diff --git a/.gitignore b/.gitignore index 10b1924..7ceac16 100644 --- a/.gitignore +++ b/.gitignore @@ -5,11 +5,19 @@ *.bcf *.blg *.dpth +*.glg +*.glo +*.gls +*.idx +*.ilg +*.ind +*.ist *.kilepr *.log *.md5 *.out *.pdf +*.slo *.synctex.gz *.toc *.xml diff --git a/generalstyle.sty b/generalstyle.sty index d16f770..af0a575 100644 --- a/generalstyle.sty +++ b/generalstyle.sty @@ -1,23 +1,38 @@ %%%%% PACKAGES -\usepackage{gitinfo2} % including git information on the title page -\usepackage{fontspec} % font stuff (needed?) \usepackage[backend=biber, style=alphabetic, dateabbrev=false, urldate=long]{biblatex} % bibliography - -\usepackage{appendix} % the appendix -\usepackage{enumitem} % for better enumerates -\usepackage{lastpage} % to be able to refer to the last page +\usepackage{fontspec} % font stuff (needed?) +\usepackage{gitinfo2} % including git information on the title page +\usepackage[original]{imakeidx} % for the index, load before hyperref + +% using EB Garamond as the mainfont, with ebgaramond-maths for maths +% (don’t confuse ebgaramond-maths with garamond-math) +\usepackage[cmbraces]{newtxmath} +\usepackage{ebgaramond-maths} +\setmainfont{EB Garamond} + +% undefine openbox from nextxmath, as amsmath defines it again +\let\openbox\undefined + +\usepackage{appendix} % the appendix +\usepackage{csquotes} +\usepackage{enumitem} % for better enumerates +\usepackage{lastpage} % to be able to refer to the last page +\usepackage{microtype} % microtypography \usepackage[automark, headsepline, autooneside=false]{scrlayer-scrpage} % header design \usepackage{mathtools, amssymb, amsthm} % general math stuff -\usepackage{arydshln} % allowing dashed lines in arrays +% \usepackage{arydshln} % allowing dashed lines in arrays \usepackage{extarrows} % \xlongequal -\usepackage{stmaryrd} % \mapsfrom +% \usepackage{stmaryrd} % \mapsfrom \usepackage{tikz-cd} % commutative diagrams \usetikzlibrary{positioning} % for better positioning of nodes relative to each other \usepackage[colorlinks=true]{hyperref} % hyperlinks -\usepackage[capitalise,noabbrev]{cleveref} % smarter referencing; load after hyperref! + +% the following are to be loaded after hyperref +\usepackage[symbols, sort=use, style=long3col]{glossaries-extra} % list of symbols +\usepackage[capitalise,noabbrev]{cleveref} % smarter referencing @@ -28,6 +43,17 @@ % the file containing the references for biblatex \bibliography{references.bib} +% index and glossary +\makeindex +\makenoidxglossaries + +% big penalties to prevent thingsk +\binoppenalty=\maxdimen % breaking of binary operators +\relpenalty=\maxdimen % breaking of relation symbols +\widowpenalty=\maxdimen % widows +\displaywidowpenalty=\maxdimen % orphans + + % fonts for titlepage \setkomafont{publishers}{\normalsize} @@ -64,6 +90,7 @@ \newtheorem{definition}[everything]{Definition} \newtheorem{example}[everything]{Example} \newtheorem{examples}[everything]{Examples} +\newtheorem{recall}[everything]{Recall} \newtheorem{remark}[everything]{Remark} % printing of counter numbers @@ -71,12 +98,28 @@ \renewcommand{\theequation}{\arabic{equation}} \renewcommand{\theclaim}{\arabic{claim}} +% style of enumerating lists +\setlist[enumerate, 1]{leftmargin=*, align=left, label=\arabic*)} +\setlist[enumerate, 2]{leftmargin=*, align=left, label=\alph*)} +\setlist[description, 1]{leftmargin=\labelwidth, font=\normalfont} +\setlist[description, 2]{leftmargin=\labelwidth, font=\normalfont} +\setlist[itemize, 1]{leftmargin=*} +\setlist[itemize, 2]{leftmargin=*, label={\textopenbullet}} +\setlist[itemize, 3]{leftmargin=*} %%%%% MATH STUFF +% text shortcuts involving nobreakdash +\newcommand{\algebra}[1]{#1\nobreakdash-algebra} +\newcommand{\bilinear}[1]{#1\nobreakdash-bilinear} +\newcommand{\liealgebra}[1]{#1\nobreakdash-Lie~algebra} +\newcommand{\linear}[1]{#1\nobreakdash-linear} +\newcommand{\subalgebra}[1]{#1\nobreakdash-subalgebra} +\newcommand{\vectorspace}[1]{#1\nobreakdash-vector space} + % numbers and fields \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} @@ -84,16 +127,17 @@ \newcommand{\R}{\mathbb{R}} \newcommand{\Cbb}{\mathbb{Cbb}} \newcommand{\F}{\mathbb{F}} +\newcommand{\kf}{\mathbf{k}} % lie algebras -\newcommand{\cl}{\mathfrak{c}} -\newcommand{\g}{\mathfrak{g}} -\newcommand{\h}{\mathfrak{h}} -\newcommand{\gl}{\mathfrak{gl}} -\newcommand{\sll}{\mathfrak{sl}} -\newcommand{\tl}{\mathfrak{t}} -\newcommand{\n}{\mathfrak{n}} -\newcommand{\Loop}{\mathcal{L}} +\newcommand{\clie}{\mathfrak{c}} +\newcommand{\glie}{\mathfrak{g}} +\newcommand{\hlie}{\mathfrak{h}} +\newcommand{\gllie}{\mathfrak{gl}} +\newcommand{\sllie}{\mathfrak{sl}} +\newcommand{\tlie}{\mathfrak{t}} +\newcommand{\nlie}{\mathfrak{n}} +\newcommand{\looplie}{\mathcal{L}} % categories \newcommand{\catname}[1]{\mathbf{#1}} @@ -104,11 +148,13 @@ % spaces and sets \DeclareMathOperator{\Aut}{Aut} +\DeclareMathOperator{\centerlie}{Z} \DeclareMathOperator{\Der}{Der} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\Hom}{Hom} -\DeclareMathOperator{\Mat}{M} \DeclareMathOperator{\im}{im} +\DeclareMathOperator{\Mat}{M} +\DeclareMathOperator{\normallie}{Z} \DeclareMathOperator{\rad}{rad} \newcommand{\Ue}{\mathop{\mathcal{U}}} \DeclareMathOperator{\vspan}{span} @@ -125,12 +171,19 @@ \DeclareMathOperator{\tr}{tr} % relations and operators +\newcommand{\defined}{\coloneqq} \newcommand{\dotcup}{\mathrel{\mathaccent\cdot\cup}} \newcommand{\inc}{\hookrightarrow} -\newcommand{\subideal}{\trianglelefteq} +\newcommand{\ideal}{\trianglelefteq} +\newcommand{\tensor}{\otimes} % delimiters \DeclarePairedDelimiter{\pair}{\langle}{\rangle} +\NewDocumentCommand{\suchthat}{s}{ + \IfBooleanTF{#1} + {\,\middle|\,} + {\mathrel{|}} +} % decorations \newcommand{\vect}[1]{\begin{pmatrix}#1\end{pmatrix}} @@ -138,3 +191,13 @@ % shortcuts \newcommand{\dd}[1]{\frac{\text{d}}{\text{d}#1}} \newcommand{\mc}[1]{\mathcal{#1}} + + + + + +%%%%% GLOSSARY LIST +% glossary uses macros defined above +% input at the very end + +\input{glossarylist} diff --git a/glossarylist.tex b/glossarylist.tex new file mode 100644 index 0000000..e87ea9b --- /dev/null +++ b/glossarylist.tex @@ -0,0 +1,6 @@ +\glsxtrnewsymbol[description={center of~$\glie$}]{center}{\ensuremath{\centerlie(\glie)}} +\glsxtrnewsymbol[description={commutator of~$X$ and~$Y$}]{commutator space}{\ensuremath{[X,Y]}} +\glsxtrnewsymbol[description={general linear Lie algebra}]{general lie matrix}{\ensuremath{\gllie_n(\kf)}} +\glsxtrnewsymbol[description={general linear Lie algebra of~$V$}]{general lie endomorphism}{\ensuremath{\gllie(V)}} +\glsxtrnewsymbol[description={Lie bracket}]{lie bracket}{\ensuremath{[-,-]}} +\glsxtrnewsymbol[description={Lie ideal}]{lie ideal}{\ensuremath{\ideal}} diff --git a/notes.tex b/notes.tex index fb52865..e3ced78 100644 --- a/notes.tex +++ b/notes.tex @@ -1,4 +1,4 @@ -\documentclass[a4paper, 10pt, oneside, openany, bibliography=totocnumbered]{scrbook} +\documentclass[a4paper, 10pt, oneside, openany, bibliography=totocnumbered, headings=standardclasses]{scrbook} \usepackage{generalstyle} @@ -36,17 +36,17 @@ \mainmatter % TODO: notations \include{sections/basics} -\include{sections/slightly_advanced_basics} -\include{sections/semisimple_Lie_algebras} -\include{sections/root_systems} +% \include{sections/slightly_advanced_basics} +% \include{sections/semisimple_Lie_algebras} +% \include{sections/root_systems} \appendix % still part of mainmatter -\include{sections/appendix} +% \include{sections/appendix} \backmatter +\printnoidxglossary[type=symbols, title={List of Symbols}] +\printindex \printbibliography -% TODO: index - \end{document} diff --git a/sections/basic_definitions.tex b/sections/basic_definitions.tex index 010a9bb..d07b25d 100644 --- a/sections/basic_definitions.tex +++ b/sections/basic_definitions.tex @@ -1,762 +1,964 @@ -\section{Basic definitions} +\section{Basic Definitions} -\subsection{Definition and examples of Lie algebras} +\subsection{Definition and Examples of Lie~Algebras} \begin{definition} - Let $\g$ be a vector space over some field $k$. A $k$-bilinear map - \[ - [\cdot, \cdot] \colon \g \times \g \to \g - \] - is called a \emph{Lie bracket} if it satisfies the following two conditions: - \begin{enumerate} - \item - $[\cdot, \cdot]$ is \emph{alternating}, i.e.