Reading Griffiths-Harris, at page 56 I find some parts that I can't understand.
1) After having proved the Poincaré Duality Theorem, Griffiths and Harris proceed to prove a weaker result, that's to say
$P: H_k(M,\mathbb{Q}) \mapsto H_{n-k}(M,\mathbb{Q})^{*}$ given by $P(A)(B)=(A.B)$ is the intersection pairing.
Why are considering the case of the homology group with coefficients in $\mathbb{Q}$ when the duality theorem is stated for homologies with coefficients in $\mathbb{Z}$?
They do the same thing below when they talk about the Kunneth formula. How can they pass from coefficients of a type to coefficients of another type so freely?
2) They also say that via the De Rham isomorphism $H_{DR}^{n-k} \mapsto H^{n-k}(M,\mathbb{C})$, this is equivalent to the assertion that for any $n-k$ cycle $B$ su $M$ holds the relation $\int_B \phi=(A.B)$, and this is ok! At the end of the proof they define the bilinear map $H_{DR}^k(M) \otimes H_{DR}^{n-k}(M) \mapsto H_{DR}^n(M) \simeq \mathbb{C}$. Is it always true that if $M$ is a complex compact $n-$dimensional manifold, it's $n^{th}$-cohomology group is always isomorphic to $\mathbb{C}$?
Can you explain me this points, please?
Note that an isomorphism $H_k\to H_{n-k}^*$ is very different than an isomorphism $H_k\to H^{n-k}$! A cohomology class determines an element of $H_{n-k}^*$, but the converse does not hold over $\mathbb{Z}$. Instead, $H^{n-k}(M,\mathbb{Z})\cong H_{n-k}(M,\mathbb{Z})^*\oplus \mathrm{Tors}(H_{n-k-1})(M,\mathbb{Z})$. Intuitively, taking $\mathbb{Q}$ coefficients instead of $\mathbb{Z}$ coefficients kills torsion but preserves the free part (one proof is by the universal coefficient theorem,) so that Poincaré duality implies $H_k(M,\mathbb{Q})\cong H_{n-k}(M,\mathbb{Q})$. Then Griffiths-Harris' proposition is just making that isomorphism explicit.