Eulrt class of normal bundle

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  • #1
lavinia
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I want to make sure of this question: If a compact oriented n manifold is smoothly embedded in another oriented manifold and is homologous to zero as a cycle in this higher dimensional manifold then the Euler class of its normal bundle is zero.

why do i think this?

since M is null homologous as an n cycle, then every closed n form on N, the ambient manifold, integrates to zero on M. by Poincare duality the Thom class of the normal bundle is zero.

so the euler class of the normal bundle of any embedding of a smooth manifold in euclidean space is zero.

yes/no?
 

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  • #2
WWGD
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Hi lavinia, sorry for the necropost; I was doing a search, and I came into this:

I don't know enough to give a definitive answer, but, AFAIK, the Euler class of ## S^{2n+1} ## is 2 :

http://en.wikipedia.org/wiki/Euler_class

Although I don't know if it is assumed there that the spheres are embedded in ## \mathbb R^{2n+2} ## ( or higher);
still, if one is talking about normal bundles of a manifold M , there must some ambient space for this to make sense, and I think this article assumes ## S^{2n+1}## is embedded in ## \mathbb R^{2n+2}## ; there is even reference of nowhere-zero sections, which I imagine live in Euclidean (2n+2)-space.
 
  • #3
lavinia
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Hi lavinia, sorry for the necropost; I was doing a search, and I came into this:

I don't know enough to give a definitive answer, but, AFAIK, the Euler class of ## S^{2n+1} ## is 2 :

http://en.wikipedia.org/wiki/Euler_class

Although I don't know if it is assumed there that the spheres are embedded in ## \mathbb R^{2n+2} ## ( or higher);
still, if one is talking about normal bundles of a manifold M , there must some ambient space for this to make sense, and I think this article assumes ## S^{2n+1}## is embedded in ## \mathbb R^{2n+2}## ; there'll is even reference of nowhere-zero sections, which I imagine live in Euclidean (2n+2)-space.

Hi WWGD

Thanks for writing. I am not sure what you are saying about the sphere. Can you restate your point? BTW: The Euler class of an odd dimensional sphere is zero. Did you mean even dimensional?

I think I now understand the answer to this post.
I
 
  • #4
WWGD
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Actually, sorry, I should learn a bit more about this before commenting; I am still at a rudimentary level, I have not done any serious reading yet.
 
  • #5
lavinia
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Hi WWGD

Thanks for writing. I am not sure what you are saying about the sphere. Can you restate your point? BTW: The Euler class of an odd dimensional sphere is zero. Did you mean even dimensional?

I think I now understand the answer to this post.
I
Where have you got to in your study of characteristic classes? How is your algebraic topology?
 

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