Let $X=\mathbb{P}^1$, $E=\mathcal{O}\oplus\mathcal{O}(1)\oplus \mathcal{O}(2)$ and let $Y=\mathbb{P}(E)$. Denote $\pi:Y\to X$ the projection map. There is a section of $\pi$ corresponding to the surjection $E\to \mathcal{O}$. Let $C$ be the image of this section in $Y$. How can we compute the normal bundle of this curve?
The case when $E$ is a rank 2 vector bundle is discussed here.Normal bundle of a section of a $\mathbb{P}^1$-bundle. However, I believe in our case, the argument cannot be directly applied.
Thanks for the help!
Let us do this more generally, so that we do not have to look at many cases. Let $E$ be a vector bundle on a scheme $X$ and assume we have an exact sequence $0\to F\to E\to L\to 0$, with $L$ a line bundle. This gives a section $D$ of $\pi:P=\mathbb{P}_X(E)\to X$, the projective bundle. Let us calculate the normal bundle of $D$ in $P$.
We have a natural map $\pi^*F\to \mathcal{O}_P(1)$ and one easily checks that the cokernel is just $\mathcal{O}_D(1)$, which can easily seen to be $L$ (identifying $D$ with $X$). So, we get a surjection $\pi^*(F\otimes L^{-1})\to I_D$, the ideal sheaf of $D\subset P$. Since codimension of $D$ is just the rank of $F$, we see that $D$ is a local complete intersection, coming from a correct rank vector bundle. Thus the normal bundle of $D$ then is just $F^*\otimes L$.