If $B_{<N}^{(\ell-1)}\neq\emptyset$, then it is the disjoint union of lower
weights, i. e.

$B_{<N}^{(\ell-1)}=\mathop{\dot{\bigcup}}_{n=0}^{N-3}B_{n}^{(\ell-1)}.$

######  Lemma 4.17.

The map

$\displaystyle\psi\colon B_{<N}^{(\ell-1)}$ $\displaystyle\rightarrow
B_{N}^{(\ell)}$ $\displaystyle\mathsf{b}(u)$
$\displaystyle\mapsto\mathsf{b}\big{(}\mathsf{2}^{r-1}\mathsf{3}u\big{)}$

where $2r=N-1-\operatorname{wt}(u)$ is a bijection.

###### Proof.

If $B_{<N}^{(\ell-1)}=\emptyset$, we also have $B_{N}^{(\ell)}=\emptyset$ and
vice versa by Remark 4.16. So we can assume without loss of generality that
$B_{<N}^{(\ell-1)}\neq\emptyset$. Let $\mathsf{b}(u)\in B_{<N}^{(\ell-1)}$.
Indeed, $\mathsf{b}\big{(}\mathsf{2}^{r-1}\mathsf{3}u\big{)}\in
B_{N}^{(\ell)}$ by the choice of $r$. Since $\ell\geq 1$, all words in
$B_{N}^{(\ell)}$ can be uniquely written as
$\mathsf{b}(\mathsf{2}^{s-1}\mathsf{3}u^{\prime})$ for some $s\in\mathbb{N}$
and a word $u^{\prime}\in\mathcal{X}^{*}$ with $\mathsf{b}(u^{\prime})\in
B_{<N}^{(\ell-1)}$. The assignment
$\mathsf{b}(\mathsf{2}^{s-1}\mathsf{3}u^{\prime})\mapsto\mathsf{b}(u^{\prime})$
clearly gives an inverse map to $\psi$. Hence $\psi$ is a bijection, since
both sets are finite. ∎

######  Corollary 4.18.

We have

$\\#B_{N}^{(\ell)}=\\#B_{<N}^{(\ell-1)}.$

∎

######  Remark 4.19.

The statement in Corollary 4.18 can also be proven directly by combinatorial
means. We again disregard the case where either set is empty (cf. Remark
4.16). Then we have $N=2m+3\ell$ for some non-negative $m\in\mathbb{Z}$ and
observe that

$\\#B_{N}^{(\ell)}=\binom{m+\ell}{\ell}=\sum_{0\leq m^{\prime}\leq
m}\binom{m^{\prime}+\ell-1}{\ell-1}=\\#B_{<N}^{(\ell-1)},$

where the second equality is the so-called hockey-stick identity of binomial
coefficients.

######  Definition 4.20.

We endow $B_{N}^{(\ell)}$ with the lexicographic order $\leq_{\text{lex}}$
with respect to the order $\mathsf{2}<\mathsf{3}$. With this we define an
order on $B_{<N}^{(\ell-1)}$ by $\mathsf{b}(u)\leq\mathsf{b}(u^{\prime})$, iff
$\psi(\mathsf{b}(u))\leq_{\text{lex}}\psi(\mathsf{b}(u)^{\prime})$.

In other words, in Definition 4.20 we require that the bijection $\psi$ is an
order preserving map. Observe we get $\mathsf{b}(u)\leq\mathsf{b}(u^{\prime})$
if and only if $\operatorname{wt}(u)<\operatorname{wt}(u^{\prime})$ or if
$\operatorname{wt}(u)=\operatorname{wt}(u^{\prime})$ and
$\mathsf{b}(u)\leq_{\text{lex}}\mathsf{b}(u^{\prime})$ The same order is used
in [bgf].

######  Remark 4.21.

Brown [br] uses a similar lexicographic order on $B_{N}^{(\ell)}$, but a
different order on $B_{<N}^{(\ell-1)}$. The order on $B_{<N}^{(\ell-1)}$ can
be recovered from the lexicographic order on $B_{N}^{(\ell)}$ via a variation
of the map $\psi\colon B_{N}^{(\ell)}\to B_{<N}^{(\ell-1)}$ from Lemma 4.17
given by $u\mapsto u\mathsf{3}\mathsf{2}^{s-1}$. This difference in Brown’s
setup is due to the inverted MZV-notation.

######  Proposition 4.22.

Let $w\in B_{N}^{(\ell)}$, $r\geq 1$. If there is a decomposition $w=\beta v$
with $\beta\in B_{2r+1}^{(1)}$, then

$\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{2r+1}(w)=c_{\beta}\,v+(\text{terms
with coefficients in $2\mathbb{Z}$}).$

Otherwise, $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{2r+1}(w)$
is given by a sum of terms with coefficients in $2\mathbb{Z}$.

###### Proof.

Recall the four types of strict subwords $w$ that contribute non-trivially to
$\partial_{2r+1}^{(\ell)}(w)$ from the proof of Theorem 4.9.
Unless $w=\beta v$, the first two cases come in pairs $\beta,\beta^{\prime}$
by Remark 4.11. They contribute with $c_{\beta}=c_{a,b}$ and
$c_{\beta^{\prime}}=-c_{b,a}$ for some non-negative $a,b\in\mathbb{Z}$. By
Lemma 4.32 2\. we have $c_{\beta}+c_{\beta^{\prime}}\in 2\mathbb{Z}$.
The last two cases contribute with the even coefficients $\pm 2$. ∎

We deduce for the linear maps $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
from Definition 4.14 the following from Proposition 4.22.

######  Proposition 4.23.

For $w\in B_{N}^{(\ell)}$ we have

$\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}(w)=\sum_{\begin{subarray}{c}w=\beta
v,\\\ \beta\in B_{2r+1}^{(1)},r\geq 1\end{subarray}}c_{\beta}\,v+(\text{terms
with coefficients in }2\mathbb{Z}).$

∎

######  Definition 4.24.

Let $N,\ell\in\mathbb{N}$ with $N=2m+3\ell$ for some non-negative
$m\in\mathbb{Z}$. We define

$M_{N}^{(\ell)}=\big{(}m_{w,w^{\prime}}\big{)}_{w\in
B_{N}^{(\ell)},\,w^{\prime}\in B_{<N}^{(\ell-1)}}$

to be the representing matrix of $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
with respect to the bases $B_{N}^{(\ell)}$ and $B_{<N}^{(\ell-1)}$ from
Definition 4.20.

Precisely, $m_{w,w^{\prime}}\in\mathbb{Q}$ are for $w\in B_{N}^{(\ell)}$
defined by the equations

$\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}(w)=\sum_{w^{\prime}\in
B_{<N}^{(\ell-1)}}m_{w,w^{\prime}}\,w^{\prime}.$

We now deduce the following from Corollary 4.18.

######  Corollary 4.25.

The matrix $M_{N}^{(\ell)}$ is quadratic. ∎

Recall that our goal is to show that the level reducing map $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
is an isomorphism. By Remark 4.16, it suffices to study the non-trivial cases
where $N=2m+3\ell$ for some non-negative $m\in\mathbb{Z}$. The key idea of the
proof is showing that the transformation matrices $M_{N}^{(\ell)}$ are
invertible via the following lemma. Therefore, recall the $p$-adic valuation
$\nu_{p}$ of a rational number also given in Definition 4.31.

######  Lemma 4.26.

Let $p$ be a prime number and $n\in\mathbb{N}$. Let $A=(a_{ij})$ be an
$n\times n$ matrix with entries in $\mathbb{Q}$ such that

  1. (i)

$\nu_{p}(a_{ij})\geq 1$ for all $i>j$ and

  2. (ii)

$\nu_{p}(a_{jj})=\min\\{\nu_{p}(a_{ij})\mid i\in\\{1,\dots,n\\}\\}\leq 0$ for
all $j\in\\{1,\dots,n\\}$.

Then $A$ is invertible.

###### Proof.

To show that $\det(A)\neq 0$ it suffices to show that $\det(A^{\prime})\neq 0$
where the matrix $A^{\prime}=(a_{ij}^{\prime})$ arises from $A$ by multiplying
the $j$-th column with $p^{-\nu_{p}(a_{jj})}$ for all $j\in\\{1,\dots,n\\}$.
Since $\nu_{p}(a_{jj})\leq 0$, condition (i) still holds for $A^{\prime}$.
Condition (ii) implies that $\nu_{p}(a_{ij}^{\prime})\geq 0$ for all entries
of $i,j\in\\{1,\dots,n\\}$. By construction of $A^{\prime}$, we have
$\nu_{p}(a_{jj}^{\prime})=0$. So in particular, the entries on the diagonal
are not zero. It follows that $A^{\prime}$ modulo $p$ is an upper triangular
matrix with non-zero entries on the diagonal and is thus invertible. ∎

######  Theorem 4.27 (Brown [br]).

The map $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
is an isomorphism of vector spaces.

###### Proof.

We show that the matrix $M_{N}^{(\ell)}$ of the operator $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$,
given in Definition 4.24, is invertible, since it satisfies the assumptions of
Lemma 4.26 for $p=2$.
Let $v\in B_{<N}^{(\ell-1)}$. Recall the bijection $\psi\colon
B_{<N}^{(\ell-1)}\to B_{N}^{(\ell)}$ from Lemma 4.17 which is order-preserving
by Definition 4.20. Let $2r=N-1-\operatorname{wt}(v)$ and let $w\in
B_{N}^{(\ell)}$ with $w=\beta v$ for some $\beta\in B_{2r+1}^{(1)}$. If
$w\neq\psi(v)$, then $\beta$ does not end in $x_{0}x_{0}x_{1}$, hence
$w>\psi(v)$. We deduce from Proposition 4.23 that the entries of
$M_{N}^{(\ell)}$ that are not in $2\mathbb{Z}$ are either on or above the main
diagonal due to the orders on $B_{N}^{(\ell)}$ and $B_{<N}^{(\ell-1)}$. This
implies condition (i) from Lemma 4.26.
The entries on the main diagonal of $M_{N}^{(\ell)}$ are given by the
coefficients of $v$ in $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}(\psi(v))$
with $r$ as above. By Lemma 4.32 3\. and Proposition 4.23, these entries have
a non-positive $2$-adic valuation and realize the minimum of this valuation
within its column. Hence $M_{N}^{(\ell)}$ satisfies condition (ii) from Lemma
4.26. ∎

We illustrate the content of this section in two examples.

######  Example 4.28.

In our first example, we compute the matrix $M_{9}^{(1)}$ corresponding to the
map $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(1)}_{<9}$.

We have

$\displaystyle B_{9}^{(1)}$
$\displaystyle=\\{\mathsf{b}(3,2,2,2),\mathsf{b}(2,3,2,2),\mathsf{b}(2,2,3,2),\mathsf{b}(2,2,2,3)\\}$
and $\displaystyle B_{<9}^{(0)}$
$\displaystyle=\\{\mathsf{b}(2,2,2),\mathsf{b}(2,2),\mathsf{b}(2),\mathbf{1}\\}.$

In order to compute, e. g., $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(1)}_{<9}(\mathsf{b}(3,2,2,2))$
we first observe that

$\displaystyle\partial_{3}^{(1)}\big{(}\mathsf{b}(3,2,2,2)\big{)}$
$\displaystyle=I(x_{1};x_{0}x_{0}x_{1};x_{0})\otimes\mathsf{b}(2,2,2)+I(x_{0};x_{0}x_{1}x_{0};x_{1})\otimes\mathsf{b}(2,2,2)$
$\displaystyle=(x_{0}x_{0}x_{1}-x_{0}x_{1}x_{0})\otimes\mathsf{b}(2,2,2).$

These terms can be depicted, respectively, as follows:

$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{0}$

Since $\phi(x_{0}x_{0}x_{1})=2$ and $\phi(x_{0}x_{1}x_{0})=-1$ (cf. Definition
4.13) we obtain that

$(\phi\otimes\operatorname{id})\circ\partial_{3}^{(1)}\big{(}\mathsf{b}(3,2,2,2)\big{)}=3\otimes\mathsf{b}(2,2,2)$

which we identify with $3\,\mathsf{b}(2,2,2)$. Similarly, one computes that

$\displaystyle\partial_{5}^{(1)}\big{(}\mathsf{b}(3,2,2,2)\big{)}=$
$\displaystyle(x_{0}x_{0}x_{1}x_{0}x_{1}-2\,(x_{0}x_{1})^{2}x_{0})\otimes\mathsf{b}(2,2),$
$\displaystyle\partial_{7}^{(1)}\big{(}\mathsf{b}(3,2,2,2)\big{)}=$
$\displaystyle(x_{0}x_{0}x_{1}(x_{0}x_{1})^{2}+2\,(x_{0}x_{1})^{3}x_{0})\otimes\mathsf{b}(2),$
$\displaystyle\partial_{9}^{(1)}\big{(}\mathsf{b}(3,2,2,2)\big{)}=$
$\displaystyle x_{0}x_{0}x_{1}(x_{0}x_{1})^{3}\otimes\mathbf{1}.$

Thus we obtain

$\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(1)}_{<9}(\mathsf{b}(3,2,2,2))=3\,\mathsf{b}(2,2,2)-\frac{15}{2}\,\mathsf{b}(2,2)+\frac{189}{16}\,\mathsf{b}(2)-\frac{223}{16}.$

Similar computations for the remaining words in $B_{9}^{(1)}$ give

$M_{9}^{(1)}=\begin{pmatrix}3&-\frac{15}{2}&\frac{189}{16}&-\frac{223}{16}\\\\[4.30554pt]
0&-\frac{15}{2}&\frac{299}{8}&-\frac{889}{16}\\\\[4.30554pt]
0&2&-\frac{291}{16}&\frac{455}{16}\\\\[4.30554pt]
-2&12&-30&\frac{641}{16}\end{pmatrix}.$

######  Example 4.29.

In our second example, we compute the matrix $M_{10}^{(2)}$. We have

$\displaystyle B_{10}^{(2)}$
$\displaystyle=\\{\mathsf{b}(3,3,2,2),\mathsf{b}(3,2,3,2),\mathsf{b}(3,2,2,3),\mathsf{b}(2,3,3,2),\mathsf{b}(2,3,2,3),\mathsf{b}(2,2,3,3)\\}$
and $\displaystyle B_{<10}^{(1)}$
$\displaystyle=\\{\mathsf{b}(3,2,2),\mathsf{b}(2,3,2),\mathsf{b}(2,2,3),\mathsf{b}(3,2),\mathsf{b}(2,3),\mathsf{b}(3)\\}.$

Similar to the previous example, one computes that

$\displaystyle\partial_{3}^{(2)}\big{(}\mathsf{b}(3,2,2,3)\big{)}$
$\displaystyle=(x_{0}x_{1}x_{0}-x_{0}x_{0}x_{1}+x_{0}x_{0}x_{1})\otimes\mathsf{b}(3,2,2)$
$\displaystyle+(x_{0}x_{0}x_{1}-x_{0}x_{1}x_{0})\otimes\mathsf{b}(2,2,3),$
$\displaystyle\partial_{5}^{(2)}\big{(}\mathsf{b}(3,2,2,3)\big{)}$
$\displaystyle=(x_{0}x_{0}x_{1}x_{0}x_{1}-(x_{0}x_{1})^{2}x_{0})\otimes\mathsf{b}(2,3)$
$\displaystyle+((x_{0}x_{1})^{2}x_{0}-x_{0}x_{0}x_{1}x_{0}x_{1}+x_{0}x_{1}x_{0}x_{0}x_{1})\otimes\mathsf{b}(3,2),$
$\displaystyle\partial_{7}^{(2)}\big{(}\mathsf{b}(3,2,2,3)\big{)}$
$\displaystyle=(x_{0}x_{0}x_{1}(x_{0}x_{1})^{2}-x_{0}x_{0}x_{1}(x_{0}x_{1})^{2}+(x_{0}x_{1})^{2}x_{0}x_{0}x_{1})\otimes\mathsf{b}(3),$
$\displaystyle\partial_{9}^{(2)}\big{(}\mathsf{b}(3,2,2,3)\big{)}$
$\displaystyle=0.$

By applying $\phi$ we obtain

$\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(2)}_{<10}\big{(}\mathsf{b}(3,2,2,3)\big{)}=-2\,\mathsf{b}(3,2,2)+3\,\mathsf{b}(2,2,3)+12\,\mathsf{b}(3,2)-\frac{15}{2}\,\mathsf{b}(2,3)-\frac{291}{16}\,\mathsf{b}(3).$

Similar computations for the remaining elements of $B_{10}^{(2)}$ yield

$M_{10}^{(2)}=\begin{pmatrix}3&0&0&-12&0&28\\\\[4.30554pt]
0&3&0&-\frac{11}{2}&0&0\\\\[4.30554pt]
-2&0&3&12&-\frac{15}{2}&-\frac{291}{16}\\\\[4.30554pt]
0&0&0&\frac{9}{2}&-10&0\\\\[4.30554pt]
0&-2&0&0&\frac{9}{2}&\frac{75}{8}\\\\[4.30554pt]
0&0&-2&0&12&-\frac{291}{16}\\\\[4.30554pt] \end{pmatrix}.$

We conclude the examples with the observations that both $M_{9}^{(1)}$ and
$M_{10}^{(2)}$ are invertible since

$\displaystyle\det(M_{9}^{(1)})$ $\displaystyle=\frac{4865}{512}$
$\displaystyle\det(M_{10}^{(2)})$ $\displaystyle=-\frac{435419}{64}$

and that all entries below the main diagonal are in $2\mathbb{Z}$,
respectively.

