Kaehler form in coordinates

Question

Consider $M^n$ a $n$-dimensional Kaehler manifold and $\omega(X,Y) := g(JX,Y)$ its Kaehler 2-form. Let $\{e_0,\dots,e_{2n-1}\}$ be a orthonormal hermitian frame, i.e, $e_{2i+1} = Je_{2i}$ and $\{\theta_i\}$ its dual. We know that $ g = \sum_{i = 0}^{2n-i}\theta_i \otimes \theta_i $ Then, I want to obtain $\omega$ in terms of $\theta_i$. So $$ \omega(X,Y) = \sum \theta_i\otimes\theta_i(JX,Y) = \sum \theta_{2i}(JX)\theta_{2i}(Y) = - \sum \theta_{2i+1}(X)\theta_{2i}(Y) $$ but I know that I should obtain $$ \omega = \sum_i \theta_{2i}\wedge \theta_{2i+1} $$ how can I obtain the exterior product and not the tensor product?

Answer 1

First notice that the complex structure acts on 1-forms with its transpose map $J^* : \theta \mapsto \theta \circ J$. Then you can easily check that, for $k = 1 \ldots n$ $$ J^* \theta_{2k-1} = -\theta_{2k}\\ J^* \theta_{2k} = \theta_{2k-1}. $$ Then, by definition $$ \begin{align*} \omega &= \sum_{k=1}^n J^* \theta_{2k-1} \otimes \theta_{2k-1} + J^* \theta_{2k} \otimes \theta_{2k}\\ &= \sum_{k=1}^n -\theta_{2k} \otimes \theta_{2k-1} + \theta_{2k-1} \otimes \theta_{2k}\\ &=\sum_{k = 1}^n \theta_{2k-1}\wedge \theta_{2k} \end{align*} $$

Note that in this computation I used indexes in $\{1, \dots, 2n\}$, if you use yours this formula changes a bit.