§.§ Proof Strategy for Elementary Moves

The discussion of the elementary moves in the subsequent sections follows a general pattern, which we outline here.

Recall that any move $\mathbf{M}$ between extruded graphs $\sigsurf$ and $\postmove{\sigsurf}$ involves linear maps

\begin{align}\label{eq_movemap}
\movemap : \nodespace \to \postmove{\nodespace} && \text{between node spaces,}\\
\Pmovemap : \preblock \to \postmove{\preblock} && \text{between pre-block spaces, and}\\
\Movemap : \blockspace \to \postmove{\blockspace} && \text{between block spaces.}
\end{align}

In order to show that an elementary move is an invariance, we consider the diagram (<ref>) below, which is a subdivision of (<ref>).

\begin{equation}\label{diag_invariance_scheme}
\begin{tikzcd}[column sep = huge]
\nodespace \arrow[r,"{\delta}"] \arrow[d, "\movemap"] & \preblock \arrow[r, "\holsurj"] \arrow[d, "\Pmovemap"] & \blockspace \arrow[d, "\Movemap"] \\
\postmove{\nodespace} \arrow[r,"{\postmove{\delta}}"]& \postmove{\preblock} \arrow[r, "\postmove{\holsurj}"] & \postmove{\blockspace}
\end{tikzcd}
\end{equation}

We will deduce the commutativity of the outer paths in (<ref>) from the commutativity of the two cells.
Commutativity of the outer paths then implies by <Ref> that we have a move of invariance at hand.

For the right-hand square in (<ref>),
this is facilitated by the fact that $\blockspace$ is a retract of $\preblock$.
Consider the following general observation.

Let $\xcat$ be a category, and let

\begin{equation}
\begin{tikzcd}
x \arrow[r,hook,"\iota"{below}] & y \arrow[l, two heads, bend right, "\pi"{above}]
\end{tikzcd}
\qquad\text{and}\qquad
\begin{tikzcd}
x' \arrow[r,hook,"\iota'"{below}] & y' \arrow[l, two heads, bend right, "\pi'"{above}]
\end{tikzcd}
\end{equation}

be retracts in $\xcat$.
Denote the corresponding idempotents by $h = \iota \circ \pi$ and $h' = \iota' \circ \pi'$.
A morphism $\varphi : y \to y'$ which satisfies

\begin{equation}\label{diag_map_of_retracts_cond}
\begin{tikzcd}
y \arrow[r, "h"] \arrow[d, "\varphi"] & y \arrow[d, "\varphi"] \\
y' \arrow[r, "h'"] & y'
\end{tikzcd}
\end{equation}

is a morphism of retracts, meaning that

\begin{equation}\label{diag_map_of_retracts}
\begin{tikzcd}
x \arrow[r, "\iota"] \arrow[d, "\psi"] & y \arrow[d, "\varphi"] \\
x' \arrow[r, "\iota'"] & y'
\end{tikzcd}
\qquad\text{and}\qquad
\begin{tikzcd}
y \arrow[r, "\pi"] \arrow[d, "\varphi"] & x \arrow[d, "\psi"] \\
y' \arrow[r, "\pi'"] & x'
\end{tikzcd}
\end{equation}

commute for $\psi = \pi' \circ \varphi \circ \iota$.
Conversely, any pair of morphisms $(\varphi,\psi)$ which satisfies (<ref>) also satisfies (<ref>), and $\psi = \pi' \circ \varphi \circ \iota$ holds.

Applied to our situation, <Ref> becomes

The commutativity of the diagram

\begin{equation}\label{diag_moveproof_hol_condition}
\begin{tikzcd}
\preblock \arrow[r, "\holidem"] \arrow[d, "\Pmovemap"{left}] & \preblock \arrow[d, "\Pmovemap"]\\
\postmove{\preblock} \arrow[r, "\postmove{\holidem}"] & \postmove{\preblock}
\end{tikzcd},
\end{equation}

involving the holonomy idempotents $\holidem$, $\postmove{\holidem}$ instead of the projectors $\holsurj$, $\postmove{\holsurj}$ implies the commutativity of the right square in (<ref>), and thus that the elementary move $\mathbf{M}$ is an invariance.

For every elementary move, we thus have to check two things: the commutativity of the left square in the diagram (<ref>), and the commutativity of the square (<ref>).
This is done in the following sections.

§.§ Elementary Moves

This is the first of two sections that comprise the proof of <Ref>.
For some moves, we have to prove the invariance property by directly using <Ref>, the definition of the evaluation procedure.
These moves are called elementary.
Other moves can be obtained as compositions of elementary moves, and we will treat them in <Ref>.

§.§.§ ${\moveOR}$ – Orientation Reversal

In the introductory discussion in <Ref>, we have already seen that the node spaces before and after the move $\moveOR$ can be identified, and we can indeed view them as equal without loss of generality, by assuming that the acting categories $\acat$ and $\bcat$ are strictly pivotal.
As a consequence of the equality of node spaces, the pre-block  spaces are also equal.
This makes the commutativity of the left square in (<ref>) particularly easy to prove, because the linear maps $\movemap$ and $\Pmovemap$ are identities in this case.

Explicitly, using (<ref>), the pre-block space is of the form

\begin{equation}
\preblock = \homsetin{\mcat}{\sweed{y}{\opcat{\mcat}}}{\sweed{x}{\mcat}} \otimes \cdots = \homsetin{\opcat{\mcat}}{\sweed{x}{\opcat{\opcat{\mcat}}}}{\sweed{y}{\opcat{\mcat}}} \otimes \cdots  = \postmove{\preblock}.
\end{equation}

We must now show that the diagram (<ref>) commutes.
Unraveling the definition of the two holonomy idempotents $\holidem$ for $\sigsurf$ and $\holidem'$ for $\sigsurf'$ reveals that $h = h'$.
This is because in the holonomy balancing (<ref>), say for the domain $\acat$, the balancing

\begin{equation}
\homsetin{\mcat}{a^*\sweed{y}{\opcat{\mcat}}}{\sweed{x}{\mcat}} \otimes \cdots \cong \homsetin{\mcat}{\sweed{y}{\opcat{\mcat}}}{a\sweed{x}{\mcat}} \otimes \cdots
\end{equation}

appears as a composition factor.
Thus balancing is the same for both $\preblock$ and $\postmove{\preblock}$.
A similar statement hold for the domain $B$.

As a consequence, the linear map $\Movemap$ between block spaces is an isomorphism.

§.§.§ ${\moveC}$ – Contraction

The second move for which we show the invariance property is the contraction move, which is depicted in <Ref>.
Following the general proof strategy outlined in <Ref>, we now need to check that the left cell of the diagram (<ref>) commutes.
To this end, consider an extruded graph $\sigsurf_T$ which contains the disk $\sigsurf$ from figure  (<ref>).
The node space $\nodespace$ of $\sigsurf_T$ is given by $\nodespace = \homset{k\deligne\opcat{m}}{U(x)} \otimes \homset{m\deligne\opcat{l}}{U(y)}\otimes\nodespace_\rest$, where $\nodespace_\rest$ is a rest term that takes into account the nodes of $\sigsurf_T$ that lie outside of $\sigsurf$.
The commutativity of (<ref>) amounts to the equality of the linear maps $\postmove{\delta} \circ \movemap$ and $\Pmovemap \circ \delta$, which we do by evaluating them on a vector $f = \sweed{f}{\kcat} \otimes \sweed{f}{\opcat{\mcat}} \otimes \sweed{f}{\mcat} \otimes \sweed{f}{\opcat{\lcat}} \otimes \sweed{f}{\rest} \in \nodespace$.
We then calculate:

\begin{align}
\begin{aligned}
\postmove{\delta} \circ \movemap (f) &= \postmove{\delta} \left( (\sweed{f}{\mcat} \circ \sweed{f}{\opcat{\mcat}}) \otimes  \sweed{f}{\kcat}\otimes \sweed{f}{\opcat{\lcat}} \otimes \sweed{f}{\rest} \right) \\
&= (\sweed{f}{\mcat} \circ \sweed{f}{\opcat{\mcat}}) \otimes \bigoplus_{\si{k},\si{l},\dots} \pdim{\si{k}}\pdim{\si{l}}\pdim{\cdots} \left( (\sweed{f}{\kcat}\otimes \sweed{f}{\opcat{\lcat}} \otimes \sweed{f}{\rest}) \circ (\basisel{\si{k}}{k} \otimes \basisel{\opcat{\si{l}}}{\opcat{l}} \otimes \cdots)  \right),
\end{aligned}
\end{align}

where the first equality uses the definition of $\movemap$ from (<ref>) and the second equality is the definition (<ref>) of $\postmove{\delta}$.
On the other hand, we have

\begin{align}
\begin{aligned}
\Pmovemap \circ \delta (f) &\eqwithref{eq_eval_delta} \bigoplus_{\si{k}, \si{m}, \si{l},\dots} \pdim{\si{k}}\pdim{\si{m}}\pdim{\si{l}}\pdim{\cdots} \Pmovemap \left( f \circ (\basisel{\si{k}}{k} \otimes  \basisel{\opcat{\si{m}}}{\opcat{m}} \otimes \basisel{\si{m}}{m} \otimes \basisel{\opcat{\si{l}}}{\opcat{l}} \otimes \cdots) \right)\\
&\hspace{-1.4cm}= \bigoplus_{\si{k}, \si{m}, \si{l},\dots} \pdim{\si{k}}\pdim{\si{m}}\pdim{\si{l}}\pdim{\cdots}\\&\hspace{-1cm}\Pmovemap \left( \left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\mcat}} \circ \basisel{\opcat{\si{m}}}{\opcat{m}}) \otimes
(\sweed{f}{\mcat} \circ \basisel{\si{m}}{m}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right)  \right)\\
&\hspace{-1.4cm}\eqwithref{eq_Pmovemap} \bigoplus_{\si{k},\si{m}, \si{l},\dots} \pdim{\si{k}}\pdim{\si{m}}\pdim{\si{l}}\pdim{\cdots} \postmove{\delta}_{\si{k},\si{l},\dots} \circ\\&\hspace{-1cm} \movemap_{\si{k},\si{l},\dots} \left( \left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\mcat}} \circ \basisel{\opcat{\si{m}}}{\opcat{m}}) \otimes
(\sweed{f}{\mcat} \circ \basisel{\si{m}}{m}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right) \right)\\
&\hspace{-1.4cm}\eqwithref{eq_gencomp_def} \bigoplus_{\si{k},\si{m}, \si{l},\dots} \pdim{\si{k}}\pdim{\si{m}}\pdim{\si{l}}\pdim{\cdots}\\&\hspace{-1cm} \postmove{\delta}_{\si{k},\si{l},\dots}  \left((\sweed{f}{\mcat} \circ \basisel{\si{m}}{m}  \circ\basisel{\opcat{\si{m}}}{\opcat{m}} \circ \sweed{f}{\opcat{\mcat}}) \otimes
\left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right) \right)\\
&\hspace{-1.4cm}\eqwithref{eq_res_of_id} \bigoplus_{\si{k}, \si{l},\dots}\pdim{\si{k}}\pdim{\si{l}}\pdim{\cdots}\\&\hspace{-1cm} \postmove{\delta}_{\si{k},\si{l},\dots}  \left((\sweed{f}{\mcat} \circ \sweed{f}{\opcat{\mcat}}) \otimes
\left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right) \right)\\
&\hspace{-1.4cm}\eqwithref{eq_eval_delta}(\sweed{f}{\mcat} \circ \sweed{f}{\opcat{\mcat}}) \otimes \bigoplus_{\si{k'}, \si{l'},\si{k}, \si{l},\dots}\pdim{\si{k}}\pdim{\si{l}}\pdim{\si{k'}}\pdim{\si{l'}}\pdim{\cdots}\\&\hspace{-1cm}
\left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right) \circ (\basisel{\si{k'}}{\si{k}} \otimes \basisel{\opcat{\si{l'}}}{\opcat{\si{l}}} \otimes \cdots) \\
&\hspace{-1.4cm}=(\sweed{f}{\mcat} \circ \sweed{f}{\opcat{\mcat}}) \otimes \bigoplus_{\si{k}, \si{l},\dots}\pdim{\si{k}}\pdim{\si{l}}\pdim{\cdots}
\left((\sweed{f}{\kcat} \circ \basisel{\si{k}}{k}) \otimes
(\sweed{f}{\opcat{\lcat}} \circ \basisel{\opcat{\si{l}}}{\opcat{l}}) \otimes
(\sweed{f}{\rest} \circ \cdots)\right).
\end{aligned}
\end{align}

In the last equality,  we used Schur's lemma.

