Hyperelliptic and trigonal curves

Let ${C}$ be a genus ${g}$ curve over the field ${\mathbb{C}}$ of complex numbers. I’ve been trying to understand a little about special linear series on ${C}$ : that is, low degree maps ${C \rightarrow \mathbb{P}^1}$ , or equivalently divisors on ${C}$ that move in a pencil. Once the degree is at least ${2g + 1}$ , any divisor will produce a map to ${\mathbb{P}^1}$ (in fact, many maps), and these fit into nice families. In degrees ${\leq 2g-2}$ , maps ${C \rightarrow \mathbb{P}^1}$ are harder to write down, and the families they form (for fixed $C$ ) aren’t quite as nice.

However, it turns out that there are varieties of special linear series—that is, varieties parametrizing line bundles of degree ${\leq 2g-2}$ with a certain number of sections, and techniques from deformation theory and intersection theory can be used to bound below and predict their dimensions (the predictions will turn out to be accurate for a general curve). For instance, one can show that any genus ${g}$ curve has a map to ${\mathbb{P}^1}$ of degree at most ${\sim \frac{g}{2}}$ , but for degrees below that, the “general” genus ${g}$ curve does not admit such a map. This is the subject of the Brill-Noether theory.

In this post, I’d just like to do a couple of low-degree examples, to warm up for more general results. Most of this material is from Arbarello-Cornalba-Griffiths-Harris’s book Geometry of algebraic curves.

1. Hyperelliptic curves

For ${d = 2}$ , we are considering hyperelliptic curves: that is, curves with a degree two map to ${\mathbb{P}^1}$ . An application of the Riemann-Hurwitz formula shows that the hyperelliptic map

$\displaystyle \phi: C \rightarrow \mathbb{P}^1$

must be branched over exactly ${2g + 2}$ points. In other words, it is a two-sheeted cover of ${\mathbb{P}^1}$ , and the sheets come together at ${2g + 2}$ points. Since it is a degree two cover, it is necessarily Galois, and ${C}$ has a hyperelliptic involution ${\iota: C \rightarrow C}$ over ${\mathbb{P}^1}$ with those branch points as its fixed points, such that

$\displaystyle C/(x \sim \iota x ) \stackrel{\phi}{\simeq} \mathbb{P}^1.$

When ${g = 1}$ , ${C}$ is an elliptic curve (once one chooses an origin on ${C}$ ), and the hyperelliptic involution can be realized as ${x \mapsto -x}$ with respect to the group law on ${C}$ . The resulting map ${C \rightarrow C/( \mathbb{Z}/2)}$ can be realized, for a Weierstrass curve ${y^2 = x^3 + A x + B}$ , by the function ${x: C \rightarrow \mathbb{P}^1}$ which respects the involution ${(x,y) \mapsto (x, -y)}$ : in analytic terms, the hyperelliptic map is given by the Weierstrass ${\wp}$ -function. The ramification points of this map are the fixed points of the involution ${x \mapsto -x}$ , or the four 2-torsion points on ${C}$ .

In this case, it’s important that the hyperelliptic map is not unique (even up to automorphisms of ${\mathbb{P}^1}$ ). Namely, the hyperelliptic map depended upon the choice of an origin ${p \in C}$ , and then was given by the divisor ${(2p)}$ —which moved in a “pencil” and thus defined a map to ${\mathbb{P}^1}$ . (In other words, the line bundle ${\mathcal{O}({2p})}$ corresponding to the divisor had a two-dimensional space of sections ${H^0( \mathcal{O}(2p))}$ .) However, a different ${q \in C \setminus \left\{p\right\}}$ would have provided a different line bundle ${\mathcal{O}(2q)}$ and a different map to ${\mathbb{P}^1}$ , except for three exceptional choices of ${q}$ .

However, in genus ${\geq 2}$ , the hyperelliptic map on a curve (if it exists) is unique. To see this, we use the following lemma, called the basepoint-free pencil trick.

Lemma 1 Let ${\mathcal{L}, \mathcal{M}}$ be line bundles on a curve ${C}$ . Let ${s_1, s_2}$ be sections of ${H^0(\mathcal{L})}$ without common zeros and consider the map

$\displaystyle s_1 H^0( \mathcal{M}) \oplus s_2 H^0( \mathcal{M}) \rightarrow H^0( \mathcal{L} \otimes \mathcal{M}) .$

Then the kernel of this map is ${H^0( \mathcal{M} \otimes \mathcal{L}^{-1})}$ .

