Calculating Christoffel symbols from riemannian metric

[MathNugget]

On [imath]\mathbb{R}^2[/imath], I have [imath]f: \mathbb{R}^2 \rightarrow \mathbb{R}[/imath] a smooth function, and the metric [imath]g_{ij}=e^{2f(x^1, x^2)}\delta_{ij}[/imath], with [imath]\delta_{ij}[/imath] being kronecker function.

Now I try to calculate [imath]\Gamma^{k}_{ij}[/imath].


I have the formula: [imath]\Gamma^{k}_{ij}=\frac{1}{2}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m})[/imath].

I have been researching it, there should be [imath]2^3[/imath] Christoffel symbols. I also suspect the formula for [imath]\Gamma^{k}_{ij}[/imath] is a summation by m (here, from m=1 to m=2). Firstly I calculate [imath]g^{ij}=\begin{pmatrix} e^{-2f(x^1, x^2)} & 0\\ 0 & e^{-2f(x^1, x^2)} \end{pmatrix}[/imath], if I am not mistaken.

Then, I compute [imath]\frac{\partial g_{mk}}{\partial x^l}[/imath], for all indexes, so I can just plug them in at the end. Here comes a first question though: when I compute [imath]g_{11}[/imath], is it equal to [imath]e^{2f(x^1, x^2)}\delta_{ij}[/imath] or [imath]e^{2f(x^1, x^1)}\delta_{ij}[/imath]?

 

[MathNugget]

I thought now, it's obvious that [imath]g_{11}=e^{2f(x^1, x^2)}[/imath]. But who are [imath]x_1, x_2[/imath]? I suspect that when we compute [imath]g(v_1, v_2)[/imath], we transport the whole thing in the same tangent space to a point, and [imath]x_1, x_2[/imath] are the coordinates of that point (aka they are the coordinates of p in [imath]g_p[/imath])