In order to check the commutativity of the  square (<ref>), we need to prove that the two composition factors of the holonomy operations $\holidem$ and $\postmove{\holidem}$ corresponding to the paths from $k$ to $l$ agree.
These paths are labeled as (1) and (2) in the picture (<ref>) below.

\begin{equation}\label{pic_moveC_pf}
\pica{1}{moveC_pf_1} \xrightarrow{\moveC} \pica{1}{moveC_pf_2}
\end{equation}

We perform the check only for the holonomy with respect to the domain $\acat$; the argument for the $\bcat$-holonomy is similar.
Specifically, the following diagram (<ref>) needs to commute, in which the left vertical column composes to the path (1) in the picture (<ref>), and the right vertical column composes to the path (2).
For convenience, only the relevant parts of the pre-block spaces are written out: the sums over the variables $\si{l}$ and $\si{k}$ are omitted in (<ref>).

\begin{equation}\label{diag_moveC_pf_diag}
\begin{tikzcd}[row sep=small]
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}a}{U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{U(x)} \arrow[d] \arrow[r, "\Pmovemap"] & \homset{\si{k} \deligne \opcat{\si{l}}a}{U(x \mitosis{\mcat} y)} \arrow[d] \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{U(y)a^*} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{U(x)} \arrow[d] \arrow[r] & \homset{\si{k} \deligne \opcat{\si{l}}}{U(x \mitosis{\mcat} y)a^*} \arrow[ddddd] \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{a^*U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{U(x)} \arrow[d]  & \\
\bigoplus_\si{n} \homset{a\si{n} \deligne \opcat{\si{l}}}{U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{U(x)}\arrow[d]  & \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}a}{U(x)} \arrow[d] &  \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{U(x)a^*} \arrow[d] & \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{U(y)} \otimes \homset{\si{k} \deligne\opcat{\si{n}}}{a^*U(x)} \arrow[d] \arrow[r] & \homset{\si{k} \deligne \opcat{\si{l}}}{a^*U(x \mitosis{\mcat} y)} \arrow[d] \\
\bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{l}}}{U(y)} \otimes \homset{a\si{k} \deligne\opcat{\si{n}}}{U(x)} \arrow[r] & \homset{a\si{k} \deligne \opcat{\si{l}}}{U(x \mitosis{\mcat} y)}
\end{tikzcd}
\end{equation}

In the above diagram, the commutativity of the top and bottom squares is trivial.
The large cell commutes by definition of the balancing on $x \mitosis{\mcat} y$, as can be seen by unraveling the definition of the module Eilenberg-Watts functors.

The proof for the one-sided case is completely analogous; however here, only a single domain is adjacent to the $(m\in \mcat)$-labeled defect line, and hence the following paths need to be compared for the holonomy operation:

\begin{equation}\label{pic_moveC_oneside_pf}
\pica{1}{moveC_oneside_pf_1} \xrightarrow{\moveC} \pica{1}{moveC_oneside_pf_2}
\end{equation}

This completes the proof that the move $\moveC$ is an invariance.

The linear map $\Movemap$ between block spaces for the $\moveC$-move also appears in [Lem. 4.39]fss-statesum, where it is proved that $\Movemap$ is an isomorphism.

§.§.§ ${\moveEF}$ – Edge Fusion

As described in <Ref>, the edge fusion move allows use to merge parallel defect lines which share start and end nodes.
We proceed to show that $\moveEF$ is a move of invariance, following the proof strategy outlined in <Ref>.
The first step is to show that the left square in the diagram (<ref>) commutes, i.e. that

\begin{equation}\label{eq_ef_lsq}
\postmove{\delta} \circ \movemap = \Pmovemap\circ \delta.
\end{equation}

To improve legibility, we perform the calculations using the diagrammatic calculus outlined in <Ref>.
The linear map $\movemap : \nodespace \to \postmove{\nodespace}$ defined in (<ref>) is composed of the adjunction isomorphisms (<ref>) and (<ref>).
Disregarding tensor factors of the node spaces that are unaffected by the move, $\movemap$ takes for our given defect line labels $m$ and $n$ the form

\begin{equation}\label{eq_EF_movemap}
\movemap =  \bigoplus_{\si{a} \in \bcat} \pica{1}{pic_ef_moma_1} \otimes \bigoplus_{\si{b}\in \bcat} \pdim{\si{b}}  \pica{1}{pic_ef_moma_2}.
\end{equation}

Implementing the definition (<ref>) of $\delta$, we obtain

\begin{equation}\label{eq_EF_lsq_1}
\postmove{\delta} \circ \movemap =  \bigoplus_{\si{k} \in \mcat \reldel \ncat} \pdim{\si{k}} \pdim{\si{b}} \sum_{\si{a}\si{b}} \pica{1}{pic_ef_lsq_1} \otimes  \pica{1}{pic_ef_lsq_2},
\end{equation}

where we stress that the $\star$-notation here denotes a sum over dual bases of Hom-spaces in the relative Deligne product, which is why the coupons are drawn in red.
Turning to the other side of equation (<ref>), we read off the expression for $\Pmovemap$ from (<ref>):

\begin{equation}\label{eq_EF_Pmovemap}
\Pmovemap =  \bigoplus_{\si{k}} \bigoplus_{\substack{\si{m}\in \mcat \\ \si{n} \in \ncat}} \pdim{\si{k}}\sum_{\si{a}\si{b}} \pdim{\si{b}}  \pica{1}{pic_ef_pmoma_1} \otimes  \pica{1}{pic_ef_pmoma_2},
\end{equation}

and calculate:

\begin{align}\label{eq_EF_lsq_2}
\begin{aligned}
\Pmovemap \circ \delta &\eqwithref{eq_eval_delta}  \bigoplus_{\si{k}\si{m}\si{n}} \pdim{\si{k}} \pdim{\si{m}}\pdim{\si{n}}\sum_{\si{a}\si{b}} \pdim{\si{b}}  \pica{1}{pic_ef_lsq_3} \otimes  \pica{1}{pic_ef_lsq_4}\\
&\hspace{-3mm}\eqwithref{eq_def_indadjr} \bigoplus_{\si{k}\si{m}\si{n}} \pdim{\si{k}} \pdim{\si{m}}\pdim{\si{n}}\sum_{\si{a}\si{b}} \pdim{\si{b}}  \pica{1}{pic_ef_lsq_5} \otimes  \pica{1}{pic_ef_lsq_6}
\eqwithref{eq_parallel_res} \bigoplus_{\si{k}} \pdim{\si{k}} \sum_{\si{a}\si{b}} \pdim{\si{b}}  \pica{1}{pic_ef_lsq_7} \otimes  \pica{1}{pic_ef_lsq_8}.
\end{aligned}
\end{align}

Comparison with (<ref>) shows that $\postmove{\delta} \circ \movemap = \Pmovemap\circ \delta$, and hence the left square in (<ref>) commutes.

We are left to show that the square (<ref>) (which is the outer square in (<ref>) below) commutes for the move $\moveEF$.
To this end, we note that the holonomy idempotent on $\sigsurf$ factorizes into commuting idempotents $\holidem = \holidem_\mathrm{Rest} \circ\holidem_\bcat$, where $\holidem_\bcat$ is the holonomy idempotent with respect to the $\bcat$-labeled domain in $\sigsurf$.
Thus, $\holidem_\bcat \circ \holidem = \holidem = \holidem_\mathrm{Rest} \circ \holidem_\bcat$.
The square (<ref>) takes, with a subdivision, the following form.

\begin{equation} \label{diag_emove_proof}
\begin{tikzcd}
\preblock \arrow[rrr, "\holidem"] \arrow[dr, "\holidem_\bcat"] \arrow[dd, "\Pmovemap"] & & & \preblock \arrow[dd, "\Pmovemap"] \arrow[dl, "\holidem_\bcat"] \\
& \preblock \arrow[r, "\holidem_\mathrm{Rest}"] & \preblock & \\
\postmove{\preblock} \arrow[rrr, "\postmove{\holidem}"] \arrow[ur, "c", dashed] & & &  \postmove{\preblock} \arrow[ul, dashed, "c"]
\end{tikzcd}
\end{equation}

We have to find an injective map $c$ such that all smaller cells in (<ref>) which involve $c$ commute.
This then implies commutativity of the outer square.
As a map between direct sums

\begin{align} \label{eq_moveEF_pdef_first}
\begin{aligned}
\postmove{\preblock} &= \bigoplus_{\si{k} \in \mcat\reldel\ncat}\homset{\opcat{\si{k}} \deligne \cdots}{\postmove{U}(x)} \otimes \homset{\si{k} \deligne \cdots}{\postmove{U}(y)}\\
& \xrightarrow{c} \bigoplus_{\si{q}\in \mcat}\bigoplus_{\si{r}\in\ncat}\homset{\opcat{\si{r}}\deligne\opcat{\si{q}}\deligne \cdots}{U(x)} \otimes \homset{\si{q}\deligne \si{r}\deligne \cdots}{U(y)} = \preblock,
\end{aligned}
\end{align}

$c$ is determined by its components $\ctc{c}{\si{k}}{\si{qr}}$, which are

\begin{align}\label{eq_moveEF_pdef_second}
\begin{aligned}
\ctc{c}{\si{k}}{\si{qr}}= \pdim{\si{q}}\pdim{\si{r}} \; 	\homset{\basisel{\sweed{\si{k}}{\ncat}}{\si{r}}\deligne \basisel{\sweed{\si{k}}{\mcat}}{\si{q}} \deligne \cdots}{U(x)} \otimes \homset{\basisel{\si{q}}{\sweed{\si{k}}{\mcat}} \deligne \basisel{\si{r}}{\sweed{\si{k}}{\ncat}} \deligne \cdots}{U(y)} \circ U\\
\end{aligned}
\end{align}

Here we used the form of Sweedler notation for the forgetful functor from (<ref>), writing $U(k) = \sweed{k}{\mcat} \deligne \sweed{k}{\ncat}$.

To see that $c : \postmove{\preblock} \to \preblock$ is injective, note that under the isomorphisms
\postmove\preblock \cong \homset{\sweed{x}{\mcat \reldel \ncat}}{\sweed{y}{\mcat\reldel\ncat}} \otimes \cdots
\preblock \cong \homset{\sweed{x}{\mcat}\deligne \sweed{x}{\ncat}}{\sweed{y}{\mcat}\deligne \sweed{y}{\ncat}} \otimes \cdots,
that we encountered in <Ref>,
the map $c$ corresponds to the subspace inclusion, meaning that the following diagram commutes:

\begin{equation}
\begin{tikzcd}
\postmove{\preblock} \arrow[d, "c"] \arrow[r, "\cong"] & \homset{\sweed{x}{\mcat \reldel \ncat}}{\sweed{y}{\mcat\reldel\ncat}} \otimes \cdots \arrow[d, hook, "\text{subspace inclusion}"{right}] \\
\preblock \arrow[r, "\cong"] & \homset{\sweed{x}{\mcat}\deligne \sweed{x}{\ncat}}{\sweed{y}{\mcat}\deligne \sweed{y}{\ncat}} \otimes \cdots
\end{tikzcd}
\end{equation}

This can be checked explicitly, using the isomorphism $\soc$ from <Ref>.
Thus $c$ is monic, and the outer square in the diagram (<ref>) commutes if all the cells commute.

To verify that the bottom square in (<ref>) commutes, recall that the holonomy idempotents $\postmove{\holidem}$ and $\holidem_\mathrm{Rest}$ are each defined as a composition of individual holonomy idempotents for every domain except $\bcat$ – these domains appear in both $\sigsurf$ and $\postmove{\sigsurf}$.
On the other hand, the linear map $c$ affects only the algebraic data associated with the domain $\bcat$.
It thus commutes with the holonomy idempotents for the other domains.

This leaves us with the checking the commutativity of the triangles on the left and the right of  (<ref>), which happen to be equal.
They represent the equation

\begin{equation}\label{eq_moveEF_trianglegoal}
c \circ \Pmovemap = \holidem_\bcat,
\end{equation}

which needs to be checked now.
Knowing the components of $c$ from (<ref>) and of $\Pmovemap$ from (<ref>), we write out the composition $c\circ \Pmovemap$ in components $\si{m},\si{q} \in \mcat, \; \si{n}, \si{r} \in \ncat$:

\begin{align}\label{eq_edgefusion_pcircpsi_0}
\begin{aligned}
&\ctc{(c\circ \Pmovemap)}{\si{mn}}{\si{qr}} = \sum_{\si{k}\,\si{a}\si{b}} \pdim{\si{k}} \pdim{\si{q}}\pdim{\si{r}} \pdim{\si{b}}  \; \pica{1}{pic_ef_rsq_1} \otimes \pica{1}{pic_ef_rsq_2}.
\end{aligned}
\end{align}

In order to compare this expression to the respective components of $\holidem_\bcat$, it is most convenient to evaluate it on morphisms $g : U(x) \to \si{m} \deligne \si{n}$ and $f : \si{m} \deligne\si{n} \to U(y)$, and then pass through the isomorphism

\begin{equation}
\soc^{-1} : \bigoplus_{\si{q}\si{r}} \homset{U(x)}{\si{q}\deligne\si{r}} \otimes \homset{\si{q}\deligne\si{r}}{U(y)} \to \homset{U(x)}{U(y)}
\end{equation}

from (<ref>).
Doing so, we obtain

\begin{align}\label{eq_EF_rsq_calc_1}
\begin{aligned}
&\soc^{-1}\left( \ct{(c\circ \Pmovemap)}{\si{mn}}(g\otimes f) \right)
= \hspace{-2mm}\sum_{\si{k}\,\si{q}\si{r}\,\si{a}\si{b}} \pdim{\si{k}} \pdim{\si{q}}\pdim{\si{r}} \pdim{\si{b}}  \hspace{-5mm} \pica{1}{pic_ef_rsq_3}
\eqwithref{eq_res_of_id}  \sum_{\si{k}\,\si{a}\si{b}} \pdim{\si{k}} \pdim{\si{b}}  \hspace{-5mm} \pica{1}{pic_ef_rsq_4}
\eqwithref{eq_res_of_id} \sum_{\si{a}} \pdim{\si{a}} \pica{1}{pic_ef_rsq_5}.
\end{aligned}
\end{align}

On the other hand, an unraveling of the definition of $\holidem_\bcat$ similar to (<ref>) reveals the following:

\begin{align}
\begin{aligned}
\ctc{(\holidem_\bcat)}{\si{mn}}{\si{qr}} &= \frac{1}{\catdim{\bcat}} \sum_{\si{i}\si{j}\,\si{a}}  \pdim{\si{a}} \pdim{\si{i}}\pdim{\si{j}}  \pica{1}{pic_ef_rsq_6} \otimes \pica{1}{pic_ef_rsq_7}
\eqwithref{eq_parallel_res} \frac{1}{\catdim{\bcat}} \sum_{\si{a}}  \pdim{\si{a}} \pica{1}{pic_ef_rsq_8} \otimes \pica{1}{pic_ef_rsq_9}.
\end{aligned}
\end{align}

Again, evaluating on $g\otimes f$ and applying $\soc^{-1}$ yields

\begin{align}
\begin{aligned}
&\soc^{-1}\left( \ct{(\holidem_\bcat)}{\si{mn}}(g\otimes f)  \right)
= \frac{1}{\catdim{\bcat}} \sum_{\si{q}\si{r}\,\si{a}} \pdim{\si{a}} \pdim{\si{q}}\pdim{\si{r}} \hspace{-5mm} \pica{1}{pic_ef_rsq_10} \eqwithref{eq_res_of_id} \frac{1}{\catdim{\bcat}} \sum_{\si{a}}\pdim{\si{a}}  \pica{1}{pic_ef_rsq_5},
\end{aligned}
\end{align}

which is identical to (<ref>).
This implies $c\circ \Pmovemap = \holidem_\bcat$, and so the outer paths in the diagram (<ref>) commute, and the move $\moveEF$ is an invariance.

