The regular representation of $U_v(\mathfrak{gl}_{m|n})$

TL;DR

This paper constructs a super representation of the quantum superalgebra $U_v(\mathfrak{gl}_{m|n})$ using quantum differential operators, providing an explicit basis and multiplication formulas for the algebra.

Contribution

It introduces a new super representation of $U_v(\mathfrak{gl}_{m|n})$ via quantum differential operators and explicitly describes its basis and multiplication rules.

Findings

Constructed a super representation on polynomial superalgebra

Extended to a formal power series algebra containing the regular representation

Provided explicit basis and multiplication formulas for $U_v(\mathfrak{gl}_{m|n})$

Abstract

Using quantum differential operators, we construct a super representation of $U_v(\mathfrak{gl}_{m|n})$ on a certain polynomial superalgebra. We then extend the representation to its formal power series algebra which contains a $U_v(\mathfrak{gl}_{m|n})$-submodule isomorphic to the regular representation of $U_v(\mathfrak{gl}_{m|n})$. In this way, we obtain a presentation of $U_v(\mathfrak{gl}_{m|n})$ by a basis together with explicit multiplication formulas of the basis elements by generators.

Equations

$$\hat{i}=\{\begin{array}{ll}0,& \text{mbox}if1\le i\le m;\\ 1,& \text{mbox}ifm+1\le i\le m+n.\end{array}$$

$${\mathbf{e}}_i\ast {\mathbf{e}}_j=({\mathbf{e}}_i,{\mathbf{e}}_j{)}_s=(-1{)}^{\hat{i}}{\delta }_{ij}.$$

$${\mathbf{\upsilon }}_h={\mathbf{\upsilon }}^{(-1{)}^{\hat{h}}}(h\in [1,m+n]),[a{]}^{!}=[1][2]\cdots [a],[i]=\frac{{\mathbf{\upsilon }}^i-{\mathbf{\upsilon }}^{-i}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}(a\in \mathbb{N}).$$

$$[X,Y]=[X,Y{]}_s=XY-(-1{)}^{\hat{X}\hat{Y}}YX.$$

$$\{\begin{array}{ll}\text{even generators: }& E_h,F_h,K_j^{\pm 1},1\le h,j\le m+n,h\ne m,m+n;\\ \text{odd generators: }& E_m,F_m.\end{array}$$

$$E_a^2{E}_b-({\mathbf{\upsilon }}_a+{\mathbf{\upsilon }}_a^{-1})E_a{E}_b{E}_a+E_b{E}_a^2=0,$$

$$F_a^2{F}_b-({\mathbf{\upsilon }}_a+{\mathbf{\upsilon }}_a^{-1})F_a{F}_b{F}_a+F_b{F}_a^2=0;$$

$$E_{m-1,m+2}$$

$$E_{m+2,m-1}$$

$$\Delta (K_i)=K_i\otimes K_i,$$

$$\Delta (E_i)=E_i\otimes {\tilde{K}}_i+1\otimes E_i,\Delta (F_i)=F_i\otimes 1+{\tilde{K}}_i^{-1}\otimes F_i,$$

$$\epsilon (K_i)=1,\epsilon (E_i)=\epsilon (F_i)=0,$$

$$S(K_i)=K_i^{-1},S(E_i)=-E_i{\tilde{K}}_i^{-1},S(F_i)=-{\tilde{K}}_i{F}_i,$$

$${\mathcal{D}}_i(x_1^{a_1}{x}_2^{a_2}\cdots x_N^{a_N})=\{\begin{array}{ll}[a_i{]}_q{x}_1^{a_1}{x}_2^{a_2}\cdots x_i^{a_i-1}\cdots x_N^{a_N},& \text{if }{a}_i\ge 1;\\ 0,& \text{otherwise}.\end{array}$$

$${\mathcal{K}}_i(x_1^{a_1}{x}_2^{a_2}\cdots x_N^{a_N})=q^{a_i}{x}_1^{a_1}{x}_2^{a_2}\cdots x_N^{a_N}.$$

$$E_i⟼{\mathcal{E}}_i,F_i⟼{\mathcal{F}}_i,K_j⟼{\mathcal{K}}_j$$

$$d_i^2=0(1\le i\le N),d_i{d}_j=-d_j{d}_i(1\le i\ne j\le N).$$

$$K_i.d^{\mathbf{a}}=q^{a_i}{d}^{\mathbf{a}},E_h.d^{\mathbf{a}}=\{\begin{array}{ll}{d}^{\mathbf{a}+{\alpha }_h},& \text{if }{a}_{h+1}>0;\\ 0,& \text{otherwise,}\end{array}{F}_h.d^{\mathbf{a}}=\{\begin{array}{ll}{d}^{\mathbf{a}-{\alpha }_h},& \text{if }{a}_h>0;\\ 0,& \text{otherwise,}\end{array}$$

$$S(V_0)\otimes \Lambda (V_1)$$

$$\Lambda (V_0)\otimes S(V_1)$$

$${\mathcal{S}}_{0|1}={\mathcal{S}}_{0|1}(V):=\mathbb{Q}(\mathbf{\upsilon })[X_1,X_2,\cdots ,X_{m+n}]\text{ with }{X}_i=x_i,X_{m+j}=d_j,$$

