The grand unified theory appeared in a 1974 paper by Howard Georgi and Sheldon Glashow . It was the first grand unified theory, and is still considered the prototypical example. As such, there are many accounts of it in the physics literature. The textbooks by Ross  and Mohapatra  both devote an entire chapter to the theory, and a lucid summary can be found in a review article by Witten , which also discusses the supersymmetric generalization of this theory.
In this section, we will limit our attention to the nonsupersymmetric version of theory, which is how it was originally proposed. Unfortunately, this theory has since been ruled out by experiment; it predicts that protons will decay faster than the current lower bound on proton lifetime . Nevertheless, because of its prototypical status and intrinsic interest, we simply must talk about the theory.
The core idea behind the grand unified theory is that because the Standard Model representation is 32-dimensional, each particle or antiparticle in the first generation of fermions can be named by a 5-bit code. Roughly speaking, these bits are the answers to five yes-or-no questions:
We can flesh out this scheme by demanding that the operation of taking antiparticles correspond to switching 0's for 1's in the code. So the code for the antiparticle of , the `antired right-handed antidown antiquark', is . This is cute: it means that being antidown is the same as being up, while being antired is the same as being both green and blue.
Furthermore, in this scheme all antileptons are `black' (the particles with no color, ending in 000), while leptons are `white' (the particles with every color, ending in 111). Quarks have exactly one color, and antiquarks have exactly two.
We are slowly working our way to the theory. Next let us bring Hilbert spaces into the game. We can take the basic properties of being up, down, red, green or blue, and treat them as basis vectors for . Let us call these vectors . The exterior algebra has a basis given by wedge products of these 5 vectors. This exterior algebra is 32-dimensional, and it has a basis labelled by 5-bit strings. For example, the bit string corresponds to the basis vector , while the bit string corresponds to .
Next we bring in representation theory. The group has an obvious representation on . And since the operation of taking exterior algebras is functorial, this group also has a representation on . In the grand unified theory, this is the representation we use to describe a single generation of fermions and their antiparticles.
Just by our wording, though, we are picking out a splitting of into : the isospin and color parts, respectively. Choosing such a splitting of picks out a subgroup of , the set of all group elements that preserve this splitting. This subgroup consists of block diagonal matrices with a block and a block, both unitary, such that the determinant of the whole matrix is 1. Let us denote this subgroup as .
Now for the miracle: the subgroup is isomorphic to the Standard Model gauge group (at least modulo a finite subgroup). And, when we restrict the representation of on to , we get the Standard Model representation!
There are two great things about this. The first is that it gives a concise and mathematically elegant description of the Standard Model representation. The second is that the seemingly ad hoc hypercharges in the Standard Model must be exactly what they are for this description to work. So, physicists say the theory explains the fractional charges of quarks: the fact that quark charges come in units the size of electron charge pops right out of this theory.
With this foretaste of the fruits the theory will bear, let
us get to work and sow the seeds. Our work will have two
parts. First we need to check that
First, the group isomorphism. Naively, one might seek to build
the theory by including
as a subgroup of .
Can this be done?
Clearly, we can include
as block diagonal
matrices in :
The first clue is that elements of must commute with the
. But the only elements of
that commute with everybody in the
diagonal, since they must separately commute with
, and the only elements doing so are
diagonal. Moreover, they must be scalars on each block. So, they have
to look like this:
So if we throw in elements of this form, do we get
? More precisely, does this map:
The answer is no: the map has a kernel,
The kernel is the set of all elements of the form
This sets up a nerve-racking test that the theory must pass for it to have any chance of success. After all, not all representations of factor through , but all those coming from representations of must do so. A representation of will factor through only if the subgroup acts trivially.
In short: the GUT is doomed unless acts trivially on every fermion. (And antifermion, but that amounts to the same thing.) For this to be true, some nontrivial relations between hypercharge, isospin and color must hold.
