6D gauge-Higgs unification on T^2/Z_N with custodial symmetry

We investigate the gauge-Higgs unification models compactified on T^2/Z_N that have the custodial symmetry. We select possible gauge groups, orbifolds and representations of the matter fermions that are consistent with the custodial symmetry, by means of the group theoretical analysis. The best candidate we found is 6D SU(3)_C x U(4) gauge theory on T^2/Z_3 and the third generation quarks are embedded into bulk fermions that are the symmetric traceless rank-2 tensor of SO(6).


Introduction
The gauge-Higgs unification [1,2,3,4] is an interesting candidate for the new physics beyond the standard model. This solves the gauge hierarchy problem because a higher dimensional gauge symmetry protects the Higgs mass against quantum corrections. Chiral fermions in four-dimensional (4D) effective theory can be obtained by compactifying the extra dimensions on an orbifold.
The current experimental constraints require θ H to be small, i.e., θ H < ∼ O(0.1). In order to realize such small values of θ H , we need some amount of fine-tuning among the model parameters, such as 5D mass parameters for fermions. This stems from the fact that the effective potential for θ H does not exist at tree level and is induced at one-loop level in 5D gauge-Higgs unification models. The one-loop potential is typically expressed as a sum of periodic functions of θ H with the period of π and 2π (or π/2), which are roughly approximated as the cosine functions [11,12]. Therefore, θ H = O(1) is realized unless the model parameters are fine-tuned. This problem can be evaded in the six-dimensional (6D) gauge-Higgs unification models. In this case, quartic terms in the Wilson line phases exist at tree level, while quadratic terms are induced at one-loop level. In the flat spacetime, for example, the effective potential has a form of V (θ H ) = − c 2 g 2 l 6 R 2 θ H gπR 2 + c 4 g 2 θ H gπR and the KK modes are estimated to be around a few TeV.
In extra-dimensional models, coupling constants in 4D effective theories generally deviate from the standard model values even at tree level due to mixing with the KK modes [13,14,15]. Unless m KK is very high, models need some mechanisms to suppress such deviations. Especially a requirement that the ρ parameter and the Z boson coupling to the left-handed bottom quark (the Zb LbL coupling) do not deviate too much often imposes severe constraints on the model building. It is known that the custodial symmetry can protect them against the corrections induced by the mixing with the KK modes [8,16].
Hence we focus on 6D gauge-Higgs unification models that has the custodial symmetry in this paper.
The purpose of this paper is to select candidates for realistic 6D gauge-Higgs unification models by means of the group theoretical analysis. The analysis is useful to investigate the gauge-Higgs unification models because the Higgs sector is determined by the gauge group structure. There are some works along this direction. 5D models are analyzed in Ref. [17], the tree-level Higgs potentials in 6D models are calculated in Ref. [18], and models in arbitrary dimensions are discussed in Ref. [19]. In these works, the custodial symmetry is not considered and the electroweak gauge symmetry SU(2) L × U(1) Y is embedded into a simple group. Thus the Weinberg angle θ W is determined only by the group structure, and they found that no simple group realizes the observed value of θ W . However, the assumption that SU(2) L × U(1) Y is embedded into a simple group is not indispensable because the color symmetry SU(3) C is not unified anyway. Besides, any brane localized terms allowed by the symmetries are not introduced in Refs. [17,19]. In fact, the realistic models constructed so far allow both an extra U(1) gauge symmetry, which is relevant to the realization of the experimental value of θ W , and various terms and fields localized at the fixed points of the orbifolds [6,7,10,11]. Therefore, we include both ingredients in our analysis. Since larger gauge groups contain more unwanted exotic particles, we consider a case that the 6D gauge group is SU(3) C × G × U(1), where G is a simple group whose rank is less than four.
The paper is organized as follows. In the next section, we explain our setup and derive conditions for zero-modes. In Sec. 3, we list the zero-modes in the bosonic sector for all the rank-two and the rank-three groups that include the custodial symmetry. In Sec. 4, we find a condition to preserve the custodial symmetry, and provide explicit expressions of the W and Z boson masses. In Sec. 5, we discuss embeddings of quarks into 6D fermions, and search for appropriate representations of G that the 6D fermions should belong to. In Sec. 6, we calculate the Higgs potential at tree level. Sec. 7 is devoted to the summary. In Appendix A, we collect formulae in the Cartan-Weyl basis of the gauge group generators. In Appendix B, general forms of the orbifold boundary conditions are shown. In Appendix C, we list irreducible decompositions of various G representations into the SU(2) L × SU(2) R multiplets.

