# Heisenberg double of the generalized quantum euclidean group and its
representations

Limeng Xia  XIA: Institute of Applied System Analysis, Jiangsu University,
XueFu Road 301, Zhenjiang 212013, Jiangsu, PR China<EMAIL_ADDRESS>
###### Abstract.

The generalized quantum Euclidean group
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ is a natural generalization of the
quantum Euclidean group ${\mathcal{O}_{q}}(\mathfrak{b}_{1,1})$. The
Heisenberg double ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ of
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ is the smash product of
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ with its Hopf dual
${\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$. In this paper, we study the weight
modules, the prime spectrum and the automorphism group of the Heisenberg
double ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$.

Keywords: Heisenberg double, generalized quantum euclidean group, weight
module, prime spectrum

Mathematics Subject Classification (2000): 17B37

##  1\. Introduction

Throughout this paper, ${\mathbb{K}}$ is an algebraically closed field of
characteristic zero, for example ${\mathbb{C}}$. Let
${\mathbb{K}}^{*}={\mathbb{K}}\setminus\\{0\\}$ and $q\in{\mathbb{K}}^{*}$ is
not a root of unity.

Assume that $\mathfrak{g}$ is an indecomposable Lie algebra of dimension $3$,
then it is one of the following:

  * •

The three dimensional simple Lie algebra $\mathfrak{sl}_{2}$.

  * •

The three dimensional Heisenberg Lie algebra.

  * •

$\mathfrak{b}_{\lambda}={\rm span}\\{h,x,y\\}$ such that
$[h,x]=x,[h,y]=\lambda y,[x,y]=0$ for some $\lambda\in{\mathbb{K}}^{*}$.

They are very important in the representation theory of Lie algebras. The
simple modules over $\mathfrak{sl}_{2}$ and the simple modules over Heisenberg
Lie algebra have been completely classified in [B]. If $\lambda$ is not
rational, the classification of simple $\mathfrak{b}_{\lambda}$-modules is not
yet done. When $\lambda$ is rational, $\mathfrak{b}_{\lambda}$ is isomorphic
to some $\mathfrak{b}_{m,n}$, which is the three dimensional indecomposable
Lie algebra with basis $\\{h,x,y\\}$ and brackets

$\displaystyle[h,x]=mx,[h,y]=ny,[x,y]=0,$

for some nonzero integers $m,n$. The simple modules over $\mathfrak{b}_{m,n}$
have been classified, and the classification has a little difference according
to ${\rm sgn}(mn)$.

In this paper, we consider the Heisenberg double of quantum groups obtained
from the quantized universal enveloping algebra of $\mathfrak{b}_{m,n}$.

######  Definition 1.1.

The Hopf algebra ${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ is generated by the
elements $a,a^{-1},b,c$ subject to the following relations

$\displaystyle ab=q^{n}ba,\;ac=q^{m}ca,\;bc=cb,\;aa^{-1}=a^{-1}a=1.$

Its comultiplication $\Delta$, counit $\varepsilon$, antipode $S$ are defined
by

$\displaystyle\Delta(a)=a\otimes a,\;\varepsilon(a)=1,\;S(a)=a^{-1},$
$\displaystyle\Delta(b)=b\otimes a^{-n}+a^{n}\otimes
b,\;\varepsilon(b)=0,\;S(b)=-q^{-n^{2}}b,$ $\displaystyle\Delta(c)=c\otimes
a^{m}+a^{-m}\otimes c,\;\varepsilon(c)=0,\;S(c)=-q^{m^{2}}c.$

######  Definition 1.2.

The Hopf algebra ${\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$ is generated by the
elements $K,K^{-1},E,F$ subject to the following relations

$\displaystyle KE=q^{2m}EK,\;KF=q^{-2n}FK,\;EF=FE,\;KK^{-1}=K^{-1}K=1.$

Its comultiplication $\Delta$, counit $\varepsilon$, antipode $S$ are defined
by

$\displaystyle\Delta(K)=K\otimes K,\;\varepsilon(K)=1,\;S(K)=K^{-1},$
$\displaystyle\Delta(E)=E\otimes K^{m}+1\otimes
E,\;\varepsilon(E)=0,\;S(E)=-EK^{-m},$ $\displaystyle\Delta(F)=F\otimes
1+K^{-n}\otimes F,\;\varepsilon(F)=0,\;S(F)=-K^{n}F.$

Note that ${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ is a quantized analogue of
$U(\mathfrak{b}_{m,n})$, at the same time
${\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$ is a quantized analogue of
$U(\mathfrak{b}_{m,-n})$. In particular,
${\mathcal{O}_{q}}(\mathfrak{b}_{1,1})$ is the quantum Euclidean group and
$\mathfrak{b}_{1,-1}$ is the Euclidean Lie algebra. Thus we call
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ the generalized quantum Euclidean
group and call $\mathfrak{b}_{m,-n}$ the generalized Euclidean Lie algebra.

######  Proposition 1.3.

${\mathcal{U}_{q}}(\mathfrak{b}_{m,-n})$ is isomorphic to a Hopf subalgebra of
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$.

###### Proof.

It is easy to verify that the following map defines an embedding of Hopf
algebras:

$\displaystyle K\mapsto a^{2},\;E\mapsto a^{m}c,\;F\mapsto ba^{n}.$

∎

There is a unique non-degenerate dual pair
$\langle\cdot,\cdot\rangle:{\mathcal{U}_{q}}(\mathfrak{b}_{m,n})\times{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})\rightarrow{\mathbb{K}}$
determined by non-vanishing pairs:

$\displaystyle\langle K,a\rangle={q}^{-1},\langle K,a^{-1}\rangle=q,\langle
E,c\rangle=1,\langle F,b\rangle=1.$

Note that there are two different non-degenerate dual pairs
$\langle\cdot,\cdot\rangle:{\mathcal{O}_{q}}(\mathfrak{b}_{m,-n})\times{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})\rightarrow{\mathbb{K}}$.

