Algebraic Curves Lecture 22

The 22nd lecture of algebraic curves by Karl Christ

Extremal Curves
Interlude: Rational normal curves and rational normal scrolls
Rational normal scrolls

I missed the previous lecture due to an unfortunate phone call.

Extremal Curves

$d < 2r$ means you've got a non-special line bundle. Not sure where this showed up last time as I wasn't there.

Interlude: Rational normal curves and rational normal scrolls

Suppose $C\subset \mathbb P^r$ is embedded by $\mathcal O_{\mathbb P^1}(r)$ ,

\begin{aligned} [s:t] \mapsto [s^r:s^{r-1}t:...:t^r]. \end{aligned}

Remember:

such maps are characterized as non-degenerate curves of degree $r$ in $\mathbb P^r$
any $r+1$ points on $C$ span $\mathbb P^r$ .
$\varphi_{K_X}(X)$ is a rational normal curve if $X$ is hyperelliptic.

Set

\begin{aligned} M = \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}

The ideal of $C$ is defined by $2\times 2$ minors of $M$ . There are $\binom{r}{2}$ linearly independent quadrics cutting out $C$ .

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

and

\begin{aligned} \operatorname{dim}(\operatorname{img}\varphi) &= \binom{r+2}{2} - \binom{r}{2} \\ &= \frac{(r+2)(r+1) - r(r-1)}{2} \\ &= \frac{r^2 + 3r + 2 - r^2 + r}{2} \\ &= 2r + 1 = h^0(C,\mathcal O_C(2)) \end{aligned}

which implies that $\varphi$ is surjective which happens if and only if $C$ is 2-normal.

Lemma: If

C\subset \mathbb P^r

is embedded by a non-special line bundle

L

(i.e.

C = \varphi_L(X)

h^0(K_X - L) = 0

then

C

is projectively normal if and only if

C

2

-normal.

Proof. Assume $C$ is $0$ , $1$ , $2$ -normal. We need to show that $C$ is $\ell$ normal for any $\ell > 3$ . Consider the exact sequence:

\begin{aligned} 0 \to \mathcal O_C(1) \xrightarrow{\cdot H} \mathcal O_C(2) \to \mathcal O_{C\cap H}(2)\to 0. \end{aligned}

As is standard, take the long exact sequence on cohomology

\begin{aligned} H^{0}(\mathbb P^r, \mathcal O_{\mathbb P^r}(2)) \to H^0(C, \mathcal O_C(2)) \to H^0(C\cap H, \mathcal O_{C\cap H}(2)) \to H^1(C, \mathcal O_C(1)). \end{aligned}

This means the points in

C\cap H

impose independent conditions on quadrics in

\mathbb P^r

which in turn implies the points also impose independent conditions on all hypersurfaces of degree

\ell \geq 2

Now consider the exact sequence of the ideal sheaf $\mathcal I_{C\cap H}$ given by

\begin{aligned} 0 \to \mathcal I_{C\cap H}(\ell) \to \mathcal O_{\mathbb P^r}(\ell) \to \mathcal O_{C\cap H}(\ell) \to 0. \end{aligned}

Then

\begin{aligned} H^1(C, \mathcal I_{C\cap H}(\ell)) = 0 \end{aligned}

for all

\ell\geq 2

, since

\mathcal O_{\mathbb P^r}(\ell) \to \mathcal O_{C\cap H}(\ell)

is surjective on global sections whenever

\ell \geq 2

. We'd like to show that

H^1(C, \mathcal I_{C\cap H}(1) = 1

too. We have that

\begin{aligned} 0 \to \mathcal I_{C\cap H}(\ell - 1) \to \mathcal I_{C\cap H}(\ell) \to \mathcal I_{C\cap H}^H(\ell) \to 0 \end{aligned}

where

\mathcal I_{C\cap H}

is the ideal sheaf of

C\cap H

inside of

\mathcal P^r

and

\mathcal I_{C\cap H}^H(\ell)

is the ideal sheaf of

C\cap H

inside of

H\cong \mathbb P^{r-1}

. This exact sequence comes from the fact that...I'm not sure. Will need to consult the notes.

\square

Thus, rational normal curves are projectively normal.

Theorem: (Griffths, Harris page 528 - 531).

Through any $r+3$ points in $\mathbb P^{r}$ in linear general position, there is a unique rational normal curve.
Fix integers $d$ and $r$ such that $r\geq 2$ and $d\geq 2r + 3$ . Let $\Gamma \subseteq \mathbb P^r$ be $d$ points in linear general position. Then $\Gamma$ lines on a rational normal curve if and only if $\Gamma$ imposes $2r + 1$ conditions on quadrics.

