# Ramified and Unramified Motivic Multiple $t$-, $T$\- and $S$-Values

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. In this paper we shall consider a few variants of the motivic
multiple zeta values of level two by restricting the summation indices in the
definition of multiple zeta values to some fixed parity patterns. These
include Hoffman’s multiple $t$-values, Kaneko and Tsumura’s multiple
$T$-values, and the multiple $S$-values studied previously by the authors. By
applying Brown and Glanois’s descent theory on the motivic versions of these
values we shall derive some criterion for when these values are ramified and
unramified. Assuming Grothendieck’s period conjecture, our results partially
confirms a conjecture of Kaneko and Tsumura about when multiple $T$-values can
be expressed as a rational linear combination of multiple zeta values (i.e.,
unramified) if their depth is less than four. Similar results are obtained for
motivic multiple $S$-values. Further, we are able to generalize a result of
Charlton to more families of unramified multiple $t$-values with unit
components (i.e. components equal to 1). We propose some more unsolved
problems at the end of the paper.

Keywords: motivic iterated integrals, (motivic) multiple zeta values,
(motivic) Euler sums, (motivic) multiple $T$-values, (motivic) multiple
$t$-values, (motivic) multiple $S$-values.

AMS Subject Classifications (2020): 11M32, 11M99, 33B15, 33B30.

###### Contents

  1. 1 Introduction
  2. 2 Notation and the motivic setup
  3. 3 Some preliminary results
  4. 4 Reduction formula for double Euler sums of odd weight
  5. 5 Motivic double $t$-, $T$\- and $S$-values
  6. 6 Unramified triple $S$-values
  7. 7 Motivic triple $T$\- and $S$-values: three special cases
  8. 8 Triple $T$\- and $S$-values: the general case
  9. 9 Ramified and unramified MtVs
  10. 10 Concluding remarks

##  1 Introduction

Let $\mathbb{N}$ be the set of positive integers and
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$. For any $d\in\mathbb{N}$ and
$(k_{1},\dots,k_{d})\in\mathbb{N}^{d}$ we define the _multiple zeta values_
(MZVs) by

$\zeta(k_{1},\dots,k_{d}):=\displaystyle\sum_{0<n_{1}<\cdots<n_{d}}\displaystyle\frac{1}{n_{1}^{k_{1}}\cdots
n_{d}^{k_{d}}}.$

We need to impose the condition $k_{d}\geq 2$ to guarantee convergence and we
call the composition ${\boldsymbol{\sl{k}}}:=(k_{1},\dots,k_{d})$ _admissible_
in this case. Conventionally, $d$ is called the _depth_ and
$|{\boldsymbol{\sl{k}}}|:=k_{1}+\cdots+k_{d}$ is called the _weight_.

MZVs and their higher level generalizations have played significant roles in
both mathematics and physics in recent years due to their far-reaching
connections with many areas such as algebra geometry, number theory, knot
theory and theoretical physics (see e.g. [3, 5, 7, 13, 17]). At level two, we
obtain the so-called _Euler sums_ (also called alternating MZVs) by
introducing alternating signs on the numerators: for any
${\varepsilon}_{1},\dots,{\varepsilon}_{d}=\pm 1$,

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

As a convention we put a bar on top of $k_{j}$ if the corresponding
${\varepsilon}_{j}=-1$. For example, $\zeta(\bar{1})=\zeta(1;-1)=-\log 2$.

In the mid 1990s, Broadhurst [4] already noticed that some of the Euler sums
actually lie in the $\mathbb{Q}$-span $\mathsf{MZV}$ of MZVs, and he called
these values honorary MZVs. Such a phenomenon is very interesting because it
reflects the important Galois descent structure studied recently by Glanois in
her thesis [8, 9]. Following her terminology, we call such values unramified.

The _multiple $T$-values_ (MTVs) are a class of infinite sums which can be
regarded as a level two variation of the multiple zeta values, first defined
by Kaneko and Tsumura [11]:

$T(k_{1},\dots,k_{d}):=\displaystyle\sum_{\begin{subarray}{c}0<n_{1}<\cdots<n_{d}\\\
n_{j}\equiv
j\pmod{2}\end{subarray}}\displaystyle\frac{2^{d}}{n_{1}^{k_{1}}\cdots
n_{d}^{k_{d}}},$

for all admissible compositions $(k_{1},\dots,k_{d})$. We again call
$|{\boldsymbol{\sl{k}}}|$ its _weight_ and $d$ its _depth_. From numerical
evidence, they propose the following conjecture (cf. [11, Conjecture 5.2]).

######  Conjecture 1.1.

1) For even weights, $T(k,l,m)$ with $p,r$: odd $\geq 3$ and $q$: even (and
their duals) are in $\mathsf{MZV}$. Together with the single $T$-values these
(and their duals) are the only unramified MTVs (i.e., contained in
$\mathsf{MZV}$).

2) For odd weights, $T(p,1,r)$ with $p,r$: even (and their duals) are in
$\mathsf{MZV}$. Together with the single and double $T$-values these (and
their duals) are the only unramified MTVs.

In [12], Murakami proved that the conditions in the above conjecture are
sufficient by using the motivic theory of Euler sums developed in [5, 9]. In
this paper, we will use this machinery to further show that, on the motivic
level, these conditions are also necessary, at least when the depth is bounded
by three. This would completely confirm the above conjecture for depth $\leq
3$ if we assume Grothendieck’s period conjecture.

######  Theorem 1.2.

Suppose ${\boldsymbol{\sl{k}}}$ has depth at most three. We have the following
necessary and sufficient conditions for
$T^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ to be unramified.

  1. (1)

If $\operatorname*{dep}({\boldsymbol{\sl{k}}})=1$, then
${\boldsymbol{\sl{k}}}=k\geq 2$.

  2. (2)

If ${\boldsymbol{\sl{k}}}=(k,l)$ then $l\geq 2$ with $k+l$ odd.

  3. (3)

If $\operatorname*{dep}({\boldsymbol{\sl{k}}})=3$ then either

    1. (i)

${\boldsymbol{\sl{k}}}=(1,1,2),(1,1,3),(1,2,2)$ or

    2. (ii)

$|{\boldsymbol{\sl{k}}}|$ is even, ${\boldsymbol{\sl{k}}}=(O_{>1},E,O_{>1})$,
or

    3. (iii)

$|{\boldsymbol{\sl{k}}}|$ is odd, $|{\boldsymbol{\sl{k}}}|=(E,1,E)$,

where $O$ (resp. $E$) may represent different odd (resp. even) numbers.

Case (1) of Thm. 1.2 is obvious. We will prove case (2) in Thm. 5.3. Proof of
case (3) will be broken into many subcases throughout §7 and §8.

Beside MTVs, there are other variants of MZVs of level two. For any admissible
composition ${\boldsymbol{\sl{k}}}$, Hoffman defines the _multiple $t$-values_
in [10] by

$\displaystyle
t({\boldsymbol{\sl{k}}}):=\displaystyle\sum_{\begin{subarray}{c}0<n_{1}<\cdots<n_{d}\\\
n_{j}:\text{ odd}\end{subarray}}\displaystyle\frac{2^{d}}{n_{1}^{k_{1}}\cdots
n_{d}^{k_{d}}}.$

The current authors defined the _multiple $S$-values_
$S({\boldsymbol{\sl{k}}})$ in [14] by

$\displaystyle
S({\boldsymbol{\sl{k}}}):=\displaystyle\sum_{\begin{subarray}{c}0<n_{1}<\cdots<n_{d}\\\
n_{j}\equiv
j+1\pmod{2}\end{subarray}}\displaystyle\frac{2^{d}}{n_{1}^{k_{1}}\cdots
n_{d}^{k_{d}}},$

All the above values are special cases of the so-called _multiple mixed
values_ first studied by the current authors in [14], the definition of which
allows all possible parity patterns in the infinite sums.

Applying the same technique for MTVs and alongside with the proof of Thm. 1.2,
we will prove the following results about multiple $S$-values in §7 and §8.

######  Theorem 1.3.

Suppose ${\boldsymbol{\sl{k}}}$ has depth at most three. We have the following
necessary and sufficient conditions for
$S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ to be unramified.

  1. (1)

If $\operatorname*{dep}({\boldsymbol{\sl{k}}})=1$, then
${\boldsymbol{\sl{k}}}=k\geq 2$.

  2. (2)

If ${\boldsymbol{\sl{k}}}=(k,l)$ then $k\geq 2$ and $l\geq 2$ with $k+l$ odd.

  3. (3)

If $\operatorname*{dep}({\boldsymbol{\sl{k}}})=3$ then either

    1. (i)

$|{\boldsymbol{\sl{k}}}|$ is even, ${\boldsymbol{\sl{k}}}=(O_{>1},O,E)$ or
${\boldsymbol{\sl{k}}}=(O_{>1},E,O_{>1})$, or

    2. (ii)

$|{\boldsymbol{\sl{k}}}|$ is odd, ${\boldsymbol{\sl{k}}}=(E,1,E)$,

where $O$ (resp. $E$) may represent different odd (resp. even) numbers.

On the other hand, there are many unramified MtVs as manifested by the results
of Murakami and Charlton. We denote by $\\{{\boldsymbol{\sl{s}}}\\}_{a}$ the
string obtained by repeating ${\boldsymbol{\sl{s}}}$ exactly $a$ times. If
${\boldsymbol{\sl{s}}}$ has just one component then we often drop the curly
brackets for simplicity.

######  Theorem 1.4.

The MtVs $t({\boldsymbol{\sl{k}}})$ are unramified if either one of the
following conditions holds:

  1. (1)

_([12, Thm. 1])_ all components of ${\boldsymbol{\sl{k}}}$ are at least $2$;

  2. (2)

_([6, Prop. 5.11])_ ${\boldsymbol{\sl{k}}}=(2_{a},1,2_{b},2n+1,2_{c})$,
$a,n\in\mathbb{N}$ and $b,c\in\mathbb{N}_{0}.$

We can extend the above to more cases of unramified MtVs in the next theorem.

######  Theorem 1.5.

The MtVs $t({\boldsymbol{\sl{k}}})$ are unramified if either one of the
following conditions holds:

  1. (1)

${\boldsymbol{\sl{k}}}=(\\{2n\\}_{e},1,{\boldsymbol{\sl{q}}})$, where
$n,e\in\mathbb{N}$ and all components of ${\boldsymbol{\sl{q}}}$ are at least
$2$;

  2. (2)

${\boldsymbol{\sl{k}}}=(2n,2m,2n,1,{\boldsymbol{\sl{q}}})$, where
$m,n\in\mathbb{N}$ and all components of ${\boldsymbol{\sl{q}}}$ are at least
$2$;

  3. (3)

${\boldsymbol{\sl{k}}}=(2,1,3,2,1,{\boldsymbol{\sl{q}}})$, where all
components of ${\boldsymbol{\sl{q}}}$ are at least $2.$

In fact, we prove the above three cases for the motivic MtVs in Thm. 9.8, Thm.
9.10, and Thm. 9.12, respectively. Then Thm. 1.5 follows immediately by
applying the period map.

Note that if a unit component (i.e. component equal to 1) appears at either
end of a motivic MtV then it is ramified (see Lemma 9.2). Our next result
provides a family of ramified motivic MtVs neither starting nor ending with a
unit component.

######  Theorem 1.6.

_(Thm. 9.13)_ Suppose $j\leq n<\ell$ such that $k_{j}$ is odd and $k_{i}$ is
even for all $1\leq i<n$ and $i\neq j$. Suppose further that $k_{n}=1$ and
$k_{i}\geq 2$ for all $n<i\leq\ell$. Then the motivic MtV
$t^{\mathfrak{m}}(k_{1},\dots,k_{\ell})$ is ramified.

We now briefly describe the organization of this paper. After introducing the
motivic setup in the next section, we will prove a few simple preliminary
results on the motive mixed values of depth one in §3. We then present a
generalization of Zagier’s formula on reducible double zeta values in terms of
Riemann zeta values to level two, i.e., Euler sums, which will be used
crucially in the derivation computation of various motivic multiple mixed
values. Next, in §5 we provide a complete description of when double mixed
values are ramified. We then study the triple MSVs and MTVs in §6-§8 and
complete the proofs of Thm. 1.2 and Thm. 1.3. The longest section §9 is
devoted to the study of motivic MtVs. By utilizing and building upon the
recent work of Murakami and Charlton, we are able to prove Thm. 1.5 and Thm.
1.6 concerning a few families of unramified/ramified MtVs. We end the paper by
posing a conjecture and a few unsolved problems involving MSVs and MtVs.

##  2 Notation and the motivic setup

We will adopt the motivic setup developed in [5, 9]. Let $\Gamma_{N}$ be the
group of $N$-th roots of 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 $0\leq j\leq w+1$. Then
one can define the so-called motivic integrals
$I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{w};a_{w+1})$ subjecting to a list of
axioms. The most important property of the motivic integrals is that there is
a period map $\operatorname{dch}$ such that

$\operatorname{dch}\big{(}I^{\mathfrak{m}}(a_{0};a_{1},\dots,a_{w};a_{w+1})\big{)}=(-1)^{\operatorname*{dep}(a_{1},\dots,a_{w})}\displaystyle\int_{a_{0}}^{a_{w+1}}\displaystyle\frac{dt}{t-a_{1}}\cdots\displaystyle\frac{dt}{t-a_{w}}$
(2.1)

as an iterated integral (we always integrate from left to right in this paper)
whenever it converges in which case we can obtain some special values of
multiple polylogarithms at some $N$th roots of unity (see [17, Ch. 14]).

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

$\zeta^{\mathfrak{m}}_{c}\left(k_{1},\dots,k_{d}\atop{\varepsilon}_{1},\dots,{\varepsilon}_{d}\right):=I^{\mathfrak{m}}(0;0_{c},\eta_{1},0_{k_{1}-1},\dots,\eta_{d},0_{k_{d}-1};1),$

where $c\in\mathbb{N}_{0}$,
$k_{j}\in\mathbb{N},{\varepsilon}_{j}\in\Gamma_{N}$,
$\eta_{j}=1/{\varepsilon}_{j}\cdots{\varepsilon}_{d}$ for all $j=1,\dots,d$.
These are the motivic MZVs when $N=1$ and are the motivic Euler sums when
$N=2$.

We now list all the important properties/axioms of the motivic integrals as
follows (cf. [5, §2.4], [9, p. 9], and [12, §2, (I1)-(I6)]):

  * (I1)

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

  * (I2)

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

  * (I3)

Shuffle product: $\forall c\in\mathbb{N}_{0}$ we have

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

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

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

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 $0<r<w$ one 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_.

Recall the following theorem which combines Glanois’s result [9, Cor. 2.4]
with Brown’s [5, 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 ${\boldsymbol{\sl{k}}}$ be a composition of
positive integers such that $a+|{\boldsymbol{\sl{k}}}|=w$. Then the weight $w$
motivic Euler sum
$\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\in{\mathcal{H}}_{k}$ if and
only $D_{1}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$ and
$D_{r}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\in{\mathcal{L}}_{r}{\otimes}{\mathcal{H}}_{w-r}$
for all odd $r<w$. Moreover, if
$D_{r}\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$ for all odd $r<w$
then
$\zeta_{a}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=c\,\zeta^{\mathfrak{m}}(k)$
for some rational number $c$.

The following result is due to Glanois. A complete proof is given in [15,
Lemma 2.2].

######  Lemma 2.2.

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

For all $a\in\mathbb{N}_{0}$ and
${\boldsymbol{\sl{k}}}=(k_{1},\dots,k_{d})\in\mathbb{N}^{d}$ we define the
motivic multiple $t$-, $T$-, $S$\- and $\tilde{\zeta}$-values by

$\displaystyle t^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\eta_{1}I^{\mathfrak{m}}(0;0_{a},\eta_{1},0_{k_{1}-1},\dots,\eta_{d},0_{k_{d}-1};1),$
(2.2) $\displaystyle T^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\eta_{1}\cdots\eta_{d}I^{\mathfrak{m}}(0;0_{a},\eta_{1},0_{k_{1}-1},\dots,\eta_{d},0_{k_{d}-1};1),$
(2.3) $\displaystyle S^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\eta_{2}\cdots\eta_{d}I^{\mathfrak{m}}(0;0_{a},\eta_{1},0_{k_{1}-1},\dots,\eta_{d},0_{k_{d}-1};1),$
(2.4) $\displaystyle\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}I^{\mathfrak{m}}(0;0_{a},\eta_{1},0_{k_{1}-1},\dots,\eta_{d},0_{k_{d}-1};1),$
(2.5)

respectively. Setting $f^{\mathfrak{m}}=f^{\mathfrak{m}}_{0}$ for $f=t,T,S$,
it is easy to see that for all admissible ${\boldsymbol{\sl{k}}}$ we have

$\operatorname{dch}T^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=T({\boldsymbol{\sl{k}}}),\quad\operatorname{dch}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=t({\boldsymbol{\sl{k}}}),\quad\operatorname{dch}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=S({\boldsymbol{\sl{k}}}).$

Further, by the motivic version of the distribution relation proved by
Charlton we see that
$\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=2^{\operatorname*{dep}({\boldsymbol{\sl{k}}})-a-|{\boldsymbol{\sl{k}}}|}\zeta^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})$
if ${\boldsymbol{\sl{k}}}$ is admissible. For future reference, we state it
below.

