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\begin{document}
\title{An Elementary Introduction to Lie Groups: Lecture notes}
\author{Kevin Iga}
\date{Summer 1995}
\maketitle
\chapter{Introduction}
These are lecture notes for an informal course I taught to fellow graduate
students at Stanford University during the summer of 1995. It was as much
to help fellow students learn this topic so central to all of mathematics as
it was for me to learn the material better.
The applicability of the study of Lie groups is quite astonishing. In
as much as groups are ubiquitous as a way of studying mathematical
objects with certain symmetries, Lie groups provide a means to
understanding mathematical objects with a ``continuous'' family of
symmetries. Since Lie groups are only a special kind of group, one
might assume that this area of study might be very specialized. But
this is not the case for two reasons: Lie groups are everywhere, and
Lie groups have so much structure that we can describe them easily. The
theory of Lie groups is easier than the study of finite groups, for example.
For this course, I assume the reader is familiar with the concept of a
group, the concept of a manifold, and a little basic measure theory
(enough to know what one is) and topology (enough to know about the
fundamental group). In some later chapters, I will use homology and
cohomology, but the reader not familiar with these can safely skip
these sections.
\chapter{Lie Groups}
\section{Foundational Definitions}
Recall that a group is a set $G$ together with a special element $e\in
G$ called the identity, a multiplication $\mu:G\times G\to G$ and an
inverse map $i:G\to G$, satisfying the properties
\begin{eqnarray}
\mu(x,\mu(y,z))&=&\mu(\mu(x,y),z)\\
\mu(x,e)&=&x=\mu(e,x)\\
\mu(x,i(x))&=&e=\mu(i(x),x)
\end{eqnarray}
Recall that a subgroup of $G$ is a subset of $G$ containing $e$,
together with the restrictions of $\mu$ and $i$, such that these
restrictions map into the subset. Note that under these operations,
a subgroup is a group.
Recall that a manifold is a Hausdorff, second-countable topological space $M$,
together with an open cover ${U_\alpha}$, and homeomorphisms $\phi_\alpha:
U_\alpha\to V_\alpha\subset \re^n$ for some $n$, such that whenever $U_\alpha$
intersects $U_\beta$, $\phi_\beta\circ \phi_\alpha^{-1}:V_alpha\to V_\beta$
is a diffeomorphism. A function between manifolds is said to be smooth if
the composition with each of the $\phi_\alpha$ are smooth, and in a similar
way we define diffeomorphism.
Recall that a submanifold of $M$ is a topological subspace $N$ of $M$
together with an open refinement $\tilde{U}_\alpha$ of the open cover,
and new smooth functions
\[\tilde{\phi}_\alpha:\tilde{U}_\alpha\to\tilde{V}_\alpha\subset
\re^n\]
such that in each open cover, $\tilde{\phi}(N)$ is a vector subspace
of $R^n$. Note that a submanifold of a manifold is a manifold.
\begin{definition}
A {\em Lie Group} is a manifold $G$, together with a group structure
$\{e,\mu:G\times G\to G,i:G\to G\}$ on the manifold considered as a set,
such that $\mu$ and $i$ are smooth maps.
\end{definition}
Examples:
Example 1: ${\re}$, with the additive group, is clearly a manifold,
and a group, and $\mu(x,y)=x+y$ and $i(x)=-x$ are smooth functions.
So this is a Lie group.
Example 2: ${\re^*}$, which as a set is $\re\setminus\{0\}$, is a manifold,
and multiplication is a group on this set. Since $\mu(x,y)=xy$, and
$i(x)=x^{-1}$ are smooth on this set, $\re^*$ is a Lie group.
Example 3: ${\co}$, with the additive operation, is a Lie group.
Example 4: ${\co^*}=\co\setminus\{0\}$, with the multiplicative operation,
is a Lie group.
Example 5: ${\re^n}$, with the additive operation, is a Lie group.
Example 6: ${\co^n}$, with the additive operation, is a Lie group.
Example 7: $\re/\zz$, meaning equivalence classes of real numbers, under
the equivalence $a\sim b$ if $a-b\in\zz$, with the additive operation,
is a Lie group. Topologically, it is a circle, so we sometimes call this
group $S^1$.
Example 8: $\mat_{m\times n}$, under the additive operation, is a Lie group,
though of course, this can be thought of as $\re^{mn}$, with the basis
vectors as matrices with a single 1 entry, and 0's everywhere else.
Example 9: $GL_n(\re)$, the general linear group on $\re$, is the set of
$n\times n$ real-valued matrices with determinant nonzero. It is a manifold,
since it is an open subset of $\mat_{n\times n}=\re^{n^2}$, so one coordinate
neighborhood $U_\alpha$, with the inclusion into $\re^{n^2}$, provides the
manifold structure, and matrix multiplication provides the group structure.
We eliminated the matrices with determinant zero in order to have inverses,
and we consider square matrices to ensure that any two matrices have a product.
Since matrix multiplication is given by a formula involving additions and
multiplications of matrix elements, this is clearly smooth. The inverse also
has a formula, Cramer's rule, which involves additions and multiplications,
although it also has one division. This division is by the determinant of
the matrix, which in our set is never zero. So this inverse function is also
smooth. Sor $GL_n(\re)$ is a Lie group. Note that we sometimes write
$GL(n,\re)$ or even $GL(n)$.
Example 10: $GL_n(\co)$ is a Lie group for the same reasons.
