# On Some Unramified Families of Motivic Euler Sums

Ce Xua, and Jianqiang Zhaob,
a. School of Mathematics and Statistics, Anhui Normal University, Wuhu 241002,
P.R. China
b. Department of Mathematics, The Bishop’s School, La Jolla, CA 92037, USA
Email<EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract. It is well known that sometimes Euler sums (i.e., alternating
multiple zeta values) can be expressed as $\mathbb{Q}$-linear combinations of
multiple zeta values (MZVs). In her thesis Glanois presented a criterion for a
motivic Euler sums (MES) to be unramified, namely, expressible as
$\mathbb{Q}$-linear combinations of motivic MZVs. By applying this criterion
we present a few families of such unramified MES in two groups. In one such
group we can further prove the concrete identities relating the MES to the
motivic MZVs, determined up to rational multiple of a motivic Riemann zeta
value by a result of Brown.

Keywords: (motivic) multiple zeta values, (motivic) Euler sums.

AMS Subject Classifications (2020): 11M32; 11M99.

##  1 Introduction

The ubiquitous nature of multiple zeta values (MZVs) has attracted many
mathematicians and theoretical physicists in recent years after the seminal
works of Zagier [13] and Hoffman [8]. Its higher level generalization is given
by the _colored multiple zeta values_ (CMZVs) of level $N$ defined as follows.
Let $\mathbb{N}$ be the set of positive integers and
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$. For any composition
$(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ and $N$-th roots of unity
$({\varepsilon}_{1},\dots,{\varepsilon}_{d})$ we define

$\zeta\left(s_{1},\dots,s_{d}\atop{\varepsilon}_{1},\dots,{\varepsilon}_{d}\right):=\displaystyle\sum_{0<k_{1}<\dots<k_{d}}\displaystyle\frac{{\varepsilon}_{1}^{k_{1}}\cdots{\varepsilon}_{d}^{k_{d}}}{k_{1}^{s_{1}}\cdots
k_{d}^{s_{d}}}.$

To guarantee convergence we impose the condition that
$(s_{d},{\varepsilon}_{d})\neq(1,1)$.

The CMZVs have played a pivotal role in the theory of mixed Tate motives over
$\mathbb{Z}[\mu_{N}][1/N]$ (resp. $\mathbb{Z}[\mu_{N}]$), where
$\mu_{N}=\exp(2\pi i/N)$, for $N=1,2,4,6,8$ (resp. $N=6$) as manifested by the
works [4, 5, 6]. In fact, they first appeared unexpectedly in the computation
of Feynman integrals in the 1990s. In particular, the level two MZVs,
sometimes also called _Euler sums_ , have been studied quite intensively in
[1, 3, 7, 9]. To save space, if ${\varepsilon}_{j}=-1$ then we conventionally
put a bar on top of $s_{j}$. If such an Euler sum can be expressed in terms of
the MZVs then we say it is an _unramified_ Euler sum because the corresponding
motivic version is unramified. These values are also called honorary MZVs by
Broadhurst. For example, we have the following beautiful unramified family of
Euler sums discovered numerically in [2] and proved by the second author [14]

$8^{\ell}\zeta(\\{1,\bar{2}\\}_{\ell})=\zeta(3_{\ell})$ (1.1)

for all $\ell\in\mathbb{N}$. Here and in the rest of the paper,
${\boldsymbol{\sl{s}}}_{n}$ means the string ${\boldsymbol{\sl{s}}}$ is
repeated $n$ times and if the string has only one number then we remove the
curly brackets to save space.

Our goal in this paper is to lift (1.1) to its motivic version and then
provide a few more families of unramified motivic Euler sums (MES) in two
groups. Denote by $\zeta^{\mathfrak{m}}({\boldsymbol{\sl{s}}})$ the motivic
version of the Euler sum $\zeta({\boldsymbol{\sl{s}}})$ (see next section for
the precise definition). Then the first group is presented below.

######  Theorem 1.1.

For all $\ell\in\mathbb{N}$, there are some rational numbers $c_{\ell}$,
$d_{\ell}$ and $e_{\ell}$ such that

$\displaystyle 2^{3\ell}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell})=$
$\displaystyle\,\zeta^{\mathfrak{m}}(3_{\ell}),$ (1.2) $\displaystyle
2^{3\ell+1}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell},\bar{2})=$
$\displaystyle\,c_{\ell}\zeta^{\mathfrak{m}}(3\ell+2)-\displaystyle\sum_{{\alpha}+{\beta}=\ell}(-1)^{\alpha}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),$
(1.3) $\displaystyle 2^{3\ell}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell})=$
$\displaystyle\,d_{\ell}\zeta^{\mathfrak{m}}(3\ell)-\zeta^{\mathfrak{m}}(3_{\ell})+2\displaystyle\sum_{{\alpha}+{\beta}=\ell-1}(-1)^{a}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),$
(1.4) $\displaystyle
2^{3\ell-1}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell},1)=$
$\displaystyle\,e_{\ell}\zeta^{\mathfrak{m}}(3\ell+1)-3\displaystyle\sum_{{\alpha}+{\beta}=\ell,2\nmid
b}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}}).$ (1.5)

After applying the period map (2.8) we see that (1.3) and (1.5) partially
confirm a conjecture by M. Hirose and N. Sato [11]. Based on this conjecture
and some further numerical evidence, we believe that all the undetermined
constants in the above theorem must vanish.

######  Conjecture 1.2.

For all $\ell\in\mathbb{N}$, we have $c_{\ell}=d_{\ell}=e_{\ell}=0$ in Theorem
1.1.

Note that the corresponding Euler sum for
$\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell},1)$ diverges. When applying the
period map (2.8) we need to use its shuffle or stuffle regularized version
(see the explanation after (2.8)).

We will further prove the following seond group of families of unramified
Euler sums. Here $\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})$ corresponds
to some shuffle regularized Euler sum for all $a\in\mathbb{N}$ (see (I3) in
the next section for precise definition).

######  Theorem 1.3.

For all integers $a,\ell\geq 0$ the MES

$\displaystyle\zeta_{a}^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell}),\qquad\zeta_{a}^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell},\bar{2}),$
(1.6)
$\displaystyle\zeta_{a}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell}),\qquad\zeta_{a}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell},3),$
(1.7)

are all unramified.

Using exactly the same approach we may generalized the first group of
unramified families to the the following form. We leave the details to the
interested reader.

######  Theorem 1.4.

For all integers $a,\ell\geq 0$ the MES

$\zeta_{a}^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell}),\quad\zeta_{a}^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell},\bar{2}),\quad\zeta_{a}^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell}),\quad\zeta_{a}^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell},1),$

are all unramified.

######  Remark 1.5.

In [6, 7] Glanois proved that an interpolated version of MES denoted by
$\zeta^{\sharp,{\mathfrak{m}}}({\boldsymbol{\sl{s}}})$ are all unramified if
${\boldsymbol{\sl{s}}}=(\\{\overline{\text{even}},\text{odd}\\}_{\ell})$ where
the even components can differ and so can the odd ones. This is unlikely true
for the MES in general. Numerial compuatation using the LLL algorithm shows
that it is unlikely that $\zeta(\bar{2},3,\bar{2},5)$ is in the weight 12
piece of the MZV space.

The main results obtained in this paper can be regarded as some good evidence
for the following conjecture.

######  Conjecture 1.6.

Let $a\in\mathbb{N}_{0}$, $\ell\in\mathbb{N}$, and $u_{j},v_{j}\in
X:=\\{1,\bar{2},3,\bar{4},\dots\\}$ for $j=1,\dots,\ell$ such that
$1\not\in\\{u_{j}:1\leq j\leq\ell\\}\displaystyle\cap\\{v_{j}:1\leq
j\leq\ell\\}$. Then

$\displaystyle\sum_{{\sigma},\tau\in{\mathfrak{S}}_{\ell}}\zeta_{a}^{\mathfrak{m}}(u_{{\sigma}(1)},v_{\tau(1)},\dots,u_{{\sigma}(\ell)},v_{\tau(\ell)})$

and

$\displaystyle\sum_{{\sigma}\in{\mathfrak{S}}_{\ell},\tau\in{\mathfrak{S}}_{\ell-1}}\zeta_{a}^{\mathfrak{m}}(u_{{\sigma}(1)},v_{\tau(1)},\dots,v_{\tau(\ell-1)},u_{{\sigma}(\ell)}).$

are both unramified, where ${\mathfrak{S}}_{\ell}$ is the symmetry group of
$\ell$ letters.

When $a=0$, after taking the period map this conjecture is reduced to a
conjecture first proposed by M. Hirose and N. Sato [11].

##  2 Motivic set-up

We adopt the motivic set-up defined by [7]. Let $\Gamma_{N}$ be the group of
$N$-th unity for any $N\in\mathbb{N}$. Let $w\in\mathbb{N}$ and suppose
$a_{j}=0$ or $a_{j}\in\Gamma_{N}$ for all $0\leq j\leq w+1$. The most
important property of the motivic integrals is that there is a period map
$\operatorname{dch}$ such that

$\operatorname{dch}\big{(}I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{w};a_{w+1})\big{)}=(-1)^{\operatorname*{dep}(a_{1},\dots,a_{w})}\displaystyle\int_{a_{0}}^{a_{w+1}}\displaystyle\frac{dt}{t-a_{1}}\cdots\displaystyle\frac{dt}{t-a_{w}}$
(2.8)

as an iterated integral (we always integrate from left to right in this paper)
whenever it converges. Here
$\operatorname*{dep}(a_{1},\dots,a_{w}):=\sharp\\{1\leq j\leq w:a_{j}\neq
0\\}$ is called the _depth_ and $w$ the _weight_ of the integral (and its
corresponding MZV). When the integral diverges then there is a way to
regularize it so that it becomes a polynomial in $\mathsf{CMZV}[T]$, where
$\mathsf{CMZV}$ is the $\mathbb{Q}$-span of all CMZVs of level $N$. By setting
$T=0$ we can define the so-called shuffle regularized CMZVs (see [15, Ch. 13]
for more details).

We now list all the important properties of the motivic integrals as follows
(cf. [4, §2.4], [7, p. 9], and [10, §2, (I1)-(I6)]):

  * (I1)

Empty word: $I^{\mathfrak{m}}(a_{0};a_{1})=1$.

  * (I2)

Trivial path: $\forall n\geq 1$,
$I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{n};a_{n+1})=0$ if $a_{0}=a_{n+1}$.

