Explicit construction of irreducible modules for U_q(gl_n)

TL;DR

This paper introduces a method to explicitly construct irreducible modules for the quantum group U_q(gl_n) using combinatorial admissible relations, extending finite-dimensional and generic modules.

Contribution

It provides a systematic construction of irreducible U_q(gl_n)-modules from admissible sets of relations, broadening the understanding of module representations.

Findings

Constructed new families of irreducible modules from admissible relations.

Unified finite-dimensional and generic modules within this framework.

Proved that each admissible set yields an irreducible module.

Abstract

We construct new families of U_q(gl_n)-modules by continuation from finite dimensional representations. Each such module is associated with a combinatorial object - admissible set of relations defined in \cite{FRZ}. More precisely, we prove that any admissible set of relations leads to a family of irreducible U_q(gl_n)-modules. Finite dimensional and generic modules are particular cases of this construction.

Equations

$$(x{)}_q=\frac{q^x-1}{q-1},[x{]}_q=\frac{q^x-q^{-x}}{q-q^{-1}}.$$

$$q^0=1,q^h{q}^{h^{\prime }}=q^{h+h^{\prime }}(h,h^{\prime }\in \mathfrak{h}),$$

$$q^h{e}_i{q}^{-h}=q^{⟨h,{\alpha }_i⟩}{e}_i,$$

$$q^h{f}_i{q}^{-h}=q^{-⟨h,{\alpha }_i⟩}{f}_i,$$

$$e_i{f}_j-f_j{e}_i={\delta }_{ij}\frac{q^{{\alpha }_i}-q^{-{\alpha }_i}}{q-q^{-1}},$$

$$e_i^2{e}_j-(q+q^{-1})e_i{e}_j{e}_i+e_j{e}_i^2=0(|i-j|=1),$$

$$f_i^2{f}_j-(q+q^{-1})f_i{f}_j{f}_i+f_j{f}_i^2=0(|i-j|=1),$$

$$e_i{e}_j=e_j{e}_i,f_i{f}_j=f_j{f}_i(|i-j|>1).$$

$$M=\underset{\mathsf{m}\in \mathrm{Specm}{\Gamma }_q}{⨁}M(\mathsf{m}),$$

$$\begin{array}{rl}{q}^{{ϵ}_k}(T(L))& =q^{a_k}T(L),a_k=\sum _{i=1}^k{l}_{k,i}-\sum _{i=1}^{k-1}{l}_{k-1,i}+k,k=1,\dots ,n,\\ e_k(T(L))& =-\sum _{j=1}^k\frac{{\prod }_i[l_{k+1,i}-l_{k,j}{]}_q}{{\prod }_{i\ne j}[l_{k,i}-l_{k,j}{]}_q}T(L+{\delta }^{kj}),\\ f_k(T(L))& =\sum _{j=1}^k\frac{{\prod }_i[l_{k-1,i}-l_{k,j}{]}_q}{{\prod }_{i\ne j}[l_{k,i}-l_{k,j}{]}_q}T(L-{\delta }^{kj}).\end{array}$$

$${\gamma }_{mk}(L)=(k{)}_{q^{-2}}!(m-k{)}_{q^{-2}}!q^{k(k+1)+\frac{m(m-3)}{2}}\sum _{\tau }{q}^{{\sum }_{i=1}^k{l}_{m\tau (i)}-{\sum }_{i=k+1}^m{l}_{m\tau (i)}}$$

$$\mathcal{R}$$

$${\mathcal{R}}^{\ge }$$

$${\mathcal{R}}^{>}$$

$${\mathcal{R}}^0$$

$$\{(i,j)\ge (i_1,j_1),(i_1,j_1)\ge (i_2,j_2),\cdots ,(i_m,j_m)\ge (r,s)\}$$

$$\begin{array}{rl}& \{(k,i)>(k+1,s)\ge (k,j),(k,i)\ge (k-1,t)>(k,j)\}\subseteq \mathcal{C},\\ & \{(k,i)>(k+1,s),(k+1,t)\ge (k,j)\}\subseteq \mathcal{C}.\end{array}$$

$$\mathcal{S}$$

$$e_{ki}(L)=\left\{\begin{array}[]{cc}0,&\text{ if }T(L)\notin\mathcal{B}_{\mathcal{C}}(T(L))\\ -\frac{\prod_{j=1}^{k+1}[l_{ki}-l_{k+1,j}]_{q}}{\prod_{j\neq i}^{k}[l_{ki}-l_{kj}]_{q}},&\text{ if }T(L)\in\mathcal{B}_{\mathcal{C}}(T(L))\end{array}\right.$$

$$f_{ki}(L)=\left\{\begin{array}[]{cc}0,&\text{ if }T(L)\notin\mathcal{B}_{\mathcal{C}}(T(L))\\ \frac{\prod_{j=1}^{k-1}[l_{ki}-l_{k-1,j}]_{q}}{\prod_{j\neq i}^{k}[l_{ki}-l_{kj}]_{q}},&\text{ if }T(L)\in\mathcal{B}_{\mathcal{C}}(T(L))\end{array}\right.$$

$$h_{k}(L)=\left\{\begin{array}[]{cc}0,&\text{ if }T(L)\notin\mathcal{B}_{\mathcal{C}}(T(L))\\ q^{2\sum_{i=1}^{k}l_{ki}-\sum_{i=1}^{k-1}l_{k-1,i}-\sum_{i=1}^{k+1}l_{k+1,i}-1},&\text{ if }T(L)\in\mathcal{B}_{\mathcal{C}}(T(L))\end{array}\right.$$

