The Missing Label of $\mathfrak {su}_3$ and Its Symmetry

Abstract

We present explicit formulas for the operators providing missing labels for the tensor product of two irreducible representations of $\mathfrak {su}_3$. The result is seen as a particular representation of the diagonal centraliser of $\mathfrak {su}_3$ through a pair of tridiagonal matrices. Using these explicit formulas, we investigate the symmetry of this missing label problem and we find a symmetry group of order 144 larger than what can be expected from the natural symmetries. Several realisations of this symmetry group are given, including an interpretation as a subgroup of the Weyl group of type $E_6$, which appeared in an earlier work as the symmetry group of the diagonal centraliser. Using the combinatorics of the root system of type $E_6$, we provide a family of representations of the diagonal centraliser by infinite tridiagonal matrices, from which all the finite-dimensional representations affording the missing label can be extracted. Besides, some connections with the Hahn algebra, Heun–Hahn operators and Bethe ansatz are discussed along with some similarities with the well-known symmetries of the Clebsch–Gordan coefficients.

Appendices

A Calculation of $X_{{\textbf {m}}}$ and $Y_{{\textbf {m}}}$

In this appendix, a basis of the multiplicity space $M_{\textbf {m}}$ is given in terms of the highest-weight vectors of weight $(m''_1,m''_2)$ in the tensor product $[m_1,m_2]\otimes [m'_1,m'_2]$. The general idea is borrowed from [24] but, let us recall that the basis constructed here is different and has the advantage of being better adapted to the symmetries. Some details on the calculation of the symmetric forms (18) of $X_{{\textbf {m}}}$ and $Y_{{\textbf {m}}}$ are also provided.

Extremal projector. As in [24], the extremal projector $P_{(m''_1,m''_2)}$ of $\mathfrak {su}_3$ from [1] is used to build highest-weight vectors. Its definition, adapted to our setting, and its necessary properties are recalled.

A basis of the irreducible representation $[m''_1,m''_2]$ consists of weight vectors, and up to a scalar, there is a unique vector of weight $(m''_1,m''_2)$ (the highest-weight vector). A projector $P_{(m''_1,m''_2)}$ acting on $[m''_1,m''_2]$ is defined by:

$$\begin{aligned} P_{(m''_1,m''_2)}(v)=\left\{ \begin{array}{ll} v &{} \text { if }v \text { is the highest-weight vector of } [m''_1,m''_2],\\ 0 &{} \text { if }v \text { is a weight vector of weight different from }(m''_1,m''_2).\end{array}\right. \end{aligned}$$

The projector $P_{(m''_1,m''_2)}$ is also set to 0 on any irreducible finite-dimensional representation of $\mathfrak {su}_3$ different from $[m''_1,m''_2]$. Thus, the element $P_{(m''_1,m''_2)}$ is defined as a linear operator on $[m_1,m_2]\otimes [m'_1,m'_2]$ (since this representation decomposes as a direct sum of irreducible finite-dimensional representation of $\mathfrak {su}_3$) and projects the whole space $[m_1,m_2]\otimes [m'_1,m'_2]$ onto the multiplicity space $M_{\textbf {m}}$.

From this definition, it is also clear that $P_{(m''_1,m''_2)}$ satisfies:

$$\begin{aligned} P_{(m''_1,m''_2)}.(e_{ij}\otimes 1+1\otimes e_{ij})=0\ \ \ \ \ \text { if }i>j, \end{aligned}$$

(65)

where the equality is meant as linear operators on $[m_1,m_2]\otimes [m'_1,m'_2]$. Moreover, recall that any element commuting with the diagonal action of $\mathfrak {su}_3$ on $[m_1,m_2]\otimes [m'_1,m'_2]$ sends a highest-weight vector on a linear combination of highest-weight vectors with the same weight. Therefore, $P_{(m''_1,m''_2)}$ commutes with any such element and, in particular, with $X_{{\textbf {m}}}$ and $Y_{{\textbf {m}}}$.

