Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

Supersymmetry. Theory, Experiment, and Cosmology

.pdf
Скачиваний:
79
Добавлен:
01.05.2014
Размер:
12.6 Mб
Скачать

The minimal supersymmetric SU(5) model 239

Σ

Σ

Σ

Σ

Σ

~

Σ

 

 

 

Σ

 

Σ

 

Σ, H

Σ, H

 

 

~

 

~

 

 

X

 

X

 

 

 

H

 

~

 

H

H

H

H

H

H

H

 

 

 

 

Fig. 9.4

 

 

 

ones have vanishing terms on the diagonal. Thus, the mixed terms (9.56) do not generate a mass term for the Hi superfields.

We still have not shown that we can generate a mass term for the triplets of Higgs while keeping the doublets light. But the preceding discussion shows that, if we succeed to do that at tree level, then we can count on supersymmetry to ensure that radiative corrections will follow the same pattern. Let us show how this may be realized in practice.

If we consider the following superpotential for H1 and H2:

W = H1 (µ + λΣ) H2,

(9.57)

the scalar potential reads

 

V = |(µ + λΣ)H2|2 + |H1(µ + λΣ)|2 + · · ·

(9.58)

where the minimization of the extra terms is supposed to fix Σ to its value (9.49).

Then,

 

 

 

 

µ + λΣ =

(µ + 2λσ)1l

3

0

,

0

(µ − 3λσ)1l2

with σ = M/Λ. Thus, the mass of the doublet is m2 = µ − 3λσ whereas the mass of the triplets is MT = µ + 2λσ. If we impose that

µ 3λσ

(9.59)

which is of the order of the grand unification scale, then one may realize a doublet– triplet splitting. This condition may seem at this point ad hoc but, as long as supersymmetry is unbroken, it is natural in the technical sense3.

It remains to see which dynamics might be responsible for the condition (9.59). In the sliding singlet approach [292,372], one introduces a chiral superfield S which

is a gauge singlet and one modifies the superpotential (9.57) as follows:

W = H1 (κS + λΣ) H2.

(9.61)

3Using the fact that the terms in the superpotential (9.57) are not renormalized and defining the wave function renormalization constant Z as Hi = Z1/2HiR , we have

µR = Zµ, λR σR = Zλσ,

(9.60)

and thus µR 3λR σR follows from (9.59).

240 Supersymmetric grand unification

The potential is now of the form (9.58) with µ replaced by κS. Its minimization leads to the conditions (κs − 3λσ)vi = 0, i = 1, 2 where s ≡ S . Hence, once SU (2) × U (1) is broken (v1 or v2 = 0), we have automatically κs − 3λσ = 0, which ensures massless doublets at this order. In other words, the vev of the gauge singlet slides in order to minimize the ground state energy, and correspondingly the mass of the doublets. Obviously, supersymmetry breaking will modify the analysis: it is thus important to make sure that the scale of supersymmetry breaking in this sector remains low.

Another line of attack when the scale of supersymmetry breaking is not small is to introduce a Higgs representation which contains triplets of SU (3) (i.e. (3, 1) under SU (3) × SU (2)) but no doublet of SU (2) (i.e. (1, 2) under SU (3) × SU (2)). Coupling this representation to H1 and H2, one thus gives a mass to the triplets but not to the doublets. This is the missing doublet mechanism [178, 208, 284].

An example is provided by the representation 50 of SU (5). Its decomposition under

SU (3) × SU (2) reads:

 

¯

¯

50 = (8, 2) + (6, 3) + (6, 1) + (3, 2) + (3, 1) + (1, 1).

