Proof. We first show that $\mathsf {read}_\tau (\mathrm {SYT}(\lambda /\mu )) \subseteq \Sigma _L(\mathsf {poset}(\tau ))$ . To do this, take any $T \in \mathrm {SYT}(\lambda /\mu )$ and $i,j \in [n]$ with $i \preceq _\tau j$ (for the definition of $\preceq _\tau $ , see Equation (2.1). Let $B_1$ and $B_2$ be the boxes in $\mathtt {yd}(\lambda /\mu )$ such that $\tau _{B_1} = i$ and $\tau _{B_2} = j$ . Since $i \preceq _\tau j$ , $B_1$ is weakly upper-left of $B_2$ in $\mathtt {yd}(\lambda /\mu )$ . This implies that

$$ \begin{align*}\mathsf{read}_\tau(T)(i) = T_{B_1} \leq T_{B_2} = \mathsf{read}_\tau(T)(j). \end{align*} $$

Therefore, $\mathsf {read}_\tau (T) \in \Sigma _L(\mathsf {poset}(\tau ))$ .

To complete the proof, let us show that $|\Sigma _L(\mathsf {poset}(\tau ))| = |\mathsf {read}_\tau (\mathrm {SYT}(\lambda /\mu ))|$ . Note that

$$ \begin{align*} \mathrm{ch}([\mathsf{M}_{\mathsf{poset}(\tau^{\lambda/\mu}_0)}]) = \sum_{\gamma \in \Sigma_L(\mathsf{poset}(\tau^{\lambda/\mu}_0))} F_{\mathrm{comp}(\mathrm{Des}_L(\gamma))^{\mathrm{c}}} \quad \text{and} \quad \mathrm{ch}([\mathsf{M}_{\mathsf{poset}(\tau^{\lambda/\mu}_0)}]) = s_{\lambda/\mu}, \end{align*} $$

where the second equality follows from Theorem 2.9(2). Putting these together with the well-known equality $s_{\lambda / \mu } = \sum _{T \in \mathrm {SYT}(\lambda / \mu )} F_{\mathrm {comp}(T)}$ , we have

(3.1) $$ \begin{align} \sum_{\gamma \in \Sigma_L(\mathsf{poset}(\tau^{\lambda/\mu}_0))} F_{\mathrm{comp}(\mathrm{Des}_L(\gamma))^{\mathrm{c}}} = \sum_{T \in \mathrm{SYT}(\lambda/\mu)} F_{\mathrm{comp}(T)}. \end{align} $$

Here, $\mathrm {comp}(T) = \mathrm {comp}(\{i \in [n-1] \mid i\text { is weakly right of }i+1\text { in }T\})$ . As a consequence of Equation (3.1), we have

$$\begin{align*}|\Sigma_L(\mathsf{poset}(\tau))| = |\Sigma_L(\mathsf{poset}(\tau^{\lambda/\mu}_0))| = |\mathrm{SYT}(\lambda/\mu)| = |\mathsf{read}_\tau(\mathrm{SYT}(\lambda/\mu))|.\\[-42pt] \end{align*}$$

Proof. To prove the ‘only if’ part, assume that P is a regular Schur labeled skew shape poset and $\lambda / \mu $ is the shape of $\tau _P$ . We claim that every connected component of $\tau _P$ is filled with consecutive integers. Take an arbitrary connected component $\mathsf {C}$ of $\tau _P$ . Let $B_1$ be the box at the top of the rightmost column of $\mathsf {C}$ and $B_2$ the box at the bottom of the leftmost column of $\mathsf {C}$ . Then, we may choose boxes $A_0:= B_1, A_2, A_3, \ldots , A_k := B_2$ satisfying that for all $1 \le i \le k-1$ , $A_{i+1}$ is weakly below and weakly left of $A_i$ and $A_{i}, A_{i+1}$ are in the same row or in the same column. Let $m \in [(\tau _P)_{B_1}, (\tau _P)_{B_2}]$ . Then, there exists a unique index $1 \leq i \leq k-1$ such that $(\tau _P)_{A_i} \leq m \leq (\tau _P)_{A_{i+1}}$ . Since P is regular, one of the following holds:

1. (i) If $(\tau _P)_{A_i} \preceq _P (\tau _P)_{A_{i+1}}$ , then $(\tau _P)_{A_i} \preceq _P m$ or $m \preceq _P (\tau _P)_{A_{i+1}}$ .

2. (ii) If $(\tau _P)_{A_{i+1}} \preceq _P (\tau _P)_{A_{i}}$ , then $(\tau _P)_{A_{i+1}} \preceq _P m$ or $m \preceq _P (\tau _P)_{A_{i}}$ .

It follows that $(\tau _P)_{A_i}$ , m and $(\tau _P)_{A_{i+1}}$ appear in the same connected component; that is, m appears in $\mathsf {C}$ . Thus, $\tau _P$ is distinguished.

Next, to prove the ‘if’ part, assume that $\tau _P$ is a distinguished Schur labeling and $\lambda / \mu $ is the shape of $\tau _P$ . Let $B_1, B_2 \in \mathtt {yd}(\lambda /\mu )$ with $(\tau _P)_{B_1} \prec _{\tau _P} (\tau _P)_{B_2}$ . By the definition of $\prec _{\tau _P}$ , $B_1$ and $B_2$ are in the same connected component. In order to establish the regularity of P, we need to prove that either $(\tau _P)_{B_1} \preceq _{\tau _P} (\tau _P)_C$ or $(\tau _P)_{C} \preceq _{\tau _P} (\tau _P)_{B_2}$ for all $C \in \mathtt {yd}(\lambda /\mu )$ satisfying $(\tau _P)_{B_1} < (\tau _P)_C < (\tau _P)_{B_2}$ or $(\tau _P)_{B_1}> (\tau _P)_C > (\tau _P)_{B_2}$ .

Assume that there exists $C \in \mathtt {yd}(\lambda /\mu )$ such that $(\tau _P)_{B_1} < (\tau _P)_C < (\tau _P)_{B_2}$ . Since $\tau _P$ is a Schur labeling and $(\tau _P)_{B_1} \prec _{\tau _P} (\tau _P)_{B_2}$ , the inequality $(\tau _P)_{B_1} < (\tau _P)_{B_2}$ implies that $B_2$ is strictly below $B_1$ . In addition, since $\tau _P$ is distinguished and $(\tau _P)_{B_1},(\tau _P)_{B_2}$ appear in the same connected component in $\tau _P$ , $(\tau _P)_{C}$ appears in the same connected component with them in $\tau _P$ . Suppose for the sake of contradiction that $(\tau _P)_{B_1} \not \preceq _{\tau _P} (\tau _P)_C$ and $(\tau _P)_{C} \not \preceq _{\tau _P} (\tau _P)_{B_2}$ . Then C satisfies one of the following conditions:

1. (i) C is strictly above $B_1$ and strictly right of $B_2$ .

2. (ii) C is strictly left of $B_1$ and strictly below $B_2$ .

