Goldstone modes in renormalizable supersymmetric SO(10) model

Abstract

We solve the Goldstone modes in the renormalizable SUSY SO(10) model with general couplings. The Goldstones are expressed by the Vacuum Expectation Values and the Clebsch–Gordan coefficients of relevant symmetries without explicit dependence on the parameters of the model.

1 Introduction

Grand Unified Theories (GUTs) are very important candidates for the physics beyond the Standard Model (SM). The Pati–Salam $G_{422} = SU(4)_C \otimes SU(2)_L \otimes SU(2)_R$ [1], the Georgi–Glashow SU(5)[2], the flipped SU(5) of $G_{51}=(SU(5) \otimes U(1))^{Flipped}$ [3], the SO(10) [4, 5] and the $E_6$ [6] models have all been studied extensively in the literature. Except SU(5), the GUT symmetries of these models are larger in ranks than the SM symmetry $G_{321} = SU(3)_C \otimes SU(2)_L \otimes U(1)_Y$. Consequently, they have several different routines to break into the SM symmetry, since the SM symmetry $G_{321}$ is not the maximal subgroup of the GUT symmetries.

The supersymmetric (SUSY) GUT models are even more important in realizing gauge coupling unification [7,8,9,10,11,12,13,14,15]. Again, except in the SU(5) models, these GUT symmetries have different routines in the symmetry breaking chains.

Spontaneously Symmetry Breaking (SSB) generates the massless Goldstone bosons corresponding to the broken generators [16,17,18]. In gauge theories, these Goldstone bosons act as the longitudinal components of the gauge bosons. In model buildings, we need to check the spectra to verify the existence of these Goldstone bosons. When more than one Higgs multiplets contribute to the SSB, we need to check that there are massless eigenstates in the mass squared matrices with correct representations.

In SUSY models, there are easier ways to get the Goldtone bosons. The Goldstinos, which are the fermionic partners of the Goldstone bosons, are also massless which mix with gauginos, the SUSY partners of the gauge bosons. Then, we can get the Goldstinos by solving the massless eigenstates of the mass matrices. Being the SUSY partners, the Goldstone bosons are determined accordingly.

Usually in building GUT models, the Higgs sectors relevant to SSB are complicated. Hereafter we will focus on SUSY SO(10) [19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36]. Even in the minimal renormalizble SUSY SO(10) model [26], Higgs in $210+126+\overline{126}$ are needed to break SO(10) into $G_{321}$. There are Goldstinos in $(1,1,0),(1,1,1),(3,1,\frac{2}{3}), (3,2, -\frac{5}{6}),(3,2,\frac{1}{6})$ and the conjugates under $G_{321}$. None of these representations has only one field in the spectrum. To verify the Goldstone modes, usually people need to calculate the eigenvalues of the mass matrices for the Goldstinos [26, 32]. In more general models with also $45+54$ in the SSB of the GUT symmetry, the Goldstino check is done numerically [23, 29].

In this work, we will study the Goldstone modes within SUSY. In [37] it has been realized that for the SSB of U(1) symmetries in SUSY models, in the Goldstone mode a component is proportional to the charge and to the Vacuum Expectation Value (VEV). Furthermore, it has been shown in the Non-Abelian cases [38], that in any Goldstone mode, the component from a representation is proportional to the VEV from the same representation and the CGCs determine the remainder dependence. Examples on the simplest SU(2) SSB are given in [38]. Here we will show test explicitly in the SUSY SO(10) model. We will give the Goldstone modes for the SSB of SO(10) into $G_{321}$ in the general renormalizable model.

In Sect. 2 we will give a general discussion on the method to solve the Goldstone modes. Then, in Sect. 3, we associate in SO(10) the Goldstones and the corresponding broken symmetries. In Sects. 4 and 5, Goldstones associated with the SSB of $G_{422}$ and SU(5) are solved, respectively, and relevant identities among the CGCs are examined. In Sect. 6, Goldstones associated with the SSB of the flipped SU(5) are solved. The relevant CGCs, which are not available in the literature, are presented and the identities are examined. The mass matrix for the Goldstones is given in the Appendix. Finally we summarize in Sect. 7.

2 General SSB in SUSY models

Following [38], we consider the SSB of $G_1 \rightarrow G_2$ in SUSY models. For models with only real fields, the general Higgs superpotential can be written as

$$\begin{aligned} W = \frac{1}{2} \sum _{I,i} m_I (I_i)^2 + \sum _{IJK,ijk} {\lambda }^{IJK} I_i J_j K_k, \end{aligned}$$

(1)

where I, J, K denotes the superfields, i, j, k denote the representations of group $G_1$. Under $G_2$, the couplings in (1) are of the forms

$$\begin{aligned} {\lambda }^{IJK} I_{i_a} J_{j_b} K_{k_c} C^{IJK}_{i_a j_b k_c}, \end{aligned}$$

(2)

where a, b, c are the representations under $G_2$, and $C^{IJK}_{i_a j_b k_c}$ is the Clebsch–Gordan coefficient (CGC). A singlet of $G_2$, denoted by $I_{i_1}$ etc., can have a VEV $\hat{I}_{i_1}$ in the SSB. SUSY requires the F-flatness conditions for the singlets,

$$\begin{aligned} 0\equiv F_{I_{i_1}} = \left\langle \frac{\partial W}{\partial I_{i_1}} \right\rangle = m_I \hat{I}_{i_1} + \sum _{JK,jk} \lambda ^{IJK} \hat{J}_{j_1} \hat{K}_{k_1} C^{IJK}_{{i_1}{j_1}{k_1}}. \end{aligned}$$

(3)

The Goldstinos in the SSB will be denoted by $\alpha ,\overline{\alpha }$ under $G_2$. A mass matrix element for the Goldstinos is

$$\begin{aligned} M^{IJ}_{{i_\alpha }{j_{\bar{\alpha }}}} = \left\langle \frac{\partial ^2 W}{\partial I_{i_\alpha } J_{j_{\bar{\alpha }}}} \right\rangle = m_I \delta _{IJ} \delta _{ij} + \sum _{K,k} \lambda ^{IJK} \hat{K}_{k_1} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}}. \end{aligned}$$

For $G^I_{i_{\bar{\alpha }}}$ denotes the element of the Goldstino corresponding to $\bar{\alpha }$, the zero-eigenvalue equation is

$$\begin{aligned} 0\equiv \sum _{J,j} M^{IJ}_{{i_\alpha }{j_{\bar{\alpha }}}} G^J_{j_{\bar{\alpha }}} = m_I G^I_{i_{\bar{\alpha }}} + \sum _{JK,jk} \lambda ^{IJK} G^J_{j_{\bar{\alpha }}} \hat{K}_{k_1} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}}. \nonumber \\ \end{aligned}$$

(4)

Eliminating $M_I$ in (3) and (4) gives

$$\begin{aligned} 0= & {} G^I_{i_{\bar{\alpha }}} \sum \limits _{JK,jk} \lambda ^{IJK} \hat{J}_{j_1} \hat{K}_{k_1} C^{IJK}_{{i_1}{j_1}{k_1}} \nonumber \\&-\, \hat{I}_{i_1} \sum \limits _{JK,jk} \lambda ^{IJK} G^J_{j_{\bar{\alpha }}} \hat{K}_{k_1} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}}. \end{aligned}$$

(5)

It follows that if a representations of $G_1$ does not contain a $G_2$ singlet, even if it may contain representations of the Goldstinos, it does not contribute to the Goldstino modes since the $M_I$ term in (4) cannot be eliminated unless multiplied by zero.

The superpotential parameters can be arbitrary so that we can focus on a specified coupling $\lambda ^{IJK}$, and the summation over different $\lambda$s is unnecessary. Furthermore, denoting

$$\begin{aligned} T^{Ii}_{\bar{\alpha }} = \frac{G^I_{i_{\bar{\alpha }}}}{\hat{I}_{i_1}}, ~\hbox {etc.}, \end{aligned}$$

(6)

(5) becomes

$$\begin{aligned} 0= & {} T^{Ii}_{\bar{\alpha }} \lambda ^{IJK} \hat{J}_{j_1} \hat{K}_{k_1} C^{IJK}_{{i_1}{j_1}{k_1}} - T^{Jj}_{\bar{\alpha }} \lambda ^{IJK} \hat{J}_{j_1} \hat{K}_{k_1} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}} \\&-\,T^{Kk}_{\bar{\alpha }} \lambda ^{IJK} \hat{J}_{j_1} \hat{K}_{k_1} C^{IJK}_{{i_\alpha }{j_1}{k_{\bar{\alpha }}}}. \end{aligned}$$

The nonzero $\lambda ^{IJK}\hat{J}_{j_1} \hat{K}_{k_1}$ can be eliminated and we can reiterate the same operation for J, K, then

$$\begin{aligned} 0 = \left( \begin{array}{ccc} - C^{IJK}_{{i_1}{j_1}{k_1}} &{} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}} &{} C^{IKJ}_{{i_\alpha }{k_{\bar{\alpha }}}{j_1}} \\ \\ C^{JIK}_{{j_\alpha }{i_{\bar{\alpha }}}{k_1}} &{} - C^{JIK}_{{j_1}{i_1}{k_1}} &{} C^{JKI}_{{j_\alpha }{k_{\bar{\alpha }}}{i_1}} \\ \\ C^{KIJ}_{{k_\alpha }{i_{\bar{\alpha }}}{j_1}} &{} C^{KJI}_{{k_\alpha }{j_{\bar{\alpha }}}{i_1}} &{} -C^{KJI}_{{k_1}{j_1}{i_1}} \end{array} \right) \left( \begin{array}{c} T^{Ii}_{\bar{\alpha }} \\ \\ T^{Jj}_{\bar{\alpha }} \\ \\ T^{Kk}_{\bar{\alpha }} \end{array} \right) . \end{aligned}$$

(7)

Similar result is given for the Goldstino in $\alpha$. (7) is followed by an identity that the determinant of the square matrix in (7) is zero, which must hold for any SSB.

Some special cases need to be clarified now.

(1) If k does not contain $\alpha , {\bar{\alpha }}$, (7) is simplified into

$$\begin{aligned} 0 = \left( \begin{array}{cc} - C^{IJK}_{{i_1}{j_1}{k_1}} &{} C^{IJK}_{{i_\alpha }{j_{\bar{\alpha }}}{k_1}} \\ \\ C^{JIK}_{{j_\alpha }{i_{\bar{\alpha }}}{k_1}} &{} - C^{JIK}_{{j_1}{i_1}{k_1}} \end{array} \right) \left( \begin{array}{c} T^{Ii}_{\bar{\alpha }} \\ \\ T^{Jj}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

(8)

which gives an identity that the determinant of the square matrix is zero, and a simple ratio between the two components. Taking all different couplings $\lambda$s setup all relations among the Goldstino components. This determines the Goldstino content up to an overall normalization.

(2) For $J = K$ and $j = k$, we have

$$\begin{aligned} 0 = \left( \begin{array}{cc} - C^{IJJ}_{{i_1}{j_1}{j_1}} &{} 2 C^{IJJ}_{{i_\alpha }{j_{\bar{\alpha }}}{j_1}} \\ \\ C^{IJJ}_{{i_{\bar{\alpha }}}{j_{\alpha }}{j_1}} &{} C^{IJJ}_{{i_1}{j_\alpha }{j_{\bar{\alpha }}}} - C^{IJJ}_{{i_1}{j_1}{j_1}} \end{array} \right) \left( \begin{array}{c} T^{Ii}_{\bar{\alpha }} \\ \\ T^{Ij}_{\bar{\alpha }} \end{array} \right) . \end{aligned}$$

(9)

(3) For $I = J = K$ and $i = j = k$, the result is

$$\begin{aligned} C^{III}_{{i_1}{i_1}{i_1}} =2 C^{III}_{i_{\alpha } i_1 i_{{\bar{\alpha }}}}, \end{aligned}$$

(10)

without information on T.

Generalization to the models with complex fields is straightforward with the general results given in [38]. We have also identities among the CGCs and equations for the Goldstinos involving the CGCs. Two types of special superpotentials might be relevant in the SO(10) study.

