5

I know that the span of the vectors in the tangent space at the identity of a Lie Group is isomorphic to the span of the left-invariant vector fields over the Lie Group. It seems to me (a physicist) so much easier to consider a Lie Algebra by this $T_{e}$ tangent space at the identity, so I'm wondering why groups resources define Lie Algebras using left-invariant vector fields at all?

My only guess is that left-invariant vector fields are required in order to uniquely specify the integral curves used in defining the exponential map?

Alex Gower
  • 717
  • 5
    Is it obvious to you that this tangent space actually has a Lie algebra structure? To me the advantage is that vector fields on a smooth manifolds automatically form a Lie algebra, and the $G$-invariant ones form a Lie subalgebra for free. – Nate Nov 23 '21 at 16:58

1 Answers1

4

To expand out Nate's comment, if we just define a Lie algebra as the tangent space of a Lie group at the identity it isn't clear why it should have the structure of a Lie bracket on it and why other manifolds don't have this property.

So we note that vector fields on a manifold have a Lie bracket structure intrinsically ($XY$ is not a vector field but $[X,Y]=XY-YX$ is where we are treating vector fields as derivations and $XY$ is their composition). Then on a Lie group we can define left invariant vector fields and the Lie bracket on all vector fields gives a natural Lie bracket on this subspace. Then we identify the tangent space at the identity with the space of left invariant vector fields and this equips the tangent space with this Lie bracket.

Callum
  • 4,321
  • So are you saying that we can always define a Lie Algebra on any differentiable manifold by using $[X,Y]$ with all vector fields on the manifold, but that to define a Lie Algebra associated with a Lie Group we must use $[X,Y]$ with only all left-invariant vector fields on the manifold (which is believable since the definition of left-invariant vector fields uses the group action)? – Alex Gower Dec 18 '21 at 11:07
  • 2
    Yes the set of vector fields in any manifold is automatically a Lie algebra (albeit an infinite dimensional one). If the manifold is actually a Lie group then the subspace of left invariant vector fields is a finite dimensional Lie algebra called the Lie algebra of the group (which we can subsequently identify with a single tangent space at the identity). – Callum Dec 18 '21 at 12:43
  • As an extra aside we can associate a group to the infinite dimensional algebra of vector fields. We should be careful here as the infinite dimensional story is not as friendly. The corresponding group is just the diffeomorphism group of the manifold. Thus restricting from vector fields to left invariant vector fields corresponds to restricting from any diffeomorphisms to those in the Lie group itself. – Callum Dec 18 '21 at 12:50