Geometric RTT realization of $U_v(gl_n)^+$

TL;DR

This paper presents a geometric realization of the positive part of the quantum group of $U_v(gl_n)$ using RTT relations, providing a new perspective on its algebraic structure.

Contribution

It introduces a BLM realization of $U_v(gl_n)^+$ based on RTT relations, which is a novel approach in quantum group theory.

Findings

Establishes a geometric realization of $U_v(gl_n)^+$

Connects BLM realization with RTT relations

Provides new tools for studying quantum groups

Abstract

In this paper, we give a BLM realization of the positive part of the quantum group of $U_v(gl_n)$ with respect to RTT relations.

Equations

$$R_{12}{R}_{13}{R}_{23}=R_{23}{R}_{13}{R}_{12}.$$

$$\mathcal{A}=\mathbb{Z}[v^{\pm 1}].$$

$$f\ast g(V,V^{\prime })=\sum _{V^{\prime \prime }\in \mathcal{X}}f(V,V^{\prime \prime })g(V^{\prime \prime },V^{\prime }),\forall V,V^{\prime }\in \mathcal{X}.$$

$$\begin{array}{rl}\mathrm{ro}(M)& ={\left(\sum _{j=1}^c{m}_{ij}\right)}_{1\le i\le c},\\ \mathrm{co}(M)& ={\left(\sum _{i=1}^c{m}_{ij}\right)}_{1\le j\le c}.\end{array}$$

$$G\backslash \mathcal{X}\times \mathcal{X}\simeq {\Theta }_d,$$

$$t_{ij}=\overline{t_{ij}}=0,1\le i<j\le n.$$

$$v^{{\delta }_{ij}}{t}_{ia}{t}_{jb}-v^{{\delta }_{ab}}{t}_{jb}{t}_{ia}=(v-v^{-1})({\delta }_{b<a}-{\delta }_{i<j})t_{ja}{t}_{ib},$$

$$v^{{\delta }_{ij}}{\overline{t}}_{ia}{\overline{t}}_{jb}-v^{{\delta }_{ab}}{\overline{t}}_{jb}{\overline{t}}_{ia}=(v-v^{-1})({\delta }_{b<a}-{\delta }_{i<j}){\overline{t}}_{ja}{\overline{t}}_{ib},$$

$$v^{{\delta }_{ij}}{\overline{t}}_{ia}{t}_{jb}-v^{{\delta }_{ab}}{t}_{jb}{\overline{t}}_{ia}=(v-v^{-1})({\delta }_{b<a}{t}_{ja}{\overline{t}}_{ib}-{\delta }_{i<j}{\overline{t}}_{ja}{t}_{ib}).$$

$$♯S=q^{m-n}.$$

$$♯S=q^{m-1}-q^{m-n-1}.$$

$$e_B\ast e_A=\sum _{\text{(j,p)}}{f}_{\text{(j,p)}}{e}_{(A+{\sum }_{l=1}^m(E_{j_{l-1},p_l}-E_{j_l,p_l}))},$$

$$e_C\ast e_A=\sum _{\text{(j,p)}}{f}_{\text{(j,p)}}^{\prime }{e}_{(A+{\sum }_{l=1}^m(E_{j_{l-1},p_l}-E_{j_l,p_l}))},$$

$$e_B\ast e_A=\sum _{\text{(j,p)}}{f}_{\text{(j,p)}}{e}_{(A+{\sum }_{1\le k\le m,1\le l\le r_k}(E_{j_{k,l-1},p_{k,l}}-E_{j_{k.l},p_{k,l}}))},$$

$$({\text{j}}_k,{\text{p}}_k)=((j_{k,1},p_{k_1}),\cdots ,(j_{k,r_k},p_{k,r_k}))$$

$$d(A)=\dim {\mathcal{O}}_A\text{mbox}andr(A)=\dim {\mathcal{O}}_B,\forall A\in {\Theta }_d,$$

$$\begin{array}{rl}\dim {\mathrm{C}}_G(V,V^{\prime })& =\sum _{i\ge k,j\ge l}{a}_{ij}{a}_{kl},\mathrm{if}(V,V^{\prime })\in {\mathcal{O}}_A,\\ \dim {\mathcal{O}}_A& =\sum _{i<k\mathrm{or}j<l}{a}_{ij}{a}_{kl},\\ d(A)-r(A)& =\sum _{i\ge k,j<l}{a}_{ij}{a}_{kl}.\end{array}$$

$${\overline{t}}_{ji}=0,\text{if}1\le i<j\le n,$$

$$\begin{array}[]{ccc}\overline{t}_{ij}&=&-(v^{-1}-v)\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-1\end{subarray}}v^{\lambda_{i}}[D_{\lambda}+E_{ij}]\ \text{if}\ 1\leq i<j\leq n,\\[28.90755pt] \overline{t}_{ii}&=&\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d\end{subarray}}v^{\lambda_{i}}[D_{\lambda}]\ \text{if}\ 1\leq i\leq n.\\ \end{array}$$

$$(R1)$$

$$(R2)$$

$$(R3)$$

$$(R4)$$

$$\begin{array}[]{ccc}\overline{t}_{ia}\overline{t}_{jb}&=&(v-v^{-1})^{2}\sum\limits_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\sum\limits_{\lambda^{\prime}}v^{\lambda_{j}}[D_{\lambda}+E_{jb}]\\[10.84006pt] &=&(v-v^{-1})^{2}\sum\limits_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\sum\limits_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{ia}}\\[10.84006pt] &=&(v-v^{-1})^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum\limits_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-1}e_{D_{\lambda}+E_{ia}+E_{jb}}.\\ \end{array}$$

$$\begin{array}[]{ccc}\overline{t}_{jb}\overline{t}_{ia}&=&(v-v^{-1})^{2}\sum\limits_{\lambda^{\prime}}v^{\lambda_{j}^{\prime}}[D_{\lambda}+E_{jb}]\sum\limits_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\\[10.84006pt] &=&(v-v^{-1})^{2}\sum\limits_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{ia}}\sum\limits_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\\[10.84006pt] &=&(v-v^{-1})^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-1}e_{D_{\lambda}+E_{ia}+E_{jb}}.\\ \end{array}$$

$${\overline{t}}_{ia}{\overline{t}}_{jb}-{\overline{t}}_{jb}{\overline{t}}_{ia}=0.$$

$${\overline{t}}_{ia}{\overline{t}}_{jb}-v^{-{\delta }_{ab}}{\overline{t}}_{jb}{\overline{t}}_{ia}=-(v^{-1}-v){\overline{t}}_{ja}{\overline{t}}_{ib}.$$

$$v^{-{\delta }_{ij}}{\overline{t}}_{ia}{\overline{t}}_{jb}-{\overline{t}}_{jb}{\overline{t}}_{ia}=(v^{-1}-v){\overline{t}}_{ja}{\overline{t}}_{ib}.$$