$$\Phi(L,z_{1},\ldots,z_{m})=\left\{\begin{array}[]{cc}1,&\text{ if }T(L+z_{1}+\ldots+z_{t})\in\mathcal{B}_{\mathcal{C}}(T(L))\text{ for any }t\\ 0,&\text{ otherwise}.\end{array}\right.$$

$$\underset{v\to l}{\lim }{e}_{kj}(v)f_{kj}(v+{\delta }^{kj})=0.$$

$$\underset{v\to l}{\lim }{f}_{kj}(v)e_{kj}(v-{\delta }^{kj})=0.$$

$$\underset{v\to l}{\lim }{e}_{kj}(v)f_{kj}(v+{\delta }^{k,j})-f_{ki}(v)e_{ki}(v-{\delta }^{k,i})=0.$$

$$\underset{v\to l}{\lim }{e}_{kj}(v)f_{kj}(v+{\delta }^{kj})=0.$$

$$\underset{v\to l}{\lim }{e}_{kj}(v)f_{kj}(v+{\delta }^{k,j+1})-f_{kj}(v)e_{kj}(v-{\delta }^{k,j})=0.$$

$$(e_i^2{e}_j-(q+q^{-1})e_i{e}_j{e}_i+e_j{e}_i^2)T(R)$$

$$=$$

$$+$$

$$-$$

$$\begin{array}{rl}(e_i{f}_j-f_j{e}_i)T(R)=& \sum _{r=1}^j\sum _{s=1}^i\Phi (R,-{\delta }^{jr})f_{jr}(R)e_{is}(R+{\delta }^{jr})T(R-{\delta }^{jr}+{\delta }^{is})\\ -& \sum _{r=1}^j\sum _{s=1}^i\Phi (R,{\delta }^{is})e_{is}(R)f_{jr}(R+{\delta }^{is})T(R-{\delta }^{jr}+{\delta }^{is}).\end{array}$$

$$\underset{v\to l}{\lim }\left(\sum _{r=1}^k\sum _{s=1}^k{f}_{kr}(v)e_{ks}(v+{\delta }^{kt})-\sum _{r=1}^k\sum _{s=1}^k{e}_{ks}(v)f_{kr}(v+{\delta }^{ks})\right)$$

$$=\underset{v\to R}{\lim }{h}_k(v)=h_k(R).$$

$$\mathcal{I}(T(R))=\{T(S)\in \mathcal{B}(T(R)):{\Omega }^{+}(T(S))={\Omega }^{+}(T(R))\},$$

$${\Omega }^{+}(T(S)):=\{(i,j,k)|s_{i,j}-s_{i-1,k}\in \frac{1(q)}{2}+{\mathbb{Z}}_{\ge 0}\}.$$

Full text

1. Introduction

Gelfand and Graev [13] proposed a method of constructing of $\operatorname{\mathfrak{gl}}_{n}$-modules which extend finite dimensional modules and admit a basis of tableaux with the standard action of the generators of the Lie algebra [14]. This construction is based on a choice of certain relations satisfied by the entries of the Gelfand-Tsetlin tableaux. Lemire and Patera [20] conjectured sufficient conditions under which the Gelfand-Graev defines in fact a module, proving it for $n=3$ and $n=4$. In [11] the authors proved this conjecture and extended the construction for a larger class of irreducible $\operatorname{\mathfrak{gl}}_{n}$-modules. The purpose of this letter is to show how to deform this construction and obtain new large families of irreducible modules for the quantum group $U_{q}(\mathfrak{gl}_{n})$. Infinite dimensional generic and finite dimensional modules are particular cases of this construction. New irreducible modules are presented explicitly with a basis consisting of certain tableaux and with explicit action of the generators of $U_{q}(\mathfrak{gl}_{n})$ generalizing the construction of finite dimensional representations [27]. Having such an explicit construction will be useful for possible applications.

Constructed modules belong to the category of Gelfand-Tsetlin modules with a diagonalizable action of the Gelfand-Tsetlin subalgebra. For $\operatorname{\mathfrak{gl}}_{n}$ the theory of Gelfand-Tsetlin modules has origin in the classical paper of Gelfand and Tsetlin [14]. It is related to many concepts arizing in Mathematics and Physics, see for example [18], [19], [15], [4], [16], [17], [1], [2], [9]. The general theory of Gelfand-Tsetlin modules for $\operatorname{\mathfrak{gl}}_{n}$ was developed in [3], [24], [10], [21], [22], [23], [5], [6], [7] and references therein. For $U_{q}(\mathfrak{gl}_{n})$ certain families of Gelfand-Tsetlin modules were constructed in [25] and [12], while the general theory was developed in [8].

Current letter provides new information about Gelfand-Tsetlin modules for

$U_{q}(\mathfrak{gl}_{n})$. The paper is organized as follows. Section 2 contains some preliminary information. In Section 3 we introduce our main technical tools - admissible sets of relations and realizable sets of relations. To any realizable set of relations we associate a family of $U_{q}(\mathfrak{gl}_{n})$-modules. We prove the main result of this letter stating that any admissible set of relations is a realizable set of relations (Theorem 3.9). A certain effective method (RR-method) of constructing the admissible relation is described in Theorem 3.11. Finally, in Section 4 we study the action of the generators of the Gelfand-Tsetlin subalgebra on modules associated with admissible sets of relations. The Gelfand-Tsetlin subalgebra $\Gamma_{q}$ is diagonalizable on all constructed modules (Theorem 4.1), moreover, it separates the basis tableaux (Proposition 4.4). Using the action of $\Gamma_{q}$ we obtain a criterion of irreducibility of constructed admissible modules: irreducible modules correspond to maximal sets of admissible relations (Theorem 4.5).

