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 ) ?

• What did you try ? 1) and 2) follows from definitions. And 3) also follows from definition as well (you need to know the existence of Hodge decomposition of course). – Nicolas Hemelsoet May 1 '18 at 14:39
• Yes, that's right Nicolas,thank you. :-) but, can you write me a detail answer about these three questions please ?. :-) Concerning question $1)$, i know we need to use the determinant of every minor of a matrix that i don't knom its form, but, i don't know what is the final formula, because it's hard for me to do this tiresome calculus which never ends. Thank you. :-) – YoYo May 1 '18 at 15:05

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.
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 $PD[S] \in H^{p,p}(M)$.
• Thank you Nicolas for your answer, but if we have a long piece $dz_1 \wedge d \overline{z}_2 \wedge \dots \wedge d \overline{z}_{m-1} \wedge dz_{m}$ for example, how to obtain the right formula without passing through doing this fastidious developpement that you use above, only using determinant of minors of a matrix for example ? :-) – YoYo May 1 '18 at 15:18