\documentclass[10pt,twoside]{article} \usepackage{amsmath,amssymb,amsthm, fullpage} \usepackage{graphics} \usepackage{graphicx} \usepackage{amsmath,amsfonts} \newcommand{\Rr}{\mathbb R} \newcommand{\Cc}{\mathbb C} \newcommand{\Zz}{\mathbb Z} \newcommand{\mg}{\mathfrak{g}} \newcommand{\mh}{\mathfrak{h}} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{prop}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newcommand{\exercise}[1]{ \vspace{9pt}\noindent {\bf Exercise #1.} } \title{18.745: Lecture 21} \author{Professor: Victor Ka\v{c} \\Scribe: Ruochuan Liu, Zhongtao Wu} \date{} \begin{document} \maketitle Let $(V$,$\triangle$) be a root system. Define a reflection $r_{\alpha}\in $End$(V)$ for every $\alpha\in\triangle$ by: $$r_{\alpha}(v)=v-\frac{2(v,\alpha)}{(\alpha,\alpha)}\alpha$$ $r_{\alpha}$ has properties: \renewcommand{\labelenumi}{(\roman{enumi})} \begin{enumerate} \item $r_{\alpha}$ fixes pointwise the hyperplane $\tau_{\alpha}=\{u\in V|(u,\alpha)=0\}$. \item $r_{\alpha}=-\alpha$. \item $r_{\alpha}^{2}=1$ and $\gamma_{\alpha}\in O(V,(\cdot,\cdot))$, i.e $r_{\alpha}$ is invertible and $(\gamma_{\alpha}(v),\gamma_{\alpha}(v))=(v,v)$. \item $r_{\alpha}(\triangle)=\triangle$. \end{enumerate} (i), (ii) and (iii) are obvious. (iv) follows from the string property. \exercise{21.1 (optional)} Show that property (iv) along with $\frac{2(\alpha, \beta)}{(\alpha, \alpha)}\in \Zz$ for all $\alpha, \beta \in \Delta$ implies the string property. \begin{proof} Denote $<\beta, \alpha> := \frac{2(\beta, \alpha)}{(\alpha,\alpha)}$. Then $<\beta, \alpha>\in \Zz$ for all $\alpha, \beta \in \Delta$, and $r_{\alpha}(\beta)=\beta-<\beta,\alpha>\alpha$. \renewcommand{\descriptionlabel}[1]{\hspace\labelsep #1:} \begin{description} \item[Step I] Let $\alpha, \beta$ be nonproportional roots. If $(\alpha, \beta)>0$, then $\alpha-\beta$ is a root. If $(\alpha,\beta)<0$, then $\alpha+\beta$ is a root. \item[Proof] If $(\alpha, \beta)>0$, then both $<\alpha, \beta>$ and $<\beta,\alpha>$ are positive. By Cauchy's inequality, $$<\alpha, \beta><\beta,\alpha>=\frac{4(\alpha, \beta)^2}{(\alpha,\alpha)(\beta, \beta)}< 4, $$ where the inequality is strict since $\alpha$ and $\beta$ are nonproportional roots. Hence, at least one of $<\alpha,\beta>$, $<\beta,\alpha>$ equals 1. If $<\alpha,\beta>=1$, then $r_\beta(\alpha)=\alpha-\beta\in \Delta$; similarly, if $<\beta, \alpha>=1$, then $\beta-\alpha\in \Delta$, hence $r_{\beta-\alpha}(\beta-\alpha)=\alpha-\beta \in \Delta$. The case $(\alpha, \beta)<0$ is similar. \item[Step II] The $\alpha$-string through $\beta$ is unbroken, i.e., if $p,q\in \Zz_+$ are the largest integers for which $\beta-p\alpha, \beta+q\alpha \in \Delta$, then $\beta+i\alpha\in \Delta$, $\forall -p\leq i \leq q$. \item[Proof] If not, we can find $-p \leq i $. \item[Proof] Since $r_\alpha$ just adds or subtracts a multiple of $\alpha$ to any root, the string is invariant