We now fix some notation and conventions. Throughout the paper we fix an integer $n\geq 2$ and $q\in\mathbb{C}$ which is not root of unity. The ground field will be ${\mathbb{C}}$. By $U_{q}$ we denote the quantum enveloping algebra of $\operatorname{\mathfrak{gl}}_{n}$. We fix the standard Cartan subalgebra $\mathfrak{h}$, the standard triangular decomposition and the corresponding basis of simple roots $\alpha_{1},\ldots,\alpha_{n-1}$. The weights of $U_{q}$ will be written as $n$-tuples $(\lambda_{1},\ldots,\lambda_{n})$. For a commutative ring $R$, by ${\rm Specm}\,R$ we denote the set of maximal ideals of $R$. For $i>0$ by $S_{i}$ we denote the $i$th symmetric group. Let $1(q)$ be the set of all complex $x$ such that $q^{x}=1$. Finally, for any complex number $x$, we set

[TABLE]

Acknowledgements. V.F. is supported in part by CNPq (301320/2013-6) and by Fapesp (2014/09310-5). J. Z. is supported by Fapesp (2015/05927-0).

2. Preliminaries

We define $U_{q}$ as a unital associative algebra generated by $e_{i},f_{i}(1\leq i\leq n)$ and $q^{h}(h\in\mathfrak{h})$ with the following relations:

[TABLE]

The quantum special linear algebra $U_{q}(sl_{n})$ is the subalgebra of $U_{q}$ generated by $e_{i},\ f_{i},\ q^{\pm\alpha_{i}}(i=1,2,\ldots,n-1)$.

We denote by $(U_{m})_{q}$ the quantum universal enveloping algebra of $\operatorname{\mathfrak{gl}}_{m}$. We have the following chain $(U_{1})_{q}\subset$ $(U_{2})_{q}\subset$ $\ldots$ $\subset(U_{n})_{q}$. Let $Z_{m}$ denotes the center of $(U_{m})_{q}$. The subalgebra of $U_{q}$ generated by $\{Z_{m}\,|\,m=1,\ldots,n\}$ will be called the Gelfand-Tsetlin subalgebra of $U_{q}$ and will be denoted by ${{\Gamma}}_{q}$ [8], [12].

Definition 2.1**.**

A finitely generated $U_{q}$-module $M$ is called a Gelfand-Tsetlin module (with respect to ${\Gamma}_{q}$) if

[TABLE]

where $M({\mathsf{m}})=\{v\in M|{\mathsf{m}}^{k}v=0\text{ for some }k\geq 0\}$.

For a vector $L=(l_{ij})$ in $\mathbb{C}^{\frac{n(n+1)}{2}}$, by $T(L)$ we will denote the following array with entries $\{l_{ij}:1\leq j\leq i\leq n\}$

$l_{n1}$ $l_{n2}$          $\cdots$           $l_{n,n-1}$ $l_{nn}$

$l_{n-1,1}$              $\cdots$               $l_{n-1,n-1}$

$\cdots$         $\cdots$         $\cdots$

$l_{21}$ $l_{22}$

$l_{11}$

such an array will be called a Gelfand-Tsetlin tableau of height $n$. For any $1\leq j\leq i\leq n-1$, the vector $\delta^{ij}\in{\mathbb{Z}}^{\frac{n(n+1)}{2}}$ is defined by $(\delta^{ij})_{ij}=1$ and all other $(\delta^{ij})_{k\ell}$ are zero. Finally, a Gelfand-Tsetlin tableau of height $n$ is called standard if $l_{ki}-l_{k-1,i}\in\mathbb{Z}_{\geq 0}$ and $l_{k-1,i}-l_{k,i+1}\in\mathbb{Z}_{>0}$ for all $1\leq i\leq k\leq n$.

Recall the quantum version of the classical result of Gelfand and Tsetlin which provides an explicit basis in the finite dimensional case.

Theorem 2.2** ([27], Theorem 2.11 and [12], Proposition 4.3).**

Let $L(\lambda)$ be the finite dimensional irreducible module over $U_{q}$ of highest weight $\lambda=(\lambda_{1},\ldots,\lambda_{n})$, where $\lambda_{i}-\lambda_{i+1}\in\mathbb{Z}_{\geq 0}$. Then there exist a basis of $L(\lambda)$ consisting of all standard tableaux $T(L)$ with fixed top row $l_{nj}=\lambda_{j}-j$. Moreover, the action of the generators of $U_{q}$ on $L(\lambda)$ is given by the Gelfand-Tsetlin formulae:

[TABLE]

Moreover, the generators $c_{mk}$ of $\Gamma_{q}$ acts on $T(L)$ as multiplication by

[TABLE]

where $\tau\in S_{m}$ is such that $\tau(1)<\cdots<\tau(k),\ \tau(k+1)<\cdots<\tau(m)$.

