Proof. For simplicity, let $\mathrm {SRT}(\alpha )^{(i)}$ be the set of $\mathrm {SRT}$ x of shape $\alpha $ such that the topmost entry in the ith column is greater than that in the $(i+1)$ st column.

We first show that $H_n(0) \cdot T^{(i)}_{\alpha }$ is included in the $\mathbb C$ -span of $\mathrm {SRT}(\alpha )^{(i)}$ , equivalently $\pi _{\sigma } \cdot T^{(i)}_{\alpha } \in \mathrm {SRT}(\alpha )^{(i)} \cup \{0\}$ for all $\sigma \in \Sigma _n$ . Suppose that there exists $\sigma \in \Sigma _n$ such that $\pi _{\sigma } \cdot T^{(i)}_{\alpha } \neq 0$ and $\pi _{\sigma } \cdot T^{(i)}_{\alpha } \notin \mathrm {SRT}(\alpha )^{(i)}$ . Let $\sigma _0$ be such a permutation with minimal length and j a left descent of $\sigma _0$ . By the minimality of $\sigma _0$ , we have $\pi _{s_j \sigma _0} \cdot T^{(i)}_{\alpha } \in \mathrm {SRT}(\alpha )^{(i)}$ , and therefore

$$\begin{align*}(\pi_{s_j \sigma_0} \cdot T^{(i)}_{\alpha})^1_{i}> (\pi_{s_j \sigma_0} \cdot T^{(i)}_{\alpha})^1_{i+1}.\end{align*}$$

By the definition of the $\pi _j$ -action on $\mathrm {SRT}(\alpha )$ , we have

$$\begin{align*}( \pi_j \cdot (\pi_{s_j \sigma_0} \cdot T^{(i)}_{\alpha}) )^1_i> ( \pi_j \cdot (\pi_{s_j \sigma_0} \cdot T^{(i)}_{\alpha}) )^1_{i+1}.\end{align*}$$

However, since $\pi _j \cdot (\pi _{s_j \sigma _0} \cdot T^{(i)}_{\alpha }) = \pi _{\sigma _0} \cdot T^{(i)}_{\alpha }$ , this contradicts the assumption that $\pi _{\sigma _0} \cdot T^{(i)}_{\alpha } \notin \mathrm {SRT}(\alpha )^{(i)}$ .

We next show the opposite inclusion $\mathrm {SRT}(\alpha )^{(i)} \subseteq H_n(0) \cdot T^{(i)}_{\alpha }$ . Our strategy is to use [Reference Huang20, Theorem 3.3], which implicitly says that for $T_1, T_2 \in \mathrm {SRT}(\alpha )$ , $T_2 \in H_n(0) \cdot T_1$ if and only if $\mathbf {w}(T_1) \preceq _L \mathbf {w}(T_2)$ . Here, $\mathbf {w}(T_i) \,(i=1,2)$ denotes the word obtained from $T_i$ by reading the entries from left to right starting with the bottom row. For each $T \in \mathrm {SRT}(\alpha )^{(i)}$ , we define $\tau _T$ to be the filling of $\mathtt {rd}({\boldsymbol {\unicode{x3b1} }}^{(i)})$ whose entries in each column are increasing from top to bottom and whose columns are given as follows: for $1 \le p \le \ell (\alpha )$ ,

(6.1) $$ \begin{align} (\tau_T)_p^{\bullet} = \begin{cases} T_i^{\bullet} \cup \{T_{i+1}^1\} & \text{if } p = i,\\ T_{i+1}^{\bullet} \setminus \{ T_{i+1}^1 \} & \text{if } p = i+1,\\ T_p^{\bullet} & \text{otherwise.}\\ \end{cases} \end{align} $$

The inequality $(\tau _T)_i^1 < (\tau _T)_{i+1}^{-1}$ shows that $\tau _T \in \mathrm {SRT}({\boldsymbol {\unicode{x3b1} }}^{(i)})$ . Combining

$$ \begin{align*} \mathbf{w}(\tau_T) = \mathbf{w}(T) s_{m_{i+1} - 1} s_{m_{i+1} - 2} \cdots s_{m_i} \end{align*} $$

with $\tau _{T^{(i)}_{\alpha }} = T_{{\boldsymbol {\unicode{x3b1} }}^{(i)}}$ (=the source tableau of $\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}}$ ) yields that $\mathbf {w}(\tau _{T^{(i)}_{\alpha }}) \preceq _L \mathbf {w}(\tau _T)$ for $T \in \mathrm {SRT}(\alpha )^{(i)}$ . Moreover, for each $m_i \le j < m_{i+1}$ , it holds that

(6.2) $$ \begin{align} s_j \in \mathrm{Des}_R(\mathbf{w}(\tau_{T^{(i)}_{\alpha}}) s_{m_i} s_{m_i + 1} \cdots s_{j-1})\cap \mathrm{Des}_R(\mathbf{w}(\tau_T) s_{m_i} s_{m_i + 1} \cdots s_{j-1}). \end{align} $$

Here, $s_{m_i} s_{m_i + 1} \cdots s_{j - 1}$ is regarded as the identity when $j = m_i$ . Finally, applying Lemma 6.1 to equation (6.2) yields that $\mathbf {w}(T^{(i)}_{\alpha }) \preceq _L \mathbf {w}(T)$ , as required.

