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\title{Introduction to Lie Groups and Lie Algebras\\August 31, 2004}
\author{Gabriel C. Drummond-Cole}
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The website is www.math.sunysb.edu/~kirillov/mat552/
The texts are "Representation Theory: a First Course," by Fulton and Harris, and "Lie Groups and algebraic groups" by Onishchik and Vinberg, which is out of print and on reserve.  I won't follow either of these exactly; sometimes I will give references to one or the other of these texts.

I want to have regular homework, a midterm, and a final.  Here is the first homework, due in two weeks.  I will give a grade, which will take homework and tests into account.

I will require some linear algebra and basic topology, including manifolds, vector fields, and so on.

I'm going to start with a baby example.
\begin{example}
Suppose you have $m$ numbers $a_1,\ldots,a_m$ arranged on a circle.  You have a transformation which replaces $a_1$ with $\frac{a_n+a_2}{2},\ a_2$ with $a_1+a_3/2,$ and so on.  If you do this sufficiently many times, will the numbers roughly equalize?  To answer this we need to look at the eigenvalues of $A.$  To get the characteristic polynomial, and then to find the roots, is too hard.  But we have a rotational symmetry.  We can call this $\mathbb{Z}_n$-symmetry.  So $BAB^{-1}=A,$ where $B$ is a rotation.  This helps us because of the following result from linear algebra:  if two operators $A$ and $B$ commute and $B$ is diagonalizable, i.e., that $V=\bigoplus_\lambda V_\lambda,\ B|_{V_\lambda}=\lambda id,$ then $AV_\lambda\subset V_\lambda.$  We can diagonalize $B$ and see that the eigenvalues of $B$ are the $m\ m^{th}$ roots of unity.  Let $\epsilon$ be a primitive $m^{th}$ root of unity.  Then the eigenvectors are $(1,\epsilon^i,\epsilon^{2i},\ldots,\epsilon^{mi}),$ where $i$ ranges from $0$ to $m-1.$
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So since each of these eigenvectors spans an $A$-invariant subspace, they are thus also eigenvectors of $A.$  So we can answer the question by looking at what $A$ does to the eigenvectors.
\end{example} 
This is the baby version of a real life problem.  Say we have $S^2\subset \mathbb{R}^3.$  We have the Laplace operator $\Delta_{sph}:C^\infty(S^2)\to C^\infty(S^2).$  I'm not going to talk about this in detail, it involves extending to $\mathbb{R}^3$ and taking second derivatives.  The question is, what are the eigenvalues and eigenfunctions of $\Delta_{sph}?$  This describes information about the hydrogen atom.  If we try to solve it generally, we run into a difficult differential equation.

So we use symmetry, this time the symmetry of $SO(3)$ acting on the sphere by rotation.  This is not a finitely generated group.  The second problem is that $SO(3)$ is noncommutative.  So the approach before is hopeless, you'll never be able to diagonalize all of $SO(3)$ at the same time, as we could with $\mathbb{Z}_m.$  We'll eventually be able to solve this by generating with 3 "infinitessimal rotations" and decomposing into invariant representations instead of eigenspaces.

