\begin{document}
\maketitle

\begin{abstract}
We extend the block filtration, defined by Brown based on the work of Charlton, to all motivic multiple zeta values, and study relations compatible with this filtration. We construct a Lie algebra describing relations among motivic multiple zeta values modulo terms of lower block degree, proving Charlton's cyclic insertion conjecture in this structure, and showing the existence of a `block shuffle' relation, a dihedral symmetry, and differential relation.
\end{abstract}

\section{Introduction}
Consider the Tannakian category $\mathcal{M}\mathcal{T}(\mathbb{Z})$ of mixed Tate motives over $\text{Spec}(\mathbb{Z})$, with Galois group $\text{G}_{\mathcal{M}\mathcal{T}(\mathbb{Z})}$. It is well known \cite{delmixedtate} that this group decomposes as a semidirect product of the multiplicative group and a pro-unipotent group, whose graded Lie algebra is non-canonically isomorphic to a free Lie algebra with generators in odd weight
$$\gmot\cong\text{Lie}[\sigma_3,\sigma_5,\ldots]$$

\noindent called the motivic Lie algebra.
There exists a non-canonical injection $i:\{\sigma_{2k+1}\}_{k\geq 1}\to\qpoly$, allowing us to consider elements of $\gmot$ as noncommutative polynomials. This Lie algebra has ties to the Grothendieck-Teichmuller group \cite{racinet}, associators \cite{drinfeld}, and multiple zeta values \cite{brownmixedtate}, and injects into Racinet's double shuffle Lie algebra $\dmr$ \cite{racinet}.

This Lie algebra is known to be dual to (motivic) multiple zeta values, in the following sense: in \cite{goncharov}, Goncharov defines a bialgebra $\mathcal{A}$ of unipotent de Rham multiple zeta values, arising from the fundamental groupoid of $\P\setminus\{0,1,\infty\}$, which act as formal analogues to multiple zeta values (modulo $\z(2)$), endowed with a coproduct. Brown \cite{brownmixedtate} extends this definition to a comodule $\mathcal{H}$ over the graded ring of affine functions over $\text{G}_{\mathcal{M}\mathcal{T}(\mathbb{Z})}$, called motivic multiple zeta values. This algebra, along with its coaction
$$\Delta:\mathcal{H}\to\mathcal{A}\otimes\mathcal{H}$$

\noindent encodes all motivic relations among multiple zeta values. We have that $\mathcal{A}=\mathcal{H}/(\zm(2))$, where $\zm(2)$ is the motivic analogue of $\z(2)$. Taking the Lie coalgebra of indecomposables of $\mathcal{A}$, we obtain the dual of $\gmot$.
\begin{remark}
We take our convention as identifying $e_i\leftrightarrow \frac{dz}{z-i}$. As such, elements of $\gmot$ describe relations among iterated integrals, rather than multiple zeta values. As such, our results are `depth-signed'.
\end{remark}

From this, we can see that $\gmot$ inherits two filtrations arising from filtrations on motivic multiple zeta values. The multiple zeta value (MZV)
$$\z(n_1,\ldots,n_r) := \sum_{1\leq k_1<k_2<\cdots<k_r}\frac{1}{k_1^{n_1}\ldots k_r^{n_r}}$$
is said to have \textit{weight} $n_1+\cdots+n_r$ and \textit{depth} $r$. Considering $\z$ instead as a function $\qpoly\to\C$ via iterated integrals, as described in \cite{waldschmidt}, we extend these notions to $\qpoly$, where the weight $|w|$ of a word $w\in\{e_0,e_1\}^\times$ is given by its length, and the depth $d(w)$ of $w$ is the number of occurrences of $e_1$.
Thus we define

\begin{equation*}
\begin{split}
&\mathcal{W}_n\qpoly:=\langle w\ :\ |w|\leq n\rangle_\Q\\
&\mathcal{D}_n\qpoly:=\langle w\ :\ d(w)\leq n \rangle_\Q
\end{split}
\end{equation*}

These induce increasing filtrations on $\mathcal{H}$ and hence decreasing filtrations on $\gmot$, given via its embedding into $\qpoly$ (viewed here as the graded dual of $\qpoly$) by


\begin{equation*}
\begin{split}
\mathcal{W}^n\qpoly&:=\langle w\ :\ |w|\geq n\rangle_\Q\\
\mathcal{D}^n\qpoly&:=\langle w\ :\ d(w)\geq n \rangle_\Q
\end{split}
\end{equation*}

It is known that weight is a grading for $\gmot$, but depth is not. In \cite{depthgraded}, Brown studies the associated bigraded Lie algebra $\dg:=\gr_\mathcal{D}\gmot$ and its embedding into the linearised double shuffle algebra $\ls$. However, while he finds that  the depth 1 part of $\sigma_{2k+1}$ define canonical representatives of $\sigma_{2k+1}$ in $\dg\to\qpoly$, there also exist relations in $\dg$, and hence `exceptional' generators are needed. These relations are shown to have a somewhat mysterious connection to modular forms by Pollack \cite{pollack}, and this has been further explored by Baumard and Schneps \cite{schneppollack}. However, it is a computationally challenging task, and suggests that `depth graded' multiple zeta values may not be the most natural choice of object to study.

In this paper, we propose an alternative graded analogue of MZVs, based on the work of Charlton \cite{charthesis}. We go on to show that Charlton's cyclic insertion conjecture holds in this setting and additionally show a new family of `block shuffle' relations hold.


\section{The block filtration}
In addition to the weight and depth filtrations, we will define a `block filtration' on (motivic) multiple zeta values, arising from the work of Charlton \cite{charthesis}. In his thesis, Charlton defines the block decomposition of a word in two letters $\{x,y\}$ as follows.

Begin by defining a word in $\{x,y\}$ to be \textit{alternating} if it is non empty and has no subsequences of the form $xx$ or $yy$. There are exactly two alternating words of any given length: one beginning with $x$ and one beginning with $y$. Charlton shows that every non-empty word $w\in\{x,y\}^\times$ can be written uniquely as a minimal concatenation of alternating words. In particular, he defines the block decomposition $w=w_1w_2\ldots w_k$ as the unique factorisation into alternating words such that the last letter of $w_i$ equals the first letter of $w_{i+1}$.

We can use this to define a degree function on words in two letters.

\begin{defn}
Let $w\in\{x,y\}^\times$ be a word of length $n$, given by $w=a_1\ldots a_n$. Define its block degree $\degb(w)$ to be one less than the number of alternating words in its block decomposition. Equivalently, define
$$\degb(w):=\#\{i : 1\leq i < n\text{ such that }a_i=a_{i+1}\}$$
\end{defn}

\begin{remark}
	Note that, unlike depth, the block degree of a word is preserved by the duality anti-homomorphism mapping $e_i\mapsto e_{1-i}$, induced by the automorphism $z\mapsto 1-z$ of $\P^1\setminus\{0,1,\infty\}$.
\end{remark}

We can then define an increasing filtration on $\qpoly$ by
$$\mathcal{B}_n\qpoly := \langle w\ :\ \degb(w)\leq n \rangle_\Q$$

\noindent which, following the suggestion of Brown \cite{francisnote}, when restricted to a filtration on $e_1\qpoly e_0$ induces a filtration on motivic multiple zeta values
$$\mathcal{B}_n\mathcal{H}:=\langle \zm(w)\ : \ w=e_1ue_0,\ \degb(w)\leq n\rangle_\Q.$$
Brown goes on to show the following.

\begin{prop}[Brown]\label{blockmotivic}
Let $G_{\mathcal{M}\mathcal{T}(\mathbb{Z})}^{dR}$ denote the de Rham motivic Galois group of the category $\mathcal{M}\mathcal{T}(\mathbb{Z})$, and let $U_{\mathcal{M}\mathcal{T}(\mathbb{Z})}^{dR}$ denote its unipotent radical. Then $\mathcal{B}_n$ is stable under the action of $G_{\mathcal{M}\mathcal{T}(\mathbb{Z})}^{dR}$, and $U_{\mathcal{M}\mathcal{T}(\mathbb{Z})}^{dR}$ acts trivially on $\text{gr}^{\mathcal{B}} \mathcal{H}$. Equivalently
$$\Delta^r(\mathcal{B}_n\mathcal{H})\subset\mathcal{O}(U_{\mathcal{M}\mathcal{T}(\mathbb{Z})}^{dR})\otimes\mathcal{B}_{n-1}\mathcal{H}$$
where $\Delta^r(x):=\Delta(x)-x\otimes 1 -1\otimes x$ is the reduced coproduct.
\end{prop}

\begin{cor}[Brown]\label{blocklevel}
The block filtration induces the level filtration on the subspace spanned by the Hoffman motivic multiple zeta values $\zm(n_1,\ldots,n_r)$, with $n_i\in\{2,3\}$, where the level is the number of indices equal to $3$.
\end{cor}
\begin{proof}
The word corresponding to $(n_1,\ldots,n_r)$, with $n_i\in\{2,3\}$ of level $m$ has exactly $m$ occurrences of the subsequence $e_0e_0$ and none of $e_1e_1$. Therefore, its block degree is exactly $m$.
\end{proof}

As a corollary to both this and Brown's proof that the Hoffman motivic multiple zeta values form a basis of $\mathcal{H}$, we obtain the following.

\begin{cor}[Brown]
Every element in $\mathcal{B}_n\mathcal{H}$ of weight $N$ can be written uniquely as a $\Q$-linear combination of motivic Hoffman elements of weight $N$ and level at most $n$. Additionally
$$\sum_{m,\ n\geq 0}\dim\ \text{gr}^\mathcal{B}_m\mathcal{H}_ns^mt^n = \frac{1}{1-t^2-st^3}$$
where $\mathcal{H}_n$ denotes the weight $n$ piece of $\mathcal{H}$.
\end{cor}

However, trying to naively extend this filtration by
$$\mathcal{B}_n\mathcal{H}=\langle\zm(w)\ :\ \degb(w)\leq n\rangle_\Q$$
we find that the associated graded $\text{gr}^\mathcal{B}\mathcal{H}$ becomes nearly trivial. If we instead extend the filtration as follows, we obtain a much more interesting structure.

\begin{defn}\label{trueblockdef}
We define the block filtration of $\qpoly$ by

$$\mathcal{B}_n\qpoly := \langle w\ : \ \degb(e_0we_1)\leq n\rangle_\Q.$$

This induces the block filtration of motivic multiple zeta values

$$\mathcal{B}_n\mathcal{H} := \langle \zm(w)\ : \ \degb(0w1)\leq n\rangle_\Q.$$
\end{defn}

\noindent This filtration agrees with our earlier definition if we restrict to $w\in e_1\qpoly e_0$, but the associated graded remains interesting.

\begin{prop}\label{blockfullmotivic}
$$\Delta^r\mathcal{B}_n\mathcal{H} \subset \sum_{k=1}^{n-1} \mathcal{B}_{k}\mathcal{A}\otimes \mathcal{B}_{n-k}\mathcal{H}$$
\end{prop}
\begin{proof}
We will in fact show a stronger statement, that $\Delta$ is graded for block degree at the level of words. Let $\mathcal{I}:=\langle \iform(0;w;1)\ :\ w\in\{0,1\}^\times\rangle_\Q$ be the vector space spanned by formal symbols, with natural projection
\begin{equation*}
\begin{split}
\mathcal{I}&\to\mathcal{H},\\
\iform(0;w;1)&\mapsto \im(0;w;1)
\end{split}
\end{equation*}
and, similarly, a natural projection $\mathcal{I}\to\mathcal{A}$.

