# Formal Computations of Euler Classes

1. Oct 2, 2014

### WWGD

Hi All,
The Euler class of the tangent bundle of a compact, oriented manifold agrees with the evaluation of the top homology class on the fundamental class (which is represented by the manifold itself), and maybe also figure out how to do actual computations using Poincare duality (to figure out the top cohomology using 0-th homology).
I am trying to both find out the formal/actual evaluation here, say for the case of the spheres S^{2n} bring together different interpretations/perspectives of the Euler class of a vector bundle , i.e., so that , for the tangent bundle of a manifold M, its Euler class is the Poincare dual of the zero-set of a generic section. Also, if possible, we can consider that the odd-dimensional spheres have E(S^{2n+1})=0 , so that the Poincare dual of a generic section is the 0-class.

Let's fix the integers $\mathbb Z$ as coefficient group.

So, first, say for M=S^n , how do we do the evaluation? By Poincare duality, $H^n = H_0$ ( with = meaning up to isomorphism ). So , since S^n is path-connected, $H_0( S^n) = \mathbb Z = H^n$ and we must evaluate this at the fundamental class of $S^n$ , which is generated by S^n itself. How is this done?

Sorry if this is too broad; any ideas welcome.

2. Oct 2, 2014

### homeomorphic

That doesn't quite make sense. The fundamental class is a homology class, so it gets eaten by cohomology classes. If you feed the fundamental class to the Euler class, it spits out the Euler characteristic. I think all you need to do is compute the Euler characteristic.

So we have <EulerClass, fundamental class> = Euler characteristic. But the fundamental class generates the top homology (by Poincare duality, for a connected manifold), so we should be okay, since we know what the Euler class will do to any top dimensional homology class. Adding boundaries won't change anything by well-definedness of the Kronecker pairing, so we also know what it does to cochains, so we should be done.

I'm a little fuzzy on the details of why the Euler class evaluates to the Euler characteristic on the fundamental class. Seems to be related to the Poincare-Hopf theorem and the fact the that the Euler class is an obstruction to the existence of a nowhere vanishing vector field.

For general vector bundles, I don't know how you go about computing the Euler characteristic.

3. Oct 3, 2014

### WWGD

That is actually the correct answer. How does the evaluation of the top cohomology class (which is Poincare dual to H_0) on the fundamental class result in the Euler characteristic? How do we even Poincare-dualize the homology class H_0 , which
here is $\mathbb Z$. I understand this is ultimately linear algebra, but I am a bit stumped by the technicalities. Thanks.

Last edited: Oct 3, 2014
4. Oct 3, 2014

### lavinia

On a smooth manifold the Euler class of a smooth oriented vector bundle is the pullback of the Poincare dual to the zero section . One may think of it as the pullback of the Thom class of the normal bundle to the zero section(see Bott and Tu). If the bundle is an n-plane bundle, for instance the tangent bundle, the Euler class is a n dimensional cohomology class and may be evaluated on the fundamental class of the manifold.

For an n dimensional bundle, if one perturbs the zero section slightly to get a manifold that is transverse to it, the intersection number of this manifold with the zero section.is the value of the Euler class on the fundamental cycle. Note that the transverse manifold is just a vector field - not necessarily tangent-with isolated zeros so that the intersection number is the index of this vector field. Also note that this works for any n dimensional vector bundle not just the tangent bundle.

If there is a non zero section then the index is zero so the value of the Euler class on the fundamental cycle is zero.

As a zero dimensional homology cycle the sum of the zeros of the vector field times their indices is Poincare dual to the Euler class. For two vector fields with isolated zeros, these cycles are homologous.

In particular given a triangulation of the manifold, one can find a tangent vector field that has index +1 on even dimensional simplices and index -1 on odd dimensional simplices. Thus the sum of the indices of a tangent vector field is the alternating sum of the number of simplices in each dimension.( see Steenrod The Topology of Fiber Bundles). This is the Euler characteristic of the manifold.

On a Riemannian manifold the Euler class of the tangent bundle may be represented by a differential form in the curvature 2 form. Its integral over the manifold is the Euler characteristc of the manifold. On surfaces, the integral can be computed using the Gauss Bonnet theorem. One can also compute it from a section of the unit circle bundle(a unit vector field with isolated singularities) by integrating the curvature 2 form over its image in the unit circle bundle. Using Stokes Theorem this integral reduces to calculating the index of the vector field. This method generalizes to arbitrary dimensions. I think there is a paper of Chern that describes this.

So in principal,the Euler class of a compact Riemannian manifold can always be evauated as the integral of a canonical differential form and one can show that this integral directly computes the index of a vector field with isolated zeros.

