Quartic Curves in Projective Space

Today as part of my qualifying exam studying I want to sketch an example which occurs in chapter 4 of the tome of algebraic geometry written by everyone’s favorite shakuhachi player.

We are interested in studying smooth proper curves over an algebraically closed field k and it turns out that an interesting invariant (especially for curves of higher genus) is whether or not a curve admits a nontrivial degree two morphism to \mathbb{P}^1. We call such curves hyperelliptic if their genus is at least 2.

One the one hand, if a curve X is hyperelliptic we get an interesting map to work with which among other things immediately tells us that the functional field K(X) is a degree two extension of k(x). On the other, if X is not hyperelliptic, then the canonical divisor K of X is very ample and so gives a canonical embedding X \hookrightarrow \mathbb{P}^{g-1}.

Lets see how this second property comes about. Recall that a divisor D on X is very ample precisely when for any two points \{P, Q\}, \mathrm{dim} | D|= \mathrm{dim}|D-P-Q| + 2. In particular, removing a point from a divisor drops the dimension of the local system by at most 1 and very ampleness is equivalent to the dimension drop being the maximum possible for every pair of points. Picking any two  P, Q \in X and applying Riemann-Roch gives

\mathrm{dim}| P + Q | - \mathrm{dim}|K-P-Q| = 3-g

Since \mathrm{dim} |K|=g-1, we see that \mathrm{dim} |P+Q| is zero for all P, Q exactly when K is very ample and that if this is not the case, we have \mathrm{dim} |P+Q|=1 for some pair of points and this gives a degree two morphism to \mathbb{P}^1.

Now, since the canonical divisor of an elliptic curve is degree 0 it’s clearly impossible for it to be very ample so all elliptic curves admit a morphism to \mathbb{P}^1. In fact, we knew this anyway because Weierstrass form gives a very explicit degree 2 extension of k(x).

For genus 2, \mathrm{dim}|K|=1 and so very ampleness would imply that \mathrm{dim}|K-P-Q|=-1 which is absurd. Thus all genus two curves are hyperelliptic.

We’d like to find a non-hyperelliptic curve and so the next place to look is genus three. Here, the situation is not immediately impossible, and in fact we will find examples. The image of a canonical embedding would have to be a quartic curve in \mathbb{P}^2 and so we consider such curves. In fact, we will see that if I: X \hookrightarrow \mathbb{P}^2 is a nonsingular quartic, then I^* \mathcal{O}(1)=\omega_X and so all such curves are non-hyperelliptic.

Since X is nonsingular, we have an exact sequence of sheaves on X:

0 \to \mathcal{J}/ \mathcal{J}^2 \to \Omega_{\mathbb{P}^2/k} \otimes \mathcal{O}_X \to \Omega_{X/ k} \to 0

Then here is a nice observation. If 0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0 is an exact sequence  of locally sheaves of rank n', n, n'' respectively, then \wedge^r \mathcal{F} splits up into a filtration \wedge^r \mathcal{F}=F_0 \supset ... \supset F_{r+1}=0 so that F_p/F_{p+1}= \wedge^{p} \mathcal{F'} \otimes \wedge^{r-p} \mathcal{F}''. Thus, if we take the top exterior power our filtration has only one nonzero term, and this gives \omega_{\mathbb{P}^2/k} \otimes \mathcal{O}_X=\wedge^2( \Omega_{\mathbb{P}^2/ k} \otimes \mathcal{O}_X) \equiv \mathcal{J}/ \mathcal{J}^2 \otimes \omega_{X/k}.

Tensoring both sides with \widehat{\mathcal{J}/ \mathcal{J}^2} gives

\omega_{\mathbb{P}^2/k} \otimes \widehat{ \mathcal{J}/ \mathcal{J}^2 } \simeq \omega_X

But the ideal sheaf \mathcal{J}_X \simeq \mathcal{L}(-X) and so \mathcal{O}_X \otimes \mathcal{L}(-X) \simeq \mathcal{O}_X \otimes \mathcal{J} \simeq \mathcal{J} / \mathcal{J}^2 and so our final formula is

\omega_X \simeq \mathcal{L}(X) \otimes \omega_{\mathbb{P}^1} \otimes \mathcal{O}_X


The right-hand side is \mathcal{O}(4) \otimes \mathcal{O}(-3) \otimes \mathcal{O}_X \simeq I^* \mathcal{O}(1) and so we get what we wanted.




Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s