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.


3 Answers 3


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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – Saigyouji
    Oct 30, 2014 at 4:05
  • 6
    $\begingroup$ 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. $\endgroup$ 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.

  • 2
    $\begingroup$ $y$ is nonpositive, so the lattermost equation does not map out an entire sphere. $\endgroup$
    – Saigyouji
    Oct 30, 2014 at 4:11

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