I'm reading the book of Andrei Moroianu, "Lectures on kahler geometry" and at the page 79 is this theorem:
I don't understand the last proposition. namely the fact that $\nabla_Z(H(\sigma))=H(\nabla_{\over{Z}}\sigma)$
Asked
Active
Viewed 38 times
1
-
The LHS being $\mathbb{C}$-linear in $Z$ so inside the $H$ of the RHS must be antilinear in $Z$, i.e., only have $\bar{Z}$ and no $Z$. Moreover we know it is $\nabla_X\sigma$ for real $X$ so it must be $\nabla_{\bar{Z}}\sigma$. – user10354138 Oct 15 '18 at 20:00