On the derivation representation of the fundamental Lie algebra of mixed elliptic motives

Abstract

Richard Hain and Makoto Matsumoto constructed a category of universal mixed elliptic motives, and described the fundamental Lie algebra of this category: it is a semi-direct product of the fundamental Lie algebra ${{\mathrm{Lie}}}\pi _1(\mathsf {MTM})$ of the category of mixed Tate motives over $\mathbb {Z}$ with a filtered and graded Lie algebra $\mathfrak {u}$. This Lie algebra, and in particular $\mathfrak {u}$, admits a representation as derivations of the free Lie algebra on two generators. In this paper we study the image $\mathscr {E}$ of this representation of $\mathfrak {u}$, starting from some results by Aaron Pollack, who determined all the relations in a certain filtered quotient of $\mathscr {E}$, and gave several examples of relations in low weights in $\mathscr {E}$ that are connected to period polynomials of cusp forms on ${{\mathrm{SL}}}_2(\mathbb {Z})$. Pollack’s examples lead to a conjecture on the existence of such relations in all depths and all weights, that we state in this article and prove in depth 3 in all weights. The proof follows quite naturally from Ecalle’s theory of moulds, to which we give a brief introduction. We prove two useful general theorems on moulds in the appendices.

Résumé

Richard Hain et Makoto Matsumoto ont construit une catégorie de motifs elliptiques mixtes universels et décrit l’algèbre de Lie fondamentale de cette catégorie: il s’agit d’un produit semi-direct de l’algèbre de Lie fondamentale $\mathrm{Lie}\,\pi _1$(MTM) de la catégorie des motifs de Tate mixtes sur $\mathbb {Z}$ par une algèbre de Lie graduée et filtrée $\mathfrak {u}$. Cette algèbre de Lie, et en particulier l’algèbre $\mathfrak {u}$, admet une représentation par dérivations sur l’algèbre de Lie libre à deux générateurs. Dans cet article nous étudions l’image $\mathscr {E}$ de cette représentation de $\mathfrak {u}$ à partir de résultats d’Aaron Pollack, qui a déterminé toutes les relations dans un certain quotient filtré de $\mathscr {E}$, et donné plusieurs exemples de relations en petit poids dans $\mathscr {E}$ qui sont liées aux polynômes de périodes de formes paraboliques sur ${{\mathrm{SL}}}_2(\mathbb {Z})$. Les exemples de Pollack conduisent à une conjecture sur l’existence de telles relations en toute profondeur et en tout poids, que nous énonçons dans cet article et prouvons en profondeur 3 en tout poids. La démonstration découle naturellement de la théorie des moules d’Écalle, à laquelle nous donnons une brève introduction. Nous démontrons deux théorèmes utiles sur les moules dans les appendices.

Appendices

Appendix 1: Proof of Theorem 3.3

This appendix is devoted to the proof of Theorem 3.3, which makes essential use of the ${{\mathrm{\mathrm{{swap}}}}}$ operator and some of the basic notions of Ecalle’s theory of moulds, in particular the ${{\mathrm{ari}}}$-bracket.

Definitions

If we break the tuple $w=(u_1,\ldots ,u_r)$ into three parts

$$\begin{aligned} a=(u_1,...,u_k),\ \ b=(u_{k+1},...,u_{\ell }), \ \ c=(u_{\ell +1},...,u_r), \end{aligned}$$

we write

$$\begin{aligned} {\left\{ \begin{array}{ll} a\lceil c=(u_1,\ldots ,u_k,u_{k+1}+\cdots +u_{\ell }+u_{\ell +1},u_{\ell +2},\ldots ,u_r)&{} \text {if }\, c\ne \varnothing , \\ a\rceil c=(u_1,\ldots ,u_k+u_{k+1}+\cdots +u_{\ell },u_{\ell +1},\ldots ,u_r) &{}\text {if }\, a\ne \varnothing . \end{array}\right. } \end{aligned}$$

If $A,B\in {{\mathrm{ARI}}}$, we define the operator ${{\mathrm{arit}}}(B)$ acting on A by

$$\begin{aligned} \bigl ({{\mathrm{arit}}}(B)\cdot A\bigr )(u_1,\ldots ,u_r)=\sum _{0\leqslant k<\ell <r} A(a\lceil c)B(b) -\sum _{1\leqslant k<\ell \leqslant r} A(a\rceil c)B(b). \end{aligned}$$

If $A,B\in \overline{{{\mathrm{ARI}}}}$ and the tuple $(v_1,\ldots ,v_r)$ breaks into pieces

$$\begin{aligned} a=(v_1,...,v_k),\ \ b=(v_{k+1},...,v_{\ell }), \ \ c=(v_{\ell +1},...,v_r), \end{aligned}$$

then setting

$$\begin{aligned} \begin{array}{lll} b\rfloor =\left\{ \begin{array}{l@{\quad }l} (u_{k+1}-u_{\ell +1},u_{k+2}-u_{\ell +1},\ldots ,u_{\ell }-u_{\ell +1}) &{}\text {if }\,{c\ne \varnothing },\\ b&{}\mathrm{otherwise,} \end{array} \right. \\ \lfloor b=\left\{ \begin{array}{l@{\quad }l} (u_{k+1}-u_k,u_{k+2}-u_k,\ldots ,u_{\ell }-u_k) &{}\text {if }\, a\ne \varnothing ,\\ b &{}\mathrm{otherwise, } \end{array}\right. \end{array} \end{aligned}$$

we let

$$\begin{aligned} \bigl ({{\mathrm{arit}}}(B)\cdot A\bigr )(v_1,\ldots ,v_r)=\sum _{0\leqslant k<\ell <r} A(ac)B(b\rfloor ) -\sum _{1\leqslant k<\ell \leqslant r} A(ac)B(\lfloor b). \end{aligned}$$

