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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.

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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.

  1. The fact that $\mathcal N_{S/\mathbb P^3}|_C=\mathscr O_C(2)$ is exactly as for the conic: $\mathcal N_{S/\mathbb P^3}=\mathscr O_S(2)$. The fact that $\mathcal N_{C/Q}=\mathscr O_C(1)$ is because any such curve is obtained as a hperplane section of $Q$, so the line bundle $\mathcal N_{C/Q}$ has to have degree $1$ on $C$.
  2. The fact that $\mathcal N_{X/\mathbb P^3}=\mathscr O_X(3)$ is again identical to the case of the conic, at the beginning. As for $\mathcal N_{C/X}=\mathscr O_C(C)$, it is something more general which happens (so you can forget about this special situation): if $Y\subset X$ is a smooth divisor, its ideal is $I=\mathscr O_X(-Y)$, so $$\mathcal N_{Y/X}=(I/I^2)^\vee=(\mathscr O_X(-Y)|_Y)^\vee=(\mathscr O_Y(-Y))^\vee=\mathscr O_Y(Y).$$
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  • $\begingroup$ Very nice answer! $\endgroup$ – Alex Youcis Mar 20 '14 at 7:19
  • $\begingroup$ Very helpful, thank you! $\endgroup$ – Fq00 Mar 20 '14 at 23:05
  • $\begingroup$ @Fra_24: You are welcome! $\endgroup$ – Brenin Mar 23 '14 at 10:15
  • $\begingroup$ A very clear and insightful answer, Brenin: congratulations! $\endgroup$ – Georges Elencwajg Nov 17 '14 at 21:43
  • $\begingroup$ @GeorgesElencwajg: thank you very much for your kind comment! $\endgroup$ – Brenin Nov 19 '14 at 0:54
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Normal bundle is O_C(C) see 2.9 here: http://math.mit.edu/~mckernan/Teaching/07-08/Autumn/18.735/l_2.pdf
In some cases, use the dual to the tangent bundle exact sequence to compute it (i.e. the exact sequence in Hartshorne chapter II.8)

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