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\begin{document}

\title{MA 792M: Special Topics in Lie Algebras}


\author{William Jeffrey Cook\\
Department of Mathematics\\
North Carolina State University\\
wjcook@unity.ncsu.edu\\
https://www4.ncsu.edu/\textasciitilde{}wjcook/ma792/}

\maketitle
\begin{abstract}
These are notes from a presentation for MA 792M (Representation Theory
for Lie Algebras) taught by Professor Kailash Misra. The goal is to
present a construction of \( G_{2} \). In the first section I will
present a summary of basic facts about \( G_{2} \). In the second
section I construct the root system of \( G_{2} \) in \( R^{3} \)
using the root system of \( B_{3} \) (which is the 7x7 orthoginal
matrix algebra). In the third section I will construct \( G_{2} \)
itself (as a subalgebra of \( o(7,C) \) ). In the final section I
will \emph{sketch} a construction of \( G_{2} \) as the derivation
algebra of the octonians. My primary source was \emph{Introduction
to Lie Algebras and Representation Theory} by James E. Humphreys pages
102-106.
\end{abstract}

\section{Preliminaries}

\vspace{0.3cm}
{\centering \resizebox*{3.5in}{1in}{\includegraphics{G2-Dynkin.jpg}} \par}
\vspace{0.3cm}

Starting with the Dynkin diagram, we see that \( G_{2} \)'s Cartan
matrix is: \[
A=(a_{ij})=\left( \begin{array}{cc}
2 & -1\\
-3 & 2
\end{array}\right) \]
Let us label our simple roots as \( \alpha  \) and \( \beta  \).
(Let \( \pi =\{\alpha ,\, \beta \} \) be our set of simple roots).
Using the algorithm outlined in Humphreys it is easy to find all the
positive roots (hence all the roots). After calculating \( \alpha  \)
and \( \beta  \) root strings, we find that \( \Phi ^{+}=\{\alpha ,\, \beta 
,\, \alpha +\beta ,\, 2\alpha +\beta ,\, 3\alpha +\beta ,\, 3\alpha +2\beta 
\} \)
are all the positive roots thus there are 12 roots all together: \[
\Phi =\{\pm \alpha ,\, \pm \beta ,\, \pm (\alpha +\beta ),\, \pm (2\alpha 
+\beta ),\, \pm (3\alpha +\beta ),\, \pm (3\alpha +2\beta )\}\]


\vspace{0.3cm}
{\centering \resizebox*{3.75in}{3in}{\includegraphics{G2-RootSystem.jpg}} \\
The Root System of \( G_{2} \)\par}
\vspace{0.3cm}

\emph{Note}: The long roots form a root system of type \( A_{2} \)
(this we be important later).

Thus a Lie algebra of type \( G_{2} \) must be \( 2+12=14 \)-dimensional.
\( G_{2} \)'s Weyl group is generated by the two reflections: \( r_{\alpha 
}(x)=x-<x,\alpha >\alpha  \)
and \( r_{\beta }(x)=x-<x,\beta >\beta  \) where \( x\, \epsilon \, E \)
(\( x \) is in the Euclidean space spanned by \( \alpha  \) and
\( \beta  \)). Now using a proposition from class (noting that \( 
a_{12}a_{21}=3 \))
we see that the order of \( r_{\alpha }r_{\beta } \) is 6. Thus the
Weyl group is isomorphic to \( D_{6} \) (dihedral group of order
12).


\section{The Root System}

Using representation theory one can show that \( G_{2} \) has an
irreducible faithful representation by 7x7 matricies (and in fact
this is the smallest irreducible faithful representation).

Consider \( o(7,C) \). Let us construct the root system in \( R^{3} \)
(this is constructed in Humphreys page~64). Let \( (e_{1},e_{2},e_{3}) \)
be the standard basis for \( R^{3} \). Then \( \{e_{1}-e_{2},\, 
e_{2}-e_{3},\, e_{3}\} \)
is also a basis for \( R^{3} \). For the short roots we have \( \pm 
e_{1},\pm e_{2},\pm e_{3} \).
For the long roots we have \( \pm (e_{1}-e_{2}),\pm (e_{1}-e_{3}),\pm 
(e_{2}-e_{3}), \)
\( \pm (e_{1}+e_{2}),\pm (e_{1}+e_{3}),\pm (e_{2}+e_{3}) \).

