2-Sphere surface coordinate dimension

Question

Ordinary sphere in $\mathbb{R}^3$ is two-dimensional object (2-sphere), i.e. it requires at least two coordinates to define point on a surface. As I notice, however, there is a catch.
If we use spherical coordinates,$(\phi,\theta)$, then there are two points where $\phi$ is not defined (polar regions). If we use stereographic coordinates, $j=x+iy$, then there is one point where both $x$ and $y$ are infinite, hence not defined. But if we use three coordinates (e.g. cartesian), then we have no such problem.
In that sense, is 2-sphere a three-dimensional object, i.e. there is some dimension function that is equal three for 2-sphere (but equal two for 2-sphere with two points excluded)? Or is there a way to use two coordinates and not to encounter such 'problematic' points?

Answer 1

some dimension function that is equal three for 2-sphere (but equal two for 2-sphere with two points excluded)?

Yes: the smallest number $n$ such that the space can be embedded into $\mathbb{R}^n$, in the sense of being diffeomorphic to a submanifold of $\mathbb{R}^n$. This is sometimes called the embedding dimension.

• The embedding dimension of $S^2$ is $3$, since it embeds into $\mathbb{R}^3$ but not into $\mathbb{R}^2$.
• The embedding dimension of $S^2$ with two points excluded is $2$, since this set is diffeomorphic to $\mathbb{R}^2$ minus a point.

(The embedding dimension of $S^2$ with one point excluded is also $2$, thanks to the stereographic projection.)

Answer 2

First, there is no way to cover the sphere with a single coordinate system. If there were, the sphere would be diffeomorphic to an open set in the plane, which it isn't. (Generally, a manifold of positive dimension can be covered by a single "coordinate chart" only if it is diffeomorphic to some open subset of a Euclidean space. Particularly, no compact manifold has this property. Some non-compact manifolds, such as the complement of a point in a torus, also cannot be covered by one coordinate system.)

Second, "ambient" Cartesian coordinates aren't suitable as "local coordinates" on a sphere because "they don't vary independently". Generally, I know of no intrinsic sense in which the unit sphere in $\mathbf{R}^{3}$ is anything but "$2$-dimensional".

Answer 3

A spherical surface in $\mathbb{R}^3$ is a two dimension manifold that is not globally diffeomorphic to a plane. But we can establish a diffeomorphism locally. This means that in a neighborhood of any point we can define a smooth function that is a diffeomorphism between the neighborhood and $\mathbb{R}^2$.The spherical coordinates $(\phi, \theta)$ are an example of how we can do this. The dimension of the manifold is, by definition, the dimension $n$ of the locally diffeomorphic space, so, for the sphere this dimension is $2$.