\documentclass[11pt]{article} \usepackage{cancel} \usepackage{amsmath} \usepackage{amssymb} \usepackage{epsfig} \usepackage{pstricks} \newcommand{\handout}[5]{ \noindent \begin{center} \framebox{ \vbox{ \hbox to 5.78in { {\bf 18.745 Introduction to Lie Algebras } \hfill #2 } \vspace{4mm} \hbox to 5.78in { {\Large \hfill #5 \hfill} } \vspace{2mm} \hbox to 5.78in { {\em #3 \hfill #4} } } } \end{center} \vspace*{4mm} } \newcommand{\lecture}[4]{\handout{#1}{#2}{#3}{Scribe: #4}{Lecture #1}} \DeclareSymbolFont{AMSb}{U}{msb}{m}{n} \DeclareMathSymbol{\N}{\mathbin}{AMSb}{"4E} \DeclareMathSymbol{\Z}{\mathbin}{AMSb}{"5A} \DeclareMathSymbol{\R}{\mathbin}{AMSb}{"52} \DeclareMathSymbol{\Q}{\mathbin}{AMSb}{"51} \DeclareMathSymbol{\I}{\mathbin}{AMSb}{"49} \DeclareMathSymbol{\C}{\mathbin}{AMSb}{"43} \DeclareMathSymbol{\F}{\mathbin}{AMSb}{"46} \newcommand{\sll}{\mbox{sl}} \newcommand{\gl}{\mbox{gl}} \newcommand{\GL}{\mbox{GL}} \newcommand{\tr}{\mbox{tr\ }} \newcommand{\Mat}{\mbox{Mat}} \newcommand{\Lie}{\mbox{Lie}} \newcommand{\Der}{\mbox{Der\ }} \newcommand{\End}{\mbox{End\ }} \newcommand{\ad}{\mbox{ad\ }} \newcommand{\im}{\mbox{im\ }} \newcommand{\Ker}{\mbox{ker\ }} \newcommand{\g}{\mathfrak{g}} \newcommand{\h}{\mathfrak{h}} \newcommand{\m}{\mathfrak{m}} \newcommand{\He}{\mathfrak{H}} \newcommand{\sk}{\vspace*{1em}} \newtheorem{defn}{Definition} \newtheorem{remark}{Remark} \newtheorem{example}{Example} \newtheorem{proof}{Proof} % 1-inch margins, from fullpage.sty by H.Partl, Version 2, Dec. 15, 1988. \topmargin 0pt \advance \topmargin by -\headheight \advance \topmargin by -\headsep \textheight 8.9in \oddsidemargin 0pt \evensidemargin \oddsidemargin \marginparwidth 0.5in \textwidth 6.5in \parindent 0in \parskip 1.5ex %\renewcommand{\baselinestretch}{1.25} \begin{document} \lecture{14}{Fall 2004}{Prof.\ Victor Ka\v{c}}{Maksim Lipyanskiy} Recall that $g = h\oplus (\oplus_{\alpha \in \Delta} g_{\alpha})$, dim $g_{\alpha}=1$ and $h$ is abelian. Let us compute the Killing form on $h$: $K(h_1,h_2) = tr_g(ad(h_1)ad(h_2))=tr_h(ad(h_1)ad(h_2))+\sum tr_{g_\alpha}(ad(h_1)ad(h_2))$ Thus, $K(h_1,h_2) = \sum_{\alpha} \alpha(h_1) \alpha(h_2)$. On $h^*$, $K(\lambda, \mu) = \sum_{\alpha} K(\lambda,\alpha)K(\mu,\alpha)$. {Theorem 4.} (a) $\Delta$ span $h^*$ over $\mathbb{F}$. \\ (b) $K(\alpha,\beta) \in \mathbb{Q}$ if $\alpha, \beta \in \Delta$ \\ (c) $ K $ on $h^* \mathbb{Q}$ is a positive-definite, $\mathbb{Q}$-valued, symmetric bilinear form. Proof. (a) In the contrary case, there exists a nonzero $h_1 \in h$ such that $\alpha(h_1)=0$ for all $\alpha \in \Delta$. Hence $[h_1,g_\alpha] =0$, and $[h_1,h]=0$ hence $h_1$ is in the center which is not possible. \\ (b) Recall that $2K(\alpha,\beta)K(\alpha,\alpha)^{-1} = p-q \in \mathbb{Z}$. But by the formula above $K(\alpha,\alpha)=\sum K(\alpha,\gamma)^2$ thus, $4K(\alpha,\alpha)^{-1} = \sum_\gamma 2K(\alpha,\gamma)K(\alpha,\alpha)^{-1} \in \mathbb{Z}$. The result follows from polarization identity.