\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{1 --- September 7, 2004}{Fall 2004}{Prof.\ Victor Ka\v{c}}{Patrick Lam} \begin{defn} (a) An \emph{algebra} is a vector space over a field $\F$, endowed with a multiplication $ab$, which is bilinear: \begin{eqnarray*} a(\lambda b+\mu c) &=& \lambda ab + \mu ac \\ (\lambda b + \mu c) a &=& \lambda ba + \mu ca \end{eqnarray*} An altebra is associative if $(ab)c = a(bc)$. (b) A \emph{Lie algebra} is an algebra $\g$ with product $[a, b]$, called the \emph{bracket} of $a$ and $b$, subject to two axioms: \begin{itemize} \item skew commutativity: $[a,a] = 0$ \item Jacobi identity: $[a, [b, c]] + [b, [c, a]] + [c, [a,b]] = 0$. \end{itemize} \begin{center} % PSTricks TeX macro % Title: l1-diag.dia % Creator: Dia v0.94 % CreationDate: Mon Sep 20 13:20:15 2004 % For: plam % \usepackage{pstricks} % The following commands are not supported in PSTricks at present % We define them conditionally, so when they are implemented, % this pstricks file will use them. \ifx\setlinejoinmode\undefined \newcommand{\setlinejoinmode}[1]{} \fi \ifx\setlinecaps\undefined \newcommand{\setlinecaps}[1]{} \fi % This way define your own fonts mapping (for example with ifthen) \ifx\setfont\undefined \newcommand{\setfont}[2]{} \fi \pspicture(5.550000,-7.350000)(9.088962,-3.350000) \scalebox{1.000000 -1.000000}{ \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \newrgbcolor{diafillcolor}{1.000000 1.000000 1.000000} \psset{fillcolor=diafillcolor} \newrgbcolor{dialinecolor}{1.000000 1.000000 1.000000} \psset{linecolor=dialinecolor} \psellipse*(7.000000,5.450000)(1.000000,1.000000) \psset{linewidth=0.050000} \psset{linestyle=solid} \psset{linestyle=solid} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psellipse(7.000000,5.450000)(1.000000,1.000000) \setfont{Helvetica}{0.800000} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \rput[l](6.750000,4.000000){\scalebox{1 -1}{a}} \setfont{Helvetica}{0.800000} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \rput[l](8.175000,6.975000){\scalebox{1 -1}{b}} \setfont{Helvetica}{0.800000} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \rput[l](5.550000,7.000000){\scalebox{1 -1}{c}} \psset{linewidth=0.050000} \psset{linestyle=solid} \psset{linestyle=solid} \setlinecaps{0} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psline(6.950000,4.253750)(6.950000,4.778750) \psset{linewidth=0.050000} \psset{linestyle=solid} \psset{linestyle=solid} \setlinecaps{0} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psline(7.600000,5.903750)(8.000000,6.303750) \psset{linewidth=0.050000} \psset{linestyle=solid} \psset{linestyle=solid} \setlinecaps{0} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psline(6.400000,6.003750)(6.050000,6.353750) \psset{linewidth=0.060000} \psset{linestyle=solid} \psset{linestyle=solid} \setlinecaps{0} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psclip{\pswedge[linestyle=none,fillstyle=none](5.429375,6.587915){4.884727}{316.263492}{336.748666}} \psellipse(5.429375,6.587915)(3.454024,3.454024) \endpsclip \psset{linewidth=0.060000} \psset{linestyle=solid} \setlinejoinmode{0} \setlinecaps{0} \newrgbcolor{dialinecolor}{0.000000 0.000000 0.000000} \psset{linecolor=dialinecolor} \psline(8.216726,4.906880)(8.626435,5.287193)(8.684860,4.731238) }\endpspicture\end{center} \end{defn} \paragraph{Remark.