Let $M^2 \subset\mathbb{R}$ be a regular surface and $ A \in M$.
Assume the gaussian curvature of M at the point A is K
and $e_1, e_2 \in T_AM$ orthonormal basis for the tanget space.
I want to find the representing matrix for the tensor $R(e_1, e_2)$ with respect to the orthonormal basis $e_1, e_2$.
Note: $R: Vec(M)^3 \to Vec(M)$
, given by, $R(X,Y)Z = \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$
, where X, Y, Z Are vector fields on M, $\nabla$ is the Levi-Civita connection and $[X,Y]$ is the commutator of X and Y.
My initial idea was to look at the embedding of $M^2$ in $\mathbb{R}^3$, by $i :M \to \mathbb{R}^3$.
We have $g_M=i^*(g_{\mathbb{R}^3)}$, so representing $g_M$ in basis $e_1, e_2$ will result in $g_M=I$.
From here, I can easily see that all of the Christoffel ($\Gamma^k_{ij}$) symbols are vanishing at the point A.
If we look at Ricci's curvature $\mathscr{R}_{ij}=R^k_{kij}=\frac{\partial\Gamma_{ij}^k} {\partial q^k} – \frac{\partial\Gamma_{kj}^k} {\partial q^i}$ (I use the notation $\partial q^i=e_i$) and using the fact that $g^{ij }\mathscr{R}_{ij}=\mathscr{K}=2K$ (This is how we formulated theorema egregium in our class)
I ended up with the following equation :
$2K=\frac{\partial\Gamma_{22}^1} {\partial q^1}+\frac{\partial\Gamma_{11}^2} {\partial q^2}-\frac{\partial\Gamma_{21}^2} {\partial q^1}-\frac{\partial\Gamma_{12}^1} {\partial q^2}$
Now
$R(e_1,e_2)_{ij} = \begin{bmatrix}R_{121}^1&R_{121}^2\\R_{122}^1&R_{122}^2\end{bmatrix}$
And generally because the Christoffel coefficients vanishes :
$R_{12k}^l= \frac{\partial\Gamma_{2k}^l} {\partial q^1}-\frac{\partial\Gamma_{1k}^l} {\partial q^2}$.
I hoped that getting to this stage things would start cancelling out, but they don't.
I would appriciate any help completing this question!
Best Answer
After few days of thinking I will expand Ted's comment a little in case anyone else would need it:
I will use the notation $R(X,Y,Z,W) = <R(X,Y,Z),W>$.
We have $R(X,Y,Y,X) = \frac {B(X,X)B(Y,Y)-B(X,Y)}{|X|^2|Y|^2-<X,Y>^2}$ and for the orthonormal basis $R(e_1,e_2,e_2,e_1)=<R(e_1,e_2)e_2,e_1>=K$, symmetric properties of R we get $<R(e_1,e_2)e_1,e_2>=<-R(e_2,e_1)e_1,e_2>=-K$, and $<R(e_1,e_2)e_2,e_2>=<R(e_1,e_2)e_1,e_1>=<R(e_1,e_1)e_1,e_2>=0$ and therefore
$R=\begin{bmatrix}0&K\\-K&0\end{bmatrix}$