I know the definition of the normal bundle $N_{C/S}$ of a curve $C$ over a surface $S$ as the cokernel of the injection $T_C \subset T_S|_C$ where $T$ is the tangent bundle.
I would like to exhibit an explicit example of the normal bundle $N_{C|S}$ over a surface $S$ of a curve $C \subset S$. $S$ is a complete intersection of $4$ quadrics $Q_i$ $i=1,...,4$ lying in the complex projective space $\mathbb{P}^6$. So $S=Q_1 \cap Q_2 \cap Q_3 \cap Q_4$. In my assumption i suppose that the curve $C=S \cap H$ where $H$ is an hyperplane of the six dimensional projective space. So, due to Bertini, $C$ is an hyperplane section.
Can someone helping with some hint or suggestion?
Perhaps an usefull relation. If i use the adjubction formula i get: $$K_C=(K_S\otimes N_{c/S})|_{C}=K_S|_C\otimes C|_C=K_S|_C\otimes H|_C$$ where $K$ is the canonical bundle.
Than $K_C=2H|_C$ so using both equation i get $H|_C=N_{C/S}$. So if i want to compute $H^0(C,N_{C/S})$ i must calculate $H^0(C,H|_C)$. Any ideas to compute it?
This is my idea to compute the dimension of the space $H^0(C,H|_{C})$. I can take the fundamental sequence of the curve $C$ i get $$ 0 \rightarrow O_S(-C)\rightarrow O_S \rightarrow O_C \rightarrow 0.$$ The exactness of the sequence is preserved if i multiply with the line bundle $H$. So i get $$ 0 \rightarrow O_S(H-C)\rightarrow O_S(H) \rightarrow O_C(H) \rightarrow 0$$.
This sequence induces the following long exact sequence in cohomology $$0 \rightarrow H^0(S,O_S(H-C))\rightarrow H^0(S,O_S(H)) \rightarrow H^0(C,O_C(H)) \rightarrow H^1(S,O_S(H-C)) \rightarrow ....$$ and so on.
I would like to compute $H^0(C,O_C(H))$. How can i compute the remaining term of the cohomology sequence to obtain $h^0(C,O_C(H))$?

  • $\begingroup$ I only know $h^0(C,O_C(H))$ is between 6 to 8... As $h^0(P^6,H)=7$, and we know $2H=K$. I would guess the number is 8 generically, but I don't know how to prove it. $\endgroup$ – lee Nov 28 '15 at 16:53

Okay, assume the characteristic is $0$ (over complex numbers). Let's stare at your last exact sequence:

$$0 \rightarrow H^0(S,O_S(H-C))\rightarrow H^0(S,O_S(H)) \rightarrow H^0(C,O_C(H)) \rightarrow H^1(S,O_S(H-C)) \rightarrow ....$$

In the surface $S$, $C=H|_S$ as divisor classes. So we know your first term has dimension 1 ($S$ is smooth and simply connected by Lefschetz's Hyperplane Section Theorem).

The fourth term has dimension $0$ as $H^1(S,O_S(H-C))=H^{1,0}(S)$, and we know $S$ is simply connected.

Now what left is we have to compute $h^0(S,O_S(H))=h^{0,2}(S)$, by adjunction formula. So we only have to compute $\chi(O(S))$. Now it follows directly by computing the Hilbert polynomial of $S$, that $\chi(O(S))=8$. Hence $h^0(S,O_S(H))=h^{0,2}(S)=7$.

So we know that the number you are interested in, i.e., $h^0(C,O_C(H))=6$, and obviously they are given by all the restriction of hyperplane sections. I find it interesting, and maybe this curve $C$ is projectively normal? I'm too sleepy to compute higher degree hyperplane sections to see if the numbers match...


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