Unifying Curvatures with Riemann Tensor

[bchui]

I always wonder how the definitions of curvatures of curves and surfaces be unified by the Riemann Tensor symbols.
For surfaces, I know R_{1,2,1,2} corresponds to the Gaussian curvature of a surface. How come R_{1,1,1,1}=0 and not corresponds to the curvature of a curve in \RE^2 or in \Re^3 ?

 

[the.bone]

Recall that:

[tex]R_{ijkl}=\dfrac{\partial \Gamma_{ijl}}{\partial u^k}- \dfrac{\partial \Gamma_{ijk}}{\partial u^l}+\Gamma^h_{ik}\Gamma_{jhl}-\Gamma^h_{il}\Gamma_{jhk}[/tex]

So that:

[tex]R_{1111}=\dfrac{\partial \Gamma_{111}}{\partial u^1}-\dfrac{\partial \Gamma_{111}}{\partial u^1}+\Gamma^h_{11}\Gamma_{1h1}-\Gamma^h_{11}\Gamma_{1h1}\\
=0[/tex]

For any finite dimensional Riemannian manifold. Moreover, we can still calculate the Gaussian curvature, [tex]K[/tex] for a 1-D curve via

[tex]K=-\dfrac{R_{1212}}{\text{det}g}[/tex]

by first viewing the curve as an embedding in a higher dimesional manifold, in particular [tex]\mathbb{R}^2[/tex] or higher--which is possible via the Nash-Kuiper theorem--allowing the appropriate number of terms to appear in the calculation.

There you have it!

 

[bchui]

How about the curvature of a curve on the plane?
Applying Gauss-Codazzi equation and we have [tex]R_{1,1,1,1}=0[/tex] and the Gaussian curvature of the plane [tex]\Re^2[/tex] is also zero!
The difference is that the curvatures of curves are the "main curvatures" [tex]k_1,k_2[/tex] while the "curvature tensors" [tex]R_{i,j,k,l}[/tex] has something to do with the Gaussian curvatures [tex]K=k_1k_2[/tex]