\vspace{0.3cm}
{\centering \resizebox*{3.75in}{3in}{\includegraphics{B3-RootSystem.jpg}} \\
The Root System of \( o(7,C) \)\par}
\vspace{0.3cm}

Now rotating and viewing this root system from the right angle we
see the root system for \( G_{2} \)

\vspace{0.3cm}
{\centering 
\resizebox*{3.75in}{3in}{\includegraphics{B3-RootSystem-View2.jpg}} \\
The Root System of \( o(7,C) \) (different view)\par}
\vspace{0.3cm}

Now to construct the root system of \( G_{2} \). We want to project
the roots of \( B_{3} \) onto the plane \( E=\{(x,y,z)\, |\, x+y+z=0\} \)
(the plane orthoginal to \( e_{1}+e_{2}+e_{3} \)). Projecting to
this plane we get:\[
\pm \{e_{1}-e_{2},\, e_{1}-e_{3},\, e_{2}-e_{3},\, 
\frac{2}{3}e_{1}-\frac{1}{3}e_{2}-\frac{1}{3}e_{3},\, 
-\frac{1}{3}e_{1}+\frac{2}{3}e_{2}-\frac{1}{3}e_{3},\, 
-\frac{1}{3}e_{1}-\frac{1}{3}e_{2}+\frac{2}{3}e_{3}\}\]


which happens to be the root system of \( G_{2} \).

\emph{Note}: The long roots (of \( G_{2} \)) form a root system of
type \( A_{2} \) and also these roots are precisely half of the long
roots of \( B_{3} \).


\section{Constructing \protect\( G_{2}\protect \) (as a subalgebra of 
\protect\( o(7,C)\protect \))}

Let's start with the classical Lie algebra \( L_{0}=o(7,C) \). The
standard basis for \( L_{0} \) in terms of matrix units \( E_{i,j} \)
(7 x 7 complex matrix with 1 in the \( i,j^{th} \) position everything
else zero):

\begin{lyxlist}{00.00.0000}
\item [\( g_{i,-j}=E_{i+1,j+1}-E_{j+4,i+4} \)]for \( 1\leq i\neq j\leq 3 \)
(6 matrices)
\item [\( d_{i}=E_{i+1,i+1}-E_{i+4,i+4} \)]for \( 1\leq i\leq 3 \) (3
matrices)
\item [\( E_{1,i+4}-E_{i+1,1} \)]and \( E_{1,i+1}-E_{i+4,1} \) for \( 1\leq 
i\leq 3 \)
(3 + 3 matrices)
\item [\( E_{i+1,j+4}-E_{j+1,i+4} \)]and \( E_{i+4,j+1}-E_{j+4,i+1} \)
for \( 1\leq i<j\leq 3 \) (3 + 3 matrices)
\end{lyxlist}
A Cartan subalgebra (CSA i.e. maximal toral subalgebra) of \( L_{0} \)
is \( H_{0}=span\{d_{1},d_{2},d_{3}\} \). For the CSA of \( L \)
(our type \( G_{2} \) Lie algebra) we will use \( 
H=span\{d_{1}-d_{2},d_{2}-d_{3}\}=\{\left. \sum a_{i}d_{i}\: \right| \sum 
a_{i}=0\} \).
Let \( L \) be the span of \( H \) together with \( g_{i,-j} \)
where \( 1\leq i\neq j\leq 3 \) (these correspond to the 6 long roots
which form a type \( A_{2} \) root system).

Now when we formed the root system of \( G_{2} \) from that of \( B_{3} \)
we had to project onto a plane and collapse 3 pairs of roots from
\( B_{3} \) into the 3 short roots of \( G_{2} \). So we need a
combination of elements from these pairs of root spaces (after some
work) the correct choices are:

\begin{lyxlist}{00.00.0000}
\item [\( g_{1}=-g_{-1}^{t}=\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}) \)]~
\item [\( g_{2}=-g_{-2}^{t}=\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5}) \)]~
\item [\( g_{3}=-g_{-3}^{t}=\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5}) \)]~
\end{lyxlist}
OR \( g_{i}=-g_{-i}^{t}=\sqrt{2}(E_{1,i+1}-E_{i+4,1})-\delta 
_{i1}(E_{3,7}-E_{4,6})+\delta _{i2}(E_{2,7}-E_{4,5})-\delta 
_{i3}(E_{2,6}-E_{3,5}) \)\\
\\
Thus \( L \) is a subspace of \( L_{0} \) of dimension 14. First
we need to show that \( L \) is closed under the commutator bracket.
Recall that \( [E_{i,j},\: E_{k,l}]=\delta _{jk}E_{i,l}-\delta _{li}E_{k,j} 
\)
where \( \delta _{ij}=\left\{ \begin{array}{cc}
1 & i=j\\
0 & i\neq j
\end{array}\right\}  \) is the Kronecker delta. We need the following 
relations \( 1\leq i,\: j,\: k,\: l\leq 3 \):

