖(𝑢)|0≤𝑖≤𝑟− 1}
𝐶𝑦(𝑤) = {𝑝𝑢𝑠ℎ𝑖(𝑢)𝑦|0≤𝑖≤𝑟− 1}.
Note that𝐶𝑦(𝑤)may be of order less than𝑟(in fact strictly dividing𝑟) when𝑤has a symmetry under
This allows us to rewrite the divergence condition (1) on an element 𝑓 ∈ 𝔨𝔯𝔳(1,1)𝑛,𝑟 as the following family of relations for all words𝑢of weight𝑛− 2and depth𝑟− 1:
This is the version of the divergence condition that we will use for comparison with the Lie algebra 𝔨𝔯𝔳𝑒𝑙𝑙.
Thus, in order to prove that𝔨𝔯𝔳𝑒𝑙𝑙 is in bijection with𝔨𝔯𝔳(1,1), it remains only to prove that the circ-constance condition onΔ−1(
𝑠𝑤𝑎𝑝(𝐹))
is equivalent to the divergence condition (5) on𝑓. Since so the circ-constance condition is given explicitly by
𝑠𝑤𝑎𝑝(𝐹)(𝑣1,…, 𝑣𝑟)
for all depths𝑟≥2. Putting this over a common denominator gives the equivalent equality The circ-constance condition on𝑠𝑤𝑎𝑝(
Δ−1(𝐹))
can thus be expressed by the family of relations for every tuple(𝑖1,…, 𝑖𝑟):
We can now translate this equality directly back into terms of the polynomial𝑓(𝑥, 𝑦)corresponding to the mould𝐹. We can write the right-hand side of the above equation as
𝐾𝑟(
by numbering the𝑣𝑖in the second term of (III.3.48) in the opposite order.
We have[𝑥, 𝑦] =𝑎𝑑(𝑥)(𝑦) =𝐶2, so[𝑥, 𝑦]𝑟=𝐶2𝑟, so the polynomial-valued mould corresponding to[𝑥, 𝑦]𝑟is given by
𝐴(𝑢1,…, 𝑢𝑟) = (−1)𝑟𝑢1⋯𝑢𝑟. The swap of this mould is given by
𝑠𝑤𝑎𝑝(𝐴)(𝑣1,…, 𝑣𝑟) = −(𝑣1−𝑣2)⋯(𝑣𝑟−1−𝑣𝑟)𝑣𝑟.
Thus the mould𝐵of (III.3.49) satisfies𝐵 = −𝑠𝑤𝑎𝑝(𝐴), so the expression (III.3.49) reformulating the RHS translates back to polynomials as to
−𝐾𝑟 Using (7) to directly translate the left-hand side of (10) in terms of the polynomial𝑓, we thus obtain the following expression equivalent to the circ-neutrality property (10):
(𝑓|𝑥𝑖1+1𝑦𝑥𝑖2𝑦…𝑦𝑥𝑖𝑟𝑦)
Now all words in the positive terms start in𝑥and end in𝑦, and all words in the negative terms start in 𝑦and end in𝑥, so we can remove these letters and write The left-hand side of this equal to
(𝑝𝑢𝑠ℎ𝑠𝑦𝑚(𝑓𝑦𝑥−𝑓𝑥𝑦)|𝑥𝑖1𝑦⋯𝑦𝑥𝑖𝑟) ,
so to the left-hand side of the divergence condition (5) for the weight𝑛− 2word𝑢=𝑥𝑖1𝑦⋯𝑦𝑥𝑖𝑟. If 𝑛 ≠ 2𝑟+ 1, then the right-hand sides of (5) and (16) are both equal to0. To prove that (5) and (16) are identical, it remains only to check that the right-hand sides are equal when𝑛 = 2𝑟+ 1, which, cancelling the factor𝐾𝑟from both sides, reduces to the following lemma.
Lemma III.3.28. For each word𝑢of depth𝑟− 1and weight2𝑟− 1, we have
where𝑢′denotes the word𝑢written backwards.
