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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?

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  • $\begingroup$ Sorry if I am ignorant, but what do you mean by the square of a curve?Thanks for clarifying. $\endgroup$
    – awllower
    Jun 3, 2012 at 16:18
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    $\begingroup$ The self-intersection $C \cdot C$. $\endgroup$
    – Evariste
    Jun 3, 2012 at 16:40
  • $\begingroup$ So the intersection is a number? It appears that I have to check out some definitions. Thanks again. $\endgroup$
    – awllower
    Jun 3, 2012 at 16:41
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    $\begingroup$ Dear @awllower: this is a long story! Given a smooth compact complex algebraic surface $S$, there is a bilinear form $Pic(S)\times Pic(S)\to \mathbb Z$, where $Pic$ denotes classes of divisors. A curve $C$ is a divisor so you can compute the value of that form on the pair $(class(C), class(C))$, and the result is (dangerously!) written $C.C$ or even worse $C^2$. It is rather technical but means intuitively that you somehow "deform" one copy of $C$ to $C'$ within $S$ and then $C.C$ is the ordinary cardinality of the set-theoretic intersection $C\cap C'$. $\endgroup$
    – Georges Elencwajg
    Jun 3, 2012 at 17:25

2 Answers 2

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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.

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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.

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  • $\begingroup$ Why submit another answer to a 9 year old problem which was answered the day it was asked saying essentially the same things as the accepted answer? $\endgroup$
    – KReiser
    Mar 30, 2022 at 9:49
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    $\begingroup$ Dear @KReiser, because I think that my answer adds something to the existing answer. Hope that's ok with you. $\endgroup$
    – Manolis C. Tsakiris
    Mar 30, 2022 at 10:42

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