The linear map $\Movemap$ between block spaces for the $\moveEF$-move also appears in [Thm. 4.43]fss-statesum, where it is proved that $\Movemap$ is an isomorphism.

§.§ Composite Moves

This section is the second and final part of the proof of <Ref>.
The remaining moves do not have to be proved by explicitly using the definition of the evaluation procedure, but can be obtained as compositions of moves we already know to be invariances.

§.§.§ ${\moveDV}$ – Silent Node

We begin by redrawing the picture (<ref>) in a slightly larger region, this time involving another node labeled by $x$.
Such a node $x$ exists because the extruded graph on the left-hand side of (<ref>) is not allowed to feature a loop without a node.

\begin{equation}\label{pic_moveDV_pf}
\pica{1}{moveDV_pf_1} \xrightarrow{\moveDV} \pica{1}{moveDV_pf_2}
\end{equation}

The invariance under the $\moveDV$-move is a consequence of the following more general lemma:

Let $\mathbf{M}_1$ be a move between regions $\sigsurf$ and $\postmove{\sigsurf}$, and let $\mathbf{M}_2$ be a move between $\postmove{\sigsurf}$ and $\postmove{\postmove{\sigsurf}}$.
Suppose that both $\mathbf{M}_2$, and the composite move $\mathbf{M}_1 \mathbf{M}_2$ between $\sigsurf$ and $\postmove{\postmove{\sigsurf}}$ are moves of invariance.
If the linear map $\Movemap_2$ between block spaces of $\postmove{\sigsurf}$ and $\sigsurf''$ is injective, then $\mathbf{M}_1$ is an invariance as well.

In the following diagram, commutativity of the left square is equivalent to $\mathbf{M}_1$ being an invariance.
This commutativity follows from the commutativity of the right square (due to the fact that $\mathbf{M}_2$ is an invariance), the commutativity of the outer paths ($\mathbf{M}_1 \mathbf{M}_2$ is an invariance), and the fact that $\Movemap_2$ is a monomorphism.

\begin{equation}
\begin{tikzcd}
\nodespace \arrow[r, "\movemap_1"] \arrow[d, "\evaluate{\sigsurf}"] & \postmove{\nodespace} \arrow[r, "\movemap_2"] \arrow[d, "\evaluate{\postmove{\sigsurf}}"] & \postmove{\postmove{\nodespace}} \arrow[d, "\evaluate{\postmove{\postmove{\sigsurf}}}"] \\
\blockspace \arrow[r, "\Movemap_1"] & \postmove{\blockspace} \arrow[r, "\Movemap_2"] & \postmove{\postmove{\blockspace}}
\end{tikzcd}
\end{equation}

To apply <Ref>  to the present situation by setting $\mathbf{M}_1 = \moveDV$ and $\mathbf{M}_2 = \moveC$,
we now show that the composite $\moveDV\moveC$ is the identity move, which is clearly an invariance.
This is obvious on the topological level.
As for the node labels, it follows immediately from the definition (<ref>) of the contraction operation $\mitosis{\mcat}$ that $\zewid{\mcat}$ is a unit for the operation:

\begin{equation}
\zewid{\mcat} \mitosis{\mcat} x \cong x,
\end{equation}

Lastly we need to verify that the linear maps $\movemap_{\moveDV}$, $\movemap_{\moveC}$ between node spaces compose to the identity.
This is done by checking that for $f\in \nodespace = \homset{m}{\sweed{x}{\mcat}}$,

\begin{equation}
\movemap_{\moveC}\left(\soc(\id_\mcat) \gencomp{m} f\right) = f,
\end{equation}

which follows from unraveling the definitions (<ref>) of $\gencomp{m}$ and (<ref>) of $\soc$.
This proves that $\moveDV$ is an invariance.

Moreover, the $\moveDV$-move's linear map $\Movemap$ between block spaces is an isomorphism, as it is the inverse of the $\moveC$-move's map $\Movemap_{\moveC}$.

§.§.§ ${\moveL}$ – Loop Move

The $\moveL$-move is obtained as the following composition of moves.

\begin{align}\label{pic_lmove_proof}
\begin{aligned}
&\pica{1}{moveDE_deloop_1} \xrightarrow{\moveDV} \pica{1}{moveDE_deloop_2} \\
&\hspace{1cm}\xrightarrow{\moveOR,\; \moveEF} \pica{1}{moveDE_deloop_3} \xrightarrow{\moveC} \pica{1}{moveDE_deloop_4}
\end{aligned}
\end{align}

First, we apply the $\moveDV$-move to the loop, which splits it into two $\mcat$-labeled defect lines and creates a new silent node.
Next, we flip the orientation of one of the $\mcat$-labeled defect lines using the $\moveOR$-move, so that both point away from the $x$-labeled node.
The $\moveEF$-move then merges the two defect lines into one, labeled by $\opcat{m} \lexreldel m \in \opcat{\mcat} \reldelov{\acat} \mcat$.
In the last step, we use the (one-sided version of the) $\moveC$-move to absorb the silent node  into the node labeled by $x$.
This results in the node label $x' = \zewcoid{\mcat} \mitosis{\opcat{\mcat}\reldel \mcat} x$, as described in (<ref>).

We still have to compute the linear map $\movemap$ between node spaces, which is a composition of the individual maps for each move in the chain (<ref>).
The first morphism is $\movemap_{\moveDV} = \soc(\id_m) \otimes \id_{\homset{\opcat{m}\deligne m}{\sweed{x}{\opcat{\mcat}\deligne\mcat}}}$.
The map $\movemap_{\moveOR}$ is the identity.
In the next step, we obtain

\begin{equation}
\movemap_{\moveEF} \circ \movemap_{\moveDV} = \frac{1}{\catdim{\acat}} \bigoplus_{\si{a},\si{b} \in \acat} \pdim{\si{b}}\; \pica{1}{pic_l_moma_2} \otimes \pica{1}{pic_l_moma_1}.
\end{equation}

Lastly, when performing the $\moveC$-move, the two tensor factors are composed:

\begin{equation}
\movemap_{\moveL} = \movemap_{\moveC} \circ \movemap_{\moveEF} \circ \movemap_{\moveDV} = \frac{1}{\catdim{\acat}} \sum_\si{a} \pdim{\si{a}} \;\pica{1}{pic_l_moma_3} = r(\soc(\id_m) \circ -),
\end{equation}

where $r$ denotes the retraction from <Ref>.
This agrees with the linear map defined in <Ref>, and so the proof is done.

This completes the proof of <Ref>.

§ DETAILS ON LOOP GRAPHS

This section forms the proof of <Ref>.
As in <Ref>, we restrict ourselves without loss of generality to the case $\bcat = \vect$.

* Let $g \in \preblock = \homset{\ewid{\mcat}}{\ewid{\mcat}}$ be a vector in the pre-block space.
We make the formula

\begin{equation}
\holsurj(g) = \sum_\si{a} \frac{\pdim{\si{a}}}{\catdim{\acat}}\; \brev{\zewid{\mcat}}{\si{a}^*} \circ (\si{a}^* g  \si{a}) \circ \cobrev{\zewid{\mcat}}{\si{a}}
\end{equation}

obtained in <Ref> for the retraction $\holsurj : \preblock \to \blockspace$ from (<ref>) more explicit.
From (<ref>) and (<ref>), we can work out how to express the morphisms $\brev{\zewid{\mcat}}{\si{a}^*}$ and $\cobrev{\zewid{\mcat}}{\si{a}}$ in components:

\begin{align}
\ete{(\brev{\zewid{\mcat}}{\si{a}^*})}{\si{l}}{\si{q}} &=\pdim{\si{l}} \; \basisel{\si{a}^*\si{l}}{\si{q}} \btimes \basisel{\si{q}}{\si{a}^*\si{l}}\;\colon\; \si{a}^*\si{l} \deligne \opcat{\si{a}^*\si{l}} \to \si{q}\deligne \opcat{\si{q}}, \label{eq_silent_brev_components}\\
\ete{(\cobrev{\zewid{\mcat}}{\si{a}})}{\si{n}}{\si{k}} &= \pdim{\si{n}} \; \basisel{\si{n}}{\si{a}^*\si{k}} \btimes \basisel{\si{a}^*\si{k}}{\si{n}} \;:\; \si{n} \deligne\opcat{\si{n}}  \to \si{a}^*\si{k} \deligne\opcat{\si{a}^*\si{k}}. \label{eq_silent_cobrev_components}
\end{align}

This allows us to state the components $\ete{\holsurj(g)}{\si{n}}{\si{q}} : \si{n} \deligne \opcat{\si{n}} \to \si{q} \deligne \opcat{\si{q}}$ of $\holsurj(g)$:

\begin{align}\label{eq_pi_component}
\begin{aligned}
\ete{\holsurj(g)}{\si{n}}{\si{q}} = \sum_{\si{a},\si{k}, \si{l}} \frac{\pdim{\si{a}}\pdim{\si{n}}\pdim{\si{l}}}{\catdim{\acat}}\; (\basisel{\si{a}^*\si{l}}{\si{q}} \btimes \basisel{\si{q}}{\si{a}^*\si{l}}) \circ (\si{a}^* \ete{g}{\si{k}}{\si{l}}  \si{a}) \circ (\basisel{\si{n}}{\si{a}^*\si{k}} \btimes \basisel{\si{a}^*\si{k}}{\si{n}}).
\end{aligned}
\end{align}

* For the evaluation of the loop graph $O$, we also need to compute the morphism $\delta$ from (<ref>).
Let $f : m \to m$ be an arbitrary endomorphism of $m \in \mcat$, as in <Ref>.
We compute

\begin{align} \label{eq_thmelempf_calc_1}
\begin{aligned}
\delta(\soc(f)) &= \bigoplus_\si{k} \pdim{\si{k}} \; \soc(f) \circ (\basisel{\si{k}}{m} \deligne \basisel{m}{\si{k}}).
\end{aligned}
\end{align}

Ultimately though, we need the components $\ete{\delta(\soc(f))}{\si{k}}{\si{l}}$.
To obtain them, we make use of the explicit form of the isomorphism $\soc$ from (<ref>), $\te{\soc(f)}{\si{l}} = (\basisel{m}{\si{l}} \circ f) \deligne \basisel{\si{l}}{m}$.
We insert this into (<ref>) to find

\begin{align}
\begin{aligned}
\ete{\delta(\soc(f))}{\si{k}}{\si{l}}
&=  \pdim{\si{k}} \; \te{\soc(f)}{\si{l}} \circ (\basisel{\si{k}}{m} \deligne \basisel{m}{\si{k}}) \\
&=  \pdim{\si{k}} \; \left((\basisel{m}{\si{l}} \circ f) \deligne \basisel{\si{l}}{m}\right) \circ (\basisel{\si{k}}{m} \deligne \basisel{m}{\si{k}}) \\
&=  \pdim{\si{k}} \; (\basisel{m}{\si{l}} \circ f \circ\basisel{\si{k}}{m})  \deligne ( \basisel{m}{\si{k}}\circ \basisel{\si{l}}{m}) \\
\end{aligned}
\end{align}

* We substitute  $\delta(\soc(f))$ for $g$ in (<ref>):

\begin{align}\label{eq_thmelempf_calc_2}
\begin{aligned}
\ete{\holsurj(\delta(\soc(f)))}{\si{n}}{\si{q}} &= \sum_{\si{a},\si{k}, \si{l}} \frac{\pdim{\si{a}}\pdim{\si{n}}\pdim{\si{l}}}{\catdim{\acat}}\; (\basisel{\si{a}^*\si{l}}{\si{q}} \btimes \basisel{\si{q}}{\si{a}^*\si{l}}) \circ (\si{a}^* \ete{\delta(\soc(f))}{\si{k}}{\si{l}}  \si{a}) \circ (\basisel{\si{n}}{\si{a}^*\si{k}} \btimes \basisel{\si{a}^*\si{k}}{\si{n}}) \\
&= \sum_{\si{a},\si{k}, \si{l}} \frac{\pdim{\si{a}}\pdim{\si{n}}\pdim{\si{l}} \pdim{\si{k}}}{\catdim{\acat}}\; \left(\basisel{\si{a}^*\si{l}}{\si{q}} \circ \si{a}^* (\basisel{m}{\si{l}} \circ f \circ\basisel{\si{k}}{m}) \circ \basisel{\si{n}}{\si{a}^*\si{k}}\right) \\ &\hspace{2cm}  \btimes \left(\basisel{\si{a}^*\si{k}}{\si{n}} \circ (\basisel{m}{\si{k}}\circ \basisel{\si{l}}{m}) \si{a} \circ  \basisel{\si{q}}{\si{a}^*\si{l}} \right).
\end{aligned}
\end{align}

