Talk:Poincaré–Hopf theorem

I added a paragraph entitled "Significance" which I hope will give the reader some concept of what index theorems are and why they are important. I hope I wasn't overly, well, enthusiastic. This theorem is a favorite of mine. The proof here is also sufficiently elementary that it seemed reasonable to briefly outline the main ideas underlying the proof. Greg Woodhouse 23:10, 20 March 2007 (UTC)[reply]

Revised Introduction

The introduction paragraph does not actually give a brief description of the theorem, formal or otherwise. I feel that a sentence or two summarising the result would somewhat improve a reader's first impressions of this article. PhysicsSean (talk) 16:07, 9 August 2017 (UTC)[reply]

Definition of Index

The definitions used in the theorem need more complete explanations. The Euler characteristic can be understood from the link, but the term index is too general, so the referenced page doesn't lead directly to the required definition. Could somebody who knows this please insert a summary of what index means in this case (winding number?)? —Preceding unsigned comment added by 146.186.131.40 (talk) 20:21, 6 October 2010 (UTC)[reply]

I agree. Either the definition of index should be more clear somewhere, or the definition should be added here. If this Theorem comprises a significant part of the usage of this meaning of "index", it should probably be here (but I'm not sure)? Franp9am (talk) 02:01, 22 August 2011 (UTC)[reply]

Sketch of Proof

"Then use the fact that the boundary of an n-dimensional manifold to an n-1-dimensional sphere, that can be extended to the whole n-dimensional manifold, is zero." — I fail to parse this phrase. What is meant? Maybe, it is about THE INDEX OF A MAP of the boundary ... to ... sphere? Boris Tsirelson (talk) 20:52, 12 October 2009 (UTC)[reply]

I think it should be read:

Then use the fact that index of a map from the boundary of an n-dimensional manifold to an n-1-dimensional sphere, that can be extended to the whole n-dimensional manifold, is zero.

(I think that) this will be true of any map to any topological space of dimension n-1 since the inclusion map will be the zero map, as the boundary of the manifold is the image of the boundary map of the entire space, and so zero in homology.

References?

I can't understand why there is a sketch of a proof when no reference is given to the complete proof. —Preceding unsigned comment added by 24.255.19.253 (talk) 07:01, 1 November 2009 (UTC)[reply]

I've added a quick reference to the Springer site. That page mentions a well-known book by John Milnor, and I presume it has a proof. Charles Matthews (talk) 22:12, 24 November 2009 (UTC)[reply]

Your presumption is wrong, as it contains only the sketch. In case you are curious, the book is "Topology from the differentiable viewpoint", and the sketch is on page 35. 72.69.194.235 (talk) 15:27, 28 November 2009 (UTC)[reply]

Notation for index

I've changed the notation for the index of a vector field v at a point x from "index_v(x)" to "index_x(v)". This is to emphasize that the index is being viewed as a property of the vector field v as opposed to of the point x. Foxjwill (talk) 23:04, 22 December 2012 (UTC)[reply]

Does the statement make any sense?

What does it mean for a vector field on an abstract diffenrentiable manifold with boundary to point in the outward normal direction on the boundary? Certainly "normal" makes no sense. As for the "outward", maybe it needs to be explained? As far as I know, a priori it only makes sense to consider vector fields that are tangent to the boundary. — Preceding unsigned comment added by 2A01:E34:EE33:210:D17E:A934:2EB2:91B8 (talk) 11:37, 14 November 2013 (UTC)[reply]

Generalisation to relax the boundary condition.

There is a generalisation due to Morse that allows one to remove the constraint that the vector field must be pointing outwards on the boundary of the manifold. A modern proof and explanation of the history of similar generalisations can be found here. 212.201.78.215 (talk) 11:12, 29 March 2023 (UTC)[reply]