Natural Vacuum Alignment from Group Theory Martin Holthausen - - PowerPoint PPT Presentation

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Natural Vacuum Alignment from Group Theory Martin Holthausen - - PowerPoint PPT Presentation

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Natural Vacuum Alignment from Group Theory Martin Holthausen based on MH, Michael A. Schmidt JHEP 1201 (2012) 126 , arXiv: 1111.1730 Why Flavour Symmetry? in SM(+Majorana neutrinos) there are a total of 28 Parameters 1 1 12 2 10 2 Why

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SLIDE 1

Natural Vacuum Alignment from Group Theory

Martin Holthausen

based on MH, Michael A. Schmidt JHEP 1201 (2012) 126 , arXiv: 1111.1730

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SLIDE 2

Why Flavour Symmetry?

in SM(+Majorana neutrinos) there are a total of 28 Parameters 12 10 2 2 1 1

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SLIDE 3

Why Flavour Symmetry?

in SM(+Majorana neutrinos) there are a total of 28 Parameters 12 10 2 2 1 1 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles

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SLIDE 4

Why Flavour Symmetry?

in SM(+Majorana neutrinos) there are a total of 28 Parameters 12 10 2 2 1 1 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles in quark sector: small mixing angles and hierarchical masses can be explained by Frogatt-Nielsen symmetry

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SLIDE 5

Why Flavour Symmetry?

in SM(+Majorana neutrinos) there are a total of 28 Parameters 12 10 2 2 1 1 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles in quark sector: small mixing angles and hierarchical masses can be explained by Frogatt-Nielsen symmetry in lepton sector: two large and one small mixing angle suggestive of non-abelian discrete symmetry

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SLIDE 6

Lepton mixing from discrete groups

complete flavour group residual symmetry of (me+me) residual symmetry of mν (most general choice if mixing angles do not depend on masses & Majorana νs) (smallest choice, but can also be continuous)

Tmem†

eT † = mem† e

T =   1 1 1  

S =   1 −1 −1   U =   1 1 1  

SmνST = mν UmνU T = mν

Ge=Z3 Gν=Z2xZ2 Gf

UP MNS = UT BM = B B @ q

2 3 1 √ 3

− 1

√ 6 1 √ 3

− 1

√ 2 1 √ 6 1 √ 3 1 √ 2

1 C C A sin2 θ12 = 1 3, sin2 θ23 = 1 2, sin2 θ13 = 0 sin2 θ12 = 0.312+0.017

−0.015,

sin2 θ23 = 0.52+0.06

−0.07,

sin2 θ13 = 0.013+0.007

−0.006

„tri-bimaximal mixing“(TBM) gives good LO description of lepton mixing

misaligned non-communting symmetries lead to

[He, Keum, Volkas ‘06; Lam’07,‘08; Altarelli,Feruglio‘05]

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SLIDE 7

SU3 332 632 36Φ 72Φ 216Φ 360Φ 60 A4 S4 168 T7

[Merle,Zwicky 1110.4891]

Candidate Groups

∆(27) ∼ = (Z3 × Z3) o Z3 T7 ∼ = Z7 o Z3 T 0 ∼ = Z2.A4 S4 ∼ = (Z2 × Z2) o S3

A4 ∼ = (Z2 × Z2) o Z3

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SLIDE 8

A4 Symmetry Group

A4 is the smallest symmetry group that can lead to TBM mixing:

A4 ⇠ = (Z2 ⇥ Z2) o Z3 ⇠ = hS, T|S2 = T 3 = (ST)3 = 1i

S T 11 1 1 12 1 ! 13 1 !2 31 B @ 1 1 1 1 C A B @ 1 1 1 1 C A 1 1

re ! = ei2π/3

(a) Character Table

1 T T 2 S 11 1 1 1 1 12 1 ! !2 1 13 1 !2 ! 1 3 3

  • 1

1-d reps. correspond to

  • reps. of Z3

3 × 3 = 11 + 12 + 13 + 3S + 3A

(ab)11 = 1 √ 3 (a1b1 + a2b2 + a3b3) (ab)12 = 1 √ 3

  • a1b1 + ω2a2b2 + ωa3b3
  • (ab)13 =

1 √ 3

  • a1b1 + ωa2b2 + ω2a3b3
  • (ab)A,3 = 1

2 B @ a2b3 − a3b2 a3b1 − a1b3 a1b2 − a2b1 1 C A (ab)S,3 = 1 2 B @ a2b3 + a3b2 a3b1 + a1b3 a1b2 + a2b1 1 C A

@ A where (a1, a2, a3), (b1, b2, b3) ∼ 3.

