Polology of Superconformal Blocks

Abstract

We systematically classify all possible poles of superconformal blocks as a function of the scaling dimension of intermediate operators, for all superconformal algebras in dimensions three and higher. This is done by working out the recently-proven irreducibility criterion for parabolic Verma modules for classical basic Lie superalgebras. The result applies to correlators for external operators of arbitrary spin, and indicates presence of infinitely many short multiplets of superconformal algebras, most of which are non-unitary. We find a set of poles whose positions are shifted by linear in $\mathcal {N}$ for $\mathcal {N}$-extended supersymmetry. We find an interesting subtlety for 3d $\mathcal {N}$-extended superconformal algebra with $\mathcal {N}$ odd associated with odd non-isotropic roots. We also comment on further applications to superconformal blocks.

Appendices

ABD Roots and Weights

In this Appendix we summarize some standard facts about the root system and the highest weights for the ABD Lie algebras. We can spare the C case (associated with Lie algebra $\mathfrak {sp}(2N)$) for the applications to SCFTs in this paper, since we can use the isomorphism of Lie algebras: $\mathfrak {sp}(2)=\mathfrak {su}(2)$, $\mathfrak {sp}(4)=\mathfrak {so}(5)$.

1.1 $A_{N-1}$

The $A_{N-1}$ root system is given by

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm (\delta _a - \delta _b) \bigg \} \ , \end{aligned}$$

(88)

where we introduced a orthonormal basis $\delta _a$

$$\begin{aligned} \begin{aligned} (\delta _a, \delta _b)=\delta _{a, b} \ , \quad a, b=1, \ldots , N. \end{aligned} \end{aligned}$$

(89)

Under an ordering $\delta _1> \ldots >\delta _{\mathcal {N}}$, the positive simple roots are given by $\alpha _a=\delta _a-\delta _{a+1}$, with $a=1, \ldots , N-1$.

The fundamental weight is given by

$$\begin{aligned} w_a:=\sum _{b=1}^{a} \delta _b -\frac{a}{N} \sum _{c=1}^{N} \delta _c , \quad \alpha _a:=\delta _a-\delta _{a+1}, \end{aligned}$$

(90)

which are determined by the conditions

$$\begin{aligned}&2\frac{ (w_a, \alpha _b)}{(\alpha _b, \alpha _b)}=\delta _{ab}, \quad a,b=1, \ldots , N, \quad \alpha _a:=\delta _a-\delta _{a+1}, \end{aligned}$$

(91)

as well as the condition that $w_a$ is in the $\mathbb {R}$-span of $\alpha _b$’s.

The highest weight for a finite-dimensional representation of the $\mathfrak {su}(N)$ algebra is given by a dominant integral weight, which is given by a $\mathbb {Z}_{\ge 0}$-span of fundamental weights:

$$\begin{aligned} \lambda =\sum _{a=1}^{N-1} \ell _a w_a =\sum _{a=1}^{N-1} \ell _a \left( \sum _{b=1}^{a} \delta _b -\frac{a}{N} \sum _{c=1}^{N} \delta _c \right) , \end{aligned}$$

(92)

where the coefficients $[\ell _1, \ldots , \ell _{\mathcal {N}}]$ are called Dynkin labels. The highest weight (92) can also be written as

$$\begin{aligned} \lambda =\sum _{a=1}^{N-1} \ell _a w_a =\sum _{a=1}^{N} \left( \lambda _a -\frac{|\lambda |}{N} \right) \delta _a , \end{aligned}$$

(93)

where we introduced $\lambda _a$ and $|\lambda |$ by

$$\begin{aligned}&\lambda _a=\ell _a+ \ell _{a+1} +\cdots +\lambda _{N-1}, \quad \text {or} \quad \ell _a=\lambda _a-\lambda _{a+1}, \end{aligned}$$

(94)

$$\begin{aligned}&|\lambda |=\sum _a \lambda _a . \end{aligned}$$

(95)

Since this $\lambda$ satisfies

$$\begin{aligned} \lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _{N-1} \ge 0, \end{aligned}$$

(96)

$\{ \lambda _a \}$ define a partition (Young diagram), where the number of boxes at height a is given by $\lambda _a$; in this language, $|\lambda |$ is the total number of boxes.

1.2 $D_{N}$

The $D_N$ root system (corresponding to the Lie algebra $\mathfrak {so}(2N)$) is given by

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \delta _a \pm \delta _b \bigg \} \ , \quad a,b=1, \ldots , N, \end{aligned}$$

(97)

and the inner product is given by (89). In dictionary ordering $\delta _1> \delta _2>\ldots >\delta _N$, the positive simple roots are given by

$$\begin{aligned} \alpha _a=\delta _a-\delta _{a+1} \quad a=1, \ldots , N-1, \quad \alpha _N=\delta _{N-1}+\delta _N. \end{aligned}$$

(98)

The fundamental weights satisfying (91) are obtained as

$$\begin{aligned} \begin{aligned} w_a&=\delta _1 + \cdots + \delta _a, \quad a=1, \ldots , N-2, \\ w_{N-1}&=\frac{\delta _1 + \cdots + \delta _{N-1}-\delta _N}{2},\\ w_{N}&=\frac{\delta _1 + \cdots + \delta _{N-1}+\delta _N}{2}.\\ \end{aligned} \end{aligned}$$

(99)

The highest weight for a finite-dimensional representation is given as

$$\begin{aligned} \begin{aligned} \lambda =\sum _{a=1}^N \ell _a w_a =\sum _{a=1}^N \lambda _a \delta _a \end{aligned} \end{aligned}$$

(100)

where $\ell _a\in \mathbb {Z}_{\ge 0}$ are integers called Dynkin labels, and we defined half-integers $\lambda _a\in \mathbb {Z}_{\ge 0}/2$ by

$$\begin{aligned} \begin{aligned}&\lambda _a = \ell _a+ \cdots \ell _{N-2}+\frac{\ell _{N-1}+\ell _N}{2} , \quad a=1, \ldots , N-2, \\&\lambda _{N-1}=\frac{\ell _{N-1}+\ell _N}{2}, \quad \lambda _{N}=\frac{-\ell _{N-1}+\ell _N}{2}. \end{aligned} \end{aligned}$$

(101)

By definition this satisfies $\lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _{N-1}\ge |\lambda _N| \ge 0$. Note that $\lambda _N$ can be negative. The $\lambda _a$’s are half-integers for a general representation. If we consider the tensor representation, however, all the $\lambda _a$’s takes values in integers; a tensor representation is labeled by a partition and a sign.

