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% Adjoint boson realization of SU($N$) and a family of structural zeros
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% J. Van der Jeugt
% J. Math. Phys. 35 (1994), 4383-4390.
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\begin{center}
{\LARGE
Adjoint boson realization of SU($N$) and a family of structural zeros
of 6$j$ coefficients}\\[2cm]
J. Van der Jeugt$^{a)}$ \\[.5cm]
{\em Laboratorium voor Numerieke Wiskunde en Informatica,\\
Universiteit Gent, Krijgslaan 281--S9, B-9000 Gent, Belgium}\\
E-mail~: Joris.VanderJeugt@rug.ac.be
\end{center}


\vspace{3cm}
\noindent {Abstract}

\begin{minipage}{14cm}
For the fundamental boson realization of SU($N$) in its principal SO(3)
basis, the generators of SU($N$) are SO(3) tensor operators constructed
out of one single type of boson, namely a boson with SO(3) multiplet label
$(N-1)/2$. Here, the adjoint boson realization of SU($N$) is given~:
this time the SO(3) tensor operators are
constructed by means of $N-1$ types of bosons (those with SO(3)
labels $1,2,\ldots,N-1$). The validity of this realization depends
upon a new identity between 6$j$ coefficients. From this realization,
or equivalently from this new identity, one can deduce the existence
of a family of structural zeros of 6$j$ coefficients.
\end{minipage}
\vskip 1cm
\noindent PACS~: 02.20b, 03.65F
\vfill
\noindent-----------------------------------\\
$^{a)}$ {\footnotesize Senior Research Associate of N.F.W.O.
(National Fund for Scientific Research of Belgium)}
\newpage

\section{Introduction}

Racah coefficients or 6-$j$ symbols$^{1,2}$ can be
identically equal to zero, even though all of the triangular relations are
satisfied. Such zeros are known in the literature as structural, nontrivial
or accidental zeros, and an extensive list of them has been given in the
book by Biedenharn and Louck$^3$.
The original motivation for the results presented here came from the
study of such nontrivial zeros and its
relations with other topics in mathematics.

A straightforward approach of such zeros leads to a study of the
structure of the analytic expression of the
6-$j$ symbol. Investigations of the polynomial part of this expression
have been carried out by Srinivasa Rao {\em et al\/}$^{4-7}$, by Bremner,
Brudno, Louck and Stein$^{8-14}$, and by Lindner$^{15}$ and
Labarthe$^{16}$. These investigations analyse integer solutions of
polynomial expressions, that is, of Diophantine equations (which can
often be related to the Pell equation).

Some of the structural zeros of 6-$j$ symbols have
been related to physical quantities, e.g.~in the quasi-spin model$^{17}$
and in atomic spectroscopy$^{18}$. Other explanations for the existence
of these zeros have been sought by trying to relate them to other mathematical
constructions. In particular it has been shown that the construction of the
generators of exceptional Lie groups by means of SO(3) tensor operators can
account for the existence of some nontrivial zeros. The first example was in
fact given by Racah$^{19}$, explaining the vanishing of $\sj{5}{5}{3}{3}{3}{3}$
by analysing the chain $\rm{SO(7)}\supset G_2\supset\rm{SO(3)}$.
In a similar fashion, Van der Jeugt, Vanden Berghe and De Meyer$^{20-23}$ have
related many structural zeros of 6-$j$ symbols to SO(3) boson realizations
of the exceptional groups $F_4$, $E_6$ and $E_7$. Again in the same spirit,
Minnaert$^{24}$ has recently explained the vanishing of
$\sj{2}{2}{2}{3/2}{3/2}{3/2}$ by means of a realization of the Lie
superalgebra gl(2/2). A closely related method using tensor products
of irreducible representations was applied more recently$^{25}$.

In the present paper, the technique used in Refs.~20--23 for exceptional
Lie groups is shown to extend in an elegant way to SU($N$). A basis for
the SU($N$) generators in terms of SO(3) tensor operators is given.
However, rather than using the fundamental representation for the
construction of these tensor operators (which does not give rise to
new relations), the adjoint representation itself is used for that
purpose. The closure condition is equivalent to a new identity,
which is proved in the Appendix. Rather than explaining isolated
zeros of 6$j$ coefficients, as in the case of similar realizations
for the exceptional Lie algebras, the present realization
is related to a one-parameter family of zeros of 6$j$ coefficients.

