Is the Riemann Curvature Tensor a Mathematical Tool or Physically Significant?

[superbat]

Can someone explain mathematically why do we say Riemann Curvature Tensor has all the information about curvature of Space
Thank You

 

[andrewkirk]

Statements like that are not mathematically precise, so I wouldn't worry about them too much. After all, the metric tensor also has all the information about curvature, since anything else can be expressed in terms of various partial derivatives of the metric.

I think it's a loose way of saying that in everyday operations of differential geometry, and in particular of general relativity, one can derive any piece of info we want about curvature from the Riemann tensor without having to take additional derivatives (which would take the order of differentiation from two to at least three) or taking fancy tensor products (which the Weyl tensor requires).

 

[superbat]

Ok Thanks
Since you mention metric, I was also wondering what does contravariant version of metric mean.
Covariant version of metric tells us about distance between 2 points. What does contravariant version of metric physically mean?

 

[andrewkirk]

Covariant version has components $g^{ab}$. It is a linear function that takes two vectors in the tangent space at the relevant manifold point as input and returns a real scalar as output.
Contravariant version has components $g_{ab}$. It is a linear function that takes two covectors (aka one-forms or dual vectors) in the cotangent space at the relevant manifold point as input and returns a real scalar as output.

 

[superbat]

Thanks
So do they have any physical significance or should I consider them as just mathematical tools