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\begin{document}
\thispagestyle{empty}
\begin{center}
{\bf \Large Lie groups and Lie algebras}
\vskip.2in
{\large Eckhard Meinrenken} 
\vskip.6in
{\Large
Lecture Notes, University of Toronto, Fall 2010}
%\footnote{WARNING: THESE NOTES ARE FULL OF MISTAKES, AND ARE 
%NOT INTENDED FOR DISTRIBUTION. 
%USE THEM AT YOUR OWN RISK!}
\vskip1in
\end{center} 




\section{Terminology and notation}
\subsection{Lie groups}
%
\begin{definition}
A Lie group is a group $G$, equipped with a manifold structure such that the 
group operations 
\[ \on{Mult}\colon G\times G\to G,\ \ (g_1,g_2)\mapsto g_1 g_2\]
\[ \on{Inv}\colon G\to G,\ \ g\mapsto g^{-1}\]
are smooth. A morphism of Lie groups $G,G'$ is a morphism of groups
$\phi\colon G\to G'$ that is smooth.   
\end{definition}
%
\begin{remark}
Using the implicit function theorem, one can show that smoothness of $\on{Inv}$ is in fact automatic. (Exercise) 
\end{remark}

The first example of a Lie group is the \emph{general linear group}
\[ \GL(n,\R)=\{A\in \on{Mat}_n(\R)|\ \det(A)\not=0\}\]
of invertible  $n\times n$ matrices. It is an open subset of $\on{Mat}_n(\R)$, hence a submanifold, and the smoothness of group multiplication follows since the product map for $\on{Mat}_n(\R)$ is obviously smooth. 

Our next example is the orthogonal group
%
\[ \on{O}(n)=\{A\in \on{Mat}_n(\R)|\ A^T A=I\}.\]
% 
To see that it is a Lie group, it suffices to show that $\on{O}(n)$ is an embedded submanifold of $\on{Mat}_n(\R)$. In order to construct submanifold charts, we use the exponential map of matrices 
%
\[ \exp\colon \on{Mat}_n(\R)\to \on{Mat}_n(\R),\ \ 
B\mapsto \exp(B)=\sum_{n=0}^\infty \f{1}{n!} B^n\]
%
(an absolutely convergent series). One has $\f{\d}{\d
t}|_{t=0}\exp(tB)=B$, hence the differential of $\exp$ at $0$ is the
identity $\on{id}_{\on{Mat}_n(\R)}$. By the inverse function theorem,
this means that there is $\eps>0$ such that $\exp$ restricts to a 
diffeomorphism from the open neighborhood $U=\{B:\ ||B||<\eps\}$ of $0$ 
onto an open neighborhood $\exp(U)$ of $I$. 
Let 
\[ \mf{o}(n)=\{B\in \on{Mat}_n(\R)|\ B+B^T=0\}.\]
%
We claim that 
\[ \exp(\mf{o}(n)\cap U)=\on{O}(n)\cap \exp(U),\]
%
so that $\exp$ gives a submanifold chart for $\on{O}(n)$ over 
$\exp(U)$. To prove the claim, let $B\in U$. Then 
\[ \begin{split} \exp(B)\in \on{O}(n) & \Leftrightarrow 
\exp(B)^T=\exp(B)^{-1}\\
& \Leftrightarrow 
\exp(B^T)=\exp(-B)\\
& \Leftrightarrow 
B^T=-B\\
& \Leftrightarrow B\in \mf{o}(n).
\end{split} 
\]
For a more general $A\in \on{O}(n)$, we use that the map 
$\on{Mat}_n(\R)\to \on{Mat}_n(\R)$ given by left multiplication is a 
diffeomorphism. Hence, $A\exp(U)$ is an open neighborhood of $A$, 
and we have 
\[ A\exp(U)\cap \on{O}(n)=A(\exp(U)\cap \on{O}(n))=A\exp(U\cap \mf{o}(n)).\]
%
Thus, we also get a submanifold chart near $A$. This proves that $\on{O}(n)$ 
is a submanifold. Hence its group operations are induced from those of 
$\GL(n,\R)$, they are smooth. Hence $\on{O}(n)$ is a Lie group. 
Notice that $\on{O}(n)$ is compact (the column vectors of an orthogonal matrix 
are an orthonormal basis of $\R^n$; hence $\on{O}(n)$ is a subset of 
$S^{n-1}\times\cdots S^{n-1}\subset \R^n\times\cdots \R^n$). 

A similar argument shows that the \emph{special linear group}
\[ \SL(n,\R)=\{A\in \on{Mat}_n(\R)|\ \det(A)=1\}\]
is an embedded submanifold of $\GL(n,\R)$, and hence is a Lie group. 
The submanifold charts are obtained by exponentiating the subspace 
\[ \mf{sl}(n,\R)=\{B\in \on{Mat}_n(\R)|\ \on{tr}(B)=0\},\]
using the identity $\det(\exp(B))=\exp(\on{tr}(B))$. 

Actually, we could have saved most of this work with $\on{O}(n),\
\SL(n,\R)$ once we have the following beautiful result of E. Cartan:
\begin{quote}
{\bf Fact:} \emph{Every closed subgroup of a Lie group is an embedded submanifold, 
hence is again a Lie group.} 
\end{quote}
%
We will prove this very soon, once we have developed some more basics
of Lie group theory. A closed subgroup of $\GL(n,\R)$ (for suitable $n$) is 
called a \emph{matrix Lie group}. Let us now give a few more examples of Lie
groups, without detailed justifications. 

\begin{examples}
\begin{enumerate}
\item Any finite-dimensional vector space $V$ over $\R$ is a Lie group, with product $\on{Mult}$ given by addition. 
\item 
Let $\A$ be a finite-dimensional associative algebra over $\R$, with
unit $1_\A$. Then the group $\A^\times$ of invertible elements is a
Lie group. More generally, if $n\in\N$ we can create the algebra
$\on{Mat}_n(\A)$ of matrices with entries in $\A$, and the
\emph{general linear group}
%
\[ \GL(n,\A):=\on{Mat}_n(\A)^\times\]
%
is a Lie group. If $\A$ is \emph{commutative}, one has a determinant map  
$\on{det}\colon \on{Mat}_n(\A)\to \A$, and $\GL(n,\A)$ is the pre-image of $\A^\times$. One may then define a \emph{special linear group}
\[ \on{SL}(n,\A)=\{g\in \GL(n,\A)|\ \det(g)=1\}.\]
\item 
We mostly have in mind the cases $\A=\R,\C,\H$. Here $\H$ is the algebra of 
\emph{quaternions} (due to Hamilton). Recall that $\H=\R^4$ as a vector space, with elements 
$(a,b,c,d)\in \R^4$ written as 
%
\[ x=a+ib+jc+kd\]
%
with imaginary units $i,j,k$. The algebra structure is determined by 
\[ i^2=j^2=k^2=-1,\ ij=k,\ jk=i,\ ki=j.\]
%
Note that $\H$ is non-commutative (e.g. $ji=-ij$), hence $\on{SL}(n,\H)$ is 
\emph{not} defined. On the other hand, one can define complex conjugates
\[ \ol{x}=a-ib-jc-kd\]
and 
\[ |x|^2:=x\ol{x}=a^2+b^2+c^2+d^2.\]
defines a norm $x\mapsto |x|$, with $|x_1x_2|=|x_1| |x_2|$ just as for 
complex or real numbers. The spaces $\R^n,\C^n,\H^n$ inherit norms, 
by putting 
\[ ||x||^2=\sum_{i=1}^n |x_i|^2,\ \ x=(x_1,\ldots,x_n).\]
%
The subgroups of $\GL(n,\R),\ \GL(n,\C),\ \GL(n,\H)$ preserving this 
norm (in the sense that $||Ax||=||x||$ for all $x$) are denoted
%
\[ \on{O}(n),\ \on{U}(n),\ \on{Sp}(n)\]
%
and are called the \emph{orthogonal, unitary, and symplectic group}, 
respectively.  Since the norms of $\C,\H$ coincide with those of 
$\C\cong \R^2$, $\H=\C^2 \cong \R^4$,  we have 
\[ \U(n)=\GL(n,\C)\cap \on{O}(2n),\ \ 
\on{Sp}(n)=\GL(n,\H)\cap \on{O}(4n).\]
%
In particular, all of these groups are compact.  
One can also define 
%
\[ \SO(n)=\on{O}(n)\cap \SL(n,\R),\ \ \SU(n)=\on{U}(n)\cap \SL(n,\C),\]
% 
these are called the \emph{special orthogonal} and \emph{special unitary}
groups. The groups $\SO(n),\ \SU(n), \on{Sp}(n)$ are often called the 
\emph{classical groups} (but this term is used a bit loosely). 
\item 
For any Lie group $G$, its universal cover $\wt{G}$ is again a Lie group. 
The universal cover $\wt{\SL(2,\R)}$ is an example of a Lie group that is not 
isomorphic to a matrix Lie group. 
\end{enumerate}
\end{examples}

\subsection{Lie algebras}
%
\begin{definition}
A Lie algebra is a vector space $\g$, together with a bilinear map 
$[\cdot,\cdot]:\,\g\times\g\to \g$ satisfying {\em anti-symmetry}
%
$$ [\xi,\eta]=-[\eta,\xi] \mbox{ for all } \xi,\eta\in\g,$$
%
and the {\em Jacobi identity},
%
$$ [\xi,[\eta,\zeta]]+[\eta,[\zeta,\xi]]+[\zeta,[\xi,\eta]]=0
\mbox{ for all }\xi,\eta,\zeta\in\g.
$$
The map $[\cdot,\cdot]$ is called the Lie bracket. A morphism of Lie algebras 
$\g_1,\g_2$ is a linear map $\phi\colon \g_1\to \g_2$ preserving brackets.
\end{definition}
\vskip.1in
%
The space 
\[ \mf{gl}(n,\R)=\on{Mat}_n(\R)\]
% 
is a Lie algebra, with bracket the commutator of matrices. 
(The notation indicates that we think of $\on{Mat}_n(\R)$ 
as a Lie algebra, not as an algebra.) 

A Lie subalgebra of  $\mf{gl}(n,\R)$, i.e. a subspace preserved under commutators,  is called a \emph{matrix Lie algebra}. For instance, 
\[ \mf{sl}(n,\R)=\{B\in\on{Mat}_n(\R)\colon \on{tr}(B)=0\}\]
and 
\[ \mf{o}(n)=\{B \in\on{Mat}_n(\R)\colon \ B^T=-B\}\]
are matrix Lie algebras (as one easily verifies). 
It turns out that every finite-dimensional real Lie algebra is
isomorphic to a matrix Lie algebra (\emph{Ado's theorem}), but the proof is
not easy. 