3. Admissible relations

Set $\mathfrak{V}:=\{(i,j)\ |\ 1\leq j\leq i\leq n\}$. We will consider relations between elements of $\mathfrak{V}$ of the form $(i,j)\geq(s,t)$ or $(i,j)>(s,t)$. More precisely, we will consider subsets of the following set of relations:

[TABLE]

where

[TABLE]

Let $\mathcal{C}$ be a subset of $\mathcal{R}$. Denote by $\mathfrak{V}(\mathcal{C})$ the set of all $(i,j)$ in $\mathfrak{V}$ such that $(i,j)\geq(\text{respectively }>,\ \leq,\ <)\ (r,s)\in\mathcal{C}$ for some $(r,s)$. Let $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ be two subsets of $\mathcal{C}$. We say that $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are disconnected if $\mathfrak{V}(\mathcal{C}_{1})\cap\mathfrak{V}(\mathcal{C}_{2})=\emptyset$, otherwise $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are connected. $\mathcal{C}$ is called decomposable if it can be decomposed into the union of two disconnected subsets of $\mathcal{R}$, otherwise $\mathcal{C}$ is called indecomposable.

Note that any subset of $\mathcal{R}$ is a union of disconnected indecomposable sets and such decomposition is unique.

Definition 3.1**.**

Let $\mathcal{C}$ be any subset of $\mathcal{R}$. Given $(i,j),\ (r,s)\in\mathfrak{V}(\mathcal{C})$ we will write:

• (i)

$(i,j)\succeq_{\mathcal{C}}(r,s)$* if, there exists $\{(i_{1},j_{1}),\ldots,(i_{m},j_{m})\}\subseteq\mathfrak{V}(\mathcal{C})$ such that*

[TABLE]

• (ii)

We write $(i,j)\succ_{\mathcal{C}}(r,s)$ if there exists $\{(i_{1},j_{1}),\ldots,(i_{m},j_{m})\}\subseteq\mathfrak{V}(\mathcal{C})$ such that in the condition 15, at least one of the inequalities is $>$.

Given another set of relations $\mathcal{C}^{\prime}$, we say that $\mathcal{C}$** implies $\mathcal{C}^{\prime}$*** if whenever we have $(i,j)\succ_{\mathcal{C}^{\prime}}(r,s)$ (respectively $(i,j)\succeq_{\mathcal{C}^{\prime}}(r,s)$) we also have $(i,j)\succ_{\mathcal{C}}(r,s)$ (respectively $(i,j)\succeq_{\mathcal{C}}(r,s)$).*

Definition 3.2**.**

Let $\mathcal{C}$ be an indecomposable subset of $\mathcal{R}$. A subset of $\mathcal{C}$ of the form $\{(k,i)\geq(k-1,t),\ (k-1,s)>(k,j)\}$ with $i<j$ and $s<t$ will be called a cross.

We will define now our main concept which is a slight modification of the definition of an admissible set in [11].

Definition 3.3**.**

Let $\mathcal{C}$ be an indecomposable set. We say that $\mathcal{C}$ is admissible if it satisfies the following conditions:

• (i)

For any $1\leq k-1\leq n$, we have $(k,i)\succ_{\mathcal{C}}(k,j)$ only if $i<j$;

• (ii)

$(n,i)\succeq_{\mathcal{C}}(n,j)$* only if $i<j$;*

• (iii)

There is not cross in $\mathcal{C}$;

• (iv)

For every $(k,i)\ (k,j)\in\mathfrak{V}(\mathcal{C})$ with $1\leq k\leq n-1$ there exists $s<t$ such that one of the following holds

[TABLE]

An arbitrary set $\mathcal{C}$ is admissible if every indecomposable subset of $\mathcal{C}$ is admissible.

Denote by $\mathfrak{F}$ the set of all indecomposable admissible subsets. Note that we obtain the same admissible sets as in [11]. In the following we will use the admissible sets to construct modules for $U_{q}$.

3.1. Tableaux realization of admissible sets of relations

In this section we will describe $\mathbb{C}$-vector spaces associated with sets of relations $\mathcal{C}$ with Gelfand-Tsetlin tableaux as a bases. We will prove that we have a structure of a $U_{q}$-module on such space with the action of the generators of $U_{q}$ given by the Gelfand-Tsetlin formulas (9).

Definition 3.4**.**

Let $\mathcal{C}$ be any subset of $\mathcal{R}$ and $T(L)$ any Gelfand-Tsetlin tableau. Recall that $1(q):=\{x\in\mathbb{C}\ |\ q^{x}=1\}$.

• (i)

We say that $T(L)$ satisfies a relation $(i,j)\geq(r,s)$ (respectively, $(i,j)>(r,s)$) if $l_{ij}-l_{st}\in\mathbb{Z}_{\geq 0}+\frac{1(q)}{2}\ (\text{respectively, }l_{ij}-l_{st}\in\mathbb{Z}_{>0}+\frac{1(q)}{2})$.

• (ii)

We say that a Gelfand-Tsetlin tableau $T(L)$ satisfies $\mathcal{C}$ if $T(L)$ satisfies all the relations in $\mathcal{C}$ and $l_{ki}-l_{kj}\in\mathbb{Z}+\frac{1(q)}{2}$ only if $(k,i)$ and $(k,j)$ in the same indecomposable subset of $\mathfrak{V}(\mathcal{C})$. In this case we call $T(L)$ a $\mathcal{C}$-realization.

• (iii)

$\mathcal{C}$* is a maximal set of relations for $T(L)$ if $T(L)$ satisfies $\mathcal{C}$ and whenever $T(L)$ satisfies a set of relations $\mathcal{C}^{\prime}$ we have that $\mathcal{C}$ implies $\mathcal{C}^{\prime}$.*

• 2.

If $T(L)$ satisfies $\mathcal{C}$ we denote by ${\mathcal{B}}_{\mathcal{C}}(T(L))$ the set of all tableaux of the form $T(L+z)$ ($z\in\mathbb{Z}^{\frac{n(n-1)}{2}}$) that satisfy $\mathcal{C}$, and by $V_{\mathcal{C}}(T(L))$ the complex vector space spanned by ${\mathcal{B}}_{\mathcal{C}}(T(L))$.

