# Torsion-freeness of a connection and anti-symmetrization

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?

## 1 Answer

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 $$d\theta_i = \sum\omega_{ij}\wedge\theta_j = \sum \omega_{ij}\otimes\theta_j - \theta_j\otimes\omega_{ij}$$.