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Natural Doublet-Triplet Splitting in SU(N)X U(1)

It is shown that natural doublet-triplet splitting can be achieved in a relatively simple way in supersymmetric SU(N)X U(1) models with N>5.

Much more attention has been paid to SO(10) unification than to unification based on SU(N) with N > 5. Part of the appeal of SO(10) is that the observed quarks and leptons fit perfectly into spinor multiplets. As pointed out long ago, however, there is also a simple explanation of the observed fermion spectrum in the context of large unitary groups. Any anomaly-free set of antisymmetric tensor multiplets of fermions in SU(N), with N > 5, will decompose under the Standard Model subgroup into families with the observed quantum numbers [1]. Interestingly, in such SU(N) models, the different families can come from tensor multiplets of different size, unlike the situation in SU (5) and SO (10), where the families arise from having three copies of the same multiplets (3 × (10 + 5) or 3 × (16), respectively). This feature may be relevant to explaining the observed mass hierarchy among the families [2].
One problem with SU(N) unification, however, is achieving doublettriplet splitting [3] in a way that is both technically natural and economical. In SU (5), the only technically natural doublet-triplet splitting is by the missing partner mechanism [4], but it requires Higgs fields in rather large tensor multiplets, specifically H [αβγ] [δǫ] = 50 and H [αβ] [δǫ] = 75. This can be generalized to SU(N), N > 5, but has the same drawback there. The "sliding singlet" mechanism can be made to work in SU(N) models with N > 5 [5], but it requires the Higgs superpotential to have a very special form [6]. There is also the so-called GIFT mechanism [7], which can be implemented in models with unitary groups larger than SU (5).
As is well known, the missing partner mechanism can be implemented in an extremely economical way in supersymmetric "flipped SU(5)" [8] (the group of which is really SU(5) × U(1)). In this paper, we show that this can be generalized to supersymmetric SU(N) × U(1).
The possibility of a "flipped" breaking of SU(5) × U(1) to the Standard Model in addition to the "Georgi-Glashow" breaking can be understood group-theoretically as arising from the fact that SU(5) × U(1) is embeddable in SO(10) in two distinct ways. Even if the SU(5) × U(1) gauge group is not actually embedded in an SO(10) gauge group, the flipped breaking requires that the U(1) charges be "SO(10)-like", i.e. that the fermion multiplets 10, 5, and 1 have U(1) charges 1, -3, and 5, just as though they came from a spinor of SO(10), and similarly for the charges of Higgs multiplets.
In a realistic SU(N) × U(1) model, with N > 5, however, the fermion multiplets are not embeddable in an SO(10) (nor, in most cases, in any larger orthogonal group). Nevertheless, in many simple SU(N) × U(1) models it happens that the U(1) charges of the quark/lepton multiplets are forced to be "SO(10)-like" by anomaly cancellation. That is enough to allow for flipped breaking of SU(N) × U(1) down to the Standard Model. With such breaking, as will be seen, an economical implementation of the missing partner mechanism is possible. Unlike the case of SU(5) × U(1), however, it seems to require that some of the known quarks and leptons obtain mass from higher-dimension effective Yukawa operators. We shall briefly review how the missing partner mechanism works in flipped SU(5)×U(1). Then we shall show explicitly how it can be generalized to SU(6) × U(1) and SU(7) × U(1) models. The further generalization to larger unitary groups is relatively straightforward.
