# Double-bosonization and Majid’s conjecture, (I): Rank-inductions of $ABCD$

Hongmei Hu†,1 †School of Mathematical Sciences, University of Science and
Technology of China, Jin Zhai Road 96, Hefei 230026, PR China
<EMAIL_ADDRESS>and  Naihong Hu∗,1 1Department of Mathematics, Shanghai Key
Laboratory of Pure Mathematics and Mathematical Practice, East China Normal
University, Minhang Campus, Dong Chuan Road 500, Shanghai 200241, PR China
<EMAIL_ADDRESS>
###### Abstract.

Majid developed in [M3] the double-bosonization theory to construct
$U_{q}(\mathfrak{g})$ and expected to generate inductively not just a line but
a tree of quantum groups starting from a node. In this paper, the authors
confirm the Majid’s first expectation (see p. 178 [M3]) through giving and
verifying the full details of the inductive constructions of
$U_{q}(\mathfrak{g})$ for the classical types, i.e., the $ABCD$ series. Some
examples in low ranks are given to elucidate that any quantum group of
classical type can be constructed from the node corresponding to
$U_{q}(\mathfrak{sl}_{2})$.

N. H., supported by the NNSFC (Grant No. 11271131).

$\ast$ corresponding author.

##  1\. Introduction

The invention of quantum groups is one of the outstanding achievements of
mathematical physics and mathematics in the late twentieth century, which
arose in the work of L. D. Faddeev with his school for solving integrable
models in $1{+}1$ dimension by the quantum inverse scattering method. In the
history of development, a major event was the discovery of quantized universal
enveloping algebras $U_{q}(\mathfrak{g})$ over the complex field by V. G.
Drinfeld [D] and M. Jimbo [Ji] independently around 1985. A striking feature
of quantum group theory is the close connections with many branches of
mathematics and physics, such as Lie groups, Lie algebras and their
representations, representation theory of Hecke algebras, link invariant
theory, conformal field theory, and so on. This attracts many mathematicians
to find some better way in a suitable frame to understand the structure of
quantum groups defined initially by generators and relations. For instance,
the first we must mention is the famous FRT-construction [FRT] of
$U_{q}(\mathfrak{g})$ for the classical types based on the $R$-matrices of the
vector representations associated to the classical simple Lie algebras
$\mathfrak{g}$, which is a natural analogue of the matrix realization of the
classical Lie algebras. It is a belief that the commutation relations in
quantum groups expressed by means of the $R$-matrices are of fundamental
importance for the theory. Afterwards, Majid rediscovered the so-called
“Radford-Majid bosonization” arising from a framework of a braided category
over a Hopf algebra (see [Ra], [RT], [M1], [M2], [M9]), and then, based on the
FRT-construction and extending the Drinfeld double to the generalized quantum
double associated to a dual pair of Hopf algebras equipped with a (not
necessarily non-degenerate) pairing, Majid [M3] in 1996 developed the double-
bosonization theory to give a direct construction of $U_{q}(\mathfrak{g})$
through viewing the Lusztig’s algebra $\mathfrak{f}$ ([L2]) as a braided group
in a special braided category $\mathfrak{M}_{H}$ (or ${}_{H}\mathfrak{M}$),
where ($H=kQ=k[K_{i}^{\pm 1}]$, $A=kQ^{\vee}$) is a weakly quasitriangular
dual pair (see p. 169 of [M3]), where $Q$ (resp. $Q^{\vee}$) is a (dual) root
lattice of $\mathfrak{g}$. An analogous construction in spirit in the Yetter-
Drinfeld category of a Hopf algebra was given independently by Sommerhäuser in
[So].

Besides the constructions above, in the early 90s, Ringel [Ri] realized the
positive part of quantum groups by quiver representations and Hall algebras,
which inspired Lusztig’s canonical bases theory [L1, L2]. Rosso [Ro] also
realized the positive part of $U_{q}(\mathfrak{g})$ by introducing the quantum
shuffle algebra in a braided category, and gave a recipe of axiomatic
inductive construction based on the quantum “shuffle” operation defined on the
(braided) tensor algebra of a braided vector space which is sometimes a bit
inconvenient for making practical calculations, in contrast with the Majid’s
double-bosonization [M3]. Bridgeland [Br] recently realized the entire quantum
groups of type $ADE$ using the Ringel-Hall algebras. From another way, Fang-
Rosso [FR] realized the whole quantum groups by means of a new theory on the
quantum quasi-symmetric algebras due to Jian-Rosso [JR]. More general, the
whole axiomatic description for the multi-parameter quantum groups as well as
the Hopf $2$-cocycle deformation under the machinery of multi-parameter quasi-
symmetric algebras have been achieved in Hu-Li-Rosso [HLR].

In this paper, let us focus on the double-bosonizaiton theory in [M3].

Associated to any mutually dual braided groups $B^{\star},B$ covariant under a
background quasitriangular Hopf algebra H, there is a new quantum group on the
tensor space $B^{\star}\otimes H\otimes B$ by double-bosonization in [M3],
consisting of $H$ extended by $B$ as additional ‘positive roots’ and its dual
$B^{\star}$ as additional ‘negative roots’. The construction is more powerful
than the quantum double since one can reach directly to the
$U_{q}(\mathfrak{g})$ (the quantum double is a bit big and one has to make
quotient). Specially, Majid viewed $U_{q}(\mathfrak{n}^{\pm})$ as the mutually
dual braided groups in braided category of right “Cartan subalgebra”
$H$-modules, then recovered $U_{q}(\mathfrak{g})$ by his double-bosonization
theory. Also, based on two examples in low ranks given in [M3, M8], Majid
claimed that many new quantum groups, as well as the inductive (by rank)
construction of $U_{q}(\mathfrak{g})$ can be obtained in principle by this
theory [M3] (we call it the Majid’s expectation), since he imaged his double-
bosonization framework allows to generate a tree of quantum groups and at each
node of the tree, there are many choices to adjoin a pair of braided groups
covariant under the corresponding quantum group at that node. Actually, it is
a combinatorial and representation-theoretical challenge to elaborate the full
tree structure, which brings us the first motivation. We want to describe in
details the inductive construction of quantum groups $U_{q}(\mathfrak{g})$ for
all complex semisimple Lie algebras $\mathfrak{g}$, which just elaborates some
main branches of the tree. We will limit ourself to consider the classical
$ABCD$ series in this paper, which is organized as follows.

In section 2, we recall some basic facts about the FRT-construction and the
Majid’s double-bosonization construction. In section 3, we analyse explicitly
the procedure of the inductive construction of $U_{q}({\mathfrak{sl}}_{n})$,
and find some tips in low rank cases. That is, although the entire
$FRT$-matrix $m^{\pm}$ is hard to know from the vector representation, we only
need to know those diagonal and minor diagonal elements. This inspires us how
to proceed the general rank-inductive construction from
$U_{q}({\mathfrak{sl}}_{n})$ to $U_{q}({\mathfrak{sl}}_{n+1})$. At the node
diagram of $U_{q}({\mathfrak{sl}}_{n})$, we can choose a pair of braided
groups corresponding to the standard $R$-matrix arising from the vector
representation. Similarly, we can do the same thing for the $BCD$ series.
These demonstrate that the Majid’s double-bosonization framework does allow to
generate four branching-lines of nodes diagram of quantum groups. Furthermore,
in the last section, we will give some examples to show how to grow the tree
(with $ABCD$-branches) of nodes diagram of quantum groups out of the same
‘root’ node at type $A$ via type-crossing construction starting from type
$A_{r}$ for $r=1,2,3$.

##  2\. FRT-construction and Majid’s double-bosonization construction

In this paper, let $k$ be the complex field, $\mathfrak{g}$ a finite-
dimensional complex simple Lie algebra with simple roots $\alpha_{i}$. Let
$\lambda_{i}$ be the fundamental weight corresponding to $\alpha_{i}$. Cartan
matrix of $\mathfrak{g}$ is $(a_{ij})$, where
$a_{ij}=\frac{2(\alpha_{i},\alpha_{j})}{(\alpha_{i},\alpha_{i})}$, and
$d_{i}=\frac{(\alpha_{i},\alpha_{i})}{2}$. Let $(H,\mathscr{R})$ be a
quasitriangular Hopf algebra, where $\mathscr{R}$ is the universal $R$-matrix,
$\mathscr{R}=\mathscr{R}^{(1)}\otimes\mathscr{R}^{(2)}$,
$\mathscr{R}_{21}=\mathscr{R}^{(2)}\otimes\mathscr{R}^{(1)}$. Denote by
$\Delta,\,\eta,\,\epsilon$, $S$ its coproduct, counit, unit, its antipode,
respectively. We shall use Sweedler’s notation: for $h\in H$,
$\Delta(h)=h_{1}\otimes h_{2}$. Write $H^{op}\ (H^{cop})$ the opposite
(co)algebra structure of $H$, respectively. Let $\mathfrak{M}_{H}\
({}_{H}\mathfrak{M}$) be the braided category consisting of right (left)
$H$-modules, respectively. If there exists a coquasitriangular Hopf algebra
$A$ such that $(H,A)$ is a weakly quasitriangular dual pair, then
$\mathfrak{M}_{H}\ ({}_{H}\mathfrak{M}$) is equivalent to the braided category
${}^{A}\mathfrak{M}\ (\mathfrak{M}^{A})$ consisting of left (right)
$A$-comodules, respectively. For the detailed description of these theories,
we left to the readers to refer to Drinfeld’s and Majid’s papers [D], [M4],
[M5], and so on. By a braided group, following [M4], we mean a braided
bialgebra or Hopf algebra in some braided category. In order to distinguish
from the ordinary Hopf algebras, denote by $\underline{\Delta},\underline{S}$
its coproduct and antipode, respectively. An invertible matrix solution of the
quantum Yang-Baxter equation (QYBE) $R_{12}R_{13}R_{23}=R_{23}R_{13}R_{12}$ is
called a $R$-matrix. For later use, one needs a certain extension
$U_{q}^{ext}(\mathfrak{g})$ of $U_{q}(\mathfrak{g})$. This is constructed by
adjoining formally certain products of the elements $K_{i}^{\pm\frac{1}{n}}$,
$i=1,\cdots,n-1$ to $U_{q}({\mathfrak{sl}}_{n})$, the elements
$K_{n}^{\pm\frac{1}{2}}$ to $U_{q}({\mathfrak{sp}}_{2n})$, and
$K_{n-1}^{\pm\frac{1}{2}}K_{n}^{\pm\frac{1}{2}}$ and
$K_{n-1}^{\pm\frac{1}{2}}K_{n}^{\mp\frac{1}{2}}$ to
$U_{q}({\mathfrak{so}}_{2n})$. The precise definitions of them can be found in
[KS]. Denote by $T_{V}$ an irreducible representation of
$U_{q}(\mathfrak{g})$, where $V$ is the corresponding module with a basis
$\\{\,x_{i}\,\\}$.

###  2.1. $FRT$-construction

One can obtain an invertible (basic) $R$-matrix from the quasitriangular Hopf
algebra $U_{q}(\mathfrak{g})$ and its (vector) representation. Conversely,
starting from an invertible (basic) $R$-matrix, if does there exist a
quasitriangular Hopf algebra to recover such a $R$-matrix through a suitable
representation? First of all, by Faddeev-Reshetikhin-Takhtajan [FRT], the
following fact is basic and well-known.

######  Definition 2.1.

Given an invertible matrix solution $R$ of the QYBE, there is a bialgebra
$A(R)$, named the $FRT$-bialgebra, which is generated by $1$ and $t^{i}_{j}$,
for $1\leq i,j\leq n$, with the relations $RT_{1}T_{2}=T_{2}T_{1}R,\
\Delta(T)=T\otimes T$ and $\epsilon(T)=I$ using standard notation in [FRT],
where the matrix $T=(t^{i}_{j})$, $T_{1}=T\otimes I$, $T_{2}=I\otimes T$.

Observe that $A(R)$ is a coquasitriangular bialgebra with
$\mathscr{R}:\,A(R)\otimes A(R)\longrightarrow k$ such that
$\mathscr{R}(t^{i}_{j}\otimes t^{k}_{l})=R^{ik}_{jl}$. Here $R^{ik}_{jl}$
denotes the entry at row $(ik)$ and column $(jl)$ in matrix $R$. Secondly, in
the dual space $A(R)^{\ast}=Hom(A(R),k)$, $\Delta$ of $A(R)$ induces the
multiplication of $A(R)^{\ast}$. In [FRT], $U_{R}$ is defined to be the
subalgebra of $A(R)^{\ast}$ generated by $L^{\pm}=(l^{\pm}_{ij})$, with
relations

$(PRP)L_{1}^{\pm}L_{2}^{\pm}=L_{2}^{\pm}L_{1}^{\pm}(PRP),\qquad(PRP)L_{1}^{+}L_{2}^{-}=L_{2}^{-}L_{1}^{+}(PRP),$

where $l^{\pm}_{ij}$ is defined by $(l^{+}_{ij},t^{k}_{l})=R^{ki}_{lj},$
$(l^{-}_{ij},t^{k}_{l})=(R^{-1})^{ik}_{jl}.$

The algebra $U_{R}$ is a bialgebra with coproduct
$\Delta(L^{\pm})=L^{\pm}\otimes L^{\pm}$ and counit
$\varepsilon(l_{ij}^{\pm})=\delta_{ij}$. Specially, when $R$ is the classical
$R$-matrix, bialgebra $A(R)$ has a quotient coquasitriangular Hopf algebra,
denoted by $Fun(G_{q})$ or $\mathcal{O}_{q}(G)$, and $U_{R}$ also has a
corresponding quotient quasitriangular Hopf algebra, which is isomorphic to
the extended quantized enveloping algebra $U_{q}^{ext}(\mathfrak{g})$.
Moreover, there exists a (non-degenerate) dual pairing $\langle,\,\rangle$
between $\mathcal{O}_{q}(G)$ and $U_{q}^{ext}(\mathfrak{g})$. The way of
getting the resulting quasitriangular algebras $U_{q}^{ext}(\mathfrak{g})$ is
the so-called FRT-construction of the quantized enveloping algebras (for the
classical types).

