Riemann--Roch for surfaces

On a surface, Riemann--Roch replaces degree by $\frac12D\cdot(D-K)$; with Serre duality, positivity turns Euler characteristic into sections.Suggest a slogan

(Riemann--Roch for curves)

Let $C$ be a smooth projective curve of genus $g$ and $D$ a divisor of degree $d$ on $C$. Then \[ h^0\qty{\OO_C(D)} - h^0\qty{\OO_C(K_C - D)} = d - g + 1 , \] equivalently $\chi\qty{\OO_C(D)} = d - g + 1$. In particular $h^0(D) - h^0(K_C - D) = \deg D - g + 1$.

(Riemann--Roch for surfaces)

Let $X$ be a smooth projective surface with canonical divisor $K$, and $D$ any divisor. Then \[ \chi\qty{\OO_X(D)} = \chi\qty{\OO_X} + \frac{1}{2} D \cdot (D - K) . \]

The curve case is the same statement with $\deg$ in place of the intersection pairing, and the surface case is the first one where the correction term is quadratic rather than linear. $\chi(\OO_X) = 1 - q + p_g$ is the constant, with $q = h^1(\OO_X)$ the irregularity.

What it is used for is existence, not counting. Serre duality turns $h^2(D)$ into $h^0(K-D)$, so \[ h^0(D) \geq \chi(\OO_X) + \tfrac{1}{2} D\cdot(D-K) - h^0(K - D) , \] and once $D$ is positive enough that $K - D$ has no sections, a large right-hand side forces $h^0(D) > 0$ and hence $D$ effective. That inequality gives a standard sufficient criterion for nonemptiness of a linear system.

State Riemann--Roch for a surface.

How do you show a divisor on a surface is effective?

Dependencies

T-COHSDSerre duality in dimension $n$uses
D-COHEULERThe Euler characteristic and the Hilbert polynomialuses

What links to this

Vanishing and duality