# Intersection of twisted cubics in $\mathbb{P}^3$

Suppose we have two twisted cubics $C_1$, $C_2$ in $\mathbb{P}^3$ such that both of them lie in some cubic surface, which means that $h^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))>0$. I want to show that in this case they intersect.

Suppose that they do not intersect. Then $\mathcal{O}_{C_1\cup C_2}=\mathcal{O}_{C_1}\oplus\mathcal{O}_{C_2}$. Twisting by 3 the exact sequence $$0\to I_{C_1\cup C_2}\to \mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_{C_1\cup C_2}\to0$$ and taking cohomologies we obtain $$0\to H^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))\to H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(3))\stackrel{f}\to H^0(\mathbb{P}^3, \mathcal{O}_{C_1\cup C_2}(3))\to H^1(\mathbb{P}^3, I_{C_1\cup C_2}(3))\to0.$$ But $h^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(3))=20$ and $h^0(\mathbb{P}^3, \mathcal{O}_{C_1\cup C_2}(3))=2h^0(\mathbb{P}^3, \mathcal{O}_{C_1}(3))=2h^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(9))=20$, so it seems that the map $f$ is an isomorphism, which contradicts the assumption $h^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))>0$.

Is it true or not that $f$ is an isomorphism? If yes, then how to prove this?

$\textbf{Edit}$

It seems that the approach I gave in the post is not the best possible.

If someone could give a canonical answer based on the different argument it would be welcomed!

$\textbf{Edit 2}$

Since $C_1$ and $C_2$ do not intersect, $I_{C_1\cup C_2}=I_{C_1}\otimes I_{C_2}$. For $I_{C_i}$ there is a resolution of the form

$$0\to\mathcal{O}(-3)^{\oplus2}\to\mathcal{O}(-2)^{\oplus3}\to I_{C_i}\to0.$$

Tensoring these two resolution and twisting by 3 we obtain the exact sequence

$$0\to\mathcal{O}(-3)^{\oplus4}\to\mathcal{O}(-2)^{\oplus6}\to\mathcal{O}(-1)^{\oplus9}\to I_{C_1}\otimes I_{C_2}(3)\to0.$$

The first three terms do not have cohomologies, so it seems that the fourth doesn't have too. How to prove this? Should I use some spectral sequence?

• You have not proved that the map $f$ is an isomorphism, but only that its domain and codomain are equidimensional Commented Jan 25, 2016 at 11:44
• @GeorgesElencwajg: hi Georges, I'm pretty interested in this question too. Is it actually possible to show that $f$ is surjective (and thus conclude the result)? To me it seems a lot to hope for to get every pair of degree 9 polynomials on the two curves from the cubic polynomials on $\mathbb{P}^3$.
– john
Commented Jan 25, 2016 at 12:42
• Your proof seems a little bit complicated to me. If you want to pursue it and show that $f$ is an isomorphism, then you want to compute $h^1({\mathbf P}^3,I_{C_1\cup C_2}(3))$ yet in another way, and find that it is zero. You can probably do so by writing down the minimal resolution of the ideal sheaf in question, but is going to be a long story, I'm afraid. Commented Jan 29, 2016 at 10:58
• @Johannes Huisman Do you know some different approach to solve the initial problem? Commented Jan 30, 2016 at 19:02
• Once you have that exact sequence, I believe you can splice it into two short exact sequences $$0 \longrightarrow \mathcal{O}(-3)^{\oplus 4} \longrightarrow \mathcal{O}(-2)^{\oplus 6} \longrightarrow \mathscr{Q} \longrightarrow 0$$ and $$0 \longrightarrow \mathscr{Q} \longrightarrow \mathcal{O}(-1)^{\oplus 9} \longrightarrow I_{C_1} \otimes I_{C_2}(3) \longrightarrow 0$$ Commented Feb 1, 2016 at 4:37

It is probably true that your map $f$ is an isomorphism if the curves are chosen "sufficiently generally". But that would be a circular argument. It may well fail in general. The best I can come up with at present is the following suggestion.
Suppose that $C_1$ and $C_2$ have $2$ common bisecant lines $L$ and $M$ that meet in a point $P = L \cap M$. (I don't see any particular reason why that should not happen—maybe you can look for an example? The bisecant lines to $C_1$ cover $\mathbb{P}^3$, and similarly $C_2$.)
Then the condition for cubics to contain $C_1$ and $C_2$ would be linearly dependent: argue first on the $4$ points $L \cap C_1$ and $L \cap C_2$ ($4$ conditions on cubics, as should be). Then argue on $4$ points $P$ and $2$ points of $M \cap C_1$ and one of the points of $M \cap C_2$; cubics through these necessarily pass through the second point of $M \cap C_2$. So the two curves $C_1$ and $C_2$ each imposes $10$ conditions on cubics, but these conditions are not linearly independent.
It is true that twisted cubics $C_1$ and $C_2$ cannot be contained in a cubic surface $S$ that is nonsingular along both. I hope you can follow intersection numbers of curves on a surface. Because (after you resolve singularities if necessary), $C_1^2 = +1$ and $C_2^2 = +1$, so that $C_1 \cdot C_2 = 0$ would contradict the algebraic index theorem. But that argument does not work if $S$ has any singularities along the $C_i$.