S4 Flavor Symmetry of Quarks and Leptons in SU(5) GUT

0.21 MB·English

by  Hajime Ishimori

S Flavor Symmetry of Quarks and Leptons
4
in SU(5) GUT
Hajime Ishimori1,, Yusuke Shimizu1,,
∗ †
Morimitsu Tanimoto2,
‡
9
0
0
1
Graduate School of Science and Technology, Niigata University,
2

n Niigata 950-2181, Japan
a
2
J Department of Physics, Niigata University, Niigata 950-2181, Japan

8

]
h
p
Abstract
-
p
e We present a S flavor model to unify quarks and leptons in the framework of the SU(5)
4
h
GUT. Three generations of 5-plets in SU(5) are assigned 3 of S while the first and
[ 1 4
the second generations of 10-plets in SU(5) are assigned to be 2 of S4, and the third
2
generation of 10-plet is to be 1 of S . Right-handed neutrinos are also assigned 2 for the
v 1 4
1 first and second generations and 1 for the third generation, respectively. Taking vacuum
1
3
alignments of relevant gauge singlet scalars, we predict the quark mixing as well as the
0
tri-bimaximal mixing of neutrino flavors. Especially, the Cabbiboangle is predicted to be
5
. 15 in the limit of the vacuum alignment. We can improve the model to predict observed
2 ◦
1 CKM mixing angles as well as the non-vanishing Ue3 of the neutrino flavor mixing.
8
0
:
v
i
X
r
a
∗E-mail address: [email protected]
†E-mail address: [email protected]
‡E-mail address: [email protected]
1 Introduction
Neutrino experimental data provide an important clue for elucidating the origin of the ob-
served hierarchies in mass matrices for quarks and leptons. Recent experiments of the neu-
trino oscillation go into a new phase of precise determination of mixing angles and mass
squared differences [1], which indicate the tri-bimaximal mixing for three flavors in the lep-
ton sector [2]. These largemixing angles are completely different fromthe quark mixing ones.
Therefore, it is very important to find a natural model that leads to these mixing patterns
of quarks and leptons with good accuracy.
Flavor symmetry is expected to explain the mass spectrum and the mixing matrix of
quarks and leptons. In particular, some predictive models with non-Abelian discrete flavor
symmetries have been explored by many authors. Among them, the tri-bimaximal mixing of
leptons has been understood based on the non-Abelian finite group S [3]-[19], A [20]-[41],
3 4
and T [42]-[47], because these symmetries provide the definite meaning of generations and
′
connects different generations. On the other hand, much attention has been devoted to the
question whether these models can be extended to describe the observed pattern of quark
masses and mixing angles, and whether these can be made compatible with the SU(5) or
SO(10) grand unified theory (GUT).
Recently, group-theoretical arguments indicate that the discrete symmetry S is the min-
4
imal flavor symmetry compatible with the tri-bimaximal neutrino mixing [48]. Actually, the
exact tri-bimaximal neutrino mixing is realized in the S flavor model [49]. Thus, the S
4 4
flavor model is attractive for the lepton sector [50]-[54]. Although an attempt to unify the
quark and lepton sector was presented towards a grand unified theory of flavor [51], mixing
angles are not predicted clearly.