\begin{enumerate}
\item \( [g_{i,-j},\: g_{j,-i}]=d_{i}-d_{j} \) for \( i\neq j \)\\
\( [g_{i,-j},\: g_{k,-l}]=\delta _{jk}g_{i,-l}-\delta _{il}g_{k,-j} \)
for \( i\neq j \), \( k\neq l \), and \( (i,j)\neq (l,k) \)
\item \( [g_{i},\: g_{-i}]=3d_{i}-(d_{1}+d_{2}+d_{3}) \)
\item \( [g_{i,-j},\: g_{k}]=-\delta _{ik}g_{j} \) \\
\( [g_{i,-j},\: g_{-k}]=\delta _{jk}g_{-i} \)
\item \( [g_{i},\: g_{-j}]=3g_{j,-i} \) for \( i\neq j \)
\item \( [g_{i},\: g_{j}]=\pm 2g_{-k} \) for \( i,\: j,\: k \) distinct\\
\( [g_{-i},\: g_{-j}]=\pm 2g_{k} \) for \( i,\: j,\: k \) distinct
(for the exact signs see the following proof)
\end{enumerate}
(1.)\\
\( [g_{i,-j},\: g_{k,-l}]=[E_{i+1,j+1}-E_{j+4,i+4},\: 
E_{k+1,l+1}-E_{l+4,k+4}]= \)
\( [E_{i+1,j+1},\: E_{k+1,l+1}]-[E_{i+1,j+1},\: E_{l+4,k+4}]-[E_{j+4,i+4},\: 
E_{k+1,l+1}]+[E_{j+4,i+4},\: E_{l+4,k41}] \)
\( =\delta _{jk}E_{i+1,l+1}-\delta _{li}E_{k+1,j+1}-0-0-\delta 
_{jk}E_{l+4,i+4}+\delta _{li}E_{k+4,j+4} \)\\
\\
\( [g_{i,-j},\: g_{j,-i}]= \)\( \delta _{jk}g_{i,-l}-\delta _{li}g_{k,-j} \)
for \( (i,j)\neq (l,k) \) and \\
\( [g_{i,-j},\: g_{j,-i}]= \) \( 
E_{i+1,i+1}-E_{j+1,j+1}-E_{i+4,i+4}+E_{j+4,j+4}= \)
\( d_{i}-d_{j} \) for \( (i,j)=(l,k) \)\\


(2.)\\
\( [g_{1},\: g_{-1}]=[\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}),\: 
-\sqrt{2}(E_{2,1}-E_{1,5})+(E_{7,3}-E_{6,4})]= \)
\( -2[E_{1,2},\: E_{2,1}]-2[E_{5,1},\: E_{1,5}]-[E_{3,7},\: 
E_{7,3}]-[E_{4,6},\: E_{6,4}]= \)
\( -2E_{1,1}+2E_{2,2}-2E_{5,5}+2E_{1,1}-E_{3,3}+E_{7,7}-E_{4,4}+E_{6,6}= 
\)\( 2d_{1}-d_{2}-d_{3}= \)
\( 3d_{1}-(d_{1}+d_{2}+d_{3}) \) \\
\\
\( [g_{2},\: g_{-2}]=[\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5}),\: 
-\sqrt{2}(E_{3,1}-E_{1,6})-(E_{7,2}-E_{5,4})]= \)
\( -2[E_{1,3},\: E_{3,1}]-2[E_{6,1},\: E_{1,6}]-[E_{2,7},\: 
E_{7,2}]-[E_{4,5},\: E_{5,4}]= \)
\( -2E_{1,1}+2E_{3,3}-2E_{6,6}+2E_{1,1}-E_{2,2}+E_{7,7}-E_{4,4}+E_{5,5}= 
\)\( 2d_{2}-d_{1}-d_{3}= \)
\( 3d_{2}-(d_{1}+d_{2}+d_{3}) \) \\
\\
\( [g_{3},\: g_{-3}]=[\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5}),\: 
-\sqrt{2}(E_{4,1}-E_{1,7})+(E_{6,2}-E_{5,3})]= \)
\( -2[E_{1,4},\: E_{4,1}]-2[E_{7,1},\: E_{1,7}]-[E_{2,6},\: 
E_{6,2}]-[E_{3,5},\: E_{5,3}]= \)
\( -2E_{1,1}+2E_{4,4}-2E_{7,7}+2E_{1,1}-E_{2,2}+E_{6,6}-E_{3,3}+E_{5,5}= 
\)\( 2d_{3}-d_{1}-d_{2}= \)
\( 3d_{3}-(d_{1}+d_{2}+d_{3}) \) \\


