TITLE: Geometric motivation for negative self-intersection
QUESTION [50 upvotes]: 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.

REPLY [24 votes]: One way to think about it is that $E^2=-1$ is necessary for the consistency of any reasonable intersection theory for divisors that descends to linear equivalence classes. Consider the blowup $\pi:X\to \mathbf P^2$ of the plane at a point $P$. We compute the self intersection $\pi^*O(1)\cdot \pi^*O(1)$ in two ways. First represent $O(1)$ by lines $l_1$ and $l_2$ in the plane that do not pass through $P$. Then $\pi^*O(1)$ is represented by the inverse images of these lines in the blow up $X$, which are just their strict transforms $\tilde l_1$ and $\tilde l_2$. Because $\pi$ is an isomorphism away from the exceptional divisor, the disired self intersection is $\tilde l_1\cdot \tilde l_2=l_1\cdot l_2=1$. On the other hand, if we represent $O(1)$ by lines $l_1'$ and $l_2'$ passing through the center of the blow up $P$, we find that $\pi^*O(1)$ is represented by the inverse images $\tilde l_1'+E$ and $\tilde l_2'+E$, where again a tilde denotes the strict transform. Thus we have
$$
1=(\tilde l_1'+E)\cdot(\tilde l_2'+E)=0+1+1+E^2,
$$
from which $E^2=-1$ follows.

Added later: Here is some heuristic motivation for an intersection theory of linear equivalence classes of divisors. Suppose $X$ is a proper variety and $D,D'\subset X$ are Cartier divisors. Let $C\subset X$ be a curve. Suppose $D$ and $D'$ are linearly equivalent. Then $D$ and $D'$ are zero loci of global sections $s$ and $s'$ of a common invertible sheaf/line bundle $L:= O_X(D)=O_X(D')$. For $t\in \mathbb A^1$, let $D_t := \{ t s+(1-t)s' = 0\}\subset X$. Thus the divisors $D_t$ interpolate between $D'$ (when $t=0$) and $D$ (when $t=1$). As $t$ varies, the intersection $D_t\cap C$ moves in $X$. As it does so, it cannot "lose points at infinity" because $X$ is proper. Because of this, the integer valued function $t\mapsto \deg D_t\cap C$ (which counts intersection points) should be constant.