2 The BLM construction of $ {{\mathbf {U}}(\mathfrak {gl}_n)} $

Following [Reference Jimbo20], we define the quantized enveloping algebra ${{\mathbf {U}}(\mathfrak {gl}_n)}$ of $\mathfrak {gl}_n$ to be the $\mathbb Q(v)$ algebra with generators

$$ \begin{align*}E_i,\ F_i\quad(1\leqslant i\leqslant n-1),\ K_j,\ K_j^{-1}\quad(1\leqslant j\leqslant n),\end{align*} $$

and relations

$\mathrm{(a)}\ K_{i}K_{j}=K_{j}K_{i},\ K_{i}K_{i}^{-1}=1;$

$\mathrm{(b)}\ K_{i}E_j=v^{\delta _{i,j}-\delta _{i,j+1}} E_jK_{i};$

$\mathrm{(c)}\ K_{i}F_j=v^{\delta _{i,j+1}-\delta _{i,j}} F_jK_{i};$

$\mathrm{(d)}\ E_iE_j=E_jE_i,\ F_iF_j=F_jF_i\ when\ |i-j|>1;$

$\mathrm{(e)}\ E_iF_j-F_jE_i=\delta _{i,j}\frac {\widetilde K_{i} -\widetilde K_{i}^{-1}}{v-v^{-1}},\ where \ \widetilde K_i =K_{i}K_{i+1}^{-1};$

$\mathrm{(f)}\ E_i^2E_j-(v+v^{-1})E_iE_jE_i+E_jE_i^2=0\ when\ |i-j|=1;$

$\mathrm{(g)}\ F_i^2F_j-(v+v^{-1})F_iF_jF_i+F_jF_i^2=0\ when\ |i-j|=1.$

Following [Reference Lusztig22], let ${U_{\mathcal Z}(\mathfrak {gl}_n)}$ be the Lusztig integral form of ${{\mathbf {U}}(\mathfrak {gl}_n)}$ generated by $E_{i}^{(m)}, F_{i}^{(m)}$ , $K_{j}^{\pm 1}$ , and $\big [{K_j;c \atop t}\big ](1 \leqslant i \leqslant n-1,1 \leqslant j \leqslant n, m, t \in \mathbb {N}, c \in \mathbb {Z})$ , where

$$ \begin{align*}E_{i}^{(m)}=\frac{E_{i}^{m}}{[m]^!}, \quad F_{i}^{(m)}=\frac{F_{i}^{m}}{[m]^!} \quad \text{and} \quad \left[{K_{j}; c \atop t}\right]=\prod_{s=1}^{t} \frac{K_{j} v^{c-s+1}-K_{j}^{-1} v^{-c+s-1}}{v^{s}-v^{-s}}, \end{align*} $$

with $[m]^{!}=[1][2] \cdots [m]$ and $[i]=\frac {v^{i}-v^{-i}}{v-v^{-1}}$ . The following result is given by Lusztig [Reference Lusztig21].

Let $\Pi (n)=\{\alpha _i\mid 1\leqslant i\leqslant n-1\}$ , where $\alpha _i=\boldsymbol {e}_i-\boldsymbol {e}_{i+1}$ with $\boldsymbol e_i=(0,\ldots ,0,\underset i1,0\cdots ,0)\in \mathbb Z^n.$ We have the following direct sum decomposition:

$$ \begin{align*} {{\mathbf{U}}(\mathfrak{gl}_n)}=\bigoplus\limits_{\nu\in\mathbb Z\Pi(n)}{{\mathbf{U}}(\mathfrak{gl}_n)}_\nu, \end{align*} $$

where ${{\mathbf {U}}(\mathfrak {gl}_n)}_\nu $ is defined by the conditions ${{\mathbf {U}}(\mathfrak {gl}_n)}_{\nu '}{{\mathbf {U}}(\mathfrak {gl}_n)}_{\nu "}\subseteq {{\mathbf {U}}(\mathfrak {gl}_n)}_{\nu '+\nu "}$ , $K_j^{\pm 1}\in {{\mathbf {U}}(\mathfrak {gl}_n)}_0$ , $E_i\in {{\mathbf {U}}(\mathfrak {gl}_n)}_{\alpha _i}$ , $F_i\in {{\mathbf {U}}(\mathfrak {gl}_n)}_{-\alpha _i}$ for all $\nu ',\nu "\in \mathbb Z\Pi (n)$ , $1\leqslant i\leqslant n-1$ and $1\leqslant j\leqslant n$ .

Following [Reference Lusztig23, 23.1], we introduce the modified quantum group $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ associated with ${{\mathbf {U}}(\mathfrak {gl}_n)}$ as follows. Let

$$ \begin{align*}\dot{{{\mathbf{U}}}}(\mathfrak{gl}_n)=\bigoplus\limits_{{\lambda},\mu\in\mathbb Z^{n}}{}_{\lambda}{{\mathbf{U}}(\mathfrak{gl}_n)}_\mu,\end{align*} $$

where

$$ \begin{align*}{}_{\lambda}{{\mathbf{U}}(\mathfrak{gl}_n)}_\mu={{\mathbf{U}}(\mathfrak{gl}_n)}/\left(\sum_{{\mathbf{j}}\in\mathbb Z^{n}}(K^{\mathbf{j}}- v^{{\lambda}\cdot{\mathbf{j}}}){{\mathbf{U}}(\mathfrak{gl}_n)}+\sum_{{\mathbf{j}}\in\mathbb Z^{n}}{{\mathbf{U}}(\mathfrak{gl}_n)}(K^{\mathbf{j}} -v^{\mu\cdot{\mathbf{j}}})\right),\end{align*} $$

and ${\lambda }\cdot {\mathbf {j}}=\sum _{1\leqslant i\leqslant n}{\lambda }_ij_i$ . Let $\pi _{{\lambda },\mu }: {{\mathbf {U}}(\mathfrak {gl}_n)}\rightarrow {}_{{\lambda }}{{\mathbf {U}}(\mathfrak {gl}_n)}_{\mu }$ be the canonical projection.

We define the product in $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ as follows. For ${\lambda }', \mu ', {\lambda }", \mu " \in \mathbb Z^{n}$ with ${\lambda }'-\mu '$ , ${\lambda }"-\mu " \in \mathbb Z \Pi (n)$ and any $t \in {{\mathbf {U}}(\mathfrak {gl}_n)}_{{\lambda }'-\mu '}$ , $s \in {{\mathbf {U}}(\mathfrak {gl}_n)}_{{\lambda }"-\mu "}$ , define

$$ \begin{align*}\pi_{{\lambda}',\mu'}(t) \pi_{{\lambda}", \mu"}(s)= \begin{cases}\pi_{{\lambda}',\mu"}(t s), & \text { if } \mu'={\lambda}" ,\\ 0 & \text { otherwise. }\end{cases} \end{align*} $$

Then $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ becomes an associative $\mathbb Q(v)$ -algebra with the above product. Moreover, the algebra $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ is naturally a ${{\mathbf {U}}(\mathfrak {gl}_n)}$ -bimodule defined by $t' \pi _{{\lambda }', {\lambda }"}(s) t"=\pi _{{\lambda }'+\nu ', {\lambda }"-\nu "}(t' s t")$ for $t' \in {{\mathbf {U}}(\mathfrak {gl}_n)}_{\nu '}, s \in {{\mathbf {U}}(\mathfrak {gl}_n)}, t" \in {{\mathbf {U}}(\mathfrak {gl}_n)}_{\nu "}$ , and ${\lambda }', {\lambda }" \in \mathbb Z^{n} .$ Let $\dot {{U}}_{\mathcal Z}(\mathfrak {gl}_n)$ be the ${\mathcal Z}$ -subalgebra of $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ generated by the elements $E_{i}^{(m)} 1_{\lambda }$ and $1_{\lambda } F_{i}^{(m)}$ for $1 \leqslant i \leqslant n-1$ and $m \in \mathbb {N}$ , where $1_{\lambda }=\pi _{\lambda , \lambda }(1)$ .

We now follow [Reference Dipper and James6] to recall the definition of q-Schur algebras as follows. The Hecke algebra ${{\mathcal H}_{\mathcal Z}(d)}$ associated with ${\mathfrak S}_d$ is the ${\mathcal Z}$ -algebra generated by $T_i$ ( $1\leqslant i\leqslant d-1$ ), with the following relations:

$$ \begin{align*}(T_i+1)(T_i-q)=0,\;\; T_iT_{i+1}T_i=T_{i+1}T_iT_{i+1},\;\;T_iT_j=T_jT_i\;(|i-j|>1). \end{align*} $$

where $q=v^2$ . Let ${\boldsymbol {\mathcal H}}(d)={{\mathcal H}_{\mathcal Z}(d)}\otimes _{\mathcal Z}\mathbb Q(v)$ . If $w=s_{i_1}s_{i_2}\cdots s_{i_m}$ is reduced let $T_w=T_{i_1}T_{i_2}\cdots T_{i_m}$ . Then the set $\{T_w\mid w\in {\mathfrak S}_d\}$ forms a ${\mathcal Z}$ -basis for ${{\mathcal H}_{\mathcal Z}(d)}$ . Let $\Lambda (n,d)=\{{\lambda }\in \mathbb N^{n}\mid \sigma ({\lambda })=d\}$ , where $\sigma ({\lambda })=\sum _{1\leqslant i\leqslant n}{\lambda }_i$ . For ${\lambda }\in \Lambda (n,d)$ , let $x_{{\lambda }}=\sum _{w\in {\mathfrak S}_{{\lambda }}}T_w$ , where ${\mathfrak S}_{{\lambda }}$ is the Young subgroup of ${\mathfrak S}_d$ . The endomorphism algebras

$$ \begin{align*}{\mathcal S}_{\mathcal Z}(n,d):=\operatorname{End}_{{{\mathcal H}_{\mathcal Z}(d)}}\bigg( \bigoplus_{{\lambda}\in\Lambda(n,d)}x_{{\lambda}}{{\mathcal H}_{\mathcal Z}(d)}\bigg),\quad {\boldsymbol{\mathcal S}}(n,d):=\operatorname{End}_{{\boldsymbol{\mathcal H}}(d)}\bigg( \bigoplus_{{\lambda}\in\Lambda(n,d)}x_{{\lambda}}{\boldsymbol{\mathcal H}}(d)\bigg),\end{align*} $$

are called q-Schur algebras over ${\mathcal Z}$ and over $\mathbb Q(v)$ , respectively.

We now recall the BLM construction of ${{\mathbf {U}}(\mathfrak {gl}_n)}$ . Let $\widetilde {\Theta }(n)$ be the set of all $n \times n$ matrices over $\mathbb {Z}$ with all off diagonal entries in $\mathbb {N}$ . Let $\Theta (n)$ be the set of all $n\times n$ matrices over $\mathbb N$ . Let ${\Theta (n,d)}$ be the set of all $n \times n$ matrices A over $\mathbb {N}$ such that $\sigma (A)=d$ , where $\sigma (A)=\sum _{1 \leqslant i, j \leqslant n} a_{i, j}$ . For $A \in \widetilde {\Theta }(n)$ , let $\operatorname {ro}(A)=(\sum _{j} a_{1, j}, \ldots , \sum _{j} a_{n, j})$ and $\operatorname {co}(A)=$ $(\sum _{i} a_{i, 1}, \ldots , \sum _{i} a_{i, n})$ .