Motivated by the work of [FRT], Majid built the theory of the weakly
quasitriangular dual pairings associated with $R$-matrices [M3] in a more
general context.

######  Remark 2.1.

The $R$-matrices used in Majid’s papers [M1, M3] are a bit different from the
standard ones as in [FRT], which are the conjugations $P\circ\cdot\circ P$ of
the ordinary $R$-matrices by the permutation matrix $P:P(u\otimes v)=v\otimes
u$. It can be checked directly that $A(P\circ R\circ P)=A(R)^{\text{op}}$,
where $(P\circ R\circ P)^{ij}_{kl}=R^{ji}_{lk}$. Note that we will use Majid’s
notation for $R$-matrices as in [M3] in the remaining sections of this paper.

###  2.2. Majid’s double-bosonization

Majid [M3] proposed the concept of a weakly quasitriangular dual pair via his
insight on more examples on matched pairs of bialgebras or Hopf algebras in
[M7]. This allowed him to establish a theory of double-bosonization in a broad
framework that generalized the FRT’s construction which was limited to the
classical types.

######  Definition 2.2.

Let $(H,A)$ be a pair of Hopf algebras equipped with a dual pairing
$\langle,\,\rangle$ and convolution-invertible algebra / anti-coalgebra maps
$\mathscr{R},\bar{\mathscr{R}}:A\rightarrow H$ obeying

$\langle\bar{\mathscr{R}}(a),b\rangle=\langle\mathscr{R}^{-1}(b),a\rangle,\quad\partial^{R}h=\mathscr{R}\ast(\partial^{L}h)\ast\mathscr{R}^{-1},\quad\partial^{R}h=\bar{\mathscr{R}}\ast(\partial^{L}h)\ast\bar{\mathscr{R}}^{-1}$

for $a,b\in A,\ h\in H$. Here $\ast$ is the convolution product in $hom(A,H)$
and $(\partial^{L}h)(a)=\langle h_{(1)},a\rangle h_{(2)},$
$(\partial^{R}h)(a)=h_{(1)}\langle h_{(2)},a\rangle$ are left, right
“differentiation operators” regarded as maps $A\rightarrow H$ for fixed $h$.

Let $C,B$ be a pair of braided groups in $\mathfrak{M}_{H}$, which are called
dually paired if there is an intertwiner $ev:C\otimes B\longrightarrow k$ such
that
$\text{ev}(cd,b)=\text{ev}(d,b_{\underline{(1)}})ev(c,b_{\underline{(2)}}),$
$\text{ev}(c,ab)=\text{ev}(c_{\underline{(2)}},a)ev(c_{\underline{(1)}},b),$
$\forall a,b\in B,c,d\in C.$ Then $C^{op/cop}$ (with opposite product and
coproduct) is a Hopf algebra in ${}_{H}\mathfrak{M}$, which is dual to $B$ in
the sense of an ordinary dual pairing $\langle~{},~{}\rangle$ with
$H$-bicovariant: $\langle h\rhd c,b\rangle=\langle c,b\lhd h\rangle$ for all
$h\in H$. Let $\overline{C}=(C^{op/cop})^{\underline{cop}}$, then
$\overline{C}$ is a braided group in ${}_{\overline{H}}\mathfrak{M}$, where
$\overline{H}$ is $(H,\mathscr{R}_{21}^{-1})$. With these, Majid gave the
following double-bosonization theorem.

######  Theorem 2.1.

$($Majid$)$ On the tensor space $\bar{C}\otimes H\otimes B$, there is a unique
Hopf algebra structure $U=U(\bar{C},H,B)$ such that $H\ltimes B$
(bosonization) and $\bar{C}\rtimes H$ (bosonization) are sub-Hopf algebras by
the canonical inclusions, with cross relation

$None$ $bh=h_{(1)}(b\lhd h_{(2)}),\quad ch=h_{(2)}(c\lhd h_{(1)}),$

here $b\in B,c\in C,h\in H$. If there exists a coquasitriangular Hopf algebra
$A$ such that $(H,A)$ is a weakly quasitriangular dual pair, and $b,c$ are
primitive elements, then some relations simplify to

$None$ $[b,c]=\mathscr{R}(b^{\overline{(1)}})\langle
c,b^{\overline{(2)}}\rangle-\langle
c^{\overline{(1)}},b\rangle\bar{\mathscr{R}}(c^{\overline{(2)}});$ $None$
$\Delta b=b^{\overline{(2)}}\otimes\mathscr{R}(b^{\overline{(1)}})+1\otimes
b,\quad\Delta c=c\otimes 1+\bar{\mathscr{R}}(c^{\overline{(2)}})\otimes
c^{\overline{(1)}}.$

Let $\widetilde{U(R)}$ be the double cross product bialgebra of
$A(R)^{\text{op}}$ in [M7] generated by $m^{\pm}$ with the bialgebra structure
given by

$None$ $Rm^{\pm}_{1}m^{\pm}_{2}=m^{\pm}_{2}m^{\pm}_{1}R,\quad
Rm_{1}^{+}m_{2}^{-}=m_{2}^{-}m_{1}^{+}R,\quad\Delta((m^{\pm})^{i}_{j})=(m^{\pm})_{j}^{a}\otimes(m^{\pm})_{a}^{i},\quad\epsilon((m^{\pm})^{i}_{j})=\delta_{ij},$

where $(m^{\pm})^{i}_{j}$ are the $FRT$-generators, $m^{\pm}$ are called
$FRT$-matrices.

######  Proposition 2.1.

$(1)$ $(\widetilde{U(R)},A(R))$ is a weakly quasitriangular dual pair with

$\langle(m^{+})^{i}_{j},t^{k}_{l}\rangle=R^{ik}_{jl},\quad\langle(m^{-})^{i}_{j},t^{k}_{l}\rangle=(R^{-1})^{ki}_{lj},\quad\mathscr{R}(T)=m^{+},\quad\bar{\mathscr{R}}(T)=m^{-}.$

Specially, when $R$ is the classical $R$-matrix, then the weakly
quasitriangular dual pair can be descended to a mutually dual pair of quotient
Hopf algebras $(U_{q}^{ext}(\mathfrak{g}),\mathcal{O}_{q}(G))$ (where $G$ is
one of connected and simply connected algebraic groups of classical types)
with the correct modification [M3]:

$\langle(m^{+})^{i}_{j},t^{k}_{l}\rangle=\lambda
R^{ik}_{jl},\quad\langle(m^{-})^{i}_{j},t^{k}_{l}\rangle=\lambda^{-1}(R^{-1}){}^{ki}_{lj}.$

Such $\lambda$ is called a quantum group normalization constant.

$(2)$ Suppose that $R^{\prime}$ is another matrix such that (i)
$R_{12}R_{13}R^{\prime}_{23}=R^{\prime}_{23}R_{13}R_{12}$,
$R_{23}R_{13}R^{\prime}_{12}$ $=R^{\prime}_{12}R_{13}R_{23}$, (ii)
$(PR+1)(PR^{\prime}-1)=0$, (iii)
$R_{21}R^{\prime}_{12}=R^{\prime}_{21}R_{12}$, where $P$ is the permutation
matrix with entries $P^{ij}_{kl}=\delta_{il}\delta_{jk}$. Then a braided-
vector algebra $V(R^{\prime},R)$ generated by generators $1$,
$\\{e^{i}~{}|~{}i=1,\cdots,n\\}$, and relations
$e^{i}e^{j}=\sum\limits_{a,b}{R^{\prime}}^{ji}_{ab}e^{a}e^{b}$ forms a braided
group with $\underline{\Delta}(e^{i})=e^{i}\otimes 1+1\otimes
e^{i},\underline{\epsilon}(e^{i})=0,\underline{S}(e^{i})=-e^{i},\Psi(e^{i}\otimes
e^{j})=\sum\limits_{a,b}R^{ji}_{ab}e^{a}\otimes e^{b}$ in braided category
${}^{A(R)}\mathfrak{M}$. Under duality $\langle
f_{j},e^{i}\rangle=\delta_{ij}$, a braided-covector algebra
$V^{\vee}(R^{\prime},R_{21}^{-1})$ generated by 1 and
$\\{f_{j}~{}|~{}j=1,\cdots,n\\}$, and relations
$f_{i}f_{j}=\sum\limits_{a,b}f_{b}f_{a}{R^{\prime}}_{ab}^{ij}$ forms another
braided group with $\underline{\Delta}(f_{i})=f_{i}\otimes 1+1\otimes f_{i}$,
$\underline{\epsilon}(f_{i})=0$, $\underline{S}(f_{i})=-f_{i}$,
$\Psi(f_{i}\otimes f_{j})=\sum\limits_{a,b}f_{b}\otimes
f_{a}R^{a}_{i}{}^{b}_{j}$ in braided category $\mathfrak{M}^{A(R)}$.

######  Remark 2.2.

In fact, by Remark 2.1, $\widetilde{U(R)}$ determined uniquely by relations
$(C4)$ with the $R$-matrix in [M3] is the opposite of $U_{R}$ constructed by
$FRT$-approach. So the quantized enveloping algebra $U_{q}(\mathfrak{g})$
recovered by Majid (Proposition 4.3 in [M3]) satisfies
$E_{i}K_{j}=q_{i}^{a_{ij}}K_{j}E_{i},$ $F_{i}K_{j}=q_{i}^{-a_{ij}}K_{j}F_{i},$
$[E_{i},F_{j}]=\delta_{ij}\frac{K_{i}-K_{i}^{-1}}{q-q^{-1}},$ where
$q_{i}=q^{d_{i}}$, which is the opposite of the ordinary
$U_{q}(\mathfrak{g})$. By abuse of notation, we still use
$U_{q}(\mathfrak{g})$ instead of $U_{q}(\mathfrak{g})^{op}$ in the remaining
sections of this paper.

With the new $\lambda R$-matrix in Proposition 2.1, in order that
$V(R^{\prime},R),V^{\vee}(R^{\prime},R_{21}^{-1})$ are still braided groups ,
then $(U_{q}^{ext}(\mathfrak{g}),\mathcal{O}_{q}(G))$ must be centrally
extended to the pair

$\Bigl{(}\widetilde{U_{q}^{ext}(\mathfrak{g})}=U_{q}(\mathfrak{g})\otimes
k[c,c^{-1}],\widetilde{\mathcal{O}_{q}(G)}=\mathcal{O}_{q}(G)\otimes
k[g,g^{-1}]\Bigr{)}$

with action $e^{i}\lhd c=\lambda e^{i}$, $f_{i}\lhd c=\lambda f_{i}$, $\langle
c,g\rangle=\lambda$. By Theorem 2.1, we have the following

######  Corollary 2.1.

$U=U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}(\mathfrak{g})},V(R^{\prime},R))$
is a new quantum group with the cross relations: $e^{i}(m^{+})^{j}_{k}=\lambda
R^{ji}_{ab}(m^{+})^{a}_{k}e^{b}$, $(m^{-})^{i}_{j}e^{k}=\lambda
R^{ki}_{ab}e^{a}(m^{-})^{b}_{j}$, $(m^{+})^{i}_{j}f_{k}=\lambda
f_{b}(m^{+})^{i}_{a}R^{ab}_{jk}$,
$f_{i}(m^{-})^{j}_{k}=\lambda(m^{-})^{j}_{b}f_{a}R^{ab}_{ik}$, $cf_{i}=\lambda
f_{i}c$, $e^{i}c=\lambda ce^{i}$, $[c,m^{\pm}]=0,$
$[e^{i},f_{j}]=\delta_{ij}\frac{(m^{+})^{i}_{j}c^{-1}-c(m^{-})^{i}_{j}}{q_{\ast}-q_{\ast}^{-1}};$
and the coproduct: $\Delta c=c\otimes c,$ $\Delta
e^{i}=e^{a}\otimes(m^{+})^{i}_{a}c^{-1}+1\otimes e^{i},$ $\Delta
f_{i}=f_{i}\otimes 1+c(m^{-})^{a}_{i}\otimes f_{a},$ $\epsilon e^{i}=\epsilon
f_{i}=0$, where one can normalize $e^{i}$ such that the factor
$q_{\ast}-q_{\ast}^{-1}$ satisfies the situation you need.

So the Majid’s double-bosonization construction can lead to new quantum
groups. After giving an example of obtaining $U_{q}({\mathfrak{sl}}_{3})$ from
$U_{q}^{ext}({\mathfrak{sl}}_{2})$ in [M3], Majid expected that the novel
resulting quantum group is the quantum group of higher-one rank in the
classical $ABCD$’s series. In the current paper, we will solve such Majid’s
expectation in the classical types that has been an open question aimed by
Majid ([M3]) since the mid of 90’s. In the next section, we will give full
details of the rank-inductive construction in the $ABCD$ series.

