I'm working on G&P problem sets to prepare for a comp. Today was day 1! All went fairly well, except this problem here:
Prove that the hyperboloid in $\mathbb{R}^3$ defined by $x^2+y^2-z^2=a$ is a manifold for $a>0$. Why isn't it a manifold when $a=0$?
I've already taken the course, so my first instinct was the following:
Proof: Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $(x,y,z)\mapsto x^2+y^2-z^2$. If $a>0$, then for each point $(x,y,z)$ satisfying $f(x,y,z)=a$ we have that $x^2+y^2=a+z^2>0$ and thus that one of $x$ or $y$ is non-zero. Hence, for each such point $(x,y,z)\in\mathbb{R}^3$, the linear map $df_{(x,y,z)}:\mathbb{R}^3\to\mathbb{R}$, computed directly as $[2x \ \ 2y \ -2z]$, is surjective; wlog, suppose $x\neq 0$ and set $L=\lbrace (t,0,0) :t\in\mathbb{R}\rbrace$. Then $df_{(x,y,z)}(L)=\mathbb{R}$. Since $df_{(x,y,z)}$ is surjective for all such values, $a$ is a regular value and $f^{-1}(a)$ is a 2-dimensional manifold (preimage theorem).$\blacksquare$
However, this problem is in the first section so I imagine they want a more constructive argument. Since showing that $S^1$ is a manifold was done in the section, I think this is what they wanted. But I'm not sure how to do it this way. I thought about projecting in the following way:
$f:X\to Y$ by $(x,y,z)\to (x,0,z)$ for $x>0$
where $Y$ is the portion grey plane that lies inside of the hyperboloid. So $f$ would only project the upper half of the hyperboloid onto this grey portion, but by symmetry this covers $X$. Here's the problem, though. Even though I think this is doable (so far), how do we know that $Y$ is diffeomorphic to $\mathbb{R}^2$?
While the preimage theorem is great, we can also use local parametrizations in the same way that G&P parametrized $S^1$. We can parametrize $H_a=\lbrace (x,y,z): x^2+y^2-z^2=a\rbrace$ with four maps. Let $g:\mathbb{R}^2\to\mathbb{R}$ be given by $g(s,t)=a^2-s^2+t^2$. Then the four maps we will use to parametrize are
$\begin{array}{ll} \phi_1(x,z)=(x,\sqrt{g(x,z)},z)\quad\text{ for $y>0$,} &\phi_2(x,z)=(x,-\sqrt{g(x,z)},z)\quad\text{ for $y<0$,} \\ \phi_3(y,z)=(\sqrt{g(y,z)},y,z)\quad\text{ for $x>0$,} & \phi_4(y,z)=(-\sqrt{g(y,z)},y,z)\quad\text{ for $x<0$,} \end{array}$
Then $\phi_1$ is defined on the region $g^{-1}((0,\infty))$ which is an open subset of $\mathbb{R}^2$ and and hence an 2-dimensional manifold. The inverse function for $\phi_1$ is $(x,y,z)\mapsto (x,z)$ and this is clearly smooth. The other functions are symmetric.