Example 3.5**.**

Set

[TABLE]

It follows from the definition that $\emptyset$ and $\mathcal{S}$ are admissible sets of relations. Moreover, the set of all tableaux satisfying $\mathcal{S}$ coincides with the set of all standard tableaux and the set of all tableaux satisfying $\emptyset$ coincide with the set of all generic tableaux.

Our goal is to show that any admissible set of relations $\mathcal{C}$ leads to a family of $U_{q}$-modules. In fact, for any $\mathcal{C}$-realization $T(L)$ we will prove that $V_{\mathcal{C}}(T(L))$ is a Gelfand-Tsetlin module with the action of the generators of $U_{q}$ given by the Gelfand-Tsetlin formulas (9). For this we will need some technical lemmas.

Set

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We will denote by $T(v)$ the tableau with variable entries $v_{ij}$.

Lemma 3.6**.**

Let $\mathcal{C}\in\mathfrak{F}$, $T(L)$ any tableau satisfying $\mathcal{C}$.

• (i)

If $T(L+\delta^{kj})\notin\mathcal{B}_{\mathcal{C}}(T(L))$ and $l_{k,i}-l_{kj}\notin 1+\frac{1(q)}{2}$ for any $i$, then

[TABLE]

• (ii)

If $T(L-\delta^{kj})\notin\mathcal{B}_{\mathcal{C}}(T(L))$ and $l_{k,j}-l_{k,i}\notin 1+\frac{1(q)}{2}$ for any i, then

[TABLE]

• (iii)

If $l_{k,i}-l_{k,j}\notin 1+\frac{1(q)}{2}$, then $T(L+\delta^{k,j}),T(L-\delta^{k,i})\notin\mathcal{B}_{\mathcal{C}}(T(L))$, and

[TABLE]

Proof.

Since $T(L+\delta^{kj})\notin\mathcal{B}_{\mathcal{C}}(T(L))$, we have $\{(k+1,s)\geq(k,j)\}\subseteq\mathcal{C}$ or $\{(k-1,t)>(k,j)\}\subseteq\mathcal{C}$. Suppose $\{(k+1,s)\geq(k,j)\}\subseteq\mathcal{C}$ and $T(L+\delta^{kj})\notin\mathcal{B}_{\mathcal{C}}(T(L))$. Then $l_{k+1,s}-l_{k,j}\notin\frac{1(q)}{2}$ and by direct computation one has $\lim\limits_{v\rightarrow l}e_{kj}(v)f_{kj}(v+\delta^{kj})=0$. Suppose $T(L+\delta^{kj})$ does not satisfies the relation $l_{k-1,t}-l_{k,j}\in\mathbb{Z}_{>0}+\frac{1(q)}{2}$. Then we have $l_{k-1,t}-l_{k,j}\in 1+\frac{1(q)}{2}$ and

[TABLE]

The proof of (ii) is similar to (i).

It is clear that $T(L-\delta^{k,j}),\ T(L+\delta^{k,j+1})\notin\mathcal{B}_{\mathcal{C}}(T(L))$ if $l_{k,j}-l_{k,j+1}\in 1+\frac{1(q)}{2}$. It is easy to see that $\#\{l_{k+1,i^{\prime}},l_{k-1,j^{\prime}}\ |\ l_{k+1,i^{\prime}}-l_{kj}\in\frac{1(q)}{2},\ l_{k-1,j^{\prime}}-l_{ki}\in\frac{1(q)}{2}\}\geq 2$. By direct computation one has

[TABLE]

∎

Lemma 3.7** ([11], Lemma 4.21).**

Let $\mathcal{C}\in\mathfrak{F}$, $z^{(1)},z^{(2)}\in\mathbb{Z}^{\frac{n(n-1)}{2}}$. Denote $I_{1}=\{(i,j)\ |\ z_{ij}^{(1)}\neq 0\}$, $I_{2}=\{(i,j)\ |\ z_{ij}^{(2)}\neq 0\}$. If $I_{1}\cap I_{2}=\emptyset$ and for any $(i_{1},j_{1})\in I_{1}$, $(i_{2},j_{2})\in I_{2}$ there is no relation between $(i_{1},j_{1})$ and $(i_{2},j_{2})$, then $T(R+z^{(1)}+z^{(2)})\neq 0$ if and only if $T(R+z^{(1)})\neq 0$ and $T(R+z^{(2)})\neq 0$.

3.2. $U_{q}$-modules defined by admissible relations

Definition 3.8**.**

Let $\mathcal{C}$ be a subset of $\mathcal{R}$. We call $\mathcal{C}$ realizable if for any tableau $T(L)$ satisfying $\mathcal{C}$, the vector space $V_{\mathcal{C}}(T(L))$ has a structure of a $U_{q}$-module, endowed with the action of $U_{q}$ given by the Gelfand-Tsetlin formulas (9).

Theorem 3.9**.**

If $\mathcal{C}$ is a union of disconnected sets from $\mathfrak{F}$ then $\mathcal{C}$ is realizable.

Proof.

Let $\mathcal{C}$ be a union of disconnected sets from $\mathfrak{F}$. It is sufficient to consider the case when $\mathcal{C}$ is a union of two disconnected subsets from $\mathfrak{F}$. Suppose $\mathcal{C}=\mathcal{C}_{1}\cup\mathcal{C}_{2}$.