(4) For $I,\overline{I},K$ contain Goldstones and K is real, if

$$\begin{aligned} W=m_I I\overline{I}+\frac{1}{2}m_K^2+\lambda I^2K+\bar{\lambda }\overline{I}^2K,\nonumber \end{aligned}$$

which gives

$$\begin{aligned} 0=\left( \begin{array}{ccc} \begin{array}{c} -C^{IIK}_{{i_1}{i_1}{k_1}} \\ \\ XC^{\bar{I}\bar{I} K}_{\bar{i}_{\bar{\alpha }} {\bar{i}_1} {k_\alpha } }\\ \\ C^{\bar{I}\bar{I} K}_{\bar{i}_{\bar{\alpha }}{\bar{i}_\alpha }{k_1}} \end{array} \begin{array}{c} C^{IIK}_{{i_{\alpha }}{i_1}{k_{\bar{\alpha }}}}\\ \\ -C^{IIK}_{{i_1}{i_1}{k_1}} \\ \\ C^{\bar{I}\bar{I} K}_{\bar{i}_{\bar{\alpha }}{\bar{i}_1}{k_\alpha }} \end{array} \begin{array}{c} C^{IIK}_{{i_{\alpha }}{i_{\bar{\alpha }}}{k_1}}\\ \\ C^{IIK}_{{i_{\bar{\alpha }}}{i_1}{k_{\alpha }}}\\ \\ -C^{\bar{I}\bar{I} K}_{\bar{i}_1{\bar{i}_1}{k_1}} \end{array} \end{array} \right) \left( \begin{array}{c} T^{\bar{I}i}_{\bar{\alpha }} \\ \\ T^{Kk}_{\bar{\alpha }}\\ \\ T^{Ii}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

(11)

where $X=C^{\overline{I}\overline{I} K}_{\bar{i}_1{\bar{i}_1}{k_1}}/C^{IIK}_{{i_1}{i_1}{k_1}}$.

(5) For $I,\overline{I},K$ contain Goldstones and K is real, if

$$\begin{aligned} W=m_I I\overline{I}+\frac{1}{2}m_K^2+\lambda I \overline{I}K, \end{aligned}$$

which gives

$$\begin{aligned} 0=\left( \begin{array}{ccc} \begin{array}{c} -C^{I\bar{I}K}_{{i_1}{\bar{i}_1}{k_1}} \\ \\ C^{I\bar{I}K}_{{i_1}{\bar{i}_{\bar{\alpha }}}{k_{\alpha }}}\\ \\ 0 \end{array} \begin{array}{c} C^{I\bar{I}K}_{{i_{\alpha }}{\bar{i}_1}{k_{\bar{\alpha }}}}\\ \\ -C^{I\bar{I}K}_{{i_1}{\bar{i}_1}{k_1}} \\ \\ C^{I\bar{I}K}_{{i_1}{\bar{i}_{\alpha }}{k_{\bar{\alpha }}}} \end{array} \begin{array}{c} 0\\ \\ C^{I\bar{I}K}_{i_{\bar{\alpha }} {\bar{i}_1}{k_{\alpha }}}\\ \\ -C^{I\bar{I}K}_{{i_1}{\bar{i}_1}{k_1}} \end{array} \end{array} \right) \left( \begin{array}{c} T^{\bar{I}i}_{\bar{\alpha }} \\ \\ T^{Kk}_{\bar{\alpha }}\\ \\ T^{Ii}_{\bar{\alpha }} \end{array} \right) . \end{aligned}$$

(12)

Furthermore, in case that $I_i$ $(\overline{I})$ contains only $\alpha$ $(\bar{\alpha })$ but not $\bar{\alpha }$ $(\alpha )$, we have

$$\begin{aligned} 0=\left( \begin{array}{cc} \begin{array}{c} -C^{I\bar{I}K}_{{i_1}{\bar{i}_1}{k_1}} \\ \\ C^{I\bar{I}K}_{{i_1}{\bar{i}_{\bar{\alpha }}}{k_{\alpha }}} \end{array} \begin{array}{c} C^{I\bar{I}K}_{{i_{\alpha }}{\bar{i}_1}{k_{\bar{\alpha }}}}\\ \\ -C^{I\bar{I}K}_{{i_1}{\bar{i}_1}{k_1}} \end{array} \end{array} \right) \left( \begin{array}{c} T^{\bar{I}i}_{\bar{\alpha }} \\ \\ T^{Kk}_{\bar{\alpha }} \end{array} \right) . \end{aligned}$$

(13)

We summarize in this section that the Goldstino and thus the Goldstone components are proportional to the VEVs, and the remaining determinations of these components need only the calculations of the CGCs. This is the approach which we will use to determine the Goldstones in the following sections, and the identities of the CGC determinants will be used to check the consistencies of these calculations.

3 SSB and Goldstones in SO(10)

We study the most general renormalizable couplings containing Higgs $H(10), D(120), \overline{\Delta }(\overline{126})+ \Delta (126), A(45), E(54)$ and $\Phi (210)$ in the SUSY SO(10) models. The most general renormalizable Higgs superpotential is [29]

$$\begin{aligned} W= & {} \frac{1}{2} m_{1} \Phi ^2 + m_{2} \overline{\Delta } \Delta + \frac{1}{2} m_{3} H^2+ \frac{1}{2} m_{4} A^2 \nonumber \\&+\, \frac{1}{2} m_{5} E^2 + \frac{1}{2} m_{6} D^2 \nonumber \\&+\, \lambda _{1} \Phi ^3 + \lambda _{2} \Phi \overline{\Delta } \Delta + \left( \lambda _3 \Delta + \lambda _4 \overline{\Delta } \right) H \Phi +\lambda _{5} A^2 \Phi \nonumber \\&-\,i \lambda _{6} A \overline{\Delta } \Delta + \frac{\lambda _7}{120} \varepsilon A \Phi ^2 \nonumber \\&+\, E ( \lambda _{8} E^2 + \lambda _{9} A^2 + \lambda _{10} \Phi ^2 + \lambda _{11} \Delta ^2\nonumber \\&+\, \lambda _{12} \overline{\Delta }^2 + \lambda _{13} H^2 ) \nonumber \\&+\,D^2 \left( \lambda _{14} E + \lambda _{15} \Phi \right) \nonumber \\&+\, D \left\{ \lambda _{16} H A + \lambda _{17} H \Phi + \left( \lambda _{18} \Delta + \lambda _{19} \overline{\Delta } \right) A \right. \nonumber \\&\left. +\,\left( \lambda _{20} \Delta + \lambda _{21} \overline{\Delta } \right) \Phi \right\} , \end{aligned}$$

(14)

When we study the SSB of a SO(10) subgroup $G_1$ into $G_2$, we firstly decompose the SO(10) representations into $G_1$ representations, and the couplings are now of the form

$$\begin{aligned} \lambda ^{IJK} IJK=\lambda ^{IJK}\sum _{ijk} C^{IJK}_{ijk} I_i J_j K_k, \end{aligned}$$

(15)

where, for example, $I_i$ stands for the representations i of $G_1$ from SO(10) superfield I. Then, under $G_2$, i, j, k are further decomposed, (15) is

$$\begin{aligned} \lambda ^{IJK} I J K= & {} \lambda ^{IJK}\sum _{ijk}\sum _{abc} C^{IJK}_{ijk}C^{ijk}_{abc} I_{i_a} J_{j_b} K_{kc}\nonumber \\\equiv & {} \lambda ^{IJK}\sum _{ijk}\sum _{abc} C^{I_i J_j K_k}_{a b c} I_{i_a} J_{j_b} K_{k_c}, \end{aligned}$$

(16)

where $I_{i_a}$ is a representations a of $G_2$ coming from $I_i$, and $C^{IJK}_{i_a j_b k_c}$ is the SO(10) CGC. Most of these CGCs have been given [29, 36] so that no separate calculations of $C^{IJK}_{ijk}$ and $C^{ijk}_{abc}$ are needed. A special kind of CGCs relevant for the SSB of the flipped SU(5) symmetry will be given later.

To study SSB of SO(10) into $G_{321}$, it is necessary to decompose the Higgs representations of SO(10) under the $G_{321}$ subgroup. There are $45 - 12 = 33$ Goldstone modes. We link the Goldstones of a SM representation with the SSB of a specific subgroup of SO(10), as is summarized in Table 1. In Table 1, the first three modes can be studied by the SO(10) CGCs using $G_{422}$ as the maximal subgroup[29], the $(3,2, -\frac{5}{6})+\mathrm{c.c.}$ modes need the CGCs using SU(5)[36], and $(3,2,\frac{1}{6})+\mathrm{c.c.}$ using $G_{51}$ CGCs which will be given below.

Table 1 The Goldstone modes and the corresponding SSB

4 Goldstone modes for SSB of $G_{422}$

Under $G_{422}$, the following representations

$$\begin{aligned} A_1\equiv & {} \widehat{A}_{(1,1,3)},\quad A_2\equiv \widehat{A}_{(15,1,1)},\quad E \equiv \widehat{E}_{(1,1,1)},\nonumber \\ v_R\equiv & {} \widehat{\Delta }_{(\overline{10},1,3)},\quad \overline{v_R}\equiv \widehat{\overline{\Delta }}_{(10,1,3)},\nonumber \\ \Phi _1\equiv & {} \widehat{\Phi }_{(1,1,1)},\quad \Phi _2\equiv \widehat{\Phi }_{(15,1,1)},\quad \Phi _3\equiv \widehat{\Phi }_{(15,1,3)}, \end{aligned}$$

(17)

contain the SM singlets (1, 1, 0) whose VEVs will be denoted in the same symbols. Obviously, E and $\Phi _1$ contain no Goldstones relevant for the $G_{422}$ breaking, $A_1$ contains Goldstone components relevant for the $SU(2)_R$ breaking, $A_2,\Phi _2$ for the $SU(4)_C$ breaking, and $v_R,\overline{v_R},\Phi _3$ for both. The needed CGCs are taken from [29] and are summarized in Tables 2, 3 and 4.

Table 2 CGCs for the SM singlets (1, 1, 0) under $G_{422}$ SSB. These SM singlets also correspond to Goldstones (1, 1, 0) in the SSB of $U(1)_{I_{3R}}\otimes U(1)_{B-L}$ into $U(1)_Y$. Hereafter, the CGCs for $A\Phi \Phi$ are understood as those for $\frac{1}{120}\epsilon A\Phi \Phi$ [29]. Here, for example, ${A}_1$ stands for $\widehat{A}_{(1,1,3)}^{(1,1,0)}$. For $X=\widehat{\overline{\Delta }}^{(1,1,0)}_{(10,1,3)}$ and $Y=\widehat{\Delta }^{(1,1,0)}_{(\overline{10},1,3)}$, $XY{A}_1$ corresponds to the CGC $C^{\overline{v_{R}}v_RA_1}_{1 ~1 ~1}$

Table 3 CGCs for the Goldstones (1, 1, 1) $+$ c.c. of $G_{422}$ SSB. Here, for example, $XY{A}_1$ for $X=\widehat{\Delta }^{(1,1,1)}_{(\overline{10},1,3)}$ and $Y=\widehat{\overline{\Delta }}^{(1,1,-1)}_{(10,1,3)}$ corresponds to the CGC $C^{\overline{v_{R}} v_R A_2}_{\bar{\alpha } ~\alpha ~1}$ where $\bar{\alpha }=(1,1,1)$ under $G_{321}$

Table 4 CGCs for the Goldstones (3, 1, $\frac{2}{3}$) $+$ c.c. of $G_{422}$ SSB

4.1 Goldstone (1, 1, 0) under $G_{321}$

This Goldstone is the result of the SSB of $U(1)_{I_{3R}}\otimes U(1)_{B-L}$ into $U(1)_Y$ and has been studied in [37]. In the renormalizable models, only ${\Delta }(\mathbf{126 }) +\overline{\Delta }(\overline{\mathbf{126 }})$ have the SM singlets with nonzero $U(1)_{I_{3R}},U(1)_{B-L}$ charges. Furthermore, if we include also fields in the spinor representations ${\Psi }(\mathbf{16 })+\overline{\Psi }(\overline{\mathbf{16 }})$ of SO(10), which are usually used in building non-renormalizable models and have halves of the charges of ${\Delta }(\mathbf{126 }) +\overline{\Delta }(\overline{\mathbf{126 }})$, the Goldstino and thus the Goldstone mode is found to be

$$\begin{aligned}&\frac{\langle \Delta \rangle }{N} \widehat{\Delta }^{(1,1,0)}_{(\overline{10},1,3)} -\frac{\langle \overline{\Delta }\rangle }{N} \widehat{\overline{\Delta }}^{(1,1,0)}_{(10,1,3)} +\frac{1}{2} \frac{\langle \Psi \rangle }{N} \widehat{\Psi }^{(1,1,0)}_{(\overline{4},1,2)} \nonumber \\&\quad -\,\frac{1}{2} \frac{\langle \overline{\Psi }\rangle }{N}\widehat{\overline{\Psi }}^{(1,1,0)}_{(4,1,2)}, \end{aligned}$$

(18)

where N is a simple normalization factor. Equation (18) tell us two important conclusions in the U(1) symmetry breaking. First, the component of a field in the Goldstone mode is proportional to its charge under the breaking U(1) and to its VEV. Second, there is no dependence on the superpotential parameters of the model, besides through the VEV determinations indirectly.