However, since $\tau _P$ is a Schur labeling and $(\tau _P)_{B_1} < (\tau _P)_C$ , C cannot satisfy (i). Similarly, since $\tau _P$ is a Schur labeling and $(\tau _P)_{C} < (\tau _P)_{B_2}$ , C cannot satisfy (ii). Therefore, $(\tau _P)_{B_1} \preceq _{\tau _P} (\tau _P)_C$ or $(\tau _P)_{C} \preceq _{\tau _P} (\tau _P)_{B_2}$ . In a similar way, one can show that if there exists $C \in \mathtt {yd}(\lambda /\mu )$ such that $(\tau _P)_{B_1}> (\tau _P)_C > (\tau _P)_{B_2}$ , then $(\tau _P)_{B_1} \preceq _{\tau _P} (\tau _P)_C$ or $(\tau _P)_{C} \preceq _{\tau _P} (\tau _P)_{B_2}$ . Thus, P is regular.

Proof. Let us show the inclusion $\{\mathsf {read}_\tau (T) \mid \tau \in \mathsf {DS}(\lambda /\mu )\} \subseteq [\mathsf {read}_{\tau _0}(T), \mathsf {read}_{\tau _1}(T)]_R$ . This can be done by proving $\mathsf {read}_{\tau _0}(T) \preceq _R \mathsf {read}_{\tau }(T)$ and $\mathsf {read}_{\tau }(T) \preceq _R \mathsf {read}_{\tau _1}(T)$ for all $\tau \in \mathsf {DS}(\lambda /\mu )$ . Since the method of proof for the latter inequality is essentially the same as that for the former one, we omit the proof for the latter inequality. Let $\tau \in \mathsf {DS}(\lambda /\mu )$ and $(i,j) \in \mathrm {Inv}_R(\mathsf {read}_{\tau _0}(T))$ . Since ${i> j}$ and $\mathsf {read}_{\tau _0}(T)^{-1}(i) < \mathsf {read}_{\tau _0}(T)^{-1}(j)$ , the box $T^{-1}(j)$ is placed strictly left and weakly below $T^{-1}(i)$ . This, together with the definition of distinguished Schur labeling, implies that $\tau _{T^{-1}(i)} < \tau _{T^{-1}(j)}$ ; equivalently, $\mathsf {read}_{\tau }(T)^{-1}(i) < \mathsf {read}_{\tau }(T)^{-1}(j)$ . It follows that $(i,j) \in \mathrm {Inv}_R(\mathsf {read}_{\tau }(T))$ . Since we chose an arbitrary $(i,j) \in \mathrm {Inv}_R(\mathsf {read}_{\tau _0}(T))$ , we have the inclusion $\mathrm {Inv}_R(\mathsf {read}_{\tau _0}(T)) \subseteq \mathrm {Inv}_R(\mathsf {read}_{\tau }(T))$ . Therefore, $\mathsf {read}_{\tau _0}(T) \preceq _R \mathsf {read}_{\tau }(T)$ .

Let us show the opposite inclusion $\{\mathsf {read}_\tau (T) \mid \tau \in \mathsf {DS}(\lambda /\mu )\} \supseteq [\mathsf {read}_{\tau _0}(T), \mathsf {read}_{\tau _1}(T)]_R$ . Since $\{\mathsf {read}_\tau (T) \mid \tau \text {is a bijective tableau of shape} \lambda / \mu \}$ is equal to $\mathfrak {S}_n$ as a set, the inclusion can be obtained by proving that $\mathsf {read}_\tau (T) \notin [\mathsf {read}_{\tau _0}(T), \mathsf {read}_{\tau _1}(T)]_R$ for all bijective tableaux $\tau $ of shape $\lambda /\mu $ with $\tau \notin \mathsf {DS}(\lambda /\mu )$ . To prove it, choose an arbitrary bijective tableau $\tau $ of shape $\lambda /\mu $ with $\tau \notin \mathsf {DS}(\lambda /\mu )$ . Since $\tau \notin \mathsf {DS}(\lambda /\mu )$ , there exists $ 1 \leq i < j \leq n$ such that i is weakly below and weakly left of j in $\tau $ . Set $x := T_{\tau ^{-1}(i)}$ and $y := T_{\tau ^{-1}(j)}$ . Then, x appears left of y in $\mathsf {read}_\tau (T)$ . However, since $\tau _0, \tau _1 \in \mathsf {DS}(\lambda /\mu )$ , we have $(\tau _0)_{\tau ^{-1}(i)}> (\tau _0)_{\tau ^{-1}(j)}$ and $(\tau _1)_{\tau ^{-1}(i)}> (\tau _1)_{\tau ^{-1}(j)}$ , which implies that x appears right of y in both $\mathsf {read}_{\tau _0}(T)$ and $\mathsf {read}_{\tau _1}(T)$ . If $x < y$ , then $(y, x) \in \mathrm {Inv}_R(\mathsf {read}_{\tau _0}(T))$ and $(y, x) \notin \mathrm {Inv}_R(\mathsf {read}_\tau (T))$ , thus $\mathrm {Inv}_R(\mathsf {read}_{\tau _0}(T)) \not \subseteq \mathrm {Inv}_R(\mathsf {read}_{\tau }(T))$ . Similarly, if $x> y$ , then $\mathrm {Inv}_R(\mathsf {read}_\tau (T)) \not \subseteq \mathrm {Inv}_R(\mathsf {read}_{\tau _1}(T))$ . Hence, $\mathsf {read}_\tau (T) \notin [\mathsf {read}_{\tau _0}(T), \mathsf {read}_{\tau _1}(T)]_R$ .

Proof. Due to Theorem 2.4 and Lemma 3.2, it suffices to show that $\mathsf {read}_{\tau _P}(T_{\lambda /\mu })$ is minimal and $\mathsf {read}_{\tau _P}(T^{\prime}_{\lambda /\mu })$ is maximal in $\mathsf {read}_{\tau _P}(\mathrm {SYT}(\lambda /\mu ))$ with respect to $\preceq _L$ . Let $T \in \mathrm {SYT}(\lambda / \mu )$ . In the case where $\tau _P = \tau _0$ , one can easily see that

(3.4) $$ \begin{align} \mathrm{Inv}_L(\mathsf{read}_{\tau_0}(T_{\lambda/\mu})) \subseteq \mathrm{Inv}_L(\mathsf{read}_{\tau_0}(T)) \subseteq \mathrm{Inv}_L(\mathsf{read}_{\tau_0}(T_{\lambda/\mu}')). \end{align} $$

In the case where $\tau _P \neq \tau _0$ , we consider the equality

(3.5) $$ \begin{align} \mathsf{read}_{\tau_P}(T) = \mathsf{read}_{\tau_0}(T) \mathsf{read}_{\tau_0}(\tau_P)^{-1} \end{align} $$

which follows from Definition 3.1. By Lemma 3.6, we have $\tau _P \in \mathsf {DS}(\lambda /\mu )$ ; therefore, combining Lemma 3.7 with Equation (3.5) yields that

$$ \begin{align*} \ell(\mathsf{read}_{\tau_P}(T)) = \ell(\mathsf{read}_{\tau_0}(T)) + \ell(\mathsf{read}_{\tau_0}(\tau_P)^{-1}). \end{align*} $$

