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quantum plane, quantum algebra,  singular magnetic field
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\begin{document}


\title{\bf  Representations of the Quantum Plane and the Quantum Algebra
$U_{q}({\rm sl}_2)$ on $L^2(\R^d)$}
\author{ Asao Arai\thanks{
Supported by the Grant-In-Aid No.17340032 for Scientific Research
from the JSPS.}\\ Department of Mathematics, 
Hokkaido University \\
Sapporo 060-0810, Japan \\ e-mail:  arai@math.sci.hokudai.ac.jp
      }

\maketitle
\begin{abstract}A class of representations on the Hilbert space
$L^2(\R^d)$ ($d\geq 2$) of the quantum plane $\C_q^2$ and the quantum
algebra $U_q({\rm sl}_2)$ is presented. The boundedness and the unboundedness of
the representations are discussed. A physically interesting 
example of the representations
is shown to appear in a two-dimensional quantum system with a 
magnetic field concentrated on an infinite lattice.  
\end{abstract}

\noindent
{\bf AMS Mathematics Subject Classification}: 81R50, 81Q99

 
\noindent
{\bf Keywords:} quantum plane, quantum algebra,  singular magnetic field.

\section{Introduction}


In the previous paper \cite{Ar98} (cf. also \cite{Ar96}), 
the author considered a quantum system
in the plane $\R^2$ which is under the influence of a perpendicular magnetic field
concentrated on an infinite discrete set $\mathbf{D}$ of $\R^2$ 
 and showed that, in this quantum system,  there appear, in a natural way, representations of the canonical commutation relations (CCR) with two degrees of freedom.
Interestingly enough, these representations are not necessarily unitarily 
 equivalent to the Schr\"odinger representation of the CCR with two degrees of freedom and each inequivalent representation physically corresponds to the occurrence
of the  Aharonov-Bohm effect
in the context of the quantum system under consideration. Moreover,  
in connection with the inequivalent representations with $\mathbf{D}$ being an infinite lattice,
bounded operator representations of 
the quantum plane $\C_q^2$ and the quantum algebra
$U_q({\rm sl}_2)$ with $|q|=1$ ($q\in \C$)
were constructed as well as their reductions to  lattice quantum systems.


From  a mathematical point of view, it would be natural to ask if there is any general
structure behind the representations of $\C_q^2$ ($|q|=1$) constructed
in \cite{Ar98}. This is one of the motivations for the present work. In this paper, we show that the answer is affirmative, presenting a general theory
of representations of $\C_q^2$ on the Hilbert space $L^2(\R^d)$ ($d\geq 2$) with
$|q|$ not necessarily equal to one. 
This general theory, of course, includes, as
examples, the previously obtained representations of $\C_q^2$ mentioned above.
A new feature of the representations of $\C_q^2$ given in this paper
is in that they may be unbounded. 



The present paper is organized as follows. Section 2 is a preliminary,
where we define a class of closed linear operators on $L^2(\R^d)$
constructed from the Schr\"odinger representation of the 
CCR with $d$ degrees of freedom.
In Section 3 we present  a general scheme for the construction of  a class of
 representations of $\C_q^2$ on $L^2(\R^d)$. 
Here we see that, to obtain an unbounded operator representation,
the constituent operators of the representation are required to have
a common invariant domain. A possible form  of  such an invariant domain is
given in Section 4 in the case $d=2$. 
In the paper \cite{Ar96}, we showed a scheme to obtain  representations
of $U_q({\rm sl}_2)$ from a representation of $\C_q^2$.
Using this scheme and the results in Section 3, we present in Section 5 a class of 
representations of $U_q({\rm sl}_2)$ on $L^2(\R^d)$. In the last section,
we consider a two dimensional quantum system
with a singular gauge potential 
determined by a meromorphic function 
of which  poles are on  an infinite lattice  and  show that
representations of $\C_q^2$  (resp. $U_q({\rm sl}_2)$)  of the type discussed in 
Section 3 (resp. Section 5) 
are realized  in this system. 


\section{Preliminary}


For a linear operator $A$ on a Hilbert space, we denote its domain by $D(A)$. 
Let $d\geq 2$ be a natural number,
$\R^d=\{\x=(x_1,\cdots,x_d)|x_j\in \R,j=1,\cdots,d\}$ and
$D_j$ be the generalized partial differential operator in $x_j$ on
$L^2(\R^d)$. Then, as is well known, the operator
\begin{equation}
p_j:=-iD_j
\end{equation}
($i$ is the imaginary unit) is self-adjoint.
We set
\begin{equation}
\p:=( p_1,\cdots, p_d).
\end{equation}
It follows that,
for each vector $\mathbf{a}=(a_1,\cdots,a_d)\in \R^d$, the operator
\begin{equation}
\mathbf{a}\cdot \p:=\sum_{j=1}^da_jp_j
\end{equation}
with $D(\mathbf{a}\cdot \p)=\cap_{j=1}^dD(p_j)$ is essentially self-adjoint and
\begin{equation}
V(\mathbf{a}):=e^{i\overline{\mathbf{a}\cdot\p}}
=e^{ia_1p_1}\cdots e^{ia_dp_d}=e^{ia_dp_d}\cdots e^{ia_1p_1},
\end{equation}
where $\overline{\mathbf{a}\cdot \p}$ denotes the closure
of  $\mathbf{a}\cdot\p$.

