I seem to have made a mistake while doing the simple exercises of calculating 2D sectional curvature of paraboloid $z=\frac a2 (x^2+y^2)$. I used polar coordinates to do this; $(r,\theta)=(\sqrt{x^2+y^2},\arctan\frac xy)$.
To calculate the sectional curvature, the connection coefficients were first calculated, and my calculation of the connection coefficients were identical to one I found on the web: http://mathserver.neu.edu/~bratus/diffgeom/sol3/sol3.htm (Ctrl+F "formula for the curvature tensor")
$\Gamma^1_{11}=\frac{a^2r}{1+a^2r}, \Gamma^2_{12}=\Gamma^2_{21}=\frac 1r, \Gamma^1_{22}=\frac{-r}{1+a^2r}$, and the rest are $0$.
And then I proceeded to calculate the curvature tensor, which gave me a wrong result! I think that the mistake I made is probably that $\nabla_1(\Gamma^1_{22} e_1) = \Gamma^1_{22,1} e_1 + \Gamma^1_{22}\nabla _1 e_1 = \Gamma^1_{22,1} + \Gamma^1_{22} \Gamma^1_{11} e_1 + \Gamma^1_{22} \Gamma^2_{22} e_2$, but I can't find any reason why this should be wrong though. The full calculation I did and the comparison with the solution is given below:
$R(e_1,e_2)e_2 = \nabla_1\nabla_2 e_2 - \nabla_2 \nabla_1 e_2 \\= \nabla_1(\Gamma^1_{22} e_1 + \Gamma^2_{22}e_2) - \nabla_2(\Gamma^1_{12}e_1 + \Gamma^2_{12}e_2) \\= \nabla_1(\Gamma^1_{22}e_1) - \nabla_2(\Gamma^2_{12}e_2)\text{ (since $\Gamma^2_{22}=\Gamma^1_{12}=0$.)} \\= \Gamma^1_{22,1} e_1 + \Gamma^1_{22} \nabla_1 e_1 - \Gamma^2_{12,2}e_2 - \Gamma^2_{12} \nabla_2 e_2\\ =\Gamma^1_{22,1} e_1 + \Gamma^1_{22} (\Gamma^1_{11} e_1 + \Gamma^2_{11} e_2) - \Gamma^2_{12,2} e_2 - \Gamma^2_{12} (\Gamma^1_{22} e_1 + \Gamma^2_{22} e_2)$
and therefore
$R^2_{212}=d\theta(R(e_1,e_2)e_2) = \Gamma^1_{22}\Gamma^2_{11} - \Gamma^2_{12,2} - \Gamma^2_{12}\Gamma^2_{22} = 0 - \frac{\partial}{\partial \theta}(\frac 1r) + 0 = 0$
However, thet solution says that $R^2_{212} = \frac 1{r^2}$. I have checked over this calculation multiple times and I can't find any problem. Can someone help me out, please?
Appended: I append some further (probably flawed) calculation results of mine:
$R^1_{212} \\=dr(R(e_1,e_2)e_2) \\=dr(\Gamma^1_{22,1} e_1 + \Gamma^1_{22} (\Gamma^1_{11} e_1 + \Gamma^2_{11} e_2) - \Gamma^2_{12,2} e_2 - \Gamma^2_{12} (\Gamma^1_{22} e_1 + \Gamma^2_{22} e_2)) \\=\Gamma^1_{22,1} + \Gamma^1_{22}\Gamma^1_{11} - \Gamma^2_{12}\Gamma^1_{22} \\=\frac{-(1+a^2r^2)+r\cdot 2a^2 r}{(1+a^2r^2)^2} - \frac{a^2r^2}{(1+a^2r^2)^2} - \frac{-1}{(1+a^2r^2)^2} \\=\frac{-1-a^2r^2+2a^2r^2 - a^2r^2 + 1}{(1+a^2r^2)^2} \\=0$
and therefore, sectional curvature is
$K(e_1,e_2) \\=\frac{\langle R(e_1,e_2)e_2,e_1\rangle}{|e_1|^2|e_2|^2-\langle e_1,e_2 \rangle^2} \\=g_{11} R^1_{212}/(r^2(1+a^2r^2)) \\=(1+a^2r^2)\cdot 0 / (r^2(1+a^2r^2))=0$
To answer your revised question:
$$ \Gamma^1_{22} \Gamma^2_{12} = \frac{-r}{1 + a^2 r^2} \frac{1}{r} = - \frac{1}{1+a^2 r^2}. $$
In your computations you wrote instead
$$ \Gamma^1_{22} \Gamma^2_{12} "=" - \frac{1}{(1+a^2 r^2)^2}$$
Fixing that you should get
$$ R^1_{212} = \frac{a^2 r^2}{(1 + a^2 r^2)^2} $$
So that
$$ R_{1212} = g_{11} R^1_{212} = \frac{a^2 r^2}{1 + a^2 r^2} $$
So that
$$ K = R_{1212} / g = \frac{a^2}{(1+ a^2 r^2)^2} $$