Compactness of Tangent Bundle: Manifold M

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SUMMARY

The discussion centers on the compactness of the tangent bundle TM of a compact manifold M. It is established that while the tangent sphere bundle of a compact manifold is compact, the tangent bundle itself is not compact. Specific examples are provided, such as the tangent bundle of the two-sphere being homeomorphic to real projective 3-space (RP^3), while the tangent bundle of the circle is homeomorphic to an open cylinder, confirming its non-compactness. The conversation also touches on the properties of the tangent bundle of the Moebius band, clarifying misconceptions about its dimensionality and structure.

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  • Understanding of manifold theory and compactness
  • Familiarity with tangent bundles and vector bundles
  • Knowledge of real projective spaces (RP^n)
  • Basic concepts of fiber bundles and local trivializations
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Mathematicians, particularly those specializing in differential geometry, topology, and manifold theory, will benefit from this discussion. It is also relevant for students and researchers interested in the properties of tangent bundles and their applications in geometry.

math6
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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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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.
 
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.
 
thnx friends. it is more clear now.
 
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].
 
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.
 
Last edited:
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
 
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?
 
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!
 

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