For a Borel measurable function $F$ on $\R^d$ almost everywhere (a.e.) finite
with respect to the Lebesgue measure on $\R^d$, 
we denote the multiplication
operator by the function $F$ on $L^2(\R^d)$ by the same symbol $F$.
For a vector $\mathbf{a}\in \R^d$, we define a function $F_\mathbf{a}$ on $\R^d$ by
\begin{equation}
F_\mathbf{a}(\x):=F(\x+\mathbf{a}), \quad \x\in \R^d.
\end{equation}
The following lemma is  well-known (or easily proved).


\begin{lemma}\label{lem2-1} For all $\mathbf{a}\in\R^d$ and each Borel measurable function $F$ on $\R^d$, the operator equality
\begin{equation}
V(\mathbf{a})FV(\mathbf{a})^*
=F_{\mathbf{a}}
\end{equation}
holds, where $V(\a)^*$ denotes the adjoint of $V(\a)$.
\end{lemma}


Let 
\begin{equation}
T(\a):=FV(\mathbf{a}).\label{T-def}
\end{equation}


Since the function $F$ is a.e. finite with respect to the Lebesgue measure on $\R^d$, the multiplication operator
$F$ is a densely defined closed
linear operator on $L^2(\R^d)$. It follows that the same holds for $T(\a)$.
But $T(\a)$ may be unbounded. We first give a characterization for the boundedness of 
$T(\a)$.




\begin{proposition}\label{prop2-1} The operator
$T(\a)$ is bounded if and only if 
the function $F$ is essentially bounded on $\R^d$.
In that case, $D(T(\a))=D(F)=L^2(\R^d)$.
\end{proposition}

{\it Proof}. Suppose that $T(\a)$ is bounded. 
Then it follows from the closedness of $T(\a)$ and the denseness of $D(T(\a))$,
as mentioned above, that $D(T(\a))=L^2(\R^d)$.
Moreover there exists a constant $C>0$ such that, for all $\psi\in L^2(\R^d)$, 
 $\|FV(\mathbf{a})\psi\|\leq C\|\psi\|$, where
$\|\cdot\|$ denotes the norm of $L^2(\R^d)$. Since $V(\mathbf{a})$
is unitary, it follows that  $D(F)=L^2(\R^d)$  and $F$ is bounded. Hence 
the function $F$ is essentially bounded on $\R^d$.
Proving the converse statement is easy.
\hfill $\Box$

\medskip

The next proposition  is concerned with unitarity of $T(\a)$.


\begin{proposition}\label{unitary}
The operator $T(\a)$ is unitary if and only if
$|F(\x)|=1, {\rm a.e.}\x\in \R^d$.
\end{proposition}


{\it Proof}. Suppose that $T(\a)$ is unitary.
Then, $F=T(\a)V(\a)^{-1}$ is unitary.
Hence $|F|$ is a unitary self-adjoint multiplication operator.
Therefore the spectrum of $|F|$ is $\{1\}$.
This implies that
$|F(\x)|=1, \,{\rm a.e.}\x\in \R^d$.


Conversely, suppose that $|F(\x)|=1, \,{\rm a.e.}\x\in \R^d$.
Then the multiplication operator  $F$ is unitary. The operator $V(\a)$ is also unitary. 
Hence $T(\a)$ is unitary.
\hfill $\Box$

\medskip



We also note the following fact:

\begin{proposition}\label{prop2-2}
Let
\begin{equation}
N_F:=\{\x\in \R^d|F(\x)=0\}.
\end{equation}
Then 
 $T(\a)$ is injective if and only if the Lebesgue measure $|N_F|$ of $N_F$ is equal to zero. In that case, 
the multiplication operator $F$  is injective and 
\begin{equation}
T(\a)^{-1}=V(-\mathbf{a})F^{-1}\label{T-inv}
\end{equation}
on $D(F^{-1})$.
\end{proposition}


{\it Proof}. Suppose that $T(\a)$ is injective.
Let $\psi\in \ker F$ so that $F\psi=0$.
Hence, putting $\phi:=V(-\mathbf{a})\psi$, we have $\phi\in D(T(\a))$ and $T(\a)\phi=0$.
Hence $\phi=0$. This implies that $\ker F=\{0\}$.
Hence $F$ is injective. On the other hand, it is easy to see that 
\begin{equation}
\ker F
=\{\psi \in L^2(\R^d)| \{\x\in K_{\psi}^c|\psi(\x)\not=0\}\subset N_F\},
\label{kerF}
\end{equation}
where $K_{\psi}$ is a null set of $\R^d$ depending on $\psi$. 
Hence $|N_F|$ must be zero.

Conversely, suppose that $|N_F|=0$. Then, by (\ref{kerF}),
 $\ker F=\{0\}$. Hence $F$ is injective. Let $\psi\in \ker T(\a)$ so that $T(\a)\psi=0$.
Then $V(\mathbf{a})\psi\in \ker F$. Hence $V(\mathbf{a})\psi=0$, implying
$\psi=0$.