One chooses to break SU (5) down to SU (3) × SU (2) × U (1) with a 75 of SU (5)

 

˜

 

 

¯

 

 

which we note Σ. Then we introduce a 50 (C) and a 50 (C); using the fact that

¯

1, we may write the superpotential as

 

50 × 75 × 5

 

 

W = ρ CΣ˜ H1 + ρ C¯Σ˜ H2 + M CC.¯

(9.62)

˜

 

 

 

 

 

 

 

 

 

 

 

 

 

Then, replacing Σ by its vev of order MX , we may write the part of the superpotential

 

 

 

 

 

 

 

¯

 

 

 

 

 

 

relevant for the triplets (T ) and antitriplets (T ) as

 

 

 

 

W = ρMX T¯H

TC + ρ MX T¯¯ TH

2

+ M T¯¯ TC

 

 

 

 

1

 

 

C

 

 

 

C

 

 

 

 

 

 

M

 

TC

M

 

TH2

M 2

(9.63)

= M T¯C¯ + ρ M T¯H1

+ ρ M

− ρρ

M T¯H1 TH2 .

 

 

 

X

 

 

 

 

 

X

 

 

X

 

Hence the triplets acquire a mass

of

order MX2 /M

whereas

the doublets

remain

massless.

 

 

 

 

 

 

 

 

 

 

 

 

 

9.3.3Fermion masses

Fermion masses arise from Yukawa interactions which are derived directly from the superpotential. Denoting, as above, by χmn and ηm the superfields in representations

¯

10 and 5, the superpotential reads:

 

W = −λd ηmχmnH1n − λu mnpqrχmnχpq H2r,

(9.64)

 

¯

¯

10 × 10 × 5

 

 

5

× 10 × 5

 

p

 

 

 

¯

where H1

and H2r are respectively the Higgs superfields in representations 5 and 5

of SU (5) and is the completely antisymmetric tensor.

One easily deduces the Yukawa couplings and, setting the Higgs fields at their

vacuum expectation values H1m = −v1δm5 and H2m = v2δm5, one obtains

 

Lm = −λdv1 Ψηm Ψχm5 + λuv2 mnpq5Ψχmn Ψχpq + h.c.

(9.65)

The minimal supersymmetric SU(5) model 241

with obvious notation. Reading in (9.4) and (9.6) the fermion content of the quark– lepton representations, we conclude that

md = me = −λdv1, mu = λuv2.

(9.66)

If we restore the three families, we conclude that SU (5) grand unification (supersymmetric or not, as a matter of fact) makes the following extra prediction as compared to the Standard Model:

mb = mτ ,

(9.67)

ms = mµ,

(9.68)

md = me.

(9.69)

These relations are valid at the scale of grand unification and must be renormalized down to low energies.

[Bottom–tau unification

We first study bottom–tau unification: the renormalization group evolution is complicated by the fact that the top Yukawa coupling is not small. In fact, we have already solved the evolution equation for the top coupling in Section 6.9.1 of Chapter 6 and we will follow the same method here. Neglecting all other Yukawa couplings than λt in the evolution, the renormalization group equations for the bottom and tau Yukawa couplings read:

 

 

µ dλb

 

1

16 2

 

2

 

7

 

 

2

 

λt2

 

 

 

 

 

 

 

 

=

 

 

 

 

g3

+ 3g2

+

 

 

g1

+

 

(9.70)

 

λb

16π2

3

 

15

16π2

 

 

µ dλτ

=

1

3g22 +

9

g12

.

 

 

 

 

 

 

 

(9.71)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λτ

16π2

5

 

 

 

 

 

 

 

Defining as in Chapter 6, equation (6.104),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λ2

 

 

 

 

λ2

 

 

 

 

 

(9.72)

 

 

 

 

 

 

 

Yb

b

 

Yt

 

t

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

,

 

 

 

 

 

 

 

 

 

 

16π2

 

16π2

 

 

 

the renormalization equation for λb is written

µ dYb

 

1

32 2

2

 

14 2

 

 

 

 

 

=

 

 

 

g3

+ 6g2

+

 

g1

+ 2Yt,

(9.73)

Yb

16π2

3

15

where Yt(µ) is given explicitly in equation (6.108) of Chapter 6. In the absence of the Yt term, this would read:

 

 

 

µ dYb

=

32 µ dg3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Yb

9

g3

which is readily solved as

 

