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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 $\mathcal O(5)^{\oplus 2}$.

qpow
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  • Have you seen this thread? (http://math.stackexchange.com/questions/569237/the-normal-bundle-of-the-twisted-cubic) Open the link in the accepted answer. I do not think this is a trivial result. Essential is that a twisted cubic lies on a smooth quadric surface. – Brenin Dec 20 '15 at 14:46
  • Actually, how to show that $T_{\mathbb{P}^3|C}=\mathcal{O}(4)^{\oplus3}$? – ARM Jan 07 '16 at 22:46

2 Answers2

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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 $\scr O(-5)\oplus \scr O(-5)$. :)

Ted Shifrin
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  • Nice answer! :-) – Brenin Dec 20 '15 at 17:36
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    Explicitly, the last change of coordinates seems to be $\tau_1' = \tau_1, \tau_2' = \tau_2 - 2/3t \cdot \tau_1$ and $\sigma_1' = \sigma_1 - 2/3 t^{-1} \sigma_2, \sigma_2' = \sigma_2$. Nice answer. – Jake Levinson Dec 20 '15 at 19:43
  • Sorry I am a bit confused, I think we view the matrix over PID $k[t^{-1}]$, the smith normal form is determined by the g.c.d. of minors of fixed rank, for example, the first elementary factor is the g.c.d. of all the minors of size $1$. Which is $t^{-4}$, not $t^{-5}$. Am I understanding wrongly? –  Jan 23 '18 at 20:08
  • @JakeLevinson Sorry I am a bit confused, I think if we think the matrix over ring $k[t^{-1}]$, then the invertible matrices used to obtain the smith normal form need to have coefficient in $k[t^{-1}]$, but $2/3t$ does not.. –  Jan 23 '18 at 20:17
  • @Qixiao First, if we were working with a different bundle, say $\mathscr O(2)\oplus\mathscr O(-3)$, both positive and negative powers of $t$ would be appearing. I haven't thought about this in years, but presumably we allow $k[t]$ column operations (change of basis in the main chart) and $k[t^{-1}]$ row operations (change of basis in the chart at infinity). – Ted Shifrin Jan 23 '18 at 21:41
  • @TedShifrin Ah, that's a good point! –  Jan 24 '18 at 16:28
  • By your method, for a rational normal curve in $\mathbb{P}^n$, the normal bundle is of form $\mathcal{O}(2n-1)^{\oplus 2}\oplus\mathcal{O}(n)^{\oplus{(n-3)}}$, cheers! –  Jan 24 '18 at 18:43
  • Interesting, @Qixiao. I've never worked this out. I'm would be it's in print somewhere, perhaps as an exercise (maybe Arbarello/Cornalba/Griffiths/Harris?). – Ted Shifrin Jan 24 '18 at 18:45
  • @TedShifrin I am again confused by a thing...I think we picked the "basis" $\mathrm{d}y-2x\mathrm{d}x$ for the conormal bundle, viewed as a subbundle of $T_{\mathbb{P}^4}|C$, but it seems this section does not lie in $T{\mathbb{P}^4}|_C$, because the tangent bundle is the kernel in the Euler sequence, however $\mathrm{d}y-2x\mathrm{d}x$ does not.. –  Jan 24 '18 at 19:28
  • @Qixiao: We're working on affine charts. The Euler sequence is not visible in this approach, although it's lurking there when you do the transition functions for the $1$-forms on the charts. – Ted Shifrin Jan 24 '18 at 19:36
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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 $\mathcal O(-2) \oplus \mathcal O$.

evgeny
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