The presentation of the fundamental boson realization in Section 2 and
of the adjoint boson realization in Section~3 will be given for
SU($2n+1$) only, but from the rest of the paper it follows that
the results are easily converted to the case of SU($2n$).

\section{Fundamental boson realization}

Let $b_{j,m}^\dagger$ (resp.\ $b_{j,m}$), with $m=-j,-j+1,\ldots,j$,
be a set of creation (resp.\ annihilation) operators satisfying
\beq
[b_{j,m}, b_{j',m'}^\dagger]=\delta_{j,j'}\delta_{m,m'} .
\eeq
With these operators, one can construct SO(3) or SU(2) tensor operators~:
\beq
T_q^k(j_1j_2)=\sum_{m_1,m_2}\langle j_1\;m_1\;j_2\;m_2\;|\;k\;q\rangle
 b_{j_1,m_1}^\dagger (-1)^{j_2+m_2}b_{j_2,-m_2} ,
\eeq
where $\langle \cdots|\cdots\rangle$ is an SO(3) or SU(2) coupling
coefficient$^2$. Then, the commutator between such tensor operators
reads as follows$^{26}$
\bea
[T_{q_1}^{k_1}(j_1j_2),T_{q_2}^{k_2}(j_3j_4)]& =&
 \sum_{k_3,q_3}\sqrt{(2k_1+1)(2k_2+1)(2k_3+1)}
 \left({k_1\atop q_1}\;{k_2\atop q_2}\;{k_3\atop -q_3}\right)
 (-1)^{j_2+j_3+q_3} \nn \\
&\times&\left( (-1)^{k_1+k_2+k_3+j_1+j_2-j_3-j_4}\sj{k_1}{k_2}{k_3}{j_4}{j_1}{j_2}
 \delta_{j_2,j_3} T_{q_3}^{k_3}(j_1j_4) \right. \nn\\
&&\left. -
 \sj{k_1}{k_2}{k_3}{j_3}{j_2}{j_1} \delta_{j_1,j_4} T_{q_3}^{k_3}(j_3j_2)
 \right) .
\label{3}
\eea
This formula will be one of the basic ingredients in this paper.

Consider the principal SO(3) subgroup of SU($2n+1$). With respect to
this subgroup, the fundamental representation of SU($2n+1$) with
Dynkin labels $(1,0,\ldots)$ decomposes
into the SO(3) multiplet $(n)$, and the conjugate representation
with Dynkin labels $(\ldots,0,1)$ also decomposes into the SO(3)
multiplet $(n)$. Since the tensor product
\beq
(1,0,\ldots)\otimes (\ldots,0,1)= (1,0,\ldots,0,1)\oplus (0,\ldots,0),
\eeq
contains the adjoint representation, it follows that the Lie algebra
basis for SU($2n+1$) can be built using the bosons with SO(3) quantum
label $n$. In other words, the operators
\beq
G^k_q = T^k_q(nn),
\eeq
with commutation relation
\bea
[G_{q_1}^{k_1},G_{q_2}^{k_2}]& =&
 \sum_{k_3,q_3}\sqrt{(2k_1+1)(2k_2+1)(2k_3+1)}
 \left({k_1\atop q_1}\;{k_2\atop q_2}\;{k_3\atop -q_3}\right)
 (-1)^{q_3} \nn \\
&\times&\bigl( (-1)^{k_1+k_2+k_3}-1\bigr)\sj{k_1}{k_2}{k_3}{n}{n}{n}
 G_{q_3}^{k_3},
\label{6}
\eea
form a realization of the SU($2n+1$) Lie algebra in the SO(3) basis.
From this well-known realization it follows that the operators $G^k_q$
with $k$ odd form the Lie subalgebra corresponding to SO($2n+1$) (or,
in the other case the subalgebra chain SU($2n$) $\supset$
Sp($2n$) $\supset$ SO(3) will become apparent).

