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% Quantum algebra embeddings~: deforming functionals and
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% J. Van der Jeugt
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\begin{document}
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\begin{center}
{\LARGE
Quantum algebra embeddings~: deforming functionals and
algebraic approach
}\\[2cm]
J. Van der Jeugt$^{a)}$ \\[.5cm]
{\em Vakgroep Toegepaste Wiskunde en Informatica,\\
Universiteit Gent, Krijgslaan 281--S9, B-9000 Gent, Belgium}\\
\end{center}
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\vspace{3cm}
\begin{abstract}
The study of subalgebras of Lie algebras arising in physical
models has been important for many applications. In the present
paper we examine the $q$-deformation of such embeddings; the
Lie algebras are then replaced by quantum algebras. Two methods
are presented~: one based upon deforming functionals, and a
direct algebraic approach. A number of examples are given, e.g.\ $su_q(2)
\oplus su_q(2) \supset su_q(2)$ and $u_q(3)\supset so_q(3)$.
For the last example, we give the $q$-boson construction, and
the relevant overlap coefficients are related to a generalized
basic hypergeometric function $_3\phi_2$.
\end{abstract}
\vspace{1cm}
\noindent
PACS numbers~: 02.20, 03.65F, 21.60F.
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\noindent-----------------------------------\\
$^{a)}$ {\footnotesize Research Associate of the N.F.W.O.
(National Fund for Scientific Research of Belgium)}
\newpage
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\section{Introduction}

For classical Lie algebras, the study of subalgebras has been of
great importance both in mathematics and in physics. Subalgebra
chains arise in physical models which possess symmetries.
For example, chains of Lie algebras $u(N)\supset so(N) \supset so(3) \supset so(2)$
make their appearance in many algebraic models in hadronic~\cite{iachello},
nuclear~\cite{elliott}, atomic~\cite{judd} and molecular
physics~\cite{iachellol}. The study of subalgebra chains
and their related mathematical quantities (branching rules, tensor products,
generator matrix elements) are fundamental and bear a direct
relation to the physical quantities of the model. It is in this
field that Prof.\ Sharp -- to whom this paper is dedicated --
made numerous important contributions, a few of which will be
mentioned and used here.

The objects treated in this paper are, however, not Lie
algebras but quantum algebras.
Since the introduction of quantum groups and quantum enveloping
algebras (quantum algebras, $q$-algebras, deformed
algebras)~\cite{kulish,drinfeld,jimbo}
these new mathematical structures are also finding
applications in various branches in physics~\cite{zachos}.
In a first approach, a quantum algebra can be seen as
a deformation of the enveloping algebra of a (simple) Lie algebra;
this structure can then be endowed with other quantities that
turns it into a quasi-triangular Hopf algebra~\cite{majid}.
The simplest quantum algebra is $su_q(2)$, an associative algebra
with generators $\{L_-,L_+,q^{\pm L_0}\}$ and relations
$q^{L_0}L_\pm=q^\pm L_\pm q^{L_0}$, $L_+L_--L_-L_+=(q^{2L_0}-
q^{-2L_0})/(q-q^{-1})\equiv[2l_0]$, where $q$ is a deformation
parameter. When $L_0$ is considered as an operator and $q^{L_0}$
as an exponential, these relations can actually be rewritten
as $[L_0,L_\pm]=\pm L_\pm$ and $[L_+,L_-]=[2L_0]$, and in the limit
$q\rightarrow 1$ this yields the relations satisfied in the
enveloping algebra of $su(2)$.

Quantum algebras have by now a well established role in
2-dimensional solvable models and inverse scattering theory~\cite{doebner}.
In models of deformed molecules, atoms or nuclei, some
research has been
performed~\cite{bonatsosrrs,bonatsosar,bonatsosdk}, 
but only to a certain extent.
In the study of such deformed models in physics, it is
mainly the quantum algebra $su_q(2)$ (or $su_q(1,1)$) which
has been used in the literature. The use of larger quantum
algebras has so far been limited by the fact that many
relevant properties of classical Lie algebras (e.g.\ subalgebra
chains, decomposition rules) do not trivially extend to
quantum algebras.

In the present paper we mention some of the problems and solutions
in the study of subalgebras of quantum algebras.
We shall highlight two methods in the quantum algebra
embedding problem~: the algebraic approach and the deforming
functional approach. All methods will be illustrated by
means of examples which not only have a pedagogical meaning,
but are also potential candidates for applications in
deformed physical models.

In the following section we illustrate by means of the
quantum algebra $u_q(3)$ the algebraic approach. Herein,
the definition of quantum subalgebra is very natural.
In Section~3 we use a well-known technique, of deforming
functionals, to treat the deformation of embeddings.
This technique has a number of shortcomings but nevertheless
it can be applied in a straighforward way. A simple example,
illustrating these shortcomings, is presented in Section~4.
Section~5 returns to the quantum subalgebra chain given
in Section~2, makes a correspondence with so-called $q$-boson
creation and annihilation operators, and makes an explicit
study of the transformation of $q$-boson states to $so_q(3)$
states labelled by the classical $(l,m)$-labels. The final
section presents some conclusions and relations with other
work.

\section{Quantum (sub)algebras~: algebraic approach}

Quantum algebras associated with simple Lie algebras are usually
defined by means of relations between the Chevalley triples
$\{h_i,e_i,f_i\}$. Considering the algebra generated by
a subset of such triples (and with the corresponding relations),
it is thus rather trivial to construct quantum subalgebras of
the original quantum algebra. One could refer to this as the
construction of subalgebras of quantum algebras in a Chevalley
basis. In this case, also the Hopf structure (comultiplication,
counit, and antipode) transfers trivially to the subalgebra.
The construction of subalgebras of quantum algebras in a non-Chevalley
basis is definitely a much harder problem, and so far only some
partial solutions have been presented in the literature.

