Global bases for quantum Borcherds–Bozec algebras

Abstract

We provide an explicit construction of global bases for quantum Borcherds–Bozec algebras and their irreducible highest weight modules. Moreover, we give a new presentation for quantum Borcherds–Bozec algebras in terms of primitive generators.

1 Introduction

The quantum Borcherds–Bozec algebras were introduced by Bozec [1, 2] in terms of generators and relations when he solved a question asked by Lusztig in [15]. Namely, if we consider a quiver with loops, the Grothendieck group arising from Lusztig sheaves on representation varieties is generated by the elementary simple perverse sheaves $F_i^{(n)}$ with all vertices i and $n\in {\mathbf {N}}$. Bozec proved an analogue of Gabber–Kac theorem for the negative part $U_q^-({\mathfrak {g}})$ of a quantum Borcherds–Bozec algebra, and showed that the above Grothendieck group is isomorphic to $U_{q}^{-}({\mathfrak {g}})$, which gives a construction of its canonical basis.

The canonical basis theory was first introduced by Lusztig in the simply-laced case in [13], arising from his geometric construction of the negative parts of quantum groups, and it has been generalized to symmetric Kac–Moody type in [14, 16]. On the other hand, Kashiwara constructed the crystal bases and global bases for quantum groups associated with symmetrizable Kac–Moody algebras in an algebraic way [11, 12]. In [5], Grojnowski and Lusztig proved that Kashiwara’s global bases coincide with Lusztig’s canonical bases. The canonical/crystal basis theory has become one of the most central themes in combinatorial and geometric representation theory of quantum groups because it provides us with very powerful geometric and combinatorial tools to investigate the structure of quantum groups and their integrable representations. In [7], Jeong, Kang, and Kashiwara developed the crystal basis theory for quantum Borcherds algebras, which were introduced in [8]. In [10], Kang and Schiffmann gave a construction of canonical basis for quantum Borcherds algebras and proved that, when all the diagonal entries of the corresponding Borcherds–Cartan matrices are non-zero, the canonical bases coincide with global bases.

Bozec’s crystal basis theory for quantum Borcherds–Bozec algbras is based on primitive generators ${\texttt {a}}_{il}$, ${{\texttt {b}}}_{il}$ $(i,l) \in I^{\infty }$, not on the Chevalley generators $e_{il}$, $f_{il}$ $(i,l) \in I^{\infty }$. The primitive generators have simpler commutation relations than Chevalley generators. Bozec defined the Kashiwara operators using primitive generators and proved several crucial theorems which are important steps for Kashiwara’s grand-loop argument [2,  Lemma 3.33, Lemma 3.34]. In this way, even though he did not check all the details, he was able to deduce that there exist unique crystal bases for quantum Borcherds–Bozec algbras and their integrable highest weight modules. Moreover, using Lusztig’s and Nakajima’s quiver varieties, he also gave a geometric construction of $\mathcal B(\infty )$, the crystal of the negative half $U_{q}^{-}({\mathfrak {g}})$, and ${\mathcal {B}}(\lambda )$, the crystal of the integrable highest weight representation $V(\lambda )$, respectively.

The main goal of this paper is to construct the global bases for quantum Borcherds–Bozec algebras and their irreducible highest weight modules. As the first step, we give an explicit description of the radical ${{\mathscr {R}}}$ of Lusztig’s bilinear form (Theorem 4). The higher order Serre relations we obtained have more general forms than those given in [1]. As a direct application, we give a new presentation of the quantum Borcherds–Bozec algebra $U_q({\mathfrak {g}})$ in terms of primitive generators (Proposition 7).

Then we set up the frame work that can be found in [7, 12]. However, we still need more preparations. In the case of quantum Borcherds–Bozec algebras, for each $I^{\text {im}}$, there are infinitely many generators with higher degrees. Thus, compared with quantum Borcherds algebras, we need to take a much more complicated approach to the construction of global bases. To overcome these difficulties, we introduce a very natural and much expanded notion of balanced triples corresponding to the compositions or partitions of each higher degree of primitive generators (Proposition 22, Corollary 23). As can be expected, to prove our assertions, the imaginary indices with higher degrees should be treated with special care. In particular, the isotropic case (i.e., when $a_{ii}=0$) requires very subtle and delicate treatments.

Now we can follow the steps given in [7, 12] and prove the existence and uniqueness of global bases (Theorem 26). The key ingredients of our proof are Proposition 22 and Corollary 23. We conjecture that our global bases coincide with (a variation of) Bozec’s canonical bases.

This paper is organized as follows. In Sect. 2, we give an explicit description of the radical ${\mathscr {R}}$ of Lusztig’s bilinear form $( \ , \ )_L$ via higher order quantum Serre relations in quantum Borcherds–Bozec algebras. In Sect. 3, we give a new presentation of quantum Borcherds–Bozec algebras in terms of primitive generators as an application of higher order quantum Serre relations. In Sect. 4, we review the crystal basis theory for quantum Borcherds–Bozec algebras and give canonical characterizations of the crystal bases $({\mathcal {L}}(\infty ),{\mathcal {B}}(\infty ))$ and $({\mathcal {L}}(\lambda ), {\mathcal {B}}(\lambda ))$, respectively. We also define the quantum Boson algebra ${\mathscr {B}}_q({\mathfrak {g}})$ for an arbitrary Borcherds–Cartan datum. In Sect. 5, we define the ${\mathbb {A}}$-forms $U_{{\mathbb {A}}}({\mathfrak {g}})$ of $U_{q}({\mathfrak {g}})$ and $V(\lambda )^{{\mathbb {A}}}$ of $V(\lambda )$, respectively. We prove that $U_{{\mathbb {A}}}({\mathfrak {g}})$ has the triangular decomposition and both $U_{{\mathbb {A}}}^-({\mathfrak {g}})$ and $V(\lambda )^{{\mathbb {A}}}$ are stable under the Kashiwara operators. In Sect. 6, we prove the existence and uniqueness of global bases. As expected, most of this section is devoted to the proof of Proposition 22 and Corollary 23.

2 Higher order quantum serre relations

Let I be an index set possibly countably infinite. An integer-valued matrix $A=(a_{ij})_{i,j\in I}$ is called an even symmetrizable Borcherds–Cartan matrix if it satisfies the following conditions:

• (i) $a_{ii}=2, 0, -2, -4, \ldots$,

• (ii) $a_{ij} \in {\mathbf {Z}}_{\le 0}$ for $i \ne j$,

• (iii) there is a diagonal matrix $D=\text {diag} (r_i \in {\mathbf {Z}}_{>0} \mid i \in I)$ such that DA is symmetric.

Let $I^{\text {re}}:=\{ i \in I \mid a_{ii}=2 \}$, $I^{\text {im}}:= \{ i \in I \mid a_{ii} \le 0 \}$, and $I^{\text {iso}}: =\{ i \in I \mid a_{ii}=0 \}$. The elements of $I^{\text {re}}$ (resp. $I^{\text {im}}$, $I^{\text {iso}}$) are called real indices (resp. imaginary indices, isotropic indices).

A Borcherds–Cartan datum consists of

• (a) an even symmetrizable Borcherds–Cartan matrix $A=(a_{ij})_{i,j \in I}$,

• (b) a free abelian group $P^{\vee }=\left( \bigoplus _{i\in I}{{\mathbf {Z}} h_i}\right) \oplus \left( \bigoplus _{i\in I}{{\mathbf {Z}} d_i}\right)$, the dual weight lattice,

• (c) ${\mathfrak {h}}={\mathbf {Q}} \otimes _{{\mathbf {Z}}}P^{\vee }$, the Cartan subalgebra,

• (d) $P=\{ \lambda \in {\mathfrak {h}}^* \mid \lambda (P^{\vee })\subseteq {\mathbf {Z}} \}$, the weight lattice,

• (e) $\Pi ^{\vee } = \{ h_i \in P^{\vee } \mid i \in I \}$, the set of simple coroots,

• (f) $\Pi = \{ \alpha _i \in P \mid i \in I \}$, the set of simple roots, which is linearly independent over ${\mathbf {Q}}$ and satisfies

  $$\begin{aligned} \alpha _j(h_i)=a_{ij},\ \alpha _j(d_i)=\delta _{ij} \ \ \text {for all} \ i, j \in I. \end{aligned}$$

• (g) for each $i \in I$, there is a $\Lambda _i \in P$, called the fundamental weight, defined by

  $$\begin{aligned} \Lambda _i(h_j)=\delta _{ij}, \ \Lambda _i(d_j)=0\ \ \text {for all} \ i, j \in I. \end{aligned}$$

We denote by $P^+$ the set $\{ \lambda \in P \mid \lambda (h_i) \ge 0 \text {~~for all~} i \in I \}$ of dominant integral weights. The free abelian group $Q:=\bigoplus _{i \in I} {\mathbf {Z}} \alpha _i$ is called the root lattice. Set $Q_{+} = \sum _{i \in I} {\mathbf {Z}}_{\ge 0} \alpha _i$ and $Q_{-} = - Q_{+}$. For $\beta = \sum k_i \alpha _i \in Q_{+}$, we define its height to be $\vert \beta \vert : = \sum k_i$.

There is a non-degenerate symmetric bilinear form $( \ , \ )$ on ${\mathfrak {h}}^*$ satisfying

$$\begin{aligned} (\alpha _i, \lambda )=r_i\lambda (h_i), \ (\Lambda _i, \lambda )=r_i \lambda (d_i) \ \ \text {for any} \ \lambda \in {\mathfrak {h}}^* \ \text {and} \ i \in I, \end{aligned}$$

and therefore we have

$$\begin{aligned} (\alpha _i, \alpha _j)=r_i a_{ij}=r_j a_{ji} \ \text {for all} \ i, j \in I. \end{aligned}$$

For $i \in I^{\text {re}}$, we define the simple reflection $\omega _{i} \in GL({\mathfrak {h}}^{*})$ by

$$\begin{aligned} \omega _{i} (\lambda ) = \lambda -\lambda (h_{i}) \alpha _{i} \end{aligned}$$

for all $\lambda \in {\mathfrak {h}}^{*}$. The subgroup W of $GL({\mathfrak {h}}^*)$ generated by $\omega _{i}$ $(i \in I^{\text {re}})$ is called the Weyl group of the Borcherds–Cartan datum given above. Note that the symmetric bilinear form $( \ , \ )$ is W-invariant.

Let $I^{\infty }:= (I^{\text {re}} \times \{1\}) \cup (I^{\text {im}} \times {\mathbf {Z}}_{>0})$. If $i \in I^{\text {re}}$, we often write i for (i, 1). Let q be an indeterminate and set

$$\begin{aligned} q_i=q^{r_i},\quad q_{(i)}=q^{\frac{(\alpha _i,\alpha _i)}{2}}. \end{aligned}$$

For each $i \in I^{\text{ re }}$ and $n \in {\mathbf {Z}}_{\ge 0}$, we define

$$\begin{aligned}{}[n]_i=\frac{q_{i}^n-q_{i}^{-n}}{q_{i}-q_{i}^{-1}},\quad [n]_i!=\prod _{k=1}^n [k]_i,\quad {\begin{bmatrix} n \\ k \end{bmatrix}}_i=\frac{[n]_i!}{[k]_i![n-k]_i!}. \end{aligned}$$

Let ${\mathscr {F}}={\mathbf {Q}}(q)\langle f_{il} \mid (i,l) \in I^{\infty } \rangle$ be the free associative algebra over ${\mathbf {Q}}(q)$ generated by the symbols $f_{il}$ for $(i,l)\in I^{\infty }$. By setting $\deg f_{il}= -l\alpha _{i}$, ${\mathscr {F}}$ become a $Q_-$-graded algebra. For a homogeneous element u in ${\mathscr {F}}$, we denote by |u| the degree of u, and for any subset $A \subseteq Q_{-}$, set ${{\mathscr {F}}}_{A}=\{ x\in {{\mathscr {F}}} \mid \vert x\vert \in A \}$.

We define a twisted multiplication on ${\mathscr {F}}\otimes {\mathscr {F}}$ by

$$\begin{aligned} (x_1\otimes x_2)(y_1\otimes y_2)=q^{-(\left| x_2\right| , \left| y_1 \right| )}x_1y_1\otimes x_2y_2, \end{aligned}$$

for all homogeneous $x_1,x_2,y_1,y_2\in {\mathscr {F}}$, and equip ${\mathscr {F}}$ with a co-multiplication $\varrho$ defined by

$$\begin{aligned} \varrho (f_{il})=\sum _{m+n=l}q_{(i)}^{-mn}f_{im}\otimes f_{in} \ \ \text {for} \ (i,l)\in I^{\infty }. \end{aligned}$$

Here, we understand $f_{i0}=1$ and $f_{il}=0$ for $l<0$.

Proposition 1

[1, 2] For any family $\nu =(\nu _{il})_{(i,l)\in I^{\infty }}$ of non-zero elements in ${\mathbf {Q}}(q)$, there exists a symmetric bilinear form $( \ , \ )_L :{\mathscr {F}}\times {\mathscr {F}}\rightarrow {\mathbf {Q}}(q)$ such that

• (a) $(x, y)_{L} =0$ if $\vert x\vert \ne \vert y\vert$,

• (b) $(1,1)_{L} = 1$,

• (c) $(f_{il}, f_{il})_{L} = \nu _{il}$ for all $(i,l) \in I^{\infty }$,

• (d) $(x, yz)_{L} = (\varrho (x), y \otimes z)_L$ for all $x,y,z \in {{\mathscr {F}}}$.

Here, $(x_1\otimes x_2,y_1\otimes y_2)_L=(x_1,y_1)_L(x_2,y_2)_L$ for any $x_1,x_2,y_1,y_2\in {{\mathscr {F}}}$.

We denote by ${\mathscr {R}}$ the radical of $( \ , \ )_L$.

Let ${\mathcal {C}}_l$ be the set of compositions ${\mathbf {c}}$ of l, and set $f_{i,{\mathbf {c}}}=f_{ic_1}\ldots f_{ic_m}$ for every $i\in I^{\text {im}}$ and every ${\mathbf {c}}=(c_1,\ldots ,c_m)\in \mathcal C_l$ . Assume ${\mathcal {C}}_l=\{{\mathbf {c}}_1, {\mathbf {c}}_2, \ldots ,{\mathbf {c}}_r\}$, we see that $f_{i,{\mathbf {c}}_1},\ldots ,f_{i,{\mathbf {c}}_r}$ form a basis of $\mathscr {F}_{-l\alpha _i}$. Hence, for any homogeneous element x in $\mathscr {F}$, $\varrho (x)$ can be written into the forms

$$\begin{aligned} \varrho (x)= & {} x_{{\mathbf {c}}_1}\otimes f_{i,{\mathbf {c}}_1}+\cdots +x_{{\mathbf {c}}_r}\otimes f_{i,{\mathbf {c}}_r}+\ \text {terms of bidegree not in}\ Q_-\times -l\alpha _i,\\ \varrho (x)= & {} f_{i,{\mathbf {c}}_1}\otimes x_{{\mathbf {c}}_1}'+\cdots +f_{i,{\mathbf {c}}_r}\otimes x_{{\mathbf {c}}_r}'+\ \text {terms of bidegree not in}\ -l\alpha _i\times Q_-. \end{aligned}$$

We denote by $\varrho _{i,l}(x), \varrho ^{i,l}(x):\mathscr {F}\rightarrow {\mathscr {F}}^r$ the ${\mathbf {Q}}(q)$-linear maps:

$$\begin{aligned} \varrho _{i,l}(x)=(x_{{\mathbf {c}}_1},\ldots ,x_{{\mathbf {c}}_r}), \ \varrho ^{i,l}(x)=(x_{{\mathbf {c}}_1}',\ldots ,x_{{\mathbf {c}}_r}'). \end{aligned}$$

If x, y are homogeneous elements such that $\varrho _{i,k}(y)=0$ for any $k>0$, then we have

$$\begin{aligned} \varrho _{i,l}(xy)=q^{l(\alpha _i,|y|)}\varrho _{i,l}(x)y\ \ \text {and} \ \ \varrho _{i,l}(yx)=y\varrho _{i,l}(x). \end{aligned}$$

Here, $\varrho _{i,l}(x)y=(x_{{\mathbf {c}}_1}y,\ldots ,x_{{\mathbf {c}}_r}y)$ and $y\varrho _{i,l}(x)=(yx_{{\mathbf {c}}_1},\ldots ,yx_{{\mathbf {c}}_r})$ if $\varrho _{i,l}(x)=(x_{{\mathbf {c}}_1},\ldots ,x_{{\mathbf {c}}_r})$. Similarly, if $\varrho ^{i,k}(y)=0$ for any $k>0$, we have

$$\begin{aligned} \varrho ^{i,l}(xy)=\varrho ^{i,l}(x)y\ \ \text {and} \ \ \varrho ^{i,l}(yx)=q^{l(\alpha _i,|y|)}y\varrho ^{i,l}(x). \end{aligned}$$

For $i\in I^{\text {re}}$, we define the ${\mathbf {Q}}(q)$-linear maps $\varrho _i,\varrho ^i:{\mathscr {F}}\rightarrow {\mathscr {F}}$ by

$$\begin{aligned} \varrho _i(1)= & {} 0, \varrho _i(f_{j,k})=\delta _{i,j}\delta _{k,1}, \ \text {and} \ \varrho _i(xy)=q^{(\alpha _i,\vert y\vert )}\varrho _i(x)y+x\varrho _i(y),\\ \varrho ^i(1)= & {} 0, \varrho ^i(f_{j,k})=\delta _{i,j}\delta _{k,1}, \ \text {and} \ \varrho ^i(xy)=\varrho ^i(x)y+q^{(\alpha _i,\vert x\vert )}x\varrho ^i(y), \end{aligned}$$

for all homogeneous x, y. Note that for any homogeneous $x\in {\mathscr {F}}$, we have

$$\begin{aligned} \varrho (x)= & {} \varrho _i(x)\otimes f_{i}+\ \text {terms of other bi-homogeneities},\\ \varrho (x)= & {} f_{i}\otimes \varrho ^i(x)+\ \text {terms of other bi-homogeneities}. \end{aligned}$$

The following lemma can be derived directly from the definitions of $\varrho _{i,l}$ and $\varrho ^{i,l}$.