This dual pair makes ${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ to be a left
${\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$-module algebra with the action defined
by

$\displaystyle u\cdot x=\sum_{x}x_{(1)}\langle u,x_{(2)}\rangle,\;\forall
u\in{\mathcal{U}_{q}}(\mathfrak{b}_{m,n}),x\in{\mathcal{O}_{q}}(\mathfrak{b}_{m,n}).$

Explicitly, the action takes the following form

$\displaystyle K\cdot a=q^{-1}a,\;K\cdot b=q^{n}b,\;K\cdot c=q^{-m}c,$
$\displaystyle E\cdot a=0,\;E\cdot b=0,\;E\cdot c=a^{-m},$ $\displaystyle
F\cdot a=0,\;F\cdot b=a^{n},\;F\cdot c=0.$

We define the Heisenberg double of the generalized quantum Euclidean group,
which is the smash product algebra

$\displaystyle{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})={\mathcal{O}_{q}}(\mathfrak{b}_{m,n})\rtimes{\mathcal{U}_{q}}(\mathfrak{b}_{m,n}).$

In this paper we are concerned with the Heisenberg double
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ and its representation theory.

######  Definition 1.4.

The Heisenberg double ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ of the
generalized quantum Euclidean group is the unital algebra generated by the
elements $a,a^{-1},b,c,K,E,F$ subject to the following relations

$\displaystyle ab=q^{n}ba,\;ac=q^{m}ca,\;bc=cb,\;aa^{-1}=a^{-1}a=1,$
$\displaystyle KE=q^{2m}EK,\;KF=q^{-2n}FK,\;EF=FE,\;KK^{-1}=K^{-1}K=1,$
$\displaystyle Ka=q^{-1}aK,\;Kb=q^{n}bK,\;Kc=q^{-m}cK,$ $\displaystyle
Ea=aE,\;Eb=bE,\;Ec=cE+a^{-m}K^{m},$ $\displaystyle
Fa=q^{n}aF,\;Fb=q^{-n^{2}}bF+a^{n},\;Fc=q^{mn}cF.$

The centrally extended Heisenberg double of $SL_{2}$ and the Heisenberg double
${\mathcal{D}_{q}}(\mathfrak{b}_{1,1})$ of the quantum Euclidean group were
studied by Tao (see [T1, T2]). For the general construction of the Heisenberg
double of a Hopf algebra, one is referred to [L].

##  2\. The weight modules

Because the map $\xi_{K}:x\rightarrow KxK^{-1}$ (resp. $\xi_{a}:x\rightarrow
axa^{-1}$) is diagonal, it is possible to study the $K$-weight (resp.
$a$-weight) modules over ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$.

######  Definition 2.1.

A ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$-module $V$ is called a $K$-weight
module provided that $V=\oplus_{\lambda\in{\mathbb{K}}^{*}}V_{\lambda}$ where
$V_{\lambda}=\\{v\in V:Kv=\lambda v\\}$. Denote by ${\rm
supp}_{K}(V)=\\{\lambda\in{\mathbb{K}}^{*}|V_{\lambda}\not=0\\}$ the set of
all $K$-weights of $V$.

######  Definition 2.2.

A ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$-module $V$ is called an $a$-weight
module provided that $V=\oplus_{\mu\in{\mathbb{K}}^{*}}V^{\mu}$ where
$V^{\mu}=\\{v\in V:av=\mu v\\}$. Denote by ${\rm
supp}_{a}(V)=\\{\mu\in{\mathbb{K}}^{*}|V^{\mu}\not=0\\}$ the set of all
$a$-weights of $V$.

Let ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{+}$ be the polynomial algebra
${\mathbb{K}}[E,b]$, let ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}$ be the
quantum torus ${\mathbb{K}}[K^{\pm 1},a^{\pm 1}]$ and let
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{-}$ be the quantum polynomial algebra
${\mathbb{K}}[F,c]$. Then as algebra, ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$
has a triangular decomposition

$\displaystyle{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})={\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{-}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{+}.$

Let
$b^{\prime}=a^{n}bK^{-n},c^{\prime}=a^{-m}cK^{-m},E^{\prime}=a^{2m}E,F^{\prime}=q^{-n^{2}}a^{-2n}FK^{n}$.
Then $\\{a^{\pm 1},K^{\pm 1},b^{\prime},c^{\prime},E^{\prime},F^{\prime}\\}$
is also a generating set of ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$. In
particular, we have

$\displaystyle
Kb^{\prime}=b^{\prime}K,\;Kc^{\prime}=c^{\prime}K,\;KE^{\prime}=E^{\prime}K,\;KF^{\prime}=F^{\prime}K,$
$\displaystyle
ab^{\prime}=b^{\prime}a,\;ac^{\prime}=c^{\prime}a,\;aE^{\prime}=E^{\prime}a,\;aF^{\prime}=F^{\prime}a,$
$\displaystyle
b^{\prime}c^{\prime}=q^{2mn}c^{\prime}b^{\prime},\;E^{\prime}b^{\prime}=q^{2mn}b^{\prime}E^{\prime},\;F^{\prime}b^{\prime}=q^{-2n^{2}}b^{\prime}F^{\prime}+1,$
$\displaystyle
E^{\prime}c^{\prime}=q^{2m^{2}}c^{\prime}E^{\prime}+1,\;F^{\prime}c^{\prime}=q^{-2mn}c^{\prime}F^{\prime},\;E^{\prime}F^{\prime}=q^{-2mn}F^{\prime}E^{\prime}.$

Let $S$ be the subalgebra
${\mathbb{K}}[b^{\prime},c^{\prime},E^{\prime},F^{\prime}]$, then we have the
following isomorphism of algebras

$\displaystyle{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ $\displaystyle\cong$
$\displaystyle{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}\otimes S.$

######  Lemma 2.3.

The $S$ has no finite dimensional modules.