In the model (14), we have

$$\begin{aligned} \overrightarrow{G}_{(1,1,0)} = v_R \widehat{\Delta }^{(1,1,0)}_{(\overline{10},1,3)} - \overline{v_R} \widehat{\overline{\Delta }}^{(1,1,0)}_{(10,1,3)}. \end{aligned}$$

(19)

up to an obvious normalization.

4.2 [(1, 1, 1) $+$ c.c.]

They are relevant for the SSB of $SU(2)_R$ into $U(1)_{I_{3R}}$. The fields contain $\bar{\alpha }=(1,1,1)$ are

$$\begin{aligned} \widehat{A}^{(1,1,1)}_{(1,1,3)}, \widehat{D}^{(1,1,1)}_{(\overline{10},1,1)}, \widehat{\Delta }^{(1,1,1)}_{(\overline{10},1,3)}, \widehat{\Phi }^{(1,1,1)}_{(15,1,3)}. \end{aligned}$$

Note that D has no VEV, the Goldstone corresponding to $\overline{\alpha }=(1,1,1)$ of $G_{321}$ is written as

$$\begin{aligned} \overrightarrow{G}_{(1,1,1)}= & {} T^{A_1}_{\overline{\alpha }} A_1 \widehat{A}^{(1,1,1)}_{(1,1,3)} + T^{v_R}_{\overline{\alpha }} v_R \widehat{\Delta }^{(1,1,1)}_{(\overline{10},1,3)}\\&+\,T^{\Phi _3}_{\overline{\alpha }} \Phi _3 \widehat{\Phi }^{(1,1,1)}_{(15,1,3)}. \end{aligned}$$

Here $A_1\dots$ are the VEVs, $\widehat{A},\dots$ are the fields and Ts are what will be solved. We will first classify the couplings and discuss their results, both on solve the Goldstones and on the identities among the CGCs.

(1) The representations $E,A_2,\Phi _1,\Phi _2$ have no $(1,1,\pm 1)$ under $G_{321}$, so that couplings of them with two same representations, such as $A_1^2 E$ or $\Phi _3^2 \Phi _1$, do not give relations among the Ts. These couplings lead to relations of CGCs such as

$$\begin{aligned} 0= & {} C^{A_1 A_1 E}_{1~1~1} - C^{A_1 A_1 E}_{\bar{\alpha }~\alpha ~1} = \frac{\sqrt{3}}{2\sqrt{5}} - \frac{\sqrt{3}}{2\sqrt{5}},\nonumber \\ 0= & {} C^{\Phi _3 \Phi _3 \Phi _1}_{1~1~1} - C^{\Phi _3 \Phi _3 \Phi _1}_{\bar{\alpha }~\alpha ~1} =\frac{1}{6\sqrt{6}}-\frac{1}{6\sqrt{6}}, \end{aligned}$$

which are trivial.

(2) Any coupling of $E,A_2,\Phi _1,\Phi _2$ with two different representations will set up a relation of the two Ts. Following (8), the coupling $A_1\Phi _3 A_2$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{cc} -C^{A_1 \Phi _3 A_2}_{1~1~1} &{} C^{A_1 \Phi _3 A_2}_{\alpha ~\bar{\alpha }~1} \\ \\ C^{A_1 \Phi _3 A_2}_{\bar{\alpha }~\alpha ~1} &{} - C^{A_1 \Phi _3 A_2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{A_1}_{\bar{\alpha }} \\ \\ T^{\Phi _3}_{\bar{\alpha }} \end{array} \right) \\= & {} \left( \begin{array}{cc} - \dfrac{1}{\sqrt{6}} &{} - \dfrac{1}{\sqrt{6}} \\ \\ - \dfrac{1}{\sqrt{6}} &{} - \dfrac{1}{\sqrt{6}} \end{array} \right) \left( \begin{array}{c} T^{\Phi _2}_{\bar{\alpha }} \\ \\ T^{\Phi _3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

where the determinant is zero, so

$$\begin{aligned} \frac{T^{A_1}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}} = \frac{C^{A_1 \Phi _3 A_2}_{\bar{\alpha }~\alpha ~1}}{C^{A_1 \Phi _3 A_2}_{1~1~1}} = \left( - \frac{1}{\sqrt{6}}\right) \Bigg / \left( \frac{1}{\sqrt{6}}\right) = - 1. \end{aligned}$$

The coupling $A_1 \Phi _3 \Phi _2$ also gives

$$\begin{aligned} \frac{T^{A_1}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}} = \frac{C^{A_1 \Phi _3 \Phi _2}_{\bar{\alpha }~\alpha ~1}}{C^{A_1 \Phi _3 \Phi _2}_{1~1~1}} = \left( - \frac{1}{\sqrt{6}}\right) \Bigg / \left( \frac{1}{\sqrt{6}}\right) = -1. \end{aligned}$$

(3) $\Phi _3 v_R \overline{v_R}$ and $A_1 v_R \overline{v_R}$ are couplings with all three representations contain the Goldstone modes. However, $v_R$ contains only $\overline{\alpha }$ while $\overline{v_R}$ contains only $\alpha$. Following (13), the coupling $\Phi _3 v_R \overline{v_R}$ gives

$$\begin{aligned} \frac{T^{\Phi _3}_{\bar{\alpha }}}{T^{v_R}_{\bar{\alpha }}} = \frac{C^{v_R \Phi _3 \overline{v_R}}_{\bar{\alpha }~\alpha ~1}}{C^{v_R \Phi _3 \overline{v_R}}_{1~1~1}} = \left( - \frac{1}{10}\right) \Bigg / \left( \frac{1}{10}\right) = - 1, \end{aligned}$$

and $A_1 v_R \overline{v_R}$ gives

$$\begin{aligned} \frac{T^{A_1}_{\bar{\alpha }}}{T^{v_R}_{\bar{\alpha }}} = \frac{C^{v_R A_1 \overline{v_R}}_{\bar{\alpha }~\alpha ~1}}{C^{v_R A_1 \overline{v_R}}_{1~1~1} } = \left( - \frac{1}{5}\right) \Bigg / \left( - \frac{1}{5}\right) = 1. \end{aligned}$$

Altogether, we have consistently $T^{A_1}_{\bar{\alpha }}: T^{v_R}_{\bar{\alpha }}:T^{\Phi _3}_{\bar{\alpha }} = 1:1: (-1)$, so the Goldstone mode is

$$\begin{aligned} \overrightarrow{G}_{(1,1,1)} = A_1 \widehat{A}^{(1,1,1)}_{(1,1,3)} + v_R \widehat{\Delta }^{(1,1,1)}_{(\overline{10},1,3)} - \Phi _3 \widehat{\Phi }^{(1,1,1)}_{(15,1,3)} \end{aligned}$$

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with an obvious normalization.

4.3 [(3, 1, $\frac{2}{3}$) $+$ c.c.]

They are the Goldstones for $SU(4)_C$ SSB into $SU(3)_C\otimes U(1)_{B-L}$. The fields corresponding to $\bar{\alpha }=(3,1,\frac{2}{3})$ are

$$\begin{aligned} \widehat{A}^{(3,1,\frac{2}{3})}_{(15,1,1)}, \widehat{D}^{(3,1,\frac{2}{3})}_{(6,1,3)}, \widehat{\overline{\Delta }}^{(3,1,\frac{2}{3})}_{(10,1,3)}, \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,1)}, \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,3)}. \end{aligned}$$

Again, D does not have a VEV, the Goldstone mode is

$$\begin{aligned} \overrightarrow{G}_{(3,1,\frac{2}{3})}= & {} T^{A_2}_{\bar{\alpha }} A_2 \widehat{A}^{(3,1,\frac{2}{3})}_{(15,1,1)} + T^{\overline{v_R}}_{\bar{\alpha }} \overline{v_R} \widehat{\overline{\Delta }}^{(3,1,\frac{2}{3})}_{(10,1,3)} \\&+\,T^{\Phi _2}_{\bar{\alpha }} \Phi _2 \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,1)} + T^{\Phi _3}_{\bar{\alpha }} \Phi _3 \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,3)}. \end{aligned}$$

(1) According to (10), $\Phi _2^3$ leads to

$$\begin{aligned} 0 = C^{\Phi _2 \Phi _2 \Phi _2}_{1~1~1} - 2 C^{\Phi _2 \Phi _2 \Phi _2}_{\bar{\alpha }~\alpha ~1}= \frac{1}{9\sqrt{2}} - 2*\frac{1}{18\sqrt{2}}. \end{aligned}$$

(2) $A_1,E,\Phi _1$ does not contain $\alpha$ or $\overline{\alpha }$ and are SU(4) singlets, their couplings with the same fields, $A_2^2 E, \Phi _3^2 E, \Phi _3^2 \Phi _1$ lead to only trivial results. In the couplings of one of them with two different fields, according to (8), they give relations between two Goldstone components. $A_2 \Phi _3 A_1$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}} = \frac{C^{A_2 \Phi _3 A_1}_{\bar{\alpha }~\alpha ~1}}{C^{A_2 \Phi _3 A_1}_{1~1~1}} = \left( - \frac{1}{\sqrt{6}}\right) \Bigg / \left( \frac{1}{\sqrt{6}}\right) = - 1, \end{aligned}$$

$A_2 \Phi _2 \Phi _1$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _2}_{\bar{\alpha }}} = \frac{C^{A_2 \Phi _2 \Phi _1}_{\bar{\alpha }~\alpha ~1}}{C^{A_2 \Phi _2 \Phi _1}_{1~1~1}} = \left( - \frac{\sqrt{2}}{5}\right) \Bigg / \left( \frac{\sqrt{2}}{5}\right) = - 1, \end{aligned}$$

and $\Phi _2 \Phi _3 A_1$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}} = \frac{C^{A_2 \Phi _3 A_1}_{\bar{\alpha }~\alpha ~1}}{C^{A_2 \Phi _3 A_1}_{1~1~1}} = \left( - \frac{1}{\sqrt{6}}\right) \Bigg / \left( \frac{1}{\sqrt{6}}\right) = - 1. \end{aligned}$$

(3) $v_R$ contains only $\overline{\alpha }$ while $\overline{v_R}$ contains only ${\alpha }$. According to (13), $v_R \overline{v_R}\Phi _2$ gives

$$\begin{aligned} \frac{T^{\Phi _2}_{\bar{\alpha }}}{T^{\overline{v_R}}_{\bar{\alpha }}} = \frac{C^{ \overline{v_R} \Phi _2 v_R}_{\bar{\alpha }~\alpha ~1}}{C^{\overline{v_R} \Phi _2 v_R}_{1~1~1}} = \left( -\frac{1}{10\sqrt{3}}\right) \Bigg / \left( \frac{1}{10\sqrt{2}}\right) = - \sqrt{\frac{2}{3}}, \end{aligned}$$