$$\sum _{r\le i,s\ge j}{a}_{rs}$$

$$\sum _{r\ge i,s\le j}{a}_{rs}$$

$$\prod _{1\le j<i\le n}{e}_{D_{ij}+a_{ij}{E}_{ij}}\ast \prod _{1\le i<j\le n}{e}_{D_{ij}+a_{ij}{E}_{ij}}={\chi }_A{e}_A+\mathrm{lower}\mathrm{terms},$$

$$A=\left(\begin{array}[]{ccc}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{array}\right)$$

$$A_{1}=\left(\begin{array}[]{ccc}a_{11}+a_{21}+a_{31}&a_{12}&0\\ 0&a_{22}+a_{32}&0\\ 0&0&a_{13}+a_{23}+a_{33}\\ \end{array}\right)$$

$$A_{2}=\left(\begin{array}[]{ccc}a_{11}+a_{21}+a_{31}+a_{12}&0&a_{13}\\ 0&a_{22}+a_{32}&0\\ 0&0&a_{23}+a_{33}\\ \end{array}\right)$$

Full text

1. Introduction

There are two ways to study quantum groups. One is algebraic, the another is geometric.

In algebriaic way, there are two methods. The first was adopted by Drinfeld [10, 12] and Jimbo [15] to define the quantum enveloping algebra $U_{q}(\mathfrak{g})$ as a $q$-deformation of the enveloping algebra $U(\mathfrak{g})$ in terms of the Chevalley generators and Serre relations based on the data coming from the corresponding Cartan matrix. The second approach to realize the quantum groups was through RTT   method. The approach was from the quantum inverse scattering method developed by the Leningrad school. Sometimes it was more available for us. For example, In [13] Faddeev, Reshetikhin and Takhtajan have shown that both the quantum enveloping algebras $U_{q}(\mathfrak{g})$ and the dual quantum groups for finite classical simple Lie algebras $\mathfrak{g}$ can be studied in the RTT method using the solutions $R$ of the Yang-Baxter equation:

[TABLE]

In geometric way, there are many results on the geometric realization of quantum groups with respect to the Chevalley generators. For example, the $\mathbb{C}$-valued $GL_{n}$ equivalence functions on the flag variety associated to the finite dimensional vector space over $\mathbb{F}_{q}$ was considered by Beilinson, Lusztig, and McPherson. It gave realization of Schur algebra of $U_{v}(gl_{n})$, and $U_{v}(gl_{n})$ was the limit algebra of the Schur algebra. Also Du and Gu used the same way give the realization of $U_{v}(gl(m|n))$[5]. Moreover, they also obtained a new basis containing the standard generators for quantum linear super group and explicit multiplication formulas between the generators and an arbitrary basis element. In affine case, Fu gave a BLM realization for $U_{\mathbb{Z}}(\widehat{gl}_{n})$[4]. By the similar way, Bao-Wang [2] and Fan-Li[3] gave the several new quantum groups and the Schur-like duality of type B/C and D. But there are few results on the geometric RTT realization. So the questions are if the the geometric realization of quantum groups with respect to the RTT relations can be given and if more information about the representation of quantum group can be given though the geometry RTT realization. In this paper, we use the BLM’s way to construct the RTT realization of $U_{v}(gl_{n})^{+}$.

The paper is organized as follows. In section 2, we recall the basic results on flag varieties and the RTT realization of $U_{v}(gl_{n})$. In section 3, we calculate generator realizations of the Schur algebra, find a subalgebra of the limit algebra of the Schur algebra and show that it is isomorphic to the positive part of $U_{v}(gl_{n})$.

2. Preliminary

In this section, let us recall some basic facts on flag varieties appear in [1] and the RTT realization of $U_{v}(gl_{n})$.

Let $\mathbb{F}_{q}$ be a finite field of $q$ elements and of odd characteristic. For positive integers $d$ and $n$, consider the set $\mathscr{X}$ of $n$-step flags $V=(V_{i})_{0\leq i\leq n}$ in $\mathbb{F}_{q}^{d}$ such that $V_{0}=0$, $V_{i}\subseteq V_{i+1}$.

Let $G=\mathrm{GL}(\mathbb{F}_{q}^{d})$, and $G$ acts naturally on set $\mathscr{X}$. Let $G$ act diagonally on the product $\mathscr{X}\times\mathscr{X}$. Set

[TABLE]

Let $\mathcal{S}_{\mathscr{X}}=\mathcal{A}(\mathscr{X}\times\mathscr{X})^{G}$ be the set of all $\mathcal{A}$-valued $G$-invariant functions on $\mathscr{X}\times\mathscr{X}$. Clearly, the set $\mathcal{S}_{\mathscr{X}}$ is a free $\mathcal{A}$-module. Moreover, $\mathcal{S}_{\mathscr{X}}$ admits an associative $\mathcal{A}$-algebra structure ‘$*$’ under a standard convolution product as discussed in  [1]. In particular, when $v$ is specialized to $\sqrt{q}$, we have

[TABLE]

Let us describe the $G$-orbits on $\mathscr{X}\times\mathscr{X}$. We start by introducing the following notations associated to a matrix $M=(m_{ij})_{1\leq i,j\leq c}$.

[TABLE]

We also write $\mathrm{ro}(M)_{i}$ and $\mathrm{co}(M)_{j}$ for the $i$-th and $j$-th component of the row vectors of $\mathrm{ro}(M)$ and $\mathrm{co}(M)$, respectively. For any pair $(V,V^{\prime})$ of flags in $\mathscr{X}$, we can assign an $n$ by $n$ matrix whose $(i,j)$-entry equal to $\dim\frac{V_{i-1}+V_{i}\cap V_{j}^{\prime}}{V_{i-1}+V_{i}\cap V_{j-1}^{\prime}}$. We have the following bijection.

[TABLE]

where $\Theta_{d}$ is the set of all matrices $\Theta_{d}$ in $\mbox{Mat}_{n\times n}(\mathbb{N})$ such that $\sum_{i,j}(\Theta_{d})_{i,j}=d$

Definition 2.0.1**.**

The algebra $U_{v}(gl_{n})$ is generated by the elements $t_{ij}$ and $\overline{t_{ij}}$ with $1\leq i,j\leq n$ subject to the relations

[TABLE]

3. BLM Realization Of $U_{v}(gl_{n})^{+}$ With Respect To RTT Relations

3.1. Calculus of the algebra $\mathcal{S}$

For simplicity, we shall denote $\mathcal{S}$ instead of $\mathcal{S}_{\mathscr{X}}$. In this section, we determine the generators for $\mathcal{S}$ and the associated multiplication formula.

