I would be pleased to have some information about coordinates in differential geometry.
A) First I would like to check whether or not the definitions I use are correct. (Mainly for the sake of clarity.)
Let us consider a smooth $n$-dimensional manifold $\cal M$.
- Local coordinates are defined by a diffeomorphism $f$ such that
$\begin{array}{ccc}
f \colon & U & \to & f(U) \subset \mathbb{R}^n \\
& p & \mapsto & (x^0, \ldots, x^n) \, ,
\end{array}$
where $U$ is a neighborhood of the point $p$.
- Global coordinates are given by a chart defined on the whole manifold, hence such that its domain $U$ can be extended to $\cal M$.
Are these definitions correct?
B) When one is able to define global coordinates on a manifold, does it imply that the manifold is flat?
(In my case of interest, I consider a four-dimensional Lorentzian manifold $\cal M$. If global coordinates can be defined, does it imply $\cal M$ is Minkowskian?)
If this is indeed the case, how can it be shown?
Thanks for your help!
Xavier
Best Answer
A) Your definitions look fine, though a lot of authors require $f(U) \simeq \mathbb R^n$.
B) No - global coordinates determine the topology of a manifold, but not the geometry - the metric (and thus the curvature) has a lot of freedom once the topology has been fixed. The most classical counterexample is probably the complete hyperbolic space $\mathbb H^n$, which is diffeomorphic to $\mathbb R^n$ but has constant negative curvature. A GR-flavoured example you might have worked with are small peturbations $\eta + \epsilon h$ of the Minkowski metric $\eta$ - these are non-flat metrics on $\mathbb R^4$. You need a global coordinate chart in which the metric has components $g_{ij} = \eta_{ij}$ in order to conclude that the manifold is Minkowskian.