# Exercise in Lee: Proving $\mathbb{S}^2$ is an embedded submanifold of $\mathbb{R}^3$

Exercise: Show that spherical coordinates form a slice chart for $\mathbb{S}^2$ in $\mathbb{R}^3$ on any open subset where they are defined.

Solution: For this post I will just show this for one portion of the sphere, for brevity. $(U, \psi)$ is a spherical coordinate chart on $\mathbb{R}^3$ where $U = (0, \infty) \times (0, \pi) \times (-\pi, \pi)$ and $\psi(\rho, \varphi, \theta) = (\rho\cos\theta \sin\psi, \rho\sin\theta\sin\varphi, \rho\cos\varphi)$.

We know that $\mathbb{S}^2 \cap U$ is just the portion of the sphere where $y > 0$ (in Cartesian coordinates). So, $\psi(\mathbb{S}^2 \cap U) = \{(\cos\theta\sin\varphi, \sin\theta\sin\varphi, \cos\varphi): -\pi < \theta < \pi, 0 < \varphi < \pi\}$. This isn't a $2$-slice of $\psi(U)$ because none of the coordinates are constant. Where am I going wrong?

What you wrote down is $\psi(U) \cap \mathbb S^2$, not $\psi(U\cap \mathbb S^2)$. The spherical coordinate map is $\psi^{-1}$, not $\psi$.
• But doesn't the co-domain of a coordinate map have to be an open subset of $\mathbb{R}^n$ where $\mathbb{R}^n$ has Cartesian coordinates? $\psi^{-1}$ has as its co-domain $\mathbb{R}^n$ with spherical coordinates. Jun 1 '17 at 0:31
• In this case, both the domain and the codomain have to be open subsets of $\mathbb R^3$. But you need a map $F$ with the property that $F(\mathbb S^2\cap U)$ is a coordinate slice. That is satisfied by $\psi^{-1}$, not by $\psi$. Jun 1 '17 at 19:33
• @ConfusedMonkey: I don't know what you mean by "its codomain does not have Cartesian coordinates." The codomain $U$ of $\psi^{-1}$ is an open subset of $\mathbb R^3$, which we always (unless otherwise specified) consider as a smooth manifold with its standard coordinates. The fact that we're calling the coordinate functions $(\rho,\phi,\theta)$ in this case instead of $(x,y,z)$ does not change the fact that they are the standard coordinates of $\mathbb R^3$. Jun 5 '17 at 22:49
• @ConfusedMonkey: But in the set $U = (0, \infty) \times (0, \pi) \times (-\pi, \pi)\subseteq\mathbb R^3$ where $\psi^{-1}$ takes its values, $\rho$ is just the first component, $\phi$ is the second component, and $\theta$ is the third component. These are Cartesian coordinates on that set! Jun 5 '17 at 23:32
• @ConfusedMonkey: $\psi^{-1}(1,0,0) = (1,\pi/2,0)$, which is just an element of $\mathbb R^3$. Jun 5 '17 at 23:48