Lemma 3.1.1**.**

Assume $V_{1},~{}V_{2},~{}V_{3}$ are the vector space over $F_{q}$, $V_{1}\overset{1}{\subset}V_{2}\subset V_{3}$, and $\mathrm{dim}V_{1}=n-1$, $\mathrm{dim}V_{2}=n$, $\mathrm{dim}V_{3}=m$. Set $S=\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{1}\subset V_{3}^{\prime},V_{2}\cap V_{3}^{\prime}\neq V_{2}\}$. Then

[TABLE]

Where $V_{1}\overset{1}{\subset}V_{2}$ means $V_{1}\subset V_{2}$ and $\mathrm{dim}V_{2}-\mathrm{dim}V_{1}=1.$

Proof.

$\sharp S=\sharp\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{1}\subset V_{3}^{\prime}\}-\sharp\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{2}\subset V_{3}^{\prime}\}$

\begin{array}[]{ccc}\sharp S&=&\sharp\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{1}\subset V_{3}^{\prime}\}-\sharp\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{2}\subset V_{3}^{\prime}\}\\ &=&\frac{q^{m-n+1}-1}{q-1}-\frac{q^{m-n}-1}{q-1}\\ &=&q^{m-n}\end{array}

∎

Lemma 3.1.2**.**

Assume $V_{1},~{}V_{2},~{}V_{3}$ are the vector space over $F_{q}$, $V_{2}\subset V_{3}$, $V_{1}\subset V_{3}$, $V_{1}\cap V_{2}=0$, $\mathrm{dim}V_{1}=1$, $\mathrm{dim}V_{2}=n$, $\mathrm{dim}V_{3}=m$. Set $S=\{V_{3}^{\prime}|V_{3}^{\prime}\overset{1}{\subset}V_{3},V_{1}\cap V_{3}^{\prime}=0,$ $\mathrm{dim}V_{2}\cap V_{3}^{\prime}=n-1\}$. Then

[TABLE]

Proof.

$\sharp S=\sharp\{V^{\prime}|V^{\prime}\overset{1}{\subset}V_{3}\}-\sharp\{V^{\prime}|V^{\prime}\overset{1}{\subset}V_{3},V_{1}\subset V^{\prime}\}-\sharp\{V^{\prime}|V^{\prime}\overset{1}{\subset}V_{3},V_{2}\subset V^{\prime}\}+\sharp\{V^{\prime}|V^{\prime}\overset{1}{\subset}V_{3},V_{1}\subset V^{\prime},V_{2}\subset V^{\prime}\}$

$=\frac{q^{m}-1}{q-1}-\frac{q^{m-1}-1}{q-1}-\frac{q^{m-n}-1}{q-1}+\frac{q^{m-n-1}-1}{q-1}$

$=q^{m-1}-q^{m-n-1}.$

∎

Lemma 3.1.3**.**

Let $A=(a_{ij})\in\Theta_{d}$.

(a)Assume $B=(b_{ij})\in\Theta_{d}$. There exist $1\leq i_{0}<i_{1}\leq n$ such that $B-E_{i_{0},i_{1}}$ is the diagonal matrices, and $\sum_{i}b_{ij}=\sum_{k}a_{jk}.$ Then

[TABLE]

where $\textbf{(j,p)}=((j_{1},p_{1}),\cdots,(j_{m},p_{m}))$ satisfied the conditions: $i_{0}=j_{0}<j_{1}<\cdots<j_{m}=i_{1}$, $1\leq p_{m}<\cdots<p_{1}\leq n$, and for any $1\leq k\leq m$, $a_{j_{k},p_{k}}\geq 1$. $f_{\textbf{(j,p)}}=f_{1}f_{2}\cdots f_{m}$, where

f_{l}=\left\{\begin{array}[]{ll}\frac{v^{2(1+\sum{j\geq p_{1}}a_{j_{0},j})}-v^{2\sum_{j>p_{1}}a_{j_{0},j}}}{v^{2}-1}\prod\limits_{k=j_{0}+1}^{k=j_{1}-1}v^{2\sum_{j\geq p_{1}}a_{k,j}}&\text{if}\ l=1;\\[10.84006pt] (v^{2\sum_{j\geq p_{l}}a_{j_{l-1},j}}-v^{2(\sum_{j>p_{l}}a_{j_{l-1},j}-1)})\prod\limits_{k=j_{l-1}+1}^{k=j_{l}-1}v^{2\sum_{j\geq p_{l}}a_{k,j}}&\text{if}\ l>1.\end{array}\right.**

(b)Assume $C=(c_{ij})\in\Theta_{d}$. There exist $n\geq i_{0}>i_{1}\geq 1$ such that $C-E_{i_{0},i_{1}}$ is the diagonal matrices, and $\sum_{i}c_{ij}=\sum_{k}a_{jk}.$ Then

[TABLE]

where $\textbf{(j,p)}=((j_{1},p_{1}),\cdots,(j_{m},p_{m}))$ satisfied the conditions: $i_{0}=j_{0}>j_{1}>\cdots>j_{m}=i_{1}$, $1\leq p_{1}<\cdots<p_{m}\leq n$, and for any $1\leq k\leq m$, $a_{j_{k},p_{k}}\geq 1$. $f_{\textbf{(j,p)}}^{\prime}=f_{1}^{\prime}f_{2}^{\prime}\cdots f_{m}^{\prime}$, where

f_{l}^{\prime}=\left\{\begin{array}[]{ll}\frac{v^{2(1+\sum_{j\leq p_{1}}a_{j_{0},j})}-v^{2\sum_{j<p_{1}}a_{j_{0},j}}}{v^{2}-1}\prod\limits_{k=j_{1}+1}^{k=j_{0}-1}v^{2\sum_{j\leq p_{1}}a_{k,j}}&\text{if}\ l=1;\\[10.84006pt] (v^{2\sum_{j\leq p_{l}}a_{j_{l-1},j}}-v^{2(\sum_{j<p_{l}}a_{j_{l-1},j}-1)})\prod\limits_{k=j_{l}+1}^{k=j_{l-1}-1}v^{2\sum_{j\leq p_{l}}a_{k,j}}&\text{if}\ l>1.\end{array}\right.**

Proof.

Assume $A^{\prime}=A+\sum_{l=1}^{m}(E_{j_{l-1},p_{l}}-E_{j_{l},p_{l}})$. Let $f=(V_{1}\subset\cdots\subset V_{i_{0}-1}\subset V_{i_{0}}\subset\cdots\subset V_{i_{1}-1}\subset V_{i_{1}}\subset\cdots\subset V_{n}),~{}f^{\prime}=(V_{1}^{\prime}\subset\cdots\subset V_{i_{0}-1}^{\prime}\subset V_{i_{0}}^{\prime}\subset\cdots\subset V_{i_{1}-1}^{\prime}\subset V_{i_{1}}^{\prime}\subset\cdots\subset V_{n}^{\prime})$ be such that $(f,f^{\prime})\in\mathcal{O}_{A^{\prime}}$. Set $V_{0}=V_{0}^{\prime}=0.$

To prove (a), we need to compute how many $f^{\prime\prime}$ such that $(f,~{}f^{\prime\prime})\in\mathcal{O}_{B}$, $(f^{\prime\prime},~{}f^{\prime})\in\mathcal{O}_{A}$. Assume $f^{\prime\prime}=(V_{1}^{\prime\prime}\subset\cdots\subset V_{i_{0}-1}^{\prime\prime}\subset V_{i_{0}}^{\prime\prime}\subset\cdots\subset V_{i_{1}-1}^{\prime\prime}\subset V_{i_{1}}^{\prime\prime}\subset\cdots\subset V_{n}^{\prime\prime}).$ For any $i<i_{0}$ or $i\geq i_{1}$, $V_{i}^{\prime\prime}=V_{i}$.