The following examples of finite-dimensional Lie algebras correspond 
to our examples for Lie groups. The origin of this correspondence will 
soon become clear. 

\begin{examples}
\begin{enumerate}
\item 
Any vector space $V$ is a Lie algebra for the zero bracket. 
\item 
Any associative algebra $\A$ can be viewed as a Lie algebra under commutator. 
Replacing $\A$ with matrix algebras over $\A$, it follows that 
$\mf{gl}(n,\A)=\on{Mat}_n(\A)$, is a Lie algebra, with bracket the commutator. 
If $\A$ is commutative, then the subspace $\mf{sl}(n,\A)\subset \mf{gl}(n,\A)$
of matrices of trace $0$ is a Lie subalgebra. 
\item 
We are mainly interested in the cases $\A=\R,\C,\H$. Define an inner product on $\R^n,\C^n,\H^n$ by putting
%
\[ \l x,y\r=\sum_{i=1}^n \ol{x}_i y_i,\]
%
and define $\mf{o}(n),\ \mf{u}(n),\ \mf{sp}(n)$ as the matrices in 
$\mf{gl}(n,\R),\ \mf{gl}(n,\C),\ \mf{gl}(n,\H)$ satisfying 
\[ \l B x,y\r=-\l x,By\r\] 
for all $x,y$. These are all Lie algebras called the (infinitesimal)
orthogonal, unitary, and symplectic Lie algebras. For $\R,\C$ one can impose the additional condition $\on{tr}(B)=0$, thus defining the special 
orthogonal and special unitary Lie algebras
$\mf{so}(n),\ \mf{su}(n)$. Actually, 
\[ \mf{so}(n)=\mf{o}(n)\]
%
since $B^T=-B$ already implies $\on{tr}(B)=0$.
\end{enumerate}
\end{examples}

\vskip1in

\begin{exercise}
Show that $\on{Sp}(n)$ can be characterized as follows. Let $J\in U(2n)$ 
be the unitary matrix
\[ \left(\begin{array}{cc}0&I_n\\ -I_n&0 
\end{array}\right)\]
where $I_n$ is the $n\times n$ identity matrix. Then 
\[ \on{Sp}(n)=\{A\in \U(2n)|\ \ol{A}=J A J^{-1}\}.\]
Here $\ol{A}$ is the componentwise complex conjugate of $A$. 
\end{exercise}

\begin{exercise}
Let $R(\theta)$ denote the $2\times 2$ rotation matrix 
\[ R(\theta)=\left(\begin{array}{cc} \cos\theta & -\sin\theta\\
\sin\theta &\cos\theta\end{array}\right).\]
Show that for all $A\in \SO(2m)$ there exists $O\in \SO(2m)$ such that 
$O A O^{-1}$ is of the block diagonal form 
\[ \left( \begin{array}{ccccc} R(\theta_1) & 0& 0&\cdots&0\\
 0& R(\theta_2)&0&\cdots &0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\
0&0&0&\cdots&R(\theta_m)\end{array}\right).\] For $A\in \SO(2m+1)$ one
has a similar block diagonal presentation, with $m$ $2\times 2$ blocks
$R(\theta_i)$ and an extra $1$ in the lower right corner. Conclude that 
$\SO(n)$ is connected. 
\end{exercise}


\begin{exercise}\label{ex:3}
Let $G$ be a connected Lie group, and $U$ an open neighborhood of the group unit $e$. Show that 
any $g\in G$ can be written as a product $g=g_1\cdots g_N$ of elements $g_i\in U$. 
\end{exercise}

\begin{exercise}
Let $\phi\colon G\to H$ be a morphism of connected Lie groups, and assume that the differential 
$\d_e\phi\colon T_eG\to T_eH$ is bijective (resp.~ surjective). 
Show that $\phi$ is a covering (resp.~ surjective).  
Hint: Use Exercise \ref{ex:3}. 
\end{exercise}

\section{The covering $\SU(2)\to \SO(3)$}
%
The Lie group $\SO(3)$ consists of rotations in 3-dimensional space. 
Let $D\subset \R^3$ be the closed ball of radius $\pi$. Any 
element $x\in D$ represents a rotation by an angle $||x||$ in the direction 
of $x$. This is a 1-1 correspondence for points in the interior of $D$, but 
if $x\in \partial D$ is a boundary point then $x,-x$ represent the same rotation. Letting $\sim$ be the equivalence relation on $D$, given by antipodal identification on the boundary, we have $D^3/\sim =\R P(3)$. Thus 
%
\[ \SO(3)=\R P(3)\]
%
(at least, topologically). With a little extra effort (which we'll make below)
one can make this into a diffeomorphism of manifolds. 
 
By definition 
\[ \SU(2)=\{A\in \on{Mat}_2(\C)|\ A^\dagger=A^{-1},\ \det(A)=1\}.\]
%
Using the formula for the inverse matrix, we see that $\SU(2)$ consists of matrices of the form 
%
\[ \SU(2)=\left\{\left(\begin{array}{cc}
z&-\ol{w}\\ w& \ol{z}
\end{array}\right)|\ |w|^2+|z|^2=1\right\}.\]
%
That is, $\SU(2)=S^3$ as a manifold. Similarly, 
\[ \mf{su}(2)=\left\{\left(\begin{array}{cc}
it &-\ol{u}\\ u& -it
\end{array}\right)|\ t\in\R,\ u\in\C\right\}\]
%
gives an identification $\mf{su}(2)=\R\oplus \C=\R^3$. 
Note that for a matrix $B$ of this form, $\det(B)=t^2+|u|^2$, so that 
$\det$ corresponds to $||\cdot||^2$ under this identification. 

The group $\SU(2)$ acts linearly on the vector space $\mf{su}(2)$, 
by matrix conjugation: $B\mapsto ABA^{-1}$. 
Since  the conjugation action preserves $\det$, we obtain a linear 
action on $\R^3$, preserving the norm. This defines a Lie group morphism 
from $\SU(2)$ into $\on{O}(3)$. Since $\SU(2)$ is connected, this must take values in the identity component:
\[ \phi\colon \SU(2)\to \SO(3).\]
The kernel of this map consists of matrices $A\in\SU(2)$ such that $A
B A^{-1}=B$ for all $B\in\mf{su}(2)$. Thus, $A$ commutes with all skew-adjoint matrices of trace $0$. 
Since $A$ commutes with multiples of the identity, it then commutes with all skew-adjoint matrices. 
But since $\on{Mat}_n(\C)=\mf{u}(n)\oplus i\mf{u}(n)$ (the decomposition into skew-adjoint and self-adjoint parts), 
it then follows that $A$ is a multiple of the identity matrix. Thus $\ker(\phi)=\{I,-I\}$ is discrete. Since $\d_e\phi$ is
an isomorphism, it follows that the map $\phi$ is a double
covering. This exhibits $\SU(2)=S^3$ as the double cover of $\SO(3)$. 


\begin{exercise}
Give an explicit construction of a double covering of $\SO(4)$ by 
$\SU(2)\times\SU(2)$. Hint: Represent the quaternion algebra $\H$ as 
an algebra of matrices $\H\subset \on{Mat}_2(\C)$, by 
\[ x=a+ib+jc+kd\mapsto x=\left(\begin{array}{cc}
a+ib & c+id\\ -c+id & a-ib
\end{array}\right).\]
Note that $|x|^2=\det(x)$, and that $\SU(2)=\{x\in \H|\ \det(x)=1\}$. 
Use this to define an action of $\SU(2)\times\SU(2)$ on $\H$ preserving 
the norm. 
\end{exercise}





\section{The Lie algebra of a Lie group}
%
\subsection{Review: Tangent vectors and vector fields}
%
We begin with a quick reminder of some manifold theory, partly just to set up our notational conventions.

Let $M$ be a manifold, and $C^\infty(M)$ its algebra of smooth real-valued 
functions.  For $m\in M$, we define the tangent space 
$T_mM$ to be the space of directional derivatives: 
%
\[ T_mM=\{v\in \Hom(C^\infty(M),\R)|\ v(fg)=v(f) g+v(g) f\}.\]
%
Here $v(f)$ is local, in the sense that $v(f)=v(f')$ if $f'-f$ vanishes on a neighborhood of $m$. 
%
\begin{example}
If $\gamma\colon J\to M$, $J\subset \R$ is a smooth curve we obtain tangent vectors to the curve,
%
\[ \dot{\gamma}(t)\in T_{\gamma(t)}M,\ \ \dot{\gamma}(t)(f)= \f{\p}{\p t}|_{t=0}f(\gamma(t)).\] 
%
\end{example}
\begin{example}
We have $T_x\R^n=\R^n$, where the isomorphism takes $a\in\R^n$ to the 
corresponding velocity vector of the curve $x+t a$. That is, 
\[ v(f)=\f{\p}{\p t}|_{t=0}f(x+ta)=\sum_{i=1}^n a_i \f{\p f}{\p x_i}.\] 
\end{example}

A smooth map of manifolds $\phi\colon M\to M'$ defines a 
\emph{tangent map}:
\[ \d_m\phi\colon T_mM\to T_{\phi(m)}M',\ \ 
(\d_m\phi(v))(f)=v(f\circ \phi).\]
%
The locality property ensures that for an open neighborhood $U\subset M$, 
the inclusion identifies $T_mU=T_mM$. In particular, a coordinate chart 
$\phi\colon U\to \phi(U)\subset \R^n$ gives an isomorphism 
%
\[ \d_m\phi\colon T_mM=T_mU\to T_{\phi(m)}\phi(U)=T_{\phi(m)}\R^n=\R^n.\]
%
Hence $T_mM$ is a vector space of dimension $n=\dim M$. 
The union $TM=\bigcup_{m\in M}T_mM$ is a vector bundle over $M$, 
called the tangent bundle. Coordinate charts for $M$ give vector 
bundle charts for $TM$. For a smooth map of manifolds  
$\phi\colon M\to M'$, the entirety of all maps $\d_m\phi$ defines a smooth vector bundle map 
\[ \d\phi\colon TM\to TM'.\]




A \emph{vector field} on $M$ is a derivation $X\colon C^\infty(M)\to C^\infty(M)$. That is, it is a linear map satisfying 
%
\[ X(fg)=X(f) g+f X(g).\]
%  
The space of vector fields is denoted $\mf{X}(M)=\on{Der}(C^\infty(M))$.  
Vector fields are local, in the sense that for any open subset $U$ there is 
a well-defined restriction $X|_U\in\mf{X}(U)$ such that $X|_U(f|_U)=(X(f))|_U$. 
For any vector field, one obtains tangent vectors $X_m\in T_mM$ by
$X_m(f)=X(f)|_m$. One can think of a vector field as an assignment of
tangent vectors, depending smoothly on $m$. More precisely, a vector
field is a smooth section of the tangent bundle $TM$. In local
coordinates, vector fields are of the form $\sum_i a_i \f{\p}{\p x_i}$ 
where the $a_i$ are smooth functions. 

It is a general fact that the commutator of derivations of an
algebra is again a derivation. Thus, $\mf{X}(M)$ is a Lie algebra for
the bracket
%
\[ [X,Y]=X\circ Y-Y\circ X.\]
%

In general, smooth maps $\phi\colon M\to M'$ of manifolds do not induce maps 
of the Lie algebras of vector fields (unless $\phi$ is a diffeomorphism). 
One makes the following definition. 
%
\begin{definition}
Let $\phi\colon M\to N$ be a smooth map. 
Vector fields $X,Y$ on $M,N$ are called $\phi$-related,  written 
$X\sim_\phi Y$, if 
\[ X(f\circ \phi)=Y(f)\circ \phi\]
for all $f\in C^\infty(M')$. 
\end{definition}
In short, $X\circ \phi^*=\phi^* \circ Y$ where $\phi^*\colon C^\infty(N)\to C^\infty(M),\ f\mapsto f\circ \phi$. 

One has $X\sim_\phi Y$ if and only if $Y_{\phi(m)}
=\d_m\phi (X_m)$. From the definitions, one checks 
\[ X_1\sim_\phi
Y_1,\ X_2\sim_\phi Y_2\  \Rightarrow\  [X_1,X_2]\sim_\phi[Y_1,Y_2].\] 
%
\begin{example}
Let $j\colon S\hra M$ be an embedded submanifold. We say that a vector field $X$ is \emph{tangent to $S$}  if $X_m\in T_mS\subset T_mM$ for all $m\in S$. We claim that if two vector fields are tangent to $S$ then so is their Lie bracket. 
That is, the vector fields on $M$ that are tangent to $S$ form a Lie subalgebra. 

Indeed, the definition means that there exists a vector field $X_S\in\mf{X}(S)$ such that $X_S\sim_j X$. Hence, if $X,Y$ are tangent to $S$, then 
$[X_S,Y_S]\sim_j [X,Y]$, so $[X_S,Y_S]$ is tangent. 