- If one knows the Z homology of a manifold,one can compute the Euler characteristic as the alternating sum of the ranks of the homology groups. For spheres this gives 2 for even dimensions, 0 for odd dimensions.

By Poincare duality, every odd dimensional compact oriented manifold without boundary has Euler characteristic 0. In terms of the Euler class, this follows because the Euler class of any odd dimensional oriented vector bundle is 2 torsion (its double is cohomologous to zero) , this because multiplication of each fiber by -1 is an orientation reversing bundle automorphism. But the Z cohomology of an oriented compact manifold without boundary is isomorphic to the integers.

-

Last edited: Oct 4, 2014
5. Oct 4, 2014

### WWGD

Thanks, both, I am trying to work out a specific example, that of the Mobius bundle, using/constructing the associated Thom space . So , here each fiber is a copy of the Reals, so that fibers compactify into an $S^1$. But we then glue
them together using a flip ( as we would do if we had a fundamental domain for the Mobius band ). So this seems like
we cut open an $S^1 \times S^1$ meridinoally (i.e., along a great circle going through both North- and south- poles, or along an inclusion of {pt} $\times S^1$ into $S^1 \times S^1$), twist one of the bounding circles and glue it to the other circle, i.e., we are gluing two copies of $S^1$ with opposite orientations. This reminds me of $\mathbb RP^2$ (right?), but then I can't see how to glue all the new "$\infty$'s " into a single space, as defined in.

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

Any ideas, here?

PS: I can see how to evaluate cochains on chains in DeRham cohomology; this is just integration of forms; I think this is also straight-forward in Simplicial, but I cant see how to do it in the above case. I know it is a matter of playing around with different types of Hom functors, but I still have trouble seeing it.

PPS: Hope I am not overloading this post (I will start a new one if necessary), but I wonder if we can show that the Mobius band is not orientable ( as a manifold, not as a bundle over the circle) by using the fact that the gluing map is just multiplication by -1 , and then the map induced on cohomology would take a class in $T_p^{*}$ to its negative; I know this needs work, but is it on the right track?

Last edited: Oct 4, 2014
6. Oct 4, 2014

### lavinia

WWGD I am not sure what you are trying to do with the Mobius bundle so let me know if this is off base.

Your idea about cutting a torus then attaching with a reflection around the vertical axis is correct. This compactifies the fibers of the Mobius band and produces the Klein bottle - not the projective plane.

But this Klein bottle is not the Thom space of the Mobius bundle because the Thom space identifies all of the end points of all of the fibers to a single point. This can be done by attaching a closed disk to the bounding circle of the Mobius band. The result is the projective plane. To see this more clearly,think about making a projective plane from a sphere by modding out by the antipodal map. A cylindrical strip around the equator becomes a Mobius band and the two polar caps become a single disk.

The Mobius bundle is not orientable so its Euler class is not defined. But it is orientable mod 2 and thus has a mod 2 analogue of the Euler class. It is called the first Stiefel Whitney class of the bundle and is a mod 2 cohomology class.

7. Oct 5, 2014

### WWGD

Thanks, lavinia, I am basically trying to find my way around characteristic classes, starting with Euler classes; please correct anything that doesn't make sense in my posts.

Actually, is it feasible to work out the Thom isomorphism here somewhat-explicitly, for the Mobius band, of course with coefficients in Z/2?

Last edited: Oct 5, 2014
8. Oct 5, 2014

### lavinia

Absolutely. Try it. Notice that the Z2 cohomology of the projective plane is Z2 in dimensions 0 1and 2 and zero in higher dimensions.

The cup product of the one dimensional class with itself is the 2 dimensional class. You can see this from the intersection of the equatorial circle with a transverse perturbation.

You should also work out the exact mod 2 cohomology sequence of the pair, ( Mobius band,bounding circle).

Last edited: Oct 5, 2014
9. Oct 5, 2014

### lavinia

Try computing the Euler characteristic of the 2 sphere using the Thom isomorphism and the exact sequence of the pair, ( Tangent disk bundle,tangent circle bundle).

You will need the topology of the tangent circle bundle. Show that it is homeomorphic to the 3 dimensional real projective space.

10. Oct 10, 2014

### WWGD

Thanks, lavinia, I will give a shot in a few days.

11. Oct 10, 2014

### lavinia

Do not hesitate to ask more questions. I will guide you through it.

12. Oct 14, 2014

### WWGD

Thanks again; I fell behind in my class work, but I'll let you know after I catch up.