Compactified space
The 6D spacetime is assumed to be flat, and the metric is given by where M, N = 0, 1, · · · , 5, η µν = diag(−1, 1, 1, 1) is the 4D Minkowski metric, and a point in the extra space (x 4 , x 5 ) is identified as x 4 x 5 ∼ x 4 where n 1 and n 2 are integers, and R 1 , R 2 > 0 and 0 < θ < π are constants. In order to obtain a 4D chiral theory at low energies, we compactify the extra space on a twodimensional orbifold. All possible orbifolds are T 2 /Z N (N = 2, 3, 4, 6) [20]. It is convenient to use a complex (dimensionless) coordinate z ≡ 1 2πR 1 (x 4 + ix 5 ). Then, the orbifold obeys the identification, where ω = e 2πi/N and τ ≡ R 2 R 1 e iθ . Note that an arbitrary value of τ is allowed when N = 2 while it must be equal to ω when N = 2.
The orbifold T 2 /Z N has the following fixed points in the fundamental domain [21,22].
4D fields or interactions are allowed to be introduced on these fixed points.

Field content
We consider a 6D gauge theory whose gauge group is SU(3) C × G × U(1) Z , where G is a simple group. Since G must include SU(2) L × SU(2) R , its rank r is greater than one.
In this paper, we investigate cases of r = 2, 3. In the following, we omit SU(3) C since it is irrelevant to the discussion. The 6D gauge fields for G and U(1) Z are denoted as A M and B Z M , and the field strengths and the covariant derivative are defined as F where g A and g Z are the 6D gauge coupling constants for G and U(1) Z , Γ M are the 6D gamma matrices, and L (z f ) are 4D Lagrangians localized at the fixed points z = z f .
The G gauge field A M is decomposed as where {H i , E α } are the generators in the Cartan-Weyl basis, i.e., H i (i = 1, · · · , r) are the Cartan generators and α runs over all the roots of G. Since A M is Hermitian, C i M are real and W −α M = (W α M ) * . In the complex coordinate (x µ , z), the extra-dimensional components of the gauge fields are expressed as

Orbifold conditions for gauge fields
As shown in Appendix B, the general orbifold boundary conditions for the gauge fields can be expressed as where P is an element of G. The orbifold conditions for 6D fermions are provided in (5.2).
Since zero-modes of the gauge fields have flat profiles over the extra dimensional space, we can see from (2.8) that B Z µ has a zero-mode while B Z z does not. Namely U(1) Z is unbroken by the orbifold conditions. The condition for A M to have zero-modes is determined by the choice of the matrix P in (2.8). It is always possible to choose the generators so that P is expressed as where p · H ≡ i p i H i and p i are real constants. Thus P H i P −1 = H i and P E α P −1 = e ip·α E α , and the relevant conditions in (2.8) to the zero-mode conditions are rewritten as This indicates that C i µ always have zero-modes while C i z do not irrespective of the choice of the matrix P . Therefore the orbifold boundary conditions cannot reduce the rank of G as pointed out in Ref. [23]. In contrast, whether W α µ and W α z have zero-modes depend on the choice of P . Since (2.10) is the Z N transformation, p i must satisfy e iN p·α = 1. Thus possible values of p · α are where n α is an integer.
In this paper, we focus on P such that the orbifold bondary conditions break G to We denote the positive roots that specify SU(2) L and SU(2) R as α L and α R , respectively. The SU(2) L and SU(2) R generators are given by (4.1). Then (2.11) is further restricted as From the last condition in (2.10), the zero-mode condition for W β z is (2.13)

Zero-modes of gauge and Higgs fields
In this section, we investigate the field content of the zero-modes from the 6D gauge fields.

