# 4 Conclusion

We have studied three different grand unified theories: the , and theories. The and theories were based on different visions about how to extend the Standard Model. However, we saw that both of these theories can be extended to the theory, which therefore unites these visions.

The theory is all about treating isospin and color on an equal footing: it combines the two isospins of with the three colors of , and posits an symmetry acting on the resulting . The particles and antiparticles in a single generation of fermions are described by vectors in . So, we can describe each of these particles and antiparticles by a binary code indicating the presence or absence of up, down, red, green and blue.

In doing so, the theory introduces unexpected relationships between matter and antimatter. The irreducible representations of

unify some particles we normally consider to be matter' with some we normally consider antimatter', as in

In the Standard Model representation, we can think of the matter-antimatter distinction as a -grading, because the Standard Model representation splits into and . By failing to respect this grading, the symmetry group fails to preserve the usual distinction between matter and antimatter.

But the Standard Model has another -grading that does respect. This is the distinction between left- and right-handedness. Remember, the left-handed particles and antiparticles live in the even grades:

while the right-handed ones live in the odd grades:

The action of automatically preserves this -grading, because it comes from the -grading on , which already respects.

This characteristic of the theory lives on in its extension to . There, the distinction between left and right is the only distinction among particles and antiparticles that knows about, because and are irreducible. This says the theory unifies all left-handed particles and antiparticles, and all right-handed particles and antiparticles.

In contrast, the theory was all about adding a fourth color', , to represent leptons, and restoring a kind of symmetry between left and right by introducing a right-handed that treats right-handed particles like the left-handed treats left-handed particles.

Unlike the theory, the theory respects both -gradings in the Standard Model: the matter-antimatter grading, and the right-left grading. The reason is that respects the -grading on , and we have:

Moreover, the matter-antimatter grading and the right-left grading are all that respects, since each of the four spaces listed is an irrep of this group.

When we extend to the theory, we identify with . Then the -grading on gives the -grading on using addition in . This sounds rather technical, but it is as simple as even + odd = odd'':

and odd + odd = even'', even + even = even'':

Recall that consists of all the fermions and antifermions that are right-handed, while consists of the left-handed ones.

Furthermore, the two routes to the theory that we have described, one going through and the other through , are compatible. In other words, this cube commutes:

So, all four theories fit together in an elegant algebraic pattern. What this means for physics--if anything--remains unknown. Yet we cannot resist feeling that it means something, and we cannot resist venturing a guess: the Standard Model is exactly the theory that reconciles the visions built into the and theories.

What this might mean is not yet precise, but since all these theories involve symmetries and representations, the reconciliation' must take place at both those levels--and we can see this in a precise way. First, at the level of symmetries, our Lie groups are related by the commutative square of homomorphisms:

Because this commutes, the image of lies in the intersection of the images of and inside . But we claim it is precisely that intersection!

To see this, first recall that the image of a group under a homomorphism is just the quotient group formed by modding out the kernel of that homomorphism. If we do this for each of our homomorphisms above, we get a commutative square of inclusions:

This implies that

as subgroups of . To make good on our claim, we must show these subgroups are equal:

In other words, our commutative square of inclusions is a pullback square'.

As a step towards showing this, first consider what happens when we pass from the spin groups to the rotation groups. We can accomplish this by modding out by an additional above. We get another commutative square of inclusions:

Here the reader may wonder why we could quotient and by without having to do the same for their respective subgroups, and . It is because intersects both of those subgroups trivially. We can see this for because we know the inclusion is just the lift of the inclusion to universal covers, so it makes this diagram commute:

But this means that intersects in only the identity. The subgroup therefore intersects trivially as well.

Now, let us show:

Theorem 8   .

Proof. We can prove this in the same manner that we showed, in Section 3.1, that

is precisely the subgroup of that preserves the splitting of .

To begin with, the group is the group of orientation-preserving symmetries of the 10-dimensional real inner product space . But is suspiciously like , a 5-dimensional complex inner product space. Indeed, if we forget the complex structure on , we get an isomorphism , a real inner product space with symmetries . We can consider the subgroup of that preserves the original complex structure. This is . If we further pick a volume form on , i.e. a nonzero element of , and look at the symmetries fixing that volume form, we get a copy of .

Then we can pick a splitting on . The subgroup of that also preserves this is

These inclusions form the top and right sides of our square:

We can also reverse the order of these processes. Imposing a splitting on yields a splitting on the underlying real vector space, . The subgroup of that preserves this splitting is : the block diagonal matrices with and orthogonal blocks and overall determinant 1. The connected component of this subgroup is .

The direct summands in came from forgetting the complex structure on . The subgroup of preserving the original complex structure is , and the subgroup of this that also fixes a volume form on is . This group is connected, so it must lie entirely in the connected component of the identity, and we get the inclusions:

These maps form the left and bottom sides of our square.

It follows that is precisely the subgroup of that preserves a complex structure on , a chosen volume form on the resulting complex vector space, and a splitting on this space. But this splitting is the same as a compatible splitting of , one in which each summand is a complex vector subspace as well as a real subspace. This means that

and since is connected,

as desired.

From this, a little diagram chase proves our earlier claim:

Theorem 9   . .

Proof. By now we have built the following commutative diagram:

where both the bottom vertical arrows are two-to-one, but the composite vertical maps and are one-to-one. Our previous theorem says that the big square with and as vertical sides is a pullback. Now we must show that the upper square is also a pullback. So, suppose we are given and with

We need to show there exists such that

Now, we know that

so since the big square is a pullback, there exists with

The second equation is half of what we need to show. So, we only need to check that the first equation implies .

The kernel of consists of two elements, which we will simply call . Since , we know

Since , we thus have

The one-to-one map sends the kernel of to the kernel of , which consists of two elements that we may again call . So, . On the other hand, since the top square commutes we know . Thus the element must actually be , so as desired.

In short, the Standard Model has precisely the symmetries shared by both the theory and the theory. Now let us see what this means for the Standard Model representation.

We can break the symmetry' of the theory in two different ways. In the first way, we start by picking the subgroup of that preserves the -grading and volume form in . This is . Then we pick the subgroup of that respects the splitting of into . This subgroup is the Standard Model gauge group, modulo a discrete subgroup, and its representation on is the Standard Model representation.

We can draw this symmetry breaking process in the following diagram:

The theory shows up as a `halfway house' here.

We can also break the symmetry of in a way that uses the theory as a halfway house. We do essentially the same two steps as before, but in the reverse order! This time we start by picking the subgroup of that respects the splitting of as . This subgroup is modulo a discrete subgroup. The two factors in this subgroup act separately on the factors of . Then we pick the subgroup of that respects the -grading and volume form on . This subgroup is the Standard Model gauge group, modulo a discrete subgroup, and its representation on is the Standard Model representation.

We can draw this alternate symmetry breaking process in the following diagram:

Will these tantalizing patterns help us understand physics beyond the Standard Model? Only time will tell.

Subsections
2010-01-11