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Does trivial cotangent bundle implies trivial tangent bundle?

by quasar987
Tags: bundle, cotangent, implies, tangent, trivial
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quasar987
#1
Mar2-10, 02:49 PM
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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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quasar987
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Mar2-10, 03:43 PM
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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]".
Ben Niehoff
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Mar2-10, 03:45 PM
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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.

quasar987
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Mar2-10, 04:26 PM
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Does trivial cotangent bundle implies trivial tangent bundle?

What happens in that case?
Ben Niehoff
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Mar2-10, 05:23 PM
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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).
zhentil
#6
Mar11-10, 11:26 AM
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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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