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Elementary questions about fibre bundles

  1. Oct 11, 2008 #1

    haushofer

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    Hi, I'm a little stuck on Nakahara's treatment about fibre bundles. I hope someone can give me a clear answer on this; they are quite elementary questions, I guess.

    We have:

    * A principal bundle P(M,G)
    * A fibre [itex]G_{p}[/itex] at [itex]p= \pi(u)[/itex]

    Then the vertical subspace [itex]V_{u}P[/itex] is defined as a subspace of the tangent space [itex]T_{u}P[/itex] which is tangent to the fibre [itex]G_{p}[/itex] at u. An element [itex]X[/itex] of this vertical subspace is called the fundamental vector field, and is defined as

    [tex]
    X f(u) = \frac{d}{dt}f(ue^{tA})|_{t=0}
    [/tex]

    in analogy with tangent vectors on a manifold M. Here f is a curve from P to the real line and A is an element of your left invariant vector field on G. It is then asked to show that (ex. 10.1)

    [tex]
    \pi_{*}X = 0
    [/tex]

    for a fundamental vector field X. I don't see this, and it's probably because I don't grasp the meaning of this vertical space. Is it only tangent to the fibre but not to the base manifold M? Is that the reason for the pushforward vector (which is defined on M) is zero (because vectors on M are only defined in the tangent space [itex]T_{p}M[/itex])?

    It is further claimed on page 378 that if you act with the Ehresmann connection on a fundamental vector field, you can use

    [tex]
    d_{P}g_{i} = \frac{dg(ue^{tA})}{dt}|_{t=0}
    [/tex]

    where d is the exterior derivative on P. Where does this come from? I hope my questions are clear and that someone can help me with this.
     
  2. jcsd
  3. Oct 11, 2008 #2

    haushofer

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    So my first question can be put as: why is

    [tex]
    T_{u}(P_{\pi(u)}) = ker[\pi_{*}|_{u}:T_{u}P \rightarrow T_{\pi(u)}M]
    [/tex]

    ?

    Another question which rises then is why the intersection of the horizontal subspace and the vertical subspace is zero.
     
  4. Oct 13, 2008 #3

    haushofer

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    No-one?
     
  5. Oct 13, 2008 #4
    The action of the group on the bundle preserves the fibers. A fundamental vector field corresponds to (the derivative of) such an action and the vector field is along the fibers. It follows that the projection to the base manifold tangent space of a fundamental vector field is zero.

    For your second question, an Ehresmann connection [itex]\omega[/itex] fulfills
    [tex]\omega (\sigma (X)) = X[/tex]
    where X is in g and [itex]\sigma (X)[/itex] is a fundamental vector field. Since no fundamental vector field is in the kernel of [itex]\omega[/itex], the vertical and horizontal vector spaces do not intersect.
     
  6. Oct 15, 2008 #5

    haushofer

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    I think I'm missing something very trivial here, but why is this projection then zero? Why can't a vector in the bundle lie in the tangent space of the fiber AND the tangent space of the basis manifold?
     
  7. Oct 15, 2008 #6
    Let [itex]\gamma(t)[/itex] be an integral curve of a fundamental vector field. The whole curve is then contained in a single fiber [itex]F_p[/itex]. The image of the curve when projecting down to the base manifold is a point [itex]\pi(\gamma)\subset \pi(F_p)=p[/itex]. A velocity vector of the curve projects down to a veclocity vector of the point. But the point does not move and the velocity vector is zero.

    [tex]\tilde{\gamma}(t) = \pi(\gamma(t)) = p[/tex]
    [tex]\pi^*(\gamma'(t)) = \tilde{\gamma}'(t) = \frac{d}{dt}\, p = 0[/tex]
     
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