Similarly, the vector fields vanishing on $S$ are a Lie subalgebra. 
\end{example}

Let $X\in\mf{X}(M)$. A curve $\gamma(t),\ t\in J\subset \R$ is called an \emph{integral curve} of $X$ 
if for all $t\in J$, 
%
\[ \dot{\gamma}(t)=X_{\gamma(t)}.\]
%
In local coordinates, this is an ODE $\f{\d x_i}{\d t}=a_i(x(t))$. 
The existence and uniqueness theorem for ODE's (applied in coordinate charts, and then patching the local solutions) 
shows that for any $m\in M$, there is a unique maximal
integral curve $\gamma(t),\ t\in J_m$ with $\gamma(0)=m$. 
\begin{definition} A vector
field $X$ is \emph{complete} if for all $m\in M$, the maximal integral curve with $\gamma(0)=m$ 
is defined for all $t\in\R$. 
\end{definition}
%
In this case, one obtains a \emph{smooth} map 
%
\[ \Phi\colon \R\times M\to M,\ (t,m)\mapsto \Phi_t(m)\]
%
such that $\gamma(t)=\Phi_{-t}(m)$ is the integral curve through $m$. 
The uniqueness property gives 
%
\[ \Phi_0=\on{Id},\ \ \Phi_{t_1+t_2}=\Phi_{t_1}\circ \Phi_{t_2}\]
%
i.e. $t\mapsto \Phi_t$ is a group homomorphism. Conversely, given such a 
group homomorphism such that the map $\Phi$ is smooth, one obtains a vector field $X$ by setting 
%
\[ X=\f{\p}{\p t}|_{t=0} \Phi_{-t}^*,\]
%
as operators on functions. That is, 
$X(f)(m)=\f{\p}{\p t}|_{t=0} f(\Phi_{-t}(m))$. 
\footnote{The minus sign is convention, but it is motivated as follows. Let $\on{Diff}(M)$ be 
the infinite-dimensional group of diffeomorphisms of $M$. It acts on 
$C^\infty(M)$ by $\Phi.f=f\circ \Phi^{-1}=(\Phi^{-1})^* f$. 
Here, the inverse is needed so that $\Phi_1.\Phi_2.f=(\Phi_1\Phi_2).f$. 
We think of vector fields as `infinitesimal flows', i.e. informally 
as the tangent space at $\on{id}$ to $\on{Diff}(M)$. 
Hence, given a curve $t\mapsto \Phi_t$ through $\Phi_0=\on{id}$, 
smooth in the sense that 
the map $\R\times M\to M,\ (t,m)\mapsto \Phi_t(m)$ is smooth, we 
define the corresponding vector field $X=\f{\p}{\p t}|_{t=0}\Phi_t$ in terms of the action on functions: 
as 
\[ X.f=\f{\p}{\p t}|_{t=0} \Phi_t.f=\f{\p}{\p t}|_{t=0} (\Phi_t^{-1})^*f.
\]
If $\Phi_t$ is a flow, we have $\Phi_t^{-1}=\Phi_{-t}$.}


The Lie bracket of vector fields measure the non-commutativity of their flows. In particular, 
if $X,Y$ are complete vector fields, with flows $\Phi_t^X,\ \Phi_s^Y$, then $[X,Y]=0$ if and 
only if 
\[\Phi_t^X\circ \Phi_s^Y =\Phi_s^Y \circ \Phi_t^X.\]
% 
In this case, $X+Y$ is again a complete vector field with flow
$\Phi_t^{X+Y}=\Phi_t^X\circ \Phi_t^Y$. (The right hand side defines a
flow since the flows of $X,Y$ commute, and the corresponding vector
field is identified by taking a derivative at $t=0$.)

\subsection{The Lie algebra of a Lie group}
%
Let $G$ be a Lie group, and $TG$ its tangent bundle.  For all $a\in
G$, the left,right translations
%
\[ L_a\colon G\to G,\ g\mapsto ag\]
\[ R_a\colon G\to G,\ g\mapsto ga\]
%
are smooth maps. Their differentials at $e$ define isomorphisms $\d_g
L_a\colon T_g G\to T_{ag} G$, and similarly for $R_a$. Let
\[ \g=T_eG\]
be the tangent space to the group unit. 

A vector field $X\in \mf{X}(G)$ is called left-invariant if 
\[ X\sim_{L_a} X\]
for all $a\in G$, i.e. if it commutes with $L_a^*$. 
The space $\mf{X}^L(G)$ of left-invariant vector fields is thus a Lie subalgebra of 
$\mf{X}(G)$. Similarly the space of right-invariant vector fields $\mf{X}^R(G)$ is a Lie subalgebra. 

%
\begin{lemma}
The map 
\[ \mf{X}^L(G)\to \g,\ X\mapsto X_e\] 
is an isomorphism of vector 
spaces.  (Similarly for $\mf{X}^R(G)$.) 
\end{lemma}
\begin{proof}
For a left-invariant vector field, $X_a=(\d_eL_a)X_e$, hence the map 
is  injective. To show that it is surjective, let $\xi\in \g$, and 
put $X_a=(\d_eL_a)\xi \in T_a G$. We have to show that the map $G\to TG,\ 
a\mapsto X_a$ is smooth. It is the composition of the map 
$G\to G\times\g,\ g\mapsto (g,\xi)$ (which is obviously smooth) 
with the map $G\times\g\to TG,\ (g,\xi)\mapsto \d_e L_g(\xi)$. 
The latter map is the restriction of $\d\on{Mult}\colon TG\times TG\to TG$ 
to $G\times\g\subset TG\times TG$, and hence is smooth. 
\end{proof}

%
We denote by
$\xi^L\in\mf{X}^L(G),\ \xi^R\in\mf{X}^R(G)$ 
the left,right invariant vector fields defined by
$\xi\in \g$. Thus
\[ \xi^L|_e=\xi^R|_e=\xi\]
%
\begin{definition}
The Lie algebra of a Lie group $G$ is the vector space $\g=T_eG$, equipped with the unique bracket such that 
%
\[ [\xi,\eta]^L=[\xi^L,\eta^L],\ \ \xi\in\g.\]
%
\end{definition}

\begin{remark}
If you use the right-invariant vector fields to define the bracket on $\g$, we get a minus sign.
Indeed, note that $\on{Inv}\colon G\to G$ takes left translations to right translations. 
Thus, $\xi^R$ is $\on{Inv}$-related to some left invariant vector field. Since 
$\d_e\on{Inv}=-\on{Id}$, we see  
$\xi^R\sim_{\on{Inv}}-\xi^L$. Consequently, 
\[ [\xi^R,\eta^R]\sim_{\on{Inv}} [-\xi^L,-\eta^L]=[\xi,\eta]^L.\]
But also $ -[\xi,\eta]^R\sim_{\on{Inv}} [\xi,\eta]^L$, hence we get 
%
\[ [\xi^R,\zeta^R]=-[\xi,\zeta]^R.\]
%
\end{remark}

The construction of a Lie algebra is compatible with morphisms.
That is, we have a \emph{functor} from Lie groups to 
finite-dimensional Lie algebras. 
%
\begin{theorem}\label{th:phirelated}
For any morphism of Lie groups $\phi\colon G\to G'$, the tangent map 
$\d_e\phi\colon \g\to \g'$ is a morphism of Lie algebras. For all 
$\xi\in\g,\ \xi'=\d_e\phi(\xi)$ one has
%
\[ \xi^L\sim_\phi (\xi')^L,\ \xi^R\sim_\phi (\xi')^R.\]
%
\end{theorem}
\begin{proof}
Suppose $\xi\in\g$, and let $\xi'=\d_e\phi(\xi)\in\g'$. 
The property $\phi(ab)=\phi(a)\phi(b)$ shows that 
$L_{\phi(a)}\circ \phi=\phi\circ L_a$. Taking the differential 
at $e$, and applying to $\xi$ we find
$(\d_e L_{\phi(a)})\xi'=(\d_a\phi)(\d_e L_a(\xi))$
hence $(\xi')^L_{\phi(a)}= (\d_a\phi)(\xi^L_a)$. That is
$\xi^L\sim_{\phi}(\xi')^L$. The proof for right-invariant 
vector fields is similar. Since the Lie brackets of two pairs of 
$\phi$-related vector fields are again $\phi$-related, it follows that 
$\d_e\phi$ is a Lie algebra morphism. 
\end{proof}
%

\begin{remark}
Two special cases are worth pointing out.  
\begin{enumerate}
\item Let $V$ be a finite-dimensional (real) vector space. 
A representation of a Lie group $G$ on $V$ is a 
Lie group morphism $G\to \GL(V)$. A representation of a 
Lie algebra $\g$ on $V$ is a Lie algebra morphism $\g\to \mf{gl}(V)$. 
The Theorem shows that the differential of any Lie group representation is a 
representation of its a Lie algebra. 
\item
An \emph{automorphism of a Lie group $G$} is a Lie group morphism
$\phi\colon G\to G$ from $G$ to itself, with $\phi$ a
diffeomorphism. An \emph{automorphism of a Lie algebra} is an
invertible morphism from $\g$ to itself.  By the Theorem, the
differential of any Lie group automorphism is an automorphism of its
Lie algebra. As an example, $\SU(n)$ has a Lie group automorphism 
given by complex conjugation of matrices; its differential 
is a Lie algebra automorphism of $\mf{su}(n)$ given again by complex 
conjugation. 
\end{enumerate}
\end{remark}


\begin{exercise}
Let $\phi\colon G\to G$ be a Lie group automorphism. Show that its 
fixed point set is a closed subgroup of $G$, hence a Lie subgroup. 
Similarly for Lie algebra automorphisms. What is the fixed point set 
for the complex conjugation automorphism of $\SU(n)$?
\end{exercise}

\section{The exponential map}
%
\begin{theorem}
The left-invariant vector fields $\xi^L$ are complete, i.e. they define a flow 
$\Phi^\xi_t$ such that 
\[ \xi^L=\f{\p}{\p t}|_{t=0}(\Phi^\xi_{-t})^*.\]
% 
Letting $\phi^\xi(t)$ denote the unique integral curve with $\phi^\xi(0)=e$. 
It has the property 
\[ \phi^\xi(t_1+t_2)=\phi^\xi(t_1)\phi^\xi(t_2),\]
and the flow of $\xi^L$ is given by right translations:
%
\[ \Phi^\xi_t(g)=g\phi^\xi(-t).\]
%
Similarly, the right-invariant vector fields $\xi^R$ are complete. 
$\phi^\xi(t)$ is an integral curve for $\xi^R$ as well, and
the flow of $\xi^R$ is  given by left translations, $g\mapsto \phi^\xi(-t)g$. 
\end{theorem}
\begin{proof}
If $\gamma(t),\ t\in J\subset \R$ is an integral curve of a
left-invariant vector field $\xi^L$, then its left translates
$a\gamma(t)$ are again integral curves. In particular, for $t_0\in J$
the curve $t\mapsto \gamma(t_0)\gamma(t)$ is again an integral
curve. Hence it coincides with $\gamma(t_0+t)$ for all $t\in J\cap
(J-t_0)$. In this way, an integral curve defined for small $|t|$ can
be extended to an integral curve for all $t$, i.e. $\xi^L$ is complete. 