\\ (c) $K(\lambda,\lambda)$, $(\lambda \in h^*) = \sum_\alpha K(\alpha,\lambda)^2 \geq 0$ with equality iff all $K(\alpha,\lambda)=0$. Hence $\lambda=0$ since $K$ on $h$ is nondegenerate. Proposition. If $g$ is semisimple and $a \subset g$ is an ideal then $K$ restricted to $a$ is nondegenerate and $g=a\oplus a^\bot$. Thus, $a$ is semisimple. Proof. $K(a \cap a^\bot,a \cap a^\bot)=0$, hence by Cartan's criterion $a \cap a^\bot$ is a solvable ideal. This contradicts the semisimplicity of $g$. Since dim$(g) \leq $ dim$(a) + $dim$(a^\bot)$ the result follows. Definition. A Lie algebra is $simple$ if it is not abelian and has no proper nonzero ideals. Corollary. $g$ is semisimple iff it is a direct sum of simple Lie algebras. Exercise 14.1 Show that decomposition is unique up to permutation and any ideal is a subsum of the ideals in decomposition. \\ Solution. It suffices to prove the second statement. So let $g=\bigoplus_i g_i$ where the $g_i$ are simple. If $a$ is an ideal $a \cap g_i$ is either zero or $g_i$. We need only establish that $a$ is homogeneous, i.e. $a=\bigoplus_i a \cap g_i$. If $a$ has a nontrivial projection to $g_i$ we have $[g_i,a]=g_i$ since $g_i$ has trivial center. But $[g_i,a]\subset a$. Let $g=g_1\oplus g_2$ be a direct sum of semisimple Lie algebras. If $h_1$, $h_2$ are Cartan subalgebras then $h_1\oplus h_2$ is a Cartan subalgebra. If $g_1 = h_1\oplus \bigoplus_{\alpha \in \Delta_1} g_{\alpha}, g_2 = h_2\oplus \bigoplus_{\beta \in \Delta_2 } g_{\beta}$ then $g$ has root space $\Delta = \Delta_1 \sqcup \Delta_2$ where we extend $\Delta_i$ by zero to the other Cartan subalgebra. This decomposition has the property that:\\ (*) If $\alpha \in \Delta_1$, $\beta \in \Delta_2$ then $\beta + \alpha \notin \Delta \cup 0$. Definition. A set of roots is $indecomposable$ if there is no nontrivial decomposition such that (*) holds. Clearly an indecomposable semisimple Lie algebra must be simple. $\Delta$ is indecomposable iff for any $\alpha, \beta \in \Delta$ there exists a sequence $\alpha = \gamma_1, . . .,\gamma_n=\beta$ such that $\gamma_i+\gamma_{i+1} \in \Delta$ or $0$. Example 1. $g=sl_n (n \geq2)$, $h=$ diagonal matrices with trace zero. Let $\epsilon_i(a_{jj})=\delta_{ij}$ and restrict $\epsilon_i$ to $h$. The eigenvectors for the Cartan subalgebra are $E_{ij}$. Note that the corresponding root is $\epsilon_i -\epsilon_j$. Hence we have root space decomposition $sl_n\mathbb{F}=h\oplus\bigoplus_{i,j}\mathbb{F}E_{ij}$. The root space $\epsilon_i-\epsilon_j$ where $i\neq j$ is indecomposable. Indeed, $(\epsilon_i-\epsilon_j) +(\epsilon_l -\epsilon_k)$ is a root or zero if $j=l$ and $(\epsilon_i-\epsilon_j) +(\epsilon_j -\epsilon_l)$, $(\epsilon_j-\epsilon_l) +(\epsilon_l -\epsilon_k)$ are roots or zero otherwise. Example 2. $g=so_n(\mathbb{F}) = A\in gl_n(\mathbb{F})$ such that $A^TB+BA=0$. The best choice of $B$ is $B_{i,n+1-i}=1$ and zero otherwise. Exercise 14.2 $A\in so_n(\mathbb{F})$ iff $A'=-A$ where $A'_{ij}=A_{n-j+1,n-i+1}$. \\ Solution. We have $(AB)_{ij}=A_{i,n+1-j}$ and $(BA)_{ij}=A_{n+1-i,j}$. From this it is easy to see that $A^TB+BA=0$ iff $A_{n+1-j,i}+A_{n+1-i,j}=0$. Make the substitution $i'=n+1-i$. We have: $A_{n+1-j,n+1-i'}+A_{i',j}=0$ which is what we wanted. In this case the Cartan subalgebra $h$ is diag$(a_1, ...,a_r,-a_r, ...,-a_1)$. The roots are $F_{ij}=E_{ij}-E_{n+1-j,n+1-i}$ where the $i,j$ are determined below. Observe that $\epsilon_i=-\epsilon_{n+1-i}$ when restricted to $h$. $F_{ij}$ is a root vector with root $\epsilon_i-\epsilon_j$. Note that a basis for the root system is formed by the vectors $e_1, ..., e_r$ where either $n=2r$ or $n=2r+1$. In the case $n=2r+1$ the roots are $\{ \epsilon_i-\epsilon_j, \epsilon_i, -\epsilon_i,\epsilon_i+\epsilon_j,-\epsilon_i-\epsilon_j \}$ where $i\neq j$ and $i,j \leq r$. This follows from the fact that $e_{r+1}=0$ and that $\epsilon_i=-\epsilon_{n+1-i}$.\\ The case $n=2r$ is almost identical. Here the roots are $\{ \epsilon_i-\epsilon_j,\epsilon_i+\epsilon_j,-\epsilon_i-\epsilon_j \}$ where $i\neq j$ and $i,j \leq r$. Exercise 14.3 Do the case $r=2n$. \\ Solution. See above. Exercise 14.4 (a) $\Delta_{so_{2r+1}}$ is indecomposable iff $r\geq 1$. Thus $so_{2r+1}$ is simple iff $r\geq 1$. \\ (b) $\Delta_{so_{2r}}$ is indecomposable iff $r\geq 3$. \\ $\Delta_{so_4}=\{\pm (\epsilon_1-\epsilon_2) \} \cup \{\pm (\epsilon_1+\epsilon_2 )\} $, Thus, it is not simple. ($so_2$ is abelian). \\ Solution. (a) Notice that each $\epsilon_i$ is connected to $\epsilon_j$. Note the obvious fact that roots are connected to their inverses. Furthermore, each $\pm \epsilon_i \pm \epsilon_j$ is connected to an $\pm \epsilon_i$ by choosing the sign of this last root judiciously. \\ (b) Note that $\pm \epsilon_i + \mu \epsilon_j$ is connected to $-\mu \epsilon_j \pm \epsilon_k$ where $i, j, k$ are distinct and $\mu =\pm 1$ ($r \geq 3$). We have shown that one can change one index of any root without leaving the component. Thus we can connect all roots by changing one index at a time. \end{document}