# How does the stereographic map from $S^{2}$ to $\mathbb R^2$ “induce” a metric on $\mathbb R^2$

Let $S^2\subset\mathbb{E}^3$ be a sphere with radius $1$ and center $(0,0,1)$ in cartesian coordinates. The northpole is point $(0,0,2)$ on the sphere and $Oxy$ is the $xy$-plane. Let $\pi:S^2\setminus N \to Oxy$ be a stereographic projection defined by $\pi(x,y,z)=(u,v)$ where $(u,v)$ is the intersection of the $xy$-plane and of a straight line through the the north pole and point $(x,y,z)$ on the sphere. I know $\pi$ is given by $$\pi(x,y,z)=(u,v)=\left(\frac{2x}{2-z},\frac{2y}{2-z}\right),$$ and its inverse $\pi^{-1}:\mathbb{R}^2\to S^2\setminus N$ by $$\pi^{-1}(u,v)=\left(\frac{4u}{u^2+v^2+4},\frac{4v}{u^2+v^2+4}, \frac{2(u^2+v^2)}{u^2+v^2+4}\right)$$ The sphere can be parametrized by $$r(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).$$

I understand how a submanifold inherits an induced metric from a manifold in this case we do not have such a relation as the $XY$ plane is not a submanifold of $S^2$. What does induced metric mean in this case? How should I go about determining it? Thanks in advance.

Consider the map $\pi^{-1}$, it is a diffeomorphism between $\mathbb{R}^2\to S^2\setminus N$.

$S^2\setminus N$ has the metric induced by $S^2$, let us call this metric $g$. So you can pull-back this metric on $\mathbb{R}^2$. The resulting metric ${\pi^{-1}}^*g$ is defined by $$({\pi^{-1}}^*g)_x (X_x,Y_x) = g_{\pi^{-1}x}(d_x\pi^{-1}X_x, d_x\pi^{-1}Y_x)$$

where $X_x,Y_x \in \mathbb{R}^2$ are tangent vector at $x\in \mathbb{R}^2$ and $d_x\pi^{-1}$ is the differential of the map $\pi^{-1}$ at $x$.

Since $\pi^{-1}$ is a diffeomorphism, its differential at $x$ is a bijection between tangent spaces, so the pull back metric has the same properties of the staring metric (for example they are both Riemannian).

• Thanks, that seems to make sense to me. I guess I got too caught up with the submanifold part of "induced" metric. – dak. cola Mar 11 '18 at 17:22

Let $\rho: S^2 \times S^2 \to \mathbb{R}_{\geq0}$ be the metric on the sphere. Define a metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}_{\geq0}$ on the plane by $d(a,b) = \rho(\pi^{-1}(a),\pi^{-1}(b))$.

• Thanks, I should have been more clear as to what type of metric, I meant a metric in the sense of @Warlock of Firetop Mountain 's answer. – dak. cola Mar 11 '18 at 17:29

Let $(u_1,v_1),\, (u_2,v_2)$ be two points in the plane. Define a metric thus: $$D\big( (u_1,v_1),\, (u_2,v_2)\big) = \text{great-circle distance between } \pi^{-1}(u_1,v_1) \text{ and } \pi^{-1}(u_2,v_2).$$ Or instead of great-circle distance you could use chord length or some other metric on the sphere. What you get is a metric on the plane induced by some metric on the sphere.

• Thanks, I should have been more clear as to what type of metric, I meant a metric in the sense of @Warlock of Firetop Mountain 's answer. – dak. cola Mar 11 '18 at 17:29