Proof: Indeed, one has an exact sequence of sheaves

$\displaystyle 0 \rightarrow \mathcal{M} \otimes \mathcal{L}^{-1} \rightarrow \mathcal{M} \oplus \mathcal{M} \stackrel{s_1, s_2}{\rightarrow} \mathcal{M} \otimes \mathcal{L} \rightarrow 0,$

where the first map sends a section ${t}$ of ${\mathcal{M} \otimes \mathcal{L}^{-1}}$ to the pair ${(ts_2, -ts_1)}$ . This is precisely a Koszul-type complex, for the regular sequence ${(s_1, s_2)}$ —regularity follows because the vanishing loci are disjoint. Taking global sections gives the desired claim. $\Box$

Let’s now suppose that ${C}$ is a hyperelliptic curve of genus ${g > 1}$ and ${\mathcal{L}, \mathcal{M}}$ are two degree two line bundles with ${h^0 = 2}$ ; that means they’re generated by their global sections (since if they had a basepoint, one would get a degree one line bundle with sections). The basepoint-free pencil trick now shows that, if ${\mathcal{L} \not\simeq \mathcal{M}}$ , then

$\displaystyle h^0( \mathcal{L} \otimes \mathcal{M}) \geq 4,$

where ${\mathcal{L} \otimes \mathcal{M}}$ has degree four. Since ${g(C) >0}$ , we must have

$\displaystyle h^0( \mathcal{L} \otimes \mathcal{M}) = 4,$

since for any line bundle ${\mathcal{N}}$ on a curve of genus ${>1}$ , we have ${h^0( \mathcal{N}) \leq \deg \mathcal{N}}$ for ${\deg \mathcal{N} > 0}$ : otherwise we could keep subtracting points of ${\mathcal{N}}$ to get a degree one map to ${\mathbb{P}^1}$ .

Now let’s apply the basepoint-free pencil trick to ${\mathcal{L}}$ and ${\mathcal{L} \otimes \mathcal{M}}$ . We get

$\displaystyle h^0( \mathcal{L}^{2} \otimes \mathcal{M}) \geq 6,$

and, once again, equality holds since ${\deg( \mathcal{L}^2 \otimes \mathcal{M}) = 6}$ . Inductively, we get

$\displaystyle h^0( \mathcal{L}^n \otimes \mathcal{M}) = 2n + 2.$

Taking ${n \gg 0}$ , we know that this has to be equal to ${2n + 2 + 1 - g}$ , so that ${g = 1}$ .

In other words, the choice of a hyperelliptic map ${C \rightarrow \mathbb{P}^1}$ is a condition, not extra data (modulo automorphisms of ${\mathbb{P}^1}$ ). For example, when ${g = 2}$ , the hyperelliptic map can be described as the canonical map: the map associated to the canonical line bundle. To specify a hyperelliptic curve of genus ${g}$ is thus equivalent to specifying ${2g + 2}$ points on ${\mathbb{P}^1}$ over which the degree two cover ${C \rightarrow \mathbb{P}^1}$ is branched, modulo automorphisms of ${\mathbb{P}^1}$ . In other words, the moduli space of hyperelliptic curves is given by

$\displaystyle \mathrm{Conf}_{2g+2}( \mathbb{P}^1)/ \mathrm{Aut}(\mathbb{P}^1),$

where ${\mathrm{Conf}_{2g+2}}$ is the configuration space of ${2g+2}$ distinct (unordered) points of ${\mathbb{P}^1}$ . In particular, the dimension is given by

$\displaystyle 2g + 2 - \dim \mathrm{Aut}(\mathbb{P}^1) = 2g - 1,$

so that hyperelliptic curves form a rather small subspace of the moduli space of curves, which has dimension ${3g-3}$ . Moreover, it is a unirational variety: it admits a dominant rational map from a projective space. This is in sharp contrast to the moduli space ${\mathcal{M}_g}$ , which is of general type for ${g \gg 0}$ by a celebrated theorem of Harris and Mumford.

2. Trigonal curves

Let’s consider the next case: that of a degree three map ${\phi: C \rightarrow \mathbb{P}^1}$ (or a trigonal curve). We will also assume that the genus of ${C}$ is at least ${3}$ .

In other words, there exists a basepoint-free line bundle ${\mathcal{L}}$ on ${C}$ of degree three, defining the map ${\phi}$ . In fact, in this case, we have

$\displaystyle h^0( \mathcal{L}) = 2,$

and so the map ${\phi: C \rightarrow \mathbb{P}^1}$ is associated to a complete linear system. One way to see this is to appeal to Clifford’s theorem, which states that:

Theorem 2 (Clifford) For a line bundle ${\mathcal{L}}$ of degree at most ${2g -2}$ on a curve ${C}$ , one has

$\displaystyle 2( h^0 ( \mathcal{L}) - 1) \leq \deg \mathcal{L}.$

Observe that ${h^0( \mathcal{L}) - 1}$ is the dimension of the complete linear system associated to ${\mathcal{L}}$ , i.e. the projective space ${\mathbb{P}( H^0(\mathcal{L})}$ . In this case, Clifford’s theorem shows that the linear system associated to a degree three line bundle on ${C}$ (if ${g(C) \geq 3}$ ) has dimension at most ${1}$ , which was our claim.

Every curve of genus ${\leq 4}$ is either hyperelliptic or trigonal. Given a genus two curve, we already know that it is hyperelliptic via the canonical map. Let’s look at the next two cases.

