$\alpha,\beta$ are as in §8.1. Again we can assume that
$\mu_{x^{\prime\prime}}+\vartheta_{\gamma}^{\vee}\npreceq\lambda$ for any
$x^{\prime\prime}\in\mathcal{S}_{\lambda,b}^{+}$ and $\gamma\in\mathcal{O}$.
If there do not exist $\gamma\in\mathcal{O}$, $1\leqslant r\leqslant 3d-1$,
and $x^{\prime}\in\mathcal{S}_{\lambda,b}^{+}$ such that
$x\overset{(\gamma,r)}{\rightarrowtail}x^{\prime}$, by [30, Lemma 8.6] the
statement follows from Lemma 8.2 and Corollary 6.8. Assume otherwise. Then
there exists $x_{i}\in\mathcal{S}_{\lambda,b}^{+}$, $\gamma_{i}\in\mathcal{O}$
and $1\leqslant r_{i}\leqslant 3d-1$ for $1\leqslant i\leqslant m$ such that
$\omega_{\mathcal{O}}=\sum_{i=1}^{m}\sum_{j=0}^{r_{i}-1}\sigma^{j}(\gamma_{i})^{\vee}\in\pi_{1}(M_{J})$
and

$x=x_{0}\overset{(\gamma_{1},r_{1})}{\to}x_{1}\overset{(\gamma_{2},r_{1})}{\to}\cdots\overset{(\gamma_{m},r_{m})}{\to}x_{m}=x.$

if $d+1\leqslant r_{i}\leqslant 2d-1$, then either Lemma 8.3 or Lemma 8.6
occurs. If for each $1\leqslant i\leqslant m$ we have either $r_{i}\leqslant
d$ or $2d\leqslant r_{i}\leqslant 3d-1$ or Lemma 8.3 (for
$(x,x^{\prime},\alpha,r)=(x_{i-1},x_{i},\gamma_{i},r_{i})$) occurs, it follows
that $\omega_{\mathcal{O}}\in\mathcal{A}_{\lambda,b}$ by Lemma 8.1, 8.2, Lemma
8.3 and Corollary 6.8. Otherwise, by the proof of [30, Proposition 6.8], there
exists $1\leqslant i\leqslant m$ such that the situation of Lemma 8.6 occurs
(for $(x,x^{\prime},\alpha,r)=(x_{i-1},x_{i},\gamma_{i},r_{i})$).

Let $x,x^{\prime},\alpha,r$ be as in Lemma 8.6. If
$\langle\sigma^{r+d}(\alpha),\mu_{x}\rangle\leqslant 0$, then
$\langle\sigma^{r}(\vartheta_{\alpha}),\mu_{x}\rangle\leqslant-1$, which
contradicts our assumption. So we have
$\langle\sigma^{r+d}(\alpha),\mu_{x}\rangle\geqslant 1$, and hence

$x\overset{(\sigma^{d+r}(\alpha),3d-r)}{\to}y:=x-\sigma^{r+d}(\alpha)^{\vee}+\sigma^{d}(\alpha)^{\vee}\overset{(\sigma^{d}(\alpha),r)}{\to}x.$

Then it suffices to show that

$\displaystyle g_{2}I\sim_{\lambda,b}g_{2}\omega_{2}I$ $\displaystyle\text{
for }g_{2}\in\mathbb{J}_{b,{\tilde{w}}_{y}};$ (a) $\displaystyle
g_{1}I\sim_{\lambda,b}g_{1}\omega_{1}I$ $\displaystyle\text{ for
}g_{1}\in\mathbb{J}_{b,{\tilde{w}}_{x}},$ (b)

where
$\omega_{1}=\sigma^{r+d}(\alpha)^{\vee}+\cdots+\sigma^{4d-1}(\alpha)^{\vee},\omega_{2}=\sigma^{d}(\alpha)^{\vee}+\cdots+\sigma^{r+d-1}(\alpha)^{\vee}\in\pi_{1}(M_{J})\cong\Omega_{J}$.

