Local coordinates for rotation manifold SO(3)

Question

My major is mechanical engineering. Recently, I am working on some subject involving three-dimensional finite rotations.

When I read the chapter titled "Charts on SO(3)" in Wikipedia and math books about smooth manifolds (e.g., Introduction to smooth manifolds, 2nd edition, by JM Lee). I got trouble to understand the following basic concept:

A coordinate chart on $M$ is a pair $(U,\phi)$, where $U$ is an open set of $M$, and $\phi: U\to \hat U$ is a homeomorphism from $U$ to an open set $\hat U=\phi (U)\subseteq \mathbb R^{n}$.

Does that mean the space $\hat U$ of parameters (local coordinates) should always be an Euclidean space $\mathbb R^{n}$? Taking SO(3) as an example, if we choose the Euler angles as the local coordinates, I can figue out how to establish the map $\phi$, however, as far as I know, the space of parameters (the set of Euler angles in this case) is not an Euclidean space. So, where am I misunderstanding?

This question bothers me for a while. Please help me, thank you very much!

Answer 1

It is not possible to cover $SO(3)$ with a single chart, at least using Euler angles. For instance, choosing $\alpha = 0$ or $\alpha= 2\pi$ gives precisely the same result. Note that the map is from the manifold to $\mathbb{R}^n$ and not the other way around. Hence for each point on the manifold you need to be able to unambiguously define the value in $\mathbb{R}^n$.

You're right in that if we choose to identify these points are equal, the resulting space of angles on which $\phi$ maps is no longer Euclidean. However, in this case it is not a chart in a topological sense anymore.

In general, to define an atlas on a closed manifold you need at least two charts. This is even true for a sphere. The stereographic projection maps the sphere onto $\mathbb{R}^n$, but leaves out a single point.