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\documentclass{article} 
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	\begin{document}   
 
	\begin{center}   
	{\bf The True Internal Symmetry Group of the Standard Model \\}   
	{\small John\ C.\ Baez, May 19 2003\\}   
	\vspace{0.3cm}   
        \end{center}

The symmetry group of the Standard Model is usually said
to be
\[     \ISpin(3,1) \times G,  \]
where $\ISpin(3,1)$ describes the {\bf 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 usually called {\bf 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 {\it do} know more than what
I've said so far about {\it 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.\ homomorphism) 
\[          \tilde \rho \maps G/N \to \End(V) \]
making the following diagram commute:
\[  
\begin{diagram}[\End(V)]
\node{G} \arrow{s} \arrow{se,t}{\rho}    \\
\node{G/N}  \arrow{e,t}{\tilde\rho}  \node{\End(V)}  
\end{diagram} 
\]
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 {\bf true internal symmetry group} of the
Standard Model.   

\vskip 2em
\centerline{\it To prepare for this problem, you need to do some math....}
\vskip 1em
\noindent
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
$\alpha$.  

\vskip 1em
\noindent
{\it 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 {\bf 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 $\alpha \in \C$.}

\vskip 1em
\noindent
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 $\alpha$.

\vskip 1em
\noindent
{\it 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.}

\vskip 1em
\noindent
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 $\alpha$.

\vskip 1em
\noindent
{\it 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.}

\vskip 1em
\noindent
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 $\alpha$.

\vskip 1em
\noindent
{\it 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.}

\vskip 1em
\noindent
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 $\exp(2\pi i/n)I$.

\vskip 1em
\noindent
{\it Hint: Use Problem 4.}

\vskip 2em
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:
\[ \begin{array}{cll}
        Z(G) &=& \Z_3 \times \Z_2 \times \U(1) \\
             &\subset& \SU(3) \times \SU(2) \times \U(1) = G 
\end{array}
\]

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
\[          g = 
\left( \begin{array}{ccc}
        g_3 & 0 & 0  \\
        0 & g_2 & 0  \\
        0 & 0   & g_1 \\
         \end{array} \right)  
\qquad \qquad g_3 \in \SU(3), g_2 \in \SU(2), g_1 \in \U(1). \]
Similarly, we can think of $x \in \G = \su(3) \oplus \su(2) \oplus 
\u(1)$ as a matrix
\[          x = 
\left( \begin{array}{ccc}
        x_3 & 0 & 0  \\
        0 & x_2 & 0  \\
        0 & 0   & x_1 \\
         \end{array} \right)  
\qquad \qquad x_3 \in \su(3), x_2 \in \su(2), x_1 \in \u(1). \]
In these terms, the adjoint representation of $G$ on $\G$ is
given by conjugation:
\[          {\rm Ad}(g) \maps x \to gxg^{-1} . \]
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
$G$:
\[ \begin{array}{cccl}
      N \subseteq & Z(G) &=& \Z_3 \times \Z_2 \times \U(1) \\
                  &  &\subset& \SU(3) \times \SU(2) \times \U(1) = G 
\end{array}
\]
\vskip 2em
\centerline{\it 
To figure out which subgroup, consider the following
chart:}
\vskip 1em

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

\vskip 1em
\noindent
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 $N$.)

\vskip 1em
\noindent   
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.

\vskip 2em
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 $\Z_3 \times \Z_2 \times
\Z_6$.  

\vskip2em
\noindent
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.

\vskip 2em
\centerline{\it And now for the miracle:}
\vskip 2em
\noindent
9.   What do you get when you multiply all 3 numbers in any row
of the above chart?   

\vskip 1em
\noindent
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.  

\vskip 1em
\noindent
{\it Hint: Problem 9 is an incredibly important clue.}

\vskip 2em
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
\[         \su(3) \oplus \su(2) \oplus \u(1) \subset \su(5) \]
giving rise to the Georgi--Glashow model with $\SU(5)$ as its
internal symmetry group, we do {\it 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 proton decay of the form
$p \to \overline{e} \pi^0$, 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 $5.5 \times 10^{32}$
years.  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.


\vfill
\eject
\end{document}

 