Proof. Observe that if ([𝑥, 𝑦]𝑟|𝑢𝑦) ≠ 0, then𝑢𝑦must satisfy theparity propertythat, writing𝑢𝑦 = 𝑢1⋯𝑢2𝑟where each 𝑢𝑖 is letter𝑥or𝑦, the pair𝑢2𝑖−1𝑢2𝑖must be either𝑥𝑦or𝑦𝑥for0 ≤ 𝑖 ≤ 𝑟. The coefficient of the word𝑢𝑦in[𝑥, 𝑦]𝑟is equal to(−1)𝑗where𝑗is the number of pairs𝑢2𝑖−1𝑢2𝑖in𝑢𝑦that
are equal to𝑦𝑥. In other words, if a word𝑤appears with non-zero coefficient in[𝑥, 𝑦]𝑟, then letting 𝑈 =𝑦𝑥and𝑉 =𝑥𝑦, we must be able to write𝑤as a word in𝑈 , 𝑉, and the coefficient of𝑤in[𝑥, 𝑦]𝑟 is(−1)𝑚 where𝑚denotes the number of times the letter𝑈occurs.
If 𝑤 = 𝑢𝑦 = 𝑉𝑟 = (𝑥𝑦)𝑟, then 𝑢′𝑦 = 𝑢𝑦. The coefficient of 𝑉𝑟 in [𝑥, 𝑦]𝑟 is equal to 1, so the right-hand side of (III.3.28) is equal to 2 if 𝑟is even and0 if𝑟is odd. For the left-hand side, 𝐶(𝑢𝑦) = {𝑉𝑟, 𝑈𝑟}and𝐶𝑦(𝑢𝑦) = {𝑉𝑟}, so|𝐶𝑦(𝑢𝑦)| = 1. The coefficient of𝑈𝑟in[𝑥, 𝑦]𝑟 is equal to (−1)𝑟, so the left-hand side is again equal to 2 if𝑟is even and0if𝑟is odd. This proves (III.3.28) in the case𝑢𝑦=𝑉𝑟.
Suppose now that𝑢𝑦≠𝑉𝑟but that it satisfies the parity property. Write𝑢𝑦= 𝑈𝑎1𝑉𝑏1⋯𝑈𝑎𝑠𝑉𝑏𝑠 in which all the𝑎𝑖, 𝑏𝑖≥1except for𝑎1, which may be0. Then𝑢′𝑦is equal to𝑥𝑈𝑏𝑠−1𝑉𝑎𝑠⋯𝑈𝑏1𝑉𝑎1𝑦. If𝑏𝑠 > 1, then the pair𝑢2(𝑏
𝑠−1)+1𝑢2(𝑏
𝑠−1)+2 is𝑥𝑥, so([𝑥, 𝑦]𝑟|𝑢′𝑦) = 0. If𝑏𝑠 = 1, then the word𝑢′𝑦 begins with𝑥𝑥and thus does not have the parity property, so again([𝑥, 𝑦]𝑟|𝑢′𝑦) = 0. This shows that if([𝑥, 𝑦]𝑟|𝑢𝑦)≠0then([𝑥, 𝑦]𝑟|𝑢′𝑦) = 0and vice versa.
This leaves us with three possibilities for𝑢𝑦≠𝑉𝑟.
Case 1: ([𝑥, 𝑦]𝑟|𝑢𝑦) ≠ 0. Then𝑢𝑦has the parity property, so we write𝑢𝑦 = 𝑈𝑎1𝑉𝑏1⋯𝑈𝑎𝑠𝑉𝑏𝑠 as above. The right-hand side of (III.3.28) is then equal to (−1)𝑗 where 𝑗 = 𝑎1 +⋯+ 𝑎𝑠. For the left-hand side, we note that the only words in the cyclic permutation class𝐶(𝑢𝑦)that have the parity property are the cyclic shifts of𝑢𝑦by an even number of letters, otherwise a pair𝑥𝑥or𝑦𝑦necessarily occurs as above. These are the same as the cyclic permutations of the word𝑢𝑦written in the letters 𝑈 , 𝑉. All these cyclic permutations obviously have the same number of occurrences𝑗of the letter𝑈. Thus, the words in𝐶(𝑢𝑦)for which[𝑥, 𝑦]𝑟has a non-zero coefficient are the cyclic permutations of the word𝑢𝑦in the letters𝑈 , 𝑉, and the coefficient is always equal to(−1)𝑗. These words are exactly half of the all the words in𝐶(𝑢𝑦), so the sum in the left-hand side is equal to(−1)𝑗|𝐶(𝑢𝑦)|∕2. But
|𝐶𝑦(𝑢𝑦)| = |𝐶(𝑢𝑦)|∕2, so the left-hand side is equal to(−1)𝑗, which proves (III.3.28) for words𝑢𝑦 having the parity property.
