Proof. The first claim follows from exactness of coinduction. (In particular, the surjection $\widehat {\mathsf {Z}}(\lambda ) \twoheadrightarrow \widehat {\mathsf {L}}(\lambda )$ factors through $\widehat {\mathsf {M}}_I(\lambda )$.) The second claim can be deduced from (3.5).

Proof. Let $\widehat {\mathsf {M}}'_I(\lambda ) = \operatorname {ind}_{P_1T}^{G_1T}\, \widehat {\mathsf {L}}_I(\lambda )$ as in Remark 3.1. The ‘dual’ of (3.5) gives a $P_1^{+}T$-module isomorphism

\[ \widehat{\mathsf{M}}'_I(\lambda) \cong \Bbbk[(U_I^{+})_1] \otimes \widehat{\mathsf{L}}_I(\lambda), \]

where, in particular, $(U_I^{+})_1$ acts via the left regular representation on the first term and trivially on the second term. Thus,

\[ \widehat{\mathsf{M}}'_I(\lambda)^{(U_I^{+})_1} \cong \Bbbk[(U_I^{+})_1]^{(U_I^{+})_1} \otimes \widehat{\mathsf{L}}_I(\lambda) \cong \widehat{\mathsf{L}}_I(\lambda), \]

where the last isomorphism follows from the fact that $\Bbbk [(U_I^{+})_1]^{(U_I^{+})_1} = \Bbbk$.

Moreover, the ‘dual’ of Lemma 3.2 gives an inclusion $\widehat {\mathsf {M}}'_I(\lambda ) \hookrightarrow \widehat {\mathsf {Z}}'(\lambda )$. Thus, there also exists an inclusion $\widehat {\mathsf {L}}(\lambda ) \hookrightarrow \widehat {\mathsf {M}}'_I(\lambda )$ since $\widehat {\mathsf {M}}'_I(\lambda )$ must contain the (simple) socle of $\widehat {\mathsf {Z}}'(\lambda )$. This implies

\[ \widehat{\mathsf{L}}(\lambda)^{(U_I^{+})_1} \subseteq \widehat{\mathsf{M}}'_I(\lambda)^{(U_I^{+})_1} \cong \widehat{\mathsf{L}}_I(\lambda). \]

Finally, since the first term is non-zero (because non-zero $(U_I^{+})_1$ invariants always exist), and the middle term is irreducible for $L_1T$, then we must have equality.

Proof. It suffices to show that $\dim \operatorname {Hom}_{G_1T}(N, \widehat {\mathsf {L}}(\mu )) = \dim \operatorname {Hom}_{L_1T}(M, \widehat {\mathsf {L}}_I(\mu ))$ for all $\mu \in \mathbf {X}$. Observe,

\[ \operatorname{Hom}_{G_1T}(N, \widehat{\mathsf{L}}(\mu)) \cong \operatorname{Hom}_{P_1^{+}T}(M, \widehat{\mathsf{L}}(\mu)) \cong \operatorname{Hom}_{L_1T}\left(M, \widehat{\mathsf{L}}(\mu)^{(U_I^{+})_1}\right), \]

where the first isomorphism follows from [Reference Jantzen7, I.8.14(4)] and the second isomorphism holds because $(U_I^{+})_1$ acts trivially on $M$, so the image of any morphism must also be $(U_I^{+})_1$-invariant. Finally, by Lemma 3.4,

\[ \operatorname{Hom}_{L_1T}\left(M, \widehat{\mathsf{L}}(\mu)^{(U_I^{+})_1}\right) \cong \operatorname{Hom}_{L_1T}(M, \widehat{\mathsf{L}}_I(\mu)). \]

Therefore, $\dim \operatorname {Hom}_{G_1T}(N, \widehat {\mathsf {L}}(\mu )) = \dim \operatorname {Hom}_{L_1T}(M, \widehat {\mathsf {L}}_I(\mu ))$.

Proof. We will use Jantzen's criterion for the simplicity of Weyl modules (cf. [Reference Jantzen7, II.8.21] or [Reference Jantzen6]). Set $\nu _i =\lambda _i + \rho$ for $i=0,\ldots,n$. It will suffice to show that for any $i = 0, \ldots, n$ and $1 \leq k< j \leq n+1$, the quantity

\[ \langle \nu_i, \epsilon_k - \epsilon_j \rangle \]

satisfies the criterion. We will proceed by dividing this problem into the following cases:

1. Case 1 $i=0$

   1. 1.1) $k = 1$, $2 \leq j \leq n+1$

   2. 1.2) $2 \leq k < j \leq n+1$

2. Case 2 $i=n$

   1. 2.1) $1\leq k \leq n$, $j=n+1$

   2. 2.2) $1 \leq k < j \leq n$

3. Case 3 $1 \leq i \leq n-1$

   1. 3.1) $1 \leq k < j \leq i$

   2. 3.2) $i+2 \leq k < j \leq n+1$

   3. 3.3) $1 \leq k \leq i$, $j=i+1$

   4. 3.4) $k = i+1$, $i+1 < j \leq n+1$

   5. 3.5) $1 \leq k \leq i < i+2 \leq j \leq n+1$.

(Note that some of these sub-cases may be empty for certain choices of $i$ and $n$.)

Let us first consider Case 1. We calculate ${\langle \nu _0, \epsilon _1 - \epsilon _{2} \rangle = 1}$, and for $2\leq k\leq n$, ${\langle \nu _0, \epsilon _k - \epsilon _{k+1} \rangle = p}$. In the situation of 1.1), first observe

\[ {\langle \nu_0, \epsilon_1 - \epsilon_{j} \rangle = 1 + (j-2)p}, \]

for $2 \leq j \leq n+1$. Following the notation in [Reference Jantzen7, II.8.21], set $a = 1$, $b = j-2$ and $s=0$. The criterion is satisfied by setting $\beta _0 = \epsilon _1 - \epsilon _{2}$ and $\beta _r = \epsilon _{r+1}-\epsilon _{r+2}$ for $r = 1,\ldots, j-2$. Similarly, in the case of 1.2), first note that for $2 \leq k < j \leq n+1$,

\[ \langle \nu_0, \epsilon_k - \epsilon_{j} \rangle = (j-k)p. \]

If we write ${j-k= ap^{s-1} + bp^{s}}$ for some $s \geq 1$ and $0 < a < p$, then ${\langle \nu _0, \epsilon _k - \epsilon _{j} \rangle } = ap^{s} + bp^{s+1}$. Finally, set ${\beta _0 = \epsilon _k - \epsilon _{k+ ap^{s-1}}}$, and ${\beta _r = \epsilon _{k+ ap^{s-1}+(r-1)p^{s}} - \epsilon _{k+ ap^{s-1} + rp^{s}}}$ for ${r= 1,\ldots, b}.$

Now we consider Case 2. Again, we calculate for $1\leq k\leq n-1$, $\langle \nu _n, \epsilon _k - \epsilon _{k+1} \rangle = p$ and ${\langle \nu _n, \epsilon _n - \epsilon _{n+1}\rangle =p-1}$. In the situation of 2.1), note that for $1\leq k \leq n$,

\[ {\langle \nu_n, \epsilon_k - \epsilon_{n+1} \rangle = (p-1) + (n-k)p}, \]

so $a = p-1$, $b = n-k$ and $s=0$. The criterion is satisfied by setting $\beta _0 = \epsilon _n - \epsilon _{n+1}$ and $\beta _r = \epsilon _{r+k-1}-\epsilon _{r+k}$ for $r = 1,\ldots, n-k$. In the situation of 2.2), we get

\[ \langle \nu_n, \epsilon_k - \epsilon_{j} \rangle = (j-k)p, \]

for $1 \leq k < j \leq n$. If we write ${j-k= ap^{s-1} + bp^{s}}$ for some $s \geq 1$ and $0 < a < p$, then ${\langle \nu _n, \epsilon _k - \epsilon _{j} \rangle } = ap^{s} + bp^{s+1}$. Now set ${\beta _0 = \epsilon _k - \epsilon _{ap^{s-1} +k}}$, and $\beta_{r} = \epsilon _{k+ ap^{s-1}+(r-1)p^{s}} - \epsilon _{k+ ap^{s-1} + rp^{s}}$ for ${r= 1,\ldots, b}$ as in 1.2).

