Assume all varieties are projective and smooth over $\Bbb{C}$.
Let $X\subset\Bbb{P}^3$ be surface and $C\subset X$ a curve on it.
The normal bundle $\mathcal{N}_{C/X}$ is the cokernel of the map $T_C\hookrightarrow T_X\big|_C$, which intuitively represent the tangent vectors on $X$ which are perpendicular to $C$.
I've read recently that the selfintersection $C^2$ equals the degree of the normal bundle $\mathcal{N}_{C/X}$, which can be interpreted as how free the curve $C$ is to move inside $X$ (in particular, if $C^2<0$ then $C$ can't move).
At first I understood this explanation to be just a way to make us more comfortable with the idea of selfintersection. But this answer made me think twice.
The answer deals in particular with the example of (I'll adapt it a little) a curve $C\subset\Bbb{P}_\Bbb{C}^3$ of degree $2$ and genus $0$. It can be proven that there is a quadric $Q\subset\Bbb{P}^3$ containing $C$. Furthermore $\mathcal{N}_{C/Q}=\mathcal{O}_C(1)$, which according to the answer is because "any such curve is obtained as a hyperplane section of $Q$, so the line bundle $\mathcal{N}_{C/Q}$ has to have degree $1$ on $C$".
I want to explore that. Does that mean that the possibilities of $C$ to move in $Q$ amount to the way we move the hyperplane cutting $Q$? If, more generally, we had $\mathcal{N}_{C/X}=\mathcal{O}_C(d)$, would this mean that $C$ is the section of $X$ by a hypersurface of degree $d$ and that $C$ moves in $X$ according to how we move the hypersurface?
I'd like to understand this as geometrically as possible.
Thank you!
Best Answer
This is correct. The normal bundle to a divisor in $X$ measures the self-intersection of the divisor as it moves in its complete linear system on $X$. So for divisors which are hypersurface sections, you can look at the system of hypersurfaces in the ambient projective space.
It's also useful to note that in this situation, you can restrict-then-intersect or vice versa. So not only can you realize the deformation of $C$ by cutting with a different hyperplane to get a different conic that meets $C$ in two points (because $\mathcal O_C(1) \cong \mathcal O_{\mathbb {P^1}}(2)$), you can also instead look at two hyperplanes, which intersect in a line, and then intersect that line with $Q$ to get two points.