The q-Schur algebra ${\mathcal S}_{\mathcal Z}(n,d)$ was reconstructed using the geometry of pairs of n-step filtrations on a d-dimensional vector space in [Reference Beilinson, Lusztig and MacPherson3]. In particular, a normalized ${\mathcal Z}$ -basis $\{[A]\}_{A \in {\Theta (n,d)}}$ for ${\mathcal S}_{\mathcal Z}(n,d)$ was constructed. Using the stabilization property of multiplication for q-Schur algebra, an important ${\mathcal Z}$ -algebra $K_{\mathcal Z}(n)$ (without $ 1 $ ), with basis $\{[A]\}_{A \in \widetilde {\Theta }(n)} $ , was constructed in [1, §4]. Let ${\mathbf K}(n)=K_{\mathcal Z}(n)\otimes _{\mathcal Z}\mathbb Q(v)$ . Following [Reference Beilinson, Lusztig and MacPherson3, 5.1], we define $\widehat {{\mathbf K}}(n)$ to be the vector space of all formal (possibly infinite) $\mathbb {Q}(v)$ -linear combinations $\sum _{A \in \widetilde {\Theta }(n)} \beta _{A}[A]$ satisfying the following property: for any ${\mathbf {x}} \in \mathbb Z^{n}$ , the sets $\{A \in \widetilde {\Theta }(n) \mid \beta _{A} \neq 0, \operatorname {ro}(A)={\mathbf {x}}\}$ and $\{A \in \widetilde {\Theta }(n) \mid \beta _{A} \neq 0, \operatorname {co}(A)={\mathbf {x}}\}$ are finite. The product of two elements $\sum _{A \in \widetilde {\Theta }(n)} \beta _{A}[A]$ , $\sum _{B \in \widetilde {\Theta }(n)} \gamma _{B}[B]$ in $\widehat {{\mathbf K}}(n)$ is defined to be $\sum _{A, B} \beta _{A} \gamma _{B}[A] \cdot [B]$ , where $[A] \cdot [B]$ is the product in $K_{\mathcal Z}(n).$ Then $\widehat {{\mathbf K}}(n)$ is an associative algebra.

Let $\Theta ^\pm (n)$ be the set of all $A \in \Theta (n)$ such that all diagonal entries are zero. For $A \in \Theta ^\pm (n)$ and ${\mathbf {j}} \in \mathbb {Z}^{n}$ , let

$$ \begin{align*}\begin{aligned} A({\mathbf{j}},d) &=\sum_{\lambda \in \Lambda(n,d-\sigma(A))} v^{\lambda \cdot {\mathbf{j}}}[A+\operatorname{diag}(\lambda)] \in {\boldsymbol{\mathcal S}}(n,d), \\ A({\mathbf{j}}) &=\sum_{\lambda \in \mathbb{Z}^{n}} v^{\lambda \cdot {\mathbf{j}}}[A+\operatorname{diag}(\lambda)] \in \widehat{{\mathbf K}}(n), \end{aligned} \end{align*} $$

where ${\lambda }\cdot{{\mathbf {j}}}=\sum _{1\leqslant i\leqslant n}{\lambda }_i{j_i}$ .

We shall denote by $\mathbf {V}(n)$ the subspace of $\widehat {{\mathbf K}}(n)$ spanned by the elements $A({\mathbf {j}})$ for $A \in \Theta ^{\pm }(n)$ and ${\mathbf {j}} \in \mathbb {Z}^{n}$ . For $1 \leqslant i, j \leqslant n$ , let $E_{i, j} \in \Theta (n)$ be the matrix whose $(i,j)$ -entry is $1$ and the other entries are $0$ . The following result was given by Beilinson–Lusztig–MacPherson [Reference Beilinson, Lusztig and MacPherson3].

Theorem 2.2. $(1)$ $\mathbf {V}(n)$ is a subalgebra of $\widehat {{\mathbf K}}(n)$ and there is an algebra isomorphism ${{\mathbf {U}}(\mathfrak {gl}_n)}\stackrel {\thicksim }{\,\rightarrow }\mathbf {V}(n)$ satisfying

$$ \begin{align*}E_h\mapsto E_{h,h+1}(\mathbf{0}),\ K_1^{j_1}K_2^{j_2}\cdots K_n^{j_n}\mapsto 0({\mathbf{j}}),\ F_h\mapsto E_{h+1,h}(\mathbf{0}).\end{align*} $$

$(2)$ There is an algebra epimorphism $\zeta _d:{{\mathbf {U}}(\mathfrak {gl}_n)}\rightarrow {\boldsymbol {\mathcal S}}(n,d)$ satisfying

$$ \begin{align*}E_h\mapsto E_{h,h+1}(\mathbf{0},d),\ K_1^{j_1}K_2^{j_2}\cdots K_n^{j_n}\mapsto 0({\mathbf{j}},d),\ F_h\mapsto E_{h+1,h}(\mathbf{0},d).\end{align*} $$

We shall identify ${{\mathbf {U}}(\mathfrak {gl}_n)}$ with $\mathbf {V}(n)$ . By [Reference Du and Fu14] we have the following result (cf. [Reference Fu17]).

Lemma 2.3. $(1)$ There is an algebra isomorphism $\varphi :\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)\rightarrow {\mathbf K}(n)$ satisfying

$$ \begin{align*}\pi_{{\lambda}\mu}(u)\mapsto[\operatorname{diag}({\lambda})]u[\operatorname{diag}(\mu)],\end{align*} $$

for all $u\in {{\mathbf {U}}(\mathfrak {gl}_n)}$ and ${\lambda },\mu \in \mathbb Z^{n}$ . Furthermore, we have $\varphi (\dot {{U}}_{\mathcal Z}(\mathfrak {gl}_n))=K_{\mathcal Z}(n)$ .

$(2)$ There is a surjective algebra homomorphism $\dot {\zeta }_d:{\mathbf K}(n)\rightarrow {\boldsymbol {\mathcal S}}(n,d)$ such that

$$ \begin{align*} { \dot{\zeta}_d([A])=\begin{cases}[A], &\text{ if } A\in{\Theta(n,d)};\\ 0, &\text{ otherwise.} \end{cases}} \end{align*} $$

We shall identify $\dot {{U}}_{\mathcal Z}(\mathfrak {gl}_n)$ with $K_{\mathcal Z}(n)$ .

3 The infinitesimal q-Schur algebra

Let . We shall denote the images of $E_{i}^{(m)}, F_{i}^{(m)}$ , etc. in by the same letters. Let (resp. ) be the subalgebra of generated by the elements $E_{i}^{(m)}$ (resp. $F_{i}^{(m)}$ ) for $1\leqslant i\leqslant n-1$ and $m\in \mathbb N$ . Let be the subalgebra of generated by the elements $K_{j}^{\pm 1}$ and $\big [{K_j;0 \atop t} \big ]$ for $1 \leqslant j \leqslant n$ and $t \in \mathbb {N}$ . Then we have . The algebras and are both $\mathbb {N}$ -graded in terms of the degrees of monomials in the $E_{i}^{(m)}$ and $F_{i}^{(m)}$ .

For $r\geqslant 1$ , let be the -subalgebra of generated by the elements $E_{i}^{(m)}$ , $F_{i}^{(m)}$ , $K_{j}^{\pm 1}$ , and $\big [{K_j;0 \atop t} \big ]$ for $1 \leqslant i \leqslant n-1$ , $1 \leqslant j \leqslant n$ , $t \in \mathbb {N}$ and $0 \leqslant m,t<l p^{r-1}$ . Furthermore, let

Clearly, the algebra is a Hopf subalgebra of . Let (resp. ) be the subalgebra of generated by the elements $E_{i}^{(m)}$ (resp. $F_{i}^{(m)}$ ) for $1\leqslant i\leqslant n-1$ and $0 \leqslant m<l p^{r-1}$ . Then we have .

Let $\Theta ^+(n)=\left \{A \in \Theta (n) \mid a_{i, j}=0, \forall i \geqslant j\right \}$ and $\Theta ^-(n)=\left \{A \in \Theta (n) \mid a_{i, j}=0, \forall i \leqslant j\right \}$ . For $A \in \widetilde \Theta (n)$ , write $A=A^{+}+\operatorname {diag}({\lambda })+A^{-}$ with $A^{+} \in \Theta ^+(n)$ , $A^{-} \in \Theta ^-(n)$ and ${\lambda }\in \mathbb Z^{n}$ . Let

$$ \begin{align*}\Theta^\pm(n)_r=\{A\in\Theta^\pm(n)\mid a_{i,j}<lp^{r-1},\,\forall i\not=j\}.\end{align*} $$

Let

(3.1) $$ \begin{align} \Theta^+(n)_r=\Theta^\pm(n)_r\cap\Theta^+(n),\ \Theta^-(n)_r=\Theta^\pm(n)_r\cap\Theta^-(n). \end{align} $$

For $A \in \Theta ^\pm (n)_r$ , let

where

$$ \begin{align*} M_{j}=M_{j}(A^{+})=E_{j-1}^{(a_{j-1, j})}(E_{j-2}^{(a_{j-2, j})} E_{j-1}^{(a_{j-2, j})}) \cdots(E_{1}^{(a_{1, j})} E_{2}^{(a_{1, j})} \cdots E_{j-1}^{(a_{1, j})}), \end{align*} $$

and

$$ \begin{align*}M_{j}^{\prime}=M_{j}^{\prime}(A^{-})=(F_{j-1}^{(a_{j, 1})} \cdots F_{2}^{(a_{j, 1})} F_{1}^{(a_{j, 1})}) \cdots(F_{j-1}^{(a_{j, j-2})} F_{j-2}^{(a_{j, j-2})}) F_{j-1}^{(a_{j, j-1})}. \end{align*} $$

For $A \in \Theta ^\pm (n)$ let

$$ \begin{align*}\operatorname{deg}(A)= \sum_{1 \leqslant i, j \leqslant n}|j-i| a_{i, j}.\end{align*} $$

Then we have $\textrm {deg}(E^{(A^+)})=\textrm {deg}(A^+)$ and $\textrm {deg}(F^{(A^-)})=\textrm { deg}(A^-)$ for $A\in \Theta ^\pm (n)$ . For ${\lambda }\in \mathbb N^{n}$ and ${\mathbf {j}}\in \mathbb Z^{n}$ let

$$ \begin{align*}K_{\lambda}=\prod_{1 \leqslant i \leqslant n}\bigg[ {K_i;0 \atop {\lambda}_i} \bigg],\,K^{{\mathbf{j}}}=\prod_{1 \leqslant i \leqslant n}K_i^{j_i}.\end{align*} $$

The following result is given in [Reference Fu18, Lem. 6.3].

