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Trivial Tangent Bundles

  1. Apr 6, 2009 #1
    In trying to understand why not all tangent bundles are trivial, I've attempted to prove that they are all trivial and see where things go wrong. Unfortunately, I finished the proof and cannot find my mistake. Here it is:

    Let M be an n-manifold with coordinate charts [tex](U_\alpha, \phi_\alpha)[/tex]. Therefore [tex](\pi^{-1}(U_\alpha), \tilde{\phi_\alpha})[/tex] are charts for TM where [tex]\pi[/tex] is the projection map and [tex]\tilde{\phi_\alpha}(p, v^i \frac{\partial}{\partial x^i}\vert_p) = (\phi(p),v^1, \ldots, v^n)[/tex]. I claim that the map F from TM to M x R^n given by [tex]F(p, v^i \frac{\partial}{\partial x^i}\vert_p) = (p, v^1, \ldots, v^n)[/tex] is a diffeomorphism. Clearly F is bijective so it is sufficient to check that F is a local diffeomorphism. Thus let [tex](p, v^i \frac{\partial}{\partial x^i}\vert_p)[/tex] be an arbitrary point in TM. [tex]p \in U_\alpha[/tex] for some [tex]\alpha[/tex] so [tex]\pi^{-1}(U_\alpha)[/tex] is an open set (indeed a chart) of TM containing [tex](p, v^i \frac{\partial}{\partial x^i}\vert_p)[/tex] and [tex]F(\pi^{-1}(U_\alpha)) = U_\alpha \times R^n[/tex] is a chart of M x R^n. But the restriction of F to [tex]\pi^{-1}(U_\alpha)[/tex] is [tex](\phi_\alpha^{-1} \times Id_{R^n}) \circ \tilde{\phi_\alpha}[/tex], which is a diffeomorphism (being a composition of diffeomorphisms).

    Where did I go wrong?
     
  2. jcsd
  3. Apr 6, 2009 #2
    The basic idea is that when you write
    [tex](p, v^i \frac{\partial}{\partial x^i}\vert_p)[/tex]
    you've already chosen an open set which is diffeomorphic to an open set in Euclidean space. Thus if your manifold can be covered by one coordinate chart, your proof is entirely correct (though manifolds with this property are not very interesting).

    Here's an idea of what can go wrong: consider [tex]S^2[/tex] minus the north and south poles. Put spherical coordinates on it, and consider the standard basis for the tangent space in spherical coordinates. What happens if you try to extend these vector fields to the north and south poles?
     
  4. Apr 6, 2009 #3
    Thanks for the quick reply! I'm pretty sure I get what you're saying: I'm basically defining the "diffeomorphism" locally and when I go to show that it is a local diffeomorphism, I'm using a specific chart for a given point but for other points in the chart, the function may have been defined using a different chart. Like in the case of the sphere, if I use spherical coordinates for everything but the poles, then whatever charts I use with the poles will overlap with the spherical coordinates.
     
  5. Apr 6, 2009 #4
    Not really. When you choose local coordinates, you are trivializing the tangent bundle over that open set - diffeomorphism induces isomorphism on the tangent bundle. But this diffeomorphism (and hence the bijection you referred to) are local in character - when you write down a basis for the tangent space, it's not guaranteed that the basis will extend to the entire manifold - in the sphere example, there's no way to extend the vector fields beyond the coordinate chart, which is the basic obstruction to trivializing the tangent bundle.
     
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