Example 11: $O_n$, the group of orthgonal matrices, is the subset of
$GL_n(\re)$ consisting of matrices $A$ preserving the natural inner
product on $\re^n$. That is, if $v, w \in \re^n$, $\langle
Av,Aw\rangle = \langle v,w\rangle$. It is easy to see that this is
equivalent to the condition $A^{-1}=A^{t}$. This is the set of
isometries, that is, rigid motions, of $\re^n$ preserving the origin,
which include rotations and reflections. This is a subgroup of
$GL_n(\re)$, as is easy to check. The fact that it is a manifold and
that the operations of multiplication and inverse are smooth on this
manifold, are harder to check, and we will use a theorem later in the
course to do this. Essentially, it turns out to apply the theorem, we
will simply need to check that it is a closed subset and is a subgroup
of a Lie group.
Example 12: $U_n$, the group of unitary matrices, is the subset of
$GL_n(\co)$ consisting of matrices $A$ preserving the natural
hermitian inner product on $\co^n$. This is equivalent to the
condition $A^{-1}=\bar{A^{t}}$. This is a subgroup of $GL_n(\co)$,
but again checking that this is a manifold and that the operations are
smooth is difficult and will be left until we have a theorem to help
us.
Example 13: $UT_n(\re)$ is the subset of $GL_n(\re)$ consisting of
upper-diagonal matrices, that is, have ones down the diagonal, and zeroes
below the diagonal, e.g.
\[\left(\begin{array}{ccccc}
1&5&2&4&9\\
0&1&4&0&-1\\
0&0&1&6&2\\
0&0&0&1&\frac{1}{2}\\
0&0&0&0&1
\end{array}\right)\]
Under multiplication, this is a subgroup, and since it is a linear subspace,
it is a manifold. The group operations are smooth for the same reasons
they were smooth on $GL_n(\re)$. So this is a Lie group.
Example 14: $SL_n(\re), SL_n(\co)$, the special linear group (over $\re$ or
$\co$) is the subgroup of $GL_n(\re)$ or $GL_n(\co)$, respectively, which
have determinant equal to 1. This is a subgroup since determinants are
multiplicative, and it is a submanifold by the Implicit Function Theorem,
since the determinant is a smooth function and 1 is a regular value, as
can be computed. Now the group operations are smooth since they are
restrictions of smooth functions to a submanifold. So they are Lie groups.
Example 15: $SO_n, SU_n$ are the intersections $O_n\cap SL_n(\re)$ and
$U_n\cap SL_n(\co)$, respectively. These are subgroups since they are
intersections of subgroups, and submanifolds since they are intersections
of submanifolds, and note that, considering an intersection of submanifolds
as a manifold, the restrictions of the group operations are still smooth.
So $SO_n$ and $SU_n$ are Lie groups. We could do the same with $UT$ but
the determinant of all its elements are already $1$.
There are many other examples we could do, but the list above will
suffice for now. In later chapters we will introduce more Lie groups as
they are needed.
\begin{definition}
A {\em Lie subgroup} of a Lie group is a subgroup which is also a submanifold
of the Lie group.
\end{definition}
\begin{proposition}
A Lie subgroup is a Lie group.
\end{proposition}
Proof: By definition, a Lie subgroup is a group, and a manifold. We only
need to check that the operations are smooth. Restricting smooth
functions to a submanifold gives rise to smooth functions on the
submanifold.\qed
If $G$ and $H$ are Lie groups, then $G\times H$ is defined considering
$G$ and $H$ as groups, and also considering $G$ and $H$ as manifolds.
The underlying set of both is the same, the cartesian product of $G$ and $H$.
So $G\times H$ has a structure of a group and as a manifold. Furthermore,
the group operations on $G\times H$ are smooth. So $G\times H$ is a Lie
group. So we have:
\begin{definition}
(Proposition) Let $G$ and $H$ be Lie groups. The {\em cartesian product} of
$G$ and $H$ is $G\times H$, considered as a group and as a manifold. This is
a Lie group.
\end{definition}
\begin{definition}
A {\em Lie group homomorphism} between Lie groups $G$ and $H$ is a map
$\phi:G\to H$ which is both a group homomorphism and a smooth map. Similarly,
a {\em Lie group isomorphism} is a map which is both a group isomorphism
and a diffeomorphism.
\end{definition}
For every group element $g\in G$, we have a map $L_g:G\to G$, defined
as $L_g(h)=\mu(g,h)$, and a map $R_g:G\to G$ with $R_g(h)=\mu(h,g)$.
These are called the {\em left-translation} and {\em right-translation}
maps. They are diffeomorphisms, and have no fixed points. For those
familiar with some algebraic topology, note that by taking $g$ to be in the
same path component of $e$ in $G$, we can find diffeomorphisms isotopic to
the identity with no fixed points. Thus, $S^2$ is not a Lie group. In
fact, a great many manifolds cannot be Lie groups for this very reason.
\section{The Lie Algebra}
The study of Lie groups, as I mentioned in the introduction, is fairly
complete. Most natural questions about Lie groups have been answered
quite satisfactorily. In fact, Lie groups are in some sense easier than
finite groups.
At first this seems inconceivable. The theory of countably infinite
groups is very difficult, and very little can be said in general about
these. Lie groups are uncountably infinite groups. How could there possibly
be any way of studying them?
What gives Lie groups extra structure is the fact that it is a manifold.
Somehow the topology puts severe limitations on what kind of group you can
get, and the group theory puts severe limitations on what kind of manifold
you can get. Thus we have a curious interplay between two areas of
mathematics: topology and algebra.
What saves us is an object called a {\em Lie algebra} which encodes the local
group structure and the local manifold structure into one algebraic object.
Thus instead of studying two areas of mathematics at once, we need only
study one area of mathematics: algebra. Studying the Lie algebra is fairly
easy, and furthermore gives deep insights in the study of Lie groups.
The Lie algebra is a real vector space, together with a kind of
``multiplication''. This multiplication is not commutative nor associative,
nor does it have an identity element. Nevertheless, it satisfies other
rules which are quite beautiful. Before we define a Lie algebra, or define
the Lie algebra associated to a Lie group, I will motivate the concept
with some examples.
