In this post, we will give some results related to the classification of irreducible projective smooth curves of small genus, and make a careful study on hyperelliptic curves. In this note, a curve means a projective smooth irreducible curve.
Curves of genus 0 and 1. We will first begin with a curve $X$ of genus $0$. By Riemann-Roch's theorem, for any divisor of the form $p\in Div(X)$, we have $l(p)=2$. This yields $p$ is a very ample divisor. Hence, the induced map from $X\to \mathbb{P}^1$ is an isomorphism. We have another way to look at this: because $l(p)=2$, there exists $f\in L(p)$ a non-constant rational function, i.e. $f$ has simple pole at $p$. Hence, $f$ defines a map from $X$ to $\mathbb{P}^1$ sending $p\mapsto \infty$. Hence, $f$ is a map of degree 1, and hence, an isomorphism.
Let $X$ be a smooth projective curve of genus 1. We will prove that in fact $X$ is a smooth cubic curve in $\mathbb{P}^2$. By using Riemann-Roch, one can see $l(np)=n$, for all $n\ge 1$. In particular, $(1,\phi)$ is the basis for $L(2p)$, $(1,\phi,\psi)$ is the basis for $L(3p)$, $(1,\phi,\psi,\phi^2)$ is the basis for $L(4p)$, $(1,\phi,\psi,\phi^2,\phi\psi)$ is the basis for $L(5p)$. From this, one can see that $(1,\phi,\psi,\phi^2,\phi\psi, \psi^2,\phi^3)$ for $L(6p)$. This yields the non-trivial relation between 7 functions:
$$a_0+a_1\phi+a_2\psi+a_3\phi^2+a_4\phi\psi+a_5\psi^2+a_6\phi^3$$
This yields
$$a_5\psi^2 + a_4\phi\psi + a_2\psi =a_6\phi^3 + a_3\phi^2 + a_1\phi+a_0$$
By change of variable, one can assume that $a_5=a_6=1$. And $(\phi,\psi)$ defines a regular map between $X$ and a cubic curve $E$. By more careful analysis, the cubic curve $E$ defined about (in terms of $\phi,\psi$) is smooth, and the map from $X$ to $E$ is an isomorphism (For reference, see Milne's "Elliptic Curves").
Hyperelliptic curves and curves of genus 2 and 3. We begin with the following
Definition 1. Let $X$ be a smooth projective irreducible curve with $g_X\ge 1$. Then $X$ is a hyperelliptic curve if there exists a degree 2 morphism $X\to \mathbb{P}^1$.
We then prove that
Proposition 2. Any curve $X$ of genus 2 is hyperelliptic.
Proof. By Riemann-Roch, we have $\deg(K)=2$, and $l(K)=2$, where $K$ is the canonical divisor of $X$. That means, there exists $f\in K(K)$, and $f$ is non-constant, and $div(f)+K\ge 2$ with $\deg(div(f)+K)=\deg(K)=2$. And hence, because $div(f)+K\sim K$, we can assume without loss of generality that $K$ is of the form $p+q$.
We will prove next that $f$ has simple pole at $P,Q$. Because $f$ is non-constant, it must have poles somewhere. And its possible poles is simple at $P$ and $Q$. If $f$ has only simple pole at $P$, then we know that $f$ will define a map to $\mathbb{P}^1$, and $\deg(f)=1$ (since $\#f^{-1}(\infty)=1$), and $f$ defines an isomorphism from a curve of genus 2 to a curve of genus 0, a contradiction. Hence, $f$ must have simple pole at $P,Q$. Hence, $div(f)=P+Q-R-S$. And one can see that $f$ defines a degree two map from $X$ to $\mathbb{P}^1$, and $X$ is a hyperelliptic curve. (Q.E.D)
It is natural to ask: with given $g\ge 0$, there exists a curve of genus $g$?
Proposition 3. There exists hyperelliptic curve of every genus $g\ge 1$.
Proof. Let us consider the affine curve $X'$ defined by $y^2=\prod_{i=1}^{2g+1}(x-a_i)$, where $a_i\in k$ are distinct values. It can be seen that $X'$ is a smooth affine curve, but for $g\ge 2$, its compactification is not smooth, because the point of infinity is a singular point. We can now assume for simplicity that $k=\mathbb{C}$, then $X'\to \mathbb{P}^1\setminus\{a_1,...,a_{2g+1},\infty\}$ defines an unbranched 2-sheeted holomorphic map. Now, using a theorem for Riemann surfaces, there exists a compact Riemann surface $X$ with a branched covering $f: X\to \mathbb{P}^1$ that extend the map $X'\to \mathbb{P}^1\setminus\{a_1,...,a_{2g+1}, \infty\}$. And because $2g+1$ is odd, the map $f$ has critical values at $a_1,...,a_{2g+1},\infty$, and hence, we have $2g+2$ branched points of $X$ with multiplicity 2 for each. Now, we use the Riemann-Hurwitz's formula: $2g_X-2=2g+2-4$. This implies $g_X=g$.