\ $[x,x] = 0$ for every $x \in \g$. - \item - $[\cdot, \cdot]$ satisfies the \emph{Jacobi identity} - \[ - [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0 - \quad - \text{for all $x,y,z \in \g$}. - \] - \end{enumerate} - A $k$-vector space $\g$ together with the Lie-bracket $[\cdot,\cdot]$ is called a \emph{$k$-Lie algebra}. + Let~$\glie$ be a vector space over some field~$\kf$. + A~{\bilinear{$\kf$}} map + \[ + \gls*{lie bracket} + \colon + \glie \times \glie + \to + \glie + \] + is called a \emph{Lie~bracket}\index{Lie!bracket} if it satisfies the following two conditions: + \begin{enumerate} + \item + $[-, -]$ is \emph{alternating}\index{alternating}, i.e.~$[x,x] = 0$ for every~$x \in \glie$. + \item + $[-, -]$ satisfies the \emph{Jacobi identity}\index{Jacobi identity} + \[ + [x,[y,z]] + [y,[z,x]] + [z,[x,y]] + = + 0 + \] + for all~$x,y,z \in \glie$. + \end{enumerate} + A~{\vectorspace{$\kf$}}~$\glie$ together with a Lie~bracket~$[-,-]$ is called a~\emph{\liealgebra{$\kf$}}\index{Lie!algebra}. \end{definition} \begin{remark} - Any Lie bracket $[\cdot, \cdot]$ is antisymmetric, i.e.\ $[y,x] = -[x,y]$ for all $x,y \in \g$, because - \[ - 0 = [x+y, x+y] = [x,x] + [x,y] + [y,x] + [y,y] = [x,y] + [y,x]. - \] + A Lie~bracket~$[-, -]$ on a vector space~$\glie$ is always antisymmetric, i.e.~$[y,x] = -[x,y]$ for all~$x,y \in \glie$, because + \[ + 0 + = + [x+y, x+y] + = + [x,x] + [x,y] + [y,x] + [y,y] + = + [x,y] + [y,x] \,. + \] \end{remark} \begin{remark} - Using the antisymmetry of the Lie bracket the Jacobi identity can be rewritten as - \[ - [x,[y,z]] = [[x,y],z] + [y,[x,z]] \quad \text{for all $x,y,z \in \g$}. - \] + By using the antisymmetry of the Lie~bracket the Jacobi identity\index{Jacobi identity} can be rewritten as + \[ + [x,[y,z]] + = + [[x,y],z] + [y,[x,z]] + \] + for all~$x, y, z \in \glie$. \end{remark} - \begin{examples} - \begin{enumerate}[leftmargin=*] - \item - Any vector space $V$ becomes a Lie algebra via - \[ - [x,y] = 0 \quad \text{for all $x,y \in V$}. - \] - \item - Any \emph{associative} $k$-algebra $A$ becomes a Lie algebra via - \[ - [a,b] = ab-ba \quad \text{for all $a,b \in A$}. - \] - Then $[\cdot, \cdot]$ is alternating and because $A$ is associative it follows that for all $a,b,c \in A$ - \begin{align*} - &\, [a,[b,c]] + [b,[c,a]] + [c,[a,b]] \\ - &= [a, (bc-cb)] + [b, (ca-ac)] + [c, (ab-ba)] \\ - &= a(bc-cb)-(bc-cb)a + b(ca-ac) - (ca-ac)b + c(ab-ba) - (ab-ba)c \\ - &= abc -acb -bca +cba +bca -bac -cab +acb +cab -cba -abc +bac \\ - &= 0. - \end{align*} - - The following two are important examples of this construction: - \begin{enumerate} + \leavevmode + \begin{enumerate} \item - The $k$-algebra of ($n \times n$)-matrices $\Mat_n(k)$ becomes a Lie algebra via - \[ - [A,B] = AB-BA \quad \text{for all $A,B \in \Mat_n(k)$}. - \] - This is called the \emph{general linear Lie algebra} and is denoted by $\gl_n(k)$. + Any vector space~$\glie$ becomes a Lie~algebra via~$[x,y] = 0$ for all~$x,y \in \glie$. \item - More generally for any $k$-vector space the $k$-algebra $\End_k(V)$ becomes a Lie algebra via - \[ - [\varphi_1, \varphi_2] \coloneqq \varphi_1 \circ \varphi_2 - \varphi_2 \circ \varphi_1 - \quad - \text{for all $\varphi_1, \varphi_2 \in \End_k$}. - \] - This is called the \emph{general linear Lie algebra for $V$} and is denoted by $\gl(V)$. - \end{enumerate} - \end{enumerate} -\end{examples} - - -\begin{definition} - Let $\g$ be a $k$-Lie algebra. A \emph{Lie subalgebra} of a $\g$ is a $k$-linear subspace $\h \subseteq \g$ such that - \[ - [x,y] \in \h \quad \text{for all $x,y \in \h$}. - \] - An \emph{ideal} inside $\g$ is a $k$-linear subspace $I \subseteq \g$ such that - \[ - [x,y] \in I \quad \text{for all $x \in \g$ and $y \in I$}. - \] - That $I$ is an ideal in $\g$ is denoted by $I \subideal \g$. -\end{definition} - - -\begin{remark} - It is not necessary to distinguish between left ideals or right ideals in a Lie algebra because the Lie bracket is antisymmetric. -\end{remark} - - -\begin{remark} - For a Lie algebra $\g$ and a subalgebra $\h \subseteq \g$ it follows that $\h$ becomes a Lie algebra by restricting the Lie bracket of $\g$ to $\h$. Every ideal inside $\g$ is also a subalgebra of $\g$. -\end{remark} - - -\begin{definition} - Let $\g$ be a Lie algebra. The \emph{center} of $\g$ is - \[ - Z(\g) \coloneqq \{x \in \g \mid \text{$[x,y] = 0$ for every $y \in \g$}\} - \] - $\g$ is called \emph{abelian} if $Z(\g) = 0$, i.e.\ if $[x,y] = 0$ for all $x,y \in \g$. -\end{definition} - - -\begin{definition} - For a Lie-algebra $\g$ over some field $k$ and subsets $X, Y \subseteq \g$ let - \[ - [X,Y] \coloneqq \vspan_k \{[x,y] \mid x \in X, y \in Y\}. - \] -\end{definition} - - -\begin{remark} - Notice that $\g$ is abelian if and only if $[\g,\g] = 0$. Also notice that $[\g,\g]$ and $Z(\g)$ are ideals inside $\g$. -\end{remark} - - -\begin{lemma} - Let $\g$ be a Lie algebra over some field $k$. - \begin{enumerate}[leftmargin=*] - \item - If $I_\lambda$, $\lambda \in \Lambda$ is a collection of ideals $I_\lambda \subideal \g$ then also $\bigcap_{\lambda \in \Lambda} I_\lambda \subideal \g$ and $\sum_{\lambda \in \Lambda} I_\lambda \subideal \g$. - \item - If $I, J \subideal \g$ then also $[I,J] \subideal \g$. - \end{enumerate} -\end{lemma} -\begin{proof} - \begin{enumerate}[leftmargin=*] - \item - This follows from direct calculation. - \item - As $[I,J]$ is spanned by the elements $[y,z]$ with $y \in I$ and $z \in J$ it is enough to show that $[x,[y,z]] \in [I,J]$ for every $x \in \g$, $y \in I$ and $z \in J$. This follows from $I, J \subideal \g$ and the Jacobi identity, because - \[ - [x,[y,z]] - = [\underbrace{[x,y]}_{\in I},z] + [y,\underbrace{[x,z]}_{\in J}] - \in [I,J]. - \qedhere - \] - \end{enumerate} -\end{proof} - - -\begin{definition} - A Lie algebra $\g$ is called \emph{linear} if $\g$ is a Lie subalgebra of $\gl(V)$ for some finite dimensional vector space $V$. -\end{definition} - - -\begin{example} - \begin{enumerate}[leftmargin=*] - \item - Let $\g = \gl_n(k)$. Then - \begin{gather*} - \sll_n(k) = \{A \in \g \mid \tr A = 0\} - \shortintertext{is an ideal in $\g$ because} - \sll_n(k) = [\g,\g]. - \end{gather*} - To see this first notice that on the one hand - \[ - \tr [A,B] - = \tr(AB-BA) - = \tr(AB) - \tr(BA) - = \tr(AB) - \tr(AB) - = 0 - \] - for all $A, B \in \g$ and therefore $[\g,\g] \subseteq \sll_n(k)$. - - On the other hand notice that $\sll_n(k)$ has a basis given by the elementary matrices $e_{ij}$ with $1 \leq i \neq j \leq n$ and $e_{11} - e_{ii}$ with $i = 2, \dotsc, n$. Each of these matrices is given as a commutator, namely $e_{ij} = [e_{ii},e_{ij}]$ for $1 \leq i \neq j \leq n$ and $e_{11} - e_{ii} = [e_{1i},e_{i1}]$ for $i = 2, \dotsc, n$. Therefore $\sll_n(k) \subseteq [\g,\g]$. - - \item - The upper triangular matrices - \[ - \tl_n(k) \coloneqq - \left\{ - \begin{pmatrix} - a_{11} & \cdots & \cdots & a_{1n} \\ - 0 & \ddots & & \vdots \\ - \vdots & \ddots & \ddots & \vdots \\ - 0 & \cdots & 0 & a_{nn} - \end{pmatrix} - \,\middle|\, - \text{$a_{ij} \in k$ for every $1 \leq i \leq j \leq n$} - \right\} - \] - are a Lie subalgebra of $\gl_n(k)$. + Any \emph{associative}~{\algebra{$\kf$}}~$A$ becomes a~{\liealgebra{$k$}} via~$[a,b] \defined ab - ba$ for all~$a, b \in A$. + Then~$[-, -]$ is alternating and it follows from the associativity of the multiplication of~$A$ that + \begin{align*} + {}& [a,[b,c]] + [b,[c,a]] + [c,[a,b]] \\ + ={}& [a, (bc-cb)] + [b, (ca-ac)] + [c, (ab-ba)] \\ + ={}& a(bc-cb)-(bc-cb)a + b(ca-ac) - (ca-ac)b + c(ab-ba) - (ab-ba)c \\ + ={}& abc - acb - bca + cba + bca - bac - cab + acb + cab - cba - abc + bac \\ + ={}& 0 + \end{align*} + for all~$a, b, c \in A$. + The element~$[a,b]$ is the \emph{commutator}\index{commutator} of~$a$ and~$b$. - \item - The strictly upper triangular matrices - \[ - \n_n(k) \coloneqq - \left\{ - \begin{pmatrix} - 0 & a_{12} & \cdots & a_{1n} \\ - \vdots & \ddots & \ddots & \vdots \\ - \vdots & & \ddots & a_{n-1,n} \\ - 0 & \cdots & \cdots & 0 - \end{pmatrix} - \,\middle|\, - \text{$a_{ij} \in k$ for every $1 \leq i < j \leq n$} - \right\} - \] - are a