Now, we have that

$$ \begin{align*} \ell(\mathsf{read}_{\tau_P}(T)) - \ell(\mathsf{read}_{\tau_P}(T_{\lambda/\mu})) & = \ell(\mathsf{read}_{\tau_0}(T)) - \ell(\mathsf{read}_{\tau_0}(T_{\lambda/\mu})) \\ & = \ell(\mathsf{read}_{\tau_0}(T) \mathsf{read}_{\tau_0}(T_{\lambda/\mu})^{-1}) & \text{by Equation }({3.4}) \\ & = \ell(\mathsf{read}_{\tau_P}(T) \mathsf{read}_{\tau_P}(T_{\lambda/\mu})^{-1}) & \text{by Definition }{3.1}. \end{align*} $$

Therefore, $\mathsf {read}_{\tau _P}(T_{\lambda /\mu }) \preceq _L \mathsf {read}_{\tau _P}(T)$ . In the same manner, we can prove that $\mathsf {read}_{\tau _P}(T) \preceq _L \mathsf {read}_{\tau _P}(T^{\prime}_{\lambda /\mu })$ .

Proof. Let $\gamma , \gamma ' \in I_1$ with $\gamma \preceq _L^c \gamma '$ . Since f is a poset isomorphism, we have $f(\gamma ) \preceq _L^c f(\gamma ')$ . Let $i,j \in [n-1]$ , satisfying $\gamma ' = s_i \gamma $ and $f(\gamma ') = s_j f(\gamma )$ . For the assertion, it suffices to show that $i = j$ . Let $D_1 := \mathrm {Des}_L(\gamma )$ and $D_2 := \mathrm {Des}_L(\gamma ')$ . Since $\gamma ' = s_i \gamma $ , we have

$$\begin{align*}\{i\} \subseteq (D_1 \cup D_2) \setminus (D_1 \cap D_2) \subseteq \{i-1, i, i+1\}. \end{align*}$$

In addition, since $D_1 = \mathrm {Des}_L(f(\gamma ))$ , $D_2 = \mathrm {Des}_L(f(\gamma '))$ and $f(\gamma ') = s_j f(\gamma )$ , we have $j \in D_2 \setminus D_1$ , and it follows that j is one of $i-1$ , i, and $i+1$ .

If $j = i-1$ , then $i-1, i \in D_2$ . It follows that

$$\begin{align*}\gamma' = \cdots \ i+1 \ \cdots \ i \ \cdots \ i-1 \ \cdots \quad \text{in one-line notation}; \end{align*}$$

equivalently,

$$\begin{align*}\gamma = \cdots \ i \ \cdots \ i+1 \ \cdots \ i-1 \ \cdots \quad \text{in one-line notation.} \end{align*}$$

This implies that $j=i-1 \in D_1$ , which is a contradiction to $j \notin \mathrm {Des}_L(f(\gamma ))$ . Therefore, $j\neq i-1$ .

In a similar manner, one can show that $j \neq i+1$ . Hence, $j = i$ , as required.

Proof. From the condition $\ell (\xi _C) = 1$ , it follows that $\xi _C = s_{i_0}$ for some $i_0 \in [n-1]$ .

First, let us prove that there exists a unique minimal element in $\overline {\mathrm {min}}(C)$ . Let

(4.1) $$ \begin{align} D_1 := \mathrm{Des}_L(\sigma) \quad \text{and} \quad D_2 := \mathrm{Des}_L(s_{i_0}\sigma) \quad \text{for any }[\sigma, s_{i_0}\sigma]_L \in C \end{align} $$

and

(4.2) $$ \begin{align} X := (D_1 \cup D_2) \setminus (D_1 \cap D_2). \end{align} $$

One sees that $\{i_0\} \subseteq X \subseteq \{i_0 - 1, i_0, i_0 + 1\}$ , and therefore, X can be one of the following:

$$\begin{align*}\{i_0-1, i_0, i_0+1\}, \ \ \{i_0\}, \ \ \{i_0-1, i_0\}, \ \ \text{and} \ \ \{i_0, i_0+1\}. \end{align*}$$

Case 1: $X = \{i_0 - 1, i_0, i_0 + 1\}$ . Since $i_0 - 1, i_0+1 \in D_1$ and $i_0 \notin D_1$ ,

$$\begin{align*}w_0(D_1) = \cdots \ i_0 \ i_0 - 1 \ \cdots \ i_0+2 \ i_0+1 \ \cdots \end{align*}$$

in one-line notation. Considering this equality, one can see that $[w_0(D_1), s_{i_0} w_0(D_1)]_L \in C$ . By Lemma 2.2, $w_0(D_1) \preceq _R \sigma $ for all $\sigma \in \mathfrak {S}_n$ with $\mathrm {Des}_L(\sigma ) = D_1$ . Thus, $w_0(D_1)$ is a unique minimal element in $\overline {\mathrm {min}}(C)$ .

Case 2: $X = \{i_0\}$ . In this case, we have

$$\begin{align*}w_0(D_2) = \cdots \ i_0 + 1 \ i_0 \cdots \ \quad \end{align*}$$

in one-line notation. Considering this equality, one can see that $[s_{i_0} w_0(D_2), w_0(D_2)]_L \in C$ . Again, by Lemma 2.2, $w_0(D_2) \preceq _R \sigma $ for all $\sigma \in \mathfrak {S}_n$ with $\mathrm {Des}_L(\sigma ) = D_2$ . Thus, $s_{i_0} w_0(D_2)$ is a unique minimal element in $\overline {\mathrm {min}}(C)$ .

Case 3: $X = \{i_0-1, i_0\}$ . When $i_0 + 1 \notin D_1$ , following the way as in Case 1, one can see that $w_0(D_1)$ is a unique minimal element in $\overline {\mathrm {min}}(C)$ .

From now on, assume that $i_0 + 1 \in D_1$ . We begin by introducing necessary notation. Let

$$\begin{align*}\mathsf{m}_1 := \min\{m \in [n - 1] \mid [m, i_0 - 1] \subseteq D_1 \} \quad \text{and} \quad \mathsf{m}_2 := \max\{m \in [n - 1] \mid [i_0 + 1, m] \subseteq D_1 \}. \end{align*}$$

And set

$$\begin{align*}p_1 := \mathsf{m}_1 - 1, \quad p_2 := \mathsf{m}_2 - i_0, \quad p_3 := i_0 - \mathsf{m}_1, \quad \text{and} \quad p_4 := n - (\mathsf{m}_2 + 1). \end{align*}$$

Let

• • $\mathbf {w}^{(1)}$ be the longest element of the subgroup $\mathfrak {S}_{D_1 \cap [p_1 - 1]}$ of $\mathfrak {S}_{p_1}$ ,

• • $\mathbf {w}^{(2)}$ be the longest element of $\mathfrak {S}_{p_2}$ ,

• • $\mathbf {w}^{(3)}$ be the longest element of $\mathfrak {S}_{p_3}$ , and

• • $\mathbf {w}^{(4)}$ be the longest element of the subgroup $\mathfrak {S}_{D_1 \cap [p_4 -1]}$ of $\mathfrak {S}_{p_4}$ .