(3.)\\
\( [g_{i,-j},\: g_{k}]=[E_{i+1,j+1}-E_{j+4,i+4},\: 
\sqrt{2}(E_{1,k+1}-E_{k+4,1})-\delta _{k1}(E_{3,7}-E_{4,6})+\delta 
_{k2}(E_{2,7}-E_{4,5})-\delta _{k3}(E_{2,6}-E_{3,5})]= \)
\( [E_{i+1,j+1}-E_{j+4,i+4},\: \sqrt{2}(E_{1,k+1}-E_{k+4,1})]= \)
\( -\delta _{ik}\sqrt{2}E_{1,j+1}+\delta _{ik}\sqrt{2}E_{j+4,1} \)
\\
\\
(\( k=1) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: -(E_{3,7}-E_{4,6})]= \)
\( -\delta _{j2}E_{i+1,7}+\delta _{j3}E_{i+1,6}-\delta _{j3}E_{3,i+4}+\delta 
_{j2}E_{4,i+4} \)
\\
(\( k=2) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: (E_{2,7}-E_{4,5})]= \)
\( \delta _{j1}E_{i+1,7}-\delta _{j3}E_{i+1,5}+\delta _{j3}E_{2,i+4}-\delta 
_{j1}E_{4,i+4} \)
\\
(\( k=3) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: -(E_{2,6}-E_{3,5})]= \)
\( -\delta _{j1}E_{i+1,6}+\delta _{j2}E_{i+1,5}-\delta _{j2}E_{2,i+4}+\delta 
_{j1}E_{3,i+4} \)
\\
\\
\( [g_{i,-j},\: g_{k}]=-\delta _{ik}\sqrt{2}(E_{1,j+1}-E_{j+4,1}) \)
\( +\delta _{1k}(\delta _{j2}(E_{4,i+4}-E_{i+1,7})+\delta 
_{j3}(E_{i+1,6}-E_{3,i+4})) \)
\( -\delta _{2k}(\delta _{j1}(E_{4,i+4}-E_{i+1,7})+\delta 
_{j3}(E_{i+1,5}-E_{2,i+4})) \)
\( +\delta _{3k}(\delta _{j1}(E_{3,i+4}-E_{i+1,6})+\delta 
_{j2}(E_{i+1,5}-E_{2,i+4}))= \)
\( -\delta _{ik}g_{j} \)\\
\\
\( [g_{i,-j},\: g_{-k}]=[E_{i+1,j+1}-E_{j+4,i+4},\: 
\sqrt{2}(E_{1,k+4}-E_{k+1,1})+\delta _{k1}(E_{7,3}-E_{6,4})-\delta 
_{k2}(E_{7,2}-E_{5,4})+\delta _{k3}(E_{6,2}-E_{5,3})]= \)
\( [E_{i+1,j+1}-E_{j+4,i+4},\: \sqrt{2}(E_{1,k+4}-E_{k+1,1})]= \)\( -\delta 
_{jk}\sqrt{2}E_{i+1,1}+\delta _{jk}\sqrt{2}E_{1,i+4} \)
\\
\\
(\( k=1) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: (E_{7,3}-E_{6,4})]= \)
\( -\delta _{i2}E_{7,j+1}+\delta _{i3}E_{6,j+1}-\delta _{i3}E_{j+4,3}+\delta 
_{i2}E_{j+4,4} \)
\\
(\( k=2) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: -(E_{7,2}-E_{5,4})]= \)
\( \delta _{i1}E_{7,j+1}-\delta _{i3}E_{5,j+1}+\delta _{i3}E_{j+4,2}-\delta 
_{i1}E_{j+4,4} \)
\\
(\( k=3) \) \( [E_{i+1,j+1}-E_{j+4,i+4},\: (E_{6,2}-E_{5,3})]= \)
\( -\delta _{i1}E_{6,j+1}+\delta _{i2}E_{5,j+1}-\delta _{i2}E_{j+4,2}+\delta 
_{i1}E_{j+4,3} \)
\\
\\
\( [g_{i,-j},\: g_{-k}]=\delta _{jk}\sqrt{2}(E_{1,i+4}-E_{i+1,1}) \)
\( -\delta _{1k}(\delta _{i2}(E_{7,j+1}-E_{j+4,4})+\delta 
_{i3}(E_{j+4,3}-E_{6,j+1})) \)
\( +\delta _{2k}(\delta _{i1}(E_{7,j+1}-E_{j+4,4})+\delta 
_{i3}(E_{j+4,2}-E_{5,j+1})) \)
\( -\delta _{3k}(\delta _{i1}(E_{6,j+1}-E_{j+4,3})+\delta 
_{i2}(E_{j+4,2}-E_{5,i+4}))= \)
\( \delta _{jk}g_{-i} \)\\


