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\begin{document}

\title{Representations of Lie Algebras, with Applications to Particle Physics}


\author{James Marrone\\
University of Chicago Mathematics REU, August 2007}

\begin{abstract}
The structure of Lie groups and the classification of their representations
are subjects undertaken by many an author of mathematics textbooks;
Lie algebras are always considered as an indispensable component of
such a study. Yet every author approaches Lie algebras differently
- some begin axiomatically, some derive the algebra from prior principles,
and some barely connect them to Lie groups at all. The ensuing problem
for the student is that the importance of the Lie algebra can only
be deduced by {}``reading between the lines,'' so to speak; the
relationship between a Lie algebra and a Lie group is not always illuminated
(in fact, there isn't always a relationship in the first place) and
so considering the representations of Lie algebras may appear redundant
or at best cloudy. The approach to Lie algebras in this paper should
clarify such problems. Lie algebras are historically a consequence
of Lie groups (namely, they are the tangent space of a Lie group at
the identity) which may be axiomatized so that they stand alone as
an algebraic concept. The important correspondence between representations
of Lie algebras and Lie groups, however, makes Lie algebras indispensable
to the study of Lie groups; one example of this is the Eight-Fold
Way of particle physics, which is actually just an 8-dimensional representation
of $\mathfrak{sl}_{3}(\mathbb{C})$ but which has the enlightening
property of corresponding to the strong nuclear interaction, giving
a physical manifestation of the action of a representation and showing
just one of the many interesting ramifications of mathematics on science
within the past 50 years.
\end{abstract}
\maketitle

\section{Lie Groups and Representations}

We begin with

\begin{defn}
A \emph{Lie group} is a group such that the map $\omega:G\times G\rightarrow G$
sending $(x,y)\mapsto xy^{-1}$ is $\mathit{C^{\infty}}$.
\end{defn}
The important consequence of this is that a Lie group is endowed with
a manifold structure. Without looking very far, we can easily find
Lie groups - just take groups of non-singular \emph{n $\times$n }matrices
over a complete metric space, such as $\mathbb{R}$ or $\mathbb{C}$.
Then this is locally equivalent to working in the finite-dimensional
space $\mathbb{R^{\mathrm{\mathit{m}}}}$ or $\mathbb{C\mathrm{^{\mathit{m}}}}$,
depending on the number of degrees of freedom characterizing the matrix
group. For example, $\mathrm{\mathit{SL_{n}}(\mathbb{R\mathrm{)}}}$,
the group of non-singular real matrices with determinant 1, has dimension
$\mathrm{\mathit{n}^{2}-1}$. These are elementary facts from linear
algebra, and we might create a complete list of all the classical
groups $\mathrm{\mathit{SO_{n}}(\mathbb{R})}$, $\mathrm{\mathit{SU_{n}}(\mathbb{R})}$,
$\mathrm{\mathit{Sp_{p,q}}(\mathbb{R\mathrm{)}}}$ and the corresponding
complex Lie groups with their corresponding dimensions. To see how
these might have a manifold structure, consider the map above as a
continous map on a vector space. For example, consider any matrix
$\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right)\in\mathrm{\mathit{SL_{2}}(\mathbb{R\mathrm{)}}}$. Since $ad-cb=1$, we have 3 free parameters but the last element
is then fixed by the other three in order to get determinant 1. This
group consists of many 'components,' one of which is the set of matrices
$\left(\begin{array}{cc}
a & b\\
c & \frac{1+cb}{a}\end{array}\right)$ such that $\mathit{a}\neq0$. We can map such a matrix to the element
$\left(a,b,c,\frac{1+cb}{a}\right)\in\mathbb{R}^{4}$. This defines
a subspace of $\mathbb{R}^{4}$ that gives the local structure that
defines a manifold; to get differentiability, the map $\omega:(x,y)\mapsto xy^{-1}$
can be defined by

\[
\left(\left(\begin{array}{cc}
a & b\\
c & \frac{1+cb}{a}\end{array}\right),\left(\begin{array}{cc}
e & f\\
g & \frac{1+fg}{e}\end{array}\right)\right)\mapsto\left(\begin{array}{cc}
a & b\\
c & \frac{1+cb}{a}\end{array}\right)\left(\begin{array}{cc}
\frac{1+fg}{e} & -f\\
-g & e\end{array}\right)=\left(\begin{array}{cc}
\frac{a+afg}{e}-bg & \frac{c+cfg}{e}-\frac{g+gcb}{a}\\
be-af & \frac{e+ecb}{a}-cf\end{array}\right)\]


corresponding to the map

\[
\omega_{\mathbb{R}}:\left((a,b,c,\frac{1+cb}{a}),(e,f,g,\frac{1+fg}{e})\right)\mapsto(\frac{a+afg}{e}-bg,\frac{c+cfg}{e}-\frac{g+gcb}{a},be-af,\frac{e+ecb}{a}-cf).\]


Since any of the six parameters $a,b,c,e,f,g$ can be continuously
varied in the reals, we can take $\frac{\partial\omega_{\mathbb{R}}}{\partial x_{i}}$
for $\mathit{x_{i}}=a,b,c,$ etc. Hence $\frac{\partial\omega_{\mathbb{R}}}{\partial c}=(0,\frac{1+fg}{e}-\frac{gb}{a},0,\frac{eb}{a}-f)$
and so forth, and we have a smooth map on the group. This map gives
a manifold structure because it provides a local coordinatization
for each element by multiplying The next important concept is

\begin{defn}
A \emph{representation} of a group $\mathit{G}$ is a smooth group
homomorphism $\varphi:G\rightarrow GL_{n}(V)$ for some finite-dimensional
vector space $\mathit{V}$.
\end{defn}
Hence, a representation assigns to each $\mathit{g\in G}$ a linear
map $\varphi_{g}:V\rightarrow V$, and one can say for short that
{}``$\mathit{V}$is a representation of $\mathit{G}$.'' Given $\mathit{V}$
a finite-dimensional vector space over a field $\mathit{F}$, it is
clear that if $\mathit{V}$ is an $\mathit{FG}$-module, $\mathit{V}$
is also a representation of $\mathit{G}$ since we can define $\varphi_{g}(v)=g\cdot v$.
Conversely, if $\mathit{V}$ is a representation of $\mathit{G}$
we can turn it into an $\mathit{FG}$-module by defining

\begin{center}
$\left(\sum\alpha_{i}g_{i}\right)\cdot v=\sum\alpha_{i}\varphi_{g_{i}}(v)$
\par\end{center}