It turns out that the Lie algebra, as a vector space, will be the tangent
space of the manifold at the identity element. There are many ways of
describing the tangent space to a manifold at a point, some of which are
quite abstract and obfuscated. Many of the more abstract definitions are
in some sense the most useful, but since I am not assuming a background in
differential geometry, I will use the most elementary notion and as a
result some of the later computations will be a bit awkward.
\subsection{Tangent spaces}
The notion of a tangent space (tangent plane, tangent line) is
familiar to calculus students. If you have a curve or surface in space,
it is a line or plane that passes through a point in the curve or surface
and has the same slope as the curve or surface at that point. It is a first
order approximation to the curve or surface.
Similarly, given an n-dimensional manifold $M$ embedded in $\re^n$,
and a point $p\in M$, the tangent space to $M$ at $p$ is the affine
space (that is, vector space translated so it may not contain the
origin) containing the point $p$ and sharing the same
first-derivatives of the embedding of $M$. That is to say, if
$\gamma$ is a smooth curve in $M$ passing through the point $p$, so
that $\gamma(0)=p$, then $p+\gamma'(0)$ is a point in the tangent
space. Furthermore, all points in the tangent space are obtained in
this way. We denote the tangent space to $M$ at $p$ as $T_pM$. By
translating $p$ to the origin, we get a linear space, which we also
call the tangent space.
It turns out that every manifold can indeed be embedded in $\re^N$ for some
$N$, and that there is some aspect of a tangent space that is intrinsic to
$M$; that is, is the same no matter what embedding you use or which $N$ you
take. This is easier to see if we use the more abstract definitions,
but for our purposes we ask that readers unfamiliar to those techniques take
our word for it, and that readers familiar to them check it for themselves.
Those advanced readers might want to translate the following discussion
into the more advanced language and see how much easier some computations
become.
As an exercise, and because we will need this information later, let us
calculate the tangent space to some Lie groups at their identity element.
Example 1: $O(n)$. We think of $O(n)$ as embedded in $\mat_{n\times n}(\re)$
which as we noted above is actually $\re^{n^2}$. In this embedding, we
will find the tangent space to $O(n)$ at the identity element $1$.
Let $\gamma$ be a curve in $O(n)$, with $\gamma(0)=1$. For each $t$,
$\gamma(t)$ is an $n\times n$ matrix with $\gamma(t)\gamma(t)^t=1$.
Taking the derivative of this equation with respect to $t$, we get:
\begin{eqnarray}
\frac{d}{dt}(\gamma\gamma^t)&=&0\\
\frac{d}{dt}(\gamma)\gamma^t+\gamma\frac{d}{dt}(\gamma^t)&=&0\\
\frac{d}{dt}(\gamma)\gamma^t+\gamma(\frac{d}{dt}\gamma)^t&=&0\\
\end{eqnarray}
For $t=0$, we have $\gamma(0)=1$, so we have
\[\gamma'(0)+(\gamma'(0))^t=0\]
So elements in the tangent space to $O(n)$ at $1$ consist of points
$1+a$ where $a$ is a matrix satisfying $a^t=-a$, that is, $a$ is
skew-symmetric. Translating this plane back to the origin, we get a
vector space $\{a:a^t=-a\}$.
Conversely, if $\gamma$ is a smooth curve with $\gamma(0)=1$, and
satisfying
\[\frac{d}{dt}(\gamma)\gamma^t+\gamma(\frac{d}{dt}\gamma)^t=0\]
then $\frac{d}{dt}\gamma\gamma^t=0$, so $\gamma\gamma^t$ takes on the
value at $t=0$, which is $1$. So then $\gamma$ lies in $O(n)$. We can
clearly take $\gamma$ to satisfy the above solution and specify any
$\gamma'(0)$ as long as it satifies the equation at $0$. So the space
$\{a:a^t=-a\}$ is the tangent space of $O(n)$ at $1$.
Curiously, this space is not preserved under matrix multiplication, but
it is preserved under the binary operation known as {\em commutation}, or
{\em bracket}. It is defined as $[A,B]=AB-BA$. It is easy to check that
if $A$ and $B$ are skew-hermitian, so is $[A,B]$. This is a foreshadowing
of things to come.
Incidentally, by virtue of the fact that the tangent space and the manifold
have the same dimension, we can calculate the dimension of $O(n)$ by
calculating the dimension of its tangent space at the identity. Calculating
the dimension of $O(n)$ directly might be tricky, but calculating the
dimension of a vector space is fairly straightforward. Every
skew-hermitian matrix has zeros down the main diagonal; other than that,
we are free to specify the entries above the diagonal, and these will
uniquely determine the matrix entries below the diagonal by the skew-symmetry.
So the dimension of $O(n)$ is $\frac{1}{2}n(n-1)$.
Example 2: $U(n)$: We can similarly do a calculation for $U(n)$, so that
$\gamma$ is a curve satisfying
\[\gamma(t)\bar{\gamma(t)^t}=1\]
and $\gamma(0)=1$. Then by differentiating the above equation and setting
$t=0$, we get
\[\gamma'(0)+\bar{\gamma'(0)^t}=0\]
so the space is $\{a:\bar{a}^t=-a\}$, the space of skew-Hermitian matrices.
Note that again, this space is closed under the commutation operation.
That is if $A$ and $B$ are skew-hermitian matrices, so is $[A,B]=AB-BA$.
Again, we can use the tangent space to show that the dimension of $U(n)$
is $n^2$. After all, the diagonal elements must be pure imaginary, and
once we specify the complex numbers above the diagonal, the elements
below the diagonal are specified, so we have $n$ real numbers for the
diagonal and $\frac{1}{2}n(n-1)$ complex numbers = $n(n-1)$ real numbers
for the off-diagonals. Adding the two gives $n^2$.
