Show that $\omega_{ab}=-\omega_{ba}$ for a Riemannian connection How can we see for the Riemannian connection, connection 1-form with its first index lowered $\omega_{ab}=\delta_{ac}{\omega^c}_b$ is antisymmetric in a, b, i.e. $\omega_{ab}=-\omega_{ba}$. Thanks.
 A: Let $\{e_a\}$ be a local orthonormal frame for the tangent bundle, so $\delta_{ab} = \langle e_a, e_b\rangle$, and let $\nabla$ be a Riemannian connection. We have $\nabla e_a = e_c\otimes {\omega^c}_a$ where ${\omega^c}_a$ are the connection one-forms.
Taking the covariant derivative of both sides of the identity $\delta_{ab} = \langle e_a, e_b\rangle$ we see that 
$$0 = \nabla\langle e_a, e_c\rangle = \langle\nabla e_a, e_c\rangle + \langle e_a, \nabla e_c\rangle$$
where the last equality holds because $\nabla$ is a Riemannian connection (i.e. $\nabla g = 0$).
Using the connection one-forms we have 
\begin{align*}
\langle \nabla e_a, e_b\rangle + \langle e_a, \nabla e_b\rangle &= \langle e_c\otimes {\omega^c}_a, e_b\rangle + \langle e_a, e_c\otimes {\omega^c}_b\rangle\\ 
&= \langle e_c, e_b\rangle{\omega^c}_a + \langle e_a, e_c\rangle{\omega^c}_b\\ 
&= \delta_{cb}{\omega^c}_a + \delta_{ac}{\omega^c}_b\\
&= \omega_{ab} + \omega_{ba}.
\end{align*}
So $\omega_{ab} + \omega_{ba} = 0$ and hence $\omega_{ab} = -\omega_{ba}$.
