## Canonical Linear Systems

Given any curve $C$, we have a natural linear system $|K|$, consisting of all the effective canonical divisors. Now, sometimes this is unhelpful. After all, if $g=0$, then the canonical system has negative degree, so $|K|=\emptyset$. If $g=1$, then the canonical system is trivial, and so $|K|=0$. However, we do have that in general $\dim |K|=g-1$, and so starting at $g=2$ we have something we can do with it. We call the rational map defined by $|K|$ $\phi_K$.

First thing we need to do is check that $|K|$ has no base points. If that’s true, then $\phi_K$ is in fact a morphism. So we assume that $p$ is a base point of $|K|$. This is equivalent to saying that $H^0(K\otimes \mathscr{O}(-p))\to H^0(K)$ is an isomorphism. That implies that $h^0(K-p)=g$. So $p$ is a base point if and only if $h^0(K-p)=g$. Now, we apply Riemann-Roch and see that $h^0(K-p)=2g-2-1-g+1+h^1=g-2+h^1(K-p)=g-2+h^0(K-(K-p))=g-2+h^0(p)$, and we know that $h^0(p)=1$ (with the one dimension being the constant functions with no poles), because otherwise there’s a rational function with a simple pole at $p$, which means that $C$ is rational, but we assumed that the genus is $\geq 2$. Thus, there are no base points, so we have a canonical morphism.

Now we claim that the canonical morphism is either an embedding or else is a two-to-one map onto $\mathbb{P}^1$ embedded into $\mathbb{P}^{g-1}$, the hyperelliptic case.

Now, a linear system is very ample, that is, defines an embedding, if and only if it has no base point and it satisfies for each $p,q\in C$, $\dim |D-p-q|=\dim |D|-2$, including the $p=q$ case. What this essentially says is that we can use the linear system to tell the points and tangents apart. If $p\neq q$, it says that the two points aren’t mapped to the same point in the image and if $p=q$, it says that we get a well defined tangent vector again. Another way to see it is that if $p\neq q$, the pair of maps $p,q:pt\to C$ with $pt=\mathrm{Spec}(\mathbb{C})$ can be distinguished after composition with the morphism given by the linear system, and if $p=q$, it says that two morphisms $\mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^2)\to C$ with image $p$ can be distinguished. With a little work it can be seen that the set of such maps is naturally isomorphic to the tangent space at the point.

So, we just need to check $\dim |K-p-q|=\dim|K|-2=g-3$ to show that the canonical map is an embedding. So we apply Riemann-Roch to $p+q$, and get $\dim|p+q|-\dim|K-p-q|=2+1-g=3-g$. So then the question is just whether or not $\dim|p+q|=0$. Now is where hyperellipticity comes in. If a curve is hyperelliptic, then there is a divisor of the form $p+q$ such that $|p+q|=1$. So in the hyperelliptic case we don’t get an embedding, but instead we get a map onto a copy of $\mathbb{P}^1$. And, if $\dim |p+q|>0$ ever, then $|p+q|$ has a linear subsystem which gives a map to $\mathbb{P}^1$, and so is hyperelliptic. Thus, in the nonhyperelliptic case, $|p+q|=0$, and so we have an embedding in the first place.

So we call the morphism $\phi_K:C\to \mathbb{P}^{g-1}$ the canonical embedding when $C$ isn’t hyperelliptic. It’s image is a curve of degree $2g-2$ and we call that the canonical curve. Now, canonical curves are rather important, because their geometry is actually intrinsic to the curve, as the canonical linear system was.

For hyperelliptic curves, we can actually very explicitly write out the canonical map, to show that the image is the Veronese embedding of $\mathbb{P}^1$ into $\mathbb{P}^{g-1}$. Now, to get there, we first need that hyperelliptic curves can all be realized as the (smooth) closures of plane curves of the form $y^2=f(x)$ where $f(x)$ has distinct roots. We can do this because for any collection of $2g+2$ points in the plane, we have $y^2=\prod_{i=1}^{2g-2}(x-x_i)$ as a smooth curve, whose completion is hyperelliptic and can be viewed as a branched cover of $\mathbb{P}^1$, branched at the $x_i$. Now, take any hyperelliptic curve at all. It is branched at $2g+2$ points (Hurwitz says that $2g-2=-2d+\deg B$, but $d=2$, so $2g+2=\deg B$) and if we delete the ramification points, then we have a legitimate topological covering space.

We, however, get another topological covering space if we take the plane hyperelliptic curve branched at these points and delete the ramification, and these covering spaces are isomorphic. The isomorphism will extend across the ramification points, and so we get the hyperelliptic curve and the plane curve as isomorphic. If we started with a curve branched at infinity, then the $f(x)$ above will be of degree $2g-1$ rather than $2g-2$.

Now, on a plane hyperelliptic curve, any 1-form (that is, global section of $K$) can be written as $p(x)dx/y$ where $p(x)$ has degree at most $g-1$. So we get a basis $dx/y, xdx/y, \ldots, x^{g-1}dx/y$. This gives the map $(1:x:x^2:\ldots:x^{g-1})$ as the canonical map for the curve, which is just the composition of the degree two map to $\mathbb{P}^1$ and the Veronese embedding, as desired.

Now, all curves of genus two are hyperelliptic, so we’ve described their canonical map. For a nonhyperlliptic curve of genus three, the canonical map is going to be a map into $\mathbb{P}^2$ as a curve of degree 4. So for one thing, this shows that every genus three curve is a plane curve. This isn’t true in general, most curves can’t be plane curves.

## About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 5 Responses to Canonical Linear Systems

1. David Sevilla says:

Thanks for this explanation, I was looking for a way to explain some people this classification of curves without mentioning schemes and Hartshorne wasn’t being completely helpful.

2. I’ve found that Hartshorne is rarely helpful. That’s actually part of why I wrote up this series – to study this material for my orals and try to go as far as possible without using schemes.

3. Joel Fine says:

Thanks very much for this post. I was writing notes for a lecture I am about to give on the canonical morphism and hyperelliptic curves and this was a very helpful account to read.

4. Today, I went to the beach front with my children. I found a sea shell and gave it to my 4 year old daughter and said “You can hear the ocean if you put this to your ear.” She put the shell to her ear and
screamed. There was a hermit crab inside and it pinched her ear.
She never wants to go back! LoL I know this is totally off topic but I had
to tell someone!