#  Braid group actions of quantum Borcherds-Bozec algebras

Zhaobing Fan  Harbin Engineering University, Harbin, China
<EMAIL_ADDRESS>and  Bolun Tong  Harbin Engineering University,
Harbin, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we construct the Lusztig symmetries for quantum Borcherds-Bozec
algebra $U_{q}(\mathfrak{g})$ and its weight module $M\in\mathcal{O}$, on
which the generators with real indices of $U_{q}(\mathfrak{g})$ act
nilpotently. We show that these symmetries satisfy the defining relations of
the braid group, associated to the Weyl group $W$ of $U_{q}(\mathfrak{g})$,
which gives a braid group action.

###### Key words and phrases:

quantum Borcherds-Bozec algebra, braid group action, PBW-basis, primitive
generator

###### 2010 Mathematics Subject Classification:

17B37, 17B67, 16G20

## Introduction

Lusztig considered in [L90] certain perverse sheaves on representation
varities of a quiver of type A,D,E, and gave a geometric approach to the half
parts of corresponding quantum groups. The canonical basis theory arose in
this setting, which is given by simple perverse sheaves. Moreover, the
algebraic and coalgebraic structures of the quantum groups are given by
induction functor and restriction functor, respectively. These results were
generalized by Lusztig in [L91] to Kac-Moody cases. Later on, Lusztig consider
more general cases, namely arbitrary quivers, possibly carrying loops in
[L93]. The obtained algebra is denoted by $U^{-}$. In [L93], Lusztig proposed
a question that if $U^{-}$ is generated by the elementary simple perverse
sheaves $F_{i}^{(n)}$ with all vertices $i$ and $n\in\mathbb{N}$ as an
algebra. The question is answered by himself in the case of the quiver with
one vertex and multiple loops, by a quadratic form criterion for a monomial to
be tight or semi-tight. Based on Kang and Schiffmann’s work for quantum
generalied Kac-Moody algebras (cf. [KS06]), Li and Lin [LL09] answered the
question when the quiver has at least two loops on each imaginary vertex. In
[B15], Bozec solved the Lusztig’s question completely. As a bialgebra, the
resulting $U^{-}$ is so called quantum Borcherds-Bozec algebra, which is the
main object we studied in the current paper.

On algebraic side, the quantum Borcherds-Bozec algebras can be treated as a
further generalization of quantum generalized Kac-Moody algebras [K95]. More
precisely, a quantum Borcherds-Bozec algebra $U_{q}(\mathfrak{g})$ has
infinitely many generators $e_{il},f_{il}$ $(l\in\mathbb{Z}_{>0})$ for each
imaginary index $i\in I^{\text{im}}$, and their degrees are $l$ multiples of
$\alpha_{i}$ and $-\alpha_{i}$, respectively. The commutation relations
between these generators are rather complicated and are higher order in some
sense (cf. [FKKT20]). Thanks to Bozec, there exists a set of primitive
generators $\mathtt{a}_{il},\mathtt{b}_{il}$ $((i,l)\in I^{\infty})$ with
better properties and simpler commutation relations. Using these generators,
Bozec constructed the Kashiwara operators. He then developed the crystal basis
theory for quantum Borcherds-Bozec algebras and their irreducible highest
weight modules [B16]. In [FKKT21], the authors and Kang and Kim constructed
the Global bases for the quantum Borcherds-Bozec algebras.

Lusztig defined in [L10] the symmetries
$T^{\prime}_{i,e},T^{\prime\prime}_{i,e}$ ($e=\pm 1$) for the quantum group
$U$ of Kac-Moody type and its integrable weight modules. He proved that these
automorphisms $T^{\prime}_{i,e},T^{\prime\prime}_{i,e}$ satisfy the braid
group relations both on $U$ and integrable $U$-modules. As an important
application, he gave some linearly independent subsets of $U$ associated to
reduced expressions of elements in the Weyl group $W$. In the case of finite
types, if one choose the longest element $\omega_{0}\in W$, the independent
set considered form the PBW-basis. He also shown how to extend as much as
possible this construction to arbitrary Cartan data, especially for affine
cases.

In this paper, we shall define the Lusztig symmetries for quantum Borcherds-
Bozec algebra $U_{q}(\mathfrak{g})$. Note that, the Weyl group of
$U_{q}(\mathfrak{g})$ is generated by the reflections associated to real
indexes, it is not enough to construct the PBW-type basis as Lusztig did. But
since the primitive generators $\mathtt{a}_{il}$ and $\mathtt{b}_{il}$ satisfy
the ‘higer order’ quantum Serre relations for $i\in I^{\text{im}}$, we could
define the Lusztig symmetries in a natural way. In the case where $I$ consists
of exact two real indexes $i\neq j$ and $a_{ij}a_{ji}$ is finite, Lusztig
proved that there is a braid group action on the integral modules (cf. [L10])
through the symmetries. In Theorem 2.6, we prove that it can be generalized to
an arbitrary Borcherds-Cartan datum by using our constructed symmetries, which
give the braid group actions on $U_{q}(\mathfrak{g})$ and
$U_{q}(\mathfrak{g})$-module $M$ in a certain category $\mathcal{O}$.

In Lusztig’s construction for PBW-basis, a crucial result is that the inner
product of $U^{+}$ is $T^{\prime\prime}_{i,1}$-invariant on the subalgebra
$U^{+}[i]$, which is generated by a set of elements of the forms similar to
the Serre-type relations but allows for smaller degrees. To verify this in
quantum Borcherds-Bozec algebra case (Theorem 3.5), we follow the framework
given in [J95, Chapter 8A] rather that Section 38.2 in [L10], since we have a
more general setting for the values of our bilinear form. We make a notice
here that, as a consequence of Theorem 3.5, one could get a lot of linearly
independent subsets of $U^{+}_{q}(\mathfrak{g})$ as in [L10, 38.2.2] by a
similar argument.

This paper is organized as follows. In Section 1, we review the definition of
quantum Borcherds-Bozec algebras and the notion of the primitive generators.
In Section 2, we define the Lusztig symmetries
$L^{\prime}_{i,e},L^{\prime\prime}_{i,e}$ ($e=\pm 1$) for these algebras and
their weight module $M\in\mathcal{O}$, and prove the braid group actions on
them. In Section 3, we investigate the relations between the symmetries and
the bilinear form $\\{\ ,\ \\}$ on $U^{+}_{q}(\mathfrak{g})\times
U^{-}_{q}(\mathfrak{g})$.

Acknowledgements.

Z. Fan was partially supported by the NSF of China grant 11671108, the NSF of
Heilongjiang Province grant JQ2020A001, and the Fundamental Research Funds for
the central universities. We would like to express our sincere gratitude to
Professor Seok-Jin Kang for his helpful discussions.

##  1\. Quantum Borcherds-Bozec algebras

Let $I$ be a finite or countably infinite index set. 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)

$a_{ii}=2,0,-2,-4,\dots$,

  * (ii)

$a_{ij}\in\mathbb{Z}_{\leq 0}$ for $i\neq j$,

  * (iii)

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

Let $I^{\text{re}}=\\{i\in I\mid a_{ii}=2\\}$ be the set of real indices. Let
$I^{\text{im}}=\\{i\in I\mid a_{ii}\leq 0\\}$ and $I^{\text{iso}}=\\{i\in
I\mid a_{ii}=0\\}$ be the set of imaginary indices and isotropic indices,
respectively.

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 $P^{\vee}=(\bigoplus_{i\in
I}\mathbb{Z}h_{i})\oplus(\bigoplus_{i\in I}\mathbb{Z}d_{i})$, the dual weight
lattice,

  * (c)

$\mathfrak{h}=\mathbb{Q}\otimes_{\mathbb{Z}}P^{\vee}$, the Cartan subalgebra,

  * (d)

$P=\\{\lambda\in\mathfrak{h}^{*}\mid\lambda(P^{\vee})\subseteq\mathbb{Z}\\}$,
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 $\mathbb{Q}$ and satisfies

$\alpha_{j}(h_{i})=a_{ij},\ \alpha_{j}(d_{i})=\delta_{ij}\ \ \text{for all}\
i,j\in I,$

  * (g)

for each $i\in I$, there is an element $\Lambda_{i}\in P$, called the
fundamental weight, defined by

$\Lambda_{i}(h_{j})=\delta_{ij},\ \Lambda_{i}(d_{j})=0\ \ \text{for all}\
i,j\in I.$

We denote by $P^{+}$ the set $\\{\lambda\in P\mid\lambda(h_{i})\geq 0\
\text{for all}\ i\in I\\}$ of dominant integral weights. The free abelian
group $Q=\bigoplus_{i\in I}\mathbb{Z}\alpha_{i}$ is called the root lattice.
Set $Q_{+}=\sum_{i\in I}\mathbb{Z}_{\geq 0}\alpha_{i}$ and $Q_{-}=-Q_{+}$. For
$\beta=\sum_{i\in I}k_{i}\alpha_{i}\in Q_{+}$, we define its hight to be
$|\beta|=\sum_{i\in I}k_{i}$.

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

$(\alpha_{i},\lambda)=s_{i}\lambda(h_{i}),\
(\Lambda_{i},\lambda)=s_{i}\lambda(d_{i})\ \ \text{for any}\
\lambda\in\mathfrak{h}^{*}\ \text{and}\ i\in I,$

and therefore we have

$(\alpha_{i},\alpha_{j})=s_{i}a_{ij}=s_{j}a_{ji}\ \ \text{for all}\ i,j\in I.$

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

$r_{i}(\lambda)=\lambda-\lambda(h_{i})\alpha_{i}\ \ \text{for}\
\lambda\in\mathfrak{h}^{*}.$

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

$W$ is a coxeter group generated by $r_{i}$ $(i\in I^{\text{re}})$ with
defining relations $r_{i}^{2}=1$ $(i\in I^{\text{re}})$ and
$(r_{i}r_{j})^{m_{ij}}=1$ $(i\neq j)$, where $m_{ij}$ is the order of
$r_{i}r_{j}$ and is related to $a_{ij}$ as follows:

$a_{ij}a_{ji}$ | $0$ | $1$ | $2$ | $3$ | $\geq 4$
---|---|---|---|---|---
$m_{ij}$ | $2$ | $3$ | $4$ | $6$ | $\infty$

We also use $r_{i}$ $(i\in I^{\text{re}})$ to denote the automorphism of
$P^{\vee}$ given by

$r_{i}(h)=h-\alpha_{i}(h)h_{i}\ \ \text{for all}\ h\in P^{\vee}.$

Let $\mathbf{i}=(i_{1},i_{2},\dots,i_{N})$ be a sequence in $I^{\text{re}}$.
Note that for any $\lambda\in\mathfrak{h}^{*}$ and $h\in P^{\vee}$, we have

$\lambda\big{(}r_{i_{1}}r_{i_{2}}\cdots{r_{i_{N}}}(h)\big{)}=\big{(}r_{i_{N}}r_{i_{N-1}}\cdots{r_{i_{1}}}(\lambda)\big{)}(h).$

Let
$I^{\infty}=(I^{\text{re}}\times\\{1\\})\cup(I^{\text{im}}\times\mathbb{Z}_{>0})$.
For simplicity, we will often write $i$ instead of $(i,1)$ when $i\in
I^{\text{re}}$. Let $q$ be an indeterminate, and set for each $i\in I$

$q_{i}=q^{s_{i}},\ q_{(i)}=q^{\frac{(\alpha_{i},\alpha_{i})}{2}}.$

For $i\in I^{\text{re}}$ and $n\in\mathbb{Z}_{\geq 0}$, we define

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

Let $\mathscr{E}=\mathbb{Q}(q)\left<e_{il}\mid(i,l)\in I^{\infty}\right>$ be
the free associative algebra over $\mathbb{Q}(q)$ generated by the symbols
$e_{il}$ for $(i,l)\in I^{\infty}$. By setting $\text{deg}e_{il}=l\alpha_{i}$,
$\mathscr{E}$ becomes a $Q_{+}$-graded algebra. For a homogeneous element $u$
in $\mathscr{E}$, we denote by $|u|$ the degree of $u$, and for any
$A\subseteq Q_{+}$, set ${\mathscr{E}}_{A}=\\{x\in{\mathscr{E}}\mid|x|\in
A\\}$.

