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\begin{document}
\begin{center}
{\large Classical Mechanics, Lecture 14 \\}
{\small February 26, 2008 \\}
{\small lecture by John Baez \\}
{\small notes by Alex Hoffnung}
\end{center}
\section{The Lie Algebra of a Lie Group}
We've seen some Lie groups already, like $\SO(n)$. Now it's time
for a formal definition:
\begin{defn}
A {\bf Lie group} is a group $G$ that is also a manifold such that
multiplication
\[m\maps G\times G\to G\]
\[(g,h)\mapsto gh\]
and inverse map
\[{\rm inv}\maps G\to G\]
\[g\mapsto g^{-1}\]
are smooth maps.
\end{defn}
Every Lie group has a Lie algebra. To get our hands on this,
given a Lie group $G$ we define a vector space
\[\g = T_1G\]
where $1\in G$ is the identity. We want to make this into a Lie algebra
and construct an `exponentiation' map
\[ \exp\maps\g\to G .\]
But we need to start slowly, and build up some basic tools first.
Recall that given a smooth map between manifolds $\psi\maps X\to N$ we
get a linear map called {\bf pushing forward along} $\psi$:
\[\psi_* = d\psi_x\maps T_xX\to T_{psi(x)}N\]
\[\textrm{picture of pushforward of tangent vector}\]
We can define $\psi_*$ by saying it sends $[\gamma]\in T_xX$ where $\gamma\maps\R\to X$ with $\gamma(0) = x$, to $[\psi\gamma]\in T_{\psi(x)}N$.\\\\
Check:
\[(1_X)_* = 1_{T_xX}\]
and if $\psi\maps M\to N$ and $\psi\maps N\to P$ then
\[\phi_*\psi_* = (\phi\psi)_*\]
This idea lets us think of the $\g$ as a space of vector fields called
`left-invariant' vector fields:
\begin{thm}
$\g$ is isomorphic to the vector space of {\bf left-invariant vector fields} on $G$, i.e.\ vector fields $v\in Vect(G)$ such that
\[(L_g)_*v(h) = v(gh),~~~\forall g,h\in G\]
where {\bf left multiplication by} $g$ is:
\[L_g\maps G\to G\]
\[h\mapsto gh.\]
The isomorphism goes as follows:
\[\{\textrm{left-invariant vector fields}\}\to\g\]
\[v\mapsto v(1)\]
\end{thm}
Before we prove this, let's draw a picture to explain it!
\[\textrm{picture of }U(1)\textrm{ in the complex plane with tangent space at identity before and after left multiplication}\]
A left-invariant vector field on $U(1)$ is one of constant length,
always pointing the same way. Given $x\in T_1G$ we get a
left-invariant vector field $V$ on $G$ by
\[v(g) = (L_g)_*x\]
Now let's prove the theorem:
\noindent {\bf Proof} - We construct an inverse map:
\[\g\to\{\textrm{left-invariant vector fields}\}\]
\[x\mapsto v^x\]
where
\[v^x(h) = (L_h)_*x\in T_hG.\]
Let's check that $v^x$ is left-invariant:
\begin{eqnarray*}
(L_g)_*v^x(h) &=& (L_g)_*(L_h)_*x\\
&=& (L_gL_h)_*x\\
&=& (L_{gh})_*x\\
&=& v^x(gh).
\end{eqnarray*}
Next check it is an inverse map. First: start with a left-invariant $w$, turn it into $w(1)\in\g$, then turn that back into a left-invariant vector field $v^{w(1)}$. Check: $w = v^{w(1)}$.
\begin{eqnarray*}
v^{w(1)}(h) &=& (L_h)_*w(1)\\
&=& w(h).\\
\end{eqnarray*}
Second: start with $x\in\g$, turn it into a left-invariant vector field $v^x$, then turn that back into $v^x(1)\in\g$. Check: $x = v^x(1)$.
\begin{eqnarray*}
v^x(1) &=& (L_1)_*x\\
&=& (1_G)_*x\\
&=& x.
\end{eqnarray*}
\[\textrm{picture of Lie-algebra element going to left-invariant vector field on the circle and vice versa}\]
We henceforth use this isomorphism to freely think of $\g$ {\it either} as $T_1G$ or as the space of all left-invariant vector fields on $G$. We use this to define a bracket operation on $\g$, using the fact that $Vect(G)$ is a Lie algebra. Using $\g\subseteq Vect(G)$ to make $\g$ into a Lie algebra we just need:
\begin{lem}
If $v,w\in Vect(G)$ are left-invariant, so is $[v,w]$.