1.3 $B_{N}$

The $B_N$ root system (corresponding to the Lie algebra $\mathfrak {so}(2N+1)$) is similar to the $D_N$ root system, but with some extra roots added:

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \delta _a \pm \delta _b, \pm \delta _a \bigg \} , \quad a,b=1, \ldots , N \end{aligned}$$

(102)

and the inner product is given by (89). In dictionary ordering $\delta _1> \delta _2>\ldots >\delta _N$, the positive simple roots are given by

$$\begin{aligned} \alpha _a=\delta _a-\delta _{a+1} \quad a=1, \ldots , N-1, \quad \alpha _N=\delta _N. \end{aligned}$$

(103)

The fundamental weights satisfying (91) are obtained as

$$\begin{aligned} \begin{aligned} w_a&=\delta _1 + \cdots + \delta _a, \quad a=1, \ldots , N-1, \\ w_{N}&=\frac{\delta _1 + \cdots + \delta _{N-1}+\delta _N}{2}.\\ \end{aligned} \end{aligned}$$

(104)

The highest weight for a finite-dimensional representation is given as

$$\begin{aligned} \begin{aligned} \lambda =\sum _{a=1}^N \ell _a w_a =\sum _{a=1}^N \lambda _a \delta _a \end{aligned} \end{aligned}$$

(105)

where $\ell _a\in \mathbb {Z}_{\ge 0}$ are Dynkin labels and we defined half-integers $\lambda _a\in \mathbb {Z}_{\ge 0}/2$ by

$$\begin{aligned} \begin{aligned}&\lambda _a = \ell _a+ \cdots \ell _{N-2}+\frac{\ell _N}{2}, \quad a=1, \ldots , N-1, \\&\lambda _{N}=\frac{\ell _N}{2} . \end{aligned} \end{aligned}$$

(106)

By definition this satisfies $\lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _N\ge 0$. In tensor representations all the $\lambda _a$’s are integers; a tensor representation is labeled by a partition.

$\mathcal {N}=0$ Results

In this appendix we present the analysis for the non-supersymmetric cases (i.e. $\mathcal {N}=0$) in spacetime dimensions 4, 5 and 6. Note that in these cases the conformal algebra contains no odd elements, and hence $\Delta =\Delta _{\bar{0}}, \Delta _1=\Delta _{1}^{+}=\varnothing$ and $\rho =\rho _0$.

Note that such an analysis (for a general spacetime dimension) was already given in [45], except there the spinor representations are not considered there. In this appendix we therefore present the full analysis including the cases of spinor representations. We present this analysis in the same notations as in the rest of this paper, to make the comparison easier.²³ We will find that while intermediate steps of the analysis changes slightly from [45], the final result is in the end the same as in [45], even for spinor representations.

1.1 4d $\mathcal {N}=0$

The conformal algebra in this case is $\mathfrak {g}=\mathfrak {su}(4)=A(3)$. Let us introduce a basis

$$\begin{aligned} \begin{aligned} (\varepsilon _i, \varepsilon _j)=\delta _{i,j} \ , \quad \end{aligned} \end{aligned}$$

(107)

with $i,j=1, \ldots , 4$. Then we have

$$\begin{aligned}&\Delta =\bigg \{ \pm ( \varepsilon _i - \varepsilon _j ) \bigg \} \ , \end{aligned}$$

(108)

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm ( \varepsilon _1 - \varepsilon _2 ), \quad \pm ( \varepsilon _3 - \varepsilon _4) \bigg \}. \end{aligned}$$

(109)

Let us choose an ordering $\varepsilon _1> \varepsilon _2>-\varepsilon _3>-\varepsilon _4$. We then have

$$\begin{aligned} \Delta ^+&=\bigg \{ \varepsilon _{1}-\varepsilon _{2} , \quad \varepsilon _{4}-\varepsilon _{3} \bigg \}, \end{aligned}$$

(110)

$$\begin{aligned} \Delta _{\mathfrak {n}}&=\bigg \{ \varepsilon _1 - \varepsilon _3 , \quad \varepsilon _1 - \varepsilon _4 , \quad \varepsilon _2 - \varepsilon _3 , \quad \varepsilon _2 - \varepsilon _4 \bigg \}. \end{aligned}$$

(111)

The highest weight vector and the Weyl vector is

$$\begin{aligned} \begin{aligned} \lambda&=-\frac{\Delta }{2}(\varepsilon _1+\varepsilon _2-\varepsilon _3-\varepsilon _4)+\frac{\ell _1}{2}(\varepsilon _1-\varepsilon _2)-\frac{\ell _2}{2}(\varepsilon _3-\varepsilon _4) \\ \rho&=\frac{1}{2} \left( 3 \varepsilon _1 +\varepsilon _2 -3\varepsilon _3 -\varepsilon _4 \right) ,\\ \lambda +\rho&=\frac{-\Delta +3}{2} (\varepsilon _1 -\varepsilon _3 ) +\frac{-\Delta +1}{2}(\varepsilon _2 -\varepsilon _4) +\frac{\ell _1}{2}(\varepsilon _1-\varepsilon _2)-\frac{\ell _2}{2}(\varepsilon _3-\varepsilon _4) \,. \end{aligned} \end{aligned}$$

(112)

Here $\ell _1$ and $\ell _2$ are the two angular spins, and take integer values.

We find

$$\begin{aligned} \begin{aligned}&n_{\varepsilon _1-\varepsilon _3}=3-\Delta +\frac{\ell _1+\ell _2}{2}, \ \ n_{\varepsilon _1-\varepsilon _4}=2-\Delta +\frac{\ell _1-\ell _2}{2}, \\&n_{\varepsilon _2-\varepsilon _3}=2-\Delta +\frac{\ell _2-\ell _1}{2}, \ \ n_{\varepsilon _2-\varepsilon _4}=1-\Delta -\frac{\ell _1+\ell _2}{2}\, \end{aligned} \end{aligned}$$

(113)

and $\Psi _\lambda ^+\ne \varnothing$ in the following cases:

$$\begin{aligned} \Delta -C_{\ell _1+\ell _2} = {\left\{ \begin{array}{ll} 3+\frac{\ell _1+\ell _2}{2}-k, \ k=\{1,\ldots \infty \},\ \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3\}\\ 2-\frac{\ell _1-\ell _2}{2}-k,\ k=\{1,2,\ldots ,1+\ell _2\},\ \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4,\varepsilon _2-\varepsilon _3\},\\ 2+\frac{\ell _1-\ell _2}{2}-k, \ k=\{1,2,\ldots \ell _1-\ell _2\}, \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4\},\\ 1-\frac{\ell _1+\ell _2}{2}-k, \ k=\{1,2,\ldots \}, \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4,\varepsilon _2-\varepsilon _3,\varepsilon _2-\varepsilon _4\},\\ \end{array}\right. } \end{aligned}$$

(114)

for $\ell _1\ge \ell _2$, and

$$\begin{aligned} \Delta -C_{\ell _1+\ell _2} ={\left\{ \begin{array}{ll}1-\frac{\ell _1+\ell _2}{2}-k, \ k=\{1,2,\ldots \}, \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4,\varepsilon _2-\varepsilon _3,\varepsilon _2-\varepsilon _4\},\\ 2+\frac{\ell _2-\ell _1}{2}-k, \ k=\{1,2,\ldots ,\ell _2-\ell _1\}, \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _2-\varepsilon _3\},\\ 2-\frac{\ell _2-\ell _1}{2}-k,\ k=\{1,2,\ldots ,1+\ell _1\} ,\ \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4,\varepsilon _2-\varepsilon _3\},\\ 3+\frac{\ell _1+\ell _2}{2}-k, \ k=\{1,\ldots \infty \},\ \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3\} \end{array}\right. } \end{aligned}$$

