What follows could be done more "abstractly", like in the note Euclidean plane, I think.
Consider the Levi-Civita connection in $\mathbb{R}^3$, $\nabla$. 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 $\Theta=0$.
This can be deduced also with the Cartan's first structural equation. We have the coframe $\{dx,dy,dz\}$ and since they are all closed
$$ 0=d(dx)=dx\wedge \Theta^1_1+dy\wedge \Theta^1_2+dz\wedge \Theta^1_3 $$ $$ 0=d(dy)=dx\wedge \Theta^2_1+dy\wedge \Theta^2_2+dz\wedge \Theta^2_3 $$ $$ 0=d(dz)=dx\wedge \Theta^3_1+dy\wedge \Theta^3_2+dz\wedge \Theta^3_3 $$Moreover, since the connection is metric and the frame is orthonormal we have that $\Theta_i^j=-\Theta_j^i$ (see here) and, since is torsion free, $\partial_i\lrcorner\Theta_{j}^k-\partial_j\lrcorner\Theta_{i}^k=-T_{ij}^k=0$ (see here). Then
$$ 0=dy\wedge \Theta^1_2+dz\wedge \Theta^1_3 $$ $$ 0=-dx\wedge \Theta^1_2+dz\wedge \Theta^2_3 $$ $$ 0=-dx\wedge \Theta^1_3-dy\wedge \Theta^2_3 $$Observe, first,
$$ \partial_1 \lrcorner \Theta^1_2=\partial_2 \lrcorner \Theta^1_1=0 $$Now, contracting $\partial_2$ in the second equation
$$ 0=dx\wedge\partial_2\lrcorner \Theta^1_2-dz\wedge \partial_2\lrcorner \Theta^2_3 $$and from here $\partial_2\lrcorner \Theta^1_2=0$.
So $\Theta^1_2$ is a multiple of $dz$. For the same reasons, $\Theta^1_3$ is a multiple of $dy$ and $\Theta^2_3$ is a multiple of $dx$.
Therefore
$$ 0=dy\wedge K_{12}dz+dz\wedge K_{13}dy $$ $$ 0=-dx\wedge K_{12}dz+dz\wedge K_{23}dx $$ $$ 0=-dx\wedge K_{12}dy-dy\wedge K_{23}dx $$and $K_{13}=K_{12}=-K_{23}=-K_{12}$ and therefore all of them are 0.
Consider now an orthonormal frame $\{e_1,e_2,e_3\}$ with dual coframe $\{\omega_1,\omega_2,\omega_3\}$. The connection form is, as usual, a matrix of 1-form $\Theta$ such that
$$ \begin{array}{l} \nabla e_1=\quad\qquad-\Theta^1_2 e_2-\Theta^1_3 e_3\\ \nabla e_2=\Theta_2^1 e_1+\qquad-\Theta_3^2e_3\\ \nabla e_3=\Theta_3^1 e_1+\Theta_3^2e_2\\ \end{array} $$where we have used that since the connection is metric and the frame is orthonormal we have that $\Theta_i^j=-\Theta_j^i$ (see here).
We have two pieces of data: the connection and the frame. And we can get $\Theta$ from them.
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Author of the notes: Antonio J. Pan-Collantes
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