Spontaneous Breaking of Gauge Groups to Discrete Symmetries

TL;DR

This paper explores how gauge theories can spontaneously break to discrete symmetries, ensuring anomaly freedom and compatibility with quantum gravity, by analyzing Higgs potentials and scalar spectra.

Contribution

It introduces methods to minimize Higgs potentials for gauge breaking to discrete groups and studies the resulting scalar mass spectra, advancing understanding of symmetry breaking mechanisms.

Findings

Successful minimization of Higgs potentials for specific gauge breakings

Identification of scalar mass spectra after symmetry breaking

Framework for connecting gauge theories with discrete symmetries

Abstract

Many models of beyond Standard Model physics connect flavor symmetry with a discrete group. Having this symmetry arise spontaneously from a gauge theory maintains compatibility with quantum gravity and can be used to systematically prevent anomalies. We minimize a number of Higgs potentials that break gauge groups to discrete symmetries of interest, and examine their scalar mass spectra.

Tables

Table 1. Table 1: Scalar mass eigenstates for the SSB pattern S ​ O ​ ( 3 ) → A 4 → 𝑆 𝑂 3 subscript 𝐴 4 SO(3)\rightarrow A_{4} using a real 7 of S ​ O ​ ( 3 ) 𝑆 𝑂 3 SO(3) .

Value	Multiplicity
0	3
$4m^2$	1
$\frac{8m^2\kappa }{5\left(3\lambda +\kappa \right)}$	3

Table 2. Table 2: Scalar mass eigenstates for the SSB pattern S ​ O ​ ( 3 ) → S 4 → 𝑆 𝑂 3 subscript 𝑆 4 SO(3)\rightarrow S_{4} using a real 9 of S ​ O ​ ( 3 ) 𝑆 𝑂 3 SO(3) .

Value	Multiplicity
0	3
$4m^2$	1
$\frac{5m^2\left(5\kappa +8\rho -2\tau \right)}{7\left(90\lambda +10\kappa +7\rho +2\tau \right)}$	3
$\frac{20m^2\left(5\kappa +8\rho -2\tau \right)}{7\left(90\lambda +10\kappa +7\rho +2\tau \right)}$	2

Table 3. Table 3: Scalar mass eigenstates for the SSB pattern S ​ O ​ ( 3 ) → A 5 → 𝑆 𝑂 3 subscript 𝐴 5 SO(3)\rightarrow A_{5} using a real 13 of S ​ O ​ ( 3 ) 𝑆 𝑂 3 SO(3) .

Value	Multiplicity
0	3
$4m^2$	1
$\frac{28m^2\left(105\kappa +196\rho +240\tau -14\nu -19\sigma +12\chi \right)}{33(420\lambda +140\kappa +84\rho +65\tau +14\nu +9\sigma -2\chi }$	5
$\frac{28m^2\left(14\rho +45\tau +14\nu -11\sigma +18\chi \right)}{33(420\lambda +140\kappa +84\rho +65\tau +14\nu +9\sigma -2\chi }$	4

Table 4. Table 4: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 2 ) → Q 6 → 𝑆 𝑈 2 subscript 𝑄 6 SU(2)\rightarrow Q_{6} using a complex 7 of S ​ U ​ ( 2 ) 𝑆 𝑈 2 SU(2) .

Value	Multiplicity
0	4
$4m^2$	1
$\frac{2m^2\kappa }{3\left(2\lambda +\kappa +\rho +\tau \right)}$	2
$\frac{4m^2\left(\kappa +\rho +\tau \right)}{2\lambda +\kappa +\rho +\tau }$	1
$\frac{-3m^2\left(2\rho +3\tau \right)}{5\left(2\lambda +\kappa +\rho +\tau \right)}$	1
$\frac{-2m^2\left(2\rho +3\tau \right)}{5\left(2\lambda +\kappa +\rho +\tau \right)}$	2
$\frac{-m^2\left(6\rho +7\tau \right)}{5\left(2\lambda +\kappa +\rho +\tau \right)}$	1
$\frac{-2m^2\left(8\rho +9\tau \right)}{15\left(2\lambda +\kappa +\rho +\tau \right)}$	2

Table 5. Table 5: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 2 ) → T ′ → 𝑆 𝑈 2 superscript 𝑇 ′ SU(2)\rightarrow T^{\prime} using a complex 7 of S ​ U ​ ( 2 ) 𝑆 𝑈 2 SU(2) .

Value	Multiplicity
0	4
$4m^2$	1
$\frac{12m^2\tau }{5\left(12\lambda +6\kappa +4\rho +3\tau \right)}$	3
$\frac{16m^2\left(2\rho +3\tau \right)}{5\left(12\lambda +6\kappa +4\rho +3\tau \right)}$	3
$\frac{4m^2\left(8\kappa +8\rho +9\tau \right)}{3\left(12\lambda +6\kappa +4\rho +3\tau \right)}$	3

Table 6. Table 6: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 2 ) → O ′ → 𝑆 𝑈 2 superscript 𝑂 ′ SU(2)\rightarrow O^{\prime} using a complex 9 of S ​ U ​ ( 2 ) 𝑆 𝑈 2 SU(2) .

Value	Multiplicity
0	4
$4m^2$	1
$\frac{-24m^2\sigma }{7\left(60\lambda +30\kappa +20\rho +15\tau +14\sigma \right)}$	2
$\frac{5m^2\left(10\rho +15\tau +16\sigma \right)}{7\left(60\lambda +30\kappa +20\rho +15\tau +14\sigma \right)}$	3
$\frac{20m^2\left(10\rho +15\tau +16\sigma \right)}{7\left(60\lambda +30\kappa +20\rho +15\tau +14\sigma \right)}$	2
$\frac{2m^2\left(25\kappa +25\rho +25\tau +24\sigma \right)}{60\lambda +30\kappa +20\rho +15\tau +14\sigma }$	3
$\frac{3m^2\left(25\tau +32\sigma \right)}{7\left(60\lambda +30\kappa +20\rho +15\tau +14\sigma \right)}$	3

Table 7. Table 7: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 2 ) → I ′ → 𝑆 𝑈 2 superscript 𝐼 ′ SU(2)\rightarrow I^{\prime} using a complex 13 of S ​ U ​ ( 2 ) 𝑆 𝑈 2 SU(2) .

Value	Multiplicity
0	4
$4m^2$	1
$\frac{28m^2\left(14\nu +35\sigma +45\chi \right)}{33\left(420\lambda +210\kappa +140\rho +105\tau +84\nu +70\sigma +65\chi \right)}$	4
$\frac{5m^2\left(49\sigma +72\chi \right)}{33\left(420\lambda +210\kappa +140\rho +105\tau +84\nu +70\sigma +65\chi \right)}$	5
$\frac{14m^2\left(210\rho +315\tau +392\nu +455\sigma +480\chi \right)}{33\left(420\lambda +210\kappa +140\rho +105\tau +84\nu +70\sigma +65\chi \right)}$	5
$\frac{m^2\left(980\kappa +980\rho +882\tau +784\nu +735\sigma +720\chi \right)}{3\left(420\lambda +210\kappa +140\rho +105\tau +84\nu +70\sigma +65\chi \right)}$	3
$\frac{4m^2\left(441\tau +882\nu +1225\sigma +1350\chi \right)}{33\left(420\lambda +210\kappa +140\rho +105\tau +84\nu +70\sigma +65\chi \right)}$	4

Table 8. Table 8: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 3 ) → A 4 → 𝑆 𝑈 3 subscript 𝐴 4 SU(3)\rightarrow A_{4} using a 15 of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	9
$4m^2$	1
$\frac{m^2\left(-2\eta +\kappa -2\rho +4\tau \right)}{3\lambda +\eta +\kappa +\rho +\tau }$	2
$\frac{-3m^2\eta }{3\lambda +\eta +\kappa +\rho +\tau }$	6
$\frac{m^2}{4\left(3\lambda +\eta +\kappa +\rho +\tau \right)}\left(5\kappa +2\rho +4\tau +\sqrt{{\left(4\tau +2\rho -3\kappa \right)}^2+16{\left(\rho +\kappa +2\eta \right)}^2}\right)$	3
$\frac{m^2}{4\left(3\lambda +\eta +\kappa +\rho +\tau \right)}\left(5\kappa +2\rho +4\tau -\sqrt{{\left(4\tau +2\rho -3\kappa \right)}^2+16{\left(\rho +\kappa +2\eta \right)}^2}\right)$	3
$\frac{m^2}{2\left(3\lambda +\eta +\kappa +\rho +\tau \right)}\left(3\kappa -5\eta -2\rho +4\tau +\frac{1}{3}\sqrt{{\left(9\eta -7\kappa +10\rho -4\tau \right)}^2+8{\left(\rho +2\kappa -4\tau \right)}^2}\right)$	3
$\frac{m^2}{2\left(3\lambda +\eta +\kappa +\rho +\tau \right)}\left(3\kappa -5\eta -2\rho +4\tau -\frac{1}{3}\sqrt{{\left(9\eta -7\kappa +10\rho -4\tau \right)}^2+8{\left(\rho +2\kappa -4\tau \right)}^2}\right)$	3

Table 9. Table 9: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 3 ) → T 7 → 𝑆 𝑈 3 subscript 𝑇 7 SU(3)\rightarrow T_{7} using a 15 of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	9
$4m^2$	1
$\frac{2\left(2\kappa -\rho +2\tau \right)m^2}{\kappa +3\lambda +\rho +\tau }$	2
$\frac{m^2}{12\left(3\lambda +\kappa +\rho +\tau \right)}\times \alpha$	6
$\frac{m^2}{12\left(3\lambda +\kappa +\rho +\tau \right)}\times \beta$	6
$\frac{m^2}{12\left(3\lambda +\kappa +\rho +\tau \right)}\times \gamma$	6

Table 10. Table 10: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 3 ) → Δ ​ ( 27 ) → 𝑆 𝑈 3 Δ 27 SU(3)\rightarrow\Delta(27) using a 10 of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	11
$4m^2$	1
$\frac{4\kappa m^2}{3\lambda +\kappa }$	2
$\frac{\kappa m^2}{3\left(3\lambda +\kappa \right)}$	6

Table 11. Table 11: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 3 ) → P ​ S ​ L ​ ( 2 , 7 ) → 𝑆 𝑈 3 𝑃 𝑆 𝐿 2 7 SU(3)\rightarrow PSL(2,7) using a 𝟏𝟓 ′ superscript 15 ′ {\bf 15^{\prime}} of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	9
$4m^2$	1
$\frac{\left(7\kappa +8\rho \right)m^2}{2\left(6\lambda +2\kappa +\rho \right)}$	8
$\frac{2\left(3-\sqrt{2}\right)\rho m^2}{3\left(6\lambda +2\kappa +\rho \right)}$	6
$\frac{2\left(3+\sqrt{2}\right)\rho m^2}{3\left(6\lambda +2\kappa +\rho \right)}$	6