Proof. For each $T \in H_n(0) \cdot T^{(i)}_{\alpha }$ , let $\tau _T$ be the filling of $\mathtt {rd}({\boldsymbol {\unicode{x3b1} }}^{(i)})$ defined in equation (6.1). The surjectivity of $\partial _1^{(i)}$ is straightforward since $\tau _T \in \mathrm {SRT}({\boldsymbol {\unicode{x3b1} }}^{(i)})$ and $L(\tau _T)=T$ . Thus, to prove our assertion, it suffices to show that

$$\begin{align*}\partial_1^{(i)}(\pi_k \cdot \tau) = \pi_k \cdot \partial_1^{(i)}(\tau) \end{align*}$$

for all $k = 1,2, \ldots , n-1$ and $\tau \in \mathrm {SRT}({\boldsymbol {\unicode{x3b1} }}^{(i)})$ .

Case 1: $\pi _k \cdot \tau = \tau $ . If $\partial _1^{(i)}(\tau ) = 0$ , then there is nothing to prove. Suppose that $\partial _1^{(i)}(\tau ) \neq 0$ : that is, $L(\tau ) \in \mathrm {SRT}(\alpha )$ . We claim that $k \notin \mathrm {Des}(L(\tau ))$ . If $k = \tau _i^1$ and $k+1 = \tau _i^2$ , then $k \in L(\tau )_{i+1}^{\bullet }$ and $k+1 \in L(\tau )_i^{\bullet }$ . If $k \in \tau _{i+1}^{\bullet }$ and $k+1 = \tau _i^1$ , then both k and $k+1$ are in $L(\tau )_{i+1}^{\bullet }$ . In the remaining cases, from the fact that k is weakly right of $k+1$ in $\tau $ , it follows that k is weakly right of $k+1$ in $L(\tau )$ . For any cases, we can see that $k \notin \mathrm {Des}(L(\tau ))$ .

Case 2: $\pi _k \cdot \tau = 0$ . If $\partial _1^{(i)}(\tau ) = 0$ , then there is nothing to prove. Suppose that $\partial _1^{(i)}(\tau ) \neq 0$ . Since k and $k+1$ are in the same row of $\tau $ , k is the top and $k+1$ is the bottom for some two consecutive columns of $\tau $ . If $k \neq \tau _i^1$ , then k and $k+1$ are still in the same row of $L(\tau )$ , so $\pi _k \cdot L(\tau ) = \pi _k \cdot \partial _1^{(i)}(\tau ) = 0$ , as required. Assume that $k = \tau _i^1$ . Note that $|\tau _i^{\bullet }| = \alpha _i +1 \geq 2$ and $\tau _i^2$ greater than both k and $k+1$ . By the definition of $L(\tau )$ , we have that $L(\tau )_i^1 = \tau _i^2> L(\tau )_{i+1}^{-1} = k+1$ . This implies that $\partial _1^{(i)}(\tau )=0$ , which contradicts our assumption $\partial _1^{(i)}(\tau ) \neq 0$ .

Case 3: $\pi _k \cdot \tau = s_k \cdot \tau $ . First, consider the case where $\partial _1^{(i)}(\tau ) = 0$ : that is, $L(\tau ) \notin \mathrm {SRT}(\alpha )$ . Then $\tau $ must satisfy either $\tau _i^2> \tau _{i+1}^{-1}$ or $\min (\tau _i^1, \tau _{i+1}^1)> \tau _{i+2}^{-1}$ . Thus, to $L(\pi _k \cdot \tau ) \in \mathrm {SRT}(\alpha )$ , either $\tau _i^2 = k+1$ and $\tau _{i+1}^{-1}=k$ or $\min (\tau _i^1, \tau _{i+1}^1)= k+1$ and $\tau _{i+2}^{-1}=k$ . However, these are absurd because k is strictly left of $k+1$ in $\tau $ .

Next, consider the case where $\partial _1^{(i)}(\tau ) \neq 0$ : that is, $L(\tau ) \in \mathrm {SRT}(\alpha )$ . Since $\pi _k \cdot \tau = s_k \cdot \tau $ , k is strictly left of $k+1$ in $\tau $ . Therefore, k is weakly left of $k+1$ in $L(\tau )$ by the definition of $L(\tau )$ . Hence if neither k and $k+1$ are in the same column in $L(\tau )$ , nor are they in the same row in $L(\tau )$ , then $\pi _k \cdot L(\tau ) = s_k \cdot L(\tau )$ . Therefore, in such case, we have that

$$\begin{align*}\pi_k \cdot \partial_1^{(i)}(\tau) = \pi_k \cdot L(\tau) = s_k \cdot L(\tau) = L(s_k \cdot \tau) = L(\pi_k \cdot \tau)= \partial_1^{(i)}(\pi_k \cdot \tau) . \end{align*}$$