Rank-two groups
First we consider a case of r = 2, i.e., G = SO(5), G 2 . In this case, the unbroken gauge group by the orbifold conditions is SU(2) L ×SU(2) R ×U(1) Z . We do not consider G =SU (3) because it does not contain SU(2) L ×SU(2) R as a subgroup. The roots of G can be expressed as linear combinations of two-dimensional basis vectors e i (i = 1, 2).

SO(5)
The roots are {±e i ± e j , ±e i } (1 ≤ i = j ≤ 2). We can choose the unbroken sub- A candidate for the Higgs fields is a bidoublet (2, 2), which consists of ±e 1 and ±e 2 . The conditions in (2.12) are now expressed as It is enough to find a solution in a range: 0 ≤ p 1 , p 2 < 2π. A solution exists when N = 3, and it is or Therefore the zero-mode condition (2.13) for (2, 2) is expressed as Namely, we have one Higgs bidoublet when N = 2, while no Higgs exists in the other cases.
Let us first consider the case of (α L , α R ) = (e 1 , e 2 / A candidate for the Higgs fields is (2, 4). The conditions in (2.12) become It is enough to find a solution in a range 0 ≤ p 1 , p 2 √ 3 < 2π. A solution exists when N = 3, and it is Therefore the zero-mode condition (2.13) for (2, 4) is expressed as Namely, we have a (2, 4) multiplet as the Higgs fields when N = 2, while no Higgs exists in the other cases.
In the case of (α L , α R ) = (e 2 / √ 3, e 1 ), the results are obtained by exchanging SU(2) L and SU(2) R in the above resuts. Hence we do not have SU(2) L -doublet Higgses.

Rank-three groups
Next we consider a case of r = 3, i.e., G =SU(4),SO(7),Sp (6). In this case, the unbroken gauge group by the orbifold conditions is SU(2) L ×SU(2) R ×U(1) X ×U(1) Z . The roots of G can be expressed as linear combinations of three-dimensional basis vectors e i (i = 1, 2, 3).
When N = 3, the scalar zero-modes we have are Nothing : (when n P = 2, 4) The results are obtained by exchanging SU(2) L and SU(2) R in the case (II). Hence we do not have SU(2) L -doublet Higgses.

Sp(6)
The roots are The U(1) X generator is The irreducible decomposition of the adjoint representation of G is Independent conditions in (2.12) are expressed as Solutions exist only when N = 4, 6. They are where n P = 0, N/2.
When N = 4, the scalar zero-modes we have are When N = 6, they are The U(1) X generator is The irreducible decomposition of the adjoint representation of G is Independent conditions in (2.12) are expressed as where n P = 0, N/2. Solutions exist only when N = 3, 4, 6. They are (3.45) When N = 3, the scalar zero-modes we have are When N = 4, they are When N = 6, they are The results are obtained by exchanging SU(2) L and SU(2) R in the case (II).

Custodial symmetry and Weinberg angle 4.1 Custodial symmetry
Here we consider a condition that the custodial symmetry is preserved after the electroweak symmetry is broken. The SU(2) L and SU(2) R generators are respectively. Thus (2.6) is rewritten as and B X µ ≡ x·Cµ |x| 2 is the U(1) X gauge field that does not exist when r = 2. The ellipsis denotes components that do not have zero-modes. Since the generators in (4.1) are normalized as the canonically normalized zero-mode gauge fields arê where A is the area of the fundamental domain of T 2 /Z N .
Since we have assumed that SU(2) R × U(1) Z is unbroken by the orbifold boundary conditions, we introduce some 4D scalar fields at one of the fixed points of T 2 /Z N in order to break it to U(1) Y . We demand that the custodial symmetry SU(2) V ⊂ SU(2) L × SU(2) R remains unbroken after the Higgs fields have VEVs. The generators of SU(2) V are Thus the conditions for SU(2) V to be unbroken are since C i z do not have zero-modes and thus C i z = 0.

