Generalized 331 models

The standard model (SM) cannot predict the number of fermion generations. If one expands the gauge group to ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$[[1, 2, 3, 4, 5]], then the number of generations come out to be 3, or any multiple of 3. In addition to this feature for which acted as a prime motivation for constructing these models, the models can provide an automatic solution to the strong CP problem [[6]] and an interesting explanation of electric charge quantization [[7, 8]]. Because of these attractive features, phenomenological implications of these models have been extensively studied [[9, 10, 11, 12, 13, 14, 15, 16]]. In addition, people have also examined the Higgs potential [[17, 18]], possible candidates for cosmological dark matter [[19, 20]] and the possibilities of embedding of these models [[21, 22, 23]] into some grand unified model.

For the sake of convenience, one often refers to the gauge group as the 331 group, as has been done in the title of this article. There are different versions of 331 models, or really, different models with the 331 gauge group. Among the two early versions in which the number of generations is predicted, right-handed neutrinos are absent in one version [[3, 4]] and present in another [[1, 2, 5]]. The common feature is the gauge group, the prediction of the number of generations, and the fact that new fermions must be introduced in order to complete the multiplets as well as to cancel gauge anomalies. The difference lies in the properties of these new fermions. In addition, it was shown later [[23]] that if one chooses, one can also construct models based on the same gauge group in which each generation of fermions is independent in the sense that all anomalies cancel within a single generation, and different generations are mere copies of one another. Of course, in this model the prediction of the number of generations is lost, but the model is nonetheless viable.

The aim of this article is to examine whether other variants of the model are possible. We show that, for the case of non-identical generations, the proposed models are specific examples of a class of models characterized by a free parameter. For the case of identical fermion generations, however, the proposed model is unique under some restrictive assumptions. However, those assumptions are by no ways a necessity, so they can be spared and more general models can be built. We show that there are other kinds of models possible, which do not resemble the earlier models in any way.

So far as the color symmetry is concerned, this group is no different from SM. The quarks will be in color triplets and the leptons will be in color singlets. Since there will be new fields in these models, let us say that anything that transforms like a triplet of ${\rm SU(3)}_{c}$ will be called quark, and any singlet of the color group will be called lepton.

The ${\rm SU(3)}_{L}$ part of the group requires obvious thoughts. We assume that all fermions transform either like the singlet or in the multiplet whose dimension equals that of the fundamental representation, as is the case in the SM. This means that the representations under ${\rm SU(3)}_{L}$ are $\mathit{1}$, $\mathit{3}$ and $\mathit{3^{*}}$. We take all right-chiral fields in ${\rm SU(3)}_{L}$ singlets. This is no loss of generality, because if we encounter any right-chiral field that is not a singlet of ${\rm SU(3)}_{L}$, we can disregard it and instead tabulate its complex conjugate which would then be a left-chiral field. With these ground rules, then, all possible multiplets would fall under one of the categories listed in Table 1. It should be commented that restriction to singlets and triplets does not follow from any fundamental principle. It is just an assumption. There are models where other representations are used [[24]], but we will not discuss them here.

We now write down the constraints coming from gauge anomaly cancellation. While reading the ensuing equations, it has to be remembered that the range of the sum on $k$ is in general different for different lettered variables, as can be seen from Table 1.

	$\displaystyle[{\rm SU(3)}_{c}]^{3}~{}:$	$\displaystyle 3(n_{3}+n_{4})-n_{6}$	$\displaystyle=0\,,$		(1a)
	$\displaystyle[{\rm SU(3)}_{L}]^{3}~{}:$	$\displaystyle n_{1}-n_{2}+3(n_{3}-n_{4})$	$\displaystyle=0\,,$		(1b)
	$\displaystyle[{\rm SU(3)}_{c}]^{2}\times[{\rm U(1)}_{X}]~{}:$	$\displaystyle\sum_{k}\Big{(}3\gamma_{k}+3\delta_{k}-\zeta_{k}\Big{)}$	$\displaystyle=0\,,$		(1c)
	$\displaystyle[{\rm SU(3)}_{L}]^{2}\times[{\rm U(1)}_{X}]~{}:$	$\displaystyle\sum_{k}\Big{(}\alpha_{k}+\beta_{k}+3\gamma_{k}+3\delta_{k}\Big{)}$	$\displaystyle=0\,,$		(1d)
	$\displaystyle[{\rm U(1)}_{X}]^{3}~{}:$	$\displaystyle\sum_{k}\Big{(}3\alpha_{k}^{3}+3\beta_{k}^{3}+9\gamma_{k}^{3}+9\delta_{k}^{3}-\xi_{k}^{3}-3\zeta_{k}^{3}\Big{)}$	$\displaystyle=0\,.$		(1e)
In addition, gravitational anomaly cancellation would require the condition
	$\displaystyle\sum_{k}\Big{(}3\alpha_{k}+3\beta_{k}+9\gamma_{k}+9\delta_{k}-\xi_{k}-3\zeta_{k}\Big{)}$	$\displaystyle=0\,.$			(1f)

