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.

**Answer**

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.

**Attribution***Source : Link , Question Author : unk220 , Answer Author : user875280*