For $A\in \widetilde \Theta (n)$ let

$$ \begin{align*}\sigma_{i,j}(A) =\begin{cases} \sum_{s\leqslant i;t\geqslant j}a_{s,t}&\text{if } i<j\\ \sum_{s\geqslant i;t\leqslant j}a_{s,t}&\text{if } i>j. \end{cases}\end{align*} $$

Following [Reference Beilinson, Lusztig and MacPherson3], for $A,B\in \widetilde \Theta (n)$ , define $B \preccurlyeq A$ if and only if $\sigma _{i,j}( B )\leqslant \sigma _{i,j}(A)$ for all $i\not =j$ . Put $ B \prec A$ if $ B \preccurlyeq A$ and $\sigma _{i,j}( B )<\sigma _{i,j}(A)$ for some $i\not =j$ .

Proof. By [Reference Beilinson, Lusztig and MacPherson3, 4.6(c)] for $A \in \Theta ^\pm (n)_r$ , we have

(3.2) $$ \begin{align} E^{(A^+)}=A^+(\mathbf{0})+f,\quad F^{(A^-)}=A^-(\mathbf{0})+g, \end{align} $$

where f is a -linear combination of $B(\mathbf {0})$ for $B\in \Theta ^+(n)$ with $B\prec A^+$ and g is a -linear combination of $C(\mathbf {0})$ for $C\in \Theta ^-(n)$ with $C\prec A^-$ . By [Reference Fu18, Lem. 6.3] we know that f must a -linear combination of $B(\mathbf {0})$ for $B\in \Theta ^+(n)_r$ with $B\prec A^+$ and g is a -linear combination of $C(\mathbf {0})$ for $C\in \Theta ^-(n)_r$ with $C\prec A^-$ . Now the assertion follows from Proposition 3.1.

Let . We shall denote the images of $E_{i}^{(m)}1_{\lambda }$ , $1_{\lambda } F_{i}^{(m)}$ , $E^{(A^+)} 1_{\lambda }$ , $1_{\lambda }F^{(A^-)}$ in by the same letters. For $A\in \widetilde \Theta (n)$ let

Let be the -subalgebra of generated by the elements $E_{i}^{(m)}1_{\lambda }$ and $1_{\lambda } F_{i}^{(m)}$ for $1 \leqslant i \leqslant n-1$ , $\lambda \in \mathbb Z^{n}$ and $0 \leqslant m<l p^{r-1}$ .

For $ A \in \widetilde \Theta (n)$ and $1 \leqslant i \leqslant n$ , let

$$ \begin{align*}{\boldsymbol\sigma}(A)=(\sigma_1(A),\sigma_2(A),\ldots,\sigma_n(A)),\end{align*} $$

where $ \sigma _{i}(A)=a_{i,i}+\sum _{1 \leqslant j<i}(a_{i j}+a_{j i})$ . Let

$$ \begin{align*} \begin{aligned} \widetilde\Theta(n)_r&=\{A\in\widetilde\Theta(n)\mid a_{i,j}<lp^{r-1},\,\forall i\not=j\}. \end{aligned} \end{align*} $$

We have the following monomial, BLM and PBW bases of .

Proof. Let be the -submodule of spanned by the elements $[A]_{\varepsilon }$ for $A\in \widetilde \Theta (n)_r$ . By [Reference Beilinson, Lusztig and MacPherson3, 4.6(a)] for $1\leqslant h\leqslant n-1$ , $0\leqslant m<lp^{r-1}$ , $A\in \widetilde \Theta (n)_r$ , we have

$$ \begin{align*}E_h^{(m)} [A]_{\varepsilon}=\sum_{{\mathbf{t}}\in\Lambda(n,m)\atop a_{h+1,u}\geqslant t_u,\,\forall u\not=h+1}\varepsilon^{\beta({\mathbf{t}})} \prod_{1\leq u\leq n}{\bigg[{a_{h,u}+t_u\atop t_u}\bigg]_{\varepsilon}}\big[A+\sum_{1\leq u\leq n}t_u(E_{h,u}-E_{h+1,u})\big]_{\varepsilon},\end{align*} $$

where $\beta ({\mathbf {t}})=\sum _{j> u}(a_{h,j}-a_{h+1,j})t_u+\sum _{u<u'}t_ut_{u'}$ . If $A+\sum _{u}t_u(E_{h,u}-E_{h+1,u})\not \in \widetilde \Theta (n)_r$ for some ${\mathbf {t}}\in \Lambda (n,m)$ , then we have $a_{h,u}+t_u\geqslant lp^{r-1}$ for some $u\not =h$ . Since $A\in \widetilde \Theta (n)_r$ , ${\mathbf {t}}\in \Lambda (n,m)$ and $m<lp^{r-1}$ , we have $a_{h,u}<lp^{r-1}$ and $t_u<lp^{r-1}$ . Hence, by [Reference Fu18, Cor. 3.4] we have $\big [{a_{h,u}+t_u\atop t_u}\big ]_{\varepsilon }=0$ . Therefore, we have

Similarly, we have

for $1\leqslant h\leqslant n-1$ and $0\leqslant m<lp^{r-1}$ . Consequently, we have

(3.3)

Furthermore, by [Reference Beilinson, Lusztig and MacPherson3, 4.6(c)] for $A \in \widetilde \Theta (n)_r$ ,

(3.4) $$ \begin{align} E^{(A^+)} 1_{{\boldsymbol\sigma}(A)} F^{(A^-)}=[A]_{\varepsilon} +f, \end{align} $$

where f is a -linear combination of $[B]_{\varepsilon }$ for $B\in \widetilde \Theta (n)$ with $B\prec A$ . By (3.3), we see that f must be a -linear combination of $[B]_{\varepsilon }$ for $B\in \widetilde \Theta (n)_r$ with $B\prec A$ . It follows from (3.2) that

(3.5) $$ \begin{align} A^+(\mathbf{0}) 1_{{\boldsymbol\sigma}(A)} A^-(\mathbf{0})=[A]_{\varepsilon} +g, \end{align} $$

where g is a -linear combination of $[B]_{\varepsilon }$ for $B\in \widetilde \Theta (n)_r$ with $B\prec A$ . Therefore, each of the sets ${\mathscr M}_r$ , ${\mathscr L}_r$ , ${\mathscr P}_r$ forms a -basis of and

Hence, by (3.3) we have . The proof is completed.

Let . By [Reference Du13] we have $\zeta _d({U_{\mathcal Z}(\mathfrak {gl}_n)})={\mathcal S}_{\mathcal Z}(n,d)$ . Therefore, the map $\zeta _d: {{\mathbf {U}}(\mathfrak {gl}_n)} \rightarrow {\boldsymbol {\mathcal S}}(n,d)$ given in Theorem 2.2 restricts to a surjective algebra homomorphism

(3.6) $$ \begin{align} \zeta_{d}: {U_{\mathcal Z}(\mathfrak{gl}_n)} \rightarrow {\mathcal S}_{\mathcal Z}(n,d). \end{align} $$

A generating set for the kernel of $\zeta _{d}:{U_{\mathcal Z}(\mathfrak {gl}_n)} \rightarrow {\mathcal S}_{\mathcal Z}(n,d)$ was given in [Reference Fu and Gao19]. The map $\zeta _{d}$ induces, upon tensoring with , a surjective algebra homomorphism

(3.7)

Let

$$ \begin{align*}\mathtt{e}_i=\zeta_{d} (E_i),\,\mathtt{f}_i=\zeta_{d}(F_i),\,\mathtt{k}_j=\zeta_{d}(K_j),\end{align*} $$

for $1 \leqslant i \leqslant n-1$ and $1\leqslant j \leqslant n$ . For $A \in \Theta (n)$ and ${\lambda }\in \mathbb N^{n} $ , let

$$ \begin{align*}{\mathtt e}^{(A^+)}=\zeta_{d}(E^{(A^+)}),\,{\mathtt f}^{(A^-)}=\zeta_{d}(F^{(A^-)}),\,\mathtt{k}_{\lambda}= \zeta_{d}(K_{\lambda}).\end{align*} $$

For $ A \in {\Theta (n,d)}$ , let

By [Reference Du and Parshall15, Cor. 5.3], we have

(3.8) $$ \begin{align} \mathtt{k}_{\lambda}=[\operatorname{diag}({\lambda})]_{\varepsilon}, \end{align} $$

for ${\lambda }\in \Lambda (n,d)$ .

Let be the infinitesimal q-Schur algebra introduced in [Reference Cox4]. The algebra is a -subalgebra of the q-Schur algebra . Let

$$ \begin{align*}\Theta(n,d)_r=\{A \in {\Theta(n,d)} \mid a_{ij} < lp^{r-1} \ \text{for all} \ i \neq j\}.\end{align*} $$

According to [Reference Cox4, 5.3.1] and the proof of [Reference Fu16, Th. 5.5], we have the following result.

By [Reference Fu18, Prop. 6.4], we have the following result.

The map $\dot {\zeta }_d: \dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)={\mathbf K}(n)\rightarrow {\boldsymbol {\mathcal S}}(n,d)$ given in Lemma 2.3 restricts to a surjective algebra homomorphism

(3.9) $$ \begin{align} \dot{\zeta}_{d}: \dot{{U}}_{\mathcal Z}(\mathfrak{gl}_n)\rightarrow {\mathcal S}_{\mathcal Z}(n,d); \end{align} $$

tensoring with , we obtain a surjective algebra homomorphism

(3.10)

Combining Lemma 2.3 with Proposition 3.3, we obtain the following result.

For ${\lambda },\mu \in \mathbb Z^{n}$ , write ${\lambda }\leqslant \mu \Leftrightarrow {\lambda }_i\leqslant \mu _i$ for $1\leqslant i\leqslant n$ . We have the following monomial and PBW bases of .

Proof. By Lemma 2.3, (3.4), (3.5), and (3.8), for $A \in \Theta ^\pm (n)_r$ , ${\lambda }\in \Lambda (n,d)$ with ${\lambda }\geqslant {\boldsymbol \sigma }(A)$ , we have

$$ \begin{align*} \begin{aligned} {\mathtt e}^{(A^+)} \mathtt{k}_{{\lambda}} {\mathtt f}^{(A^-)} &=\dot{\zeta}_{d}(E^{(A^+)} 1_{{\lambda}}F^{(A^-)})=[A+{\lambda}-{\boldsymbol\sigma}(A)]_{\varepsilon} +f,\\ A^+(\mathbf{0},d) \mathtt{k}_{{\lambda}} A^-(\mathbf{0},d) &=[A+{\lambda}-{\boldsymbol\sigma}(A)]_{\varepsilon} +g,\\ \end{aligned} \end{align*} $$

where . Now the assertion follows from Lemma 3.4.

