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{\small
\ni{Lectures by Robert Geroch}
\ni{{\TeX}ed and supplemented by R. Salgado}
}

\ni{\bf LIE ALGEBRAS}
\bigskip

A {\bf lie algebra}\marginpar{\bf lie algebra} is a real vector space $L$,
together with a mapping $L\times L \stackrel{\beta}{\rightarrow} L$
(always written, rather than $\beta(l,l')$, as $[l,l']$)
, a rule that assigns, for any two elements $l$ and $l'$ of $L$,
a third element, $[l,l']$), satisfying the following three conditions:
     \begin{enumerate}
          \item (Linearity) For any $l$, $l'$, and $l'' \in L$ and
               $a,b\in\Re{}$, $[l,al'+bl'']=a[l,l']+b[l,l'']$, and
               $[al+bl',l'']=a[l,l'']+b[l',l''].$
          \item (Antisymmetry) For any $l, l' \in L$, $[l,l']=-[l',l].$
          \item (Jacobi relation) For any $l, l', l'' \in L$,
               $[[l,l'],l'']+[[l',l''],l]+[[l'',l],l']=0.$
     \end{enumerate}

Thus, to specify a lie algebra one must say what the vector space $L$ is;
say, for any $l, l' \in L$, what element of $L$ $[l,l']$ is;
and check the three conditions above.

\ni EXAMPLES

Any vector space $L$, with $[l,l']=0$ for all $l,l'\in L$, is a lie algebra.
     \tech{The three conditions follow immediately when all ``$[\ ,\ ]$''
     vanish.}

The vector space $L$ of all (bounded) operators on a Hilbert space, with,
for $A$ and $B$ such operators, $[A,B]=AB-BA$, is a lie algebra.
     \tech{The first two conditions are immediate.
     A direct computation will demonstrate the thrid condition.}\\
Similarly, the vector space $L$ of all bounded Hermitian operators
on a Hilbert space, with $[A,B]=i(AB-BA)$, is a lie algebra.
     \tech{The ``$i$'' is needed here to guarantee that, for $A$ and $B$
          Hermitian, $[A,B]$ is Hermitian.}

The vector space $L$ of all smooth vector fields on a manifold (\eg $\Re{3}$),
with, for $\vec v, \vec v'$ such vector fields,
$[\vec v,\vec v']=(\vec v\cdot\vec\nabla)\vec v'-
                         (\vec v'\cdot\vec\nabla)\vec v$,
is a lie algebra.

The vector space $L$ of all antisymmetric, real, $n\times n$ matrices, with
for $A$ and $B$ such matrices, $[A,B]=AB-BA$, is a lie algebra.
     \tech{The right side is indeed an antisymmetric matrix, using the fact
          that $(AB)^t=B^t A^t$.}

The vector space $L$ consisting of ordinary $3$-dimensional Euclidean space,
with, for $\vec v, \vec v' \in L$, $[\vec v,\vec v']=\vec v \times \vec v'$,
is a lie algebra.
     \tech{The first two conditions are immediate, while the Jacobi relation
          follows from a standard vector identity.}


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