\noindent so that there is a one-to-one correspondence between finite-dimensional
representations of $\mathit{G}$ and $\mathit{FG}$-modules. This
is useful in creating a conceptual basis for the idea of a representation,
especially since finite-dimensional representations are the only ones
in which we will be interested. Without going into too many details,
two important examples are stated, as these will reappear later:

\begin{example}
(The Standard Representation): Let $\mathit{G}$ be a permutation
group, i.e. $\mathit{G\cong S_{n}}$ for some $\textit{n}$. Then
$\mathit{g\in G}$ acts on elements of $\mathit{F^{n}}$: let $\mathit{e_{1},e_{2},\ldots,e_{n}}$
be a basis; then $\mathit{g\cdot e_{i}=e_{g(i)}}$. Then the vector
space is a representation of $\mathit{G}$, as each element of the
group is associated with a linear map on the vector space.
\end{example}
-

\begin{example}
(The Dual Representation): If $\mathit{V}$ is an arbitrary vector
space, take $\psi:V\rightarrow V$ with its dual $\psi^{*}:V^{*}\rightarrow V^{*}$
defined by $\psi^{*}(\rho)=\rho\circ\psi$ in the normal way for a
vector space. In matrix form, $\psi^{*}$ is $\mathit{^{t}\psi}$
and since $\mathit{^{t}(\psi\cdot\phi)=^{t}\phi\cdot^{t}\psi}$ for
any two matrices $\psi$, $\phi$, we have $(\psi\cdot\phi)^{*}(\rho)=\phi^{*}(\psi^{*}(\rho))$.
If $\mathit{V}$ is a representation of a group $\mathit{G}$, we
have $\varphi_{g}$for each $\mathit{g\in G}$and we see $\varphi_{g\cdot h}^{*}=\varphi_{h}^{*}\circ\varphi_{g}^{*}$so
this is an anti-homomorphism. By defining, in the case of group representations,
$\varphi_{g}^{*}=^{t}\varphi_{g^{-1}}$, we get
\end{example}
\begin{center}
$\varphi_{g\cdot h}^{*}=^{t}\varphi_{(g\cdot h)^{-1}}=^{t}\left(\varphi_{h^{-1}}\cdot\varphi_{g^{-1}}\right)=^{t}\varphi_{g^{-1}}\cdot^{t}\varphi_{h^{-1}}=\varphi_{g}^{*}\circ\varphi_{h}^{*}$
\par\end{center}

\noindent as desired. We have thus described $\mathit{V^{*}}$as a
representation of $\mathit{G}$.

We now turn away from the group structure of a Lie group and explore
the implications of its manifold structure.


\section{Lie Algebras}


\subsection{The Tangent Space at the Identity}

Since a Lie group is a finite-dimensional manifold, given a point
$\mathit{g_{o}}$ on its surface we may find the tangent space at
$\mathit{g_{o}}$. That is, for $\mathfrak{\mathit{G}}$ a Lie group
we have a space of all analytic functions (not necessarily linear)
$\mathfrak{\mathit{G}\rightarrow\mathit{G}}$. We may take the partial
derivatives of any of these functions at our specified point $\mathit{g_{o}}$.
We do this by fixing a chart $(\mathit{U},\varphi)$ and considering
the space of operators $\left\{ \xi^{i}\frac{\partial}{\partial x_{i}}\right\} $,
applying them to analytic functions at $\mathit{g_{o}}$. The direction
of the partial derivative is determined by $\xi^{i}$ (or, specifically,
the ratio of $\xi^{i}$ to $\xi^{j}$ for various coordinates $\mathit{i,}j$),
and the space of such operators analyzed at $\mathit{g_{o}}$is the
tangent space at $\mathit{g_{o}}$, $\mathfrak{\mathit{T}_{\mathit{g_{o}}}\mathit{G}}$.
If we have a tangent vector at each point of the manifold, we have
a vector field. We wish to invest the space of analytic vector fields
$A$ with the structure of a vector space, like its corresponding
Lie group $\mathit{G}$, and so we need to have a multiplication.
Merely taking two tangent vectors $\xi^{i}\frac{\partial}{\partial x_{i}}$
and $\eta^{i}\frac{\partial}{\partial x_{i}}$ at a point $\mathit{g_{o}}$
and composing them as operators yields

\[
\xi^{i}\frac{\partial}{\partial x_{i}}\left(\eta^{j}\frac{\partial}{\partial x_{j}}\right)=\xi^{i}\eta^{j}\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}+\xi^{i}\frac{\partial\eta^{j}}{\partial x_{i}}\frac{\partial}{\partial x_{j}}\]


which is not a tangent vector because of the double partial derivative.
However, if we take two such operators as before, we see that

\[
\left(\xi^{i}\frac{\partial}{\partial x_{i}}\right)\left(\eta^{j}\frac{\partial}{\partial x_{j}}\right)-\left(\eta^{j}\frac{\partial}{\partial x_{j}}\right)\left(\xi^{i}\frac{\partial}{\partial x_{i}}\right)=\left(\xi^{i}\frac{\partial\eta^{j}}{\partial x_{j}}-\eta^{j}\frac{\partial\xi^{i}}{\partial x_{j}}\right)\frac{\partial}{\partial x_{i}}\]


which is itself another tangent vector. Hence, we can define multiplication
of two vector fields $\mathsf{X,Y}\in\mathfrak{\mathit{A}}$ to be
$\mathsf{\left[X,Y\right]=XY-YX}$, which is also known as the \emph{commutator}.

