I Fiber bundle homeomorphism with the fiber

[WWGD]

A general question on tangent and cotangent bundle. Are they always trivial bundle regardless of the base space topological manifold ?

They're only guanteed to be locally , only at times globally trivial. IIRC, the latter is true only for product spaces. The term " Trivial Bundle" wouldn't have been coined.

 

[cianfa72]

Have you systematically worked through a book like Tu's Introduction to Manifolds?

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

 

[martinbn]

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

Isomorphisms of the bundles not $M$.

 

[cianfa72]

Isomorphisms of the bundles not $M$.

However, in general, bundles do not have a vector space structure.

Edit: eventually the term isomorphism is not restricted to just vector spaces (e.g. in the context of differentiable manifolds a diffeomorphism is actually an isomorphism).

 

[WWGD]

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

Maybe the manifolds themselves only have a topological structure, not a differentiable one?

 

[cianfa72]

Maybe the manifolds themselves only have a topological structure, not a differentiable one?

Yes, in that case (topological manifolds alone) the isomorphism is actually an homeomorphism.

Btw up to dimension 3, every topological manifold admits a differentiable structure and all these structures are equivalent.

 

[jbergman]

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

The cylinder and the mobius strip are two different vector bundles over the base manifold of the circle. The cylinder is a trivial bundle and the Mobius strip is not. They are not isomorphic. I would work through Lee's chapter on vector bundles because it goes more in depth.

Basically, different vector bundles can be twisted in different ways over the base space, but locally it is like a product manifold. I think you need to build up intuition as to what twisting means here. Essentially it means that our patches above open sets on M where we can define local trivializations are different. With the cylinder we can define a trivialization over the open set U which is the entire circle. With the Mobius strip we can only define trivializations on patches above proper subsets of the circle.

 

[WWGD]

Yes, in that case (topological manifolds alone) the isomorphism is actually an homeomorphism.

Btw up to dimension 3, every topological manifold admits a differentiable structure and all these structures are equivalent.

You may want to look up some more into Stieffel-Whitney classes.

 

[cianfa72]

As far as I can tell, the topology assigned on the tangent bundle is not the same as the initial topology from the canonical projection map $\pi$ on the first factor (see Tu's book section 12.1).

 

[martinbn]

As far as I can tell, the topology assigned on the tangent bundle is not the same as the initial topology from the canonical projection map $\pi$ on the first factor (see Tu's book section 12.1).

Yes, the initial topology will contain whole fibers. The bundle topology also has open subsets of fibers.

Why!

 

[cianfa72]

Why!

Yes, open sets in initial topology contain whole fibers, however each fiber $\pi^{-1} \{p \}$ will be open in the initial topology iff the base space B has the discrete topology.

 

[martinbn]

Yes, open sets in initial topology contain whole fibers, however each fiber $\pi^{-1} \{p \}$ will be open in the initial topology iff the base space B has the discrete topology.

No, i meant why do you make these statements/questions?

 

[cianfa72]

No, i meant why do you make these statements/questions?

Just to double check my understanding (also sometimes I've problem with reading english).

 

[fresh_42]

Let's leave it for now as it becomes quite general. A Google search on "introduction to vector bundles" returned good results, and for those who like it fast, see
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/#Vector-Bundles

Thank you all for your participation.

This thread is closed.