\section{Adjoint boson realization}

The decomposition of the adjoint representation $(1,0,\ldots,0,1)$
of SU($2n+1$) into SO(3) multiplets reads~:
\beq
(1,0,\ldots,0,1) \longrightarrow (1)\oplus(2)\oplus\cdots\oplus(2n).
\eeq
The tensor product of $(1,0,\ldots,0,1)$ with itself contains
(note that this is now a self-conjugate representation)
the irrep $(1,0,\ldots,0,1)$ twice, one corresponding to the
so-called symmetric coupling and the other to the
so-called antisymmetric coupling. It follows that the Lie algebra
of SU($2n+1$) can also be constructed using SO(3) bosons
with quantum numbers $1,2,\ldots,2n$. In the realization
related to the symmetric coupling, the Lie algebra basis takes
the following form~:
\beq
G^a_q = \sum_{\ss j,k \atop \ss {\cal O}(j+k+a)} g^a_{jk} T^a_q(jk),
\label{8}
\eeq
where $j,k\in\{1,2,\ldots,2n\}$ and ${\cal O}(j+k+a)$ restricts
the summation over $j$ and $k$ to values such that $j+k+a$ is odd.
In this expression, the $g^a_{jk}$ can be considered as the unknowns.
The equations for the $g^a_{jk}$ follow from substituting (\ref{8}) in (\ref{6}),
using (\ref{3}). Thus expressing that $[G^a_q,G^b_{q'}]$ yields the correct
contribution in $G^c_{q''}$ implies that the following relation
should hold (only when $a+b+c$ is odd, because of the factor
$ (-1)^{a+b+c} -1 $)~:
\beq
\sum_{k,{\cal O}(j+k+a)} g^a_{jk}g^b_{kl}\sj{a}{b}{c}{l}{j}{k}
+ \sum_{k,{\cal O}(j+k+b)} g^a_{kl}g^b_{jk}\sj{a}{b}{c}{j}{l}{k}
= -2 (-1)^c g^c_{jl}\sj{a}{b}{c}{n}{n}{n}.
\label{9}
\eeq
This is a complicated set of non-linear equations, and in
general it requires a lot of computations to solve (see, for example,
similar constructions for boson realizations of the exceptional
Lie groups)$^{20-23}$. In the present case, however, these $g$-coefficients
have a very simple solution. To present this solution, we shall
explicitly construct the coefficients $g^1_{jk}$ and $g^2_{jk}$.
From their expression, a general form can be postulated. Then,
it will be shown (in the Appendix) that this general form is
a genuine solution.

