Algebraic Curves Lecture 5: Corollaries to Riemann-Roch

The 5th lecture of algebraic curves by Karl Christ.

More Riemann-Roch

More Riemann-Roch

Recall that for a divisor $D$ on a curve $X$ , if we write

$\ell(D) = h^0(X,\mathcal O_X(D))$
$i(D) = h^0(X,K_X\otimes \mathcal O_X(D)^{-1}) = h^1(X,\mathcal O_X(D))$ (here $h^0(X,K_X\otimes \mathcal O_X(D)^{-1})$ is called the "index of speciality")

Then Riemann-Roch says $\ell(D) - i(D) = d - g + 1$ where $d$ is the degree of the line bundle. Today we'll collect some easy Corollaries of Riemann Roch.

Definition:

\mathcal O_X(D)/D

is special if

\ell(D) > 0

and

i(D) > 0

Cor: Let

K_X

be the canonical sheaf. Then

h^0(K_X) = g

and

\deg(K_X) = 2g - 2

Proof. Apply Riemann-Roch to

\mathcal O_X

. Then

h^0(\mathcal O_X) = 1

since the only global regular functions on

X

are constant,

h^0(K_X\otimes \mathcal O_X^{-1}) = h^0(K_X)

and

d = 0

since the degree of

\mathcal O_X

0

. Then

\begin{aligned} h^0(K_X) = g. \end{aligned}

Now applying Riemann-Roch to

K_X

gives us

\deg(K_X) = 2g - 2

\square

Cor: If

\deg(L) < 0

\deg(L) > 2g- 2

then

L

is not special and

\begin{aligned} h^0(L) = \begin{cases} 0 & \text{if} \deg(L) < 0 \\ d - g + 1 &\text{if} \deg(L) > 2g - 2 \end{cases} \end{aligned}

Proof. The first case where $\deg(L) < 0$ was already established, so assume $\deg(L) > 2g - 2$ . Then $h^0(L) - h^0(K_X - L) = d - g + 1$ .

Karl says that $h^0(K_X - L) = 0$ , I'm not sure why, but this gives you the result.

\square

Cor: The Hilbert Polynomial of a degree d subcurve of

\mathbb P^n

h_X(m) = d\cdot m - g + 1

Proof. By definition the Hilbert function is the function

\begin{aligned} f_X(m) = h^0(X,\mathcal O_X(m)). \end{aligned}

Applying Riemman-Roch gives us

\begin{aligned} h^0(X,\mathcal O_X(m)) - h^1(X,\mathcal O_X(m)) = dm - g + 1. \end{aligned}

For

m \gg 0

h^1(X,\mathcal O_X(m)) = 0

\square

Cor:

X

is rational (i.e. isomorphic to

\mathcal P^1

) if and only if

g = 0

. That is, there is only one genus

0

curve up to isomorphism.

Proof. Suppose $g=0$ . We now want to construct a map to $\mathbb P^1$ , Pick $p\in X$ and consider $\mathcal O_X(p) = L$ . By Riemann-Roch, $h^0(X,L) \geq d - g + 1 = 2$ since $g = 0$ and $d = \deg(\mathcal O_X(p) = 1$ . This gives us a degree 1 map $X \dashrightarrow \mathcal P^1$ , which implies $X\cong \mathbb P^1$ .

Now suppose that $X\cong \mathbb P^1$ . Then $h_X(m) = h^0(\mathcal O_{\mathbb P^1}(m)) = m + 1.$ Then the previous corollary concerning the Hilbert polynomial means $d\cdot m - g + 1 = m + 1$ , and since $d = 1$ (viewing $X$ as a curve with a degree 1 embedding in $\mathbb P^1$ ) we have $g = 0$ .

\square

Cor: For any line bundle

L

h^0(L) \leq \deg(L) + 1

. With equality if and only if

X

is rational.

Proof. Suppose $h^0(L) > \deg(L) + 1$ . Let $D$ be an effective divisor of degree $d + 1$ . Then $h^0(L(-D))\geq 0$ . But $0 > \deg(L(-D))$ , which is a contradiction.

If $X$ is rational, $L$ has degree d. Then $h^0(L) \geq d + 1$ . Then $h^0(L)\geq d + 1$ since $h^1(L) = 0$ . Conversely, suppose there is a line bundle $L$ of degree $d > 0$ and space of global sections $h^0(L) = d + 1$ .

\square

Now let $D$ be an effective divisor of degree $d - 1$ . Then $h^0(L(-D)) \geq 2$ and the degree of $L(-D)$ is $1$ (since $L$ is degree $d$ and $-D$ is of degree $1 - d$ ). This again gives a map $X\dashrightarrow \mathbb P^1$ of degree $1$ and hence $X\cong \mathbb P^1$ .

Now for the most important corollary:

Cor:

Any line bundle of degree $2g$ is base point free
Any line bundle of degree $2g + 1$ is very ample.

Proof. If $\deg(L) \geq 2g$ , then $h^0(L) = d - g + 1$ and $h^0(L(-p)) = d - 1$ for all $p$ . This implies there are no base points.

If $\deg(L)\geq 2g + 1$ then $h^0(L) = h^0(L(-p-q)) + 2$ for all $p,q$ .

\square