Is $S$ a regular submanifold of $\Bbb R^{3}$?

Question

$$S=\{(x,y,z) \mid x^{2}+y^{2}=z^{2}\}$$

$g: \Bbb R^{3}\to \Bbb R$, $S=g^{-1}(0)$

Is $S$ a regular submanifold of $\Bbb R^{3}$?

I'd be grateful for a clear and explicit explanation of why this is regular submanifold.

Answer 1

No it's not. As you say, define $g(x,y,z) := x^2+y^2-z^2$ so that $S=g^{-1}(0)$.

Consider the Jacobian matrix of $g : \mathbb{R}^3 \to \mathbb{R}$. This will be a $1 \times 3$ matrix, namely:

$$J_g = \left( \frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}\right) = \left(2x,2y,-2z\right).$$

The critical points of $g$ are given by the points for which $J_g$ has less than maximal rank, i.e. rank less than one. In this case, we need $x=y=z=0$. The only critical point of $g$ is $(0,0,0)$.

Next we check for critical values. We look at the image of the critical points under $g$. Sadly $g(0,0,0) = 0$, and so $0$ is a critical value of $g$. That means that $0$ is not a regular value of $g$ and hence $g^{-1}(0)$ is not a regular submanifold.

Geometrically, $S$ is a circular cone. The cross-sections, for fixed $z$, are all circles. For example, fixing $z=1$ gives the unit circle $x^2+y^2=1$ sitting in the $z=1$ plane. The vertex of this cone is the origin $(0,0,0)$. (That's why $(0,0,0)$ was a critical point of $g$.)