Example 3: $GL(n)$. Since $GL(n)$ is an open subset of $\mat_{n\times n}$,
the tangent space in this embedding is the whole space $\mat_{n\times n}$.
Of course, this is closed under commutation. We can easily compute the
dimension to be $n^2$ even without the tangent space argument but simply
since $GL(n)$ is an open subset of $\mat_{n\times n}$.
Example 4: $SL(n)$. Again we take a curve $\gamma$ with $\gamma(0)=1$,
and:
\[\det(\gamma(t))=\sum_{\sigma\in S_n} (-1)^\sigma \prod_{i=1}^n
\gamma_{i,\sigma(i)}(t)=1\]
Differentiating the second equation with respect to time, we get a
mess.
But when we set $t=0$, we know that $\gamma(0)=1$, which is a matrix
with zeros everywhere except at the diagonals, where the value is 1.
The derivative of the determinant, in terms of its matrix entries,
is a sum of terms, each of which involves one derivative of one of
the matrix entries of $\gamma$ multiplied by some other matrix entries
of $\gamma$. The term is zero when at least one of the other matrix
entries is zero. In other words, when there are at least two
matrix entries in the term which are off-diagonal. Permutations other
than the identity change at least two elements, so the only nonzero
terms come from the trivial permutation:
\[\frac{d}{dt}\det(\gamma(t))|_{t=0}=\sum_{i=1}^n\gamma'_{i,i}(0)=
\tr(\gamma'(0))\]
So the condition $\det(A)=1$ differentiates to $\tr(A)=0$. Similarly,
if the trace is zero, it is the tangent vector to a curve in $SL(n)$.
This is closed under commutation as well. This is because the trace of
$AB$ is the same as the trace of $BA$, so the trace of any commutator
is zero.
The dimension of $SL(n,\re)$ is $n^2-1$, since $\tr=0$ poses only one
linear constraint, and it is nonsingular. Similarly, the dimension
of $SL(n,\co)$ is $2n^2-2$.
Example 5: $SO(n)$: The intersection of two submanifolds has, as a tangent
space to a point in the intersection, the intersection of the two
tangent spaces. So the tangent space to $SO(n)$ at the origin consists of
skew-symmetric matrices with trace=0. Since skew-symmetric matrices have
only zeros on the diagonal anyway, the ``trace=0'' condition does not
change the tangent space. So the tangent space to $SO(n)$ at the identity
is the space of skew-symmetric matrices. As before the dimension is
$\frac{1}{2}n(n-1)$.
Example 6: $SU(n)$: The tangent space should again be the intersection
of the two tangent spaces $T_1U(n)$ and $T_1SL(n)$. These are skew-hermitian
traceless matrices. If the tangent space to $U(n)$ and $SL(n)$ are
preserved under the commutation relation, then the intersection is as well.
The dimension is $n^2-1$.
Example 7: $UT(n)$: Since this is an affine space to begin with, by subtracting
the identity element we get the tangent space. These are the purely-upper
triangular matrices, that is, matrices with zeros on the diagonal and lower.
The commutator sends these purely-upper triangular elements to others; in
fact, not only the main diagonal but the next higher diagonal is forced
to be zero.
The dimension of the tangent space is clearly $\frac{1}{2}n(n-1)$.
End of Examples.\qed
\subsection{Commutators}
There is a good reason for the commutator to be preserved. That is
because it turns up as the infinitessimal version of the group
commutator $ABAA^{-1}B^{-1}$.
\begin{theorem}
Let $G$ be a Lie subgroup of $GL_n$. Then if $A$ and $B$ are in the
tangent space to $G$ translated back to the origin, then so is
$[A,B]=AB-BA$.
\end{theorem}
Proof:
Let $A(t)$ and $B(t)$ be curves in $G$, with $A(0)=B(0)=1$. Then consider
$C(t)=A(t)B(t)A(t)^{-1}B(t)^{-1}$. We will be taking derivatives of this,
so we need to know how to take derivatives of matrix inverses.
\begin{lemma}
$frac{d}{dt}A(t)^{-1}|_{t=0}=-A'(0)$
\end{lemma}
Proof:
\[AA^{-1}=1.\]
Taking derivatives of both sides we get
\[A'(t)A(t)^{-1}+A(t)\frac{d}{dt}A(t)^{-1}=0.\]
Setting $t=0$ and using $A(0)=1$ we get
\[\frac{d}{dt}A(t)|_{t=0}=-A'(0)\]
\qed
Return to proof of theorem:
Taking the derivative of $C$ at $t=0$, we get
$C'(0)=A'(0)1+1B'(0)1+1(-A'(0))1+1(-B'(0))=0$
So $C'(0)=0$.
Now we take second derivatives of $C$. In order to do this we need
second derivatives of inverses.
