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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?

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up vote 8 down vote accepted

$\omega \wedge \eta = (-1)^{ab}\eta \wedge \omega$ 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.

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I don't follow, If $\eta_A=dx_1 \wedge dx_2$, then in $\eta_A \wedge \eta_A=dx_1 \wedge dx_2 \wedge dx_1 \wedge dx_2 =0.$ –  ymo Sep 17 '10 at 6:40
3  
If $\eta=dx_1\wedge dx_2+dx_3\wedge dx_4$ then $\eta\wedge\eta = 2dx_1\wedge x_2\wedge x_3\wedge x_4\ne0$. –  Robin Chapman Sep 17 '10 at 7:28
    
oh, I see. Is there a way to see using differential forms that when you blowup the $\mathbb{P}^2$ at a point, the exceptional divisor has self intersection -1? –  ymo Sep 17 '10 at 7:47
    
I frequently find algebraic geometry terminology disorienting, but I take it the exceptional divisor of a blow-up is the $\mathbb CP^1$ you have to blow down to get your original variety? Then by Mayer-Vietoris your question reduces to the self-intersection of $\mathbb CP^1$ in $\mathbb CP^2$ which I'm pretty certain is written up in the language of differential forms in Bott and Tu's book. –  Ryan Budney Sep 17 '10 at 23:20

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