1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $\psi$.

$\endgroup$
9
  • $\begingroup$ 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. $\endgroup$ Jun 1 '17 at 0:31
  • 2
    $\begingroup$ 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 $\psi$. $\endgroup$
    – Jack Lee
    Jun 1 '17 at 19:33
  • 1
    $\begingroup$ @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 $\mathbb R^3$. $\endgroup$
    – Jack Lee
    Jun 5 '17 at 22:49
  • 1
    $\begingroup$ @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! $\endgroup$
    – Jack Lee
    Jun 5 '17 at 23:32
  • 1
    $\begingroup$ @ConfusedMonkey: $\psi^{-1}(1,0,0) = (1,\pi/2,0)$, which is just an element of $\mathbb R^3$. $\endgroup$
    – Jack Lee
    Jun 5 '17 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.