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

$(x_{1}\otimes x_{2})(y_{1}\otimes
y_{2})=q^{(|x_{2}|,|y_{1}|)}x_{1}y_{1}\otimes x_{2}y_{2}$

for all homogeneous elements $x_{1},x_{2},y_{1},y_{2}\in\mathscr{E}$, and
equip $\mathscr{E}$ with a comultiplication $\rho$ defined by

$\rho(e_{il})=\sum_{m+n=l}q_{(i)}^{mn}e_{im}\otimes e_{in}\ \ \text{for}\
(i,l)\in I^{\infty}.$

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

######  Proposition 1.1.

[B15, B16] For a family $\nu=(\nu_{il})_{(i,l)\in I^{\infty}}$ of non-zero
elements in $\mathbb{Q}(q)$, there exists a symmetric bilinear form $\\{\ ,\
\\}:\mathscr{E}\times\mathscr{E}\rightarrow\mathbb{Q}(q)$ such that

  * (a)

$\\{x,y\\}=0$ if $|x|\neq|y|$,

  * (b)

$\\{1,1\\}=1$,

  * (c)

$\\{e_{il},e_{il}\\}=\nu_{il}$ for all $(i,l)\in I^{\infty}$,

  * (d)

$\\{x,yz\\}=\\{\rho(x),y\otimes z\\}$ for all $x,y,z\in{\mathscr{E}}$.

Here, $\\{x_{1}\otimes x_{2},y_{1}\otimes
y_{2}\\}=\\{x_{1},y_{1}\\}\\{x_{2},y_{2}\\}$ for any
$x_{1},x_{2},y_{1},y_{2}\in{\mathscr{E}}$.

Let $\mathcal{C}_{n}$ be the set of compositions $\mathbf{c}$ of $n$, and
$e_{i,\mathbf{c}}=e_{ic_{1}}\cdots e_{ic_{m}}$ for each $i\in I^{\text{im}}$,
$\mathbf{c}=(c_{1},\cdots,c_{m})\in\mathcal{C}_{n}$. It is clear that
$\\{e_{i,\mathbf{c}}\mid\mathbf{c}\in\mathcal{C}_{n}\\}$ form a basis of
$\mathscr{E}_{l\alpha_{i}}$.

Assume that $i\in I^{\text{re}}$, $j\in I$ and $i\neq j$. Let
$m\in\mathbb{Z}_{>0}$, $n\in\mathbb{Z}_{\geq 0}$ with $m>-a_{ij}n$, then for
any $\mathbf{c}\in\mathcal{C}_{n}$, the following element of $\mathscr{E}$
belongs to the radical $\mathcal{J}$ of the form $\\{\ ,\ \\}$

(1.1) $\mathcal{F}_{i,j,n,m,\mathbf{c},\pm 1}=\sum_{r+s=m}(-1)^{r}q_{i}^{\pm
r(-a_{ij}n-m+1)}e_{i}^{(r)}e_{j,\mathbf{c}}e_{i}^{(s)}$

Here, if $j\in I^{\text{re}}$, we set $e_{j,\mathbf{c}}=e_{j}^{(n)}$ as the
divided power of $e_{j}$.

Moreover, if $(i,k),(j,l)\in I^{\infty}$ such that $a_{ij}=0$, one can show
that the element $e_{ik}e_{jl}-e_{jl}e_{ik}$ belongs to $\mathcal{J}$.

From now on, we assume that

(1.2) $\nu_{il}\in 1+q^{-1}\mathbb{Z}_{\geq 0}[[q^{-1}]]\ \ \text{for all}\
(i,l)\in I^{\infty}.$

Under this assumption, the bilinear form $\\{\ ,\ \\}$ is non-degenerate on
$\mathscr{E}(i)=\bigoplus_{l\geq 1}{\mathscr{E}}_{l\alpha_{i}}$ for $i\in
I^{\rm{im}}\backslash I^{\rm{iso}}$. Moreover, its radical is generated by a
simpler set consisting of

$\sum_{r+s=1-la_{ij}}(-1)^{r}e_{i}^{(r)}e_{jl}e_{i}^{(s)}\ \ \text{for}\ i\in
I^{\text{re}},(j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l),$

and $e_{ik}e_{jl}-e_{jl}e_{ik}$ for all $(i,k),(j,l)\in I^{\infty}$ with
$a_{ij}=0$ (cf. [B15, Proposition 14]).

######  Definition 1.2.

Given a Borcherds-Cartan datum $(A,P,P^{\vee},\Pi,\Pi^{\vee})$, the quantum
Borcherds-Bozec algebra $U_{q}(\mathfrak{g})$ is the associative algebra over
$\mathbb{Q}(q)$ with $\mathbf{1}$ generated by the elements $q^{h}$ $(h\in
P^{\vee})$ and $e_{il},f_{il}$ $((i,l)\in I^{\infty})$, subjecting to

$\displaystyle q^{0}=\mathbf{1},\quad q^{h}q^{h^{\prime}}=q^{h+h^{\prime}}\ \
\text{for}\ h,h^{\prime}\in P^{\vee}$ $\displaystyle
q^{h}e_{jl}q^{-h}=q^{l\alpha_{j}(h)}e_{jl},\ \
q^{h}f_{jl}q^{-h}=q^{-l\alpha_{j}(h)}f_{jl}\ \ \text{for}\ h\in
P^{\vee},(j,l)\in I^{\infty},$
$\displaystyle\sum_{r+s=1-la_{ij}}(-1)^{r}{e_{i}}^{(r)}e_{jl}e_{i}^{(s)}=0\ \
\text{for}\ i\in I^{\text{re}},(j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l),$
$\displaystyle\sum_{r+s=1-la_{ij}}(-1)^{r}{f_{i}}^{(r)}f_{jl}f_{i}^{(s)}=0\ \
\text{for}\ i\in I^{\text{re}},(j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l),$
$\displaystyle e_{ik}e_{jl}-e_{jl}e_{ik}=f_{ik}f_{jl}-f_{jl}f_{ik}=0\ \
\text{for}\ a_{ij}=0,$ $\displaystyle e_{ik}f_{jl}=f_{jl}e_{ik}\ \ \text{for}\
(i,k),(j,l)\in I^{\infty}\ \text{and}\ i\neq j,$
$\displaystyle\sum_{\begin{subarray}{c}m+n=k\\\
n+s=l\end{subarray}}q_{(i)}^{n(m-s)}\nu_{in}K_{i}^{-n}f_{is}e_{im}=\sum_{\begin{subarray}{c}m+n=k\\\
n+s=l\end{subarray}}q_{(i)}^{n(s-m)}\nu_{in}K_{i}^{n}e_{im}f_{is}\ \
\text{for}\ (i,k),(i,l)\in I^{\infty}.$

Here, $K_{i}=q_{i}^{h_{i}}$ for all $i\in I$. We extend the grading by setting
$|q^{h}|=0$ and $|f_{il}|=-l\alpha_{i}$. For each $\beta=\sum
n_{i}\alpha_{i}\in Q$, we set $K_{\beta}=\prod_{i}K_{i}^{n_{i}}$.

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 $h\in P^{\vee}$. Then the
quantum Borcherds-Boec algebra $U_{q}(\mathfrak{g})$ has the triangular
decomposition

$U_{q}{(\mathfrak{g})}\cong U^{-}_{q}(\mathfrak{g})\otimes
U^{0}_{q}(\mathfrak{g})\otimes U^{+}_{q}(\mathfrak{g}).$

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.

The algebra $U$ is endowed with a comultiplication $\Delta:U\rightarrow
U\otimes U$ given by

(1.3) $\displaystyle\Delta(q^{h})=q^{h}\otimes q^{h},$
$\displaystyle\Delta(e_{il})=\sum_{m+n=l}q_{(i)}^{mn}e_{im}K_{i}^{n}\otimes
e_{in},$ $\displaystyle\Delta(f_{il})=\sum_{m+n=l}q_{(i)}^{-mn}f_{im}\otimes
K_{i}^{-m}f_{in}.$

We shall give the relations between $\rho:U^{+}\rightarrow U^{+}\otimes U^{+}$
and $\Delta$. Let $x\in U^{+}$ be a homogeneous element such that
$\rho(x)=\sum x_{1}\otimes x_{2}$, then we have

(1.4) $\Delta(x)=\sum x_{1}K_{|x_{2}|}\otimes x_{2}.$

If $y\in U^{-}$ is a homogeneous element with $\rho(\omega(y))=\sum
y_{1}\otimes y_{2}$, where $\omega$ is the involution of $U$ such that
$\omega(q^{h})=q^{-h},\ \omega(e_{il})=f_{il},\ \omega(f_{il})=e_{il}$ for
$h\in P^{\vee}$ and $(i,l)\in I^{\infty}$. We have

(1.5) $\Delta(y)=\sum q^{-(|y_{1}|,|y_{2}|)}\omega(y_{2})\otimes
K_{-|y_{2}|}\omega(y_{1}).$

######  Proposition 1.3.

[B15, B16] For any $i\in I^{\text{im}}$ and $l\geq 1$, there exist unique
elements $\mathtt{a}_{il}\in U^{+}_{l\alpha_{i}}$ and
$\mathtt{b}_{il}=\omega(\mathtt{a}_{il})$ such that

  * (1)

$\mathbb{Q}(q)\left<e_{il}\mid l\geq
1\right>=\mathbb{Q}(q)\left<\mathtt{a}_{il}\mid l\geq 1\right>$ and
$\mathbb{Q}(q)\left<f_{il}\mid l\geq
1\right>=\mathbb{Q}(q)\left<\mathtt{b}_{il}\mid l\geq 1\right>$,

  * (2)

$(\mathtt{a}_{il},z)_{L}=0$ for all
$z\in\mathbb{Q}(q)\left<e_{i1},\cdots,e_{il-1}\right>$,
$(\mathtt{b}_{il},z)_{L}=0$ for all
$z\in\mathbb{Q}(q)\left<f_{i1},\cdots,f_{il-1}\right>$,

  * (3)

$\mathtt{a}_{il}-e_{il}\in\mathbb{Q}(q)\left<e_{ik}\mid k<l\right>$ and
$\mathtt{b}_{il}-f_{il}\in\mathbb{Q}(q)\left<f_{ik}\mid k<l\right>$,

  * (4)

$\overline{\mathtt{a}}_{il}=\mathtt{a}_{il},\
\overline{\mathtt{b}}_{il}=\mathtt{b}_{il}$,

  * (5)

$\rho(\mathtt{a}_{il})=\mathtt{a}_{il}\otimes 1+1\otimes\mathtt{a}_{il},\
\rho(\mathtt{b}_{il})=\mathtt{b}_{il}\otimes 1+1\otimes\mathtt{b}_{il}$.