Table 12. Table 12: Numerical results where cubic terms are included for the scalar mass eigenstates of the SSB pattern S ​ U ​ ( 3 ) → P ​ S ​ L ​ ( 2 , 7 ) → 𝑆 𝑈 3 𝑃 𝑆 𝐿 2 7 SU(3)\rightarrow PSL(2,7) using a 𝟏𝟓 ′ superscript 15 ′ {\bf 15^{\prime}} of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	8
$4.002$	1
$.006003$	1
$0.323647$	6
$0.125244$	6
$0.838169$	8

Table 13. Table 13: Scalar mass eigenstates for the SSB pattern S ​ U ​ ( 3 ) → P ​ S ​ L ​ ( 2 , 7 ) → 𝑆 𝑈 3 𝑃 𝑆 𝐿 2 7 SU(3)\rightarrow PSL(2,7) using a 28 of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

Value	Multiplicity
0	16
$4m^2$	1
$\frac{4\left(7\rho +9\tau \right)m^2}{5\left(42\lambda +14\kappa +7\rho +6\tau \right)}$	7
$\frac{\left(21\kappa +20\rho +18\tau \right)m^2}{42\lambda +14\kappa +7\rho +6\tau }$	8
$\frac{1}{200{\left(42\lambda +14\kappa +7\rho +6\tau \right)}^2}\times A$	6
$\frac{1}{200{\left(42\lambda +14\kappa +7\rho +6\tau \right)}^2}\times B$	6
$\frac{1}{200{\left(42\lambda +14\kappa +7\rho +6\tau \right)}^2}\times C$	6
$\frac{1}{200{\left(42\lambda +14\kappa +7\rho +6\tau \right)}^2}\times D$	6

Table 14. Table 14: Scalar mass eigenstates for the SSB patterns S ​ O ​ ( 3 ) → A 4 , → 𝑆 𝑂 3 subscript 𝐴 4 SO(3)\rightarrow A_{4}, S 4 subscript 𝑆 4 S_{4} and A 5 subscript 𝐴 5 A_{5} using real irreps of S ​ O ​ ( 3 ) 𝑆 𝑂 3 SO(3) .

SSB pattern	decomposition
$SO\left(3\right)\to A_4$	$\mathrm{𝟕}\to \mathrm{𝟏}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟑}$
$SO\left(3\right)\to S_4$	$\mathrm{𝟗}\to \mathrm{𝟏}+\mathrm{𝟐}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟑}$
$SO\left(3\right)\to A_5$	$\mathrm{𝟏𝟑}\to \mathrm{𝟏}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟒}+\mathrm{𝟓}$

Table 15. Table 15: Scalar mass eigenstates for the SSB patterns S ​ U ​ ( 2 ) → Q 6 , → 𝑆 𝑈 2 subscript 𝑄 6 SU(2)\rightarrow Q_{6}, T ′ superscript 𝑇 ′ T^{\prime} , O ′ superscript 𝑂 ′ O^{\prime} and I ′ superscript 𝐼 ′ I^{\prime} using complexified irreps of S ​ O ​ ( 3 ) 𝑆 𝑂 3 SO(3) .

SSB pattern	decomposition
$SU\left(2\right)\to Q_6$	${\mathrm{𝟕}}_{𝐜}\to \mathrm{𝟏}+{\mathrm{𝟏}}^{\prime }+{\left({\mathrm{𝟏}}^{\prime }+{\mathrm{𝟐}}^{\prime }\right)}_{{}_{𝐆}}+{\mathrm{𝟐}}^{\prime }+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+{\mathrm{𝟏}}^{\prime }+{\mathrm{𝟏}}^{\prime }+{\mathrm{𝟐}}^{\prime }+{\mathrm{𝟐}}^{\prime }$
$SU\left(2\right)\to T^{\prime }$	${\mathrm{𝟕}}_{𝐜}\to \mathrm{𝟏}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟑}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟑}+\mathrm{𝟑}$
$SU\left(2\right)\to O^{\prime }$	${\mathrm{𝟗}}_{𝐜}\to \mathrm{𝟏}+\mathrm{𝟐}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟑}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟐}+\mathrm{𝟑}+\mathrm{𝟑}$
$SU\left(2\right)\to I^{\prime }$	${\mathrm{𝟏𝟑}}_{𝐜}\to \mathrm{𝟏}+{\mathrm{𝟑}}_{{}_{𝐆}}+\mathrm{𝟒}+\mathrm{𝟓}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟑}+\mathrm{𝟒}+\mathrm{𝟓}$

Table 16. Table 16: Scalar mass eigenstates for the SSB patterns S ​ U ​ ( 3 ) → A 4 , → 𝑆 𝑈 3 subscript 𝐴 4 SU(3)\rightarrow A_{4}, T 7 subscript 𝑇 7 T_{7} , Δ ​ ( 27 ) Δ 27 \Delta(27) and P ​ S ​ L ​ ( 2 , 7 ) 𝑃 𝑆 𝐿 2 7 PSL(2,7) using various complex irreps of S ​ U ​ ( 3 ) 𝑆 𝑈 3 SU(3) .

SSB pattern	decomposition
$SU\left(3\right)\to A_4$	$\mathrm{𝟏𝟓}\to \mathrm{𝟏}+{\left({\mathrm{𝟏}}^{\prime }+{\mathrm{𝟏}}^{\prime \prime }+\mathrm{𝟑}+\mathrm{𝟑}\right)}_{{}_{𝐆}}+{\left(\mathrm{𝟑}+\mathrm{𝟑}\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+{\left({\mathrm{𝟏}}^{\prime }+{\mathrm{𝟏}}^{\prime \prime }\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}+\mathrm{𝟑}+\mathrm{𝟑}+\mathrm{𝟑}+\mathrm{𝟑}$
$SU\left(3\right)\to T_7$	$\mathrm{𝟏𝟓}⟶\mathrm{𝟏}+{\left({\mathrm{𝟏}}^{\prime }+{\mathrm{𝟏}}^{\prime \prime }+{\mathrm{𝟑}}^{\prime }+{\mathrm{𝟑}}^{\prime \prime }\right)}_{{}_{𝐆}}+{\left({\mathrm{𝟑}}^{\prime }+{\mathrm{𝟑}}^{\prime \prime }\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}$
	$+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+{\left({\mathrm{𝟏}}^{\prime }+{\mathrm{𝟏}}^{\prime \prime }\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}+{\left({\mathrm{𝟑}}^{\prime }+{\mathrm{𝟑}}^{\prime \prime }\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}+{\left({\mathrm{𝟑}}^{\prime }+{\mathrm{𝟑}}^{\prime \prime }\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}$
$SU\left(3\right)\to \Delta \left(27\right)$	$\mathrm{𝟏𝟎}\to \mathrm{𝟏}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+{\left({𝚺}_{𝐧=\mathrm{𝟐}}^{\mathrm{𝟗}}\right)}_{{}_{𝐆}}+\mathrm{𝟑}+{\left(\mathrm{𝟏}+\mathrm{𝟏}\right)}_{{}_{\mathrm{𝐩𝐆𝐁}}}+{\left({\mathrm{𝟏}}_{\mathrm{𝟐}}+{\mathrm{𝟏}}_{\mathrm{𝟑}}\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}+{\left({𝚺}_{𝐧=\mathrm{𝟒}}^{\mathrm{𝟗}}\right)}_{{}_{\mathrm{𝐝𝐞𝐠}.}}$
$SU\left(3\right)\to PSL(2,7)$	${\mathrm{𝟏𝟓}}^{\prime }\to \mathrm{𝟏}+\mathrm{𝟔}+{\mathrm{𝟖}}_{{}_{𝐆}}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟔}+\mathrm{𝟖}$
$SU\left(3\right)\to PSL(2,7)$	$\mathrm{𝟐𝟖}\to \mathrm{𝟏}+\mathrm{𝟔}+\mathrm{𝟔}+\mathrm{𝟕}+{\mathrm{𝟖}}_{{}_{𝐆}}+{\mathrm{𝟏}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟔}+\mathrm{𝟔}+{\mathrm{𝟕}}_{{}_{\mathrm{𝐩𝐆𝐁}}}+\mathrm{𝟖}$

Table 17. Table 17: S ​ O ​ ( 3 ) → A 4 → 𝑆 𝑂 3 subscript 𝐴 4 SO(3)\rightarrow A_{4}

Dimension	${\mathrm{𝟏}}_{\mathrm{𝟏}}$	${\mathrm{𝟏}}_{\mathrm{𝟐}}$	${\mathrm{𝟏}}_{\mathrm{𝟑}}$	$\mathrm{𝟑}$
2	0	0	0	0
3	0	0	0	1
4	0	0	0	0
5	0	1	1	1
6	0	0	0	0
7	1	0	0	2
8	0	0	0	0
9	1	1	1	2
10	0	0	0	0
11	0	1	1	3

Table 18. Table 19: S ​ O ​ ( 3 ) → A 5 → 𝑆 𝑂 3 subscript 𝐴 5 SO(3)\rightarrow A_{5}

Dimension	$\mathrm{𝟏}$	$\mathrm{𝟑}$	$\mathrm{𝟑}$	$\mathrm{𝟒}$	$\mathrm{𝟓}$
2	0	0	0	0	0
3	0	0	1	0	0
4	0	0	0	0	0
5	0	0	0	0	1
6	0	0	0	0	0
7	0	1	0	1	0
8	0	0	0	0	0
9	0	0	0	1	1
10	0	0	0	0	0
11	0	1	1	0	1
12	0	0	0	0	0
13	1	0	1	1	1

Table 19. Table 21: S ​ U ​ ( 2 ) → T ′ → 𝑆 𝑈 2 superscript 𝑇 ′ SU(2)\rightarrow T^{\prime}

Dimension	${\mathrm{𝟏}}_{\mathrm{𝟏}}$	${\mathrm{𝟏}}_{\mathrm{𝟐}}$	${\mathrm{𝟏}}_{\mathrm{𝟑}}$	${\mathrm{𝟐}}_{\mathrm{𝟏}}$	${\mathrm{𝟐}}_{\mathrm{𝟐}}$	${\mathrm{𝟐}}_{\mathrm{𝟑}}$	$\mathrm{𝟑}$
2	0	0	0	1	0	0	0
3	0	0	0	0	0	0	1
4	0	0	0	0	1	1	0
5	0	1	1	0	0	0	1
6	0	0	0	1	1	1	0
7	1	0	0	0	0	0	2
8	0	0	0	2	1	1	0
9	1	1	1	0	0	0	2
10	0	0	0	1	2	2	0
11	0	1	1	0	0	0	3

Table 20. Table 23: S ​ U ​ ( 2 ) → I ′ → 𝑆 𝑈 2 superscript 𝐼 ′ SU(2)\rightarrow I^{\prime}