Case 2: ([𝑥, 𝑦]𝑟|𝑢′𝑦) ≠ 0. In this case it is 𝑢′𝑦that has the parity property, and the right-hand side of (III.3.28) is equal to (−1)𝑟+𝑗′ where 𝑗′ is the number of occurrences of 𝑈 in the word 𝑢′𝑦 = 𝑈𝑎1𝑉𝑏1⋯𝑈𝑎𝑠𝑉𝑏𝑠, i.e.𝑗′ = 𝑎1+⋯+𝑎𝑠. We have𝑢𝑦 = 𝑥𝑈𝑏𝑠−1𝑉𝑎𝑠⋯𝑈𝑏1𝑉𝑏1𝑦. The word𝑤 = 𝑈𝑏𝑠−1𝑉𝑎𝑠⋯𝑈𝑏1𝑉𝑏1𝑈 then occurs in𝐶(𝑢𝑦), and the number of occurrences of the letter𝑈 in𝑢𝑦is equal to𝑗=𝑏1+⋯+𝑏𝑠−1+𝑏𝑠. Since𝑎1+𝑏1+⋯+𝑎𝑠+𝑏𝑠=𝑟, we have𝑗+𝑗′=𝑟so𝑗′=𝑟−𝑗and the right-hand side of (III.3.28) is equal to(−1)𝑗. The number of words in𝐶(𝑢𝑦)that have non-zero coef-ficient in[𝑥, 𝑦]𝑟is|𝐶(𝑢𝑦)|∕2 =|𝐶𝑦(𝑢𝑦)|as above, these words being exactly the cyclic permutations of𝑤written in𝑈 , 𝑉, and the coefficient is always equal to(−1)𝑗. So the left-hand side of (III.3.28) is equal to(−1)𝑗, which proves (III.3.28) in the case where𝑢′𝑦has the parity property.
Case 3: ([𝑥, 𝑦]𝑟|𝑢𝑦) = ([𝑥, 𝑦]𝑟|𝑢′𝑦) = 0. The right-hand side of (III.3.28) is zero. For the left-hand side, consider the words in𝐶(𝑢𝑦). If there are no words in𝐶(𝑢𝑦)whose coefficient in[𝑥, 𝑦]𝑟is non-zero, then the left-hand side of (III.3.28) is also zero and (III.3.28) holds. Suppose instead that there is a word𝑤 ∈ 𝐶(𝑢𝑦) whose coefficient in[𝑥, 𝑦]𝑟 is non-zero. Then as we saw above, 𝑤is a cyclic shift of𝑢𝑦by an odd number of letters, and since all cyclic shifts of𝑤by an even number of letters then have the same coefficient in[𝑥, 𝑦]𝑟as𝑤, we may assume that𝑤is the cyclic shift of𝑢𝑦 by one letter, i.e. taking the final𝑦and putting it at the beginning. Since𝑤has non-zero coefficient in[𝑥, 𝑦]𝑟, we can write 𝑤 = 𝑈𝑎1𝑉𝑏1⋯𝑈𝑎𝑠𝑉𝑏𝑠, where 𝑎1 > 0 since𝑤now starts with 𝑦, but𝑏𝑠 may be equal to 0 since 𝑤may end with 𝑥. Then 𝑢𝑦 = 𝑥𝑈𝑎1−1𝑉𝑏1⋯𝑈𝑎𝑠𝑉𝑏𝑠𝑦, so we can write 𝑢′𝑦 = 𝑈𝑏𝑠𝑉𝑎𝑠⋯𝑈𝑏1𝑉𝑎1−1𝑥𝑦 = 𝑈𝑏𝑠𝑉𝑎𝑠⋯𝑈𝑣1𝑉𝑎1. But then𝑢′𝑦satisfies the parity property, so its coefficient in[𝑥, 𝑦]𝑟is non-zero, contradicting our assumption. Thus under the assumption, all words in𝐶(𝑢𝑦)have coefficient zero in[𝑥, 𝑦]𝑟, which completes the proof of the Lemma.