Lie subalgebra of $\tl_n(k)$ and therefore also of $\gl_n(k)$. It is even an ideal in $\tl_n(k)$ because $\n_n(k) = [\tl_n(k), \tl_n(k)]$. - \end{enumerate} -\end{example} - - -\begin{definition} - If $\g$ is a Lie algebra and $U \subseteq \g$ a linear subspace then - \[ - N_\g(U) \coloneqq \{x \in \g \mid \text{$[x,y] \in \g$ for every $y \in U$}\} - \] - is called the \emph{normalizer} of $U$ in $\g$ and - \[ - Z_\g(U) \coloneqq \{x \in \g \mid \text{$[x,y] = 0$ for every $y \in U$}\} - \] - is called the \emph{centralizer} of $U$ in $\g$. For a single element $x \in \g$ the centralizer of $x$ in $\g$ defined as - \[ - Z_\g(x) \coloneqq \{y \in \g \mid [x,y] = 0\}. - \] -\end{definition} - - -\begin{lemma} - Let $\g$ be a Lie algebra and $U \subseteq \g$ a linear subspace. Then $N_\g(U)$ and $Z_\g(U)$ are Lie subalgebras of $\g$. $Z_\g(x)$ is a Lie subalgebra of $\g$ for every $x \in \g$. -\end{lemma} -\begin{proof} - If $x,y \in N_\g(U)$ then by the Jacobi identity it follows for every $z \in U$ that - \[ - [[x,y],z] - = -[z,[x,y]] - = -[[z,x],y]-[x,[z,y]] - = \in \h - \] - and therefore $[x,y] \in N_\g(U)$. In the same way it follows for all $x,y \in \g$ and $z \in U$ that - \[ - [[x,y],z] - = -[[z,x],y]-[x,[z,y]] - = 0 - \] - and therefore $[x,y] \in Z_\g(U)$. For every $x \in \g$ the span $kx \subseteq \g$ is a linear subspace with $Z_\g(x) = Z_\g(kx)$, which is why $Z_\g(x)$ is a Lie subalgebra of $\g$. -\end{proof} - - -\begin{remark} - Let $\g$ be a Lie algebra and $L \subseteq \g$ a linear subspace. Then $L$ is a Lie subalgebra if and only if $L \subseteq N_\g(L)$. Then $L$ is not only contained in $N_\g(L)$ but $N_\g(L)$ is the maximal subalgebra of $\g$ which contains $L$ as an ideal. In particular $L$ is an ideal in $\g$ if and only if $N_\g(L) = \g$. -\end{remark} - - - - - - - - - -\subsection{Homomorphisms of Lie algebras} - + The following two are important examples of this construction: + \begin{enumerate} + \item + The~{\algebra{$\kf$}} of~($n \times n$)-matrices~$\Mat_n(\kf)$ becomes a Lie~algebra via + \[ + [A,B] + \defined + AB - BA + \] + for all~$A, B \in \Mat_n(\kf)$. + This Lie~algebra is called the \emph{general linear Lie~algebra}\index{general linear Lie algebra} and is denoted by~\gls*{general lie matrix}. + \item + More generally for any~{\vectorspace{$\kf$}} the~{\algebra{$\kf$}}~$\End_k(V)$ becomes a Lie~algebra via + \[ + [\phi_1, \phi_2] + \defined + \phi_1 \circ \phi_2 - \phi_2 \circ \phi_1 + \] + for all~$\phi_1, \phi_2 \in \End_k$. + This Lie~algebra is called the \emph{general linear Lie~algebra of~$V$}\index{general linear Lie algebra} and is denoted by~\gls*{general lie endomorphism}. + \end{enumerate} + \end{enumerate} +\end{examples} -\begin{definition} - Given Lie algebras $\g_1$ and $\g_2$ over the same field $k$ a \emph{homomorphism of Lie algebras} $\g_1 \to \g_2$ is a $k$-linear map $f \colon \g_1 \to \g_2$ such that - \[ - f([x,y]) =[f(x),f(y)] \quad \text{for all $x,y \in \g_1$}. - \] -\end{definition} +\begin{recall} + For calculations in~$\gllie_n(\kf)$ it is often useful to remember that + \[ + E_{ij} E_{kl} + = + \begin{cases} + E_{il} & \text{if~$j = k$} \,, \\ + 0 & \text{otherwise} \,, + \end{cases} + \] + where~$E_{ij}$ with~$1 \leq i,j \leq n$ denotes the standard basis of~$\gllie_n(\kf)$. + One may think about the matrix~$E_{ij}$ as \enquote{going from~$j$ to~$i$}. + The composition~$E_{ij} E_{kl}$ does then \enquote{go from~$l$ to~$i$} if the positions~$j$ and~$k$ fit together; + if they do not fit, then the composition is simply zero. + + This intuition can be formalized by observing that + \[ + E_{ij} e_k + = + \begin{cases} + e_i & \text{if~$j = k$} \,, \\ + 0 & \text{otherwise} \,. + \end{cases} + \] + This means that the matrix~$E_{ij}$ maps one of the standard basis vectors~$e_1, \dotsc, e_n$ (namely~$e_j$) to another standard basis vector (namely~$e_i$), but filters out all other standard basis vectors. +\end{recall} -\begin{examples}\label{expls: homomorphisms of Lie algebras} - \begin{enumerate}[leftmargin=*] - \item - For any Lie algebra $\g$ the identity $\id_\g \colon \g \to \g$ is a Lie algebra homomorphism. - \item - Given Lie algebras $\g_1$, $\g_2$ and $\g_3$ and Lie algebra homomorphisms $f_1 \colon \g_1 \to \g_2$ and $f_2 \colon \g_2 \to \g_3$ the composition $f_2 \circ f_1 \colon \g_1 \to \g_3$ is also a homomorphism of Lie algebras. - \item - If $\g$ is a Lie algebra and $\h \subseteq \g$ a Lie subalgebra then the inclusion $\h \inc \g$ is a homomorphism of Lie algebras. - \item - Given two abelian Lie algebras $\g_1$ and $\g_2$ any linear map $\g_1 \to \g_2$ is already a homomorphism of Lie algebras. - \item - Let $\g$ ba a Lie algebra over an arbitrary fielid $k$. Then for every $x \in \g$ let - \[ - \ad(x) \colon \g \to \g \quad \text{mit} \quad \ad(x)(y) = [x,y] - \quad \text{for every $y \in \g$}. - \] - Then the map $\ad \colon \g \to \gl(\g)$ is a homomorphism of Lie algebras. This follows from the Jacobi identity because for all $x,y,z \in \g$ - \begin{align*} - \ad([x,y])(z) - &= [[x,y],z] - = -[z,[x,y]] - = -[[z,x],y] -[x,[z,y]] - = [x,[y,z]] - [y,[x,z]] \\ - &= \ad(x)\ad(y)(z) - \ad(y)\ad(x)(z) - = [\ad(x),\ad(y)](z). - \end{align*} - \item - If $A_1$ and $A_2$ are associative $k$-algebras and $f \colon A_1 \to A_2$ a homomorphism of $k$-algebras then it is also a homomorphism of Lie algebras because - \[ - f([a,b]) = f(ab-ba) = f(a)f(b)-f(b)f(a) = [f(a),f(b)] - \quad \text{for all $a,b \in A_1$}. - \] - \item - Let $\g$ be a Lie algebra over an arbitary field $k$. If $\phi \colon \sll_2(k) \to \g$ is a homomorphism of Lie algebras then the images - \[ - E \coloneqq \phi(e), \quad H \coloneqq \phi(h), \quad F \coloneqq \phi(f) - \] - satisfy the relations - \[ - [H,E] = 2E, \quad [H,F] = 2F, \quad [E,F] = H. - \] - On the other hand every triple $(E', H', F')$ of elements satisfying the relations above (with $X$ replaced by $X'$ for $X \in \{E,H,F\}$) gives rise to a unique homomorphism of Lie algebras $\phi' \colon \sll_2(k) \to \g$ with - \[ - \phi'(E) = E', \quad \phi'(H) = H', \quad \phi'(F) = F'. - \] - - Hence there is a bijection between Lie algebra homomorphisms $\sll_2(k) \to \g$ and triples as above. Such triples will play an important part later on. - \end{enumerate} -\end{examples} \begin{definition} - Let $\g_1, \g_2$ be Lie algebras over the same field $k$. A homomorphism of Lie algebras $f \colon \g_1 \to \g_2$ is called an \emph{isomorphism of $k$-Lie algebras} if $f$ is bijective. + Let~$\glie$ be a~{\liealgebra{$\kf$}}. + \begin{enumerate} + \item + A \emph{Lie~subalgebra}\index{Lie!subalgebra} of~$\glie$ is a~{\linear{$\kf$}} subspace~$\hlie \subseteq \glie$ such that~$[x,y] \in \hlie$ for all~$x, y \in \hlie$. + \item + A \emph{Lie~ideal}\index{Lie!ideal}, or simply~\emph{ideal} in~$\glie$ is a~{\linear{$\kf$}} subspace~$I \subseteq \glie$ such that~$[x,y] \in I$ for all~$x \in \glie$ and~$y \in I$. + That~$I$ is an ideal in~$\glie$ is denoted by~$I \mathrel{\gls*{lie ideal}} \glie$. + \end{enumerate} \end{definition} -\begin{lemma} - If $f \colon \g_1 \to \g_2$ is an isomorphism of $k$-Lie algebras, then the $k$-linear map $f^{-1} \colon \g_2 \to \g_1$ is also a homomorphism of Lie-algebras and therefore also an isomorphism. -\end{lemma} -\begin{proof} - For all $x,y \in \g_2$ - \begin{align*} - f^{-1}( [x,y] ) - &= f^{-1}( [ f(f^{-1}(x)), f(f^{-1}(y)) ] ) \\ - &= f^{-1}(f( [f^{-1}(x) , f^{-1}(y)] )) \\ - &= [f^{-1}(x), f^{-1}(y)]. - \qedhere - \end{align*} -\end{proof} - - -\begin{remark} - It follows that Lie algebras over the same field $k$ together with homomorphisms of Lie algebras and their usual composition form a category, which will be denoted by $\cLie{k}$. An \emph{isomorphism of $k$-Lie algebras} is the same as an isomorphism in $\cLie{k}$. -\end{remark} - - -\begin{example}[Classification of one- and two-dimensional Lie algebras] - Let $k$ be any field. - - As every linear map between abelian Lie algebras is already a homomorphism of Lie algebras it follows that there exists up to isomorphism exactly one $n$-dimensional abelian Lie algebra over $k$ for every $n \in \N$. - - If $\g$ is a one-dimensional Lie