(115)

for $\ell _1< \ell _2$. Here we defined

$$\begin{aligned} C_{\ell _1+\ell _2}= {\left\{ \begin{array}{ll} 0 &{} (\ell _1+\ell _2\text { even})\\ \frac{1}{2} &{} (\ell _1+\ell _2 \text { odd}). \end{array}\right. } \end{aligned}$$

(116)

The final step is to check whether there exists $\beta \in \Delta _\mathfrak {n}$ such that $(\lambda +\rho ,\beta )=0$. This happens for

$$\begin{aligned} \begin{aligned}&\Delta =3+\frac{\ell _1+\ell _2}{2}, \ n_{\varepsilon _{1}-\varepsilon _{3}}=0, \ \Psi _\lambda ^+=\varnothing \,.\\&\Delta =2+\frac{\ell _1-\ell _2}{2}, \ n_{\varepsilon _{1}-\varepsilon _{4}}=0, \ \Psi _\lambda ^+={\left\{ \begin{array}{ll}\{\varepsilon _1-\varepsilon _3,\varepsilon _2-\varepsilon _3\},\ell _2>\ell _1\\ \{\varepsilon _1-\varepsilon _3\},\ell _2<\ell _1\end{array}\right. }\,.\\&\Delta =2+\frac{\ell _2-\ell _1}{2}, \ n_{\varepsilon _{2}-\varepsilon _{3}}=0,\ \Psi _\lambda ^+={\left\{ \begin{array}{ll}\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4\},\ell _1>\ell _2\\ \{\varepsilon _1-\varepsilon _3\},\ell _1<\ell _2\end{array}\right. }\,.\\&\Delta =1-\frac{\ell _1+\ell _2}{2},\ n_{\varepsilon _{2}-\varepsilon _{4}}=0, \ \Psi _\lambda ^+=\{\varepsilon _1-\varepsilon _3,\varepsilon _1-\varepsilon _4,\varepsilon _2-\varepsilon _3\}\,. \end{aligned} \end{aligned}$$

(117)

The first line is trivial since $\Psi _\lambda ^+=\varnothing$. For the remaining cases, we need to apply the final condition (25). In most cases the condition (25) does not hold. The first exception is the obvious case of $\Delta =3+\frac{\ell _1+\ell _2}{2}$, when $\Psi _\lambda ^+$ is empy. The other exception, assuming $\ell _1>\ell _2$, happens for $\Delta =2+\frac{\ell _1-\ell _2}{2}$. In this case $\lambda +\rho$ is in the hyperplane orthogonal to $\varepsilon _1 -\varepsilon _4$, and hence $s_{\varepsilon _1-\varepsilon _3}(\lambda +\rho )$ is orthogonal to $\varepsilon _3 -\varepsilon _4$. This means that $s_{\varepsilon _1-\varepsilon _3}(\lambda +\rho )$ is fixed by an element of the Weyl group $W_{\mathfrak {l}}$ exchanging $\varepsilon _3$ and $\varepsilon _4$. The case of $\ell _1<\ell _2$ is similar, and we find the reducible points in four dimensions comprises of,

$$\begin{aligned} \Delta =\bigg \{3+\frac{\ell _1+\ell _2}{2}-\mathbb {Z}_{>0}\bigg \}\bigg \backslash \bigg \{2+\frac{|\ell _1-\ell _2|}{2}\bigg \}\, . \end{aligned}$$

(118)

In terms of partitions $\lambda _1, \lambda _2\in \mathbb {Z}$ with $\lambda _1\ge |\lambda _2|\ge 0$ (see Appendix A), this becomes

$$\begin{aligned} \Delta =\bigg \{\frac{ \lambda _1}{2}+3-\mathbb {Z}_{>0}\bigg \} \setminus \bigg \{ \frac{|\lambda _2|}{2}+2 \bigg \}. \end{aligned}$$

(119)

1.2 5d $\mathcal {N}=0$

For this case, we use the root system for $\mathfrak {so}_7=B(3)$:

$$\begin{aligned}&\Delta =\bigg \{ \pm \beta _D \pm \beta _{J_1}, \quad \pm \beta _D \pm \beta _{J_2}, \quad \pm \beta _{J_1} \pm \beta _{J_2}, \quad \pm \beta _D , \quad \pm \beta _{J_1}, \quad \pm \beta _{J_2} \bigg \} \ , \end{aligned}$$

(120)

with the inner product $(\beta _i, \beta _j)=\delta _{i,j}$ Under an ordering $\beta _D> \beta _{J_1}>\beta _{J_2}$,

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \beta _{J_1} \pm \beta _{J_2}, \quad \pm \beta _{J_1}, \quad \pm \beta _{J_2} \bigg \}, \end{aligned}$$

(121)

$$\begin{aligned}&\Delta _{\mathfrak {n}}=\bigg \{ \beta _D \pm \beta _{J_1}, \quad \beta _D \pm \beta _{J_2}, \quad \beta _D \bigg \}. \end{aligned}$$

(122)

We have

$$\begin{aligned}&\lambda =-\Delta \beta _D + \left( \ell _1 +\frac{\ell _2}{2}\right) \beta _{J_1}+\frac{\ell _2}{2} \beta _{J_2} + k\delta , \end{aligned}$$

(123)

$$\begin{aligned}&\rho =\frac{5}{2}\beta _D+\frac{3}{2}\beta _{J_1}+\frac{1}{2}\beta _{J_2}+\frac{1}{2}\delta , \end{aligned}$$

(124)

$$\begin{aligned}&\lambda +\rho =\left( -\Delta +\frac{5}{2}\right) \beta _D+\left( \ell _1+\frac{\ell _2}{2}+\frac{3}{2}\right) \beta _{J_1}+\left( \frac{\ell _2}{2}+\frac{1}{2}\right) \beta _{J_2}+ \left( k+\frac{1}{2} \right) \delta . \end{aligned}$$

(125)

In Step 1 we have

$$\begin{aligned} \begin{aligned}&n_{\beta _D}=5-2\Delta ,\\&n_{\beta _D+\beta _{J_1}}=4-\Delta +\ell _1+\frac{\ell _2}{2}, \\&n_{\beta _D-\beta _{J_1}}=1-\Delta -\ell _1-\frac{\ell _2}{2},\\&n_{\beta _D+\beta _{J_2}}=3-\Delta +\frac{\ell _2}{2},\\&n_{\beta _D-\beta _{J_2}}=2-\Delta -\frac{\ell _2}{2}\,. \end{aligned} \end{aligned}$$

(126)