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[SS] A.Salerno, L.Schneps, Mould theory and the double shuffle Lie algebra structure, preprint (2015), available at https://arxiv.org/abs/1510.05535
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Appendices
A Appendix : Proof of Lemma III.3.22 Let us recall the statement of the technical lemma 31.
Lemma 31. For𝑛 > 1 and any constant𝑐 ≠ 0, let𝑇𝑐𝑛 be the homogeneous polynomial mould of degree𝑛defined by
𝑇𝑐𝑛(𝑣1,…, 𝑣𝑟) = 𝑐 𝑟𝑃𝑟𝑛,
where 𝑃𝑛𝑟 is the sum is over all monomials of degree 𝑛−𝑟in the variables 𝑣1,…, 𝑣𝑟. Then 𝑇𝑐𝑛 is circ-constant and𝑔𝑎𝑛𝑖𝑡(𝑝𝑜𝑐)⋅𝑝𝑎𝑟𝑖(𝑇𝑐𝑛)is circ-neutral.
Proof. The mould 𝑇𝑐𝑛 is trivially circ-constant. Consider the proof of Proposition III.3.21 with 𝑀 = 𝑇𝑐𝑛,𝑁 = 𝑝𝑎𝑟𝑖(𝑀). In order to show that𝑔𝑎𝑛𝑖𝑡(𝑝𝑜𝑐)⋅𝑁 is circ-neutral, we start by recalling from the proof of Proposition III.3.21 that for each𝑟 >1, the cyclic sum
𝑔𝑎𝑛𝑖𝑡(𝑝𝑜𝑐)⋅𝑁(𝑣1,…, 𝑣𝑟) +⋯+𝑔𝑎𝑛𝑖𝑡(𝑝𝑜𝑐)⋅𝑁(𝑣𝑟, 𝑣1,…, 𝑣𝑟−1)
is equal to the expression (III.3.40). Thus we need to show that (III.3.40) is equal to zero for all𝑟 >1. To show this, we will break up the sum
∑
𝐛={𝐛1,…,𝐛𝑠}
(−1)𝑙𝐚𝑝𝑜𝑐(⌊𝐛1)⋯𝑝𝑜𝑐(⌊𝐛𝑠) ∑
𝑤∈𝐖𝐚𝑛
𝑤 (.0.51)
into parts that are simpler to express.
We need a little notation. Let us write𝑉𝑗= {1,…, 𝑗}. If𝐵 ⊂ 𝑉𝑟, let𝑃𝐵𝑑 denote the sum of of all monomials of degree𝑑in the variables𝑣𝑖∈𝐵. We write𝑃0
𝐵 = 1for all𝐵.
We will break up the sum (.0.51) as the sum of partial sums𝑆0+⋯+𝑆𝑟, where𝑆0 is the term of (.0.51) corresponding to the empty set and 𝑆𝑖 is the sum over the 𝐛-parts containing 𝑣𝑖 but not 𝑣𝑖+1,…, 𝑣𝑟, for each𝑖 ∈ {1,…, 𝑟}. Notice that the𝐛-parts containing𝑣𝑖but not𝑣𝑖+1,…, 𝑣𝑟are in bijection with the2𝑖−1subsets𝐵 ⊂ 𝑉𝑖−1, by taking𝐛to be the set𝐵′ =𝐵∪ {𝑣𝑖}, divided into chunks consisting of consecutive integers. For example, if𝑖= 5and𝐵 = {1,3}then𝐵′ = {1,3,5}and the associated𝐛-part is(𝑣1)(𝑣3)(𝑣5); if𝐵 = {1,2}then𝐵′ = {1,2,5}and the𝐛-part is(𝑣1, 𝑣2)(𝑣5), and if𝐵 = {1,4}then𝐵′= {1,4,5}and the𝐛-part is(𝑣1)(𝑣4, 𝑣5).
