1
$\begingroup$

I'm interested in the first basic case of excess intersection in intersection theory:

Let $X$ be a smooth projective 4-fold and let $S,T$ be two surfaces in $X$. Assume that the intersection $S\cap T$ contains an effective 1-cycle $D$ as its 1-dimensional part. In other words, $Z$ defines a Cartier divisor on $S$.

Is there some way of extracting information about $D$ as a divisor on $S$ (e.g., the self-intersection $D^2$) given the intersection number $S\cdot T$ and the normal bundles $N_S,N_T$?

$\endgroup$
2
  • $\begingroup$ In what sense are the normal bundles given? On $S$ and $T$, or on $Z$, or what? Does the intersection contain $Z$ or it is $Z$? Do you assume some kind of regularity (whatever this may mean in this situation)? $\endgroup$ Feb 18, 2015 at 19:56
  • $\begingroup$ The normal bundle is a vector bundle on $S$, and say, I know the Chern classes. The 1-dimensional part of the intersection is $Z$. Feel free to assume that everything is smooth. $\endgroup$ Feb 19, 2015 at 0:10

1 Answer 1

4
+100
$\begingroup$

The answer is NO. Take $X=\mathbb{P}^4$, $S,T$ two smooth quadrics. Then $(S\cdot T)=4$, the normal bundles are $N_S=\mathcal{O}_S(1)\oplus \mathcal{O}_S(2)$, and same for $T$. If $S$ and $T$ are general, $Z=0$. If they are given by $$S:\ X=0\ ,\ YU+TV=0\quad;\quad T:\ Y=0\ ,\ XV+TU=0$$(coordinates $(X,Y,T,U,V)$ on $\mathbb{P}^4$), then $Z$ is the line $X=Y=T=0$ on $S$, with $Z^2=0$. If $S$ and $T$ lie in the same hyperplane, then $Z$ is a degree 4 elliptic curve in $S$, linearly equivalent to 2 times the hyperplane class $H$, and $Z^2=4H^2=8$.

$\endgroup$
5
  • $\begingroup$ Cheers! So there is in general not much hope to say anything about the divisor $Z$, e.g., when it is nef? $\endgroup$ Feb 22, 2015 at 11:32
  • $\begingroup$ No. You could play the same game with two cubic surfaces (contained in a hyperplane), so that your 1-cycle may be a line (of square $-1$, hence not nef) or a triple of the hyperplane section. $\endgroup$
    – abx
    Feb 22, 2015 at 11:39
  • $\begingroup$ So does @abx collect that bounty?? $\endgroup$
    – j0equ1nn
    Feb 24, 2015 at 2:59
  • $\begingroup$ @B.Wellington — If you think these examples are good; please consider accepting this answer, by clicking the check mark to the left of the answer (below the votes). $\endgroup$
    – jmc
    Feb 24, 2015 at 14:05
  • $\begingroup$ Thanks, I'll wait a bit, in case there are other answers too. $\endgroup$ Feb 24, 2015 at 16:30

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.