# Geometric motivation for negative self-intersection

Consider the blow-up of the plane in one point. Let $$E$$ the exceptional divisor. We know that $$(E,E)=-1$$. What is the geometrical reason for which the auto-intersection of $$E$$ is $$-1$$? In general, what does it mean, geometrically, that a divisor has a negative self-intersection or that the “right” number of divisors gives rise to a negative intersection?

Thanks.

If you take a line (isomorphic to $$\mathbb P^1$$) in $$\mathbb P^2$$, its self-intersection is $$1$$. This is a manifestation of the fact that if you slightly move it, the unmoved copy will intersect the moved copy in $$1$$ point. Great you say, I’ll do the same with the exceptional divisor $$E$$ in the blow-up. Let’s see: I’ll slightly move $$E$$ and the moved copy will intersect the fixed copy in…-1 points?! But this is utter nonsense! Yes, it is: the way out of this absurdity is to realize that you can’t move $$E$$ !. In slightly more technical terms, the normal bundle of $$E$$ in the blown-up plane has degree (-1) and this shows that it cannot be moved.So one intuition could be: negative self-intersection= rigidity.
Another intuition could be that negative self-intersection smells of blow-up. A basic result in that direction is Castelnuevo’s criterion: if $$S$$ is a projective surface containing a curve $$E$$ isomorphic to $$\mathbb P^1$$ of self-intersection -1, then $$E$$ can be blown down to a point $$e$$. This means that there exists a surface $$S_0$$ containing a point $$e$$, which when blown-up in $$S_0$$ will become the curve $$E$$ in $$S$$. Grauert and others have proved very profound generalizations of Castelnuevo’s theorem.