The intersection pairing between two divisors on a nonsingular algebraic surface over a field is defined thanks to the following theorem (the reference is Hartshorne's book): enter image description here enter image description here

One can define a pairing for any couple of invertible sheaves $\mathcal L,\mathcal M\in\operatorname{Pic}(X)$ as follows:

$$\mathcal L.\mathcal M:=\chi(\mathcal O_X)-\chi(\mathcal L^{-1})-\mathcal (M^{-1})-\chi( \mathcal L^{-1}\otimes \mathcal M^{-1})\quad\quad (\ast)$$

By using the well known isomorphism between $\operatorname{Pic}(X)$ and the group of divisors up linear equivalence, one can clearly define:

$$C.D:=\mathcal O_X(C).\mathcal O_X(D)$$

and the final step is to show that this definition satisfies properties (1)-(4) of the above theorem.

So everything is very clear, but I don't understand what is the meaning of the definition $(\ast)$. It seems to me that this pairing for invertible sheaves appears out of the blue. Can you give any intuitive motivation about its nature? Why do we need the Poincare characteristics? Why are we taking the inverse sheaves?

  • $\begingroup$ Are you asking about (*), which seems a very intuitive notion, when you intersect 2 curves, more or less you are counting the number of points common? Or, are you asking about the proof using line bundles? $\endgroup$ – Mohan Feb 24 '17 at 1:20
  • 2
    $\begingroup$ I'm asking why $(\ast)$ is the intersection pairing. From its definitions by means of Poincare characteristic I don't see any geometry and any "more or less (...) counting the number of points common" $\endgroup$ – notsure Feb 24 '17 at 1:32

Here is a simple way to connect ($\star$) to your geometric intuition.

Suppose $C$ and $D$ are two curves on your surface. There exists an exact sequence, $$ 0 \to \mathcal O_{X}(-C-D) \to \mathcal O_X(-C)\oplus \mathcal O_X(-D) \to O_X \to \mathcal O_{C \cap D} \to 0,$$ where $\mathcal O_{C \cap D}$ is the skyscraper sheaf supported on the intersection points of $C$ and $D$ with stalks of dimension equal to the intersection multiplicity at each intersection point.

[To spell out the above morphisms more explicitly, suppose that ${\rm Spec \ } A \subset X$ is a open affine on which $C$ and $D$ are the vanishing loci of $f$ and $g$ in $A$ respectively. Then the associated exact sequence of modules over ${\rm Spec \ } A$ is $$ 0 \to A \overset{(-g,f)}{\to} A^{\oplus 2}\overset{(f,g)}{\to} A \to A/(f,g) \to 0.$$ Indeed, the stalk at an intersection point $x \in C \cap D$ is $A_x/(f,g)$, whose dimension is the standard definition of the intersection multiplicity of $C$ and $D$ at $x$. ]

Anyway, your intuitive notion of intersection pairing is that $C.D$ is the number of points in $C\cap D$ counted with multiplicity. But this is exactly the same thing as the number of global sections of $\mathcal O_{C \cap D}$. Since $\mathcal O_{C \cap D}$ is a skyscraper sheaf, its higher cohomologies vanish, so this is the same as the Euler characteristic $\chi(\mathcal O_{C \cap D})$. But then, by the above exact sequence, this Euler characteristic is equal to $\chi(\mathcal O_X) - \chi(\mathcal O_X(-C)) - \chi(\mathcal O_X(-D)) + \chi(\mathcal O_X(-C-D)),$ which agrees with your $(\star)$.

  • $\begingroup$ Great answer, Kenny: bravo and +1 (I have taken the liberty to correct a typo in your third line, replacing $ \mathcal O_x(-D)$ by $ \mathcal O_X(-D)$). $\endgroup$ – Georges Elencwajg Feb 24 '17 at 19:31
  • $\begingroup$ Nice answer! I will wait a few days and then I'll think about its acceptance. $\endgroup$ – notsure Feb 24 '17 at 20:30

Your questions are a frequent reaction to the "rabbit-out-of-a-hat" type of proof. The reason for doing it that way is a matter of exposition. A more natural approach would be step-by-step. First, for two nonsingular curves $C$ and $D$, meeting transversally, define $C.D$ to the the number of intersection points. Next, keeping $C$ fixed, let's show this depends only on the linear equivalence class of $D$. This is because if $D \sim D'$ on the surface, then $C.D \sim C.D'$ as divisors on $C$. And then we know that linearly equivalent divisors on a curve have the same degree. This will still work if $D$ becomes singular, as long as its intersection with $C$ is a finite set of points.

Maybe you can work out the rest for yourself. If you start with arbitrary divisors, any such is a difference of effective curves, and using Bertini's theorem, you can take them to be nonsingular. This gives a definition for any two divisors, but you have to show it is independent of the choices made. By the time you work this all out in complete detail, perhaps you can appreciate the choice of Hartshorne to take the more efficient "rabbit-out-of-a-hat" method.

Of course Kenny's answer to your question is quite excellent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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