My question concern the definition, geometric meaning, and usage of the normal bundle in algebraic geometry.
Let $X$ be a nonsingular variety over an algebraically closed field $k$, and $Y\subseteq X$ a nonsingular closed subvariety. Let $\mathcal{I}$ be the sheaf of ideals defined by the closed embedding $i: Y \hookrightarrow X$, and consider the sheaf $\mathcal{C}_{Y/X}=: \mathcal{I}/\mathcal{I}^{2}$. We define the normal sheaf of $Y$ in $X$ to be $\mathcal{N}_{Y/X}=:\mathsf{Hom}(\mathcal{C}_{Y/X},\mathcal{O}_{Y})$.
This definition can be found on Hartshorne's "Algebraic Geometry". Recently I came across some concrete examples of normal bundles that I cannot understand.
$\mathbf{(1)}$ Let $C$ be a nonsingular curve of degree $2$ and genus $0$ in $\mathbb{P}^{3}_{k}$. with $k$ an algebraically closed field. It can be proven that for any such curve, there exists a quadric $Q$ in $\mathbb{P}^{3}_{k}$ such that $C$ lies on $Q$. Then $\mathcal{N}_{C/Q}=\mathcal{O}_{C}(1)$, and $\mathcal{N}_{S}|_{C}=\mathcal{O}_{C}(2)$.
$\mathbf{(2)}$ Let $X$ a nonsingular irreducible cubic surface in $\mathbb{P}^{3}_{k}$, with $k$ an algebraically closed field. Let $H$ the hyperplane section of $X$. It can be proven that there exists a nonsingualr irreducible curve $C\in |4H+2L|$ of degree $14$ and genus $24$, for a line $L$ on $X$. Then $\mathcal{N}_{C/X}=\mathcal{O}_{C}(C)$, and $\mathcal{N}_{X}=\mathcal{O}_{X}(3)$.
The geometrical meaning of the twisting sheaf, and the Picard group construction are very clear to me. I don't understand how can normal sheaves be computed in the examples presented above. However, rather than focusing on those two cases, I would like to understand the nature of the normal sheaf – i.e. what does it say about the varieties? Is there a more agile definition? How do normal sheaves relate to twisting sheaves? How can they be computed? I am mosty interested in the case of regular projecive schemes of low dimension, over an algebraically closed field. Simple examples are also welcome.
Best Answer
First of all, a warning (to myself first, because I use to get confused). The sheaf $I/I^2$ is a sheaf on $X$. This is not the conormal sheaf. Of course, it is supported on $Y$, but the conormal sheaf is the sheaf $i^\ast(I/I^2)$, and this is really on $Y$. That said, I will write $I/I^2$ instead of $i^\ast(I/I^2)$.
Example. Let us compute the normal bundle of a plane conic $C\subset\mathbb P^2$. The ideal of the conic is $I=\mathscr O_{\mathbb P^2}(-2)$, so $$I/I^2=I|_C=\mathscr O_{\mathbb P^2}(-2)|_C=\mathscr O_{C}(-2)\,\Rightarrow\,\mathcal N_{C/\mathbb P^2}=\mathscr O_C(2).$$ Note that this line bundle has actually a $5$-dimensional space of sections, as $$\mathscr O_C(2)\cong \mathscr O_{\mathbb P^1}(4).$$
This "$5$" is the dimension of the Hilbert scheme of plane conics, which is the smooth space $\mathbb P^5$. Actually, the vector space $H^0(C,\mathcal N_{C/\mathbb P^2})$ computes the tangent space of this $\mathbb P^5$ at $[C]$, and the latter is the space of deformations of the conic in the plane.
This is why I think it is useful to have in mind the following association: $$\textrm{(Sections of the) Normal bundle }\mathcal N_{Y/X}\,\,\longleftrightarrow\,\,\textrm{Deformations of } Y\textrm{ inside }X.$$ I will try to explain why, before coming to your examples.
In the case that interests you, namely that with $Y$ and $X$ both smooth, there is an exact sequence (called the conormal exact sequence): $$0\to I/I^2\to \Omega_X|_Y\to \Omega_Y\to 0.$$ Taking the dual, we find $$0 \to T_Y\to T_X|_Y\to \mathcal N_{Y/X}\to 0.$$ The normal bundle appears as cokernel of the map which identifies a tangent vector on $Y$ with a tangent vector on $X$, restricted to $Y$: those in the cokernel are tangent vectors restricted from $X$, up to those coming from $Y$. I cannot draw pictures here, but the idea is that a section of the normal bundle should be the datum of a family of vectors, orthogonal (normal!) to the tangent spaces, and these normal vectors draw for you a "nearby $Y$" inside $X$, i.e. a deformation of $Y$ inside $X$. The zeros of a section should then be the points $y\in Y$ where the vector was the zero vector: that particular $y$ did not contribute to the deformation. Formally, if $H$ is the Hilbert scheme of $X$, you have $$H^0(Y,\mathcal N_{Y/X})=T_{[Y]}H=\{\textrm{Deformations of }Y\textrm{ in }X\}.$$
Now for your examples.