4 The algebra

For $d\in \mathbb N$ let

where $I_{d,r}$ is the two-sided ideal of generated by the elements $1-\sum _{\mu \in \Lambda (n,d)} K_{\mu }$ , $K_{i} K_{\lambda }-\varepsilon ^{\lambda _{i}} K_{\lambda }$ and $\big [{K_i;0 \atop t}\big ] K_{\lambda }-\big [{{\lambda }_i \atop t}\big ]_{\varepsilon } K_{\lambda }$ for $1 \leqslant i \leqslant n$ , $t \in \mathbb {N}$ and $\lambda \in \Lambda (n,d)$ . For $1\leqslant i\leqslant n-1$ , $t\in \mathbb N$ and $0\leqslant m<lp^{r-1}$ let

$$ \begin{align*}\mathbf{e}_i^{(m)}=E_i^{(m)}+I_{d,r},\,\mathbf{f}_i^{(m)}=F_i^{(m)}+I_{d,r}.\end{align*} $$

Furthermore, for $1\leqslant j\leqslant n$ , $c\in \mathbb Z$ , $t\in \mathbb N$ and ${\lambda }\in \mathbb N^{n}$ let

$$ \begin{align*}\mathbf{k}_j=K_j+I_{d,r},\, \bigg[{\mathbf{k}_j;c \atop t}\bigg]=\bigg[{K_j;c \atop t}\bigg]+I_{d,r},\,\mathbf{k}_{\lambda}=K_{\lambda}+I_{d,r}.\end{align*} $$

We will prove in Theorem 4.10 that the algebra is isomorphic to the infinitesimal q-Schur algebra .

Proof. For $1\leqslant i\leqslant n$ and $t\in \mathbb N$ , we have $\big [{\mathbf {k}_i;0 \atop t}\big ]\mathbf {k}_\mu = \big [{ \mu _i\atop t}\big ]_{\varepsilon } \mathbf {k}_\mu $ . It follows that

$$ \begin{align*}\mathbf{k}_{\lambda} \mathbf{k}_\mu = \bigg[{ \mu\atop{\lambda}}\bigg]_{\varepsilon} \mathbf{k}_\mu.\end{align*} $$

If $\big [{ \mu \atop {\lambda }}\big ]_{\varepsilon }\not =0$ , then we have $\mu \geqslant {\lambda }$ . This implies that $\mu ={\lambda }$ since ${\lambda },\mu \in \Lambda (n,d)$ . Therefore, we have $\mathbf {k}_{\lambda }\mathbf {k}_\mu =\delta _{{\lambda },\mu }\mathbf {k}_{\lambda }$ . Furthermore, since $1=\sum _{{\gamma } \in \Lambda (n,d)}\mathbf {k}_{\gamma }$ and $\sigma (\nu )>d$ , we have

$$ \begin{align*}\mathbf{k}_\nu= \sum_{{\gamma} \in \Lambda(n,d)}\mathbf{k}_\nu\mathbf{k}_{\gamma} =\sum_{{\gamma} \in \Lambda(n,d)}\bigg[{ {\gamma}\atop\nu}\bigg]_{\varepsilon} \mathbf{k}_{\gamma}=\sum_{{\gamma} \in \Lambda(n,d)\atop\sigma({\gamma})\geqslant\sigma(\nu)>d}\bigg[{ {\gamma}\atop\nu}\bigg]_{\varepsilon} \mathbf{k}_{\gamma}=0.\end{align*} $$

The proof is completed.

For $a,b\in \mathbb Z$ , we have

(4.1) $$ \begin{align} \begin{aligned} \bigg[{b+a\atop t}\bigg]_{\varepsilon}&=\sum_{0\leqslant j\leqslant t}\varepsilon^{a(t-j)-bj}\bigg[{a\atop j}\bigg]_{\varepsilon}\bigg[{b\atop t-j}\bigg]_{\varepsilon}. \end{aligned} \end{align} $$

Proof. Assume $c\geqslant 0$ . By [Reference Lusztig22, 2.3 (g9), (g10)], we have

$$ \begin{align*}\left[{\mathbf{k}_i;\pm c\atop t}\right]=\sum_{0\leqslant j\leqslant t}\varepsilon^{c(t-j)}\left[{\pm c\atop j}\right]_{\varepsilon}\mathbf{k}_i^{\mp j}\left[{\mathbf{k}_i;0\atop t-j}\right]. \end{align*} $$

Hence, by (4.1), we have

$$ \begin{align*} \left[{\mathbf{k}_i;\pm c\atop t}\right]\mathbf{k}_{\lambda} =\sum_{0\leqslant j\leqslant t}\varepsilon^{c(t-j)\mp j{\lambda}_i}\left[{\pm c\atop j}\right]_{\varepsilon}\left[{{\lambda}_i\atop t-j}\right]_{\varepsilon}\mathbf{k}_{\lambda}=\bigg[{{\lambda}_i\pm c\atop t}\bigg]_{\varepsilon}\mathbf{k}_{\lambda}. \end{align*} $$

The proof is completed.

By the definition of , we have the following result.

Lemma 4.3. There is an algebra anti-automorphism $\tau _d$ on such that

$$ \begin{align*}\tau_d(\mathbf{e}_i^{(m)})=\mathbf{f}_i^{(m)},\quad \tau_d(\mathbf{f}_i^{(m)})=\mathbf{e}_i^{(m)}, \quad \tau_d(\mathbf{k}_j)=\mathbf{k}_j,\quad\tau_d\bigg(\bigg[{\mathbf{k}_j;0\atop t}\bigg]\bigg)=\bigg[{\mathbf{k}_j;0\atop t}\bigg], \end{align*} $$

for $1\leqslant i \leqslant n-1$ , $1\leqslant j \leqslant n$ , $t\in \mathbb N$ and $0\leqslant m<lp^{r-1}$ .

Proof. By Lemma 2.1 and 4.2, we have

$$ \begin{align*} \begin{aligned} \mathbf{e}_i^{(a)}\mathbf{k}_{\lambda}&=\mathbf{e}_i^{(a)}\prod_{s\not=i,i+1} \left[{\mathbf{k}_s;0\atop{\lambda}_s}\right]\left[{\mathbf{k}_i;a\atop{\lambda}_i+a}\right]\mathbf{k}_{\lambda} =\mathbf{k}_{{\lambda}+a\boldsymbol{e}_i-{\lambda}_{i+1}\boldsymbol{e}_{i+1}}\mathbf{e}_i^{(a)}\mathbf{k}_{\lambda},\\ \mathbf{k}_\mu\mathbf{e}_j^{(a)}&=\mathbf{k}_\mu\prod_{s\not=j,j+1} \left[{\mathbf{k}_s;0\atop\mu_s}\right]\left[{\mathbf{k}_{j+1};a\atop\mu_{j+1}+a}\right]\mathbf{e}_j^{(a)} =\mathbf{k}_\mu\mathbf{e}_j^{(a)}\mathbf{k}_{{\lambda}+a\boldsymbol{e}_{j+1}-\mu_j\boldsymbol{e}_j}. \end{aligned} \end{align*} $$

Hence, by Lemma 4.1 we have $\mathbf {k}_{{\lambda }+a\boldsymbol {e}_i-{\lambda }_{i+1}\boldsymbol {e}_{i+1}}=0$ and $\mathbf {k}_{{\lambda }+a\boldsymbol {e}_{j+1}-\mu _j\boldsymbol {e}_j}=0$ , since ${\lambda }_{i+1}<a$ , $\mu _j<a$ and ${\lambda },\mu \in \Lambda (n,d)$ . Therefore we have $\mathbf {e}_i^{(a)}\mathbf {k}_{\lambda }=\mathbf {k}_\mu \mathbf {e}_j^{(a)}=0$ . Consequently, we have $\mathbf {k}_{\lambda }\mathbf {f}_i^{(a)}=\tau _d(\mathbf {e}_i^{(a)}\mathbf {k}_{\lambda })=0$ and $\mathbf {f}_j^{(a)}\mathbf {k}_\mu =\tau _d(\mathbf {k}_\mu \mathbf {e}_j^{(a)})=0$ .

Proof. By Lemma 2.1 and 4.2, we have

$$ \begin{align*} \begin{aligned} \mathbf{e}_i^{(a)}\mathbf{k}_{\lambda} &=\prod_{j\not=i,i+1}\left[{\mathbf{k}_j;0\atop{\lambda}_j}\right] \left[{\mathbf{k}_i;-a\atop{\lambda}_i}\right]\left[{\mathbf{k}_{i+1};a\atop{\lambda}_{i+1}}\right] \mathbf{e}_i^{(a)}\\ &= \sum_{\mu\in\Lambda(n,d)} \prod_{j\not=i,i+1}\left[{\mathbf{k}_j;0\atop{\lambda}_j}\right] \left[{\mathbf{k}_i;-a\atop{\lambda}_i}\right]\left[{\mathbf{k}_{i+1};a\atop{\lambda}_{i+1}}\right] \mathbf{k}_\mu \mathbf{e}_i^{(a)}. \end{aligned} \end{align*} $$

Hence, by Lemma 4.4, we have

$$ \begin{align*} \mathbf{e}_i^{(a)}\mathbf{k}_{\lambda} =\sum_{\mu\in\Lambda(n,d)\atop\mu_i\geqslant a}\left[{\mu-a\alpha_i\atop{\lambda}}\right]_{\varepsilon} \mathbf{k}_\mu \mathbf{e}_i^{(a)} =\sum_{\mu\in\Lambda(n,d),\,\mu-a\alpha_i\geqslant{\lambda}}\left[{\mu-a\alpha_i\atop{\lambda}}\right]_{\varepsilon} \mathbf{k}_\mu \mathbf{e}_i^{(a)}=\mathbf{k}_{{\lambda}+a\alpha_i} \mathbf{e}_i^{(a)}. \end{align*} $$

Therefore, we have $\mathbf {k}_{\lambda }\mathbf {f}_i^{(a)}=\tau _d(\mathbf {e}_i^{(a)}\mathbf {k}_{\lambda })= \tau _d(\mathbf {k}_{{\lambda }+a\alpha _i} \mathbf {e}_i^{(a)} )=\mathbf {f}_i^{(a)}\mathbf {k}_{{\lambda }+a\alpha _i}$ .

For simplicity, we set $\mathbf {k}_{\lambda }=0$ if ${\lambda }\not \in \mathbb N^{n}$ with $\sigma ({\lambda })=d$ , where $\sigma ({\lambda })=\sum _{1\leqslant i\leqslant n}{\lambda }_i$ . Then, by Lemma 4.4 and 4.5, we have the following result.

For $A\in \Theta ^\pm (n)$ let $\mathbf {e}^{(A^+)}=E^{(A^+)}+I_{d,r}$ and $\mathbf {f}^{(A^-)}=F^{(A^-)}+I_{d,r}$ .

Proof. By Lemma 4.6, we have $ \mathbf {e}^{(A^+)}\mathbf {k}_{\lambda }=\mathbf {k}_\mu \mathbf {e}^{(A^+)}$ and $\mathbf {k}_{\lambda }\mathbf {f}^{(A^-)}=\mathbf {f}^{(A^-)}\mathbf {k}_\nu $ where

$$ \begin{align*}\mu={\lambda}+\sum_{2\leqslant j\leqslant n}\sum_{1\leqslant k<j}a_{k,j}\sum_{k\leqslant s<j}\alpha_s={\lambda}+\sum_{2\leqslant j\leqslant n}\sum_{1\leqslant k<j}a_{k,j}(\boldsymbol{e}_k-\boldsymbol{e}_j)={\lambda}-\textrm{co}(A^+)+\textrm{ro}(A^+), \end{align*} $$

and

$$ \begin{align*}\nu={\lambda}+\sum_{2\leqslant j\leqslant n}\sum_{1\leqslant k<j}a_{j,k}\sum_{k\leqslant s<j}\alpha_s ={\lambda}+\sum_{2\leqslant j\leqslant n}\sum_{1\leqslant k<j}a_{j,k}(\boldsymbol{e}_k-\boldsymbol{e}_j)={\lambda}+\textrm{co}(A^-)-\textrm{ro}(A^-).\end{align*} $$

The proof is completed.

