Does trivial cotangent bundle implies trivial tangent bundle?

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In summary: 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.
  • #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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  • #2
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]".
 
  • #3
Yeah, if a matrix is invertible, then so is its inverse. :D

The subtlety only happens when things are infinite-dimensional.
 
  • #4
What happens in that case?
 
  • #5
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).
 
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  • #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.
 

1. What is a trivial cotangent bundle?

A trivial cotangent bundle is a mathematical object that is used in differential geometry to describe the space of all possible tangent vectors at each point on a manifold. It is called "trivial" because it is isomorphic to the product of the manifold with the dual of its tangent space, meaning that all the tangent spaces are essentially the same.

2. What is a trivial tangent bundle?

A trivial tangent bundle is a mathematical object that is used in differential geometry to describe the space of all possible tangent vectors at each point on a manifold. It is called "trivial" because it is isomorphic to the product of the manifold with its tangent space, meaning that all the tangent spaces are essentially the same.

3. What does it mean for a cotangent bundle to imply a tangent bundle?

The statement "a trivial cotangent bundle implies a trivial tangent bundle" means that if the cotangent bundle of a manifold is trivial (isomorphic to the product of the manifold with the dual of its tangent space), then the tangent bundle of the manifold must also be trivial (isomorphic to the product of the manifold with its tangent space). In other words, if the tangent vectors at each point on the manifold are all essentially the same, then the cotangent vectors at each point must also be all essentially the same.

4. What is the significance of a trivial cotangent bundle implying a trivial tangent bundle?

The statement "a trivial cotangent bundle implies a trivial tangent bundle" has significant implications in differential geometry. It means that if we can show that the cotangent bundle of a manifold is trivial, then we automatically know that the tangent bundle is also trivial. This can greatly simplify calculations and proofs in certain situations.

5. Can a trivial cotangent bundle imply a non-trivial tangent bundle?

No, a trivial cotangent bundle cannot imply a non-trivial tangent bundle. This is because the statement "a trivial cotangent bundle implies a trivial tangent bundle" is a logical implication, meaning that if the first statement is true, then the second statement must also be true. If the cotangent bundle is trivial, then the tangent bundle must also be trivial.

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