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SLIDE 9

An A4 Prototype model

(A4,Z4) charge assignments: l∼ (3,i), ec∼ (11,-i), 휇c∼ (12,-i), 휏c∼ (13,-i) ,휒∼(3,1), Φ∼(3,-1), 휉∼(1,-1) auxiliary Z4 separates neutral and charged lepton sectors at LO

A4

〈휒〉∼(1,1,1) 〈Φ〉∼(1,0,0)

Z3=〈T〉 Z2=〈S〉 (ll)3Φ+(ll)1휉 (L휒)ec

B @ ˜ a ˜ a ˜ d ˜ d ˜ a 1 C A , v Λ B @ ye yµ yτ ye !yµ !2yτ ye !2yµ !yτ 1 C A

me∼ m휈∼ TBM Vacuum alignment crucial!

(Z2xZ2 symmetry accidental)

[e.g. Ma,Rajasekaran’01, Babu, Ma, Valle ’03, Altarelli,Feruglio, ’05,’06]

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SLIDE 10

Can Vacuum Alignment be realized?

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SLIDE 11

Can Vacuum Alignment be realized?

SSB: Minimum Energy State exhibits less symmetry than full Lagrangian

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SLIDE 12

Can Vacuum Alignment be realized?

Vχ = m2

0 (χχ)11 + λ1 (χχ)11(χχ)11 + λ2 (χχ)12(χχ)13

Vsoft,Z2 = m2

Aχ2 1 + m2 Bχ2 2 + m2 Cχ2χ3

SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: has minima (1,1,1) and (1,0,0). Effect of breaking to Z2 in another sector can be included by adding:

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SLIDE 13

Can Vacuum Alignment be realized?

Vχ = m2

0 (χχ)11 + λ1 (χχ)11(χχ)11 + λ2 (χχ)12(χχ)13

Vsoft,Z2 = m2

Aχ2 1 + m2 Bχ2 2 + m2 Cχ2χ3

SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: has minima (1,1,1) and (1,0,0). Effect of breaking to Z2 in another sector can be included by adding: Minimization conditions then give:

0 =  ∂V ∂χ1

  • χi=v0 =

2 √ 3 ⇣ m2

0 +

√ 3m2

A

⌘ v0 + 4λ1v03 0 =  ∂ ∂χ2 V − ∂ ∂χ3 V

  • χi=v0

= 2 m2

B v0

0 =  ∂ ∂χ1 V − ∂ ∂χ3 V

  • χi=v0 =
  • 2 m2

A − m2 C

  • v0
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SLIDE 14

Can Vacuum Alignment be realized?

Vχ = m2

0 (χχ)11 + λ1 (χχ)11(χχ)11 + λ2 (χχ)12(χχ)13

Vsoft,Z2 = m2

Aχ2 1 + m2 Bχ2 2 + m2 Cχ2χ3

SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: has minima (1,1,1) and (1,0,0). Effect of breaking to Z2 in another sector can be included by adding: Minimization conditions then give: This thus requires mA= mB= mC=0, i.e. all non-trivial contractions between Φ and 휒 have to vanish in the potential.

0 =  ∂V ∂χ1

  • χi=v0 =

2 √ 3 ⇣ m2

0 +

√ 3m2

A

⌘ v0 + 4λ1v03 0 =  ∂ ∂χ2 V − ∂ ∂χ3 V

  • χi=v0

= 2 m2

B v0

0 =  ∂ ∂χ1 V − ∂ ∂χ3 V

  • χi=v0 =
  • 2 m2

A − m2 C

  • v0
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SLIDE 15

Can Vacuum Alignment be realized?

Vχ = m2

0 (χχ)11 + λ1 (χχ)11(χχ)11 + λ2 (χχ)12(χχ)13

Vsoft,Z2 = m2

Aχ2 1 + m2 Bχ2 2 + m2 Cχ2χ3

SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: has minima (1,1,1) and (1,0,0). Effect of breaking to Z2 in another sector can be included by adding: Minimization conditions then give: This thus requires mA= mB= mC=0, i.e. all non-trivial contractions between Φ and 휒 have to vanish in the potential. Breaking to the same subgroup of A4 can be realized. The non-trivial couplings, i.e. (Φ Φ)3(휒휒)3 thus force breaking of group to the same subgroup. the couplings cannot be forbidden by an internal symmetry that commutes with A4, as e.g.(Φ† Φ)3 is invariant under the commuting symmetry.