A generating set of $M_{{\textbf {m}}}$. Let $v_0=v_{(m_1,m_2)}\otimes v_{(m'_1,m'_2)}$, where $v_{(m_1,m_2)}$ (respectively, $v_{(m'_1,m'_2)}$) is a highest-weight vector of $[m_1,m_2]$ (respectively, $[m'_1,m'_2]$). Since $[m_1,m_2]$ and $[m'_1,m'_2]$ are highest-weight representations, any vector of $[m_1,m_2]\otimes [m'_1,m'_2]$ is a linear combination of vectors of the form:

$$\begin{aligned} (e_{31}^{a'}e_{21}^{b'}e_{32}^{c'}\otimes e_{31}^ae_{21}^be_{32}^c)\cdot v_0\,. \end{aligned}$$

(66)

Recall that we are looking for the image of the projector $P_{(m''_1,m''_2)}$ which satisfies (65). Therefore it is enough to consider the set of vectors:

$$\begin{aligned} P_{(m''_1,m''_2)}\cdot \left( 1\otimes e_{31}^ae_{21}^be_{32}^c\right) \cdot v_0 ,\ \ \ \ \ \ \ \ a,b,c\in \mathbb {Z}_{\ge 0}\ . \end{aligned}$$

An easy calculation shows that $e_{31}e_{21}^b=\frac{1}{b+1}(e_{32}e_{21}^{b+1}-e_{21}^{b+1}e_{32})$ and, by induction, any of the above vectors can be rewritten as a linear combination of the following ones:

$$\begin{aligned} P_{(m''_1,m''_2)}\cdot \left( 1\otimes e_{32}^ae_{21}^be_{32}^c\right) \cdot v_0 ,\ \ \ \ \ \ \ \ a,b,c\in \mathbb {Z}_{\ge 0}\ . \end{aligned}$$

Since $P_{(m''_1,m''_2)}$ sends to 0 any vector of weight different from $(m''_1,m''_2)$, restrict a, b, c such that the vector, before the application of $P_{(m''_1,m''_2)}$, is of weight $(m''_1,m''_2)$. This implies that $b=n-1$ and that $a+c=\ell -1$. So, with a choice of normalisation that will be convenient later, we conclude that the space $M_{{\textbf {m}}}$ is spanned by the following vectors:

$$\begin{aligned} \omega _m:=P_{(m''_1,m''_2)}\cdot \left( 1\otimes \frac{e_{32}^m\,e_{21}^{n-1}\,e_{32}^{\ell -1-m}}{m!\,(n-1)!\,(\ell -1-m)!}\right) \cdot v_0 ,\ \ \ \text { where }m\in \{0,\dots ,\ell -1\}.\nonumber \\ \end{aligned}$$

(67)

We can also start with the opposite order on the generators of $\mathfrak {su}_3$ in (66) and use a similar reasoning exchanging the role of $e_{21}$ and $e_{32}$. For some reasons, it will prove convenient to put this other spanning set of $M_{{\textbf {m}}}$ in the following form:

$$\begin{aligned} \omega '_m:=P_{(m''_1,m''_2)}\cdot \left( \frac{e_{21}^m\,e_{32}^{\ell -1}\,e_{21}^{n-1-m}}{m!\,(\ell -1)!\,(n-1-m)!}\otimes 1\right) \cdot v_0 ,\ \ \text { where }m\in \{0,\dots ,n-1\}.\nonumber \\ \end{aligned}$$

(68)

We will use the conventions $\omega _m:=0$ if $m\notin \{0,\dots ,\ell -1\}$ and $\omega '_m:=0$ if $m\notin \{0,\dots ,n-1\}$.

A basis of $M_{{\textbf {m}}}$. Let $v\in [m_1,m_2]\otimes [m'_1,m'_2]$ such that $(1\otimes e_{23})\cdot v=0$ and v is an eigenvector of $1\otimes h_{2}$ with eigenvalues $N\in \mathbb {Z}_{\ge 0}$. Then v is the highest-weight vector of a copy of the irreducible representation of dimension $N+1$ of the subalgebra generated by $1\otimes e_{32},1\otimes h_{2},1\otimes e_{23}$ isomorphic to $\mathfrak {su}_2$. In particular, we have that $(1\otimes e_{32})^{N+1}\cdot v=0$. We will also use this fact for the other $\mathfrak {su}_2$-subalgebras. We deduce that:

• $(1\otimes e_{32}^{\ell -1-m})\cdot v_0=0$ unless $\ell -1-m<m'_2$, that is, unless $m>\ell -1-m'_2$.

• $(e_{32}^{m}\otimes 1)\cdot v_0=0$ unless $m<m_2$.

