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


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


(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


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



\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


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




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


or more generally


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


[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). Freely available at;from=objects;id=1112


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