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.

Solution 1:

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.

Solution 2:

I'll just add a remark to Georges's (excellent) answer.

The real plane $\mathbb R^2$ has two freely substitutable orientations but the complex plane $\mathbb C$ has a canonical one (as a real vector space). Essentially, when you chose $i$ and not $-i$ as your favorite root of $-1$, you chose at the same time an orientation for the real plane. As a consequence, when two (complex) lines intersect in the (complex) plane, all the orientations are compatible and intersection is positive. This is more generally always the case when two complex manifolds intersect. (My Geometric Algebrish isn't that good, so I won't try to translate this statement for algebraic varieties, but I'm confident a simple translation exists).

Now, when Georges say: "the way out of this absurdity is to realize that you can't move E", he is merely admitting that algebraists are powerless. I'm a topologist and I can move E. But I only obtain two real surfaces $S_1 = E$ and $S_2 = E_{\text{slightly moved}}$ in a real 4-manifold $W$ having a total intersection number of -1. That means that at least one of the intersection points $x$ comes with a negative sign: putting together an orientation of direct basis of $T_x S_1$ and a direct basis of $T_x S_2$ gives an indirect basis of $T_x W$. And the facts I stated above about intersections of complex submanifolds prove that the deformation I've just made has absolutely no counterpart in the complex/algebraic world.

Solution 3:

Let me complement PseudoNeo's excellent answer by actually writing down a motion of $E$ and showing that it intersects itself with multiplicity $-1$. Recall that $\mathbb{C}^2$ blown up at $(0,0)$ can be thought of as the space of ordered pairs $(z, \ell)$ where $z$ is a point of $\mathbb{C}^2$ and $\ell$ is a line through $z$ and through $(0,0)$. The exceptional fiber $E$ is the set of pairs of the form $((0,0),\ \ell)$ where $\ell$ can be any line through $(0,0)$.

In order to perturb this, I want to move to a set of the form $\{ \zeta(\ell), \ell \}$ where $\zeta$ is some continuous function which, to a line through the origin in $\mathbb{C}^2$, assigns a point on that line. We'll write $\ell$ in homogenous coordinates as $(x : y)$. So we need a continuous function $\zeta$ which, given $(x, y) \in \mathbb{C}^2 \setminus \{ (0,0) \}$, chooses some point on the line through $(x,y)$, so that $\zeta(x, y) = \zeta(\lambda x, \lambda y)$ for any nonzero $\lambda$.

Some experimentation produces $$\zeta(x,y) := \left( \frac{x \overline{x}}{|x|^2+|y|^2},\ \frac{y \overline{x}}{|x|^2+|y|^2} \right).$$ Let $E_{moved}$ be the set of points $\left( \zeta(x,y), (x:y) \right)$. (If you want to reassure yourself that $E_{moved}$ is homotopic to $E$, consider the homotopy $\left( t\cdot \zeta(x,y), (x:y) \right)$, as $t$ goes from $1$ to $0$.)

Now $E$ intersects $E_{moved}$ when $\zeta(x,y)=(0,0)$, which happens when $(x:y) = (0:1)$. Near $\left( (0,0), (0:1) \right)$, local complex coordinates are $u$ and $v$, corresponding to the point $\left( (uv,u), (v:1) \right)$ in the blow up. In this chart, $E$ is given by $u=0$ and $E_{moved}$ is given by $$u = \frac{\overline{v}}{1+|v|^2}.$$

The intersection takes place at $(u,v) = (0,0)$. Writing $u = u_1 + i u_2$ and $v = v_1 + i v_2$, the Jacobian of $v \mapsto \overline{v}/(1+|v|^2)$ is $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$. The fact that this has determinant $-1$ indicates that the intersection with $u=0$ counts with sign $-1$.

Solution 4:

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.