If $F$ is injective, then 
(\ref{T-inv}) easily follows from (\ref{T-def}).
\hfill $\Box$

\medskip


For each vector $\b\in \R^d$, we define a unitary operator
\begin{equation}
U(\b):=e^{i\b\cdot \x}
\end{equation}
(the multiplication operator
by the function $:\x\mapsto e^{i\b\cdot \x}$)
on $L^2(\R^d)$.
It is easy to see that
\begin{equation}
V(\a)U(\b)=e^{i\a\cdot \b}U(\b)V(\a), \quad \a,\b\in \R^d.\label{UV}
\end{equation}
Therefore, $V(\a)$ and $U(\b)$ commute
if and only if $\a\cdot \b=2\pi n$ with some $n\in \Z$ (the set of integers).





\begin{proposition}\label{prop2-3} 
For all $\a,\b\in \R^d$, $U(\b)D(T(\a))=D(T(\a))$ and the operator equality
\begin{equation}
U(\b)T(\a)=e^{-i\a\cdot\b}T(\a)U(\b) \label{UT}
\end{equation}
holds. 
In particular, if  $\a\cdot \b \in 2\pi \Z:=\{2\pi n|n\in \Z\}$.
then 
$$
U(\b)T(\a)=T(\a)U(\b).
$$
\end{proposition}

{\it Proof}. By definition,  $\psi\in L^2(\R^d)$ is in $D(T(\a))$ if and only if  
$$
\int_{\R^d}|F(\x)|^2|(V(\a)\psi)(\x)|^2d\x<\infty.
$$ 
Let $\psi\in D(T(\a))$ and $\phi=U(\b)\psi$. Then, by (\ref{UV}), we have
$$
V(\a)\phi=e^{i\a\cdot\b}U(\b)V(\a)\psi.
$$
Hence 
$$
\int_{\R^d}|F(\x)|^2|(V(\a)\phi)(\x)|^2d\x=
\int_{\R^d}|F(\x)|^2|(V(\a)\psi)(\x)|^2d\x<\infty.
$$
 Therefore $\phi \in D(T(\a))$. Thus $U(\b)D(T(\a))\subset D(T(\a))$. This implies that 
$$
D(T(\a))\subset U(-\b)D(T(\a)) \subset D(T(\a)).
$$
Hence it follows that  $U(\b)D(T(\a))=D(T(\a))$. 
By this result, we have $D(U(\b)T(\a))=D(e^{-\a\cdot\b}T(\a)U(\b))$.
It is straightforward to see that 
$$
U(\b)T(\a)\psi=e^{-I\a\cdot\b}T(\a)U(\b)\psi, \quad \psi \in D(T(\a)).
$$
\ \hfill $\Box$

\medskip

\begin{proposition}\label{prop2-4} For all $\a\in \R^d$, 
$$
T(\a)^*=(F^*)_{-\a}V(-\a).
$$
In particular, $D(F)\subset D(T(\a)^*)$.
\end{proposition}

{\it Proof}. By Lemma \ref{lem2-1}, we have $T(\a)=V(\a)F_{-\a}$.
Since $V(\a)$ is bounded, it follows that 
$$
T(\a)^*=F_{-\a}^*V(\a)^*=
(F^*)_{-\a}V(-\a).
$$ 

Let $\psi \in D(F)$. Then 
$$
\int_{\R^d}|(F^*)_{-\a}(\x)|^2|(V(-\a)\psi)(\x)|^2d\x
=\int_{\R^d}|F(\x)|^2|\psi(\x)|^2d\x< \infty.
$$
Hence $\psi\in D(T(\a)^*)$.
\hfill $\Box$

\medskip



\section{A Class of  Representations of
the Quantum Plane on $L^2(\R^d)$}


Let $\mathbf{w}_1$ and $\mathbf{w}_2$
be linearly independent vectors in $\R^d$.
Suppose that there exist non-zero Borel measurable 
functions $F_j:\R^d\to \C$ a.e. finite with respect to
the Lebesgue measure on $\R^d$  and a constant
$q\in \C\setminus\{0\}$
satisfying
\begin{equation}
F_1(\x)F_2(\x+\mathbf{w}_1)=q
F_1(\x+\mathbf{w}_2)F_2(\x), \quad{\rm a.e.}\x\in \R^d. \label{U}
\end{equation}

A simple example for such $F_1$ and $F_2$ is given as follows:

\begin{example}{\rm Let $\u,\v\in \R^d, \alpha,\beta\in \C$ and 
$$
q:=e^{\beta\w_1\cdot \v-\alpha\w_2\cdot \u}.
$$
Let
$$
F_1(\x):=e^{\alpha \x\cdot\u}, \quad F_2(\x):=e^{\beta\x\cdot\v}.
$$
Then it is easily checked that these $F_1$ and $F_2$ satisfy (\ref{U}).
}
\end{example}



We define
\begin{equation}
T_j:=F_jV(\w_j), \quad j=1,2.
\end{equation}
A simple application of Proposition \ref{prop2-1}
gives the following fact:

\begin{proposition}\label{prop3-1}
For each $j=1,2$, $T_j$ is bounded if and only if the function
$F_j$ is essentially bounded on $\R^d$. In that case, $D(T_j)=D(F_j)=
L^2(\R^d)$.
\end{proposition}

Proposition \ref{prop2-2} yields the following result:

\begin{proposition}\label{prop3-2} For each $j=1,2$, $T_j$ is injective if and only if
$|N_{F_j}|=0$. In that case, $F_j$ is injective and 
$T_j^{-1}=V(-\w_j)F_j^{-1}$ on $D(F_j^{-1})$.
\end{proposition}