13. Oct 16, 2014

### WWGD

Hi, Lavinia , let me see if I have the right notion, at least informally : say we have a fiber bundle \pi: E-->B . Then
we consider B embedded as the 0 section S in E. Then we perturb this 0 section into S' (meaning, I think, that
we select a vector field V on S, and we map s-->v(s) . Now this set {v(s)} represents a homology class h in E.
The Euler class of the bundle \pi is the Poincare dual PD[ {v(s)}]. If we can define a nowhere-zero vector field v(s) on S , then we can push S of itself, and then the euler class is zero. Am I on the right path?

WWGD: What Would Gauss Do?

14. Oct 17, 2014

### lavinia

Yes you have the right idea.

A few things to take further

- The vector field cannot be arbitrary. It must have isolated zeros. Otherwise its intersection with the manifold (viewed as the zero section of the bundle) will not be zero dimensional. Further, as a submanifold of the vector bundle, it must be transverse to the zero section. (It is a theorem that a small perturbation of the zero section will be transverse).

- The zero dimensional class depends on two orientations, the orientation of the bundle and the orientation of the manifold. If the bundle is the tangent bundle,then they are the same.

- You should really learn first about the Thom class of an oriented vector bundle. It leads directly to the Euler class and its Poincare dual. The Thom class is a cohomology class of the vector bundle with compact supports along the fibers and can be defined as the unique differential form that integrates to exactly 1 on each fiber.Its pullback by the zero section is the Euler class.

- The Thom class of the normal bundle of an embedded oriented (closed) submanifold is Poincare dual to the submanifold.

The zero section of a vector bundle over a manifold is an embedded submanifold (of the vector bundle) and its normal bundle is just the vector bundle itself. So the Thom class of the normal bundle is actually the Thom class of the vector bundle.

Last edited: Oct 18, 2014
15. Oct 21, 2014

### WWGD

Lavinia, thanks again. A couple of questions to see how well I am doing; I will look into the Thom class of an oriented vector bundle when I get a chance:

1) How does one usually go about computing Poincare duals ( in this case to a submanifold ). This would be a cocycle. I am aware of the P.Duality theorem, but, is there a practical way of dualizing submanifolds? Is there a "natural format"
for expressing either (co)homology class that allows us to Poincare-dualize the class?

2) This came about in my effort to figure some things out: say f: M -->N is smooth between an m-manifold and an n-manifold, and c is any regular value. Are then all the classes [f^{-1}(c)] in H_1(M) ( which are (m-n) submanifolds of M ) equal ( so that their duals H^{n-1}(M) are also equal) ? I think this is true, maybe using Morse functions, since in this case the homology changes only when there is a critical value to the Morse function/Morse homology?

Thanks.

16. Oct 22, 2014

### lavinia

I am not sure what a general method for calculating Poincare duals would be. One would need to know the homology of the manifold. Can you tell me more what you have in mind?

In simplicial homology, one can in principal calculate duals with cap products but again one would need to know the simplicial structure of the manifold.

For an embedded smooth submanifold one gets a cohomology class by taking transverse intersections with submanifolds of complementary dimension. In principal it calculates the value of the Poincare dual to the submanifold on manifolds of complementary dimension.

If the manifold is triangulated, then there is a dual complex. Not sure of a reference.

- The general theorem that I know is that when a smooth function is transverse to a submanifold, then the Poincare dual to the inverse image manifold is the pullback of the Poincare dual. The proof is straightforward using Thom classes.

However if the inverse image manifold is not connected, I doubt- but have not proved - that the components are necessarily homologous. But let's think about this more.

Last edited: Oct 23, 2014
17. Oct 22, 2014

### WWGD

Sorry, I meant to say the inverse images are not necessarily homologous; otherwise every pair of regular images would have to co-bound; pretty sure this does not always happen.

18. Oct 23, 2014

### lavinia

Here is the definition of the Thom class and the Euler class in Algebraic Topology.

If E is an oriented n plane bundle, then the Thom isomorphism says that the n'th relative cohomology group, H^n(E,E-zero section;Z)) is isomorphic to Z and it generator pulls back to the oriented generator of H^n(F,F-0) of each fiber. Note that this is the algebraic topological equivalent of the Thom class. The Euler class is the pullback of the Thom class via two maps: H^n(E,E-zero section) -> H^n(E) -> H^n(B) where B is the base space of the bundle. The first map is the pullback of the inclusion and the second is the pullback of the zero section.

Note that by the exact sequence of the pair, (E,E-zero section), that if the bundle has a non zero section, the Euler class is zero.

Last edited: Oct 23, 2014
19. Oct 23, 2014

### WWGD

Thanks, sorry I have been too fuzzy . Will look into it some more.

20. Oct 26, 2014

### homeomorphic

Munkres Elements of Algebraic Topology.