######  Theorem 2.3.

_(cf.[6, Cor. 5.4])_ Let $a\in\mathbb{N}_{0}$ and ${\boldsymbol{\sl{k}}}$ be
an admissible composition of positive integers. Then we have

$\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=2^{\operatorname*{dep}({\boldsymbol{\sl{k}}})-a-|{\boldsymbol{\sl{k}}}|}\zeta^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}}).$
(2.6)

We will use the next technical lemma of [12] a number of times in this paper.

######  Lemma 2.4.

_(cf.[12, Lemma 9])_ Suppose $a_{0},\dots,a_{d}\geq 1$. For any fixed
${\alpha},{\beta}\in\\{\pm 1\\}$ we have

$\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}I^{\mathfrak{l}}({\alpha};0_{a_{0}-1},\eta_{1},0_{a_{1}-1},\dots,\eta_{d},0_{a_{d}-1};{\beta})=0.$

##  3 Some preliminary results

To save space, we extend $I^{\mathfrak{l}}(a;--;b)$ and
$I^{\mathfrak{m}}(a;--;b)$ $\mathbb{Q}$-linearly to all finite
$\mathbb{Q}$-linear combinations of strings of 0’s and $\pm 1$’s. For example,

$I^{\mathfrak{m}}(1;011+0\bar{1}0;0)=I^{\mathfrak{m}}(1;011;0)+I^{\mathfrak{m}}(1;0\bar{1}0;0).$

The next lemma follows from definition (I3) in section 2 immediately.

######  Lemma 3.1.

For all integers $a\geq 0$ and $k\geq 2$ we have

$\displaystyle\zeta^{\mathfrak{m}}_{a}(k)=$
$\displaystyle\,(-1)^{a}\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k),$
$\displaystyle\zeta^{\mathfrak{m}}_{a}(\overline{k})=$
$\displaystyle\,(-1)^{a}(2^{1-a-k}-1)\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k).$

######  Lemma 3.2.

Suppose $a\in\mathbb{N}_{0}$, $k\geq 1$, and $a+k\geq 2$. For motivic single
$t$-, $T$\- and $S$-values we have

$\displaystyle t^{\mathfrak{m}}(1)=T^{\mathfrak{m}}(1)=$
$\displaystyle\,\log^{\mathfrak{m}}2,\qquad
S^{\mathfrak{m}}(1)=-\log^{\mathfrak{m}}2,$ $\displaystyle\quad
t^{\mathfrak{m}}_{a}(k)=T^{\mathfrak{m}}_{a}(k)=$
$\displaystyle\,(-1)^{a}(2-2^{1-a-k})\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k),$
$\displaystyle S^{\mathfrak{m}}_{a}(k)=$
$\displaystyle\,(-1)^{a}2^{1-a-k}\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k).$

###### Proof.

By definition

$\displaystyle t^{\mathfrak{m}}(1)=T^{\mathfrak{m}}(1)=$
$\displaystyle\,I^{\mathfrak{m}}(0;1;1)-I^{\mathfrak{m}}(0;\bar{1};1)=\log^{\mathfrak{m}}2,$
$\displaystyle S^{\mathfrak{m}}(1)=$
$\displaystyle\,I^{\mathfrak{m}}(0;1;1)+I^{\mathfrak{m}}(0;\bar{1};1)=-\log^{\mathfrak{m}}2,$

and for all $k\geq 2$

$\displaystyle t^{\mathfrak{m}}_{a}(k)=T^{\mathfrak{m}}_{a}(k)=$
$\displaystyle\,I^{\mathfrak{m}}(0;0_{a-1}10_{k-1}-0_{a-1}\bar{1},0_{k-1};1)=\zeta_{a}^{\mathfrak{m}}(k)-\zeta_{a}^{\mathfrak{m}}(\overline{k})$
$\displaystyle=$
$\displaystyle\,(-1)^{a}(2-2^{1-a-k})\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k),$
$\displaystyle S^{\mathfrak{m}}_{a}(k)=$
$\displaystyle\,\zeta_{a}^{\mathfrak{m}}(k)+\zeta_{a}^{\mathfrak{m}}(\overline{k})(-1)^{a}2^{1-a-k}\binom{a+k-1}{a}\zeta^{\mathfrak{m}}(a+k)$

by Lemma 3.1. ∎

######  Lemma 3.3.

Let ${\boldsymbol{\sl{k}}}=(k_{1},\dots,k_{p})$ be an admissible composition.

  1. (1)

$D_{1}T^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$.

  2. (2)

$D_{1}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=-2\delta_{k_{1}=1}\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(k_{2},\dots,k_{p})$.
Consequently, $S^{\mathfrak{m}}(1,{\boldsymbol{\sl{l}}})$ are ramified for all
nontrivial ${\boldsymbol{\sl{l}}}$.

###### Proof.

Part (1) of the lemma is [12, Lemma 43] by Murakami. By the same argument, we
see that in general

$D_{1}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=\delta_{k_{1}=1}\displaystyle\sum_{\eta_{1},\dots,\eta_{p}=\pm
1}(\eta_{2}-\eta_{1})\eta_{3}\dots\eta_{p}I^{\mathfrak{l}}(0;\eta_{1},\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{k_{2}-1},\dots,\eta_{p},0_{k_{p}-1};1)$

for ${\boldsymbol{\sl{k}}}=(k_{1},\dots,k_{p})$. Thus
$D_{1}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$ if $k_{1}\geq 2$. On the
other hand, if $k_{1}=1$ then we see that

$D_{1}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=-2\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(k_{2},\dots,k_{p})\neq
0$

which implies that $S^{\mathfrak{m}}(1,k_{2},\dots,k_{p})$ is ramified by Thm.
2.1. This proves (2). ∎

##  4 Reduction formula for double Euler sums of odd weight

In this section, we will express every motivic double Euler sum of odd weight
as a $\mathbb{Q}$-linear combination of products of Riemann zeta values (after
setting $\zeta^{\mathfrak{m}}(0)=-1/2$). As corollaries, we can write all the
elements $t^{\mathfrak{l}}(k,l)$, $T^{\mathfrak{l}}(k,l)$, and
$S^{\mathfrak{l}}(k,l)$ in ${\mathcal{L}}$ as rational multiples of
$\zeta^{\mathfrak{l}}(k+l)$ when the weight $k+l$ is odd.

First, we generalize a result of Zagier [16, Prop. 7] concerning double zetas
to double Euler sums. The non-motivic version can be found in [2, (74)].

######  Proposition 4.1.

Let $m,n\in\mathbb{N}$ and $\eta_{1},\eta_{2}=\pm 1$. Put $\zeta(0;\pm
1)=-1/2$. Then the double zeta value
$\zeta^{\mathfrak{m}}(m,n;\eta_{1},\eta_{2})$ of weight $m+n=W=2K+1$ is given
by

$\displaystyle\zeta^{\mathfrak{m}}(m,n;\eta_{1},\eta_{2})=$
$\displaystyle\,-\displaystyle\frac{1}{2}\zeta^{\mathfrak{m}}(w;\eta_{1}\eta_{2})+(-1)^{m}\displaystyle\sum_{s=0}^{K-1}\bigg{[}\binom{w-2s-1}{m-1}\zeta^{\mathfrak{m}}(w-2s;\eta_{1})\zeta^{\mathfrak{m}}(2s;\eta_{1}\eta_{2})$
$\displaystyle\,+\binom{w-2s-1}{n-1}\zeta^{\mathfrak{m}}(w-2s;\eta_{2})\zeta^{\mathfrak{m}}(2s;\eta_{1}\eta_{2})\bigg{]}+{\delta}_{2|n}\,\zeta^{\mathfrak{m}}(m;\eta_{1})\zeta^{\mathfrak{m}}(n;\eta_{2}).$

###### Proof.

We will follow Zagier idea in the proof of [16, Prop. 7]. For
$\eta_{1},\eta_{2}=\pm 1$ and $w\geq 3$ we define

$\displaystyle P^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=$
$\displaystyle\,\displaystyle\sum_{r,s\in\mathbb{N},r+s=w}\zeta^{\mathfrak{m}}\binom{r}{\eta_{1}}\zeta^{\mathfrak{m}}\binom{s}{\eta_{2}}x^{r-1}y^{s-1},$
$\displaystyle D^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=$
$\displaystyle\,\displaystyle\sum_{r,s\in\mathbb{N},r+s=w}\zeta^{\mathfrak{m}}\binom{r\,,\,s}{\eta_{1},\eta_{2}}x^{r-1}y^{s-1}.$

Then we can derive

$\displaystyle P^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=$
$\displaystyle\,D^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)+D^{\mathfrak{m}}_{\eta_{2},\eta_{1}}(y,x)+\zeta^{\mathfrak{m}}(w;\eta_{1}\eta_{2})F_{w}(x,y)$
$\displaystyle=$
$\displaystyle\,D^{\mathfrak{m}}_{\eta_{1}\eta_{2},\eta_{2}}(x,x+y)+D^{\mathfrak{m}}_{\eta_{1}\eta_{2},\eta_{1}}(y,x+y),$

where $F_{w}(x,y)=\displaystyle\frac{x^{w-1}-y^{w-1}}{x-y}$. Setting
$(\eta_{1},\eta_{2})=(-1,-1)$ and $(-1,1)$

$\displaystyle P^{\mathfrak{m}}_{-,-}(x,y)=$
$\displaystyle\,D^{\mathfrak{m}}_{-,-}(x,y)+D^{\mathfrak{m}}_{-,-}(y,x)+\zeta^{\mathfrak{m}}(w)F_{w}(x,y)$
$\displaystyle=$
$\displaystyle\,D^{\mathfrak{m}}_{+,-}(x,x+y)+D^{\mathfrak{m}}_{+,-}(y,x+y),$
$\displaystyle P^{\mathfrak{m}}_{-,+}(x,y)=$
$\displaystyle\,D^{\mathfrak{m}}_{-,+}(x,y)+D^{\mathfrak{m}}_{+,-}(y,x)+\zeta^{\mathfrak{m}}(\overline{w})F_{w}(x,y)$
$\displaystyle=$
$\displaystyle\,D^{\mathfrak{m}}_{-,-}(x,x+y)+D^{\mathfrak{m}}_{-,+}(y,x+y)$

and repeatedly applying the above two equations, we get (setting $z=x-y$)

$\displaystyle\,D^{\mathfrak{m}}_{-,+}(x,y)$ $\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(x,y)-D^{\mathfrak{m}}_{+,-}(y,x)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)-P^{\mathfrak{m}}_{-,-}(y,z)+D^{\mathfrak{m}}_{+,-}(z,x)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)-P^{\mathfrak{m}}_{-,-}(y,z)+P^{\mathfrak{m}}_{-,+}(x,z)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(z,x)-D^{\mathfrak{m}}_{-,+}(x,z)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)-P^{\mathfrak{m}}_{-,-}(y,z)+P^{\mathfrak{m}}_{-,+}(x,z)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(z,x)$
$\displaystyle\,-P^{\mathfrak{m}}_{-,+}(x,-y)+D^{\mathfrak{m}}_{-,-}(-y,z)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)-P^{\mathfrak{m}}_{-,-}(y,z)+P^{\mathfrak{m}}_{-,+}(x,z)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(z,x)$
$\displaystyle\,-P^{\mathfrak{m}}_{-,+}(x,-y)+P^{\mathfrak{m}}_{-,-}(-y,z)-\zeta^{\mathfrak{m}}(w)F_{w}(-y,z)-D^{\mathfrak{m}}_{-,-}(z,-y)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)-P^{\mathfrak{m}}_{-,-}(y,z)+P^{\mathfrak{m}}_{-,+}(x,z)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(z,x)$
$\displaystyle\,-P^{\mathfrak{m}}_{-,+}(x,-y)+P^{\mathfrak{m}}_{-,-}(-y,z)-\zeta^{\mathfrak{m}}(w)F_{w}(-y,z)-P^{\mathfrak{m}}_{-,+}(-x,z)+D^{\mathfrak{m}}_{-,+}(-x,-y)$
$\displaystyle=$
$\displaystyle\,P^{\mathfrak{m}}_{-,+}(x,y)-\zeta(\overline{w})F_{w}(x,y)+P^{\mathfrak{m}}_{-,-}(-y,y-x)+P^{\mathfrak{m}}_{-,+}(x,z)-\zeta^{\mathfrak{m}}(\overline{w})F_{w}(z,x)$
$\displaystyle\,+P^{\mathfrak{m}}_{-,+}(-x,y)+P^{\mathfrak{m}}_{-,-}(-y,z)-\zeta^{\mathfrak{m}}(w)F_{w}(-y,z)+P^{\mathfrak{m}}_{-,+}(x,y-x)-D^{\mathfrak{m}}_{-,+}(x,y)$
$\displaystyle=$
$\displaystyle\,Q^{\mathfrak{m}}_{-,+}(x,y)+Q^{\mathfrak{m}}_{+,-}(z,x)+Q^{\mathfrak{m}}_{-,-}(z,-y)-D^{\mathfrak{m}}_{-,+}(x,y),$

where

$Q^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=P^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)+P^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(-x,y)-\zeta^{\mathfrak{m}}(w,\eta_{1}\eta_{2})F_{w}(x,y).$

Similarly,

$\displaystyle 2D^{\mathfrak{m}}_{+,-}(x,y)=$
$\displaystyle\,Q^{\mathfrak{m}}_{+,-}(x,y)+Q^{\mathfrak{m}}_{+,-}(z,-y)+Q^{\mathfrak{m}}_{-,-}(z,x).$

In general

$\displaystyle 2D^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=$
$\displaystyle\,Q^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)+Q^{\mathfrak{m}}_{\eta_{1},\eta_{1}\eta_{2}}(z,-y)+Q^{\mathfrak{m}}_{\eta_{2},\eta_{1}\eta_{2}}(z,x).$

Hence, putting $c_{m}=(1-(-1)^{m})$ for all $m$ we get

$\displaystyle[x^{m-1}y^{n-1}]Q^{\mathfrak{m}}_{\eta_{1},\eta_{2}}(x,y)=$
$\displaystyle\,c_{m}\zeta^{\mathfrak{m}}(m,\eta_{1})\zeta^{\mathfrak{m}}(n,\eta_{2})-\zeta^{\mathfrak{m}}(w,\eta_{1}\eta_{2}),$
$\displaystyle[x^{m-1}y^{n-1}]Q^{\mathfrak{m}}_{\eta_{1},\eta_{1}\eta_{2}}(z,-y)=$
$\displaystyle\,\displaystyle\sum_{r=m}^{w-1}\binom{r-1}{m-1}(-1)^{n}\Big{[}\zeta^{\mathfrak{m}}(w,\eta_{2})-c_{r}\zeta^{\mathfrak{m}}(r,\eta_{1})\zeta^{\mathfrak{m}}(w-r,\eta_{1}\eta_{2})\Big{]},$
$\displaystyle[x^{m-1}y^{n-1}]Q^{\mathfrak{m}}_{\eta_{2},\eta_{1}\eta_{2}}(z,x)=$
$\displaystyle\,\displaystyle\sum_{r=n}^{w-1}\binom{r-1}{n-1}(-1)^{n}\Big{[}\zeta^{\mathfrak{m}}(w,\eta_{1})-c_{r}\zeta^{\mathfrak{m}}(r,\eta_{2})\zeta^{\mathfrak{m}}(w-r,\eta_{1}\eta_{2})\Big{]}.$

Therefore, the proposition follows by the substitution $r=w-2s$ and the
identity

$\displaystyle\sum_{r=m}^{w-1}\binom{r-1}{m-1}=\binom{w-1}{m}=\binom{w-1}{n-1},\qquad\displaystyle\sum_{r=n}^{w-1}\binom{r-1}{n-1}=\binom{w-1}{n}=\binom{w-1}{m-1}.$

∎

######  Corollary 4.2.

Let $m,n\in\mathbb{N}$ with $m+n=W=2K+1$ is given by

$\displaystyle\zeta^{\mathfrak{l}}(m,n;\eta_{1},\eta_{2})$
$\displaystyle\,=-\displaystyle\frac{1}{2}\zeta^{\mathfrak{l}}(w;\eta_{1}\eta_{2})-\displaystyle\frac{1}{2}(-1)^{m}\bigg{[}\binom{w-1}{m-1}\zeta^{\mathfrak{l}}(w;\eta_{1})+\binom{w-1}{n-1}\zeta^{\mathfrak{l}}(w;\eta_{2})\bigg{]},$
$\displaystyle\tilde{\zeta}^{\mathfrak{l}}(m,n)$
$\displaystyle\,=-\bigg{(}1+(-1)^{m}\binom{w}{m}\bigg{)}\tilde{\zeta}^{\mathfrak{l}}(w),\quad
t^{\mathfrak{l}}(m,n)=-t^{\mathfrak{l}}(w),$ $\displaystyle
T^{\mathfrak{l}}(m,n)$
$\displaystyle\,=(-1)^{n}\binom{w-1}{m-1}T^{\mathfrak{l}}(w),\quad
S^{\mathfrak{l}}(m,n)=(-1)^{n}\binom{w-1}{n-1}S^{\mathfrak{l}}(w).$

######  Remark 4.3.

The formula for $\zeta^{\mathfrak{l}}(m,n;\eta_{1},\eta_{2})$ is depth two
case of a more general reduction formula of Glanois’s in [8, Cor. 4.2.4].