Lemma 2

• (a) If $i\in I^{\text {re}}$, then for any $x,y \in {\mathscr {F}}$, we have

  $$\begin{aligned} (y f_{i},x)_L=(f_i,f_i)_L(y,\varrho _i(x))_L, \ (f_{i}y,x)_L=(f_i,f_i)_L(y,\varrho ^i(x))_L. \end{aligned}$$

• (b) If $i\in I^{\text {im}}$, let $x\in {\mathscr {F}}$ with $\varrho _{i,l}(x)=(x_{{\mathbf {c}}_1},\cdots ,x_{{\mathbf {c}}_r})$ and $\varrho ^{i,l}(x)=(x_{{\mathbf {c}}_1}',\cdots ,x_{{\mathbf {c}}_r}')$. Then for any $x\in {\mathscr {F}}$, we have

  $$\begin{aligned} (yf_{il},x)_L= & {} (f_{il},f_{i,{\mathbf {c}}_1})_L(y,x_{{\mathbf {c}}_1})_L+\cdots +(f_{il},f_{i,{\mathbf {c}}_r})_L(y,x_{{\mathbf {c}}_r})_L,\\ (f_{il}y,x)_L= & {} (f_{il},f_{i,{\mathbf {c}}_1})_L(y,x_{{\mathbf {c}}_1}')_L+\cdots +(f_{il},f_{i,{\mathbf {c}}_r})_L(y,x_{{\mathbf {c}}_r}')_L. \end{aligned}$$

• (c) Let $x\in {\mathscr {F}}$ be a homogeneous element with $\vert x\vert \ne 0$, we have

• (i) if $\varrho _{i,l}(x)\in {\mathscr {R}}$ for any $(i,l)\in I^{\infty }$, then $x\in {\mathscr {R}}$,

• (ii) if $\varrho ^{i,l}(x)\in {\mathscr {R}}$ for any $(i,l)\in I^{\infty }$, then $x\in {\mathscr {R}}$.

  Here, if $i\in I^{\text {im}}$, $\varrho ^{i,l}(x)\in {\mathscr {R}}$ means each component of $\varrho ^{i,l}(x)$ belongs to ${\mathscr {R}}$.

For any $i\in I^{\text {re}}$ and $n \in {\mathbf {N}}$, we set

$$\begin{aligned} f_i^{(n)}=\frac{f_i^{n}}{[n]_i!}. \end{aligned}$$

By a similar argument in [16,  1.4.2], we can prove:

Lemma 3

We have

$$\begin{aligned} \varrho (f_i^{(n)})=\sum _{p+p'=n}q_i^{-pp'}f_i^{(p)}\otimes f_i^{(p')} \end{aligned}$$

(1)

for any $i\in I^{\text {re}}$ and $n \in {\mathbf {N}}$.

Theorem 4

Assume that $i\in I^{\text {re}}$, $j\in I$ and $i\ne j$. Let $m\in {\mathbf {Z}}_{>0}$, $n\in {\mathbf {Z}}_{\ge 0}$ with $m>-a_{ij}n$. Then for any ${\mathbf {c}}\in {\mathcal {C}}_n$, the following element of ${\mathscr {F}}$

$$\begin{aligned} {\texttt {F}}_{i,j,m,n,{\mathbf {c}},\pm 1}=\sum _{r+s=m}(-1)^rq_i^{\pm r(-a_{ij}n-m+1)}f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s)} \end{aligned}$$

(2)

belongs to ${\mathscr {R}}$. Here, we put $f_{j,{\mathbf {c}}}=f_j^{n}$ for $j \in I^{\text {re}}$.

Proof

If $n=0$, then

$$\begin{aligned} {\texttt {F}}_{i,j,m,0,{{\mathbf {c}}},\pm 1}=\sum _{r+s=m}(-1)^rq_i^{\pm r(1-m)}f_i^{(r)}f_i^{(s)}. \end{aligned}$$

Since $\sum _{r+s=m}(-1)^rq_i^{\pm r(1-m)}\begin{bmatrix} m \\ r \end{bmatrix}_i=0$, we have ${\texttt {F}}_{i,j,m,0,{{\mathbf {c}}},\pm 1}=0$.

We shall only show it for $j\in I^{\text {im}}$, and the case $j\in I^{\text {re}}$ can be shown similarly. By Lemma 2 (c), it is enough to show each component of $\varrho ^{i',k}({\texttt {F}}_{i,j,m,n,{{\mathbf {c}}},\pm 1})$ belongs to ${\mathscr {R}}$. If $i'\ne i, j$, there is nothing to show.

We first show it for $i'=j$. For $0<k\le n$ and ${\mathbf {c}}=(n_1,\ldots ,n_t)\in {\mathcal {C}}_n$, we have

$$\begin{aligned} \begin{aligned}&\varrho ^{j,k}(f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s)})= \varrho ^{j,k}(f_i^{(r)}f_{j,{\mathbf {c}}})f_i^{(s)}=q^{-(r\alpha _i,k\alpha _j)}f_i^{(r)}\varrho ^{j,k}(f_{j,{\mathbf {c}}})f_i^{(s)}\\&=q^{-(r\alpha _i,k\alpha _j)}f_i^{(r)}\left( \beta _{a_1,\ldots ,a_t} f_{j,(n_1-a_1,\ldots ,n_t-a_t)}\right) _{\begin{array}{c} a_1\le n_1,\ldots ,a_t\le n_t \\ a_1+\cdots +a_t=k \end{array}}f_i^{(s)}, \end{aligned} \end{aligned}$$

(3)

where

$$\begin{aligned} \beta _{a_1,\ldots ,a_t}=q_{(j)}^{\sum _{h=1}^ta_h(a_h-n_h)}q_{(j)}^{2\sum _{1\le p< q\le t}(a_p-n_p)a_q}. \end{aligned}$$

Note that $m>-a_{i,j}n\ge -a_{i,j}(n-k)$ and

$$\begin{aligned} q_i^{ -r(-a_{ij}n-m+1)}q^{-(r\alpha _i,k\alpha _j)}=q_i^{ -r[-a_{ij}(n-k)-m+1]}. \end{aligned}$$

Therefore each component of $\varrho ^{j,k}({\texttt {F}}_{i,j,m,n,{\mathbf {c}},-1})$ is a scalar multiple of ${\texttt {F}}_{i,j,m,n-k,{\mathbf {c}}',-1}$ for some ${\mathbf {c}}'\in \mathcal C_{n-k}$.

We now show it for $i'= i$. Since $i\in I^{\text {re}}$, we have

$$\begin{aligned} \begin{aligned} \varrho ^{i}(f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s)})&=\varrho ^i(f_i^{(r)}f_{j,{\mathbf {c}}})f_i^{(s)}+q^{-(r\alpha _i+n\alpha _j,\alpha _i)}q_i^{1-s}f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s-1)}\\&=q_i^{1-r}f_i^{(r-1)}f_{j,{\mathbf {c}}}f_i^{(s)}+q^{-(r\alpha _i+n\alpha _j,\alpha _i)}q_i^{1-s}f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s-1)}. \end{aligned} \end{aligned}$$

(4)

Hence

$$\begin{aligned} \begin{aligned} \varrho ^{i}({\texttt {F}}_{i,j,m,n,{\mathbf {c}},-1})&=\sum _{r+s=m}(-1)^rq_i^{-r(-a_{ij}n-m+1)}q_i^{1-r}f_i^{(r-1)}f_{j,{\mathbf {c}}}f_i^{(s)}\\&\qquad +\sum _{r+s=m}(-1)^rq_i^{-r(-a_{ij}n-m+1)}q^{-(r\alpha _i+n\alpha _j,\alpha _i)}q_i^{1-s}f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s-1)}. \end{aligned} \end{aligned}$$

(5)

Note that the coefficient of $f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s-1)}$ is

$$\begin{aligned} \begin{aligned}&q_i^{-(r+1)(-a_{ij}n-m+1)}q_i^{-r}-q_i^{-r(-a_{ij}n-m+1)}q^{-(r\alpha _i+n\alpha _j,\alpha _i)}q_i^{1-m+r}\\&\quad =q_i^{-(r+1)(-a_{ij}n-m+1)}q_i^{-r}-q_i^{-r(-a_{ij}n-m+1)}q_i^{-2r-na_{ij}}q_i^{1-m+r}\\&\quad =q_i^{-r(-a_{ij}n-m+2)}q_i^{a_{ij}n+m-1}(1-q_i^{-2m-2na_{ij}+2}). \end{aligned} \end{aligned}$$

(6)

Therefore

$$\begin{aligned} \varrho ^{i}({\texttt {F}}_{i,j,m,n,{\mathbf {c}},-1})&=(1-q_i^{-2m-2na_{ij}+2})q_i^{a_{ij}n+m-1}\\&\quad \cdot \sum _{r+s=m-1}(-1)^rq_i^{-r(-a_{ij}n-m+2)}f_i^{(r)}f_{j,{\mathbf {c}}}f_i^{(s)} \\&= {\left\{ \begin{array}{ll} \beta {\texttt {F}}_{i,j,m-1,n,{\mathbf {c}},-1} &{} \text {if} \ \ m>-a_{ij}n+1,\\ 0 &{} \text {if} \ \ m=-a_{ij}n+1. \end{array}\right. }\\ \end{aligned}$$

Here $\beta =(1-q_i^{-2m-2na_{ij}+2})q_i^{a_{ij}n+m-1}$ is a constant.

By the induction and Lemma 2(c), theorem follows. $\square$

In particular, when $m=1-la_{ij}, n=l$ and ${\mathbf {c}}=(l)$, by Theorem 4, we conclude

$$\begin{aligned} {\texttt {F}}_{i,j,m,n,{\mathbf {c}},\pm 1}={\left\{ \begin{array}{ll}\sum _{r+s=1-la_{ij}}(-1)^rf_i^{(r)}f_{j}^{l}f_i^{(s)} &{}\text {if} \ j\in I^{\text {re}},\\ \sum _{r+s=1-la_{ij}}(-1)^rf_i^{(r)}f_{jl}f_i^{(s)} &{}\text {if} \ j\in I^{\text {im}},\end{array}\right. } \end{aligned}$$

belongs to ${\mathscr {R}}$.

Lemma 5

Let $(i,k),(j,l)\in I^{\infty }$ such that $a_{ij}=0$. Set $X=f_{ik}f_{jl}-f_{jl}f_{ik}$, Then $X\in {\mathscr {R}}$.

Proof

Note that if $i,j\in I^{\text {re}}$, then $X=f_{i}f_{j}-f_{j}f_{i}$. Since i and j cannot be equal, we have $X=-{\texttt {F}}_{i,j,1,1,(1),\pm 1}\in {\mathscr {R}}$.

If $i\in I^{\text {re}}$ and $j\in I^{\text {im}}$, we have $X=f_{i}f_{jl}-f_{jl}f_{i}=-{\texttt {F}}_{i,j,1,l,{\mathbf {c}}=(l), \pm 1}\in {\mathscr {R}}$.

We now assume that $i,j\in I^{\text {im}}$ and $i=j\in I^{\text {iso}}$. Note for any $0<s\le k+l$, we have

$$\begin{aligned}&\varrho ^{i,s}(X)=\varrho ^{i,s}(f_{ik}f_{il}-f_{il}f_{ik})\\&=\left( f_{i,k-a_1}f_{i,l-a_2}-f_{i,l-a_2}f_{i,k-a_1}\right) _{\begin{array}{c} a_1\le k,a_2\le l \\ a_1+a_2=s \end{array}}. \end{aligned}$$

By induction and Lemma 2, $X\in {\mathscr {R}}$.

Finally, if $i,j\in I^{\text {im}}$ and $i\ne j$, then for any $0<s\le k$ and $0<t\le l$, we have

$$\begin{aligned}&\varrho ^{i,s}(X)=\varrho ^{i,s}(f_{ik}f_{jl}-f_{jl}f_{ik})=q_{(i)}^{-s(k-s)} (f_{i,k-s}f_{jl}-f_{jl}f_{i,k-s}),\\&\varrho ^{j,t}(X)=\varrho ^{j,t}(f_{ik}f_{jl}-f_{jl}f_{ik})=q_{(j)}^{-t(l-t)} (f_{ik}f_{j,l-t}-f_{j,l-t}f_{ik}). \end{aligned}$$

By induction and Lemma 2, $X\in {\mathscr {R}}$. $\square$

3 Quantum Borcherds–Bozec algebras

From now on, we always assume that

$$\begin{aligned} \nu _{il} \in 1+q{\mathbf {Z}}_{\ge 0}[[q]]\ \ \text {for all} \ (i,l)\in I^{\infty }. \end{aligned}$$

(7)

Under this assumption, the bilinear form $( \ , \ )_L$ is non-degenerate on ${\mathscr {F}}(i)=\bigoplus _{l\ge 1} {\mathscr {F}}_{-l \alpha _i}$ for $i\in I^{\mathrm{im}} \backslash I^{\mathrm{iso}}$. Moreover, it has been showed in [1,  Proposition 14] that the two-sided ideal ${\mathscr {R}}$ is generated by

$$\begin{aligned} \sum _{r+s=1-la_{ij}}(-1)^rf_i^{(r)}f_{jl}f_i^{(s)} \ \ \text {for} \ i\in I^{\text {re}},(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \end{aligned}$$

and $f_{ik}f_{jl}-f_{jl}f_{ik}$ for all $(i,k),(j,l)\in I^{\infty }$ with $a_{ij}=0$

Given a Borcherds–Cartan datum $(A, P,P^{\vee },\Pi , \Pi ^{\vee })$, we denote by $\widehat{U}$ the associative algebra over ${\mathbf {Q}}(q)$ with ${\mathbf {1}}$, generated by the elements $q^h$ $(h\in P^{\vee })$ and $e_{il}, f_{il}$ $((i,l) \in I^{\infty })$ with defining relations

$$\begin{aligned} \begin{aligned}&q^0={\mathbf {1}},\quad q^hq^{h'}=q^{h+h'} \ \ \text {for} \ h,h' \in P^{\vee } \\&q^h e_{jl}q^{-h} = q^{l\alpha _j(h)} e_{jl}, \ \ q^h f_{jl}q^{-h} = q^{-l\alpha _j(h)} f_{jl}\ \ \text {for} \ h \in P^{\vee }, (j,l)\in I^{\infty }, \\&\sum _{r+s=1-la_{ij}}(-1)^r {e_i}^{(r)}e_{jl}e_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \\&\sum _{r+s=1-la_{ij}}(-1)^r {f_i}^{(r)}f_{jl}f_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \\&e_{ik}e_{jl}-e_{jl}e_{ik} = f_{ik}f_{jl}-f_{jl}f_{ik} =0 \ \ \text {for} \ a_{ij}=0. \end{aligned} \end{aligned}$$

(8)

We extend the grading by setting $\vert q^h\vert =0$ and $\vert e_{il}\vert = l \alpha _{i}$.

The algebra $\widehat{U}$ is endowed with a co-multiplication $\Delta :\widehat{U} \rightarrow \widehat{U} \otimes \widehat{U}$ given by

$$\begin{aligned} \begin{aligned}&\Delta (q^h) = q^h \otimes q^h, \\&\Delta (e_{il}) = \sum _{m+n=l} q_{(i)}^{mn}e_{im}\otimes K_{i}^{-m}e_{in}, \\&\Delta (f_{il}) = \sum _{m+n=l} q_{(i)}^{-mn}f_{im}K_{i}^{n}\otimes f_{in}, \end{aligned} \end{aligned}$$

(9)

where $K_i=q^{r_ih_i}$ $(i \in I)$.

