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% Tensor product of group representations and structural zeros
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% J. Van der Jeugt
% J. Math. Phys. 33 (1992), 2417-2421.
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\begin{center}
{\LARGE Tensor product of group representations and structural zeros
of Racah coefficients}\\[2cm]
J. Van der Jeugt$^{a)}$ \\[.5cm]
{\em Laboratorium voor Numerieke Wiskunde en Informatica,\\
Universiteit Gent, Krijgslaan 281--S9, B-9000 Gent, Belgium}\\
\end{center}


\vspace{3cm}
\noindent \underline{Abstract}

\begin{minipage}{14cm}
It is shown that structural (or nontrivial) zeros of Racah coefficients
(6-$j$ symbols) can be related to decompositions of tensor products of
group representations reduced to SO(3). This extends a previous relation
between structural zeros of 6-$j$ symbols and boson realizations of
exceptional Lie groups. In the present case, both classical and exceptional
groups may appear, and examples are given for SU(3), SO(7), SO(9) and $F_4$.
\end{minipage}

\vfill
\noindent-----------------------------------\\
$^{a)}$ {\footnotesize Research Associate of N.F.W.O.
(National Fund for Scientific Research of Belgium)}
\newpage

\section{Introduction}

It is well known that certain Racah coefficients or 6-$j$ symbols$^{1,2}$
are identically 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$^3$. A simple explanation of these zeros
resides of course in 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 known
polynomial expressions, that is, of Diophantine equations. This approach
not only shows that the number of structural zeros is infinite, it also
provides elegant algorithms to generate the zeros of weight 1 or 2.
It does, however, not suggest any physical applications, nor does it
provide a relation with other mathematical structures.

In the literature, 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$^{20-22}$ have
related many structural zeros of 6-$j$ symbols to SO(3) boson realizations
of the exceptional groups $F_4$ and $E_6$. Again in the same spirit,
Minnaert$^{23}$ 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).

The number of exceptional Lie groups is very limited and although their
use to explain zeros of 6-$j$ symbols has in no way been exhausted it
seems very unrealistic to believe that they could account for all
nontrivial zeros. In the present paper a new but closely related method
is proposed which relates structural zeros to tensor products of
irreducible representations (irreps) of simple Lie groups, both classical
and exceptional. The method uses a construction by means of SO(3)
tensor operators not only for the Lie group generators, but also for
the basis vectors of an irrep. By expressing that the Lie group action
on an irrep should not produce vectors outside the same irrep, one
can relate vanishings of structural zeros of 6-$j$ symbols to these irreps.

In Section~2 we describe the general method. In the following section
examples are given for SU(3), SO(7), SO(9) and $F_4$. Finally, we
discuss some conclusions.

\section{General construction}

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$^{24}$
\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) .
\eea
This formula will be the basic ingredient in this paper.

Let $G$ be a (simple) Lie group and let SO(3) (or SU(2)) be a fixed
subgroup of $G$. Suppose that the representation $\rho$ of $G$
decomposes to SO(3) as $j_1\oplus j_2\oplus j_3 \cdots$. Denote by
$\rho^*$ the representation contragredient to $\rho$. The basis vectors
of $\rho^*$ and $\rho$ can be identified with the creation and annihilation
operators $b_{j,m}^\dagger$ and $b_{j,m}$ ($j\in\{j_1,j_2,j_3,\ldots\}$).
Consider now the tensor product $\rho^*\otimes\rho$, and its decomposition
in irreps of $G$~:
\beq
\rho^*\otimes\rho \rightarrow \gamma\oplus\lambda\oplus\cdots
\eeq
Here, we shall assume that the tensor product decomposition (4) contains
the adjoint representation $\gamma$ as a component. Let the decomposition
of $\gamma$ to SO(3) be $k_1\oplus k_2\oplus k_3\cdots$. Then the basis
elements $G_q^k$ of the Lie algebra of $G$ can be written in terms of
the tensor operators $T_q^k(j j')$ with $j,j'\in\{j_1,j_2,j_3,\ldots\}$
and $k\in\{k_1,k_2,k_3,\ldots\}$. If $\lambda$ denotes another
irreducible representation of $G$ appearing in (4), with SO(3)
components $l_1\oplus l_2\oplus l_3\cdots$, then the basis vectors
$L_q^l$ of this irrep can also be constructed in terms of the tensor
operators $T_q^l(j j')$ with $l\in\{l_1,l_2,l_3,\ldots\}$. Once the
adjoint representation and the basis of $\lambda$ have been constructed,
the action of the adjoint representation on $\lambda$ follows from (3).
Obviously, this action should remain within $\lambda$. Expressing by
means of (3) that the adjoint action on $\lambda$ should not produce
vectors outside $\lambda$, can lead to strong conditions on 6-$j$
symbols, and in particular it can imply the vanishing of certain
6-$j$ symbols.