First we show (a). Note that
$\langle\sigma^{r}(\alpha),\mu_{y}\rangle=\langle\sigma^{r}(\alpha),\mu_{x}\rangle\leqslant-1$.
We have

$y\overset{(\sigma^{d}(\alpha),r-d)}{\to}y-\sigma^{d}(\alpha)^{\vee}+\sigma^{r}(\alpha)^{\vee}\overset{(\sigma^{r}(\alpha),d)}{\to}x,$

and (a) follows from Lemma 8.1 and Corollary 6.8.

Now we show (b). If $\langle\alpha,\mu_{x}\rangle\leqslant-1$, the statement
follows from that

$x\overset{(\sigma^{r+d}(\alpha),2d-r)}{\to}x-\sigma^{r+d}(\alpha)+\alpha^{\vee}\overset{(\alpha,d)}{\to}y.$

So we can assume $\langle\alpha,\mu_{x}\rangle=0$. If
$\langle\sigma^{i}(\alpha),\mu_{x}\rangle=0$ for $r+d+1\leqslant i\leqslant
3d-1$, it follows from Lemma 8.3. Otherwise, let $k=\max\\{r+d+1\leqslant
i\leqslant 3d-1;\langle\sigma^{i}(\alpha),\mu_{x}\rangle\neq 0\\}$. If
$\langle\sigma^{k}(\alpha),\mu_{x}\rangle=-1$, then
$\langle\sigma^{k-d}(\alpha),\mu_{x}\rangle\geqslant 1$ since
$\langle\sigma^{k}(\vartheta_{\alpha}),\mu_{x}\rangle\geqslant 0$, which means
$x\overset{(\sigma^{k-d}(\delta),2d)}{\to}x_{1}:=x+\sigma^{k}(\alpha)^{\vee}-\sigma^{k-d}(\alpha)^{\vee}$
and
$y\overset{(\sigma^{k-d}(\delta),2d)}{\to}y_{1}:=y+\sigma^{k}(\alpha)^{\vee}-\sigma^{k-d}(\alpha)^{\vee}$.
By Lemma 7.2, we have

$g_{1}\sim_{\lambda,b}g_{1}\omega^{\prime}I\text{ for
}g_{1}\in\mathbb{J}_{b,{\tilde{w}}_{x}},\
g_{2}\sim_{\lambda,b}g_{2}\omega^{\prime}I\text{ for
}g_{2}\in\mathbb{J}_{b,{\tilde{w}}_{y}},$

where
$\omega^{\prime}=\sigma^{k-d}(\delta^{\vee})+\cdots+\sigma^{k+d}(\delta^{\vee})\in\pi_{1}(M_{J})\cong\Omega_{J}$.
So we can replace the pair $(x,y)$ with $(x_{1},y_{1})$ so that
$\langle\sigma^{k}(\alpha),\mu_{x}\rangle\geqslant 1$. Then

$x\overset{(\sigma^{k}(\alpha),4d-k)}{\to}x-\sigma^{k}(\alpha)^{\vee}+\sigma^{d}(\alpha)^{\vee}\overset{(\sigma^{r+d}(\alpha),k-r-d)}{\to}y,$

and (b) follows from Lemma 8.3, Lemma 8.1 and Corollary 6.8. ∎

##  Appendix A Distinct elements in ${\rm{Adm}}(\lambda)$

In this Appendix, we study the distinct elements introduced in [4].

### A.1.

First we recall the following lemmas.

######  Lemma A.1.

Let $s,s^{\prime}\in\SS^{a}$ and ${\tilde{w}}\in\tilde{W}$ such that
$\ell(s{\tilde{w}})=\ell({\tilde{w}}s^{\prime})$ and
$\ell(s{\tilde{w}}s^{\prime})=\ell({\tilde{w}})$. Then
${\tilde{w}}=s{\tilde{w}}s^{\prime}$.