Rank-two groups
Let us first consider the rank-two groups. We introduce the following Lagrangian at z = 0. 3 and V (φ) is a potential that force φ to have a nonvanishing VEV. After φ gets a VEV, SU(2) R ×U(1) Z is broken to U(1) Y , and the corresponding massless gauge field is expressed where a mixing angle θ Z is determined by tan After W β z have nonvanishing VEVs, SU(2) L × U(1) Y is broken to the electromagnetic symmetry U(1) em . Since W β z is U(1) Z neutral and only U(1) em neutral W β z can have nonvanishing VEVs, the root β must satisfy if W β z = 0. Thus the second condition in (4.7) is automatically satisfied. The roots that satisfy (4.11) are ±e 2 ∈ (2, 2) in SO (5), and ± e 1 2 − e 2 2 √ 3 ∈ (2, 4) in G 2 . Then, from the first condition in (4.7), we obtain a condition, for SO (5), while no nonvanishing VEV is allowed for G 2 .

Rank-three groups
Next consider the rank-three groups. Since the unbroken gauge symmetry by the orbifold conditions is SU(2) L × SU(2) R × U(1) X × U(1) Z , let us first assume that φ in (4.8) also has a nonzero U(1) X charge in order to obtain SU(2) L ×U(1) Y at low energies. Then the U(1) Y gauge field B Y µ becomes a linear combination of W 3 Rµ , B X µ and B Z µ , and the hypercharge is identified as Thus the condition (4.11) now becomes From this and the second condition in (4.7), both (4.11) and x · β = 0 must be satisfied if W β z = 0. Such roots do not exist among the zero-modes listed in Sec. 3.2. Therefore we introduce two complex scalar fields φ 1 and φ 2 instead of φ on the fixed point, where φ 1 and φ 2 are complex scalars belonging to (1, 2) 0,+1/2 and (1, 1) +1,0 respectively Hence the hypercharge is identified as (4.10). The U(1) X charges are no longer relevant to the U(1) Y and U(1) em charges because U(1) X is completely broken by a VEV of another scalar φ 2 .
Thus the U(1) Y gauge field is given by (4.9). In this case, the U(1) em neutral condition becomes (4.11), which is consistent with the second condition in (4.7). As a result, possible nonvanishing VEVs are as follows. (4), where the double signs correspond.
In summary, fields that can have nonzero VEVs are the neutral components of a bidoublet (2, 2) or a singlet (1, 1). The above conditions indicate that a bidoublet H a must have a VEV: where v a > 0, if we redefine a phase of each field component appropriately.

Weinberg angle and weak gauge boson masses
In the approximation that the W and Z bosons have constant profiles over the extra dimensions, the 4D SU(2) L and U(1) Y gauge coupling constants are read off from couplings to the matter zero-modes, and are identified as .

(4.18)
Thus the Weinberg angle is calculated as .

(4.19)
We can obtain the experimental value tan 2 θ W ≃ 0.30 by tuning the ratio g Z /g A .
Next we derive the expressions of the W and Z boson masses. From (4.5) and (4.9), the expression (4.2) becomes From the results in the previous subsections, the only components that contribute to the W and Z boson masses are the neutral components of the bidoublets. Since the roots that form a bidoublet are expressed as where a labels the bidoublets, (4.21) are rewritten as where γ a is the T 3 L = T 3 R = −1/2 component of the zero-mode bidoublets H a . We have used that |N −α L ,γa+α L | 2 = |N α L ,γa+α R | 2 = |α L | 2 /2, and ζ ≡ arg(N −α L ,γa+α L ) and η ≡ arg(N α L ,γa+α R ). Thus the relevant terms in 6D Lagrangian are calculated as At the last step, we have used that (4.5), and (g: 4D SU(2) L gauge coupling), which follows from (4.12) or (4.16). We obtain the W and Z boson mass terms by integrating (4.24) over the extra dimensions, and their masses are read off as From these and (4.19), we find that ρ ≡ m 2 W /(m 2 Z cos 2 θ W ) = 1. This is expected because we have assumed that only SU(2) L doublets and singlets have nonzero VEVs and neglected the z-dependence of the mode functions for the W and Z bosons. The custodial symmetry will play a crucial role when such z-dependence is taken into account.

