Does trivial cotangent bundle implies trivial tangent bundle?

[quasar987]

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 \alpha^i(X_j)=\delta_{ij} 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 \alpha^i(X_j)=\delta_{ij}" because such a vector field might not exist. And if locally, \alpha^i=\sum_j\alpha^i_jdx^j, then defining a (global) vector field by setting X_i:=\sum_j\alpha^i_j\partial_j is inconsistent because the coefficients \alpha^i_j do no transform correctly.

 

[quasar987]

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 \alpha^i(X_j)=\delta_{ij}".

 

[Ben Niehoff]

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

The subtlety only happens when things are infinite-dimensional.

 

[quasar987]

What happens in that case?

 

[Ben Niehoff]

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

Df = \frac{df}{dx}

and an integral operator J as

Jf = \int_a^x f(x') \; dx'

Then D has a right inverse, given by J:

DJf = f

but no left inverse, because

JDf \neq f

for all possible f (specifically, all the constant functions are mapped to 0 by D, and J maps 0 to 0).

 

[zhentil]

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.