##  5 Motivic double $t$-, $T$\- and $S$-values

In this section, we present the precise criterion for when a motivic double
mixed value is unramified. Obviously, there are only three kinds of double
mixed values that are not automatically MZVs: double $t$-, $T$\- and
$S$-values.

First, we recall the following result of Murakami.

######  Lemma 5.1.

_(cf.[12, Thm. 42])_ For $m,k\geq 1$, $l\geq 2$ with $k+l+m$ odd, the motivic
$T$-value $T^{\mathfrak{m}}_{m}(k,l)$ is unramified.

######  Proposition 5.2.

For $k,l\in\mathbb{N}$ the motivic double $t$-value is unramified if and only
if $k,l\geq 2$.

###### Proof.

This follow immediately from Prop. 9.3 and Murakami’s result in Lemma 9.1. ∎

######  Theorem 5.3.

Let $k,l\in\mathbb{N}$. The motivic double $T$-value $T^{\mathfrak{m}}(k,l)$
is unramified if and only if $k\geq 1$ and $l\geq 2$ with odd weight $k+l$.
The motivic double $S$-value $S^{\mathfrak{m}}(k,l)$ is unramified if and only
if $k,l\geq 2$ with odd weight $k+l$.

We break the proof of theorem into the next two propositions according the
weight parity.

######  Proposition 5.4.

Suppose $k,l\in\mathbb{N}$ have different parities. Then we have

  1. (1)

$T^{\mathfrak{m}}(k,1)$, $S^{\mathfrak{m}}(k,1)$ and $S^{\mathfrak{m}}(1,l)$
are ramified for all even $k\in\mathbb{N}$, and

  2. (2)

$T^{\mathfrak{m}}(k,l)$ is unramified if $l\geq 2$ and $S^{\mathfrak{m}}(k,l)$
is unramified if $k,l\geq 2$.

###### Proof.

The claims about motivic double $t$-values follows from Lemma 9.1 and Prop.
9.3.

Let $a=k-1$ and $b=l-1$. We will prove the rest of the proposition by
computing the derivations $D_{r}$ for all odd $r<k+l$.

I. First we consider $D_{1}$.

(i). If $a=0$ then $b>0$. Then by Lemma 3.3

$D_{1}T^{\mathfrak{m}}(1,l)=0,\quad\text{and}\quad
D_{1}S^{\mathfrak{m}}(1,l)=-2\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(l)\neq
0.$ (5.7)

Thus $S^{\mathfrak{m}}(1,l)$ is ramified.

(ii). If $a\neq 0$. Then there are no nontrivial cuts for
$D_{1}T^{\mathfrak{m}}(k,l)$ and $D_{1}S^{\mathfrak{m}}(k,l)$ except when
$b=0$. If $b=0$ then

$\displaystyle D_{1}T^{\mathfrak{m}}(k,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a};1)=\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(k)\neq
0,$ (5.8) $\displaystyle D_{1}S^{\mathfrak{m}}(k,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}}\eta_{2}I^{\mathfrak{l}}(0;\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a};1)=\log^{\mathfrak{l}}2{\otimes}S^{\mathfrak{m}}(k)\neq
0.$ (5.9)

Thus $T^{\mathfrak{m}}(k,1)$ and $S^{\mathfrak{m}}(k,1)$ are both ramified.

II. Suppose $a=0,b\geq 1$, $3\leq r\leq b$, and both $b$ and $r$ are odd. Then
there is only one possible cut for $D_{r}T^{\mathfrak{m}}(1,b+1)$ shown in the
following picture.

$0;\eta_{1},\eta_{2},0_{r-1},0,0_{b-r};1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Only one possible cut of
$D_{r}T^{\mathfrak{m}}(1,b+1)$, $b\geq 1$.

${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}=\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(\eta_{1};\eta_{2},0_{r-1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{b-r};1)=-T^{\mathfrak{l}}_{r-1}(1){\otimes}t^{\mathfrak{m}}(b-r+1)$

which is unramified by Lemma 3.2. Hence, $T^{\mathfrak{m}}(1,l)$ is unramified
by Thm. 2.1.

III. Suppose $a,b\geq 1$ and $r\leq\displaystyle\min\\{a,b\\}.$ Clearly there
is no nontrivial cuts for $D_{r}T^{\mathfrak{m}}(k,l)$.

IV. Suppose $a,b\geq 1$ and $r=a+s$ is odd, where $1\leq s\leq b.$ Then all
possible cuts are shown in the following picture.

$0;\eta_{1},0_{b-s},0,0_{a+s-b-1},\eta_{2},0_{s-1},0,0_{b-s};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 $D_{a+s}T^{\mathfrak{m}}(a+1,b+1)$,
$a\neq 0$.

Here ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}\,$ appears if $s=1$ and $a$ is even while
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,$ appears if $a+s>b$. In this case

$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;\eta_{1},0_{a};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{b};1)=T^{\mathfrak{l}}(a+1){\otimes}S^{\mathfrak{m}}(b+1),$
$\displaystyle
S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(0;\eta_{1},0_{a};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{b};1)=S^{\mathfrak{l}}(a+1){\otimes}t^{\mathfrak{m}}(b+1).$

For general $1\leq s\leq b$ we always have (after putting $c=b-s+1$)

$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{s-1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{c};1)=-T^{\mathfrak{l}}_{s-1}(a+1){\otimes}S^{\mathfrak{m}}(c+1),$
$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;0_{a+s-b-1},\eta_{2},0_{b};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{c};1)=T^{\mathfrak{l}}_{a-c}(b+1){\otimes}T^{\mathfrak{m}}(c+1),$
$\displaystyle
S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{s-1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{c};1)=-T^{\mathfrak{l}}_{s-1}(a+1){\otimes}T^{\mathfrak{m}}(c+1),$
$\displaystyle
S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,)=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(0;0_{a+s-b-1},\eta_{2},0_{b};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{c};1)=T^{\mathfrak{l}}_{a-c}(b+1){\otimes}S^{\mathfrak{m}}(c+1).$

All of the above are clearly unramified by Lemma 3.2. By Thm. 2.1 both
$T^{\mathfrak{m}}(k,l)$ and $S^{\mathfrak{m}}(k,l)$ are unramified if $k,l\geq
2$. This completes the proof of the proposition. ∎

######  Proposition 5.5.

Let $k,l\in\mathbb{N}$ have the same parity. Then the motivic double
$T$-values $T^{\mathfrak{m}}(k,l)$ and double $S$-values
$S^{\mathfrak{m}}(k,l)$ are ramified.

###### Proof.

Suppose $a=k-1,b=l-1$ have the same parity. Let $w=k+l$. If $b=0$ and $a=0$
then

$\displaystyle D_{1}T^{\mathfrak{m}}(1,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}\Big{(}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2};1)+I^{\mathfrak{l}}(\eta_{1};\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,-\log^{\mathfrak{l}}2{\otimes}\log^{\mathfrak{m}}2+\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\bar{1};\eta_{2};1){\otimes}\log^{\mathfrak{m}}2=-\log^{\mathfrak{l}}2{\otimes}\log^{\mathfrak{m}}2$
(5.10)

by Lemma 2.4. Thus $T^{\mathfrak{m}}(1,1)$ is ramified by Thm. 2.1. Similarly,

$D_{1}S^{\mathfrak{m}}(1,1)=-\log^{\mathfrak{l}}2{\otimes}\log^{\mathfrak{m}}2.$
(5.11)

Hence, $S^{\mathfrak{m}}(1,1)$ is ramified

If $b=0$ and $a>0$ is even then

$\displaystyle D_{1}T^{\mathfrak{m}}(a+1,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a};1)=\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(a+1),$
$\displaystyle D_{w-1}S^{\mathfrak{m}}(a+1,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{2}\Big{(}I^{\mathfrak{l}}(0;\eta_{1},0_{a};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2};1)+I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1};1)\Big{)}$
$\displaystyle=S^{\mathfrak{m}}(a+1)$
$\displaystyle\,{\otimes}\log^{\mathfrak{m}}2-T_{a}^{\mathfrak{m}}(1){\otimes}\log^{\mathfrak{m}}2+T^{\mathfrak{m}}(a+1){\otimes}\log^{\mathfrak{m}}2=S^{\mathfrak{m}}(a+1){\otimes}\log^{\mathfrak{m}}2$

which implies that both $T^{\mathfrak{m}}(k,1)$ and $S^{\mathfrak{m}}(k,1)$
are ramified by Thm. 2.1.

If $b>0$ then

$\displaystyle\,D_{w-1}T^{\mathfrak{m}}(a+1,b+1)=\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}D_{w-1}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b};1)$
$\displaystyle=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2}=\pm
1}\eta_{1}\eta_{2}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1};1)$
$\displaystyle=$ $\displaystyle\,-\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\bar{1};0_{a},\eta_{2},0_{b};1){\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$ $\displaystyle\,-\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}\Big{(}I^{\mathfrak{l}}(0;0_{a},\eta_{2},0_{b};1)-I^{\mathfrak{l}}(0;0_{b},\eta_{2},0_{a};\bar{1})\Big{)}{\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$
$\displaystyle\,-\Big{(}T_{a}^{\mathfrak{l}}(b+1)+T_{b}^{\mathfrak{l}}(a+1)\Big{)}{\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$
$\displaystyle\,-\big{(}(-1)^{a}+(-1)^{b}\big{)}\binom{a+b}{a}T^{\mathfrak{l}}(w-1){\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$
$\displaystyle\,-(-1)^{a}2\binom{a+b}{a}T^{\mathfrak{l}}(w-1){\otimes}\log^{\mathfrak{m}}2\not\in{\mathcal{L}}_{w-1}{\otimes}{\mathcal{H}}_{1}$
(5.12)

by Lemma 3.2, which implies that $T^{\mathfrak{m}}(k,l)$ is ramified by Thm.
2.1. Similarly, if $b>0$ then

$D_{w-1}S^{\mathfrak{m}}(a+1,b+1)=-D_{w-1}T^{\mathfrak{m}}(a+1,b+1)=(-1)^{a}2\binom{a+b}{a}T^{\mathfrak{l}}(w-1){\otimes}\log^{\mathfrak{m}}2$
(5.13)

is ramified and therefore $S^{\mathfrak{m}}(k,l)$ is ramified. ∎

##  6 Unramified triple $S$-values

In this section, we show that the conditions in Thm. 1.3(3) are sufficient,
which is rephrased in the next propositions.

######  Proposition 6.1.

If ${\boldsymbol{\sl{k}}}=(k,l,m)=(O_{>1},O,E)$, or $(O_{>1},E,O_{>1})$, or
$(E,1,E)$ then $S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ is unramified.

###### Proof.

First, it follows immediately from Lemma 3.3 that
$D_{1}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$. By Thm. 2.1 we now only need
to show that the images of the motivic MSVs in the proposition under under
$D_{r}$ are all unramified for odd $r$ such that
$1<r<|{\boldsymbol{\sl{k}}}|$.

If $r>1$ is odd then the same argument as in the proof of [12, Thm. 41] is
mostly still valid and therefore
$D_{r}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ is unramified, at least when no
component is 1. However, there are some subtle differences when $1$ appears.
We now consider three cases.

(I) $1<r<k+l+m-1$.

(I.1) When the weight is even, then both left and right factor of
$D_{r}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ are of odd weight $>1$ and
depth at most two. Thus both factors are unramified by Lemma 3.2 and Prop. 4.1
if no component in the factors is 1. If the left factor of $D_{r}$ has a
component 1 (this can occur only when
${\boldsymbol{\sl{k}}}=(k,l,m)=(O_{>1},1,E)$) then it must have depth two and
therefore the right factor has depth one and consequently the cut is
unramified. If the right factor has a component 1 and depth two, then the cut
must straddle over exactly one of the $\eta$’s and it must has the following
form:

  1. (i)

$I^{\mathfrak{l}}(0;0_{s},\eta_{i},0_{a_{i}};\eta_{i+1}){\otimes}I^{\mathfrak{m}}(0;\dots,\eta_{i-1},0_{a_{i-1}-s},\eta_{i+1},\dots;1)\Longrightarrow
k_{\ell}=1$ for some $\ell\neq i-1,i$.

  2. (ii)

$I^{\mathfrak{l}}(\eta_{i-1};0_{a_{i-1}-1},\eta_{i},0_{s};0){\otimes}I^{\mathfrak{m}}(0;\dots,\eta_{i-1},0_{a_{i}-s},\eta_{i+1},\dots;1)\Longrightarrow
k_{\ell}=1$ for some $\ell\neq i-1,i$.

  3. (iii)

$I^{\mathfrak{l}}(\eta_{i};0_{a_{i}},\eta_{i+1},0_{a_{i+1}};\eta_{i+2}){\otimes}I^{\mathfrak{m}}(0;\dots,\eta_{i},\eta_{i+1},\dots;1)$.

Only ${\boldsymbol{\sl{k}}}=(O_{>1},O,E)$ need to be considered which involves
both (i) and (iii). Let $k=a+1,m=c+1$. Then

$\displaystyle\,D_{k}S^{\mathfrak{m}}(k,1,m)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{1},0_{a};\eta_{2}){\otimes}I^{\mathfrak{l}}(0;\eta_{2},\eta_{3},0_{c};1)+I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{l}}(0;\eta_{1},\eta_{3},0_{c};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\Big{(}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{1},0_{a};1){\otimes}I^{\mathfrak{l}}(0;\eta_{2},\eta_{3},0_{c};1)$
$\displaystyle\,+\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(1;0_{a},\eta_{2};0){\otimes}I^{\mathfrak{l}}(0;\eta_{1},\eta_{3},0_{c};1)+\eta_{2}I^{\mathfrak{l}}(0;0_{a},\eta_{2};1){\otimes}I^{\mathfrak{l}}(0;\eta_{1},\eta_{3},0_{c};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\Big{(}S^{\mathfrak{l}}(a+1)-T^{\mathfrak{l}}(a+1)\Big{)}{\otimes}T^{\mathfrak{m}}(1,c+1)-T^{\mathfrak{l}}_{a}(1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(1,a+1)$

which is unramified by Lemma 5.1 and Thm. 2.3. Thus
$D_{r}S^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ is unramified if
${\boldsymbol{\sl{k}}}=(k,l,m)=(O_{>1},O,E)$ or $(O_{>1},E,O_{>1})$ with even
weight and $1<r<k+l+m-1$ with $r$ odd.

(I.2) If the weight is odd then ${\boldsymbol{\sl{k}}}=(k,1,m)=(E,1,E)$. The
computation is almost exactly the same as that in the proof of [12, Lemma 41]
except that we replace the factor $\eta_{1}\eta_{2}\eta_{3}$ by
$\eta_{2}\eta_{3}$ inside all the sums for
$D_{r}T^{\mathfrak{m}}(k_{1},1,k_{3})=D_{r}T^{\mathfrak{m}}(k,1,m)$ there (the
overall sign difference is because our choices of definition of the motivic
iterated integral have such a difference). Thus among the five terms on the
antepenultimate page of [12], the first, fourth and fifth terms are all
unramified because the second factors are all of depth 1 and weight at least
2. The second and the third terms have the form

$\delta_{r<k}T^{\mathfrak{l}}(r){\otimes}\tilde{\zeta}(k-r+1,m),\qquad\delta_{r<m}T^{\mathfrak{l}}_{r-1}(1){\otimes}\tilde{\zeta}(k,m-r+1)$

both of which are unramified since $k-r+1,m,k,m-r+1$ are all $\geq 2$.

(II) $r=k+l+m-1$. Then we have

$\displaystyle\,D_{r}S^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1};1)$
$\displaystyle=$ $\displaystyle\,\displaystyle\sum_{\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\bar{1};0_{a},\eta_{2},0_{b},\eta_{3},0_{c};1){\otimes}\log^{\mathfrak{m}}2=0$

by [12, Lemma 9]. Consequently, $S^{\mathfrak{m}}(k,l,m)$ is unramified by
Thm. 2.1. ∎

##  7 Motivic triple $T$\- and $S$-values: three special cases

We consider three special cases in this section which have either a leading 1
or an ending 1 in the motivic triple $T$-values.