0 =  ∂V ∂χ1

  • χi=v0 =

2 √ 3 ⇣ m2

0 +

√ 3m2

A

⌘ v0 + 4λ1v03 0 =  ∂ ∂χ2 V − ∂ ∂χ3 V

  • χi=v0

= 2 m2

B v0

0 =  ∂ ∂χ1 V − ∂ ∂χ3 V

  • χi=v0 =
  • 2 m2

A − m2 C

  • v0
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SLIDE 16

Solutions in the Literature

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings.

Altarelli, Feruglio 2005

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SLIDE 17

Solutions in the Literature

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment.

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

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SLIDE 18

Solutions in the Literature

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Babu and Gabriel(2010) proposed the flavour group (S3)4⋊A4, which has the properties leptons transform only under A4 subgroup if one takes Φ∼16, vacuum alignment possible as V=V(Φ)+V(휒)+(Φ Φ)1(휒휒)1 neutrino masses then generated by coupling to〈Φ4〉∼(1,0,0)

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

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SLIDE 19

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Babu and Gabriel(2010) proposed the flavour group (S3)4⋊A4, which has the properties leptons transform only under A4 subgroup if one takes Φ∼16, vacuum alignment possible as V=V(Φ)+V(휒)+(Φ Φ)1(휒휒)1 neutrino masses then generated by coupling to〈Φ4〉∼(1,0,0)

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

Problems with the Solutions

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SLIDE 20

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Babu and Gabriel(2010) proposed the flavour group (S3)4⋊A4, which has the properties leptons transform only under A4 subgroup if one takes Φ∼16, vacuum alignment possible as V=V(Φ)+V(휒)+(Φ Φ)1(휒휒)1 neutrino masses then generated by coupling to〈Φ4〉∼(1,0,0)

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

Problems with the Solutions

needs high flavour scale, hard to test

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SLIDE 21

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Babu and Gabriel(2010) proposed the flavour group (S3)4⋊A4, which has the properties leptons transform only under A4 subgroup if one takes Φ∼16, vacuum alignment possible as V=V(Φ)+V(휒)+(Φ Φ)1(휒휒)1 neutrino masses then generated by coupling to〈Φ4〉∼(1,0,0)

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

Problems with the Solutions

Model is fine-tuned/needs special UV completion: different mass entries in neutrino mass matrix stem from operators of very different mass dimensions (ll)3Φ4+(ll)1 non-minimal(size: 15552), needs large representations

needs high flavour scale, hard to test

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SLIDE 22

In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Babu and Gabriel(2010) proposed the flavour group (S3)4⋊A4, which has the properties leptons transform only under A4 subgroup if one takes Φ∼16, vacuum alignment possible as V=V(Φ)+V(휒)+(Φ Φ)1(휒휒)1 neutrino masses then generated by coupling to〈Φ4〉∼(1,0,0)

Altarelli, Feruglio 2005 Altarelli, Feruglio 2006

Problems with the Solutions

Model is fine-tuned/needs special UV completion: different mass entries in neutrino mass matrix stem from operators of very different mass dimensions (ll)3Φ4+(ll)1 non-minimal(size: 15552), needs large representations

needs high flavour scale, hard to test

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SLIDE 23

Group extensions and vacuum alignment

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SLIDE 24

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)].

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SLIDE 25

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H

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SLIDE 26

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H there therefore exist representations ρi, which are directly related to the representations ρHi of H by, via ρi ≡ ρH ◦ ξ.

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SLIDE 27

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H there therefore exist representations ρi, which are directly related to the representations ρHi of H by, via ρi ≡ ρH ◦ ξ. to keep the same flavour structure as within H, we assign SM fermions to these representations, also flavon 휒 ∼ 3H ◦ ξ.

slide-28
SLIDE 28

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H there therefore exist representations ρi, which are directly related to the representations ρHi of H by, via ρi ≡ ρH ◦ ξ. to keep the same flavour structure as within H, we assign SM fermions to these representations, also flavon 휒 ∼ 3H ◦ ξ. We further demand the existence of an irreducible representation Φ, whose product Φn contains the triplet representation that the leptons transform under.

slide-29
SLIDE 29

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H there therefore exist representations ρi, which are directly related to the representations ρHi of H by, via ρi ≡ ρH ◦ ξ. to keep the same flavour structure as within H, we assign SM fermions to these representations, also flavon 휒 ∼ 3H ◦ ξ. We further demand the existence of an irreducible representation Φ, whose product Φn contains the triplet representation that the leptons transform under. Furthermore, we demand that the renormalizable scalar potential formed out of such a triplet flavon 휒 and Φ exhibits an accidental symmetry G×H, i.e. V=V(Φ) +V(휒)+(ΦΦ)1(휒휒)1.

