Beyond the Standard Model: the minimal \(SU(5)\) GUT
The standard model is incredibly exhaustive and accurate, to the point that it’s been a challenge finding empirical facts that would directly contradict it. Yet, for the past 50 years, the quest for Grand Unification has been the overarching goal of fundamental physics.
Despite the incredible success of the standard model, there are reasons to think it does not offer a complete description of reality. Those are the most compelling arguments:
Charge quantization: How come that protons and electrons have exactly opposite electric charges? From the standard model perspective, it is just a concidence that the electron’s charge is \(-1\) and the quarks’ are \(+2/3\) and \(-1/3\). Yet we know that it’s not an approximation: this has been measured to an extremely high accuracy. I may add it’s rather consequential, since there could be no matter at all without this "coincidence": atoms couldn’t exist if a set of quarks couldn’t exactly counterbalance the proton’s electric charge.
Free parameters: There are 18 free parameters in the standard model (interactions’ mixing parameters, Yukawa couplings (which determine fermion masses), Higgs potential parameters...).
Generations: Three full generations of fermions have been observed, and further data seems to prove there cannot be fourth one. Yet the presence of three identical families (except for their mass) is completely sui generis in the Standard Model.
The hierarchy problem: Why is the weak interaction so much stronger than gravity? One way to look at this is to consider that gravitation’s strength is characterized by the Planck mass \(m_p\) while for the weak force the relevant quantity is the mass of its bosons \(m_W\) or \(m_Z\), which depend directly on the Higgs boson’s mass \(m_H\). In fact, the value of \(m_H\) is not a prediction of the standard model, it is an observation we’ve made. The issue is that in QFT, the renormalization procedure that helps us go from the "bare masses" of the lagrangian to the "physical mass" that we observe, relies on the cancellation of certain terms that require fine-tuning. The form of Higgs renormalization makes the cancellation terms proportional to \(\Lambda / m_H\), where \(\Lambda\) is roughly the ambient energy level. So if we wanted the standard model to be valid up to energies comparable to the Planck mass, then we would need terms \(10^34\) times larger than the Higgs mass to "magically" cancel out and give us the right result. This is mathematically acceptable but appears as a "coincidence" which could only be explained by the anthropic principle (only a universe where that coincidence happens could have us in it, so we must be in that one), or this procedure can be avoided thanks to a higher-level theory.
Matter-antimatter asymmetry: The imbalance between matter and antimatter in our universe is difficult to square with the symmetries of the standard model (where fermions and anti-fermions only get produced, or annihilate, in pairs)/
Neutrino mass: Observations now strongly suggest that neutrinos have a small mass, which means that right-handed neutrinos could exist. That is not ruled out but unlike right-handed electrons, right-handed neutrinos would need to be completely separate from the other fermions: not part of any representation of the basic gauge group.
A great way to explain the first two items is through unification of the interactions, which as we’ve seen, also fulfils a long-held aesthetic preference that has also borne fruit in spectacular fashion in physics history.
We know that unification cannot happen in the context of the \(SU(3)\times SU(2) \times U(1)\) group. Since that group is a product of three simple groups, there can be three different coupling constants (interaction strength, which we called \(g_s\), \(g\) and \(g'\) above). We can compute their dependence to temperature based on the standard model. When we increase it (or equivalently, when we "turn back time" closer to the Big Bang), there is no point when the three constants would be equal. So, unification is a miss.
One way to avoid this is to build a theory around one simple Lie group, which by definition has only one coupling constant (say, \(g_{GUT}\)). That symmetry can then be broken into \(SU(3) \times SU(2) \times U(1)\), at which point we retrieve the standard model.
The strategy should then be as follows:
Find a simple Lie group \(G\) that contains the Standard Model group and has complex representations (lest it couldn’t account for the difference between left-handed and right-handed particles which we observe in the Standard Model).
Find representations matching the known particle content of the standard model.
Find a symmetry breaking procedure that leads to the Standard Model group.
Check that the predictions are not inconsistent with our observations (in particular, extra particles should all be heavy enough to explain why they were not observed before).
Luckily, the set of possible groups is very small. First let’s remember that the classification of all simple Lie groups is itself parsimonious.
The special unitary groups: \(A_n = SU(n+1)\)
The "odd-dimensional" special orthogonal groups: \(B_n = SO(2n+1)\)
The symplectic groups: \(C_n = Sp(n)\)
The "even-dimensional" special orthogonal groups: \(D_n = SO(2n)\)
\(E_6\),\(E_7\),\(E_8\),\(F_4\),\(G_2\)
That list covers all the possible simple Lie groups. This can be proved very elegantly using Dynkin diagrams, which we can only show here for illustration.