  * (I3)

Shuffle product: for any $s_{1},\dots,s_{n}\in\mathbb{N}$ and
${\varepsilon}_{1},\dots,{\varepsilon}_{n}\in\Gamma_{N}$, define

$\zeta_{k}^{\mathfrak{m}}\left(s_{1},\dots,s_{n}\atop{\varepsilon}_{1},\dots,{\varepsilon}_{n}\right):=I^{\mathfrak{m}}\big{(}0;0^{k},\eta_{1},0^{s_{1}-1},\eta_{2},0^{s_{2}-1},\dots,\eta_{n},0^{s_{n}-1};1\big{)}$

where $\eta_{j}=1/{\varepsilon}_{j}\cdots{\varepsilon}_{d}$ for all
$j=1,\dots,d$. Then

$\zeta_{k}^{\mathfrak{m}}\left({s_{1},\cdots,s_{p}\atop{\varepsilon}_{1},\cdots,{\varepsilon}_{p}}\right)=(-1)^{k}\displaystyle\sum_{\begin{subarray}{c}i_{1}+\cdots+i_{p}=k\\\
i_{1},\dots,i_{p}\geq
0\end{subarray}}\left(\prod_{j=1}^{p}\binom{s_{j}+i_{j}-1}{i_{j}}\right)\zeta^{\mathfrak{m}}\left({s_{1}+i_{1},\cdots,s_{p}+i_{p}\atop{\varepsilon}_{1}\
\ ,\cdots,\ \ {\varepsilon}_{p}}\right).$

  * (I4)

Regularization: If $a_{1}=\cdots=a_{n}\in\\{0,1\\}$, then
$I^{\mathfrak{m}}(0;a_{1},\cdots,a_{n};1)=0.$

  * (I5)

Path reversal:
$I^{\mathfrak{m}}(a_{0};a_{1},\cdots,a_{n};a_{n+1})=(-1)^{n}I^{\mathfrak{m}}(a_{n+1};a_{n},\cdots,a_{1};a_{0}).$

  * (I6)

Homothety: $\forall\alpha\in\Gamma_{N},I^{\mathfrak{m}}(0;\alpha
a_{1},\cdots,\alpha a_{n};\alpha
a_{n+1})=I^{\mathfrak{m}}(0;a_{1},\cdots,a_{n};a_{n+1})$.

  * (I7)

Change of variable $t\to 1-t:\forall a_{1},\cdots,a_{n}\in\\{0,1\\},$

$I^{\mathfrak{m}}(0;a_{1},\cdots,a_{n};1)=I^{\mathfrak{m}}(0;1-a_{n},\cdots,1-a_{1};1).$

  * (I8)

Path composition: $\forall a,b,x\in\Gamma_{N}\cup\left\\{0\right\\}$,

$I^{\mathfrak{m}}(a;a_{1},\cdots,a_{n};b)=\displaystyle\sum_{i=0}^{n}I^{\mathfrak{m}}(a;a_{1},\cdots,a_{i};x)I^{\mathfrak{m}}(x;a_{i+1},\cdots,a_{n};b).$

Let ${\mathcal{H}}^{N}$ be $\mathbb{Q}$-vector space spanned by the motivic
CMZVs of the form

$\zeta^{\mathfrak{m}}_{a}\left(n_{1},\dots,n_{d}\atop{\varepsilon}_{1},\dots,{\varepsilon}_{d}\right):=I^{\mathfrak{m}}(0;0_{a},\eta_{1},0_{n_{1}-1},\dots,\eta_{d},0_{n_{d}-1};1),\quad
a\in\mathbb{N}_{0},$

where $n_{j}\in\mathbb{N},{\varepsilon}_{j}\in\Gamma_{N}$, and $\eta_{j}$’s
are defined in (I3) for all $j$. Note that $a+n_{1}+\dots+n_{d}$ is its weight
which is denoted by $|{\boldsymbol{\sl{n}}}|$ when $a=0$. If
${\varepsilon}_{j}=1$ (resp. ${\varepsilon}_{j}=\pm 1$) for all $j$ then we
get the motivic MZV (resp. MES) $\zeta^{\mathfrak{m}}_{a}(n_{1},\dots,n_{d})$.
Set
${\mathcal{A}}^{N}={\mathcal{H}}^{N}/\zeta^{\mathfrak{m}}(2){\mathcal{H}}^{N}$.
Denote by ${\mathcal{H}}^{N}_{w}$ and ${\mathcal{A}}_{w}^{N}$ the weight $w$
part for all $w\geq 0$. Let
${\mathcal{L}}^{N}={\mathcal{A}}^{N}_{>0}/{\mathcal{A}}^{N}_{>0}\cdot{\mathcal{A}}^{N}_{>0}$.
For any weight $w$ and odd $r$ such that $r<w$ we can define a derivation as
part of a coaction

$D_{r}:{\mathcal{H}}_{w}^{N}\to{\mathcal{L}}_{r}^{N}{\otimes}{\mathcal{H}}_{w-r}^{N}$

by sending $I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{w};a_{w+1})$ to

$\displaystyle\sum_{p=0}^{w-r}I^{\mathfrak{l}}(a_{p};a_{p+1},\dots,a_{p+r};a_{p+r+1}){\otimes}I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{p},a_{p+r+1},\dots,a_{w};a_{w+1}).$

The sequence in the left motivic integral is called a _subsequence_ of
$(a_{0};a_{1},\dots,a_{w};a_{w+1})$ while that in the right factor is called a
_quotient sequence_. Each such a choice is called a _cut_.

Now we set $N=2$ and recall the following theorem which combines Glanois’s
result [7, Cor. 2.4] and Brown’s [4, Thm. 3.3]. Set
${\mathcal{H}}={\mathcal{H}}^{1}$ and ${\mathcal{L}}={\mathcal{L}}^{1}$.

######  Theorem 2.1.

Let $a\in\mathbb{N}_{0}$ and ${\boldsymbol{\sl{s}}}$ be a composition of
positive integers such that $a+|{\boldsymbol{\sl{s}}}|=k$. Then the weight $k$
MES $\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})\in{\mathcal{H}}_{k}$ if
and only $D_{1}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})=0$ and
$D_{r}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})\in{\mathcal{L}}_{r}{\otimes}{\mathcal{H}}_{k-r}$
for all odd $r<k$. Moreover, if
$D_{r}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})=0$ for all odd $r<k$
then
$\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})=c\,\zeta^{\mathfrak{m}}(k)$
for some rational number $c$.

Set $\eta_{j}=\prod_{i=j}^{d}{\varepsilon}_{i}$ for all $j=1,\dots,d$ as above
(note that $\eta_{j}=1/\eta_{j}$) and define

$\rho\left(n_{1},\dots,n_{d}\atop{\varepsilon}_{1},\dots,{\varepsilon}_{d}\right):=(0;\eta_{1},0_{n_{1}-1},\dots,\eta_{d},0_{n_{d}-1};1).$

We often use the bar notation instead of the two-row one for Euler sums.

The following simple fact was already noticed by Glanois [6] without proof. We
provide one here for completeness.

######  Lemma 2.2.

If $\bar{1}$ does not appear in ${\boldsymbol{\sl{s}}}$ then
$D_{1}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{s}}})=0$ for all
$a\in\mathbb{N}_{0}$.

###### Proof.

If $|s_{j}|>1$ for all components of ${\boldsymbol{\sl{s}}}$ then no two
consecutive nonzero terms can appear in $\rho({\boldsymbol{\sl{s}}})$ so that
$D_{1}=0$ clearly. If $s_{j}=1$ then the corresponding sign
${\varepsilon}_{j}=1$ and therefore $\eta_{j}=\eta_{j+1}$. Namely, whenever
two consecutive nonzero terms appear they must have the same sign. Thus if
there are more than one nonzero numbers in the left factor
$I^{\mathfrak{l}}(a;b;c)$ of $D_{1}$ then it must look like
$I^{\mathfrak{l}}(0;\eta;\eta)$, or $I^{\mathfrak{l}}(\eta;\eta;0)$,or
$I^{\mathfrak{l}}(\eta;0;\eta)$ or $I^{\mathfrak{l}}(\eta;\eta;\eta)$ all of
which must be 0. All the other cases are clearly 0, too. ∎

Recall that the Bernoulli numbers $B_{n}$ are defined by the generating
function

$\displaystyle\frac{e^{t}}{e^{t}-1}=\displaystyle\sum_{n=0}^{\infty}B_{n}\displaystyle\frac{t^{n}}{n!}.$

######  Proposition 2.3.

For any positive integers $m\geq 2$ and $d\geq 1$ we have

$\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{d})\in{\mathcal{H}}_{md}.$ (2.9)

Moreover, if $m=2s$ is even we have

$\zeta^{\mathfrak{m}}(\\{\overline{2s}\\}_{d})=c\,\zeta^{\mathfrak{m}}(2sd)$
(2.10)

where

$c=\displaystyle\frac{-2(2sd)!}{16^{sd}B_{2sd}}\displaystyle\sum_{\begin{subarray}{c}\Sigma_{j=1}^{s}n_{j}=sd\\\
n_{j}\in\mathbb{N}_{0}\ \forall
j\end{subarray}}\left(\prod_{j=1}^{s}\displaystyle\frac{\big{(}1+e^{\tfrac{\pi
i}{2s}}\big{)}^{2n_{j}+1}+\big{(}1-e^{\tfrac{\pi
i}{2s}}\big{)}^{2n_{j}+1}}{2(2n_{j}+1)!}e^{\tfrac{\pi
i}{s}\underset{\scriptscriptstyle j=1}{\overset{\scriptscriptstyle
m}{\Sigma}}(2j-1)n_{j}}\right)\in\mathbb{Q}.$

###### Proof.

When $d=1$ we see that $D_{r}\zeta^{\mathfrak{m}}(\bar{m})=0$ for all $r<m$
since all cuts for $D_{r}$ either have 0 on both ends or have all 0’s inside.
Thus (2.10) follows directly from Theorem 2.1.

Suppose now $d>1$. To save space we let $m=n+1$ and use $\bar{1}$ to denote
$-1$ . Then

$\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{2\ell})=I^{\mathfrak{m}}(0;\\{1,0_{n},\bar{1},0_{n}\\}_{\ell};1),\quad\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{2\ell+1})=I^{\mathfrak{m}}(0;\bar{1},0_{n},\\{1,0_{n},\bar{1},0_{n}\\}_{\ell};1).$

For every odd integer $r<m$ we clearly have
$D_{r}\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{d})=0$.