In the following section we shall give a number of examples of this
technique. There we shall explain how to construct the elements of the adjoint
representation and of a representation $\lambda$. For the determination of
the tensor product (4), one can use standard techniques, tables$^{25}$ or
programs$^{26}$. For the decomposition of irreps to SO(3) or SU(2), we
have used the tables of McKay and Patera$^{27}$, whose notation for
representations by means of Dynkin labels we shall adopt here.

\section{Examples}
\subsection{$G = {\rm SU(3)} $}
Let SO(3) correspond to the principal subalgebra of the Lie algebra of
SU(3), and take for $\rho$ the 8-dimensional adjoint representation itself,
with Dynkin labels (1,1) and with $\rho^*=\rho$. The $j$-values are 1
and 2, since
\[
(1,1) \rightarrow 1\oplus 2 .
\]
The tensor product decomposition (4) reads
\[
(1,1)\otimes(1,1) \rightarrow (2,2)\oplus(3,0)\oplus(0,3)\oplus 2(1,1)
 \oplus (0,0) ,
\]
with (1,1) appearing with multiplicity 2. Thus, the tensor product contains
again the adjoint representation, which means that the Lie algebra of SU(3)
can be constructed out of the tensor operators $T_q^k(j_1j_2)$ with
$j_1,j_2\in\{1,2\}$. The most general form for the SU(3) generators is
\beas
G_q^1 &=& aT_q^1(11)+bT_q^1(22)+cT_q^1(12)+dT_q^1(21) , \\
G_q^2 &=& eT_q^1(11)+fT_q^1(22)+gT_q^1(12)+hT_q^1(21) .
\eeas
Expressing that $[G_q^1,G_{q'}^1]$ can be written in terms of $G_{q''}^1$
only leads by means of (3) to a number of equations in the unknowns,
with solution $a=1$, $b=\sqrt{5}$ and $c=d=0$ (up to a common
proportionality factor). Similarly, expressing that the $k=1$ part in
$[G_q^2,G_{q'}^2]$ must be proportional to $T_{q''}^1(11)+
\sqrt{5}\:T_{q''}^1(22)$ implies $e=f=0$. Thus, the SU(3) generators can
be taken as follows~:
\beas
G^1 &=& T^1(11)+\sqrt{5}\:T^1(22) , \\
G^2 &=& T^2(12)+T^2(21) ,
\eeas
where we have dropped the projection number $q$ in order to simplify
the notation.

In this case, we shall take for $\lambda$ again the adjoint representation.
Then the action condition is equivalent to the closure condition for
the commutators. Looking at the commutator $[G^2_q,G^2_{q'}]$, it follows
from (3) that this contains a term $T_{q''}^3(22)$ with factor
$\sj{2}{2}{3}{2}{2}{1}$. Thus the closure condition implies
\[
\sj{2}{2}{3}{2}{2}{1} = 0,
\]
relating this structural zero to SU(3) representations.

\subsection{$G = {\rm SO(7)} $}

Let SO(3) again correspond to the principal subalgebra of the Lie algebra
of SO(7), which is not maximal since ${\rm SO(7)}\supset G_2 \supset
{\rm SO(3)}$, and take for $\rho$ the 21-dimensional adjoint representation
with Dynkin labels $\rho=(010)$, and decomposition
\[
(010) \rightarrow 1\oplus 3\oplus 5 .
\]
The tensor product decomposition is
\[
(010)\otimes (010)\rightarrow (020)\oplus(102)\oplus(200)\oplus(002)
 \oplus(010)\oplus(000).
\]
It contains the adjoint representation again, thus the Lie algebra of SO(7)
can be constructed out of SO(3) tensor operators with $j=1,3,5$. Without
going into the details of the construction, which is similar to the
case of SU(3) but involves rather lengthy calculations, we simply give the
result here~:
\beas
G^1&=&T^1(11)+\sqrt{2\cdot 7}\:T^1(33)+\sqrt{5\cdot 11}\:T^1(55) ,\\
G^3&=&T^3(33)+\sqrt{3\over 7}\: [T^3(13)+T^3(31)]+\sqrt{11\over 7}\:
 [T^3(35)+T^3(53)] ,\\
G^5&=&T^3(33) -\sqrt{13\over 7}\:T^5(55)+\sqrt{3\cdot 5 \over 2\cdot 7}\:
 [T^5(15)+T^5(51)].
\eeas
It should be noted that there is usually some freedom of phase-factors
in the expressions, but here we have made a fixed choice.