algebra over $k$ then the Lie bracket of $\g$ is zero because it is alternating, which is why $\g$ is abelian. Hence there is up to isomorphism exactly one one-dimensional Lie algebra over $k$. - - Up to isomorphism there exists exactly one two-dimensional abelian $k$-Lie algebra - - Suppose that $\g$ is a two-dimensional, non-abelian Lie algebra over $k$. Let $\tilde{x}$,$\tilde{y}$ be a basis of $\g$. Because $[\g,\g]$ is non-abelian it follows that $[\g,\g] \neq 0$ and on the other hand $[\g,\g]$ is spanned by $[\tilde{x},\tilde{y}]$ because the Lie bracket is alternating, so $[\g,\g] = k [\tilde{x},\tilde{y}]$ with $[\tilde{x},\tilde{y}] \neq 0$. Let $x \coloneqq [\tilde{x},\tilde{y}]$ and $y \in \g$ such that $x,y$ is a basis of $\g$. Then $[x,y] \neq 0$ and $[x,y] \in k x$. By rescaling $y$ it can be assumed that $[x,y] = x$. This that up to isomorphism there is at most one two-dimensional, non-abelian Lie algebra $\g$ over $k$. - - Such an Lie algebra also exists. It can be realized as a subalgebra of $\gl_2(k)$ by choosing - \begin{gather*} - x \coloneqq \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = e_{12} - \quad\text{and}\quad - y \coloneqq \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = e_{22} - \shortintertext{because} - [x,y] = [e_{12}, e_{22}] = e_{12} e_{22} - e_{22} e_{12} = e_{12} = x. - \end{gather*} - Hence there are up to isomorphism exactly two two-dimensional Lie algebras over $k$. -\end{example} - - -\begin{proposition}[Homomorphism theorem] - Let $\g_1$ and $\g_2$ be Lie algebras and $f \colon \g_1 \to \g_2$ a homomorphism of Lie algebras. - \begin{enumerate}[leftmargin=*] - \item - $\ker f \subideal \g_1$ is an ideal. - \item - $\im f \subseteq \g_2$ is a Lie subalgebra. - \item - If $I \subideal \g_1$ is an ideal with $\ker f \subseteq I$ then there exists a unique homomorphism of Lie algebras $\tilde{f} \colon \g_1/I \to \g_2$ with $f = \tilde{f} \circ \pi$ where $\pi \colon \g_1 \to \g_1/I$ is the canonical projection. - \[ - \begin{tikzcd} - \g_1 - \arrow{dr}[above right]{f} - \arrow{d}[left]{d} - & - {} - \\ - \g_1/I - \arrow[dashed]{r}[below]{\exists! f} - & - \g_2 - \end{tikzcd} - \] - - \item - If $I, J \subideal \g$ are subideals with $I \subseteq J$ then $J/I \subideal \g/I$ and the natural isomorphism of vector spaces - \[ - (\g/I)/(J/I) \to \g/I, \quad (x+I)+(J/I) \mapsto x+I - \] - is already a natural isomorphism of Lie algebras. - \item - If $I, J \subideal \g$ are subideals then the natural isomorphism of vector spaces - \begin{gather*} - (I + J)/J \to I/(I \cap J) - \shortintertext{defined by} - (x+J)+I \mapsto x + (I \cap J) \quad \text{for every $x \in I$} - \end{gather*} - is already a natural isomorphism of Lie algebras. - \end{enumerate} -\end{proposition} - - \begin{remark} - For a Lie algebra $\g$ the ideal $[\g,\g]$ is the minimal ideal inside $\g$ such that $\g/[\g,\g]$ is abelian. Furthermore given any abelian Lie algebra $\h$ any homomorphism of Lie algebras $\g \to \h$ factorizes through a unique homomorphism of Lie algebras $\g/[\g,\g] \to \h$. - \[ - \begin{tikzcd} - \g - \arrow{rr} - \arrow{dr} - & - {} - & - \h - \\ - {} - & - \g / [\g,\g] - \arrow[dashed]{ur}[below right]{\exists!} - & - {} - \end{tikzcd} - \] + It is not necessary to distinguish between left ideals or right ideals in a Lie~algebra because the Lie~bracket is antisymmetric. \end{remark} - - -\subsection{New Lie algebras from old ones} - - -\begin{definition} - Let $\g_1$ and $\g_2$ be Lie algebras over the same field $k$. Then the \emph{product} of $\g_1$ and $\g_2$ is defined as the $k$-vector space $\g_1 \times \g_2$ together with the Lie bracket - \[ - [(x_1, y_1), (x_2, y_2)] - = ([x_1, x_2], [y_1, y_2]) - \quad - \text{for all $(x_1, y_1), (x_2, y_2) \in \g_1 \times \g_2$}. - \] -\end{definition} - - -\begin{definition} - Let $\g$ be a Lie algebra and $I \subideal \g$. Then the induced Lie algebra structure on the quotient vector space $\g/I$ is given by - \[ - [x+I, y+I] = [x,y] + I \quad \text{for all $x,y \in \g$}. - \] -\end{definition} - - \begin{remark} - The Lie algebra structure on the quotient $\g/I$ is well-defined: If $x,y, x',y' \in \g$ with $x+I = x'+I$ and $y+I = y'+I$ then $x-x' \in I$ and $y-y' \in I$ and thus - \begin{align*} - [x,y] + I - &= [x' + x-x', y' + y-y'] + I \\ - &= [x',y'] + \underbrace{[x', y-y']}_{\in I} + \underbrace{[x-x', y']}_{\in I} + \underbrace{[x-x', y-y']}_{\in I} + I - = [x', y'] + I. - \end{align*} - The additional properties of a Lie bracket follows from the Lie bracket of $\g$ safisfying them. + \leavevmode + \begin{enumerate} + \item + For a Lie~algebra~$\glie$ and a subalgebra~$\hlie \subseteq \glie$ it follows that~$\hlie$ becomes a Lie~algebra by restricting the Lie~bracket of~$\glie$ to~$\hlie$. + \item + Every ideal in~$\glie$ is also a subalgebra of~$\glie$. + \end{enumerate} \end{remark} -\begin{lemma} - \begin{enumerate}[leftmargin=*] - \item - If $\g_1$ and $\g_2$ are Lie algebras then the canonical projections - \[ - \pi_i \colon \g_1 \times \g_2 \to \g_i, \quad (x_1, x_2) \mapsto x_i - \quad \text{for $i = 1, 2$} - \] - are homomorphisms of Lie-algebras. - \item - If $\g$ is a Lie algebra and $I \subideal \g$ then the canonical projection $\pi \colon \g \to \g/I, x \mapsto [x]$ is a homomorphism of Lie algebras. - \end{enumerate} -\end{lemma} - - - -\begin{lemma}\label{lem: quasi extension of scalars for Lie algebras} -Let $\g$ be a Lie algebra over $k$ and $A$ an associative, commutative $k$-algebra. Then $A \otimes_k \g$ is a Lie algebra over $k$ via -\[ - [a \otimes x, b \otimes y] = (ab) \otimes [x,y] - \quad - \text{for all $a,b \in A$ and $x,y \in \g$}. -\] -Similarly $\g \otimes_k A$ carries the structure of a Lie algebra over $k$ via -\[ - [x \otimes a, y \otimes b] = [x,y] \otimes (ab) - \quad - \text{for all $x,y \in \g$ and $a,b \in A$}. -\] -\end{lemma} - - -\begin{example} - If $L/k$ is a field extension and $\g$ a Lie algebra over $k$, then $L \otimes_k \g$ is a Lie algebra over $k$ via - \[ - [\lambda \otimes x, \mu \otimes y] = (\lambda \mu) \otimes [x,y] - \quad - \text{for alle $\lambda, \mu \in k$ and $x,y \in \g$}. - \] - $L \otimes_k \g$ also carries the structure of an $L$-vector space via extension of scalars, i.e. - \[ - \lambda \cdot (\mu \otimes x) = (\lambda \mu) \otimes x - \quad - \text{for alle $\lambda, \mu \in k$ and $x \in \g$}, - \] - and the Lie bracket is not only $k$-bilinear, but also $L$-bilinear. Hence the structure of a $k$-Lie algebra on $L \otimes_k \g$ can be extended to the structure of a Lie algebra over $L$. (Notice that the Jacobi-Identity is independent of the ground field.) -\end{example} - - \begin{definition} - Let $\g$ be a Lie algebra and $A = k[t,t^{-1}]$ be the algebra of Laurent polynomials over $k$. Then - \[ - \Loop(\g) \coloneqq \g \otimes_k A - \] - with the Lie bracket as in Lemma~\ref{lem: quasi extension of scalars for Lie algebras} is called the \emph{loop (Lie) algebra} of $\g$. -\end{definition} - - -Another example for constructing new Lie algebras out of old ones are \emph{central extensions}: Let $\g$ be any $k$-Lie algebra. -\[ - \tilde{\g} - \coloneqq \g \oplus k - = \{x + \lambda c \mid x \in \g, \lambda \in k\}, -\] -where we understand $c$ as a formal variable. Suppose that $\kappa \colon \g \times \g \to k$ is a $k$-bilinear map satisfying the following properties: -\begin{enumerate} - \item - $\kappa$ is antisymmetric, i.e.\ $\kappa(x,y) = -\kappa(y,x)$ for all $x,y \in \g$. - \item - $\kappa$ satisfies the $2$-cocycle condition + Let~$\glie$ be a Lie~algebra. + The \emph{center}\index{center} of~$\glie$ is \[ - \kappa([x,y],z) + \kappa([y,z],x) + \kappa([z,x],y) = 0 - \quad - \text{for all $x,y,z \in \g$}. + \gls*{center} + \defined + \{ + x \in \glie + \suchthat + \text{$[x,y] = 0$ for every~$y \in \glie$} + \} \,. \] -\end{enumerate} -Then $\tilde{\g}$ becomes a Lie algebra via -\[ - [x + \lambda c, y + \mu c] \coloneqq [x,y] + \kappa(x,y) \lambda \mu c - \quad \text{for all $x,y \in \g$ and $\lambda, \mu \in k$.