Suppose that k and $k+1$ are in the same column in $L(\tau )$ . This is possible only the case where $k = \tau _i^1$ and $k+1 \in \tau _{i+1}^{\bullet }$ since k is strictly left of $k+1$ in $\tau $ . Moreover, $k+1 \neq \tau _{i+1}^{-1}$ since $\pi _k \cdot \tau = s_k \cdot \tau $ . Hence $k+1 = (\pi _k \cdot \tau )_{i}^1$ and $k \in (\pi _k \cdot \tau )_{i+1}^{\bullet }$ , which implies that $L(\tau ) = L(\pi _k \cdot \tau )$ . Therefore, we have

$$\begin{align*}\pi_k \cdot \partial_1^{(i)}(\tau) = \pi_k \cdot L(\tau) = L(\tau) = L(\pi_k \cdot \tau)= \partial_1^{(i)}(\pi_k \cdot \tau) . \end{align*}$$

Here the second equality follows from the assumption that k and $k+1$ are in the same column in $L(\tau )$ .

Suppose that k and $k+1$ are in the same row in $L(\tau )$ . Then $\pi _k \cdot L(\tau ) = 0$ . In addition, since $\pi _k \cdot \tau = s_k \cdot \tau $ , we have that either $L(\tau _{i+1}^1) = k$ and $L(\tau )_{i+2}^{-1} = k+1$ , or $L(\tau )_i^1 = k$ and $L(\tau )_{i+1}^{-1} = k+1$ . In the case where $L(\tau )_{i+1}^1 = k$ and $L(\tau )_{i+2}^{-1} = k+1$ , the assumption $\pi _k \cdot \tau = s_k \cdot \tau $ implies that $L(\pi _k \cdot \tau )_{i+1}^1 = k+1$ and $L(\pi _k \cdot \tau )_{i+2}^{-1} = k$ . Thus, $L(\pi _k \cdot \tau ) \notin \mathrm {SRT}(\alpha )$ : that is, $\partial _1^{(i)}(\pi _k \cdot \tau ) = 0$ as desired. In the case where $L(\tau )_i^1 = k$ and $L(\tau )_{i+1}^{-1} = k+1$ , one can easily see that $L(\pi _k \cdot \tau ) \notin \mathrm {SRT}(\alpha )$ . Thus $\pi _k \cdot \partial _1^{(i)}(\tau ) = 0 = \partial _1^{(i)}(\pi _k \cdot \tau )$ .

Proof. If $i = \ell (\alpha )-1$ , then ${\boldsymbol {\unicode{x3b1} }}^{(i)}$ is a composition. Therefore, $\mathrm {rad}(\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}})$ is the $\mathbb C$ -span of $\mathrm {SRT}({\boldsymbol {\unicode{x3b1} }}^{(i)})\setminus \{ \boldsymbol {\tau }_{(i)} \}$ . Since $\partial _1^{(i)}(\boldsymbol {\tau }_{(i)}) \neq 0$ , this implies that $\ker (\partial _1^{(i)}) \subseteq \mathrm {rad}(\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}} )$ .

Suppose that $i \neq \ell (\alpha )-1$ . Let

(6.3) $$ \begin{align} \begin{aligned} \beta^{(1)} & = (\alpha_1, \alpha_2, \ldots, \alpha_{i-1}, \alpha_i +1, \alpha_{i+1} - 1, \alpha_{i+2} , \alpha_{i+3}, \ldots, \alpha_{\ell(\alpha)}), \\ \beta^{(2)} & = (\alpha_1, \alpha_2, \ldots, \alpha_{i-1}, \alpha_i +1, \alpha_{i+1} - 1 + \alpha_{i+2} , \alpha_{i+3}, \ldots, \alpha_{\ell(\alpha)}). \end{aligned} \end{align} $$

To ease notation, we denote the source tableaux of $\mathbf {P}_{\beta ^{(1)}}$ and $\mathbf {P}_{\beta ^{(2)}}$ by $\tau ^{(1)}$ and $\tau ^{(2)}$ , respectively. By Theorem 2.3, we may choose an $H_n(0)$ -module isomorphism

$$\begin{align*}f:\mathbf{P}_{{\boldsymbol{\unicode{x3b1}}}^{(i)}} \rightarrow \mathbf{P}_{\beta^{(1)}} \oplus \mathbf{P}_{\beta^{(2)}}. \end{align*}$$

Let

$$ \begin{align*}f(\boldsymbol{\tau}_{(i)}) = \sum_{\tau \in \mathrm{SRT}(\beta^{(1)})} c_{\tau} \tau + \sum_{\tau \in \mathrm{SRT}(\beta^{(2)})} d_{\tau} \tau \quad \text{for } c_{\tau}, d_{\tau} \in {\mathbb C}. \end{align*} $$

Since $f(\boldsymbol {\tau }_{(i)})$ is a generator of $\mathbf {P}_{\beta ^{(1)}} \oplus \mathbf {P}_{\beta ^{(2)}}$ , $c_{\tau ^{(1)}}$ and $d_{\tau ^{(2)}}$ are nonzero.