Let $r\geq m$ be an odd integer. Suppose first that $m=2s$ is an even integer.
If $d=2\ell$ is even let $r=m+i+2mp<md$ for some integer $i,p\geq 0$. Then all
nonzero terms of $D_{r}$ can be paired as shown below:

$0;\\{1,0_{n},\bar{1},0_{n}\\}_{j},1,0_{n-i-1},0,0_{i},\\{\bar{1},0_{n},1,0_{n}\\}_{p},\bar{1},0_{i},0,0_{n-i-1},1,0_{n},,\bar{1},0_{n},\\{1,0_{n},\bar{1},0_{n}\\}_{k};$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{r}\zeta^{\mathfrak{m}}(\\{\overline{n+1}\\}_{d}),2j+2p+2k+4=d$ is even.

Strictly speaking there are also cut pairs such as
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{1}$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{2}$}}}\hskip
1.07639pt{}}{}$ for which 1 and $\bar{1}$ are exchanged inside
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ in the above graph. Every such pair have the same quotient
sequence but reversed subsequences of each other. Thus the pair cancel each
other by path reversal (I5). The argument for odd $d=2\ell+1$ is exactly the
same. Thus, by Theorem 2.1 we see that
$\zeta^{\mathfrak{m}}(\\{\overline{2s}\\}_{d})=c\zeta^{\mathfrak{m}}(2sd)$ for
some rational number $c$. Applying the period map (2.8) we can verify (2.10)
by [12, Thm. 2].

Suppose now that $m$ is an odd integer. Then the above proof still works
unless $r=mp$ for some odd $p<d$. Now, if a cut starts at a 0 immediately
before a $\pm 1$ then the term will be canceled by the cut starting from the
next $\pm 1$, unless the cut itself ends with the last 1 which yields the
following single nonzero term

$D_{r}\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{d})=\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{p}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{m}\\}_{d-p})\in{\mathcal{L}}_{mp}{\otimes}{\mathcal{H}}_{m(d-p)}$

by induction. Hence (2.9) follows by Theorem 2.1. This completes the proof of
the proposition. ∎

##  3 Proof of (1.2) in Theorem 1.1

In this section, we first prove (1.2) due to its simplicity and repeated use
throughout the paper. The proof technique is also very typical for this kind
of claims on unramified MES. As pointed out by S. Charlton, the equation (1.2)
follows from the original proof of (1.1) in [14] which uses the double shuffle
relations and the distribution relations all of which are motivic. For
convenience we restate (1.2) as follows: for all $\ell\in\mathbb{N}$

$8^{\ell}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell})=\zeta^{\mathfrak{m}}(3_{\ell}).$
(3.11)

We now prove by (3.11) by induction on $\ell$.

###  3.1 The base case

If $\ell=1$ then we see easily that $D_{1}\zeta^{\mathfrak{m}}(1,\bar{2})=0$
by Lemma 2.2. Thus by Theorem 2.1
$\zeta^{\mathfrak{m}}(1,\bar{2})=c\zeta^{\mathfrak{m}}(3)$ for some
$c\in\mathbb{Q}$. Applying the period map (2.8) and comparing with (1.1) we
get $c=1/8$ immediately.

We now assume that (3.11) holds for all positive integers $\ell<k$. We need to
show that

$8^{k}D_{r}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k})=D_{r}\zeta^{\mathfrak{m}}(3_{k})$
(3.12)

for all positive integers $r=6n+3,6n+5,6n+7<3k$ where $n\in\mathbb{Z}$.

###  3.2 Inductive step

####  3.2.1 $r=6n+3$

We have the following picture when $k$ is even (odd $k$ case can be dealt with
similarly).

$0;$$1,$$1,$$0,$$\bar{1},$$\bar{1},$$\cdots\\!,$$0,$$1,$$1,$$\\{0\bar{1}\bar{1}011\\}_{n}$$0,$$\bar{1},$$\bar{1},$$0,$$1,$$1,\cdots\\!,$$0,$$\bar{1},$$\bar{1},$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k})$

Suppose a cut starts in a 011-block. It is easy to see that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$ since they have opposite subsequences but the same
quotient one. Further, the subsequence is anti-symmetric so that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}=0$ by path reversal (I5) and homothety with $\alpha=-1$ in (I6).
If a cut starts in a $0\bar{1}\bar{1}$-block then the same argument works.
However, if a cut of type
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{1}$}}}\hskip
1.07639pt{}}$ ends at the last 1 then it cannot be canceled. Thus

$8^{k}D_{6n+3}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k})=8^{k}\zeta^{\mathfrak{l}}(\\{1,\bar{2}\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k-2n-1})=\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1})$

by induction. Similarly, by the picture

$0;$$1,$$0,$$\cdots\\!,$$0,$$1,$$0,$$\\{010010\\}_{n}$$0,$$1,$$0,$$0,$$1,$$0,\cdots\\!,$$0,$$1,$$0,$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(3_{k})$

we see that ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=0$ unless the cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ ends at the last 1 so that it cannot be canceled. Thus (3.12)
holds for $r=6n+3$.

####  3.2.2 $r=6n+5$

It is obvious that $D_{6n+5}=0$ on both sides of (3.12) since both
$\rho(3_{k})$ and $\rho(\\{1,\bar{2}\\}_{k})$ have period 6 so that every cut
of $D_{6n+5}$ starts and ends with the same number.

####  3.2.3 $r=6n+7$

We have the following pictures of possible cuts.

$0;$$1,$$1,$$0,$$\bar{1},$$\bar{1},$$\cdots\\!,$$0,$$1,$$1,$$\\{0\bar{1}\bar{1}011\\}_{n}$$0,$$\bar{1},$$\bar{1},$$0,$$1,$$1,$$0,$$\bar{1},$$\bar{1},\cdots\\!,$$0,$$\bar{1},$$\bar{1},$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+7}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k})$ (3.13)

$0,$$1,$$0,$$\cdots\\!,$$0,$$1,$$0,$$\\{010010\\}_{n}$$0,$$1,$$0,$$0,$$1,$$0,$$0,$$1,$$0,\cdots\\!,$$0,$$1,$$0,$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+7}\zeta^{\mathfrak{m}}(3_{k})$ (3.14)

It is clear that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ has an anti-symmetric subsequence so that it disappears and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=0$ in both pictures. This implies that (3.12) holds
for $r=6n+7$.

By combining all the cases for different $r$’s above, we see that (3.12) holds
for all odd $r<k$. By Theorem 2.1 there is some $c\in\mathbb{Q}$ such that
$8^{k}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k})=\zeta^{\mathfrak{m}}(3_{k})+c\,\zeta^{\mathfrak{m}}(3k)$.
Applying the period map (2.8) and using formula (1.1) we finally see that
$c=0$ which implies (1.2) (or its restatement (3.11)) immediately.

###  3.3 Proof of base cases of Theorem 1.3

We first prove Theorem 1.3 when $a=0$ and $a=1$ as part of the base cases. We
will then prove the full Theorem 1.3 by induction in the next section.

######  Lemma 3.1.

For all integers $\ell\geq 0$ the MES

$\displaystyle\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell}),\qquad\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell},\bar{2}),$
(3.15)
$\displaystyle\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell}),\qquad\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell},\bar{2}),$
(3.16)
$\displaystyle\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell}),\qquad\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell},3),$
(3.17)
$\displaystyle\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell}),\qquad\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{\ell},3)$
(3.18)

are all unramified.

We originally only planned to prove the pair of families in (3.15) are both
unramified. When we tried the induction we found that we had to use a
“quadruple” induction to prove all the four pairs in Lemma 3.1 are unramified.
This is what will do below.

By simple computation or using the datamine [1] we get

$\displaystyle\zeta(\bar{2},3)=$
$\displaystyle\,-\tfrac{1}{8}\zeta(3,2)-\tfrac{3}{32}\zeta(5),$
$\displaystyle\zeta(\bar{2},3,\bar{2})=$
$\displaystyle\,\tfrac{7}{640}\zeta(7)-\tfrac{21}{640}\zeta(3,4),$
$\displaystyle\zeta_{1}(\bar{2},3)=$
$\displaystyle\,-\big{(}2\zeta(\bar{3},3)+3\zeta(\bar{2},4)\big{)}=\tfrac{3}{8}\zeta(3)^{2}+\tfrac{5}{64}\zeta(6),$
$\displaystyle\zeta_{1}(\bar{2},3,\bar{2})=$
$\displaystyle\,-\big{(}2\zeta(\bar{3},3,\bar{2})+3\zeta(\bar{2},4,\bar{2})+2\zeta(\bar{2},3,\bar{3})\big{)}$
$\displaystyle=$
$\displaystyle\,\tfrac{41}{32}\zeta(6,2)-\tfrac{185}{64}\zeta(1,7)-\displaystyle\frac{1}{8}\zeta(2,3,3)-\tfrac{215}{256}\zeta(8),$
$\displaystyle\zeta(3,\bar{2})=$
$\displaystyle\,\tfrac{7}{32}\zeta(5)-\tfrac{1}{4}\zeta(2,3),$
$\displaystyle\zeta(3,\bar{2},3)=$
$\displaystyle\,\tfrac{3}{64}\zeta(6,2)-\tfrac{3}{16}\zeta(3,5)+\tfrac{3}{16}\zeta(3,3,2)-\tfrac{1}{48}\zeta(8),$
$\displaystyle\zeta_{1}(3,\bar{2})=$
$\displaystyle\,-\big{(}2\zeta(3,\bar{3})+3\zeta(4,\bar{2})\big{)}=\tfrac{147}{64}\zeta(6)-\tfrac{9}{8}\zeta(3)^{2},$
$\displaystyle\zeta_{1}(3,\bar{2},3)=$
$\displaystyle\,-\big{(}3\zeta(4,\bar{2},3)+2\zeta(3,\bar{3},3)+3\zeta(3,\bar{2},4)\big{)}$
$\displaystyle=$
$\displaystyle\,\tfrac{153}{64}\zeta(7,2)-\tfrac{21}{4}\zeta(6,3)-\tfrac{387}{64}\zeta(3,6)+\tfrac{11}{8}\zeta(9).$

This convinced us that the base cases are all correct even though we need to
do this on the motivic level. First, we can see easily that $D_{1}=0$ for all
the MES in Theorem 1.3 by Lemma 2.2.

