Algebraic Curves Lecture 25

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:

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 \(Q_1, Q_2, Q_3\).

1. 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 \(X_{1,2}\).
2. Otherwise, \(C\) is a complete intersection of the 3 quadrics: \(C = Q_1\cap Q_2 \cap Q_3\).

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

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

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 \(g^1_3\).)

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

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

The domain has dimension \(10\). The dimension of the domain is bounded by \(10\) by Clifford:

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

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

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 \(11\).

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

1. 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 \(h^0(\mathcal O_C(2)) = 9\).
2. 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.