##  3\. Rank-inductive construction of quantum groups for classical types

In order to explore the structure of the resulting quantum group in Corollary
2.1, we need to know how to get the explicit form of the $FRT$-matrix
$m^{\pm}$, which can be obtained by the following lemma.

######  Lemma 3.1.

Corresponding to the invertible matrix $R$ obeying the QYBE, we have
$S(l_{ij}^{\pm})=(m^{\pm})^{i}_{j}$ in the quotient Hopf algebra, $S$ is the
corresponding antipode.

###### Proof.

If there exists a quotient Hopf algebra of $\widetilde{U(R)}$, according to
Remark 2.2, we obtain the following relations in this quotient Hopf algebra:
$RL_{2}^{\pm}L_{1}^{\pm}=L_{1}^{\pm}L_{2}^{\pm}R,$
$RL_{2}^{-}L_{1}^{+}=L_{1}^{+}L_{2}^{-}R,$ and $\Delta(L^{\pm})=L^{\pm}\otimes
L^{\pm}$. We can describe these relations in view of the entries in the matrix
$R$ and $L^{\pm}$. For example, for any fixed $i,j,k,l$, we have
$(RL_{2}^{\pm}L_{1}^{\pm})^{ij}_{kl}=(L_{1}^{\pm}L_{2}^{\pm}R)^{ij}_{kl}$,
then we obtain the following equality on the left hand side

$\left.\begin{array}[]{rl}(RL_{2}^{\pm}L_{1}^{\pm})^{ij}_{kl}&=R^{ij}_{ab}(I\otimes
L^{\pm})^{ab}_{mn}(L^{\pm}\otimes
I)^{mn}_{kl}=R^{ij}_{ab}\delta_{am}(l^{\pm})_{bn}(l^{\pm})_{mk}\delta_{nl}\\\
&=R^{ij}_{mb}(l^{\pm})_{bl}(l^{\pm})_{mk}=R^{ij}_{mn}(l^{\pm})_{nl}(l^{\pm})_{mk},\end{array}\right.$

and the equality by the right hand side

$(L_{1}^{\pm}L_{2}^{\pm}R)^{ij}_{kl}=(L^{\pm}\otimes I)^{ij}_{ab}(I\otimes
L^{\pm})^{ab}_{mn}R^{mn}_{kl}=(l^{\pm})_{ia}\delta_{jb}\delta_{am}(l^{\pm})_{bn}R^{mn}_{kl}=(l^{\pm})_{im}(l^{\pm})_{jn}R^{mn}_{kl}.$

So
$R^{ij}_{mn}(l^{\pm})_{nl}(l^{\pm})_{mk}=(l^{\pm})_{im}(l^{\pm})_{jn}R^{mn}_{kl}.$
Taking the antipode $S$ on both sides, we obtain

$R^{ij}_{mn}S((l^{\pm})_{mk})S((l^{\pm})_{nl})=S((l^{\pm})_{jn})S((l^{\pm})_{im})R^{mn}_{kl}.$

Under the notation $(m^{\pm})^{i}_{j}=S(l_{ij}^{\pm})$, we have

$R^{ij}_{mn}(m^{\pm})^{m}_{k}(m^{\pm})^{n}_{l})=(m^{\pm})^{j}_{n})(m^{\pm})^{i}_{m})R^{mn}_{kl},\
\text{i.e.},\ Rm_{1}^{\pm}m_{2}^{\pm}=m_{2}^{\pm}m_{1}^{\pm}R.$

We also obtain $Rm_{1}^{+}m_{2}^{-}=m_{2}^{-}m_{1}^{+}R$ by a similar
argument. On the other hand,

$\begin{split}\Delta((m^{\pm})^{i}_{j})&=\Delta(S(l^{\pm}_{ij}))=P\circ(S\otimes
S)\Delta(l^{\pm}_{ij})=P\circ(S\otimes S)(l^{\pm}_{ia}\otimes l^{\pm}_{aj})\\\
&=P((m^{\pm})^{i}_{a}\otimes(m^{\pm})^{a}_{j})=(m^{\pm})^{a}_{j}\otimes(m^{\pm})^{i}_{a}.\end{split}$

So, the generators $(m^{\pm})^{i}_{j}$ satisfy relation $(C4)$. This completes
the proof. ∎

###  3.1. Inductive construction of $U_{q}({\mathfrak{sl}}_{n})$

Since $m^{\pm}$ and $R$-matrix will become larger and larger with the growing
of rank, it is a challenge to describe explicitly the procedure of general
inductive construction. We want to find some tips on some concrete examples.
Majid already described explicitly the case of
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{2})},V(R^{\prime},R))\simeq
U_{q}({\mathfrak{sl}}_{3})$ in [M3]. First of all, in order to capture much
more hints coming from the theory of Majid’s double-bosonization construction,
in what follows, we will describe in detail an example of rank $2$ case:
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{3})},V(R^{\prime},R))\simeq
U_{q}({\mathfrak{sl}}_{4})$.

######  Example 3.1.

Let us start with $R$-matrix datum

$R=\left(\begin{array}[]{ccccccccc}q^{2}&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0\\\
0&~{}q&~{}0&~{}q^{2}-1&~{}0&~{}0&~{}0&~{}0&~{}0\\\
0&~{}0&~{}q&~{}0&~{}0&~{}0&~{}q^{2}-1&~{}0&~{}0\\\
0&~{}0&~{}0&~{}q&~{}0&~{}0&~{}0&~{}0&~{}0\\\
0&~{}0&~{}0&~{}0&~{}q^{2}&~{}0&~{}0&~{}0&~{}0\\\
0&~{}0&~{}0&~{}0&~{}0&~{}q&~{}0&~{}q^{2}-1&~{}0\\\
0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}q&~{}0&~{}0\\\
0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}q&~{}0\\\
0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}0&~{}q^{2}\end{array}\right).$

Take $R^{\prime}=q^{-2}R$, then $R,R^{\prime}$ give braided groups
$V^{\vee}(R^{\prime},R_{21}^{-1})=k\,\langle f_{i}\mid i=1,2,3\rangle$ and
$V(R^{\prime},R)=k\,\langle e^{i}\mid i=1,2,3\rangle$. Identify
$e^{3},f_{3},(m^{+})^{3}_{3}c^{-1}$ with the additional simple root vectors
$E_{3},F_{3}$ and group-like element $K_{3}$, then the resulting quantum group
$(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{3})}$,
$V(R^{\prime},R))$ is exactly the quantum group $U_{q}({\mathfrak{sl}}_{4})$
with $K_{i}^{\frac{1}{3}},\ i=1,2$ adjoined.

###### Proof.

The quantum group normalization constant for $R$ needed for the weakly
quasitriangular structure on $U_{q}^{ext}({\mathfrak{sl}}_{3})$ is
$\lambda=q^{-\frac{4}{3}}$ obtained by the facts in [FRT]. Moreover, the
$m^{\pm}$-matrix corresponding to the vector representation can be obtained by
Lemma 3.1.

$\left.\begin{array}[]{rcl}m^{+}=\left(\begin{array}[]{ccc}K^{\frac{2}{3}}_{1}K^{\frac{1}{3}}_{2}&~{}~{}(q-q^{-1})E_{1}K^{-\frac{1}{3}}_{1}K^{\frac{1}{3}}_{2}&~{}~{}q^{-1}(q-q^{-1})[E_{1},E_{2}]_{q^{-1}}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}\\\
0&K^{-\frac{1}{3}}_{1}K^{\frac{1}{3}}_{2}&(q-q^{-1})E_{2}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}\\\
0&0&K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}\end{array}\right)\end{array}\right.,$
$\left.\begin{array}[]{rcl}m^{-}=\left(\begin{array}[]{ccc}K^{-\frac{2}{3}}_{1}K^{-\frac{1}{3}}_{2}&0&0\\\
(q-q^{-1})K^{\frac{1}{3}}_{1}K^{-\frac{1}{3}}_{2}F_{1}&K^{\frac{1}{3}}_{1}K^{-\frac{1}{3}}_{2}&0\\\
q(q-q^{-1})K^{\frac{1}{3}}_{1}K^{\frac{2}{3}}_{2}[F_{2},F_{1}]_{q}&~{}~{}(q-q^{-1})K^{\frac{1}{3}}_{1}K^{\frac{2}{3}}_{2}F_{2}&~{}~{}K^{\frac{1}{3}}_{1}K^{\frac{2}{3}}_{2}\end{array}\right)\end{array}\right.,$

where $[E_{1},E_{2}]_{q^{-1}}=E_{1}E_{2}-q^{-1}E_{2}E_{1}$,
$[F_{2},F_{1}]_{q}=F_{2}F_{1}-qF_{1}F_{2}$. By Corollary 2.1, we get

$[e^{3},f_{3}]=\frac{(m^{+})^{3}_{3}c^{-1}-c(m^{-})^{3}_{3}}{q-q^{-1}},\Longrightarrow[E_{3},F_{3}]=\frac{K_{3}-K_{3}^{-1}}{q-q^{-1}}.$

$E_{3}K_{3}=e^{3}(m^{+})^{3}_{3}c^{-1}=\lambda
R^{33}_{ab}(m^{+})^{a}_{3}e^{b}c^{-1}=R^{33}_{ab}(m^{+})^{a}_{3}c^{-1}e^{b}=R^{33}_{33}(m^{+})^{3}_{3}c^{-1}e^{3}=q^{2}K_{3}E_{3}.$
From the expression of $(m^{+})^{i}_{i}$, we have
$(m^{+})^{2}_{2}K_{1}=(m^{+})^{1}_{1},(m^{+})^{3}_{3}K_{2}=(m^{+})^{2}_{2}.$
Associating with the cross relation $e^{3}(m^{+})^{i}_{i}=\lambda
R^{i3}_{ab}(m^{+})^{a}_{i}e^{b},$ we obtain

$\left.\begin{array}[]{l}e^{3}(m^{+})^{1}_{1}=q^{-\frac{1}{3}}(m^{+})^{1}_{1}e^{3},\\\
e^{3}(m^{+})^{2}_{2}=q^{-\frac{1}{3}}(m^{+})^{2}_{2}e^{3},\\\
e^{3}(m^{+})^{3}_{3}=q^{\frac{2}{3}}(m^{+})^{3}_{3}e^{3},\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}e^{3}K_{1}=K_{1}e^{3},\\\
e^{3}K_{2}=q^{-1}K_{2}e^{3}.\end{array}\right.\Longleftrightarrow\left\\{\begin{array}[]{l}E_{3}K_{1}=K_{1}E_{3},\\\
E_{3}K_{2}=q^{-1}K_{2}E_{3}.\end{array}\right.$

In order to explore the relations between $F_{3}$ and $K_{i},i=1,2,3$, we have
$K_{3}F_{3}=(m^{+})^{3}_{3}c^{-1}f_{3}=\frac{1}{\lambda}(m^{+})^{3}_{3}f_{3}c^{-1}=\frac{1}{\lambda}\lambda
f_{3}(m^{+})^{3}_{3}R^{33}_{33}c^{-1}=f_{3}(m^{+})^{3}_{3}c^{-1}R^{33}_{33}=q^{2}F_{3}K_{3}$
through the cross relation $(m^{+})^{i}_{i}f_{3}=\lambda
f_{b}(m^{+})^{i}_{a}R^{ab}_{i3}.$ Similarly,

$\left.\begin{array}[]{l}(m^{+})^{i+1}_{i+1}K_{i}=(m^{+})^{i}_{i},\ i=1,2,\\\
(m^{+})^{i}_{i}f_{3}=q^{-\frac{1}{3}}f_{3}(m^{+})^{i}_{i},\ i=1,2,\\\
(m^{+})^{3}_{3}f_{3}=q^{\frac{2}{3}}f_{3}(m^{+})^{3}_{3}.\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}f_{3}K_{1}=K_{1}f_{3},\\\
f_{3}K_{2}=qK_{2}f_{3}.\end{array}\right.\Longleftrightarrow\left\\{\begin{array}[]{l}F_{3}K_{1}=K_{1}F_{3},\\\
F_{3}K_{2}=qK_{2}F_{3}.\end{array}\right.$

Then the relations between $K_{3}$ and $E_{i},F_{i},i=1,2$ are given as
follows

$E_{1}K_{3}=E_{1}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}=q^{-\frac{2}{3}}q^{\frac{2}{3}}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}E_{1}c^{-1}=K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}E_{1}=K_{3}E_{1},$
$E_{2}K_{3}=E_{2}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}=q^{\frac{1}{3}}q^{-\frac{4}{3}}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}E_{2}c^{-1}=q^{-1}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}E_{2}=q^{-1}K_{3}E_{2},$
$F_{1}K_{3}=F_{1}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}=q^{\frac{2}{3}}q^{-\frac{2}{3}}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}F_{1}c^{-1}=K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}F_{1}=K_{3}F_{1},$
$F_{2}K_{3}=F_{2}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}=q^{-\frac{1}{3}}q^{\frac{4}{3}}K^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}F_{2}c^{-1}=qK^{-\frac{1}{3}}_{1}K^{-\frac{2}{3}}_{2}c^{-1}F_{2}=qK_{3}F_{2}.$