Matter field
We consider a case that quarks and leptons live in the bulk. This case is interesting because the hierarchical structure of the Yukawa coupling constants can be realized by the wave function localization [24,25], and the generation structure can also be obtained by a background magnetic flux [22]. In the following, we focus on the quark sector, but a similar argument is also applicable to the lepton sector.
As pointed out in Ref. [26], the generations and the hierarchy among the Yukawa couplings can be obtained by introducing an extra gauge symmetry G F and assuming a magnetic flux on T 2 /Z N and the Wilson line phases for G F . The zero-modes are contained in Ψ χ 6 ,χ 4 as where µ runs over the weights of the zero-mode states, 4 and the ellipsis denotes the nonzero KK modes. The number of the zero-modes j max is determined by the magnetic flux [22].
The zero-mode functions f (j)µ χ 6 (z) are determined so that (5.3) satisfies the first two conditions in (5.2). From the last condition in (5.2), we obtain Namely, the zero-mode is an eigenvector of ω − χ 4 χ 6 2 e iϕω P with an eigenvalue 1. Denote the highest weight of a representation R that Ψ χ 4 ,χ 6 belongs to as µ max . Then µ is expressed where k i are non-negative integers, and α i are the simple roots. Since P N |µ = e iN p·µ |µ = e iN p·µmax |µ , 5 the phase ϕ ω is determined by (B.4) as ϕ ω = π N (2m ω + 1) − p · µ max , where m ω = 0, 1, · · · , N − 1. Thus we find that Namely, the zero-mode condition for the state |µ is

Zb LbL coupling
When the quarks live in the bulk, the Zb LbL coupling often receives a large correction induced by mixing with the KK modes. The authors of Ref. [16] pointed out that the custodial symmetry plays an important role to suppress the deviation of this coupling from the standard model value. The Zb LbL coupling is protected if the theory has a parity symmetry P LR that exchanges SU(2) L and SU(2) R , and b L is the component of (2,2). Since the Higgs fields also belong to (2, 2), the right-handed quarks should belong to (1, 1) or (1, 3) + (3, 1). 4 Do not confuse it with the 4D Lorentz index. 5 We have used (2.11) at the second equality.
Cases in which the bosonic sector has the parity symmetry P LR and a scalar bidoublet are SO(5), SU(4) and SO(7) (I) in Sec. 3. In Appendix C, we list the irreducible representations of these groups whose dimensions are less than 30, and their decomposition into the SU(2) L × SU(2) R (×U(1) X ) multiplets. There is no (1, 3) + (3, 1) multiplets included in the list. Hence the left-handed and the right-handed quarks should be embedded into (2, 2) and (1, 1), respectively.

Yukawa couplings 5.3.1 General expression
The Yukawa couplings originate from the 6D minimal couplings in the kinetic term, where g is the SU(2) L gauge coupling constant. (See (6.5).) Then the Yukawa couplings in 4D effective Lagrangian are expressed as where R H , R L and R R are irreducible representations of SU(2) L ×SU(2) R (×U(1) X )×U(1) Z that |β , |µ L and |µ R = |µ L + χ 6 β belong to, and Note that these coupling constants only depend on the representations {R H , R L , R R }, and take common values for all β ∈ R H and µ L ∈ R L .
Exponentially small Yukawa couplings can be obtained by using the wave function localization in the extra dimensions [24,25]. For the third generation, we assume that the overlap integrals in (5.9) do not provide any suppression factors, i.e., equal one. Then the Yukawa couplings are determined only by the group-theoretical factors. In the following, we focus on the third generation quarks.

