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\title{Introduction to Lie Groups and Lie Algebras\\November 9, 2004}
\author{Gabriel C. Drummond-Cole (based on the notes of Jaimie Thind)}
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Recall: $\la{g}$ is a semisimple complex Lie algebra, and $\la{h}\in\la{g}$ is a Cartan subalgebra so that $\la{g}=\la{h}\oplus\bigoplus_{\alpha\in R}\la{g}_\alpha.$  Then $R\in\la{h}^*$ has a number of properties:
\begin{defi}
Let $E$ be a finite dimensional real Euclidean space.  A finite subset $R\subset E\backslash \{0\}$ is called a root system if it satisfies
\begin{enumerate}
\item $R$ spans $E.$
\item If $\alpha,\beta\in R$ then $S_\alpha(\beta)=\beta-\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\alpha=\beta-\langle \alpha^\vee,\beta\rangle\alpha\in R.$  This is reflection in the hyperplane perpendicular to $\alpha.$  Here, also, $\alpha^\vee\in E^*$ is defined by the above by $\langle \alpha^\vee,\lambda\rangle=\frac{2(\alpha,\lambda)}{(\lambda,\lambda)}.$
\item If $\alpha,\beta\in R$ then $\langle \alpha^\vee,\beta\rangle=\frac{2(\alpha,\beta)}{(\alpha,\alpha)}\in\mathbb{Z}.$
\end{enumerate}
$R$ is called reduced if $\alpha\in R$ implies $2\alpha\notin R.$
\end{defi}
Some remarks:
\begin{enumerate}
\item If $R_1\subset E_1,R_2\subset E_2$ are root systems then $R_1\sqcup R_2\subset E_1\oplus E_2$ is a root system.
\item If we rescale a root system $\alpha\mapsto c\alpha$ for $c\in\mathbb{R}_*$ it will again be a root system.
\item The third part of the definition means that the projection of $\beta$ on $\alpha$ is an integer multiple of $\frac{\alpha}{2}.$
\end{enumerate}
\begin{example}
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(Pictures)
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Assume we are working with reduced root systems.
Back to Lie algebras, let $\la{g}$ be as above.  So $\la{h}^*=\la{h}^*_\mathbb{R}\oplus i\la{h}^*_\mathbb{R}.$  Then $(\, ,\,)|_{\la{h}^*_\mathbb{R}}$ is positive definite and $R\subset\la{h}^*_\mathbb{R}$ is a reduced root system.
\begin{example}
Take $\la{g}=\la{sl}(n+1,\mathbb{C}).$  Then take $\la{h}$ to be traceless diagonal matrices $$\left\{\left(\begin{array}{ccc}\lambda_1&&\\&\ddots&\\&&\lambda_{n+1}\end{array}\right)|\sum \lambda_i=0\right\}\subset \mathbb{C}^{n+1}.$$

We'll index the roots by $\alpha_{ij},i\ne j.$  Let $\la{g}_{\alpha_{ij}}=\mathbb{C}E_{ij}.$  Then $[h,E_{ij}]=(\lambda_i-\lambda_j)E_{ij}=(\epsilon_i-\epsilon_j)(h)E_{ij},$ where $\epsilon_i(\left(\begin{array}{ccc}\lambda_1&&\\&\ddots&\\&&\lambda_{n+1}\end{array}\right))=\lambda_i.$

Then $\la{h}^*=\mathbb{C}\epsilon_1\oplus\cdots\oplus\mathbb{C}\epsilon_{n+1}/\sum \epsilon_i=0.$

So the root system is $\{\epsilon_i-\epsilon_j,i\ne j\}.$
\\$\la{h}_\mathbb{R}=\{\left(\begin{array}{ccc}\lambda_1&&\\&\ddots&\\&&\lambda_{n+1}\end{array}\right)|\lambda_i\in\mathbb{R},\sum\lambda_i=0\}.$