 

 

 

 

 

 

 

 

Yb(µ) =

1

Eb(µ) with Eb(µ) =

g3(µ)

 

αb

g3(µ0)

 

6

µ dg2

14 µ dg1

(9.74)

g2

 

99

 

g1

 

 

 

32/9

 

g2(µ) 6

 

g1(µ)

14/99

 

 

 

 

 

. (9.75)

 

 

g2(µ0)

g1(µ0)

5g2
1 + 3
12π2

242 Supersymmetric grand unification

The complete equation is solved by turning the constant αb into a function αb(µ) which satisfies (µ/αb)(b/dµ) = 2Yt. Using (6.108) of Chapter 6, this is solved as:

αb(µ) = αb(µ0) [1 12Yt(µ0)Ft(µ)]1/6

where Ft(µ) is given in equation (6.109) of the same chapter. Hence

Yb(µ0)Eb(µ)

Yb(µ) = [1 12Yt(µ0)Ft(µ)]1/6 .

Choosing µ0 = MU , we thus have for the bottom mass

g3(µ) mb(µ) = mb(MU ) g3(MU )

7/99

×g1(µ)

g1(MU )

16/9 3

g2(µ)

g2(MU )

1

[1 3λ2t (MU )Ft(µ)/(4π2)]1/12

(9.76)

(9.77)

(9.78)

and we would obtain along the same lines from (9.71)

mτ (µ) = mτ (MU )

g2(µ)

3

 

g1(µ)

3/11

(9.79)

 

 

 

 

.

g2(MU )

 

g1(MU )

Using the unification constraint mb(MU ) = mτ (MU ), we deduce

 

 

mb(mb) = mτ (mb)

g3(mb)

16/9

 

g1(mb)

20/99

 

 

1

 

 

 

 

 

 

 

 

.

g3(MU )

 

 

g1(MU )

 

 

 

[1 3λt2(MU )Ft(mb))/(4π2)]1/12

In this equation, we can safely approximate mτ (mb) with the physical mass mτ

(9.80)

= 1.78

GeV. On the other hand, QCD corrections introduce a significant di erence between the running bottom quark mass and its physical mass which we will denote by Mb. More precisely, we have at one loop in the DR renormalization scheme4:

Mb = mb(Mb)DR . (9.82)

For a physical mass Mb = 4.7 GeV, this gives a current mass mb(Mb)DR = 4.14 GeV

or mb(Mb)MS = 4.24 GeV (we use α3(Mb) = 0.2246).

Putting these numbers in (9.80) shows that the correction due to the top Yukawa coupling (i.e. the last factor) should reduce the ratios of gauge coupling constants by an approximate factor of 2. It is thus necessary to take λt(MU ) rather large. As discussed in Section 6.9.1 of Chapter 6, this drives the low energy evolution of the top coupling towards the quasi-infrared fixed points.]

4As explained above, we are working in the dimensional reduction scheme DR. The translation to the standard MS scheme is given by:

 

 

 

 

1

g2

(µ)

.

 

mb(µ)

 

= mb(µ)

 

3

 

 

(9.81)

DR

MS

3π

The minimal supersymmetric SU(5) model 243

First and second family unification

The same procedure can be applied to the first two families, following (9.68) and (9.69). The formulas obtained are similar to (9.80) without the correction due to the top Yukawa coupling. This gives a s quark mass of order 500 MeV, which is di cult to reconcile with estimates based on chiral symmetry breaking.

This becomes even worse when one realizes that one may get rid of most of the

renormalization e ects by considering the ratios:

 

 

 

 

md

 

me

1

 

 

(9.83)

 

 

 

 

=

 

,

 

 

ms

mµ

207

 

which is in gross contradiction with the current algebra value of m

2

2

m2 ) = 1/24 [176].

 

 

 

 

 

d/ms mπ /(2mK

π

 

 

 

 

 

 

 

 

The latter problem has led to reconsider partially the problem of fermion masses in SU (5). One attitude is to consider that d being a light quark, its mass is sensitive to nonrenormalizable interactions which arise from the fundamental theory behind the grand unified model (string, gravity, etc.).