####  3.3.1 The base cases

Lemma 3.1 clearly holds for $\ell=0$. To see the basic ideas we now prove
$\ell=1$ case. To save space, we often suppress the commas in the sub- and
quotient sequence in the rest of the paper. Using the left picture below we
can check that

$\displaystyle D_{3}\zeta^{\mathfrak{m}}(\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};010;0){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)+I^{\mathfrak{l}}(0;100;1){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)$
$\displaystyle=$
$\displaystyle\,(\zeta^{\mathfrak{l}}(3)+\zeta_{1}^{\mathfrak{l}}(\bar{2})){\otimes}\zeta^{\mathfrak{m}}(\bar{2})=(\zeta^{\mathfrak{l}}(3)-2\zeta^{\mathfrak{l}}(\bar{3})){\otimes}\zeta^{\mathfrak{m}}(\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{2},$

by (I3) and Prop. 2.3, which shows that
$\zeta^{\mathfrak{m}}(\bar{2},3)\in{\mathcal{H}}_{5}$ by Theorem 2.1.

$0;$$\bar{1},$$0,$$1,$$0,$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(\bar{2},3)$$0;$$1,$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip
0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})$$0;$$1,$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
0$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{1}$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{2}$}}}\hskip
0.75346pt{}}$$D_{5}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})$

To compute $D_{3}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})$ we can use the
middle picture above to see that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=0$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$ by path reversal. Hence
$D_{3}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})=0$. Further, using the right
picture above we get

$\displaystyle D_{5}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;10\bar{1}00;\bar{1}){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)+\big{(}I^{\mathfrak{l}}(1;0\bar{1}00\bar{1};0)+I^{\mathfrak{l}}(0;\bar{1}00\bar{1}0;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;10;1)$
$\displaystyle=$
$\displaystyle\,I^{\mathfrak{l}}(0;\bar{1}0100;1){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)=\zeta^{\mathfrak{l}}(\bar{2},3){\otimes}\zeta^{\mathfrak{m}}(\bar{2})$

by homothety and path reversal. Thus by the above proof that
$\zeta^{\mathfrak{m}}(\bar{2},3)\in{\mathcal{H}}_{5}$ and Prop. 2.3 we get
$D_{5}\zeta^{\mathfrak{m}}(\bar{2},3,\bar{2})\in{\mathcal{L}}_{5}{\otimes}{\mathcal{H}}_{2}.$

The above computation shows that both MES in (3.15) are unramified if $\ell=1$
by Theorem 2.1. One can similarly check that
$\zeta_{1}^{\mathfrak{m}}(\bar{2},3),\zeta_{1}^{\mathfrak{m}}(\bar{2},3,\bar{2}),\zeta^{\mathfrak{m}}(3,\bar{2}),\zeta^{\mathfrak{m}}(3,\bar{2},3)$
are all unramified. Indeed, by similar computation as above, we get

$\displaystyle D_{3}\zeta_{1}^{\mathfrak{m}}(\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(0;0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;100;1)+\big{(}I^{\mathfrak{l}}(\bar{1};010;0)+I^{\mathfrak{l}}(0;100;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;0\bar{1}0;1)$
$\displaystyle=$
$\displaystyle\,\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta^{\mathfrak{m}}(3)+\big{(}\zeta^{\mathfrak{l}}(3)-\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta_{1}^{\mathfrak{m}}(\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{3},$
$\displaystyle D_{3}\zeta_{1}^{\mathfrak{m}}(\bar{2},3,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;010;\bar{1}){\otimes}I^{\mathfrak{m}}(0;\bar{1}00\bar{1}0;1)+\big{(}I^{\mathfrak{l}}(1;0\bar{1}0;0)+I^{\mathfrak{l}}(0;\bar{1}00;\bar{1})\big{)}{\otimes}I^{\mathfrak{m}}(0;010\bar{1}0;1)$
$\displaystyle\,+\big{(}I^{\mathfrak{l}}(\bar{1};00\bar{1};0)+I^{\mathfrak{l}}(0;0\bar{1}0;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;010\bar{1}0;1)$
$\displaystyle=$
$\displaystyle\,\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta^{\mathfrak{m}}(3,\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{5},$
$\displaystyle D_{5}\zeta_{1}^{\mathfrak{m}}(\bar{2},3,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;10\bar{1}00;\bar{1}){\otimes}I^{\mathfrak{m}}(0;0\bar{1}0;1)+\big{(}I^{\mathfrak{l}}(1;0\bar{1}00\bar{1};0)+I^{\mathfrak{l}}(0;\bar{1}00\bar{1}0;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;010;1)$
$\displaystyle=$
$\displaystyle\,\zeta^{\mathfrak{l}}(\bar{2},3){\otimes}\zeta_{1}^{\mathfrak{m}}(\bar{2})\in{\mathcal{L}}_{5}{\otimes}{\mathcal{H}}_{3}$

by essentially the same proof of unramification of
$\zeta^{\mathfrak{m}}(\bar{2},3)$ above.

$0;$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0;$$1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$$D_{3}\zeta_{1}^{\mathfrak{m}}(3,\bar{2})$$0;$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1,$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
5$}}}\hskip
0.75346pt{}}$$D_{3}\zeta_{1}^{\mathfrak{m}}(3,\bar{2},3)$$0;$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1,$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip 0.75346pt{}}$$D_{5}\zeta_{1}^{\mathfrak{m}}(3,\bar{2},3)$

Moreover, using the above pictures
(${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ always cancel) one checks easily that

$\displaystyle D_{3}\zeta_{1}^{\mathfrak{m}}(3,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;0\bar{1}0;1)=\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta_{1}^{\mathfrak{m}}(\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{3},$
$\displaystyle D_{5}\zeta_{1}^{\mathfrak{m}}(3,\bar{2})=$ $\displaystyle\,0,$
$\displaystyle D_{3}\zeta_{1}^{\mathfrak{m}}(3,\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(0;0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;0\bar{1}0100;1)+\big{(}I^{\mathfrak{l}}(\bar{1};010;0)+I^{\mathfrak{l}}(0;100;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;0\bar{1}00\bar{1}0;1)$
$\displaystyle=$
$\displaystyle\,\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta_{1}^{\mathfrak{m}}(\bar{2},3)+\big{(}\zeta^{\mathfrak{l}}(3)-\zeta_{1}^{\mathfrak{l}}(\bar{2})\big{)}{\otimes}\zeta_{1}^{\mathfrak{m}}(3,\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{6},$
$\displaystyle D_{5}\zeta_{1}^{\mathfrak{m}}(3,\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(0;\bar{1}0100;1){\otimes}I^{\mathfrak{m}}(0;0\bar{1}00;1)=\zeta_{1}^{\mathfrak{l}}(\bar{2},3){\otimes}\zeta_{1}^{\mathfrak{m}}(\bar{3})\in{\mathcal{L}}_{5}{\otimes}{\mathcal{H}}_{4},$
$\displaystyle D_{7}\zeta_{1}^{\mathfrak{m}}(3,\bar{2},3)=$ $\displaystyle 0,$

using the known unramified MES. Similarly,

$\displaystyle D_{3}\zeta^{\mathfrak{m}}(3,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)=\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{2},$
$\displaystyle D_{3}\zeta^{\mathfrak{m}}(3,\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(0;0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;\bar{1}0100;1)+\big{(}I^{\mathfrak{l}}(\bar{1};010;0)+I^{\mathfrak{l}}(0;100;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;\bar{1}00\bar{1}0;1)$
$\displaystyle=$
$\displaystyle\,\zeta_{1}^{\mathfrak{l}}(\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\bar{2},3)+\big{(}\zeta^{\mathfrak{l}}(3)-\zeta_{1}^{\mathfrak{l}}(\bar{2})\big{)}{\otimes}\zeta^{\mathfrak{m}}(3,\bar{2})\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{5},$
$\displaystyle D_{5}\zeta^{\mathfrak{m}}(3,\bar{2},3)=$
$\displaystyle\,I^{\mathfrak{l}}(0;\bar{1}0100;1){\otimes}I^{\mathfrak{m}}(0;\bar{1}00;1)=\zeta^{\mathfrak{l}}(\bar{2},3){\otimes}\zeta^{\mathfrak{m}}(\bar{3})\in{\mathcal{L}}_{5}{\otimes}{\mathcal{H}}_{3}.$

The above computation shows that all the MES in Lemma 3.1 are unrmified when
$\ell=1$.

We now assume that $k\geq 2$ and that the MES in (3.15) are unramified for all
$\ell<k$.

###  3.4 The inductive step for $\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$

As the sequence $\rho(\\{\bar{2},3\\}_{k})$ has a period of 10, it is clear
that $D_{9}=0$ and the pattern of $D_{10+r}$ is basically the same as that of
$D_{r}$ for any $r>1$, with one more of $\\{\bar{2},3\\}_{2}$ in the
subsequence and one less of $\\{\bar{2},3\\}_{2}$ in the quotient. Thus it
suffices for us to show that $D_{r}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$
are stable derivations for $r=10n+3,10n+5,10n+7,10n+11$, meaning that all its
factors are of similar forms which are then unramified by induction.

####  3.4.1 $r=10n+3$

We have the following picture when $k$ is odd and we only need to modify the
left end by exchange 1 and $\bar{1}$ when $k$ is even.

$0;$$\bar{1},$$0,$$1,$$0,\ldots,$$0,$$1,$$0,$$\bar{1},$$0,$$\\{S_{\bar{1}},S_{1}\\}_{n}$$0,$$\bar{1},$$0,$$1,$$0,\ldots,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{10n+3}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$

We call a subsequence $S_{\bar{1}}=(0\bar{1}010)$ a $S_{\bar{1}}$-block, and a
subsequence $S_{1}=(010\bar{1}0)$ a $S_{1}$-block. For general $k$, the
sequence $\rho(\\{\bar{2},3\\}_{k})$ always starts with a $S_{(-1)^{k}}$-block
(here we identify $-1$ with $\bar{1}$) and ends with a $S_{\bar{1}}$-block
followed by $0,1$.

If a subsequence starts in a $S_{1}$-block (as shown in the picture above)
then there are only four possible nonzero cuts with the same quotient
sequence:

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(1;0,\bar{1},0,\\{S_{\bar{1}},S_{1}\\}_{n};0){\otimes}I^{\mathfrak{m}}(0;\bar{1},\dots,0;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;\bar{1},0,\\{S_{\bar{1}},S_{1}\\}_{n},0;\bar{1}){\otimes}I^{\mathfrak{m}}(0;\bar{1},\dots,0;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};0,\\{S_{\bar{1}},S_{1}\\}_{n},0,\bar{1};0){\otimes}I^{\mathfrak{m}}(0;\bar{1},\dots,0;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;\\{S_{\bar{1}},S_{1}\\}_{n},0,\bar{1},0;1){\otimes}I^{\mathfrak{m}}(0;\bar{1},\dots,0;1).$

By path reversal and homothety, we see immediately that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip
1.07639pt{}}={\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$.