First of all, we should note that the rank of the standard model group is \(4\). The rank can be defined as the largest number of generators that commute with each other. We know that \(SU(3)\) has rank \(2\), \(SU(2)\) has rank \(1\), and \(U(1)\) only has one generator. By construction of the group product, generators of distinct simple groups commute, hence the rank of \(SU(3)\times SU(2) \times U(1)\) ought to be \(2+1+1=4\).
The nomenclature we’ve presented above uses the rank as an index. So the simple groups of rank 4 are \(A_4\),\(B_4\),\(C_4\),\(D_4\) and \(F_4\). But Lie group theory (an aesthetically delighting discipline) tells us exactly which groups fulfill our basic criteria (to embed the Standard Model, and to admit complex representations). They are: the \((A_n)_{n \geq 4}\), the \((D_{2n+1})_{n\geq 2}\), and \(E_6\).
For rank \(4\), we know that the only eligible group is \(A_4=SU(5)\). At rank 5, there are \(A_5=SU(6)\) or \(D_5=SO(10)\). At rank 6 the eligible groups are \(A_6=SU(7)\) and the exceptional group \(E_6\).
SU(5) has a particular status, being the minimal supergroup of the Standard Model: the smallest supergroup that contains it. Building a gauge theory, thanks to all the rules inherited from group theory, is a lot like Meccano at this stage! So let’s get on with it...
In theories we’ve considered so far, fermions have been placed in the fundamental representation and its anti-fundamental counterpart. For \(SU(N)\) groups, these were alway \(N\)-dimensional. No difference here, the fundamental representation is indeed \(\mathbf{5}\), and the anti-fundamental one is \(\bar{\mathbf{5}}\).
What are the generators of \(SU(5)\). Remember that they form a basis of the Lie algebra \(\mathfrak{su}(N)\), which is the space of \(N\times N\) traceless Hermitean matrices. We already know that the basis for \(\mathfrak{su}(3)\) is the family of 8 Gell-Mann matrices \((\lambda^i)\), and for \(\mathfrak{su}(2)\) is the family of 3 Pauli matrices \((\tau^i)\).
These are already traceless Hermitean matrices in their own space, so it’s easy to see that those matrices also form a free family of \(\mathfrak{su}(5)\):
\[\begin{pmatrix} \lambda^i & 0 \\ 0 & 0 \end{pmatrix}_{i=1\dots8} \quad , \quad \begin{pmatrix} 0 & 0 \\ 0 & \tau^i \end{pmatrix}_{i=1\dots3}\]
This means that if we divide the quintuplet representation into a triplet (on top) and a doublet (on the bottom), those will be respectively \(SU(3)\)- and \(SU(2)\)-invariant. It makes sense, based on the way we’ve modelized fermions to fill out the quintuplet with a right-handed quark triplet and a left-handed lepton doublet.
If we continue to explore the generators of \(SU(5)\), we can show there is a diagonal matrix among them: \(\text{diag}(2,2,2,-3,-3)\). From the independent perspectives of a triplet and of a doublet, this is acting as a simple scalar (a multiple of the identity). We can then identify it with \(\tilde{Y}\), the operator corresponding to the weak hypercharge \(Y\).
The difference is that in the standard model, the generator of \(U(1)_Y\) does not need to be a traceless matrix, and indeed it is not. Here, the matrix is a generator of \(SU(5)\) and thus had to be traceless, so it could not be a multiple of the identity. The fact is traceless ensures that the sum of hypercharges within the quintuplet. Since the electric charge matrix \(Q\) will ultimately result from the same kind of symmetry breaking as within the Standard Model and will be some linear combination of \(\tau^3\) (also traceless) and \(\tilde{Y}\). Therefore \(Q\) will be traceless and the sum of electric charges in the quintuplet will be nil.
This means that \(3q_Q + q_e=0\) (because quarks of each color have the same electric charge, and we know the neutrino’s charge is nil). Furthermore, the quark has positive hypercharge and the leptons negative hypercharge. Based on our standard model study, this means the quark triplet is actually an antiquark triplet (since only antifermions carry positive hypercharge). The charged lepton is then an electron (not an antiparticle). We end up with \(q_e=-1\) and then \(q_Q=+1/3\), which we then identify with an anti-down quark.
But most importantly, this relationship is imposed by the theory! Right off the bat, \(SU(5)\) solves the puzzle of charge quantization.