First, note that $T(k,l,1)$ all diverge and among the convergent triple
$T$-values of the form $T(1,k,l)$ only $T(1,1,2)=T(4)$, $T(1,1,3)=T(1,4)$ and
$T(1,2,2)=T(2,3)$ (by duality) are unramified by the Kaneko-Tsumura
Conjecture. It is straight-forward to show that $T^{\mathfrak{m}}(1,1,2)$,
$T^{\mathfrak{m}}(1,1,3)$ and $T^{\mathfrak{m}}(1,2,2)$ are unramified by
considering $D_{1}$ and $D_{3}$. This takes care of Thm. 1.2(3)(i).

We now deal with all the other special cases by the next three propositions.

######  Proposition 7.1.

For any $k,l\in\mathbb{N}$, the motivic triple $S$\- and $T$-values
$S^{\mathfrak{m}}(k,l,1)$ and $T^{\mathfrak{m}}(k,l,1)$ are both ramified.

###### Proof.

Let $a=k-1$ and $b=l-1$. If $b=0$ then we have

$\displaystyle\,D_{1}T^{\mathfrak{m}}(k,1,1)=\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}D_{1}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},\eta_{3};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{3};1)+I^{\mathfrak{l}}(\eta_{2};\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2};1)\Big{)}$
$\displaystyle=$ $\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{3}=\pm
1}\eta_{1}\log^{\mathfrak{l}}2{\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{3};1)+\displaystyle\sum_{\eta_{1},\eta_{3}=\pm
1}I^{\mathfrak{l}}(\bar{1};\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\bar{1};1)$
$\displaystyle=$
$\displaystyle\,\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(a+1,1)\neq 0$

since $\displaystyle\sum_{\eta_{3}=\pm
1}I^{\mathfrak{l}}(\bar{1};\eta_{3};1)=0$ by Lemma 2.4. Similarly,

$D_{1}S^{\mathfrak{m}}(k,1,1)=\log^{\mathfrak{l}}2{\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+1,1)\neq
0.$

Thus both $S^{\mathfrak{m}}(k,1,1)$ and $T^{\mathfrak{m}}(k,1,1)$ are ramified
by Thm. 2.1.

If $b\geq 1$ then we have only one cut for $D_{1}$:

$\displaystyle D_{1}T^{\mathfrak{m}}(k,l,1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}D_{1}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b},\eta_{3};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b};1)$
$\displaystyle=$
$\displaystyle\,\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(a+1,b+1)\neq 0.$

Similarly,

$D_{1}S^{\mathfrak{m}}(k,l,1)=\log^{\mathfrak{l}}2{\otimes}S^{\mathfrak{m}}(a+1,b+1)\neq
0.$

Hence, both $S^{\mathfrak{m}}(k,l,1)$ and $T^{\mathfrak{m}}(k,l,1)$ are
ramified by Thm. 2.1. This completes the proof of the proposition. ∎

######  Proposition 7.2.

For any $k\geq 4$ the motivic triple $T$-values $T^{\mathfrak{m}}(1,1,k)$ is
ramified.

###### Proof.

Write $a=k-1\geq 3$. Then

$\displaystyle\,D_{3}T^{\mathfrak{m}}(1,1,a+1)=\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}D_{3}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},\eta_{3},0_{a};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(\eta_{1};\eta_{2},\eta_{3},0;0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-1};1)+I^{\mathfrak{l}}(\eta_{2};\eta_{3},0,0;0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{a-2};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{3}\Big{(}\eta_{2}I^{\mathfrak{l}}(0;0,\eta_{3},\eta_{2};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-1};1)+I^{\mathfrak{l}}(0;0,0,\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{a-2};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,-T^{\mathfrak{l}}_{1}(1,1){\otimes}T^{\mathfrak{m}}(a)-T^{\mathfrak{l}}_{2}(1){\otimes}t^{\mathfrak{m}}(1,a-1)$
$\displaystyle=$
$\displaystyle\,\Big{(}T^{\mathfrak{l}}(2,1)+T^{\mathfrak{l}}(1,2)\Big{)}{\otimes}(2-2^{1-a})\zeta^{\mathfrak{m}}(a)-T^{\mathfrak{l}}(3){\otimes}t^{\mathfrak{m}}(1,a-1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\frac{7}{4}(2-2^{1-a})\zeta^{\mathfrak{l}}(3){\otimes}\zeta^{\mathfrak{m}}(a)-\displaystyle\frac{7}{4}\zeta^{\mathfrak{l}}(3){\otimes}t^{\mathfrak{m}}(1,a-1)$

by Lemma 3.2 and Cor. 4.2. In the above and throughout the rest of the paper,
we often use substitutions such as $\eta_{3}\to\eta_{3}\eta_{1}$ to convert
the motivic integrals to the standard forms of motivic multiple $T$-, $t$\-
and $S$-values. Since $t^{\mathfrak{m}}(1,a-1)$ is ramified by Prop. 9.3
$D_{3}T^{\mathfrak{m}}(1,1,a+1)\not\in{\mathcal{L}}_{3}{\otimes}{\mathcal{H}}_{a}$
and therefore $T^{\mathfrak{m}}(1,1,a+1)$ is ramified by Thm. 2.1. ∎

######  Proposition 7.3.

For any $k\in\mathbb{N}$ the motivic triple $S$-values
$S^{\mathfrak{m}}(1,1,k)$ is ramified.

###### Proof.

Write $a=k-1$. First, we assume $a\geq 1$. Then

$\displaystyle\,D_{1}S^{\mathfrak{m}}(1,1,a+1)=\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}D_{1}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},\eta_{3},0_{a};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},\eta_{3},0_{a};1)$
$\displaystyle\,+I^{\mathfrak{l}}(\eta_{1};\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{a};1)+I^{\mathfrak{l}}(\eta_{2};\eta_{3};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{a};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},\eta_{3},0_{a};1)-I^{\mathfrak{l}}(0;\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{a};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}(\eta_{2}-\eta_{1})\eta_{3}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},\eta_{3},0_{a};1)$
$\displaystyle=$
$\displaystyle\,-2\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(1,a+1)\neq 0.$

Hence, $S^{\mathfrak{m}}(1,1,k)$ is ramified by Thm. 2.1. If $a=0$ then

$\displaystyle\,D_{1}S^{\mathfrak{m}}(1,1,1)=\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}D_{1}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},\eta_{3};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},\eta_{3};1)$
$\displaystyle\,+I^{\mathfrak{l}}(\eta_{1};\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{a};1)+I^{\mathfrak{l}}(\eta_{2};\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\Big{(}(\eta_{2}-\eta_{1})\eta_{3}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},\eta_{3};1)+\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,-\log^{\mathfrak{l}}2{\otimes}\Big{(}2T^{\mathfrak{m}}(1,1)-S^{\mathfrak{m}}(1,1)\Big{)}\neq
0$

since
$D_{1}T^{\mathfrak{m}}(1,1)=D_{1}S^{\mathfrak{m}}(1,1)=-\log^{\mathfrak{l}}2{\otimes}\log^{\mathfrak{m}}2\neq
0$ by (5.10) and (5.11). Hence, $S^{\mathfrak{m}}(1,1,k)$ is ramified by Thm.
2.1. ∎

######  Proposition 7.4.

For all $k,l\geq 2$ the motivic triple $T$\- and $S$-values
$T^{\mathfrak{m}}(1,k,l)$ (if $(k,l)\neq(2,2)$) and $S^{\mathfrak{m}}(1,k,l)$
are ramified.

###### Proof.

Let $a=k-1$ and $b=l-1$. We break the lemma into three cases:

I. $a$ and $b$ are even; II. $a$ is even, $b$ odd; III. $a$ is odd.

Case I. Suppose $a,b\geq 2$ and $a$ and $b$ are both even. Since $a+b\geq 4$
then we have the following possible cuts for $D_{a+1}$.

$0;\eta_{1},\eta_{2},0_{b-1},0,0_{a-b},\eta_{3},0,0_{b-1};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
$D_{a+1}T^{\mathfrak{m}}(1,a+1,b+1)$, $a$ even and $b\geq 2$.

First, since $a$ is even we observe that

$I^{\mathfrak{m}}(\eta_{1};\eta_{2},0_{a};0)=-\zeta^{\mathfrak{m}}_{a}(1;\eta_{1}\eta_{2})=-\zeta^{\mathfrak{m}}(a+1;\eta_{1}\eta_{2})=-I^{\mathfrak{m}}(0;\eta_{2},0_{a};\eta_{1}).$

We get

$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}\,)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{1};\eta_{2},0_{a};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{b};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{2},0_{a};\eta_{3})-I^{\mathfrak{l}}(0;\eta_{2},0_{a};\eta_{1})\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{b};1),$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\Big{(}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;\eta_{2},0_{a};1)-\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{2},0_{a};1)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{b};1),$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}(a+1){\otimes}\Big{(}t^{\mathfrak{m}}(1,b+1)-S^{\mathfrak{m}}(1,b+1)\Big{)},$
$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{2};0_{a},\eta_{3};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{b};1)=-T^{\mathfrak{l}}(a+1){\otimes}t^{\mathfrak{m}}(1,b+1).$

Since $b$ is even, by Lemma 3.2 and Prop. 5.5 we see that

$T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}\,)=\left\\{\begin{array}[]{ll}-T^{\mathfrak{l}}(a+1){\otimes}S^{\mathfrak{m}}(1,b+1)\in{\mathcal{L}}_{a+1}{\otimes}({\mathcal{H}}_{b+1}^{2}\setminus{\mathcal{H}}_{b+1}),\phantom{\displaystyle\frac{1}{2}}&\hbox{if
$a>0$;}\\\
-\log^{\mathfrak{l}}2{\otimes}S^{\mathfrak{m}}(1,b+1)\in({\mathcal{L}}_{1}^{2}\setminus{\mathcal{L}}_{1}){\otimes}({\mathcal{H}}_{b+1}^{2}\setminus{\mathcal{H}}_{b+1}),\phantom{\displaystyle\frac{1}{2}}&\hbox{if
$a=0$.}\end{array}\right.$

If $a\geq b$ then we need to consider the additional term

$\displaystyle
T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{a-b},\eta_{3},0_{b};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{b};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{a-b}(b+1){\otimes}T^{\mathfrak{m}}(1,b+1)=\binom{a}{b}T^{\mathfrak{l}}(a+1){\otimes}T^{\mathfrak{m}}(1,b+1)$

by Lemma 3.2 since $(-1)^{a-b}=1$. Thus

$T({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip
1.07639pt{}}\,)=T^{\mathfrak{l}}(a+1){\otimes}\left(\binom{a}{b}T^{\mathfrak{m}}(1,b+1)-S^{\mathfrak{m}}(1,b+1)\right).$

If $a\neq b$, by (5.13) we have

$D_{b+1}\left(\binom{a}{b}T^{\mathfrak{m}}(1,b+1)-S^{\mathfrak{m}}(1,b+1)\right)=2\left(\binom{a}{b}+1\right)T^{\mathfrak{l}}(b+1){\otimes}\log^{\mathfrak{m}}2$

which is ramified. Thus $T^{\mathfrak{m}}(1,k,l)$ is ramified in this case.

For $S^{\mathfrak{m}}(1,k,l)$, if $a+b\geq 4$ then similarly we have by Prop.
5.5

$S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}\,)=\left\\{\begin{array}[]{ll}-T^{\mathfrak{l}}(a+1){\otimes}T^{\mathfrak{m}}(1,b+1)\in{\mathcal{L}}_{a+1}{\otimes}({\mathcal{H}}_{b+1}^{2}\setminus{\mathcal{H}}_{b+1}),\phantom{\displaystyle\frac{1}{2}}&\hbox{if
$a>0$;}\\\
-\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(1,b+1)\in({\mathcal{L}}_{1}^{2}\setminus{\mathcal{L}}_{1}){\otimes}({\mathcal{H}}_{b+1}^{2}\setminus{\mathcal{H}}_{b+1}),\phantom{\displaystyle\frac{1}{2}}&\hbox{if
$a=0$.}\end{array}\right.$

If $a\geq b$ then we need to consider the additional term

$\displaystyle
S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}\,)=\binom{a}{b}T^{\mathfrak{l}}(a+1){\otimes}S^{\mathfrak{m}}(1,b+1).$

Thus

$S({\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip
1.07639pt{}}\,)=T^{\mathfrak{l}}(a+1){\otimes}\left(\binom{a}{b}S^{\mathfrak{m}}(1,b+1)+T^{\mathfrak{m}}(1,b+1)\right).$

Similar to the above we see this is ramified and therefore so is
$S^{\mathfrak{m}}(1,k,l)$.

Case II. Suppose $a,b\geq 1$, $a$ is even and $b$ is odd. Then

$\displaystyle\,D_{w-1}T^{\mathfrak{m}}(1,a+1,b+1)$ $\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{1};\eta_{2},0_{a},\eta_{3},0_{b};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1};1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\bar{1};\eta_{2},0_{a},\eta_{3},0_{b};1){\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{2},0_{a},\eta_{3},0_{b};1)-I^{\mathfrak{l}}(0;0_{b},\eta_{3},0_{a},\eta_{2};\bar{1})\Big{)}{\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=$
$\displaystyle\,\Big{(}T^{\mathfrak{l}}(a+1,b+1)-T^{\mathfrak{l}}_{b}(a+1,1)\Big{)}{\otimes}\log^{\mathfrak{m}}2$

by the substitutions $\eta_{2}\to-\eta_{2},\eta_{3}\to-\eta_{3}$ in
$I^{\mathfrak{l}}(0;0_{b},\eta_{3},0_{a},\eta_{2};\bar{1})$. By Cor. 4.2

$\displaystyle T^{\mathfrak{l}}_{b}(a+1,1)=$
$\displaystyle\,-\displaystyle\sum_{i=0}^{b}\binom{a+i}{i}T^{\mathfrak{l}}(a+i+1,b+1-i)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{i=0}^{b}(-1)^{i}\binom{a+i}{i}\binom{a+b+1}{a+i}T^{\mathfrak{l}}(a+b+2)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{i=0}^{b}(-1)^{i}\displaystyle\frac{(a+b+1)!}{a!i!(b+1-i)!}T^{\mathfrak{l}}(a+b+2)$
$\displaystyle=$
$\displaystyle\,\binom{a+b+1}{a}\displaystyle\sum_{i=0}^{b}(-1)^{i}\binom{b+1}{i}T^{\mathfrak{l}}(a+b+2)=0.$

Thus

$T^{\mathfrak{l}}(a+1,b+1)-T^{\mathfrak{l}}_{b}(a+1,1)=(-1)^{b+1}\binom{a+b+1}{a}T^{\mathfrak{l}}(a+b+2)\neq
0.$

Similar computation shows that
$D_{w-1}S^{\mathfrak{m}}(1,a+1,b+1)=-D_{w-1}T^{\mathfrak{m}}(1,a+1,b+1)$ which
is ramified, and therefore $S^{\mathfrak{m}}(1,a+1,b+1)$ is ramified by Thm.
2.1.

Case III. Suppose $a,b\geq 1$ and $a$ is odd. It is easy to see that

$\displaystyle D_{1}S^{\mathfrak{m}}(1,a+1,b+1)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{a},\eta_{3},0_{b};1)$
$\displaystyle\,\qquad+I^{\mathfrak{l}}(\eta_{1};\eta_{2};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{3},0_{b};1)\Big{)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}(\eta_{2}-\eta_{1})\eta_{3}I^{\mathfrak{l}}(0;\eta_{1};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{a},\eta_{3},0_{b};1)$
$\displaystyle=$
$\displaystyle\,-\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(a+1,b+1)$

which is ramified. Thus $S^{\mathfrak{m}}(1,a+1,b+1)$ is ramified.

Now we turn to $T^{\mathfrak{m}}(1,a+1,b+1)$. We divide into two cases. (A)
$b=1$ (B) $b\geq 2$.

(A) If $b=1$ then $a\geq 3$ and

$\displaystyle
D_{1}T^{\mathfrak{m}}(1,a+1,2)=\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{3};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{a};1)=\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{l}}(1,a+1)$

yielding that $T^{\mathfrak{m}}(1,a+1,2)$ is ramified in this case.

(B) Assume $b\geq 2$. We have the following possible cuts for
$D_{a+2}T^{\mathfrak{m}}(1,a+1,b+1)$.

$0;\eta_{1},\eta_{2},0_{b-2},0,0_{a-b+1},\eta_{3},0,0,0_{b-2};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{}}$Possible cuts of
$D_{a+2}T^{\mathfrak{m}}(1,a+1,b+1)$, $a$ odd and $b\geq 2$.

