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    Riemann--Roch for surfaces

    T-SRFRR

    Surface Riemann--Roch turns Euler characteristic into intersection theory: $\chi(\OO_X(D))-\chi(\OO_X)=\tfrac12D\cdot(D-K_X)$.Suggest a slogan

    For a divisor \(D\) on a smooth projective surface \(X\), \[ \chi(\OO_X(D)) = \chi(\OO_X) + \tfrac{1}{2} D \cdot (D - K_X) . \]

    (Noether's formula)

    \[ \chi(\OO_X) = \tfrac{1}{12}\left( K_X^2 + c_2(X) \right) , \] where \(c_2(X)\) is the topological Euler characteristic over \(\CC\).

    The surface formula parallels the curve formula, with \(\deg D\) replaced by \(\tfrac{1}{2}D\cdot(D-K)\) and \(1-g\) replaced by \(\chi(\OO_X)\); adjunction makes these substitutions consistent.

    The typical use is not to compute \(\chi\) but to force a section. Serre duality gives \(h^2(D) = h^0(K-D)\), so \[ h^0(D) \geq \chi(\OO_X) + \tfrac{1}{2}D\cdot(D-K) , \] and if in addition \(h^0(K-D) = 0\) — for instance because \(D \cdot H > K \cdot H\) for an ample \(H\) — then \(D^2\) large forces \(D\) or \(-D\) to be effective. This inequality is a standard mechanism for proving that a divisor on a surface has sections and hence moves in a linear system.

    Noether’s formula is the constraint relating \(K^2\) and the topology: on \(\PP^2\), \(K^2 = 9\) and \(c_2 = 3\), and \(\tfrac{1}{12}(9+3) = 1 = \chi(\OO)\).

    State Riemann--Roch for surfaces.
    What is Noether's formula?
    How do you show a divisor on a surface is effective?

    Dependencies

    • T-SRFADJThe adjunction formula on a surfaceuses
    • D-SRFINTThe intersection pairing on a surfaceuses

    Backlinks

    • D-VARTODDThe Todd class and the Todd genusrelated-to

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