Recall the sets $\Theta ^+(n)_r$ and $\Theta ^-(n)_r$ defined in (3.1).

Proof. If $A \in \Theta ^+(n)_r$ and ${\lambda }_i<\sigma _i(A)$ for some i, then by Lemma 4.6 we have

$$ \begin{align*} \mathbf{e}^{(A)}\mathbf{k}_{\lambda}=\mathbf{m}_n\mathbf{m}_{n-1}\cdots \mathbf{m}_2\mathbf{k}_{\lambda}=\mathbf{m}_n\mathbf{m}_{n-1}\cdots\mathbf{m}_{i+1}\mathbf{k}_\mu\mathbf{m}_i\mathbf{m}_{i-1}\cdots\mathbf{m}_2, \end{align*} $$

where $\mathbf {m}_j=\mathbf {e}^{(a_{j-1,j})}_{j-1}(\mathbf {e}_{j-2}^{(a_{j-2, j})} \mathbf {e}_{j-1}^{(a_{j-2, j})}) \cdots (\mathbf {e}_{1}^{(a_{1, j})} \mathbf { e}_{2}^{(a_{1, j})} \cdots \mathbf {e}_{j-1}^{(a_{1, j})})$ and

$$ \begin{align*}\mu={\lambda}+\sum_{2\leqslant j\leqslant i}\sum_{1\leqslant k<j}a_{k,j}\sum_{k\leqslant s<j}\alpha_s={\lambda}+\sum_{2\leqslant j\leqslant i}\sum_{1\leqslant k<j}a_{k,j}(\boldsymbol{e}_k-\boldsymbol{e}_j). \end{align*} $$

Since $\mu _i={\lambda }_i-\sigma _i(A)<0$ we have $\mathbf {k}_\mu =0$ . Hence, we have $\mathbf {e}^{(A)}\mathbf {k}_{\lambda }=0$ . Assume now that $A \in \Theta ^-(n)_r$ and ${\lambda }_i<\sigma _i(A)$ for some i. Then, we have $\mathbf {k}_{\lambda }\mathbf {f}^{(A)}=\tau _d(\mathbf {e}^{({}^tA)}\mathbf {k}_{\lambda })=0$ . The proof is completed.

For $A \in \Theta ^\pm (n)_r$ and ${\lambda } \in \Lambda (n,d)$ , let

$$ \begin{align*}\mathbf{m}^{(A,{\lambda})}=\mathbf{e}^{(A^+)} \mathbf{k}_{\lambda} \mathbf{f}^{(A^-)}.\end{align*} $$

Proof. By the definition of $I_{d,r}$ we have $\mathbf {k}_i=\sum _{{\lambda }\in \Lambda (n,d)}\varepsilon ^{{\lambda }_i}\mathbf {k}_{\lambda }$ and $\mathbf {k}_{\lambda }=\sum _{\mu \in \Lambda (n,d)} \big [{\mu \atop {\lambda }}\big ]_{\varepsilon }\mathbf {k}_\mu $ for ${\lambda }\in \Lambda (n,d)$ and $1\leqslant i\leqslant n$ . Hence, by Proposition 3.1, we see that the algebra is spanned by the elements $\mathbf {e}^{(A^+)} \mathbf {k}_{\lambda } \mathbf {f}^{(A^-)}$ for $A\in \Theta ^\pm (n)_r$ and $\lambda \in \Lambda (n,d)$ . Therefore, to prove the proposition, we have to show that if ${\lambda }_i< \sigma _i(A)$ for some i, then $\mathbf {m}^{(A,{\lambda })}$ lies in the span of $\mathbf {M}_{d,r}$ .

We argue by induction on $\textrm {deg}(A)$ . The result follows from Lemma 4.4 in the cases where $\textrm {deg}(A)=1$ . Assume now that $\textrm {deg}(A)>1$ , and suppose ${\lambda }_i<\sigma _i(A)$ for some $1\leqslant i\leqslant n$ . For $2\leqslant j\leqslant n$ let

$$ \begin{align*} \begin{aligned} \mathbf{m}_j&=\mathbf{e}^{(a_{j-1,j})}_{j-1}(\mathbf{e}_{j-2}^{(a_{j-2, j})} \mathbf{e}_{j-1}^{(a_{j-2, j})}) \cdots(\mathbf{e}_{1}^{(a_{1, j})} \mathbf{ e}_{2}^{(a_{1, j})} \cdots \mathbf{e}_{j-1}^{(a_{1, j})}),\\ \mathbf{m}_j'&=(\mathbf{f}_{j-1}^{(a_{j, 1})} \cdots \mathbf{f}_{2}^{(a_{j, 1})} \mathbf{ f}_{1}^{(a_{j, 1})}) \cdots(\mathbf{f}_{j-1}^{(a_{j, j-2})} \mathbf{f}_{j-2}^{(a_{j, j-2})}) \mathbf{f}_{j-1}^{(a_{j, j-1})}. \end{aligned} \end{align*} $$

Then, we have $\mathbf {e}^{(A^+)} =\mathbf {m}_n\mathbf {m}_{n-1}\cdots \mathbf {m}_2$ and $\mathbf {f}^{(A^-)} =\mathbf {m}_2^{\prime }\mathbf {m}_{3}^{\prime }\cdots \mathbf {m}_n'$ . Let $A_i$ be the submatrix of A consisting of the first i rows and columns, and write $\mathbf {e}^{(A^+)}=\mathbf {x}_1 \mathbf {e}^{(A_i^{+})}, \mathbf { f}^{(A^-)}=\mathbf {f}^{(A_i^-)}\mathbf {x}_1^\prime $ . Then,

$$ \begin{align*}\mathbf{m}^{(A,{\lambda})}=\mathbf{x}_1 \mathbf{e}^{(A_i^{+})}\mathbf{k}_{\lambda}\mathbf{f}^{(A_i^-)}\mathbf{x}_1^\prime,\end{align*} $$

where $\mathbf {x}_1=\mathbf {m}_n\mathbf {m}_{n-1}\cdots \mathbf {m}_{i+1}$ and $\mathbf {x}_1^{\prime }=\mathbf {m}_{i+1}^{\prime }\mathbf {m}_{i+2}^{\prime }\cdots \mathbf {m}_{n}^{\prime }$ . By Lemma 4.8, we may assume that ${\lambda }_i\geqslant \sigma _i(A_i^{+})=\sigma _i(A^{+})$ . Furthermore, by Lemma 4.7, we have

$$ \begin{align*}\mathbf{m}^{(A,{\lambda})}=\mathbf{x}_1 \mathbf{k}_{{\lambda}'} \mathbf{e}^{(A_i^{+})}\mathbf{f}^{(A_i^-)}\mathbf{x}_1', \end{align*} $$

where ${\lambda }^\prime ={\lambda }-\textrm {co}(A_i^{+})+\textrm {ro}(A_i^+)$ . By Lemma 2.1,

$$ \begin{align*}\mathbf{e}^{(A_i^{+})}\mathbf{f}^{(A_i^-)}=\mathbf{f}^{(A_i^-)}\mathbf{e}^{(A_i^{+})}+f, \end{align*} $$

where f is a -linear combination of $\mathbf {x}_j^{\mathbf {e}}\mathbf {h}_j \mathbf {x}^{\mathbf {f}}_j$ with and $\textrm {deg}(\mathbf {x}_j^{\mathbf {e}})+\textrm {deg} (\mathbf {x}_j^{\mathbf { f}}) < \textrm {deg}(A_i)$ . Here, $\mathbf {x}_j^{\mathbf {e}}$ (resp. $\mathbf {x}_j^{\mathbf {f}}$ ) denotes a monomial in the $\mathbf {e}_i^{(a)}$ (resp. $\mathbf {f}_i^{(a)}$ ). Thus, $\textrm {deg}(\mathbf {x}_1)+\textrm {deg}(\mathbf {x}_j^{\mathbf {e}})+\textrm {deg}(\mathbf {x}^{\mathbf {f}}_j)+\textrm {deg}(\mathbf {x}_1')<\textrm {deg}(A)$ . Since ${\lambda }_i<\sigma _i(A)$ , we have ${\lambda }_i^{\prime }={\lambda }_i-\sigma _i(A_i^+)<\sigma _i(A)-\sigma _i(A_i^+)=\sigma _i(A_i^-)$ . It follows from Lemma 4.7 that $ \mathbf {k}_{{\lambda }'}\mathbf {f}^{(A_i^-)}=0$ . Hence, we have

$$ \begin{align*}\mathbf{m}^{(A,{\lambda})}=\mathbf{x}_1\mathbf{k}_{{\lambda}'}f\mathbf{x}_1'.\end{align*} $$

Furthermore, by Proposition 3.1, we see that each $\mathbf {x}_1\mathbf {x}_j^{\mathbf {e}}$ is a -linear combination of $\mathbf { e}^{(B)}$ with $B\in \Theta ^+(n)_r$ , $\textrm {deg}(B)=\textrm {deg}(\mathbf {x}_1\mathbf {x}_j^{\mathbf {e}})$ and each $\mathbf {x}_j^{\mathbf {f}} \mathbf {x}_1^\prime $ is a -linear combination of $\mathbf {f}^{(C)}$ with $C\in \Theta ^-(n)_r$ , $\textrm {deg}(C)=\textrm { deg}(\mathbf {x}_j^{\mathbf {f}} \mathbf {x}_1^\prime )$ . Therefore, by Lemma 4.7 each $\mathbf {x}_1 \mathbf {k}_{{\lambda }'} \mathbf {x}_j^{\mathbf {e}}\mathbf {h}_j \mathbf {x}^{\mathbf {f}}_j\mathbf {x}_1^\prime $ is a -linear combination of $\mathbf {m}^{(A',\mu )}$ with $\textrm {deg}(A')<\textrm {deg}(A)$ , since $\textrm {deg}(\mathbf {x}_1)+\textrm {deg}(\mathbf {x}_j^{\mathbf {e}})+\textrm {deg}(\mathbf {x}^{\mathbf {f}}_j)+\textrm {deg}(\mathbf {x}_1')<\textrm {deg}(A)$ . Consequently, by induction, we have . The proof is completed.

