# Non-proper intersection of surfaces

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 $N_S,N_T$?

• 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)? – Alex Degtyarev Feb 18 '15 at 19:56
• 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. – Walter Neff Feb 19 '15 at 0:10

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 $Z^2=4H^2=8$.
• Cheers! So there is in general not much hope to say anything about the divisor $Z$, e.g., when it is nef? – Walter Neff Feb 22 '15 at 11:32
• 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. – abx Feb 22 '15 at 11:39