What we mean by the algebraic approach for the construction
of quantum subalgebras shall be illustrated in an example.
Consider the quantum algebra $u_q(3)$~: this is the
associative algebra
with generators $e_1$, $e_2$, $f_1$, $f_2$, $k_i^{\pm}=
q^{\pm N_i}$ ($i=0,1,2$), subject to the relations~\cite{jimbo246}~:
\begin{eqnarray}
&&\hbox{all }k_i^\pm \hbox{ and }k_j^\pm\hbox{ commute for }i\ne j,\nn\\
&&k_i^+k_i^-=1=k_i^-k_i^+ ,\nn\\
&&k_i^\pm e_j = q^{\pm(\de_{i,j-1}-\de_{ij})} e_j k_i^\pm,\nn\\
&&k_i^\pm f_j = q^{\mp(\de_{i,j-1}-\de_{ij})} f_j k_i^\pm, \label{rel} \\
&&e_if_j-f_je_i=\de_{ij} (k_{i-1}^+k_i^- - k_{i-1}^-k_i^+)/(q-q^{-1}),\nn\\
&&e_i^2e_j -(q+q^{-1})e_ie_je_i+e_je_i^2 =0 \quad (i\ne j),\nn\\
&&f_i^2f_j -(q+q^{-1})f_if_jf_i+f_jf_i^2 =0 \quad (i\ne j). \nn
\end{eqnarray}
Herein, the parameter $q$ is taken to be generic (not a root of unity).
In the limit $q\rightarrow 1$, this reduces to the universal enveloping
algebra of $u(3)$. The quantum enveloping algebra $u_q(3)$ as defined
here can be made into a Hopf algebra, by defining a co-product, a
co-unit and an antipode; however, for the purposes of this paper only
the $q$-deformed algebra as determined by (\ref{rel}) is needed. In
fact, the subalgebra structure introduced here
will only be a $q$-deformed subalgebra structure (and not a Hopf
subalgebra).

To illustrate the remarks of the first paragraph of this section,
it is easy to see that the triples $\{ e_i, f_i, h_i=N_{i-1}-N_i\}$
($i=1,2$), together with the corresponding relations, generate
an $su_q(2)$ subalgebra of $u_q(3)$. These are two examples of
a subalgebra in the Chevalley basis. Another example
is~\cite{dobrev,proc_sal}~:
\beq
\begin{array}{l}
E=e_1e_2-qe_2e_1 ,\\
F=f_2f_1-q^{-1}f_1f_2 ,\\
H=N_0-N_2 .
\end{array}
\label{EFH}
\eeq
This example is less trivial but still simple since it corresponds
to a regular $su(2)$ subalgebra of $u(3)$ in the nondeformed case.
From (\ref{EFH}) it is also clear what we mean by the algebraic
approach to the embedding problem~: a subalgebra of a quantum
algebra ${\cal G}$ is a subalgebra of ${\cal G}$ as an associative algebra.
When this subalgebra is generated by a number of generators
with relations, such that it is itself a quantum algebra, we
speak of a quantum algebra embedding. For instance, the three
elements of (\ref{EFH}) belong to $u_q(3)$, and they satisfy
\beq
[H,E]=2E,\quad [H,F]=-2F,\quad [E,F]=[H]={q^H-q^{-H}\over q-q^{-1}} ,
\label{commEFH}
\eeq
the defining relations for $su_q(2)$.
Hence, we have identified a quantum algebra embedding $u_q(3)
\supset su_q(2)$.

Another more important set of elements of $u_q(3)$ has been
constructed in~\cite{proc_sal,princ}~:
\beq
\begin{array}{l}
H=2N_0-2N_2 ,\\
E=\sqrt{2}q^{2N_2}e_1+e_2 q^{-N_1}(q^{N_2}+q^{-N_2})/\sqrt{2} ,\\
F=f_1q^{-N_1}(q^{N_0}+q^{-N_0})/\sqrt{2}+\sqrt{2} q^{2N_0} f_2 .
\end{array}
\label{soq3}
\eeq
These elements are of interest because in the limit $q\rightarrow 1$
they reduce to the generators of the principal $so(3)$ subalgebra
of $u(3)$; it is therefore convenient to label the above elements
as the generators of $so_q(3)$. However, the elements (\ref{soq3})
satisfy the last relation in (\ref{commEFH}) {\em only} when acting on
the class of totally symmetric representations of $u_q(3)$. In this
sence, we do not have a genuine quantum algebra embedding, and
the embedding $u_q(3)\supset so_q(3)$ is valid {\em for symmetric
representations only}~\cite{princ}.
Nevertheless, the expressions (\ref{soq3})
are important and in Section~5 we shall show that they have a
natural relation with so-called $q$-boson operators. Indeed,
if a set of relations for generators is valid only for a certain class of
representations, it means that there are in fact more relations than in the
original set, or that the generators themselves can be expressed in terms
of simpler objects (also satisfying some relations); here these simpler
objects will be $q$-boson creation and annihilation operators.