Recall that the motivic coaction is given by the formula 

\begin{equation*}
\begin{split}
\Delta I^\mathfrak{m}(a_0;a_1,\ldots,a_n;a_{n+1}):&= \\
\sum_{0=i_0<i_1<\ldots<i_k<i_{k+1}=n+1}\prod_{p=0}^k I^\mathfrak{a}(a_{i_p};a_{i_p+1},\ldots,a_{i_{p+1}-1};a_{i_{p+1}})&\otimes I^\mathfrak{m}(a_0;a_{i_1},\ldots,a_{i_k};a_{n+1})
\end{split}
\end{equation*}
where $0\leq k \leq n$. The infinitesimal coactions are given by 

\begin{equation*}
\begin{split}
D_{2r+1}:\mathcal{H}_N \to \mathcal{L}_{2r+1}&\otimes\mathcal{H}_{N-2r-1}\\
I^\mathfrak{m}(a_0;a_1,\ldots,&a_N;a_{N+1})\mapsto \\
\sum_{p=0}^{N-2r-1}I^\mathfrak{a}(a_p;a_{p+1},\ldots,a_{p+2r+1};a_{p+2r+2})&\otimes I^\mathfrak{m}(a_0;a_1,\ldots,a_p,a_{p+2r+2},\ldots,a_N;a_{N+1})
\end{split}
\end{equation*}
where $I^\mathfrak{a}$ is taken to be its projection into $\mathcal{L}:=\mathcal{A}_{>0}/\mathcal{A}_{>0}\mathcal{A}_{>0}$. Note that these lift to coactions $\mathcal{I}\to \mathcal{I}\otimes\mathcal{I}$, by calculating these purely symbolically.

Define $\mathcal{I}_n:= \langle \iform(0;w;1)\ : \ \degb(0w1)=n\rangle_\Q$. It is sufficient to show that $\Delta\mathcal{I}_n\subset \sum_{i=0}^n\mathcal{I}_i\otimes\mathcal{I}_{n-i}$, as the result follows upon composition with the necessary projections. In fact, it suffices to show that 
$$\D_{2r+1}\mathcal{I}_n\subset \sum_{i=0}^n\mathcal{I}_i\otimes\mathcal{I}_{n-i}.$$
Now, consider $\iform(0;w;1)$, $w$ a word in $\{0,1\}$ such that $\degb(0w1)= n$. Then we can decompose $0w1=b_1b_2\ldots b_{n+1}$ into alternating blocks, and consider the action of $D_{2n+1}$ on $\iform(b_1\ldots b_{n+1})$. All terms in $D_{2n+1}\iform(b_1\ldots b_{n+1})$ will be of the form
$$\iform(x;b_i''b_{i+1}\ldots b_{i+j}';y)\otimes \iform(b_1\ldots b_{i-1}b_i'xyb_{i+j}''b_{i+j+1}\ldots b_{n+1})$$
\noindent
for some $1\leq i \leq {n+1}$, where $b_i = b_i'xb_i''$, $b_{i+j}=b_{i+j}'yb_{i+j}''$. For the left hand term to be non-zero, we must have $x\neq y$, and so we see
\begin{equation*}
\begin{split}
\degb(xb_i''b_{i+1}\ldots b_{i+j}'y) &= j,\\
\degb(b_1\ldots b_i'xyb_{i+j}''\ldots b_{n+1}) &= n - j,
\end{split}
\end{equation*}
\noindent
by counting the blocks. Thus, we get that the total block degree of any term in the coproduct is $n$, and the result follows.
\end{proof}


\begin{cor}\label{coradical}
The block filtration on $\qpoly$ induces the coradical filtration on $\mathcal{H}$.
\end{cor}

\begin{cor}\label{iharagraded}
The (linearised) Ihara action $\circ:\text{Lie}[e_0,e_1]_{\geq 2}\otimes \qpoly \to \qpoly$ is graded for block degree, where $\text{Lie}[e_0,e_1]_{\geq 2}$ is the space of Lie polynomials of degree at least 2.
\end{cor}
\begin{proof}
The Ihara action is dual to the motivic coaction. As this proof shows the coaction to be, at the level of words, graded for block degree, the claim follows immediately. One can also show this directly via the recursive formula \cite{anatomy} for the linearised Ihara action.
\end{proof}

We also recall a short observation due to Charlton \cite{charthesis}.

\begin{lem}\label{blockparity}
Let $w=w_1\ldots w_n$ be a word in $\{0,1\}^\times$ of length $n$, with $\degb(w)=b$. Then $\im(w)=0$ if $b\equiv w+1\text{ (mod 2)}$.
\end{lem}


\begin{remark}
This provides a natural analogue of the depth parity theorem \cite{depthgraded}.
\begin{prop}
Suppose $\sigma\in\mathfrak{ls}$ is of weight $N$ and depth $d$. Then, if $N$ and $d$ are of opposite parity, $\sigma=0$. That is, there are no non-trivial solutions to the linearised double shuffle equations with weight and depth of opposite parity.
\end{prop}
With Proposition \ref{canonicalgen}, we obtain a similar corollary to the final conclusion of the following corollary.
\begin{cor}
For a solution to the double shuffle equations mod products $\phi\in\dmr$, of weight  $N$, the depth $d+1\not\equiv N\text{ (mod 2)}$ components are uniquely determined by the lower depths. In particular, $\sigma_{2n+1}$ is uniquely determined in depths 1 and 2.
\end{cor}
Specifically, $\sigma_{2n+1}$ is uniquely determined in block degree 1 and 2.
\end{remark}

\section{Block-graded multiple zeta values and an encoding of relations}
As the block filtration is motivic and invariant under the duality arising from the symmetry $z \mapsto 1-z$ of $\P^1\setminus\{0,1,\infty\}$, we can consider the associated graded algebra $\gr^\mathcal{B}\mathcal{A}:=\bigoplus_{n=0}^\infty \mathcal{B}_n\mathcal{A}/\mathcal{B}_{n-1}\mathcal{A}$. We follow the example of Brown's depth graded multiple zeta values \cite{depthgraded}.

\begin{remark}
In the following it is important to keep in mind that we are identifying $e_i\leftrightarrow \frac{dz}{z-i}$, and, as such, elements of $\gmot$ describe relations among iterated integrals, rather than multiple zeta values. As such, our results are `depth signed' compared to standard notation.
\end{remark}

\begin{defn}
Define $\mathcal{B}^n\qpoly:=\langle w\ : \ \degb(e_0we_1)\geq n\rangle_\Q$ and define $$\gr_\mathcal{B}\gmot := \bigoplus_{n=0}^\infty \mathcal{B}^n\gmot/\mathcal{B}^{n+1}\gmot$$ where we identify $\mathcal{B}^n\gmot/\mathcal{B}^{n+1}\gmot$ with its image in $\mathcal{B}^n\Q\langle e_0,e_1\rangle/\mathcal{B}^{n+1}\Q\langle e_0,e_1\rangle$, equipped with the block graded Ihara bracket.
\end{defn}

\begin{defn}
If  $\degb(e_0we_1)=n$, define $\ib(0;w;1)$ to be the image of $\ia(0;w;1)$ in $\mathcal{B}_n\mathcal{A}/\mathcal{B}_{n-1}\mathcal{A}$. Similarly, define $\ibl(0;w;1)$ to be the image of $\il(0;w;1)$ in $\mathcal{B}_n\mathcal{L}/\mathcal{B}_{n-1}\mathcal{L}$. Define $\zb$ and $\zbl$ similarly.
\end{defn}

\begin{defn}
Fix an embedding of $\{\sigma_3,\sigma_5,\ldots\}\hookrightarrow \qpoly$. We define the block graded generators $\{p_{2k+1}\}_{k\geq 1}$ to be the image of the generators $\{\sigma_{2k+1}\}_{k\geq 1}$ of $\gmot$ in $\mathcal{B}^1\qpoly/\mathcal{B}^2\qpoly$. We define the bigraded Lie algebra $\bg$ to be the Lie algebra generated by $p_{2k+1}$ and the Ihara bracket.
\end{defn}

One of the challenges in studying $\gmot$ is that we have an ambiguity in our representation of the generators: $\sigma_{2k+1}$ is unique only up to addition of another element of weight $2k+1$. Its depth one part is canonical, so Brown's depth graded Lie algebra avoids this issue. We find similar success here.

\begin{prop}\label{canonicalgen}
The generators $p_{2k+1}$ of $\bg$ are canonical, i.e. independent of our choice of embedding of generators $\{\sigma_{2k+1}\}\hookrightarrow \qpoly$.
\end{prop}
\begin{proof}
Let $\sigma_{2k+1}$, $\sigma_{2k+1}^\prime\in \qpoly$ be two choices of generator for $\gmot$ in weight $2k+1$. We must have
$$\sigma_{2k+1}-\sigma_{2k+1}^\prime\in \{\gmot,\gmot\}.$$
Proposition \ref{iharagraded} tells us that the Ihara action is compatible with the block filtration, and so
$$ \{\gmot,\gmot\}\subset \mathcal{B}^2\gmot$$
and therefore
$$p_{2k+1}-p_{2k+1}^\prime= 0.$$
\end{proof}


Note that we can still define a concept of depth on $\bg$ as before. We define the depth of a word $w$ to be $d(w)$, and induce a decreasing filtration on $\bg$ via its embedding $\bg\hookrightarrow\qpoly$. It is interesting here that depth grading gives canonical generators in depth 1, while block grading gives $p_{2k+1}$ consisting only of terms of depth $k$ or $k+1$.

