Premise: I know this post exixts Manifold diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$., but I'm asking for something more.

I need to prove that $A=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2-z^2=1 \}$ is diffeomorphic to $S^1 \times \mathbb{R}$. $A$ is a rotational hyperboloid. $$F: A \rightarrow S^1 \times \mathbb{R}, \quad (x,y,z) \mapsto (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},z)$$ with inverse $$F^{-1}: S^1 \times \mathbb{R} \rightarrow A, \quad (\xi,\eta,z) \mapsto (\xi \sqrt{z^2+1},\eta \sqrt{z^2+1},z) $$ looks like a good candidate.

Now, both $A$ (at glance) and $S^1 \times \mathbb{R}$ are $2$ dimensional smooth manifolds. My firt approach would be to exhibit explicit atlases $\mathscr{A}=\{(U_i,\phi_i)\}$ for $A$, $\mathscr{B}=\{(V_i,\psi_i)\}$ for $S^1 \times \mathbb{R}$; then for each chart check that $\tilde{F_i}= \psi_i \cdot F \cdot \phi_i^{-1}$ between open sets in $\mathbb{R}^2$ is smooth. Charts in $\mathscr{B}$ are canonical; for $\mathscr{A}$ I was thinking about something like this:

Atlas for $A$

$A$ without a "meridian" $M_+=\{(x,y,z) : y=0, \quad x^2-z^2=1, \quad x>0 \}$ is open in $A$ (with the induced topology, of course), so a homeomorphism between $A \setminus M_+$ and $\mathbb{R}^2$ may be given by the projection of $(x,y,z) \in A \setminus M_+ $ from $(\sqrt{z^2+1},0,z)$ to the plane $x=0$ ; something like a generalized stereographic projection. This is a chart; the second one is of course obtained using $M_-$ and projecting the same way. This is just and idea, I haven't done any calculation.

Still the question is: is this necessary? In the mentioned post Manifold diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$. the author briefly concludes with

Both f and g are smooth since their components are smooth. So f is a diffeomorphism.

And of course this is a lucky case with only a bunch of charts, but I believe there must be of course a way of checking whether $F$ is a diffeomorphism without showing explicits atlases.

How can this be done? Any help is appreciated!

  • 1
    $\begingroup$ If $M, N$ are smooth manifolds and $f:M\to N$ is a diffeomorphism, and if $\Sigma$ is a submanifold of $M$, then by pushing definitions you see that $f|_\Sigma$ gives a diffeomorphism of $\Sigma$ to its image. So, embed $\mathbb{S}^1\times \mathbb{R}$ in $\mathbb{R}^3$ in the trivial way, and you can write an explicit diffeomorphism of $\mathbb{R}^3$ to itself that maps the embedded copy of $\mathbb{S}^1\times\mathbb{R}$ to the hyperboloid. $\endgroup$ – Willie Wong Sep 2 '16 at 15:45
  • $\begingroup$ @WillieWong let me see if I got it: if I find a diffeomorphism $f$ of $\mathbb{R}^3$ to itself such that the image of the (trivial) embedding of $S^1 \times \mathbb{R}$ in $\mathbb{R}^3$ (read cylinder $C$) under $f$ is the hyperboloid, then $f|_C$ is the wanted diffeomorphism. Of course this is not the case of $F$ as written in the question, so I'd need a hint to write this new diffeomorphism. Plus, I still don't know how to show $F$ as above is a diffeomorphism without exhibiting charts. $\endgroup$ – DavideL Sep 2 '16 at 17:58
  • $\begingroup$ In terms of $f$: take $f$ such that for every constant $z$ slice it is a rescaling. e.g. $$ f(x,y,z) = \left( \sqrt{1 + z^2} x, \sqrt{1 + z^2} y, z\right) $$ then the cylinder $\{ x^2 + y^2 = 1\}$ gets mapped to the hyperboloid. $f$ is manifestly a diffeomorphism since $f$ is smooth and its jacobian = $1 + z^2$ is non-vanishing everywhere. $\endgroup$ – Willie Wong Sep 3 '16 at 14:50
  • $\begingroup$ The restriction of $f$ to the cylinder is in fact your $F^{-1}$... $\endgroup$ – Willie Wong Sep 3 '16 at 14:51

As I can't continue the discussion started in the comments (reputation got me here...), I'll try to explain what Willie Wong meant (or, at least, what I understood).

Define $f$ as the function : $$ f : \mathbb{R}^3 - \{0\}\longrightarrow \mathbb{R}^3 , \qquad (x,y,z) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, z\right) $$

Taking your definitions of atlases, you need to show that $\psi_i \circ F \circ \phi_i^{-1}$ is a diffeomorphism. With the definitions just above, we have : $$ \psi_i \circ F \circ \phi_i^{-1} = \psi_i \circ f \circ \phi_i^{-1} $$ where $\phi_i^{-1}$ is defined. This is because $f|_A = F$ and $A \subset \mathbb{R}^3$ and $f(A) \subset \mathbb{S}^1 \times \mathbb{R}$.

Recalling that $\psi_i$, $\phi_i$ and $f$ are diffeomorphisms, we conclude that $\psi_i \circ F \circ \phi_i^{-1}$ is smooth because $\psi_i \circ f \circ \phi_i^{-1}$ is smooth.


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