Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
(Missed Monday due to flight)
General and uniform positions
Theorem
Let \(C\subseteq \mathbb P^r\), \(r\geq 3\), \(C\) is non-degenerate irreducible of degree \(d\). Let \(H\) be a general hyperplane in \(\mathbb P^r\). Then any \(k\leq r\) points in \(H\cap C\) will span a linear space of dimension \(k-1\).
Example: Take \(C\subseteq \mathbb P^3\). Then no three points in \(C\cap H\) are colinear for a general plane \(H\subseteq \mathbb P^3\).
Slogan: For a general hyperplane \(H\) the subsets containing \(k\) elements in \(H\cap C\) are indistinguishable
Let’s let
\begin{align*} I = \{(p,H) | p\in H \cap C\} \subseteq C\times U \end{align*}where \(U\subseteq (\mathbb P^r)^*\) is the set of hyperplanes in \(\mathbb P^r\) meeting \(C\) transversely. Now consider the map \(\pi_2:I\to U\), it is a \(d\)-sheeted covering of \(U\). Fix \(H_0\in U\) and consider the action of \(\pi_1(U,H_0)\) (the fundamental group of \(U\) based at \(H_0\)) on \(\pi^{-1}_2(H_0)\).
Lemma
\(\pi_1(U,H_0)\) acts as the full symmetric group \(S_{d}\) on \(\pi^{-1}_2(H_0)\).
Proof
Introduce the sets
\begin{align*} \tilde{I}(2) = \{(p_1,p_2,H) ~\mid~ p_1\neq p_2, p_1,p_2 \in C\cap H\}\subset C\times C\times (\mathbb P^r)^* \end{align*}and
\begin{align*} I(2) := \tilde{I}(2) \cap (C\times C\times U). \end{align*}Note that passing from \(\tilde I(2)\) to \(I(2)\) simply requires the plane to meet the curve transversely. Now let \(\pi_1:\tilde I(2)\to (C\times C)\setminus \Delta\) be the projection map (here we remove the diagonal since we required that \(p_1\neq p_2\) for a tuple in \(\tilde I(2)\).
The set \(\pi_1^{-1}(p_1,p_2)\) is the set of all hyperplanes passing through \(p_1\) and \(p_2\), and it is isomorphic to \(\mathbb P^{r-2}\).
Claim: both \(\tilde I(2)\) and \(I(2)\) are both irreducible and are in particular connected.
> Side Remark: if \(f:X\to Y\) is some dominant map of varieties and you know that \(X\) is irreducible, you can immediately conclude that \(Y\) is irreducible. However, if \(Y\) is irreducible, concluding that \(X\) is irreducible is much harder. A sufficient condition is the following: (1) \(Y\) is irreducible (2) the fibers of \(f\) are all irreducible (3) the fibers have the same dimension and (4) \(f\) is proper.
The space \(C\times C\setminus \Delta\) is irreducible – the diagonal is a codimension 2 subset in this case. Furthermore, \(\pi_1\) satisfies (1)-(4) in the remark above, so \(\tilde I(2)\) is irreducible. A similar argument tells me that \(I(2)\) is irreducible. Now we return our attention to the map and fiber
\begin{align*} \pi_2:I\to U, \hspace{1cm} \pi_2^{-1}(H_0). \end{align*}This means \(\pi_1(U,H_0)\) acts 2-transitively on \(\pi^{-1}_2(H_0)\). Let \(G\subseteq S_d\) be the transitive subgroup of \(S_d\) given as the image of the fundamental group \(\pi_1(U,H_0)\to \Aut(\pi_2^{-1}(H_0)) = S_d\).
To show that \(G = S_d\) it suffices to show that \(G\) contains a transposition. Indeed, if it has a transposition \(\tau \in G\), then
- I can get any other transposition \(\tau'\) by conjugation with a permutation \(\sigma\) such that \((\tau_1,\tau_2) \mapsto (\tau_1', \tau_2')\). Then \(\sigma\tau\sigma^{-1} = \tau'.\)
- The transpositions generate \(S_d\), so \(G = S_d\).