6
$\begingroup$

How to understand (maybe, informally) why the intersection of two quadrics in general position in $\mathbb{CP}^3$ is an elliptic curve?

It is obvious that it is a compact 2-manifold, i.e. a sphere with handles, but how to find its genus? I suppose one can apply Mayer-Vietoris sequence, but the union of two quadrics seems to be hard to realise.


Update: I was told that it is possible to use Riemann-Hurwitz formula. Let us choose two lines $a$ and $l$ and project all points from $a$ to $l$. Than it will be branched covering, with 4 sheets by Bezout's theorem. It shoould be $$0=\chi(\text{intersection})=\chi(l=\mathbb{CP}^1)*\text{sheets}-\text{ramification}=2*4-8,$$ but is there an easy way to see these ramification points?

$\endgroup$
3
$\begingroup$

Your comments about Mayer-Vietoris seem off kilter a bit (you would need to write the intersection $X=Q\cap Q'$ as a union of two good sets). Although there may be an ad hoc approach, the standard approach is to use the adjunction formula. We compute the canonical bundle in two steps: $$K_Q \cong K_{\Bbb P^3}\otimes\mathscr O_{\Bbb P^3}(2)\big|_Q \cong\mathscr O_Q(-2), \text{ and } K_X \cong K_Q\otimes \mathscr O_Q(2)\big|_X \cong \mathscr O_X.$$ The genus formula for smooth curves tells us that trivial canonical bundle is equivalent to $g=1$.

EDIT: Cool argument you've added with Riemann-Hurwitz. One elementary way to get the $8$ is this: Say the line $\ell$ is given by homogeneous linear equations $L_1=L_2=0$ and the two quadrics are $Q_1=0$ and $Q_2=0$, respectively. You get another quadric surface by taking the equation $$D=\left|\begin{matrix} - & dL_1 & - \\- & dL_2 & - \\- & dQ_1 & - \\- & dQ_2 & - \end{matrix}\right| = 0.$$ The exercise is to show that points of $X\cap D$ correspond to points of the curve $X=Q_1\cap Q_2$ where the projection $\pi_a$ from $X$ to $\ell$ ramifies. By Bezout, since $X$ has degree $4$ and $D$ has degree $2$, we get $8$ such points.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.