slide-30
SLIDE 30

Group extensions and vacuum alignment

To solve the vacuum alignment problem, we extend the flavour group H [e.g. the successfull groups H=A4,T7,S4,T′ or ∆(27)]. demand that there should be a a surjective homomorphism ξ : G → H there therefore exist representations ρi, which are directly related to the representations ρHi of H by, via ρi ≡ ρH ◦ ξ. to keep the same flavour structure as within H, we assign SM fermions to these representations, also flavon 휒 ∼ 3H ◦ ξ. We further demand the existence of an irreducible representation Φ, whose product Φn contains the triplet representation that the leptons transform under. Furthermore, we demand that the renormalizable scalar potential formed out of such a triplet flavon 휒 and Φ exhibits an accidental symmetry G×H, i.e. V=V(Φ) +V(휒)+(ΦΦ)1(휒휒)1.

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SLIDE 31

Scan for Small Groups

using the computer algebra system GAP and its SmallGroups catalogue, we have checked all groups with size smaller than 1000 (11,758,814 groups) and we have found a number of candidates:

Subgroup H Order of G GAP Structure Description Z(G) A4 96 204 Q8 o A4 Z2 288 860 T 0 o A4 Z2 384 617, 20123 ((Z2 × Q8) o Z2) o A4 Z2 576 8273 (Z2.S4) o A4 Z2 768 1083945 (Z4.Z2

4) o A4

Z4 1085279 ((Z2 × Q16) o Z2) o A4 Z2 S4 192 1494 Q8 o S4 Z2 384 18133, 20092 (Z2 × Q8) o S4 Z2 20096 ((Z4 × Z2) o Z2) o S4 Z4 576 8282 T 0 o S4 Z2 8480 (Z3 × Q8) o S4 Z6 768 1086052, 1086053 ((Z2 × Q8) o Z2) o S4 Z2 960 11114 (Z5 × Q8) o S4 Z10 T 0 192 1022 Q8 o T 0 Z2

2

648 533 ∆(27) o T 0 Z3 768 1083573, 1085187 ((Z2 × Q8) o Z2) o T 0 Z2

2

no candidates for T7 or Δ(27), maybe because here 3 is complex and there are more couplings that have to be forbidden (also smaller number of possible extensions) all candidates in list have non- trivial centre(=element that commute with

all other elements), not necessary true

for all groups(see e.g. (S3)4⋊A4 studied in

Babu/Gabriel 2010)

Groups of the Structure G ⋍N⋊H, H is subgroup of G

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SLIDE 32

Scan for Small Groups

using the computer algebra system GAP and its SmallGroups catalogue, we have checked all groups with size smaller than 1000 (11,758,814 groups) and we have found a number of candidates: no candidates for T7 or Δ(27), maybe because here 3 is complex and there are more couplings that have to be forbidden (also smaller number of possible extensions) all candidates in list have non- trivial centre(=element that commute with

all other elements), not necessary true

for all groups(see e.g. (S3)4⋊A4 studied in

Babu/Gabriel 2010)

Quotient Group H Order of G GAP Structure Description A4 96 201 Z2.(Z2

2 × A4)

144 127 Z2.(A4 × S3) 192 1017 Z2.(D8 × A4) S4 96 67, 192 Z4.S4 144 121, 122 Z6.S4 192 187, 963 Z8.S4 192 987, 988 Z2.((Z2

2 × A4) o Z2)

192 1483,1484 Z2.(Z2

2 × S4)

192 1492 Z2.((Z4

2 o Z3) o Z2)

T 0 192 1007 Z2

2.(Z2 2 × A4)

Groups for which H is not a subgroup of G

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SLIDE 33

Smallest Group

The smallest candidate group that contains A4 as a subgroup is the semidirect product of the quaternion group Q8 ⌦ X, Y |X4 = 1, X2 = Y 2, Y −1XY = X−1↵ with A4 ⌦ S, T|S2 = T 3 = (ST)3 = 1 ↵ defined by the additional relations (⇔ the homomorphism 휑:H→Aut(N) introduced earlier) SXS−1 = X, SY S−1 = Y −1, TXT −1 = Y X, TY T −1 = X . Representations:

1 T SY X SY Y 2 T 2 TY S SX X STY T 11 1 1 1 1 1 1 1 1 1 1 1 12 1 ω 1 1 1 ω2 ω 1 1 1 ω2 13 1 ω2 1 1 1 ω ω2 1 1 1 ω 31 3 .