\begin{lem}\label{generatordepth}
$p_{2k+1}$ contains only depth $k$ and $k+1$ terms.
\end{lem}
\begin{proof}
Suppose $w$ is a word of block degree 1 and weight $2k+1$. Then $e_0we_1$ has two blocks and hence contains exactly one of $e_0^2$ or $e_1^2$. In the first case, the number of $e_1$ must be exactly half $2k+1 - 1$, i.e. $k$. In the second case, the number of $e_0$ must similarly be $k$ and hence the number of $e_1$ is $k+1$.
\end{proof}

\begin{thm}\label{gradedisbg}
$\bg$ is freely generated by $\{p_{2k+1}\}_{k\geq 1}$ as a Lie algebra.
\end{thm}
\begin{proof}
We have a bijection between the generators of $\gmot$ and of $\bg$, and Corollary \ref{iharagraded} tells us that the Ihara action is graded for block degree. Thus, we can write an element $$\{p_{2k_1+1},\{\ldots,\{p_{2k_{b-1}+1},p_{2k_b+1}\},\ldots\}\}$$ as the image of $$\{\sigma_{2k_1+1},\{\ldots,\{\sigma_{2k_{b-1}+1},\sigma_{2k_b+1}\},\ldots\}\}$$ in $\mathcal{B}^b\gmot/\mathcal{B}^{b+1}\gmot$. Hence, we have a relation in $\bg$ if and only if the corresponding sum of terms is 0 in $\text{gr}_{\mathcal{B}}\gmot$. Indeed, we have an injective Lie algebra homomorphism $\bg\hookrightarrow\text{gr}_\mathcal{B}\gmot$ induced by the bijection $\{\sigma_{2k+1}\}_{k\geq 1}\leftrightarrow\{p_{2k+1}\}_{k\geq 1}$. 
Now, as $\text{gr}_{\mathcal{B}}\gmot$ is dual to $\text{gr}^{\mathcal{B}}\mathcal{L}$, the existence of relations in $\bg$ implies the existence of additional relations in $\text{gr}^{\mathcal{B}}\mathcal{L}$. To be precise, we must have that $$\dim\ \text{gr}^\mathcal{B}_n\mathcal{L}_N<\dim\ \langle \il(w) | \degb(w)=n,\ |w|=N\rangle_Q.$$
Then, by the proof of Theorem 7.4 in \cite{brownmixedtate}, we know that the right hand side has is spanned by $\{\za(k_1,\ldots,k_r)\}$, where $k_i\in\{3,2\}$,  and $k_i=3$ exactly $n$ times and $k_1+\cdots+k_r=N$. In particular, it has a basis given by $\za(k_1,\ldots,k_r)$ such that $(k_1,\ldots,k_r)$ is a Lyndon word with respect to the order $3<2$. This basis, called the Hoffman-Lyndon basis, forms a spanning set for $\text{gr}^\mathcal{B}_n\mathcal{L}_N$. Thus,
$$\dim\ \text{gr}^\mathcal{B}_n\mathcal{L}_N<\dim\ \langle \ia(w) | \degb(w)=n,\ |w|=N\rangle_Q$$
which implies that there is a sum of Hoffman-Lyndon elements of weight $N$ with $n$ threes that can be written as a sum of Hoffman-Lyndon elements of weight $N$ with fewer threes. However, the Hoffman-Lyndon elements of weight $N$ form a basis of $\mathcal{L}_N$, and, so, no such relation can exist. Thus, we must have that $\mathcal{L}\equiv\text{gr}^\mathcal{B}\mathcal{L}$ as they have equal dimensions, and hence, $\text{gr}_\mathcal{B}\gmot\equiv \gmot$. This implies $\text{gr}_\mathcal{B}\gmot$ and $\bg$ are both freely generated and isomorphic.
\end{proof}

\begin{remark}
While both Brown's $\mathfrak{dg}$ and our $\bg$ have canonical generators, Theorem \ref{gradedisbg} tells us that $\bg$ is free, while there exist relations in $\dg$, and hence `exceptional' generators are needed, first appearing in depth four. These relations are shown to have a somewhat mysterious connection to modular forms by Pollack \cite{pollack}, and this has been further explored by Baumard and Schneps \cite{schneppollack}. However, it is a computationally challenging task, and suggests that `depth graded' multiple zeta values may not be the most natural choice of object to study.
\end{remark}

\section{Polynomial representations}
We now reframe this Lie algebra in terms of commutative polynomials, similarly to Brown \cite{anatomy}\cite{zeta3} and \'Ecalle \cite{ecalleassoc}, as follows.

Recall that Charlton shows that every word $w\in\{ e_0,e_1\}^\times$ can be written uniquely as a sequence of alternating blocks \cite{charthesis}. In doing so, he establishes a bijection
\begin{equation*}
\begin{split}
\text{bl}:\{e_0,e_1\}^\times\setminus\{\emptyset\} &\to \cup_{n=1}^\infty \{0,1\}\times \mathbb{N}^n\\
w&\mapsto (\epsilon;l_1,l_2\ldots,l_n)
\end{split}
\end{equation*}
where $\epsilon$ defines the first letter of $w$, and $l_1,\ldots,l_n$ describe the length of the alternating blocks.

\begin{example}
\begin{equation*}
\begin{split}
e_0e_1e_0e_0e_1e_0e_1e_1&\mapsto (0;3,4,1),\\
e_1e_1e_0e_1e_0e_1e_1e_0e_0&\mapsto (1;1,5,2,1)
\end{split}
\end{equation*}
\end{example}

We can use this to define an injection of  vector spaces by
\begin{equation}\label{isomorph}
\begin{split}
\pi_\text{bl}:\qpoly\setminus\{\Q\cdot 1\}&\to \bigoplus_{n=1}^\infty x_1\Q[x_1,\ldots,x_n]x_n\\
w&\mapsto x_1^{l_1}\ldots x_n^{l_n}
\end{split}
\end{equation}
where $\text{bl}(e_0we_1)=(0;l_1,\ldots,l_n)$.

In this formulation, a word of block degree $n$ and weight $N\geq 1$ is  represented by a polynomial in $n+1$ variables of degree $N+2$. From this point on, we shall freely identify elements of $\bg$ with their images under this injection.

\begin{prop}\label{generatordef}
The projections of the depth-signed $\sigma_{2k+1}\in\gmot$ onto their block degree one part are given by
$$p_{2k+1}(x_1,x_2) = q_{2k+1}(x_1,x_2) - q_{2k+1}(x_2,x_1)$$
where
\begin{equation*}
q_{2k+1}(x_1,x_2)=\sum_{i=1}^{k} \left[\binom{2k}{2i} - \left(1-\frac{1}{2^{2k}}\right)\binom{2k}{2k+1-2i}\right]x_1^{2i+1}x_2^{2k+2-2i} -  x_1x_2^{2k+2}
\end{equation*}
and $\sigma_{2k+1}$ have been normalised to correspond to $\frac{(-1)^k}{2}\zeta(2k+1)$.
\end{prop}

\begin{proof}
We will compute the block degree 1 part of $\sigma_{2k+1}$ consisting of terms containing an $e_0^2$. This will give $q_{2k+1}$. That $p_{2k+1}(x_1,x_2)=q_{2k+1}(x_1,x_2)-q_{2k+1}(x_2,x_1)$ follows from duality. In terms of $e_0,\ e_1$, we have
$$q_{2k+1}= \sum_{i=0}^k c_i(e_1e_0)^ie_0(e_1e_0)^{k-i},$$
where $\zm(\{2\}^{i-1},3,\{2\}^{k-i}) = \alpha c_i\zm(2k+1) \text{ (mod }\zm(2)\text{)}$, for $i>0$ and some $\alpha\in\Q$, and $c_0$ is obtained via shuffle regularisation \cite{brownmixedtate}.

Shuffle regularisation of $e_0e_1\ldots e_0$ tells us that
$$c_0 + 2\sum_{i=1}^k c_i = 0.$$
Next, from the work of Zagier \cite{zaghoff},
$$\zeta(\{2\}^{a},3,\{2\}^b) = 2\sum_{r=1}^{a+b+1}(-1)^r\left[\binom{2r}{2a+2}-(1-\frac{1}{2^{2r}})\binom{2r}{2b+1}\right]\zeta(\{2\}^{a+b-r+1})\zeta(2r+1).$$
Brown then shows in \cite{brownmixedtate} Theorem 4.3 that this lifts to an identity among motivic multiple zeta values. Considered modulo $\zm(2)$, we find

$$\zm(\{2\}^{i-1},3,\{2\}^{k-i})  = 2(-1)^{k}\left[\binom{2k}{2i} - \left(1-\frac{1}{2^{2k}}\right)\binom{2k}{2k+1-2i}\right]\zm(2k+1),$$
and thus, we can take $c_i=\left[\binom{2k}{2i} - (1-\frac{1}{2^{2k}})\binom{2k}{2k+1-2i}\right]$ for $i>0$. The result then follows.
\end{proof}

Computing these sums explicitly, we obtain the following theorem.

\begin{thm}\label{summedgen}
$$p_{2k+1}(x_1,x_2)= x_1x_2(x_1-x_2)\left(\frac{(1-2^{2k+1})(x_1+x_2)^{2k}-(x_1-x_2)^{2k}}{2^{2k}}\right).$$
\end{thm}

With this in mind, we can provide a characterisation of these generators in terms of polynomial equations.

\begin{cor}\label{newgendef}
The polynomial $p_{2k+1}(x_1,x_2)$ is, up to rescaling, the unique homogeneous polynomial $p(x_1,x_2)$ of degree $2k+3$ such that
$$p(x_1,0)=p(0,x_2)=p(x_1,x_2)+p(x_2,x_1)=0,$$
and, defining $r(x_1,x_2):=\frac{p(x_1,x_2)}{x_1x_2(x_1-x_2)}$, satisfying
$$r(0,x)=2r(x,-x),$$
and
$$\left(\partialx{1}\right)^2 r(x_1,x_2)=\left(\partialx{2}\right)^2r(x_1,x_2).$$
\end{cor}

\begin{proof}
The condition $p(x_1,0)=p(0,x_2)=p(x_1,x_2)+p(x_2,x_1)=0$ suggests we can write $p(x_1,x_2)=x_1x_2(x_1-x_2)r(x_1,x_2)$. Letting $u=x_1+x_2$, and $v=x_1-x_2$, we can rewrite 

$$\left(\partialx{1}\right)^2 r(x_1,x_2)=\left(\partialx{2}\right)^2r(x_1,x_2)\Leftrightarrow \frac{\partial^2 r}{\partial u\partial v}(u,v)=0,$$
which has polynomial solution, homogeneous of degree $(2k+3)-3=2k$
$$r(u,v)=\alpha u^{2k} + \beta v^{2k}$$
which is to say
$$r(x_1,x_2)=\alpha(x_1+x_2)^{2k}+\beta(x_1-x_2)^{2k}.$$
Finally, the condition
$$r(0,x)=2r(x,-x)$$
gives
$$(\alpha + \beta)x^{2k} = 2^{2k+1}\beta x^{2k},$$
and hence
$$\alpha = -(1-2^{2k+1})\beta,$$
giving the desired result.
\end{proof}

