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.