To simplify this expression, we use that

\begin{align}\label{eq_thmelempf_calc_3}
\begin{aligned}
&\sum_{\si{l}} \pdim{\si{l}}\; (\basisel{\si{a}^*\si{l}}{\si{q}} \circ \si{a}^* \basisel{m}{\si{l}})  \otimes  (\basisel{\si{l}}{m} \si{a} \circ  \basisel{\si{q}}{\si{a}^*\si{l}} ) = \basisel{\si{a}^*m}{\si{q}} \otimes \basisel{\si{q}}{\si{a}^*m}, \quad \text{and} \\
&\sum_{\si{k}} \pdim{\si{k}}\; (\si{a}^* \basisel{\si{k}}{m} \circ \basisel{\si{n}}{\si{a}^* \si{k}} ) \otimes ( \basisel{\si{a}^*\si{k}}{\si{n}} \circ \basisel{m}{\si{k}}\si{a} ) = \basisel{\si{n}}{\si{a}^* m} \otimes \basisel{\si{a}^* m}{\si{n}}.
\end{aligned}
\end{align}

Using (<ref>) to continue the calculation (<ref>), we obtain

\begin{align}\label{eq_thmelempf_calc_4}
\begin{aligned}
\ete{\holsurj(\delta(\soc(f)))}{\si{n}}{\si{q}}
&= \sum_{\si{a}} \frac{\pdim{\si{a}}\pdim{\si{n}}}{\catdim{\acat}}\; \left(\basisel{\si{a}^*m}{\si{q}} \circ \si{a}^*f \circ \basisel{\si{n}}{\si{a}^* m} \right)  \btimes \left( \basisel{\si{a}^* m}{\si{n}} \circ \basisel{\si{q}}{\si{a}^*m} \right).
\end{aligned}
\end{align}

* The cells of the following diagram commute:

\begin{equation}\label{eq_thmelempf_calc_5}
\begin{tikzcd}
\homsetin{\raycat}{\zewid{\mcat}}{\zewid{\mcat}} \arrow[r, hook] \arrow[d, "\trace{\raycat}"] & \homsetin{\nodecat}{\ewid{\mcat}}{\ewid{\mcat}} = \bigoplus_\si{n} \homset{\si{n} \deligne \opcat{\si{n}}}{\ewid{\mcat}} \arrow[r, "\bigoplus_\si{n} \soc^{-1}"] \arrow[d, "\trace{\nodecat}"] & \bigoplus_\si{n} \homsetin{\mcat}{\si{n}}{\si{n}} \arrow[d, "\sum_\si{n} \pdim{\si{n}} \trace{\mcat}"] \\
\field \arrow[r, "\catdim{\acat}"] & \field \arrow[r, "\id"] & \field.
\end{tikzcd}
\end{equation}

Commutativity of the left cell follows from <Ref>; the right square can be checked explicitly using the explicit form of the isomorphism $\soc$ form (<ref>).

* We compute $\bigoplus_\si{n} \soc^{-1}(\et{\holsurj(\delta(\soc(f)))}{\si{n}})$, starting from (<ref>).

\begin{align}\label{eq_thmelempf_calc_6}
\begin{aligned}
\soc^{-1}(\et{\holsurj(\delta(\soc(f)))}{\si{n}})
&=  \sum_{\si{a}\si{q}} \frac{\pdim{\si{a}}\pdim{\si{n}}}{\catdim{\acat}}\; \left(\basisel{\si{a}^* m}{\si{n}} \circ \basisel{\si{q}}{\si{a}^*m} \circ \basisel{\si{a}^*m}{\si{q}} \circ \si{a}^*f \circ \basisel{\si{n}}{\si{a}^* m} \right)\\
&= \sum_{\si{a}} \frac{\pdim{\si{a}}}{\catdim{\acat}}\; \left(\basisel{\si{a}^* m}{\si{n}} \circ \si{a}^*f \circ \basisel{\si{n}}{\si{a}^* m} \right)
\end{aligned}
\end{align}

To arrive at the first equality, we used the explicit form of the isomorphism $\soc$.
For the second equality, we recognized a resolution of the identity. To perform this step, it is necessary to replace the dimension factor $\pdim{\si{n}}$ with $\pdim{\si{q}}$, which is possible thanks to Schur's lemma.

* We are now in a position to compute the evaluation:

\begin{align}\label{eq_thmelempf_calc_7}
\begin{aligned}
\left(\theta, \evaluate{O}\soc(f)\right)
&\eqwithref{eq_total_eval} \theta \circ \pi \circ \delta \circ \soc(f) \eqwithref{eq_dualoftheta} \catdim{\acat} \; \trace{\raycat}\left(\holsurj \circ \delta \circ \soc(f) \right) \\
&\eqwithref{eq_thmelempf_calc_5} \sum_\si{n} \pdim{\si{n}} \trace{\mcat} \left(\soc^{-1}(\et{\holsurj (\delta (\soc(f)))}{\si{n}})\right) \\
&= \frac{1}{\catdim{\acat}} \sum_{\si{n}, \si{a}} \pdim{\si{n}} \pdim{\si{a}} \trace{\mcat}  \left(\basisel{\si{a}^* m}{\si{n}} \circ \si{a}^*f \circ \basisel{\si{n}}{\si{a}^* m}\right) \\
&\eqdesc{(*)} \frac{1}{\catdim{\acat}} \sum_{\si{a}} \pdim{\si{a}} \trace{\mcat} \left(  \si{a}^*f \right) =  \trace{\mcat} \left( f \right).
\end{aligned}
\end{align}

The equality $(*)$ uses the fact that for any endomorphism $\alpha : x \to x$ in a Calabi-Yau category $\xcat$, $\trace{\xcat}(\alpha) = \sum_\si{y} \pdim{\si{y}} \trace{\xcat}(\basisel{x}{\si{y}} \circ \alpha \circ \basisel{\si{y}}{x})$.
To check this, one uses symmetry of the trace and a resolution of the identity.

It is easy to see that the result (<ref>) is equivalent to the statement

\begin{equation}
\theta \circ \evaluate{O} = \trace{\mcat} \circ \soc^{-1}
\end{equation}

of <Ref>.
The claim also holds true in the case where $\bcat$ is not necessarily $\vect$. Thus, the proof is done.

Missing 'biblatex' package
The bibliography requires the 'biblatex' package.

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By Theorem <ref> and Proposition <ref>, we define automorphisms $\TT_i^{\pm1}$ on $\hA_{q}(\n)$
by the following:

\begin{equation} \label{eq:brd1pY}
\begin{aligned}
\TT_{i}(\sfy_{j,m}) =
\bc
\sfy_{j,m+\delta_{i,j}}  &  \text{if} \  \sfc_{i,j} \ge 0, \\[1ex]
\dfrac{ \left( \displaystyle \sum_{k=0}^{- \sfc_{i,j}} (-1)^{k} q_i^{ - \sfc_{i,j}/2 -k  } \sfy_{i,m}^{(k)}\sfy_{j,m}  \sfy_{i,m}^{(- \sfc_{i,j}-k)}  \right)}{ (q_i -q_i^{-1})^{- \sfc_{i,j}}  }  &\text{if} \  \sfc_{i,j} < 0,  \ec
\end{aligned}
\end{equation}

\begin{equation}  \label{eq:invbrd1pY}
\begin{aligned}
\TT_{i}^{-1}(\sfy_{j,m}) =
\bc
\sfy_{j,m-\delta_{i,j}}  &  \text{if} \  \sfc_{i,j} \ge 0, \\[1ex]
\dfrac{ \left( \displaystyle \sum_{k=0}^{-\sfc_{i,j}} (-1)^{k} q_i^{ -\sfc_{i,j}/2 -k  } \sfy_{i,m}^{(-\sfc_{i,j}-k)} \sfy_{j,m}  \sfy_{i,m}^{(k)}  \right)}{ (q_i -q_i^{-1})^{-\sfc_{i,j} }  }  &\text{if} \  \sfc_{i,j} < 0,
\ec
\end{aligned}
\end{equation}

\sfy_{i,m}^{(k)}  \seteq \dfrac{\sfy_{i,m}^{k} }{[k]_{i}!}.

Now we are in a position to state our final main result of this paper.

The automorphisms $\{ \TT^{\pm 1}_i \ | \ i \in I\}$ give a braid group $B(\Dynkin)$ action on $\hA_{q}(\n)$ and hence on $\frakK_q$.

We give the proof of Theorem <ref> in Section <ref>.

The rest of this subsection will be devoted to present consequences of Theorem <ref> and its proof.

For $m \in \Z$, we set

$$\hA_{q}(\n)_m \seteq\lan \sfy_{i,m} \ | \ i \in \Dynkin_0 \ran \subset \hA_{q}(\n).$$

Note that $\hA_{q}(\n)_m$ is isomorphic to $\calU_q^-(\g)$ via the $\Q(q^{1/2})$-algebra isomorphism $\varsigma_m$ such that  $\varsigma_m(\sfy_{i,m})=f_i$.
From the proofs of the propositions in Section <ref>, we obtain the following corollary:

For $i,j \in I$ and $0 \le p < h_{i,j}$, we have

\underbrace{\cdots \TT_{i}\TT_{j}}_{\text{$p$-times}}(\sfy_{i}), \ \underbrace{\cdots \TT_{j}\TT_{i}}_{\text{$p$-times}}(\sfy_{j}) \in   \lan \sfy_{i,m},\sfy_{j,m}  \ran \subseteq
\hA_{q}(\n)_m.

For $w \in \weyl$ with $(i_1, \dots, i_r) \in I(w)$, set

\begin{align*}
\TT^{\pm 1}_w \seteq \TT^{\pm 1}_{i_1} \dots \TT^{\pm 1}_{i_r}.
\end{align*}

Note that $\TT^{\pm 1}_w$ is well-defined by Theorem <ref>.

For $w \in \weyl$ and $\ell(ws_{i})=\ell(w)+1$. Then we have the following:

*  $  \TT_{w}( \sfy_{i,m} ) \in \hA_{q}(\n)_m.$

*  There exists $\bfc \in \Q[[q^{1/2}]] \setminus \{0\}$ such that
$\varsigma_m(\TT^{\pm}_{w}( \sfy_{i,m} )) = \bfc T^{\pm}_w(f_i).$

By the standard argument, there exists $j \in I$ such that $\ell(ws_j)=\ell(w)-1$ and a unique $w' \in \weyl$ such that

$$\ell(w's_j)=\ell(w')+1,\quad \ell(w's_i)=\ell(w')+1 \quad \text{ and } \quad w = w'y$$

whether either

* $y=\underbrace{s_is_j \cdots s_j}_{\text{$(h_{i,j}> p \ge 2)$-factors}}$ and $\ell(w)=\ell(w')+p$, or

* $y=\underbrace{s_js_i \cdots s_j}_{\text{$(h_{i,j}> p \ge 1)$-factors}}$ and $\ell(w)=\ell(w')+p$.

With this observation, the first assertion now follows from an induction argument on $\ell(w)$ and Corollary <ref> (see <cit.>).
Next, Remark <ref> implies that the second assertion holds for $w$ with $\ell(w)=1$. In Propositions <ref>, <ref>
and, <ref>, we will see that

\varsigma_m(\TT^{\pm}_{y}( \sfy_{i,m} )) = \bfc T^{\pm}_y(f_i).

Then the second assertion also follows from an induction argument on the length and (<ref>).

For a reduced expression $\uw=s_{i_1}\cdots s_{i_{r-1}}s_{i_r}$ of $w$, if $s_{i_1}\cdots s_{i_{r-1}}(\al_{i_r})=\al_j$, then

$$  \TT^\pm_{i_1}\cdots \TT^\pm_{i_{r-1}}( \sfy_{i_r,m} ) = \sfy_{j,m}.$$

*  For $w_\circ \in \weyl$, we have

\TT^\pm_{w_\circ}(\sfy_{i,m}) =\sfy_{i^*,m\pm1} \in \hA_{q}(\n)_{m\pm1}.

In this proof, we only consider $\TT_i$-case, since the proof for  $\TT^{-}_i$-case is similar.

(<ref>) Assigning $(-1)^{m}\al_i$ for the weight of $\sfy_{i,m}$, the algebra $\hA_{q}(\n)$ is a $\Phi$-graded algebra. With this $\Phi$-grading, the automorphism $\TT_i$ sends a $\be$-homogeneous element to a $s_i(\be)$-homogeneous element. Since $(-1)^{m}\al_j$-graded space in $ \hA_{q}(\n)_m$ is $1$-dimensional and spanned by $\sfy_{j,m}$, our assertion follows.

(<ref>) Note that there exists $\ii_\circ = (i_1,\ldots, i_\ell)\in I(w_\circ)$ such that $i_\ell=i$. Then we have $s_{i_1} \cdots s_{i_{\ell-1}}(\al_i)=\al_{i^*}$ and hence

$$  \TT_{i_1} \cdots \TT_{i_{\ell-1}}(\sfy_{i,m}) = \sfy_{i^*,m}  $$

by (<ref>). Thus

$$  \TT_{i_1} \cdots \TT_{i_{\ell-1}}\TT_{i_\ell}(\sfy_{i,m})=\TT_{i_1} \cdots \TT_{i_{\ell-1}}(\sfy_{i,m+1}) =\sfy_{i^*,m+1},$$

as desired.