Inourwork, wepresent aS flavor modeltounifythequarksandleptonsintheframework
4
of the SU(5) GUT. The group S has irreducible representations 1 , 1 , 2, 3 , and 3 . Three
4 1 2 1 2
generations of 5-plets in SU(5) are assigned 3 of S while the first andthe second generations
1 4
of 10-plets in SU(5) are assigned to be 2 of S , and the third generation of 10-plet is to be
4
1 of S . These assignments of S for 5 and 10 lead to the completely different structure of
1 4 4
quark and lepton mass matrices. Right-handed neutrinos, which are SU(5) gauge singlets,
are also assigned 2 for the first and second generations, and 1 for the third generation,
1
respectively. These assignments are essential to realize the tri-bimaximal mixing of neutrino
flavors. Taking vacuum alignments of relevant gauge singlet scalars, we predict the quark
mixing as well as the tri-bimaximal mixing of leptons. Especially, the Cabbibo angle is
predicted to be 15 in the limit of the vacuum alignment. We improve the model to predict
◦
observed CKM mixing angles as well as the non-vanishing U of the neutrino flavor mixing.
e3
The paper is organized as follows. We present the prototype of the S flavor model of
4
quarks and leptons in SU(5) GUT in section 2, and discuss the lepton sector in section 3, and
the quark sector in section 4. In section 5, we present the improved model with additional
scalar in order to study detail of the model. Section 6 is devoted to the summary. In the
appendix, wepresent themultiplicationrulesofS ,andthescalarpotentialanalysis. Vacuum
4
alignments and magnitude of VEVs are also summarized in the appendix.
1
2 Prototype of S flavor model in SU(5) GUT
4
We present the prototype of the S flavor model in the SU(5) GUT to understand the essence
4
of our model clearly. We consider the supersymmetric GUT based on SU(5). The flavor
symmetry of quarks and leptons is the discrete group S in our model. The group S has ir-
4 4
reducible representations 1 , 1 , 2, 3 ,and3 . The multiplication rules of S aresummarized
1 2 1 2 4
in appendix.
T (T ,T ) (F ,F ,F ) (Nc,Nc) Nc H H
3 1 2 1 2 3 e µ τ 5 ¯5
SU(5) 10 10 ¯5 1 1 5 ¯5
S 1 2 3 2 1 1 1
4 1 1 1 1 1
Z ω2 ω3 ω 1 1 1 1
4
χ (χ ,χ ) (χ ,χ ) (χ ,χ ,χ ) (χ ,χ ,χ ) (χ ,χ ,χ )
1 2 3 4 5 6 7 8 9 10 11 12 13 14
SU(5) 1 1 1 1 1 1
S 1 2 2 3 3 3
4 1 1 1 1
Z ω2 ω2 1 ω3 1 ω
4
Table 1: Assignments of SU(5), S , and Z representations, where the phase factor ω is i.
4 4
Let us present the model of the quark and lepton flavor with the S group in SU(5)
4
GUT. In SU(5), matter fields are unified into a 10(q ,uc,ec) and a ¯5(dc,l ) dimensional
1 L e L
representations. Three generations of ¯5, which are denoted by F , are assigned by 3 of S .
i 1 4
On the other hand, the third generation of the 10-dimensional representation is assigned by
1 of S , so that the top quark Yukawa coupling is allowed in tree level. While, the first
1 4
and the second generations are assigned 2 of S . These 10-dimensional representations are
4
denoted by T and (T ,T ), respectively. Right-handed neutrinos, which are SU(5) gauge
3 1 2
singlets, are also assigned 1 and 2 for Nc and (Nc,Nc), respectively 1.
1 τ e µ
We introduce new scalars χ in addition to the 5-dimensional and ¯5-dimensional Higgs