Dimension	${\mathrm{𝟏}}_{\mathrm{𝟏}}$	${\mathrm{𝟐}}_{\mathrm{𝟐}}$	${\mathrm{𝟐}}_{\mathrm{𝟑}}$	${\mathrm{𝟑}}_{\mathrm{𝟏}}$	${\mathrm{𝟑}}_{\mathrm{𝟐}}$	${\mathrm{𝟒}}_{\mathrm{𝟏}}$	${\mathrm{𝟒}}_{\mathrm{𝟐}}$	$\mathrm{𝟓}$	$\mathrm{𝟔}$
2	0	1	0	0	0	0	0	0	0
3	0	0	0	0	1	0	0	0	0
4	0	0	0	0	0	0	1	0	0
5	0	0	0	0	0	0	0	1	0
6	0	0	0	0	0	0	0	0	1
7	0	0	0	1	0	1	0	0	0
8	0	0	1	0	0	0	0	0	1
9	0	0	0	0	0	1	0	1	0
10	0	1	0	0	0	0	1	0	1
11	0	0	0	1	1	0	0	1	0
12	0	1	0	0	0	0	1	0	1
13	1	0	0	0	1	1	0	1	0

Table 21. Table 25: S ​ U ​ ( 3 ) → T 7 → 𝑆 𝑈 3 subscript 𝑇 7 SU(3)\rightarrow T_{7}

Dimension	${\mathrm{𝟏}}_{\mathrm{𝟏}}$	${\mathrm{𝟏}}_{\mathrm{𝟐}}$	${\mathrm{𝟏}}_{\mathrm{𝟑}}$	${\mathrm{𝟑}}_{\mathrm{𝟏}}$	${\mathrm{𝟑}}_{\mathrm{𝟐}}$
3	0	0	0	1	0
6	0	0	0	1	1
8	0	1	1	1	1
10	1	0	0	1	2
15	1	1	1	2	2
15′	1	1	1	2	2
21	1	1	1	3	3
24	1	1	1	4	3
27	1	1	1	4	4

Table 22. Table 27: S ​ U ​ ( 3 ) → P ​ S ​ L ​ ( 2 , 7 ) → 𝑆 𝑈 3 𝑃 𝑆 𝐿 2 7 SU(3)\rightarrow PSL(2,7)

Dimension	$\mathrm{𝟏}$	${\mathrm{𝟑}}_{\mathrm{𝟐}}$	${\mathrm{𝟑}}_{\mathrm{𝟐}}$	$\mathrm{𝟔}$	$\mathrm{𝟕}$	$\mathrm{𝟖}$
3	0	1	0	0	0	0
6	0	0	0	1	0	0
8	0	0	0	0	0	1
10	0	0	1	0	1	0
15	0	0	0	0	1	1
15′	1	0	0	1	0	1
21	0	1	1	0	1	1
24	0	1	0	1	1	1
27	0	0	0	2	1	1
28	1	0	0	2	1	1

Equations

$$3\times 3\times 3=1+3\cdot 3+2\cdot 5+7$$

$$V_7=-m^2{T}_{ijk}{T}_{ijk}+\lambda (T_{ijk}{T}_{ijk}{)}^2+\kappa T_{ijm}{T}_{ijn}{T}_{kln}{T}_{klm}$$

$$\begin{array}{rl}{V}_9=-& m^2{T}_{ijkl}{T}_{ijkl}+\lambda (T_{ijkl}{T}_{ijkl}{)}^2+\kappa T_{ijkl}{T}_{ijkp}{T}_{mnop}{T}_{mnol}\\ +& \rho T_{ijkl}{T}_{ijop}{T}_{mnop}{T}_{mnkl}+\tau T_{ijkl}{T}_{ijmn}{T}_{kmop}{T}_{lnop}\end{array}$$

$$\begin{array}{rl}{V}_{13}=-& m^2{T}_{ijklmn}{T}_{ijklmn}+\lambda (T_{ijklmn}{T}_{ijklmn}{)}^2+\kappa T_{ijklmn}{T}_{ijklmt}{T}_{opqrsn}{T}_{opqrst}\\ +& \rho T_{ijklmn}{T}_{ijklst}{T}_{opqrmn}{T}_{opqrst}+\tau T_{ijklmn}{T}_{ijkrst}{T}_{opqlmn}{T}_{opqrst}\end{array}$$

$$\begin{array}{rl}{V}_{7_c}=-& m^2{T}_{ijklmn}{T}^{ijklmn}+\lambda (T_{ijklmn}{T}^{ijklmn}{)}^2+\kappa T_{ijklmn}{T}^{ijklmt}{T}_{opqrst}{T}^{opqrsn}\\ +& \rho T_{ijklmn}{T}^{ijklst}{T}_{opqrst}{T}^{opqrmn}+\tau T_{ijklmn}{T}^{ijkrst}{T}_{opqrst}{T}^{opqlmn}\end{array}$$

$$\begin{array}{rl}{V}_{9_c}=-& m^2{T}_{ijklmnop}{T}^{ijklmnop}+\lambda (T_{ijklmnop}{T}^{ijklmnop}{)}^2+\kappa T_{ijklmnop}{T}^{ijklmnox}{T}_{qrstuvwx}{T}^{qrstuvwp}\\ +& \rho T_{ijklmnop}{T}^{ijklmnwx}{T}_{qrstuvwx}{T}^{qrstuvop}+\tau T_{ijklmnop}{T}^{ijklmvwx}{T}_{qrstuvwx}{T}^{qrstunop}\\ +& \sigma T_{ijklmnop}{T}^{ijkluvwx}{T}_{qrstuvwx}{T}^{qrstmnop}\end{array}$$

$$\begin{array}{rl}{V}_{13_c}=& -m^2{T}_{abcdefghijkl}{T}^{abcdefghijkl}+\lambda (T_{abcdefghijkl}{T}^{abcdefghijkl}{)}^2\\ +& \kappa T_{abcdefghijkl}{T}^{abcdefghijkx}{T}_{mnopqrstuvwx}{T}^{mnopqrstuvwl}+...\end{array}$$

$$T_{12a}\cdot T^{12a}=T_{abcdefghijkl}{T}_{abcdefghijkl},$$

$$(T_{11a}\cdot T^{11a}{)}_c^b(T_{11a}\cdot T^{11a}{)}_b^c=T_{abcdefghijkl}{T}^{abcdefghijkx}{T}_{mnopqrstuvwx}{T}^{mnopqrstuvwl},$$

$$\begin{array}{rl}{V}_{13_c}=& -m^2{T}_{12a}\cdot T^{12a}+\lambda (T_{12a}\cdot T^{12a}{)}^2+\kappa (T_{11a}\cdot T^{11a}{)}_c^b(T_{11a}\cdot T^{11a}{)}_b^c\\ +& \rho (T_{10a}\cdot T^{10a}{)}_{2c}^{2b}(T_{10a}\cdot T^{10a}{)}_{2b}^{2c}+\tau (T_{9a}\cdot T^{9a}{)}_{3c}^{3b}(T_{9a}\cdot T^{9a}{)}_{3b}^{3c}+\nu (T_{8a}\cdot T^{8a}{)}_{4c}^{4b}(T_{8a}\cdot T^{8a}{)}_{4b}^{4c}\\ +& \sigma (T_{7a}\cdot T^{7a}{)}_{5c}^{5b}(T_{7a}\cdot T^{7a}{)}_{5b}^{5c}+\chi (T_{6a}\cdot T^{6a}{)}_{6c}^{6b}(T_{6a}\cdot T^{6a}{)}_{6b}^{6c}\end{array}$$

$$\begin{array}{rl}{V}_{15}=& -m^2{T}_{ij}^k{T}_k^{ij}+\lambda (T_{ij}^k{T}_k^{ij}{)}^2+\kappa T_{jm}^i{T}_i^{jn}{T}_{ln}^k{T}_k^{lm}\\ & +\rho T_{jm}^i{T}_i^{jn}{T}_{kl}^m{T}_n^{kl}+\tau T_{ij}^m{T}_n^{ij}{T}_{kl}^n{T}_m^{kl}+\nu T_{jm}^i{T}_{in}^j{T}_l^{km}{T}_k^{ln}\end{array}$$

$$V_{10}=-m^2{T}_{ijk}{T}^{ijk}+\lambda (T_{ijk}{T}^{ijk}{)}^2+\kappa T_{ijm}{T}^{ijn}{T}_{kln}{T}^{klm}$$

$$3\times 3\times 3\times 3=3\cdot 3+2\cdot \bar{6}+3\cdot 15+15{}^{\prime }$$

$$\begin{array}{rl}{V}_{15^{\prime }}=-& m^2{T}_{ijkl}{T}^{ijkl}+\lambda (T_{ijkl}{T}^{ijkl}{)}^2\\ +& \kappa T_{ijkl}{T}^{ijkm}{T}_{mnop}{T}^{lnop}+\rho T_{ijkl}{T}^{ijmn}{T}_{mnop}{T}^{klop}\end{array}$$

$$\begin{array}{rl}{V}_{28}=-& m^2{T}_{ijklmn}{T}^{ijklmn}+\lambda (T_{ijklmn}{T}^{ijklmn}{)}^2\\ +& \kappa T_{ijklmn}{T}^{ijklmt}{T}_{opqrst}{T}^{opqrsn}+\rho T_{ijklmn}{T}^{ijklst}{T}_{opqrst}{T}^{opqrmn}\\ +& \tau T_{ijklmn}{T}^{ijkrst}{T}_{opqrst}{T}^{opqlmn}\end{array}$$

$$\begin{array}{rl}& |1⟩=\frac{1}{\sqrt{2}}(|11⟩-|22⟩)\\ & |2⟩=\frac{1}{\sqrt{6}}(|11⟩+|22⟩-2\cdot |33⟩)\\ & |3⟩=\frac{1}{\sqrt{2}}(|12⟩+|21⟩)\\ & |4⟩=\frac{1}{\sqrt{2}}(|13⟩+|31⟩)\\ & |5⟩=\frac{1}{\sqrt{2}}(|23⟩+|32⟩)\end{array}$$

$$T_{ij}=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}|1⟩+\frac{1}{\sqrt{6}}|2⟩& \frac{1}{\sqrt{2}}|3⟩& \frac{1}{\sqrt{2}}|4⟩\\ \frac{1}{\sqrt{2}}|3⟩& -\frac{1}{\sqrt{2}}|1⟩+\frac{1}{\sqrt{6}}|2⟩& \frac{1}{\sqrt{2}}|5⟩\\ \frac{1}{\sqrt{2}}|4⟩& \frac{1}{\sqrt{2}}|5⟩& -\sqrt{\frac{2}{3}}|2⟩\end{array}\right)$$

$$\begin{array}{rl}& |1⟩=\frac{1}{\sqrt{2}}(x^2-y^2)\\ & |2⟩=\frac{1}{\sqrt{6}}(x^2+y^2-2z^2)\\ & |3⟩=\frac{1}{\sqrt{2}}(xy+yx)=\sqrt{2}xy\\ & |4⟩=\sqrt{2}xz\\ & |5⟩\sqrt{2}yz\end{array}$$

$$\mathcal{I}(x,y,z)=\frac{1}{|\mathcal{R}(H)|}\sum _{h\in \mathcal{R}(H)}f(h\circ \left(\begin{array}{c}x\\ y\\ z\end{array}\right))$$

