[Math] Coordinate systems on manifolds

Question

I am fairly new to differential geometry and something I can't get my head around is, if an $n$-dimensional manifold is locally homeomorphic to $\mathbb{R}^{n}$, i.e. Euclidean space, then isn't it possible to cover any manifold with a collection of coordinate charts whose coordinates are just the usual Cartesian coordinates of Euclidean space? Why does one need to even consider more general, cuvilinear coordinate systems, other than that they may simplify the problem at hand?

For example, the 2-sphere $S^{2}$ can be locally described (perhaps most easily) by spherical polar coordinates $(\theta , \phi)$ that can be mapped to local Cartesian coordinates, $x^{1}=\sin (\theta)\cos (\phi),\; x^{2}=\sin (\theta)\sin (\phi),\; x^{3}=\cos (\theta)$. Couldn't one equally just start from the definition of $S^{2}=\lbrace (x^{1},x^{2},x^{3})\in\mathbb{R}^{3}\;\vert\; (x^{1})^{2}+(x^{2})^{2}+(x^{3})^{2}=1\rbrace$ and just use Cartesian coordinates (forgoing curvilinear coordinates altogether)?

However, I have read that, in general, curved manifolds cannot be described even locally by Cartesian coordinates. I'm confused how this is the case when supposedly all manifolds are locally homeomorphic to Euclidean space?

Answer 1

A couple points:

1. the coordinate chart (or inversely a patch) does not approximately describe the manifold. It is exactly on the manifold. For example, at the top of the unit-sphere in $\mathbb{R}^3$ near the point $(0,0,1)$ it is true that $z=1$ locally approximates the sphere (it is the tangent plane), however, it is certainly not true that $\Phi(x,y) = (x,y,1)$ provides a patch of the sphere near $(0,0,1)$. We could use $\Psi ( x, y) = (x,y, \sqrt{1-x^2-y^2} )$ as the image of $\Psi$ is on the sphere.
2. there are abstract examples of manifolds formed by sets of matrices, or projective spaces. Such examples have points which are not even in $\mathbb{R}^n$, thus, without some fine print, it is clearly impossible to use coordinates in $\mathbb{R}^n$ as coordinates for such manifolds. But...
3. item 2. is not quite as imposing as it appears because it is usually possible to find a model of the abstract manifold which fits inside some copy of $\mathbb{R}^n$. Moreover, Whitney's Embedding Theorem and Nash's Embedding Theorem show that we can find a set $S$ inside $\mathbb{R}^k$ for $k$ sufficiently large to represent an abstract $n$-dimensional manifold $\mathcal{M}$ in such a way that $S$ has the same structure as $\mathcal{M}$. That structure could involve the metric, or just the topology, it depends on the type of manifold and theorem we wish to invoke. I point you to the links.
4. what is distance on a manifold ? This would seem to be part of your current confusion. For a given point-set, there are multiple structures we can place. For example, the plane can be given a metric which gives it spherical or hyperbolic geometry (angles add up to more or less than 180 degrees in a triangle). Of course, those metrics are not induced from the ambient Euclidean metric in $\mathbb{R}^2$. Likewise, for manifolds, the metric need not be induced from the metric on the larger space on which it is embedded. We develop a theory of geometry for Riemannian manifolds which is completely based on the abstract structure of the manifold itself. The intrinsic geometry of a manifold is independent of the details of its embedding. This is a bit of a mind-bender when you first come across the idea. In the classical differential geometry we have some mixture of intrinsic and extrinsic quantities for surfaces. For example, the mean curvature is extrinsic whereas the Gaussian curvature is intrinsic.