Left-Right Gauge Model in Nonassociative Geometry

arXiv:hep-ph/9907558v1 30 Jul 1999

UC-ph/99-50

1

Left-Right Gauge Model in Nonassociative Geometry A. Benslama and N. Mebarki D´epartement de Physique Th´eorique. Universit´e Mentouri. Constantine. Algeria. May 1999 Abstract We reformulate the left-right gauge model of Pati-Mohapatra using nonassociative geometry approach. At the tree level we obtain the mass relations MW = 21 mt , MH = 23 mt and the mixing angles are sin2 θW = 38 and sin2 θs = 53 , which are identical to the ones obtained in SO(10) GUT .

2

1

Introduction

One of the greatest achievement of noncommutative geometry (abbreviated NCG hereafter) is its geometrization of the standard model ([1],[2],[3]). NCG provides a framework where the Higgs boson H may be introduced on the same level as W ± and Z bosons. In this approach we introduce additional discrete dimensions to the four-dimensional space-time. If the gauge bosons are associated to the continuous directions, the Higgs boson results from gauging the discrete directions. However one of the shortcomings of NCG is its non accessibility to grand unified theories GUT ([4],[5]). The other formulation of NCG due to Coquereaux [6] does not suffer from this problem ([7]-[10]). There are some variant of Connes’ theory ([11],[12]) where we use an auxiliary Hilbert space to fit GUT. R. Wulkenhaar has recently succeeded in formulating another type of geometry which shares a lot of common points with NCG `a la Connes [13]. The theory was baptized ”nonassociative geometry” NAG. The main difference with the two theories is that NAG is based on unitary Lie algebra instead of unital associative ⋆-algebra in the case of NCG. Its application to several physical models has been successful ([14]-[17]). We try using this approach to reformulate the left-right model LRM ([18],[19]). One of the features of NAG (and NCG) is its geometric explanation of the spontaneous breakdown of gauge symmetry. Our aim is to investigate whether NAG could encompass also the parity violation. In section 2 we present the main elements to formulate gauge theories using NAG. In section 3 we derive the bosonic action of left-right model and finally we discuss our results.

2

Some elements of Nonassociative Geometry

NAG is based on the L-cycle (g,H, D, π, Γ) [13] g =C ∞ (X) ⊗ a

(1)

where C ∞ (X) is the algebra of smooth functions on the manifold X and a is the matrix Lie algebra. H is the Hilbert space H =L2 (X, S) ⊗ CF 3

(2)

where L2 (X, S) is the Hilbert space of spinors. D is the total Dirac operator given by: D = D ⊗ 1F + γ 5 ⊗ M

(3)

where D is the Dirac operator associated with the continuous algebra C ∞ (X) and M is associated to the discrete algebra a. The representation of g on H is given by π = 1⊗ˆ π,

(4)

where π ˆ is the representation of a on CF . Finally, the graded operator Γ, acting on H, is given by: ˆ Γ = γ 5 ⊗ Γ, ˆ is the grading operator acting on CF . where Γ ˆ 1 a is generated by the elements of the type The space Ω X ω1 = [az , ...[a1 , da0 ]...] ai ∈ a.

(5)

(6)

z

ˆ 1 a as: The representation π ˆ acts on the space Ω ˆ 1 a −→MF (C) π ˆ:Ω τ 1 = πˆ (ω 1 ) :=

X

[ˆ π (az ), ...[ˆ π (a1 ), [−iM, π ˆ (a0 )]]...]

(7)

z

We define also the mapping: ˆ 1 a −→MF (C) σ ˆ:Ω σ ˆ (ω 1 ) :=

X

[ˆ π (az ), ...[ˆ π (a1 ), [M2, π ˆ (a0 )]]...]

(8)

z

For n> 2 we have

n o ˆ k g∩ ker(π) . π(J k+1 g) = σ(ω k ); ω k ∈ Ω 4

(9)

The main difference between NCG and NAG is that in NAG the connection ρ and the curvature θ are not, in general, elements of Ω1 g and Ω2 g. To construct connection and curvature we need the spaces r0 a , r1 a ⊂ MF (C) defined by the conditions:

and

r0 (a) 0  r (a), π ˆ (a) 0  r (a), π ˆ (Ω1 a)  0 r (a), π ˆ (a) 0  r (a), π ˆ (Ω1 a)

= ⊂ ⊂ ⊂ ⊂

ˆ 0 (a)Γ ˆ Γr πˆ (a) πˆ (Ω1 a) {ˆ π (a), π ˆ (a)} + π ˆ (Ω2 a) {ˆ π (a), π ˆ (Ω1 a)} + π ˆ (Ω3 a)