(4.)\\
\( [g_{1},\: g_{-2}]=[\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}),\: 
-\sqrt{2}(E_{3,1}-E_{1,6})-(E_{7,2}-E_{5,4})]= \)
\( -2[E_{1,2},\: E_{3,1}]-2[E_{5,1},\: E_{1,6}]+[E_{3,7},\: 
E_{7,2}]+[E_{4,6},\: E_{5,4}]= \)
\( 2E_{3,2}-2E_{5,6}+E_{3,2}-E_{5,6} \) \( =3g_{2,-1} \) \\
\\
\( [g_{1},\: g_{-3}]=[\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}),\: 
-\sqrt{2}(E_{4,1}-E_{1,7})+(E_{6,2}-E_{5,3})]= \)
\( -2[E_{1,2},\: E_{4,1}]-2[E_{5,1},\: E_{1,7}]+[E_{4,6},\: 
E_{6,2}]+[E_{3,7},\: E_{5,3}]= \)
\( 2E_{4,2}-2E_{5,7}+E_{4,2}-E_{5,7} \) \( =3g_{3,-1} \) \\
\\
\( [g_{2},\: g_{-3}]=[\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5}),\: 
-\sqrt{2}(E_{4,1}-E_{1,7})+(E_{6,2}-E_{5,3})]= \)
\( -2[E_{1,3},\: E_{4,1}]-2[E_{6,1},\: E_{1,7}]+[E_{2,7},\: 
E_{6,2}]+[E_{4,5},\: E_{5,3}]= \)
\( 2E_{4,3}-2E_{6,7}-E_{6,7}+E_{4,3} \) \( =3g_{2,-3} \)\\
\\
Now note that \( [g_{i},\: 
g_{-j}]^{t}=(g_{i}g_{-j})^{t}-(g_{-j}g_{i})^{t}=g^{t}_{-j}g^{t}_{i}-g^{t}_{i}g^{t}_{-j} 
\)
\( =(-g_{j})(-g_{-i})-(-g_{-i})(-g_{j}) \) \( 
=g_{j}g_{-i}-g_{-i}g_{j}=[g_{j},\: g_{-i}] \).
Thus we have that: \( [g_{2},\: g_{-1}]=[g_{1},\: 
g_{-2}]^{t}=3g^{t}_{2,-1}=3g_{1,-2} \),
\( [g_{3},\: g_{-1}]=[g_{1},\: g_{-3}]^{t}=3g^{t}_{3,-1}=3g_{1,-3} \),
and \( [g_{3},\: g_{-2}]=[g_{2},\: g_{-3}]^{t}=3g^{t}_{3,-2}=3g_{2,-3} \)\\


(5.)\\
\( [g_{1},\: g_{2}]=[\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}),\: 
\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5})]= \)
\( -2[E_{1,2},\: E_{6,1}]-2[E_{5,1},\: E_{1,3}]+\sqrt{2}[E_{1,2},\: 
E_{2,7}]+\sqrt{2}[E_{5,1},\: E_{4,5}] \)
\( -\sqrt{2}[E_{3,7},\: E_{1,3}]-\sqrt{2}[E_{4,6},\: E_{6,1}]= \)
\( 
2E_{6,2}-2E_{5,3}+\sqrt{2}E_{1,7}-\sqrt{2}E_{4,1}+\sqrt{2}E_{1,7}-\sqrt{2}E_{4,1}= 
\)
\( 2g_{-3} \) \\
\\
\( [g_{1},\: g_{3}]=[\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6}),\: 
\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5})]= \)
\\
\( -2[E_{1,2},\: E_{7,1}]-2[E_{5,1},\: E_{1,4}]-\sqrt{2}[E_{1,2},\: 
E_{2,6}]+\sqrt{2}[E_{5,1},\: E_{3,5}] \)
\( -\sqrt{2}[E_{3,7},\: E_{7,1}]-\sqrt{2}[E_{4,6},\: E_{1,4}]= \)
\( 
2E_{7,2}-2E_{5,4}-\sqrt{2}E_{1,6}+\sqrt{2}E_{3,1}-\sqrt{2}E_{1,6}+\sqrt{2}E_{3,1}= 
\)
\( -2g_{-2} \) \\
\\
\( [g_{2},\: g_{3}]=[\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5}),\: 
\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5})]= \)
\\
\( -2[E_{1,3},\: E_{7,1}]-2[E_{6,1},\: E_{1,4}]+\sqrt{2}[E_{1,3},\: 
E_{3,5}]+\sqrt{2}[E_{6,1},\: E_{2,6}] \)
\( -\sqrt{2}[E_{2,7},\: E_{7,1}]-\sqrt{2}[E_{4,5},\: E_{1,4}]= \)
\( 
2E_{7,3}-2E_{6,4}+\sqrt{2}E_{1,5}-\sqrt{2}E_{2,1}-\sqrt{2}E_{2,1}+\sqrt{2}E_{1,5}= 
\)
\( 2g_{-1} \) \\
\\
Now note that \( [g_{i},\: 
g_{j}]^{t}=(g_{i}g_{j})^{t}-(g_{j}g_{i})^{t}=g^{t}_{j}g^{t}_{i}-g^{t}_{i}g^{t}_{j} 
\)
\( =(-g_{-j})(-g_{-i})-(-g_{-i})(-g_{-j}) \) \( 
=g_{-j}g_{-i}-g_{-i}g_{-j}=[g_{-j},\: g_{-i}] \).
Thus we have that: \( [g_{-1},\: g_{-2}]=[g_{2},\: 
g_{1}]^{t}=-2g^{t}_{-3}=2g_{3} \),
\( [g_{-1},\: g_{-3}]=[g_{3},\: g_{1}]^{t}=2g^{t}_{-2}=-2g_{2} \),
and \( [g_{-2},\: g_{-3}]=[g_{3},\: g_{2}]^{t}=-2g^{t}_{-1}=2g_{1} \)\\
\\


The only relations left to check are those involving \( d_{1}-d_{2} \)
and \( d_{2}-d_{3} \). First note that: \( [d_{1}-d_{2},\: d_{2}-d_{3}]=0 \)
because \( d_{i} \)'s are diagonal (hence commute).