Following (\ref{8}), we write~:
\beq
G^1_q = \sum_{j=1}^{2n} g^1_{jj} T^1_q(jj).
\eeq
Using (\ref{3}) with $k_1=k_2=k_3=1$, it follows that the coefficients
$g^1_{jj}$ satisfy~:
\beq
g^1_{jj} = \sj{1}{1}{1}{n}{n}{n} {\huge /} \sj{1}{1}{1}{j}{j}{j}\quad (j=1,2,\ldots,2n),
\eeq
or, explicitly,
\beq
g^1_{jj} = \left( j(j+1)(2j+1)\over n(n+1)(2n+1)\right)^{1/2}
\quad (j=1,2,\ldots,2n).
\label{12}
\eeq
The expression for $G^2_q$ reads~:
\beq
G^2_q = \sum_{j=1}^{2n-1} g^2_{j,j+1} \left( T^2_q(j,j+1) + \alpha
 T^2_q(j+1,j) \right),
\eeq
where $\alpha$ is some phase factor to be determined. One can now
require that the commutator $[G^2_q,G^2_{q'}]$ yields the correct
contribution in $G^1_{q''}$, which has already been constructed.
Using (\ref{3}) and (\ref{6}), this gives rise to the following equations~:
\bea
\alpha (g^2_{j,j+1})^2&=&\left(g^1_{j+1,j+1} \sj{2}{2}{1}{n}{n}{n}\right.\nn\\
&-& \left.\alpha (g^2_{j+1,j+2})^2 \sj{2}{2}{1}{j+1}{j+1}{j+2} \right) {\huge /}
\sj{2}{2}{1}{j+1}{j+1}{j+1},
\label{14}
\eea
where $j=0,1,2,\ldots,2n-1$; note that for $j=0$ one puts $g^2_{01}=0$
and for $j=2n-1$ one puts $g^2_{2n,2n+1}=0$. Thus one obtains a set
of $2n$ linear equations in the $2n-1$ unknowns $(g^2_{j,j+1})^2$.
This set has a positive solution provided $\alpha=-1$. Using the explicit
forms of the 6$j$ coefficients in (\ref{14}), and backward substitution, one
finds~:
\beq
(g^2_{j,j+1})^2 = {3j(j+1)(j+2)(2n-j)(2n+j+2)\over n(n+1)(2n-1)(2n+1)(2n+3)}.
\label{15}
\eeq
This suggest a relation between the $g$-coefficients and Racah coefficients,
since
\beq
\sj{j}{j+1}{2}{n}{n}{n}=(-1)^j\left( 3j(j+1)(j+2)(2n-j)(2n+j+2) \over
 (2j+1)(2n+3)(2n+3)(2n+2)(2n+1)(2n)(2n-1) \right)^{1/2}.
\label{16}
\eeq
Similarly, the $g^1$-coefficient can be related to
\beq
\sj{j}{j}{1}{n}{n}{n} = (-1)^{j+1} \left( j(j+1)\over (2j+1)(2n)(2n+1)(2n+2)
\right)^{1/2}.
\label{17}
\eeq
Comparing (\ref{12}) with (\ref{17}) and (\ref{15}) with (\ref{16})
suggest the following
expression~:
\beq
g^a_{jk} = (-1)^{a+j} 2\sqrt{(2j+1)(2k+1)} \sj{j}{k}{a}{n}{n}{n}.
\label{18}
\eeq
In other words, the relation to prove is
\bea
&& (-1)^b\sum_{k,{\cal O}(j+k+a)} (2k+1)\sj{j}{k}{a}{n}{n}{n}
 \sj{k}{l}{b}{n}{n}{n}\sj{a}{b}{c}{l}{j}{k} \nn\\
&& +(-1)^a\sum_{k,{\cal O}(j+k+b)} (2k+1)\sj{k}{l}{a}{n}{n}{n}
 \sj{j}{k}{b}{n}{n}{n}\sj{a}{b}{c}{j}{l}{k} \nn\\
&& = (-1)^j \sj{j}{l}{c}{n}{n}{n} \sj{a}{b}{c}{n}{n}{n},
\eea
for $a+b+c$ odd and $j+l+c$ odd.

This relation is indeed valid. In fact, it is a special case of the
following general relation (valid for all $a,b,c,d,e$ and integer or
half-integer $n$)~:
\bea
&&\sj{a}{b}{c}{n}{n}{n} \sj{c}{d}{e}{n}{n}{n} = \nn\\
&& \sum_{k,{\cal O}(a+d+k)}(-1)^{a+b+c+d+e+k} (2k+1)\sj{a}{d}{k}{n}{n}{n}
 \sj{e}{b}{k}{n}{n}{n}\sj{a}{d}{k}{e}{b}{c} \nn\\
&&+ \sum_{k,{\cal E}(b+c+e+k)}(-1)^{a+b+c+d+e+k} (2k+1)\sj{a}{e}{k}{n}{n}{n}
 \sj{d}{b}{k}{n}{n}{n}\sj{a}{e}{k}{d}{b}{c}.
\label{20}
\eea
Herein, ${\cal E}(b+c+e+k)$ in the second summation restricts the sum
to $k$-values with $b+c+e+k$ even.
This relation~(\ref{20}), reminiscent of but quite different from the
Biedenharn-Elliott identity, will be proved in the Appendix.