$v_R \overline{v_R}\Phi _3$ gives

$$\begin{aligned} \frac{T^{\Phi _3}_{\bar{\alpha }}}{T^{\overline{v_R}}_{\bar{\alpha }}} = \frac{C^{ \overline{v_R} \Phi _3 v_R}_{\bar{\alpha }~\alpha ~1}}{C^{\overline{v_R} \Phi _3 v_R}_{1~1~1}} = \left( -\frac{1}{5\sqrt{6}}\right) \Bigg / \left( \frac{1}{10}\right) = - \sqrt{\frac{2}{3}}, \end{aligned}$$

and $v_R \overline{v_R}A_2$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\overline{v_R}}_{\bar{\alpha }}} = \frac{C^{ \overline{v_R} A_2 v_R}_{\bar{\alpha }~\alpha ~1}}{C^{\overline{v_R} A_2 v_R}_{1~1~1}} = \left( -\frac{1}{5}\right) \Bigg / \left( - \frac{\sqrt{3}}{5\sqrt{2}}\right) = \sqrt{\frac{2}{3}}. \end{aligned}$$

(4) According to (9), $\Phi _2\Phi _3^2$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{cc} -C^{\Phi _2 \Phi _3 \Phi _3}_{1~1~1} &{} 2 C^{\Phi _2 \Phi _3 \Phi _3}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\Phi _3 \Phi _2 \Phi _3}_{\bar{\alpha }~\alpha ~1} &{} C^{\Phi _3 \Phi _3 \Phi _2}_{\bar{\alpha }~\alpha ~1} - C^{\Phi _3 \Phi _3 \Phi _2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{\Phi _2}_{\bar{\alpha }} \\ \\ T^{\Phi _3}_{\bar{\alpha }} \end{array} \right) \\= & {} \left( \begin{array}{cc} - \dfrac{1}{9\sqrt{2}} &{} 2*\dfrac{1}{18\sqrt{2}} \\ \\ \dfrac{1}{18\sqrt{2}} &{} \dfrac{1}{9\sqrt{2}} - \dfrac{1}{18\sqrt{2}} \end{array} \right) \left( \begin{array}{c} T^{\Phi _2}_{\bar{\alpha }} \\ \\ T^{\Phi _3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

so

$$\begin{aligned} \frac{T_{\Phi _3}}{T_{\Phi _2}}= & {} \frac{C^{\Phi _3 \Phi _2 \Phi _3}_{\bar{\alpha }~\alpha ~1}}{C^{\Phi _3 \Phi _3 \Phi _2}_{1~1~1} - C^{\Phi _3 \Phi _3 \Phi _2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{1}{18\sqrt{2}}\right) \Bigg / \left( \frac{1}{9\sqrt{2}} - \frac{1}{18\sqrt{2}}\right) = 1. \nonumber \\ or{:}\ \frac{T_{\Phi _3}}{T_{\Phi _2}}= & {} \frac{C^{\Phi _2 \Phi _3 \Phi _3}_{1~1~1}}{2 C^{\Phi _2 \Phi _3 \Phi _3}_{\bar{\alpha }~\alpha ~1}} = \left( \frac{1}{9\sqrt{2}}\right) \Bigg / \left( 2*\frac{1}{18\sqrt{2}}\right) = 1. \end{aligned}$$

Similarly, $\Phi _2 A_2^2$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _2}_{\bar{\alpha }}}= & {} \frac{C^{A_2 \Phi _2 A_2}_{\bar{\alpha }~\alpha ~1}}{C^{A_2 A_2 \Phi _2}_{1~1~1} - C^{A_2 A_2 \Phi _2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( - \frac{1}{3\sqrt{2}}\right) \Bigg /\left( \frac{\sqrt{2}}{3} - \frac{1}{3\sqrt{2}}\right) = - 1. \nonumber \\ or{:}\ \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _2}_{\bar{\alpha }}}= & {} \frac{C^{\Phi _2 A_2 A_2}_{1~1~1}}{2 C^{\Phi _2 A_2 A_2}_{\bar{\alpha }~\alpha ~1}} = \left( \frac{\sqrt{2}}{3}\right) \Bigg / \left( 2*\frac{-1}{3\sqrt{2}}\right) = -1. \end{aligned}$$

$A_2 \Phi _3^2$ gives

$$\begin{aligned} \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}}= & {} \frac{2 C^{A_2 \Phi _3 \Phi _3}_{\bar{\alpha }~\alpha ~1}}{C^{A_2 \Phi _3 \Phi _3}_{1~1~1}} = \left( - \frac{2*\sqrt{2}}{5\sqrt{3}}\right) \Bigg / \left( \frac{2\sqrt{2}}{5\sqrt{3}}\right) = -1, \nonumber \\ or{:}\ \frac{T^{A_2}_{\bar{\alpha }}}{T^{\Phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\Phi _3 \Phi _3 A_2}_{1~1~1} - C^{\Phi _3 \Phi _3 A_2}_{\bar{\alpha }~\alpha ~1}}{C^{\Phi _3 A_2 \Phi _3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{2\sqrt{2}}{5\sqrt{3}} - \frac{\sqrt{2}}{5\sqrt{3}}\right) \Bigg / \left( -\frac{\sqrt{2}}{5\sqrt{3}}\right) = - 1. \end{aligned}$$

Table 5 CGCs for the SM singlets (1, 1, 0) under SU(5). Here, for example, $a_2$ stands for $\widehat{A}_{(24)}^{(1,1,0)}$ etc.

All together we have $T_A = \sqrt{\tfrac{2}{3}} T^{\overline{v_R}}_{\bar{\alpha }} = - T^{\Phi _2}_{\bar{\alpha }} = - T^{\Phi _3}_{\bar{\alpha }}$, then the Goldstone is

$$\begin{aligned} \overrightarrow{G}_{(3,1,\frac{2}{3})}= & {} A_2 \widehat{A}^{(3,1,\frac{2}{3})}_{(15,1,1)} + \sqrt{\frac{3}{2}} \overline{v_R} \widehat{\overline{\Delta }}^{(3,1,\frac{2}{3})}_{(10,1,3)}\nonumber \\&-\, \Phi _2 \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,1)} - \Phi _3 \widehat{\Phi }^{(3,1,\frac{2}{3})}_{(15,1,3)}. \end{aligned}$$

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5 Goldstone modes for SSB of SU(5): [(3, 2, $-\frac{5}{6}$) $+$ c.c.]

In study SSB of SU(5), the fields need to be decomposed into SU(5) representations and use the CGCs calculated in [36]. Under SU(5), the following representations

$$\begin{aligned} {a_1}\equiv & {} \widehat{A}_{(1)},\quad {a_2} \equiv \widehat{A}_{(24)},\quad {S} \equiv \widehat{E}_{(24)}, \nonumber \\ {V_R}\equiv & {} \widehat{\Delta }_{(1)},\quad {\overline{V_R}} \equiv \widehat{\overline{\Delta }}_{(1)},\nonumber \\ {\phi }_1\equiv & {} \widehat{\Phi }_{(1)},\quad {\phi }_2 \equiv \widehat{\Phi }_{(24)},\quad {\phi }_3 \equiv \widehat{\Phi }_{(75)}. \end{aligned}$$

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contain the SM singlets whose VEVs will be denoted by the same symbols. Here the subscripts are the representations under SU(5).

Since the SU(5) singlets do not contribute to the Goldstone modes, the fields contains $\bar{\alpha }=(3,2,-\frac{5}{6})$ are

$$\begin{aligned} \widehat{A}^{(3,2,-\frac{5}{6})}_{(24)}, \widehat{E}^{(3,2,-\frac{5}{6})}_{(24)}, \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(24)}, \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(75)}. \end{aligned}$$

The Goldstone corresponding to $(3,2,-\frac{5}{6})$ is denoted as

$$\begin{aligned} \overrightarrow{G}_{(3,2,-\frac{5}{6})}= & {} T^{a_2}_{\bar{\alpha }} a_2 \widehat{A}^{(3,2,-\frac{5}{6})}_{(24)} + T^{S}_{\bar{\alpha }} S \widehat{E}^{(3,2,-\frac{5}{6})}_{(24)} \\&+\, T^{\phi _2}_{\bar{\alpha }} \phi _2 \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(24)} + T^{\phi _3}_{\bar{\alpha }} \phi _3 \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(75)}. \end{aligned}$$

The needed CGCs are taken from [36] and summarized in Tables 5 and 6.

Table 6 CGCs for the Goldstones (3, 2, $- \frac{5}{6}$) $+$ c.c. of SU(5) SSB

(1) According to (10), the coupling $\phi _2^3$ gives

$$\begin{aligned} 0 = C^{\phi _2 \phi _2 \phi _2}_{1~1~1} - 2 C^{\phi _2 \phi _2 \phi _2}_{\bar{\alpha }~\alpha ~1} =\left( - \frac{7}{9\sqrt{15}}\right) - 2*\left( - \frac{7}{18\sqrt{15}}\right) , \end{aligned}$$

$\phi _3^3$ gives

$$\begin{aligned} 0 =C^{\phi _3 \phi _3 \phi _3}_{1~1~1} - 2 C^{\phi _3 \phi _3 \phi _3}_{\bar{\alpha }~\alpha ~1} = \frac{8}{9\sqrt{3}} - 2*\frac{4}{9\sqrt{3}}, \end{aligned}$$

and $S^3$ gives

$$\begin{aligned} 0 =C^{SSS}_{111} - 2 C^{SSS}_{\bar{\alpha } \alpha 1} =\sqrt{\frac{3}{5}} - 2 *\frac{\sqrt{3}}{2\sqrt{5}}. \end{aligned}$$

(2) $\phi _1, a_1$ contain neither Goldstones, the couplings $\phi _2^2 \phi _1, \phi _3^2 \phi _1, a_2^2 \phi _1, \phi _2^2 a_1$ and $\phi _3^2 a_1$ lead to only trivial results. The coupling $a_2 \phi _2 a_1$, according to (8), gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}} = \frac{C^{a_2 \phi _2 a_1}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 \phi _2 a_1}_{1~1~1}} = \left( -\sqrt{\frac{2}{5}}\right) \Bigg / \left( \sqrt{\frac{2}{5}}\right) = -1, \end{aligned}$$

$a_2 \phi _2 \phi _1$ gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}} = \frac{C^{a_2 \phi _2 \phi _1}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 \phi _2 \phi _1}_{1~1~1}} = \left( \frac{2\sqrt{3}}{5\sqrt{5}}\right) \Bigg / \left( -\frac{2\sqrt{3}}{5\sqrt{5}}\right) = -1, \end{aligned}$$

$S a_2 a_1$ gives

$$\begin{aligned} \frac{T^{S}_{\bar{\alpha }}}{T^{a_2}_{\bar{\alpha }}} = \frac{C^{S a_2 a_1}_{\bar{\alpha }~\alpha ~1}}{C^{S a_2 a_1}_{1~1~1}} = \left( -\sqrt{\frac{2}{5}}\right) \Bigg / \left( \sqrt{\frac{2}{5}}\right) = -1, \end{aligned}$$

and $S \phi _2 \phi _1$ gives

$$\begin{aligned} \frac{T^{S}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}} = \frac{C^{S \phi _2 \phi _1}_{\bar{\alpha }~\alpha ~1}}{C^{S \phi _2 \phi _1}_{1~1~1}} = \left( \frac{\sqrt{3}}{2\sqrt{5}} \right) \Bigg / \left( \frac{\sqrt{3}}{2\sqrt{5}} \right) = 1. \end{aligned}$$