  • 1
  • 1

3 . .

  • 1
  • 1

3 . 32 3 . 3

  • 1

3 . .

  • 1
  • 1
  • 1

. 33 3 .

  • 1

3 3 . .

  • 1
  • 1
  • 1

. 34 3 .

  • 1
  • 1

3 . . 3

  • 1
  • 1

. 35 3 .

  • 1
  • 1

3 . .

  • 1

3

  • 1

. 41 4 1 . .

  • 4

1

  • 1

. . .

  • 1

42 4 ω2 . .

  • 4

ω

  • ω2

. . .

  • ω

43 4 ω . .

  • 4

ω2

  • ω

. . .

  • ω2

unfaithful A4 reps for leptons, 휒 faithful rep for Φ

X2 1 X X3 YX3 YX2 Y YX

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SLIDE 34

Smallest Group

The smallest candidate group that contains A4 as a subgroup is the semidirect product of the quaternion group Q8 ⌦ X, Y |X4 = 1, X2 = Y 2, Y −1XY = X−1↵ with A4 ⌦ S, T|S2 = T 3 = (ST)3 = 1 ↵ defined by the additional relations (⇔ the homomorphism 휑:H→Aut(N) introduced earlier) SXS−1 = X, SY S−1 = Y −1, TXT −1 = Y X, TY T −1 = X . Representations:

X2 1 X X3 YX3 YX2 Y YX

A4 reps

3i ⇥ 3i = 11 + 12 + 13 + 3iS + 3iA 3i ⇥ 3j =

5

X

k=1

k6=i,j

3k (i 6= j) 3i ⇥ 4j = 41 + 42 + 43 41 ⇥ 41 = 11S + 31A + 32S + 33S + 34S + 35A 41 ⇥ 42 = 12S + 31A + 32S + 33S + 34S + 35A

S T X Y 11 1 1 1 1 12 1 ω 1 1 13 1 ω2 1 1 31   1 −1 −1     1 1 1     1 1 1     1 1 1   41     1 1 −1 −1         1 1 1 1         −1 1 1 −1         1 −1 1 −1     Φ

faithful representation Φ is what we were looking for. (Φ Φ) only contains non-trivial contraction of the A4 subgroup.

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SLIDE 35

The model

particle SU(3)c SU(2)L U(1)Y Q8 o A4 Z4 ` 1 2

  • 1/2

31 i ec + µc + ⌧ c 1 1 1 11 + 12 + 13 −i H 1 2 1/2 11 1

  • 1

1 31 1 1 1 1 41 1 2 1 1 41 −1

slide-36
SLIDE 36

The model

particle SU(3)c SU(2)L U(1)Y Q8 o A4 Z4 ` 1 2

  • 1/2

31 i ec + µc + ⌧ c 1 1 1 11 + 12 + 13 −i H 1 2 1/2 11 1

  • 1

1 31 1 1 1 1 41 1 2 1 1 41 −1

hχi = (v0, v0, v0)T , hφ1i = 1 p 2(a, a, b, b)T , hφ2i = 1 p 2(c, c, d, d)T VEVs: h(φ1φ2)31i = 1 2(bc ad, 0, 0)T h(φ1φ2)11i = 1 2(ac + bd)

slide-37
SLIDE 37

The model

particle SU(3)c SU(2)L U(1)Y Q8 o A4 Z4 ` 1 2

  • 1/2

31 i ec + µc + ⌧ c 1 1 1 11 + 12 + 13 −i H 1 2 1/2 11 1

  • 1

1 31 1 1 1 1 41 1 2 1 1 41 −1

L(5)

e

= ye(`)11ec ˜ H/Λ + yµ(`)13µc ˜ H/Λ + yτ(`)12⌧ c ˜ H/Λ + h.c. , LO charged lepton masses: hχi = (v0, v0, v0)T , hφ1i = 1 p 2(a, a, b, b)T , hφ2i = 1 p 2(c, c, d, d)T VEVs: h(φ1φ2)31i = 1 2(bc ad, 0, 0)T h(φ1φ2)11i = 1 2(ac + bd)

v Λ B @ ye yµ yτ ye !yµ !2yτ ye !2yµ !yτ 1 C A

me∼

slide-38
SLIDE 38

The model

particle SU(3)c SU(2)L U(1)Y Q8 o A4 Z4 ` 1 2

  • 1/2

31 i ec + µc + ⌧ c 1 1 1 11 + 12 + 13 −i H 1 2 1/2 11 1

  • 1

1 31 1 1 1 1 41 1 2 1 1 41 −1

L(5)

e

= ye(`)11ec ˜ H/Λ + yµ(`)13µc ˜ H/Λ + yτ(`)12⌧ c ˜ H/Λ + h.c. , L(7)