$$\begin{array}{rl}& |1⟩=\frac{1}{2}(|111⟩-|122⟩-|212⟩-|221⟩)\\ & |2⟩=\frac{1}{\sqrt{60}}(3\cdot |111⟩+|122⟩+|212⟩+|221⟩-4\cdot |133⟩-4\cdot |313⟩-4\cdot |331⟩)\\ & |3⟩=\frac{1}{2}(|222⟩-|112⟩-|121⟩-|211⟩)\\ & |4⟩=\frac{1}{\sqrt{60}}(3\cdot |222⟩+|112⟩+|121⟩+|211⟩-4\cdot |233⟩-4\cdot |323⟩-4\cdot |332⟩)\\ & |5⟩=\frac{1}{2}(|333⟩-|113⟩-|131⟩-|311⟩)\\ & |6⟩=\frac{1}{\sqrt{60}}(3\cdot |333⟩+|113⟩+|131⟩+|311⟩-4\cdot |223⟩-4\cdot |232⟩-4\cdot |322⟩)\\ & |7⟩=\frac{1}{\sqrt{6}}(|123⟩+|132⟩+|213⟩+|231⟩+|312⟩+|321⟩)\end{array}$$

$$v=[0,0,0,0,0,0,1]$$

$$\begin{array}{rl}& |1⟩=Y_4^0;|2⟩=\frac{i}{\sqrt{2}}(Y_4^1+Y_4^{-1});|3⟩=\frac{1}{\sqrt{2}}(Y_4^1-Y_4^{-1});|4⟩=\frac{1}{\sqrt{2}}(Y_4^2+Y_4^{-2});\\ & |5⟩=\frac{i}{\sqrt{2}}(Y_4^2-Y_4^{-2});|6⟩=\frac{i}{\sqrt{2}}(Y_4^3+Y_4^{-3});|7⟩=\frac{1}{\sqrt{2}}(Y_4^3-Y_4^{-3});\\ & |8⟩=\frac{1}{\sqrt{2}}(Y_4^4+Y_4^{-4});|9⟩=\frac{i}{\sqrt{2}}(Y_4^4-Y_4^{-4}).\end{array}$$

$$v=[\sqrt{\frac{7}{5}},0,0,0,0,0,0,1,0]$$

$$\begin{array}{rl}& |1⟩=Y_6^0;|2⟩=\frac{i}{\sqrt{2}}(Y_6^1+Y_6^{-1});|3⟩=\frac{1}{\sqrt{2}}(Y_6^1-Y_6^{-1});|4⟩=\frac{1}{\sqrt{2}}(Y_6^2+Y_6^{-2});\\ & |5⟩=\frac{i}{\sqrt{2}}(Y_6^2-Y_6^{-2});|6⟩=\frac{i}{\sqrt{2}}(Y_6^3+Y_6^{-3});|7⟩=\frac{1}{\sqrt{2}}(Y_6^3-Y_6^{-3});\\ & |8⟩=\frac{1}{\sqrt{2}}(Y_6^4+Y_6^{-4});|9⟩=\frac{i}{\sqrt{2}}(Y_6^4-Y_6^{-4});|10⟩=\frac{i}{\sqrt{2}}(Y_6^5+Y_6^{-5});\\ & |11⟩=\frac{1}{\sqrt{2}}(Y_6^5-Y_6^{-5})|12⟩=\frac{1}{\sqrt{2}}(Y_6^6+Y_6^{-6});|13⟩=\frac{i}{\sqrt{2}}(Y_6^6-Y_6^{-6}).\end{array}$$

$$v=[1,0,0,-\sqrt{\frac{21}{2}},0,0,0,-\sqrt{7},0,0,0,\sqrt{\frac{105}{22}},0]$$

$$v=[1,1,0,0,0,0,0,0,0,0,0,0,0,0]$$

$$\begin{array}{rl}|1⟩=& |111111⟩\\ |2⟩=& \frac{1}{\sqrt{6}}(|111112⟩+|111121⟩+|111211⟩+|112111⟩+|121111⟩+|211111⟩)\\ |3⟩=& \frac{1}{\sqrt{15}}(|111122⟩+|111212⟩+|111221⟩+|112112⟩+|112121⟩+|112211⟩+\\ & |121112⟩+|121121⟩+|121211⟩+|122111⟩+|211112⟩+|211121⟩+|211211⟩+|212111⟩+|221111⟩)\\ |4⟩=& \frac{1}{\sqrt{20}}(|111222⟩+|112122⟩+|112212⟩+|112221⟩+|121122⟩+|121212⟩+|121221⟩+|122112⟩+\\ & |122121⟩+|122211⟩+|211122⟩+|211212⟩+|211221⟩+|212112⟩+|212121⟩+|212211⟩+\\ & |221112⟩+|221121⟩+|221211⟩+|222111⟩)\\ |5⟩=& \frac{1}{\sqrt{15}}(|112222⟩+|121222⟩+|122122⟩+|122212⟩+|122221⟩+|211222⟩+|212122⟩+|212212⟩+\\ & |212221⟩+|221122⟩+|221212⟩+|221221⟩+|222112⟩+|222121⟩+|222211⟩\\ |6⟩=& \frac{1}{\sqrt{6}}(|122222⟩+|212222⟩+|221222⟩+|222122⟩+|222212⟩+|222221⟩\\ |7⟩=& |222222⟩\end{array}$$

$$v=[0,-1,0,0,0,1,0,0,0,0,0,0,0,0]$$

$$\begin{array}{rl}& |1⟩=\frac{}{}|11111111⟩;|2⟩=\frac{}{}|22222222⟩;|3⟩=\frac{1}{\sqrt{8}}(|11111112⟩+perms);\\ & |4⟩=\frac{1}{\sqrt{8}}(|22222221⟩+perms);|5⟩=\frac{1}{\sqrt{28}}(|11111122⟩+perms);\\ & |6⟩=\frac{1}{\sqrt{28}}(|22222211⟩+perms);;|7⟩=\frac{1}{\sqrt{56}}(|11111222⟩+perms);\\ & |8⟩=\frac{1}{\sqrt{56}}(|22222111⟩+perms);|9⟩=\frac{1}{\sqrt{70}}(|11112222⟩+perms),\end{array}$$

$$v=[1,1,0,0,0,0,0,0,\frac{14}{\sqrt{70}},0,0,0,0,0,0,0,0,0]$$

Full text

I Introduction

The standard model (SM) does not explain quark and lepton masses, nor does it explain how quarks and leptons mix. The most studied and perhaps the most aesthetic approach to parameterizing the masses and mixing angles is to extend the SM by a discrete group $\Gamma$ into whose irreducible representations (irreps) the standard model particles are assigned. Many choices of this discrete flavor symmetry have been tried. As expected, larger groups can typically provide a fuller description of flavor physics, but there are examples of relatively small nonabelian discrete groups like $A_{4}$ and $T^{\prime}$ that are somewhat more economical. Here we take an agnostic approach as to the choice of discrete group and study a representative set of examples that have been used in model building.

Notable early extensions of the standard model with discrete symmetries include the work of Pakvasa and Sugawara Pakvasa:1977in who used $\Gamma=S_{3}$ and focused on the quark sector, as well as Ma and collaborators Ma:2001dn ; Babu:2002dz who used $\Gamma=A_{4}$ to describe the lepton sector. Many other choices for $\Gamma$ have been used in model building, several of which will be discussed below. For an early brief review of possible discrete groups that can be used for SM extensions see Frampton:1994rk . Recent extensive reviews with more complete and up to date bibliographies are also available. See for instance Altarelli:2010gt ; Ishimori:2010au ; King:2013eh ; King:2014nza .

Extending the SM by a discrete group is not without its perils. Global discrete symmetries are violated by gravity Krauss:1988zc , the discrete group can be anomalous Luhn:2008sa , unwanted cosmic defects can be produced Hindmarsh:1994re , etc. To avoid as many of these problems as possible the most expedient approach is to gauge the discrete symmetry, i. e., extend the SM by a continuous group $G$ in such a way that no chiral anomalies are produced. Then one breaks this gauge group to the desired discrete group, $G\rightarrow\Gamma$, where now $\Gamma$ is effectively anomaly free and avoids problems with gravity.

Various examples of Lie groups breaking to discrete groups have been discussed in the literature, but only in a few cases have the details of the minimization of the scalar potential and the extraction of the scalar spectrum been investigated. Here we plan to include these important details for many of the discrete groups of interest via the following procedure:

(i) First we provide irreps of $G$ that contain trivial $\Gamma$ singlets. These results are summarized in the Appendix.

(ii) Next we set up scalar potentials $V$ with scalars in one of these irreps.

(iii) Then we find a vacuum expectation value (VEV) via the Reynolds operator Reynolds (similar to the perhaps more familiar Molien series Molien ) that can break $G$ to $\Gamma$.

(iv) Next we minimize $V$ to show that the VEV indeed does properly break the symmetry.

(v) Finally, we provide the spectrum of scalar masses at the $\Gamma$ level after the breaking. Our calculations are carried out with Mathematica and checked by hand where practical.

Many of the methods we employ were developed in work by Luhn Luhn:2011ip and by Merle and Zwicky Merle:2011vy , where some of the results summarized here can be found. We believe our results will be of interest to many model builders, since it will allow them to include the minimal set of scalars necessary to break a gauge symmetry to a discrete symmetry of interest. A few examples that go beyond the minimal set of scalars are also included, where the symmetry breaking is carried out from a nonminimal $G$ irrep or a non-minimal $G$.

II Lie Group Invariant Potentials

Our task in this section is to construct Higgs potentials invariant under Lie groups $G$ for specific irreps. But first we must see which irreps are suitable for spontaneous symmetry breaking (SSB), i.e., irreps whose decompositions include a trivial singlet of the desired subgroup $\Gamma\subset G$ to which we hope to break. Using the Mathematica package decomposeLGreps Fallbacher:2015pga along with GAP to generate the groups GAP4 , one can easily produce tables of branching rules from Lie group irreps to subgroup irreps and find such singlets. We have done this for a number of cases and have included them in a short appendix for convenience and to make the paper self contained.

II.1 Gauge group irreps containing discrete gauge singlets

The discrete groups $\Gamma$ we will discuss and the gauge groups where they can be minimally embedded are $A_{4},S_{4},A_{5}\subset SO(3)$; $Q_{6},T^{\prime},O^{\prime},I^{\prime}\subset SU(2)$; and $T_{7},\Delta(27),PSL(2,7)\subset SU(3)$.

These discrete groups can also be embedded non-minimally. For example, we include the case $A_{4}\subset SU(3)$. Minimal and non-minimal embedding of other discrete groups can be handled in a way similar to what is discussed here, and we hope that the examples we discuss are sufficiently informative to aid in other cases.