ˆ 1 (a)Γ ˆ r1 (a) = −Γr  1 1 r (a), π ˆ (a) ⊂ π ˆ (Ω a) 1 1 r (a), π ˆ (Ω a) ⊂ π ˆ (Ω2 a) + {ˆ π(a), π ˆ (a)}

(10)

(11)



ˆ is the grading operator. where Γ The connection is given by: X ρ= (c1α ⊗ m0α + c0α γ 5 ⊗ m1α )

(12)

α

where c1α ∈ Λ1 , c0α ∈ Λ0

(13)

m0α ∈ r0 (a) m1α ∈ r1 (a)

(14)

Λk is the space of k-forms represented by gamma matrices, and

To construct the curvature θ we need the spaces j0 a, j1 a and j2 a defined as: j0 a =c0 a

(15)

ˆ c0 a c0 a = c0 a Γ c0 a .ˆ π (a) = 0 0 c a .ˆ π (Ω1 a) = 0

(16)

where

5

and j1 a =c1 a

(17)

ˆ c1 a c1 a = −c1 a Γ c1 a .ˆ π (a) = 0 1 c a .ˆ π (Ω1 a) = 0

(18)

j2 a =c2 a + π ˆ (J2 a) + {ˆ π (a), π ˆ (a)}

(19)

where

and with

ˆ c2 a c2 a = c2 a Γ  2  c a ,π ˆ (a) = 0  2  c a ,π ˆ (Ω1 a) = 0

(20)

θ = dρ + ρ2 − i{γ 5 ⊗ M, ρ} + σ ˆ g(ρ)γ 5 + J2 g

(21)

J2 g =(Λ2 ⊗ j 0 a)⊕(Λ1 γ 5 ⊗ j 1 a)⊕(Λ0 ⊗ j 2 a)

(22)

ε(θ) = dρ + ρ2 − i{γ 5 ⊗ M, ρ} + σˆ g(ρ)γ 5 + j

(23)

The curvature θ is given by:

where We have to choose the representative of the curvature θ , ε(θ), orthogonal to J2 g,given by:

where

Z

X

dxtr(ε(θ)j ′ ) = 0 ∀j ′ ∈ J2 g

The bosonic and the fermionic actions are given by: Z 1 SB = dxtr(ε(θ)2 ) NF g02 X and SF =

Z

Ψ∗ (D + iρ)Ψ

X

where g0 is a coupling constant.

6

(24)

(25)

(26)

3

The construction

The discrete L-cycle (a,H, M) consists of the Lie algebra given by: a = su(3) ⊕ su(2) ⊕ su(2) ⊕ u(1)

(27)

∋ {a3 , a2 , a′2 , a1 }

The Hilbert space H is C48 labelled by the elements (uL , dL , uR , dR , ν L , eL , ν R , eR )T where uL , dL , uR , dR ∈ C3 ⊗ C3 and ν L , eL , ν R , eR ∈ C3 . The Lie algebra a acts on H via the representation:   0 π ˆ q ((a1 , a2 , a′2 , a3 )) ′ π ˆ ((a1 , a2 , a2 , a3 )) = 0 π ˆ l ((a1 , a2 , a′2 , a3 ))

(28)

where q and l label, respectively, the quarkionic and the leptonic sectors, with the representation of the quark sector given by:  1  0 π ˆ q ((a1 , a2 , a′2 , a3 )) ′ π ˆ q ((a1 , a2 , a2 , a3 )) = (29) 0 π ˆ 2q ((a1 , a2 , a′2 , a3 )) where (30) π ˆ 1q ((a1 , a2 , a′2 , a3 )) = if0 diag(α13 ⊗ 13 , β13 ⊗ 13 ) +   (a3 + if31 13 ) ⊗ 13 i(f11 − if21 )13 ⊗ 13 i(f11 + if21 )13 ⊗ 13 (a3 − if31 13 ) ⊗ 13 and (31) π ˆ 2q ((a1 , a2 , a′2 , a3 )) = if0 diag(γ13 ⊗ 13 , δ13 ⊗ 13 ) +   (a3 + if32 13 ) ⊗ 13 i(f12 − if22 )13 ⊗ 13 i(f12 + if22 )13 ⊗ 13 (a3 − if32 13 ) ⊗ 13 and the representation of the leptonic sector:   1 π ˆ l ((a1 , a2 , a′2 , a3 )) 0 ′ π ˆ l ((a1 , a2 , a2 , a3 )) = 0 π ˆ 2l ((a1 , a2 , a′2 , a3 )) where 7

(32)

π ˆ 1l ((a1 , a2 , a′2 , a3 )) = if0 diag(υ13 , ε13 ) +   1 if3 ⊗ 13 i(f11 − if21 ) ⊗ 13 i(f11 + if21 ) ⊗ 13 −if31 ⊗ 13

