2 The i-quantum group ${\mathbf{U}}^{\jmath}(n)$ and its BLM-type realisation

We first recall the definition of the i-quantum group ${\mathbf U^\jmath (n)}$ from [Reference Bao, Kujawa, Li and Wang1, Section 4.3]. Here, $n\geqslant 1$ denotes the rank of the i-quantum group and we always fix n in the following discussion.

The algebra ${\mathbf {U}}^{{\jmath }}(n)$ admits an involutory automorphism $\omega $ defined by

$$ \begin{align*} \omega(e_i) = f_i,\quad \omega(f_i) = e_i,\quad \omega(d_i) = d_i^{-1}\ (1 \leqslant i \leqslant n),\quad \omega(d_{n+1}) = v^{-1}d_{n+1}^{-1}. \end{align*} $$

Now, we follow [Reference Du and Wu8] to define the Lusztig (integral) form of ${\mathbf {U}}^{{\jmath }}(n)$ . Let $U^{\jmath }_{\mathcal Z}(n)$ be the ${\mathcal Z}$ -subalgebra of ${\mathbf {U}}^{{\jmath }}(n)$ generated by the elements

$$ \begin{align*}e_{i}^{(m)}:=\frac{e_i^{m}}{[m]^!},\quad f_{i}^{(m)}:=\frac{f_i^{m}}{[m]^!},\quad d_i^{\pm 1},\quad\bigg[ {d_{i};0\atop t}\bigg],\quad d_{n+1}^{\pm 1},\quad \bigg[{d_{n+1};0\atop t}\bigg]_{\upsilon^2}\end{align*} $$

for all $m,t\in \mathbb N$ and $1\leqslant i\leqslant n$ , where

(2-1) $$ \begin{align} \bigg[ {d_{i};0\atop t}\bigg]=\prod_{1\leqslant s\leqslant t}\frac{d_i\upsilon^{-s+1}-d_i^{-1}\upsilon^{s-1}}{\upsilon^s-\upsilon^{-s}},\quad \bigg[ {d_{n+1};0\atop t}\bigg]_{\upsilon^2}=\prod_{1\leqslant s\leqslant t}\frac{d_{n+1}\upsilon^{-2s+2}-d_{n+1}^{-1}\upsilon^{2s-2}}{\upsilon^{2s}-\upsilon^{-2s}}. \end{align} $$

Let $\text {Mat}_{{2n+1}}(\mathbb N)$ be the ${2n+1\times 2n+1}$ matrices with entries in $\mathbb N$ and introduce the following index sets:

$$ \begin{align*} &\Xi_{2n+1}=\{A=(a_{i,j})\in \text{Mat}_{2n+1}(\mathbb{N}) \mid a_{i,j}=a_{{2n+2}-i,{2n+2}-j} \text{ for all } i,j\in [1,{2n+1}]\},\\ &\Xi_{2n+1}^{0\operatorname{diag}}=\{A-\operatorname{diag}(a_{1,1},a_{2,2},\ldots,a_{{2n+1},{2n+1}})\mid A=(a_{i,j})\in\Xi_{2n+1}\},\\ &\Xi_{2n+1,2r+1}=\bigg\{A=(a_{i,j})\in\Xi_{2n+1}\mid |A|:=\sum_{i,j}a_{i,j}=2r+1\bigg\}. \end{align*} $$

Denote $E_{i,j}$ as the standard unit matrix in $\text {Mat}_{2n+1}(\mathbb N)$ . Note that the matrix $E_{i,j}^\theta :=E_{i,j}+E_{2n+2-i,2n+2-j}$ is in $\Xi _{2n+1}$ .

Let ${\mathcal S}^{\jmath }_{\mathcal Z}(n,r)$ be the q-Schur algebra of type B over ${\mathcal Z}$ introduced in [Reference Bao, Kujawa, Li and Wang1, Section 2.3] (see also [Reference Du and Wu8, (3.0.3)]). It has a normalised (or standard) ${\mathcal Z}$ -basis $\{[A]\}_{A\in \Xi _{2n+1,2r+1}}$ . Let

$$ \begin{align*} {{\boldsymbol{{\mathcal S}}}^\jmath(n,r)}:={{\mathcal S}^\jmath_{\mathcal Z}(n,r)}\otimes_{\mathcal Z}\mathbb Q(\upsilon). \end{align*} $$

Let

$$ \begin{align*} \Lambda(n+1,r)=\{\lambda=(\lambda_1,\ldots, \lambda_{n+1})\in \mathbb N^{n+1}\mid {|\lambda|:=}\lambda_1+\cdots+\lambda_{n+1}=r\}. \end{align*} $$

Define the bijection

(2-2) $$ \begin{align} \widetilde{\ }:\Lambda(n+1,r) & \longrightarrow{\widetilde\Lambda}(n+1,r)\subseteq\Lambda(2n+1,2r+1),\nonumber\\ \lambda &\longmapsto\widetilde\lambda:= (\lambda_{1},\ldots,\lambda_n,2\lambda_{n+1}+1,\lambda_n,\ldots,\lambda_{1}), \end{align} $$

where ${\widetilde \Lambda }(n+1,r)$ is the image of $\Lambda (n+1,r)$ under the map $\widetilde {\ }\,$ . In addition, let $\boldsymbol {e}_i=(0,\ldots ,0,\underset {(i)}1,0,\ldots ,0)\in \mathbb N^{2n+1}$ and set $\boldsymbol {e}^\theta _i=\boldsymbol {e}_i+\boldsymbol {e}_{2n+2-i}$ .

We record the fact that for $\lambda \in \Lambda (n+1,r)$ and $A\in \Xi _{2n+1,2r+1}$ ,

$$ \begin{align*} \ [\tilde{\lambda}][A]= \delta_{\tilde{\lambda},\mathrm{ro}(A)}[A] \quad \text{and}\quad [A][\tilde{\lambda}]=\delta_{\mathrm{co}(A),\tilde{\lambda}}[A], \end{align*} $$

where $\mathrm {ro}(A)=(\sum _ja_{1,j},\ldots ,\sum _ja_{2n+1,j})$ and $\mathrm {co}(A)=(\sum _ia_{i,1},\ldots ,\sum _ia_{i,2n+1})$ are the sequences of row and column sums of A. Here, ${\widetilde \lambda }$ in $[{\widetilde \lambda }]$ denotes the diagonal matrix $\text {diag}({\widetilde \lambda })$ for simplicity.

Define

$$ \begin{align*} (\ )^\circ&:\mathbb Z^{n+1}\longrightarrow\mathbb Z^{2n+1}, {\mathbf{i}}\longmapsto{\mathbf{i}}^\circ:=({\mathbf{i}},0^n),\\ (\ )^{\bullet}&:\mathbb Z^{n+p}\longrightarrow\mathbb Z^{n}, {\mathbf{i}}\longmapsto{\mathbf{i}}^{\bullet}:=(i_1,\ldots,i_n). \end{align*} $$

Here, $\circ $ indicates expansion by n zeros and $\bullet $ means a ‘wiping-out’ (or blacking-out) of the last p components. We regard $\mathbb Z^{n+1}$ as a subset of $\mathbb Z^{2n+1}$ via the injection ${(\ )}^\circ : \mathbb Z^{n+1}\rightarrow \mathbb Z^{2n+1}$ .

For $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ , ${\mathbf {i}}=(i_1,i_2,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ , recall the elements $A({\mathbf {i}},r)$ defined in [Reference Du and Wu8, (4.1.1)]:

(2-3) $$ \begin{align} A({{\mathbf{i}}},r):=A({{\mathbf{i}}}^\circ,r)=\begin{cases}\displaystyle \sum_{\mu\in\Lambda(n+1,r-{|A|}/2)}\upsilon^{{\widetilde \mu}\cdot{\mathbf{i}}^\circ}[A+{\widetilde\mu}]&\text{ if }|A|\leqslant 2r,\\ 0\qquad&\text{ if }|A|>2r,\end{cases} \end{align} $$

where ${\widetilde \mu }\cdot {\mathbf {i}^{\circ }}=\sum _{k=1}^{n}\mu _k i_k+(2\mu _{n+1}+1){i}_{n+1}$ .

According to [Reference Du and Wu8, Corollary 6.2], for all $r\geqslant 0$ , there are algebra epimorphisms

$$ \begin{align*} \phi^{\jmath}_r:{\mathbf U^\jmath(n)} \,-\!\!\!\twoheadrightarrow{{\boldsymbol{{\mathcal S}}}^\jmath(n,r)} \end{align*} $$

satisfying, for all $1\leqslant h\leqslant n$ ,

$$ \begin{align*}e_{h}\mapsto E^{\theta}_{h,h+1}(\mathbf{0},r),\;f_{h}\mapsto E^{\theta}_{h+1,h}(\mathbf{0},r),\;d_{h}^{\pm1}\mapsto {O}({\pm {\boldsymbol{e}}_h},r),\;d^{\pm1}_{n+1}\mapsto \upsilon^{\mp1}{ O}({\pm {\boldsymbol{e}}_{n+1}},r),\end{align*} $$

where O denotes the zero matrix in $\Xi _{2n+1}^{\mathrm {0\ diag}}$ . Thus, by [Reference Du and Wu8, Theorem 7.1], there is an algebra monomorphism