We can provide an exact polynomial formula for the Ihara action. Recall that we have chosen $\gmot$ to differ from Brown's by sending $e_1\mapsto -e_1$, and so this is only accurate for `depth-signed' elements. We delay the proof of this until later.

\begin{thm}\label{Iharanice}
For (depth-signed) elements of the motivic Lie algebra, the Ihara action is given at the level of block-polynomials by

\begin{equation}\label{prettyIhara}
\begin{split}
(f\circ g) (x_1,\ldots,x_{m+n-1}) &= (-1)^{(m+1)(n+1)}\sum_{i=1}^{n}\frac{f (x_i,x_{i+1},\ldots,x_{i+m-1})}{x_i-x_{i+m-1}}\\
&\times\left( \frac{1}{x_i}g (x_1,\ldots,x_{i-1},x_i,x_{i+m},\ldots,x_{m+n-1})\right.\\
&-\left.\frac{1}{x_{i+m-1}}g (x_1,\ldots,x_{i-1},x_{i+m-1},\ldots,x_{m+n-1})\right).
\end{split}
\end{equation}

\end{thm}
\section{Relations arising in the polynomial representation}
We find several relations arising naturally in the polynomial representation which are preserved by the Ihara action, and dual to relations in $\gr^\mathcal{B}\mathcal{L}$. We prove these by inducting on block degree. To illustrate this method, we will reprove the duality identity.

\begin{prop}\label{duality}
For all $f(x_1,\ldots,x_n)\in\bg$,
$$f(x_1,\ldots,x_n)=(-1)^{n+1}f(x_n,\ldots,x_1).$$ 
\end{prop}
\begin{proof}
It suffices to show that this holds for $p_{2k+1}$, and that, if this holds for $f,g\in \bg$, then it holds for $f\circ g$. The former holds by definition of $p_{2k+1}$. To see the latter, note that
\begingroup
\allowdisplaybreaks
\begin{align*}
(f\circ g) (x_{m+n-1},\ldots,x_{1}) &= (-1)^{(m+1)(n+1)}\sum_{i=1}^{n}\frac{f (x_{m+n-i},x_{m+n-i-1},\ldots,x_{n+1-i})}{x_{m+n-i}^2-x_{n+1-i}^2}\\
&\times\left((1 + \frac{x_{n+1-i}}{x_{m+n-i}})g (x_{m+n-1},\ldots,x_{m+n-i+1},x_{m+n-i},x_{n-i},\ldots,x_{1})\right.\\
&-\left. (1+\frac{x_{m+n-i}}{x_{n+1-i}})g (x_{m+n-1},\ldots,x_{m+n-i+1},x_{n+1-i},\ldots,x_{1})\right)\\
&= (-1)^{(m+1)(n+1)}\sum_{i=1}^{n}(-1)^m\frac{f (x_{n+1-i},x_{n+2-1},\ldots,x_{m+n-i})}{x_{n+1-i}^2-x_{m+n-i}^2}\\
&\times\left((-1)^{n+1}(1 + \frac{x_{n+1-i}}{x_{m+n-i}})g (x_{1},\ldots,x_{n-i},x_{m+n-i},x_{m+n-i},\ldots,x_{m+n-1})\right.\\
&-\left. (-1)^{n+1}(1+\frac{x_{m+n-i}}{x_{n+1-i}})g (x_{1},\ldots,x_{n+1-i},x_{m+n-i+1},\ldots,x_{m+n-1})\right)\\
&=(-1)^{m+n} (-1)^{(m+1)(n+1)}\sum_{i=1}^{n}\frac{f (x_{n+1-i},x_{n+2-1},\ldots,x_{m+n-i})}{x_{n+1-i}^2-x_{m+n-i}^2}\\
&\times\left( (1+\frac{x_{m+n-i}}{x_{n+1-i}})g (x_{1},\ldots,x_{n+1-i},x_{m+n-i+1},\ldots,x_{m+n-1})\right.\\
&-\left.(1 + \frac{x_{n+1-i}}{x_{m+n-i}})g (x_{1},\ldots,x_{n-i},x_{m+n-i},x_{m+n-i},\ldots,x_{m+n-1})\right)\\
&=(-1)^{m+n}(f\circ g)(x_1,\ldots,x_{m+n-1}),
\end{align*}
\endgroup
and hence, the duality relation is preserved by the Ihara bracket.
\end{proof}

We can similarly prove Charlton's cyclic insertion conjecture, up to terms of lower block degree. While this has been verified in upcoming work due to Hirose-Sato, in this formulation, it is merely a consequence of the Ihara action, allowing for a significantly simpler proof. We will instead show that a more general relation holds, of which cyclic insertion is a corollary. These are the `block shuffle' relations.


\begin{defn}
For any $1\leq r \leq n$, define the shuffle set

$$\text{Sh}_{n,r}=\{\sigma\in\mathcal{S}_n | \sigma^{-1}(1)<\ldots<\sigma^{-1}(r);\sigma^{-1}(r+1)<\ldots<\sigma^{-1}(n)\}.$$

Then, for any $f\in\Q[x_1,\ldots,x_n]$, define

$$f(x_1\ldots x_r\sh x_{r+1}\ldots x_n):=\sum_{\sigma\in \text{Sh}_{n,r}}f(x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)}).$$
\end{defn}


\begin{thm}\label{blockshuffle}
For any $f(x_1,\ldots,x_n)\in\bg$, and any $1\leq r <n$, we have 
$$f(x_1x_2\ldots x_r\sh x_{r+1}\ldots x_n) = 0.$$
\end{thm}
\begin{proof}
For $p_{2k+1}$, this is equivalent to $p(x_1,x_2)+p(x_2,x_1)=0$, given by Proposition \ref{duality}. Then, as the Ihara action is associative, it in fact suffices to show that
$$(f\circ g)(x_1\ldots x_r\sh x_{r+1}\ldots x_{n+1}) = 0$$
for all $f=p_{2k+1}(x_1,x_2)$, $g(x_1,\ldots,x_n)\in\bg$.

 We write $(f\circ g)(x_1\ldots x_r\sh x_{r+1}\ldots x_{n+1})$ as

\begin{equation*}
\begin{split}
&\sum_{\sigma\in \text{Sh}_{n+1,r}}\sum_{i=1}^n\frac{f(x_{\sigma(i)},x_{\sigma(i+1)})}{x_{\sigma(i)}-x_{\sigma(i+1)}}\times\\
&\left(\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i)},x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i)}} -\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_{\sigma(i+1)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i+1)}}\right).
\end{split}
\end{equation*}

This sum splits as follows

\begin{equation*}
\begin{split}
&\sum_{\sigma\in \text{Sh}_{n+1,r}}\sum_{i=1}^{r-1}\frac{f(x_{\sigma(i)},x_{\sigma(i+1)})}{x_{\sigma(i)}-x_{\sigma(i+1)}}\times\\
&\left(\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i)},x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i)}} -\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_{\sigma(i+1)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i+1)}}\right)\\  
&+\sum_{\sigma\in \text{Sh}_{n+1,r}}\sum_{i=r+1}^n\frac{f(x_{\sigma(i)},x_{\sigma(i+1)})}{x_{\sigma(i)}-x_{\sigma(i+1)}}\times\\
&\left(\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i)},x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i)}} -\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_{\sigma(i+1)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(i+1)}}\right)\\
&+\sum_{\substack{\sigma\in \text{Sh}_{n+1,r}\\
\text{such that}\\
\{\sigma(r),\sigma(r+1)\}\neq\{r,r+1\}}}\frac{f(x_{\sigma(r)},x_{\sigma(r+1)})}{x_{\sigma(r)}-x_{\sigma(r+1)}}\times\\
&\left(\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(r)},x_{\sigma(r+2)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(r)}} -\frac{g(x_{\sigma(1)},\ldots,x_{\sigma(r-1)},x_{\sigma(r+1)},\ldots,x_{\sigma(n+1)})}{x_{\sigma(r+1)}}\right).
\end{split}
\end{equation*}


Denote the first sum by $A$, the second by $B$, and the third by $C$. Now, this sum can be written uniquely as
$$\sum_{1\leq k < l \leq n+1}\frac{f(x_k,x_l)}{x_k-x_l}\left(\frac{G_{k,l}}{x_k}-\frac{H_{k,l}}{x_l}\right)$$
where $G_{k,l}$, $H_{k,l}$ are polynomials related by swapping $x_k\leftrightarrow x_l$.
We have 4 cases to consider
\begin{enumerate}
    \item $l\leq r$,
    \item $k\geq r+1$,
    \item $k<r<r+1<l$,
    \item $k=r=l-1$.
\end{enumerate}

In the first case, both $A$ and $B$ only contribute non-zero terms if $l=k+1$, while $C$ only contributes if $l>k+1$. Thus, denoting by $\Phi_k(\sigma,i)$ the condition $\{\sigma(i)=k,\ \sigma(i+1)=k+1\}$, we have
\begin{equation*}
    \begin{split}
        G_{k,k+1}=&\sum_{k\leq i<r}\sum_{\substack{ \sigma\in \text{Sh}_{n+1,r}\\
\text{such that } \Phi_k(\sigma,i)}} g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_k,x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)})\\
        &+\sum_{i>r}\sum_{\substack{ \sigma\in \text{Sh}_{n+1,r}\\
\text{such that }\Phi_k(\sigma,i)}} g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_k,x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)})\\
        &+\sum_{\substack{\sigma\in \text{Sh}_{n+1,r}\\
\text{such that } \Phi_k(\sigma,r)}} g(x_{\sigma(1)},\ldots,x_{\sigma(i-1)},x_k,x_{\sigma(i+2)},\ldots,x_{\sigma(n+1)}).\\
    \end{split}
\end{equation*}
Let $P(\sigma,r)$ denote the condition
$$\{\sigma^{-1}(1)<\ldots<\sigma^{-1}(r);\sigma^{-1}(r+1)<\ldots<\sigma^{-1}(n+1)\}.$$
Then, this is a sum over the set of permutations
\begin{equation*}
        \cup_{k\leq i <r}\{\sigma|\Phi_k(\sigma,i)\text{ and } P(\sigma,r)\} \bigcup\cup_{i>r}\{\sigma|\Phi_k(\sigma,i)\text{ and } P(\sigma,r)\}
\end{equation*}
which is clearly in bijection with a set of shuffles of $[n+1]\setminus\{k+1\}$, and so the contribution is $0$ by induction.