Setting𝑣0=𝑣𝑟, this means that𝑆0 =𝑃𝑉𝑛−𝑟
𝑟 and for1≤𝑖≤𝑟, 𝑆𝑖= ∑
𝐵⊆{1,…,𝑖−1}
(−1)𝑟−|𝐵′|𝑃𝑛−𝑟+|𝐵′|
𝑉𝑟⧵𝐵′
∏
𝑗∈𝐵′(𝑣𝑗−1−𝑣𝑗) . (.0.52)
In order to prove that (.0.51) is zero, we will give simplified expressions for𝑆1,…, 𝑆𝑟−1in Claim 1, a simplified expression for𝑆𝑟in Claim 2, and then show how to sum them up in Claim 3.
Claim 1. For1≤𝑖≤𝑟− 1, we have𝑣𝑖+1,…, 𝑣𝑟. Let𝑣0=𝑣𝑟. Then we have 𝑆𝑖=
(−1)𝑟−𝑖𝑃𝑛−𝑟+𝑖
{𝑖−1,𝑖+1,…,𝑟−1}
(𝑣𝑟−𝑣1)(𝑣1−𝑣2)⋯(𝑣𝑖−1−𝑣𝑖). (.0.53) Proof. We will use the following trivial but useful identity. Let 𝐵 ⊊ 𝑉𝑟, let𝑣𝑗 ∉ 𝐵, and let𝐵′ = 𝐵∪ {𝑣𝑗}. Then
𝑃𝐵𝑑′ =𝑃𝐵𝑑+𝑣𝑗𝑃𝐵𝑑−1′ . (.0.54)
Multiplying by the common denominator, we write (.0.52) as induction on𝑘. Let us do the base case𝑘= 1by showing that the right-hand side of (.0.55) is equal to𝑄1. We start by breaking the right-hand side of (.0.55) into𝑣1∈𝐵and𝑣1 ∉𝐵, and compute
which is exactly𝑄1. The last equality is obtained by using (.0.54) twice on the right-hand factor. This proves the base case𝑘= 1.
then setting𝐶 =𝐵⧵{𝑣𝑘+1},𝐶𝑘+1 =𝐶∪ {𝑣𝑘+1} =𝐵,𝐶𝑘′ =𝐶∪ {𝑣𝑘+1, 𝑣𝑖} =𝐵′in the first sum, right-hand side of (.0.55) equals𝑄𝑖−1as desired.
Unfortunately, the expression in Claim 1 for𝑆𝑖does not work for𝑖=𝑟due to the fact that when 𝑖=𝑟in (.0.55), the subset𝐵 =𝑉𝑟−1occurs in the sum and the corresponding polynomial𝑃𝑉
𝑟⧵𝐵′ = 0. It turns out that the expression for𝑆𝑟is actually simpler.
Claim 2. The term𝑆𝑟is given by
starting from the equality (.0.55) for𝑖=𝑟, slightly rewritten as
∏𝑟
We have𝑃∅= 0, and we may as well sum over the subsets𝐶, so it becomes
This equality suffices to prove the desired result (.0.57). Indeed, taking𝑖=𝑟− 1, we see that𝑅𝑛𝑟−1 is equal to the right-hand side of (.0.60), so
∏𝑟
= (𝑣𝑟−𝑣𝑖−1)𝑣𝑛−1𝑖−1 + (𝑣𝑖−1−𝑣𝑖)𝑣𝑛−1𝑖 − (𝑣𝑟−𝑣𝑖−1)(𝑣𝑖−1−𝑣𝑖)
∑𝑛−2 𝑘=0
𝑣𝑘𝑖𝑣𝑛−𝑘−2𝑖−1
= (𝑣𝑟−𝑣𝑖−1)𝑣𝑛−1𝑖−1 + (𝑣𝑖−1−𝑣𝑖)𝑣𝑛−1𝑖 − (𝑣𝑟−𝑣𝑖−1)(𝑣𝑛−1𝑖−1 −𝑣𝑛−1𝑖 )
= (𝑣𝑟−𝑣𝑖−1)𝑣𝑛−1𝑖 + (𝑣𝑖−1−𝑣𝑖)𝑣𝑛−1𝑖
= (𝑣𝑟−𝑣𝑖)𝑣𝑛−1𝑖 .