(3) According to (9), $\phi _2 \phi _3^2$ gives

$$\begin{aligned} \frac{T^{\phi _2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{2 C^{\phi _2 \phi _3 \phi _3}_{\bar{\alpha }~\alpha ~1}}{C^{\phi _2 \phi _3 \phi _3}_{1~1~1}} \\= & {} 2* \left( - \frac{\sqrt{2}}{9\sqrt{3}}\right) \Bigg / \left( -\frac{8}{9\sqrt{15}}\right) = \sqrt{\frac{5}{8}},\\ or{:}\ \frac{T^{\phi _2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _3 \phi _3 \phi _2}_{1~1~1} - C^{\phi _3 \phi _3 \phi _2}_{\bar{\alpha }~\alpha ~1}}{C^{\phi _3 \phi _2 \phi _3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{-8}{9\sqrt{15}} - \frac{-11}{18\sqrt{15}}\right) \Bigg / \left( - \frac{\sqrt{2}}{9\sqrt{3}}\right) = \sqrt{\frac{5}{8}}, \end{aligned}$$

$\phi _2^2 \phi _3$ gives

$$\begin{aligned} \frac{T^{\phi _2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _2 \phi _3 \phi _2}_{\bar{\alpha }~\alpha ~1}}{C^{\phi _2 \phi _2 \phi _3}_{1~1~1} - C^{\phi _2 \phi _2 \phi _3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{\sqrt{5}}{9\sqrt{6}}\right) \Bigg / \left( \frac{5}{18\sqrt{3}} - \frac{1}{18\sqrt{3}}\right) = \sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{\phi _2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _3 \phi _2 \phi _2}_{1~1~1} }{2 C^{\phi _3 \phi _2 \phi _2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{5}{18\sqrt{3}}\right) \Bigg / \left( 2*\frac{\sqrt{5}}{9\sqrt{6}}\right) = \sqrt{\frac{5}{8}}, \end{aligned}$$

$a_2^2 \phi _2$ gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{C^{a_2 \phi _2 a_2}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 a_2 \phi _2}_{1~1~1} - C^{a_2 a_2 \phi _2}_{\bar{\alpha }~\alpha ~1}}\\= & {} \left( \frac{1}{3\sqrt{15}}\right) \Bigg / \left( \frac{-2}{3\sqrt{15}} - \frac{-1}{3\sqrt{15}}\right) = -1, \\ or{:}\ \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{C^{\phi _2 a_2 a_2}_{1~1~1}}{2 C^{\phi _2 a_2 a_2 }_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( -\frac{2}{3\sqrt{15}}\right) \Bigg / 2*\left( \frac{1}{3\sqrt{15}}\right) = -1, \end{aligned}$$

$a_2^2 \phi _3$ gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{a_2 \phi _3 a_2}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 a_2 \phi _3}_{1~1~1} - C^{a_2 a_2 \phi _3}_{\bar{\alpha }~\alpha ~1}}\\= & {} \left( - \frac{\sqrt{10}}{3\sqrt{3}}\right) \Bigg / \left( \frac{5}{3\sqrt{3}} - \frac{1}{3\sqrt{3}}\right) = -\sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _3 a_2 a_2}_{1~1~1}}{2 C^{\phi _3 a_2 a_2 }_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{5}{3\sqrt{3}}\right) \Bigg / \left( 2*(-\frac{\sqrt{10}}{3\sqrt{3}})\right) = -\sqrt{\frac{5}{8}}, \end{aligned}$$

$\phi _2^2 a_2$ gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{2 C^{a_2 \phi _2 \phi _2}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 \phi _2 \phi _2}_{1~1~1}} \\= & {} 2*\left( -\frac{4}{15\sqrt{15}}\right) \Bigg /\left( \frac{8}{15\sqrt{15}}\right) = -1, \\ or{:}\ \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{C^{\phi _2 \phi _2 a_2}_{1~1~1} - C^{\phi _2 \phi _2 a_2}_{\bar{\alpha }~\alpha ~1} }{C^{\phi _2 a_2 \phi _2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{8}{15\sqrt{15}}- \frac{4}{15\sqrt{15}}\right) \Bigg / \left( - \frac{4}{15\sqrt{15}}\right) = -1, \end{aligned}$$

$\phi _3^2 a_2$ gives

$$\begin{aligned} \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{2 C^{a_2 \phi _3 \phi _3}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 \phi _3 \phi _3}_{1~1~1}} \\= & {} 2*\left( \frac{4\sqrt{2}}{15\sqrt{3}}\right) \Bigg /\left( - \frac{32}{15\sqrt{15}}\right) = -\sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{a_2}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _3 \phi _3 a_2}_{1~1~1} - C^{\phi _3 \phi _3 a_2}_{\bar{\alpha }~\alpha ~1} }{C^{\phi _3 a_2 \phi _3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{-32}{15\sqrt{15}} - \frac{-22}{15\sqrt{15}}\right) \Bigg / \left( \frac{4\sqrt{2}}{15\sqrt{3}}\right) = -\sqrt{\frac{5}{8}}, \end{aligned}$$

$S a_2^2$ gives

$$\begin{aligned} \frac{T^{S}_{\bar{\alpha }}}{T^{a_2}_{\bar{\alpha }}}= & {} \frac{2 C^{S a_2 a_2}_{\bar{\alpha }~\alpha ~1}}{C^{S a_2 a_2}_{1~1~1}}\\= & {} 2*\left( - \frac{1}{2\sqrt{15}} \right) \Bigg / \left( \frac{1}{\sqrt{15}}\right) = -1, \nonumber \\ or{:}\ \frac{T^{S}_{\bar{\alpha }}}{T^{a_2}_{\bar{\alpha }}}= & {} \frac{C^{a_2 a_2 S}_{1~1~1} - C^{a_2 a_2 S}_{\bar{\alpha }~\alpha ~1}}{C^{a_2 S a_2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{1}{\sqrt{15}} - \frac{1}{2\sqrt{15}}\right) \Bigg /\left( -\frac{1}{2\sqrt{15}}\right) = -1, \end{aligned}$$

$S \phi _2^2$ gives

$$\begin{aligned} \frac{T^{S}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{2 C^{S \phi _2 \phi _2}_{\bar{\alpha }~\alpha ~1}}{C^{S \phi _2 \phi _2}_{1~1~1}} \\= & {} 2*\left( \frac{1}{12\sqrt{15}}\right) \Bigg / \left( \frac{1}{6\sqrt{15}}\right) = 1, \nonumber \\ or{:}\ \frac{T^{S}_{\bar{\alpha }}}{T^{\phi _2}_{\bar{\alpha }}}= & {} \frac{C^{\phi _2 \phi _2 S}_{1~1~1} - C^{\phi _2 \phi _2 S}_{\bar{\alpha }~\alpha ~1}}{C^{\phi _2 S \phi _2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{1}{6\sqrt{15}} - \frac{1}{12\sqrt{15}}\right) \Bigg /\left( \frac{1}{12\sqrt{15}}\right) = 1, \end{aligned}$$

and $S \phi _3^2$ gives

$$\begin{aligned} \frac{T^{S}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{2 C^{S \phi _3 \phi _3}_{\bar{\alpha }~\alpha ~1}}{C^{S \phi _3 \phi _3}_{1~1~1}} \\= & {} 2*\left( \frac{1}{3\sqrt{6}}\right) \Bigg / \left( \frac{4}{3\sqrt{15}}\right) = \sqrt{\frac{5}{8}}, \nonumber \\ or{:}\ \frac{T^{S}_{\bar{\alpha }}}{T^{\phi _3}_{\bar{\alpha }}}= & {} \frac{C^{\phi _3 \phi _3 S}_{1~1~1} - C^{\phi _3 \phi _3 S}_{\bar{\alpha }~\alpha ~1}}{C^{\phi _3 S \phi _3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{4}{3\sqrt{15}} - \frac{11}{12\sqrt{15}}\right) \Bigg /\left( \frac{1}{3\sqrt{6}}\right) = \sqrt{\frac{5}{8}}. \end{aligned}$$

(4) According to (7), $a_2 \phi _2 \phi _3$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{ccc} - C^{a_2 \phi _2 \phi _3}_{1~1~1} &{} C^{a_2 \phi _2 \phi _3}_{\bar{\alpha }~\alpha ~1} &{} C^{a_2 \phi _3 \phi _2}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\phi _2 a_2 \phi _3}_{\bar{\alpha }~\alpha ~1} &{} - C^{\phi _2 a_2 \phi _3}_{1~1~1} &{} C^{\phi _2 \phi _3 a_2 }_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\phi _3 a_2 \phi _2 }_{\bar{\alpha }~\alpha ~1} &{} C^{\phi _3 \phi _2 a_2 }_{\bar{\alpha }~\alpha ~1} &{} -C^{\phi _3 a_2 \phi _2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{A_2}_{\bar{\alpha }} \\ \\ T^{\phi _2}_{\bar{\alpha }} \\ \\ T^{\phi _3}_{\bar{\alpha }} \end{array} \right) \\= & {} \left( \begin{array}{ccc} - \dfrac{2}{3\sqrt{3}} &{} - \dfrac{2}{15\sqrt{3}} &{} -\dfrac{2\sqrt{2}}{3\sqrt{15}} \\ \\ - \dfrac{2}{15\sqrt{3}} &{} - \dfrac{2}{3\sqrt{3}} &{} \dfrac{2\sqrt{2}}{3\sqrt{15}} \\ \\ -\dfrac{2\sqrt{2}}{3\sqrt{15}} &{} \dfrac{2\sqrt{2}}{3\sqrt{15}} &{} -\dfrac{2}{3\sqrt{3}} \end{array} \right) \left( \begin{array}{c} T^{A_2}_{\bar{\alpha }} \\ \\ T^{\phi _2}_{\bar{\alpha }} \\ \\ T^{\phi _3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

so that

$$\begin{aligned} T^{A_2}_{\bar{\alpha }} {:} T^{\phi _2}_{\bar{\alpha }} {:} T^{\phi _3}_{\bar{\alpha }} = 1 {:} -1 {:} \left( -\sqrt{\frac{8}{5}}\right) , \end{aligned}$$

and $S \phi _2 \phi _3$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{ccc} - C^{S \phi _2 \phi _3}_{1~1~1} &{} C^{S \phi _2 \phi _3}_{\bar{\alpha }~\alpha ~1} &{} C^{S \phi _3 \phi _2}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\phi _2 S \phi _3}_{\bar{\alpha }~\alpha ~1} &{} - C^{\phi _2 S \phi _3}_{1~1~1} &{} C^{\phi _2 \phi _3 S}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\phi _3 S \phi _2 }_{\bar{\alpha }~\alpha ~1} &{} C^{\phi _3 \phi _2 S }_{\bar{\alpha }~\alpha ~1} &{} -C^{\phi _3 S \phi _2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{S}_{\bar{\alpha }} \\ \\ T^{\phi _2}_{\bar{\alpha }} \\ \\ T^{\phi _3}_{\bar{\alpha }} \end{array} \right) \\= & {} \left( \begin{array}{ccc} - \dfrac{5}{6\sqrt{3}} &{} \dfrac{1}{6\sqrt{3}} &{} \dfrac{\sqrt{5}}{3\sqrt{6}} \\ \\ \dfrac{1}{6\sqrt{3}} &{} - \dfrac{5}{6\sqrt{3}} &{} \dfrac{\sqrt{5}}{3\sqrt{6}} \\ \\ \dfrac{\sqrt{5}}{3\sqrt{6}} &{} \dfrac{\sqrt{5}}{3\sqrt{6}} &{} - \dfrac{5}{6\sqrt{3}} \end{array} \right) \left( \begin{array}{c} T^{S}_{\bar{\alpha }} \\ \\ T^{\phi _2}_{\bar{\alpha }} \\ \\ T^{\phi _3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

so that

$$\begin{aligned} T^{S}_{\bar{\alpha }} {:} T^{\phi _2}_{\bar{\alpha }} {:} T^{\phi _3}_{\bar{\alpha }} = 1 {:} 1 {:} \sqrt{\frac{8}{5}}. \end{aligned}$$

Altogether, the Goldstone mode is

$$\begin{aligned} \overrightarrow{G}_{(3,2,-\frac{5}{6})}= & {} a_2 \widehat{A}^{(3,2,-\frac{5}{6})}_{(24)} - S \widehat{E}^{(3,2,-\frac{5}{6})}_{(24)} \nonumber \\&-\,\phi _2 \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(24)} - \sqrt{\frac{8}{5}} \phi _3 \widehat{\Phi }^{(3,2,-\frac{5}{6})}_{(75)}. \end{aligned}$$

