Flipped SO(10) model

We show that as in the flipped SU(5) models, doublet-triplet splitting is realized by the missing partner mechanism in the flipped SO(10) models. The gauge group $SO(10)_F\times U(1)_{V'_F}$ includes $SU(2)_E$ gauge symmetry, that plays an important role in solving supersymmetric flavor problem by introducing non-abelian horizontal gauge symmetry and anomalous $U(1)_A$ gauge symmetry. The gauge group can be broken into the standard model gauge group by VEVs of only spinor fields, such models may be easier than $E_6$ models to be derived from the superstring theory.


Introduction
In the previous paper [1], one of the authors shows that the SUSY flavor problem can be solved in E 6 unification by using non-abelian horizontal gauge symmetry and anomalous U(1) A gauge symmetry [2], whose anomaly is cancelled by the Green-Schwarz mechanism [3], even if large neutrino mixing angles are obtained. It is essential that the fundamental representation 27 of E 6 has two5 fields of SU (5). Actually 27 is decomposed as 27 → [10 (1,1) +5 (1,−3) + 1 (1,5) ] If we introduce three 27 fields Ψ i (i = 1, 2, 3) for three generation quarks and leptons, three of six5 fields become massive with three 5 fields after breaking E 6 into SU (5), and the remaining fields (3 ×5) remain massless. In the 6 × 3 mass matrix for5 and 5 fields, it is natural to expect that the elements for the third generation field Ψ 3 become larger to realize larger Yukawa couplings than the first and second generation fields Ψ 1 and Ψ 2 . Therefore, all the three massless modes of5 come mainly from the first two generation fields Ψ 1 and Ψ 2 . This structure is interesting because it can explain larger mixing angles of lepton sector than of quark sector as discussed in Ref. [4]. Moreover, if we introduce non-abelian horizontal symmetry SU(2) H and take the first two generation fields as doublet, then all three generation5 fields have degenerate sfermion masses, which are very important to suppress flavor changing neutral current (FCNC) processes with large neutrino mixing angles as discussed in Ref. [1].
In the above arguments, E 6 gauge group plays an important role. Actually, it is essential that a single field includes two5 fields to realize large neutrino mixing angles with suppressing FCNC processes. However, in order to break E 6 into the standard model (SM) gauge group SU(3) C × SU(2) L × U(1) Y , adjoint Higgs fields 78 are required, which may not be so easily realized in the framework of superstring models. To avoid the adjoint Higgs, it is a simple way to adopt non-simple group as a unification group. Which kinds of non-simple group do not spoil the above interesting features? The answer is simple. In order to satisfy the essential point that two fields with the same quantum number under the SM gauge group are included in a single multiplet, SU(2) E , which is a subgroup of E 6 group and rotates (5 (1,−3) ,5 (−2,2) ) and (1 (1,5) , 1 (4,0) ) as doublets, is sufficient. Therefore, it is interesting to consider the unification group which include SU(2) E . The SU(3) 3 ⊂ E 6 is in the case and we know that realistic SU(3) 3 model can be straightforwardly constructed [5], in which doublet-triplet splitting problem is solved and realistic quark and lepton mass matrices are obtained including large neutrino mixing angles. Therefore, if we introduce non-abelian horizontal symmetry in addition to SU(3) 3 , FCNC processes can be naturally suppressed with large neutrino mixing angles. In this paper, we consider another non-simple gauge group, SO(10) F × U(1) V ′ F , which can include SU(2) E because of the unusual embedding of the SM gauge group. We show that in this model, doublet-triplet splitting is realized by missing partner mechanism. The original missing partner mechanism was introduced in SU(5) unification group [6], but it requires several large dimensional representation Higgs fields. To avoid the large dimensional Higgs fields, flipped SU(5) [7] has been considered. It is known that the gauge group SU(5) F × U(1) X cannot be unified into SO(10) without spoiling the missing partner mechanism, but we show that SO(10) F × U(1) V ′ F ⊂ E 6 can embed the flipped SU(5) without spoiling the missing partner mechanism. As noted in the above, the flipped SO(10) gauge group includes SU(2) E , that is important to solve the SUSY flavor problem by introducing non-abelian horizontal gauge symmetry and anomalous U(1) A gauge symmetry.