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

Set
$\tau_{il}=\\{\mathtt{a}_{il},\mathtt{a}_{il}\\}=\\{\mathtt{b}_{il},\mathtt{b}_{il}\\}$.
We have the following commutation relations in $U$ derived from the Drinfeld
double process,

(1.6)
$\mathtt{a}_{il}\mathtt{b}_{jk}-\mathtt{b}_{jk}\mathtt{a}_{il}=\delta_{ij}\delta_{lk}\tau_{il}(K_{i}^{-l}-K_{i}^{l}).$

The $\mathtt{a}_{il}$’s and $\mathtt{b}_{il}$’s are called the primitive
generators of $U_{q}(\mathfrak{g})$.

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

$\mathcal{D}_{i,l}=\begin{cases}\mathcal{C}_{l}&\text{if}\ i\in
I^{\text{im}}\backslash I^{\text{iso}},\\\ \mathcal{P}_{l}&\text{if}\ i\in
I^{\text{iso}}.\end{cases}$

and $\mathcal{D}_{i}=\bigsqcup_{l\geq 0}\mathcal{D}_{i,l}$. Let
$\mathbf{c}=(c_{1},\cdots,c_{t})\in\mathcal{D}_{i,l}$, we set

$\mathtt{a}_{i,\mathbf{c}}=\mathtt{a}_{ic_{1}}\cdots\mathtt{a}_{ic_{t}},\
\mathtt{b}_{i,\mathbf{c}}=\mathtt{b}_{ic_{1}}\cdots\mathtt{b}_{ic_{t}}\
\text{and}\ \tau_{i,\mathbf{c}}=\tau_{ic_{1}}\cdots\tau_{ic_{t}}.$

Note that $\\{\mathtt{a}_{i,\mathbf{c}}\mid\mathbf{c}\in\mathcal{D}_{i,l}\\}$
forms a basis of $U^{+}_{l\alpha_{i}}$. For each $i\in I^{\text{re}}$, we set
$\mathtt{a}_{i1}=e_{i1}$, $\mathtt{b}_{i1}=f_{i1}$, and write $\mathtt{a}_{i}$
(resp. $\mathtt{b}_{i}$) instead of $\mathtt{a}_{i1}$ (resp.
$\mathtt{b}_{i1}$) in this case for simplicity.

######  Example 1.4.

$\lambda\in\mathcal{P}_{l}$ can be written as the form
$\lambda=1^{\lambda_{1}}2^{\lambda_{2}}\cdots l^{\lambda_{l}}$, 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

$\mathtt{a}_{il}=e_{il}-\sum_{\lambda\in\mathcal{P}_{l}\backslash(l)}\frac{1}{\prod_{i=1}^{l}\lambda_{i}!}\mathtt{a}_{i,\lambda}.$

######  Example 1.5.

For $i\in I^{\text{iso}}$ and
$\mathbf{c},\mathbf{c}^{\prime}\in\mathcal{P}_{l}$, we have
$\\{\mathtt{a}_{i,\mathbf{c}},\mathtt{a}_{i,\mathbf{c}^{\prime}}\\}=0$ when
$\mathbf{c}\neq\mathbf{c}^{\prime}$. For $i\in I^{\text{im}}\backslash
I^{\text{iso}}$ and $\mathbf{c},\mathbf{c}^{\prime}\in\mathcal{C}_{l}$, if the
partitions obtained by rearranging $\mathbf{c}$ and $\mathbf{c}^{\prime}$ are
not equal, we also have
$\\{\mathtt{a}_{i,\mathbf{c}},\mathtt{a}_{i,\mathbf{c}^{\prime}}\\}=\\{\mathtt{b}_{i,\mathbf{c}},\mathtt{b}_{i,\mathbf{c}^{\prime}}\\}=0$.

######  Example 1.6.

Let $\mathbf{c}=(c_{1},c_{2},\cdots,c_{r})$ be a composition of $l$, we denote
by $\widetilde{\mathbf{c}}$ the reverse of $\mathbf{c}$, i.e.
$\widetilde{\mathbf{c}}=(c_{r},c_{r-1},\cdots,c_{1})$. Assume that
$\mathbf{c},\mathbf{c}^{\prime}$ are two compositions of $l$, which determine
the same partition, say it is $\mathbf{p}=(p_{1},p_{2},\cdots,p_{r})$. For
$i\in I^{\text{im}}\backslash I^{\text{iso}}$, if
$\\{\mathtt{a}_{i,\mathbf{c}},\mathtt{a}_{i,\mathbf{c}^{\prime}}\\}=\gamma\tau_{i,\mathbf{p}}$
for some $\gamma\in\mathbb{Q}(q)$, then we have

$\\{\mathtt{a}_{i,\mathbf{c}},\mathtt{a}_{i,\widetilde{\mathbf{c}}^{\prime}}\\}=q_{(i)}^{m}\overline{\gamma}\tau_{i,\mathbf{p}}.$

Here, $m=2\sum_{r<s}c_{r}c_{s}=2\sum_{r<s}c_{r}^{\prime}c_{s}^{\prime}$ and -
be the involution of $\mathbb{Q}(q)$ mapping $q$ to $q^{-1}$. In particular,
we see that
$\\{\mathtt{a}_{i,\widetilde{\mathbf{c}}},\mathtt{a}_{i,\widetilde{\mathbf{c}}^{\prime}}\\}=\\{\mathtt{a}_{i,\mathbf{c}},\mathtt{a}_{i,\mathbf{c}^{\prime}}\\}$.

######  Definition 1.7.

For every $(i,l)\in I^{\infty}$, we define the linear maps
$\delta_{i,l},\delta^{i,l}:U^{+}\rightarrow U^{+}$ by

$\delta_{i,l}(1)=0,\ \delta_{i,l}(\mathtt{a}_{jk})=\delta_{ij}\delta_{lk}\
\text{and}\
\delta_{i,l}(xy)=q^{l(|y|,\alpha_{i})}\delta_{i,l}(x)y+x\delta_{i,l}(y),$
$\delta^{i,l}(1)=0,\ \delta^{i,l}(\mathtt{a}_{jk})=\delta_{ij}\delta_{lk}\
\text{and}\
\delta^{i,l}(xy)=\delta^{i,l}(x)y+q^{l(|x|,\alpha_{i})}x\delta^{i,l}(y),$

for any homogeneous elements $x,y$ in $U^{+}$.

Let $(i,l)\in I^{\infty}$ and $z\in U^{+}$, one deduce from the Drinfeld
double process that (cf. [B16, Proposition 3.10])

(1.7)
$\left[\mathtt{a}_{il},\omega(z)\right]=\tau_{il}\\{\omega(\delta_{i,l}(z))K_{i}^{-l}-K_{i}^{l}\omega(\delta^{i,l}(z))\\}.$

Applying the involution $\omega$ to both sides of (1.7), we obtain the
following commutation relation

(1.8)
$\left[\mathtt{b}_{il},z\right]=\tau_{il}\\{\delta_{i,l}(z)K_{i}^{l}-K_{i}^{-l}\delta^{i,l}(z)\\}.$

##  2\. Lusztig symmetries and the braid group actions

Let $U_{q}(\mathfrak{g})$ be a quantum Borcherds-Bozec algebra, we denote by
$\mathcal{O}$ the category of $U_{q}(\mathfrak{g})$-module $M$ satisfying the
following two conditions

  * (i)

$M$ has a weight space decomposition

$M=\bigoplus_{\mu\in P}M_{\mu},\ \text{where}\ M_{\mu}=\\{z\in M\mid
q^{h}z=q^{\mu(h)}z\\},$

  * (ii)

$\mathtt{a}_{i}$ and $\mathtt{b}_{i}$ act locally nilpotent on $M$ for all
$i\in I^{\text{re}}$.

For $i\in I^{\text{re}}$, we set
$B_{i}=\mathtt{b}_{i}\big{/}\tau_{i}(q_{i}^{-1}-q_{i})$, then the commutation
relation between $\mathtt{a}_{i}$ and $\mathtt{B}_{i}$ is given by

$\mathtt{a}_{i}\mathtt{B}_{i}-\mathtt{B}_{i}\mathtt{a}_{i}={(K_{i}-K_{i}^{-1})}/{(q_{i}-q_{i}^{-1})}.$

Since every element $M$ in $\mathcal{O}$ is a weight module by (i), so it has
a direct sum decomposition as follow

$M=\bigoplus_{n\in\mathbb{Z}}M^{n},\ \text{where}\ M^{n}=\\{z\in M\mid
K_{i}z=q_{i}^{n}z\\}.$

We see that $\mathtt{a}_{i}(M^{n})\subseteq M^{n+2}$,
$\mathtt{B}_{i}(M^{n})\subseteq M^{n-2}$ for all $n$, and
$\mathtt{a}_{i}\mathtt{B}_{i}-\mathtt{B}_{i}\mathtt{a}_{i}$ acts on $M^{n}$ is
the multiplication by $[n]_{i}$ for all $n$, these properties yield $M$
becomes an object of the category $\mathcal{C}^{\prime}_{i}$ defined in [L10,
5.1.1]. Hence the standard argument in [L10, Chapter 5] can be applied to our
case. We define the $\mathbb{Q}(q)$-linear maps
$L^{\prime}_{i,e},L^{\prime\prime}_{i,e}:M\rightarrow M$ $(e=\pm 1)$ for $i\in
I^{\text{re}}$ and $M\in\mathcal{O}$ by

$L^{\prime}_{i,e}(z)=\sum_{a,b,c;a-b+c=n}(-1)^{b}q_{i}^{e(-ac+b)}\mathtt{B}_{i}^{(a)}\mathtt{a}_{i}^{(b)}\mathtt{B}_{i}^{(c)}z,$
$L^{\prime\prime}_{i,e}(z)=\sum_{a,b,c;-a+b-c=n}(-1)^{b}q_{i}^{e(-ac+b)}\mathtt{a}_{i}^{(a)}\mathtt{B}_{i}^{(b)}\mathtt{a}_{i}^{(c)}z,$

where $z\in M^{n}$ and $e=\pm 1$. $L^{\prime}_{i,e},L^{\prime\prime}_{i,e}$
are called symmetries on $M$. According to [L10, 5.2.3], we have
$L^{\prime}_{i,e}L^{\prime\prime}_{i,-e}=L^{\prime\prime}_{i,-e}L^{\prime}_{i,e}={\text{id}}:M\rightarrow
M$.

Given $i\in I^{\text{re}}$, $j\in I$ and $i\neq j$. Let
$m,n\in\mathbb{Z}_{\geq 0}$. Along the notations in (1.1), we set

(2.1)
$\mathcal{F}_{i,j,n,m,e}=\sum_{r+s=m}(-1)^{r}q_{i}^{er(-a_{ij}n-m+1)}\mathtt{a}_{i}^{(r)}\mathtt{a}_{jn}\mathtt{a}_{i}^{(s)}\in
U^{+}.$

Here $a_{jn}=1$ if $n=0$. Otherwise, $\mathtt{a}_{jn}$ are the generators of
$U^{+}$ corresponding to $(j,n)\in I^{\infty}$ when $j\in I^{\text{im}}$, and
$\mathtt{a}_{jn}=a_{j}^{(n)}$ when $j\in I^{\text{re}}$.

We shall denote $\mathcal{F}_{i,j,n,m,e}$ by $\mathcal{F}_{n,m,e}$ for
simplicity if there is no risk of confusion, and set $\beta=-a_{ij}$,
$\beta^{\prime}=-a_{ji}$.

We shall give several equations about $\mathcal{F}_{n,m,e}$ that can be proved
by similar inductive processes in [L10, Chapter 7].