One of the main results of this section is the following theorem:
 


\begin{The}\label{th3-1} 
For all $\psi\in D(T_1^*)\cap D(T_2^*)$ and $\phi
\in D(T_1)\cap D(T_2)$,
\begin{equation}
\lang T_1^*\psi, T_2\phi\rang=q\lang T_2^*\psi,T_1\phi\rang,\label{com1}
\end{equation}
where $\lang \,\cdot\,, \,\cdot \,\rang$ denotes the inner product of $L^2(\R^d)$.
\end{The}


{\it Proof}. 
By Proposition \ref{prop2-4}, we have 
\begin{eqnarray*}
&&\lang T_1^*\psi, T_2\phi\rang
=\int_{\R^d} F_1(\x-\w_1)F_2(\x)\psi(\x-\w_1)^*
\phi(\x+\w_2)d\x,\\
&&\lang T_2^*\psi, T_1\phi\rang
=\int_{\R^d} F_2(\x-\w_2)F_1(\x)\psi(\x-\w_2)^*
\phi(\x+\w_1)d\x.
\end{eqnarray*}
By change of variables and condition (\ref{U}),
one sees that (\ref{com1}) holds.
\hfill $\Box$

\medskip

Theorem \ref{th3-1} and Proposition \ref{prop2-1} imply the following
result:


\begin{corollary} Suppose that $F_1$ and $F_2$ are essentially bounded on $\R^d$.
Then $T_1$ and $T_2$ are bounded with $D(T_j)=L^2(\R^d)$ ($j=1,2$) and
\begin{equation}
T_1T_2=qT_2T_1.
\end{equation}
\end{corollary}

Thus, in the case where $F_1$ and $F_2$ are essentially bounded,
the set $\{T_1,T_2\}$ of operators yields a bounded representation
of the quantum plane $\C_q^2$ with $q\not=0,\pm 1$ \cite{KS}. 

In the  case where at least one of $F_1$ and $F_2$ is not essentially bounded on $\R^d$, however, we need some condition for $\{T_1,T_2\}$ to give a representation
of $\C_q^2$:



\begin{corollary}\label{cor3-2} 
Suppose that there exists a dense subspace $D$ in $L^2(\R^d)$ such that,
for $j=1,2$, 
$V(\w_j)D\subset D$ and $F_jD\subset D$.
Then $T_jD\subset D$ ($j=1,2$) and 
\begin{equation}
T_1T_2=qT_2T_1 \label{com2}
\end{equation}
on $D$.
\end{corollary}

{\it Proof}. The property $T_jD\subset D$ follows from the present assumption and
the definition of $T_j$. 
By Proposition \ref{prop2-4}, 
$D\subset D(T_1^*)\cap D(T_2^*)$. Hence, by Theorem \ref{th3-1},
we have that, for all $\psi,\phi\in D$, $\lang \psi, T_1T_2\phi\rang=\lang \psi, qT_2T_1\phi\rang$. Since $D$ is dense, we obtain (\ref{com2}) on $D$.
\hfill $\Box$

\medskip
 


Suppose that the assumption of Corollary \ref{cor3-2} holds. 
Then we can define the algebra
generated by $T_1|D$ (the restriction of $T_1$ to $D$) and $T_2|D$.
We denote it 
$\mathscr{O}_{\w_1,\w_2}(D)$. 
We  define
\begin{equation}
\mathscr{O}_{\w_1,\w_2}(D)':=\{B\in \mathfrak{B}(L^2(\R^d))
|\lang B^*\phi,T_j\psi\rang=\lang T_j^*\phi, B\psi\rang, 
\phi,\psi \in D,j=1,2\},
\end{equation}
where $\mathfrak{B}(L^2(\R^d))$ denotes the Banach space of all
bounded linear operators $B$ on $L^2(\R^d)$ with $D(B)=L^2(\R^d)$.

We can apply Proposition \ref{prop2-3} to obtain the following
fact:


\begin{proposition}
Let $\a\in \R^d$ be such that $\a\cdot \w_j\in 2\pi \Z, j=1,2$ and
$U(\a)D\subset D$.
Then $U(\a)\in\mathscr{O}_{\w_1,\w_2}(D)'$.
\end{proposition}




\section{Invariant Domains in the Case $d=2$}


We consider the case $d=2$ and suppose that
$\w_1\times \w_2>0$. In this case, 
a possible invariant domain for $T_1$ and $T_2$ can be 
constructed as follows. 
We first introduce  a subset  $\L_j$ of $\R^d$ ($j=1,2$) by
\begin{equation}
\L_1:=\cup_{n\in \Z}\{t\w_1+n\w_2|t\in \R\}, \quad
 \L_2:=\cup_{m\in \Z}\{m\w_1+t\w_2|t\in \R\}.
\end{equation}
Note that  
\begin{equation}
\L_1\cap \L_2=\L_{\w_1,\w_2}:=\{m\w_1+n\w_2|m,n\in \Z\},\label{lattice}
\end{equation}  
a two-dimensional lattice. 
The set
\begin{equation}
\Omega:=\L_1^c\cap \L_2^c=(\L_1\cup \L_2)^c
\end{equation}
is an open set in $\R^d$. It is easy to see that
\begin{equation}
\Omega\pm \w_j=\Omega, \quad j=1,2. \label{omega}
\end{equation}
Hence $\Omega$ has the translation symmetry with vectors $\pm \w_j$ ($j=1,2$).

