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\title{Introduction to Lie Groups and Lie Algebras\\September 21, 2004}
\author{Gabriel C. Drummond-Cole}
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\begin{document}
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From now on all of my Lie groups are connected.  If $G$ is a Lie group 
then $\mathfrak{g}$ is the Lie algebra corresponding to $G,$ also 
written $T_1G$ and $Lie(G).$

There is a bracket $[,],$ and I won't remind you of how it's defined, 
but it's bilinear, skew-symmetric, and satisfies the Jacobi identity.  
There are relations to the group multiplication.  There is the $Ad$ 
action $Ad\ e^{tx}=1+t ad\ x+\ldots$ for $x\in\la{g},$ which defines the 
$ad\ x$ action $\la{g}\to\la{g}$ by $ad\ x(y)=[x,y].$

You have the formula $e^xe^y=e^f(x,y).$  Here 
$f=x+y+\frac{1}{2}[x,y]+\ldots,$ where the other terms have $x,\ y,$ and 
the commutator.  The coefficients are universal.  This is called the 
Campbell Hausdorff formula and we haven't actually proved it.

Now, what next?  I want to continue the story about the relation between 
Lie groups and Lie algebras.  Say we have $\phi:G_1\to G_2;$ then there 
is a corresponding map $\phi_*:\la{g}_1\to\la{g}_2$ which preserves 
$[,].$  That is, $Hom(G_1,G_2)\injects Hom(\la{g}_1,\la{g}_2).$  Is this 
equality?  It's pretty clear that a lift would be unique.

If you know the map at the level of Lie algebras, it uniquely defines 
the map in a neighborhood of the identity in the Lie group, which then 
generates the group, giving uniqueness.

In general, though, not everything can be lifted.  Let 
$G_1=S^1,G_2=\mathbb{R}.$  Then the Lie algebras are $\mathbb{R}$ with 
zero bracket.  There is a map (the identity) on the level of Lie 
algebras, but a Lie group morphism (which must be induced by 
$t\to \alpha t:\mathbb{R}\to\mathbb{R}$) should satisfy $2\pi\alpha=0.$  
Then it's the zero morphism.

Is there any way to fix this?
\begin{thm}
If $G_1$ is simply connected then $Hom(G_1,G_2)=Hom(\la{g}_1,\la{g}_2).$
\end{thm}
The reason we're trying to do this is because Lie algebras are easier to 
work with.  The other type of question is the question of subgroups.  
Namely, suppose $H\subset G$ (a Lie subgroup, I don't consider any 
others).  Then of course you can immediately form the corresponding Lie 
algebras, and you have an inclusion $\la{h}\subset\la{g}.$  This is not 
only an embedding of vector spaces, but also preserves the commutator.  

Think of inclusion as a group morphism; then this induces an algebra 
morphism.  Can we do it the other way around?  Does a Lie subalgebra 
correspond to a Lie subgroup?

The answer again is unfortunately no.  Not every Lie subalgebra can be 
related to a Lie subgroup.  Say your $G$ is $\mathbb{R}^2,$ and $\la{h}$ 
is a subspace spanned by $(1,\alpha).$  Then $H=\mathbb{R}(1,a).$

Look instead at $T^2,$ where $\mathfrak{g}=\mathbb{R}^2$ with the 
trivial bracket.  Then let $\mathfrak{h}=\mathbb{R}(1,a)$ with $a$ 
irrational.  At the level of Lie groups, this covers an everywhere dense 
subset of the torus.  This is then not a closed one-dimensional 
submanifold.

To accomodate for this kind of problem we have the definition
\begin{defi}
An immersed (analytic, etc.) subgroup in $G$ is a subset $H\subset G$ 
such that \begin{enumerate}
\item $H$ is a subgroup of $G$
\item $H$ is almost a submanifold; it is the image of an immersion: 
$H=Im(\tilde{H}\injects G)$ for an immersion $i$ of a Lie group 
$\tilde{H}.$  That is, it is a map of manifolds.
\end{enumerate}
\end{defi}
So that's almost a Lie subgroup, but not quite.  So you should be 
talking about these more general subgroups.  Does this solve the 
problem?  The answer is yes.  Here is a theorem:
\begin{thm}
There is a bijection between connected immersed subgroups $H\subset G$ 
and 
subalgebras $\la{h}\subset\la{g}.$  The correspondence comes from $H\to 
\la{h}=Lie(H)=T_1H.$
\end{thm}
There is also a third fundamental theorem which we'll discuss next time.  
Let me note that theorem one easily follows from theorem two, and the 
other way around as well.

The third theorem will tell us that an abstract Lie algebra is the Lie 
algebra of some Lie group.

Here's the proof of the first theorem from the second.  Suppose 
$f:\la{g}_1\to\la{g}_2.$  We want to show it lifts to $\phi:G_1\to 
G_2$ with $\phi_*=f.$

Define $G=G_1\times G_2.$  Then the Lie algebra of $G$ is 
$\la{g}_1\times\la{g}_2.$  Then this contains $\la{h}=\{(x,f(x))\}.$  I 
claim that this is a subalgebra.  It's a subspace, and 
$[(x,f(x)),(y,f(y))]=([x,y],[f(x),f(y)])=([x,y],f([x,y])).$  So that's a 
very simple but quite useful trick.   Then there is a corresponding 
subgroup $H\subset G_1\times G_2.$  It's the image of some embedding 
$Im(\tilde{H}\injects G_1\times G_2).$  Look at the projection 
$p:G_1\times G_2\to G_1.$  Apply this projection to $H$ or if you prefer 
to $\tilde{H}.$  Then I claim that $\tilde{H}\injects G_1\times G_2\to 
G_1$ is a covering map, that is, at the level of tangent spaces is an 
isomorphism.  This follows from the fact that projection 
$\la{g}_1\to\la{g}_1\times\la{g}_2\to\la{g}_1$ is an isomorphism.  