###### Proof.

Note that $S$ has two subalgebras $S_{1}={\mathbb{K}}[E^{\prime},c^{\prime}]$
and $S_{2}={\mathbb{K}}[F^{\prime},b^{\prime}]$, both of them are generalized
Weyl algebras. It is known that a generalized Weyl algebra has no finite
dimensional modules. ∎

Our first main result is the following theorem.

######  Theorem 2.4.

(i) Let $V$ be a simple $K$-weight
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$-module. Then ${\rm
supp}_{K}(V)=\\{q^{i}\lambda|i\in{\mathbb{Z}}\\}$ for some $\lambda$ and any
$K$-weight space of $V$ is a simple $S$-module. In particular, any $K$-weight
space of $V$ is infinite dimensional.

(ii) Let $W$ be a simple $a$-weight
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$-module. Then ${\rm
supp}_{a}(W)=\\{q^{i}\mu|i\in{\mathbb{Z}}\\}$ for some $\mu$ and any
$a$-weight space of $W$ is a simple $S$-module. In particular, any $a$-weight
space of $W$ is infinite dimensional.

###### Proof.

We only prove part (i), the proof for part (ii) is very similar.

Let $V$ be a simple $K$-weight module with a weight $\lambda\in{\rm
supp}_{K}(V)$. Because $a,a^{-1}$ are invertible and $KaK^{-1}=q^{-1}a$, we
have

$\displaystyle aV_{\mu}\subseteq V_{q^{-1}\lambda},\;a^{-1}V_{\mu}\subseteq
V_{q\lambda}.$

By induction, it holds ${\rm
supp}_{K}(V)\supseteq\\{q^{i}\lambda|i\in{\mathbb{Z}}\\}$. Because $[K,S]=0$,
it implies

$\displaystyle S\cdot V_{q^{i}\lambda}\subseteq V_{q^{i}\lambda}.$

So ${\rm supp}_{K}(V)=\\{q^{i}\lambda|i\in{\mathbb{Z}}\\}$ and any weight
space is an $S$-module. If $V_{q^{i_{0}}\lambda}$ contains a proper submodule
$V^{\prime}$ over $S$, then

$\displaystyle\sum_{i\in{\mathbb{Z}}}a^{i}V^{\prime}$

is a proper $K$-weight submodule over ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$.
Thus any weight space is a simple $S$-module. By Lemma 2.3, any $K$-weight
space of $V$ is infinite dimensional. ∎

##  3\. Some simple $S$-modules

In this section, we construct some classes of simple $S$-modules. In
particular, all of these modules have GKdim $2$.

######  Proposition 3.1.

Let $J_{1}(\sigma,\tau)$ be the left $S$-ideal generated by
$b^{\prime}-\sigma,c^{\prime}-\tau$ for all $\sigma,\tau\in{\mathbb{K}}$ such
that $\sigma\tau=0$. Then $S/J_{1}(\sigma,\tau)$ is a simple $S$-module of
GKdim $2$.

###### Proof.

It is obvious that
$S/J_{1}(\sigma,\tau)\cong{\mathbb{K}}[E^{\prime},F^{\prime}]v$ with
$b^{\prime}v=\sigma v,c^{\prime}v=\tau v$.

If $\sigma=0$, for any
$x=\sum_{i=0}\varphi_{i}(E^{\prime})(F^{\prime})^{i}\not=0$, we have

$\displaystyle
b^{\prime}xv=\sum_{i=1}^{I}\varphi_{i}(q^{-2mn}E^{\prime})c_{i}(F^{\prime})^{i-1}v$

for some nonzero constants $c_{i},i\geq 1$. Induction on $I$, we may obtain a
nonzero element $\varphi(E^{\prime})v$. It is known that
${\mathbb{K}}[E^{\prime}]v$ is a simple $S_{1}$-module, so this lemma holds.

If $\tau=0$, the proof is similar. ∎

######  Proposition 3.2.

Let $J_{2}(\sigma,\tau)$ be the left $S$-ideal generated by
$b^{\prime}-\sigma,E^{\prime}-\tau$ for all $\sigma,\tau\in{\mathbb{K}}$ such
that $\sigma\tau=0$. Then $S/J_{2}(\sigma,\tau)$ is a simple $S$-module of
GKdim $2$.

###### Proof.

First we have $S/J_{2}(\sigma,\tau)\cong{\mathbb{K}}[c^{\prime},F^{\prime}]v$
with $b^{\prime}v=\sigma v,E^{\prime}v=\tau v$.

If $\sigma=0$, for any
$x=\sum_{i=0}\varphi_{i}(c^{\prime})(F^{\prime})^{i}\not=0$, we have

$\displaystyle
b^{\prime}xv=\sum_{i=1}^{I}\varphi_{i}(q^{2mn}c^{\prime})c_{i}(F^{\prime})^{i-1}v$

for some nonzero constants $c_{i},i\geq 1$. Induction on $I$, we may obtain a
nonzero element $\varphi(c^{\prime})v$. Because ${\mathbb{K}}[c^{\prime}]v$ is
also a simple $S_{1}$-module, so this lemma holds.

If $\tau=0$, the proof is similar. ∎

Similar to above, we state the following propositions without proof.

######  Proposition 3.3.

Let $J_{3}(\sigma,\tau)$ be the left $S$-ideal generated by
$F^{\prime}-\sigma,c^{\prime}-\tau$ for all $\sigma,\tau\in{\mathbb{K}}$ such
that $\sigma\tau=0$. Then $S/J_{3}(\sigma,\tau)$ is a simple $S$-module of
GKdim $2$.

######  Proposition 3.4.

Let $J_{4}(\sigma,\tau)$ be the left $S$-ideal generated by
$F^{\prime}-\sigma,E^{\prime}-\tau$ for all $\sigma,\tau\in{\mathbb{K}}$ such
that $\sigma\tau=0$. Then $S/J_{4}(\sigma,\tau)$ is a simple $S$-module of
GKdim $2$.