This is a repetition of results in differential geometry, but there
are some useful results for Lie groups. In particular, as with any
group, we consider the action of a Lie group on itself by right (or
left) multiplication, i.e. for $\mathit{g\in G}$ consider the map
$\mathit{\Phi_{g}(g')=g\cdot g'}$. Any homomorphism of groups $\rho:\mathit{G\rightarrow H}$
must preserve this action, since

\[
\rho(g\cdot g')=\rho(g)\cdot\rho(g')=\rho(g)\cdot\rho(g')=\Phi_{\rho(g)}(g').\]


That a map respects multiplication on the right or left is equivalent
to saying that it is a homomorphism. Hence, finding group representations
amounts in some sense to finding maps between groups that are compatible
with right multiplication. Since right multiplication is everywhere
continuous, we can take its differential at some point $\mathit{g_{o}}$.
Let $\mathit{X=\xi^{i}\frac{\partial}{\partial x_{i}}}$ be a tangent
vector at $\mathit{g_{o}}$ and consider $\left.\mathit{X\Phi_{g}(v)=\xi^{i}\frac{\partial(\varphi\circ\Phi_{g})(v)}{\partial v_{i}}}\right|_{v=g_{o}}$.
Letting $\left.\frac{\partial((\varphi\circ\Phi_{g})(v))_{k}}{\partial v_{i}}\right|_{v=g_{o}}=\Phi_{g}^{k,i}(g_{o})$,
we can define, for an arbitrary analytic function $\mathit{f}$, $\mathit{X_{g}f(v)=X(f\circ\Phi_{g})(v)=Xf(g\cdot v)}$
which gives

\[
\mathit{\left.\xi^{i}\frac{\partial f(g\cdot v)}{\partial v_{i}}\right|_{v=g_{o}}=\xi^{i}\left[\frac{\partial f(g\cdot v)}{\partial(g\cdot v)_{k}}\frac{\partial(g\cdot v)_{k}}{\partial v_{i}}\right]_{v=g_{o}}}=\xi^{i}\Phi_{g}^{k,i}(g_{o})\left.\frac{\partial f(g\cdot v)}{\partial(g\cdot v)_{k}}\right|_{v=g_{o}}\]


which as a tangent vector $\xi^{i}\Phi_{g}^{k,i}(g_{o})\frac{\partial}{\partial(g\cdot v)}$
is an element an element of $\mathfrak{\mathit{T}\mathit{_{g\cdot g_{o}}G}}$.
We get a function between tangent spaces defined by this {}``infinitesimal''
right translation. We can map any tangent space $\mathfrak{\mathit{T}\mathit{_{g_{o}}G}}$to
$\mathfrak{\mathit{T}\mathit{_{e}G}}$ by defining a map $\omega$
such that

\[
\omega(X)=X_{g_{o}^{-1}}\in\mathfrak{\mathit{T}\mathit{_{g_{o}\cdot g_{o}^{-1}}G=\mathfrak{\mathit{T}\mathit{_{e}G}}}}\]


so that we have {}``pulled back'' the tangent vector to lie in the
tangent space to the identity. Suppose $\mathit{\mathsf{X}}$ were
a right-invariant vector field; then $\mathit{\omega}$ must map every
tangent vector to the same element, namely to $\mathit{\mathsf{X}(e)}$.
But given a tangent vector at the identity, we can construct a right-invariant
vector field by right-translating it, so these vector fields are in
one-to-one correspondence with elements of $\mathfrak{\mathit{T}\mathit{_{e}}\mathit{G}}$.
Further, the map $\omega$ preserves the commutator of two right-invariant
vector fields:

\[
\omega(\left[\mathsf{X,Y}\right]_{i})=\omega(\mathsf{X}_{i}\frac{\partial\mathsf{Y}_{j}}{\partial x_{i}}-\mathsf{Y}_{i}\frac{\partial\mathsf{X}_{j}}{\partial x_{i}})=\mathsf{X}(e)\frac{\partial\mathsf{Y}(e)_{j}}{\partial x_{i}}-\mathsf{Y}(e)_{i}\frac{\partial\mathsf{X}(e)_{j}}{\partial x_{i}}=[\omega(\mathsf{X}),\omega(\mathsf{Y})]_{i}\]
.

We have proved:

\begin{prop}
The tangent space at the identity $\mathit{T_{e}G}$ is isomorphic
to the space of right-invariant vector fields.
\end{prop}
To see that $T_{e}G$ encodes many desired properties of $\mathit{G}$,
we will use this isomorphism to move from elements of the tangent
space to the manifold itself.


\subsection{The Exponential Map}

Given $\mathit{X\in T_{e}G}$ let $\mathsf{X}$ be the associated
right-invariant vector field, i.e. $\mathsf{X}(e)=X$. Then the results
of differential equations allow us to integrate over $\mathsf{X}$,
getting a map $\phi_{X}:U\subset\mathbb{R}\rightarrow G$ such that
$\phi_{X}(0)=g_{o}$ for some chosen point $\mathit{g_{o}}$ and $\phi_{X}'(t)=\mathsf{X}(\phi(t))$
for $\mathit{t\in U}$, where $\mathit{U}$ is some open ball around
the origin. Essentially, we are solving the differential equation

\[
\frac{d\phi_{X}}{dt}=\mathsf{X}(\phi_{X}(t))\]


given $\phi_{X}(0)=g_{o}$, so we are using the principle of least
action and {}``following the tangent vectors.'' By the invariance
of $\mathsf{X}$ and the definition of $\phi_{X}$ it follows that
$\phi_{X}(s+t)=\phi_{X}(s)\phi_{X}(t)$ so $\phi_{X}$ is a homomorphism:
Fix $\mathit{s}$ and define $\alpha_{X}(t)=\phi_{X}(s+t)$, $\beta_{X}(t)=\phi_{X}(s)\phi_{X}(t)$;
take the differential of both sides to see that, by invariance, $\mathsf{X}(\alpha(t))=\alpha'(t)$
and $\mathsf{X}(\beta(t))=\beta'(t)$. Since $\alpha(0)=\beta(0)$
their differentials must be the same at the origin, so by the uniqueness
of the integral curve they are equal for all $\mathit{t}$. Then $\phi_{X}$
is a homomorphism of Lie groups and since it is defined around the
origin, it extends uniquely to all of $\mathbb{R\mathit{^{n}}}$.
As there is one such map for each tangent vector $\mathit{X}$ , one
might assume that they will somehow be useful. They do get their own
special name:

\begin{defn}
The homomorphisms $\phi_{X}:\mathbb{R}\rightarrow G$ are the \emph{one-parameter
subgroups of G}.
\end{defn}
Now let $\mathit{g_{o}=e}$ and define a map that will take elements
of $\mathit{T_{e}G}$ to the manifold:

\begin{defn}
The \emph{exponential map} of a Lie group $\mathit{G}$ is the map
sending elements $\mathit{X}\in T_{e}G$ to the corresponding one-parameter
subgroup of $G$.

