4
$\begingroup$

I'm trying to show that the hyperboloid $x^2+y^2-z^2=a$ ($a > 0$) is a manifold in the sense that every point on it has a neighborhood diffeomorphic to an open subset of $R^n$ (or $R^n$ itself).

First let us show that all points of the hyperboloid with $z > 0$ (call this set $H_{z > 0}$) have the desired property. Let $U$ be the exterior of the closed ball of radius $a$ centered at the origin in $R^2$. Define the parametrization $\phi_1: U\rightarrow H_{z > 0}$ by $$(u,v)\mapsto (u,v,\sqrt{u^2+v^2-a}).$$

The image lies in $H_{z > 0}$: Every point in $U$ is of the form $be^{i\theta},\ b> \sqrt a, \theta \in [0, 2\pi)$, so $(u,v)=(b\cos\theta,b\sin \theta)\mapsto (b\cos\theta,b\sin\theta,\sqrt{b^2-a})$. Because $b> \sqrt{a}, b^2> a$, so the 3rd coordinate is $> 0$.

This map is injective because if $(u,v,\sqrt{u^2+v^2-a})=(s,t,\sqrt{s^2+t^2-a})$, then $u=s,v=t$. It is surjective for the following reason. Every point on $H_{z > 0}$ is of the form $(b\cos \theta, b\sin\theta, z)$ with $z> 0, b> \sqrt a. \theta \in [0, 2\pi)$. Its preimage is $(b\cos\theta, b\sin\theta)$.

This map is smooth and its inverse, $(u,v,\sqrt{u^2+v^2-a})\mapsto (u,v)$, is also smooth. (So the argument about bijectivity above is not necessary since there is an explicit inverse.)

The parametrization of $H_{z < 0}$ is defined similarly.

It remains to show that every point on $H_{z = 0}$ has the desired property. For this I guess I should use something like $(u,v)\mapsto (\pm\sqrt{a+u^2-v^2},u,v)$, or $(u,\pm\sqrt{a+u^2-v^2},v)$. But in this case I cannot determine which set a neighborhood of $z=0$ in the hyperboloid is diffeomorphic to.

So, my questions are:

  • Is what I've written so far correct? (In particular, are there any gaps?)
  • How to proceed with $H_{z=0}$?
$\endgroup$
6
  • $\begingroup$ Why does $a$ not have a sign? If $a=0,$ not a manifold. If $a > 0,$ one sheet. If $a < 0,$ two sheets. $\endgroup$
    – Will Jagy
    Jan 16 '18 at 0:42
  • $\begingroup$ $a$ is assumed to be positive, fixed. $\endgroup$
    – user500094
    Jan 16 '18 at 0:44
  • $\begingroup$ Do you know what a hyperboloid looks like? Can you see that it is diffeomorphic to a cylinder $\mathbb{R} \times S^1$? $\endgroup$ Jan 16 '18 at 5:49
  • $\begingroup$ @QiaochuYuan Yes I know how it looks like, and on intuitive level it is clear that it is diffeomorphic to a cylinder. But I haven't tried to prove this rigorously. How can this be helpful? $\endgroup$
    – user500094
    Jan 17 '18 at 1:21
  • $\begingroup$ There is a better way to do this by using the regular value theorem: for $f(x,y,z)=x^2+y^2 -z^2$ show that $\nabla f(P)\ne 0$ for each $P\ne 0$. $\endgroup$ Jan 17 '18 at 1:50
1
$\begingroup$

For $z=0$, the goal is to map the neighbourhood around the circle to a subset of $\mathbb{R}^2$. Obtain four semicircles from this circle so that no point $(x,y,0)$ is left out (like picking one for each half of $x$ axis, $y$ axis respectively). Pick an open disc of radius a: $\mathbb{B}_a(0,0)$ in $\mathbb{R}^2$. Define the mapping from $\mathbb{B}_a(0,0)$ to the neighbourhood for each semi-circle as:

$(u,v) \to (\sqrt{a+ v^2 - u^2},u,v)$

$(u,v) \to (-\sqrt{a+ v^2 - u^2},u,v)$

$(u,v) \to (u,\sqrt{a+ v^2 - u^2},v)$

$(u,v) \to (u,-\sqrt{a+ v^2 - u^2},v)$

With this you have a smooth mapping for every point on the hyperboloid.

$\endgroup$

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.