Can anyone provide a diffeomorphism between these "spheres": $\mathbb{S}^2$ and $\{(x,y,z)\in \mathbb{R}^3: x^4+y^2+z^2=1\}$?

PS: If you know a result that can solve this problem, I would be glad to know.


The problem is $sign(x)\sqrt{x}$ and $sign(x)x^2$ are not strictly speaking diffeomorphisms from $[-1,1]$ to $[-1,1]$ since the differential at $x=0$ either does not exist or vanish.

This is fixable by partition of unity. Take the suggested map $g: (x,y,z)\mapsto(sign(x)\sqrt{x},y,z)$ on the domain where $|x|>1/4$.

And on $|x|<1/2$ the projection map $h: (x,r,\theta)\mapsto(x,r'=\sqrt{x^2-x^4+r^2},\theta)$ (here the obvious $(y,z)$ into polar coordinates $(r,\theta)$ is used.)

These two maps can be patched together using a partition of unity $f=\lambda g + (1-\lambda) h$ where $\lambda$ is a smooth function $\lambda:\mathbb{R}^3\rightarrow \mathbb{R}$ such that $\lambda=1$ for $|x|\ge 1/2$, $\lambda=0$ for $|x|\le 1/4$.

  • $\begingroup$ Could you be clearer it is a bit confuse in first glance! $\endgroup$ – checkmath May 31 '12 at 2:04
  • $\begingroup$ The idea is to define two different maps and patch them together where they do not agree. The map $h$ define a diffeomorphism near the equator, that is, the domain where $|x|<1/2$. The map $g$ define a diffeomorphism around the two poles (away from the equator) where $|x|>1/4$. The smooth function $\lambda$ is then used to 'patch' $g$ and $h$ together and make a complete diffeomorphism $f$ between the two spheres. One can think of $f$ as a weighted average of $h$ and $g$. Hope that helps :) $\endgroup$ – S.L.H May 31 '12 at 12:03
  • $\begingroup$ I got conffused can you write down editing the answer:? $\endgroup$ – checkmath Jun 2 '12 at 20:13

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