(33)

and π ˆ 2l ((a1 , a2 , a′2 , a3 )) = if0 diag(ζ13 , η13 ) +   2 if3 ⊗ 13 i(f12 − if22 ) ⊗ 13 i(f12 + if22 ) ⊗ 13 −if32 ⊗ 13

(34)

where we have used the fact that elements of su(2) algebra have the representation:   if3 i(f1 − if2 ) . (35) a2 = i(f1 + if2 ) −if3 The Dirac operator is given by:   Mq 0 M= 0 Ml

with

and

 0 0 13 ⊗ Mu 0  0 0 0 13 ⊗ Md   Mq =  ∗   13 ⊗ Mu 0 0 0 ∗ 0 13 ⊗ Md 0 0 

 0 0 Mν 0  0 0 0 Me   Ml =   Mν∗ 0 0 0  0 Me∗ 0 0 

(36)

(37)

(38)

where Mu , Md , Mν , Me ∈ M3 (C) are the mass matrices of the fermions. Remark that we have chosen the coefficients associated with u(1) algebra arbitrary, we will see when calculating the fermionic action that in fact they corresponds to the hypercharges. The grading operator is given by: ˆ = diag(−13 ⊗ 13 , −13 ⊗ 13 , 13 ⊗ 13 , 13 ⊗ 13 , −13 , −13 , 13 , 13 ) Γ 8

(39)

The space π ˆ (Ω1 a) is generated by elements of the type X     τ1 = π ˆ (azα ), ... πˆ (a1α ), −iM, π ˆ (a0α ) ...

(40)

α,z

aiα

(ai1 , ai2 , ai′2 , ai3 )

where = . Simple calculation gives the following decomposition:  1  τq 0 1 τ = 0 τ 1l

(41)

where for the quark sector we have   0 τ 11q 1 τq = τ 12q 0

(42)

with τ 11q

=i

¯b2 13 ⊗ Md + c¯2 13 ⊗ Mu b1 13 ⊗ Md + c1 13 ⊗ Mu ¯ −b1 13 ⊗ Mu − c¯1 13 ⊗ Md b2 13 ⊗ Mu + c2 13 ⊗ Md



b2 13 ⊗ Md∗ + c2 13 ⊗ Mu∗ −b1 13 ⊗ Mu∗ − c1 13 ⊗ Md∗ ¯b1 13 ⊗ M ∗ + c¯1 13 ⊗ M ∗ ¯b2 13 ⊗ M ∗ + c¯2 13 ⊗ M ∗ u u d d





(43)

and τ 12q

=i



.

(44)

For the leptonic sector we have τ 1l where τ 11l

=i

and τ 12l

=i





=



0 τ 11l τ 12l 0



(45)

¯b2 ⊗ Me + c¯2 ⊗ Mν b1 ⊗ Me + c1 ⊗ Mν −¯b1 ⊗ Mν − c¯1 ⊗ Me b2 ⊗ Mν + c2 ⊗ Me



b2 ⊗ Mν∗ + c2 ⊗ Me∗ −b1 ⊗ Mν∗ − c1 ⊗ Me∗ ¯b1 ⊗ M ∗ + c¯1 ⊗ M ∗ ¯b2 ⊗ M ∗ + c¯2 ⊗ M ∗ e ν e ν



(46)

.

(47)

We can show that the form τ 1 is antihermetian provided that: α+β =γ+δ

(48)

υ + ε = ζ + η.

(49)

and

9

Similarly, straightforward calculations give:  2  τq 0 2 2 1 1 π ˆ (Ω a) ∋ τ = {τ , τ } = 0 τ 2l with

τ l2 11 l2  τ 21 τ 2l =   0 0 

τ l2 12 τ l2 22 0 0

0 0 τ l2 33 τ l2 43

where the diagonal elements are:

 0 0    τ l2 34 l2 τ 44

2 2 2 2 ∗ ∗ − τ l2 11 = 2(|b1 | + |c2 | ) ⊗ Me Me + 2(|c1 | + |b2 | ) ⊗ Mν Mν +2(b1 c¯1 + c¯2 b2 ) ⊗ Me Mv∗ + 2(¯b1 c1 + c2¯b2 ) ⊗ Mν Me∗ 2 2 2 2 ∗ ∗ −τ l2 22 = 2(|c1 | + |b2 | ) ⊗ Me Me + 2(|b1 | + |c2 | ) ⊗ Mν Mν +2(b1 c¯1 + c¯2 b2 ) ⊗ Me Mv∗ + 2(¯b1 c1 + c2¯b2 ) ⊗ Mν Me∗ 2 2 2 2 ∗ ∗ −τ l2 33 = 2(|c2 | + |c1 | ) ⊗ Me Me + 2(|b1 | + |b2 | ) ⊗ Mν Mν +2(b1 c¯1 + c¯2 b2 ) ⊗ Me Mv∗ + 2(¯b1 c1 + c2¯b2 ) ⊗ Mν Me∗ 2 2 2 2 ∗ ∗ −τ l2 44 = 2(|b1 | + |b2 | ) ⊗ Me Me + 2(|c1 | + |c2 | ) ⊗ Mν Mν +2(b1 c¯1 + c¯2 b2 ) ⊗ Me Mv∗ + 2(¯b1 c1 + c2¯b2 ) ⊗ Mν Me∗