Since $\xi^L$ is left-invariant, so is its flow $\Phi^\xi_t$. Hence 
\[ \Phi^\xi_t(g)=\Phi^\xi_t \circ L_g(e)=L_g\circ \Phi^\xi_t(e)
=g \Phi^\xi_t(e)=g\phi^\xi(-t).\] The property 
$\Phi^\xi_{t_1+t_2}=\Phi^\xi_{t_1}\Phi^\xi_{t_2}$ shows that 
$\phi^\xi(t_1+t_2)=\phi^\xi(t_1)\phi^\xi(t_2)$. Finally, since 
$\xi^L\sim_{\on{Inv}}-\xi^R$, the image 
\[ \on{Inv}(\phi^\xi(t))=\phi^\xi(t)^{-1}=\phi^\xi(-t)\]
is an integral curve of $-\xi^R$. Equivalently, $\phi^\xi(t)$ is an integral curve of $\xi^R$. 
\end{proof}
%
Since left and right translations commute, it follows in particular that 
\[[\xi^L,\eta^R]=0.\]

\begin{definition}
A 1-parameter subgroup of $G$ is a group homomorphism $\phi\colon \R\to G$.
\end{definition}
%
We have seen that every $\xi\in\g$ defines a 1-parameter group, by taking 
the integral curve through $e$ of the left-invariant vector field $\xi^L$. 
Every 1-parameter group arises in this way: 

\begin{proposition}
If $\phi$ is a 1-parameter subgroup of $G$, then $\phi=\phi^\xi$ where 
$\xi=\dot{\phi}(0)$. One has 
%
\[ \phi^{s\xi}(t)=\phi^\xi(st).\]
%
The map 
\[ \R\times\g\to G,\ (t,\xi)\mapsto \phi^\xi(t)\]
is smooth. 
\end{proposition}
\begin{proof}
Let $\phi(t)$ be a 1-parameter group. 
Then $\Phi_t(g):=g\phi(-t)$
defines a flow. Since this flow commutes with left translations, it is
the flow of a left-invariant vector field, $\xi^L$. Here $\xi$ is
determined by taking the derivative of $\Phi_{-t}(e)=\phi(t)$ at
$t=0$: $\xi=\dot{\phi}(0)$.  This shows $\phi=\phi^\xi$. As an application, 
since $\psi(t)=\phi^{\xi}(st)$ is a 1-parameter group with 
$\dot{\psi}^{\xi}(0)=s\dot{\phi}^\xi(0)=s\xi$, we have 
$\phi^\xi(st)=\phi^{s\xi}(t)$. Smoothness of the map $(t,\xi)\mapsto
\phi^\xi(t)$ follows from the smooth dependence of solutions of ODE's
on parameters.
\end{proof}


%
\begin{definition} 
The \emph{exponential map} for the Lie group $G$ is the smooth map defined by
%
\[ \exp\colon \g\to G,\ \xi\mapsto \phi^\xi(1),\]
% 
where $\phi^\xi(t)$ is the 1-parameter subgroup with $\dot{\phi}^\xi(0)=\xi$. 
\end{definition}
%
\begin{proposition}
We have 
\[ \phi^\xi(t)=\exp(t\xi).\] 
%
If $[\xi,\eta]=0$ then 
%
\[ \exp(\xi+\eta)=\exp(\xi)\exp(\eta).\]
%
\end{proposition}
\begin{proof}
By the previous Proposition, $\phi^\xi(t)=\phi^{t\xi}(1)=\exp(t\xi)$. 
 For the second claim, note that 
$[\xi,\eta]=0$ implies that $\xi^L,\eta^L$ commute. Hence their flows $\Phi^\xi_t,\ \Phi^\eta_t$, and 
$\Phi^\xi_t \circ \Phi^\eta_t$ is the flow of $\xi^L+\eta^L$. Hence it coincides with 
$\Phi^{\xi+\eta}_t$. Applying to $e$, we get $\phi^\xi(t)\phi^\eta(t)=\phi^{\xi+\eta}(t)$. Now put $t=1$. 
\end{proof}

In terms of the exponential map, we may now write the flow of $\xi^L$ as 
$\Phi^\xi_t(g)=g\exp(-t\xi)$, and similarly for the flow of $\xi^R$. 
That is, 
\[ \xi^L=\f{\p}{\p t}|_{t=0}R_{\exp(t\xi)}^*,\ \ 
\xi^R=\f{\p}{\p t}|_{t=0}L_{\exp(t\xi)}^*.\]








\begin{proposition}
The exponential map is functorial with respect to Lie group homomorphisms
$\phi\colon G\to H$. That is, we have a commutative diagram 
\[ \begin{CD} G @>{\phi}>> H \\
@A{\exp}AA @AA{\exp}A \\
\g @>>{\d_e\phi}> \h
\end{CD}\]
%
\end{proposition}
\begin{proof}
$t\mapsto \phi(\exp(t\xi))$ is a 1-parameter subgroup of $H$, with differential at $e$ given by 
\[ \f{d}{d t}\Big|_{t=0} \phi(\exp(t\xi))=\d_e\phi(\xi).\]
Hence $ \phi(\exp(t\xi))=\exp(t \d_e\phi(\xi))$. Now put $t=1$. 
\end{proof}

\begin{proposition}
%
Let $G\subset \GL(n,\R)$ be a matrix Lie group, and $\g\subset \mf{gl}(n,\R)$ its Lie algebra. 
Then $\exp\colon \g\to G$ is just the exponential map for matrices, 
\[ \exp(\xi)=\sum_{n=0}^\infty \f{1}{n!}\xi^n.\]
Furthermore, the Lie bracket on $\g$ is just the commutator of matrices.
\end{proposition}
\begin{proof}
By the previous Proposition, applied to the inclusion of $G$ in
$\GL(n,\R)$, the exponential map for $G$ is just the restriction of
that for $\GL(n,\R)$.  Hence it suffices to prove the claim for
$G=\GL(n,\R)$.  The function $\sum_{n=0}^\infty \f{t^n}{n!}\xi^n$ is a
1-parameter group in $\GL(n,\R)$, with derivative at $0$ equal to
$\xi\in \mf{gl}(n,\R)$.  Hence it coincides with $\exp(t\xi)$. Now put
$t=1$.
\end{proof}

\begin{proposition}
For a matrix Lie group $G\subset \GL(n,\R)$, the Lie bracket on $\g=T_IG$ is just the commutator of matrices.
\end{proposition}
\begin{proof}
It suffices to prove for $G=\GL(n,\R)$. Using 
$\xi^L=\f{\p}{\p t}\Big|_{t=0} R_{\exp(t\xi)}^*$
we have
\[ \begin{split}
\lefteqn{\f{\p}{\p s}\Big|_{s=0}\f{\p}{\p t}\Big|_{t=0} 
(R_{\exp(-t\xi)}^*R_{\exp(-s\eta)}^* 
R_{\exp(t\xi)}^* R_{\exp(s\eta)}^*)}\\
&=\f{\p}{\p s}\Big|_{s=0}(R_{\exp(-s\eta)}^* \xi^L R_{\exp(s\eta)}^*-\xi^L)\\
&= \xi^L\eta^L-\eta^L\xi^L\\
&=[\xi,\eta]^L.
\end{split}\]
On the other hand, write 
\[ R_{\exp(-t\xi)}^*R_{\exp(-s\eta)}^* 
R_{\exp(t\xi)}^* R_{\exp(s\eta)}^*
=R_{\exp(-t\xi)\exp(-s\eta)\exp(t\xi)\exp(s\eta)}^*.\]
Since the Lie group exponential map for $\GL(n,\R)$ coincides with the 
exponential map for matrices, we may use Taylor's expansion, 
\[\exp(-t\xi)\exp(-s\eta)\exp(t\xi)\exp(s\eta)=I+st(\xi\eta-\eta\xi)+\ldots
=\exp(st(\xi\eta-\eta\xi))+\ldots
\]
where $\ldots$ denotes terms that are cubic or higher in $s,t$. 
Hence 
\[ R_{\exp(-t\xi)\exp(-s\eta)\exp(t\xi)\exp(s\eta)}^*=R_{\exp(st (\xi\eta-\eta\xi)}^*+\ldots\]
and consequently
\[ \f{\p}{\p s}\Big|_{s=0}\f{\p}{\p t}\Big|_{t=0} R_{\exp(-t\xi)\exp(-s\eta)\exp(t\xi)\exp(s\eta)}^*
=\f{\p}{\p s}\Big|_{s=0}\f{\p}{\p t}\Big|_{t=0} R_{\exp(st (\xi\eta-\eta\xi))}^*
=(\xi\eta-\eta\xi)^L.\]
We conclude that $[\xi,\eta]=\xi\eta-\eta\xi$. 
\end{proof}

\begin{remark}
Had we defined the Lie algebra using right-invariant vector fields, we would have 
obtained \emph{minus} the commutator of matrices. Nonetheless, some authors use that convention. 
\end{remark}

The exponential map gives local coordinates for the group $G$ on a
neighborhood of $e$:
%
\begin{proposition}
The differential of the exponential map at the origin is $\d_0\exp=\on{id}$. As a consequence, 
there is an open neighborhood $U$ of $0\in\g$ such that the exponential map restricts to a diffeomorphism 
$U\to \exp(U)$. 
\end{proposition}
\begin{proof}
Let $\gamma(t)=t\xi$. Then $\dot{\gamma}(0)=\xi$ 
since $\exp(\gamma(t))=\exp(t\xi)$ is the 1-parameter group, we have 
\[ (\d_0\exp)(\xi)=\f{\p}{\p t}|_{t=0} \exp(t\xi)=\xi.\] 
\end{proof}
%
%
\begin{exercise}
Show hat the exponential map for $\SU(n),\ \SO(n)\  \on{U}(n)$ are surjective. 
(We will soon see that the exponential map for any compact, connected Lie group is 
surjective.)
\end{exercise}
\begin{exercise}
A matrix Lie group $G\subset \GL(n,\R)$ is called \emph{unipotent} if for all $A\in G$, 
the matrix $A-I$ is nilpotent (i.e. $(A-I)^r=0$ for some $r$). The prototype of such a group are 
the upper triangular matrices with $1$'s down the diagonal. Show that for a connected 
unipotent matrix Lie group, the exponential map is a diffeomorphism. 
\end{exercise}
\begin{exercise} Show that $\exp\colon \mf{gl}(2,\C)\to \GL(2,\C)$ is 
surjective. More generally, show that the exponential map for 
$\GL(n,\C)$ is surjective. (Hint: First conjugate the given matrix into 
Jordan normal form). 
\end{exercise}
\begin{exercise}
Show that $\exp\colon \mf{sl}(2,\R)\to \SL(2,\R)$ is not surjective, by proving that 
the matrices
$\left(\begin{array}{cc} -1&\pm 1\\0&-1\end{array}\right)\in \SL(2,\R)$ are not in the image. 
(Hint: Assuming these matrices are of the form $\exp(B)$, what would the 
eigenvalues of $B$ have to be?) Show that these two matrices represent 
\emph{all} conjugacy classes of elements that are not in the image of $\exp$. 
 (Hint: Find a classification of the conjugacy classes of $\SL(2,\R)$, e.g. in terms of eigenvalues.) 
%Consider the exponential map $\exp\colon \mf{sl}(2,\R)\to \SL(2,\R)$ for the 
%group $\SL(2,\R)$. Let $\mf{sl}(2,\R)_0,\ \mf{sl}(2,\R)_{\on{re}},\  \mf{sl}(2,\R)_{\on{im}}$ 
%denote the subset of $\mf{sl}(2,\R)$ consisting of matrices with zero, non-zero real, or 
%non-zero imaginary eigenvalues.  
%Prove: 
%\begin{itemize}
%\item[(i)] Any $B\in \mf{sl}(2,\R)$ is in one of these three sets. 
%\item[(ii)] $B\in \mf{sl}(2,\R)_{\on{re}} \Rightarrow \on{tr}(\exp(B))>2$. 
%\item[(iii)] $B\in \mf{sl}(2,\R)_{0} \Rightarrow \on{tr}(\exp(B))=2$. 
%\item[(iv)] $B\in \mf{sl}(2,\R)_{\on{im}} \Rightarrow -2\le\on{tr}(\exp(B))\le 2$. 
%\item[(v)] The matrix $A=\left(\begin{array}{cc} -1&1\\0&-1\end{array}\right)\in \SL(2,\R)$ does not lie in the image of %the exponential map for $\SL(2,\R)$. 
%\end{itemize}
%Encore: Can you find all conjugacy classes of elements that are not in the image of $\exp$? Try to visualize the 
%set of conjugacy classes of $\SL(2,\R)$ (is it a Hausdorff space?). 
\end{exercise}










\section{Cartan's theorem on closed subgroups}
%
Using the exponential map, we are now in position to prove Cartan's theorem on closed subgroups. 
%
\begin{theorem}
Let $H$ be a closed subgroup of a Lie group $G$. Then $H$ is an embedded submanifold, and hence is a Lie subgroup.  
\end{theorem}
We first need a Lemma. 
Let $V$ be a Euclidean vector space, and $S(V)$ its unit sphere. For $v\in V\backslash\{0\}$, let 
$[v]=\f{v}{||v||}\in S(V)$.   

\begin{lemma}
Let $v_n,v\in V\backslash\{0\}$ with $\lim_{n\to \infty}v_n=0$. Then 
\[ \lim_{n\to \infty} [v_n]=[v] \Leftrightarrow \exists a_n\in\N\colon \lim_{n\to \infty} a_nv_n=v.\]
\end{lemma}
\begin{proof} The implication $\Leftarrow$ is obvious. For the opposite direction, 
suppose $\lim_{n\to \infty} [v_n]=[v] $. 
Let $a_n\in\N$ be defined by $a_n-1<\f{||v||}{||v_n||}\le a_n$. Since $v_n\to 0$, we have 
$\lim_{n\to \infty} a_n  \f{||v_n||}{||v||}=1$, and  
\[ a_n v_n =\left(a_n  \f{||v_n||}{||v||}\right) [v_n]\, ||v||\to [v]\,||v||=v.\qedhere\]
%   
\end{proof}

\begin{proof}[Proof of E. Cartan's theorem]
It suffices to construct a submanifold chart near $e\in H$. (By left 
translation, one then obtains submanifold charts near arbitrary $a\in H$.) 
Choose an inner product on $\g$. 