Let $\widehat{U}^{\le 0}$ be the subalgebra of $\widehat{U}$ generated by $f_{il}$ and $q^h$, for all $(i,l) \in I^{\infty }$ and $h\in P^{\vee }$, and $\widehat{U}^+$ be the subalgebra generated by $e_{il}$ for all $(i,l) \in I^{\infty }$. We extend $( \ , \ )_L$ to a symmetric bilinear form $( \ , \ )_L$ on $\widehat{U}^{\le 0}$ and on $\widehat{U}^+$ by setting

$$\begin{aligned} \begin{aligned}&(q^h,1)_L=1,\ (q^h,f_{il})_L=0, \\&(q^h,K_j)_L=q^{-\alpha _j(h)},\\&(x,y)_L=(\omega (x),\omega (y))_L \ \ \text {for all} \ x,y\in \widehat{U}^+, \end{aligned} \end{aligned}$$

(10)

where $\omega :\widehat{U}\rightarrow \widehat{U}$ is the involution defined by

$$\begin{aligned} \omega (q^h)=q^{-h},\ \omega (e_{il})=f_{il},\ \omega (f_{il})=e_{il}\ \ \text {for}\ h \in P^{\vee },\ (i,l)\in I^{\infty }. \end{aligned}$$

For any $x\in \widehat{U}$, we shall use the Sweedler’s notation, and write

$$\begin{aligned} \Delta (x)=\sum x_{(1)}\otimes x_{(2)}. \end{aligned}$$

Definition 1

Following the Drinfeld double process, we define the quantum Borcherds–Bozec algebra $U_q({\mathfrak {g}})$ associated with a given Borcherds–Cartan datum $(A, P,P^{\vee }, \Pi , \Pi ^{\vee })$ as the quotient of $\widehat{U}$ by the relations

$$\begin{aligned} \sum (a_{(1)},b_{(2)})_L\omega (b_{(1)})a_{(2)}=\sum (a_{(2)},b_{(1)})_La_{(1)}\omega (b_{(2)})\ \ \text {for all}\ a,b \in \widehat{U}^{\le 0}. \end{aligned}$$

(11)

Let $U^+_q({\mathfrak {g}})$ (resp. $U^-_q({\mathfrak {g}})$) be the subalgebra of $U_q({\mathfrak {g}})$ generated by $e_{il}$ (resp. $f_{il}$) for $(i,l)\in I^{\infty }$, and $U^{0}_q({\mathfrak {g}})$ the subalgebra of $U_q({\mathfrak {g}})$ generated by $q^h$ for $h\in P^{\vee }$. We shall denote by U (resp. $U^+$ and $U^-$) for $U_q({\mathfrak {g}})$ (resp. $U^+_q({\mathfrak {g}})$ and $U^-_q({\mathfrak {g}})$) for simplicity. Then U has the following triangular decomposition $U\cong U^-\otimes U^0 \otimes U^+$.

Proposition 6

[1, 2] For any $i\in I^{\text{ im }}$ and $l\ge 1$, there exist unique elements ${\texttt {b}}_{il}\in U^-_{-l \alpha _{i}}$ and $\texttt {a}_{il}=\omega ({\texttt {b}}_{il})$ such that

• (1) ${\mathbf {Q}} (q) \langle f_{il} \mid l\ge 1\rangle ={\mathbf {Q}} (q) \langle {\texttt {b}}_{il} \mid l\ge 1\rangle$ and ${\mathbf {Q}} (q) \langle e_{il} \mid l\ge 1\rangle ={\mathbf {Q}} (q) \langle \texttt {a}_{il} \mid l\ge 1\rangle$,

• (2) $({\texttt {b}}_{il},z)_L=0$ for all $z\in {\mathbf {Q}} (q) \langle f_{i1} ,\ldots ,f_{i,l-1}\rangle$, $(\texttt {a}_{il},z)_L=0$ for all $z\in {\mathbf {Q}} (q) \langle e_{i1} ,\ldots ,e_{i,l-1}\rangle$,

• (3) ${\texttt {b}}_{il}-f_{il}\in {\mathbf {Q}} (q) \langle f_{ik} \mid k<l \rangle$ and $\texttt {a}_{il}-e_{il}\in {\mathbf {Q}} (q) \langle e_{ik} \mid k<l \rangle$,

• (4) $\overline{{\texttt {b}}}_{il}={\texttt {b}}_{il},\ \overline{\texttt {a}}_{il}=\texttt {a}_{il}$,

• (5) $\varrho ({\texttt {b}}_{il})={\texttt {b}}_{il}\otimes 1+ 1\otimes {\texttt {b}}_{il}, \ \varrho (\texttt {a}_{il})=\texttt {a}_{il}\otimes 1+1\otimes \texttt {a}_{il}$,

• (6) $\Delta ({\texttt {b}}_{il})={\texttt {b}}_{il}\otimes 1+K_i^l\otimes {\texttt {b}}_{il}, \ \Delta (\texttt {a}_{il})=\texttt {a}_{il}\otimes K_i^{-l}+ 1\otimes \texttt {a}_{il}$,

• (7) $S({\texttt {b}}_{il})=-K_i^{-l}{\texttt {b}}_{il}, \ S(\texttt {a}_{il})=-\texttt {a}_{il}K_i^{l}$.

Here, S is the antipode of U, and $^-:U^{\pm }\rightarrow U^{\pm }$ is the ${\mathbf {Q}}$-algebra involution defined by $\overline{e}_{il}=e_{il},\ \overline{f}_{il}=f_{il}$ and $\overline{q}=q^{-1}$.

Set $\tau _{il}=(\texttt {a}_{il},\texttt {a}_{il})_L=({\texttt {b}}_{il},{\texttt {b}}_{il})_L$, we have the following commutation relations in $U_q({\mathfrak {g}})$

$$\begin{aligned} \texttt {a}_{il}{\texttt {b}}_{jk}-{\texttt {b}}_{jk}\texttt {a}_{il}=\delta _{ij}\delta _{lk}\tau _{il}(K_i^{l}-K_i^{-l}). \end{aligned}$$

(12)

The elements $\texttt {a}_{il}$’s and ${\texttt {b}}_{il}$’s are called the primitive generators. Let ${\mathcal {C}}_{l}$ (resp. ${\mathcal {P}}_{l}$) be the set of compositions (resp. partitions) of l. For $i\in I^{\text {im}}$, we define

$$\begin{aligned} {\mathscr {C}}_{i,l}={\left\{ \begin{array}{ll} {\mathcal {C}}_{l}&{} \text {if} \ i\in I^{\text {im}}\backslash I^{\text {iso}}, \\ {\mathcal {P}}_{l}&{} \text {if} \ i\in I^{\text {iso}}.\end{array}\right. } \end{aligned}$$

and ${\mathscr {C}}_i=\bigsqcup _{l\ge 0}{\mathscr {C}}_{i,l}$. For $i\in I^{\text {re}}$, we just put ${\mathscr {C}}_{i,l}=\{l\}$.

Assume that $i\in I^{\text {im}}$. Let ${\mathbf {c}} =(c_1,\ldots ,c_t)\in {\mathscr {C}}_{i,l}$ and set

$$\begin{aligned} {\texttt {b}}_{i,{\mathbf {c}}}={\texttt {b}}_{ic_1}\cdots {\texttt {b}}_{ic_t} , \ \texttt {a}_{i,{\mathbf {c}}}=\texttt {a}_{ic_1}\ldots \texttt {a}_{ic_t} \ \text {and} \ \tau _{i,{\mathbf {c}}}=\tau _{ic_1}\ldots \tau _{ic_t} . \end{aligned}$$

Note that $\{ {\texttt {b}}_{i,{\mathbf {c}}} \mid {\mathbf {c}} \in \mathscr {C}_{i,l}\}$ forms a basis of $U^-_{-l \alpha _{i}}$. For each $i\in I^{\text{ re }}$, we put ${\texttt {b}}_{i}=f_{i}$, $\texttt {a}_{i}=e_{i}$ and $\tau _{i}=\nu _{i}$.

Example 1

• (1) Each $\lambda \in {\mathcal {P}}_l$ can be written as the form $\lambda =1^{\lambda _1}2^{\lambda _2}\ldots l^{\lambda _l}$, where $\lambda _k$ are non-negative integers such that $\lambda _1+2\lambda _2+\cdots +l\lambda _l=l$. For $i\in I^{\text {iso}}$, we have

  $$\begin{aligned} {\texttt {b}}_{il}=f_{il}-\sum _{\lambda \in {\mathcal {P}}_l\backslash (l)} \frac{1}{\prod _{k=1}^{l} \lambda _k!} {\texttt {b}}_{i,\lambda }. \end{aligned}$$

  Note that assumption (7) implies $\nu _{il}\equiv 1 \ (\text {mod} \ q)$, hence we have $\tau _{il}\equiv \frac{1}{l} \ (\text {mod} \ q)$ by the following equation

  $$\begin{aligned} \sum _{\lambda \in {\mathcal {P}}_l} \frac{1}{\prod _{k=1}^{l} k^{\lambda _k}\lambda _k!}=1. \end{aligned}$$

• (2) Under the assumption (7), if $i\in I^{\text {im}}\backslash I^{\text {iso}}$, it was shown in [2,  Lemma 3.32] that $\tau _{il}\equiv 1 \ (\text {mod} \ q)$ for all $l\ge 1$. Moreover, $\tau _{il} \in 1+q{\mathbf {Z}}[[q]]$.

Let ${\mathcal {U}}$ be the associative algebra over ${\mathbf {Q}}(q)$ with ${\mathbf {1}}$ generated by the elements $\texttt {t}_{il}, \texttt {w}_{il}$ $((i,l) \in I^{\infty })$ and $q^h$ $(h\in P^{\vee })$ with defining relations

$$\begin{aligned} \begin{aligned}&q^0={\mathbf {1}},\quad q^hq^{h'}=q^{h+h'} \ \ \text {for} \ h,h' \in P^{\vee } \\&q^h \texttt {t}_{jl}q^{-h} = q^{l\alpha _j(h)} \texttt {t}_{jl}, \ \ q^h \texttt {w}_{jl}q^{-h} = q^{-l\alpha _j(h)} \texttt {w}_{jl}\ \ \text {for} \ h \in P^{\vee }, (j,l)\in I^{\infty }, \\&\texttt {t}_{il}\texttt {w}_{jk}-\texttt {w}_{jk}\texttt {t}_{il}=\delta _{ij}\delta _{lk}\tau _{il}(K_i^{l}-K_i^{-l}),\\&\sum _{r+s=1-la_{ij}}(-1)^r {\texttt {t}_i}^{(r)}\texttt {t}_{jl}\texttt {t}_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \\&\sum _{r+s=1-la_{ij}}(-1)^r {\texttt {w}_i}^{(r)}\texttt {w}_{jl}\texttt {w}_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \\&\texttt {t}_{ik}\texttt {t}_{jl}-\texttt {t}_{jl}\texttt {t}_{ik} = \texttt {w}_{ik}\texttt {w}_{jl}-\texttt {w}_{jl}\texttt {w}_{ik} =0 \ \ \text {for} \ a_{ij}=0. \end{aligned} \end{aligned}$$

(13)

Theorem 7

There exists a ${\mathbf {Q}}(q)$-algebra isomorphism $\Phi :U\xrightarrow {\sim } {\mathcal {U}}$ mapping $\texttt {a}_{il}$ to $\texttt {t}_{il}$, ${\texttt {b}}_{il}$ to $\texttt {w}_{il}$, and $q^h$ to $q^h$.

Proof

Recall that, for $j\in I^\text {im}$, $e_{jl}$ (resp. $f_{jl}$) in U can be written as a homogeneous polynomials in $\texttt {a}_{jk}$’s (resp. ${\texttt {b}}_{jk}$’s) for $1\le k\le l$. We may write

$$\begin{aligned} e_{jl}=\sum _{{\mathbf {c}}\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}}\texttt {a}_{j,{\mathbf {c}}}, \ f_{jl}=\sum _{{\mathbf {c}}\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}}{\texttt {b}}_{j,{\mathbf {c}}}, \end{aligned}$$

and let $\Phi :U\rightarrow {\mathcal {U}}$ be an algebra homomorphism sending $q^h\mapsto q^h$, $\texttt {e}_{i}\mapsto \texttt {t}_{i}$, ${\texttt {b}}_{i}\mapsto \texttt {w}_{i}$ if $i \in I^\text {re}$ and

$$\begin{aligned} \Phi (e_{jl})= & {} \sum _{{\mathbf {c}}\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}}\texttt {t}_{j,{\mathbf {c}}}, \\ \Phi (f_{jl})= & {} \sum _{{\mathbf {c}}\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}}\texttt {w}_{j,{\mathbf {c}}},\ \mathrm{if}\ j\in I^\text {im}. \end{aligned}$$

We shall show $\Phi$ is well-defined. For each ${\mathbf {c}}\in \mathscr {C}_{j,l}$, we have $q^h \texttt {t}_{j,{\mathbf {c}}}q^{-h} = q^{l\alpha _j(h)} \texttt {t}_{j,{\mathbf {c}}}$ and $q^h \texttt {w}_{j,{\mathbf {c}}}q^{-h} = q^{-l\alpha _j(h)} \texttt {w}_{j,{\mathbf {c}}}$ in ${\mathcal {U}}$. Hence

$$\begin{aligned} \Phi (q^h e_{jl}q^{-h} - q^{l\alpha _j(h)} e_{jl})=\Phi (q^h f_{jl}q^{-h} - q^{-l\alpha _j(h)} f_{jl})=0. \end{aligned}$$

For the Serre-type relations, if $a_{ij} \ne 0$ we have

$$\begin{aligned} \Phi \left( \sum _{r+s=1-la_{ij}}(-1)^r {e_i}^{(r)}e_{jl}e_i^{(s)}\right) =\sum _{{\mathbf {c}}\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}}\sum _{r+s=1-la_{ij}}(-1)^r {\texttt {t}_i}^{(r)}\texttt {t}_{j,{\mathbf {c}}}\texttt {t}_i^{(s)}=0. \end{aligned}$$

If $a_{ij}=0$, we have

$$\begin{aligned} \Phi ( e_{ik}e_{jl} )=\sum _{{\mathbf {c}}\in {\mathscr {C}}_{i,k}}\alpha _{{\mathbf {c}}}\texttt {t}_{i,{\mathbf {c}}} \sum _{{\mathbf {c}}'\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}'}\texttt {t}_{j,{\mathbf {c}}'} = \sum _{{\mathbf {c}}'\in {\mathscr {C}}_{j,l}}\alpha _{{\mathbf {c}}'}\texttt {t}_{j,{\mathbf {c}}'}\sum _{{\mathbf {c}}\in {\mathscr {C}}_{i,k}}\alpha _{{\mathbf {c}}}\texttt {t}_{i,{\mathbf {c}}} = \Phi (e_{jl} e_{ik}). \end{aligned}$$

The other half Serre-type relations can be shown similarly.

For the commutation relations, we first claim that $\Phi (\texttt {a}_{il})=\texttt {t}_{il}$ and $\Phi ({\texttt {b}}_{il})=\texttt {w}_{il}$ for any $(i,l)\in I^{\infty }$. If $i \in I^{\mathrm{re}}$, there is nothing to show. If $i \in I^{\mathrm{im}}$, we could show it by induction on l. If $l=1$, it is obvious since $e_{i1} = a_{i1}$. Assume the claim is true for all $k<l$. Since $\texttt {a}_{il}= e_{il} - \sum _{{\mathbf {c}}\in {\mathscr {C}}_{i,l}, {\mathbf {c}} \ne (l)}\alpha _{{\mathbf {c}}}\texttt {a}_{i,{\mathbf {c}}}$, we have $\Phi (\texttt {a}_{il}) = \sum _{{\mathbf {c}}\in \mathscr {C}_{i,l}}\alpha _{{\mathbf {c}}}\texttt {t}_{i,{\mathbf {c}}} - \sum _{{\mathbf {c}}\in {\mathscr {C}}_{i,l}, {\mathbf {c}}\ne (i)}\alpha _{{\mathbf {c}}}\texttt {t}_{i,{\mathbf {c}}}= \texttt {t}_{il}$. And $\Phi ({\texttt {b}}_{il})=\texttt {w}_{il}$ can be shown similarly.

Now by the Drinfeld double process, the commutation relation in (11) is equivalent to the one in (12) (cf. [17,  Lemma 3.2]). Moreover, $\Phi (\texttt {a}_{il}{\texttt {b}}_{jk}-{\texttt {b}}_{jk}\texttt {a}_{il}-\delta _{ij}\delta _{lk}\tau _{il}(K_i^{l}-K_i^{-l}))=0.$ This shows that $\Phi$ is well-defined.