(50)

(51)

(52)

and the off-diagonal ones: − τ l2 12 −τ l2 21 −τ l2 34 −τ l2 43

= = = =

2(¯ c2 c1 − b1¯b2 ) ⊗ Mνe 2(c2 c¯1 − ¯b1 b2 ) ⊗ Mνe 2(c1 b2 − b1 c2 ) ⊗ Mνe 2(¯ c1¯b2 − ¯b1 c¯2 ) ⊗ Mνe

(53)

with Mνe = Mν Mν∗ − Me Me∗ M{νe} = Mν Mν∗ + Me Me∗

(54)

and similarly τ q2 11 q2  τ 21 τ 2q =   0 0 

τ q2 12 τ q2 22 0 0 10

0 0 τ q2 33 τ q2 43

 0 0    τ q2 34 τ q2 44

(55)

with 2 2 2 2 ∗ ∗ − τ q2 11 = 2(|b1 | + |c2 | )13 ⊗ Md Md + 2(|c1 | + |b2 | )13 ⊗ Mu Mu (56) +2(b1 c¯1 + c¯2 b2 )13 ⊗ Md Mu∗ + 2(¯b1 c1 + c2¯b2 )13 ⊗ Mu Md∗ 2 2 2 2 ∗ ∗ −τ l2 22 = 2(|c1 | + |b2 | )13 ⊗ Md Md + 2(|b1 | + |c2 | )13 ⊗ Mu Mu +2(b1 c¯1 + c¯2 b2 )13 ⊗ Md Mu∗ + 2(¯b1 c1 + c2¯b2 )13 ⊗ Mu Md∗ 2 2 2 2 ∗ ∗ −τ l2 33 = 2(|c2 | + |c1 | )13 ⊗ Md Md + 2(|b1 | + |b2 | )13 ⊗ Mu Mu +2(b1 c¯1 + c¯2 b2 )13 ⊗ Md Mu∗ + 2(¯b1 c1 + c2¯b2 )13 ⊗ Mu Md∗ 2 2 2 2 ∗ ∗ −τ l2 44 = 2(|b1 | + |b2 | )13 ⊗ Md Md + 2(|c1 | + |c2 | )13 ⊗ Mu Mu +2(b1 c¯1 + c¯2 b2 )13 ⊗ Md Mu∗ + 2(¯b1 c1 + c2¯b2 )13 ⊗ Mu Md∗

and − τ q2 12 −τ q2 21 −τ q2 34 −τ q2 43

= = = =

2(¯ c2 c1 − b1¯b2 )13 ⊗ Mud 2(c2 c¯1 − ¯b1 b2 )13 ⊗ Mud 2(c1 b2 − b1 c2 )13 ⊗ Mud 2(¯ c1¯b2 − ¯b1 c¯2 )13 ⊗ Mνud

(57)

where Mud = Mu Mu∗ − Md Md∗ M{ud} = Mu Mu∗ + Md Md∗ .

(58)

We can show that: 1

σ ˆ (ω ) =



σˆ q (ω 1 ) 0 0 σ ˆ l (ω 1 )



(59)

where 1

σ ˆ q (ω ) = with

σ ˆ 2q = i

σ ˆ 1q 0 0 σ ˆ 2q



(60)



if31 13 ⊗ Mud i(f11 − if21 )13 ⊗ Mud i(f11 + if21 )13 ⊗ Mud −if31 13 ⊗ Mud



(61)



if32 13 ⊗ Mud i(f12 − if22 )13 ⊗ Mud i(f12 + if22 )13 ⊗ Mud −if32 13 ⊗ Mud



(62)

σ ˆ 1q = i and



11

.

For the leptonic sector we have 1

σ ˆ l (ω ) = with

σ ˆ 2l = i

σ ˆ 1l 0 0 σ ˆ 2l



if31 ⊗ Mνe i(f11 − if21 ) ⊗ Mνe 1 1 i(f1 + if2 ) ⊗ Mνe −if31 ⊗ Mνe



(64)



if32 ⊗ Mνe i(f12 − if22 ) ⊗ Mνe i(f12 + if22 ) ⊗ Mνe −if32 ⊗ Mνe



(65)

After trivial calculation we can show that

with and

(63)



σ ˆ 1l = i and



.