######  Lemma 2.1.

  * (i)

$-q_{i}^{e(\beta
n-2m)}\mathtt{a}_{i}\mathcal{F}_{n,m,e}+\mathcal{F}_{n,m,e}\mathtt{a}_{i}={[m+1]}_{i}\mathcal{F}_{n,m+1,e}$,

  * (ii)

$\mathtt{a}_{i}^{(p)}\mathcal{F}_{n,m,e}=\sum_{p^{\prime}=0}^{p}(-1)^{p^{\prime}}q_{i}^{e(2pm-\beta
pn+pp^{\prime}-p^{\prime})}{\left[\begin{subarray}{c}m+p^{\prime}\\\
p^{\prime}\end{subarray}\right]}_{i}\mathcal{F}_{n,m+p^{\prime},e}\mathtt{a}_{i}^{(p-p^{\prime})}$,

  * (iii)

$-\mathtt{B}_{i}\mathcal{F}_{n,m,e}+\mathcal{F}_{n,m,e}\mathtt{B}_{i}={[\beta
n-m+1]}_{i}K_{-ei}\mathcal{F}_{n,m-1,e}$,

  * (iv)

$\mathtt{B}_{i}^{(p)}\mathcal{F}_{n,m,e}=\sum_{p^{\prime}=0}^{p}(-1)^{p^{\prime}}q_{i}^{-e(pp^{\prime}-p^{\prime})}{\left[\begin{subarray}{c}\beta
n-m+p^{\prime}\\\
p^{\prime}\end{subarray}\right]}_{i}K_{-ep^{\prime}i}\mathcal{F}_{n,m-p^{\prime},e}\mathtt{B}_{i}^{(p-p^{\prime})}$.

######  Lemma 2.2.

  * (i)

If $j\in I^{\text{re}}$ and $m=1+\beta n$, then

$\mathtt{B}_{j}\mathcal{F}_{n,m,e}-\mathcal{F}_{n,m,e}\mathtt{B}_{j}=K_{-j}\frac{q_{j}^{n-1}}{q_{j}-q_{j}^{-1}}\mathcal{F}_{n-1,m,1}-K_{j}\frac{q_{j}^{1-n}}{q_{j}-q_{j}^{-1}}\mathcal{F}_{n-1,m,-1}.$

  * (ii)

If $j\in I^{\text{im}}$, $(j,l)\in I^{\infty}$ and $m=1+\beta n$, then

$\mathtt{b}_{jl}\mathcal{F}_{n,m,e}-\mathcal{F}_{n,m,e}\mathtt{b}_{jl}=0.$

###### Proof.

In the case of $j\in I^{\text{re}}$, the proof is the same as [L10, Lemma
7.1.4]. We now prove (ii). Note that

$\displaystyle\mathtt{b}_{jl}\mathcal{F}_{n,m,e}-\mathcal{F}_{n,m,e}\mathtt{b}_{jl}=\sum_{r+s=m}(-1)^{r}q_{i}^{er(\beta
n-m+1)}\mathtt{a}_{i}^{(r)}\big{(}\mathtt{b}_{jl}\mathtt{a}_{jn}-\mathtt{a}_{jn}\mathtt{b}_{jl}\big{)}\mathtt{a}_{i}^{(s)}$
$\displaystyle\phantom{\mathtt{b}_{jl}\mathcal{X}_{n,m,e}}=\delta_{ln}\tau_{jn}\sum_{r+s=m}(-1)^{r}q_{i}^{er(\beta
n-m+1)}\mathtt{a}_{i}^{(r)}(K_{j}^{n}-K_{j}^{-n})\mathtt{a}_{i}^{(s)}$
$\displaystyle\phantom{\mathtt{b}_{jl}\mathcal{X}_{n,m,e}}=\delta_{ln}\tau_{jn}\sum_{r+s=m}(-1)^{r}q_{i}^{er(\beta
n-m+1)}(q_{i}^{nr\beta}K_{j}^{n}-q_{i}^{-nr\beta}K_{j}^{-n})\mathtt{a}_{i}^{(r)}\mathtt{a}_{i}^{(s)}.$

Since $m=1+\beta n$, we see that the right hand side of the above equality
equals to

$\delta_{ln}\tau_{jn}\sum_{r+s=m}(-1)^{r}(q_{i}^{r(m-1)}K_{j}^{n}-q_{i}^{r(1-m)}K_{j}^{-n})\frac{1}{[r]_{i}![m-r]_{i}!}\mathtt{a}_{i}^{m},$

and our assertion follows by the identity $\sum_{r=0}^{m}(-1)^{r}q_{i}^{\pm
r(1-m)}{\begin{bmatrix}m\\\ r\end{bmatrix}}_{i}=0$ for all $m>0$. ∎

The following lemma is an analogue of [L10, Lemma 3.5.4]. One just need to
note the fact that if $i\in I^{\text{re}}$, $(j,l)\in I^{\infty}$ and $i\neq
j$, then

$\mathtt{a}_{i}^{m}\mathtt{a}_{jl}\in\sum_{k=0}^{-la_{ij}}\mathbb{Q}(q)\mathtt{a}_{i}^{k}\mathtt{a}_{jl}\mathtt{a}_{i}^{m-k}$

for all $m\geq 1-la_{ij}$.

######  Lemma 2.3.

Assume that $I^{\text{re}}\neq\emptyset$. Let $u$ be an element of $U$ such
that $u$ annihilates all $M\in\mathcal{O}$. Then $u=0$.

Given an $i\in I^{\text{re}}$, we define the symmetries
$L^{\prime}_{i,e}:U\rightarrow U$ $(e=\pm 1)$ on the generators of $U$ as
follows

$\displaystyle L^{\prime}_{i,e}(\mathtt{a}_{i})=-K_{ei}\mathtt{B}_{i},\
L^{\prime}_{i,e}(\mathtt{B}_{i})=-\mathtt{a}_{i}K_{-ei};$ $\displaystyle
L^{\prime}_{i,e}(\mathtt{a}_{jl})=\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{er}\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(s)}\
\ \text{for}\ (j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l);$ $\displaystyle
L^{\prime}_{i,e}(\mathtt{b}_{jl})=\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{-er}\mathtt{B}_{i}^{(s)}\mathtt{b}_{jl}\mathtt{B}_{i}^{(r)}\
\ \text{for}\ (j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l);$ $\displaystyle
L^{\prime}_{i,e}(q^{h})=q^{r_{i}(h)}=q^{h-\alpha_{i}(h)h_{i}},$

and define the symmetries $L^{\prime\prime}_{i,-e}$ $(e=\pm 1)$ by

$\displaystyle
L^{\prime\prime}_{i,-e}(\mathtt{a}_{i})=-\mathtt{B}_{i}K_{-ei},\
L^{\prime\prime}_{i,-e}(\mathtt{B}_{i})=-K_{ei}\mathtt{a}_{i};$ $\displaystyle
L^{\prime\prime}_{i,-e}(\mathtt{a}_{jl})=\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{er}\mathtt{a}_{i}^{(s)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(r)}\
\ \text{for}\ (j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l);$ $\displaystyle
L^{\prime\prime}_{i,-e}(\mathtt{b}_{jl})=\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{-er}\mathtt{B}_{i}^{(r)}\mathtt{b}_{jl}\mathtt{B}_{i}^{(s)}\
\ \text{for}\ (j,l)\in I^{\infty}\ \text{and}\ i\neq(j,l);$ $\displaystyle
L^{\prime\prime}_{i,-e}(q^{h})=q^{h-\alpha_{i}(h)h_{i}}.$

Let $j\in I$ with $j\neq i$, and let $m,n\in\mathbb{Z}_{\geq 0}$. We set
$\mathcal{F}_{i,j,n,m,e}=\mathcal{F}_{n,m,e}$ as in (2.1) and

(2.2)
$\displaystyle\mathcal{F}^{\prime}_{i,j,n,m,e}=\mathcal{F}^{\prime}_{n,m,e}=\sum_{r+s=m}(-1)^{r}q_{i}^{er(\beta
n-m+1)}\mathtt{a}_{i}^{(s)}\mathtt{a}_{jn}\mathtt{a}_{i}^{(r)};$
$\displaystyle\mathcal{G}_{i,j,n,m,e}=\mathcal{G}_{n,m,e}=\sum_{r+s=m}(-1)^{r}q_{i}^{-er(\beta
n-m+1)}\mathtt{B}_{i}^{(s)}\mathtt{b}_{jn}\mathtt{B}_{i}^{(r)};$
$\displaystyle\mathcal{G}^{\prime}_{i,j,n,m,e}=\mathcal{G}^{\prime}_{n,m,e}=\sum_{r+s=m}(-1)^{r}q_{i}^{-er(\beta
n-m+1)}\mathtt{B}_{i}^{(r)}\mathtt{b}_{jn}\mathtt{B}_{i}^{(s)}.$

Here, $\mathtt{b}_{jn}$ has the same interpretation as $\mathtt{a}_{jn}$ and
$\beta=-a_{ij}$.

Using Lemma 2.1 and Lemma 2.3, we can verify the following statements step by
step according to [L10, Chapter 37], we leave it to readers. Precisely, thanks
to Lemma 2.1, one can treat the generator $\mathtt{a}_{jn}$ as the
$\mathtt{a}_{j}^{(n)}$ in form when we work with (i) in the following
proposition, which is a counterpart of Lemma 37.2.2 in loc.cit.. Then the
proof of (ii) is entirely similar to Lemma 37.2.3 in [L10] with the help of
(i) and Lemma 2.3.

######  Proposition 2.4.

  * (i)

Let $M\in\mathcal{O}$ and $z\in M$, then we have

$\displaystyle L^{\prime\prime}_{i,-e}(\mathcal{F}_{n,\beta
n,e}z)=\mathtt{a}_{jn}L^{\prime\prime}_{i,-e}(z);\ \
L^{\prime}_{i,e}(\mathcal{F}^{\prime}_{n,\beta
n,e}z)=\mathtt{a}_{jn}L^{\prime}_{i,e}(z);$ $\displaystyle
L^{\prime\prime}_{i,-e}(\mathcal{G}_{n,\beta
n,e}z)=\mathtt{b}_{jn}L^{\prime\prime}_{i,-e}(z);\ \
L^{\prime}_{i,e}(\mathcal{G}^{\prime}_{n,\beta
n,e}z)=\mathtt{b}_{jn}L^{\prime}_{i,e}(z).$

  * (ii)

$L^{\prime}_{i,e}$ and $L^{\prime\prime}_{i,-e}$ are automorphisms of $U$
sending the $\alpha$-root space onto the $r_{i}(\alpha)$-root space, they are
the inverse of each other. For any $M\in\mathcal{O}$, let $z\in M$ and $u\in
U$, we have

$L^{\prime}_{i,e}(uz)=L^{\prime}_{i,e}(u)L^{\prime}_{i,e}(z)\ \text{and}\
L^{\prime\prime}_{i,-e}(uz)=L^{\prime\prime}_{i,-e}(u)L^{\prime\prime}_{i,-e}(z).$

Furthermore, the operators $L^{\prime}_{i,e}$ and $L^{\prime\prime}_{i,-e}$ of
$U$ are uniquely determined by these properties.