} In a Lie algebra, one has $[b, a] = -[a,b]$. \paragraph{Proof.} $0 = [a+b, a+b] = [a, b] + [b, a] + \cancel{[a, a]} + \cancel{[b, b]}$ \hfill $\Box$ \paragraph{Examples.} \begin{enumerate} \item $\g$ a vector space with bracket $[a,b] = 0$. This is called an \emph{abelian Lie algebra}. \item $\R^3$ with vector multiplication $\times$ (cross product). \item If $A$ is an associative algebra, then $[a,b] = ab - ba$ satisfies the two identities. This Lie algebra is denoted by $A_-$. \sk\noindent {\bf Exercise 1.1}. Check the Jacobi identity on $[a, b] = ab-ba$. Moreover, this is true if $A$ is only quasi-associative, \emph{i.e.} $(ab)c-a(bc)$ is symmetric in $a, b$ $(= (ba)c-b(ac))$. \paragraph{Solution.} First, we show that associativity implies quasi-associativity. Let $(ab)c = a(bc)$. Then $(ab)c - a(bc) = 0 = (ba)c - b(ac)$. Hence we only need to show that $[a,b]$ satisfies the Jacobi identity if $A$ is quasi-associative. Here are some consequences of quasi-associativity. \begin{eqnarray*} (ab)c-a(bc)-(ba)c+b(ac) &=& 0 \\ (cb)a-c(ba)-(bc)a+b(ca) &=& 0 \\ (ac)b-a(cb)-(ca)b+c(ab) &=& 0 \\ (ba)c-b(ac)-(ab)c+a(bc) &=& 0 \\ \end{eqnarray*} We expand the Jacobi identity, group terms, and apply quasi-associativity: \begin{eqnarray*} && [a, [b, c]] + [b, [c, a]] + [c, [a, b]] \\ &=& a(bc-cb) - (bc-cb)a + b(ca-ac) - (ca-ac)b + c(ab-ba) - (ab-ba)c\\ &=& \quad [(cb)a-c(ba)-(bc)a+b(ca)] \\ && +\; [(ac)b-a(cb)-(ca)b+c(ab)] \\ && +\; [(ba)c-b(ac)-(ab)c+a(bc)]\\ &=& 0 \mbox{ (since all these terms $= 0$ by quasi-associativity)} \end{eqnarray*} $\Box$ A special case is $A = \End V$, then $A_- = \gl_V$ is called the \emph{general linear Lie algebra}. In particular, $A = \Mat_n \F$, then $A_- = \gl_n(\F)$. \item Any subalgebra of a Lie algebra is a Lie algebra. Notation: for subsets $M, N$ of $\g$ we denote $[M,N]$ the span of all commutators $[m, n]$, where $m \in M$ and $n \in N$. For example, subspace $\h$ is a subalgebra of $\g$ if $[\h, \h'] \in \h$. \paragraph{Example.} $\sll_n(\F) = \{ a \in \gl_n(\F) \mid \tr a = 0 \}$ \sk\noindent {\bf Exercise 1.2}. Show that $\mbox{tr} [a,b] = 0$ when $a, b \in \gl_n(\F)$. Also show that if $f:\gl_n(\F) \rightarrow \F$ is a linear function such that $f([a, b]) = 0, a, b \in \gl_n(\F)$, then $f(k) = c \cdot \tr k$. \paragraph{Solution.} We have \[ \tr [a, b] = \tr ab - \tr ba = \sum_i \sum_j a_{ji} b_{ij} - \sum_i \sum_j b_{ji} a_{ij} = 0 \] Now, any matrix $e_{ij}, i \neq j$ can be expressed as a commutator $[a,b]$ where $a, b \in \gl_n(\F)$, because $e_{ij}e_{jj} - e_{jj}e_{ij} = e_{ij}$. Hence $f(e_{ij}) = 0$ for all such matrices. But $f(e_{ii}) = f(e_{jj}) = c$ for all $i, j$ because $f(e_{ii})-f(e_{jj}) = f(e_{ii}) - f(e_{jj}) = f(e_{ij}e_{ji} - e_{ji}e_{ij}) = 0$. Any $x \in \gl_n(\F)$ can be split into $x' + x''$ where $x'$ is the sum of $e_{ij}, i \neq j$, and $x''$ is of the form $\sum_i \lambda_i e_{ii}$. By linearity, \[f(x) = 0 + f(x'') = c \sum_i \lambda_i = c \cdot \mbox{tr}.\] $\Box$ \begin{defn} An \emph{ideal} $\m \subset \g$ is a subspace such that $[\m, \g] \subset \m$. \end{defn} \paragraph{Example.