τ 2 = diag(K ′ ⊗ 13 , K ′ ⊗ 13 , K ′ ⊗ 13 , K ′ ⊗ 13 , K, K, K, K) mod(ˆ σ (Ω1 a))

(66)

K=α ˆ M{νe} + βˆ Me Mv∗ + γˆ Mν Me∗

(67)

K′ = α ˆ M{ud} + βˆ Md Mu∗ + γˆ Mu Md∗

(68)

α ˆ = |b1 |2 + |b2 |2 + |c1 |2 + |c2 |2 βˆ = b1 c¯1 + b2 c¯2

(69)

where

¯ ∧

γˆ = β= ¯b1 c1 + ¯b2 c2 . Using the conditions given by eqs (10,11), we can show that: r0 (a) = π ˆ (a) 1 r (a) = π ˆ (Ω1 a)

(70)

Let us adopt the following parametrization X α

c1α ⊗ a1α =

X α

′ ˜ ∈ Λ1 ⊗ u(1) = i g γ µ Cµ c1α ⊗ if0α = iA 2

12

(71)

X α

X α

ˆ ∈ Λ1 ⊗ su(3) c1α ⊗ a3α = A

(72)

c1α ⊗ a2α = A ∈ Λ1 ⊗ su(2)  P

α =  P α

=



(73)

c1α ⊗ if3α c1α ⊗ i(f1α + if2α )

P

α P α

iA3 i(A1 − iA2 ) i(A1 + iA2 ) −iA3

c1α ⊗ i(f1α − if2α ) c1α ⊗ (−if3α ) 

 

and Φi = − Ξi = −

X

c0α ⊗ biα ∈ Λ0 ⊗ C

α

X

(74)

c0α ⊗ ciα ∈ Λ0 ⊗ C

α

The connection ρ is given by ρ=



ρq 0 0 ρl



(75)

where ρl =



1 1 ρA ρH l l 2 ρH2 ρA l l



(76)

with 1 ρA l

=

2 = ρA l

1 = ρH l

ρH2 = l

 i(υ A˜ + A3 ) ⊗ 13 i(A1 − iA2 ) ⊗ 13 (77) i(A1 + iA2 ) ⊗ 13 i(εA˜ − A3 ) ⊗ 13   i(ζ A˜ + A′3 ) ⊗ 13 i(A′1 − iA′2 ) ⊗ 13 i(A′1 + iA′2 ) ⊗ 13 i(η A˜ − A′3 ) ⊗ 13   ¯ 2 ⊗ Me − iγ 5 Ξ ¯ 2 ⊗ Mν −iγ 5 Φ1 ⊗ Me − iγ 5 Ξ1 ⊗ Mν −iγ 5 Φ ¯ 1 13 ⊗ Mν + iγ 5 Ξ ¯ 1 ⊗ Me −iγ 5 Φ2 ⊗ Mν − iγ 5 Ξ2 ⊗ Me iγ 5 Φ   −iγ 5 Φ2 ⊗ Me∗ − iγ 5 Ξ2 ⊗ Mν∗ iγ 5 Φ1 ⊗ Mν∗ + iγ 5 Ξ1 ⊗ Me∗ ¯ 1 ⊗ Me∗ − iγ 5 Ξ ¯ 1 ⊗ Mν∗ −iγ 5 Φ ¯ 2 ⊗ Mν∗ − iγ 5 Ξ ¯ 2 ⊗ Me∗ −iγ 5 Φ 

13

and ρq = with 1 ρA q

2 ρA q

1 ρH q

2 ρH q

= =





=



1 1 ρA ρH q q H2 A2 ρq ρq



(78)

(Aˆ + i(αA˜ + A3 )13 ) ⊗ 13 i(A1 − iA2 )13 ⊗ 13 i(A1 + iA2 )13 ⊗ 13 (Aˆ + i(β A˜ − A3 )13 ) ⊗ 13

=





(Aˆ + i(γ A˜ + A′3 )13 ) ⊗ 13 i(A′1 − iA′2 )13 ⊗ 13 i(A′1 + iA′2 )13 ⊗ 13 (Aˆ + i(δ A˜ − A′3 )13 ) ⊗ 13



 (79)

¯ 2 13 ⊗ Md − iγ 5 Ξ ¯ 2 13 ⊗ Mu −iγ 5 Φ1 13 ⊗ Md − iγ 5 Ξ1 13 ⊗ Mu −iγ 5 Φ 5¯ 5¯ iγ Φ1 13 ⊗ Mu + iγ Ξ1 13 ⊗ Md −iγ 5 Φ2 13 ⊗ Mu − iγ 5 Ξ2 13 ⊗ Md

