Algebraic Curves Lecture 23

The 23rd (probably?) lecture of algebraic curves by Karl Christ

More extremal curves

More extremal curves

We return to the case that $d > 2r$ . Assume that $C\subseteq \mathbb P^r$ is extremal, i.e.

non-degenerate
$g(C) = \pi(d,r)$ , where $\pi(d,r)$ is the bound from Castelnuova's bound.

Set $H$ to be a general hyperplane and let $D = C\cap H$ as always. Then

\begin{aligned} \alpha_1 = h^0(C, \mathcal O_C(D)) = r + 1 \end{aligned}

and on the level of quadrics

\begin{aligned} \alpha_2 = h^0(C, \mathcal O_C(2D)) = (r+1) + (2r - 1) = 3r. \end{aligned}

So for an extremal curve, I exactly know the dimension of the restriction of quadrics to $C$ : we've got the map

\begin{aligned} \varphi:H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(2)) \twoheadrightarrow H^0(C, \mathcal O_C(2)) \end{aligned}

where the domain has dimension $\binom{r+2}{2}$ and the codomain has dimension $3r$ , so

\begin{aligned} \operatorname{dim}(\operatorname{ker}\varphi) = \frac{(r+2)(r+1)}{2}-3r = \frac{r^2 - 3r + 2}{2} = \binom{r-1}{2}. \end{aligned}

This is the number of quadrics containing a rational normal scroll.

New questions: What is $\rho = \bigcap Q_i$ , where the $Q_i$ are the $\binom{r-1}{2}$ linearly independent quadrics containing $C$ ?

Pick a general hyperplane $H$ and consider $D = C\cap H$ . Then $|D| = d \geq 2r + 1$ . The points in $D$ are in linearly general position and impose $2r - 1$ conditions on quadrics. By Castelnuova's $2r + 3$ lemma, these is a rational normal curve $C_D \subseteq H$ containing $D$ .

Claim: $C_D\subseteq S$ . Suppose there is $Q_i$ such that $C_D \not\subseteq Q_i$ . Then

\begin{aligned} C_D . Q_i = 2\cdot (r - 1) \geq d = \deg (D) \end{aligned}

because $D\subseteq C_D\cap Q_i$ . The count above is simply Bezout's theorem. This is a contradiction however, since $d > 2r$ , so we can't have $2\cdot (r - 1) \geq d$ . Thus $C_D\not\subseteq Q_i$ implies that $C_d\cap Q_i$ has dimension $0$ .

Conversely: $C_D$ is cut out by $\binom{r - 1}{2}$ quadrics in $H\cong \mathbb P^{r-1}$ which implies $S\cap H = C_D$ . A rational normal curve in $\mathbb P^r$ is cut out by the $2\times 2$ minors of a matrix of the form

\begin{aligned} \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{r-1} \\ x_1 & x_2 & x_3 & \dots & x_r \end{pmatrix}, \end{aligned}

of which there are $\binom{r}{2}$ . Because $C_D$ is a rational normal curve in $H\cong \mathbb P^{r-1}$ , it is therefore cut out by $\binom{r-1}{2}$ quadrics. We know already that $C_D$ vanishes on the $Q_i$ , of which there are $\binom{r-1}{2}$ , so there are no degrees of freedom left and hence $S\cap H \supset C_D$ . This gives us both inclusions and hence $S\cap H = C_D$ .

Summary: We get that $\bigcap Q_i = S$ gives a rational normal curve of degree $r-1$ when we intersect with a general hyperplane:

\begin{aligned} C_D := S\cap H. \end{aligned}

This implies $S$ is a surface of degree $r-1$ , and hence that $C$ lies on a rational normal scroll or on $\mathbb P^2 \to \mathbb P^5$ (the Veronese embedding $[x:y:z] \mapsto [x^2:y^2:z^2:xy:xz:yz:]$ ).

Suppose $S = X_{a_1, a_2}$ with $a_1\neq 0$ is a smooth, rational normal scroll. We saw (or claimed without proof? Can't remember) last time that $\operatorname{Pic}(S) = \langle L, H\rangle$ with $L.L = 0$ , $L.H = 1$ and $H.H = r-1$ .

Recall if you have a curve $C$ with bidegree $(d_1,d_2)$ then $2g - 2 = K_S. C + C.C$ and $C = \alpha L + \beta H$ . Then

\begin{aligned} K_S = K_1\cdot L + K_2\cdot H. \end{aligned}

From $L$ you have

\begin{aligned} -2 = K_S\cdot L + L^2 = K_2 \end{aligned}

and for $H$

\begin{aligned} -2 = K_S H + H^2 = K_1 + K_2(r-1) + (r-1) = K_1 - (r-1). \end{aligned}

This implies $K_1 = r-3$ . Hence $K_S = (r-3)L - 2H$ , $2g(C) - 2 = K_S.C + C.C$ and so

\begin{aligned} g(C) = \frac{(\alpha - 1)(2m - \alpha)}{2}(r-1) + \epsilon\cdot (\alpha - 1), \end{aligned}

hence $d-1 = m(r-1) + \epsilon$ . So $g$ is maximal for $\alpha = m+1$ , $\beta = \epsilon + 2 - r$ for any $\epsilon$ or $\alpha = m$ and $\beta = 1$ and $\epsilon = 0$ . For these values, equality in Castelneuovo's bound is achieved.

Theorem: Let $r\geq 3$ and $d \geq 2r + 1$ . Then extremal curves of degree $d$ exist and are of the following forms:

The image of a smooth degree $k$ curve in $\mathbb P^2$ under the Veronese embedding $\mathbb P^2 \to \mathbb P^5$ (so $d = 2k$ and $r = 5$ )
A non-singular member of $|mH + L|$ on a rational normal scroll
A non-singular member of the linear system $|(m+1)H - (r - \epsilon - 2)L|$ on a rational normal scroll.

Cor: In case (3),

C

admits a

g_{m+1}^1

cut out by

L