# Poincaré-Hopf Theorem

1. Dec 4, 2009

### Jamma

Hi. I've been going through the proof of this theorem; if you don't know the statement, see:

http://en.wikipedia.org/wiki/Poincaré–Hopf_theorem

However, I'm getting confused due to inaccuracies of the way this theorem is written down in different sources.

On the wiki article, it says that the theorem applies to compact, ORIENTABLE, DIFFERENTIABLE manifolds.

However, the reference from this page says that it must be just compact and smooth. Further still, in "Topology from a differentiable viewpoint" by Milnor (I need to read through this sketch proof a bit more rigourously), we seem to only require that M be a compact manifold (actually, just read that the vector field needs to be smooth, so I'm guessing that the original manifold needs to be smooth to even have the notion of a smooth vector field?).
So do we need only differentiability of the manifold? Or do we need either smoothness OR orientability?

My problem is that the sketch proof on wikipedia seems to make use of the fact that the manifold is orientable since we define the Gauss map from its boundary, which can only be defined if the boundary is orientable (the degree of a map between manifolds is only defined when the top homology group is Z, implying orientability), which will be implied by the original manifold being orientable.

I would like to be able to prove the most general result possible (I need to present the proof in class); maybe we don't need the manifold to be orientable, and when forming our new manifold in the higher dimensional Euclidean space, our boundary must now be orientable? - If this is true, I must be missing something.

My other annoyance with this sketch proof is that at the end it makes use of a triangulation of our manifold M. But can we be sure that a triangulation exists? It certainly doesn't for every manifold, but again, maybe when it is differentiable, orientable and compact, (or smooth and compact?!) then there will be a triangulation.

Sorry for the length of this post, I shall summarise:
1) In the hypotheses for this proof, can we get away with M just being differentiable and compact? Or do we require that it be smooth and compact OR differentiable, orientable and compact?

2) In the sketch proof, if we had not assumed orientability, can we assume that the boundary of our new manifold in the higher dimensional Euclidean space has an orientable boundary?

3) Will a smooth and compact (or [differentiable and compact] or [differentiable, compact and orientable]) always have a triangulation?

(I assume that the answers to 2) and 3) will be negative and that we actually need a more elaborate proof for the general theorem than the badly written on on wiki).

If you managed to read all that, thanks! Any answers greatly appreciated :)

2. Dec 4, 2009

### zhentil

If your manifold is not orientable, indices of vector fields are only defined modulo two. I am not sure if a mod two version of the P-H theorem holds in this case, but it certainly doesn't hold over the integers, since one side of the equation is not defined.

You definitely need some idea of differentiability to talk about vector fields. It's possible that the theorem may be true for C^1 vector fields, but again, I'm not sure.

3. Dec 5, 2009

### wofsy

I am glad you asked this question since it has forced me to think carefully about this theorem.

I am not sure of the answer but it seems that the sum of the indices of a vector field does not require the manifold to be orientable. Please refute the arguments.

First of all, at any isolated zero of the field, the index is defined in a small coordinate ball as the degree of the map of the bounding sphere into the standard sphere in Euclidean space. This degree is independent of the coordinate ball and in particular does not depend upon whether the coordinate map is is orientation preserving or reversing.

This means that the sum of the indices of a vector field on any manifold is well defined.

If the manifold is non-orientable pull the vector field back to an orientable 2 fold cover, do the theorem there then divide by 2. Since the coordinate projection will not change the index of any particular zero, the theorem follows because zeros are identified in pairs with no change of local index and the Euler Characteristic of the non-orientable base is just half the Euler characteristic of its 2 fold covering space.

In the Poincare-Hopf theorem one first embeds the manifold in Euclidean space and then extends the vector fields to a tubular neighborhood making sure that each local index is preserved and that the new vector field points outward along the boundary. A simple argument ( Stokes theorem ) shows that the sum of the indices is the degree of the Gauss map on the boundary of the tube. The Gauss map is well defined because the boundary of the tube is orientable (since any closed hypersurface of Euclidean space is orientable). I do not see where orientability of the original manifold comes in here.

Any smooth manifold can be triangulated. If one chooses a vector field that follows the contours of a barycentric subdivision of the triangulation a simple counting argument shows that the sum of the indices is the Euler characteristic. Again,I do not see where orientability comes in here.

Lastly it seems to me that the vector field only needs to be continuous because the idea of degree of a mapping only requires continuity.

On the other hand, if one wants to apply the Gauss Bonnet theorem to prove the index theorem the vector field would have to be at least C^1.

Last edited: Dec 5, 2009
4. Dec 5, 2009

### zhentil

Hi Wofsy, I believe you're correct (except for one minor point). One needs only to specify that the lift of the vector field to the orientable double cover is unique, and everything else follows easily.