We begin with a candidate for the Lie algebra of $H$. Let $W\subset
\g$ be the subset such that $\xi\in W$ if and only if either $\xi=0$, or
$\xi\not=0$ and there exists $\xi_n\not=0$ with
%
\[ \exp(\xi_n)\in H,\ \ \xi_n\to 0,\ \ [\xi_n]\to [\xi].\] 
%
We will now show the following: 
\begin{enumerate}
\item[(i)]  $\exp(W)\subset H$,
\item[(ii)] $W$ is a subspace of $\g$,
\item[(iii)] There is an open neighborhood $U$ of $0$ and a diffeomorphism 
$\phi\colon U\to \phi(U)\subset G$ with $\phi(0)=e$ such that 
\[ \phi(U\cap W)=\phi(U)\cap H.\]
(Thus $\phi$ defines a submanifold chart near $e$.) 
\end{enumerate}
Step (i). Let  $\xi\in W\backslash\{0\}$, with  sequence $\xi_n$ as in 
the definition of $W$. By the Lemma, there are $a_n\in\N$ with 
$a_n\xi_n\to \xi$. Since $\exp(a_n\xi_n)=\exp(\xi_n)^{a_n}\in H$, and $H$ is closed, it follows that  
\[ \exp(\xi)=\lim_{n\to \infty} \exp(a_n\xi_n)\in H.\] 

Step (ii). Since the subset $W$ is invariant under scalar
multiplication, we just have to show that it is closed under
addition. Suppose $\xi,\eta\in W$. To show that $\xi+\eta\in W$, we
may assume that $\xi,\eta,\xi+\eta$ are all non-zero.  For $t$
sufficiently small, we have
\[ \exp(t \xi)\exp(t \eta)=\exp(u(t))\] 
for some smooth curve 
$t\mapsto u(t)\in\g$ with $u(0)=0$.  Then $\exp(u(t))\in H$ and 
\[ \lim_{n\to \infty} n\, u(\f{1}{n})=
\lim_{h\to 0} \f{u(h)}{h}=\dot{u}(0)=\xi+\eta.\]
hence $u(\f{1}{n})\to 0,\ \exp(u(\f{1}{n})\in H,\  
[u(\f{1}{n})]\to [\xi+\eta]$. This shows $[\xi+\eta]\in W$, 
proving (ii). 

Step (iii). Let $W'$ be a complement to $W$ in $\g$, and define 
\[ \phi\colon \g\cong W\oplus W'\to G,\ \ \phi(\xi+\xi')=\exp(\xi)
\exp(\xi').\]
%
Since $\d_0\phi$ is the identity, there is an open neighborhood
$U\subset\g$ of $0$ such that $\phi\colon U\to \phi(U)$ is a
diffeomorphism.  It is automatic that $\phi(W\cap U)\subset
\phi(W)\cap \phi(U)\subset H\cap \phi(U)$. We want to show that we can
take $U$ sufficiently small so that we also have the opposite
inclusion
% 
\[  H\cap \phi(U)\subset \phi(W\cap U).\]
%
Suppose not. Then, any neighborhood $U_n\subset \g=W\oplus W'$
of $0$ contains an element $(\eta_n,\eta_n')$ such that 
\[ \phi(\eta_n,\eta_n')=\exp(\eta_n)\exp(\eta_n')\in H\] 
(i.e. $\exp(\eta_n')\in H$) but $(\eta_n,\eta_n')\not\in W$ (i.e. 
$\eta_n'\not=0$). Thus, taking $U_n$ to be a nested sequence
of neighborhoods with intersection $\{0\}$,  we could construct a sequence 
$\eta_n'\in W'-\{0\}$ with $\eta_n'\to 0$ 
and $\exp(\eta_n')\in H$. Passing to a subsequence we may assume that 
$[\eta_n']\to [\eta]$ for some $\eta\in W'\backslash \{0\}$. On the 
other hand, such a convergence would mean $\eta\in W$, by definition of $W$. 
Contradiction. 
\end{proof}


As remarked earlier, Cartan's theorem is very useful in practice. 
For a given Lie group $G$, the term `closed subgroup' is often used 
as synonymous to `embedded Lie subgroup'. 
 
\begin{examples}
\begin{enumerate}
\item The matrix groups $G=\on{O}(n), \on{Sp}(n), \SL(n,\R),\ldots$ are 
all closed subgroups of some $\GL(N,\R)$, and hence are Lie groups. 
\item 
Suppose that $\phi\colon G\to H$ is a morphism of Lie groups. 
Then $\ker(\phi)=\phi^{-1}(e)\subset G$ is a closed subgroup. Hence 
it is an embedded Lie subgroup of $G$. 
\item 
The center $Z(G)$ of a Lie group $G$ is the set of all $a\in G$ such
that $ag=ga$ for all $a\in G$. It is a closed subgroup, and hence an
embedded Lie subgroup.
\item 
Suppose $H\subset G$ is a closed subgroup. Its \emph{normalizer} 
$N_G(H)\subset G$  is the set of all $a\in G$ such that $aH=Ha$. 
(I.e. $h\in H$ implies $aha^{-1}\in H$.) This is a closed subgroup, hence 
a Lie subgroup. The \emph{centralizer} $Z_G(H)$ is the set of all 
$a\in G$ such that $ah=ha$ for all $h\in H$, it too is a closed 
subgroup, hence a Lie subgroup. 
\end{enumerate}
\end{examples}
The E. Cartan theorem is just one of many `automatic smoothness' results in Lie theory. 
Here is another. 
\begin{theorem}
Let $G,H$ be Lie groups, and $\phi\colon G\to H$ be a continuous group morphism. Then 
$\phi$ is smooth.  
\end{theorem}
As a corollary, a given topological group carries at most one smooth structure for which 
it is a Lie group. For profs of these (and stronger) statements, see the book by 
Duistermaat-Kolk. 
%\begin{proof}
%Let $\on{gr}(\phi)\subset G\times H$ be the graph of $\phi$. It is a closed subset, since $\phi$ is 
%continuous.  It is also a subgroup of $G\times H$. Hence it is an embedded submanifold. The projection 
%$G\times H\to H$ restricts to a smooth map $\alpha\colon \on{gr}(\phi)\to H$. On the other hand, projection to the first 
%factor gives a smooth bijection $\beta\colon \on{gr}(\phi)\to G$. 
% with invertible differential.\footnote{The kernel of the differential of the projection map is the intersection of $G\times TH\subset T(G\times H)$ with $T\on{gr}(\phi)$.  
%Thus, we have to show that the graph does not contain vectors of the form $(0,w)$. 
%Hence $\beta$ is a diffeomorphism and, $\alpha\circ \beta^{-1}$ is smooth. 
%\end{proof}


\section{The adjoint representations}
%
\subsection{Automorphisms}

The group $\on{Aut}(\g)$ of automorphisms of a Lie algebra $\g$ is closed in the
group $\End(\g)^\times$ of vector space automorphisms, hence it is a
Lie group. To identify its Lie algebra, let $D\in \End(\g)$ be such that $\exp(tD)\in \Aut(\g)$ for 
$t\in\R$. Taking the derivative of the defining condition 
$\exp(tD)[\xi,\eta]=[\exp(tD)\xi,\exp(tD)\eta]$, we obtain the property 
\[ D[\xi,\eta]=[D\xi,\eta]+[\xi,D\eta]\]
saying that $D$ is a \emph{derivation} of the Lie algebra. Conversely, if 
$D$ is a derivation then 
\[ D^n[\xi,\eta]=\sum_{k=0}^n \binom{n}{k} [D^k \xi,\,D^{n-k}\eta]\]
by induction, which then shows that $\exp(D)=\sum_n \f{D^n}{n!}$ is an automorphism. 
Hence the Lie algebra of $\on{Aut}(\g)$ is the Lie algebra $\on{Der}(\g)$ of derivations 
of $\g$. 
\begin{exercise} 
Using similar arguments, verify that the Lie algebra of $\SO(n),\SU(n),\on{Sp}(n),\ldots$
are $\mf{so}(n),\ \mf{su}(n),\ \mf{sp}(n),\ldots$.  
\end{exercise}

\subsection{The adjoint representation of $G$}
%
Recall that an automorphism of a Lie group $G$ is an invertible morphism 
from $G$ to itself. The automorphisms form a group $\on{Aut}(G)$. 
Any $a\in G$ defines an `inner' automorphism $\Ad_a\in \on{Aut}(G)$ 
by conjugation: 
%
\[ \Ad_a(g)=aga^{-1}\]
%
Indeed, $\Ad_a$ is an automorphism since $\Ad_a^{-1}=\Ad_{a^{-1}}$ and 
% 
\[ \Ad_a(g_1g_2)= ag_1g_2 a^{-1}=ag_1 a^{-1}ag_2
a^{-1}=\Ad_a(g_1)\Ad_a(g_2).\] 
%
Note also that $\Ad_{a_1 a_2}=\Ad_{a_1}\Ad_{a_2}$, thus we have a 
group morphism 
\[ \Ad\colon G\to \on{Aut}(G)\] 
%
into the group of automorphisms. The kernel of this morphism is the center $Z(G)$, 
the image is (by definition) the subgroup $\on{Int}(G)$ of inner 
automorphisms. Note that for any $\phi\in \on{Aut}(G)$, and any $a\in G$, 
\[ \phi\circ \Ad_a\circ \phi^{-1}=\Ad_{\phi(a)}.\]
That is, $\on{Int}(G)$ is a \emph{normal} subgroup of $\on{Aut}(G)$. 
(I.e. the conjugate of an inner automorphism by any automorphism is inner.) 
It follows that $\on{Out}(G)=\on{Aut}(G)/\on{Int}(G)$ inherits a group 
structure; it is called the \emph{outer automorphism group}.  
\begin{example}
If $G=\SU(2)$ the complex conjugation of matrices is an inner automorphism, but for $G=\SU(n)$ with $n\ge 3$ it 
cannot be inner (since an inner automorphism has to preserve the 
spectrum of a matrix). Indeed, one know that $\on{Out}(\SU(n))=\Z_2$ for $n\ge 3$. 
\end{example}

The differential of the automorphism $\Ad_a\colon G\to G$ is a Lie algebra automorphism, 
denoted by the same letter: $\Ad_a=\d_e\Ad_a\colon\g\to \g$. 
The resulting map 
\[ \Ad\colon G\to \on{Aut}(\g)\] 
is called the \emph{adjoint representation of $G$}.
%
%By Theorem \ref{th:phirelated} we have 
%$ \xi^L\sim_{\Ad_a} (\Ad_a\xi)^L,\ \ \xi^R\sim_{\Ad_a} (\Ad_a\xi)^R$. 
Since the $\Ad_a$ are Lie algebra/group morphisms, they are 
compatible with the exponential map, 
%
\[ \exp(\Ad_a\xi)=\Ad_a\exp(\xi).\]
%
 
\begin{remark}
If $G\subset \GL(n,\R)$ is a matrix Lie group, then 
$\Ad_a\in \on{Aut}(\g)$ is the conjugation of matrices
%
\[ \Ad_a(\xi)= a\xi a^{-1}.\] 
%
This follows by taking the derivative of $\Ad_a(\exp(t\xi))=a\exp(t\xi)a^{-1}$, using that $\exp$ is just the exponential series for matrices. 
\end{remark}



\subsection{The adjoint representation of $\g$}
%
Let $\on{Der}(\g)$ be the Lie algebra of derivations of the Lie algebra $\g$. 
There is a Lie algebra morphism, 
\[ \ad\colon \g\to \on{Der}(\g),\ \ \xi\mapsto [\xi,\cdot].\]
%
The fact that $\ad_\xi$ is a derivation follows from the Jacobi identity; 
the fact that $\xi\mapsto \ad_\xi$ it is a Lie algebra morphism is again the 
Jacobi identity. 
The kernel of $\ad$ is the center of the Lie algebra $\g$, i.e. elements 
having zero bracket with all elements of $\g$, while the image is the 
Lie subalgebra $\on{Int}(\g)\subset \on{Der}(\g)$ of \emph{inner} derivations. 
It is a normal Lie subalgebra, i.e $[\on{Der}(\g),\on{Int}(\g)]\subset \on{Int}(\g)$, and the quotient Lie algebra $\on{Out}(\g)$ are the 
\emph{outer automorphims}. 