Then, if $l>k+1$, we find that the non-zero terms in

$$\frac{f(x_k,x_l)}{x_k-x_l}\left(\frac{G_{k,l}}{x_k}-\frac{H_{k,l}}{x_l}\right)$$ 
due to permutations with $\sigma(r)=k,\sigma(r+1)=l$ cancel with those due to $\sigma(r)=l,\sigma(r+1)=k$. Thus, in this case, 
$$\frac{f(x_k,x_l)}{x_k-x_l}\left(\frac{G_{k,l}}{x_k}-\frac{H_{k,l}}{x_l}\right)=0.$$ 

The second case, $k\geq r+1$, is similar. In the third case, every term due to a permutation with $\sigma(i)=k,\sigma(i+1)=l$ cancels with the term due to the permutation $\tau_{k,l}\circ\sigma$, where $\tau_{k,l}$ is the transposition $(k,l)$.

Finally, in the fourth case, our sum splits into a sum over the following sets
$$\cup_{i<r}\{\sigma\in \text{Sh}_{n+1,r} | \sigma(i)=r,\sigma(i+1)=r+1\},$$
$$\cup_{i>r}\{\sigma\in \text{Sh}_{n+1,r} | \sigma(i)=r,\sigma(i+1)=r+1\},$$
$$\cup_{i<r}\{\sigma\in \text{Sh}_{n+1,r} | \sigma(i)=r+1,\sigma(i+1)=r\},$$
$$\cup_{i>r}\{\sigma\in \text{Sh}_{n+1,r} | \sigma(i)=r+1,\sigma(i+1)=r\}.$$

All of these must be empty due to the order preserving property of shuffle permutations. Thus,

$$(f\circ g)(x_1\ldots x_r\sh x_{r+1}\ldots x_{n+1}) = 0.$$
\end{proof}

\begin{cor}
For any finite sequence of integers $l_1,\ldots,l_n$, and any $1\leq r < n$, we have
$$\sum_{\sigma\in\text{Sh}_{n,r}}\ibl((l_{\sigma(1)},\ldots,l_{\sigma(n)})=0$$
when considered modulo products.
\end{cor}

\begin{proof}
Using Theorem \ref{gradedisbg}, we can consider $\bg$ as the dual Lie algebra to the graded Lie coalgebra of indecomposables $\gr^\mathcal{B}\mathcal{L}$, and hence, relations among the coefficients of elements of $\bg$ induce relations among elements of $\gr^\mathcal{B}\mathcal{L}$. Specifically, we define a $\Q$-linear pairing
$$\langle \ibl(l_1,\ldots,l_n) | x_1^{k_1}\ldots x_m^{k_m}\rangle := \delta_{l_1,k_1}\ldots\delta_{l_n,k_n}$$
where $\ibl(l_1,\ldots,l_n)$ is the image of $\ib(l_1,\ldots,l_n)$ in $\gr_\mathcal{B}\mathcal{L}$. We have that $R$ is a relation in $\gr^\mathcal{B}\mathcal{L}$ if and only if $\langle R | f\rangle=0$ for all $f\in\bg$. Hence, as $f(x_1x_2\ldots x_r\sh x_{r+1}\ldots x_n) = 0$ for all $f\in\bg$, we must have that

$$\sum_{\sigma\in\text{Sh}_{n,r}}\ibl((l_{\sigma(1)},\ldots,l_{\sigma(n)})=0.$$
\end{proof}

\begin{cor}[Block graded cyclic insertion]\label{cyclic}
The cyclic sum
$$\sum_{\sigma\in\text{C}_n}\ibl(l_{\sigma(1)},l_{\sigma(2)},\ldots,l_{\sigma(n)}) = 0.$$
\end{cor}
\begin{proof}
It suffices to show that
$$\sum_{\sigma\in\text{C}_n} f(x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)})=0$$
for all $f\in\bg$.

Suppose $f\in\bg$. Then, Theorem \ref{blockshuffle} implies that the image of $f$ under the following vector space isomorphism
\begin{equation}\label{zalphabet}
\begin{split}
\bigoplus_{n=0}^\infty \Q[x_1,\ldots,x_n] &\xrightarrow{\sim} \Q\langle z_1,z_2,z_3,\ldots\rangle\\
x_1^{i_1} x_2^{i_2}\ldots x_n^{i_n} &\mapsto z_{i_1}z_{i_2}\ldots z_{i_n}
\end{split}
\end{equation}
lies in $\text{Lie}[z_1,z_2,\ldots]$. In particular, the image lies in the span of elements of degree at least 2. Now, we define a linear map $\mathcal{C}:\Q\langle z_1,z_2,\ldots\rangle\to \Q\langle z_1,z_2,\ldots\rangle$ by
$$\mathcal{C}(z_{i_1}z_{i_2}\ldots z_{i_n}) = \sum_{\sigma\in\text{C}_n}z_{i_{\sigma(1)}}z_{i_{\sigma(2)}}\ldots z_{i_{\sigma(n)}}$$
for a word of length $n$. Thus, it suffices to show that $\mathcal{C}(Z)=0$ for all $Z\in\text{Lie}[z_1,z_2,\ldots]$ of degree at least 2.

Note, for any monomials $X,\ Y$ in $\{z_1,z_2,\ldots\}$ of degree $k,\ n-k$ respectively, we have $[X,Y] = XY - \sigma(XY)$, for some $\sigma\in\text{C}_n$ acting by cyclic rotations on words of length $n$. Thus,
$$\mathcal{C}([X,Y])=\mathcal{C}(XY)-\mathcal{C}(\sigma(XY))=\mathcal{C}(XY)-\mathcal{C}(XY)=0$$
and so the image of  any element of degree at least two in $\text{Lie}[z_1,z_2,\ldots]$ is zero, and hence 

$$\sum_{\sigma\in\text{C}_n} f(x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)})=0$$

\end{proof}

\begin{remark}
As in this proof, it can be useful to consider $\bg$ as a subspace of the Hopf $ \Q\langle z_1,z_2,z_3,\ldots\rangle$, with the standard concatenation product, and a coproduct given by $\Delta z_i =z_i\otimes 1 + 1\otimes z_i$. For example, Theorem \ref{blockshuffle} implies elements of $\bg$ are primitive for this coproduct, we immediately obtain Proposition \ref{duality} as a corollary, by considering the antipode map, i.e. the antihomomorphism $z_i\mapsto -z_i$. This is an idea explored further in the author's thesis
\end{remark}

\section{Shuffle Regularisation}


The double shuffle relations among iterated integrals are not, in general, compatible with the block filtration. However, the regularisation relation obtained by shuffling with an element of weight 1, does respect the block filtration.

\begin{thm}\label{weight1shuff}
Let $\pi_1:\qpoly\to\Q e_0\oplus\Q e_1$ denote the projection map onto weight 1 , and let $\Delta:\qpoly\to\qpoly\otimes\qpoly$ be the coproduct defined by $\Delta(e_i)=e_i\otimes 1 + 1\otimes e_i$. The map $\Delta_1:=(\pi_1\otimes\text{id})\Delta$ is compatible with the block filtration:
$$\Delta_1\mathcal{B}^n\qpoly\subset\mathcal{B}^1\qpoly\otimes\mathcal{B}^{n-1}\qpoly.$$
\end{thm}
\begin{proof}
For $w\in\qpoly$ every term in $\Delta_1(w)$ is of the form $e_i\otimes \bar{w}$ for $i\in\{0,1\}$, where $\bar{w}$ is obtained from $w$ by omitting a letter. The left hand side is of block degree 1. The right hand side is of higher block degree, if the omitted letter was internal to a block, and of block degree 1 lower than $w$, if the omitted letter was at the beginning or end of a block.
\end{proof}

Thus, we can take the associated graded map of $\Delta_1$.
\begin{cor}\label{gradedshuffle}
$\gr_\mathcal{B}(\Delta_1)(\bg)=0.$
\end{cor}
\begin{proof}
This follows from the work of Brown \cite{brownmixedtate} and Racinet \cite{racinet}, as any element $\psi\in\gmot$ satisfies $\Delta(\psi)=0$.
\end{proof}

In low degree, we can translate this to a statement about elements of $\bg$ considered as polynomials.

\begin{example}\label{singleshuffexplicit}
For $f(x_1,x_2)\in\bg$ and $g(x_1,x_2,x_3)\in\bg$, we have
\begin{equation*}
\begin{split}
x\frac{\partial f}{\partial x_1}(0,x) &= f(x,-x),\\
yz\left(\frac{\partial g}{\partial x_1}(0,y,z) - \frac{\partial g}{\partial x_1}(0,y,-z)\right)&=y\left(g(y,z,-z)+g(-y,z,-z)\right)\\
&+z\left(g(-y,y,-z)-g(-y,y,z)\right),\\
yz\left(\frac{\partial g}{\partial x_1}(0,y,z) + \frac{\partial g}{\partial x_1}(0,y,-z)+\frac{\partial g}{\partial x_2}(y,0,z) + \frac{\partial g}{\partial x_2}(y,0,-z)\right)&=y\left(g(y,z,-z)-g(-y,z,-z)\right)\\
&-z\left(g(-y,y,-z)+g(-y,y,z)\right).
\end{split}
\end{equation*}
\end{example}

In order to better describe elements of $\bg$, we use the following lemma to transform our polynomial representation.