With respect to continuous vector fields, the answer is clearly true if the manifold has a smooth structure, since any continuous (or C^1) vector field is homotopic to a smooth vector field. The question I had in mind is whether indices of vector fields can be defined in a meaningful way, say, for PL manifolds and continuous vector fields. If one loses the notion of a tangent space, what does it mean for a vector field to be transverse to the zero section?

5. Dec 7, 2009

### wofsy

I don't know the PL theory but suspect that there are analogues of vector bundles. I would be willing to do some reading with you. I could ask around for papers if you like.

6. Dec 8, 2009

### Jamma

Thanks Wofsy, that certainly is very helpful; much more so than the wikipedia article.

You wouldn't mind explaining what you mean by a 2-fold cover do you? Its not something that I have ever encountered before, although I assume it is something that is quite intuitive- an example with a Mobius band maybe? And yes, we would need to know that this construction (whatever it is!) is unique.

So yes, orientability is not an issue. However, we still need the manifold to be smooth, so that we can be sure to have a triangulation (I was not aware that all smooth manifolds can be triangulated; where can I find this statement?). I would assume that the theorem will still hold for just a differentiable manifold though; maybe we need a slightly more elaborate version of the contruction on the vector field?

7. Dec 8, 2009

### wofsy

- You do not need the manifold to be smooth only C^1 then I think all of the arguments work. Take a look at Milnor's proof and see if smoothness is actually used. I don't think so.

- In practice mathematicians only work with smooth manifolds. This is because they want to take as many derivatives as they like. There are theorems about when a C^k manifold has a smooth structure but I am not sure what they say. I suspect that if k is large enough say 2 or greater then it is true. I will look it up. It is an interesting question. I think it was first studied by Whitney and I have the feeling that he completely solved the problem.

- Every smooth manifold not only has a triangulation but actually has a smooth triangulation. This means that the maps of the simplices in the triangulation are themselves smooth. This means that one can extend the maps of the standard simplices in Euclidean space into the manifold to an open neighborhood so that the maps are smooth in the neighborhood and so the the maps line up smoothly(differentiably )on the intersections of the simplices. This is called a smoothing of the triangulation - I think. I think this is a theorem of Whitehead and I do not know the proof. It would be fun to try to figure it out for ourselves.

This issue has other interesting aspects to it. There are triangulations of manifolds that are not smoothable. And there are manifolds which can not be triangulated. I do not know this theory.

- A two fold covering space is another manifold that wraps around the first twice. This wrapping or covering is a local homeomorphism and is exactly 2 to 1. It is a standard fact that a non-orientable manifold has an orientable 2 fold covering. There are a zillion books on algebraic topology that discuss this. I think the best introductory book to topology and geometry from the modern point of view is Singer and Thorpe's Lecture Notes on Elementary Topology and Geometry.

- In the case of Moebius band the two fold cover is the cylinder. Try to figure out the covering map. When you finish that try figuring out how the torus is a two fold cover of the Klein bottle.

The sphere is a two fold cover of projective space. Mod out by the anitpodal map (the map that sends opposite poles to each other). In the case of even dimensional sphere the projective space is non-orientable. For odd spheres it is orientable. Try to prove this.

Notice that in the case of the two sphere you can see the cylinder covering the Moebius band.

Last edited: Dec 8, 2009
8. Dec 10, 2009

### Jamma

Ahha, so you can just take your standard "box with identifications on it" and double it in one direction. So yes, I see how you would get the Klein bottle and torus, this is interesting.

How this works in higher dimensions I'm sure is not so simple, but sounds interesting. I will make sure to take a look at this subject, when I have more time.

As for Milnor's proof, despite starting out by saying that we just need a manifold, but a smooth vector field (so we would need a smooth manifold too), at the end he starts deducing that we don't need such strong differentiability conditions. Its more of a sketch and I don't really have time to go in to it too much, but he doesn't seem to use the Whitney embedding theorem, but uses vector fields with non-degenerate zeros, and later goes on to say that this wasn't really a restriction.

For the Whitney embedding theorem, on wikipedia it says that it applies to smooth manifolds, but then that they can be smoothly embedded. So I'm guessing that we don't need a smooth manifold? (although then obviously, we will just have an embedding?).

Last edited: Dec 10, 2009
9. Dec 22, 2009

### wofsy

This thread has got me reading. In the section on the Poincare-Hopf theorem, Milnor's book mentions Morse functions on manifolds. These are real valued functions whose gradients have only non-degenerate zeros.

