# Submanifold diffeomorphic to sphere

1. Jun 9, 2017

### ConfusedMonkey

1. The problem statement, all variables and given/known data
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 $\mathbb{S}^2$.

2. Relevant equations

3. The attempt at a solution

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 $\mathbb{R}^4$. Quick question: Did I calculate the differential properly?

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?

2. Jun 10, 2017

### andrewkirk

I think if we instead set $\psi(x,y,s,t)=(y\ \mathrm{sign}(x),s,t)$ we would have $S$ a perfect sphere, rather than just a spheroid.

I can't see a way of avoiding local coordinates. Perhaps there's a theorem that can be used to shortcut it, but none pops to mind.

Coordinate maps to use might be:
for $S$:
$\ (y,s,t)\mapsto (s,t)$, with inverse map $(s,t)\mapsto (\pm\sqrt{1-s^2-t^2},s,t)$, if $s^2+t^2\leq 0.9$ (two disjoint coordinate patches), otherwise:
$(y,s,t)\mapsto (y,s)$, with inverse map $(y,s)\mapsto (y,s,\pm\sqrt{1-s^2-y^2})$, if $y^2+s^2\leq 0.9$ (two disjoint coordinate patches), otherwise:
$(y,s,t)\mapsto (y,t)$, with inverse map $(y,t)\mapsto (y,\pm\sqrt{1-y^2-t^2},t)$, (two disjoint coordinate patches)
where the $\pm$ has a different sign for each of the two patches in a pair.

and for the level set:
$\ (x,y,s,t)\mapsto (s,t)$, with inverse map $(s,t)\mapsto (\pm(1-s^2-t^2)^\frac14,-\sqrt{1-s^2-t^2},s,t)$, if $s^2+t^2\leq 0.9$ (two disjoint coordinate patches), otherwise:
$(x,y,s,t)\mapsto (x,s)$, with inverse map $(x,s)\mapsto (x,-x^2,s,\pm\sqrt{1-s^2-x^4})$, if $y^2+s^2\leq 0.9$ (two disjoint coordinate patches), otherwise:
$(x,y,s,t)\mapsto (x,t)$, with inverse map $(x,t)\mapsto ((x,-x^2,\pm\sqrt{1-x^4-t^2},t)$.

3. Jun 11, 2017

### ConfusedMonkey

I can solve it like this:

I will keep $\psi$ defined how it is in the original post. Let $M = \phi^{-1}(0,1)$. I've already proven that $M$ is a submanifold and so the inclusion map $\iota_M: M \rightarrow \mathbb{R}^4$ is smooth. The projection map $\pi: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ given by $\pi(x,y,s,t) = (x,s,t)$ is smooth. Because $\psi = \pi \circ \iota_M$, then $\psi$ is smooth.

We can prove the smoothness of $\psi^{-1}$ in a similar manner: Let $\iota_S: S \hookrightarrow \mathbb{R}^3$ be an inclusion map - it is smooth as $S$ is an embedded submanifold of $\mathbb{R}^3$. Let $\iota: \mathbb{R}^3 \rightarrow \mathbb{R}^4$ be an inclusion map - it too is smooth. And finally let $F: \mathbb{R}^4 \rightarrow M$ be given by $F(x,y,s,t) = (x,-x^2,s,t)$, which is obviously smooth. Then $\psi^{-1} = F \circ \iota \circ \iota_S$ is the composition of smooth maps and hence also smooth.

Note: I know that $S$ is a submanifold of $\mathbb{R}^3$, and hence $\iota_S$ is smooth, because $S$ is diffeomorphic to $\mathbb{S}^2$ (constructing a diffeomorphism is very easy).