(2-4) $$ \begin{align} \phi^\jmath:{\mathbf U^\jmath(n)}&\longrightarrow {\widehat{{\boldsymbol{{\mathcal S}}}}^\jmath(n)}:=\prod_{r\geqslant 0}{{\boldsymbol{{\mathcal S}}}^\jmath(n,r)},\nonumber\\ e_{h}\mapsto E^{\theta}_{h,h+1}(\mathbf{0}),\quad f_{h}\mapsto E^{\theta}_{h+1,h}(\mathbf{ 0}),&\quad d_{h}^{\pm1}\mapsto {O}({\pm {\boldsymbol{e}}_h}),\quad d^{\pm1}_{n+1}\mapsto \upsilon^{\mp1}{ O}({\pm {\boldsymbol{e}}_{n+1}}), \end{align} $$

where $E^{\theta }_{h,h+1}(\mathbf {0}):=(E^{\theta }_{h,h+1}(\mathbf {0},r))_{r \geqslant 0}$ , and so forth. Moreover, putting $A({\mathbf {i}})=(A({\mathbf {i}},r))_{r\geqslant 0}$ , $\phi ^\jmath $ has the image

(2-5) $$ \begin{align} {\boldsymbol{\mathfrak{A}}^\jmath(n)}:={\mathrm{im}}(\phi^\jmath)=\text{span}_{\mathbb Q(\upsilon)}\{A({{\mathbf{i}}})\mid A\in\Xi^{\mathrm{0\ diag}}_{2n+1}, {\mathbf{i}}=(i_1,i_2,\ldots,i_{n+1})\in\mathbb{Z}^{n+1}\}. \end{align} $$

We end this section with the triangular relation within ${{\boldsymbol {{\mathcal S}}}^\jmath (n,r)}$ . We linearly order the set

$$ \begin{align*} \mathscr T_{2n+1} =\{(i,h,j)\mid 2n+1\geqslant i>h\geqslant j\geqslant1\}, \end{align*} $$

by the following semi-lexicographic order (see [Reference Bao, Kujawa, Li and Wang1, Theorem 3.10]):

$$ \begin{align*} (i,h,j)\angle(i',h',j')\iff i<i'\text{ or }i=i',j<j'\text{ or }i=i',j=j', h>h'. \end{align*} $$

Denote the resulting sequence by $\mathscr T_{2n+1}^\angle $ . Alternatively, consider a lower triangular pseudomatrix M whose $(i,j)$ -entry with $i>j$ is given by the sequence $\{(i,i-1,j), (i,i-2,j),\ldots ,(i,j,j)\}$ , then the sequence $\mathscr T_{2n+1}^\angle $ is obtained by reading each row of M from left to right, down successive rows.

Moreover, as in [Reference Bao, Kujawa, Li and Wang1] or [Reference Beilinson, Lusztig and MacPherson3, 3.5], we define a preorder $\preceq $ on ${\Xi _{2n+1}}$ as follows:

$$ \begin{align*} A\preceq B \Longleftrightarrow \sum_{r\leqslant i;s\geqslant j}a_{rs}\leqslant \sum_{r\leqslant i;s\geqslant j}b_{rs} \quad\text{for all }1\leqslant i<j\leqslant 2n+1. \end{align*} $$

Clearly, $A\preceq B\Longleftrightarrow \sum _{r\geqslant i;s\leqslant j}a_{rs}\leqslant \sum _{r\geqslant i;s\leqslant j}b_{rs}\text { for all }i>j$ . We write $A\prec B$ if $A\preceq B$ and $B\not \preceq A$ . Note that $({\Xi _{2n+1}^{0\operatorname {diag}}},\preceq )$ is a poset.

Then comes the following result, which has been proved in [Reference Du and Wu8, Lemma 5.1].

Proposition 2.2. For each $A\in {\Xi _{2n+1}^{0\operatorname {diag}}}$ ,

$$ \begin{align*} \prod _{(i,h,j)\in\mathscr T_{2n+1}^\angle }(a_{i,j}E^{\theta}_{h+1,h})(\mathbf{0})=A({\mathbf{0}})+\sum_{B\in{\Xi_{2n+1}^{0\operatorname{diag}}},{\mathbf{j}}\in\mathbb{Z}^{n+1}\atop B\prec A}g_{A,B,{\mathbf{j}}}B({\mathbf{j}}), \end{align*} $$

where the product is taken over the sequence $\mathscr T_{2n+1}^\angle $ and $g_{A,B,{\mathbf {j}}}\in \mathbb {Q}(\upsilon )$ .

3 Some integral multiplication formulae in q-Schur algebras of type B

Following [Reference Fu10], we introduce in this section new elements $A({\mathbf {i}},\lambda ,r)$ in ${{\mathcal S}^\jmath _{\mathcal Z}(n,r)}$ that are more complicated than the elements $A({\mathbf {i}},r)$ defined in (2-3). More importantly, we derive certain useful multiplication formulae for the elements $A({\mathbf {i}},\lambda ):=(A({\mathbf {i}},\lambda ,r))_{r\geqslant 0}$ in the limiting algebra ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ . Of course, such multiplication formulae for $A({\mathbf {i}},\lambda )$ are derived from the corresponding formulae for $A({\mathbf {i}},\lambda ,r)$ in the q-Schur algebras ${{\mathcal S}^\jmath _{\mathcal Z}(n,r)}$ of type B.

Recall the ring automorphism $\bar \ :{\mathcal Z}\rightarrow {\mathcal Z}$ defined by $\bar \upsilon =\upsilon ^{-1}$ and some fundamental multiplication formulae in [Reference Bao, Kujawa, Li and Wang1].

Proposition 3.2 [Reference Bao, Kujawa, Li and Wang1, 3.7]

Suppose that $1\leqslant h\leqslant n$ and $m\in \mathbb N$ .

$(1)$ For $A,B\in \Xi _{2n+1,2r+1}$ such that $B-mE^{\theta }_{h,h+1}$ is diagonal and $\mathrm {co}(B)=\mathrm {ro}(A)$ ,

$$ \begin{align*} [B]\cdot[A] =\sum_{{\mathbf{t}}\in\Lambda(2n+1,m)}\upsilon^{\beta_{\mathbf{t}}(A,h)} \bigg(\prod_{p\in[1,2n+1]}\overline{\bigg[\!\bigg[{a_{h,p}+t_p\atop t_p}\bigg]\!\bigg]}\bigg)\, \bigg[ A+\sum_{p\in[1,2n+1]}t_p(E^{\theta}_{h,p}-E^{\theta}_{h+1,p})\bigg], \end{align*} $$

where $\beta _{\mathbf {t}}(A,h)\ =\ \sum _{j\leqslant l}a_{h,l}t_j\ -\ \sum _{j<l}a_{h+1,l}t_j\ +\ \sum _{j<l}t_jt_l\ +\ \delta _{h,n}( \sum _{j<l\atop j+l<2n+2}t_jt_l\ +\ \sum _{j<n+1} ({t_j(t_j+1)}/{2})).$

$(2)$ For $A,C\in \Xi _{2n+1,2r+1}$ such that $C-mE^{\theta }_{h+1,h}$ is diagonal and $\mathrm {co}(C)=\mathrm {ro}(A)$ ,

$$ \begin{align*} [C]\cdot[A] & = \sum_{{\mathbf{t}}\in\Lambda(2n+1,m)}\upsilon^{\beta^{\prime}_{{\mathbf{t}}}(A,h)} \prod_{p\in[1,2n+1]}\overline{\bigg[\!\bigg[{a_{h+1,p}+t_p\atop t_p}\bigg]\!\bigg]} \\ & \quad\cdot \bigg[ A-\sum_{p\in[1,2n+1]}t_p(E^{\theta}_{h,p}-E^{\theta}_{h+1,p})\bigg] \quad\text{for all } h<n, \end{align*} $$

where $\beta _{{\mathbf {t}}}'(A,h)=\sum _{j\geqslant l}a_{h+1,l}t_j-\sum _{j>l}a_{h,l}t_j+\sum _{j>l}t_jt_{l}$ and, for $h=n$ ,

$$ \begin{align*} \begin{aligned} [C]\cdot[A]=& \sum_{{\mathbf{t}}\in\Lambda(2n+1,m)}\upsilon^{\beta^{\prime\prime}_{{\mathbf{t}}}(A,n)} \prod_{p<n+1}\overline{\bigg[\!\bigg[{a_{n+1,p}+t_p+t_{2n+2-p}\atop t_p}\bigg]\!\bigg]} \prod_{p>n+1}\overline{\bigg[\!\bigg[{a_{n+1,p}+t_p\atop t_p}\bigg]\!\bigg]}\\ &\cdot\prod_{i=0}^{t_{n+1}-1}\frac{\overline{[\![a_{n+1,n+1}+1+2i]\!]}} {\overline{[\![i+1]\!]}} \bigg[A-\sum_{p\in[1,2n+1]}t_p(E^{\theta}_{n,p}-E^{\theta}_{n+1,p})\bigg], \end{aligned} \end{align*} $$

where

$$ \begin{align*} \beta_{{\mathbf{t}}}"(A,n)= \sum_{j\geqslant l}a_{n+1,l}t_j-\sum_{j>l}a_{n,l}t_j+\sum_{j>l}t_jt_{l} -\sum_{j<l\atop j+l<2n+2}t_jt_l-\sum_{j<n+1}\frac{t_j(t_j-1)}{2}+\frac{m(m-1)}{2}. \end{align*} $$

For a possible construction of a certain integral form for ${\boldsymbol {\mathfrak {A}}^\jmath (n)}$ , we modify the definition given in [Reference Fu10, Section 2] (see also [Reference Du and Fu5, (3.0.6)]) to define some new elements in ${{\mathcal S}^\jmath _{\mathcal Z}(n,r)}$ as follows: for $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ , ${\mathbf {i}}=(i_1,i_2,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ and $\lambda \in \mathbb {N}^{n+1}$ ,

