May 19, 2003

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The symmetry group of the Standard Model is usually said
to be

where describes the

describes the rest of the symmetries. These other symmetries are often called

Although there are many theories,
we don't know why the internal symmetry group of the
Standard Model is what it is. But we *do* know more than what
I've said so far about *what it really is*. Whenever we have any
representation of any group :

we can think of it as a representation of the quotient group where is the kernel of , since there is a unique representation (i.e.)

making the following diagram commute:

In this situation one can argue that is the `true' symmetry group. After all, the representation sends elements of to the identity transformation of , so they only act as symmetries in a trivial sort of way. This means we don't lose anything by modding out by them!

The goal of this homework is to determine
where
and is
the direct sum of all representations
corresponding to elementary particles in the Standard Model.
Elements of act as the identity
on all particles, so deserves to be
called the **true internal symmetry group** of the
Standard Model.

1. Let be the algebra of complex matrices. Show that commutes with all iff for some complex number .

*Hint: given , assume that
for all
, where is
the elementary matrix with a in the entry lying
on the th row and th column, and all other entries equal
to . Get enough equations to show
for some .*

2. Let be the Lie algebra of skew-adjoint complex matrices. Show that commutes with all iff for some complex number .

*Hint: Show that any matrix is of the
form where
. Use this to reduce
Problem 2 to Problem 1.*

3. Let be the Lie algebra of skew-adjoint traceless complex matrices. Show that commutes with all iff for some complex number .

*Hint: Show that any is of the form
where
and . Use this to
reduce Problem 3 to Problem 2.*

4. Let be the Lie group of unitary matrices with determinant 1. Show that commutes with all iff for some complex number .

*Hint: We've seen that if
then
for all . Show that if commutes with for
all then commutes with . Use this to reduce
Problem 4 to Problem 3.*

5. Show that the center of consists of all elements of that are of the form for some complex number . Show the center is generated by the matrix .

*Hint: Use Problem 4.*

Since
is an th root of unity, it
follows that that the center of is isomorphic
to ! It's easy to see that the center of a product
of groups is the product of their centers. So, the center
of
is isomorphic
to
. Or, for short:

Now let us work out the subgroup consisting
of elements that act as the identity on all elementary
particles in the Standard Model.
An element will be in this subgroup
iff it acts trivially on the fermion rep , the gauge
boson rep **G**,
and the Higgs rep .

To get started, note that acts trivially on
**G**
precisely when is in the center of . The key to
seeing this is remembering that the gauge boson rep is
the adjoint representation of !
We can think of as a block diagonal matrix

Similarly, we can think of as a matrix

In these terms, the adjoint representation of on

By Problem 3, together with the fact that matrices all commute, it follows that maps every to itself iff and are multiples of the identity. By Problem 5 this happens precisely when lies in the center of .

It follows that must be a subgroup of the center of
:

type of particle | action of | action of | action of |

6. By Problem 5, the center of is generated by the element . Fill out the first column of the above chart by saying how this element acts on each irrep appearing in the Higgs and fermion reps. In each case this element acts as multiplication by some number, so just write down this number. For example, the Higgs boson lives in the trivial rep of , and acts as multiplication by on this rep, so you can write down `' for the Higgs.

(We have not listed the fermions of the second and third generations. Since these transform in the same representations of as the fermions of the first generation, they are irrelevant to the problem of finding the group .)

7. By Problem 5, the center of is generated by the element . Fill out the second column of the above chart by saying how this element acts on each irrep in the Higgs and fermion reps. In each case this element acts as multiplication by some number, so just write down this number. For example, the Higgs boson lives in the defining rep of on , and acts as multiplication by on this rep, so you can write down `' for the Higgs.

If you do Problems 6 and 7 correctly, you should see that every number in the first two columns is a sixth root of unity. So, to find elements of that act trivially on all reps in the Standard Model, we only need to consider elements of that are sixth roots of unity. In other words, the subgroup must be be contained in .

8. Every sixth root of unity is a power of . So, fill out the third column of the above chart by saying how this element acts on each irrep in the Higgs and fermion reps. In each case this element acts as multiplication by some number, so just write down this number. For example, the Higgs boson lives in the hypercharge- rep of . In the hypercharge- rep, each element acts as multiplication by . Thus, for the Higgs, the element acts as multiplication by . So, you can write down `' for the Higgs.

9. What do you get when you multiply all 3 numbers in any row of the above chart?

10. Determine the group consisting of all elements that act as the identity on the Higgs and fermion reps.

*Hint: Problem 9 is an incredibly important clue.*

The precise nature of the subgroup turns out to be
crucial in setting up grand unified theories of particle physics, because
while we have

giving rise to the Georgi-Glashow model with as its internal symmetry group, we do

Instead, we just have

It's the miracle in Problem 9 that makes this possible. If you think about it, you'll see this miracle relies on the the `coincidence' between the 3 in and the fact that quark charges are multiples of . So, people often say the Georgi-Glashow model `explains' why quarks have fractional charge.

Unfortunately, this model predicts that a proton will eventually decay
into a positron and neutral pion,
with the mean lifetime of the proton
being somewhere between 10^{26} and 10^{30} years.
Experiments
have shown that the proton lifetime is at least 5.5 x 10^{32}
years -- no proton decay has ever been seen.
So, people don't believe in the Georgi-Glashow model.
Another beautiful theory killed by an ugly fact! But, this model
serves as a basis for most other grand unified theories, so it's
worth understanding.

** Next:** The
Eightfold Way
** Previous:** Hypercharge and Weak Isospin

© 2003 John Baez - all rights reserved.

baez@math.removethis.ucr.andthis.edu