\section{Discussion}

We have given a new and interesting realization for the SU($2n+1$)
Lie algebra, namely~:
\beq
G^a_q=\sum_{j,k,{\cal O}(j+k+a)}(-1)^{a+j} 2\sqrt{(2j+1)(2k+1)}
 \sj{j}{k}{a}{n}{n}{n} T^a_q(jk).
\label{21}
\eeq
Since the generators are constructed by means of
boson operators corresponding to the adjoint representation itself,
this has been called the adjoint boson realization. The validity
of this realization depends upon a new identity, relation (\ref{20}). Since
this relation holds for half-integer $n$-values as well, it is clear
that the present construction can immediately be transferred to the
case of SU($2n$), thus giving the adjoint boson realization for
all SU($N$) Lie algebras.

The original motivation for such new realizations was the hope of
finding relations between classical Lie algebras and structural
zeros of Racah coefficients. In the present case, such a relation
can indeed be seen, as we shall now illustrate for the case
of SU($2n+1$). Consider the commutator of the type $[G^{2n}_q,
G^{2n}_{q'}]$, with $G^{2n}$ given in (\ref{8}). Working out this commutator
by means of (\ref{3}), it appears to have a contribution in $T^{2n+1}_{q''}(2,2n)$
with coefficient proportional to $\sj{2n}{2n}{2n+1}{2n}{2}{2n-1}$.
Since the Lie algebra consists of tensor operators $G^{k}$ with
$k=1,2,\ldots,2n$ only, this contribution has to be zero and hence
the vanishing of the above 6$j$ coefficient follows.

More directly, this vanishing can also be seen from relation
(\ref{20}). If we substitute $a=b=2n$, $c=2n+1$, $d=2$ and $e=2n$ in (\ref{20}),
the left hand side reduces to
\beq
\sj{2n}{2n}{2n+1}{n}{n}{n} \sj{2n+1}{2}{2n}{n}{n}{n}
\eeq
which is trivially zero because of triangular conditions.
On the other hand, the right hand side becomes~:
\beq
\sum_{k,{\cal O}(k)}2(2k+1)\sj{2n}{2}{k}{n}{n}{n}
 \sj{2n}{2n}{k}{n}{n}{n}\sj{2n}{2}{k}{2n}{2n}{2n+1};
\eeq
the triangular conditions imply that only one sum remains and we obtain
\beq
\sj{2n}{2}{2n-1}{n}{n}{n}
 \sj{2n}{2n}{2n-1}{n}{n}{n}\sj{2n}{2}{2n-1}{2n}{2n}{2n+1}=0.
\eeq
Since the first two 6$j$ coefficients in this last expression are
stretched or single term coefficients (there is no summation part
in the 6$j$ coefficient), it follows that the third coefficient
must vanish.

A similar analysis has shown the relation between exceptional Lie
algebras and structural zeros of 6$j$ coefficients$^{20-23}$; in fact, at
that time it was believed that only an isolated number of zeros
could be explained by means of such a technique.
Here, we have shown that a whole family of zeros,
\beq
\sj{a}{a}{a+1}{a}{2}{a-1}=0,
\eeq
can be explained by such an approach and be related to Lie
algebra realizations.

\section*{Acknowledgements}
It is a pleasure to thank Prof.\ K.~Srinivasa Rao (IMSC, Madras,
India) for stimulating discussions, and Ms.\ Sangita Pitre (Ghent)
for numerical verification of the identities appearing in this paper.
This work was partially supported by the E.E.C.\ (contract
No.~CI1*-CT92-0101).

\section*{Appendix}

The relations that will be used for the proof of (\ref{20}) are the
two orthogonality properties for 6$j$ coefficients, the expression for
the 9$j$ coeffiecient, and the Biedenharn-Elliott
identity, resp.\ eqs.\ (3), (4), (5) and (6) on page 305 of Varshalovich
{\em et al}~$^{27}$~:
\beq
\sum_X (2X+1)\sj{a}{b}{X}{c}{d}{p}\sj{a}{b}{X}{c}{d}{q} = \delta_{pq}/(2p+1),
\label{29}
\eeq
\beq
\sum_X (-1)^{p+q+X}(2X+1)\sj{a}{b}{X}{c}{d}{p}\sj{a}{b}{X}{d}{c}{q} =
 \sj{a}{c}{q}{b}{d}{p}, \label{30}
\eeq
\beq
\sum_X (-1)^{2X}(2X+1)\sj{a}{b}{X}{c}{d}{p}\sj{c}{d}{X}{e}{f}{q}
 \sj{e}{f}{X}{a}{b}{r} =
 \left\{ \begin{array}{ccc} a&f&r\\ d&q&e\\ p&c&b \end{array} \right\},
\label{31}
\eeq
\beq
\sum_X (-1)^{R+X}(2X+1)\sj{a}{b}{X}{c}{d}{p}\sj{c}{d}{X}{e}{f}{q}
 \sj{e}{f}{X}{b}{a}{r} =
 \sj{p}{q}{r}{e}{a}{d}\sj{p}{q}{r}{f}{b}{c}, \label{32}
\eeq
where $R=a+b+c+d+e+f+p+q+r$.