You also need to check that it's surjective.  This is because it's a 
group morphism covering a neighborhood of the identity.  So it's a 
covering map.  So leaving apart some stupid topological problem, a 
surjective map which is an isomorphism at the level of covering spaces 
is a covering map.  Since $G_1$ is simply connected this is just a 
diffeomorphism.  Then we can follow the inverse map and then the other 
projection.  

So define $\phi$ as composition of $G_1\to H\subset G_1\times 
G_2\to^{p_2} G_2.$  You have to check that it is a group morphism, but 
that basicly follows because $H$ is a subgroup.

I still haven't proved theorem 2.  To do this I need to ask how many of 
you are familiar with distribution.  Not all, so I'll give you the 
method but I won't go into detail.  

Let me start by the following remark.  Assume you have a Lie group which 
corresponds to the subalgebra $\la{h}.$  Then at every point $p\in H,\ 
T_pH=(T_1H)p=\la{h}\cdot p=(R_{p^{-1}})_*\la{h}.$

Now at every point in $G,$ define $\mathscr{D}_p=\la{h}\cdot p\subset 
T_pG.$  This is called a distribution, and generalizes the picture of a 
differential equation as a direction field.  This simple argument shows 
that the Lie group should have the property that the tangent space is 
$(T_1H)p.$  If $H$ exists, then $T_pH=\mathscr{D}_p.$  So $H$ is called 
an integral manifold of the distribution $\mathscr{D}_p.$
\begin{lemma}
For every point $g\in G,\ $ there is loally an integral manifold 
containing $g,$ namely $H^0\cdot g,$ where $H^0=\exp$ in some 
neighborhood of $0$ in $\la{h}.$
\end{lemma}
After we prove this we're basically done.  There are at least two or 
three proofs.  I know two.
\begin{enumerate}
\item You can use the Frobenius thoerem, that tells you that 
$\mathscr{D}_p$ is integrable.  It's generated by right-invariant vector 
fields corresponding to elements of $\la{h},$ and $\la{h}$ is closed 
under $[,]$ If I take $ge^{tx}$ for $x\in\la{h},$ then this curve is the 
integral curve for a right invariant vector field corresponding to $g.$  
In particular, its tangent vector at every point will lie in this 
distribution.

Thus, $e^{tx}\cdot g$ lies in an integral manifold containing $g.$  Now 
it shows that elements of form $H^0\cdot g$ are in the integral manifold 
containing $g.$  Comparing dimensions you find out they are the same.
\item There is an alternative proof which doesn't use Frobenius at all, 
but instead uses the Campbell Hausdorff formula.  They are actually 
closely related but I'd rather not spend time on this.
\end{enumerate}
Now that I've proved the lemma, you have $G,\ \la{g},$ and $\la{h}.$  
You create out of it a distribution.  Locally there is always a 
manifold.  In particular near the identity this is $\exp(\la{h}).$  So 
just take an integral manifold and continue it as far as you can.  Now 
take $H$ to be the maximal connected integral manifold containing $1.$  
This doesn't have to be a submanifold in $G.$  But by arguments having 
nothing to do with Lie theory you can show that this is the image of an 
injective map between manifolds.  I'm not going to show this because it 
is so believable.

The only thing which remains to be 
shown is why this thing is a subgroup.  The points on this integral 
manifold will be exactly of for $\exp(t_1x_1).$    This is how you 
extend.  Then an element will be in $H$ only if it is the product of 
finitely many things of form $\exp(t_ix_i)$ for $x_i\in\la{h}.$  So $H$ 
is a subgroup, and that is basically the end.

So that essentially proves these two theorems.  In the last two minutes 
let me give you some obvious corrolaries.  There are obvious ways to get 
morphisms from a Lie group to another, if the first is simply connected.  
For example, where you're supposed to show the map $SU(2)\to SO(3)$ is a 
covering map, instead of using a messy fromula it suffices to define 
$\phi_*:\la{su}(2)\to\la{so}(3)$ and check that this is a morphism of 
Lie algebras.  

For example, if the basis is
$$(
\sigma_x=\left(\begin{array}{cc}
0&1\\
-1&0
\end{array}\right)
,\sigma_y=\left(\begin{array}{cc}
0&i\\
i&0
\end{array}\right)
,\sigma_z=\left(\begin{array}{cc}
i&0\\
0&-i
\end{array}\right)$$

Then you take $\sigma_i\to 2T_i,$ where these latter are the 
infinitessimal rotations.

There is a similar statement for normal groups.
\begin{thm}
Say $H$ is a connected Lie subgroup of $G.$  Then $H$ is normal if and 
only if $ad\ x h\in\la{h}$ for all $x\in\la{g}.$  That is, 
$[x,h]\in\la{h}.$  Such an $\la{h}$ is called an ideal in $\la{g}.$

If $H$ is normal then $Lie(G/H)=\la{g}/\la{h}.$
\end{thm}
This theorem is relatively easy and the time is up so let's stop here.


\end{document}
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