3.4 The Route to Spin(10) via Pati-Salam

In the last section, we showed how the Pati-Salam model answers two questions about the Standard Model:

Why are quarks and leptons so similar? Why are left and right so different?

We were able to describe leptons as a fourth color of quark, `white', and treat right-handed and left-handed particles on a more equal footing. Neither of these ideas worked on its own, but together, they made a full-fledged extension of the Standard Model, much like and , but based on seemingly different principles.

Yet thinking of leptons as `white' should be strangely familiar, not just from the Pati-Salam perspective, but from the binary code that underlies both the and the theories. There, leptons were indeed white: they all have color .

Alas, while hints that leptons might be a fourth color, it does not
deliver on this. The quark colors

lie in a different irrep of than does . So, leptons in the theory are white, but unlike the Pati-Salam model, this theory does not unify leptons with quarks.

Yet theory is not the only game in town when it comes to the binary
code. We also have
, which acts on the same vector space as
. As a representation of
,
breaks up into just
two irreps: the even grades,
, which contain the left-handed
particles and antiparticles:

and the odd grades , which contain the right handed particles and antiparticles:

Unlike , the GUT really does unify with the colors , and , because they both live in the irrep .

In short, it seems that the GUT, which we built as an extension of the GUT, somehow managed to pick up this feature of the Pati-Salam model. How does relate to Pati-Salam's gauge group , exactly? In general, we only know there is a map , but in low dimensions, there is much more, because some groups coincide:

What really stands out is this:

This brings out an obvious relationship between the Pati-Salam model and the
theory, because the inclusion
lifts to the universal covers, so we get a homomorphism

A word of caution is needed here. While is the lift of an inclusion, it is not an inclusion itself: it is two-to-one. This is because the universal cover of is a four-fold cover, being a double cover on each factor.

So we can try to extend the symmetries
to
, though this can only work if the kernel of acts
trivially on the Pati-Salam representation.
What is this representation like? There is an obvious
representation of
that extends to a
rep.
Both and have Dirac spinor representations, so their
product
has a representation on
. And in fact, the obvious map

given by

is an isomorphism compatible with the actions of on these two spaces. More concisely, this square:

commutes.

We will prove this in a moment. First though, we must check that
this representation of
is secretly
just another name for the Pati-Salam representation of
on the space we discussed in
Section 3.3:

Checking this involves choosing an isomorphism between and . Luckily, we can choose one that works:

Proof. We can prove this in pieces, by separately finding
a unitary operator

that makes this square commute:

and a unitary operator

that make this square:

commute.

First, the piece. It suffices to show that the Dirac spinor rep
of
on
is isomorphic to
as a rep of
. We start with the action of on
. This breaks up
into irreps:

called the

Since these reps are dual, it suffices just to consider one of them, say .

Passing to Lie algebras, we have a homomorphism

Homomorphic images of semisimple Lie algebras are semisimple, so the image of must lie entirely in . In fact is simple, so this nontrivial map must be an injection

and because the dimension is 15 on both sides, this map is also onto. Thus is an isomorphism of Lie algebras, so is an isomorphism of the simply connected Lie groups and :

Furthermore, under this isomorphism

as a representation of . Taking duals, we obtain an isomorphism

Putting these together, we get an isomorphism that makes this square commute:

which completes the proof for .

Next, the piece. It suffices
to show that the spinor rep
of
is isomorphic to
as a rep of
. We start with the action of on
. This
again breaks up into irreps:

Again we call these representations

First consider
. Passing to Lie algebras, this gives
a homomorphism

Homomorphic images of semisimple Lie algebras are semisimple, so the image of must lie entirely in . Similarly, also takes to :

and we can combine these maps to get

which is just the derivative of 's representation on . Since this representation is faithful, the map of Lie algebras is injective. But the dimensions of and agree, so is also onto. Thus it is an isomorphism of Lie algebras. This implies that is an isomorphism of of the simply connected Lie groups and

under which acts on . The left factor of acts irreducibly on , which the second factor is trivial on. Thus as a rep of . Similarly, as a rep of this group. Putting these together, we get an isomorphism that makes this square commute:

which completes the proof for .

In the proof of the preceding theorem, we merely showed that there
exists an isomorphism

making the square commute. We did not say exactly what was. In the proof, we built it after quietly choosing three unitary operators, giving these isomorphisms:

Since the remaining map is determined by duality, these three operators determine , and they also determine the Lie group isomorphism via the construction in our proof.

There is, however, a specific choice for these unitary operators that we prefer, because this choice makes the particles in the Pati-Salam representation look almost exactly like those in the representation .

First, since
is spanned by and
, the (left-handed) isospin states of the Standard Model, we really ought
to identify the left-isospin states and of the Pati-Salam model
with these. So, we should use this unitary operator:

Next, we should use this unitary operator for right-isospin states:

Why? Because the right-isospin up particle is the right-handed neutrino , which corresponds to in the theory, but in the Pati-Salam model. This suggests that and should be identified.

Finally, because
is spanned by the colors , and , while
is spanned by the colors , , and , we really ought
to use this unitary operator:

Dualizing this, we get the unitary operator

These choices determine the unitary operator

With this specific choice of , we can combine the commutative squares built in Theorems 3 and 4:

to obtain the following result:

The map , which tells us how to identify and
, is given by applying to the `Pati-Salam code' in
Table 5. This gives a binary code for the
Pati-Salam model:

We have omitted wedge product symbols to save space. Note that if we apply the obvious isomorphism

given by

then the above table does more than merely resemble Table 4, which gives the binary code for the theory. The two tables become identical!

This fact is quite intriguing. We will explore its meaning in the next section. But first, let us start by relating the Pati-Salam model to the theory:

Proof. At the Lie algebra level, we have the inclusion

by block diagonals, which is also just the differential of the inclusion at the Lie group level. Given how the spinor reps are defined in terms of creation and annihilation operators, it is easy to see that

commutes, because is an intertwining operator between representations of . That is because the part only acts on , while the part only acts on .

But these Lie algebras act by skew-adjoint operators, so really

commutes. Since the 's and their direct sums are semisimple, so are their images. Therefore, their images live in the semisimple part of the unitary Lie algebras, which is just another way of saying the special unitary Lie algebras. We get that

commutes, and this gives a commutative square in the world of simply connected Lie groups:

This completes the proof.

This result shows us how to reach the theory, not through the theory, but through the Pati-Salam model. For physics texts that treat this issue, see for example Zee [40] and Ross [31].

2010-01-11