(23)

6 Goldstone modes for SSB of $G_{51}$: [(3, 2, $\frac{1}{6}$) $+$ c.c.]

In studying the SSB of the flipped SU(5), the fields need to be decomposed into $G_{51}=(SU(5) \otimes U(1))^{Flipped}$ representations and use the CGCs accordingly. Under $G_{51}$, the following representations

$$\begin{aligned} \widetilde{a}_1\equiv & {} \widehat{A}_{(1,0)},\quad \widetilde{a}_2 \equiv \widehat{A}_{(24,0)},\quad \widetilde{S} \equiv \widehat{E}_{(24,0)}, \nonumber \\ {\delta }\equiv & {} \widehat{\Delta }_{(\overline{50},-2)},\quad {\overline{\delta }} \equiv \widehat{\overline{\Delta }}_{(50,2)},\nonumber \\ \widetilde{\phi }_1\equiv & {} \widehat{\Phi }_{(1,0)},\quad \widetilde{\phi }_2 \equiv \widehat{\Phi }_{(24,0)},\quad \widetilde{\phi }_3 \equiv \widehat{\Phi }_{(75,0)}. \end{aligned}$$

(24)

contain the SM singlets whose VEVs will be denoted by the same symbols. Here the subscripts are the representations under $G_{51}$. These VEVs are related to those under $G_{422}$ and SU(5) by

$$\begin{aligned} \widetilde{S}=E= & {} S,\quad {\delta } = v_R =V_R, \quad {\overline{\delta }} = \overline{v_R} =\overline{V_R}, \end{aligned}$$

(25)

$$\begin{aligned} \left( \begin{array}{c} \widetilde{a}_1\\ \\ \widetilde{a}_2 \end{array}\right)= & {} \left[ \begin{array}{cc} \sqrt{\dfrac{2}{5}} &{} - \sqrt{\dfrac{3}{5}} \\ \sqrt{\dfrac{3}{5}} &{} \sqrt{\dfrac{2}{5}} \end{array} \right] \left( \begin{array}{c} A_1 \\ \\ A_2 \end{array}\right) \nonumber \\= & {} \left[ \begin{array}{cc} -\dfrac{1}{5} &{} \dfrac{2\sqrt{6}}{5} \\ \dfrac{2\sqrt{6}}{5} &{} \dfrac{1}{5} \end{array} \right] \left( \begin{array}{c} a_1 \\ \\ a_2 \end{array}\right) , \end{aligned}$$

(26)

and

$$\begin{aligned} \left( \begin{array}{c} \widetilde{\phi }_1 \\ \\ \widetilde{\phi }_2 \\ \\ \widetilde{\phi }_3 \end{array}\right)= & {} \left[ \begin{array}{ccc} \sqrt{\dfrac{1}{10}} &{} \sqrt{\dfrac{3}{10}} &{} -\sqrt{\dfrac{3}{5}} \\ \sqrt{\dfrac{2}{5}} &{} - \sqrt{\dfrac{8}{15}} &{} - \sqrt{\dfrac{1}{15}} \\ \sqrt{\dfrac{1}{2}} &{} \sqrt{\dfrac{1}{6}} &{} \sqrt{\dfrac{1}{3}} \end{array} \right] \left( \begin{array}{c} \Phi _1 \\ \\ \Phi _2 \\ \\ \Phi _3 \end{array}\right) \nonumber \\= & {} \left[ \begin{array}{ccc} -\dfrac{1}{5} &{} -\dfrac{2}{5} &{} \dfrac{2\sqrt{5}}{5} \\ -\dfrac{2}{5} &{} \dfrac{13}{15} &{} \dfrac{2\sqrt{5}}{15} \\ \dfrac{2\sqrt{5}}{5} &{} \dfrac{2\sqrt{5}}{15} &{} \dfrac{1}{3} \end{array} \right] \left( \begin{array}{c} \phi _1 \\ \\ \phi _2 \\ \\ \phi _3 \end{array}\right) . \end{aligned}$$

(27)

The fields contain to $\bar{\alpha }=(3,2,\frac{1}{6})$ are

$$\begin{aligned}&\widehat{A}^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{E}^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{D}^{(3,2,\frac{1}{6})}_{(10,-6)}, \widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{50},-2)}, \widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(\overline{45}, -2)},\\&\quad \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(75,0)}, \end{aligned}$$

they are related to the fields under $G_{422}$ and SU(5) by

$$\begin{aligned}&\widehat{A}^{(3,2,\frac{1}{6})}_{(24,0)} = \widehat{A}^{(3,2,\frac{1}{6})}_{(6,2,2)} = \widehat{A}^{(3,2,\frac{1}{6})}_{(10, 4)}, \\&\widehat{E}^{(3,2,\frac{1}{6})}_{(24,0)} = \widehat{E}^{(3,2,\frac{1}{6})}_{(6,2,2)} =\widehat{E}^{(3,2,\frac{1}{6})}_{(15,4)}, \\&\widehat{D}^{(3,2,\frac{1}{6})}_{(10,-6)}= \widehat{D}^{(3,2,\frac{1}{6})}_{(15,2,2)} =\widehat{D}^{(3,2,\frac{1}{6})}_{(10, -6)}, \\&\widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{50},-2)} = \widehat{\Delta }^{(3,2,\frac{1}{6})}_{(15,2,2)} =\widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{15}, 6)}, \\&\widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(\overline{45}, -2)} = \widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(15,2,2)} =\widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(15,-6)}, \end{aligned}$$

$$\begin{aligned} \left( \begin{array}{c} \Phi ^{(3,2,\frac{1}{6})}_{(24,0)} \\ \\ \Phi ^{(3,2, \frac{1}{6})}_{(75,0)} \end{array}\right)= & {} \left[ \begin{array}{cc} \sqrt{\dfrac{2}{3}} &{} \sqrt{\dfrac{1}{3}} \\ - \sqrt{\dfrac{1}{3}} &{} \sqrt{\dfrac{2}{3}} \end{array} \right] \left( \begin{array}{c} \Phi ^{(3,2,\frac{1}{6})}_{(6,2,2)} \\ \\ \Phi ^{(3,2,\frac{1}{6})}_{(10,2,2)} \end{array}\right) \nonumber \\= & {} \left[ \begin{array}{cc} \dfrac{2\sqrt{2}}{3} &{} \dfrac{1}{3} \\ \dfrac{1}{3} &{} -\dfrac{2\sqrt{2}}{3} \end{array} \right] \left( \begin{array}{c} \Phi ^{(3,2,\frac{1}{6})}_{(\overline{10},0)} \\ \\ \Phi ^{(3,2,\frac{1}{6})}_{(\overline{40},0)} \end{array}\right) . \end{aligned}$$

(28)

Note that $\widehat{D}_{(10,-6)}$ and $\widehat{\overline{\Delta }}{(\overline{45}, -2)}$ contain no VEV, so the Goldstone mode can be denoted as

$$\begin{aligned} \overrightarrow{G}_{(3,2,\frac{1}{6})}= & {} T^{\widetilde{a}_2}_{\bar{\alpha }} \widetilde{a}_2 \widehat{A}^{(3,2,\frac{1}{6})}_{(24,0)} + T^{\widetilde{S}}_{\bar{\alpha }} \widetilde{S} \widehat{E}^{(3,2,\frac{1}{6})}_{(24,0)} + T^{\delta }_{\bar{\alpha }} \delta \widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{50},-2)} \\&+\, T^{\overline{\delta }}_{\bar{\alpha }} \overline{\delta } \widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(\overline{45}, -2)} + T^{\widetilde{\phi }_2}_{\bar{\alpha }} \widetilde{\phi }_2 \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(24,0)} + T^{\widetilde{\phi }_3}_{\bar{\alpha }} \widetilde{\phi }_3 \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(75,0)}. \end{aligned}$$

The needed CGCs can be transformed from those under $G_{422}$ through (25)–(28), and are summarized in Tables 7 and 8. For completeness, the full mass matrix for [(3, 2, $\frac{1}{6}$) $+$ c.c.] can be also transformed through (25)–(28) and is given in the Appendix, for the readers to check the results of this section.

Table 7 CGCs for the SM singlets (1, 1, 0) through $G_{51}$. Here, for example, $\widetilde{\phi }_3$ stands for $\widehat{\Phi }_{(75,0)}^{(1,1,0)}$

Table 8 CGCs for the Goldstone (3, 2, $\frac{1}{6}$) $+$ c.c. of $G_{51}$ SSB

(1) According to (10), the coupling $\widetilde{\phi }_2^3$ gives

$$\begin{aligned} 0 = C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{\phi }_2}_{1~1~1} - 2 C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1} = \left( - \frac{7}{9\sqrt{15}}\right) - 2*\left( - \frac{7}{18\sqrt{15}}\right) , \end{aligned}$$

$\widetilde{\phi }_3^3$ gives

$$\begin{aligned} 0 = C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{\phi }_3}_{1~1~1} - 2 C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1} = \frac{8}{9\sqrt{3}} - 2*\frac{4}{9\sqrt{3}}, \end{aligned}$$

and $\widetilde{S}^3$ gives

$$\begin{aligned} 0 =C^{\widetilde{S}\widetilde{S}\widetilde{S}}_{111} - 2 C^{\widetilde{S}\widetilde{S}\widetilde{S}}_{\bar{\alpha } \alpha 1} = \sqrt{\frac{3}{5}} - 2 *\frac{\sqrt{3}}{2\sqrt{5}}. \end{aligned}$$

(2) $\widetilde{\phi }_1, \widetilde{a}_1$ contain neither Goldstones, the couplings $\widetilde{\phi }_2^2 \widetilde{\phi }_1 \widetilde{\phi }_3^2 \widetilde{\phi }_1 \widetilde{a}_2^2 \widetilde{\phi }_1 \widetilde{\phi }_2^2 \widetilde{a}_1 \widetilde{\phi }_3^2 \widetilde{a}_1$ lead to only trivial results. The coupling $\delta \overline{\delta } \widetilde{\phi }_2$, according to (8), gives

$$\begin{aligned} \frac{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}{T^{\delta }_{\bar{\alpha }}} = \frac{C^{ \delta \widetilde{\phi }_2 \overline{\delta }}_{\bar{\alpha }\alpha 1}}{C^{ \delta \widetilde{\phi }_2 \overline{\delta }}_{111}} = \left( -\frac{1}{30}\right) \Bigg / \left( - \frac{1}{5\sqrt{15}}\right) = \frac{\sqrt{15}}{6}, \end{aligned}$$

$\delta \overline{\delta } \widetilde{\phi }_3$ gives

$$\begin{aligned} \frac{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}{T^{\delta }_{\bar{\alpha }}} = \frac{C^{\delta \widetilde{\phi }_3 \overline{\delta }}_{\bar{\alpha }\alpha 1}}{C^{ \delta \widetilde{\phi }_3 \overline{\delta }}_{111}} = \left( -\frac{\sqrt{2}}{15}\right) \Bigg / \left( \frac{1}{5\sqrt{3}}\right) = - \frac{\sqrt{6}}{3}, \end{aligned}$$

$\delta \overline{\delta } \widetilde{a}_2$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\delta }_{\bar{\alpha }}} = \frac{C^{\delta \widetilde{a}_2 \overline{\delta }}_{\bar{\alpha }\alpha 1}}{C^{\delta \widetilde{a}_2 \overline{\delta }}_{111}} = \left( -\frac{1}{5}\right) \Bigg / \left( -\frac{2\sqrt{3}}{5\sqrt{5}}\right) = \frac{\sqrt{15}}{6}, \end{aligned}$$

$\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{a}_1$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}} = \frac{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{a}_1}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{a}_1}_{1~1~1}} = \left( \sqrt{\frac{2}{5}}\right) \Bigg / \left( \sqrt{\frac{2}{5}}\right) = 1, \end{aligned}$$

$\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_1$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}} = \frac{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_1}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_1}_{1~1~1}} = \left( \frac{2\sqrt{3}}{5\sqrt{5}}\right) \Bigg / \left( \frac{2\sqrt{3}}{5\sqrt{5}}\right) = 1, \end{aligned}$$