This means that we have $\Psi _\lambda ^+\ne \varnothing$ when

$$\begin{aligned} \ell _2 \text { even}, \ \ \Delta ={\left\{ \begin{array}{ll} 1-\ell _1-\frac{\ell _2}{2}-k, \ \Psi _\lambda ^+=\{\beta _D,\beta _D\pm \beta _{J_1},\beta _D\pm \beta _{J_2}\}\\ 2-\frac{\ell _2}{2}-k, \ k=\{1,\ldots , 1+\ell _1\}\ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D\pm \beta _{J_2}\}\\ \frac{5}{2}-k, \ k=\{\frac{3}{2},\ldots ,\frac{\ell _2+1}{2}\}\ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\\ 3+\frac{\ell _2}{2}-k, \ k=\{1,\ldots ,\frac{1+\ell _2}{2}\}\ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\\ 4+\ell _1+\frac{\ell _2}{2}-k, \ k=\{1,\ldots ,1+\ell _1\}\ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1}\} \\ \frac{5}{2}-k, \ k=\{1,\ldots ,\frac{\ell _2}{2}\}\ \Psi _\lambda ^+=\{\beta _D \}\\ \end{array}\right. } \end{aligned}$$

(127)

and,

$$\begin{aligned} \ell _2\text { odd}, \ \ \Delta ={\left\{ \begin{array}{ll} 1-\ell _1-\frac{\ell _2}{2}-k, \ \Psi _\lambda ^+=\{\beta _D,\beta _D\pm \beta _{J_1},\beta _D\pm \beta _{J_2}\}\\ 2-\frac{\ell _2}{2}-k, \ k=\{\frac{1}{2},\ldots , \ell _1+\frac{1}{2}\}\ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D\pm \beta _{J_2}\}\\ \frac{5}{2}-k, \ k=\{\frac{1}{2},\ldots ,\frac{\ell _2}{2}\}\ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\\ 3+\frac{\ell _2}{2}-k, \ k=\{\frac{1}{2},\ldots ,\frac{\ell _2}{2}\}\ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\\ 4+\ell _1+\frac{\ell _2}{2}-k, \ k=\{\frac{1}{2},\ldots ,\ell _1+\frac{1}{2}\}\ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1}\}. \end{array}\right. } \end{aligned}$$

(128)

The next step is to identify the walls for which there exists $\beta \in \Delta _\mathfrak {n}$ such that $(\lambda +\rho ,\beta )=0$,

$$\begin{aligned}&n_{\beta _D}=0, \ \Delta =\frac{5}{2}, \ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\nonumber \\&n_{\beta _D+\beta _{J_1}}=0, \ \Delta =4+\ell _1+\frac{\ell _2}{2}, \ \Psi _\lambda ^+=\varnothing \nonumber \\&n_{\beta _D-\beta _{J_1}}=0, \ \Delta =1-\ell _1-\frac{\ell _2}{2}, \ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D\pm \beta _{J_2}\}\\&n_{\beta _D+\beta _{J_2}}=0,\ \Delta =3+\frac{\ell _2}{2}, \ \Psi _\lambda ^+=\{\beta _D+\beta _{J_1}\}\nonumber \\&n_{\beta _D-\beta _{J_2}}=0, \ \Delta =2-\frac{\ell _2}{2}, \ \Psi _\lambda ^+=\{\beta _D,\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}\}\,.\nonumber \end{aligned}$$

(129)

Finally we need to check the condition (25) for each of the above cases. In most cases the condition is not satisfied and the representation is reducible. The exception in the case $n_{\beta _D}=0$, when $\Psi _\lambda ^+$ has two elements $\beta _D+\beta _{J_1},\beta _D+\beta _{J_2}$, and $s_{\beta _D+\beta _{J_1}}(\lambda +\rho )$ and $s_{\beta _D+\beta _{J_2}}(\lambda +\rho )$ are each in the hyperplane orthogonal to $\beta _{J_1}$ and $\beta _{J_2}$. Since these two vectors can be rotated by an element of $W_{\mathfrak {l}}$ (rotation symmetry) we have an irreducible representation. Hence we get the following set of reducible points,

$$\begin{aligned} \Delta =\bigg (\bigg \{4+\ell _1+\frac{\ell _2}{2}-\mathbb {Z}_{>0}\bigg \}\bigg \backslash \bigg \{3+\frac{\ell _2}{2},2-\frac{\ell _2}{2},1-\ell _1-\frac{\ell _2}{2}\bigg \}\bigg )\bigcup \bigg (\frac{5}{2}-\mathbb {Z}_{>0}\bigg )\,. \end{aligned}$$

(130)

In terms of $\lambda _1 \ge \lambda _2 \ge 0$ this can be written as

$$\begin{aligned} \begin{aligned}&\Delta =\left( \bigg \{ \lambda _1+4-\mathbb {Z}_{>0}\bigg \} \setminus \bigg \{\lambda _2+3, -\lambda _2+2 , -\lambda _1+1\bigg \}\right) \bigcup \left( \frac{5}{2}-\mathbb {Z}_{>0}\right) . \end{aligned} \end{aligned}$$

(131)

1.3 6d $\mathcal {N}=0$

The root system for $\mathfrak {g}=\mathfrak {so}(8)=D(4)$ is

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \alpha _i \pm \alpha _j, \bigg \} \quad (i,j=D,1,2,3)\; . \end{aligned}$$

(132)

Under an ordering $\alpha _D>\alpha _1>\alpha _2>\alpha _3$,

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \alpha _1 \pm \alpha _2, \quad \pm \alpha _1 \pm \alpha _3, \quad \pm \alpha _2 \pm \alpha _3 \bigg \}, \end{aligned}$$

(133)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \alpha _D\pm \alpha _{1}, \quad \alpha _D\pm \alpha _{2}, \quad \alpha _D\pm \alpha _{3} \bigg \}\; \end{aligned}$$

(134)

with the constraint $s_1 s_2 s_3=1$.

$$\begin{aligned}&\lambda = -\Delta \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2} \right) \alpha _1+\frac{\ell _2+\ell _3}{2}\alpha _2+\frac{-\ell _2+\ell _3}{2} \alpha _3+k \beta , \end{aligned}$$

(135)

$$\begin{aligned}&\rho =3\alpha _D+2 \alpha _1+\alpha _2+\frac{1}{2}\beta , \end{aligned}$$

(136)

$$\begin{aligned}&\lambda +\rho =( -\Delta +3) \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2} +2\right) \alpha _1+\left( \frac{\ell _2+\ell _3}{2} +1\right) \alpha _2\nonumber \\&\qquad \qquad + \frac{-\ell _2+\ell _3}{2} \alpha _3+\left( k+\frac{1}{2}\right) \beta . \end{aligned}$$

(137)

We compute

$$\begin{aligned} \begin{aligned}&n_{\alpha _D+\alpha _1}=5-\Delta +\ell _1+\frac{\ell _2+\ell _3}{2},\\&n_{\alpha _D-\alpha _1}=1-\Delta -\ell _1-\frac{\ell _2+\ell _3}{2},\\&n_{\alpha _D+\alpha _2}=4-\Delta +\frac{\ell _2+\ell _3}{2},\\&n_{\alpha _D-\alpha _2}=2-\Delta -\frac{\ell _2+\ell _3}{2},\\&n_{\alpha _D+\alpha _3}=3-\Delta -\frac{\ell _2-\ell _3}{2},\\&n_{\alpha _D-\alpha _3}=3-\Delta +\frac{\ell _2-\ell _3}{2}\,. \end{aligned} \end{aligned}$$