By Lemma 3.5, we have Therefore, the map given in (3.7) restricts to a surjective algebra homomorphism

(4.2)

By (3.8), we have

$$ \begin{align*} \zeta_{d,r}(K_{\lambda})=[\operatorname{diag}({\lambda})]_{\varepsilon}, \end{align*} $$

for ${\lambda }\in \Lambda (n,d)$ . So, we have $\zeta _{d,r} (I_{d,r})=0$ . Hence, the map induces a surjective algebra homomorphism

5 The algebra

For $d\in \mathbb N$ , let

where $J_{d,r}$ is the two-sided ideal of generated by the elements $1_{\lambda }$ for $\lambda \notin \Lambda (n,d)$ . For ${\lambda }\in \mathbb Z^{n}$ , let

$$ \begin{align*}\mathfrak{k}_{\lambda}=1_{\lambda}+J_{d,r}.\end{align*} $$

Then, we have $1=\sum _{{\lambda }\in \Lambda (n,d)}\mathfrak {k}_{\lambda }$ . For $1\leqslant i\leqslant n-1$ and $0\leqslant m<lp^{r-1}$ , let

$$ \begin{align*}\mathfrak{e}_i^{(m)}=\sum_{{\lambda}\in\Lambda(n,d)}E_i^{(m)}1_{\lambda}+J_{d,r},\ \mathfrak{f}_i^{(m)}=\sum_{{\lambda}\in\Lambda(n,d)}1_{\lambda} F_i^{(m)}+J_{d,r}.\end{align*} $$

For $A\in \Theta ^\pm (n)_r$ , let. Furthermore for $1\leqslant i\leqslant n$ , $1\leqslant j\leqslant n-1$ , $c\in\mathbb{Z}$ and $t\in\mathbb{N}$ let

$$ \begin{align*}\mathfrak{k}_i=\sum_{\lambda\in\Lambda(n,d)}K_i1_\lambda+J_{d,r},\ \bigg[{\mathfrak{k}_i;c\atop t}\bigg]=\sum_{\lambda\in\Lambda(n,d)}\bigg[{K_i;c\atop t}\bigg]1_\lambda+J_{d,r},\ \bigg[{\widetilde{\mathfrak{k}}_j;c\atop t}\bigg]=\sum_{\lambda\in\Lambda(n,d)}\bigg[{\widetilde K_j;c\atop t}\bigg]1_\lambda+J_{d,r}.\end{align*} $$

$$ \begin{align*}\mathfrak{e}^{(A^+)}=\sum_{{\lambda}\in\Lambda(n,d)}E^{(A^+)} 1_{\lambda}+J_{d,r},\ \mathfrak{f}^{(A^-)}=\sum_{{\lambda}\in\Lambda(n,d)}1_{\lambda}F^{(A^-)} +J_{d,r}.\end{align*} $$

We will prove in Theorem 5.5 that the algebra is isomorphic to the infinitesimal q-Schur algebra .

By [Reference Beilinson, Lusztig and MacPherson3, Lem. 3.10 and Prop. 4.2], we have the following result.

Clearly we have the following result.

Lemma 5.2. Let $1 \leqslant i \leqslant n, 1\leqslant j\leqslant n-1$ , $c\in \mathbb Z$ , $t\in \mathbb N$ and $\lambda \in \mathbb {Z}^{n}$ . The following formulas hold in

$$ \begin{align*}\mathfrak{k}_{i}\mathfrak{k}_{\lambda}=\varepsilon^{\lambda_{i}}\mathfrak{k}_{\lambda},\ \bigg[{\mathfrak{k}_i;c\atop t}\bigg]\mathfrak{k}_{\lambda}=\bigg[{{\lambda}_i+c \atop t}\bigg]_{\varepsilon}\mathfrak{k}_{\lambda},\ \bigg[{\widetilde{\mathfrak{k}}_j;c \atop t}\bigg]\mathfrak{k}_{\lambda}=\bigg[{{\lambda}_j-{\lambda}_{j+1}+c \atop t}\bigg]_{\varepsilon} \mathfrak{k}_{\lambda}.\end{align*} $$

Recall from (3.1) that $\Theta ^\pm (n)_r=\{A\in \Theta ^\pm (n) \mid 0 \leqslant a_{ij} < lp^{r-1}, \forall i \neq j\} $ , $\Theta ^+(n)_r=\{A\in \Theta ^+(n) \mid 0 \leqslant a_{ij} < lp^{r-1}, \forall i < j \}$ and $\Theta ^-(n)_r=\{A\in \Theta ^-(n) \mid 0 \leqslant a_{ij} < lp^{r-1}, \forall i> j \}$ .

Proof. Assume $A\in \Theta ^+(n)_r$ and ${\lambda }_i<\sigma _i(A)$ for some i. For $\mu \in \mathbb Z^{n}, 1 \leqslant j \leqslant n-1$ , we have $\mathfrak {e}_j\mathfrak {k}_\mu =\mathfrak {k}_{\mu +\alpha _j} \mathfrak {e}_j$ . This implies that

$$ \begin{align*}\mathfrak{m}_i(A)\mathfrak{m}_{i-1}(A)\cdots \mathfrak{m}_2(A)\mathfrak{k}_{\lambda}=\mathfrak{k}_{{\lambda}+\nu}\mathfrak{m}_i(A)\mathfrak{m}_{i-1}(A)\cdots \mathfrak{m}_2(A),\end{align*} $$

where $\mathfrak {m}_{j} =\mathfrak {e}_{j-1}^{(a_{j-1, j})}(\mathfrak {e}_{j-2}^{(a_{j-2, j})} \mathfrak {e}_{j-1}^{(a_{j-2, j})}) \cdots (\mathfrak {e}_{1}^{(a_{1, j})} \mathfrak {e}_{2}^{(a_{1, j})} \cdots \mathfrak {e}_{j-1}^{(a_{1, j})})$ and

$$ \begin{align*}\nu=\sum_{2\leqslant j\leqslant i}\bigg(\sum_{1\leqslant s\leqslant j-1}a_{s,j}\boldsymbol{e}_s-(\sum_{1\leqslant s\leqslant j-1}a_{s,j})\boldsymbol{e}_j\bigg).\end{align*} $$

Thus, we have $ {\mathfrak {e}^{(A)}\mathfrak {k}_{\lambda }}= {\mathfrak {m}_n \mathfrak {m}_{n-1} \cdots \mathfrak {m}_{i+1} \mathfrak {k}_{{\lambda }+\nu } \mathfrak {m}_i \mathfrak {m}_{i-1} \cdots \mathfrak {m}_2 }.$ Since ${\lambda }_i<\sigma _i(A)$ , we have ${\lambda }+\nu \not \in \Lambda (n,d)$ . It follows that $\mathfrak {k}_{{\lambda }+\nu }=0$ . Therefore, we have $\mathfrak {e}^{(A)}\mathfrak {k}_{\lambda }=0$ . Applying $\dot \tau _d$ to the identity in (1) gives that in (2).

Proof. By Proposition 3.3, we have

Thus, it is enough to prove that if ${\lambda } \in \Lambda (n,d)$ and $ {\lambda }_i <\sigma _i(A) $ for some i, . We apply induction on $\textrm {deg}(A)$ . If $\textrm {deg}(A)=1$ , then by Lemma 5.3 we have $\mathfrak {e}^{(A^+)}\mathfrak {k}_{\lambda }\mathfrak {f}^{(A^-)}=0$ . Now suppose $\textrm {deg}(A)>1$ . For $2\leqslant j\leqslant n$ let

$$ \begin{align*} \begin{aligned} \mathfrak{m}_{j}&=\mathfrak{e}_{j-1}^{(a_{j-1, j})}(\mathfrak{e}_{j-2}^{(a_{j-2, j})} \mathfrak{e}_{j-1}^{(a_{j-2, j})}) \cdots(\mathfrak{e}_{1}^{(a_{1, j})} \mathfrak{e}_{2}^{(a_{1, j})} \cdots \mathfrak{e}_{j-1}^{(a_{1, j})}) \\ \mathfrak{m}_j'&=(\mathfrak{f}_{j-1}^{(a_{j, 1})} \cdots \mathfrak{f}_{2}^{(a_{j, 1})} \mathfrak{f}_{1}^{(a_{j, 1})}) \cdots(\mathfrak{f}_{j-1}^{(a_{j, j-2})} \mathfrak{f}_{j-2}^{(a_{j, j-2})}) \mathfrak{f}_{j-1}^{(a_{j, j-1})}. \end{aligned} \end{align*} $$

Then, we have

$$ \begin{align*}\mathfrak{e}^{(A^+)}\mathfrak{k}_{\lambda}\mathfrak{f}^{(A^-)} =X_1(X_2\mathfrak{k}_{\lambda})Y_1 Y_2=X_1(\mathfrak{k}_{{\lambda}'}X_2)Y_1 Y_2,\end{align*} $$

where $X_1=\mathfrak {m}_n\mathfrak {m}_{n-1}\cdots \mathfrak {m}_{i+1}$ , $X_2=\mathfrak {m}_i\mathfrak {m}_{i-1}\cdots \mathfrak {m}_2$ , $Y_1=\mathfrak {m}_2^{\prime }\mathfrak {m}_3^{\prime }\cdots \mathfrak {m}_i^{\prime }$ , $Y_2=\mathfrak {m}_{i+1}^{\prime }\mathfrak {m}_{i+2}^{\prime }\cdots \mathfrak {m}_{n}^{\prime }$ and

$$ \begin{align*}{\lambda}'={\lambda}+\sum_{2\leqslant j\leqslant i}\left(\sum_{1\leqslant s\leqslant j-1}a_{s,j}\boldsymbol{e}_s-\left(\sum_{1\leqslant s\leqslant j-1}a_{s,j}\right)\boldsymbol{e}_j\right).\end{align*} $$

By Lemma 2.1 and 5.2, we have $\mathfrak {k}_{{\lambda }'}X_2Y_1=\mathfrak {k}_{{\lambda }'}Y_1X_2+\mathfrak {k}_{{\lambda }'}f_1f_2$ where $f_1$ is a -linear combination of monomials $f_{1,k}$ in the $\mathfrak {e}_s^{(a)}$ , $f_{2}$ is a -linear combination of monomials $f_{2,k}$ in the $\mathfrak {f}_s^{(a)}$ , and $\textrm {deg}(f_{1,k})+\textrm {deg}(f_{2,k})<\textrm {deg}(X_2)+\textrm {deg}(Y_1)$ . Since ${\lambda }_i<\sigma _i(A)$ , we have ${\lambda }_i^{\prime }<\sigma _i(A^-)$ . Hence by Lemma 5.3, we have $\mathfrak {k}_{{\lambda }'}Y_1=0$ . This implies that

$$ \begin{align*}\mathfrak{e}^{(A^+)}\mathfrak{k}_{\lambda}\mathfrak{f}^{(A^-)}=X_1\mathfrak{k}_{{\lambda}'}f_1f_2Y_2 =\mathfrak{k}_{{\lambda}"}X_1f_1f_2Y_2,\end{align*} $$

where

$$ \begin{align*}{\lambda}"={\lambda}+\sum_{2\leqslant j\leqslant n}\left(\sum_{1\leqslant s\leqslant j-1}a_{s,j}\boldsymbol{e}_s-\left(\sum_{1\leqslant s\leqslant j-1}a_{s,j}\right)\boldsymbol{e}_j\right).\end{align*} $$

By Proposition 3.1, we have and . Thus,

By induction, we have for $B\in \Theta ^\pm (n)_r$ with $\textrm {deg}(B)<\textrm {deg}(A)$ . Therefore, .

By Lemma 3.6, we have Therefore, the map given in (3.10) restricts to a surjective algebra homomorphism

(5.1)

Since $\dot {\zeta }_{d,r} (J_{d,r})=0$ , the map induces a surjective algebra homomorphism

A presentation of was given in [Reference Fu16, Th. 3.9] in the case where $r=1$ , is a field and $l'$ is odd. We now generalize this result to the general case.