Example 1 Given a genus three curve ${C}$ , if it is not hyperelliptic, the canonical map imbeds ${C}$ as a smooth plane quartic in ${\mathbb{P}^2}$ , and projection from a point on ${C}$ is a degree three map from ${C}$ to ${\mathbb{P}^1}$ .

Example 2 Given a genus four curve, if it is not hyperelliptic, the canonical map ${C \rightarrow \mathbb{P}^3}$ imbeds ${C}$ as the (degree six) complete intersection of a quadric ${Q}$ and a cubic ${S}$ in ${\mathbb{P}^3}$ . Let’s admit this, and see how to produce the degree three map ${C \rightarrow \mathbb{P}^1}$ .

We need to produce three points on ${C}$ which move in a pencil. The Riemann-Roch theorem, in its “geometric” form, states that this is equivalent to finding ${p,q,r \in C}$ such that the images of ${p,q,r}$ via the canonical imbedding live inside a line, a ${\mathbb{P}^1 \subset \mathbb{P}^3}$ . In other words, ${p,q,r}$ impose one less than the expected number of linear conditions on differential 1-forms in ${C}$ : we have

$\displaystyle h^0( K - (p + q + r)) = h^0(K) - 2,$

rather than ${h^0(K) - 3}$ .

To produce these three points, observe that a quadric ${Q \subset \mathbb{P}^3}$ always contains a ${\mathbb{P}^1}$ : in fact, lots of copies of ${\mathbb{P}^1}$ . Now take the three points of intersection between ${S}$ and ${\mathbb{P}^1}$ given by Bezout’s theorem; they are in ${C}$ and live on a line, and so move in a pencil. (This is the sort of argument that Brill-Noether theory does very efficiently, in higher genera when one doesn’t have as clear a picture of curves.)

Once again, trigonal divisors on a curve ${C}$ — degree three divisors that move in a pencil without base points — are very special divisors for large ${g}$ , and we should expect them to be in short supply.

Proposition 3 A curve ${C}$ of genus ${\geq 3}$ cannot be both hyperelliptic and trigonal.

Proof: To see this, suppose given a degree two map ${\phi: C \rightarrow \mathbb{P}^1}$ and a degree three map ${\psi: C \rightarrow \mathbb{P}^1}$ . Equivalently, suppose given line bundles ${\mathcal{L}, \mathcal{M}}$ on ${C}$ of degrees ${2,3}$ with ${h^0(\mathcal{L}) = h^0(\mathcal{M}) = 2}$ .

The basepoint-free pencil trick now implies that

$\displaystyle h^0( \mathcal{L} \otimes \mathcal{M}) \geq h^0(\mathcal{M}) + h^0( \mathcal{M}) = 4.$

If the genus is at least four, then ${\mathcal{L} \otimes \mathcal{M}}$ is a degree five special divisor (of degree at most ${2g -2}$ ) whose ${h^0}$ contradicts Clifford’s theorem. $\Box$

If the genus is four, then the result fails. Namely, we can take a smooth two-dimensional quadric surface (i.e., a ${\mathbb{P}^1 \times \mathbb{P}^1}$ ), and take a smooth divisor of type ${(3, 3)}$ . Given a smooth curve in of type ${(a,b)}$ in ${\mathbb{P}^1 \times \mathbb{P}^1}$ , the genus is given by ${(a-1) (b-1)}$ , so if ${a = b = 3}$ the genus is four. Such a curve comes with two natural degree three maps to ${\mathbb{P}^1}$ , which must be distinct since the curve is imbedded in ${\mathbb{P}^1 \times \mathbb{P}^1}$ . In fact, it follows from this — since every (edit: not quite, some of these live on singular quadrics and are trigonal in only one way) nonhyperelliptic genus four curve is given by a ${(3,3)}$ -curve in ${\mathbb{P}^1 \times \mathbb{P}^1}$ — that the general curve of genus four has (at least) two distinct maps ${C \rightarrow \mathbb{P}^1}$ .

Proposition 4 A curve of genus ${\geq 5}$ cannot be trigonal in two different ways.

Proof: Similarly, suppose there exist two different line bundles ${\mathcal{L}, \mathcal{M}}$ of degree ${3}$ and with ${h^0 = 2}$ . In this case, we can use the basepoint-free pencil trick (again!) to get

$\displaystyle h^0( \mathcal{L} \otimes \mathcal{M}) \geq 4,$

and that contradicts the equality case of Clifford’s theorem: ${\mathcal{L} \otimes \mathcal{M}}$ is special and has degree too small to be the canonical divisor. $\Box$

It’s interesting that this pattern does not persist: a curve (of high genus) can be tetragonal in infinitely many ways. To construct such examples, consider bi-elliptic curves: that is, curves ${C}$ with a degree two map ${C \rightarrow E}$ for ${E}$ an elliptic curve. By increasing the branching, we can make ${C}$ of genus as high as we want. Then there are lots of degree four maps