i
of the SU(5), H and H which are assigned 1 of S . These new scalars are supposed
5 ¯5 1 4
to be SU(5) gauge singlets. The χ scalar is assigned 1 , (χ ,χ ), and (χ ,χ ) are 2,
1 1 2 3 4 5
(χ ,χ ,χ ), (χ ,χ ,χ ), and (χ ,χ ,χ ) are 3 of the S representations, respectively.
6 7 8 9 10 11 12 13 14 1 4
The χ and (χ ,χ ) scalars are coupled with the up type quark sector, (χ ,χ ) are cou-
1 2 3 4 5
pled with the right-handed Majorana neutrino sector, (χ ,χ ,χ ) are coupled with the Dirac
6 7 8
neutrino sector, (χ ,χ ,χ ) and (χ ,χ ,χ ) are coupled with the charged lepton and
9 10 11 12 13 14
down type quark sector, respectively. We also add Z symmetry in order to obtain relevant
4
couplings. The particle assignments of SU(5), S , and Z are summarized Table 1.
4 4
We can now write down the superpotential at the leading order in terms of the cut off
scale Λ, which is taken to be the Planck scale. The SU(5) invariant superpotential of the
1
The similar assignments of right-handed neutrinos were presented in the first version of Ref.[49].
2
Yukawa sector respecting S and Z symmetries is given as
4 4
w(0) = yu(T ,T ) (T ,T ) χ H /Λ+yu(T ,T ) (T ,T ) (χ ,χ ) H /Λ
SU(5) 1 1 2 ⊗ 1 2 ⊗ 1 ⊗ 5 2 1 2 ⊗ 1 2 ⊗ 2 3 ⊗ 5
+yuT T H +M (Nc,Nc) (Nc,Nc)+M Nc Nc
3 3 ⊗ 3 ⊗ 5 1 e µ ⊗ e µ 2 τ ⊗ τ
+yN(Nc,Nc) (Nc,Nc) (χ ,χ )
e µ ⊗ e µ ⊗ 4 5
+yD(Nc,Nc) (F ,F ,F ) (χ ,χ ,χ ) H /Λ
1 e µ ⊗ 1 2 3 ⊗ 6 7 8 ⊗ 5
+yDNc (F ,F ,F ) (χ ,χ ,χ ) H /Λ
2 τ ⊗ 1 2 3 ⊗ 6 7 8 ⊗ 5
+y (F ,F ,F ) (T ,T ) (χ ,χ ,χ ) H /Λ
1 1 2 3 1 2 9 10 11 ¯5
⊗ ⊗ ⊗
+y (F ,F ,F ) T (χ ,χ ,χ ) H /Λ, (1)
2 1 2 3 3 12 13 14 ¯5
⊗ ⊗ ⊗
where M and M are mass parameters for right-handed Majorana neutrinos, and Yukawa
1 2
coupling constants ya andy arecomplex ingeneral. By decomposing this superpotential into
i i
the quark sector and the lepton sector, we can discuss mass matrices of quarks and leptons
in following sections.
3 Lepton sector
(0)
At first, we begin to discuss the lepton sector of the superpotential w . Denoting Higgs
SU(5)
doublets as h and h , the superpotential of the Yukawa sector respecting the S Z sym-
u d 4 4
×
metry is given for charged leptons as
ec µc
w = y (l χ l χ )+ ( 2l χ +l χ +l χ ) h /Λ
l 1 µ 10 τ 11 e 9 µ 10 τ 11 d
√2 − √6 −
(cid:20) (cid:21)
+y τc(l χ +l χ +l χ )h /Λ. (2)
2 e 12 µ 13 τ 14 d
For right-handed Majorana neutrinos, the superpotential is given as
w = M (NcNc +NcNc)+M NcNc
N 1 e e µ µ 2 τ τ
+yN (NcNc +NcNc)χ +(NcNc NcNc)χ , (3)
e µ µ e 4 e e − µ µ 5
(cid:2) (cid:3)
and for neutrino Yukawa couplings, the superpotential is
Nc Nc
w = yD e (l χ l χ )+ µ( 2l χ +l χ +l χ ) h /Λ
D 1 √2 µ 7 − τ 8 √6 − e 6 µ 7 τ 8 u
(cid:20) (cid:21)
+yDNc(l χ +l χ +l χ )h /Λ. (4)
2 τ e 6 µ 7 τ 8 u
Higgs doublets h ,h and gauge singlet scalars χ , are assumed to develop their vacuum
u d i
expectation values (VEVs) as follows:
h = v , h = v , (χ ,χ ) = (u ,u ), (χ ,χ ,χ ) = (u ,u ,u ),
u u d d 4 5 4 5 6 7 8 6 7 8
h i h i h i h i
(χ ,χ ,χ ) = (u ,u ,u ), (χ ,χ ,χ ) = (u ,u ,u ), (5)
9 10 11 9 10 11 12 13 14 12 13 14
h i h i
3
which are supposed to be real. Then, we obtain the mass matrix for charged leptons as
0 α /√2 α /√2 0 0 0
10 11
−
M = y v 2α /√6 α /√6 α /√6 +y v 0 0 0 , (6)
l 1 d− 9 10 11  2 d 
0 0 0 α α α
12 13 14
   