(138)

This means that if

$$\begin{aligned} \Delta =5+\ell _1+\frac{\ell _2+\ell _3}{2}-\mathbb {Z}_{>0}, \end{aligned}$$

(139)

then we will have $\Psi _\lambda ^+\ne \varnothing$ with varying elements depending on what $n_\beta$ are non-zero. We will directly go to the next step which is to analyze the wall condition for $\beta \in \Delta _\mathfrak {n}$ such that $(\lambda +\rho ,\beta )=0$:

$$\begin{aligned} \begin{aligned}&n_{\alpha _D+\alpha _1}=0,\ \Delta =5+\ell _1+\frac{\ell _2+\ell _3}{2}, \ \Psi _\lambda ^+=\varnothing ,\\&n_{\alpha _D-\alpha _1}=0,\ \Delta =1-\ell _1-\frac{\ell _2+\ell _3}{2}, \ \Psi _\lambda ^+=\{\alpha _D+\alpha _1,\alpha _D\pm \alpha _2,\alpha _D\pm \alpha _3\},\\&n_{\alpha _D+\alpha _2}=0,\ \Delta =4+\frac{\ell _2+\ell _3}{2}, \ \Psi _\lambda ^+=\{\alpha _D+\alpha _1\},\\&n_{\alpha _D-\alpha _2}=0,\ \Delta =2-\frac{\ell _2+\ell _3}{2}, \ \Psi _\lambda ^+=\{\alpha _D+\alpha _1,\alpha _D+\alpha _2,\alpha _D\pm \alpha _3\},\\&n_{\alpha _D+\alpha _3}=0,\ \Delta =3-\frac{\ell _2-\ell _3}{2},\ \Psi _\lambda ^+={\left\{ \begin{array}{ll}\{\alpha _D+\alpha _1,\alpha _D+\alpha _2,\alpha _D-\alpha _3\}, \ \ell _2>\ell _3\\ \{\alpha _D+\alpha _1,\alpha _D+\alpha _2\}, \ \ell _2<\ell _3\end{array}\right. },\\&n_{\alpha _D-\alpha _3}=0,\ \Delta =3+\frac{\ell _2-\ell _3}{2},\ \Psi _\lambda ^+={\left\{ \begin{array}{ll}\{\alpha _D+\alpha _1,\alpha _D+\alpha _2\}, \ \ell _2>\ell _3\\ \{\alpha _D+\alpha _1,\alpha _D+\alpha _2,\alpha _D+\alpha _3\},\ \ell _2<\ell _3\end{array}\right. }\,. \end{aligned} \end{aligned}$$

(140)

After working out the condition (25) we find that the reducible points are

$$\begin{aligned} \Delta =\bigg (\bigg \{5+\ell _1+\frac{\ell _2+\ell _3}{2}-\mathbb {Z}_{>0}\bigg \}\bigg \backslash \bigg \{4+\frac{\ell _2+\ell _3}{2},3+\frac{|\ell _2-\ell _3|}{2} \bigg \}\bigg )\,. \end{aligned}$$

(141)

In the language of the parametrization $\lambda _1\ge \lambda _2\ge |\lambda _3|\ge 0$ in Appendix A this becomes

$$\begin{aligned} \Delta =\bigg \{ \lambda _1+5-\mathbb {Z}_{>0}\bigg \} \setminus \bigg \{\lambda _2+4, |\lambda _3|+3 \bigg \}. \end{aligned}$$

(142)

Root System for SCAs

1.1 3d $\mathcal {N}$ Even

The root system for $\mathfrak {g}=D\left( \frac{\mathcal {N}}{2}, 2\right)$ is given by

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \beta _D , \quad \pm \beta _J, \quad \pm \beta _D \pm \beta _J , \quad \pm \delta _i \pm \delta _j \bigg \} \ , \end{aligned}$$

(143)

$$\begin{aligned}&\Delta _{\bar{1}}=\bigg \{ \pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J \pm \delta _i \bigg \} \ , \end{aligned}$$

(144)

with $i=1, \ldots , \frac{\mathcal {N}}{2}$ and the inner product is given by

$$\begin{aligned} (\beta _a, \beta _b)=2\delta _{ab}, \quad (\delta _i, \beta _a)=0, \quad (\delta _i, \delta _j)=-\delta _{ij}, \end{aligned}$$

(145)

with $a,b=D,J$. In dictionary ordering $\beta _D> \beta _J> \delta _1> \delta _2>\ldots >\delta _{\frac{\mathcal {N}}{2}}$, we find

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \beta _J, \quad \pm \delta _i \pm \delta _j \bigg \}, \end{aligned}$$

(146)

$$\begin{aligned}&\Delta _{1}^{+}=\overline{\Delta }_{1}^{+}=\bigg \{ \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J\pm \delta _i \bigg \}, \end{aligned}$$

(147)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \beta _D+\beta _J, \quad \beta _D, \quad \beta _D-\beta _J \bigg \}. \end{aligned}$$

(148)

1.2 3d $\mathcal {N}$ Odd

We have $\mathfrak {g}=B\left( \frac{\mathcal {N}-1}{2}, 2\right)$. We have the root system

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \beta _D , \quad \pm \beta _J, \quad \pm \beta _D \pm \beta _J,\quad \pm \delta _i \pm \delta _j, \quad \pm \delta _i \bigg \} \ , \end{aligned}$$

(149)

$$\begin{aligned}&\Delta _{\bar{1}}=\bigg \{ \pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J \pm \delta _i, \quad \pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J \bigg \} \ , \end{aligned}$$

(150)

with $i=1,\ldots , \frac{\mathcal {N}-1}{2}$. For $\mathcal {N}=1$ the vectors $\delta _i$’s are absent, and correspondingly we disregard those roots containing these vectors. The inner product is given by (145), and with dictionary ordering $\beta _D> \beta _J> \delta _1> \delta _2>\cdots >\delta _{\frac{\mathcal {N}-1}{2}}$ we obtain

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \beta _J, \quad \pm \delta _i \pm \delta _j, \quad \pm \delta _i, \bigg \}, \end{aligned}$$

(151)

$$\begin{aligned}&\overline{\Delta }_{1}^{+}=\bigg \{\pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J \pm \delta _i \bigg \}, \end{aligned}$$

(152)

$$\begin{aligned}&\Delta _{1}^{+}\setminus \overline{\Delta }_{1}^{+}=\bigg \{ \pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _J \bigg \}, \end{aligned}$$

(153)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \beta _D+\beta _J, \quad \beta _D, \quad \beta _D-\beta _J \bigg \}. \end{aligned}$$

(154)