ν

= xa(`H`H)11(12)11/Λ3 + xd(`H`H)31 · (12)31/Λ3 + h.c. . LO charged lepton masses: LO neutral lepton masses: hχi = (v0, v0, v0)T , hφ1i = 1 p 2(a, a, b, b)T , hφ2i = 1 p 2(c, c, d, d)T VEVs: h(φ1φ2)31i = 1 2(bc ad, 0, 0)T h(φ1φ2)11i = 1 2(ac + bd)

B @ ˜ a ˜ a ˜ d ˜ d ˜ a 1 C A , v Λ B @ ye yµ yτ ye !yµ !2yτ ye !2yµ !yτ 1 C A

me∼ m휈∼ TBM

(Z2xZ2 symmetry accidental)

slide-39
SLIDE 39

The model

particle SU(3)c SU(2)L U(1)Y Q8 o A4 Z4 ` 1 2

  • 1/2

31 i ec + µc + ⌧ c 1 1 1 11 + 12 + 13 −i H 1 2 1/2 11 1

  • 1

1 31 1 1 1 1 41 1 2 1 1 41 −1

L(5)

e

= ye(`)11ec ˜ H/Λ + yµ(`)13µc ˜ H/Λ + yτ(`)12⌧ c ˜ H/Λ + h.c. , L(7)

ν

= xa(`H`H)11(12)11/Λ3 + xd(`H`H)31 · (12)31/Λ3 + h.c. . LO charged lepton masses: LO neutral lepton masses: hχi = (v0, v0, v0)T , hφ1i = 1 p 2(a, a, b, b)T , hφ2i = 1 p 2(c, c, d, d)T VEVs: h(φ1φ2)31i = 1 2(bc ad, 0, 0)T h(φ1φ2)11i = 1 2(ac + bd) additional 41 necessary to get correct symmetry breaking (otherwise only breaking to A4) same # of d.o.f. as in case of complex triplet and singlet, no additional driving fields necessary low flavour symmetry breaking scale possible, testable

slide-40
SLIDE 40

Scalar Potential & Vacuum Alignment

Vφ(φ1, φ2) =µ2

1(φ1φ1)11 + α1(φ1φ1)2

11 + X

i=2,3

αi(φ1φ1)3i · (φ1φ1)3i +µ2

2(φ2φ2)11 + β1(φ2φ2)2

11 + X

i=2,3

βi(φ2φ2)3i · (φ2φ2)3i +γ1(φ1φ1)11(φ2φ2)11 + X

i=2,3,4

γi(φ1φ1)3i · (φ2φ2)3i Vχ(χ) = µ2

3(χχ)11 + ρ1(χχχ)11 + λ1(χχ)2

11 + λ2(χχ)12(χχ)13 Vmix(χ, φ1, φ2) = ζ13(φ1φ1)11(χχ)11 + ζ23(φ2φ2)11(χχ)11 The most general scalar potential invariant under the flavour symmetry is given by V (χ, φ1, φ2) = Vχ(χ) + Vφ(φ1, φ2) + Vmix(χ, φ1, φ2) with [(Q8 o A4) × A4] × Z4 Potential has an accidental symmetry invariant under independent transformations of Φ and 휒 note that couplings such as are forbidden by the auxiliary Z4 symmetry that separates the charged and neutral lepton sectors χ · (φ1φ2)31

slide-41
SLIDE 41

Scalar Potential & Vacuum Alignment

slide-42
SLIDE 42

Scalar Potential & Vacuum Alignment

Characterization of Minima If there is a minimum, in which the symmetry generator Q∈G is left unbroken, i.e. Q〈Φi〉=〈Φi〉,there are degenerate minima〈Φ‘i〉=g 〈Φ‘i〉that leave gQg -1 unbroken, with g∈G. The physically distinct minima are therefore characterized by the conjugacy class G Q={ gQg-1: g∈G} Only conjugacy classes with an eigenvalue +1 can lead to a non-trivial little group.

