First of all, the metric tensor is one additional piece of structure one inserts on a smooth manifold to measure lenghts and angles. The metric is indeed not present in all applications of Differential Geometry to Physics (see e.g. Lagrangian Mechanics). In that case, it is important to know also how to deal with manifolds without metric tensors.

Now, about the coordinate systems the point is that indeed usually manifolds require more than one to be covered. Take the two sphere $S^2$ for example, you need at least two stereographic projections to cover it all. The idea, however is not to piece two coordinate systems to get a global one.

The idea is that given a point, around it there is some coordinate system that works and if that you have any overlapping one, you can be sure that the results and definitions don't depend on which coordinate system you use. One more readily seem example is cartesian and spherical coordinates in $\mathbb{R}^3$: you can use any one of them.

If $(x,U)$ and $(y,V)$ are two coordinate systems, on the overlap $U\cap V$, if you make sure results independ on the coordinate system you can think of them as intrinsic to $M$ and yet use coordinates to carry down calculations.

You can't assume that there is just one coordinate system because if you look at examples you find objects which you certainly want to consider as manifolds which cannot be covered by just one coordinate system.

To clarify those points I recommend you take a look in this two books

- Modern Differential Geometry for Physicists - C. J. Isham
- A Comprehensive Introduction to Differential Geometry Vol. 1 - Michael Spivak

Book 2 is more technical and it's for mathematicians, but it's very good. I recommend you first look at 1 and then look at some things in 2 to see some more detailed constructions.

**Edit:** One counterexample might help you out, so I decided to give one. If $M$ is a smooth manifold and if $(x,U)$ is a coordinate system, then $x : U\subset M\to \mathbb{R}^n$ is a homeomorphism. If there exists one global coordinate system $(x,M)$ then $M$ is homeomorphic to $\mathbb{R}^n$ itself. This is problematic because many manifolds one encounters (not just in Math but in Physics as well) have a more complicated topology.

Let's show that the sphere $S^2$ cannot have a global coordinate system.

$S^2$ is compact: in truth, $S^2$ is endowed with the subspace topology and because of that, it suffices to show $S^2$ regarded as a subset of $\mathbb{R}^3$ is closed and bounded. Bounded it's easy, if $p\in S^2$ then $|p|=1$, hence $|p|< 2$ so that $S^2\subset B(0, 2)$ where $B(a,r)$ is the ball centered in $a$ with radius $r$. Closed is also easy: $S^2 = \{(a,b,c)\in \mathbb{R}^3 : a^2+b^2+c^2 = 1\}$ so that if we set $f : \mathbb{R}^3\to \mathbb{R}$ by $f(a,b,c)=a^2+b^2+c^2$ then $S^2 = f^{-1}(1)$, but $f$ is continuous and $\{1\}$ is a closed set so that $S^2$ is closed. Since $S^2$ is closed and bounded, $S^2$ is compact.

Suppose now that $S^2$ has a global coordinate system $(x,S^2)$ then $x: S^2\to \mathbb{R}^2$ is a homeomorphism, but since $S^2$ is compact, then $\mathbb{R}^2$ is compact which is obviously wrong. So we are forced to conclude $S^2$ has no global coordinate system.

So your procedure gives one $n$-tuple of numbers for each point of the manifold, but it doesn't respect the topological structure. If we suppose a global coordinate system for the sphere, you get an absurd unless you accept coordinates which are not continuous and those are not really interesting.

Now regarding the metric tensor: I'll say again, the metric is not something you deduce from coordinates. The metric is something you postulate. In GR in particular it is the solution to Einstein's Equations. The formula you say about is just a way to relate coordinate representations of the metric tensor in different coordinate systems, not a way to deduce it from coordinates. For example if $M = \{(a,b)\in \mathbb{R}^2 : b > 0\}$ and if you use cartesian coordinates $(x,y)$ you can define $g$ in coordinates by

$$g = \dfrac{dx\otimes dx + dy\otimes dy}{y^2}$$

This is the coordinate representation of $g$ in **this** coordinate system. If you choose any other coordinates, $g$ will transform it's representation according to the formula you gave. Also, see that I postulated the metric, instead of deducing it.

This post imported from StackExchange Mathematics at 2015-01-20 11:58 (UTC), posted by SE-user user1620696