To spontaneously break $G$ to $\Gamma$ with some irrep $R$ of $G$, it is necessary that $R$ contains a trivial $\Gamma$ singlet. It is straightforward to look at the decomposition of $R$ from $G$ to $\Gamma$ to make this determination. The decomposition can be carried out by standard techniques starting from character tables. Since it is the character tables that are usually provided in the literature, we here provide an appendix with the tables of decompositions of the first few irreps of $SO(3)$, $SU(2)$, and $SU(3)$ to discrete groups of interest. For example, as one can see in Table 18 of the Appendix, the 7 and 9 dimensional irreps of $SO(3)$ have trivial $A_{4}$ singlets, therefore these irreps are candidates for the scalar potential that allows the spontaneous symmetry breaking $SO(3)\rightarrow A_{4}$.

II.2 $SO(3)$ Potentials

We will begin our study of SSB by starting with relatively simple examples and then proceed to more sophisticated cases. But first, a note on cubic terms in the potential; a general renormalizable potential has quadratic, cubic, and quartic terms, but the cubic terms tend to significantly complicate the analysis. We exclude these terms for simplicity by imposing a $Z_{2}$ symmetry (or, like in some cases, they vanish upon summation), so the following potentials are actually $SO(3)\times Z_{2}$ invariant. (The $Z_{2}$ symmetry can be avoided by including the cubic terms or by gauging it too.) The effect of including the cubic terms is studied for some cases where the analysis is tractable in Section IV. We now proceed to our first example, the breaking pattern $SO(3)\rightarrow A_{4}$.

II.2.1 $A_{4}$

We begin by constructing an $SO(3)$ invariant potential 333Group Theory Comments: The tetrahedral group $A_{4}\subset SO(3)$ has double-valued representations that correspond to single-valued representations of the binary (double) tetrahedral group $T^{\prime}\subset SU(2)$. As $SO(3)$ is not a subgroup of $SU(2)$, likewise $A_{4}$ is not a subgroup of $T^{\prime}$ Frampton:2009pr . Hence, besides the irreps of $T^{\prime}$ that are coincident with those of $A_{4}$, it has three additional spinor doublet-like irreps. The relationships between $S_{4}$ and $O^{\prime}$ and between $A_{5}$ and $I^{\prime}$ are similar.. As stated above, which irrep

we use depends on the discrete subgroup of interest. For example, if we want to break to the tetrahedral group $A_{4}$, which has been used to describe the tri-bimaximal neutrino mixing pattern Altarelli:2010gt Berger:2009tt and co-bimaximal mixing Ma:2015fpa , we look at Table 18 and see that the lowest dimensional irrep we can use is the 7. For references to other recent work with $A_{4}$ models see Altarelli:2008bg ; King:2011ab ; Ferreira:2013oga . In terms of the fundamental 3 of $SO(3)$, we obtain a 7 as a direct product of three 3s.

[TABLE]

This product gives a generic rank 3 tensor with 27 independent components. To isolate the 7, we take only the totally symmetric part, which reduces the number of components from 27 to 10, giving the symmetric tensor $S_{ijk}$. Then, using the fact that the Kroenecker delta $\delta_{ij}$, is an invariant of the fundamental irrep of $SO$ groups (for a discussion of Lie group invariant tensors see Cvitanovic:1976am ) we subtract off the three traces, $\sum_{j}^{3}\delta_{jk}S_{ijk}$, i=1,2,3 , to obtain the traceless symmetric tensor $T_{ijk}$, which is our 7 dimensional $SO(3)$ irrep. As mentioned above, the most general renormalizable potential is constructed from the independent quadratic, cubic, and quartic contractions of this tensor. In this case there are two quartic terms, but notice that all the cubic terms, which necessarily include the anti-symmetric Levi-Civita Tensor, $\epsilon_{ijk}$ vanish upon summation. Hence the potential for the 7 is

[TABLE]

In subsequent sections we find a vector (in a particular basis) pointing in the $A_{4}$ direction, then minimize the potential and find the mass eigenstates and show that they can all be positive which implies the minimum is stable. Minimization implies certain constraints on the coupling constants must be satisfied as will be discussed. We proceed in analogous fashion for other $G\rightarrow\Gamma$ cases, but first we will collect all the potentials we need for the purpose.

II.2.2 $S_{4}$

To break to the octahedral group, $S_{4}$, we see from Table 18 that the lowest irrep we can use is the 9. From examining Kroenecker products, we see that we must begin with the direct product of four 3s. Similar to the results in the previous subsection, we take the symmetric part of this rank 4 tensor, $S_{ijkl}$, which reduces the number of components to 15. We then subtract off the six trace elements, $\,\sum\delta_{kl}S_{ijkl}$, to obtain the desired 9-component tensor. The associated potential is

[TABLE]

For examples where the octahedral group has been used to build models see Hagedorn:2006ug ; Hagedorn:2010th .

II.2.3 $A_{5}$

Another subgroup of interest, which has been used in a number of recent models Everett:2008et ; Feruglio:2011qq ; Ding:2011cm ; Chen:2010ty , is $A_{5}$. From Table 20 we see that the 13 is the lowest irrep that contains a trivial $A_{5}$ singlet. Again starting from the fundamental $SO(3)$ triplet one can show that the Kroenecker product of six 3s is needed to get an irrep of this dimension. The symmetric part of this rank 6 tensor, $S_{ijklmn}$ has 28 independent components, which is then reduced to 13 by subtracting off the 15 trace elements, $\,\sum\delta_{mn}S_{ijklmn}$. The potential is constructed in a fashion similar to the $A_{4}$ case.

[TABLE]

II.3 $SU(2)$ Potentials

We now proceed in a similar vein to construct $SU(2)$ invariant potentials. In fact, for the odd dimensional (real) representations, invariants must be constructed from triplets which furnish an unfaithful representation of $SU(2)$. As such the true symmetry of the theory is not given by the potential alone and must be determined from the specifics of the model, i.e., from the full Lagrangian. In the following cases, the omission of the cubic terms means the potentials have a $SU(2)\times U(1)$ symmetry, where the $U(1)$ is a phase. This phase can also be gauged and then broken if necessary to avoid problems with global symmetries, or in some cases cubic terms can be added that do not respect the $U(1)$.

II.3.1 $Q_{6}$

If we want to break to $Q_{6}$ we see from Table 20 that the lowest dimensional irrep we can use is the 5. However, as explained in Frampton:1994xm ; Frampton:1999hk , this irrep will actually break to the continuous subgroup $Pin(2)$. So we must look at the next lowest irrep with a trivial $SU(2)$ singlet, the 7. We cannot break with a real 7 as in Eq.(2) because there are no triplet representations of $Q_{6}$ that can be used to find a VEV in the unfaithful SO(3) representation. Thus we must use the complex 7, which has the same potential as needed for the $T^{\prime}$ case which is given in Eq.(5) below.

II.3.2 $T^{\prime}$

To break from $SU(2)$ to $T^{\prime}$, the binary tetrahedral group, we see from Table 22 that the smallest $SU(2)$ irrep we can use is the 7. Since we must construct it from triplets the potential is the same as in equation (2). The VEVs will also be the same.

Another possibility is to do the breaking to $T^{\prime}$ with a complex 7, which can be thought of as a pair of real 7s. We can now build our representation out of the fundamental doublets of $SU(2)$, where we get the 7 by taking the direct product of six 2s and isolating the tensor symmetric on all indices. The potential is

[TABLE]

where the indices now run from 1 to 2. All cubic terms have vanished upon summation. $T^{\prime}$ models are economical and have been used to explain both quark and lepton sector parameters Frampton:1994rk ; Aranda:1999kc ; Chen:2007afa ; Frampton:2007et ; Frampton:2007et ; Frampton:2008bz ; Frampton:2010uw ; Natale:2016xob ; Carone:2016xsi . A more complete set of recent $T^{\prime}$ model references can be found in Carone:2016xsi .

II.3.3 $O^{\prime}$

To break to from $SU(2)$ to $O^{\prime}$, the binary octahedral group, we see from Table 22 that the smallest $SU(2)$ irrep we can use is the 9. As in the $S_{4}$ example, we can construct our potential from triplets so the potential is the same as in equation (3) and the VEVs will again be the same.

We can also consider the case of a complex 9 and build the representation out of $SU(2)$ doublets. We obtain the 9 through the symmetric product of eight 2s. The potential is

[TABLE]

$O^{\prime}$ is maximal in $SU(2)$, so the proper SSB is assured for a VEV that is $O^{\prime}$ invariant.

II.3.4 $I^{\prime}$

The final $SU(2)$ breaking case we consider is $I^{\prime}$, the binary icosahedral group, which has been used in both three and four family extensions of the SM Everett:2010rd ; Chen:2011dn . Here the lowest $SU(2)$ irrep we can use is the real 13, which yields the same potential as we used for $A_{5}$ (Eq. (4)).

Alternatively for the case of a complex 13 we see that it is given by the symmetric product of twelve 2s. The potential has seven quartic invariants, and the first few terms are

[TABLE]

Potentials for higher tensors can be cumbersome to write, so let us introduce a new notation to deal with them. For instance for the potential for the 13, let us define

[TABLE]

and

[TABLE]

etc. Specifically we write $na$ for the collection of indices $a_{1}a_{2}a_{3}...a_{n}$, etc. Then the full potential is for the complex 13 takes the form

[TABLE]

This notation is consistent when the tensor $T$ is totally symmetric on all of its indices 444We could write an even more compact notation in generalized dyadic form, e.g., the $\nu$ term would be $\nu(T:^{8}T):^{4}(T:^{8}T)$ which again defines how the tensor contractions are to be made, but we find this form unnecessary here, but it could be useful for expressions involving more complicated group invariants. Cvitanovic’s “Bird Track” notationCvitanovic:1976am can also be useful for this purpose.

Again, since $I^{\prime}$ is maximal in $SU(2)$, the proper SSB is assured for an $I^{\prime}$ invariant VEV.

II.4 $SU(3)$ potentials

Similar to the previous section, the omission of cubic terms means that the following potentials have an $SU(3)\times U(1)$ symmetry, where the $U(1)$ can be dealt with as described above.

II.4.1 $A_{4}$ and $T_{7}$

In addition to SO(3), $A_{4}$ can originate from a broken $SU(3)$ symmetry. Looking at Table 24 we see that the lowest dimensional irrep containing a trivial $A_{4}$ singlet is the 6, but as explained in Luhn:2011ip , neither the 6, 10, nor $\bf{15^{\prime}}$ will break $SU(3)$ uniquely to $A_{4}$, i.e., giving these irreps an $A_{4}$ VEV will necessarily leave a group larger than $A_{4}$ unbroken. This leaves us with the 15 as the smallest irrep that will uniquely break to an $A_{4}$ subgroup, and the same logic applies to $T_{7}$. (A variety of $T_{7}$ models have been proposed, see Luhn:2007sy ; Kile:2014kya ; Vien:2015koa ; Vien:2016qbb .) To obtain a useful form of the $\bf{15}$ we first take the product $\mathbf{3}\times\mathbf{3}\times\mathbf{{\bar{3}}}$ in $SU(3)$; then by specifying the part that is symmetric on 2 indices, $S_{ij}^{k}$, we reduce the number of independent components from 27 to 18. Finally, subtracting off the three traces: $\sum_{j}^{3}\delta_{jk}S_{ij}^{k},\,i=1,2,3\,$, gives us the desired 15 component tensor. The associated potential Luhn:2011ip is

[TABLE]

II.4.2 $\Delta(27)$

From Table 26 we see that we can use the 10 to spontaneously break from $SU(3)$ to $\Delta(27)$. We can get to this irrep by taking the product of three triplets and specifying the fully symmetric part of the resulting tensor, which reduces to the desired ten independent components. The potential is

[TABLE]

where the cubic terms have vanished upon summation. This result can also be found in Luhn:2011ip . Examples where $\Delta(27)$ has been used are Ferreira:2012ri ; Vien:2016tmh .