###  4.3 Some properties of the numbers $c_{a,b}^{r}$

In this subsection we present some of the arithmetic properties of the numbers
$c_{a,b}^{r}$ introduced in Definition 4.13. Later in Section 6.2 these
numbers occur in Zagier’s Theorem and will play a crucial role in the main
result of these notes.

######  Definition 4.30.

For integers $a,b,r\geq 0$ we set

$\displaystyle A_{a,b}^{r}$ $\displaystyle=(1-2^{-2r})\binom{2r}{2a+1},$
$\displaystyle B_{a,b}^{r}$ $\displaystyle=\binom{2r}{2b+2}.$

Then, we get immediately

$c_{a,b}^{r}=2\cdot(-1)^{r}(B_{a,b}^{r}-A_{a,b}^{r}).$

######  Definition 4.31.

Let $p$ be a prime number and $q\in\mathbb{Q}^{\times}$ non-zero. The _$p$
-adic valuation_ $\nu_{p}(q)$ of $q$ is the integer $n\in\mathbb{Z}$ such that
$q=p^{n}\frac{a}{b}$ with $a,b$ relatively prime to $p$. We further set
$\nu_{p}(0)=\infty$.

It is not hard to verify that $p$-adic valuations satisfy the following basic
properties for $q_{1},q_{2}\in\mathbb{Q}$

$\displaystyle\nu_{p}(q_{1}\cdot q_{2})$
$\displaystyle=\nu_{p}(q_{1})+\nu_{p}(q_{2})$ (32)
$\displaystyle\nu_{p}(q_{1}+q_{2})$
$\displaystyle\geq\min\\{\nu_{p}(q_{1}),\nu_{p}(q_{2})\\}$ (33)

with equality in (33) if $\nu_{p}(q_{1})\neq\nu_{p}(q_{2})$.

######  Lemma 4.32.

For all integers $a,b\geq 0$ we define

$c_{a,b}=c_{a,b}^{a+b+1}.$

These numbers satisfy

  1. 1.

$c_{a,b}\in\mathbb{Z}[\frac{1}{2}]$,

  2. 2.

$c_{a,b}-c_{b,a}\in 2\mathbb{Z}$ and

  3. 3.

$\nu_{2}(c_{a+b,0})=\nu_{2}(c_{0,a+b})\leq\nu_{2}(c_{a,b})\leq 0$.

###### Proof.

Write $n=a+b+1$. The first claim follows immediately from the definition

$c_{a,b}=2\cdot(-1)^{n}\left(\binom{2n}{2b+2}-(1-2^{-2n})\binom{2n}{2a+1}\right).$

The second claim follows from (34), i. e. $A_{a,b}^{n}-A_{b,a}^{n}=0$ hence

$c_{a,b}-c_{b,a}=2\cdot(-1)^{n}(B_{a,b}^{n}-B_{b,a}^{n})\in 2\mathbb{Z}.$

It remains to prove the third claim. We first show that $\nu_{2}((2n)!)<2n$.
By (32) we have

$\nu_{2}((2n)!)=\sum_{i=1}^{2n}\nu_{2}(i).$

We rearrange this sum and obtain

$\displaystyle\sum_{i=1}^{2n}\nu_{2}(i)$
$\displaystyle=\sum_{k=1}^{\lfloor\log_{2}(2n)\rfloor}\\#\big{\\{}i\in\\{1,\dots,2n\\}\mid\text{$i$
is divisible by $2^{k}$}\big{\\}}$
$\displaystyle=n+\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor+\dots+1<2n.$

It follows that

$\nu_{2}\left(2^{-2n}\cdot\binom{2n}{2a+1}\right)<0.$

On the other hand, we clearly have
$\nu_{2}\big{(}\binom{2n}{2b+2}-\binom{2n}{2a+1}\big{)}\geq 0$. In particular,
(33) becomes an equality for $q_{1}=2^{-2n}\cdot\binom{2n}{2a+1}$ and
$q_{2}=\binom{2n}{2b+2}-\binom{2n}{2a+1}$, hence

$\displaystyle\nu_{2}(c_{a,b})$
$\displaystyle=1+\nu_{2}\left(2^{-2n}\cdot\binom{2n}{2a+1}+\binom{2n}{2b+2}-\binom{2n}{2a+1}\right)$
$\displaystyle=1+\nu_{2}\left(2^{-2n}\cdot\binom{2n}{2a+1}\right)\leq 0.$

This proves the last inequality in the third claim. By writing
$\binom{2n}{2a+1}=\frac{2n}{2a+1}\binom{2n-1}{2a}$ we further obtain from (32)
that

$\nu_{2}(c_{a,b})=1+\nu_{2}\left(2^{-2n}\cdot\frac{2n}{2a+1}\binom{2n-1}{2a}\right)=2-2n+\nu_{2}(n)+\nu_{2}\left(\binom{2n-1}{2a}\right).$

For $n\in\mathbb{N}$ fixed, this is minimal for $a=0$ and $a=n-1$. The latter
case is equivalent to $b=0$. ∎

Later in Subsection 6.2 we need the following lemma.

######  Lemma 4.33.

For $a,b\geq 0$ and $1\leq r\leq a+b+1$ we have

$c_{a,b}^{r}=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\beta,\alpha}^{r}+2\cdot(-1)^{r}\big{(}\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r)\big{)}.$

Here, $\mathbb{I}$ denotes the indicator function, i.e. $\mathbb{I}(A)=1$ if
$A$ is a true statement and $\mathbb{I}(A)=0$ else.

###### Proof.

Using the quantities $A_{a,b}^{r}$ and $B_{a,b}^{r}$, it suffices to show that

$B_{a,b}^{r}-A_{a,b}^{r}=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}(B_{\alpha,\beta}^{r}-A_{\alpha,\beta}^{r})-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}(B_{\beta,\alpha}^{r}-A_{\beta,\alpha}^{r})+(\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r))$

for $a,b\geq 0$ and $1\leq r\leq a+b+1$.

We first show for all $\alpha,\beta\geq 0$ and $r=\alpha+\beta+1$ that

$\displaystyle A_{\alpha,\beta}^{r}$ $\displaystyle=A_{\beta,\alpha}^{r}$ (34)
$\displaystyle B_{\alpha,\beta}^{r}$ $\displaystyle=B_{\beta+1,\alpha-1}^{r}.$
(35)

Recall that we have for $0\leq k\leq n$

$\binom{n}{k}=\binom{n}{n-k}.$

Identity (34) follows immediately for $n=2r$ and $k=2\alpha+1$. Identity (35)
follows for $n=2r$ and $k=2\beta+2$ from

$2(\alpha+\beta+1)-(2\beta+2)=2\alpha=2(\alpha-1)+2.$

It suffices to show that

$\displaystyle A_{a,b}^{r}$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq
b\\\
\alpha+\beta+1=r\end{subarray}}A_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\ \alpha+\beta+1=r\end{subarray}}A_{\beta,\alpha}^{r}$ (36)
$\displaystyle B_{a,b}^{r}$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq
b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}B_{\beta,\alpha}^{r}+\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r).$ (37)

We first observe that $a\geq r$ is equivalent to

  1. 1.

$A_{a,\beta}^{r}$ vanishes for all $\beta\geq 0$,

  2. 2.

the summand for $\alpha=a$ does not appear in the first sum since
$\alpha+\beta+1=r$ would imply $\beta<0$ and

  3. 3.

the summand for $\beta=0$ appears in the second sum since then $\alpha=r-1<a$.

We deduce (36) from the first two observations since

$\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}A_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}A_{\beta,\alpha}^{r}=A_{a,b}^{r}+\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}(A_{\alpha,\beta}^{r}-A_{\beta,\alpha}^{r})\overset{\eqref{eq:final_lemma_claim_A}}{=}A_{a,b}^{r}.$

We similarly observe for identity (37) that $b\geq r$ is equivalent to

  1. 1.

$B_{\alpha,b}^{r}$ vanishes for all $\alpha\geq 0$,

  2. 2.

the summand for $\beta=b$ does not appear in the first sum since
$\alpha+\beta+1=r$ would imply $\alpha<0$ and

  3. 3.

the summand for $\alpha=0$ appears in the first sum since then $\beta=r-1<b$.

So we have

$\displaystyle\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq
b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}B_{\beta,\alpha}^{r}+\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r)$
$\displaystyle\overset{\eqref{eq:final_lemma_claim_B}}{=}B_{a,b}^{r}+\sum_{\begin{subarray}{c}0\leq\alpha\leq
a\\\ 0\leq\beta<b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha+1,\beta-1}^{r}+\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r)$
$\displaystyle=B_{a,b}^{r}+\sum_{\begin{subarray}{c}0<\alpha\leq a\\\
0\leq\beta<b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha,\beta}^{r}+\mathbb{I}(b\geq
r)-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\ 0<\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}B_{\alpha+1,\beta-1}^{r}-\mathbb{I}(a\geq r)$
$\displaystyle\quad+\mathbb{I}(a\geq r)-\mathbb{I}(b\geq r)$
$\displaystyle=B_{a,b}^{r}+\sum_{\begin{subarray}{c}0<\alpha\leq a\\\
0\leq\beta<b\\\
\alpha+\beta+1=r\end{subarray}}\Big{(}B_{\alpha,\beta}^{r}-B_{\alpha,\beta}^{r}\Big{)}$
$\displaystyle=B_{a,b}^{r},$

where the second equality follows from the third observations above for $a\geq
r$ and $b\geq r$, respectively, and the third equality follows from a shift of
indices. ∎

##  5 Formal multiple zeta values

###  5.1 Multiple zeta values and their general structure

We provide a short basic introduction in the theory of multiple zeta values,
which will serve as the motivation for formal multiple zeta values. For a
detailed exposition we refer to [ba_notes], [bgf], [ikz].

######  Definition 5.1.

To integers $k_{1}\geq 2,\ k_{2},\ldots,k_{d}\geq 1$, associate the _multiple
zeta value_

$\zeta(k_{1},\ldots,k_{d})=\sum_{n_{1}>\dots>n_{d}>0}\frac{1}{n_{1}^{k_{1}}\dots
n_{d}^{k_{d}}}\in\mathbb{R}.$

Denote the $\mathbb{Q}$-vector space spanned by all multiple zeta values by

$\operatorname{\mathcal{Z}}=\operatorname{span}_{\mathbb{Q}}\\{\zeta(k_{1},\ldots,k_{d})\mid
d\geq 0,\ k_{1}\geq 2,\ k_{2},\ldots,k_{d}\geq 1\\},$

where $\zeta(\emptyset)=1$. For an index
$(k_{1},\ldots,k_{d})\in\mathbb{N}^{d}$, define the _weight_ and _depth_ by

$\displaystyle\operatorname{wt}(k_{1},\ldots,k_{d})=k_{1}+\dots+k_{d},\qquad\operatorname{dep}(k_{1},\ldots,k_{d})=d.$

For simplicity, we will also refer to these numbers as the weight and depth of
$\zeta(k_{1},\ldots,k_{d})$.

Numerical experiments have led to the following dimension conjectures for
$\operatorname{\mathcal{Z}}$.

######  Conjecture 5.2.

([zag, p. 509])
1) The vector space $\operatorname{\mathcal{Z}}$ is graded with respect to the
weight, i.e.,

$\operatorname{\mathcal{Z}}=\bigoplus\limits_{w\geq
0}\operatorname{\mathcal{Z}}_{w},$

where $\operatorname{\mathcal{Z}}_{w}$ is spanned by all multiple zeta values
of weight $w$.
2) The dimensions of the homogeneous subspaces
$\operatorname{\mathcal{Z}}_{w}$ are given by

$H_{\operatorname{\mathcal{Z}}}(x)=\sum_{w\geq
0}\dim_{\mathbb{Q}}(\operatorname{\mathcal{Z}}_{w})x^{w}=\frac{1}{1-x^{2}-x^{3}}.$

This conjecture implies that the numbers
$d_{w}=\dim_{\mathbb{Q}}(\operatorname{\mathcal{Z}}_{w})$ should satisfy the
recursion

$\displaystyle d_{w}=d_{w-2}+d_{w-3}$

with initial values $d_{1}=0$ and $d_{2}=d_{3}=1$.

It is well-known that $\operatorname{\mathcal{Z}}$ is not graded with respect
to the depth, e.g., there is Euler’s relation

$\displaystyle\zeta(2,1)=\zeta(3).$ (38)

In these lectures we just want to mention the Broadhurst-Kreimer conjecture
([brkr, (7)]), which is a refinement of Zagier’s dimension conjecture that in
addition relies also on the depth filtration.

There exists also a suggestion for an explicit basis for
$\operatorname{\mathcal{Z}}$. We set

$\displaystyle\mathcal{H}=\operatorname{span}_{\mathbb{Q}}\\{\zeta(k_{1},\ldots,k_{d})\mid
k_{i}\in\\{2,3\\}\text{ for
}i=1,\ldots,d\\}\subset\operatorname{\mathcal{Z}}.$

######  Conjecture 5.3.

([h2, Conjecture C]) A basis for $\operatorname{\mathcal{Z}}$ is given by the
Hoffman elements $\zeta(k_{1},\ldots,k_{d})$, $k_{i}\in\\{2,3\\}$. In
particular, we have

$\mathcal{H}=\operatorname{\mathcal{Z}}.$

This conjecture would imply Zagier’s dimension conjecture 5.2.

It is quite obvious, that there exists more structure on the space
$\operatorname{\mathcal{Z}}$.

######  Proposition 5.4.

The space $\operatorname{\mathcal{Z}}$ equipped with the usual multiplication
of real numbers is an algebra. ∎

There are two ways of expressing the product of multiple zeta values, called
the _stuffle_ and the _shuffle product_. The stuffle product comes from the
combinatorics of multiplying infinite nested sums. E.g., for $k_{1},k_{2}\geq
2$, there is the simple calculation

$\displaystyle\zeta(k_{1})\zeta(k_{2})$
$\displaystyle=\left(\sum_{m>0}\frac{1}{m^{k_{1}}}\right)\left(\sum_{n>0}\frac{1}{n^{k_{2}}}\right)=\left(\sum_{m>n>0}+\sum_{n>m>0}+\sum_{m=n>0}\right)\frac{1}{m^{k_{1}}n^{k_{2}}}$
$\displaystyle=\zeta(k_{1},k_{2})+\zeta(k_{2},k_{1})+\zeta(k_{1}+k_{2}).$

The shuffle product is obtained from expressing multiple zeta values as
iterated integrals ([bgf, Theorem 1.108.]). E.g., in depth $2$ the shuffle
product reads for $k_{1},k_{2}\geq 2$

$\zeta(k_{1})\zeta(k_{2})=\sum_{j=2}^{k_{1}+k_{2}-1}\left(\binom{j-1}{k_{1}-1}+\binom{j-1}{k_{2}-1}\right)\zeta(j,k_{1}+k_{2}-j).$

###  5.2 Extended double shuffle relations

To describe these two product expressions of multiple zeta values in general,
we will use Hoffman’s quasi-shuffle Hopf algebras (Subsection 2.2). The
comparison of these two product formulas lead us then to the (extended) double
shuffle relations.

The shuffle product of the multiple zeta values can be described in terms of
the previously studied shuffle Hopf algebra
$(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle,\Delta_{\operatorname{dec}})$.
Denote by $\mathfrak{h}^{0}$ the subspace of
$\mathbb{Q}\langle\mathcal{X}\rangle$ generated by $\mathbf{1}$ and all words
starting in $x_{0}$ and ending in $x_{1}$, so

$\mathfrak{h}^{0}=\mathbb{Q}\mathbf{1}+x_{0}\mathbb{Q}\langle\mathcal{X}\rangle
x_{1}.$

######  Theorem 5.5.

The map

$\displaystyle\zeta:(\mathfrak{h}^{0},\shuffle)$
$\displaystyle\to(\operatorname{\mathcal{Z}},\cdot),$ $\displaystyle
x_{0}^{k_{1}-1}x_{1}\cdots x_{0}^{k_{d}-1}x_{1}$
$\displaystyle\mapsto\zeta(k_{1},\ldots,k_{d})$

is a surjective algebra morphism compatible with notions of weight and depth
for words and indices. ∎

Recall that the _weight_ of a word $w\in\mathbb{Q}\langle\mathcal{X}\rangle$
is the number of its letters, and by the _depth_ of a word
$w\in\mathbb{Q}\langle\mathcal{X}\rangle$ we mean the number of the letter
$x_{1}$ in $w$.

To describe the stuffle product of multiple zeta values, we introduce a new
alphabet.

######  Definition 5.6.

Consider the infinite alphabet $\mathcal{Y}=\\{y_{1},y_{2},\ldots\\}$. For a
word in $\mathbb{Q}\langle\mathcal{Y}\rangle$, define the _weight_ and _depth_
by

$\displaystyle\operatorname{wt}(y_{k_{1}}\cdots
y_{k_{d}})=k_{1}+\dots+k_{d},\qquad\operatorname{dep}(y_{k_{1}}\cdots
y_{k_{d}})=d.$

Let the _stuffle product_ $\ast$ on $\mathbb{Q}\langle\mathcal{Y}\rangle$ be
the quasi-shuffle product corresponding to

$y_{i}\diamond y_{j}=y_{i+j}\qquad\text{ for }i,j\geq 1.$

From Theorem 2.19 we obtain the following.