From above propositions, we obtain various simple $K$-weight modules (resp.
$a$-weight modules) of GKdim $3$:

$\displaystyle\sum_{i\in{\mathbb{Z}}}a^{i}S/J_{k}(\sigma,\tau),\quad\left(\hbox{resp.}\sum_{i\in{\mathbb{Z}}}K^{i}S/J_{k}(\sigma,\tau)\right)\quad
k=1,2,3,4.$

In particular, when $m=n=1$, the
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$-modules

$\displaystyle\sum_{i\in{\mathbb{Z}}}a^{i}S/J_{k}(0,0),\quad\left(\hbox{resp.}\sum_{i\in{\mathbb{Z}}}K^{i}S/J_{k}(0,0)\right)\quad
k=1,2,3,4.$

cover those modules constructed in [T2].

##  4\. The prime and primitive spectra of $S$

Let $\mathfrak{p}$ be a proper ideal of a ring $R$. Then for all ideals
$\mathfrak{i},\mathfrak{j}$, if $\mathfrak{ij}\subseteq\mathfrak{p}$ then
either $\mathfrak{i}\subseteq\mathfrak{p}$ or
$\mathfrak{j}\subseteq\mathfrak{p}$. In particular, if $R/\mathfrak{p}$ is a
domain, the ideal $\mathfrak{p}$ must be prime. Let ${\rm Spec}(S)$ be the set
of all prime ideals of $S$.

Define symbols

$\displaystyle\phi_{1}=E^{\prime}c^{\prime}-c^{\prime}E^{\prime},\;\phi_{2}=F^{\prime}b^{\prime}-b^{\prime}F^{\prime}.$

######  Lemma 4.1.

$\displaystyle\phi_{1}\phi_{2}=\phi_{2}\phi_{1},\;\phi_{1}F^{\prime}=F^{\prime}\phi_{1},\;\phi_{1}b^{\prime}=b^{\prime}\phi_{1},\;E^{\prime}\phi_{2}=\phi_{2}E^{\prime},\;c^{\prime}\phi_{2}=\phi_{2}c^{\prime},$
$\displaystyle\phi_{1}E^{\prime}=q^{-2m^{2}}E^{\prime}\phi_{1},\;\phi_{1}c^{\prime}=q^{2m^{2}}c^{\prime}\phi_{1},\;F^{\prime}\phi_{2}=q^{-2n^{2}}\phi_{2}F^{\prime},\;b^{\prime}\phi_{2}=q^{2n^{2}}\phi_{2}b^{\prime}.$

###### Proof.

It follows by directly computing. ∎

Assume $d=(m,n)$. Set
$(0)=\\{0\\},I_{1}=S\phi_{1},I_{2}=S\phi_{2},I_{3}=I_{1}+I_{2}$. If $mn>0$, we
set

$\displaystyle J_{1}(z)=S(\phi_{1}^{|n|/d}-zb^{\prime|m|/d})+S\phi_{2},\quad
J_{2}(z)=S\phi_{1}+S(\phi_{2}^{|m|/d}-zc^{\prime|n|/d}).$

If $mn<0$, we set

$\displaystyle J_{1}(z)=S(\phi_{1}^{|n|/d}-zF^{\prime|m|/d})+S\phi_{2},\quad
J_{2}(z)=S\phi_{1}+S(\phi_{2}^{|m|/d}-zE^{\prime|n|/d}).$

######  Proposition 4.2.

The following statements hold.

1\. Both $S/I_{1}$ and $S/I_{2}$ are domains.

2\. $S/I_{3}$ is simple.

3\. Both $S/J_{1}(z)$ and $S/J_{2}(z)$ are domains.

###### Proof.

1\. In the quotient algebra $S/S\phi_{1}$, we have
$E^{\prime}c^{\prime}=c^{\prime}E^{\prime}$, then
$E^{\prime}c^{\prime}=q^{2m^{2}}c^{\prime}E^{\prime}+1$ implies
$E^{\prime}c^{\prime}\in{\mathbb{K}}^{*}$. Consequently,

$\displaystyle
S/I_{1}\cong{\mathbb{K}}[E^{\prime},E^{\prime-1}][F^{\prime},b^{\prime}]\cong{\mathbb{K}}[c^{\prime},c^{\prime-1}][F^{\prime},b^{\prime}],$

which is a domain. In a similar way, $S/I_{2}$ also is a domain.

2\. The quotient algebra $S/I_{3}$ is isomorphic to the quantum torus
${\mathbb{K}}[E^{\prime\pm 1},F^{\prime\pm 1}]$, which is simple.

3\. $S/J_{1}(z)$ is isomorphic to a finite extension of $S_{1}$. Explicitly,

$\displaystyle S/J_{1}(z)\cong
S_{1}+S_{1}\sqrt[|m|]{\phi_{1}^{|n|}}+S_{1}\sqrt[|m|]{\phi_{1}^{2|n|}}+\cdots+S_{1}\sqrt[|m|]{\phi_{1}^{(|m|/d-1)|n|}},$

which is a domain. In special, when $m=\pm n$, $S/J_{1}(z)\cong S_{1}$ is
simple. The proof for $S/J_{2}(z)$ is very similar. ∎

######  Lemma 4.3.

Let $\Omega=\\{zb^{\prime i}c^{\prime
j}|z\in{\mathbb{K}}^{*},i,j\in\mathbb{N}\\}$. If $\mathfrak{p}$ is a prime
ideal of $S$, then $\Omega\cap\mathfrak{p}=\emptyset$.