(3-1) $$ \begin{align} A({\mathbf{i}},\lambda,r)= \begin{cases} {\displaystyle\sum_{\mu\in\Lambda(n+1,r-{|A|}/{2})}\upsilon^{{\widetilde \mu}{\cdot}{\mathbf{i}}^\circ} \bigg[{\mu^{\bullet}\atop \lambda^{\bullet}}\bigg] \bigg[{\mu_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}}[A+\tilde{\mu}]& \text{ if }|A|\leqslant 2r,\\ 0 &\text{ if }|A|>2r. \end{cases} \end{align} $$

We also define the limiting (or completion) algebra over ${\mathcal Z}$ :

$$ \begin{align*} {\widehat{{\mathcal S}}^\jmath_{\mathcal Z}(n)}:=\prod_{r\geqslant 0}{{\mathcal S}^\jmath_{\mathcal Z}(n,r)}, \end{align*} $$

and consider elements in ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ : $A({\mathbf {i}},\lambda ):=(A({\mathbf {i}},\lambda ,r))_{r\geqslant 0}$ . Here we modified the notion used in [Reference Du and Wu8] by putting a hat on ${{\boldsymbol {{\mathcal S}}}^\jmath (n)}$ over $\mathcal {Z}$ . Note that the elements $A({\mathbf {i}},r)$ defined in (2-3) have the form $A({\mathbf {i}},r)=A({\mathbf {i}},{\mathbf {0}},r)$ , where ${\mathbf {0}}=(0,\ldots ,0)\in \mathbb {N}^{n+1}$ . We prove in Section 4 that these elements span a ${\mathcal Z}$ -subalgebra of ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ .

First, we derive the integral version for [Reference Du and Wu8, Theorem 4.4(1)], which is much simpler.

Proof. We only prove Item (1); the proof for Item (2) is similar.

We compute Item (1) component-wise. Clearly, we may assume $r\geqslant |A|/2$ (as otherwise, both sides are zero). By the definition (3-1),

(3-2) $$ \begin{align} &O({\mathbf{j}},\mu,r)A({\mathbf{i}},\lambda,r) \nonumber\\ &\quad=\sum_{\alpha\in\Lambda(n+1,r-{|A|}/{2})}\!\!\upsilon^{ (\mathrm{ro}(A)+\tilde\alpha ) \cdot{\mathbf{j}}^\circ+\tilde\alpha{\cdot}{\mathbf{i}}^\circ} \bigg[{(\mathrm{ro}(A)+\alpha)^{\bullet}\atop\mu^{\bullet}}\bigg]\bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}+\alpha_{n+1}\atop \mu_{n+1}}\bigg]_{\upsilon^2}\! \bigg[{\alpha^{\bullet}\atop \lambda^{\bullet}}\bigg]\bigg[{\alpha_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}[A+\tilde\alpha]. \end{align} $$

We want to show that this is a linear combination of some $A({\mathbf {j}},\nu ,r)$ . By (1-1),

$$ \begin{align*} \begin{aligned} &\bigg[{(\text{ ro}(A)+\alpha)^{\bullet}\atop\mu^{\bullet}}\bigg]\bigg[{\alpha^{\bullet}\atop \lambda^{\bullet}}\bigg]\cdot\bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}+\alpha_{n+1}\atop\mu_{n+1}}\bigg]_{\upsilon^2} \bigg[{\alpha_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}\\&\quad= \sum_{ {\beta }\in\mathbb N^{n+1}_{\mu }} \upsilon^{\wp} \bigg[{\mathrm{ro}(A)^{\bullet}\atop {\beta^{\bullet}}}\bigg]\cdot \bigg[{\alpha^{\bullet}\atop\mu^{\bullet}- {\beta^{\bullet}}}\bigg] \bigg[{\alpha^{\bullet}\atop\lambda^{\bullet}}\bigg] \cdot \bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}\atop {\beta_{n+1}}}\bigg]_{\upsilon^2}\cdot \bigg[{\alpha_{n+1}\atop\mu_{n+1}- {\beta_{n+1}}}\bigg]_{\upsilon^2} \bigg[{\alpha_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}\\&\quad= \sum_{ {\beta }\in\mathbb{N}^{n+1}_{\mu }} \upsilon^{\wp}\bigg[{\mathrm{ro}(A)^{\bullet}\atop {\beta^{\bullet}}}\bigg] \bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}\atop {\beta_{n+1}}}\bigg]_{\upsilon^2}\\&\quad\quad \cdot\bigg(\sum_{ {\gamma }\in\mathbb{N}^{n+1}_{\lambda \cap(\mu - {\beta })}} \upsilon^{\wp'} \bigg[{\alpha^{\bullet}\atop\lambda^{\bullet}+\mu^{\bullet}- {\beta^{\bullet}}- {\gamma^{\bullet}}}\bigg] \bigg[{\lambda^{\bullet}+\mu^{\bullet}- {\beta^{\bullet}}- {\gamma^{\bullet}}\atop {\gamma^{\bullet}},\,\lambda^{\bullet}- {\gamma^{\bullet}},\,\mu^{\bullet}-\beta^{\bullet}- {\gamma^{\bullet}}}\bigg] \\&\quad\quad \quad \cdot \bigg[{\alpha_{n+1}\atop\lambda_{n+1}+\mu_{n+1}- {\beta_{n+1}}- {\gamma_{n+1}}}\bigg]_{\upsilon^2} \bigg[{\lambda_{n+1}+\mu_{n+1}- {\beta_{n+1}}- {\gamma_{n+1}}\atop {\gamma_{n+1}},\,\lambda_{n+1}- {\gamma_{n+1}},\,\mu_{n+1}-\beta_{n+1}- {\gamma_{n+1}}}\bigg]_{\upsilon^2}\bigg), \end{aligned} \end{align*} $$

where

$$ \begin{align*} \wp&=\mathrm{ro}(A)^{\bullet}\cdot(\mu^{\bullet}- {\beta^{\bullet}})- {\beta^{\bullet}}\cdot\alpha^{\bullet}+2\bigg(\frac{\mathrm{ro}(A)_{n+1}}{2}(\mu_{n+1}- {\beta_{n+1})-\beta_{n+1} \alpha_{n+1}\bigg)}\\ &=\text{ ro}(A)\cdot(\mu-\beta)^\circ-{\beta}\cdot\alpha {-\beta_{n+1}\alpha_{n+1}}\\ \wp'&=\lambda^{\bullet}\cdot(\mu^{\bullet}- {\beta^{\bullet}})-\alpha^{\bullet}\cdot {\gamma^{\bullet}}+2(\lambda_{n+1}(\mu_{n+1}- {\beta_{n+1}})-\alpha_{n+1}{\gamma_{n+1}})\\&={\widetilde\lambda}\cdot(\mu- {\beta})^\circ-\mu_{n+1}-\alpha\cdot {\gamma} {+\beta_{n+1}-\alpha_{n+1}{\gamma}_{n+1}}. \end{align*} $$

Thus, substituting into (3-2) and noting the exponents of $\upsilon $

$$ \begin{align*}(\mathrm{ro}(A)+\tilde\alpha ) \cdot{\mathbf{j}}^\circ+\tilde\alpha\cdot{\mathbf{i}}^\circ+\wp+\wp'=\wp"+\widetilde\alpha\cdot({\mathbf{j}}+{\mathbf{i}}- \beta-{\gamma})^\circ,\end{align*} $$

where $\wp "=\text { ro}(A)\cdot ({\mathbf {j}}+\mu -\beta )^\circ +\tilde \lambda \cdot (\mu -\beta )^\circ + {(2\beta +{\gamma }-\mu )}\cdot {\boldsymbol {e}}_{n+1}$ , we have, for $n'=n+1$ ,

$$ \begin{align*} \begin{aligned} & O({\mathbf{j}},\mu,r)A({\mathbf{i}},\lambda,r)\\ &\quad =\sum_{\beta,{\gamma}\in\mathbb{N}_\mu^{n+1}\atop {\gamma}\leqslant\lambda\cap(\mu-\beta)} \upsilon^{\wp"}\bigg[{\mathrm{ro}(A)^{\bullet}\atop \beta^{\bullet}}\bigg] \bigg[{(\lambda+\mu-\beta-{\gamma})^{\bullet}\atop {\gamma}^{\bullet},\,(\lambda-{\gamma})^{\bullet},\,(\mu-\beta-{\gamma})^{\bullet}}\bigg]\\ & \qquad \bigg[{\frac{\mathrm{ro}(A)_{n'}}{2}\atop \beta_{n'}}\bigg]_{\upsilon^2} \bigg[{\lambda_{n'}+\mu_{n'}-\beta_{n'}-{\gamma}_{n'}\atop {\gamma}_{n'},\,\lambda_{n'}-{\gamma}_{n'},\,\mu_{n'}-\beta_{n'}-{\gamma}_{n'}}\bigg]_{\upsilon^2}\\ &\qquad\times \sum_{\alpha\in\Lambda(n+1,r-{|A|}/{2})}\upsilon^{\widetilde\alpha\cdot({\mathbf{j}}+{\mathbf{i}}- \beta-{\gamma})^\circ}\bigg[{\alpha^{\bullet}\atop(\lambda+\mu-\beta-{\gamma})^{\bullet}}\bigg] \bigg[{\alpha_{n'}\atop\lambda_{n'}+\mu_{n'}-\beta_{n'}-{\gamma}_{n'}}\bigg]_{\upsilon^2}[A+\tilde\alpha]\\ &\quad=\sum_{\nu{ \in\mathbb{N}^{n+1}_\mu}}f_\nu A({\mathbf{j}}+{\mathbf{i}}-\nu,\mu+\lambda-\nu,r). \end{aligned} \end{align*} $$

Since $f_\nu $ is independent of r, Item (1) follows.