Embedding of quarks
As we will see in Sec. 6, the Higgs potential at tree level only contains quartic terms. The electroweak symmetry breaking occurs at one-loop level, and the top Yukawa coupling provides a dominant contribution to the one-loop Higgs potential. In general, such oneloop potential breaks SU(2) L × SU(2) R , and thus the Higgs VEVs are not aligned as (4.17).
Namely the custodial symmetry is broken. A simple way to avoid this difficulty is to assume that the quark fields couple to the Higgs fields only through a combination H a +H a . This is achieved when y Specifically, consider a case that ν L = −µ L and 4D fermions ζ R ∈ (2, 2) and η L ∈ (1, 1) are localized at a fixed point, which transform as ζ R → −ζ R and η L → −η L under Ψ ± → −Ψ ∓ . L yukawa = y λ 2 a tr Q L H a +H a λ R + h.c. + · · · , (5.12) 6 Notice that Q where y λ ≡ y Now we will see how the quark fields should be embedded into 6D fields. For simplicity, we consider a case that there is one Higgs bidoublet H as a zero-mode for a while. We introduce two 6D Dirac fermions Ψ (2/3) = Ψ L , H 1 } are SU(2) L doublets whose T 3 R eigenvalues are −1/2 and 1/2, respectively. Then the Yukawa couplings in the form of (5.12) are expressed as where θ q is a mixing angle. The orthogonal combination and Q L are exotic fields that must be decoupled at low energies. Hence we need to introduce 4D localized fermions that couple with those exotic components. As a result, the following Yukawa couplings are obtained at low energies.
. 7 In such a case, m t is calculated as where v is defined as H 1 = H 2 = (0, v/2) t (see (4.17)). We have used that cos θ q ≃ 1, (4.25) and (5.11). Therefore, the observed top quark mass is obtained if |N β,µ L | = √ 2. 8 We can extend this result to the two-Higgs-bidoublet case straightforwardly.

Available representations for matter fermions
In summary, the quark multiplets should be embedded into two 6D Dirac fermions Ψ (2/3) and Ψ (−1/3) whose U(1) Z charges are 2/3 and −1/3, respectively. Irreducible representations R which they belong to must satisfy the following conditions.
We will search for R that satisfies these conditions from the list in Appendix C. We focus on the cases of G = SO(5), SU(4), SO(7)(I), which have the P LR symmetry.

SO(5)
There is no irreducible representation that satisfies the condition 1 among the list in Appendix C.1.

SU(4)
Only 20 ′ satisfies the condition 1 among the list in Appendix C.2. The weights of 7 In contrast to the mixing between Q L , the mixing angle θ q can take arbitrary values because there is no symmetry to fix it. 8 A small deviation from the observed value of m t is expected to be explained by quantum correction.
(See Ref. [27].) 20 ′ that form (2, 2) and (1, 1) are where the double signs correspond. Notice that the weights that form bidoublets are the same as the roots that form the Higgs bidoublets.

SO(7) (I)
The irreducible representations that satisfy the condition 1 among the list in Appendix C.3 are 21 and 27. These also satisfy the condition 2, but they cannot satisfy the condition 3 for any choice of (N, n P , m ω ).

Higgs potential
In contrast to the 5D gauge-Higgs unification model, we have quartic couplings of the Higgs fields at tree level. The relevant terms in the 6D Lagrangian are (6.1) In this section, we calculate the classical Higgs potential V tree focusing on the Higgs bidoublets, which are relevant to the electroweak symmetry breaking. In the previous section, we have shown that only a model of G =SU(4) compactified on T 2 /Z 3 has required zero-mode spectrum for the quarks. For the sake of completeness, however, we will also calculate V tree in the other cases that have Higgs bidoublets. We have one Higgs bidoublet in the cases of SO (5)