\begin{lem}\label{reducable}
For $f(x_1,x_2,\ldots,x_n)\in\bg$, we can write
$$f(x_1,\ldots,x_n)=x_1\ldots x_n(x_1-x_n)r(x_1,\ldots,x_n)$$
for some polynomial $r\in\Q[x_1,\ldots,x_n]$.
\end{lem}

\begin{proof}
We induct on the number of variables. For $n=2$, this follows from Theorem \ref{summedgen}. Now, suppose this factorisation holds for $f(x_1,x_2),\ g(x_1,\ldots,x_n)\in\bg$. We have
\begin{equation*}
\begin{split}
\{f,g\}&= \sum_{i=1}^n \frac{f(x_i,x_{i+1})}{x_i-x_{i+1}}\times\\
&\left(\frac{1}{x_i}g(x_1,\ldots,x_i,x_{i+2}\ldots,x_{n+1})-\frac{1}{x_{i+1}}g(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1})\right)\\
&-g(x_1,\ldots,x_n)\left(\frac{1}{x_1}f(x_1,x_{n+1})-\frac{1}{x_n}f(x_n,x_{n+1})\right)\\
&-g(x_2,\ldots,x_{n+1}\left(\frac{1}{x_2}f(x_1,x_2)-\frac{1}{x_{n+1}}f(x_1,x_{n+1})\right).
\end{split}
\end{equation*}
Applying our induction hypothesis, we find
\begingroup
\allowdisplaybreaks
\begin{align*}
\{f,g\}&= x_1\ldots x_{n+1}r_f(x_1,x_2)\times\\
&\left((x_1-x_{n+1})r_g(x_1,x_3,\ldots,x_{n+1}) - (x_2-x_{n+1})r_g(x_2,\ldots,x_{n+1})\right)\\
+&\sum_{i=2}^{n-1} x_1\ldots x_{n+1}(x_1-x_{n+1})r_f(x_i,x_{i+1})\times\\
&\left(r_g(x_1,\ldots,x_i,x_{i+2}\ldots,x_{n+1})-r_g(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1})\right)\\
&+ x_1\ldots x_{n+1}r_f(x_n,x_{n+1})\times\\
&\left((x_1-x_{n})r_g(x_1,\ldots,x_{n})-(x_1-x_{n+1})r_g(x_1,\ldots,x_{n-1},x_{n+1})\right)\\
&-x_1\ldots x_{n+1}r_g(x_1,\ldots,x_n)\left((x_1-x_{n+1})r_f(x_1,x_{n+1})-(x_n-x_{n+1})r_f(x_n,x_{n+1})\right)\\
&-x_1\ldots x_{n+1}r_g(x_2,\ldots,x_{n+1})\left((x_1-x_2)r_f(x_1,x_2)-(x_1-x_{n+1})r_f(x_1,x_{n+1})\right).
\end{align*}
\endgroup

Considering only the terms not immediately divisible by $x_1\ldots x_{n+1}(x_1-x_{n+1})$, we reduce the problem to showing that

\begin{equation*}
\begin{split}
& -x_1\ldots x_{n+1}(x_2-x_{n+1})r_f(x_1,x_2) r_g(x_2,\ldots,x_{n+1})\\
&+ x_1\ldots x_{n+1}(x_1-x_n)r_f(x_n,x_{n+1})r_g(x_1,\ldots,x_n)\\
&+x_1\ldots x_{n+1}(x_n-x_{n+1})r_f(x_n,x_{n+1})r_g(x_1,\ldots,x_n)\\
&-x_1\ldots x_{n+1}(x_1-x_2)r_f(x_1,x_2)r_g(x_2,\ldots,x_{n+1})\\
&=-x_1\ldots x_{n+1}(x_1-x_{n+1})r_f(x_1,x_2) r_g(x_2,\ldots,x_{n+1})\\
&+ x_1\ldots x_{n+1}(x_1-x_{n+1})r_f(x_n,x_{n+1})r_g(x_1,\ldots,x_n)
\end{split}
\end{equation*}

is divisible by $x_1\ldots x_{n+1}(x_1-x_{n+1})$. This is clear, so we are done.

\end{proof}

\begin{defn}
For $f(x_1,\ldots,x_n)\in\bg$, define the reduced block polynomial to be 

$$r(x_1,\ldots,x_n):=\frac{f(x_1,\ldots,x_n)}{x_1\ldots x_n(x_1-x_n)}.$$

Define $\mathfrak{rbg}$ to be the bigraded $\Q$-vector space of reduced block polynomials.
\end{defn}

\begin{remark}
It may be useful to recall how the various degrees we assign to motivic iterated integrals relate to the reduced block polynomials. A reduced block polynomial $r(x_1,x_2,\ldots,x_n)$ of degree $N$ corresponds to elements of weight $N+n-1$ and block degree $n-1$.
\end{remark}

\section{The dihedral action}
As an immediate corollary to Proposition \ref{duality} we obtain:

\begin{lem}\label{reflection}
For all $r(x_1,\ldots,x_n)\in\mathfrak{rbg}$,
$$r(x_n,\ldots,x_1) = (-1)^nr(x_1,\ldots,x_n).$$
\end{lem}

\begin{defn}
We define a Lie algebra structure on $\mathfrak{rbg}$ via the Lie bracket

$$\{r_1,r_2\}(x_1,\ldots,x_{m+n-1}):=\frac{\{f_1,f_2\}(x_1,\ldots,x_{m+n-1})}{x_1\ldots x_{m+n-1}(x_1-x_{m+n-1})}$$

\noindent for $r_1(x_1,\ldots,x_m)=\frac{f_1(x_1,\ldots,x_m)}{x_1\ldots x_m(x_1-x_m)}, r_2(x_1,\ldots,x_n)=\frac{f_2(x_1,\ldots,x_n)}{x_1\ldots x_n(x_1-x_n)}\in\mathfrak{rbg}$. We call this the reduced Ihara bracket. It produces a polynomial of degree $\text{deg}(r_1)+\text{deg}(r_2)$.
\end{defn}

We can explicitly compute this, and in the case of $r_1=r_1(x_1,x_2)$, we obtain a particularly nice formula.

\begin{prop}\label{reducedihara}
For $r(x_1,x_2),\ q(x_1,\ldots,x_{n-1})\in\mathfrak{rbg}$, the reduced Ihara bracket is given by

$$\{r,q\}(x_1,\ldots,x_n)= \sum_{i=1}^n r(x_i,x_{i+1})(q(x_1,\ldots,x_i,x_{i+2},\ldots,x_n)-q(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n))$$
where we consider indices modulo $n$.
\end{prop}

\begin{cor}\label{cyclicinvariance}
$$r(x_1,\ldots,x_n)=r(x_2,\ldots,x_n,x_1)$$
for $r(x_1,\ldots,x_n)\in\mathfrak{rbg}$.
\end{cor}
\begin{proof}
This follows from a simple induction argument, using Lemma \ref{reflection} as our base case, and the natural cyclic symmetry in Proposition \ref{reducedihara}.
\end{proof}

\begin{remark}
Corollary \ref{cyclic} follows as an immediate corollary to this invariance.
\end{remark}

With this cyclic invariance, we can write down the general case of the reduced Ihara bracket quite succinctly.
\begin{cor}\label{generalreducedihara}
For $r(x_1,\ldots,x_m),q(x_1,\ldots,x_n)\in\rbg$, the reduced Ihara bracket is given by
\begin{equation*}
\begin{split}
\{r,q\}(x_1,\ldots,x_{m+n-1})=\sum_{i=1}^{m+n-1}r(x_i,\ldots,x_{i+m-1})&\left(q(x_{i+m},\ldots,x_{m+n-1},x_1,\ldots,x_i)\right.\\
&-\left. q(x_{i+m-1},\ldots,x_{m+n-1},x_1,\ldots,x_{i-1})\right)
\end{split}
\end{equation*}
where the indices are considered modulo $m+n-1$.
\end{cor}

Thus, we have an action of the dihedral group on $\rbg$, restricting to either the trivial or sign representation on the block graded parts. 

\section{A differential relation}

We additionally obtain a differential relation, generalising the differential relation defining the generators of $\bg$.

\begin{defn}\label{differentialdefinition}
For $n\geq 2$, define the differential operator
$$\D_n:\Q[x_1,\ldots,x_n] \to \Q[x_1,\ldots,x_n]$$ by
$$\D_n := \prod_{i_1,\ldots,i_{n-1}\in\{0,1\}}\left(\partialx{1}+(-1)^{i_1}\partialx{2}+\cdots +(-1)^{i_{n-1}}\partialx{n}\right).$$
\end{defn}

\begin{thm}\label{differential}
$$\D_n r(x_1,\ldots,x_n)=0$$
for all $r(x_1,\ldots,x_n)\in\mathfrak{rbg}.$
\end{thm}

\begin{proof}
We induct on $n$. For $n=2$, this follows from Corollary \ref{newgendef}. Suppose this holds for $q(x_1,\ldots,x_{n})\in\mathfrak{rbg}$.

Next define 
$$I_n:=\{M\in M_n(\mu_2) \ |\ M_{i,i}=1,\ \frac{M_{i+1,j}}{M_{i,j}}=\frac{M_{i+1,j+1}}{M_{i,j+1}}\},$$
and 
$$L_M:= \sum_{i=1}^n M_{1,i}\partialx{i} = \pm \sum_{i=1}^n M_{j,i}\partialx{i}.$$

Note that $\D_n = \prod_{M\in I_n}L_M$, and thus we have, for $r(x_1,x_2), q(x_1,\ldots,x_{n})\in\mathfrak{rbg}$,

\begin{equation*}
\begin{split}
\D_{n+1}\{r,q\}(x_1,\ldots,x_{n+1}) &= \sum_{i=1}^n \D_{n+1}(r(x_i,x_{i+1})q(x_i,x_{i+2},\ldots,x_{i+n})-r(x_i,x_{i+1})q(x_{i+1},x_{i+2},\ldots,x_{i+n})\\
&=\sum_{i=1}^n\sum_{S\subset I_{n+1}}(\prod_{M\in S}L_M )r(x_i,x_{i+1})(\prod_{M\in I_{n+1}\setminus S}L_M)q(x_i,x_{i+2},\ldots,x_{i+n})\\
&-\sum_{i=1}^n\sum_{S\subset I_{n+1}}(\prod_{M\in S}L_M) r(x_i,x_{i+1})(\prod_{M\in I_{n+1}\setminus S}L_M)q(x_{i+1},x_{i+2},\ldots,x_{i+n})
\end{split}
\end{equation*}

where we have used the cyclic invariance of $\mathfrak{rbg}$ and considering indices modulo $n+1$.

Next denote by $M[i_1,\ldots,i_k]$ the submatrix of $M$ obtained by restricting to rows and columns $i_1,\ldots,i_k$. We see that $L_M f(x_{i_1},\ldots,x_{i_k}) = L_{M[i_1,\ldots,i_k]}f(x_{i_1},\ldots,x_{i_k})$. 

