Let $e_1, \ldots, e_n$ be a basis in an Lie algebra $\mathfrak{g}$, regarded as a left-invariant movable frame, and $w^i -$ a dual movable left-invariant frame of 1-forms on $G$. $C_{i\, j}^k$ are the structure constants of the Lie algebra
$$ \left[e_i, e_j \right] = \sum C^k_{i\, j} e_k $$
Let $\nabla$ be a connection on a Lie group $G$ with a bi-invariant metric.
$$ \nabla_X e_j= \sum w^i_j \left( X \right) e_i $$
How to prove that then $ w^i_j \left( X \right) = \dfrac{1}{2} \sum C^i_{k\, j} w^k $?
Before this exercise, there were some tasks. And I proved that
$$ d w^i = - \dfrac{1}{2} \sum C^i_{j\, k} w^j \wedge w^k$$
$$ \left( \left[X,Y \right], Z \right)_g + \left(Y, \left[X,Z \right] \right)_g = 0 \quad \implies C^i_{j\, k} + C^k_{j \, i} $$
Then i know how to prove that $ \nabla_X Y = \dfrac{1}{2} \left[X,Y \right]$ for left-invariant vector fields. But this is only in the case of the Levy-Chevita connection.
Thank you so much!