Algebraic Curves Lecture 10: Hurwitz bound and hyperelliptic curves.

The 10th lecture of algebraic curves with Karl Christ

Proof of Hurwitz Bound
Hyperelliptic curves

Proof of Hurwitz Bound

Last time we ended with the Hurwitz bound:

Cor: (Hurwitz bound). Suppose

X

has genus

\geq 2

and finitely many automorphisms. Then

|\operatorname{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 $|\operatorname{Aut}(X)| = f_ir_i$ . Let $h$ denote the genus of $X/\operatorname{Aut}(X)$ . Riemann-Hurwitz says that

\begin{aligned} 2g(X) - 2 &= |\operatorname{Aut}(X)| \cdot (2h - 2) + \sum_{i,j}(e_{i,j} - 1) \\ &= |\operatorname{Aut}(X)| \cdot (2h - 2) + \sum_{i=1}^bf_i(r_i - 1) &= |\operatorname{Aut}(X)| \cdot \left(2h - 2 - \sum_{i=1}^b(1 - \frac{1}{r_i})\right). \end{aligned}

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 $|\operatorname{Aut}(X)| \leq g - 1$ .

Case 2: $h = 1$ . Then $b \geq 1$ (remember, $b$ is the number of branch points) and $c \geq \frac{1}{2}$ . Thus $|\operatorname{Aut}(X)|\leq 4(g - 1)$ .

Case 3: $h = 0$ .

Case 3.1: If $b\geq 5$ , then $c \geq \frac{1}{2}$ and so $|\operatorname{Aut}(X) \leq 4(g - 1)$ .

Case 3.2: If $b = 4$ then not all $r_i$ can be equal to $2$ . Then

\begin{aligned} c \geq -2 \frac{3}{2} = \frac{g + 4 - 12}{6} = \frac{1}{6} \implies |\operatorname{Aut}(X)| \leq 12(g - 1). \end{aligned}

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{aligned} c \geq -2 + \frac{2}{3} + \frac{2}{3} + \frac{3}{4} = \frac{-24 + 16 + 9}{12} = \frac{1}{12}, \end{aligned}

from which it follows that

|\operatorname{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{aligned} c \geq -2 + \frac12 + \frac34 + \frac 45 = \frac{-40 + 10 + 15 + 16}{20} = \frac{1}{20} \end{aligned}

|\operatorname{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{aligned} c \leq -2 + \frac{1}{2} + \frac34 + \frac67 = \frac{1}{42} \end{aligned}

|\operatorname{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 $|\operatorname{Aut}(X)| \leq 84(g - 1)$ .

\square

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{aligned} h^0(L) - h^0(K_X - L) = 2g - 2 - g + 1 = g - 1. \end{aligned}

\square

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{aligned} \varphi^*_L\mathcal O_{\mathbb P^1}(1)\cong L \end{aligned}

and

\begin{aligned} \mathcal O_{\mathbb P^1}(g - 1) \cong \mathcal O_{\mathbb P^1}(1)^{\otimes g - 1}. \end{aligned}

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

\square

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{aligned} h^0(K_X - p) - h^0(\mathcal O_X(p)) = 2g - 3 - g + 1 = g - 2. \end{aligned}

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 $h^0(\mathcal O_X O(p + q) = 1$
$\begin{aligned}h^0(K_x - p - q) - &h^0(\mathcal O_X O(p + q)) = 2g - 4 - g + 1 = g-2 \\ &\implies h^0(K_x - p - q) = g - 2.\end{aligned}$
This was the previous proposition.

\square

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

X

and it defines this cover. This implies the

g^1_2

is unique.

\square