How to calculate the Euler class of a sphere bundle?

1. Dec 11, 2011

kakarotyjn

I have read the section about sphere bundle in Differential Forms in Algebraic Topology,but I still don't understand the Euler class very clear.I don't know how to calculate it for a sphere bundle,for example the sphere bundle of S^2.

And I can't work out the exercise at the end of the section(the exercise said the Euler class for sphere bundle with even dimensional fiber is zero).I always read the book and understand some,but when came to the exercise,I don't have any idea.Could somebody tell me what should I do to learn it better?Thank you!

2. Dec 12, 2011

lavinia

I do not know of a general computation for the Euler class of a sphere bundle. But here are a few things that might help you.

- the Euler class is not defined if the bundle is not orientable.
- the Euler class of a trivial bundle is zero.
- the Euler class of an induced bundle with induced orientation is the pull back of the Euler class.

A good exercise to see how this works is to take a smooth surface embedded in 3 space and look at the Euler class of the bundle induced by the sphere map (Gauss map).

- If the orientation changes sign then the Euler class changes sign. From this you should be able to figure out that the Euler class of an odd dimensional vector bundle is a 2 torsion class. In the case of a n sphere bundle over an orientable n manifold this means that it is actually zero.

- For an n plane bundle over an orientable n manifold I think that the Euler class is Poincare dual to the zero section. This would imply that the Euler class can be calculated from the self intersection number of the zero section.

I do know that if the bundle has a non-zero section then the Euler class is zero. So for instance the tangent bundle of the 2 sphere has non-zero Euler class since its tangent bundle has no non-zero section. The Euler class of the tangent bundle of a torus is zero since it does have a non-zero section.

- For the tangent bundle of an orientable manifold the Euler class is the Euler characteristic multiplied by the fundamental top dimensional cohomology class. Since the Euler characteristic is a combinatorial invariant, it can be computed from a triangulation of the manifold.

- The Euler class of a Whitney sum of two oriented vector bundles is the cup product of their Euler classes. This gives another proof that the tangent bundle of the 2 sphere has non non-zero section. It also shows that if a n plane bundle is the Whitney sum of two odd dimensional subbundles then its Euler class is zero.

- If a vector bundle over a smooth orientable manifold is compatible with a Riemannian metric then the Euler class can be computed from a universal differential form constructed from the curvature 2 form. Generally one needs a metric where the curvature is simple enough to be able to compute the integral. For instance, for the standard 2 sphere of radius 1, the Euler class is (1/2pi)times the volume form. For a Riemannian manifold with a flat metric, the Euler class is zero. So since the torus can be given a flat Riemannian metric, its Euler class is zero.

If the connection is not compatible with a Riemannian metric, then the Euler class may not be computable from the curvature 2 from. There are elementary examples of this.

- If an orientable manifold is the boundary of another manifold then its Euler characteristic is even. This is a good exercise.

There is of course much more of this type of stuff.

Perhaps you could explain to me the derivation of the global angular form for an oriented 2 plane bundle that is described in Bott and Tu.

What about the Euler characteristic of a closed hypersurface of Euclidean space?

Last edited: Dec 12, 2011
3. Dec 12, 2011

zhentil

There are two points to clear up. The first is that the Euler class of a sphere bundle is just the Euler class of the corresponding vector bundle. The second is that the Euler class of the corresponding vector bundle is defined using the Thom isomorphism.

For your other question, the Euler class of a vector bundle with odd-dimensional fibers is zero using the following: the intersection of the zero section with a transverse section represents a homology class. Repeating the construction with the negative of the section also yields a homology class. Because the fibers are odd-dimensional, it represents the negative of the first. However, the sections are homotopic, hence the homology class of the intersection must be the same - i.e. zero. The euler class is Poincare dual to the homology class of the intersection, hence also zero.

4. Dec 13, 2011

kakarotyjn

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$$\frac{d\theta_\alpha}{2\pi}-\pi^* \ksi_\alpha=\frac{d \theta_\beta}{2\pi}-\pi^*\ksi_\beta$$ where $$\theta$$ is angular coordinates and $$\ksi_\alpha$$ is a 1 form on $$U_\alpha$$,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!

5. Dec 13, 2011

lavinia

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.

6. Dec 13, 2011

lavinia

I looked at Bott and Tu and I think I get the idea of the intersection argument.

Map the manifold into the vector bundle so that two things are true: 1)The map is transverse to the zero section: 2) the map is transverse to a tubular neighborhood of the zero section.

In this case, the inverse image of the tube is a tube around the transverse intersection.
So then Thom class of the vector bundle pulls back to the Thom class of the normal bundle to the transverse intersection.

But the pull back of the Thom class is the Euler class by definition so the Euler class is Poincare dual to the transverse intersection.

7. Dec 14, 2011

kakarotyjn

Thank you lavinia! I can get a sketchy understand of the proof.There are some basic knowledge of transverse intersection I don't know.I will look at it.