In the context of the SM gauge group, it was shown [[25]] that the anomaly cancellation equations, aided by some reasonable assumptions like the presence of a mass of the charged fermions, are enough to determine the charges of the fermion fields. This will not be the case in the context of the 331 models. There are too many parameters, and one will need so many assumptions to reach the goal that finally it would seem that the goals were pre-determined and the assumptions were tailored to obtain them. This statement will be supported by the discussions in the rest of this article, where we will see that there is enough freedom in the construction which allows many models to be constructed.

Note that the $[{\rm SU(3)}_{c}]^{3}$ anomaly cancellation condition, Eq. (1a), demands that the number of left-chiral triplets of color must equal the number of right-chiral triplets. That requirement is automatically satisfied if we assume that all quarks are massive. Similarly, Eq. (1b) implies that the number of triplets and antitriplets of ${\rm SU(3)}_{L}$ must be equal.

Many of the other equations become trivial if we impose the requirement that all charged particles must be massive. In the mass term, a left-chiral fermion field pairs with a right-chiral field. Therefore, this requirement can be summarized by saying that all non-trivial representations of the final unbroken group of ${\rm SU(3)}_{c}\times{\rm U(1)}_{Q}$ should be vectorlike [[25]], the latter factor being the gauge group of quantum electrodynamics. Because the electric charge generator is a linear combination of ${\rm SU(3)}_{L}$ generators and the ${\rm U(1)}_{X}$ generator, Eqs. (1a) and (1c) together imply that the $[{\rm SU(3)}_{c}]^{2}\times{\rm U(1)}_{Q}$ anomalies must also cancel. This will produce a condition that is similar to Eq. (1c), except that there will be sums over the electric charges instead of the $X$ quantum numbers. But that equation is automatically obeyed, because for each left-chiral quark field there must be a right-chiral quark with the same charge. For the same reason, the quark terms in Eqs. (1e) and (1f) must also vanish among themselves. The same argument applies for the lepton fields. If there is a left-chiral field with a certain non-zero charge, either there would be a type-5 right-chiral field of the same charge present in the model, or a left-chiral field of opposite charge must be present in the type-1 fields. To summarize, the anomaly conditions of Eqs. (1e) and (1f) need not be considered separately if we arrange the right-chiral fields in such a way that all left-chiral fields of nonzero charge can obtain masses.

Eqs. (1b) and (1d) are then the important equations in the context of model building. Note that all $n$’s must be non-negative integers by definition. We can always take

$\displaystyle n_{1}\geq n_{2}$

(2)

by adjusting our definition of the $\mathit{3}$ and $\mathit{3^{*}}$ representations of ${\rm SU(3)}_{L}$. Eq. (1b) then implies that

$\displaystyle n_{4}\geq n_{3}\,.$

(3)

We will also impose

$\displaystyle n_{1}\geq 1,\qquad n_{4}\geq 1\,,$

(4)

because otherwise the model would lack either leptons, or quarks, or both. Next, we note that Eq. (1b) shows that $n_{1}-n_{2}$ must be a multiple of 3. We can explore how this equation can be satisfied, taking small values of all $n$’s. Here are some examples. Corresponding to each pair of values of $n_{3}$ and $n_{4}$, we choose the minimum values of $n_{1}$ and $n_{2}$, subject to the conditions of Eqs. (2) and (4), that will satisfy Eq. (1b).

$\displaystyle\begin{array}[]{ccccc}\mbox{Choice}&n_{3}&n_{4}&n_{1}&n_{2}\\ \hline\cr{\rm A}&0&1&3&0\\ {\rm B}&1&1&1&1\\ {\rm C}&1&2&3&0\end{array}$

(9)

We will see that Choice 9 will give us the sequential model [[23]], whereas Choice 9 will give us models in which the generations are non-sequential, and one needs all three generations for all gauge anomalies to cancel [[1, 2, 3, 4, 5]], which can be called the entangled models. An intermediate case, corresponding to Choice 9, provides a new kind of 331 model which has not been explored so far.