Note that, for $\ii_\circ =(i_1,i_2,\ldots,i_\ell) \in I(w_\circ)$, the set

$$ \bbP_{\ii_\circ,m} \seteq \{ \TT_{i_1}\cdots  \TT_{i_{k-1}}(\sfy_{i_k,m}) \ | \ 1 \le k \le \ell \} \text{ is linearly independent and

For $\ii_\circ=(j_1,j_2,\ldots,j_\ell) \in I(w_\circ)$, we define $\widetilde{\ii}_0 = (i_k)_{k \in \Z_{>0}}$ as follows:

$$   i_k = j_k  \text{ for } 1\le k \le \ell \quad  \text{ and } \quad j_{k+\ell}=j_k^*.$$

Then Corollary <ref> implies that

$$ \{ \TT_{i_1}\cdots  \TT_{i_{k-1}}(\sfy_{i_k,m}) \ | \  k \in \Z_{>0}  \}  \sqcup  \{ \TT^-_{i_1}\cdots  \TT^-_{i_{k-1}}(\sfy_{i_k,m-1}) \ | \  k \in \Z_{>0}  \}    \text{ is linearly independent}$$

for any $m\in\Z$.

§.§ Proof of Theorem <ref>

We investigate the braid relations (<ref>) on $\left\{  \TT_i \, | \, i \in I \,\right\}$ in the next five propositions.

Assume $\sfc_{i,j}=0$. Then we have

$$\TT_i\TT_j   = \TT_j\TT_i.$$

We verify $\TT_i\TT_j(\sfy_{k,m}) = \TT_j\TT_i(\sfy_{k,m})$ for any pair $(k,m) \in I \times \Z$.
In the case of $\sfc_{i,k}\ge 0$ and $\sfc_{j,k} \ge 0$,
it is straightforward to check it.
Let us consider the other case, that is, $\sfc_{i,k}<0$ or $\sfc_{j,k}<0$.
Since $\sfc_{i,j}=0$, it allows us to take a Dynkin quiver $Q$ with sources $i$ and $j$.
By Theorem <ref> and Proposition <ref>, we have

\begin{align*}
\tUptheta_Q \circ \TT_i\TT_j (\sfy_{k,m}) = L^Q_q(s_is_j(\al_k),m) =  \tUptheta_Q \circ \TT_j\TT_i(\sfy_{k,m}).
\end{align*}

Since $\tUptheta_Q$ is invertible, we conclude $\TT_i\TT_j (\sfy_{k,m}) = \TT_j\TT_i(\sfy_{k,m})$.
This completes the proof.

For a simply-laced $\g$, assume $\sfc_{i,j}=-1$. Then we have

$$\TT_i\TT_j\TT_i   = \TT_j\TT_i\TT_j.$$

We claim that $\TT_i\TT_j\TT_i(\sfy_{k,m})   = \TT_j\TT_i\TT_j(\sfy_{k,m})$ for any pair $(k,m) \in I \times \Z$.
If $i \not\sim k$ and $j \not\sim k$, that is, $k$ is not adjacent to both $i$ and $j$ on the Dynkin diagram, then we have $\TT_i\TT_j\TT_i(\sfy_{k,m})   = \TT_j\TT_i\TT_j(\sfy_{k,m})$ by definition.
Let us consider the other two cases:

Case 1. Assume $k=i$ or $k=j$.
We prove the case of $k=i$ here, since the proof for the other case is quite similar.
Let $Q$ be a Dynkin quiver with  source $i$.
By similar computation to the proof of Proposition <ref>, we have

\begin{equation*}
\begin{aligned}
\tUptheta_Q \circ \TT_i\TT_j (\sfy_{i,m}) & =  \tUptheta_Q \circ  \TT_i  \left(   \dfrac{ q^{1/2}\sfy_{i,m}\sfy_{j,m} -q^{-1/2} \sfy_{j,m} \sfy_{i,m} }{q-q^{-1}}  \right) \\
& \hspace{-9ex}= \left(   \dfrac{ q^{1/2}L_q^{Q}(\al_i,m+1)*L_q^{Q}(\al_i+\al_j,m) -q^{-1/2} L_q^{Q}(\al_i+\al_j,m)* L_q^{Q}(\al_i,m+1) }{q-q^{-1}}  \right) \\
& \hspace{-9ex} = \left(   \dfrac{ q^{1/2}L_q^{s_i Q}(\al_i,m)*L_q^{s_iQ}(\al_j,m) -q^{-1/2} L_q^{s_i Q}(\al_j,m) *L_q^{s_i Q}(\al_i,m) }{q-q^{-1}}  \right)  \\
&\hspace{-9ex}   = L_q^{s_iQ}(\al_i+\al_j,m) = L_q^{Q}(\al_j,m),
\end{aligned}
\end{equation*}

where $\tUptheta_Q \circ  \TT_i (\sfy_{i,m}) = L_q^Q(\al_i, m+1) = L_q^{s_i Q}(\al_i,m)$ and $\tUptheta_Q \circ  \TT_i (\sfy_{j,m}) = L_q^Q(\al_i+\al_j,m) = L_q^{s_iQ}(\al_j,m)$.
This implies

\begin{align} \label{eq: ij(i,m)=(j,m)}
\TT_i\TT_j (\sfy_{i,m}) = \sfy_{j,m},
\end{align}

which prove the current case because (<ref>) yields

\begin{align*}
& \TT_i\TT_j\TT_i (\sfy_{i,m}) = \TT_i\TT_j(\sfy_{i,m+1}) = \sfy_{j,m+1}, \qquad  \TT_j\TT_i\TT_j (\sfy_{i,m}) = \TT_j(\sfy_{j,m})= \sfy_{j,m+1}.
\end{align*}

Case 2.
Assume $i \sim k$,   $j \not\sim k$ or $i \not\sim k$, $j \sim k$.
We only consider the former case, since the other case is proved in a similar way.
Under the assumption of $i \sim k$ and  $j \not\sim k$,
we have

\begin{align}
\TT_i\TT_j\TT_i (\sfy_{k,m})
& =
\left( \dfrac{q^{1/2}\TT_i(\sfy_{k,m}) \sfy_{j,m} - q^{-1/2}\sfy_{j,m} \TT_i(\sfy_{k,m})}{q-q^{-1}} \right),  \label{eq: st1} \allowdisplaybreaks\\
\TT_j\TT_i\TT_j (\sfy_{k,m}) &%=  \TT_j\TT_i (\sfy_{k,m})
= \left( \dfrac{ q^{1/2} \sfy_{k,m}\TT_j(\sfy_{i,m}) - q^{1/2}\TT_j(\sfy_{i,m})\sfy_{k,m} }{q-q^{-1}} \right), \label{eq: st2}
\end{align}

where (<ref>) follows from (<ref>).

To see the equality between  (<ref>) and (<ref>),
it is enough to consider the case of $\g=A_3$, $k=1$, $i=2$, $j=3$, and $m=0$.
Set $y_t\seteq \sfy_{t,0}$ for $t=1,2,3$.
Under this setting,
(<ref>) and (<ref>) become

\begin{align*}
& \dfrac{q^{1/2}\TT_2(y_{1}) y_{3} - q^{-1/2}y_{3} \TT_2(y_{1})}{q-q^{-1}}
=  \left(\dfrac{ qy_{1}y_{2}y_{3}- y_{2}y_{1}y_{3}- y_{3}y_{1}y_{2}+ q^{-1}y_{3}y_{2}y_{1} }{(q-q^{-1})^2} \right), \\
& \dfrac{q^{1/2} y_{1}\TT_3(y_{2}) - q^{1/2}\TT_3(y_{2})y_{1} }{q-q^{-1}}
= \left(       \dfrac{qy_{1}y_{2}y_{3}-y_{1}y_{3}y_{2} -y_{2}y_{3}y_{1}+q^{-1}y_{3}y_{2}y_{1} }{(q-q^{-1})^2} \right),
\end{align*}

Since $y_1y_3=y_3y_1$ by the defining relation of $\hA_{q}(\n)_m$,
we prove our claim.

For $\g$  of type $B_n$, $C_n$ or $F_4$, assume $\sfc_{j,i}\sfc_{i,j}=1$. Then we have

$$\TT_{i}\TT_{j}\TT_{i}   = \TT_{j}\TT_{i}\TT_{j}.$$

Since the proofs for $B_n$, $C_n$ and $F_4$ are all similar, we only give the proof when $\g=C_n$.
The case of $i, j \in I \setminus \{ n-2, n-1 \}$ with $\sfc_{j,i}\sfc_{i,j}=1$ follows from Proposition <ref>, so
it is enough to prove that

We see that

\begin{equation} \label{eq: stepp0}
\begin{aligned}
& \TT_{n-2}\TT_{n-1}\TT_{n-2}(\sfy_{n,m}) =  \TT_{n-2}\TT_{n-1}(\sfy_{n,m}) \\
& =
\dfrac{ q\sfy_{n,m}  \TT_{n-2}(\sfy_{n-1,m})^2 - (q+q^{-1})\TT_{n-2}(\sfy_{n-1,m})\sfy_{n,m} \TT_{n-2}(\sfy_{n-1,m}) + q^{-1} \TT_{n-2}(\sfy_{n-1,m})^2\sfy_{n,m}  }
\end{aligned}
\end{equation}

\begin{equation} \label{eq: stepp1}
\begin{aligned}
&\TT_{n-1}\TT_{n-2}\TT_{n-1}(\sfy_{n,m}) = \TT_{n-1}\TT_{n-2} \left( \dfrac{q\sfy_{n,m}\sfy_{n-1,m}^{(2)} - \sfy_{n-1,m}\sfy_{n,m}\sfy_{n-1,m} + q^{-1}\sfy_{n-1,m}^{(2)}\sfy_{n,m}}{(q-q^{-1})^2}\right)\\
& \overset{\dagger}{=}
\dfrac{q\TT_{n-1}(\sfy_{n,m})\sfy_{n-2,m}^{ 2} -(q+q^{-1}) \sfy_{n-2,m}\TT_{n-1}(\sfy_{n,m})\sfy_{n-2,m} + q^{-1}\sfy_{n-2,m}^{2}\TT_{n-1}(\sfy_{n,m})}
{(q-q^{-1})(q^2-q^{-2}) },
\end{aligned}
\end{equation}

where $\overset{\dagger}{=}$ follows from (<ref>).

To see the equality between  (<ref>) and (<ref>), it suffices to assume that $n=3$ and $m=0$. Set $y_i\seteq \sfy_{i,0}$ for $i=1,2,3$ and recall that  $q_1=q_2=q$, and  $q_3=q^2$.
By multiplying the numerators of (<ref>) and (<ref>) by $(q-q^{-1})(q^2-q^{-2})$, the first one becomes

\begin{align} \label{eq: stepp00}
\begin{split}
&  \big(qy_{3}  \TT_{1}(y_{2})^2 - (q+q^{-1})\TT_{1}(y_{2})y_{3} \TT_{1}(y_{2}) + q^{-1} \TT_{1}(y_{2})^2y_{3}  \big) \times (q-q^{-1})(q^2-q^{-2})  \\
& = \left\{   q^2y_{3}y_{2}y_{1}y_{2}y_{1}- qy_{3}y_{1}y_{2}^2y_{1} - qy_{3}y_{2}y_{1}^2y_{2}  +y_{3}y_{1}y_{2}y_{1}y_{2}       \right.   \allowdisplaybreaks \\
&\hspace{5ex}   -(q+q^{-1})         (q  y_{2}y_{3}y_{1}y_{2}y_{1} -  y_{1}y_{2}y_3y_{2}y_{1}    -  y_{2}y_{1}^2y_3y_{2}+q^{-1}y_{1}y_{2}y_{1}y_{3}y_{2})    \allowdisplaybreaks \\
&\hspace{15ex} \left. +   y_{2}y_{1}y_{2}y_{1}y_{3} -q^{-1}  y_{1}y_{2}^2y_{1}y_{3} - q^{-1}y_{2}y_{1}^2y_{2}y_{3} + q^{-2}y_{1}y_{2}y_{1}y_{2}y_{3}  \right\}  \times   (q+q^{-1})
\allowdisplaybreaks \\
\end{split}
\end{align}

and the second one becomes

\begin{align} \label{eq: stepp11}
\begin{split}
& \big( q\TT_{2}(y_{3})y_{1}^{2} -(q+q^{-1}) y_{1}\TT_{2}(y_{3})y_{1} + q^{-1}y_{1}^{2}\TT_{2}(y_{3}) \big) \times  (q-q^{-1})(q^2-q^{-2})  \\
& = \left\{ q^2y_{3}y_{2}^{2}y_{1}^{2} - (q^2+1)y_{2}y_{3}y_{2}y_{1}^{2} + y_{2}^{2}y_{3}y_{1}^{2} \right.  \allowdisplaybreaks\\
&\hspace{5ex}   - (q^2+1)y_{1}y_{3}y_{2}^{2}y_{1} +(q^2+2+q^{-2})y_{1}y_{2}y_{3}y_{2}y_{1} - (1+q^{-2})y_{1}y_{2}^{2}y_{3}y_{1}    \allowdisplaybreaks\\
&\hspace{15ex} \left.  +y_{1}^2y_{3}y_{2}^{2} - (1+q^{-2})y_{1}^2y_{2}y_{3}y_{2} + q^{-2}y_{1}^2y_{2}^{2}y_{3} \right\}.  \allowdisplaybreaks\\
\end{split}
\end{align}

Now, we claim that  (<ref>) and  (<ref>) coincide with each other.
By the quantum Serre relations (<ref>), we have

\begin{eqnarray} &&
\parbox{90ex}{
\bna
\item  \label{it: C12}  $y_1^2y_2 - (q+q^{-1})y_1y_2y_1 + y_2y_1^2 = 0 \iff (q+q^{-1})y_1y_2y_1 =  y_1^2y_2+ y_2y_1^2 $,
\item \label{it: C21}  $y_2^2y_1 - (q+q^{-1})y_2y_1y_2 + y_1y_2^2 = 0 \iff (q+q^{-1})y_2y_1y_2 = y_2^2y_1+ y_1y_2^2$.
\ee
}\label{eq: C3 q-serre}
\end{eqnarray}

