Proof. If ${\tilde {w}} s \notin {}^K \tilde {W}$, then there exits $s' \in K$ such that $s' {\tilde {w}} s < {\tilde {w}} s$, that is, $s{\tilde {w}}^{-1}({\tilde {\alpha }}') \in \tilde {\Phi }^-$, where ${\tilde {\alpha }}' \in \tilde {\Phi }^+$ is the simple affine root of $s'$. Note that ${\tilde {w}}^{-1}({\tilde {\alpha }}') \in \tilde {\Phi }^+$ (since ${\tilde {w}} \in {}^K \tilde {W}$) and that $s$ is a simple reflection, it follows that ${\tilde {w}}^{-1}({\tilde {\alpha }}')$ is the affine simple root of $s$. Thus $s' {\tilde {w}} = {\tilde {w}} s$ and statement (1) is proved. The statement (2) is proved in [Reference HeHe07, Corollary 2.6].

Proof. As $I(K, x)$ is a finite set, there exists $n \in \mathbb {Z}_{\geqslant 1}$ such that $(x\sigma )^n = t^{n \nu _x}$ fixes each element of $I(K, x)$, that is, $p(s)(\nu _x) = \nu _x$ for $s \in I(K, x)$. Here $p: \tilde {W} \rtimes \langle \sigma \rangle \to W_0 \rtimes \langle \sigma \rangle$ is the natural projection. Thus, $I(K, x) \subseteq W_{\nu _x}$ as desired.

Proof. Note that $\nu _{z {\tilde {w}} \sigma (z)^{-1}} = p(z)(\nu _{\tilde {w}})$ and, hence, $z(\tilde {\Phi }_{\nu _{\tilde {w}}}) = \tilde {\Phi }_{\nu _{z {\tilde {w}} \sigma (z)^{-1}}}$, where $p: \tilde {W} \rtimes \langle \sigma \rangle \to W_0 \rtimes \langle \sigma \rangle$ is the natural projection. Thus, $z (\tilde {\Phi }_{\nu _{{\tilde {w}}}}^\pm ) = \tilde {\Phi }_{\nu _{z {\tilde {w}} \sigma (z)^{-1}}}^\pm$, and part (1) follows by definition.

Let $z' \in W_0^{J_{\bar {\nu }_{{\tilde {w}}}}}$ such that $z'(\bar {\nu }_{{\tilde {w}}}) = \nu _{{\tilde {w}}}$. Let ${\tilde {w}}' = {z'}^{-1} {\tilde {w}} \sigma (z')$. Note that $z'(\tilde {\Phi }_{\bar {\nu }_{{\tilde {w}}}}^+) = \tilde {\Phi }_{\nu _{{\tilde {w}}}}^+$. So ${\tilde {w}}' \in \mathcal {S}^+$ by (1). Assume there is another ${\tilde {w}}'' \in \mathcal {S}^+$ such that ${\tilde {w}} = w {\tilde {w}}'' \sigma (w)^{-1}$ for some $w \in W_0$. Write $w = z'' u$ with $z'' \in W_0^{J_{\bar {\nu }_{{\tilde {w}}}}}$ and $u \in W_{J_{\bar {\nu }_{{\tilde {w}}}}}$. Then $\nu _{{\tilde {w}}'} = \nu _{{\tilde {w}}''} = \bar {\nu }_{{\tilde {w}}}$ and ${z'}^{-1} z'' \in W_{J_{\bar {\nu }_{{\tilde {w}}}}}$. Thus, $z' = z'' \in W_0^{J_{\bar {\nu }_{{\tilde {w}}}}}$, and ${\tilde {w}}'', {\tilde {w}}' \in \Omega _{J_{\bar {\nu }_{\tilde {w}}}}$ are $\sigma$-conjugate by $W_{J_{\bar {\nu }_{\tilde {w}}}}$. Thus, ${\tilde {w}}' = {\tilde {w}}''$ as desired.

To prove part (3) we can assume ${\tilde {w}} \to _s {\tilde {w}}'$ for some $s \in \mathbb {S}^a$ and ${\tilde {w}} \neq {\tilde {w}}'$. Thus, either $s {\tilde {w}} < {\tilde {w}}$ or ${\tilde {w}} \sigma (s) < {\tilde {w}}$. In view of part (1) it suffices to show $s(\tilde {\Phi }_{\nu _{{\tilde {w}}}}^+) \subseteq \tilde {\Phi }^+$. Otherwise, the simple affine root of $s$ lies in $\tilde {\Phi }_{\nu _{{\tilde {w}}}}^+$. Hence, $s {\tilde {w}}, {\tilde {w}} \sigma (s) > {\tilde {w}}$ (since ${\tilde {w}}\sigma (\tilde {\Phi }_{\nu _{{\tilde {w}}}}^+) = \tilde {\Phi }_{\nu _{{\tilde {w}}}}^+$), contradicting our assumption.

Note that $\mathbb {J}_{\tilde {w}} \subseteq M_{\nu _{{\tilde {w}}}}$. Then part (4) follows from that ${}^{{\tilde {w}}\sigma } I_{M_{\nu _{\tilde {w}}}} = I_{M_{\nu _{\tilde {w}}}}$, ${}^{{\tilde {w}}\sigma } \tilde {W}_{M_{\nu _{\tilde {w}}}} = \tilde {W}_{M_{\nu _{\tilde {w}}}}$, and the Bruhat decomposition $M_{\nu _{\tilde {w}}}({\breve F}) = I_{M_{\nu _{\tilde {w}}}} \tilde {W}_{M_{\nu _{\tilde {w}}}} I_{M_{\nu _{\tilde {w}}}}$.

Assume ${\tilde {w}} \in \mathcal {S}^+$, that is, $\nu _{\tilde {w}} = \sigma (\nu _{\tilde {w}})$ is dominant. As $p({\tilde {w}}\sigma )(\nu _{\tilde {w}}) = \nu _{{\tilde {w}}}$, we have $p({\tilde {w}})(\nu _{\tilde {w}}) = \nu _{\tilde {w}}$ and, hence, ${\tilde {w}} \in \tilde {W}_{\nu _{\tilde {w}}} = \tilde {W}_{J_{\nu _{\tilde {w}}}}$. Moreover, ${\tilde {w}}(\tilde {\Phi }_{J_{\nu _{\tilde {w}}}}^+) = {\tilde {w}}\sigma (\tilde {\Phi }_{J_{\nu _{\tilde {w}}}}^+) = \tilde {\Phi }_{J_{\nu _{\tilde {w}}}}^+$ by definition. This means that ${\tilde {w}} \in \Omega _{J_{\nu _{\tilde {w}}}}$ and part (5) is proved.