Note that, so far we have not used Eq. (1d) at all. It will prove useful when we will try to construct the models explicitly.

The SM lepton doublet must lie within one of the $(1,3,\alpha_{k})$ representations. We can choose the triplet, for the first generation of left-chiral lepton fields, to be

$\displaystyle\left(\begin{array}[]{c}\nu_{e}\\ e\\ f\end{array}\right)_{L}\,,$

(13)

where $f_{L}$ is a lepton field that has to be added to complete the triplet. It might be a field that is already present in the SM, or might be a new field — the choice is left open for now.

The electric charge generator must be of the form

$\displaystyle Q=T_{L}+X\,,$

(14)

where $T_{L}$ is a diagonal generator of ${\rm SU(3)}_{L}$, and $X$ is the generator of the ${\rm U(1)}_{X}$ part of the gauge group. If one uses the standard representations of SU(3) generators in which $T_{3}$ and $T_{8}$ are the diagonal generators, then $T_{L}$ is a linear combination of those two. For the $\mathit{3}$ representation of ${\rm SU(3)}_{L}$, this means that the charges of its components will be given by

$\displaystyle Q_{(\mathit{3})}=\mathop{\rm diag}\Big{(}c_{1}+x,c_{2}+x,c_{3}+x\Big{)}\,,$

(15)

where $x$ is the ${\rm U(1)}_{X}$ quantum number, and

$\displaystyle c_{1}+c_{2}+c_{3}=0$

(16)

because they come from the ${\rm SU(3)}_{L}$ generators which are traceless. We note that, since the representation of the generators in $\mathit{3}$ and $\mathit{3^{*}}$ are different, in the $\mathit{3^{*}}$ representation the electric charge will be given by

$\displaystyle Q_{(\mathit{3^{*}})}=\mathop{\rm diag}\Big{(}{-}c_{1}+x,-c_{2}+x,-c_{3}+x\Big{)}\,,$

(17)

with the same values of $c_{1}$, $c_{2}$ and $c_{3}$ that appear in Eq. (15).

If the $X$ quantum number is taken to be $\alpha_{1}$ for the multiplet shown in Eq. (13), the charges of the first two components would imply the relations

$\displaystyle c_{1}+\alpha_{1}=0\,,\qquad c_{2}+\alpha_{1}=-1\,.$

(18)

If we take the charge of the field $f$ to be $q$, Eq. (16) would imply the relation

$\displaystyle\alpha_{1}={q-1\over 3}\,.$

(19)

So now we can find $c_{1}$, $c_{2}$ and $c_{3}$ in terms of $q$ by using Eqs. (16) and (18), and thereby rewrite Eqs. (15) and (17) in the form

	$\displaystyle Q_{(\mathit{3})}$	$\displaystyle=$	$\displaystyle\mathop{\rm diag}\Big{(}{1-q\over 3}+x,\;-{2+q\over 3}+x,\;{1+2q\over 3}+x\Big{)}\,,$		(20a)
	$\displaystyle Q_{(\mathit{3^{*}})}$	$\displaystyle=$	$\displaystyle\mathop{\rm diag}\Big{(}{q-1\over 3}+x,\;{2+q\over 3}+x,\;-{1+2q\over 3}+x\Big{)}\,.$		(20b)

Let us now look at the quark sector. We see from Eq. (9) that there must be at least one $(3,3^{*})$ multiplet of ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$. Let us denote its $X$ quantum number by $\delta_{1}$. It must contain one of the usual quark doublets, and they must occur in the first two components of the ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$ representation. However, note that Eqs. (18) and (20b) tell us that, for a $\mathit{3^{*}}$ representation of ${\rm SU(3)}_{L}$, the first component has lower charge than the second one. Therefore, we should identify the first component as $d_{L}$ and the second one as $u_{L}$. These identifications imply

$\displaystyle{q-1\over 3}+\delta_{1}=-\frac{1}{3}\,,$

(21)

or

$\displaystyle\delta_{1}=-{q\over 3}\,.$

(22)

From this, we can find the charges of all three components of the multiplet. In particular, the third component will have a charge $-\frac{1}{3}-q$. Exactly similarly, we can argue that, for a $(3,3)$ representation of ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$, if the first two components are $u$-type and $d$-type quarks, the third component would have a charge $\frac{2}{3}+q$. We will now use these results to construct the full models. We will not arrange our construction in the order of the choices given in Eq. (9), Rather, we start with the possibility $n_{2}=0$, and will later go up to $n_{2}=1$.

