- #1

- 431

- 0

## Main Question or Discussion Point

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 :)

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 :)