Suppose now that $G$ is a Lie group, with Lie algebra $\g$.  
We have remarked above that the Lie algebra of $\on{Aut}(\g)$ is $\on{Der}(\g)$. 
Recall that the differential of any $G$-representation is a 
$\g$-representation. In particular, we can consider the differential 
of $G\to \on{Aut}(\g)$. 
%
\begin{theorem}
If $\g$ is the Lie algebra of $G$, then the adjoint representation
$\ad\colon \g\to \on{Der}(\g)$ is the differential of the adjoint 
representation $\Ad\colon G\to \on{Aut}(\g)$. 
One has the equality of operators 
\[ \exp(\ad_\xi)=\Ad(\exp\xi)\]
for all $\xi\in\g$. 
\end{theorem}
\begin{proof}
For the first part we have to show 
$\f{\p}{\p t}\big|_{t=0}\Ad_{\exp(t\xi)}\eta=\ad_\xi\eta$. 
This is easy if $G$ is a matrix Lie group: 
\[ \f{\p}{\p t}\Big|_{t=0}\Ad_{\exp(t\xi)}\eta=\f{\p}{\p t}\Big|_{t=0}\exp(t\xi)\eta\exp(-t\xi)=\xi\eta-\eta\xi=[\xi,\eta].\]
For general Lie groups we compute, using 
\[\exp(s\Ad_{\exp(t\xi)}\eta)=\Ad_{\exp(t\xi)}\exp(s\eta)
=\exp(t\xi)\exp(s\eta)\exp(-t\xi),\]
\[ \begin{split}
\f{\p}{\p t}\Big|_{t=0} (\Ad_{\exp(t\xi)}\eta)^L
&=\f{\p}{\p t}\Big|_{t=0} \f{\p}{\p s}\Big|_{s=0}
R_{\exp(s\Ad_{\exp(t\xi)}\eta)}^*
\\
&=\f{\p}{\p t}\Big|_{t=0} \f{\p}{\p s}\Big|_{s=0}
R_{\exp(t\xi)\exp(s\eta)\exp(-t\xi)}^*\\
&=\f{\p}{\p t}\Big|_{t=0} \f{\p}{\p s}\Big|_{s=0}
R_{\exp(t\xi)}^*R_{\exp(s\eta)}^*R_{\exp(-t\xi)}^*\\
&=\f{\p}{\p t}\Big|_{t=0} R_{\exp(t\xi)}^*\ \eta^L\ R_{\exp(-t\xi)}^*\\
&=[\xi^L,\eta^L]\\
&=[\xi,\eta]^L=(\ad_\xi\eta)^L. 
\end{split}\] 
This proves the first part. The second part is the commutativity of the diagram
\[ \begin{CD} G @>{\on{Ad}}>> \on{Aut}(\g) \\
@A{\exp}AA @AA{\exp}A \\
\g @>>{\on{ad}}> \on{Der}(\g)
\end{CD}\]
%
which is just a special case of the functoriality property of $\exp$
with respect to Lie group morphisms.   
\end{proof}

\begin{remark}
As a special case, this formula holds for matrices. That is, for 
$B,C\in \on{Mat}_n(\R)$, 
\[ e^B\, C\, e^{-B}=\sum_{n=0}^\infty \f{1}{n!} [B,[B,\cdots [B,C]\cdots]].\]
The formula also holds in some other contexts, e.g. if $B,C$ are elements 
of an algebra with $B$ nilpotent (i.e. $B^N=0$ for some $N$). In this case, both the exponential series for 
$e^B$ and the series on the right hand side are finite. (Indeed, $[B,[B,\cdots [B,C]\cdots]]$ with $n$ $B$'s is a 
sum of terms $B^j C B^{n-j}$, and hence must vanish if $n\ge 2N$.)  
\end{remark}





\section{The differential of the exponential map}
%
We had seen that $\d_0\exp=\on{id}$. More generally, one can derive a formula for 
the differential of the exponential map at arbitrary points $\xi\in\g$, 
\[ \d_\xi\exp\colon \g=T_\xi\g\to T_{\exp\xi}G.\] 
%
Using left translation, we can move $T_{\exp\xi}G$ back to $\g$, and obtain an endomorphism 
of $\g$. 
%
\begin{theorem}
The differential of the exponential map $\exp\colon \g\to G$ at
$\xi\in \g$ is the linear operator $\d_\xi\exp\colon\g\to T_{\exp(\xi)}\g$ given
by the formula,
%
\[  \d_\xi\exp=(\d_e L_{\exp\xi})\circ \f{1-\exp(-\ad_\xi)}{\ad_\xi}.\]
%
\end{theorem}
%
Here the operator on the right hand side is defined to be the result of substituting $\ad_\xi$ in the 
entire holomorphic function $\f{1-e^{-z}}{z}$. Equivalently, it may be written as an integral
\[ \f{1-\exp(-\ad_\xi)}{\ad_\xi}=\int_0^1 \d s\  \exp(-s\ad_\xi).\]
%
\begin{proof}
We have to show that for all $\xi,\eta\in\g$, 
%
\[ (\d_\xi\exp)(\eta)\circ L_{\exp(-\xi)}^*=\int_0^1 \d s\ (\exp(-s\ad_\xi)\eta)\]
%
as operators on functions  $f\in C^\infty(G)$.
To compute the left had side, write
%
\[ (\d_\xi\exp)(\eta)\circ L_{\exp(-\xi)}^*(f)=\f{\p}{\p t}\Big|_{t=0} 
(L_{\exp(-\xi)}^*(f))(\exp(\xi+t\eta))=\f{\p}{\p t}\Big|_{t=0} f(\exp(-\xi)\exp(\xi+t\eta)).\]
%
We think of this as the value of $\f{\p}{\p t}\Big|_{t=0} R_{\exp(-\xi)}^* R_{\exp(\xi+t\eta)}^* f$ 
at $e$, and compute as follows: \footnote{We will use the identities 
$\f{\p}{\p s} R_{\exp(s\zeta)}^*= R_{\exp(s\zeta)}^*\ \zeta^L=\zeta^L\ R_{\exp(s\zeta)}^*$
for all $\zeta\in\g$. 
Proof: $\f{\p}{\p s} R_{\exp(s\zeta)}^*=\f{\p}{\p u}|_{u=0} 
R_{\exp((s+u)\zeta)}^*=\f{\p}{\p u}|_{u=0}R_{\exp(u\zeta)}^* R_{\exp(s\zeta)}^*=
\zeta^L R_{\exp(s\zeta)}^*$.}
% 
\[ \begin{split}
\f{\p}{\p t}\Big|_{t=0} R_{\exp(-\xi)}^* R_{\exp(\xi+t\eta)}^* 
&=\int_0^1 \d s\ \f{\p}{\p t}\Big|_{t=0} \f{\p}{\p s} R_{\exp(-s\xi)}^*R_{\exp(s(\xi+t\eta)}^* \\
&=\int_0^1 \d s\ \f{\p}{\p t}\Big|_{t=0} R_{\exp(-s\xi)}^*  (t\eta)^L R_{\exp(s(\xi+t\eta)}^*\\
&=\int_0^1 \d s\ R_{\exp(-s\xi)}^*\ \eta^L\  R_{\exp(s(\xi))}^*\\
&=\int_0^1 \d s\ (\Ad_{\exp(-s\xi)}\eta)^L\\
&=\int_0^1 \d s\ (\exp(-s\ad_\xi)\eta)^L.
\end{split}\]
Applying this result to $f$ at $e$, we obtain 
$\int_0^1 \d s\ (\exp(-s\ad_\xi)\eta)(f)$ as desired.
\end{proof}
%
\begin{corollary}
The exponential map is a local diffeomorphism near $\xi\in\g$ if and only if
$\ad_\xi$ has no eigenvalue in the set $2 \pi i \Z\backslash\{0\}$.
\end{corollary}
\begin{proof}
$\d_\xi\exp$ is an isomorphism if and only if $\f{1-\exp(-\ad_\xi)}{\ad_\xi}$ is invertible, i.e. has 
non-zero determinant. The determinant is given in terms of the eigenvalues of $\ad_\xi$ as a product, 
$\prod_{\lambda} \f{1-e^{-\lambda}}{\lambda}$. This vanishes if and only if there is a non-zero eigenvalue 
$\lambda$ with $e^\lambda=1$. 
\end{proof}

As an application, one obtains a version of the 
\emph{Baker-Campbell-Hausdorff formula}. 
%
Let $g\mapsto \log(g)$ be the inverse function to $\exp$, defined for $g$ close to $e$. 
For $\xi,\eta\in\g$ close to $0$, the function 
\[ \log(\exp(\xi)\exp(\eta))\] 
The BCH formula gives the Taylor series expansion of this function. 
The series starts out with 
\[ \log(\exp(\xi)\exp(\eta))=\xi+\eta+\hh [\xi,\eta]+\cdots\]
but gets rather complicated.  To derive the formula, introduce a $t$-dependence, and let 
$f(t,\xi,\eta)$ be defined by 
$\exp(\xi)\exp(t\eta)=\exp(f(t,\xi,\eta))$ (for $\xi,\eta$ sufficiently small). Thus 
\[ \exp(f)=\exp(\xi)\exp(t\eta)\]
We have, on the one hand, 
\[ (\d_eL_{\exp(f)})^{-1}\f{\p}{\p t}\exp(f)=
\d_e L_{\exp(t\eta)}^{-1}\f{\p}{\p t}\exp(t\eta)
=\eta.\]
On the other hand, by the formula for the differential of $\exp$, 
\[ (\d_eL_{\exp(f)})^{-1}\f{\p}{\p t}\exp(f)
=(\d_eL_{\exp(f)})^{-1}(\d_f\exp) (\f{\p f}{\p t})
= \f{1-e^{-\ad_f}}{\ad_f}(\f{\p f}{\p t}).\]
Hence 
%
\[ \f{\d f}{\d t}=\f{\ad_f}{1-e^{-\ad_f}}\eta.\]
%
Letting $\chi$ be the function, holomorphic near $w=1$,
%
\[ \chi(w)=\f{\log(w)}{1-w^{-1}}=1+\sum_{n=1}^\infty \f{(-1)^{n+1} }{n(n+1)}(w-1)^n,\] 
we may write the right hand side as $\chi(e^{\ad_f})\eta$. 
By Applying $\Ad$ 
to the defining equation for $f$ 
we obtain $e^{\ad_f}=e^{\ad_\xi}e^{t\ad_\eta}$. 
Hence 
%
\[ \f{\d f}{\d t}=\chi(e^{\ad_\xi}e^{t\ad_\eta}) \eta.\]
%
Finally, integrating from $0$ to $1$ and using $f(0)=\xi,\ f(1)=\log(\exp(\xi)\exp(\eta))$, we find:
%
\[ \log(\exp(\xi)\exp(\eta))=\xi+\Big(\int_0^1 \chi(e^{\ad_\xi}e^{t\ad_\eta})\d t\Big) \eta.\]
%
To work out the terms of the series, one puts 
\[ w-1=e^{\ad_\xi} e^{t\ad_\eta}-1=\sum_{i+j\ge 1} \f{t^j}{i! j!}\ad_\xi^i \ad_\eta^j \]
in the power series expansion of $\chi$, and integrates the resulting series in $t$. We arrive at: 