Finally, we consider Case 3. For $1\leq k\leq i-1$ and $i+2 \leq k \leq n+1$, we have $\langle \nu _i, \epsilon _k - \epsilon _{k+1} \rangle = p,$ ${ \langle \nu _i, \epsilon _i - \epsilon _{i+1} \rangle = p-1}$ and ${ \langle \nu _i, \epsilon _{i+1} - \epsilon _{i+2} \rangle = 1}$. To handle 3.1) and 3.2), note that in both instances

\[ \langle \nu_i, \epsilon_k - \epsilon_{j} \rangle = (j-k)p. \]

These cases then follow by writing ${j-k= ap^{s-1} + bp^{s}}$ with $0 < a < p$, and defining $\beta _0$ and $\beta _r$ for $r=1,\ldots, b$ as in 1.2) and 2.2), respectively. Similarly, the verifications of 3.3) and 3.4) are identical to the verifications of 2.1) and 1.1), respectively.

Thus, we only need to consider 3.5). We first observe

\[ \langle \nu_i, \epsilon_k - \epsilon_{j} \rangle = (j-k-1)p, \]

and write ${j-k-1= ap^{s-1} + bp^{s}}$ where $0 < a < p$. Now there are two further sub-cases:

1. a) $k + ap^{s-1} \geq i+1$,

2. b) $k + ap^{s-1} \leq i$.

In the situation of a), we set $\beta _0 = \epsilon _k - \epsilon _{k+1+ap^{s-1}}$ and $\beta _r = \epsilon _{k+ 1+ ap^{s-1}+(r-1)p^{s}} - \epsilon _{k+ 1+ ap^{s-1} + rp^{s}}$ for ${r= 1,\ldots, b}$. In the situation of b), we set ${\beta _0 = \epsilon _k - \epsilon _{k+ap^{s-1}}}$ and

\[ \beta_r = \begin{cases} \epsilon_{k+ ap^{s-1}+(r-1)p^{s}} - \epsilon_{k+ ap^{s-1} + rp^{s}} & \text{if }k+ ap^{s-1} + rp^{s} \leq i\\ \epsilon_{k+ ap^{s-1}+(r-1)p^{s}} - \epsilon_{k+ 1+ap^{s-1} + rp^{s}} & \text{if }k+ ap^{s-1} + (r-1)p^{s} \leq i \text{ and} \\ & k+ ap^{s-1} + rp^{s} \geq i+1\\ \epsilon_{k+ 1+ ap^{s-1}+(r-1)p^{s}} - \epsilon_{k+ 1+ ap^{s-1} + rp^{s}} & \text{otherwise}\\ \end{cases} \]

for $r = 1,\ldots, b$.

Proof. It will be enough to prove (1), since (2) will follow from the exact same arguments. For notational simplicity, set $\nu _i = \lambda _i + \rho$ for $i=0,\ldots, n-1$. By Corollary 4.3, we can apply the Weyl dimension formula, which gives

\[ \dim_{\Bbbk} \mathsf{L}(\lambda_i) = \frac{\prod_{1 \leq k < j \leq n+1} \langle \nu_i, \epsilon_k - \epsilon_j\rangle }{\prod_{1 \leq k < j \leq n+1} \langle \rho, \epsilon_k - \epsilon_j\rangle}, \]

where

\[ \prod_{1 \leq k < j \leq n+1} \langle \rho, \epsilon_k - \epsilon_j\rangle = n!(n-1)!\cdots 2!1!. \]

By (3.4), the description of the weight basis of $\mathsf {D}_I$ in (3.3), and the analogue of Corollary 4.3 for the Levi factor $L_I$ (applied to $\mathsf {L}_I(\lambda _i)$), we can deduce

\[ \dim_{\Bbbk}\mathsf{M}_I(\lambda_i) = \frac{p^{n}\prod_{1 \leq k < j \leq n} \langle \nu_i, \epsilon_k - \epsilon_j\rangle}{(n-1)!\cdots 2!1!}. \]

Thus, the equation in the statement of the lemma is equivalent to

(4.4)\begin{equation} \prod_{1 \leq k < j \leq n+1} \langle \nu_i, \epsilon_k - \epsilon_j\rangle + \prod_{1 \leq k < j \leq n+1} \langle \nu_{i+1}, \epsilon_k - \epsilon_j\rangle = n!p^{n}\prod_{1 \leq k < j \leq n} \langle \nu_i, \epsilon_k - \epsilon_j\rangle, \end{equation}

for $i=0,\ldots, n-1$. Notice that if we treat $p$ as an indeterminate variable, then the expressions

\[ \langle \nu_i, \epsilon_{i+1}-\epsilon_{n+1}\rangle = (n-1-i)p + 1, \quad \langle \nu_{i+1}, \epsilon_{1} - \epsilon_{i+2} \rangle = (i+1)p - 1 \]

are exclusive to the first and second terms appearing in (4.4), respectively. So it will be helpful to introduce the notation

\[ \varGamma = \{ (k,j) \mid 1 \leq k < j \leq n+1\}. \]

Claim. For $i=0,\ldots, n-1$,

\[ \prod_{\Gamma\backslash \{(i+1,n+1)\}} \langle \nu_i, \epsilon_k-\epsilon_j\rangle = \prod_{\Gamma\backslash \{(1,i+2)\}}\langle \nu_{i+1}, \epsilon_k-\epsilon_j\rangle. \]

Suppose for now that this claim holds, then by the observation immediately preceding the claim,

\[ \prod_{1 \leq k < j \leq n+1} \langle \nu_i, \epsilon_k - \epsilon_j\rangle + \prod_{1 \leq k < j \leq n+1} \langle \nu_{i+1}, \epsilon_k - \epsilon_j\rangle = np \prod_{\Gamma\backslash \{(i+1,n+1)\}} \langle \nu_i, \epsilon_k-\epsilon_j\rangle. \]

Combining this with the following identity:

\begin{align*} \prod_{\Gamma\backslash \{(i+1,n+1)\}} \langle \nu_i, \epsilon_k-\epsilon_j\rangle & = \prod_{1\leq k \leq n, \, k\neq i+1} \langle \nu_i, \epsilon_k-\epsilon_{n+1} \rangle \prod_{1 \leq k < j \leq n} \langle \nu_i, \epsilon_k - \epsilon_j\rangle \\ & = (n-1)!p^{n-1} \prod_{1 \leq k < j \leq n} \langle \nu_i, \epsilon_k - \epsilon_j\rangle, \end{align*}

verifies (4.4).

