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Let $C\subset \mathbb P^3$ be the twisted cubic given by the ideal $I=(xz-y^2,yw-z^2,xw-yz)$. I want to compute the normal bundle $N_{C/\mathbb P^3}$, i.e. the dual of $\mathcal I/\mathcal I^2=(I/I^2)^\sim$. My goal is to find $h^0(N_{C/\mathbb P^3})$, so I would like to write $N_{C/\mathbb P^3}$ in such a way that its $h^0$ can be easily computed.

I tried to compute $I^2$, hoping to be able to write down $I/I^2$. But I did not succeed (too many relations), and I do not even know if this is the right track. What do you think?

Afterwards, I wrote $$\mathcal I/\mathcal I^2=\mathcal I\otimes_{\mathcal O_{\mathbb P^3}}\mathcal O_{\mathbb P^3}/\mathcal I=\mathcal I\otimes_{\mathcal O_{\mathbb P^3}}\mathcal O_C=\mathcal I|_C.$$ But even there I got stuck, as I cannot compute that restriction. It would be much easier if $C$ were a complete intersection.

Any help would be greatly appreciated. Thank you.

Brenin
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  • See http://math.stackexchange.com/questions/1583326/normal-bundle-of-twisted-cubic/1603525#1603525 – evgeny Jan 07 '16 at 18:47

2 Answers2

4

We have for the normal bundle the splitting $N_{C/\mathbb P^3}=\mathcal O_C(5)\oplus \mathcal O_C(5)$.
This is proved in Proposition 6 of the article On the Normal Bundles of Smooth Rational Space Curves by Eisenbud-Van de Ven.
So your goal is fulfilled: $h^0(C, N_{C/\mathbb P^3})=6+6=12$

The authors (at the end of their article) sketch a second, very computational proof, of their Proposition 6 based on the fact the cubic curve lies on the smooth quadric $xw-yz=0$.

Georges Elencwajg
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3

Let $R$ be the the coordinate ring of $\mathbb P^3$ with variables $x,y,z,w$, and $S=R/I$. Let $Y$ be the twisted cubic inside $Proj R$. We than have an exact sequence $$ 0 \to R(3)^2 \xrightarrow{\alpha} R(-2)^3 \to I \to 0,$$ where $\alpha$ is the matrix $\begin{bmatrix}-z & w \\ y & -z \\ -x & y \end{bmatrix}$. Tensoring the above sequence with $ \otimes_R R/I$ gives a presentation of $I/I^2$ :

$$ A(-3)^2 \to A(-2)^3 \to I/I^2 \to 0 .$$

Homming is left-exact, so applying $Hom_B(-,B)$ gives $$0 \to Hom(I/I^2,B) \to B(2)^3 \to B(3) ^2.$$

Sheafifying gives the exact sequence of $\mathcal O _Y$-modules: $$ 0 \to \mathcal N_{Y/X} \to \mathcal O_Y(2)^3 \xrightarrow{\alpha^T} \mathcal O_Y(3)^2.$$

Thus a basis for $H^0(\mathcal N_{Y/X})$ is given by the degree zero piece of $\ker \alpha^T $.

I did the computation in Macaulay2, but it shouldn't be unbearable by hand. The difficulty is keeping track of gradings. 

>ker C.dd_2
>o114 = image {-2} | z y  0 x  0 0 |
              {-2} | w 0  z 0  y x |
              {-2} | 0 -w w -z z y |
                                3
>o114 : S-module, submodule of S
>matrix basis(0,oo)
>o115 = {-1} | w 0 0 0 0 0 0 0 0 0 0 0 |
        {-1} | 0 w 0 0 0 0 0 0 0 0 0 0 |
        {-1} | 0 0 w 0 0 0 0 0 0 0 0 0 |
        {-1} | 0 0 0 x y z w 0 0 0 0 0 |
        {-1} | 0 0 0 0 0 0 0 w 0 0 0 0 |
        {-1} | 0 0 0 0 0 0 0 0 x y z w |
               6       12
o115 : Matrix S  <--- S
Fredrik Meyer
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