Now, if $\{M[i,i+1]\ |\ M\in S\}=I_2$, then $(\prod_{M\in S}L_M )r(x_i,x_{i+1})=0$. Otherwise, we must have $M[i,i+1]=\bigl( \begin{smallmatrix}1 & 1\\ 1 & 1\end{smallmatrix}\bigr)$ for all $M\in S$, or $M[i,i+1]=\bigl( \begin{smallmatrix}1 & -1\\ -1 & 1\end{smallmatrix}\bigr)$ for all $M\in S$. In the first case, we must have all $M\in I_{n+1}$ with $M[i,i+1]=\bigl( \begin{smallmatrix}1 & -1\\ -1 & 1\end{smallmatrix}\bigr)$ contained in $I_{n+1}\setminus S$. The second case is similar. In either case, this implies that
$$\{M[i,i+2,\ldots,i+n]\ |\ M\in I_{n+1}\setminus S\}=\{M[i+1,\ldots,i+n]\ |\ I_{n+1}\setminus S\}=I_n$$
and so 

$$(\prod_{M\in I_{n+1}\setminus S}L_M)q(x_{i},x_{i+2},\ldots,x_{i+n})=(\prod_{M\in I_{n+1}\setminus S}L_M)q(x_{i+1},x_{i+2},\ldots,x_{i+n})=0$$

Thus $D_{n+1}\{r,q\}$=0.
\end{proof}

\begin{remark}
Note that, in sufficiently high degree, $r(x_1,\ldots,x_n)\in\text{ker}\D_n$ is equivalent to $r(x_1,\ldots,x_n)\in\sum_{M\in I_n}\text{ker}L_M$. This second condition clearly holds for $n=2$, and can easily be shown to be preserved by the Ihara bracket. Hence, we can equivalently state Theorem \ref{differential} as the following:
$$r(x_1,\ldots,x_n)\in\sum_{M\in I_n}\text{ker}L_M\text{ for all }r(x_1,\ldots,x_n)\in\rbg.$$
\end{remark}

\begin{remark}\label{wildmassguessing}
We have shown that, in block degree 1, $\bg$ is isomorphic as a vector space to the bigraded vector space of homogeneous polynomials satisfying Theorem \ref{blockshuffle}, Example \ref{singleshuffexplicit}, Lemma \ref{reducable}, and whose reduced forms satisfy Corollary \ref{cyclicinvariance} and Theorem \ref{differential}. Note also that, as all these properties are preserved by the Ihara bracket, $\bg$ is a Lie subalgebra of the Lie algebra of homogeneous polynomials satisfying these properties. However, in block degree $b$, and weight $w$, we can only show that the dimension of the bigraded piece of the vector space of homogeneous polynomials satisfying these constraints is bounded above by $Cw^{b-1}$ for some constant $C$.
\end{remark}


\section{Deriving the Ihara action formula}
For elements of the double shuffle Lie algebra, the (linearised) Ihara action is given by the following \cite{zeta3}:

\begin{prop}
For $\sigma\in\text{Lie}[e_0,e_1]$, $u\in\{e_0,e_1\}^\times$, the linearised Ihara action is given recursively by
\begin{equation}\label{ihararecursive}
\sigma\circ e_0^ne_1u := e_0^n\sigma e_1u - e_0^ne_1 \sigma^* u + e_0^n e_1(\sigma\circ u)
\end{equation}
where $(a_1\ldots a_n)^*:=(-1)^na_n\ldots a_1$.
\end{prop}


Translating the linearised Ihara action into the language of commutative variables, we find the following.

\begin{thm}\label{uglyIhara}
Let $f(x_1,\ldots,x_m)$ be the image of the block degree $m-1$ part of $\sigma\in\text{Lie}[e_0,e_1]$, and $g\in\Q[x_1,\ldots,x_n]$. Then the linearised Ihara action is given by
\begin{equation*}
\begin{split}
(f\circ g)(x_1,\ldots,x_{m+n-1}) &= \sum_{i=1}^{n}(-1)^{(m+1)(i-1)}\frac{f(x_i,x_{i+1},\ldots,x_{i+m-1})}{x_i^2-x_{i+m-1}^2}\\
&\times\left((1 +(-1)^{m+1} \frac{x_{i+m-1}}{x_i})g(\bar x_1,\ldots, \bar x_i,x_{i+m},\ldots,x_{m+n-1})\right.\\
&\left.-(1+(-1)^{m+1}\frac{x_i}{x_{i+m-1}})g(\bar x_1,\ldots, \bar x_{i-1},x_{i+m-1},\ldots,x_{m+n-1})\right)
\end{split}
\end{equation*}

where we define $\bar x_i := (-1)^{m+1}x_i$.
\end{thm}
\begin{proof}
We start by writing, for $u=u_1\ldots u_n\in\{e_0,e_1\}^\times$,

$$\sigma\circ u_1\ldots u_n = \epsilon_0 \sigma u_1\ldots u_n + \sum_{i=1}^n \epsilon_i u_1\ldots u_i \sigma u_{i+1}\ldots u_n$$

where $\epsilon_i\in\{0,\pm 1\}$ for each $i$. We first claim that $\epsilon_i=0$ if $u_i=u_{i+1}$. We take here $u_0=e_0$ and $u_{n+1}=e_1$.

If $u_i=u_{i+1}=e_0$, then our recursive formula (\ref{ihararecursive}) shows $\epsilon_i=0$, as $\sigma$ does not `insert' between adjacent $e_0$. If $u_i=u_{i+1}=e_1$, then our recursion gives us terms of the form

$$\cdots + u_1\ldots u_{i-1} e_1 \sigma^* e_{1} u_{i+2}\ldots u_{n} + u_1\ldots u_{i-1}e_1 \sigma e_1 u_{i+2}\ldots u_n + u_1\ldots u_{i-1} e_1e_1 \sigma^* u_{i+2}\ldots u_n + \cdots$$

As, for $\sigma\in\text{Lie}[e_0,e_1]$, $\sigma+\sigma^*=0$, the terms corresponding to $u_1\ldots u_i \sigma  u_{i+1}\ldots u_n$ cancel, giving us that $\epsilon_i=0$. Hence, our block-polynomial formula will consist of a sum over the blocks of $u$, each corresponding to the insertion of $\sigma$ into a single block.

We will induct on the number of blocks in $u$. If $u$ consists of a single block, $u=(e_1e_0)^k$, and 

\begin{equation*}
\begin{split}
\sigma\circ u &= \sigma (e_1e_0)^k + e_1\sigma^* e_0 (e_1e_0)^{k-1} + e_1e_0\sigma (e_1e_0)^{k-2} +e_1e_0e_1\sigma^*e_0(e_1e_0)^{k-3}+\ldots\\
&=\sum_{i=0}^k [(e_1e_0)^i\sigma (e_1e_0)^{k-i} + (e_1e_0)^i e_1 \sigma^* e_0 (e_1e_0)^{k-1-i}].
\end{split}
\end{equation*}


Letting $f(x_1,\ldots,x_m)$ be the polynomial representing the block degree $n$ part of $\sigma$ and $g(x_1)=x_1^{2k+2}$ be the polynomial representing $u$, this is equivalent to the statement that

\begin{equation*}
\begin{split}
(f\circ g)(x_1,\ldots,x_m)&= \sum_{i=0}^k (\frac{x_1}{x_m})^{2i}\frac{f(x_1,\ldots,x_m)g(x_m)}{x_m^2} + (-1)^{m+1}\frac{x_1}{x_m}\sum_{i=0}^{k-1}(\frac{x_1}{x_m})^{2i}\frac{f(x_1,\ldots,x_m)g(m)}{x_m^2}\\
&=\frac{f(x_1,\ldots,x_m)}{x_1^2-x_m^2}\left((\frac{x_1}{x_m})^{2k+2} - 1 +(-1)^{m+1} (\frac{x_1}{x_m})^{2k+1} - (-1)^{m+1}\frac{x_1}{x_m}\right)g(x_m)\\
&=\frac{f(x_1,\ldots,x_m)}{x_1^2-x_m^2}\left(g(x_1)-g(x_m)+(-1)^{m+1}\frac{x_m}{x_1}g(x_1)-(-1)^{m+1}\frac{x_1}{x_m}g(x_m)\right),
\end{split}
\end{equation*}

which is precisely the result given by the formula.

Now suppose our formula is correct for words consisting of $n-1$ blocks, and let $e_0ue_1$ be a word consisting of $n$ blocks, i.e. $e_0ue_1=b_1\ldots b_n$, represented by the monomial $g(x_1,\ldots,x_n)$. As we have merely appended a block onto the end of a word, the first $n-2$ terms of $(f\circ g)$ will be given by our formula, by our induction hypothesis. To see this, consider $(f\circ g_{alt})$, where $g_{alt}$ is the polynomial  corresponding to the word  $e_0u_{alt}e_1=b_1\ldots b_{n-1}^\prime$. Here $b_{n-1}^\prime$ is the smallest block extending $b_{n-1}$ and ending on $e_1$. The Ihara action of any $\sigma\in\text{Lie}[e_0,e_1]$ on $u$ and $u_{alt}$ will produce terms that are identical upon swapping $b_{n-1}b_n \leftrightarrow b_{n-1}^\prime$ up to those terms in which $\sigma$ inserts into $b_{n-1}b_n$. Indeed, they will agree under this swapping until we consider terms in which $\sigma$ inserts beyond the end of $b_{n-1}$. Thus, it suffices to show that the formula holds for a word $e_0ue_1=b_1b_2$ of block degree 1.

We have 2 cases: the repeated letter in $e_0ue_1$ is $e_0$, or it is $e_1$.
In the first case, $u=(e_1e_0)^ke_0(e_1e_0)^l$ and

\begin{equation*}
\begin{split}
\sigma\circ u &=  \sum_{i=0}^k (e_1e_0)^i\sigma (e_1e_0)^{k-i}e_0(e_1e_0)^l + \sum_{i=0}^{k-1}(e_1e_0)^ie_1\sigma^* e_0(e_1e_0)^{k-1-i}e_0(e_1e_0)^l\\
&+\sum_{i=0}^l(e_1e_0)^ke_0(e_1e_0)^i\sigma (e_1e_0)^{l-i} + \sum_{i=0}^{l-1}(e_1e_0)^ke_0(e_1e_0)^ie_1\sigma^*e_0(e_1e_0)^{l-1-i}
\end{split}
\end{equation*}

In terms of commutative polynomials, after summing the geometric series, we obtain

\begin{equation*}
\begin{split}
(f\circ g)&(x_1,\ldots,x_{m+1}) = \frac{f(x_1,\ldots,x_m)}{x_1^2-x_m^2}\left((\frac{x_1}{x_m})^{2k} - 1 + (-1)^{m+1}(\frac{x_1}{x_m})^{2k+1}-(-1)^{m+1}\frac{x_1}{x_m}\right)g(x_m,x_{m+1})\\
&+\frac{f(x_2,\ldots,x_{m+1})}{x_2^2-x_{m+1}^2}\left((\frac{x_2}{x_{m+1}})^{2l+2} - 1 + (-1)^{m+1}(\frac{x_2}{x_{m+1}})^{2l+1}-(-1)^{m+1}\frac{x_2}{x_{m+1}}\right)g(x_1,x_{m+1}).
\end{split}
\end{equation*}


