The normal bundle of the twisted cubic 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.
 A: 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 $xw-yz=0$.
A: 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