We get

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{1},\eta_{2},0_{a};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{3},0_{b};1)=T^{\mathfrak{l}}(1,a+1){\otimes}T^{\mathfrak{m}}(b+1),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{1};\eta_{2},0_{a},\eta_{3};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{b};1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{1},0_{a},\eta_{2};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{3},0_{b};1)=-T^{\mathfrak{l}}(a+1,1){\otimes}T^{\mathfrak{m}}(b+1).$

Since $a$ is odd so that both $T^{\mathfrak{l}}(1,a+1)$ and
$T^{\mathfrak{l}}(a+1,1)$ have odd weight 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$}}}\hskip
1.07639pt{}}=\Big{(}T^{\mathfrak{l}}(1,a+1)-T^{\mathfrak{l}}(a+1,1)\Big{)}{\otimes}T^{\mathfrak{m}}(b+1)\in{\mathcal{L}}_{a+2}{\otimes}{\mathcal{H}}_{b+1}$
(7.14)

by Lemma 5.1. Further

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{2};0_{a},\eta_{3},0;0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{b-1};1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{3}I^{\mathfrak{l}}(0;0,\eta_{2}\eta_{3},0_{a};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{b-1};1)$
$\displaystyle=$
$\displaystyle\,-T^{\mathfrak{l}}_{1}(a+1){\otimes}t^{\mathfrak{m}}(1,b)=(a+1)T^{\mathfrak{l}}(a+2){\otimes}t^{\mathfrak{m}}(1,b)$
(7.15)

by Lemma 3.2, which is ramified by Lemma 9.3.

(1) If $b\geq 2$ and $a<b-1$ then
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}\,$ does not appear and we see that
$D_{a+2}T^{\mathfrak{m}}(1,a+1,b+1)={\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}\,$ is ramified by (7.14) and (7.15).

(2) If $a\geq b-1\geq 1$ then
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}\,$ appears and we get

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{a-b+1},\eta_{3},0_{b};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{2},0_{b-1};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{a-b+1}(b+1){\otimes}T^{\mathfrak{m}}(1,b)=(-1)^{b}\binom{a+1}{b}T^{\mathfrak{l}}(a+2){\otimes}T^{\mathfrak{m}}(1,b).$

By combining ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}\,$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}\,$ we get

${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip
1.07639pt{}}=T^{\mathfrak{l}}(a+2){\otimes}\bigg{(}(a+1)t^{\mathfrak{m}}(1,b)+(-1)^{b}\binom{a+1}{b}T^{\mathfrak{m}}(1,b)\bigg{)}.$

Since $D_{1}T^{\mathfrak{m}}(1,b)=0$ we see that by Lemma 9.2

$D_{1}\left((a+1)t^{\mathfrak{m}}(1,b)+(-1)^{b}\binom{a+1}{b}T^{\mathfrak{m}}(1,b)\right)=2(a+1)\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(b)\neq
0.$

Thus ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip
1.07639pt{}}\in{\mathcal{L}}_{a+2}{\otimes}({\mathcal{H}}_{b+1}^{2}\setminus{\mathcal{H}}_{b+1})$.
Combining this with (7.14) we find that
$D_{a+2}T^{\mathfrak{m}}(1,a+1,b+1)\not\in{\mathcal{L}}_{a+2}{\otimes}{\mathcal{H}}_{b+1}$.

Putting (A) and (B) together we can conclude that if $a+b\geq 3$ and $a$ is
odd then $T^{\mathfrak{m}}(1,a+1,b+1)$ is ramified by Thm. 2.1. ∎

##  8 Triple $T$\- and $S$-values: the general case

We now consider motivic triple $T$-values $T^{\mathfrak{m}}(k,l,m)$ and
motivic triple $S$-values $S^{\mathfrak{m}}(k,l,m)$. In view of Prop. 7.1 and
Prop. 7.4 we may assume that $a=k-1\geq 1,c=m-1\geq 1$ and $b=l-1\geq 0$ such
that $w=k+l+m$. Since $T^{\mathfrak{m}}(O_{>1},E,O_{>1})$ are all unramified
by Murakami (see [12, Thm. 41]), $S^{\mathfrak{m}}(E,O_{>1},O_{>1})$ and
$S^{\mathfrak{m}}(O_{>1},E,O_{>1})$ are all unramified by Prop. 6.1, we only
consider the following three cases for triple $T^{\mathfrak{m}}$-values and
two cases for triple $S^{\mathfrak{m}}$-values when $w$ is even:

$f^{\mathfrak{m}}(E,E,E),\quad f^{\mathfrak{m}}(E,O,O_{>1}),\quad
T^{\mathfrak{m}}(O_{>1},O,E),$

where $f=T$ or $S$. When $w$ is odd, since $T^{\mathfrak{m}}(E,1,E)$ are all
unramified by Murakami (see [12, Thm. 41]) and $S^{\mathfrak{m}}(E,1,E)$ are
all unramified by Prop. 6.1 we only consider the following three cases for
triple $T^{\mathfrak{m}}$-values and three cases for triple
$S^{\mathfrak{m}}$-values when $w$ is even:

$f^{\mathfrak{m}}(E,E,O_{>1}),\quad f^{\mathfrak{m}}(O_{>1},E,E),\quad
f^{\mathfrak{m}}(E,O_{>1},E)$,

where $f=T$ or $S$.

Case I. Suppose $a$ and $b$ have the same parity. Then we have the following
possible cuts for $D_{a+b+1}T^{\mathfrak{m}}(k,l,m)$.

$0;\eta_{1},\underbrace{0,\dots,0}_{\scriptstyle\text{$a$
times}},\eta_{2},\underbrace{0,\dots,0}_{\scriptstyle\text{$b$
times}},\eta_{3},0_{a},0,0_{c-a-1};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{}}$Possible cuts of $D_{a+b+1}f^{\mathfrak{m}}(k,l,m)$
($f=T$ or $S$), $a\equiv b\pmod{2}$.

We have

$\displaystyle D_{a+b+1}T^{\mathfrak{m}}(k,l,m)=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\bigg{[}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{c};1)$
$\displaystyle\,+\delta_{c>a+1}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},0_{a};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{c-a};1)$
$\displaystyle\,+\delta_{a=c}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},0_{a};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2};1)$
$\displaystyle\,+\delta_{c+1>a}I^{\mathfrak{l}}(0;0_{b-c+a},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{c-a};1)$
$\displaystyle\,+\delta_{c+1<a}I^{\mathfrak{l}}(0;0_{a-c-1},\eta_{2},0_{b},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{c+1};1)\bigg{]}.$

Since $a$ and $b$ have the same parity, we have

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;0_{a},\eta_{2},0_{b};\eta_{3})+I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b};0)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{c};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\Big{(}\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;0_{a},\eta_{2},0_{b};1)+\eta_{2}\eta_{3}I^{\mathfrak{l}}(1;0_{a},\eta_{2},0_{b};0)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},\eta_{3},0_{c};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{a}(b+1){\otimes}\Big{(}t^{\mathfrak{m}}(1,c+1)-S^{\mathfrak{m}}(1,c+1)\Big{)}$
$\displaystyle=$
$\displaystyle\,(-1)^{a}\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}\Big{(}t^{\mathfrak{m}}(1,c+1)-S^{\mathfrak{m}}(1,c+1)\Big{)},$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,-\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{3}I^{\mathfrak{l}}(0;0_{a},\eta_{3},0_{b};1){\otimes}\eta_{1}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{c-a};1)$
$\displaystyle=$
$\displaystyle\,-T^{\mathfrak{l}}_{a}(b+1){\otimes}t^{\mathfrak{m}}(a+1,c-a+1)=-(-1)^{a}\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}t^{\mathfrak{m}}(a+1,c-a+1),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{3}=\pm
1}\eta_{1}\eta_{3}\Big{(}I^{\mathfrak{l}}(0;0_{b},\eta_{3},0_{a};1)+\eta_{2}I^{\mathfrak{l}}(1;0_{b},\eta_{3},0_{a};0)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{b}(a+1){\otimes}t^{\mathfrak{m}}(a+1,1)-T^{\mathfrak{l}}_{a}(b+1){\otimes}T^{\mathfrak{m}}(a+1,1)$
$\displaystyle=$
$\displaystyle\,-(-1)^{a}2\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}\Big{(}t^{\mathfrak{m}}(a+1,1)-T^{\mathfrak{m}}(a+1,1)\Big{)},$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{b-c+a},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{c-a};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{b-c+a}(c+1){\otimes}T^{\mathfrak{m}}(a+1,c-a+1)=(-1)^{c}\binom{a+b}{c}T^{\mathfrak{l}}(a+b+1){\otimes}T^{\mathfrak{m}}(a+1,c-a+1),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
5$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,T^{\mathfrak{l}}_{a-c-1}(b+1,c+1){\otimes}T^{\mathfrak{m}}(c+2)$
$\displaystyle=$
$\displaystyle\,(-1)^{a-c-1}\binom{a+b}{c}T^{\mathfrak{l}}(a+b-c,c+1){\otimes}T^{\mathfrak{m}}(c+2).$

Note that ${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
5$}}}\hskip 1.07639pt{}}\,$ is clearly unramified while
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}\,$ is unramified by Lemma 9.1 since $a\geq 1$ and $c-a+1\geq 2$
if $c\geq a+1$. Thus by putting the ramified contributions together we get
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=T^{\mathfrak{l}}(a+b+1){\otimes}X$ where

$\displaystyle X:=$
$\displaystyle\,(-1)^{a}\binom{a+b}{a}\Big{(}t^{\mathfrak{m}}(1,c+1)-S^{\mathfrak{m}}(1,c+1)\Big{)}$
$\displaystyle\,-(-1)^{a}2\binom{a+b}{a}\delta_{a=c}\Big{(}t^{\mathfrak{m}}(a+1,1)-T^{\mathfrak{m}}(a+1,1)\Big{)}$
$\displaystyle\,+(-1)^{c}\binom{a+b}{c}\delta_{c+1>a}T^{\mathfrak{m}}(a+1,c-a+1).$

By Lemma 9.2 and Lemma 3.3(2)

$D_{1}t^{\mathfrak{m}}(1,c+1)=2\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(c+1),\qquad
D_{1}S^{\mathfrak{m}}(1,c+1)=-2\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(c+1).$
(8.16)

If $c\neq a$ then
$D_{1}t^{\mathfrak{m}}(a+1,c-a+1)=D_{1}T^{\mathfrak{m}}(a+1,c-a+1)=0$ by Lemma
9.2 and Lemma 3.3(1). Hence,

$\displaystyle
D_{1}X=(-1)^{a}(2+2)\binom{a+b}{a}\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(a+1)\neq
0.$

If $a=c$, then we have

$\displaystyle
X=(-1)^{a}\binom{a+b}{a}\Big{(}3T^{\mathfrak{m}}(a+1,1)+t^{\mathfrak{m}}(1,a+1)-S^{\mathfrak{m}}(1,a+1)-2t^{\mathfrak{m}}(a+1,1)\Big{)}.$

Thus if $a=c$ then by (8.16) and (5.10)

$\displaystyle
D_{1}X=9(-1)^{a}\binom{a+b}{a}\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(a+1)\neq
0.$

In any case we see that $D_{a+b+1}T^{\mathfrak{m}}(k,l,m)$ is ramified. Hence,
$T^{\mathfrak{m}}(k,l,m)$ is ramified when $k\equiv l\pmod{2}$.

By the same argument we see that if $(k,l,m)=(E,E,O)$ then

$\displaystyle D_{a+b+1}S^{\mathfrak{m}}(k,l,m)=$
$\displaystyle\,(-1)^{a}\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}\Big{(}\tilde{\zeta}^{\mathfrak{m}}(1,c+1)-T^{\mathfrak{m}}(1,c+1)\Big{)}$
$\displaystyle\,-(-1)^{a}\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}\delta_{c>a+1}\tilde{\zeta}^{\mathfrak{m}}(a+1,c-a+1)$
$\displaystyle\,-(-1)^{a}2\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}\delta_{a=c}\Big{(}\tilde{\zeta}^{\mathfrak{m}}(a+1,1)-S^{\mathfrak{m}}(a+1,1)\Big{)}$
$\displaystyle\,+(-1)^{c}\binom{a+b}{c}T^{\mathfrak{l}}(a+b+1){\otimes}\delta_{c+1>a}S^{\mathfrak{m}}(a+1,c-a+1)$
$\displaystyle\,+(-1)^{a-c-1}\binom{a+b}{c}T^{\mathfrak{l}}(a+b-c,c+1){\otimes}\delta_{c+1<a}S^{\mathfrak{m}}(c+2),$

where the second and the fifth terms are again unramified. Now that $a$ is odd
and $c$ is even the third term cannot appear. By Thm. 2.3he only ramified
terms are produce by
$(-1)^{a}\binom{a+b}{a}T^{\mathfrak{l}}(a+b+1){\otimes}Y$, where

$Y=\binom{a+b}{a}T^{\mathfrak{m}}(1,c+1)+\binom{a+b}{c}\delta_{c+1>a}S^{\mathfrak{m}}(a+1,c-a+1)$

for which by (5.12) and (5.13) we have

$\displaystyle
D_{c+1}Y=\bigg{(}2\binom{a+b}{a}-\delta_{c>a}2\binom{a+b}{c}\binom{c}{a}\bigg{)}T^{\mathfrak{l}}(c+1){\otimes}\log^{\mathfrak{m}}2$
$\displaystyle=2\binom{a+b}{a}\bigg{(}1-\delta_{c>a}\binom{b}{c-a}\bigg{)}T^{\mathfrak{l}}(c+1){\otimes}\log^{\mathfrak{m}}2\neq
0$

since $b\neq c-a$ as they have different parities. Thus
$S^{\mathfrak{m}}(k,l,m)$ is ramified in this case, too.

Case II. Suppose $a\not\equiv b\equiv c\pmod{2}$. Then we have the following
possible cuts for $D_{b+c+1}T^{\mathfrak{m}}(k,l,m)$.

$0;\eta_{1},0_{a-c-1},0,0_{c},\eta_{2},0_{b-a+c},0,0_{c-a-1},\eta_{3},0_{c-a-1},0,0_{a};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{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
5$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
6$}}}\hskip 0.75346pt{}}$Possible cuts of $D_{b+c+1}f^{\mathfrak{m}}(k,l,m)$
($f=T$ or $S$), $a\not\equiv b\equiv c\pmod{2}$.

Thus

$\displaystyle\,D_{b+c+1}T^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}\bigg{[}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2};1)$
$\displaystyle\,+{\delta}_{c\geq
a+1}\,I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b},\eta_{3},0_{c-a-1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a+1};1)$
$\displaystyle\,+{\delta}_{c<a}\,I^{\mathfrak{l}}(0;0_{c},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{c\leq a\leq
b+c}\,I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{b-a+c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{a+1=c}\,I^{\mathfrak{l}}(0;\eta_{1},0_{a},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{a=b+c,2|a}\,I^{\mathfrak{l}}(0;\eta_{1},0_{a};\eta_{2}){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{b},\eta_{3},0_{c};1)\bigg{]}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\bigg{[}\Big{(}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{b},\eta_{3},0_{c};1)+\eta_{1}\eta_{3}I^{\mathfrak{l}}(1;0_{b},\eta_{3},0_{c};0)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2};1)$
$\displaystyle\,-{\delta}_{c\geq
a+1}\,\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{c-a-1},\eta_{3},0_{b},\eta_{2},0_{a};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a+1};1)$
$\displaystyle\,+{\delta}_{c<a}\,\eta_{1}\eta_{2}I^{\mathfrak{l}}(0;0_{c},\eta_{2},0_{b};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{c<a\leq
b+c}\,\eta_{2}\eta_{3}I^{\mathfrak{l}}(1;0_{a},\eta_{2},0_{b-a+c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{a+1=c}\,\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;\eta_{1}\eta_{3},0_{a},\eta_{2}\eta_{3},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{a=b+c,2|a}\,\eta_{1}\eta_{3}I^{\mathfrak{l}}(0;\eta_{1},0_{a};1){\otimes}I^{\mathfrak{m}}(0;\eta_{2},0_{b},\eta_{3},0_{c};1)\bigg{]}$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{b}(c+1){\otimes}T^{\mathfrak{m}}(a+1,1)-T^{\mathfrak{l}}_{c}(b+1){\otimes}t^{\mathfrak{m}}(a+1,1)$
$\displaystyle\,-{\delta}_{c\geq
a+1}\,T^{\mathfrak{l}}_{c-a-1}(b+1,a+1){\otimes}T^{\mathfrak{m}}(a+2)$
$\displaystyle\,+{\delta}_{c<a}\,T_{c}(b+1){\otimes}t^{\mathfrak{m}}(a-c+1,c+1)$
$\displaystyle\,-{\delta}_{c<a\leq
b+c}\,T_{b-a+c}(a+1){\otimes}S^{\mathfrak{m}}(a-c+1,c+1)$
$\displaystyle\,+{\delta}_{a+1=c}\,T^{\mathfrak{l}}(a+1,b+1){\otimes}T^{\mathfrak{m}}(a+2)$
$\displaystyle\,+{\delta}_{a=b+c,2|a}\,T^{\mathfrak{l}}(a+1){\otimes}S^{\mathfrak{m}}(b+1,c+1)$
$\displaystyle\equiv$
$\displaystyle\,(-1)^{b}T^{\mathfrak{l}}(b+c+1){\otimes}\bigg{[}\binom{b+c}{b}\Big{(}T^{\mathfrak{m}}(a+1,1)-t^{\mathfrak{m}}(a+1,1)\Big{)}$
$\displaystyle\,\qquad\qquad\qquad+\bigg{(}(-1)^{b}{\delta}_{a=b+c,2|a}-{\delta}_{c<a\leq
b+c}\,\binom{b+c}{a}\bigg{)}S^{\mathfrak{m}}(a-c+1,c+1)\bigg{]}$

modulo the unramified terms $\cdots{\otimes}T(a+2)$ by Lemma 3.2 and Lemma
5.4, unramified term $T_{c}(b+1){\otimes}t^{\mathfrak{m}}(a-c+1,c+1)$ by Lemma
9.1. Note that if $c<a$ then $a-c+1,c+1\geq 2$ and therefore
$D_{1}S^{\mathfrak{m}}(a-c+1,c+1)=0$ by Lemma 2.2. But

$D_{1}\Big{(}T^{\mathfrak{m}}(a+1,1)-t^{\mathfrak{m}}(a+1,1)\Big{)}=2\log^{\mathfrak{l}}2{\otimes}T^{\mathfrak{m}}(a+1)\neq
0.$

This implies that $T^{\mathfrak{m}}(k,l,m)$ is ramified in this case.

