Algebraic Curves Lecture 25

The 25th lecture of algebraic curves by Karl Christ

Last bit about extremal curves
1. Last time
2. Final example of extremal curves
Curves in $\mathbb P^3$ and examples

Last bit about extremal curves

Last time

We saw that $\varphi_{K_X}(X)$ , the image curves, is cut out by quadrics when $X$ is trigonal or isomorphic to a plane quintic (Enriques-Babbage). Have one more remark about this:

Remark: What the argument showed is that its set theoretically cut out by quadrics. You can actually show that it is scheme-theoretically cut out by quadrics. That is, $\varphi_{K_X}(X)$ is cut out by quadrics as a scheme; the defining ideal of $\varphi_{K_X}(X)$ is generated by the intersection of the quadrics.

In general, $\varphi_{K_X}(X) \subset \mathbb P^r$ is cut out by quadrics and cubics (Petri's theorem).

Final example of extremal curves

Our final example of extremal curves will be $g = 5$ curves. In this case, $C\subseteq \mathbb P^4$ is canonically embedded of degree $8$ and there are $3$ linearly independent quadrics vanishing on $C$ . Call them $Q_1, Q_2, Q_3$ .

If $Q_1 \cap Q_2\cap Q_3 \neq C$ , then $C$ lies on $X_{1,2}$ and $C$ is trigonal with $g^1_3$ cut out by the ruling on $X_{1,2}$ .
Otherwise, $C$ is a complete intersection of the 3 quadrics: $C = Q_1\cap Q_2 \cap Q_3$ .

Curves in $\mathbb P^3$ and examples

The following image from Hartshorne IV Section 6 summarizes what we know about the existence of curves of degree $d$ and genus $g$ in $\mathbb P^3$ :

All possible values of $(d, g)$ below the line $g = d - 3$ exist, and they will be embedded in $\mathbb P^3$ by a non-special line bundle. We know everything in degree $d \leq 6$ using what we've discussed thus far in class.

The first case in this image that isn't covered by the methods thus far discussed in class is the point $(7,5)$ .

Example: is there a degree $7$ , genus $5$ curve?

First try: suppose it lives on a quadric. Then it has bidegree $(d_1,d_2)$ and $d_1 + d_2 = 7$ and $(d_1 - 1)(d_2 - 1) = 5$ . This implies $d_1 = 2$ and $d_2 = 6$ .

Second try: take $X$ an abstract curve of genus $5$ . We want $L$ on $X$ such that $\deg(L) = 7$ and $h^0(X, L)\geq 4$ (we're forgetting about the geometry of an embedding into $\mathbb P^3$ for a moment). Then by Riemann-Roch

\begin{aligned} 4 = 7 - 5 + 1 + h^0(K_X - L) \implies h^0(K_X - L) = 1, \end{aligned}

meaning $L - K_X(-p)$ and $h^0(X, L) = 4$ . We also want this line bundle to be very ample so that it embedds $X$ as a smooth curve into $\mathbb P^3$ .

If $L$ is very ample, then $h^0(K_X(-p-q-r)) = 2$ for all $q,r\in X$ . This comes from our condition of very ampleness; subtracting any two points must lower the dimension of the space of global sections by $2$ . Thus

\begin{aligned} 5 - 5 + 1 +h^0(\mathcal O_X(p+q+r)) = 2 \iff h^0(\mathcal O_X(p + q + r)) = 1. \end{aligned}

This is the case if and only if $X$ is not trigonal and is not hyperelliptic. (A curve is trigonal if it admits a basepoint free $g^1_3$ .)

Example: is there a $d = 9$ and $g = 11$ curve?

Now suppose $d = 9$ and $g = 11$ . Such a curve would need to lie on a quadric:

\begin{aligned} H^0(\mathbb P^3, \mathcal O_{\mathbb P^3}(2)) &\to H^0(C, \mathcal O_C(2)). \end{aligned}

The domain has dimension $10$ . The dimension of the domain is bounded by $10$ by Clifford:

\begin{aligned} h^0(C, \mathcal O_C(2)) \leq \frac{d}{2} + 1 = 10 \end{aligned}

and if we have equality then $\mathcal O_C(2) = 9\cdot g^1_2$ , which is not very ample, so $h^0(C, \mathcal O_C(2)) \leq 9$ . We thus have a linear map from something $10$ dimensional to something $9$ dimensional, meaning we have nontrivial kernel and hence $C$ lies on a quadric in $\mathbb P^3$ . This in turn implies that $C$ has bidegree $(d_1,d_2)$ , so

\begin{aligned} d_1 + d_2 = 9 \hspace{1em}\text{and}\hspace{1em} (d-1)(d-2) = 11 \end{aligned}

and since $11$ is prime, $d_1 = 2$ and $d_2 = 12$ . These don't sum to $9$ , however. Thus there exists no curve of degree $9$ and genus $11$ .

Theorem: Let $C\subseteq \mathbb P^3$ be a smooth curve of degree $d$ and genus $g$ .

If $C\subseteq \mathbb P^2$ then $g = \frac{(d-1)(d-2)}{2}$ and for any such values $(d,g)$ there exists a curve
If $C\subseteq Q_2$ , some quadric in $\mathbb P^3$ , then there exists integers $d_1,d_2\in \mathbb N$ such that $d = d_1 + d_2,$ $g = (d_1 - 1)(d_2 - 1)$ .
If $C$ is not contained in a quadric, then $g \leq \frac{1}{6}(d(d-3)) + 1$ , and in this case a curve exists.

Example: a $d = 9$ and $g = 10$ curve

There are a few ways to construct curves of this type.

Take a curve of bidegree $(d_1, d_2) = (3, 6)$ on a quadric surface in $\mathbb P^3$ . Then $d_1 + d_2 = 9$ and $(d_1 - 1)(d_2 - 1) = 10$ , as desired.
- $h^0(I_C(2)) > 0$
- $h^0(\mathcal O_C(2))$ can be calculated via the short exact sequence
  $0 \to \mathcal O_Q(-3-6) \xrightarrow{\cdot C} \mathcal O_Q \to \mathcal O_C\to 0$ and twisting by $2$ . You'll get that $h^0(\mathcal O_C(2)) = 9$ .
Take $C = Q_1\cap Q_2$ to be a complete intersection of two cubics. Then $g = \frac{1}{2}(d_1 + d_2 - 4) + 1$ , taking $d_1 = d_2 = 3$ gives $10$ as needed.