Moreover, $\Delta(E_{3})=E_{3}\otimes K_{3}+1\otimes E_{3}$,
$\Delta(F_{3})=F_{3}\otimes 1+K_{3}^{-1}\otimes F_{3}$ can be obtained by
Corollary 2.1. The most important relations are the $q$-Serre relations. Since
the generators $E_{i}$’s belong to $(m^{+})^{i}_{i+1},i=1,2$, we will consider
the following equalities

$\left.\begin{array}[]{l}(m^{+})^{1}_{2}=(q-q^{-1})E_{1}(m^{+})^{2}_{2},\\\
e^{3}(m^{+})^{1}_{2}=\lambda R^{13}_{ab}(m^{+})^{a}_{2}e^{b}=\lambda
q(m^{+})^{1}_{2}e^{3},\\\ e^{3}(m^{+})^{2}_{2}=\lambda
R^{23}_{ab}(m^{+})^{a}_{2}e^{b}=\lambda
q(m^{+})^{2}_{2}e^{3}.\end{array}\right\\}\Longrightarrow
e^{3}E_{1}=E_{1}e^{3},\Longrightarrow E_{3}E_{1}=E_{1}E_{3}.$
$\left.\begin{array}[]{l}(m^{+})^{2}_{3}=(q-q^{-1})E_{2}(m^{+})^{3}_{3},\\\
e^{3}(m^{+})^{2}_{3}=\lambda R^{23}_{ab}(m^{+})^{a}_{3}e^{b}=\lambda
q(m^{+})^{2}_{3}e^{3}+\lambda(q^{2}-1)(m^{+})^{3}_{3}e^{2},\\\
e^{3}(m^{+})^{3}_{3}=\lambda R^{33}_{ab}(m^{+})^{a}_{2}e^{b}=\lambda
q^{2}(m^{+})^{3}_{3}e^{3}.\end{array}\right\\}\Longrightarrow
e^{2}=e^{3}E_{2}-q^{-1}E_{2}e^{3}.$

So, we need to explore the relation between $e^{2}$ and $e^{3}$. Note that
$e^{2}e^{3}=R^{\prime}{}^{32}_{ab}e^{a}e^{b}=q^{-2}R^{32}_{ab}e^{a}e^{b}=q^{-2}R^{32}_{32}e^{3}e^{2}=q^{-1}e^{3}e^{2}$,
then combining with $e^{2}=e^{3}E_{2}-q^{-1}E_{2}e^{3}$, we obtain

$(E_{3})^{2}E_{2}-(q+q^{-1})E_{3}E_{2}E_{3}+E_{2}(E_{3})^{2}=0.$

On the other hand, we need to know another relation between $e^{2}$ and
$E_{2}$, which can be explored by the following cross relations

$\left.\begin{array}[]{l}(m^{+})^{2}_{3}=(q-q^{-1})E_{2}(m^{+})^{3}_{3},\\\
e^{2}(m^{+})^{2}_{3}=\lambda q^{2}(m^{+})^{2}_{3}e^{2},\\\
e^{2}(m^{+})^{3}_{3}=\lambda q(m^{+})^{3}_{3}e^{3},\\\
e^{2}=e^{3}E_{2}-q^{-1}E_{2}e^{3}.\end{array}\right\\}\Longrightarrow(E_{2})^{2}E_{3}-(q+q^{-1})E_{2}E_{3}E_{2}+E_{3}(E_{2})^{2}=0.$

The $q$-Serre relation of $F_{i},1\leq i\leq 3$ can be obtained similarly.
Then the resulting quantum group is just the quantized enveloping algebra
$U_{q}({\mathfrak{sl}}_{4})$, where $e^{i},f_{i},i=1,2$ can be expressed by
the $q$-commutators with generators $E_{i},F_{i},i=1,2,3.$ ∎

######  Remark 3.1.

In the above concrete example, we find that it is not necessary to know the
entire $m^{\pm}$-matrix for determining the structure of resulting quantum
group, except for the diagonal and minor diagonal entries. At least,
corresponding to the vector representation of $U_{q}(\mathfrak{g})$, all the
simple root vectors $E_{i},F_{i}$ and group-like elements $K_{i}$ are included
in the diagonal and minor diagonal entries. Other entries in $m^{\pm}$ are
filled by non-simple root vectors generated by $q$-commutators with generators
$E_{i},F_{i}$ and $K_{i}$. Moreover, the explicit expressions of the diagonal
and minor diagonal entries in $m^{\pm}$ can be easily obtained by Lemma 3.1.

For the $A$ series, the data $R,R^{\prime},m^{\pm}$ as above are deduced from
vector representation, which inspires us to start with the vector
representation when considering the general rank-inductive construction. The
$R_{VV}$-matrix of vector representation $T_{V}$ satisfies the quadratic
equation $(PR_{VV}-q^{\frac{n-1}{n}}I)(PR_{VV}+q^{-\frac{n+1}{n}}I)=0$. So
setting $R=q^{\frac{n+1}{n}}R_{VV},R^{\prime}=q^{-2}R,$ then we have
$(PR+I)(PR^{\prime}-I)=0,$ and
$R^{ij}_{kl}=qq^{\delta_{ij}}\delta_{ik}\delta_{jl}+(q^{2}-1)\delta_{il}\delta_{jk}\theta(j-i),\
$ where

$\theta(k)=\left\\{\begin{array}[]{lcl}1&{}{}\hfil&k>0,\\\ 0&{}{}\hfil&k\leq
0.\end{array}\right.$

On the other hand, according to Lemma 3.1, we obtain the following

######  Lemma 3.2.

Corresponding to the vector representation, the diagonal and minor diagonal
entries in $FRT$-matrix $m^{\pm}$ of $U_{q}^{ext}({\mathfrak{sl}}_{n})$ are
given by

$\displaystyle(m^{+})^{i}_{i+1}=(q-q^{-1})E_{i}K^{-\frac{1}{n}}_{1}K^{-\frac{2}{n}}_{2}\cdots
K^{-\frac{i-1}{n}}_{i-1}K^{-\frac{i}{n}}_{i}K^{\frac{n-(i+1)}{n}}_{i+1}\cdots
K^{\frac{n-(n-1)}{n}}_{n-1},\quad 1\leq i\leq n-1,$
$\displaystyle(m^{+})^{i}_{i}=K^{-\frac{1}{n}}_{1}K^{-\frac{2}{n}}_{2}\cdots
K^{-\frac{i-1}{n}}_{i-1}K^{\frac{n-i}{n}}_{i}\cdots
K^{\frac{n-(n-1)}{n}}_{n-1},\quad 1\leq i\leq n.$
$\displaystyle(m^{-})^{i+1}_{i}=(q-q^{-1})K^{\frac{1}{n}}_{1}K^{\frac{2}{n}}_{2}\cdots
K^{\frac{i-1}{n}}_{i-1}K^{\frac{i}{n}}_{i}K^{-\frac{n-(i+1)}{n}}_{i+1}\cdots
K^{-\frac{n-(n-1)}{n}}_{n-1}F_{i},\quad 1\leq i\leq n-1,$
$\displaystyle(m^{-})^{i}_{i}=K^{\frac{1}{n}}_{1}K^{\frac{2}{n}}_{2}\cdots
K^{\frac{i-1}{n}}_{i-1}K^{-\frac{n-i}{n}}_{i}\cdots
K^{-\frac{n-(n-1)}{n}}_{n-1},\quad 1\leq i\leq n.$

Obviously, we observe that $(m^{+})^{i}_{i}K_{i}^{-1}=(m^{+})^{i+1}_{i+1}$ for
$1\leq i\leq n-1$ by the above lemma. With these, we have the following

######  Theorem 3.1.

For type $A$, let $\lambda=q^{-\frac{n+1}{n}}$, and identify
$e^{n},f_{n},(m^{+})^{n}_{n}c^{-1}$ with the additional simple root vectors
$E_{n},F_{n}$ and group-like element $K_{n}$. Then the resulting quantum group
$U(V^{\vee}(R_{21}^{-1},R^{\prime}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{n})},V(R,R^{\prime}))$
is exactly the $U_{q}({\mathfrak{sl}}_{n+1})$ with $K_{i}^{\pm\frac{1}{n}}$
adjoined.

###### Proof.

$(m^{+})^{n}_{n}=K^{-\frac{1}{n}}_{1}K^{-\frac{2}{n}}_{2}\cdots
K^{-\frac{n-1}{n}}_{n-1}$ can be obtained by Lemma 3.2.
$[E_{n},F_{n}]=\frac{K_{n}-K_{n}^{-1}}{q-q^{-1}},$ $\Delta(E_{n})=E_{n}\otimes
K_{n}+1\otimes E_{n}$, and $\Delta(F_{n})=F_{n}\otimes 1+K_{n}^{-1}\otimes
F_{n}$ can be deduced easily from Corollary 2.1. On the other hand, we have
the following cross relations by Corollary 2.1

$\left.\begin{array}[]{l}(m^{-})^{i+1}_{i}=(q-q^{-1})(m^{-})^{i+1}_{i+1}F_{i},\\\
(m^{-})^{i+1}_{i}e^{n}=\lambda R^{n,i+1}_{n,i+1}e^{n}(m^{-})^{i+1}_{i},\\\
(m^{-})^{i+1}_{i+1}e^{n}=\lambda
R^{n,i+1}_{n,i+1}e^{n}(m^{-})^{i+1}_{i+1},\end{array}\right\\}\Longrightarrow
F_{i}e^{n}=e^{n}F_{i},\Longrightarrow[E_{n},F_{i}]=0,\ 1\leq i\leq n-1.$
$\left.\begin{array}[]{l}(m^{+})^{i}_{i+1}=(q-q^{-1})E_{i}(m^{+})^{i+1}_{i+1},\\\
(m^{+})^{i}_{i+1}f_{n}=\lambda f_{n}(m^{+})^{i}_{i+1}R^{i+1,n}_{i+1,n},\\\
(m^{+})^{i+1}_{i+1}f_{n}=\lambda
f_{n}(m^{+})^{i+1}_{i+1}R^{i+1,n}_{i+1,n},\end{array}\right\\}\Longrightarrow
E_{i}f_{n}=f_{n}E_{i},\Longrightarrow[E_{i},F_{n}]=0,\ 1\leq i\leq n-1.$

With these, we get

$None$
$[E_{n},F_{i}]=\delta_{ni}\frac{K_{n}-K_{n}^{-1}}{q-q^{-1}},\quad[E_{i},F_{n}]=\delta_{in}\frac{K_{n}-K_{n}^{-1}}{q-q^{-1}}.$

The relations between $E_{n},\,F_{n}$ and $K_{i}$, $1\leq i\leq n$ also can be
obtained by the cross relations in Corollary 2.1,

$None$
$\left.\begin{array}[]{l}(m^{+})^{i}_{i}K_{i}^{-1}=(m^{+})^{i+1}_{i+1},\ 1\leq
i\leq n-1,\\\ e^{n}(m^{+})^{j}_{j}=\lambda
R^{j}_{j}{}^{n}_{n}(m^{+})^{j}_{j}e^{n},\\\ R^{jn}_{jn}=q,\ 1\leq j\leq n-1,\
R^{nn}_{nn}=q^{2}.\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}E_{n}K_{j}=K_{j}E_{n},\
1\leq j\leq n-2,\\\ E_{n}K_{n-1}=q^{-1}K_{n-1}E_{n}.\end{array}\right.$ $None$
$\left.\begin{array}[]{l}(m^{+})^{i}_{i}K_{i}^{-1}=(m^{+})^{i+1}_{i+1},\ 1\leq
i\leq n-1,\\\ (m^{+})^{i}_{i}f_{n}=\lambda f_{n}(m^{+})^{i}_{i}R^{in}_{in},\\\
R^{jn}_{jn}=q,\ 1\leq j\leq n-1,\
R^{nn}_{nn}=q^{2}.\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}F_{n}K_{j}=K_{j}F_{n},\
1\leq j\leq n-2,\\\ F_{n}K_{n-1}=qK_{n-1}F_{n}.\end{array}\right.$

In the $U_{q}({\mathfrak{sl}}_{n})$, $E_{i}K_{i}=q^{2}K_{i}E_{i},E_{i}K_{i\pm
1}=q^{-1}K_{i\pm 1}E_{i},E_{i}K_{j}=K_{j}E_{i},j\neq i\pm
1;F_{i}K_{i}=q^{-2}K_{i}F_{i},F_{i}K_{i\pm 1}=qK_{i\pm
1}F_{i},F_{i}K_{j}=K_{j}F_{i},j\neq i\pm 1.$ Then we get the relations between
$K_{n}$ and $E_{i},F_{i}$, where$1\leq i\leq n-2$.

