Algebraic Curves Lecture 4: Riemann Roch

The fourth lecture of the algebraic curves class

Riemann Roch

Last time we stopped with the following characterization of "very ampleness":

Proposition: Let $L$ be a line bundle on a curve $X$ . Then

$L$ is base point free if and only if $h^0(X,L) = h^0(X,L(-p)) + 1$ for all $p\in X$ .
$L$ is very ample if $h^0(X,L) = h^0(X,L(-p-q))+2$ for all $p,q \in X$ .

Remark: The same is true for linear series, and the definitions (of base point, of very ample) are the same as well.

Example: Take a map $\mathbb P^1 \to \mathbb P^2$ given by $(s:t) \mapsto (s^3:st^2:t^3)$ . Take the linear series $V = \mathrm{span}\{s^3,st^2,t^3\} \subseteq H^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(3))$ . In the affine chart $r\mapsto (r^2, r^3)$ this is given by the equation $x^3 - y^2$ ; it's the cuspoidal cubic. The linear series $V$ is NOT very ample because at the singularity the differential is not injective: the sections in $V$ that vanish at $0$ are of the form $a\cdot r^2 + b\cdot t^3 = r^2\cdot (a+b\cdot r)$ , meaning

\begin{aligned} \operatorname{dim}(V(-0)) = \operatorname{dim}(V(-2\cdot 0)). \end{aligned}

Here

0

is the origin

(0,0)

, the divisor.

Here are other ways to see that $(0,0)$ is singular in the the image:

$k\mapsto (2r \cdot k, 3r^2 \cdot k)$
Jacobi's criterion
If you have a plane curve you can directly read off the equation of the Zariski tangent space at the origin – the equation of the Zariski tangent space at the origin is the linear term in the equation. In the case of $x^3 - y^2 = 0$ we don't have a linear term, hence singular at the origin.
You can also use the definition of a singularity; show that the dimension of the Zariski tangent space is larger than the dimension of the ring. In our case,

\begin{aligned} \operatorname{dim}_{\mathbb C}\left(\frac{(x,y)}{(x^2,xy,y^2)} = 2 \neq 1.\right) \end{aligned}

Note that the equation for the tangent cone is given by the lowest order terms. The tangent cone to an affine variety $X$ given by an ideal $I$ at the origin is the Zariski closed subset corresponding to the ideal $in(I)$ , the ideal generated by all the lowest degree terms of elements in $I$ .

Riemann Roch

Lemma:

If $\deg(L) < 0$ then $h^0(X,L) = 0$ .
If $\deg(L) = 0$ and $h^0(X,L) > 0$ then $L\cong \mathcal O_X$ .

Proof.

Assume $h^0(X,L) > 0$ . We know that $L\cong \mathcal O_X(D)$ for some $D\in \textrm{Div}(X)$ . If $h^0(X,L) > 0$ , then $D\sim D'$ with $D'$ effective which then implies that $\deg(D') = \deg(L) \geq 0$ .
This follows since the only effective divisor of degree $0$ is 0 (all coefficients equal to 0 in the divisor).

\square

Definition: We denote by

K_X

the canonical sheaf on

X

. Analytically, the sheaf of holomorphic one forms.

Theorem: (Serre Duality).

\begin{aligned} H^1(X,L) \cong H^0(X,K_X\otimes L^{-1}). \end{aligned}

Note that we call $K_X\otimes L^{-1}$ the residual of $\mathbf{L}$ .

Theorem: (Riemann Roch).

\begin{aligned} h^0(X,L) - h^0(X,K_X\otimes L^{-1}) = d - g + 1. \end{aligned}

Proof. By Serre Duality, we need to show that

\begin{aligned} h^0(X,L) - h^1(X,L) = d - g + 1. \end{aligned}

We prove this by induction, and we write

h^i(-)

for

h^i(X,-)

to simplify notation a bit.
Base case: The claim is true for

L = \mathbb O_X

\begin{aligned} h^0(\mathcal O_X) = 1, ~h^1(\mathcal O_X) = g, ~d = 0. \end{aligned}

We show that equality holds for

L

if and only if it holds for

L(-p)

.
We have a short exact sequence

\begin{aligned} 0\to L(-p)\to L\to L|_p \to 0 \end{aligned}

where

L|_p

denotes the skyscraper sheaf of

L

p

. I can then pass to the long exact sequence of cohomology to get

\begin{aligned} 0 \to H^0(L(-p)) \to H^0(L) \xrightarrow{\mathrm{ev}} K \to H^1(L(-p)) \to H^1(L) \to 0 \end{aligned}

noting that

H^0(L|_p) = K

and

H^1(L|_p) = 0

. From this long exact sequence we have that

\begin{aligned} h^0(L) = h^0(L(-p)) + \operatorname{dim}(\operatorname{ker} \mathrm{ev}) \end{aligned}

and

\begin{aligned} h^1(L(-p)) &= h^1(L) + \textrm{codim}(\operatorname{ker} \mathrm{ev}) \\ &= h^1(L) + 1 - \operatorname{dim}(\operatorname{ker}\textrm{ev}). \end{aligned}

This implies

\begin{aligned} h^0(L) - h^0(L(-p)) = h^1(L) + 1 - h^1(L(-p)) \end{aligned}

and then

\begin{aligned} h^0(L) - h^1(L) = h^0(L(-p)) - h^1(L(-p)) + 1. \end{aligned}

\square