In Milnor's book,Morse Theory, he shows that any manifold has a Morse function and that its gradient vector field not-only tells you the Euler Charateristic of the manifold but also gives you a cell decomposition. One cell is added for each zero of the gradient and the dimension of the cell is computable from the Hessian of the Morse function. Also the index of the vector fields at a zero is computable from the Hessian - it is just its determinant. A negative determinant means a cell of odd dimension is attached, a positive determinant means a cell of even dimension is attached. This means that the sum of the indices of the gradient vector field equals the Euler Characteristic of the manifold since its equals the alternating sum of the number of cells in each dimension.

So these special gradient vector fields seem to tell you a lot more that just the Euler Charateristic. They tell you the homotopy type of the manifold. It might be interesting to investigate what the total information in a vector field is in general.

One last thought. Milnor's book, Topology from the Differentiable Viewpoint, only proves that the sum of the indices of a vector field is the same for all vector fields. He does not prove that it is the Euler Characteristic. (Technically, he only shows it for vector fields with non-degenerate zeros.) This he does in Morse theory but also the example I mentioned above that uses a triangulation works.

Last edited: Dec 22, 2009
10. Dec 22, 2009

### Jamma

Very interesting, thanks Wofsy.

11. Dec 22, 2009

### zhentil

In general, one can recover the homology of the manifold from a single Morse function. And his construction does prove that every smooth manifold has the homotopy type of a CW complex.

You can actually push this stuff quite far. It can be used to prove Bott Periodicity and Poincare's conjecture in dimensions greater than 4. You can also generalize from a real-valued function to a closed one-form and get the basic setup of Floer homology.

I think your question about the information that a vector field encodes is interesting. This stuff barely scratches the surface (i.e. a signed count of the zeros). I think some of the guys in the '60s looked into the information one could obtain about the maximum number of linearly independent sections of vector bundles, but I think this got gradually subsumed by characteristic classes and obstruction theory. One completely different angle is periodic orbits of vector fields on manifolds, but the problem with this is that one needs more rigidity to make anything interesting happen (i.e. a symplectic/contact/Kahler structure).

12. Dec 23, 2009

### wofsy

This stuff seems really amazing. Can you give references. I have good books on charateristic classes but how about Floer homology and the Poincare conjecture?

Our discussion still leaves some open questions as well. When you first thought that the manifold needed to be orientable for the index theorem to work I thought that was right but couldn't reconcile it with some simple examples e.g. the projective plane.

But the vector bundle does need to be orientable in order for there to be an Euler class. This is true on any vector bundle including the tangent bundle and this I think was your original thought. It seems that the tangent bundle is an exception and for a while this had me worried. I suspect that the tangent bundle is special because it has a differential. Other bundles do not. Milnor's proof uses the differential of coordinate charts and the Morse Theory argument uses the differential of a function. I am still uneasy about this.

If the vector bundle and the manifold are both orientable and have the same dimension then the Euler class is Poincare dual to the zero section so by the transversality argument you mentioned, the sum of the degrees of a vector field must equal the Euler number of the bundle. It seems that if the manifold is not orientable this transversality argument only works mod 2 but I am not sure. Anyway this again seems to be your original thought.

13. Dec 23, 2009

### zhentil

My original thought was basically that intersection numbers can only be defined mod 2. In my original line of thinking, I took the index of a vector field to be the intersection number of the section with the zero section.

As for the other stuff, I'd look either at Floer's original papers or (particularly) their motivation, Witten's paper on Supersymmetry and Morse Theory. For the Poincare conjecture stuff, you can't go wrong with Milnor's book on the h-cobordism theorem, of which the high-dimensional poincare conjecture is a corollary.

14. Dec 23, 2009

### quasar987

I read that Floer's motivation for inventing Floer homology was that he saw a way to prove Arnold's conjecture. Is there something else relating to Witten's paper that he had in mind?

15. Dec 24, 2009

### wofsy

Thanks for the references. I looked quickly at the Milnor book. It looks difficult.

I think the intersection number of the manifold with itself is the Poincare dual to the Euler class for an oriented n-plane bundle over an oriented n-manifold. The Euler class seems to be just the Thom class of the zero section so the cup product of the Euler class with itself is dual to the transverse intersection of the zero section with itself (the Thom class of the transverse intersection is the wedge product of the Thom classes). So it seems but I am not sure that the index theorem works by this transversality argument for any oriented vector bundle over an oriented manifold of the same dimension.

If the vector bundle is of lower dimension transverse intersection gives a submanifold. What can we say about this submanifold? I also still don't see that the manifold needs to be orientable if the vector bundle is orientable.

Last edited: Dec 24, 2009
16. Dec 26, 2009

### quasar987

The h-cobordism theorem is also presented in the book Differential manifolds of Kosinski. (It's in dover!)

17. Dec 26, 2009

### wofsy

Thaanks. What is the S-cobordism Theorem?

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