$${\mathcal{S}}_{1|0}={\mathcal{S}}_{1|0}(V):=\mathbb{Q}(\mathbf{\upsilon })[X_1,X_2,\cdots ,X_{m+n}]\text{ with }{X}_i=d_i,X_{m+j}=x_j,$$

$$X^{(\mathbf{a})}=X_1^{(a_1)}{X}_2^{(a_2)}\cdots X_{m+n}^{(a_{m+n})},$$

$$K_i.X^{(\mathbf{a})}$$

$$E_h.X^{(\mathbf{a})}$$

$$F_h.X^{(\mathbf{a})}$$

$$(E_m{F}_m+F_m{E}_m).X^{(\mathbf{a})}=F_m{E}_m.X^{(\mathbf{a})}=[a_m+1]F_m{X}^{(\mathbf{a}+{\alpha }_m)}=[a_m+1]X^{(\mathbf{a})}.$$

$$\frac{K_m{K}_{m+1}^{-1}-K_m^{-1}{K}_{m+1}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}.X^{(\mathbf{a})}=\frac{{\mathbf{\upsilon }}^{a_m+1}-{\mathbf{\upsilon }}^{-a_m-1}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}.X^{(\mathbf{a})}=[a_m+1]X^{(\mathbf{a})}.$$

$$(E_m{F}_m+F_m{E}_m).X^{(\mathbf{a})}$$

$$=\frac{K_m{K}_{m+1}^{-1}-K_m^{-1}{K}_{m+1}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}.X^{(\mathbf{a})}.$$

$$(E_m{F}_m+F_m{E}_m).X^{(\mathbf{a})}=0=\frac{K_m{K}_{m+1}^{-1}-K_m^{-1}{K}_{m+1}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}.X^{(\mathbf{a})}.$$

$$(E_m{F}_m+F_m{E}_m).X^{(\mathbf{a})}=\frac{K_m{K}_{m+1}^{-1}-K_m^{-1}{K}_{m+1}}{\mathbf{\upsilon }-{\mathbf{\upsilon }}^{-1}}.X^{(\mathbf{a})},$$

$$E_m{E}_{m-1,m+2}.X^{(\mathbf{a})}$$

$$-{\mathbf{\upsilon }}^{-1}{E}_m{E}_m{E}_{m+1}{E}_{m-1}+E_m{E}_{m+1}{E}_m{E}_{m-1})X^{(\mathbf{a})}$$

$$=(-\mathbf{\upsilon }{E}_m{E}_{m-1}{E}_{m+1}{E}_m+E_m{E}_{m+1}{E}_m{E}_{m-1})X^{(\mathbf{a})}$$

$$=(-\mathbf{\upsilon }[a_m+1][a_{m-1}+1][a_m+1]+[a_{m-1}+1][a_m][a_m+1])X^{({\mathbf{a}}^{\prime })},$$

$$E_{m-1,m+2}{E}_m.X^{(\mathbf{a})}$$

Full text

The second author would like to thank UNSW for its hospitality during his one year visit and thank the Jiangsu Provincial Department of Education for financial support.

1. Introduction

Arising from the natural representation $V$ of the quantum supergroup $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$, the investigation on the tensor products $V^{\otimes r}$ for all $r\geq 0$ has recently produced interesting outcomes. For example, the root-of-unity theory resulted in a new proof for the quantum Mullineux conjecture (see [9]). On the other hand, the generic theory on ${\boldsymbol{\upsilon}}$-Schur superalgebras, which are homomorphic images of the representations $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})\to\text{End}(V^{\otimes r})$, gives rise to a new construction for $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ itself (see [7]). This latter work extends the geometric realisation of quantum $\mathfrak{gl}_{n}$, given by Beilinson–Lusztig–MacPherson (BLM) in [1], to the super case. The BLM work has also been generalised to the quantum affine case [3, 5] and the case for the other classical types [2, 10].

Furthernore, in the nonsuper case, there are other representations of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{n})$ arising from the symmetric and exterior algebras $S(V)$ and $\Lambda(V)$ of the natural representation $V$; see, e.g., [11] and [12, §§5A.6-7], where the module actions are defined by using certain quantum differential operators. Can these representations be used to determine the structure of a quantum supergroup? We will provide an affirmative answer in this paper.

We will start with the natural super representations $V=V_{0}\oplus V_{1}$ of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n}).$ We first introduce two types of symmetric superalgebras $\mathcal{S}_{0|1}(V)=S(V_{0})\otimes\Lambda(V_{1})$ and $\mathcal{S}_{1|0}(V)=\Lambda(V_{0})\otimes S(V_{1})$ and their mixed tensor product ${{\mathcal{S}^{m|n}}}(V)$. The supermodule structure on each of them is defined via quantum differential (super) operators. We then extend the supermodule structure to the formal power series algebra ${\widetilde{\mathcal{S}}^{m|n}}(V)$ which is naturally a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-module. We will extract a submodule from ${\widetilde{\mathcal{S}}^{m|n}}(V)$ which naturally possesses a supermodule structure. We prove in the main theorem (Theorem 5.3) that this supermodule is isomorphic to the regular representation of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$. Thus, we obtain a new presentation for $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ (cf. Lemma 5.1). Surprising enough, this presentation from the regular representation of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ coincides with the one from [1, Lemma 5.3] when $n=0$ and with the one as given in [7, Thm 8.4] in general, both of which were obtained either by a geometric method involving quantum Schur algebras or by an algebraic method involving quantum Schur superalgebras.

