Fiber bundle homeomorphism with the fiber

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SUMMARY

The discussion centers on the homeomorphism properties of fiber bundles, specifically the relationship between the preimage of a point in the base space and the fiber itself. It establishes that for a fiber bundle defined by a continuous onto map ##\pi## from total space ##E## to base space ##B##, the local trivialization maps ##\varphi: \pi^{-1}(U) \rightarrow U \times F## are homeomorphisms. The participants confirm that the homeomorphism between ##\pi^{-1}(\{p\})## and the fiber ##F## is valid under the subspace topology derived from ##\pi^{-1}(U)##. Furthermore, it is clarified that the homeomorphism condition holds only for specific open neighborhoods in ##B##, not universally.

PREREQUISITES
  • Understanding of fiber bundles and their definitions
  • Knowledge of continuous maps and homeomorphisms in topology
  • Familiarity with local trivialization maps and subspace topology
  • Basic concepts of topological spaces and manifolds
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  • Study the properties of continuous maps in topology, focusing on homeomorphisms
  • Explore the concept of local trivializations in fiber bundles
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Mathematicians, particularly those specializing in topology, differential geometry, and anyone studying fiber bundles and their applications in theoretical frameworks.

  • #31
cianfa72 said:
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.
 
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  • #32
jbergman said:
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.
 
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  • #33
cianfa72 said:
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##.
 
  • #34
martinbn said:
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).
 
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  • #35
cianfa72 said:
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?
 
  • #36
WWGD said:
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.
 
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  • #37
cianfa72 said:
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.
 
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  • #38
cianfa72 said:
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.
 
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  • #39
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).
 
  • #40
cianfa72 said:
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!
 
  • #41
martinbn said:
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.
 
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  • #42
cianfa72 said:
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?
 
  • #43
martinbn said:
No, i meant why do you make these statements/questions?
Just to double check my understanding (also sometimes I've problem with reading english).
 
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