######  Lemma A.2 ([4, Lemma 1.8 & 1.9], [7, Lemma 4.5]).

Let $s\in\SS^{a}$ and ${\tilde{w}}\in{\rm{Adm}}(\lambda)$ with $\lambda\in Y$
such that ${\tilde{w}}<s{\tilde{w}}$. Then we have

(1) ${\tilde{w}}s\in{\rm{Adm}}(\lambda)$ if ${\tilde{w}}s<s{\tilde{w}}s$;

(2) ${\tilde{w}}s=s{\tilde{w}}$ if ${\tilde{w}}s\notin{\rm{Adm}}(\lambda)$;

(3) $s{\tilde{w}}s\in{\rm{Adm}}(\lambda)$ if
$\ell(s{\tilde{w}}s)=\ell({\tilde{w}})$.

######  Lemma A.3.

Let ${\tilde{w}}\notin{\rm{Adm}}(\lambda)$ and $s\in\SS^{a}$ such that
${\tilde{w}}s>{\tilde{w}}$. Then $s{\tilde{w}}s\notin{\rm{Adm}}(\lambda)$.

###### Proof.

Assume $s{\tilde{w}}s\in{\rm{Adm}}(\lambda)$, then
$s{\tilde{w}}s<{\tilde{w}}s$ and hence
$\ell(s{\tilde{w}}s)=\ell({\tilde{w}})$. By Lemma A.2 (3), we have
${\tilde{w}}\in{\rm{Adm}}(\lambda)$, contradicting the assumption that
${\tilde{w}}\notin{\rm{Adm}}(\lambda)$. ∎

### A.2.

Fix $\lambda\in Y^{+}$. Let $R\subseteq\SS_{0}$ and
${\tilde{w}}\in{\rm{Adm}}(\lambda)$. We say ${\tilde{w}}$ is left $R$-distinct
(resp. right $R$-distinct) if $s{\tilde{w}}\notin{\rm{Adm}}(\lambda)$ (resp.
${\tilde{w}}s\notin{\rm{Adm}}(\lambda)$) for all $s\in R$. Let $w_{R}$ denote
the longest element of $W_{R}$.

For a reflection $s\in W_{0}$ we denote by $\alpha_{s}\in\Phi^{+}$ the
corresponding simple root.

######  Lemma A.4.

Let $R=\\{s,s^{\prime}\\}\subseteq\SS_{0}$. Let
${\tilde{w}}\in{\rm{Adm}}(\lambda)$ be right $R$-distinct. Let
$u,u^{\prime}\in W_{R}$ with $\ell(u^{\prime})\leqslant\ell(u)$. Then
$u^{\prime}{\tilde{w}}u^{-1}\in{\rm{Adm}}(\lambda)$ if and only if
$u=u^{\prime}$. As a consequence,
$w_{R}{\tilde{w}}w_{R}\in{\rm{Adm}}(\lambda)$ is left $R$-distinct.

###### Proof.

First we notice that ${\tilde{w}}\in\tilde{W}^{R}$, see §1.1. Without loss of
generality, we can assume $s\neq s^{\prime}$ and
$ss^{\prime}s=s^{\prime}ss^{\prime}$.

First we show the “only if” part. By symmetry it suffices to consider the
following cases.

Suppose $s{\tilde{w}}s^{\prime}\in{\rm{Adm}}(\lambda)$. Then
$s{\tilde{w}}s^{\prime}<{\tilde{w}}s^{\prime}$ and
$s{\tilde{w}}s^{\prime}s\notin{\rm{Adm}}(\lambda)$ (see Lemma A.3). By Lemma
A.2 we have $s{\tilde{w}}s^{\prime}(\alpha_{s})=\alpha_{s}$, that is,
${\tilde{w}}(\alpha_{s}+\alpha_{s^{\prime}})=-\alpha_{s}$. This is impossible
since ${\tilde{w}}\in\tilde{W}^{R}$.