II.4.3 $PSL(2,7)$

Another group that has garnered considerable interest as a flavor symmetry is $PSL(2,7)$ Chen:2014wiw . Looking at Table 27 we see that the lowest dimensional irrep of $SU(3)$ we can use to break to $PSL(2,7)$ is the 15′, (Dynkin label [4 0]). To get to a 15′ we take the product of four fundamental triplets

[TABLE]

The generic rank 4 tensor has 81 independent components, requiring it be symmetric on all four indices reduces it to $\bf{15^{\prime}}$ as required. The associated potential is

[TABLE]

Also of interest is the next lowest irrep suitable for breaking from $SU(3)$ to $PSL(2,7)$, the 28. We build this irrep by taking the symmetric product of six triplets, giving a fully symmetric rank 6 tensor with 28 components. The associated potential is

[TABLE]

III Vaccuum Alignments for Spontaneous Symmetry Breaking

III.1 Vacuua for $SO(3)$ Potentials

The invariant tensors from the previous section can be written in terms of a $d$-dimensional orthonormal bases, where $d$ is the number of independent tensor components. To illustrate this consider the 5 of $SO(3)$ which is a second rank symmetric traceless tensor $T_{ij}$. It has a basis

[TABLE]

Where $\ket{ij}$ is the $ij^{th}$ component of the tensor. Using this basis the matrix form of $T_{ij}$ is

[TABLE]

With an explicit basis, it now makes sense to look for a d-component vacuum alignment that minimizes the potential and is invariant under the desired discrete subgroup. How do we find this specified direction? First, note that we can express our basis above in polynomial form, assigning component 1 to $x$, 2 to $y$, and 3 to $z$:

[TABLE]

So if we find a polynomial that is invariant under the desired subgroup we can convert it into a vacuum alignment by expressing it as a vector in terms of these basis functionsMerle:2011vy . To find a polynomial, $\mathcal{I}(x,y,z)$, invariant under a group $H$, one employs the Reynolds Operator Reynolds

[TABLE]

Where $\mathcal{R}(H)$ is a representation of the group, $|\mathcal{R}(H)|$ is the number of elements in the group, and $f(h\circ\begin{pmatrix}x\\ y\\ z\end{pmatrix})$ signifies the result of a group element $h$ acting on the vector $(x,y,z)$ and then input into a trial function $f(x,y,z)$. Trial polynomials of the form $x^{n}y^{m}z^{d-n-m}$ will typically be most useful in finding invariants of degree $d$. Note we have specified polynomials in three variables here, but we can use the same procedure to find invariants in terms of any number of variables, real or complex. E.g., in two real dimensions we can find an invariant $\mathcal{I}(x,y)$ with a trial function $f(x,y)$.

III.1.1 $A_{4}$

As an initial practical example lets examine the symmetry breaking pattern $SO(3)\rightarrow A_{4}$. The irrep of interest is a $\bf{7}$ which is the symmetric, traceless part of ${\bf 3\times 3\times 3}$. Expressed it in terms of 7 independent components we have

[TABLE]

Using $xyz$ as a trial polynomial in equation (16), ($d=3,\,n=m=1$) gives us back $xyz$ as our invariant polynomial. Expressed in terms of this basis our $A_{4}$ invariant vacuum alignment is remarkably simple:

[TABLE]

The VEV for spontaneous breaking will be this unit vector multiplied by a constant which minimizes the potential. We must show that this VEV is unique to $A_{4}$. The gauge group will spontaneously break to the largest subgroup which leaves that VEV invariant. So $G$ will only break to a desired subgroup, $H$, if there is no other group, $H^{\prime}$, which is invariant under the specified VEV and satisfies $H\subset H^{\prime}\subset G$. It is difficult to systematically determine which subgroup will be left invariant for a given breaking, and in particular if there is a higher invariance than the desired discrete group, so each case must be considered individually. For the present case we start with the fact that the only groups that contain $A_{4}$ and are subgroups of $SO(3)$ are $S_{4}$ and $A_{5}$. Examining the branching rules for both these groups, one sees that a 7 of $SO(3)$ does not break to a trivial singlet of either $S_{4}$ or $A_{5}$, and thus the largest group left invariant by this VEV must be $A_{4}$. Hence we have obtained the desired result for the case at hand.

III.1.2 $S_{4}$

For the 9 of SO(3), it is more convenient to express our basis in terms of spherical harmonics:

[TABLE]

We find that the polynomial, $x^{4}+y^{4}+z^{4}$ is $S_{4}$ invariant. Expressed in terms of our basis this is

[TABLE]

$S_{4}$ is also a maximal subgroup of SO(3), so we can be certain our alignment breaks SO(3) uniquely to $S_{4}$.

III.1.3 $A_{5}$

As mentioned previously, to break from $SO(3)$ to $A_{5}$ the irrep of interest is the totally symmetric traceless tensor with 13 independent components contained in ${\bf 3\times 3\times 3\times 3\times 3\times 3}$. In this case it is again easier (and yields equivalent results) to express the components in terms of spherical harmonics555One can also use this method for the $A_{4}$ case, see Koca:2003jy . of degree l = 6,  $Y_{6}^{m}\>(\text{where}\>m=-6,-5...0...5,6)$. In order to get real basis vectors, we define them as

[TABLE]

We find that a degree six invariant polynomial is $(\frac{(1+\sqrt{5})^{2}}{4}\,x^{2}-y^{2})(\frac{(1+\sqrt{5})^{2}}{4}\,y^{2}-z^{2})(\frac{(1+\sqrt{5})^{2}}{4}\,z^{2}-x^{2})$ Merle:2011vy . The associated VEV is proportional to

[TABLE]

Because $A_{5}$ is a maximal subgroup of SO(3), i.e., there is no group $H^{\prime}$ that nontrivially satisfies $A_{5}\subset H^{\prime}\subset SO(3)$ for any VEV of the $\bf{13}$, and again we can be sure the VEV in eq. (22) breaks $SO(3)$ uniquely to $A_{5}$.

III.2 Vacuua for SU(2) Potentials

III.2.1 $Q_{6}$

For the breaking $SU(2)\rightarrow Q_{6}$ we use the same basis as with $T^{\prime}$ above. We find the polynomial $\frac{1}{2}(x^{6}+y^{6})$ is left invariant by $Q_{6}$, and this leads to a VEV proportional to

[TABLE]

To make sure we have broken to $Q_{6}$ and not any larger subgroups, we first note that the 7 does not break to any $Q_{n}$ with $n>6$ (see page 6 of Frampton:1999hk ). The only other larger $SU(2)$ subgroup that can be spontaneously broken with a 7 is $T^{\prime}$, but we find that $T^{\prime}$ has only one degree six invariant which is given in the subsection above. Therefore, the VEV in Eq. (23) is the result we were seeking.

III.2.2 $T^{\prime}$

Because $SU(2)$ breaks to $T^{\prime}$ from the same real seven dimensional irrep that breaks $SO(3)$ to $A_{4}$, the potentials are the same and the basis will be the same as in the $A_{4}$ section above. In addition, the Reynolds operator yields the same polynomial invariant $xyz$, so the VEV is identical. On the other hand the complex 7 has a different basis666Because this is a complex irrep there are actually 14 basis states; the basis states listed are the 7 real parts of the tensor components, while bases 8 through 14 are the imaginary parts. These conjugate components have been suppressed here since they will always be set to zero at vacuum in order to have a real VEV. This will be the case for most of the complex irreps we consider., specifically that of the symmetric tensor with 6 indices.

[TABLE]

We find that the polynomial $\frac{1}{2}(xy^{5}-yx^{5})$ is left invariant for this representation and the associated VEV is proportional to

[TABLE]

To make sure we have broken to T*′* we must show that this VEV does not break $SU(2)$ to any larger group. The only $SU(2)$ subgroups that contain T*′* as a subgroup are I*′*, the binary icosahedral group, and $O^{\prime}$, the binary octahedral group. Looking at tables of branching rules we see that the 7 of $SU(2)$ does not contain a trivial singlet of either of these groups, so we can be sure the breaking is to $T^{\prime}$ as desired.

III.2.3 $O^{\prime}$

Like the other double cover groups, the basis and vacuum direction for the breaking of $O^{\prime}$ with a real 9 of $SU(2)$ will be the same as its $SO(3)\rightarrow S_{4}$ counterpart above.

The complex 9 arises from the basis of the symmetric tensor with 8 doublet indices:

[TABLE]

where here and in what follows ‘$+perms$’ means we include all permutations of tensor indices.

Here the $O^{\prime}$ invariant polynomial is $x^{8}+y^{8}+14x^{4}y^{4}$, which leads to a VEV proportional to

[TABLE]

Where $\ket{1}=\ket{2}$ and $\ket{9}=\frac{14}{\sqrt{70}}\ket{1}$.

$O^{\prime}$ is a maximal subgroup of $SU(2)$, so we can be certain our alignment breaks SU(2) uniquely to $O^{\prime}$.

III.2.4 $I^{\prime}$

Similar to the spontaneous symmetry breaking behavior of the $T^{\prime}$ case relative to the $A_{4}$ case with a real $\bf{7}$, the basis for the symmetry breaking to $I^{\prime}$ with the real 13 will be the same as for $A_{5}$ above. Additionally, both groups have the same invariant polynomial so the vacuum directions will be the same.

On the other hand, a complex 13 arises from the basis of the symmetric tensor with 12 doublet indices:

[TABLE]

Here the $I^{\prime}$ invariant polynomial is $x^{11}y+11x^{6}y^{6}-y^{11}x$, which leads to a VEV proportional to

[TABLE]

Where clearly $\ket{4}=-\ket{3}$ and $\ket{13}=\sqrt{\frac{11}{12}}\cdot\ket{3}$.

$I^{\prime}$ is a known maximal subgroup of SU(2), so we can be certain our alignment breaks SU(2) uniquely to $I^{\prime}$.

III.3 Vacuua for $SU(3)$ Potentials

First let us show that we can get discrete subgroups from continuous groups in a nonminimal way. For this purpose we use the example $SU(3)\rightarrow A_{4}$ where we break with a 15 of $SU(3)$. Then we find vacuua for the minimal cases discussed above. Then finally, for $PSL(2,7)$ we give both a minimal case with a VEV for the ${\bf 15^{\prime}}$ of $SU(3)$ and a nonminimal breaking via a 28 of $SU(3)$ using the potential given in eq. (13).