Note that the $(E,F,H)$-basis of (\ref{commEFH}) is the one
used mainly by mathematicians; in the physics literature one
identifies $(E,F,H)$ with $(L_+,L_-,2L_0)$ and then the relations
become~\cite{sklyanin,jimbo}
\beq
[L_0,L_\pm]=\pm L_\pm,\quad  [L_+,L_-]=[2L_0]={q^{2L_0}-q^{-2L_0}\over q-q^{-1}} ,
\label{commL}
\eeq

In the present section we have defined the concept of quantum
subalgebras. A few examples have been given, from which it
is clear that the problem of finding the ``$q$-analog'' of a
Lie algebra embedding is a highly nontrivial task, in particular
if the embedding is nonregular. In the following section we
study the embedding problem by means of a technique which is
commonly used in quantum algebras.

\section{Deforming functionals}

In a number of papers, deforming functionals have been introduced
and studied for the quantum enveloping algebra $su_q(2)$ and for
the quantum plane~\cite{curtrightz,curtrightgz,fairliez}.
In the present section, we shall study
how such deforming functionals can be used to treat the quantum
subalgebra embedding problem. In the following section we shall
give an example.

In the framework of deforming functionals, explicit invertible functionals
of the generators of $su(2)$ are given, which satisfy the defining
relations of the $q$-deformed algebra $su_q(2)$. Let $l_0$, $l_+$
and $l_-$ be the basis of $su(2)$ satisfying
\beq
[l_0,l_\pm]=\pm l_\pm,\qquad [l_+,l_-]=2l_0.
\label{su2}
\eeq
The Casimir operator is $I=l_+l_-+l_0^2-l_0\equiv l(l+1)$; then the
positive operator $l$ is defined as $l=(-1+\sqrt{1+4I})/2$. Clearly,
$l$ does not belong to the Lie algebra $su(2)$ nor to its enveloping
algebra $U(su(2))$. However, $l$ acts as a scalar on all $su(2)$ modules $V$
generated by a vector $v$, i.e.~$V=U(su(2))v$, hence $l$ is well
defined on such modules. Consider the category ${\cal C}$ of modules
of the form $U(su(2))v$ with $v$ a weight vector with respect to $l_0$.
Let $q$ be a generic real number, and denote again $(q^x-q^{-x})/(q-q^{-1})$
by $[x]$. The following deforming functionals~\cite{curtrightz}
are well defined on modules belonging to ${\cal C}$~:
\beq
L_0=l_0,\qquad L_\pm=\sqrt{ [l\pm l_0][l\mp l_0+1]\over
 (l\pm l_0)(l\mp l_0+1)} l_\pm .
\label{deform}
\eeq
Since the elements $L_0$, $L_+$, $L_-$ satisfy the standard relations
(\ref{commL}) of the quantum enveloping algebra $su_q(2)$,
the map (\ref{deform}) is said to be the deforming functional.
For finite dimensional irreducible modules of $su(2)$ with basis $|l,m\rangle$
($m=l,l-1,\ldots,-l$; $2l$ integer) and action $l_0|l,m\rangle=
m|l,m\rangle$,
$l_\pm|l,m\rangle=\sqrt{(l\mp m)(l\pm m+1)}|l,m\pm 1\rangle$, one can
introduce the following basis~:
\beq
|n_1,n_2\rangle,\qquad\hbox{with}\qquad n_1=l+m,\;n_2=l-m,
\eeq
and denote the $su(2)$ generators by~:
\beq
l_0={1\over 2}(e_{11}-e_{22}),\quad
l_+=e_{12},\quad l_-=e_{21} ,
\eeq
with usual commutation relations
\beq
[e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{il}e_{kj}.
\label{comm}
\eeq
The deforming functional (\ref{deform}) now takes the following form~:
\beq
E_{ii}=e_{ii}\;(i=1,2),\quad
E_{12}=\sqrt{[e_{11}][e_{22}+1]\over (e_{11})(e_{22}+1)} e_{12},\quad
E_{21}=\sqrt{[e_{22}][e_{11}+1]\over (e_{22})(e_{11}+1)} e_{21}.
\eeq
These expressions suggest a way to define deforming functionals for
$u(N)$ or $su(N)$. The basis of $u(N)$ is $e_{ij}$ with $i,j=1,\ldots,N$,
and the commutation relations are given by~(\ref{comm}). The actual
generators to be deformed are $e_{ii}$ and $e_{ij}$ with $j=i\pm 1$.
The deformed generators, $E_{ii}$ and $E_{ij}$ with $j=i\pm 1$, of
$u_q(N)$ should satisfy~\cite{drinfeld,jimbo,sklyanin}~:
\begin{eqnarray}
&&[E_{ii},E_{jj}]=0, \label{rel1} \\
&&[E_{ii},E_{jk}]=(\delta_{ij}-\delta_{ik})E_{jk}, \label{rel2} \\
&&[E_{i,i+1},E_{j+1,j}]=\delta_{ij} [E_{ii}-E_{i+1,i+1}], \label{rel3}
\end{eqnarray}
and the Serre relations, which will not be repeated here. If one
defines~:
\beq
E_{ii}=e_{ii},\quad
E_{ij}=\sqrt{[e_{ii}][e_{jj}+1]\over (e_{ii})(e_{jj}+1)} e_{ij},\quad
(j=i\pm 1),
\eeq
then (\ref{rel1}) and (\ref{rel2}) are satisfied, but it is easily verified
that~(\ref{rel3}) is in general not valid. However, if all the relations
are restricted to the class ${\cal S}$ of totally symmetric representations
of $u_q(N)$,
then~(\ref{rel3}) and also the Serre relations are automatically satisfied.
An irreducible representation of ${\cal S}$ is labelled by an integer $n$,
and has basis states $|n_1,n_2,\ldots,n_N\rangle$ with $\sum_{i=1}^N n_i=n$;
the action of the $u_q(N)$ generators is~:
\begin{eqnarray}
E_{ii}|n_1,n_2,\ldots,n_N\rangle&=& n_i|n_1,n_2,\ldots,n_N\rangle,\nn\\
E_{i,i+1}|n_1,n_2,\ldots,n_N\rangle&=&
\sqrt{[n_i+1][n_{i+1}]}|n_1,\ldots,n_i+1,n_{i+1}-1,\ldots,n_N\rangle,
 \label{action}\\
E_{i+1,i}|n_1,n_2,\ldots,n_N\rangle&=&
\sqrt{[n_i][n_{i+1}+1]}|n_1,\ldots,n_i-1,n_{i+1}+1,\ldots,n_N\rangle.\nn
\end{eqnarray}
In the following, we shall assume that we are always working in the
class ${\cal S}$. Then the deforming functionals for $u(N)$ or $su(N)$
are well defined and the corresponding generators of $u_q(N)$ or
$su_q(N)$ satisfy the defining relations for these quantum enveloping
algebras. As previously, the fact that we are working in the class of
symmetric representations means that the deforming functionals
are in fact those defined for $q$-boson operators of $u_q(N)$.