2. The quantum supergroup $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ and differential operators

For fixed non-negative integers $m,n$ with $m+n>0$, let $[1,m+n]:=\{1,2,\cdots,m+n\},$ and define the parity function $\widehat{\ }:[1,m+n]\to\mathbb{Z}_{2},i\mapsto\widehat{i}$ by

[TABLE]

We will always regard $\mathbb{Z}_{2}=\{0,1\}$ as a subset of $\mathbb{N}$ unless it is used for the grading of a super structure. For any superspace $V$ and a homogeneous element $v\in V$, we often use $\widehat{v}$ to denote its parity.

Let $\{{\boldsymbol{e}}_{1},{\boldsymbol{e}}_{2},\cdots,{\boldsymbol{e}}_{m+n}\}$ be the standard basisfor $\mathbb{Z}^{m+n}$, and define the “super dot product” on $\mathbb{Z}^{m+n}$ by

[TABLE]

Let $\mathbb{Q}({\boldsymbol{\upsilon}})$ be the field of rational functions in indeterminate ${\boldsymbol{\upsilon}}$ and let

[TABLE]

Let $[i]_{q}$ denote the value at $q$.

Define the super (or graded) commutator on the homogeneous elements $X,Y$ for an (associative) superalgebra by

[TABLE]

The following quantum enveloping superalgebra $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ is defined in [13].

Definition 2.1**.**

(1) The quantum enveloping superalgebra $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ over $\mathbb{Q}({\boldsymbol{\upsilon}})$ is generated by

[TABLE]

These elements are subject to the following relations:

1. (QG1)

$K_{a}K_{b}=K_{b}K_{a},\ K_{a}K_{a}^{-1}=K_{a}^{-1}K_{a}=1;$ 2. (QG2)

$K_{a}E_{b}={\boldsymbol{\upsilon}}^{(\varepsilon_{a},\alpha_{b})_{s}}E_{b}K_{a},\ K_{a}F_{b}={\boldsymbol{\upsilon}}^{(\varepsilon_{a},-\alpha_{b})_{s}}F_{b}K_{a};$ 3. (QG3)

$[E_{a},F_{b}]=\delta_{a,b}\frac{K_{a}K_{a+1}^{-1}-K_{a}^{-1}K_{a+1}}{{\boldsymbol{\upsilon}}_{a}-{\boldsymbol{\upsilon}}_{a}^{-1}};$ 4. (QG4)

$E_{a}E_{b}=E_{b}E_{a},\ F_{a}F_{b}=F_{b}F_{a},$ if $|a-b|>1;$ 5. (QG5)

For $a\neq m$ and $|a-b|=1,$

[TABLE] 6. (QG6)

$E_{m}^{2}=F_{m}^{2}=[E_{m},E_{m-1,m+2}]=[F_{m},E_{m+2,m-1}]=0,$ where

[TABLE]

(2) Let ${\mathbbm{k}}$ be a field and let $q\in{\mathbbm{k}}$, $q\neq 0$ and $q^{2}\neq 1$. Then, with ${\boldsymbol{\upsilon}}$ replaced by $q$, we may similarly define the quantum group $U_{q}(\mathfrak{gl}_{m|n})$ over ${\mathbbm{k}}$ (see [12, 13]).

Note that, if $n=0$, then (QG1)–(QG5) form a presentation for the quantum group $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m})$.

A Hopf algebra structure on $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ is defined (see [13, Section II]) by:

[TABLE]

where ${\widetilde{K}}_{i}=K_{i}K_{i+1}^{-1}.$

Representations of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ have been investigated in [13] (see also [9] for representations of its hyperalgebra at roots of unity). We will need two special $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodules in the next section for our construction. They are built on the following two $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{N})$-modules defined by quantum differential operators.

Example 2.2**.**

Let $V$ be a vector space over a field ${\mathbbm{k}}$ of dimension $N$ and let ${\mathbbm{k}}[x_{1},x_{2},\cdots,x_{N}]$ be the polynomial algebra over ${\mathbbm{k}}$ in indeterminates $x_{1},\ldots,x_{N}$.

(1) Let $S(V)$ be the symmetric algebra on $V$, identified as $S(V)={\mathbbm{k}}[x_{1},x_{2},\cdots,x_{N}]$. Following [12, 5A.6], we define quantum differential operators $\mathcal{D}_{i}:S(V)\to S(V)$ by

[TABLE]

We also introduce algebra automorphism $\mathcal{K}_{i}:S(V)\to S(V)$ by setting

[TABLE]

Let $\mathcal{E}_{i}=x_{i}\circ\mathcal{D}_{i+1}$ and $\mathcal{F}i=x_{i+1}\circ\mathcal{D}_{i}$. Then, by [12, Prop. 5A.6], the following map

[TABLE]

for all $1\leq i,j\leq N$ ($i\neq N$) defines an algebra homomorphism from $U_{q}(\mathfrak{gl}_{N})$ to the endomorphism algebra of $S(V)$. Hence, $S(V)$ becomes a $U_{q}(\mathfrak{gl}_{N})$-module (cf. [11, Thm 4.1(A)]).