  * (iii)

For any $m\in\mathbb{Z}$, we have

$L^{\prime}_{i,e}(\mathcal{F}^{\prime}_{n,m,e})=\mathcal{F}_{n,\beta n-m,e}\
\text{and}\
L^{\prime\prime}_{i,-e}(\mathcal{F}_{n,m,e})=\mathcal{F}^{\prime}_{n,\beta
n-m,e}.$

It is more convenient to twist the involution $\omega$. We shall denote by
$\varpi$ the automorphism of $U$ such that

$\varpi(\mathtt{a}_{i})=\mathtt{B}_{i},\
\varpi(\mathtt{B}_{i})=\mathtt{a}_{i},\ \varpi(q^{h})=q^{-h},\
\varpi(\mathtt{a}_{jl})=\mathtt{b}_{jl}\ \text{and}\
\varpi(\mathtt{b}_{jl})=\mathtt{a}_{jl}\ \text{for}\ (j,l)\neq i.$

Let $*$ be the anti-automorphism of $U$ given by

$*(q^{h})=q^{-h},\ *(\mathtt{a}_{jl})=\mathtt{a}_{jl}\ \text{and}\
*(\mathtt{b}_{jl})=\mathtt{b}_{jl}\ \ \text{for}\ h\in P^{\vee},\ (i,l)\in
I^{\infty}.$

Checking for the generators of $U$, we obtain $\varpi
L^{\prime}_{i,e}=L^{\prime\prime}_{i,e}\varpi:U\rightarrow U$ and
$*L^{\prime}_{i,e}=L^{\prime\prime}_{i,-e}*:U\rightarrow U$. Moreover, if
$u\in U$ such that $K_{i}uK_{i}^{-1}=q_{i}^{n}u$, we have

(2.3) $L^{\prime\prime}_{i,e}(u)=(-1)^{n}q_{i}^{en}L^{\prime}_{i,e}(u).$

######  Lemma 2.5.

[L10, Lemma 39.4.1] Assume that $I=I^{\text{re}}$ consists of exact two
elements $i\neq j$ such that $m_{ij}=2,3,4$ or $6$, and assume
$M\in\mathcal{O}$ in this case, we have

(2.4)
$L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}\cdots=L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots:M\rightarrow
M$

with both sides have $m_{ij}$ terms.

We shall generalize it to an arbitrary Borcherds-Cartan datum by the meothod
in [L10, Lemma 39.4.3].

######  Theorem 2.6.

Given a Borcherds-Cartan datum $(A,P,P^{\vee},\Pi,\Pi^{\vee})$ with
$I^{\text{re}}\neq\emptyset$, let $U$ be the associated quantum Borcherds-
Bozec algebra. For any $i\neq j$ such that $m_{ij}<\infty$, we have the
following equalities of automorphisms of $U$ and any $U$-module $M$ in
$\mathcal{O}$

(2.5)
$L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}\cdots=L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots,$
(2.6)
$L^{\prime}_{i,e}L^{\prime}_{j,e}L^{\prime}_{i,e}\cdots=L^{\prime}_{j,e}L^{\prime}_{i,e}L^{\prime}_{j,e}\cdots,$

where all the products have $m_{ij}$ factors.

###### Proof.

By the preceding argument, equality (2.5) for $M\in\mathcal{O}$ can be shown
by considering $M$ as a $U^{\prime}$-module, where $U^{\prime}$ is the
subalgebra of $U$ generated by
$\mathtt{a}_{i},\mathtt{b}_{i},\mathtt{a}_{j},\mathtt{b}_{j}$ and $U^{0}$.

Let $u\in U$, we set
$u_{1}=(L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}\cdots)(u)$
and
$u_{2}=(L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots)(u)$.For
any $M\in\mathcal{O}$ and $z\in M$, we have by using Proposition 2.4 (ii) that

$\displaystyle
u_{1}((L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots)z)=u_{1}((L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}\cdots)z)$
$\displaystyle=(L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}\cdots)(uz)=(L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots)(uz)$
$\displaystyle=u_{2}((L^{\prime\prime}_{j,-e}L^{\prime\prime}_{i,-e}L^{\prime\prime}_{j,-e}\cdots)z).$

It follows that $u_{1}-u_{2}$ annihilates all $M\in\mathcal{O}$. Hence we get
$u_{1}=u_{2}$ by Lemma 2.3, this proves (2.5) for $U$. Finally, the equality
for $L^{\prime}$ follows by taking inverse. ∎

The braid group associated to a Borcherds-Cartan datum is the group generated
by $r_{i}$ $(i\in I^{\text{re}})$ with defining relations

$r_{i}r_{j}r_{i}\cdots=r_{j}r_{i}r_{j}\cdots\quad\text{for}\ i\neq j\
\text{and}\ m_{ij}<\infty,$

where both sides have $m_{ij}$ factors. Theorem 2.6 yields the braid group
actions on $U$ and $U$-module $M\in\mathcal{O}$.

If $r_{i_{1}}r_{i_{2}}\cdots r_{i_{N}}$ and
$r_{i^{\prime}_{1}}r_{i^{\prime}_{2}}\cdots r_{i^{\prime}_{N}}$ are two
reduced expressions of $r\in W$, then we have the equality
$r_{i_{1}}r_{i_{2}}\cdots r_{i_{N}}=r_{i^{\prime}_{1}}r_{i^{\prime}_{2}}\cdots
r_{i^{\prime}_{N}}$ in the braid group. Hence the following definition is
valid

$L^{\prime\prime}_{r,e}=L^{\prime\prime}_{i_{1},e}L^{\prime\prime}_{i_{2},e}\cdots
L^{\prime\prime}_{i_{N},e},\
L^{\prime}_{r,e}=L^{\prime}_{i_{1},e}L^{\prime}_{i_{2},e}\cdots
L^{\prime}_{i_{N},e},$

where $r=r_{i_{1}}r_{i_{2}}\cdots r_{i_{N}}$ is a reduced expression of $r\in
W$. From the definition, we have

$L^{\prime\prime}_{rr^{\prime},e}=L^{\prime\prime}_{r,e}L^{\prime\prime}_{r^{\prime},e},\
L^{\prime}_{rr^{\prime},e}=L^{\prime}_{r,e}L^{\prime}_{r^{\prime},e}$

if $r,r^{\prime}\in W$ such that $l(rr^{\prime})=l(r)+l(r^{\prime})$. The
almost same calculate in [L10, Chapter 40] gives the following lemma.

######  Lemma 2.7.

Let $r\in W$ and let $i\in I^{\text{re}}$ with $l(rr_{i})=l(r)+1$, we have
$L^{\prime\prime}_{r,e}(\mathtt{a}_{i})\in U^{+}$ and
$L^{\prime}_{r,e}(\mathtt{a}_{i})\in U^{+}$ for $e=\pm 1$.

######  Corollary 2.8.

Let $r_{i_{1}}r_{i_{2}}\cdots r_{i_{N}}$ be a reduced expression for some
$r\in W$. Then

  * (i)

$L^{\prime\prime}_{i_{1},e}L^{\prime\prime}_{i_{2},e}\cdots
L^{\prime\prime}_{i_{N-1},e}(\mathtt{a}_{i_{N}})\in U^{+}$ and
$L^{\prime}_{i_{1},e}L^{\prime}_{i_{2},e}\cdots
L^{\prime}_{i_{N-1},e}(\mathtt{a}_{i_{N}})\in U^{+}.$

  * (ii)

For any $j\in I^{\text{im}}$, we have
$L^{\prime\prime}_{r,e}(\mathtt{a}_{jl})\in U^{+}$ and
$L^{\prime}_{r,e}(\mathtt{a}_{jl})\in U^{+}$.

###### Proof.

The first assertion follows from the previous lemma directly. We prove (ii) by
induction on $l(r)$. Let $r^{\prime}=r_{i_{1}}r_{i_{2}}\cdots r_{i_{N-1}}$ and
$i=i_{N}$, we have $r=r^{\prime}r_{i}$. Since $l(r^{\prime})=l(r)-1$, we have
$L^{\prime\prime}_{r,e}=L^{\prime\prime}_{r^{\prime},e}L^{\prime\prime}_{i,e}$
and $L^{\prime\prime}_{r^{\prime},e}(\mathtt{a}_{jl})\in U^{+}$ by the
induction hypothesis. Applying $L^{\prime\prime}_{r,e}$ to $\mathtt{a}_{jl}$,
we have

$\displaystyle
L^{\prime\prime}_{r,e}(\mathtt{a}_{jl})=L^{\prime\prime}_{r^{\prime},e}L^{\prime\prime}_{i,e}(\mathtt{a}_{jl})=L^{\prime\prime}_{r^{\prime},e}(\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{-er}\mathtt{a}_{i}^{(s)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(r)})$
$\displaystyle\phantom{L^{\prime\prime}_{r,e}(\mathtt{a}_{jl})}=\sum_{r+s=-la_{ij}}(-1)^{r}q_{i}^{-er}L^{\prime\prime}_{r^{\prime},e}(\mathtt{a}_{i}^{(s)})L^{\prime\prime}_{r^{\prime},e}(\mathtt{a}_{jl})L^{\prime\prime}_{r^{\prime},e}(\mathtt{a}_{i}^{(r)}).$

Hence by (i), we deduce that $L^{\prime\prime}_{r,e}(\mathtt{a}_{jl})\in
U^{+}$. Similarly, we have $L^{\prime}_{r,e}(\mathtt{a}_{jl})\in U^{+}$. ∎

##  3\. Link with the bilinear form on $U$

Fix $i\in I^{\text{re}}$. For any $(j,l)\in I^{\infty}$ with $i\neq(j,l)$ and
any $m\in\mathbb{Z}$, we set

  * (i)

$f(i,(j,l),m)=\sum_{k=0}^{m}(-1)^{k}q_{i}^{k(la_{ij}+m-1)}\mathtt{a}_{i}^{(k)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(m-k)};$

  * (ii)

$f^{\prime}(i,(j,l),m)=\sum_{k=0}^{m}(-1)^{k}q_{i}^{k(la_{ij}+m-1)}\mathtt{a}_{i}^{(m-k)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(k)};$

  * (iii)

$g(i,(j,l),m)=\sum_{k=0}^{m}(-1)^{k}q_{i}^{k(-la_{ij}-m+1)}\mathtt{B}_{i}^{(m-k)}\mathtt{b}_{jl}\mathtt{B}_{i}^{(k)};$

  * (iv)

$g^{\prime}(i,(j,l),m)=\sum_{k=0}^{m}(-1)^{k}q_{i}^{k(-la_{ij}-m+1)}\mathtt{B}_{i}^{(k)}\mathtt{b}_{jl}\mathtt{B}_{i}^{(m-k)}.$

Note that $f(i,(j,l),m)=\mathcal{F}_{i,j,l,m,-1}$,
$f^{\prime}(i,(j,l),m)=\mathcal{F}^{\prime}_{i,j,l,m,-1}$,
$g(i,(j,l),m)=\mathcal{G}_{i,j,l,m,-1}$ and
$g^{\prime}(i,(j,l),m)=\mathcal{G}^{\prime}_{i,j,l,m,-1}$ by using the
notations in (2.2). Sometimes, we will simply write
$f_{m},f^{\prime}_{m},g_{m},g^{\prime}_{m}$ for convenience.

Let $U^{+}[i]$ (resp. $U^{+}_{*}[i]$, $U^{-}[i]$ and $U^{-}_{*}[i]$) be the
subalgebra of $U$ generated by the elements $f(i,(j,l),m)$ (resp.
$f^{\prime}(i,(j,l),m)$, $g(i,(j,l),m)$ and $g^{\prime}(i,(j,l),m)$) for all
$(j,l)\neq i$ and $m\in\mathbb{Z}$. We have $*(U^{+}[i])=U^{+}_{*}[i]$ and
$*(U^{-}[i])=U^{-}_{*}[i]$.