Using (<ref>)  (<ref>),  (<ref>) and the relation $y_1y_2^2y_1 = y_2y_1^2y_2$,
the equality of (<ref>) and (<ref>) follows from the computation as shown below:

\begin{align*}
&\left\{   q^2y_{3}y_{2} \underline{y_{1}y_{2}y_{1}}- qy_{3}y_{1}y_{2}^2y_{1} - qy_{3}y_{2}y_{1}^2y_{2}  +y_{3}y_{1} \underline{y_{2}y_{1}y_{2}}       \right.   \nonumber\allowdisplaybreaks\\
&\hspace{5ex}   -(q+q^{-1})         (q  y_{2}y_{3} \underline{y_{1}y_{2}y_{1}} -  y_{1}y_{2}y_3y_{2}y_{1}    -  y_{2}y_{1}^2y_3y_{2}+q^{-1} \underline{y_{1}y_{2}y_{1}}y_{3}y_{2})    \nonumber\allowdisplaybreaks \\
&\hspace{15ex} \left. +   y_{2} \underline{y_{1}y_{2}y_{1}}y_{3} -q^{-1}  y_{1}y_{2}^2y_{1}y_{3} - q^{-1}y_{2}y_{1}^2y_{2}y_{3} + q^{-2} \underline{y_{1}y_{2}y_{1}}y_{2}y_{3}  \right\}  \times  \underline{(q+q^{-1})}  \\
&=    q^2y_{3}y_{2}^2y_{1}^2 - (q^2+1) y_{2}y_{3}y_{2}y_{1}^2 +     y_{2}^2y_{3}y_{1}^2            \nonumber\allowdisplaybreaks\\
&\hspace{5ex}  -  (q^2 +1)y_{1}y_{3}y_{2}^2y_{1}  + (q^2+2+q^{-2})y_{1}y_{2}y_{3}y_{2}y_{1}  -(1+q^{-2})  y_{1}y_{2}^2y_{3}y_{1}         \nonumber\allowdisplaybreaks \\
& \hspace{15ex}  +y_{1}^2y_{3}y_{2}^2 - (1+q^{-2}) y_{1}^2y_{2}y_{3}y_{2} +  q^{-2}y_{1}^2y_{2}^2y_{3}+y_{3} (y_{1}y_{2}^2y_{1}   - y_{2}y_{1}^2y_{2})     \nonumber\allowdisplaybreaks
\end{align*}

where all underlined monomials in $y_i$ $(1 \le i \le 3)$ with the factor $(q+q^{-1})$ are replaced by (<ref>), and then one can see that the last expression is equal to (<ref>).
This completes the proof.

For $\g$ of type $B_n$ or $C_n$ $($resp. $F_4$$)$, we have

$$\TT_{n}\TT_{n-1}\TT_{n}\TT_{n-1}   = \TT_{n-1}\TT_{n}\TT_{n-1}\TT_{n}  \qquad \text{$($resp. } \TT_2\TT_3\TT_2\TT_3=\TT_3\TT_2\TT_3\TT_2)$$

Since the proofs for $B_n$, $C_n$ and $F_4$ are all similar, we only give the proof when $\g=B_n$.
We claim that

\TT_{n} \TT_{n-1}\TT_{n}\TT_{n-1}(\sfy_{k,m}) =  \TT_{n-1}\TT_{n}\TT_{n-1}\TT_{n}(\sfy_{k,m})

for all $1 \le k \le n$. The case when $k<n-2$ is obvious by definition.
Let us verify three non-trivial cases as follows, which prove our claim.

Case 1.
Let $k=n-1$.  For the time being, we omit $m+1$ for simplicity of notation to write $\sfy_i = \sfy_{i,m+1}$ and $L_q^{\widetilde{Q}}(\be) = L_q^{\widetilde{Q}}(\be,m+1)$ for a Dynkin quiver $\widetilde{Q}$.
By taking a Dynkin quiver $Q$ with source at $n-1$,

\begin{align}
& (q-q^{-1})  \tUptheta_{Q} \circ  \TT_{n-1}\TT_{n}\TT_{n-1}(\sfy_{n-1,m})=   (q-q^{-1}) \tUptheta_{Q} \circ \TT_{n-1}\TT_{n}(\sfy_{n-1})  \nonumber\allowdisplaybreaks \\
& = (q-q^{-1}) \tUptheta_{Q} \circ \TT_{n-1} \left(\dfrac{  q \sfy_{n-1}  \sfy_{n}^{(2)} - \sfy_{n}  \sfy_{n-1} \sfy_{n} +  q^{-1}\sfy_{n}^{(2)} \sfy_{n-1} }{(q-q^{-1})^{2}} \right)  \nonumber \allowdisplaybreaks\\
\left(\dfrac{q  L_q^Q(\al_{n-1},m+2) *L_q^Q( \al_{n-1}+\al_n)  - q^{-1}L_q^Q( \al_{n-1}+\al_n) * L_q^Q(\al_{n-1},m+2)}{q^2-q^{-2}} \right) * L_q^Q( \al_{n-1}+\al_n)
\nonumber \allowdisplaybreaks\\
& \,\, -   L_q^Q( \al_{n-1}+\al_n) *
\left(
\dfrac{  q  L_q^Q(\al_{n-1},m+2) *L_q^Q( \al_{n-1}+\al_n)  - q^{-1}L_q^Q( \al_{n-1}+\al_n) * L_q^Q(\al_{n-1},m+2) }{(q^2-q^{-2})}
\right) \nonumber \allowdisplaybreaks \\
& =
\left(\dfrac{q  L_q^{s_{n-1}Q}(\al_{n-1}) *L_q^{s_{n-1}Q}( \al_n)  - q^{-1}L_q^{s_{n-1}Q}(\al_n) * L_q^{s_{n-1}Q}(\al_{n-1})}{q^2-q^{-2}} \right) * L_q^{s_{n-1}Q}( \al_n)
\nonumber \allowdisplaybreaks\\
& \,\, -   L_q^{s_{n-1}Q}( \al_n)
* \left(
\dfrac{  q  L_q^{s_{n-1}Q}(\al_{n-1}) *L_q^{s_{n-1}Q}(\al_n)  - q^{-1}L_q^{s_{n-1}Q}(\al_n) * L_q^{s_{n-1}Q}(\al_{n-1}) }{(q^2-q^{-2})}
\right)
\nonumber \allowdisplaybreaks \\
&  \overset{\ddagger}{=}
L_q^{s_{n-1}Q}(\al_{n-1}+\al_{n})*L_q^{s_{n-1}Q}(\al_{n}) - L_q^{s_{n-1}Q}(\al_{n})*L_q^{s_{n-1}Q}(\al_{n-1}+\al_{n})
\nonumber \allowdisplaybreaks \\
&  \overset{\star}{=}  (q-q^{-1}) L_q^{s_{n-1}Q}(  \al_{n-1}+2\al_n) = (q-q^{-1}) L_q^{Q}(  \al_{n-1}+2\al_n)
\nonumber \allowdisplaybreaks\\
& \overset{\dagger}{=}
L_q^{Q}(   \al_n)*L_q^{Q}(   \al_{n-1}+ \al_n)  -L_q^{Q}(   \al_{n-1}+ \al_n)* L_q^{Q}(   \al_n),\label{eq: rstep1}
\end{align}

where $ \overset{\ddagger}{=}$, $\overset{\star}{=}$, and
$ \overset{\dagger}{=}$ follow from  (<ref>).
Since (<ref>) tells us that

\begin{equation}\label{eq: rep}
\begin{aligned}
& L_q^{Q}(   \al_{n-1}+ \al_n,m+1) \\
& \hspace{5ex}=  \dfrac{q L_q^{Q}(   \al_{n},m+1)*L_q^{Q}(   \al_{n-1},m+1) - q^{-1}L_q^{Q}(   \al_{n-1},m+1)* L_q^{Q}(   \al_{n},m+1)}{q^2-q^{-2}},
\end{aligned}
\end{equation}

we have

\begin{equation}\label{eq: 4move n-1 0}
\begin{aligned}
\TT_{n-1}\TT_{n}(\sfy_{n-1,m+1})  & =
\dfrac{q \sfy_{n,m+1}^{(2)} \sfy_{n-1,m+1} - \sfy_{n,m+1} \sfy_{n-1,m+1} \sfy_{n,m+1} +q^{-1}  \sfy_{n-1,m+1} \sfy_{n,m+1}^{(2)} }{(q-q^{-1})^2} \\
& = \TT_{n}^{-1}(\sfy_{n-1,m+1}),
\end{aligned}
\end{equation}

by replacing $L_q^{Q}(   \al_{n-1}+ \al_n )$ in (<ref>) with (<ref>).
By (<ref>), we have

\begin{equation}\label{eq: 4move n-1}
\begin{aligned}
& \TT_{n}\TT_{n-1}\TT_{n}\TT_{n-1}(\sfy_{n-1,m})= \TT_{n}\TT_{n-1}\TT_{n}(\sfy_{n-1,m+1}) = \TT_{n}\TT_{n}^{-1}(\sfy_{n-1,m+1}) = \sfy_{n-1,m+1}.
\end{aligned}
\end{equation}

\begin{align*}
\TT_{n-1}\TT_{n}\TT_{n-1}\TT_{n}(\sfy_{n-1,m})
&=   \TT_{n-1}\TT_{n}\TT_{n}^{-1}(\sfy_{n-1,m}) =   \TT_{n-1}(\sfy_{n-1,m}) =\sfy_{n-1,m+1}.
\end{align*}

Case 2.
Let us consider $k=n$. Take a Dynkin quiver $Q$ with source $n$.
By a similar argument as in Case 1 with (<ref>), we have

\begin{align*}
& \tUptheta_{Q} \circ\TT_{n}\TT_{n-1}\TT_{n}(\sfy_{n,m})   = \tUptheta_{Q} \circ\TT_{n}\TT_{n-1}(\sfy_{n,m+1})   \allowdisplaybreaks\\
& =\tUptheta_{Q} \circ\TT_{n} \left( \dfrac{q \sfy_{n,m+1}\sfy_{n-1,m+1} - q^{-1} \sfy_{n-1,m+1} \sfy_{n,m+1} }{q^2-q^{-2}} \right) \allowdisplaybreaks \\
& =\dfrac{ q  L_q^Q(\al_{n},m+2) *L_q^Q(\al_{n-1}+2\al_n,m+1) - q^{-1}  L_q^Q(\al_{n-1}+2\al_n,m+1)* L_q^Q(\al_{n},m+2) }{q^2-q^{-2}}\allowdisplaybreaks \\
& =\dfrac{ q  L_q^{s_nQ}(\al_{n},m+1) * L_q^{s_nQ}(\al_{n-1},m+1) - q^{-1}  L_q^{s_nQ}(\al_{n-1},m+1)* L_q^{s_nQ}(\al_{n},m+1) }{q^2-q^{-2}}\allowdisplaybreaks \\
&  = L_q^{s_nQ}(\al_{n-1}+\al_{n},m+1)   = L_q^{Q}(\al_{n-1}+\al_{n},m+1)  \\
&=   \dfrac{q L_q^Q(\al_{n-1},m+1)*L_q^Q(\al_{n},m+1) - q^{-1}L_q^Q(\al_{n},m+1)* L_q^Q(\al_{n-1},m+1) }{q^2-q^{-2}} \\
&= \tUptheta_{Q} \circ \TT_{n-1}^{-1} (\sfy_{n,m+1}).
\end{align*}

Hence, it yields

\TT_{n-1}\TT_{n}\TT_{n-1}\TT_{n}(\sfy_{n,m}) = \sfy_{n,m+1},

\begin{align*}
\TT_{n}\TT_{n-1}\TT_{n}\TT_{n-1}(\sfy_{n,m}) &=   \TT_{n}\TT_{n-1}\TT_{n}\TT_{n-1}\TT_{n}(\sfy_{n,m-1}) = \TT_{n}(\sfy_{n,m}) = \sfy_{n,m+1}.
\end{align*}

Case 3.
Let us consider $k=n-2$.

We have

\begin{align}
& \TT_{n}\TT_{n-1}  \TT_{n} \TT_{n-1}(\sfy_{n-2,m})
=  \TT_{n} \TT_{n-1}  \TT_{n} \left( \dfrac{q\sfy_{n-2,m}\sfy_{n-1,m} - q^{-1}\sfy_{n-1,m}\sfy_{n-2,m} }{q^2-q^{-2}} \right) \allowdisplaybreaks \nonumber \\
&\overset{\star}{=}    \left( \dfrac{q\TT_{n} \TT_{n-1}  (\sfy_{n-2,m}) \sfy_{n-1,m} - q^{-1}\sfy_{n-1,m}\TT_{n} \TT_{n-1}  (\sfy_{n-2,m}) }{q^2-q^{-2}} \right). \label{eq: goal LHS}
\end{align}

Here $ \overset{\star}{=}$ follows from  (<ref>).

On the other hand,

\begin{align}
&\TT_{n-1} \TT_{n}\TT_{n-1}  \TT_{n} (\sfy_{n-2,m})    =   \TT_{n-1}  \TT_{n} \TT_{n-1} (\sfy_{n-2,m})  \allowdisplaybreaks \nonumber \\
&    =   \TT_{n-1}  \TT_{n} \left( \dfrac{q\sfy_{n-2,m}\sfy_{n-1,m} - q^{-1}\sfy_{n-1,m}\sfy_{n-2,m} }{q^2-q^{-2}} \right) \allowdisplaybreaks \nonumber \\
&\overset{\dagger}{=}  \left( \dfrac{q \TT_{n-1} (\sfy_{n-2,m}) \TT_{n}^{-1} (\sfy_{n-1,m}) - q^{-1}\TT_{n}^{-1}(\sfy_{n-1,m} )\TT_{n-1} (\sfy_{n-2,m}) }{q^2-q^{-2}} \right).  \label{eq: goal RHS}
\end{align}

Here $ \overset{\dagger}{=}$ follows from  (<ref>).