In the simplest supersymmetric flipped SU(5) model, the quarks and leptons of a family are in 10 1 + 5 −3 + 1 5 of SU(5) × U(1) X and the Higgs multiplets include 10 1 H +10  Table I represent the fact that these multiplets are eaten by the Higgs mechanism when H 12 and H 12 obtain vacuum expectation values (VEVs). One sees that the doublet fields H i and H i do not acquire mass as they have no weak-doublet, color-singlet "partners" in the 10 H +10 H . These are the doublets of the MSSM and couple to the quark and lepton multiplets through the usual terms (10 1 10 1 ) Let us now see how this mechanism can be generalized to the gauge groups SU(6) × U(1). The details which we now present are more involved than in flipped SU (5), simply because the group has higher rank and there is another step of symmetry breaking required, and also because the multiplets of SU(6) contain more Standard Model fields, which have to be kept track of. This will be even more the case when we turn to SU(7) × U(1). But the essence of what is going on is the same as in flipped SU (5), and is basically simple: the missing-partner mechanism of flipped SU (5) is being embedded in larger unitary groups. The matter (quark and lepton) content in each case is essentially minimal, and the structure of the Higgs superpotentials (in Eqs. (2) and (4)) is simple. We now go through the details carefully to demonstrate that, indeed, the mechanism can be implented in a realistic way and no obstruction arises for these larger groups.
A family in SU (6) consists of ψ AB +ψ A +ψ A = 15+6+6. (Capital letters A, B = 1, 2, ..., 6 denote SU (6) indices.) Up to an overall normalization, there is only one way to assign U(1) charges to the quarks and leptons in a family-independent way that is consistent with anomaly cancellation, namely 15 0 + 6 1 + 6 −1 . (The contribution to the anomalies of the Higgsinos would vanish, if the Higgs multiplets are in conjugate pairs under the gauge group as would generally be required to avoid D-term breaking of supersymmetry.) The U (1)  H and the corresponding vacuum expectation values in the conjugate Higgs multiplets. (We use the superscripts (±1) to distinguish the two anti-fundamental Higgs multiplets based on their U(1) charge.) The first of these VEVs breaks SU (6) (1, 1, 1, 1, 1, −5) and T 1 is the generator of the original U(1) as normalized above, then the generator of the unbroken U(1) X is given by X = 1 2 (T 6 + 5T 1 ). This is clear, since H (−1) 6 has T 6 = 5 and T 1 = −1. The field H 12 ∈ 15 0 H then does the usual "flipped" breaking of SU(5) × U(1) X down to the Standard Model group. One can see that it has the correct SU(5) × (1) X charges to do this. It has T 6 = 2, T 1 = 0, and thus X = 1. It also has Y 5 /2 = 1, where Y 5 /2 is the SU (6) generator , which is also a generator of the SU(5) subgroup. Therefore, its VEV breaks SU (5) . The decomposition of the quarks-lepton and Higgs multiplets is given in Table II.  Note that the X charges are exactly those that would result if SU(5)×U(1) X were embedded in an SO(10), even though obviously SU(6) × U(1) is not a subgroup of SO(10). This can be understood in two ways: (1) in terms of the structure of E 6 , or (2) as the result of anomaly cancellation. The argument based on E 6 is as follows. E 6 contains the subgroups E 6 ⊃ SU(6) × SU(2) ⊃ SU(6) × U(1) under which 27 −→ (15, 1) + (6, 2) −→ 15 0 + 6 1 + 6 −1 . Note that these SU(6) ×U(1) charges assignments are just those assumed in Table  II. But also E 6 contains the subgroups E 6 ⊃ SO(10) ⊃ SU(5) × U(1) X . The fact that the flipped SU(5) × U(1) X is a subgroup of the SU(6) × U(1) in Table II can therefore be understood as a consequence of the group theory of E 6 . This explanation is special to SU(6) × U(1), however, and does not generalize to SU(N) × U(1) with N > 6.
The anomaly cancellation argument is more general. The point is that if SU(N) × U(1) is broken to SU(5) × U(1) X and the light families end up as 10 + 5 + 1 then X has to have the "SO(10)-like" assignments 10 1 + 5 −3 + 1 5 because of anomaly freedom. (This is so, even if each light family contains several extra SU(5) singlets.) This is the crucial point that makes it possible to have "flipped" breaking and to implement the missing partner mechanism in an economical way in SU(N) × U(1) models, even if they have particle content that cannot be understood in terms of orthogonal or exceptional groups.