Review of flipped SU (5) model
We briefly review the flipped SU(5) model and the reason why the flipped SU(5) model cannot be embedded in SO(10) GUT.
It is well-known that one family standard model fermions Q(3, 2) 1 , and E c (1, 1) 1 plus the right-handed neutrino N c (1, 1) 0 under the SM gauge group SU(3) C × SU(2) L × U(1) Y are unified into an SO(10)spinorial 16 superfield: where the decomposition is specified into SU(5) × U(1) V . The matter content of the flipped SU(5) models can be obtained from the corresponding assignment of the standard SU(5) GUT model by means of the "flipping" U c ↔ D c , N c ↔ E c : It is important that if 10 1 representation Higgs 10 C is introduced, SU(5)×U(1) X can be broken into the standard model gauge group SU(3) C × SU(2) L × U(1) Y by the vacuum expectation value (VEV) of the component of N c . Here, the hypercharge operator is written where Y ′ is the generator of SU(5) F which commutes with SU(3) C × SU(2) L . Then the SO(10)-vectorial 10 superfield decomposed as where D c′ and L ′ have the same quantum number of SM gauge group as D c and L, respectively. If we introduce interactions in the superpotential as only the triplet HiggsD c′ and D c′ can be superheavy with D c in 10 C andD c in 10C, respectively, by developing the VEVs of 10 C and 10C, but the doublet Higgs L ′ andL ′ have no partner and remain massless. This is essential of the missing partner mechanism in the flipped SU(5) model. Unfortunately, this missing partner mechanism in the flipped SU(5) model cannot be extended to SO(10) unification. In SO(10) unification the interactions (2.6) are included in the SO(10) symmetric interactions C(16)C(16)H(10) and C(16)C(16)H (10), which include also Through these interactions, the doublet Higgs (L ′ ) H and (L ′ ) H become superheavy with L C and (L * )C, respectively, by developing the VEVs of 10C and 10C.
(In this paper, X * is a component of 16 of SO (10) and denotes the complex conjugate representation of X which is a component of 16 of SO(10). ) Therefore, doublet-triplet splitting is spoiled by this extension. 1 In the next section, we show that the missing partner mechanism of the flipped SU(5) model can be embedded in SO (10) 3 Flipped SO(10) model As noted in the introduction, 27 of E 6 is decomposed as There are two ways to embed the flipped SU(5) matters 10 Ψ = (Q, D c , N c ),5 Ψ = (U c , L) and 1 Ψ = E c in the above decomposition of 27 of E 6 into SO(10) × U(1) V ′ . As discussed in the previous section, the usual embedding SU(5) F × U(1) X in SO (10), where 5 H = (D c′ , L ′ ),5 H = (D c′ ,L ′ ) and 1 S is singlet under SU(5) F × U(1) X , spoils the missing partner mechanism. The other embedding can be obtained by means of the "flipping"5 Ψ ↔5 H and 1 Ψ ↔ 1 S : In this embedding, if 1 S component of 16 1 field have non-vanishing VEV, SO (10) Here, the operator X is obtained as L) and (N c , S), which has the same quantum number of SM gauge group, is included into a single multiplet 16 1 , 10 −2 and 16 1 , respectively. This means that SU(2) E is embedded in SO(10) F . We introduce two pairs of Higgs fields [Φ(16 1 ),Φ(16 −1 )] and [C(16 1 ),C(16 −1 )] to break SO (10) (1) X , the components 10 Φ and 10Φ are absorbed by the Higgs mechanism. The VEVs | C | = | C | break SU(5) F × U(1) X into the SM gauge group, and the components Q and N c are absorbed by the Higgs mechanism. All the remaining components5 Φ , 5Φ,5 C , 5C, (D c ) C and (D c * )C must be massive except a pair of doublets. For example, through the interactions in the superpotential, W SO(10) =ΦΦCC +CCΦΦ, (3.6) which include the interactions (2.6) after developing the VEVs If we introduce the mass term for C andC, then only (L ′ ) Φ and (L ′ * )Φ remain massless, namely, doublet-triplet splitting is realized. There are several interactions which unstabilize the doublet-triplet splitting. For example, the termsΦΦF (CC,ΦΦ) give directly the doublet Higgs mass, so they must be forbidden. (We will return to this subject lator in a concrete model.) We assume that three generation matter fields Ψ i (27) = 16 Ψ i + 10 Ψ i + 1 Ψ i (i = 1, 2, 3) respect E 6 symmetry. It is an easy way to guarantee the cancellation of gauge anomaly. Among the three generation matter fields Ψ i , there are six fields which have the same quantum number under the SM gauge group as (D c , L). Only three linear combinations of these fields become quarks and leptons, and other modes become superheavy with the three (D c′ ,L ′ ) fields through the interactions 16 Ψ i 10 Ψ j Φ and 16 Ψ i 10 Ψ j C by developing the VEVs of Φ and C. It is interesting that up-type Yukawa coupling can be obtained from the renormalizable interactions 16 Ψ i 10 Ψ j Φ, because O(1) top Yukawa coupling can be naturally realized. But Yukawa couplings of down quark sector and of charged lepton sector are obtained from the higher dimensional interactions 16 Ψ i 16 Ψ jCΦ and 10 Ψ i 1 Ψ jCΦ , respectively. Because we have six singlets N c i and S i in the matter sector, the mass matrix for right-handed neutrino becomes 6 × 6 matrix which are obtained from the interactions Ψ i Ψ jΦΦ , Ψ i Ψ jΦC and Ψ i Ψ jCC . Yukawa couplings of Dirac neutrino sector are obtained from the interactions 16 Ψ i 10 Ψ j Φ. Therefore, the mass terms of all quarks and leptons can be obtained in this scenario.
Unfortunately, as in the flipped SU(5) model, this missing partner mechanism in the flipped SO(10) model cannot be extended to E 6 unification. In give mass terms to doubet Higgs by taking non-vanishing VEVs | C | = | C |. Therefore, doublet-triplet splitting is spoiled in this extension.