\[
\mathrm{exp:}X\mapsto\phi_{X}(1).\]

\end{defn}
If the Lie group is a real or complex matrix group, the 'usual' properties
of the exponential hold (see below). However, for any group, the exp
map gives us the useful result

\begin{prop}
The map $\exp$ is natural, i.e. for any homomorphism of groups $\Psi$with
differential $\psi:T_{e}G\rightarrow T_{e}G$ the following diagram
commutes:

\[
\begin{array}{ccc}
G & {\Psi\atop \longrightarrow} & H\\
{\scriptstyle exp}\uparrow &  & {\scriptstyle exp}\uparrow\\
T_{e_{G}}G & {\psi\atop \longrightarrow} & T_{e_{H}}H\end{array}.\]

\end{prop}
\begin{proof}
Let $\mathit{X\in T_{e_{G}}G}$ and take $\phi_{X}$; then $\Psi\circ\phi_{X}:\mathbb{R}\rightarrow H$
is the one-parameter subgroup corresponding to $\psi(X)\in T_{e_{H}}H$,
so by definition $\exp(\psi(X))=(\Psi\circ\phi_{X})(1)=\Psi\circ\exp(X)$.
\end{proof}
Note that this means $\exp$ lifts the tangent space to give a neighborhood
of the identity in $G$; if $G$ is connected then we generate the
whole group from this neighborhood. The connectedness of Lie groups
will be important for the construction of the Lie algebra (see Theorem
1 below). If our Lie group is indeed a real or complex matrix group,
the exponential can be written

\[
\mathrm{exp(}X)=I+X+\frac{X^{2}}{2}+\frac{X^{3}}{6}+\cdots\]


as with the 'usual' definition of exponential.%
\footnote{Since $\mathbb{R}$ is itself a Lie group, the 'usual' exponential
might be seen as a special case of the map described here.%
} This will always converge; if we are dealing with an $n\times n$
matrix $\mathit{X}$ whose largest value is $m$ then the elements
of $\mathit{X^{i}}$ will be at most $\left((n^{i-1}m^{i})\right)$
and since $m\sum_{i=1}^{\infty}\frac{(nm)^{i-1}}{i!}$ will converge
for any number $\mathit{nm}$, the infinite sum $\mathit{I+X+\frac{X^{2}}{2}+\cdots}$
will always converge. One can additionally check that if $\left[X,Y\right]=0$
then $\exp(X)\exp(Y)=\exp(X+Y)$ on a sufficiently small neighborhood
of the identity. Because $\exp$ is injective on a neighborhood of
the identity, we can introduce an inverse on this neighborhood. It
will map from a subset of $\mathit{GL_{n}}(\mathbb{R})$ to the tangent
space $\mathit{T_{e}GL_{n}}(\mathbb{R})$:

\begin{defn}
Define on a neighborhood of $e\in GL_{n}(\mathbb{R})$ the map

\[
\log:g\mapsto(g-I)-\left(\frac{(g-I)^{2}}{2}\right)+\left(\frac{(g-I)^{3}}{6}\right)-\cdots\]


and on a neighborhood of 0 in $T_{e}GL_{n}(\mathbb{R})$ the binary
operation $*$ such that

\[
\mathit{X*Y=\log\mathrm{(exp(\mathit{X})\cdot exp(}Y})).\]


This is the \emph{Campbell-Hausdorff Formula}.
\end{defn}
Careful computation gives the first few terms of the Campbell-Hausdorff
formula:

\[
\mathit{X*Y}=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]+\cdots\]


The Campbell-Hausdorff formula provides a way to expand the binary
operation above without the use of exp or log, i.e. it relies only
on the algebraic relations of the elements of the tangent space. We
observe that this series always converges, so that $\exp(X)\cdot\exp(Y)$
must always be in the Lie group if $X,Y$ are in the tangent space.
The formula shows how $\mathit{T_{e}G}$ encodes the local group structure
of the manifold while simplifying the topology via a {}``linearization.''

As we said earlier, finding homomorphisms of groups is our main goal,
and we employed right translation to aid us in our search. Before
we can completely characterize homomorphisms of Lie groups, we prove
a lemma.

\begin{lem}
Let $\mathit{G}$ be a real matrix Lie group with $\mathit{\mathfrak{h}\leq T_{e}G}$
a subspace of the tangent space at the identity of $\mathit{G}$.
Then the subgroup of $\mathit{G}$ obtained via $\exp(\mathfrak{h})$
is an immersed%
\footnote{The term \emph{immersed} borrows from \cite{key-1} and corresponds
to what is normally termed a Lie subgroup, not necessarily closed.%
} subgroup $\mathit{H}$ with tangent space $\mathit{T_{e}H=\mathfrak{h}}$.
\end{lem}
\begin{proof}
Let $\mathit{h\in H}$ and consider the map $\rho_{g}:H\rightarrow T_{e}H$
taking $\mathit{y}\mapsto\log(g^{-1}\cdot y)$. This gives an atlas
of local coordinatization of elements of $\mathit{H}$ because log
is defined only on a neighborhood of the identity. Hence only those
elements for which $\mathit{g^{-1}}$ is sufficiently near $\mathit{y}$
will map under $\rho_{g}$ to the tangent space, and so we have identified
a neighborhood of $y$ in $G$.
\end{proof}
Now we can prove the final result, showing a correspondence between
homomorphisms of Lie groups and homomorphisms of their tangent spaces
at the identity.