For two moulds A and B, let ${{\mathrm{mu}}}(A,B)$ be the product defined by

$$\begin{aligned} {{\mathrm{mu}}}(A,B)(u_1,\ldots ,u_r)=\sum _{i=0}^r A(u_1,\ldots ,u_i)B(u_{i+1},\ldots ,u_r). \end{aligned}$$

When $A={{\mathrm{ma}}}(F)$ and $B={{\mathrm{ma}}}(G)$ for polynomials F, G in a, b, the multiplication ${{\mathrm{mu}}}$ coincides with ordinary multiplication of polynomials  (cf. [22, (3.2.13)]):

$$\begin{aligned} {{\mathrm{mu}}}({{\mathrm{ma}}}(F),{{\mathrm{ma}}}(G))={{\mathrm{ma}}}(FG). \end{aligned}$$

Let

$$\begin{aligned} {{\mathrm{lu}}}(A,B)={{\mathrm{mu}}}(A,B)-{{\mathrm{mu}}}(B,A), \end{aligned}$$

so ${{\mathrm{lu}}}$ is the corresponding Lie bracket. For $A,B\in {{\mathrm{ARI}}}$ or $\overline{{{\mathrm{ARI}}}}$, we set

$$\begin{aligned} {{\mathrm{ari}}}(A,B)={{\mathrm{arit}}}(B)\cdot A-{{\mathrm{arit}}}(A)\cdot B+{{\mathrm{lu}}}(A,B)\text{. } \end{aligned}$$

(5.1)

Recall that  is the mould in the $u_i$’s defined by

Let ; explicitly,

For any mould $A\in \overline{{{\mathrm{ARI}}}}$, let $\check{A}$ be the mould defined by

Let $\overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$ denote the space of alternal rational-valued moulds A in the variables $v_i$ such that $\check{A}$ is polynomial-valued.

The heart of the proof of Theorem 3.3 [1, lemme 4.40] consists in the following proposition.

Proposition 5.1

The space $\overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$ is a Lie algebra under the ${{\mathrm{ari}}}$-bracket.

We first need a lemma.

Lemma 5.2

Let M be an element of $\overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$. Then the mould $\check{M}$ satisfies the relation

$$\begin{aligned} \check{M}(0,v_2,\ldots ,v_{r}) = \check{M}(v_2,\ldots ,v_r,0)\text{. } \end{aligned}$$

Proof

Let us write the simplest of the alternality relations, which gives

$$\begin{aligned} 0&= \sum _{w\in {{\mathrm{\text {sha}}}}(v_1,v_2\cdots v_r)}M(w) \\&= \sum _{i=0}^{r}M(v_2,\ldots ,v_i,v_1,v_{i+1},\ldots ,v_r) \\&= \sum _{i=1}^{r-1}\frac{\check{M}(v_2,\ldots ,v_i,v_1,v_{i+1},\ldots ,v_r)}{v_2\,(v_2-\cdots )\cdots (\cdots -v_r)\,v_r} +\frac{\check{M}(v_1,\ldots ,v_r)}{v_1\,(v_1-v_2)\cdots (v_{r-1}-v_r)\,v_r }\\&\quad + \frac{\check{M}(v_2,\ldots ,v_{r},v_1)}{v_2\,(v_2-v_3)\cdots (v_{r-1}-v_{r})(v_r-v_1)\,v_1}\cdot \\ \end{aligned}$$

Hence, multiplying by $v_1$ and evaluating at $v_1=0$, we get

$$\begin{aligned} 0\;&=\; \frac{\check{M}(0,v_2,\ldots ,v_r)}{(-v_2)(v_2-v_3)\cdots (v_{r-1}-v_r)\,v_r } + \frac{\check{M}(v_2,\ldots ,v_{r},0)}{v_2\,(v_2-v_3)\cdots (v_{r-1}-v_{r})v_r}, \end{aligned}$$

which is well-defined since both numerators at play are polynomials by the hypothesis on M. The result follows immediately.

Proof of Proposition 5.1

Let $M,N\in \overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$. By linearity, we may assume that M and N are respectively concentrated in depths $r\geqslant 1$ and $s\geqslant 1$. Now, alternality is preserved by the ${{\mathrm{ari}}}$-bracket (cf. [19, Theorem 3.1] for a complete proof), so it only remains to show that ${{\mathrm{ari}}}(M,N)$ is an element of $\overline{{{\mathrm{ARI}}}}^{\text {sing}}$.