\begin{lemma}
$\frac{d^2}{dt^2}A(t)^{-1}|_{t=0}=-A''(0)-2A'(0)^2$
\end{lemma}
Proof:
Again, we take $AA^{-1}=1$ and take derivatives, but this time, we take
two derivatives:
\begin{eqnarray}
AA^{-1}&=&1\\
A'A^{-1}+AA^{-1'}&=&0\\
A''A^{-1}+A'A^{-1'}+A'A^{-1'}+AA^{-1''}&=&0\\
A''(0)+A'(0)(-A'(0))+A'(0)(-A'(0))+A(0)^{-1''}&=&0\\
A(0)^{-1''}=-A''(0)+2A'(0)^2
\end{eqnarray}
\qed
Return to proof of theorem:
Taking the second derivative of $C$ at $t=0$, we get:
\begin{eqnarray}
C''(0)&=&A''(0)+B''(0)+A^{-1''}(0)+B^{-1''}(0)+2A'(0)B'(0)+2A'(0)A^{-1'}(0)
+2A'(0)B^{-1'}(0)\\
&&+2B'(0)A^{-1'}(0)+2B'(0)B^{-1'}(0)+2A^{-1'}(0)B^{-1'}(0)\\
&=&A''(0)+B''(0)-A''(0)+2A'(0)^2-B''(0)+2B'(0)^2+2A'(0)B'(0)-2A'(0)^2
-2A'(0)B'(0)\\
&&-2B'(0)A'(0)-2B'(0)^2+2A'(0)B'(0)\\
&=&2A'(0)B'(0)-2B'(0)A'(0)\\
&=&2[A'(0),B'(0)]
\end{eqnarray}
Something interesting has happened. We usually expect the first
derivative of a product to depend on the first derivatives of the
multiplicands, but here it was always zero. We usually expect the
second derivative of a product to depend on the second derivatives of
the multiplicands, but here it depended on the first derivatives
of the multiplicands. In general, the $n$th derivatives of $C$ will
involve $n-1$ derivatives of $A$ and $B$. But just from our
calculations to second order, we have:
\[C(t)=1+[A'(0),B'(0)]t^2+O(t^3)\]
This is enough to allow us to reparametrize with $t=\sqrt{s}$.
Then we get
\[C(s)=1+[A'(0),B'(0)]s+o(s)\]
So we get a curve passing through 1 which has as its tangent vector
the commutator of $A'(0)$ and $B'(0)$. (This curve, of course,
is a curve from $[0,\infty)$, but the tangent vector still exists since
the formula for $C(s)$ above allows the derivative with respect to $s$ to
be taken for all $s>0$, and shows that the limit as $s\to 0$ exists.)
\qed
\subsection{Commutators of general Lie Groups}
This proof illuminates how we might generalize the notion of commutator
to general Lie groups. After all, in a Lie group, we have no notion
of what a commutator of two elements of the tangent space is, but
we know how to form $ABA^{-1}B^{-1}$.
If we define the commutator $[a,b]$ to be the tangent vector to the
curve $C(s)=A(\sqrt{s})B(\sqrt{s})A(\sqrt{s})^{-1}B(\sqrt{s})^{-1}$ at
$s=0$, where $A(s)$ and $B(s)$ are curves with $A(0)=B(0)=e$ and
$A'(0)=a$ and $B'(0)=b$, and show that this limit as $s\to 0$ exists,
and is independent of the choice of curves $A$ and $B$, then we will
have a definition of commutator that works for a general Lie group
which gives the ordinary commutator when applied to Lie subgroups of
$GL_n$. This is the goal of the next few lemmas.
Note that all along, we assume $G$ is a submanifold of $\re^N$. Then
the multiplication map $\mu$ and inverse map $i$ are defined on $G$ in
terms of coordinates in $\re^N$. Those familiar with more abstract
ways of dealing with derivatives on manifolds may find it good practice
to prove the following lemmas using the machinery of differential geometry,
and see how much easier they become.
\begin{lemma}
Let $G$ be a Lie group with identity $e$, multiplication map $\mu$ and
inverse map $i$. Let $A$ and $B$ be curves in $G$ such that
$A(0)=B(0)=e$. Let $L_g(h)=\mu(g,h)$ and $R_g(h)=\mu(h,g)$.
Then at $h=e$, the Jacobians $dL_e=dR_e=1$, the identity map.
\end{lemma}
Proof:
$L_e(h)=h=R_e(h)$. Differentiation gives the identity map.
\qed
\begin{lemma}
Using the hypotheses from the previous lemma, $iA(0)'=-A'(0)$.
\end{lemma}
Proof:
We take a coordinate system $\{x_1,...,x_n\}$ on a subset of $G$ around
the identity, and write $\mu$ in terms of the coordinates
$\{x_1,...,x_n,y_1,...,y_n\}$.
Take $\mu(A,i(A))=e$, and differentiate both sides. Then set $t=0$, so
$A=e$.
\begin{eqnarray}
\sum_{j=1}^n \frac{\partial}{\partial x_j}\mu|_{(A,i(A))}(A_j')+
\sum_{j=1}^n \frac{\partial}{\partial y_j}\mu|_{(A,i(A))}(i(A)_j')&=&0\\
dR_{i(A)}(A)(A')+dL_A(i(A))(i(A)')&=&0\\
dR_{e}(e)(A')+dL_e(e)(i(A)')&=&0\\
A'+i(A)'&=&0
\end{eqnarray}
\qed
\begin{lemma}
Using the hypotheses from the previous lemma, $(i(A))''(0)+A''(0)$ only
depends on $A'(0)$. \label{lem:twoinv}
\end{lemma}
Proof:
Again we take $\mu(A,i(A))=1$ and differentiate twice, then set $t=0$.
First, since $dL_e(A)$ is the identity, taking another derivative of it
will give 0, and similarly with $dR_e$. With this in mind, we write
\begin{eqnarray}
\sum_{j=1}^n \frac{\partial}{\partial x_j}\mu|_{(A,i(A))}(A_j')+
\sum_{j=1}^n \frac{\partial}{\partial y_j}\mu|_{(A,i(A))}(i(A)_j')&=&0\\
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j^2}\mu|_{(A,i(A))}(A_j'A_k')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j\partial y_j}\mu|_{(A,i(A))}(A_j'i(A)_k')+
\sum_{j=1}^n \frac{\partial}{\partial x_j}\mu|_{(A,i(A))}(A_j'')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j\partial y_j}\mu|_{(A,i(A))}(A_j'i(A)_k')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial y_j^2}\mu|_{(A,i(A))}(i(A)_j'i(A)_k')+
\sum_{j=1}^n \frac{\partial}{\partial y_j}\mu|_{(A,i(A))}(i(A)_j'')&=&0\\
\mbox{terms involving $A'$ }+ dR_e(e)(A''(0)) + dL_e(e)((iA)''(0))&=&0\\
(iA)''&=&-A''+\mbox{ terms involving $A'$ }
\end{eqnarray}
\qed
\begin{lemma}
Assume the hypotheses in the previous lemma. If $K(t)=\mu(A,B)$, then
then $K'(0)=A'(0)+B'(0)$.