The rational normal scrolls will be surfaces which contain our extremal curves. Fix $a_1$ and $a_2$ such that $a_1\geq a_2$ , $a_1\geq 0$ $a_2\geq 0$ and $a_1 + a_2 > r - 1$ . Fix subspaces $L_1$ and $L_2$ such that $L_1$ and $L_2$ span $\mathbb P^r$ and $\operatorname{dim}(L_i) = a_i$ . Let $C_i \subseteq C_i \subseteq L_i$ be a rational normal curve and fix an isomorphism $\varphi:C_1\to C_2$ . Then

\begin{aligned} X_{a_1, a_2} = \bigcup_{p\in C_1} \{\text{line connecting } P \text{ to } \varphi(P)\}. \end{aligned}

Example:

Take $r = 2$ , then $X_{0,1}\cong \mathbb P^2$ .
Take $r = 3$ . Then $X_{0,2} -$ cone over a conic, $X_{1,1} \cong \mathbb P^1\times \mathbb P^1$ .
Take $r = 4$ . Then $X_{1,2} \operatorname{BL}_p\mathbb P^2$ .

An Observation: Up to linear automorphisms of $\mathbb P^r$ this construction does not depend on the choices made other than $a_1$ and $a_2$ .

Lemma:

$X_{a_1,a_2}$ is a non-degenerate surface
The degree of $X_{a_1,a_2}$ is $a_1+a_2 = r - 1$
$X_{a_1,a_2}$ is non-singular if $a_1 \neq 0$ .

Proof.

First, $\operatorname{span}(c_i) = L_i$ and $\operatorname{span}(L_1\cup L_2) = \mathbb P^r$ . Any linear subspace $H\supset X_{a_1, a_2}$ needs to contain the $C_i \implies L_i \subset H$ , hence $\operatorname{span}(L_1\cup L_2) \subset H \implies H = \mathbb P^r$ .
$X_{a_1,a_2} \cap H = C_2 + \sum_{i=1}^{a_1}L_i$ . Thus the degree of $X_{a_1,a_2}$ is equal to the degree of $X_{a_1,a_2}\cap H$ which is $a_1 + a_2$ .
$X_{a_1, a_2} \cap H_p$ . For any $p$ , I can choose $H_p$ such that $X_{a_1,a_2} \cap H_p$ is smooth at $p$ as long as $a_1\neq 0$ . Then

\begin{aligned} \operatorname{deg}(H\cap X_{a_1,a_2}) = r-1 \end{aligned}

which implies

X_{a_1,a_2} \cap H\subset H

is a rational normal curve if smooth.

\square

Theorem:

Any surface in $\mathbb P^r$ that is non-degenerate has degree at least $r - 1$ .
If it has degree $r - 1$ , it is either a rational normal scroll or $\mathbb P^2 \to \mathbb P^5$ .

What about the Picard group of $X_{a_1,a_2}$ ? This is generated freely by the class $L$ of a line in the ruling and by $H$ the class of a hyperplane section. Note that you can write $H = C_1 + a_2 \cdot L = C_2 + a_1 \cdot L$ , so you can choose the class of one of the curves instead of $H$ if you prefer. Some intersection pairings:

$L.L = 0$
$L.H = 1$
$H.H = a_1 + a_2 = r - 1$

Remarks: Remember

\begin{aligned} M = \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}

Set

\begin{aligned} M' = \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{a_1-1} & y_0 & y_1 & \dots & y_{a_2 - 1} \\ x_1 & x_2 & x_3 & \dots & x_{a_1} & y_0 & y_1 & \dots & y_{a_2} \end{pmatrix}. \end{aligned}

$X_{a_1, a_2}$ will be cut out by $2\times 2$ minors of $M'$ , $\binom{r-1}{2}$ quadrics.
The surface $X_{a_1,a_2}$ is projectively normal:

0 \to I_{X_{a_1,a_2}} \xrightarrow{\cdot H} I_{X_{a_{1},a_2}}(\ell) \to I_{X_{a_1,a_2}\cap H}^H(\ell) \to 0.

$X_{a_1,a_2}\cong \mathbb P(\mathcal O_{\mathbb P^1}(a_1)\oplus \mathcal O_{\mathbb P^1}(a_2))\cong \mathbb F_{|a_2 - a_1|}$ where $\mathbb F_{|a_2 - a_1|}$ is a Hirzebruch surface.

Algebraic Curves Lecture 22

Extremal Curves

Interlude: Rational normal curves and rational normal scrolls

Rational normal scrolls