# intersection form of $CP^2$

I am trying to understand why the intersection form of $CP^2$ is <1>. First we introduce {[x:y:z], x=0} as a generator of second homology and then we say that it has one intersection with {[x:y:z], y=0}. The part I do not understand is why the intersection number is 1 and not -1. Similarly, why for $CP^2$ with the reverse orientation the intersection number is -1. So altogether I want to know how we determine the sign of the intersection.

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Complex manifolds are always orientable. – user641 Mar 13 '13 at 15:43

All of these operations are only up to sign. Once you pick the sign though, everything is fixed. It is similar to integration: you pick a generator $\omega \in H^n(M, \mathbb R)$ and this gives you a good notion of integration. But $-\omega$ works just as well and will give you consistently opposite results. Basically it all boils down to picking one of two equivalent options and sticking to it for the rest of your life.
I see, you want a concrete way from picking an orientation to the sign. Very well. Suppose the orientation is picked. Now, since the submanifolds (let's call them $M$ and $N$) meet transversally at every point $p$ of intersection, their tangent spaces at $p$ generate the full tangent space $T_p X \cong T_p M \oplus T_p N$ (where $X$ is the ambient manifold, in your case $CP^2$). Since we have orientation, we can tell whether this isomorphism is orientation-preserving or not and this is where the sign comes from.