1.3 4d $\mathcal {N}\ge 1$

For $\mathfrak {g}=\mathfrak {su}(4|\mathcal {N})$ ($\mathfrak {g}=\mathfrak {psu}(4|4)$ for $\mathcal {N}=4$), it is useful to introduce a basis

$$\begin{aligned} \begin{aligned} (\varepsilon _i, \varepsilon _j)=\delta _{i,j} \ , \quad (\varepsilon _i, \delta _b)=0 \ , \quad (\delta _a, \delta _b)=-\delta _{a, b} \ , \end{aligned} \end{aligned}$$

(155)

with $i,j=1, \ldots , 4$ and $a, b=1, \ldots , \mathcal {N}$. Then we have

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm ( \varepsilon _i - \varepsilon _j ), \quad \pm (\delta _a - \delta _b) \bigg \} \ , \quad \Delta _{\bar{1}}=\bigg \{ \pm ( \varepsilon _i + \delta _a) \bigg \}, \end{aligned}$$

(156)

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm ( \varepsilon _1 - \varepsilon _2 ), \quad \pm ( \varepsilon _3 - \varepsilon _4), \quad \pm (\delta _a - \delta _b) \bigg \}. \end{aligned}$$

(157)

Following [19, 41] we choose an ordering

$$\begin{aligned} \varepsilon _1> \varepsilon _2>-\varepsilon _3>-\varepsilon _4>-\delta _1> \cdots >-\delta _{\mathcal {N}}. \end{aligned}$$

(158)

We have positive simple roots

$$\begin{aligned}&\bigg \{ \varepsilon _{1}-\varepsilon _{2} , \quad \varepsilon _{4}-\varepsilon _{3} , \quad \varepsilon _{2}+\delta _{1} , \quad -\varepsilon _{4}-\delta _{\mathcal {N}}, \quad -\delta _{a}+\delta _{a+1} \quad (a=1, \ldots , \mathcal {N}-1) \bigg \}. \end{aligned}$$

(159)

We have

$$\begin{aligned}&\Delta _{1}^{+}=\overline{\Delta }_{1}^{+}=\bigg \{ \varepsilon _1 - \delta _a , \varepsilon _2 - \delta _a , -\varepsilon _3 + \delta _a , -\varepsilon _4 + \delta _a \bigg \}, \end{aligned}$$

(160)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \varepsilon _1 - \varepsilon _3 , \quad \varepsilon _1 - \varepsilon _4 , \quad \varepsilon _2 - \varepsilon _3 , \quad \varepsilon _2 - \varepsilon _4 \bigg \}. \end{aligned}$$

(161)

The highest weight vector is

$$\begin{aligned} \begin{aligned} \lambda&=-\frac{\Delta }{2}(\varepsilon _1+\varepsilon _2-\varepsilon _3-\varepsilon _4)+\frac{\ell _1}{2}(\varepsilon _1-\varepsilon _2)-\frac{\ell _2}{2}(\varepsilon _3-\varepsilon _4)\\&\quad + \frac{\mathcal {N}-4}{8 \mathcal {N}} R \left( \sum _{i=1}^4 \varepsilon _i- \sum _{a=1}^{\mathcal {N}} \delta _a \right) -\sum _{b=1}^{\mathcal {N}} \left( \lambda _b -\frac{|\lambda |}{\mathcal {N}} \right) \delta _b \,. \end{aligned} \end{aligned}$$

(162)

Here $\ell _1$ and $\ell _2$ are the two angular spins, and take integer values. The set of integers $\{\lambda _a\}$ define a partition, see Appendix A.1. Note the factor with R is absent for the special case of $\mathcal {N}=4$, where there is a reduction of $\mathfrak {u}(1)$ symmetry, from $\mathfrak {sl}(4|4)$ into $\mathfrak {psl}(4|4)$.

The Weyl vector is given by

$$\begin{aligned}&\rho _0=\frac{1}{2} \left( 3 \varepsilon _1 +\varepsilon _2 -3\varepsilon _3 -\varepsilon _4 \right) -\sum _{a=1}^{\mathcal {N}}\frac{\mathcal {N}+1-2a}{2}\delta _a, \end{aligned}$$

(163)

$$\begin{aligned}&\rho _1=\frac{\mathcal {N}}{2}\left( \varepsilon _1+ \varepsilon _2 - \varepsilon _3- \varepsilon _4 \right) , \end{aligned}$$

(164)

$$\begin{aligned}&\rho = \left( \frac{3-\mathcal {N}}{2} \varepsilon _1 +\frac{1-\mathcal {N}}{2}\varepsilon _2 -\frac{3-\mathcal {N}}{2}\varepsilon _3 -\frac{1-\mathcal {N}}{2}\varepsilon _4 \right) -\sum _{a=1}^{\mathcal {N}} \frac{\mathcal {N}+1-2a}{2} \delta _a\,. \end{aligned}$$

(165)

We therefore obtain

$$\begin{aligned} \begin{aligned} \lambda +\rho&=\frac{-\Delta +3-\mathcal {N}}{2} \varepsilon _1 +\frac{-\Delta +1-\mathcal {N}}{2}\varepsilon _2 -\frac{-\Delta +3-\mathcal {N}}{2}\varepsilon _3 -\frac{-\Delta +1-\mathcal {N}}{2}\varepsilon _4 \\&\quad +\frac{\ell _1}{2}(\varepsilon _1-\varepsilon _2)-\frac{\ell _2}{2}(\varepsilon _3-\varepsilon _4)\\&\quad + \frac{\mathcal {N}-4}{8 \mathcal {N}} R \left( \sum _{i=1}^4 \varepsilon _i- \sum _{a=1}^{\mathcal {N}} \delta _a \right) -\sum _{a=1}^{\mathcal {N}} \frac{2\lambda _b -2\frac{|\lambda |}{\mathcal {N}} +\mathcal {N}+1-2a}{2} \delta _a\,. \end{aligned} \end{aligned}$$

(166)

For Step 1, we compute

$$\begin{aligned} \begin{aligned} (\lambda +\rho , \varepsilon _1)&=\frac{1}{2} \left( -\Delta +\ell _1+3-\mathcal {N}+\frac{\mathcal {N}-4}{4\mathcal {N}} R\right) , \\ (\lambda +\rho , \varepsilon _2)&=\frac{1}{2} \left( -\Delta -\ell _1+1-\mathcal {N}+\frac{\mathcal {N}-4}{4\mathcal {N}} R\right) , \\ (\lambda +\rho , \varepsilon _3)&=\frac{1}{2} \left( \Delta -\ell _2-3+\mathcal {N}+\frac{\mathcal {N}-4}{4\mathcal {N}} R\right) , \\ (\lambda +\rho , \varepsilon _4)&=\frac{1}{2} \left( \Delta +\ell _2-1+\mathcal {N}+\frac{\mathcal {N}-4}{4\mathcal {N}} R\right) , \\ (\lambda +\rho , \delta _a)&=\frac{1}{2} \left( \frac{\mathcal {N}-4}{4\mathcal {N}}R+2\lambda _a -2\frac{|\lambda |}{\mathcal {N}} +\mathcal {N}+1-2a \right) . \end{aligned} \end{aligned}$$

(167)