(2) Let ${\Lambda}(V)$ be the exterior algebra on $V$. In this case, we may identify ${\Lambda}(V)$ with the Grassman superalgebra ${\Lambda}(d_{1},\ldots,d_{N})$ with odd generators $d_{1},\ldots,d_{N}$ and relations

[TABLE]

Thus, ${\Lambda}(V)$ has a basis $d^{{\bf a}}:=d_{1}^{a_{1}}\cdots d_{N}^{a_{N}}$, ${{\bf a}}\in\mathbb{Z}_{2}^{N}$. Define a $U_{q}(\mathfrak{gl}_{N})$-action on ${\Lambda}(V)$ by

[TABLE]

for all $1\leq h,i\leq N$, $h\neq N$. It is direct to check that all relations (QG1-5) are satisfied. Hence, ${\Lambda}(V)$ becomes a $U_{q}(\mathfrak{gl}_{N})$-module (cf. [11, §§2,4]).

3. The polynomial superalgebra ${{\mathcal{S}^{m|n}}}(V)$ as a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule

We generalize the constructions of the module structures on symmetric and exterior algebras to the supergroup $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n}).$

Consider the natural representation on the superspace $V=V_{0}\oplus V_{1}$ of $\mathfrak{gl}_{m|n}({\mathbbm{k}})$ where $\dim V_{0}=m$ and $\dim V_{1}=n$. We will consider two superalgebras in the notation of Example 2.2:

[TABLE]

These are known as polynomial superalgebras with even generators $x_{i}$ and odd generators $d_{j}$. By Example 2.2, both algebras are also $U_{q}(\mathfrak{gl}_{m})\otimes U_{q}(\mathfrak{gl}_{n})$-modules

We now assume ${\mathbbm{k}}=\mathbb{Q}({\boldsymbol{\upsilon}})$. In order to introduce supermodule structure for $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$, we set

[TABLE]

where $1\leq i\leq m,1\leq j\leq n$. We use divided powers to denote their monomial bases:

[TABLE]

where ${{\bf a}}=(a_{1},\cdots,a_{m+n})\in\mathbb{N}^{m}\times\mathbb{Z}_{2}^{n}$ for $\mathcal{S}_{0|1}$, ${{\bf a}}=(a_{1},\cdots,a_{m+n})\in\mathbb{Z}_{2}^{m}\times\mathbb{N}^{n}$ for $\mathcal{S}_{1|0}$, and $X_{i}^{(a_{i})}=\frac{X_{i}^{a_{i}}}{[a_{i}]!}$.

For the superspace structure, we have, for $i\in\mathbb{Z}_{2}$, ${X^{({{\bf a}})}}\in(\mathcal{S}_{0|1})_{i}$ if and only if $\widehat{{\bf a}}:=\sum_{j=1}^{n}a_{m+j}\equiv i(\text{mod}\;2)$, while ${X^{({{\bf a}})}}\in(\mathcal{S}_{1|0})_{i}$ if and only if $\widehat{{\bf a}}:=\sum_{j=1}^{m}a_{j}\equiv i(\text{mod}\;2)$.

As algebras, both $\mathcal{S}_{0|1}$ and $\mathcal{S}_{1|0}$ have a graded structure $\mathcal{S}_{0|1}(r)$ and $\mathcal{S}_{1|0}(r)$ for all $r\geq 0$, where $\mathcal{S}_{0|1}(r)$ (resp., $\mathcal{S}_{1|0}(r)$) is the $r$-th homogeneous component spanned by all $X^{({{\bf a}})}$ with $\deg(X^{({{\bf a}})})=r$. Here $\deg(X^{({{\bf a}})}):=\sum_{i=1}^{m+n}a_{i}$.

We now define the following actions on $\mathcal{S}_{0|1}$ and $\mathcal{S}_{1|0}$ by the same rules:

[TABLE]

where $1\leq h,i\leq m+n$, $h\neq m+n$, and $\alpha_{h}={\boldsymbol{e}}_{h}-{\boldsymbol{e}}_{h+1}$ are “simple roots”. Note that, for the even generators, the action above (on quantum divided powers) coincides with those given in Example 2.2. Thus, both $\mathcal{S}_{0|1}$ and $\mathcal{S}_{1|0}$ are $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m})\otimes U_{{\boldsymbol{\upsilon}}^{-1}}(\mathfrak{gl}_{n})$-modules under the action above.

Lemma 3.1**.**

Both $\mathcal{S}_{0|1}$ and $\mathcal{S}_{1|0}$ are $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodules under the actions above. In particular, their homogeneous components $\mathcal{S}_{0|1}(r),\mathcal{S}_{1|0}(r),r\geq 0$ are all subsuperbmodules.

Proof.

We only need to verify the defining relations that involve the odd generators.

We only prove the case for $\mathcal{S}_{0|1}.$ It is easy to verify the relations (QG2) and (QG4). Note that the actions of $E_{m},F_{m}$ is consistent with those for even generators, so (QG5) holds. It remains to check (QG3) and (QG6).

The relations $[E_{m},E_{b}]=0=[E_{b},F_{m}]$ with $m\neq b$ in (QG3) are clear. Assume now $a=b=m$. Let ${{\bf a}}=(a_{1},\cdots,a_{m+n})\in\mathbb{N}^{m}\times\mathbb{Z}_{2}^{n}.$ If $a_{m+1}=1$ then

[TABLE]

If $a_{m+1}=0,a_{m}>0$ then

[TABLE]

If $a_{m+1}=0,a_{m}=0$ then

[TABLE]

So, in all three cases, we obtain

[TABLE]

for all ${{\bf a}}\in\mathbb{N}^{m}\times\mathbb{Z}_{2}^{n}$, proving (QG3).