\end{lemma}
Proof:
\begin{eqnarray}
K'(0)&=&\sum_{j=1}^n \frac{\partial}{\partial x_j}\mu(A,B)(A_j'(0))+
\sum_{j=1}^n \frac{\partial}{\partial y_j}\mu(A,B)(B_j'(0))\\
&=&dR_e(e)(A'(0))+dL_e(e)(B'(0))\\
&=&A'(0)+B'(0)
\end{eqnarray}
\qed
\begin{lemma}
Assume the hypotheses in the previous lemma. Then $K''(0)-A''(0)-B''(0)$
only depends on $A'(0)$ and $B'(0)$. \label{lem:twod}
\end{lemma}
Proof:
\begin{eqnarray}
K'(0)&=&\sum_{j=1}^n\frac{\partial}{\partial x_j}\mu|_{(A,B)}(A_j'(0))+
\sum_{j=1}^n\frac{\partial}{\partial y_j}\mu|_{(A,B)}(B_j'(0))\\
K''(0)&=&\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j^2}\mu|_{(A,B)}(A_j'A_k')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j\partial y_j}\mu|_{(A,B)}(A_j'B_k')+
\sum_{j=1}^n \frac{\partial}{\partial x_j}\mu|_{(A,B)}(A_j'')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial x_j\partial y_j}\mu|_{(A,B)}(A_j'B_k')+
\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2}{\partial y_j^2}\mu|_{(A,B)}(B_j'B_k')+
\sum_{j=1}^n \frac{\partial}{\partial y_j}\mu|_{(A,B)}(B_j'')\\
\mbox{terms involving $A'$ }+ dR_e(e)(A''(0)) + dL_e(e)(B''(0))&=&0\\
(iA)''&=&-A''+\mbox{ terms involving $A'$ }
\end{eqnarray}
\qed
\begin{lemma}
Assume the hypotheses in the previous lemma. If $C(t)=ABA^{-1}B^{-1}$, then
$C'(0)=0$.
\end{lemma}
Proof:
$C'(0)=A'(0)+B'(0)-A'(0)-B'(0)=0$.
\qed
\begin{lemma}
Assume the hypotheses in the previous lemma. Then $C''(0)$ only depends
on $A'(0)$ and $B'(0)$.
\end{lemma}
Taking the second derivative, and using lemma \ref{lem:twod}, we get
$C''(0)-A''(0)-B''(0)-(iA)''(0)-(iB)''(0)$ depending only on $A'(0)$
and $B'(0)$. Similarly, by lemma \ref{lem:twoinv}, $(iA)''(0)+A''(0)$
and $(iB)''(0)+A''(0)$ depend only on $A'(0)$ and $B'(0)$,
respectively. So $C''(0)$ depends on $A'(0)$ and $B'(0)$.
\qed
\begin{theorem}
Using the hypotheses above, $C(\sqrt{s})$ is a curve with a derivative
at $s=0$ that depends only on $A'(0)$ and $B'(0)$.
\end{theorem}
By Taylor's theorem, $C(\sqrt{s})=1+sf(s)$ for some function $f(s)$, which
is defined and continuous on $[0,\infty)$. As in the matrix case,
$f(0)=\frac{1}{2}C''(0)$, which only depends on $A'(0)$ and $B'(0)$.
\qed
\begin{definition}
Using the hypotheses above, we define the bracket of $A'(0)$ and
$B'(0)$, written $[A'(0),B'(0)]$, to be $C'(\sqrt{s})|_{s=0}$.
\label{def:bracket}
\end{definition}
\subsection{Lie Algebras}
The commutator on matrices satisfies the following three identities:
\begin{enumerate}
\item bilinearity: $a[X,Y]+b[W,Y]=[aX+bW,Y]$ and $a[X,Y]+b[X,W]=[X,aY+bW]$.
\item skew symmetry: $[X,Y]=-[Y,X]$.
\item Jacobi Identity: $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.
\end{enumerate}
The first two are immediately obvious. The third is a computation.
The same properties hold true for the general commutator on a general
Lie group, but as they are awkward to obtain in our limited notation,
we leave the task of verifying these to those familiar with a more
appropriate notation.
This leads to a more general definition:
\begin{definition}
Aa {\em Lie Algebra} is a vector space $V$ (over $\re$, over $\co$)
together with a bilinear operation $[,]: V\times V\to V$ satisfying
\[a[X,Y]+b[W,Y]=[aX+bW,Y]\mbox{ and }a[X,Y]+b[X,W]=[X,aY+bW]\]
\[ [X,Y]=-[Y,X] \]
\[[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0\]
\end{definition}
Although we do not define a Lie algebra to be finite dimensional,
we will only work with finite-dimensional Lie algebras here.
\begin{definition}
The {\em Lie Algebra associated to a Lie group} is the tangent space
at the identity to the Lie group, together with the commutator described
in Definition \ref{def:bracket}.
\end{definition}
\chapter{Relationships between the Lie Group and Lie Algebra}
In the last chapter, we defined a Lie group, and showed how another
object, the Lie algebra, arose from it. As I mentioned earlier,
the Lie algebra is easier to deal with, and holds the key to understanding
the Lie group. In this chapter we try to address the issue of to what
extent a Lie algebra gives rise to a Lie group. We will not answer this
question until Chapter ..., but along the way, we will lay the groundwork
for the study of Lie groups.