Now consider \( [d_{i},\: g_{j,-k}]= \) \( [E_{i+1,i+1}-E_{i+4,i+4},\: 
E_{j+1,k+1}-E_{k+4,j+4}]= \)
\( [E_{i+1,i+1},\: E_{j+1,k+1}]-[E_{i+4,i+4},\: E_{j+1,k+1}]-[E_{i+1,i+1},\: 
E_{k+4,j+4}]+[E_{i+4,i+4},\: E_{k+4,j+4}]= \)
\( \delta _{ij}E_{i+1,k+1}-\delta _{ik}E_{j+1,i+1}+\delta 
_{ik}E_{i+4,j+4}-\delta _{ij}E_{k+4,i+4}= \)
\( \delta _{ij}g_{i,-k}-\delta _{ik}g_{j,-i} \)

Thus we have that \( [d_{1}-d_{2},\: g_{j,-k}]= \) \( \delta 
_{1j}g_{j,-k}-\delta _{1k}g_{j,-k}-\delta _{2j}g_{j,-k}+\delta _{2k}g_{j,-k} 
\)
\( =(\delta _{1j}-\delta _{1k}-\delta _{2j}+\delta _{2k})g_{j,-k} \)
which is non-zero since \( j \) and \( k \) are distinct. Likewise
\( [d_{2}-d_{3},\: g_{j,-k}]=Cg_{i,-k} \) where \( C \) is non-zero.
We have only \( g_{i} \)'s left to check.\\
\\
\( [d_{1}-d_{2},\: g_{1}]=[E_{2,2}-E_{5,5}-E_{3,3}+E_{6,6},\: 
\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6})] \)
\( =\sqrt{2}[E_{2,2},\: E_{1,2}]+\sqrt{2}[E_{5,5},\: E_{5,1}]+[E_{3,3},\: 
E_{3,7}]+[E_{6,6},\: E_{4,6}] \)
\( =-\sqrt{2}E_{1,2}+\sqrt{2}E_{5,1}+E_{3,7}-E_{4,6} \) \( =-g_{1} \)\\
\\
\( [d_{2}-d_{3},\: g_{1}]=[E_{3,3}-E_{6,6}-E_{4,4}+E_{7,7},\: 
\sqrt{2}(E_{1,2}-E_{5,1})-(E_{3,7}-E_{4,6})] \)
\( =-[E_{3,3},\: E_{3,7}]-[E_{6,6},\: E_{4,6}]-[E_{4,4},\: 
E_{4,6}]-[E_{7,7},\: E_{3,7}] \)
\( =-E_{3,7}+E_{4,6}-E_{4,6}+E_{3,7} \) \( =0 \)\\
\\
\( [d_{1}-d_{2},\: g_{2}]=[E_{2,2}-E_{5,5}-E_{3,3}+E_{6,6},\: 
\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5})] \)
\( =[E_{2,2},\: E_{2,7}]+[E_{5,5},\: E_{4,5}]-\sqrt{2}[E_{3,3},\: 
E_{1,3}]-\sqrt{2}[E_{6,6},\: E_{6,1}] \)
\( =\sqrt{2}E_{1,3}-\sqrt{2}E_{6,1}+E_{2,7}-E_{4,5} \) \( =g_{2} \)\\
\\
\( [d_{2}-d_{3},\: g_{2}]=[E_{3,3}-E_{6,6}-E_{4,4}+E_{7,7},\: 
\sqrt{2}(E_{1,3}-E_{6,1})+(E_{2,7}-E_{4,5})] \)
\( =\sqrt{2}[E_{3,3},\: E_{1,3}]-\sqrt{2}[E_{6,6},\: E_{6,1}]+[E_{4,4},\: 
E_{4,5}]+[E_{7,7},\: E_{2,7}] \)
\( =-\sqrt{2}E_{1,3}+\sqrt{2}E_{6,1}+E_{4,5}-E_{2,7} \) \( =-g_{2} \)\\
\\
\( [d_{1}-d_{2},\: g_{3}]=[E_{2,2}-E_{5,5}-E_{3,3}+E_{6,6},\: 
\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5})] \)
\( =-[E_{2,2},\: E_{2,6}]-[E_{5,5},\: E_{3,5}]-[E_{3,3},\: 
E_{3,5}]-[E_{6,6},\: E_{2,6}] \)
\( =-E_{2,6}+E_{3,5}-E_{3,5}+E_{2,6} \) \( =0 \)\\
\\
\( [d_{2}-d_{3},\: g_{3}]=[E_{3,3}-E_{6,6}-E_{4,4}+E_{7,7},\: 
\sqrt{2}(E_{1,4}-E_{7,1})-(E_{2,6}-E_{3,5})] \)
\( =[E_{3,3},\: E_{3,5}]+[E_{6,6},\: E_{2,6}]-\sqrt{2}[E_{4,4},\: 
E_{1,4}]-\sqrt{2}[E_{7,7},\: E_{7,1}] \)
\( =\sqrt{2}E_{1,4}-\sqrt{2}E_{7,1}-E_{2,6}+E_{3,5} \) \( =g_{3} \)\\
\\
Because \( d_{i} \)'s are diagonal we have that \( [d_{i},\: 
g_{-j}]^{t}=(d_{i}g_{-j}-g_{-j}d_{i})^{t}=g^{t}_{-j}d_{i}^{t}-d_{i}^{t}g^{t}_{-j} 
\)
\( =-g_{j}d_{i}+d_{i}g_{j}=[d_{i},\: g_{j}] \) thus \( [d_{1}-d_{2},\: 
g_{-1}]=g_{-1} \),
\( [d_{2}-d_{3},\: g_{-1}]=0 \), \( [d_{1}-d_{2},\: g_{-2}]=-g_{-2} \),
\( [d_{2}-d_{3},\: g_{-2}]=g_{-2} \), \( [d_{1}-d_{2},\: g_{-3}]=0 \),
and \( [d_{2}-d_{3},\: g_{-3}]=-g_{-3} \).