Having defined the deforming functionals, we shall now consider the
problem of constructing quantum subalgebras by means of these deforming
maps. Let ${\cal G}=u(N)$ or $su(N)$, and let ${\cal G}'\subset{\cal G}$
be a subalgebra of ${\cal G}$, also of type $u(K)$ or $su(K)$. In particular,
with a view on defining deforming functionals, we shall assume that in
the reduction of an irreducible symmetric representation of ${\cal G}$
with respect to ${\cal G}'$, only symmetric representations of ${\cal G}'$
appear (this is certainly the case if ${\cal G}'=su(2)$). Let ${\cal S}$
be the class of symmetric representations of ${\cal G}$, and $V\in{\cal S}$.
The standard (orthogonal) basis vectors $|n_1,n_2,\ldots,n_N\rangle$ of
$V$ shall be denoted as
$v_\la$, and the action of the generators $g_i$ of ${\cal G}$, which is
well known, will be denoted by~:
\beq
g_i v_\la=\sum_\mu (g_i)_\la^\mu v_\mu.
\label{gi}
\eeq
The generators $G_i$ of the quantum enveloping algebra ${\cal G}_q$ are
then defined in terms of the $g_i$ by means of the deforming functionals.
Hence, $V$ is a representation for ${\cal G}_q$ and the action of $G_i$
on the basis $v_\la$ follows from~(\ref{gi}) and the explicit form
of the deforming functionals. Here, this action will be denoted by~:
\beq
G_i v_\la=\sum_\mu (G_i)_\la^\mu v_\mu.
\label{Gi}
\eeq
Next, consider the branching ${\cal G}\rightarrow{\cal G}'$. In this
reduction, the symmetric representations $W^{(l)}$ of ${\cal G}'$
will be labelled by the symbol $l$, and the internal labels for the
(orthogonal) basis vectors will be denoted by $m$. Let the reduction of $V$
in ${\cal G}\rightarrow{\cal G}'$ be
\beq
V \rightarrow \oplus_l W^{(l)},
\label{decomp}
\eeq
where multiplicities can occur. We now assume that for the representation
$V$ the decomposition coefficients are known explicitly, i.e.~:
\beq
v_\la=\sum_{l,m} T_\la^{lm} w_{lm}.
\label{vw}
\eeq
Since we are dealing with orthogonal bases, it also follows that
\beq
w_{lm} = \sum_\la T_\la^{lm} v_\la.
\label{wv}
\eeq
The action of the generators $g_k'$ of ${\cal G}'$ on the representations
$W^{(l)}$ are known, and since the deforming functionals for ${\cal G}'$
are also known, the action of the generators $G_k'$ of ${\cal G}_q'$
follows~:
\beq
G_k' w_{lm}= \sum_{m'} (G_k')^{lm'}_{lm} w_{lm'} .
\label{Gk}
\eeq
Next, we use (\ref{vw}), (\ref{Gk}) and (\ref{wv})~:
\begin{eqnarray}
G_k' v_\la
&=& \sum_{l,m} T_\la^{lm} G_k' w_{lm} \nn\\
&=& \sum_{l,m} \sum_{m'} T_\la^{lm} (G_k')^{lm'}_{lm} w_{lm'} \nn\\
&=& \sum_\mu \left(\sum_{l,m} \sum_{m'} T_\la^{lm} (G_k')^{lm'}_{lm}
     T_\mu^{lm'}\right) v_\mu . \label{result}
\end{eqnarray}
This last equation gives the action of the elements $G_k'$ of ${\cal G}_q'$
on the standard basis $v_\la$ of the representations $V$ of ${\cal G}_q$.
From such a relation it is, at least in principle, possible to
express the $G_k'$ in terms of the $G_i$. However, it does not
necessarily follow that ${\cal G}_q'$ is a subalgebra of
${\cal G}_q$ (seen as an associative algebra). This will be clear
from a simple example treated in the following section.