Suppose $s{\tilde{w}}ss^{\prime}\in{\rm{Adm}}(\lambda)$. Then
$s{\tilde{w}}ss^{\prime}<s{\tilde{w}}s^{\prime}$ (as
$s{\tilde{w}}s^{\prime}\notin{\rm{Adm}}(\lambda)$), that is,
$s{\tilde{w}}s^{\prime}(s^{\prime}(\alpha_{s}))=s{\tilde{w}}(\alpha_{s})\in\tilde{\Phi}^{+}$.
Since ${\tilde{w}}(\alpha_{s})\in\tilde{\Phi}^{-}$ (as
${\tilde{w}}\in\tilde{W}^{R}$), we have ${\tilde{w}}(\alpha_{s})=\alpha_{s}$.
This means
$s{\tilde{w}}ss^{\prime}={\tilde{w}}s^{\prime}\notin{\rm{Adm}}(\lambda)$, a
contradiction. Notice that $s{\tilde{w}}s^{\prime}s\notin{\rm{Adm}}(\lambda)$
by Lemma A.3.

Suppose $ss^{\prime}{\tilde{w}}ss^{\prime}\in{\rm{Adm}}(\lambda)$. Then
$ss^{\prime}{\tilde{w}}ss^{\prime}<s^{\prime}{\tilde{w}}ss^{\prime}$. If
$s^{\prime}{\tilde{w}}ss^{\prime}<s^{\prime}{\tilde{w}}ss^{\prime}s$, then
$ss^{\prime}{\tilde{w}}ss^{\prime}s\notin{\rm{Adm}}(\lambda)$ by Lemma A.3.
Otherwise, by Lemma A.1 we have
$s^{\prime}{\tilde{w}}ss^{\prime}s={\tilde{w}}ss^{\prime}$ (since
${\tilde{w}}ss^{\prime}<{\tilde{w}}ss^{\prime}s$) and hence
$ss^{\prime}{\tilde{w}}ss^{\prime}s=s{\tilde{w}}ss^{\prime}\notin{\rm{Adm}}(\lambda)$.
So we always have
$ss^{\prime}{\tilde{w}}ss^{\prime}s\notin{\rm{Adm}}(\lambda)$. By Lemma A.2 we
have $ss^{\prime}{\tilde{w}}ss^{\prime}(\alpha_{s})=\alpha_{s}$, that is,
${\tilde{w}}(\alpha_{s^{\prime}})=-(\alpha_{s}+\alpha_{s^{\prime}})$, which is
impossible as ${\tilde{w}}\in\tilde{W}^{R}$.

Suppose $ss^{\prime}{\tilde{w}}ss^{\prime}s\in{\rm{Adm}}(\lambda)$. Then
$ss^{\prime}{\tilde{w}}ss^{\prime}s<s^{\prime}{\tilde{w}}ss^{\prime}s$. Since
$ss^{\prime}{\tilde{w}}ss^{\prime}\notin{\rm{Adm}}(\lambda)$, by Lemma A.2 we
have $ss^{\prime}{\tilde{w}}ss^{\prime}s(\alpha_{s})=\alpha_{s}$, that is,
${\tilde{w}}(\alpha_{s^{\prime}})=\alpha_{s}+\alpha_{s^{\prime}}$. This means
$ss^{\prime}{\tilde{w}}ss^{\prime}s=s^{\prime}{\tilde{w}}ss^{\prime}\in{\rm{Adm}}(\lambda)$,
a contradiction.

