# Diffeomorphism from level set onto $S^2$

I'm given a map $\phi : R^4 \rightarrow R^2$ defined by $\phi(x,y,s,t) = (x^2 + y, x^2+y^2+s^2+t^2+y)$.

It's easy to show that the level set $C = \phi^{-1}(0,1)$ is a smooth submanifold of $R^4$ with dimension $2$ by showing that $(0,1)$ is a regular value.

I'm trying to show that this level set is diffeomorphic to $S^2$, the unit sphere in $R^3$. The only thing I can think of is to construct a smooth embedding from $S^2$ into $R^4$ whose image is $C$, since smooth embeddings are diffeomorphisms onto their images.

It's easy enough to get that $C$ is diffeomorphic to $\{(x,y,z) \mid x^4 + y^2 + z^2 = 1\}$, but from there I'm stuck. The map $(x, y, z) \rightarrow (\text{sign}(x) x^2, y, z)$ is a differentiable homeomorphism from $S^2$ onto $C$, but it's not even $C^2$ let alone smooth.

Once you're working with the surface $C=\{ (x,y,z) : x^4+y^2+z^2=1 \}$, how about the map $$(x,y,z) \mapsto \frac{(x,y,z)}{ \sqrt{x^2+y^2+z^2}} ?$$

$C$ is a ball-type surface centered at the origin, so why not just project to the sphere in this simple way?

• You seem to suggest that this is a very natural thing to try, but I have no intuition about how you came up with this. Commented Oct 30, 2014 at 4:05
• C is roughly a spheroid surface. Draw all the radial rays out of the origin. They each intersect C in one point and $S^2$ in one point. This map takes each point on C, and puts it on the point on $S^2$ that's on the same ray, by dividing by its norm. Commented Oct 31, 2014 at 0:09

Note: The theorems referenced in the answer are from John Lee's Introduction to Smooth Manifolds (2nd Edition).

First, we note the following fact:

Let $$S$$ be an embedded submanifold of a manifold $$M$$. Let $$N$$ be another manifold and let $$\varphi: S\rightarrow N$$ be a bijective map that is a "diffeomorphism" in the extended sense (i.e. both $$\varphi$$ and $$\varphi^{-1}$$ have local smooth extensions). Then $$T = \varphi(S)$$ is an embedded submanifold of $$N$$ and $$\varphi: S\rightarrow T$$ is a diffeomorphism (with the smooth structure on $$T$$).

Proof: Extend $$\varphi$$ to an open neighborhood $$U$$ of $$S$$ (and call the extension $$\Phi$$). Similarly, extend $$\varphi^{-1}$$ to an open neighborhood $$V$$ of $$T$$ (and call the extension $$\Psi$$). Now using Theorem 5.29, we may regard $$S$$ as an embedded submanifold of $$U$$ and thus $$\varphi = \Phi|_{S}\in C^{\infty}(S)$$ (i.e. it is a smooth map with respect to the smooth structure on $$S$$). Since $$\Psi\circ\varphi = \text{id}_{S}$$, using the chain rule we conclude that $$\varphi$$ is a smooth immersion. Moreover, $$\varphi$$ is a homeomorphism between $$S$$ and $$T$$ and is thus a smooth embedding from $$S$$ to $$N$$. The result now follows from Proposition 5.2.

This result helps us to define the necessary diffeomorphisms since it is easier to specify the local smooth extensions on Euclidean space.

Let $$C = \{(x, y, z)\in\mathbb{R}^{3}: x^{4}+y^{2}+z^{2}\}$$. Then $$\phi^{-1}(0, 1)$$ is diffeomorphic to $$C$$ via the map $$(x, y, s, t)\mapsto (x, s, t)$$ whose inverse is given by $$(x, y, z)\mapsto (x, -x^{2}, y, z).$$ Now, to show that $$C$$ is diffeomorphic to $$S^{2}$$ consider the map $$(x, y, z)\mapsto \frac{(x, y, z)}{\sqrt{x^{2}+y^{2}+z^{2}}}.$$ To define the inverse map, consider the intersection of the ray $$\{\lambda(x, y, z): \lambda > 0\}$$ with $$C$$. Solving for $$\lambda$$ we get $$\lambda(x, y, z) = \begin{cases}\sqrt{\frac{\sqrt{(y^{2}+z^{2})^{2}+4x^{4}}-(y^{2}+z^{2})}{2x^{4}}}, & x\neq 0\\ \frac{1}{\sqrt{y^{2}+z^{2}}}, & x = 0\end{cases}.$$ To show that $$\lambda$$ is smooth on $$\mathbb{R}^{3}\setminus\{0\}$$ consider $$f: (\mathbb{R}^{3}\setminus\{0\})\times (0, \infty)\rightarrow \mathbb{R}$$ defined by $$f(x, y, z, \lambda) = x^{4}\lambda^{4}+(y^{2}+z^{2})\lambda^{2}-1.$$ Using the inverse function theorem, we conclude that $$\lambda(x, y, z)$$ is locally smooth and hence smooth on its entire domain. Thus the inverse map may be defined as $$(x, y, z)\mapsto \lambda(x, y, z)(x, y, z).$$

The level set is traced out by the intersection of x^2+y=0 and x^2 + y^2 +s^2 +t^2 + y =1. Combine the equations and get y^2 + s^2 + t^2 = 1.

• $y$ is nonpositive, so the lattermost equation does not map out an entire sphere. Commented Oct 30, 2014 at 4:11