• $(1\otimes e_{21}^{n-1}\,e_{32}^{\ell -1-m})\cdot v_0=0$ unless $n-1<m'_1+\ell -1-m$, that is, unless $m<m'_1+\ell -n$.

• $(e_{21}^{n-1}\,e_{32}^{m}\otimes 1)\cdot v_0=0$ unless $n-1<m_1+m$, that is, unless $m>n-m_1-1$.

Using the defining formula (67) for the vectors $\omega _m$ together with the property (65) for the projector $P_{(m''_1,m''_2)}$, we deduce that:

$$\begin{aligned} \omega _m=0\ \ \ \ \text { unless max}\{\ell -m'_2 ,\ n-m_1 ,\ 0\}\le m< \text {min}\{\ell ,\ m_2 ,\ m'_1+\ell -n\}. \end{aligned}$$

So the number of non-zero vectors $\omega _m$ is at most the difference between the min and the max. If $\ell \le n$, this number is the dimension of $M_{{\textbf {m}}}$. We conclude that if $\ell \le n$, the non-zero vectors $\omega _m$ form a basis of $M_{{\textbf {m}}}$.

A similar reasoning with the vectors $\omega '_m$ if $n\le \ell$ exchanges the role of the parameters as follows: $\ell \leftrightarrow n$ and $(m_1,m_2)\leftrightarrow (m'_2,m'_1)$. Therefore, a basis of $M_{{\textbf {m}}}$ is

$$\begin{aligned} \{\omega _m\}_{\text {max}\{\ell -m'_2 ,\ n-m_1 ,\ 0\}\le m< \text {min}\{\ell ,\ m_2 ,\ m'_1+\ell -n\}}\ \ \ \ \ \ (\text {if }\ell \le n) , \end{aligned}$$

(69)

$$\begin{aligned} \{\omega '_m\}_{\text {max}\{\ell -m'_2 ,\ n-m_1 ,\ 0\}\le m< \text {min}\{n ,\ m'_1 ,\ m_2+n-\ell \}}\ \ \ \ \ \ (\text {if }n\le \ell )\,. \end{aligned}$$

(70)

Matrix for $X_{{\textbf {m}}}$. Two equivalent expressions for X are:

$$\begin{aligned} X=\frac{1}{2}(T^{(1,1,2)}-T^{(1,2,2)})+\frac{1}{3}(\ell _1-\ell _2)=T^{(1,1,2)}+\frac{1}{6}(3l_1-l_2-l_3)+\frac{3}{4}(k_3-k_2-k_1)\,. \end{aligned}$$

Recall that X acts on the space $M_{{\textbf {m}}}$, the resulting operator being denoted $X_{{\textbf {m}}}$. We are going to calculate the action of X on the vectors $\omega _m$ defined in (67). In fact, due to the second expression of X, it would be enough to calculate the action of $T^{(1,1,2)}$, since the remaining terms are Casimir elements which act on $M_{{\textbf {m}}}$ simply by numbers given in Remark 1.2. The strategy presented below is equally applicable for any elements of the centraliser, so for simplicity, we present it for X, but we mention that the actual technical computations are simpler to perform on $T^{(1,1,2)}$.

We are going to use the following rules:

$$\begin{aligned} {\left\{ \begin{array}{ll} P_{(m''_1,m''_2)}\,(e_{ij}\otimes 1)=-P_{(m''_1,m''_2)}\, (1\otimes e_{ij}) \quad \text { if }i>j,\\ (e_{ij}\otimes 1)\,(1\otimes \gamma _m)\cdot v_0=0 \qquad \text { if }i<j,\\ (h_1^ah_2^b\otimes h_1^ch_2^d)\,(1\otimes \gamma _m)\cdot v_0=(m_1-1)^a(m_2-1)^b\\ (m'_1+\ell -2n)^c(m'_2+n-2\ell )^d (1\otimes \gamma _m)\cdot v_0,\qquad \end{array}\right. } \end{aligned}$$

(71)

where $\gamma _m:=0$ if $m\notin \{0,\dots ,\ell -1\}$ and $\gamma _m:= \frac{e_{32}^m\,e_{21}^{n-1}\,e_{32}^{\ell -1-m}}{m!\,(n-1)!\,(\ell -1-m)!}$, otherwise. The first rule is (65), the second rule follows from the fact that $e_{ij}\otimes 1$ will hit $v_0$ which is the tensor product of two highest-weight vectors, and the third rule comes from recalling that $\gamma _m\cdot v_0$ is an eigenvector of the Cartan generators and from calculating the eigenvalues.