Let $T(L)$ be any $\mathcal{C}$-realization. In order to prove that $V_{\mathcal{C}}(T(L))$ is a $U_{q}$-module one needs to verify all the defining relations (2–7) for any $T(R)\in{\mathcal{B}}_{\mathcal{C}}(T(L))$.

First we show that $(e_{i}^{2}e_{j}-(q+q^{-1})e_{i}e_{j}e_{i}+e_{j}e_{i}^{2})T(R)=0\,(|i-j|=1)$.

[TABLE]

Now we consider the coefficients by nonzero tableau $T(R+\delta^{jr}+\delta^{is}+\delta^{it})$.

Let $s=t$.

• (a)

Suppose there is no relation between $(i,s)$ and $(j,r)$. Then $\Phi(R,\delta^{jr},\delta^{is})=\Phi(R,\delta^{is},\delta^{is})=\Phi(R,\delta^{is},\delta^{jr})=1$ by Lemma 3.7. Then the coefficient of $T(R+\delta^{jr}+2\delta^{is})$ is the limit of the coefficient of $T(v+\delta^{jr}+2\delta^{is})$ when $v\rightarrow R$ (here $T(v)$ again is a tableau with variable entries). Thus the coefficient of $T(R+\delta^{jr}+2\delta^{is})$ is zero.

• (b)

Suppose there exists a relation between $(i,s)$ and $(j,r)$. Without loss of generality we assume that this relation is $\mathcal{C^{\prime}}=\{(i,s)\geq(j,r)\}$, Let $T(v^{\prime})$ be the tableau with $v^{\prime}_{s^{\prime}t^{\prime}}=l_{s^{\prime}t^{\prime}}$ if $(s^{\prime},t^{\prime})=(i,s)\text{ or }(j,r)$, and variable entries otherwise. Then $T(v^{\prime})$ is a $\mathcal{C^{\prime}}$-realization and $V_{\mathcal{C}^{\prime}}(T(v^{\prime}))$ is a module for arbitrary generic values of free variables in $v^{\prime}$. Let $z^{(1)},z^{(2)}\in\{\delta^{jr},\delta^{is}\}$. Then $\Phi(R,z^{(1)},z^{(2)})=\Phi(v,z^{(1)},z^{(2)})$ where $z^{(1)}=z^{(2)}$ only if $z^{(1)}=z^{(2)}=\delta^{is}$. Therefore the coefficient of $T(R+\delta^{jr}+2\delta^{is})$ is the limit of the coefficient of $T(v+\delta^{jr}+2\delta^{is})$ when $v\rightarrow R$, hence, it is zero.

Suppose $s\neq t$. Then there is no relation between $(i,s)$ and $(i,t)$.

• (a)

Suppose there is no relation between $(j,r)$ and $(i,s)$ or between $(j,r)$ and $(i,t)$. Then the value of the function $\Phi$ that appears along with $T(R+\delta^{jr}+\delta^{is}+\delta^{it})$ is $1$ by Lemma 3.7. Thus the coefficient of $T(R+\delta^{jr}+\delta^{is}+\delta^{it})$ is zero similarly to (a) in (i).

• (b)

Suppose there is a relation between $(j,r)$ and one of $\{(i,s),(i,t)\}$. Similarly to (b) in (i), one has that the coefficient of $T(R+\delta^{jr}+\delta^{is}+\delta^{it})$ is zero.

• (c)

Suppose there exist relations between $(j,r)$ and both $\{(i,s),(i,t)\}$. In this case $(j,r),(i,s),(i,t)$ are in the same indecomposable set. If $r_{is}-r_{it}=1$ then $r_{jr}=r_{it}$ and there exists $r^{\prime}$ such that $\{(i,s)\geq(i-1,r^{\prime})>(i,t))\}\subseteq\mathcal{C}$ and $r_{i-1,r^{\prime}}=r_{is}$. It contradicts with $T(R+\delta^{jr}+\delta^{is}+\delta^{it})$ nonzero. Therefore $r_{is}-r_{it}\in\frac{1(q)}{2}+\mathbb{Z}_{>0}$. Then $r_{is}-r_{jr}\in\frac{1(q)}{2}+\mathbb{Z}_{>0}$ or $r_{jr}-r_{it}\in\frac{1(q)}{2}+\mathbb{Z}_{>0}$. Without loss of generality we assume that $r_{jr}-r_{it}\in\frac{1(q)}{2}+\mathbb{Z}_{>0}$. Let $\mathcal{C^{\prime}}=\{(i,s)\geq(j,r)\}$ and $T(v^{\prime})$ the tableau with $v^{\prime}_{s^{\prime}t^{\prime}}=l_{s^{\prime}t^{\prime}}$ if $(s^{\prime},t^{\prime})=(i,s)\text{ or }(j,r)$ and variable entries otherwise. Then $T(v^{\prime})$ is a $\mathcal{C^{\prime}}$-realization and $V_{\mathcal{C}^{\prime}}(T(v^{\prime}))$ is a module. Let $z^{(1)},z^{(2)}\in\{\delta^{jr},\delta^{is},\delta^{it}\}$. One has that $\Phi(R,z^{(1)},z^{(2)})=\Phi(v,z^{(1)},z^{(2)})$ whenever $z^{(1)}\neq z^{(2)}$. Therefore the coefficient of $T(R+\delta^{jr}+2\delta^{is})$ is the limit of the coefficient of $T(v+\delta^{jr}+2\delta^{is})$ when $v\rightarrow R$, which is zero.