In the SU(6) × U(1) model, the "flipped" breaking of the SU(5) × U(1) X subgroup down to the Standard Model group is done (as in ordinary flipped SU (5)) by H 12 and H 12 . As shown in Table II, these are contained in the 15 0 + 15 0 of SU(6) × U(1). The doublet-triplet splitting can be achieved through a Higgs superpotential that contains terms of the following form: where Z and Z ′ are gauge singlets that are integrated out to give nonzero superlarge VEVs to the components that break SU (6) (2), since such terms can be eliminated simply by shifting the singlet superfields Z and Z ′ . (Or one could forbid such terms with a discrete symmetry.) The color-triplet Higgs(ino) fields in 15 0 H and 15 0 H obtain mass from the first terms in Eq. (2), which contain ǫ 12abc6 H 12 H bc H a6 +ǫ 12abc6 H 12 H bc H a6 . The doublet-triplet splitting works essentially as shown in Eq. (1). There are altogether three triplet Higgs multiplets with Standard Model quantum numbers (3, 1, − 1 3 ), which are H a6 ∈ 15 0 H , H bc ∈ 15 0 H , and H (−1)a ∈ 6 −1 H ; and there are three anti-triplet Higgs in (3, 1, + 1 3 ) that are conjugate to these. The mass matrix of these colored Higgs(ino) fields with Y /2 = ± 1 3 is (3) All these triplets acquire superlarge mass. There are also a pair of triplets with Standard Model quantum numbers (3, 1, 2 ) that are conjugate representations to these. The mass matrix of these doublet Higgs is Two pairs of doublet Higgs fields are left massless by the mass matrix in Eq.  (2), which was discussed above. That is to say, it is due to the fact that Z = 0 and Z ′ = 0.) Of these two massless pairs of doublets, one pair consists of goldstone fields that get eaten by the Higgs mechanism, and the other pair are the two doublet Higgs multiplets of the MSSM. The MSSM doublet with (1,  0 H that are eaten by the Higgs mechanism. The absence of uneaten goldstone Higgs fields implies that this minimum is not continuously connected to other degenerate minima, which is a condition for satisfactory breaking of SU(6) × U(1) to the Standard Model.
Although we arrived at this set-up by generalizing flipped SU(5) × U(1) to SU(6) × U(1), it may not be obvious from examining the mass matrices in Eqs. (3) and (4) that the doublet-triplet splitting is being accomplished here by the missing partner mechanism. It can be seen more readily if we note that the 15 0 H contains both a (3, 1, − 1 3 ) (namely H a6 ) and a (3, 1, + 1 3 ) (namely ǫ abc H bc ), and these are "partners" of each other in a mass term of the form ǫ 12abc6 H a6 H bc H 12 . On the other hand, in the 15 0 H there is a (1, 2, − 1 2 ) (namely H i6 ), but no (1, 2, + 1 2 ). That is, the "partner" of the doublet is "missing". The same is true for the 15 are not connected to each other by a mass term (because Z = 0), but that does not matter: the triplets in them are eaten and the extra pair of doublets in them is needed, since a pair of doublets must be eaten by the Higgs mechanism when SU(6) breaks to SU (5).
Turning now to the spectrum of quarks and leptons, the extra vectorlike quarks and leptons in 5 −2 + 5 2 (called "heavy" in Table I)  Note that unlike the usual flipped SU(5) models, the mass matrix of the can be absorbed by a shift in the singlet superfields Z and Z ′ . Note that there has to be more than one pair of Higgs fields of type 7 0 H + 7 The vanishing of the lower-right 2-by-2 block is due to the absence of a mass term of the form M(7 0 H 7 0 H ) (or, equivalently, the vanishing of the VEV of Z ′′ ). There is one zero eigenvalue of this matrix, corresponding to a pair of