According to Proposition 2.4(iii) and equality (2.3), for any
$m\in\mathbb{Z}$, we have

$L^{\prime\prime}_{i,1}(f_{m})=(-1)^{n}q_{i}^{n}L^{\prime}_{i,1}(f_{m})=f^{\prime}_{l\beta-m},$

where $n=2m-l\beta$, and we obtain by using $*$ that

$L^{\prime}_{i,-1}(f^{\prime}_{m})=(-1)^{n}q_{i}^{n}L^{\prime\prime}_{i,-1}(f^{\prime}_{m})=f_{l\beta-m}.$

Note that

$\varpi f_{m}=(-1)^{m}q_{i}^{m(la_{ij}+m-1)}g_{m},\ \varpi
f^{\prime}_{m}=(-1)^{m}q_{i}^{m(la_{ij}+m-1)}g^{\prime}_{m},$

so we obtain

$\displaystyle
L^{\prime\prime}_{i,1}(g_{m})=(-1)^{m}q_{i}^{m(l\beta-m+1)}L^{\prime\prime}_{i,1}(\varpi
f_{m})$
$\displaystyle\phantom{L^{\prime\prime}_{i,1}}=(-1)^{m}q_{i}^{m(l\beta-m+1)}\varpi
L^{\prime}_{i,1}(f_{m})$
$\displaystyle\phantom{L^{\prime\prime}_{i,1}}=(-1)^{m+n}q_{i}^{m(l\beta-m+1)-n}\varpi
f^{\prime}_{l\beta-m}=g^{\prime}_{l\beta-m},$

and $L^{\prime}_{i,-1}(g^{\prime}_{m})=g_{l\beta-m}$ by taking inverse. Hence
we have

$\varpi U^{+}[i]=U^{-}[i],\ \varpi U^{+}_{*}[i]=U^{-}_{*}[i],$

and

$L^{\prime\prime}_{i,1}U^{+}[i]=U^{+}_{*}[i];\
L^{\prime\prime}_{i,1}U^{-}[i]=U^{-}_{*}[i];$
$L^{\prime}_{i,-1}U^{+}_{*}[i]=U^{+}[i];\
L^{\prime}_{i,-1}U^{-}_{*}[i]=U^{-}[i].$

######  Lemma 3.1.

We have

  * (i)

$U^{+}=\bigoplus_{t\geq 0}\mathtt{a}_{i}^{t}U^{+}[i]=\bigoplus_{t\geq
0}U^{+}[i]\mathtt{a}_{i}^{t}$;

  * (ii)

$U^{+}=\bigoplus_{t\geq 0}\mathtt{a}_{i}^{t}U^{+}_{*}[i]=\bigoplus_{t\geq
0}U^{+}_{*}[i]\mathtt{a}_{i}^{t}$.

###### Proof.

Note that (ii) follows from (i) by applying $*$. To prove (i), it suffices to
show that $U^{+}[i]=\text{ker}\delta^{i}$. By Lemma 2.1(i), we see that
$\mathtt{a}_{i}$ and $f_{m}$ interchange by the following formula

$-q_{i}^{2m-l\beta}\mathtt{a}_{i}f(i,(j,l),m)+f(i,(j,l),m)\mathtt{a}_{i}={[m+1]}_{i}f(i,(j,l),m+1).$

Note that $U^{+}$ is generated by $\mathtt{a}_{i}$ and $U^{+}[i]$, we have the
following decomposition

$U^{+}=\sum_{t\geq 0}\mathtt{a}_{i}^{t}U^{+}[i].$

Since $L^{\prime\prime}_{i,1}U^{+}[i]\subseteq U^{+}$, one can show by the
exposition in [L10, Lemma 38.1.4] that $U^{+}[i]$ is contained in
$\text{ker}\delta^{i}$. Let $x\in\text{ker}\delta^{i}$, we write $x$ into the
form $x=\sum_{t\geq 0}\mathtt{a}_{i}^{t}x_{t}$, where $x_{t}\in
U^{+}[i]\subseteq\text{ker}\delta^{i}$. Note that $U^{+}=\bigoplus_{t\geq
0}\mathtt{a}_{i}^{t}\text{ker}\delta^{i}$, we have $x=x_{0}\in U^{+}[i]$. This
completes the proof. ∎

Using $\varpi$, we have the decompositions for $U^{-}$

$U^{-}=\bigoplus_{t\geq 0}\mathtt{B}_{i}^{t}U^{-}_{*}[i]=\bigoplus_{t\geq
0}U^{-}_{*}[i]\mathtt{B}_{i}^{t}=\bigoplus_{t\geq
0}\mathtt{B}_{i}^{t}U^{-}_{*}[i]=\bigoplus_{t\geq
0}U^{-}_{*}[i]\mathtt{B}_{i}^{t}.$

We now assume that $x\in U^{+}$ and $L^{\prime\prime}_{i,1}(x)\in U^{+}$, and
write $x=\sum_{\alpha\in Q^{+}}x_{\alpha}$. Since $L^{\prime\prime}_{i,1}$
sends the different root spaces into the different root spaces, we have
$L^{\prime\prime}_{i,1}(x_{\alpha})\in U^{+}$ for each $\alpha\in Q^{+}$. Note
that
$L^{\prime\prime}_{i,1}(x_{\alpha})=(-1)^{\alpha(h_{i})}q_{i}^{\alpha(h_{i})}L^{\prime}_{i,1}(x_{\alpha})$,
we deduce that $L^{\prime}_{i,1}(x_{\alpha})\in U^{+}$ and therefore
$L^{\prime}_{i,1}(x)\in U^{+}$. Thus, the following four subspaces of $U^{+}$
coincide

$U^{+}[i];\ \text{ker}\delta^{i};\ \\{x\in U^{+}\mid
L^{\prime\prime}_{i,1}(x)\in U^{+}\\};\ \text{and}\ \\{x\in U^{+}\mid
L^{\prime}_{i,1}(x)\in U^{+}\\}.$

By using $*$ , we have the following equal subspaces

$U^{+}_{*}[i];\ \text{ker}\delta_{i};\ \\{x\in U^{+}\mid
L^{\prime\prime}_{i,-1}(x)\in U^{+}\\};\ \text{and}\ \\{x\in U^{+}\mid
L^{\prime}_{i,-1}(x)\in U^{+}\\},$

and by using $\varpi$, we obtain

$U^{-}[i]=\\{x\in U^{-}\mid L^{\prime\prime}_{i,1}(x)\in U^{-}\\}=\\{x\in
U^{-}\mid L^{\prime}_{i,1}(x)\in U^{-}\\},$ $U_{*}^{-}[i]=\\{x\in U^{-}\mid
L^{\prime\prime}_{i,-1}(x)\in U^{-}\\}=\\{x\in U^{-}\mid
L^{\prime}_{i,-1}(x)\in U^{-}\\}.$

Let $f_{m}=f(i,(j,l),m)$ and $f^{\prime}_{m}=f^{\prime}(i,(j,l),m)$. By direct
calculation, we can get the following equations in $U^{+}$ easily as in [L10,
Lemma 38.1.7],

(3.1) $\displaystyle\rho(f_{m})=1\otimes
f_{m}+\sum_{t=0}^{m}\prod_{h=0}^{m-t-1}(1-q_{i}^{2m-2h-2l\beta-2})q_{i}^{t(m-t)}f_{t}\otimes\mathtt{a}_{i}^{(m-t)},$
$\displaystyle\rho(f^{\prime}_{m})=f^{\prime}_{m}\otimes
1+\sum_{t=0}^{m}\prod_{h=0}^{m-t-1}(1-q_{i}^{2m-2h-2l\beta-2})q_{i}^{t(m-t)}\mathtt{a}_{i}^{(m-t)}\otimes
f^{\prime}_{t}.$

Hence we have $\delta^{i}(f_{m})=0$,
$\delta_{i}(f_{m})=(1-q_{i}^{2m-2l\beta-2})q_{i}^{m-1}f_{m-1}$ and

(3.2)
$\delta^{j,l}(f_{m})=\prod_{h=0}^{m-1}(1-q_{i}^{2m-2h-2l\beta-2})\mathtt{a}_{i}^{(m)}=\prod_{h=1}^{m}(1-q_{i}^{-2(l\beta+1-h)})\mathtt{a}_{i}^{(m)}.$

Since $*\delta^{k,s}=\delta_{k,s}*$ for any $(k,s)\in I^{\infty}$, so we
obtain

$\delta_{i}(f^{\prime}_{m})=0,\
\delta^{i}(f^{\prime}_{m})=(1-q_{i}^{2m-2l\beta-2})q_{i}^{m-1}f^{\prime}_{m-1},$

and

$\delta_{j,l}(f^{\prime}_{m})=\prod_{h=1}^{m}(1-q_{i}^{-2(l\beta+1-h)})\mathtt{a}_{i}^{(m)}.$

######  Lemma 3.2.

Let $f_{m}=f(i,(j,l),m)$ and $g_{m}=g(i,(j,l),m)$, we have

$\\{f_{m},g_{m}\\}=\\{L^{\prime\prime}_{i,1}f_{m},L^{\prime\prime}_{i,1}g_{m}\\}.$

###### Proof.

Note that $\\{\mathtt{B}_{i}U^{-},f_{m}\\}=0$, so we have by (3.2) that

$\displaystyle\\{g_{m},f_{m}\\}=(-1)^{m}q_{i}^{m(l\beta-m+1)}\\{\mathtt{b}_{jl}\mathtt{B}_{i}^{(m)},f_{m}\\}$
$\displaystyle\phantom{\\{g_{m},f_{m}\\}}=(-1)^{m}q_{i}^{m(l\beta-m+1)}\tau_{jl}\prod_{h=1}^{m}(1-q_{i}^{-2(l\beta+1-h)})\\{\mathtt{B}_{i}^{(m)},\mathtt{a}_{i}^{(m)}\\}.$

Since
$\\{\mathtt{a}_{i}^{(m)},\mathtt{a}_{i}^{(m)}\\}=\tau_{i}^{m}q_{i}^{m(m-1)/2}[m]_{i}^{-1}$,
we have

$\\{\mathtt{B}_{i}^{(m)},\mathtt{a}_{i}^{(m)}\\}=q_{i}^{m(m-1)/2}(q_{i}^{-1}-q_{i})^{-m}{[m]_{i}}^{-1}.$

Hence

$\displaystyle\\{g_{m},f_{m}\\}=\tau_{jl}{\begin{bmatrix}l\beta\\\
m\end{bmatrix}}_{i}=\\{g_{l\beta-m},f_{l\beta-m}\\}=\\{*g_{l\beta-m},*f_{l\beta-m}\\}$
$\displaystyle\phantom{\\{g_{m},f_{m}\\}}=\\{g^{\prime}_{l\beta-m},f^{\prime}_{l\beta-m}\\}=\\{L^{\prime\prime}_{i,1}g_{m},L^{\prime\prime}_{i,1}f_{m}\\}.$

The lemma is proved. ∎

Let $n>0$ and $i\in I^{\text{re}}$, we define the linear maps
$\delta_{ni},\delta^{ni}:U^{+}\rightarrow U^{+}$ by

$\displaystyle\rho(x)=x\otimes
1+\sum_{n>0}\delta_{ni}(x)\otimes\mathtt{a}_{i}^{(n)}+\text{terms of bidegree
not in}\ Q_{+}\times\mathbb{N}\alpha_{i},$ $\displaystyle\rho(x)=1\otimes
x+\sum_{n>0}\mathtt{a}_{i}^{(n)}\otimes\delta^{ni}(x)+\text{terms of bidegree
not in}\ \mathbb{N}\alpha_{i}\times Q_{+},$

where $x$ is a homogeneous element of $U^{+}$. Let $|x|=\mu$ and let $(j,l)\in
I^{\infty}$ with $i\neq(j,l)$, we have

