# Global and local coordinates on a manifold, and their relations to curvature

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?

Xavier

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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.

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Like the hyperbolic space example. +1! (Used to work in GR a long time ago, but a nice and classic example!) – Sanath K. Devalapurkar Jun 13 '14 at 0:29
Thanks for your answer Anthony. So one is able to define global coordinates on $\mathbb{H}^n$? I also have in mind the general covariance of GR which states (hoping I'm correct) that two diffeomorphic manifolds are physically equivalent. Hence if a diffeormorphism exists between $\mathbb{H}^n$ and $\mathbb{R}^n$, this property of general covariance would imply that everything on $\mathbb{H}^n$ happens as if $\mathbb{H}^n$ was flat? – Xavier Jun 13 '14 at 10:03
@Xavier: yes, the Poincare Disc and Half-Space models give global coordinates for $\mathbb H^n$. You're misunderstanding the "general covariance" - this principle states that the laws of physics should be coordinate-independent. For full equivalence of physical data in GR you need an isometry. – Anthony Carapetis Jun 13 '14 at 10:12
Thanks for your answer Anthony. – Xavier Jun 18 '14 at 17:49

You are confusing two different notions here: In part (A) both definitions are correct and refer to the differential topology of the manifold. In part (B) the notion of flatness is not topological, it depends on a metric or a connection. However, it is true that if a manifold is diffeomorphic to an open subset of $R^n$, then it admits a flat Riemannian metric as well as a flat Lorentzian metric. However, this metric on $M$ is fact from unique (up to isometry). Just consider $M=R^2$ and $M$ equal to the open 2d disk: The induced flat Riemannian (or Lorentzian) metrics on $M$ are not isometric to each other. Furthermore, even if $M$ is an open disk, you can define, say, the hyperbolic metric on $M$, which is not flat.

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Thanks studiosus. I realize I am probably confused indeed. You are saying that "if a manifold is diffeomorphic to an open subset of $\mathbb{R}^n$, then it admits a [flat metric]". Taking the example of Anthony written above, can we say that $\mathbb{H}^n$ is flat then? I understand with what you wrote that, on the open 2D disk, one can write a flat metric. I am not sure to see that, or to understand it. Can you please elaborate or give precise examples? – Xavier Jun 13 '14 at 10:10
@Xavier: "It admits" means "there exists": An open subset of $R^n$ has many Riemannian (and Lorentzian) metrics. Some of them will be flat, but most (like the hyperbolic one) will not be flat. Again, the point is that such open subset is just a smooth manifold, you have to specify the metric to talk about flatness. – studiosus Jun 13 '14 at 15:25
Thanks for your answer studiosus. Perhaps, my confusion stems from the assumption that a manifold is always given with a metric, which is the case in GR. This being said, it makes me think of other questions, which may start to be off-topic. I'll probably open a new post then. – Xavier Jun 18 '14 at 17:55