For example, consider the left-handed electron
Or, consider the right-handed quark:
For to work, though, has to act trivially on every fermion. There are 16 cases to check, and it is an awful lot to demand that hypercharge, the most erratic part of the Standard Model representation, satisfies 16 relations.
Or is it? In general, for a fermion with hypercharge , there are four distinct possibilities:
|Nontrivial , nontrivial|
|Nontrivial , trivial|
|Trivial , nontrivial|
|Trivial , trivial|
By this analysis, we have shown that
acts trivially on the Standard Model
rep, so it is contained in the kernel of this rep. It is better than just a
is the entire kernel. Because of this, we could say
is the `true' gauge group of the Standard Model. And because
we now know that
Of course, we still need to find a representation of that extends the Standard Model representation. Luckily, there is a very beautiful choice that works: the exterior algebra . Since acts on , it has a representation on . Our next goal is to check that pulling back this representation from to using , we obtain the Stadard model representation
As we do this, we will see another fruit of the theory ripen. The triviality of already imposed some structure on hypercharges, as outlined in above in Table 3. As we fit the fermions into , we will see this is no accident: the hypercharges have to be exactly what they are for the theory to work.
To get started, our strategy will be to use the fact that, being representations of compact Lie groups, both the fermions and the exterior algebra are completely reducible, so they can be written as a direct sum of irreps. We will then match up these irreps one at a time.
The fermions are already written as a direct sum of irreps, so we need
to work on
. Now, any element
acts as automorphisms
of the exterior algebra
Let us see how this works, starting with the easiest cases. and are both trivial irreps of . There are two trivial irreps in the Standard Model representation, namely and its dual , where we use angle brackets to stand for the Hilbert space spanned by a vector or collection of vectors. So, we could select and , or vice versa. At this juncture, we have no reason to prefer one choice to the other.
Next let us chew on the next piece: the first exterior power,
. We have
In short, as a rep of
, we have
Now this is problematic, because another glance at Table 1
reveals that there is no left-handed lepton with
hypercharge 1. The only particles with hypercharge 1 are the right-handed
antileptons, which span the representation
Now we can use our knowledge of the first exterior power to
compute the second exterior power, by applying the formula
So, let us calculate! As reps of
Consider the first summand,
. As a rep of this space is just
which is the one-dimensional trivial rep, .
As a rep of it is also trivial. But as a rep of
, it is nontrivial. Inside it we are juxtaposing
two particles with hypercharge 1. Hypercharges add, just like charges,
so the composite particle, which consists of one particle and
the other, has hypercharge 2. So, as a representation of the Standard
Model gauge group we have
Next consider the second summand:
Finally, the third summand in
In summary, the following pieces of the Standard Model rep
We are almost done. Because preserves the canonical volume form on
, taking Hodge duals gives an isomorphism between
How does all this look in terms of the promised binary code? Remember, a 5-bit code is short for a wedge product of basis vectors . For example, 01101 corresponds to . And now that we have found an isomorphism , each of these wedge products corresponds to a fermion or antifermion. How does this correspondence go, exactly?
First consider the grade-one part
This has basis vectors called and . We have seen that
, spanned by and , corresponds to
Next consider the grade-two part:
To work out the other grades, note that Hodge duality corresponds to switching 0's and 1's in our binary code. For instance, the dual of 01101 is 10010: or written in terms of basis vectors, the dual of is . Thus given the binary codes for the first few exterior powers:
Now we can see a good, though not decisive, reason to choose . With this choice, and not the other, we get left-handed particles in the even grades, and right-handed particles in the odd grades. We choose to have this pattern now, but later on we need it.
Table 4 defines a linear isomorphism in terms of the basis vectors, so the equations in this table are a bit of a exaggeration. When we write say, , we really mean . This map is an isomorphism between representations of . It tells us how these representations are the `same'.
More precisely, we mean these representations are the same when we
using the isomorphism
induced by . In general,
we can think of a unitary representation as a Lie group homomorphism
Indeed, we just showed this! We have seen there exists a
unitary operator from the Standard Model rep to