The 17th lecture of algebraic curves by Karl Christ
We stopped in the middle last time: we have
\[\begin{aligned}
I = \{(p,H) ~\mid~ p\in H\} \subset C\times U
\end{aligned}\]
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
\[\begin{aligned}
\varphi:\pi_1(U,H_0) \hookrightarrow \operatorname{Aut}(\rho_2^{-1}_2(H_0))
\end{aligned}\]
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
\[\begin{aligned}
I(m) = \{(p_1,...,p_m,H) ~\mid~ p_i \in H\} \subset C^m \times U.
\end{aligned}\]
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.
\(\square\)
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
\[\begin{aligned}
J_K = \{(p_1,...,p_m,H) ~\mid~ r(V-\sum p_i)\} \subseteq I(m).
\end{aligned}\]
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 \(U\).
\(\square\)
Cor: (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.
\(\square\)
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 \(d \leq 2g - 2\)
\[\begin{aligned}
\operatorname{rank}(L) = d - g \leq \frac{d}{2}.
\end{aligned}\]
Thus we can assume \(L\) is special, so \(L\) and \(K_X - L\) are effective. For any two divisors \(D, D'\):
\[\begin{aligned}
\operatorname{rank}(D) + \operatorname{rank}(D') \leq \operatorname{rank}(D + D').
\end{aligned}\]
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 \(\operatorname{rank}(L) + \operatorname{rank}(K_X - L) \leq \operatorname{rank}(K_X) = g - 1\).
\(\square\)