The strategy is to start by rewriting X as a linear combination of monomials in the generators of $U(\mathfrak {su}_3)\otimes U(\mathfrak {su}_3)$ organized (in each factor) in the following order: $e_{21},e_{31},e_{32},e_{12},e_{13},e_{23},h_1,h_2$. Then we use that X commutes with P and write:

$$\begin{aligned} X\cdot \omega _m{} & {} =XP_{(m''_1,m''_2)}\,(1 \otimes \gamma _m)\cdot v_0=P_{(m''_1,m''_2)}X\,(1 \otimes \gamma _m)\cdot v_0\nonumber \\{} & {} = P_{(m''_1,m''_2)}\,\sum _{W\in S} x_W (1\otimes W\gamma _m)\,\cdot v_0\ , \end{aligned}$$

(72)

where $S=\{1,\, e_{21}e_{12},\, e_{32}e_{23},\, e_{31}e_{13},\, e_{32}e_{21}e_{13},\, e_{31}e_{23}e_{12}\}$ and $x_W$ are numbers. The last equality has been obtained by using only the rules (71). Then, for each $W\in S$, $W\gamma _m$ is put in the same order as above and applied on $v_0$. Finally, after a tedious but straightforward calculation, one can check that:

$$\begin{aligned} X\cdot \omega _m=\alpha _m\,\omega _{m-1}+\beta _m\,\omega _m+\delta _m\,\omega _{m+1} , \end{aligned}$$

(73)

with $E_{m,k}=\sum _{i=1}^6(\xi _i-m-\frac{1}{2})^k$,

$$\begin{aligned}{} & {} \alpha _m=(m-\xi _1)(m-\xi _2)(m-\xi _3)\ ,\ \ \ \ \ \ \ \ \delta _m=(m+1)(m+1-\xi _4)(m+1-\xi _5) ,\\{} & {} \beta _m=-\frac{1}{108}\Bigl (\frac{7}{2}E_{m,1}^3-18 E_{m,1}E_{m,2}+18 E_{m,3}\Bigr )-\frac{1}{24}E_{m,1}\Bigl (\Lambda ^2+2\Bigr )\ , \end{aligned}$$

and $(\xi _1,\,\xi _2,\,\xi _3,\,\xi _4,\,\xi _5,\,\xi _6)$ and $\Lambda$ defined in Sect. 1.3 for the situation $\ell \le n$. In the case $\ell \le n$, we have proved that a basis is given by the subset of vectors $\omega _m$ with indices $\text {max}\{\ell -m'_2 ,\ n-m_1 ,\ 0\}\le m< \text {min}\{\ell ,\ m_2 ,\ m'_1+\ell -n\}$, and moreover that all other vectors $\omega _m$ are 0. So it remains only to extract the diagonal block of X corresponding to the restricted set of indices to obtain the matrix for $X_{{\textbf {m}}}$. This proves the formula given in Sect. 1.3 for $X_{{\textbf {m}}}$ in the case $\ell \le n$.

In the situation $n\le \ell$, we have to consider the vectors $\omega '_m$ instead of $\omega _m$. We see, comparing the definitions (67) and (68), that $\omega '_m$ is obtained from $\omega _m$ by applying an automorphism on what is between the projector and $v_0$. This automorphism is the composition of the exchange of the two factors and the automorphism $\sigma$ Defined in (13) (in fact this gives $\omega '_m$ up to a global sign not depending on m). Then the key fact is that X is invariant under the same automorphism as indicated in Sect. 1.2. From this it is immediate to deduce that the action of X on $\omega '_m$ is as in (73), namely:

$$\begin{aligned} X\cdot \omega '_m=\alpha '_m\,\omega '_{m-1}+\beta '_m\,\omega '_m+\delta '_m\,\omega '_{m+1} , \end{aligned}$$

where $\alpha '_m,\beta '_m,\delta '_m$ are obtained from $\alpha _m,\beta _m,\delta _m$ with the following exchange of the parameters: $\ell \leftrightarrow n$ and $(m_1,m_2)\leftrightarrow (m'_2,m'_1)$. So the action of X on the vectors $\omega '_m$, $m=0,\dots ,\ell -1$, is also by a tridiagonal matrix and in this case, similarly as above, it remains only to extract the square submatrix corresponding to the indices $\text {max}\{\ell -m'_2 ,\ n-m_1 ,\ 0\}\le m< \text {min}\{n ,\ m'_1 ,\ m_2+n-\ell \}$. This concludes the verification of the formulas give in Sect. 1.3 for $X_{{\textbf {m}}}$ in all cases.

