Connection form for LV in R3

What follows could be done more "abstractly", like in the note Euclidean plane, I think.

With the standard coordinate frame

Consider the Levi-Civita connection in $R^{3}$ , $\nabla$ . If we use the standard coordinates $(x, y, z)$ we have the orthonormal frame ${\partial_{1}, \partial_{2}, \partial_{3}}$ and we know that

\nabla_{\partial_{i}} \partial_{j} = 0

so in this case the connection form (Christoffel symbols, indeed) is $Θ = 0$ .

This can be deduced also with the Cartan's first structural equation. We have the coframe ${d x, d y, d z}$ and since they are all closed

0 = d (d x) = d x \land Θ_{1}^{1} + d y \land Θ_{2}^{1} + d z \land Θ_{3}^{1}

0 = d (d y) = d x \land Θ_{1}^{2} + d y \land Θ_{2}^{2} + d z \land Θ_{3}^{2}

0 = d (d z) = d x \land Θ_{1}^{3} + d y \land Θ_{2}^{3} + d z \land Θ_{3}^{3}

Moreover, since the connection is metric and the frame is orthonormal we have that $Θ_{i}^{j} = - Θ_{j}^{i}$ (see here) and, since is torsion free, $\partial_{i} ⌟ Θ_{j}^{k} - \partial_{j} ⌟ Θ_{i}^{k} = - T_{i j}^{k} = 0$ (see here). Then

0 = d y \land Θ_{2}^{1} + d z \land Θ_{3}^{1}

0 = - d x \land Θ_{2}^{1} + d z \land Θ_{3}^{2}

0 = - d x \land Θ_{3}^{1} - d y \land Θ_{3}^{2}

Observe, first,

\partial_{1} ⌟ Θ_{2}^{1} = \partial_{2} ⌟ Θ_{1}^{1} = 0

Now, contracting $\partial_{2}$ in the second equation

0 = d x \land \partial_{2} ⌟ Θ_{2}^{1} - d z \land \partial_{2} ⌟ Θ_{3}^{2}

and from here $\partial_{2} ⌟ Θ_{2}^{1} = 0$ .
So $Θ_{2}^{1}$ is a multiple of $d z$ . For the same reasons, $Θ_{3}^{1}$ is a multiple of $d y$ and $Θ_{3}^{2}$ is a multiple of $d x$ .

Therefore

0 = d y \land K_{12} d z + d z \land K_{13} d y

0 = - d x \land K_{12} d z + d z \land K_{23} d x

0 = - d x \land K_{12} d y - d y \land K_{23} d x

and $K_{13} = K_{12} = - K_{23} = - K_{12}$ and therefore all of them are 0.

With a general orthonormal frame

Consider now an orthonormal frame ${e_{1}, e_{2}, e_{3}}$ with dual coframe ${ω_{1}, ω_{2}, ω_{3}}$ . The connection form is, as usual, a matrix of 1-form $Θ$ such that

\begin{array}{l} \nabla e_{1} = - Θ_{2}^{1} e_{2} - Θ_{3}^{1} e_{3} \\ \nabla e_{2} = Θ_{2}^{1} e_{1} + - Θ_{3}^{2} e_{3} \\ \nabla e_{3} = Θ_{3}^{1} e_{1} + Θ_{3}^{2} e_{2} \end{array}

where we have used that since the connection is metric and the frame is orthonormal we have that $Θ_{i}^{j} = - Θ_{j}^{i}$ (see here).

We have two pieces of data: the connection and the frame. And we can get $Θ$ from them.