slide-43
SLIDE 43

Scalar Potential & Vacuum Alignment

Characterization of Minima If there is a minimum, in which the symmetry generator Q∈G is left unbroken, i.e. Q〈Φi〉=〈Φi〉,there are degenerate minima〈Φ‘i〉=g 〈Φ‘i〉that leave gQg -1 unbroken, with g∈G. The physically distinct minima are therefore characterized by the conjugacy class G Q={ gQg-1: g∈G} Only conjugacy classes with an eigenvalue +1 can lead to a non-trivial little group. For 41 there are 5 such classes

There are three physically distinct minima of φ1, that preserve a Z2 sub- group:

  • hφ1i =

1 √ 2(a, a, b, b)T results in the little group hSi,

  • hφ1i = (0, a, b, 0)T in hSY i and
  • hφ1i =

1 √ 2(a, b, a, b)T in hSY Xi .

In addition, there is one preserving a Z3 subgroup:

  • hφ1i =

1 √ 2(a, a, a, b)T preserves hTi (as well as

⌦ T 2↵ = hTi).

slide-44
SLIDE 44

Scalar Potential & Vacuum Alignment

Characterization of Minima If there is a minimum, in which the symmetry generator Q∈G is left unbroken, i.e. Q〈Φi〉=〈Φi〉,there are degenerate minima〈Φ‘i〉=g 〈Φ‘i〉that leave gQg -1 unbroken, with g∈G. The physically distinct minima are therefore characterized by the conjugacy class G Q={ gQg-1: g∈G} Only conjugacy classes with an eigenvalue +1 can lead to a non-trivial little group. For 41 there are 5 such classes

There are three physically distinct minima of φ1, that preserve a Z2 sub- group:

  • hφ1i =

1 √ 2(a, a, b, b)T results in the little group hSi,

  • hφ1i = (0, a, b, 0)T in hSY i and
  • hφ1i =

1 √ 2(a, b, a, b)T in hSY Xi .

In addition, there is one preserving a Z3 subgroup:

  • hφ1i =

1 √ 2(a, a, a, b)T preserves hTi (as well as

⌦ T 2↵ = hTi).

⇒

h(φ1φ2)31i = 1

2(bc ad, 0, 0)T

h(φ1φ2)11i = 1

2(ac + bd)

(a,b) replaced by (c,d) in Φ2

slide-45
SLIDE 45

Scalar Potential & Vacuum Alignment

Minimum Conditions

a

  • α+
  • a2 + b2

+ α

  • a2 − b2

+ γ+

  • c2 + d2

+ γ

  • c2 − d2

+ U1

  • + Γbcd = 0

b

  • α+
  • a2 + b2

− α

  • a2 − b2

+ γ+

  • c2 + d2

− γ

  • c2 − d2

+ U1

  • + Γacd = 0

c

  • β+
  • c2 + d2

+ β

  • c2 − d2

+ γ+

  • a2 + b2

+ γ

  • a2 − b2

+ U2

  • + Γabd = 0

d

  • β+
  • c2 + d2

− β

  • c2 − d2

+ γ+

  • a2 + b2

− γ

  • a2 − b2

+ U2

  • + Γabc = 0

v0 ⇣ 4 √ 3λ1v02 + 3ρ1v0 + 2µ2

3 + ζ13(a2 + b2) + ζ23(c2 + d2)

⌘ = 0

Ui = 1 2 ⇣ µ2

i +

√ 3ζi3 v02⌘ with for〈S〉, similar relations for 〈SY〉, 〈SYX〉

eleven minimization conditions reduce to these 5 equations for 5 VEVs there is therefore generally a solution note that e.g. a=b=0 or c=d=0 is also a solution, here the singlet and triplet contraction vanishes we have performed a numerical study to show that there is finite region of parameter space where 〈S〉 is the global minimum

ξ+ = ξ1

2 , ξ− = ξ2+ξ3 2 √ 3 for ξ = α, β

γ+ =

√ 3γ1+γ4 4 √ 3

, γ− = γ2+γ3

4 √ 3

and Γ = γ4

√ 3

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SLIDE 46

Higher Order Corrections

NLO Corrections to vacuum potential V (5) =

2

X

L,M=1 4

X

i,j=2

δ(LM)

ij

Λ χ · ⇢ (φLφL)3i · (φMφM)3j

  • 31

+ +χ3 Λ ⇣ δ(3)

1 χ2 + δ(3) 2 (φ1φ1)11 + δ(3) 3 (φ2φ2)11

⌘ leads to shifts in VEVs

hχi = (v0 + δv0

1, v0 + δv0 2, v0 + δv0 2)T ,

hφ1i = 1 p 2(a + δa1, a + δa2, b + δa3, b + δa4)T , hφ2i = 1 p 2(c + δb1, c + δb2, d + δb3, d + δb4)T