We claim that $[\boldsymbol {\tau }_{(i)}, \hat {\boldsymbol {\tau }}_{(i)}]^{\mathrm {c}} \subset \mathrm {rad} (\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}})$ . Take any $\tau \notin [\boldsymbol {\tau }_{(i)},\hat {\boldsymbol {\tau }}_{(i)}]$ . To get $\tau $ from $\boldsymbol {\tau }_{(i)}$ , there should exist an $H_n(0)$ -action switching two entries such that at least one of them lies apart from the $(i+1)$ st and $(i+2)$ nd columns. Thus there exist $\sigma , \rho \in \Sigma _n$ and $k \notin [m_i+2, m_{i+2} - 1]$ such that

$$ \begin{align*}\tau = \pi_{\sigma} \pi_k \pi_{\rho} \cdot \boldsymbol{\tau}_{(i)}, \quad \pi_{\rho} \cdot \boldsymbol{\tau}_{(i)} \in [\boldsymbol{\tau}_{(i)},\hat{\boldsymbol{\tau}}_{(i)}] \quad \text{and} \quad \pi_k \pi_{\rho} \cdot \boldsymbol{\tau}_{(i)} = s_k \cdot (\pi_{\rho} \cdot \boldsymbol{\tau}_{(i)}). \end{align*} $$

Ignoring the columns filled with entries $[m_i+2, m_{i+2}]$ , we can see that all $\pi _{\rho } \cdot \boldsymbol {\tau }_{(i)}$ , $\tau ^{(1)}$ and $\tau ^{(2)}$ are the same. This implies that $\pi _k \cdot \tau ^{(j)} = s_k \cdot \tau ^{(j)}$ for $j = 1,2$ . In all, we have

$$ \begin{align*} f(\tau) &= \pi_{\sigma} \pi_k \pi_{\rho} \cdot f(\boldsymbol{\tau}_{(i)})\\ &= \pi_{\sigma} \pi_k \pi_{\rho} \cdot \left( \sum_{\tau \in \mathrm{SRT}(\beta^{(1)})} c_{\tau} \tau + \sum_{\tau \in \mathrm{SRT}(\beta^{(2)})} d_{\tau} \tau \right) \\ &= \sum_{\substack{\tau \in \mathrm{SRT}(\beta^{(1)}) \\ \tau> \tau^{(1)} }} c^{\prime}_{\tau} \tau + \sum_{\substack{\tau \in \mathrm{SRT}(\beta^{(2)}) \\ \tau > \tau^{(2)} }} d^{\prime}_{\tau} \tau \end{align*} $$

for some $c^{\prime }_{\tau }, d^{\prime }_{\tau } \in \mathbb C$ . This implies that $f(\tau ) \in \mathrm {rad}(\mathbf {P}_{\beta ^{(1)}} \oplus \mathbf {P}_{\beta ^{(2)}})$ , and hence $\tau \in \mathrm {rad}(\mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}})$ .

By virtue of the above discussion, to complete our assertion, it is enough to show that $\ker (\partial _1^{(i)}) \subseteq \mathbb C [\boldsymbol {\tau }_{(i)}, \hat {\boldsymbol {\tau }}_{(i)}]^{\mathrm {c}}$ , or equivalently, $L(\tau ) \in \mathrm {SRT}(\alpha )$ for every $\tau \in [\boldsymbol {\tau }_{(i)},\hat {\boldsymbol {\tau }}_{(i)}]$ . But this is obvious since $L(\tau )_i^1 = \tau _i^2 = m_{i-1}+2$ , $L(\tau )_{i+1}^1 = \tau _{i}^{1} = m_{i-1}+1$ and $L(\tau )_{i+1}^{-1}, L(\tau )_{i+2}^{-1} \in [m_i+2, m_{i+2}]$ .

Proof of Theorem 3.3.

(a) As mentioned after the proof of Lemma 6.3, $\partial _1: \bigoplus _{i \in \mathcal {I}(\alpha )} \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}} \rightarrow \Omega (\mathcal {V}_{\alpha })$ is a surjective $H_n(0)$ -module homomorphism. Therefore, we only need to check $\ker (\partial _1) \subseteq \mathrm {rad} \left (\bigoplus _{i \in \mathcal {I}(\alpha )} \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}} \right )$ to complete the proof of the assertion. Let

$$ \begin{align*}\mathbf{T} := \bigoplus_{i \in \mathcal{I}(\alpha)} \mathbb C[\boldsymbol{\tau}_{(i)},\hat{\boldsymbol{\tau}}_{(i)}] \quad \text{and} \quad \mathbf{B} := \bigoplus_{i \in \mathcal{I}(\alpha)} \mathbb C[\boldsymbol{\tau}_{(i)}, \hat{\boldsymbol{\tau}}_{(i)}]^{\mathrm{c}}. \end{align*} $$

In the proof of Lemma 6.4, we see that $[\boldsymbol {\tau }_{(i)}, \hat {\boldsymbol {\tau }}_{(i)}]^{\mathrm {c}} \subseteq \mathrm {rad} \, \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}}$ for $i \in \mathcal {I}(\alpha )$ and thus $\mathbf {B} \subseteq \mathrm {rad} \left (\bigoplus _{i \in \mathcal {I}(\alpha )} \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}} \right )$ .