\section{Example~: $su_q(2)\oplus su_q(2)$}

Note that the construction of the previous section
is also valid for direct sums
of $su(N)$. Consider now the case ${\cal G}=su(2)\oplus su(2)$
and ${\cal G}'=su(2)$ (the diagonal $su(2)$). The basis elements
of $su(2)\oplus su(2)$ are $\{s_i,t_i (i=0,+1,-1)\}$,
with commutation relations~:
\begin{eqnarray*}
&&[s_i,t_j]=0,\\
&&[s_0,s_\pm]=\pm s_\pm,\qquad [s_+,s_-]=2s_0,\\
&&[t_0,t_\pm]=\pm t_\pm,\qquad [t_+,t_-]=2t_0.
\end{eqnarray*}
The irreducible representations $V^{(s,t)}$ are labelled by two half-integers,
and have basis states $|s,t,m_s,m_t\rangle$, with
$m_s=s,s-1,\ldots,-s$ and $m_t=t,t-1,\ldots,-t$.
The subalgebra has generators $l_0=s_0+t_0$ and $l_\pm=s_\pm+t_\pm$,
with commutation relations~(\ref{su2}).
Here, the decomposition~(\ref{decomp}) is given by~:
\[
V^{(s,t)}\rightarrow W^{(s+t)}\oplus W^{(s+t-1)}\oplus
 \cdots\oplus W^{(|s-t|)}.
\]
The decomposition coefficients (\ref{vw}) are the well known $su(2)$ coupling
coefficients~\cite{edmonds}~:
\[
|s,t,m_s,m_t\rangle=\sum_{l,m}\langle l\,m|s\,t\,m_s\,m_t\rangle\;
|l,m\rangle,
\]
or inversely~:
\[
|l,m\rangle=\sum_{m_s,m_t}\langle l\,m|s\,t\,m_s\,m_t\rangle\;
|s,t,m_s,m_t\rangle.
\]
Consider now the quantum algebras ${\cal G}_q$ and ${\cal G}_q'$
with generators $\{S_i, T_i (i=0,+1,-1)\}$ and $\{L_i (i=0,+1,-1)\}$
respectively, obtained from the nondeformed algebras $su(2)\oplus su(2)$
and $su(2)$ by means of the deforming functionals. For the generator
$L_+$, for instance, it follows that the action~(\ref{Gk}) is given
by
\[
L_+|l,m\rangle = \sqrt{[l-m][l+m+1]} |l,m+1\rangle.
\]
Then (\ref{result}) implies that
\begin{eqnarray}
&&L_+|s,t,m_s,m_t\rangle = \nn\\
&&\sum_{m_s',m_t'} \left(\sum_{l,m}\langle l\,m|s\,t\,m_s\,m_t\rangle
\sqrt{[l-m][l+m+1]}\langle l\,m+1|s\,t\,m_s'\,m_t'\rangle \right)
|s,t,m_s',m_t'\rangle.\nn \\ &&
\label{Lplus}
\end{eqnarray}
This result is remarkable for the following reasons. When the nondeformed
generator $l_+$ acts on $|s,t,m_s,m_t\rangle$, it yields only two
contributions, namely in
\beq
|s,t,m_s+1,m_t\rangle\qquad\hbox{and}\qquad|s,t,m_s,m_t+1\rangle.
\label{2states}
\eeq
When the deformed generator $L_+$ acts on $|s,t,m_s,m_t\rangle$, it
follows from~(\ref{Lplus}) that in general it gives contributions in
\beq
|s,t,s,m_s+m_t-s+1\rangle,\;|s,t,s-1,m_s+m_t-s+2\rangle,\ldots,
|s,t,m_s+m_t-t+1,t\rangle.
\label{allstates}
\eeq
The coefficients accompanying these states are $q$-dependent.
In the limit $q\rightarrow 1$ most of the coefficients of the
states~(\ref{allstates}) tend to zero, such that only contributions
in~(\ref{2states}) remain. One can now try to express the $L_+$
action in terms of the generators of ${\cal G}_q$. It follows
that $L_+$ would take the following form~:
\begin{eqnarray}
L_+&=&a_1S_+ + a_2S_+^2T_- + a_3S_+^3T_-^2 + a_4S_+^4T_-^3 + \cdots\nn\\
&&+b_1T_+ + b_2S_+T_-^2 + b_3S_+^2T_-^3 + b_4S_+^3T_-^4 + \cdots,
\label{series}
\end{eqnarray}
where $a_k$ and $b_k$ are coefficients still depending upon the
diagonal generators $S_0$ and $T_0$. This implies that the ${\cal G}_q'$
algebra obtained this way is not a genuine subalgebra of ${\cal G}_q$
considered as an associative algebra.

For the present example, it is rather surprising that the
deforming functionals approach gives rise to such complicated
objects. This is in strong contrast to
the algebraic approach which for this case can also successfully
be applied, and yields very simple expressions.
The quantum algebra ${\cal G}_q$ of this example does indeed contain
a quantum subalgebra of type $su_q(2)$, namely,
\beq
L_0=S_0+T_0,\qquad L_\pm=S_\pm q^{T_0}+T_\pm q^{-S_0}.
\label{sub}
\eeq
This is a genuine subalgebra which tends in the limit $q\rightarrow 1$
to ${\cal G}'$, just as ${\cal G}_q'$ does.
It may be surprising that the technique of quantum deformations
does not give rise to the subalgebra~(\ref{sub}). The underlying reason
is because we are dealing with a nonregular embedding ${\cal G}
\supset{\cal G}'$.

Recently, the deforming functionals approach has also been
applied for the (symmetric representations) of
$u_q(3)\supset so_q(3)$~\cite{feng,delsol}.
In both papers, the treatment falls within the framework of
our present Section~3. The conclusions are similar as in the
example presented here, i.e.~one does not find a genuine subalgebra.

