Riemann-Roch 05: The Theorem

Riemann-Roch notes / V

The theorem, read carefully

The formula is short, but it is safer to read it as a comparison between an expected count and a dual correction term.

Let \(X\) be a compact Riemann surface of genus \(g\), let \(D\) be a divisor, and let \(K\) be a canonical divisor. Riemann-Roch says

Riemann-Roch

\(\ell(D)-\ell(K-D)=\deg(D)+1-g.\)

It is often more useful to solve for \(\ell(D)\):

\[\ell(D)=\deg(D)+1-g+\ell(K-D).\]

The term \(\deg(D)+1-g\) is the expected dimension. The term \(\ell(K-D)\) is the special contribution.

The nonspecial range

Since \(\deg(K)=2g-2\),

\[\deg(K-D)=2g-2-\deg(D).\]

If \(\deg(D)>2g-2\), then \(\deg(K-D)<0\), and hence \(\ell(K-D)=0\). In that case

\[\ell(D)=\deg(D)+1-g.\]

Divisors for which the correction term vanishes are called nonspecial.

A basic existence use

Constants always lie in \(L(D)\) when \(D\) is effective. A nonconstant meromorphic function appears once \(\ell(D)\ge 2\). In the nonspecial range this becomes a numerical condition:

\[\deg(D)+1-g\ge 2.\]

Thus sufficiently many allowed poles force the existence of a nonconstant meromorphic function.

Two substitutions that test the formula

On \(\mathbb P^1\), take \(D=2\infty+0\). We already computed

\[L(D)=\operatorname{span}\{z^{-1},1,z,z^2\},\]

so \(\ell(D)=4\). Riemann-Roch gives the same value because \(g=0\), \(\deg(D)=3\), and \(\ell(K-D)=0\):

\[\ell(D)=3+1=4.\]

For a less spherical test, take the genus two hyperelliptic curve

\[C:y^2=x^5-x+1\]

with its unique point \(O\) at infinity, and put \(D=3O\). Here \(g=2\) and \(\deg(D)=3>2g-2=2\), so the correction term vanishes. Riemann-Roch gives

\[\ell(3O)=3+1-2=2.\]

This agrees with the explicit pole orders: \(x\) has pole order \(2\) at \(O\), while \(y\) has pole order \(5\). Thus

\[L(3O)=\operatorname{span}\{1,x\}.\]

The example shows why the theorem is more than degree counting: it predicts the absence of a third function even though the divisor has degree three.

Where a proof comes from

The modern proof is best remembered through the line bundle \(\mathcal O(D)\). One proves

\[\chi(\mathcal O(D))=\deg(D)+1-g,\]

where

\[\chi=h^0-h^1.\]

Serre duality identifies

\[H^1(X,\mathcal O(D))^*\cong H^0(X,\mathcal O(K-D)).\]

Substituting \(h^0=\ell(D)\) and \(h^1=\ell(K-D)\) gives the classical formula.

Practical reading

Most computations are not proofs of Riemann-Roch. They are uses of it: compute the degree, decide whether \(K-D\) has sections, and then read off \(\ell(D)\).