In the following link, pages : $133$ and $65$, we find the following paragraph translated from spanish to english :
Let $M$ be a compact kahler manifold
Let $S \subset M $ be a complex submanifold of ( complex ) dimension $ n-p $
We have : $PD[S] \in H^{2p} ( M , \mathbb{Z} ) \cap H^{p,p} (M) $ :
Indeed :
Obviously : $PD[S] \in H^{Zp} ( M , \mathbb{Z} )$
$H^{2p}(M,\mathbb{C} ) = \displaystyle \bigoplus_{i+j=2p} H^{i,j} (M) $
Let $ \alpha \in H^{q_{1} , q_{2}} (M) $ , $ q_1 + q_2 = 2n-2p $.
$ \alpha \in \sum f_{I,J} dz_{i_{1}} \wedge \dots \wedge dz_{i_{q_{1}}} \wedge d \overline{z}_{j_{1}} \wedge \dots \wedge d\overline{z}_{j_{q_{2}}} $
$ \langle \alpha , PD[S] \rangle = \int_S \alpha $
Locally, $ S = \{ \ z_1 = 0 , \dots , z_p = 0 \ \} $
If $ q_1 > n-p $ then $ dz_{i_{1}} \wedge \dots \wedge dz_{i_{q_{1}}} = 0 $
If $ q_2 > n-p $ then $ d \overline{z}_{j_{1}} \wedge \dots \wedge d \overline{z}_{j_{q_{2}}} = \overline{dz_{j_{1}} \wedge \dots \wedge dz_{j_{q_{2}}}} = 0 $.
If $ (q_1 , q_2) \neq (n-p , n-p)$, then $ \int_S \alpha = 0 $
$ \Longrightarrow PD[S] \in H^{p,p} (M) $
My questions are :
$ 1) $ According to page : $65$, if $ \begin{cases} dz_j = dx_j + i dy_j \\ d \overline{z}_j = dx_j - i dy_j \end{cases} $, how do we go from $ dz_{i_{1}} \wedge \dots \wedge dz_{i_{q_{1}}} \wedge d \overline{z}_{j_{1}} \wedge \dots \wedge d\overline{z}_{j_{q_{2}}} $ and developpe it, to obtain a formula in terms of $ dx_i $ and $ dy_j $ and the symbol : $ \wedge $ ?
$ 2) $ Why is $ d \overline{z}_{j_{1}} \wedge \dots \wedge d \overline{z}_{j_{q_{2}}} = \overline{dz_{j_{1}} \wedge \dots \wedge dz_{j_{q_{2}}}} $.
$ 3) $ Why does the fact $ \ \ \big( \ \ $ if $ (q_1 , q_2) \neq (n-p , n-p) $, then $ \int_S \alpha = 0 \ \ \big) \ \ $ involve that $ \Longrightarrow \ \ PD[S] \in H^{p,p} (M) $ ( explicitly, please ) ?
Thanks in advance for your help.
As I said in comments everything follows from definitions.
1) e.g we have $$dz_1 \wedge d \overline z_2 = (dx_1 + idy_1) \wedge (dx_2 - idy_2) = dx_1 \wedge dx_2 + dy_1 \wedge dy_2 - (dx_1 \wedge dy_2 + dy_1 \wedge dx_2)$$ and you can immediately generalize.
2) This is how you define conjugaison of differential forms.
3) By Hodge decomposition, write $\alpha = \lambda_{0,2p}\alpha^{0,2p} + \dots + \lambda_{2p,0} \alpha^{2p,0} $. Integration against $S$ shows that all the $\lambda$'s are zero excepted $\lambda_{p,p}$, this exactly means that $PD[S] \in H^{p,p}(M)$.