Since Theorem 4 yields the following relations in U

$$\begin{aligned} \begin{aligned}&\sum _{r+s=1-la_{ij}}(-1)^r {\texttt {a}_i}^{(r)}\texttt {a}_{jl}\texttt {a}_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},~(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l), \\&\sum _{r+s=1-la_{ij}}(-1)^r {{\texttt {b}}_i}^{(r)}{\texttt {b}}_{jl}{\texttt {b}}_i^{(s)}=0 \ \ \text {for} \ i\in I^{\text {re}},~(j,l)\in I^{\infty } \ \text{ and } \ i \ne (j,l).\\ \end{aligned} \end{aligned}$$

We see that $\Phi$ has an obvious inverse $\Psi :{\mathcal {U}} \rightarrow U$ given by

$$\begin{aligned} \Psi (\texttt {t}_{il})=\texttt {a}_{il},\ \Psi (\texttt {w}_{il})={\texttt {b}}_{il}, \ \Psi (q^h)=q^h, \end{aligned}$$

for $(i,l)\in I^{\infty }$ and $h\in P^\vee$. $\square$

We note that Theorem 7 provides a new presentation of U with primitive generators. Now, Proposition 6 provides a Hopf algebra structure on ${\mathcal {U}}$:

$$\begin{aligned} \begin{aligned} \Delta (q^h)&= q^h \otimes q^h, \\ \Delta (\texttt {t}_{il})&=\texttt {t}_{il}\otimes K_i^{-l}+ 1\otimes \texttt {t}_{il},\\ \Delta (\texttt {w}_{il})&=\texttt {w}_{il}\otimes 1+K_i^l\otimes \texttt {w}_{il}, \\ S(\texttt {t}_{il})&=-\texttt {t}_{il}K_i^{l},\ S(\texttt {w}_{il})=-K_i^{-l}\texttt {w}_{il}, \ S(q^h)=q^{-h},\\ \epsilon (\texttt {t}_{il})&=\epsilon (\texttt {w}_{il})=0,\ \epsilon (q^h)=1, \end{aligned} \end{aligned}$$

(14)

where $\Delta ,S$, and $\epsilon$ denote the co-multiplication, antipode and counit, respectively.

4 Crystal bases and polarization

Definition 2

For $i\in I^{\text {im}}$ and ${\mathbf {c}}\in {\mathscr {C}}_i$, we define the linear maps $\delta _{i,{\mathbf {c}}}, \delta ^{i,{\mathbf {c}}}:U^{-}\rightarrow U^{-}$ by

$$\begin{aligned} \varrho (x)=\sum _{{\mathbf {c}}\in {\mathscr {C}}_{i}}\delta _{i,{\mathbf {c}}}(x)\otimes {\texttt {b}}_{i,{\mathbf {c}}}+\ \text {terms of bidegree not in}\ Q_-\times -{\mathbf {N}} \alpha _i, \end{aligned}$$

$$\begin{aligned} \varrho (x)=\sum _{{\mathbf {c}}\in {\mathscr {C}}_{i}}{\texttt {b}}_{i,{\mathbf {c}}}\otimes \delta ^{i,{\mathbf {c}}}(x)+\ \text {terms of bidegree not in}\ -{\mathbf {N}} \alpha _i\times Q_-, \end{aligned}$$

where x is a homogeneous element in $U^-$.

Let $i\in I^{\text {im}}$, $l>0$, then for any homogeneous $x,y,z\in U^-$ and ${\mathbf {c}}=(c_1,\ldots ,c_t)\in {\mathscr {C}}_i$, we have the following equations

$$\begin{aligned}&\gamma ^{i,l}(xy)=\gamma ^{i,l}(x)y+q^{l(\alpha _i,\vert x\vert )}x\gamma ^{i,l}(y), \end{aligned}$$

(15)

$$\begin{aligned}&\quad \gamma ^{i,l}({\texttt {b}}_{i,{\mathbf {c}}})=\sum _{k:c_k=l} q_{(i)}^{-2l\sum _{j<k}c_j}{\texttt {b}}_{i,{\mathbf {c}}\backslash c_k}, \end{aligned}$$

(16)

$$\begin{aligned}&\quad \left[ {\texttt {a}}_{il},z\right] =\tau _{il} \left( \gamma _{i,l}(z)K_i^l-K_i^{-l}\gamma ^{i,l}(z)\right) , \end{aligned}$$

(17)

where ${\mathbf {c}}\backslash c_k=(c_1,\ldots ,\widehat{c}_k,\ldots ,c_r)$ means removing $c_k$ from ${\mathbf {c}}$. We will denote the operator $\gamma ^{i,l}$ by $e'_{i,l}$ in the following.

Recall from [2] that every $u\in U^-$ can be written uniquely as

$$\begin{aligned} u=\sum _{{\mathbf {c}}\in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}}, \end{aligned}$$

where $e'_{i,l}\, u_{{\mathbf {c}}} = 0$ for all $l\ge 1$ and ${\mathbf {c}} \in {\mathscr {C}}_i$. Moreover, if u is homogeneous, then every $u_{{\mathbf {c}}}$ is homogeneous. Then the Kashiwara operators are defined by

$$\begin{aligned} \begin{aligned}&{\widetilde{e}}_{il}u={\left\{ \begin{array}{ll}\displaystyle \sum _{{\mathbf {c}}:c_1=l}{\texttt {b}}_{i,{\mathbf {c}}\backslash c_1}u_{{\mathbf {c}}}\qquad &{} \text {if}\ i\notin I^{\text {iso}},\\ \displaystyle \sum _{{\mathbf {c}}\in {\mathscr {C}}_i}\sqrt{\frac{m_l({\mathbf {c}})}{l}}{\texttt {b}}_{i,{\mathbf {c}}\backslash l}u_{{\mathbf {c}}}&{} \text {if}\ i\in I^{\text {iso}},\end{array}\right. }\\&{\widetilde{f}}_{il}u={\left\{ \begin{array}{ll}\displaystyle \sum _{{\mathbf {c}}\in {\mathscr {C}}_i}{\texttt {b}}_{i,\mathbf (l,c)}u_{{\mathbf {c}}}\qquad \qquad &{} \text {if}\ i\notin I^{\text {iso}},\\ \displaystyle \sum _{{\mathbf {c}}\in \mathscr {C}_i}\sqrt{\frac{l}{m_l({\mathbf {c}})+1}}{\texttt {b}}_{i,{\mathbf {c}}\cup l}u_{{\mathbf {c}}}&{} \text {if}\ i\in I^{\text {iso}},\end{array}\right. } \end{aligned} \end{aligned}$$

where $m_{l}({\mathbf {c}}) = \# \{k \mid c_k = l\}$.

Remark 1

Note that the square roots appear in the above definition. So we need to consider an extension ${\mathbf {F}}$ of ${\mathbf {Q}}$ that contains all the necessary square roots (see [2,  Remark 3.12]).

Let ${\mathbb {A}}_0=\{f\in {\mathbf {F}}(q) \mid f \ \text {is regular at} \ q=0\}$, and let ${\mathcal {L}}(\infty )$ be the $\mathbb A_0$-submodule of $U^-$ generated by the elements $\widetilde{f}_{i_1,l_1}\ldots \widetilde{f}_{i_r,l_r} {\mathbf {1}}$ for $r\ge 0$ and $(i_k,l_k)\in I^{\infty }$, where the Kashiwara operators $\widetilde{f}_i$ for $i\in I^{\text {re}}$ have been defined in [12]. Set

$$\begin{aligned} {\mathcal {B}}(\infty )=\{\widetilde{f}_{i_1,l_1}\ldots \widetilde{f}_{i_r,l_r} {\mathbf {1}} \ \text {mod} \ q{\mathcal {L}}(\infty ) \mid r \ge 0, (i_k, l_k) \in I^{\infty } \}\subseteq {{\mathcal {L}}}(\infty )/ q{{\mathcal {L}}}(\infty ), \end{aligned}$$

then $(\mathcal L(\infty ),{\mathcal {B}}(\infty ))$ is the crystal base of $U^-$.

By [2,  Lemma 3.33] and [12,  Proposition 5.1.2], we have the following proposition.

Proposition 8

• (i) $({\mathcal {L}}(\infty ),{\mathcal {L}}(\infty ))_L\subseteq {\mathbb {A}}_0.$

  Let $( \ , \ )_L^0$ denote the ${\mathbf {F}}$-valued inner product on ${\mathcal {L}}(\infty )/ q{\mathcal {L}} (\infty )$ induced by $( \ , \ )_L \vert _{q=0}$ on ${\mathcal {L}}(\infty )$.

• (ii) $(\widetilde{e}_{il} u,v)_L^0=(u,\widetilde{f}_{il}v)_L^0$ for $u,v\in {\mathcal {L}}(\infty )/ q{\mathcal {L}}(\infty )$ and $(i,l)\in I^{\infty }$.

• (iii) ${\mathcal {B}}(\infty )$ is an orthonormal base of $( \ , \ )_L^0$. In particular, $( \ , \ )_L^0$ is positive definite.

• (iv) ${\mathcal {L}}(\infty )=\{u\in U^- \mid (u,{\mathcal {L}}(\infty ))_L\subseteq {\mathbb {A}}_0\}=\{u\in U^- \mid (u,u)_L\in {\mathbb {A}}_0\}$.

Let $\lambda \in P^+$, and let $V(\lambda )$ be the irreducible highest weight $U_q({\mathfrak {g}})$-module with highest weight $\lambda$ and highest weight vector $v_{\lambda }$. Then we have a $U^-_q({\mathfrak {g}})$-module isomorphism (cf. [3, 9])

$$\begin{aligned} \begin{aligned} V(\lambda )\simeq U^-_q({\mathfrak {g}}) \bigg / \left( \sum _{i\in I^{\text {re}}}U^-_q({\mathfrak {g}})f_i^{\lambda (h_i)+1}+\sum _{\begin{array}{c} i\in I^{\text {im}}, \lambda (h_i)=0\\ (i,l)\in I^{\infty } \end{array}}U^-_q({\mathfrak {g}})f_{il}\right) . \end{aligned} \end{aligned}$$

(18)

Recall from [2] that, for any $i\in I^{\text {im}}$ and $\lambda \in P^+$, $v\in V(\lambda )_\mu$ has a decomposition of the following form

$$\begin{aligned} v=\sum _{{\mathbf {c}} \in {{\mathscr {C}}}_i}{\texttt {b}}_{i,{\mathbf {c}}}v_{{\mathbf {c}}}, \end{aligned}$$

where $v_{{\mathbf {c}}}\in V(\lambda )_{\mu + \vert {\mathbf {c}} \vert \alpha _i}$ and $e_{il} v_{{\mathbf {c}}}=0$ for all $l\ge 1$ and ${\mathbf {c}} \in {\mathscr {C}}_i$. Moreover, if we omit the terms $\texttt {b}_{i,{\mathbf {c}}} v_{{\mathbf {c}}}$ with $\vert {\mathbf {c}} \vert \ne 0$ and $(\mu + \vert {\mathbf {c}} \vert \alpha _i,\alpha _i)=0$, which are equal to zero trivially, then the decomposition of v is unique.

Define the Kashiwara operators on $V(\lambda )$ by

$$\begin{aligned} \begin{aligned}&{\widetilde{e}}_{il}v={\left\{ \begin{array}{ll}\displaystyle \sum _{{\mathbf {c}}:c_1=l}{\texttt {b}}_{i,{\mathbf {c}}\backslash c_1}v_{{\mathbf {c}}}&{} \text {if}\ i\notin I^{\text {iso}},\\ \displaystyle \sum _{{\mathbf {c}}\in {\mathscr {C}}_i}\sqrt{\frac{m_l({\mathbf {c}})}{l}}{\texttt {b}}_{i,{\mathbf {c}}\backslash l}v_{{\mathbf {c}}}&{} \text {if}\ i\in I^{\text {iso}},\end{array}\right. }\\&{\widetilde{f}}_{il}v={\left\{ \begin{array}{ll}\displaystyle \sum _{{\mathbf {c}}\in {\mathscr {C}}_i}{\texttt {b}}_{i,\mathbf (l,c)}v_{{\mathbf {c}}}\qquad \qquad &{} \text {if}\ i\notin I^{\text {iso}},\\ \displaystyle \sum _{{\mathbf {c}}\in \mathscr {C}_i}\sqrt{\frac{l}{m_l({\mathbf {c}})+1}}{\texttt {b}}_{i,{\mathbf {c}}\cup l}v_{{\mathbf {c}}}&{} \text {if}\ i\in I^{\text {iso}}.\end{array}\right. } \end{aligned} \end{aligned}$$

Let ${\mathcal {L}}(\lambda )=\sum _{\iota _1,\ldots ,\iota _s\in I^{\infty }}\mathbb A_0\widetilde{f}_{\iota _1}\ldots \widetilde{f}_{\iota _s}v_{\lambda }$ be an ${\mathbb {A}}_0$-submodule of $V(\lambda )$ and let

$$\begin{aligned} {\mathcal {B}}(\lambda )=\{\widetilde{f}_{\iota _1}\ldots \widetilde{f}_{\iota _s}v_{\lambda }\mid \iota _k\in I^{\infty }\}\backslash \{0\}\subseteq {{\mathcal {L}}}(\lambda )/ q{{\mathcal {L}}}(\lambda ), \end{aligned}$$

then $({\mathcal {L}}(\lambda ),{\mathcal {B}}(\lambda ))$ is the crystal base of $V(\lambda )$.

There exists a unique symmetric bilinear form $\{ -, - \}$ on $V(\lambda )$ such that

$$\begin{aligned} \begin{aligned}&\{v_{\lambda },v_{\lambda }\}=1, \\&\{q^hv,v'\}=\{v,q^hv'\},\\&\{{\texttt {b}}_{il}v,v'\}=-\{v, K_i^{l}\texttt {a}_{il}v'\} \ \ \text {if} \ i\in I^{\text {im}},\\&\{{\texttt {b}}_{i}v,v'\}=\frac{1}{q_i^{2}-1}\{v, K_i\texttt {a}_{i}v'\} \ \ \text {if} \ i\in I^{\text {re}},\\ \end{aligned} \end{aligned}$$

for every $v,v'\in V(\lambda )$ and $(i,l)\in I^{\infty }$.

Similarly, by [2,  Lemma 3.34] and [12,  Proposition 5.1.1], we have the following proposition.

Proposition 9

• (i) $\{{\mathcal {L}}(\lambda ),{\mathcal {L}}(\lambda )\}\subseteq {\mathbb {A}}_0.$

  Let $\{ \ , \ \}_0$ denote the ${\mathbf {F}}$-valued inner product on ${{\mathcal {L}}}(\lambda )/ q{{\mathcal {L}}}(\lambda )$ induced by $\{ \ , \ \}\vert _{q=0}$ on ${{\mathcal {L}}}(\lambda )$.

• (ii) $\{\widetilde{e}_{il}u,v\}_0=\{u,\widetilde{f}_{il}v\}_0$ for $u,v\in {\mathcal {L}}(\lambda )/ q{\mathcal {L}}(\lambda )$ and $(i,l)\in I^{\infty }$.

• (iii) ${\mathcal {B}}(\lambda )$ is an orthonormal base of $\{ \ , \ \}_0$. In particular, $\{ \ , \ \}_0$ is positive definite.

• (iv) ${\mathcal {L}}(\lambda )=\{v\in V(\lambda ) \mid \{v,{\mathcal {L}}(\lambda )\}\subseteq {\mathbb {A}}_0\}=\{v\in V(\lambda ) \mid \{v,v\}\in {\mathbb {A}}_0\}$.

The following proposition follows from Kashiwara’s grand-loop argument, which describes the relations between ${\mathcal {B}}(\infty )$ and ${\mathcal {B}}(\lambda )$.

Proposition 10

Let $\pi _\lambda :U^-_q({\mathfrak {g}})\rightarrow V(\lambda )$ be the $U^-_q({\mathfrak {g}})$-module homomorphism given by $P\mapsto Pv_{\lambda }$, then we have

• (i) $\pi _\lambda ({\mathcal {L}}(\infty ))={\mathcal {L}}(\lambda )$, hence $\pi _{\lambda }$ induces the surjective homomorphism

  $$\begin{aligned} \overline{\pi }_{\lambda }:{\mathcal {L}}(\infty )/ q{\mathcal {L}}(\infty )\rightarrow {\mathcal {L}}(\lambda )/ q{\mathcal {L}}(\lambda ). \end{aligned}$$

• (ii) $\{b\in {\mathcal {B}}(\infty )\mid \overline{\pi }_{\lambda }(b)\ne 0 \}$ is isomorphic to ${\mathcal {B}}(\lambda )$ under the map $\overline{\pi }_{\lambda }$.

• (iii) If $b\in {\mathcal {B}}(\infty )$ satisfies $\overline{\pi }_{\lambda }(b)\ne 0$, then $\widetilde{e}_{il}\overline{\pi }_{\lambda }(b)=\overline{\pi }_{\lambda }({\widetilde{e}}_{il}b)$.

• (iv) $\widetilde{f}_{il}\circ \overline{\pi }_{\lambda }=\overline{\pi }_{\lambda }\circ \widetilde{f}_{il}$.