It remains to show that any compact Riemann surface is an algebraic curve. But it is not a trivial fact. We can sketch the proof here. First, the Riemann-Roch's theorem can be applied for compact Riemann surfaces, and using it, we can deduce that there is a non-constant meromorphic function on $X$. This function defines a branched covering map from $X$ to $\mathbb{P}^1$, and it is a finite map, since $X$ is compact. And then, one can deduce that the meromorphic function field of $X$, denoted $M(X)$ is an extension of $M(\mathbb{P}^1)$ of degree $n$, and hence, there exists an irreducible polynomial $F(w)$ of degree $n$ such that $M(\mathbb{P}^1)[w]/(F(w))\cong M(X)$. Now, it yields $X$ is biholomorphic with the Riemann surfaces defined by $F(w)\in M(\mathbb{P}^1)[w]$, i.e. all coefficients of $F$ are meromorphic functions on $\mathbb{P}^1$. But meromorphic functions on $\mathbb{P}^1$ are just quotient of polynomials. This yields $X$ is in fact defined by an algebraic equation, and $X$ is an algebraic curve. (Q.E.D)
There is an useful criterion that help us to detect hyperelliptic curves via divisors.
Proposition 4. Let $X$ be a curve of genus $g\ge 1$, then $X$ is a hyperelliptic curve iff there exists a divisor $D$ on $X$ such that $\deg D=2, l(D)=2$.
Proof. We already saw in our previous results that the statement holds for $g=1,2$. Assume that $X$ is a hyperelliptic curve of genus $g\ge 3$, i.e. there exists a map $\phi: X\to \mathbb{P}^1$ and $\deg \phi=2$. This yields $\phi^*: k(\mathbb{P}^1)\to k(X)$ is a field embedding of degree 2. We know that there exists a non-constant rational function $f$ on $\mathbb{P}^1$, such that $div(f)=p-q$, where $p,q\in\mathbb{P}^1$, i.e. $q$ is the only simple pole of $f$. This yields $\phi^*f$ have exactly two poles $q_1,q_2\in X$, which lie on the fiber of $q$ via $\phi$. Now, let us consider $D:=q_1+q_2\in Div(X)$. It can be seen that $\deg(D)=2$, and $l(D)\ge 2$. If $l(D)=3$, then $D$ is a very ample divisor, since $l(D-p_1-p_2)\le 1$, for all $p_1,p_2\in X$. And hence, $X$ can be embedded into $\mathbb{P}^2$ via $D$. Let $F$ be the homogeneous equation for $X$, then due to the genus degree formula, $\deg f=d$, such that $g=\frac{(d-1)(d-2)}{2}$, and because $g\ge 3, d\ge 4$. But then, the function field $k(X)$ is an extension field of $k(\mathbb{P}^1)$ of degree 4. It is a contradiction. Hence, $D$ is not a very ample divisor, and $l(D)=2$. The converse is already done by our earlier arguments. (Q.E.D)
This result will lead to the classification of curves of genus $3$. Let $X$ be a curve of genus $3$. It can be seen that the canonical divisor $K$ has degree $3$. We will prove that it is actually base-point free.
Proposition 5. Let $X$ be a curve of genus $g\ge 1$. Then the canonical divisor $K$ is base point free.
Proof. It can be seen that $l(K)=g$, and $K$ is base point free iff $l(K-p)=l(K)-1$, for all point $p\in X$. By Riemann-Roch's theorem, we have
$$l(p)-l(K-p)=2-g$$
And it is sufficient for us to prove that $l(p)=1$, for all point $p\in X$. We have $1\le l(p)\le 2$. If $l(p)=2$, then it is a very ample divisor, and one can embedded $X$ into $\mathbb{P}^1$, a contradiction. So, $l(p)=1$ for all point $p\in X$, and hence $K$ is a base point free divisor (Q.E.D)
If $K$ is a base point free divisor, we then have a canonical map $\phi: X\to \mathbb{P}^{g-1}$ defined by $K$. In the case $g=3$, we will have a canonical map $\phi: X\to \mathbb{P}^2$. If furthermore, $K$ is base point free, then the map above is actually an embedding. And in this case $X$ is given by a polynomial of degree 4, by the genus degree formula.
