Can we interpret spherical polars as covering spaces?

29 Views Asked by At

My understanding of polar coordinates is that we are implicitly working on a covering space of the punctured plane, given by:

$p: \mathbb{R}^+ \times \mathbb{R} \to \mathbb{R}^2 \setminus \{0\} $

$ p(r,\theta) = (r \cos(\theta), r \sin(\theta))$

Is there a similar interpretation of spherical polars?

1

There are 1 best solutions below

0
On

No, because the mapping at the poles is not a local homeomorphism.

But if, as in your example, you remove the polar axes (i.e., map $R^{+} \times R \times R \to R^3 - \{(0, y, 0) : y \in R \}$) via $$ (r, \theta, \phi) \mapsto r (\cos \theta \cos \phi, \sin \theta, \cos \theta \sin \phi), $$ then you indeed do get a covering space.

In your example, you can actually include the origin, and you get what's called a "branched cover." Arguably, spherical polar coordinates are a generalization of that idea.