# Hyperboloid is a manifold

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 $H_{z=0}$?
• Why does $a$ not have a sign? If $a=0,$ not a manifold. If $a > 0,$ one sheet. If $a < 0,$ two sheets. Jan 16 '18 at 0:42
• $a$ is assumed to be positive, fixed. Jan 16 '18 at 0:44
• Do you know what a hyperboloid looks like? Can you see that it is diffeomorphic to a cylinder $\mathbb{R} \times S^1$? Jan 16 '18 at 5:49
• @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? Jan 17 '18 at 1:21
• 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 $P\ne 0$. Jan 17 '18 at 1:50

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