The remainder of the proof will be devoted to proving the preceding claim. Let

\[ X = \{(k,j) \in \Gamma \, \mid \, j = i+1,i+2, \text{ or } k = i+1, i+2\}. \]

It can be checked that $|X| = 2n-1$ and that $X$ is the subset of $\Gamma$ consisting of all $(k,j)$ satisfying

\[ \langle \nu_i, \epsilon_k - \epsilon_j \rangle \neq \langle \nu_{i+1}, \epsilon_k - \epsilon_j \rangle. \]

In particular, the sets ${X\backslash \{(i+1,n+1)\}}$ and ${X \backslash \{(1,i+2)\} }$ have precisely $2n-2$ elements. We get immediately that

\[ \prod_{\Gamma\backslash X} \langle \nu_i, \epsilon_k-\epsilon_j\rangle = \prod_{\Gamma\backslash X}\langle \nu_{i+1}, \epsilon_k-\epsilon_j\rangle, \]

and we only have to check the $(2n-2)$-fold products

\[ \prod_{X \backslash \{(i+1,n+1)\}} \langle \nu_i, \epsilon_k-\epsilon_j\rangle, \quad \prod_{X \backslash \{(1,i+2)\}}\langle \nu_{i+1}, \epsilon_k-\epsilon_j\rangle. \]

If $i=0$,

\begin{align*} & \prod_{X \backslash \{(1,n+1)\}} \langle \nu_0, \epsilon_k-\epsilon_j\rangle \\ & \quad = \bigg( \big(p\big)\big(2p\big)\cdots \big((n-1)p\big)\bigg) \bigg(\big(p+1\big)\big(2p+1\big)\cdots \big((n-2)p+1\big)\bigg)\\ & \quad = \prod_{X \backslash \{(1,2)\}} \langle \nu_1, \epsilon_k-\epsilon_j\rangle, \end{align*}

and for $i=n-1$,

\begin{align*} & \prod_{X \backslash \{(n,n+1)\}} \langle \nu_{n-1}, \epsilon_k-\epsilon_j\rangle \\ & \quad = \bigg(\big((n-1)p-1\big)\big((n-2)p-1\big)\cdots \big(p-1\big)\bigg) \\ & \qquad\times \bigg( \big((n-1)p\big)\big((n-2)p\big)\cdots \big(p\big)\bigg) \\ & \quad = \prod_{X \backslash \{(1,n+1)\}} \langle \nu_n, \epsilon_k-\epsilon_j\rangle. \end{align*}

Finally, suppose that $1 \leq i \leq n-2$, then

\begin{align*} & \prod_{X \backslash \{(i+1,n+1)\}}\langle \nu_i, \epsilon_k-\epsilon_j\rangle \\ & \quad =\bigg(\big(ip-1\big)\big((i-1)p-1\big)\cdots \big(p-1\big)\bigg) \bigg( \big(ip\big)\big((i-1)p\big)\cdots \big(p\big)\bigg) \\ & \qquad\times \bigg( \big(p+1\big)\big(2p+1\big)\cdots \big((n-2-i)p+1\big)\bigg) \\ & \qquad \times \bigg(\big(p\big)\big(2p\big)\cdots \big( (n-1-i)p\big)\bigg) \\ & \quad = \prod_{X \backslash \{(1,i+2)\}}\langle \nu_{i+1}, \epsilon_k-\epsilon_j\rangle. \end{align*}

Proof. The proofs of (1) and (2) are identical, so we will only prove (1). It will be sufficient to show that the lowest weights of $\widehat {\mathsf {M}}_I(\lambda _i)$ and ${\widehat {\mathsf {L}}(\lambda _{i+1}-p\varpi _n)}$ coincide. This is because Lemma 4.4 will then imply that the modules $\widehat {\mathsf {L}}(\lambda _{i+1} - p\varpi _n)$ and $\widehat {\mathsf {L}}(\lambda _{i})$ account for the complete set of composition factors (including multiplicity) of $\widehat {\mathsf {M}}_I(\lambda _i)$. The fact that $\widehat {\mathsf {L}}(\lambda _{i})$ is the unique simple quotient of $\widehat {\mathsf {M}}_I(\lambda _i)$ by Lemma 3.2 will then force $\widehat {\mathsf {L}}(\lambda _{i+1} - p\varpi _n)$ to be a submodule.

By Lemma 3.2 and (4.1), the lowest weight of $\widehat {\mathsf {M}}_I(\lambda _{i})$ is

\[ -(p-1)(n+1)\varpi_n + w_I(\lambda_{i}). \]

Likewise, the lowest weight of $\widehat {\mathsf {L}}(\lambda _{i+1}-p\varpi _n)$ is

\[ w_0(\lambda_{i+1}) - p\varpi_n. \]

Hence, the result will follow if we can prove that

(4.5)\begin{equation} -(p-1)(n+1)\varpi_n + w_I(\lambda_{i}) - w_0(\lambda_{i+1}) ={-}p\varpi_n. \end{equation}

(Recall the definitions (2.1), (4.2), and (4.3).)

To verify this identity, let us first define

\[ \rho_I = \frac{1}{2}\sum_{\alpha \in \varPhi_I^{+}}\alpha \in \frac{1}{2}\mathbf{X}, \]

then

\begin{align*} \rho & = \rho_I + \frac{1}{2}(\epsilon_1 - \epsilon_{n+1} + \epsilon_2 - \epsilon_{n+1} + \cdots + \epsilon_n - \epsilon_{n+1}) \\ & = \rho_I + \frac{(n+1)}{2}\varpi_n. \end{align*}

Thus, $w_0(\rho ) = -\rho _I - \frac {(n+1)}{2}\varpi _n$ and $w_I(\rho )= -\rho _I + \frac {(n+1)}{2}\varpi _n.$ We also observe that for $i=1,\ldots, n-1$, ${w_I(\varpi _{i}) = \varpi _n - \varpi _{n-1-i}}$ and, by recalling the $\mu _i$ from (2.10),

\begin{align*} w_I(\mu_i) & = \epsilon_{n-i} + \rho_I - \frac {(n+1)}{2}\varpi_n, \\ w_0(\mu_{i+1}) & = \epsilon_{n-i} + \rho_I + \frac{(n+1)}{2}\varpi_n, \end{align*}

for $i=0,\ldots, n-1$. Finally, we verify (4.5) by substituting the preceding identities into (2.11).

Proof. Without loss of generality, we can assume $\nu =0$. Also, it will be enough to prove (1), since (2) will follow from an identical argument.

The description of $\widehat {\mathsf {M}}_I(\lambda _i)$ for $i=0,\ldots, n-1$ follows immediately from Lemmas 4.4 and 4.5. On the other hand, the description of $\widehat {\mathsf {M}}_I(\lambda _n)$ is a consequence of [Reference Jantzen7, Lemma II.11.8]. Namely, since ${\langle \lambda _n + \rho, \alpha ^{\vee }\rangle = p}$ for all $\alpha \in I$, then it follows that $\widehat {\mathsf {L}}_I(\lambda _n) \cong \widehat {\mathsf {Z}}_{I}(\lambda _n).$ By transitivity of coinduction, we obtain

\[ \widehat{\mathsf{M}}_I(\lambda_n) \cong \operatorname{coind}_{(P_I^{+})_1T}^{G_1T}\operatorname{coind}_{B_1^{+}T}^{(P_I^{+})_1T}\lambda_n \cong \widehat{\mathsf{Z}}(\lambda_n). \]

Proof. We shall perform induction with respect to $n$, where $G = SL_{n+1}$. For the base case, when $n=1$, we have

\[ \lambda_0 = 0, \quad \lambda_1 = p-2. \]

In this case, $\mathcal {C}(\lambda _0)$ is actually the regular block and the claim can be verified through explicit computation (eg. [Reference Jantzen7, II.9.10]). Now suppose $n \geq 2$ and that the formula holds for $SL_{r+1}$ with $r \leq n-1$. We set $G' = [L_I, L_I]$ and note that $G' \cong SL_{n}$. Similarly, set $B' = B \cap G' \subset B \cap L_I$ and $T' = T \cap G'$, where $B'$ is the (lower) Borel subgroup of $G'$ and $T'$ is the torus of $G'$.