In the following, we will prove $\ker (\partial _1) \subseteq \mathbf {B}$ , which is obviously a stronger inclusion than necessary. We begin by collecting the following properties, which were shown in the proof of Lemma 6.4: For all $i \in \mathcal {I}(\alpha )$ , $1\le j < i$ and $\tau \in [\boldsymbol {\tau }_{(i)},\hat {\boldsymbol {\tau }}_{(i)}]$ ,

$$ \begin{align*} &\ker(\partial_1^{(i)}) \subseteq \mathbb C[\boldsymbol{\tau}_{(i)}, \hat{\boldsymbol{\tau}}_{(i)}]^{\mathrm{c}},\\ &\partial_1^{(i)}(\tau)_i^1 = m_{i-1}+2\, \text{ and} \\ &\partial_1^{(i)}(\tau)_j^1 = m_{j-1} +1. \end{align*} $$

Therefore, for any $i, j \in \mathcal {I}(\alpha )$ with $j < i$ , if $\tau \in [\boldsymbol {\tau }_{(i)},\hat {\boldsymbol {\tau }}_{(i)}] \subset \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}}$ and $\tau ' \in [\boldsymbol {\tau }_{(j)},\hat {\boldsymbol {\tau }}_{(j)}] \subset \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(j)}}$ , then $\partial _1(\tau )_j^1 = \partial _1^{(i)} (\tau )_j^1 = m_{j-1} + 1$ and $\partial _1(\tau ')_j^1 = \partial _1^{(j)} (\tau ')_j^1 = m_{j-1} + 2$ : that is, $\partial _1(\tau ) \neq \partial _1(\tau ')$ . This implies that that the set $\{\partial _1(\tau ) \mid \tau \in [\boldsymbol {\tau }_{(i)},\hat {\boldsymbol {\tau }}_{(i)}] \text { for } i \in \mathcal {I}(\alpha )\}$ is linearly independent, hence every $\mathbf {x} \in \ker (\partial _1) \setminus \{0\}$ is decomposed as $\mathbf {x} = \mathbf {x}^{(1)} + \mathbf {x}^{(2)}$ for some $\mathbf {x}^{(1)} \in \mathbf {T}$ and $\mathbf {x}^{(2)} \in \mathbf {B} \setminus \{0\}$ .

We claim that $\mathbf {x}^{(1)} = 0$ . Suppose on the contrary that $\mathbf {x}^{(1)} \neq 0$ . Let

$$\begin{align*}\partial_1(\mathbf{x}^{(1)}) = \sum_{T \in \mathrm{SRT}(\alpha) \cap \Omega(\mathcal{V}_{\alpha}) }c_T T \quad \text{and} \quad \partial_1(\mathbf{x}^{(2)}) = \sum_{T \in \mathrm{SRT}(\alpha) \cap \Omega(\mathcal{V}_{\alpha}) }d_T T. \end{align*}$$

Since $\partial _1(\mathbf {x}^{(1)}) \neq 0$ , there exists $T \in \mathrm {SRT}(\alpha ) \cap \Omega (\mathcal {V}_{\alpha })$ such that $c_{T} \neq 0$ . In addition, since $\mathrm {SRT}(\alpha ) \cap \Omega (\mathcal {V}_{\alpha }) $ is linearly independent and $\partial _1(\mathbf {x}) = 0$ , we have $c_{T} = -d_{T}$ . Therefore, there exist $i,j \in \mathcal {I}(\alpha )$ , $\tau _{\mathbf {T}} \in [\boldsymbol {\tau }_{(i)}, \hat {\boldsymbol {\tau }}_{(i)}]$ and $\tau _{\mathbf {B}} \in [\boldsymbol {\tau }_{(j)}, \hat {\boldsymbol {\tau }}_{(j)}]^{\mathrm {c}}$ such that $\partial _1(\tau _{\mathbf {T}}) = T = \partial _1(\tau _{\mathbf {B}})$ . Since $\{\partial _1(\tau ) \mid \tau \in \mathrm {SRT}({\boldsymbol {\unicode{x3b1} }}^{(i)})\} \setminus \{0\}$ is linearly independent, we have $i \neq j$ . Note that $\partial _1(\tau _{\mathbf {B}}) = \partial _1^{(j)}(\tau _{\mathbf {B}}) \in H_n(0)\cdot T^{(j)}_{\alpha }$ . By Lemma 6.2, we have $T_j^1> T_{j+1}^1$ . On the other hand, since $T = \partial _1^{(i)}(\tau _{\mathbf {T}})$ and $\tau _{\mathbf {T}} \in [\boldsymbol {\tau }_{(i)}, \hat {\boldsymbol {\tau }}_{(i)}]$ , T is equal to $T^{(i)}_{\alpha }$ except for the $(i+1)$ st and $(i+2)$ nd columns. Note that the $(i+1)$ st and $(i+2)$ nd columns of them are filled with $\{ (\tau _{\mathbf {T}})_i^1 \} \cup [m_i+2, m_{i+2}]$ and $T_{i+1}^1 = \partial _1^{(i)}(\tau _{\mathbf {T}})_{i+1}^1 = m_{i-1} + 1$ . This shows that $T_j^1 < T_{j+1}^1$ , which is absurd. Hence $\mathbf {x}^{(1)} = 0$ , and it follows that $\ker (\partial _1) \subseteq \mathbf {B}$ , as required.