Simplifying, and noting that $g(x_1,x_2)= x_1^{2k+1}x_2^{2l+2}$, we obtain

\begin{equation*}
\begin{split}
(f\circ g)(x_1,\ldots,x_{m+1})&= \frac{f(x_1,\ldots,x_m)}{x_1^2-x_m^2}\left(\frac{x_m}{x_1}g(x_1,x_m+1) -g(x_m,x_m+1)\right.\\
&\left.+(-1)^{m+1}g(x_1,x_{m+1})-(-1)^{m+1}\frac{x_1}{x_m}g(x_m,x_{m+1})\right)\\
&+\frac{f(x_2,\ldots,x_{m+1})}{x_2^2-x_{m+1}^2}\bigg(g(x_1,x_2)-g(x_1,x_{m+1})\\
&\left.+(-1)^{m+1}\frac{x_{m+1}}{x_2}g(x_1,x_2) - (-1)^{m+1}\frac{x_2}{x_{m+1}}g(x_1,x_{m+1})\right).
\end{split}
\end{equation*}

Considering parity, and defining $\bar x_i:=(-1)^{m+1}x_i$, we can rewrite this as

\begin{equation*}
\begin{split}
(f\circ g)(x_1,\ldots,x_{m+1})&= (-1)^{(0)(m+1)}\frac{f(x_1,\ldots,x_m)}{x_1^2-x_m^2}\left(g(\bar x_1,x_{m+1})+(-1)^{m+1}\frac{x_m}{x_1}g(\bar x_1,x_{m+1})\right.\\
&\left. - g(x_m,x_{m+1}) + (-1)^{m+1}\frac{x_m}{x_{m+1}}g(x_m,x_{m+1})\right)\\
&+(-1)^{m+1}\frac{f(x_2,\ldots,x_{m+1})}{x_2^2-x_{m+1}^2}\left(g(\bar x_1,\bar x_2) - (-1)^{m+1}\frac{x_{m+1}}{x_2}g(\bar x_1,\bar x_2)\right.\\
&\left. -g(\bar x_1,x_{m+1})+(-1)^{m+1}\frac{x_2}{x_{m+1}}g(\bar x_1,x_{m+1})\right)
\end{split}
\end{equation*}
\noindent giving the desired formula. The second case follows similarly.

Hence, our general formula is $(f\circ g)(x_1,\ldots,x_{m+n-1}) =A\pm B\pm C$, where
\begin{equation*}
\begin{split}
A &= \sum_{i=1}^{n-2}(-1)^{(m+1)(i-1)}\frac{f(x_i,x_{i+1},\ldots,x_{i+m-1})}{x_i^2-x_{i+m-1}^2}\\
&\times\left((1 +(-1)^{m+1} \frac{x_{i+m-1}}{x_i})g(\bar x_1,\ldots, \bar x_i,x_{i+m},\ldots,x_{m+n-1})\right.\\
&\left.-(1+(-1)^{m+1}\frac{x_i}{x_{i+m-1}})g(\bar x_1,\ldots, \bar x_{i-1},x_{i+m-1},\ldots,x_{m+n-1})\right),\\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
B &=(-1)^{(m+1)(n-2)}\frac{f(x_{n-1},\ldots,x_{m+n-2})}{x_{n-1}^2-x_{n+m-2}^2}\\
&\times\left((1 +(-1)^{m+1} \frac{x_{n+m-2}}{x_{n-1}})g(\bar x_1,\ldots, \bar x_{n-1},x_{m+n-1})\right.\\
&\left.-(1+(-1)^{m+1}\frac{x_{n-1}}{x_{n+m-2}})g(\bar x_1,\ldots, \bar x_{n-2},x_{n+m-2},x_{m+n-1})\right),\\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
C &=(-1)^{(m+1)(n-1)}\frac{f(x_{n},\ldots,x_{m+n-1})}{x_{n}^2-x_{m+n-1}^"}\\
&\times\left((1 +(-1)^{m+1} \frac{x_{n+m-1}}{x_{n}})g(\bar x_1,\ldots, \bar x_{n})\right.\\
&\left.-(1+(-1)^{m+1}\frac{x_{n}}{x_{n+m-1}})g(\bar x_1,\ldots, \bar x_{n-1},x_{m+n-1})\right),\\
\end{split}
\end{equation*}

\noindent and the signs of $B$ and $C$ agree. To fix this sign, we need only to consider the sign of the term corresponding to $(\frac{x_{n-1}}{x_{m+n-2}})^2\frac{f(x_{n-1},\ldots,x_{m+n-2})g(x_1,\ldots,x_{n-2},x_{m+n-2},x_{m+n-1})}{x_{m+n-2}^2}$. This corresponds to inserting $\sigma$ after the first two letters of the $(n-1)^{th}$ block. The sign will be positive if this block starts with an $e_1$ and must have the same sign as $(-1)^{m+1}$ otherwise. Let $g(x_1,\ldots,x_n)=x_1^{d_1}\ldots x_n^{d_n}$. Then by Lemma \ref{blockparity} the $(n-1)^{th}$ block starts with $e_0$ if $d_1+\ldots+d_{n-2}\equiv n-2 \text{ (mod 2)}$, and $e_1$ otherwise. Thus, the sign of the term corresponding to $(\frac{x_{n-1}}{x_{m+n-2}})^2\frac{f(x_{n-1},\ldots,x_{m+n-2})g(x_1,\ldots,x_{n-2},x_{m+n-2},x_{m+n-1})}{x_{m+n-2}^2}$ is $(-1)^{(m-1)(1 + d_1+\ldots + d_{n-2} - n +2)}$. Comparing this with our formula, we see that the final two terms must appear with a positive sign, giving the desired result.


\end{proof}

To obtain (\ref{prettyIhara}), we must translate this across into the `depth-signed' convention. Specifically, we must find the action of the map $e_1\mapsto -e_1$ in terms of commutative variables.

\begin{lem}\label{depthsigned}
The automorphism $\qpoly\to\qpoly$ given by $e_1\mapsto -e_1$, is equivalent under the isomorphism (\ref{isomorph}) to the map
\begin{equation}\label{depthsigning}
\begin{split}
\Q[x_1,\ldots,x_n]&\to\Q[x_1,\ldots,x_n]\\
f(x_1,\ldots,x_n)&\mapsto(-1)^{\lceil\frac{l}{2}\rceil}f(-x_1,x_2,\ldots,(-1)^nx_n)
\end{split}
\end{equation}
for $f$ a homogeneous polynomial of degree $l+2$.
\end{lem}


\begin{proof}
Note that it suffices to show that, for a word $w$ of length $l$ and depth $d$, with $\pi_{\text{bl}}(w)=x_1^{d_1}\ldots x_n^{d_n}$, this congruence holds

$$d\equiv \lceil\frac{l}{2}\rceil + d_1 + d_3+\cdots\text{ (mod 2)}.$$

We will induct on the number of blocks in $e_0we_1$. If $e_0we_1$ consists of a single block, then $e_0we_1 = (e_0e_1)^{\frac{l}{2}+1}$, and so $d= \frac{l}{2}$, and $d_1=l+2$. Thus the result holds.

Suppose the result holds for $w$ such that $e_0we_1$ consists of $n$ blocks. Let $e_0we_1=w^\prime w_{d_{n+1}}$ be a word of length $l+2$ and depth $d+1$,consisting of $n+1$ blocks, where $w_{d_{n+1}}$ is a single block of length $d_{n+1}$ and $w^\prime$ is a word of length $l^\prime$ and depth $d^\prime$. Suppose $\pi_{\text{bl}}(w)=x_1^{d_1}\ldots x_n^{d_n}x_{n+1}^{d_{n+1}}$.

If $l^\prime$ is even, then $w^\prime=e_0 u e_1$ consists of $n\equiv 1\text{ (mod 2)}$ blocks, and $d_{n+1}$ must be odd. So, by induction,

$$d^\prime - 1 \equiv \frac{l^\prime-2}{2}+\sum_{1\leq 2i+1\leq n}d_{2i+1}\text{ (mod 2)}.$$

\noindent Thus 

\begin{equation*}
\begin{split}
d &= d^\prime + \lceil\frac{d_{n+1}}{2}\rceil - 1\\
&\equiv  \frac{l^\prime-2}{2}+\sum_{1\leq 2i+1\leq n}d_{2i+1}\text{ (mod 2)}+ \lceil\frac{d_{n+1}}{2}\rceil\text{ (mod 2)}\\
&\equiv \lceil\frac{l^\prime + d_{n+1}-2}{2}\rceil +\sum_{1\leq 2i+1\leq n+1}d_{2i+1}\text{ (mod 2)}\\
&\equiv \lceil\frac{l}{2}\rceil +\sum_{1\leq 2i+1\leq n+1}d_{2i+1}\text{ (mod 2)},
\end{split}
\end{equation*}

\noindent and so the result holds. Similar considerations for $l^\prime$ odd prove the result in general.
\end{proof}


Applying this transformation, and simplifying, we obtain Proposition \ref{Iharanice}, giving the formula

\begin{equation*}
\begin{split}
(f\circ g)(x_1,\ldots,x_{m+n-1}) &= \sum_{i=1}^n \frac{f(x_i,\ldots,x_{i+m-1})}{x_i-x_{i+m-1}}\left (\frac{1}{x_i}g(x_1,\ldots,x_i,x_{i+m},\ldots,x_{m+n-1})\right.\\
&\left. -\frac{1}{x_{i+m-1}}g(x_1,\ldots,x_{i-1},x_{i+m-1},\ldots,x_{m+n-1})\right).
\end{split}
\end{equation*}


\section{Further Remarks}
As $\bg$ is a free Lie algebra, we have a non-canonical isomorphism $\gmot\cong \bg$, and hence, we should be able to lift relations such as cyclic insertion and the block shuffle relations to $\gmot$ . One would hope that we could follow the example of Brown in \cite{anatomy}, in which he lifts solutions from $\ls$ to $\dmr$, but it is as yet unclear how this should work. Progress on this has been made by Hirose and Sato \cite{damnsato}, who provide an ungraded version of the block shuffle relations, but it is an area ripe for further consideration. Then, as the collection of relations presented in this paper, alongside block graded double shuffle relations, should completely describe all relations among block graded motivic multiple zeta values, a successful lifting of these relations could provide a complete description of all relations among motivic multiple zeta values, and give an approach to tackling questions about the completeness of the associator and double shuffle relations. The construction of genuine relations, and the connection between these block graded relations and known relations is explored further in the author's doctoral thesis.


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