We denote by $C_0^{\infty}(\Omega)$ the set
of infinitely differentiable functions on $\R^d$ with compact support in
$\Omega$.


\begin{proposition}\label{prop3-8} Suppose that each
$F_j$ ($j=1,2$) is infinitely differentiable on $\Omega$.
Then $T_jC_0^{\infty}(\Omega)\subset C_0^{\infty}(\Omega)$
and (\ref{com2}) holds on $C_0^{\infty}(\Omega)$.
\end{proposition}


{\it Proof}. Let $\psi \in C_0^{\infty}(\Omega)$. We have
\begin{equation}
(V(\w_j)\psi)(\x)=\psi(\x+\w_j), \quad \x\in \R^d.
\end{equation}
This relation and (\ref{omega}) imply that $V(\w_j)\psi\in C_0^{\infty}(\Omega)$. 
It is obvious from the present assumption for $F_j$
that $F_jC_0^{\infty}(\Omega)\subset C_0^{\infty}(\Omega)$.
Thus the desired result follows.
\hfill $\Box$

\medskip

\section{A Class of Representations of $U_q({\rm sl}_2)$ on $L^2(\R^d)$}


For a complex number $q\in \C\setminus\{0,-1,1\}$, the quantum algebra
$U_q({\rm sl}_2)$ is defined to be the complex associative algebra with unit $1$ 
 generated by four elements
$E,F,K,K^{-1}$ subject to the following relations \cite{KS}:
\begin{eqnarray}
&&KK^{-1}=K^{-1}K=1,\\
&& KEK^{-1}=q^2E, \quad KFK^{-1}=q^{-2}F,\\
&& [E,F]=\frac{K-K^{-1}}{q-q^{-1}},
\end{eqnarray}
where $[X,Y]:=XY-YX$.
As is shown in Lemma 4.1 and Lemma 5.1 in \cite{Ar96}, 
there is a general scheme to construct 
a representation of $U_q({\rm sl}_2)$ 
from a representation of the quantum plane $\C_q^2$.

Let $T_1$ and $T_2$ be as in Section 3.

\begin{The}\label{th4-1} Assume that $|N_{F_j}|=0, \,j=1,2$.
Suppose that there exists a dense subspace $D$ in $L^2(\R^d)$ such that,
for $j=1,2$, 
$V(\w_j)D\subset D$ and $F_jD\subset D, F_j^{-1}D\subset D$.
Let $a,b,a',b'\in \C$ be constants satisfying
\begin{equation}
abq=\frac{a'b'}{q}=-\frac 1{\left(q-q^{-1}\right)^2}.
\end{equation}
and
\begin{eqnarray}
&&E:=T_2(aT_2+a'T_2^{-1})T_1^{-1},\\
&& F:=T_1(bT_2+b'T_2^{-1})T_2^{-1},\\
&& K:=T_2^2, \quad K^{-1}:=(T_2^{-1})^2.
\end{eqnarray}
Then the set $\{E,F, K, K^{-1}\}$ generates
a representation of $U_q({\rm sl}_2)$ on the vector space $D$.
\end{The}


{\it Proof}. We need only to apply Lemma 4.1 in \cite{Ar96} to the representation $\{T_1,T_2\}$
of $\C_q^2$.
\hfill $\Box$

\medskip


We denote  
by $\Pi_{a,b;a',b'}$ the representation of $U_q({\rm sl}_2)$ given in Theorem \ref{th4-1}.



\begin{proposition}\label{prop4-1} There is no
finite dimensional subspace $M$
of $D$ which reduces the representation $\Pi_{a,b;a',b'}$ and
in which the highest weight of $\Pi_{a,b;a',b'}|M$ is not equal to $-a'/aq^2$.
\end{proposition}


{\it Proof}. Suppose that there is a finite dimensional subspace $M$ of $D$
which reduces the representation $\Pi_{a,b;a',b'}$.
By a general fact on  finite dimensional
representations of $U_q({\rm sl}_2)$ (\cite[Proposition VI.3.3]{Ka}),
The representation $\Pi_{a,b;a',b'}|M$ has a
highest weight vector $\psi\in M\setminus\{0\}$:
$E\psi=0$ and $K\psi=\lambda \psi$ with highest weight 
$\lambda\in \C\setminus\{0\}$. Then
\begin{eqnarray}
&& T_2(aT_2+a'T_2^{-1})T_1^{-1}\psi=0,\label{E}\\
&&
F_2(\x)F_2(\x+\w_2)\psi(\x+2\w_2)=\lambda \psi(\x), \quad {\rm a.e.}\x\in \R^d.
\label{K}
\end{eqnarray}
By (\ref{E}) and the easily derived relation
$$
T_2T_1^{-1}=qT_1^{-1}T_2,
$$
we obtain $(aq^2\lambda+a')T_1^{-1}\psi=0$.
Hence $aq^2\lambda+a'\not=0$. Therefore $\lambda=-a'/aq^2$.
This implies the desired assertion.
\hfill $\Box$

\medskip


\begin{The} Suppose that the following (i) or (ii) hold:
\begin{list}{}{}
\item{(i)} For a constant $c>0$, $|F_2(\x)|=c, \ {\rm a.e.}\x\in \R^d$.
\item{(ii)} $|F_2(\x)|^2 \geq |a'|/|a||q|^2, \ {\rm a.e.}\x\in \R^d$.
\end{list}
Then there is no finite dimentional subspace  $M$ of $D$ which reduces
the representation $\Pi_{a,b;a',b'}$.
\end{The}