$None$
$\left\\{\begin{array}[]{l}E_{i}K_{n}=q^{\frac{i-1}{n}}q^{-\frac{2i}{n}}q^{\frac{i+1}{n}}K^{-\frac{1}{n}}_{1}\cdots
K^{-\frac{n-1}{n}}_{n-1}c^{-1}E_{i}=K_{n}E_{i},\\\
F_{i}K_{n}=q^{-\frac{i-1}{n}}q^{\frac{2i}{n}}q^{-\frac{i+1}{n}}K^{-\frac{1}{n}}_{1}\cdots
K^{-\frac{n-1}{n}}_{n-1}c^{-1}F_{i}=K_{n}F_{i},\\\
E_{n-1}K_{n}=q^{\frac{n-2}{n}}q^{-\frac{2(n-1)}{n}}K^{-\frac{1}{n}}_{1}\cdots
K^{-\frac{n-2}{n}}_{n-2}K^{-\frac{n-1}{n}}_{n-1}c^{-1}E_{n-1}=q^{-1}K_{n}E_{n-1},\\\
F_{n-1}K_{n}=q^{-\frac{n-2}{n}}q^{\frac{2(n-1)}{n}}K^{-\frac{1}{n}}_{1}\cdots
K^{-\frac{n-2}{n}}_{n-2}K^{-\frac{n-1}{n}}_{n-1}c^{-1}F_{n-1}=qK_{n}F_{n-1}.\end{array}\right.$

We want to explore the relations between $e^{n}$ and $E_{i},1\leq i\leq n-1,$
and observe that $E_{i}$ just belongs to the entry $(m^{+})^{i}_{i+1}$, so

$\left.\begin{array}[]{l}(m^{+})^{i}_{i+1}=(q-q^{-1})E_{i}(m^{+})^{i+1}_{i+1},\\\
e^{n}(m^{+})^{i}_{i+1}=\lambda R^{in}_{in}(m^{+})^{i}_{i+1}e^{n},1\leq i\leq
n-2,\\\ e^{n}(m^{+})^{n-1}_{n}=\lambda
R^{n-1,n}_{n-1,n}(m^{+})^{n-1}_{n}e^{n}+\lambda
R^{n-1,n}_{n,n-1}(m^{+})^{n}_{n}e^{n-1},\\\ e^{n}(m^{+})^{i+1}_{i+1}=\lambda
R^{i+1,n}_{i+1,n}(m^{+})^{i+1}_{i+1}e^{n},\\\ R^{jn}_{jn}=q,1\leq j\leq
n-1,R^{nn}_{nn}=q^{2},R^{n-1,n}_{n,n-1}=q^{2}-1.\end{array}\right\\}\Rightarrow\left\\{\begin{array}[]{l}E_{n}E_{i}=E_{i}E_{n},1\leq
i\leq n-2,\\\ e^{n-1}=e^{n}E_{n-1}-q^{-1}E_{n-1}e^{n}.\end{array}\right.$
$\left.\begin{array}[]{l}(m^{-})^{i+1}_{i}=(q-q^{-1})(m^{-})^{i+1}_{i+1}F_{i},\\\
f_{n}(m^{-})^{i+1}_{i}=\lambda R^{ni}_{ni}(m^{-})^{i+1}_{i}f_{n},1\leq i\leq
n-2,\\\ f_{n}(m^{-})^{n}_{n-1}=\lambda
R^{n,n-1}_{n,n-1}(m^{-})^{n}_{n-1}f_{n}+\lambda
R^{n-1,n}_{n,n-1}(m^{-})^{n}_{n}f_{n-1},\\\ f_{n}(m^{-})^{i+1}_{i+1}=\lambda
R^{n,i+1}_{n,i+1}(m^{-})^{i+1}_{i+1}f_{n},\\\
\end{array}\right\\}\Rightarrow\left\\{\begin{array}[]{l}F_{n}F_{i}=F_{i}F_{n},1\leq
i\leq n-2,\\\ f_{n-1}=qf_{n}F_{n-1}-F_{n-1}f_{n}.\end{array}\right.$

Note that
$e^{n-1}e^{n}=R^{\prime}{}^{n,n-1}_{ab}e^{a}e^{b}=q^{-2}R^{n,n-1}_{n,n-1}e^{n}e^{n-1}=q^{-1}e^{n}e^{n-1}$
and
$f_{n-1}f_{n}=f_{b}f_{a}R^{\prime}{}^{ab}_{n-1,n}=q^{-2}f_{n}f_{n-1}R^{n-1,n-1}_{n-1,n}=q^{-1}f_{n}f_{n-1}.$
Then we obtain

$None$
$\left\\{\begin{array}[]{l}(E_{n})^{2}E_{n-1}-(q+q^{-1})E_{n}E_{n-1}E_{n}+E_{n-1}(E_{n})^{2}=0,\\\
(F_{n})^{2}F_{n-1}-(q+q^{-1})F_{n}F_{n-1}F_{n}+F_{n-1}(F_{n})^{2}=0.\end{array}\right.$

On the other hand,

$\left.\begin{array}[]{l}(m^{+})^{n-1}_{n}=(q-q^{-1})E_{n-1}(m^{+})^{n}_{n},\\\
e^{n-1}(m^{+})^{n-1}_{n}=\lambda q^{2}(m^{+})^{n-1}_{n}e^{n-1},\\\
e^{n-1}(m^{+})^{n}_{n}=\lambda
q(m^{+})^{n}_{n}e^{n-1}.\end{array}\right\\}\Longrightarrow
e^{n-1}E_{n-1}=qE_{n-1}e^{n-1}.$
$\left.\begin{array}[]{l}(m^{-})^{n}_{n-1}=(q-q^{-1})(m^{-})^{n}_{n}F_{n-1},\\\
f_{n-1}(m^{-})^{n}_{n-1}=\lambda q^{2}(m^{-})^{n}_{n-1}f_{n-1},\\\
f_{n-1}(m^{-})^{n}_{n}=\lambda
q(m^{-})^{n}_{n}f_{n-1}.\end{array}\right\\}\Longrightarrow
f_{n-1}F_{n-1}=qF_{n-1}f_{n-1}.$

Combining with $e^{n-1}=e^{n}E_{n-1}-q^{-1}E_{n-1}e^{n}$ and
$f_{n-1}=qf_{n}F_{n-1}-F_{n-1}f_{n}$, we get

$None$
$\left\\{\begin{array}[]{l}(E_{n-1})^{2}E_{n}-(q+q^{-1})E_{n-1}E_{n}E_{n-1}+E_{n}(E_{n-1})^{2}=0,\\\
(F_{n-1})^{2}F_{n}-(q+q^{-1})F_{n-1}F_{n}F_{n-1}+F_{n}(F_{n-1})^{2}=0.\end{array}\right.$

The other elements $e^{i},f_{j}$ can be identified with non-simple root
vectors generated by $q$-commutators with $E_{i},F_{i},K_{i},1\leq i\leq n$.
With the above equalities $(A1)$—$(A6)$, we prove that the resulting quantum
groups is $U_{q}({\mathfrak{sl}}_{n+1})$. ∎

###  3.2. Inductive construction of $U_{q}(\mathfrak{g})$ for the $BCD$ series

The rank-inductive construction of $U_{q}({\mathfrak{sl}}_{n})$ gives us the
confidence to consider the $BCD$ series. Their Dynkin diagrams are given
respectively by the following diagrams, and the arrow is point to the shorter
of the two roots in the diagrams.

$<$$1$$2$$n-1$$n$$:B_{n}\ (n\geq 2)$

$>$$1$$2$$n-1$$n$$:C_{n}\ (n\geq 3)$

$1$$2$$3$$n-1$$n$$:D_{n}\ (n\geq 4)$

Corresponding to their vector representations, the matrix $PR_{VV}$ satisfies
the following cubic equation [FRT, KS]:

$(PR_{VV}+q^{-1}I)(PR_{VV}-qI)(PR_{VV}-\epsilon q^{\epsilon-N}I)=0.$

Then set $R=qR_{VV},R^{\prime}=RPR-(\epsilon
q^{\epsilon-N+1}+q^{2})R+(\epsilon q^{\epsilon-N+3}+1)P,$ we get
$(PR+I)(PR^{\prime}-I)=0$. Moreover, there is a unique formula for the matrix
entries of $R$,

$R^{ij}_{kl}=qq^{\delta_{ji}-\delta_{ji^{\prime}}}\delta_{ik}\delta_{jl}+(q^{2}-1)\theta(j-l)(\delta_{il}\delta_{jk}-K^{ij}_{lk}).$

Here $K^{ij}_{lk}=\epsilon C^{i}_{j}C^{l}_{k},$ and
$C^{m}_{t}=\epsilon_{m}\delta_{mt^{\prime}}q^{-\rho_{m}},$ where
$i^{\prime}=N+1-i$, let $N=2n$ if $N$ is even, and $N=2n+1$ if $N$ is odd.
$\rho_{i}=\frac{N}{2}-i$ if $i<i^{\prime}$; $\rho_{i^{\prime}}=-\rho_{i}$ if
$i\leq i^{\prime}$, $\epsilon=\epsilon_{1}=\cdots=\epsilon_{N}=1$ for
$g={\mathfrak{so}}_{N},$ and
$\rho_{i}=\frac{N}{2}+1-i,\rho_{i^{\prime}}=-\rho_{i}$ if $i<i^{\prime}$,
$\epsilon_{1}=\cdots=\epsilon_{n}=1,\epsilon=\epsilon_{n+1}=\cdots=\epsilon_{N}=-1$
for $g={\mathfrak{sp}}_{N}$.

For their vector representations, the diagonal and minor diagonal entries we
need in $m^{\pm}$ can be obtained by Lemma 3.1.

######  Lemma 3.3.

$(1)$ For $U_{q}^{ext}({\mathfrak{so}}_{2n+1}):$

$\displaystyle(m^{+})^{i}_{i}=K_{1}K_{2}\cdots
K_{n-i}K_{n+1-i},\quad(m^{+})^{n+1}_{n+1}=1,\quad 1\leq i\leq n,$
$\displaystyle(m^{+})^{i}_{i+1}=-(q{-}q^{-1})E_{n+1-i}K_{1}K_{2}\cdots
K_{n-i},\quad 1\leq i\leq n{-}1,\quad(m^{+})^{n}_{n+1}=-c_{0}E_{1},$
$\displaystyle(m^{-})^{i+1}_{i}=(q{-}q^{-1})K_{1}^{-1}K_{2}^{-1}\cdots
K_{n-i}^{-1}F_{n+1-i},\quad
m^{-(n+1)}_{n}=c_{0}F_{1},\quad(m^{+})^{i}_{i}(m^{-})^{i}_{i}=1,$

where
$c_{0}=(q^{\frac{1}{2}}+q^{-\frac{1}{2}})^{\frac{1}{2}}(q^{\frac{1}{2}}-q^{-\frac{1}{2}}).$

$(2)$ For $U_{q}^{ext}({\mathfrak{sp}}_{2n}):$

$\displaystyle(m^{+})^{i}_{i}=K_{1}^{\frac{1}{2}}K_{2}\cdots
K_{n+1-i},\quad(m^{+})^{i^{\prime}}_{i^{\prime}}(m^{+})^{i}_{i}=1,\quad 1\leq
i\leq n,$
$\displaystyle(m^{+})^{i}_{i+1}=-(q{-}q^{-1})E_{n+1-i}K_{1}^{\frac{1}{2}}K_{2}\cdots
K_{n-i},\quad 1\leq i\leq
n{-}1,\quad(m^{+})^{n}_{n+1}=-(q^{2}{-}q^{-2})E_{1}K_{1}^{-\frac{1}{2}},$
$\displaystyle(m^{-})^{i}_{i}=K_{1}^{-\frac{1}{2}}K_{2}^{-1}\cdots
K_{n+1-i}^{-1},\quad(m^{-})^{i^{\prime}}_{i^{\prime}}(m^{-})^{i}_{i}=1,\quad
1\leq i\leq n,$
$\displaystyle(m^{-})^{i+1}_{i}=(q{-}q^{-1})K_{1}^{-\frac{1}{2}}K_{2}^{-1}\cdots
K_{n-i}^{-1}F_{n+1-i},\quad 1\leq i\leq
n{-}1,\quad(m^{-})^{n+1}_{n}=(q^{2}{-}q^{-2})K_{1}^{\frac{1}{2}}F_{1}.$

$(3)$ For $U_{q}^{ext}({\mathfrak{so}}_{2n}):$

$\displaystyle(m^{+})^{i}_{i}=(K_{1}^{\frac{1}{2}}K_{2}^{\frac{1}{2}})K_{3}\cdots
K_{n+1-i},\quad(m^{+})^{i^{\prime}}_{i^{\prime}}(m^{+})^{i}_{i}=1,\quad 1\leq
i\leq n{-}2,$
$\displaystyle(m^{+})^{n-1}_{n-1}=K_{1}^{\frac{1}{2}}K_{2}^{\frac{1}{2}},\quad(m^{+})^{n}_{n}=K_{1}^{\frac{1}{2}}K_{2}^{-\frac{1}{2}},\quad(m^{+})^{n-1}_{n+1}=-(q{-}q^{-1})E_{1}(K_{1}^{-\frac{1}{2}}K_{2}^{\frac{1}{2}}),$
$\displaystyle(m^{+})^{i}_{i+1}=-(q{-}q^{-1})E_{n+1-i}(K_{1}^{\frac{1}{2}}K_{2}^{\frac{1}{2}})K_{3}\cdots
K_{n-i},\quad 1\leq i\leq n{-}1,$
$\displaystyle(m^{-})^{i}_{i}=(K_{1}^{-\frac{1}{2}}K_{2}^{-\frac{1}{2}})K_{3}^{-1}\cdots
K_{n+1-i}^{-1},\quad(m^{-})^{i^{\prime}}_{i^{\prime}}(m^{-})^{i}_{i}=1,\quad
1\leq i\leq n{-}2,$
$\displaystyle(m^{-})^{n-1}_{n-1}=K_{1}^{-\frac{1}{2}}K_{2}^{-\frac{1}{2}},\quad(m^{-})^{n}_{n}=K_{1}^{-\frac{1}{2}}K_{2}^{\frac{1}{2}},\quad(m^{-})^{n+1}_{n-1}=(q{-}q^{-1})(K_{1}^{\frac{1}{2}}K_{2}^{-\frac{1}{2}})F_{1},$
$\displaystyle(m^{-})^{i+1}_{i}=(q{-}q^{-1})(K_{1}^{-\frac{1}{2}}K_{2}^{-\frac{1}{2}})K_{3}^{-1}\cdots
K_{n-i}^{-1}F_{n+1-i},\quad 1\leq i\leq n{-}1.$

With these, we have the following

######  Theorem 3.2.

With quantum group normalization constant $\lambda=q^{-1}$.

