Let $(M,J,g)$ be an almost Hermitian manifold, and $\{e_i\}$ be $(1,0)$-vector field basis, $\{\theta^i\}$ be its dual basis. We have $$g=g_{i\bar{j}}\theta^i\otimes\bar{\theta^j}$$
If connection $D$ satisfies $$DJ=0~~Dg=0$$ then $$d\theta^i=-\theta_j^i\wedge\theta^j+\Theta^i$$ and $$\theta_j^i+\bar{\theta_i^j}=0$$
I am confused about the notion of $\bar{\theta^j}$ and $\bar{\theta_i^j}$. Let $f$ be one-form, define $\bar{f}(v):=\bar{f(v)}$. $\bar{f}$ is not a one-form because $$\bar{f}(\lambda v)=\bar{\lambda}\bar{f}(v)$$ That is to say, $\bar{\theta^j}$ and $\bar{\theta_i^j}$ are not one-form. How are they calculated in $\otimes$ and $\wedge$?
Any advice is helpful. Thank you.