Let $(i,l)\in I^{\infty }$ and let $P\in U^-$, then there exist unique $Q,R\in U^-$ such that

$$\begin{aligned}{}[\texttt {a}_{il},P]=\tau _{il}(K_i^{l}Q-K_i^{-l}R). \end{aligned}$$

Note that $e'_{i,l}(P)=R$ by (17). If we set $e''_{i,l}(P)=Q$, then we have

$$\begin{aligned} \begin{aligned}&e'_{i,l}{\texttt {b}}_{jk}=\delta _{ij}\delta _{kl}+q_i^{-kla_{ij}}{\texttt {b}}_{jk}e'_{i,l},\\&e''_{i,l}{\texttt {b}}_{jk}=\delta _{ij}\delta _{kl}+q_i^{kla_{ij}}{\texttt {b}}_{jk}e''_{i,l}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} e'_{i,l}e''_{j,k}=q_i^{kla_{ij}}e''_{j,k}e'_{i,l}. \end{aligned}$$

Definition 3

Let ${\mathscr {B}}_q({\mathfrak {g}})$ be the algebra over ${\mathbf {F}}(q)$ generated by $e'_{i,l}$, ${\texttt {b}}_{il}$ $(i,l)\in I^{\infty }$ with defining relations

$$\begin{aligned} \begin{aligned}&e'_{i,l}{\texttt {b}}_{jk}=\delta _{ij}\delta _{kl}+q_i^{-kla_{ij}}{\texttt {b}}_{jk}e'_{i,l},\\&\sum _{r=0}^{1-la_{ij}}(-1)^r\begin{bmatrix} 1-la_{ij}\\ r \end{bmatrix}_i {e'_i}^{1-la_{ij}-r}e'_{j,l}{e'}_i^{r}=0 \ \ \text {for} \ i\in I^{\text {re}} \ \text{ and } \ i \ne (j,l), \\&\sum _{r=0}^{1-la_{ij}}(-1)^r\begin{bmatrix} 1-la_{ij}\\ r \end{bmatrix}_i {{\texttt {b}}_i}^{1-la_{ij}-r}{\texttt {b}}_{j,l}{\texttt {b}}_i^{r}=0 \ \ \text {for} \ i\in I^{\text {re}} \ \text{ and } \ i \ne (j,l), \\&e'_{i,k}e'_{j,l}-e'_{j,l}e'_{i,k} = {\texttt {b}}_{ik}{\texttt {b}}_{jl}-{\texttt {b}}_{jl}{\texttt {b}}_{ik} =0 \ \ \text {for} \ a_{ij}=0. \end{aligned} \end{aligned}$$

We call ${\mathscr {B}}_q({\mathfrak {g}})$ the quantum boson algebra associated with ${\mathfrak {g}}$. One can show that $\mathscr {B}_q({\mathfrak {g}})$ is a left $U^-_q({\mathfrak {g}})$-module by the standard argument in [12]. Furthermore, we have

$$\begin{aligned} U^-_q({\mathfrak {g}})\cong {\mathscr {B}}_q({\mathfrak {g}})\bigg /\sum _{(i,l)\in I^{\infty }}{\mathscr {B}}_q({\mathfrak {g}})e'_{i,l}. \end{aligned}$$

Lemma 11

For all $P, Q\in U^-$ and $(i,l)\in I^{\infty }$, we have

$$\begin{aligned} (P{\texttt {b}}_{il}, Q)_L=\tau _{il}(P,K_i^{l}e''_{i,l}QK_i^{-l})_L. \end{aligned}$$

Proof

By (17), we have $K_i^{l}e''_{i,l}Q=\delta _{i,l}(Q)K_i^{l}$ and hence $K_i^{l}e''_{i,l}QK_i^{-l}=\delta _{i,l}(Q)$. Thus we obtain

$$\begin{aligned} (P{\texttt {b}}_{il}, Q)_L=\tau _{il}(P,\delta _{i,l}(Q))_L=\tau _{il}(P,K_i^{l}e''_{i,l}QK_i^{-l})_L \end{aligned}$$

as desired. $\square$

Let $*:U_q({\mathfrak {g}})\rightarrow U_q({\mathfrak {g}})$ be the ${\mathbf {F}}(q)$-linear anti-involution given by

$$\begin{aligned} e^*_{il}=e_{il}, \quad f_{il}^*=f_{il},\quad (q^h)^*=q^{-h}. \end{aligned}$$

Note that $**=id$ and $*-=-*$ on $U^\pm$, and ${\texttt {a}}_{il}, {\texttt {b}}_{il}$ are stable under $*$ for any $(i,l)\in I^{\infty }$.

Lemma 12

For any $P, Q\in U^-$, we have

$$\begin{aligned} (P^*,Q^*)_L=(P,Q)_L. \end{aligned}$$

Proof

Note that $e''_{i,l}(Q^*)=K_i^{-l}(e'_{i,l}Q)^*K_i^{l}$ and $e'_{i,l}(Q^*)=K_i^{l}(e''_{i,l}Q)^*K_i^{-l}$. We shall prove this lemma by induction on $\vert P\vert$. If $P=1$, our assertion is clear. By Lemma 11 and the inductive hypothesis, we have

$$\begin{aligned} \begin{aligned} ((P{\texttt {b}}_{il})^*, Q^*)_L&=({\texttt {b}}_{il}P^*,Q^*)_L=\tau _{il}(P^*,e'_{i,l}(Q^*))_L\\&=\tau _{il}(P^*,K_i^{l}(e''_{i,l}Q)^*K_i^{-l})_L\\&=\tau _{il}(P,K_i^{l}e''_{i,l}QK_i^{-l})_L\\&=(P{\texttt {b}}_{il}, Q)_L, \end{aligned} \end{aligned}$$

which proves our claim. $\square$

The following corollary is an immediate consequence of Lemma 12 and Proposition 8.

Corollary 13

${\mathcal {L}}(\infty )^*={\mathcal {L}}(\infty )$.

Proposition 14

Let $P, Q\in U^-_q({\mathfrak {g}})_{-\beta }$ for $\beta \in Q_+$. If $\lambda \gg 0$, we have

$$\begin{aligned} \{Pv_\lambda ,Qv_\lambda \}\equiv c(P,Q)_L \pmod {q{\mathbb {A}}_0} \end{aligned}$$

for some $c\in {\mathbb {A}}_0 \backslash q{\mathbb {A}}_0$.

Proof

We use the induction on $\vert \beta \vert$. If $i\in I^{\text {im}}$, we have

$$\begin{aligned} \begin{aligned} \{{\texttt {b}}_{il}Pv_\lambda ,Qv_\lambda \}&=-\{Pv_{\lambda },K_i^{l}{\texttt {a}}_{il}Qv_\lambda \}\\&=-\{Pv_{\lambda }, K_i^{l}(Q{\texttt {a}}_{il}+\tau _{il}(K_i^{l}e''_{i,l}Q-K_i^{-l}e'_{i,l}Q))v_\lambda \}\\&=-\tau _{il}\{Pv_{\lambda }, K_i^{2l}e''_{i,l}Qv_\lambda -e'_{i,l}Qv_\lambda \}\\&=-\tau _{il}\{Pv_{\lambda }, q_i^{2l(\lambda -\beta )(h_i)}e''_{i,l}Qv_\lambda \}+\tau _{il}\{Pv_\lambda ,e'_{i,l}Qv_{\lambda }\}, \end{aligned} \end{aligned}$$

where $P\in U^-_{-\beta }$ and $Q\in U^-_{-\beta -l\alpha _i}$. Hence

$$\begin{aligned} \{{\texttt {b}}_{il}Pv_\lambda ,Qv_\lambda \}\equiv \tau _{il}\{Pv_\lambda ,e'_{il}Qv_{\lambda }\} \equiv c\tau _{il}(P,e'_{i,l}Q)_L=c({\texttt {b}}_{il}P, Q)_L \pmod {q{\mathbb {A}}_0}. \end{aligned}$$

if $i\in I^{\text {re}}$, we have

$$\begin{aligned} \begin{aligned} \{{\texttt {b}}_{i}Pv_\lambda ,Qv_\lambda \}&=\frac{1}{q_i^2-1}\{Pv_\lambda ,K_i{\texttt {a}}_iQv_\lambda \} \\&=\frac{1}{q_i^2-1}\{Pv_\lambda ,K_i\tau _i(K_ie''_iQ-K_i^{-1}e'_iQ)v_\lambda \}\\&=\frac{1}{q_i^2-1}\tau _i\{Pv_\lambda , q_i^{2(\lambda -\beta )(h_i)}e''_{il}Qv_\lambda \}+\frac{1}{q_i^2-1}\tau _i\{Pv_\lambda ,e'_iQv_\lambda \}, \end{aligned} \end{aligned}$$

where $P\in U^-_{-\beta }$ and $Q\in U^-_{-\beta -\alpha _i}$. Hence

$$\begin{aligned} \begin{aligned} \{{\texttt {b}}_{i}Pv_\lambda ,Qv_\lambda \}&\equiv \frac{1}{q_i^2-1}\tau _i\{Pv_\lambda ,e'_iQv_\lambda \} \equiv \frac{1}{q_i^2-1}c\tau _i(P,e'_{i}Q)_L\\&=\frac{1}{q_i^2-1}c({\texttt {b}}_{i}P, Q)_L \pmod {q{\mathbb {A}}_0}, \end{aligned} \end{aligned}$$

which completes the proof. $\square$

Corollary 15

If $\lambda \gg 0$ and $Pv_\lambda \in {\mathcal {L}}(\lambda )$, then $P^*v_\lambda \in {\mathcal {L}}(\lambda )$.

Proof

If $Pv_\lambda \in {\mathcal {L}}(\lambda )$, then $\{Pv_\lambda ,Pv_\lambda \}\in {\mathbb {A}}_0$ by Proposition 9. Since $\{Pv_\lambda ,Pv_\lambda \}\equiv c(P,P)_L \pmod {q{\mathbb {A}}_0}$ for some $c \in {\mathbb {A}}_{0} \setminus q \, {\mathbb {A}}_{0}$, we have $(P,P)_L\in {\mathbb {A}}_0$. Hence $P\in {\mathcal {L}}(\infty )$ and $P^*\in {\mathcal {L}}(\infty )$ by Proposition 8 and Corollary 13. Now Proposition 10 yields $\pi _{\lambda }(\mathcal L(\infty ))={\mathcal {L}}(\lambda )$. Thus we get $P^*v_{\lambda }\in {\mathcal {L}}(\lambda )$ by applying $\pi _\lambda$. $\square$

5 ${\mathbb {A}}$-form of $U^-_q({\mathfrak {g}})$

Let ${\mathbb {A}}={\mathbf {F}}[q,q^{-1}]$ and ${\mathbb {A}}_\infty =\{f\in {\mathbf {F}}(q) \mid f \ \text {is regular at} \ q=\infty \}$. We denote by $U^-_{{\mathbb {A}}}({\mathfrak {g}})$ the ${\mathbb {A}}$-subalgebra of $U_q({\mathfrak {g}})$ generated by ${\texttt {b}}_i^{(n)}$ $(i\in I^{\text {re}}, n\ge 0)$ and ${\texttt {b}}_{il}$ $(i\in I^{\text {im}}, l\ge 1)$.

For each $i\in I^{\text {re}}$, set

$$\begin{aligned} A_i={\texttt {a}}_i/\tau _i(q_i-q_i^{-1}), \end{aligned}$$

(19)

which yields the following commutation relation

$$\begin{aligned} A_i{\texttt {b}}_i-{\texttt {b}}_iA_i=\frac{K_i-K_i^{-1}}{q_i-q_i^{-1}}. \end{aligned}$$

(20)

For $i\in I^{\text {im}}$ and $l\ge 1$, set $A_{il}={\texttt {a}}_{il}/\tau _{il}$. Then we have

$$\begin{aligned} A_{il}{\texttt {b}}_{il}-{\texttt {b}}_{il}A_{il}=K_i^{l}-K_i^{-l}. \end{aligned}$$

(21)

Let $U_{{\mathbb {A}}}({\mathfrak {g}})$ be the ${\mathbb {A}}$-subalgebra of $U_q({\mathfrak {g}})$ generated by $A_i^{(n)},{\texttt {b}}_i^{(n)}$ $(i\in I^{\text {re}}, n\ge 0)$, $A_{il}, {\texttt {b}}_{il}$ $(i\in I^{\text {im}}, l\ge 1)$ and $q^h$ $(h \in P^{\vee })$, $\left\{ \begin{matrix}K_iq_i^n \\ m \end{matrix}\right\} _i$ $(i \in I^{\text {re}}, m\in {\mathbf {Z}}_{\ge 0}, n\in {\mathbf {Z}})$, where

$$\begin{aligned} \left\{ \begin{matrix}K_iq_i^n \\ m \end{matrix}\right\} _i=\frac{1}{[m]_i!}\prod _{s=1}^m\frac{K_iq_i^{n+1-s}-K_i^{-1}q_i^{-n-1+s}}{q_i-q_i^{-1}}. \end{aligned}$$

(22)

Let $U_{{\mathbb {A}}}^+({\mathfrak {g}})$ (resp. $U_{\mathbb A}^0({\mathfrak {g}})$) be the ${\mathbb {A}}$-subalgebra of $U_q({\mathfrak {g}})$ generated by $A_i^{(n)}$ $(i\in I^{\text {re}})$ and $A_{il}$ $(i\in I^{\text {im}}, l \ge 1)$ (resp. $q^h, \left\{ \begin{matrix}K_iq_i^n \\ m \end{matrix}\right\} _i$ for $h\in P^{\vee }, m\in {\mathbf {Z}}_{\ge 0}, n\in {\mathbf {Z}} \ \text {and} \ i\in I^{\text {re}}$ ). Then using the commutations relations (20), (21) and the definition (22), one can prove that $U_{{\mathbb {A}}}({\mathfrak {g}})$ has the triangular decomposition (see also [12,  Section 1], [6,  Exercise 3.6])

$$\begin{aligned} U_{{\mathbb {A}}}({\mathfrak {g}})\cong U^-_{{\mathbb {A}}}({\mathfrak {g}})\otimes U^0_{{\mathbb {A}}}({\mathfrak {g}})\otimes U^+_{{\mathbb {A}}}({\mathfrak {g}}). \end{aligned}$$

Let $\lambda \in P^{+}$ and consider an ${\mathbf {F}}$-linear automorphism $^-:V(\lambda )\rightarrow V(\lambda )$ given by $Pv_\lambda \mapsto \overline{P}v_\lambda$ for $P\in U_q({\mathfrak {g}})$. Set ${\mathcal {L}}(\lambda )^-=\overline{\mathcal L(\lambda )}$. Then ${\mathcal {L}}(\lambda )$ (resp. $\mathcal L(\lambda )^-$) is a free ${\mathbb {A}}_0$-lattice (resp. free $\mathbb A_\infty$-lattice) of $V(\lambda )$.

Since

$$\begin{aligned} \left\{ \begin{matrix}K_iq_i^n \\ m \end{matrix}\right\} _iv_\lambda =\begin{bmatrix}\lambda (h_i)+n\\ m\end{bmatrix}_iv_\lambda \in {\mathbf {Z}}[q,q^{-1}]v_\lambda , \end{aligned}$$

we get $U^0_{\mathbb {A}}({\mathfrak {g}})v_\lambda =\mathbb Av_\lambda$. This leads us to give the following definition

$$\begin{aligned} V(\lambda )^{{\mathbb {A}}}:=U_{{\mathbb {A}}}({\mathfrak {g}})v_\lambda =U^-_{{\mathbb {A}}}({\mathfrak {g}})v_\lambda . \end{aligned}$$

Note that $\overline{{\texttt {b}}}_{il}={\texttt {b}}_{il}$ for all $(i,l)\in I^{\infty }$. Hence $U^-_{{\mathbb {A}}}({\mathfrak {g}})$ and $V(\lambda )^{{\mathbb {A}}}$ are stable under −. Also, since $U^-_{{\mathbb {A}}}({\mathfrak {g}})$ is graded by $Q_-$, we have $V(\lambda )^{{\mathbb {A}}}=\bigoplus _{\mu \le \lambda }V(\lambda )_{\mu }^{{\mathbb {A}}}$, where $V(\lambda )_{\mu }^{{\mathbb {A}}}=V(\lambda )^{{\mathbb {A}}}\cap V(\lambda )_\mu$.

Fix $i \in I$. In [2], Bozec proved that every $u \in U_{q}^{-}({\mathfrak {g}})$ has the following decomposition.

$$\begin{aligned} u = {\left\{ \begin{array}{ll} \sum _{n\ge 0}{\texttt {b}}_i^{(n)}u_n &{} \text {with}\ i\in I^{\text {re}} \ \text {and}\ e'_iu_n=0 \ \text {for all} \ n\ge 0,\\ \sum _{{\mathbf {c}} \in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}} &{} \text {with}\ i\in I^{\text {im}} \ \text {and}\ e'_{il}u_{{\mathbf {c}}}=0 \ \text {for all} \ l>0, {\mathbf {c}}\in {\mathscr {C}}_i. \end{array}\right. } \end{aligned}$$

(23)

Lemma 16

For each $i \in I$ and $u \in U_{q}^{-}({\mathfrak {g}})$, consider the decomposition (23). If $u\in U^-_{\mathbb A}({\mathfrak {g}})$, then all $u_n, u_{{\mathbf {c}}} \in U^-_{\mathbb A}({\mathfrak {g}})$.

Proof

We first prove that $e'_{i,l}U^-_{{\mathbb {A}}}({\mathfrak {g}})\subseteq U^-_{{\mathbb {A}}}({\mathfrak {g}})$ for all $(i,l)\in I^{\infty }$.

