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:

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

So that:

$$R_{1111}=\dfrac{\partial \Gamma_{111}}{\partial u^1}-\dfrac{\partial \Gamma_{111}}{\partial u^1}+\Gamma^h_{11}\Gamma_{1h1}-\Gamma^h_{11}\Gamma_{1h1}\\<br /> =0$$

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

$$K=-\dfrac{R_{1212}}{\text{det}g}$$

by first viewing the curve as an embedding in a higher dimesional manifold, in particular $$\mathbb{R}^2$$ 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 $$R_{1,1,1,1}=0$$ and the Gaussian curvature of the plane $$\Re^2$$ is also zero!
The difference is that the curvatures of curves are the "main curvatures" $$k_1,k_2$$ while the "curvature tensors" $$R_{i,j,k,l}$$ has something to do with the Gaussian curvatures $$K=k_1k_2$$