If a subsequence starts in a $S_{\bar{1}}$-block, then the argument is
completely similar. We only need to exchange $1$ and $\bar{1}$ and therefore
the four terms still cancel each other.

However, when a subsequence starts in the final $S_{\bar{1}}$-block, we only
get two terms

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{1}$}}}\hskip
1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};0,1,0,\\{S_{1},S_{\bar{1}}\\}_{n};0){\otimes}I^{\mathfrak{m}}(0;\\{S_{\bar{1}},S_{1}\\}_{k/2-n},\bar{1},0;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{2}$}}}\hskip
1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;1,0,\\{S_{1},S_{\bar{1}}\\}_{n},0;1){\otimes}I^{\mathfrak{m}}(0;\\{S_{\bar{1}},S_{1}\\}_{k/2-n},\bar{1},0;1)$

when $k$ is odd and

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};0,1,0,\\{S_{1},S_{\bar{1}}\\}_{n};0){\otimes}I^{\mathfrak{m}}(0;S_{1},\\{S_{\bar{1}},S_{1}\\}_{k/2-n-1},\bar{1},0;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;1,0,\\{S_{1},S_{\bar{1}}\\}_{n},0;1){\otimes}I^{\mathfrak{m}}(0;S_{1},\\{S_{\bar{1}},S_{1}\\}_{k/2-n-1},\bar{1},0;1)$

when $k$ is even. In both cases

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,-\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n},\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2n-1},\bar{2})\in{\mathcal{L}}_{10n+3}{\otimes}{\mathcal{H}}_{5k-10n-3},$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,\zeta^{\mathfrak{l}}(\\{3,\bar{2}\\}_{2n},3){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2n-1},\bar{2})\in{\mathcal{L}}_{10n+3}{\otimes}{\mathcal{H}}_{5k-10n-3}$

by induction.

####  3.4.2 $r=10n+5$

We have the following picture when $k$ is odd and we only need to modify the
left end by exchange 1 and $\bar{1}$ when $k$ is even.

$0;$$\bar{1},$$0,$$1,$$0,\ldots,$$0,$$1,$$0,$$\bar{1},$$0,$$\\{S_{\bar{1}},S_{1}\\}_{n}$$0,$$\bar{1},$$0,$$1,$$0,\ldots,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
0$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{10n+5}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$

If a subsequence starts in a $S_{1}$-block as shown above then there are only
four possible nonzero terms with the same quotient sequence in
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}$:

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
0$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;1,0,\bar{1},0,\\{S_{\bar{1}},S_{1}\\}_{n},0;\bar{1}){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},0\bar{1}010,S_{1},\dots;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(1;0,\bar{1},0,\\{S_{\bar{1}},S_{1}\\}_{n},0,\bar{1};0){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},01010,S_{1},\dots;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;\bar{1},0,\\{S_{\bar{1}},S_{1}\\}_{n},0,\bar{1},0;1){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},01010,S_{1},\dots;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};0,\\{S_{\bar{1}},S_{1}\\}_{n},0,\bar{1},0,1;0){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},010\bar{1}0,S_{1},\dots;1).$

By path reversal and homothety, we see easily that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=0$. But we don’t see the immediate cancelation of
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 0$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$. However, all possible cuts of
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 0$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{0}$}}}\hskip
1.07639pt{}}\,$ together yield

$\zeta^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+1}){\otimes}\displaystyle\sum_{j=0}^{k-2n-1}\zeta^{\mathfrak{m}}(\\{2,\bar{3}\\}_{k-2n-1-j},\\{\bar{2},3\\}_{j})$

while all possible cuts of
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{3}$}}}\hskip
1.07639pt{}}\,$ together yield

$-\zeta^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+1}){\otimes}\displaystyle\sum_{j=0}^{k-2n-2}\zeta^{\mathfrak{m}}(\\{2,\bar{3}\\}_{k-2n-1-j},\\{\bar{2},3\\}_{j})$

by path reversal. Note that there is one extra term for
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 0$}}}\hskip
1.07639pt{}}\,$ when a subsequence ends at the last 1. Therefore

$D_{5}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})=\zeta^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2n-1})\in{\mathcal{L}}_{10n+5}{\otimes}{\mathcal{H}}_{5k-10n-5}.$

####  3.4.3 $r=10n+7$

We have the following picture when $k$ is odd.

$0;$$\bar{1},$$0,$$1,$$0,\ldots,$$0,$$1,$$0,$$\bar{1},$$0,$$\\{S_{\bar{1}},S_{1}\\}_{n}$$0,$$\bar{1},$$0,$$1,$$0,$$0,$$1,$$0,$$\bar{1},$$0,\ldots,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{10n+7}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$

If a subsequence starts in a $S_{1}$-block then there are only three possible
nonzero terms:

$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(0;10\bar{1}0,\\{S_{\bar{1}},S_{1}\\}_{n},0\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},0,1,0,S_{1},\dots;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(1;0\bar{1}0,\\{S_{\bar{1}},S_{1}\\}_{n},0\bar{1}01;0){\otimes}I^{\mathfrak{m}}(0;\dots,S_{\bar{1}},0,1,0,S_{1},\dots;1),$
$\displaystyle{{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}}=$
$\displaystyle\,I^{\mathfrak{l}}(\bar{1};0,\\{S_{\bar{1}},S_{1}\\}_{n},0\bar{1}0100;1){\otimes}I^{\mathfrak{m}}(0;\cdots;1).$

Then ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$ since its subsequence is anti-symmetric. The
quotient sequence in
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ are the same. By path reversal, we see immediately that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=0$.

If a subsequence starts in a $S_{\bar{1}}$-block, then the argument is
completely similar so that all the terms will be canceled.

In conclusion we see that
$D_{10n+7}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})=0$.

####  3.4.4 $r=10n+11$

We notice that any such cut would produce the same type of quotient sequence
with $n+1$ blocks of $(S_{1},S_{\bar{1}})$ missing. So the right factor is
unramified by induction. The left factor must have one of the following forms:

$\displaystyle
I^{\mathfrak{l}}(0;\\{S_{1},S_{\bar{1}}\\}_{n+1},0;1)=I^{\mathfrak{l}}(0;\\{S_{\bar{1}},S_{1}\\}_{n+1},0;\bar{1})=$
$\displaystyle\,\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+2})\in{\mathcal{L}}_{10n+11},$
$\displaystyle
I^{\mathfrak{l}}(1;0,\\{S_{1},S_{\bar{1}}\\}_{n+1};0)=I^{\mathfrak{l}}(\bar{1};0,\\{S_{\bar{1}},S_{1}\\}_{n+1};0)=$
$\displaystyle\,-\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+2})\in{\mathcal{L}}_{10n+11}.$

by inductive assumption.

Combining all the findings in §3.4.1-§3.4.4 we see that
$D_{1}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})=0$ and
$D_{r}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})\in{\mathcal{L}}_{r}{\otimes}{\mathcal{H}}_{5k-r}$
(i.e., $D_{r}$ is a stable derivation) for all odd $r<5k$. Consequently,
$\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$ is unramified by Theorem 2.1.

###  3.5 The inductive step for
$\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k},\bar{2})$

The computation of $D_{r}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell},\bar{2})$
can be carried out in a similar manner as that for
$\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell})$. We can attach $0;\bar{1}$ to
the right end of the picture in §3.4.1 and then exchange $\bar{1}$ and 1. By
the same argument there, $D_{10n+r}$ are all stable for $r=3,5,7,9,11$. For
brevity, we only show the computation when $r=7$ in which case the induction
assumption for MES $\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell})$ is
essentially used. In this case, all the computation will be the same as that
of $D_{7}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{\ell})$ except for the only one
extra possible cut starting at the beginning of the last $S_{\bar{1}}$-block
for which we have the following picture.

$\dots,$$0,$$1,$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{1}$}}}\hskip
0.75346pt{}}$

which produces

${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\text{$\bar{1}$}$}}}\hskip
1.07639pt{}}=I^{\mathfrak{l}}(0;10\bar{1}00\bar{1}0;1){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-1})=\zeta^{\mathfrak{l}}(\bar{2},3,\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-1})\in{\mathcal{L}}_{7}{\otimes}{\mathcal{H}}_{5k-5}$

by the initial step and the inductive assumption.

By Theorem 2.1 we have now proved that both families of MES in (3.15) are
unramified when $\ell=k$.

###  3.6 The inductive step for $\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$
and $\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$

We now prove that the pair of families in (3.17) are both unramified. It is
mostly the same as that of (3.15) except that in the inductive step we need to
use the validity of unramification of (3.15).

When $r=3$ (or similarly for $r=10n+3<5k$) we can modify the picture in 3.4.1
at the two ends if $k$ is odd. We just add $0;\bar{1}0$ to the left end and
remove $0,0;1$ from the right end (no right-end change needed for
$\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$). So we consider the following
picture for even $k>1$:

$0;$$\bar{1},$$0,$$0,$$\bar{1},$$0,$$1,$$0,$$0,$$1,$$0,$$\bar{1},$$0,\ldots,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1\phantom{,}$
for
$\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$$0,$$\bar{1},$$0;$$1\phantom{,}$
for
$\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{2}$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{3}$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\bar{4}$}}}\hskip
0.75346pt{}}$

Similar to what we have shown previously we see easily that
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{2}$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{3}$}}}\hskip
1.07639pt{}}=0$. But
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{4}$}}}\hskip
1.07639pt{}}\,$ from the first block cannot be canceled by the missing
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{1}$}}}\hskip
1.07639pt{}}$ in front of it. All the other cuts in
$D_{3}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$ will either cancel each other
or vanish themselves (see §3.4.1). For general $D_{10n+3}$ the argument is the
same and therefore

$D_{10n+3}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})=\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n},\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2n-1},\bar{2})\in{\mathcal{L}}_{10n+3}{\otimes}{\mathcal{H}}_{5k-10n-3}$

by inductive assumption. Similarly, we see that

$D_{10n+3}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)=\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n},\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2n})+\zeta^{\mathfrak{l}}(\\{3,\bar{2}\\}_{2n},3){\otimes}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k-2n})$

also lies in ${\mathcal{L}}_{10n+3}{\otimes}{\mathcal{H}}_{5k-10n}$ by
induction, where the second term comes from the cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ ending at the last 1.