Otherwise, $K$ is not a base point free divisor, and there exists two points $p,q$ such that $l(K-p-q)=l(K)-1$, i.e. there exists a divisor $D$ of degree 2, and $l(D)=2$. This yields by our previous proposition that $X$ is a hyperelliptic curve. In conclusion, we have
Proposition 6. Let $X$ be a curve of genus 3. Then either $X$ is hyperelliptic, or $X$ is a plane curve, given by a polynomial of degree 4.
Maps from hyperelliptic curve to $\mathbb{P}^1$. We can now deduce that if $X$ is a hyperelliptic curve, there exists only one map from $X$ to $\mathbb{P}^1$, by recalling some knowledge about morphisms to projective space.
We recall that from a curve $X$ to $\mathbb{P}^n$ is uniquely characterized by $n+1$-tuples of meromorphic function $\lambda f_0,...,\lambda f_n$ on $X$, where $\lambda\in k(X)^\times$, and $f_0,..,f_n$ do not vanish simultaneously at any point $p\in X$, and if we define a divisor $D\in Div(X)$, where $-D(p)$ is the minimum of $ord_p(f_i)$, then it can be seen that $D$ (up to linearly equivalent) is independent to the choice of $f_i$. One can see that in fact $f_i\in L(D)$. And hence, the vector space spanned by $(f_0,...,f_n)$ is a subspace of $L(D)$.
In the case $X$ is a hyperelliptic curve, one has the map $\phi$ from $X\to\mathbb{P}^1$, which is characterized by 2-tuples of meromorphic functions $f_0,f_1$ on $X$. The condition that $f_i$ do not vanish simultaneously at any point $p\in X$ implies that $f_0,f_1$ are linearly independent. And hence, $(f_0,f_1)$. In an open subset of $X$, where $f_0$ does not vanish, we have $\phi$ is given by $(1:f_1/f_0)$. Let us denote $f:=f_1/f_0$. Because $\phi$ is of degree 2, $f$ is not defined at exactly 2 points $q_1,q_2$, which are the fiber of $\infty\in \mathbb{P}^1$. Hence, the map $\phi:X\to\mathbb{P}^1$ is exactly $p\mapsto (1:f(p))$, for $p\ne q_i$ and $q_i\mapsto \infty$. It can be seen that the divisor $D$ associated to $\phi$ defined as above in this case is $q_1+q_2$, which has degree 2, and due to Lemma 4, $l(D)=2$. This yields by our previous argument that $(1,f)$ is the basis for $L(D)$. In conclusion, we have
Proposition 7. There exists only one map of degree 2 from a hyperelliptic curve to $\mathbb{P}^1$.
We now conclude this section by considering the canonical morphism from a hyperelliptic curve $X\to\mathbb{P}^{g-1}$ defined by the canonical divisor. It is quite surprising: $X$ will be mapped to the rational normal curve in $\mathbb{P}^{g-1}$.
Lemma 8. The canonical divisor on a hyperelliptic curve $X$ is of the form $(g-1)p+(g-1)q$, for some points $p,q\in X$.
Proof. We begin with a divisor $D=p+q$ on $X$ such that $l(D)=2$. This yields $(1,\phi)$ is the basis for $L(D)$, where $\phi$ is a non-constant meromorphic function. This yields $(1,\phi,...\phi^{g-1})$ are linearly independent in $L((g-1)D)$. Hence, $l((g-1)D)\ge g$. By Riemann-Roch, because $\deg (g-1)D=\deg K=2g-2$, one has $l((g-1)D)=l(K-(g-1)D)+g-1$. But we know that for any divisor $E$ with $\deg(E)=0$, then $l(E)\le 1$, and $l(E)=1$ iff $E$ is a principal divisor. This yields $l((g-1)D)=g$, and hence, $K\sim (g-1)D$. (Q.E.D)
Using this, we have
Proposition 9. The canonical morphism from a hyperelliptic curve $X$ maps $X$ into a rational normal curve in $\mathbb{P}^{g-1}$, and it is a map of degree 2.
Proof. One can see by the previous lemma that there exists $D=p+q\in Div(X)$, and $l(D)=2$, and $K=(g-1)D$. This yields, if $(1,\phi)$ is a basis for $l(D)$, we have $(1,\phi,...,\phi^{g-1})$ is a basis for $K$. From this, the canonical morphism is defined by $p\mapsto (1:\phi(p):...:\phi^{g-1}(p))$, and it is exactly the equation for the rational normal curve in $\mathbb{P}^{g-1}$. Because rational normal curves are isomorphic to $\mathbb{P}^1$, this induces a morphism from $X$ to $\mathbb{P}^1$, given by $p\mapsto (1:\phi(p))$. And it is exactly the map in Proposition 7, i.e. it is a degree 2 map. Hence, the canonical map from $X$ to the rational normal curve is of degree 2. (Q.E.D)
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