Unit sphere difficult isometry

Question

Let $S^2$ be the unit sphere $x^2+y^2+z^2=1$ and let $f:\mathbb{R^2} \to S^2$ , $(u,v) \to (x,y,z)$. The point $(x,y,z)$ is the second point where the line that passes through $(u,v,0)$ and $(0,0,1)$ intersects $S^2$, and give $ \mathbb{R^2}$ the usual metric.

I found that the point $$(x,y,z)= ({2u \over u^2+v^2+1} , {2v \over u^2+v^2+1}, 1-{2 \over u^2+v^2+1})$$

so $$f(u,v)=({2u \over u^2+v^2+1} , {2v \over u^2+v^2+1}, 1-{2 \over u^2+v^2+1})$$

The question is : What metric should we give $S^2$ so that $f$ is an isometry?

I think that the usual metric of $\mathbb {R^3}$ doesn´t work but I can´t think of another metric such that $f$ is an isometry

Any hints or suggestions would be highly appreciated

Answer 1

A broad hint, but not a complete solution

At the point $f(u, v)$, can you find a basis for the tangent space, a pair of vectors $t_1$ and $t_2$ that span the space? Hint: you could take a (standard) orthonormal basis of the tangent space to $\Bbb R^2$ and push it forward by $Df(u, v)$ to get $t_1, t_2$.

Now you know that in your new metric, you need $g(t_1, t_1) = g(t_2, t_2) = 1$ and $g(t_1, t_2) = 0$. That completely defines $g$! For if $p$ and $q$ are any two tangent vectors at $f(u, v)$, you can write $$ p = at_1 + b t_2\\ q = c t_1 + d t_2 $$ and then work out $q(p, q)$ via bilinearity.

Now you're probably hoping to consume tangent vectors at $f(u, v)$ that are written in xyz-coordinates, rather than in terms of $t_1, t_2$. So you'll need a change of basis from xyz-coords to coords in the $t$-basis. Put all that together, and you're done.

Answer 2

By definition $$f : (\Bbb R^2, \bar g) \to S^2 - \{N\}$$ (here we denote $N := (0, 0, 1)$) is an isometry if and only if for any point $x \in \Bbb R^2$ and any vectors $x, y \in T_x \Bbb R^2$ we have $$\bar g(X, Y) = g(T_x f \cdot X, T_x f \cdot Y),$$ where $T_x f : T_x M \to T_{f(x)} (S^2 \setminus \{N\})$ is the pushforward (differential) map. By definition, the expression on the r.h.s. is the pullback $f^* g .$ of $g$ by $f$, and so $$\bar g = f^* g ,$$ or just as well $$g = (f^{-1})^* \bar g .$$ But we can compute $f^{-1}$ explicitly (in fact, as Christian Blatter points out in his very good answer, this map is exactly stereographic projection from $N$), and we already know the coordinate representation of $\bar g$, namely, $\bar g = du^2 + dv^2$.

Answer 3

Note that the word isometry is used in two different contexts: "isometry of metric spaces" and "isometry of Riemannian manifolds". It is not clear which of the two you have in mind. The answers of Travis and John Hughes refer to the isometry of Riemannian manifolds. I shall deal with the first interpretation.

It has not yet been said that your map $f$ is well known. It is the inverse of the stereographic projection $$\sigma\!: \>\dot S^2\to{\mathbb R}^2,\qquad(x,y,z)\mapsto (u,v):=\left({x\over1-z}, \ {y\over 1-z}\right)\ .\tag{1}$$ Here $\dot S^2$ denotes the sphere $S^2$, punctured at $(0,0,1)$. The formula $(1)$ (which follows easily from your formula for $f$) allows to transport the euclidean metric $$d\bigl((u_1,v_1),(u_2,v_2)\bigr):=\sqrt{(u_1-u_2)^2+(v_1-v_2)^2}\tag{2}$$ from the plane ${\mathbb R}^2$ to $\dot S^2$. If $${\bf r}_i=(x_i,y_i,z_i)\in\dot S^2\qquad(i=1,2)$$ are two given points on the sphere we simply define $$\hat d({\bf r}_1,{\bf r}_2):=d\bigl(\sigma({\bf r}_1),\sigma({\bf r}_2)\bigr)$$and obtain, by plugging $(1)$ into $(2)$: $$\hat d({\bf r}_1,{\bf r}_2)=\sqrt{\left({x_1\over1-z_1}-{x_2\over1-z_2}\right)^2+ \left({y_1\over1-z_1}-{y_2\over1-z_2}\right)^2}\ .$$