Alternatively, one may envisage enlarging the scalar structure of the model. Fol-

lowing Georgi and Jarlskog [182], one introduces a representation 45, Φmnp, m, n, p = 1, . . . , 5, which satisfies Φmnp = Φmpn and Φmmp = 0, and a 45, Φnpm of SU (5). If the

corresponding scalar fields develop a SU (3) × U (1)Y invariant vacuum expectation

value:

Φnm5 = a δnm 4δ4mδn4 ,

(9.84)

 

 

n5

 

 

 

and similarly for Φm

, this generates mass terms for quarks and leptons through the

Yukawa couplings:

 

 

 

 

 

 

 

 

 

 

 

 

 

W = λ

ηmχ

 

 

np

+ λ

mnprsχ

 

 

 

Φq .

(9.85)

np

Φ

mn

χ

pq

¯

d

 

m

u

 

 

rs

 

 

 

 

 

 

 

 

 

 

 

 

 

 

× 10 × 45

 

10 × 10 × 45

 

 

 

5

 

 

 

 

Now, if we restrict our attention to the charged leptons and charge 1/3 quarks of the first two families, we may use discrete symmetries to keep only the following terms in the superpotential:

η(1)m (2) n

+ η(2)mχ(1)

Hn

 

+ λ

 

η(2)m

(2)

 

np

h.c.

 

 

W |family (1) and (2) = λd

χmnH1

mn

1

 

d

 

χnp

Φm

+(9.86)

This gives a matrix structure of the form:

 

 

 

 

 

 

 

 

 

 

0

λdv1

for (e, µ),

 

 

 

 

 

 

 

 

λdv1 3λda

 

 

 

 

 

 

 

 

 

0 λdv1

for (d, s),

 

 

 

 

 

 

 

 

λdv1

λda

 

 

 

 

 

 

 

 

which naturally yields memµ mdms whereas mµ

me mµ and md ms

md 9 me ms mµ

which is in much better agreement with estimates.

+ me = 3(ms + md). Hence, since

(9.87)

244 Supersymmetric grand unification

Table 9.2 Charges under U (1)Y and U (1)χ of the low energy fields.

 

Q L

U c

Dc

Ec

H1

H2

y

1/3

1

4/3

2/3

2

1

1

yχ

1

3

1

3

1

2

2

(2y − yχ)/5

1/3

1

1/3

1/3

1

0

0

Global symmetries

Before closing this section, let us note that the superpotential which consists of (9.57) and (9.64) has a global abelian symmetry U (1)χ with charges:

yχ(Σ) = 0, yχ(H1) = 2, yχ(H2) = +2, yχ(η) = 3, yχ(χ) = 1.

(9.88)

Such a symmetry seems to lead to a disaster since it is broken by H1 and H2 . However, the combination (2y − yχ)/5 remains unbroken: considering H1 for example, its doublet component, which acquires a nonzero vev, has y = 1. Using (9.4) and (9.6), we deduce from the charges (9.88) the value of (2y − yχ)/5 for the low energy fields (see Table 9.2).

We recognize in the last line of Table 9.2 the quantum number B −L. Hence the SU (5) model has a global B −L symmetry.

9.3.4Proton decay

Since quarks and leptons are in the same representations of the unified group, one expects quark–lepton transitions and thus the possibility of proton decay. Writing the coupling of fermions (of the first family for the time being) to the SU (5) vector bosons Xαµ or Yαµ (α = 1, 2, 3 is a color index)

g

2

g

+

2

X

 

 

 

 

u¯

c

γ

µ

u

 

¯

γ

µ

e

c

e¯ γ

µ

d

c

 

 

 

 

 

 

 

 

+ d

 

 

 

 

 

 

 

 

 

αµ

 

αβγ

c

 

µ

 

 

 

 

µ

cL

L

µ

 

c

(9.89)

Yαµ αβγ u¯γ

 

 

d− u¯γ

 

 

eL + ν¯Lγ

 

 

d,

one infers that the following decays are allowed

5

: X

¯

+

, Y

→ ud, ue¯

+

.

