The True Internal Symmetry Group of the Standard Model
John C. Baez

May 19, 2003

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

$$\ISpin (3,1) \times G,$$

where $\ISpin (3,1)$ describes the geometrical symmetries of this theory -- that is, those coming from the symmetries of spacetime -- while

$$G = \SU (3) \times \SU (2) \times \U (1)$$

describes the rest of the symmetries. These other symmetries are often called internal symmetries, because we only know about them from experiments on elementary particles, so in some vague sense they describe the `inner workings' of matter and the forces of nature.

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 $\rho$ of any group $G$:

$$\rho \maps G \to \End (V)$$

we can think of it as a representation of the quotient group $G/N$ where $N$ is the kernel of $\rho$, since there is a unique representation (i.e.)

$$\tilde \rho \maps G \to \End (V)$$

making the following diagram commute:

$$% latex2html id marker 357 \begin{diagram}[\End (V)] \node... ...de{G/N} \arrow{e,t}{\tilde\rho} \node{\End (V)} \end{diagram}$$

In this situation one can argue that $G/N$ is the `true' symmetry group. After all, the representation $\rho$ sends elements of $N$ to the identity transformation of $V$, 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 $N = \ker \rho$ where $G = \SU (3) \times \SU (2) \times \U (1)$ and $\rho$ is the direct sum of all representations corresponding to elementary particles in the Standard Model. Elements of $N$ act as the identity on all particles, so $G/N$ deserves to be called the true internal symmetry group of the Standard Model.

To prepare for this problem, you need to do some math....

1. Let $\C[n]$ be the algebra of $n \times n$ complex matrices. Show that $X \in \C[n]$ commutes with all $Y \in \C[n]$ iff $X = \alpha I$ for some complex number $\alpha$.

Hint: given $X \in \C[n]$, assume that $Xe(ij) = e(ij)X$ for all $1 \le i,j \le n$, where $e(ij)$ is the elementary matrix with a $1$ in the entry lying on the $i$th row and $j$th column, and all other entries equal to $0$. Get enough equations to show $X_{ij} = \alpha \delta_{ij}$ for some $\alpha \in \C$.

2. Let $\u (n)$ be the Lie algebra of $n \times n$ skew-adjoint complex matrices. Show that $X \in \C[n]$ commutes with all $A \in \u (n)$ iff $X = \alpha I$ for some complex number $\alpha$.

Hint: Show that any matrix $Y \in \C[n]$ is of the form $A+iB$ where $A,B \in \u (n)$. Use this to reduce Problem 2 to Problem 1.

3. Let $\su (n)$ be the Lie algebra of $n \times n$ skew-adjoint traceless complex matrices. Show that $X \in \C[n]$ commutes with all $A \in \su (n)$ iff $X = \alpha I$ for some complex number $\alpha$.

Hint: Show that any $A \in \u (n)$ is of the form $A_0 + \alpha I$ where $A_0 \in \su (n)$ and $\alpha \in \C$. Use this to reduce Problem 3 to Problem 2.

4. Let $\SU (n)$ be the Lie group of $n \times n$ unitary matrices with determinant 1. Show that $X \in \C[n]$ commutes with all $U \in \SU (n)$ iff $X = \alpha I$ for some complex number $\alpha$.

Hint: We've seen that if $A_0 \in \su (n)$ then $U(t) = e^{tA_0} \in \SU (n)$ for all $t \in \R$. Show that if $X$ commutes with $U(t)$ for all $t$ then $X$ commutes with $A_0$. Use this to reduce Problem 4 to Problem 3.

5. Show that the center of $\SU (n)$ consists of all elements of $\SU (n)$ that are of the form $\alpha I$ for some complex number $\alpha$. Show the center is generated by the matrix $\exp(2\pi i/n)I$.

Hint: Use Problem 4.