$\widetilde{S} \widetilde{a}_2 \widetilde{a}_1$ gives

$$\begin{aligned} \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{a}_2}_{\bar{\alpha }}} = \frac{C^{\widetilde{S} \widetilde{a}_2 \widetilde{a}_1}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{S} \widetilde{a}_2 \widetilde{a}_1}_{1~1~1}} = \left( \sqrt{\frac{2}{5}}\right) \Bigg / \left( \sqrt{\frac{2}{5}}\right) = 1, \end{aligned}$$

and $\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_1$ gives

$$\begin{aligned} \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}} = \frac{C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_1}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_1}_{1~1~1}} = \left( \frac{\sqrt{3}}{2\sqrt{5}}\right) \Bigg / \left( \frac{\sqrt{3}}{2\sqrt{5}}\right) = 1. \end{aligned}$$

(3) According to (9), $\widetilde{\phi }_2 \widetilde{\phi }_3^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{\phi }_2 \widetilde{\phi }_3 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{\phi }_2 \widetilde{\phi }_3 \widetilde{\phi }_3}_{1~1~1}} \\= & {} 2* \left( \frac{\sqrt{2}}{9\sqrt{3}}\right) \Bigg / \left( -\frac{8}{9\sqrt{15}}\right) = - \sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{\phi }_2}_{1~1~1} - C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{\phi }_3 \widetilde{\phi }_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{-8}{9\sqrt{15}} - \frac{-11}{18\sqrt{15}}\right) \Bigg / \left( \frac{\sqrt{2}}{9\sqrt{3}}\right) = - \sqrt{\frac{5}{8}}, \end{aligned}$$

$\widetilde{\phi }_2^2 \widetilde{\phi }_3$ gives

$$\begin{aligned} \frac{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_2 \widetilde{\phi }_3 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{\phi }_3}_{1~1~1} - C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( -\frac{\sqrt{5}}{9\sqrt{6}}\right) \Bigg / \left( \frac{5}{18\sqrt{3}} - \frac{1}{18\sqrt{3}}\right) = - \sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_3 \widetilde{\phi }_2 \widetilde{\phi }_2}_{1~1~1} }{2 C^{\widetilde{\phi }_3 \widetilde{\phi }_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{5}{18\sqrt{3}}\right) \Bigg / 2*\left( - \frac{\sqrt{5}}{9\sqrt{6}}\right) = - \sqrt{\frac{5}{8}}, \end{aligned}$$

$\widetilde{a}_2^2 \widetilde{\phi }_2$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{\phi }_2}_{1~1~1} - C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}}\\= & {} \left( - \frac{1}{3\sqrt{15}}\right) \Bigg / \left( \frac{-2}{3\sqrt{15}} - \frac{-1}{3\sqrt{15}}\right) = 1, \\ or{:}\ \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_2 \widetilde{a}_2 \widetilde{a}_2}_{1~1~1}}{2 C^{\widetilde{\phi }_2 \widetilde{a}_2 \widetilde{a}_2 }_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( -\frac{2}{3\sqrt{15}}\right) \Bigg / 2*\left( -\frac{1}{3\sqrt{15}}\right) = 1, \end{aligned}$$

$\widetilde{a}_2^2 \widetilde{\phi }_3$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{a}_2 \widetilde{\phi }_3 \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{\phi }_3}_{1~1~1} - C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}}\\= & {} \left( - \frac{\sqrt{10}}{3\sqrt{3}}\right) \Bigg / \left( \frac{5}{3\sqrt{3}} - \frac{1}{3\sqrt{3}}\right) = -\sqrt{\frac{5}{8}},\\ or{:}\ \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_3 \widetilde{a}_2 \widetilde{a}_2}_{1~1~1}}{2 C^{\widetilde{\phi }_3 \widetilde{a}_2 \widetilde{a}_2 }_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{5}{3\sqrt{3}}\right) \Bigg / 2*\left( -\frac{\sqrt{10}}{3\sqrt{3}}\right) = -\sqrt{\frac{5}{8}}, \end{aligned}$$

$\widetilde{a}_2 \widetilde{\phi }_2^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_2}_{1~1~1}} \\= & {} 2*\left( -\frac{4}{15\sqrt{15}}\right) \Bigg /\left( - \frac{8}{15\sqrt{15}}\right) = 1,\\ or{:}\ \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{a}_2}_{1~1~1} - C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1} }{C^{\widetilde{\phi }_2 \widetilde{a}_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( - \frac{8}{15\sqrt{15}} + \frac{4}{15\sqrt{15}}\right) \Bigg / \left( - \frac{4}{15\sqrt{15}}\right) = 1, \end{aligned}$$

$\widetilde{a}_2 \widetilde{\phi }_3^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{a}_2 \widetilde{\phi }_3 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{\phi }_3 \widetilde{\phi }_3}_{1~1~1}} \\= & {} 2*\left( - \frac{4\sqrt{2}}{15\sqrt{3}}\right) \Bigg /\left( \frac{32}{15\sqrt{15}}\right) = -\sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{\widetilde{a}_2}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{a}_2}_{1~1~1} - C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1} }{C^{\widetilde{\phi }_3 \widetilde{a}_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{32}{15\sqrt{15}} - \frac{22}{15\sqrt{15}}\right) \Bigg / \left( - \frac{4\sqrt{2}}{15\sqrt{3}}\right) = -\sqrt{\frac{5}{8}}, \end{aligned}$$

$\widetilde{S} \widetilde{a}_2^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{a}_2}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{S} \widetilde{a}_2 \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{S} \widetilde{a}_2 \widetilde{a}_2}_{1~1~1}}\\= & {} 2*\left( \frac{1}{2\sqrt{15}}\right) \Bigg / \left( \frac{1}{\sqrt{15}}\right) = 1, \\ or{:}\ \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{a}_2}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{S}}_{1~1~1} - C^{\widetilde{a}_2 \widetilde{a}_2 \widetilde{S}}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{a}_2 \widetilde{S} \widetilde{a}_2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{1}{\sqrt{15}} - \frac{1}{2\sqrt{15}} \right) \Bigg /\left( \frac{1}{2\sqrt{15}}\right) = 1, \end{aligned}$$

$\widetilde{S} \widetilde{\phi }_2^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_2}_{1~1~1}}\\= & {} 2*\left( \frac{1}{12\sqrt{15}}\right) \Bigg / \left( \frac{1}{6\sqrt{15}}\right) = 1, \\ or{:}\ \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{\phi }_2}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{S}}_{1~1~1} - C^{\widetilde{\phi }_2 \widetilde{\phi }_2 \widetilde{S}}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{\phi }_2 \widetilde{S} \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{1}{6\sqrt{15}} - \frac{1}{12\sqrt{15}}\right) \Bigg /\left( \frac{1}{12\sqrt{15}}\right) = 1, \end{aligned}$$

$\widetilde{S} \widetilde{\phi }_3^2$ gives

$$\begin{aligned} \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{2 C^{\widetilde{S} \widetilde{\phi }_3 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{S} \widetilde{\phi }_3 \widetilde{\phi }_3}_{1~1~1}} \\= & {} 2*\left( - \frac{1}{3\sqrt{6}}\right) \Bigg / \left( \frac{4}{3\sqrt{15}}\right) = - \sqrt{\frac{5}{8}}, \\ or{:}\ \frac{T^{\widetilde{S}}_{\bar{\alpha }}}{T^{\widetilde{\phi }_3}_{\bar{\alpha }}}= & {} \frac{C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{S}}_{1~1~1} - C^{\widetilde{\phi }_3 \widetilde{\phi }_3 \widetilde{S}}_{\bar{\alpha }~\alpha ~1}}{C^{\widetilde{\phi }_3 \widetilde{S} \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1}} \\= & {} \left( \frac{4}{3\sqrt{15}} - \frac{11}{12\sqrt{15}}\right) \Bigg /\left( -\frac{1}{3\sqrt{6}}\right) = - \sqrt{\frac{5}{8}}. \end{aligned}$$

(4) According to (7), $\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_3$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{ccc} - C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_3}_{1~1~1} &{} C^{\widetilde{a}_2 \widetilde{\phi }_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1} &{} C^{\widetilde{a}_2 \widetilde{\phi }_3 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\widetilde{\phi }_2 \widetilde{a}_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1} &{} - C^{\widetilde{\phi }_2 \widetilde{a}_2 \widetilde{\phi }_3}_{1~1~1} &{} C^{\widetilde{\phi }_2 \widetilde{\phi }_3 \widetilde{a}_2 }_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\widetilde{\phi }_3 \widetilde{a}_2 \widetilde{\phi }_2 }_{\bar{\alpha }~\alpha ~1} &{} C^{\widetilde{\phi }_3 \widetilde{\phi }_2 \widetilde{a}_2 }_{\bar{\alpha }~\alpha ~1} &{} -C^{\widetilde{\phi }_3 \widetilde{a}_2 \widetilde{\phi }_2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{\widetilde{a}_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_3}_{\bar{\alpha }} \end{array} \right) \\= & {} \left( \begin{array}{ccc} \dfrac{2}{3\sqrt{3}} &{} - \dfrac{2}{15\sqrt{3}} &{} \dfrac{2\sqrt{2}}{3\sqrt{15}} \\ \\ - \dfrac{2}{15\sqrt{3}} &{} \dfrac{2}{3\sqrt{3}} &{} \dfrac{2\sqrt{2}}{3\sqrt{15}} \\ \\ \dfrac{2\sqrt{2}}{3\sqrt{15}} &{} \dfrac{2\sqrt{2}}{3\sqrt{15}} &{} \dfrac{2}{3\sqrt{3}} \end{array} \right) \left( \begin{array}{c} T^{\widetilde{a}_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

so that

$$\begin{aligned} T^{\widetilde{a}_2}_{\bar{\alpha }} {:} T^{\widetilde{\phi }_2}_{\bar{\alpha }} {:} T^{\widetilde{\phi }_3}_{\bar{\alpha }} = 1 {:} 1 {:} \left( -\sqrt{\frac{8}{5}}\right) , \end{aligned}$$

and $\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_3$ gives

$$\begin{aligned} 0= & {} \left( \begin{array}{ccc} - C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_3}_{1~1~1} &{} C^{\widetilde{S} \widetilde{\phi }_2 \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1} &{} C^{\widetilde{S} \widetilde{\phi }_3 \widetilde{\phi }_2}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\widetilde{\phi }_2 \widetilde{S} \widetilde{\phi }_3}_{\bar{\alpha }~\alpha ~1} &{} - C^{\widetilde{\phi }_2 \widetilde{S} \widetilde{\phi }_3}_{1~1~1} &{} C^{\widetilde{\phi }_2 \widetilde{\phi }_3 \widetilde{S}}_{\bar{\alpha }~\alpha ~1} \\ \\ C^{\widetilde{\phi }_3 \widetilde{S} \widetilde{\phi }_2 }_{\bar{\alpha }~\alpha ~1} &{} C^{\widetilde{\phi }_3 \widetilde{\phi }_2 \widetilde{S} }_{\bar{\alpha }~\alpha ~1} &{} -C^{\widetilde{\phi }_3 \widetilde{S} \widetilde{\phi }_2}_{1~1~1} \end{array} \right) \left( \begin{array}{c} T^{\widetilde{S}}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_3}_{\bar{\alpha }} \end{array} \right) \\ {}= & {} \left( \begin{array}{ccc} - \dfrac{5}{6\sqrt{3}} &{} \dfrac{1}{6\sqrt{3}} &{} - \dfrac{\sqrt{5}}{3\sqrt{6}} \\ \\ \dfrac{1}{6\sqrt{3}} &{} - \dfrac{5}{6\sqrt{3}} &{} - \dfrac{\sqrt{5}}{3\sqrt{6}} \\ \\ - \dfrac{\sqrt{5}}{3\sqrt{6}} &{} - \dfrac{\sqrt{5}}{3\sqrt{6}} &{} - \dfrac{5}{6\sqrt{3}} \end{array} \right) \left( \begin{array}{c} T^{\widetilde{S}}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_2}_{\bar{\alpha }} \\ \\ T^{\widetilde{\phi }_3}_{\bar{\alpha }} \end{array} \right) , \end{aligned}$$

so that

$$\begin{aligned} T^{\widetilde{S}}_{\bar{\alpha }} {:} T^{\widetilde{\phi }_2}_{\bar{\alpha }} {:} T^{\widetilde{\phi }_3}_{\bar{\alpha }} = 1 {:} 1 {:} \left( - \sqrt{\frac{8}{5}}\right) . \end{aligned}$$

Altogether, the Goldstino is consistently

$$\begin{aligned} \overrightarrow{G}_{(3,2,\frac{1}{6})}= & {} \widetilde{a}_2 \widehat{A}^{(3,2,\frac{1}{6})}_{(24,0)} + \widetilde{S} \widehat{E}^{(3,2,\frac{1}{6})}_{(24,0)} + \frac{2\sqrt{15}}{5} \delta \widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{50},-2)}\nonumber \\&+\,\widetilde{\phi }_2 \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(24,0)} -\sqrt{\frac{8}{5}} \widetilde{\phi }_3 \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(75,0)}. \end{aligned}$$

(29)

7 Summary

We have studied the Goldstone modes of SSB in the SUSY SO(10) model with general renormalizable couplings. VEVs and CGCs determine the contents of these Goldstone modes while the parameters of the model are irrelevant. Identities among the CGCs are examined to be in accord with the general conclusions in [38].