By part (2) there exits $z \in {}^{J_{\bar {\nu }_{\tilde {w}}}} W_0$ and ${\tilde {w}}' \in \mathcal {S}^+$ such that ${\tilde {w}}' = z^{-1} {\tilde {w}} \sigma (z)$. Let $u \in W_{\nu _{\tilde {w}}}$. Then $z^{-1} u z \in W_{J_{\bar {\nu }_{\tilde {w}}}}^a$ since $z(\nu _{\tilde {w}}) = \nu _{\bar {\nu }_{\tilde {w}}}$. By part (5) we have ${\tilde {w}}'' \in \Omega _{J_{\nu _{\tilde {w}}}}$ and, hence, ${\tilde {w}}'' \leq (z^{-1} u z)$. It follows from [Reference Chen and NieCN19, Lemma 1.3] that

\[ {\tilde{w}} = z {\tilde{w}}' \sigma(z)^{-1} \leq z (z^{-1} u z) {\tilde{w}}' \sigma(z)^{-1} = u {\tilde{w}} \]

and part (b) follows.

Proof. Let $s \in R$. If $s {\tilde {w}} < {\tilde {w}}$, then $s{\tilde {w}} \in {\rm {Adm}}(\lambda )$ since ${\tilde {w}} \in {\rm {Adm}}(\lambda )$, which is a contradiction. Thus, $s {\tilde {w}} > {\tilde {w}}$ and, hence, ${\tilde {w}} \in {}^R \tilde {W}$.

To show part (1) we can assume $d = 1$. Then one checks that $R$ is either commutative or is of type $A_2$. Thus, part (1) follows from Lemma A.4.

Now we show part (2). Let $\alpha \in \mathcal {P}_{\tilde {w}}$ and set $\alpha ^i = ({\tilde {w}}\sigma )^i(\alpha ) \in \tilde {\Phi }$ for $i \in \mathbb {Z}$. Let

\[ n_\alpha = \min \{i \in \mathbb{Z}_{\geqslant 0}; \alpha^{-i} \notin \Phi_R^+\} \leqslant m_{\alpha, {\tilde{w}}}. \]

It suffices to show that $w_R(\alpha ^{-n_\alpha }) \in \mathcal {P}_{w_R {\tilde {w}} w_R}$. First we check that

(a)$$\begin{align*} \alpha^{-n_\alpha} \in \Phi^+ \text{ and } w_R {\tilde{w}} w_R \sigma(s_{w_R(\alpha^{-n_\alpha})}) = w_R {\tilde{w}} \sigma(s_{\alpha^{-n_\alpha}}) w_R \in {\rm{Adm}}(\lambda). \end{align*}$$

If $n_\alpha = 0$, then $\alpha ^{-n_\alpha } = \alpha \in \Phi ^+ \setminus \Phi _R$, and part (a) follows from Corollary A.6. Otherwise, $\alpha ^{-n_\alpha +1} \in \Phi _R^+ \subseteq \tilde {\Phi }^-$. Noting that ${\tilde {w}} \in {}^R \tilde {W}$ (since ${\tilde {w}} \in {\rm {Adm}}(\lambda )$ is left $R$-distinct), by Lemma 1.1 we have $\alpha ^{-n_\alpha } = ({\tilde {w}}\sigma )^{-1} (\alpha ^{-n_\alpha +1}) \in \tilde {\Phi }^-$. Hence, $\alpha ^{-n_\alpha } \in \Phi ^+$ since $n_\alpha \leqslant m_{\alpha, {\tilde {w}}}$ and $\alpha ^{-m_{\alpha, {\tilde {w}}}} \in \tilde {\Phi }^+$. Moreover,

\[ w_R {\tilde{w}} w_R \sigma(w_R(\alpha^{-n_\alpha})) = w_R(\alpha^{-n_\alpha+1}) \in \Phi_R^- \subseteq \tilde{\Phi}^+. \]

By Lemma 1.1, $w_R {\tilde {w}} w_R \sigma (s_{w_R(\alpha ^{-n_\alpha })}) \leq w_R {\tilde {w}} w_R \in {\rm {Adm}}(\lambda )$, and part (a) follows.

Note the following three facts: $(w_R {\tilde {w}} w_R)^i(w_R(\alpha ^{-n_\alpha })) = w_R(\alpha ^{-n_\alpha +i})$ for $i \in \mathbb {Z}$; $w_R \in W_0$ preserves $\tilde {\Phi } \setminus \Phi$; and $n_\alpha < n_{\alpha, {\tilde {w}}}$ (since $\alpha ^{-n_\alpha } \in \Phi$). Then it follows, by definition, that $n_{w_R(\alpha ^{-n_\alpha }), w_R {\tilde {w}} w_R} = n_{\alpha, {\tilde {w}}} - n_\alpha$ and

\[ (w_R {\tilde{w}} w_R)^{-n_{w_R(\alpha^{-n_\alpha}), w_R {\tilde{w}} w_R}}(w_R(\alpha^{-n_\alpha})) = w_R(\alpha^{-n_{\alpha, {\tilde{w}}}}) \in \tilde{\Phi}^+ \setminus \Phi, \]

where the inclusion follows from $\alpha ^{-n_{\alpha, {\tilde {w}}}} \in \tilde {\Phi }^+ \setminus \Phi$. Therefore, we have $w_R(\alpha ^{-n_\alpha }) \in \mathcal {P}_{w_R {\tilde {w}} w_R}$ as desired.

Proof. By assumption, there exist $s' \in R$ and $0 \leqslant i \leqslant |R|-1$ such that $\sigma ^{-i}(s') {\tilde {w}} < {\tilde {w}}$ and ${\tilde {w}} \sigma (s') \in {\rm {Adm}}(\lambda )$. Thus, we can define

\[ k = \min\{0 \leqslant i \leqslant |R|-1; \sigma^{-i}(s') {\tilde{w}} < {\tilde{w}}, {\tilde{w}} \sigma(s') \in {\rm{Adm}}(\lambda) \text{ for some } s' \in R \}. \]

Choose $s \in R$ such that $\sigma ^{-k}(s) {\tilde {w}} < {\tilde {w}}$ and ${\tilde {w}}\sigma (s) \in {\rm {Adm}}(\lambda )$. Let $\alpha \in \Phi ^+$ be the simple root of $s$. Set $\gamma ^i = ({\tilde {w}}\sigma )^i(\gamma ) \in \tilde {\Phi }$ for $\gamma \in \Phi$ and $i \in \mathbb {Z}$. We claim that

(a)$$\begin{align*} \alpha^{-i} = \sigma^{-i}(\alpha) \text{ for } 0 \leqslant i \leqslant k, \text{ and hence } m_{\alpha, {\tilde{w}}} \geqslant k+1. \end{align*}$$

Let $0 \leqslant i \leqslant k-1$. By the choice of $k$ we have ${\tilde {w}} < \sigma ^{-i}(s) {\tilde {w}}$ and ${\tilde {w}} \sigma ^{-i}(s) \notin {\rm {Adm}}(\lambda )$, which means that ${\tilde {w}}\sigma ^{-i}(s) = \sigma ^{-i}(s) {\tilde {w}}$ by Lemma A.2, that is, $\sigma ^{-i}(\alpha ) = {\tilde {w}} \sigma ^{-i}(\alpha )$ (since ${\tilde {w}} < \sigma ^{-i}(s) {\tilde {w}}$). Thus,

\[ \alpha^{-i-1} = (\sigma^{-1} {\tilde{w}}^{-1})^{i+1}(\alpha) = (\sigma^{-1}{\tilde{w}}^{-1})^i \sigma^{-1}(\alpha) = \cdots = (\sigma^{-1}{\tilde{w}}^{-1}) \sigma^{-i}(\alpha) = \sigma^{-i-1}(\alpha). \]

Thus, part (a) is proved.