Flipped SO(10) model with anomalous U (1) A
It is important to find a concrete flipped SO(10) model in which doublet-triplet splitting is realized with generic interactions and to examine whether the realistic quark and lepton mass matrices are realized or not. In a series of papers [1,4,5,9,10,11], we have pointed out that anomalous U(1) A symmetry plays an important role in solving various problems in SUSY grand unified theory (GUT) with generic interactions. This is mainly because the SUSY zero mechanism (holomorphic zero) 2 can control various terms which must be forbidden.
In this section, we present a concrete flipped SO(10) model with generic interaction by introducing anomalous U(1) A symmetry.

Higgs sector
The Higgs contents are listed in Table I.
Following the general discussion on the determination of VEVs of the models with anomalous U(1) A charges, only the negatively charged fields can have nonvanishing VEVs [4,9,10,11]. The scale of these VEVs are determined by the anomalous U(1) A charges as where λ is the ratio of the VEV of Froggatt-Nielsen field Θ, which is essentially determined by the Fayet-Illiopoulos D-term parameter, to the cutoff Λ. In this paper, we take λ as around the Cabbibo angle sin θ W ∼ 0.22. If the 1 (1,5) component of Φ and the 1 (−1,−5) component ofΦ have non-vanishing VEVs, (1,1) of Φ and 10 (−1,−1) ofΦ are absorbed by the Higgs mechanism at that time. Moreover, if the 10 (1,1) component of C and the 10 (−1,−1) component ofC have non-vanishing VEVs, SU(5) F × U(1) X is broken into the SM gauge group. Then the Q component of 10 (1,1) of C and theQ component of 10 (−1,−1) ofC are absorbed by the Higgs mechanism. Therefore, the remaining negatively charged fields except singlets under the SM gauge group are the5 (1,−3) components of Φ and C, the D c component of C, and the mirror components ofΦ andC. Among these negatively charged fields, no mass term appears because of the SUSY zero (holomorphic zero) mechanism. In order to make them massive, we have to take account of the positively charged fields Φ ′ i andΦ ′ i . Note that in a 16 1 field, there are two colored Higgs D c and D c ′ because of SU(2) E symmetry, but only one doublet L ′ . Therefore, the colored Higgs mass matrix becomes 7 × 7 matrix M T which is given by where ∆ ≡ 1 2 (φ − φ −c + c). The rank becomes seven for the charge assignment in Table I. On the other hand, the mass matrix for doublet Higgs becomes 4 × 4 matrix. The charges in Table I lead It is obvious that the rank is reduced to three, and therefore one pair of doublet Higgs appears in this model. The massless modes are written where H u and H d are the doublet Higgs for up-quark sector and for down-quark sector, respectively.