Similarly, for $S^{\mathfrak{m}}(k,l,m)$ by assuming $b$ and $c$ are both odd
we can compute to get

$\displaystyle\,D_{b+c+1}S^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{b}(c+1){\otimes}S^{\mathfrak{m}}(a+1,1)-T^{\mathfrak{l}}_{c}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+1,1)$
$\displaystyle\,-{\delta}_{c\geq
a+1}\,T^{\mathfrak{l}}_{c-a-1}(b+1,a+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+2)$
$\displaystyle\,+{\delta}_{c<a}\,T_{c}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a-c+1,c+1)$
$\displaystyle\,-{\delta}_{c<a\leq
b+c}\,T_{b-a+c}(a+1){\otimes}T^{\mathfrak{m}}(a-c+1,c+1)$
$\displaystyle\,+{\delta}_{a=c+1}\,S^{\mathfrak{l}}(a+1,b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(c+1)$
$\displaystyle\,+{\delta}_{a=b+c,2|a}\,T^{\mathfrak{l}}(a+1){\otimes}T^{\mathfrak{m}}(b+1,c+1)$
$\displaystyle\equiv$
$\displaystyle\,(-1)^{b}T^{\mathfrak{l}}(b+c+1){\otimes}\bigg{[}\binom{b+c}{b}\Big{(}S^{\mathfrak{m}}(a+1,1)-\tilde{\zeta}^{\mathfrak{m}}(a+1,1)\Big{)}$
$\displaystyle\,\qquad\qquad-{\delta}_{c<a\leq
b+c}\,\binom{b+c}{a}T^{\mathfrak{m}}(a-c+1,c+1)+(-1)^{b}{\delta}_{a=b+c,2|a}\,T^{\mathfrak{m}}(b+1,c+1)\bigg{]}$

modulo all the unramified terms. We can see that $S^{\mathfrak{m}}(k,l,m)$ is
ramified since

$D_{1}\Big{(}S^{\mathfrak{m}}(a+1,1)-\tilde{\zeta}^{\mathfrak{m}}(a+1,1)\Big{)}=2\log^{\mathfrak{l}}2{\otimes}S^{\mathfrak{m}}(a+1)\neq
0$

using (5.9) and Lemma 9.7. Note that in this case, $a-c+1,c+1,b+1\geq 2$.

Case III. Suppose $b\not\equiv a\equiv c\equiv 1\pmod{2}$ and $b\geq 2$.

(i) First we consider $D_{a+2}$ when $a\leq c$.

$0;\eta_{1},0_{b-2},0,0_{a-b+1},\eta_{2},0,0,0_{b-2},\eta_{3},0_{a-b+1},0,0_{c-a+b-2};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{}}$Possible cuts of $D_{a+2}f^{\mathfrak{m}}(k,l,m)$
($f=T$ or $S$), $a\leq c$, $b\not\equiv a\equiv c\equiv 1\pmod{2}$.

Thus

$\displaystyle\,D_{a+2}T^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0;0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{b-1},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{b\leq
a+1}\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{a-b+1},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{b-1},\eta_{3},0_{c};1)$
$\displaystyle\,+{\delta}_{b\leq a+1\leq
b+c}\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},0_{a-b+1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{c-a+b-1};1)$
$\displaystyle\,+{\delta}_{c\leq a+1\leq
b+c}\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{a-c+1},\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b-a+c-1};1)$
$\displaystyle=$
$\displaystyle\,-T^{\mathfrak{l}}_{1}(a+1){\otimes}S^{\mathfrak{m}}(b,c+1)+{\delta}_{b\leq
a+1}\,T^{\mathfrak{l}}_{a-b+1}(b+1){\otimes}t^{\mathfrak{m}}(b,c+1)$
$\displaystyle\,-{\delta}_{b\leq a+1\leq
b+c}\,T^{\mathfrak{l}}_{a-b+1}(b+1){\otimes}t^{\mathfrak{m}}(a+1,c-a+b)+{\delta}_{c=a}\,T^{\mathfrak{l}}_{1}(c+1){\otimes}T^{\mathfrak{m}}(a+1,b)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}(a+2){\otimes}\bigg{[}(a+1)S^{\mathfrak{m}}(b,c+1)+{\delta}_{b\leq
a+1}\,\binom{a+1}{b}t^{\mathfrak{m}}(b,c+1)$
$\displaystyle\,\qquad-{\delta}_{b\leq a+1\leq
b+c}\,\binom{a+1}{b}t^{\mathfrak{m}}(a+1,c-a+b)-{\delta}_{c=a}\,(a+1)T^{\mathfrak{m}}(c+1,b)\bigg{]}$

by Lemma 3.2 since $a\leq c$. But

$\displaystyle\,D_{b+c}S^{\mathfrak{m}}(b,c+1)=\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\bar{1};0_{b-1},\eta_{2},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\bar{1};1)$
$\displaystyle=$
$\displaystyle\,\Big{(}T^{\mathfrak{l}}_{c}(b)+T^{\mathfrak{l}}_{b-1}(c+1)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\bar{1};1)=-2\binom{b+c-1}{c}T^{\mathfrak{l}}(b+c){\otimes}\log^{\mathfrak{m}}2\neq
0$

while $D_{b+c}t^{\mathfrak{m}}(b,c+1)=D_{b+c}t^{\mathfrak{m}}(a+1,c-a+b)=0$ by
Lemma 9.4 since $c+1\geq 2$ and $c-a+b\geq 2$. If $a=c$ we need to compute one
extra contribution:

$\displaystyle\,D_{b+c}T^{\mathfrak{m}}(c+1,b)=-\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\bar{1};0_{c},\eta_{2},0_{b-1};1){\otimes}I^{\mathfrak{m}}(0;\bar{1};1)$
$\displaystyle=$
$\displaystyle\,-\Big{(}T^{\mathfrak{l}}_{c}(b)+T^{\mathfrak{l}}_{b-1}(c+1)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\bar{1};1)=2\binom{b+c-1}{c}T^{\mathfrak{l}}(b+c){\otimes}\log^{\mathfrak{m}}2$

so that

$\displaystyle
D_{b+c}\Big{(}S^{\mathfrak{m}}(b,c+1)-T^{\mathfrak{m}}(c+1,b)\Big{)}=-4\binom{b+c-1}{c}T^{\mathfrak{l}}(b+c){\otimes}\log^{\mathfrak{m}}2\neq
0.$

Thus $D_{a+2}T^{\mathfrak{m}}(k,l,m)$ is ramified in this case.

Similarly,

$\displaystyle\,D_{a+2}S^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,-T^{\mathfrak{l}}_{1}(a+1){\otimes}T^{\mathfrak{m}}(b,c+1)+{\delta}_{b\leq
a+1}\,T^{\mathfrak{l}}_{a-b+1}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(b,c+1)$
$\displaystyle\,-{\delta}_{b\leq a+1\leq
b+c}\,T^{\mathfrak{l}}_{a-b+1}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+1,c-a+b)+{\delta}_{c=a}\,T^{\mathfrak{l}}_{1}(c+1){\otimes}S^{\mathfrak{m}}(a+1,b)$
$\displaystyle\equiv$
$\displaystyle\,T^{\mathfrak{l}}(a+2){\otimes}\bigg{[}(a+1)\Big{(}T^{\mathfrak{m}}(b,c+1)-{\delta}_{c=a}\,S^{\mathfrak{m}}(a+1,b)\Big{)}-{\delta}_{a+1=b+c}\,\binom{a+1}{b}\tilde{\zeta}^{\mathfrak{m}}(a+1,1)\bigg{]}$

modulo unramified terms. This is clearly ramified if $c\neq a$ and $a+1\neq
b+c$ by Prop. 5.5. If $c=a$ then $a+1\neq b+c$ since $b\geq 2$, and therefore
by (5.12) and (5.13)

$\displaystyle
D_{a+b}\Big{(}T^{\mathfrak{m}}(b,a+1)-S^{\mathfrak{m}}(a+1,b)\Big{)}=4\binom{a+b-1}{a}T^{\mathfrak{l}}(a+b){\otimes}\log^{\mathfrak{m}}2$

which is ramified. If $a+1=b+c$ then $a\neq c$ and therefore

$\displaystyle
D_{1}\Big{(}(a+1)T^{\mathfrak{m}}(b,a+1)-\binom{a+1}{b}\tilde{\zeta}^{\mathfrak{m}}(a+1,1)\Big{)}=\binom{a+1}{b}S^{\mathfrak{m}}(a+1){\otimes}\log^{\mathfrak{m}}2\neq
0.$

Consequently, $D_{a+2}S^{\mathfrak{m}}(k,l,m)$ is always ramified in this
case.

(ii) Next we consider $D_{c+2}$ when $c\leq a-2$.

$0;\eta_{1},0_{a+b-c-2},0,0_{c-b+1},\eta_{2},0_{c-a+1},0,0_{a+b-c-2},\eta_{3},0_{c-b+1},0,0_{b-2};1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
3$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
2$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip 0.75346pt{}}$Possible cuts of $D_{c+2}T^{\mathfrak{m}}(k,l,m)$,
$c\leq a-2$, $b\not\equiv a\equiv c\equiv 1\pmod{2}$.

We have

$\displaystyle\,D_{c+2}T^{\mathfrak{m}}(k,l,m)$ $\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0,\eta_{3},0_{c};1){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b-1};1)$
$\displaystyle\,+{\delta}_{c\geq
b-1}\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},0_{c-b+1};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b-1};1)$
$\displaystyle\,+{\delta}_{a\geq c-b+1\geq
0}\displaystyle\sum_{\eta_{1},\eta_{2},\eta_{3}=\pm
1}\eta_{1}\eta_{2}\eta_{3}I^{\mathfrak{l}}(0;0_{c-b+1},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a+b-c-1},\eta_{3},0_{c};1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}_{1}(c+1){\otimes}T^{\mathfrak{m}}(a+1,b)-{\delta}_{c\geq
b-1}\,T^{\mathfrak{l}}_{c-b+1}(b+1){\otimes}t^{\mathfrak{m}}(a+1,b)$
$\displaystyle\,+{\delta}_{a\geq c-b+1\geq
0}\,T^{\mathfrak{l}}_{c-b+1}(b+1){\otimes}t^{\mathfrak{m}}(a+b-c,c+1)$
$\displaystyle=$
$\displaystyle\,T^{\mathfrak{l}}(c+2){\otimes}\bigg{[}-(c+1)T^{\mathfrak{m}}(a+1,b)-{\delta}_{c\geq
b-1}\,\binom{c+1}{b}t^{\mathfrak{m}}(a+1,b)$
$\displaystyle\,\qquad\qquad+{\delta}_{a\geq c-b+1\geq
0}\,\binom{c+1}{b}t^{\mathfrak{m}}(a+b-c,c+1)\bigg{]}$

by Lemma 3.2 since $c\leq a-2$. But

$\displaystyle D_{a+b}T^{\mathfrak{m}}(a+1,b)=$
$\displaystyle\,-\displaystyle\sum_{\eta_{2}=\pm
1}\eta_{2}I^{\mathfrak{l}}(\bar{1};0_{a},\eta_{2},0_{b-1};1){\otimes}I^{\mathfrak{m}}(0;\bar{1};1)$
$\displaystyle=$
$\displaystyle\,-\Big{(}T^{\mathfrak{l}}_{a}(b)+T^{\mathfrak{l}}_{b-1}(a+1)\Big{)}{\otimes}I^{\mathfrak{m}}(0;\bar{1};1)$
$\displaystyle=$
$\displaystyle\,2\binom{a+b-1}{a}T^{\mathfrak{l}}(a+b){\otimes}\log^{\mathfrak{m}}2\neq
0$

while $D_{a+b}t^{\mathfrak{m}}(a+1,b)=D_{a+b}t^{\mathfrak{m}}(a+b-c,c+1)=0$.
Thus $D_{c+2}T^{\mathfrak{m}}(k,l,m)$ is ramified in this case.

Similarly,

$\displaystyle D_{c+2}S^{\mathfrak{m}}(k,l,m)=$
$\displaystyle\,T^{\mathfrak{l}}_{1}(c+1){\otimes}S^{\mathfrak{m}}(a+1,b)-{\delta}_{c\geq
b-1}\,T^{\mathfrak{l}}_{c-b+1}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+1,b)$
$\displaystyle\,+{\delta}_{a\geq c-b+1\geq
0}\,T^{\mathfrak{l}}_{c-b+1}(b+1){\otimes}\tilde{\zeta}^{\mathfrak{m}}(a+b-c,c+1)$

whose first term is the only ramified term so the expression as a whole is
ramified.

Combining (i) and (ii) we see that both $T^{\mathfrak{m}}(k,l,m)$ and
$S^{\mathfrak{m}}(k,l,m)$ are ramified by Thm. 2.1.

We have now exhausted all the possibilities for the triple $T$-values and
triple $S$-values, which complete the proof of Thm. 1.2 and Thm. 1.3.

##  9 Ramified and unramified MtVs

In this section, we shall prove Thm. 1.5 and Thm. 1.6 using Glanois’s descent
theory. We start with the following result due to Murakami.

######  Lemma 9.1.

_(cf.[12, Thm. 8])_ If positive integers $k_{1}\geq 2,\dots,k_{d}\geq 2$ then
the motivic multiple $t$-value $t^{\mathfrak{m}}(k_{1},\dots,k_{d})$ is
unramified.

In [6], Charlton improved some of Murakami’s results and proved the following.
For ${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{d})\in\mathbb{N}^{d}$, we put
${\boldsymbol{\sl{k}}}_{i,j}=(k_{i},\ldots,k_{j})$ and set
${\boldsymbol{\sl{k}}}_{j,i}$ to be the reversal of
${\boldsymbol{\sl{k}}}_{i,j}$ for all $1\leq i\leq j\leq d$.

######  Lemma 9.2.

_(cf.[6, Prop. 5.8])_ For every composition
${\boldsymbol{\sl{k}}}=(k_{1},\dots,k_{d})$ of positive integers

$D_{1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=\log^{\mathfrak{l}}2{\otimes}\Big{(}2\delta_{k_{1}=1}\,t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{2,d})-\delta_{k_{d}=1}\,t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,d-1})\Big{)},$
(9.17)

where ${\delta}_{\bullet}=1$ if $\bullet$ is true and $0$ otherwise.

######  Proposition 9.3.

For integers $a,b$ with $a+b\geq 1$ the motivic $t$-values
$t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b})$ are ramified for all
compositions ${\boldsymbol{\sl{k}}}$.

###### Proof.

Without loss of generality we may assume ${\boldsymbol{\sl{k}}}$ neither
starts or ends with 1. If ${\boldsymbol{\sl{k}}}$ is empty and $b=0$ then from
Lemma 3.2 we may assume $a\geq 2$. By [6, Prop. 5.8] we get

$\displaystyle D_{1}t^{\mathfrak{m}}(1_{a})=$
$\displaystyle\,\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}(1_{a-1})\neq 0.$

Thus the proposition follows from Thm. 2.1.