\begin{thm}
Let G, H be Lie groups with G simply connected. Let $\mathit{T_{e_{G}}G},T_{e_{H}}H$
be the corresponding tangent spaces. Then $\Psi:G\rightarrow H$ is
a (differentiable) group homomorphism iff $d\Psi_{e_{G}}=\psi:T_{e_{G}}G\rightarrow T_{e_{H}}H$
is a differentiable homomorphism of the tangent spaces (that is, it
preserves the commutator).
\end{thm}
\begin{proof}
Consider the Lie group $G\times H$, which has tangent space $\mathit{T_{e_{G}}G\times T_{e_{H}}H}$.
Then if $\psi$ is a homomorphism as hypothesized, the graph $j$
of $\psi$ is a subspace of $\mathit{T_{e_{G}}G\times T_{e_{H}}H}$.
By Lemma 1, this would imply that we have an immersed subgroup $J\subset G\times H$
with tangent space $\mathit{T_{e_{G\times H}}J}$. Consider the projection
map on the first factor $\pi_{1}:J\rightarrow G$, which has differential
$d\pi_{1}:j\mathrm{\rightarrow}T_{e}G$ that is an isomorphism by
hypothesis. But since $\mathit{G}$ is simply connected, $\pi_{1}$
is also an isomorphism. Hence the projection on the second factor
$\pi_{2}:J\rightarrow H$ must be a homomorphism with differential
at the identity equal to $\psi$. The other direction follows from
the diagram above; that is, if $\Psi$ is a homomorphism then its
differential is also.
\end{proof}
We used the fact that $\mathit{G}$ is simply connected in order to
show that the map $\Psi$ is defined on all of $\mathit{G}$ while
preserving its differentiability. Hopefully it is clear now that the
tangent space at the identity is quite important, because as Theorem
1 shows, any representation of a Lie group will induce a representation
of the tangent space. Although we have seen before now that the tangent
space has the structure of an algebra (bilinearity in its commutator),
its importance was not illuminated until now, and so we finally give
the tangent space a special name.

\begin{defn}
The tangent space at the identity of a Lie group, $T_{e}G$, is called
the \emph{Lie algebra} of the group and is labeled $\mathfrak{g}$.
\end{defn}
Hence we can now name the Lie algebras of the classical Lie groups:
for the special linear group $\mathit{SL}(n)$ we have $\mathfrak{sl}_{n}$,
for the symplectic group $\mathit{Sp}(2n)$ we have $\mathfrak{sp}_{2n}$,
etc.

Note that we may axiomatize Lie algebras:

\begin{defn}
A \emph{Lie algebra} is a vector space $\mathfrak{g}$ over a field
$\mathit{F}$ with a map $\left[\cdot,\cdot\right]:\mathfrak{g\rightarrow g}$
satisfying three axioms:

1. Bilinearity: $\left[aX+bY,Z\right]=a\left[X,Z\right]+b\left[Y,Z\right]$

2. Anticommutativity: $\left[X,Y\right]=-\left[Y,X\right]$

3. Jacobi identity: $\left[X,\left[Y,Z\right]\right]+\left[Y,\left[Z,X\right]\right]+\left[Z,\left[X,Y\right]\right]=0$
\end{defn}
Property 3 was not verified for the tangent space of a Lie group,
but it holds. Definition 7 would imply that perhaps there exist Lie
algebras without an associated Lie group - indeed there do, but they
must be infinite-dimensional. On the other hand, we could construct
an 'abstract' Lie algebra by taking generators of any vector space
such as a Lie group and applying commutation relations to them to
determine what the commutators are. This is the method we use in Section
3 to develop the Eight-Fold Way. Other instances of a Lie algebra
being developed 'from scratch' rather than from a Lie group would
be the Heisenberg group:

\[
\mathfrak{h}_{3}=\left\langle x,y,z:\left[x,y\right]=z,\left[x,z\right]=0=\left[y,z\right]\right\rangle \]


which has for elements the upper-triangular matrices with diagonal
elements equal to 1. This has additional applications to quantum mechanics.

Now that Lie algebras have been developed and their relationships
to Lie groups and Lie subgroups have been discussed,%
\footnote{Note: One may equivalently develop Lie groups using the adjoint representation
of a Lie group. This is perfectly equivalent to the method outlined
above but seems rather artificial because one doesn't know \emph{a
priori} that the adjoint operator will be useful. Hence, examining
the tangent space, which is a very normal endeavor when dealing with
manifolds, seems to be the most natural method of describing the Lie
group, and is also closer to what Sophus Lie originally did.%
} we describe an example in which representations of a Lie algebra
prove useful in examining a Lie group. Lie algebras will be important
in the following exposition because they are endowed with a commutator.
Lacking any other information about how to find a representation,
we know that any representation of a Lie group must preserve the commutation
relations of the Lie algebra, so we can construct an 'abstract' Lie
algebra using the generators of the Lie group and find the commutation
relations between them, thereby gaining some structure for the action
of the representation.


\section{The Eight-Fold Way}


\subsection{Historical Background}

One common method for predicting the potential for certain interactions
to occur in physics is to determine what observable quantities must
be conserved during the reaction; if a given interaction conserves
all of these quantities, then it is considered {}``possible'' and
can be looked for in an experiment in order to confirm the theory.
Most people are familiar with the concepts of conservation of energy
and momentum, but there are other quantities that must be conserved,
depending on the reaction. Two of these observable are known as isospin
and strangeness. As science began to uncover the inner structure of
the atom, it became clear that some force must be binding the protons
and neutrons together in the nucleus. This force must be stronger
over short distances than the electromagnetic force but ultimately
weaker at far distances (where {}``far'' can mean billionths of
a meter). This so-called {}``strong'' force is charge-independent,
since neutron and proton do not have the same charge but act the same
under the force. Heisenberg postulated in 1932 that these two particles
are just different states of a single particle, the nucleon. In fact,
this would imply that they could be transformed from one to the other
as matrices might be transformed under the action of a symmetry group.
Heisenberg assumed that the associated group was $\mathit{SU}(2)$
and that the two eigenstates of the nucleon were $\left|n\right\rangle =\left(\begin{array}{c}
0\\
1\end{array}\right)$ and $\left|p\right\rangle =\left(\begin{array}{c}
1\\
0\end{array}\right)$ (the similarity to the spin-up/spin-down states of an electron caused
this nucleon matrix group to be known as isospin, with each particle
having an observable {}``isospin,'' and the total isospin must be
conserved in any interaction).%
\footnote{The $\left|\cdot\right\rangle ${}``bra-ket'' notation is a product
of Paul Dirac and indicates that the corresponding matrix represents
a particle's state, in this case the state of the neutron and proton,
respectively.%
} Note that the choice of $\mathit{SU}(2)$ was probably due to the
fact that there are generators of $\mathit{SU}(2)$ which take $\left(\begin{array}{c}
1\\
0\end{array}\right)$ to $\left(\begin{array}{c}
0\\
1\end{array}\right)$ and vice-versa. In addition, a few other particles involved in the
strong interaction (baryons, including the proton and neutron, and
mesons, such as pions and kaons) must be related by matrix relations
of the same group, as they are involved in interactions with each
other. Then the transformation between the states of the nucleon would
be equivalent to the action of the two-dimensional representation
of $\mathit{SU}(2)$. We won't go into the details of this representation
- finding generators and so forth - because some of the reactions
that were hypothesized to occur were not found in high-energy accelerator
experiments. Such interactions conserved all the supposedly necessary
observables required by the strong interaction. Because these reactions
did not occur, Murray Gell-Mann and Nishijima hypothesized that there
must be an additional observable quantity which these {}``unseen''
reactions did not conserve. Because this quantity was theretofore
unknown, they called it {}``strangeness.'' By normalizing the strangeness
of the proton and neutron to zero, the strangeness of other particles
could be deduced from their production or decay reactions. A new quantum
number, the hypercharge, was introduced and related to the strangeness
$S$ and the baryon number $B$:

\begin{quote}
\[
Y=S+B\]
.
\end{quote}
After trying alternatives, a model for the group of strong interactions
was developed which described the reactions as seen in experiment.
It is called the {}``flavor'' model and is the foundation for the
quark model of subatomic particles. By introducing these unobservable
states called {}``quarks,'' observed mesons and baryons could be
built. This all corresponds to a representation of the group $SU(3)$,
which was chosen over the smaller group $\mathit{SU}(2)$ to account
for the new quantum number $Y$. Diagrams displaying the quantum numbers
$\mathit{I_{3}}$ (isospin) and $\mathit{Y}$ are shown in Figure
1: the vertical axis shows the isospin and the horizontal axis shows
the strangeness, with the particles' names written at the intersection
of their quantum numbers.

%
\begin{figure}
\caption{Quantum Numbers of Mesons and Baryons \cite{key-2}}
\includegraphics{\string"S,I chart\string".eps}
\end{figure}



\subsection{The Mathematics}

$SU(3)$ is the space of unitary $3\times3$ matrices with determinant
1. These are all traceless, as can be seen thus:

This space has dimension eight, and the standard example of a basis
is the set of Gell-Mann $\lambda$-matrices:

\[
\lambda_{1}=\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 0\end{array}\right),\lambda_{2}=\left(\begin{array}{ccc}
0 & -i & 0\\
i & 0 & 0\\
0 & 0 & 0\end{array}\right),\lambda_{3}=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 0\end{array}\right)\]
(the Pauli matrices, which generate $SU(2)$, with extra row of zeroes)

\[
\lambda_{4}=\left(\begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
1 & 0 & 0\end{array}\right),\lambda_{5}=\left(\begin{array}{ccc}
0 & 0 & -i\\
0 & 0 & 0\\
i & 0 & 0\end{array}\right),\lambda_{6}=\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\end{array}\right),\lambda_{7}=\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & -i\\
0 & i & 0\end{array}\right),\lambda_{8}=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -2\end{array}\right).\]


Without actually finding the corresponding Lie algebra, we can construct
an {}``abstract'' version by implementing the fact that elements
of the Lie algebra obey the commutation relation (in fact, the Lie
algebra of $SU(3)$ is isomorphic to that of $\mathit{SL}_{3}(\mathbb{C})$,
so we can call it $\mathfrak{sl}_{3}(\mathbb{C})$). Out of eventual
convenience (and partly to reflect the structure of the basis of $SU(2)$
that we didn't consider) we define some new matrices (the names are
historical artifices):

\[
T_{x}=\frac{1}{2}\lambda_{1},T_{y}=\frac{1}{2}\lambda_{2},T_{z}=\frac{1}{2}\lambda_{3},V_{x}=\frac{1}{2}\lambda_{4},V_{y}=\frac{1}{2}\lambda_{5},U_{x}=\frac{1}{2}\lambda_{6},U_{y}=\frac{1}{2}\lambda_{7},Y=\frac{1}{\sqrt{3}}\lambda_{8}\]


and

\[
T_{\pm}=T_{x}\pm iT_{y},V_{\pm}=V_{x}\pm iV_{y},U_{\pm}=U_{x}\pm iU_{y}\,\mathrm{(the\, so-called\,``shift''\, operators).}\]


The names should give some hint of what the matrices mean; $\mathit{Y}$,
for instance, corresponds to the hypercharge and accordingly, $\mathit{T_{z}}$,
the only other matrix not encoded in a {}``$\pm$'' relation, corresponds
to isospin (note that these are the only two diagonalizable matrices,
and so the only ones with eigenvectors). The {}``convenience'' comes
when we consider the commutation relations (in Table 1), so that all
commutators are just scalar multiples of other matrices in the basis.
We construct an abstract Lie algebra by assuming elements $\mathit{t_{x}},t_{y},t_{z},$
etc. whose commutators are the same as the commutators of the matrices
above.