} -\] -Note that $c$ is central in $\tilde{\g}$ in the sense that $[x,c] = 0$ for all $x \in \g$. - - -\begin{example} - Let $\g = \gl_n(k)$. We define a symmetric bilinear form on $\g$ via - \[ - (A,B)_{\tr} = \tr(AB) \quad \text{for all $A,B \in \g$}. - \] - We define a bilinear form - \[ - \Loop(\g) \times \Loop(\g) \to k[t,t^{-1}], \quad - (x \otimes p, y \otimes q) \mapsto (x,y)_{\tr}\ pq - \] - We now get a $2$-cocycle $\kappa \colon \Loop(\g) \times \Loop(\g) \to k$ via - \[ - \kappa(a,b) \coloneqq \mathrm{Res}\left(\frac{\partial a}{\partial t}, b\right). - \] - $\kappa$ is also antisymmetric: Let $a = x \otimes t^i$ and $b = y \otimes t^j$ with $x,y \in \g$ and $i,j \in \Z$. Then - \begin{align*} - \kappa(x \otimes t^i, y \otimes t^j) - = \mathrm{Res}(i x \otimes t^{i-1}, y \otimes t^j) - &= \mathrm{Res}(i t^{i+j-1} (x,y)_{\tr}) \\ - &= - \begin{cases} - i (x,y)_{\tr} & \text{if $i+j = 0$}, \\ - 0 & \text{otherwise}. - \end{cases} - \end{align*} - In the same way we find that - \[ - \kappa(y \otimes t^j, x \otimes t^i) = - \begin{cases} - j (x,y)_{\tr} & \text{if $i+j = 0$}, \\ - 0 & \text{otherwise}. - \end{cases} - \] - Since $(\cdot,\cdot)_{\tr}$ is symmetric we find that - \begin{align*} - \kappa(x \otimes t^i, y \otimes t^j) - &= - \begin{cases} - i (x,y)_{\tr} & \text{if $i+j = 0$}, \\ - 0 & \text{otherwise}, - \end{cases} \\ - &= - \begin{cases} - -j (x,y)_{\tr} & \text{if $i+j = 0$}, \\ - 0 & \text{otherwise}, - \end{cases} \\ - &= - -\kappa(y \otimes t^j, x \otimes t^i). - \end{align*} -\end{example} - - - - - -\subsection{Derivations} - - -\begin{definition} - Let $A$ be a $k$-algebra (not necessarily unitary of even associative). A \emph{derivation of $A$} is a $k$-linear map $d \colon A \to A$ such that - \[ - d(ab) = d(a)b + ad(b) \quad \text{for all $a,b \in A$}. - \] - We set - \[ - \Der(A) \coloneqq \{d \colon A \to A \mid \text{$d$ is a derivation of $A$} \}. - \] \end{definition} -\begin{remark} - $\Der(A)$ is a $k$-linear subspace of $\End_k(A)$. -\end{remark} - - -\begin{example} - Let $A$ be a $k$-algebra. It follows from direct calculation that for all $d, d' \in \Der(A)$ the commutator $[d,d'] = d \circ d' - d' \circ d$ is again a derivation $\Der(A)$. Hence $\Der(A)$ is a Lie subalgebra of $\gl(A)$. -\end{example} - - -\begin{lemma}\label{lem: Lie algebras act adjoint by derivations} - Let $\g$ be a Lie algebra. Then for any $x \in \g$ the map - \[ - \ad(x) \colon \g \to \g, \quad y \mapsto [x,y] - \] - is a derivation of $\g$. -\end{lemma} -\begin{proof} - By the Jacobi identity - \begin{align*} - \ad(x)([y,z]) - &= [x,[y,z]] - = [[x,y],z] + [y,[x,z]] \\ - &= [\ad(x)(y),z] + [y,\ad(x)(z)] - \end{align*} - for all $y,z \in \g$. -\end{proof} - - \begin{definition} - Let $\g$ be a Lie algebra. A derivation of $\g$ is called \emph{inner} if it is of the form $\ad(x)$ for some $x \in \g$. -\end{definition} - - -\begin{lemma}\label{lem: inner derivations are in ideal} - If $\g$ is a Lie algebra then the inner derivations form an ideal inside of $\Der(\g)$. -\end{lemma} -\begin{proof} - Let $I \coloneqq \im \ad \subseteq \Der(\g)$ be the linear subspace of inner derivations. For any $\delta \in \Der$ and $x \in \g$ it follows that for any $y \in \g$ - \begin{align*} - &\,[\delta, \ad(x)](y) - = (\delta \ad(x) - \ad(x) \delta(x))(y) \\ - &= \delta([x,y]) - [x,\delta(y)] - = [\delta(x),y] + [x,\delta(y)] - [x,\delta(y)] \\ - &= [\delta(x),y] - = \ad(\delta(x))(y). - \end{align*} - Hence $[\delta, \ad(x)] = \ad(\delta(x)) \in I$. -\end{proof} - - - - - -\subsection{Simple Lie algebras} - - -\begin{definition} - A Lie algebra $\g$ is \emph{simple} if $0$ and $\g$ are the only ideals inside $\g$ and $\g$ is not abelian. -\end{definition} - - -\begin{lemma} - Let $\g$ be a simple Lie algebra. Then $[\g,\g] = \g$ and $Z(\g)=0$. -\end{lemma} -\begin{proof} - Because $\g$ is simple it is not abelian. Therefore $[\g,\g] \neq 0$ and $Z(\g) \neq \g$. Since $[\g,\g]$ and $Z(\g)$ are ideals inside $\g$ it follows that $[\g,\g] = \g$ and $Z(\g) = 0$. By the homomorphism theorem $\g$ is isomorphic to its image $\ad \g$ and hence to a linear Lie algebra. -\end{proof} - - -\begin{corollary} - Let $\g$ be simple. Then the homomorphism $\ad \colon \g \to \gl(\g), x \mapsto \ad(x)$ is injective. In particular $\g$ can be realized as a linear Lie algebra. -\end{corollary} -\begin{proof} - As part of Examples~\ref{expls: homomorphisms of Lie algebras} has already been shown that $\ad$ is a homomorphism of Lie algebras. That it is injective follows directly from $\ker \ad = Z(\g) = 0$. -\end{proof} - - -It can be shown that every finite dimensional Lie algebra can be realized as a linear Lie algebra. This will not be proven in this lecture and is by far not trivial. - - -\begin{theorem}[Ado] - Every finite dimensional Lie algebra $\g$ is isomorphic to a linear Lie algebra. -\end{theorem} - - -\begin{examples} - \begin{enumerate}[leftmargin=*] - \item - Since $[\gl_n(k),\gl_n(k)] = \sll_n(k) \neq \gl_n(k)$ we find that $\gl_n(k)$ is not simple. - \item - Let $\g = \sll_2(k)$. Then $\g$ is simple if and only if $\chara k \neq 2$. To see this consider the basis $(e,h,f)$ of $\sll_2(k)$ consisting of the matrices - \[ - e = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \quad - h = \vect{1 & 0 \\ 0 & -1}, \quad - f = \vect{0 & 0 \\ 1 & 0}. - \] - of $\sll_2(k)$. Then - \[ - [h,e] = 2e, \quad - [h,f] = -2f, \quad - [e,f] = h. - \] - If $\chara k = 2$ then $h$ spans a $1$-dimensional ideal, thus $\sll_2(k)$ is not simple. Suppose that $\chara k \neq 2$ and let $I \subseteq \sll_2(k)$ be an ideal with $I \neq 0$. From the above relations it follows that if $I$ contains one of the basis vectors $e$, $h$ or $f$ then already $I = \sll_2(k)$. Let $x \in I$ with $x \neq 0$ and write $x = \alpha e + \beta h + \gamma f$. Then - \[ - [e,x] = -2 \beta e + \gamma h \quad \text{and} \quad [e,[e,x]] = -2 \gamma e. - \] - Since $\gamma = 0$ or $\gamma \neq 0$ we find that $e \in I$. - \end{enumerate} -\end{examples} - - -\begin{definition} - Let $k$ be any field. The basis - \[ - e = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \quad - h = \vect{1 & 0 \\ 0 & -1}, \quad - f = \vect{0 & 0 \\ 1 & 0}. - \] - of $\sll_2(k)$ is called the \emph{standard basis} of $\sll_2(k)$. + For a Lie-algebra~$\glie$ over some field~$\kf$ and all subsets~$X, Y \subseteq \glie$ the subspace + \[ + \gls*{commutator space} + \defined + \vspan_k + \{ + [x,y] + \suchthat + x \in X, + y \in Y + \} + \] + is the \emph{commutator}\index{commutator} of~$X$ and~$Y$. \end{definition} - - -\begin{remark} - If $\chara k = 0$ then $\sll_n(k)$ is simple for all $n \geq 2$. -\end{remark} +% +% +% \begin{definition} +% A Lie~algebra~$\glie$ is \emph{abelian} if~$[x,y] = 0$ for all~$x, y \in \glie$. +% \end{definition} +% +% +% \begin{remark} +% Note that a Lie~algebra~$\glie$ is abelian if and only if~$\centerlie(\glie) = 0$, if and only if~$[\glie, \glie] = 0$. +% \end{remark} +% +% +% \begin{example} +% A~{\algebra{$\kf$}}~$A$ is commutative if and only if it is abelian as a Lie~algebra. +% \end{example} +% +% +% \begin{lemma} +% Let~$\glie$ be a Lie~algebra over some field~$\kf$. +% \begin{enumerate} +% \item +% If~$I_\lambda$,~$\lambda \in \Lambda$ is a collection of ideals~$I_\lambda \subideal \glie$ then their intersection~$\bigcap_{\lambda \in \Lambda} I_\lambda$ and their sum~$\sum_{\lambda \in \Lambda} I_\lambda$ are again ideals in~$\glie$. +% \item +% If $I$ and~$J$ are ideals in~$\glie$ then their commutator~$[I,J]$ is again an ideal in~$\glie$. +% \end{enumerate} +% \end{lemma} +% +% +% \begin{proof} +% \leavevmode +% \begin{enumerate} +% \item +% This follows from direct calculation. +% \item +% As~$[I,J]$ is spanned by the elements~$[y,z]$ with~$y \in I$ and~$z \in J$ it sufficies to show that~$[x,[y,z]] \in [I,J]$ for all~$x \in \glie$,~$y \in I$ and~$z \in J$. +% This follows from $I, J \subideal \glie$ and the Jacobi identity, because +% \[ +% [x,[y,z]] +% = +% [[x,y], z] + [y, [x,z]] +% \subseteq +% [I, z] + [y, I] +% \subseteq +% [I, J] + [J, I] +% = +% [I,J] \,. +% \] +% Here we used that~$[J,I] = [I,J]$ by the antisymmetry of the Lie bracket~$[-,-]$. +% \qedhere +% \end{enumerate} +% \end{proof} +% +% +% \begin{definition} +% A Lie~algebra $\glie$ is~\emph{linear} if~$\glie$ is a Lie subalgebra