First, we need to count how many $V_{i_{0}}^{\prime\prime}$ there exist. Consider the set $Z_{i_{0}}$ of all subspace $U_{i_{0}}$ of $V$ such that $V_{i_{0}-1}\subset U_{i_{0}}\overset{1}{\subset}V_{i_{0}}$, $V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}-1}^{\prime}\subset U_{i_{0}}$, and$(V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}}^{\prime})\cap U_{i_{0}}\neq V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}}^{\prime}$.

$\sharp Z_{i_{0}}=\sharp\{U_{i_{0}}|U_{i_{0}}\supset V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}-1}^{\prime}\}-\sharp\{U_{i_{0}}|U_{i_{0}}\supset V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}}^{\prime}\}$

$=\frac{v^{2(1+\sum_{j\geq p_{1}}a_{j_{0},j})}-v^{2\sum_{j>p_{1}}a_{j_{0},j}}}{v^{2}-1}.$

Second, we fix $U_{i_{0}}$. Consider the set $Z_{i_{0}+1}$ of all subspace $U_{i_{0}+1}$ of $V$ such that $U_{i_{0}}\subset U_{i_{0}+1}\overset{1}{\subset}V_{i_{0}+1}$, $V_{i_{0}}\cap U_{i_{0}+1}=U_{i_{0}}$(since $B_{i_{0},~{}i_{0}+1}=0$), $U_{i_{0}}+V_{i_{0}+1}\cap V_{p_{1}-1}\subset U_{i_{0}+1}$. That is, $V_{i_{0}+1}\cap V_{p_{1}-1}^{\prime}+U_{i_{0}}\subset U_{i_{0}+1},V_{i_{0}}\cap U_{i_{0}+1}=U_{i_{0}}.$

$\sharp Z_{i_{0}+1}=\sharp\{U_{i_{0}+1}|U_{i_{0}+1}\supset V_{i_{0}+1}\cap V_{p_{1}-1}^{\prime}+U_{i_{0}}\}-\sharp\{U_{i_{0}+1}|U_{i_{0}+1}\supset V_{i_{0}}+V_{i_{0}+1}\cap V_{p_{1}-1}^{\prime}\}$

$=\frac{v^{2(1+\sum_{j\geq p_{1}}a_{j_{0}+1,j})}-v^{2\sum_{j\geq p_{1}}a_{j_{0}+1,j}}}{v^{2}-1}$

$=v^{2\sum_{j\geq p_{1}}a_{j_{0}+1,j}}.$

Similarly, as above, for any $i_{0}+1<k<j_{1}$, $\sharp Z_{k}=v^{2\sum_{j\geq p_{1}}a_{i_{k},j}}$. Then we get the coefficient of $f_{1}$.

Finally, we fix $U_{k},~{}i_{0}\leq k\leq j_{1}-1.$ We need to count how many $V_{j_{1}}^{\prime\prime}$ exist. Consider the set $Z_{j_{1}}$ of all subspace $U_{j_{1}}$ of $V$ such that $U_{j_{1}-1}\subset U_{j_{1}}\overset{1}{\subset}V_{j_{1}}$, $V_{i_{0}}\cap U_{j_{1}}=V_{i_{0}}\cap U_{j_{1}-1}=U_{i_{0}}$ (since $b_{j_{0},j_{1}}=0$), $U_{j_{1}-1}+V_{j-1}\cap V_{P_{2}-1}^{\prime}\subset U_{j_{1}}$, $(U_{j_{1}-1}+V_{j-1}\cap V_{P_{2}}^{\prime})\cap U_{j_{1}}\neq U_{j_{1}-1}+V_{j-1}\cap V_{P_{2}}^{\prime}.$ By the lemma3.1.2. $\sharp Z_{j_{1}}=q^{\sum_{j\geq p_{2}}a_{j_{1},j}}-q^{\sum_{j>p_{2}}a_{j_{1},j}-1}$.

All other $Z_{k}\ (j_{1}+1\leq k\leq i_{1}-1)$ can be counted by the similar way as above. Then $(a)$ follows.

To prove (b), we need to compute how many $f^{\prime\prime}$ such that $(f,~{}f^{\prime\prime})\in\mathcal{O}_{B}$, $(f^{\prime\prime},~{}f^{\prime})\in\mathcal{O}_{A}$. Assume $f^{\prime\prime}=(V_{1}^{\prime\prime}\subset\cdots\subset V_{i_{0}-1}^{\prime\prime}\subset V_{i_{0}}^{\prime\prime}\subset\cdots\subset V_{i_{1}-1}^{\prime\prime}\subset V_{i_{1}}^{\prime\prime}\subset\cdots\subset V_{n}^{\prime\prime}).$ For any $i\geq i_{0}$ or $i<i_{1}$, $V_{i}^{\prime\prime}=V_{i}$.

First, we need to count how many $V_{i_{0}-1}^{\prime\prime}$ exist. Consider the set $Z_{i_{0}-1}$ of all subspace $U_{i_{0}-1}$ of $V$ such that $V_{i_{0}-1}\overset{1}{\subset}U_{i_{0}-1}\subset V_{i_{0}}$, $(V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}-1}^{\prime})\cap U_{i_{0}-1}\neq U_{i_{0}-1}$, and $U_{i_{0}-1}\subset V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}}^{\prime}$.