Appendix: Mass matrix for [(3, 2, \(\frac{1}{6}\)) \(+\) c.c.]

For the SSB of $G_{422}$ and SU(5), the mass matrices for the Goldstone representations can be found in [29, 36], respectively. Here we give the mass matrix for [(3, 2, $\frac{1}{6}$) + c.c.] using $G_{51}$ as the maximal subgroup of SO(10). This matrix is transformed from [29] through (25)–(28). $[\mathbf{{(3,2}}, \frac{1}{6}) +c.c.]$

$$\begin{aligned}&\mathbf{c}:\widehat{A}^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{E}^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{D}^{(3,2,\frac{1}{6})}_{(10, -6)}, \widehat{\Delta }^{(3,2,\frac{1}{6})}_{(\overline{50},-2)},\\&\quad \widehat{\overline{\Delta }}^{(3,2,\frac{1}{6})}_{(\overline{45},-2)}, \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(24,0)}, \widehat{\Phi }^{(3,2,\frac{1}{6})}_{(75,0)}\\&\mathbf{r}:\widehat{A}^{(\overline{3},2,-\frac{1}{6})}_{(24,0)}, \widehat{E}^{(\overline{3},2,-\frac{1}{6})}_{(24,0)}, \widehat{D}^{(\overline{3},2,-\frac{1}{6})}_{(\overline{10}, 6)}, \widehat{\overline{\Delta }}^{(\overline{3},2,-\frac{1}{6})}_{(50,2)},\\&\quad \widehat{\Delta }^{(\overline{3},2,-\frac{1}{6})}_{(45,2)}, \widehat{\Phi }^{(\overline{3},2,-\frac{1}{6})}_{(24,0)}, \widehat{\Phi }^{(\overline{3},2,-\frac{1}{6})}_{(75,0)} \end{aligned}$$

$$\begin{aligned} \left( \begin{array}{ccccccc} m_{11} &{} \sqrt{\frac{2}{5}} \widetilde{a}_1 \lambda _9+\frac{\widetilde{a}_2 \lambda _9}{2 \sqrt{15}} &{} -\frac{i {\overline{\delta }} \lambda _{19}}{\sqrt{10}} &{} -\frac{{\overline{\delta }} \lambda _6}{5} &{} 0 &{} m_{16} &{} m_{17} \\ \sqrt{\frac{2}{5}} \widetilde{a}_1 \lambda _9+\frac{\widetilde{a}_2 \lambda _9}{2 \sqrt{15}} &{} m_5+\frac{1}{2} \sqrt{\frac{3}{5}} \widetilde{S} \lambda _8 &{} 0 &{} 0 &{} -\frac{2 {\delta } \lambda _{12}}{5} &{} m_{26} &{} m_{27} \\ \frac{i {\delta } \lambda _{18}}{\sqrt{10}} &{} 0 &{} m_{33} &{} m_{34} &{} m_{35} &{} -\frac{{\delta } \lambda _{20}}{6 \sqrt{10}} &{} -\frac{{\delta } \lambda _{20}}{3 \sqrt{5}} \\ -\frac{{\delta } \lambda _6}{5} &{} 0 &{} m_{43} &{} m_{44} &{} \frac{\widetilde{S} \lambda _{12}}{\sqrt{15}} &{} -\frac{{\delta } \lambda _2}{30} &{} -\frac{\sqrt{2}}{15} {\delta } \lambda _2 \\ 0 &{} -\frac{2 {\delta }\lambda _{11}}{5} &{} m_{53} &{} \frac{\widetilde{S} \lambda _{11}}{\sqrt{15}} &{} m_{55} &{} 0 &{} 0 \\ m_{16} &{} m_{26} &{} -\frac{\overline{{\delta }} \lambda _{21}}{6 \sqrt{10}} &{} -\frac{\overline{{\delta }} \lambda _2}{30} &{} 0 &{} m_{66} &{} m_{67} \\ m_{17} &{} m_{27} &{} -\frac{\overline{{\delta }} \lambda _{21}}{3 \sqrt{5}} &{} -\frac{\sqrt{2}}{15} \overline{{\delta }} \lambda _2 &{} 0 &{} m_{67} &{} m_{77} \\ \end{array} \right) , \end{aligned}$$

where:

$$\begin{aligned} m_{11}= & {} m_4+\frac{{\delta } \lambda _9}{2 \sqrt{15}}-\frac{\lambda _5\widetilde{\phi }_1}{\sqrt{15}}-\frac{\lambda _5 \widetilde{\phi }_2}{3 \sqrt{15}}+\frac{\lambda _5 \widetilde{\phi }_3}{3 \sqrt{3}}, \\ m_{16}= & {} \sqrt{\frac{2}{5}} \widetilde{a}_1 \lambda _5-\frac{\widetilde{a}_2 \lambda _5}{3 \sqrt{15}}+\frac{2}{5} \sqrt{\frac{3}{5}} \lambda _7 \widetilde{\phi }_1-\frac{4 \lambda _7 \widetilde{\phi }_2}{15 \sqrt{15}}-\frac{2 \lambda _7 \widetilde{\phi }_3}{15 \sqrt{3}},\\ m_{17}= & {} -\frac{1}{3} \sqrt{\frac{10}{3}} \widetilde{a}_2 \lambda _5+\frac{2}{3} \sqrt{\frac{2}{15}} \lambda _7 \widetilde{\phi }_2-\frac{4}{15} \sqrt{\frac{2}{3}} \lambda _7 \widetilde{\phi }_3, \\ m_{26}= & {} \frac{1}{2} \sqrt{\frac{3}{5}} \lambda _{10} \widetilde{\phi }_1+\frac{\lambda _{10} \widetilde{\phi }_2}{12 \sqrt{15}}+\frac{\lambda _{10} \widetilde{\phi }_3}{6\sqrt{3}},\\ m_{27}= & {} -\frac{1}{3} \sqrt{\frac{5}{6}} \lambda _{10} \widetilde{\phi }_2-\frac{\lambda _{10} \widetilde{\phi }_3}{3 \sqrt{6}},\\ m_{33}= & {} m_6-\frac{{\delta } \lambda _{14}}{3 \sqrt{15}}-\frac{\lambda _{15} \widetilde{\phi }_1}{3 \sqrt{15}}-\frac{2 \lambda _{15}\widetilde{\phi }_2}{3 \sqrt{15}}+\frac{\lambda _{15}\widetilde{\phi }_3}{3 \sqrt{3}},\\ m_{34}= & {} -\frac{i \widetilde{a}_2 \lambda _{18}}{2 \sqrt{6}}+\frac{\lambda _{20} \widetilde{\phi }_2}{12 \sqrt{6}}-\frac{1}{3} \sqrt{\frac{2}{15}} \lambda _{20} \widetilde{\phi }_3, \\ m_{35}= & {} -\frac{1}{5} i \widetilde{a}_1 \lambda _{19}-\frac{i \widetilde{a}_2 \lambda _{19}}{10 \sqrt{6}}+\frac{\lambda _{21} \widetilde{\phi }_1}{10\sqrt{6}}-\frac{13 \lambda _{21} \widetilde{\phi }_2}{60 \sqrt{6}}-\frac{\lambda _{21} \widetilde{\phi }_3}{6 \sqrt{30}}, \\ m_{43}= & {} \frac{i \widetilde{a}_2 \lambda _{19}}{2 \sqrt{6}}+\frac{\lambda _{21} \widetilde{\phi }_2}{12 \sqrt{6}}-\frac{1}{3} \sqrt{\frac{2}{15}} \lambda _{21} \widetilde{\phi }_3, \\ m_{44}= & {} m_2+\frac{\widetilde{a}_1 \lambda _6}{5 \sqrt{10}}-\frac{7 \widetilde{a}_2 \lambda _6}{10 \sqrt{15}}-\frac{\lambda _2 \widetilde{\phi }_1}{10 \sqrt{15}}-\frac{7 \lambda _2 \widetilde{\phi }_2}{60 \sqrt{15}}+ \frac{\lambda _2 \widetilde{\phi }_3}{15 \sqrt{3}}, \\ m_{53}= & {} \frac{1}{5} i \widetilde{a}_1 \lambda _{18}+\frac{i \widetilde{a}_2 \lambda _{18}}{10 \sqrt{6}}+\frac{\lambda _{20} \widetilde{\phi }_1}{10\sqrt{6}}-\frac{13 \lambda _{20} \widetilde{\phi }_2}{60 \sqrt{6}}-\frac{\lambda _{20} \widetilde{\phi }_3}{6 \sqrt{30}}, \\ m_{55}= & {} m_2-\frac{\widetilde{a}_1 \lambda _6}{5 \sqrt{10}}+\frac{7 \widetilde{a}_2 \lambda _6}{10 \sqrt{15}}-\frac{\lambda _2 \widetilde{\phi }_2}{12 \sqrt{15}}+\frac{\lambda _2 \widetilde{\phi }_3}{30 \sqrt{3}}, \\ m_{66}= & {} m_1+\frac{4}{5} \sqrt{\frac{2}{5}} \widetilde{a}_1 \lambda _7-\frac{4 \widetilde{a}_2 \lambda _7}{15 \sqrt{15}}+\frac{{\delta } \lambda _{10}}{12 \sqrt{15}}+\frac{\lambda _1 \widetilde{\phi }_1}{2 \sqrt{15}}\\&-\,\frac{7 \lambda _1 \widetilde{\phi }_2}{18 \sqrt{15}} +\frac{\lambda _1 \widetilde{\phi }_3}{18 \sqrt{3}}, \\ m_{67}= & {} \frac{2}{3} \sqrt{\frac{2}{15}} \widetilde{a}_2 \lambda _7-\frac{1}{3} \sqrt{\frac{5}{6}} \widetilde{S} \lambda _{10}-\frac{1}{9} \sqrt{\frac{5}{6}} \lambda _1 \widetilde{\phi }_2+\frac{1}{9} \sqrt{\frac{2}{3}} \lambda _1 \widetilde{\phi }_3, \\ m_{77}= & {} m_1-\frac{2}{5} \sqrt{\frac{2}{5}} \widetilde{a}_1 \lambda _7+\frac{22 \widetilde{a}_2 \lambda _7}{15 \sqrt{15}}+\frac{11 {\delta } \lambda _{10}}{12 \sqrt{15}}\\&-\,\frac{\lambda _1 \widetilde{\phi }_1}{\sqrt{15}}-\frac{11 \lambda _1 \widetilde{\phi }_2}{18 \sqrt{15}}+\frac{4 \lambda _1 \widetilde{\phi }_3}{9 \sqrt{3}}. \end{aligned}$$