To see the equality between  (<ref>) and (<ref>), it suffices to assume that $n=3$ and $m=0$. Set $y_i\seteq \sfy_{i,0}$ for $i=1,2,3$ and recall that  $q_1=q_2=q^2$, and $q_3=q$.
By direct computation, we have

\begin{equation} \label{eq: should be zero}
\begin{aligned}
& \dfrac{
(q\TT_{3} \TT_{2}  (y_{1})y_2  - q^{-1}y_2\TT_{3} \TT_{2}  (y_{1})) - (q\TT_{2} (y_{1}) \TT_{3}^{-1} (y_{2})-q^{-1}\TT_{3}^{-1} (y_{2})\TT_{2} (y_{1}))}{(q^2-q^{-2})} \\
& =   \left\{ (q^3+q)(y_{1}y_{2}y_{3}y_{2}y_{3} -  y_{1}y_{3}y_{2}y_{3}y_{2} )   + (q- q^{-1})( y_{3}^2y_{2}y_{1}y_{2}   - y_{2}y_{1}y_{2}y_{3}^2 )  \right.\\
&  \left. \hspace{4ex}  +(q^{-1}+q^{-3})(y_{3}y_{2}y_{3}y_{2}y_{1}   - y_{2}y_{3}y_{2}y_{3}y_{1})  +  qy_{3}^2y_{1}y_{2}^2 +  q^{-1}   y_{2}^2y_{1}y_{3}^2     -qy_{1}y_{2}^2y_{3}^2      -q^{-1}y_{3}^2y_{2}^2y_{1} \right\} \\
&   \hspace{12ex} \times (q-q^{-1})^{-1}(q^2-q^{-2})^{-2}.
\end{aligned}
\end{equation}

Now we claim that (<ref>) vanishes.
Note that

\begin{eqnarray} &&
\parbox{90ex}{
\bna
\item  \label{it: 21}  $y_2^2y_1 - (q^2+q^{-2})y_2y_1y_2 + y_1y_2^2 = 0 \iff (q^2+q^{-2})y_2y_1y_2 =  y_2^2y_1+ y_1y_2^2 $,
\item \label{it: 23}  $y_2^2y_3 - (q^2+q^{-2})y_2y_3y_2 + y_3y_2^2 = 0 \iff (q^2+q^{-2})y_2y_3y_2 = y_2^2y_3+ y_3y_2^2$.
\ee
}\label{eq: B3 q-serre}
\end{eqnarray}

Then we have

\begin{align*}
& \underline{(q^2+q^{-2})}  \times \left\{ (q^3+q)(y_{1}\underline{y_{2}y_{3}y_{2}}y_{3} -  y_{1}y_{3}\underline{y_{2}y_{3}y_{2}} )
+ (q- q^{-1})( y_{3}^2\underline{y_{2}y_{1}y_{2}}   - \underline{y_{2}y_{1}y_{2}}y_{3}^2 ) \right.\\
& \hspace{4ex} \left. +(q^{-1}+q^{-3})(y_{3}\underline{y_{2}y_{3}y_{2}}y_{1}   - \underline{y_{2}y_{3}y_{2}}y_{3}y_{1})
+  qy_{3}^2y_{1}y_{2}^2 +  q^{-1}   y_{2}^2y_{1}y_{3}^2     -qy_{1}y_{2}^2y_{3}^2      -q^{-1}y_{3}^2y_{2}^2y_{1} \right\} \\
& =
(q^3+q) \left( y_{1}y_2^2y_3^2 - y_{1}y_{3}^2y_2^2   \right)  \quad (\because ~\eqref{eq: B3 q-serre}~\eqref{it: 23})\\
& \hspace{3ex}+ (q-q^{-1})  \left(  y_{3}^2y_2^2y_1+ y_{3}^2y_1y_2^2    -y_2^2y_1y_{3}^2- y_1y_2^2y_{3}^2  \right)
\quad (\because ~\eqref{eq: B3 q-serre}~\eqref{it: 21})\\
& \hspace{6ex}+ (q^{-1}+q^{-3})  \left(  y_3^2y_2^2y_{1}    -   y_2^2y_3^2y_{1}   \right) \quad (\because ~\eqref{eq: B3 q-serre}~\eqref{it: 23}) \\
& \hspace{9ex} +  (q^3+q^{-1})y_{3}^2y_{1}y_{2}^2 +  (q+q^{-3})   y_{2}^2y_{1}y_{3}^2     - (q^3+q^{-1})y_{1}y_{2}^2y_{3}^2      - (q+q^{-3})y_{3}^2y_{2}^2y_{1} \\
& =0,
\end{align*}

as desired,
where all underlined monomials in $y_i$ $(1 \le i \le 3)$ with the factor $(q^2+q^{-2})$ are replaced by (<ref>).

For $\g=G_2$, we have

$$\TT_1\TT_2\TT_1\TT_2\TT_1\TT_2  = \TT_2\TT_1\TT_2\TT_1\TT_2\TT_1.$$

It suffices to show that

$$\TT_1\TT_2\TT_1\TT_2\TT_1\TT_2(\sfy_{i,m}) = \TT_2\TT_1\TT_2\TT_1\TT_2\TT_1 (\sfy_{i,m}) \qquad (i=1,2).

In this proof, we omit $*$-operation and $m+1$ frequently for simplicity of notation. Note that $q_1=q$ and $q_2=q^3$.
Let us take a Dynkin quiver $Q'$ with source $2$ and $\xi_2=5$ whose $\Gamma^{Q'}$ is described as

\Gamma^{Q'} = \raisebox{3em}{ \scalebox{0.9}{\xymatrix@!C=4ex@R=2ex{
(i\setminus p)  & 0 & 1 & 2&3&4&5\\
1&   \al_1  \ar@{=>}[dr]  && 2\al_1+\al_2  \ar@{=>}[dr]     &&\al_1+\al_2 \ar@{=>}[dr] \\
2&&  3\al_1+\al_2  \ar@{-}[ul]\ar@{->}[ur]  && 3\al_1+2\al_2 \ar@{-}[ul]\ar@{->}[ur]  &&\al_2 \ar@{-}[ul]

and set $Q\seteq s_2Q'$.
We remark that the height function $\xi$ of $Q$ is given by $\xi_1 = 4$ and $\xi_2 = 3$
(cf. (<ref>) below for $\Gamma^Q$).
Then we have

\begin{align} \label{eq:Q' T2T1T2y2 in G2}
\begin{split}
&\tUptheta_{Q'} \circ\TT_2\TT_1\TT_2(\sfy_{2,m}) =\tUptheta_{Q'} \circ\TT_2\TT_1 (\sfy_{2})  \allowdisplaybreaks\\
&=\tUptheta_{Q'} \circ \TT_2 \left(  \dfrac{   q^{3/2} \sfy_{2}\sfy_{1}^3 -q^{1/2}[3]_1\sfy_{1}\sfy_{2}\sfy_{1}^2 + q^{-1/2}[3]_1\sfy_{1}^2\sfy_{2}\sfy_{1} - q^{-3/2} \sfy_{1}^3\sfy_{2} }{(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})}  \right) \allowdisplaybreaks\\
& = \left\{ q^{3/2}L^{Q'}_q(\al_2,m+2)L^{Q'}_q( \al_1+\al_2)^3 - q^{1/2}[3]_1L^{Q'}_q( \al_1+\al_2)L^{Q'}_q(\al_2,m+2)L^{Q'}_q( \al_1+\al_2)^2 \right. \allowdisplaybreaks\\
& \hspace{6ex} \left. - q^{-1/2}[3]_1L^{Q'}_q(\al_1+\al_2)^2L^{Q'}_q(\al_2,m+2)L^{Q'}_q( \al_1+\al_2) +q^{-3/2} L^{Q'}_q( \al_1+\al_2)^3L^{Q'}_q(\al_2,m+2) \right\} \allowdisplaybreaks\\
& \hspace{54ex}  \times (q-q^{-1})^{-1}(q^2-q^{-2})^{-1}(q^3-q^{-3})^{-1} \allowdisplaybreaks\\
& = \left\{ q^{3/2}L^{Q}_q(\al_2)L^{Q}_q(  \al_1)^3 - q^{1/2}[3]_1L^{Q}_q( \al_1)L^{Q}_q(\al_2)L^{Q}_q( \al_1)^2 \right. \allowdisplaybreaks\\
& \hspace{32ex} \left. - q^{-1/2}[3]_1L^{Q}_q(\al_1)^2L^{Q}_q(\al_2)L^{Q}_q( \al_1) +q^{-3/2} L^{Q}_q( \al_1)^3L^{Q}_q(\al_2) \right\} \allowdisplaybreaks\\
& \hspace{54ex}  \times (q-q^{-1})^{-1}(q^2-q^{-2})^{-1}(q^3-q^{-3})^{-1} \allowdisplaybreaks\\
& = L^{Q}_q(3\al_1+\al_2) = L^{Q'}_q(3\al_1+2\al_2)  \allowdisplaybreaks\\
& =  \dfrac{ q^{-1/2}  L^{Q'}_q(2\al_1+\al_2)  L^{Q'}_q( \al_1+\al_2) - q^{1/2} L^{Q'}_q( \al_1+\al_2) L^{Q'}_q(2\al_1+\al_2)   }{q-q^{-1}},
\end{split}
\end{align}

by (<ref>).

\begin{align*}
&  L^{Q'}_q( \al_1+\al_2) =  \dfrac{q^{3/2} L^{Q'}_q( \al_1) L^{Q'}_q( \al_2) - q^{-3/2}  L^{Q'}_q( \al_2) L^{Q'}_q( \al_1)}{q^3-q^{-3}} \allowdisplaybreaks\\
& \hspace{18ex}  \iff    \tUptheta_{Q'}^{-1}(L^{Q'}_q( \al_1+\al_2)) = \dfrac{q^{3/2} \sfy_{1}\sfy_{2}- q^{-3/2}  \sfy_{2}\sfy_{1}}{q^3-q^{-3}}, \allowdisplaybreaks\\
& L^{Q'}_q( 2\al_1+\al_2)  =\dfrac{q^{1/2} L_q^{Q'}( \al_1) L_q^{Q'}(\al_1+\al_2)    -  q^{-1/2}L_q^{Q'}(\al_1+\al_2)  L_q^{Q'}( \al_1) }{q^2-q^{-2}}\allowdisplaybreaks \\
& \hspace{18ex}  \iff   \tUptheta_{Q'}^{-1}(L^{Q'}_q( 2\al_1+\al_2)) = \left( \dfrac{q^{2} \sfy_{1}^2\sfy_{2}- (q+q^{-1}) \sfy_{1}  \sfy_{2}\sfy_{1} + q^{-2}  \sfy_{2}\sfy_{1}^2}{(q^3-q^{-3})(q^2-q^{-2})} \right),
\end{align*}

the equality (<ref>) implies that

\begin{align} \label{eq:T2T1y2}
\begin{split}
&\TT_2\TT_1\TT_2(\sfy_{2,m}) =\TT_2\TT_1 (\sfy_{2}) \allowdisplaybreaks\\
& =\left( q^{-1/2} (q^2\sfy_{1}^2\sfy_{2}-(q+q^{-1})\sfy_{1}\sfy_{2}\sfy_{1} +q^{-2}\sfy_{2}\sfy_{1}^2  ) (q^{3/2}\sfy_{1}\sfy_{2}-q^{-3/2}\sfy_{2}\sfy_{1})  \right. \allowdisplaybreaks\\
& \hspace{5ex} \left.  - q^{1/2} (q^{3/2}\sfy_{1}\sfy_{2}-q^{-3/2}\sfy_{2}\sfy_{1}) (q^2\sfy_{1}^2\sfy_{2}-(q+q^{-1})\sfy_{1}\sfy_{2}\sfy_{1} +q^{-2}\sfy_{2}\sfy_{1}^2  )  \right) \allowdisplaybreaks\\
& \hspace{45ex} \times (q-q^{-1})^{-1} (q^2-q^{-2})^{-1}(q^3-q^{-3})^{-2} \allowdisplaybreaks\\
& =\left\{   q^3\sfy_{1}^2\sfy_{2}\sfy_{1}\sfy_{2}-q^2(q^2+1+q^{-2})\sfy_{1}\sfy_{2}\sfy_{1}^2\sfy_{2} +(q+q^{-1})\sfy_{2}\sfy_{1}^3\sfy_{2}     \right. \allowdisplaybreaks\\
& \hspace{5ex}
+ q^2(q+q^{-1})\sfy_{1}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}  -(1+q^{-2}+q^{-4})\sfy_{2}\sfy_{1}^2\sfy_{2}\sfy_{1} +q^{-3}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}^2  \allowdisplaybreaks\\
& \hspace{10ex} \left. -  \sfy_{1}( \sfy_{2}^2\sfy_{1}- (q^3+q^{-3})\sfy_{2}\sfy_{1}\sfy_{2} +\sfy_{1}\sfy_{2}^2 ) \sfy_{1}
\right\} \times (q-q^{-1})^{-1} (q^2-q^{-2})^{-1}(q^3-q^{-3})^{-2}  \allowdisplaybreaks\\
& \overset{\dagger}{=}\left\{    q^3\sfy_{1}^2\sfy_{2}\sfy_{1}\sfy_{2}-q^2(q^2+1+q^{-2})\sfy_{1}\sfy_{2}\sfy_{1}^2\sfy_{2} +(q+q^{-1})\sfy_{2}\sfy_{1}^3\sfy_{2}     \right. \allowdisplaybreaks\\
& \hspace{5ex} \left.
+ q^2(q+q^{-1})\sfy_{1}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}  -(1+q^{-2}+q^{-4})\sfy_{2}\sfy_{1}^2\sfy_{2}\sfy_{1} +q^{-3}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}^2 \right\} \allowdisplaybreaks\\
& \hspace{45ex}  \times  (q-q^{-1})^{-1} (q^2-q^{-2})^{-1}(q^3-q^{-3})^{-2}  \allowdisplaybreaks \\
& = \dfrac{
q^3\sfy_{1}^2\sfy_{2}\sfy_{1}\sfy_{2}-q^2 [3]_1\sfy_{1}\sfy_{2}\sfy_{1}^2\sfy_{2} +[2]_1\sfy_{2}\sfy_{1}^3\sfy_{2}   -q^{-2}[3]_1 \sfy_{2}\sfy_{1}^2\sfy_{2}\sfy_{1}  + q^2[2]_1\sfy_{1}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}   +q^{-3}\sfy_{2}\sfy_{1}\sfy_{2}\sfy_{1}^2
}{ (q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})^{2}},
\end{split}
\end{align}

where $\overset{\dagger}{=}$ follows from (<ref>).
Therefore we obtain

\begin{align*}
&\tUptheta_{Q'} \circ \TT_2\TT_1 (\sfy_{2}) = L^{Q'}_q(3\al_1+2\al_2)\\
\dfrac{q^3 \calK^2\calR\calK\calR-q^2[3]_1\calK\calR\calK^2\calR +[2]_1\calR\calK^3\calR   -q^{-2}[3]_1 \calR\calK^2\calR\calK  + q^2[2]_1\calK\calR\calK\calR\calK   +q^{-3}\calR\calK\calR\calK^2
}{ (q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})^{2}}
\end{align*}

