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Edit File: CHAPT24.TEX

\section{Representations of $q$-relations}

\subsection{Finitely-dimensional representations of $q$-relations}

When stydying representations of $q$-relations,
\begin{align*}
(VI_0), (VII_0) \ \frac 1i [A,B]& = \alpha (A^2 \pm B^2), \quad \alpha
>0,
\\
(VI_1), (VII_1) \ \frac 1i [A,B]& = \alpha (A^2 \pm B^2) +I, \quad \alpha
\in \mathbb{R}, \alpha \ne 0,
\end{align*}
it would be nice to have a kind of theorem of Kleinecke-Shirokov type
for general relations of the form
\[
\frac1i [A,B] = f(A) + \phi(B),
\]
where $f(\cdot)$, $\phi(\cdot)$ are real functions of $t \in
\mathbb{R}^1$. Indeed, in the finite-dimensional case ($\dim H <\infty$)
we have the following statement.

\begin{proposition}
If $A$, $B$ are self-adjoint operators in a finite-dimensional space
$H$, and
\[
[A, B] =T+S,
\]
where
\[
[A,T] =[B,S] =0
\]
then $[A, B] = 0$.
\end{proposition}

\begin{proof}
Indeed, since
\[
[A,B]^2 = (T+S) [A, B] = T(AB - BA) + S(AB - BA) =
A(TB) - (TB) A + (SA)B - B(SA) = [A, TB] + [SA,B],
\]
we conclude that $\tr [A,B]^2 = 0$. But $[A,B]$ is a skew-adjoint
operator, therefore, $\tr[A,B]^2 =0$ implies $[A,B]=0$.
\end{proof}

\begin{corollary}
Irreducible finite-dimensional representations of relations
$(VI_0)$--$(VII_1)$ are one-dimensional, $A = \lambda$, $B =\mu$,
($\lambda$, $\mu \in M_{(\cdot)}(\alpha)$, where
\begin{align*}
M_{(VI_0)}(\alpha)& = \{(\lambda, \mu) \in \mathbb{R}^2 \mid \lambda =0,
\mu =0\}, \quad \text{for all $\alpha >0$,}
\\
M_{(VI_1)}(\alpha)& = \emptyset, \quad \alpha >0,
\\
M_{(VI_1)}(\alpha)& = \{(\lambda, \mu) \in \mathbb{R}^2 \mid \lambda^2
+ \mu^2 = -\frac1\alpha\}, \quad \alpha <0,
\\
M_{(VII_0)}(\alpha)& = \{(\lambda, \mu) \in \mathbb{R}^2 \mid \lambda^2 
=\mu^2\}, \quad \text{for all $\alpha >0$,}
\\
M_{(VI_1)}(\alpha)& = \{(\lambda, \mu) \in \mathbb{R}^2 \mid \lambda^2-
\mu^2 =-\frac1\alpha\}, \quad \text{for all $\alpha \in
\mathbb{R}^2$, $\alpha \ne 0$.}
\end{align*}
\end{corollary}

\begin{remark}
In the finite-dimensional situation, the corresponding generalization
of the Jackobson theorem is not true. Indeed, even for $\dim H= 3$,
there exist matrices $A$, $B$, $T$, $S$, such that $[P,T] = [Q,S]=0$,
but the matix $[P,Q]$ is not nilpotent.

\begin{example}(V. S. Guba).
Let
\begin{gather*}
A= \begin{pmatrix}2&0&0\\0&1&0\\0&0&0\end{pmatrix}, \quad B
=\begin{pmatrix}2&4&4\\6&-4&4\\-3&6&2\end{pmatrix},
\\
T=\begin{pmatrix}4&0&0\\0&16&0\\0&0&4\end{pmatrix}, \quad S =
\begin{pmatrix} 4&4&8\\-6&16&4\\6&-6&4\end{pmatrix} = B^2 /4;
\end{gather*}
then $[A,B] =T+S$, $[T,A]= [S,B]=0$, but $\sigma([A,B]) \ne \{0\}$.
\end{example}
\end{remark}

Similar to Proposition~\ref{???} fact holds for operators in
infinite-dimensional $H$, but under the additional assumption that the
operators $T$ and $S$ are compact.

