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I know that a diffeomorphsim between the sphere $S^2 = \{(x, y, z) \; | \; x^2+y^2 +z^2= 1\}$ and the ellipse $E^2 = \{(x, y, z) \; | \; (\frac{x}{a})^2+(\frac{y}{a})^2 +(\frac{z}{a})^2= 1\}$ is given by

$$f:S^2\to E^2 \;\;\ni\;\; (x,y,z) \mapsto (ax, by, cz)$$

I want to know how to actually argue that this map is indeed a diffeomorphism? Please see my arguments below

Argument #1.

Clearly $f$ is injective with $f(S^2) = E^2$, so $f$ is bijective with inverse $f^{-1} :E^2 \to S^2$ given by $ (x,y,z) \mapsto (x/a,y/b,z/c)$. So all that is left to show is that $f$ and $f^{-1}$ are smooth.

Note that $$g:\mathbb{R}^3\to \mathbb{R}^3 \;\;\ni\;\; (x,y,z) \mapsto (ax, by, cz)$$ is a smooth map because the partial derivatives of all orders of its component functions exists on $\mathbb{R}^3$. Now we have $f = g \big|_{S^2}$ so again $f$ is smooth.

Also note that $$g^{-1}:\mathbb{R}^3\to \mathbb{R}^3 \;\;\ni\;\; (x,y,z) \mapsto (x/a, y/b, z/c)$$ is a smooth map because the partial derivatives of all orders of its component functions exists on $\mathbb{R}^3$. Now we have $f^{-1} = g^{-1} \big|_{S^2}$ so again $f^{-1}$ is smooth.

This proves that $f$ is diffeomorphism.

Argument #2 (A more rigorous way to prove that $f$ is diffeomorphism)

Let

$\phi : (0,2\pi)\times(0,\pi) \to S^2 $ given by $(u,v)\mapsto (\cos u \sin v, \sin u \sin v, \cos v)$ and

$\Gamma : (0,2\pi)\times(0,\pi) \to E^2 $ given by $(u,v)\mapsto (a\cos u \sin v, b\sin u \sin v, c\cos v)$.

Then, $\phi and \Gamma$ are parametrization of $S^2$ and $E^2$ respectively. So, in order to show that $f$ is diffeomorphism, I must show that both $\Gamma^{-1}\circ f\circ \phi$ and $\phi^{-1}\circ f^{-1}\circ \Gamma$ are smooth. I am not going to show this, but it would be correct to say that the rigorous approach to proving diffeomorphism would be this way, if am correct.

Are these arguments correct?

chesslad
  • 2,533
  • I believe you would also need to show that another chart that overlaps the points where $\phi$ and $\Gamma$ are ill-defined is also a diffeomorphism. I think your original argument works pretty well, though you've essentially shown that this linear transformation $f$ on $\mathbb{R}^3$ is a diffeomorphism which you could then use to reason that $E^2$ and $S^2$'s embeddings are diffeomorphic. If $\Lambda: S^2\to \mathbb{R}^3$ is an embedding of $S^2$ and $\Psi: E^2 \to \mathbb{R}^3$ is an embedding of $E^2$ then $\Psi^{-1}\circ f \circ \Lambda$ is a diffeomorphism. – J.V.Gaiter Dec 04 '20 at 01:17
  • $f = g \big|_{S^2}$ is a homeomorphism, but it does not make sense to ask about differentibility. Think about how you would apply the definition and see where it fails. One of the reasons manifold theory was developed was to deal with calculus on surfaces. The other comment shows that you do need to use either parameterizations (or charts) to get defintions that make sense. – Matematleta Dec 04 '20 at 01:23
  • @J.V.Gaiter I think the OP's parameterization covers the whole sphere, only he/she needs to half- close and close the intervals defining $u$ and $v$. I have used $x = \cos(v)\cos(u), y = \cos(v) \sin(u); z = \sin(v) ; u = [0, 2\pi);\ v = [-\pi/2,\pi/2]$ and this covers the whole sphere. – Matematleta Dec 04 '20 at 01:30
  • OP's parametrization does cover the entire sphere, but it can only define a chart on the sphere if the intervals are open. – J.V.Gaiter Dec 04 '20 at 01:37

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