When $c=d=n$, one can take appropriate combinations of (\ref{29})
and (\ref{30}) to find~:
\beq
\sum_{\ss {\cal E}(X) \atop \ss {\cal O}(X) }2(2X+1)(2p+1)
\sj{a}{b}{X}{n}{n}{p}\sj{a}{b}{X}{n}{n}{q} = \delta_{pq}\pm(-1)^{p+q}(2p+1)
 \sj{a}{n}{q}{b}{n}{p},       \label{33}
\eeq
where the plus sign is taken when the sum ranges over all even $X$-values,
and the minus sign when it ranges over all odd $X$-values.
Consider now the first part in the right hand side of (\ref{20})~:
\beq
\sum_{k,{\cal O}(a+d+k)}(-1)^{a+b+c+d+e+k} (2k+1)\sj{a}{d}{k}{n}{n}{n}
 \sj{e}{b}{k}{n}{n}{n}\sj{a}{d}{k}{e}{b}{c}
\eeq
Applying the Biedenharn-Elliott identity (\ref{32}) to
$\sj{e}{b}{k}{n}{n}{n}\sj{e}{b}{k}{a}{d}{c}$ yields~:
\beq
\sum_{k,{\cal O}(a+d+k)}\sum_X (-1)^{3n+X} (2k+1)(2X+1)\sj{a}{d}{k}{n}{n}{n}
 \sj{n}{d}{X}{c}{n}{e}\sj{c}{n}{X}{n}{a}{b}\sj{n}{a}{X}{d}{n}{k}.
\eeq
Using (\ref{33}), the summation over $k$ can be performed~:
\beq
\sum_X (-1)^{3n+X}
 \sj{n}{d}{X}{c}{n}{e}\sj{c}{n}{X}{n}{a}{b} {1\over 2}
 \left(\delta_{nX} - (-1)^{a+d+n+X}(2X+1)\sj{a}{n}{n}{d}{n}{X}\right).
\eeq
Finally, using (\ref{31}) this reduces to~:
\beq
{1\over 2} \sj{n}{d}{n}{c}{n}{e}\sj{c}{n}{n}{n}{a}{b}- {1\over 2}
(-1)^{a+d} \left\{ \begin{array}{ccc} n&a&n\\ n&b&n\\ e&c&d
\end{array} \right\}. \label{37}
\eeq

For the second part of (\ref{20}), one uses the Biedenharn-Elliott identity
on the product $\sj{b}{d}{k}{n}{n}{n}\sj{b}{d}{k}{e}{a}{c}$,
then one can again use
(\ref{33}) and (\ref{31}); this part then reduces to
\beq
{1\over 2} \sj{n}{a}{n}{c}{n}{b}\sj{c}{n}{n}{n}{e}{d}+ {1\over 2}
(-1)^{b+c+e} \left\{ \begin{array}{ccc} n&e&n\\ n&d&n\\ b&c&a
\end{array} \right\}.   \label{38}
\eeq
Due to the symmetries of 9$j$ coefficients, the contributions in (\ref{37}) and
(\ref{38}) cancel each other, and we find for the sum~:
\beq
\sj{a}{b}{c}{n}{n}{n}\sj{c}{d}{e}{n}{n}{n},
\eeq
thus proving (\ref{20}).

\newpage
\section*{References}
{\small
\addtolength{\baselineskip}{-3mm}
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\end{enumerate}
 }
\end{document}