III.3.1 $A_{4}$

The complex 15 dimensional basis needed to break SU(3) to $A_{4}$ is that of the traceless $3\times 3\times\bar{3}$ tensor that is symmetric on the first two indices Luhn:2011ip .

[TABLE]

Because this tensor is symmetric on only two indices we find that the invariant should be of degree 2 in the variables $x,y,z$ and degree 1 in the conjugate variables, $x^{*},y^{*},z^{*}$. Inputting the trial polynomial $xyz^{*}$ into the Reynolds operator produces the invariant: $xyz^{*}+yzx^{*}+xzy^{*}$. In this basis the VEV is proportional to

[TABLE]

i.e., where $\ket{13}=\ket{14}=\ket{15}$ with all other components zero.

One can examine the generators of $A_{4}$ and SU(3) to see that this VEV breaks SU(3) uniquely to $A_{4}$, see Luhn:2011ip .

III.3.2 $T_{7}$

The invariant tensor object and therefore our basis for $T_{7}$ is the same as for $A_{4}$ above. The invariant polynomial in this case is $x^{2}y^{*}+y^{2}z^{*}+z^{2}x^{*}$ and the corresponding VEV is proportional to

[TABLE]

where $\ket{7}=\ket{9}=\ket{11}$.

Similarly to the $A_{4}$ case, one can verify this VEV uniquely breaks $SU(3)$ to $T_{7}$ by examining how the $T_{7}$ generators operate on $v$, see Luhn:2011ip .

III.3.3 $\Delta(27)$

For $\Delta(27)$, the relevant invariant tensor is the fully symmetric part of $3\times 3\times 3$ with 10 independent components

[TABLE]

The invariant polynomial is $x^{3}+y^{3}+z^{3}$, which gives us a VEV proportional to

[TABLE]

Again explicit forms of the generators can be examined in order to verify the uniqueness of this VEV for breaking from $SU(3)$ to $\Delta(27)$ Luhn:2011ip .

III.3.4 $PSL(2,7)$

Our basis for the $\mathbf{15^{\prime}}$ is that of the fully symmetric $3\times 3\times 3\times 3$ tensor

[TABLE]

The relevant invariant polynomial is $x^{3}z+y^{3}x+z^{3}y$ Merle:2011vy , which gives a VEV proportional to

[TABLE]

where the nonvanishing vacuum components are $\ket{5}=\ket{6}=\ket{9}$. We can be sure we have broken to the correct subgroup 777Luhn C.Luhn has shown that the VEV in eq. (36) has a $Z_{28}$ symmetry and the vacuum of the potential $V_{15^{\prime}}$ in eq. (12) is also symmetric under this symmetry. However, other terms in the Lagrangian will violate this $Z_{28}$, e.g., the Yukawa terms. As it is a discrete symmetry, its breaking can not lead to a pseudo Goldstone boson, but there could be other phenomenological consequences of this $Z_{28}$ that would be interesting to explore. because $PSL(2,7)$ is known to be a maximal in $SU(3)$.

Finally, for the 28 of $SU(3)$ we have the basis for the fully symmetric ${\bf 3}^{6}$ tensor of the form

[TABLE]

The necessary invariant polynomial is $x^{5}y+y^{5}z+z^{5}x-5x^{2}y^{2}z^{2}$Merle:2011vy , which gives real components with VEV proportional to

[TABLE]

i.e., where $\ket{4}=\ket{7}=\ket{8}$, and $\ket{28}=-\sqrt{\frac{5}{3}}\cdot\ket{4}$ and where we recall that all conjugate components (29–56) are set to zero.

IV Vacuum expectation values and Mass Spectra

Thus far, we have discussed how to set up potentials corresponding to specific gauge group representations and then found vacuum alignments that can be used to break the gauge symmetry to desired subgroups. In this section we minimize the scalar potentials and show where symmetry breaking in the desired directions are allowed. We will find the scale of the symmetry breaking and resulting tree level scalar mass states in terms of the coupling constants of the potential. As usual, the minimization conditions of the potential will lead to constraints on the values of these constants.

IV.1 $SO(3)$ Cases

IV.1.1 $A_{4}$ scalar spectrum

We found earlier that a VEV in the direction (18) will break $SO(3)$ to $A_{4}$. The actual VEV is proportional to this direction vector, with the constant of proportionality being the scale of the breaking. To determine this scale one must minimize the potential (2). To achieve this we compute the first derivative with respect to each basis state, insert the alignment from (18), and set this equal to zero. This alignment (and all of our alignments below) will give an equation in terms of one basis state (or one linear combination of basis states). For the present case we solve for $\ket{7}$ and take the positive solution to obtain the VEV

[TABLE]

As for any non-trivial stable vacuum, $m^{2}$ must be positive. To have a real value for our breaking scale $3\lambda+\kappa$ must also be positive. We find the scalar mass states by calculating the matrix of second derivatives (the Hessian), inserting the VEV from above, and computing the eigenvalues of the matrix. The resulting values and their multiplicities are given in Table 1.

Looking at Table 18 in the Appendix, we see that the multiplicities of the eigenvalues match up with the branching of the 7 for $SO(3)\rightarrow A_{4}$, as expected. We see that there are three zero eigenvalues as expected corresponding to the three Goldstone boson from the breaking of all the generators of $SO(3)$. Constraints on the coupling constants arise from the requirement that at a minimum of the potential, the eigenvalues must all positive or zero. Since $m^{2}$ and $3\lambda+\kappa$ must be positive, requiring the third eigenvalue to be positive leads to the constraint $\kappa>0$ in this case.

IV.1.2 $S_{4}$

For $S_{4}$ we minimize the potential from (3) using the alignment (20). We obtain a VEV

[TABLE]

A real value for our breaking scale requires $90\lambda+10\kappa+7\rho+2\tau>0$.

The scalar mass states are found in Table 2 888In order to normalize the eigenvalues for $S_{4}$ to those in other cases when we use the spherical harmonic basis, we have multiplied all quadratic terms by a factor of $\frac{1}{8}$ and quartic terms by a factor of $\frac{1}{64}$. and are all non-negative if

$5\kappa+8\rho-2\tau>0.$

The three zeros correspond to the broken $SO(3)$ generators.

IV.1.3 $A_{5}$ scalar spectrum

For $A_{5}$, we minimize the potential from (4) using the alignment (22). We obtain a VEV

[TABLE]

A real value for our breaking scale requires $420\lambda+140\kappa+84\rho+65\tau+14\nu+9\sigma-2\chi>0$. For the scalar mass states 999Again by expressing our states in terms of spherical harmonics, we obtain different normalizations for our basis states which lead to a different normalization scale for the VEV scale and scalar mass states. To correct this for $A_{5}$ we have multiplied the quadratic term by a factor of $\frac{5}{352}$ and the quartic terms by $(\frac{5}{352})^{2}$ so that our states are now normalized the same way as our other breakings. are given in Table 3.

We see we have the three zeros corresponding to the broken $SO(3)$ generators and must satisfy the constraints

$105\kappa+196\rho+240\tau-14\nu-19\sigma+12\chi>0$

$14\rho+45\tau+14\nu-11\sigma+18\chi>0.$

IV.2 $SU(2)$ Cases

IV.2.1 $Q_{6}$ scalar spectrum

We break the symmetry of the potential given in Eq.(5) with the alignment in Eq.(23) to obtain a VEV

[TABLE]

Thus we require $\kappa+2\lambda+\rho+\tau>0$. The eigenvalues of the Hessian are given in Table 4.

The constraints from these mass eigenvalues are

$\kappa>0$

$\kappa>-(\rho+\tau)$

$2\rho+3\tau,\>6\rho+7\tau,\>8\rho+9\tau<0$

There are clearly stable minima when $\lambda>0$, $\kappa>0$, $\rho<0$ and $\tau<0$. The extra zero eigenvalue comes from breaking an accidental $U(1)$ phase symmetry. This gives rise to a pseudo-goldstone boson that can gain a mass through quantum corrections.

IV.2.2 $T^{\prime}$ scalar spectrum

The potential and the vacuum alignment of the breaking of of SU(2) to T*′* with a real 7 are the same as for SO(3)$\rightarrow A_{4}$. Therefore the breaking scale and the mass states will be exactly the same, as the two models can only be differentiated by the non-scalar part of the Lagrangian.

For a the breaking with a complex 7 we minimize the potential in Eq.(5) but this time using the alignment Eq.(25) to obtain the VEV

[TABLE]

which leads to the constraint that $12\lambda+6\kappa+4\rho+3\tau>0$. The eigenvalues of the Hessian are shown in Table 5.

From the requirement of positive eigenvalues we deduce the constraints

$\tau>0$

$\rho>-\frac{3}{2}\tau$

$\frac{3}{8}\tau>\kappa>-8\rho-\frac{9}{8}\tau$

As in the $Q_{6}$ example, the extra zero eigenvalue is a result of breaking the accidental $U(1)$ phase symmetry in the potential.

IV.2.3 $O^{\prime}$ scalar spectrum

The breaking scale and scalar mass spectrum of $SU(2)$ to $O^{\prime}$ with a real $\bf 9$ is exactly the same as that for $SO(3)$ to $S_{4}$, where differences between two models would come from the non-scalar part of the Lagrangian.

For a complex 9 we minimize the potential in Eq.(6) using the alignment Eq.(27) and obtain a VEV

[TABLE]

Thus $60\lambda+30\kappa+20\rho+15\tau+14\sigma$ must be $>0$. The eigenvalues of the Hessian (see Table 6) are all real and positive semidefinite for positive scalar quartic couplings, while more detailed constraints on the scalar quartics can clearly be extracted from the individual mass eigenvalues. There are 3 zeros corresponding to the 3 broken $SU(2)$ generators, as well as an extra zero from breaking the $U(1)$ phase symmetry.

We have the additional constraints

$\sigma<0$

$10\rho+15\tau+16\sigma>0$

$25\kappa+25\rho+25\tau+24\sigma>0$

and

$25\tau+32\sigma>0.$

IV.2.4 $I^{\prime}$ scalar spectrum

The breaking of $SU(2)$ to $I^{\prime}$ and $SO(3)$ to $A_{5}$ with a real 13, are completely analogous to the breakings of $SU(2)$ and $SO(3)$ to $T^{\prime}$ and $A_{4}$ respectively with a real 7.

For a complex 13 we minimize the potential of Eq.(8) using the alignment Eq.(29) and obtain a VEV

[TABLE]

Thus $420\lambda+210\kappa+140\rho+105\tau+84\nu+70\sigma+65\chi$ must be $>0$. The eigenvalues of the Hessian (see Table 7) are all real and positive semidefinite for positive scalar quartic couplings. (More detailed constraints on the scalar quartics can clearly be extracted from the individual mass eigenvalues.)

There are 3 zeros corresponding to the 3 broken $SU(2)$ generators, as well as an extra zero from breaking the $U(1)$ phase symmetry.