Quark and lepton sector
In this subsection, we use the standard definition of5 ≡ (D c , L) field. If we introduce three generation matter fields Ψ i (27) = 16 Ψ i + 10 Ψ i + 1 Ψ i (i = 1, 2, 3) with their charges (ψ 1 , ψ 2 , ψ 3 ) = (4,3,1) in addition to the Higgs sector in Table  I, the massless modes of5 fields, where we have used the usual definition for5, become5 where5 ′ ≡ (D c′ , L ′ ) and we fix the three bases of the massless modes (5 1 ,5 2 ,5 3 ) to . These are obtained from the mass matrix of three 5 fields and six 5 fields which are given from the interactions Ψ i Ψ j ΦZ and Ψ i Ψ j C by developing the VEVs of Φ, C and Z. Then we can estimate the Yukawa couplings of quarks and leptons.
The Yukawa couplings of up quark sector are obtained as from the interactions λ ψ i +ψ j +c 16 Ψ i 10 Ψ j C. The Yukawa couplings of down quark sector and of charged lepton sector are given as 4.8) from the higher dimensional interactions λ ψ i +ψ j +2c 16 Ψ i 16 Ψ jCC and λ ψ i +ψ j +2c 10 Ψ i 1 Ψ jCC , respectively. Note that only5 ′ fields can have non-vanishing Yukawa couplings through the interactions. This is because the interactions 16 Ψ i 16 Ψ jCΦ and 10 Ψ i 1 Ψ jCΦ are forbidden by Z 2 -parity. The above mass matrices give almost good values for masses and mixings for quark sector and charged lepton sector. The Yukawa couplings for the Dirac neutrino are given as through the interactions λ ψ i +ψ j +c 10 Ψ i 16 Ψ j C. The vanishing component is caused by SUSY zero (holomorphic zero). The right-handed neutrino mass matrix becomes Here vanishing components are caused by the SUSY zero (holomorphic zero) mechanism. Then the neutrino mass matrix is given by where η is a renormalization factor. This gives bi-large neutrino mixings but to realize the mass scale for the neutrino, we have to take the cutoff Λ ∼ 10 13 GeV if we take H u η ∼ 200 GeV. Such a small cutoff scale leads to too short neucleon life-time via dimension six operators. Therefore, the charge assignment in Table  I looks unrealistic. However, because the neutrino scale is determined by the anomalous U(1) A charges as there may be other realistic models with other charge assignments. To obtain larger value of l, smaller c and/or largerc is needed. Because C includes H u , the charge c is determined as c = −2ψ 3 = −2n so that the top Yukawa coupling becomes O(1). Here we take ψ i = δ i + n [(δ 1 , δ 2 , δ 3 ) = (3, 2, 0)] to obtain realistic Cabbibo-Kobayashi-Maskawa matrix. To realize bi-large neutrino mixings (i.e., 5 fields in Eq. (4.6)), we must take (4.14) Once we fix the mixing structure of5 fields, the Yukawa couplings for down quarks are proportional to λ ψ i +ψ j +2c C ∼ λ δ i +δ j + 3 2 (c−c) . Therefore, roughly speaking, tan β ≡ Hu H d is proportional to λ 3 2 (c−c) . Then, smaller c and/or largerc lead to smaller tan β. For fixed tan β, smaller c andc lead to larger l. However, unless the condition c − 2c ≤ 2 (4.15) is satisfied, (Y d ) 33 vanishes by the SUSY zero mechanism. A compromised charge assignment is (φ,φ, c,c, φ ′ i ,φ ′ i ,z i , z, s ′ ) = (−1, −1, −4, −3, 8, 8, −1, −6, 12). Then l becomes −5, so the cutoff scale can be larger than the 10 15 GeV. Actually, the running gauge couplings of SU(3) C and SU(2) L , which should meet at the cutoff scale in this flipped SO(10) scenario, meet around the scale in this charge assignment. And the Yukawa coupling of bottom quark becomes λ 3.5 which can be realistic although the large ambiguity of O(1) coefficients is required.

Summary
In this paper, we have shown that the missing partner mechanism in flipped SU(5) model can be embedded in flipped SO(10) model whose gauge group is SO(10) F × U(1) V ′ F ⊂ E 6 . It is interesting that the gauge group includes SU(2) E , that plays an important role in solving SUSY flavor problem by the horizontal gauge symmetry and anomalous U(1) A gauge symmetry. As a proof of existence of a concrete model, we build a flipped SO(10) model by introducing anomalous U(1) A gauge symmetry.