In the following we show that $(e_{i}f_{j}-f_{j}e_{i})T(R)=\delta_{ij}\frac{q^{\alpha_{i}}-q^{-\alpha_{i}}}{q-q^{-1}}T(R)$. We have

[TABLE]

Now we consider the coefficients of nonzero tableaux $T(L-\delta^{jr}+\delta^{is})$. If $(i,r)\neq(j,s)$ then the coefficient of $T(L-\delta^{jr}+\delta^{is})$ is zero similarly to the above case and, hence, $[e_{i},f_{j}]T(R)=0$ if $i\neq j$.

Suppose $i=j=k$. The coefficient of $T(R-\delta^{ir}+\delta^{is})$ is zero if $r\neq s$.

By Corollary 3.6, the coefficient of $T(R)$ is

[TABLE]

Hence $(e_{i}f_{j}-f_{j}e_{i})T(R)=\delta_{ij}\frac{q^{\alpha_{i}}-q^{-\alpha_{i}}}{q-q^{-1}}T(R)$.

All other relations can be verified similarly. Thus $\mathcal{C}$ is realizable.

∎

Remark 3.10**.**

The realizable sets of relations for $\operatorname{\mathfrak{gl}}_{n}$ were all obtained in [11] (Theorem 4.22). Here we only deform the definition of a tableau satisfying a set of relations, i.e. replace $\mathbb{Z}$ by $\mathbb{Z}+\frac{1(q)}{2}$. As in the non quantum case, if we only consider irreducible modules, all realizable sets of relations are admissible. Thus the converse of Theorem 3.9 holds.

An effective method of constructing of realizable sets of relations was introduced in [11], called relations removal method (RR-method for short). We will show that the same method can be applied to construct admissible sets of relations (and hence realizable subsets of $\mathcal{R}$ by Theorem 16,) in the quantum case.

Let $\mathcal{C}$ be any realizable subset of $\mathcal{R}$ and $T(L)$ a tableau satisfying $\mathcal{C}$. Fix $(k,i)\in\mathfrak{V}(\mathcal{C})$ and suppose that $T(L+m\delta^{ki})$ is a $\mathcal{C}$-realization for infinitely many choices of $m\in\mathbb{Z}$. Denote by $\widetilde{\mathcal{C}_{ki}}$ the set of relations obtained from $\mathcal{C}$ by removing all relations that involve $(k,i)$. We say that $\widetilde{\mathcal{C}}\subsetneq\mathcal{C}$ is obtained from $\mathcal{C}$ by the RR-method if it is obtained by a sequence of such removing of relations for different indexes.

Theorem 3.11**.**

Let $\mathcal{C}_{1}$ be any realizable subset of $\mathcal{R}$. If $\mathcal{C}_{2}$ is obtained from $\mathcal{C}_{1}$ by the RR-method then $\mathcal{C}_{2}$ is realizable.

Proof.

Analogous to the proof of Theorem 4.24 in [11]. ∎

We immediately obtain the following statement for generic modules which was shown in [12], Theorem 5.2 (cf. [28], Theorem 2).

Corollary 3.12**.**

Let $T(L)$ be a generic Gelfand-Tsetlin tableau of height $n$. Then $V_{\emptyset}(T(L))$ has a structure of a $U_{q}$-module with the action of the generators of $U_{q}$ given by the Gelfand-Tsetlin formulas (9).

Proof.

By Theorem 16 the set $\mathcal{S}$ is realizable and applying the RR-method to $\mathcal{S}$, after finitely many steps we can remove all the relations in $\mathcal{S}$, then $\emptyset$ is realizable by Theorem 3.11. ∎

We call $V_{\mathcal{C}}(T(L))$ admissible Gelfand-Tsetlin module associated with the admissible set of relations $\mathcal{C}$. Note that $V_{\mathcal{C}}(T(L))$ is infinite dimensional if $\mathcal{C}$ does not imply $\mathcal{S}$.

4. Action of Gelfand-Tsetlin subalgebra

From now on we will assume that $\mathcal{C}$ is an admissible subset of $\mathcal{R}$ and consider the $U_{q}$-module $V_{\mathcal{C}}(T(L))$. We will analyze the action of the Gelfand-Tsetlin subalgebra $\Gamma_{q}$ on modules $V_{\mathcal{C}}(T(L))$.

The action of $U_{q}$ on this irreducible module is given by the Gelfand-Tsetlin formulas (9). Moreover, the action of $\Gamma_{q}$ is given by (10). Then as in the non quantum case, we have the following Theorem:

Theorem 4.1**.**

For any admissible $\mathcal{C}$ the module $V_{\mathcal{C}}(T(L))$ is a Gelfand-Tsetlin module with diagonalisable action of the generators of the Gelfand-Tsetlin subalgebra given by the formula (10).

Proof.

Essentially repeats the proof of Theorem 5.2 in [11]. ∎

Remark 4.2**.**

Note that for any $x\in 1(q)$ we have $\gamma_{mk}(L)=\gamma_{mk}(L+x\delta^{ij})$ for any $m,\ k$ and any $1\leq j\leq i\leq n$. In particular, the tableaux $T(L)$ and $T(L+x\delta^{ij})$ define the same Gelfand-Tsetlin character.

Remark 4.3**.**

Let $\mathcal{C}$ be any realizable set of relations and $T(L)$ satisfying $\mathcal{C}$. Then for any ${x\in 1(q)}$ the tableau $T(L+x\delta^{ij})$ is also a $\mathcal{C}$-realization and the Gelfand-Tsetlin modules $V_{\mathcal{C}}(T(L+x\delta^{ij}))$ and $V_{\mathcal{C}}(T(L))$ are isomorphic (see Remark 4.2). On the other hand, when $\frac{x}{2}\notin 1(q)$, the two modules are not isomorphic. In general, every admissible set of relations defines infinitely many nonisomorphic modules.

