Normal bundle of twisted cubic.

Question

Let $C$ be a twisted cubic in $\mathbb P^3$. I'd like to compute the splitting type of normal bundle $N_{C/\mathbb P^3}$? I understood that $T_{\mathbb P^3}|_C=\mathcal O(4)^{\oplus 3}.$ So we have an short exact sequence $$ 0 \to \mathcal O(2) \to \mathcal O(4)^{\oplus 3} \to N_{C/\mathbb P^3} \to 0. $$ So $N_{C/\mathbb P^3} =\mathcal O(4) \oplus \mathcal O(6)$ or $N_{C/\mathbb P^3} =\mathcal O(5)^{\oplus 2}.$ But I don't know how to prove that this normal bundle is actually $\mathcal O(5)^{\oplus 2}$.

Answer 1

My favorite way to do this is to work with the conormal bundle instead, seeing very concretely what its transition functions are. Here is a bit to get you started. We have $$0\to N^*_{C/\Bbb P^3}\to T^*\Bbb P^3\big|_C \overset{\phi}\to T^*C \to 0.$$ In the "main" chart $[1,x,y,z]$, where $C$ is parametrized by $(t,t^2,t^3)$, we have \begin{align*}\phi(dx)&=dt \\ \phi(dy)&=2t\,dt \\ \phi(dz)&=3t^2\,dt, \end{align*} and in the chart "at infinity," $[X,Y,Z,1]$, where $C$ is parametrized by $(s^3,s^2,s)$, we have \begin{align*} \phi(dX)&=3s^2\,ds \\ \phi(dY)&=2s\,ds \\ \phi(dZ)&=ds. \end{align*} Thus, on the respective charts we have the relations $$dy-2x\,dx = dz-3x^2\,dx = 0 \qquad dX -3Z^2\,dZ = dY - 2Z\,dZ = 0.$$ Now, on the overlap of the two charts, the frame for the conormal bundle in the first chart is given by the two sections $$\sigma_1=-2t\,dx + dy \quad\text{and}\quad \sigma_2=-3t^2dx + dz,$$ and performing the change of coordinates (here is where you have to do some work), the frame in the second chart is given by the two sections $$\tau_1 = -3t^{-5}dy + 2t^{-6}dz \quad\text{and}\quad \tau_2=t^{-3}dx-2t^{-4}dy+t^{-5}dz.$$ Thus, we have \begin{align*} \tau_1 &= -3t^{-5}\sigma_1+2t^{-6}\sigma_2 \\ \tau_2 &=-2t^{-4}\sigma_1+t^{-5}\sigma_2.\end{align*} Now, the usual row- and column-operation (Smith normal form) game shows that, after change of basis, we obtain frames satisfying \begin{align*} \tau_1' &= -3t^{-5}\sigma_1' \\ \tau_2' &= -\frac13t^{-5}\sigma_2',\end{align*} from which we see that the conormal bundle is $\scr O(-5)\oplus \scr O(-5)$. :)

Answer 2

You can use an ad hoc representation theory trick: twisting by $\mathcal O_{\mathbb P^1}(-6)$ you get

$$0 \to \mathcal O(-4) \to \mathcal O(-2)^{\oplus 3} \to N_{C/\mathbb P^3}(-6) \to 0,$$

$$0 \to H^0(N_{C/\mathbb P^3}(-6)) \to H^1(\mathcal O(-4)).$$

If Veronese embedding is $\mathbb P(W) \to \mathbb P(S^3W)$ for 2-dimensional $W$, then $H^1(\mathcal O(-4)) = S^2W$ is irreducible $GL(W)=GL_2$-representation, so $H^0(N_{C/\mathbb P^3}(-6))$ can be trivial, as for $\mathcal O(-1)^{\oplus 2}$, but not 1-dimensional, as for $\mathcal O(-2) \oplus \mathcal O$.