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

Algebraic Curves Lecture 24

Last case of extremal curves: d=2rd = 2rd=2r

Last case of extremal curves: $d = 2r$