−iγ 5 Φ2 13 ⊗ Md∗ − iγ 5 Ξ2 13 ⊗ Me∗ iγ 5 Φ1 13 ⊗ Mu∗ + iγ 5 Ξ1 13 ⊗ Md∗ ¯ 1 13 ⊗ M ∗ − iγ 5 Ξ ¯ 1 13 ⊗ M ∗ −iγ 5 Φ ¯ 2 13 ⊗ M ∗ − iγ 5 Ξ ¯ 2 13 ⊗ M ∗ −iγ 5 Φ e u d d



The interaction fermions-bosons is given by

LF −B = iΨ∗ ρΨ = LlB + LqB + LlH + LqH ,

(80)

where LlB , LqB ; LlH , LqH are ,respectively, the Lagrangians of interaction leptons-bosons, quarks-bosons, leptons-Higgs and quark-Higgs, and Ψ = (uL , dL , uR , dR , ν L , eL , ν R , eR )T .

(81)

Using the parametrization ′

˜ = i g γ µ Cµ A 2 A=i A′ = i and

3 gL X a µ W γ ⊗ σa 2 a=1 µ

3 gR X ′a µ W γ ⊗ σa 2 a=1 µ

ˆ =i gs A 2

3 X a=1

Gaµ γ µ ⊗ λa

14

(82)

(83)

(84)

(85)



where g ′ , gL, gR , gs are respectively the coupling constants of u(1)L−B , su(2)L, su(2)R and su(3) algebras. Let [20] Bµ = cos(θW )Aµ − sin(θ W )Zµ Wµ3 = sin(θW )Aµ + cos(θW )Zµ

(86)

Cµ = cos(θS )Bµ − sin(θS )Zµ′

(87)

and

Wµ′3

= sin(θS )Bµ +

cos(θS )Zµ′ ,

where the field Bµ is associated to the group U(1)Y , and θW and θ S are the mixing angles. Once performing a Wick rotation and by identification with the fermionic action provided by the classical model we obtain [22]: ε = η = υ = ζ = −1

(88)

which are the leptons hypercharges in the left-right model [21]. The same treatment, applied to the quarkionic sector, gives α=β=γ=δ=

1 3

(89)

which are the quark hypercharges. Using the conditions given by eqs. (16, 18 and 20) we can show that j0 a = 0

(90)

j1 a = 0

(91)

j2 a =diag(RI36 , RI12 ) + πˆ (J2 a) + {ˆ π(a), π ˆ (a)} .

(92)

and Let the anticommutator ˆ ψ(a) = {ˆ π (a), π ˆ (a)} = diag({ˆ π q (a), π ˆ q (a)} , {ˆ π l (a), π ˆ l (a)}) where 15

(93)

{ˆ π q (a), π ˆ q (a)} ∋ Aq + ∆q with Aq = where



A1q 0 0 A2q



(94)

⊗I 3

(95)

= i

"

# ˆ 1 − iλ ˆ 2 )I3 λ01 a3 + iθ 01 I3 (̟ 1 − i̟ 2 )a3 + i(λ ˆ 1 + iλ ˆ 2 )I3 λ0 a3 − iθ 0 I3 (̟1 + i̟ 2 )a3 + i(λ 2 1 (96)

A2q = i

"

# ˆ 3 − iλ ˆ 4 )I3 (̟ 3 − i̟ 4 )a3 + i(λ λ03 a3 + iθ 02 I3 ˆ 3 + iλ ˆ 4 )I3 λ0 a3 − iθ 0 I3 (̟3 + i̟ 4 )a3 + i(λ 4 2 (97)

A1q

and ∆q = diag(λ1 I3 , λ1 I3 , λ2 I3 , λ2 I3 ) ⊗ I3

(98)

λ01 + λ02 = λ03 + λ04

(99)

{ˆ π l (a), π ˆ l (a)} ∋ Al + ∆l

(100)

with the condition and

with 

  Al = i  

iκ01 I3 ˇ 1 + iλ ˇ 2 )I3 i(λ 0 0

ˇ 1 − iλ ˇ 2 )I3 i(λ −iκ01 I3 0 0

0 0 iκ02 I3 ˇ 3 + iλ ˇ 4 )I3 i(λ

and

∆l = diag(χI3 , χI3 , χI3 , χI3 ) i

ˇi, λ ˆ , θ0 , κ0 , ̟i , χ (i = 1..4, j = 1..2) ∈ R. with λi , λ0i , λ j j 16

0 0 ˇ 3 − iλ ˇ 4 )I3 i(λ −iκ02 I3



   (101)  (102)

Trivial calculations give [22]

where

j2 a = π ˆ (J 2 a) ⊕ diag(Aq , Al ) ⊕ diag(Kq , Kl )

(103)

Kq = diag(ϕ1 I3 , ϕ1 I3 , ϕ2 I3 , ϕ2 I3 ) ⊗ I3 Kl = diag(ϕ3 I3 , ϕ3 I3 , ϕ3 I3 , ϕ3 I3 )