So $\la{h}^*_\mathbb{R}=\mathbb{R}\epsilon_1\oplus\cdots\oplus\mathbb{R}\epsilon_{n+1}/\sum\epsilon_i=0.$
\end{example}
\begin{thm}
Let $\alpha,\beta\in R$ and $\alpha\ne c\beta.$ Then, up to interchanging $\alpha,\beta,$ we must have one of the following.
\begin{enumerate}
\item $\alpha\perp\beta$ (like $A_1\times A_1).$
\item\begin{enumerate}\item $|\alpha|=|\beta|,\phi=\pi/3.$
\item $|\alpha|=|\beta|,\phi=2\pi/3.$
\end{enumerate} These cases correspond to $A_2.$
\item\begin{enumerate}\item $|\alpha|=\sqrt{2}|\beta|,\phi=\pi/4.$
\item $|\alpha|=\sqrt{2}|\beta|,\phi=3\pi/4.$
\end{enumerate} These cases correspond to $B_2.$
\item\begin{enumerate}\item $|\alpha|=\sqrt{3}|\beta|,\phi=\pi/6.$
\item $|\alpha|=\sqrt{3}|\beta|,\phi=5\pi/6.$
\end{enumerate} These cases correspond to $G_2.$
\end{enumerate}\end{thm}
Look at $n_{\alpha\beta}=\langle b^\vee,\alpha\rangle=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\in\mathbb{Z}.$  This is $\frac{2|\alpha||\beta|}{|\alpha|^2}\cos\phi=\frac{2|\beta|}{|\alpha|}\cos\phi.$  
Then $n_{\alpha\beta}n_{\beta\alpha}=4\cos^2\phi.$  The options are 
\begin{itemize}
\item $n_{\alpha\beta}=n{\beta\alpha}=0.$ This is case one.
\item $n_{\alpha\beta}=\pm 1, n{\beta\alpha}=\pm 1$ so that $\cos^2\phi=\frac{1}{4}$ and $\cos\phi=\pm\frac{1}{2}.$  This is case two above.
\item $n_{\alpha\beta}=\pm 1, n{\beta\alpha}=\pm 2$ so that $\cos^2\phi=\frac{1}{2}$ and $\cos\phi=\pm\frac{1}{\sqrt{2}}.$  This is case three above.
\item $n_{\alpha\beta}=\pm 1, n{\beta\alpha}=\pm 3$ so that $\cos^2\phi=\frac{3}{4}$ and $\cos\phi=\pm\frac{\sqrt{3}}{2}.$  This is case four above.\end{itemize}
\begin{thm} The only rank two root systems are $A_1\times A_1,A_2,B_2,G_2.$\end{thm}
Let $R$ be a rank two root system.  Take $\alpha,\beta\in R$ such that the angle between them is the smallest possible.  Then we must be in one of the cases $1,2a,3a,4a$ (to eliminate the $b$ cases reflect to get a root with smaller angle).

If we are in the case $2a,$ by reflections it must contain all of the six roots in $A_2;$ if it contained anything else we'd get a pair of roots with smaller angle.  Similarly for $3a,4a.$

Now let $t\in E$ such that $(t,\alpha)\ne 0$ for all $\alpha\in R.$  Then $R=R_+\sqcup R_-,$ where $R_+=\{\alpha\in R|(\alpha,t)>0\}$ and $R_-=\{\alpha\in R|(\alpha,t)<0\}.$
\begin{example} Look at the root system $A_n$ of $\la{sl}(n+1,\mathbb{C}).$  Take $t\in \la{h}^*_\mathbb{R}$ as $t=(t_1,\cdots,t_{n+1})$ with $t_1>\cdots>t_{n+1}.$  Then $(t,\alpha_{ij})=(t,\epsilon_i-\epsilon_j)=t_i-t_j,$ so this is positive when $i<j,$ negative when $i>j.$

Then $R_+=\{\alpha_{ij}|i<j\};R_-=\{\alpha_{ij}|i>j\}.$  So $\oplus_{\alpha\in R_+}\la{g}_\alpha$ is the set of strictly upper triangular matrices, and  $\oplus_{\alpha\in R_-}\la{g}_\alpha$ is the set of strictly lower triangular matrices.
\end{example}
\begin{defi} A root $\alpha\in R_+$ is simple if it cannot be written as a sum of positive roots.
\end{defi}
Let $\Pi=\{\alpha_1,\cdots,\alpha_k\}$ be the set of simple roots.
\begin{lemma} Any positive root can be written as $\alpha=\sum n_i\alpha_i$ for $n_i\in\mathbb{Z}_+.$
\end{lemma}
\begin{lemma}$(\alpha_i,\alpha_j)\le 0.$
\end{lemma}\begin{lemma}$\{\alpha_i,\ldots,\alpha_k\}$ are linearly independent
\end{lemma}
The proof is an exercise.

In $\la{sl}(n+1),$ the simple roots are $\epsilon_1-\epsilon_2,\cdots,\epsilon_n-\epsilon_{n+1}.$

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