} $\sll_n(\F)$ is an ideal of $\gl_n(\F)$ by Exercise 1.2. \item Factor algebras: If $\g$ is a Lie algebra and $\m$ is an ideal, then $\g / \m$ is a Lie algebra with bracket $[a+\m, b+\m] = [a, b]+\m$. \item Direct sum of two (Lie) algebras $\g_1 \oplus \g_2$: $[(a, b), (a_1, b_1)] = ([a, a_1], [b, b_1])$ where $a, a_1 \in \g_1, b, b_1 \in \g_2$. \end{enumerate} More examples of subalgebras of $\gl_V$: Let $B$ be a bilinear ($\F$-valued) form on a vector space $V$ over $\F$, define $o_{V,B} = \{ a \in \gl_V \mid B(a(u), v) + B(u, a(v)) = 0 \}$. \sk\noindent {\bf Exercise 1.3}. Let $B$ be a bilinear $\F$-valued form on a vector space $V$ over $\F$. Define \[ o_{V,B} = \{a \in \gl_V \ \mid \ B(a(u), v) + B(u, a(v)) = 0 \] Check that this is a subalgebra of the Lie algebra $\gl_V$. \paragraph{Solution.} To show that $o_{V,B}$ is a subalgebra (it is clearly a subspace), we need only show that $o_{V,B}$ is closed under the bracket. Let $x, y \in o_{V,B}$. Consider \[ B(x(y(u)), v) = -B(y(u), x(v)) = B(u, y(x(v))) \] But since $B$ is bilinear and $yx = [y,x]-xy$, $B(u, y(x(v))) = B(u, [x, y] - xy) = -B(u, x(y(v))) + B(u, [x, y])$ where $B(u, [x, y])$ is 0 from above, implying that $B(x(y(u)), v) = -B(u, x(y(v)))$, giving closure: \[ B(x(y(u)), v) + B(u, x(y(v))) = 0 \] $\Box$ Important special cases: $\dim V < \infty$, $B$ is non-degenerate (\emph{i.e.} det of the matrix of $B$ in some basis is non-zero). \begin{itemize} \item case 1: $B$ is symmetric. $B(a,b) = B(b,a)$, then $o_{V,B}$ is called the orthogonal Lie algebra, notation $so_{V,B}$. \item case 2: $B$ is skew-symmetric. $B(a,b) = -B(b,a)$, then $o_{V,B}$ is called the symplectic Lie algebra, notation $sp_{V,B}$. \end{itemize} \sk\noindent {\bf Exercise 1.4}. Suppose $\dim V = n$, choose a basis of $V$, let $so_{V, B}$ and $sp_{V, B} \subset \gl_n$. Let $B$ be the matrix of the bilinear form. Show \begin{eqnarray*} so_{V,B} &=& \{ a \in \gl_n(\F) \ \mid \ a^T B + Ba = 0 \}\\ sp_{V,B} &=& \{ a \in \gl_n(\F) \ \mid \ a^T B + Ba = 0 \} \end{eqnarray*} \paragraph{Solution.} Recall that $\gl_n$ is associative, and that $so_{V,B}$ is the set of $a \in A$ such that $B(u, a(v)) + B(a(u), v) = 0$ and $B$ symmetric; similarly, $sp_{V,B}$ is the corresponding set when $B$ is skew-symmetric. For $so_{V,B}$, we expand the definition of the bilinear form to get \[ u^T B a(v) + (a(u))^T B v = 0 \] and expressing $a$ as matrix multiplication, \[ u^T (Ba) v + (au)^T Bv = u^T (Ba) v + u^T a^TB v = 0 \] which is equivalent to the matrix condition \[ a^T B + Ba = 0 \] when $B$ is a symmetric matrix; similarly, for a skew-symmetric $B$, we expand to get \[ u^T B a(v) + (a(u))^T B v = 0 \] and expressing $a$ as matrix multiplication, \[ u^T (Ba) v + (au)^T Bv = u^T (Ba) v + u^T a^TB v = 0 \] which is equivalent to the matrix condition \[ a^T B + Ba = 0. \] $\Box$ \begin{defn} The \emph{derived algebra} of $\g$ is $[\g, \g]$. Obviously, this is an ideal, and hence a subalgebra. \end{defn} We now classify Lie algebras in dimensions 1 and 2. \paragraph{dim 1.} $\g = \F a, [a, a] = 0$. Only the abelian one. \paragraph{dim 2.