Finally, we prove the four relations in (QG6). As $a_{m+1}\leq 1$, we have $E_{m}^{2}.X^{({{\bf a}})}=0=F_{m}^{2}.X^{({{\bf a}})}$ for all ${{\bf a}}$. For the other two relations, if $a_{m+1}=1,a_{m+2}=1$ then

[TABLE]

where ${{\bf a}}^{\prime}={{\bf a}}+\alpha_{m-1}+2\alpha_{m}+\alpha_{m+1}={{\bf a}}+{\boldsymbol{e}}_{m-1}+{\boldsymbol{e}}_{m}-{\boldsymbol{e}}_{m+1}-{\boldsymbol{e}}_{m+2}$. On the other hand,

[TABLE]

Since $[a_{m}]+[a_{m}+2]-({\boldsymbol{\upsilon}}+{\boldsymbol{\upsilon}}^{-1})[a_{m}+1]=0$, it follows that

[TABLE]

If $a_{m+1}=0$ or $a_{m+2}=0$ then $E_{m-1,m+2}E_{m}.X^{({{\bf a}})}=0=E_{m}E_{m-1,m+2}.X^{({{\bf a}})}$ by the definition of the actions. The last case can be proved similarly. This proves (QG6). ∎

The following result is a super analog of a result stated at the end of [12, 5A.7] (see also [11, Thms 4.1(A), 4.2]). Recall from, say, [9] that irreducible weight $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-modules are indexed by

[TABLE]

Corollary 3.2**.**

Let $\Delta(r{\boldsymbol{e}}_{1})$ (resp., $\nabla(r{\boldsymbol{e}}_{m+n})$) be the irreducible weight $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-module of highest (resp., lowest) weight $r{\boldsymbol{e}}_{1}$ (resp., $r{\boldsymbol{e}}_{m+n}$). Then, there are $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-module isomorphisms:

[TABLE]

Proof.

Let $\lambda=r{\boldsymbol{e}}_{1}\in\mathbb{N}^{m|n}_{++}$. Then $X^{(\lambda)}\in\mathcal{S}_{0|1}(r)$ is a highest weight vector, since, for any ${{\bf a}}=(a_{1},\cdots,a_{m+n})\in\mathbb{N}^{m}\times\mathbb{Z}_{2}^{n}$ with $|{{\bf a}}|=r$, $r{\boldsymbol{e}}_{1}-{{\bf a}}=a_{2}({\boldsymbol{e}}_{1}-{\boldsymbol{e}}_{2})+a_{3}({\boldsymbol{e}}_{1}-{\boldsymbol{e}}_{3})+\cdots+a_{m+n}({\boldsymbol{e}}_{1}-{\boldsymbol{e}}_{m+n})$ and

[TABLE]

Hence, $\mathcal{S}_{0|1}(r)$ is generated by an highest weight vector. On the other hand, a reversed sequence in the $E_{i}^{(a_{i})}$’s send $X^{({{\bf a}})}$ back to $X^{({\lambda})}$. Thus, $\mathcal{S}_{0|1}(r)$ is irreducible. The proof for $\mathcal{S}_{1|0}(r)$ is similar. ∎

Consider the tensor product

[TABLE]

where $X_{i,j}$ denotes the $i$th generator of the $j$-th tensor factor. Thus, we may regard ${{\mathcal{S}^{m|n}}}$ as the polynomial superalgebra as indicated by the right hand side of (3.2.1), which has even generators $X_{i,j}$ for all $i,j$ with $\hat{i}+\hat{j}=0$ and odd generators $X_{i,j}$ for all $i,j$ with $\hat{i}+\hat{j}=1$. In particular, we may describe the monomial basis for ${{\mathcal{S}^{m|n}}}$ in terms of the following matrix set:

[TABLE]

For $A=(a_{i,j})\in M(m|n),$ let

[TABLE]

be the $i$-th column of $A$ and let

[TABLE]

The parity of $X^{[A]}$ is given by $\widehat{A}:=\sum_{\widehat{i}+\widehat{j}=1}a_{i,j}$.

Via the coalgebra structure (2.1.1) of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$, ${{\mathcal{S}^{m|n}}}$ becomes a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-module (see the lemma below). Recall also the sign rule: for supermodules $V_{1},V_{2}$ over a superlagbera $U$, if $u_{1},u_{2}\in U,v_{i}\in V_{i}$ with $u_{2},v_{1}$ homogeneous, then

[TABLE]

For $A\in M(m|n),$ $i\in[1,m+n]$, let

[TABLE]

and

[TABLE]

Lemma 3.3**.**

The set $\{X^{[A]}\,|\,A\in M(m|n)\}$ forms a $\mathbb{Q}({\boldsymbol{\upsilon}})$-basis for the $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule ${{\mathcal{S}^{m|n}}}$ which has the following actions:

[TABLE]

where

[TABLE]

Proof.