\begin{proposition}
If $A= A^*$, $B=B^* \in L(H)$, and $[A,B] = T+S$, where $T$, $S$, are
compact operators in $H$ such that $[A,T] = [B,S]=0$, then $[A,B]=0$.
\end{proposition}

\begin{proof}
Since $[A,B]$ is skew-adjoiint, then
\begin{equation}\label{skew1}
[A,B] = (T_1 + S_1),
\end{equation}
where compact operators $T_1 = (T- T^*)/2$, and $S_1 = (S- S^*)/2$ are
skew-adjoint. By the Fuglede-Putnam-Rosenblum theorem (for its
formulation and proof see below Section~\ref{sec_243}) we have
\[
[T_1,A] = S_1, B] =0.
\]
Further, let $\|T_1 \| \ge \|S_1\| >0$. Choose an eigenvalue $\mu$ of
the operator $T_1$ such that $|\mu | = \|T_1\|$, and write $H_\mu = \{
f\in H \colon T_1 f = \mu f\}$. The space $H_\mu $ is finte-dimensional
and invariant with respect to $A$. If $e\in H_\mu$, $\|e\| =1$, and $Ae
= \lambda e$, $\lambda \in \mathbb{R}$, then due to \eqref{skew1} we
have
\[
(A - \lambda I)Be = \mu e + S_1 e,
\]
which implies $(S_1 e, e) = -\mu$, and therefore, $\|S_1\|\ge |\mu|$.
But since $\|S_1\| \le \|T_1\| = |\mu|$, we have $\|S_1 \| =|\mu|$,
whicch yields $S_1 e = - \mu e$. Since $T_1$ is compact, we have $S_1 =
-T-1$ on all eigenspaces of the operator $T$, and, therefore, on the
whole $H$.
\end{proof}

In Section~\ref{sec_242}, investigating relation $(V1_1)$,
we will see that the corresponding statement
for arbitrary bounded operators does not hold.

\subsection{Hermitean $q$-plane and $q$-CCR}

We will consequently study representations of relations 
$(VI_0)$, $(VI_1)$, $(VII_0)$, $(VII_1)$ by bounded self-adjoint
operators.

1. $(VI_0)$. Consider the pairs of bounded self-adjoint operators
satisfying the relation
\[
[A,B] = i\alpha (A^2 + B^2), \quad \alpha >0.
\]

\begin{proposition}
If a pair of bounded self-adjoint operators $A$, $B$, satisfies $(VI_0)$, 
then $A= B=0$.
\end{proposition}

\begin{proof}
Introduce the operators $X = A+ iB$, $X^* = A- iB$. Then the operators
$X$ and $X^*$ satisfy the Hermitean $q$-plane relation:
\[
(1-\alpha ) \, XX^* = (1+\alpha) X^* X,
\]
but since $\alpha>0$, putting $q= \frac{1-\alpha}{1+\alpha}$, we get
\begin{equation} \label{q-plane}
X^* X = q XX^*
\end{equation}
For $q \le 0$ ($\alpha \ge 1$), this equation possesses only zero
solution $X= X^* = 0$, since for $q \le 0$, the non-negative operator
$X^*X$ should be equal to the non-positive one $qXX^*$. But then
$A=B=0$.

We will do more detailed investigation of the case $1>q>0$ ($0<\alpha <
1$).

\begin{lemma}
If $X$, $X^* \in L(H)$, and \eqref{q-plane} holds, then $\ker X = \ker
X^*$.
\end{lemma}

\begin{proof}
Indeed, we have $\ker X = \ker X^*X = \ker XX^* = \ker X^*$.
\end{proof}

The representation space of of relation \eqref{q-plane} now has the 
form $H = H_0 \oplus H_1$, where $H_0$ and $H_1$ are invariant with
respect to $X$, $X^*$ subspaces, on $H_0 = \ker X = \ker X^*$ we have
$X= X^* =0$, and on $H_1 = H_0^\perp$ these operators are
non-degenerated.

Consider representations of \eqref{q-plane} on $H_1$. For the polar
decomposition of the operator $X= UC$, $U$ is unitary, $C>0$, we have
\begin{equation}\label{q-plane_ii}
C^2 U = U (qC^2), \quad \text{and}\quad C^2 U^* = U^*(\frac1q
c^2).
\end{equation}
But then if $\lambda >0$ belongs to the spectrum $\sigma(C^2)$ of the
operator $C^2$, then $\sigma(C^2) \supset \bigcup_{k\in \mathbb{Z}} q^k
\lambda$. But since the set $\bigcup_{k\in \mathbb{Z}} q^k
\lambda$ is unbounded, for bounded representations of $(VI_0)$ we have
$A= B=0$.
\end{proof}

\begin{remark}
The arguments above enable us to write down explicit formula for a
family of irreducible \em{unbounded} representations of $(VI_0)$,
\[
X = \begin{pmatrix} \ddots&\ddots\\& \sqrt{q^{-1} \lambda}&0&&&0\\
&&\sqrt{\lambda}& 0 \\ &&&\sqrt{q\lambda}&0\\0&&&\sqrt{q^2\lambda}&0\\
 &&&&\ddots&\ddots\end{pmatrix}, \quad \lambda \in  [1,q),
\]
if we define the correct meaning of the relation for unbounded operators
(see Chapter~3).

