0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.