Next, we consider as representation $\lambda$ the 27-dimensional
irrep with Dynkin labels (200). This decomposes to SO(3) as
\[
(200)\rightarrow 2\oplus 4\oplus 6.
\]
In order to construct the representation $\lambda$ in terms of tensor
operators, one starts again from the most general form, e.g.\
\[
L^2=T^2(11)+aT^2(33)+bT^2(55)+c[T^2(13)+T^2(31)]+d[T^2(35)+T^2(53)],
\]
and one expresses by means of (3) that $[G^5_q,L^2_{q'}]$ should only
give terms proportional to $L^2_{q''},L^4_{q''}$ and $L^6_{q''}$. Using
this and similar conditions, one finds~:
\beas
L^2&=&T^2(11)+{\sqrt{2\cdot 7}\over 3\cdot 3}\:T^2(33)
 - {\sqrt{11\cdot 13}\over 3\cdot 3}\:T^2(55)
 +{2 \over \sqrt{3}}\:[T^2(13)+T^2(31)]\\
 &&+{2\sqrt{2\cdot 11}\over 3\cdot 3} \:[T^2(35)+T^2(53)], \\
L^4&=&T^4(33) + 2\sqrt{13\over 7} \:T^4(55)
 - \sqrt{3\cdot 11\over 7}\: [T^4(13)+T^4(31)] \\
 &&- \sqrt{2\cdot 3\cdot 5\over 7}\: [T^4(15)+T^4(51)]
 + \sqrt{5\cdot 13\over 7}\: [T^4(35)+T^4(53)],\\
L^6&=&T^6(33)+\sqrt{5\cdot 17\over 7}\: T^6(55)
 + 3\sqrt{3\over 2\cdot 7}\: [T^6(15)+T^6(51)]
 - \sqrt{2\cdot 7}\: [T^6(35)+T^6(53)].
\eeas

Now, in the commutator of $G^3_q$ and $L^6_{q'}$, there appears a term
in $T^8_{q''}(55)$ with factor $\sj{3}{6}{8}{5}{5}{3}$. Since the
decomposition of (200) contains only 2, 4 and 6, and no 8, it follows
that this term should vanish, hence
\[
\sj{3}{6}{8}{5}{5}{3} = 0.
\]
Similarly, the commutator of $G^5_q$ and $L^6_{q'}$ has a term
proportional to $\sj{5}{6}{3}{3}{1}{5}[T^3(13)-T^3(31)]$. This
implies
\[
\sj{5}{6}{3}{3}{1}{5} = 0.
\]

\subsection{$G = {\rm SO(9)} $}

SO(3) corresponds again to the principal subalgebra of the Lie algebra
of SO(9), and take for $\rho$ the 16-dimensional spin representation
with Dynkin labels (0001) and decomposition
\[
(0001)\rightarrow 2\oplus 5 .
\]
The tensor product $\rho^*\otimes\rho$ is
\[
(0001)\otimes(0001)\rightarrow
 (0002)\oplus(0010)\oplus(0100)\oplus(1000)\oplus(0000).
\]
The right hand side contains the adjoint representation (0100) with
decomposition
\[
(0100)\rightarrow 1\oplus 3\oplus 5\oplus 7.
\]
The basis elements of the Lie algebra of SO(9) are constructed as
described in the previous examples, and are given by
\beas
G^1&=&T^1(22)+\sqrt{11}\:T^1(55),\\
G^3&=&T^3(22)-\sqrt{13\over 3\cdot 11}\:T^3(55)
 - {5\sqrt{7}\over\sqrt{3\cdot 3\cdot 11\cdot 17}}\:[T^3(25)-T^3(52)],\\
G^5&=&T^5(55)+{5\cdot 7\over 2\cdot\sqrt{17}}\:[T^5(25)-T^5(52)],\\
G^7&=&T^7(55)+\sqrt{7\over 2\cdot 17}\:[T^7(25)-T^7(52)].
\eeas
As representation $\lambda$ we take the 9-dimensional standard
representation (1000) with decomposition
\[
(1000)\rightarrow 4.
\]
It is rather simple to construct the tensor product operator basis
for (1000)~:
\[
L^4=T^4(22)+\sqrt{11\cdot 13\over 5}\:T^4(55)
 +\sqrt{11}\:[T^4(25)-T^4(52)].
\]
The action of the adjoint representation on $L^4$ should give terms
only in $L^4$; in particular, the action $[G^5_q,L^4_{q'}]$ has
a term proportional to $\sj{5}{4}{2}{2}{2}{5} T^2_{q''}(22)$.
Hence,
\[
\sj{5}{4}{2}{2}{2}{5}=0.
\]