 

→ uu, de

 

 

Thus, the exchange of X or Y , as in Fig. 9.5, provides an e ective interaction of

dimension 6:

2MX2

 

 

 

 

 

 

 

 

 

 

 

 

 

H

 

 

 

 

 

 

 

 

 

 

 

 

g2

 

c

µ

 

 

 

 

¯c

 

c

µ

 

 

 

 

e =

 

2

αβγ

u¯γ

 

u

 

dγµeL

 

e¯Lγ

 

d

 

 

 

 

g

 

c

 

 

µ

 

 

¯c

 

c

 

 

µ

 

 

 

 

αβγ u¯γ

 

d

dγµνL − e¯Lγ

 

u,

(9.90)

 

2MY2

 

 

which contributes to the decay channels p → e+π0 and n → e+π.

The corresponding amplitude includes: (i) a renormalization factor due to the fact that the e ective interaction (9.89) must be renormalized from MZ down to the

5One may note that, whereas the final states have various values of B and L, they have the same value B −L = 2/3; thus we can define B −L for X and Y .

The minimal supersymmetric SU(5) model 245

u

e+

u

e+

 

X

 

Y

u

d

d

u

Fig. 9.5 Example of amplitude contributing to the decay p → e+π0 (left) and n → e+π(right).

proton mass, say 1 GeV, (ii) a flavor factor, and (ii) a hadronization factor due to the fact that the amplitude is measured between hadrons, not between quarks. One obtains:

A

α3(1 GeV) 2/9

 

α3

(mc)

6/25

 

α3(mb) 6/23

1

, (9.91)

 

 

 

 

 

 

 

Fflavorαlat

 

α3(mc)

α3

(mb)

 

α3(MZ )

MX,Y2

where αlat = 0 αβγ dαR proton) is computed on

uβR uγL p /NL (NL is is the wave function of the left-handed the lattice. This yields:

τ p → e+π0 8 × 1034 years

0.015 GeV3

2

 

MX,Y

2

 

 

(9.92)

 

 

 

.

αlat

 

1016 GeV

This lies approximately one order of magnitude beyond the present experimental limit.

However, we have seen in Section 5.4 that, even when assuming R-parity, one remains with the possibility of dangerous B and L violating processes. In the context of grand unification, the exchange of color triplet fermions of mass MT generates the following dimension-5 operators (see Fig. 9.6):

1

αβγ

1

QQQLl + C5ijklR Uc Dc Uc Elc ,

 

W =

 

 

C5ijklL

(9.93)

MT

2

with α, β, γ color indices and i, j, k, l family indices.

This is why such color triplet supermultiplets must have a superheavy mass MT .

For example, the e ective operator

 

1

Q1Q1Q2Li (i = 1, 2, 3) generates the ∆B =

 

 

 

 

 

 

˜

 

 

 

 

 

 

 

 

M

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c

c c

c

L = 1 process u˜L dL

→ ν¯is¯ with coupling of order 1/MT , whereas

 

 

U1 U3 D1E3

MT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˜

. Through

 

 

 

 

 

 

 

 

 

 

 

 

the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

 

 

 

 

 

generates ud → τ˜R

R

gaugino or Higgsino exchange, such as in +

ν¯

amplitudes0 given in Fig.

9.7, this

induces, respectively,

the transitions

p

K

 

+

ν¯τ .

 

 

 

 

 

 

 

 

 

 

 

i

(or n → K ν¯i) and p → K

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The amplitude now reads

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

α3(1 GeV) 2/9

 

α3(mc)

6/25

 

α3(mb) 6/23

Fflavor Floopβlat

 

1

 

,

(9.94)

α3(mc)

 

 

 

α3(mb)

 

 

 

 

α3(MZ )

 

MT

 

246 Supersymmetric grand unification

di

 

el

di

 

 

el

~

X

 

 

 

X

 

~

 

 

~

~

 

H1

H2

~

~

H1

H2

~

~

 

 

 

dLj

 

uLk

dRj

 

 

uRk

Fig. 9.6 Dimension-5 operators generated by color triplet exchange.

