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	\begin{document}

       \begin{center}
      {\large Classical Mechanics, Lecture 13 \\}  
      {\small February 21, 2008 \\}  
      {\small lecture by John Baez \\}  
      {\small notes by Alex Hoffnung}  
      \end{center}

\section{Lie Algebras}

Lie groups are groups that are also manifolds.  These let us describe
continuous symmetries in physics.  A classic example is $\SO(n)$ - the
group of rotations of $\R^n$.  Every Lie group has a `Lie algebra',
which lets us do calculations more easily.  The Lie algebra is a
vector space, so it lets us use linear algebra to study Lie
groups.\\\\ 

For example: $\SO(n)$ has a Lie algebra
\[\so(n) = \{A : A \textrm{ is an } n\times n \textrm{ real  matrix with } A^* = -A\}\]
The point is that given $A\in\so(n)$,
\[\exp(tA) = \sum_{k=0}^\infty\frac{(tA)^k}{k!}\in\SO(n),~~~(t\in\R)\]
In fact, any Lie group $G$ has a Lie algebra $\g$ and there is a map called exponentiation
\[\exp\maps\g \to G\]
such that
\[\exp((s+t)A) = \exp(sA)\exp(tA).\]
For example, if $\g = \R$ (a vector spce) and $G = \R^+$ (the Lie group of positive real numbers, with multiplication as the group operation), then
\[\exp:\R\to\R^+\]
is the usual exponential function and I am just saying $\exp(s+t) = \exp(s)\exp(t)$.  Here a slide rule (or table of logarithms) converts problems in the Lie group into problems in the Lie algebra.

\begin{defn}
A {\bf Lie algebra} is a (real) vector space  $\g$ equipped with an operation
\[[\cdot,\cdot]\maps\g\times\g\to\g\]
called the {\bf Lie bracket} which has the following properties:
\begin{enumerate}
\item bilinearity
\item antisymmetry
\item the Jacobi identity
\end{enumerate}
\end{defn}

We've seen one type of Lie algebra so far: if $X$ is a Poisson
manifold, $C^\infty(X)$ is a Lie algebra with the Poisson bracket
as its bracket.  But in fact, {\em every} manifold gives a Lie
algebra, in a different way:

\begin{thm}
Given any manifold $X$, let $\Vect(X)$ be the set of vector fields on $X$.  
This becomes a Lie algebra by:
\[(\alpha v)f = \alpha(v(f))\]
\[(v+w)f = vf+wf\]
\[[v,w]f = v(wf)-w(vf)\]
where $\alpha\in\R$ and $f\in C^\infty(X)$.
\end{thm}
\noindent Sketch of proof:\\
Just calculate to check the Lie algebra axioms.  For example: a vector field $v\in \Vect(X)$ is a linear map
\[v\maps C^\infty(X)\to C^\infty(X)\]
satisfying $v(fg) = v(f)g + fv(g)$.\\\\
Why does $v,w\in \Vect(X)\Rightarrow [v,w]\in \Vect(X)$?\\
\begin{eqnarray*}
[v,w](fg) &=& vw(fg) - wv(fg)\\
				&=& v(w(f)g + fw(g)) - w(v(f)g + fv(g))\\
				&=& (vw(f))g + w(f)v(g) + v(f)w(g) + f(vw(g)) - wv(f)g - v(f)w(g) - w(f)v(g) - f(wv(g))\\
				&=& ([v,w]f)g + f([v,w]g).
\end{eqnarray*}
\\\\
\noindent Why does $[\cdot,\cdot]$ satisfy the Jacobi identity?
\[[u,[v,w]] = [[u,v],w] + [v,[u,w]]\]
\begin{eqnarray*}
([[u,v],w] + [v,[u,w]])f &=& (uvw - vuw -wuv +wvu +vuw - vwu - uwv + wuv)f\\
									&=& [u,[v,w]]f
\end{eqnarray*}
Etc.\ldots\\\\

This Lie algebra $\Vect(X)$ is very related to the Poisson algebra
$C^\infty(X)$ when $X$ is a Poisson manifold:

\begin{thm}
If $X$ is a Poisson manifold and $f,g\in C^\infty(X)$ then
\[[v_f,v_g] = v_{\{f,g\}}.\]
\end{thm}

\noindent Proof - For any $h\in C^\infty(X)$,
\begin{eqnarray*}
v_{\{f,g\}}h &=& \{\{f,g\},h\}\\
						&=& \{f\{g,h\}\} + \{\{f,h\},g\}\\
						&=&\{f,\{g,h\}\} - \{g,\{f,h\}\}\\
						&=& v_fv_gh - v_gv_fh\\
						&=& [v_f,v_g]h
\end{eqnarray*}

Later we'll define a `homomorphism' between Lie algebras, and the
theorem we just proved will say that 
\[          v \mapsto v_f \]
is a homomorphism from the Lie algebra $C^\infty(X)$ to the Lie
algebra $\Vect(X)$.

\end{document}
