$SU(n)$ symmetry breaking by rank three and rank two antisymmetric tensor scalars

We study $SU(n)$ symmetry breaking by rank three and rank two antisymmetric tensor fields. Using tensor analysis, we derive branching rules for the adjoint and antisymmetric tensor representations, and explain why for general $SU(n)$ one finds the same $U(1)$ generator mismatch that we noted earlier in special cases. We then compute the masses of the various scalar fields in the branching expansion, in terms of parameters of the general renormalizable potential for the antisymmetric tensor fields.


I. INTRODUCTION
The most familiar case of symmetry breaking for grand unified theories, such as minimal SU (5) ⊃ SU (2) × SU (3) × U (1), utilizes a scalar field in the adjoint representation, with a gauge singlet component with U (1) generator zero that receives a vacuum expectation.The symmetry breaking mechanism is then straightforward: since the gauge fields and the symmetry breaking scalar are both in the adjoint representation, the same representations appear in their branching expansions.As a consequence, the massless gauge fields that pick up masses, and the scalars that supply their longitudinal components, have the same group theoretic quantum numbers.
We recently noted [1,2] that when the symmetry breaking scalar is in a totally antisymmetric representation, the situation is more complicated.Using as explicit examples SU (8) broken by a rank three antisymmetric tensor scalar, and SU (5) broken by a rank two antisymmetric tensor scalar, we showed that there is a mismatch between the U (1) generator values of the massless gauge fields that obtain masses, and the scalars that supply their longitudinal components.We noted that this mismatch is related to the fact that the gauge singlet component of the antisymmetric tensor field that receives a vacuum expectation has a nonzero U (1) generator N , requiring a modular ground state that is periodic in integer divisors p of N .
The purpose of this paper is twofold.First, we show that the mismatch found in [1,2] appears in the case of general SU (n), and can be traced to the fact that invariant tensors lying in SU (3) * Electronic address: adler@ias.eduor SU (2) subgroups are available to lower subgroup indices.This analysis is given in Sec. 2, where we use tensor methods to compute the relevant branching expansions and U (1) generator values.
The second aim of this paper is to calculate the masses of the various scalar field components in the branching expansions, obtained by expanding the general renormalizable scalar field potential around the generic symmetry breaking minimum.This analysis is given in Sec. 3, and a brief summary of our results follows in Sec. 4.
Our notation is to define the upper index totally antisymmetric tensor with R components to be a basis for the representation R, and the corresponding lower index tensor to be a basis for the conjugate representation R. Thus in SU (n) the tensor φ α , α = 1, ..., n is a basis for the fundamental representation n, and φ α is a basis for the conjugate representation n.In SU (3), the tensor φ α is a basis for the 3, and since the totally antisymmetric tensor ǫ αβγ is invariant and can be used to lower indices, both the tensors φ α and φ [αβ] give a basis for the 3, and the tensor φ [αβγ] ∝ ǫ αβγ is a singlet.Similarly, in SU (2), since the invariant tensor ǫ αβ can be used to lower indices the representations 2 and 2 are equivalent, and can be represented by either φ α or φ α , and the tensor φ [αβ] ∝ ǫ αβ is a singlet [3].We assume that SU (n) is broken by the ground state expectation of a single component φ [123] = a = 0, corresponding to the simplest case considered by Cummins and King [4], which applies for all n.The conditions on the scalar potential for this case to apply will be given in Sec.
To get the needed branching expansions, we have to enumerate the possibilities for tensor indices to belong to these two classes.We use the notation R SU (3) , R SU (n−3) (g), with g the U (1) generator eigenvalue.Writing the U (1) generator G as We begin by deriving the branching expansion for the SU (n) rank three antisymmetric tensor representation n(n − 1)(n − 2)/6, represented by the tensor φ [αβγ] , enumerating cases as follows.

2 indices in
As a check on the counting, we note the identity For the case n = 8 discussed in [1], the SU (5) three upper index antisymmetric tensor is equivalent, by use of the invariant tensor ǫ αβγδǫ , to the SU (5) two lower index antisymmetric tensor, and so represents a 10 rather than a 10.Thus we get the expansion This agrees with the expansion given in [1] and the Slansky tables [5], apart from the fact that in this paper we have chosen the opposite sign convention for the U (1) generator G.For n < 8, one makes similar conversions of upper index tensors to lower index ones in Eq. ( 3) , when the number of lower indices can be made smaller than the number of upper indices, with the corresponding replacement of the representation R by R. We also note that when n − 3 is divisible by 3, the U ( generator values can all be divided by 3, and this is the convention that is used in the Slansky tables see e.g. the expansion for the 20 of SU (6) .
We turn next to the branching expansion for the SU (n) adjoint representation n 2 −1, represented by the tensor φ α β , with α φ α α = 0, again enumerating cases.
2. upper index and lower index both in A, traceless part.This corresponds to the representation 8, 1 (0).
3. upper index and lower index both in B, traceless part.This corresponds to the representation 1, (n − 3) 2 − 1 (0).Thus we have the branching expansion