while the right-handed Majorana neutrino mass matrix is given as
M +yNα Λ yNα Λ 0
1 5 4
M = yNα Λ M yNα Λ 0 , (7)
N 4 1 5
 − 
0 0 M
2
 
and the Dirac mass matrix of neutrinos is
0 α /√2 α /√2 0 0 0
7 8
−
M = yDv 2α /√6 α /√6 α /√6 +yDv 0 0 0 , (8)
D 1 u− 6 7 8  2 u 
0 0 0 α α α
6 7 8
   
where we denote α u /Λ.
i i
≡
Let us discuss lepton masses and mixing angles by considering mass matrices in Eqs.(6),
(7) and (8). In order to get the left-handed mixing of charged leptons, we investigate Ml†Ml:
Ml†Ml = vd2×
2 y 2α2 + y 2α2 1 y 2α α + y 2α α 1 y 2α α + y 2α α
3| 1| 9 | 2| 12 −3| 1| 9 10 | 2| 12 13 −3| 1| 9 11 | 2| 12 14
1 y 2α α + y 2α α 2 y 2α2 + y 2α2 1 y 2α α + y 2α α .
−3| 1| 9 10 | 2| 12 13 3| 1| 10 | 2| 13 −3| 1| 10 11 | 2| 13 14
1 y 2α α + y 2α α 1 y 2α α + y 2α α 2 y 2α2 + y 2α2
−3| 1| 9 11 | 2| 12 14 −3| 1| 10 11 | 2| 13 14 3| 1| 11 | 2| 14
 (9)
If we can take vacuum alignments (u ,u ,u ) = (u ,u ,0) and (u ,u ,u ) = (0,0,u ),
9 10 11 9 10 12 13 14 14
that is α = α = α = 0, we obtain
11 12 13
2 y 2α2 1 y 2α α 0
3| 1| 9 −3| 1| 9 10
Ml†Ml = vd2−31|y1|2α9α10 23|y1|2α120 0 , (10)
0 0 y 2α2
| 2| 14
 
which gives θl = θl = 0, where θl denote left-handed mixing angles to diagonalize the
13 23 ij
charged lepton mass matrix. Since the electron mass is tiny compared with the muon mass,
we expect α α and then we get the mixing angle θl as,
9 ≪ 10 12
α
tanθl 9 , (11)
12 ≈ −2α
10
and charged lepton masses,
1 2 1 2
m2 y 2α2v2 , m2 y 2α2 v2 + y 2α2v2 y 2α2 v2 , m2 = y 2α2 v2 . (12)
e ≈ 2| 1| 9 d µ ≈ 3| 1| 10 d 6| 1| 9 d ≈ 3| 1| 10 d τ | 2| 14 d
Therefore, the mixing of θl is estimated as
12
1 m
tanθl e 2.8 10 3, (13)
| 12| ≈ √3m ≈ × −
µ
4
which is negligibly small. The mixing angle θl is at most (m /m ) even if we take into
13 O e τ
account of non-zero α . These tiny θl and θl hardly affect the magnitude of the lepton
12 12 13
mixing matrix element U , which will be discussed later.
e3
It is noticed that one can take at the leading order the vacuum alignment (u ,u ,u ) =
9 10 11
(0,u ,0) in order to guarantee α α , in which the electron mass vanishes. In conclusion,
10 9 10
≪
we find the charged lepton mass matrix to be almost diagonal one.
Taking vacuum alignments (u ,u ) = (0,u ) and (u ,u ,u ) = (u ,u ,u ) in Eq.(7), the
4 5 5 6 7 8 6 6 6
Majorana mass matrix of neutrinos turns to
M +yNα Λ 0 0
1 5
M = 0 M yNα Λ 0 , (14)
N 1 5
 − 
0 0 M
2
 
and the Dirac mass matrix of neutrinos turns to
0 α /√2 α /√2 0 0 0
6 6
−
M = yDv 2α /√6 α /√6 α /√6 +yDv 0 0 0 . (15)
D 1 u− 6 6 6  2 u 
0 0 0 α α α
6 6 6
   
By using the seesaw mechanism M = MTM 1M , the left-handed Majorana neutrino mass
ν D N− D
matrix is written as
a+ 2b a 1b a 1b
3 − 3 − 3
M = a 1b a+ 1b+ 1c a+ 1b 1c , (16)
ν  − 3 6 2 6 − 2 
a 1b a+ 1b 1c a+ 1b+ 1c
− 3 6 − 2 6 2
 
where
(yDα v )2 (yDα v )2 (yDα v )2
a = 2 6 u , b = 1 6 u , c = 1 6 u . (17)
M M yNα Λ M +yNα Λ
2 1 5 1 5
−
The neutrino mass matrix is decomposed as
1 0 0 1 1 1 1 0 0
b+c 3a b b c
M = 0 1 0 + − 1 1 1 + − 0 0 1 . (18)
ν
2   3   2  
0 0 1 1 1 1 0 1 0
     