The essential step of the proof consists in determining the poles of ${{\mathrm{arit}}}(M)\cdot N$ by reducing their values modulo the subspace of polynomials in the $v_i$’s. We will use the simplifying notation . The fact that M is concentrated in depth r and N in depth s also simplifies the defining formula for ${{\mathrm{arit}}}(M)\cdot N$, as follows:

$$\begin{aligned}&\bigl ({{\mathrm{arit}}}(M)\cdot N\bigr )(v_1\ldots v_{r+s})= \sum _{0\leqslant i< s}N(v_1\cdots \,v_i\, v_{i+r+1}\cdots \,v_{r+s})\,M(v_{i+1}\cdots v_{i+r}\rfloor )\\&\qquad -\sum _{0< i\leqslant s}N(v_1\cdots \,v_i\, v_{i+r+1}\ldots v_{r+s})\,M(\lfloor v_{i+1}\cdots \,v_{i+r})\\&\quad = N(v_{r+1}\cdots \,v_{r+s})\,M(v_1-v_{r+1},\ldots ,v_r-v_{r+1})\\&\quad \quad +\sum _{i=1}^{s-1} N(v_1\cdots \,v_i\, v_{i+r+1}\cdots v_{r+s})\\&\quad \quad \times \Bigl (M(v_{i+1}-v_{i+r+1},\ldots ,v_{i+r}-v_{i+r+1}) - M(v_{i+1}-v_i,\ldots ,v_{i+r}-v_i)\Bigr )\\&\qquad -\, N(v_1\cdots \,v_s)\,M(v_{s+1}-v_s,\ldots v_{r+s}-v_s). \end{aligned}$$

Consequently,

(5.2)

with

$\square$

or, equivalently,

(5.3)

In light of equalities (5.2) and (5.3), there are three types of poles which can occur in the rational function :

• (1) the poles $\frac{1}{v_{r+1}}$ and $\frac{1}{v_{s}}$, which come from only one term and thus do not immediately cancel out;

• (2) the poles of the form $\tfrac{1}{v_i-v_{i+r+1}}$;

• (3) the poles of the form $\tfrac{1}{v_i-v_{i+r}}\cdot$

Let us deal first with Case (2). The corresponding pole $\tfrac{1}{v_i-v_{i+r+1}}$ can only appear in the term ; we then want to check that this pole is compensated by the corresponding difference of the $\check{M}$’s. Let us compute this difference for

$$\begin{aligned} v_i=v_{i+r+1}=u_i=u_{i+r+1}=a; \end{aligned}$$

it gives

$$\begin{aligned} \displaystyle \frac{a-v_{i+1}}{v_{i+1}-a}\,\check{M}(v_{i+1}-a,\ldots ,v_{i+r}-a)+ \displaystyle \frac{v_{i+r}-a}{v_{i+r}-a}\,\check{M}(v_{i+1}-a,\ldots ,v_{i+r}-a), \end{aligned}$$

which is indeed zero, hence the compensation.

For Case (3), we need to distinguish three subcases, according to whether $i=1$, or $i=s$, or i belongs to $\mathopen {[\![} 2,s-1 \mathclose {]\!]}$:

(a) $i=1$: the pole is multiplied by

$$\begin{aligned}&\displaystyle \frac{v_1}{v_{r+1}}\,\check{N}(v_{r+1}v_{r+2}\cdots \,v_{r+s})\,\check{M}(v_1-v_{r+1},\ldots ,v_r-v_{r+1})\\&\quad -\displaystyle \frac{1}{v_1-v_{r+2}}\,\check{N}(v_1\, v_{r+2}\cdots \,v_{r+s})\cdot (v_{r+1}-v_{r+2})\, \check{M}(v_{2}-v_1,\ldots ,v_{r+1}-v_1); \end{aligned}$$

if $v_1=v_{r+1}=a$, this gives

$$\begin{aligned}&\frac{a}{a}\,\check{N}(a,v_{r+2},\ldots , v_{r+s})\,\check{M}(a-a,v_2-a,\ldots ,v_r-a)\\&\qquad -\frac{1}{a-v_{r+2}}\,\check{N}(a, v_{r+2},\cdots , v_{r+s})\cdot (a-v_{r+2})\, \check{M}(v_{2}-a,\ldots ,v_{r}-a,a-a) \\&\quad = \check{N}(a,v_{r+2},\ldots , v_{r+s})\,\check{M}(0,v_2-a,\ldots ,v_r-a)\\&\qquad -\check{N}(a, v_{r+2},\cdots , v_{r+s})\, \check{M}(v_{2}-a,\ldots ,v_{r}-a,0), \end{aligned}$$

which is zero thanks to Lemma 5.2.