%
\begin{table}


\caption{Commutation Relations for Elements of Abstract Lie Algebra}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
 & $\mathbf{t}_{+}$ & $\mathbf{t}_{-}$ & $\mathbf{t_{z}}$ & $\mathbf{v_{+}}$ & $\mathbf{v_{-}}$ & $\mathbf{u}_{+}$ & $\mathbf{u}_{-}$ & $\mathbf{y}$\tabularnewline
\hline
$\mathbf{t}_{+}$ & 0 & $2t_{z}$ & $-t_{+}$ & 0 & $-u_{-}$ & $v_{+}$ & 0 & 0\tabularnewline
\hline
$\mathbf{t_{-}}$ & $-2t_{z}$ & 0 & $t_{-}$ & $u_{+}$ & 0 & 0 & $-v_{-}$ & 0\tabularnewline
\hline
$\mathbf{t}_{z}$ & $t_{+}$ & $-t_{-}$ & 0 & $\frac{1}{2}v_{+}$ & $-\frac{1}{2}v_{-}$ & $-\frac{1}{2}u_{+}$ & $\frac{1}{2}u_{z}$ & 0\tabularnewline
\hline
$\mathbf{v}_{+}$ & 0 & $-u_{+}$ & $-\frac{1}{2}v_{+}$ & 0 & $\frac{3}{2}y+t_{z}$ & 0 & $t_{-}$ & $-v_{+}$\tabularnewline
\hline
$\mathbf{v_{-}}$ & $-u_{-}$ & 0 & $\frac{1}{2}v_{-}$ & $-\frac{3}{2}y-t_{z}$ & 0 & $-t_{-}$ & 0 & $v_{-}$\tabularnewline
\hline
$\mathbf{u}_{+}$ & $-v_{+}$ & 0 & $\frac{1}{2}u_{+}$ & 0 & $t_{-}$ & 0 & $\frac{3}{2}y-t_{z}$ & $-u_{+}$\tabularnewline
\hline
$\mathbf{u_{-}}$ & 0 & $v_{-}$ & $-\frac{1}{2}u_{z}$ & $-t_{-}$ & 0 & $-\frac{3}{2}y+t_{z}$ & 0 & $u_{-}$\tabularnewline
\hline
$\mathbf{y}$ & 0 & 0 & 0 & $v_{+}$ & $-v_{-}$ & $u_{+}$ & $-u_{-}$ & 0\tabularnewline
\hline
\end{tabular}
\end{table}


We see that $y$ and $t_{z}$ commute, so they are simultaneously
diagonalizable; hence we might write simultaneous eigenvectors as
particle states in the Dirac notation $\left|\lambda,\mu\right\rangle $
where $\lambda,\mu$ are eigenvalues of the matrices on that vector.
Then the notation $\left|\lambda,\mu\right\rangle $ would correspond
to the state of a particle with isospin $\lambda$ and strangeness
$\mu$. By examination we see that $\mathit{T_{+}}$ shifts the eigenstate
to $\left|\lambda+1,\mu\right\rangle $, $\mathit{U_{+}}$ shifts
it to $\left|\lambda-\frac{1}{2},\mu+1\right\rangle $, etc. so that
the shift operators (hence the reason for the subscripts) form a diagram
with vectors $\mathit{T_{+}}=(1,0),\, U_{+}=(-\frac{1}{2},1)$, etc.
The shift operators, then, actually transform \emph{between particles},
which are themselves \emph{eigenstates of the isospin and strangeness
operators}. Figure 2 shows the six vectors (note that in the text
from which this came, the operators $\mathit{T}$ are called $\mathit{I}$;
someone with previous background might recognize this chart as the
root diagram $\mathit{A_{2}}$, which corresponds to the Lie algebra
we are considerinig). A representation of this Lie algebra would be
any map from the generators $\mathit{t_{+}},\, t_{-}$, etc. to a
vector space while preserving the commutator relations.

%
\begin{figure}


\caption{Root Diagram for Shift Operators$\left(\mathrm{A_{2}}\right)$ \cite{key-2}}


\includegraphics{\string"Vector Chart\string".eps}
\end{figure}


Beginning with the standard representation from Example 1 above, with
the vector space $V=\mathbb{C}^{3}$, we have basis vectors $e_{1}=\left(\begin{array}{c}
1\\
0\\
0\end{array}\right),\, e_{2}=\left(\begin{array}{c}
0\\
1\\
0\end{array}\right),\, e_{3}=\left(\begin{array}{c}
0\\
0\\
1\end{array}\right)$. The Lie group acts on these by multiplying on the left. We (suggestively)
write these $e_{1}=u\,(\mathrm{up\, quark)},\, e_{2}=d\,(\mathrm{down\, quark)},\, e_{3}=s\,(\mathrm{strange\, quark)}$.
Using the definition of the dual representation as stated in Example
2 above, the basis vectors are $u^{*}=\left(\begin{array}{ccc}
-1 & 0 & 0\end{array}\right),\, d^{*}=\left(\begin{array}{ccc}
0 & -1 & 0\end{array}\right),\, s^{*}=\left(\begin{array}{ccc}
0 & 0 & -1\end{array}\right)$ which are anti-up, anti-down, and anti-strange. The Lie group acts
on these by multiplying on the right by a matrix. For example,

\[
u^{*}V_{+}=\left(\begin{array}{ccc}
-1 & 0 & 0\end{array}\right)\left(\begin{array}{ccc}
0 & 0 & -1\\
0 & 0 & 0\\
0 & 0 & 0\end{array}\right)=\left(\begin{array}{ccc}
0 & 0 & -1\end{array}\right)=s^{*}\]
.

Then taking the tensor product $V\otimes V^{*}$ yields a representation
with 9 generators (hence 9-dimensional). Given the Lie algebra relations
in Table 1 and the shift operator relations in Figure 2, we can determine
the eigenstate corresponding to each vector $\mathit{u},\, d,\, s$,
etc. by finding their eigenvalues under the action by $\mathit{T_{z}}$
and $\mathit{Y}$. For instance, $\mathit{T_{z}u}=\frac{1}{2}u$ and
$\mathit{Yu}=\frac{1}{3}u$ so $\mathit{u}=\left|\frac{1}{2},\frac{1}{3}\right\rangle $
as an eigenstate. The diagram formed by charting these is called a
\emph{weight diagram}; the weight diagrams for the standard and dual
representations are shown in Figures 3 and 4.