Note that the closure of the symmetric matrix $X+X^*$ from the set of
finite vectors, is not a self-adjoint operator; its deficiency indices
are (1,1).
\end{remark}

$(VI_1)$ Consider pairs of bounded self-adjoint operators satisfying the
relation
\[
\frac1i [A,B] = \alpha (A^2 + b^2) + I, \quad \alpha \in \mathbb{R}^1
\setminus \{0\}.
\]
Making a change of variables $X = A+ iB$, $X^* = A- iB$, we get the
following
\begin{equation}\label{alpha-ccr}
(1-\alpha) XX^* = (1+\alpha)X^*X + 2I.
\end{equation}

For $\alpha \ge 1$ this equation does not have solutions, since in the
left-hand side we have non-positive operator, and in the right-hand side
we have strictly positive one.

Assuming $\alpha < 1$, we have
\begin{equation} \label{alpha-ccr2}
XX^* =  \frac{1+\alpha}{1-\alpha} X^*X + \frac 2{1-\alpha} I = q X^*X +
(q+1) I,
\end{equation}
where $q = \frac{1+\alpha}{1-\alpha} > -1$. To rewrite the latter in the
form used in the literature, introduce operator $a = \sqrt{q+1} X$, $a^*
= \sqrt{q=1} X^*$. Then we have
\begin{equation} \label{q-ccr}
aa^* = q a^* a + I
\end{equation}

\begin{remark}
Equations \eqref{alpha-ccr} and \eqref{q-ccr} are not equivalent. For
$q= -1$ equation \eqref{q-ccr}, $\{ a, a^*\} = I$ possesses irreducible
one-dimensional and two-dimensional solutions. For $q<1$, replacing $a$
with $a^*$ we get from \eqref{q-ccr} relation of the same form, but with
$-1 < q <0$.
\end{remark}

Further, select on the interval $q>-1$ points $q=0$ ($\alpha = -1$), and
$q=1$ ($\alpha =0$), corresponding to co-isometry $aa^* = I$, and CCR:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$q$ & $1<q$ & $q=1$ & $0<q<1$ & $q=0$ & $-1 <q<0$ \\
\hline
$\alpha$ & $0<\alpha < 1$ & $\alpha =0$ & $-1 < \alpha < 0$ & $ \alpha =
-1$ & $\alpha < -1$ \\
\hline
\end{tabular}
\end{center}

For $q>1$, the operator $a^*$ is non-degenerated. Let $a=UC$ be its
polar decomposition such that $\ker U = \ker C$, and $U$ is a
co-isometry; $C\ge0$. Then
\[
UC^2 U^* = q CU^*UC + I = q c^2 +I,
\]
which gives
\begin{equation} \label{ds-qccr}
C^2 U^* = U^* (qC^2 +I) = U^* f(C^2), \quad \text{and} \quad C^2 U  = U
q^{-1}( C^2 -I) = f^{-1}(C^2).
\end{equation}
If $\lambda \in \sigma(C^2)$, then points $f(\lambda)= q\lambda +1$,
$f(f(\lambda)) = q^2 \lambda +q +1$,~\dots, also  belong to
$\sigma(C^2)$. But this set of points is unbounded, i.e., {\em for $q>1$
there are no representations by bounded operators}.

\begin{remark}
For $q>1$ one can consider a formal unbounded solution of \eqref{q-ccr}
given by the following Jacobi matrix
\[
a= \begin{pmatrix}
0&0&\\
1&0& & & 0\\
 & \sqrt{1+q} & 0 \\
 &            & \ddots & \ddots \\
&0& &\sqrt{\frac{1-q^n}{1-q}}&0 &\\
 &     &  &                        & \ddots &\ddots
\end{pmatrix}
\]
But in this case, the problem with introducing unbounded operators is
not quite trivial. For example, the closure of the following Jacobi
matrix
\[
a= \begin{pmatrix}
0&1&\\
1&0& \sqrt{1+q} & & 0\\
 & \sqrt{1+q} & 0 &\ddots\\ 
 &            & \ddots & \ddots & \sqrt{\frac{1-q^n}{1-q}}\\
&0& &\sqrt{\frac{1-q^n}{1-q}}&0 &\ddots\\
 &     &  &                        & \ddots &\ddots
\end{pmatrix}
\]
definded on the set of finite vectors, is not self-adjoint, but has
deficiency indices (1,1).
\end{remark}