6 The classical case

Let ${\mathcal U}(\mathfrak {gl}_n)$ be the $\mathbb Q$ -algebra defined by the generators

$$ \begin{align*}\bar E_i,\ \bar F_i\quad(1\leqslant i\leqslant n-1),\ H_j \quad(1\leqslant j\leqslant n),\end{align*} $$

and the relations

1. (a) $H_iH_j=H_jH_i;$

2. (b) $H_{i}\bar E_j-\bar E_jH_i=(\delta _{i,j}-\delta _{i,j+1}) \bar E_j;$

3. (c) $H_{i}\bar F_j-\bar F_jH_i=(-\delta _{i,j}+\delta _{i,j+1}) \bar F_j;$

4. (d) $ \bar E_i\bar E_j=\bar E_j\bar E_i,\ \bar F_i\bar F_j=\bar F_j\bar F_i\ when\ |i-j|>1;$

5. (e) $\bar E_i\bar F_j-\bar F_j\bar E_i=\delta _{i,j}H_i;$

6. (f) $ \bar E_i^2\bar E_j-2\bar E_i\bar E_j\bar E_i+\bar E_j\bar E_i^2=0\ when\ |i-j|=1;$

7. (g) $ \bar F_i^2\bar F_j-2\bar F_i\bar F_j\bar F_i+\bar F_j\bar F_i^2=0\ when\ |i-j|=1.$

Then, ${\mathcal U}(\mathfrak {gl}_n)$ is the universal enveloping algebra of $\mathfrak {gl}_n$ . Let ${\mathcal U}_{\mathbb Z}(\mathfrak {gl}_n)$ be the $\mathbb Z$ -subalgebra of ${\mathcal U}(\mathfrak {gl}_n)$ generated by $\bar E_i^{(m)}$ , $\bar F_i^{(m)}$ , and $\big ({H_j\atop t}\big )$ for $1\leqslant i\leqslant n-1$ , $1\leqslant j\leqslant n$ and $m,t\in \mathbb N$ , where

$$ \begin{align*}\bar E_i^{(m)}=\frac{\bar E_i^m}{m!},\, \bar F_i^{(m)}=\frac{\bar F_i^m}{m!},\, \left({H_j\atop t}\right)=\frac{H_j(H_{j}-1)\cdots (H_{j}-t+1)}{t!}.\end{align*} $$

Let ${U_{\mathbb Z}(\mathfrak {gl}_n)}={U_{\mathcal Z}(\mathfrak {gl}_n)}\otimes _{\mathcal Z}\mathbb Z$ , where $\mathbb Z$ is viewed as ${\mathcal Z}$ -modules by specializing v to $1$ . Let ${\bar U_{\mathbb Z}(\mathfrak {gl}_n)}={U_{\mathbb Z}(\mathfrak {gl}_n)}/\langle K_i-1\mid 1\leqslant i\leqslant n\rangle $ . We shall denote the images of $E_i^{(m)}$ , $F_i^{(m)}$ , etc. in ${\bar U_{\mathbb Z}(\mathfrak {gl}_n)}$ by the same letters. By [Reference Lusztig22, 6.7(c)], there is an algebra isomorphism

(6.1) $$ \begin{align} \theta:{\bar U_{\mathbb Z}(\mathfrak{gl}_n)}\rightarrow{\mathcal U}_{\mathbb Z}(\mathfrak{gl}_n), \end{align} $$

such that $\theta (E_i^{(m)})=\bar E_i^{(m)}$ , $\theta (F_i^{(m)})=\bar F_i^{(m)}$ , $\theta (\big [{K_j;0\atop t}\big ])=\big ({H_j\atop t}\big )$ for $1\leqslant i\leqslant n-1$ , $1\leqslant j\leqslant n$ , $m,t\in \mathbb N$ . We will identify ${\bar U_{\mathbb Z}(\mathfrak {gl}_n)}$ with ${\mathcal U}_{\mathbb Z}(\mathfrak {gl}_n)$ .

Let ${\mathcal S}_{\mathbb Z}(n,d)={\mathcal S}_{\mathcal Z}(n,d)\otimes _{\mathcal Z}\mathbb Z$ where $\mathbb Z$ is viewed as ${\mathcal Z}$ -modules by specializing v to $1$ . The map $\zeta _{d}$ given in (3.6) induces, upon tensoring with $\mathbb Z$ , a surjective algebra homomorphism

$$ \begin{align*}\xi_{d}:{U_{\mathbb Z}(\mathfrak{gl}_n)}\rightarrow{\mathcal S}_{\mathbb Z}(n,d).\end{align*} $$

Since $\xi _{d}(K_i)=1$ , the map $\xi _{d}$ induces a surjective algebra homomorphism

$$ \begin{align*}\xi_{d}:{\mathcal U}_{\mathbb Z}(\mathfrak{gl}_n)={\bar U_{\mathbb Z}(\mathfrak{gl}_n)}\rightarrow{\mathcal S}_{\mathbb Z}(n,d).\end{align*} $$

In the remainder of this section, we assume that $l'=l=1$ . Let

We shall denote the images of $\bar E_{i}^{(m)}$ , $\bar F_{i}^{(m)}$ , etc. in by the same letters. For $A\in {\Theta (n,d)}$ , let $[A]_1$ be the image of $[A]$ in . The map $\xi _{d}$ induces, upon tensoring with , a surjective algebra homomorphism

Let be the infinitesimal Schur algebra introduced in [Reference Doty, Nakano and Peters11]. By [Reference Doty, Nakano and Peters11, (5.3.4)], the set $\{[A]_1\mid A\in {\Theta (n,d)},\,a_{i,j}<p^r,\,\forall i,j\}$ forms a -basis for . Hence, we have

Let be the -subalgebra of generated by the elements $\bar E_i^{(m)}$ , $\bar F_i^{(m)}$ , $\big ({H_j\atop t}\big )$ for $1\leqslant i\leqslant n-1$ , $1\leqslant j\leqslant n$ , $t\in \mathbb N$ and $0\leqslant m<p^{r}$ . Then, by (6.1), we have

Therefore, by Lemma 3.5, we have . Hence, by restricting the map $\xi _{d}$ to , we obtain a surjective algebra homomorphism

By Theorem 4.10, we obtain the following result.

Let

$$ \begin{align*}\dot{\mathcal U}(\mathfrak{gl}_n):=\bigoplus_{{\lambda},\mu\in\mathbb Z^{n}}{}_{\lambda}{\mathcal U}(\mathfrak{gl}_n)_\mu,\end{align*} $$

where

$$ \begin{align*} {}_{\lambda}{\mathcal U}(\mathfrak{gl}_n)_\mu={\mathcal U}(\mathfrak{gl}_n)/\left(\sum_{{\mathbf{j}}\in\mathbb Z^{n}}\big(H^{\mathbf{j}}-{\lambda}^{{\mathbf{j}}}\big) {\mathcal U}(\mathfrak{gl}_n)+\sum_{{\mathbf{j}}\in\mathbb Z^{n}}{\mathcal U}(\mathfrak{gl}_n)\big(H^{\mathbf{j}}-\mu^{{\mathbf{j}}})\right), \end{align*} $$

$H^{\mathbf {j}}=\prod _{1\leqslant i\leqslant n} H_i^{j_i}$ and ${\lambda }^{{\mathbf {j}}}=\prod _{1\leqslant i\leqslant n}{\lambda }_i^{j_i}$ . Let $\bar \pi _{{\lambda },\mu }:{\mathcal U}(\mathfrak {gl}_n)\rightarrow {}_{\lambda }{\mathcal U}(\mathfrak {gl}_n)_\mu $ be the canonical projection. Let $\bar 1_{\lambda }=\bar \pi _{{\lambda },{\lambda }}(1)$ . As in the case of $\dot {{{\mathbf {U}}}}(\mathfrak {gl}_n)$ , there is a natural associative $\mathbb Q$ -algebra structure on $\dot {\mathcal U}(\mathfrak {gl}_n)$ inherited from that of ${\mathcal U}(\mathfrak {gl}_n)$ , and $\dot {\mathcal U}(\mathfrak {gl}_n)$ is naturally a ${\mathcal U}(\mathfrak {gl}_n)$ -bimodule. Let $\dot {\mathcal U}_{\mathbb Z}(\mathfrak {gl}_n)$ be the $\mathbb Z$ -subalgebra of $\dot {\mathcal U}(\mathfrak {gl}_n)$ generated by the elements $\bar E_i^{(m)}\bar 1_{\lambda }$ and $\bar 1_{\lambda }\bar F_i^{(m)}$ for $1\leqslant i\leqslant n-1$ , ${\lambda }\in \mathbb Z^{n}$ and $m\in \mathbb N$ .

Let ${\dot U_{\mathbb Z}(\mathfrak {gl}_n)}=\dot {{U}}_{\mathcal Z}(\mathfrak {gl}_n)\otimes _{\mathcal Z}\mathbb Z$ , where $\mathbb Z$ is viewed as a ${\mathcal Z}$ -module by specializing v to $1$ . By (6.1), we have

(6.2) $$ \begin{align} {\dot U_{\mathbb Z}(\mathfrak{gl}_n)}\cong\dot{\mathcal U}_{\mathbb Z}(\mathfrak{gl}_n). \end{align} $$

We will identify ${\dot U_{\mathbb Z}(\mathfrak {gl}_n)}$ with $\dot {\mathcal U}_{\mathbb Z}(\mathfrak {gl}_n)$ . The map $\dot {\zeta }_{d}$ given in (3.9) induces, upon tensoring with $\mathbb Z$ , a surjective algebra homomorphism

$$ \begin{align*}\dot\xi_{d}:\dot{\mathcal U}_{\mathbb Z}(\mathfrak{gl}_n)\rightarrow{\mathcal S}_{\mathbb Z}(n,d).\end{align*} $$

Let . We shall denote the images of $\bar E_{i}^{(m)}\bar 1_{\lambda }$ , $\bar 1_{\lambda }\bar F_{i}^{(m)}$ in by the same letters. The map $\dot \xi _{d}$ induces, upon tensoring with , a surjective algebra homomorphism

Let be the -subalgebra of generated by the elements $\bar E_i^{(m)}\bar 1_{\lambda }$ , $\bar 1_{\lambda }\bar F_i^{(m)}$ for $1\leqslant i\leqslant n-1$ , ${\lambda }\in \mathbb Z^{n}$ and $0\leqslant m<p^{r}$ . Then, by (6.2), we have

By Lemma 3.6, we have . Hence, by restricting the map $\dot {\xi }_{d}$ to , we obtain a surjective algebra homomorphism

By Theorem 5.5, we obtain the following result.

7 Borel subalgebras of the infinitesimal q-Schur algebra

In this section, we investigate Borel subalgebras of the infinitesimal q-Schur algebra . In what follows, we focus entirely on the quantum case as the corresponding results for the classical cases are essentially the same.