under $r_\alpha$. In particular, $r_\alpha(\beta+q\alpha)=\beta-p\alpha$. The left hand side is also equal to $\beta-<\beta,\alpha>\alpha-q\alpha$. Hence, $p-q=<\beta,\alpha>$. \end{description} Thus, we proved the string property for $\Delta$. \end{proof} \begin{definition} The {\em Weyl group} of a root system is a subgroup $W$ of $O(V,(\cdot,\cdot))$ generated by all reflections $\gamma_{\alpha}$ for $\alpha\in\triangle$. This is a finite group since it permutes elements of a finite set which spans $V$. \end{definition} \begin{example} $\triangle_{A_{r}}=\{\varepsilon_{i}-\varepsilon_{j}|1\leq i,j\leq r+1, i\neq j\}$. Let $\alpha=\varepsilon_{i}-\varepsilon_{j}$, then $$ r_{\alpha}(\varepsilon_{k})=\left\{ \begin{array}{r@{\quad:\quad}l} \varepsilon_{k} & k\neq i,j\\ \varepsilon_{j}& k=i \\ \varepsilon_{i} & k=j \end{array} \right\} $$ Hence $r_{\alpha}=(ij)$, i.e., it just permutes $\varepsilon_{i}$ and $\varepsilon_{j}$. Thus the Weyl group of $A_{r}$ is $S_{r+1}$. \end{example} \exercise{21.2} Compute the Weyl group of the root systems $B$, $C$ and $D$. \begin{proof}\ \renewcommand{\labelenumi}{(\alph{enumi})} \begin{enumerate} \item Root system $\Delta_{B_r}=\{\pm\epsilon_i\pm\epsilon_j \ ( i\neq j), \ \pm\epsilon_i \}$. \begin{itemize} \item When $\alpha=\epsilon_i-\epsilon_j$, $r_\alpha(\epsilon_k)=\begin{cases} \epsilon_k & \text{if $k\neq i,j$;}\\ \epsilon_j & \text{if $k=i$;}\\ \epsilon_i & \text{if $k=j$.} \end{cases}$ \item When $\alpha=\epsilon_i+\epsilon_j$, $r_\alpha(\epsilon_k)=\begin{cases} \epsilon_k & \text{if $k\neq i,j$;}\\ -\epsilon_j & \text{if $k=i$;}\\ -\epsilon_i & \text{if $k=j$.} \end{cases}$ \item When $\alpha=\epsilon_i$, $r_\alpha(\epsilon_k)=\begin{cases} \epsilon_k & \text{if $k\neq i$;}\\ -\epsilon_i & \text{if $k=i$.} \end{cases}$ \end{itemize} Hence, the Weyl group is generated by all permutations of the set $\{ \epsilon_1, \epsilon_2,\dotsc, \epsilon_r \}$ and the operations of sign changes. In terms of group, it is isomorphic to $S_r \times \Zz_2^r$. \item Root system $\Delta_{C_r}=\{\pm\epsilon_i\pm\epsilon_j \ ( i\neq j), \ \pm 2\epsilon_i \}$. We can check that $r_\alpha(\epsilon_k)$ has exactly the same form as the previous case, hence its Weyl group is also $S_r \ltimes \Zz_2^r$. \item Root system $\Delta_{D_r}=\{\pm\epsilon_i\pm\epsilon_j, \ i\neq j \}$. In this case, the Weyl group is generated by the first two types of reflections in (a). Also, notice that $$r_{\epsilon_i-\epsilon_j}\circ r_{\epsilon_i+\epsilon_j}(\epsilon_k)=\begin{cases} \epsilon_k & \text{if $k\neq i,j$;}\\ -\epsilon_i & \text{if $k=i$;}\\ -\epsilon_j & \text{if $k=j$.