Since $\exp(2\pi i/n)$ is an $n$th root of unity, it follows that that the center of $\SU (n)$ is isomorphic to $\Z_n$! It's easy to see that the center of a product of groups is the product of their centers. So, the center of $G = \SU (3) \times \SU (2) \times \U (1)$ is isomorphic to $\Z_3 \times \Z_2 \times \U (1)$. Or, for short:

$$% latex2html id marker 358 \begin{array}{cll} Z(G) &=& \Z_... ...&\subset& \SU (3) \times \SU (2) \times \U (1) = G \end{array}$$

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

To get started, note that $g \in G$ acts trivially on G precisely when $g$ is in the center of $G$. The key to seeing this is remembering that the gauge boson rep is the adjoint representation of $G$! We can think of $g \in G$ as a block diagonal $6 \times 6$ matrix

$$% latex2html id marker 359 g = \left( \begin{array}{ccc} ... ...qquad \qquad g_3 \in \SU (3), g_2 \in \SU (2), g_1 \in \U (1).$$

Similarly, we can think of $x \in \G = \su (3) \oplus \su (2) \oplus \u (1)$ as a matrix

$$% latex2html id marker 360 x = \left( \begin{array}{ccc} ... ...qquad \qquad x_3 \in \su (3), x_2 \in \su (2), x_1 \in \u (1).$$

In these terms, the adjoint representation of $G$ on G is given by conjugation:

$$% latex2html id marker 361 {\rm Ad}(g) \maps x \to gxg^{-1} .$$

By Problem 3, together with the fact that $1 \times 1$ matrices all commute, it follows that ${\rm Ad}(g)$ maps every $x \in \G$ to itself iff $g_3 \in \SU (3)$ and $g_2 \in \SU (2)$ are multiples of the identity. By Problem 5 this happens precisely when $g$ lies in the center of $G$.

It follows that $N$ must be a subgroup of the center of $G$:

$$% latex2html id marker 362 \begin{array}{cccl} N \subseteq... ...&\subset& \SU (3) \times \SU (2) \times \U (1) = G \end{array}$$

To figure out which subgroup, consider the following chart:

type of particle	action of	action of	action of
	$e^{\frac{2\pi i}{3}}I \in \SU (3)$	$-I \in \SU (2)$	$e^{\frac{\pi i}{3}}I \in \U (1)$
$(H^+,H^0)$
$(\nu_e^L,e^L)$
$\nu_e^R$
$e^R$
$(u_r^L,u_g^L,u_b^L,d_r^L,d_g^L,d_b^L)$
$(u_r^R,u_g^R,u_b^R)$
$(d_r^R,d_g^R,d_b^R)$

6. By Problem 5, the center of $\SU (3)$ is generated by the element $\exp(2\pi i/3) I$. 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 $\SU (3)$, and $\exp(2\pi i/3) I$ acts as multiplication by $1$ on this rep, so you can write down `$1$' for the Higgs.

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

7. By Problem 5, the center of $\SU (2)$ is generated by the element $\exp(2\pi i/2)I = -I$. 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 $\SU (2)$ on $\C$, and $-I$ acts as multiplication by $-1$ on this rep, so you can write down `$-1$' 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 $Z(G) = \Z_3 \times \Z_2 \times \U (1)$ that act trivially on all reps in the Standard Model, we only need to consider elements of $\U (1)$ that are sixth roots of unity. In other words, the subgroup $N \subseteq Z(G)$ must be be contained in $\Z_3 \times \Z_2 \times \Z_6$.

8. Every sixth root of unity is a power of $\exp(\pi i/3) \in \U (1)$. 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-$1$ rep of $\U (1)$. In the hypercharge-$y$ rep, each element $\alpha \in \U (1)$ acts as multiplication by $\alpha^{3y}$. Thus, for the Higgs, the element $\exp(\pi i/3) \in \U (1)$ acts as multiplication by $-1$. So, you can write down `$-1$' for the Higgs.

And now for the miracle:

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

10. Determine the group $N \subseteq \Z_3 \times \Z_2 \times \Z_6 \subset G$ 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 $N$ turns out to be crucial in setting up grand unified theories of particle physics, because while we have

$$% latex2html id marker 363 \su (3) \oplus \su (2) \oplus \u (1) \subset \su (5)$$

giving rise to the Georgi-Glashow model with $\SU (5)$ as its internal symmetry group, we do not have

$$\SU (3) \times \SU (2) \times \U (1) \subset \SU (5) .$$

Instead, we just have

$$(\SU (3) \times \SU (2) \times \U (1))/N \subset \SU (5).$$

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 $\SU (3)$ and the fact that quark charges are multiples of $\frac{1}{3}$. 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 10²⁶ and 10³⁰ years. Experiments have shown that the proton lifetime is at least 5.5 x 10³² 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.

© 2003 John Baez - all rights reserved.
baez@math.removethis.ucr.andthis.edu

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