######  Proposition 5.7.

The tuple
$(\mathbb{Q}\langle\mathcal{Y}\rangle,\ast,\Delta_{\operatorname{dec}})$ is a
weight-graded commutative Hopf algebra. The complete dual Hopf algebra with
respect to the pairing in (5) is given by
$(R\langle\langle\mathcal{Y}\rangle\rangle,\operatorname{conc},\Delta_{\ast})$,
where the coproduct $\Delta_{\ast}$ is defined on the generators by

$\Delta_{\ast}(y_{i})=y_{i}\otimes\mathbf{1}+\mathbf{1}\otimes
y_{i}+\sum_{j=1}^{i-1}y_{j}\otimes y_{i-j},\quad i=1,2,\ldots.$

Denote by $\mathbb{Q}\langle\mathcal{Y}\rangle^{0}$ the subspace of
$\mathbb{Q}\langle\mathcal{Y}\rangle$ generated by all words, which do not
start in $y_{1}$.

######  Theorem 5.8.

The map

$\displaystyle\zeta:(\mathbb{Q}\langle\mathcal{Y}\rangle^{0},\ast)$
$\displaystyle\to(\operatorname{\mathcal{Z}},\cdot),$ $\displaystyle
y_{k_{1}}\cdots y_{k_{d}}$ $\displaystyle\mapsto\zeta(k_{1},\ldots,k_{d})$

is a surjective algebra morphism compatible with the notions of weight and
depth for words and indices. ∎

Comparing the shuffle and stuffle product formulas for multiple zeta values
(Theorem 5.5, 5.8) gives the _(finite) double shuffle relations_ among
multiple zeta values. Euler’s relation given in (38) is not covered by the
finite double shuffle relations, since there is no product decomposition in
weight $3$. To get these kind of relations we will introduce regularizations.

######  Proposition 5.9.

Let $T$ be a commutative variable and extend the stuffle product $\ast$ by
$\mathbb{Q}[T]$-linearity to $\mathbb{Q}\langle\mathcal{Y}\rangle^{0}[T]$. We
have an algebra isomorphism

$\displaystyle\operatorname{reg}_{\ast}:\mathbb{Q}\langle\mathcal{Y}\rangle^{0}[T]\to\mathbb{Q}\langle\mathcal{Y}\rangle,$
$\displaystyle wT^{n}\mapsto w\ast y_{1}^{\ast n}.$

∎

For any $w\in\mathbb{Q}\langle\mathcal{Y}\rangle$, set

$\displaystyle\zeta_{\ast}^{T}(w)=\zeta(\operatorname{reg}_{\ast}^{-1}(w))\in\operatorname{\mathcal{Z}}[T],$

where we extend also the map
$\zeta:\mathbb{Q}\langle\mathcal{Y}\rangle^{0}\to\operatorname{\mathcal{Z}}$
by $\mathbb{Q}[T]$-linearity to $\mathbb{Q}\langle\mathcal{Y}\rangle^{0}[T]$.
We call

$\displaystyle\zeta_{\ast}(w)=\zeta_{\ast}^{T=0}(w)=\zeta(\operatorname{reg}_{\ast}^{-1}(w)|_{T=0})\in\operatorname{\mathcal{Z}}$

the _stuffle-regularized multiple zeta values_. An immediate consequence of
Proposition 5.9 is the following.

######  Theorem 5.10.

The map
$\zeta_{\ast}:\mathbb{Q}\langle\mathcal{Y}\rangle\to\operatorname{\mathcal{Z}}$
given by $w\mapsto\zeta_{\ast}(w)$ is the unique map satisfying

  * (i)

$\zeta_{\ast}(w)=\zeta(w)$ for all
$w\in\mathbb{Q}\langle\mathcal{Y}\rangle^{0}$,

  * (ii)

$\zeta_{\ast}(y_{1})=0$,

  * (iii)

$\zeta_{\ast}(u)\zeta_{\ast}(v)=\zeta_{\ast}(u\ast v)$ for all
$u,v\in\mathbb{Q}\langle\mathcal{Y}\rangle$.

Similarly, there also exists a regularization with respect to the shuffle
product for multiple zeta values.

######  Proposition 5.11.

Let $T,U$ be a commutative variables and extend the shuffle product by
$\mathbb{Q}[T,U]$-linearity to $\mathfrak{h}^{0}[T,U]$. There is an algebra
isomorphism

$\displaystyle\operatorname{reg}:\mathfrak{h}^{0}[T,U]$
$\displaystyle\to\mathbb{Q}\langle\mathcal{X}\rangle,$ $\displaystyle
wT^{n}U^{m}$ $\displaystyle\mapsto w\shuffle x_{1}^{\shuffle n}\shuffle
x_{0}^{\shuffle m}.$

∎

For any $w\in\mathbb{Q}\langle\mathcal{X}\rangle$, set

$\displaystyle\zeta^{T,U}(w)=\zeta(\operatorname{reg}^{-1}(w))\in\operatorname{\mathcal{Z}}[T,U],$

where the map $\zeta:\mathfrak{h}^{0}\to\operatorname{\mathcal{Z}}$ needs to
be extended by $\mathbb{Q}[T,U]$-linearity to $\mathfrak{h}^{0}[T,U]$. We set

$\displaystyle\zeta^{T}(w)=\zeta(\operatorname{reg}^{-1}(w)|_{U=0})\in\operatorname{\mathcal{Z}}[T],$

and, moreover, call

$\displaystyle\zeta(w)=\zeta^{T=U=0}(w)=\zeta(\operatorname{reg}_{\ast}^{-1}(w)|_{T=U=0})\in\operatorname{\mathcal{Z}}$

the _shuffle-regularized multiple zeta values_. One derives the following from
Proposition 5.11.

######  Theorem 5.12.

The map
$\zeta:\mathbb{Q}\langle\mathcal{X}\rangle\to\operatorname{\mathcal{Z}},\
w\mapsto\zeta(w)$ is the unique map satisfying

  * (i)

$\zeta(w)=\zeta(w)$ for all $w\in\mathfrak{h}^{0}$,

  * (ii)

$\zeta(x_{1})=\zeta(x_{0})=0$,

  * (iii)

$\zeta(u)\zeta(v)=\zeta(u\shuffle v)$ for all
$u,v\in\mathbb{Q}\langle\mathcal{X}\rangle$.

We want to relate these two regularizations of multiple zeta values. Define
the $\operatorname{\mathcal{Z}}$-linear map
$\rho:\operatorname{\mathcal{Z}}[T]\to\operatorname{\mathcal{Z}}[T]$ by

$\displaystyle\rho\left(\frac{T^{m}}{m!}\right)=\sum_{i=0}^{m}\gamma_{i}\frac{T^{m-i}}{(m-i)!},\qquad
m=0,1,2,\ldots,$ (39)

where the coefficients $\gamma_{i}\in\operatorname{\mathcal{Z}}$ are defined
by $\sum\limits_{i\geq 0}\gamma_{i}u^{i}=\exp\left(\sum\limits_{n\geq
2}\frac{(-1)^{n}}{n}\zeta(n)u^{n}\right)$.

######  Theorem 5.13.

([ikz, Theorem 1]) For all $k_{1},\ldots,k_{d}\geq 1$, one has

$\rho\big{(}\zeta_{\ast}^{T}(y_{k_{1}}\cdots
y_{k_{d}})\big{)}=\zeta^{T}(x_{0}^{k_{1}-1}x_{1}\cdots x_{0}^{k_{d}-1}x_{1}).$

∎

Combining the stuffle product formula for the stuffle-regularized multiple
zeta values and the shuffle product formula for the shuffle-regularized
multiple zeta values together with Theorem 5.13 gives the _extended double
shuffle relations_ among multiple zeta values.

######  Conjecture 5.14.

([ikz, Conjecture 1]) All algebraic relations in the algebra
$\operatorname{\mathcal{Z}}$ of multiple zeta values are a consequence of the
extended double shuffle relations.

In particular, Conjecture 5.14 would imply that the algebra
$\operatorname{\mathcal{Z}}$ is graded by weight, since the stuffle and the
shuffle product are both homogeneous for the weight.

######  Example 5.15.

We calculate

$\displaystyle\zeta_{\ast}(y_{1})\zeta_{\ast}(y_{2})$
$\displaystyle=\zeta_{\ast}(y_{1}y_{2})+\zeta_{\ast}(y_{2}y_{1})+\zeta_{\ast}(y_{3}),$
$\displaystyle\zeta(x_{1})\zeta(x_{0}x_{1})$
$\displaystyle=\zeta(x_{1}x_{0}x_{1})+2\zeta(x_{0}x_{1}x_{1})$

By Theorem 5.13, we have $\zeta_{\ast}(y_{1}y_{2})=\zeta(x_{1}x_{0}x_{1})$,
and, moreover, $\zeta_{\ast}(y_{1})=\zeta(x_{1})=0$. Thus, the above equations
reduce to

$\displaystyle\zeta_{\ast}(y_{2}y_{1})+\zeta_{\ast}(y_{3})=2\zeta(x_{0}x_{1}x_{1}).$

By Theorem 5.10, 5.12, this is equivalent to

$\displaystyle\zeta(3)=\zeta(2,1).$

So we recover Euler’s relations from the extended double shuffle relations.

######  Remark 5.16.

There are various kinds of relations obtained for multiple zeta values in the
literature, and several of them are expected to give all algebraic relations
in $\operatorname{\mathcal{Z}}$. A graphical overview is given by H. Bachmann
in [ba_notes, Section 3.3].

###  5.3 Formal multiple zeta values and some properties

We have a canonical embedding

$\displaystyle\iota:\mathbb{Q}\langle\mathcal{Y}\rangle$
$\displaystyle\hookrightarrow\mathbb{Q}\langle\mathcal{X}\rangle,$
$\displaystyle y_{k_{1}}\cdots y_{k_{d}}$ $\displaystyle\mapsto
x_{0}^{k_{1}-1}x_{1}\cdots x_{0}^{k_{d}-1}x_{1},$

which allows to transfer the stuffle product $\ast$ on
$\mathbb{Q}\langle\mathcal{Y}\rangle$ (Definition 5.6) to the algebra
$\mathfrak{h}^{1}=\mathbb{Q}\mathbf{1}+\mathbb{Q}\langle\mathcal{X}\rangle
x_{1}$. So motivated by Conjecture 5.14, we can reformulate the extended
double shuffle relations purely algebraically in
$\mathbb{Q}\langle\mathcal{X}\rangle$.

######  Definition 5.17.

The algebra $\mathcal{Z}^{\mathsf{f}}$ of _formal multiple zeta values_ is
given by

$\mathcal{Z}^{\mathsf{f}}={\mathchoice{\raisebox{3.75pt}{$\displaystyle{(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle)}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-3.41666pt}{$\displaystyle{\operatorname{R}_{\operatorname{EDS}}}$}}{\raisebox{3.75pt}{$\textstyle{(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle)}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-3.41666pt}{$\textstyle{\operatorname{R}_{\operatorname{EDS}}}$}}{\raisebox{3.75pt}{$\scriptstyle{(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle)}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-2.39166pt}{$\scriptstyle{\operatorname{R}_{\operatorname{EDS}}}$}}{\raisebox{3.75pt}{$\scriptscriptstyle{(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle)}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-1.70833pt}{$\scriptscriptstyle{\operatorname{R}_{\operatorname{EDS}}}$}}},$

where $\operatorname{R}_{\operatorname{EDS}}$ is the ideal generated by the
extended double shuffle relations. Following [ikz],
$\operatorname{R}_{\operatorname{EDS}}$ is for example generated by $x_{0}$,
$x_{1}$ and

$u\shuffle v-\iota\Big{(}\iota^{-1}(u)\ast\iota^{-1}(v)\Big{)},\qquad
u\in\mathfrak{h}^{0},v\in\mathfrak{h}^{1}.$

######  Remark 5.18.

Since the stuffle product $\ast$ as well as the shuffle product are graded for
the weight, the ideal $\operatorname{R}_{\operatorname{EDS}}$ is weight-
homogeneous. Therefore, the algebra $\mathcal{Z}^{\mathsf{f}}$ is weight-
graded.

######  Remark 5.19.

We won’t compare the formal multiple zeta values to the motivic multiple zeta
values, which are formalisations that depend on a different set of relations.
Their construction is requires a profound knowledge of modern arithmetic
geometry, for detail we refer to [bgf]. See also Remark 5.16.

Let us denote the canonical projection by

$\zeta^{\mathsf{f}}:\mathbb{Q}\langle\mathcal{X}\rangle\to\mathcal{Z}^{\mathsf{f}},$
(40)

thus $\zeta^{\mathsf{f}}(w)$ is the class of
$w\in\mathbb{Q}\langle\mathcal{X}\rangle$ in the quotient algebra
$\mathcal{Z}^{\mathsf{f}}$. The space $\mathcal{Z}^{\mathsf{f}}$ is a weight-
graded algebra spanned by the elements $\zeta^{\mathsf{f}}(w)$,
$w\in\mathcal{X}^{*}$, which satisfy exactly the extended double shuffle
relations. In particular, we have for all $k\geq 1$ that

$\displaystyle\zeta^{\mathsf{f}}(x_{0}^{k})=\zeta^{\mathsf{f}}(x_{1}^{k})=0.$
(41)

If we have a word $w=x_{0}^{k_{1}-1}x_{1}\cdots
x_{0}^{k_{d}-1}x_{1}\in\mathfrak{h}^{0}$, we often write instead of
$\zeta^{\mathsf{f}}(w)$ also $\zeta^{\mathsf{f}}(k_{1},\ldots,k_{d})$. By
Proposition 5.11 and (41) these are also a spanning set for
$\mathcal{Z}^{\mathsf{f}}$. Additionally, the
$\zeta^{\mathsf{f}}(k_{1},\ldots,k_{d})$ with $k_{1}\geq 2$ also satisfy the
usual stuffle product formulas.

By construction, the _evaluation map_

$\displaystyle\mathcal{Z}^{\mathsf{f}}$
$\displaystyle\to\operatorname{\mathcal{Z}},$ (42)
$\displaystyle\zeta^{\mathsf{f}}(w)$ $\displaystyle\mapsto\zeta(w)$

is a surjective algebra morphism. By Conjecture 5.14, this map should be an
algebra isomorphism.

######  Proposition 5.20.

For $k\geq 0$, let $B_{k}$ be the $k$-th Bernoulli number and set

$b_{n}=(-1)^{n+1}B_{2n}\frac{24^{n}}{2(2n)!}.$

Then we have for each $n\in\mathbb{N}$

$\displaystyle\zeta^{\mathsf{f}}(2n)$
$\displaystyle=b_{n}\zeta^{\mathsf{f}}(2)^{n},$
$\displaystyle\zeta^{\mathsf{f}}(\\{2\\}^{n})$
$\displaystyle=\frac{6^{n}}{(2n+1)!}\zeta^{\mathsf{f}}(2)^{n}.$

Here we use the notation
$\zeta^{\mathsf{f}}(\\{2\\}^{n})=\zeta^{\mathsf{f}}(\underbrace{2,\ldots,2}_{n\text{
times}})$.

###### Proof.

Euler showed by analytic means his famous formula

$\zeta(2n)=(-1)^{n+1}\frac{B_{2n}}{2(2n)!}(2\pi)^{2n}=b_{n}\zeta(2)^{n}.$

For the formal multiple zeta values we deduce this formula from the extended
double shuffle as follows: Considering the generating series for the formal
multiple zeta values in depth two we deduce as Gangl, Kaneko and Zagier in
[gkz] the identity

$\sum_{\begin{subarray}{c}r,s\text{ even}\\\
r+s=k\end{subarray}}\zeta^{\mathsf{f}}(r,s)=\frac{3}{4}\zeta^{\mathsf{f}}(k).$

Now using
$\zeta^{\mathsf{f}}(r)\zeta^{\mathsf{f}}(s)=\zeta^{\mathsf{f}}(r,s)+\zeta^{\mathsf{f}}(s,r)+\zeta^{\mathsf{f}}(r+s)$
we get for the sum of products

$\sum_{\begin{subarray}{c}r,s\text{ even}\\\
r+s=k\end{subarray}}\zeta^{\mathsf{f}}(r)\zeta^{\mathsf{f}}(s)=\frac{k+1}{2}\zeta^{\mathsf{f}}(k).$

This yields a recursion for the numbers $b_{n}$ in the equation
$\zeta^{\mathsf{f}}(2n)=b_{n}\zeta^{\mathsf{f}}(2)^{n}$, whose first terms are

$\displaystyle\zeta^{\mathsf{f}}(4)$
$\displaystyle=\frac{2}{5}\zeta^{\mathsf{f}}(2)^{2},$
$\displaystyle\zeta^{\mathsf{f}}(6)$
$\displaystyle=\frac{2}{7}\Big{(}\zeta^{\mathsf{f}}(2)\zeta^{\mathsf{f}}(4)+\zeta^{\mathsf{f}}(4)\zeta^{\mathsf{f}}(2)\Big{)}=\frac{8}{35}\zeta^{\mathsf{f}}(2)^{3}.$

Now using a simple manipulation of the generating series for Bernoulli
numbers333It is not difficult to check that
$-\frac{1}{2}\left(\sqrt{6}x\operatorname{cot}(\sqrt{6}x)-1\right)$ is the
generating series for the rational numbers $b_{n}$. This approach to obtain
this formula of Euler in the formal setting is elaborated in [BaIt_fMES,
Corollary 2.12].  or using the evaluation map and Euler’s formula for the even
zeta values we deduce

$b_{n}=(-1)^{n+1}B_{2n}\frac{24^{n}}{2(2n)!}.$

In their paper [hi] on quasi-shuffle algebras Hofmann and Ihara showed that
for all $k\geq 2$ we have the identity of generating series

$\displaystyle\exp\left(\sum_{i\geq
1}\frac{(-1)^{i-1}}{i}\zeta^{\mathsf{f}}(ik)t^{i}\right)=\sum_{n=0}^{\infty}\zeta^{\mathsf{f}}(\\{k\\}^{n})\,t^{n}.$

If we specialize this to $k=2$ we see that $\zeta^{\mathsf{f}}(\\{2\\}^{n})$
is a polynomial in even formal zeta values, however by the algebraic Euler
formula this equals a rational multiple of $\zeta^{\mathsf{f}}(2)^{n}$. As
before we can determine this proportionality factor either by a direct
calculation or by a comparison with the analytic identity

$\zeta(\\{2\\}^{n})=\frac{\pi^{2n}}{(2n+1)!}=\frac{6^{n}}{(2n+1)!}\zeta(2)^{n}.$

due to Hoffman and Zagier. ∎

######  Remark 5.21.

In the above proof we reduced identities for formal multiple zeta values to
the determination of a rational number, which we computed by means of the
evaluation map. Actually, we could even had calculated it without referring to
the previous known analytic identities. In Subsection 6.2 we see a much more
elaborate lift of another set of identities satisfied by multiple zeta values
to the formal setting. Those formulae due to Zagier can be seen as a
refinement of the following proposition and so far no algebraic proof without
referring to analytical identities is known.