$\displaystyle\delta_{ni}(\mathtt{a}_{i}x)=\mathtt{a}_{i}\delta_{ni}(x)+[n]_{i}q^{(\alpha_{i},\mu-(n-1)\alpha_{i})}\delta_{(n-1)i}(x),$
$\displaystyle\delta^{ni}(x\mathtt{a}_{i})=\delta^{ni}(x)\mathtt{a}_{i}+[n]_{i}q^{(\alpha_{i},\mu-(n-1)\alpha_{i})}\delta^{(n-1)i}(x),$

and

$\delta_{ni}(\mathtt{a}_{jl}x)=\mathtt{a}_{jl}\delta_{ni}(x),\
\delta^{ni}(x\mathtt{a}_{jl})=\delta^{ni}(x)\mathtt{a}_{jl}.$

Note that, we have by induction on $n$ that

$\displaystyle(\delta_{i})^{n}(\mathtt{a}_{i}x)=\mathtt{a}_{i}(\delta_{i})^{n}(x)+[n]_{i}q^{(\alpha_{i},\mu)}q_{i}^{-(n-1)}(\delta_{i})^{n-1}(x),$
$\displaystyle(\delta^{i})^{n}(x\mathtt{a}_{i})=(\delta^{i})^{n}(x)\mathtt{a}_{i}+[n]_{i}q^{(\alpha_{i},\mu)}q_{i}^{-(n-1)}(\delta^{i})^{n-1}(x),$

and

$(\delta_{i})^{n}(\mathtt{a}_{jl}x)=\mathtt{a}_{jl}(\delta_{i})^{n}(x),\
(\delta^{i})^{n}(x\mathtt{a}_{jl})=(\delta^{i})^{n}(x)\mathtt{a}_{jl}.$

It follows that

(3.3) $(\delta_{i})^{n}=q_{i}^{n(n-1)/2}\delta_{ni}\ \text{and}\
(\delta^{i})^{n}=q_{i}^{n(n-1)/2}\delta^{ni}.$

Fix $i\in I^{\text{re}}$, let $(j,l)\in I^{\infty}$ with $(j,l)\neq i$. Note
that, the following set of elements are linearly independent in $U^{+}$

$\\{\mathtt{a}_{i}^{(n)};\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(s)}\mid
n,r,s\geq 0\ \text{and}\ r+s\leq-la_{ij}\\}.$

It can be extended to be a basis of $U^{+}$, by adding a set
$\\{u_{w}\\}_{w\in S}$ of monomials in $U^{+}$.

Hence for any $x\in U^{+}$, we can write $\rho(x)$ into

(3.4) $\rho(x)=1\otimes
x+\sum_{n>0}\mathtt{a}_{i}^{(n)}\otimes\delta^{ni}(x)+\sum_{r=0}^{l\beta}\sum_{s=0}^{l\beta-r}\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(s)}\otimes\delta^{ri;(j,l);si}(x)+\sum_{w\in
S}u_{w}\otimes x_{w}.$

Here $\beta=-a_{ij}$ as usual.

######  Lemma 3.3.

Let $x\in U^{+}$ and $y\in U^{-}$. For any $m\in\mathbb{Z}$, we have

  * (i)

$\\{g_{m}y,x\\}=\\{g_{m},f_{m}\\}\\{y,\delta^{(j,l);mi}(x)\\}.$

  * (ii)

$\\{g^{\prime}_{m}y,x\\}=\\{g^{\prime}_{m},f^{\prime}_{m}\\}\\{y,\delta^{mi;(j,l)}(x)\\}.$

###### Proof.

We shall prove (i) first. Since $f_{m}=g_{m}=0$ when $m>l\beta$ or $m<0$, we
may assume $0\leq m\leq l\beta$. Let $u$ be a monomial in
$U^{+}_{m\alpha_{i}+l\alpha_{j}}$. If $u$ contains no $\mathtt{a}_{jl}$, then
we can write $u=\mathtt{a}_{i}^{t}\mathtt{a}_{jk}u^{\prime}$ for some $t\geq
0$, $k<l$ and a monomial $u^{\prime}$ of
$U^{+}_{(m-t)\alpha_{i}+(l-k)\alpha_{j}}$. Note that

$\rho(\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(m-r)})=(\sum_{a+b=r}q_{i}^{ab}\mathtt{a}_{i}^{(a)}\otimes\mathtt{a}_{i}^{(b)})(\mathtt{a}_{jl}\otimes
1+1\otimes\mathtt{a}_{jl})(\sum_{a^{\prime}+b^{\prime}=m-r}q_{i}^{a^{\prime}b^{\prime}}\mathtt{a}_{i}^{(a^{\prime})}\otimes\mathtt{a}_{i}^{(b^{\prime})}),$

and the right hand side is in

$\sum_{\begin{subarray}{c}a+b=r\\\
a^{\prime}+b^{\prime}=m-r\end{subarray}}\mathbb{Q}(q)\mathtt{a}_{i}^{(a)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(a^{\prime})}\otimes\mathtt{a}_{i}^{b+b^{\prime}}+\sum_{\begin{subarray}{c}a+b=r\\\
a^{\prime}+b^{\prime}=m-r\end{subarray}}\mathbb{Q}(q)\mathtt{a}_{i}^{a+a^{\prime}}\otimes\mathtt{a}_{i}^{(b)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(b^{\prime})}.$

Hence we obtain

$\\{\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(m-r)},u\\}=\\{\rho(\mathtt{a}_{i}^{(r)}\mathtt{a}_{jl}\mathtt{a}_{i}^{(m-r)}),\mathtt{a}_{i}^{t}\mathtt{a}_{jk}\otimes
u^{\prime}\\}=0,$

and therefore $\\{g_{m},u\\}=0$ for such $u$.

Since $\\{g_{m},\mathtt{a}_{i}U^{+}\\}=0$, we get
$\\{g_{m},f_{m}\\}=\\{g_{m},\mathtt{a}_{jl}\mathtt{a}_{i}^{(m)}\\}$. On the
other hand, by the above argument, we have $\\{g_{m},u_{w}\\}=0$ for any $w\in
S$, which implies

$\\{g_{m}y,x\\}=\\{g_{m}\otimes
y,\mathtt{a}_{jl}\mathtt{a}_{i}^{(m)}\otimes\delta^{(j,l);mi}(x)\\}=\\{g_{m},f_{m}\\}\\{y,\delta^{(j,l);mi}(x)\\}.$

The proof of (ii) is similar. ∎

Let $x\in U^{+}_{\mu}$ and $x^{\prime}\in U^{+}_{\mu^{\prime}}$. Let $0\leq
n\leq l\beta$. The expression (3.4) yields

(3.5)
$\displaystyle\delta^{(j,l);ni}(xx^{\prime})=\delta^{(j,l);ni}(x)x^{\prime}+q^{(\mu,l\alpha_{j}+n\alpha_{i})}x\delta^{(j,l);ni}(x^{\prime})$
$\displaystyle\phantom{\delta}+\sum_{t=0}^{n-1}q^{(\mu-l\alpha_{j}-t\alpha_{i},(n-t)\alpha_{i})}{\begin{bmatrix}n\\\
t\end{bmatrix}}_{i}\delta^{(j,l);ti}(x)\delta^{(n-t)i}(x^{\prime}),$

and

(3.6)
$\displaystyle\delta^{ni;(j,l)}(xx^{\prime})=\delta^{ni;(j,l)}(x)x^{\prime}+q^{(\mu,n\alpha_{i}+l\alpha_{j})}x\delta^{ni;(j,l)}(x^{\prime})$
$\displaystyle\phantom{\delta}+\sum_{t=0}^{n-1}q^{(\mu-(n-t)\alpha_{i},t\alpha_{i}+l\alpha_{j})}{\begin{bmatrix}n\\\
t\end{bmatrix}}_{i}\delta^{(n-t)i}(x)\delta^{ti;(j,l)}(x^{\prime}).$

In particular, if $x^{\prime}\in U^{+}[i]$, we have
$\delta^{ni}(x^{\prime})=0$ for all $n>0$, which implies

(3.7)
$\delta^{(j,l);ni}(xx^{\prime})=\delta^{(j,l);ni}(x)x^{\prime}+q^{(\mu,l\alpha_{j}+n\alpha_{i})}x\delta^{(j,l);ni}(x^{\prime}).$

Let $f_{m}=f(i,(j,l),m)$ and $f^{\prime}_{m}=f^{\prime}(i,(j,l),m)$. By (3.1),
we have

$\delta^{(j,l);ni}(f_{m})=\begin{cases}\gamma_{mn}\mathtt{a}_{i}^{(m-n)}&\text{if}\
n\leq m,\\\ 0&\text{if}\ n>m,\end{cases}\ \text{and}\ \
\delta^{ni;(j,l)}(f^{\prime}_{m})=\begin{cases}1\ \ \text{if}\ n=m,\\\ 0\ \
\text{otherwise},\end{cases}$

where

$\gamma_{mn}=\prod_{h=0}^{m-n-1}(1-q_{i}^{2m-2h-2l\beta-2})q_{i}^{n(m-n)}.$

Denote by $P_{i}:U^{+}\rightarrow U^{+}[i]$ the projection obeys the
decomposition $U^{+}=\mathtt{a}_{i}U^{+}\oplus U^{+}[i]$ given in Lemma 3.1.
We have by the definition

$P_{i}(xx^{\prime})=P_{i}(P_{i}(x)x^{\prime})\ \text{for any}\ x,x^{\prime}\in
U^{+}.$

In particular, we have a more simpler formula
$P_{i}(xx^{\prime})=P_{i}(x)x^{\prime}$ when $x^{\prime}\in U^{+}[i]$.

######  Lemma 3.4.

Let $x\in U^{+}[i]$ with $|x|=\mu$ and let $n>0$, we have

$P_{i}(x\mathtt{a}_{i}^{n})=\frac{q^{n(\mu,\alpha_{i})}q_{i}^{n(n+1)}}{(q_{i}-q_{i}^{-1})^{n}}L^{\prime}_{i,-1}(\delta^{i})^{n}L^{\prime\prime}_{i,1}(x).$

###### Proof.

We use induction on $n$. Assume that $n=1$. Since
$L^{\prime\prime}_{i,1}(x)\in U^{+}_{*}[i]$, we have
$\delta_{i}(L^{\prime\prime}_{i,1}(x))=0$ and so

$L^{\prime\prime}_{i,1}(x)\mathtt{B}_{i}-\mathtt{B}_{i}L^{\prime\prime}_{i.1}(x)=-K_{i}^{-1}\delta^{i}(L^{\prime\prime}_{i,1}(x))/(q_{i}-q_{i}^{-1}).$

Applying $L^{\prime}_{i,-1}$ to both sides, we get

$-x\mathtt{a}_{i}K_{i}+\mathtt{a}_{i}K_{i}x=-K_{i}L^{\prime}_{i,-1}\delta^{i}L^{\prime\prime}_{i,1}(x)/(q_{i}-q_{i}^{-1}),$

and by applying $K^{-1}$, we obtain

$-q^{-(\mu+\alpha_{i},\alpha_{i})}x\mathtt{a}_{i}+q^{-(\alpha_{i},\alpha_{i})}\mathtt{a}_{i}x=-L^{\prime}_{i,-1}\delta^{i}L^{\prime\prime}_{i,1}(x)/(q_{i}-q_{i}^{-1}).$

It follows that

$x\mathtt{a}_{i}=q^{(\mu,\alpha_{i})}\mathtt{a}_{i}x+\frac{q^{(\mu+\alpha_{i},\alpha_{i})}}{q_{i}-q_{i}^{-1}}L^{\prime}_{i,-1}\delta^{i}L^{\prime\prime}_{i,1}(x).$

Note that, according to the second formula in (3.1), we have
$\rho(U^{+}_{*}[i])\subseteq U^{+}\otimes U^{+}_{*}[i]$, which implies
$\delta^{i}L^{\prime\prime}_{i,1}(x)\in U^{+}_{*}[i]$. Hence

$P_{i}(x\mathtt{a}_{i})=\frac{q^{(\mu+\alpha_{i},\alpha_{i})}}{q_{i}-q_{i}^{-1}}L^{\prime}_{i,-1}\delta^{i}L^{\prime\prime}_{i,1}(x).$

This proves our assertion for $n=1$. For the induction step, assume the lemma
is true for $n$, then we have

$\displaystyle
P_{i}(x\mathtt{a}_{i}^{n+1})=P_{i}(P_{i}(x\mathtt{a}_{i}^{n})\mathtt{a}_{i})=\frac{q^{(\mu+(n+1)\alpha_{i},\alpha_{i})}}{q_{i}-q_{i}^{-1}}L^{\prime}_{i,-1}\delta^{i}L^{\prime\prime}_{i,1}(P_{i}(x\mathtt{a}_{i}^{n}))$
$\displaystyle\phantom{P_{i}(x\mathtt{a}_{i}^{n+1})}=\frac{q^{(n+1)(\mu,\alpha_{i})}q_{i}^{(n+1)(n+2)}}{(q_{i}-q_{i}^{-1})^{n+1}}L^{\prime}_{i,-1}(\delta^{i})^{n+1}L^{\prime\prime}_{i,1}(x)$

as desired. ∎

Now, we shall prove our main theorem by using the method given in [J95,
Chapter 8A]. Note that there is no special restriction on the values of
$\tau_{i}$’s for $i\in I^{\text{re}}$, but only ask them to take values in
$1+q^{-1}\mathbb{Z}_{\geq 0}[[q^{-1}]]$, hence the argument in [L10, Lemma
38.2.1] is not available for our case.