\subsection{$G = F_4 $}

We take for $\rho$ the 26-dimensional representation (0001) with
decomposition
\[
(0001)\rightarrow 4\oplus 8
\]
into the principal SO(3). The tensor product is
\[
(0001)\otimes(0001) \rightarrow (0002)\oplus(0010)\oplus
(1000)\oplus(0001)\oplus(0000),
\]
thus it contains the adjoint representation (1000), which decomposes
to SO(3) as
\[
(1000)\rightarrow 1\oplus 5\oplus 7\oplus 11.
\]
The present form of the Lie algebra of $F_4$ has been constructed
before$^{20}$~:
\beas
G^1&=&T^1(44)+\sqrt{2\cdot 17\over 5}\:T^1(88),\\
G^5&=&T^5(44)+\sqrt{2\cdot 19\over 3\cdot 17}\:T^5(88)
 + 2\sqrt{2\cdot 5\over 3\cdot 17}\:[T^5(48)+T^5(84)],\\
G^7&=&T^7(44)-\sqrt{17\cdot 23\over 11\cdot 19}\:T^7(88)
 + 3\sqrt{2\cdot 7\cdot 17\over 3\cdot 11\cdot 19}\:[T^7(48)+T^7(84)],\\
G^{11}&=& T^{11}(88) - \sqrt{11\over 2\cdot 3\cdot 5}\:[T^{11}(48)
 +T^{11}(84)].
\eeas
In fact, this construction of the exceptional Lie algebra has itself
been related to structural zeros of 6-$j$ symbols$^{20}$.

Now, if we take $\lambda=(0001)$, then the basis vectors of
$\lambda$ are found to be
\beas
L^4&=&T^4(44)-\sqrt{17\cdot 19\over 2\cdot 7\cdot 11}\:T^4(88)
 + {\sqrt{5\cdot 17}\over 2\sqrt{3\cdot 7}}\:[T^4(48)+T^4(84)],\\
L^8&=&T^8(44)-{\sqrt{19\cdot 23}\over 3\sqrt{11}}\:T^8(88)
 - \sqrt{2\cdot 3\cdot 19 \over 5\cdot 11}\:[T^8(48)+T^8(84)].
\eeas
The action of $G^{11}$ on $L^8$ should only give terms in $L^4$
and $L^8$, in particular it should contain no terms in $T^6(44)$.
But $[G^{11}_q,L^8_{q'}]$ has a term proportional to
$\sj{11}{8}{6}{4}{4}{8}T^6(44)_{q''}$, hence
\[
\sj{11}{8}{6}{4}{4}{8}=0.
\]
Note that this symbol has previously been related$^{22}$ to a boson
realization of $E_6$.

\section{Conclusions}

Using group theoretical methods, i.e.~tensor products and branching
rules, we have been able to explain a number of structural zeros of
6-$j$ symbols. Only a few examples have been given but no doubt
many others exist. With a previous group theoretical technique$^{20-22}$
to explain structural zeros of Racah coefficients, boson realizations
were used and there only the exceptional Lie groups seemed to be related
to structural zeros. In the present framework, both exceptional and
classical groups can be used. This certainly enlarges the application
of group theoretical methods to explain zeros of Racah coefficients.
Whether all structural zeros of 6-$j$ symbols can be given a group
theoretical explanation is a rather academical question. In view of
this, we should point out that only a few examples were presented
here but more exist. For example, here only principal SO(3)
embeddings were considered, but one can just as well consider
non-principal SO(3) embeddings, or branchings to SO(3)$\otimes$SO(3)
etc. We believe that the point is not to give as many examples as possible
or to relate all structural zeros to group irreps -- the purpose of this
paper is only to illustrate that there exists a relationship between
certain structural zeros of 6-$j$ symbols on the one hand and the
decomposition of tensor products of Lie group representations on the
other hand.

Finally, note that due to Regge symmetries the structural zeros
given here give rise to many other zeros appearing in the table
of Biedenharn and Louck$^3$.

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