This proves (.0.61) and thus completes the proof of Claim 2.
We can now prove that the expression (.0.51) is equal to zero by showing that𝑆0+⋯+𝑆𝑟= 0. Claim 3. We have𝑆0+⋯+𝑆𝑟= 0.
Proof.The key point is the following computation of partial sums for𝑖 < 𝑟: 𝑆0+⋯+𝑆𝑖=
(−1)𝑟−𝑖𝑃𝑣𝑛−𝑟+𝑖
𝑖,…,𝑣𝑟−1
(𝑣𝑟−𝑣1)(𝑣1−𝑣2)⋯(𝑣𝑖−1−𝑣𝑖). (.0.62) We prove it by induction on 𝑖. The base case𝑖 = 0 is just given by the formula for 𝑆0 (with 𝑣0 =𝑣𝑟). Assume (.0.62) up to𝑖− 1. Then
𝑆0+⋯+𝑆𝑖= (𝑆0+⋯+𝑆𝑖−1) +𝑆𝑖 (.0.63)
=
(−1)𝑟−𝑖+1𝑃𝑣𝑛−𝑟+𝑖−1
𝑖−1,…,𝑣𝑟−1
(𝑣𝑟−𝑣1)⋯(𝑣𝑖−2−𝑣𝑖−1)+
(−1)𝑟−𝑖𝑃𝑣𝑛−𝑟+𝑖
𝑖−1,𝑣𝑖+1,…,𝑣𝑟−1
(𝑣𝑟−𝑣1)⋯(𝑣𝑖−1−𝑣𝑖), (.0.64) Using (.0.54) and multiplying (.0.63) by the denominator, we find
(−1)𝑟−𝑖(𝑣𝑟−𝑣1)⋯(𝑣𝑖−1−𝑣𝑖)(𝑆0+⋯+𝑆𝑖) =𝑃𝑣𝑛−𝑟+𝑖
𝑖−1,𝑣𝑖+1,…,𝑣𝑟−1− (𝑣𝑖−1−𝑣𝑖)𝑃𝑣𝑛−𝑟+𝑖−1
𝑖−1,…,𝑣𝑟−1
=𝑃𝑣𝑛−𝑟+1
𝑖−1,𝑣𝑖+1,…,𝑣𝑟−1+𝑣𝑖𝑃𝑣𝑛−𝑟+𝑖−1
𝑖−1,…,𝑣𝑟−1−𝑣𝑖−1𝑃𝑣𝑛−𝑟+𝑖−1
𝑖−1,…,𝑣𝑟−1
=𝑃𝑣𝑛−𝑟+1
𝑖−1,…,𝑣𝑟−1−𝑣𝑖−1𝑃𝑣𝑛−𝑟+𝑖−1
𝑖−1,…,𝑣𝑟−1
=𝑃𝑣𝑛−𝑟+1
𝑖,…,𝑣𝑟−1.
Now, taking this equality for𝑖=𝑟− 1yields 𝑆0+⋯+𝑆𝑟−1=
−𝑃𝑣𝑛−1
𝑟−1
(𝑣𝑟−𝑣1)(𝑣1−𝑣2)⋯(𝑣𝑟−2−𝑣𝑟−1) = −𝑣𝑛−1𝑟−1
(𝑣𝑟−𝑣1)⋯(𝑣𝑟−2−𝑣𝑟−1), which is equal to−𝑆𝑟by Claim 2. This proves Claim 3.
Since𝑆0+⋯+𝑆𝑟is equal to (.0.51), we have finally shown that whatever the value of𝑐,𝑔𝑎𝑛𝑖𝑡(𝑝𝑜𝑐)⋅
𝑁is circ-neutral, completing the proof of Proposition III.3.21.