For $0<q<1$, the operator $a^*$ is also non-degenerated, in the polar
decomposition $a= UC$, $U$ is a co-isometry, and equality
\eqref{ds-qccr} holds.

If $\lambda \in \sigma(C^2)$, then all points $f(\lambda)$,
$f(f(\lambda))$, \dots, should belong to to the spectrum $\sigma(C^2)$,
too. To the point $\lambda = (1-q)^{-1}$ (stationary point of this
mapping) there corresponds a circle of one-dimensional operators $a =
(1-q)^{-1/2} e^{i\phi}$, $\phi \in S^1$. If $\lambda > (1-q)^{-1}$, the
corresponding solutions are unbounded.

\begin{remark}
There exist a series of such inequivalent unbounded representations of
\eqref{q-ccr} depending on a parameter $\lambda \in (\lambda_0,
q\lambda_0 +1]$, $\lambda_0 > (1-q)^{-1}$.
\end{remark}

If $(1-q) > \lambda >0$, then $\lambda \in \sigma(C^2)$ implies that
$\sigma(C^2)$ contains all points $f(\lambda)$, $f(f(\lambda))$, \dots,
tending to $(1-q)^{-1}$; also $\sigma(C^2)$ contains all points
$f^{-1}(\lambda)$, $f^{-1}(f^{-1}(\lambda))$,~\dots, unless this
sequence contains the zero point. But if this would not happen, then the
spectrum $\sigma(C^2)$ contatins negative points, which is impossible.
This means that equation \eqref{q-ccr} has {\em unique} irreducible
infinite-dimensional representation by bounded operators
\[
a= \begin{pmatrix}
0&0&\\
1&0& & & 0\\
 & \sqrt{1+q} & 0 \\
 &            & \ddots & \ddots \\
&0& &\sqrt{\frac{1-q^n}{1-q}}&0 &\\
 &     &  &                        & \ddots &\ddots
\end{pmatrix}
\]
with vacuum vector $e_0$ such that $ae_0 =0$ (Fock representation).

Notice that the spectrum of bounded self-adjoint operator
\[
a=a^* = \begin{pmatrix}
0&1\\
1&0&\sqrt{1+q} \\
&\sqrt{1+q} &0 &\ddots \\
&&\ddots & \ddots
\end{pmatrix}
\]
is concentrated on the interval $[-2/(1-q), 2 /(1-q)]$.

For $q=0$, we have $aa^* =I$, i.e., $a$ is a co-isometry. Irreducible
representations are the following: a circle of one-dimensional unitary
operators $a = e^{i\phi}$, $\phi \in S^1$, and a single operator, adjoint
to the unilateral shift operator in $l_2(\mathbb{Z}_+)$. In this case,
any isometry in  $H$ gives rise to unique  decomposition of $H$ into
invariant with respect to $a$ and $a^*$ subspaces, $H= H_{\text{uni}}
\oplus H_{\text{shift}}$ such that $a|_{H_{\text{uni}}}$ is unitary
operator, and $a|_{H_{\text{shift}}}$ is unitarily equivalent to
multiple of the operator adjoint to the unilateral shift (H. Wold's
decomposition). One can easily get this decomposition putting
\[
H_{\text{shift}} = \bigcup_{k=0}^\infty (a^*)^k (H \ominus a^* H).
\]

For $-1 , q < 0$ we have the following:

a) a circle of one-dimensional representations $a = (1-q)^{-1/2}
e^{i\phi}$, $\phi \in S^1$;

b) Fock representation by bounded operators
\[
a= \begin{pmatrix}
0&0&\\
1&0& & & 0\\
 & \sqrt{1+q} & 0 \\
 &            & \ddots & \ddots \\
&0& &\sqrt{\frac{1-q^n}{1-q}}&0 &\\
 &     &  &                        & \ddots &\ddots
\end{pmatrix}
\]
There are no representations by unbounded operators, since for each
$0<\lambda \in \sigma(C^2)$ the point $f^{-1}(\lambda)$ also belongs to
$\sigma(C^2)$.

