Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
Proof of Hurwitz Bound
Last time we ended with the Hurwitz bound:
Corollary
(Hurwitz bound). Suppose \(X\) has genus \(\geq 2\) and finitely many automorphisms. Then \(|\Aut(X)| \leq 84(g - 1)\).
Note that if \(X\) is genus \(0\) or \(1\) then it will never have finitely many automorphisms.
Proof
let \(f_i\) be the number of points of ramification index \(r_i\) over a point \(p_i\) in the base. Orbit stabilizer theorem says that \(|\Aut(X)| = f_ir_i\). Let \(h\) denote the genus of \(X/\Aut(X)\). Riemann-Hurwitz says that
\begin{align*} 2g(X) - 2 &= |\Aut(X)| \cdot (2h - 2) + \sum_{i,j}(e_{i,j} - 1) \\ &= |\Aut(X)| \cdot (2h - 2) + \sum_{i=1}^bf_i(r_i - 1) &= |\Aut(X)| \cdot \left(2h - 2 - \sum_{i=1}^b(1 - \frac{1}{r_i})\right). \end{align*}Set \(c = \left(2h - 2 - \sum_{i}(1 - \frac{1}{r_i})\right)\). We’ll come up with a lower bound on \(c\) by examining a few cases:
Case 1: \(h \geq 2\). Then \(c \geq 2\) and \(|\Aut(X)| \leq g - 1\).
Case 2: \(h = 1\). Then \(b \geq 1\) (remember, \(b\) is the number of branch points) and \(c \geq \frac{1}{2}\). Thus \(|\Aut(X)|\leq 4(g - 1)\).
Case 3: \(h = 0\).
Case 3.1: If \(b\geq 5\), then \(c \geq \frac{1}{2}\) and so \(|\Aut(X) \leq 4(g - 1)\).
Case 3.2: If \(b = 4\) then not all \(r_i\) can be equal to \(2\). Then
\begin{align*} c \geq -2 \frac{3}{2} = \frac{g + 4 - 12}{6} = \frac{1}{6} \implies |\Aut(X)| \leq 12(g - 1). \end{align*}Case 3.3: The case that \(b \leq 2\) is not possible since \(g\) is at least \(2\).
Case 3.4: If \(b = 3\) then suppose \(r_1\leq r_2 \leq r_3\) without loss of generality. Now MORE subcases:
Case 3.4.1: If \(r_1\geq 3\), then not all \(r_i\) can be equal to \(3\), so
\begin{align*} c \geq -2 + \frac{2}{3} + \frac{2}{3} + \frac{3}{4} = \frac{-24 + 16 + 9}{12} = \frac{1}{12}, \end{align*}from which it follows that \(|\Aut(X)| \leq 24(g - 1)\).
Case 3.4.2: If \(r_1 = 2\) then \(r_2 > 4\) and \(r_3 \geq 5\). Then
\begin{align*} c \geq -2 + \frac12 + \frac34 + \frac 45 = \frac{-40 + 10 + 15 + 16}{20} = \frac{1}{20} \end{align*}so \(|\Aut(X)| \leq 40(g - 1)\).
Case 3.4.3: If \(r_1 = 2\), \(r_2 = 3\) and then \(r_3 \geq 7\). Then
\begin{align*} c \leq -2 + \frac{1}{2} + \frac34 + \frac67 = \frac{1}{42} \end{align*}so \(|\Aut(X)| \leq 84(g - 1)\), which is exactly the Hurwitz bound.
So in all cases, if \(X\) is a genus \(\geq 2\) curve, then \(|\Aut(X)| \leq 84(g - 1)\).
Hyperelliptic curves
A curve \(X\) is called hyperelliptic if it admits a \(g^1_2\), that is, a degree \(2\) map \(X\to \mathbb P^1\).
- If \(g = 0\) (meaning \(X \cong \mathbb P^1\)) then \(h^0(\mathcal O_{\mathbb P^1}(2)) = 3\). Any 2-dimensional subspace \(V\subset H^0(\mathcal O_{\mathbb P^1}(2))\) gives a \(g^1_2\).
- If \(g \geq 1\) and \(X\) is hyperelliptic, then the \(g^1_2\) needs to be complete.
- If \(g = 1\), then any degree 2 line bundle gives a \(g^1_2\) (Riemann-Roch).
- If \(g = 2\), then the canonical divisor has degree equal to \(2g - 2 = 2\), and its space of global sections is \(h^0(K_X) = g = 2\). From Riemann-Roch we can deduce that this is the unique \(g^1_2\).
As a reminder:
Lemma
If \(X\) has genus \(g \geq 1\) and \(L\) has degree \(2g - 2\) then \(h^0(L) \leq g\) and equality holds if and only if \(L\cong K_X\).
Proof
\begin{align*} h^0(L) - h^0(K_X - L) = 2g - 2 - g + 1 = g - 1. \end{align*}Proposition
If \(X\) is a hyperelliptic curve with \(g\geq 2\) then \(K_X\cong L^{\otimes g - 1}\) where \(L\) is a \(g^1_2\).
Proof
\(L\) induces a map \(\varphi_L:X\to \mathbb P^1\). Consider the Veronese embedding of \(\mathbb P^1\), i.e. the map \(\mathbb P^1 \to \mathbb P^{g - 1}\) induced by \(\mathcal O_{\mathbb P^1}(g - 1)\): \([x,y] \mapsto [x^3, x^2y, xy^2, y^3]\). Then
\begin{align*} \varphi^*_L\mathcal O_{\mathbb P^1}(1)\cong L \end{align*}and
\begin{align*} \mathcal O_{\mathbb P^1}(g - 1) \cong \mathcal O_{\mathbb P^1}(1)^{\otimes g - 1}. \end{align*}Thus \(\varphi\circ \varphi_L:X\to \mathbb P^{g - 1}\) given by \(L^{\otimes g - 1}\). In particular, \(h^0(L^{\otimes g-1})\) and \(\deg(L^{\otimes g-1}) = 2g -2\), hence \(L^{g-1}\cong K_X\).
Theorem
Let \(g \geq 2\). Then \(K_X\) is base point free and
- very ample if \(X\) is not hyperelliptic
- a \(2:1\) cover of a rational normal curve in \(\mathbb P^{g-1}\) if \(X\) is hyperelliptic.
Proof
Let \(p\in X\).
\begin{align*} h^0(K_X - p) - h^0(\mathcal O_X(p)) = 2g - 3 - g + 1 = g - 2. \end{align*}We previously saw that \(h^0(\mathcal O_X(p)) = 1\) in this case (I forgot why this is true) so \(h^0(K_X - p) = g - 1 = h^0(K_X) - 1\).
- Since \(h^0(\mathcal O_X O(p + q) = 1\) \begin{align*}h^0(K_x - p - q) - &h^0(\mathcal O_X O(p + q)) = 2g - 4 - g + 1 = g-2 \\ &\implies h^0(K_x - p - q) = g - 2.\end{align*}
- This was the previous proposition.
Proposition
If \(X\) is a hyperelliptic curve of genus \(\geq 2\) then the \(g^1_2\) is unique.
Proof
Consider \(\varphi_{K_X}:X\to \mathbb P^{g -1}\). By the previous proposition, this is a \(2:1\) cover of its image which is a rational curve. Conversely, give any \(g^1_2\) on \(X\) and it defines this cover. This implies the \(g^1_2\) is unique.