Since $e'_{i,l}{\texttt {b}}_{jk}=\delta _{ij}\delta _{kl}+q_i^{-kla_{ij}}{\texttt {b}}_{jk}e'_{i,l},$ we have

$$\begin{aligned} e'_i{\texttt {b}}_i=1+q_i^{-2}{\texttt {b}}_ie'_i \ \ \text {for} \ i\in I^{\text {re}}. \end{aligned}$$

It follows that

$$\begin{aligned} e'_i{\texttt {b}}_i^{(n)}=q_i^{1-n}{\texttt {b}}_i^{(n-1)}+q_i^{-2n}{\texttt {b}}_i^{(n)}e'_i. \end{aligned}$$

Furthermore, by a direct calculation, we have

$$\begin{aligned} e'^{n}_i{\texttt {b}}_i^{(m)}=\sum _{k=0}^{n}q_i^{-2nm+(m+n)k-k(k-1)/2}\begin{bmatrix}n\\ k\end{bmatrix}_i{\texttt {b}}_i^{(m-k)}e'^{n-k}_i, \end{aligned}$$

where ${\texttt {b}}_i^{(r)}=0$ if $r<0$. These imply our assertion.

For $i\in I^{\text {re}}$, let

$$\begin{aligned} P=\sum _{n\ge 0}(-1)^nq_i^{-n(n-1)/2}{\texttt {b}}_i^{(n)}e'^{n}_i. \end{aligned}$$

Then we obtain (cf. [12,  Section 3.2]):

• (i) $P{\texttt {b}}_i=0$,   $e'_iP=0$,

• (ii) $\sum _{n\ge 0}q_i^{n(n-1)/2} \, {\texttt {b}}_i^{(n)}Pe'^{n}_i=1$,

• (iii) $Pe'^{n}_iu=q_i^{-n(n-1)/2}u_n$ for $u\in U^-_q({\mathfrak {g}})$.

Hence, if $u\in U^-_{{\mathbb {A}}}({\mathfrak {g}})$, then $u_n\in U^-_{{\mathbb {A}}}({\mathfrak {g}})$ for all $n\ge 0$.

For $i\in I^{\text {im}}$, we use a similar argument in [1,  Proposition 3.11]. Assume that $u\in U^-_{\mathbb A}({\mathfrak {g}})$ has the form $u=m{\texttt {b}}_{i,{\mathbf {c}}}m'$ for some ${\mathbf {c}}\in {\mathscr {C}}_i$ and homogeneous elements $m, m'\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$, where ${\mathcal {K}}_i=\bigcap _{l>0}\text {ker}\,e'_{i,l}$. We shall show that u can be written into the form

$$\begin{aligned} u=\sum _{{\mathbf {c}}'\in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}'}u_{{\mathbf {c}}'} \ \ \text {with} \ u_{{\mathbf {c}}'}\in {\mathcal {K}}_i\cap U^-_{\mathbb A}({\mathfrak {g}}). \end{aligned}$$

If $\vert {\mathbf {c}}\vert =0$, then $u=mm'\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$. Otherwise, we have

$$\begin{aligned} u=(m{\texttt {b}}_{ic_1}-q^{c_1(|m|,\alpha _i)}{\texttt {b}}_{ic_1}m){\texttt {b}}_{i,{\mathbf {c}}\backslash c_1}m'+q^{c_1(|m|,\alpha _i)}{\texttt {b}}_{ic_1}m{\texttt {b}}_{i,{\mathbf {c}}\backslash c_1}m', \end{aligned}$$

where $m{\texttt {b}}_{ic_1}-q^{c_1(|m|,\alpha _i)}{\texttt {b}}_{ic_1}m\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$. Now our claim follows by using the induction on $\vert {\mathbf {c}}\vert$.

We next show that if $u\in U^-_{{\mathbb {A}}}({\mathfrak {g}})$, then u can be written into the form

$$\begin{aligned} u=\sum _{{\mathbf {c}} \in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}} \ \text { with} \ u_{{\mathbf {c}}}\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}}). \end{aligned}$$

We will use the induction on $-|u|$.

Assume that u is a monomial in $U^-_{{\mathbb {A}}}({\mathfrak {g}})$. Then there exists some monomial $u'\in U^-_{\mathbb A}({\mathfrak {g}})$ such that $u={\texttt {b}}_j^{(n)}u'$ for some $j\in I^{\text {re}}$ or $u={\texttt {b}}_{jl}u'$ for some $j\in I^{\text {im}}$. By induction hypothesis, $u'=\sum _{{\mathbf {c}} \in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}}$ with $u_{{\mathbf {c}}}\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$. If $j\ne i$, then $u=\sum _{{\mathbf {c}} \in {\mathscr {C}}_i}{\texttt {b}}_j^{(n)}{\texttt {b}}_{i,{\mathbf {c}}} \, u_{{\mathbf {c}}}$ or $u=\sum _{{\mathbf {c}} \in \mathscr {C}_i}{\texttt {b}}_{jl}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}}$ is of the form $m{\texttt {b}}_{i,{\mathbf {c}}}m'$ with $m, m'\in {\mathcal {K}}_i\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$. If $i=j$, then $u=\sum _{{\mathbf {c}} \in {\mathscr {C}}_i}{\texttt {b}}_{i,(l,{\mathbf {c}})}u_{{\mathbf {c}}}$ is already in the form we wanted.

Thus, our assertion follows from the uniquesness of the decomposition. $\square$

Define

$$\begin{aligned}&({\texttt {b}}_i^nU_q^-({\mathfrak {g}}))^{\mathbb {A}}:={\texttt {b}}_i^nU_q^-({\mathfrak {g}})\cap U^-_{{\mathbb {A}}}({\mathfrak {g}}) \quad \text {for} \ i\in I^{\text {re}} \ \text {and}\ n\ge 1,\\&({\texttt {b}}_{i,{\mathbf {c}}}U_q^-({\mathfrak {g}}))^{\mathbb {A}}:={\texttt {b}}_{i,{\mathbf {c}}}U_q^-({\mathfrak {g}})\cap U^-_{{\mathbb {A}}}({\mathfrak {g}}) \quad \text {for} \ i\in I^{\text {im}} \ \text {and}\ {\mathbf {c}} \in {\mathscr {C}}_i\backslash \{0\}. \end{aligned}$$

By the above lemma, $U^-_{{\mathbb {A}}}({\mathfrak {g}})$ is stable under the Kashiwara operators $\widetilde{e}_{il}$ and $\widetilde{f}_{il}$. Moreover, we have the following corollary.

Corollary 17

• (i) For $i\in I^{\text {re}}$, $({\texttt {b}}_i^nU_q^-({\mathfrak {g}}))^{\mathbb {A}}=\sum _{k\ge n}{\texttt {b}}_i^{(k)}U^-_{{\mathbb {A}}}({\mathfrak {g}})=\bigoplus _{k\ge n} {\texttt {b}}_i^{(k)} (U^-_{{\mathbb {A}}}({\mathfrak {g}})\cap \text {ker}\, e'_i)$. For $i\in I^{\text {im}}$, $({\texttt {b}}_{i,{\mathbf {c}}}U_q^-({\mathfrak {g}}))^{\mathbb {A}}= b_{i,{\mathbf {c}}}U^-_{\mathbb A}({\mathfrak {g}})=\bigoplus _{{\mathbf {c}}'\in {\mathscr {C}}_i}{\texttt {b}}_{i,({\mathbf {c}},{\mathbf {c}}')}(U^-_{{\mathbb {A}}}({\mathfrak {g}})\cap {\mathcal {K}}_i)$.

• (ii) For $i\in I^{\text {re}}$, $({\texttt {b}}_i^nV(\lambda ))^{{\mathbb {A}}}:=({\texttt {b}}_i^nU_q^-({\mathfrak {g}}))^\mathbb Av_\lambda =\sum _{k\ge n}{\texttt {b}}_i^{(k)}V(\lambda )^{{\mathbb {A}}}$. For $i\in I^{\text {im}}$, $({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{\mathbb {A}}:=({\texttt {b}}_{i,{\mathbf {c}}}U_q^-({\mathfrak {g}}))^\mathbb Av_\lambda ={\texttt {b}}_{i,{\mathbf {c}}}V(\lambda )^{{\mathbb {A}}}$.

6 Existence of global bases

Let V be a finite-dimensional vector space over ${\mathbf {F}}(q)$, M be an ${\mathbb {A}}$-submodule of V, and $L_0$ (resp. $L_\infty$) be a free ${\mathbb {A}}_0$-submodule (resp. free $\mathbb A_\infty$-submodule) of V such that $V\cong {\mathbf {F}}(q)\otimes _{{\mathbb {A}}_0}L_0 \cong {\mathbf {F}}(q)\otimes _{{\mathbb {A}}_\infty }L_\infty$.

Definition 4

The triple $(V^{{\mathbb {A}}}, L_{0}, L_{\infty })$ is called an balanced triple for V if the canonical linear map $V^{{\mathbb {A}}} \cap L_{0} \cap L_{\infty } \longrightarrow L_{0} \big / qL_{0}$ is an isomorphism.

Lemma 18

[12,  Lemma 7.1.1] Let $V, M, L_0, L_\infty$ be as above.

• (i) Assume that the canonical map $M\cap L_0 \cap L_\infty \rightarrow M\cap L_0/M\cap qL_0$ is an isomorphism. Then

  $$\begin{aligned} M\cap L_0&\cong {\mathbf {F}}[q]\otimes _{{\mathbf {F}}}(M\cap L_0 \cap L_\infty ),\\ M\cap L_\infty&\cong {\mathbf {F}}[q^{-1}]\otimes _{{\mathbf {F}}}(M\cap L_0 \cap L_\infty ),\\ M&\cong {\mathbb {A}}\otimes _{{\mathbf {F}}}(M\cap L_0 \cap L_\infty ),\\ M\cap L_0 \cap L_\infty&\cong (M\cap L_\infty )/(M\cap q^{-1}L_\infty ),\\ M\cap L_0 \cap L_\infty&\cong ({\mathbf {F}}(q)\otimes _{\mathbb A}M)\cap L_0/({\mathbf {F}}(q)\otimes _{{\mathbb {A}}}M)\cap qL_0. \end{aligned}$$

• (ii) Let E be an ${\mathbf {F}}$-vector space and $\varphi :E\rightarrow M\cap L_0 \cap L_\infty$ a homomorphism. Assume that $M={\mathbb {A}}\varphi (E)$ and $E\rightarrow L_0/qL_0, E\rightarrow L_\infty /q^{-1}L_\infty$ are injective. Then $E\rightarrow M\cap L_0 \cap L_\infty \rightarrow M\cap L_0/M \cap qL_0$ is an isomorphism.

Lemma 19

[12,  Lemma 7.1.2] Let $V, M, L_0, L_\infty$ be as above and let N be an $\mathbb A$-submodule of M. Assume that

• (1) $N\cap L_0\cap L_\infty \cong N\cap L_0/N\cap qL_0$.

• (2) There exist an ${\mathbf {F}}$-vector space F and a homomorphism $\varphi :F\rightarrow M\cap (L_0+N)\cap (L_\infty +N)$ such that

• (a) $M={\mathbb {A}}\varphi (F)+N$,

• (b) the induced homomorphisms $\varphi _0:F\rightarrow (L_0+N) / (qL_0+N)$ and $\varphi _\infty :F\rightarrow (L_\infty +N) / (q^{-1}L_\infty +N)$ are injective.

Then the following statements hold.

• (i) $M\cap L_0\cap L_\infty \rightarrow M\cap L_0/M\cap qL_0$ is an isomorphism.

• (ii) $M\cap L_0/M\cap qL_0 \cong F\oplus (N\cap L_0/N\cap qL_0)$.

For $r\ge 0$, set

$$\begin{aligned} Q_+(r)=\{\alpha \in Q_+\mid \vert \alpha \vert \le r\}. \end{aligned}$$

We will prove the following inductive statements on $r\ge 0$.

A(r): For any $\alpha \in Q_+(r)$, we have the following canonical isomorphism

$$\begin{aligned} U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^- \xrightarrow {\sim } \frac{U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap {\mathcal {L}}(\infty )}{U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q{\mathcal {L}}(\infty )}\xrightarrow {\sim } {\mathcal {L}}(\infty )_{-\alpha }/q{\mathcal {L}}(\infty )_{-\alpha }. \end{aligned}$$

We denote by $G_\infty$ the inverse of this isomorphism.

B(r): For any $\alpha \in Q_+(r)$ and $\lambda \in P^{+}$, we have the following canonical isomorphism

$$\begin{aligned} V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^- \xrightarrow {\sim } \frac{V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )}{V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )}\xrightarrow {\sim } {\mathcal {L}}(\lambda )_{\lambda -\alpha }/q{\mathcal {L}}(\lambda )_{\lambda -\alpha }. \end{aligned}$$

We denote by $G_\lambda$ the inverse of this isomorphism.

C(r): For $\alpha \in Q_+(r)$, $(i,l)\in I^{\infty }$, and $n\ge 0$, assume that $b\in \widetilde{f}_{il}^n \, {\mathcal {B}}(\infty )_{-\alpha +l n\alpha _i}$. Then we have

$$\begin{aligned} G_\infty (b)\in {\texttt {b}}_{il}^n \, U_q^-({\mathfrak {g}}). \end{aligned}$$

If $r=0$, our assertions are obvious. Now we assume that A($r-1$), B($r-1$) and C($r-1$) are true. Then Lemma 18 and Proposition 10 imply the following result.

Lemma 20

For $\alpha \in Q_+(r-1)$, we have

$$\begin{aligned} \begin{aligned} U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap \mathcal L(\infty )&=\bigoplus _{b\in \mathcal B(\infty )_{-\alpha }}{\mathbf {F}}[q]G_\infty (b), \quad U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }=\bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }}\mathbb AG_\infty (b),\\ V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap \mathcal L(\lambda )&=\bigoplus _{b\in \mathcal B(\lambda )_{\lambda -\alpha }}{\mathbf {F}}[q]G_\lambda (b), \quad V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }=\bigoplus _{b\in \mathcal B(\lambda )_{\lambda -\alpha }}\mathbb AG_\lambda (b), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} G_\infty (b)v_\lambda =G_\lambda (\overline{\pi }_\lambda (b)). \end{aligned}$$

Lemma 21

For $\alpha \in Q_+(r-1)$, $b\in {\mathcal {B}}(\infty )_{-\alpha }$ (resp. $b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha }$), we have $\overline{G_\infty (b)}=G_\infty (b)$ (resp. $\overline{G_\lambda (b)}=G_\lambda (b)$).

Proof

Let $Q=(G_\infty (b)-\overline{G_\infty (b)})/(q-q^{-1})$. Then we have $Q\in U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q\mathcal L(\infty )\cap {\mathcal {L}}(\infty )^-$ since $1/(q-q^{-1})\in q\mathbb A_0$. $\square$

Let $i\in I^{\text {iso}}$, $\lambda \in P^{+}$ and $\alpha \in Q_{+}$. For each partition ${\mathbf {c}}=(1^{l_1}2^{l_2}\cdots k^{l_k}\cdots )$, we define

$$\begin{aligned}&({{\texttt {b}}}_{i, {\mathbf {c}}} * U_{q}^{-}({\mathfrak {g}}))^{\mathbb A}_{-\alpha } := \sum _{k \ge 1} ({{\texttt {b}}}_{i,k}^{l_k} U_{q}^{-}({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha } = \sum _{k \ge 1} {{\texttt {b}}}_{i,k}^{l_k} (U_{{\mathbb {A}}}^{-}({\mathfrak {g}})_{-\alpha + k l_{k} \alpha _{i}}), \\&({{\texttt {b}}}_{i, {\mathbf {c}}} * V(\lambda ))^{{\mathbb {A}}}_{\lambda - \alpha } : = \sum _{k \ge 1} ({{\texttt {b}}}_{i, k}^{l_k} V(\lambda ))^{{\mathbb {A}}}_{\lambda - \alpha } =\sum _{k \ge 1} {{\texttt {b}}}_{i, k}^{l_k}(V(\lambda )^{{\mathbb {A}}}_{\lambda - \alpha + k l_k \alpha _{i}}). \end{aligned}$$

Here $({\texttt {b}}_{i,k}^{l_k}U^-_q({\mathfrak {g}}))^{\mathbb A}_{-\alpha } = ({{\texttt {b}}}_{i,k}^{l_k} V(\lambda ))^{\mathbb A}_{\lambda - \alpha } = 0$ if $l_k=0$.

Proposition 22

Let $\alpha \in Q_+(r)$ and $\lambda \in P^+$.