As well known, the neutrino mass matrix with the tri-bimaximal mixing is expressed in terms
of neutrino mass eigenvalues m , m and m as
1 2 3
1 0 0 1 1 1 1 0 0
m +m m m m m
1 3 2 1 1 3
M = 0 1 0 + − 1 1 1 + − 0 0 1 . (19)
ν
2   3   2  
0 0 1 1 1 1 0 1 0
     
Therefore, our neutrino mass matrix M gives the tri-bimaximal mixing matrix U and
ν tri-bi
mass eigenvalues as follows:
2 1 0
√6 √3
U = 1 1 1 , m = b , m = 3a , m = c . (20)
tri-bi −√6 √3 −√2 1 2 3
1 1 1
−√6 √3 √2 
 
5
We remind ourselves that the flavor mixing from the charged lepton sector is negligibly small.
Defining parameters µ = v /Λ, λ = M /Λ and λ = M /Λ, and taking yD = yD, the
0 u 1 1 2 2 1 2
observed values ∆m2 and ∆m2 are expressed as
atm sol
4yNα λ 9(λ2 yNα )2 λ2
∆m2 = 5 1 (yDα )4µ2v2, ∆m2 = 1 − 5 − 2 (yDα )4µ2v2. (21)
atm −(λ2 yN2α2)2 1 6 0 u sol λ2(λ yNα )2 1 6 0 u
1 − 5 2 1 − 5
PuttingΛ = 2.43 1018GeVandexperimental
×
values of ∆m2 = (2.1 2.8) 10 3eV2 and
atm − × −
∆m2 = (7.1 8.3) 10 5eV2 [1], wecanestimate 1
sol − × −
magnitudesofα andα inthecaseofthenormal 0.5
5 6
neutrino mass hierarchy. We show the numerical
0.2
2
result in Figure 1, where we fix yN = 1 and 0
1 0.1
− ´
yD = 1. We find α 0.1, which is much larger
1 6 ≥ Α50.05
than α 10 4 10 2.
5 − −
≈ − 0.02
Let us discuss a possible case of the non-
vanishing U , which is the deviation from the
e3 0.1 0.2 0.3 0.4 0.5
tri-bimaximal mixing. If α = 0, which corre- Α
4 6
6
sponds to the non-vanishing u , the left-handed
4
Majorana neutrino mass matrix deviates from Figure 1: The allowed region on the α
5
−
Eq.(16). After rotating by the tri-bimaximal α plane.
6
mixing matrix (Mˆ = UT M U ), we obtain
ν tri-bi ν tri-bi
off-diagonalelementsintheneutrinomassmatrix
due to the non-zero α as follows:
4
y1D2(M1+yNα5Λ) 0 y1D2yNα4Λ
Mˆν = α62vu2 M12−yN2(0α24+α25)Λ2 3My2D22 −M12−yN20(α24+α25)Λ2. (22)
−M12−yy1DN22y(Nα24α+4Λα25)Λ2 0 My121D−2y(MN21(−αy24N+αα525Λ)Λ)2 
Then the mixing angle δθν ,which diagonalizes this mass matrix, is givenas
13
α
tan2δθν = 4, (23)
13 α
5
which leads to
2 1 α 1 1 1 α
U = δθν 4 , U = , U + 4 . (24)
| e3| √6| 13| ≈ √6 α | e2| √3 | µ3| ≈ √2 2√6α
(cid:12) 5(cid:12) (cid:12) 5(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Thus, the magnitude of Ue3 i(cid:12)s de(cid:12)termined by the non-vanishing ra(cid:12)tio α4/α5. (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
4 Quark sector
(0)
In this section, we discuss quark sector of the superpotential w . For up type quarks, the
SU(5)
superpotential of the Yukawa sector with S Z is given as
4 4
×
w = yu(ucq +ccq )χ h /Λ
u 1 1 2 1 u
+yu[(ucq +ccq )χ +(ucq ccq )χ ]h /Λ+yutcq h . (25)
2 2 1 2 1 − 2 3 u 3 3 u
6
For down type quarks, we can write the superpotential as follows:
1 1
w = y (scχ bcχ )q + ( 2dcχ +scχ +bcχ )q h /Λ
d 1 10 11 1 9 10 11 2 d
√2 − √6 −
(cid:20) (cid:21)
+y (dcχ +scχ +bcχ )q h /Λ. (26)
2 12 13 14 3 d
We assume that scalar fields, χ , develop their VEVs as follows:
i
χ = u , (χ ,χ ) = (u ,u ),
1 1 2 3 2 3
h i h i
(χ ,χ ,χ ) = (u ,u ,u ), (χ ,χ ,χ ) = (u ,u ,u ). (27)
9 10 11 9 10 11 12 13 14 12 13 14
h i h i
Then, the mass matrix for up type quarks is given as
yuα +yuα yuα 0
1 1 2 3 2 2
M = v yuα yuα yuα 0 , (28)
u u 2 2 1 1 − 2 3 
0 0 yu
3
 