Observe the coefficients $f_\nu $ for $A=O$ , the zero matrix. If $\beta>{\mathbf {0}}$ , then

$$ \begin{align*}\bigg[{\mathrm{ro}(A)^{\bullet}\atop \beta^{\bullet}}\bigg]\cdot\bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}\atop \beta_{n+1}}\bigg]_{\upsilon^2}=0. \end{align*} $$

If $\beta ={\mathbf {0}}$ , then the product is 1 and $\nu \leqslant \lambda $ . Thus, we have the following corollary.

Corollary 3.4. Maintain the notation in the above lemma. We have

$$ \begin{align*} O({\mathbf{i}},\lambda)O({\mathbf{j}},\mu) & =\sum_{\nu\in\mathbb{N}^{n+1}_{\lambda\cap\mu}} \upsilon^{\lambda{\cdot}\mu+\lambda_{n+1}\mu_{n+1}+\nu{\cdot}{\boldsymbol{e}}_{n+1}} \bigg[{(\lambda+\mu-\nu)^{\bullet}\atop \nu^{\bullet},\,(\lambda-\nu)^{\bullet},\,(\mu-\nu)^{\bullet}}\bigg]\\ &\quad\,\times \bigg[{\lambda_{n+1}+\mu_{n+1}-\nu_{n+1}\atop \nu_{n+1},\,\lambda_{n+1}-\nu_{n+1},\,\mu_{n+1}-\nu_{n+1}}\bigg]_{\upsilon^2}O({\mathbf{i}}+{\mathbf{j}}-\nu,\lambda+\mu-\nu). \end{align*} $$

In particular, $O({\mathbf {i}},\lambda )O({\mathbf {j}},{\mathbf {0}})=O({\mathbf {i}}+{\mathbf {j}},\lambda )=O({\mathbf {j}},{\mathbf {0}})O({\mathbf {i}},\lambda )$ .

The next lemma gives the integral version for [Reference Du and Wu8, Theorems 4.4(2) and (3)]. We use Convention 3.1 to assume that, for a matrix $B\in \text {Mat}_{2n+1}(\mathbb Z)$ , $B({\mathbf {i}},\lambda ,r)= 0=B({\mathbf {i}},\lambda )$ if $b_{i,j}< 0$ for some $i\not =j$ .

Lemma 3.5. Let $A=(a_{i,j})\in \Xi _{2n+1}^{\mathrm {0\ diag}}$ , ${\mathbf {i}}=(i_1,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ , $\lambda =(\lambda _1,\ldots ,\lambda _{n+1})\in \mathbb {N}^{n+1}$ , $m\in {\mathbb N}$ and $1\leqslant h\leqslant n$ . The following multiplication formulae hold in ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ :

$$ \begin{align*} \begin{aligned} (1)\;\;&(mE^{\theta}_{h,h+1})({\mathbf{0}})A({\mathbf{i}},\lambda)\\ &\quad=\sum_{{\mathbf{t}}\in\Lambda(2n+1,m)\atop {{j\in\mathbb{N}_{\lambda_h},k\in\mathbb N_{\lambda_{h+1}}\atop { c\in\mathbb N_{j\cap t_h}}} }} \upsilon^{b^{h,{\mathbf{t}}}_{j,k}}f^{h,{\mathbf{t}}}_{j,c,k} \bigg(A+\sum_{p\not=h}t_pE^{\theta}_{h,p}-\sum_{p\not=h+1}t_pE^{\theta}_{h+1,p}\bigg) ({\mathbf{i}}+\alpha^{{\mathbf{t}}}_{j,c,k},\lambda+\alpha^{\prime\;{\mathbf{t}}}_{j,c,k}) \end{aligned} \end{align*} $$

for some $b^{h,{\mathbf {t}}}_{j,k}\in \mathbb Z$ , $f^{h,{\mathbf {t}}}_{j,c,k}\in {\mathcal Z}$ and $\alpha ^{{\mathbf {t}}}_{j,c,k},\alpha ^{\prime \;{\mathbf {t}}}_{j,c,k}\in \mathbb Z^{n+1}$ .

$$ \begin{align*} \begin{aligned} (2)\;\;&(mE^{\theta}_{h+1,h})({\mathbf{0}})A({\mathbf{i}},\lambda)\\ &\quad=\sum_{{\mathbf{t}}\in\Lambda(2n+1,m)\atop {{j\in\mathbb{N}_{\lambda_{h+1}},k\in\mathbb N_{\lambda_{h}}\atop {c\in\mathbb N_{j\cap t_{h+1}}}} }} \upsilon^{{c}^{h,{\mathbf{t}}}_{j,k}}{ {g}}^{h,{\mathbf{t}}}_{j,c,k} \bigg(A-\sum_{p\not=h}t_pE^{\theta}_{h,p}+\sum_{p\not=h+1}t_pE^{\theta}_{h+1,p}\bigg) ({\mathbf{i}}+{\beta}^{{\mathbf{t}}}_{j,c,k},\lambda+{\beta}^{\prime\;{\mathbf{t}}}_{j,c,k}) \end{aligned} \end{align*} $$

for some $c^{h,{\mathbf {t}}}_{j,k}\in \mathbb Z$ , ${g}^{h,{\mathbf {t}}}_{j,c,k}\in {\mathcal Z}$ and $\beta ^{{\mathbf {t}}}_{j,c,k},\beta ^{\prime \;{\mathbf {t}}}_{j,c,k}\in \mathbb Z^{n+1}$ .

The proof of the lemma is quite technical and explicit formulae for these specified coefficients $\upsilon ^{b^{{\mathbf {t}},h}_{j,k}},f^{{\mathbf {t}},h}_{j,c,k}$ and $\upsilon ^{{c}^{{\mathbf {t}},h}_{j,k}},{ {g}}^{{\mathbf {t}},h}_{j,c,k}$ at level r are given in Appendix A. We need such explicit expressions to show the fact that the coefficients are independent of r. This fact is crucial to Theorem 4.6.

4 The main results

We are now ready to give a realisation for the Lusztig form $U^{{\jmath }}_{{\mathcal Z}}(n)$ by investigating its image in ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ . Meanwhile, we show several ${\mathcal Z}$ -bases for $U^{{\jmath }}_{{\mathcal Z}}(n)$ via this new realisation.

Let $\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ be the ${\mathcal Z}$ -submodule of ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ spanned by

$$ \begin{align*} \mathfrak{A}^{{\jmath}}_{{\mathcal Z}}(n)=\text{span}_{\mathcal Z}\{A({\mathbf{i}},\lambda)\mid A\in\Xi^{\mathrm{0\ diag}}_{2n+1},\,{\mathbf{i}}\in\mathbb{Z}^{n+1},\,\lambda\in\mathbb{N}^{n+1}\}. \end{align*} $$

We see in Proposition 4.5 below that this is a ${\mathcal Z}$ -subalgebra generated by ‘divided powers’. Note that if we use the spanning set in (2-5), then the ${\mathcal Z}$ -submodule of ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}$ :

$$ \begin{align*}\text{span}_{{\mathcal Z}}\{A({{\mathbf{i}}})\mid A\in\Xi^{\mathrm{0\ diag}}_{2n+1}, {\mathbf{i}}=(i_1,i_2,\ldots,i_{n+1})\in\mathbb{Z}^{n+1}\}\end{align*} $$

is not closed under the multiplication by the generators $e_i,f_i$ ; see [Reference Du and Wu8, Theorem 4.4].

Let ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}^0$ be the ${\mathcal Z}$ -subalgebra of ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}$ generated by $O(\pm {\boldsymbol {e}}_i,{\mathbf {0}})$ and $O({\mathbf {0}}, m{\boldsymbol {e}}_i)$ for all $i\in [1,n+1],m\in \mathbb {N}$ . Recall the notation $\Bbbk _{i}$ defined in [Reference Du and Wu8, Section 6] (see (2-4)):

$$ \begin{align*} \Bbbk_{i}:=\phi^{\jmath}(d_i)= \begin{cases} O({\boldsymbol{e}}_i) &\text{ if }1\leqslant i\leqslant n,\\ \upsilon^{-1}O({\boldsymbol{e}}_{n+1})&\,\,\text{if }i=n+1, \end{cases} \end{align*} $$

and define $\Bbbk _{i,r}=\pi _r(\Bbbk _i)$ , where $\pi _r$ is the canonical projection map ${\boldsymbol {\mathfrak {A}}^\jmath (n)}\to {{\boldsymbol {{\mathcal S}}}^\jmath (n,r)}$ . Then, for all $i\in [1,n]$ ,

$$ \begin{align*} \bigg[{\Bbbk_{i,r};0\atop m}\bigg] =\sum_{\mu\in\Lambda(n+1,r)}\bigg[{\Bbbk_{i,r};0\atop m}\bigg][\tilde\mu] =\sum_{\mu\in\Lambda(n+1,r)}\bigg[{\mu_i\atop m}\bigg][\tilde{\mu}]=O({\mathbf{0}},m{\boldsymbol{e}}_i,r). \end{align*} $$

Similarly, for $i=n+1$ , replacing $O({\boldsymbol {e}}_i,r)$ by $\upsilon ^{-1}O({\boldsymbol {e}}_{n+1},r)$ and $\upsilon $ by $\upsilon ^2$ in the above proof gives rise to $[{\Bbbk _{n+1,r};0\atop m}]_{\upsilon ^2}=O({\mathbf {0}},m{\boldsymbol {e}}_{n+1},r)$ . The above two equations hold independent of r; thus, we get that

$$ \begin{align*} O({\mathbf{0}},m\boldsymbol{e}_i)= \begin{cases} \bigg[{\Bbbk_{i};0\atop m}\bigg] & \text{if } i\in[1,n],\\ \bigg[{\Bbbk_{n+1};0\atop m}\bigg]_{\upsilon^2}&\text{if }i=n+1.\\ \end{cases} \end{align*} $$