Applying the same idea we find that all of $D_{10n+5}$, $D_{10n+7}$ and
$D_{10n+11}$ are stable derivations. By Theorem 2.1 we know
$\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$ and
$\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$ are both unramified. We leave
the details to the interested reader.

###  3.7 The inductive step for
$\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$ and
$\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$

We first consider the case $r=10n+9<5k$. There is only one cut in $D_{r}$ that
is nonzero.

$0;$$0,$$\bar{1},$$0,$$0,$$\bar{1},$$0,$$1,$$0,$$\\{S_{1},S_{\bar{1}}\\}_{n},$$0,$$1,$$0,$$\bar{1},$$0,\ldots,$$0,$$\bar{1},$$0,$$1,$$0,$$0;$$1\phantom{,}$
for
$\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)$$0,$$\bar{1},$$0;$$1\phantom{,}$
for
$\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
9$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-4.52083pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1\\!1$}}}\hskip 0.75346pt{}}$

This occurs when the subsequence starts at the very beginning which produces
the terms

$\displaystyle
D_{r}\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})={\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
9$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\zeta_{1}(\\{3,\bar{2}\\}_{2n+1},3){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2-2n},\bar{2})\in{\mathcal{L}}_{10n+9}{\otimes}{\mathcal{H}}_{5k-10n-8},$
$\displaystyle
D_{r}\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)={\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
9$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\zeta_{1}(\\{3,\bar{2}\\}_{2n+1},3){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-1-2n})\in{\mathcal{L}}_{10n+9}{\otimes}{\mathcal{H}}_{5k-10n-5}$

by induction and the proceeding section. Similarly, if $r=10n+11<5k$ then the
only nontrivial cut in $D_{r}$ occurs when the subsequence starts at the very
beginning, which produces the terms

$\displaystyle
D_{r}\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})={\bigcirc\hskip-8.61108pt{\raisebox{0.0pt}{\text{$\scriptstyle
11$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\zeta_{1}(\\{3,\bar{2}\\}_{2n+2}){\otimes}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k-2-2n})\in{\mathcal{L}}_{10n+11}{\otimes}{\mathcal{H}}_{5k-10n-10},$
$\displaystyle
D_{r}\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k},3)={\bigcirc\hskip-8.61108pt{\raisebox{0.0pt}{\text{$\scriptstyle
11$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\zeta_{1}(\\{3,\bar{2}\\}_{2n+2}){\otimes}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k-2-2n},3)\in{\mathcal{L}}_{10n+11}{\otimes}{\mathcal{H}}_{5k-10n-7}$

by induction and the proceeding section. For all other odd $r$’s the proof is
essentially the same as the starting $0$ does not contribute to $D_{r}$. By
Theorem 2.1 we see that $\zeta_{1}^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k})$ is
unramified.

###  3.8 The inductive step for
$\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$ and
$\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{k},\bar{2})$

We now consider the pair of families of MES in (3.16). We first consider the
case $r=10n+3$. Adding one 0 to the left of the picture in §3.4.1 we see that
there is only one extra cut in $D_{r}$ compared to the (3.17) case when the
subsequence starts at the very beginning. This yields the terms

${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\bar{4}$}}}\hskip
1.07639pt{}}=\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n},\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k-1-2n},3)\quad\text{or}\quad\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n},\bar{2}){\otimes}\zeta^{\mathfrak{m}}(\\{3,\bar{2}\\}_{k-2n})$

By induction and the proceeding section (if $n=0$) we see that all the factors
are unramified. Similarly, if $r=10n+11$ then the only nontrivial cut in
$D_{r}$ occurs when the subsequence starts at the very beginning which
produces the terms

$\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2-2n})\quad\text{or}\quad\zeta_{1}^{\mathfrak{l}}(\\{\bar{2},3\\}_{2n+2}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k-2-2n},\bar{2}).$

Again, all the factors are unramified by induction. For other cuts when
$r=10n+3$ as well as for other odd $r$’s the proof is essentially the same as
that for $\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$ and
$\zeta^{\mathfrak{m}}(\\{\bar{2},3\\}_{k},\bar{2})$ since the starting $0$
does not contribute to $D_{r}$. By Theorem 2.1 we see that both
$\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{k})$ and
$\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},3\\}_{k},\bar{2})$ are unramified.

We have now completed the proof of Lemma 3.1.

##  4 Proof of Theorem 1.3

We first handle another set of base cases for arbitrary $a$.

######  Lemma 4.1.

For any integer $a\geq 0$ the MES

$\zeta_{a}^{\mathfrak{m}}(\bar{2}),\quad\zeta_{a}^{\mathfrak{m}}(\bar{2},3),\quad\zeta_{a}^{\mathfrak{m}}(\bar{2},3,\bar{2}),\quad\zeta_{a}^{\mathfrak{m}}(3,\bar{2}),\quad\zeta_{a}^{\mathfrak{m}}(3,\bar{2},3)$

are all unramified.

###### Proof.

The proof is again a standard application of Theorem 2.1 using the stable
derivations $D_{r}$. We leave the details to the interested reader. ∎

We now use induction to prove Theorem 1.3. Assume $a\geq 2$, $\ell\geq 2$ and
Theorem 1.3 holds if $a$ or $k$ or both are replaced by any smaller
nonnegative integers.

For any odd integer $r\geq 3$ if a cut of $D_{r}$ starting after the first
nonzero $\pm 1$ is not canceled by another cut nor does it vanish by itself
then by the proof of Lemma 3.1 the left factor must have the form treated in
Lemma 3.1 and the right factor one of the forms in Theorem 1.3 but with
smaller $\ell$. By induction assumption $D_{r}$ must be stable.

If a cut starts somewhere within the first block of $a+1$ 0’s then its left
factor must have one of the forms in Theorem 1.3 but with $\ell<k$ and its
right factor must have the form treated in Lemma 3.1. Thus $D_{r}$ must be
stable by induction assumption again. This completes the proof of the theorem.

##  5 Proof of (1.3) and (1.4) in Theorem 1.1

We now show that for all $\ell\in\mathbb{N}$, there are some rational numbers
$c_{\ell}$ and $d_{\ell}$ such that

$\displaystyle
2^{3\ell+1}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell},\bar{2})=$
$\displaystyle\,c_{\ell}\zeta^{\mathfrak{m}}(3\ell+2)-\displaystyle\sum_{{\alpha}+{\beta}=\ell}(-1)^{\alpha}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),$
(5.19) $\displaystyle 2^{3\ell}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{\ell})=$
$\displaystyle\,d_{\ell}\zeta^{\mathfrak{m}}(3\ell)-\zeta^{\mathfrak{m}}(3_{\ell})+2\displaystyle\sum_{{\alpha}+{\beta}=\ell-1}(-1)^{a}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),$
(5.20)

by induction on $\ell$.

###  5.1 Proof of base cases of (5.19) and (5.20)

If $\ell=1$ then we see easily that $D_{1}=0$ for all MES involved by Lemma
2.2. Then by the regularized stuffle relation of Euler sums

$\zeta_{*}(\bar{2},1)=-\zeta(1,\bar{2})-\zeta(\bar{3})=-\displaystyle\frac{1}{8}\zeta(3)+\displaystyle\frac{3}{4}\zeta(3)=\displaystyle\frac{5}{8}\zeta(3)$

using Prop. 2.3 and (1.2). Thus (5.20) follows from Theorem 2.1 when $\ell=1$:

$8\zeta^{\mathfrak{m}}(\bar{2},1)=-\zeta^{\mathfrak{m}}(3)+2\zeta_{1}^{\mathfrak{m}}(2)=5\zeta^{\mathfrak{m}}(3).$
(5.21)

Using the pictures (5.22)

$0;$$1,$$0,$$\bar{1},$$\bar{1},$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(\bar{2},1,\bar{2})$
$0;$$1,$$0,$$0,$$1,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(3,2)$
$0;$$1,$$0,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(2,3)$
$0,$$\bar{1},$$\bar{1},$$0,$$1;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(1,\bar{2},1)$
$0,$$1,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(1,3)$ (5.22)

together with the fact that
$\zeta^{\mathfrak{m}}(\bar{2})=-\zeta^{\mathfrak{m}}(2)/2$ by Prop. 2.3 and
(5.21), we have

$\displaystyle D_{3}\zeta^{\mathfrak{m}}(\bar{2},1,\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;10\bar{1};\bar{1}){\otimes}I^{\mathfrak{m}}(0;\bar{1}0;1)=-\displaystyle\frac{1}{2}\zeta^{\mathfrak{l}}(\bar{2},1){\otimes}\zeta^{\mathfrak{m}}(2)=-\displaystyle\frac{5}{16}\zeta^{\mathfrak{l}}(3){\otimes}\zeta^{\mathfrak{m}}(2),$
$\displaystyle D_{3}\zeta^{\mathfrak{m}}(3,2)=$
$\displaystyle\,I^{\mathfrak{l}}(0;010;1){\otimes}I^{\mathfrak{m}}(0;10;1)=\zeta_{1}^{\mathfrak{l}}(2){\otimes}\zeta^{\mathfrak{m}}(2)=-2\zeta^{\mathfrak{l}}(3){\otimes}\zeta^{\mathfrak{m}}(2),$
$\displaystyle D_{3}\zeta^{\mathfrak{m}}(2,3)=$
$\displaystyle\,\big{(}I^{\mathfrak{l}}(1;010;0)+I^{\mathfrak{l}}(0;100;1)\big{)}{\otimes}I^{\mathfrak{m}}(0;10;1)=3\zeta^{\mathfrak{l}}(3){\otimes}\zeta^{\mathfrak{m}}(2).$

Hence

$16D_{3}\zeta^{\mathfrak{m}}(\bar{2},1,\bar{2})=D_{3}\big{(}\zeta^{\mathfrak{m}}(3,2)-\zeta^{\mathfrak{m}}(2,3)\big{)}.$

For (6.32) it is easy to see from the last two pictures in (5.22) that

$D_{r}\zeta^{\mathfrak{m}}(1,\bar{2},1)=D_{r}\zeta^{\mathfrak{m}}(1,3)=0\quad\text{for
}r=1,3.$

Hence, the $\ell=1$ case of (5.19) and (5.20) follows from Theorem 2.1.

We now assume (5.19) and (5.20) holds for all positive integers $\ell<k$.

###  5.2 Inductive proof of (5.19)

We first show that

$2^{3\ell+1}D_{r}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k},\bar{2})=-\displaystyle\sum_{{\alpha}+{\beta}=k}(-1)^{\alpha}D_{r}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
(5.23)

for all positive integers $r=6n+3,6n+5,6n+7<3k+2$.