(104)

with ϕ1 , ϕ2 et ϕ3 ∈ R. The representative ε({τ 1 , τ 1 }) of {τ 1 , τ 1 } + j2 a orthogonal to j2 a is given by:  ˜ ′ ⊗ 13 , K ˜ ′ ⊗ 13 , K ˜ ′ ⊗ 13 , K ˜ ′ ⊗ 13 , ε( τ 1 , τ 1 ) = diag(K ˜ K, ˜ K, ˜ K) ˜ mod(ˆ K, σ (Ω1 a))

(105)

with ∽

∽

∽

∽

˜ = α ˜ {νe} + βˆ Me M ∗ +ˆ K ˆM γ Mν Me∗ v

(106)

˜′ = α ˜ {ud} + βˆ Md M ∗ +ˆ K ˆM γ Mu Md∗ u where

˜ {νe} = Mν M ∗ + Me M ∗ − 1 tr(Mν M ∗ + Me M ∗ )13 , M ν e ν e 3 ∽ 1 Me Mv∗ = Me Mv∗ − tr(Me Mv∗ )13 , 3 ∽ 1 Mν Me∗ = Mν Me∗ − tr(Mν Me∗ )13 , 3

˜ {ud} = = Mu M ∗ + Md M ∗ − 1 tr(Mu M ∗ + Md M ∗ )13, M u d u d 3 ∽ 1 Md Mu∗ = Md Mu∗ − tr(Md Mu∗ )13, 3

and

∽ 1 Mu Md∗ = Mu Md∗ − tr(Mu Md∗ )13 . 3

17

(107)

Remark also that M2 =

1 ˜ {ud} , 13 ⊗ M ˜ {ud} , 13 ⊗ M ˜ {ud} , 13 ⊗ M ˜ {ud} , diag(13 ⊗ M 2 ˜ {νe} , M ˜ {νe} , M ˜ {νe} , M ˜ {νe} ) mod(ˆ M σ (Ω1 a)) (108)

The bosonic action is given by Z SB = (L0 + L1 + L2 )dx

(109)

X

where L0 =

1 ˜ 2 + 4M ˜ 2 ) × (|Φ1 |2 + |Φ2 |2 + |Ξ1 |2 + |Ξ2 + 1|2 − 1)2 {tr(12M {ud} {νe} 2 96g0 ∽

∽

+tr(12(Md Mu∗ )2 + 4(Me Mν∗ )2 ) × (Φ1 Ξ1 +Φ2 (Ξ2 +1)) ∽

∽

+ tr(12(Mu Md∗ )2 + 4(Mν Me∗ )2 ) × (Φ1 Ξ1 + Φ2 (Ξ2 + 1))}

(110)

is the Higgs Lagrangian, and

L2 =

o n 1 ˆ A ˆ )2 + 1 trc ((d A+ 1 {A, A})2 ) (111) ˆ 1 A, tr (d A+ c 4g02 2 4g02 2 1 1 1 ˜ )2 )) + 2 trc (d A′ + {A′ , A′ })2 + 2 trc ((d A 4g0 2 3g0

is the Yang-Mills Lagrangian. Since L1 , the Lagrangian of interaction Higgs-bosons, is quite long it is not displayed here in its totality [22]. The parts of L1 which give rise to the masses of WR and WL bosons are respectively: L11 =

1 2 m trc |i(A1 − iA2 )(Ξ2 + 1)|2 8g02

(112)

L21 =

1 2 2 m trc |i(A′1 − iA′2 )(Ξ2 + 1)| , 8g02

(113)

and

18

where

1 1 m2 = tr( Me Me∗ + Mν Mν∗ + Mu Mu∗ + Md Md∗ ). 3 3 Let us make the following rescaling Ξ2 =

g¯φ2 , m

where g¯ is a coupling constant which is specified later. Since gL † µ γ i(A1 − iA2 ) = i √ WµL 2 gL † µ i(A′1 − iA′2 ) = i √ WµR γ 2

(114)

(115)

(116)

and trc (γ µ γ ν ) = 4δ µν

(117)

then 1 gL 1 gL 1 g¯gL 2 † † † ) WµL WLµ φ22 + ( )2 m¯ g WµL WLµ φ2 + ( )2 m2 WµL WLµ (118) L11 = ( 4 g0 2 g0 4 g0 The last term gives: 1 gL MWL = ( )m 2 g0

(119)

1 gR MWR = ( )m 2 g0

(120)

gL MWL = MWR gR

(121)

We can show also that:

Hence

In the case of left-right symmetric model gL = gR,

(122)

MWL = MWR .

