2
$\begingroup$

The Wikipedia entry on intersection theory contains the following statement:

[for C a curve, on a surface] "the self-intersection points of C is the generic point of C, taken with multiplicity C · C."

This statement is intriguing and rather plausible. But I don't know how to make it rigorous, in terms of the standard presentation of intersection theory. (And neither does an algebraic-geometer friend, who points out various issues, for instance, "... how it generalizes to higher dimensions: would the self-intersection of a surface in a threefold be a generic curve on that surface?")

So, is any rigorous version of this statement available? Failing this, does there even exist any discussion or mention of this heuristic, other than in that Wikipedia article?

$\endgroup$
1
  • $\begingroup$ I think it is time for someone to clean up this Wikipedia page! $\endgroup$ Jun 15, 2016 at 15:10

1 Answer 1

2
$\begingroup$

I don't like this statement. I want this self-intersection to be (at least formally) zero-dimensional, not one-dimensional like the generic point of the curve.

I regard Fulton's as the now-standard presentation of intersection theory. There, the intersection of $C$ and $D$ inside $X$ is defined for $C$ regularly embedded (which I think of as a cap product, $C$ in cohomology and $D$ in homology): one replaces $D$ by the normal cone $N_{D\cap C} D \subseteq N_C X$, and the intersection is the class $[N_{D\cap C} D] \in A(N_C X) \cong A(C) \to A(X)$, where the (Gysin) isomorphism used is only available because $C$ was regularly embedded (meaning, its normal cone is a vector bundle). That isomorphism changes degree by ${\mathrm {codim}}\ C$.

The canonical, geometric object here representing the intersection is the scheme $N_{D\cap C} D$, defining an effective class on $N_C X$. But we want a cycle in $C$, and we may not be able to get one, because the Gysin isomorphism may take this effective class to a possibly ineffective one on $C$. That's life, of course -- your $C\cdot C$ may be negative. But anyway, under the standard presentation the resulting class is then a negative multiple of the ordinary, not generic, point class on $C$.

$\endgroup$
2
  • $\begingroup$ There is no "ordinary point class" in $C$. As you say, you get a class in the Chow group $A_0(C)$ (a.k.a $CH_0(C)$), which is very far from being trivial. $\endgroup$
    – abx
    Feb 12, 2015 at 6:12
  • 1
    $\begingroup$ Oh, very good point! (No pun intended.) I'm too used to $C = {\mathbb P}^1$. $\endgroup$ Feb 12, 2015 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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