and the down type quark mass matrix is given as
0 2α /√6 0 0 0 α
9 12
−
M = y v α /√2 α /√6 0 +y v 0 0 α . (29)
d 1 d 10 10 2 d 13
   
α /√2 α /√6 0 0 0 α
11 11 14
−
   
Let us discuss masses and mixing of the quark sector. For up type quarks, if we take
α = 0, yuα = yuα , (30)
3 1 1 2 2
which will be reexamined to get observed CKM mixing angles in section 5.2, then, we have
yuα yuα 0
1 1 1 1
M = v yuα yuα 0 , (31)
u u 1 1 1 1 
0 0 yu
3
 
which is diagonalized by the orthogonal matrix U
u
cos45 sin45 0
◦ ◦
U = sin45 cos45 0 . (32)
u ◦ ◦
− 
0 0 1
 
The up type quark masses are given as
m = 0, m = 2yuv α , m = yuv . (33)
u c 1 u 1 t 3 u
For down type quarks, putting α = α = α = 0, which is the condition in the charged
11 12 13
lepton sector, we have
1 y 2α2 1 y 2α2 0
2| 1| 10 2√3| 1| 10
Md†Md = vd22√13|y1|2α120 16|y1|2(4α92 +α120) 0 . (34)
0 0 y 2α2
| 2| 14
 
7
Then, the mass matrix is diagonalized by the orthogonal matrix U as
d
cos60 sin60 0
◦ ◦
U = sin60 cos60 0 , (35)
d ◦ ◦
− 
0 0 1
 
where the small α is neglected. The down type quark masses are given as
9
1 2
m2 y 2α2v2 , m2 y 2α2 v2 , m2 y 2α2 v2 , (36)
d ≈ 2| 1| 9 d s ≈ 3| 1| 10 d b ≈ | 2| 14 d
which are the same ones of charged lepton masses in Eq.(12).
Now, we get the CKM matrix as follows:
cos15 sin15 0
◦ ◦
VCKM = U U = sin15 cos15 0 . (37)
u† d − ◦ ◦ 
0 0 1
 
