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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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Sorry if I am ignorant, but what do you mean by the square of a curve?Thanks for clarifying. – awllower Jun 3 '12 at 16:18
The self-intersection $C \cdot C$. – Evariste Jun 3 '12 at 16:40
So the intersection is a number? It appears that I have to check out some definitions. Thanks again. – awllower Jun 3 '12 at 16:41
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 $C\cap C'$. – Georges Elencwajg Jun 3 '12 at 17:25
up vote 6 down vote accepted

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 this book.

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Thank you Georges. – Evariste Jun 3 '12 at 20:59

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