\section{Elementary concepts}
\begin{definition}
A {\em Lie subalgebra} of a Lie algebra is a vector subspace which is
closed under the commutation operation of the Lie algebra.
\end{definition}
\begin{proposition}
A Lie subalgebra is a Lie algebra.
\end{proposition}
Proof:
It is a vector space, and the restriction of the commutation operation
to it gives an operation, and the axioms of Lie algebras are still
preserved.
\qed
\begin{proposition}
If $G$ is a Lie group, and $H$ a Lie subgroup, the Lie algebra of $H$ is
a Lie subalgebra of the Lie algebra of $G$.
\end{proposition}
Proof:
The tangent space is a vector subspace, and the commutator is defined
in terms of tangent vectors to a curve in the subspace. So the
tangent subspace is closed under the commutator.
\qed
\begin{definition}
A {\em Lie algebra homomorphism} between Lie algebras $V$ and $W$ is a
linear map $f:V\to W$ such that $f([x,y]_V)=[f(x),f(y)]_W$, where the
subscripts to the brackets indicate which Lie algebra's bracket we are
using. Similarly, a bijective Lie algebra homomorphism is a {\em Lie
algebra isomorphism}.
\end{definition}
\begin{proposition}
If $\rho:G\to H$ is a Lie group homomorphism, then the Jacobian
$d\rho:T_eG\to T_eH$ is a Lie algebra homomorphism.
\end{proposition}
Proof:
The fact that this is a linear map is standard. That the brackets
are preserved follows from the formula in terms of the derivative
of $A((\sqrt{s}))B(\sqrt{s})A((\sqrt{s}))^{-1}B(\sqrt{s})^{-1}$, and
that $\rho$ preserves multiplications and inverses.
\qed
\section{One-Parameter subgroups}
The Lie algebra in a certain sense captures the ``infinitessimal'' group
at the identity. We have already seen that a curve in the Lie group
through the identity gives rise to a vector in the Lie algebra. We can
try to go the other way around, by ``integrating'' the vector back out
again.
Intuitively, to find the group element corresponding to a vector $v$ in the
Lie algebra, we do the following: first embed the Lie group in $\re^N$.
Then look at $v$ translated so its tail lies at the identity element.
Its head should be approximately where the Lie group element ought to be.
But this is only approximate. It might not even land inside the Lie group!
We can take the closest point in the Lie group to this, but still this is
approximate.
So we divide $v$ in two. That is, we start at $e$, go forward by
$v/2$, and find the closest point in the group $g$. Then we go by $v/2$ again.
Only this second time, instead of using $v/2$ which was calculated at
the origin, we use the appropriate version of $v/2$ at $g$. To find out
what this is, we use the left-translation function $L_g$. The Jacobian
$dL_g$ maps tangent vectors at $e$ to tangent vectors at $g$. So
$dL_g(v/2)$ is the vector we want to use.
Now it is clear how to generalize this process. We divide $v$ in $k$
pieces, and each step of the way we use $dL_g(v/k)$ where $g$ is the point
in the Lie group we reached in the last step.
A slicker way to do this is to translate $v$ all over the group to begin
with, then find paths which follow the resulting vector field.
\begin{definition}
A {\em Left-invariant vector field} $v$ is a vector field on $G$ such that
$dL_g(v)=v$.
\end{definition}
\begin{proposition}
Left invariant vector fields are in one-to-one correspondence with vectors
in the tangent space at the identity.
\end{proposition}
Note: clearly the role of the identity is not special in this proposition.
Any other element of the group would do.
Proof:
Given a left-invariant vector field, the restriction to the identity gives
aa vector at the identity. Given a vector at the identity, $dL_g$ gives
aa vector at every other point in the group. These two maps are inverses.
\qed
For those familiar with the concept of a Lie bracket of vector fields, note
that the Lie bracket of two left-invariant vector fields is a left-invariant
vector field, and that this Lie bracket agrees with the commutator
operation defined earlier.
\begin{definition}
A {\em one-parameter subgroup} of a Lie group $G$ is a curve
$\gamma:\re\to G$ such that $\gamma(s+t)=\mu(\gamma(s),\gamma(t))$.
\end{definition}
Note: A one-parameter subgroup is not a subgroup. It is a homomorphism.
But its image is not even a Lie subgroup. For instance, if $G$ is
the torus $S^1\times S^1$, and we take the curve $\gamma(t)=(\sqrt{2}t,
\sqrt{3}t)$, modulo $\zz$, this is a one-parameter subgroup, but it
densely fills $G$, and is definitely not a submanifold of $G$. The
notation is confusing, but it is a historical usage that has been
ingrained in the literature.
\begin{proposition}
A one-parameter subgroup $\gamma$ satisfies the equations $\gamma(0)=e$,
and $\gamma'(t)=dL_{\gamma(t)}(\gamma'(0))$.
\end{proposition}
Proof:
The first equation arises by setting $s=0$ in $\gamma(t+s)=
\mu(\gamma(t),\gamma(s))$. The second is obtained from the same
equation by differentiation with respect to $s$ and setting $s=0$.
\qed
\begin{theorem}
(Uniqueness and Existence of solutions to Ordinary Differential Equations)
Given a vector field $v$ on a manifold $M$, and a point $p$, there is
an $\epsilon>0$ such that there exists a unique curve $\gamma:(-\epsilon,
\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(t)=v(\gamma(t))$.
\end{theorem}
For the proof of this fact, see ....?