of~$\gllie(V)$ for some finite dimensional vector space~$V$, or a Lie subalgebra of~$\gllie_n(\kf)$ for some~$n$. +% \end{definition} +% +% +% \begin{examples}[Linear Lie~Algebras] +% \leavevmode +% \begin{enumerate} +% \item +% Let $\glie = \gllie_n(\kf)$. +% Then +% \[ +% \sllie_n(\kf) +% \defined +% \{ +% A \in \glie +% \suchthat +% \tr A = 0 +% \} +% \] +% is an ideal in~$\glie$, namely +% \[ +% \sllie_n(\kf) +% = +% [\glie,\glie] \,. +% \] +% +% Indeed, we have for all~$A, B \in \glie$ that~$\tr(AB) = \tr(BA)$ and hence +% \[ +% \tr( [A,B] ) +% = \tr(AB-BA) +% = \tr(AB) - \tr(BA) +% = \tr(AB) - \tr(AB) +% = 0 \,. +% \] +% This shows that~$[\glie, \glie] \subseteq \sllie_n(\kf)$. +% Observe on the other hand that~$\sllie_n(\kf)$ has a basis given by the matrices $E_{ij}$ with $1 \leq i \neq j \leq n$ and $E_{11} - E_{ii}$ with $i = 2, \dotsc, n$. +% Each of these matrices is given as a commutator, namely +% \[ +% E_{ij} +% = +% E_{ij} E_{jj} - \underbrace{ E_{jj} E_{ij} }_{=0} +% = +% [E_{ij}, E_{jj}] +% \] +% for $1 \leq i \neq j \leq n$ and +% \[ +% E_{11} - E_{ii} +% = +% E_{1i} E_{i1} - E_{i1} E_{1i} +% = +% [E_{1i}, E_{i1}] +% \] +% for $i = 2, \dotsc, n$. +% Therefore $\sllie_n(\kf) \subseteq [\glie,\glie]$. +% +% The Lie~algebra~$\sllie_n(\kf)$ is the \emph{special Lie~algebra}. +% +% If~$V$ is any finite dimensional~{\vectorspace{$\kf$}} then +% \[ +% \sllie(V) +% \defined +% [\gllie(V), \gllie(V)] +% = +% \{ +% f \in \gllie(V) +% \suchthat +% \tr(f) = 0 +% \} +% \] +% is the \emph{special Lie~algebra} of~$V$. +% +% \item +% The upper triangular matrices +% \[ +% \tlie_n(\kf) +% \defined +% \left\{ +% \begin{pmatrix} +% a_{11} +% & \cdots +% & \cdots +% & a_{1n} +% \\ +% 0 +% & \ddots +% & {} +% & \vdots +% \\ +% \vdots +% & \ddots +% & \ddots +% & \vdots +% \\ +% 0 +% & \cdots +% & 0 +% & a_{nn} +% \end{pmatrix} +% \suchthat* +% \text{$a_{ij} \in \kf$ for all~$1 \leq i \leq j \leq n$} +% \right\} +% \] +% form a Lie~subalgebra of $\gllie_n(\kf)$. +% This holds because the upper triangular matrices form a~{\subalgebra{$\kf$}} of~$\Mat_n(\kf)$, and are hence closed under the commutator~$[A,B] = AB - BA$. +% +% \item +% The strictly upper triangular matrices +% \[ +% \nlie_n(\kf) +% \defined +% \left\{ +% \begin{pmatrix} +% 0 +% & a_{12} +% & \cdots +% & a_{1n} +% \\ +% \vdots +% & \ddots +% & \ddots +% & \vdots +% \\ +% \vdots +% & {} +% & \ddots +% & a_{n-1,n} +% \\ +% 0 +% & \cdots +% & \cdots +% & 0 +% \end{pmatrix} +% \suchthat* +% \text{$a_{ij} \in \kf$ for all~$1 \leq i < j \leq n$} +% \right\} +% \] +% form a Lie~subalgebra of~$\tlie_n(\kf)$ and therefore also of $\gllie_n(\kf)$. +% It is even an ideal in $\tlie_n(\kf)$, namely $\nlie_n(\kf) = [\tlie_n(\kf), \tlie_n(\kf)]$: +% +% For any two matrices~$A, B \in \tlie_n(\kf)$ their products~$AB$ and~$BA$ are again upper triangular and both products have the same diagonal entries. +% The commutator~$[A,B] = AB - BA$ is therefore a strictly upper triangular matrix. +% Hence~$[\tlie_n(\kf), \tlie_n(\kf)] \subseteq \nlie_n(\kf)$. +% We know on the other hand that~$\nlie_n(\kf)$ has as a basis the matrices~$E_{ij}$ with~$1 \leq i < j \leq n$, and each of those matrices can be written as a commutator +% \[ +% E_{ij} +% = +% E_{ii} E_{ij} - \underbrace{ E_{ij} E_{ii} }{= 0} +% \] +% with~$E_{ii}, E_{ij} \in \tlie_n(\kf)$. +% Therefore~$\nlie_n(\kf) \subseteq \tlie_n(\kf)$. +% \end{enumerate} +% \end{examples} +% +% +% \begin{definition} +% If~$\glie$ is a Lie~algebra and~$U \subseteq \glie$ any linear subspace then +% \[ +% \normallie_\glie(U) +% \defined +% \{ +% x \in \glie +% \suchthat +% \text{$[x,y] \in \glie$ for every~$y \in U$} +% \} +% \] +% is the \emph{normalizer} of~$U$ in~$\glie$, and +% \[ +% \centerlie_\glie(U) +% \defined +% \{ +% x \in \glie +% \suchthat +% \text{$[x,y] = 0$ for every $y \in U$} +% \} +% \] +% is the \emph{centralizer} of~$U$ in~$\glie$. +% For a single element~$x \in \glie$ the centralizer of~$x$ in~$\glie$ is +% \[ +% \centerlie_\glie(x) +% \defined +% \{ +% y \in \glie +% \suchthat +% [x,y] = 0 +% \} \,. +% \] +% \end{definition} +% +% +% \begin{lemma} +% Let $\glie$ be a Lie~algebra and $U \subseteq \glie$ a linear subspace. Then $\normallie_\glie(U)$ and $\centerlie_\glie(U)$ are Lie subalgebras of $\glie$. +% $\centerlie_\glie(x)$ is a Lie subalgebra of $\glie$ for every $x \in \glie$. +% \end{lemma} +% +% +% \begin{proof} +% If $x,y \in \normallie_\glie(U)$ then by the Jacobi identity it follows for every $z \in U$ that +% \[ +% [[x,y],z] +% = +% -[z,[x,y]] +% = +% -[[z,x],y]-[x,[z,y]] +% = +% \in \hlie +% \] +% and therefore $[x,y] \in \normallie_\glie(U)$. In the same way it follows for all $x,y \in \glie$ and $z \in U$ that +% \[ +% [[x,y],z] +% = -[[z,x],y]-[x,[z,y]] +% = 0 +% \] +% and therefore $[x,y] \in Z_\glie(U)$. For every $x \in \glie$ the span $kx \subseteq \glie$ is a linear subspace with $Z_\glie(x) = Z_\glie(kx)$, which is why $Z_\glie(x)$ is a Lie subalgebra of $\glie$. +% \end{proof} +% +% +% \begin{remark} +% Let $\glie$ be a Lie~algebra and $L \subseteq \glie$ a linear subspace. Then $L$ is a Lie subalgebra if and only if $L \subseteq \normallie_\glie(L)$. Then $L$ is not only contained in $\normallie_\glie(L)$ but $\normallie_\glie(L)$ is the maximal subalgebra of $\glie$ which contains $L$ as an ideal. In particular $L$ is an ideal in $\glie$ if and only if $\normallie_\glie(L) = \glie$. +% \end{remark} +% +% +% +% +% +% +% +% +% +% \subsection{Homomorphisms of Lie~algebras} +% +% +% \begin{definition} +% Given Lie~algebras $\g_1$ and $\g_2$ over the same field $\kf$ a \emph{homomorphism of Lie~algebras} $\g_1 \to \g_2$ is a $\kf$-linear map $f \colon \g_1 \to \g_2$ such that +% \[ +% f([x,y]) =[f(x),f(y)] \quad \text{for all $x,y \in \g_1$}. +% \] +% \end{definition} +% +% +% \begin{examples}\label{expls: homomorphisms of lie algebras} +% \begin{enumerate}[leftmargin=*] +% \item +% For any Lie~algebra $\glie$ the identity $\id_\g \colon \g \to \glie$ is a Lie~algebra homomorphism. +% \item +% Given Lie~algebras $\g_1$, $\g_2$ and $\g_3$ and Lie~algebra homomorphisms $f_1 \colon \g_1 \to \g_2$ and $f_2 \colon \g_2 \to \g_3$ the composition $f_2 \circ f_1 \colon \g_1 \to \g_3$ is also a homomorphism of Lie~algebras. +% \item +% If $\glie$ is a Lie~algebra and $\hlie \subseteq \glie$ a Lie subalgebra then the inclusion $\hlie \inc \glie$ is a homomorphism of Lie~algebras. +% \item +% Given two abelian Lie~algebras $\g_1$ and $\g_2$ any linear map $\g_1 \to \g_2$ is already a homomorphism of Lie~algebras. +% \item +% Let $\glie$ ba a Lie~algebra over an arbitrary fielid $\kf$. Then for every $x \in \glie$ let +% \[ +% \ad(x) \colon \g \to \g \quad \text{mit} \quad \ad(x)(y) = [x,y] +% \quad \text{for every $y \in \glie$}. +% \] +% Then the map $\ad \colon \g \to \gl(\g)$ is a homomorphism of Lie~algebras. This follows from the Jacobi identity because for all $x,y,z \in \glie$ +% \begin{align*} +% \ad([x,y])(z) +% &= [[x,y],z] +% = -[z,[x,y]] +% = -[[z,x],y] -[x,[z,y]] +% = [x,[y,z]] - [y,[x,z]] \\ +% &= \ad(x)\ad(y)(z) - \ad(y)\ad(x)(z) +% = [\ad(x),\ad(y)](z). +% \end{align*} +% \item +% If $A_1$ and $A_2$ are associative $\kf$-algebras and $f \colon A_1 \to A_2$ a homomorphism of $\kf$-algebras then it is also a homomorphism of Lie~algebras because +% \[ +% f([a,b]) = f(ab-ba) = f(a)f(b)-f(b)f(a) = [f(a),f(b)] +% \quad \text{for all $a,b \in A_1$}. +% \] +% \item +% Let $\glie$ be a Lie~algebra over an arbitary field $\kf$. If $\phi \colon \sll_2(k) \to \glie$ is a homomorphism of Lie~algebras then the images +% \[ +% E \coloneqq \phi(e), \quad H \coloneqq \phi(h), \quad F \coloneqq \phi(f) +% \] +% satisfy the relations +% \[ +% [H,E] = 2E, \quad [H,F] = 2F, \quad [E,F] = H. +% \] +% On the other hand every triple $(E', H', F')$ of elements satisfying the relations above (with $X$ replaced by $X'$ for $X \in \{E,H,F\}$) gives rise to a unique homomorphism of Lie~algebras $\phi' \colon \sll_2(k) \to \glie$ with +% \[ +% \phi'(E) = E', \quad \phi'(H) = H', \quad \phi'(F) = F'. +% \] +% +% Hence there is a bijection between Lie~algebra homomorphisms $\sll_2(k) \to \glie$ and triples as above. Such triples will