To keep things as simple as possible, we will first assume, along with $n_{2}=0$ as mentioned, that all $\alpha_{k}$’s are equal to $\alpha_{1}$ found in Eq. (19), and similarly all $\gamma_{k}$’s and all $\delta_{k}$’s are also equal. We will denote this common value without any subscripted index. Then, Eq. (1d) will imply

$\displaystyle n_{1}\alpha+3n_{3}\gamma+3n_{4}\delta=0\,.$

(23)

The values of $\alpha$ and $\delta$ must be those given in Eqs. (19) and (22). Then,

$\displaystyle n_{3}\gamma=-\frac{1}{3}n_{1}\alpha-n_{4}\delta={(1-q)n_{1}+3n_{4}q\over 9}\,.$

(24)

This equation will be inconsistent if we take Choice 9 of Eq. (9), where $n_{3}=0$. Therefore, the next simplest solution is Choice 9, i.e., $n_{1}=3$, $n_{3}=1$ and $n_{4}=2$. This will mean that the three left-chiral quark doublets of SM appear in two different types of representations of the 331 group. The generations are not identical, and all anomalies also do not cancel within a single generation of fermions, which is why we call these models ‘entangled’.

Eq. (24) now gives

$\displaystyle\gamma={q+1\over 3}\,.$

(25)

Remarkably, this value coincides exactly with the value of the $X$ quantum number obtained from the requirement that a (3,3) representation of ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$ contains the usual quark doublet. Therefore we find that the generalized 331 model will have the following multiplet structure for the fermions:

$\displaystyle\begin{array}[]{c@{\hspace{7mm}}c@{\hspace{7mm}}c@{\hspace{7mm}}l}\hline\cr\mbox{${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$}\hfil\hskip 19.91692pt&\hbox{\multirowsetup\mbox{Chirality}}\hfil\hskip 19.91692pt&\mbox{Number of}\hfil\hskip 19.91692pt&\mbox{electric}\\ \mbox{representation}\hfil\hskip 19.91692pt&\hfil\hskip 19.91692pt&\mbox{copies}\hfil\hskip 19.91692pt&\mbox{charges}\\ \hline\cr\Big{(}1,3,{q-1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&3\hfil\hskip 19.91692pt&0,-1,q\\ \Big{(}3,3,{q+1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&\frac{2}{3},-\frac{1}{3},\frac{2}{3}+q\\ \Big{(}3,3^{*},-{q\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&2\hfil\hskip 19.91692pt&-\frac{1}{3},\frac{2}{3},-\frac{1}{3}-q\\ \lx@intercol\hfil\mbox{and R-chiral fields to match their charges.}\hfil\lx@intercol\\ \hline\cr\end{array}$

(32)

It can now be easily checked that all anomaly cancellation conditions of Eq. (1) are obeyed for any value of $q$ as long as the right chiral fields contain the same charges as the left chiral fields, as argued earlier in Sec. 2.

Clearly, there are two distinguished values of $q$ that need to be discussed. One is $q=+1$, and the other is $q=0$. For each of these values, there arises the possibility that the new field can pair with one of the two other fields of the same multiplet to form a mass term for the fermion. Interestingly, these values give the variants of the 331 models discussed in the literature: the $q=+1$ case gives the model of Pisano, Pleitez and Frampton [[3, 4]], whereas the model with $q=0$ was presented by Singer, Valle, Schechter, Foot, Long and Tran [[1, 2, 5]]. To the best of our knowledge, models with any other value of $q$ have not been studied or proposed.

In this case, we deal with the solution with $n_{2}=n_{3}=0$ that was presented as Choice 9 in Eq. (9). Since $n_{1}=3$ for $n_{4}=1$, there will be three lepton multiplets corresponding to each quark multiplet. We can therefore use the quark multiplet to be the marker of a fermion generation, and construct the field content of a single generation, which can be repeated arbitrary number of times. We can now write $\delta$ instead of $\delta_{1}$ since it is the only parameter of its kind. Clearly, we cannot take all $\alpha$’s to be the same because, if we do that, then Eq. (1d), or equivalently Eq. (24), is not obeyed with the values obtained in Eqs. (19) and (22).