###### Proof.

If $zb^{\prime i}c^{\prime j}\in\mathfrak{p}$, we have $b^{\prime i}c^{\prime
j}\in\mathfrak{p}$. When $i>0$,

$\displaystyle F^{\prime}b^{\prime i}c^{\prime j}-q^{-2mnj-2n^{2}i}b^{\prime
i}c^{\prime j}F^{\prime}\in{\mathbb{K}}^{*}b^{\prime i-1}c^{\prime j},$

thus $b^{\prime i-1}c^{\prime j}\in\mathfrak{p}$. By induction, we infer that
$c^{\prime j}\in\mathfrak{p}$. Similarly by
$E^{\prime}c^{\prime}=q^{2m^{2}}c^{\prime}E^{\prime}+1$, we obtain
$1\in\mathfrak{p}$ and $\mathfrak{p}=S$, a contradiction. ∎

######  Theorem 4.4.

The prime spectrum of the algebra $S$ is given below,

$\displaystyle{\rm Spec}(S)$ $\displaystyle=$
$\displaystyle\\{(0),I_{1},I_{2},I_{3}\\}\cup\\{J_{1}(z)|z\in{\mathbb{K}}^{*}\\}\cup\\{J_{2}(z)|z\in{\mathbb{K}}^{*}\\}.$

All the containments of the prime ideals of $S$ are shown in the following
diagram

$\textstyle{\\{J_{1}(z)|z\in{\mathbb{K}}^{*}\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\\{J_{1}(z)|z\in{\mathbb{K}}^{*}\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(0)}$

###### Proof.

We only prove for $mn>0$.

By Lemma 6.3, the spectrum of $S$ is homeomorphic to the spectrum of its Ore
extension $S\Omega^{-1}$. Now we consider the Ore extension
$\mathbb{S}=S\Omega_{1}^{-1}$ with Ore set

$\displaystyle\Omega_{1}=\\{zb^{\prime i}c^{\prime
j}\phi_{1}^{k}\phi_{2}^{l}|z\in{\mathbb{K}}^{*},i,j,k,l\in\mathbb{N}\\}.$

In particular, $\mathbb{S}$ is the localization of $S\Omega^{-1}$ obtained by
inverting the elements $\phi_{1},\phi_{2}$.

The algebra $\mathbb{S}$ is a quantum torus with the structural matrix

$\displaystyle Q=\left[\begin{array}[]{cccc}1&q^{2m^{2}}&1&1\\\
q^{-2m^{2}}&1&q^{-2mn}&1\\\ 1&q^{2mn}&1&q^{2n^{2}}\\\
1&1&q^{-2n^{2}}&1\end{array}\right],$

associated to the variables $\phi_{1},c^{\prime},b^{\prime},\phi_{2}$.

Let $\mathfrak{p}_{1}$ be a prime ideal of $\mathbb{S}$ and given a nonzero
element $x\in\mathfrak{p}_{1}$ such that

$\displaystyle x=\sum_{i,j,k,l}z_{i,j,k,l{\mathbb{Z}}}b^{\prime i}c^{\prime
j}\phi_{1}^{k}\phi_{2}^{l},$

with finitely many nonzero coefficients. Because

$\displaystyle\phi_{1}x-q^{2m^{2}j_{0}}x\phi_{1}=\sum_{i,j,k,l\in{\mathbb{Z}}}(1-q^{2m^{2}(j_{0}-j)})z_{i,j,k,l}b^{\prime
i}c^{\prime j}\phi_{1}^{k}\phi_{2}^{l},$

we can choose $j_{0}$ such that $z_{i,j_{0},k,l}$ for some $i,k,l$. Either
$z_{i,j,k,l}=0$ for all $j\not=j_{0}$ or we obtain a nonzero element with less
nonzero coefficients.

Consider the powers $i,k,l$ in the same way and by induction on the number of
nonzero coefficients, we can find a monomial $zb^{\prime i}c^{\prime
j}\phi_{1}^{k}\phi_{2}^{l}\in\mathfrak{p}_{1}$. Therefore
$1\in\mathfrak{p}_{1}$ and it is impossible. This implies $\mathbb{S}$ is
simple.

Consequently, for any nonzero prime ideal $\mathfrak{p}$ of $S$ (equivalent to
consider the prime ideal $\mathfrak{p}\Omega^{-1}$ of $S\Omega^{-1}$), either
$I_{1}\subseteq\mathfrak{p}$ or $\phi_{2}\subseteq\mathfrak{p}$.

When $I_{3}\in\mathfrak{p}$, since $S/I_{3}$ is isomorphic to a quantum torus,
which is simple, we obtain $\mathfrak{p}=I_{3}$.

Assume $I_{1}\subseteq\mathfrak{p}\not=I_{3}$. Because
$S\Omega^{-1}/I_{1}\Omega^{-1}\cong{\mathbb{K}}[b^{\prime},b^{\prime-1},c^{\prime},c^{\prime-1},\phi_{2}]$,
there exists an irreducible polynomial

$\displaystyle y=\sum_{i,j\in{\mathbb{Z}},k\in\mathbb{N}}z_{i,j,k}b^{\prime
i}c^{\prime j}\phi_{2}^{k}\in\mathfrak{p}\Omega^{-1}.$

Similar to above approach, by considering the number of nonzero coefficients
of

$\displaystyle\phi_{2}y-q^{-2n^{2}i_{0}}y\phi_{2},$

we may assume $z_{i,j,k}=0$ except $i=i_{0}$ for some $i_{0}$. Multiple by
$b^{\prime-i_{0}}$, we obtain a nonzero irreducible element

$\displaystyle y_{1}=\sum_{j\in{\mathbb{Z}},k\in\mathbb{N}}z_{j,k}c^{\prime
j}\phi_{2}^{k}\in\mathfrak{p}\Omega^{-1},$

and hence some

$\displaystyle
y_{2}=\sum_{j=0}^{r}f_{j}(c^{\prime})\phi_{2}^{j}\in\mathfrak{p},$

with $r>0$ and $f_{0}\not=0$. Because

$\displaystyle b^{\prime
i}y_{2}=\sum_{j=0}^{r}q^{2n^{2}ij}f_{j}(q^{2mni}c^{\prime})\phi_{2}^{j}b^{\prime
i},$

it implies

$\displaystyle\sum_{j=0}^{r}q^{2n^{2}ij}f_{j}(q^{2mni}c^{\prime})\phi_{2}^{j}\in\mathfrak{p},$

for all $i\in\mathbb{N}.$ Therefore, we can assume

$\displaystyle y_{2}=\sum_{j=0}^{r}z_{j}c^{\prime
i_{j}}\phi_{2}^{j}\in\mathfrak{p},$

such that $nj-mi_{j}$ is a constant. Such irreducible element of the
polynomial ring ${\mathbb{K}}[c^{\prime},\phi_{2}]$ has to be
$\phi_{2}^{|m|/d}-zc^{\prime|n|/d}$ for some nonzero constant $z$. So
$J_{1}(z)\subseteq\mathfrak{p}$. Because $S/J_{1}(z)$ is simple, we have
$\mathfrak{p}=J_{1}(z)$.