From this we can easily see that the module is reducible at²⁴

$$\begin{aligned} \Delta = {\left\{ \begin{array}{ll} \ell _1+\frac{\mathcal {N}-4}{2\mathcal {N}} R +2 \lambda _a-2\frac{|\lambda |}{\mathcal {N}} -2a +4&{} (\varepsilon _1-\delta _a),\\ -\ell _1+\frac{\mathcal {N}-4}{2\mathcal {N}}R +2 \lambda _a-2\frac{|\lambda |}{\mathcal {N}} -2a+2&{}(\varepsilon _2-\delta _a),\\ -\ell _2-\frac{\mathcal {N}-4}{2\mathcal {N}} R -2 \lambda _a+2\frac{|\lambda |}{\mathcal {N}} +2a+2-2\mathcal {N}&{} (\varepsilon _3-\delta _a),\\ \ell _2-\frac{\mathcal {N}-4}{2\mathcal {N}} R -2 \lambda _a+2\frac{|\lambda |}{\mathcal {N}} +2a-2\mathcal {N}&{}(\varepsilon _4-\delta _a).\\ \end{array}\right. } \end{aligned}$$

(168)

We next come to Step 2$^{\prime }$.

A care is needed in this step since in the expression for $\lambda +\rho$ in (166) the coefficients of the $\varepsilon _i$ do depend non-trivially on the R-charge R. However, such a R-dependence drops out when we consider irreducibility in this step, since $\sum _i \varepsilon _i -\sum _a \delta _a$ is orthogonal to all the roots corresponding to momentum generators. Indeed, we can compute

$$\begin{aligned} \begin{aligned}&n_{\varepsilon _1-\varepsilon _3}= -\Delta +\frac{\ell _1-\ell _2}{2}+3 -\mathcal {N},\quad n_{\varepsilon _1-\varepsilon _4}= -\Delta +\frac{\ell _1+\ell _2}{2}+2 -\mathcal {N},\\&n_{\varepsilon _2-\varepsilon _3}= -\Delta +\frac{-\ell _1-\ell _2}{2}+2 -\mathcal {N},\quad n_{\varepsilon _2-\varepsilon _4}= -\Delta +\frac{-\ell _1+\ell _2}{2}+1 -\mathcal {N}\,. \end{aligned} \end{aligned}$$

(169)

From this we can work out when the set $\Psi _{\lambda , \mathrm{non-iso}}$ is non-empty. We again learn that only the effect of $\mathcal {N}$ in the rest of the analysis is to shift $\Delta \rightarrow \Delta +\mathcal {N}$.

The result is then obtained by combining (168) and (118), where $\Delta$ in the latter is shifted by $\mathcal {N}$.

1.4 5d $\mathcal {N}=1$

For this case, we use the root system for $\mathfrak {f}_4$:

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \beta _D \pm \beta _{J_1}, \quad \pm \beta _D \pm \beta _{J_2}, \quad \pm \beta _{J_1} \pm \beta _{J_2}, \quad \pm \beta _D , \quad \pm \beta _{J_1}, \quad \pm \beta _{J_2}, \quad \pm \delta \bigg \} \ , \end{aligned}$$

(170)

$$\begin{aligned}&\Delta _{\bar{1}}=\bigg \{ \pm \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _{J_1} \pm \frac{1}{2}\beta _{J_2}\pm \frac{1}{2}\delta \bigg \} \ , \end{aligned}$$

(171)

with the inner product

$$\begin{aligned} \begin{aligned} (\beta _i, \beta _j)=\delta _{i,j}, \quad (\beta _i, \delta )=0, \quad (\delta , \delta )=-3. \end{aligned} \end{aligned}$$

(172)

Under an ordering $\beta _D> \beta _{J_1}>\beta _{J_2}>\delta$,

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \beta _{J_1} \pm \beta _{J_2}, \quad \pm \beta _{J_1}, \quad \pm \beta _{J_2}, \quad \pm \delta \bigg \}, \end{aligned}$$

(173)

$$\begin{aligned}&\Delta _{1}^{+}=\overline{\Delta }_{1}^{+}=\bigg \{ \frac{1}{2}\beta _D \pm \frac{1}{2}\beta _{J_1} \pm \frac{1}{2}\beta _{J_2} \pm \frac{1}{2}\delta \bigg \}, \end{aligned}$$

(174)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \beta _D \pm \beta _{J_1}, \quad \beta _D \pm \beta _{J_2}, \quad \beta _D \bigg \}. \end{aligned}$$

(175)

We have

$$\begin{aligned}&\lambda =-\Delta \beta _D + \left( \ell _1 +\frac{\ell _2}{2}\right) \beta _{J_1}+\frac{\ell _2}{2} \beta _{J_2} + k\delta , \end{aligned}$$

(176)

$$\begin{aligned}&\rho _0=\frac{5}{2}\beta _D+\frac{3}{2}\beta _{J_1}+\frac{1}{2}\beta _{J_2}+\frac{1}{2}\delta ,\quad \rho _1=2 \beta _D, \end{aligned}$$

(177)

$$\begin{aligned}&\rho =\frac{1}{2}\beta _D+\frac{3}{2}\beta _{J_1}+\frac{1}{2}\beta _{J_2}+\frac{1}{2}\delta , \end{aligned}$$

(178)

$$\begin{aligned}&\lambda +\rho =\left( -\Delta +\frac{1}{2}\right) \beta _D+\left( \ell _1+\frac{\ell _2}{2}+\frac{3}{2}\right) \beta _{J_1}+\left( \frac{\ell _2}{2}+\frac{1}{2}\right) \beta _{J_2}+ \left( k+\frac{1}{2} \right) \delta . \end{aligned}$$

(179)

Step 1 gives

$$\begin{aligned} \Delta =\frac{1}{2}+s_1\left( \ell _1+\frac{\ell _2}{2}+\frac{3}{2}\right) +s_2\left( \frac{\ell _2}{2}+\frac{1}{2}\right) -3\sigma \left( k+\frac{1}{2} \right) , \end{aligned}$$

(180)

with $s_1, s_2, \sigma =\pm 1$. In Step 2$^{\prime }$ the shift of the value of $\Delta$ is 2, as originating from the coefficient of $\beta _D$ in $\rho _1$. Hence the module is reducible at these values, as well at at (130) with shift of $\Delta$ by 2.