$(1)$ Type $B$

Identify $e^{2n+1},f_{2n+1},(m^{+})^{2n+1}_{2n+1}c^{-1}$ with the additional
simple root vectors $E_{n+1}$, $F_{n+1}$ and the group-like element $K_{n+1}$.
Then the resulting quantum group
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{so}}_{2n+1})}$,
$V(R^{\prime},R))$ is the quantum group $U_{q}({\mathfrak{so}}_{2n+3})$.

$(2)$ Type $C$

Identify $e^{2n},f_{2n},(m^{+})^{2n}_{2n}c^{-1}$ with the additional simple
root vectors $E_{n+1},F_{n+1}$ and the group-like element $K_{n+1}$. Then the
resulting quantum group
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sp}}_{2n})},V(R^{\prime},R))$
is the quantum group $U_{q}({\mathfrak{sp}}_{2n+2})$ with
$K_{1}^{\pm\frac{1}{2}}$ adjoined.

$(3)$ Type $D$

Identify $e^{2n},f_{2n},(m^{+})^{2n}_{2n}c^{-1}$ with the additional simple
root vectors $E_{n+1},F_{n+1}$ and the group-like element $K_{n+1}$. Then the
resulting quantum group
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{so}}_{2n})},V(R^{\prime},R))$
is the quantum group $U_{q}({\mathfrak{so}}_{2n+2})$ with
$K_{1}^{\pm\frac{1}{2}}K_{2}^{\pm\frac{1}{2}}$,
$K_{1}^{\pm\frac{1}{2}}K_{2}^{\mp\frac{1}{2}}$ adjoined.

###### Proof.

The proof of Theorem 3.1 means that the relations of negative part can be
obtained in a similar way, so we only focus on the relations of the positive
part.

$(1)$ $(m^{+})^{2n+1}_{2n+1}c^{-1}=K_{1}^{-1}\cdots K_{n-1}^{-1}K_{n}^{-1}$
follows from Lemma 3.3. From the identification in the above theorem, it is
easily deduced from Corollary 2.1 that
$[E_{n+1},F_{n+1}]=\frac{K_{n+1}-K_{n+1}^{-1}}{q-q^{-1}}$,
$\Delta(E_{n+1})=E_{n+1}\otimes K_{n+1}+1\otimes E_{n+1}$, and
$\Delta(F_{n+1})=F_{n+1}\otimes 1+K_{n+1}^{-1}\otimes F_{n+1}$.

On the other hand, we have
$E_{n+1}K_{n+1}=e^{2n+1}(m^{+})^{2n+1}_{2n+1}c^{-1}=\lambda
R^{2n+1}_{a}{}^{2n+1}_{b}(m^{+})^{a}_{2n+1}e^{b}c^{-1}$
$=R^{2n+1}_{2n+1}{}^{2n+1}_{2n+1}(m^{+})^{2n+1}_{2n+1}c^{-1}e^{2n+1}=q^{2}K_{n+1}E_{n+1}.$
The relations between the additional simple root vector $e^{2n+1}$ and other
$K_{i},1\leq i\leq n$ can be deduced from $e^{2n+1}(m^{+})^{p}_{p}=\lambda
R^{p,2n+1}_{ab}(m^{+})^{a}_{p}e^{b}$ $=\lambda
R^{p,2n+1}_{p,2n+1}(m^{+})^{p}_{p}e^{2n+1},\ 1\leq p\leq n+1$. Combining with
$(m^{+})^{i+1}_{i+1}K_{n+1-i}=(m^{+})^{i}_{i},\ 1\leq i\leq n,$ and
$R^{1}_{1}{}^{2n+1}_{2n+1}=1,\ R^{p}_{p}{}^{2n+1}_{2n+1}=q,\ 2\leq p\leq n+1,$
we obtain $e^{2n+1}K_{n}=q^{-1}K_{n}e^{2n+1},\ e^{2n+1}K_{i}=K_{i}e^{2n+1}$,
for $1\leq i\leq n-1,$ namely,

$E_{n+1}K_{n}=q^{-1}K_{n}E_{n+1},\quad E_{n+1}K_{i}=K_{i}E_{n+1},\quad 1\leq
i\leq n-1.$

We observe that $F_{i}$ belongs to $(m^{-})^{i+1}_{i}$, so the relations
between $E_{n+1}$ and $F_{i}$ can be obtained by the equality
$(m^{-})^{i}_{j}e^{k}=\lambda R^{k}_{a}~{}{}^{i}_{b}e^{a}(m^{-})^{b}_{j}$ in
Corollary 2.1.

$\left.\begin{array}[]{l}(m^{-})^{i+1}_{i}=(q-q^{-1})(m^{-})^{i+1}_{i+1}F_{i},\
1\leq i\leq n-1,\\\ (m^{-})^{i+1}_{i}e^{2n+1}=\lambda
R^{2n+1}_{a}{}^{i+1}_{b}e^{a}(m^{-})^{b}_{i}=e^{2n+1}(m^{-})^{i+1}_{i},1\leq
i\leq n,\\\ (m^{-})^{p}_{p}e^{2n+1}=\lambda
R^{2n+1}_{a}{}^{p}_{b}e^{a}(m^{-})^{b}_{p}=e^{2n+1}(m^{-})^{p}_{p},\ 2\leq
p\leq n,\end{array}\right\\}\Longrightarrow[E_{n+1},F_{i}]=0,\ 1\leq i\leq n.$

We will explore the $q$-Serre relations of the positive part. We also observe
that $E_{n+1-i}$ belongs to $(m^{+})^{i}_{i+1},\ 1\leq i\leq n,$ so

$\left\\{\begin{array}[]{l}e^{2n+1}(m^{+})^{1}_{2}=\lambda
R^{1}_{a}{}{}^{2n+1}_{b}(m^{+})^{a}_{2}e^{b}=\lambda
R^{1}_{1}{}{}^{2n+1}_{2n+1}(m^{+})^{1}_{2}e^{2n+1}+\lambda
R^{1}_{2}{}{}^{2n+1}_{2n}(m^{+})^{2}_{2}e^{2n},\\\
e^{2n+1}(m^{+})^{i}_{i+1}=\lambda
R^{i}_{a}{}{}^{2n+1}_{b}(m^{+})^{a}_{i+1}e^{b}=\lambda
R^{i}_{i}{}{}^{2n+1}_{2n+1}(m^{+})^{i}_{i+1}e^{2n+1}=\lambda
q(m^{+})^{i}_{i+1}e^{2n+1},\ 2\leq i\leq n.\end{array}\right.$

Putting the expression of $(m^{+})^{i}_{i+1},\ 1\leq i\leq n$ into the above
equalities, we get

$\left\\{\begin{array}[]{l}e^{2n+1}E_{j}=E_{j}e^{2n+1},\ 1\leq j\leq n-1,\\\
e^{2n}=e^{2n+1}E_{n}-q^{-1}E_{n}e^{2n+1}.\end{array}\right.$

So, we need to know the relations between $e^{2n}$ and $e^{2n+1},E_{n}$. We
have $e^{2n+1}e^{2n}=R^{\prime}{}^{2n}_{a}{}^{2n+1}_{b}e^{a}e^{b}=-(q+\epsilon
q^{\epsilon-N+2})e^{2n}e^{2n+1}+(\epsilon q^{\epsilon-N+1}+2)e^{2n+1}e^{2n}$,
so $e^{2n+1}e^{2n}=qe^{2n}e^{2n+1}$. Combining with
$e^{2n}=e^{2n+1}E_{n}-q^{-1}E_{n}e^{2n+1},$ we get
$(e^{2n+1})^{2}E_{n}-(q+q^{-1})e^{2n+1}E_{n}e^{2n+1}+E_{n}(e^{2n+1})^{2}=0,$
namely,

$(E_{n+1})^{2}E_{n}-(q+q^{-1})E_{n+1}E_{n}E_{n+1}+E_{n}(E_{n+1})^{2}=0.$

On the other hand, according to the equality $e^{2n}(m^{+})^{1}_{2}=\lambda
R^{1}_{a}{}^{2n}_{b}(m^{+})^{a}_{2}e^{b}=\lambda
R^{1}_{1}{}^{2n}_{2n}(m^{+})^{1}_{2}e^{2n}=(m^{+})^{1}_{2}e^{2n}$ and
$e^{2n}(m^{+})^{2}_{2}=\lambda
R^{2}_{2}{}^{2n}_{2n}(m^{+})^{2}_{2}e^{2n}=q^{-1}(m^{+})^{2}_{2}e^{2n},$ we
get $e^{2n}E_{n}=qE_{n}e^{2n}$. Combining with
$e^{2n}=e^{2n+1}E_{n}-q^{-1}E_{n}e^{2n+1}$ again, we obtain

$(E_{n})^{2}E_{n+1}-(q+q^{-1})E_{n}E_{n+1}E_{n}+E_{n+1}(E_{n})^{2}=0.$

With these relations, we prove that the resulting quantum group is
$U_{q}({\mathfrak{so}}_{2n+3})$. $(2)$ and $(3)$ can be proved in a similar
way. ∎

##  4\. Type-crossing constructions of types $BCD$ starting from type $A$

Up to now, we have got the general inductive constructions of the classical
quantum groups within the same type as in the above section. However, Majid
claimed that his double-bosonization allows to create not only a line of nodes
diagram but also a tree of nodes diagram of quantum groups. At each node of
the tree, we have more choices to adjoin suitable braided groups to obtain
possible different new quantum groups of higher rank one. In this section, we
will give some examples to demonstrate this fact. As we known, at the source
node corresponding to $U_{q}({\mathfrak{sl}}_{2})$, Majid [M3] chose a pair of
braided groups generated by the vector representation of
$U_{q}({\mathfrak{sl}}_{2})$ to give $U_{q}({\mathfrak{sl}}_{3})$ as above. In
what follows, we will give $3$ kinds of type-crossing constructions: from type
$A_{1}$ to type $B_{2}$, from type $A_{2}$ to type $C_{3}$, and from type
$A_{3}$ to type $D_{4}$.

Corresponding to an irreducible representation of $U_{q}(\mathfrak{g})$, the
matrix $R_{VV}$ is induced by $R_{VV}=B_{VV}\circ(T_{V}\otimes
T_{V})(\mathfrak{R})$. Here
$\mathfrak{R}=\sum\limits_{r_{1},\cdots,r_{n}=0}^{\infty}\prod\limits_{j=1}^{n}\frac{(1-q_{\beta_{j}}^{-2})^{r_{j}}}{[r_{j}]_{q_{\beta_{j}}}!}q_{\beta_{j}}^{\frac{r_{j}(r_{j}+1)}{2}}E_{\beta_{j}}^{r_{j}}\otimes
F_{\beta_{j}}^{r_{j}}$ is the main part of the universal $R$-matrix of
$U_{h}(\mathfrak{g})$. $B_{VV}$ denotes the linear operator on $V\otimes V$
given by $B_{VV}(v\otimes w):=q^{(\mu,\mu^{\prime})}v\otimes w$ for $v\in
V_{\mu},$ $w\in V_{\mu^{\prime}}$.

###  4.1. Type-crossing construction from type $A_{1}$ to type $B_{2}$

Now, we still start from the node diagram of $U_{q}({\mathfrak{sl}}_{2})$ and
choose other braided groups to give $U_{q}({\mathfrak{so}}_{5})$ in the
following.

######  Example 4.1.

Starting from a $3$-dimensional representation $T_{V}$ of
$U_{q}({\mathfrak{sl}}_{2})$, which is given by
$E_{1}(x_{i})=[2]_{q}x_{i+1},E_{1}(x_{3})=0,F_{1}(x_{i+1})=x_{i},F_{1}(x_{1})=0,i=1,2,$
where $x_{1},x_{2},x_{3}$ is the base of $V$ with corresponding weights
$-\alpha_{1},0,\alpha_{1}$. Then we get a $9\times 9$ $R_{VV}$-matrix datum

$R_{VV}=\left(\begin{array}[]{ccccccccc}q^{2}&0&0&0&0&0&0&0&0\\\
0&1&0&q^{2}{-}q^{-2}&0&0&0&0&0\\\
0&0&q^{-2}&0&q^{2}{-}q^{-2}&0&(1{-}q^{-2})(q^{2}{-}q^{-2})&0&0\\\
0&0&0&1&0&0&0&0&0\\\ 0&0&0&0&1&0&1{-}q^{-4}&0&0\\\
0&0&0&0&0&1&0&q^{2}{-}q^{-2}&0\\\ 0&0&0&0&0&0&q^{-2}&0&0\\\
0&0&0&0&0&0&0&1&0\\\ 0&0&0&0&0&0&0&0&q^{2}\end{array}\right).$

Clearly, $R_{VV}$ is invertible, and according to the submodules decomposition
of module $V^{\otimes 2}$, it is easy to see that $PR_{VV}$ obeys the minimal
polynomial

$(PR_{VV}+q^{-2}I)(PR_{VV}-q^{2}I)(PR_{VV}-q^{-4}I)=0.$

Setting $R=q^{2}R_{VV},R^{\prime}=RPR-q^{-2}R-q^{4}R+(q^{2}+1)P$, then

$(PR+I)(PR^{\prime}-I)=0.$

With these $R$ and $R^{\prime}$, choose a pair of braided groups
$V^{\vee}(R^{\prime},R_{21}^{-1}),V(R^{\prime},R)$, and $\lambda=q^{-2}$.
Identify $e^{3},f_{3},(m^{+})^{3}_{3}c^{-1}$ with the additional simple root
vectors $E_{2},F_{2}$ and the group-like element $K_{2}$. Then the resulting
quantum group $U(V^{\vee}(R^{\prime},R_{21}^{-1})$,
$\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{2})},V(R^{\prime},R))$ is the quantum
group $U_{q}({\mathfrak{so}}_{5})$.