(b) $i=s$: the pole is multiplied by

$$\begin{aligned}&-\displaystyle \frac{v_{r+s}}{v_s}\,\check{N}(v_1\cdots \,v_s)\,\check{M}(v_{s+1}-v_s,\ldots , v_{r+s}-v_s)+ \frac{1}{v_{s-1}-v_{r+s}}\,\check{N}(v_1,\ldots , v_{s-1}, v_{r+s})\\&\quad \cdot (v_{s-1}-v_{s})\, \check{M}(v_{s}-v_{r+s},\ldots ,v_{r+s-1}-v_{r+s}); \end{aligned}$$

if $v_s=v_{r+s}=a$, this gives

$$\begin{aligned}&-\displaystyle \frac{a}{a}\,\check{N}(v_1,\ldots ,v_{s-1}, a)\,\check{M}(v_{s+1}-a,\ldots , v_{r+s-1}-a,a-a)\\&\qquad + \displaystyle \frac{1}{v_{s-1}-a}\,\check{N}(v_1,\ldots , v_{s-1}, a)\cdot (v_{s-1}-a)\, \check{M}(a-a,v_{s+1}-a,\ldots ,v_{r+s-1}-a) \\&\quad = -\check{N}(v_1,\ldots ,v_{s-1}, a)\,\check{M}(v_{s+1}-a,\ldots , v_{r+s-1}-a,0)\\&\qquad + \check{N}(v_1,\ldots , v_{s-1}, a)\check{M}(0,v_{s+1}-a,\ldots ,v_{r+s-1}-a) , \end{aligned}$$

which is zero for the same reason as above.

(c) $i\in \mathopen {[\![} 2,s-1 \mathclose {]\!]}$: the pole $\tfrac{1}{v_i-v_{i+r}}$ comes from , and is multiplied by

$$\begin{aligned}&\displaystyle \frac{v_{i-1}-v_{i}}{v_{i-1}-v_{i+r}}\,\check{N}(v_1,\ldots , v_{i-1}, v_{i+r},\cdots , v_{r+s})\cdot \check{M}(v_{i}-v_{i+r},\ldots ,v_{i-1+r}-v_{i+r})\\&\quad - \displaystyle \frac{v_{i+r}-v_{i+r+1}}{v_i-v_{i+r+1}}\,\check{N}(v_1,\ldots , v_i, v_{i+r+1},\ldots , v_{r+s})\cdot \check{M}(v_{i+1}-v_i,\ldots ,v_{i+r}-v_i); \end{aligned}$$

if $v_i=v_{i+r}=a$, this gives

$$\begin{aligned}&\check{N}(v_1,\ldots , v_{i-1}, a,v_{i+r+1},\cdots , v_{r+s})\cdot \check{M}(a-a,v_{i+1}-a,\ldots ,v_{i-1+r}-a)\\&\quad \quad -\check{N}(v_1,\ldots ,v_{i-1}, a, v_{i+r+1},\ldots , v_{r+s})\cdot \check{M}(v_{i+1}-a,\ldots ,v_{i+r-1}-a,a-a) \\&\quad = \check{N}(v_1,\ldots , v_{i-1}, a,v_{i+r+1},\cdots , v_{r+s})\cdot \check{M}(0,v_{i+1}-a,\ldots ,v_{i-1+r}-a)\\&\quad \quad - \check{N}(v_1,\ldots ,v_{i-1}, a, v_{i+r+1},\ldots , v_{r+s})\cdot \check{M}(v_{i+1}-a,\ldots ,v_{i+r-1}-a,0), \end{aligned}$$

which is zero again.

In consequence, the only remaining poles in  are those in $\frac{1}{v_{r+1}}$ and $\frac{1}{v_s}$ from Case (1), and more precisely we can write

It remains to show that the above expression cancels in the bracket ${{\mathrm{ari}}}(M,N)$, using the definition by equation (5.1). The possible poles for the rational function come from the sum

$$\begin{aligned}&\displaystyle \frac{v_1}{v_{s+1}\,(v_1-v_{s+1})}\,\check{M}(v_{s+1},\ldots , v_{s+r})\,\check{N}(v_1-v_{s+1},\ldots ,v_s-v_{s+1}) \\&\quad +\frac{v_{s+r}}{v_r\,(v_{s+r}-v_r)}\,\check{M}(v_1,\ldots , v_r)\,\check{N}(v_{r+1}-v_r,\ldots , v_{s+r}-v_r) \\&\quad -\displaystyle \frac{v_1}{v_{r+1}\,(v_1-v_{r+1})}\,\check{N}(v_{r+1},\ldots , v_{r+s})\,\check{M}(v_1-v_{r+1},\ldots ,v_r-v_{r+1}) \\&\quad +\displaystyle \frac{v_{r+s}}{v_s\,(v_{r+s}-v_s)}\,\check{N}(v_1,\ldots , v_s)\,\check{M}(v_{s+1}-v_s,\ldots v_{r+s}-v_s) \\&\quad +\displaystyle \frac{v_r-v_{r+1}}{v_r\,v_{r+1}}\,\check{M}(v_1,\ldots ,v_r)\,\check{N}(v_{r+1},\ldots ,v_{r+s}) \\&\quad -\displaystyle \frac{v_s-v_{s+1}}{v_s\,v_{s+1}}\,\check{N}(v_1,\ldots ,v_s)\,\check{M}(v_{s+1},\ldots ,v_{r+s}); \end{aligned}$$

paying attention to the exchange of r and s when we switch M and N. The discussion above shows that it suffices to check that the poles $\frac{1}{v_{r}}$ and $\frac{1}{v_s}$, $\frac{1}{v_{r+1}}$ and $\frac{1}{v_{s+1}}$ cancel out.