play an important part later on. +% \end{enumerate} +% \end{examples} +% +% +% \begin{definition} +% Let $\g_1, \g_2$ be Lie~algebras over the same field $\kf$. A homomorphism of Lie~algebras $f \colon \g_1 \to \g_2$ is called an \emph{isomorphism of $\kf$-Lie~algebras} if $f$ is bijective. +% \end{definition} +% +% +% \begin{lemma} +% If $f \colon \g_1 \to \g_2$ is an isomorphism of $\kf$-Lie~algebras, then the $\kf$-linear map $f^{-1} \colon \g_2 \to \g_1$ is also a homomorphism of Lie-algebras and therefore also an isomorphism. +% \end{lemma} +% \begin{proof} +% For all $x,y \in \g_2$ +% \begin{align*} +% f^{-1}( [x,y] ) +% &= f^{-1}( [ f(f^{-1}(x)), f(f^{-1}(y)) ] ) \\ +% &= f^{-1}(f( [f^{-1}(x) , f^{-1}(y)] )) \\ +% &= [f^{-1}(x), f^{-1}(y)]. +% \qedhere +% \end{align*} +% \end{proof} +% +% +% \begin{remark} +% It follows that Lie~algebras over the same field $\kf$ together with homomorphisms of Lie~algebras and their usual composition form a category, which will be denoted by $\cLie{k}$. An \emph{isomorphism of $\kf$-Lie~algebras} is the same as an isomorphism in $\cLie{k}$. +% \end{remark} +% +% +% \begin{example}[Classification of one- and two-dimensional Lie~algebras] +% Let $\kf$ be any field. +% +% As every linear map between abelian Lie~algebras is already a homomorphism of Lie~algebras it follows that there exists up to isomorphism exactly one $n$-dimensional abelian Lie~algebra over $\kf$ for every $n \in \N$. +% +% If $\glie$ is a one-dimensional Lie~algebra over $\kf$ then the Lie~bracket of $\glie$ is zero because it is alternating, which is why $\glie$ is abelian. Hence there is up to isomorphism exactly one one-dimensional Lie~algebra over $\kf$. +% +% Up to isomorphism there exists exactly one two-dimensional abelian $\kf$-Lie~algebra +% +% Suppose that $\glie$ is a two-dimensional, non-abelian Lie~algebra over $\kf$. Let $\tilde{x}$,$\tilde{y}$ be a basis of $\glie$. Because $[\g,\g]$ is non-abelian it follows that $[\g,\g] \neq 0$ and on the other hand $[\g,\g]$ is spanned by $[\tilde{x},\tilde{y}]$ because the Lie~bracket is alternating, so $[\g,\g] = k [\tilde{x},\tilde{y}]$ with $[\tilde{x},\tilde{y}] \neq 0$. Let $x \coloneqq [\tilde{x},\tilde{y}]$ and $y \in \glie$ such that $x,y$ is a basis of $\glie$. Then $[x,y] \neq 0$ and $[x,y] \in k x$. By rescaling $y$ it can be assumed that $[x,y] = x$. This that up to isomorphism there is at most one two-dimensional, non-abelian Lie~algebra $\glie$ over $\kf$. +% +% Such an Lie~algebra also exists. It can be realized as a subalgebra of $\gl_2(k)$ by choosing +% \begin{gather*} +% x \coloneqq \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = e_{12} +% \quad\text{and}\quad +% y \coloneqq \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = e_{22} +% \shortintertext{because} +% [x,y] = [e_{12}, e_{22}] = e_{12} e_{22} - e_{22} e_{12} = e_{12} = x. +% \end{gather*} +% Hence there are up to isomorphism exactly two two-dimensional Lie~algebras over $\kf$. +% \end{example} +% +% +% \begin{proposition}[Homomorphism theorem] +% Let $\g_1$ and $\g_2$ be Lie~algebras and $f \colon \g_1 \to \g_2$ a homomorphism of Lie~algebras. +% \begin{enumerate}[leftmargin=*] +% \item +% $\ker f \subideal \g_1$ is an ideal. +% \item +% $\im f \subseteq \g_2$ is a Lie subalgebra. +% \item +% If $I \subideal \g_1$ is an ideal with $\ker f \subseteq I$ then there exists a unique homomorphism of Lie~algebras $\tilde{f} \colon \g_1/I \to \g_2$ with $f = \tilde{f} \circ \pi$ where $\pi \colon \g_1 \to \g_1/I$ is the canonical projection. +% \[ +% \begin{tikzcd} +% \g_1 +% \arrow{dr}[above right]{f} +% \arrow{d}[left]{d} +% & +% {} +% \\ +% \g_1/I +% \arrow[dashed]{r}[below]{\exists! f} +% & +% \g_2 +% \end{tikzcd} +% \] +% +% \item +% If $I, J \subideal \glie$ are subideals with $I \subseteq J$ then $J/I \subideal \g/I$ and the natural isomorphism of vector spaces +% \[ +% (\g/I)/(J/I) \to \g/I, \quad (x+I)+(J/I) \mapsto x+I +% \] +% is already a natural isomorphism of Lie~algebras. +% \item +% If $I, J \subideal \glie$ are subideals then the natural isomorphism of vector spaces +% \begin{gather*} +% (I + J)/J \to I/(I \cap J) +% \shortintertext{defined by} +% (x+J)+I \mapsto x + (I \cap J) \quad \text{for every $x \in I$} +% \end{gather*} +% is already a natural isomorphism of Lie~algebras. +% \end{enumerate} +% \end{proposition} +% +% +% \begin{remark} +% For a Lie~algebra $\glie$ the ideal $[\g,\g]$ is the minimal ideal inside $\glie$ such that $\g/[\g,\g]$ is abelian. Furthermore given any abelian Lie~algebra $\hlie$ any homomorphism of Lie~algebras $\g \to \hlie$ factorizes through a unique homomorphism of Lie~algebras $\g/[\g,\g] \to \hlie$. +% \[ +% \begin{tikzcd} +% \g +% \arrow{rr} +% \arrow{dr} +% & +% {} +% & +% \h +% \\ +% {} +% & +% \g / [\g,\g] +% \arrow[dashed]{ur}[below right]{\exists!} +% & +% {} +% \end{tikzcd} +% \] +% \end{remark} +% +% +% +% +% \subsection{New Lie~algebras from old ones} +% +% +% \begin{definition} +% Let $\g_1$ and $\g_2$ be Lie~algebras over the same field $\kf$. Then the \emph{product} of $\g_1$ and $\g_2$ is defined as the $\kf$-vector space $\g_1 \times \g_2$ together with the Lie~bracket +% \[ +% [(x_1, y_1), (x_2, y_2)] +% = ([x_1, x_2], [y_1, y_2]) +% \quad +% \text{for all $(x_1, y_1), (x_2, y_2) \in \g_1 \times \g_2$}. +% \] +% \end{definition} +% +% +% \begin{definition} +% Let $\glie$ be a Lie~algebra and $I \subideal \glie$. Then the induced Lie~algebra structure on the quotient vector space $\g/I$ is given by +% \[ +% [x+I, y+I] = [x,y] + I \quad \text{for all $x,y \in \glie$}. +% \] +% \end{definition} +% +% +% \begin{remark} +% The Lie~algebra structure on the quotient $\g/I$ is well-defined: If $x,y, x',y' \in \glie$ with $x+I = x'+I$ and $y+I = y'+I$ then $x-x' \in I$ and $y-y' \in I$ and thus +% \begin{align*} +% [x,y] + I +% &= [x' + x-x', y' + y-y'] + I \\ +% &= [x',y'] + \underbrace{[x', y-y']}_{\in I} + \underbrace{[x-x', y']}_{\in I} + \underbrace{[x-x', y-y']}_{\in I} + I +% = [x', y'] + I. +% \end{align*} +% The additional properties of a Lie~bracket follows from the Lie~bracket of $\glie$ safisfying them. +% \end{remark} +% +% +% \begin{lemma} +% \begin{enumerate}[leftmargin=*] +% \item +% If $\g_1$ and $\g_2$ are Lie~algebras then the canonical projections +% \[ +% \pi_i \colon \g_1 \times \g_2 \to \g_i, \quad (x_1, x_2) \mapsto x_i +% \quad \text{for $i = 1, 2$} +% \] +% are homomorphisms of Lie-algebras. +% \item +% If $\glie$ is a Lie~algebra and $I \subideal \glie$ then the canonical projection $\pi \colon \g \to \g/I, x \mapsto [x]$ is a homomorphism of Lie~algebras. +% \end{enumerate} +% \end{lemma} +% +% +% +% \begin{lemma}\label{lem: quasi extension of scalars for lie algebras} +% Let $\glie$ be a Lie~algebra over $\kf$ and $A$ an associative, commutative $\kf$-algebra. Then $A \otimes_k \glie$ is a Lie~algebra over $\kf$ via +% \[ +% [a \otimes x, b \otimes y] = (ab) \otimes [x,y] +% \quad +% \text{for all $a,b \in A$ and $x,y \in \glie$}. +% \] +% Similarly $\g \otimes_k A$ carries the structure of a Lie~algebra over $\kf$ via +% \[ +% [x \otimes a, y \otimes b] = [x,y] \otimes (ab) +% \quad +% \text{for all $x,y \in \glie$ and $a,b \in A$}. +% \] +% \end{lemma} +% +% +% \begin{example} +% If $L/k$ is a field extension and $\glie$ a Lie~algebra over $\kf$, then $L \otimes_k \glie$ is a Lie~algebra over $\kf$ via +% \[ +% [\lambda \otimes x, \mu \otimes y] = (\lambda \mu) \otimes [x,y] +% \quad +% \text{for alle $\lambda, \mu \in k$ and $x,y \in \glie$}. +% \] +% $L \otimes_k \glie$ also carries the structure of an $L$-vector space via extension of scalars, i.e. +% \[ +% \lambda \cdot (\mu \otimes x) = (\lambda \mu) \otimes x +% \quad +% \text{for alle $\lambda, \mu \in k$ and $x \in \glie$}, +% \] +% and the Lie~bracket is not only $\kf$-bilinear, but also $L$-bilinear. Hence the structure of a $\kf$-Lie~algebra on $L \otimes_k \glie$ can be extended to the structure of a Lie~algebra over $L$. (Notice that the Jacobi-Identity is independent of the ground field.) +% \end{example} +% +% +% \begin{definition} +% Let $\glie$ be a Lie~algebra and $A = k[t,t^{-1}]$ be the algebra of Laurent polynomials over $\kf$. Then +% \[ +% \Loop(\g) \coloneqq \g \otimes_k A +% \] +% with the Lie~bracket as in Lemma~\ref{lem: quasi extension of scalars for lie algebras} is called the \emph{loop (Lie) algebra} of $\glie$. +% \end{definition} +% +% +% Another example for constructing new Lie~algebras out of old ones are \emph{central extensions}: Let $\glie$ be any $\kf$-Lie~algebra. +% \[ +% \tilde{\g} +% \coloneqq \g \oplus k +% = \{x + \lambda c \mid x \in \g, \lambda \in k\}, +% \] +% where we understand $c$ as a formal variable. Suppose that $\kappa \colon \g \times \g \to k$ is a $\kf$-bilinear map satisfying the following properties: +% \begin{enumerate} +% \item +% $\kappa$ is antisymmetric, i.e.