SO(5) case
First we consider the SO(5) case. In this case, the roots that form the bidoublet are From (6.1), the kinetic terms of the zero-modes W β z in the 4D effective Lagrangian are We have used (A.4), and A is the area of T 2 /Z N . Thus the canonically normalized Higgs bidoublet is defined as Then it follows that where we have used (A.2). Hence, from (6.1), V tree is calculated as where H 2 ≡ (H 1 2 , H 2 2 ) t and H 1 ≡ (H 1 1 , H 2 1 ) t are the SU(2) L doublets with the hypercharge Y = 1/2. We have used that (4.18) with |α L | 2 = 2, and |N e 1 ,e 2 | 2 = |N e 1 ,−e 2 | 2 = 1.
The above result agrees with Eq.(7) in Ref. [18]. The final expression in (6.6) is manifestly invariant under the transformation: H → U L HU † R (U L ∈ SU(2) L and U R ∈ SU(2) R ).

Cases of rank-three groups
Next we consider the cases of the rank-three groups. In these cases, the candidates for the zero-mode Higgs bidoublets consist of the following roots.
where γ = − e 1 √ 2 − e 2 √ 2 + e 3 for SU(4), γ = −e 1 + e 3 for SO(7) (I), and γ = −e 2 − e 3 for Sp(6) (II) or (III). The canonically normalized Higgs bidoublets are defined as where the signs in the suffixes denote the signs of the U(1) X charges. Then it follows that where γ L ≡ γ + α L , γ R ≡ γ + α R and γ LR ≡ γ + α L + α R , and the ellipses denote fields belonging to other multiplets, if any. After some calculations, we obtain We have used that U L H ± U R (U L ∈ SU(2) L and U R ∈ SU(2) R ). Except for the case of SU(4) on T 2 /Z 2 , one of the bidoublets H ± is absent due to the orbifold boundary conditions. In such cases, the model becomes a two-Higgs-doublet model. In contrast to the SO(5) case, the potential (6.10) with H + = 0 or H − = 0 does not agree with (7) of Ref. [18]. This is because they have assumed γ + α L + α R = −γ, which only holds in the SO(5) case.
Finally we comment on the Higgs mass. We consider a case of SU(4) on T 2 /Z 3 . The tree-level Higgs potential (6.10) becomes where m ± > 0, and the ellipsis denotes terms involving the charged components. We can always redefine the phase of fields so that h + > 0. Then, from the minimization condition for the potential, we obtain Therefore, the alignment (4.17) is actually achieved. Note that the one-loop induced quadratic terms do not have the SU(2) L × SU(2) R symmetry. Thus the custodial symmetry is broken in the Higgs sector. This does not cause a problem for the protection of the ρ parameter at tree level as long as the relevant terms to the W and Z boson masses in (4.24) have the custodial symmetry.
The mass of the lightest neutral Higgs boson is where v is defined as H 2 1 = H 2 2 = v/2 > 0, and we have used (4.25) at the last equality. We expect that the deviation from the observed value m H ≃ 125 GeV is explained by quantum corrections 9 .

Summary
We have investigated 6D gauge-Higgs unification models compactified on T 2 /Z N (N = 2, 3, 4, 6) that have the custodial symmetry. The gauge group is assumed to be SU(3) C × G × U(1) Z , where G is a simple group. Since G includes SU(2) L × SU(2) R , its rank must be more than one. The Higgs fields originate from the extra-dimensional components of the G gauge field. In contrast to 5D models [8,9,10], we have at least two Higgs doublets.
Thus their VEVs need to be aligned as (4.17) to preserve the custodial symmetry. This severely constrains the structure of models.
In order to select candidates for viable models, we demanded the following requirements.
• The model has a scalar bidoublet zero-mode as the Higgs fields.
• The bosonic sector has a symmetry under a parity P LR that exchanges SU(2) L and SU(2) R in order to protect the Zb LbL coupling against a large deviation induced by mixing with the KK modes.
• The quark fields are embedded into 6D fermions in such a way that they couple to the Higgs bidoublet H only through a combination H + σ 2 H * σ 2 .
• The representation R that the 6D fermions belong to provides a right size group factor to realize the top Yukawa coupling constant. 9 Although the quantum correction to the Higgs quartic couplings is divergent, the ratio m H /m W is proven to be finite and calculable in Ref. [28]. (4) SO (7) Sp (6) (I) (II), (III) (I) (II), (III)