Let us note that the alternality of ${{\mathrm{ari}}}(M,N)$, together with the basic fact that any alternal mould M satisfies

$$\begin{aligned} M(u_1,\ldots ,u_r)=(-1)^{r-1} M(u_r,\ldots ,u_1) \end{aligned}$$

[22, Lemma 2.5.3], imply that it is enough to deal with the case of $\frac{1}{v_{r}}$ and $\frac{1}{v_{s}}$, the other two being deduced from them by applying the involution corresponding to the symmetry around $\tfrac{r+s+1}{2}$. And the pole $\frac{1}{v_s}$ is deduced from $\frac{1}{v_{r}}$ since the total expression for ${{\mathrm{ari}}}(M,N)$ is antisymmetric in M and N, and thus in r and s.

The only case left to check is that of the pole $\frac{1}{v_{r}}$. The corresponding factor is

$$\begin{aligned}&\displaystyle \frac{v_{s+r}}{v_{s+r}-v_r}\,\check{M}(v_1,\ldots , v_r)\,\check{N}(v_{r+1}-v_r,\ldots , v_{s+r}-v_r)\\&\quad +\displaystyle \frac{v_r-v_{r+1}}{v_{r+1}}\,\check{M}(v_1,\ldots ,v_r)\,\check{N}(v_{r+1},\ldots ,v_{r+s}), \end{aligned}$$

which is clearly zero when $v_r=0$. This concludes the proof of Proposition 5.1.

We can now complete the proof of Theorem 3.3. We use the following elementary result of mould theory [22, Lemma 2.4.1]:

$$\begin{aligned} {{\mathrm{ari}}}\bigl ({{\mathrm{\mathrm{{swap}}}}}(A),{{\mathrm{\mathrm{{swap}}}}}(B)\bigr )={{\mathrm{\mathrm{{swap}}}}}\bigl ({{\mathrm{ari}}}(A,B)\bigr )\qquad \forall \ A,B\in {{\mathrm{ARI}}}_{{{\mathrm{\mathrm{{push}}}}}}\text{. } \end{aligned}$$

(5.4)

Let $A,B\in {{\mathrm{ARI}}}^{\text {sing}}_{\underline{\text {al}}/\underline{\text {al}}}$. Then

$$\begin{aligned} C={{\mathrm{ari}}}(A,B)\in {{\mathrm{ARI}}}_{\underline{\text {al}}/\underline{\text {al}}} \end{aligned}$$

by the part (v) of Theorem 3.1. Consider the mould ${{\mathrm{\mathrm{{swap}}}}}(C)\in \overline{{{\mathrm{ARI}}}}$. Again by part (v) of Theorem 3.1, ${{\mathrm{ARI}}}_{\underline{\text {al}}/\underline{\text {al}}}$ is contained in ${{\mathrm{ARI}}}_{{{\mathrm{\mathrm{{push}}}}}}$, so the moulds A, B and C are all ${{\mathrm{\mathrm{{push}}}}}$-invariant. Thus (5.4) holds, i.e.,

$$\begin{aligned} {{\mathrm{\mathrm{{swap}}}}}(C)={{\mathrm{ari}}}({{\mathrm{\mathrm{{swap}}}}}(A),{{\mathrm{\mathrm{{swap}}}}}(B)). \end{aligned}$$

Both ${{\mathrm{\mathrm{{swap}}}}}(A)$ and ${{\mathrm{\mathrm{{swap}}}}}(B)$ lie in $\overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$ by definition. Thus by Proposition 5.1, we also have ${{\mathrm{\mathrm{{swap}}}}}(C)\in \overline{{{\mathrm{ARI}}}}^{\text {sing}}_{\text {al}}$ and thus $C\in {{\mathrm{ARI}}}^{\text {sing}}$. But we also have

$$\begin{aligned} C={{\mathrm{ari}}}(A,B)\in {{\mathrm{ARI}}}_{\underline{\text {al}}/\underline{\text {al}}}, \end{aligned}$$

so $C\in {{\mathrm{ARI}}}^{\text {sing}}_{\underline{\text {al}}/\underline{\text {al}}}$, which concludes the proof of Theorem 3.3. $\square$

Appendix 2: Proof of Theorem 3.5

This section is devoted to the proof of Theorem 3.5, which is stated as Proposition 6.2 below.

We use the notation and terminology of Sect. 3, with one further definition: for any polynomial $F\in {{\mathrm{Lie}}}[a,b]$ of homogeneous depth r, we define ${{\mathrm{da}}}(F)$ to be the mould given by

$$\begin{aligned} {{\mathrm{da}}}(F)(u_1,\ldots ,u_r)=\frac{{{\mathrm{ma}}}(F)}{u_1\cdots u_r}\cdot \end{aligned}$$

As we did for ${{\mathrm{Da}}}(F)$, if $F=\sum _r F^r$ is any Lie polynomial broken up into its depth-graded parts, we define

$$\begin{aligned} {{\mathrm{da}}}(F)=\sum _r {{\mathrm{da}}}(F^r). \end{aligned}$$