We have successfully identified the fundamental representation with known fermion content: \[\begin{aligned} \bar{\mathbf{5}} &\rightarrow (\bar{\mathbf{3}},\mathbf{1})_{2/3} \oplus (\mathbf{1},\mathbf{2})_1\\ & = \overline{d_R} \oplus L_L = \begin{pmatrix} \overline{d_R^b} & \overline{d_R^r} & \overline{d_R^y} & \nu_L & e_L \end{pmatrix} \end{aligned}\]
However, we’re still missing lots of our fermions. To find them we can go to the next smallest representation, the ten-dimensional \(\mathbf{10}\), which we can evidence by taking the product of two fundamental representations.
\[\mathbf{5} \otimes \mathbf{5} = \mathbf{10} \oplus \mathbf{15}\]
\(\mathbf{10}\) is the anti-symmetric part of that product. It can be represented as an anti-symmetric \(5\times 5\) matrix, which indeed has 10 degrees of freedom. One way to think about this product is that each component of \(\mathbf{10}\) is the "sum" of two distinct components of \(\mathbf{5}\) (or rather, the particle whose conserved charges are the sum of the two components’ respective charges).
\[\begin{aligned} \mathbf{10} &\rightarrow (\bar{\mathbf{3}},\mathbf{1})_{4/3} \oplus (\mathbf{3},\mathbf{2})_{1/3} \oplus (\bar{\mathbf{1}},\mathbf{1})_{2} \\ & = \overline{u_R} \oplus Q_L \oplus \overline{e_R}\\ & = \begin{pmatrix} 0 & \overline{u_R^b} & -\overline{u_R^y} & u_L^r & d_L^r\\ - \overline{u_R^b} & 0 & \overline{u_R^r} & u_L^y & d_L^y\\ \overline{u_R^y} & -\overline{u_R^r} & 0 & u_L^b & d_L^b\\ -u_L^r & -u_L^y & -u_L^b & 0 & \overline{e_R}\\ -d_L^r & -d_L^y & -d_L^b & -\overline{e_R} & 0\\ \end{pmatrix} \end{aligned}\]
Between \(\mathbf{5}\), \(\bar{\mathbf{5}}\),\(\mathbf{10}\) and \(\bar{\mathbf{10}}\), we’ve retrieved the 15 fermions and 15 anti-fermion of the Standard Model!
What about the bosons? As usual, they occupy the adjoint representation, which is 24-dimensional (\(5^2-1\)).
\[\begin{aligned} \mathbf{5} \otimes \bar{\mathbf{5}} &= \mathbf{24} \\ & \rightarrow (\mathbf{8},\mathbf{1})_0 \oplus (\mathbf{3},\mathbf{1})_0 \oplus (\mathbf{1},\mathbf{1})_0 \oplus (\mathbf{3},\mathbf{2})_{-10/3} \oplus (\bar{\mathbf{3}},\mathbf{2})_{10/3} \end{aligned}\]
This matches or analysis of the generators. In this sum we find:
8 gluons \(G\)
3 weak bosons \(W^\pm\) and \(W^3\)
1 hypercharge boson \(B\)
12 \(SU(5)\)-specific bosons called \(X\) and \(Y\)
The \(X\) and \(Y\) bosons have charges correspond to generators that govern interactions between quarks and leptons. That’s why they’re sometimes called leptoquarks. They carry weak, electric and color charges.
Among the reactions allowed by the presence of the \(X\) and \(Y\), some involve the decay of these boson in two fermions or two anti-fermions. Therefore, in the context of a very fast cooling of the universe, we are much closer to explaining the matter-antimatter asymmetry!
One important task remains: we have a wonderful symmetry, but as for all unbroken local gauge symmetries, it only allows massless fermions and bosons.
As for the Standard Model, we’ll use the wonderful Higgs mechanism, which has abundantly proven its worth. At this stage it is very important to highlight the difference between GUT symmetry breaking and the electroweak symmetry breaking.
The electroweak Higgs process breaks the symmetry this way: \[SU(3) \times SU(2) \times U(1)_Y \rightarrow SU(3) \times U(1)_{EM}\]
On both sides we have a product of simple Lie groups, each one can have its own coupling constant (interaction strength). The symmetry breaking has caused the coupling of the Higgs field with other particles which gave mass to the \(W^\pm\), the \(Z\), the fermions and the Higgs boson. But it has not caused the divergence of the \(SU(2)\) and \(U(1)_Y\) coupling constants: they were already distinct. In that sense, it is an exaggeration when we speak of the "electroweak" force. And the symmetry breaking process is much better described as a phase transition than as a "force de-unification".