$\sharp Z_{i_{0}-1}=\sharp\{U_{i_{0}-1}|U_{i_{0}-1}\subset V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}}^{\prime}\}-\sharp\{U_{i_{0}}|U_{i_{0}}\subset V_{i_{0}-1}+V_{i_{0}}\cap V_{p_{1}-1}^{\prime}\}$

$=\frac{v^{2(1+\sum_{j\leq p_{1}}a_{j_{0},j})}-v^{2\sum_{j<p_{1}}a_{j_{0},j}}}{v^{2}-1}.$

Second, we fix $U_{i_{0}-1}$. Consider the set $Z_{i_{0}-2}$ of all subspace $U_{i_{0}-2}$ of $V$ such that $V_{i_{0}-2}\overset{1}{\subset}U_{i_{0}-2}\subset U_{i_{0}-1}$, $V_{i_{0}-1}+U_{i_{0}-2}=U_{i_{0}-1}$(since $c_{i_{0},i_{0}-1}=0$), $U_{i_{0}-2}\subset V_{i_{0}-2}+U_{i_{0}-1}\cap V_{p_{1}}^{\prime}$. That is, $U_{i_{0}-2}\subset V_{i_{0}-2}+U_{i_{0}-1}\cap V_{p_{1}}^{\prime},U_{i_{0}-2}\cap(V_{i_{0}-2}+V_{i_{0}-1}\cap V_{p_{1}}^{\prime})\neq U_{i_{0}-2}.$

$\sharp Z_{i_{0}-2}=\sharp\{U_{i_{0}-2}|U_{i_{0}-2}\subset V_{i_{0}-2}+U_{i_{0}-1}\cap V_{p_{1}}^{\prime}\}-\sharp\{U_{i_{0}-2}|U_{i_{0}-2}\subset V_{i_{0}-2}+V_{i_{0}-1}\cap V_{p_{1}}^{\prime}\}$

$=\frac{v^{2(1+\sum_{j\leq p_{1}}a_{j_{0}-1,j})}-v^{2\sum_{j\leq p_{1}}a_{j_{0}-1,j}}}{v^{2}-1}$

$=v^{2\sum_{j\leq p_{1}}a_{j_{0}-1,j}}.$

Similarly, as above, for any $j_{1}\leq k<j_{0}-2$, $\sharp Z_{k}=v^{2\sum_{j\leq p_{1}}a_{k+1,j}}$. Then we get the coefficient of $f_{1}$

Finally, we fix $U_{k},~{}j_{1}\leq k\leq j_{0}-1.$ We need to count how many $V_{j_{1}-1}^{\prime\prime}$ exist. Consider the set $Z_{j_{1}-1}$ of all subspace $U_{j_{1}-1}$ of $V$ such that $V_{j_{1}-1}\overset{1}{\subset}U_{j_{1}-1}\subset U_{j_{1}}$, $U_{j_{1}}=V_{j_{1}}+U_{j_{1}-1}$, $U_{j_{1}-1}\subset V_{j_{1}-1}+U_{j_{1}}\cap V_{p_{2}}^{\prime}$, $U_{j_{1}-1}\overset{\bullet}{\subset}V_{j_{1}-1}+U_{j_{1}}\cap V_{p_{2}-1}^{\prime}$, where $U_{j_{1}-1}\overset{\bullet}{\subset}V_{j_{1}-1}+U_{j_{1}}\cap V_{p_{2}-1}^{\prime}$ means $U_{j_{1}-1}\subset V_{j_{1}-1}+U_{j_{1}}\cap V_{p_{2}-1}^{\prime}$ and $U_{j_{1}-1}\cap(V_{j_{1}-1}+V_{j_{1}}\cap V_{p_{2}-1}^{\prime})\neq U_{j_{1}-1}$.

$\sharp Z_{j_{1}-1}=\frac{q^{(1+\sum_{k\leq p_{2}})}-1}{q-1}-\frac{q^{\sum_{k<p_{2}}}-1}{q-1}-(\frac{q^{\sum_{k\leq p_{2}}}-1}{q-1}-\frac{q^{(\sum_{k<p_{2}}-1)}-1}{q-1})$

$=q^{\sum_{k\leq p_{2}}}-q^{(\sum_{k<p_{2}}-1)}.$

All other $Z_{k}\ (j_{m}\leq k\leq j_{1}-2)$ can be counted by the similar way as above. Then $(b)$ follows.

∎

Proposition 3.1.4**.**

Let $A=(a_{ij})\in\Theta_{d}$. Assume $B=(b_{ij})\in\Theta_{d}$. There exist $1\leq i_{0}<i_{1}<\cdots<i_{m-1}<i_{m}\leq n$ such that $B-\sum_{k=1}^{m}E_{i_{k-1},i_{k}}$ is the diagonal matrices, and $\sum_{i}b_{ij}=\sum_{k}a_{jk}.$ Then

[TABLE]

where (j,p) runs over $((\textbf{j}_{1},~{}\textbf{p}_{1}),~{}\cdots,~{}(\textbf{j}_{m},~{}\textbf{p}_{m}))$. For any $1\leq k\leq m$,

[TABLE]

satisfied the conditions: $i_{k-1}=j_{k,0}<j_{k,1}<\cdots<j_{k,r_{k}}=i_{k}$, $1\leq p_{k,r_{k}}<\cdots<p_{k,1}\leq n$, and for any $1\leq l\leq r_{k}$, $a_{j_{k,l},p_{k,l}}\geq 1$. $f_{\textbf{(j,p)}}=\prod_{1\leq k\leq m,1\leq l\leq r_{k}}f_{k,l}.$

f_{k,l}=\left\{\begin{array}[]{ll}\frac{v^{2(1+\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2\sum_{j>p_{k,1}}a_{j_{k,0},j}}}{v^{2}-1}\prod\limits_{\xi=j_{k,0}+1}^{\xi=j_{k,1}-1}v^{2\sum_{j\geq p_{k,1}}a_{\xi,j}}&\text{if}\ l=1,p_{k-1,r_{k-1}}<p_{k,1};\\[10.84006pt] \frac{v^{2(\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2(-1+\sum_{j>p_{k,1}}a_{j_{k,0},j}})}{v^{2}-1}\prod\limits_{\xi=j_{k,0}+1}^{\xi=j_{k,1}-1}v^{2\sum_{j\geq p_{k,1}}a_{\xi,j}}&\text{if}\ l=1,p_{k-1,r_{k-1}}>p_{k,1};\\[10.84006pt] \frac{v^{2(\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2(\sum_{j>p_{k,1}}a_{j_{k,0},j}})}{v^{2}-1}\prod\limits_{\xi=j_{k,0}+1}^{\xi=j_{k,1}-1}v^{2\sum_{j\geq p_{k,1}}a_{\xi,j}}&\text{if}\ l=1,p_{k-1,r_{k-1}}=p_{k,1};\\[10.84006pt] (v^{2\sum_{j\geq p_{k,l}}a_{j_{k,l-1},j}}-v^{2\sum_{j>p_{k,l}}a_{j_{k,l-1},j}})\prod\limits_{\xi=j_{k,l-1}+1}^{k=j_{k,l}-1}v^{2\sum_{j\geq p_{k,l}}a_{\xi,j}}&\text{if}\ l>1.\end{array}\right.**

Proof.