Now suppose ${\boldsymbol{\sl{k}}}$ is nonempty. By (9.17) we get

$D_{1}t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b})=\log^{\mathfrak{l}}2{\otimes}\Big{(}2\delta_{a>0}t^{\mathfrak{m}}(1_{a-1},{\boldsymbol{\sl{k}}},1_{b})-\delta_{b>0}t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b-1})\Big{)}.$

Defining $D_{1}^{(0)}=D_{1}$ and
$D_{1}^{(\ell)}=\text{id}{\otimes}D_{1}^{(\ell-1)}$ for all $\ell\geq 1$ we
get

$D_{1}^{(\ell)}t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b})=(\log^{\mathfrak{l}}2)^{{\otimes}(\ell-1)}{\otimes}\displaystyle\sum_{\begin{subarray}{c}i+j=\ell\\\
i,j\geq 0\end{subarray}}{\delta}_{a\geq i,b\geq
j}\,C_{i,j}^{(\ell)}t^{\mathfrak{m}}(1_{a-i},{\boldsymbol{\sl{k}}},1_{b-j}),$

where the coefficients $C_{i,j}^{(\ell)}$ satisfy the recurrence relation

$C_{i,j}^{(\ell)}=2C_{i-1,j}^{(\ell-1)}-C_{i,j-1}^{(\ell-1)}.$

With substitution $C_{i,j}^{(\ell)}=2^{i}(-1)^{j}B_{i,j}^{(\ell)}$ the
recurrence becomes

$B_{i,j}^{(\ell)}=B_{i-1,j}^{(\ell-1)}+B_{i,j-1}^{(\ell-1)}$

so that $B_{i,j}^{(\ell)}=\binom{\ell}{i}$ are just the binomial coefficients.
Hence, $C_{i,j}^{(\ell)}=2^{i}(-1)^{j}\binom{\ell}{i}$ and therefore

$\displaystyle
D_{1}^{(a+b)}t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b})=(\log^{\mathfrak{l}}2)^{{\otimes}(a+b-1)}{\otimes}2^{a}(-1)^{b}\binom{a+b}{a}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\neq
0.$

Repeatedly using Thm. 2.1 we see that none of the $D_{1}^{(\ell)}$
($\ell<a+b$) vanishes so that
$t^{\mathfrak{m}}(1_{a},{\boldsymbol{\sl{k}}},1_{b})$ is ramified. ∎

The following fact will be useful in our computations later.

######  Lemma 9.4.

For any admissible composition ${\boldsymbol{\sl{k}}}$ of weight $w$ we have
$D_{w-1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$.

###### Proof.

Suppose ${\boldsymbol{\sl{k}}}=(n_{1}+1,\dots,n_{d}+1)$, $n_{d}\geq 1$. Then

$\displaystyle D_{w-1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=$
$\displaystyle\,\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\eta_{1}D_{w-1}I^{\mathfrak{m}}(0;\eta_{1},0_{n_{1}},\dots,,\eta_{d},0_{n_{d}};1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{\eta_{2},\dots,\eta_{d}=\pm
1}I^{\mathfrak{l}}(\bar{1};0_{n_{1}},\eta_{2},0_{n_{2}},\dots,\eta_{d},0_{n_{d}};1){\otimes}\log^{\mathfrak{m}}2=0$

by Lemma 2.4. ∎

In fact, Charlton proved a detailed computation of the derivation $D_{r}$ on
motivic MtVs as follows. Lemma 9.2 above is only one of its corollaries.

######  Theorem 9.5.

_(cf.[6, Prop. 5.6])_ Let
${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{d})\in\mathbb{N}^{d}$. Then

$\displaystyle
D_{r}\big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\big{)}=\displaystyle\sum_{1\leq
j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{1,j}|=r}t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,j}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{j+1,d})$
(9.18) $\displaystyle+\\!\\!\displaystyle\sum_{1\leq i<j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{i+1,j}|\leq
r<|{\boldsymbol{\sl{k}}}_{i,j}|-1}\binom{\zeta^{\mathfrak{l}}_{r-|{\boldsymbol{\sl{k}}}_{i+1,j}|}({\boldsymbol{\sl{k}}}_{i+1,j})}{-\delta_{r=1}\log^{\mathfrak{l}}2}{\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d})$
(9.19) $\displaystyle-\\!\\!\displaystyle\sum_{1\leq i<j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{i,j-1}|\leq
r<|{\boldsymbol{\sl{k}}}_{i,j}|-1}\binom{\zeta^{\mathfrak{l}}_{r-|{\boldsymbol{\sl{k}}}_{i,j-1}|}({\boldsymbol{\sl{k}}}_{j-1,i})}{-\delta_{r=1}\log^{\mathfrak{l}}2}{\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d}).$
(9.20)

The next result will be crucial in our proof of some families of unramified
MtVs.

######  Proposition 9.6.

Suppose ${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{d})$ where
$k_{1},\ldots,k_{d}\geq 2$. Then

$t^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2,\quad\tilde{\zeta}^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+\tilde{\zeta}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2,\quad\tilde{\zeta}^{\mathfrak{m}}(1,{\boldsymbol{\sl{k}}},1)+\tilde{\zeta}^{\mathfrak{m}}(1,{\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2$

are all unramified.

###### Proof.

Let’s consider the MtVs first. By Glonois’s result Thm. 2.1 we only need to
show that
$D_{r}\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2\Big{)}$
is unramified for odd $1<r\leq|{\boldsymbol{\sl{k}}}|$ and vanished for $r=1$.
As $D_{r}$’s are derivations, by Lemma 9.2 we clearly have

$D_{1}\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2\Big{)}=-\log^{\mathfrak{l}}2{\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})+D_{1}\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2\Big{)}=(1{\otimes}\log^{\mathfrak{m}})D_{1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$

by Lemma 9.1.

Now assume $r$ is odd and $1<r\leq|{\boldsymbol{\sl{k}}}|$. Let $k_{d+1}=1$.
Then by Thm. 9.5 we see that (after setting
${\boldsymbol{\sl{k}}}_{d+1,d}=\emptyset$ and $t^{\mathfrak{m}}(\emptyset)=1$)

$\displaystyle
D_{r}\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2\Big{)}=\displaystyle\sum_{1\leq
j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{1,j}|=r}t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,j}){\otimes}\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{j+1,d},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{j+1,d})\log^{\mathfrak{m}}2\Big{)}$
$\displaystyle+\\!\\!\displaystyle\sum_{1\leq i<j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{i+1,j}|\leq
r<|{\boldsymbol{\sl{k}}}_{i,j}|-1}\zeta^{\mathfrak{l}}_{r-|{\boldsymbol{\sl{k}}}_{i+1,j}|}({\boldsymbol{\sl{k}}}_{i+1,j}){\otimes}$
$\displaystyle\hskip
56.9055pt\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d})\log^{\mathfrak{m}}2\Big{)}$
$\displaystyle-\\!\\!\displaystyle\sum_{1\leq i<j\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{i,j-1}|\leq
r<|{\boldsymbol{\sl{k}}}_{i,j}|-1}\zeta^{\mathfrak{l}}_{r-|{\boldsymbol{\sl{k}}}_{i,j-1}|}({\boldsymbol{\sl{k}}}_{j-1,i}){\otimes}$
$\displaystyle\hskip
56.9055pt\Big{(}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d},1)+t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,j}|-r,{\boldsymbol{\sl{k}}}_{j+1,d})\log^{\mathfrak{m}}2\Big{)}$
$\displaystyle+\\!\\!\displaystyle\sum_{1\leq i\leq
d}\delta_{|{\boldsymbol{\sl{k}}}_{i+1,d+1}|\leq
r<|{\boldsymbol{\sl{k}}}_{i,d}|}\zeta^{\mathfrak{l}}_{r-|{\boldsymbol{\sl{k}}}_{i+1,d+1}|}({\boldsymbol{\sl{k}}}_{i+1,d+1}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},|{\boldsymbol{\sl{k}}}_{i,d}|+1-r).$

Every term in the first three summands is unramified by induction while the
last sum is also unramified by Lemma 9.1.

The proof that
$\tilde{\zeta}^{\mathfrak{m}}({\boldsymbol{\sl{k}}},1)+\tilde{\zeta}^{\mathfrak{m}}({\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2$
and
$\tilde{\zeta}^{\mathfrak{m}}(1,{\boldsymbol{\sl{k}}},1)+\tilde{\zeta}^{\mathfrak{m}}(1,{\boldsymbol{\sl{k}}})\log^{\mathfrak{m}}2$
are unramified is completely similar using next lemma so we leave it to the
interested reader. ∎

######  Lemma 9.7.

For any nontrivial composition of positive integers ${\boldsymbol{\sl{k}}}$ we
have

$D_{1}\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}},1)=-\log^{\mathfrak{l}}2{\otimes}\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}}).$

###### Proof.

Observe that nontrivial cuts of
$D_{1}\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})$ can appear when
${\boldsymbol{\sl{k}}}$ has unit components. First, assume such a component
appears in the following block:

$\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}D_{1}I^{\mathfrak{m}}(0;{\boldsymbol{{\alpha}}},0,\eta_{i},\eta_{i+1},\dots,\eta_{j-1},\eta_{j},0,{\boldsymbol{{\beta}}};1)$

for some strings ${\boldsymbol{{\alpha}}}$ and ${\boldsymbol{{\beta}}}$. We
observe that nontrivial cuts produced from the above block
$0,\eta_{i},\dots,\eta_{j},0$ lead to

$\displaystyle\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\Big{(}I^{\mathfrak{l}}(0;\eta_{i};\eta_{i+1}){\otimes}I^{\mathfrak{m}}(0;{\boldsymbol{{\alpha}}},0,\eta_{i+1},\eta_{i+2},\dots,\eta_{j-1},\eta_{j},0,{\boldsymbol{{\beta}}};1)$
$\displaystyle+I^{\mathfrak{l}}(\eta_{j-1};\eta_{j};0){\otimes}I^{\mathfrak{m}}(0;{\boldsymbol{{\alpha}}},0,\eta_{i},\eta_{i+1},\dots,\eta_{j-2},\eta_{j-1},0,{\boldsymbol{{\beta}}};1)\Big{)}$
$\displaystyle+\displaystyle\sum_{\eta_{1},\dots,\eta_{d}=\pm
1}\displaystyle\sum_{\ell=i+1}^{j-1}I^{\mathfrak{l}}(\eta_{\ell-1};\eta_{\ell};\eta_{\ell+1}){\otimes}I^{\mathfrak{m}}(0;{\boldsymbol{{\alpha}}},0,\eta_{i},\dots,\widehat{\eta_{\ell}},\dots,\eta_{j},0,{\boldsymbol{{\beta}}};1).$

The first two terms cancel if we apply to the second term the substitution
$\eta_{i}\to\eta_{i+1}\to\cdots\to\eta_{j-1}\to\eta_{j}\to\eta_{j}\eta_{i}/\eta_{i+1}$.
Further, for each fixed $\ell$, by Lemma 2.4 the sum

$\displaystyle\sum_{\eta_{\ell}=\pm
1}I^{\mathfrak{l}}(\eta_{\ell-1};\eta_{\ell};\eta_{\ell+1})=0.$

Therefore we see that
$D_{1}\tilde{\zeta}^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=D_{1}\zeta^{\mathfrak{m}}_{a}({\boldsymbol{\sl{k}}})=0$.

Now if a unit components at the end then the above argument shows that all but
one term is left after cancelation and the lemma follows easily. ∎

We are now ready to prove the following theorem.

######  Theorem 9.8.

Let $n,e\in\mathbb{N}$ and all components of ${\boldsymbol{\sl{q}}}$ are at
least $2$. Then the MtV
$t^{\mathfrak{m}}(\\{2n\\}_{e},1,{\boldsymbol{\sl{q}}})$ is unramified. Hence,
$t(\\{2n\\}_{e},1,{\boldsymbol{\sl{q}}})$ is unramified.

###### Proof.

By Lemma 9.2 we see that $D_{1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$.
Assume now $1<r<|{\boldsymbol{\sl{k}}}|$ is odd. Then by Lemma 2.4 we only
need to consider the cuts of $D_{r}$ either starting at or ending at 0:

$\
\hskip-147.95424pt0;\dots,\eta_{i},0,\dots,0,\eta_{i+1},\dots,\eta_{j},0,\dots,0,\eta_{j+1},\dots,\eta_{e+1},\eta_{e+2},\dots,\eta_{g},0,\dots,0,\eta_{g+1},\dots$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
1$}}}\hskip
0.75346pt{}}$${}^{\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{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
5$}}}\hskip
0.75346pt{}}$$\dots,\eta_{g},0,\dots,0,\eta_{g+1},\dots,\eta_{h},0,\dots,0,\eta_{h+1},\dots;1$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
6$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
7$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{r}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$.

We now show by induction on the depth of ${\boldsymbol{\sl{k}}}$ that
$D_{r}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$ is unramified. Let
${\boldsymbol{\sl{q}}}=(q_{1},\dots,q_{d})$ and
${\boldsymbol{\sl{k}}}=(\\{2n\\}_{e},1,{\boldsymbol{\sl{q}}})=(k_{1},\dots,k_{\ell})$
with $\ell=e+1+d$. Put $\eta_{\ell+1}=1$ and $a_{m}=k_{m}-1$ for all $1\leq
m\leq\ell$. Then for all $e+2\leq g\leq\ell$ by substitution
$\eta_{m}\to\eta_{m}\eta_{g}$ for $1\leq m<g$ we get

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{|{\boldsymbol{\sl{k}}}_{1,g-1}|=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;\eta_{1},\dots;\eta_{g}){\otimes}I^{\mathfrak{m}}(0;\eta_{g},\dots;1)$
$\displaystyle=$
$\displaystyle\,{\delta}_{|{\boldsymbol{\sl{k}}}_{1,g-1}|=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}\eta_{g}I^{\mathfrak{l}}(0;\eta_{1},\dots;1){\otimes}I^{\mathfrak{m}}(0;\eta_{g},\dots;1)$
$\displaystyle=$
$\displaystyle\,{\delta}_{|{\boldsymbol{\sl{k}}}_{1,g-1}|=r}\,t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,g-1}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{g,\ell}).$

By induction, $t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,g-1})$ is unramified
(if $g=e+2$ then this follows from Prop. 9.6). By Lemma 9.1,
$t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{g,\ell})$ is unramified. Next, for
any fixed $1\leq i<j\leq e$

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{b=0}^{a_{j}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i,j-1}|+b=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{i};0_{a_{i}},\dots,\eta_{j},0_{b};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\dots,\eta_{i},0_{a_{j}-b},\eta_{j+1},\dots;1),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{b=0}^{a_{i}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i+1,j}|+b=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{b},\eta_{i+1},\dots;\eta_{j+1}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\dots,\eta_{i},0_{a_{i}-b},\eta_{j+1},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{b=0}^{a_{i}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i+1,j}|+b=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{j+1};\dots,\eta_{i+1},0_{b};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\dots,\eta_{i},0_{a_{i}-b},\eta_{j+1},\dots;1).$

Since $a_{i}=a_{j}$ we get
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 2$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=0$. Next, for any fixed $1\leq i\leq e+1,e+2\leq
g\leq\ell$

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{b=0}^{a_{g}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i,g-1}|+b=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{i};0_{a_{i}},\dots,\eta_{g},0_{b};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\dots,\eta_{i},0_{a_{g}-b},\eta_{g+1},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{b=0}^{a_{g}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i,g-1}|+b=r}\,\tilde{\zeta}_{b}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{g-1,i}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},k_{g}-b,{\boldsymbol{\sl{k}}}_{g+1,\ell})$

which is unramified by Thm. 2.3 (if $i=e+1$ and $g>e+2$ then by Prop. 9.6) and
Lemma 9.1, except for the case $i=e+1=g-1$ and $b=0$ when
$\tilde{\zeta}_{b}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{g-1,i})=\tilde{\zeta}(1)$
which is ramified. Similarly,

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
5$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{b=0}^{a_{i}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i+1,g-1}|+b=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{b},\eta_{i+1},\dots;\eta_{g}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},\dots,\eta_{i},0_{a_{i}-b},\eta_{g},\dots;1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{b=0}^{a_{i}-1}{\delta}_{|{\boldsymbol{\sl{k}}}_{i+1,g-1}|+b=r}\,\tilde{\zeta}_{b}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i+1,g-1}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{1,i-1},k_{i}-b,{\boldsymbol{\sl{k}}}_{g,\ell})$

which is unramified by Thm. 2.3 (if $g=e+2$ and $i<e$ then by Prop. 9.6) and
Lemma 9.1, except for the case $i=e=g-2$ and $b=0$ when
$\tilde{\zeta}_{b}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i+1,g-1})=\tilde{\zeta}(1)$
which is ramified. But this ramified term is canceled exactly by the
exceptional ramified term of
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 4$}}}\hskip
1.07639pt{}}$ .