For ${\mathbf {i}}=(i_1,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ and $\lambda =(\lambda _1,\ldots ,\lambda _{n+1})\in \mathbb {N}^{n+1}$ , let

$$ \begin{align*} \Bbbk^{\mathbf{i}}=\Bbbk_1^{i_1}\Bbbk_2^{i_2}\cdots \Bbbk_{n+1}^{i_{n+1}}\quad\text{and}\quad\bigg[{\Bbbk\atop \lambda}\bigg] =\prod_{i=1}^n\bigg[{\Bbbk_i;0\atop \lambda_i}\bigg]\cdot \bigg[{\Bbbk_{n+1};0\atop \lambda_{n+1}}\bigg]_{\upsilon^2}. \end{align*} $$

Lemma 4.1. For any $\lambda =(\lambda _1,\ldots ,\lambda _{n+1})\in \mathbb N^{n+1}$ and ${\mathbf {i}}=(i_1,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ , we have in ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ :

$$ \begin{align*}\Bbbk^{\mathbf{i}}\bigg[{\Bbbk\atop \lambda}\bigg]=\upsilon^{-i_{n+1}}O({\mathbf{i}})\prod_{i=1}^{n+1}O({\mathbf{0}},\lambda_i{\boldsymbol{e}}_i)=\upsilon^{-i_{n+1}}O({\mathbf{i}},\lambda).\end{align*} $$

Proof. By the definition of $A({\mathbf {i}},\lambda ,r)$ , we easily see that

$$ \begin{align*} \prod_{i=1}^{n+1}O({\mathbf{0}},\lambda_i{\boldsymbol{e}}_i,r)&=\prod_{i=1}^n \bigg(\sum_{\mu\in\Lambda(n+1,r)}\bigg[{\mu_i\atop \lambda_i}\bigg][\tilde{\mu}]\bigg)\sum_{\mu\in\Lambda(n+1,r)}\bigg[{\mu_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}[\tilde{\mu}]\\ &=\sum_{\mu\in\Lambda(n+1,r)}\prod_{i=1}^n\bigg[{\mu_i\atop \lambda_i}\bigg]\bigg[{\mu_{n+1}\atop \lambda_{n+1}}\bigg]_{\upsilon^2}[\tilde{\mu}]\\ &=O({\mathbf{0}},\lambda,r). \end{align*} $$

The assertion follows from Corollary 3.4.

Proof. This follows from Lemma 4.1, and the facts that $\mathbb {Q}(\upsilon )[\Bbbk _{1}^{\pm 1},\ldots ,\Bbbk _{n+1}^{\pm 1}]$ is a Laurent polynomial algebra and its ${\mathcal Z}$ -subalgebra generated by $\Bbbk _{1}^{\pm 1},\ldots ,\Bbbk _{n+1}^{\pm 1}$ and $\big [{\Bbbk _{i};0\atop m}\big ], \big [{\Bbbk _{n+1};0\atop m}\big ]_{\upsilon ^2}$ for $i\in [1,n]$ and $m\in \mathbb {N}$ has a basis

$$ \begin{align*} \bigg\{\Bbbk^{\mathbf{i}}\bigg[{\Bbbk\atop \lambda}\bigg]\mid {\mathbf{i}}\in\mathbb{N}_1^{n+1},\lambda\in\mathbb{N}^{n+1}\bigg\} \end{align*} $$

(see, for example, [Reference Deng, Du, Parshall and Wang4, Section 6.9, Corollary 14.19]). Note that the last factor $\big [{\Bbbk _{n+1}\atop \lambda _{n+1}}\big ]_{\upsilon ^2}$ in $\big [{\Bbbk \atop \lambda }\big ]$ does not affect the statement; see [Reference Deng, Du, Parshall and Wang4, Lemma 6.47].

Proof. The proof below is a modification of the proof for [Reference Fu10, Lemma 4.2]. We first prove Assertion (1). For $\lambda \in \mathbb {N}^{n+1}$ , $\mu ,{\mathbf {i}}\in \mathbb {Z}^{n+1}$ , we first observe that

$$ \begin{align*} \text{if }k\in[1,n],\quad \upsilon^{i_k\mu_k}\bigg[{\mu_k\atop\lambda_k}\bigg] &=\upsilon^{\lambda_k}(\upsilon^{\lambda_k+1}-\upsilon^{-\lambda_k-1})\upsilon^{(i_k-1)\mu_k} \bigg[{\mu_k\atop \lambda_k+1}\bigg]+\upsilon^{2\lambda_k+(i_k-2)\mu_k} \bigg[{\mu_k\atop\lambda_k}\bigg]\quad\text{and}\\ \upsilon^{i_{n+1}{\widetilde\mu}_{n+1}}\bigg[{\mu_{n+1}\atop\lambda_{n+1}}\bigg]_{\upsilon^2} &=\upsilon^{2\lambda_{n+1}+1}(\upsilon^{2\lambda_{n+1}+2}-\upsilon^{-2\lambda_{n+1}-2}) \upsilon^{({\mathbf{i}}_{n+1}-1){\widetilde \mu}_{n+1}}\bigg[{\mu_{n+1}\atop \lambda_{n+1}+1}\bigg]_{\upsilon^2}\\ &\quad +\upsilon^{4\lambda_{n+1}+2+({\mathbf{i}}_{n+1}-2){\widetilde \mu}_{n+1}}\bigg[{\mu_{n+1}\atop\lambda_{n+1}}\bigg]_{\upsilon^2}. \end{align*} $$

Thus, replacing the k th factor of the product $\prod _{k=1}^n\upsilon ^{i_k\mu _k}\big [{\mu _k\atop \lambda _k}\big ]\cdot \upsilon ^{i_{n+1}{\widetilde \mu }_{n+1}}\big [{\mu _{n+1}\atop \lambda _{n+1}}\big ]_{\upsilon ^2}$ in $A({\mathbf {i}},\lambda )$ by the right-hand side above yields

$$ \begin{align*} A({\mathbf{i}},\lambda)=\begin{cases}\upsilon^{\lambda_k}(\upsilon^{\lambda_k+1}-\upsilon^{-\lambda_k-1})A({\mathbf{i}}-{\boldsymbol{e}}_k,\lambda+{\boldsymbol{e}}_k) +\upsilon^{2\lambda_k}A({\mathbf{i}}-2{\boldsymbol{e}}_k,\lambda) \quad\text{ for }k\in[1,n];&\\ \upsilon^{2\lambda_{n+1}+1}(\upsilon^{2\lambda_{n+1}+2}-\upsilon^{-2\lambda_{n+1}-2})A({\mathbf{i}}-{\boldsymbol{e}}_{n+1},\lambda+{\boldsymbol{e}}_{n+1}) +\upsilon^{4\lambda_{n+1}+2}A({\mathbf{i}}-2{\boldsymbol{e}}_{n+1},\lambda).&\end{cases} \end{align*} $$

With the notation ${\widetilde \lambda }$ in (2-2), we have ${\widetilde \lambda }_k=\lambda _k$ for all $k\in [1,n]$ and ${\widetilde \lambda }_{n+1}=2\lambda _{n+1}+1$ . Thus, the formula above can be combined as follows:

(4-1) $$ \begin{align} A({\mathbf{i}},\lambda)=\upsilon^{{\widetilde\lambda}_k}(\upsilon^{{\widetilde\lambda}_k+1}-\upsilon^{-{\widetilde\lambda}_k-1})A({\mathbf{i}}-{\boldsymbol{e}}_k,\lambda+{\boldsymbol{e}}_k) +\upsilon^{2{\widetilde\lambda}_k}A({\mathbf{i}}-2{\boldsymbol{e}}_k,\lambda) \end{align} $$

for all $k\in [1,n+1]$ . Further replacing ${\mathbf {i}}$ by ${\mathbf {i}}+2{\boldsymbol {e}}_k$ in (4-1) and solving for $A({\mathbf {i}},\lambda )$ gives

(4-2) $$ \begin{align} A({\mathbf{i}},\lambda) & =\upsilon^{-2{\widetilde\lambda}_k}A({\mathbf{i}}+2{\boldsymbol{e}}_k,\lambda) -\upsilon^{-{\widetilde\lambda}_k}(\upsilon^{{\widetilde\lambda}_k+1}-\upsilon^{-{\widetilde\lambda}_k-1})A({\mathbf{i}}+{\boldsymbol{e}}_k,\lambda+{\boldsymbol{e}}_k)\nonumber\\ & \quad\text{for all }k\in[1,n+1]. \end{align} $$

Thus, if ${\mathbf {i}}$ has a k th component $>1$ (respectively $<0$ ), by repeatedly applying (4-1) (respectively (4-2)), $A({\mathbf {i}},\lambda )$ is a linear combination of $A({\mathbf {j}},\mu )$ such that the k th component of ${\mathbf {j}}$ is in $\mathbb N_1$ . Hence, doing this for every such k shows that $A({\mathbf {i}},\lambda )$ is in the span of $\mathfrak B_1$ , proving that $\mathfrak B_1$ spans $\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ .

It remains to prove $\mathfrak B_1$ is linearly independent. Suppose $\sum _{A\in {\mathcal A}\atop {\mathbf {i}}\in {\mathcal B},\lambda \in {\mathcal C}}f_{A,{\mathbf {i}},\lambda }A({\mathbf {i}},\lambda )=0$ , where $f_{A,{\mathbf {i}},\lambda }\in {\mathcal Z}$ and ${\mathcal A}\subset {\Xi _{2n+1}^{0\operatorname {diag}}}$ , ${\mathcal B}\subset \mathbb N_1^{n+1}$ , ${\mathcal C}\subset \mathbb N^{n+1}$ are all finite subsets. Then, $\sum _{A\in {\mathcal A}\atop {\mathbf {i}}\in {\mathcal B},\lambda \in {\mathcal C}}f_{A,{\mathbf {i}},\lambda }A({\mathbf {i}},\lambda ,r)=0$ for all $r\geqslant {|A|}/2$ . Thus, for all r satisfying $r-{|A|}/2\geqslant 0$ ,

$$ \begin{align*} \sum_{A\in {\mathcal A}\atop\mu\in\Lambda(n+1,r-{|A|}/2)}\bigg (\sum_{{\mathbf{i}}\in {\mathcal B},\lambda\in {\mathcal C}}f_{A,{\mathbf{i}},\lambda} \upsilon^{{\widetilde\mu}{\cdot}{\mathbf{i}}^\circ}\bigg[{\mu^{\bullet}\atop\lambda^{\bullet}}\bigg] \bigg[{\mu_{n+1}\atop\lambda_{n+1}}\bigg]_{\upsilon^2}\bigg)[A+\tilde\mu]=0. \end{align*} $$

Hence,

$$ \begin{align*} \sum_{{\mathbf{i}}\in {\mathcal B},\lambda\in {\mathcal C}}f_{A,{\mathbf{i}},\lambda} \upsilon^{{\widetilde\mu}{\cdot}{\mathbf{i}}^\circ}\bigg[{\mu^{\bullet}\atop\lambda^{\bullet}}\bigg] \bigg[{\mu_{n+1}\atop\lambda_{n+1}}\bigg]_{\upsilon^2}=0 \end{align*} $$

for all $A\in {\mathcal A}$ and $\mu \in \mathbb {N}^{n+1}$ . It follows that

$$ \begin{align*}\sum_{{\mathbf{i}}\in {\mathcal B},\lambda\in {\mathcal C}}f_{A,{\mathbf{i}},\lambda}O({\mathbf{i}},\lambda,r)=\sum_{\mu\in\Lambda(n+1,r)}\sum_{{\mathbf{i}}\in {\mathcal B},\lambda\in {\mathcal C}}f_{A,{\mathbf{i}},\lambda} \upsilon^{{\widetilde\mu}{\cdot}{\mathbf{i}}^\circ}\bigg[{\mu^{\bullet}\atop\lambda^{\bullet}}\bigg] \bigg[{\mu_{n+1}\atop\lambda_{n+1}}\bigg]_{\upsilon^2}[{\widetilde \mu}]=0 \end{align*} $$

for $r\geqslant {|A|}/2$ . Consequently, $\sum _{{\mathbf {i}}\in {\mathcal B},\lambda \in {\mathcal C}}f_{A,{\mathbf {i}},\lambda }O({\mathbf {i}},\lambda )=0$ for all $A\in {\mathcal A}$ . Thus, by Corollary 4.2, we conclude that $f_{A,{\mathbf {i}},\lambda }=0$ for all $A,{\mathbf {i}},\lambda $ . This shows that the set ${\frak B}_1$ is linearly independent and hence forms a ${\mathcal Z}$ -basis for ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}$ , proving Assertion (1).

Note that $\mathfrak B_1$ and $\mathfrak B_2$ are triangularly related,

$$ \begin{align*} O({\mathbf{i}},\lambda)A({\mathbf{0}})&= \upsilon^{\mathrm{ro}(A){\cdot} ({\mathbf{i}}+\lambda)^\circ}A({\mathbf{i}},\lambda)\nonumber\\ &\quad + \sum_{\nu\in\mathbb{N}^{n+1}\atop 0< \nu\leqslant \lambda}\upsilon^{\mathrm{ro}(A){\cdot}({\mathbf{i}}+\lambda-\nu)^\circ+\nu_{n+1}} \bigg[{\mathrm{ro}(A)^{\bullet}\atop\nu^{\bullet}}\bigg]\cdot\bigg[{\frac{\mathrm{ro}(A)_{n+1}}{2}\atop \nu_{n+1}}\bigg]_{\upsilon^2}A({\mathbf{i}}-\nu,\lambda-\nu). \end{align*} $$

Thus, we can deduce Assertion (2) from the above relation.

The proof for $\mathfrak B_3$ in Assertion (3) is similar.

To show that $\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ is actually a ${\mathcal Z}$ -subalgebra, now we give an integral version of the triangular relation in Proposition 2.2.

Proposition 4.4. For each $A\in {\Xi _{2n+1}^{0\operatorname {diag}}}$ ,

$$ \begin{align*}\prod _{(i,h,j)\in\mathscr T_{2n+1}^\angle }(a_{i,j}E^{\theta}_{h+1,h})(\mathbf{ 0})=A({\mathbf{0}})+\sum_{B\in{\Xi_{2n+1}^{0\operatorname{diag}}}, B\prec A\atop {\mathbf{i}}\in\mathbb{N}_1^{n+1},\nu\in\mathbb{N}^{n+1}}f^A_{B,{\mathbf{i}},\nu}B({\mathbf{i}},\nu),\end{align*} $$

where $f^A_{B,{\mathbf {i}},\nu }\in {\mathcal Z}$ .

Proof. Lemmas 3.3 and 3.5 give that the left-hand side is a ${\mathcal Z}$ -linear combination of $B({\mathbf {i}},\nu )$ for some ${\mathbf {i}}\in \mathbb Z^{n+1},\nu \in \mathbb N^{n+1}$ . Proposition 2.2 tells us that the coefficient of the leading term $A({\mathbf {0}})$ is $1$ . Hence,

$$ \begin{align*}\text{Left-hand side}=A({\mathbf{0}})+\sum_{B\in\Xi^{\mathrm{0\ diag}}_{2n+1},B\prec A\atop{\mathbf{i}}\in\mathbb Z^{n+1},\nu\in\mathbb N^{n+1}}f^A_{B,{\mathbf{i}},\nu}B({\mathbf{i}},\nu) \quad\text{for some } f^A_{B,{\mathbf{i}},\nu}\in{\mathcal Z}.\end{align*} $$

Furthermore, (the proof of) Lemma 4.3(1) implies that we may choose ${\mathbf {i}}\in \mathbb N_1^{n+1}$ .

Proposition 4.5. The ${\mathcal Z}$ -submodule $\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ is a subalgebra of ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ that is generated by the following elements:

$$ \begin{align*}(mE^{\theta}_{h,h+1})({\mathbf{0}}),\ (mE^{\theta}_{h+1,h})({\mathbf{0}}),\ O({\mathbf{i}},\lambda)\end{align*} $$

for $m\in \mathbb N$ , $1\leqslant h\leqslant n$ , ${\mathbf {i}}\in \mathbb {Z}^{n+1}$ and $\lambda \in \mathbb {N}^{n+1}$ .

Proof. Let $\mathfrak {A}^{\prime }_{{\mathcal Z}}$ be the ${\mathcal Z}$ -subalgebra of ${\widehat {{\mathcal S}}^\jmath _{\mathcal Z}(n)}$ generated by $(mE^{\theta }_{h,h+1})({\mathbf {0}})$ , $(mE^{\theta }_{h+1,h})({\mathbf {0}})$ and $O({\mathbf {i}},\lambda )$ for $m\in \mathbb N$ , $1\leqslant h\leqslant n$ , ${\mathbf {i}}\in \mathbb {Z}^{n+1}$ and $\lambda \in \mathbb {N}^{n+1}$ . From Lemmas 3.3 and 3.5, we see that $\mathfrak {A}^{\prime }_{{\mathcal Z}}\subseteq \mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n).$

We now prove the converse inclusion $\mathfrak {A}^{\prime }_{{\mathcal Z}}\supseteq \mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ . Lemma 4.3(2) implies that it suffices to prove that all $O({\mathbf {j}},\mu )A({\mathbf {0}})\in \mathfrak {A}^{\prime }_{{\mathcal Z}}$ . We now prove this by induction on $|\!|A|\!|$ , where

$$ \begin{align*}|\!|A|\!|=\sum_{1\leqslant i<j\leqslant 2n+1}\frac{(j-i)(j-i+1)}{2}(a_{ij}+a_{ji}).\end{align*} $$

If $|\!|A|\!| = 0$ , then A is the zero matrix. In this case, $O({\mathbf {j}},\mu )A({\mathbf {0}})=O({\mathbf {j}},\mu )O({\mathbf {0}})= O({\mathbf {j}},\mu )\in \mathfrak {A}^{\prime }_{{\mathcal Z}}$ . Now, we assume that $|\!|A|\!|>0$ and $O({\mathbf {j}},\mu )B({\mathbf {0}})\in \mathfrak {A}^{\prime }_{{\mathcal Z}}$ (and hence $B({\mathbf {j}},\mu )\in \mathfrak {A}^{\prime }_{{\mathcal Z}}$ ) for all B with $0\leqslant |\!|B|\!|<|\!|A|\!|$ . By Proposition 4.4,

$$ \begin{align*} \begin{aligned} \quad\,&O({\mathbf{j}},\mu)\prod _{(i,h,j)\in\mathscr T_{2n+1}^\angle }(a_{i,j}E^{\theta}_{h+1,h})(\mathbf{0})=O({\mathbf{j}},\mu)A({\mathbf{0}})+f, \end{aligned}\end{align*} $$

where f is a ${\mathcal Z}$ -linear combination of $B({\gamma },\nu )$ with $B\prec A$ , $\gamma \in \mathbb Z^{n+1}$ and $\nu \in \mathbb N^{n+1}$ . The left-hand side of the above equation is obviously in $\mathfrak {A}^{\prime }_{{\mathcal Z}}$ . By induction, the summation f on the right-hand side belongs to $\mathfrak {A}^{\prime }_{{\mathcal Z}}$ . Hence, $O({\mathbf {j}},\mu )A({\mathbf {0}})\in \mathfrak {A}^{\prime }_{{\mathcal Z}}.$

Recall (2-1). For ${\mathbf {i}}=(i_1,\ldots ,i_{n+1})\in \mathbb {Z}^{n+1}$ and $\lambda =(\lambda _1,\ldots ,\lambda _{n+1})\in \mathbb {N}^{n+1}$ , let

$$ \begin{align*} d^{\mathbf{i}}=d_1^{i_1}d_2^{i_2}\cdots d_{n+1}^{i_{n+1}}\quad\text{and}\quad \bigg[{d\atop\lambda}\bigg]=\prod_{k=1}^n\bigg[{d_k;0\atop \lambda_k}\bigg]\bigg[{d_{n+1};0\atop \lambda_{n+1}}\bigg]_{\upsilon^2}. \end{align*} $$

Summarising the above results gives the following integral form of (2-4).