The proof for $I_{2}\subseteq\mathfrak{p}\not=I_{3}$ is similar. ∎

######  Theorem 4.5.

The prime spectrum of the algebra ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ is
given below,

$\displaystyle{\rm
Spec}(S)=\\{(0),{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{1},{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{2},{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{3}\\}\cup\\{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}J_{1}(z),{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}J_{2}(z)|z\in{\mathbb{K}}^{*}\\}.$

All the containments of the prime ideals of $S$ are shown in the following
diagram

---
$\textstyle{\\{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}J_{1}(z)|z\in{\mathbb{K}}^{*}\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\\{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}J_{1}(z)|z\in{\mathbb{K}}^{*}\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}I_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(0)}$

###### Proof.

Because ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}$ is simple,
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}\cap\mathfrak{p}=0$ for any prime
ideal of ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$. Assume $\mathfrak{p}$ is a
prime ideal of ${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$. Given a nonzero
element

$\displaystyle x=\sum_{i,j\in{\mathbb{Z}}}x_{i,j}K^{i}a^{j}\in\mathfrak{p}$

for finitely many nonzero elements $x_{i,j}\in S$. Then

$\displaystyle K^{r}a^{s}xa^{-s}K^{-r}=\sum_{i,j\in{\mathbb{Z}}}q^{is-
jr}x_{i,j}K^{i}a^{j}\in\mathfrak{p}.$

So it is easy to see that $x_{i,j}\in\mathfrak{p}$ for all $i,j$. Hence
$\mathfrak{p}={\mathcal{D}_{q}}(\mathfrak{b}_{m,n})(\mathfrak{p}\cap S)$. The
proof is finished. ∎

For $m=n=1$, this agrees [T2, Theorem 3.6].

##  5\. The automorphism groups

In this section, we investigate the automorphisms of algebras
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$,
${\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$ and
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$.

######  Theorem 5.1.

1\. For $m=\pm n$, let $\tau\in{\rm Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$
be defined by

$\displaystyle\tau$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}a\mapsto a,b\mapsto c,c\mapsto
b,&m=n,\\\ a\mapsto a^{-1},b\mapsto c,c\mapsto b,&m=-n.\end{array}\right.$

Then $\\{{\rm id},\tau\\}\cong{\mathbb{Z}}_{2}$.

2\. For any $i\in{\mathbb{Z}}$, let $\xi_{i}\in{\rm
Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ be defined by

$\displaystyle\xi_{i}:a\mapsto a,b\mapsto a^{\frac{in}{(m,n)}}b,c\mapsto
a^{\frac{im}{(m,n)}}c.$

Then $\\{\xi_{i}|i\in{\mathbb{Z}}\\}\cong{\mathbb{Z}}$.

3\. For all $z,z_{1},z_{2}\in{\mathbb{K}}^{*}$, let
$\zeta_{z,z_{1},z_{2}}\in{\rm Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ be
defined by

$\displaystyle\zeta_{z,z_{1},z_{2}}:a\mapsto za,b\mapsto z_{1}b,c\mapsto
z_{2}c.$

Then
$\\{\zeta_{z,z_{1},z_{2}}|z,z_{1},z_{2}\in{\mathbb{K}}^{*}\\}\cong({\mathbb{K}}^{*})^{3}$.

4.

$\displaystyle{\rm Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$
$\displaystyle\cong$
$\displaystyle\left\\{\begin{array}[]{ll}({\mathbb{K}}^{*})^{3}\rtimes({\mathbb{Z}}\times{\mathbb{Z}}_{2}),&m=\pm
n,\\\ ({\mathbb{K}}^{*})^{3}\rtimes{\mathbb{Z}},&m\not=\pm
n.\end{array}\right.$

###### Proof.

It is clear that the statements 1-3 hold. Moreover, the subgroup $G$ generated
by
$\tau,\xi_{i}(i\in{\mathbb{Z}}),\zeta_{z,z_{1},z_{2}}(z,z_{1},z_{2}\in{\mathbb{K}}^{*})$
is isomorphic to
$({\mathbb{K}}^{*})^{3}\rtimes({\mathbb{Z}}\times{\mathbb{Z}}_{2})$ and the
subgroup $G_{1}$ generated by
$\xi_{i}(i\in{\mathbb{Z}}),\zeta_{z,z_{1},z_{2}}(z,z_{1},z_{2}\in{\mathbb{K}}^{*})$
is isomorphic to $({\mathbb{K}}^{*})^{3}\rtimes{\mathbb{Z}}$.

We prove statement 4. Assume $\rho\in{\rm
Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$. Obviously, the group
$\\{za^{i}|z\in{\mathbb{K}}^{+},i\in{\mathbb{Z}}\\}={\mathbb{K}}^{*}\times\langle
a\rangle$ is the set of all invertible elements in
${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$, which can be generated by
${\mathbb{K}}^{*}$ and $a$. So this group also can be generated by
${\mathbb{K}}^{*}$ and $\rho(a)$. This implies that $\rho(a)=za^{\pm 1}$ for
some $z\in{\mathbb{K}}^{*}$.