By the remark following [Reference Jantzen7, Proposition I.8.20],

\[ \mathsf{Z}_I(\lambda)|_{G_1'} \cong \operatorname{coind}_{(B'^{+})_1}^{G'_1}(\lambda|_{T'}) \]

and by [Reference Jantzen7, II.2.10(2)]

\[ \mathsf{L}_I(\lambda)|_{G'_1} \cong \mathsf{L}(\lambda|_{T'}). \]

For $i=0,\ldots, n-1$, set $\lambda '_i = \lambda _i|_{T'}$. Applying the inductive hypothesis to $G'$ yields

\[ [\mathsf{Z}_I(\lambda_i)|_{G_1'}: \mathsf{L}_I(\lambda_j)|_{G_1'}] = [\operatorname{coind}_{(B'^{+})_1}^{G'_1}(\lambda'_i): \mathsf{L}(\lambda'_j)] = {n-1\choose j} \]

for $0 \leq i,j \leq n-1$. The modules $\mathsf {L}_I(\lambda _0),\ldots, \mathsf {L}_I(\lambda _{n-1})$ form the entire set of irreducibles of the $\mathsf {Rep}((L_I)_1)$ block $\mathcal {C}_I(\lambda _0)$. Now since $\mathsf {Z}_I(\lambda _i)$ is an object of $\mathcal {C}_I(\lambda _0)$ and for each $i=0,\ldots, n-1$, $\mathsf {L}_I(\lambda _i)$ is the only irreducible of $\mathcal {C}_I(\lambda _0)$ which satisfies $\mathsf {L}_I(\lambda _i)|_{G_1'}\cong \mathsf {L}(\lambda '_i)$, then we must also have

\[ [\mathsf{Z}_I(\lambda_i): \mathsf{L}_I(\lambda_j)] = {n-1\choose j}. \]

If we take any Jordan–Hölder filtration of $\mathsf {Z}_I(\lambda _i)$ for $0 \leq i \leq n-1$ and apply the exact functor $\operatorname {coind}_{(P_I)^{+}_1}^{G_1}(-)$, we will get a filtration whose layers are of the form $\mathsf {M}_I(\lambda _j)$ for $0 \leq j \leq n-1$. Thus,

(4.6)\begin{equation} [\mathsf{Z}(\lambda_i): \mathsf{M}_I(\lambda_j)] = [\mathsf{Z}_I(\lambda_i): \mathsf{L}_I(\lambda_j)] = {n-1\choose j}, \end{equation}

where $[\mathsf {Z}(\lambda _i): \mathsf {M}_I(\lambda _j)]$ denotes the filtration multiplicity.

By Proposition 4.6, each $\mathsf {M}_I(\lambda _j)$ contributes a single copy of $\mathsf {L}(\lambda _j)$ and $\mathsf {L}(\lambda _{j+1})$. Thus, $[\mathsf {Z}(\lambda _i): \mathsf {L}(\lambda _0)] = 1$ and $[\mathsf {Z}(\lambda _i): \mathsf {L}(\lambda _n)] = 1$ since by (4.6), $[\mathsf {Z}(\lambda _i): \mathsf {M}_I(\lambda _0)] =1$ and $[\mathsf {Z}(\lambda _i): \mathsf {M}_I(\lambda _{n-1})] =1$. Likewise, for $1 \leq j \leq n-1$ the multiplicity

\[ [\mathsf{Z}(\lambda_i): \mathsf{L}(\lambda_j)] = {n-1\choose j} + {n-1\choose j-1} \]

arises from the ${n-1\choose j}$ copies of $\mathsf {M}_I(\lambda _j)$ and the ${n-1\choose j-1}$ copies of $\mathsf {M}_I(\lambda _{j-1})$. The proposition now follows from the well-known identity

\[ {n\choose j} = {n-1 \choose j} + {n-1 \choose j-1}. \]

Thus, we have verified the formula for $\mathsf {Z}(\lambda _i)$ when $0 \leq i \leq n-1$. The $\mathsf {Z}(\lambda _n)$ case can be verified by replacing $I$ with $J$ and repeating the same arguments.

Proof. By BGG reciprocity,

\begin{align*} {[\mathsf{Q}(\lambda_i):\mathsf{L}(\lambda_j)]} & = \sum_{k=0}^{n}[\mathsf{Q}(\lambda_i):\mathsf{Z}(\lambda_k)][\mathsf{Z}(\lambda_k):\mathsf{L}(\lambda_j)] \\ & = \sum_{k=0}^{n}[\mathsf{Z}(\lambda_k):\mathsf{L}(\lambda_i)][\mathsf{Z}(\lambda_k):\mathsf{L}(\lambda_j)] \\ & = (n+1){n \choose i}{n \choose j}, \end{align*}

where the last equality follows from Proposition 4.7.

Proof. First recall that for any $\lambda, \mu \in \mathbf {X}$,

(5.4)\begin{equation} [\overline{\text{rad}}_1\, \widehat{\mathsf{Q}}(\lambda):\widehat{\mathsf{L}}(\mu)] = \dim\operatorname{Ext}^{1}_{G_1T}(\widehat{\mathsf{L}}(\lambda), \widehat{\mathsf{L}}(\mu)), \end{equation}

which we obtain by applying $\operatorname {Hom}_{G_1T}(-, \widehat {\mathsf {L}}(\mu ))$ to the short exact sequence

\[ 0 \rightarrow \text{rad}^{1}\, \widehat{\mathsf{Q}}(\lambda) \rightarrow \widehat{\mathsf{Q}}(\lambda) \rightarrow \widehat{\mathsf{L}}(\lambda) \rightarrow 0. \]

On the other hand, if we combine Theorem 5.1 with (5.1), then we can deduce that the dimensions

\[ \dim\operatorname{Ext}^{1}_{G_1T}(\widehat{\mathsf{L}}(\lambda_i), \widehat{\mathsf{L}}(\lambda_j -p\nu)), \]

for $0 \leq i,j \leq n$ and $\nu \in \mathbf {X}$ are given by the appropriate weight multiplicities of $V$ and $V^{*}$.

Proof. The case for $n=1$ follows from [Reference Jantzen7, II.9.10] and the $n=2$ case follows from [Reference Xi12, Theorems 2.4-2.5]. Now suppose $n > 2$ and that the statement of the lemma holds for $SL_{r+1}$ whenever $1 \leq r < n$. By the same argument as in the proof of Proposition 4.7, we can assume the statement also holds for $\mathsf {Z}_I(\lambda _i)$ with $0 \leq i \leq n-1$ (respectively, for $\mathsf {Z}_J(\lambda _i)$ with $1 \leq i \leq n$).