With this notation, we define $\mathbf {w}_0$ to be the permutation given by

$$\begin{align*}\mathbf{w}_0(k) := \begin{cases} \mathbf{w}^{(1)}(k) & \text{if }k \in [p_1], \\ \mathbf{w}^{(2)}(k - p_1) + (i_0 + 1) & \text{if }k \in [p_1 +1, p_1 + p_2], \\ i_0 & \text{if }k = p_1 + p_2 + 1, \\ \mathbf{w}^{(3)}(k - (p_1 + p_2 + 1)) + (\mathsf{m}_1 - 1) & \text{if }k \in [p_1 + p_2 + 2, p_1 + p_2 + p_3 + 1], \\ i_0 + 1 & \text{if }k = p_1 + p_2 + p_3 + 2, \\ \mathbf{w}^{(4)}(k - (p_1 + p_2 + p_3 + 2)) + (\mathsf{m}_2 + 1) & \text{if }k \in [p_1 + p_2 + p_3 + 3, n]. \end{cases} \end{align*}$$

It should be remarked that

$$ \begin{align*} \mathbf{w}_0([1,p_1]) &= [1,\mathsf{m}_1-1], \\ \mathbf{w}_0([p_1+1,p_1+p_2]) & = [i_0+2,\mathsf{m}_2+1], \\ \mathbf{w}_0(p_1+p_2+1) &= i_0, \\ \mathbf{w}_0([p_1+p_2+2,p_1+p_2+p_3+1]) &= [\mathsf{m}_1, i_0-1], \\ \mathbf{w}_0(p_1+p_2+p_3+2) &= i_0+1, \quad \text{and} \\ \mathbf{w}_0([p_1+p_2+p_3+3,n]) &= [\mathsf{m}_2+2, n]. \end{align*} $$

From the definition of $\mathbf {w}_0$ , it follows that $[\mathbf {w}_0, s_{i_0} \mathbf {w}_0]_L \in C$ ; equivalently, $\mathbf {w}_0 \in \overline {\mathrm {min}}(C)$ . We claim that $\mathbf {w}_0$ is a unique minimal element in $\overline {\mathrm {min}}(C)$ . This can be verified by showing that every minimal element in $\overline {\mathrm {min}}(C)$ is equal to $\mathbf {w}_0$ . Let $\sigma _0$ be a minimal element in $\overline {\mathrm {min}}(C)$ . Set

$$ \begin{align*} \begin{array}{l} \mathcal{I}_{\mathrm{L}} := \{\sigma_0(k) \mid 1 \le k < \sigma_0^{-1}(i_0)\}, \\[.5ex] \mathcal{I}_{\mathrm{C}} := \{\sigma_0(k) \mid \sigma_0^{-1}(i_0) < k < \sigma_0^{-1}(i_0+1) \}, \\[.5ex] \mathcal{I}_{\mathrm{R}} := \{\sigma_0(k) \mid \sigma_0^{-1}(i_0 + 1) < k \le n\}. \end{array} \end{align*} $$

To begin with, we establish the equalities

(4.3) $$ \begin{align} \mathcal{I}_{\mathrm{L}} = [\mathsf{m}_1 -1] \cup [i_0+2, \mathsf{m}_2+1], \quad \mathcal{I}_{\mathrm{C}} = [\mathsf{m}_1, i_0 - 1], \quad \text{and} \quad \mathcal{I}_{\mathrm{R}} = [\mathsf{m}_2 + 2,n]. \end{align} $$

These equalities are derived by verifying the following claims.

Claim 1. $\mathcal {I}_{\mathrm {C}} = [\mathsf {m}_1, i_0 - 1]$ . Let us show $\mathcal {I}_{\mathrm {C}} \subseteq [\mathsf {m}_1, i_0 - 1]$ . First, to prove $\mathcal {I}_{\mathrm {C}} \subseteq [i_0 - 1]$ , we assume that there exists $i \in \mathcal {I}_{\mathrm {C}}$ such that $i \geq i_0$ . Let

$$ \begin{align*} k_1 & := \mathrm{max}\{k \in [n] \mid \sigma_0(k) \in \mathcal{I}_{\mathrm{C}} \ \text{and} \ \sigma_0(k) \ge i_0 \}. \end{align*} $$

By the definition of $\mathcal {I}_{\mathrm {C}}$ , $i_0,i_0+1 \notin \mathcal {I}_{\mathrm {C}}$ . If $i_0 + 2 \in \mathcal {I}_{\mathrm {C}}$ , then $i_0 + 1 \in D_1 \setminus D_2$ , which contradicts the assumption that $i_0 +1 \notin X$ . Since $\sigma _0(k_1) \in \mathcal {I}_{\mathrm {C}}$ , it follows that $\sigma _0(k_1) \geq i_0 + 3$ . And, by the choice of $k_1$ , we have $\sigma _0(k_1 + 1) \leq i_0+1$ . Putting these together yields that $[\sigma _0 s_{k_1}, s_{i_0} \sigma _0s_{k_1}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k_1} \prec _R \sigma _0$ . This contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ ; therefore, $\mathcal {I}_{\mathrm {C}} \subseteq [i_0 - 1]$ . Next, to prove $\mathcal {I}_{\mathrm {C}} \subseteq [\mathsf {m}_1, n]$ , we assume that there exists $i \in \mathcal {I}_{\mathrm {C}}$ such that $i < \mathsf {m}_1$ . Let

$$\begin{align*}k_2 := \mathrm{min}\{ k \in [n] \mid \sigma_0(k) \in \mathcal{I}_{\mathrm{C}} \ \text{and} \ \sigma_0(k) < \mathsf{m}_1\}. \end{align*}$$

By the choice of $k_2$ , we have $\sigma _0(k_2) + 1 \le \mathsf {m}_1 \le \sigma _0(k_2 - 1)$ . In addition, if $\sigma _0(k_2) + 1 = \mathsf {m}_1 = \sigma _0(k_2 - 1)$ , then $\mathsf {m}_1 - 1 \in \mathrm {Des}_L(\sigma _0)$ , which cannot happen by the definition of $\mathsf {m}_1$ . Therefore, $\sigma _0(k_2) + 1 < \sigma _0(k_2 - 1)$ , which implies that $[\sigma _0 s_{k_2 - 1}, s_{i_0} \sigma _0 s_{k_2 - 1}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k_2 - 1} \prec _R \sigma _0$ . This contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ . Thus, $\mathcal {I}_{\mathrm {C}} \subseteq [\mathsf {m}_1, i_0 - 1]$ .

