Let $\nabla$ denote the Levi-Civita connection on the following manifold in $\mathbb{R}^3$ with Riemannian metric $g$ as follows:
\begin{equation} \mathcal{H}^3=\lbrace (x,y,z)\in \mathbb{R}^3 \mid z>0 \rbrace \end{equation} \begin{equation} g=\frac{(dx)^2+(dy)^2+(dz)^2}{z^2} \end{equation} \begin{equation} A= z\partial_x \, , \, B=z\partial_y \, , \, C= z\partial_z \end{equation}
I want to compute $\nabla_X Y$ for all $X,Y \in \lbrace A,B,C \rbrace$. I know that it should be easiest to do this via Koszul's formula, but I don't know exactly how to use it. Koszul's formula reads as on wikipedia:
\begin{equation} 2g(\nabla_X Y, Z)=\partial_X(g(Y,Z))+\partial_Y(g(X,Z))-\partial_Z(g(X,Y))+g([X,Y],Z)-g([X,Z],Y)-g([Y,Z],X) \end{equation}
I don't know exactly how to interpret this formula. First, what is $Z$? is it a dummy variable in this case or what? and second, what does the term $\partial_X$ and the others mean? Since it seems that I need to take derivatives with respect to derivatives? All the other terms I know how to handle, but the others mentioned are really bugging me.
Okay, if I try to let $X=A$ and $Y=B$, then the first term in Koszul's formula is $$g(Y,Z)=\frac{dx(B)dx(Z)+dy(B)dy(Z)+dz(B)dz(Z)}{z^2}= \frac{dx(z\partial_y)dx(Z)+dy(z\partial_y)dy(Z)+dz(z\partial_y)dz(Z)}{z^2}=\frac{zdx(\partial_y)dx(Z)+zdy(\partial_y)dy(Z)+zdz(\partial_y)dz(Z)}{z^2}=\frac{z \cdot 0 \cdot dx(Z)+z \cdot 1 \cdot dy(Z)+z \cdot 0 \cdot dz(Z)}{z^2}=\frac{z\cdot 1 \cdot dy(Z)}{z^2}$$
Or am I totally off track here?