The inductive hypothesis gives

(5.8)\begin{equation} \overline{\text{rad}}_1\, \mathsf{Z}_I(\lambda_i) =\mathsf{L}_I(\lambda_{i-1})^{{\oplus} i} \oplus \mathsf{L}_I(\lambda_{i+1})^{{\oplus} n-1-i} \end{equation}

for $0 \leq i \leq n-1$, and

\[ \overline{\text{rad}}_1\, \mathsf{Z}_J(\lambda_i) =\mathsf{L}_J(\lambda_{i-1})^{{\oplus} i-1} \oplus \mathsf{L}_J(\lambda_{i+1})^{{\oplus} n-i} \]

for $1 \leq i \leq n$. Now we coinduce to get

\[ \overline{\text{F}}_I^{0}(\lambda_i) = \operatorname{coind}_{(P_I^{+})_1}^{G_1}(\overline{\text{rad}}_0\, \mathsf{Z}_I(\lambda_i)) = \mathsf{M}_I(\lambda_i), \]

and

(5.9)\begin{equation} \overline{\text{F}}_I^{1}(\lambda_i) = \operatorname{coind}_{(P_I^{+})_1}^{G_1}(\overline{\text{rad}}_1\, \mathsf{Z}_I(\lambda_i)) =\mathsf{M}_I(\lambda_{i-1})^{{\oplus} i} \oplus \mathsf{M}_I(\lambda_{i+1})^{{\oplus} n-1-i} \end{equation}

for $0 \leq i \leq n-1$. (The formulas for $\overline {\text {F}}_J^{0}(\lambda _i)$ and $\overline {\text {F}}_J^{1}(\lambda _i)$ are similar.)

Let us focus on $I$ for now. By Proposition 4.6,

(5.10)\begin{equation} \overline{\text{rad}}_0\, \overline{\text{F}}_I^{0}(\lambda_i) = \mathsf{L}(\lambda_i), \quad \overline{\text{rad}}_1\, \overline{\text{F}}_I^{0}(\lambda_i) = \mathsf{L}(\lambda_{i+1}). \end{equation}

We also have

(5.11)\begin{equation} \overline{\text{rad}}_0\, \text{F}_I^{1}(\lambda_i) =\mathsf{L}(\lambda_{i-1})^{{\oplus} i} \oplus \mathsf{L}(\lambda_{i+1})^{{\oplus} n-1-i} = \overline{\text{rad}}_0\, \overline{\text{F}}_I^{1}(\lambda_i) \end{equation}

for $0 \leq i \leq n-1$, where the first isomorphism is a consequence of (5.8) and Proposition 3.5 and the second isomorphism is deduced from (5.9) and Proposition 4.6. It then follows that the surjective map

\[ \overline{\text{rad}}_0\, \text{F}_I^{1}(\lambda_i) \twoheadrightarrow \frac{\text{F}_I^{1}(\lambda_i)}{\text{rad}^{1}\, \text{F}_I^{1}(\lambda_i) + \text{F}_I^{2}(\lambda_i)} = \overline{\text{rad}}_0\, \overline{\text{F}}_I^{1}(\lambda_i) \]

is an isomorphism since the left- and right-hand sides have the same dimension. Consequently,

(5.12)\begin{equation} \text{F}_I^{2}(\lambda_i) \subseteq \text{rad}^{1} \text{F}_I^{1}(\lambda_i). \end{equation}

Now let $M = \text {rad}^{1}\, \mathsf {Z}(\lambda _i) / \text {rad}^{1}\text {F}_I^{1}(\lambda _i)$ and observe

\begin{align*} [M] & = [\text{rad}^{1}\, \mathsf{Z}(\lambda_i)] - [\text{rad}^{1}\text{F}_I^{1}(\lambda_i)] = [\mathsf{Z}(\lambda_i)] - [\mathsf{L}(\lambda_i)]-[\text{F}_I^{1}(\lambda_i)] + [\overline{\text{rad}}_0\, \text{F}_I^{1}(\lambda_i)] \\ & =[\overline{\text{F}}_I^{0}(\lambda_i)] - [\mathsf{L}(\lambda_i)] + [\overline{\text{rad}}_0\, \overline{\text{F}}_I^{1}(\lambda_i)]. \end{align*}

Thus, if we combine this with (5.10) and (5.11), we deduce the formula

\[ [M] = i[\mathsf{L}(\lambda_{i-1})] + (n-i)[\mathsf{L}(\lambda_{i+1})]. \]

Now observe that $\text {rad}^{1}\,\text {F}_I^{1}(\lambda _i) \subseteq \text {rad}^{2}\, \mathsf {Z}(\lambda _i)$ since $\text {F}_I^{1}(\lambda _i) \subseteq \text {rad}^{1}\, \mathsf {Z}(\lambda _i)$. As a consequence,

\[ \text{rad}^{1}\, M = \frac{\text{rad}^{2}\, \mathsf{Z}(\lambda_i) + \text{rad}^{1}\, \text{F}_I^{1}(\lambda_i)}{\text{rad}^{1}\,\text{F}_I^{1}(\lambda_i)} = \frac{\text{rad}^{2}\, \mathsf{Z}(\lambda_i)}{\text{rad}^{1}\,\text{F}_I^{1}(\lambda_i)}, \]

and hence,

\[ \overline{\text{rad}}_0\, M = \frac{\text{rad}^{1}\, \mathsf{Z}(\lambda_i)}{\text{rad}^{1}\,\text{F}_I^{1}(\lambda_i)} \bigg / \frac{\text{rad}^{2}\, \mathsf{Z}(\lambda_i)}{\text{rad}^{1}\,\text{F}_I^{1}(\lambda_i)} = \overline{\text{rad}}_1\, \mathsf{Z}(\lambda_i). \]

Moreover, $\overline {\text {rad}}_0\, \text {F}_I^{1}(\lambda _i) = \text {F}_I^{1}(\lambda _i)/\text {rad}^{1}\text {F}_I^{1}(\lambda _i) \hookrightarrow M.$ Thus by (5.11), there exists an injective map

\[ \iota: \mathsf{L}(\lambda_{i-1})^{{\oplus} i} \oplus \mathsf{L}(\lambda_{i+1})^{{\oplus} n-1-i} \hookrightarrow M, \]

whose image must account for every irreducible factor of $M$ except for a single copy of $\mathsf {L}(\lambda _{i+1})$. Consequently, $M$ fits into a short exact sequence of the form

\[ 0 \longrightarrow \mathsf{L}(\lambda_{i-1})^{{\oplus} i} \oplus \mathsf{L}(\lambda_{i+1})^{{\oplus} n-1-i} \xrightarrow{\iota} M\longrightarrow \mathsf{L}(\lambda_{i+1}) \longrightarrow 0. \]

Now let $N = \iota (\mathsf {L}(\lambda _{i-1})^{\oplus i})$ so that $[M/N] = (n-i)[\mathsf {L}(\lambda _{i+1})]$. From [Reference Jantzen7, Proposition II.12.9], we know that any $G_1$-module with precisely one isotypic component must be semisimple. It follows that $M/N \cong \mathsf {L}(\lambda _{i+1})^{\oplus n-i}$. In particular, we get a surjection

\[ \text{rad}^{1}\, \mathsf{Z}(\lambda_i) \twoheadrightarrow M/N \cong \mathsf{L}(\lambda_{i+1})^{{\oplus} n-i}. \]

Finally, since every map from $\text {rad}^{1}\, \mathsf {Z}(\lambda _i)$ to a semisimple module factors through $\overline {\text {rad}}_1\, \mathsf {Z}(\lambda _i)$ (recall that the latter is the head of the former), we get a surjection

(5.13)\begin{equation} \overline{\text{rad}}_1\, \mathsf{Z}(\lambda_i) \twoheadrightarrow \mathsf{L}(\lambda_{i+1})^{{\oplus} n-i}. \end{equation}

(Recall that $\overline {\text {rad}}_0\, M = \overline {\text {rad}}_1\, \mathsf {Z}(\lambda _i)$.) Observe now that if $i=0$, then the preceding map must be an isomorphism since every possible factor of $\overline {\text {rad}}_1\, \mathsf {Z}(\lambda _i)$ has been accounted for and we are done.

