Fiber Bundle Basics: An Overview of Penrose's Twisted Bundles

  • #31
quasar987 said:
I'm trying to read another meaning into what you wrote, but it really looks like you're saying that every rank 1 vector bundle is trivial, which is of course false.

What are you saying here?

I have to think about it. Quite possible I was wrong. E\otimes E is then trivial.
 

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  • #32
Rasalhague said:
(By the way, Penrose is talking about a slightly different pair of shapes, where F is the R1 vector space, rather than just a finite interval of it.)
I haven't followed the thread so I don't know if it's relevant, but topologically R1 and a finite (open) interval are homeomorphic spaces.
 
  • #33
@quasar987
Thanks. Now I remember: the story was that in real vector bundles there is always a global cross section, but not that it is nowhere vanishing. Next time I will check better before posting anything.
 
  • #34
arkajad said:
I have to think about it. Quite possible I was wrong. E\otimes E is then trivial.

Well, the topic of the thread is the Mobius bundle, which is a nontrivial rank 1 vector bundle over the circle. Also, the tautological line bundle over RP^n is another example of a nontrivial rank 1 bundle over RP^n that has also been mentioned by lavinia earlier.

I do not see the relevance of the lemma you quote. It talks about the possibility of extending globally a locally extendable section defined on a closed set. But it does not say that the extension will be nonvanishing outside of the closed set S...so I don't see why it is relevant.
 
  • #35
arkajad said:
@quasar987
Thanks. Now I remember: the story was that in real vector bundles there is always a global cross section, but not that it is nowhere vanishing. Next time I will check better before posting anything.

No problem! I am too often guilty of that myself. :-p
 
  • #36
If I understand this characteristic class stuff, then it also seems that the tangent bundle of the 2 sphere can not have a 1 dimensional subbundle. For if so the tangent bundle would decompose into a Whitney sum of two line bundles and each would have zero Euler class because the sphere is simply connected. The Whitney sum formula for the Euler class would then imply that the Euler class of the 2 sphere is also zero which can not be because its Euler characteristic is 2.

More generally from the same kind of reasoning, it would seem that the tangent bundle of an even dimensional sphere does not have any proper subbundle.

I think that Arkajad was thinking that the Whitney sum of the Mobius line bundle over the circle is trivial. In this bundle, parallel translation around the circle brings a vector back to its negative. So one can not get a section through parallel translation, I guess. However, if one allows the vector to rotate 180 degrees as one moves around the circle once, you get a section. What does this mean about the bundle?
 

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