Let us show $\mathcal {I}_{\mathrm {C}} \supseteq [\mathsf {m}_1, i_0 - 1]$ . Assume for the sake of contradiction that there exists $i \in [\mathsf {m}_1, i_0 - 1]$ such that $i \notin \mathcal {I}_{\mathrm {C}}$ . Let j be the maximal element in $[\mathsf {m}_1, i_0 - 1]$ such that $j \notin \mathcal {I}_{\mathrm {C}}$ . Since $i_0 - 1 \in X$ , we have $i_0 - 1 \in \mathcal {I}_{\mathrm {C}}$ . It follows that $j < i_0 - 1$ , so $j+1 \in [\mathsf {m}_1,i_0-1]$ . Combining this with the maximality of j, we have $j+1 \in \mathcal {I}_{\mathrm {C}}$ . And, by the definition of $\mathsf {m}_1$ , we have $j \in D_1$ . Putting these together yields that $j \in \mathcal {I}_{\mathrm {R}}$ . Let

$$\begin{align*}k_3 := \mathrm{min}\{ k \in [n] \mid \sigma_0(k) \in \mathcal{I}_{\mathrm{R}} \ \text{and} \ \sigma_0(k) \leq j \}. \end{align*}$$

If $\sigma _0(k_3 -1) \le \sigma _0(k_3) + 1$ , then $\sigma _0(k_3 -1) \le j + 1 < i_0$ . So, $\sigma _0(k_3 - 1) \neq i_0 + 1$ , which implies $\sigma _0(k_3 -1) \in \mathcal {I}_{\mathrm {R}}$ . This, together with the minimality of $k_3$ , yields that $j+1 \le \sigma _0(k_3 -1)$ . It follows that $\sigma _0(k_3 - 1) = j + 1$ , which is a contradiction because $\sigma _0(k_3 -1) \in \mathcal {I}_{\mathrm {R}}$ , but $j+1 \in \mathcal {I}_{\mathrm {C}}$ . Therefore, we have $\sigma _0(k_3) + 1 < \sigma _0(k_3-1)$ . In addition, since $j < i_0 - 1$ , we have $\sigma _0(k_3) < i_0 - 1$ . Putting these together yields that $[\sigma _0 s_{k_3 - 1}, s_{i_0} \sigma _0 s_{k_3 - 1}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k_3 - 1} \prec _R \sigma _0$ , which contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ . Thus, $\mathcal {I}_{\mathrm {C}} \supseteq [\mathsf {m}_1, i_0 - 1]$ .

Claim 2. $[\mathsf {m}_1 -1] \cup [i_0+2, \mathsf {m}_2+1] \subseteq \mathcal {I}_{\mathrm {L}}$ . By the definition of $\mathsf {m}_2$ , we have $[i_0 + 1, \mathsf {m}_2] \subseteq \mathrm {Des}_L(\sigma _0)$ . Since $i_0+2 \notin \mathcal {I}_{\mathrm {C}}$ , we have $[i_0+2, \mathsf {m}_2+1] \subseteq \mathcal {I}_{\mathrm {L}}$ . To prove $[\mathsf {m}_1 - 1] \subseteq \mathcal {I}_{\mathrm {L}}$ , suppose that there exists $i \in [\mathsf {m}_1 - 1]$ such that $i \notin \mathcal {I}_{\mathrm {L}}$ . Let

$$\begin{align*}k_4 := \mathrm{min}\{k \in [n] \mid \sigma_0(k) \in [\mathsf{m}_1 - 1] \ \text{and} \ \sigma_0(k) \notin \mathcal{I}_{\mathrm{L}}\}. \end{align*}$$

Since $\sigma _0(k_4) \notin \mathcal {I}_{\mathrm {L}}$ and $\sigma _0(k_4) < \mathsf {m}_1$ , we have $\sigma _0(k_4) \in \mathcal {I}_{\mathrm {R}}$ . This implies that $\sigma _0(k_4-1) \notin \mathcal {I}_{\mathrm {L}} \cup \{i_0\} \cup \mathcal {I}_{\mathrm {C}}$ . In addition, the minimality of $k_4$ gives $\sigma _0(k_4 - 1) \geq \mathsf {m}_1$ . Since $[\mathsf {m}_1, i_0-1] = \mathcal {I}_{\mathrm {C}}$ , we have $\sigma _0(k_4 - 1) \geq i_0+1$ . Putting the above inequalities together, we have

$$\begin{align*}\sigma_0(k_4) < \mathsf{m}_1 \le i_0-1 < i_0 + 1 \le \sigma_0(k_4-1), \end{align*}$$

and so $\sigma _0(k_4) + 2 < \sigma _0(k_4 - 1)$ . It follows that $[\sigma _0 s_{k_4 - 1}, s_{i_0} \sigma _0 s_{k_4 - 1}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k_4 - 1} \prec _R \sigma _0$ . This contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ ; thus, $[\mathsf {m}_1 - 1] \subseteq \mathcal {I}_{\mathrm {L}}$ .

Claim 3. $[\mathsf {m}_2 + 2,n] \subseteq \mathcal {I}_{\mathrm {R}}$ . Suppose that there exists $i \in [\mathsf {m}_2 + 2,n]$ such that $i \notin \mathcal {I}_{\mathrm {R}}$ . Let

$$\begin{align*}k_5 := \mathrm{max}\{k \in [n] \mid \sigma_0(k) \notin \mathcal{I}_{\mathrm{R}} \ \ \text{and} \ \ \sigma_0(k) \in [\mathsf{m}_2 + 2,n]\}. \end{align*}$$

Since $k_5 \notin \mathcal {I}_{\mathrm {R}}$ and $\sigma _0(k_5)> i_0 + 1$ , we have $\sigma _0(k_5) \in \mathcal {I}_{\mathrm {L}}$ , which implies that $\sigma _0(k_5 + 1) \notin \mathcal {I}_{\mathrm {R}}$ . By the maximality of $k_5$ , we have $\sigma _0(k_5+1) \le \mathsf {m}_2 + 1$ . If $\sigma _0(k_5+1) < \mathsf {m}_2 + 1$ , then

$$\begin{align*}\sigma_0(k_5+1) +1 < \mathsf{m}_2 + 2 \le \sigma_0(k_5). \end{align*}$$

If $\sigma _0(k_5+1) = \mathsf {m}_2 + 1$ , then $\sigma _0^{-1}(\mathsf {m}_2 + 2)> k_5+1$ due to the maximality of $\mathsf {m}_2$ , so $\sigma _0(k_5) \neq \mathsf {m}_2 + 2$ , which implies

$$\begin{align*}\sigma_0(k_5+1) + 1 < \sigma_0(k_5). \end{align*}$$

Putting these together with the inequalities $\sigma _0(k_5) \ge \mathsf {m}_2 + 2>i_0 + 2$ yields that $[\sigma _0 s_{k_5}, s_{i_0} \sigma _0 s_{k_5}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k_5} \prec _R \sigma _0$ . This contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ ; thus, $[\mathsf {m}_2 + 2, n] \subseteq \mathcal {I}_{\mathrm {R}}$ .