1.5 6d $\mathcal {N}=(0,1)$

The root system for $\mathfrak {g}=\mathfrak {osp}(8|2)=D(4,1)$ is

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \alpha _i \pm \alpha _j, \quad \pm \beta , \bigg \} \quad (i,j=D,1,2,3), \end{aligned}$$

(181)

$$\begin{aligned}&\Delta _{\bar{1}}=\bigg \{ \frac{s_D}{2}\alpha _D + \frac{s_1}{2}\alpha _1 + \frac{s_2}{2}\alpha _2 + \frac{s_3}{2}\alpha _3 + \frac{\sigma }{2}\beta \bigg \} \ , \end{aligned}$$

(182)

where $s_D, s_1, \ldots , s_3, \sigma =\pm 1$ with the constraint $s_D s_1 s_2 s_3=1$, and we have

$$\begin{aligned} (\alpha _i, \alpha _j)=\delta _{ij}, \quad (\beta , \beta )=-2. \end{aligned}$$

(183)

Under an ordering $\alpha _D>\alpha _1>\alpha _2>\alpha _3>\beta$,

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \alpha _1 \pm \alpha _2, \quad \pm \alpha _1 \pm \alpha _3, \quad \pm \alpha _2 \pm \alpha _3, \quad \pm \beta \bigg \}, \end{aligned}$$

(184)

$$\begin{aligned}&\Delta _{1}^{+}=\overline{\Delta }_{1}^{+}=\bigg \{ \frac{1}{2}\alpha _D +\frac{s_1}{2}\alpha _1 + \frac{s_2}{2}\alpha _2 +\frac{s_3}{2}\alpha _3 +\frac{\sigma }{2}\beta \bigg \}, \end{aligned}$$

(185)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \alpha _D\pm \alpha _{1}, \quad \alpha _D\pm \alpha _{2}, \quad \alpha _D\pm \alpha _{3} \bigg \}, \end{aligned}$$

(186)

with the constraint $s_1 s_2 s_3=1$.

$$\begin{aligned}&\lambda = -\Delta \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2} \right) \alpha _1+\frac{\ell _2+\ell _3}{2}\alpha _2+\frac{-\ell _2+\ell _3}{2} \alpha _3+\frac{k}{2} \beta , \end{aligned}$$

(187)

$$\begin{aligned}&\rho =\alpha _D+2 \alpha _1+\alpha _2+\frac{1}{2}\beta , \end{aligned}$$

(188)

$$\begin{aligned}&\lambda +\rho =( -\Delta +1) \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2} +2\right) \alpha _1+\left( \frac{\ell _2+\ell _3}{2} +1\right) \alpha _2\nonumber \\&\qquad \qquad \,\, + \frac{-\ell _2+\ell _3}{2} \alpha _3+\left( \frac{k}{2}+\frac{1}{2}\right) \beta . \end{aligned}$$

(189)

Then $(\lambda +\rho , \alpha )=0$ for an odd isotropic root $\alpha$ from (185) gives

$$\begin{aligned} \Delta&=1+\left( \ell _1 +\frac{\ell _2+\ell _3}{2} +2\right) s_1+\left( \frac{\ell _2+\ell _3}{2} +1\right) s_2\nonumber \\&\quad +\frac{-\ell _2+\ell _3}{2} s_3 +4\sigma \left( \frac{k}{2}+\frac{1}{2}\right) . \end{aligned}$$

(190)

The module is reducible either at these values or at values (141), with $\Delta$ shifted by 2.

1.6 6d $\mathcal {N}=(0,2)$

We have $\mathfrak {g}=\mathfrak {osp}(8|4)=D(4,2)$, which has the root system

$$\begin{aligned}&\Delta _{\bar{0}}=\bigg \{ \pm \alpha _i \pm \alpha _j, \quad \pm \beta _1 \pm \beta _2, \quad \pm \beta _1 , \quad \pm \beta _2 \bigg \} \ , \quad (i,j=D,1,2,3) \end{aligned}$$

(191)

$$\begin{aligned}&\Delta _{\bar{1}}=\bigg \{ \frac{s_D}{2}\alpha _D + \frac{s_1}{2}\alpha _1 + \frac{s_2}{2}\alpha _2 + \frac{s_3}{2}\alpha _3 + \frac{\sigma _1}{2}\beta _1 +\frac{\sigma _2}{2}\beta _2 \bigg \}, \end{aligned}$$

(192)

where $s_D, s_1, \ldots , s_3, \sigma _1, \sigma _2=\pm 1$ with the constraint $s_D s_1 s_2 s_3=1$, and with

$$\begin{aligned} (\alpha _i, \alpha _j)=\delta _{ij}, \quad (\beta _a, \beta _b)=-\delta _{ij}. \end{aligned}$$

(193)

Under an ordering $\alpha _D>\alpha _1>\alpha _2>\alpha _3>\beta _1>\beta _2$,

$$\begin{aligned}&\Delta _{\mathfrak {l}}=\bigg \{ \pm \alpha _1 \pm \alpha _2, \quad \pm \alpha _1 \pm \alpha _3, \quad \pm \alpha _2 \pm \alpha _3, \quad \pm \beta _1 \pm \beta _2, \quad \pm \beta _1 , \quad \pm \beta _2\bigg \}, \end{aligned}$$

(194)

$$\begin{aligned}&\Delta _{1}^{+}=\overline{\Delta }_{1}^{+}=\bigg \{ \frac{1}{2}\alpha _D + \frac{s_1}{2}\alpha _1 + \frac{s_2}{2}\alpha _2 +\frac{s_3}{2}\alpha _3 + \frac{\sigma _1}{2}\beta _ 1+ \frac{\sigma _2}{2}\beta _ 2 \bigg \}, \end{aligned}$$

(195)

$$\begin{aligned}&\Delta _{\mathfrak {n}}\cap \overline{\Delta }_0=\Delta _{\mathfrak {n}}=\bigg \{ \alpha _D\pm \alpha _{1}, \quad \alpha _D\pm \alpha _{2}, \quad \alpha _D\pm \alpha _{3} \bigg \}. \end{aligned}$$

(196)

with the constraint $s_1 s_2 s_3=1$. We have

$$\begin{aligned}&\lambda = -\Delta \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2} \right) \alpha _1+\frac{\ell _2+\ell _3}{2} \alpha _2+\frac{-\ell _2+\ell _3}{2} \alpha _3+k_1 \beta _1+k_2 \beta _2, \end{aligned}$$

(197)

$$\begin{aligned}&\rho =-\alpha _D+2 \alpha _1+\alpha _2+\frac{3}{2}\beta _1+\frac{1}{2}\beta _2, \end{aligned}$$

(198)

$$\begin{aligned}&\lambda +\rho =( -\Delta -1) \alpha _D +\left( \ell _1 +\frac{\ell _2+\ell _3}{2}+2\right) \alpha _1+ \left( \frac{\ell _2+\ell _3}{2}+1 \right) \alpha _2+\frac{-\ell _2+\ell _3}{2} \alpha _3 \nonumber \\&\qquad \qquad +\left( k_1 +\frac{3}{2}\right) \beta _1+\left( k_2 +\frac{1}{2}\right) \beta _2. \end{aligned}$$

(199)

Then $(\lambda +\rho , \alpha )=0$ for an odd isotropic root $\alpha$ from (195) gives

$$\begin{aligned} \Delta&=-1+\left( \ell _1 +\frac{\ell _2+\ell _3}{2}+2\right) s_1+ \left( \frac{\ell _2+\ell _3}{2}+1 \right) s_2+\frac{-\ell _2+\ell _3}{2} s_3 \nonumber \\&\quad +2\sigma _1 \left( k_1 +\frac{3}{2}\right) +2\sigma _2 \left( k_2 +\frac{1}{2}\right) . \end{aligned}$$

(200)

The module is reducible either at these values or at (141) with $\Delta$ shifted by minus 4.