I am reading a paper and there I have the following:
$F:=Y\times \mathbb P^1$ where $Y$ is a sextic hypersurface in the weighted projective space $\mathbb P(1^3,2,3)$.
(1) Let $X\subset F$ be a Fano threefold that is a (1,1)-section of
$F$.
The anticanonical line bundle
(2) $\omega_X^{\vee}\cong (\mathcal O_F(2,2)\otimes \mathcal O_F(-X))|_X\cong \mathcal O_F(1,1)|_X$
The conormal sequence for the inclusion $X\hookrightarrow F$ twisted by the anticanonical line bundle.
(3) $0\rightarrow \mathcal O_X\rightarrow\Omega_F^1(1,1)|_X\rightarrow \Omega_X^1(1,1)\rightarrow 0$
In (1), what does "$X$ is a (1,1)-section of $F$" mean?
In (2), How can I prove the last isomorphism? I know that the anticanonical line bundle is $\omega_X^{\vee}\cong \omega_F^{\vee}\otimes I/I^2\cong \mathcal O_F(2,2)\otimes \mathcal O_F(-X))$, since $\omega_F^{\vee}=\pi_1^*\mathcal O(6-(3+2+3))|_Y\otimes \pi_2^*\mathcal O(-2)=\mathcal O_F(2,2)$ according with the notation of the paper. Also I know that $I/I^2=\mathcal O_F(-X)$ but I don't understand how to prove the last isomorphism.
In (3), I know that the the conormal sequence for the inclusion $X\hookrightarrow F$ is $0\rightarrow I/I^2\rightarrow\Omega_F^1|_X\rightarrow \Omega_X^1\rightarrow 0$ but, why after twist it by $\omega_X^{\vee}$ the firts sheaf is $\mathcal O_X$? I can't conclude that from (2).
Best Answer
I guess $\mathcal{O}_Y(1)$ denotes the ample generator of the Picard group of $Y$ --- this is the restriction of the reflexive sheaf $\mathcal{O}(1)$ from the weighted projective space. Note that by adjunction formula $$ \omega_Y \cong \mathcal{O}_Y(-2). $$ Further, I think $X \subset F = Y \times \mathbb{P}^1$ is defined as the hypersurface from the linear system $\mathcal{O}_F(1,1) = \mathcal{O}_Y(1) \boxtimes \mathcal{O}_{\mathbb{P}^1}(1)$ (this answers (1)).
Furthermore, (2) follows from the standard isomorphism $$ I/I^2 \cong \mathcal{O}_F(-1,-1)\vert_X = \mathcal{O}_X(-1,-1) $$ and (3) is obtained be tensoring the conormal sequnce with $\mathcal{O}_X(1,1)$.