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\begin{document}

\lecture{1 --- Basic Definitions (I)}{September 9, 2010}{Prof.\
Victor Kac}{John Smith}


\setcounter{section}{1}

\begin{defn}
An algebra is a vector space over a field $\mathbb{F}$, endowed with
a bilinear product.
\end{defn}

\begin{exmp}
All $n\times n$ matrices with the matrix multiplication is an
associative algebra: $(ab)c=a(bc)$.
\end{exmp}

\begin{defn}
Lie algebra is an algebra with product $[a,b]$ (usually called
bracket), satisfying the following two axioms:

\begin{enumerate}
  \item [1](skew-commutativity) $[a,b]=-[b,a]$
  \item [2](Jacobi identity) $[a,[b,c]]+[b,[c,a]]+[c,[a,b]]=0$
\end{enumerate}
\end{defn}
\begin{exmp}
\begin{enumerate}
  \item [1)]Take a vector space $\g$ with bracket $[a,b]=0$.
  This is called an abelian Lie algebra;
  \item [2)]$\g=\mathbb{R}^{3}$,$[a,b]=a\times b$(cross
  product);
  \item [3)]Let $A$ be an associative algebra.
  Then the space $A$ with the bracket $[a,b]=ab-ba$ is a Lie algebra, denoted by $A_{\_}$. (skew commutativity is obvious.)
\end{enumerate}
\end{exmp}
\begin{exercise}
Check the Jacobi identity in Example $3)$.
\end{exercise}
\begin{proof}

\end{proof}
Special cases of this example:

$A=End V$, where $V$ is a vector space over a field $\mathbb{F}$,
with product being the composition of endomorphisms. Then $gl_V=(End
V)_{\_}$ is a Lie algebra, called the general Lie algebra. In the
case $V=\mathbb{F}^{n}$, we denote $gl_{V}=gl_n(\mathbb{F})$, the
set of all $n\times n$ matrices with the bracket $[a,b]=ab-ba$.

\begin{exercise}
Show that $sl_n(\mathbb{F})=\{$ all traceless $n\times n$ matrices
$\}$is a Lie algebra, which is called the special Lie algebra.
(Hint: derive this from the following property of the trace:
$tr[a,b]=0$, for any $a,b\in Mat_n(\mathbb{F})$.
\end{exercise}
\begin{proof}

\end{proof}
\begin{defn}
Let $\g$ be a Lie algebra. A subspace $\h\subset\mathcal {G}$ is
called a subalgebra if $[a,b]\in \h$ for any $a,b\in \h$. A subspace
$\h$ of $\g$ is called an ideal if $[a,b]\in\h$ for any $a\in \g$,
$b\in \h$.
\end{defn}
Notation: If $M,N\subset \g$ are subspaces,
 the span of all the elements $[m,n]$, where $m\in M$, $n\in
N$, is denoted by $[M,N]$.
\begin{itemize}
  \item $\h$ is a subalgebra means that
$[\h,\h]\subset \h$;
  \item $\h$ is an ideal means that
$[\g,\h]\subset \h$. Note that $[\g,\h]=[\h,\g]$ (due to the skew
commutativity).
\end{itemize}


Notice that by the property that $tr[a,b]=0$, for any $a,b\in
Mat_n$, $sl_n(\mathbb{F})$ is not only a subalgebra of
$gl_n(\mathbb{F})$, but also an ideal of  $gl_n(\mathbb{F})$.

\begin{exercise}
Show that if $f$ is a linear function on $Mat_n(\mathbb{F})$, such
that $f([a,b])=0$, for any $a,b\in Mat_n(\mathbb{F})$, then
$f(a)=$constant$\cdot tr(a)$, for all $a\in Mat_n(\mathbb{F})$.
\end{exercise}
\begin{proof}

\end{proof}
\begin{exmp}
Let $B:V\times V\rightarrow \mathbb{F}$ be a bilinear form on the
vector space $V$. Show that $$o_{V,B}=\{a\in
gl_V|B(a(u),v)=-B(v,a(u)), u,v\in V\}$$ is a subalgebra of the Lie
algebra $gl_n(\mathbb{F})$.
\end{exmp}
\begin{exercise}
Prove this fact.
\end{exercise}
\begin{proof}

\end{proof}
\begin{exercise}
Suppose $dim V=n$. Choose a basis of $V$, and let $B$ be the matrix
of the bilinear form under this basis. Then
$gl_n(\mathbb{F}))\supset o_{V,B}=\{a\in
gl_n(\mathbb{F})|a^{T}B+Ba=0\}$, where $a^{T}$ stands for the
transpose of $a$.
\end{exercise}
\begin{proof}
\end{proof}
The most important examples are:

\begin{exmp}Let $B$ be a non-degenerate symmetric bilinear form. Then $o_{V,B}$ is called the orthogonal Lie algebra, denoted by
$so_{n,B}(\mathbb{F})$.
\end{exmp}
\begin{exmp}
Let $B$ be a non-degenerate skew symmetric bilinear form (which is
possible only if  $n$ is even). Then $o_{V,B}$ is called the
symplectic Lie algebra, denoted by $sp_{n,B}(\mathbb{F})$.
\end{exmp}

The three most important series of Lie algebras, called the
classical series, are $sl_{n}(\mathbb{F})$, $so_{n,B}(\mathbb{F})$
and $sp_{n,B}(\mathbb{F})$. All of them are subalgebras of
$gl_{n}(\mathbb{F})$, but only $sl_{n}(\mathbb{F})$ is an ideal.

\begin{defn}[derived algebra] The derived algebra of a Lie algebra $\g$
is $[\g,\g]$.
\end{defn}

Now we consider the classification of Lie algebras of dimension $1$
and $2$.

\end{document}
