0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Nicolas Hemelsoet May 1 '18 at 14:39
  • $\begingroup$ 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. :-) $\endgroup$ – YoYo May 1 '18 at 15:05
1
$\begingroup$

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 $PD[S] \in H^{p,p}(M)$.

$\endgroup$
  • $\begingroup$ 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 ? :-) $\endgroup$ – YoYo May 1 '18 at 15:18
  • $\begingroup$ Ok, thank you, in understand now. :-) $\endgroup$ – YoYo May 1 '18 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.