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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!

rmdmc89
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1 Answers1

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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.

Tabes Bridges
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    Maybe this is beyond the scope of this answer (feel free to say so!) but is there a particular story behind why the normal bundle is the thing that measures this here and why degree is the correct way to count this? If I squint, I can see an analogy with how the tangent space is heavily involved in deformation theory, but I haven't learned as much intersection theory as I probably should have. – KReiser Sep 15 '20 at 07:36
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    @KReiser I can certainly try...if we start with the idea of a tangent vector field on $X$, then it's fairly standard that we think of the field as the derivative/1st order approximation of an automorphism of $X$. But of course your typical automorphism will not fix a given subvariety. Now the normal bundle sequence of a divisor essentially says that a normal vector field is a tangent vector field restricted to $D$ with the "part that fixes $D$" (the tangent bundle to $D$) killed off. So a section of the normal bundle... – Tabes Bridges Sep 15 '20 at 17:34
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    ...is basically a 1st order deformation of the embedding $D \to X$. The zeroes of this section are precisely those points that do not move under the deformation, i.e. the points of self-intersection, and the degree of course counts zeroes of a generic section. In the case where the degree of the normal bundle is negative, I have always thought of this in terms of zeroes minus poles: if $\deg(N_{D/X}) = -n$ (and $D$ is a curve), I need to delete $n$ points from $D$ before I'm able to deform what remains. – Tabes Bridges Sep 15 '20 at 17:34
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    Ah, excellent! That's a very satisfying description, thank you. – KReiser Sep 15 '20 at 23:32
  • @TabesBridges, does the number of zeroes have something to do with the base points of the linear system $|C|$ in $X$? I ask this because I know $C^2=0\Rightarrow $ then $|C|$ is base-point free. – rmdmc89 Sep 18 '20 at 00:57
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    @rmdmc89 Hmm... Certainly in some cases, but not always. For instance your fact should be an if and only if statement in the case that $|C|$ is a pencil. But on the other hand a degree $d$ curve in $\mathbb P^2$ has $C^2 = d^2$, but every (nontrivial) complete linear system on $\mathbb P^2$ is base-point free. – Tabes Bridges Sep 18 '20 at 01:27
  • @TabesBridges, just one last question. From the intuition I got so far, it seems reasonable to say that if $P$ is a base point in $|C|$, then $P$ doesn't move under deformation, so it will contribute with at least $1$ to the value of $C^2$. In other words, $C^2\geq #(\text{base points of }|C|)$. – rmdmc89 Sep 18 '20 at 19:26
  • @TabesBridges, this works for the example I gave and for the one you gave, at least – rmdmc89 Sep 18 '20 at 19:37
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    @rmdmc89 I think that works. I think the precise statement you want is something like $C^2$ is equal to the number of base points of a general sub-pencil of $|C|$ (as long as $|C|$ has no $1$-dimensional fixed part [I've been assuming we're still on a surface this whole time]). – Tabes Bridges Sep 18 '20 at 23:22