upper index in
As a check on the counting, we note the identity We now note the phenomenon discussed in the n = 8 case in [1,2], that the U (1) generator of the (3, n − 3) is −n in the branching expansion for the adjoint, whereas it is 2n − 9 in the branching expansion for the rank three antisymmetric tensor.The difference between these two , which is just the U (1) generator of the singlet (1, 1) in the expansion of Eq. ( 3).This is a direct result of the fact that the 3 is represented by a two upper index antisymmetric tensor in the expansion of Eq. ( 3), and by a one lower index tensor in the expansion of Eq. ( 6), so the difference in U (1) generator values is 2 − (−1) (n − 3) = 3(n − 3).
When we discuss the scalar potential in Sec. 3, we will see that the complex states (3, n − 3) in Eq. ( 3) are zero mass Goldstone modes.When the rank three antisymmetric tensor is used to break the SU (n) symmetry, the Goldstone modes are absorbed as longitudinal parts of the In this case we shall assume that SU (n) is broken by the ground state expectation of a single component φ [12] = a = 0, corresponding to the case studied by Li [6].We now define the index classes by Since the enumeration of cases parallels that in the rank three case, we go directly to the results.
For the rank two antisymmetric tensor, we have for n > 5 with the three terms corresponding, respectively, to zero, one, and two upper indices in B. As a check on the counting, we note the identity For the case n = 5, since the SU (3) two upper index antisymmetric tensor represents a 3, we get the expansion 10 = (1, 1)( 6) + (2, 3)(1) + (1, 3)(−4) , in agreement with the expansion given in the Slansky tables [5].When n − 2 is divisible by 2, the U (1) generator values can all be divided by 2, and this is the convention used in the Slansky tables see, e.g., the expansion for the 15 of SU (6).
For the adjoint representation n 2 − 1 of SU (n), we get the branching expansion and as a check on counting We again see the mismatch discussed in [2] in the n=5 case.This results from the fact that the 2 is represented by a one upper index tensor in the expansion of Eq. ( 10), and by a one lower index tensor in the expansion of Eq. ( 13), with a resulting difference of U (1) generator values 1 − (−1) (n − 2) = 2(n − 2).When we discuss the scalar potential in Sec.
3, we will see that the complex states (2, n − 2) in Eq. ( 10) are zero mass Goldstone modes.When the rank two antisymmetric tensor is used to break the SU (n) symmetry, the Goldstone modes are absorbed as longitudinal parts of the (2, n − 2) + (2, n − 2) in the adjoint.This is possible, despite the U (1) generator mismatch, because for the (1, 1)(2n−4) to get a ground state expectation value, the ground state must have a periodic structure modulo an integer divisor of 2n − 4, and so the mismatch of the U (1) generator values is equivalent to zero.