Let $\Delta^{(N)}=(\Delta\otimes\underbrace{1\otimes\cdots\otimes}_{N-1}1)\circ\cdots\circ(\Delta\otimes 1)\circ\Delta$. Then, for $N=m+n-1$,

[TABLE]

Thus, by (3.0.2) and the sign rule (and, for $h=m$, noting ${\boldsymbol{\upsilon}}_{m}={\boldsymbol{\upsilon}},{\boldsymbol{\upsilon}}_{m+1}={\boldsymbol{\upsilon}}^{-1}$),

[TABLE]

The actions of $F_{h},K_{i}$ can be proved similarly. ∎

Remark 3.4**.**

We remark that these module formulas are easily obtained, but are the key to the determination of the regular representation of $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$. As a comparison, analogous formulas for quantum Schur superalgebras are certain multiplication formuas (see [7, Props 4.4-5]) which are obtained by rather lengthy calculations.

4. The formal power series algebra ${\widetilde{\mathcal{S}}^{m|n}}(V)$

We now extend the module structure on ${{\mathcal{S}^{m|n}}}$ to its formal power series algebra and then focus on a submodule which has a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule structure. We will displayed explicitly the actions on a basis.

Recall from (3.2.1) the polynomial superalgebra ${{{\mathcal{S}^{m|n}}}}$ and its basis $\{X^{[A]}\}_{\mid A\in M(m|n)}$. By turning the direct sum of all $\mathbb{Q}({\boldsymbol{\upsilon}})X^{[A]}$ into a direct product, we obtain the formal power series algebra:

[TABLE]

For clarity of the $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-actions below, we continue to write the elements in ${{\widetilde{\mathcal{S}}^{m|n}}}$ by infinite series in $X^{[A]}$’s. Natrually, the $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-action on ${{\mathcal{S}^{m|n}}}$ extends to ${\widetilde{\mathcal{S}}^{m|n}}$ so that ${\widetilde{\mathcal{S}}^{m|n}}$ becomes a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-module. We now construct a submodule on which a natural super structure can be built.

Let

[TABLE]

For $\lambda\in\mathbb{N}^{m+n},A\in M(m|n)^{\mp}$, let $A+{\lambda}=A+\operatorname{diag}({\lambda})$ and, for ${{\bf j}}\in\mathbb{Z}^{m+n}$, define

[TABLE]

Let $\mathcal{U}(m|n)$ be the subspace of ${\widetilde{\mathcal{S}}^{m|n}}$ spanned by $A({{\bf j}})$ for all $A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{Z}^{m+n}$. Since every $X^{[A+\lambda]}$ in $A({{\bf j}})$ has parity $\widehat{A}$, $\mathcal{U}(m|n)$ has a natural superspace structure $\mathcal{U}(m|n)=\mathcal{U}(m|n)_{0}\oplus\mathcal{U}(m|n)_{1}$. In the rest of the section, we will prove that $\mathcal{U}(m|n)$ is a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule.

Let $\alpha_{h}={\boldsymbol{e}}_{h}-{\boldsymbol{e}}_{h+1},\beta_{h}={\boldsymbol{e}}_{h}+{\boldsymbol{e}}_{h+1}$, $\sigma_{h,m}(i)=\sigma_{h,m}(i,A)$, $f(i)=f(i,A),$ and $g(i)=g(i,A)$ (see (3.2.4) and (3.3.1)).

Theorem 4.1**.**

The superspace $\mathcal{U}(m|n)$ is a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-submodule of ${\widetilde{\mathcal{S}}^{m|n}}$ with basis $\{A({{\bf j}})\mid A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{Z}^{m+n}\}$ and the following explicit actions of $E_{h},F_{h},K_{i}$: for $A=(a_{s,t}),{{\bf j}}=(j_{s})$, $1\leq i\leq m+n$, and $1\leq h<m+n$,

[TABLE]

[TABLE]

where $\blacktriangleright$ (resp., $\blacktriangleleft$) is 0 if $a_{h+1,h}=0$ (resp., $a_{h,h+1}=0$), and is 1 otherwise.

Moreover, it is a $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule.

Proof.

The proof of linear independence is similar to that of [6, Prop. 4.1(2)].

By Lemma 3.3(1),

[TABLE]

Similarly, by Lemma 3.3(2), and noting $\sigma(i,A)=\sigma(i,A+{\lambda})$,

[TABLE]

where

[TABLE]

By (3.2.4), for $i\geq h+1$, we have $f(i,A+{\lambda})=f(i,A)$. Thus,

[TABLE]

and

[TABLE]

Similarly, for $i<h$, $f(i,A+{\lambda})=f(i,A)+{\lambda}_{h}-(-1)^{\delta_{h,m}}{\lambda}_{h+1}$. So, by (2.0.1),

[TABLE]

Finally, for $\Sigma_{3}$ when $a_{h+1,h}>0$, since $f(h,A+{\lambda})=f(h,A)-(-1)^{\delta_{h,m}}\lambda_{h+1}$ and

[TABLE]

it follows that

[TABLE]

proving (4.1.1). (Notice a cancellation for the terms associated to those ${\lambda}$ with ${\lambda}_{h}=0$ when expanding the numerator of the last expression.)

The proof for the action of $F_{h}$ is similar. Finally, the supermodule assertion follows easily from the action formulas. ∎

5. The main result

We are now ready to prove the main result of the paper by the following.

