Algebraic Curves Lecture 24

The 24th (probably?) lecture of algebraic curves by Karl Christ

Last case of extremal curves: \(d = 2r\)

Last case of extremal curves: \(d = 2r\)

\[\begin{aligned} \pi(d,r) = \frac{m(m-1)}{2}(r-1) + m\cdot \epsilon \end{aligned}\]

Then \(2r - 1 = d - 1 = (r-1)m + \epsilon\). That \(d = 2r\) means \(m = 2\) and \(\epsilon = 1\). We then calculate \(g = (r - 1) + 2 = r + 1\), \(r = g - 1\), \(d = 2r = 2g - 2\). It follows that \(C\) is canonically embedded. Conversely, canonical curves are extremal.

Theorem: (Max Noether's theorem). The map

\[\begin{aligned} \operatorname{Sym}^\ell H^0(C, K_C) = H^0(\mathbb P^{g-1}, \mathcal O_{\mathbb P^{g-1}}(\ell)) \to H^0(C, \mathcal O_C(\ell)) = H^0(C, K^{\otimes \ell}_C) \end{aligned}\]

is surjective for all \(\ell\) and \(C\) is canonically embedded.

Let \(C\subseteq \mathbb P^{g-1}\) be canonically embedded.

\[\begin{aligned} H^0(\mathbb P^{g-1}, \mathcal O_{\mathbb P^{g-1}}(2)) &\twoheadrightarrow H^0(C, \mathcal O_C(2)) \\ \operatorname{dim}: \hspace{1em} \binom{g+1}{2}, &\phantom{\twoheadrightarrow} 4g - h - g + 1 = 3g - 3. \end{aligned}\]

Then

\[\begin{aligned} \operatorname{dim}\operatorname{ker}\varphi = \frac{(g+1)g}{2}-3g + 3 = \frac{g^2 - 5g + 6}{2} = \frac{(g-2)(g-3)}{2}. \end{aligned}\]

Example: In \(g = 4\), \(C\to \mathbb P^3\), \(Q_2 \cap Q_3\). Every nonhyperelliptic curve in genus 4 is trigonal, and an extremal curve with \(d = 2r\) must not be hyperelliptic since it admits a canonical embedding. If \(C\) is trigonal, \(\varphi_{K_C}(C)\) cannot be cut out by quadrics, since by geometric Riemann-Roch, for any \(D\in |g^1_3|\) the three points in \(\varphi_{K_C}(D)\) need to be colinear.

\[\begin{aligned} r(D) = d - 1 - \operatorname{dim}(\overline{\varphi_{K_C}(D)}). \end{aligned}\]

Example: If \(g = 3\) then \(C \hookrightarrow \mathbb P^2\), if \(g = 2\) then \(C\xrightarrow{2:1}\mathbb P^2\).

If \(Q\supseteq C\), then \(Q\) contains \(D\). Let \(L\) be the copy of \(\mathbb P^1\) passing through \(D\). Then the defining equation of \(Q\) restricted to \(L\) gives a degree \(2\) polynomial vanishing at three points. This implies it needs to vanish along all of \(L\) which happens if and only if \(L\subseteq Q\).

Theorem: (Enriques-Babbage). If \(C\subseteq \mathbb P^{g-1}\) is canonically embedded, then either \(C\) is cut out by quadrics, or \(C\) is trigonal or \(C\) is isomorphic to a plane quintic and then \(g = \frac{(5 - 1)(5 - 2)}{2} = 6\).

Remember: Last time: used that through any \(\geq 2r + 3\) points in linear general position, that impose \(2r + 1\) cconditions on quadrics, there is a unique rational normal curve (in \(\mathbb P^r\)).

Remark: The bound \(2r + 3\) is strict. For example: \(Q_1, Q_2, Q_3 \subseteq \mathbb P^3\), \(Q_1\cap Q_2 \cap Q_3 = 8 = 2r_2\) points (and \(D\) is the divisor given by those \(8\) points).

\[\begin{aligned} H^0(\mathbb P^3, \mathcal_{\mathbb P^3}(2)) \to H^0(D, \mathcal O_{\mathbb P^2}(D)), \end{aligned}\]

the \(8\) points in \(D\) impose \(7\) conditions on quadrics. There is no twisted cubic which intersects all 8 of these points.

Lemma: Suppose the intersection \(S\) of all quadrics containing a canonically embedded curve \(C\subseteq \mathbb P^{g-1}\) of genus \(g \geq 4\) contains a point \(p\not\in C\). Then \(C\) lies either on the image of \(\mathbb P^2\to \mathbb P^5\) (\(g=6\)) or on a rational normal scroll, which is smooth if \(g\geq 5\).

Proof. (Sketch of proof.) First, we show that we may assume \(p\) lies on finitely many secants of \(C\).

Second, notice this implies that projection \(\mathbb P^{g - 1} \to \mathbb P^{g-2}\) maps \(C\) birationally to a non0-degenerate curve \(C\) in \(\mathbb P^{g - 2}\).

Third, let \(D = (C\cap H) \cup \{p\}\subseteq H\cong \mathbb P^{r-1}\) (noting that \((C\cap H) \cup \{p\}\) consists of \(2r + 1\) points). Then check that the points in \(D\) are in linear general position.

Fourth, calculate the number of conditions imposed on quadrics. The result of this calculation will show there exists a rational normal curve \(C_D\) through \(D\) in \(H\). \(C_D\) needs to be contained in \(S\), and this implies \(S\) is either a rational normal scroll or \(\mathbb P^2 \to \mathbb P^5\).

\(\square\)