Compactness of Tangent Bundle: Manifold M

In summary, the conversation discusses the compactness of the tangent bundle and whether it is necessarily compact if the underlying manifold is compact. It is mentioned that the tangent unit sphere bundle of a compact manifold is always compact, but the tangent bundle itself is never compact. A couple of counterexamples are given to support this claim.
  • #1
math6
67
0
hello friends :smile:
I have a question about the compactness of the tangent bundle: assume that the manifold M is compact, does it make necessarily TM compact ? if not TM, a submanifold of TM (precisely a submanifold of vector norm equal to 1) can be compact?
 
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  • #2
TM is basically one copy of R^n glued on each point of M. Since R^n is not compact, how can TM be? (Find a rigorous version of this argument).

I'm not sure what you're asking in your second question.
 
  • #3
math6 said:
hello friends :smile:
I have a question about the compactness of the tangent bundle: assume that the manifold M is compact, does it make necessarily TM compact ? if not TM, a submanifold of TM (precisely a submanifold of vector norm equal to 1) can be compact?

The tangent unit sphere bundle of a compact manifold is compact. It is not hard to prove this.
 
  • #4
thnx friends. it is more clear now.
 
  • #5
Indeed, Lavinia is correct, the tangent bundle of the two sphere is real projective 3 space, which is compact. However, obviously the tangent bundle of the circle is homeomorphic to the open cylinder (this is easy to see) and hence is not compact.

It doesn't seem that there is any relation to the compactness of a manifold and its tangent bundle (unless its tangent bundle is compact iff its tangent bundle is non-trivial. This I highly doubt to be true).[edit- the tangent bundle of the moebius band would be the moebius band again wouldn't it? So this would be a counter-example to this obviously wrong claim].
 
  • #6
Jamma said:
Indeed, Lavinia is correct, the tangent bundle of the two sphere is real projective 3 space, which is compact. However, obviously the tangent bundle of the circle is homeomorphic to the open cylinder (this is easy to see) and hence is not compact.

It doesn't seem that there is any relation to the compactness of a manifold and its tangent bundle (unless its tangent bundle is compact iff its tangent bundle is non-trivial. This I highly doubt to be true).[edit- the tangent bundle of the moebius band would be the moebius band again wouldn't it? So this would be a counter-example to this obviously wrong claim].

The tangent bundle is never compact. The tangent sphere bundle of a compact manifold is always compact.

The tangent bundle of the 2 sphere is not RP^3. The tangent circle bundle is RP^3.

The tangent bundle to the Moebius band is not the Moebius band. It is a 4 dimensional manifold. The Moebius band is a surface.
 
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  • #7
Wow, I feel mega stupid.

Of course, you are right. I think that I read somewhere that the tangent sphere bundle of a sphere is RP^3 and my mind left out the "sphere" part (I must have somehow convinced myself that the bundle twists in such a way that it becomes non-compact :S). And the tangent bundle of a manifold of dimension n is a 2n dimensional manifold, so God knows where the Moebius band thing came from- sorry about that! :D
 
  • #8
Quasar, could you give a more rigourous argument to what you mentioned? Obviously because the tangent space is a locally trivial fibre bundle, you can take a local trivialisation, cover that in a canonical way and finish it off to an open covering of TM which has no finite subcovering. Is this the method you had in mind?
 
  • #9
Not really. Simply the fact that in the tangent bundle (or in any vector bundle for that matter) TM-->M, the fiber over a point is
a) closed
b) homeomorphic to R^n
Assume TM were compact and derive a contradiction.
 
  • #10
Ok, yeah, that's pretty obvious now :/ Thanks!
 

1. What is the definition of compactness in the context of a tangent bundle on a manifold?

The compactness of a tangent bundle on a manifold refers to the property of the bundle being finite-dimensional and having a “nice” topology that allows for smooth transition functions between tangent spaces. In simpler terms, it means that the tangent bundle is well-behaved and can be easily studied and manipulated.

2. Why is the compactness of a tangent bundle important in mathematical and scientific research?

The compactness of a tangent bundle is important because it allows for a more structured and systematic approach to studying the manifold. This can lead to a better understanding of the properties and behavior of the manifold, which can then be applied to various fields such as physics, engineering, and computer science.

3. How is the compactness of a tangent bundle related to the concept of smoothness?

The compactness of a tangent bundle is closely related to the concept of smoothness. In fact, compactness is a necessary condition for a manifold to be considered smooth. This means that a manifold must have a compact tangent bundle in order to be considered a smooth, well-behaved mathematical object.

4. Can a tangent bundle on a manifold be non-compact?

Yes, a tangent bundle on a manifold can be non-compact. This means that the bundle is infinite-dimensional and does not have a nice topology that allows for smooth transition functions. Non-compact tangent bundles can still be studied and used in mathematical and scientific research, but they may be more difficult to work with compared to compact tangent bundles.

5. Are there any practical applications of the concept of compactness in tangent bundles on manifolds?

Yes, there are many practical applications of compactness in tangent bundles on manifolds. For example, compactness allows for the study of differential equations and geometric structures on manifolds, which can be applied to fields such as robotics, computer graphics, and image processing. Additionally, compact tangent bundles play a crucial role in the theory of fiber bundles, which has applications in physics and engineering.

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