\ $\kappa(x,y) = -\kappa(y,x)$ for all $x,y \in \glie$. +% \item +% $\kappa$ satisfies the $2$-cocycle condition +% \[ +% \kappa([x,y],z) + \kappa([y,z],x) + \kappa([z,x],y) = 0 +% \quad +% \text{for all $x,y,z \in \glie$}. +% \] +% \end{enumerate} +% Then $\tilde{\g}$ becomes a Lie~algebra via +% \[ +% [x + \lambda c, y + \mu c] \coloneqq [x,y] + \kappa(x,y) \lambda \mu c +% \quad \text{for all $x,y \in \glie$ and $\lambda, \mu \in k$.} +% \] +% Note that $c$ is central in $\tilde{\g}$ in the sense that $[x,c] = 0$ for all $x \in \glie$. +% +% +% \begin{example} +% Let $\g = \gllie_n(\kf)$. We define a symmetric bilinear form on $\glie$ via +% \[ +% (A,B)_{\tr} = \tr(AB) \quad \text{for all $A,B \in \glie$}. +% \] +% We define a bilinear form +% \[ +% \Loop(\g) \times \Loop(\g) \to k[t,t^{-1}], \quad +% (x \otimes p, y \otimes q) \mapsto (x,y)_{\tr}\ pq +% \] +% We now get a $2$-cocycle $\kappa \colon \Loop(\g) \times \Loop(\g) \to k$ via +% \[ +% \kappa(a,b) \coloneqq \mathrm{Res}\left(\frac{\partial a}{\partial t}, b\right). +% \] +% $\kappa$ is also antisymmetric: Let $a = x \otimes t^i$ and $b = y \otimes t^j$ with $x,y \in \glie$ and $i,j \in \Z$. Then +% \begin{align*} +% \kappa(x \otimes t^i, y \otimes t^j) +% = \mathrm{Res}(i x \otimes t^{i-1}, y \otimes t^j) +% &= \mathrm{Res}(i t^{i+j-1} (x,y)_{\tr}) \\ +% &= +% \begin{cases} +% i (x,y)_{\tr} & \text{if $i+j = 0$}, \\ +% 0 & \text{otherwise}. +% \end{cases} +% \end{align*} +% In the same way we find that +% \[ +% \kappa(y \otimes t^j, x \otimes t^i) = +% \begin{cases} +% j (x,y)_{\tr} & \text{if $i+j = 0$}, \\ +% 0 & \text{otherwise}. +% \end{cases} +% \] +% Since $(\cdot,\cdot)_{\tr}$ is symmetric we find that +% \begin{align*} +% \kappa(x \otimes t^i, y \otimes t^j) +% &= +% \begin{cases} +% i (x,y)_{\tr} & \text{if $i+j = 0$}, \\ +% 0 & \text{otherwise}, +% \end{cases} \\ +% &= +% \begin{cases} +% -j (x,y)_{\tr} & \text{if $i+j = 0$}, \\ +% 0 & \text{otherwise}, +% \end{cases} \\ +% &= +% -\kappa(y \otimes t^j, x \otimes t^i). +% \end{align*} +% \end{example} +% +% +% +% +% +% \subsection{Derivations} +% +% +% \begin{definition} +% Let $A$ be a $\kf$-algebra (not necessarily unitary of even associative). A \emph{derivation of $A$} is a $\kf$-linear map $d \colon A \to A$ such that +% \[ +% d(ab) = d(a)b + ad(b) \quad \text{for all $a,b \in A$}. +% \] +% We set +% \[ +% \Der(A) \coloneqq \{d \colon A \to A \mid \text{$d$ is a derivation of $A$} \}. +% \] +% \end{definition} +% +% +% \begin{remark} +% $\Der(A)$ is a $\kf$-linear subspace of $\End_k(A)$. +% \end{remark} +% +% +% \begin{example} +% Let $A$ be a $\kf$-algebra. It follows from direct calculation that for all $d, d' \in \Der(A)$ the commutator $[d,d'] = d \circ d' - d' \circ d$ is again a derivation $\Der(A)$. Hence $\Der(A)$ is a Lie subalgebra of $\gl(A)$. +% \end{example} +% +% +% \begin{lemma}\label{lem: Lie algebras act adjoint by derivations} +% Let $\glie$ be a Lie~algebra. Then for any $x \in \glie$ the map +% \[ +% \ad(x) \colon \g \to \g, \quad y \mapsto [x,y] +% \] +% is a derivation of $\glie$. +% \end{lemma} +% \begin{proof} +% By the Jacobi identity +% \begin{align*} +% \ad(x)([y,z]) +% &= [x,[y,z]] +% = [[x,y],z] + [y,[x,z]] \\ +% &= [\ad(x)(y),z] + [y,\ad(x)(z)] +% \end{align*} +% for all $y,z \in \glie$. +% \end{proof} +% +% +% \begin{definition} +% Let $\glie$ be a Lie~algebra. A derivation of $\glie$ is called \emph{inner} if it is of the form $\ad(x)$ for some $x \in \glie$. +% \end{definition} +% +% +% \begin{lemma}\label{lem: inner derivations are in ideal} +% If $\glie$ is a Lie~algebra then the inner derivations form an ideal inside of $\Der(\g)$. +% \end{lemma} +% \begin{proof} +% Let $I \coloneqq \im \ad \subseteq \Der(\g)$ be the linear subspace of inner derivations. For any $\delta \in \Der$ and $x \in \glie$ it follows that for any $y \in \glie$ +% \begin{align*} +% &\,[\delta, \ad(x)](y) +% = (\delta \ad(x) - \ad(x) \delta(x))(y) \\ +% &= \delta([x,y]) - [x,\delta(y)] +% = [\delta(x),y] + [x,\delta(y)] - [x,\delta(y)] \\ +% &= [\delta(x),y] +% = \ad(\delta(x))(y). +% \end{align*} +% Hence $[\delta, \ad(x)] = \ad(\delta(x)) \in I$. +% \end{proof} +% +% +% +% +% +% \subsection{Simple Lie~algebras} +% +% +% \begin{definition} +% A Lie~algebra $\glie$ is \emph{simple} if $0$ and $\glie$ are the only ideals inside $\glie$ and $\glie$ is not abelian. +% \end{definition} +% +% +% \begin{lemma} +% Let $\glie$ be a simple Lie~algebra. Then $[\g,\g] = \glie$ and $Z(\g)=0$. +% \end{lemma} +% \begin{proof} +% Because $\glie$ is simple it is not abelian. Therefore $[\g,\g] \neq 0$ and $Z(\g) \neq \glie$. Since $[\g,\g]$ and $Z(\g)$ are ideals inside $\glie$ it follows that $[\g,\g] = \glie$ and $Z(\g) = 0$. By the homomorphism theorem $\glie$ is isomorphic to its image $\ad \glie$ and hence to a linear Lie~algebra. +% \end{proof} +% +% +% \begin{corollary} +% Let $\glie$ be simple. Then the homomorphism $\ad \colon \g \to \gl(\g), x \mapsto \ad(x)$ is injective. In particular $\glie$ can be realized as a linear Lie~algebra. +% \end{corollary} +% \begin{proof} +% As part of Examples~\ref{expls: homomorphisms of lie algebras} has already been shown that $\ad$ is a homomorphism of Lie~algebras. That it is injective follows directly from $\ker \ad = Z(\g) = 0$. +% \end{proof} +% +% +% It can be shown that every finite dimensional Lie~algebra can be realized as a linear Lie~algebra. This will not be proven in this lecture and is by far not trivial. +% +% +% \begin{theorem}[Ado] +% Every finite dimensional Lie~algebra $\glie$ is isomorphic to a linear Lie~algebra. +% \end{theorem} +% +% +% \begin{examples} +% \begin{enumerate}[leftmargin=*] +% \item +% Since $[\gllie_n(\kf),\gllie_n(\kf)] = \sllie_n(\kf) \neq \gllie_n(\kf)$ we find that $\gllie_n(\kf)$ is not simple. +% \item +% Let $\g = \sll_2(k)$. Then $\glie$ is simple if and only if $\chara k \neq 2$. To see this consider the basis $(e,h,f)$ of $\sll_2(k)$ consisting of the matrices +% \[ +% e = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \quad +% h = \vect{1 & 0 \\ 0 & -1}, \quad +% f = \vect{0 & 0 \\ 1 & 0}. +% \] +% of $\sll_2(k)$. Then +% \[ +% [h,e] = 2e, \quad +% [h,f] = -2f, \quad +% [e,f] = h. +% \] +% If $\chara k = 2$ then $h$ spans a $1$-dimensional ideal, thus $\sll_2(k)$ is not simple. Suppose that $\chara k \neq 2$ and let $I \subseteq \sll_2(k)$ be an ideal with $I \neq 0$. From the above relations it follows that if $I$ contains one of the basis vectors $e$, $h$ or $f$ then already $I = \sll_2(k)$. Let $x \in I$ with $x \neq 0$ and write $x = \alpha e + \beta h + \gamma f$. Then +% \[ +% [e,x] = -2 \beta e + \gamma h \quad \text{and} \quad [e,[e,x]] = -2 \gamma e. +% \] +% Since $\gamma = 0$ or $\gamma \neq 0$ we find that $e \in I$. +% \end{enumerate} +% \end{examples} +% +% +% \begin{definition} +% Let $\kf$ be any field. The basis +% \[ +% e = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \quad +% h = \vect{1 & 0 \\ 0 & -1}, \quad +% f = \vect{0 & 0 \\ 1 & 0}. +% \] +% of $\sll_2(k)$ is called the \emph{standard basis} of $\sll_2(k)$. +% \end{definition} +% +% +% \begin{remark} +% If $\chara k = 0$ then $\sllie_n(\kf)$ is simple for all $n \geq 2$. +% \end{remark} diff --git a/sections/basics.tex b/sections/basics.tex index c05d4ac..847294f 100644 --- a/sections/basics.tex +++ b/sections/basics.tex @@ -1,7 +1,7 @@ \chapter{The Basics} \input{sections/basic_definitions} -\input{sections/representations} +% \input{sections/representations}