%
\begin{figure}


\caption{Weight Diagram for Standard Representation $(V)$ \cite{key-2}}
\includegraphics{\string"Weight Chart V\string".eps}


\end{figure}


%
\begin{figure}


\caption{Weight Diagram for Dual Representation $(V^{*})$ \cite{key-2}}
\includegraphics{\string"Weight Chart Dual\string".eps}


\end{figure}


We can see by comparison to Figure 1 that there are no particles corresponding
to these eigenvectors; hence, particles and quarks are distinct and
the particles are actually \emph{tensor products of} \emph{quarks.
}To\emph{ }actually find the elements of the tensor product, we compute
them in the normal way: if $\mathit{L_{1}},\, L_{2}$ are generators
for two representations $\mathit{V}_{1},\, V_{2}$ then $\mathit{L}(v_{1}\otimes v_{2})=L_{1}(v_{1})\otimes v_{2}+v_{1}\otimes L_{2}(v_{2})$.
For example,

\[
T_{z}(ud^{*})=T_{z}(u)d^{*}+uT(d^{*})=\frac{1}{2}ud^{*}+\frac{1}{2}ud^{*}=1\cdot ud^{*}\]


\[
Y(ud^{*})=Y(u)d^{*}+uY(d^{*})=0\]


so $\mathit{ud^{*}}$ corresponds to the state $\left|1,0\right\rangle $
which, as one can see from Diagram 1, is the same as that of the pion
$\pi^{+}$. Further, we can get relationships using the shift operators.
For example,

$\mathit{U}_{+}ud^{*}=us^{*}=\left|\frac{1}{2},1\right\rangle $ which
is the kaon $K^{+}$.

>From these relationships, we can generate every meson; however, the
particle $\eta=\frac{uu^{*}+dd^{*}+ss^{*}}{\sqrt{6}}=\left|0,0\right\rangle $
is invariant under the action (one can check that $\mathit{uu*},\, dd^{*},\, ss^{*}$
go to zero under the shift operators) so it cannot be transformed
into any other particle. Hence the representation $\mathit{V\otimes V^{*}}$
reduces to the direct sum of two irreducible representations of degrees
8 and 1.%
\footnote{Actually, the representation of degree 8 is the adjoint representation
mentioned in footnote 2; one may actually begin with this representation
as teh basis for the eight-fold way, but it makes more sense to me
to start with the standard and dual representations, as their actions
will end up corresponding to the actions of quarks and antiquarks,
respectively.%
} The tensor product of the standard representation and its dual, then,
accurately explain the observed interactions of mesons, and the group
of eight mesons generated by the shift operators gives the Eight-Fold
Way its name. The meson octet is shown in Figure 5 (by historical
convention, $\pi^{+}=-ud^{*}$ instead of $ud^{*}$ as we determined).

%
\begin{figure}


\caption{The Meson Octet \cite{key-2}}
\includegraphics{MesonOctet}
\end{figure}



\subsection{Physical Re-Interpretation}

As stated above, quarks are the basis vectors for the standard representation
of $SU(3)$ and anti-quarks are the basis vectors for the dual representation
(although today there are six quarks in the model of the universe).
The tensor product has generators consisting of quark-anti-quark pairs,
and these correspond to the mesons. Therefore, the elementary particles
are the generators for a vector space into which there is a homomorphism
from a Lie group. That Lie group has eight elements for generators;
two of them have the particles as eigenvectors, and we can uniquely
identify the particles from their corresponding two eigenvalues. The
other six generators of the Lie group act on the mesons by transforming
them, turning them into other mesons. Therefore, one might say that
\emph{all mesons are basis vectors for $\mathbb{C}^{3}\otimes\mathbb{C}^{3*}$
and meson interactions are characterized by the action of linear combinations
of generators of the Lie group $SU(3)$.} Why this should be - why
Lie groups should be woven into the background of the universe's fabric
- is a question I simply can't answer.

We have shown that all mesons are composed of one or more quark-antiquark
pairs. The baryons (protons, neutrons, etc.), it turns out, correspond
to the triple tensor product representation $V\otimes V\otimes V$
so each of them have 3 quarks (the proton, for instance, is up-up-down).
This splits into a 10-dimensional irreducible representation, two
8-dimensional ones (the same as the 8-dimensional one above) and a
1-dimensional representation. The benefit of this mathematical model
is that given enough information from the mathematics, one can tell
when the physics is incomplete. In 1964, using a theoretical prediction
from the representation theory used here, the baryon $\Omega^{-}$
was discovered. While the number of known elementary particles has
grown to several hundred, some are still only theoretical predictions.
That the mathematics of homomorphisms between groups should help in
the endeavor to ascertain the structure of the universe is exciting
and gives a unique perspective on just what the interactions between
matter really are.

Lie algebras were indispensable here because, although we did all
our work with the matrices of the Lie group itself, we used the fact
that the Lie algebra is endowed with a commutator, and since homomorphisms
between groups must correspond to homomorphisms between algebras,
we use the information of the commutators to guarantee that we are
finding workable representations of the groups at hand. Indeed, in
this case one might not even know what vector space $V$ to use as
a representation; what ended up happening was that the Lie algebra
itself acted as that vector space, as we found an 8-dimensional representation
corresponding to a map into the Lie algebra.%
\footnote{A greater elaboration on the adjoint representation would demonstrate
immediately that this is true. In short, if $\Theta_{g}$ is conjugation
by $g\in G$ then $\mathrm{Ad}(g)=(d\Theta_{g})_{e}$ is a map from
the Lie group to the space of automorphisms of its Lie algebra, Aut($\mathfrak{g}$).%
} Hence the algebraic structure of Lie algebras are a vital tool in
the study of representations of Lie groups.

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\bibitem{key-1}Adams, J. Frank. \emph{Lectures on Lie Groups}. Chicago:
University of Chicago Press, 1969.

\bibitem{key-4}Cahn, Robert N. \emph{Semisimple Lie Algebras and
Their Representations}. Accessed online: http://www-physics.lbl.gov/\textasciitilde{}rncahn/book.html.

\bibitem{key-3}Cohn, P.M. \emph{Lie Groups}. Cambridge: Cambridge
University Press, 1957.

\bibitem{key-2}Duistermat, J.J. and J.A.C. Kolk. \emph{Lie Groups}.
New York: Springer-Verlag, 2000.

\bibitem{key-1}Dummit, David S. and Richard M. Foote. \emph{Abstract
Algebra}. Hoboken, NJ: John Wiley and Sons, Inc., 2004.

\bibitem{key-1}Fulton, William and Joe Harris. \emph{Representation
Theory: A First Course}. New York: Springer Verlag, 1991.

\bibitem{key-2}Sattinger, D. H. and O. L. Weaver. \emph{Lie Groups
and Algebras with Applications to Physics, Geometry, and Mechanics}.
New York: Springer-Verlag, 1986.
\end{thebibliography}

\end{document}