Now, we are ready to show that $\sigma _0 = \mathbf {w}_0$ . Let

$$\begin{align*}\mathcal{I}^{(1)}_{\mathrm{L}} := \{ \sigma_0(k) \in \mathcal{I}_{\mathrm{L}} \mid 1 \le k \le \mathsf{m}_1-1 \} \quad \text{and} \quad \mathcal{I}^{(2)}_{\mathrm{L}} := \{ \sigma_0(k) \in \mathcal{I}_{\mathrm{L}} \mid \mathsf{m}_1 \le k <\sigma_0^{-1}(i_0) \}. \end{align*}$$

We claim that $\mathcal {I}^{(1)}_{\mathrm {L}} = [\mathsf {m}_1-1]$ and $\mathcal {I}^{(2)}_{\mathrm {L}} = [i_0+2, \mathsf {m}_2+1]$ . We may assume that $\mathsf {m}_1> 1$ ; otherwise, the claim is obvious. To prove our claim, suppose that there exists $i \in \mathcal {I}^{(1)}_{\mathrm {L}}$ such that $i \in [i_0+2,\mathsf {m}_2+1]$ . Then, there exists $1 \le k < \sigma _0^{-1}(i_0) - 1$ such that $\sigma _0(k) \in [i_0+2, \mathsf {m}_2 + 1]$ and $\sigma _0(k + 1) \in [\mathsf {m}_1 - 1]$ . It follows that $[\sigma _0 s_{k}, s_{i_0} \sigma _0 s_{k}]_L \overset {D}{\simeq } [\sigma _0, s_{i_0}\sigma _0]_L$ and $\sigma _0 s_{k} \prec _R \sigma _0$ . Again, this contradicts the minimality of $\sigma _0$ in $\overline {\mathrm {min}}(C)$ , so

(4.4) $$ \begin{align} \mathcal{I}^{(1)}_{\mathrm{L}} = [\mathsf{m}_1-1] \quad \text{and} \quad \mathcal{I}^{(2)}_{\mathrm{L}} = [i_0+2, \mathsf{m}+1]. \end{align} $$

Putting Lemma 2.2, Equation (4.3), Equation (4.4) and the minimality of $\sigma _0$ together, we conclude that $\sigma _0 = \mathbf {w}_0$ .

Case 4: $X = \{i_0, i_0 + 1\}$ . Take $[\sigma , s_{i_0} \sigma ]_L \in C$ and let $C'$ be the equivalence class of $[\sigma ^{w_0}, (s_{i_0} \sigma )^{w_0}]_L$ . By mimicking Equation (4.1) and Equation (4.2), we define

$$\begin{align*}D^{\prime}_1 := \mathrm{Des}_L(\sigma^{w_0}), \quad D^{\prime}_2 := \mathrm{Des}_L((s_{i_0}\sigma)^{w_0}), \quad \text{and} \quad X' := (D^{\prime}_1 \cup D^{\prime}_2) \setminus (D^{\prime}_1 \cap D^{\prime}_2). \end{align*}$$

Since $D^{\prime}_1 = \{ n-i \mid i \in D_1\}$ and $D^{\prime}_2 = \{ n-i \mid i \in D_2\}$ , we have

$$\begin{align*}X' = \{n-i_0, (n-i_0)+1\}. \end{align*}$$

Following the proof of Case 3, we see that $\overline {\mathrm {min}}(C')$ has a unique minimal element $\mathbf {w}^{\prime}_0$ . And one can easily see that the map $f: \overline {\mathrm {min}}(C) \rightarrow \overline {\mathrm {min}}(C')$ , $\gamma \mapsto \gamma ^{w_0}$ is a well-defined bijection and that for $\gamma _1, \gamma _2 \in \overline {\mathrm {min}}(C)$ , $\gamma _1 \preceq _R \gamma _2$ if and only if $f(\gamma _1) \preceq _R f(\gamma _2)$ . Thus, $(\mathbf {w}^{\prime}_0)^{w_0}$ is a unique minimal element in $\overline {\mathrm {min}}(C')$ .

Second, we will show that $\overline {\mathrm {min}}(C)$ has a unique maximal element. Recall that we take $[\sigma , s_{i_0} \sigma ]_L \in C$ . Let $C"$ be the equivalence class of $[s_{i_0} \sigma w_0, \sigma w_0]_L$ . Due to the previous arguments, we know that there is a unique minimal element $\gamma _0$ in $\overline {\mathrm {min}}(C")$ . One can easily see that the map $g: \overline {\mathrm {min}}(C) \rightarrow \overline {\mathrm {min}}(C")$ , $\gamma \mapsto \gamma w_0$ is a well-defined bijection and that for $\gamma _1, \gamma _2 \in \overline {\mathrm {min}}(C)$ , $\gamma _1 \preceq _R \gamma _2$ if and only if $g(\gamma _1) \succeq _R g(\gamma _2)$ . Therefore, $\gamma _0 w_0$ is the unique maximal element in $\overline {\mathrm {min}}(C)$ .

Finally, we will show that $\overline {\mathrm {min}}(C)$ is a right weak Bruhat interval in $(\mathfrak {S}_n, \preceq _R)$ . Let $\sigma _0$ and $\sigma _1$ be the minimal and maximal elements in $\overline {\mathrm {min}}(C)$ , respectively. Let $\gamma \in [\sigma _0, \sigma _1]_R$ . Since $\mathrm {Des}_L(\sigma _0) = \mathrm {Des}_L(\sigma _1)$ , we have $\mathrm {Des}_L(\gamma ) = \mathrm {Des}_L(\sigma _0)$ by Lemma 2.2. Next, let us examine $\mathrm {Des}_L(s_{i_0}\gamma )$ . Since $\mathrm {Des}_L(\sigma _0) = \mathrm {Des}_L(\gamma )$ , it follows that $\gamma \preceq _L s_{i_0} \gamma $ . By Lemma 2.1, we have that $s_{i_0} \gamma \in [s_{i_0}\sigma _0, s_{i_0}\sigma _1]_{R}$ . Since $\mathrm {Des}_L(s_{i_0} \sigma _0) =\mathrm {Des}_L(s_{i_0} \sigma _1)$ , we have $\mathrm {Des}_L(s_{i_0}\gamma ) = \mathrm {Des}_L(s_{i_0}\sigma _0)$ by Lemma 2.2. Thus, ${\gamma \in \overline {\mathrm {min}}(C)}$ .

Proof. It is well known that $(\mathfrak {S}_n, \preceq _R)$ is a lattice; that is, every two-element subset $\{\gamma _1,\gamma _2\} \subseteq \mathfrak {S}_n$ has the least upper bound and greatest lower bound (for example, see [Reference Björner and Brenti6, Section 3.2]). Combining this with the fact $|\mathfrak {S}_n| < \infty $ , we derive the desired result.

Proof. Note that $\sigma \preceq _L \xi _C \sigma $ for any $\sigma \in \overline {\mathrm {min}}(C)$ and that $\overline {\mathrm {max}}(C) = \xi _C \cdot \overline {\mathrm {min}}(C)$ . If we prove that $\overline {\mathrm {min}}(C)$ is a right weak Bruhat interval, then Lemma 2.1 implies that $\overline {\mathrm {max}}(C)$ is also a right weak Bruhat interval. So we will only prove that $\overline {\mathrm {min}}(C)$ is a right weak Bruhat interval.