######  Proposition 5.22.

For all $n\in\mathbb{N}$ we have the identity

$\displaystyle\zeta^{\mathsf{f}}((x_{0}x_{1})^{n}x_{0})$
$\displaystyle=-2\sum_{i=0}^{n-1}\zeta^{\mathsf{f}}(\\{2\\}^{i},3,\\{2\\}^{n-1-i})$
(43)
$\displaystyle=2\sum_{i=1}^{n}(-1)^{i}\zeta^{\mathsf{f}}(2i+1)\,\zeta^{\mathsf{f}}(\\{2\\}^{n-i}).$

###### Proof.

For the first equality we consider the shuffle product identity in
$\mathbb{Q}\langle\mathcal{X}\rangle$

$\displaystyle
x_{0}\shuffle(x_{0}x_{1})^{n}=(x_{0}x_{1})^{n}x_{0}+\sum_{i=0}^{n-1}2\,(x_{0}x_{1})^{i}x_{0}x_{0}x_{1}(x_{0}x_{1})^{n-1-i}$

and this amounts in the quotient algebra $\mathcal{Z}^{\mathsf{f}}$ to

$\zeta^{\mathsf{f}}(x_{0})\zeta^{\mathsf{f}}(\\{2\\}^{n})=\zeta^{\mathsf{f}}((x_{0}x_{1})^{n}x_{0})+\sum_{i=0}^{n-1}2\,\zeta^{\mathsf{f}}(\\{2\\}^{i},3,\\{2\\}^{n-1-i}).$

The left-hand side vanishes since $\zeta^{\mathsf{f}}(x_{0})=0$ and the first
identity follows.

For the second equality we use that the multiplication of formal zeta values
with admissible indices satisfy the stuffle relation, we thus get

$\displaystyle\zeta^{\mathsf{f}}(3)\zeta^{\mathsf{f}}(\\{2\\}^{n-1})$
$\displaystyle=\sum_{i=0}^{n-1}\zeta^{\mathsf{f}}(\\{2\\}^{i},3,\\{2\\}^{n-1-i})+\sum_{i=0}^{n-2}\zeta^{\mathsf{f}}(\\{2\\}^{i},5,\\{2\\}^{n-2-i})$
$\displaystyle\zeta^{\mathsf{f}}(5)\zeta^{\mathsf{f}}(\\{2\\}^{n-2})$
$\displaystyle=\sum_{i=0}^{n-2}\zeta^{\mathsf{f}}(\\{2\\}^{i},5,\\{2\\}^{n-2-i})+\sum_{i=0}^{n-3}\zeta^{\mathsf{f}}(\\{2\\}^{i},7,\\{2\\}^{n-3-i})$
$\displaystyle\;\vdots$
$\displaystyle\zeta^{\mathsf{f}}(2n-1)\zeta^{\mathsf{f}}(2)$
$\displaystyle=\zeta^{\mathsf{f}}(2n-1,2)+\zeta^{\mathsf{f}}(2,2n-1)+\zeta^{\mathsf{f}}(2n+1).$

By taking the alternating sum we obtain

$-\sum_{i=0}^{n-1}\zeta^{\mathsf{f}}(\\{2\\}^{i},3,\\{2\\}^{n-1-i})=\sum_{i=1}^{n}(-1)^{i}\zeta^{\mathsf{f}}(2i+1)\zeta^{\mathsf{f}}(\\{2\\}^{n-i})$

and the claim follows. ∎

######  Remark 5.23.

The identity (43) implies for the space $\mathcal{B}^{1}$ defined in Theorem
4.9 that $\zeta^{\mathsf{f}}\big{(}\mathcal{B}^{1}\big{)}$ is contained in the
vector space generated by those $\zeta^{\mathsf{f}}(k_{1},...,k_{d})$, where
one $k_{i}$ equals $3$ and all other $k_{j}$ equal $2$.

###  5.4 Racinet’s approach and Ecalle’s theorem

We explain Racinet’s approach to formal multiple zeta values in terms of non-
commutative power series ([ra]). This means, we assign an affine group scheme
and a Lie algebra to the formal multiple zeta values. This has two deep,
structural consequences for the algebra $\mathcal{Z}^{\mathsf{f}}$. For more
detailed expositions we refer to [bu], [enr].

######  Definition 5.24.

Let $\mathcal{A}$ be an alphabet, $R$ be some $\mathbb{Q}$-algebra, and
$\varphi:\mathbb{Q}\langle\mathcal{A}\rangle\to R$ be a $\mathbb{Q}$-linear
map. Then the non-commutative generating series associated to $\varphi$ is

$\operatorname{Gen}_{\varphi}(\mathcal{A})=\sum_{w\in\mathcal{A}^{*}}\varphi(w)w\in
R\langle\langle\mathcal{A}\rangle\rangle.$

We assume that $(\mathbb{Q}\langle\mathcal{A}\rangle,\ast_{\diamond})$ is any
graded quasi-shuffle algebra with $\deg(a)\geq 1$ for all $a\in\mathcal{A}$.
Then we can compute the dual coproduct $\Delta_{\ast_{\diamond}}$ given in
(6).

######  Proposition 5.25.

A $\mathbb{Q}$-linear map $\varphi:\mathbb{Q}\langle\mathcal{A}\rangle\to R$
is an algebra morphism with respect to $\ast_{\diamond}$ if and only if
$\operatorname{Gen}_{\varphi}(\mathcal{A})$ is grouplike for
$\Delta_{\ast_{\diamond}}$,

$\Delta_{\ast_{\diamond}}(\operatorname{Gen}_{\varphi}(\mathcal{A}))=\operatorname{Gen}_{\varphi}(\mathcal{A})\otimes\operatorname{Gen}_{\varphi}(\mathcal{A}).$

∎

Motivated by this proposition, we consider the following sets.

######  Definition 5.26.

For any commutative $\mathbb{Q}$-algebra $R$ with unit, let
$\operatorname{DM}(R)$ be the set of all non-commutative power series $\phi\in
R\langle\langle\mathcal{X}\rangle\rangle$ satisfying

(i) | $(\phi|x_{0})\ =\ (\phi|x_{1})$ | $=$ |  $0$,
---|---|---|---
(ii) | $\Delta(\phi)$ | $=$ |  $\phi\hat{\otimes}\phi$,
(iii) | $\Delta_{\ast}(\phi_{\ast})$ | $=$ |  $\phi_{\ast}\hat{\otimes}\phi_{\ast}$,

where

$\phi_{\ast}=\exp\left(\sum_{n\geq
2}\frac{(-1)^{n-1}}{n}(\Pi_{\mathcal{Y}}(\phi)|y_{n})y_{1}^{n}\right)\Pi_{\mathcal{Y}}(\phi)\in
R\langle\langle\mathcal{Y}\rangle\rangle$

and $\Pi_{\mathcal{Y}}$ is the $R$-linear extension of the projection

$\displaystyle\mathbb{Q}\langle\mathcal{X}\rangle$
$\displaystyle\to\mathbb{Q}\langle\mathcal{Y}\rangle,$ $\displaystyle
x_{0}^{k_{1}-1}x_{1}\cdots x_{0}^{k_{d}-1}x_{1}x_{0}^{k_{d+1}-1}$
$\displaystyle\mapsto\begin{cases}y_{k_{1}}\cdots y_{k_{d}},&\quad
k_{d+1}=1\\\ 0&\quad\textsf{else}\end{cases}.$

For each $\lambda\in R$, denote by $\operatorname{DM}_{\lambda}(R)$ the set of
all $\phi\in\operatorname{DM}(R)$, which additionally satisfy

$\text{(iv)}\qquad(\phi|x_{0}x_{1})=\lambda.$

By Theorems 5.10, 5.12, 5.13 and Proposition 5.25, the non-commutative
generating series of the shuffle regularized multiple zeta values

$\displaystyle\operatorname{Gen}_{\zeta}(\mathcal{X})=\sum_{w\in\mathcal{X}^{*}}\zeta(w)w$

is an element in $\operatorname{DM}_{\pi^{2}/6}(\operatorname{\mathcal{Z}})$.

######  Theorem 5.27.

([ra2, Theorem I]) For each commutative $\mathbb{Q}$-algebra $R$ and
$\lambda\in R$, the set $\operatorname{DM}_{\lambda}(R)$ is non-empty. ∎

From [dr] and [fu] one deduces that there also exist elements $\phi$ in
$\operatorname{DM}_{\lambda}(R)$ additionally satisfying

$\displaystyle(\phi\mid x_{0}^{k}x_{1})=0\quad\text{ for }k\geq 1\text{
even}.$

The sets $\operatorname{DM}(R)$ and $\operatorname{DM}_{\lambda}(R)$ give rise
to affine schemes represented by (quotient algebras of)
$\mathcal{Z}^{\mathsf{f}}$.

######  Proposition 5.28.

([ra, p. 107])
(i) The functor
$\operatorname{DM}_{\lambda}:\mathbb{Q}\operatorname{-}\mathsf{Alg}\to\mathsf{Sets}$
is an affine scheme represented by the algebra $\mathcal{Z}^{\mathsf{f}}$ of
formal multiple zeta values. In particular, for
$R\in\mathbb{Q}\operatorname{-}\mathsf{Alg}$ there is a bijection

$\displaystyle\operatorname{Hom}_{\mathbb{Q}\operatorname{-}\mathsf{Alg}}(\mathcal{Z}^{\mathsf{f}},R)$
$\displaystyle\to\operatorname{DM}(R),$ $\displaystyle\varphi$
$\displaystyle\mapsto\sum_{w\in\mathcal{X}^{*}}\varphi(\zeta^{\mathsf{f}}(w))w.$

(ii) The functors
$\operatorname{DM}_{\lambda}:\mathbb{Q}\operatorname{-}\mathsf{Alg}\to\mathsf{Sets}$
are an affine schemes represented by the quotient algebras
${\mathchoice{\raisebox{4.38887pt}{$\displaystyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.44444pt}{$\displaystyle{(\zeta^{\mathsf{f}}(2)-\lambda)}$}}{\raisebox{4.38887pt}{$\textstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.44444pt}{$\textstyle{(\zeta^{\mathsf{f}}(2)-\lambda)}$}}{\raisebox{3.0861pt}{$\scriptstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-3.12498pt}{$\scriptstyle{(\zeta^{\mathsf{f}}(2)-\lambda)}$}}{\raisebox{2.40277pt}{$\scriptscriptstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-2.43054pt}{$\scriptscriptstyle{(\zeta^{\mathsf{f}}(2)-\lambda)}$}}}$.
∎

$\operatorname{DM}_{0}$ is often called the _double shuffle group_. To figure
out the group structure for the affine scheme $\operatorname{DM}_{0}$, one
needs to consider first the corresponding linearized space.

######  Definition 5.29.

For any $R\in\mathbb{Q}\operatorname{-}\mathsf{Alg}$, let $\mathfrak{dm}(R)$
be the $R$-vector space of all non-commutative polynomials $\psi\in
R\langle\mathcal{X}\rangle$ satisfying

(i) | $(\psi\mid x_{0})\ =\ (\psi\mid x_{1})$ | $=$ |  $0$,
---|---|---|---
(ii) | $\Delta(\psi)$ | $=$ |  $\psi\otimes\mathbf{1}+\mathbf{1}\otimes\psi$,
(iii) | $\Delta_{\ast}(\psi_{\ast})$ | $=$ |  $\psi_{\ast}\otimes\mathbf{1}+\mathbf{1}\otimes\psi_{\ast}$,

where

$\psi_{\ast}=\Pi_{\mathcal{Y}}(\psi)+\sum\limits_{n\geq
2}\frac{(-1)^{n-1}}{n}(\Pi_{\mathcal{Y}}(\psi)\mid y_{n})y_{1}^{n}\in
R\langle\mathcal{Y}\rangle$

and $\Pi_{\mathcal{Y}}$ is the $R$-linear extension of the canonical
projection
$\mathbb{Q}\langle\mathcal{X}\rangle\to\mathbb{Q}\langle\mathcal{Y}\rangle$.
By $\mathfrak{dm}_{0}(R)$ denote the subspace of all $\psi\in\mathfrak{dm}(R)$
additionally satisfying

$\text{(iv)}\qquad(\psi\mid x_{0}x_{1})=0.$

Denote $\mathfrak{dm}_{0}=\mathfrak{dm}_{0}(\mathbb{Q})$. Then one has
$\mathfrak{dm}_{0}(R)=\mathfrak{dm}_{0}\otimes R$. The space
$\mathfrak{dm}_{0}$ is often called the _double shuffle Lie algebra_ , the
name will be justified in Theorem 5.32.

######  Example 5.30.

The space $\mathfrak{dm}_{0}$ can be computed algorithmically in small
weights, see e.g. [enr], [bu_master]. One obtains the following elements in
$\mathfrak{dm}_{0}$ up to weight $5$

$\displaystyle\xi(3)$
$\displaystyle=[x_{0},[x_{0},x_{1}]]+[[x_{0},x_{1}],x_{1}],$
$\displaystyle\xi(5)$
$\displaystyle=[x_{0},[x_{0},[x_{0},[x_{0},x_{1}]]]]+2[[x_{0},[x_{0},[x_{0},x_{1}]]],x_{1}]+\frac{1}{2}[[x_{0},[x_{0},x_{1}]],[x_{0},x_{1}]]$
$\displaystyle\hskip
14.22636pt+2[x_{1},[x_{1},[x_{0},[x_{0},x_{1}]]]]-\frac{3}{2}[[x_{0},x_{1}],[[x_{0},x_{1}],x_{1}]]+[[[[x_{0},x_{1}],x_{1}],x_{1}],x_{1}].$

######  Proposition 5.31.

([ra, IV, Proposition 2.2]) For each $k\geq 2$ even and
$\psi\in\mathfrak{dm}_{0}(R)$, one has

$(\psi\mid x_{0}^{k-1}x_{1})=0.$

∎

The Lie bracket on the spaces $\mathfrak{dm}_{0}(R)$ is given by the Ihara
bracket $\\{-,-\\}$ from Definition 3.12.

######  Theorem 5.32.

([ra, IV, Proposition 2.28., Corollary 3.13.]) Let $R$ be a commutative
$\mathbb{Q}$-algebra with unit.
1) The space $\mathfrak{dm}_{0}(R)$ is a Lie subalgebra of
$(\mathfrak{f}_{2}\otimes R,\\{-,-\\})$.
2) For all $\phi_{1},\phi_{2}\in\operatorname{DM}_{\lambda}(R)$, there exists
a unique element $\psi$ in the completed Lie algebra
$\widehat{\mathfrak{dm}_{0}}(R)$ such that

$\exp(s_{\psi})(\phi_{1})=\phi_{2}.$

∎

Here for any $f\in R\langle\mathcal{X}\rangle$,
$s_{f}:R\langle\mathcal{X}\rangle\to R\langle\mathcal{X}\rangle$ is the
$R$-linear map defined by $s_{f}(w)=d_{f}(w)+fw$. Note that this allows to
write

$\displaystyle\\{f,g\\}=s_{f}(g)-s_{g}(f),\qquad f,g\in
R\langle\mathcal{X}\rangle.$

A more detailed proof of Theorem 5.32 1) is given in [fu, Appendix A] and also
in the first author’s master thesis [bu_master].

From Theorem 5.32 2), one obtains natural bijections

$\displaystyle\widehat{\mathfrak{dm}_{0}}(R)$
$\displaystyle\to\operatorname{DM}_{0}(R),$ (44) $\displaystyle\psi$
$\displaystyle\mapsto\exp(s_{\psi})(\mathbf{1}).$

We deduce the following.

######  Corollary 5.33.

([ra]) The functor $\operatorname{DM}_{0}$ is a pro-unipotent affine group
scheme with Lie algebra functor

$\widehat{\mathfrak{dm}_{0}}:\mathbb{Q}\operatorname{-Alg}\to\operatorname{Lie-
Alg},\ R\mapsto\widehat{\mathfrak{dm}_{0}}(R).$

∎

The group multiplication on $\operatorname{DM}_{0}(R)$ is given in Theorem
3.18 or Remark 3.21, so $\operatorname{DM}_{0}(R)$ is a subgroup of
$(\operatorname{Grp}(\widehat{\mathcal{U}}(\mathfrak{f}_{2}\otimes
R)),\circledast)$.