\end{lem}
\noindent {\bf Proof} - For this we will use a general fact: if $\phi\maps M\to N$ is a diffeomorphism, then given $v\in Vect(M)$
\[\textrm{picture of push-forward of vector}\]
If $\phi$ is a diffeomorphism there is a unique $x\in M$ mapping to any $y\in N$ (namely $\phi^{-1}(y)$), so we can define a vector field $\phi_*v\in Vect(N)$ by:
\[(\phi_*v)(y) = \phi_*v(\phi^{-1}(y))\]
In fact, if $\phi\maps M\to N$ is a diffeomorphism and $v,w\in Vect(M)$ then:
\[\phi_*[v,w] = [\phi_*v,\phi_*w].\]
In particular, if $v$ and $w$ are left-invariant vector fields on $G$,
\[(L_g)_*v = v\]
\[(L_g)_*w = w\]
so
\begin{eqnarray*}
(L_g)_*[v,w] &=& [(L_g)_*v,(L_g)_*w]\\
&=& [v,w]
\end{eqnarray*}
so $[v,w]$ is left-invariant.\\\\
Now that $\g$ is a Lie algebra let's define the exponential map
\[exp\maps\g\to G.\]
We will use a sort of hard fact
\begin{thm}
If $G$ is a Lie group, every left-invariant $v\in Vect(G)$ is integrable.
\end{thm}
(By the way: If $M$ is a compact manifold, then every $v\in Vect(M)$ is integrable.)\\\\
Given this, any $v\in\g$ thought of as a left-invariant vector field, generates a flow:
\[\phi \maps\R\times G\to G\]
\[(t,g)\mapsto \phi_t(g)\]
and we define:
\[\exp(tv) = \phi_t(1)\in G\]
{\bf Example}: $G = \U(1)$
\[\textrm{picture of circle with left-invariant vector field}\]
This generates a flow where $\phi_t\maps U(1)\to U(1)$ is rotation by the angle $\frac{tv}{i}\in\R$. Note: rotation by $\frac{tv}{i}$ is the same as multiplication by $e^{tv}\in U(1)$. So
\[\exp(tv) = \phi_t(1) = e^{tv}1 = e^{tv}\in U(1)\]
Whew!\\\\
{\bf Example}: $G = \SO(n)$, rotation group of $\R^n$.\\
This is a group of matrices, with matrix multiplication as the group operation. So $\SO(n)$ sits in the vector space of $n\times n$ matrices:
\[\textrm{artist's depiction of }\SO(n) \textrm{ with its Lie algebra } \so(n)\]
In this case the Lie algebra
\[\so(n) = \{A\maps\R^n\to\R^n : A \textrm{ linear and } A^*=-A\}\]
and
\[\exp\maps\so(n)\to\so(n)\]
is given by
\[\exp(A) = \sum_{k=0}^\infty\frac{A^k}{k!}\]
With this definition of $exp\maps\g\to G$ we can check:
\[\exp(0) = 1\in G\]
(easy, since a flow has $\phi_0 = 1$) and:
\[\exp((s+t)v) = exp(sv)exp(tv)\]
(trickier, but we use $\phi_{s+t} = \phi_s\phi_t$ and some more).\\\\
\section{Actions of Lie Groups}
What are Lie groups good for? They `act' on manifolds!
\begin{defn}
If $G$ is a Lie group and $X$ is some manifold, an {\bf action} of $G$ on $X$ is a smooth map
\[\phi\maps G\times X\to X\]
\[(g,x)\mapsto\phi(g)x\]
such that:
\[\phi(gh)x = \phi(g)\phi(h)x,~~~\forall g,h\in G, \forall x\in X\]
and:
\[\phi(1)x = x.\]
(These imply: $\phi(g^{-1}) = \phi(g)^{-1}$).
\end{defn}
{\bf Example}: A flow is the same as an action of $G = \R$.\\\\ {\bf
Example}: The Euclidean group $E(n)$ is a Lie group and it acts on
$\R^n$ (``space").\\\\
{\bf Example}: The Galilei group $G(n+1)$ is a
Lie group and it acts on $\R^{n+1}$ (``spacetime"). We also saw how
$G(n+1)$ acts on the phase space of a free particle in $\R^n$, $X =
\R^n\times\R^n\ni(q,p)$.\\\\
We have seen how symmetries in $G(n+1)$
are related to conserved quantities, certain functions on the Poisson
manifold $X$. We have seen that (almost) any single function on a
Poisson manifold generates a flow, i.e. an action of $\R$. (The
vector field might fail to be integrable.)\\\\
When does a collection
of functions on a Poisson manifold $X$ give rise to an action of a Lie
group $G$ on $X$? Or conversely, which group actions on $X$ give rise
to a bunch of functions on $X$? These are important questions that
we'll begin to tackle next time.
\end{document}