 

 

 

 

 

 

 

dL

 

 

τR

 

 

uL

ν

uR

 

ντ

 

 

χ

 

 

 

 

χ

u

 

 

 

 

 

 

 

 

dL

 

tR

 

 

L

 

s

dR

 

s

Fig. 9.7 Amplitude contributing to the decay p → K+ν¯eµτ

or n → K0ν¯eµτ (left) and

p → K+ν¯τ (right) through dimension-5 operators.

where the loop factor is of order M1/2/m02 for M1/2 m0 and βlat= 0 αβγ dLαuLβ uLγ 0 /

NL. This gives a lower limit on the mass of the Higgs triplet, which

reads typically,

for tan β

5:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MT

 

 

τ (p → K+ν¯)

 

1/2

 

βlat

1 TeV

tan β

2

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5.0 × 1017 GeV

5.5 × 1032 years

 

0.003 GeV3

 

mf˜

! 10

 

 

 

 

 

 

(9.95) The negative search performed by the Super-Kamiokande Cherenkov detector sets a limit of 6.7 × 1032 years (90% confidence level) for the partial lifetime of the proton in the decay channel K+ν¯ [138]. This puts MT in a mass range incompatible with the data on the gauge coupling unification (9.53). It thus rules out the minimal SU (5) model [203].

This, however, does not exclude a more general SU (5) unification. One may for example include more superheavy multiplets in order to push gauge unification to higher scales through threshold corrections. Alternatively, one may try to suppress the dimension-5 operators, as in the flipped SU (5) model. Finally, one may enlarge the grand unification symmetry, as we will now see.

9.4The SO(10) model

The SU (5) model does not truly unify the matter fields since one has to advocate

¯

two representations: 5 and 10. It turns out that, if we view SU (5) as a subgroup of

¯

SO(10), these two representations make a single one of SO(10): 16 = 5 + 10 + 1. The SU (5) singlet that is required to make the sixteenth field is singlet under SU (3) × SU (2) × U (1): it is interpreted as a right-handed neutrino. This provides a natural explanation for the cancellation of gauge anomalies: as explained in Section 9.1.2, anomalies naturally cancel for representations of SO(10).

The SO(10) model 247

Since SO(10) has rank 5, we expect to be able to define an extra gauge quantum number. In fact, the global B − L symmetry of SU (5) is promoted to the status of

a gauge symmetry in the context of SO(10). Noting that Tr(B − L)|¯5 = 3 and Tr(B − L)|10 = 2, we see that the presence of the right-handed neutrino gives the missing contribution to ensure a vanishing total contribution for Tr(B − L)|16, as is fit for a gauge generator. Moreover, SO(10) incorporates a natural left-right symmetry and thus a spontaneous breaking of parity may be implemented. We will see below that the two issues are somewhat linked.

All these properties, compared with the mixed success of the SU (5) model make SO(10) unification a very interesting candidate for a grand unified theory.

9.4.1Symmetry breaking

Among the subgroups of SO(10), we first focus on SU (5) × U (1). One can easily see why, on general grounds, SU (n) × U (1) is a subgroup of SO(2n). A transformation of SU (n) leaves invariant the scalar product of two complex n-component vectors U and V :

U · V = Re U · Re V + Im U · Im V + i Re U · Im V − i Im U · Re V. (9.96)

Using instead the real 2n-component fields U = (Re U, Im U ) and V = (Re V, Im V ), one sees that the following scalar product is conserved:

U · V = Re U · Re V + Im U · Im V.