Moreover, it is shown by Racinet that the affine schemes
$\operatorname{DM}_{\lambda}$ are $\operatorname{DM}_{0}$-torsors.

######  Theorem 5.34.

([ra, Section IV, Corollary 3.13]) The affine group scheme
$\operatorname{DM}_{0}$ acts freely and transitively on the affine schemes
$\operatorname{DM}_{\lambda}$ by left multiplication. ∎

In other words, for each $R\in\mathbb{Q}\operatorname{-}\mathsf{Alg}$ we
obtain a map

$\displaystyle\circledast:\operatorname{DM}_{0}(R)\times\operatorname{DM}_{\lambda}(R)$
$\displaystyle\to\operatorname{DM}_{\lambda}(R),$ $\displaystyle(\Phi,G)$
$\displaystyle\mapsto\Phi\circledast G,$

such that $-\circledast
G:\operatorname{DM}_{0}(R)\to\operatorname{DM}_{\lambda}(R)$ is bijective for
any $G\in\operatorname{DM}_{\lambda}(R)$. Combining those maps gives rise to
natural maps

$\displaystyle\circledast:\operatorname{DM}_{0}(R)\times\operatorname{DM}(R)\to\operatorname{DM}(R).$
(45)

An application of Yoneda’s Lemma to the isomorphism
$\exp:\widehat{\mathfrak{dm}_{0}}\to\operatorname{DM}_{0}$ of affine schemes
given in (44), yields the following main result.

######  Corollary 5.35.

([ra], Chapter IV, Corollary 3.14) There is an algebra isomorphism

$\mathcal{Z}^{\mathsf{f}}\simeq\mathbb{Q}[\zeta^{\mathsf{f}}(2)]\otimes_{\mathbb{Q}}\mathcal{U}(\mathfrak{dm}_{0})^{\vee}.$

By (7), we have proved Ecalle’s theorem [ecalle:tale].

######  Theorem 5.36.

The algebra $\mathcal{Z}^{\mathsf{f}}$ of formal multiple zeta values is a
free polynomial algebra.

###  5.5 Goncharov coproduct and Goncharov-Brown coaction

Let

$\mathcal{A}^{\mathsf{f}}=\mathcal{Z}^{\mathsf{f}}/(\zeta^{\mathsf{f}}(2))$

be the quotient algebra of $\mathcal{Z}^{\mathsf{f}}$ by the principal ideal
generated by $\zeta^{\mathsf{f}}(2)$ and write

$\mathsf{a}:\mathcal{Z}^{\mathsf{f}}\to\mathcal{A}^{\mathsf{f}}$

for the natural projection. We define a coproduct on
$\mathcal{A}^{\mathsf{f}}$ by

$\displaystyle\Delta_{\operatorname{Gon}}(a)=\Big{(}\mathsf{a}\circ\zeta^{\mathsf{f}}\otimes\,\mathsf{a}\circ\zeta^{\mathsf{f}}\Big{)}\big{(}\Delta_{\operatorname{Gon}}(w)\big{)},$
(46)

where $w\in\mathbb{Q}\langle\mathcal{X}\rangle$ is a lift for
$a\in\mathcal{A}^{\mathsf{f}}$, i.e. $w$ is any element such that
$\mathsf{a}(\zeta^{\mathsf{f}}(w))=a$.

######  Theorem 5.37.

The triple
$\big{(}\mathcal{A}^{\mathsf{f}},\cdot,\Delta_{\operatorname{Gon}}\big{)}$ is
a weight-graded Hopf algebra.

###### Proof.

In Corollary 5.33, we have seen that $\operatorname{DM}_{0}$ is a subgroup
scheme of the affine group scheme given in (25) induced by the grouplike
elements of $(\widehat{\mathcal{U}}(\mathfrak{f}_{2}),\Delta)$. Since
$\operatorname{DM}_{0}$ is represented by the algebra
$\mathcal{A}^{\mathsf{f}}$ (Proposition 5.28), and the affine group scheme in
(25) is represented by the Hopf algebra
$(\mathbb{Q}\langle\mathcal{X}\rangle,\shuffle,\Delta_{\operatorname{Gon}})$
(Lemma 3.25, Proposition 3.26), also $\mathcal{A}^{\mathsf{f}}$ must be a Hopf
algebra. The coproduct on $\mathcal{A}^{\mathsf{f}}$ is induced by the
Goncharov coproduct under the projection

$\mathsf{a}\circ\zeta^{\mathsf{f}}:\mathbb{Q}\langle\mathcal{X}\rangle\to\mathcal{A}^{\mathsf{f}}.\qed$

######  Remark 5.38.

In particular we deduce from the proof of the above theorem that the coproduct
(46) is well-defined. We like to point to the fact that our approach to the
coproduct relies on the general theory of post-Lie algebras together with the
work of Racinet, whereas the original definition of Goncharov in [gon] was
based on topological considerations for the path algebra. The latter is
directly related to the representation of multiple zeta values by iterated
integrals.

Following Proposition 2.29, the Goncharov coproduct
$\Delta_{\operatorname{Gon}}$ induces a Lie cobracket $\delta$ on the space of
indecomposables $\mathcal{L}^{\mathsf{f}}$ of the quotient algebra
$\mathcal{A}^{\mathsf{f}}$. By Proposition 2.37, there is a canonical
isomorphism of Lie algebras

$\mathfrak{dm}_{0}\simeq(\mathcal{L}^{\mathsf{f}})^{\vee}.$

Summarizing the previous results leads to the following diagram, which should
be seen as a special case of the diagram (9)

${\big{(}\mathcal{A}^{\mathsf{f}},\cdot,\Delta_{\operatorname{Gon}}\big{)}}$${\big{(}\mathcal{U}(\mathfrak{dm}_{0}),\circledast,\Delta\big{)}}$${\big{(}\operatorname{DM}_{0},\circledast\big{)}}$${\big{(}\mathcal{L}^{\mathsf{f}},\delta\big{)}}$${\big{(}\mathfrak{dm}_{0},\\{-,-\\}\big{)}}$$\scriptstyle{\sim}$graded
dual$\scriptstyle{1:1}$Prop 5.28$\scriptstyle{\sim}$graded
dual$\scriptstyle{1:1}$Cor 5.33 (47)

######  Definition 5.39.

Mimicking the construction of Brown [br] we define a coaction

$\Delta_{\operatorname{Gon}}:\mathcal{Z}^{\mathsf{f}}\rightarrow\mathcal{A}^{\mathsf{f}}\otimes\mathcal{Z}^{\mathsf{f}}$

on the whole algebra $\mathcal{Z}^{\mathsf{f}}$ of formal multiple zeta values
by setting

$\Delta_{\operatorname{Gon}}(\xi)=\Big{(}\mathsf{a}\circ\zeta^{\mathsf{f}}\otimes\,\zeta^{\mathsf{f}}\Big{)}\big{(}\Delta_{\operatorname{Gon}}(w)\big{)},$

where $w\in\mathbb{Q}\langle\mathcal{X}\rangle$ is a lift for $\xi$, i.e. $w$
is any element such that $\zeta^{\mathsf{f}}(w)=\xi$.

######  Theorem 5.40.

The coaction
$\Delta_{\operatorname{Gon}}:\mathcal{Z}^{\mathsf{f}}\to\mathcal{A}^{\mathsf{f}}\otimes\mathcal{Z}^{\mathsf{f}}$
is well-defined.

###### Proof.

By (45), we have a morphism of affine schemes

$\displaystyle\operatorname{DM}_{0}\times\operatorname{DM}$
$\displaystyle\to\operatorname{DM}$ $\displaystyle(\Phi,G)$
$\displaystyle\mapsto\Phi\circledast G.$

Applying Yoneda’s Lemma (Theorem 2.32) yields an algebra morphism

$\displaystyle\mathcal{Z}^{\mathsf{f}}$
$\displaystyle\to\mathcal{A}^{\mathsf{f}}\otimes\mathcal{Z}^{\mathsf{f}},$
$\displaystyle\xi$
$\displaystyle\mapsto\Big{(}\mathsf{a}\circ\zeta^{\mathsf{f}}\otimes\zeta^{\mathsf{f}}\Big{)}\big{(}\Delta_{\operatorname{Gon}}(w)\big{)},$

where $w\in\mathbb{Q}\langle\mathcal{X}\rangle$ satisfies
$\zeta^{\mathsf{f}}(w)=\xi$. The group structure of $\operatorname{DM}_{0}$
gives rise to the coassociativity and the counitarity of this morphism. ∎

By Example 3.28, we have

$\displaystyle\Delta_{\operatorname{Gon}}(x_{0}x_{1})$
$\displaystyle=x_{0}x_{1}\otimes\mathbf{1}+\mathbf{1}\otimes x_{0}x_{1},$

and hence we obtain

$\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2))=1\otimes\zeta^{\mathsf{f}}(2)\in\mathcal{A}^{\mathsf{f}}\otimes\mathcal{Z}^{\mathsf{f}}.$
(48)

This coaction is an extension of $\Delta_{\operatorname{Gon}}$ on
$\mathcal{A}^{\mathsf{f}}$ given in (46) to $\mathcal{Z}^{\mathsf{f}}$. In
order to give an explicit formula for it we set for
$\varepsilon_{0},\varepsilon_{n+1}\in\mathcal{X}$ and a word
$\varepsilon_{1}\cdots\varepsilon_{n}\in\mathcal{X}^{*}$

$\zeta^{\mathsf{f}}(\varepsilon_{0};\varepsilon_{1}\cdots\varepsilon_{n};\varepsilon_{n+1})=\zeta^{\mathsf{f}}(I(\varepsilon_{0};\varepsilon_{1}\cdots\varepsilon_{n};\varepsilon_{n+1})),$

and then by Definition 3.24 for the Goncharov coproduct we get the explicit
formula

$\displaystyle\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(\varepsilon_{1}\cdots\varepsilon_{n}))=\sum_{\begin{subarray}{c}0\leq
k\leq n\\\ 0<i_{1}<\dots<i_{k}<n+1\end{subarray}}\
\prod_{p=0}^{k}\mathsf{a}\big{(}\zeta^{\mathsf{f}}(\varepsilon_{i_{p}};\varepsilon_{i_{p}+1}\cdots\varepsilon_{i_{p+1}-1};\varepsilon_{i_{p+1}})\big{)}\otimes\zeta^{\mathsf{f}}(\varepsilon_{i_{1}}\cdots\varepsilon_{i_{k}})$
(49)

where $i_{0}=0,\,i_{k+1}=n+1$ and
$\varepsilon_{0}=x_{1},\,\varepsilon_{n+1}=x_{0}$.

######  Remark 5.41.

In Definition 5.39, it is necessary to consider $\mathcal{A}^{\mathsf{f}}$
instead of $\mathcal{Z}^{\mathsf{f}}$ in the first tensor product factor.
Recall from Example 3.28 that

$\displaystyle\Delta_{\operatorname{Gon}}(x_{0}x_{1})$
$\displaystyle=x_{0}x_{1}\otimes\mathbf{1}+\mathbf{1}\otimes x_{0}x_{1}$ and
therefore on the one hand side we get
$\displaystyle\Delta_{\operatorname{Gon}}(x_{0}x_{1}\shuffle x_{0}x_{1})$
$\displaystyle=\Delta_{\operatorname{Gon}}(x_{0}x_{1})^{\shuffle
2}=(x_{0}x_{1}\otimes\mathbf{1}+\mathbf{1}\otimes x_{0}x_{1})^{\shuffle 2}$
$\displaystyle=(x_{0}x_{1}\shuffle
x_{0}x_{1})\otimes\mathbf{1}+2\,x_{0}x_{1}\otimes
x_{0}x_{1}+\mathbf{1}\otimes(x_{0}x_{1}\shuffle x_{0}x_{1}).$ On the other
side we have $\displaystyle\Delta_{\operatorname{Gon}}(x_{0}x_{0}x_{0}x_{1})$
$\displaystyle=x_{0}x_{0}x_{0}x_{1}\otimes\mathbf{1}-2x_{0}x_{0}\otimes
x_{0}x_{1}+\mathbf{1}\otimes x_{0}x_{0}x_{0}x_{1}.$

Since $\frac{2}{5}\zeta^{\mathsf{f}}(2)^{2}=\zeta^{\mathsf{f}}(4)$ as shown in
Proposition 5.20, both expressions should coincide up to the scalar
$\frac{2}{5}$ after passing to formal multiple zeta values. However, the terms
in the middle differ as
$\zeta^{\mathsf{f}}(x_{0}x_{1})=\zeta^{\mathsf{f}}(2)\neq 0$ but
$\zeta^{\mathsf{f}}(x_{0}x_{0})=0$.

######  Example 5.42.

We want to compute $\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(5))$ by
using the graphical interpretation of ”eating worms” (cf. Figure 2). Since
$\zeta^{\mathsf{f}}(5)=\zeta^{\mathsf{f}}(x_{0}^{4}x_{1})$ we consider
subwords of $x_{0}^{4}x_{1}$. The terms in (49) corresponding to $\mathbf{1}$
and $x_{0}^{4}x_{1}$ are visualized, respectively, as

and we obtain the summands
$\mathsf{a}(\zeta^{\mathsf{f}}(5))\otimes\mathbf{1}$ and
$\mathbf{1}\otimes\zeta^{\mathsf{f}}(5)$, respectively. Now consider the terms
that are depicted with a ”red worm” ending on the right bound $x_{0}$, e. g.

The factor that arises from this ”red worm” vanishes as the left bound is also
$x_{0}$ and thus the summand does not contribute to
$\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(5))$. Similarly, the factors
that correspond to a ”red worm” starting in any $x_{0}$ vanish, e. g.

Note that if the worm ends in $x_{1}$ the factor is
$(-1)^{i}\zeta^{\mathsf{f}}(x_{0}^{i})$ for some $i\leq 3$ which vanishes by
(41). Hence only the first two terms contribute non-trivially and we obtain

$\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(5))=\mathsf{a}(\zeta^{\mathsf{f}}(5))\otimes\mathbf{1}+\mathbf{1}\otimes\zeta^{\mathsf{f}}(5)$

This argument can be generalized to show the following lemma.

######  Lemma 5.43.

For $n\in\mathbb{N}$ we have

$\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2n+1))=\mathsf{a}(\zeta^{\mathsf{f}}(2n+1))\otimes\mathbf{1}+\mathbf{1}\otimes\zeta^{\mathsf{f}}(2n+1).$

###### Proof.

Since $\zeta^{\mathsf{f}}(2n+1)=\zeta^{\mathsf{f}}(x_{0}^{2n}x_{1})$ each
summand in $\Delta_{\operatorname{Gon}}\zeta^{\mathsf{f}}(2n+1)$ corresponds
to a subword of $x_{0}^{2n}x_{1}$ (cf. Definition 5.39). Similar to Example
5.42, the subwords $\mathbf{1}$ and $x_{0}^{2n}x_{1}$ contribute with
$\mathsf{a}(\zeta^{\mathsf{f}}(2n+1))\otimes\mathbf{1}$ and
$\mathbf{1}\otimes\zeta^{\mathsf{f}}(2n+1)$, respectively. It thus suffices to
show that all other subwords of $x_{0}^{2n}x_{1}$ contribute trivially. So let
$k\in\\{1,\dots,2n\\}$ and let $\varepsilon_{i_{1}}\cdots\varepsilon_{i_{k}}$
be a subword of $x_{0}^{2n}x_{1}$. If $i_{k}<2n+1$, then
$\varepsilon_{i_{1}}\cdots\varepsilon_{i_{k}}=x_{0}^{k}$ and the term vanishes
since $\zeta^{\mathsf{f}}(x_{0}^{k})=0$ by (41). Thus, we restrict to the case
$i_{k}=2n+1$, i. e. $\varepsilon_{i_{k}}=x_{1}$. If $k=1$ then the summand
vanishes since $\zeta^{\mathsf{f}}(x_{1})=0$. If $k>1$ then
$\varepsilon_{i_{1}}\cdots\varepsilon_{i_{k}}=x_{0}^{k-1}x_{1}$ and the
product in the formula for $\Delta_{\operatorname{Gon}}$ contains the factor
$\zeta^{\mathsf{f}}(x_{0}^{l})=0$ for some $l>0$. Hence all other subwords
contribute trivially. ∎

###  5.6 On the free odd generation conjecture

Central for this notes is the following well-known conjecture for
$\mathfrak{dm}_{0}$, which is motivated by conjectures of Deligne ([de]) and
Y. Ihara ([ih, p. 300]) in the context of certain Galois actions and of
Drinfeld [dr] on his Grothendieck-Teichmüller Lie algebra. By work of Furusho
[fu], we know that the Grothendieck-Teichmüller Lie algebra embedds into
$\mathfrak{dm}_{0}$.

######  Conjecture 5.44.

The double shuffle Lie algebra $\mathfrak{dm}_{0}$ is a free Lie algebra with
exactly one generator in each odd weight $w\geq 3$, i.e.

$\mathfrak{dm}_{0}\simeq\operatorname{Lie}(S)$

where $S=\\{s_{3},s_{5},\ldots,s_{2n+1},\ldots\\}$.

We call this conjecture the _free odd generation conjecture._

######  Remark 5.45.

Of course the Lie algebra $\operatorname{Lie}(S)$ is also in the heart of the
theory of motivic multiple zeta values ([br],[de],[gon], [degon]). It occurs
as the Lie algebra of the motivic fundamental group of
$\mathbb{P}^{1}\setminus\\{0,1,\infty\\}$ and in fact Brown proved that it
also equals the Lie algebra of the motivic Galois group attached to the
category of mixed Tate motives.

