Calculating Christoffel symbols from riemannian metric

[MathNugget]

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

Now I try to calculate $\Gamma^{k}_{ij}$.


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

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

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

 

[MathNugget]

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

 

[shahar]

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

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