####  5.2.1 $r=6n+3$

For $r=6n+3$ we have the following picture when $k$ is even (the case of odd
$k$ can be dealt with similarly).

$0;$$\bar{1},$$0,$$1,$$1,$$0,$$\bar{1},$$\bar{1},$$0,$$\cdots\\!,$$1,$$1,$$0,$$\\{\bar{1}\bar{1}0110\\}_{n}$$\bar{1},$$\bar{1},$$0,$$1,$$1,$$0,\cdots\\!,$$\bar{1},$$\bar{1},$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k},\bar{2})$ (5.24)

Note that in general
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ is antisymmetric so that they all disappear, except for the
following special case that cannot be canceled: a type
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ cut starts at the initial $0$. Thus

$\displaystyle D_{6n+3}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k},\bar{2})=$
$\displaystyle\,I^{\mathfrak{l}}(0;10\\{\bar{1}\bar{1}0110\\}_{n}\bar{1};\bar{1}){\otimes}I^{\mathfrak{m}}(0;10\bar{1}\bar{1}0\\{110\bar{1}\bar{1}0\\}_{k/2-n-1};1)$
$\displaystyle=$
$\displaystyle\,\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k-2n-1},\bar{2}).$
(5.25)

We note that there are two kinds of cuts for
$D_{r}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$: (i) those that don’t
involve the component $\rho(2)$ and (ii) those that do.

$0;$$1,$$0,$$0,$$1,$$0,$$0,$$\cdots\\!,$$1,$$0,$$0,$$\\{100100\\}_{n}$$1,$$0,$$0,$$1,$$0,$$0,\cdots\\!,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Type (i) cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$ (5.26)

Consider the picture above for cuts of type (i). Then
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ is always canceled by
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ except when
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ ends at the last 1. Type (i) has only one nontrivial
contribution

$\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},2,3_{{\beta}}).$
(5.27)

For type (ii) cuts shown below

$0;$$1,$$0,$$0,$$1,$$0,$$0,$$\cdots\\!,$$1,$$0,$$0,$$\\{1,0,0\\}_{i},$$1,$$0,$$\\{1,0,0\\}_{j},$$1,$$0,$$0,\cdots\\!,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip 0.75346pt{}}$Type (ii) cuts
$D_{6n+3}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),i+j=2n$, starting
before $\rho(2)$ (5.28)

we see that ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ contribute to

$\displaystyle\sum_{i+j=2n}\big{(}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})-\zeta_{1}^{\mathfrak{l}}(3_{j-1},2,3_{i+1})\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-i-1},2,3_{{\beta}-j}).$

When $j=0$ the second term inside the left factor means

$-\zeta_{1}^{\mathfrak{l}}(3_{-1},2,3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},2,3_{{\beta}}):=-\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},2,3_{{\beta}}),$

which is canceled by (5.27). There are two more special cases when a type (ii)
cut starts in $\rho(2)$ (consider $j=2n+1$ in (5.28))

${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip
1.07639pt{}}=\big{(}\zeta^{\mathfrak{l}}(3_{2n+1})-\zeta_{1}^{\mathfrak{l}}(3_{2n},2)\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}-2n-1}).$

Putting all the above together we obtain

$\displaystyle D_{6n+3}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})=$
$\displaystyle\,\displaystyle\sum_{\begin{subarray}{c}i+j=2n\\\ a-1\geq i\geq
0,b\geq j\geq
1\end{subarray}}\big{(}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})-\zeta_{1}^{\mathfrak{l}}(3_{j-1},2,3_{i+1})\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-i-1},2,3_{{\beta}-j})$
$\displaystyle\,+{\delta}_{{\alpha}\geq
2n+1}\zeta_{1}^{\mathfrak{l}}(3_{2n},2){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},2,3_{{\beta}})$
$\displaystyle\,+{\delta}_{{\beta}\geq
2n+1}\big{(}\zeta^{\mathfrak{l}}(3_{2n+1})-\zeta_{1}^{\mathfrak{l}}(3_{2n},2)\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}-2n-1}).$

Therefore

$\displaystyle\,\displaystyle\sum_{{\alpha}+{\beta}=k}(-1)^{\alpha}D_{6n+3}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\begin{subarray}{c}{\alpha}+{\beta}=k,i+j=2n\\\
{\alpha}-1\geq i\geq 0,{\beta}\geq j\geq
1\end{subarray}}(-1)^{\alpha}\big{(}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})-\zeta_{1}^{\mathfrak{l}}(3_{j-1},2,3_{i+1})\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-i-1},2,3_{{\beta}-j})$
$\displaystyle\,+\displaystyle\sum_{{\alpha}+{\beta}=k}(-1)^{\alpha}{\delta}_{{\alpha}\geq
2n+1}\zeta_{1}^{\mathfrak{l}}(3_{2n},2){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},2,3_{{\beta}})$
$\displaystyle\,+\displaystyle\sum_{{\alpha}+{\beta}=k}(-1)^{\alpha}{\delta}_{{\beta}\geq
2n+1}\big{(}\zeta^{\mathfrak{l}}(3_{2n+1})-\zeta_{1}^{\mathfrak{l}}(3_{2n},2)\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}-2n-1})$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\begin{subarray}{c}i+j=2n,i\geq 0,j\geq
1\\\
{\alpha}+{\beta}=k-2n-1\end{subarray}}(-1)^{a+i+1}\big{(}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})-\zeta_{1}^{\mathfrak{l}}(3_{j-1},2,3_{i})\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle\,+\displaystyle\sum_{{\alpha}+{\beta}=k-2n-1}(-1)^{\alpha}\big{(}\zeta^{\mathfrak{l}}(3_{2n+1})-2\zeta_{1}^{\mathfrak{l}}(3_{2n},2)\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,-2\displaystyle\sum_{\begin{subarray}{c}i+j=2n,i\geq 0,j\geq
1\\\
{\alpha}+{\beta}=k-2n-1\end{subarray}}(-1)^{a+i}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j}){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle\,+\displaystyle\sum_{{\alpha}+{\beta}=k-2n-1}(-1)^{\alpha}\big{(}\zeta^{\mathfrak{l}}(3_{2n+1})-2\zeta_{1}^{\mathfrak{l}}(3_{2n},2)\big{)}{\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,\left(\zeta^{\mathfrak{l}}(3_{2n+1})-2\displaystyle\sum_{i+j=2n}(-1)^{i}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})\right){\otimes}\displaystyle\sum_{{\alpha}+{\beta}=k-2n-1}(-1)^{\alpha}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k-2n-1},\bar{2})$

by inductive assumption. By comparing with (5.25) we see that (5.23) holds for
$r=6n+3$.

####  5.2.2 $r=6n+5$

It is obvious that if $r=6n+5$ then the left-hand side of (5.23) vanishes
since $\rho(\\{\bar{2},1\\}_{k},\bar{2})$ has period 6 so that every cut of
$D_{6n+5}$ starts and ends with the same number.

Similar to the case $r=6n+3$, there are two kinds of cuts for
$D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$: (i) those involve
the component $\rho(2)$ and (ii) those don’t. All cuts in (ii) together with
the cut starting at the 0 in $\rho(2)$ clearly vanish by periodicity. For type
(ii) cuts we consider the following picture

$0;$$1,$$0,$$0,$$1,$$0,$$0,$$\cdots\\!,$$1,$$0,$$0,$$\\{1,0,0\\}_{i},$$1,$$0,$$\\{1,0,0\\}_{j},$$1,$$0,$$0,\cdots\\!,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Type (i) cuts of
$D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}}),i+j=2n+1$, starting
before 0 in $\rho(2)=10$

Then we see immediately that cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ with index $(i,j)$ shown as in the picture is canceled by the
cut ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}\,$ with index $(j,i)$ by path reversal. Note that we
allow ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}\,$ to start at the number 1 in $\rho(2)=10$ but not
after.

To summarize, (5.23) holds (actually vanishes on both sides) for $r=6n+5$.

####  5.2.3 $r=6n+7$

By periodicity, each cut of $D_{6n+7}$ straddling over $\rho(2)$ must start
and end with the same number and must vanish. For all the other cut we may use
the pictures (3.13) and (3.14) to conclude that all terms in (5.20) vanish
under the derivation $D_{6n+7}$.

In summary, the equation (5.23) holds for all odd $r<3\ell$. This completes
our inductive proof of (5.20) by Theorem 2.1.

###  5.3 Inductive proof of (5.20)

We now prove that

$2^{3\ell}D_{r}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k})=-D_{r}\zeta^{\mathfrak{m}}(3_{k})+2\displaystyle\sum_{{\alpha}+{\beta}=k-1}(-1)^{a}D_{r}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
(5.29)

for all positive integers $r=6n+3,6n+5,6n+7<3k$.

####  5.3.1 $r=6n+3$

For $r=6n+3$ we get the following picture when $k$ is even (the case of odd
$k$ can be dealt with similarly) by inserting another 1 to the left of the
last 1 and removing the $\bar{1}01$ to the right of the initial 0 in picture
(5.24):

$0;$$1,$$0,$$\bar{1},$$\bar{1},$$0,$$\cdots\\!,$$1,$$1,$$0,$$\\{\bar{1}\bar{1}0110\\}_{n}$$\bar{1},$$\bar{1},$$0,$$1,$$1,$$0,\cdots\\!,$$\bar{1},$$\bar{1},$$0,$$1;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k})$ (5.30)

Using the same argument as for (5.24) we get

$D_{6n+3}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k})=\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{\bar{2},1\\}_{k-2n-1}).$
(5.31)

On the other hand, by (3.12)

$D_{6n+3}\zeta^{\mathfrak{m}}(3_{k})=\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1}).$

Further, for any fixed ${\alpha}+{\beta}=k-1$,
$D_{6n+3}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$ can be computed
similarly as in §5.2.1 according to whether the cuts involve the component
$\rho(2)$ or not. By adding one more 0 at the beginning of pictures (5.26) and
(5.28) we find that there will be one more term of cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ starting at the very first 0 giving rise to the additional
term

$\zeta_{1}^{\mathfrak{l}}(3_{{\alpha}},2,3_{2n-a}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1}).$

Therefore, by induction assumption.

