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Dec13-11, 06:10 AM
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Quote Quote by kakarotyjn View Post
Thank you lavinia and zhentil!

To lavinia,I'm not very clear about some property of Euler class you listed,my foundation of topology is quite insufficient.The derivation of the global angular form for an oriented 2 plane bundle :by[tex]\frac{d\theta_\alpha}{2\pi}-\pi^* \ksi_\alpha=\frac{d \theta_\beta}{2\pi}-\pi^*\ksi_\beta[/tex] where [tex]\theta[/tex] is angular coordinates and [tex]\ksi_\alpha[/tex] is a 1 form on [tex]U_\alpha[/tex],these forms piece together to give a global angular form on E^0.

To zhentil,oh I'm sorry there are too many thing I don't know.what is transverse section?Why the intersection of the zero section with a transverse section represents a homology class?What is negative of the section ?

Thank you again!
Thanks. I will read this through in the book.

I don't totally understand the intersection argument but I think it is related to this.

The manifold is embedded in the vector bundle as the set of zero vectors in each fiber. Its normal bundle( as an embedded submanifold) is just the vector bundle itself. Thus the Thom class of the normal bundle,i.e. the differential form that integrates to 1 along each fiber of the normal bundle and which is zero outside of a tubular neighborhood of the manifold (as described in Bott and Tu), is just the Thom class of the vector bundle. Thus the manifold viewed as a cycle in the vector bundle is Poincare dual to the Thom class of the vector bundle.

The pull back of the Thom class to the manifold is the Euler class of the bundle. Its Poincare dual in the homology of the manifold ( not the vector bundle) is the intersection that Zhentil describes. I am not sure how to prove that the intersection is the Poincare dual. Let's think it through.

The Thom isomorphism is key to defining Euler classes topologically. The properties that I described such as naturality and the Whitney sum formula fall out of it easily. You would benefit from leaning about it. There is also a Thom isomorphism for unorientable vector bundles. In this case differential forms can not be used since the cohomology uses Z/2Z coefficients. Nevertheless the construction is much the same.