Let ${{\mathrm{lu}}}(A,B)$ be as in the beginning of Appendix 1. For any polynomial U, let ${{\mathrm{Darit}}}_U$ be the operator on moulds defined by

$$\begin{aligned} {{\mathrm{Darit}}}_U\cdot A=-{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot A-{{\mathrm{lu}}}\bigl (A, {{\mathrm{Da}}}(U)\bigr ). \end{aligned}$$

Proposition 6.1

Let U be a push-invariant polynomial in ${{\mathrm{Lie}}}[a,b]$, and let $D_U$ be the associated derivation. Then for any Lie polynomial F in a and b, we have

$$\begin{aligned} -{{\mathrm{da}}}\bigl (D_U(F)\bigr )={{\mathrm{Darit}}}_U\cdot {{\mathrm{da}}}(F).\end{aligned}$$

(6.1)

Proof

By additivity, we may assume that U is of homogeneous depth r and F is of homogeneous depth s. We will use induction on the depth of F to prove the proposition; the only difficult part is the base case $s=1$.

So assume that F is of depth 1, i.e., $F=C_n={{\mathrm{\text {ad}}}}(a)^{n-1}(b)$. Here, since $D_U([a,b])$ is zero, we have the explicit formula

$$\begin{aligned} D_U(F)=\sum _{i=0}^{n-2} {{\mathrm{\text {ad}}}}(a)^i{{\mathrm{\text {ad}}}}(U){{\mathrm{\text {ad}}}}(a)^{n-i-2}(b) =\sum _{i=0}^{n-2} {{\mathrm{\text {ad}}}}(a)^i(UC_{n-i-1}-C_{n-i-1}U). \end{aligned}$$

Applying ${{\mathrm{ma}}}$ to both sides of this formula, and using the facts that ${{\mathrm{ma}}}(C_{n-i-1})$ is equal to $(-1)^{n-i-2}u_1^{n-i-2}$ and that for a Lie polynomial P of homogeneous depth r we have

$$\begin{aligned} {{\mathrm{ma}}}([a,P])=-(u_1+\cdots +u_r){{\mathrm{ma}}}(P) \end{aligned}$$

by the part (3) of Theorem 3.1, and using the identity

$$\begin{aligned} {{\mathrm{ma}}}(UC_{n-i-1}-C_{n-i-1}U)&= (-1)^{n-i-2}{{\mathrm{ma}}}(U)(u_1,\ldots ,u_r)u_{r+1}^{n-i-2}\\&\ \ \ \ -(-1)^{n-i-2}u_1^{n-i-2}{{\mathrm{ma}}}(U)(u_2,\ldots ,u_{r+1}), \end{aligned}$$

we find that

$$\begin{aligned} {{\mathrm{ma}}}\bigl (D_U(F)\bigr )&= \sum _{i=0}^{n-3} (-1)^{n-1}(u_1+\cdots +u_{r+1})^i\\&\quad \times \Bigl ( {{\mathrm{ma}}}(U)(u_1,\ldots ,u_r)u_{r+1}^{n-i-2}-u_1^{n-i-2}{{\mathrm{ma}}}(U)(u_2,\ldots ,u_{r+1})\Bigr ), \end{aligned}$$

so, with a little rewriting of indices, we find that the left-hand side of (6.1) equals

or, adding up the sums to obtain a closed expression,

(6.2)

Let us now compute the right-hand side of (6.1). We have ${{\mathrm{ma}}}(F)=(-1)^nu_1^{n-1}$, so ${{\mathrm{da}}}(F)=(-1)^nu_1^{n-2}$; thus

$$\begin{aligned} {{\mathrm{lu}}}\left( {{\mathrm{da}}}(F),{{\mathrm{Da}}}(U)\right)&= (-1)^n\left( {\displaystyle \frac{u_1^{n-2}{{\mathrm{ma}}}(U)(u_2,\ldots ,u_{r+1})}{u_2\cdots u_{r+1}(u_2+\cdots + u_{r+1})}}\right) \nonumber \\&\ \ \ \ -(-1)^n \left( \displaystyle \frac{{{\mathrm{ma}}}(U)(u_1,\ldots ,u_r)u_{r+1}^{n-2}}{u_1\cdots u_r(u_1+ \cdots +u_r)}\right) \text{. } \end{aligned}$$

(6.3)

Also, we have

$$\begin{aligned} {{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot {{\mathrm{da}}}(F)&= \displaystyle \sum _{w=abc, c\ne \varnothing } {{\mathrm{da}}}(F)(a\lceil c){{\mathrm{Da}}}(U)(b) -\displaystyle \sum _{w=abc, a\ne \varnothing } {{\mathrm{da}}}(F)(a\rceil c){{\mathrm{Da}}}(U)(b)\nonumber \\&=(-1)^n(u_1+\cdots +u_{r+1})^{n-2}{{\mathrm{Da}}}(U)(u_1,\ldots ,u_r)\nonumber \\&\quad -(-1)^n(u_1+\cdots +u_{r+1})^{n-2}{{\mathrm{Da}}}(U)(u_2,\ldots ,u_{r+1})\nonumber \\&=(-1)^n(u_1+\cdots +u_{r+1})^{n-2}\; \nonumber \\&\quad \times \left( {\displaystyle \frac{{{\mathrm{ma}}}(U)(u_1,\ldots ,u_r)}{u_1\cdots u_r (u_1+\cdots +u_r)}} - {\displaystyle \frac{{{\mathrm{ma}}}(U)(u_2,\ldots ,u_{r+1})}{u_2\cdots u_{r+1} (u_2+\cdots +u_{r+1})}}\right) , \end{aligned}$$