Assume $A^{\prime}=A+\sum\limits_{\begin{subarray}{c}1\leq k\leq m\\ 1\leq l\leq r_{k}\end{subarray}}(E_{j_{k,l-1},p_{k,l}}-E_{j_{k.l},p_{k,l}})$. Let $f=(V_{1}\subset\cdots\subset V_{i_{0}-1}\subset V_{i_{0}}\subset\cdots\subset V_{i_{m}-1}\subset V_{i_{m}}\subset\cdots\subset V_{n}),~{}f^{\prime}=(V_{1}^{\prime}\subset\cdots\subset V_{i_{0}-1}^{\prime}\subset V_{i_{0}}^{\prime}\subset\cdots\subset V_{i_{m}-1}^{\prime}\subset V_{i_{m}}^{\prime}\subset\cdots\subset V_{n}^{\prime})$ be such that $(f,f^{\prime})\in\mathcal{O}_{A^{\prime}}$. Set $V_{0}=V_{0}^{\prime}=0.$ We need to compute how many $f^{\prime\prime}$ we have such that $(f,~{}f^{\prime\prime})\in\mathcal{O}_{B}$, $(f^{\prime\prime},~{}f^{\prime})\in\mathcal{O}_{A}$. Assume $f^{\prime\prime}=(V_{1}^{\prime\prime}\subset\cdots\subset V_{i_{0}-1}^{\prime\prime}\subset V_{i_{0}}^{\prime\prime}\subset\cdots\subset V_{i_{m}-1}^{\prime\prime}\subset V_{i_{m}}^{\prime\prime}\subset\cdots\subset V_{n}^{\prime\prime})$. For any $i<i_{0}$ or $i\geq i_{m}$, $V_{i}^{\prime\prime}=V_{i}$. The proof of this proposition is almost similar to the lemma 3.1.3. The only difference is that when $k>1,~{}l=1$, how many $V_{j_{k,0}}^{\prime\prime}$ exist. We need to count it in the following three case.

First case. $p_{k-1,r_{k-1}}<p_{k,1}$. Consider the set $Z_{j_{k,0}}$ of all subspace $U_{j_{k,0}}$ of $V$ such that $V_{j_{k,0}-1}\subset U_{j_{k,0}}\overset{1}{\subset}V_{j_{k,0}}$, $U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\subset U_{j_{k,0}}$, and $(U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime})\cap U_{j_{k,0}}\neq U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}$.

$\sharp Z_{j_{k,0}}=\sharp\{U_{j_{k,0}}|U_{j_{k,0}}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\}-\sharp\{U_{j_{k,0}}|U_{j_{k,0}}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}\}$

$=\frac{v^{2(1+\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2\sum_{j>p_{k,1}}a_{j_{k,0},j}}}{v^{2}-1}.$

Second case. $p_{k-1,r_{k-1}}=p_{k,1}$. Consider the set $Z_{j_{k,0}}^{\prime}$ of all subspace $U_{j_{k,0}}^{\prime}$ of $V$ such that$V_{j_{k,0}-1}\subset U_{j_{k,0}}^{\prime}\overset{1}{\subset}V_{j_{k,0}}$, $U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\subset U_{j_{k,0}}^{\prime}$, and $(U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime})\cap U_{j_{k,0}}^{\prime}\neq U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}$.

$\sharp Z_{j_{k,0}}=\sharp\{U_{j_{k,0}}^{\prime}|U_{j_{k,0}}^{\prime}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\}-\sharp\{U_{j_{k,0}}^{\prime}|U_{j_{k,0}}^{\prime}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}\}$

$=\frac{v^{2(\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2\sum_{j>p_{k,1}}a_{j_{k,0},j}}}{v^{2}-1}.$

Third case. $p_{k-1,r_{k-1}}>p_{k,1}$. Consider the set $Z_{j_{k,0}}^{\prime\prime}$ of all subspace $U_{j_{k,0}}^{\prime\prime}$ of $V$ such that$V_{j_{k,0}-1}\subset U_{j_{k,0}}^{\prime\prime}\overset{1}{\subset}V_{j_{k,0}}$, $U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\subset U_{j_{k,0}}^{\prime\prime}$, and $(U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime})\cap U_{j_{k,0}}^{\prime\prime}\neq U_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}$.

$\sharp Z_{j_{k,0}}^{\prime\prime}=\sharp\{U_{j_{k,0}}^{\prime\prime}|U_{j_{k,0}}^{\prime\prime}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}-1}^{\prime}\}-\sharp\{U_{j_{k,0}}^{\prime\prime}|U_{j_{k,0}}^{\prime\prime}\supset V_{j_{k,0}-1}+V_{j_{k,0}}\cap V_{p_{k,1}}^{\prime}\}$

$=\frac{v^{2(\sum_{j\geq p_{k,1}}a_{j_{k,0},j})}-v^{2\sum_{j>p_{k,1}}(-1+a_{j_{k,0},j}})}{v^{2}-1}.$

Then the proposition follows. ∎

We assume that the ground field is an algebraic closure $\overline{\mathbb{F}}_{q}$ of $\mathbb{F}_{q}$ when we talk about the dimension of a $G$-orbit or its stabilizer. Set

[TABLE]

where $B=(b_{ij})$ is the diagonal matrix such that $b_{ii}=\sum_{k}a_{ik}$. Denote by ${\rm C}_{G}(V,V^{\prime})$ the stabilizer of $(V,V^{\prime})$ in $G$. From [1], we have known the following fact. If $A\in\Theta_{d}$, we have

[TABLE]

For any $A\in\Theta_{d},$ let $[A]=v^{-(d(A)-r(A))}e_{A}$, where $e_{A}$ stand for the characteristic function of $G$-obits associated to $A$.

Define

[TABLE]

[TABLE]

Proposition 3.1.5**.**

The functions $t_{ij},~{}\overline{t}_{ij}$ in $\mathcal{S}$, for any $i,~{}j\in[1,~{}n]$, satisfy the following relations.

[TABLE]

Proof.

We show the identity R1.

We will show it in the following several cases. By the first part of lemma 3.1.3, the following identities can be obtained by directly computing.

First case. $1\leq i<j<b<a\leq n$,

[TABLE]

[TABLE]

Hence

[TABLE]

Second case. $i<j\ \text{and}\ a\leq b$,

\overline{t}_{ia}\overline{t}_{jb}\\ =(v^{-1}-v)^{2}\sum\limits_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\sum\limits_{\lambda^{\prime}}v^{\lambda_{j}}[D_{\lambda}+E_{jb}]\\ =(v^{-1}-v)^{2}\sum\limits_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\sum\limits_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{ia}}\\ =\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}+1}e_{D_{\lambda}+E_{ia}+E_{jb}}&\text{if}\ i<j<a=b;\\[28.90755pt] (v^{-1}-v)^{2}(\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}\\ +\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-2}(v^{2}-1)e_{D_{\lambda}+E_{ib}+E_{ja}})&\text{if}\ i<j<a<b;\\[28.90755pt] (v^{-1}-v)^{2}(\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}\\ +\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-1\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ib}})&\text{if}\ i<j=a<b;\\[28.90755pt] (v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}&\text{if}\ i<a<j<b.\end{array}\right.