The \(SU(2)\) and \(U(1)_Y\) coupling constants vary with the energy level, and they do meet at some point, but that is a different one, at a much higher energy than the Higgs boson mass (which is where the electroweak phase trsnsition happens). The goal of the GUT builders is to have the three couplings meeting at the same energy level, so there’s need for only one GUT breaking. And the running couplings’ calculation, after incorporating all the \(SU(5)\)-specific terms, shows them almost converging!
The \(SU(5)\) model then uses two Higgs field, one in the adjoint representation (made of 24 real scalar fields) breaks \(SU(5)\) directly into the SM group. It gives mass to the 12 \(X\) and \(Y\) bosons, and to 12 Higgs bosons. Another Higgs field in the fundamental representation is a complex quintuplet, with the lower doublet being the usual SM Higg field, and with at the top a color-charged Higgs triplet, resulting into 5 massive Higgs boson (why this doesn’t give mass to the gluon is a problem called triplet-doublet splitting that introduces significant fine-tuning in the theory).
Let’s look at our GUT checklist. \(SU(5)\) has successfully explained charge quantization, and majorly improved the situation regarding matter/antimatter asymmetry, and coupling unification. It has also established relationships between parameters, however it also introduces new free ones...
Despite this overall success, \(SU(5)\) GUT as such is not enough: it is contradicted by experience, the most striking one being the half-life of the proton, which has to be much higher than what it predicts (based on our observations). It also makes a number of wrong predictions, for instance about the ratios between fermion masses.
From \(SO(10)\) to \(E_8\)
\(SU(5)\) was our only option at rank \(4\). But not to worry, we have plenty of higher-rank ones. At rank \(5\), it is \(SO(10)\) that seems to offer the best prospects at this stage.
First off, the fundamental representation of that group is sixteen-dimensional: \(\mathbf{16}\). We can soon verify that the whole SM fermion content can fit in that representation (all fifteen of them) plus a particle we can identify with the righ-handed neutrino! This is considered more elegant since all the matter and antimatter content can fit in the fundamental and anti-fundamental representations.
The adjoint representation is 45-dimensional, which means that we should find the 24 bosons from \(SU(5)\), plus 21 new ones.
\(SU(5)\) is a subgroup of \(SO(10)\), and sure enough, the following symmetry breaking route is available:
\[\begin{aligned} SO(10) & \rightarrow SU(5) \times U(1) \\ &\rightarrow SU(3)_C \times SU(2)_L \times U(1)_Y \end{aligned}\]
This means there will be two, instead of one, GUT breaking scales. At least this route allows us to retrieve the familiar \(SU(5)\) multiplets, accompanied by the new fermion (the RH neutrino) and the new bosons.
\[\mathbf{16} \rightarrow \mathbf{10} \oplus \bar{\mathbf{5}} \oplus \mathbf{1}\]
\[\mathbf{45} \rightarrow \mathbf{24} \oplus \bar{\mathbf{10}} \oplus \mathbf{10} \oplus \mathbf{1}\]
However there is another route:
\[\begin{aligned} SO(10) & \rightarrow SU(4)_C \times SU(2)_L \times SU(2)_R \\ &\rightarrow SU(3)_C \times SU(2)_L \times SU(2)_R \times U(1)_{B-L} \\ &\rightarrow SU(3)_C \times SU(2)_L \times U(1)_R \times U(1)_{B-L} \\ &\rightarrow SU(3)_C \times SU(2)_L \times U(1)_{Y} \\ \end{aligned}\]
This is obviously a more complex scheme that involves four symmetry breaking scales. \(SU(4)_C \times SU(2)_L \times SU(2)_R\) corresponds to the Pati-Salam theory, which pre-dates even the \(SU(5)\) Georgi-Glashow model. It posits a four-color symmetry, where all leptons would have the same color, called "liliac". The two \(SU(2)\) groups correspond to a left-right symmetry, which imposes the existence of the right-handed neutrino. The breakdown of the \(SU(4)\) color group yields the QCD group and a \(B-L\) symmetry that preserves the Baryon-minus-Lepton number. The last step to reach the Standard Model gives the neutrino its mass and mixes the two \(U(1)\) symmetries into the familiar hypercharge one.
The \(SO(10)\) representations’ branching rules when it breaks into the Pati-Salam group are as follows:
\[\mathbf{16} \rightarrow (\mathbf{4},\mathbf{2},\mathbf{1}) \oplus (\bar{\mathbf{4}},\mathbf{1},\mathbf{2})\]
\[\mathbf{45} \rightarrow (\mathbf{15},\mathbf{21},\mathbf{1}) \oplus (\mathbf{6},\mathbf{2},\mathbf{2}) \oplus (\mathbf{1},\mathbf{3},\mathbf{1}) \oplus (\mathbf{1},\mathbf{1},\mathbf{3})\]
The Higgs field content and mechanism depends of course on the route chosen. It typically involves scalars in at least the \(\mathbf{16}\) and \(\mathbf{45}\) representations.