When $\ell (\xi _C) = 0$ , the assertion follows from Lemma 2.2. From now on, assume that $\ell (\xi _C) \geq 1$ . We will prove the assertion by using mathematical induction on $\ell (\xi _C)$ . When $\ell (\xi _C) = 1$ , the assertion is true by Lemma 4.3. Let k be an arbitrary positive integer and suppose that the assertion holds for every equivalence class $C \in \mathscr {C}(n)$ with $\ell (\xi _C) \leq k$ . Let $C \in \mathscr {C}(n)$ with $\ell (\xi _C) = k+1$ . Set

$$ \begin{align*}\mathcal{A} := \{i \in [n-1] \mid s_i \in [\mathrm{id}, \xi_C]_L \}. \end{align*} $$

Given $i \in \mathcal {A}$ and $\sigma \in \overline {\mathrm {min}}(C)$ , note that

$$ \begin{align*} [\sigma, s_i \sigma]_L \overset{D}{\simeq} [\sigma', s_i \sigma']_L \quad \text{and} \quad [s_i \sigma, \xi_C \sigma]_L \overset{D}{\simeq} [s_i\sigma', \xi_C \sigma']_L \end{align*} $$

for all $\sigma ' \in \overline {\mathrm {min}}(C)$ . This says that the equivalence classes of $[\sigma , s_i \sigma ]_L$ and $[s_i\sigma , \xi _C \sigma ]_L$ do not depend on $\sigma \in \overline {\mathrm {min}}(C)$ . For each $i \in \mathcal {A}$ , we set

$$ \begin{align*} E_i & := \text{the equivalence class of }[\sigma, s_i \sigma]_L, \\ E^{\prime}_i & := \text{the equivalence class of }[s_i \sigma, \xi_C \sigma]_L \end{align*} $$

for any $\sigma \in \overline {\mathrm {min}}(C)$ . Then, we set

$$ \begin{align*} J_i := \overline{\mathrm{max}}(E_i) \cap \overline{\mathrm{min}}(E^{\prime}_i) \quad \text{for }i \in \mathcal{A}, \quad \text{and} \quad J := \bigcap_{i \in \mathcal{A}}s_i \cdot J_i. \end{align*} $$

Now, the desired assertion can be achieved by proving the following claims:

1. (i) $\overline {\mathrm {min}}(C) = J$ .

2. (ii) J is a right weak Bruhat interval.

First, let us prove $\overline {\mathrm {min}}(C) = J$ . By the definition of $J_i$ , we have $s_i \sigma \in J_i$ for all $i \in \mathcal {A}$ and $\sigma \in \overline {\mathrm {min}}(C)$ . It follows that $\overline {\mathrm {min}}(C) \subseteq J$ . To prove the opposite inclusion $\overline {\mathrm {min}}(C) \supseteq J$ , take $\sigma \in J$ . By the definition of J, we have

$$ \begin{align*} [\sigma, s_i \sigma]_L \in E_i \quad \text{and} \quad [s_i \sigma ,\xi_C \sigma]_L \in E^{\prime}_i \quad \text{for all }i \in \mathcal{A}. \end{align*} $$

And Lemma 2.1 implies that

$$ \begin{align*}\{s_i \sigma \mid i \in \mathcal{A} \} = \{\gamma \in [\sigma, \xi_C\sigma]_L \mid \sigma \prec^{\mathrm{c}}_L \gamma \}. \end{align*} $$

Putting these together yields that $[\sigma , \xi _C \sigma ]_L \in C$ ; therefore, $\sigma \in \overline {\mathrm {min}}(C)$ .

Next, let us prove that J is a right weak Bruhat interval. Due to Lemma 4.5, it suffices to show that $s_i \cdot J_i$ is a right weak Bruhat interval for $i \in \mathcal {A}$ . Let us fix $i \in \mathcal {A}$ . Since $\ell (\xi _{E_i}) = 1$ and $\ell (\xi _{E^{\prime}_i}) = k$ , $\overline {\mathrm {max}}(E_i)$ and $\overline {\mathrm {min}}(E^{\prime}_i)$ are right weak Bruhat intervals by the induction hypothesis. Combining this with Lemma 4.5 yields that $J_i$ is a right weak Bruhat interval. In addition, we have $s_i \gamma \preceq _L \gamma $ for all $\gamma \in J_i$ . Therefore, by Lemma 2.1, $s_i \cdot J_i$ is a right weak Bruhat interval.

Proof. Let $\lambda /\mu = \mathrm {sh}(\tau _P)$ . Combining Equation (3.3) with Theorem 3.9 yields that

$$\begin{align*}\{\Sigma_L(Q) \mid Q \in \mathsf{RSP}_n\text{ with }\mathrm{sh}(\tau_Q) = \lambda / \mu \} = \{[\mathsf{read}_{\tau}(T_{\lambda/\mu}), \mathsf{read}_{\tau}(T^{\prime}_{\lambda/\mu})]_L \mid \tau \in \mathsf{DS}(\lambda/\mu) \}. \end{align*}$$

Therefore, for the assertion, we have only to show the equality

$$ \begin{align*} C = \{[\mathsf{read}_{\tau}(T_{\lambda/\mu}), \mathsf{read}_{\tau}(T^{\prime}_{\lambda/\mu})]_L \mid \tau \in \mathsf{DS}(\lambda/\mu) \}. \end{align*} $$

First, let us show that $\{[\mathsf {read}_{\tau }(T_{\lambda /\mu }), \mathsf {read}_{\tau }(T^{\prime}_{\lambda /\mu })]_L \mid \tau \in \mathsf {DS}(\lambda /\mu ) \} \subseteq C$ . This can be done by proving that for $\tau \in \mathsf {DS}(\lambda /\mu )$ , the map

$$ \begin{align*} f_{P;\tau}: [\mathsf{read}_{\tau_P}(T_{\lambda/\mu}), \mathsf{read}_{\tau_P}(T^{\prime}_{\lambda/\mu})]_L & \rightarrow [\mathsf{read}_{\tau}(T_{\lambda/\mu}), \mathsf{read}_{\tau}(T^{\prime}_{\lambda/\mu})]_L \\ \mathsf{read}_{\tau_P}(T) &\mapsto \mathsf{read}_{\tau}(T) \quad (T\in \mathrm{SYT}(\lambda / \mu)) \end{align*} $$

is a descent-preserving isomorphism. Let us fix $\tau \in \mathsf {DS}(\lambda /\mu )$ . The definition of $\tau $ -reading implies that for any $T_1,T_2 \in \mathrm {SYT}(\lambda /\mu )$ ,

$$ \begin{align*}\mathsf{read}_{\tau}(T_1) \preceq_L^{\mathrm{c}} \mathsf{read}_{\tau}(T_2) \quad \text{if and only if} \quad \mathsf{read}_{\tau_P}(T_1) \preceq_L^{\mathrm{c}} \mathsf{read}_{\tau_P}(T_2), \end{align*} $$

