Can someone help in computing curvature tensor of a surface?

Question

I have a problem, I've been thinking about all day. Came across this while browsing some lecture notes online. So, I have a surface in space say, described as $z= f(x,y)$ and I want to find it's sectional curvature.

I think we can do this by first finding Riemannian-metric $(g_{ij})$ of this manifold described by this surface, then find christoffel's symbols, from there one can find local expression for curvature tensor, which is sufficient to find sectional curvature.

I think that's it but with some surfaces this is taking just so much time to compute, and I don't think is the best method around. Can someone here help me figure out an alternate (better) way to do this?

Thanks and Cheers!

Answer 1

I have come to this solution. I hope it's correct

In case of a surface embedded in $\mathbb{R^3}$, say described as $z= f(x,y)$ it's sectional curvature is the same as it's Gaussian curvature. And we know that the latter is given by $$ K = \frac{f_{xx} f_{yy} - (f_{xy})^2 }{(1+(f_x)^2 +(f_y)^2)^2}$$

Answer 2

In $\mathbb{R^3}$ Monge form surface $z= f(x,y)$ has sectional curvature same as the Gaussian curvature K. Using Classical approach like with Christoffel symbols ( Diff Geom texts Barret O'Neill, DJ Struik, Ted Shifrin .. ) and with notation

$$ r = f_{xx}, t= f_{yy} , s= f_{xy}, p= f_x, q= f_y ; $$

we obtain

$$ K= \frac{(rt-s^2)}{(1+p^2+q^2)^2}$$

Answer 3

One alternative strategy for computing the Gaussian curvature of a surface equipped with a Riemannian metric is using Cartan's moving frames. Namely, if $(E_1,E_2)$ is a frame defined on an open subset of the surface and $(\theta^1,\theta^2)$ is the dual coframe, one defines the connection $1$-forms $\omega^i_{~j}$ by the relations $\nabla_XE_j = \sum_i \omega^i_{~j}(X)E_j$. If $(E_1,E_2)$ is an orthonormal frame, one can show that ${\rm d}\omega^1_{~2} = K\,\theta^1\wedge \theta^2$.

To compute $\omega^i_{~j}$ in concrete cases, one uses the structure equations ${\rm d}\theta^i = \sum_j\theta^j \wedge \omega^i_{~j}$ and the metric compatibility $\omega_{ij}+\omega_{ji} = {\rm d}g_{ij}$, where $(g_{ij})$ are the components of the metric with respect to the frame $(E_1,E_2)$ and $\omega_{ij}$ is obtained by lowering $i$ in $\omega^i_{~j}$ using the metric, that is, $\omega_{ij} = \sum_kg_{ik}\omega^k_{~j}$. You can see proofs that everything works in these notes.

In the particular example for graphs, these computations (and all others, to be honest) are a bit more complicated, because the coordinate vector fields associated to the Monge chart are not orthogonal (so Gram-Schmidt is required if one wants to employ moving frames).