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\title{Introduction to Lie Groups and Lie Algebras\\November 11, 2004}
\author{Gabriel C. Drummond-Cole}
\newtheorem{lemma}{Lemma}
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\begin{document}
\noindent
\maketitle  
Recall: $R=R_+\cup R_-\subset E$ and $r=\dim E.\ \Pi=\{\alpha_1,\cdots,\alpha_r\}\subset R_+$ is the set of simple roots.  For any $\alpha\in R_+,\ \alpha=\sum n_i\alpha_i$ for $n_i\ge 0.$
\begin{example}
$R=\{e_i-e_j,i\ne j\}\subset \mathbb{R}^{n+1}/(1,\cdots,1).
\\R_+=\{e_i-e_j,i<j\}.
\\\Pi=\{\alpha_1=e_1-e_2,\cdots,\alpha_n=e_n-e_{n+1}\}.$
\end{example}
Can one recover $R$ from $\Pi,$ and for different choices of polarization do we get ``different'' $\Pi?$

Note that $R=R_+\cup R_-$ depends only on the signs of $(t,\alpha).$  It changes when $(t,\alpha)=0,$ i.e., $t$ crosses the hyperplane $H_\alpha=\{\lambda|(\alpha,\lambda)=0\}.$
\begin{defi}
A Weyl chamber is a connected component of $E\backslash \cup H_\alpha.$
\end{defi}
Each Weyl chamber is the intersection of half-spaces so it is a convex polygonal cone.  There are only finitely many.
\begin{lemma}
$C\to R=R_+^C\cup R_-^C,$ where $R_+^C=\{\alpha:\alpha(C)>0\}$ is a bijection between Weyl chambers and polarizations.
\end{lemma}
The inverse map is $R=R_+\cup R_-$ to $C_+=\{t\in E|(\alpha,t)>0$ for all $\alpha\in R_+\}=\{t\in E|(t,\alpha_i)>0$ for all $\alpha_i\in \Pi\}.$

If $(t,\alpha_i)>0$ for all $\alpha_i\in \Pi$ then for all $\alpha\in R_+\ (t,\alpha)=\sum n_i(t,\alpha_i)>0.$

Every Weyl chamber is bounded by exactly $r$ hyperplanes, called the walls of the Weyl chamber.
\begin{defi}The Weyl group $W$ of $R\subset GL(E)$ is the group generated by $S_\alpha,\alpha\in R.$
\end{defi}
\begin{example}
Let $R=A_n,$ the root system of $\la{sl}(n+1).$  Then $\alpha_{ij}=(e_i-e_j)$ just as $s_{ij}$ transposes $i$ and $j.$  So $S_{\alpha_{ij}}$ sends $(t_1,\cdots,t_i,\cdots,t_j,\cdots,t_{n+1}$ to $(t_1,\cdots,t_j,\cdots,t_i,\cdots,t_{n+1}.$  So $W=S_{n+1}.$
\end{example}
Some properties are that $W\subset O(E)$ and $|W|>\infty.$

The first is obvious, the second is because $W\subset S(R).$
\begin{thm}
\begin{enumerate}
\item $W$ acts transitively on the set of Weyl chambers.
\item Fix $R=R_+\cup R_-.$  Then $W$ is generated by $S_i=S_{\alpha_i}$ for $\alpha_i\in\Pi.$
\item For every root $\alpha$ there exists $w\in W$ such that $w(\alpha)\in\Pi$ so $W\Pi=R.$
\end{enumerate}
\end{thm}
\begin{cor}
\begin{enumerate}
\item $R$ can be recovered from $\Pi.$
\item Say $R=R_+\cup R_-=R_+'\cup R_-',$ with simple root sets $\Pi,\Pi'.$  Then $\Pi'=w\Pi$ for some $w\in W.$
\end{enumerate}
\end{cor}
\begin{lemma}
Fix $R=R_+\cup R_-$ and let $W'=\langle S_i\rangle,$ the group generated by the simple reflections.  Then $W'$ acts transitively on the set of Weyl chambers.
\end{lemma}
It suffices to prove that for a Weyl chamber $C,$ there exists $w\in W'$ such that $w(C_+)=C.$  Let $\ell$ be the number of hyperplanes seperating $C$ and $C_+.$  Take a segment connecting a point in $C$ with a point in $C_+.$  It will intersect all these $\ell$ hyperplanes and only them at isolated points.  Then $C_1=S_{\beta_1}(C_0),C_2=S_{\beta_2}(C_1),$ and so on to $C=C_l=S_{\beta_\ell}(C_{\ell-1}).$
...
Proof of theorem from lemma:

\begin{enumerate}
\item Obvious
\item Let $\beta\in R$ and $C$ be a Weyl chamber such that $H_\beta$ is a wall of $C.$  Then $C=w(C_+);\ H_\beta=w($wall of $C_+)=w(H_{\alpha_i})$ for some $\alpha_i\in \Pi$ and $w\in W'.$  So $\beta=\pm w\alpha_i.$  Then $S_\beta=ws_iw^{-1}\in W'.$
\item We already know $\beta=\pm w(\alpha_i).$  If $\beta=-w\alpha_i$ then $\beta=wS_i(\alpha_i).$
\end{enumerate}
\end{document}