(b) For all $\beta \models n$ , it is known that

$$\begin{align*}\mathrm{Ext}_{H_n(0)}^1(\mathcal{V}_{\alpha},\mathbf{F}_{\beta})=\mathrm{Hom}_{H_n(0)}(P_1,\mathbf{F}_{\beta}) \end{align*}$$

with $P_1:=\displaystyle { \bigoplus _{i \in \mathcal {I}(\alpha )} \mathbf {P}_{{\boldsymbol {\unicode{x3b1} }}^{(i)}}}$ (for instance, see [Reference Benson3, Corollary 2.5.4]). In the case with projective indecomposable modules, one has that $\dim \mathrm {Hom}_{H_n(0)}(\mathbf {P}_{\gamma },\mathbf {F}_{\gamma '})=\delta _{\gamma ,\gamma '}$ for all $\gamma , \gamma ' \models n$ (see [Reference Benson3, Lemma 1.7.5]). This tells us that $\dim \mathrm {Ext}_{H_n(0)}^1(\mathcal {V}_{\alpha },\mathbf {F}_{\beta })$ counts the multiplicity of $\mathbf {P}_{\beta }$ in the decomposition of $P_1$ into indecomposables. The indecomposables that occur in the decomposition are precisely $\mathbf {P}_{\beta }$ with $\beta \in \mathcal {J}(\alpha )$ . We claim that all of them are multiplicity-free. For $i \in \mathcal {I}(\alpha )$ , note that $[{\boldsymbol {\unicode{x3b1} }}^{(i)}]=\{\beta ^{(1)}, \beta ^{(2)}\}$ with $\beta ^{(1)}, \beta ^{(2)}$ in equation (6.3). Obviously $\beta ^{(1)}$ and $\beta ^{(2)}$ are distinct. Furthermore, for $i<j$ , $[{\boldsymbol {\unicode{x3b1} }}^{(i)}]$ and $[{\boldsymbol {\unicode{x3b1} }}^{(j)}]$ are disjoint since the ith entry of the compositions in the former is $\alpha _i+1$ , whereas that of the compositions in the latter is $\alpha _i$ . Hence the claim is verified, which completes the proof.

Proof. First, we claim that $\overline {\pi }_{w_0(\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}})} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}$ is stabilised under the action of $\pi _i$ for all $i \in \mathrm {Des}(T_{\beta }^{\leftarrow })^{\mathrm {c}}$ . Note that $\overline {\pi }_{w_0(\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}})} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}$ is of the form

(6.4) $$ \begin{align} \sum_{T \in [{T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}}, T_{\underline{{\boldsymbol{\unicode{x3b1}}}}}^{\leftarrow}]} c_T T \quad \text{for some } c_T \in {\mathbb Z}. \end{align} $$

But from the definitions of ${T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}$ and $T_{\underline {{\boldsymbol {\unicode{x3b1} }}}}^{\leftarrow }$ , it follows that $\pi _i \cdot T = T$ for $i \in \mathrm {Des}(T_{\beta }^{\leftarrow })^{\mathrm {c}}$ . Thus our claim is verified.

Next, we claim that $\pi _i \cdot (\overline {\pi }_{w_0(\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}})} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}) = 0$ for all $i \in \mathrm {Des}(T_{\beta }^{\leftarrow })$ . Take any $i \in \mathrm {Des}(T_{\beta }^{\leftarrow })$ . Note that $T(\mathtt {S}_{k_0}) = \{1,2,\ldots ,\ell (\alpha )\}$ for any $T \in [{T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}, T_{\underline {{\boldsymbol {\unicode{x3b1} }}}}^{\leftarrow }]$ . Therefore, if $1 \leq i < \ell (\alpha )$ , then $\pi _i \overline {\pi }_{w_0(\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}})} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}} = 0$ by equation (6.4). In the case where $i \ge \ell (\alpha )$ , $i \in \mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}$ and thus $\pi _i \overline {\pi }_{w_0(\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}})}= 0$ by Lemma 6.5.

Proof of Theorem 4.1.

We first claim that $\epsilon : \mathcal {V}_{\alpha } \to \mathbf {P}_{\underline {{\boldsymbol {\unicode{x3b1} }}}}$ is an $H_n(0)$ -module homomorphism: that is,

$$ \begin{align*} \epsilon(\pi_i \cdot \mathscr{T}) = \pi_i \cdot \epsilon(\mathscr{T}) \quad\text{ for } i=1,2,\ldots,n-1 \text{ and }\mathscr{T} \in \mathrm{SIT}(\alpha). \end{align*} $$

Let us fix $1 \le i \le n-1$ and $\mathscr {T} \in \mathrm {SIT}(\alpha )$ . Let $0 \leq x, y \leq m$ be integers satisfying that $i \in \mathscr {T}(\mathtt {S}_{k_x})$ and $i+1 \in \mathscr {T}(\mathtt {S}_{k_y})$ .