On the other hand, if $1 \leq i \leq n$, then by replacing $I$ with $J$ and repeating the same arguments, we also obtain a surjection

(5.14)\begin{equation} \overline{\text{rad}}_1\, \mathsf{Z}(\lambda_i) \twoheadrightarrow \mathsf{L}(\lambda_{i-1})^{{\oplus} i}. \end{equation}

Therefore, the lemma follows by combining (5.13) and (5.14) which account for every possible factor of $\overline {\text {rad}}_1\, \mathsf {Z}(\lambda _i)$.

Proof. The $n=1$ case follows from [Reference Jantzen7, II.9.10], and the $n=2$ case is given in [Reference Xi12, Theorems 2.4-2.5]. Suppose now that $n>2$, and that the statement holds for $SL_{r+1}$ with $2 \leq r < n$. The inductive hypothesis can be applied to $L_I$ and $L_J$ as in the proof of Proposition 4.7. More precisely, for $L_I$ and $i=0,\ldots, n-1$, [Reference Jantzen7, Lemma II.9.2(3)] implies that the highest weight of every composition factor of $\widehat {\mathsf {Z}}_I(\lambda _i)$ is of the form $\lambda _i - \gamma$ for various $\gamma \in \mathbb {Z} I$.

If we set $G' = [L_I, L_I]\cong SL_n$, and $T' = T\cap G'$, we get

\[ \widehat{\mathsf{Z}}_I(\lambda_i)|_{G'_1T'} \cong \widehat{\mathsf{Z}}(\lambda_i|_{T'}), \quad\widehat{\mathsf{L}}(\lambda_i - \gamma)|_{G'_1T'} \cong \widehat{\mathsf{L}}(\lambda_i|_{T'} - \gamma|_{T'}), \]

with ${[\widehat {\mathsf {Z}}_I(\lambda _i): \widehat {\mathsf {L}}(\lambda _i - \gamma )] = [\widehat {\mathsf {Z}}(\lambda _i|_{T'}): \widehat {\mathsf {L}}(\lambda _i|_{T'} - \gamma |_{T'})] }$. Furthermore,

(5.15)\begin{equation} \gamma|_{T'} = \sum_{i =1}^{n-1} a_i (\epsilon_i - \epsilon_{i+1})|_{T'} \iff \gamma = \sum_{i =1}^{n-1} a_i (\epsilon_i - \epsilon_{i+1}). \end{equation}

So the inductive hypothesis is applied to $L_I$ by first expressing the irreducibles occurring in the inductive hypothesis for $SL_{n}$ as $\widehat {\mathsf {L}}(\lambda _i|_{T'} - \gamma |_{T'})$ for various uniquely determined $\gamma$, and then employing (5.15) to obtain the corresponding formulas for $\overline {\text {rad}}_1\, \widehat {\mathsf {Z}}_I(\lambda _i)$. (The case for $L_J$ with $i=1,\ldots, n$ is similar.)

Thus, the inductive hypothesis gives

\begin{align*} \overline{\text{rad}}_1\, \widehat{\mathsf{Z}}_I(\lambda_0) & = \widehat{\mathsf{L}}_I(\lambda_1 - p\varpi_{n-1} + p\varpi_{n}) \oplus \widehat{\mathsf{L}}_I(\lambda_1 - p\varpi_{n-2}+p\varpi_{n-1})\\ & \quad \oplus \cdots \oplus \widehat{\mathsf{L}}_I(\lambda_1 - p\varpi_1 + p\varpi_2). \end{align*}

So the formula for $\overline {\text {rad}}_1\, \widehat {\mathsf {Z}}(\lambda _0)$ follows from Proposition 4.6 and the proof of Lemma 5.3. For $i=1,\ldots, n-1$, the hypothesis also gives

\begin{align*} \overline{\text{rad}}_1\, \widehat{\mathsf{Z}}_I(\lambda_i) & =\left(\bigoplus_{k=1}^{i}\, \widehat{\mathsf{L}}_I(\lambda_{i-1} -p\varpi_k + p\varpi_{k-1})\right) \\ & \oplus \left(\bigoplus_{k=1}^{n-1-i}\, \widehat{\mathsf{L}}_I(\lambda_{i+1} -p\varpi_{n+1-k} + p\varpi_{n+2-k})\right). \end{align*}

Now once again, the formula for $\overline {\text {rad}}_1\, \widehat {\mathsf {Z}}(\lambda _i)$ is obtained by applying Proposition 4.6 and then proceeding as in the proof of Lemma 5.3.

Finally, the formula for $\overline {\text {rad}}_1\, \widehat {\mathsf {Z}}(\lambda _n)$ is verified by first applying the inductive hypothesis to $L_J$, which gives

\begin{align*} \overline{\text{rad}}_1\, \widehat{\mathsf{Z}}_J(\lambda_n) & =\widehat{\mathsf{L}}_J(\lambda_{n-1} - p\varpi_2+p\varpi_1) \oplus \widehat{\mathsf{L}}_J(\lambda_{n-1} - p\varpi_{3}+p\varpi_2)\\ & \quad \oplus \cdots \oplus \widehat{\mathsf{L}}_J(\lambda_{n-1} - p\varpi_{n} + p\varpi_{n-1}), \end{align*}

and then proceeding as above.

Proof. By definition, $E$ is an indecomposable length 2 module with head $\widehat {\mathsf {L}}(\lambda )$ and socle $\widehat {\mathsf {L}}(\mu )$. In particular, $E$ is a cyclic module for $\widehat {\mathcal {U}}$ which is generated by some $\lambda$-weight vector $v_{\lambda }$. Now every weight $\gamma$ occurring with non-zero multiplicity in $\widehat {\mathsf {L}}(\mu )$ satisfies $\gamma \leq \mu$, and hence, $\lambda \not \leq \gamma$. On the other hand, any weight $\gamma$ occurring with non-zero multiplicity in $\widehat {\mathsf {L}}(\lambda )$ with $\gamma \neq \lambda$ must also satisfy $\gamma < \lambda$, and thus, $\lambda + \alpha _i \not \leq \gamma$ for $i=1,\ldots,n$. Therefore, $\lambda + \alpha _i$ cannot occur as a weight of $E$ which forces $X_{\alpha _i}\cdot v_{\lambda } =0$ for all $i$. As a result, the surjective map

(5.16)\begin{equation} \begin{aligned} \widehat{\mathcal{U}} & \twoheadrightarrow E\\ X & \mapsto X\cdot v_{\lambda} \end{aligned} \end{equation}

factors through the ideal $\widehat {I}_{\lambda }$, and so it follows that $E$ is a quotient of ${\widehat {\mathsf {Z}}(\lambda ) \cong \widehat {\mathcal {U}}/\widehat {I}_{\lambda }}$.