(6.4)

since ${{\mathrm{da}}}(F)$ is a depth 1 mould and therefore the only possible decomposition $w=abc$ with $c\ne \varnothing$ is $a=\varnothing$, $b=(u_1,\ldots ,u_r)$ and $c=u_{r+1}$, and the only possible decomposition with $a\ne \varnothing$ is $a=u_1$, $b=(u_2,\ldots ,u_r)$ and $c=\varnothing$. We add (6.3) and (6.4) to get the right-hand side of (6.1), obtaining

(6.5)

which is equal to (6.2). This settles the base case where F is of depth 1.

Now assume that (6.1) holds up to depth $s-1$, and let F be a Lie polynomial of depth s. Then F is a linear combination of Lie brackets, so by additivity, we may assume that F is a single Lie bracket; thus $F=[G,H]$ for Lie brackets G, H that are of depth $<s$. We saw in Appendix 1 that

$$\begin{aligned} {{\mathrm{ma}}}(FG)={{\mathrm{mu}}}({{\mathrm{ma}}}(F),{{\mathrm{ma}}}(G)), \end{aligned}$$

so by definition of ${{\mathrm{da}}}$ we also have

$$\begin{aligned} {{\mathrm{mu}}}({{\mathrm{da}}}(F),{{\mathrm{da}}}(G))={{\mathrm{da}}}(FG). \end{aligned}$$

Furthermore, ${{\mathrm{Darit}}}_U$ is a derivation for the ${{\mathrm{lu}}}$-bracket since both ${{\mathrm{arit}}}(B)$ and  are — this is obvious for ${{\mathrm{lu}}}$ but difficult for ${{\mathrm{arit}}}$, cf. [22, Prop. 2.2.1] —, so using the induction hypothesis for G and H, we have

$$\begin{aligned}-{{\mathrm{da}}}\bigl (D_U(F)\bigr )&= -{{\mathrm{da}}}([D_U(G),H]+[G,D_U(H)])\\&=-{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(D_U(G)),{{\mathrm{da}}}(H)\bigr )-{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(G),{{\mathrm{da}}}(D_U(H))\bigr )\\&={{\mathrm{lu}}}\bigl ({{\mathrm{Darit}}}_U({{\mathrm{da}}}(G))),{{\mathrm{da}}}(H)\bigr )+{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(G),{{\mathrm{Darit}}}_U({{\mathrm{da}}}(H)))\bigr )\\&={{\mathrm{Darit}}}_U\cdot {{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(G)),{{\mathrm{da}}}(H)\bigr )\\&={{\mathrm{Darit}}}_U\cdot {{\mathrm{da}}}([G,H])\\&={{\mathrm{Darit}}}_U\cdot {{\mathrm{da}}}(F). \end{aligned}$$

This proves the proposition. $\square$

Proposition 6.2

The map

$$\begin{aligned} {{\mathrm{Der}}}^0{{\mathrm{Lie}}}[a,b]\rightarrow {{\mathrm{ARI}}}\end{aligned}$$

given by $D_U\mapsto {{\mathrm{Da}}}(U)$ is a Lie algebra morphism, i.e.,

$$\begin{aligned}{}[D_U,D_V]\mapsto {{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\text{. } \end{aligned}$$

(6.6)

Proof

It is easily seen that

$$\begin{aligned} {{\mathrm{arit}}}(B)\circ {{\mathrm{arit}}}(A)-{{\mathrm{arit}}}(A)\circ {{\mathrm{arit}}}(B)={{\mathrm{arit}}}\bigl ({{\mathrm{ari}}}(A,B)\bigr )\text{. } \end{aligned}$$

(6.7)

