Let $F:U \subset R^2 \rightarrow R^3$ by given by $F(u,v) = (u\sin{(\alpha)}\cos{(v)},u\sin{(\alpha)}\sin{(v)},u\cos{(\alpha}))$,
$(u,v) \in U = \{(u,v) \in R^2 \space; u > 0\}$ with $\alpha$ a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and $2\alpha$ as the angle of the vertex.
- From Do Carmo, problem 4.2.1
I was thinking of using the Inverse function theorem. $F$ is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.
Recall that the Inverse function theorem says if we have a function from $R^n \rightarrow R^n$ that is continuously differentiable on some open set containing a point $p$, and that the determinate of $Jf(a) \ne 0$ then there is some open set $W$ containing $f(a)$ such that $f:V \rightarrow W$ has a continuous inverse $f^{-1}:W \rightarrow V$ which is differentiable for all $y \in W$.
Basically it says $f$ is a diffeomorphism from $V$ to $W$ given it satisfies the above hypothesis.
Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point $p$ in the domain of F, there exists an open set $U \subset \mathrm{dom}(F)$ containing $p$ such that $f:U \rightarrow f(U)$ is a diffeomorphism.
Best Answer
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)\in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $z\gt0, x^2+y^2=z^2\tan ^2(\alpha)$ of $\mathbb R^3$.
2) This subset is indeed the intersection $C\subset \mathbb R^3$ of the upper half-space $z\gt 0$ with the cone of half-angle $\alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $\mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:U\to F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line $\{c\}\times \mathbb R\subset U \;(c\gt 0)$ is the universal covering map of the horizontal circle $C\cap\{z=c\cos \alpha\} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...