Lemma 5.1**.**

Let $U$ be an algebra over a field ${\mathbbm{k}}$ with generators $g_{i}$, $1\leq i\leq N$. Suppose $Uv$ is a cyclic $U$-module with basis $b_{j}=u_{j}.v,\;j\in J$ ($u_{j}\in U$), and trivial annihilator $\text{\rm ann}_{U}(v)=0$. Then the matrix representations $g_{i}.b_{j}=\sum_{k\in J}{\lambda}_{i,j,k}b_{k}$ of the generators give rise to a presentation of $U$ by basis $\{u_{j}\mid j\in J\}$ and the multiplication formulas:

[TABLE]

Proof.

Since the $U$-module homomorphism $\phi:U\to Uv,u\mapsto u.v$ is an isomorphism, the basis claim is clear and so are the multiplication formulas. ∎

For $A=(a_{i,j})\in M(m|n)$, let

[TABLE]

Following [1, §3.5] or [7, (8.0.1)], define a preorder relation on $M(m|n)$:

[TABLE]

Note that this is a partial order relation on $M(m|n)^{\mp}$. The $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-actions in Theorem 4.1 satisfy certain “triangular relations” relative to $\preceq$. The “lower terms” below means a linear combination of $B({{\bf j}}^{\prime})$ with $B\prec$ the leading matrix.

Lemma 5.2**.**

Let $A=(a_{i,j})\in M(m|n)^{\mp}$, ${{\bf j}}\in\mathbb{Z}^{m+n}$, and $h,k\in[1,m+n]$.

1. (1)

If $h<k$, $A^{\textsf{L}}_{h+1,k+1}=0$, $a_{h,k}=0$, and $a_{h+1,k}\geq a>0$, then, for some $b\in\mathbb{Z}$,

[TABLE] 2. (2)

If $h+1>k$, $A^{\neg}_{h,k-1}=0$, $a_{h+1,k}=0$, and $a_{h,k}\geq a>0$, then, for some $c\in\mathbb{Z}$,

[TABLE]

Proof.

This follows easily from repeatedly applying the actions in Theorem 4.1. For example, the first summation in $E_{h}.A(\mathbf{0})$ contains only the terms $(A-E_{h,i}+E_{h+1,i})(\mathbf{0})$, for some $h+1<i\leq k$, and $A+E_{h,k}-E_{h+1,k}\succ A+E_{h,i}-E_{h+1,i}$ for all $i<h$ or $h+1<i<k$ if it occurs in the first two summations. One sees also $A+E_{h,k}-E_{h+1,k}\succ A-E_{h+1,h},A+E_{h,h+1}$. Hence, $E_{h}.A(\mathbf{0})=\pm{\boldsymbol{\upsilon}}^{*}(A+E_{h,k}-E_{h+1,k})(\mathbf{0})+(\text{lower terms}).$ Inductively, $E_{h}^{a}.A(\mathbf{0})=\pm{\boldsymbol{\upsilon}}^{b}[a]^{!}(A+aE_{h,k}-aE_{h+1,k})(\mathbf{0})+(\text{lower terms}).$ Hence, the desired formulas follows. ∎

Theorem 5.3**.**

The $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-supermodule $\mathcal{U}(m|n)$ is a cyclic module generated by $O(\bf 0)$, where $O\in M(m|n)$ and $\bf 0\in\mathbb{N}^{m+n}$ are the zero elements, and the module homomorphism

[TABLE]

is an isomorphism.

Proof.

By Lemma 5.2, we may use an argument similar to that for [1, Proposition 3.9]). Consider reduced expressions of the longest elements in the symmetric groups ${\mathfrak{S}}_{\{1,2,\ldots,j\}}$ for $j=2,3,\ldots,m+n$ and ${\mathfrak{S}}_{\{k,k+1,\ldots,m+n\}}$ for $k=1,2,\ldots,m+n-1$:

[TABLE]

For any $A=(a_{i,j})\in M(m|n)^{\mp}$ and ${{\bf j}}\in\mathbb{Z}^{m+n}$, let

[TABLE]

and let $\mathfrak{m}^{A,{{\bf j}}}=\mathfrak{m}^{-}_{\bullet}K^{{\bf j}}\mathfrak{m}^{+}_{\bullet},$ where $\mathfrak{m}^{-}_{\bullet}=\mathfrak{m}^{-}_{1}\mathfrak{m}^{-}_{2}\cdots\mathfrak{m}^{-}_{m+n-1}$, $K^{{\bf j}}=K_{1}^{j_{1}}\cdots K_{m+n}^{j_{m+n}}$, and $\mathfrak{m}^{+}_{\bullet}=\mathfrak{m}^{+}_{m+n}\cdots\mathfrak{m}^{+}_{3}\mathfrak{m}^{+}_{2}$. For example, if $m=2,n=2,$ $A\in M(2|2)^{\pm},{{\bf j}}\in\mathbb{Z}^{4}$, then

[TABLE]

Repeatedly applying Lemma 5.2, we obtain

[TABLE]

In fact, $\mathfrak{m}^{+}_{2}.O(\bf 0)$ has the leading term $(a_{1,2}E_{1,2})(\bf 0)$, $\mathfrak{m}^{+}_{3}\mathfrak{m}^{+}_{2}.O(\bf 0)$ has the leading term $(a_{1,2}E_{1,2}+a_{1,3}E_{1,3}+a_{2,3}E_{2,3})(\bf 0)$, … , $\mathfrak{m}^{+}_{m+n}\cdots\mathfrak{m}^{+}_{3}\mathfrak{m}^{+}_{2}.O(\bf 0)$ has the leading term $A^{+}(\bf 0)$, where $A^{+}$ is the upper triangular part of $A$. Similarly, $K^{{\bf j}}\mathfrak{m}^{+}_{\bullet}.O(\bf 0)$ has the leading term $A^{+}({{\bf j}})$, $\mathfrak{m}^{-}_{m+n-1}K^{{\bf j}}\mathfrak{m}^{+}_{\bullet}.O(\bf 0)$ has the leading term $(A^{+}+a_{m+n,m+n-1}E_{m+n,m+n-1})({{\bf j}})$, and so on.