By part (a) we have $\alpha ^{-k} \in \Phi ^+$. Thus, $\alpha ^{-k-1} = ({\tilde {w}}\sigma )^{-1} (\alpha ^{-k}) \in \tilde {\Phi }^+$ since $\sigma ^{-k}(s) {\tilde {w}} < {\tilde {w}}$. As ${\tilde {w}} \in \mathcal {S}$, it follows that $\alpha ^{-k} \notin \Phi _{\nu _{\tilde {w}}}$ and, hence, $\alpha ^i \notin \Phi _{\nu _{\tilde {w}}}$ for $i \in \mathbb {Z}$. If $\alpha \notin \mathcal {P}_{\tilde {w}}$, then $\alpha ^{-m_{\alpha, {\tilde {w}}}} \in \tilde {\Phi }^- \setminus \Phi$ by definition. Thus, $\alpha ^{-k-1} \in \tilde {\Phi }^+ \cap \Phi = \Phi ^-$ since $k+1 \leqslant m_{\alpha, {\tilde {w}}}$ by part (a). Let $\beta = -\alpha ^{-k-1} \in \Phi ^+ \setminus \Phi _{\nu _{\tilde {w}}}$. Then $\beta ^{-m_{\beta, {\tilde {w}}}} = -\alpha ^{-m_{\alpha, {\tilde {w}}}} \in \tilde {\Phi }^+ \setminus \Phi$, and ${\tilde {w}}\sigma (s_\beta ) < {\tilde {w}} \in {\rm {Adm}}(\lambda )$ since ${\tilde {w}}\sigma (\beta ) = -\alpha ^{-k} \in \Phi ^-$. Thus, $\beta \in \mathcal {P}_{\tilde {w}}$ as desired.

Proof. By Theorem 1.4, there exist unique ${\tilde {w}}' \in {}^K \tilde {W}$ and some $u \in I(K, {\tilde {w}}')$ such that ${\tilde {w}} \to _K u {\tilde {w}}'$. Thus, $\Phi _{I(K, {\tilde {w}}')} \subseteq \Phi _{\nu _{{\tilde {w}}'}}$ (by Lemma 1.3) and $\ell (u {\tilde {w}}') = \ell (u) + \ell ({\tilde {w}}')$. As ${\tilde {w}} \in \mathcal {S}$, by Lemma 1.5(3) and (6) we have $u {\tilde {w}}' \in \mathcal {S}$ and $u{\tilde {w}}' \leq u^{-1} u{\tilde {w}}' = {\tilde {w}}'$. Thus, $u = 1$, and the first statement follows. The second statement is proved in [Reference Chen and NieCN19, Lemma 6.11].

Proof. We argue by induction on $|K|$. If $K = \emptyset$, the statement is trivial. Assume $|K| \geqslant 1$. Suppose there exists such an infinite sequence. As ${\tilde {w}} \in \mathcal {S}$, ${\tilde {w}}_i \in \mathcal {S}$ for $i \in \mathbb {Z}_{\geqslant 0}$ by Lemma 1.5(3). Noting that $\ell ({\tilde {w}}_0) \geqslant \ell ({\tilde {w}}_1) \geqslant \cdots$, we can assume that: (a) $\ell ({\tilde {w}}_0) = \ell ({\tilde {w}}_1) = \cdots$. Write ${\tilde {w}}_i = u_i y_i$ with $u_i \in W_K$ and $y_i \in {}^K \tilde {W}$. Then $s_{i+1} u_i < u_i$ and, hence, $y_i \sigma (s_{i+1}) > y_i$. By Lemma 1.2(1), for each $i \in \mathbb {Z}_{\geqslant 0}$ we have $y_i \leq y_{i+1}$, and more precisely, either $y_{i+1} = y_i \sigma (s_{i+1}) > y_i$ or $y_{i+1} = y_i$ and $y_i \sigma (s_{i+1}) y_i^{-1} \in K$. Thus, we can assume further that $y_0 = y_1 = \cdots$, that is: (b) there exists $y \in {}^K \tilde {W}$ such that ${\tilde {w}}_i \in W_K y$ and $y \sigma (s_i) y^{-1} \in K$ for $i \in \mathbb {Z}_{\geqslant 1}$. Let $K' = \{s_i; i \in \mathbb {Z}_{\geqslant 1}\} \subseteq K$. If $K' = K$, then by part (b) we have $K = I(K, y) \subseteq W_{\nu _{\tilde {w}}}$, see § 1.5. Thus, ${\tilde {w}}_i = y \in {}^K \tilde {W}$ since ${\tilde {w}}_i \in \mathcal {S}$, which is impossible. Otherwise, $K' \subsetneq K$ and it contradicts the induction hypothesis for the proper subset $K'$.

Proof. Assume otherwise. Then by Lemma 3.5 there is an infinite sequence

\[ {\tilde{w}} = {\tilde{w}}_0 \rightharpoonup_{R_0} {\tilde{w}}_1 \rightharpoonup_{R_1} \cdots, \]

where ${\tilde {w}}_{i+1} \in {}^{R_i} \tilde {W}$ and $R_i$ is some $\sigma$-orbit of $\mathbb {S}_0$ for $i \in \mathbb {Z}_{\geqslant 0}$. This contradicts Lemma 3.5. Thus, the statement follows.

Proof. Note that $\mu, \eta$ are conjugate by $W_0$. By the choice of $M \geqslant 2$ we have

\[ \biggl|\frac{\langle \alpha, p({\tilde{w}}\sigma)^n(\mu) \rangle}{M^n}\biggr| > \frac{2A}{M^{n+1}} > \frac{A}{M^{n+1}} \sum_{i=n+1}^{N-1} \frac{1}{M^{i-n-1}} \geqslant \sum_{i=n+1}^{N-1} \biggl|\frac{\langle \alpha, p({\tilde{w}}\sigma)^i(\mu) \rangle}{M^i}\biggr|. \]

Thus, the statement follows.

