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?

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    $\begingroup$ It depends on what you ask for your coordinate systems. If you want only them to reflect the topology of the manifold, then you're right, it's always locally possible and then a collection of charts will cover your whole manifold. The thing you've read, on the other hand, deals with more that mere topology. If you want your chart to reflect geometric properties of your space (angles, lengths, areas...) then it is in general not possible, even locally. It's a thing cartographers learned a long time ago: you can represent angles nicely, areas nicely, but not both at the same time. $\endgroup$ – PseudoNeo Jun 10 '15 at 13:43
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    $\begingroup$ In what sense are the coordinates you give for $S^2$ not "curvilinear"? Your question is not a precise one but I suspect your confusion is in the distinction between (smooth) topology and geometry - curved manifolds cannot be given coordinates so that the metric takes the form $dx^2 + dy^2 + \cdots$. Remember that when dealing with spherical polar coordinates you have to use special formulae for distances, areas, etc, so the geometry is not that of the plane. $\endgroup$ – Anthony Carapetis Jun 10 '15 at 13:44
  • $\begingroup$ @PseudoNeo So if one endows the manifold with a (non-Euclidean) metric does that automatically imply that when one maps locally to $\mathbb{R}^{n}$ one cannot use Cartesian coordinates as the geometry is not flat, even locally? $\endgroup$ – Will Jun 10 '15 at 14:06
  • $\begingroup$ @AnthonyCarapetis I put spherical polar coordinates as an example, but didn't phrase it very well (have updated it to try and make things a little clearer). What I meant was, couldn't one forgo the curvilinear coordinates altogether in this case and just use local Cartesian coordinates like $x^{1}=\sqrt{1-(x^{2})^{2}-(x^{3})^{2}}$ (with appropriate restrictions), why do curvilinear coordinates even need to be introduced? $\endgroup$ – Will Jun 10 '15 at 14:06
  • $\begingroup$ @Will: I would call those "graph coordinates", and they are often quite ugly to work with - certainly they do not give you the Euclidean metric (unless the surface is locally just a plane). If you allow high enough codimension they do exist in general - any manifold can be isometrically embedded in a large enough $\mathbb R^n$, and we can just align our Cartesian axes with the tangent plane at a point of interest to get local graph coordinates. $\endgroup$ – Anthony Carapetis Jun 10 '15 at 14:16

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.
  • $\begingroup$ Thanks for the detailed answer. In part 1. of your answer are $\Phi$ and $\Psi$ maps from (a subset of) $M$ to (a subset of) $\mathbb{R}^{n}$? Part of what has confused me in reading some texts on diff. geom. is when they give an example of $S^{2}$ they define a coordinate chart such as $\phi :U\subset M\rightarrow V\subset\mathbb{R}^{n},\; \phi : (\theta, \phi)\mapsto (\sin (\theta)\cos (\phi),\sin (\theta)\sin (\phi),\cos (\theta)$. For me this confuses things as it makes it seem as if $(\theta, \phi)$ are points on the manifold, rather than coordinates $\mathbb{R}^{2}$?! $\endgroup$ – Will Jun 10 '15 at 16:00
  • $\begingroup$ No my $\Phi$ and $\Psi$ are "patches". The inverse maps are the coordinate "charts" on the sphere. Let me pick on an easier example, for the $z=1$ plane we have patch $\Phi(u,v) = (u,v,1)$. On the other hand, $\phi^{-1}(x,y,z) = (x,y)$ for all $(x,y,z)$ on the plane. Part of the confusion stems from the fact that we use $x,y,z$ as both fixed points and coordinate functions. I can write $\phi^{-1} = (x,y)$ this indicates $\phi^{-1}(a,b,c) = (x(a,b,c),y(a,b,c)) = (a,b)$. So, $(x,y)$ is a chart on the $z=1$ plane, but, understand, this views $x,y$ as real-valued functions with domain $z=1$. $\endgroup$ – James S. Cook Jun 10 '15 at 17:22
  • $\begingroup$ To answer your question more abstractly, a patch is like the parametrizations we used in calculus III to set-up flux or surface integrals. A pair of parameters $(u,v)$ gives us a point $(x(u,v),y(u,v),z(u,v))$ on the surface. So, $(u,v)$ is not "on" the surface, but, they do set-up a coordinate system on the surface in a natural way. A chart takes its domain as some subset of the manifold and its image is in $\mathbb{R}^n$. When I talk about a system of coordinates on a manifold, I'm talking about a chart. On $\mathbb{R}^3$ we have three usual charts we use in calc. III. $\endgroup$ – James S. Cook Jun 10 '15 at 17:27
  • $\begingroup$ JamesS.Cook So is the statement "locally Euclidean" specifically saying that locally points on the manifold can be represent by $n$-tuples of real numbers in $\mathbb{R}^{n}$ and says nothing about whether the coordinate space is flat or that the coordinates are Cartesian?! Is the point that coordinates are just labels to keep track of where all the points on the manifold are, so within a given patch we are free to choose any coordinate system we like (although in practice we would choose one that suited the problem at hand), not just Cartesian or spherical polar etc.? $\endgroup$ – Will Jun 10 '15 at 20:41
  • $\begingroup$ Also, for the example of the sphere, is it that we represent points in a given cooordinate patch by spherical polar coordinates (for example), such that a given point has a coordinate representation $(\theta, \phi)\in\mathbb{R}^{2}$. As the manifold is locally flat within a sufficiently small neighbourhood of each point we can then map these coordinate representations to coordinates in the ambient 3-dimensional Euclidean space, whose coordinates are Cartesian, i.e. $(\theta, \phi)\mapsto (x, y,z)\in\mathbb{R}^{3}$? $\endgroup$ – Will Jun 10 '15 at 20:48

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