1
$\begingroup$

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!

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.