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Does trivial cotangent bundle implies trivial tangent bundle?

  1. Mar 2, 2010 #1

    quasar987

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    If the tangent bundle is trivial, then the cotangent bundle is trivial. To see this, consider (X_i) a global frame for TM. Then define a global frame (\alpha^i) for T*M by setting [itex]\alpha^i(X_j)=\delta_{ij}[/itex] and extend by linearity.

    Does trivial cotangent bundle implies trivial tangent bundle? A similar argument based on global frames does not seem to work in this direction: given a global frame (\alpha^i) for T*M, how do you define a global frame for TM? It does not makse sense to say "Let X_i be the vector field such that [itex]\alpha^i(X_j)=\delta_{ij}[/itex]" because such a vector field might not exist. And if locally, [itex]\alpha^i=\sum_j\alpha^i_jdx^j[/itex], then defining a (global) vector field by setting [itex]X_i:=\sum_j\alpha^i_j\partial_j[/itex] is inconsistent because the coefficients [itex]\alpha^i_j[/itex] do no transform correctly.
     
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  3. Mar 2, 2010 #2

    quasar987

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    I may have spoken too fast. It appears now that it does make sense to say "Let X_i be the vector field such that [itex]\alpha^i(X_j)=\delta_{ij}[/itex]".
     
  4. Mar 2, 2010 #3

    Ben Niehoff

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    Yeah, if a matrix is invertible, then so is its inverse. :D

    The subtlety only happens when things are infinite-dimensional.
     
  5. Mar 2, 2010 #4

    quasar987

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    What happens in that case?
     
  6. Mar 2, 2010 #5

    Ben Niehoff

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    The dual space V* of an infinite-dimensional vector space V can be strictly larger than V, and so they can't be isomorphic. For example, on a space of functions, the dual space includes distributions, which are not in the original space.

    Also, on an infinite-dimensional vector space, a linear operator might have a right inverse, but no left inverse (or vice versa). For example, on an L^2 function space in an interval [a,b], define the derivative operator D as

    [tex]Df = \frac{df}{dx}[/tex]

    and an integral operator J as

    [tex]Jf = \int_a^x f(x') \; dx'[/tex]

    Then D has a right inverse, given by J:

    [tex]DJf = f[/tex]

    but no left inverse, because

    [tex]JDf \neq f[/tex]

    for all possible f (specifically, all the constant functions are mapped to 0 by D, and J maps 0 to 0).
     
    Last edited: Mar 2, 2010
  7. Mar 11, 2010 #6
    In general, you can just use a metric to set up an isomorphism between them, i.e. any nondegenerate pairing < , > on a finite dimensional vector space gives an isomorphism with its dual. The problem of infinite dimensionality of course arises since we're using the equivalence of nondegeneracy and isomorphism.
     
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