(i) If $m=n$, ${\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ has a decomposition of
$a$-weight spaces

$\displaystyle{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$ $\displaystyle=$
$\displaystyle\bigoplus_{i=0}^{\infty}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})_{q^{im}},$

under the action $x\mapsto axa^{-1}$, where

$\displaystyle{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})_{q^{im}}$ $\displaystyle=$
$\displaystyle\bigoplus_{r=0}^{i}{\mathbb{K}}[a,a^{-1}]b^{i-r}c^{r}.$

So either $\rho(b)=xb,\rho(c)=yc$ for some $x,y\in{\mathbb{K}}[a,a^{-1}]$ or
$\rho(b)=x^{\prime}c,\rho(c)=y^{\prime}b$ for some
$x^{\prime},y^{\prime}\in{\mathbb{K}}[a,a^{-1}]$, and in this case
$\rho(a)=za$. Moreover, the elements $x,y,x^{\prime},y^{\prime}$ are
invertible. By the identity $\rho(b)\rho(c)=\rho(c)\rho(b)$, we also infer
that $xy^{-1},x^{\prime}y^{\prime-1}\in{\mathbb{K}}^{*}$. Explicitly, there
exists $z,z_{1},z_{2}\in{\mathbb{K}}^{*}$ and $i\in{\mathbb{Z}}$ such that
either

$\displaystyle\rho(a)=za,\;\rho(b)=z_{1}a^{i}b,\;\rho(c)=z_{2}a^{i}c,$

or

$\displaystyle\rho(a)=za,\;\rho(b)=z_{1}a^{i}c,\;\rho(c)=z_{2}a^{i}b.$

Then $\rho\in G$, the statement holds.

(ii) If $m=-n$, similar to (i), there exists
$z,z_{1},z_{2}\in{\mathbb{K}}^{*}$ and $i\in{\mathbb{Z}}$ such that either

$\displaystyle\rho(a)=za,\;\rho(b)=z_{1}a^{i}b,\;\rho(c)=z_{2}a^{-i}c,$

or

$\displaystyle\rho(a)=za^{-1},\;\rho(b)=z_{1}a^{-i}c,\;\rho(c)=z_{2}a^{i}b.$

Then $\rho\in G$, the statement holds.

(iii) If $m\not=\pm n$, we have $\rho(b)\in{\mathbb{K}}[a,a^{-1}]b$,
$\rho(c)\in{\mathbb{K}}[a,a^{-1}]c$. Then similar to above, there exists
$z,z_{1},z_{2}\in{\mathbb{K}}^{*}$ and $i\in{\mathbb{Z}}$ such that either

$\displaystyle\rho(a)=za,\;\rho(b)=z_{1}a^{\frac{in}{(m,n)}}b,\;\rho(c)=z_{2}a^{\frac{im}{(m,n)}}c.$

Then $\rho\in G_{1}$, the statement holds. ∎

######  Theorem 5.2.

$\displaystyle{\rm Aut}{\mathcal{U}_{q}}(\mathfrak{b}_{m,n})$
$\displaystyle\cong$
$\displaystyle\left\\{\begin{array}[]{ll}({\mathbb{K}}^{*})^{3}\rtimes({\mathbb{Z}}\times{\mathbb{Z}}_{2}),&m=\pm
n,\\\ ({\mathbb{K}}^{*})^{3}\rtimes{\mathbb{Z}},&m\not=\pm
n.\end{array}\right.$

###### Proof.

It is straightforward by the isomorphism of algebras

$\displaystyle{\mathcal{U}_{q}}(\mathfrak{b}_{m,n})\cong{\mathcal{O}_{q}}(\mathfrak{b}_{2m,-2n}).$

∎

######  Theorem 5.3.

1\. For any $z_{1},z_{2}\in{\mathbb{K}}^{*}$, let $\zeta_{z_{1},z_{2}}\in{\rm
Aut}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ be defined by

$\displaystyle K\mapsto z_{1}K,a\mapsto z_{2}a,\rho|_{S}=\hbox{\rm id}_{S}.$

Then
$\\{\zeta_{z_{1},z_{2}}|z_{1},z_{2}\in{\mathbb{K}}^{*}\\}\cong({\mathbb{K}}^{*})^{2}$.

2\. For any $A=(A_{i,j})\in SL_{2}({\mathbb{Z}})$, let $\rho_{A}\in{\rm
Aut}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ be defined by

$\displaystyle K\mapsto K^{A_{1,1}}a^{A_{2,1}},a\mapsto
K^{A_{1,2}}a^{A_{2,2}},\rho|_{S}=\hbox{\rm id}_{S}.$

The subgroup generated by
$\zeta_{z_{1},z_{2}}(z_{1},z_{2}\in{\mathbb{K}}^{*}),\rho_{A},(A\in
SL_{2}({\mathbb{Z}}))$ is isomorphic to $({\mathbb{K}}^{*})^{2}\rtimes
SL_{2}({\mathbb{Z}})$.

3\. For any $z_{3},z_{4}\in{\mathbb{K}}^{*}$, let $\xi_{z_{3},z_{4}}\in{\rm
Aut}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ be defined by

$\displaystyle K\mapsto K,a\mapsto a,E^{\prime}\mapsto
z_{3}E^{\prime},F^{\prime}\mapsto z_{4}F^{\prime},c^{\prime}\mapsto
z_{3}^{-1}c^{\prime},b\mapsto z_{4}^{-1}b^{\prime}.$

Then
$\\{\xi_{z_{3},z_{4}}|z_{3},z_{4}\in{\mathbb{K}}^{*}\\}\cong({\mathbb{K}}^{*})^{2}$.

4.

$\displaystyle{\rm Aut}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$
$\displaystyle\cong$ $\displaystyle(({\mathbb{K}}^{*})^{2}\rtimes
SL_{2}({\mathbb{Z}}))\times({\mathbb{K}}^{*})^{2}.$

###### Proof.

The statements 1 and 3 are clear.