Proof. The $G/G_1$-module $M$ is non-zero if and only if $M_{p\nu } \neq 0$ for some $\nu \in \mathbf {X}^{+}$. By Lemma 5.3, we know that there does not exist an extension $E \in M$ which is isomorphic to a quotient of $\mathsf {Z}(\lambda _i)$. Hence, there are no extensions $E \in M_{p\nu } = \operatorname {Ext}^{1}_{G_1T}(\widehat {\mathsf {L}}(\lambda _i), \widehat {\mathsf {L}}(\lambda _j - p\nu ))$ which occur as a quotient of $\widehat {\mathsf {Z}}(\lambda _i)$. Thus, by Lemma 5.5, we must have $\lambda _i \leq \lambda _j - p\nu$ if $M_{p\nu } \neq 0$.

Proof. We begin by noting that the $i=j$ case immediately follows from [Reference Jantzen7, Proposition II.12.9]. So we only have to consider the case when $|i-j| \geq 2$. By the comments immediately preceding this lemma, it suffices to show that there exist no dominant weights $\nu$ such that

\[ p\nu \leq \lambda_j - \lambda_i, \]

whenever $|i-j| \geq 2$. Applying (2.11), we compute

\[ \lambda_j - \lambda_i = \begin{cases} \epsilon_{j+1} - \epsilon_{i+1} + p(-\varpi_{j+1}+\varpi_{i+1}) & \text{ if }0\leq i,j \leq n-1, \\ \epsilon_{j+1} - \epsilon_{n+1} - p\varpi_{j+1} & \text{ if }0 \leq j \leq n-1 \text{ and }i = n, \\ \epsilon_{n+1} - \epsilon_{i+1} + p\varpi_{i+1} & \text{ if }j = n \text{ and }0\leq i \leq n-1. \end{cases} \]

First suppose that $j>i$, then we can see that

(5.17)\begin{equation} \lambda_j - \lambda_i < \begin{cases} p(-\varpi_{j+1} + \varpi_{i+1}) & \text{ if }j< n, \\ p\varpi_{i+1} & \text{ if }j = n. \end{cases} \end{equation}

The $j=n$ case is now obvious since $\varpi _{i+1}$ is minuscule, and thus $p\nu \leq \lambda _n - \lambda _i$ implies $p\nu < p\varpi _{i+1}.$ Hence by (5.3), $\nu < \varpi _{i+1}$ which is impossible for $\nu \in \mathbf {X}^{+}$. On the other hand, for $i < j < n$,

\[ \lambda_j - \lambda_i < p(-\varpi_{j+1} + \varpi_{i+1}) < p\varpi_{n+1-(j-i)}, \]

where the rightmost inequality comes from the fact that $\varpi _{n+1-(j-i)} = w(-\varpi _{j+1} + \varpi _{i+1})$ for some $w \in W$. Now by applying (5.3) as above, we can see that if $p\nu \leq \lambda _j - \lambda _i$, then $\nu < \varpi _{n+1-(j-i)}$, which is also impossible for $\nu \in \mathbf {X}^{+}$.

Suppose now that $j < i$. When $i=n$, we can see that $\lambda _j-\lambda _n \not \in p\mathbf {X}$ so if $p\nu < \lambda _j - \lambda _n$ (and thus $(\lambda _j - \lambda _n) - p\nu \in \mathbb {Z}_{\geq 0}\varPhi ^{+}$), then

\[ (\lambda_j - \lambda_n) - p\nu = (\epsilon_{j+1}-\epsilon_{n+1}) + ({-}p\varpi_{j+1} - p\nu). \]

If we set $\gamma = -\varpi _{j+1} - \nu$ and then compare both sides of the preceding equation, we can deduce that $p\gamma \in p\mathbf {X}\cap \mathbb {Z}\varPhi = p\mathbb {Z}\varPhi$, where the equality follows from (5.2). So we can write $p\gamma = \sum _{k=1}^{n} pc_k\alpha _k$ where $c_k \in \mathbb {Z}$ for all $k$. Moreover, $\gamma \in \mathbb {Z}_{\geq 0}\varPhi ^{+}$ since

\[ (\epsilon_{j+1}-\epsilon_{n+1}) + p\gamma = \sum_{k=1}^{j} pc_k\alpha_k + \sum_{k=j+1}^{n} (pc_k + 1)\alpha_k, \]

and thus if $c_k < 0$ for some $k$, then $pc_k +1 <0$. But this contradicts the assumption that $(\lambda _j - \lambda _n) - p\nu \in \mathbb {Z}_{\geq 0}\varPhi ^{+}$.

Hence,

\[ p\nu \leq{-}p\varpi_{j+1} < w_0({-}p\varpi_{j+1}) = p\varpi_{n-j} \]

and so by (5.3), $\nu < \varpi _{n-j}$, which is impossible for $\nu \in \mathbf {X}^{+}$. For $j < i < n$, the same reasoning shows that if $p\nu \leq \lambda _j - \lambda _i$, then $p\nu \leq p(-\varpi _{j+1} + \varpi _{i+1}) < p\varpi _{i-j-1}$. This forces $\nu < \varpi _{i-j-1}$, which again is impossible for $\nu \in \mathbf {X}^{+}$.

Proof. By the observation immediately preceding the lemma, we know that if $M_{\nu } \neq 0$ and $M_{\nu '} \neq 0$, then both $p\nu$ and $p\nu '$ are in the same root coset since they are both comparable to the weight $\mu ^{0} - \lambda$. Thus by (5.2),

\[ p\nu - p\nu' \in p\mathbf{X}\cap \mathbb{Z}\varPhi = p\mathbb{Z}\varPhi. \]

Therefore, $\nu - \nu ' \in \mathbb {Z}\varPhi$ and we are done.

Proof. Let $\nu = \varpi _1 + \sum _{i=1}^{n} a_i\alpha _i$, where $a_i \in \mathbb {Z}$ is arbitrary for all $i$. The lemma will follow if we can show that

\[ \nu \in \mathbf{X}^{+} \implies a_i \geq 0 \text{ for all }i. \]

(Note that if $n=1$, then the claim is immediate since $\nu = \varpi _1 + a_1\alpha _1 = (2a_1+1)\varpi _1$, where $2a_1 +1 \geq 0$ implies $a_1 \geq 0$ because $a_1$ is an integer.)

In general, write $\nu = c_1\varpi _1 + c_2\varpi _2 + \cdots +c_n \varpi _n$, and observe that $c_1 = 2a_1-a_2+1$, $c_n = -a_{n-1} + 2a_n$ and $c_i = -a_{i-1} + 2a_i -a_{i+1}$ for $i=2, \ldots, n-1$. Thus, the condition $\nu \in \mathbf {X}^{+}$ is given by

\begin{align*} 0 & \leq{-}a_{n-1} + 2a_n \\ 0 & \leq{-}a_{n-2} + 2a_{n-1} - a_n \\ & \quad \vdots \\ 0 & \leq{-}a_{n-(k+1)} + 2a_{n-k} - a_{n-(k-1)}\\ & \quad \vdots \\ 0 & \leq{-}a_1 + 2a_2 - a_3 \\ -1 & \leq 2a_1 - a_2. \end{align*}

From the above list of inequalities, we can deduce that $a_n \geq \frac {1}{2}a_{n-1}$, $a_{n-1} \geq \frac {2}{3}a_{n-2}$, and generally, that

\begin{align*} a_{n-k} & \geq \frac{k+1}{k+2}a_{n-(k+1)} \quad \text{for }k = 0, \ldots, n-2,\\ a_1 & \geq \frac{-n}{n+1}. \end{align*}

Now since $a_1 \in \mathbb {Z}$ and $-1 < \frac {-n}{n+1} \leq a_1$, we get $a_1 \geq 0$. But then $a_1 \geq 0$ implies $a_2 \geq \frac {n-1}{n}a_1 \geq 0$, which then implies $a_3 \geq \frac {n-2}{n-1}a_2 \geq 0$. Proceeding in this way, we conclude that $a_i \geq 0$ for all $i$.

