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The following is a question from Spivak's Differential Geometry text:

enter image description here

Not really sure what he's going for here. Any ideas?

Eric Auld
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  • The easiest attempts at "combing the hairs on a ball", i.e. to construct an everywhere non-zero vector field on the sphere ends up with two zero points. What he's getting at here is that it is possible to do it with just one zero point. That being said, I think the arrows in the lower middle, pointing upwards should be pointing downwards. – Arthur Aug 21 '15 at 07:59
  • @Arthur Indeed, it confused me that it seems like this vector field is continuous at $(0,0,1)$ – Eric Auld Aug 21 '15 at 18:47
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    It's continuous there, but not nonzero. –  Aug 22 '15 at 04:37

1 Answers1

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A way to explicitly construct such kind of vector field is to use the stereographic projection to transplant a constant vector field like $\frac{\partial}{\partial x}$ from the $(x,y)$ plane.

Using the projection the sphere is parameterized as $$r(x,y)=(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}, \frac{-1+x^2+y^2}{1+x^2+y^2})$$

Then $\frac{\partial}{\partial x}$ is transplanted to $$\frac{\partial r}{\partial x}=2(\frac{1-x^2+y^2}{(1+x^2+y^2)^2}, -\frac{2xy}{(1+x^2+y^2)^2}, \frac{2x}{(1+x^2+y^2)^2})$$

This is the vector field you're looking for. To verify, note that $$\left\|\frac{\partial r}{\partial x}\right\|=\frac{2}{1+x^2+y^2}$$ which never vanish everywhere else except at $(0,0,1)$, corresponding to $(x,y)\rightarrow \infty$.

Xipan Xiao
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