Proof. Statement (1) follows from Lemma 4.1 and the definition of $\nu _{{\tilde {w}}}^\flat$.

Suppose there exists $\alpha \in \Phi _{\nu _{{\tilde {w}}}}^+ \subseteq \tilde {\Phi }^-$ such that $\langle \alpha, \nu _{{\tilde {w}}}^\flat \rangle < 0$. By Lemma 4.1, there exists $n \in \mathbb {Z}_{\geqslant 0}$ such that $\langle \alpha, p({\tilde {w}}\sigma )^n(\mu ) \rangle < 0$ and $\langle \alpha, p({\tilde {w}}\sigma )^i(\mu ) \rangle = 0$ for $0 \leqslant i \leqslant n-1$. In particular, we have $({\tilde {w}}\sigma )^{-i}(\alpha ) = p({\tilde {w}}\sigma )^{-i}(\alpha )$ for $1 \leqslant i \leqslant n$ and $({\tilde {w}}\sigma )^{-n-1}(\alpha ) \in \tilde {\Phi }^+ \setminus \Phi$, contradicting that ${\tilde {w}} \in \mathcal {S}$. Thus, statement (2) follows.

Statement (3) follows by definition.

By statement (1) we have $\tilde {\Phi }_{\nu _{{\tilde {w}}}^\flat } = {\tilde {w}}\sigma (\tilde {\Phi }_{\nu _{{\tilde {w}}}^\flat }) \subseteq \tilde {\Phi }_{\nu _{{\tilde {w}}}}$. As ${\tilde {w}} \in \mathcal {S}$, we have ${\tilde {w}}\sigma (\tilde {\Phi }_{\nu _{{\tilde {w}}}}^\pm ) = \tilde {\Phi }_{\nu _{{\tilde {w}}}}^\pm$. Thus, statement (4) follows from that $\tilde {\Phi }_{\nu _{\tilde {w}}^\flat }^\pm = \tilde {\Phi }_{\nu _{\tilde {w}}^\flat } \cap \tilde {\Phi }_{\nu _{\tilde {w}}}^\pm$.

Let $\alpha \in \mathcal {P}_{\tilde {w}}$. Set $m = m_{\alpha, {\tilde {w}}}$ and $\alpha ^i = ({\tilde {w}}\sigma )^i(\alpha )$ for $i \in \mathbb {Z}$. By definition, $\langle \alpha ^{1-m}, \mu \rangle < 0$, $\alpha ^{-i} = p({\tilde {w}}\sigma )^{-i}(\alpha )$, and $\langle \alpha ^{1-i}, \mu \rangle = \langle \alpha, p({\tilde {w}}\sigma )^{i-1}(\mu ) \rangle = 0$ for $1 \leqslant i \leqslant m-1$. Thus, it follows from Lemma 4.1 that $\langle \alpha ^i, \nu _{\tilde {w}}^\flat \rangle < 0$ for $1-m \leqslant i \leqslant 0$. Suppose $\sum _{i=0}^{1-m} c_i \alpha ^i = 0$, where the coefficients $c_i \in \mathbb {R}$ are not all zero. Let $i_0 = \min \{1-m \leqslant i \leqslant 0; c_i \neq 0\}$. Then

\[ 0 = \biggl \langle p({\tilde{w}}\sigma)^{1-m - i_0}\biggl(\sum_{i=0}^{1-m} c_i \alpha^i\biggr), \mu \biggr \rangle = \sum_{i=0}^{i_0} c_i \langle \alpha^{1-m-i_0+i}, \mu \rangle = c_{i_0} \langle \alpha^{1-m}, \mu \rangle \neq 0, \]

which is a contradiction. Thus, statement (5) follows.

Proof. Assume $\nu _{\tilde {w}}^\flat$ is dominant. Let $\mu \in Y$ such that ${\tilde {w}} \in t^\mu W_0$. Then $\mu$ is dominant by Lemma 4.1. We show ${\tilde {w}} < s_\alpha {\tilde {w}}$ for $\alpha \in \Phi ^+$. If $\langle \alpha, \nu _{\tilde {w}}^\flat \rangle > 0$, then either $\langle \alpha, \mu \rangle > 0$, or $\langle \alpha, \mu \rangle = 0$ and $\langle p({\tilde {w}}\sigma )^{-1}(\alpha ), \nu _{\tilde {w}}^\flat \rangle > 0$ (hence, $p({\tilde {w}}\sigma )^{-1}(\alpha ) \in \Phi ^+$) by Corollary 4.2, which means ${\tilde {w}} < s_\alpha {\tilde {w}}$ as desired. Suppose $\langle \alpha, \nu _{\tilde {w}}^\flat \rangle = 0$, that is, $\alpha \in \Phi _{\nu _{\tilde {w}}^\flat }^+ \subseteq \tilde {\Phi }^-$. Thus, $({\tilde {w}}\sigma )^{-1}(\alpha ) \in \tilde {\Phi }_{\nu _{\tilde {w}}^\flat }^-$ by Corollary 4.2(4), which also means ${\tilde {w}} < s_\alpha {\tilde {w}}$ as desired.

Hypothesis 4.1 Recall that $\mathbb {F}_q$ is the residue field of $F$. Assume that $q^d > 2$ (respectively, ${q^d > 3}$) if some/any connected component of $\mathbb {S}_0$ is non-simply-laced except of type $G_2$ (respectively, is of type $G_2$).

Proof. By assumption, we have

(a) \begin{align*} {}^{({\tilde{w}}\sigma)^i} U_\gamma(z) = U_{\gamma^i}(c_i z^{q^i}) \text{ with } c_i \in \mathcal{O}_{\breve F}^\times \text{ for } 1-m \leqslant i \leqslant 0. \end{align*}

For $\alpha \neq -\alpha ' \in \Phi$ there exist constants $c_{\alpha, \alpha ', i, j} \in \mathcal {O}_{\breve F}$ for $i, j \in \mathbb {Z}_{\geqslant 1}$ such that

(b) \begin{align*} U_\alpha(x) U_{\alpha'}(y) U_\alpha(-x) =U_{\alpha'}(y) \prod_{i, j \in \mathbb{Z}_{\geqslant 1}} U_{i\alpha + j\alpha'}(c_{\alpha, \alpha', i, j} x^i y^j). \end{align*}

Now we argue by induction on $m$. If $m = 0$, the statement is trivial. Assume $m \geqslant 1$. If $\gamma \in \Phi ^-$, then $U_\gamma (z) \in I$ and, hence, $ {g}(\infty ) = {g}_{1, \gamma ^{-1}, {\tilde {w}}, m-1}(\infty )$, from which the statement follows by induction hypothesis. Otherwise, we have

\[ {g}(z) = {}^{({\tilde{w}}\sigma)^{1-m}} U_\gamma(z) \cdots {}^{({\tilde{w}}\sigma)^{-1}} U_\gamma(z) U_{-\gamma}(z^{-1}) s_\gamma I \text{ for } z \neq 0. \]

