Algebraic Curves Lecture 18

The 18th lecture of algebraic curves by Karl Christ

Continuing the Petri Map

Continuing the Petri Map

The Petri map is

\begin{aligned} \mu_0:H^0(X,L)\otimes H^0(X,K_X - L) \to H^0(X,K_X) \end{aligned}

defined by $s_1 \otimes s_2 \mapsto s_1s_2$ . We saw that when $X$ has genus 3, then $\mu_0$ is injective if and only if $L\not\cong g^1_2$ . Let's expand a bit more on this example:

Proposition: (Base point free pencil trick). Let

L

be a line bundle on

X

and

s_1, s_2 \in H^0(X,L)

be linearly independent global sections. Denote by

V

the span

\{s_1,s_2\}

and let

F

be another locally free sheaf. Then the kernel of the tensor product map

\begin{aligned} V\otimes H^0(X,F)\to H^0(X,L\otimes F) \end{aligned}

H^0(X,F\otimes L^{-1}(B))

where

B

is the base locus of

V

Proof. Pick an element $s_1\otimes t_2 - s_2\otimes t_1 \in \operatorname{ker}(\varphi)$ . I can write any element of the kernel like this because $s_1$ and $s_2$ form a basis for $V$ . Let $s_i = s\cdot r_i$ with $s \in H^0(X,\mathcal O_X(B))$ a global section of $L$ vanishing on $V$ with $r_1 \in V(-B)\subseteq L(-B)$ . This is confusing, so here's a disambiguation:

$r_i$ is an element of $V(-B)$
$D_i$ is the vanishing locus of $r_i$ , it's a divisor on $X$
$L(-B) \cong \mathcal O_X(D_i)$ and
$s \in H^0(X, \mathcal O_X(B))$ . So
$s_i \in H^0(X,L)$ , and the divisor given by the vanishing of $s_i$ is $(s_i)_0 = B+D_i$ .

By assumption, $s_1\cdot t_2 - s_2 \cdot t_1 = 0$ . This implies $r_1\cdot t_2 - r_2 \cdot t_1 = 0$ where $r_1$ and $r_2$ do not have common zeroes. This in turn means $t_i$ vanishes along $D_i$ .

Write $\tau_i = \frac{t_i}{r_i}$ . Since $t_i \in H^0(X,F)$ and $1/r_i \in H^0(X,L^{-1}(B))$ , $\tau_i$ is a rational section of $F\otimes L^{-1}(B)$ (the inverse of $L(-B)$ is $L^{-1}(B)$ ). In fact, it is a regular section, because the zeros of $r_i$ and $t_i$ cancel each other out. This means $\tau_i \in H^0(X, F\otimes L^{-1}(B))$ .

I now have sections $\tau_1$ and $\tau_2$ , and I want to show they are equal. This follows since $t_i = r_i\tau_i$ and hence

\begin{aligned} &\implies r_1\cdot \underbrace{r_2 \cdot \tau_2}_{t_2} - r_2\cdot \underbrace{r_1\cdot \tau_1}_{t_1} = 0 \\ &\implies \tau_1 = \tau_2 =: \tau. \end{aligned}

In summary,

\begin{aligned} s_1\otimes t_2 - s_2 \otimes t_1 = s\cdot \tau \cdot (r_1\otimes r_2 - r_2 \otimes r_1) \end{aligned}

where

\tau \in H^0(X,F\otimes L^{-1}(B))

Conversely, if we're given a $\tau \in H^0(X,F\otimes L^{-1}(B))$ , then set $t_i = \tau\cdot r_i$ . Then

\begin{aligned} s_1\otimes t_2 - s_2 \otimes t_1 = s\cdot \tau(r_1\otimes r_2 - r_2 \otimes r_1), \end{aligned}

which is clearly in the kernel of

\varphi

\square

Using this trick we get the following corollary regarding the injectivity of the Petri map:

Cor: If

(L,V)

is a

g^1_d

with

2d < g - 2

, then the Petri map

\mu_0

for

(X,L)

is not injective.

Proof. We may assume that

V

is basepoint free by subtracting base points. The map

\begin{aligned} \varphi:V\otimes H^0(X,K_X - L) \to H^0(X,K_X). \end{aligned}

The kernel of

\varphi

H^0(X,K_X - 2L)

by the base point free pencil trick. The degree of

K_X - 2L

2g - 2 - 2d > 2g - 2 - g + 2 = g

. By Riemann-Roch,

\begin{aligned} h^0(L) - h^0(K_X - L) = d - g + 1 \implies h^0(L) \geq d - g + 1, \end{aligned}

so as soon as

d > g

, we know that there is a global section of

L

. Hence

K_X - 2L

has a global section, so

\operatorname{ker}(\varphi) = H^0(X,K_X - 2L) \neq 0

\square

Remark: The dimension of the locus of curves admitting a

g^1_d

\begin{aligned} &\phantom{\implies}2g - 2 = d\cdot (2h - 2) + b \\ &\implies b = 2g - 2 + 2d. \end{aligned}

Then

b - 3 - 2g - 5 + 2d \geq 3g - 3

and so

2d \geq g + 2