Thus we have shown that \( L \) is closed under bracket (thus a subalgebra
of \( L_{0} \)) and furthermore, we have shown that the elements
of \( L \) are common eigenvectors of \( ad(H) \). We have that
\( N_{L}(H)=\{x\, \epsilon \, L\, |\, [x,\, H]\subseteq H\}=H \)
(\( H \) is self-normalizing), since all \( x\, \epsilon \, L \)
which aren't in \( H \) are eigenvectors with non-zero eigenvalues
for some \( h\, \epsilon \, H \). Now \( [H,\, H]=\{0\} \) (\( H \)
is abelian) thus \( H \) is nilpotent, hence a CSA (self-normalizing
and nilpotent).

We now have have a 14-dimensional Lie algebra \( L \) with a 2-dimensional
CSA. Thus if \( L \) is semi-simple, by the classification theorem,
\( L \) must be of type \( G_{2} \).

\begin{proposition}
Let \( L\subseteq sl(V) \) (where \( V \) is finite dimensional)
be a non-zero Li algebra acting irreducibly on \( V \), then \( L \)
is semi-simple.
\end{proposition}
\begin{proof}
See Humphreys page 102.
\end{proof}

Consider \( L \) acting on \( C^{7} \)(by regular matrix multiplication).
Let \( (e_{i})^{7}_{i=1} \) be the standard basis for \( C^{7} \).
\( 
(d_{1}-d_{2})+3(d_{2}-d_{3})=E_{2,2}+2E_{3,3}-3E_{4,4}-E_{5,5}-2E_{6,6}+3E_{7,7}\, 
\epsilon \, H \)
this matrix has distinct eigenvalues which implies that any non-zero
subspace of \( C^{7} \) invariant under \( L \) must contain at
least one of the standard basis vectors. Now consider the following:\\
\( g_{-1}e_{1}=\sqrt{2}e_{2} \), \( g_{2,-1}e_{2}=e_{3} \), \( 
g_{3,-2}e_{3}=e_{4} \),
\( g_{-2}e_{4}=e_{5} \), \( g_{1,-2}e_{5}=-e_{6}, \) \( g_{2,-3}e_{6}=-e_{7} 
\),
and \( g_{-3}e_{7}=2e_{1} \). \\
Thus if one basis vector is in a subspace then all of the basis vectors
are in that subspace. Thus the only invariant subspaces are zero and
\( L \) itself. Therefore, \( L \) acts irreducibly on \( C^{7} \).

Finally, a quick examination of the basis for \( L \) shows that
each matrix has trace zero thus \( L\subseteq sl(7,C) \). Therefore
by the above proposition \( L \) must be semi-simple, hence of type
\( G_{2} \).

\vspace{0.3cm}
{\centering 
\resizebox*{3.25in}{2.5in}{\includegraphics{G2-RootSpaceDecomposition.jpg}} 
\\
Root Space Decomposition\par}
\vspace{0.3cm}


\section{Constructing \protect\( G_{2}\protect \) (as the derivation algebra
of the octonians)}

We now sketch the construction of \( G_{2} \) as the derivation algebra
of the octonians. First we must construct the octonians and review
the definition of a derivation.

