Charaterization of the component of exterior bundle Let $(M,J)$ be an almost complex manifold,consider the complexified tangent bundle it has two subbundle $T^{(0,1)}M$ and $T^{(1,0)}M$,where $T^{(0,1)}M$ is the eigenbundle of eigen value $i$ associcated to $J$.
Then we can consider the decomposition of the exterior bundle $$\Lambda^{1,0} M:=\left\{\xi \in \Lambda_{\mathbb{C}}^{1} M \mid \xi(Z)=0 \forall Z \in T^{0,1} M\right\}\\\Lambda^{0,1} M:=\left\{\xi \in \Lambda_{\mathbb{C}}^{1} M \mid \xi(Z)=0 \forall Z \in T^{1,0} M\right\}$$
Furthure more we have,$\Lambda^{p,q} = \Lambda^{p, 0} \otimes \Lambda^{0, q}$
Then I need to prove the following charaterization (which appears on Moroianu's Kahler geometry note):

*

*$\left.\omega \text { is a section of } \Lambda^{k, 0} M \text { if and only if } Z\right\lrcorner \omega=0 \text { for all } Z \in T^{0,1} M \text {. }$

*$\omega$ is a section of $\Lambda^{p, q} M$ if and only if it vanishes whenever applied to $p+1$ vectors from $T^{1,0} M$ or to $q+1$ vectors from $T^{0,1} M$

*given $\omega\in \Lambda^{1,0}M$ the $(0,2)$ component of $d\omega$ vanish if and only if for all $Z,W \in T^{0,1}M$ ,$d\omega (Z,W) = 0$
My attempt:If we write under the local coordinate the only if part is easy to check,I don't know how to check the if part.
 A: I realize what confuse me before, that is there are two different charaterizations.Since it really does not invovle any geometry I state it in terms of language of vector space.
Given $(V,J)$ we have $T^{(1,0)}V = \{v\in V_{\Bbb{C}}\mid Jv = iv\}$, similarily the $$T^{(1,0)}V^* = \{\xi\in V^*_{\Bbb{C}}\mid i\xi(v) = \xi(Jv)\}\tag{1}$$ that is the set of complex linear form with complex structure given as $$\xi:(V,J)\to (\Bbb{C},i)$$
Note that this is really restriction of $\xi $ to $V\subset V_{\Bbb{C}}$. As $\xi \in V^{*}_{\Bbb{C}}$ is naturally is a complex linear map
$$\xi:(V_{\Bbb{C}},i)\to (\Bbb{C},i)$$

We have second chaterization as annihilator of the other component:
$$T^{(1,0)}V^* = \{\xi \in V^*_{\Bbb{C}}\mid \xi(Z) = 0 \text{ for any } Z\in T^{(0,1)}V\}\tag{2}$$

To see they are equivalent:
given any $Z \in T^{(0,1)}V$ we have $JZ =-iZ $ therefore if we pick some $\xi \in (1)$ we will have $$i\xi(Z) = \xi(iZ) = \xi(-JZ) = - \xi (JZ) = -i\xi (Z)$$
where the first equality due to $\xi$ is complexified linear form, the second equality due to we choose $Z \in T^{(0,1)}V$, the forth equality due to we choose $\xi \in (1)$.
We see then $\xi \in (2)$.
The other direction is similar.

Finally, there is a story about local representation when talking about geometry.
We know for an almost complex manifold, we can choose the real basis for the tangent space as $$\{\frac{\partial}{\partial x^i}, I(\frac{\partial}{\partial x^i})\}$$
As a bundle isomorphism $I:TM\to TM$, the $I(\frac{\partial}{\partial x^i})$ is also a local vector field. You can show these are linear independent.And finally expanded to the real frame for $TM$.
For the complexified tangent space.As notation, we can denote(you don't really know if it is induced from the coordinate, or as coordinate vector field just treated it as a notation) $$e_i = 1/2(\frac{\partial}{\partial x^i}- i I\frac{\partial}{\partial x^i})\\ \bar{e_i} = 1/2(\frac{\partial}{\partial x^i}+ i I\frac{\partial}{\partial x^i})$$
You can check this is the complex local frame for the complexified tangent bundle(You should be careful this needs not to be induced from coordinate as complex manifold did)
similarly, for the complexified cotangent bundle.
Then you can consider differential calculus, you can complexify the exterior differential, simply change it to complex linear and remain the exterior product rule , and chain complex property the same.
That how the differential calculus is really perform on almost complex manifold.

There is one thing should be caveat, notices that the local frame for $T^{(1,0)}X$ is $$dx + i I(dx)$$ is not clear $I(dx) = dy$(i.e induced from coordinate) or not , on the complex manifold however, the complex structure on the complex manifold induced from the coordinate chart gurantee that $I(dx) = dy$, which means $d(dx + i I(dx)) = 0+ i dI(dx) = i ddy = 0$ which is closed(in fact it's exact). That's one of the main difference between almost complex manifold and complex manifold.
