Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
Continuing the Petri Map
The Petri map is
\begin{align*} \mu_0:H^0(X,L)\otimes H^0(X,K_X - L) \to H^0(X,K_X) \end{align*}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{align*} V\otimes H^0(X,F)\to H^0(X,L\otimes F) \end{align*}is \(H^0(X,F\otimes L^{-1}(B))\) where \(B\) is the base locus of \(V\).
Proof
Pick an element \(s_1\otimes t_2 - s_2\otimes t_1 \in \ker(\varphi)\). 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{align*} &\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{align*}In summary,
\begin{align*} 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{align*}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{align*} s_1\otimes t_2 - s_2 \otimes t_1 = s\cdot \tau(r_1\otimes r_2 - r_2 \otimes r_1), \end{align*}which is clearly in the kernel of \(\varphi\).
Using this trick we get the following corollary regarding the injectivity of the Petri map:
Corollary
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{align*} \varphi:V\otimes H^0(X,K_X - L) \to H^0(X,K_X). \end{align*}The kernel of \(\varphi\) is \(H^0(X,K_X - 2L)\) by the base point free pencil trick. The degree of \(K_X - 2L\) is \(2g - 2 - 2d > 2g - 2 - g + 2 = g\). By Riemann-Roch,
\begin{align*} h^0(L) - h^0(K_X - L) = d - g + 1 \implies h^0(L) \geq d - g + 1, \end{align*}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 \(\ker(\varphi) = H^0(X,K_X - 2L) \neq 0\).
Remark: The dimension of the locus of curves admitting a \(g^1_d\) is
\begin{align*} &\phantom{\implies}2g - 2 = d\cdot (2h - 2) + b \\ &\implies b = 2g - 2 + 2d. \end{align*}Then \(b - 3 - 2g - 5 + 2d \geq 3g - 3\) and so \(2d \geq g + 2\).