###### Proof.

Corresponding to this representation, we get the $m^{\pm}$-matrix as follows

$m^{+}=\left(\begin{array}[]{ccc}K_{1}&-(q{-}q^{-1})E_{1}&\frac{q(1{-}q^{-2})^{2}}{[2]_{q}}E_{1}^{2}K_{1}^{-1}\\\
0&1&-(q{-}q^{-1})E_{1}K_{1}^{-1}\\\ 0&0&K_{1}^{-1}\end{array}\right),$
$m^{-}=\left(\begin{array}[]{ccc}K_{1}^{-1}&0&0\\\ (q^{4}{-}1)F_{1}&1&0\\\
(q^{4}{-}q^{2})(q^{4}{-}1)K_{1}F_{1}^{2}&(q^{2}{-}q^{-2})K_{1}F_{1}&K_{1}\end{array}\right).$

With the identification as above,
$[E_{2},F_{2}]=\frac{K_{2}-K_{2}^{-1}}{q^{2}-q^{-2}},$
$\Delta(E_{2})=E_{2}\otimes K_{2}+1\otimes E_{2}$, and
$\Delta(F_{2})=F_{2}\otimes 1+K_{2}^{-1}\otimes F_{2}$ can be deduced from
Corollary 2.1.

$E_{2}K_{2}=e^{3}(m^{+})^{3}_{3}c^{-1}=\lambda
R^{3}_{3}{}^{3}_{3}(m^{+})^{3}_{3}e^{3}c^{-1}=R^{3}_{3}{}^{3}_{3}(m^{+})^{3}_{3}c^{-1}e^{3}=q^{4}K_{2}E_{2},$
$E_{2}K_{1}=q^{-2}K_{1}E_{2}$ can be deduced from
$e^{3}(m^{+})^{1}_{1}=\lambda
R^{1}_{1}{}^{3}_{3}(m^{+})^{1}_{1}e^{3}=q^{-2}(m^{+})^{1}_{1}e^{3}.$ On the
other hand,
$E_{1}K_{2}=E_{1}(m^{+})^{3}_{3}c^{-1}=E_{1}K_{1}^{-1}c^{-1}=q^{-2}K_{1}^{-1}c^{-1}E_{1}=q^{-2}K_{2}E_{1}$.
According to the equality $(m^{-})^{2}_{1}e^{3}=\lambda
R^{3}_{3}{}^{2}_{2}e^{3}(m^{-})^{2}_{1}=e^{3}(m^{-})^{2}_{1}$, we obtain
$e^{3}F_{1}=F_{1}e^{3}$ combining with $(m^{-})^{2}_{1}=(q^{4}-1)F_{1}$,
namely, $[E_{2},F_{1}]=0.$

We will explore the $q$-Serre relation between $e^{3}$ and $E_{1}$. The
equality $(q+q^{-1})e^{2}=q^{-2}E_{1}e^{3}-e^{3}E_{1}$ can be given by
$e^{3}(m^{+})^{1}_{2}=\lambda
R^{13}_{ab}(m^{+})^{a}_{2}e^{b}=q^{-2}(m^{+})^{1}_{2}e^{3}+(q^{2}-q^{-2})(m^{+})^{2}_{2}e^{2}.$
Combining with $e^{3}e^{2}=q^{2}e^{2}e^{3}$, which is deduced from
$e^{2}e^{3}=R^{\prime}{}^{32}_{ab}e^{a}e^{b}=(q^{4}+q^{2}+1)e^{2}e^{3}-(q^{2}+1)e^{3}e^{2}$,
we obtain

$(E_{2})^{2}E_{1}-(q^{2}+q^{-2})E_{2}E_{1}E_{2}+E_{1}(E_{2})^{2}=0.$

On the other hand, we need to know the relation between $e^{2}$ and $E_{1}$.

$\left.\begin{array}[]{l}e^{2}(m^{+})^{1}_{2}=\lambda
R^{12}_{ab}(m^{+})^{a}_{2}e^{b}=(m^{+})^{1}_{2}e^{2}+(q^{2}{-}q^{-2})(m^{+})^{2}_{2}e^{1},\\\
e^{1}(m^{+})^{1}_{2}=\lambda R^{11}_{ab}(m^{+})^{a}_{2}e^{b}=\lambda
R^{11}_{11}(m^{+})^{1}_{2}e^{1}=q^{2}(m^{+})^{1}_{2}e^{1}.\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}e^{1}=\frac{1}{q+q^{-1}}(E_{1}e^{2}-e^{2}E_{1}),\\\
e^{1}E_{1}=q^{2}E_{1}e^{1}.\end{array}\right.$

Combining with $(q+q^{-1})e^{2}=q^{-2}E_{1}e^{3}-e^{3}E_{1}$, we obtain

$(E_{1})^{3}E_{2}-\left[\begin{array}[]{c}3\\\
1\end{array}\right]_{q}(E_{1})^{2}E_{2}E_{1}+\left[\begin{array}[]{c}3\\\
2\end{array}\right]_{q}E_{1}E_{2}(E_{1})^{2}-E_{2}(E_{1})^{3}=0.$

With these relations, the Cartan matrix of the resulting quantum group is
$\left(\begin{array}[]{cc}2&-2\\\ -1&2\end{array}\right)$.

So the resulting quantum group is $U_{q}({\mathfrak{so}}_{5})$, and the proof
is complete. ∎

###  4.2. Type-crossing construction from type $A_{2}$ to type $C_{3}$

In Example 3.1, we get $U_{q}({\mathfrak{sl}}_{4})$ starting from
$U_{q}({\mathfrak{sl}}_{3})$ by choosing the braided groups generated by a
$3$-dimensional vector representation via the Majid’s double-bosonization
construction. In the example below, we will choose another pair of braided
groups generated by a $6$-dimensional irreducible module to give
$U_{q}({\mathfrak{sp}}_{6})$ starting from the node diagram of
$U_{q}({\mathfrak{sl}}_{3})$.

######  Example 4.2.

The pair of braided groups we want to have is obtained from the
$6$-dimensional irreducible representation $T_{V}$ of
$U_{q}({\mathfrak{sl}}_{3})$, which is defined by

$E_{1}\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\
x_{3}\end{array}\right)=\left(\begin{array}[]{c}x_{2}\\\ (q+q^{-1})x_{4}\\\
x_{5}\end{array}\right)$ , $E_{2}\left(\begin{array}[]{c}x_{2}\\\ x_{4}\\\
x_{5}\end{array}\right)=\left(\begin{array}[]{c}x_{3}\\\ x_{5}\\\
(q+q^{-1})x_{6}\end{array}\right),$

$F_{1}\left(\begin{array}[]{c}x_{2}\\\ x_{4}\\\
x_{5}\end{array}\right)=\left(\begin{array}[]{c}(q+q^{-1})x_{1}\\\ x_{2}\\\
x_{3}\end{array}\right)$ , $F_{2}\left(\begin{array}[]{c}x_{3}\\\ x_{5}\\\
x_{6}\end{array}\right)=\left(\begin{array}[]{c}x_{2}\\\ (q+q^{-1})x_{4}\\\
x_{5}\end{array}\right),$

where $\\{\,x_{i}\mid 1\leq i\leq 6\,\\}$ is a basis of $V$ with the
corresponding weights
$-2\lambda_{1},-2\lambda_{1}+\alpha_{1},-2\lambda_{1}+\alpha_{1}+\alpha_{2},-2\lambda_{1}+2\alpha_{1},-2\lambda_{1}+2\alpha_{1}+\alpha_{2},-2\lambda_{1}+2\alpha_{1}+2\alpha_{2}$,
respectively. We can obtain a $36\times 36$ matrix $R_{VV}$ corresponding to
the $6$-dimensional representation, and the $PR_{VV}$ obeys the minimal
polynomial

$(PR_{VV}-q^{\frac{8}{3}}I)(PR_{VV}+q^{-\frac{4}{3}}I)(PR_{VV}-q^{-\frac{10}{3}}I)=0.$

Setting $R=q^{\frac{4}{3}}R_{VV},R^{\prime}=RPR-(q^{-2}+q^{4})R+(q^{2}+1)P$,
we have

$(PR+I)(PR^{\prime}-I)=0,$

and $\lambda=q^{-\frac{4}{3}}$. Identify $e^{6},f_{6},(m^{+})^{6}_{6}c^{-1}$
with the additional simple root vectors $E_{3},F_{3}$ and the group-like
element $K_{3}$, then the resulting quantum group
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{3})},V(R^{\prime},R))$
is the quantum group $U_{q}({\mathfrak{sp}}_{6})$ with $K_{i}^{\frac{1}{3}}\
(i=1,2)$ adjoined.

###### Proof.

The elements in matrix $m^{\pm}$ we need can be obtained by Lemma 3.1, which
are given by the following equalities

$\left\\{\begin{array}[]{l}(m^{+})^{1}_{2}=-(q-q^{-1})E_{1}K_{1}^{\frac{1}{3}}K_{2}^{\frac{2}{3}},\quad(m^{+})^{2}_{2}=K_{1}^{\frac{1}{3}}K_{2}^{\frac{2}{3}},\\\
(m^{+})^{5}_{6}=-(q^{2}-q^{-2})E_{2}K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}},\quad(m^{+})^{6}_{6}=K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}},\quad(m^{+})^{4}_{4}=K_{1}^{-\frac{2}{3}}K_{2}^{\frac{2}{3}},\\\
(m^{-})^{5}_{3}=q(q-q^{-1})K_{1}^{\frac{2}{3}}K_{2}^{\frac{1}{3}}F_{1},\quad(m^{-})^{5}_{5}=K_{1}^{\frac{2}{3}}K_{2}^{\frac{1}{3}},\\\
(m^{-})^{6}_{5}=(q-q^{-1})K_{1}^{\frac{2}{3}}K_{2}^{\frac{4}{3}}F_{2},\quad(m^{-})^{6}_{6}=K_{1}^{\frac{2}{3}}K_{2}^{\frac{4}{3}}.\end{array}\right.$

We only focus on the relations of the positive part.

Note that $E_{3}K_{3}=e^{6}(m^{+})^{6}_{6}c^{-1}=\lambda
R^{66}_{66}(m^{+})^{6}_{6}e^{6}c^{-1}$
$=R^{66}_{66}(m^{+})^{6}_{6}c^{-1}e^{6}=q^{4}(m^{+})^{6}_{6}c^{-1}e^{6}=q^{4}K_{3}E_{3}.$
Combining with $e^{6}(m^{+})^{i}_{i}=\lambda
R^{i}_{i}{}^{6}_{6}(m^{+})^{i}_{i}e^{6}$, we have

$\left.\begin{array}[]{l}(m^{+})^{6}_{6}K_{2}^{2}=(m^{+})^{4}_{4},\\\
e^{6}(m^{+})^{6}_{6}=q^{\frac{8}{3}}(m^{+})^{6}_{6}e^{6},\\\
e^{6}(m^{+})^{4}_{4}=q^{-\frac{4}{3}}(m^{+})^{4}_{4}e^{6}.\end{array}\right\\}\Longrightarrow\left\\{\begin{array}[]{l}e^{6}K_{1}=K_{1}e^{6},\\\
e^{6}K_{2}=q^{-2}K_{2}e^{6}.\end{array}\right.\Longrightarrow\left\\{\begin{array}[]{l}E_{3}K_{1}=K_{1}E_{3},\\\
E_{3}K_{2}=q^{-2}K_{2}E_{3}.\end{array}\right.$

On the other hand,
$E_{2}K_{3}=E_{2}K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}}c^{-1}=q^{\frac{2}{3}}q^{-\frac{8}{3}}K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}}c^{-1}E_{2}=q^{-2}K_{3}E_{2},$
$E_{1}K_{3}=E_{1}K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}}c^{-1}=q^{-\frac{4}{3}}q^{\frac{4}{3}}K_{1}^{-\frac{2}{3}}K_{2}^{-\frac{4}{3}}c^{-1}E_{1}=K_{3}E_{1}.$
We will explore the $q$-Serre relations. Since $E_{1},E_{2}$ belong to
$(m^{+})^{1}_{2},\ (m^{+})^{5}_{6}$, respectively, then

$\left.\begin{array}[]{l}(m^{+})^{1}_{2}=-(q-q^{-1})E_{1}(m^{+})^{2}_{2},\\\
e^{6}(m^{+})^{1}_{2}=q^{-\frac{4}{3}}(m^{+})^{1}_{2}e^{6},\\\
e^{6}(m^{+})^{2}_{2}=q^{-\frac{4}{3}}(m^{+})^{2}_{2}e^{6}.\end{array}\right\\}\Longrightarrow
e^{6}E_{1}=E_{1}e^{6},\Longrightarrow E_{3}E_{1}=E_{1}E_{3}.$
$\left.\begin{array}[]{l}(m^{+})^{5}_{6}=-(q^{2}-q^{-2})E_{2}(m^{+})^{6}_{6},\\\
e^{6}(m^{+})^{5}_{6}=q^{\frac{2}{3}}(m^{+})^{5}_{6}e^{6}+(q^{2}-q^{-2})q^{\frac{2}{3}}(m^{+})^{6}_{6}e^{5},\\\
e^{6}(m^{+})^{6}_{6}=q^{\frac{8}{3}}(m^{+})^{6}_{6}e^{6}.\end{array}\right\\}\Longrightarrow
e^{5}=q^{-\frac{4}{3}}(q^{-2}E_{2}e^{6}-e^{6}E_{2}).$

So we need to know the relation between $e^{5}$ and $e^{6}$.
$e^{6}e^{5}=q^{2}e^{5}e^{6}$ is obtained by $e^{i}e^{j}=R^{\prime
ji}_{ab}e^{a}e^{b}$ and
$R^{\prime}{}^{5}_{5}{}^{6}_{6}=-q^{2}-1,R^{\prime}{}^{5}_{6}{}^{6}_{5}=2+q^{-2}$,
then we have

$(E_{3})^{2}E_{2}-(q^{2}+q^{-2})E_{3}E_{2}E_{3}+E_{2}(E_{3})^{2}.$

On the other hand, we need to obtain the relation between $e^{5}$ and $E_{2}$.