• (i) For $i\in I^{\text {re}}$ and $n\ge 1$, we have

  $$\begin{aligned} ({\texttt {b}}_i^nV(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^-\xrightarrow {\sim }\frac{({\texttt {b}}_i^nV(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )}{({\texttt {b}}_i^nV(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )}\cong \bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap {\widetilde{f}}_i^n{\mathcal {B}}(\lambda )}{\mathbf {F}} b. \end{aligned}$$

• (ii) For $i\in I^{\text {im}}\backslash I^{\text {iso}}$ and any composition ${\mathbf {c}}$ with $\vert {\mathbf {c}}\vert \ne 0$, we have

  $$\begin{aligned} ({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^-\xrightarrow {\sim }\frac{({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )}{({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )}\cong \bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap {\widetilde{f}}_{i,{\mathbf {c}}}{\mathcal {B}}(\lambda )}{\mathbf {F}} b. \end{aligned}$$

• (iii) For $i\in I^{\text {iso}}$ and any partition ${\mathbf {c}}=1^{l_1}2^{l_2}\cdots k^{l_k}\cdots$, we have

  $$\begin{aligned}&({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^-\xrightarrow {\sim }\frac{({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )}{({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )}\\&\cong \bigoplus _{b\in B(\lambda )_{\lambda -\alpha }\cap ({\widetilde{f}}_{i,{\mathbf {c}}}*\mathcal B(\lambda ))}{\mathbf {F}} b, \end{aligned}$$

  where ${\widetilde{f}}_{i,{\mathbf {c}}}*\mathcal B(\lambda ):=\bigcup _{k\ge 1} {{\widetilde{f}}}_{i,k}^{l_k}\mathcal B(\lambda )$.

Proof

Our assertion (i) has been proved in [12,  Proposition 7.4.1].

Assume $i\in I^{\text {im}}\backslash I^{\text {iso}}$. Let ${\mathbf {c}}\in {{\mathscr {C}}}_i$ such that $\vert {\mathbf {c}}\vert =n>0$. Recall that $({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{\mathbb A}_{\lambda -\alpha }={\texttt {b}}_{i,{\mathbf {c}}}(V(\lambda )^{\mathbb A}_{\lambda -\alpha +n\alpha _i})$. If $(\lambda -\alpha +n\alpha _i)(h_{i})=0$, then $({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }=0$, and hence our assertion is trivial. Thus we may assume that $(\lambda -\alpha +n\alpha _i)(h_{i})>0$. In this case, for any $b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}$, we have $\widetilde{f}_{i,{\mathbf {c}}}b\ne 0$.

By B($r-1$), we have $V(\lambda )^{\mathbb A}_{\lambda -\alpha +n\alpha _i}=\bigoplus _{b\in \mathcal B(\lambda )_{\lambda -\alpha +n\alpha _i}}\mathbb AG_\lambda (b)$. Hence

$$\begin{aligned} ({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }=\sum _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}}{\mathbb {A}}{\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b). \end{aligned}$$

Let $F=\sum _{b\in \mathcal B(\lambda )_{\lambda -\alpha +n\alpha _i}}{\mathbf {F}} {\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b)$. We first show that F is a direct sum. Assume that

$$\begin{aligned} \sum _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}}\beta _b {\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b)=0\quad \text {for some} \ \beta _b\in {\mathbf {F}}. \end{aligned}$$

Since ${\widetilde{f}}_{i,{\mathbf {c}}}G_\lambda (b)={\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b)$ and $G_\lambda (b)\equiv b \ (\text {mod}\ q\mathcal L(\lambda ))$ for any $b\in \mathcal B(\lambda )_{\lambda -\alpha +n\alpha _i}$, we obtain

$$\begin{aligned} \sum _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}}\beta _b {\widetilde{f}}_{i,{\mathbf {c}}}b=0. \end{aligned}$$

By applying ${\widetilde{e}}_{i,\widetilde{{\mathbf {c}}}}$, we get $\sum _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}}\beta _b b=0$, which implies $\beta _b=0$ for all $b\in \mathcal B(\lambda )_{\lambda -\alpha +n\alpha _i}$.

Let $N=0$, $M=({\texttt {b}}_{i,{\mathbf {c}}}V(\lambda ))^{\mathbb A}_{\lambda -\alpha }$, $L_0={\mathcal {L}}(\lambda )_{\lambda -\alpha }$ and $L_\infty ={\mathcal {L}}(\lambda )_{\lambda -\alpha }^-$. Set $\varphi :F\rightarrow M\cap L_0\cap L_\infty$ be the ${\mathbf {F}}$-linear map given by

$$\begin{aligned} {\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b) \longmapsto {\texttt {b}}_{i,{\mathbf {c}}}G_\lambda (b)={\widetilde{f}}_{i,{\mathbf {c}}} G_\lambda (b). \end{aligned}$$

Then, it is easy to check $F,N,M,L_0,L_\infty$ and $\varphi$ satisfy the conditions in Lemma 19, and hence we get

$$\begin{aligned} M\cap L_0\cap L_\infty \xrightarrow {\sim }M\cap L_0/M\cap qL_0\cong \bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +n\alpha _i}}{\mathbf {F}}{\widetilde{f}}_{i,{\mathbf {c}}}b=\bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap {\widetilde{f}}_{i,{\mathbf {c}}}{\mathcal {B}}(\lambda )}{\mathbf {F}} b \end{aligned}$$

as desired.

Now, we shall prove (iii). Let $i\in I^{\text {iso}}$. If $l_k$ is sufficient large, then $({\texttt {b}}_{ik}^{l_k}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }=0$. Hence we can use descending induction on $N = \sum _{k\ge 1}l_k$. Without loss of generality, we may assume that $l_1\ne 0$ and $(\lambda -\alpha )(h_{i})>0$. Then, by A($r-1$) and B($r-1$), we have

$$\begin{aligned} ({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }= {\texttt {b}}_{i,1}^{l_1}V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha +l_1\alpha _i}+\sum _{k\ge 2}({\texttt {b}}_{i,k}^{l_k}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha } \end{aligned}$$

and

$$\begin{aligned} V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha +l_1\alpha _i}=\bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +l_1\alpha _i}}\mathbb AG_\lambda (b)=\bigoplus _{\begin{array}{c} b\in {\mathcal {B}}(\infty )_{-\alpha +l_1\alpha _i}\\ \overline{\pi }_\lambda (b)\ne 0 \end{array}}\mathbb AG_\infty (b)v_\lambda . \end{aligned}$$

Let $b\in {\mathcal {B}}(\infty )_{-\alpha +l_1\alpha _i}$ with $\widetilde{e}_{i1}b\ne 0$. Then $b\in {\widetilde{f}}_{i1}\mathcal B(\infty )\cap {\mathcal {B}}(\infty )_{-\alpha +l_1\alpha _i}$, which implies $G_\infty (b)\in {\texttt {b}}_{i1}U^-_q({\mathfrak {g}})\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$ by C($r-1$). Hence ${\texttt {b}}_{i1}^{l_1}G_\infty (b)v_\lambda \in ({\texttt {b}}_{i1}^{l_1+1}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }$. For $k\ne 1$ with $l_k\ne 0$, if $b\in {\widetilde{f}}_{ik}^{l_k}\mathcal B(\infty )$, then $G_\infty (b)\in {\texttt {b}}_{ik}^{l_k}U^-_q({\mathfrak {g}})\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})$ by C($r-1$). Therefore ${\texttt {b}}_{i,1}^{l_1}G_\infty (b)v_\lambda \in ({\texttt {b}}_{i,k}^{l_k}V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }$. Hence we have

$$\begin{aligned} ({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }=\sum _{b\in S}{\mathbb {A}}{\texttt {b}}_{i,1}^{l_1}G_\infty (b)v_\lambda +({\texttt {b}}_{i,{\mathbf {c}}\cup \{1\}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} S&=\left\{ b\in {\mathcal {B}}(\infty )_{-\alpha +l_1\alpha _i}\mid \overline{\pi }_\lambda (b)\ne 0,\ {\widetilde{e}}_{i,1}b=0,\ b\notin \bigcup _{k\ge 2}{\widetilde{f}}_{i,k}^{l_k}{\mathcal {B}}(\infty )\right\} \\&\xrightarrow {\begin{array}{c} \overline{\pi }_\lambda \\ \sim \end{array}} \left\{ b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha +l_1\alpha _i}\mid {\widetilde{e}}_{i,1}b=0,\ b\notin \bigcup _{k\ge 2}{\widetilde{f}}_{i,k}^{l_k}{\mathcal {B}}(\lambda )\right\} \\&\xrightarrow {\begin{array}{c} {\widetilde{f}}_{i,1}^{l_1}\\ \sim \end{array}} {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap \left( \widetilde{f}_{i,1}^{l_1}{\mathcal {B}}(\lambda )\backslash ({\widetilde{f}}_{i,{\mathbf {c}}\cup \{1\}}*{\mathcal {B}}(\lambda ))\right) . \end{aligned} \end{aligned}$$

(24)

The last isomorphism follows from the fact that $\widetilde{f}_{il}{\widetilde{e}}_{il'}={\widetilde{e}}_{il'}{\widetilde{f}}_{il}$ and ${\widetilde{f}}_{il}{\widetilde{f}}_{il'}={\widetilde{f}}_{il'}\widetilde{f}_{il}$ for any $l,l'\ge 1$ with $l\ne l'$.

Let $V=V(\lambda )_{\lambda -\alpha }$, $M=({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }$, $N=({\texttt {b}}_{i,{\mathbf {c}}\cup \{1\}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }$, $L_0={\mathcal {L}}(\lambda )_{\lambda -\alpha }$, $L_\infty =\mathcal L(\lambda )^-_{\lambda -\alpha }$ and

$$\begin{aligned} F=\sum _{b\in S}{\mathbf {F}} {\texttt {b}}_{i1}^{l_1}G_\infty (b)v_\lambda . \end{aligned}$$

For $b\in S$, we have $b=G_\infty (b)+q{\mathcal {L}}(\infty )$. Assume that $G_\infty (b)$ has the decomposition

$$\begin{aligned} G_\infty (b)=\sum _{{\mathbf {c}}\in {\mathscr {C}}_i}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}}\in U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha +l_1\alpha _i}\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-. \end{aligned}$$

Then we have

$$\begin{aligned} {\widetilde{f}}_{i1}\widetilde{e}_{i1}G_\infty (b)=G_\infty (b)-\sum _{{\mathbf {c}}\in {\mathscr {C}}_i; 1\notin {\mathbf {c}}}{\texttt {b}}_{i,{\mathbf {c}}}u_{{\mathbf {c}}}=G_\infty (b)-u_b\in q{\mathcal {L}}(\infty ). \end{aligned}$$

Hence we obtain

• (i) ${\texttt {b}}_{i1}^{l_1}G_\infty (b)\equiv {\texttt {b}}_{i1}^{l_1}u_b\ \text {mod}\ ({\texttt {b}}_{i1}^{l_1+1}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}$, which implies

  $$\begin{aligned} {\texttt {b}}_{i1}^{l_1}G_\infty (b)v_\lambda \in M\cap (N+L_0)\cap (N+L_\infty ).\end{aligned}$$

  (25)

• (ii) ${\widetilde{f}}_{i1}^{l_1}b=\beta _b{\texttt {b}}_{i1}^{l_1}u_b+q{\mathcal {L}}(\infty )$ for some $\beta _b\in {{\mathbf {F}}}^*$, which implies

  $$\begin{aligned} \overline{\pi }_\lambda ({\widetilde{f}}_{i1}^{l_1}b)=\beta _b{\texttt {b}}_{i1}^{l_1}u_bv_\lambda +q{\mathcal {L}}(\lambda ).\end{aligned}$$

  (26)

Set

$$\begin{aligned} H:= (N+L_0) /(N+qL_0) \cong \frac{L_0/qL_0}{N \cap L_0 / N\cap qL_0}. \end{aligned}$$

By induction hypothesis, we have $H\cong \bigoplus _{b\in \mathcal B(\lambda )_{\lambda -\alpha }\backslash ({\widetilde{f}}_{i,{\mathbf {c}}\cup \{1\}}*{\mathcal {B}}(\lambda ))}{\mathbf {F}} b.$ Hence (24), (25) and (26) imply that the following canonical maps are injective:

$$\begin{aligned} \begin{aligned}&\varphi _0:F\xrightarrow {\varphi } M\cap (L_0+N)\cap (L_\infty +N)\rightarrow \frac{N+L_0 }{N+qL_0}\xrightarrow {\sim } H \\&b_{i1}^{l_1}G_\infty (b)v_\lambda \mapsto {\texttt {b}}_{i1}^{l_1}G_\infty (b)v_\lambda \mapsto {\texttt {b}}_{i1}^{l_1}u_bv_\lambda +(N+qL_0) \mapsto \frac{1}{\beta _b}\overline{\pi }_\lambda ({\widetilde{f}}_{i1}^{l_1}b) \end{aligned} \end{aligned}$$

(27)

By taking −, the following canonical map is injective

$$\begin{aligned} \varphi _\infty :F\xrightarrow {\varphi } M\cap (L_0+N)\cap (L_\infty +N)\rightarrow \frac{N+L_\infty }{N+q^{-1}L_\infty }. \end{aligned}$$

Note that $M={\mathbb {A}}\varphi (F)+N$. Hence Lemma 19 yields

$$\begin{aligned} M\cap L_0\cap L_\infty \xrightarrow {\sim } (M\cap L_0) / (M\cap qL_0) \cong F\oplus (N\cap L_0/N\cap qL_0), \end{aligned}$$

where

$$\begin{aligned}&F\oplus (N\cap L_0/ N\cap qL_0)\cong \bigoplus _{S\sqcup \left( {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap {\widetilde{f}}_{i,{\mathbf {c}}\cup \{1\}}*{\mathcal {B}}(\lambda )\right) }{\mathbf {F}} b\\&=\bigoplus _{\mathcal B(\lambda )_{\lambda -\alpha }\cap \left( {\widetilde{f}}_{i,{\mathbf {c}}}*{\mathcal {B}}(\lambda )\right) }{\mathbf {F}} b. \end{aligned}$$

This completes the proof. $\square$

Corollary 23

Let $\alpha \in Q_+(r)$.

• (i) For $i\in I^{\text {re}}$ and $n\ge 1$, we have

  $$\begin{aligned}&({\texttt {b}}_i^nU^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-\xrightarrow {\sim }\frac{({\texttt {b}}_i^nU^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )}{({\texttt {b}}_i^nU^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap q{\mathcal {L}}(\infty )}\\&\cong \bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap \widetilde{f}_i^n{\mathcal {B}}(\infty )}{\mathbf {F}} b. \end{aligned}$$

• (ii) For $i\in I^{\text {im}}\backslash I^{\text {iso}}$ and $\vert {\mathbf {c}}\vert \ne 0$, we have

  $$\begin{aligned}&({\texttt {b}}_{i,{\mathbf {c}}}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-\xrightarrow {\sim }\frac{({\texttt {b}}_{i,{\mathbf {c}}}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )}{({\texttt {b}}_{i,{\mathbf {c}}}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap q{\mathcal {L}}(\infty )}\\&\cong \bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap \widetilde{f}_{i,{\mathbf {c}}}{\mathcal {B}}(\infty )}{\mathbf {F}} b. \end{aligned}$$

• (iii) For $i\in I^{\text {iso}}$ and any partition ${\mathbf {c}}=(1^{l_1}2^{l_2}\cdots k^{l_k}\cdots )$, we have

  $$\begin{aligned}&({\texttt {b}}_{i,{\mathbf {c}}}*U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-\xrightarrow {\sim }\frac{({\texttt {b}}_{i,{\mathbf {c}}}*U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )}{({\texttt {b}}_{i,{\mathbf {c}}}*U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap q{\mathcal {L}}(\infty )}\\&\cong \bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap (\widetilde{f}_{i,{\mathbf {c}}}*{\mathcal {B}}(\infty ))}{\mathbf {F}} b, \end{aligned}$$

  where ${\widetilde{f}}_{i,{\mathbf {c}}}*{\mathcal {B}}(\infty )=\bigcup _{k\ge 1}{\widetilde{f}}_{ik}^{l_k}{\mathcal {B}}(\infty )$.

Proof

We shall prove (iii) only. The proof of (i) and (ii) are similar. For $\lambda \gg 0$, we have

$$\begin{aligned} U^-_q({\mathfrak {g}})_{-\alpha }\xrightarrow {\sim }V(\lambda )_{\lambda -\alpha },\quad ({\texttt {b}}_{i,{\mathbf {c}}}*U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\xrightarrow {\sim }({\texttt {b}}_{i,{\mathbf {c}}}*V(\lambda ))^{{\mathbb {A}}}_{\lambda -\alpha }, \end{aligned}$$

$$\begin{aligned} {\mathcal {L}}(\infty )_{-\alpha }\xrightarrow {\sim }{\mathcal {L}}(\lambda )_{\lambda -\alpha },\quad {\mathcal {L}}(\infty )^-_{-\alpha }\xrightarrow {\sim }{\mathcal {L}}(\lambda )^-_{\lambda -\alpha }, \end{aligned}$$

and

$$\begin{aligned} \bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap ({\widetilde{f}}_{i,{\mathbf {c}}}*{\mathcal {B}}(\infty ))}{\mathbf {F}} b\xrightarrow {\sim } \bigoplus _{b\in {\mathcal {B}}(\lambda )_{\lambda -\alpha }\cap ({\widetilde{f}}_{i,{\mathbf {c}}}*{\mathcal {B}}(\lambda ))}{\mathbf {F}} b. \end{aligned}$$

Hence our assertion follows immediately. $\square$

For $\alpha \in Q_+(r)$ and $(i,l)\in I^{\infty }$, let us denote by $G_{il}$ the inverse of the isomorphism

$$\begin{aligned} ({\texttt {b}}_{il}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-\xrightarrow {\sim } \bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap {\widetilde{f}}_{il}{\mathcal {B}}(\infty )}{\mathbf {F}} b. \end{aligned}$$

Then Corollary 23 implies $({\texttt {b}}_{il}^nU^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }=\bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap {\widetilde{f}}_{il}^n\mathcal B(\infty )}\mathbb AG_{il}(b)$ for any $n\ge 1$.