As the roots $\gamma ^i$ for $1-m \leqslant i \leqslant 0$ are linearly independent, it follows by parts (a) and (b) that

\[ {}^{{}^{({\tilde{w}}\sigma)^{1-m}} U_\gamma(z) \cdots {}^{({\tilde{w}}\sigma)^{-1}} U_\gamma(z)} U_{-\gamma}(z^{-1})= \prod_{(\beta, a_\bullet)} U_\beta(c_{a_\bullet} z^{n_{a_\bullet}}), \]

where $a_\bullet = (a_i)_{0 \leqslant i \leqslant m-1} \in (\mathbb {Z}_{\geqslant 0})^m$ with $a_0 \geqslant 1$, $\beta = -a_0 \gamma + \sum _{i=1}^{m-1} a_i \gamma ^{-i} \in \Phi$, $c_{a_\bullet } \in \mathcal {O}_{\breve F}$ and $n_{a_\bullet } = -a_0 + \sum _{i=1}^{m-1} a_i q^{-i}$. Note that $a_i = 0$ unless $i \in d\mathbb {Z}$ since $\beta \in \Phi$. Moreover, $a_{jd} / a_0 \leqslant 1$ (respectively, $a_{j d} / a_0 \leqslant 2$; respectively, $a_{jd} / a_0 \leqslant 3$) for $j \geqslant 1$ if some/any connected component of $\mathbb {S}_0$ is simply-laced (respectively, is non-simply-laced except of type $G_2$; respectively, is of type $G_2$). Thus, by Hypothesis 4.1 we have $a_{jd} / a_0 \leqslant q^d-1$ for $j \geqslant 1$, which implies that $n_{a_\bullet } < 0$ and, hence,

\[ \lim_{z \to \infty} {}^{{}^{({\tilde{w}}\sigma)^{1-m}} U_\gamma(z) \cdots {}^{({\tilde{w}}\sigma)^{-1}} U_\gamma(z)} U_{-\gamma}(z^{-1}) = 1. \]

Then we have

\[ {g}(\infty) = s_{\gamma} {g}_{1, s_\gamma(\gamma^{-1}), s_\gamma {\tilde{w}} \sigma(s_\gamma), m-1}(\infty). \]

By the induction hypothesis, there exist a sequence $2-m \leqslant j_r < \cdots < j_1 \leqslant 0$ of integers such that

\begin{align*} & {g}_{1, s_\gamma(\gamma^{-1}), s_\gamma {\tilde{w}} \sigma(s_\gamma), m-1}(\infty) \\ &\qquad = s_{(s_\gamma {\tilde{w}}\sigma s_\gamma)^{j_r}(s_\gamma(\gamma^{-1}))} \cdots s_{(s_\gamma {\tilde{w}}\sigma s_\gamma)^{j_1}(s_\gamma(\gamma^{-1}))} I = s_\gamma s_{\gamma^{j_r-1}} \cdots s_{\gamma^{j_1-1}} s_\gamma I, \end{align*}

and for $1 \leqslant k \leqslant r$,

\begin{align*} & s_{(s_\gamma {\tilde{w}}\sigma s_\gamma)^{j_1}(s_\gamma(\gamma^{-1}))} \cdots s_{(s_\gamma {\tilde{w}}\sigma s_\gamma)^{j_{k-1}}(s_\gamma(\gamma^{-1}))} ((s_\gamma {\tilde{w}}\sigma s_\gamma)^{j_k}(s_\gamma(\gamma^{-1})))\\ &\qquad = s_\gamma s_{\gamma^{j_1-1}} \cdots s_{\gamma^{j_{k-1}-1}} (\gamma^{j_k - 1}) \in \Phi^+. \end{align*}

Take $i_0 = 0$ and $i_k = j_k - 1$ for $1 \leqslant k \leqslant r$. Then one checks directly that the first statement is true.

Set $\beta _k = s_{\gamma ^{i_0}} \cdots s_{\gamma ^{i_{k-1}}} (\gamma ^{i_k}) \in \Phi ^+$ and $v_k = s_{\gamma ^{i_0}} \cdots s_{\gamma ^{i_k}}(v)$ for $0 \leqslant k \leqslant r$. As $\langle \gamma ^{i_k}, v \rangle < 0$ we have

\[ v_k = s_{\beta_k}(v_{k-1}) = v_{k-1} - \langle \beta_k, v_{k-1} \rangle \beta_k^\vee = v_{k-1} - \langle \gamma^{i_k}, v \rangle \beta_k^\vee > v_{k-1}. \]

Thus, the ‘Moreover’ part follows.

Proof. Write $y z = z' \omega ^{-1} u$ with $u \in W_J^a$. Let $\delta = u x \sigma (u)^{-1} x^{-1} \in W_J^a$. Then $x \leq _J \delta x$ (see § 1.5) since $x \in \Omega _J$. As $z' \omega ^{-1} (\tilde {\Phi }_J^+) \subseteq \tilde {\Phi }^+$, it follows from [Reference Chen and NieCN19, Lemma 1.3] that

\[ {\tilde{w}}' = (z'\omega^{-1}) x \sigma(z'\omega^{-1})^{-1} \leq (z'\omega^{-1}) \delta x \sigma(z'\omega^{-1})^{-1} = y {\tilde{w}} \sigma(y)^{-1} \in {\rm{Adm}}(\lambda). \]

Thus, ${\tilde {w}}' \in {\rm {Adm}}(\lambda ) \cap \mathcal {S}$ by Lemma 1.5. Note that ${\tilde {w}}' \in W_0 t^{\mu _{x'}} W_0$ and $x' \in \Omega _J$. Then $\mu _{x'} \preceq \lambda$ and $x' \leq _J t^{\mu _{x'}} \in {\rm {Adm}}(\lambda )$, which means $x' \in {\rm {Adm}}(\lambda )$ and $x' \in \mathcal {S}_{\lambda, b}^+$. Thus, part (1) follows.

Note that $J = J_{\nu _x} = J_{\nu _{x'}}$. By definition $y {\tilde {w}} \sigma (y)^{-1}, {\tilde {w}}'$ are $\sigma$-conjugate by

\[ z'\omega^{-1} u \omega {z'}^{-1} \in z' W_J^a {z'}^{-1} = z' W_{\nu_{x'}}^a {z'}^{-1} = W_{\nu_{{\tilde{w}}'}}^a. \]

Moreover, if $y \in W_0$, then $\omega =1$, $u \in W_J$ and hence $z'\omega ^{-1} u \omega {z'}^{-1} \in W_{\nu _{{\tilde {w}}'}}$. Thus, part (2) follows.