III. RESIDUAL SCALAR MASSES
In this section we analyze the residual scalar masses arising from SU (n) symmetry breaking with a general renormalizable scalar potential, first for a rank three antisymmetric tensor scalar, and then for a rank two antisymmetric tensor.
A. Residual scalar masses for SU (n) symmetry breaking by a rank three antisymmetric tensor The most general SU (n) invariant fourth degree potential formed from φ [αβγ] , where the indices all range from 1 to n, has the form [4] V We assume µ 2 > 0, so that the origin is a local maximum, and consider the case λ 2 < 0 studied in [4], for which the potential is bounded from below, for all n, when 3λ This lower bound is attained when only one component of φ is nonzero, and as in our branching analysis we take the nonvanishing component to be φ [123] = a = 0, where We will derive Eqs. ( 16) and ( 17) shortly.
Continuing to follow [4], we note that the potential of Eq. ( 15) can be rewritten in terms of which obeys (θ τ γ ) * = θ γ τ , as To expand the potential around its minimum, we substitute where φ [αβγ] = aǫ αβγ is nonzero only when its tensor indices are some permutation of 1, 2, 3.For θ τ γ we find We consider first the term γ<τ θ τ γ θ γ τ in Eq. ( 19).The term in θ τ γ that is quadratic in φ must have γ = τ , and so does not contribute to this sum over γ < τ .Hence the term in θ τ γ that is quadratic in σ makes a contribution to this sum that is third order in σ, and can be dropped in calculating the potential to second order in σ.Thus we get where in the final line we have used the fact that when α, β, τ are permutations of 1, 2, 3, then when τ = 1 there is no γ obeying γ < τ , and when τ = 2 or τ = 3, the γ obeying γ < τ must equal either α or β, and so the factor σ * Since σ [αβτ ] in this equation has α ∈ A, β ∈ A, and τ ∈ B, it belongs to the representation (3, n − 3), and so we can rewrite Eq. ( 23) as The remaining terms in Eq. ( 19) all involve the diagonal element θ γ γ , which from Eq. ( 21) is given by From this, substituting φ [αβγ] = aǫ αβγ , splitting sums on γ into disjoint sums γ∈A and γ∈B , and dropping terms of higher order than quadratic in σ, one finds Substituting Eqs. ( 24) and (26) into Eq.( 19), and combining the first order terms in σ, we get which when equated to zero gives Eq. ( 17).Using this value of |a| 2 , we find the lower bound of Eq. ( 16) for the value of the potential at the minimum.Splitting the sum αβ γ∈B |σ [αβγ] | 2 into three pieces, and relabeling σ [αβγ] in terms of the representations appearing in the branching expansion of Eq.
(3), we get as the final result for the expansion of the potential near the minimum through second order terms, The remarks made in Sec. 2 about using the rank n − 3 epsilon tensor to replace upper index tensors by lower index tensors in conjugate representations, when this reduces the number of indices, applies here.We see that as noted in Sec. 2, the Goldstone modes, with mass 0, are in the representation (3, n − 3), which has a U (1) generator mismatch with respect to the corresponding representation in the expansion of the adjoint representation.
B. Residual scalar masses for SU (n) symmetry breaking by a rank two antisymmetric tensor The most general SU (n) invariant fourth degree potential formed from the rank two antisymmetric tensor scalar φ [αβ] , where the indices all range from 1 to n, has the form [6] for n > 4,1 Since the method of analysis parallels that used in the rank three case, we state only the final results.We assume that λ 2 < 0 and 2λ 1 + λ 2 > 0, and as in our branching analysis we take the nonvanishing component of φ [αβ] to be φ [12] = a = 0.The potential minimum is at and the value of the potential at the minimum is For the expansion of the potential through second order terms, we find ( The remarks made in Sec. 2 about using the rank n − 2 epsilon tensor to replace upper index tensors by lower index tensors in conjugate representations, when this reduces the number of II. BRANCHING RULES FOR THE SU (n) ANTISYMMETRIC TENSOR AND ADJOINT REPRESENTATIONS A. Branching under SU (n) ⊃ SU (3) × SU (n − 3) × U (1) for the rank three antisymmetric tensor and adjoint representations 3. Let us now divide the tensor indices into two classes,A ={1, 2, 3} , B ={4, ..., n} .

) with n − 3 entries − 3 ,
the U (1) generator value g is simply n − 3 times the number of upper indices in A plus −3 times the number of upper indices in B; for lower indices the U (1) contributions are reversed in sign.Since the overall normalization of the U (1) generator is arbitrary, normalization-independent statements refer only to relative values of the U (1) generators for different representations.
A, lower index in B. This corresponds to the representation 3, n − 3 (n).5. lower index in A, upper index in B. This corresponds to the representation 3, n − 3 (−n).

( 3 ,
n − 3) + (3, n − 3) in the adjoint.This is possible, even though the U (1) generators do not match, because for the (1, 1)(3n − 9) to get a ground state expectation value, the ground state must have a periodic structure modulo an integer divisor of 3n − 9, and so the mismatch of the U (1) generator values is equivalent to zero.B. Branching under SU (n) ⊃ SU (2) × SU (n − 2) × U (1) for the rank two antisymmetric tensor and adjoint representations and use the notation (R SU (2) , R SU (n−2) )(g), with g the U (1) generator.Writing the U (1) generator G as G = Diag(n − 2, n − 2, −2, −2, ..., −2) (9) with n − 2 entries −2, the U (1) generator value g is simply n − 2 times the number of upper indices in A plus −2 times the number of upper indices in B; for lower indices the U (1) contributions are reversed in sign.Again, since the overall normalization of the U (1) generator is arbitrary, normalization-independent statements refer only to relative values of the U (1) generators for different representations.
The U (1) generator of the (2, n − 2) is −n in the branching expansion of the adjoint, whereas it is n − 4 in the branching expansion for the rank two antisymmetric tensor.The difference between these two U (1) values is n − 4 − (−n) = 2n − 4 = 2(n − 2), which is the U (1) generator of the singlet (1, 1) in the expansion of Eq. (10).