{\it Proof}.
By a general fact on  finite dimensional
representations of $U_q({\rm sl}_2)$ (\cite[Proposition VI.3.3]{Ka}),
we need only to show that the representation $\Pi_{a,b;a',b'}$ has no
highest weight vector. Let $\psi$ be a  vector in $D$ such that
$E\psi=0$ and $K\psi=\lambda \psi$ with
$\lambda\in \C\setminus\{0\}$. Then
(\ref{E}) and
(\ref{K}) hold.
By Proposition \ref{prop4-1}, we need only to consider the case $\lambda=-a'/aq^2\not=0$.
Then, by (\ref{K}), we have $|F_2(\x)| |F_2(\x+\w_2)||\psi(\x+2\w_2)|=|a'| |\psi(\x)/|a||q|^2$.

Let condition (i) be satisfied. Then we have $k|\psi(\x+2\w_2)|=|\psi(\x)|$
with $k=|a||q|^2c^2/|a'|$. Since $\psi$ is in $L^2(\R^d)$, it follows that $k=1$.
Then $|\psi(\x+2\w_2)|=|\psi(\x)|$. But this implies $\int_{\R^d}|\psi(\x)|^2dx=\infty$
if $\psi\not=0$. Thus $\psi=0$.

We next consider the case where condition (ii) holds.
In this case we have $|\psi(\x+2\w_2)|\leq |\psi(\x)$. 
This inequality and condition $\psi\in L^2(\R^d)$ imply that $\psi=0$.
 \hfill $\Box$

\medskip

The element  $C$ defined by
\begin{equation}
C:=\frac{qK-2+q^{-1}K^{-1}}{(q-q^{-1})^2}+FE
\end{equation}
is 
called the Casimir element of $U_q({\rm sl}_2)$ and 
commutes with $E,F,K$ and $K^{-1}$.


In the representation $\Pi_{a,b;a',b'}$, we have 
\begin{equation}
\Pi_{a,b;a',b'}(C)=a'b+ab'-\frac{2}{(q-q^{-1})^2},\label{Casimir}
\end{equation}
which is a scalar.
   
\begin{The} Let $\Pi_{a,b;a',b'}$ and
 $\Pi_{c,d;c',d'}$ be two representations of $U_q({\rm sl}_2)$
of the form given in Theorem \ref{th4-1}. 
Suppose that
$a'b+ab'\not=c'd+cd'$. Then
$\Pi_{a,b;a',b'}$ is not unitarily equivalent to
 $\Pi_{c,d;c',d'}$. 
\end{The}

{\it Proof}. If $\Pi_{a,b;a',b'}$ is unitarily equivalent to
 $\Pi_{c,d;c',d'}$ with a unitary operator $U$, then 
$U\Pi_{a,b;a',b'}(C)U^{-1}=
\Pi_{c,d;c',d'}(C)$. Hence, by (\ref{Casimir}), we have
$a'b+ab'=c'd+cd'$. But this is a contradiction.
\hfill $\Box$

\medskip




\begin{proposition} Let $F_1$ and $F_2$ be essentially bounded and
$\mathfrak{A}_q$ be the $*$-algebra  generated by
$E,F,K,K^{-1}$ in Theorem \ref{th4-1}. Then
$\mathfrak{A}_q$ is not irreducible.
\end{proposition}


{\it Proof}. 
Let $W$ be the two-dimensional subspace generated by $\w_1$ and $\w_2$ and
$\{{\bf f}_1,{\bf f}_2\}$
be an orthonormal  basis of $W$. We expand $\w_j$ ($j=1,2$) as
$\w_j=w_{j1}{\bf f}_1+w_{j2}{\bf f}_2$ ($w_{j1},w_{j2}\in \C$) and define a
matrix 
\begin{equation}
T:=\left(
\begin{array}{cc}
w_{11} & w_{12} \\
w_{21} & w_{22}
\end{array}
\right)
\end{equation}
Then $T$ is regular. For each $\n=(n_1,n_2)\in \Z\times \Z$, we define 
$(a_{\n1},a_{\n2})\in \C^2$ by
\begin{equation}
\left(
\begin{array}{c}
a_{\n1} \\
a_{\n2}
\end{array}
\right):=2\pi T^{-1}\n.
\end{equation}
Then the vector
$\a_\n:=a_{\n1}{\bf f}_1+a_{\n2}{\bf f}_2$
satisfies $\a_\n\cdot\w_j=2\pi n_j
\in 2\pi\Z$.
Hence $U(\pm \a_\n)$ commute with $E,F,K$ and $K^{-1}$, and hence all those of 
$\mathfrak{A}_q$ (note that $U(\a_\n)^*=U(-\a_\n)$).
Therefore the commutant of $\mathfrak{A}_q$ is not the set of scalar operators.
Thus $\mathfrak{A}_q$ is not irreducible.
\hfill $\Box$

\medskip

\section{Representations of $\C_q^2$ and
$U_q({\rm sl}_2)$
in a quantum system with a singular gauge potential}


In this section we consider a two-dimensional quantum system with a perpendicular
magnetic field concentrated on the infinite lattice $\L_{\w_1,\w_2}$
defined by (\ref{lattice}).

For a vector $\a=(a_1,a_2)$, we denote its corresponding
complex number by $a=a_1+ia_2$.

