Algebraic Curves Lecture 11

The 11th lecture of algebraic curves by Karl Christ

Riemann existence theorem
Low genus examples

Last week, we saw

there exists a unique $g^1_2$ on any hyperelliptic curve with $g \geq 2$ .
$K_X$ is very ample if and only if $X$ is not hyperelliptic.

Proposition: If

C\subseteq \mathbb P^n

is a complete intersection curve and is smooth of genus at least 2, then

C

is not hyperelliptic.

Proof. $N_{C/\mathbb P^n} = \bigoplus \mathcal O_{\mathbb P^n}(d)$ , $\bigwedge^{n-1} N_{C/\mathbb P^1} = \mathcal O_{\mathbb P^n}(\sum d_i)|_C$ . The canonical bundle of $C$ is given by

\begin{aligned} K_C = \left(K_{\mathbb P^n} \otimes \bigwedge^{n-1} N_{C/\mathbb P^n}\right)|_C = \mathcal O_{\mathbb P^n}(\sum d_i - n - 1)|_C, \end{aligned}

where we've used adjunction and the fact that

K_{\mathbb P^n} = \mathcal O_{\mathbb P^n}(-n-1)

. Since

g(C) \geq 2

, we need to have

\sum d_i - n - 1 > 0

Through any pair of points $p$ and $q$ on $C$ , we can find a hypersurface of degree $\sum d_i- n - 1$ which passes through $p$ but not $q$ (this is true as long as $\sum d_i -n - 1 > 0$ ). This works for double points as well -- we can find such a hypersurface which vanishes to order 1 at a specified point. Since $K_C$ is cut out by hypersurfaces, this means $K_C$ is very ample, hence $C$ is not hyperelliptic curve by (2) above.

\square

Riemann existence theorem

Question: If given a $\mathbb P^1$ with specified branch points $p_1,...,p_b\in \mathbb P^1$ , can you find a corresponding cover $X\to \mathbb P^1$ whose branch points are precisely these $p_i$ ?

First attempt at constructing this: You want a cover of $\mathbb P^1$ that is unramified away from the $p_i$ . So consider first $\mathbb P^1 \setminus \{p_1,...,p_b\}$ topologically. The fundamental group of this is generated by loops $\gamma_i$ based at $p_0$ and going around $p_i$ . They are subject to the relation $\prod \gamma_i = \operatorname{id}$ . This means that interchanging any of these two generators corresponds to permuting the sheets of the cover.

For any cover $X\xrightarrow{d:1} \mathbb P^1$ branched at the $p_i$ , then we have $d$ sheets and therefore we get a group homomorphism

\begin{aligned} \pi_1(\mathbb P^1\setminus \{p_1,...,p_b\}, p_0) \to S_d. \end{aligned}

Thus we get permutations $\tau_i\in S_d$ such that

\begin{aligned} \prod_{i=1}^d\tau_i = \operatorname{id}. \end{aligned}

These permutations additionally must generate a transitive subgroup of $S_d$ , if we want $X$ to be connected.

Note 1 at this point: We haven't prescribed ramificaiton indices to the points

p_i

, but they are encoded in this picture as well. As constructed, the ramification index of

p_i

corresponds to the "number of sheets" in the cover of

\mathbb P^1\set minus \{p_1,...,p_b\}

which meet at

p_i

, and this number is equal to the number of cycles in the cycle decomposition of

\tau_i

. To see this, consider the analytic picture: locally near

p_i

, the map will be given

z \mapsto z^{r_i}

where

r_i

is the ramification index of

p_i

Note 2 at this point: We're trying to construct

X

and a map to

\mathbb P^1

by first constructing a topological cover

U\to \mathbb P^1\set minus \{p_1,...,p_b\}

and then gluing our branch points back in. One way to construct this cover is to take

U = \widetilde U \times \{p_1,...,p_b\}/G

, where

\widetilde{U}

is the universal cover and

G = \pi_1(\mathbb P^1\set minus \{p_1,...,p_b\}, p_0)

acts diagonally by deck transformation.

This leads us to a theorem that says constructing a cover in this way is always possible.

