Genus/Adjunction formula for a singular curve on a surface

Question

For a smooth curve $C$ on a smooth, projective surface $S$ over $\mathbb{C}$, we have the genus formula:

$g(C) = 1 + \frac12(C^2 + C \cdot K_S)$

where $K_S$ is the canonical divisor. Is this formula still true for singular (e.g. reducible) curves on $S$ if one uses the arithmetic genus in the left hand side instead of the geometric genus?

Answer 1

Yes, the formula is still true if $C\subset S$ is reduced, irreducible but not smooth.

The arithmetic genus is to be defined as $p_a(C)=\dim_{\mathbb C}H^1(C,\mathcal O_C),$ and we then have $$p_a(C)= 1+\frac {\deg[(\mathcal K_S\otimes \mathcal O_S(C))\mid C]}{2}. $$

You can find a proof in chapter II of Compact complex surfaces by Wolf Barth, C. Peters and Antonius Ven.

Answer 2

Proposition. Let $X$ be a nonsingular projective surface over an algebraically closed field. Let $D$ be an effective divisor on $X$ and let $p_a(D) = 1 - \chi(\mathscr{O}_D)$ be its arithmetic genus, where $\chi(\mathscr{O}_D)$ is the Euler characteristic. Then $2p_a-2 = D.(D+K)$.

Proof. Since $D$ is an effective divisor, we have a short exact sequence $$0 \rightarrow \mathscr{L}(-D) \rightarrow \mathscr{O}_X \rightarrow \mathscr{O}_D \rightarrow 0.$$ Since the Euler characteristic is additive on short exact sequences, we have $$\chi(\mathscr{L}(-D)) = \chi(\mathscr{O}_X) - \chi(\mathscr{O}_D).$$ On the other hand, the Riemann-Roch Theorem for surfaces (e.g., Theorem V.1.6 in Hartshorne) states that for any divisor $E$ on $X$ we have $$ \chi(\mathscr{L}(E)) = \frac{1}{2} E.(E-K) +1 + p_a(X),$$ where $K$ is the canonical divisor on $X$. Now apply this with $E=-D$.

Remark. This is Exercise V.1.3 in Hartshorne. It is also interesting to note that the adjunction formula for nonsingular curves (Proposition V.1.5 in Hartshorne) is used in the proof of the Riemman-Roch Theorem.