Theorem 4.6. The restriction of the $\mathbb Q(\upsilon )$ -algebra monomorphism $\phi ^\jmath $ in (2-4) to $U^{{\jmath }}_{{\mathcal Z}}(n)$ induces a ${\mathcal Z}$ -algebra monomorphism

(4-3) $$ \begin{align} \phi^\jmath_{{\mathcal Z}}:U^{{\jmath}}_{{\mathcal Z}}(n)\longrightarrow\widehat{\mathcal S}^{{\jmath}}_{{\mathcal Z}}(n) \end{align} $$

such that $e_h^{(m)}\mapsto (mE^{\theta }_{h,h+1})({\mathbf {0}}), f_h^{(m)}\mapsto (mE^{\theta }_{h+1,h})({\mathbf {0}})$ and $ d^{{\mathbf {i}}}\big [{d\atop \lambda }\big ]\mapsto {\Bbbk ^{\mathbf {i}}\big [{\Bbbk \atop \lambda }\big ]=\upsilon ^{-i_{n+1}}O({\mathbf {i}},\lambda )}$ for $m\in \mathbb N$ , $1\leqslant h\leqslant n$ , ${\mathbf {i}}\in \mathbb {Z}^{n+1},\lambda \in \mathbb {N}^{n+1}$ . In particular, $U^{{\jmath }}_{{\mathcal Z}}(n)\cong \mathrm {im}(\phi ^\jmath _{{\mathcal Z}})=\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ is a free ${\mathcal Z}$ -module.

We end this section with a construction of some monomial bases for ${U^\jmath _{\mathcal Z}(n)}$ .

Let $f_{2n+1-i}=e_i$ for $i=1,2,\ldots ,n$ . Then, $f_{2n}\cdots f_{n+1}f_n\cdots f_1=e_1\cdots e_nf_n\cdots f_1$ . For $A\in {\Xi _{2n+1}^{0\operatorname {diag}}}$ and $N=2n+1$ , define

$$ \begin{align*} {\mathcal M}^{A,{\mathbf{0}}}&=\prod_{1\leqslant i\leqslant N-1}\bigg(\prod_{1\leqslant j\leqslant i}(f_i^{(a_{i+1,j})}\cdots f_j^{(a_{i+1,j})})\bigg)=f_1^{(a_{2,1})}\cdot f_2^{(a_{3,1})}f_1^{(a_{3,1})}f_2^{(a_{3,2})}\cdots\\ &\quad\,\cdot e_1^{(a_{N,1})}\cdots e_n^{({a_{N,1}})}f_n^{(a_{N,1})}\cdots f_1^{(a_{N,1})}e_1^{{(a_{N,2})}}\cdots e_n^{{(a_{N,2})}}f_n^{(a_{N,2})}\cdots f_2^{(a_{N,2})}\cdots e_1^{(a_{N,N-1})}. \end{align*} $$

This element can be obtained from the element ${\mathfrak m}^{A,{\mathbf {0}}}:=\prod _{(i,h,j)\in \mathscr T_{2n+1}^\angle }(a_{i,j}E^{\theta }_{h+1,h})(\mathbf {0})$ by replacing each factor of the $(a_{i,j}E^\theta _{h+1,h})({\mathbf {0}})$ for $h\leqslant n$ by $f_h^{(a_{i,j})}$ and replacing $(a_{i,j}E^\theta _{h+1,h})({\mathbf {0}})$ for $h> n$ by $e_{2n+1-h}^{(a_{i,j})}$ . In other words, ${\mathfrak m}^{A,{\mathbf {0}}}=\phi ^\jmath _{\mathcal Z}({\mathcal M}^{A,{\mathbf {0}}})$ . Then, ${\mathcal M}^{A,{\mathbf {0}}}\in {U^\jmath _{\mathcal Z}(n)}$ .

Furthermore, for $\lambda \in \mathbb {N}^{n+1}$ , define the elements in ${U^\jmath _{\mathcal Z}(n)}$ , ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}$ , respectively:

$$ \begin{align*} {\mathcal M}^{A,\lambda}:=\bigg[{d\atop\lambda}\bigg]{\mathcal M}^{A,{\mathbf{0}}},\quad{\mathfrak m}^{A,\lambda}=\phi^\jmath_{\mathcal Z}({\mathcal M}^{A,\lambda})=\bigg[{\Bbbk\atop\lambda}\bigg]{\mathfrak m}^{A,{\mathbf{0}}} \end{align*} $$

(see [Reference Du and Wu8, (6.4.1)]).

Theorem 4.8. The set

$$ \begin{align*} {\mathfrak M}:=\{d^{\mathbf{i}}{\mathcal M}^{A,\lambda}\mid A\in{\Xi_{2n+1}^{0\operatorname{diag}}},\lambda\in\mathbb{N}^{n+1},{\mathbf{i}}\in\mathbb{N}_1^{n+1}\} \end{align*} $$

forms a ${\mathcal Z}$ -basis for the integral form ${U^\jmath _{\mathcal Z}(n)}$ .

Proof. By Proposition 4.4 and Theorem 4.6,

$$ \begin{align*} \phi^\jmath_{{\mathcal Z}}(d^{{\mathbf{i}}}{\mathcal M}^{A,\lambda})=\Bbbk^{\mathbf{i}}{\mathfrak m}^{A,\lambda}= \upsilon^{-i_{n+1}}O({\mathbf{i}},\lambda)A({\mathbf{0}})+\,\text{lower\ terms}. \end{align*} $$

Then, the assertion follows from Lemma 4.3(2).

We call ${\mathfrak M}$ a monomial basis for ${U^\jmath _{\mathcal Z}(n)}$ . By identifying ${U^\jmath _{\mathcal Z}(n)}$ with ${\mathfrak {A}^\jmath _{\mathcal Z}(n)}$ , we call the basis corresponding to the basis $\mathfrak B_1$ in Lemma 4.3 the BLM basis for ${U^\jmath _{\mathcal Z}(n)}$ .

5 The hyperalgebra in positive quantum characteristic

Let be a field containing an l th primitive root $\varepsilon $ of $1$ with $l\geqslant 1$ odd. Specialising $\upsilon $ to $\varepsilon $ , is viewed as a ${\mathcal Z}$ -module. We denote the image of $[{\mu \atop \lambda }]$ in by $[{\mu \atop \lambda }]_\varepsilon $ . Let

This is called the i-quantum group at a root of 1 (or an i-quantum hyperalgebra). Recall from [Reference Du and Wu8, Theorem 6.5] the epimorphism $\phi ^{{\jmath }}_{r,{\mathcal Z}}:U_{\mathcal Z}^{{\jmath }}(n)\rightarrow {\mathcal S}_{\mathcal Z}^{{\jmath }}(n,r)$ that induces an algebra epimorphism

Note that the q-Schur algebra ${\mathcal S}_{\mathcal Z}^{{\jmath }}(n,r)$ satisfies the base-change property.

Since , induces an algebra epimorphism

(5-1)

where and . Hence, by taking the direct product, induces an algebra homomorphism

(5-2)

This homomorphism can also be induced from (4-3) via a base change from ${\mathcal Z}$ to . We now show that the homomorphism gives rise to a realisation of .

For simplicity, let , and .

Theorem 5.1. The algebra homomorphism defined in (5-2) is injective. Furthermore, the set

forms a -basis for .

Proof. Throughout the proof, we identify $U^{{\jmath }}_{{\mathcal Z}}(n)$ with $\mathfrak {A}^{{\jmath }}_{{\mathcal Z}}(n)$ via the map $\phi ^{{\jmath }}_{{\mathcal Z}}$ defined in Theorem 4.6. Thus, by Corollary 4.2,

By [Reference Lusztig14, Lemma 6.4(b)] which continues to hold with $\upsilon $ replaced by $\upsilon ^2$ , for the factor involving $d_{n+1}$ ,

Hence, by Lemma 4.3(3),

Since $\overline {O(l{\mathbf {i}})_\varepsilon }=1$ in , it follows that the images $\overline {A({\mathbf {0}})}_\varepsilon \cdot \overline {O(-\lambda ,\lambda )}_\varepsilon $ in of the elements $A({\mathbf {0}})_\varepsilon \cdot O(l{\mathbf {i}})_\varepsilon \cdot O(-\lambda ,\lambda )_\varepsilon $ form a spanning set

$$ \begin{align*}\mathcal B=\{\overline{A({\mathbf{0}})}_\varepsilon\cdot \overline{O(-\lambda,\lambda)}_\varepsilon\mid A\in\Xi_{2n+1}^{\mathrm{0\ diag}},\ \lambda\in\mathbb{N}^{n+1}\}.\end{align*} $$

In other words, we have . Clearly, . To prove the theorem, it suffices to prove that is a linearly independent set (which forces $\mathcal B$ to form a basis for and is injective).