Using only this identity, the Jacobi relation and (6.1), we compute

$$\begin{aligned}&-{{\mathrm{da}}}\bigl ([D_U,D_V](F)\bigr )\\&\quad =-{{\mathrm{da}}}\bigl (D_U(D_V(F))\bigr ) +{{\mathrm{da}}}\bigl ({{\mathrm{ma}}}(D_V(D_U(F))\bigr )\\&\quad ={{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot {{\mathrm{da}}}(D_V(F)) +{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(D_V(F)),{{\mathrm{Da}}}(U)\bigr )\\&\qquad +{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(V)\bigr )\cdot {{\mathrm{da}}}(D_U(F)) +{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(D_U(F)),{{\mathrm{Da}}}(V)\bigr )\\&\quad =-{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot \Bigl ({{\mathrm{arit}}}({{\mathrm{Da}}}(V))\cdot {{\mathrm{da}}}(F)+{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(F),{{\mathrm{Da}}}(V)\bigr )\Bigr )\\&\qquad -{{\mathrm{lu}}}\bigl ({{\mathrm{arit}}}({{\mathrm{Da}}}(V))\cdot {{\mathrm{da}}}(F),{{\mathrm{Da}}}(U)\bigr ) -{{\mathrm{lu}}}\bigl ({{\mathrm{lu}}}({{\mathrm{da}}}(F),{{\mathrm{Da}}}(V)),{{\mathrm{Da}}}(U)\bigr )\\&\qquad +{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(V)\bigr )\cdot \Bigl ({{\mathrm{arit}}}({{\mathrm{Da}}}(U))\cdot {{\mathrm{da}}}(F)+{{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(F),{{\mathrm{Da}}}(U)\bigr )\Bigr )\\&\qquad +{{\mathrm{lu}}}\bigl ({{\mathrm{arit}}}({{\mathrm{Da}}}(U))\cdot {{\mathrm{da}}}(F),{{\mathrm{Da}}}(V)\bigr ) -{{\mathrm{lu}}}\bigl ({{\mathrm{lu}}}({{\mathrm{da}}}(F),{{\mathrm{Da}}}(U)),{{\mathrm{Da}}}(V)\bigr )\\&\quad ={{\mathrm{arit}}}\bigl ({{\mathrm{ari}}}({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V))\bigr )\cdot {{\mathrm{da}}}(F)\qquad \qquad \qquad \qquad (\text {by }(6.7))\\&\qquad -{{\mathrm{lu}}}\Bigl ({{\mathrm{da}}}(F),{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot {{\mathrm{Da}}}(V)\Bigr ) +{{\mathrm{lu}}}\Bigl ({{\mathrm{da}}}(F),{{\mathrm{arit}}}({{\mathrm{Da}}}(V))\cdot {{\mathrm{Da}}}(U)\Bigr )\\&\qquad -{{\mathrm{lu}}}\Bigl ({{\mathrm{lu}}}({{\mathrm{da}}}(F),{{\mathrm{Da}}}(V)),{{\mathrm{Da}}}(U)\Bigr ) +{{\mathrm{lu}}}\Bigl ({{\mathrm{lu}}}\bigl ({{\mathrm{da}}}(F),{{\mathrm{Da}}}(U)\bigr ),{{\mathrm{Da}}}(V)\Bigr )\\&\quad ={{\mathrm{arit}}}\Bigl ({{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr )\cdot {{\mathrm{da}}}(F)\\&\qquad -{{\mathrm{lu}}}\Bigr ({{\mathrm{da}}}(F),{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot {{\mathrm{Da}}}(V)\Bigl ) +{{\mathrm{lu}}}\Bigl ({{\mathrm{da}}}(F),{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(V)\bigr )\cdot {{\mathrm{Da}}}(U)\Bigr )\\&\qquad +{{\mathrm{lu}}}\Bigl ({{\mathrm{da}}}(F),{{\mathrm{lu}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr ) \qquad \qquad \qquad \qquad \mathrm{(by\ Jacobi)}\\&\quad ={{\mathrm{arit}}}\Bigl ({{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr )\cdot {{\mathrm{da}}}(F)\\&\qquad +{{\mathrm{lu}}}\Bigr ({{\mathrm{da}}}(F),-{{\mathrm{arit}}}\bigl ({{\mathrm{Da}}}(U)\bigr )\cdot {{\mathrm{Da}}}(V) +{{\mathrm{arit}}}({{\mathrm{Da}}}(V))\cdot {{\mathrm{Da}}}(U) +{{\mathrm{lu}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr )\\&\quad ={{\mathrm{arit}}}\Bigl ({{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr )\cdot {{\mathrm{da}}}(F) +{{\mathrm{lu}}}\Bigr ({{\mathrm{da}}}(F),{{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr )\Bigr )\\&\quad ={{\mathrm{Darit}}}_W\cdot {{\mathrm{da}}}(F), \end{aligned}$$

where W is the polynomial such that

$$\begin{aligned} {{\mathrm{Da}}}(W)={{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr ) \end{aligned}$$

(obtained by multiplying the right-hand mould by and inverting the injective map  ${{\mathrm{ma}}}$). However, the above calculation shows that

$$\begin{aligned} {{\mathrm{Darit}}}_W\cdot {{\mathrm{da}}}(F)= -{{\mathrm{da}}}\bigl ([D_U,D_V](F)\bigr )={{\mathrm{Darit}}}_P\cdot {{\mathrm{da}}}(F), \end{aligned}$$

where $P=D_U(V)-D_V(U)$, that is $D_P=[D_U,D_V]$. Thus the two derivations ${{\mathrm{Darit}}}_W$ and ${{\mathrm{Darit}}}_P$ take the same values on all elements F of ${{\mathrm{Lie}}}[a,b]$, so $W=P$, i.e.,

$$\begin{aligned}{}[D_U,D_V]=D_P=D_W, \end{aligned}$$

so under the map $D_U\mapsto {{\mathrm{Da}}}(U)$ of (6.6), we have

$$\begin{aligned}{}[D_U,D_V]=D_W\mapsto {{\mathrm{Da}}}(W)={{\mathrm{ari}}}\bigl ({{\mathrm{Da}}}(U),{{\mathrm{Da}}}(V)\bigr ). \end{aligned}$$

This proves Proposition 6.2. $\square$