######  Theorem 3.5.

For any $x\in U^{+}[i],y\in U^{-}[i]$, we have

$\\{L^{\prime\prime}_{i,1}(x),L^{\prime\prime}_{i,1}(y)\\}=\\{x,y\\}.$

###### Proof.

Assume that our assertion holds for a given $y$ in $U^{-}[i]$ and arbitrary
$x$ in $U^{+}[i]$. By Lemma 3.3(ii), we have

$\displaystyle\\{L^{\prime\prime}_{i,1}(g_{m}y),L^{\prime\prime}_{i,1}(x)\\}=\\{g^{\prime}_{l\beta-m}L^{\prime\prime}_{i,1}(y),L^{\prime\prime}_{i,1}(x)\\}$
$\displaystyle=\\{g^{\prime}_{l\beta-m},f^{\prime}_{l\beta-m}\\}\\{L^{\prime\prime}_{i,1}(y),\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)\\}$
$\displaystyle=\\{g^{\prime}_{l\beta-m},f^{\prime}_{l\beta-m}\\}\\{y,L^{\prime}_{i,-1}\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)\\},$

where the last equality follows from the fact that
$\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)\in U^{+}_{*}[i]$. On the
other hand, by Lemma 3.2 and Lemma 3.3(i), we have

$\\{g_{m}y,x\\}=\\{g_{m},f_{m}\\}\\{y,\delta^{(j,l);mi}(x)\\}=\\{g^{\prime}_{l\beta-m},f^{\prime}_{l\beta-m}\\}\\{y,\delta^{(j,l);mi}(x)\\}.$

Since $\\{y,\mathtt{a}_{i}U^{+}\\}=0$. To show
$\\{L^{\prime\prime}_{i,1}(g_{m}y),L^{\prime\prime}_{i,1}(x)\\}=\\{g_{m}y,x\\}$,
it is enough to show

$L^{\prime}_{i,-1}\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)\equiv\delta^{(j,l);mi}(x)\
\text{mod}\ \mathtt{a}_{i}U^{+}$

for all $x\in U^{+}[i]$. This is equivalent to

(3.8)
$\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)=L^{\prime\prime}_{i,1}P_{i}\delta^{(j,l);mi}(x).$

for all $x\in U^{+}[i]$.

The identity (3.8) holds for $x=1$, since both sides equal $0$ in this case.
Assume (3.8) holds for $x\in U^{+}[i]$ with $|x|=\mu$, and let
$x^{\prime}=f(i,(j^{\prime},l^{\prime}),s)$. If
$(j^{\prime},l^{\prime})\neq(j,l)$, then the first equality in (3.1) yields
$\delta^{(j,l);mi}(x^{\prime})=0$. By (3.7), we have
$\delta^{(j,l);mi}(xx^{\prime})=\delta^{(j,l);mi}(x)x^{\prime}$, which implies

$P_{i}\delta^{(j,l);mi}(xx^{\prime})=P_{i}(\delta^{(j,l);mi}(x))x^{\prime},$

and so

$L^{\prime\prime}_{i,1}P_{i}\delta^{(j,l);mi}(xx^{\prime})=L^{\prime\prime}_{i,1}P_{i}\delta^{(j,l);mi}(x)L^{\prime\prime}_{i,1}(x^{\prime}).$

On the other hand, by (3.6), we have

$\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(xx^{\prime})=\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)L^{\prime\prime}_{i,1}(x^{\prime}),$

hence (3.8) holds for $xx^{\prime}$ in this case.

If $(j^{\prime},l^{\prime})=(j,l)$, then $x^{\prime}=f_{s}=f(i,(j,l),s)$. By
using (3.7), we obtain

$\delta^{(j,l);mi}(xf_{s})=\delta^{(j,l);mi}(x)f_{s}+X,$

where $X=0$ when $m>s$, and
$X=q^{(\mu,l\alpha_{j}+m\alpha_{i})}\gamma_{sm}x\mathtt{a}_{i}^{(s-m)}$ when
$m\leq s$.

By (3.6), we have

$\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(xf_{s})=\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)f^{\prime}_{l\beta-s}+Y,$

where $Y=0$ when $m>s$, and when $m\leq s$,

$Y=q^{(r_{i}\mu-(s-m)\alpha_{i},(l\beta-s)\alpha_{i}+l\alpha_{j})}{\left[\begin{subarray}{c}l\beta-m\\\
l\beta-s\end{subarray}\right]}_{i}\delta^{(s-m)i}L^{\prime\prime}_{i,1}(x).$

Here we understand
$\delta^{0i}(L^{\prime\prime}_{i,1}(x))=L^{\prime\prime}_{i,1}(x)$. According
to our assumption, we have

$L^{\prime\prime}_{i,1}\left(P_{i}\delta^{(j,l);mi}(x)f_{s}\right)=\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(x)f^{\prime}_{l\beta-s}.$

Hence in order to show

$\delta^{(l\beta-m)i;(j,l)}L^{\prime\prime}_{i,1}(xf_{s})=L^{\prime\prime}_{i,1}P_{i}\delta^{(j,l);mi}(xf_{s}),$

it suffices to prove that $L^{\prime\prime}_{i,1}P_{i}(X)=Y$ in the case of
$m\leq s$. In this case, using the equality given Lemma 3.4 and in (3.3), we
get

$\displaystyle
L^{\prime\prime}_{i,1}P_{i}(X)=q^{(\mu,l\alpha_{j}+m\alpha_{i})}\gamma_{sm}L^{\prime\prime}_{i,1}P_{i}(x\mathtt{a}_{i}^{(s-m)})$
$\displaystyle\ \
=\frac{1}{[s-m]_{i}!}\frac{q^{(s-m)(\mu,\alpha_{i})}q^{(\mu,l\alpha_{j}+m\alpha_{i})}}{(q_{i}-q_{i}^{-1})^{(s-m)}}q_{i}^{(s-m)(3(s-m)+1)/2}\gamma_{sm}\delta^{(s-m)i}L^{\prime\prime}_{i,1}(x).$

Note that, we have

$\gamma_{sm}=q_{i}^{m(s-m)+(s-l\beta)(s-m)-(s-m)(s-m+1)/2}\prod_{h=1}^{s-m}(q_{i}^{l\beta-s+h}-q_{i}^{-l\beta+s-h}).$

Thus,

$L^{\prime\prime}_{i,1}P_{i}(X)=q^{(\mu,s\alpha_{i}+l\alpha_{j})}q_{i}^{(2s-l\beta)(s-m)}{\left[\begin{subarray}{c}l\beta-m\\\
l\beta-s\end{subarray}\right]}_{i}\delta^{(s-m)i}L^{\prime\prime}_{i,1}(x)=Y.$

The theorem is proved. ∎

Let $P^{\prime}_{i}:U^{+}\rightarrow U^{+}_{*}[i]$ be the projection alongs
the decomposition $U^{+}=U^{+}_{*}[i]\oplus U^{+}\mathtt{a}_{i}$. We have the
following result as a direct corollary of the previous theorem.

######  Corollary 3.6.

Let $x\in U^{+}[i]$, we have

(3.9) $(P^{\prime}_{i}\otimes\text{id})\circ\rho\circ
L^{\prime\prime}_{i,1}(x)=(L^{\prime\prime}_{i,1}\otimes
L^{\prime\prime}_{i,1})\circ(\text{id}\otimes P_{i})\circ\rho(x).$

###### Proof.

Note that both sides in the equation (3.9) belong to $U^{+}_{*}[i]\otimes
U^{+}_{*}[i]$. Let $y,y^{\prime}\in U^{-}[i]$, we have by the definition

$\displaystyle\\{yy^{\prime},x\\}=\\{y\otimes
y^{\prime},\rho(x)\\}=\\{y\otimes y^{\prime},(\text{id}\otimes
P_{i})\circ\rho(x)\\}$
$\displaystyle\phantom{\\{yy^{\prime},x\\}}=\\{L^{\prime\prime}_{i,1}(y)\otimes
L^{\prime\prime}_{i,1}(y^{\prime}),(L^{\prime\prime}_{i,1}\otimes
L^{\prime\prime}_{i,1})\circ(\text{id}\otimes P_{i})\circ\rho(x)\\},$

where the last equality follows from Theorem 3.5. On the other hand, since
$yy^{\prime}\in U^{-}[i]$, we obtain by using Theorem 3.5 directly

$\displaystyle\\{yy^{\prime},x\\}=\\{L^{\prime\prime}_{i,1}(y)L^{\prime\prime}_{i,1}(y^{\prime}),L^{\prime\prime}_{i,1}(x)\\}=\\{L^{\prime\prime}_{i,1}(y)\otimes
L^{\prime\prime}_{i,1}(y^{\prime}),\rho L^{\prime\prime}_{i,1}(x)\\}$
$\displaystyle\phantom{\\{yy^{\prime},x\\}}=\\{L^{\prime\prime}_{i,1}(y)\otimes
L^{\prime\prime}_{i,1}(y^{\prime}),(P^{\prime}_{i}\otimes\text{id})\circ\rho\circ
L^{\prime\prime}_{i,1}(x)\\}.$

Let $u=(P^{\prime}_{i}\otimes\text{id})\circ\rho\circ
L^{\prime\prime}_{i,1}(x)-(L^{\prime\prime}_{i,1}\otimes
L^{\prime\prime}_{i,1})\circ(\text{id}\otimes P_{i})\circ\rho(x)$. The above
argument implies that

$\\{U^{-}_{*}[i]\otimes U^{-}_{*}[i],u\\}=0.$

Since $U^{-}=U^{-}_{*}[i]\oplus U^{-}\mathtt{B}_{i}$ and
$\\{U^{-}\mathtt{B}_{i},u\\}=0$, thus we deduce that $\\{U^{-}\otimes
U^{-},u\\}=0$. Note that the bilinear form $\\{\ ,\ \\}$ is non-degenerated on
$U^{+}\otimes U^{+}$, hence $u=0$ and our assertion follows. ∎

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