So at least two $\alpha_{k}$’s will be different. Rather than considering the possibility that they are all different, let us consider the scenario where

$\displaystyle\alpha_{1}=\alpha_{2}\neq\alpha_{3}\,.$

(33)

We now look at Eq. (1d). With the values of $\alpha_{1}$ and $\delta$ from Eqs. (19) and (22), we find

$\displaystyle\alpha_{3}={2+q\over 3}\,.$

(34)

We can now write the electric charges of all left-chiral fields in a single generation:

$\displaystyle\begin{array}[]{c@{\hspace{7mm}}c@{\hspace{7mm}}c@{\hspace{7mm}}l}\hline\cr\mbox{${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$}\hfil\hskip 19.91692pt&\hbox{\multirowsetup\mbox{Chirality}}\hfil\hskip 19.91692pt&\mbox{Number of}\hfil\hskip 19.91692pt&\mbox{electric}\\ \mbox{representation}\hfil\hskip 19.91692pt&\hfil\hskip 19.91692pt&\mbox{copies}\hfil\hskip 19.91692pt&\mbox{charges}\\ \hline\cr\Big{(}1,3,{q-1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&2\hfil\hskip 19.91692pt&0,-1,q\\ \Big{(}1,3,{q+2\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&1,0,q+1\\ \Big{(}3,3^{*},-{q\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&-\frac{1}{3},\frac{2}{3},-\frac{1}{3}-q\\ \lx@intercol\hfil\mbox{and R-chiral fields to match their charges.}\hfil\lx@intercol\\ \hline\cr\end{array}$

(41)

Here also, one can take any arbitrary value of $q$ and introduce a right chiral field corresponding to every left chiral charged field and obtain anomaly cancellation thereby. But there is a more economical possibility, which was found by Deppish, Hati, Patra, Sarkar and Valle [[23]]. They entertained the possibility that each charged left chiral field has a partner in the form of a left chiral field of opposite charge. Together, they can form mass terms. This means that there is no need for the singlets of ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$. If this viewpoint is adopted, there is a unique value of $q$ that becomes acceptable. This value comes from the fact that there are two fields of charge $-1$ in the two copies of of the first kind of multiplet shown in Eq. (41), and therefore we need two fields with charge $+1$. This can be achieved by putting $q=0$, so that both fields of charge $+1$ can appear in the second kind of multiplet shown in Eq. (41). Setting $q=0$ and adding the right chiral quark fields in ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$ singlets, one obtains a single generation of fermions from Eq. (41). Other generations are exact copies, so far as the gauge transformation properties are concerned. The model thus obtained is then exactly the model suggested in Ref. [23], except for a difference in the convention: what we call a triplet of ${\rm SU(3)}_{L}$ was called an antitriplet in Ref. [23], and vice versa.

In order to exhaust all scenarios in which at least two $\alpha_{k}$’s are equal, we should also examine the possibility

$\displaystyle\alpha_{1}\neq\alpha_{2}=\alpha_{3}\,.$

(42)

Now, putting Eqs. (19) and (22) into Eq. (1d), we obtain

$\displaystyle\alpha_{2}=\alpha_{3}={1+2q\over 6}\,.$

(43)

With these values of the $\alpha_{k}$’s, the following multiplets result.

$\displaystyle\begin{array}[]{c@{\hspace{7mm}}c@{\hspace{7mm}}c@{\hspace{7mm}}l}\hline\cr\mbox{${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$}\hfil\hskip 19.91692pt&\hbox{\multirowsetup\mbox{Chirality}}\hfil\hskip 19.91692pt&\mbox{Number of}\hfil\hskip 19.91692pt&\mbox{electric}\\ \mbox{representation}\hfil\hskip 19.91692pt&\hfil\hskip 19.91692pt&\mbox{copies}\hfil\hskip 19.91692pt&\mbox{charges}\\ \hline\cr\Big{(}1,3,{q-1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&0,-1,q\\ \Big{(}1,3,{1+2q\over 6}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&2\hfil\hskip 19.91692pt&\frac{1}{2},-\frac{1}{2},\frac{1}{2}+q\\ \Big{(}3,3^{*},-{q\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&-\frac{1}{3},\frac{2}{3},-\frac{1}{3}-q\\ \lx@intercol\hfil\mbox{and R-chiral fields to match their charges.}\hfil\lx@intercol\\ \hline\cr\end{array}$

(50)

Note that these models necessarily contain fractionally charged leptons. Also note that there is no value of $q$ for which the ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$ singlets can be avoided. However, it should be emphasized that, by adding suitable right chiral fields, one can definitely build a viable model of this sort.