$\left.\begin{array}[]{l}(m^{+})^{5}_{6}=-(q^{2}-q^{-2})E_{2}(m^{+})^{6}_{6},\\\
e^{5}(m^{+})^{5}_{6}=q^{\frac{2}{3}}(m^{+})^{5}_{6}e^{5}+(q-q^{-1})q^{-\frac{4}{3}}(m^{+})^{6}_{6}e^{4},\\\
e^{4}(m^{+})^{6}_{6}=q^{-\frac{4}{3}}(m^{+})^{6}_{6}e^{4}.\end{array}\right\\}\Longrightarrow
e^{4}=-(q+q^{-1})(E_{2}e^{5}-e^{5}E_{2}).$

Then the relation between $e^{5}$ and $E_{2}$ must be deduced from the
relation between $e^{4}$ and $E_{2}$, which is given by the following cross
relations

$\left.\begin{array}[]{l}(m^{+})^{5}_{6}=-(q^{2}{-}q^{-2})E_{2}(m^{+})^{6}_{6},\\\
e^{4}(m^{+})^{5}_{6}=q^{\frac{2}{3}}(m^{+})^{5}_{6}e^{4},\\\
e^{4}(m^{+})^{6}_{6}=q^{-\frac{4}{3}}(m^{+})^{6}_{6}e^{4}.\end{array}\right\\}\Longrightarrow
e^{4}E_{2}=q^{2}E_{2}e^{4}\Longrightarrow(E_{2}e^{5}-e^{5}E_{2})E_{2}=q^{2}E_{2}(E_{2}e^{5}-e^{5}E_{2}).$

Combining with $e^{5}=q^{-\frac{4}{3}}(q^{-2}E_{2}e^{6}-e^{6}E_{2})$, we
obtain
$(E_{2})^{3}e^{6}-(q^{2}+1+q^{-2})(E_{2})^{2}e^{6}E_{2}+(q^{2}+1+q^{-2})E_{2}e^{6}(E_{2})^{2}-e^{6}(E_{2})^{3}=0,$
namely,

$(E_{2})^{3}E_{3}-\left[\begin{array}[]{c}3\\\
1\end{array}\right]_{q}(E_{2})^{2}E_{3}E_{2}+\left[\begin{array}[]{c}3\\\
2\end{array}\right]_{q}E_{2}E_{3}(E_{2})^{2}-E_{3}(E_{2})^{3}=0.$

From these relations, it follows that the resulting quantum group is
$U_{q}({\mathfrak{sp}}_{6})$. ∎

###  4.3. Type-crossing construction from type $A_{3}$ to type $D_{4}$

In the following example, we can choose the different braided groups generated
by a $6$-dimensional $U_{q}({\mathfrak{sl}}_{4})$-module to give
$U_{q}({\mathfrak{so}}_{8})$ based on the node diagram of
$U_{q}({\mathfrak{sl}}_{4})$.

######  Example 4.3.

There is a $6$-dimensional irreducible representation $T_{V}$ of
$U_{q}({\mathfrak{sl}}_{4})$, given by

$\displaystyle E_{1}\left(\begin{array}[]{c}x_{2}\\\
x_{3}\end{array}\right)=\left(\begin{array}[]{c}x_{4}\\\
x_{5}\end{array}\right),\quad E_{2}\left(\begin{array}[]{c}x_{1}\\\
x_{5}\end{array}\right)=\left(\begin{array}[]{c}x_{2}\\\
x_{6}\end{array}\right),\quad E_{3}\left(\begin{array}[]{c}x_{2}\\\
x_{4}\end{array}\right)=\left(\begin{array}[]{c}x_{3}\\\
x_{5}\end{array}\right);$ $\displaystyle F_{1}\left(\begin{array}[]{c}x_{4}\\\
x_{5}\end{array}\right)=\left(\begin{array}[]{c}x_{2}\\\
x_{3}\end{array}\right),\quad F_{2}\left(\begin{array}[]{c}x_{2}\\\
x_{6}\end{array}\right)=\left(\begin{array}[]{c}x_{1}\\\
x_{5}\end{array}\right),\quad F_{3}\left(\begin{array}[]{c}x_{3}\\\
x_{5}\end{array}\right)=\left(\begin{array}[]{c}x_{2}\\\
x_{4}\end{array}\right).$

Here $\\{x_{i}\mid 1\leq i\leq 6\\}$ is a basis of $V$ with corresponding
weights
$-2\lambda_{1}+\alpha_{1},-2\lambda_{1}+\alpha_{1}+\alpha_{2},-2\lambda_{1}+\alpha_{1}+\alpha_{2}+\alpha_{3},-2\lambda_{1}+2\alpha_{1}+\alpha_{2},-2\lambda_{1}+2\alpha_{1}+\alpha_{2}+\alpha_{3},-2\lambda_{1}+2\alpha_{1}+2\alpha_{2}+\alpha_{3}$,
respectively. Then we get another $36\times 36$ matrix $R_{VV}$, and the
$PR_{VV}$ matrix obeys the minimal polynomial

$(PR_{VV}+q^{-1}I)(PR_{VV}-q^{-1}I)(PR_{VV}-qI)=0.$

Setting $R=qR_{VV}$, $R^{\prime}=RPR-(q^{2}+1)R+(q^{2}+1)P$, then we have

$(PR+I)(PR^{\prime}-I)=0,$

and $\lambda=q^{-1}$. Identify $e^{6},f_{6},(m^{+})^{6}_{6}c^{-1}$ with the
additional simple root vectors $E_{4},\,F_{4}$ and the group-like element
$K_{4}$. Then the resulting quantum group
$U(V^{\vee}(R^{\prime},R_{21}^{-1}),\widetilde{U_{q}^{ext}({\mathfrak{sl}}_{4})},V(R^{\prime},R))$
is the quantum group $U_{q}({\mathfrak{so}}_{8})$ with $K_{i}^{\frac{1}{2}},\
1\leq i\leq 3$ adjoined.

###### Proof.

The entries in matrix $m^{\pm}$ we need are listed by the following equalities

$\left\\{\begin{array}[]{l}(m^{+})^{1}_{2}=-(q-q^{-1})E_{2}K_{1}^{\frac{1}{2}}K_{3}^{\frac{1}{2}},\quad(m^{+})^{2}_{2}=K_{1}^{\frac{1}{2}}K_{3}^{\frac{1}{2}},\\\
(m^{+})^{2}_{3}=-(q-q^{-1})E_{3}K_{1}^{\frac{1}{2}}K_{3}^{-\frac{1}{2}},\quad(m^{+})^{3}_{3}=K_{1}^{\frac{1}{2}}K_{3}^{-\frac{1}{2}},\\\
(m^{+})^{2}_{4}=-(q-q^{-1})E_{1}K_{1}^{-\frac{1}{2}}K_{3}^{\frac{1}{2}},\quad(m^{+})^{4}_{4}=K_{1}^{-\frac{1}{2}}K_{3}^{\frac{1}{2}},\quad(m^{+})^{6}_{6}=K_{1}^{-\frac{1}{2}}K_{2}^{-1}K_{3}^{-\frac{1}{2}},\\\
(m^{-})^{2}_{1}=q(q-q^{-1})K_{1}^{-\frac{1}{2}}K_{3}^{-\frac{1}{2}}F_{2},\quad(m^{-})^{2}_{2}=K_{1}^{-\frac{1}{2}}K_{3}^{-\frac{1}{2}},\\\
(m^{-})^{3}_{2}=q(q-q^{-1})K_{1}^{-\frac{1}{2}}K_{3}^{\frac{1}{2}}F_{3},\quad(m^{-})^{3}_{3}=K_{1}^{-\frac{1}{2}}K_{3}^{\frac{1}{2}},\\\
(m^{-})^{5}_{3}=q(q-q^{-1})K_{1}^{\frac{1}{2}}K_{3}^{\frac{1}{2}}F_{1},\quad(m^{-})^{5}_{5}=K_{1}^{\frac{1}{2}}K_{3}^{\frac{1}{2}}.\end{array}\right.$

The cross relations can be easily obtained as above, so we only describe the
$q$-Serre relations of the positive part. $E_{1},E_{3}$ belong to
$(m^{+})^{2}_{4},\ (m^{+})^{2}_{3}$, respectively, so we have

$\left.\begin{array}[]{l}(m^{+})^{2}_{4}=-(q-q^{-1})E_{1}(m^{+})^{4}_{4},\\\
e^{6}(m^{+})^{2}_{4}=\lambda
R^{2}_{2}{}^{6}_{6}(m^{+})^{2}_{4}e^{6}=(m^{+})^{2}_{4}e^{6},\\\
e^{6}(m^{+})^{4}_{4}=(m^{+})^{4}_{4}e^{6}.\end{array}\right\\}\Longrightarrow
e^{6}E_{1}=E_{1}e^{6},\Longrightarrow E_{4}E_{1}=E_{1}E_{4}.$
$\left.\begin{array}[]{l}(m^{+})^{2}_{3}=-(q-q^{-1})E_{3}(m^{+})^{3}_{3},\\\
e^{6}(m^{+})^{2}_{3}=\lambda
R^{2}_{2}{}^{6}_{6}(m^{+})^{2}_{3}e^{6}=(m^{+})^{2}_{3}e^{6},\\\
e^{6}(m^{+})^{3}_{3}=(m^{+})^{3}_{3}e^{6}.\end{array}\right\\}\Longrightarrow
e^{6}E_{3}=E_{3}e^{6},\Longrightarrow E_{4}E_{3}=E_{3}E_{4}.$

$E_{2}$ belongs to $(m^{+})^{1}_{2},$ then we obtain

$\left.\begin{array}[]{l}(m^{+})^{1}_{2}=-(q-q^{-1})E_{2}(m^{+})^{2}_{2},\\\
e^{6}(m^{+})^{1}_{2}=q^{-1}(m^{+})^{1}_{2}e^{6}+q^{-1}(q-q^{-1})(m^{+})^{2}_{2}e^{5},\\\
e^{6}(m^{+})^{2}_{2}=(m^{+})^{2}_{2}e^{6},\
e^{5}(m^{+})^{2}_{2}=q^{-1}(m^{+})^{2}_{2}e^{5}.\end{array}\right\\}\Longrightarrow
e^{5}=q^{-1}E_{2}e^{6}-e^{6}E_{2}.$

Combining with $e^{6}e^{5}=qe^{5}e^{6},$ which is obtained by
$e^{i}e^{j}=R^{\prime}{}^{ji}_{ab}e^{a}e^{b}$ and
$R^{\prime}{}^{5}_{5}{}^{6}_{6}=-2q,R^{\prime}{}^{5}_{6}{}^{6}_{5}=3$, we have

$E_{2}(E_{4})^{2}-(q+q^{-1})E_{4}E_{2}E_{4}+(E_{4})^{2}E_{2}=0.$

On the other hand, combining with $e^{5}E_{2}=qE_{2}e^{5}$ deduced from
$e^{5}(m^{+})^{1}_{2}=\lambda R^{1}_{1}{}^{5}_{5}(m^{+})^{1}_{2}e^{5}$, we
obtain $E_{4}(E_{2})^{2}-(q+q^{-1})E_{2}E_{4}E_{2}+(E_{2})^{2}E_{4}=0.$ ∎

######  Remark 4.1.

Observing the constructions of $C_{3}$ and $D_{4}$, we claim that $C_{n+1}$
and $D_{n+1}$ can be constructed directly from the node diagram $A_{n}$. The
pairs of braided groups or the $R$-matrices data $R,R^{\prime}$ we should
choose will be obtained by the ‘symmetric square’ and the second exterior
power of the vector representation of $A_{n}$. The verification of the claim
will rely on some skills, full details of it will be developed in a sequel.

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