An immediate consequence of the free odd generation conjecture would be the
truth of Zagier’s dimension conjecture 5.2 for formal multiple zeta values.

######  Proposition 5.46.

Under the assumption of the free odd generation conjecture, one obtains

$H_{\mathcal{Z}^{\mathsf{f}}}(x)=\sum_{w\geq
0}\dim_{\mathbb{Q}}\mathcal{Z}^{\mathsf{f}}_{w}x^{w}=\frac{1}{1-x^{2}-x^{3}}.$

###### Proof.

Under the assumption of the free odd generation conjecture 5.44, one obtains
the following Hilbert-Poincare series for the universal enveloping algebra of
$\mathfrak{dm}_{0}$

$\displaystyle
H_{\mathcal{U}(\mathfrak{dm}_{0})}(x)=\frac{1}{1-x^{3}-x^{5}-x^{7}-\dots}.$

By Corollary 5.35, there is an isomorphism
$\mathcal{Z}^{\mathsf{f}}\simeq\mathbb{Q}[\zeta^{\mathsf{f}}(2)]\otimes\mathcal{U}(\mathfrak{dm}_{0})^{\vee}$.
Hence, we deduce

$\displaystyle
H_{\mathcal{Z}^{\mathsf{f}}}(x)=H_{\mathbb{Q}[\zeta^{\mathsf{f}}(2)]}(x)H_{\mathcal{U}(\mathfrak{dm}_{0})}(x)=\frac{1}{1-x^{2}}\frac{1}{1-x^{3}-x^{5}-x^{7}-\cdots}=\frac{1}{1-x^{2}-x^{3}}.$

∎

Recall from Example 2.43, that we have for the Lie algebra in the free odd
generation conjecture 5.44

$\mathcal{U}(\operatorname{Lie}(S))=(\mathbb{Q}\langle
s_{3},s_{5},\ldots\rangle,\operatorname{conc},\Delta),$

and the graded dual is given by

$\mathcal{U}^{\vee}=(\mathbb{Q}\langle
s_{3},s_{5},s_{7},\dots\rangle,\shuffle,\Delta_{\operatorname{dec}}).$

In Definition 2.44 we defined
$\mathcal{U}^{\mathsf{f}}=\mathcal{U}^{\vee}\otimes\mathbb{Q}[s_{2}]$ and
extended the deconcatenation coproduct on $\mathcal{U}^{\vee}$ to a coaction

$\Delta_{\operatorname{dec}}:\mathcal{U}^{\mathsf{f}}\to\mathcal{U}^{\vee}\otimes\mathcal{U}^{\mathsf{f}}$

via $\Delta_{\operatorname{dec}}(s_{2})=\mathbf{1}\otimes s_{2}$. We now also
set $s_{2n}=b_{n}s_{2}^{n}$, where the rational numbers $b_{n}$ are given in
Proposition 5.20.

######  Theorem 5.47.

Assume the free odd generation conjecture 5.44 for $\mathfrak{dm}_{0}$. There
is an isomorphism of algebras with coaction

$\Phi\colon(\mathcal{Z}^{\mathsf{f}},\cdot,\Delta_{\operatorname{Gon}})\overset{\sim}{\longrightarrow}(\mathcal{U}^{\mathsf{f}},\shuffle,\Delta_{\operatorname{dec}})$
(50)

satisfying for each $N\geq 3$ odd

$\Phi(\zeta^{\mathsf{f}}(N))=s_{N}.$

###### Proof.

Assume the free odd generation conjecture 5.44 for $\mathfrak{dm}_{0}$, then
from Theorem 3.8 applied to the Ihara bracket, we get a Hopf algebra
isomorphism

$(\mathcal{U}(\mathfrak{dm}_{0}),\circledast,\Delta))\cong\mathcal{U}(\operatorname{Lie}(S))=(\mathbb{Q}\langle
s_{3},s_{5},\ldots\rangle,\operatorname{conc},\Delta).$

By Corollary 5.35 and by dualization we have a Hopf algebra isomorphism

$\displaystyle\Big{(}{\mathchoice{\raisebox{4.38887pt}{$\displaystyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.44444pt}{$\displaystyle{(\zeta^{\mathsf{f}}(2))}$}}{\raisebox{4.38887pt}{$\textstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.44444pt}{$\textstyle{(\zeta^{\mathsf{f}}(2))}$}}{\raisebox{3.0861pt}{$\scriptstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-3.12498pt}{$\scriptstyle{(\zeta^{\mathsf{f}}(2))}$}}{\raisebox{2.40277pt}{$\scriptscriptstyle{\mathcal{Z}^{\mathsf{f}}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-2.43054pt}{$\scriptscriptstyle{(\zeta^{\mathsf{f}}(2))}$}}},\
\cdot\
,\Delta_{\operatorname{Gon}}\Big{)}\simeq(\mathcal{U}^{\vee},\shuffle,\Delta_{\operatorname{dec}}).$
(51)

It is a direct consequence of Corollary 5.35 and the compatibily of the
construction of the coaction on $\mathcal{Z}^{\mathsf{f}}$ via (48) and that
on $\mathcal{U}^{\mathsf{f}}$ by Definition 2.44 that we can extend (51) to an
isomorphism of algebras with coaction
$\Phi:(\mathcal{Z}^{\mathsf{f}},\cdot,\Delta_{\operatorname{Gon}})\xrightarrow{\sim}(\mathcal{U}^{\mathsf{f}},\shuffle,\Delta_{\operatorname{dec}})$.

Since we have for any odd $N\geq 3$

$\displaystyle\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(N))$
$\displaystyle=\zeta^{\mathsf{f}}(N)\otimes\mathbf{1}+\mathbf{1}\otimes\zeta^{\mathsf{f}}(N),$
$\displaystyle\Delta_{\text{dec}}(s_{N})$
$\displaystyle=s_{N}\otimes\mathbf{1}+\mathbf{1}\otimes s_{N},$

the isomorphism $\Phi$ can be chosen such that
$\Phi(\zeta^{\mathsf{f}}(N))=s_{N}$. ∎

######  Corollary 5.48.

Assume the free odd generation conjecture 5.44 for $\mathfrak{dm}_{0}$. The
formal zeta values $\zeta^{\mathsf{f}}(2)$ and $\zeta^{\mathsf{f}}(2r+1)$,
$r\in\mathbb{N}$, are non-zero modulo products and algebraically independent.

###### Proof.

Evidently, the letters $s_{2r+1}$ are nonzero modulo products and
algebraically independent in $\mathcal{U}^{\vee}$. So the same must hold for
their preimages $\zeta^{\mathsf{f}}(2r+1)=\Phi^{-1}(s_{2r+1})$ under the
isomorphism (50). By construction of
$\mathcal{U}^{\mathsf{f}}=\mathcal{U}^{\vee}\otimes\mathbb{Q}[s_{2}]$ and the
convention $\zeta^{\mathsf{f}}(2)=\Phi^{-1}(s_{2})$, the claim follows. ∎

######  Remark 5.49.

By work of Drinfeld [dr], Brown [br], and Furusho [fu], we have inclusions

$\operatorname{Lie}(S)\subset\mathfrak{grt}_{1}\subset\mathfrak{dm}_{0}.$

This implies
$\mathcal{U}(\operatorname{Lie}(S))\subset\mathcal{U}(\mathfrak{dm}_{0})$, and
hence $\Phi:\mathcal{Z}^{\mathsf{f}}\to\mathcal{U}^{\mathsf{f}}$ is
surjective. In particular, we obtain Corollary 5.48 without assuming the free
odd generation conjecture.

###  5.7 The Kernel conjecture

Recall that in Subsection 2.4 we introduced the derivations $D_{w}$ and their
extension to algebras with particular coaction. With the notation from
Subsection 3.6 and 5.5 we get the commutative diagram

|  |  |  |
---|---|---|---|---
$\textstyle{D_{2r+1}\colon}$$\textstyle{\mathbb{Q}\langle\mathcal{X}\rangle\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Delta_{\operatorname{Gon}}^{\prime}}$$\scriptstyle{\zeta^{\mathsf{f}}}$$\textstyle{\bigoplus_{w\geq
1}\mathbb{Q}\langle\mathcal{X}\rangle_{w}\otimes\mathbb{Q}\langle\mathcal{X}\rangle\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{2r+1}\otimes\operatorname{id}}$$\scriptstyle{\mathsf{a}\circ\zeta^{\mathsf{f}}\otimes\zeta^{\mathsf{f}}}$$\textstyle{\mathcal{L}_{2r+1}\otimes\mathbb{Q}\langle\mathcal{X}\rangle\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\overline{\mathsf{a}\circ\zeta^{\mathsf{f}}}\otimes\zeta^{\mathsf{f}}}$$\textstyle{\mathcal{Z}^{\mathsf{f}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Delta_{\operatorname{Gon}}^{\prime}}$$\textstyle{\bigoplus_{w\geq
1}\mathcal{A}^{\mathsf{f}}_{w}\otimes\mathcal{Z}^{\mathsf{f}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{2r+1}\,\otimes\operatorname{id}}$$\textstyle{\mathcal{L}^{\mathsf{f}}_{2r+1}\otimes\mathcal{Z}^{\mathsf{f}}}$
(52)

where similar to the previous

$\displaystyle\pi_{2r+1}:\bigoplus_{w\geq 1}\mathcal{A}^{\mathsf{f}}_{w}$
$\displaystyle\to\mathcal{L}^{\mathsf{f}}_{2r+1},$
$\displaystyle\overline{\mathsf{a}\circ\zeta^{\mathsf{f}}}\colon\mathcal{L}$
$\displaystyle\rightarrow\mathcal{L}^{\mathsf{f}},$

are the canonical projections.

Observe as in Lemma 2.42, the maps
$D_{2r+1}:\mathcal{Z}^{\mathsf{f}}\to\mathcal{L}^{\mathsf{f}}_{2r+1}\otimes\mathcal{Z}^{\mathsf{f}}$
in the lower line are again derivations.

######  Proposition 5.50.

Assume the free odd generation conjecture 5.44 for $\mathfrak{dm}_{0}$. For
each $r\in\mathbb{N}$ the element

$\zeta^{\mathsf{f}}_{2r+1}=\pi_{2r+1}(\mathsf{a}(\zeta^{\mathsf{f}}(2r+1)))\in\mathcal{L}^{\mathsf{f}}_{2r+1}$

is non-zero.

###### Proof.

This claim is just a reformulation of Corollary 5.48. ∎

######  Lemma 5.51.

Let $r,n,k\geq 1$ be natural numbers, then we have

$D_{2n+1}(\zeta^{\mathsf{f}}(2r+1)\cdot\zeta^{\mathsf{f}}(2)^{k})=\delta_{r,n}\cdot\zeta^{\mathsf{f}}_{2r+1}\otimes\zeta^{\mathsf{f}}(2)^{k}.$
(53)

where $\delta_{r,n}$ is the Kronecker delta.

###### Proof.

Recall
$\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2)^{k})=\mathbf{1}\otimes\zeta^{\mathsf{f}}(2)^{k}$
and

$\displaystyle\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2r+1))=\mathsf{a}(\zeta^{\mathsf{f}}(2r+1))\otimes\mathbf{1}+\mathbf{1}\otimes\zeta^{\mathsf{f}}(2r+1).$

We then compute

$\displaystyle\Delta_{\operatorname{Gon}}^{\prime}(\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(2)^{k})$
$\displaystyle=\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(2)^{k})-\mathbf{1}\otimes\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(2)^{k}$
$\displaystyle=\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2r+1))\Delta_{\operatorname{Gon}}(\zeta^{\mathsf{f}}(2)^{k})-\mathbf{1}\otimes\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(2)^{k}$
$\displaystyle=\Big{(}\mathsf{a}(\zeta^{\mathsf{f}}(2r+1))\otimes\mathbf{1}+\mathbf{1}\otimes\zeta^{\mathsf{f}}(2r+1)\Big{)}\Big{(}\mathbf{1}\otimes\zeta^{\mathsf{f}}(2)^{k}\Big{)}-\mathbf{1}\otimes\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(2)^{k}$
$\displaystyle=\mathsf{a}(\zeta^{\mathsf{f}}(2r+1))\otimes\zeta^{\mathsf{f}}(2)^{k},$

which in turn yields

$\displaystyle D_{2n+1}$
$\displaystyle(\zeta^{\mathsf{f}}(2r+1)\zeta(2)^{k})=\pi_{2n+1}(\mathsf{a}(\zeta^{\mathsf{f}}(2r+1)))\otimes\zeta^{\mathsf{f}}(2)^{k}.\qed$

######  Conjecture 5.52 (Kernel conjecture).

Define

$D_{<N}=\bigoplus_{3\leq 2r+1<N}D_{2r+1},$ (54)

then we have
$\ker(D_{<N})\cap\mathcal{Z}^{\mathsf{f}}_{N}=\mathbb{Q}\,\zeta^{\mathsf{f}}(N)$
for all $N\geq 2$.

######  Remark 5.53.

Observe that $\zeta^{\mathsf{f}}(N)\in\ker(D_{<N})$ for all $N\geq 2$. This
follows from Proposition 5.20 for even $N$ and from Lemma 5.43 together with
Corollary 5.48 for odd $N$.

######  Theorem 5.54.

The free odd generation conjecture 5.44 for $\mathfrak{dm}_{0}$ implies the
Kernel conjecture 5.52, i.e., we have for all $N\geq 2$

$\ker(D_{<N})\cap\mathcal{Z}^{\mathsf{f}}_{N}=\mathbb{Q}\zeta^{\mathsf{f}}(N).$

###### Proof.

Recall that $\zeta^{\mathsf{f}}(N)\in\ker(D_{<N})$. For the other inclusion we
employ the isomorphism of algebras with coaction

$\Phi\colon(\mathcal{Z}^{\mathsf{f}},\cdot,\Delta_{\operatorname{Gon}})\overset{\sim}{\longrightarrow}(\mathcal{U}^{\mathsf{f}},\shuffle,\Delta_{\text{dec}})$

from Proposition 5.47. In particular this yields for all $N\in\mathbb{N}$ a
commutative diagram

$\textstyle{\mathcal{Z}^{\mathsf{f}}_{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{D_{<N}}$$\scriptstyle{\Phi}$$\textstyle{\mathcal{U}^{\mathsf{f}}_{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{D_{<N}}$$\textstyle{\bigoplus\limits_{1<2r+1<N}\\!\\!\\!\mathcal{L}^{\mathsf{f}}_{2r+1}\otimes\mathcal{Z}^{\mathsf{f}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\overline{\Phi}\otimes\Phi}$$\textstyle{\bigoplus\limits_{1<2r+1<N}\\!\\!\\!L_{2r+1}\otimes\mathcal{U}^{\mathsf{f}}}$
(55)

Hence, if $\xi\in\ker(D_{<N})\cap\mathcal{Z}^{\mathsf{f}}_{N}$, then (55)
implies $\Phi(\xi)\in\ker(D_{<N})$. So by the Proposition 2.48, we obtain that
$\Phi(\xi)=\alpha\,s_{N}$ for some $\alpha\in\mathbb{Q}$. Hence
$\xi=\alpha\,\Phi^{-1}(s_{N})=\alpha\,\zeta^{\mathsf{f}}(N)$. ∎

######  Example 5.55.

The direct sum in (54) is empty for $N=3$, hence $D_{<3}\equiv 0$ is the zero
map. Conjecture 5.52 would thus imply that
$\mathcal{Z}^{\mathsf{f}}_{3}=\mathbb{Q}\,\zeta^{\mathsf{f}}(3)$, so in
particular $\zeta^{\mathsf{f}}(2,1)$ would be a rational multiple of
$\zeta^{\mathsf{f}}(3)$ (possibly zero). In fact, in Example 5.15 we saw that
Euler’s identity $\zeta(3)=\zeta(2,1)$ follows from the extended double
shuffle relations, hence $\zeta^{\mathsf{f}}(3)=\zeta^{\mathsf{f}}(2,1)$.

######  Example 5.56.

We give here some potential applications of the Kernel conjecture. Let
$n\in\mathbb{N}$ be a positive integer.

  1. 1.

We compute $D_{<2n}(\zeta^{\mathsf{f}}(\\{2\\}^{n}))$. Let $r\in\mathbb{N}$
with $2r+1<2n$. Each term in $D_{2r+1}(\zeta^{\mathsf{f}}(\\{2\\}^{n}))$
corresponds to a consecutive subword of $(x_{0}x_{1})^{n}\in\mathcal{X}^{*}$
of weight $2r+1$. Note that such a subword always starts and ends in the same
letter for parity reasons. This can be depicted on a semicircle as:

$x_{1}$$x_{0}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$x_{0}$$\dots$

So we have $\zeta^{\mathsf{f}}(\\{2\\}^{n})\in\ker(D_{2r+1})$ for all
$2r+1<2n$, hence $\zeta^{\mathsf{f}}(\\{2\\}^{n})\in\ker(D_{<2n})$. We
therefore expect by Conjecture 5.52 that
$\zeta^{\mathsf{f}}(\\{2\\}^{n})\in\mathbb{Q}\,\zeta^{\mathsf{f}}(2n)$. In
fact, this holds by Proposition 5.20. A generalisation of this method can be
found in [C15].

  2. 2.

