- #1
ConfusedMonkey
- 42
- 15
I feel embarrassed to ask this, but I may have a misunderstanding in my understanding of some basics.
I was told that ##\psi: U \rightarrow \psi(U)##, where ##U = (0, \infty) \times (0, \pi) \times (-\pi, \pi)## and ##\psi(\rho, \varphi, \theta) = (\rho\cos\theta\sin\varphi, \rho\sin\theta\sin\varphi, \rho\cos\varphi)##, is not a coordinate map, but that ##\psi^{-1}## is a coordinate map.
But doesn't the co-domain of a coordinate map have to be an open subset of ##\mathbb{R}^n## with Cartesian coordinates? The co-domain of ##\psi^{-1}## is a subset of ##\mathbb{R}^3## with spherical coordinates so how could ##\psi^{-1}## be a coordinate map?
EDIT: I should add that I am using the convention that the domain of my coordinate maps are open subsets of my manifold, and the person who corrected me on the ##\psi^{-1}## thing is aware of this.
I was told that ##\psi: U \rightarrow \psi(U)##, where ##U = (0, \infty) \times (0, \pi) \times (-\pi, \pi)## and ##\psi(\rho, \varphi, \theta) = (\rho\cos\theta\sin\varphi, \rho\sin\theta\sin\varphi, \rho\cos\varphi)##, is not a coordinate map, but that ##\psi^{-1}## is a coordinate map.
But doesn't the co-domain of a coordinate map have to be an open subset of ##\mathbb{R}^n## with Cartesian coordinates? The co-domain of ##\psi^{-1}## is a subset of ##\mathbb{R}^3## with spherical coordinates so how could ##\psi^{-1}## be a coordinate map?
EDIT: I should add that I am using the convention that the domain of my coordinate maps are open subsets of my manifold, and the person who corrected me on the ##\psi^{-1}## thing is aware of this.