Case 1: $\pi _i \cdot \mathscr {T} = \mathscr {T}$ . First we handle the case where $x=0$ . Then i will be placed in the top row in ${T^{\mathscr {T}}}$ . In view of the given condition $\pi _i \cdot \mathscr {T} = \mathscr {T}$ , one sees that $x \neq y$ . This implies that $i+1$ is strictly below i in ${T^{\mathscr {T}}}$ . Next we handle the case where $x>0$ . The condition $\pi _i \cdot \mathscr {T} = \mathscr {T}$ says that $0 < x \leq y$ ; thus $i+1$ is strictly below i in ${T^{\mathscr {T}}}$ . In either case, it is immediate from equation (2.1) that $\pi _i \cdot {T^{\mathscr {T}}} = {T^{\mathscr {T}}}$ .

Case 2: $\pi _i \cdot \mathscr {T} = 0$ . From equation (2.2), it follows that i and $i+1$ are in the first column in $\mathscr {T}$ : that is, $x = y = 0$ . Hence, in ${T^{\mathscr {T}}}$ , both of them will appear in ${T^{\mathscr {T}}}(\mathtt {S}_{k_0})$ . As in Case 1, one can derive from equation (2.1) that $\pi _i \cdot {T^{\mathscr {T}}} = 0$ .

Case 3: $\pi _i \cdot \mathscr {T} = s_i \cdot \mathscr {T}$ . We claim that $\epsilon (s_i \cdot \mathscr {T}) = s_i \cdot {T^{\mathscr {T}}}$ . Observe that i appears strictly above $i+1$ in $\mathscr {T}$ . If $i+1 \in \mathscr {T}(\mathtt {S}_{k_0})$ , then we see that $i \notin \mathscr {T}(\mathtt {S}_{k_0})$ , which means i appears strictly left of $i+1$ in ${T^{\mathscr {T}}}$ . Otherwise, we also see that $i \notin \mathscr {T}(\mathtt {S}_{k_0})$ . More precisely, if $i+1 \notin \mathscr {T}(\mathtt {S}_{k_0})$ and $i \in \mathscr {T}(\mathtt {S}_{k_0})$ , then $\mathscr {T}$ is not an SIT since the entries in the row containing $i+1$ of $\mathscr {T}$ do not increase from left to right. It follows from the construction of ${T^{\mathscr {T}}}$ that i is strictly below $i+1$ in ${T^{\mathscr {T}}}$ . In either case, it holds that $T^{s_i \cdot \mathscr {T}} = s_i \cdot {T^{\mathscr {T}}}$ . Thus we conclude that

$$ \begin{align*}\pi_i \cdot \epsilon(\mathscr{T}) = \pi_i \cdot {T^{\mathscr{T}}} = T^{s_i \cdot \mathscr{T}} = \epsilon(s_i \cdot \mathscr{T}) = \epsilon(\pi_i \cdot \mathscr{T}). \end{align*} $$

We next claim that $\mathbf {P}_{\underline {{\boldsymbol {\unicode{x3b1} }}}}$ is an essential extension of $\epsilon (\mathcal {V}_{\alpha })$ . To do this, we see that $\mathrm {soc}(\mathbf {P}_{\underline {{\boldsymbol {\unicode{x3b1} }}}}) \subset \epsilon (\mathcal {V}_{\alpha })$ . Note that

$$ \begin{align*} \mathrm{soc}(\mathbf{P}_{\underline{{\boldsymbol{\unicode{x3b1}}}}}) \cong \mathrm{soc} \Big(\bigoplus_{\beta \in [\underline{{\boldsymbol{\unicode{x3b1}}}}]} \mathbf{P}_{\beta} \Big) \cong \bigoplus_{\beta \in [\underline{{\boldsymbol{\unicode{x3b1}}}}]} \mathbb C T^{\leftarrow}_{\beta}. \end{align*} $$

In view of Lemma 6.6, one sees that

(6.5) $$ \begin{align} \mathrm{soc}(\mathbf{P}_{\underline{{\boldsymbol{\unicode{x3b1}}}}}) = \bigoplus_{\beta \in [\underline{{\boldsymbol{\unicode{x3b1}}}}]} \mathbb C\left( \overline{\pi}_{w_0(\mathtt{J}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}})} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}}\right). \end{align} $$

Choose any $\beta \in [\underline {{\boldsymbol {\unicode{x3b1} }}}]$ . Then

$$ \begin{align*} \overline{\pi}_{w_0(\mathtt{J}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}})} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}} = \sum_{\sigma \in (\Sigma_n)_{\mathtt{J}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}}} (-1)^{\ell(w_0(\mathtt{J}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}})) - \ell(\sigma)} \pi_{\sigma} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}}. \end{align*} $$

For $\sigma \in (\Sigma _n)_{\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}$ , since $(\pi _{\sigma } \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}})(\mathtt {S}_{k_0}) = \{1,2, \ldots , \ell (\alpha )\}$ , we have

$$ \begin{align*} (\pi_{\sigma} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}})^1_{m+k_j-1} < \begin{cases} (\pi_{\sigma} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}})^{1}_j & \text{if } 1 \leq j < m, \\[1ex] (\pi_{\sigma} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}}})^{2}_j & \text{if } j = m. \end{cases} \end{align*} $$

It means $\pi _{\sigma } \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}} \in \epsilon (\mathcal {V}_{\alpha })$ for all $\sigma \in (\Sigma _n)_{\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}}}$ . Combining this with equation (6.5) yields that $\mathrm {soc}(\mathbf {P}_{\underline {{\boldsymbol {\unicode{x3b1} }}}}) \subset \epsilon (\mathcal {V}_{\alpha })$ .