We compute $D_{<n+2}(\zeta^{\mathsf{f}}(2,\\{1\\}^{n}))$. Let $r\in\mathbb{N}$
with $2r+1<n+2$. Since
$\zeta^{\mathsf{f}}(2,\\{1\\}^{n})=\zeta^{\mathsf{f}}(x_{0}x_{1}^{n+1})$ there
are only two kind of strict subwords whose boundaries are not both $x_{1}$.
They can be depicted as

Both corresponding terms vanish since $\zeta^{\mathsf{f}}(x_{1}^{2r+1})=0$ by
(41). So we have $\zeta^{\mathsf{f}}(2,\\{1\\}^{n})\in\ker(D_{2r+1})$ for all
$2r+1<n+2$, hence $\zeta^{\mathsf{f}}(2,\\{1\\}^{n})\in\ker(D_{<n+2})$. By the
Kernel conjecture 5.52 we expect that
$\zeta^{\mathsf{f}}(2,\\{1\\}^{n})\in\mathbb{Q}\,\zeta^{\mathsf{f}}(n+2)$.
Using the evaluation map would imply

$\zeta^{\mathsf{f}}(2,\\{1\\}^{n})=\zeta^{\mathsf{f}}(n+2).$

as $\zeta(2,\\{1\\}^{n})=\zeta(n+2)$. Remarkably, there are two independent
ways to show that multiple zeta values satisfy this identity. At first it is
an example of _duality relations_ , i. e. $\zeta(x_{0}^{k_{1}-1}x_{1}\cdots
x_{0}^{k_{d}-1}x_{1})=\zeta(x_{0}x_{1}^{k_{d}-1}\cdots x_{0}x_{1}^{k_{1}-1})$
for integers $k_{1}>1$ and $k_{2},\dots,k_{d}\geq 1$. Alternatively, as shown
in [k], this identity also follows from the extended double shuffle relations
and therefore we also get a unconditional proof without refering to the Kernel
conjecture in this case.

##  6 Brown’s theorem for formal multiple zeta values

In this section we are interested in formulas for the multiple zeta values
with level zero and one (cf Definition 4.1). We already understand
$H(n)=\zeta^{\mathsf{f}}(\\{2\\}^{n})$, since by Proposition 5.20 we have
explicit formulae for

$\displaystyle\zeta^{\mathsf{f}}(2k)\cdot\,\zeta^{\mathsf{f}}(\\{2\\}^{n-k})\in\mathbb{Q}\cdot\zeta^{\mathsf{f}}(2)^{n}.$

######  Theorem 6.1 (Zagier, 2012).

Let $a,b\geq 0$. Then we have an identity of multiple zeta values

$\zeta(\\{2\\}^{a},3,\\{2\\}^{b})=\sum_{r=1}^{a+b+1}c_{a,b}^{r}\,\zeta(2r+1)\zeta(\\{2\\}^{a+b+1-r})$

with $c_{a,b}^{r}$ from Definition 4.30, i.e.

$c_{a,b}^{r}=2\cdot(-1)^{r}\Bigg{(}\binom{2r}{2b+2}-(1-2^{-2r})\binom{2r}{2a+1}\Bigg{)}.$

Brown showed that this formula also holds for motivic MZVs. We will prove the
analogue statement for formal MZVs, assuming the Kernel conjecture 5.52.

###  6.1 Proof of Zagier’s formula

Zagier’s theorem is a vast refinement of Proposition 5.22, i.e. of the
identity

$\displaystyle\sum_{a+b=n}\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})=\sum_{i=0}^{n}(-1)^{i}\zeta^{\mathsf{f}}(2i+3)\zeta^{\mathsf{f}}(\\{2\\}^{n-i}).$

Unfortunately, by now only analytical proofs of his result are known. We
follow in our presentation the paper [LaiLupuOrr].

######  Theorem 6.2.

Let $a,b\geq 0$ be integers, then the following three quantities are equal

$\displaystyle H(a,b)$ $\displaystyle=\zeta(\\{2\\}^{a},3,\\{2\\}^{b})$ (56)
$\displaystyle I(a,b)$
$\displaystyle=\frac{\pi^{2a}2^{2b+3}}{(2a+1)!(2b+2)!}\int\limits_{0}^{\pi/2}x^{2b+2}\Big{(}1-\frac{2x}{\pi}\Big{)}^{2a+1}\cot(x)dx$
(57) $\displaystyle\hat{H}(a,b)$
$\displaystyle=\sum_{r=1}^{a+b+1}c_{a,b}^{r}\,\zeta(2r+1)\zeta(\\{2\\}^{a+b+1-r}).$
(58)

Essential ingredients for the proof are the following three lemmas.

######  Lemma 6.3.

For any polynomial $P(x)\in\mathbb{C}[x]$ with $P(0)=P(1)=0$, we have

$\displaystyle\int\limits_{0}^{1}P(x)\cot\big{(}\frac{\pi x}{2}\big{)}dx$
$\displaystyle=2\sum_{k=1}^{\lfloor\deg(P)/2\rfloor}(-1)^{k}\Big{(}\big{(}1-2^{-2k}\big{)}P^{(2k)}(1)+P^{(2k)}(0)\Big{)}\frac{\zeta(2k+1)}{\pi^{2k+1}},$
(59)

where $P^{(2k)}(x)$ denotes the $(2k)$-th derivative of $P(x)$.

###### Proof.

This is a special case of [LaiLupuOrr, Lemma 2.3]. ∎

######  Lemma 6.4.

Let $r\in\mathbb{N}$, then we have

$\displaystyle\frac{\arcsin(x)^{2r}}{(2r)!}=\frac{1}{4^{r}}\sum_{n=1}^{\infty}\frac{4^{n}}{n^{2}\binom{2n}{n}}\,x^{2n}\sum\limits_{n_{1}<n_{2}<...<n_{r-1}<n}\frac{1}{n_{1}^{2}n_{2}^{2}...n_{r-1}^{2}}.$
(60)

###### Proof.

For a proof we refer to [LaiLupuOrr, Eq. (2)]. ∎

######  Lemma 6.5.

Let $n\geq 1$, $a\geq 0$, then we have

$\displaystyle\int\limits_{0}^{1}x^{2n-1}\frac{(2\arccos(x))^{2a+1}}{(2a+1)!}dx=\frac{\binom{2n}{n}\pi}{2^{2n+1}n}\sum_{n=1}^{\infty}\sum\limits_{n<m_{1}<m_{2}<...<m_{a}}\frac{1}{m_{1}^{2}m_{2}^{2}...m_{a}^{2}},$
(61)

where the sum is set to be $1$ if $a=0$.

###### Proof.

For a proof we refer to [LaiLupuOrr, Lemma 3.1]. ∎

We are now prepared for the proof of Zagier’s theorem.

###### Proof of Theorem 6.2.

At first we observe that via the change of variables $x\mapsto\dfrac{\pi
x}{2}$ we obtain

$\displaystyle I(a,b)$
$\displaystyle=\frac{\pi^{2a}2^{2b+3}}{(2a+1)!(2b+2)!}\int\limits_{0}^{\pi/2}x^{2b+2}\Big{(}1-\frac{2x}{\pi}\Big{)}^{2a+1}\cot(x)dx$
$\displaystyle=\frac{\pi^{2a+2b+2}2}{(2a+1)!(2b+2)!}\int\limits_{0}^{1}x^{2b+2}\big{(}1-x\big{)}^{2a+1}\cot\big{(}\frac{\pi
x}{2}\big{)}dx$

Applying Lemma 6.3 with $P(x)=x^{2b+2}(1-x)^{2a+1}$ yields

$\displaystyle\int\limits_{0}^{1}x^{2b+2}\big{(}1-x\big{)}^{2a+1}\cot\big{(}\frac{\pi
x}{2}\big{)}dx$ $\displaystyle=(2a+1)!(2b+2)!$ $\displaystyle\cdot
2\sum_{k=0}^{a+b+1}(-1)^{k}\Big{[}\binom{2k}{2b+2}-\big{(}1-2^{-2k}\big{)}\binom{2k}{2a+1}\Big{]}\frac{\zeta(2k+1)}{(2a+2b+3-2k)!\pi^{2k+1}}$

Now using Proposition 5.20 the equality $I(a,b)=\hat{H}(a,b)$ follows by
rearranging the above equalities.

To obtain the second equality we start with the change of variables
$x\mapsto\arcsin(x)$. Using in addition that $\arccos(x)=\pi/2-\arcsin(x)$ we
get

$\displaystyle I(a,b)$
$\displaystyle=\frac{2^{2a+2b+4}}{(2a+1)!(2b+2)!\pi}\int\limits_{0}^{\pi/2}x^{2b+2}\Big{(}\frac{\pi}{2}-x\Big{)}^{2a+1}\cot(x)dx$
$\displaystyle=\frac{2}{\pi}\int\limits_{0}^{1}\frac{(2\arcsin(x))^{2b+2}}{(2b+2)!}\frac{(2\arccos(x))^{2a+1}}{(2a+1)!}\frac{1}{x}dx.$

If we replace the first factor under the integral by its Taylor expansion
given in Lemma 6.4, then the resulting sum of integrals are exactly those of
6.5, therefore we get

$\displaystyle\int\limits_{0}^{1}\frac{(2\arcsin(x))^{2b+2}}{(2b+2)!}\frac{(2\arccos(x))^{2a+1}}{(2a+1)!}\frac{1}{x}dx$
$\displaystyle=\sum_{n=1}^{\infty}\frac{4^{n}}{n^{2}\binom{2n}{n}}\sum\limits_{n_{1}<n_{2}<...<n_{b}<n}\frac{1}{n_{1}^{2}n_{2}^{2}\cdots
n_{b}^{2}}\int\limits_{0}^{1}x^{2n-1}\frac{(2\arccos(x))^{2a+1}}{(2a+1)!}dx$
$\displaystyle=\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\left(\sum\limits_{n_{1}<n_{2}<...<n_{b}<n}\frac{1}{n_{1}^{2}n_{2}^{2}...n_{b}^{2}}\right)\cdot\left(\sum\limits_{n<m_{1}<m_{2}<\cdots<m_{a}}\frac{1}{m_{1}^{2}m_{2}^{2}\cdots
m_{a}^{2}}\right)$
$\displaystyle=\frac{\pi}{2}\sum\limits_{n_{1}<n_{2}<...<n_{b}<n<m_{1}<m_{2}<...<m_{a}}\frac{1}{n_{1}^{2}n_{2}^{2}...n_{b}^{2}n^{3}m_{1}^{2}m_{2}^{2}...m_{a}^{2}}$
$\displaystyle=\frac{\pi}{2}H(a,b).\qed$

###  6.2 Lifting Zagier’s formula

We will show in a sequence of lemmas that both sides of Zagier’s formula
$H(a,b)=\hat{H}(a,b)$ in the formal setting have the same image under the map
$D_{<2a+2b+3}$. Thus by Theorem 5.54 their difference is a multiple of
$\zeta^{\mathsf{f}}(2a+2b+3)$. We determine this multiple via the canonical
map $\mathcal{Z}^{\mathsf{f}}\to\operatorname{\mathcal{Z}}$.

######  Lemma 6.6.

For integers $a,b\geq 0$ and $1\leq r\leq a+b$ we have

$D_{2r+1}\big{(}\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})\big{)}=\pi_{2r+1}(\mathsf{a}(\xi_{a,b}^{r}))\otimes\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r})$

where

$\displaystyle\xi_{a,b}^{r}$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq
b\\\
\alpha+\beta+1=r\end{subarray}}\zeta^{\mathsf{f}}(\\{2\\}^{\alpha},3,\\{2\\}^{\beta})-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}\zeta^{\mathsf{f}}(\\{2\\}^{\beta},3,\\{2\\}^{\alpha})$
$\displaystyle+2\,(\mathbb{I}(a\geq r)-\mathbb{I}(b\geq
r))\sum_{i=1}^{r}(-1)^{i}\zeta^{\mathsf{f}}(2i+1)\zeta^{\mathsf{f}}(\\{2\\}^{r-i})$

where $\mathbb{I}$ denotes the indicator function (i. e. $\mathbb{I}(A)=1$ if
$A$ is a true statement and $\mathbb{I}(A)=0$ else).

###### Proof.

We have

$\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})=\zeta^{\mathsf{f}}((x_{0}x_{1})^{a}x_{0}x_{0}x_{1}(x_{0}x_{1})^{b}).$

By previous considerations, each summand in
$D_{2r+1}(\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b}))$ corresponds to a
consecutive subsequence of length $2r+3$ (including boundaries). Let
$\varepsilon_{0}\dots\varepsilon_{2r+2}$ be such a subsequence.
Cases 1 + 2: $\varepsilon_{2r+2}$ is to the left or $\varepsilon_{0}$ is to
the right of $x_{0}x_{0}$. Then, by parity,
$\varepsilon_{0}=\varepsilon_{2r+2}$ and the sequence does not contribute. The
general observation here is that the right tensor factor must be
$\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r})$ since $x_{0}x_{0}x_{1}$ must be
contained $\varepsilon_{0}\dots\varepsilon_{2r+2}$.
Cases 3 + 4: The sequence $x_{0}x_{0}x_{1}$ is contained in
$\varepsilon_{1}\dots\varepsilon_{2r+2}$. In this case, we obtain factors of
the form

$\displaystyle\zeta^{\mathsf{f}}(x_{1};(x_{0}x_{1})^{\alpha}x_{0}x_{0}x_{1}(x_{0}x_{1})^{\beta};x_{0})\vspace{-.3cm}$
$\displaystyle\alpha\leq a,\beta\leq b$
$\displaystyle\zeta^{\mathsf{f}}(x_{0};(x_{1}x_{0})^{\alpha}x_{1}x_{0}x_{0}(x_{1}x_{0})^{\beta};x_{1})$
$\displaystyle\alpha<a,\beta\leq b$

such that $\alpha+\beta+1=r$ and $\alpha,\beta\geq 0$. Observe that these
factors are $\zeta^{\mathsf{f}}(\\{2\\}^{\alpha},3,\\{2\\}^{\beta})$ and
$-\zeta^{\mathsf{f}}(\\{2\\}^{\beta},3,\\{2\\}^{\alpha})$, respectively.
Cases 5 + 6: Assume that $\varepsilon_{0}\dots\varepsilon_{2r+2}$ starts or
ends in $x_{0}x_{0}$. This is only possible, if $b\geq r$ or $a\geq r$. The
corresponding factors are $\zeta^{\mathsf{f}}(x_{1};w;x_{0})$ and
$\zeta^{\mathsf{f}}(x_{0};w;x_{1})$ where $w=(x_{0}x_{1})^{r}x_{0}$ is
palindromic. Hence
$\zeta^{\mathsf{f}}(x_{0};w;x_{1})=-\zeta^{\mathsf{f}}(x_{1};w;x_{0})$. The
claim now follows from Proposition 5.22, i. e.

$\zeta^{\mathsf{f}}((x_{0}x_{1})^{r}x_{0})=2\sum_{i=1}^{r}(-1)^{i}\zeta^{\mathsf{f}}(2i+1)\,\zeta^{\mathsf{f}}(\\{2\\}^{r-i}).\qed$

######  Lemma 6.7.

Assume the Kernel conjecture 5.52, and let $a,b\geq 0$ be given integers. Then
there exist unique coefficients $\alpha_{1},\dots,\alpha_{n}\in\mathbb{Q}$
such that

$\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})=\sum_{i=1}^{n}\alpha_{i}\,\zeta^{\mathsf{f}}(2i+1)\zeta^{\mathsf{f}}(\\{2\\}^{n-i})$

with $n=a+b+1$.

###### Proof.

Let $w=(x_{0}x_{1})^{a}x_{0}x_{0}x_{1}(x_{0}x_{1})^{b}$. We proof the
statement by induction on $\operatorname{wt}(w)$. The claim is trivial for
$\operatorname{wt}(w)=3$. Let $w=\\{2\\}^{a}3\\{2\\}^{b}$ with
$\operatorname{wt}(w)=2N+1$ and assume the claim holds for all $r<N$. Now
Lemma 6.6 implies for all $1\leq r<N$

$D_{2r+1}(\zeta^{\mathsf{f}}(w))=\pi_{2r+1}(\mathsf{a}(\xi_{a,b}^{r}))\otimes\zeta^{\mathsf{f}}(\\{2\\}^{N-r}).$

Recall also from Lemma 6.6 the explicit formula for $\xi_{a,b}^{r}$ and
observe that $\operatorname{wt}(\xi_{a,b}^{r})\leq\operatorname{wt}(w)-2$ for
all $r\leq a+b$. So we can apply the induction hypothesis to all summands in
$\xi_{a,b}^{r}$. Therefore,
$\pi_{2r+1}(\mathsf{a}(\xi_{a,b}^{r}))=\alpha_{r}\,\zeta^{\mathsf{f}}_{2r+1}$.
Hence

$D_{2r+1}(\zeta^{\mathsf{f}}(w))=\alpha_{r}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\zeta^{\mathsf{f}}(\\{2\\}^{N-r}).$

In particular, we obtain numbers $\alpha_{r}\in\mathbb{Q}$ for each $1\leq
r\leq N-1$ such that

$D_{<2N+1}(\zeta^{\mathsf{f}}(w))=\sum_{r=1}^{N-1}\alpha_{r}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\zeta^{\mathsf{f}}(\\{2\\}^{N-r}).$

On the other hand, we deduce from (53) that

$D_{2r+1}\Big{(}\sum_{i=1}^{N-1}\alpha_{i}\zeta^{\mathsf{f}}(2i+1)\zeta^{\mathsf{f}}(\\{2\\}^{N-i})\Big{)}=\alpha_{r}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\zeta^{\mathsf{f}}(\\{2\\}^{N-r}),$

because $D_{2r+1}$ is a derivation. Therefore,

$D_{<2N+1}\Big{(}\zeta^{\mathsf{f}}(w)-\sum_{i=1}^{N-1}\alpha_{i}\,\zeta^{\mathsf{f}}(2i+1)\zeta^{\mathsf{f}}(\\{2\\}^{N-i})\Big{)}=0.$

By the assumption
$\ker(D_{<2N+1})\cap\mathcal{Z}^{\mathsf{f}}_{2N+1}=\mathbb{Q}\zeta^{\mathsf{f}}(2N+1)$
it follows that both sides of the claim differ by
$\alpha_{n}\,\zeta^{\mathsf{f}}(2N+1)$ for some $\alpha_{n}\in\mathbb{Q}$ and
the claim follows. ∎

######  Remark 6.8.

Note that the linear combination

$\Phi(\zeta^{\mathsf{f}}(w))=\sum_{i=1}^{n}\alpha_{i}\,s_{2i+1}s_{2}^{n-i}$

is unique where
$\Phi\colon\mathcal{Z}^{\mathsf{f}}\overset{\sim}{\longrightarrow}\mathcal{U}^{\mathsf{f}}$
is the conjectured isomorphism from (50). This may give an alternative proof.