It is also obvious that the map $\rho_{A}$ is injective. By the definition,
one can verify that $\rho_{A}\circ\rho_{B}=\rho_{AB}\circ\zeta_{z_{1},z_{2}}$
for some $z_{1},z_{2}\in{\mathbb{K}}^{*}$. This implies $\rho_{A}$ is an
isomorphism since
$\rho_{A}\circ\rho_{A^{-1}}\circ\zeta_{z_{1}^{-1},z_{2}^{-1}}={\rm id}$.
Moreover, the subgroup generated by
$\zeta_{z_{1},z_{2}}(z_{1},z_{2}\in{\mathbb{K}}^{*}),\rho_{A}(A\in
SL_{2}({\mathbb{Z}}))$ is isomorphic to $({\mathbb{K}}^{*})^{2}\rtimes
SL_{2}({\mathbb{Z}})$, this proves statement 2.

Obviously, the subgroup $G$ generated by
$\zeta_{z_{1},z_{2}}(z_{1},z_{2}\in{\mathbb{K}}^{*}),\rho_{A}(A\in
SL_{2}({\mathbb{Z}})),\xi_{z_{3},z_{4}}(z_{3},z_{4}\in{\mathbb{K}}^{*})$ is
isomorphic to $(({\mathbb{K}}^{*})^{2}\rtimes
SL_{2}({\mathbb{Z}}))\times({\mathbb{K}}^{*})^{2}$.

We prove statement 4. Assume $\rho\in{\rm
Aut}{\mathcal{O}_{q}}(\mathfrak{b}_{m,n})$.

Clearly, the group $\\{zK^{i}a^{j}|z\in{\mathbb{K}}^{*},i,j\in{\mathbb{Z}}\\}$
is the set of all invertible elements in
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$. So
$\rho(K)=z_{1}K^{i}a^{j},\rho(a)=z_{2}K^{r}a^{s}$ for some
$z_{1},z_{2}\in{\mathbb{K}}^{*}$ and $i,j,r,s\in{\mathbb{Z}}$. Moreover,
$\rho(K)\rho(a)=q^{-1}\rho(a)\rho(K)$ implies

$\displaystyle\det\left(\begin{array}[]{cc}i&r\\\ j&s\end{array}\right)=1.$

So $\rho|_{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}}\in{\rm
Aut}{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}$ and

$\displaystyle\rho|_{{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}}=\rho_{A}\circ\zeta_{z_{1},z_{2}},\;A=\left(\begin{array}[]{cc}i&r\\\
j&s\end{array}\right).$

Since
$\rho({\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0})={\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}$,
we also have $\rho(S)=S$ by the fact that $S$ is the centralizer of
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})^{0}$. Next we consider the
automorphisms of $S$.

Because $\rho$ maps prime ideals to prime ideals, it must holds

$\displaystyle\\{\rho(\phi_{1}),\rho(\phi_{2})\\}=\\{x\phi_{1},y\phi_{2}\\}$

for some invertible elements $x,y$, so $x,y\in{\mathbb{K}}^{*}$. Up to an
automorphism $\zeta_{z_{3},z_{4}}$, we may assume $x=y=1$.

When $\rho(\phi_{1})=\phi_{1},\rho(\phi_{2})=\phi_{2}$. Because
$\rho(c^{\prime})\rho(\phi_{1})=q^{-2m^{2}}\rho(\phi_{1})\rho(c^{\prime})$ and
$\rho(E^{\prime})\rho(\phi_{1})=q^{2m^{2}}\rho(\phi_{1})\rho(E^{\prime})$, we
have

$\displaystyle\rho(c^{\prime})=f_{1}(\phi_{1},b^{\prime},F^{\prime})c^{\prime},\;\rho(E^{\prime})=E^{\prime}f_{2}(\phi_{1},b^{\prime},F^{\prime})$

and

$\displaystyle{\mathbb{K}}^{*}\phi_{1}+{\mathbb{K}}\ni\rho(c^{\prime})\rho(E^{\prime})=f_{1}(\phi_{1},b^{\prime},F^{\prime})c^{\prime}E^{\prime}f_{2}(\phi_{1},b^{\prime},F^{\prime}),$

this implies
$f_{1}(\phi_{1},b^{\prime},F^{\prime})f_{2}(\phi_{1},b^{\prime},F^{\prime})$
is a constant. Moreover,
$\rho(c^{\prime})\rho(E^{\prime})=c^{\prime}E^{\prime}$ forces

$\displaystyle\rho(c^{\prime})=zc^{\prime},\rho(E^{\prime})=z^{-1}E^{\prime}$

for some $z\in{\mathbb{K}}^{*}$. Similarly,
$\rho(b^{\prime})=z^{\prime}c^{\prime},\rho(E^{\prime})=z^{\prime-1}E^{\prime}$
for some $z^{\prime}\in{\mathbb{K}}^{*}$.

When $\rho(\phi_{1})=\phi_{2},\rho(\phi_{2})=\phi_{1}$. The proof is similar.
Hence

$\displaystyle\\{\rho|_{S}\mid\rho\in{\mathcal{D}_{q}}(\mathfrak{b}_{m,n})\\}=\\{\xi_{z_{3},z_{4}}|_{S}\mid
z_{3},z_{4}\in{\mathbb{K}}^{*}\\}={\rm Aut}(S)\cong({\mathbb{K}}^{*})^{2}.$

The proof is finished. ∎

######  Remark 5.4.

The structure of the automorphism group of
${\mathcal{D}_{q}}(\mathfrak{b}_{m,n})$ is independent on the choice of $m,n$.
When $m=n=1$, Theorem 5.3 corrects the result of [T2, Theorem 4.5], in which
the automorphism group is $({\mathbb{K}}^{*})^{4}\rtimes{\mathbb{Z}}$.

## Acknowledgments

The author gratefully acknowledge partial financial supports from the National
Natural Science Foundation of China (11871249, 12171155).

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