Assume, for some finitely many ,

$$ \begin{align*} \sum_{A\in\Xi_{2n+1}^{\mathrm{0\ diag}},\,\lambda\in\mathbb{N}^{n+1}}f_{A,\lambda} (A_\varepsilon({\mathbf{0}})O_\varepsilon(-\lambda,\lambda))=0. \end{align*} $$

Then, for any $r\geqslant 0$ , the sum at level r is zero. More precisely, in ,

$$ \begin{align*} \sum_{A\in\Xi_{2n+1}^{\mathrm{0\ diag}} \atop \mu\in\Lambda(n+1,r-{|A|}/{2})}\bigg(\sum_{\lambda\in\mathbb{N}^{n+1}}f_{A,\lambda} \varepsilon^{-\lambda^\circ{\cdot} (\tilde\mu+\mathrm{co}(A))}\bigg[{(\mu+\mathrm{co}(A))^{\bullet}\atop \lambda^{\bullet}}\bigg]_\varepsilon\bigg[{\mu_{n+1}+\frac{\mathrm{co}(A)_{n+1}}{2}\atop \lambda_{n+1}}\bigg]_{\varepsilon^2}\bigg)[(A+\tilde\mu)]_\varepsilon =0. \end{align*} $$

Thus, for any $A\in \Xi _{2n+1}^{\mathrm {0\ diag}}$ with $r\geqslant {|A|}/{2}$ and $\mu \in \Lambda (n+1,r-{|A|}/{2})$ ,

(5-3) $$ \begin{align} \sum_{\lambda\in\mathbb{N}^{n+1}}f_{A,\lambda} \varepsilon^{-\lambda^\circ{\cdot} (\tilde\mu+\mathrm{co}(A))}\bigg[{(\mu+\mathrm{co}(A))^{\bullet}\atop\lambda^{\bullet}}\bigg]_\varepsilon\bigg[{\mu_{n+1}+\frac{\mathrm{co}(A)_{n+1}}{2}\atop \lambda_{n+1}}\bigg]_{\varepsilon^2}=0. \end{align} $$

We claim that, for any $A\in \Xi _{2n+1}^{\mathrm {0\ diag}}$ and $\mu ,\alpha \in \mathbb {N}^{n+1}$ , the following equation holds:

(5-4) $$ \begin{align} g_{A,\alpha,\mu}:=\sum_{\lambda\in\mathbb{N}^{n+1},\,\lambda\geqslant\alpha}f_{A,\lambda} \varepsilon^{-\lambda^\circ{\cdot}(\tilde\mu+\mathrm{co}(A))}\bigg[{(\mu+\mathrm{co}(A))^{\bullet}\atop(\lambda-\alpha)^{\bullet}}\bigg]_\varepsilon\bigg[{\mu_{n+1}+\frac{\mathrm{co}(A)_{n+1}}{2}\atop \lambda_{n+1}-\alpha_{n+1}}\bigg]_{\varepsilon^2}=0. \end{align} $$

We proceed by induction on $|\alpha |$ . If $|\alpha |=0$ , then the claim follows from (5-3). Now, we assume $|\alpha |>0$ . Then there exists some $\beta \in \mathbb {N}^{n+1}$ such that $\alpha =\beta +\boldsymbol {e}_i$ .

Case 1. If $i\in [1,n]$ , for $\lambda \in \mathbb {N}^{n+1}$ with $\lambda \geqslant \beta $ ,

$$ \begin{align*} \bigg[{(\mu+\boldsymbol{e}_i+\text{ co}(A))^{\bullet}\atop(\lambda-\beta)^{\bullet}}\bigg]_\varepsilon=\varepsilon^{\lambda_i-\beta_i}\bigg[{(\mu+\text{ co}(A))^{\bullet}\atop (\lambda-\beta)^{\bullet}}\bigg]_\varepsilon+ \varepsilon^{\lambda_i-\beta_i-1-\mu_i-\mathrm{co}(A)_{i}}\bigg[{(\mu+\text{ co}(A))^{\bullet}\atop(\lambda-\beta-\boldsymbol{e}_i)^{\bullet}}\bigg]_\varepsilon. \end{align*} $$

Thus, by the induction hypothesis, we conclude that

$$ \begin{align*} 0=g_{A,\beta,\mu+\boldsymbol{e}_i}=\varepsilon^{-\beta_i}g_{A,\beta,\mu}+\varepsilon^{-\beta_i-1-\text{ co}(A)_{i}-\mu_i}g_{A,\alpha,\mu}=\varepsilon^{-\beta_i-1-\text{ co}(A)_{i}-\mu_i}g_{A,\alpha,\mu} \end{align*} $$

for $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ and $\mu \in \mathbb {N}^{n+1}$ . It follows that $g_{A,\alpha ,\mu }=0$ for $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ and $\mu \in \mathbb {N}^{n+1}$ , proving (5-4).

Case 2. If $i=n+1$ , for $\lambda \in \mathbb {N}^{n+1}$ with $\lambda \geqslant \beta $ ,

$$ \begin{align*} \bigg[{\mu_{n+1}+1+\frac{\mathrm{co}(A)_{n+1}}{2}\atop\lambda_{n+1}-\beta_{n+1}}\bigg]_{\varepsilon^2} &=\varepsilon^{2(\lambda_{n+1}-\beta_{n+1})}\bigg[{\mu_{n+1}+({\mathrm{co}(A)_{n+1}}/{2})\atop \lambda_{n+1}-\beta_{n+1}}\bigg]_{\varepsilon^2}\\ &\quad + \varepsilon^{2(\lambda_{n+1}-\beta_{n+1}-1-\mu_{n+1}-({\mathrm{co}(A)_{n+1}}/{2}))}\bigg[{\mu_{n+1}+ ({\mathrm{co}(A)_{n+1}}/{2})\atop\lambda_{n+1}-\beta_{n+1}-1}\bigg]_{\varepsilon^2}. \end{align*} $$

Thus, by the induction hypothesis, we conclude that

$$ \begin{align*} \begin{aligned} 0=g_{A,\beta,\mu+\boldsymbol{e}_i}&=\varepsilon^{-2\beta_{n+1}}g_{A,\beta,\mu}+ \varepsilon^{2(-\beta_{n+1}-1-({\mathrm{co}(A)_{n+1}}/{2})-\mu_{n+1})}g_{A,\alpha,\mu}\\ &=\varepsilon^{2(-\beta_{n+1}-1-({\mathrm{co}(A)_{n+1}}/{2})-\mu_{n+1})}g_{A,\alpha,\mu} \end{aligned} \end{align*} $$

for $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ and $\mu \in \mathbb {N}^{n+1}$ . It follows that $g_{A,\alpha ,\mu }=0$ for $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ and $\mu \in \mathbb {N}^{n+1}$ , proving (5-4).

Let ${\mathcal X}=\{\lambda \in \mathbb {N}^{n+1}\mid f_{A,\lambda }\not =0\quad \text {for some }A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}\}$ . If ${\mathcal X}\not =\emptyset $ , we may choose a maximal element $\nu $ in ${\mathcal X}$ with respect to $\leqslant $ . Then, by (5-4),

$$ \begin{align*} f_{A,\nu}= \varepsilon^{\nu^\circ{\cdot} (\tilde\mu+\mathrm{co}(A))} {g_{A,\nu,\mu}=0\quad \text{ for all } \ A\in\Xi^{\mathrm{0\ diag}}_{2n+1}.} \end{align*} $$

This is a contradiction. Thus, $f_{A,\lambda }=0$ for all $A\in \Xi ^{\mathrm {0\ diag}}_{2n+1}$ and $\lambda \in \mathbb {N}^{n+1}$ . This completes the proof of the theorem.

Remark 5.2. We first observe that, by the algebra epimorphism in (5-1), the module category -mod of -modules is a full subcategory of that of . Thus, the various bases for would play important roles in the study of representations of . Next, we observe that we may link representations of to those of finite orthogonal groups.

Let $\widehat G=\text {GL}_{2r+1}(q)$ be the finite general linear group over a field of q elements. For a composition $\mu $ of $2r+1$ , let $\widehat P_\mu $ be the parabolic subgroup of $\widehat G$ associated with $\mu $ (see [Reference Deng, Du, Parshall and Wang4, Section 13.4]). Then, the finite orthogonal group $G:=O_{2r+1}(q)$ can be regarded as a fixed-point subgroup of $\widehat G$ . For $\lambda \in \Lambda (n+1,r)$ , $P_\lambda :=\widehat P_{{\widetilde \lambda }}\cap G$ , where $ {\widetilde \lambda }=(\lambda _1,\ldots ,\lambda _n,2\lambda _{n+1}+1,\lambda _n,\ldots ,\lambda _1)$ forms a parabolic subgroup of G.

Then, for any field with char, the isomorphism (see [Reference Bao, Kujawa, Li and Wang1] or [Reference Du and Wu8, Theorem 3.2])

establishes a link between representations of the i-hyperalgebra and the orthogonal group G. For example, for $n\geqslant r$ , the decomposition matrix of ${\mathcal S}_{\mathcal O}^{\jmath }(n,r)$ is a submatrix of the decomposition matrix of ${ {\mathcal O} G}$ (see [Reference Du, Parshall and Scott7, Theorem 5.3.1]). As an application of the theory established in [Reference Du, Parshall and Scott7], we may also conclude that the above mentioned decomposition matrices are unitriangular for any n and r (see [Reference Du, Parshall and Scott7, Theorem 5.1.4]).