} $\g = \F a + \F b, [\g, \g] = \F[a,b]$. case 1. $[a,b] = 0$. Then $\g$ abelian. case 2. $[a,b] = c \neq 0$. So $\g' = \F c$. Take $d \not\in \g'$ such that $d \neq 0$. Then $[d, c] = \alpha c$, since $\F c$ is an ideal, and $\alpha \neq 0$, since $\g$ nonabelian. Replacing $d$ by $\frac{1}{\alpha}d$, we get $[d, c] = c$. Thus we have a unique non-abelian Lie algebra, $[a, b] = b$. \sk\noindent The two most important ways to construct Lie algebras: \begin{enumerate} \item as a subalgebra (of $\gl_n$); \item by structure constants: choose a basis $e_1, \ldots, e_n$ of $\g$; then $[e_i, e_j] = \sum_{k=1}^n c_{ij}^k e_k$. The scalars $c_{ij}^k$ are called \emph{structure constants}. Of course, $c_{ii}^k = 0$ and $c_{ij}^k = -c_{ij}^k$ by skew-commutativity and a quadratic equation which is the Jacobi identity. \end{enumerate} \begin{remark} The non-abelian 2-dimensional Lie algebra is \[ \left\{ \left( \begin{array}{cc} \alpha & \beta \\ 0 & 0 \end{array} \right) \right\} \subset \gl_2(\F) \] If $a = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$ and $b = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$, then $[a,b] = b$. \end{remark} \begin{example} The Heisenberg Lie algebra $\He_n$ has basis $p_i, q_i (i = 1, \ldots, n), c$. ($\dim \He_n = 2n+1$) where $[p_i, q_j] = \delta_{ij} c, [c, p_i] = 0, [c, q_i] = 0, [p_i, p_j] = 0, [q_i, q_j] = 0$. Jacobi trivially holds. Realization by operators: $p_i = \frac{\partial}{\partial x}, q_i = x_i, c = 1$ on $\C[x_1, \ldots, x_n]$. \end{example} The first important meaning of the Jacobi identity: Rewrite it as follows: \begin{equation} \label{jac1} [a, [b, c]] = [[a, b], c] + [b, [a, c]]. \end{equation} \begin{defn} For any algebra $A$, an endomorphism $D$ is called a \emph{derivation} if the Leibniz rule holds, \emph{i.e.} \[D(ab) = (Da)b + a(Db).\] \end{defn} \begin{defn} Given an element $a \in \g$, define the operator $\ad a$ (adjoint) on $\g$ by: \[ (\ad a) b = [a, b]. \] \end{defn} Equation \ref{jac1} means that $\ad a$ is a derivation of the Lie algebra $\g$. It is called an \emph{inner derivation}. Notation: Given an algebra $A$, denote by $\Der A (\subset \End A)$ the space of derivations of $A$. \sk\noindent {\bf Exercise 1.5}. a) $\Der A$ is closed under the bracket in $\End A$ i.e. bracket of two derivations is a derivation, or $\Der A$ is a subalgebra of $\gl_A$. (b) If $A = \g$ is a Lie algebra, then $[D, \ad a] = \ad(D(a))$ for any derivation $F \in \Der \g$ and $a \in \g$. Hence inner derivations form an ideal of the Lie algebra $\Der \g$. \paragraph{Solution.} a) Let $D_1, D_2$ be derivations. Consider (by parts) $[D_1, D_2]$: \begin{eqnarray*} D_1D_2(ab) &=& D_1((D_2a)b + a(D_2b)) \\ &=& D_1((D_2a)b) + D_1(a(D_2b)) \\ &=& (D_1(D_2a))b+(D_2a)(D_1b)+(D_1a)(D_2b) + a(D_1D_2b) \\ D_2D_1(ab) &=& (D_2(D_1a))b+(D_1a)(D_2b)+(D_2a)(D_1b) + a(D_2D_1b) \\ D_1D_2(ab) - D_2D_1(ab) &=& (D_1D_2a-D_2D_1a)b + a(D_1D_2b-D_2D_1b) \\ &=& ((D_1D_2-D_2D_1)a)b + a((D_1D_2-D_2D_1)b) \end{eqnarray*} showing that the bracket is a derivation. \sk\noindent b) Consider \begin{eqnarray*} [D, \ad a] &=& (D(\ad a))b - ((\ad a)D)b \\ &=& D[a,b] - [a, D(b)]\\ &=& D(ab) - D(ba) - aD(b) + D(b)a\\ &=& D(a)b + \cancel{aD(b)} - \cancel{D(b)a} - bD(a) - \cancel{aD(b)} + \cancel{D(b)a} \\ &=& [D(a), b]\\ &=& \ad D(a) \end{eqnarray*} which shows that inner derivations form an ideal of the Lie algebra $\Der \g$. $\Box$ \end{document}