Proof. Let $1 \leq j \leq m$ and $\beta \in [\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}]$ . If $\min (\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}})> \ell (\alpha )$ , then one can prove the assertion in the same way as in Lemma 6.6. We now assume that $\min (\mathtt {J}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}) \leq \ell (\alpha )$ . We first show that

$$ \begin{align*} \pi_i \cdot (\overline{\pi}_{\boldsymbol{w_0}(\beta;j)} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}}) = \overline{\pi}_{\boldsymbol{w_0}(\beta;j)} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}} \end{align*} $$

for $i \notin \mathrm {Des}(T^{\leftarrow }_{\beta })$ . Since

$$ \begin{align*} \overline{\pi}_{\boldsymbol{w_0}(\beta;j)} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}} = \sum_{T \in [{T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}}, T_{\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}^{\leftarrow}]} c_T T \quad \text{for some } c_T \in \mathbb Z, \end{align*} $$

it suffices to show that $\pi _i \cdot T = T$ for $i \notin \mathrm {Des}(T_{\beta }^{\leftarrow })$ and $T \in [{T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}}, T_{\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}^{\leftarrow }]$ . Since $\{1,2,\ldots , \ell (\alpha )\}\subseteq \mathrm {Des}(T^{\leftarrow }_{\beta })$ by definition, we only consider that $i \geq \ell (\alpha )+1$ . If $i=\ell (\alpha )+1$ , then the assertion follows from the fact that $T(\mathtt {S}^{\prime }_{k_0})\cup T(\mathtt {S}^{\prime }_{k_{-1}}) = \{1,2,\ldots ,\ell (\alpha )+1\}$ . Otherwise, from the definitions of ${T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}}$ and $T_{\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}^{\leftarrow }$ , it follows that $\pi _i \cdot T = T$ for $i \notin \mathrm {Des}(T_{\beta }^{\leftarrow })$ . Thus our claim is verified.

We next show that $\pi _i \cdot (\overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}}) = 0$ for $i \in \mathrm {Des}(T^{\leftarrow }_{\beta })$ . Take any $i \in \mathrm {Des}(T_{\beta }^{\leftarrow })$ . If $i> \ell (\alpha ) + 1$ , then $i \in \widehat {\mathtt {J}}_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}$ . Therefore, by Lemma 6.5, we have $\pi _i \overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} = 0$ , which implies $\pi _i \overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} \cdot T^{\leftarrow }_{\beta ,\underline {{\boldsymbol {\unicode{x3b1} }}}} = 0$ . Suppose that $i \le \ell (\alpha ) + 1$ . Since $\ell (\alpha ) + 1 \notin \mathrm {Des}(T^{\leftarrow }_{\beta })$ , we have that $1 \le i \le \ell (\alpha )$ . If $i \in \mathrm {Des}_L(\boldsymbol {w_0}(\beta ;j))$ , then $\pi _i \overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} = 0$ . Thus, $\pi _i \overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} \cdot T^{\leftarrow }_{\beta ,\underline {{\boldsymbol {\unicode{x3b1} }}}} = 0$ . Otherwise, we have $s_i w_0([\ell (\alpha )];|\mathtt {S}^{\prime }_{k_0}|) = \sigma s_{i'}$ for some $\sigma \in (\Sigma _{n})_{[\ell (\alpha )]}^{(|\mathtt {S}^{\prime }_{k_0}|)}$ and $1\le i' \le \ell (\alpha )$ with $i' \neq |\mathtt {S}^{\prime }_{k_0}|$ since $w_0([\ell (\alpha )];|\mathtt {S}^{\prime }_{k_0}|)$ is the unique longest element in $(\Sigma _{n})_{[\ell (\alpha )]}^{(|\mathtt {S}^{\prime }_{k_0}|)}$ . Combining this with [Reference Jung, Kim, Lee and Oh21, Lemma 1], we have that $\pi _i \overline {\pi }_{\boldsymbol {w_0}(\beta ;j)} = \mathbf {h} \pi _{i'}$ for some $\mathbf {h} \in H_n(0)$ and $1 \le i' \le \ell (\alpha )$ with $i' \neq |\mathtt {S}^{\prime }_{k_0}|$ . Since $\pi _{i'} \cdot {T^{\leftarrow }_{\beta ;\underline {{\boldsymbol {\unicode{x3b1} }}}_{(j)}}} = 0$ for all $1 \le i' \le \ell (\alpha )$ with $i' \neq |\mathtt {S}^{\prime }_{k_0}|$ , it follows that

$$ \begin{align*} \pi_i \cdot (\overline{\pi}_{\boldsymbol{w_0}(\beta;j)} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}}) = \mathbf{h} \pi_{i'} \cdot {T^{\leftarrow}_{\beta;\underline{{\boldsymbol{\unicode{x3b1}}}}_{(j)}}} = 0.\\[-3.4pc] \end{align*} $$