# Lie Groups/Algebras, Notes 4: Lie Groups

Lie Groups:

Definition 1.  A Lie group is a manifold and group whose binary operation is a $C^\omega$ map between the manifolds $G\times G$ and $G.$

We will assume our Lie groups are Banach manifolds unless otherwise stated (locally homeomorphic to a Banach space $E$).  It turns out a Lie group is metrizable, and the metric space is complete.  Lie subgroups are subgroups which are also submanifolds and Lie groups.

Proposition 2.  Let $G$ be an $E$-Lie group (that is, a Lie group which is an $E$-manifold).  Then

1. If $E\in\{\mathbb{R},\mathbb{C}\},$ then $G$ is locally connected.
2. If $E\notin\{\mathbb{R},\mathbb{C}\},$ then $\dim(G)=\dim_F(E)=0.$
3. Suppose $E$ is locally compact, then $G$ is locally compact iff $\dim(G)<\infty.$
4. If $G$ is generated by a subspace whose topology admits a countable base, then the topology on $G$ admits a countable base.

Now suppose $X$ is a $C^\omega$-manifold with an analytic associative binary operation with identity.  Then the subset $G\subseteq X$ of invertible elements of $X$ is open in $X$ and is a Lie group.

Definition 3.  A Lie group morphism between Lie groups $G$ and $H$ is a $C^\omega$ group homomorphism $f:G\to H.$  $Aut(G)$ will denote the group of Lie group automorphisms of $G.$  A Lie group representation on a Banach space $E$ is a Lie group morphism $\rho:G\to Aut(E).$  (Note if $E$ is a Banach space over the field $F$, then $Aut(E)$ is a Lie group which is a manifold over $F$)

Definition 4.  Let $G$ be a topological group and $X$ be an analytic manifold.  We say that $X$ is a homogeneous space for $G$ if $X$ is a $G$-space (that is, continuous action) and $(g_1g_2)x=g_1(g_2x).$

This induces an equivalence relation on $X$ into orbits under the action of $G$ and a corresponding quotient space $X/G,$ called the orbit space of $X$ by $G.$

Proposition 5.  Let $X$ be Lie group and $G$ be a Lie subgroup.  Then:

1. The orbit space $X/G$ has a unique analytic manifold structure such that the canonical map $\pi:X\to X/G$ is a submersion.
2. If $G\unlhd X,$ then $X/G$ is a Lie group and $\pi$ is a Lie group morphism.

In case 1 above, we call the quotient $X/G$ the Lie homogeneous space of $X$ by $G,$ and in case 2 we call the quotient the Lie quotient group.

Let $G$ and $H$ be Lie groups.  Then there is a canonical correspondence $\tau$ between $T(G)\times T(H)$ and $T(G\times H).$  In particular, suppose we have two tangent vectors $u\in T_x(G),$ $v\in T_y(G),$ and $f\in C^\omega(G).$  Let us then define

$(uv)(f)=\tau(u,v)(f,0)+\tau(u,v)(0,f).$

It follows that $uv\in T_{xy}(G)$ for

$\begin{array}{lcl} (uv)(fg)&=&\tau(u,v)(fg,0)+\tau(u,v)(0,fg)\\&=&\tau(u,v)(f,0)\cdot g(xy)+f(xy)\tau(u,v)(g,0)+\tau(u,v)(0,f)\cdot g(xy)+f(xy)\tau(u,v)(0,g)\\&=&g(xy)\left(\tau(u,v)(f,0)+\tau(u,v)(0,f)\right)+f(xy)\left(\tau(u,v)(g,0)+\tau(u,v)(0,g)\right)\\&=&(uv)(f)g(xy)+f(xy)(uv)(g).\end{array}$

Thus we have a product on the tangent bundle of $G$ (note that $G$ being a group makes $xy$ an arbitrary element of $G$).  Hence $T(G)$ is a Lie group.

The Lie Algebra of a Lie Group:

Recall a vector field on a manifold $M$ is section of the vector fiber bundle $T(M)$ with $M$ as its base.  Hence it is a map $X:M\to T(M)$ satisfying $\pi\circ X=1$ where $\pi$ is the projection of the vector fiber bundle to its base.  Now suppose $X$ and $Y$ are two vector fields on $M,$ then we can define the vector fields

$(X+Y)(p)(f)=(X(p)+Y(p))(f)$

$(XY)(p)(f)=X(p)[Y\circ f]-Y(p)[X\circ f]$

where $(Y\circ f)(x)=Y(x)(f).$  This turns $V(M),$ the set of vector fields of $M,$ into a Lie algebra.

Now let $G$ be a Lie group and $\lambda_g:G\to G$ be left multiplication by $g\in G.$  These are certainly diffeomorphisms of $G$

Definition 6.  A vector field $X$ is left-invariant if

$(\lambda_g)_*(X)=X$

for all $g\in G$ where $(\lambda_g)_*(X)(x)=X(gx).$

Proposition 7.  The set $L(G)$ of left-invariant vector fields of $G$ is a Lie subalgebra of $V(G).$

Proof.  Let $X$ and $Y$ be left-invariant and $g\in G$.  Then

$\begin{array}{lcl}(\lambda_g)_*(X+Y)(x)(f)&=&(\lambda_g)_*X(x)(f)+(\lambda_g)_*Y(x)(f)\\&=&X(gx)(f)+Y(gx)(f)\\&=&(X+Y)(gx)(f)\end{array}$

and

$\displaystyle \begin{array}{lcl}(\lambda_g)_*(XY)(x)(f)&=&(\lambda_g)_*\left(X(x)[Y\circ f]-Y(x)[X\circ f]\right)\\&=&X(gx)[Y\circ f]-Y(gx)[X\circ f]\\&=&(XY)(gx)(f)\end{array}.$

The Lie subalgebra $L(G)$ of $V(G)$ is called the Lie algebra of $G.$  Note an obvious consequence is that if $v\in T_1(G),$ $X\in L(G),$ and $X(1)=v.$  Then

$X(g)=(\lambda_g)_*(X)(1).$

Hence it follows that left-invariant vector fields of $G$ are completely determined by where they send the identity element.  In this sense, we obtain the equivalence of the vector spaces $L(G)$ and $T_1(G).$

Lie Group of a Lie Algebra:

The standard convention is to discuss an “inverse” map from the Lie algebra of a Lie group back to the Lie group.  Let $X\in L(G).$  Then this left-invariant vector field corresponds to a tangent vector $v\in T_1(G).$  Let $\gamma_X:\mathbb{R}\to G$ be a path defined by

$\gamma_X(0)=1$

$X(\gamma_X(0))=X(1)=v$

$\gamma_X(t_1+t_2)=\gamma_X(t_1)\gamma_X(t_2).$

Note these uniquely define $\gamma_X,$ called a one-parameter subgroup of $G.$  We in turn define the map $\exp:L(G)\to G$ by

$\exp(X)=\gamma_X(1)$

or more generally

$\exp(tX)=\gamma_X(t).$

Hence $\exp(0)=1.$ It follows that $L(\exp(L(G)))=L(G).$

Suppose $G=GL_n(\mathbb{R}).$  Then $L(G)=M_n(\mathbb{R})$ and

$\displaystyle\exp(X)=\sum_{n=0}^\infty\frac{X^n}{n!}.$

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

[2]  Robert Milson, Thomas Foregger, Mike Fikes. “Lie group” (version 12). PlanetMath.org. Freely available at http://planetmath.org/?op=getobj;from=objects;id=1112