\subsection{Real quantum plane and real quantum hyperboloid}

1. Consider, finally, pairs of self-adjoint operators that satisfy
relations $(VII_0)$ and $(VII_1)$,
\begin{align*}
[A,B]& = i\alpha(A^2 - B^2), \quad \alpha >0,
\\
{}[A,B] & = i\alpha (A^2 - B^2) + iI, \quad \alpha \in \mathbb{R}
\setminus \{0\}.
\end{align*}

Representations of relations $(VII_0)$ and $(VII_1)$ by bounded
self-adjoint operators can be obtained from the following theorem.

\begin{theorem}
(B. Fuglede, C.R.Putnam, M.Rosenblum).
Let $M$, $N$, $T\in L(H)$, and the operators $M$ and $N$ are normal. If
$MT = TN$, then $M^* T = TN^*$ as well.
\end{theorem}

\begin{proof}
The proof is based on the unitarity of the operator-valued functions
$\exp (zM^* - \bar z M)$, and $\exp(-zN^* + \bar z N)$. Since $M^kT =
TN^k$ for all $k=1$, 2,~\dots, and $\exp (\bar zM) T = T \exp(\bar z N)$
$\forall z \in \mathbb{C}$, we conclude that for the bounded entire
operator-valued function
\begin{align*}
F(z) &= \exp (zM^* - \bar z M) T \exp (\bar z N
- zN^*)
\\
&= \exp(zM^*) (\exp (-\bar z M)) T \exp (\bar z N)) \exp (-zN^*)
\\
&= F(0) = T;
\end{align*}
therefore, $\exp(zM^*) T = T \exp (zN^*)$ for all $z \in
\mathbb{C}$. Thus, $M^* T = TN^*$.
\end{proof}

2. Relation $(VII_0)$,
\[
[A,B] = i\alpha (A^2 - B^2), \quad \alpha >0,
\]
can be rewritten in the form
\[
XY = q YX, \quad q = \frac{1+ i \alpha}{1-i\alpha} \ne 1, \quad |q| =1
\]
(real quantum sphere $\mathbb{R}_q^2$); here, $X = A+B$, $Y= A-B$. For
bounded self-adjoint operators, tue to the Fuglede-Putnam-Rosenblum
theorem we also have
\[
XY = \bar q YX,
\]
which implies, since $q \ne \bar q$, that $XY = YX =0$, i.e., $[A,B]=0$,
and $A^2 = B^2$.

Therefore, all irreducible representations of $(VII_0)$ by bounded
operators are one-dimensional (see Section~\ref{sec241}).

3. Now consider bounded pairs, satisfying $(VII_1)$,
\[
\frac 1i [A,B] = \alpha (A^2 - B^2)+I, \quad \alpha \in \mathbb{R}, \
\alpha \ne 1.
\]
For the self-adjoint operators $X = A+B$, $Y = A-B$, we have
\begin{equation}\label{hyperb}
XY = qYX + (q+1)I, \quad q = \frac{1+ \alpha i}{1-\alpha i} \ne 1, \quad
|q| = 1.
\end{equation}
$*$-Algebra generated by \eqref{hyperb} is called real quantum
hyperboloid.

For bounded operators \eqref{hyperb} implies
\[
X(XY + \frac1\alpha I) = q (XY + \frac1\alpha I)X,
\]
and, due to the Fuglede-Putnam-Rosenblum theorem we also have
\[
X(XY + \frac1\alpha I) = \bar q (XY + \frac 1\alpha I)X,
\]
which is possible only if
\begin{equation}\label{hyperb1}
XYX = X^2 Y = -\frac 1\alpha X.
\end{equation}
By \eqref{hyperb1}, $H_0 = \ker X$ is invariant under $A$, $B$; but then
by \eqref{hyperb}, we get $H_0 = \{0\}$. On the subspace $H_0^\perp$,
the operator $X$ is invertible, and $XY = YX$. Thus we have $[A,B]=0$,
and $\alpha (A^2 - B^2) +I =0$.

Therefore, irreducible representations of $(VII_1)$ by bounded operators
are one-dimensional (see Section~\ref{sec241}).

\begin{remark}
For possible definition of integrable pairs of unbounded pairs of
self-=adjoint operators, satisfying $(VII_0)$ and $(VII_1)$, and the
structure of such pairs see Chapter III.
\end{remark}