Lemma 24

Let $(i,l),(j,s)\in I^{\infty }$, $\alpha \in Q_+(r)$ and $b\in {\widetilde{f}}_{il}{\mathcal {B}}(\infty )\cap {\widetilde{f}}_{js}\mathcal B(\infty )\cap {\mathcal {B}}(\infty )_{-\alpha }$. Then we have

$$\begin{aligned} G_{il}(b)=G_{js}(b). \end{aligned}$$

Proof

Let us write $b=\widetilde{f}_{il}\ldots \widetilde{f}_{km}\cdot 1$, where $(k,m)\in I^{\infty }$. If $k\in I^{\text {re}}$, then our claim was proved in [12]. So we will assume that $k\in I^{\text {im}}$. Take $\lambda \in P^+$ with $\lambda (h_{k})=0$ and $\lambda (h_{j}) \gg 0$ for all $j\in I\backslash \{k\}$. Then (18) yields

$$\begin{aligned} V(\lambda )_{\lambda -\alpha }\simeq U^-_q({\mathfrak {g}})_{-\alpha } \bigg /\sum _{n\ge 1}U^-_q({\mathfrak {g}})_{-\alpha +n\alpha _k}{\texttt {b}}_{kn}. \end{aligned}$$

The same argument in [12,  Lemma 7.5.1] shows that

$$\begin{aligned} Q=G_{il}(b)-G_{js}(b)\in \left( \sum _{n\ge 1}U^-_q({\mathfrak {g}})_{-\alpha +n\alpha _k}{\texttt {b}}_{kn}\right) \cap U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q{\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-. \end{aligned}$$

Then Corollary 13 implies

$$\begin{aligned} Q^*\in \left( \sum _{n\ge 1}{\texttt {b}}_{kn}U^-_q({\mathfrak {g}})_{-\alpha +n\alpha _k}\right) \cap U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q{\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-. \end{aligned}$$

If $k\in I^{\text {im}}\backslash I^{\text {iso}}$, we assume that $Q^*={\texttt {b}}_{k1}u_1+\cdots +{\texttt {b}}_{kt}u_t$. Since $\overline{Q^*}=\overline{Q}^*=Q^*$, we have $Q^*={\texttt {b}}_{k1}\overline{u}_1+\cdots +{\texttt {b}}_{kt}\overline{u}_t$. Note that for each $1\le j\le t$, ${\texttt {b}}_{kj}u_j=\widetilde{f}_{kj}{\widetilde{e}}_{kj}Q^*\in {\texttt {b}}_{kj}U^-_q({\mathfrak {g}})\cap U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q{\mathcal {L}}(\infty )$ and $u_j=\overline{u}_j={\widetilde{e}}_{kj}Q^*$. Hence $b_{kj}u_j\in ({\texttt {b}}_{kj}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }\cap q{\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-$ and Corollary 23 (ii) implies $Q^*=0$.

If $k\in I^{\text {iso}}$, since $Q^*\in (\sum _{n\ge 1}{\texttt {b}}_{kn} U^-_q({\mathfrak {g}})_{-\alpha +n\alpha _k})\cap U^-_{\mathbb A}({\mathfrak {g}})_{-\alpha }$, the decomposition of $Q^*$ can be expressed as the form $Q^*={\texttt {b}}_{k1}u_1+\cdots +{\texttt {b}}_{kt}u_t$ with

$$\begin{aligned} u_j=\sum _{\begin{array}{c} {\mathbf {c}}\in {\mathscr {C}}_k\ \text {and} \\ {\mathbf {c}}\ \text {contains no} \ j+1,\ldots , s \end{array}}{\texttt {b}}_{k,{\mathbf {c}}}u_{{\mathbf {c}}}. \end{aligned}$$

For every $1\le j\le t$, we have

$$\begin{aligned} {\texttt {b}}_{kj}u_{j}={\widetilde{f}}_{kj}{\widetilde{e}}_{kj}(Q^*-\sum _{j<p\le s}b_{kp}u_p)\in ({\texttt {b}}_{kj}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }. \end{aligned}$$

Hence $Q^*\in ({\texttt {b}}_{k,(1^12^1\ldots t^1)}*U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }$, and Corollary 23 (iii) implies $Q^*=0$. $\square$

Thus we can define

$$\begin{aligned} G:{\mathcal {L}}(\infty )_{-\alpha }/q{\mathcal {L}}(\infty )_{-\alpha }\rightarrow U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^- \end{aligned}$$

by

$$\begin{aligned} b\mapsto G_{il}(b) \ \ \text {for} \ b\in {\widetilde{f}}_{il}{\mathcal {B}}(\infty )\cap {\mathcal {B}}(\infty )_{-\alpha }, (i,l) \in I^{\infty }. \end{aligned}$$

Then we have $b=G(b)+q{\mathcal {L}}(\infty )$ and

$$\begin{aligned} ({\texttt {b}}_{il}^nU^-_q({\mathfrak {g}}))^{\mathbb A}_{-\alpha }=\bigoplus _{b\in {\mathcal {B}}(\infty )_{-\alpha }\cap {\widetilde{f}}_{il}^n{\mathcal {B}}(\infty )}\mathbb AG(b) \end{aligned}$$

(28)

for any $n\ge 1$. Since $U^-_{\mathbb A}({\mathfrak {g}})_{-\alpha }=\sum _{(i,l)\in I^{\infty }}({\texttt {b}}_{il}U^-_q({\mathfrak {g}}))^{{\mathbb {A}}}_{-\alpha }$, we obtain

$$\begin{aligned} U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }=\sum _{b\in \mathcal B(\infty )_{-\alpha }}{\mathbb {A}} G(b). \end{aligned}$$

Let $E=\mathcal L(\infty )_{-\alpha }/q{\mathcal {L}}(\infty )_{-\alpha }$ and $M=U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }$. Then by Lemma 18(ii), we deduce that

$$\begin{aligned} {\mathcal {L}}(\infty )_{-\alpha }/q{\mathcal {L}}(\infty )_{-\alpha }\xrightarrow {G} U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap {\mathcal {L}}(\infty )\cap {\mathcal {L}}(\infty )^-\rightarrow \frac{U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap {\mathcal {L}}(\infty )}{U^-_{{\mathbb {A}}}({\mathfrak {g}})_{-\alpha }\cap q{\mathcal {L}}(\infty )} \end{aligned}$$

is an isomorphism, which proves A(r). Now C(r) follows from (28). Finally, we shall prove B(r).

Lemma 25

Let $\alpha \in Q_+(r)$, $b\in {\mathcal {B}}(\infty )_{-\alpha }$ and $\lambda \in P^+$. If $\overline{\pi }_\lambda (b)=0$, then $G(b)v_\lambda =0$.

Proof

Take $(i,l)\in I^{\infty }$ with ${\widetilde{e}}_{il}b\ne 0$. Then $G(b)v_\lambda \in ({\texttt {b}}_{il}V(\lambda ))^{\mathbb A}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )\cap \mathcal L(\lambda )^-$ by Proposition 22. $\square$

By this lemma, we have $V(\lambda )^{\mathbb A}_{\lambda -\alpha }=\sum _{\begin{array}{c} b\in \mathcal B(\infty )_{-\alpha }\\ \overline{\pi }_\lambda (b)\ne 0 \end{array}}{\mathbb {A}} G(b)v_\lambda$. Let

$$\begin{aligned} E=\sum _{\begin{array}{c} b\in \mathcal B(\infty )_{-\alpha }\\ \overline{\pi }_\lambda (b)\ne 0 \end{array}}{\mathbf {F}} G(b)v_\lambda \subseteq V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^-. \end{aligned}$$

Since $\{b\in {\mathcal {B}}(\infty )_{-\alpha }\mid \overline{\pi }_\lambda (b)\ne 0\}\xrightarrow {\sim }\mathcal B(\lambda )_{\lambda -\alpha }$, we have

$$\begin{aligned} E\xrightarrow {\sim } {\mathcal {L}}(\lambda )_{\lambda -\alpha }/q{\mathcal {L}}(\lambda )_{\lambda -\alpha } \end{aligned}$$

given by

$$\begin{aligned} G(b)v_\lambda \longmapsto G(b)v_\lambda +q{\mathcal {L}}(\lambda )=\overline{\pi }_\lambda (b). \end{aligned}$$

By Lemma 18(ii), we get

$$\begin{aligned} E\xrightarrow {\sim } V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )\cap {\mathcal {L}}(\lambda )^- \xrightarrow {\sim } \frac{V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap {\mathcal {L}}(\lambda )}{V(\lambda )^{{\mathbb {A}}}_{\lambda -\alpha }\cap q{\mathcal {L}}(\lambda )}\cong {\mathcal {L}}(\lambda )_{\lambda -\alpha }/q{\mathcal {L}}(\lambda )_{\lambda -\alpha }, \end{aligned}$$

which proves B(r).

To summarize, we obtain the main goal of this paper.

Theorem 26

There exist canonical isomorphisms

$$\begin{aligned} U^-_{{\mathbb {A}}}({\mathfrak {g}})\cap {\mathcal {L}}(\infty )\cap \mathcal L(\infty )^-&\xrightarrow {\sim } {\mathcal {L}}(\infty )/q {\mathcal {L}}(\infty ),\\ V(\lambda )^{{\mathbb {A}}}\cap {\mathcal {L}}(\lambda )\cap \mathcal L(\lambda )^-&\xrightarrow {\sim } {\mathcal {L}}(\lambda )/q \mathcal L(\lambda ) \ \ \ (\lambda \in P^{+}). \end{aligned}$$

Definition 5

• (a) ${{\mathbf {B}}} := \{ G(b) \mid b \in {\mathcal {B}}(\infty ) \}$ is called the global basis of $U_{{\mathbb {A}}}^{-}({\mathfrak {g}})$ corresponding to ${\mathcal {B}}(\infty )$.

• (b) ${{\mathbf {B}}}^{\lambda } := \{ G^{\lambda }(b) \mid b \in \mathcal B(\lambda ) \}$ is called the global basis of $V(\lambda )^{{\mathbb {A}}}$ corresponding to ${\mathcal {B}}(\lambda )$.

Remark 2

The global bases ${{\mathbf {B}}}$ and ${{\mathbf {B}}}^{\lambda }$ are unique because they are stable under the bar involution.

We conjecture that our global bases coincide with (a variation of) Bozec’s canonical bases. The following proposition would be a key ingredient of the proof.

Proposition 27

Let $i \in I^{\text {im}} \setminus I^{\text {iso}}$, $\alpha \in Q_{+}$, $l \ge 0$ and $\lambda \in P^{+}$. Define

$$\begin{aligned} \begin{aligned} B_{\alpha , i, \ge l}&:= \bigcup _{\vert {\mathbf {c}}\vert = l} \widetilde{f}_{i, {\mathbf {c}}} ({\mathcal {B}}(\infty )_{-\alpha }), \ \ \ B_{\alpha , i, l} := B_{\alpha , i, \ge l} \setminus B_{\alpha , i, \ge l+1},\\ B^{\lambda }_{\alpha , i, \ge l}&:= \bigcup _{\vert {\mathbf {c}}\vert = l} \widetilde{f}_{i, {\mathbf {c}}} ({\mathcal {B}}(\lambda )_{\lambda - \alpha }), \ \ \ \ B^{\lambda }_{\alpha , i, l} = B^{\lambda }_{\alpha , i, \ge l} \setminus B^{\lambda }_{\alpha , i, \ge l+1}. \end{aligned} \end{aligned}$$

(29)

• (a) For any $b \in B_{\alpha , i, l}$, there exist an element $b_{0} \in B_{\alpha - l \alpha _{i}, i, 0}$ and a composition ${\mathbf {c}}$ of l such that

  $$\begin{aligned} {{\texttt {b}}}_{i, {\mathbf {c}}} G(b_{0}) - G(b) \in \bigoplus _{b' \in B_{\alpha , i, \ge l+1}} {{\mathbb {A}}} G(b'). \end{aligned}$$

• (b) For any $b \in B^{\lambda }_{\alpha , i, l}$, there exist an element $b_{0} \in B^{\lambda }_{\alpha - l \alpha _{i}, i, 0}$ and a composition ${\mathbf {c}}$ of l such that

  $$\begin{aligned} {{\texttt {b}}}_{i, {\mathbf {c}}} G^{\lambda }(b_{0}) - G^{\lambda }(b) \in \bigoplus _{b' \in B^{\lambda }_{\alpha , i, \ge l+1}} {{\mathbb {A}}} G^{\lambda }(b'). \end{aligned}$$

Proof

We will prove (a) only. The proof of (b) is similar. Recall that

$$\begin{aligned} U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) = \bigoplus _{{\mathbf {c}}} {{\texttt {b}}}_{\i , {\mathbf {c}}} {{\mathcal {K}}}_{i}, \ \ \text {where} \ \ {{\mathcal {K}}}:= \bigcap _{l>0} \ker e_{il}'. \end{aligned}$$

Let $P_{i}: U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \longrightarrow {\mathcal K}_{i}$ be the projection. Then, for $u \in U_{{\mathbb {A}}}$, we have

$$\begin{aligned} \widetilde{f}_{il}(u) = {{\texttt {b}}}_{i,l} P_{i} (u) \ \ \ \text {mod} \left( \sum _{\vert {\mathbf {c}}'\vert \ge l+1} {{\texttt {b}}} _{i, {\mathbf {c}}'} \, U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \right) . \end{aligned}$$

(30)

By the crystal basis theory, for any $b \in B_{\alpha ,i, l}$, there exist an element $b_{0} \in B_{\alpha - l \alpha _{i}, i, 0}$ and a composition ${\mathbf {c}}$ of l such that $\widetilde{f}_{i, {\mathbf {c}}} b_{0} = b$. Thus we have

$$\begin{aligned} \begin{aligned} {{\texttt {b}}}_{i, {\mathbf {c}}} G(b_{0}) =&a G(b) + a_{1} G(b_{1}) + \cdots + a_{r}G(b_{r}) \\&\text {mod} \left( \sum _{\vert {\mathbf {c}}'\vert \ge l + 1} {{\texttt {b}}}_{i, {{\mathbf {c}}'}} U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \right) , \end{aligned} \end{aligned}$$

(31)

where $a, a_{1}, \ldots , a_{r} \in {{\mathbb {A}}}$, $b_{1}, \ldots , b_{r} \in B_{\alpha - l \alpha _{i}, i,0}$.

Since the left-hand side and right-hand side of (31) are invariant under the bar involution, we have

$$\begin{aligned} \overline{a} = a, \, {\overline{a}}_{1} = a_{1}, \ldots , {\overline{a}}_{r} = a_{r}. \end{aligned}$$

We know

$$\begin{aligned} \widetilde{f}_{i, {\mathbf {c}}} \, G(b_{0}) = G(b) \mod q \mathcal L(\infty ). \end{aligned}$$

(32)

On the other hand, as in (31),we have

$$\begin{aligned} \begin{aligned} \widetilde{f}_{i, {\mathbf {c}}} G(b_{0})&= {{\texttt {b}}}_{i, {\mathbf {c}}} (P_{i} G(b_{0}) \ \ \text {mod} \left( \sum _{\vert {\mathbf {c}}'\vert \ge l + 1} {{\texttt {b}}}_{i, {{\mathbf {c}}'}} U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \right) \\&= {{\texttt {b}}}_{i, {\mathbf {c}}} G(b_{0}) \ \ \text {mod} \left( \sum _{\vert {\mathbf {c}}'\vert \ge l + 1} {{\texttt {b}}}_{i, {{\mathbf {c}}'}} U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \right) \\&= (a G(b) + a_{1} G(b_{1}) + \cdots + a_{r}G(b_{r}) \ \ \text {mod} \left( \sum _{\vert {\mathbf {c}}'\vert \ge l + 1} {{\texttt {b}}}_{i, {{\mathbf {c}}'}} U_{{\mathbb {A}}}^{-}({\mathfrak {g}}) \right) . \end{aligned} \end{aligned}$$

(33)

Comparing with (32), we get

$$\begin{aligned} a = 1, a_1 = \cdots = a_{r} = 0 \ \ \mod q {\mathcal {L}}(\infty ). \end{aligned}$$

Since $a, a_{1}, \ldots , a_{r}$ are all bar-invariant, we conclude $a = 1$, $a_{1} = \cdots = a_{r} = 0$ in ${\mathbb {A}}$.

Thus we have finished our proof. $\square$