Show that $M=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2=z^2, z>0\}$ is a smooth 2-d manifold.

Question

Let $M=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2=z^2, z>0\}$. How do I show that M is a smooth 2-dimension manifold?

Answer 1

By the Implicit Function Theorem, a level set $f(x,y,z)=0$ is locally diffeomorphic to $\mathbb{R}^2$ at any point where the gradient $\nabla f(x,y,z)$ is nonzero.

Here, the relevant $f$ is $f(x,y,z)=x^2+y^2-z^2$, which has gradient $(2x,2y,-2z)$. That gradient is nonzero everywhere except the origin. In particular, it's nonzero for $z>0$, and $M$ is locally diffeomorphic to $\mathbb{R}^2$ everywhere. That makes it a smooth $2$-manifold.

The IFT also tells us that, since the $z$-coordinate of the gradient is always nonzero, we can build smooth charts of the form $z=g(x,y)$. Solving that, we get $$z=\sqrt{x^2+y^2}\quad\text{for}\quad(x,y)\neq (0,0)$$ That's a smooth global diffeomorphism between $M$ and the punctured plane.