Torsion-freeness of a connection and anti-symmetrization

Question

Let $M$ be an Hermite manifold, and $\nabla$ be the Levi-Civita connection on $TM$ and extend it to $\Lambda^*_{\mathbb{C}}(M)$. Then $\nabla$ is torsion-free by definition. But I read from a paper that the torsion-freeness implies $d\theta_i = \text{Alt}(\nabla\theta_i)$ where $\{\theta_i\}$ is a local frame of $\Lambda^{1,0}(M)$. As far as I know, the torsion-fressness means the torsion tensor $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]\equiv 0$, so how can we deduce $d\theta_i = \text{Alt}(\nabla\theta_i)$ from this?

Answer 1

This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $\theta_i$ and saying $$d\theta_i = \sum \omega_{ij}\wedge\theta_j$$ with $\omega_{ij}=-\omega_{ji}$. (Ordinarily, you'd have $d\theta_i = \sum\omega_{ij}\wedge\theta_j + \tau_i$, where $\tau$ gives the torsion.) Here $\omega_{ij}$ gives the connection form.

Then you can check that $\nabla \theta_i = \sum \omega_{ij}\otimes\theta_j$, and the rest is immediate, since $d\theta_i = \sum\omega_{ij}\wedge\theta_j = \sum \omega_{ij}\otimes\theta_j - \theta_j\otimes\omega_{ij}$.