Matrix for $Y_{{\textbf {m}}}$. The strategy used above for calculating the action of X on the vectors $\omega _m$ and $\omega '_m$ is also applicable for the computation of Y rewritten as follows to simplify the computation:

$$\begin{aligned} Y = T^{(1,1,2,2)}-\frac{1}{2}(l_1+l_2+l_3)-\frac{1}{48}(k_3-k_1-k_2)^2-\frac{5}{12}k_1k_2-(k_3-k_1-k_2)\ . \end{aligned}$$

The calculations are more complicated but leads similarly to all the formulas given in Sect. 1.3.

B Parameters z of Proposition 1.4

The parameters z in relations (27a) and (27b) are given by

$$\begin{aligned} z_2= & {} \frac{\xi _3+\xi _4}{2}-\frac{\xi _1+\xi _2+\xi _5+\xi _6}{4} ,\quad z_3 =\frac{\xi _3 -\xi _4}{2}\,,\quad z_1=z_3^2 -\frac{\Lambda ^2}{4}\ , \quad z_4=\frac{1}{2}\ , \nonumber \\ \end{aligned}$$

(74a)

$$\begin{aligned} z_0= & {} \frac{A_3}{4} -\frac{z_2}{3}\left( \frac{\Lambda ^2}{4} +z_3^2-\frac{2z_2^2}{9}-A_2-\frac{1}{2} \right) , \end{aligned}$$

(74b)

$$\begin{aligned} z_5= & {} \frac{z_3(A_2\Lambda +z_3)}{3} -\frac{z_2A_3}{3} -\frac{A_2(\Lambda -2z_3+4)(\Lambda -2z_3+2)}{24}-\frac{(\Lambda ^2-4-4z_3^2)^2}{192}\nonumber \\{} & {} -\frac{A_3(\xi _1+\xi _2+\xi _5+\xi _6)}{8}-\frac{(2A_2-1)(\xi _1+\xi _2+\xi _5+\xi _6)^2+(\xi _1^2+\xi _2^2+\xi _5^2+\xi _6^2)^2}{48} \nonumber \\{} & {} -\frac{\xi _1^4+\xi _2^4+\xi _5^4+\xi _6^4}{12} , \end{aligned}$$

(74c)

$$\begin{aligned} z_6= & {} \frac{(4z_3^2-\Lambda ^2)z_2}{6}-\frac{A_3}{2} ,\quad z_7=\frac{(\Lambda +1)(2z_3-1)}{4} -\frac{z_2^2}{3}\ , z_8= \frac{( 3\Lambda -2z_3 )z_2}{6} ,\quad \nonumber \\{} & {} z_9=-\frac{2z_2}{3} , \end{aligned}$$

(74d)

$$\begin{aligned} z_{10}= & {} \frac{(2z_3-\Lambda )(\Lambda +2-2z_3)}{4} ,\quad z_{11}=\frac{\Lambda }{2}-z_3 ,\quad z_{12}=z_3-\frac{\Lambda }{2}-1\, \end{aligned}$$

(74e)

where $A_2$ and $A_3$ are given by (3.2) with $\eta _i$ replaced by $\xi _i$.

C Root poset of $E_6$

The vertices of the graph are all the positive roots of $E_6$. The edges with label i indicate the action of the simple reflection $s_i$. The notation $1^{c_1}...6^{c_6}$ for roots is a short-hand notation for $\sum _{i=1,\dots ,6}c_i\alpha _i$.

In the top left subset of 15 roots, there are 9 roots between $\alpha _3$ and 12345 which contains 3. These nine roots are sent “as a block” to the right by $s_6$. Similarly the two blocks of underlined roots are exchanged by $s_3$. When the action of a simple reflection $s_i$ on a root is not indicated, it means that it leaves this root invariant with the exception of the action of $s_i$ on $\alpha _i$, which gives $s_i(\alpha _i)=-\alpha _i$.

With this diagram, it is straightforward to calculate the action of any word in the generators of $W(E_6)$ on any root.