Now we consider the following closed affine Deligne–Lusztig variety

\[ X_{\leq_J \delta x}^{M_J}(x) = \{m \in M_J({\breve F}) / I_{M_J}; m^{-1} x \sigma(m) \in \cup_{\delta' \leq_J \delta} I_{M_J} \delta' x I_{M_J}\!\}. \]

Note that $u^{-1} I_{M_J} \in X_{\leq _J \delta x}^{M_J}(x)$. As $x \in \Omega _J$, by [Reference He and ZhouHZ20, Theorem 4.1] (respectively, [Reference Chen and NieCN20, Lemma 6.13]), there exists $h \in \ker (\eta _{M_J}) \cap \mathbb {J}_x$ (respectively, $h \in H_x \cap \mathbb {J}_x$ if $y \in W_0$) such that $u^{-1} I_{M_J}, h I_{M_J}$ are connected in $X_{\leq _J \delta x}^{M_J}(x)$. For $g \in \mathbb {J}_{b, {\tilde {w}}}$ there is an embedding

\[ X_{\leq_J \delta x}^{M_J}(x) \hookrightarrow X(\lambda, b), \quad m I_{M_J} \mapsto g z m \omega {z'}^{-1} I, \]

from which we have $g y^{-1} I = g z u^{-1} \omega {z'}^{-1} I \sim _{\lambda, b} g z h \omega {z'}^{-1} I$ as desired.

Proof. By Proposition 3.1, there exists $\alpha \in \mathcal {P}_{{\tilde {w}}}$. Set $m = m_{\alpha, {\tilde {w}}}$ and $\alpha ^i = ({\tilde {w}}\sigma )^i(\alpha )$ for $i \in \mathbb {Z}$. Let $ {g} = {g}_{g, \alpha, {\tilde {w}}, m}$ for $g \in \mathbb {J}_{b, {\tilde {w}}}$. Let $A \subseteq I$ be the kernel of the natural reduction map $G(\mathcal {O}_{\breve F}) \overset {t \mapsto 0} \to G(\boldsymbol {k})$. Since $\alpha ^{-m} \in \tilde {\Phi }^+ \setminus \Phi$ and $\alpha ^i \in \Phi$ for $1-m \leqslant i \leqslant 0$, we have ${}^{({\tilde {w}}\sigma )^{-m}} U_\alpha = U_{\alpha ^{-m}} \subseteq A$ and ${}^{({\tilde {w}}\sigma )^i} U_\alpha = U_{\alpha ^i} \subseteq G(\mathcal {O}_{\breve F})$ for $1-m \leqslant i \leqslant 0$. Thus, for $z \in \boldsymbol {k}$ we have

\[ F(z):= {}^{({}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(z) \cdots U_\alpha(z))^{-1} ({\tilde{w}}\sigma)^{-m}} U_\alpha(-z)) \in A \subseteq I. \]

Now one computes that

\begin{align*} {g}(z)^{-1} b \sigma {g}(z) &= U_\alpha(-z) \cdots {}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(-z) {\tilde{w}}\sigma {}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(z) \cdots U_\alpha(z) \\ &= U_\alpha(-z) \cdots {}^{({\tilde{w}}\sigma)^{2-m}} U_\alpha(-z) {\tilde{w}}\sigma {}^{({\tilde{w}}\sigma)^{-m}} U_\alpha(-z) {}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(z) \cdots U_\alpha(z) \\ &= U_\alpha(-z) \cdots {}^{({\tilde{w}}\sigma)^{2-m}} U_\alpha(-z) {\tilde{w}}\sigma {}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(z) \cdots U_\alpha(z) F(z)\\ &\subseteq U_\alpha(-z) \cdots {}^{({\tilde{w}}\sigma)^{2-m}} U_\alpha(-z) {\tilde{w}}\sigma {}^{({\tilde{w}}\sigma)^{1-m}} U_\alpha(z) \cdots U_\alpha(z) I \\ &= {\tilde{w}} \sigma U_\alpha(z) I \subseteq I \{{\tilde{w}}\sigma, {\tilde{w}}\sigma s_\alpha\} I \subseteq I {\rm{Adm}}(\lambda)\sigma I. \end{align*}

By Hypothesis 4.1 and Corollary 4.2(5), the conditions in Lemma 4.4 are satisfied (for $(\gamma, m, v) = (\alpha, m_{{\tilde {w}}, \alpha }, \nu _{\tilde {w}}^\flat )$). Thus, by Lemma 4.4 we have $g I = {g}(0) \sim _{\lambda, b} {g}(\infty ) = g y^{-1} I$ for some $y \in W_0$ such that $y(\nu _{\tilde {w}}^\flat ) > \nu _{\tilde {w}}^\flat$. Then $y {\tilde {w}} \sigma (y)^{-1} \in {\rm {Adm}}(\lambda )$ and $\nu _{{\tilde {w}}}^\flat < y(\nu _{{\tilde {w}}}^\flat ) = \nu _{y {\tilde {w}} \sigma (y)^{-1}}^\flat$. Let $h \in H_x$, ${\tilde {w}}' \in \mathcal {S}_{\lambda, b, x}$, and $z' \in W_0^J$ be as in Lemma 4.7 such that $g I \sim _{\lambda, b} g y^{-1} I \sim _{\lambda, b} g z h {z'}^{-1} I$. Then $y {\tilde {w}} \sigma (y)^{-1}$ and ${\tilde {w}}'$ are $\sigma$-conjugate by $W_{\nu _{{\tilde {w}}'}}$, and, hence, $\nu _{{\tilde {w}}'}^\flat$ and $\nu _{y {\tilde {w}} \sigma (y)^{-1}}^\flat$ are conjugate by $W_{\nu _{{\tilde {w}}'}}$. By Corollary 4.2(2), $\nu _{{\tilde {w}}'}^\flat$ is dominant for $\Phi _{\nu _{{\tilde {w}}'}}^+$, which means $\nu _{{\tilde {w}}}^\flat < \nu _{y {\tilde {w}} \sigma (y)^{-1}}^\flat \leqslant \nu _{{\tilde {w}}'}^\flat$ as desired.

Proof. Note that the statement follows from Theorem 0.2, which is proved in [Reference Chen and NieCN19] when $G$ is split. Thus, we can assume $G$ is not split and Hypothesis 4.1 holds. If ${\tilde {w}} \notin {}^{\mathbb {S}_0} \tilde {W}$, by Lemma 4.8, there exist $h \in H_x \cap \mathbb {J}_x$ and ${\tilde {w}}' \in \mathcal {S}_{\lambda, b, x}$ such that $\nu _{\tilde {w}}^\flat < \nu _{{\tilde {w}}'}^\flat$ and $g I \sim _{\lambda, b} g z h {z'}^{-1} I$ for $g \in \mathbb {J}_{b, {\tilde {w}}}$, where $z' \in W_0^J$ such that ${\tilde {w}}' = z' x \sigma (z')^{-1}$. Thus, the statement follows by repeating this process.

