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Math Help - Tangent bundle of a Lie Group is trivial

  1. #1
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    Tangent bundle of a Lie Group is trivial

    Hello,

    Let G be a Lie Group and g his Lie Algebra, that is the tangent space at e \in G I have to show that the map \phi: G x g -> TG, \phi(h,X)=dL_h[X], (whereas
    dL_h[X](f)=X(f \circ L_h) with L_h left multiplication )
    is a diffeomorphism and a linear isomorphism on each fibre.

    I'm really hopeless! I don't know how i can show this property's of \phi.

    Do you know some books or links, where i can read the proof? Or can you please explain it to me?

    Regards
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  2. #2
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    Since from construction the tangent bundle is locally diffeomorphic to the product of the base manifold G and the identity fiber g, you need only to show that the map is globally one to one. This is easily verified by the fact that dL_h \cdot  dL_{h^{-1}} = id.
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  3. #3
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    Hello,

    thank you for your help! I could show that the map is bijective. But i don't see why it is diffeomorphic, even locally diffeomorphic?

    Now i have to show that \phi is a linear isomorphism on each fiber.
    I suppose that fiber doesn't mean \phi^{-1}(V_h), because \phi is bijective. So what does fiber means here?


    Regards
    Last edited by Sogan; January 17th 2011 at 02:14 PM.
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  4. #4
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    A bundle is DEFINED to be locally trivial, that is, locally diffeomorphic to the product of the base space and any of the fiber spaces( since all the fibers are isomorphic). In your case, the base space is G and the fiber at identity is g, the tangent plane. So \phi is DEFINED to be locally diffeomorphic.
    To see it is a isomorphism between two fibers T_a and T_b, you only need to show that dL_a * dL_b^{-1} is an isomorphism. Intuitively, you can always translate a vector from T_b to g( the tangent plane at identity), then translate to T_a.
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