In the models discussed in Sec. 4, anomalies do not cancel unless we take all three generations of fermions together. In the models of Sec. 5, anomalies cancel within a single generation. We can now explore the intermediate scenario in which one needs two generations to cancel the gauge anomalies. This means that we want $n_{3}+n_{4}=2$. Among the integral solutions to this equation subject to Eq. (3), the solution $n_{4}=2,\ n_{3}=0$ merely duplicates the field content of the sequential model. Thus, we are left with the only other solution, i.e., both $n_{3}$ and $n_{4}$ equals 1. This means that $n_{1}=n_{2}$, according to Eq. (1b). If we take $n_{1}=n_{2}=0$ then there will be no lepton in the model. In order to accommodate leptons, we take the next smallest solution, i.e., Choice 9 of Eq. (9). In order to accommodate the SM doublets of quarks in the $\mathit{(3,3)}$ and the $\mathit{(3,3^{*})}$ representations of ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}$, and also to accommodate an SM lepton doublet in the $\mathit{(1,3)}$ representation, we need the values of $\alpha$, $\gamma$ and $\delta$ as shown in Eq. (32). Eq. (1d) will then give the value of $\beta$. We summarize the information.

$\displaystyle\begin{array}[]{c@{\hspace{7mm}}c@{\hspace{7mm}}c@{\hspace{7mm}}l}\hline\cr\mbox{${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$}\hfil\hskip 19.91692pt&\hbox{\multirowsetup\mbox{Chirality}}\hfil\hskip 19.91692pt&\mbox{Number of}\hfil\hskip 19.91692pt&\mbox{electric}\\ \mbox{representation}\hfil\hskip 19.91692pt&\hfil\hskip 19.91692pt&\mbox{copies}\hfil\hskip 19.91692pt&\mbox{charges}\\ \hline\cr\Big{(}1,3,{q-1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&0,-1,q\\ \Big{(}1,3^{*},-{q+2\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&-1,0,-1-q\\ \Big{(}3,3,{q+1\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&\frac{2}{3},-\frac{1}{3},\frac{2}{3}+q\\ \Big{(}3,3^{*},-{q\over 3}\Big{)}\hfil\hskip 19.91692pt&L\hfil\hskip 19.91692pt&1\hfil\hskip 19.91692pt&-\frac{1}{3},\frac{2}{3},-\frac{1}{3}-q\\ \lx@intercol\hfil\mbox{and R-chiral fields to match their charges.}\hfil\lx@intercol\\ \hline\cr\end{array}$

(58)

The discussion does not imply that we can have only two generations and no more if we follow this path. It only means that all gauge anomalies cancel among the two generations of fermions whose specifications have been given in Eq. (58). In order to confront phenomenology and include three generations of quarks, one can always add a sequential generation as what has been discussed in Sec. 5. This means that the complete model, with three generations of fermions in it, will have

$\displaystyle n_{1}=4,\qquad n_{2}=1,\qquad n_{3}=1,\qquad n_{4}=2.$

(59)

We have identified many ways of extending the SM to a model based on the gauge group ${\rm SU(3)}_{c}\times{\rm SU(3)}_{L}\times{\rm U(1)}_{X}$. For the entangled models where the model has a prediction for the number of fermion generations, we show that there is a whole family of models, characterized by a parameter $q$, which are anomaly-free. This family includes the models which have been studied in detail in the literature [[1, 2, 3, 4, 5]]. In contrast, there are also sequential models, including one which has been studied in some detail [[23]], in which generations are marked by quark fields, and are copies of one another. We have shown that there can be an intermediate class of models in which there is one sequential generation and two more generations which are entangled through anomalies. Many of these models have not been studied at all.

It need not be said that one can make more complicated models with the same gauge group. It is possible to construct models with larger number of fermion fields by adding any number of vectorlike fermions, or gauge singlets, or multiple copies of the entire collection. We have only identified models which are minimal corresponding to some set of assumptions.

These minimal models can be compared on the basis of their field content. In Table 2, we summarize the total number of triplets and antitriplets of ${\rm SU(3)}_{L}$ in each kind of model. Each of these representations contain a doublet of the standard electroweak model. We see that the completely entangled models have no SM doublet other than the ones which are already present in the three generations of the SM. The sequential models need a lot of new doublets, whereas the intermediate models are intermediate in this aspect as well. The number of singlets of ${\rm SU(3)}_{L}$ are different in the individual models of each kind, and are not included in the table.