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Why is the tangent space of a lie group manifold at the origin the lie algebra?

  1. Apr 14, 2009 #1
    Question is in the title. Seems a lot of people throw that statement around as if its obvious, but it isnt obvious to me.

    I can kind of see how it might be true. If you take a group element, differentiate it wrt the group parameters to pull down the generators, and then evaluate this expression at the origin of the group manifold, ie the identity element, you are left with just the generator. So by differentiating a group element at the origin you get a generator. But this is not quite the same as differentiating a curve in the group manifold at the origin and getting a generator.
     
  2. jcsd
  3. Apr 14, 2009 #2
    The bracket of two left invariant vector fields is itself left invariant. Thus the bracket product turns left invariant vector fields into a Lie Algebra. But left invariant vector fields are completely determined by their value at the identity.
     
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