computing intersection number via differential forms

Question

Let $M$ be a $2n$ dimensional manifold, and let $A$ be a $n$-dimensional cycle on $M.$ I want to compute the self-intersection $(A.A)$ of $A$ with itself. Let $\eta_A$ be the form in $H^n(M, \mathbb{R})$ given by the Poincare duality isomorphism, i.e. $\int_B \eta_A=(A.B)$ for an $n$-cycle $B.$ Then by the rule that intersection of cycles corresponds to taking wedge product, we have $\int_M \eta_A \wedge \eta_B=(A.B)$. Taking $A=B,$ since $\eta_A \wedge \eta_A =0$, I get $(A.A)=0$, which I don't think is the case. Where am I going wrong?

Answer 1

$\omega \wedge \eta = (-1)^{ab}\eta \wedge \omega$ where $a$ is the dimension of $\omega$ and $b$ is the dimension of $\eta$.

So your "since $\eta_A \wedge \eta_A=0$" comment is only true when $n$ is odd.