Now we can show that $\Gamma_{q}$ separates basis tableaux in all constructed modules $V_{\mathcal{C}}(T(L))$ and hence, in their irreducible quotients.

Proposition 4.4**.**

For any ${\mathsf{m}}\in\operatorname{Specm}\Gamma_{q}$ from the Gelfand-Tsetlin support of $V_{\mathcal{C}}(T(L))$, the Gelfand-Tsetlin multiplicity of ${\mathsf{m}}$ is one.

Proof.

The action of $\Gamma_{q}$ is given by the formulas (10), and hence determined by the values of symmetric polynomials on the entries of the rows of the tableaux. Given two Gelfand-Tsetlin tableaux $T(R_{1})$ and $T(R_{2})$ in $\mathcal{B}_{\mathcal{C}}(T(L))$, we have $c_{rs}(T(R_{1}))=c_{rs}(T(R_{2}))$ for any $1\leq s\leq r\leq n$ if and only if $R_{1}=\sigma(R_{2})$ for some $\sigma\in G$. But if $\sigma\neq 1$ then $T(R_{1})$ and $T(R_{2})$ have different order among the entries in $r$th row. Since both $T(R_{1})$ and $T(R_{2})$ satisfy $\mathcal{C}$ and $T(R_{1})=T(R_{2}+z)$ for some $z\in\mathbb{Z}^{\frac{n(n-1)}{2}}$ we come to a contradiction. ∎

Therefore, we have an explicit basis of $V_{\mathcal{C}}(T(L))$, and of its irreducible quotients, parametrized by different Gelfand-Tsetlin tableaux with an explicit basis of the generators of $U_{q}$ and of $\Gamma_{q}$.

It was proved In [12] that the irreducible module containing generic tableau $T(R)$ has a basis of tableaux

[TABLE]

where

[TABLE]

The following statement is a generalization of this result. Recall that $\mathcal{C}$ is a maximal set of relations satisfied by $T(L)$ if $T(L)$ is a $\mathcal{C}$-realization and for any admissible set of relations $\mathcal{C}^{\prime}$ satisfied by $T(L)$, $\mathcal{C}$ implies $\mathcal{C}^{\prime}$.

Theorem 4.5**.**

Admissible Gelfand-Tsetlin module $V_{\mathcal{C}}(T(L))$ is irreducible if and only if $\mathcal{C}$ is a maximal set of relations satisfied by $T(L)$.

Proof.

Let $T(R)$ be any tableau in $\mathcal{B}_{\mathcal{C}}(T(L))$ and $\mathcal{C}$ a maximal set of relations satisfied by $T(L)$. One can show easily that $U_{q}T(R)\subseteq V_{\mathcal{C}}(T(L))$. If $T(R^{\prime})$ is another tableau in $V_{\mathcal{C}}(T(L))$ then there exist $\{(i_{s},j_{s})\}\subseteq\{(i,j)\ |\ 1\leq j\leq i\leq n-1)\}$ and $t$ such that, for any $k\leq t$, $T(R_{k})=T(R+\sum_{s=1}^{k}p_{s}\delta^{i_{s},j_{s}})\in\mathcal{B}_{\mathcal{C}}(T(L))$, where $p_{i}\in\{1,-1\}$ and $T(R_{0})=T(R)$, $T(R_{t})=T(R^{\prime})$. It is sufficient to show that we can obtain $T(R_{s})$ from $T(R_{s-1})$. If $p_{i}=1$ (resp. $p_{i}=-1$) then acting by $e_{i_{s}}$ (resp. $f_{i_{s}}$) on $T(R_{s-1})$ the coefficient of $T(R_{s})$ in the image is not zero. By Theorem 4.1, there exists an element in $\Gamma_{q}$ which annihilates all other tableaux except $T(R_{s})$. We conclude that $V_{\mathcal{C}}(T(L))\subseteq U_{q}T(R)$.

Conversely, assume $T(L)$ satisfies $\mathcal{C}$ and $\mathcal{C}$ is not maximal. Let $\mathcal{C}^{\prime}$ be the maximal set of relations satisfied by $T(L)$. Then $V_{\mathcal{C}^{\prime}}(T(L))$ is a subquotient of $V_{\mathcal{C}}(T(L))$ and $V_{\mathcal{C}^{\prime}}(T(L))\neq V_{\mathcal{C}}(T(L))$. It contradicts the irreducibility of $V_{\mathcal{C}}(T(L))$. ∎

We conclude with an example of the family of highest weight modules that can be realized as $V_{\mathcal{C}}(T(L))$ for some admissible set of relations $\mathcal{C}$.

Proposition 4.6**.**

Set $\lambda=(\lambda_{1},\ldots,\lambda_{n})$. The irreducible highest weight module $L(\lambda)$ is admissible Gelfand-Tsetlin module if $\lambda_{i}-\lambda_{j}\notin\mathbb{Z}$ or $\lambda_{i}-\lambda_{j}>i-j$ for any $1\leq i<j\leq n-1$.

Proof.

Let $T(L)$ be a tableau such that $l_{ij}=\lambda_{i}$, $\mathcal{C}$ be the maximal set of relations satisfied by $T(L)$. Then $\mathcal{C}$ is admissible and $V_{\mathcal{C}}(T(L))$ is irreducible admissible module. Moreover, $T(L)$ is a highest weight vector and $V_{\mathcal{C}}(T(L))$ is isomorphic to $L(\lambda)$. ∎