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?
Best Answer
You already did all the hard parts. Smoothness is actually easy comparing to what you've already done on your own.
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.