So how can we deal with these two problems?  The trick is that the group $SO_3$ is more than a group, it is also a smooth manifold.
\begin{defi}
A Lie group is a set $G$ with two structures: $G$ is a group and $G$ is a (smooth, real) manifold.  These structures agree in some reasonable sense, i.e., multiplication and inversion are smooth maps.  A Lie group morphism is a smooth homomorphism.
\end{defi} 
I should say what I mean by smooth.  For a Lie group, it turns out that $C^1$ is sufficient to give you real analytic.  This is a highly nontrivial result which was one of Hilbert's problems.  But the level of smoothness is unimportant here.
\begin{enumerate}
\item $\mathbb{R}^n,+$
\item $\mathbb{R}^*,\times$
\\ $\mathbb{R}_+,\times$
\item $S^1=\{z\in\mathbb{C}:|z|=1\},\times$
\item $GL(n,\mathbb{R})\subset\mathbb{R}^{n^2}.$  A lot of the groups we'll consider will be subgroups of $GL(n)$
\item $SU(2)=\{\left(\begin{array}{cc}\alpha&\beta\\-\bar{\beta}&\bar{\alpha}\end{array}\right):\alpha,\beta\in\mathbb{C},|\alpha|^2+|\beta|^2=1\}.$  This can be seen to be $S^3\subset \mathbb{R}^4$ by $(\Re \alpha,\Im\alpha,\Re\beta,\Im\beta).$
\item In fact, all classical groups in linear algebra, such as \\$GL(n,\mathbb{R}),SL(n,\mathbb{R}),O(n,\mathbb{R}),U(n),SO(n,\mathbb{R}),SU(n),Sp(2n,\mathbb{R}),$ are Lie groups.
\end{enumerate}
I never said whether my groups were connected as manifolds.  So any finite group satisfies this as a 0-dimensional manifold.  So we seperate the finite group part and the continuous part as follows.  If $G$ is a Lie group, we can denote by $G^0$ the connected component of unity.  
\begin{thm}
This is a normal subgroup of $G$ and is a Lie group itself.  $G/G^0$ is discrete.
\end{thm}
For completeness' sake, it is a group.  You have to show that it is closed under the operations of multiplication and inversion.  The continuous (inversion) map $i$ must take $G^0$ to one component of $G,$ that which contains $i(e)=e,$ namely $G^0.$  I leave it to you to show that multiplication is closed.
\\
Now, how do you check that this is a normal subgroup?  If $g\in G$ and $h\in G^0,$ then we must show that $ghg^{-1}\in G^0.$  Conjugation by $g$ is continuous will take $G^0$ to some component; since it fixes $e$ this component is $G^0.$
\\
We'll put off for a moment the proof that the quotient is discrete.  We're not going to pay much attention to the discrete case.
\begin{thm}
If $G$ is a connected Lie group then its universal cover $\tilde{G}$ has a canonical structure of a Lie group such that the covering map $p:\tilde{G}\to G$ is a morphism of Lie groups, i.e., that it agrees with the group structure.  Then $\ker\ p=\pi_1(G).$  You can prove this.
\end{thm}
\begin{defi}
A Lie subgroup $H$ of a Lie group $G$ is a subgroup which is also a submanifold.
\end{defi} 
\begin{thm}
\begin{enumerate}
\item (easy) Any Lie subgroup is a closed submanifold
\item (hard) Any closed subgroup of a Lie group is a Lie subgroup.
\end{enumerate}
\end{thm}
A torus is a Lie group, since the product structures of groups and manifolds respect one another.  So if the torus is $G,$ and $H$ is $\mathbb{R},$ then the map $H\to G$ given by $t\to (t,\sqrt(2)t)$ is not a submanifold.
\begin{cor}
\begin{enumerate}
\item if $G$ is a connected Lie group and $U$ is a neighborhood of $e,$ then $U$ generates $G.$
\item Say $f:G_1\to G_2,\ G_2$ is connected, and $f_*:T_eG_1\to T_eG_2$ is surjective, then $f$ is surjective.
\end{enumerate}
\end{cor}
Proofs:
\begin{enumerate}
\item Let $H$ be the subgroup generated by $U.$  Then $H$ is open in $G.$  That's because multaplication of $h\in H$ by $U$ yields a neighborhood of $h$ in $G.$  Since it's an open subset of a manifold, it is a submanifold, so that it is a Lie subgroup.  Therefore it is closed, and is nonempty, so is all of $G.$  
\item Given the assumption, the implicit function theorem says that $f$ is surjective on some neighborhood $U$ of $e,$ and then because $U$ generates $G_2,$ we can pull back to preimages to see that $G_1$ covers $G_2.$
\end{enumerate}
That's all I wanted to talk about today.
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