Proof. As in the proof of Lemma 5.3, we will proceed by induction on $n\geq 2$. The base case again follows from the explicit formulas given in [Reference Xi12, Theorems 2.4-2.5]. Suppose $n > 2$ and assume the statement of the lemma holds for all $SL_{r+1}$ with $2 \leq r < n$. The argument in the proof of Proposition 4.7 implies that the statement also holds for the Levi factor $L_I$ with $\mathsf {Z}_I(\lambda _i)$ and $0 \leq i \leq n-1$ (respectively, $L_J$ with $\mathsf {Z}_J(\lambda _i)$ and $1 \leq i \leq n$).

For simplicity, let us begin by fixing $i \in \{0,\ldots, n-1\}$. The inductive hypothesis gives

\[ \overline{\text{rad}}_j\, \mathsf{Z}_I(\lambda_i) = \bigoplus_{0 \leq k \leq n-1} \mathsf{L}_I(\lambda_k)^{{\oplus} m^{j}_{ik}}, \]

where $m^{j}_{ik} \neq 0$ implies $k \equiv i + j \mod 2$. Thus,

\[ \overline{\text{F}}_I^{j}(\lambda_i) = \bigoplus_{0 \leq k \leq n-1} \mathsf{M}_I(\lambda_k)^{{\oplus} m^{j}_{ik}}. \]

By Proposition 4.6,

(6.4)\begin{equation} \overline{\text{rad}}_0\, \overline{\text{F}}_I^{j}(\lambda_i) = \bigoplus_{0 \leq k \leq n-1} \mathsf{L}(\lambda_k)^{{\oplus} m^{j}_{ik}},\quad \overline{\text{rad}}_1\, \overline{\text{F}}_I^{j}(\lambda_i) = \bigoplus_{0 \leq k \leq n-1} \mathsf{L}(\lambda_{k+1})^{{\oplus} m^{j}_{ik}}. \end{equation}

From the inductive hypothesis, we can see that (6.1) will hold on the factors of $\overline {\text {rad}}_j\, \mathsf {Z}(\lambda _i)$, provided we verify (6.2). We will proceed by induction on $j\geq 0$. The base case, $j=0$, is obvious since ${\overline {\text {rad}}_0\, \mathsf {Z}(\lambda _i) = \mathsf {L}(\lambda _i)}$. Also, the $j=1$ case follows from Lemma 5.3. Now assume $j\geq 2$ and that (6.2) holds for $0 \leq l < j$.

The inductive hypothesis for $j$ gives $\text {F}_I^{j-1}(\lambda _i) \subseteq \text {rad}^{j-1}\, \mathsf {Z}(\lambda _i)$, and thus,

\[ {\text{rad}^{1}\, \text{F}_I^{j-1}(\lambda_i) \subseteq \text{rad}^{j}\, \mathsf{Z}(\lambda_i)}. \]

Now if we apply the inductive hypothesis for $L_I$ and reason as we did in the portion of the proof of Lemma 5.3 between (5.11) and (5.12), we can deduce

\[ \text{F}_I^{j}(\lambda_i) \subseteq \text{rad}^{1}\, \text{F}_I^{j-1}(\lambda_i) \subseteq \text{rad}^{j}\, \mathsf{Z}(\lambda_i), \]

and hence, the first claim of (6.2). Moreover, by imitating the arguments immediately following (5.12), we can also show that $\overline {\text {rad}}_j\, \mathsf {Z}(\lambda _i)$ is the head of the module

\[ M = \frac{\text{rad}^{j}\, \mathsf{Z}(\lambda_i)}{\text{rad}^{1}\, \text{F}_I^{j}(\lambda_i)}, \]

which fits into a short exact sequence of the form

\[ 0 \longrightarrow \overline{\text{rad}}_0\, \overline{\text{F}}_I^{j}(\lambda_i) \longrightarrow M \longrightarrow \overline{\text{rad}}_1\, \overline{\text{F}}_I^{j-1}(\lambda_i)\longrightarrow 0. \]

But now, by the inductive hypothesis and (6.4), we can see that every factor $\mathsf {L}(\lambda _k)$ of $M$ occurring with non-zero multiplicity must satisfy $k \equiv i + j \mod 2$. In particular, if $\mathsf {L}(\lambda _s)$ and $\mathsf {L}(\lambda _t)$ are two non-zero factors of $M$, then $|s-t| \neq 1$. Thus, Theorem 5.1 implies the preceding short exact sequence is split, and hence (6.2) holds for all $j\geq 0$.

So we have verified (6.1) and (6.2) for $0 \leq i \leq n-1$ and $j\geq 0$. On the other hand, if we replace $I$ with $J$ and fix any $i \in \{1,\ldots, n\}$, then the same argument as above also verifies (6.1) and (6.3) for all $j\geq 0$.

Proof. By (2.9), we are reduced to determining the radical layers of $\mathsf {Z}(\lambda _i)$. As usual, we will prove (6.6) by induction on $n\geq 1$. The base case, $n=1$, is trivial. Now assume by induction that the formula holds for $SL_{r+1}$ with $1 \leq r < n$ and apply this to $L_I$. If we fix $i \in \{0,\ldots, n-1\}$, then by Lemma 6.1 and Proposition 4.6,

\begin{align*} \overline{\text{rad}}_j\, \mathsf{Z}(\lambda_i) & = \overline{\text{rad}}_0\, \overline{\text{F}}_I^{j}(\lambda_i) \oplus \overline{\text{rad}}_1\, \overline{\text{F}}^{j-1}_I(\lambda_i)\\ & = \bigoplus_{0 \leq k \leq i} \mathsf{L}(\lambda_{i+j - 2k})^{{\oplus} {i \choose k}{n-i-1 \choose j-k}} \oplus \bigoplus_{0 \leq k \leq i} \mathsf{L}(\lambda_{i+j - 2k})^{{\oplus} {i \choose k}{n-1-i \choose j-k-1}}\\ & = \bigoplus_{0 \leq k \leq i} \mathsf{L}(\lambda_{i+j - 2k})^{{\oplus} {i \choose k}\left({n-i-1 \choose j-k} +{n-1-i \choose j-k-1}\right)}\\ & = \bigoplus_{0 \leq k \leq i} \mathsf{L}(\lambda_{i+j - 2k})^{{\oplus} {i \choose k}{n-i \choose j-k}}. \end{align*}

Similarly, we can verify (6.6) for $i \in \{1,\ldots,n\}$ by applying the inductive hypothesis to $L_J$.

Proof. We first note that the Lemmas occurring in [Reference Andersen and Kaneda3, 3.5 and 3.6] can be adapted to our setting. This is because their proofs essentially consist of the same types of arguments employed in the proof of [Reference Jantzen7, Proposition II.11.2], as well as certain general results on socle filtrations of modules, and on the basic properties of $\widehat {\mathsf {Z}}'(\lambda )$ (e.g. the highest weight structure). In particular, there is no dependence on the $p$-regularity of $\lambda \in \mathbf {X}$, or even on the prime $p$. The proof of our result follows by applying the more general versions of these lemmas to imitate the proof of [Reference Andersen and Kaneda3, Proposition 3.7].