Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X/k$ be a surface nonsingular and proper over an algebraically closed field $k$. Let $C \subset X$ be a nonsingular curve. Then it is clear that the self-intersection $(C \cdot C)_X$ is $\textrm{deg}_C ( \mathcal{N}_{X/C} )$ , basically a matter of definition in intersection theory. More generally, if $X/k$ is a proper variety of dimension $k$, and $Y \subset X$ is a cartier divisor, the the class $[Y\cdot Y] \in A_{k-2}(Y)$ is the class of the line bundle $\mathcal{O}_X(Y) \vert_{Y} = \mathcal{N}_{Y/X}$. Both of these results are fairly easy to prove. I'm asking for something a little different:

$\textbf{Question:}$ I imagine these results are "intuitively clear" at some level to geometers. Let's stick to complex algebraic varieties. In the setting of surfaces, can one explain why the normal bundle controls the number of points that divisors linearly equivalent to $C$ meet $C$? I want to say that this "follows" because we can consider the normal bundle as a "tubular neighborhood", but I don't know how to do this precisely, or how to finish the argument. How about in the higher dimensional case?

share|improve this question
    
Something can be intuitively clear without being precise. –  Will Sawin Nov 4 '12 at 16:28
    
Hi Will, First of all, let me say thanks for your helpful answers and comments on my previous questions! What I'm asking for here is very "humble". It is clear (from working with the relevant machinery) that the normal bundle to a divisor controls it's self-intersection. I suspect that this is because the normal bundle is similar to a tubular neighborhood, and hence may control "moving" the divisor. What I don't know, is the the extent to which this is or can be made precise. I am curious if I should think of this as I currently understand (above) or if there is more. It certainly seems so. :) –  LMN Nov 4 '12 at 18:21
4  
There are a couple of things that may be helpful here. One is the notion of "deformation to the normal cone." Another is the following: a section of the normal bundle $\mathcal N_{X/C}$ is precisely a first-order deformation of $C$, i.e., a tangent vector to the point $[C]$ in the space [Hilbert scheme] $H$ of closed subschemes of $X$. Whenever you can move $C$ via a smooth, one-dimensional family of curves in $X$, this corresponds to a curve in $H$ passing through $[C]$. The tangent vector at $[C]$ corresponds to a section of $\mathcal N_{X/C}$, and the zeros of this section are precisely... –  Charles Staats Nov 4 '12 at 18:54
10  
Here's a non rigorous intuitive explanation. To define $C^2$, you hold one copy of $C$ fixed, move the other, say $C'$, into general position and count the number of points of $C\cap C'$. Infinitesimally, the deformation $C'$ is given by a normal vector field $v$ on $C$. $v$ determines the directions in which to move the points of $C$, so the zeros of $v$ are the points won't get moved. So the degree of $N_C$, which is the number of zeros of $v$, ought to match the number of points on the $C\cap C'$. –  Donu Arapura Nov 4 '12 at 18:55
    
...the points that are "infinitesimally fixed" by the deformation, i.e., not moved off of $C$. This is not an unusual situation: what is true locally in topology, is true infinitesimally in algebraic geometry. –  Charles Staats Nov 4 '12 at 18:58
show 3 more comments

1 Answer

up vote 7 down vote accepted

I'd like to expand a bit on the excellent comments of Charles Staats and Donu Arapura. They both suggest understanding the self-intersection number of a curve as the number of fixed points of an infinitesimal deformation of the curve, which is manifestly the degree of the normal bundle when such a deformation exists. Here's a slightly more pedestrian route, which I think has the benefit of being rigorous and almost as intuitive.

Suppose we have two curves in a surface: $$\iota_C: C\hookrightarrow X, \iota_D: D\hookrightarrow X.$$ If $C\cap D$ has dimension zero, the intersection number should manifestly be $$C\cdot D:=\dim\Gamma(C\cap D, \mathcal{O}_{C\cap D})=\dim\Gamma(C, \iota_C^*\mathcal{O}_D).$$ We'd like to write this as an Euler characteristic, to make it constant if we vary $C$ or $D$ in a flat family. But this is easy; since $\mathcal{O}_{C\cap D}$ has zero-dimensional support, it has no higher cohomology, so its Euler characteristic equals $C\cdot D$ as defined above. Line bundles are nice (and more importantly, are acyclic with respect to restriction), so we use the short exact sequence $$0\to \mathcal{O}_X\to \mathcal{O}_X(D)\to \mathcal{O}_D\to 0$$ to rewrite this Euler characteristic as $$\chi(\mathcal{O}_X(D)|_C)-\chi(\mathcal{O}_C)=\operatorname{deg}(\mathcal{O}_X(D)|_C).$$

I think this is a reasonably intuitive motivation for the definition of the intersection number. So to fully answer your question, one should give an intuitive reason for why $\mathcal{O}_X(D)|_D$ is $\mathcal{N}_{D/X}.$ Of course, this is just the definition of the normal bundle, but let's motivate the definition. First, why is the conormal bundle $I/I^2$, for $I$ the ideal sheaf of a closed subvariety $V\subset X$? Well, an element of $I/I^2$ is precisely a function on $X$ vanishing at $V$, but ignoring higher-order terms. A section to the normal bundle precisely takes functions $f$ defined in a neighborhood of $Y$ and differentiates them--but the partial derivative should depend only on the first-order part of $f$. So the normal bundle should be precisely $(I/I^2)^\vee$. This is another name for $\mathcal{O}_X(D)|_D.$

I hope that was some reasonable intuition/motivation.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.