and therefore, $f_{P;\tau }$ is a poset isomorphism. To show that $f_{P;\tau }$ is descent-preserving, choose arbitrary $T \in \mathrm {SYT}(\lambda /\mu )$ and $i \in \mathrm {Des}_L(\mathsf {read}_{\tau _P}(T))$ . Combining the conditions $T \in \mathrm {SYT}(\lambda /\mu )$ and $\tau _P \in \mathsf {DS}(\lambda /\mu )$ with $i \in \mathrm {Des}_L(\mathsf {read}_{\tau _P}(T))$ yields that $i+1$ appears weakly above and strictly right of i in T. It follows that $i \in \mathrm {Des}_L(\mathsf {read}_{\tau }(T))$ , so $\mathrm {Des}_L(\mathsf {read}_{\tau _P}(T)) \subseteq \mathrm {Des}_L(\mathsf {read}_{\tau }(T))$ . In the same manner, one can show that $\mathrm {Des}_L(\mathsf {read}_{\tau }(T)) \subseteq \mathrm {Des}_L(\mathsf {read}_{\tau _P}(T))$ . Therefore, $f_{P;\tau }$ is a descent-preserving isomorphism.

Next, let us show $C \subseteq \{[\mathsf {read}_{\tau }(T_{\lambda /\mu }), \mathsf {read}_{\tau }(T^{\prime}_{\lambda /\mu })]_L \mid \tau \in \mathsf {DS}(\lambda /\mu ) \}$ . In the previous paragraph, we prove that $[\mathsf {read}_{\tau }(T_{\lambda /\mu }), \mathsf {read}_{\tau }(T^{\prime}_{\lambda /\mu })]_L \in C$ for any $\tau \in \mathsf {DS}(\lambda /\mu )$ . This implies that $\mathsf {read}_{\tau }(T^{\prime}_{\lambda /\mu }) = \xi _C \mathsf {read}_{\tau }(T_{\lambda /\mu })$ , and so it suffices to show that

$$ \begin{align*}\overline{\mathrm{min}}(C) \subseteq \{\mathsf{read}_\tau(T_{\lambda/\mu}) \mid \tau \in \mathsf{DS}(\lambda / \mu)\}. \end{align*} $$

Due to Lemma 3.7, this inclusion can be obtained by proving

$$ \begin{align*}\gamma \in [\mathsf{read}_{\tau_0}(T_{\lambda/\mu}), \mathsf{read}_{\tau_1}(T_{\lambda/\mu})]_R \quad \text{for any }\gamma \in \overline{\mathrm{min}}(C). \end{align*} $$

Let $\gamma \in \overline {\mathrm {min}}(C)$ . Since $\mathsf {read}_{\tau _0}(T_{\lambda /\mu }) \in \overline {\mathrm {min}}(C)$ , we have $\mathrm {Des}_L(\mathsf {read}_{\tau _0}(T_{\lambda /\mu })) = \mathrm {Des}_L(\gamma )$ . In addition, by the definitions of $\tau _0$ and $T_{\lambda /\mu }$ , we have $\mathsf {read}_{\tau _0}(T_{\lambda /\mu }) = w_0(\alpha ^{\mathrm {c}})$ , where $\alpha = (\lambda _1 - \mu _1, \lambda _2 - \mu _2, \ldots , \lambda _{\ell (\lambda )} - \mu _{\ell (\lambda )})$ . Putting these equalities together with Lemma 2.2 yields that $\mathsf {read}_{\tau _0}(T_{\lambda /\mu }) \preceq _{R} \gamma $ . Similarly, we have

$$\begin{align*}\mathrm{Des}_L(\mathsf{read}_{\tau_1}(T^{\prime}_{\lambda/\mu})) = \mathrm{Des}_L(\xi_C\gamma) \quad \text{and} \quad \mathsf{read}_{\tau_1}(T^{\prime}_{\lambda/\mu}) = w_0(\beta^{\mathrm{c}})w_0, \end{align*}$$

where $\beta = (\lambda ^{\mathrm {t}}_1 - \mu ^{\mathrm {t}}_1, \lambda ^{\mathrm {t}}_2 - \mu ^{\mathrm {t}}_2, \ldots , \lambda ^{\mathrm {t}}_{\ell (\lambda ^{\mathrm {t}})} - \mu ^{\mathrm {t}}_{\ell (\lambda ^{\mathrm {t}})})$ . This, together with Lemma 2.2, yields that $\xi _C\gamma \preceq _{R} \mathsf {read}_{\tau _1}(T^{\prime}_{\lambda /\mu })$ . Since $\mathsf {read}_{\tau _1}(T_{\lambda /\mu }) \preceq _L \mathsf {read}_{\tau _1}(T^{\prime}_{\lambda /\mu }) = \xi _C \mathsf {read}_{\tau _1}(T_{\lambda /\mu })$ , we have $\gamma \preceq _{R} \mathsf {read}_{\tau _1}(T_{\lambda /\mu })$ . Therefore, $\gamma \in [\mathsf {read}_{\tau _0}(T_{\lambda /\mu }), \mathsf {read}_{\tau _1}(T_{\lambda /\mu })]_R$ , as desired.

Proof. By the definition of ${\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\lambda /\mu )$ , the assertion (1) is clear. In addition, one can easily see that $\mathrm {Des}_R(\mathsf {read}_{\tau _0}(T^{\prime}_{\lambda /\mu })) = \mathrm {set}({\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\lambda /\mu )_{\odot }^{\mathrm {c}})$ . So, by Lemma 2.2, the assertion (2) follows.

Proof. The ‘if’ part follows from Proposition 4.1 and Theorem 4.7. To prove the ‘only if’ part, suppose that $\mathsf {M}_{P} \cong \mathsf {M}_{Q}$ . For simplicity, let $\lambda /\mu = \mathrm {sh}(\tau _{P})$ and $\nu /\kappa = \mathrm {sh}(\tau _{Q})$ . By Lemma 5.4, $\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\lambda /\mu )} \cong \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\nu /\kappa )}$ and $\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\lambda /\mu )} \cong \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\nu /\kappa )}$ , and therefore, ${\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\lambda /\mu ) = {\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\nu /\kappa )$ and ${\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\lambda /\mu ) = {\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\nu /\kappa )$ . Since ${\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\lambda /\mu ) = {\boldsymbol {\unicode{x3b1} }}_{\mathrm {proj}}(\nu /\kappa )$ , the number of boxes in the same row of $\mathtt {yd}(\lambda /\mu )$ and $\mathtt {yd}(\nu /\kappa )$ are the same. Similarly, since ${\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\lambda /\mu ) = {\boldsymbol {\unicode{x3b1} }}_{\mathrm {inj}}(\nu /\kappa )$ , the number of boxes in the same column of $\mathtt {yd}(\lambda /\mu )$ and $\mathtt {yd}(\nu /\kappa )$ are same. Thus, we have $\lambda /\mu = \nu /\kappa $ .