Further, if $\operatorname*{dep}({\boldsymbol{\sl{q}}})=1$ then neither
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 6$}}}\hskip
1.07639pt{}}$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 7$}}}\hskip
1.07639pt{}}$ can appear and therefore we have proved the base case of the
induction. If $\operatorname*{dep}({\boldsymbol{\sl{q}}})>1$ then both
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 6$}}}\hskip
1.07639pt{}}$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 7$}}}\hskip
1.07639pt{}}$ have the shape of
$\tilde{\zeta}_{b}^{\mathfrak{l}}({\boldsymbol{\sl{s}}}){\otimes}t^{\mathfrak{m}}(\\{2n\\}_{e},1,{\boldsymbol{\sl{p}}})$
where $b\geq 0$,
$\operatorname*{dep}({\boldsymbol{\sl{p}}})<\operatorname*{dep}({\boldsymbol{\sl{q}}})$,
and every component of ${\boldsymbol{\sl{s}}}$ and ${\boldsymbol{\sl{p}}}$ is
at least two. By induction and Thm. 2.3 we see that these are all unramified.

We have now completed the proof of the theorem by Thm. 2.1.. ∎

######  Lemma 9.9.

Suppose all components of ${\boldsymbol{\sl{q}}}$ are at least two. For any
positive odd number $O>1$ and even number $E$ the difference
$A=t^{\mathfrak{m}}(E,O,1,{\boldsymbol{\sl{q}}})-t^{\mathfrak{m}}(O,E,1,{\boldsymbol{\sl{q}}})$
is unramified.

###### Proof.

Clearly $D_{1}A=0$ by Lemma 9.2. Assume $r>1$ is odd. For simplicity we set
$\ell=3+\operatorname*{dep}({\boldsymbol{\sl{q}}})$,
${\boldsymbol{\sl{k}}}=(E,O,1,{\boldsymbol{\sl{q}}})$, $a=E-1$ and $b=O-1$.
Using Lemma 2.4 we only need to consider the following cuts for
$D_{r}t^{\mathfrak{m}}(E,O,1,{\boldsymbol{\sl{q}}})$ either starting at or
ending at 0. Note that if a cut starts after $\eta_{3}$ (so
$\operatorname*{dep}({\boldsymbol{\sl{q}}})>1$) then it is unramified by
induction and Lemma 9.1.

$0;\eta_{1},0_{a-c-1},0,0_{c},\eta_{2},0_{c},0,0_{b-c-1},\eta_{3},\eta_{4},\dots,\eta_{i},\dots,0,\dots,\eta_{i+1},\dots;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{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
5$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
6$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
7$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
8$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
9$}}}\hskip 0.75346pt{}}$Possible cuts of
$D_{r}t^{\mathfrak{m}}(E,O,1,{\boldsymbol{\sl{q}}})$.

We have

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{E+O=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;\eta_{1},0_{a},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{3},\eta_{4},\dots;1)$
$\displaystyle=$
$\displaystyle\,{\delta}_{E+O=r}\,t^{\mathfrak{l}}(E,O){\otimes}t^{\mathfrak{m}}(1,{\boldsymbol{\sl{q}}})=-{\delta}_{E+O=r}\,t^{\mathfrak{l}}(E+O){\otimes}t^{\mathfrak{m}}(1,{\boldsymbol{\sl{q}}})\quad\text{by
Cor.~{}\ref{cor:DBreduce};}$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;\eta_{1},\dots;\eta_{i}){\otimes}I^{\mathfrak{m}}(0;\eta_{i},\dots;1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|=r}\,t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,i-1}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{i,\ell})$
unramified by induction (or Prop. 9.6 if $i=4$) and Lemma 9.1;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{c=0}^{O-2}{\delta}_{E+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{1};0_{a},\eta_{2},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{b-c},\eta_{3},\eta_{4},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{c=0}^{O-2}{\delta}_{E+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(E){\otimes}t^{\mathfrak{m}}(O-c,1,{\boldsymbol{\sl{q}}})$
unramified by Thm. 2.3 and Thm. 9.8 since $c$ is odd;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{1};\dots,\eta_{i},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{k_{i}-c-1},\eta_{i+1},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,1}){\otimes}t^{\mathfrak{m}}(k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell})$
unramified by Thm. 2.3 (or Prop. 9.6 if $i=4$) and Lemma 9.1;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
5$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{c=0}^{E-2}{\delta}_{c+O=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{c},\eta_{2},0_{b};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{3},\eta_{4},\dots;1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{c=0}^{E-2}{\delta}_{c+O=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(O){\otimes}t^{\mathfrak{m}}(E-c,1,{\boldsymbol{\sl{q}}})$
unramified by Thm. 2.3 and Thm. 9.8 since $c$ is even;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
6$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{E-2}{\delta}_{c+|{\boldsymbol{\sl{k}}}_{2,i-1}|=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{c},\eta_{2},0_{b},\eta_{3},\dots;\eta_{i}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a-c},\eta_{i},\dots;1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{E-2}{\delta}_{c+|{\boldsymbol{\sl{k}}}_{2,i-1}|=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{2,i-1}){\otimes}t^{\mathfrak{m}}(E-c,{\boldsymbol{\sl{k}}}_{i,\ell})$
unramified by Thm. 2.3 (or Prop. 9.6 if $i=4$) and Lemma 9.1;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
7$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{2,i-1}|+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{2};0_{b},\eta_{3},\dots,\eta_{i},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{k_{i}-c-1},\eta_{i+1},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{2,i-1}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,2}){\otimes}t^{\mathfrak{m}}(E,k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell})$
unramified by Thm. 2.3 and Lemma 9.1;
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
8$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{O-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{3,i-1}|+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{c},\eta_{3},\dots;\eta_{i}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b-c},\eta_{i},\dots;1)$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{O-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{3,i-1}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{3,i-1}){\otimes}t^{\mathfrak{m}}(E,O-c,{\boldsymbol{\sl{k}}}_{i,\ell})$
unramified by Thm. 2.3 and Thm. 9.8 (note that $r>1$ so if $c=0$ then $i>4$);
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
9$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{4,i-1}|+c+1=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{3};\eta_{4},\dots,\eta_{i},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{a},\eta_{2},0_{b},\eta_{3},0_{k_{i}-c-1},\eta_{i+1},\dots;1)$
$\displaystyle=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{4,i-1}|+c+1=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,4},1){\otimes}t^{\mathfrak{m}}(E,O,k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell})$

which is unramified by Prop. 9.6 (if $i>4$) and Lemma 9.1. If $i=4$ then $c>0$
so
$\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,4},1)=\tilde{\zeta}_{c}^{\mathfrak{l}}(1)$
is unramified.

The computation for $D_{r}t^{\mathfrak{m}}(O,E,1,{\boldsymbol{\sl{q}}})$ is
completely similar such that all but
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}$ are unramified. But the contribution from
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 1$}}}\hskip
1.07639pt{}}$ is the same for
$D_{r}t^{\mathfrak{m}}(O,E,1,{\boldsymbol{\sl{q}}})$ and
$D_{r}t^{\mathfrak{m}}(E,O,1,{\boldsymbol{\sl{q}}})$ by Cor. 4.2 and therefore
$D_{r}A$ is always unramified. This completes the proof of the lemma by Thm.
2.1.. ∎

######  Theorem 9.10.

Suppose $m,n\in\mathbb{N}$ and all components of ${\boldsymbol{\sl{q}}}$ are
at least $2$. Then the MtVs
$t^{\mathfrak{m}}(2n,2m,2n,1,{\boldsymbol{\sl{q}}})$ is unramified.

###### Proof.

Let ${\boldsymbol{\sl{k}}}=(2n,2m,2n,1,{\boldsymbol{\sl{q}}})$ and
$\ell=4+\operatorname*{dep}({\boldsymbol{\sl{q}}})$. First,
$D_{1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$ by Lemma 9.2. Suppose $r>1$
is odd. Similar to the argument in Lemma 9.9 we only need to consider the
following four types of cuts for
$D_{r}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$.

$0;\eta_{1},0_{2n-c-2},0,0_{c},\eta_{2},0_{2m-1},\eta_{3},0_{c},0,0_{2n-c-2},\eta_{4},\eta_{5},\dots;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{}}$Possible cuts of
$D_{r}t^{\mathfrak{m}}(2n,2m,2n,1,{\boldsymbol{\sl{q}}})$.

We have, for $0\leq c\leq 2m-2$

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{2n+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{1};0_{2n-1},\eta_{2},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{2m-c-1},0_{2n-1},\eta_{4},\eta_{5},\dots;1)$
$\displaystyle=$
$\displaystyle\,-{\delta}_{2n+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(2n){\otimes}t^{\mathfrak{m}}(2m-c,2n,1,{\boldsymbol{\sl{q}}}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{2n+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{c},\eta_{3},0_{2n-1};\eta_{4}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{2n-1},\eta_{2},0_{2m-c-1},\eta_{4},\eta_{5},\dots;1)$
$\displaystyle=$
$\displaystyle\,{\delta}_{2n+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(2n){\otimes}t^{\mathfrak{m}}(2n,2m-c,1,{\boldsymbol{\sl{q}}}),$

while for $0\leq c\leq 2n-2$

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{2m+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(0;0_{c},\eta_{2},0_{2m-1};\eta_{3}){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{2n-c-1},\eta_{3},0_{2n-1},\eta_{4},\eta_{5},\dots;1)$
$\displaystyle=$
$\displaystyle\,{\delta}_{2m+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(2m){\otimes}t^{\mathfrak{m}}(2n-c,2n,1,{\boldsymbol{\sl{q}}}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{2m+c=r}\displaystyle\sum_{\eta_{1},\dots,\eta_{\ell}=\pm
1}\eta_{1}I^{\mathfrak{l}}(\eta_{2};0_{2m-1},\eta_{3},0_{c};0){\otimes}I^{\mathfrak{m}}(0;\eta_{1},0_{2n-1},\eta_{2},0_{2n-c-1},\eta_{4},\eta_{5},\dots;1)$
$\displaystyle=$
$\displaystyle\,-{\delta}_{2m+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}(2m){\otimes}t^{\mathfrak{m}}(2n,2n-c,1,{\boldsymbol{\sl{q}}}).$

Hence, by Lemma 9.9 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$}}}\hskip 1.07639pt{}}$ and
${\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle 3$}}}\hskip
1.07639pt{}}+{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip 1.07639pt{}}$ are both unramified. This completes the proof of the
theorem by Thm. 2.1. ∎

######  Lemma 9.11.

Suppose all components of ${\boldsymbol{\sl{q}}}$ are at least $2$. Then

$3t^{\mathfrak{m}}(3,2,1,{\boldsymbol{\sl{q}}})+t^{\mathfrak{m}}(2,1,2,1,{\boldsymbol{\sl{q}}})$

is unramified.

###### Proof.

Let ${\boldsymbol{\sl{k}}}=(3,2,1,{\boldsymbol{\sl{q}}})$. Then clearly
$D_{1}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})=0$ by Lemma 9.2. Assume now
$r>1$ is odd. By Lemma 2.4 we only need to consider the following cuts of
$D_{r}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}})$:

$\hskip-39.83368pt0;\eta_{1},0,0,\eta_{2},0,\eta_{3},\eta_{4},\dots,\eta_{i},\dots,0,\dots,\eta_{i+1},\dots;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{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
5$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
6$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
7$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
8$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle
9$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\hskip-1.2pt1\\!0$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\hskip-1.2pt1\\!1$}}}\hskip
0.75346pt{}}$${}^{\bigcirc\hskip-3.46594pt{\raisebox{-0.42677pt}{\text{$\scriptscriptstyle\hskip-1.2pt1\\!2$}}}\hskip
0.75346pt{}}$Possible cuts of
$D_{r}t^{\mathfrak{m}}(3,2,1,{\boldsymbol{\sl{q}}})$.

By tedious but straightforward computation we can see that

$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
1$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{r=3}\,t^{\mathfrak{l}}(3){\otimes}t^{\mathfrak{m}}(2,1,{\boldsymbol{\sl{q}}}),\quad{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
2$}}}\hskip
1.07639pt{}}=-{\delta}_{r=3}\,\tilde{\zeta}^{\mathfrak{l}}(3){\otimes}t^{\mathfrak{m}}(2,1,{\boldsymbol{\sl{q}}}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
3$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,{\delta}_{r=3}\,\tilde{\zeta}_{1}^{\mathfrak{l}}(2){\otimes}t^{\mathfrak{m}}(2,1,{\boldsymbol{\sl{q}}}),\quad{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
4$}}}\hskip
1.07639pt{}}=-{\delta}_{r=3}\,\tilde{\zeta}^{\mathfrak{l}}(2,1){\otimes}t^{\mathfrak{m}}(3,{\boldsymbol{\sl{q}}}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
5$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{i-1,2}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,2}){\otimes}t^{\mathfrak{m}}(3,k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
6$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=5}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{3,i-1}|=r}\,\tilde{\zeta}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{3,i-1}){\otimes}t^{\mathfrak{m}}(3,2,{\boldsymbol{\sl{k}}}_{i,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
7$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{3,i-1}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,3}){\otimes}t^{\mathfrak{m}}(3,2,k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle
9$}}}\hskip 1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=5}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|=r}\,t^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{1,i-1}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{k}}}_{i,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\hskip-1.2pt1\\!0$}}}\hskip
1.07639pt{}}=$
$\displaystyle\,-\displaystyle\sum_{i=4}^{\ell}\displaystyle\sum_{c=0}^{k_{i}-2}{\delta}_{|{\boldsymbol{\sl{k}}}_{1,i-1}|+c=r}\,\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{i-1,1}){\otimes}t^{\mathfrak{m}}(k_{i}-c,{\boldsymbol{\sl{k}}}_{i+1,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\hskip-1.2pt1\\!1$}}}\hskip
1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=5}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{2,i-1}|+1=r}\,\tilde{\zeta}_{1}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{2,i-1}){\otimes}t^{\mathfrak{m}}(2,{\boldsymbol{\sl{k}}}_{i,\ell}),$
$\displaystyle{\bigcirc\hskip-6.88889pt{\raisebox{0.0pt}{\text{$\scriptstyle\hskip-1.2pt1\\!2$}}}\hskip
1.07639pt{}}=$
$\displaystyle\,\displaystyle\sum_{i=5}^{\ell}{\delta}_{|{\boldsymbol{\sl{k}}}_{2,i-1}|=r}\,\tilde{\zeta}^{\mathfrak{l}}({\boldsymbol{\sl{k}}}_{2,i-1}){\otimes}t^{\mathfrak{m}}(3,{\boldsymbol{\sl{k}}}_{i,\ell}).$

All terms
$\tilde{\zeta}_{c}^{\mathfrak{l}}({\boldsymbol{\sl{l}}}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{s}}})$
above in $D_{r}t^{\mathfrak{m}}(3,2,1,{\boldsymbol{\sl{q}}})$ fall into three
categories: either ${\boldsymbol{\sl{s}}}=(2,1,{\boldsymbol{\sl{q}}})$ or all
components of ${\boldsymbol{\sl{s}}}$ are at least two or
${\boldsymbol{\sl{s}}}=(3,2,1,{\boldsymbol{\sl{q}}}^{\prime})$ with
$\operatorname*{dep}({\boldsymbol{\sl{q}}}^{\prime})<\operatorname*{dep}({\boldsymbol{\sl{q}}})$.
Thus, all these $t^{\mathfrak{m}}({\boldsymbol{\sl{s}}})$ are unramified by
Thm. 9.8, Lemma 9.1 and induction, respectively. The first factor
$\tilde{\zeta}_{c}({\boldsymbol{\sl{l}}})$ in the above is always unramified
by Thm. 2.3 or Prop. 9.6.

On the other hand, if the terms have the shape
$t^{\mathfrak{l}}({\boldsymbol{\sl{l}}}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{s}}})$
then ${\boldsymbol{\sl{l}}}={\boldsymbol{\sl{k}}}_{1,i}$ for some $i$ and
$|{\boldsymbol{\sl{l}}}|=r$ is odd, then these terms must have the form (i)
$t^{\mathfrak{l}}(3){\otimes}t^{\mathfrak{m}}(2,1,{\boldsymbol{\sl{q}}})$, or
(ii)
$t^{\mathfrak{l}}(3,2){\otimes}t^{\mathfrak{m}}(1,{\boldsymbol{\sl{q}}})$, or
(iii)
$t^{\mathfrak{l}}(3,2,1,{\boldsymbol{\sl{q}}}^{\prime}){\otimes}t^{\mathfrak{m}}({\boldsymbol{\sl{q}}}^{\prime\prime})$
for some nontrivial decomposition
${\boldsymbol{\sl{q}}}=({\boldsymbol{\sl{q}}}^{\prime},{\boldsymbol{\sl{q}}}^{\prime\prime})$.
The first form is unramified by Thm. 9.8.

Similar computation is valid almost verbatim for
$D_{r}t^{\mathfrak{m}}(2,1,2,1,{\boldsymbol{\sl{q}}})$. Thus we only need to
consider the terms of the forms (i)’
$t^{\mathfrak{l}}(2,1){\otimes}t^{\mathfrak{m}}(2,1,{\boldsymbol{\sl{q}}})$,
or (ii)’
$t^{\mathfrak{l}}(2,1,2){\otimes}t^{\mathfrak{m}}(1,{\boldsymbol{\sl{q}}})$,
or (iii)’