Therefore, in our prototype model of SU(5) GUT with the S flavor symmetry, the quark
4
sector has a non-vanishing mixing angle 15 only between the first and second generations
◦
while the lepton flavor mixing is tri-bimaximal. In order to get the non-vanishing but small
mixing angles VCKM and VCKM, we improve the prototype model in the next section.
cb ub
5 Improved S flavor model in SU(5) GUT
4
We improve the prototype model to get the observed quark and lepton mass spectra and the
CKM mixing matrix. We introduce the SU(5) 45-dimensional Higgs h , which is required
45
to get the difference between the charged lepton mass spectrum and the down type quark
mass spectrum. Moreover, we add a S doublet (χ ,χ ) and a S triplet (χ ,χ ,χ ), which
4 ′2 ′3 4 ′9 ′10 ′11
are SU(5) gauge singlet scalars. These assignments of SU(5), S , and Z are summarized
4 4
Table 2. Since the additional scalars do not contribute to the neutrino sector, the result of
the neutrino sector in the prototype model is not changed. Therefore, we discuss only the
charged lepton sector and the quark sector in this section.
h (χ ,χ ) (χ ,χ ,χ )
45 ′2 ′3 ′9 ′10 ′11
SU(5) 45 1 1
S 1 2 3
4 1 1
Z ω2 ω3 ω2
4
Table 2: Assignments of additional scalars in SU(5), S , and Z representations.
4 4
The superpotential of the Yukawa sector respecting the SU(5), S and Z symmetries is
4 4
given as
(0) (1)
w = w +w , (38)
SU(5) SU(5) SU(5)
where we denote
w(1) = yu(T ,T ) T (χ ,χ ) H
SU(5) 4 1 2 ⊗ 3 ⊗ ′2 ′3 ⊗ 5
+y (F ,F ,F ) (T ,T ) (χ ,χ ,χ ) h . (39)
1′ 1 2 3 ⊗ 1 2 ⊗ ′9 ′10 ′11 ⊗ 45
8
5.1 Improved lepton sector
Let us discuss the improved lepton sector of the superpotential w . The superpotential
SU(5)
of the charged lepton sector with S Z is given as
4 4
×
ec µc
w = y (l χ l χ )+ ( 2l χ +l χ +l χ ) h /Λ
l 1 µ 10 τ 11 e 9 µ 10 τ 11 d
√2 − √6 −
(cid:20) (cid:21)
ec µc
3y (l χ l χ )+ ( 2l χ +l χ +l χ ) h /Λ
− 1′ √2 µ ′10 − τ ′11 √6 − e ′9 µ ′10 τ ′11 45
(cid:20) (cid:21)
+y τc(l χ +l χ +l χ )h /Λ. (40)
2 e 12 µ 13 τ 14 d
We denote their VEVs as follows:
h = v , (χ ,χ ,χ ) = (u ,u ,u ),
45 45 9 10 11 9 10 11
h i h i
(χ ,χ ,χ ) = (u ,u ,u ), (χ ,χ ,χ ) = (u ,u ,u ). (41)
h ′9 ′10 ′11 i ′9 ′10 ′11 h 12 13 14 i 12 13 14
Then, we obtain the mass matrix for charged leptons:
0 α /√2 α /√2 0 0 0
10 11
−
M = y v 2α /√6 α /√6 α /√6 +y v 0 0 0
l 1 d 9 10 11 2 d
−   
0 0 0 α α α
12 13 14
 0 α /√2 α /√2  
1′0 − 1′1
3y v 2α /√6 α /√6 α /√6 , (42)
− 1′ 45− 9′ 1′0 1′1 
0 0 0
 
where we denote α = u /Λ and α = u /Λ. It is noticed that the third matrix in the right
i i j′ ′j
hand side is the additional one compared with the mass matrix of the prototype model in
Eq.(6).
Masses andmixing anglesofthechargedleptonsector aresimilartothoseoftheprototype
model in Eqs.(12) and (13). If we can take the vacuum alignments (u ,u ,u ) = (u ,u ,0),
9 10 11 9 10
(u ,u ,u ) = (u ,u ,0) and (u ,u ,u ) = (0,0,u ), that is α = α = α = α = 0,
′9 ′10 ′11 ′9 ′10 12 13 14 14 11 1′1 12 13
we obtain charged lepton mass matrix as follow:
0 (y α 3y¯ α )/√2 0
1 10 − 1 1′0
M = v 2(y α 3y¯ α )/√6 (y α 3y¯ α )/√6 0 , (43)
l d− 1 9 − 1 9′ 1 10 − 1 1′0 
0 0 y α
2 14
 
where we replace y v with y¯ v . Since we have
1′ 45 1 d
Ml†Ml = vd2 ×
2 y α 3y¯ α 2 1(y α 3y¯ α )(y α 3y¯ α ) 0
3| 1 9 − 1 9′| −3 1∗ 9 − 1∗ 9′ 1 10 − 1 1′0
1(y α 3y¯ α )(y α 3y¯ α ) 2 y α 3y¯ α 2 0 ,
−3 1 9 − 1 9′ 1∗ 10 − 1∗ 1′0 3| 1 10 − 1 1′0| 
0 0 y 2α2
| 2| 14
 (44)
9