Embedded submanifold of $\mathbb{S}^3$ Let's consider $f:\mathbb{S}^3\to \mathbb{R}$ with $f(x,y,z,t)=x^2+y^2-z^2-t^2$
(1) show that $M_0:= f^{-1}(0)$ is an embedded submanifold of $\mathbb{S}^3$.
(2) show that $M_0$ is diffeomorphic to the 2-torus.
(3) find all points at which $f$ is a submersion.
I have some problem with this exercise because I want to use the regular value theorem but proving that $f$ is a submersion at all points of $f^{-1}(0)$, with charts of $\mathbb{S}^3$, needs way too many calculations. How can I solve smartly this problem?
My attempt for points (1) and (3):
I can consider the natural extension of $f$ to $\mathbb{R}^4$ given by $\widetilde{f}: \mathbb{R}^4\to \mathbb{R}$ with $\widetilde{f}(x,y,z,t)=x^2+y^2-z^2-t^2$. Clearly $\widetilde{f}$ is a submersion at all points $p\neq 0$, so also at every $p\in \mathbb{S}^3$. Therefore, since $f=\widetilde{f}\circ \iota\mid_{\mathbb{S}^3}$, by chain rule I get, for $p\in \mathbb{S}^3$,
$$(df)_p=(d\widetilde{f})_p\circ (d\iota\mid_{\mathbb{S}^3})_p$$
where I know that $(d\widetilde{f})_p$ is surjective and $(d\iota\mid_{\mathbb{S}^3})_p$ is injective. It seems to be almost done but I don't know how to conclude.
 A: Hint:Let $G:\mathbb{R}^3\to\mathbb{R}$ denote the function $G(x,y,z,t)=x^2+y^2+z^2+t^2-1$.  We have the following:


*

*$T_p\mathbb{S}^3=\ker dG_p$ for $p\in\mathbb{S}^3$.

*$p\in\mathbb{S}^3$ is a critical point of $f$ if and only if $T_p\mathbb{S}^3\subset\ker d\tilde{f}_p$. By dimensional consideration, this is true if and only if $T_p\mathbb{S}^3=\ker d\tilde{f}_p$.

*Two linear functionals on a vector space are constant multiples of each other if they have the same kernel. 

A: Here are a few ideas:
For $(1)$, I think your argument is fine. There is a (smooth) embedding $e$ of $S^3$ in $R^4$, so using your idea of extending $f$ to $F$ defined on all of $\mathbb R^4$, one has $f=F\circ e$ and now the chain rule shows that $0$ is a regular value for $f$ and the result follows from the regular level set theorem. (Because $\textit{now}\ \mathcal J(F)$ is easy to calculate.)
For $(2)$, use the fact that $(r,s) \mapsto \frac{1}{\sqrt{2}}(\sin(2\pi r), \cos(2\pi r), \sin(2\pi s), \cos(2\pi s))$ maps $S^1\times S^1$ diffeomorphically into $S^3$ and show that the image is exactly $f^{-1}(0).$
For $(3)$, the points $(p,f(p)=q)$ at which $f$ is a submersion will be those for which the subset $S$ of $\mathbb R^4$ defined by $x^2+y^2-z^2-t^2=q$ and $x^2+y^2+z^2+t^2=1$ is a regular submanifold of $\mathbb R^4$ so if we define $F(x,y,z,t)=(x^2+y^2-z^2-t^2,x^2+y^2+z^2+t^2-1)$ then $S=F^{-1}(q,0)$. Now, 
$$\mathcal J(F)=\begin{pmatrix}
2x &2y  &-2z  &-2t \\ 
 2x&2y  &2z  &2t 
\end{pmatrix}$$
The critical points of $F$ are the points $(x, y, z,t)$ where 
this matrix has rank $< 2$. That is, where all $2 \times 2$ minors 
are zero. So, we get the conditions 
$\tag1 xz=0,\ yz=0,\ xt=0,\ yt=0$ 
We also have 
$\tag 2 x^2+y^2+z^2+t^2-1=0$ 
If $x\neq 0,$ we get $x^3+xy^2-x=0\ $ so 
$\tag 3 x^2+y^2=1\ $ 
If $y\neq 0$ we get the same equation. 
Similarly, if either $t$ or $z\neq 0,$ then 
$\tag4 z^2+t^2=1.$
Since $(3)$ and $(4)$ cannot both hold (because of $(2)$), if $(3)$ holds then $z=t=0$, while if $(4)$ holds then $x=y=0.$ 
But we also have 
$\tag5 x^2+y^2-z^2-t^2=q$ 
so if $(3)$ holds then $q=1$ and if $(4)$ holds, then $q=-1.$
So, the critical points of $f$ are $A=B\cup C:=\{(x,y,z,t):x^2+y^2=1;\ t=z=0\}\cup\{t^2+z^2=1;\ x=y=0\}$ with critical values $1$ and $-1$, respectively.
edit: the proof that $f$ is a submersion goes as follows:
Let $p$ be a point on the sphere in the complement of $A$, and without loss of generality, parameterize the sphere by $\phi(x,y,z)=(x,y,z,\sqrt{1-x^2-y^2-z^2})$ in some sufficiently small open set $\phi^{-1}(p)\in U$  so that $\phi$ is a smooth bijection in $U$. (the other $2$ obvious parameterizations cover the other possibilities). Then, 
$f\circ\phi(x,y,z)=x^2+y^2-z^2-(1-x^2-y^2-z^2)=-1+2x^2+2y^2$
and
$\mathcal J(f\circ \phi)=4\begin{pmatrix}
 x&y  &0
\end{pmatrix}.$
If $x=y=0$ then $t^2+z^2= 1$ and $p\in B$ so $p$ is critical. We conclude that $\mathcal J(f\circ \phi)$ has rank $1$ whenever $p\notin A$, and then so does $\mathcal J(f)$ because $\mathcal J(\phi)$ is an isomorphism. 