$\displaystyle\,\displaystyle\sum_{{\alpha}+{\beta}=k-1}(-1)^{\alpha}D_{6n+3}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,\left(\zeta^{\mathfrak{l}}(3_{2n+1})-2\displaystyle\sum_{i+j=2n}(-1)^{i}\zeta_{1}^{\mathfrak{l}}(3_{i},2,3_{j})\right){\otimes}\displaystyle\sum_{{\alpha}+{\beta}=k-2n-2}(-1)^{\alpha}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle\,+\displaystyle\sum_{{\alpha}+{\beta}=2n}(-1)^{\alpha}\zeta_{1}^{\mathfrak{l}}(3_{{\alpha}},2,3_{{\beta}}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1})$
$\displaystyle=$
$\displaystyle\,-8^{2n+1}\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\displaystyle\sum_{{\alpha}+{\beta}=k-2n-2}(-1)^{\alpha}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})+\displaystyle\sum_{{\alpha}+{\beta}=2n}(-1)^{\alpha}\zeta_{1}^{\mathfrak{l}}(3_{{\alpha}},2,3_{{\beta}}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1}).$

Thus by induction again

$\displaystyle\,-D_{6n+3}\zeta^{\mathfrak{m}}(3_{k})+2\displaystyle\sum_{{\alpha}+{\beta}=k-1}(-1)^{\alpha}D_{6n+3}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})$
$\displaystyle=$
$\displaystyle\,8^{2n+1}\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\left(-2\displaystyle\sum_{{\alpha}+{\beta}=k-2n-2}(-1)^{\alpha}\zeta_{1}^{\mathfrak{m}}(3_{{\alpha}},2,3_{{\beta}})\right)+8^{2n+1}\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{k-2n-1})$
$\displaystyle=$
$\displaystyle\,8^{k}\zeta^{\mathfrak{l}}(\\{\bar{2},1\\}_{2n+1}){\otimes}\zeta_{1}^{\mathfrak{m}}(\\{\bar{2},1\\}_{k-2n-1}).$

Comparing to (5.31) we see immediately that (5.29) holds for $r=6n+3$.

####  5.3.2 $r=6n+5$

Exactly the same argument as in §5.2.2 works almost word-for-word, implying
that both sides of (5.29) vanish if $r=6n+5$.

####  5.3.3 $r=6n+7$

By modifying the pictures (3.13) and (3.14) we find that (5.29) holds for
$r=6n+7$.

Combining all the results in §5.3.1-§5.3.3 we see that (5.20) holds by Theorem
2.1.

##  6 Proof of (1.5) in Theorem 1.1

We finally show that for all $\ell\in\mathbb{N}$, there is some rational
number $e_{\ell}$ such that

$2^{3\ell-1}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{\ell},1)=e_{\ell}\zeta^{\mathfrak{m}}(3\ell+1)-3\displaystyle\sum_{{\alpha}+{\beta}=\ell,2\nmid{\beta}}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$
(6.32)

by using induction on $\ell$.

###  6.1 The base case

If $\ell=1$ then clearly
$D_{1}\zeta^{\mathfrak{m}}(1,\bar{2},1)=D_{1}\zeta^{\mathfrak{m}}(1,3)=0$ by
Lemma 2.2. To consider the action of the derivation $D_{3}$ on (6.32) we may
look at the following pictures:

$0;$$\bar{1},$$\bar{1},$$0,$$1;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$$D_{3}\zeta^{\mathfrak{m}}(1,\bar{2},1)=0$$0;$$1,$$1,$$0,$$0;$$1\phantom{,}$$D_{3}\zeta^{\mathfrak{m}}(1,3)=0$

We find that $D_{3}\zeta^{\mathfrak{m}}(1,3)=0$ since there is no nonzero cut,
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}=0$ by anti-symmetry and

$D_{3}\zeta^{\mathfrak{m}}(1,\bar{2},1)={\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip
1.07639pt{}}=I^{\mathfrak{l}}(0;\bar{1}\bar{1}0;1){\otimes}I^{\mathfrak{m}}(0;1;1)=0.$

Hence by Theorem 2.1 there is some $e_{1}\in\mathbb{Q}$ such that
$4\zeta^{\mathfrak{m}}(1,\bar{2},1)=e_{1}\zeta^{\mathfrak{m}}(4)-3\zeta^{\mathfrak{m}}(1,3)$
and thus (6.32) holds for $\ell=1$.

In fact, by the table of values contained in [15, Appendix E.1],
$\zeta(1,3)=-16A-4B$ where $A=\zeta(1,1,\bar{2})$ and $B=\zeta(2,\bar{2})$.
Further, the stuffle regularized value

$\zeta_{*}(1,\bar{2},1)=-2\zeta(1,1,\bar{2})-\zeta(2,\bar{2})-\zeta(1,\bar{3})=-12A-3B.$

Hence

$4\zeta_{*}(1,\bar{2},1)=-3\zeta(1,3).$ (6.33)

By applying the period and using (6.33) we get $e_{1}=0$ .

###  6.2 The inductive step

We now assume that (6.32) holds for $\ell<k$ for some $k\geq 2$. We will now
prove

$2^{3\ell-1}D_{r}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k},1)=-3\displaystyle\sum_{{\alpha}+{\beta}=k,2\nmid{\beta}}D_{r}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$
(6.34)

for all positive integers $r=6n+3,6n+5,6n+7<3k$.

####  6.2.1 $r=6n+3$

For $r=6n+3$ we get the following picture when $k$ is even (the case of odd
$k$ can be dealt with similarly) by inserting another 1 to the right of the
initial 0 in picture (5.30):

$0;$$1,$$1,$$0,$$\bar{1},$$\bar{1},$$0,$$\cdots\\!,$$1,$$1,$$0,$$\\{\bar{1}\bar{1}0110\\}_{n}$$\bar{1},$$\bar{1},$$0,$$1,$$1,$$0,\cdots\\!,$$\bar{1},$$\bar{1},$$0,$$1;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{6n+3}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k},1)$ (6.35)

By the same argument as for (5.24)
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}=0$ by anti-symmetry and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$ by path reversal, except for the cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ starting at the initial 0. Hence

$D_{6n+3}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k},1)=\zeta^{\mathfrak{l}}(\\{1,\bar{2}\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k-2n-1},1).$
(6.36)

We note that there are two kinds of cuts for
$D_{r}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$: (i) those that don’t
involve the component $\rho(1)$ and (ii) those that do. By periodicity, each
type (ii) cut starts and ends with the same number so that it vanishes.
Further, as shown in (5.26) type (i) has only one nontrivial contribution

$\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},1,3_{{\beta}}).$
(6.37)

Thus

$\displaystyle-3\displaystyle\sum_{{\alpha}+{\beta}=k,2\nmid{\beta}}D_{6n+3}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})=$
$\displaystyle\,\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\left(-3\displaystyle\sum_{{\alpha}+{\beta}=k,2\nmid{\beta}}\zeta^{\mathfrak{m}}(3_{{\alpha}-2n-1},1,3_{{\beta}})\right)$
$\displaystyle=$
$\displaystyle\,2^{3(k-2n-1)-1}\zeta^{\mathfrak{l}}(3_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k-2n-1},1)$
$\displaystyle=$
$\displaystyle\,2^{3k-1}\zeta^{\mathfrak{l}}(\\{1,\bar{2}\\}_{2n+1}){\otimes}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k-2n-1},1)$

by inductive assumption and (1.2). Comparing to (6.36) we see that (6.34)
holds for $r=6n+3$.

####  6.2.2 $r=6n+5$

It is obvious that the left-hand side of (6.34) vanishes for $r=6n+5$ since
$\rho(\\{1,\bar{2}\\}_{k},1)$ have period 6 so that every cut of $D_{6n+5}$
starts and ends with the same number.

Similar to the case $r=6n+3$, there are two kinds of cuts for
$D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$: (i) those involve
the component $\rho(1)$ and (ii) those don’t. All cuts in (ii) clearly vanish
by periodicity. For type (ii) cuts we consider the following picture:

$0;$$1,$$0,$$0\cdots,$$1,$$0,$$0,$$1,$$0,$$0,$$\\{1,0,0\\}_{i},$$1,$$\\{1,0,0\\}_{j},$$1,$$0,$$0,$$1,$$0,$$0,\cdots\\!,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip 0.75346pt{}}$Type (ii) cuts of
$D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}}),i+j=2n$, starting
before/at $\rho(1)$

We see immediately that cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ with index $(i,j)$ shown as in the picture is canceled by the
cut ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}\,$ with index $(j,i)$ by path reversal. The two
special cases denoted by
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}\,$ with $i=j=2n$ cancel each other by path reversal. Thus both
sides of (6.34) vanish if $r=6n+5$.

####  6.2.3 $r=6n+7$

By attaching one more 1 to the right end of the picture (3.13) and (3.14)
exactly the same argument shows that
$D_{6n+7}\zeta^{\mathfrak{m}}(\\{1,\bar{2}\\}_{k},1)=0$ and all cuts for
$D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$ that do not involve
$\rho(1)$ cancel each other.

For cuts of $D_{6n+5}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}})$ that
involve the component $\rho(1)$ we consider the following picture:

$0;1,0,0,\cdots\\!,1,0,0,$$1,$$0,$$0,$$1,$$0,$$0,$$\\{1,0,0\\}_{i},$$1,$$\\{1,0,0\\}_{j},$$1,$$0,$$0,$$1,$$0,$$0,$$1,$$0,$$0,\cdots\\!,$$1,$$0,$$0;$$1\phantom{,}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
4$}}}\hskip 0.75346pt{}}$Type (ii) cuts of
$D_{6n+7}\zeta^{\mathfrak{m}}(3_{{\alpha}},1,3_{{\beta}}),i+j=2n$, starting
before/at $\rho(1)$

We see immediately that cut
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,$ with index $(i,j)$ shown as in the picture is canceled by the
cut ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}\,$ with index $(j,i)$ by path reversal. The two
special cases denoted by
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}\,$ with $i=j=2n$ cancel each other by path reversal. Thus both
sides of (6.34) vanish if $r=6n+7$.

Combining all the results in §6.2.1-§6.2.3 we see that (6.32) holds for
$\ell=k$ which is the last statement in Theorem 1.1. Therefore we have
completed the proof of Theorem 1.1.

Acknowledgments. Ce Xu is supported by the National Natural Science Foundation
of China (Grant No. 12101008), the Natural Science Foundation of Anhui
Province (Grant No. 2108085QA01) and the University Natural Science Research
Project of Anhui Province (Grant No. KJ2020A0057). Jianqiang Zhao is supported
by the Jacobs Prize from The Bishop’s School. Both authors would like to thank
Prof. F. Xu at the Capital Normal University and Prof. C. Bai at the Chern
Institute of Mathematics for their warm hospitality.

Declaration of interest. The authors have no competing interests to declare.

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