Let $f$ be a meromorphic function on
$\C\setminus\{nw_1+mw_2|m,n\in \Z\}$ with possible poles
on the points $nw_1+mw_2$ ($m,n\in \Z$) ($w_j=w_{j1}+iw_{j2}$).
Then one can define a gauge  potential 
$\A$ on 
\begin{equation}
\M:=\R^2\setminus \L_{\w_1,\w_2}
\end{equation} 
by
\begin{equation}
\A(\x)=(A_1(\x),A_2(\x))
\end{equation}
with
\begin{equation}
A_1(\x):=\Im f(x_1+ix_2), \quad A_2(\x):=\Re f(x_1+ix_2), \quad \x=(x_1,x_2)
\in \M.
\end{equation}
The magnetic field $B$ is defined as a distribution on $\R^2$
by
\begin{equation}
B:=D_1A_2-D_2A_1.
\end{equation}

Each component $A_j$ ($j=1,2$) is infinitely differentiable
on  $\M$. By
the Cauchy-Riemann equation for $f$, we have
\begin{eqnarray}
&&B(\x)=0,\quad  \partial_1A_1(\x)+\partial_2A_2(\x)=0, \quad \x\in \M.
\end{eqnarray}
Hence the magnetic field $B$ is concentrated, as  a distribution, on
the set $\L_{\w_1,\w_2}$. 


Let
\begin{equation}
F_j^{\A}(\x):=e^{\alpha I_\A^{(j)}(\x)}, \quad \x \in \R^2\setminus \L_j
\end{equation}
with $\alpha \in \C\setminus \{0\}$ and 
\begin{equation}
I_\A^{(j)}(\x):=\int_{\x}^{\x+\w_j}\A(\x')\cdot d\x', \quad \x \in \R^2\setminus \L_j,
\end{equation}
where the integral $\int_{\x}^{\x+\w_j}$ means the line integral
along the straightline : $t \mapsto \x+t\w_j, \  t\in [0,1]$.

Let
\begin{equation}
\Omega_{m,n}:=mw_1+nw_2.
\end{equation} and
${\rm Res}(\Omega_{m,n},f)$ be the residue
of $f$ at $z=\Omega_{m,n}$.

{\it In what follows, we assume that 
\begin{equation}
c_f:=\Re ({\rm Res}(\Omega_{m,n},f)),
\end{equation}
the real part of the residue ${\rm Res}(\Omega_{m,n},f)$, 
is a non-zero constant independently of $(m,n)\in \Z^2$}.

We introduce a constant:
\begin{equation}
q_{\alpha}:=e^{2\pi\alpha c_f}
\end{equation}

\begin{lemma} For all $\x\in \Omega$, 
\begin{equation}
F_1^{\A}(\x)F_2^{\A}(\x+\w_1)=q_{\alpha}F_1^{\A}(\x+\w_2)F_2^{\A}(\x).
\end{equation}
\end{lemma}

{\it Proof}. It is sufficient to show that, for each $\x\in \Omega$,
\begin{equation}
I_{\A}^{(1)}(\x)+I_{\A}^{(2)}(\x+\w_1)
-I_\A^{(1)}(\x+\w_2)
-I_\A^{(2)}(\x))=2\pi c_f.
\end{equation}
Let $L$ be the left hand side of this equation.
It is easy to see that 
$L=
\Im \int_Cf(z)dz$, where $C$ is  the closed curve starting and ending at $\x$
which is a composition of four straight lines 
going as $\x\to \x+\w_1\to \x+\w_1+\w_2\to \x+\w_2 \to \x$,
forming a parallelogram. There exists a unique $(m,n)\in \Z^2$
such that $\Omega_{m,n}$ is in the interior of $C$.
Hence, by the residue theorem, we have $\int_Cf(z)dz=2\pi i {\rm Res}(\Omega_{m,n},f)$.
Thus $L=2\pi  c_f$.
\hfill $\Box$

\medskip

We define
\begin{equation}
T_j^{\A}:=F_j^{\A}V(\w_j), \quad j=1,2.
\end{equation} 

\begin{The} Each $T_j^\A$ leaves $C_0^{\infty}(\Omega)$ invariant and
\begin{equation}
T_1^{\A}T_2^{\A}=q_{\alpha}T_2^{\A}T_1^{\A}
\end{equation}
holds on $C_0^{\infty}(\Omega)$.
\end{The}

{\it Proof}. It is easy to see that the function $F_j^\A$ is
infinitely differentiable on $\Omega$.
Hence we can apply Proposition   \ref{prop3-8} with
$F_j=F_j^{\A}$ to obtain the desired result.
\hfill $\Box$

\medskip
 
Thus $\{T_1^{\A},T_2^{\A}\}$ gives
a representation of the quantum plane $\C_{q_{\alpha}}^2$ on $C_0^{\infty}(\Omega)$.
We next consider the boundedness of this representation.

\begin{The} Let $\Re \alpha =0$.
Then each $T_j^{\A}$ is unitary.
\end{The}


{\it Proof}. In the present case we have
$|F_j^{\A}|=1$. Hence, by Proposition \ref{unitary},
$T_j^{\A}$   is unitary.
\hfill $\Box$

\medskip

\begin{remark}{\rm If $\Re\alpha=0$, then $q_{\alpha}
=e^{2\pi i(\Im \alpha)c_f}$. Hence $|q_{\alpha}|=1$.
}
\end{remark}