δ(LM)

ij

= 0 for i ≥ j

0.00 0.05 0.10 0.15 0.20 0.25 0.00 0.05 0.10 0.15 0.20 0.25 maxk u k

  • maxi ∆u i

maxk u k

δu u ∼ u Λ hχ2i hχ3i = O(1/Λ2) generic size of shifts generic size of shifts for scalar potential parameters of order one

VEV alignment not destroyed!

slide-47
SLIDE 47

Higher Order Corrections

[Lin’10, Shimizu,Tanimoto, Watanabe‘11,Luhn,King’11]

sin2훳13≈.1 as suggested by T2K can be accomodated at NLO

  • r by introducing additional non-trivial singlet field ξ ∼ (12,i)[does not

destroy VEV alignment]

slide-48
SLIDE 48

UV Completion

`α `β hHi h1i h2i hHi N S2 S3 N

For Seesaw UV completion, introduce fermionic singlets N∼ (31,-i), S2∼ (42,i), S3∼ (43,-i): generates singlet masses (N): m⌫ = x2

`Nv2m−1 N

L = x`N`HN + xN2NS21 + xN3NS32 + mS2S3 + x23S2S3 + h.c. ,

U T

ν mνUν = diag(

1 B + A, 1 A, 1 B − A)

the light neutrino mass matrix mN = xN2xN3 m   A A B B A   with A = −2(ac+bd) and B = i √ 3(bc−ad) . is of TBM form Accidental degeneracy of m1 and m3 is lifted by introduction of additional S2 or S3.

slide-49
SLIDE 49

Mathematica Package Discrete

We have developed a Mathematica Package that can be used to facilitate model building using discrete groups. It has the features: has access to groups catalogue of GAP, which contains all groups one would ever want to use

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SLIDE 50

Mathematica Package Discrete

We have developed a Mathematica Package that can be used to facilitate model building using discrete groups. It has the features: has access to groups catalogue of GAP, which contains all groups one would ever want to use calculate Kronecker products, Clebsch-Gordon coefficients, covariants formed out of product of any representation etc.

slide-51
SLIDE 51

Mathematica Package Discrete

We have developed a Mathematica Package that can be used to facilitate model building using discrete groups. It has the features: has access to groups catalogue of GAP, which contains all groups one would ever want to use calculate Kronecker products, Clebsch-Gordon coefficients, covariants formed out of product of any representation etc. reduce set covariants to a smaller set of independent covariants calculate flavon potentials

[see also SUtree, Merle Zwicky]

slide-52
SLIDE 52

Mathematica Package Discrete

We have developed a Mathematica Package that can be used to facilitate model building using discrete groups. It has the features: has access to groups catalogue of GAP, which contains all groups one would ever want to use calculate Kronecker products, Clebsch-Gordon coefficients, covariants formed out of product of any representation etc. reduce set covariants to a smaller set of independent covariants calculate flavon potentials available at http:/ /projects.hepforge.org/discrete/

slide-53
SLIDE 53

Conclusions

slide-54
SLIDE 54

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models

slide-55
SLIDE 55

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models We have identified the minimal set of symmetries needed to extend the smallest flavor groups(A4,T‘,S4,...)

slide-56
SLIDE 56

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models We have identified the minimal set of symmetries needed to extend the smallest flavor groups(A4,T‘,S4,...) We have presented a model based on Q8⋊A4, the smallest extension of A4

slide-57
SLIDE 57

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models We have identified the minimal set of symmetries needed to extend the smallest flavor groups(A4,T‘,S4,...) We have presented a model based on Q8⋊A4, the smallest extension of A4

vacuum alignment natural ⇒ TBM predicted at LO

slide-58
SLIDE 58

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models We have identified the minimal set of symmetries needed to extend the smallest flavor groups(A4,T‘,S4,...) We have presented a model based on Q8⋊A4, the smallest extension of A4

vacuum alignment natural ⇒ TBM predicted at LO low symmetry-breaking scale possible

slide-59
SLIDE 59

Conclusions

Group Extensions may be used to solve the vacuum alignment problem in flavor models We have identified the minimal set of symmetries needed to extend the smallest flavor groups(A4,T‘,S4,...) We have presented a model based on Q8⋊A4, the smallest extension of A4

vacuum alignment natural ⇒ TBM predicted at LO low symmetry-breaking scale possible