\section{Boson realization of $u_q(3)\supset so_q(3)$}

In this last section we comment further on the embedding discussed
in Section~2. Since this embedding holds only for the class of symmetric
representations, it is natural that there is a link with
a boson realization. On the other hand, we are dealing with
deformed structures and the bosons that appear are not ordinary
bosons but so-called $q$-bosons~\cite{biedenharn,macfarlane}.

The quantum enveloping algebra $u_q(3)$  can be realized by
means of three independent $q$-boson operators $(N_i,b_i,b^\d_i)$,
where $i=+,0,-$. These satisfy the relations~:
\beq
[N_i,b^\d_i]=b^\d_i,\qquad
[N_i,b_i]=-b_i,\qquad, b_ib_i^\d-q^{\pm 1}b_i^\d b_i= q^{\mp N_i},
\eeq
and the remaining commutators are zero.
The orthonormal $n$-boson states are of the form~\cite{princ}
\beq
|n_+,n_0,n_-\rangle = { (b^\d_+)^{n_+} (b^\d_0)^{n_0} (b^\d_-)^{n_-} \over
 [n_+]![n_0]![n_-]!} |0\rangle,
\label{nstate}
\eeq
where the vacuum $|0\rangle$ is defined by $b_i|0\rangle=0$ for $i=+,0,-$,
and $[x]!=[x][x-1]\cdots[1]$.
An orthonormal basis for the symmetric $u_q(3)$ representation
labelled by $n$ consists of the states $|n_+,n_0,n_-\rangle$ with
$n_++n_0+n_-=n$. It is common to introduce the operator $N=
N_++N_0+N_-$ with eigenvalue $n$ on the states (\ref{nstate}).

The Chevalley generators
of $u_q(3)$ are given by $N_i$ (strictly speaking $q^{\pm N_i}$),
$e_1=b^\d_+b_0$, $e_2=b^\d_0b_-$, $f_1=b^\d_0b_+$ and $f_2=b^\d_-b_0$,
satisfying the usual relations (\ref{rel}).
The $so_q(3)$ subalgebra can then also be written in this
realization by means of (\ref{soq3}). Note that, if we accept square
roots of operators which are acting diagonally on representations,
the operators (\ref{soq3}) can be written in a more symmetric form and
one obtains the following~:
\begin{eqnarray}
L_0 &=& N_{+}-N_{-}, \label{l0}\\
L_{+}&=&q^{N_{-}-{1\over 2}N_0} \sqrt{q^{N_{+}}+q^{-N_{+}}}\; b^\d_{+}b_0 +
  b^\d_0b_{-} q^{N_{+}-{1\over 2}N_0} \sqrt{q^{N_{-}}+q^{-N_{-}}} ,\label{l+} \\
L_{-}&=&b^\d_0b_{+} q^{N_{-}-{1\over 2}N_0} \sqrt{q^{N_{+}}+q^{-N_{+}}}  +
   q^{N_{+}-{1\over 2}N_0} \sqrt{q^{N_{-}}+q^{-N_{-}}}\; b^\d_{-}b_0 .
   \label{l-}
\end{eqnarray}
These operators, defined in terms of the $u_q(3)$ realization, satisfy
the usual $so_q(3)$ relations (\ref{commL}).
It is possible to simplify the expressions for $L_i$ by making the
following transformation~\cite{quesne,def-u3}~:
\beq
\tilde b^\d_i=\left(q^{N_i}+q^{-N_i}\over q+q^{-1}\right)^{1/2} b^\d_i,\qquad
\tilde b_i=b_i\left(q^{N_i}+q^{-N_i}\over q+q^{-1}\right)^{1/2}
\hbox{ for } i=+,-.
\eeq
This map transforms a $q$-boson into what could be called a $q^2$-boson,
since
\beq
[N_i,\tilde b^\d_i]=\tilde b^\d_i,\qquad
[N_i,\tilde b_i]=-\tilde b_i,\qquad,
\tilde b_i\tilde b_i^\d-q^{\pm 2}\tilde b_i^\d\tilde b_i= q^{\mp 2N_i}.
\eeq
Note that we make this transformation only for $i=+$ and $i=-$.
In terms of the two $q^2$-bosons and the single $q$-boson operators,
the expressions for $L_i$ take the simple form~:
\begin{eqnarray}
L_0 &=& N_{+}-N_{-}, \\
L_{+}&=&\sqrt{[2]}\left(q^{N_{-}-{1\over 2}N_0} \tilde b^\d_{+}b_0 +
  b^\d_0\tilde b_{-} q^{N_{+}-{1\over 2}N_0}\right)  ,\label{nl+}\\
L_{-}&=&\sqrt{[2]}\left(b^\d_0\tilde b_{+} q^{N_{-}-{1\over 2}N_0}  +
   q^{N_{+}-{1\over 2}N_0} \tilde b^\d_{-}b_0 \right).  \label{nl-}
\end{eqnarray}
So, in a sense one could say that the most natural embedding of
$so_q(3)$ in $u_q(3)$ occurs for the case when $b_0^{(+)}$ is a $q$-boson
and $b_+^{(+)}$ and $b_-^{(+)}$ are $q^2$-bosons.

Finally, we shall explicitly study the decomposition of $u_q(3)$
boson representations in the $so_q(3)$ basis. In the classical
case of $u(3)\supset so(3)$, the symmetric representations
labelled by $n$ decompose into $so(3)$ representations $(l)$ with
$l=n,n-2,\ldots,1$ or $0$. In the $q$-generalized case, this
turns out to be exactly the same decomposition.