(123)

we have

19

and the left-handed and the right-handed gauge bosons acquire the same masses, leading to the conclusion that parity violation does not occur for such models ([23],[24]). The same conclusion has been obtained by [25] when discussing the U(2)L × U(2)R model in the framework of noncommutative geometry `a la Connes. For left-right symmetric model the nonassociative geometry approach does not accommodate parity violation to be contrasted to Coquereaux approach [21]. Remark that if we neglect the mass of fermions in respect of the mass of the top quark mt we obtain

and

1 gL MWL = ( )mt 2 g0

(124)

1 gR MWR = ( )mt . 2 g0

(125)

The masses Z et Z ′ bosons come from L31 =

1 2 m trc |i(A0 − iA3 )(Ξ2 + 1)|2 2 8g0

(126)

L41 =

1 2 2 m trc |i(A0 − iA′3 )(Ξ2 + 1)| . 8g02

(127)

1 m gL ( ) . 2 g0 cos(θW )

(128)

and

The calculations give MZ = Since

1 gL MWL = ( )m 2 g0

(129)

MWL , cos(θW )

(130)

then MZ =

which is the same relation as in the standard model. Y. Okumura [21] has found the same relation when applying noncommutative geometry `a la Coquereaux to left-right gauge model.

20

Similarly, we obtain m g′ ( ) sin(θ s ). 2 g0

Mz ′ =

(131)

From eq (110) the free Lagrangian associated with the field Ξ2 is given by: 1 2 Υ (|Ξ2 + 1|2 − 1)2 2 8g0  2  2 1 g¯ Υ = φ22 + higher order terms 2 g0 m

2 = LΞf ree

where Hence

˜ 2 + 1M ˜ 2 ). Υ2 = tr(M {ud} 3 {νe}   Υ g¯ . mφ2 = ( ) g0 m

If we take g¯ = g0 , then v u ˜ 2 + 1M ˜2 ) u tr(M {ud} {νe} 3 t mφ2 = . 1 ∗ ∗ tr(Mu Mu + Md Md + 3 Me Me∗ + 31 Mν Mν∗ )

(132)

(133) (134)

(135)

Neglecting the mass of fermions in respect to the mass of the top quark gives: s ˜2 ) 3 tr(M {ud} mφ2 ≈ = mt (136) tr(Mu Mu∗ ) 2 which is the same relation found by Wulkenhaar for the neutrinos mass matrix Mν 6= 0 [14]. Let us now turn to the mixing angles, eq (111) looks like its counterpart in classical model provided that: gs = gL = gR = g0

(137)

r

(138)

and ′

g =

21

3 g0 2

which give: sin2 (θW ) =

3 8

(139)

and

3 (140) 5 which are the same angles given by SO(10) GUT. However du to eq.(138) the grand unification for coupling constants is not achieved. sin2 (θS ) =

4

Conclusion

We reformulate the left-right model in the framework of nonassociative geometry. We have found that the left-right symmetric model does not exhibit the parity violation, however NAG gives the mixing angles predicted by SO(10) GUT and at the tree level we have the mass relations 1 gL MWL = ( )mt 2 g0

(141)

1 gR MWR = ( )mt 2 g0

(142)

MZ = M

Z′

MWL cos(θW )

(143)

g ′ sin(θS ) MWR = gR

(144)

3 MH = mt 2

(145)

and

where mt is the top mass. The unification of the coupling constants is not achieved using this reformulation. The main result of this work is that nonassociative geometry, like noncommutative geometry, does not provide a geometrical explanation of the parity violation. Acknowledgments

22

One of us (A. B.) would like to express his sincere thanks to Dr. R. Wulkenhaar and Dr. F.Cuypers for private communications. This work was supported by the Algerian Minist`ere de l’Enseignement Sup´erieur et La Recherche Scientifique under the contract D2501/01/05/97.

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[16] R. Wulkenhaar, hep-th/9804046. [17] R. Wulkenhaar, Int. J. Mod. Phys. A 13 (1998) 2627, hep-th/9607137. [18] R. Mohapatra and J. Pati, Phys. Rev. D 11 (1975) 566. [19] R.Mohapatra and G. Senjanovic, Phys. Rev. D 21 (1981) 165. [20] F. Cuypers, hep-ph/9706255. [21] Y. Okumura, Nuovo-Cimento, 110 A (1997) 267. [22] A. Benslama, Ph.D thesis, University of Constantine, Algeria, 1999. [23] A. Benslama and N. Mebarki, Parity violation and SU(2)L × SU(2)R model in the framework of nonassociative geometry, preprint May 1999, Constantine University. [24] A. Benslama in the Proceeding of the 4th Workshop on Mathematical Physics, Constantine University, June 1999. [25] B. Iochum and T. Sch¨ ucker, Lett. Math. Phys. 32 (1994) 153.

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