Recall the numbers

$c_{a,b}^{r}=2\cdot(-1)^{r}\Bigg{(}\binom{2r}{2b+2}-(1-2^{-2r})\binom{2r}{2a+1}\Bigg{)}$

from Definition 4.30 that we studied in Section 4.3.

######  Theorem 6.9.

Assume the Kernel conjecture 5.52, and let $a,b\geq 0$ be given integers. Then
we have

$\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})=\sum_{r=1}^{a+b+1}c_{a,b}^{r}\,\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r}).$
(62)

###### Proof.

Induction on the weight. Let
$w=(x_{0}x_{1})^{a}x_{0}x_{0}x_{1}(x_{0}x_{1})^{b}$ and $N=a+b$. By Lemma 6.7
we have

$\zeta^{\mathsf{f}}(\\{2\\}^{\alpha},3,\\{2\\}^{\beta})=\sum_{r=1}^{\alpha+\beta+1}\alpha_{r}\,\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(\\{2\\}^{\alpha+\beta+1-r})$

We assume that $\alpha_{r}=c_{\alpha,\beta}^{r}$ holds for all
$\alpha,\beta\geq 0$ such that $\alpha+\beta<N$. By Lemma 6.6 we have

$D_{2r+1}(\zeta^{\mathsf{f}}(w))=\pi_{2r+1}(\mathsf{a}(\xi_{a,b}^{r}))\otimes\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r})$

for all $1\leq r\leq N$. Applying the induction hypothesis to the explicit
formula for $\xi_{a,b}^{r}$ given in Lemma 6.6 yields modulo products

$\pi_{2r+1}(\mathsf{a}(\xi_{a,b}^{r}))=\Big{(}\sum_{\begin{subarray}{c}0\leq\alpha\leq
a\\\ 0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\beta,\alpha}^{r}+2\cdot(-1)^{r}\big{(}\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r)\big{)}\Big{)}\zeta^{\mathsf{f}}_{2r+1}.$

By Lemma 4.33 we have

$c_{a,b}^{r}=\sum_{\begin{subarray}{c}0\leq\alpha\leq a\\\ 0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\alpha,\beta}^{r}-\sum_{\begin{subarray}{c}0\leq\alpha<a\\\
0\leq\beta\leq b\\\
\alpha+\beta+1=r\end{subarray}}c_{\beta,\alpha}^{r}+2\cdot(-1)^{r}\big{(}\mathbb{I}(a\geq
r)-\mathbb{I}(b\geq r)\big{)}.$ (63)

Hence for all $1\leq r\leq N$ we have

$D_{2r+1}\Big{(}\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})\Big{)}=c_{a,b}^{r}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r}).$

This implies with formula (53) applied to the products on the right hand side
of the claimed identity (62) that

$D_{<2N+3}\Big{(}\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})-\sum_{r=1}^{a+b+1}c_{a,b}^{r}\,\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r})\Big{)}=0.$

By assumption, there is some $\alpha\in\mathbb{Q}$ such that

$\alpha\,\zeta^{\mathsf{f}}(2N+3)=\zeta^{\mathsf{f}}(\\{2\\}^{a},3,\\{2\\}^{b})-\sum_{r=1}^{a+b+1}c_{a,b}^{r}\,\zeta^{\mathsf{f}}(2r+1)\zeta^{\mathsf{f}}(\\{2\\}^{a+b+1-r}).$
(64)

Applying the surjective morphism
$\mathcal{Z}^{\mathsf{f}}\twoheadrightarrow\mathcal{Z}$ from (42) to (64)
yields $\alpha=0$, because we have Zagier’s theorem. ∎

Applying the projection modulo products to (62), we deduce the following for
the sets $\mathcal{B}^{1}$ and $\mathcal{B}_{2,3}$ from Definition 4.8 and
4.1.

######  Corollary 6.10.

Assuming the Kernel conjecture 5.52, we have

$\pi_{2r+1}\big{(}\mathsf{a}\big{(}\zeta^{\mathsf{f}}(\mathcal{B}^{1})\big{)}\big{)}=\mathbb{Q}\,\zeta^{\mathsf{f}}_{2r+1},$

and in particular

$\pi_{2r+1}\big{(}\mathsf{a}\big{(}\zeta^{\mathsf{f}}(\operatorname{gr}_{1}^{F}(\mathcal{B}_{2,3}))\big{)}\big{)}=\mathbb{Q}\,\zeta^{\mathsf{f}}_{2r+1}.$

###  6.3 The final step in Brown’s proof

######  Definition 6.11.

We set

$\mathcal{Z}^{\mathsf{f}}_{2,3}=\operatorname{span}_{\mathbb{Q}}\\{\zeta^{\mathsf{f}}(k_{1},\dots,k_{d})\mid
d\geq 0,k_{i}\in\\{2,3\\}\\}.$

On our way to prove that the free odd generation conjecture implies
$\mathcal{Z}^{\mathsf{f}}_{2,3}=\mathcal{Z}^{\mathsf{f}}$ the next ingredient
is a detailed study of the coaction on $\mathcal{Z}^{\mathsf{f}}_{2,3}$. We
have $\zeta^{\mathsf{f}}(\mathcal{B}_{2,3})=\mathcal{Z}^{\mathsf{f}}_{2,3}$
and thus by Lemma 4.3 $\Delta_{\operatorname{Gon}}$ restricts to

$\Delta_{\operatorname{Gon}}\colon\mathcal{Z}^{\mathsf{f}}_{2,3}\rightarrow\mathcal{A}^{\mathsf{f}}\otimes\mathcal{Z}^{\mathsf{f}}_{2,3},$

According to Definition 4.1, we are interested in the following filtration.

######  Definition 6.12.

We define a _level filtration_ on $\mathcal{Z}^{\mathsf{f}}_{2,3}$ by setting

$F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}=\operatorname{span}_{\mathbb{Q}}\Big{\\{}\zeta^{\mathsf{f}}(k_{1},\dots,k_{d})\in\mathcal{Z}^{\mathsf{f}}_{2,3}\;\Big{|}\;\deg_{3}(x_{0}^{k_{1}-1}x_{1}\cdots
x_{0}^{k_{d}-1}x_{1})\leq\ell\Big{\\}}$

for all $\ell\in\mathbb{N}_{0}$.

As a corollary of Lemma 4.3 we get

$\Delta_{\operatorname{Gon}}\colon
F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}\rightarrow\mathcal{A}^{\mathsf{f}}\otimes
F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}.$

and Proposition 4.12 1\. implies

$D_{2r+1}(F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3})\subseteq\mathcal{L}^{\mathsf{f}}_{2r+1}\otimes
F_{\ell-1}\mathcal{Z}^{\mathsf{f}}_{2,3}.$ (65)

Again we follow Definition 4.7, and introduce the following.

######  Definition 6.13.

For all $\ell\in\mathbb{N}$ we set

$\operatorname{gr}_{\ell}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})={\mathchoice{\raisebox{4.38887pt}{$\displaystyle{F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.38887pt}{$\displaystyle{F_{\ell-1}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}}{\raisebox{4.38887pt}{$\textstyle{F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-4.38887pt}{$\textstyle{F_{\ell-1}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}}{\raisebox{3.0861pt}{$\scriptstyle{F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-3.0861pt}{$\scriptstyle{F_{\ell-1}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}}{\raisebox{2.40277pt}{$\scriptscriptstyle{F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}\mkern-5.0mu\diagup\mkern-4.0mu\raisebox{-2.40277pt}{$\scriptscriptstyle{F_{\ell-1}\mathcal{Z}^{\mathsf{f}}_{2,3}}$}}}$

and $\operatorname{gr}_{0}^{F}=F_{0}\mathcal{Z}^{\mathsf{f}}_{2,3}$. For all
$\ell\in\mathbb{N}$ we denote by $\pi_{\ell}^{F}\colon
F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}\twoheadrightarrow\operatorname{gr}_{\ell}^{F}\mathcal{Z}^{\mathsf{f}}_{2,3}$
the natural projections.

By (65) the map
$D_{2r+1}^{(\ell)}=(\operatorname{id}\otimes\pi_{\ell-1}^{F})\circ
D_{2r+1}\rvert_{F_{\ell}\mathcal{Z}^{\mathsf{f}}_{2,3}}$ induces a map

$D_{2r+1}^{(\ell)}\colon\operatorname{gr}_{\ell}^{F}\mathcal{Z}^{\mathsf{f}}_{2,3}\rightarrow\mathcal{L}^{\mathsf{f}}_{2r+1}\otimes\operatorname{gr}_{\ell-1}^{F}\mathcal{Z}^{\mathsf{f}}_{2,3}$
(66)

for all $r,\ell\geq 1$. We have the following refinement of (66), that
combines Theorem 4.9 with the lift of Zagier’s formula (62).

######  Theorem 6.14.

Assume the Kernel conjecture 5.52 holds. Let $r,\ell\geq 1$. Then for each
weight $N\geq 2r+1$, we have

$D_{2r+1}^{(\ell)}\colon\operatorname{gr}_{\ell}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}\rightarrow\mathbb{Q}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N-2r-1}.$

###### Proof.

Recall the derivations

$D_{2r+1}^{(\ell)}\colon\operatorname{gr}_{\ell}^{F}(\mathcal{B}_{2,3})_{N}\rightarrow\pi_{2r+1}(\mathcal{B}^{1})\otimes\operatorname{gr}_{\ell-1}^{F}(\mathcal{B}_{2,3})_{N-2r-1}$

from Proposition 4.12. As a consequence of lifting Zagier’s formula (62), we
observed in Corollary 6.10 that
$\pi_{2r+1}\big{(}\mathsf{a}\big{(}\zeta^{\mathsf{f}}(\mathcal{B}^{1})\big{)}\big{)}=\mathbb{Q}\,\zeta^{\mathsf{f}}_{2r+1}$.
The claim now follows by applying the maps $\zeta^{\mathsf{f}}$ and
$\overline{\mathsf{a}\circ\zeta^{\mathsf{f}}}$ to the dervations above since
we have
$\overline{\mathsf{a}\circ\zeta^{\mathsf{f}}}(\pi_{2r+1}(\mathcal{B}^{1}))\subseteq\pi_{2r+1}(\mathsf{a}(\zeta^{\mathsf{f}}(\mathcal{B}^{1})))$.
∎

Assuming the free odd generation conjecture 5.44, Proposition 5.50 states that
$\zeta^{\mathsf{f}}_{2r+1}$ is non-zero. Therefore we get the identifications

$\rho_{r}\colon\mathbb{Q}\,\zeta^{\mathsf{f}}_{2r+1}\otimes\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}\overset{\sim}{\longrightarrow}\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}$

induced by $\zeta^{\mathsf{f}}_{2r+1}\mapsto 1$ for all
$r,\ell,N\in\mathbb{N}$.

######  Definition 6.15.

Assuming the free odd generation conjecture, we define for all $N,\ell\geq 1$
the maps

$D_{<N}^{(\ell)}\colon\operatorname{gr}_{\ell}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}\rightarrow\bigoplus_{3\leq
2r+1\leq
N}\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N-2r-1}$

via

$D_{<N}^{(\ell)}=\bigoplus_{3\leq 2r+1\leq N}\rho_{r}\circ D_{2r+1}^{(\ell)}.$

Observe that

$\bigoplus_{3\leq 2r+1\leq
N}\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N-2r-1}=\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{<N-1}$

since the weight gives a grading on $\mathcal{Z}^{\mathsf{f}}$.

######  Lemma 6.16.

If free odd generation conjecture 5.44 holds, then the diagram

$\textstyle{\operatorname{gr}_{\ell}^{F}(\mathcal{B}_{2,3})_{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\zeta^{\mathsf{f}}}$$\scriptstyle{\mathchoice{\kern
2.33566pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
2.33566pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
2.44876pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
2.59741pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}}$$\textstyle{\bigoplus\limits_{3\leq
2r+1\leq
N}\operatorname{gr}_{\ell-1}^{F}(\mathcal{B}_{2,3})_{N-2r-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bigoplus\limits_{3\leq
2r+1\leq
N}\zeta^{\mathsf{f}}}$$\textstyle{\operatorname{gr}_{\ell}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{D_{<N}^{(\ell)}}$$\textstyle{\bigoplus\limits_{3\leq
2r+1\leq
N}\operatorname{gr}_{\ell-1}^{F}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N-2r-1}}$
(67)

is well-defined and commutative.

###### Proof.

Recall $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
is given in Definition 4.14 and by Theorem 4.27, the upper row is an
isomorphism. The lower horizontal map is given by Definition 6.15 and it is
well-defined by the discussion before its definition. Since $\mathchoice{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.33667pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.49823pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}{\kern
3.71059pt\mathop{}\mathopen{\vphantom{\partial}}^{\mathmakebox[0pt][l]{\mathmakebox[r]{\phi}}}}\partial^{(\ell)}_{<N}$
and $D_{<N}^{(\ell)}$ are compatible by Remark 4.5, it suffices to show for
$w\in\mathcal{B}^{1}$ that

$\phi(w)\zeta^{\mathsf{f}}_{2r+1}=\pi_{2r+1}(\mathsf{a}(\zeta^{\mathsf{f}}(w))).$

But now, observing that projecting the identities (43) and (62) modulo
products and weight gives exactly the coefficients $\phi(w)$ from Definition
4.13. ∎

######  Remark 6.17.

We may change the assumptions for Lemma 6.16 a little bit, namely by imposing
that the Kernel conjecture 5.52 holds and that $\zeta^{\mathsf{f}}_{2r+1}\neq
0\in\mathcal{L}^{\mathsf{f}}_{2r+1}$ for all $r\in\mathbb{N}$.

######  Theorem 6.18.

Assume the free odd generation conjecture 5.44. Then the elements

$\\{\zeta^{\mathsf{f}}(k_{1},\ldots,k_{d})\mid k_{i}\in\\{2,3\\}\\}$

are linearly independent.

###### Proof.

Recall that diagram (67) commutes by Lemma 6.16 because of our assumption of
the free odd generation conjecture. By Theorem 4.27, the upper row is an
isomorphism and obviously the vertical maps are epimorphisms. We prove by
induction on the level $\ell\in\mathbb{N}$ that they are in fact isomorphisms.
We start with the case $\ell=1$ and $N\in\mathbb{N}$ arbitrary. Since
$\mathcal{Z}^{\mathsf{f}}$ is a weight-graded algebra, the elements
$\zeta^{\mathsf{f}}(\\{2\\}^{n})$ are linearly independent for all $n\geq 0$.
We deduce that the right-hand side of (67) is an isomorphism. Thus the left-
hand side must also be an isomorphism.
Clearly, the same lines of argument would hold for all $\ell>1$, thus (67) is
a commutative diagram of isomorphisms. Now, there are no relations in
different weights and if there were a relation in different level, then it
must induce a relation in the corresponding graded pieces. But we have shown
that this is impossible, thus the claim follows. ∎

######  Theorem 6.19.

Assume the free odd generation conjecture 5.44, then

$\mathcal{Z}^{\mathsf{f}}=\mathcal{Z}^{\mathsf{f}}_{2,3}.$

###### Proof.

Under the assumption of the free odd generation conjecture, Theorem 6.18
implies that $\\{\zeta(k_{1},\ldots,k_{d})\mid k_{i}\in\\{2,3\\}\\}$ forms a
basis of $\mathcal{Z}^{\mathsf{f}}_{2,3}$. Counting the number of such indices
in a given weight $N$ yields

$\sum_{N\geq
0}\dim\big{(}(\mathcal{Z}^{\mathsf{f}}_{2,3})_{N}\big{)}\,x^{N}=\frac{1}{1-x^{2}-x^{3}}.$

This agrees with the computed dimensions for $\mathcal{Z}^{\mathsf{f}}$ in
Proposition 5.46. We deduce that the two spaces $\mathcal{Z}^{\mathsf{f}}$ and
$\mathcal{Z}^{\mathsf{f}}_{2,3}$ must agree. ∎

######  Remark 6.20.

Finally we observe that Brown’s proof for the generators also works in the
formal setup: Assume the free odd generation conjecture holds, then any
element in $\mathcal{Z}^{\mathsf{f}}$ can be written uniquely as a polynomial
of elements $\zeta^{\mathsf{f}}(w)$, where $w$ is a Lyndon word in
$\mathsf{2}$ and $\mathsf{3}$.

## References