Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
We stopped in the middle last time: we have
\begin{align*} I = \{(p,H) ~\mid~ p\in H\} \subset C\times U \end{align*}with a projection \(\rho:I\to U = \{U \in (\mathbb P^r)^* ~\mid~ H \text{ intersects } C \text{ in } d \text{points}\}\). We then had a map
\begin{align*} \varphi:\pi_1(U,H_0) \hookrightarrow \Aut(\rho_2^{-1}_2(H_0)) \end{align*}with \(\operatorname{img}(\varphi) = G\subseteq S_d\).
We claim that \(G = S_d\). We showed that \(G\) acts 2-transitively, and because \(U\) is irreducible we concluded that \(I(2)\) was irreducible. It was left to show that \(G\) contained a transposition, from which we can conclude that \(G\) contains all transpositions and hence generates the whole group. So we need to find one tranposition in \(G\).
Let \(H\) be a hyperplane such that \(H\) has a simple tangency at one point of \(C\) and intersects \(C\) transversely at all other points.
In an analytic neighborhood \(W\) of \(H\) in \((\mathbb P^r)^*\) the projection \(I\to (\mathbb P^r)^*\) is a \(d\)-sheeted covering with \(2\) branches coming together over \(H\) and all other branches staying separate. This corresponds to a transposition in \(G\).
Lemma
\(I(m)\) is irreducible where
\begin{align*} I(m) = \{(p_1,...,p_m,H) ~\mid~ p_i \in H\} \subset C^m \times U. \end{align*}Proof
\(I(m)\) is connected if and only if \(G\) acts \(m\)-transitively. We have a projection \(\pi_i:I(m)\to I\) for each \(1\leq i\leq m\) and another projection \(\pi_{m+1}:I(m) \to U\) which commutes with the map \(\rho:I\to U\). Since \(G = S_d\), it acts \(m\)-transitively and hence \(I(m)\) is connected. Since \(I(m)\to U\) is a finite covering, it being connected implies that \(I(m)\) is irreducible.
Theorem
(Uniform position theorem.) Let \(C\subset \mathbb P^r\) for \(r\geq 3\) be a non-degenerate curve. Let \(H\subset (\mathbb P^r)^\times\) be a general hyperplane with respect to some linear series \(V\) on \(C\). Then any choice of \(m\)-points in \(C\cap H\) imposes the same number of conditions on \(V\). That is, \(\{p_1,...,p_m\} \subset C\cap H\) and \(\operatorname{rank}(V - \sum p_i)\) does not depend on \(p_1,...,p_m\).
Proof
Set
\begin{align*} J_K = \{(p_1,...,p_m,H) ~\mid~ r(V-\sum p_i)\} \subseteq I(m). \end{align*}Then \(J_K \subseteq I(m)\) is closed. Let \(K_0\) be the largest \(K\) such that \(J_K\subsetneq I(m)\). Then \(\pi_2:I(m) \to U\), then also \(\pi_2(J_{K_0})\) will be a proper closed subset. Pick \(H\not\in \pi_2(J_{K_0})\).
We need that \(I(m)\) is an irreducible set to ensure that it doesn’t have an irreducible component that dominates \(U\).
Corollary
(General position theorem) Any \(m\leq r\) points in \(C\cap H\) are linearly independent.
Proof
Apply theorem above to \(V = \mathcal O_{\mathbb P^r}(1)|_C\). Thus any \(m\) points impose the same number of linear conditions. For \(H\) general, some \(m\)-tuple of points is linearly independence, hence all.
Example: Let \(X\) be a genus \(4\) curve, not hyperelliptic. \(X\rightarrow \mathbb P^3\), \(X = Q_2 \cap Q_3 \cong \mathbb P^1 \times \mathbb P^1\)… didn’t finish
Theorem
(Clifford’s theorem) Let \(L\) be a line bundle on \(X\) of degree \(0\leq d\leq 2g - 2\). Then
- \(\operatorname{rank}(L) = r \leq \frac{d}{2}\)
- Equality holds if and only if \(L = \mathcal O_X, K_X\) or if \(L = (L')^{\otimes r}\) where \(L'\) is a \(g^1_2\) (and hence \(X\) is hyperelliptic).
Proof
We can prove at least part (1) today. Recall that \(L\) is special if \(h^0(L) > 0, h^1(L) > 0\). If \(L\) is not special, then \(h^0(L) = \min\{0,d-g+1\}\). If \(h^0(L) = d - g + 1\) then since \(d \leq 2g - 2\)
\begin{align*} \operatorname{rank}(L) = d - g \leq \frac{d}{2}. \end{align*}Thus we can assume \(L\) is special, so \(L\) and \(K_X - L\) are effective. For any two divisors \(D, D'\):
\begin{align*} \operatorname{rank}(D) + \operatorname{rank}(D') \leq \operatorname{rank}(D + D'). \end{align*}Since \(\operatorname{rank}(D) \geq s \iff D - E\) is effective for any divisor of degree \(s\). Applied to \(L\) and \(K_X - L\) we get \(\operatorname{rank}(L) + \operatorname{rank}(K_X - L) \leq \operatorname{rank}(K_X) = g - 1\).