# Level set diffeomorphic to $\mathbb{S}^2$

Consider the map $\phi: \mathbb{R}^4 \rightarrow \mathbb{R}^2$ defined by $\phi(x,y,s,t) = (x^2 + y, x^2 + y^2 + s^2 + t^2 + y)$.

Show that $(0,1)$ is a regular value of $\phi$, and that the level set $\phi^{-1}(0,1)$ is diffeomorphic to $\mathbb{S}^2$.

I get two equations describing the level set:

(1) $x^2 + y = 0 \implies y = -x^2$

(2) $x^2 + y^2 + s^2 + t^2 + y = 1 \implies y^2 + s^2 + t^2 = 1$

So $\phi^{-1}(0,1) = \{(x,y,s,t) \in \mathbb{R}^4: y = -x^2 \hspace{0.1cm} \mathrm{ and } \hspace{0.1cm} y^2 + s^2 + t^2 = 1\}$.

I need to show that $d\phi(x,y,s,t)$ is surjective for all $(x,y,s,t) \in \phi^{-1}(0,1)$. I calculate:

$d\phi(x,y,s,t) = \begin{pmatrix} 2x & 1 & 0 & 0 \\ 2x & 2y + 1 & 2s & 2t \end{pmatrix}$

It is easy to show that this matrix has rank $2$ for all $(x,y,s,t) \in \phi^{-1}(0,1)$ and so $\phi^{-1}(0,1)$ is an embedded submanifold of $\mathbb{R}^4$.

Now, I need to show that this level set is diffeomorphic to the unit sphere. I can kind of see that it may be diffeomorphic to a spheroid, and I know I can show that the spheroid is diffeomorphic to the sphere. My only problem is coming up with this diffeomorphism from the level set onto the spheroid.

I imagine I can define a map $\psi: \phi^{-1}(0,1) \rightarrow S$ by $\psi(x,y,s,t) = (x,s,t)$, where $S = \psi(\phi^{-1}(0,1))$. It is easy to see that this map is invertible and its inverse is given by $\psi^{-1}(x,s,t) = (x, -x^2, s, t)$ since $y = -x^2$. My only trouble is showing that these maps are both smooth. Both the domain and codomain are submanifolds of $\mathbb{R}^4$ and $\mathbb{R}^3$, respectively, so I need to express $\psi$ in appropriate local coordinates before differentiating, but coming up with these local coordinates seems like I am making things overly complicated. What would you suggest I do?

• Try the map $\psi(x, y, s, t) = (y, s, t)$. That's clearly nice, and its differential at each point (as a map from $\Bbb R^4$ to $\Bbb R^3$) contains a $3 \times 3$ identity, which is a pretty useful start to proving that when restricted to the domain it has maximal rank. Jun 11, 2017 at 3:46
• But that map isn't invertible. $\psi(1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}) = \psi(-1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$. Jun 11, 2017 at 4:10

Let $M$ denote $\phi^{-1}(0,1)$. You define $$\psi:M\to S,\quad(x,y,s,t)\mapsto(x,s,t).$$ As $M$ is a submanifold, the inclusion $M\hookrightarrow\mathbb{R}^4$ is smooth. The projection $\mathbb{R}^4\to\mathbb{R}^3$ given by $(x,y,s,t)\mapsto(x,s,t)$ is also smooth. Your map $\psi$ is just the composition of these two smooth maps.
• Ah, that was simple. I believe I can prove the smoothness of $\psi^{-1}$ in the same spirit. Please correct me if I am wrong: Let $\iota_S: S \rightarrow \mathbb{R}^3$ be the inclusion map - it is smooth as $S$ is an embedded submanifold of $\mathbb{R}^3$ (I can prove this by easily showing that $S$ is diffeomorphic to $\mathbb{S}^2$). Let $\iota: \mathbb{R}^3 \rightarrow \mathbb{R}^4$ be the inclusion map embedding $\mathbb{R}^3$ into $\mathbb{R}^4$, and let $F: \mathbb{R}^4 \rightarrow M$ be given by $F(x,y,s,t) = (x, -x^2, s, t)$. Then $\psi^{-1} = F \circ \iota \circ \iota_S$. Jun 11, 2017 at 5:59
• Each map in the composition is smooth and hence $\psi^{-1}$ is also smooth. Is this right? Jun 11, 2017 at 6:03
• @ConfusedMonkey Yes. The only step that isn't immediate is showing that $S$ is an embedded submanifold of $\mathbb{R}^3$. If you're OK with that, then you're good. Jun 11, 2017 at 6:06
• The diffeomorphism between $S$ and $\mathbb{S}^2$ is given here: math.stackexchange.com/questions/547857/… May 1, 2021 at 7:35