SO(5) G 2 SU
There is only one candidate that satisfies the above requirements if we restrict ourselves to the cases that rank G ≤ 3 and dim R < 30. It is the case of G =SU(4), N = 3 and R = 20 ′ . Namely, the model is 6D SU(3) C × U(4) gauge theory compactified on T 2 /Z 3 , and the top and bottom quarks are embedded into the symmetric traceless rank-2 tensor of SO(6). 10 We have focused on the third generation quarks to restrict G, N and R. Embeddings of other fermions are much less constrained.
There are many issues that we have not discussed in this paper. We have approximated the mode functions of the W and Z bosons as constants. However, after the electroweak symmetry is broken, they are no longer constants and depend on z. This z-dependence causes the deviation of the ρ parameter and the Zb LbL coupling from the standard model values. We have to check that the custodial symmetry actually suppresses these deviations by using the exact mode functions. We should also calculate the one-loop effective potential to check that the vacuum alignment (4.17) is actually achieved, and to evaluate the Higgs mass spectrum. The moduli stabilization in the gauge-Higgs unification is also an important subject [12,29]. It is interesting to investigate this subject in the presence of a background magnetic flux on T 2 /Z N . All these issues are left for our future works.

Acknowledgements
This work was supported in part by Grant-in-Aid for Scientific Research (C) No.25400283 from Japan Society for the Promotion of Science (Y.S.).

A Cartan-Weyl basis
The generators of a simple group G whose rank is r in the Cartan-Weyl basis are H i (i = 1, · · · , r) and E α , which satisfy where α, β are the root vectors, and α = −β. A complex constant N α,β is nonzero only when α + β is a root, and satisfies the following equations.

B Orbifold boundary conditions
The orbifold T 2 /Z N is defined by identifying points of R 2 by a discrete group Γ which is generated by three descrete transformations O 1 : z → z + 1, O τ : z → z + τ and O ω : z → ωz.
Field values of a 6D field at Γ-equivalent points must be related to each other through gauge transformations 11 in order for the Lagrangian to be single-valued on T 2 /Z N . Thus the most general orbifold boundary conditions are given by A M (x, z + 1) = T 1 A M (x, z)T −1 1 , B Z M (x, z + 1) = B Z M (x, z), Ψ χ 6 (x, z + 1) = e iϕ 1 T 1 Ψ χ 6 (x, z), (B.1) for the translation O 1 , for the translation O τ , and for the Z N twist O ω . Matrices T 1 , T τ and P are elements of G, and ϕ 1 and ϕ τ are the Scherk-Schwarz phases. A factor ω −1 and ω − χ 4 χ 6 2 in (B.3) appears because A z , B Z z and Ψ χ 4 ,χ 6 are charged under the rotation in the extra-dimensional space. Since (ω − χ 4 χ 6 2 ) N = −1, the phase ϕ ω is determined so that e iN ϕω P N = −1. (B.4) The matrices T 1 , T τ and P satisfy the relations, [T 1 , T τ ] = 0, P N = 1, Here we perform a gauge transformation, where U(z) ≡ exp − Im (τz) Im τ ln T 1 − Im z Im τ ln T τ , (B.7) Using (B.5), we can show that U(z + 1) = U(z)T −1 1 , U(z + τ ) = U(z)T −1 τ , P −1 U(ωz)P = U(z), P −1 iU∂ z U −1 P = ω −1 iU∂ z U −1 . where α i are the simple roots, and l i are numbers such that i l i α i are positive roots. We focus on irreducible representations whose dimensions are less than 30 in the following. 12