I'm asked to show that $F(x,y,z)=(3x,2y,5z)$ is an diffeomorphism between $\mathbb{S}^{2}(1)$ and the ellipsoid $(\frac{x}{3})^{2}+(\frac{y}{2})^{2}+(\frac{z}{5})^{2}=1$. For this aim what I have seen is that (I) its class is infinity (each component is) (II) its inverse is given by $F^{-1}(x,y,z)=(\frac{x}{3},\frac{y}{2},\frac{z}{5})$ and (III) it's infinity class. This is in the general case because I haven't used yet that I define this map between the sphere and the ellipsoid. So, what I have done in this case is well let be $(u,v)\in(0,\pi)\times(0,2\pi)$ and let be $x(u,v)=(sin(u)cos(v),sin(u)sin(v),cos(u))$ and $y(u,v)=(\frac{sin(u)cos(v)}{3},\frac{sin(u)sin(v)}{2},\frac{cos(u)}{5})$ parametrizations of the sphere and the ellipsoid respectively. Then, $F\circ y$ carries points of the ellipsoid to the sphere and $F^{-1}\circ x$ carries points of the sphere to the ellipsoid. Is this correct?
Then I have to calculate its differential in any point of the sphere and it's inmmediate. \begin{equation} \begin{pmatrix} 3 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 5 \end{pmatrix} \end{equation}
I'm asked if this differential conserves angles.
The last thing is to determine if this map conserves de norm of vectors.
For both things before what I have to do is prove that this map is or no a local isometry. So for doing this I have calculated the normal vector $N\circ x(u,v)=x(u,v)$ so $T_{p}S=\{(x,y,z)\mathbb{R}^{3}:<(x,y,z),x(u,v)>=0\}$. But my problem is that with this expresion how to proceed to prove that $<DF(u),DF(v)>=<u,v>$ for all vectors in $T_{p}S$ (where $S=\mathbb{S}^{2}(1))$? I'm lost here.
Any hint to continue is appreciated!
This is one particular scenario where observing $S^2$ and the ellipsoid as surfaces embedded in $\mathbb{R}^{3}$ is actually much more helpful, compared to the usual intrinsic approach in case of manifolds.You just need to observe that a function $f$ defined on a subset $K$ of $\mathbb{R}^{n}$ is differentiable at a point $u$ of $K$ if it is the restriction of a differentiable map defined on an open(in $\mathbb{R}^{n}$) neighborhood $U$ of $u$ to $U\cap K$.Using this, it's easy to observe that $F$ in your query is actually the restriction of an invertible linear map $T$(i.e the non uniform scaling you have mentioned) to $S^2$.Hence, it's a diffeomorphism because the aforementioned linear map is. Next, this is not conformal.The simple reason behind this is, the $xy$-plane, $yz$-plane and the $zx$-planes are all invariant under $T$, and each of them arises as tangent planes of $S^2$ at $(0,0,1)$, $(1,0,0)$ and $(0,1,0)$ respectively.But on each of these planes $T$(and hence, $dF$) actually happens to be a non uniform scaling and hence, cannot be conformal.This is because in two dimensions, the only conformal linear maps are the scalar multiples of isometries.