In the basis (\ref{nstate}) the states of the $u_q(3)$ symmetric representation
labelled by $n$ are given by $|n_-,n_0,n_+\rangle$ with
$n_-+n_0+n_+=N$. In the $so_q(3)$ basis we shall use the
classical labels $l$ and $m$, where $m$ is the eigenvalue
of $L_0$ and $[l][l+1]$ is the eigenvalue of the $so_q(3)$
Casimir operator $L_+L_-+[L_0][L_0-1]$; the states are then
labelled by $w(n,l,m)$ (plus additional labels, if necessary).
Consider first the state $|l,0,0\rangle$. Since this state
has $L_0$ eigenvalue $l$ and $N$ eigenvalue $n$, and
$L_+|l,0,0\rangle=0$, it follows that $w(l,l,l)=|l,0,0\rangle$.
Next, one uses the lowering operator $L_-$ to obtain states
with quantum number $m<l$. One obtains~:
\beq
w(l,l,m) = \sqrt{[l+m]!\over [2l]![l-m]!} (L_-)^{l-m} w(l,l,l),
\label{411}
\eeq
where $m=l,l-1,\ldots,-l$. These are genuine orthonormal
$so_q(3)$ states as one can verify that~:
\beq
\begin{array}{l}
L_0 w(l,l,m) = m \;w(l,l,m) , \\[3mm]
L_\pm w(l,l,m) = \sqrt{[l\mp m][l\pm m+1]} \;w(l,l,m\pm 1).
\end{array}
\label{412}
\eeq
In Reference~\cite{princ}, the following operator was introduced~:
\beq
s=(b_0^\d)^2 q^{N_++N_-+1}
 - \sqrt{ {[2N_+]\over[N_+]}{[2N_-]\over[N_-]} } b_+^\d b_-^\d
 q^{-N_0-{1\over 2}} .
\label{41}
\eeq
This operator commutes with the action of the $so_q(3)$ generators;
on the other hand it raises the number of bosons by two when acting
on a boson-state. Consider the state $s^k w(l,l,m)$; this has the
same $so_q(3)$ labels as $w(l,l,m)$ (since $s$ commutes with $so_q(3)$),
and $N s^k w(l,l,m) = (l+2k) s^k w(l,l,m)$. Hence the state under
consideration has $N$ eigenvalue $n=l+2k$, and one can define~:
\beq
w(n,l,m)= {\cal N} s^{n-l\over 2} w(l,l,m),
\label{39}
\eeq
where ${\cal N}$ is a normalization constant. To find this
normalization constant, one can put $m=l$. From the explicit
action of $s$, namely,
\bea
s|n_+,n_0,n_-\rangle &=&
 q^{n_++n_-+1}\sqrt{[n_0+1][n_0+2]}|n_+,n_0+2,n_-\rangle\nn\\
&&- q^{-(n_0+1/2)}\sqrt{[2n_++2][2n_-+2]}|n_++1,n_0,n_-+1\rangle,
\label{saction}
\eea
one can deduce after a number of steps that
\bea
&&s^k|n_+,n_0,n_-\rangle = \sum_{y=0}^k (-1)^y q^{(k-y)(n_++n_-+1)-y(n_0+1/2)}
 {[2k]!! \over [2y]!! [2k-2y]!!} \nn\\
&\times& \left( {[2n_++2y]!!\over [2n_+]!!} {[n_0+2k-2y]!\over[n_0]!}
 {[2n_-+2y]!!\over [2n_-]!!} \right)^{1/2} |n_++y,n_0+2k-2y,n_-+y\rangle,
\label{skaction}
\eea
where $[t]!=[t][t-1]\cdots[1]$ and $[2t]!!=[2t][2t-2]\cdots[2]$.
The constant ${\cal N}$ can then, for example, be determined from
the sum of the squares of the coefficients of the boson states
in the explicit expression of $s^{n-l\over 2}|l,0,0\rangle$. This
yields~:
\beq
{\cal N}= \left( [2l+1]! [n+l]!! \over [n-l]!![2l]!![n+l+1]! \right)^{1/2}.
\label{N}
\eeq
Similarly as for $s^k$, one can from the explicit action of $L_-$,
namely
\bea
L_-|n_+,n_0,n_-\rangle&=&
 q^{n_--n_0/2}\sqrt{[2n_+][n_0+1]}|n_+-1,n_0+1,n_-\rangle\nn\\
&&+ q^{n_+-n_0/2+1/2}\sqrt{[n_0][2n_-+2]}|n_+,n_0-1,n_-+1\rangle,
\label{laction}
\eea
deduce an expression for the action of $L_-^k$, and in particular
one finds~:
\bea
L_-^{l-m}|l,0,0\rangle &=& \sum_{x=\max(0,m)}^{\lfloor{l+m\over 2}\rfloor}
q^{(2l-1)x/2-(l+m)(l+m-1)/4}
\left( [2l]!!([l-m]!)^2 \over [2x]!![2x-2m]!![l+m-2x]! \right)^{1/2}\nn\\
&&\qquad\qquad\times|x,l+m-2x,x-m\rangle.
\label{lkaction}
\eea
From (\ref{411}), (\ref{39}), (\ref{skaction}), (\ref{N}) and
(\ref{lkaction}), one obtains the following result~:
\bea
\lefteqn{w(n,l,m)= q^{-\{l(l+1)+(m-1)(m+2n)\}/4}
\left( [n+l]!![n-l]!![l+m]![l-m]![2l+1]/[n+l+1]!\right)^{1/2} } \nn\\
&\times& \sum_{x=\max(0,m)}^{\lfloor(l+m)/2\rfloor} \sum_{y=0}^{(n-l)/2}
(-1)^y q^{x(n-{1\over 2})-y(l+{3\over 2})}
\left( [2x]!![2y]!![2x-2m]!![l+m-2x]![n-l-2y]!! \right)^{-1} \nn\\
&\times& \left([m+n-2x-2y]![2x+2y]!![2x+2y-2m]!!\right)^{1/2}
|x+y,n+m-2x-2y,x+y-m\rangle . \nn \\
\label{34}
\eea
In the nondeformed case, the expressions of the states $w(n,l,m)$ --
to which (\ref{34}) reduces in the limit $q\rightarrow 1$ -- were
originally given by Sharp {\em et al}~\cite{sharp,moshinsky}.

