Compact manifold

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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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  • #2
quasar987
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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.
 
  • #3
lavinia
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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?
The tangent unit sphere bundle of a compact manifold is compact. It is not hard to prove this.
 
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thnx friends. it is more clear now.
 
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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
lavinia
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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
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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
 
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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
quasar987
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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
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Ok, yeah, that's pretty obvious now :/ Thanks!
 

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