B A Mould bestiary
A moulds bestiary
Who ? What? Tell me more ! Lives in...
alternal(ity) property equivalent to being a Lie polynomial, related to
shuf-fle II.1.39
alternil(ity) property related to stuffle II.2.7
ARI Lie algebra (with
𝑎𝑟𝑖bracket) moulds in variables𝑢𝑖with𝑀(∅) = 0 II.1.18
𝐴𝑅𝐼 Lie algebra (with
𝑎𝑟𝑖bracket) moulds in variables𝑣𝑖with𝑀(∅) = 0 II.1.18
amit(M) operator derivation of𝐴𝑅𝐼𝑙𝑢 II.1.5
anit(M) operator derivation of𝐴𝑅𝐼𝑙𝑢 II.1.4
arit(M) operator derivation of𝐴𝑅𝐼𝑙𝑢 II.1.6
𝑎𝑟𝑖 Lie bracket defined with flexions, closely related to the Poisson
bracket II.1.27
𝐴𝑅𝐼𝑎∕𝑏 space moulds in ARI having property𝑎whose swap have
property𝑏
𝐴𝑅𝐼𝑎∗𝑏 space moulds in ARI having property𝑎whose swap have
property𝑏up to adding on a constant-valued mould 𝐴𝑅𝐼𝑎∕𝑏 space subspace of 𝐴𝑅𝐼𝑎∕𝑏 of moulds that are even
func-tions of𝑢1in depth 1
𝐵𝐴𝑅𝐼 space bimoulds with𝐵(∅) = 0 II.1.18
𝐵𝐼 𝑀 𝑈 space all bimoulds II.1.3
Dari Lie bracket another bracket used on𝐴𝑅𝐼 III.3.7
Darit (M) operator derivation of𝐴𝑅𝐼𝑙𝑢 III.3.5
dar operator multiply by𝑢1…𝑢𝑟, corresponds to the
transforma-tion𝑦→[𝑦, 𝑥]in𝑓(𝑥, 𝑦) ∈ℚ⟨𝐶⟩. III.2.1 𝑑𝑢𝑝𝑎𝑙 mould in𝐴𝑅𝐼 used for the recursive construction of𝑝𝑎𝑙 II.2.11
𝑔𝑎𝑛𝑖𝑡(𝑄) operator automorphism of𝐺𝐴𝑅𝐼𝑙𝑢 III.3.28
GARI group moulds in variables𝑢𝑖with𝑀(∅) = 1 II.1.35
𝐺𝐴𝑅𝐼 group moulds in variables𝑣𝑖with𝑀(∅) = 1 II.1.35
𝑖𝑛𝑣𝑚𝑢 operation takes the inverse of a mould for the𝑚𝑢multiplication
𝑖𝑛𝑣𝑝𝑎𝑙 mould in𝐺𝐴𝑅𝐼 inverse of𝑝𝑎𝑙for the group law𝑔𝑎𝑟𝑖 II.2.2 𝑙𝑖𝑚𝑢 Lie bracket commutator for the multiplication𝑚𝑢. Also denoted
𝑙𝑢. II.1.18
𝑙𝑢 Lie bracket see𝑙𝑖𝑚𝑢
𝑚𝑢 binary operation multiplication of moulds II.1.15
𝑝𝑎𝑙 mould in𝐺𝐴𝑅𝐼 key to Ecalle’s fundamental identity II.2.2
pari operator multiplies by(−1)𝑟 III.3.21
pic mould III.3.33
𝑝𝑖𝑙 mould =𝑠𝑤𝑎𝑝(𝑝𝑎𝑙)
poc mould III.3.34
senary relation key to the construction of the map𝔨𝔯𝔳↪𝔨𝔯𝔳𝑒𝑙𝑙 III.3.23 symmetral(ity) property 𝑒𝑥𝑝𝑎𝑟𝑖(𝑀) ∈𝐺𝐴𝑅𝐼 is symmetral if𝑀 is alternal II.1.39
teru operator main ingredient of the senary relation III.3.22