Showing that the cone $x^2-y^2-z^2=0, x\geq 0$ is not an embedded submanifold I came into the exercise that the cone $x^2-y^2-z^2=0, x\geq 0$ is not an embedded submanifold.
I think I know how to do this but the hint says to use Hadamard's lemma which I don't see how it is applicable. Am I missing something?
 A: I'm not entirely sure how to construct a proof using Hadamard's lemma either. Here's how I would argue:
I will look at the manifold $z^2-y^2-x^2=0$, $z\geq 0$, which is the same as the one you gave up to rotation. For a $k$-dimensional embedded submanifold $S\subset\mathbb{R}^n$, each point $p\in S$ has a neighborhood $U$ such that there are $k$ smooth functions $g_1,\dotsc, g_k:\mathbb{R}^n\to\mathbb{R}$ with linearly independent gradients such that
$$S\cap U=\{x\in U: g_1(x)=\dotsm =g_k(x)=0\}.$$
Now observe that $\nabla g_1(0,0,0)$ of your problem must be orthogonal to the vectors $(1,1,1)$, $(1,-1,1)$ and $(-1,1,1)$. Indeed, $\frac{1}{h}(g_1(\pm h, \mp h, |h|)-g_1(0,0,0))=0$, so the corresponding directional derivatives $\nabla g_1\cdot (\pm 1,\pm 1, 1)/\sqrt{3} = 0$ and the claim follows. This is absurd though, since it would imply the existence of four linearly independent vectors in $\mathbb{R}^3$.
The problem is even easier when the cone is two-sheeted:
Recall that an embedded submanifold of $\mathbb{R}^3$ will be a manifold with respect to the subspace topology. In particular, there are (connected) neighborhoods $U\subset\mathbb{R}^3$ containing the origin $(0,0,0)\in C$ and $W\subset\mathbb{R}^2$ such that $f: U\cap C\to W$ is a homeomorphism, where $C$ is the two-sheeted cone of your problem. Such a homeomorphism is impossible, since if we remove the origin from the domain, $U\cap C$ splits into two connected components, but $W$ will remain connected upon removal of a single point.
This is essentially the same argument commonly used to show that the figure-8 is not an embedded submanifold of $\mathbb{R}^2$.