In~(\ref{34}), there is a double summation (over $x$ and $y$), but
in the boson states themselves, only the combination $x+y$ appears.
Putting $x+y=t$, one can rewrite this as a summation over $x$ and $t$.
Then, the summation over $t$ is the actual summation over the
boson states, and the summation over $x$ simply reduces to
\beq
\sum_x (-1)^x q^{x(n+l+1)} \left( [2x]!![2t-2x]!![2x-2m]!![l+m-2x]!
[n-l-2t+2x]!!\right)^{-1}.
\eeq
Such an expression can be written in terms of a basic generalized
hypergeometric series of type $_3\phi_2$, where
\beq
_3\phi_2\left( {a\quad b\quad c \atop d\quad e};q;z\right)=
\sum_{n=0}^\infty {(a;q)_n(b;q)_n (c;q)_n \over (d;q)_n (e;q)_n (q;q)_n} z^n .
\eeq
Herein, $(a;q)_n$ is the $q$-shifted factorial~\cite{rahman}~:
\beq
(a;q)_n = \left\{ \begin{array}{ll}
 1\qquad &\hbox{if } k=0,\\
 (1-a)(1-aq)\ldots(1-aq^{k-1}) \qquad &\hbox{if } k=1,2,\ldots.
 \end{array}\right.
\eeq
The transition coefficients can then be written in terms of
such a series, and we found~:
\beq
w(n,l,m) = \sum_t C(n,l,m,t) |t,n+m-2t,t-m\rangle,
\label{52}
\eeq
where
\bea
C(n,l,m,t)&=& (-1)^{t+m}q^{-l(l+1)/4-(m-1)(m+2n)/4+m(n+l+1)-t(l+3/2)}\nn\\
&\times& \left([n+l]!![n-l]!![l+m]![2l+1][m+n-2t]![2t]!!\right)^{1/2}\nn\\
&\times&\left([n+l+1]![l-m]![2t-2m]!!\right)^{-1/2}
 \left([2m]!![n-l-2t+2m]\right)^{-1}\nn\\
&\times&_3\phi_2\left( {q^{4(m-t)}\quad q^{4-2l-2m}\quad q^{8-2l-2m}
 \atop q^{4(m+1)}\quad q^{4-4t+2n-2l} };q^4;z=q^{2(n+l+m+2)}\right).
\eea
From (\ref{39}) it follows that the reduction to $so_q(3)$
of the $u_q(3)$ representations labelled by $n$ is the same as in
the nondeformed case, i.e.\ $l=n,n-2,\ldots,1$ or $0$.

Note that here another difference between the deforming functional
and the algebraic approach is apparent~: for the first approach
it follows from~(\ref{wv}) that the reduction coefficients
remain undeformed (they are the same as in the classical
nondeformed case); for the second approach the
reduction coefficients are given by~(\ref{52}) are really
$q$-deformed.

\section{Discussion}

The problem of determining deformations of classical embeddings
turns out to be difficult in the case of a nonregular embedding.
Because of its importance and potential use in physical models,
this problem is worth studying and various approaches should
be considered and compared~\cite{princ,quesne,sciarrino}.

In this paper, we have introduced an algebraic approach and
a technique based upon deforming functionals. Both methods
have been illustrated by means of simple examples, which
have a relevant meaning in physics, such as $u_q(3)\supset so_q(3)$
which can be used to describe the deformed 3-dimensional
harmonic oscillator~\cite{3-d-osc} and for which the nondeformed embedding
has often been applied.
For both techniques, there are also a number of disadvantages.
The deforming functional approach is closest related to the
nondeformed embedding since it uses the same basis states and
reductions; however, one is restricted to symmetric representations
and one does not obtain a genuiune subalgebra.
Finding a proper solution in the algebraic approach is
usually very difficult, and often not a complete solution can
be obtained, such as for the deformation of $u(3)\supset so(3)$.
In this case, the subalgebra generators are finite expressions in
the original generators, but one is again restricted to symmetric
representations. Nevertheless, the formulation in terms of
$q$-bosons, which is interesting on its own and has been studied
in detail here, also allows one to write down more general relations
which remove the restriction to symmetric representations~\cite{def-u3}.

\section*{Acknowledgements}
It is a pleasure to thank Dr.\ C.~Quesne (ULB, Brussels), Prof.\ T.D.~Palev
and Dr.\ N.I.~Stoilova (Institute for Nuclear Research and Nuclear Energy,
Sofia) for stimulating discussions.

This work was partially supported by the E.E.C.\ (contract
No.~CI1*-CT92-0101).

\addtolength{\baselineskip}{-1mm}
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\end{document}

