3.2 Riemann-Roch for surfaces

Theorem 3.4

Suppose \(X\) is a smooth surface with \(K=K_X\) and \(D\) is a Cartier divisor on \(X\) with \(L=\mathcal{O}_X(D)\). Then

\[ \chi (L) = \frac{1}{2} D\cdot (D - K) + \chi (\mathcal{O}_X). \]

Proof ▼

This is [ 4 , Theorem V.1.6 ] . Here’s the proof Hartshorne gives.

Using Bertini, we can write \(D=E-F\) where \(E\) and \(F\) are two smooth very ample divisors on \(X\). We can even assume \(E\) and \(F\) are connected.

Then we get two exact sequences

\begin{align*} & 0\to \mathcal{O}_X(C-E)=L \to \mathcal{O}_X(C)\to \mathcal{O}_X(C)_{|E}\to 0\\ & 0\to \mathcal{O}_X \to \mathcal{O}_X(C)\to \mathcal{O}_X(C)_{|C}. \end{align*}

These two exact sequences give us that

\[ \chi (L) + \chi (\mathcal{O}_X(C)_{|E} = \chi (\mathcal{O}_X) + \chi (\mathcal{O}_X(C)_{|C}. \]

So,

\begin{equation} \label{chiLt} \chi (L) = \chi (\mathcal{O}_X) + \chi (\mathcal{O}_X(C)_{|C} - \chi (\mathcal{O}_X(C)_{|E}. \end{equation}

3.5

Using the Riemann-Roch for curves along with the genus formula, we get that

\begin{align} \chi (\mathcal{O}_X(C)_{|E}) & = C\cdot E + 1 - g(E)\\ & = C\cdot E - \frac{E\cdot (E+K)}{2}\label{gfe}\\ \chi (\mathcal{O}_X(C)_{|C}) & = C\cdot C + 1 - g(C)\\ & = C\cdot C - \frac{C\cdot (C+K)}{2} \label{gfc}. \end{align}

So, substituting, we get that

\begin{align*} \chi (L) & = \chi (\mathcal{O}_X) + C^2 - C\cdot E + \frac{1}{2}\left(E^2 + E\cdot K - C^2 -C\cdot K\right)\\ & = \chi (\mathcal{O}_X) + \frac{1}{2}(C-E)\cdot (C-E -K) \end{align*}

(once you multiply it all out) as desired.

Definition 3.10

The arithmetic genus of a smooth surface \(X\) is \(p_a = p_a(X)=\chi (O_X) -1\).