Global diffeomorphism with tangent bundle

[Monocles]

I am terribly confused on the issue of trivial tangent bundles. I understand intuitively why some tangent bundles are trivial and others are not, but I'm having trouble figuring out how to show it.

Even the most trivial example, show that T\mathbb{R}^n is diffeomorphic to \mathbb{R}^{2n} I am not seeing how to show. Showing that they are locally diffeomorphic is very easy, but every tangent bundle is locally diffeomorphic to the product space of the manifold with the appropriate Euclidean space. I am new to this topic so a geometrical route is preferred. For example, I know that if there exists a vector field with no zero vectors then the tangent bundle is trivial, but I don't know how to show why that is true, so that result does not help me.

Thanks!

 

[arkajad]

<br /> \mathbb{R}^n<br />

has global coordinates. Use them to define natural global coordinates for the tangent bundle. Then notice that they define a global trivialization.

 

[Monocles]

Thanks, that gave me an idea. So, if I want to show that the tangent bundle on TS^1 is trivial, should I just find an atlas for the bundle, show that the coordinate charts are locally diffeomorphic to S^1 \times \mathbb{R}, and then show that the transition functions between the coordinate charts are smooth? I feel like that must still be missing a step somewhere, since the transition functions are smooth by definition.

 

[arkajad]

To prove that the tangent bundle of an n-dimensional manifold is trivial (and to find its trivialization) it is enough to find n vector fields that are linearly independent at every point. For the circle it is enough to find just one nowhere vanishing vector field (which should be easy).

Once you convince yourself that this is the case - it will help you in the future.

 

[lavinia]

I am terribly confused on the issue of trivial tangent bundles. I understand intuitively why some tangent bundles are trivial and others are not, but I'm having trouble figuring out how to show it.

Even the most trivial example, show that T\mathbb{R}^n is diffeomorphic to \mathbb{R}^{2n} I am not seeing how to show. Showing that they are locally diffeomorphic is very easy, but every tangent bundle is locally diffeomorphic to the product space of the manifold with the appropriate Euclidean space. I am new to this topic so a geometrical route is preferred. For example, I know that if there exists a vector field with no zero vectors then the tangent bundle is trivial, but I don't know how to show why that is true, so that result does not help me.

Thanks!

It is not true that a single vector field without zeros means that the tangent bundle is trivial. This is only true for oriented surfaces and for 1 manifolds like the circle. For instance the Klein bottle has a non-zero vector field but it's tangent bundle is not trivial.

I think that triviality formally means that there is a homeomorphism from the vector bundle into BxF ,where B is the base space of the bundle and F is a vector space, that is linear on each fiber and covers the identity map on B.

 

[arkajad]

It is not true that a single vector field without zeros means that the tangent bundle is trivial.

Of course. You need, as I wrote, n linearly independent vector fields.

This is only true for oriented surfaces and for 1 manifolds like the circle. For instance the Klein bottle has a non-zero vector field but it's tangent bundle is not trivial.

Of course, because Klein's bottle is not 1-dimensional.
A simple and useful http://www.math.uchicago.edu/~womp/2001/vbex.pdf" :

Exercise 3. Show that an n-dimensional vector bundle E -> M is trivial if and only if there are n sections s1,..., sn which, in each fiber, are linearly independent. Show that all bundles have local systems of n linearly independent sections.​

 

[lavinia]

Of course. You need, as I wrote, n linearly independent vector fields.
Of course, because Klein's bottle is not 1-dimensional.
A simple and useful http://www.math.uchicago.edu/~womp/2001/vbex.pdf" :

Exercise 3. Show that an n-dimensional vector bundle E -> M is trivial if and only if there are n sections s1,..., sn which, in each fiber, are linearly independent. Show that all bundles have local systems of n linearly independent sections.​

I was just responding to the direct words of the writer and giving an example - not disagreeing with you or for that matter not not understanding the definition of trivial. Perhaps you can explain what you are getting at.

BTW: For an oriented surface (Riemannian manifold) it suffices to have a single non-zero vector field. The orientation provides the other.

 

[arkajad]

BTW: For an oriented surface (Riemannian manifold) it suffices to have a single non-zero vector field. The orientation provides the other.

Right.