Now we show the “if” part, that is, $u{\tilde{w}}u^{-1}\in{\rm{Adm}}(\lambda)$
for $u\in W_{R}$. We argue by induction on the length $u$. If $u=1$, the
statement is true. Let $u=su_{1}>u_{1}$ with $u_{1}\in W_{R}$ and $s\in R$. We
assume $u_{1}{\tilde{w}}u_{1}^{-1}\in{\rm{Adm}}(\lambda)$ by induction
hypothesis. It remains to show that
$u{\tilde{w}}u^{-1}\in{\rm{Adm}}(\lambda)$. Otherwise, we have
$\ell(u{\tilde{w}}u^{-1})=\ell(u_{1}{\tilde{w}}u_{1}^{-1})+2$ and
$u_{1}{\tilde{w}}u^{-1}\in{\rm{Adm}}(\lambda)$ by Lemma A.2 (1) & (3), which
contradicts the “only if” part. ∎

######  Lemma A.5.

Let ${\tilde{w}}\in{\rm{Adm}}(\lambda)$ and $s\in\SS_{0}$ such that
$s{\tilde{w}}s\in{\rm{Adm}}(\lambda)$ and
$s{\tilde{w}}\notin{\rm{Adm}}(\lambda)$. Let
$\alpha\in\Phi^{+}\setminus\\{\alpha_{s}\\}$ such that
${\tilde{w}}s_{\alpha}\in{\rm{Adm}}(\lambda)$. Then
$s{\tilde{w}}s_{\alpha}s\in{\rm{Adm}}(\lambda)$

###### Proof.

Suppose $s{\tilde{w}}s_{\alpha}s\notin{\rm{Adm}}(\lambda)$, then
$s{\tilde{w}}s_{\alpha}\in{\rm{Adm}}(\lambda)$ by Lemma A.2. As
$s{\tilde{w}}\notin{\rm{Adm}}(\lambda)$, we have
$s{\tilde{w}}(\alpha)\in\tilde{\Phi}^{+}$. On the other hand, as
$s(\alpha)\in\Phi^{+}$, $s{\tilde{w}}s_{\alpha}s\notin{\rm{Adm}}(\lambda)$ and
$s{\tilde{w}}s\in{\rm{Adm}}(\lambda)$, we have
$s{\tilde{w}}(\alpha)\in\tilde{\Phi}^{-}$, which is a contradiction. ∎

######  Corollary A.6.

Let $R=\\{s,s^{\prime}\\}\subseteq\SS_{0}$. Let
${\tilde{w}}\in{\rm{Adm}}(\lambda)$ be left $R$-distinct. Let
$\alpha\in\Phi^{+}\setminus\Phi_{R}$ such that
${\tilde{w}}s_{\alpha}\in{\rm{Adm}}(\lambda)$. Then
$u{\tilde{w}}s_{\alpha}u^{-1}\in{\rm{Adm}}(\lambda)$ for $u\in W_{R}$.

###### Proof.

We argue by induction on $\ell(u)$,. If $u=1$, the statement follows by
assumption. Supposing it is true for $u_{1}$, that is,
$u_{1}{\tilde{w}}u_{1}^{-1}s_{u_{1}(\alpha)}=u_{1}{\tilde{w}}s_{\alpha}u_{1}^{-1}\in{\rm{Adm}}(\lambda)$,
we show it is also true for $u=su_{1}>u_{1}$ with $s\in R$. By Lemma A.4 we
have
$u_{1}{\tilde{w}}u_{1}^{-1},su_{1}{\tilde{w}}u_{1}^{-1}s\in{\rm{Adm}}(\lambda)$
and $su_{1}{\tilde{w}}u_{1}^{-1}\notin{\rm{Adm}}(\lambda)$. Moreover, we have
$u_{1}(\alpha)\neq\alpha_{s}$ since $\alpha\in\Phi^{+}\setminus\Phi_{R}$. Thus
$u{\tilde{w}}s_{\alpha}u^{-1}=su_{1}{\tilde{w}}u_{1}^{-1}s_{u_{1}(\alpha)}s\in{\rm{Adm}}(\lambda)$
by Lemma A.5. ∎

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