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I am confused in one question in general relativity, why we can always express a space-time geometry only by metric. It means a metric, which is just about distance in tangent space, can tell us all the information about the manifold.

I know there are standard proofs, for instance, we can express connection by metric, and so the Riemann curvature. However, I am not very satisfied by these answers. I still want a more direct reason for that.

To my understanding, a metric just defines the distance, the length of tangent vectors, however, Riemann curvature, in my eyes, tell us more, for instance, how a line differ from a direct line and how a vector travel along a closed path.

I believe there must be some neat and beautiful argument show that metric is enough, is everything.

This question is quite vague, so please just feel like a chatting.

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It's hard to understand your reasoning. In the second paragraph, you write that you know that the Riemann curvature tensor may be written in terms of the metric (and its derivatives). In the third paragraph, you say that the Riemann curvature tells you more. These two sentences directly contradict one another, don't they? One can consider more general "geometries" than ones given by the metric tensor configuration. But you would first have to define what you mean by such a generalized "geometry", and then we could discuss it (add torsion or anything else). –  Luboš Motl Dec 16 '12 at 9:29
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Sorry, I admit I am talking about this in a very confusing way. By writing Riemann curvature tensor in terms of metric, I believe the metric tells everything. However, I try to understand this because it is not very natural to me that "a local measurement of length" can tell all the properties of a manifold. –  Yingfei Gu Dec 16 '12 at 9:42
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Maybe things are clearer if you note that angles too can be expressed in terms of the metric, i.e. the angle $\theta$ between $x$ and $y$ is given by $\cos \theta = g_{\mu \nu} x^\mu y^\nu/[x^2 y^2]^{1/2},$ where I didn't bother to write out the denominator in terms of $g$. –  Vibert Dec 16 '12 at 10:10
    
keep in mind that the metric actually does not uniquely define the connection - you'll need the additional condition that the connection should be free of torsion –  Christoph Dec 16 '12 at 11:48
    
Are you familiar with the <a href="en.wikipedia.org/wiki/Tetrad_formalism">tetrad formalism</a>? The tetrad gives you a direct way to calculate the metric, and it also gives you a very physical picture of the things you are talking about. –  Jerry Schirmer Dec 16 '12 at 18:34

2 Answers 2

This is roughly how I think about it. A manifold is defined as a set of dimension $n$ with an open neighborhood at every point that has a 1-1 and continuous map to a Euclidean Space,$E^n$; such that it is locally "like" $E^n$. A single map is a Chart. The set of Charts that cover the entire manifold is an Atlas.(Shutz, Bernard. Geometrical Methods of Mathematical Physics) When the charts are constructed for a set, like a sphere, the metric for the sphere can be derived from the transformation from Euclidean coordinates and the fact that the metric is a tensor. So the metric measures the deviation from Euclidean space. And, if you know exactly how a space differs from a Euclidean space at every point, then you know "everything" about that space.

But, there are topologically distinct spaces with the same metric -a cylinder is flat. So it doesn't really tell you everything.

*Edited, to reflect the comment by gns-ank. Replaced an incorrect or imprecise definition of manifold with a rigorous definition.

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+1 fot the cylinder example. It is often ignored that the metric only tells you about the local geometry. –  ungerade Dec 16 '12 at 12:22
    
However, one should note, that the manifold is not defined as something that is locally euclidean. It's just the case, that pseudo-Riemaniann manifolds are defined as such and they are used in general and special relativity theories. –  gns-ank Dec 16 '12 at 14:10
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@gns-ank I've replaced what I wrote about the manifold with the definition of manifold found in Shutz. Geometrical Methods –  MadScientist Dec 16 '12 at 18:27
    
+1 for the cylinder example, I also ignored that metric is not everything. –  Yingfei Gu Dec 16 '12 at 20:02

It is a misconception that a Riemannian metric only specifies a norm on the tangent spaces: The inner product also adds the notion of angles and geodesic distance (and thus an actual metric on the manifold).

Personally, I do not find it surprising that distances and angles are enough to specify the geometry - what else should there be? (Torsion, of course - but in general relativity we traditionally ignore this additional degree of freedom...)

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