$\overline{t}_{jb}\overline{t}_{ia}\\ =(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{\lambda_{j}}[D_{\lambda}+E_{jb}]\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\\ =(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{ia}}\sum_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\\ =(v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}.$

\overline{t}_{ja}\overline{t}_{ib}=\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ib}+E_{ja}}&\text{if}\ i<j<a=b;\\[10.84006pt] (v^{-1}-v)^{2}\sum_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-1}e_{D_{\lambda}+E_{ib}+E_{ja}}&\text{if}\ i<j<a<b;\\[10.84006pt] -(v^{-1}-v)\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-1\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ib}}&\text{if}\ i<j=a<b;\\[10.84006pt] 0&\text{if}\ i<a<j<b.\end{array}\right.

Thus,

[TABLE]

Third case. $b<a\ \text{and}\ i\geq j$,

$\overline{t}_{ia}\overline{t}_{jb}\\ =(v^{-1}-v)^{2}\sum_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\sum_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{ia}}\\ =(v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}.$

\overline{t}_{jb}\overline{t}_{ia}=(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{\lambda_{j}}[D_{\lambda}+E_{jb}]\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\\ =(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda^{\prime}}+E_{jb}}\sum_{\lambda}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}}\\ =\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{jb}+E_{ia}}&\text{if}\ j<b<i<a;\\[28.90755pt] (v^{-1}-v)^{2}(\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}\\ +\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-1\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ja}})&\text{if}\ j<b=i<a;\\[28.90755pt] (v^{-1}-v)^{2}(\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ia}+E_{jb}}\\ +\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-2}(v^{2}-1)e_{D_{\lambda}+E_{ib}+E_{ja}})&\text{if}\ j<i<b<a;\\[28.90755pt] (v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}+1}e_{D_{\lambda}+E_{jb}+E_{ia}}&\text{if}\ j=i<b<a.\end{array}\right.

\overline{t}_{ja}\overline{t}_{ib}=\left\{\begin{array}[]{ll}0&\text{if}\ j<b<i<a;\\[10.84006pt] -(v^{-1}-v)\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-1\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ja}}&\text{if}\ j<b=i<a;\\[10.84006pt] (v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}-1}e_{D_{\lambda}+E_{ib}+E_{ja}}&\text{if}\ j<i<b<a;\\[10.84006pt] (v^{-1}-v)^{2}\sum\limits_{\begin{subarray}{c}\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})\\ s.t\sum_{k=1}^{n}\lambda_{k}=d-2\end{subarray}}v^{-\sum_{k=i+1}^{a-1}\lambda_{k}-\sum_{k=j+1}^{b-1}\lambda_{k}}e_{D_{\lambda}+E_{ja}+E_{ib}}&\text{if}\ j=i<b<a.\end{array}\right.

Therefore,

[TABLE]

Forth case. $j\leq i<a\leq b$. it is easy to know that when $j=i<a=b$, the identity R1 is equal. From the first case, when$j<i<a<b$, the identity R1 is also equal. From the second and the third case. When $j<i<a=b$ and $j=i<a<b$, the identity is right.

Thus, the identity R1 is equal.

The other identities can be shown similarly to Proposition 4.1.1 in [17]. ∎

Recall the partial order $``\leq"$ on $\Theta_{d}$ by $A\leq B$ if $\mathcal{O}_{A}\subset\overline{\mathcal{O}}_{B}$[1]. For any $A=(a_{ij})$ and $B=(b_{ij})$ in $\Theta_{\mathbf{d}}$, we say that $A\preceq B$ if and only if the following two conditions hold.

[TABLE]

The relation $``\preceq"$ defines a second partial order on $\Theta_{d}$.

Theorem 3.1.6**.**

For any $A=(a_{ij})\in\Theta_{\mathbf{d}}$. The following identity holds in $\mathcal{S}$

[TABLE]

where $\chi_{A}\in\mathcal{A}\setminus\{0\}$. The factors in the first product are taken in the following order: $(i,~{}j)$ comes before $(i^{\prime},~{}j^{\prime})$ if either $j<j^{\prime}$ or $j=j^{\prime},~{}i<i^{\prime}$. The factors in the second product are taken in the following order: $(i,~{}j)$ comes before $(i^{\prime},~{}j^{\prime})$ if either $j>j^{\prime}$ or $j=j^{\prime},~{}i>i^{\prime}$. The matrices $D_{i,j}$ are diagonal with entries in $\mathbb{N}$,which are uniquely determined.

Proof.

Assume n = 3 so that

[TABLE]

We define

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By the Lemma 3.1.3, we have

[TABLE]

Similarly, we can prove the proposition for the general case.

∎

3.2. Stabilization

Let $I$ be the identity matrix. We set ${}_{p}A=A+pI$. Let $\widetilde{\Theta}$ be the set of all $n\times n$ matrix with integer entries such that the entries off diagonal are $\geq 0$.

Let

[TABLE]

where the notation $[A]$ is a formal symbol. Let $v^{\prime}$ be a independent indeterminates, and $\mathfrak{R}$ be the ring $\mathbb{Q}(v)[v^{\prime}]$.

From [1], we have known the following results.

Proposition 3.2.1**.**

Suppose that $A_{1},~{}A_{2},~{}\cdots,~{}A_{r}\ (r\geq 2)$ are matrices in $\widetilde{\Theta}$ such that ${\rm co}(A_{i})={\rm ro}(A_{i+1})$ for $1\leq i\leq r-1$. There exist $Z_{1},~{}\cdots,~{}Z_{m}\in\widetilde{\Theta}$, $G_{j}(v,v^{\prime})\in\mathfrak{R}$ and $p_{0}\in\mathbb{N}$ such that in $\mathcal{S}_{d}$ for some $d$, we have

[TABLE]

By specialization $v^{\prime}$ at $v^{\prime}=1$, there is a unique associative $\mathcal{A}$-algebra structure on $\mathcal{K}$ without unit, where the product is given by

[TABLE]

if $A_{1},\cdots,A_{r}$ are as in Proposition 3.2.1.

Let $\hat{\mathcal{K}}$ be the $\mathbb{Q}(v)$-vector space of all formal sum $\sum\limits_{A\in\tilde{\Theta}}\xi_{A}[A]$ with $\xi_{A}\in\mathbb{Q}(v)$ and a locally finite property, that is, for any ${\mathbf{t}}\in\mathbb{Z}^{n}$, the sets $\{A\in\tilde{\Theta}~{}|~{}{\rm ro}(A)={\mathbf{t}},~{}\xi_{A}\neq 0\}$ and $\{A\in\widetilde{\Theta}~{}|~{}{\rm co}(A)={\mathbf{t}},~{}\xi_{A}\neq 0\}$ are finite. The space $\hat{\mathcal{K}}$ becomes an associative algebra over $\mathbb{Q}(v)$ which equipped with the following multiplication:

[TABLE]

where the product $[A]\cdot[B]$ is taken in $\mathcal{K}$.