Let and . These algebras are called Borel subalgebras of . Furthermore, let (resp. ) be the -subalgebra of generated by the elements $E_{i}^{(m)}1_{\lambda }$ (resp. $1_{\lambda } F_{i}^{(m)}$ ) for $1\leqslant i\leqslant n-1$ , ${\lambda }\in \mathbb Z^{n}$ and $0\leqslant m<lp^{r-1}$ .

Let (resp. ) be the -subalgebra of generated by $\mathtt {e}_i^{(m)}$ (resp. $\mathtt {f}_i^{(m)}$ ) and $\mathtt {k}_\lambda $ for $1 \leqslant i \leqslant n-1$ , $0\leqslant m<lp^{r-1}$ and $\lambda \in \Lambda (n,d)$ . These algebras are called Borel subalgebras of .

Let (resp. ) be the -subalgebra of generated by $\mathtt {e}_i^{(m)}$ (resp. $\mathtt {f}_i^{(m)}$ ) for $1\leqslant i\leqslant n-1$ and $0\leqslant m<lp^{r-1}$ .

Proof. By [Reference Doty and Giaquinto8, Prop. 8.2], we have $1=\sum _{{\lambda }\in \Lambda (n,d)}\mathtt {k}_{\lambda }$ . Hence, by [Reference Du and Parshall15, Lem. 4.10], we have

(7.1) $$ \begin{align} {\mathtt e}^{(A)}=\sum_{{\lambda}\in\Lambda(n,d) }{\mathtt e}^{(A)}\mathtt{k}_{\lambda}=\sum_{{\lambda}\in\Lambda(n,d)\atop{\lambda}\geqslant{\boldsymbol\sigma}(A)}{\mathtt e}^{(A)}\mathtt{k}_{\lambda}, \end{align} $$

for $A\in \Theta ^+(n)_r$ . Furthermore, we have

(7.2) $$ \begin{align} A(\mathbf{0},d)=\sum_{{\lambda}\in\Lambda(n,d)}A(\mathbf{0},d)[\operatorname{diag}({\lambda})]_{\varepsilon}=\sum_{{\lambda}\in\Lambda(n,d)\atop{\lambda}\geqslant{\boldsymbol\sigma}(A)} A(\mathbf{0},d)[\operatorname{diag}({\lambda})]_{\varepsilon}, \end{align} $$

for $A\in \Theta ^+(n)_r$ , since $1=\sum _{{\lambda }\in \Lambda (n,d)}[\operatorname {diag}({\lambda })]_{\varepsilon }$ . Therefore, we have

$$ \begin{align*}{\mathtt e}^{(A)}=A(\mathbf{0},d)=0,\end{align*} $$

for $A\in \Theta ^+(n)_r$ with $\sigma (A)>d$ . Hence, by Proposition 3.1 and 3.2, we conclude that . Furthermore, by Proposition 3.7, the sets ${\mathscr M}_{d,r}^+$ and ${\mathscr P}_{d,r}^+$ are both linearly independent. Our assertion follows.

Let be the quotient of by $I_{d,r}^{\geqslant 0}$ , where $I_{d,r}^{\geqslant 0}$ is the two-sided ideal of generated by the elements $1-\sum _{\mu \in \Lambda (n,d)} K_{\mu }, K_{i} K_{\lambda }-\varepsilon ^{\lambda _{i}} K_{\lambda }$ and $\big [{K_i;0 \atop t}\big ] K_{\lambda }-\big [{{\lambda }_i \atop t}\big ]_{\varepsilon } K_{\lambda }$ for $1 \leqslant i \leqslant n$ , $t \in \mathbb {N}$ and $\lambda \in \Lambda (n,d)$ .

By restricting the map $\zeta _{d,r}$ given in (4.2) to , we obtain a surjective algebra homomorphism . Since $\zeta _{d,r}(I_{d,r}^{\geqslant 0})=0$ , the map $\zeta _{d,r}$ induces an epimorphism

Proof. Using an argument similar to the proof of Proposition 4.9, we can show that the algebra is spanned as a -module by the elements $E^{(A)} K_{\lambda }+I_{d,r}^{\geqslant 0}$ for $A \in \Theta ^+(n)_r$ , $\lambda \in \Lambda (n,d)$ and $\lambda \geqslant {\boldsymbol \sigma }(A)$ . Furthermore, by Lemma 7.2, the set

$$ \begin{align*}\{\zeta_{d,r}(E^{(A)} K_{\lambda}+I_{d,r}^{\geqslant 0})\mid A \in \Theta^+(n)_r,\, \lambda \in \Lambda(n,d),\, \lambda \geqslant {\boldsymbol\sigma}(A)\},\end{align*} $$

forms a -basis for . Hence, $\zeta _{d,r}' $ is an algebra isomorphism.

Let be the quotient of by $J_{d,r}^{\geqslant 0}$ , where $J_{d,r}^{\geqslant 0}$ is the two-sided ideal of generated by the elements $1_{\lambda }$ for $\lambda \not \in \Lambda (n,d)$ .

By restricting the map $\dot {\zeta }_{d,r}$ given in (5.1) to , we obtain a surjective algebra homomorphism . Since $\dot {\zeta }_{d,r}(J_{d,r}^{\geqslant 0})=0$ , the map $\dot {\zeta }_{d,r}$ induces an epimorphism

Proof. Using an argument similar to the proof of Proposition 5.4, we can show that the algebra is spanned as a -module by the elements $E^{(A)}1_{\lambda }+J_{d,r}^{\geqslant 0}$ for $A\in \Theta ^+(n)_r$ , ${\lambda } \in \Lambda (n,d)$ and ${\lambda } \geqslant {\boldsymbol \sigma }(A)$ . Furthermore, by Lemma 7.2, the set

$$ \begin{align*}\{\dot{\zeta}_{d,r}'(E^{(A)}1_{\lambda}+J_{d,r}^{\geqslant 0})\mid A\in \Theta^+(n)_r,\,\lambda \in \Lambda(n,d),\,\lambda \geqslant {\boldsymbol\sigma}(A)\},\end{align*} $$

forms a -basis of . Therefore, $\dot {\zeta }_{d,r}'$ is an algebra isomorphism.

8 Irreducible -modules

In this section, we assume that is a field, $p>0$ and $l'=l$ is odd. Let $X=\mathbb Z^{n}$ and $X^+=\{{\lambda }\in X\mid {\lambda }_1\geqslant {\lambda }_{2}\geqslant \cdots \geqslant {\lambda }_n\}$ . For ${\lambda }\in X^+$ , let $L({\lambda })$ be the simple integrable -module of highest weight ${\lambda }$ . Let $\text {Ind}^{U_2}_{U_1}(-)= H^0(U_2/U_1,-)$ be the induction functor for quantized enveloping algebras defined in [Reference Andersen, Polo and Wen1], [Reference Andersen, Polo and Wen2].

For ${\lambda }\in X$ , let

Let $P_r=\{{\lambda }\in \mathbb N^{n}\mid 0\leqslant {\lambda }_i-{\lambda }_{i+1}<lp^{r-1}\text { for }1\leqslant i\leqslant n \}$ , where ${\lambda }_{n+1}=0$ . The following result was given in [Reference Drupieski12, Ths. 3.4.1 & 3.4.3].

If M is a -module and ${\lambda }\in X$ let

$$ \begin{align*}M_{\lambda}=\{w\in M\mid K_iw=\varepsilon^{{\lambda}_i}w,\,\left[{K_i;0\atop t}\right]w=\left[{{\lambda}_i\atop t}\right]_{\varepsilon} w \text{ for }1\leqslant i\leqslant n,\,t\in\mathbb N\}.\end{align*} $$

Let ${\Gamma }_r=P_r+lp^{r-1}\mathbb N^{n}$ and ${\Gamma }_r^d=\big \{{\lambda }\in {\Gamma }_r\big |\sum _{i=1}^n{\lambda }_i=d\big \}.$ For ${\lambda },\mu \in \mathbb Z^{n}$ with $\sum _{1\leqslant i\leqslant n}{\lambda }_i=\sum _{1\leqslant i\leqslant n}\mu _i$ we write ${\lambda }\trianglelefteq \mu $ if $\sum _{1\leqslant s\leqslant i}{\lambda }_s\leqslant \sum _{1\leqslant s\leqslant i}\mu _s$ for $1\leqslant i\leqslant n$ .

The irreducible modules for infinitesimal q-Schur algebras were classified in [Reference Cox4, Sec. 5.1]. We now use Theorem 4.10 to give a classification of irreducible -modules.

Proof. Let ${\lambda }\in {\Gamma }_r^d$ . By Lemma 8.2, we have $\widehat L_r({\lambda })=\oplus _{\mu \in \Lambda (n,d)}\widehat L_r({\lambda })_\mu $ . Let $w_\mu $ be a nonzero vector in $\widehat L_r({\lambda })_\mu $ for some $\mu \in \Lambda (n,d)$ . Since $\mu \in \Lambda (n,d)$ , we have $K_\alpha w_\mu =\big [{\mu \atop \alpha }\big ]_{\varepsilon } w_\mu =\delta _{\alpha ,\mu }w_\mu $ for $\alpha \in \Lambda (n,d)$ . It follows that

$$ \begin{align*} \begin{aligned} \sum_{\beta\in\Lambda(n,d)}K_\beta w_\mu&=\sum_{\beta\in\Lambda(n,d)}\delta_{\beta,\mu} w_\mu=w_\mu, \\ (K_i-\varepsilon^{\alpha_i})K_\alpha w_\mu&=\delta_{\alpha,\mu}K_iw_\mu-\varepsilon^{\alpha_i}\delta_{\alpha,\mu}w_\mu=0,\\ \bigg(\bigg[{K_i;0\atop t}\bigg] -\bigg[{\alpha_i\atop t}\bigg]\bigg)K_\alpha w_\mu&= \delta_{\alpha,\mu}\bigg[{K_i;0\atop t}\bigg] w_\mu-\bigg[{\alpha_i\atop t}\bigg]\delta_{\alpha,\mu} w_\mu=0, \end{aligned} \end{align*} $$

for $\alpha \in \Lambda (n,d)$ , $1\leqslant i\leqslant n$ and $t\in \mathbb N$ . Thus, by Theorem 4.10, we conclude that $\widehat L_r({\lambda })$ can be regarded as a -module.

On the other hand, let L be an irreducible -module. By Theorem 8.1, we conclude that $L\cong \widehat L_r(\nu +lp^{r-1}\delta )\cong L(\nu )\otimes lp^{r-1}\delta $ for some $\nu \in P_r$ and $\delta \in \mathbb Z^{n}$ . Hence, since L is a -module, we have $(\nu _{w(1)},\nu _{w(2)},\ldots ,\nu _{w(n)})+ lp^{r-1}\delta \in \Lambda (n,d)$ for any w in the symmetric group ${\mathfrak S}_n$ . It follows that $\nu _n+lp^{r-1}\delta _j\geqslant 0$ for $1\leqslant j\leqslant n$ . Furthermore, since $\nu \in P_r$ , we have $0\leqslant \nu _n<lp^{r-1}$ . Therefore, we have $\delta _j\geqslant 0$ for $1\leqslant j\leqslant n$ . Consequently, we have $\nu +lp^{r-1}\delta \in {\Gamma }_r^d$ . The proof is completed.