\begin{theorem}[Baker-Campbell-Hausdorff series] Let $G$ be a Lie group, with exponential map $\exp\colon\g\to G$. For $\xi,\eta\in\g$ sufficiently small we have the following formula
\[  \log(\exp(\xi)\exp(\eta))=\xi+\eta+\sum_{n=1}^\infty \f{(-1)^{n+1} }{n(n+1)}\Big(\int_0^1\d t\ \Big(\sum_{i+j\ge 1} \f{t^j}{i! j!}\ad_\xi^i \ad_\eta^j \Big)^n\Big)\eta.\]
\end{theorem}
%
An important point is that the resulting Taylor series in $\xi,\eta$ 
is a \emph{Lie series}: all terms of the series are of the form of a constant times $\ad_\xi^{n_1}\ad_\eta^{m_2}\cdots \ad_\xi^{n_r}\eta$. 
The first few terms read, 
\[  \log(\exp(\xi)\exp(\eta))=\xi+\eta+\hh [\xi,\eta]+\f{1}{12}[\xi,[\xi,\eta]]-\f{1}{12}[\eta,[\xi,\eta]]+
\f{1}{24} [\eta,[\xi,[\eta,\xi]]]+
\ldots.\]
%
\begin{exercise}
Work out these terms from the formula. 
\end{exercise}
%
There is a somewhat better version of the BCH formula, due to Dynkin.
A good discussion can be found in the book by Onishchik-Vinberg, Chapter I.3.2.  


\section{Actions of Lie groups and Lie algebras}
\subsection{Lie group actions}
%
\begin{definition}
An {\em action of a Lie group $G$ on a manifold $M$} is a group 
homomorphism 
%
$$\A\colon G\to \on{Diff}(M),\ g\mapsto \ca{A}_g$$ 
%
into the group of diffeomorphisms
on $M$, such that the {\em action map} 
%
$$  G\times M\to M,\ \ (g,m)\mapsto \A_g(m)$$
%
is smooth.
\end{definition}
%
We will often write $g.m$ rather than $\A_g(m)$. With this notation, 
$g_1.(g_2.m)=(g_1g_2).m$ and $e.m=m$. A map $\Phi\colon M_1\to M_2$ between $G$-manifolds is called 
$G$-equivariant if $g.\Phi(m)=\Phi(g.m)$ for all $m\in M$, i.e. the following diagram commutes:
\[ \begin{CD}
G\times M_1 @>>> M_1\\
@VV{\on{id}\times \Phi}V @VV{\Phi}V\\
G\times M_2 @>>> M_2
\end{CD}\]
where the horizontal maps are the action maps. 

\begin{examples}
\begin{enumerate}
\item 
An $\R$-action on $M$ is the same thing as a global flow. 
\item 
The group $G$ acts $M=G$ by right multiplication, $\A_g=R_{g^{-1}}$, left multiplication, 
$\A_g=L_g$, and by conjugation, $\A_g=\Ad_g=L_g\circ R_{g^{-1}}$. The left and right action commute, hence 
they define an action of $G\times G$. The conjugation action can be regarded as the action of the diagonal 
subgroup $G\subset G\times G$. 
\item 
Any $G$-representation $G\to \on{End}(V)$ can be regarded as a $G$-action, by viewing $M$ as a manifold.  
\item 
For any closed subgroup $H\subset G$, the space of right cosets $G/H=\{gH|\ g\in G\}$ has a unique manifold structure 
such that the quotient map $G\to G/H$  is a smooth submersion, and the action of $G$ by left multiplication on $G$ 
descends to a smooth $G$-action on $G/H$. (Some ideas of teh proof will be explained below.)  
\item 
The defining represenation of the orthogonal group $\on{O}(n)$ on $\R^n$ restricts to an action on the unit 
sphere $S^{n-1}$, which in turn descends to an action on the projective space $\R P(n-1)$. One also has 
actions on the Grassmann manifold $\on{Gr}_\R(k,n)$ of $k$-planes in $\R^n$, or on the flag manifold
$\on{Fl}(n)$ (consisting of sequences $\{0\}=V_0\subset V_1\subset \cdots V_{n-1}\subset V_n=\R^n$ with 
$\dim V_i=i$). These examples are all of the form  $\on{O}(n)/H$ for various choices of $H$. (E.g, 
for $\on{Gr}(k,n)$ one takes $H$ to be the subgroup preserving $\R^k\subset \R^n$.) 
\end{enumerate}
\end{examples}

\subsection{Lie algebra actions}

\begin{definition}
An {\em action of a finite-dimensional Lie algebra $\g$} on $M$ is a 
Lie algebra homomorphism $\g\to \mf{X}(M),\ \xi\mapsto \A_\xi$ 
such that the action map
%
$$ \g\times M\to TM,\ \ (\xi,m)\mapsto \A_\xi|_m$$ 
%
is smooth. 
 
\end{definition}
We will often write $\xi_M=:\A_\xi$ for the vector field corresponding to $\xi$. Thus, 
$[\xi_M,\eta_M]=[\xi,\eta]_M$ for all $\xi,\eta\in\g$.  A smooth map 
$\Phi\colon M_1\to M_2$ between to $\g$-manifolds is called equivariant if 
$\xi_{M_1}\sim_\Phi \xi_{M_2}$ for all $\xi\in\g$. 

\begin{examples}
\begin{enumerate}
\item 
Any vector field $X$ defines an action of the Abelian Lie algebra $\R$, by $\lambda\mapsto \lambda X$. 
\item 
Any Lie algebra representation $\phi\colon \g\to \on{gl}(V)$ may be viewed as a
Lie algebra action. Indeed, if $f\in C^\infty(V)$ we have $\d_vf\in V^*$, and 
\[ (\A_\xi f)(v)=\f{\d}{\d t}|_{t=0} f(v-t\xi.v)\]
defines a $\g$-action. (Exercise)
\item 
For any Lie group $G$, we have actions of its Lie algebra $\g$ by $\A_\xi=\xi^L,\ \A_\xi=-\xi^R$ and 
$\A_\xi=\xi^L-\xi^R$.
\item 
Given a closed subgroup $H\subset G$, the vector fields $-\xi^R\in\mf{X}(G),\ \xi\in\g$ are 
invariant under the right multiplication, hence they are related under the quotient map 
to vector fields on $G/H$. That is, there is a unique $\g$-action on $G/H$ such that 
the quotient map $G\to G/H$ is equivariant.  
\end{enumerate} 
\end{examples}

\begin{definition}
Let $G$ be a Lie group with Lie algebra $\g$. 
Given a $G$-action $g\mapsto \A_g$ on $M$, one defines its \emph{generating vector fields} by 
%
$$ \A_\xi=\f{d}{d t}\Big|_{t=0} \A_{\exp(-t\xi)}^*.$$
%
\end{definition}

\begin{example}
The generating vector field for the action by right multiplication 
\[ \A_a=R_{a^{-1}}\in \on{Diff}(G)\]
are the left-invariant vector fields,
%
\[ \A(\xi)=\f{\p}{\p t}|_{t=0} R_{\exp(t\xi)}^* =\xi^L.\]
% 
Similarly, the generating vector fields for the action by left multiplication are $-\xi^R$, and those for the conjugation action are $\xi^L-\xi^R$. 
\end{example}

Observe that if $\Phi\colon M_1\to M_2$ is an equivariant map of $G$-manifolds, then the 
generating vector fields for the action are $\Phi$-related.

\begin{theorem}
The generating vector fields of any $G$-action $g\to \A_g$ define a $\g$-action $\xi\to \A_\xi$. 
\end{theorem}
\begin{proof}
Write $\xi_M:=\A_\xi$ for the generating vector fields of a $G$-action on $M$. We have to show 
that $\xi\mapsto \xi_M$ is a Lie algebra morphism. Note that the map
%
\[ \Phi\colon G\times M\to M,\ (a,m)\mapsto a^{-1}.m\]
%
is $G$-equivariant if we take the $G$-action on $G\times M$ to be $g.(a,m)=(ag^{-1},m)$. Hence 
$\xi_{G\times M}\sim_\Phi \xi_M$. But $\xi_{G\times M}=\xi^L$ (viewed as vector fields on 
the product $G\times M$), hence $\xi\mapsto \xi_{G\times M}$ is a Lie algebra morphism. 
It follows that 
\[ 0=[(\xi_1)_{G\times M}, (\xi_1)_{G\times M}]-([\xi_1,\xi_2])_{G\times M}
\sim [(\xi_1)_M,(\xi_2)_M]-[\xi_1,\xi_2]_M.\]
%
Since $\Phi$ is a surjective submersion (i.e. the differential $\d\Phi\colon T(G\times M)\to TM$ is surjective), 
this shows that $[(\xi_1)_M,(\xi_2)_M]-[\xi_1,\xi_2]_M=0$. 
\end{proof}





\subsection{Integrating Lie algebra actions}
%
Let us now consider the inverse problem: For a Lie group $G$ with Lie algebra $\g$, 
integrating a given $\g$-action to a $G$-action. 
The construction will use some facts about \emph{foliations}. 

Let $M$ be a manifold.  A \emph{$k$-dimensional distribution} on $M$ is a linear subspace 
$\mf{R}\subset \mf{X}(M)$ of the space of vector fields such that at any point $m\in M$, the subspace $E_m\subset T_mM$ spanned by all $X_m,\ X\in \mf{R}$ is of dimension $k$. The subspaces $E_m$ define a rank $k$ 
vector bundle  $E\subset TM$ with $\mf{R}=\Gamma(E)$, hence a distribution is equivalently given by this subbundle $E$. 
An integral submanifold of the distribution $\mf{R}$ is a $k$-dimensional submanifold $S$ such that all $X\in\mf{R}$ are 
tangent to $S$. In terms of $E$, this means that $T_mS=E_m$ for all $m\in S$. 
The distribution is called \emph{integrable} if for all $m\in M$ there exists an integral submanifold containing
$m$. In this case, there exists a maximal such submanifold, $\ca{L}_m$. The decomposition of $M$ into maximal integral submanifolds is called a
$k$-dimensional foliation of $M$, the maximal integral submanifolds
themselves are called the \emph{leaves} of the foliation. 