Let \( \Theta  \) be a 8-dimensional complex vector space whose elements
are 2x2 matrices with complex entries down the diagonal and 3-vectors
on the off diagonals. That is \[
\Theta =\left\{ \left. \left( \begin{array}{cc}
a & v\\
w & b
\end{array}\right) \, \right| \, a,b\, \epsilon \, C\, and\, v,w\, \epsilon 
\, C^{3}\right\} \]


We give \( \Theta  \) its vector space structure by defining scalar
multiplication and vector addition componentwise. As for vector 
multiplication
we take the following: \[
\left( \begin{array}{cc}
a & v\\
w & b
\end{array}\right) \left( \begin{array}{cc}
c & y\\
z & d
\end{array}\right) =\left( \begin{array}{cc}
ac-v\cdot z & ay+dv+w\times z\\
cw+bz+v\times y & bd-w\cdot y
\end{array}\right) \]


where \( v\cdot z \) is the standard dot product and \( v\times y \)
is the standard cross product. \( \Theta  \) is a non-associative
algebra called the \emph{Cayley} or \emph{octonian} algebra. Fixing
a basis and examining the multiplication table for \( \Theta  \)
(see Humphreys page 104-5), we can see that \( [\Theta ,\Theta ]=\Theta _{0} 
\)
is the span of \[
\left\{ \left( \begin{array}{cc}
1 & 0\\
0 & -1
\end{array}\right) ,\left( \begin{array}{cc}
0 & e_{1}\\
0 & 0
\end{array}\right) ,\left( \begin{array}{cc}
0 & e_{2}\\
0 & 0
\end{array}\right) ,\left( \begin{array}{cc}
0 & e_{3}\\
0 & 0
\end{array}\right) ,\left( \begin{array}{cc}
0 & 0\\
e_{1} & 0
\end{array}\right) ,\left( \begin{array}{cc}
0 & 0\\
e_{2} & 0
\end{array}\right) ,\left( \begin{array}{cc}
0 & 0\\
e_{3} & 0
\end{array}\right) \right\} \]
(trace 0 elements). The complement is spanned by \( \left\{ \left( 
\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \right\}  \).

To review, a derivation is a linear transformation which satisfies
the Leibnitz rule: \( \varphi (xy)=\varphi (x)y+x\varphi (y) \) (product
rule). Let \( L=Der(\Theta ) \) (the set of all derivations). It
is easy to verify that a linear combination of derivations is still
a derivation thus \( L \) is a complex vector space. Also one can
easily verify that the commutator of two derivations is still a derivation.
Thus \( L \) is a Lie algebra with commutator bracket.

Consider \( \varphi \, \epsilon \, L \) and \( x=\left( \begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right)  \) (the muliplicative identity of the octonians) then \( 
\varphi (x)=\varphi (xx)=\varphi (x)x+x\varphi (x)=2\varphi (x) \)
thus \( \varphi (x)=0 \) (so {}``derivations kill the constants'').
Derivations leave \( \Theta _{0} \) invariant, thus \( L \) acts
faithfully on \( \Theta _{0} \) and trivially on its complement.
Let \( \phi :L\, \rightarrow \, gl(7,C) \) be the associated matrix
representation of \( L \) (using the above basis).

We need to show that \( L \) is simple and has a 2-dimensional CSA,
but we also need that \( L \) is sufficiently large (to rule out
\( A_{2} \), \( B_{2} \), \( C_{2} \), and \( D_{2} \)). To do
this first consider the following transformations: \( \delta (x)\left( 
\begin{array}{cc}
a & v\\
w & b
\end{array}\right) =\left( \begin{array}{cc}
0 & x(v)\\
-x^{t}(w) & 0
\end{array}\right)  \) where \( x\, \epsilon \, sl_{3}(C) \). Each \( \delta 
(x) \) is
a derivation in \( L \). Thus we have a non-trivial hence faithful
representation of \( sl_{3}(C) \). Let \( M=\{\delta (x)\, |\, x\, \epsilon 
\, sl_{3}(C)\} \).
Let \( H \) be the image of the diagonal subalgebra of \( sl_{3}(C) \).
A simple calculation shows that \( H \) is its own centralizer in
\( L \). We know that \( Z(M)=\{0\} \) since \( M \) is simple
thus we also have that \( Z(L)=\{0\} \). \( L \) is simple and has
a CSA, \( H \), which is 2-dimensional. To show that \( L \) must
be of type \( G_{2} \) we still need to show it is bigger than the
other rank 2 simple Lie algebras. To do this we note that \( \Theta  \)
is an alternative algebra, this is it satifies: \[
x^{2}y=x(xy)\]
\[
yx^{2}=(yx)x\]
Let \( D_{a,b} \) be a linear transformation defined by:\[
D_{a,b}=[l_{a},\, l_{b}]+[l_{a},\, r_{b}]+[r_{a},\, r_{b}]\]
where \( a,b\, \epsilon \, \Theta  \) and \( l_{a} \), \( r_{a} \)
denote left and right multiplcation respectively. Then \( D_{a,b} \)
is a derivation. These provide the remaining derivations needed to
show that \( L \) is of type \( G_{2} \). For more details see:
\emph{Lie Algebras} by Nathan Jacobson pages 142-145 and \emph{An
Introduction to Nonassociative Algebras} by Richard D. Schafer pages
75-90.
\end{document}