Since $\{A({{\bf j}})\mid A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{N}^{m+n}\}$ forms a basis for $\mathcal{U}(m|n)$ by Theorem 4.1, the triangular relation above implies that $\{\mathfrak{m}^{A,{{\bf j}}}.O({\bf 0})\mid A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{N}^{m+n}\}$ are linearly independent. Hence, the module homomorphism (5.3.1) must be an isomorphism. ∎

The theorem above gives immediately a presentation for $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$.

Corollary 5.4**.**

The supergroup $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ contains a basis

[TABLE]

such that $E_{h}=E_{h,h+1}(\bf 0)$ $F_{h}=E_{h+1,h}(\bf 0)$, and $K_{i}=O({\boldsymbol{e}}_{i})$, and the $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$-action formulas given in Theorem 4.1 become the multiplication formulas of the basis elements $A({{\bf j}})$ by the generators.

Proof.

By the module isomorphism (5.3.1) (and by abuse of notation), let $A({{\bf j}}):=f^{-1}A({{\bf j}})$. Since $E_{h}.O(\mathbf{0})=E_{h,h+1}(\bf 0)$, $F_{h}.O(\mathbf{0})=E_{h+1,h}(\bf 0)$, and $K_{i}.O(\mathbf{0})=O({\boldsymbol{e}}_{i})$, we have $E_{h}=E_{h,h+1}(\bf 0)$, $F_{h}=E_{h+1,h}(\bf 0)$, and $K_{i}=O({\boldsymbol{e}}_{i})$. The assertion now follows from Lemma 5.1. ∎

Remark 5.5**.**

The presentation above for $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ coincides with the one from [1, Lemma 5.3] (or [4, Theorem 14.8]) in the quantum $\mathfrak{gl}_{n}$ case and with the one in [7, Thm 8.4] in general after a sign modification given below.

For any $A=(a_{i,j})\in M(m|n)$, let111This number $\overline{A}$ is different from the number $\bar{A}$ defined in [7, (5.0.1)], where the super grading structure on the tensor space is under consideration.

[TABLE]

Lemma 5.6**.**

For $\lambda\in\mathbb{N}^{m+n},A=(a_{i,j})\in M(m|n)^{\mp}$ and $1\leq h,k\leq m+n$ with $h<m+n$, then

• (1)

$\overline{A+\lambda}=\overline{A}$;

• (2)

\overline{A}+\delta_{h,m}\sigma(k,A)=\overline{A+E_{h,k}-E_{h+1,k}}+{\delta_{h,m}\displaystyle\bigg{(}\sum_{{\begin{subarray}{c}i>m\\ j\leq min\{k-1,m\}\end{subarray}}}a_{i,j}-{\delta^{>}_{k,m}\sum_{{\begin{subarray}{c}i\leq m\\ j>k\end{subarray}}}a_{i,j}\bigg{)}}}.**

Here, $\delta^{>}_{k,m}=1$ if $k>m$ and [math] otherwise.

Proof.

If we write $A={X\;\;Q\choose Q^{\prime}\;\;Y}$ in blocks as in (3.2.2), then the entries involved in $\overline{A}$ are all in $Q$. Thus, (1) and (2) for $h\neq m$ or $h=m,k\leq m$ are all clear. Assume now $h=m,k>m$. Then, by definition,

[TABLE]

as desired. ∎

Let ${\underline{A}}({{\bf j}})=(-1)^{\bar{A}}A({{\bf j}})$ for all $A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{Z}^{m+n}$.

Theorem 5.7**.**

Modifying the multiplication formulas in Theorem 4.1 by using the basis $\{{\underline{A}}({{\bf j}})\mid A\in M(m|n)^{\mp},{{\bf j}}\in\mathbb{Z}^{m+n}\}$ for the supergroup $U_{\boldsymbol{\upsilon}}(\mathfrak{gl}_{m|n})$ yields exactly the same formulas as given in [7, Thm 8.4].

Proof.

We first observe that the generators $E_{h}=E_{h,h+1}({\bf 0})=\underline{E_{h,h+1}}(\bf 0)$, etc. are part of the new basis. After multiplying both sides of the multiplication formulas in Theorem 4.1 by $(-1)^{\overline{A}}$ and applying Lemma 5.6, the sign term becomes $(-1)^{s(h,i)}$ with

[TABLE]

This number $s(h,i)$ is exactly the same number $\varepsilon_{h,h+1}\sigma(i)$ defined in [7, (5.5.1-2)] and used in the multiplication formulas in [7, Thm 8.4]. ∎

Acknowledgement. The authors would like to thank the referee for a correction on the parity computation involved in Lemma 3.3. This eventually led to a significant improvement of the paper.