For any nonzero matrix $A\in\widetilde{\Theta}$, let $\hat{A}$ be the matrix obtained by replacing diagonal entries of $A$ by zeroes. We set $\Theta^{0}=\{\hat{A}|A\in\widetilde{\Theta}\}.$

For any $\hat{\mathfrak{a}}$ in $\Theta^{0}$ and ${\mathbf{j}}=(j_{1},\cdots,j_{n})\in\mathbb{Z}^{n}$, we define

[TABLE]

where the sum runs through all $\lambda=(\lambda_{i})\in\mathbb{Z}^{n}$ such that $\hat{\mathfrak{a}}+D_{\lambda}\in\widetilde{\Theta}$, where $D_{\lambda}$ is the diagonal matrices with diagonal entries $(\lambda_{i}).$

Define

[TABLE]

[TABLE]

where $\underline{i}\in\mathbb{N}^{n}$ is the vector whose $i$-th entry is 1 and 0 elsewhere.

Let $\mathcal{U}$ be the subalgebra of $\hat{\mathcal{K}}$ generated by $\overline{t}_{ij},\overline{t}_{ii}$ for all $i,j\in[1,n]$ and $\mathbf{j}\in\mathbb{Z}^{n}$.

Proposition 3.2.2**.**

The generators $t_{ij},~{}\overline{t}_{ij}$ in $\mathcal{U}$, for any $i,~{}j\in[1,~{}n]$, satisfy the following relations.

[TABLE]

Proof.

The following identities can be obtained by directly computing by the first part of Lemma 3.1.3 and Proposition 3.2.1.

First case. $1\leq i<j<b<a\leq n$,

[TABLE]

[TABLE]

Hence

[TABLE]

Second case. $i<j\ \text{and}\ a\leq b$,

\overline{t}_{ia}\overline{t}_{jb}\\ =(v^{-1}-v)^{2}\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\sum_{\lambda^{\prime}}v^{\lambda_{j}^{\prime}}[D_{\lambda}+E_{jb}]\\ =\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}v(E_{ia}+E_{jb})(\underline{i}+\underline{j}),&\text{if}\ i<j<a=b;\\[10.84006pt] (v^{-1}-v)^{2}((E_{ia}+E_{jb})(\underline{i}+\underline{j})+(v-v^{-1})(E_{ib}+E_{ja})(\underline{i}+\underline{j}))&\text{if}\ i<j<a<b;\\[10.84006pt] (v^{-1}-v)^{2}((E_{ia}+E_{jb})(\underline{i}+\underline{j})+(E_{ib})(\underline{i}+\underline{j}))&\text{if}\ i<j=a<b;\\[10.84006pt] (v^{-1}-v)^{2}(E_{ia}+E_{jb})(\underline{i}+\underline{j}).&\text{if}\ i<a<j<b.\end{array}\right.

$\overline{t}_{jb}\overline{t}_{ia}\\ =(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{\lambda_{j}^{\prime}}[D_{\lambda}+E_{jb}]\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\\ =(v^{-1}-v)^{2}(E_{ia}+E_{jb})(\underline{i}+\underline{j}).$

\overline{t}_{ja}\overline{t}_{ib}=\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}(E_{ia}+E_{jb})(\underline{i}+\underline{j}).&\text{if}\ i<j<a=b;\\[10.84006pt] (v^{-1}-v)^{2}(E_{ib}+E_{ja})(\underline{i}+\underline{j})&\text{if}\ i<j<a<b;\\[10.84006pt] -(v^{-1}-v)E_{ib}(\underline{i}+\underline{j})&\text{if}\ i<j=a<b;\\[10.84006pt] 0&\text{if}\ i<a<j<b.\end{array}\right.

Thus,

[TABLE]

Third case. $b<a\ \text{and}\ i\geq j$,

$\overline{t}_{ia}\overline{t}_{jb}\\ =(v^{-1}-v)^{2}\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\sum_{\lambda^{\prime}}v^{\lambda_{j}^{\prime}}[D_{\lambda}+E_{jb}]\\ =(v^{-1}-v)^{2}v^{\delta_{ij}}(E_{ia}+E_{jb})(\underline{i}+\underline{j})$

\overline{t}_{jb}\overline{t}_{ia}\\ =(v^{-1}-v)^{2}\sum_{\lambda^{\prime}}v^{\lambda_{j}^{\prime}}[D_{\lambda}+E_{jb}]\sum_{\lambda}v^{\lambda_{i}}[D_{\lambda}+E_{ia}]\\ =\left\{\begin{array}[]{ll}(v^{-1}-v)^{2}(E_{ia}+E_{jb})(\underline{i}+\underline{j}),&\text{if}\ j<b<i<a;\\[10.84006pt] (v^{-1}-v)^{2}((E_{ia}+E_{jb})(\underline{i}+\underline{j})+E_{ja}(\underline{i}+\underline{j}))&\text{if}\ j<b=i<a;\\[10.84006pt] (v^{-1}-v)^{2}((E_{ia}+E_{jb})(\underline{i}+\underline{j})+(v-v^{-1})(E_{ib}+E_{ja})(\underline{i}+\underline{j}))&\text{if}\ j<i<b<a;\\[10.84006pt] (v^{-1}-v)^{2}v^{2}(E_{jb}+E_{ia})(\underline{i}+\underline{j})&\text{if}\ j=i<b<a.\end{array}\right.

\overline{t}_{ja}\overline{t}_{ib}\\ =\left\{\begin{array}[]{ll}0&\text{if}\ j<b<i<a;\\[10.84006pt] -(v^{-1}-v)E_{ja}(\underline{i}+\underline{j})&\text{if}\ j<b=i<a;\\[10.84006pt] (v^{-1}-v)^{2}(E_{ib}+E_{ja})(\underline{i}+\underline{j})&\text{if}\ j<i<b<a;\\[10.84006pt] (v^{-1}-v)^{2}v(E_{ja}+E_{ib})(\underline{i}+\underline{j})&\text{if}\ j=i<b<a.\end{array}\right.

Therefore

[TABLE]

Forth case. $j\leq i<a\leq b$. it is easy to know that when $j=i<a=b$, the identity is equal. From the first case, when$j<i<a<b$, the identity is also equal.From the second and the third case. When $j<i<a=b$ and $j=i<a<b$, the identity is right. Thus, The proposition follows.

∎

The previous proposition shows that $\mathbf{U}$ satisfied the defining relations of the positive part of $U_{v}(gl_{n})$ with respect to the RTT relations. In fact , $\mathbf{U}\cong U_{v}(gl_{n})^{+}$. That is, we give the realization of the $U_{v}(gl_{n})^{+}$ with respect to the RTT relations.

Acknowledgements: This work is supported by NSFC 11571119 and NSFC 11475178.