} \end{cases}$$ This is the same as changing signs for a pair $(\epsilon_i, \epsilon_j)$. Hence, each element in the Weyl group is a permutation of set $\{ \epsilon_1, \epsilon_2,\dotsc, \epsilon_r \}$, composed with an even number of sign changes. In terms of group, it is isomorphic to $S_r \ltimes \Zz_2^{r-1}$. \end{enumerate} \end{proof} \exercise{21.3} If $A\in O(V,(\cdot,\cdot))$, then $Ar_\alpha A^{-1}=r_{A(\alpha)}$. \begin{proof} Let $AT_\alpha :=\{A(u)| \ (\alpha,u)=0 \}$. Since $A\in O(V,(\cdot,\cdot))$, we have $(A(u), A(\alpha))=(u,\alpha)=0$ for all $A(u)\in AT_\alpha$. Hence $r_{A(\alpha)}(AT_\alpha)=AT_\alpha$. Also, $Ar_\alpha A^{-1}(AT_\alpha)=Ar_\alpha (T_\alpha)=AT_\alpha$. So both reflections $Ar_\alpha A^{-1}$ and $r_{A(\alpha)}$ fix the hyperplane $AT_\alpha$, hence $Ar_\alpha A^{-1}=r_{A(\alpha)}$. \end{proof} Recall that, given a choice of $f\in V^{*}$ such that $f(\alpha)\neq 0$ for all $\alpha\in \triangle$, we get a subset $\triangle_{+}=\{\alpha\in \triangle|f(\alpha)>0\}$ and $\Pi=\{\alpha_{1},\alpha_{2},\ldots,\alpha_{r}\}$ of simple roots such that any $\alpha\in \triangle_{+}$ is of the form $\alpha=\Sigma_{i}k_{i}\alpha_{i}$, $k_{i}\in \mathbb{Z}_{\geq 0}$. The reflections $s_{i}=r_{\alpha_{i}}$ are called {\em simple reflections}. We use the notation $\operatorname{height}(\alpha):=\sum_{i}k_{i}$. \begin{theorem}\ \renewcommand{\labelenumi}{(\alph{enumi})} \begin{enumerate} \item $\triangle_{+}\backslash \{\alpha_{i}\}$ is $s_{i}$-invariant. \item If $\alpha\in \triangle_{+}\backslash \Pi$, there is a $s_{i}$ such that height$(s_{i}(\alpha))<$height$(\alpha)$. \item If $\alpha\in \triangle_{+}\backslash \Pi$, we can choose a sequence of simple reflections $s_{i_{1}},\ldots,s_{i_{k}}$ such that $s_{i_{1}}\ldots s_{i_{k}}(\alpha)\in \Pi$ and $s_{i_{j}}\ldots s_{i_{k}}(\alpha)\in \triangle_{+}$ for each $1\leq j\leq k$. \item $W$ is generated by simple reflections. \end{enumerate} \end{theorem} \begin{proof}\ \renewcommand{\labelenumi}{(\alph{enumi})} \begin{enumerate} \item Applying simple reflection $s_{i}$ changes sign of at most one coefficient $k_{i}$ of $\alpha\in\triangle_{+}$. If $k_{i}$ changes to negative, then $s_{i}(\alpha)$ wouldn't be a root, hence $s_{i}(\alpha)\in \triangle_{+}$. \item If $\operatorname{height}(s_{i}(\alpha))$ doesn't decrease for all $s_{i}$, then from $s_{i}(\alpha)=\alpha-\frac{2(\alpha,\alpha_{i})}{(\alpha_{i},\alpha_{i})}\alpha_{i}$ we get $(\alpha,\alpha_{i})\leq0$ for all $i$. Hence $(\alpha,\alpha)=\Sigma_{i}k_{i}(\alpha,\alpha_{i})\leq0$, a contradiction! \item follows from $(b)$ and $(a)$. \item Denote $W^{'}$ the subgroup of $W$ generated by simple reflections. By $(c)$, for any $\alpha\in\triangle_{+}$, there exists $w\in W^{'}$ such that $w(\alpha)=\alpha_{i}\in \Pi$ for some $i$. By Ex21.3, $wr_{\alpha}w^{-1}=r_{w(\alpha)}$, hence $r_{\alpha}=w^{-1}s_{i}w \in W^{'}$. \end{enumerate} \end{proof} Consider $V\backslash \bigcup_{\alpha\in\triangle_{+}}T_{\alpha}=\coprod_{j}C_{j}$, where $C_{j}$ are connected components of this set. $C_{j}$'s are called open chambers, $\overline{C_{j}}$'s are called (closed) chambers. Also, define the {\em fundamental chamber}: $\overline{C}=\{v\in V|(\alpha_{i},v)\geq 0$ $i=1,\ldots,r\}$. \exercise{21.4} Show that $T_\alpha \cap C=\varnothing$, where $C=\{ v\in V | \ (\alpha_i, v)>0, i=1,\dotsc,r\}$ is the open fundamental chamber. Hence the fundamental chamber is a chamber. \begin{proof} Suppose $v\in T_\alpha \cap C$, then $(v,\alpha)=0$ and $(v,\alpha_i)>0$ for $i= 1,\dotsc, r$. But $\alpha=\sum_{i=1}^{r} k_i\alpha_i$, where $k_i\in \Zz_+$. Hence, $k_i=0$ for all $i$, and consequently $v=0$. So we proved $T_\alpha \cap C=\varnothing$. Since $C$ is connected, $C \subset C_j$ for some $j$. On the other hand, $(\alpha_i,v)\neq 0$, $\forall i, v\in C_j$ by definition. And since the inner product is a continuous function of $v$, we conclude that $(\alpha_i, v)>0$ $\forall v\in C_j$. Hence $C_j\subset C$. Thus, $\bar{C}=\bar{C_j}$, i.e, the fundamental chamber is a chamber. \end{proof} \begin{theorem}\ \renewcommand{\labelenumi}{(\alph{enumi})} \begin{enumerate} \item $W$ permutes all chambers transitively, i.e for any chamber $\overline{C_{1}}$ and $\overline{C_{2}}$, there exists $w\in W$ such that $w\overline{C_{1}}=\overline{C_{2}}$. \item Let $\triangle_{+}$ and $\triangle_{+}^{'}$ be subsets of positive roots of $\triangle$ defined by $f$ and $f^{'}$ respectively . Then there exists $w\in W$ such that $w(\triangle_{+})=\triangle_{+}^{'}$. In particular, the Cartan matrix is independent of the choice of $f$. \end{enumerate} \end{theorem} \begin{proof}\ \renewcommand{\labelenumi}{(\alph{enumi})} \begin{enumerate} \item Choose $P_{i}\in C_{i}$ ($i=1,2$) such that the interval $[P_{1},P_{2}]$ doesn't intersect any of $\tau_{\alpha}\cap\tau_{\beta}$, where $\alpha$, $\beta\in\triangle_{+}$ and $\alpha\neq\beta$. Now move along the interval $[P_{1},P_{2}]$ until we hit a hyperplane $\tau_{\alpha}$. Apply reflection $r_{\alpha}$ to $\overline{C_{1}}$. Keep moving and applying reflections until we reach $\overline{C_{2}}$. \item $\triangle_{+}$ and $\triangle_{+}^{'}$ define fundamental chamber $\overline{C}$ and $\overline{C^{'}}$ respectively. By $(a)$, there exists $w\in W$ such that $w(\overline{C})=\overline{C^{'}}$. Hence $w(\triangle_{+})=w(\triangle_{+}^{'})$, since $\overline{C}=\{v\in V|(\alpha_{i},v)\geq 0$ $i=1,\ldots,r\}$. \end{enumerate} \end{proof} \begin{definition} Fix $\Pi\subset\triangle_{+}\subset\triangle$, then we have simple reflections $s_{1},\ldots,s_{r}\in W$. Given $w\in W$, an expression $w=s_{i_{1}}\ldots s_{i_{l}}$ is called {\em reduced} if $l$ is minimal possible. We let $l=l(w)$ called the length of $w$. Note that $\det w=(-1)^{l(w)}$ since $\det s_{i}=-1$. \end{definition} \begin{example} $l(s_{i})=1$, $l(s_{i}s_{j})=2$ if $i\neq j$, but $l(s_{i}^{2})=0$. \end{example} \begin{lemma} [Exchange Lemma] Suppose that $s_{i_{1}}\ldots s_{i_{t-1}}(\alpha_{i_{t}})\in \triangle_{-}$, then the expression $w=s_{i_{1}}\ldots s_{i_{t}}$ is not reduced. Namely, there exists $1\leq r