Not every distribution is integrable. Recall that if two vector fields are tangent to a submanifold, then so 
is their Lie bracket. Hence, a \emph{necessary} condition for integrability of a distribution is that 
$\mf{R}$ is a Lie subalgebra. Frobenius' theorem gives the converse: 
%
\begin{theorem}[Frobenius theorem] A rank $k$ distibution $\mf{R}\subset \mf{X}(M)$ 
is integrable if and only if $\mf{R}$ is a Lie subalgebra. 
\end{theorem}
%
The idea of proof is to show that if $\mf{R}$ is a Lie subalgebra, then  
$\mf{R}$ is spanned, near any $m\in M$, by $k$ \emph{commuting} vector fields. 
one then uses the flow of these vector fields to construct integral submanifold. 
%
\begin{exercise}
Prove Frobenius' theorem for distributions $\mf{R}$ of rank $k=2$. (Hint: If $X\in\mf{R}$ with 
$X_m\not=0$, one can choose local coordinates such that $X=\f{\p}{\p x_1}$. Given a second vector field 
$Y\in \mf{R}$, such that 
$[X,Y]\in \mf{R}$ and $X_m,Y_m$ are linearly independent, show that one can replace $Y$ by 
some $Z=aX+bY\in \mf{R}$ such that $b_m\not=0$ and $[X,Z]=0$ on a neighborhood of $m$.)  
\end{exercise}

Given a Lie algebra of dimension $k$ and an \emph{effective} $\g$-action on $M$ (i.e. $\xi_M=0$ implies $\xi=0$), 
one obtains an integrable rank $k$ distribution $\mf{R}$ as the span (over $C^\infty(M)$) of the $\xi_M$'s. 
We use this to prove: 

\begin{theorem}
Let $G$ be a connected, simply connected Lie group with Lie algebra $\g$. 
A Lie algebra action $\g\to \mf{X}(M),\ \xi\mapsto \xi_M$ integrates to an 
action of $G$ if and only if the vector fields $\xi_M$ are all complete.  
\end{theorem}

\begin{proof}[Proof of the theorem]
The idea of proof is as follows. Let $\wh{M}=G\times M$, and $\pr_1,\pr_2$ the projections to the 
two factors. A $G$-action on $M$ defines a foliation of $\wh{M}=G\times M$, with leafs 
the orbits of the diagonal action (where $G$ acts on itself by left multiplication). Equivalently, the 
leaves are the fibers of the map $\wh{M}\to M,\ (g,m)\mapsto g^{-1}.m$. Hence they are indexed by 
the elements of $m$, as follows  
% 
\[ \ca{L}_m=\{(g,g.m)|\,g\in G\}.\] 
%
$\pr_1$ restricts to diffeomorphisms $\pi_m\colon \ca{L}_m\to G$, and we recover the action as 
\[ g.m= \pr_2(\pi_m^{-1}(g)).\]
%
Given a $\g$-action, our plan is to construct the foliation from an integrable distribution. 

Consider the Lie algebra action on $\wh{M}=G\times M$, given by 
\[ \xi_{\wh{M}}=(-\xi^R,\xi_M)\in\mf{X}(G\times M).\]
%
Note that the vector fields $\xi_{\wh{M}}$ are complete, since $\xi_M$ are by assumption complete: If $\Phi_t^\xi$ is the flow of $\xi_M$, the flow of $\xi_{\wh{M}}=(-\xi^R,\xi_M)$ is given by 
%
\[ \wh{\Phi}_t^\xi=(L_{\exp(t\xi)},\Phi_t^\xi)\in\on{Diff}(G\times M).\] 
%  
The action $\xi\mapsto \xi_{\wh{M}}$ is effective, hence it defines an integable $\dim G$-dimensional distribution 
$\mf{R}\subset \mf{X}(\wh{M})$.  Let $\ca{L}_m\hra G\times M$ be the unique leaf
containing the point $(e,m)$. Projection to the first factor induces
a smooth map $\pi_m:\ \ca{L}_m\to G$. 
%

The map $\pi_m$ is \emph{surjective}: Given $g\in G$ write $g=g_r\ldots g_1$ where 
$g_i=\exp(\xi_i)$.
The path $\wh{\Phi}_{t}^{\xi_1}(e,m),\ t\in [0,1]$ lies in $\ca{L}_m$, and has end point $(g_1,m_1)$ 
where $m_1=\Phi_1^{\xi_1}(m)$. Concatenation with the path 
$\wh{\Phi}_{t}^{\xi_2}(g_1,m_1),\ t\in [0,1]$ gives a (piecewise smooth) path from $(m,e)$ to 
$(g_2g_1,m_2)$ where $m_2=\Phi_1^{\xi_2}\Phi_1^{\xi_1}(m)$. Proceeding in this manner, 
we obtain a piecewise smooth path in $\ca{L}_m$ from $(e,m)$ to $(g_r\cdots g_1,m_r)=(g,m_r)$. 
This shows $\pi_m^{-1}(g)\not=\emptyset$. 

For any $(g,x)\in\ca{L}_m$ the tangent map $\d_{(g,x)} \pi_m$ is an isomorphism. 
Hence $\pi_m\colon \ca{L}_m\to G$ is a (surjective) covering map. Since $G$ is simply connected 
by assumption, we conclude that $\pi_m\colon \ca{L}_m\to G$ is a diffeomorphism.  
We now define $\A_g(m)= \pr_2(\pi_m^{-1}(g))$. Concretely, the construction above shows that if $g=\exp(\xi_r)\cdots \exp(\xi_1)$ then 
%
\[ \A_g(m)=(\Phi^{\xi_r}_{1}\circ \cdots \circ \Phi^{\xi_1}_{1})(m).\]
%
From this description it is clear that $\A_{gh}=\A_g\circ \A_h$. 
\end{proof}

Let us remark that, in general, one cannot drop the assumption that $G$ is simply connected. Consider for example 
$G=\SU(2)$, with $\mf{su}(2)$-action $\xi\mapsto-\xi^R$. This exponentiates  to an action of $\SU(2)$ by left multiplication. But $\mf{su}(2)\cong \mf{so}(3)$ as Lie algebras, and the action does not exponentiate to 
an action of the group $\SO(3)$.
 
As an important special case, we obtain: 
%
\begin{theorem}
Let $H,G$ be Lie groups, with Lie algebras $\h\to \g$. If $H$ is connected and simply connected, then any Lie algebra 
morphism $\phi\colon \h\to \g$ integrates uniquely to a Lie group morphism $\psi\colon H\to G$.
\end{theorem}
\begin{proof}
Define an $\h$-action on $G$ by $\xi\mapsto -\phi(\xi)^R$. Since the right-invariant vector fields are complete, this action integrates to a Lie group action $\A\colon H\to \on{Diff}(G)$. 
This action commutes with the action of $G$ by right multiplication. Hence, $\A_h(g)=\psi(h)g$ where 
$\psi(h)=\A_h(e)$. The action property now shows $\psi(h_1)\psi(h_2)=\psi(h_1h_2)$, so that 
$\psi\colon H\to G$ is a Lie group morphism integrating $\phi$. 
\end{proof}
%
\begin{corollary}
Let $G$ be a connected, simply connected Lie group, with Lie algebra $\g$. Then any 
$\g$-representation on a finite-dimensional vector space $V$ 
integrates to a $G$-representation on $V$. 
\end{corollary}
\begin{proof}
A $\g$-representation on $V$ is a Lie algebra morphism $\g\to \End(V)$, hence it integrates to a 
Lie group morphism $G\to \End(V)^\times$. 
\end{proof}

By a Lie subgroup of a Lie group $H$, we mean a Lie group $G$ together with an injective Lie group morphism  
$G\hra H$. That is, the inclusion map need not be an embedding.  
%
\begin{lemma}
Let $\g\subset \h$ be a Lie subalgebra of a finite-dimensional Lie algebra, and $H$ a Lie group integrating 
$\h$. Then there exists a unique connected Lie subgroup $G\subset H$ integrating $\g$. 
\end{lemma}
\begin{proof}
Consider the distribution on $H$ spanned by the vector fields $-\xi^R,\ \xi\in\g$. It is integrable, 
hence it defines a foliation of $H$. The leaves of any foliation carry a unique manifold structure such that the 
inclusion map is an immersion. Take $G\subset H$ to be the leaf through $e\in H$, with this manifold structure. 
Explicitly, $G$ consists of products $\exp(\xi_r)\cdots \exp(\xi_1)$ where $\xi_i\in\g$. From this description it 
follows that $G$ is a Lie group.  
\end{proof}
%
By Ado's theorem, any finite-dimensional Lie algebra $\g$ is isomorphic to 
a matrix Lie algebra. We will skip the proof of this important (but relatively deep) result, since  
it involves a considerable amount of structure theory of Lie algebras. Given such a presentation 
$\g\subset \mf{gl}(n,\R)$, the Lemma gives a Lie subgroup $G\subset \GL(n,\R)$ integrating 
$\g$. Replacing  $G$ with its universal covering, this proves:
%
\begin{theorem} [Lie's third theorem]
For any finite-dimensional real Lie algebra $\g$, there exists a connected,  
simply connected Lie group $G$, unique up to isomorphism, having $\g$ as its Lie algebra. 
\end{theorem} 
%
The book by Duistermaat-Kolk contains a different, more conceptual proof of Cartan's theorem. This new 
proof has found important generalizations to the integration of \emph{Lie algebroids}. In conjunction with the previous Theorem, Lie's third theorem gives an equivalence between the categories of finite-dimensional Lie algebras $\g$
and connected, simply-connected Lie groups $G$. 





\subsection{Proper actions}

Let us quickly list some terminology for Lie group actions $\A\colon G\to \on{Diff}(M)$. 
For any $m\in M$, the set $G.m:=\{(g,m)\,g\in G\}$ is called the 
{\em orbit} of $m$. The space $M/G=\{G.m|\,m\in M\}$ is called the \emph{orbit space} 
for the given action. It inherits a topology as a quotient space of $M$, but can be 
a very singular space. The action $\A$ is called {\em transitive} if there is only one orbit, i.e. 
$M/G=\pt$. In this case, $M$ is called a \emph{homogeneous space}. 

The subgroup $G_m=\{g\in G|\,g.m=m\}$ is 
called the {\em stabilizer} of $m$. From the definition, it is clear that stabilizer subgroups are closed 
subgroups of $G$, hence are embedded Lie subgroups. In particular, the orbit $G/G_m$ inherits a manifold 
structure. The inclusion of the orbit is smooth relative to this manifold structure.  
For any $g\in G$, the stabilizers of a point $m$ and of its translate $g.m$ are 
related by the adjoint action: 
%
$$ G_{g.m}=\Ad_g(G_m).$$
%
The action is {\em free} if all stabilizers $G_m$ 
are trivial. For instance, the actions of $G$ by left or right multiplication on $G$ are both 
free, but the conjugation action is not. The action $\A$ is  {\em effective} if $\ker(\A)=\{e\}$, i.e. 
$\A_g=\on{id}_M$ implies $g=e$. For instance, the conjugation action of $G$ on itself is 
effective if and only if the center of $G$ is trivial. 

The action $\A$ is called {\em proper} if the action map $G\times M\to M$ is proper (i.e. pre-images of 
compact sets are compact). For example, the left or right actions of $G$ on itself are proper.
Note that for a proper $G$-action, the action of any closed subgroup $H\subset G$ is still proper.  
Also, for $G$ compact any $G$-action is proper. 

For a proper action, the stabilizer groups $G_m$ are \emph{compact} since $G_m$ may be viewed as the 
intersection of the closed subspace $G\times\{m\}\subset G\times M$ with the 
preimage of $\{m\}\in M$ under the action map. One can use this fact to construct \emph{slices} 
for the action, i.e. $G_m$-invariant embedded submanifolds $S\subset M$ with $m\in S$ such that 
$G.S$ is an open neighborhood of the orbit $G.m$, and such that $gS\cap S\not=\emptyset
\Leftrightarrow g\in G_m$. Slices give models a neighborhood of $G.m$ 
in the orbit space, since $(G.S)/G=S/G_m$. In particular, if $G_m$ is trivial, we see that 
a neighborhood of $G.m\subset M/G$ is a manifold (modeled by $S$).
% 
\begin{theorem}
For a free, proper action on a manifold $M$, the orbit space $M/G$ inherits a manifold structure 
such that the quotient map $M\to M/G$ is a smooth submersion. Given an $H$-action on $M$ that commutes 
 with the $G$-action, the orbit space $M/G$ inherits an $H$-action.  
\end{theorem}

\begin{example}
Let $H$ be a closed subgroup of $G$, acting on $G$ by right multiplication. This action is proper, hence
$G/H$ is a manifold. The action of $G$ by left multiplication commutes with the actuion of $H$, hence it 
descends to a smooth action on $G/H$.  
\end{example}


\begin{exercise}
Show that the conjugation action of $G$ is proper if and only if $G$ is compact. 
\end{exercise} 











\section{Universal
  cover} \section{The universal enveloping algebra} \section{Tori}
\end{document}