In the case where $\Re \alpha \not=0$, for each $j=1,2$,  
we introduce two real-valued functions  $\R^2\setminus \L_j$ as follows:
\begin{equation}
K_j(\x):=
(\Re \alpha)\Im \int_{x}^{x+w_j}f(z)dz, \quad \x\in \R^2\setminus \L_j.
\end{equation}

\begin{proposition} For each $j=1,2$, 
$T_j^{\A}$ is bounded
if and only if the function $K_j$ is bounded above on $\R^2\setminus \L_j$.
\end{proposition}

{\it Proof}. By Proposition \ref{prop2-1},
$T_1^{\A}$ and $T_2^{\A}$ are bounded if and only if
$F_1^{\A}$ and $F_2^{\A}$
are essentially bounded on $\R^2$.
Note that
$$
I_{\A}^{(j)}(\x)=\Im \int_x^{x+w_j}f(z)dz,\quad \x\in \R^2\setminus \L_j.
$$
Hence  $|F_j^{\A}(\x)|=e^{K_j(\x)}, \ \x\in \R^2\setminus \L_j$.
Obviously each $K_j$ is continuous on $\Omega$.   
Therefore
$F_j^{\A}$ is essentially bounded on $\R^2$ if and only if
$K_j$ is bounded above on $\R^2\setminus \L_j$.
\hfill $\Box$

\medskip
We can apply Theorem \ref{th4-1} to the representation
$\{T_1^{\A},T_2^{\A}\}$ of the quantum plane $\C_{q_{\alpha}}^2$
to obtain the following result:

\begin{corollary} 
Let $a,b,a',b'\in \C$ be constants satisfying
\begin{equation}
abq_{\alpha}=\frac{a'b'}{q_{\alpha}}=-\frac 1{\left(q_{\alpha}-q_{\alpha}^{-1}\right)^2}.
\end{equation}
and
\begin{eqnarray}
&&E_{\A}:=T^{\A}_2(aT^{\A}_2+a'{(T^{\A}_2)}^{-1}){(T^{\A}_1)}^{-1},\\
&& F_{\A}:=T^{\A}_1(bT^{\A}_2+b'{(T^{\A}_2)}^{-1}){(T^{\A}_2)}^{-1},\\
&& K_{\A}:={(T^{\A}_2)}^2, \quad K_{\A}^{-1}:={(T^{\A}_2)}^{-2}.
\end{eqnarray}
Then the set $\{E_{\A},F_{\A}, K_{\A}, K_{\A}^{-1}\}$ generates
a representation of $U_{q_{\alpha}}({\rm sl}_2)$ on the vector space $C_0^{\infty}(\Omega)$.
\end{corollary}

\medskip


\begin{example}{\rm Let $\zeta$ be
the Weierstrass Zeta function :
$$
\zeta(z):=\frac 1{z}+\sum_{(m,n)\not=(0,0)}\left\{
\frac 1{z-\Omega_{m,n}}+\frac 1{\Omega_{m,n}}+\frac z{\Omega_{m,n}^2}\right\}
$$
and $P$ be a polynomial of $z$.
We define
$$
f(z)=\zeta(z)+P'(z), \quad z\in M:=\C\setminus\{\Omega_{m,n}|m,n\in \Z\}.
$$
It is easy to see that $f$ is holomorphic on $M$ with poles
at $z=\Omega_{m,n}, (m,n)\in \Z^2$ 
and
$$
c_f=1.
$$ 
Hence 
$$
q_{\alpha}=e^{2\pi\alpha}.
$$
Therefore, if $\Re \alpha\not=0$, then $|q_{\alpha}|\not=1$.

It is well known that the function
$$
\sigma(z):=z\prod_{(m,n)\not=(0,0)}\left\{\left(1-\frac{z}{\Omega_{m,n}}\right)
\exp{\left(\frac{z}{\Omega_{m,n}}+\frac {z^2}{2\Omega_{m,n}^2}\right)}\right\}
$$
has the following properties \cite[Chapter XX]{WW}:
\begin{list}{}{}
\item{(i)} 
$$
\frac{d}{dz}\log \sigma(z)=\zeta(z).
$$
\item{(ii)} 
$$
\sigma(z+w_j)=-e^{\eta_j(2z+w_j)}\sigma(z), \quad j=1,2,
$$
where $\eta_j:=\zeta(w_j/2)$.
\end{list}
Hence
$$
\int_x^{x+w_j}f(z)dz=\pi i
+\eta_j(2x+w_j)+P(x+w_j)-P(x).
$$
Hence, in the present case, we have 
$$
K_j(\x)=(\Re \alpha)
(\pi +2(x_1\Im \eta_j+x_2\Re \eta_j)+\Im (\eta_jw_j) +\Im (P(x+w_j)-P(x)).
$$
Therefore, if $\Re \alpha\not=0$ and the degree of $P$ is one or
more than $2$, then
$K_j$ is neither bounded above nor below on $\R^2\setminus \L_j$ .
Thus, in this case, the operators $T_1^{\A}$ and $T_2^{\A}$ are unbounded.
}
\end{example}



\section*{Acknowledgement}

Part of this work was done during the author's stay at the Erwin Schr\"odinger International Institute for Mathematical Physics (ESI) in June, 2007.
He would like to  thank the ESI for their support.

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\end{document}
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