We apply this theorem to left-invariant vector fields, and we get
a one-parameter subgroup, by definition. Actually, we get a one-parameter
subgroup restricted to $(-\epsilon,\epsilon)$. To show there exists a
one-parameter subgroup, we use integer powers as follows:
\begin{theorem}
Given a vector $v$ in $T_eG$, there is a unique one-parameter subgroup
$\gamma$ so that $\gamma'(0)=v$. \label{thm:onepar}
\end{theorem}
Proof:
Extend $v$ to the unique left-invariant vector field on $G$. Use the
above theorem to find a curve $\gamma$ satisfying $\gamma(0)=e$, and
$\gamma'(t)=v(\gamma(t))$. Then $\gamma'(0)=v$, and
$\gamma'(t)=L_{\gamma(t)}(\gamma'(0))$. If $s<\epsilon$, we can find
another curve $\gamma_1$ with $\gamma_1(0)=\gamma(s)$ and
$\gamma_1'(t)= v(\gamma_1(t))$. By uniqueness,
$\gamma_1(t)=\gamma(s+t)$. Now we can multiply $\gamma$ on the left
with the constant $\gamma(s)$, and we again get a solution to the
differential equation. So $\gamma(s+t)=\mu(\gamma(s),\gamma(t))$.
Now to define $\gamma(t)$ for $t>\epsilon$, find any integer $N$ so that
$N>t/\epsilon$. Let $\gamma(t)=\gamma(t/N)^N$, where the power of $N$
means $n$-fold iteration using the multiplication map $\mu$. This is
well-defined because if $N_1$ and $N_2$ are two such $N$, then looking
at $\gamma(t/(N_1N_2))$, and applying uniqueness, we see they agree.
This is smooth since all the operations are smooth.
\qed
Example 1: $\re$, with addition: Left translation is the identity map, so
the differential equation is
\[\frac{d}{dt}\gamma(t)=v, \gamma(0)=0.\]
so the one parameter subgroup is
\[\gamma(t)=vt.\]
Example 2: $\re^*$, with multiplication: Left translation by $x$ is
multiplication by $x$, so the differential equation is
\[\frac{d}{dt}\gamma(t)=v\gamma(t), \gamma(0)=1.\]
so the one parameter subgroup is
\[\gamma(t)=e^{vt}.\]
Example 3: $GL_n(\re)$: Left translation by $A$ is matrix multiplication
on the left by $A$ so the differential equation is
\[\frac{d}{dt}\gamma(t)=\gamma(t)v, \gamma(0)=1.\]
This differential equation can be solved in power series in matrices, and
the result is
\[\gamma(t)=\exp(vt)=\sum_{n=0}^\infty\frac{(vt)^n}{n!}.\]
\section{The Exponential Map}
The theorem we proved (Theorem \ref{thm:onepar}) shows that given a
vector $v$ in the Lie algebra and a real number $t$, we have an
element $\gamma_v(t)$ of the Lie group. Furthermore, $\gamma_v(s+t)=
\mu(\gamma_v(s),\gamma_v(t))$, and $\gamma_{sv}(t)=\gamma_v(st)$. The
last equation can be proven by uniqueness, as before. Thus
$\gamma_v(t)=\gamma_{tv}(1)$, and so there is a well-defined function
$f$ from the Lie algebra to the Lie group so that $f(tv)=\gamma_v(t)$.
In light of the last two examples, we call this function the {\em
exponential map} and write it $\exp$.
\begin{definition}
The {\em Exponential map} from the Lie algbra to the Lie group is defined
as follows: if $v$ is an element of the Lie algebra, and $\gamma_v(t)$ its
one-parameter subgroup, then $\exp(v)=\gamma_v(1)$.
\end{definition}
Note that $\exp$ is not a homomorphism in general. Although $\exp(sv+tv)
=\mu(\exp(sv),\exp(tv))$, it is not true in general that $\exp(sv+tw)=
\mu(\exp(sv),\exp(tw))$. For instance, in $GL_2(\re)$, taking
$s=t=1$ and $v=\left(\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}\right),
w=\left(\begin{array}{cc}
0 & 0 \\
1 & 0
\end{array}\right)$, we see using the power series that
$\exp(v)=\left(\begin{array}{cc}
1 & 1 \\
0 & 1
\end{array}\right),
\exp(w)=\left(\begin{array}{cc}
1 & 0 \\
1 & 1
\end{array}\right),$
and $\exp(v)\exp(w)=\left(\begin{array}{cc}
2 & 1 \\
1 & 1
\end{array}\right)$. On the other hand,
$\exp(v+w)=\left(\begin{array}{cc}
\cosh 1 & \sinh 1 \\
\sinh 1 & \cosh 1
\end{array}\right).$
\qed
\begin{proposition}
The Jacobian of the exponential map $d\exp:T_eG\to T_eG$ is the identity map.
\end{proposition}
Proof:
Let $A$ be a curve in $T_eG$ with $A(0)=0$. Then there is a one-parameter
subgroup $\gamma(t)$ such that $\gamma'(0)=A'(0)$. The Jacobian
assigns to $A'(0)$ the vector $\gamma'(0)$.
\qed
\begin{corollary}
The exponential map is a local diffeomorphism (that is, there is an open
neighborhood $V$ of $T_eG$ and an open neighborhood $U$ of $G$ such that
$\exp$ maps $V$ onto $U$ diffeommorphically.
\end{corollary}
Proof:
This is an application of the Jacobian theorem to the exponential map, using
the proposition to show that the Jacobian is nonsingular.
\qed
Note 1: This makes $\exp$ a coordinate function.
Note 2: The exponential map is not in general a global diffeomorphism.
Otherwise, all Lie groups would be diffeomorphic to some $R^n$. But
we have $S^1$, for example. Here the exponential map is just reduction
modulo $\zz$, which is surjective but not injective. There are examples
of Lie groups for which the map is not surjective (for instance,
$SL_2(\re)$.)
\end{document}