(9.97)

Hence, a transformation of SU (n) is also a transformation of SO(2n). Moreover, under the U (1) transformation U → eU , which commutes with SU (n), U = (Re U, Im U ) transforms into (Re U cos φ − Im U sin φ, Re U sin φ +Im U cos φ). This transformation preserves the scalar product (9.97). One concludes that SU (n) × U (1) is indeed a subgroup of SO(2n).

It is thus possible that, at some scale larger than the scale obtained earlier for SU (5) unification, SO(10) is broken into SU (5) × U (1)χ. Quarks, leptons and their supersymmetric partners form NF chiral supermultiplets in 16 of SO(10), which are decomposed under SU (5) × U (1)χ as:

¯

+ 101 + 15

(9.98)

16 = 53

(Dc, E, N ) (D, U, U c, Ec) (N c)

where we have indicated as subscript for each SU (5) representation its charge yχ under U (1)χ. Similarly, gauge supermultiplets transform as 45 with the decomposition:

 

 

 

 

45 = 240 + 10 + 104 + 104,

(9.99)

where we recognize in the first places the gauge bosons of SU (5) and U (1)χ.

The first breaking (SO(10) to SU (5)) may be realized through a 16 of Higgs, whereas the second breaking (SU (5) to the Standard Model) uses a 45, since 45 includes the necessary 24 of SU (5). There is no constraint on the scale of U (1)χ breaking and the corresponding gauge boson Zχ could thus be a low energy field.

248 Supersymmetric grand unification

Alternatively, since the group SO(m + n) contains SO(m) × SO(n), and SO(6) SU (4) whereas SO(4) SU (2) × SU (2), we may consider the breaking of SO(10) to SU (4) × SU (2) × SU (2). One of the SU (2) symmetries may be interpreted as the SU (2)L symmetry of the Standard Model and the other one is identified as its parity counterpart SU (2)R. Moreover, SU (4) naturally incorporates the color SU (3) and the abelian group U (1)B−L associated with the B − L quantum number.

The 16 and 45 of SO(10) transform under SU (4) × SU (2)L × SU (2)R as

 

16

= (4, 2, 1)

+

¯

 

(9.100)

(4, 1, 2),

 

 

D

, E

U c

, N c

 

 

U

N

Dc

Ec

 

 

45

= (15, 1, 1) + (1, 3, 1) + (1, 1, 3) + (6, 2, 2).

(9.101)

The breaking of SO(10) is realized through a 54 = (6, 2, 2)+(20 , 1, 1)+(1, 3, 3)+ (1, 1, 1) and the final breaking to the Standard Model uses 16 + 16 or 126 + 126.

We note the following relation:

q = tL3 + tR3 +

B − L

,

(9.102)

2

 

 

 

which shows that B − L is a generator of SO(10) (since the others are). This provides also a nice interpretation of hypercharge, which had a somewhat mysterious origin in the context of the Standard Model:

 

y = 2tR3 + B − L.

(9.103)

We note also the relation with the charge yχ introduced earlier:

 

yχ = 2

5tR3 + 3(tL3 − q) = 2y − 5(B − L).

(9.104)

9.4.2Fermion masses

Since Yukawa couplings are trilinear, one expects to find the Higgs representations (or rather their conjugates) in the product 16 × 16 = (10 + 126)s + 120a (the subscripts refer, respectively, to the symmetric and the antisymmetric combination). The Higgs are thus searched for in 10, 120, or 126 (the first two are self-conjugates).

[We note that 16 is the spinor representation of SO(10) constructed in Section B.2.1 of Appendix B. We thus see that matter supermultiplets are in spinor representations of SO(10) whereas Higgs and gauge supermultiplets are in tensor representations (they appear in even products of spinor representations). This allows us to define a matter parity: 1 for spinors and +1 for tensors.]

If we take the Higgs in the representation 10H which decomposes under SU (5) ×

¯

U (1)χ as 10 = 52 + 52, the trilinear coupling yields the following decomposition under SU (5):

¯

¯

¯

(16 × 16) × 10 (5

× 10) × 5H + (10

× 10) × 5H + (1 × 5) × 5H . (9.105)