Theorem: (Riemann Existence). There is a bijection between degree d covers

X\to \mathbb P^1

simply branched over

p_1,...,p_b

up to isomorphism and transpositions

\tau_i\in S_d

(

1\leq i\leq b

) such that

\prod \tau_i = \operatorname{id}

and the group generated by the

\tau_i

acts transitively on

\{1,...,d\}

up to simultaneous conjugation by an element in

S_d

Example: For any even number

b

of branch points there is a unique hyperelliptic curve

X

with

X\rightarrow{2:1} \mathbb P^1

is branched over the

p_i

. Riemann-Hurwitz says

\begin{aligned} [2g - 2 = d_i (2h - 2) + \deg(R)] \end{aligned}

\begin{aligned} 2g - 2 = -4 + b \iff b = 2g + 2. \end{aligned}

The dimension of the space of hyperelliptic curves of genus

g

2g + 2 - 3 = 2g - 1

Note that this theorem only works when the base field is $\mathbb C$ . Moving from the topological construction of the cover $U\xrightarrow{d:1} \mathbb P^1\setminus \{p_1,...,p_b\}$ as we sketched above to an algebraic cover $X\to \mathbb P^1$ requires that there is a unique complex structure we can impose on $X$ . The analytic picture is therefore quite essential in this case.

Example: (Degree 3 simply ramified covers.) The relevant permutation group is

S_3

. Then

2g - 2 = -6 + b \iff b = 2g +4

. If I choose

2g + 3

transpositions, their composition is again a transposition. Then

\frac{3^{2g + 3} - 3}{6}

is the number of simpy ramified degree 3 covers

X\to \mathbb P^1

with

g(X) = g

and branched over fixed points

p_1,...,p_b

up to isomorphism.

Example: For a fixed ramification profile, there does not need to exist a cover of that profile, even if the numerics of Riemann-Hurwitz are satisfied: for

d = 4

\{(2)(2), (2)(2), (3)\}

. Then

2g - 2 = -8 + 6 = -2

which implies

g = 0

. But there are no elements

\tau_1,\tau_2,\tau_3 \in S_4

of types

(2)(2), (2)(2)

and

(3)

such that

\prod \tau_i = \operatorname{id}

.

It's a very open question to determine which covers are allowed and which are not.

Low genus examples

Genus 2: Any genus 2 curve is hyperelliptic, and the unique $g^1_2$ is the canonical sheaf. The number of branch points allowed by a map $X\to \mathbb P^1$ is $2 = -4 + b \iff b = 6$ . Morally, genus $2$ curves are paramterized by $(\mathbb P^1)^3$ . We have an action of $S_6$ on $\{(0,1,\infty)\}\times (\mathbb P^1)^3$ given on $(0,1,\infty, p_1,p_2,p_3)$ given by applying the permutation naturally on the 6-tuple and then applying an automorphism of $\mathbb P^1$ to each coordinate so that the first three coordinates are again $0$ , $1$ and $\infty$ . Removing $0,1$ and $\infty$ from each copy of $\mathbb P^1$ as the diagonal from the product $(\mathbb P^1)^3$ makes this action free, allowing us to take a quotient giving us the moduli space $M_2$ :

\begin{aligned} M_2 = (\mathbb A^1 \setminus \{0,1,\}^3 \setminus \Delta / S^6, \hspace{1cm} \operatorname{dim} = 3 = 2g - 1 = 3g - 3. \end{aligned}

Genus $3$ : If $X$ is not hyperelliptic then $\varphi_{K_X}$ realizes $X$ as a degree $2g - 2 = 4$ curve in $\mathbb P^{g - 1} = \mathbb P^2$ . Conversely, any smooth plane quartic is a canonically embedded genus $2$ curve. Indeed, by adjunction,

\begin{aligned} K_C = (\mathcal O_{\mathbb P^2}(-3 + 4))|_C = \mathcal O_{\mathbb P^2}(1)|_C. \end{aligned}

Suppose $C$ is hyperelliptic with $g^1_2$ with $g^1_2 = \mathcal O_C(p+q)$ . Then $2 = h^0(\mathcal O_C(p+q)$ . Then

\begin{aligned} 2 = h^0(\mathcal O_C(p+q)) = 2 - 3 + 1 + h^0(K_C - p - q). \end{aligned}

But for geometric reasons $h^0(K_c - p - q) = 1$ . The dimension of the space of non-hyperelliptic genus 3 curves thus is $h^0(\mathcal O_{\mathbb P^2}(4)) - 1 - 8 = \binom{6}{2} - g = 15 - g = 6$ . As before, the space of hyperelliptic curves has dimension $2g - 1 = 5$ .