\(SO(10)\) theories are then richer in terms of the phenomenology they may explain, but introduces even more new heavy particles. Unlike \(SU(5)\), \(SO(10)\) has not been ruled out by observations, but none of these extra particles have been observed at experimental energy levels.
The next step is usually the exceptional \(E_6\) group. It contains \(SO(10)\) and can break its way:
\[\begin{aligned} E_6 \rightarrow SO(10)\times U(1) \end{aligned}\]
The fundamental representation has 27 components. We then end up with many additional fermions to which we need to give high mass to explain why they don’t pop up now. The adjoint representation has 78 bosons.
\[\mathbf{27} = \mathbf{16} \oplus \bar{\mathbf{10}} \oplus \mathbf{1}\]
\[\mathbf{78} = \mathbf{45 } \oplus\mathbf{16} \oplus \bar{\mathbf{16}} \mathbf{1}\]
\(E_6\) is interesting because it has as a maximal subgroup \(SU(3)^3\), called the trinification group. Although that is by no means fleshed out, it is thought that group can implement flavorgenesis, and at last explain the existence of three fermion generations as the result of a specific symmetry. It can then break into the same left-right group we’ve seen above as an outcome of Pati-Salam symmetry breaking.
\[\begin{aligned} E_6 & \rightarrow SU(3)_C \times SU(3)_L \times SU(3)_R &\rightarrow SU(3)_C \times SU(2)_L \times SU(2)_R \times U(1)_{B-L} \\ \end{aligned}\]
\[\mathbf{27} = (\mathbf{3},\mathbf{3},\mathbf{1}) \oplus (\bar{\mathbf{3}},\bar{\mathbf{3}},\mathbf{1}) \oplus (\mathbf{1},\bar{\mathbf{3}},\mathbf{3})\]
\[\mathbf{78} = (\mathbf{8},\mathbf{1},\mathbf{1}) \oplus (\mathbf{1},\mathbf{8},\mathbf{1}) \oplus (\mathbf{1},\mathbf{1},\mathbf{8}) \oplus (\mathbf{3},\bar{\mathbf{3}},\bar{\mathbf{3}}) \oplus (\bar{\mathbf{3}},\mathbf{3}),\mathbf{3})\]
The Higgs field content to achieve these complex breaking chains from bigger and bigger groups gets overabundant, with a 351-dimensional representation involved in most models.
All in all \(E_6\) almost seems oversized, despite the potential added value from trinification. Unlike \(SO(10)\), it entails the addition of new, unobserved fermions. The main attractive factor may then be aesthetic. \(E_6\), being an exceptional group, appears like the end of the road:
\[SU(3) \times SU(2) \times U(1) \supset SU(5) \supset SO(10) \supset E_6\]
No Lie group other than \(E_7\) and \(E_8\) contain \(E_6\) as a subgroup, and we know their lack of complex representations make them unlikely candidates to be the GUT group. So picking \(E_6\) offers the advantage of singularity: it is not just "some group".
It is worth noting, however that the picture changes once superstrings come into play. We have seen that superstring theory, at a "low" energy , can be approximated by a Lie-gauge-group invariant QFT, the likes of which we’ve been considering here ("low energy" for superstrings means the high-end of the range for every GUT theory).
And it so happens that one of the only gauge groups compatible with the superstring framework is \(E_8\times E_8\), a rank-16 gauge group acting over strings from a 10D spacetime. Now, of course, we’ve seen that \(E_8\), lacking real representations, is ineligible for GUTs... but that’s only true in 4D spacetimes! As it turns out, the compactification to 4D and the use of "stringy" procedures would allow to retrieve, as an approximation, the gauge group \(E_6 \times SU(3)\) for our GUT. Alternately, \(E_8\) can break into \(SO(10) \times SU(4)\).
This connection to superstrings makes the study of \(E_6\) and \(SO(10)\) GUTs particularly attractive. \(E_8\) is also interesting to physicists since it is a unique case where the fundamental and adjoint representations have the same dimensionality (in this case, 248). Since they typically house the fermion and boson contents respectively, this makes any \(E_8\) theory a natural for supersymmetry!
From a purely aesthetic perspective, \(E_8\) is really the end of the line, since it is not a subgroup of any simple Lie group.