$\calK \seteq  L_q^{Q'}(\al_1)$ and $\calR \seteq  L_q^{Q'}(\al_2)$.
Here one can observe that

\begin{equation} \label{eq:shift by -2 in G2}
\begin{split}
L_q^Q(\al_1,m+2) &= L_q^{s_2Q'}(\al_1,m+2) = L_q^{Q'}(\al_1+\al_2,m+2) = \scrS_{-2}(L_q^{Q'}(\al_1)), \\
L_q^Q(3\al_1+\al_2) &= L_q^{s_2Q'}(3\al_1+\al_2) = L_q^{Q'}(3\al_1+2\al_2) = \scrS_{-2}(L_q^{Q'}(\al_2)).
\end{split}
\end{equation}

where $\scrS_{-2}$ is defined in (<ref>).
By (<ref>) and (<ref>), we have

\begin{align*}
&  \tUptheta_{Q} \circ \TT_1\TT_2\TT_1\TT_2(\sfy_{2,m}) =\tUptheta_{Q} \circ \TT_1(\TT_2\TT_1 (\sfy_{2}) )  \allowdisplaybreaks\\
& =
\dfrac{
q^3 K^2RKR-q^2[3]_1KRK^2R +[2]_1RK^3R   -q^{-2}[3]_1 RK^2RK  + q^2[2]_1KRKRK   +q^{-3}RKRK^2
}{ (q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})^{2}},
\end{align*}

$K \seteq L_q^{Q}(\al_1,m+2)=\scrS_{-2}(L_q^{Q'}(\al_1))$ and $R \seteq L_q^{Q}(\al_2+3\al_1)=\scrS_{-2}(L_q^{Q'}(\al_2))$.
Thus we have

\begin{equation} \label{eq:Q T1T2T1y2 in G2}
\tUptheta_{Q} \circ \TT_1\TT_2\TT_1 (\sfy_{2})=\scrS_{-2}( L_q^{Q'}(3\al_1+2\al_2)) = L_q^{Q}(3\al_1+2\al_2).
\end{equation}

On the other hand, we have

\begin{align} \label{eq:Q T2inv T1inv y2 in G2}
\begin{split}
&\tUptheta_{Q}  \circ\TT_2^{-1}\TT_1^{-1}(\sfy_{2}) \allowdisplaybreaks \\
& = \tUptheta_{Q}  \circ \TT^{-1}_2 \left(  \dfrac{   q^{3/2} \sfy_{1}^3\sfy_{2} -q^{1/2}[3]_1\sfy_{1}^2\sfy_{2}\sfy_{1} + q^{-1/2}[3]_1\sfy_{1}\sfy_{2,m}\sfy_{1}^2 - q^{-3/2} \sfy_{2}\sfy_{1}^3 }{(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})}  \right) \allowdisplaybreaks \\
& = \left\{ q^{3/2}  L_q^{Q}(\al_1+\al_2)^3 L_q^{Q}(\al_2,m) - q^{1/2}[3]_1L_q^{Q}(\al_1+\al_2)^2 L_q^{Q}(\al_2,m) L_q^{Q}(\al_1+\al_2) \right.\allowdisplaybreaks\\
&   \hspace{6ex}\left. + q^{-1/2}[3]_1 L_q^{Q}(\al_1+\al_2) L_q^{Q}(\al_2,m) L_q^{Q}(\al_1+\al_2)^2 - q^{-3/2}   L_q^{Q}(\al_2,m) L_q^{Q}(\al_1+\al_2)^3 \right\} \allowdisplaybreaks \\
& \hspace{52ex}  \times (q-q^{-1})^{-1}(q^2-q^{-2})^{-1}(q^3-q^{-3})^{-1}  \allowdisplaybreaks\\
& = \left\{ q^{3/2}  L_q^{Q'}(\al_1)^3 L_q^{Q'}(\al_2) - q^{1/2}[3]_1L_q^{Q'}(\al_1)^2 L_q^{Q'}(\al_2) L_q^{Q'}(\al_1) \right. \allowdisplaybreaks \\
&  \hspace{16ex} \left. + q^{-1/2}[3]_1 L_q^{Q'}(\al_1) L_q^{Q'}(\al_2) L_q^{Q'}(\al_1)^2 - q^{-3/2}   L_q^{Q'}(\al_2) L_q^{Q'}(\al_1)^3 \right\} \allowdisplaybreaks \\
& \hspace{52ex}  \times (q-q^{-1})^{-1}(q^2-q^{-2})^{-1}(q^3-q^{-3})^{-1}  \allowdisplaybreaks\\
&  = L_q^{Q'}(3\al_1+\al_2) = L_q^{Q}(3\al_1+2\al_2).
\end{split}
\end{align}

By combining (<ref>) with (<ref>), we obtain

\begin{align*}
\TT_1\TT_2\TT_1\TT_2\TT_1\TT_2(y_{2,m}) = y_{2,m+1}.
\end{align*}

This implies that

\begin{align*}
\TT_2\TT_1\TT_2\TT_1\TT_2\TT_1(y_{2,m}) =\TT_2\TT_1\TT_2\TT_1\TT_2\TT_1\TT_2(y_{2,m-1})=\TT_2(y_{2,m})=y_{2,m+1}.
\end{align*}

Similarly, one can see that

\begin{align*}
\TT_1\TT_2\TT_1\TT_2\TT_1\TT_2(y_{1,m}) = y_{1,m+1}=\TT_2\TT_1\TT_2\TT_1\TT_2\TT_1(y_{1,m}),
\end{align*}

as desired. We complete the proof.

In Propositions <ref>–<ref>, we have shown that

\begin{align*}
\underbrace{\TT_i\TT_j \cdots}_{\text{ $h_{i,j}$-times}} & =\underbrace{\TT_j\TT_i \cdots}_{\text{ $h_{i,j}$-times}},
\end{align*}

which amounts to a proof of the theorem.

§.§ Ending Remark
Note that the braid group action $\TT_i$ on $ \hA_{q}(\n)$ is local in the sense that $\TT_i$ affects only the variables $\sfy_{j,m}$ with $d(i,j)=1$.  Thus we can extend the action $\TT_i$ to boson-extended quantum group$\hA_{q}(\n)$ defined as follows:  Let $I$ be an index set and $\sfA=(a_{i,j})_{i,j \in I}$ a generalized Cartan matrix which is symmetrizable with $\sfD={\rm diag}(d_{i} \in \Z_{\ge1} \ | \ i\in I)$; i.e., $\sfD\sfA$ is a symmetric matrix. We write $\g$ the Kac-Moody algebra
associated with $\sfA$.

The boson-extended quantum group $\hA_{q}(\n)$ is  the associative $\Q(q^{1/2})$-algebra  generated by $\sfy_{i,k}$ $((i,k) \in I \times \Z)$ satisfying the following relations:

\begin{equation} \label{eq: bex quantum relations}
\begin{aligned}
&\sum_{s=0}^{1-a_{i,j}} (-1)^s \left[ \begin{matrix} 1-a_{i,j} \\ s  \end{matrix}\right]_{q_i} \sfy_{i,k}^{1-a_{i,j}-s}  \sfy_{j,k} \sfy_{i,k}^s =0,  \\
&\sfy_{i,k}\sfy_{j,k+1}  = q_i^{-a_{i,j}} \sfy_{j,k+1}\sfy_{i,k} + (1-q_i^{-2})\delta_{i,j}, \\
&\sfy_{i,k}\sfy_{j,l} = q_i^{(-1)^{k+l}a_{i,j}} \sfy_{j,l}\sfy_{i,k},
\end{aligned}
\end{equation}

for $i,j \in I$ and $k,l \in \Z$ with $l>k+1$.  Here $q_i \seteq q^{d_i}$

By defining $\TT_i^{\pm 1}$ action on   $\hA_{q}(\n)$ as (<ref>) and (<ref>), we can see that $\TT_i^{\pm1}$ are automorphisms of $\hA_{q}(\n)$  and  $\{ \TT^{\pm 1}_i \ | \ i\in I \}$ satisfies the braid relations also.

I realized that I did not prove that $\TT_i$ is an automorphism when $\g$ is not of finite type. Thus it would be better to remove it. But I think it must be an automorphism.

§ INTERPRETATION OF $\TT_1(\SFY_{2,M})$ WHEN $\G=G_2$

Recall that $q_1=q$, $q_2=q^3$ and

\TT_1(\sfy_{2,m}) = \dfrac{   q^{3/2} \sfy_{2,m}\sfy_{1,m}^3 -q^{1/2}[3]_1\sfy_{1,m}\sfy_{2,m}\sfy_{1,m}^2 + q^{-1/2}[3]_1\sfy_{1,m}^2\sfy_{2,m}\sfy_{1,m} - q^{-3/2} \sfy_{1,m}^3\sfy_{2,m} }{(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})}.

Let us take the Dynkin quiver $Q$ with source $1$ and $\xi_1=3$, whose $\Gamma^Q$ is described as follows:

\begin{align}\label{eq: G2 Q}
\Gamma^Q= \raisebox{2.3em}{ \scalebox{0.8}{\xymatrix@!C=4ex@R=2ex{
(i\setminus p)  & -2 & -1 & 0 & 1 & 2&3  \\
1&&   \al_1+\al_2  \ar@{=>}[dr]  && 2\al_1+\al_2  \ar@{=>}[dr]     &&\al_1  \\
2&\al_2 \ar@{->}[ur] &&  3\al_1+2\al_2  \ar@{-}[ul]\ar@{->}[ur]  && 3\al_1+\al_2 \ar@{-}[ul]\ar@{->}[ur]
\end{align}

From the viewpoint of Corollary <ref>,
the relation (<ref>) tells us that

*  $\dfrac{q^{-1/2}L_q^Q(2\al_1+\al_2,m)  * L_q^Q( \al_1,m)   -  q^{1/2} L_q^Q( \al_1,m)*L_q^Q(2\al_1+\al_2,m)}{q-q^{-1}} =L_q^Q(3\al_1+\al_2,m)$,

*  $\dfrac{q^{1/2}L_q^Q(\al_1+\al_2,m)  * L_q^Q( \al_1,m)   -  q^{-1/2} L_q^Q( \al_1,m)*L_q^Q(\al_1+\al_2,m)}{q^2-q^{-2}} =L_q^Q(2\al_1+\al_2,m)$,

*  $\dfrac{q^{3/2}L_q^Q(\al_2,m)  * L_q^Q( \al_1,m)   -  q^{-3/2} L_q^Q( \al_1,m)*L_q^Q(\al_2,m)}{q^3-q^{-3}} =L_q^Q(\al_1+\al_2,m)$,

* $p_{2\al_1+\al_2,\al_1}=2$ and $(2\al_1+\al_2,\al_1)=1$,

* $p_{\al_1+\al_2,\al_1}=1$ and $(\al_1+\al_2,\al_1)=-1$,

* $p_{\al_2,\al_1}=0$ and $(\al_2,\al_1)=-3$,
respectively. Replacing $L_q^Q(\al_1+\al_2,m)$ in (<ref>) with the LHS of (<ref>) and then replacing $L_q^Q(2\al_1+\al_2,m)$ in (<ref>) with the already-replaced one, we have

\begin{align*}
&L_q^Q(3\al_1+\al_2,m) \\
&= \dfrac{   q^{3/2} \sfx^Q_{2,m}(\sfx^Q_{1,m})^3 -q^{1/2}[3]_1\sfx^Q_{1,m}\sfx^Q_{2,m}(\sfx^Q_{1,m})^2 + q^{-1/2}[3]_1(\sfx^Q_{1,m})^2\sfx^Q_{2,m}\sfx^Q_{1,m} - q^{-3/2} (\sfx^Q_{1,m})^3\sfx^Q_{2,m} }{(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})}
\end{align*}

and hence

$$  \tUptheta_{Q} \circ \TT_1(\sfy_{2,m})   = L_q^Q(3\al_1+\al_2,m).$$

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