Proof. By Proposition 3.6 and Lemma 1.5(2), there exists $z' \in W_0$ such that $z' x \sigma (z')^{-1} \in \mathcal {S} \cap {}^{\mathbb {S}_0} \tilde {W}$. By Corollary 4.9, there exist $h_1, h_2 \in H_x \cap \mathbb {J}_x$ such that $g I \sim _{\lambda, b} g h_1 {z'}^{-1} I$ and $g z^{-1} I \sim _{\lambda, b} g h_2 {z'}^{-1} I$ for $g \in \mathbb {J}_{b, x}$. Then we have

\[ g h z^{-1} I = j g z^{-1} I \sim_{\lambda, b} j g h_2 {z'}^{-1} I = g h_1 {z'}^{-1} I \sim_{\lambda, b} g I, \]

where $h = h_1 h_2^{-1} \in H_x \cap \mathbb {J}_x$ and $j =g h_1 h_2^{-1} g^{-1} \in \mathbb {J}_b$.

Proof. It follows from Lemma 4.7 and Proposition 4.10.

Proof. Let $\gamma \in \mathcal {O}$. By definition, $\langle \gamma, \nu _G(b) \rangle = \langle \gamma, {\rm pr}_J(\mu _x)^\diamond \rangle > 0$, where ${\rm pr}_J(\mu _x)^\diamond$ is the $\sigma$-average of ${\rm pr}_J(\mu _x)$. Thus, part (1) follows since

\[ \sum_{\alpha \in \mathcal{O}} \langle \alpha, {\rm pr}_J(\mu_x) \rangle = \sum_{\alpha \in \mathcal{O}} \langle \alpha, {\rm pr}_J(\mu_x)^\diamond \rangle = |\mathcal{O}| \langle \gamma, \nu_G(b) \rangle > 0. \]

By part (1), there exists $\beta \in \mathcal {O}$ such that $\langle \beta, {\rm pr}_J(\mu _x) \rangle > 0$. As $w_J(\beta )$ is $J$-dominant and $\mu _x - {\rm pr}_J(\mu _x) \in \mathbb {R}_{\geqslant 0} (\Phi _J^+)^\vee$, we have

\[ \langle w_J(\beta), \mu_x \rangle \geqslant \langle w_J(\beta), {\rm pr}_J(\mu_x) \rangle = \langle \beta, {\rm pr}_J(\mu_x) \rangle > 0. \]

Thus, part (2) follows.

Proof. Note that parts (1) and (2) were proved in [Reference Chen, Kisin and ViehmannCKV15, Lemma 4.4.6] and [Reference Chen and NieCN20, Lemma 1.5], respectively. To show part (3) we claim that

\begin{align*} \text{there exists }\eta \in W_K(\mu)\text{ such that }\eta - \gamma^\vee + \sigma^r(\gamma^\vee)\text{ is }K\text{-minuscule}. \end{align*}

Indeed, let $\eta$ be a $W_K$-conjugate of $\mu$ such that $\eta - \gamma ^\vee + \sigma ^r(\gamma ^\vee )$ is minimal under the partial order $\preceq$. If $\eta - \gamma ^\vee + \sigma ^r(\gamma ^\vee )$ is not $K$-minuscule, then there exists $\alpha \in \Phi _K$ such that $\langle \alpha, \eta - \gamma ^\vee + \sigma ^r(\gamma ^\vee ) \rangle \geqslant 2$. As $\eta$ is $K$-minuscule, and $\gamma ^\vee, \sigma ^r(\gamma ^\vee )$ are $K$-dominant and $K$-minuscule, we deduce that $\langle \alpha, \eta \rangle = 1$. Let $\eta ' = s_\alpha (\eta ) = \eta - \alpha ^\vee$. Then we have

\[ \eta' - \gamma^\vee + \sigma^r(\gamma^\vee) = \eta - \gamma^\vee + \sigma^r(\gamma^\vee) - \alpha^\vee \prec \eta - \gamma^\vee + \sigma^r(\gamma^\vee), \]

which contradicts the choice of $\eta$. Thus, part (a) is proved.

Let $w = p({\tilde {w}}) \in W_K$. By parts (1) and (a), $\eta - \gamma ^\vee + \sigma ^r(\gamma ^\vee )$, $\mu - \gamma ^\vee + w \sigma ^r(\gamma ^\vee )$ are conjugate by $W_K$. In particular, $\eta - \gamma ^\vee + \sigma ^r(\gamma ^\vee ) \preceq \lambda$. Then part (3) follows from that

\[ s_{\tilde \gamma} {\tilde{w}} s_{\sigma^r(\tilde \gamma)} \leq s_{\tilde \gamma} t^\eta s_{\sigma^r(\tilde \gamma)} = s_\gamma t^{\eta - \gamma^\vee + \sigma^r(\gamma^\vee)} s_{\sigma^r(\gamma)} \leq t^{\eta - \gamma^\vee + \sigma^r(\gamma^\vee)} \in {\rm{Adm}}(\lambda), \]

where the first $\leq$ follows from [Reference Chen and NieCN19, Lemma 1.3], and the second $\leq$ follows from that

\begin{align*} \langle \gamma, \sigma^r(\gamma^\vee) \rangle &\leqslant 0 \text{ since } \gamma \neq \sigma^r(\gamma); \\ \langle \gamma, \eta - \gamma^\vee + \sigma^r(\gamma^\vee) \rangle &\leqslant \langle \gamma, \mu \rangle - 2 \leqslant -1; \\ \langle \sigma^r(\gamma), \eta - \gamma^\vee + \sigma^r(\gamma^\vee) \rangle &\geqslant \langle w\sigma^r(\gamma), \mu \rangle + 2 \geqslant 1. \end{align*}

The proof is finished.

