Does the Poincaré-Hopf index theorem imply that on the sphere every tangent vector field has a singularity?

Question

We know that $\chi(M) = 2 = \sum_{x_0 \in \text{Sing}(X)}I(x_o,X)$, where $X$ is any tangent vector and $x_0$ is a singularity. I mean, the sum is over all singularities of $X$, an arbitrary tangent vector field.

Then, once this sum is equal $2$ for every vector field $X$, all of these vectors has at least a point where vanishes.

Is this right?

By "any" vector field, you presumably mean continuous and with finitely many zeros. Those are common "standing hypotheses" in topology, but it's important to be aware you're not really speaking of arbitrary vector fields. — Andrew D. Hwang, Aug 13 '16 at 19:43
yes, I got it @AndrewD.Hwang, I would have to be more clear. Thank you! — L.F. Cavenaghi, Aug 13 '16 at 20:14

Michael Albanese · Accepted Answer · 2016-08-30T02:58:58.440

3

I think it makes more sense to say 'zero' rather than 'singularity', but yes. If $X$ is a vector field on $M$ which has isolated singularities, and $\chi(M) \neq 0$, then $X$ must have at least one singularity. Otherwise, the sum over singularities is empty and hence zero which is impossible as $\chi(M) \neq 0$.

edited Aug 30 '16 at 02:58

answered Aug 13 '16 at 16:52

Michael Albanese

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Does the Poincaré-Hopf index theorem imply that on the sphere every tangent vector field has a singularity?

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