IV.3 SU(3) cases

IV.3.1 $A_{4}$ scalar spectrum

For the nonminimal breaking $SU(3)\rightarrow A_{4}$ we minimize the potential Eq.(9) and use the alignment Eq.(31) to get the VEV Luhn:2011ip

[TABLE]

Thus $3\lambda+\eta+\kappa+\rho+\tau$ must be $>0$. The eigenvalues of the Hessian are are shown in Table 8.

We expect eight zeros corresponding to the broken generators of $SU(3)$, but again an extra zero eigenvalue arises from breaking the accidental $U(1)$ phase symmetry. As for constraints, we can readily see that

$\eta<0,$

$5\kappa+2\rho+4\tau>\sqrt{(4\tau+2\rho-3\kappa)^{2}+16(\rho+\kappa+2\eta)^{2}},$

and

$3\kappa-5\eta-2\rho+4\tau>\frac{1}{3}\sqrt{(9\eta-7\kappa+10\rho-4\tau)^{2}+8(\rho+2\kappa-4\tau)^{2}}$

are required. An example of where all these constraints can be satisfied is

$2\rho=3\kappa$, $\rho+\kappa=-2|\eta|$, and $5\kappa+3\rho>0$, where $\kappa,\rho,$ and $\tau>0.$

IV.3.2 $T_{7}$ scalar spectrum

For this breaking we again minimize Eq.(9), now using the alignment Eq.(32) to obtain the VEV Luhn:2011ip

[TABLE]

Thus $3\lambda+\kappa+\rho+\tau$ must be $>0$. The eigenvalues of the Hessian are shown in Table 9,

where $\alpha,\,\beta,\,\gamma$ are the three roots of the polynomial $10368\eta^{2}(\rho-\kappa-\tau)+3888\eta\rho^{2}-15552\eta\kappa\tau+(648\eta^{2}+972\eta\kappa-648\eta\rho-180\rho^{2}+1296\eta\tau+720\kappa\tau)x+(6\rho-54\eta-21\kappa-36\tau)x^{2}+x^{3}$. We have the constraints

$2\kappa-\rho+2\tau>0$

and

$\alpha,\,\beta,\,\gamma>0.$

The extra zero is once again due to breaking an accidental $U(1)$ symmetry. We cannot remedy this by including cubic terms this time, because we need the couplings on those terms to vanish in order to have a stable minimum. Numerical studies show that there is a range of scalar quartic coupling constant values where the minimum is stable. An example of such numerical analysis will be discussed below.

IV.3.3 $\Delta(27)$ scalar spectrum

Minimizing the potential of Eq.(10) with the alignment Eq.(34) we obtain a VEV

[TABLE]

giving the constraint $3\lambda+\kappa>0$. The eigenvalues of the Hessian are in Table 10,

from which we see that $\kappa>0$ is required. Again we have an extra zero from an accidental $U(1)$, and in this case the cubic terms vanish upon summation. The two extra zeros are the result of an additional $\Delta(27)$ singlet within the 10. (For a detailled explanation see Luhn:2011ip .)

IV.3.4 $PSL(2,7)$ scalar spectrum

Minimizing the potential of Eq.(12) with the alignment Eq.(36) we obtain a VEV

[TABLE]

Thus $6\lambda+2\kappa+\rho>0$. The eigenvalues of the Hessian are found in Table 11

from which we get the constraints

$\rho>0$       and       $7\kappa+8\rho>0$.

We once again have an extra zero, but this time it is possible to include cubic terms to break the $U(1)$ phase. The two cubic terms we can include are

[TABLE]

which are Hermitian conjugates and are included in the potential with the same real coupling constant, $\zeta$. The VEV scale for the potential including the cubic is now

[TABLE]

Notice that there may be two possible solutions. The constraint that must hold in both cases is $9\zeta^{2}+4m^{2}(2\kappa+6\lambda+\rho)\geq 0$.

Calculating the eigenvalues of the Hessian produces solutions involving the roots of very large polynomial which is much too large to display, but it is notable that it does produce 8 zeros rather than 9. Furthermore, following the usual procedure, but this time numerically where for simplicity setting all quartic coupling constants to unity, the quadratic coupling to -1 and the cubic to .001 (these values are selected to ensure a stable minimum) produces a VEV scale of approximately 0.3335 and eigenvalues

whose multiplicities match the branching rules SU(3)$\rightarrow PSL(2,7)$, as shown in Table 12. The degeneracy of the pseudo-Goldstone mass with that of the true Goldstones is lifted as expected, as can be seen in Table 12. Finally note that although we have set all the coupling constants except for the cubic to integer values, we can easily rescale them to smaller values to be sure we are in the perturbative regime of the theory without disturbing the stability of the result. Specifically, while the scalar quartic couplings in the numerical example are not in the perturbative range, we can rescale all the quartics by a factor $s$ and $\zeta$ by a factor $\sqrt{s}$. This leaves the eigenvalues unchanged and puts us into the perturbative regime.

Moving on to the 28, we minimize the potential in Eq.(13) with the alignment Eq.(38) to obtain a VEV

[TABLE]

Thus $42\lambda+14\kappa+7\rho+6\tau>0$. The eigenvalues of the Hessian are given in Table 13

Further constraints are

$7\rho+9\tau>0$

$21\kappa+20\rho+18\tau>0$

$A,\,B,\,C,\,D>0$

where A, B, C, and D are the roots of a very large quartic polynomial. Numerical work shows that all four roots can be positive, simultaneously leading to all positive eigenvalues in Table 13 and a stable minimum when the other constraints are also satisfied. We see that there are eight zeros from the broken generators of SU(3), and one zero from breaking the broken U(1) phase. But unique to this breaking we have seven extra zeros, which implies that there are seven more broken generators from an accidental symmetry of the Lagrangian that we have so far been unable to identify, leading to a total of 8 pseudo-Goldstone bosons.

IV.4 Symmetry Breaking Summary

Let us briefly summarize our results. We have shown that we can break from $G$ to $\Gamma$ for the gauge and discrete groups listed in the introduction. The minima can be stable since none of the eigenvalues of the scalars are negative for allowed regions of parameter space. Zero eigenvalues correspond to Goldstone bosons in each case and to additional pseudo-Goldstone bosons in several cases. Specifically for the cases we have studied of $SO(3)$ breaking to a discrete symmetry the results are summarized in Table 14. The $G$ subscript indicates the Goldstones. In each case the masses of the particles in different discrete group irreps are all different, so the initial degeneracy of the scalar masses is lifted to the extent allowed by the discrete group.

For the cases of $SU(2)$ breaking to discrete symmetries, the results are summarized in Table 15. Again all the discrete group irreps correspond to different masses except for the zero eigenvalue states where we have indicated the true Goldstones and the pseudo-Goldstones (by subscripts pGB) due to breaking of the phase symmetry on the potentials. The subscript $c$ indicates that the irreps are complexified and the decompositions are written in terms of real components. The results begin to become more complicated for the $SU(3)$ cases we have investigated, and this can be seen in Table 16. Now some irreps masses have become degenerate and we have indicated these cases by collecting those discrete group irreps with parentheses and labeling the collection with a $deg.$ subscript. All the cases have a pseudo-Goldstones associated with breaking of phase invariance. The breaking to $T_{7}$ with a ${\bf 10}$ leads to two additional pseudo-Goldstones as discussed in Luhn:2011ip and the breaking to $PSL(2,7)$ with a ${\bf 28}$ has seven additional pseudo-Goldstones. Since the ${\bf 28}$ was derived from ${\bf 3^{6}}$ one could conjecture that the potential has a $Spin(6)\sim SU(4)$ accidental symmetry that contains the gauged $SU(3)$, and that the VEV breaks all 15 $SU(4)$ plus the phase to give a total of 16 massless states. Finally, recall that for the breaking to $PSL(2,7)$ with a ${\bf 15^{\prime}}$ we have shown that phase symmetry can be avoided if we add cubic terms, hence there is no pseudo-Goldstone after SSB in that case, see Table 12.

V Discussion and Conclusion

The standard model includes 28 unspecified parameters, some of which describe fermion masses and mixing angles. Consequently, we do not know why the quark and lepton masses and mixings are what they are. To fix these parameters, a standard approach has been to extend the SM by a discrete symmetry, but this approach is not without its difficulties as discussed above. What would seem more natural would be to increase the gauge group to $SU(3)\times SU(2)\times U(1)\times G$ and extend the scalar sector. Then this model can be of the same general type as the SM, i. e., an anomaly-free gauge theory with fermions that gets spontaneously broken by VEVs of scalar fields. If the SSB of $G$ results in a discrete subgroup $\Gamma$ then we arrive at a $SU(3)\times SU(2)\times U(1)\times\Gamma$ via a route that avoids the problems just mentioned, without choosing an ad hoc discrete group for extending the SM.

Here, based on the techniques of LuhnLuhn:2011ip and Merle and R. ZwickyMerle:2011vy , we have demonstrated that we can carry out the $G\rightarrow\Gamma$ SSB in many cases of interest, specifically breaking to $A_{4},S_{4},A_{5},Q_{6},T^{\prime},O^{\prime},I^{\prime},T_{7},\Delta(27)$ and $PSL(2,7)$. Other cases can be handled by the same techniques. Many other discrete groups have been occasionally used to extend the SM, e.g., $D_{4},D_{5},D_{7},D_{14},\Delta(54),\Delta(96),$ and $\Sigma(81)$ have all appeared in the literature Hagedorn:2008bc ; King:2012in ; Vien:2014ica . For a discussion of breaking $SO(3)$ to dihedral groups see Koca:2003jy . Further information about the classification of the discrete subgroups of $SU(3)$ can be found in Ludl:2010bj ; Grimus:2011fk ; Ludl:2011gn ; Merle:2011vy . In addition products of discrete groups are often employed, where the products often contain $Z_{n}$ factors. To gauge these cases we can start with a product gauge group and break to the desired discrete group, $G_{1}\times G_{2}\times...\rightarrow\Gamma_{1}\times\Gamma_{2}\times...$. As long as there are no cross terms in the scalar potential, then we can proceed as above. In some cases the cross terms can destabilize the minima, so they must either be eliminated, or dealt with by other means. If the fundamental charge of a $U(1)$ gauge group is $q$, then by breaking a $U(1)$ with scalar particle of charge $nq$ one arrives at $Z_{n}$. Results given here could be applied to extend recent work on gauging two Higgs doublet models Huang:2015wts . Using our results to extend models currently in the literature can solve some existing problems, and the inclusion of new scalars in the spectrum may be of interest since some may be detectable either directly or indirectly depending on the details of the model. Such phenomenological investigations need to proceed on a model by model basis, and we plan to look at some specific examples in future work.

VI Acknowledgements

We have benefited greatly from discussions and correspondences with Christoph Luhn, Alex Merle and Pierre Ramond.

Appendix A Branching Rules

In this Appendix we present the branching rules for the embeddings of discrete groups into Lie groups used in the paper. The vertical axes label the dimensions of the Lie Group reps, and the horizontal the dimensions of the discrete group representations.