Proof. Let $w = p({\tilde {w}}) \in W_K$. First we claim that

(a)$$\begin{align*} \Psi := \Phi \cap (\mathbb{Z} \gamma + \mathbb{Z} w\sigma^r(\gamma))\text{ is of type }A_2,\text{ or }A_1 \times A_1,\text{ or }A_1. \end{align*}$$

Otherwise, then $\Psi$ is of type $B_2$ or $G_2$. In particular, $\gamma = \sigma ^r(\gamma )$ (since $\sigma ^d = {\rm id}$ with $d$ the number of connected components of $\mathbb {S}_0$), $\gamma \neq w\sigma ^r(\gamma ) = w(\gamma )$ are short roots, and, hence, $K \neq \emptyset$. If $\Psi$ is of $B_2$, then $\gamma \pm w\sigma ^r(\gamma ) \in \Phi$ and $\langle \gamma, w\sigma ^r(\gamma ^\vee ) \rangle = 0$ since $\gamma, w\sigma ^r(\gamma )$ are of the same length. Thus, $\gamma - w\sigma ^r(\gamma ) \in \Phi _K$ and $\langle \gamma - w\sigma ^r(\gamma ), \gamma ^\vee \rangle = 2$, contradicting that $\gamma ^\vee$ is $K$-minuscule. Thus, $\Psi$ is of type $G_2$. As $\gamma \neq w\sigma ^r(\gamma )$ are short roots and $\gamma ^\vee$ is strongly $K$-minuscule, we deduce that $K$ consists of long simple roots. This contradicts that $\gamma ^\vee$ is $K$-minuscule. Thus, part (a) is proved.

Now we claim that

(b) $$\begin{align*} \text{$U_{-\tilde \gamma} {\tilde{w}} U_{-\sigma^r(\tilde \gamma)} \subseteq I {\rm{Adm}}(\lambda) I$ if one of the following holds:} \end{align*}$$

(b1) $$\begin{align*} \text{either $\langle \gamma, \mu \rangle \geqslant 2$ or $\langle \gamma, \mu \rangle = 1$ and $\langle \gamma, w\sigma(\gamma^\vee) \rangle \geqslant 0$;} \end{align*}$$

(b2) $$\begin{align*} \text{either $\langle w\sigma^r(\gamma), \mu \rangle \leqslant -2$ or $\langle w\sigma^r(\gamma), \mu \rangle = -1$ and $\langle \gamma, w\sigma^r(\gamma^\vee) \rangle \geqslant 0$.} \end{align*}$$

By symmetry we may assume (b1) occurs. By part (a) we have

\[ U_{-{\tilde{w}}^{-1}(\tilde \gamma)}, [U_{-{\tilde{w}}^{-1}(\tilde \gamma)}, U_{-\sigma^r(\tilde \gamma)}] \subseteq I, \]

where $[g, g'] = g g' g^{-1} {g'}^{-1}$ denotes the commutator of $g, g' \in G({\breve F})$. Thus,

\[ U_{-\tilde \gamma} {\tilde{w}} U_{-\sigma^r(\tilde \gamma)} \subseteq {\tilde{w}} U_{-\sigma^r(\tilde \gamma)} I \subseteq I \{{\tilde{w}}, {\tilde{w}} s_{\sigma^r(\tilde \gamma)}\} \subseteq I {\rm{Adm}}(\lambda) I, \]

where the last inclusion follows from Lemma 5.3(2). Thus, part (b) is proved.

Suppose $U_{-\tilde \gamma } {\tilde {w}} U_{-\sigma ^r(\tilde \gamma )} \nsubseteq I {\rm {Adm}}(\lambda ) I$. Then $-\langle w \sigma ^r(\gamma ), \mu \rangle, \langle \gamma, \mu \rangle \leqslant 1$ by (b), which implies that ${\tilde {w}}^{-1}(\gamma ) \neq \sigma ^r(\gamma )$. Assume $\langle \gamma, \mu \rangle \leqslant 0$. We claim that

\[ U_{{\tilde{w}}^{-1}(\tilde \gamma)}, [U_{{\tilde{w}}^{-1}(\tilde \gamma)}, U_{-\sigma^r(\tilde \gamma)}] \subseteq I. \]

The first inclusion follows from that ${\tilde {w}}^{-1}(\tilde \gamma ) = w^{-1}(\gamma ) + 1 - \langle \gamma, \mu \rangle \in \tilde {\Phi }^+$. Note that $[U_{{\tilde {w}}^{-1}(\tilde \gamma )}, U_{-\sigma ^r(\tilde \gamma )}] = U_{{\tilde {w}}^{-1}(\tilde \gamma ) - \sigma ^r(\tilde \gamma )}$ by part (a). Thus, we can assume that $U_{{\tilde {w}}^{-1}(\tilde \gamma ) - \sigma ^r(\tilde \gamma )}$ is nontrivial, that is,

\[ {\tilde{w}}^{-1}(\tilde \gamma) - \sigma^r(\tilde \gamma) = w^{-1}(\gamma) - \sigma^r(\gamma) - \langle \gamma, \mu \rangle \in \tilde{\Phi}. \]

As $\gamma$ is $K$-dominant and $w \in W_K$, $w^{-1}(\gamma ) - \gamma \in \mathbb {Z}_{\geqslant 0} \Phi _K^-$. Thus, the $\sigma$-average of $w^{-1}(\gamma ) - \sigma ^r(\gamma )$, which equals the $\sigma$-average of $w^{-1}(\gamma ) - \gamma$, lies in $\mathbb {R}_{\geqslant 0} \Phi ^-$. This means that $w^{-1}(\gamma ) - \sigma ^r(\gamma ) \in \Phi ^-$ and, hence, ${\tilde {w}}^{-1}(\tilde \gamma ) - \sigma ^r(\tilde \gamma ) \in \tilde {\Phi }^+$ (since $\langle \gamma, \mu \rangle \leqslant 0$). Thus, the second inclusion follows, and the claim is proved.

Thus, by Lemma 5.3 we compute that

\[ U_{-\tilde \gamma} {\tilde{w}} U_{-\sigma^r(\tilde \gamma)} \subseteq I s_{\tilde \gamma} {\tilde{w}} U_{-\sigma^r(\tilde \gamma)} I \subseteq I \{ s_{\tilde \gamma}{\tilde{w}}, s_{\tilde \gamma} {\tilde{w}} s_{\sigma^r(\tilde \gamma)} \} I \subseteq I {\rm{Adm}}(\lambda) I, \]

which contradicts our assumption. Thus, $\langle \gamma, \mu \rangle = 1$, and $\langle w\sigma ^r(\gamma ), \mu \rangle = -1$ by symmetry. Moreover, we have $\langle \gamma, w\sigma ^r(\gamma ^\vee ) \rangle = -1$ by parts (b) and (a).

Write ${\tilde {w}}' = t^{\mu '} w' \in \Omega _K$ with $\mu ' \in Y$ and $w' \in W_K$. Applying Lemma 5.3(1) to ${\tilde {w}}$ and ${\tilde {w}}'$ we deduce that

\[ \mu', \mu - \gamma^\vee + w\sigma^r(\gamma^\vee), \mu' - \sigma^r(\gamma^\vee), \mu - \gamma^\vee, \mu' + w'(\gamma^\vee), \mu + w\sigma^r(\gamma^\vee) \]

are $K$-minuscule, and, hence, are conjugate by $W_K$. Since