Degree of the Normal Bundle of a Line in a Projective Variety

Question

Let $S\subset \mathbb{P}^3$ a smooth surface of degree $d$, and $L$ a line in $S$. I would like to compute the self intersection of $L$, which I know to be $L^2=deg(N_{L/S})$, where the last one is the normal bundle of $L$ in $S$. Also, I could arrive to $\chi(O_S(L))=\chi(O_S)+deg(N_{L/S})+1$.

Question: apart from the use in the above problem, how do I generally compute the degree of that normal bundle (or even in this simpler case)? By now I just know that I have an exact sequence: $$0\to T_L\to T_S|_L\to N_{L/S}\to 0 $$ but even I really don't know how to associate a divisor to the line bundle $N_{L/S}$.

Answer 1

Here is a more detailed answer that my previous one, for references I would suggest this expository paper for definitions and some examples, a more complete reference is the book 3264 and all that by Eisenbud and Harris. The canonical reference for this stuff would be Intersection theory by Fulton but there are less details.

First $N_{L/S}$ is a line bundle on $\Bbb P^1$ so it is on the form $\mathcal O(a)$ for some $a \in \Bbb Z$ : we just need to find $a$, we will do it with Chern classes.

Here is everything you need to know about them : $c(E) \in A(X)$, the Chow ring of $X$ (you can think of this ring to be the cohomology ring of $X$ for $X = \Bbb P^n$ for example). Elements of the Chow ring will be subvarieties (quotiented by "rational equivalence, which is an "algebraic" version of being homologuous) and the product is the intersection of the subvarieties.

Here are the key properties : 1) If $E = O_X(D)$ then $c(E) = 1 + D$. 2) If a sequence $0 \to E' \to E \to E'' \to 0$ is exact then $c(E) = c(E')c(E'')$.

Now, the tangent bundle of $\Bbb P^1$ is $O(2p)$ where $p \in \Bbb P^1$ so we get $c(T_L) = 1 + 2p$. Now, we need to compute $c(T_S)$. Applying 2) to the Euler exact sequence gives you $c(T_{\Bbb P^3}) = (1+h)^4$ where $h$ is the class of an hyperplane. On the other hand, we have $N_{S/ \Bbb P^3} \cong \mathcal O_S(d)$ (I believe it is a consequence of the adjunction formula) so we obtain using 2) with the sequence $0 \to T_S \to {T_{\Bbb P^3}}_{|S} \to N_{S/ \Bbb P^3} \to 0$ shows that $c(T_S) = 1 + (4-d)h + \dots$ where $h$ denote the class of an hyperplane section.

Using the exact sequence you wrote we get $(1+2p)(1+ap) = 1 + (4-d) + \dots $ so $a = 2-d$.

This coincide with everything we know : for $d=1$ the self-intersection of a line with itself is $1$ as we can move the line, for $d=2$ a quadric surface is $\Bbb P^1 \times \Bbb P^1$ and a line can be deformed away from itself, for $d=3$ this is also well-known that the $27$ lines on the cubic surfaces have self-intersection $-1$.